Search is not available for this dataset
url
string | text
string | date
timestamp[s] | meta
dict |
|---|---|---|---|
https://math.stackexchange.com/questions/1546098/spiral-of-archimedes-area-and-sketch-in-polar-coordinates | Spiral of Archimedes area and sketch in polar coordinates
This is an exercise from Apostol's Calculus, Volume 1. It asks us to sketch the graph in polar coordinates and find the area of the radial set for the function:
$$f(\theta) = \theta$$
On the interva $0 \leq \theta \leq 2 \pi$. I think to find the area we should just integrate $\theta \ d\theta$ from 0 to $2\pi$ like any other function? Is that right? Also I'm not sure how to think about sketching a function in polar coordinates.
The problem is the book gives the answer as $4\pi^3/3$ which is not what I get if I just integrate the function.
• In the 2nd edition of this book (of which I have a copy), this is exercise 5 on page 111. You should review the previous three or four pages, especially the section about "Polar coordinates" and the section about "The integral for area in polar coordinates" (or whatever those sections are called in the edition you're using). If those pages do not answer your question, perhaps you can identify something in those pages that did not make sense to you, and add that to your question. – David K Nov 25 '15 at 17:15
• By the way, I think the [calculus] tag was fine for this question. – David K Nov 25 '15 at 17:15
• Did you find the theorem yet that says, in part, "Let $R$ denote the radial set of a nonnegative function $f$ ... the area of $R$ is ..."? It gives you the formula. In my edition of the book, this is on the page just before the exercise. – David K Nov 25 '15 at 18:55
First, to sketch such a graph, you want to consider the distance from the origin as the angle from the $x$-axis changes. Just like when sketching the graph of a function in rectangular coordinates it is good to evaluate at particular values of $x$ and see the height of the function, when sketching a curve in polar coordinates, evaluate the function are a few values of the angle and find the radius, i.e., the distance from the origin at the angle. So, doing that we obtain the following graph:
Then, to calculate the area of the radial set, you must integrate $\frac{1}{2} r^2$, where the radius is the value of the function. So we have, \begin{align*} \text{Area} &= \frac{1}{2}\int_0^{2 \pi} \theta^2 \, d\theta \\ &= \left. \frac{1}{2} \cdot \frac{\theta^3}{3} \right|_0^{2 \pi} \\ &= \frac{(2 \pi)^3}{6}\\ &= \frac{4 \pi^3}{3}. \end{align*}
Sketching this in polar coordinates is pretty straightforward. Draw your axes and you know that the radial value is equal to the angle, so your curve would start at the origin when $\theta$ is zero, it would be $\frac{\pi}{2}$ at 90 degrees, etc.
As for calculating the area, you are correct that you integrate it, but I'm not sure what it means physically because the curve does not close on itself since $f(0) \neq f(2\pi)$ (perhaps this is the insight they are trying to get from you).
• The "radial set" is defined in the book. Graphically, it is formed by the line segments from the origin to each point on the curve. All the points those segments pass through are in the area to be measured. As long $f(\theta)$ is never negative, it's clear what region's area we're supposed to measure. – David K Nov 25 '15 at 17:32
commenting on Daves's answer "but I'm not sure what it means physically because the curve does not close on itself since f(0)≠f(2π)f(0)≠f(2π)"
The area calculated is the area bounded by the curve and the terminal vector. ie the curve is subtended by the sweeping vector, which in this case is at 2pi- the positive x axis. . | 2021-05-12T11:04:25 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/1546098/spiral-of-archimedes-area-and-sketch-in-polar-coordinates",
"openwebmath_score": 0.9936886429786682,
"openwebmath_perplexity": 215.99766148976502,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9706877675527112,
"lm_q2_score": 0.8824278757303677,
"lm_q1q2_score": 0.8565619447189919
} |
https://mymathforum.com/threads/how-can-i-find-the-work-on-a-block-when-it-is-pulled-up-in-a-curve.347508/ | # How can I find the work on a block when it is pulled up in a curve?
#### Chemist116
The problem is as follows:
The figure from below shows a force acting on block as it is pulled upwards in a curve from point $A$ to $B$. It is known that the block is pulled by a force which its modulus is $100\,N$. Find the work (in Joules) of such force between the points indicated. Consider that the angle given is with respect of the vertical with the floor.
The alternatives given in my book are:
$\begin{array}{ll} 1.&143\,J\\ 2.&312\,J\\ 3.&222\,J\\ 4.&98\,J\\ 5.&111\,J\\ \end{array}$
I attempted to decompose the force given as such:
$F\cos 37^{\circ} \times d = W$
But the result doesn't seem to yield an adequate result:
$W= 100\times \cos 37^{\circ} \times 2.1= 100 \times \frac{4}{5}\times 2.1=168$
Assuming the gravity does negative work?
$W= - 100 \times \sin 37^{\circ}= - 100 \times \frac{3}{5}\times 1.2= -72$
Anyways the sum doesn't yield the result which supposedly is option $5$. Can somebody help me here? :help:
#### DarnItJimImAnEngineer
Treat $\vec{F}$ as two separate forces. Horizontal force $F \sin 37°$ acts over a horizontal distance of 2.1 m. The vertical force acts over a vertical distance of 1.2 m. Try finding the work from each and adding them.
Last edited by a moderator:
2 people
#### Chemist116
Treat $\vec{F}$ as two separate forces. Horizontal force $F \sin 37°$ acts over a horizontal distance of 2.1 m. The vertical force acts over a vertical distance of 1.2 m. Try finding the work from each and adding them.
I attempted to do such thing.
Let's see:
$F\sin 37^{\circ}\cdot 2.1 = 100 \cdot \frac{3}{5} \cdot 2.1 = 126 J$
$F\cos 37^{\circ}\cdot 1.2 = 100 \cdot \frac{4}{5} \cdot 1.2 = 96 J$
The summing both would give me:
$W_{1}+W_{2}=126+96=222\,J$
which supposedly is the third option. I'm not doing any sort of vector sum here as work is scalar (thanks to romsek for that remark). Mind to confirm whether what I'm doing is what you intended to say?
I'm sorry if I mentioned that the answer is $111\,J$. I checked with the answers sheet and it lists it to be $222\,J$. I feel so dumb now, as when I compared with what I did in my notes I got an error in the summation. Learning, yikes!
But I'm still confused, by comparing it to a similar problem which I posted before this, why shouldn't I subtract the "angle" which could result by joining $AB$? Is it because it's a curve? Can you help me with this part please? :help:
Last edited by a moderator: | 2019-12-05T20:22:16 | {
"domain": "mymathforum.com",
"url": "https://mymathforum.com/threads/how-can-i-find-the-work-on-a-block-when-it-is-pulled-up-in-a-curve.347508/",
"openwebmath_score": 0.8642544746398926,
"openwebmath_perplexity": 365.674266728435,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9706877700966099,
"lm_q2_score": 0.8824278556326344,
"lm_q1q2_score": 0.856561927455175
} |
https://math.stackexchange.com/questions/1200560/matrix-of-t-over-p-2 | # Matrix of $T$ over $P_2$
I'm having a bit of trouble following the logic my professor used to construct the matrix of a given transformation $T$ in class, and was wondering if anyone could share any further intuition or insight.
Given $P_2$, the set of polynomials of degree $\leq2$, define a linear transformation $T:P_2\rightarrow \mathbb{R}^3$ such that $T\big(p(x)\big) = \begin{pmatrix} p(0)\\p(1)\\p(2)\end{pmatrix}$.
The matrix of $T$ with respect to the basis $\mathcal{B}=\left\{1,x,x^2\right\}\\$
is given by $T(1)=\begin{pmatrix} 1\\1\\1\end{pmatrix}$, $T(x)=\begin{pmatrix} 0\\1\\2\end{pmatrix}$, and $T(x^2)=\begin{pmatrix} 0\\1\\4\end{pmatrix}$,
i.e. $\,A=\begin{pmatrix}1&0&0\\1&1&1\\1&2&4\end{pmatrix}$.
I get the whole business of the coefficients on this particular transformation being the "trivial relation" such that $c_1=c_m=0$, and that $ImT=\mathbb{R}^{3}$ such that $T$ is an isomorphism.
What I'm struggling with, quite frankly, is how he got the coefficients for $A$. The coefficients for $p(x)=1$ in particular are a bit counter-intuitive. Is the idea that, if we define $p(x)=1$, then no matter what, $p(0)=p(1)=p(2)=1$?
The notation for expressing the various matrices associated with an isomorphism is also a bit nebulous for me. My textbook defines a $\mathcal{B}$-coordinate transformation as
$T^{-1}\begin{bmatrix}c_1\\\vdots\\c_n\end{bmatrix}=c_1f_1+\cdots + c_nf_n$.
What would the inverse of the transformation $T\big(p(x)\big) = \begin{pmatrix} p(0)\\p(1)\\p(2)\end{pmatrix}$ be?
What would the matrix associated with the inverse be? Or would it be a set of three different polynomials (linear equations)?
For those wondering, we're using Bretscher's Linear Algebra with Applications, 5th edition. Thanks!
• $A = [[T(1)]_B|[T(x)]_B|[T(x^2)]_B]$ where $B=\{ e_1,e_2,e_3 \}$ is just the standard basis for $\mathbb{R}^3$. More generally, you just find the coordinate vectors of the image of each vector in the domain basis and glue them together. There are good reasons for doing this, but you ought to have seen them in lecture I think... – James S. Cook Mar 22 '15 at 4:42
• My silly notation $[v]_B = v$ in this case since the coordinate system in the codomain is just the plain-old-standard Cartesian coordinate system. – James S. Cook Mar 22 '15 at 4:43
To give you a much more complicated example. Let us consider $T(f(x)) = f'(x)$ where we consider $T:P_2 \rightarrow P_2$ and to be perverse let's use $\beta = \{ 1,x,x^2 \}$ as the domain basis, but $\gamma = \{ x^2,x,1 \}$ as the codomain basis. To understand $T$ I like to consider $f(x)=a+bx+cx^2$ and see what happens: $$T(f(x)) = b+2cx$$ Then the coordinate vector of the image is easy enough to see: $$[T(f(x))]_{\gamma}= [b+2cx]_{\gamma} = [0(x^2)+2c(x)+b(1)]_{\gamma} = [0,2c,b]^T.$$ Then, the matrix $[T]_{\beta,\gamma}$ is the matrix which when multiplied on $$[f(x)]_{\beta} = [a+bx+cx^2]_{\beta} = [a,b,c]^T$$ yields $[T(f(x))]_{\gamma}=[0,2c,b]^T$. A moments reflection yields: $$[T]_{\beta, \gamma} = \left[ \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 2 \\ 0 & 1 & 0\end{array}\right].$$ Alternatively, just use: $$[T]_{\beta, \gamma} = [ [T(1)]_{\gamma} | [T(x)]_{\gamma}| [T(x^2)]_{\gamma} ] = [[0]_{\gamma}|[1]_{\gamma}|[2x]_{\gamma}] = \left[ \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 2 \\ 0 & 1 & 0\end{array}\right].$$ Some books also use $[T]_{\beta}^{\gamma}$ to reflect the differing roles the domain and codomain bases play especially in regard to coordinate change. Perhaps your text does that.
• The text doesn't explicitly use the terms "domain" and "codomain", but I think I understand. Follow-up question: if we define some other basis $\mathcal{C}=\left\{x^2,xy,y^2\right\}$ and $T(x)=\begin{pmatrix}f(1,0)\\f(0,1)\\f(1,1)\end{pmatrix}$, does the computation still move along similar lines? see here for more. – Benjamin Loya Mar 22 '15 at 5:05 | 2019-12-09T05:31:48 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/1200560/matrix-of-t-over-p-2",
"openwebmath_score": 0.9328702092170715,
"openwebmath_perplexity": 202.4662831874779,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9706877717925421,
"lm_q2_score": 0.8824278540866547,
"lm_q1q2_score": 0.8565619274510493
} |
https://www.physicsforums.com/threads/integration-of-a-rational-function.409211/ | # Integration of a rational function
stripes
## Homework Statement
find
$$\int\frac{x^{2}-2x-1}{(x-1)^{2}(x^{2}+1)}dx$$
None
## The Attempt at a Solution
$$\frac{x^{2}-2x-1}{(x-1)^{2}(x^{2}+1)} = \frac{A}{x-1} + \frac{B}{(x-1)^{2}} + \frac{Cx + D}{x^{2}+1}$$
$$x^{2}-2x-1 = A(x-1)(x^{2}+1) + B(x^{2}+1) + (Cx + D)(x-1)^{2}$$
$$x^{2}-2x-1 = Ax^{3}-Ax^{2}+Ax-A+Bx^{2}+B+Cx^{3}-2Cx^{2}+Cx+Dx^{2}-2Dx+D$$
$$x^{2}-2x-1 = x^{3}(A+B+C) + x^{2}(-A+B-2C+D) + x(A+C-2D) - A + B +D$$
so A+B+C = 0, -A+B-2C+D = 1, A+C-2D=-2 and -A+B+D=-1
solving for coefficients, we get
A = 5/3
B = -2/3
C = -1
D = 4/3
so $$\frac{x^{2}-2x-1}{(x-1)^{2}(x^{2}+1)} = \frac{5/3}{x-1} - \frac{2/3}{(x-1)^{2}} - \frac{x-(4/3)}{x^{2}+1}$$
but apparently, the last statement is not correct. Did I break down the rational function incorrectly?
Tedjn
Check A+B+C=0.
Homework Helper
i think that's ok? but -A+B-2C+D = (-5 -2 +6-4)/3 = -5/3?
Last edited:
stripes
so did I get all my coefficients messed up? Arghhh...
The Chaz
Count Iblis
Cheating to the third power
1) Obviously the Prof. doesn't want students to use Wolfram alpha.
2) The "show steps" feature adds insult to injury. Just knowing the correct answer is going to give away nontrivial information to the student.
3) And Wolfram alpha is lying to you when it shows the steps. Not that there is anything wrong in these steps, but these are not the steps Wolfram alpha uses itself when it computed the answer. If you ask Wolfram alpha to show the steps, what it does is crank up the high school algorithm that gives the same answer, as it is most likely that this will yield the desired steps (I mean, who else other than students having difficulties with their assignments would want to see this?).
Wolfram alpha uses series expansion methods to compute the answer, which is way more efficient than the usual high school method. The general rule of thumb is that methods that involve solving equations for variables are less efficient compared to methods that don't involve solving equations.
Mentor
so $$\frac{x^{2}-2x-1}{(x-1)^{2}(x^{2}+1)} = \frac{5/3}{x-1} - \frac{2/3}{(x-1)^{2}} - \frac{x-(4/3)}{x^{2}+1}$$
but apparently, the last statement is not correct. Did I break down the rational function incorrectly?
Finding the constants is tedious and error-prone, but checking them is relatively simple, and you should do this. If you get a common denominator for what you have on the right, you should end up with what you have on the left. If you do, then you have the right constants.
stripes
I ended up finding the right constants, they were 1, -1, 1, and -1 I think. Did the question again and got it right!
And as for the Wolfram Alpha thing...I use it when I am completely stumped. It gives me hints and guides me in the right direction. And this "high school method" you speak of makes me sound like a high-schooler, which I am most definitely not lolll.
Staff Emeritus
Homework Helper
One technique I like to use when solving for the coefficients is the Heaviside coverup method. When you get to this point
$$x^{2}-2x-1 = A(x-1)(x^{2}+1) + B(x^{2}+1) + (Cx + D)(x-1)^{2}$$
Substitute a value for x to eliminate some terms. In this case, x=1 works. This leaves you with $-2 = 2B$, and you can immediately see that B=-1. Ideally, you can use different values to solve to isolate different coefficients quickly. Alas, that's not the case in this problem. If you plug in the value you found for B, you get
$$2x^{2}-2x = A(x-1)(x^{2}+1) + (Cx + D)(x-1)^{2}$$
so you only have three equations and three unknowns to solve.
Count Iblis
I see! Well, this is how I would do this problem. I would start by writing:
x^2 - 2 x - 1 = x^2 - 2 x + 1 - 2 = (x-1)^2 - 2
So, we have:
(x^2 - 2 x - 1)/[(x-1)^2 (x^2 + 1)] =
1/(x^2+1) - 2/[(x-1)^2 (x^2 + 1)]
Expanding the last term about the point x = 1 and keeping only the singular terms yields:
- 2/[(x-1)^2 (x^2 + 1)] = -2/(x-1)^2 [expansion of 1/(x^2+1) around x = 1]
Put x = 1 + t:
1/(x^2+1) = 1/[(1+t)^2 + 1] = 1/(2 + 2 t + t^2) =
1/2 1/[1+t+t^2/2] = 1/2 [1-t +t^2/2 + ...]
So, we have the expansion:
- 2/[(x-1)^2 (x^2 + 1)] =
-1/(x-1)^2 + 1/(x-1) + nonsingular terms
If we now do the same at the singularities of 1/(x^2+1) and keep the singular terms, then the sum of all the singular terms from all the expansions, S(x), is the partial fraction expansion. This is because the difference between the rational function R(x) and S(x) will be a function that has no singularities, therefore it is a polynomial. But since both R(x) and S(x) tend to zero at infinity, we have that
R(x) = S(x).
So, we could now proceed to find the singular terms in the expansion around the singularities of 1/(x^2+1) at x = ±i. However, we can skip that as the partial fraction expansion is already determined by the above terms. This is because the large x behavior of
- 2/[(x-1)^2 (x^2 + 1)]
is -2/x^4, while the part of the partial fracton expansion we have so far is
-1/(x-1)^2 + 1/(x-1)
So, clearly the partial fraction expansion due to the singular terms coming from 1/(x^2+1) must cancel the asymptotic 1/x and 1/x^2 behavior of the above two terms. We know that the the remaining terms of the partial fraction expansion will be of the form:
(A x + B)/(x^2 +1)
Expanding around x = infinity gives:
(A x + B) 1/x^2 1/(1+1/x^2) =
A/x + B/x^2 + O(1/x^3)
Expanding the two terms we got so far about infinity gives:
-1/(x-1)^2 + 1/(x-1) =
-1/x^2 + 1/x 1/(1 - 1/x) + O(1/x^3) =
-1/x^2 + 1/x + 1/x^2 + O(1/x^3) =
1/x + O(1/x^3)
So, A = -1 and B = 0 and we have:
(x^2 - 2 x - 1)/[(x-1)^2 (x^2 + 1)] =
(1-x)/(x^2+1) -1/(x-1)^2 + 1/(x-1)
stripes
count iblis, B ≠ 0, B = -1
Count Iblis
count iblis, B ≠ 0, B = -1
Your B is my 1/(x-1)^2 coefficient.
Staff Emeritus
Homework Helper
I see! Well, this is how I would do this problem.
I wonder what the grader would think if stripes were to turn in his homework using that solution.
Count Iblis
I wonder what the grader would think if stripes were to turn in his homework using that solution.
It's not that different from the Heaviside coverup method you mentioned. If we have a rational function R(x) which has a linear factor in the denominator 1/(x-a)^n, then we can take that factor out and write:
R(x) = P(x) 1/(x-a)^n
So, we see that P(a) is the coefficient of 1/(x-a)^n (to do this "by the book", you would write out the equations as you did and then put x = a). To find the coeficient of 1/(x-a)^(n-1), using the Heaviside coverup method, one would subtract p(a)/(x-a)^n from R(x), simplify the rational function; you know that the numerator will contain a factor x-a that cancels against the denominator (you can use synthetic division to divide the numerator by x - a), so we can write:
R2(x) = R(x) - p(a)/(x-a)^n = P2(x)/(x-a)^(n-1)
And then we can repeat this by subtracting P2(a)/(x-a)^(n-1) from R2(x), simplify the rational function, divide numerator and denominator by x-a to obtain:
R3(x) = R2(x) - p2(a)/(x-a)^(n-1) = P3(x)/(x-a)^(n-2)
etc. etc.
But this whole process is equivalent to finding the first part of the Laurent expansion of R(x) around x = a, so you could just as well do that using any method that is most convenient, not necessarily using the above iterative process...
Count Iblis
So, in the way you wrote it down the next step would be to factor the left hand side of the last equation
2x^2 - 2x = 2x(x-1)
and cancel (x-1) on both sides:
2x = A(x^2+1) + (Cx+D) (x-1) | 2023-03-21T02:02:52 | {
"domain": "physicsforums.com",
"url": "https://www.physicsforums.com/threads/integration-of-a-rational-function.409211/",
"openwebmath_score": 0.7525143027305603,
"openwebmath_perplexity": 721.8147524644543,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9706877667047451,
"lm_q2_score": 0.8824278571786139,
"lm_q1q2_score": 0.8565619259627626
} |
https://www.physicsforums.com/threads/definition-of-average-inclination.745604/ | # Definition of 'average inclination'?
1. Mar 27, 2014
### dilloncyh
It's not a homework problem, just a problem that suddenly popped out of my mind.
1. The problem statement, all variables and given/known data
So, my question is : how to calculate, or how to define 'average inclination'? Suppose I am given an equation y=f(x) that resembles the shape of a section of a hill, and I want to calculate the average inclination (something I always see when I watch cycling or alpine skiing race), how do I do that? Let's use f(x) = x^2 as an example for the following discussion.
2. Relevant equations
I know the average of a function is defined as:
So I suppose the correct equation should be similar, with f(x) the function that gives the shape of the hill I am calculating?
3. The attempt at a solution
Here comes the problem: Do I use dx for ds? And for (b-a) in the image, I need to replace it with the actual length of the function from x=a to x=b, right?
Since I don't really know I want to calculate (to find the slope at each point of the function and add them together, and then divide it by the total length of the slope?), my question may seem very silly, but please give me some idea.
thanks
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
2. Mar 27, 2014
### Simon Bridge
The average slope is usually just the rise-over-run for two positions on the slope.
$$\frac{f(x+\Delta x)-f(x)}{\Delta x}$$ ... that would be what they did on the ski races.
You want to be a bit more detailed than that:
If $y(x)$ is the height of the slope at position $x$,
Then $$\bar y = \frac{1}{b-a}\int_a^b y(x)\;\text{d}x$$ would be the average height of the hill in a<x<b.
The slope (gradient) function would be g(x)=dy/dx ... tells you the gradient at position x.
3. Mar 27, 2014
### dilloncyh
I'm still a bit confused.
average slope = detla y / delta x seems very legit, but take y=x^2 and y=x as example.
If I want to calculate the average slope between x=0 and x=1, then by calculating the difference in height and horizontal displacement of the two end points, then both should give the same result (slope=1), but the length of the curve of y=x^2 is obviously longer than y=x (which is just sq root of 2), so by calculating the the sine of the 'triangle' if I straighten the curve section of y=x^2, I will get difference result.
4. Mar 27, 2014
### haruspex
It's a matter of how you choose to define the average. The usual would be to define it as average over horizontal distance, but as you note you could instead chose to define it as average over path distance.
5. Mar 28, 2014
### Simon Bridge
I'm with haruspex - there is no reason that two different averaging methods should produce the same value.
They are just producing a different kind of average. | 2017-12-12T17:53:26 | {
"domain": "physicsforums.com",
"url": "https://www.physicsforums.com/threads/definition-of-average-inclination.745604/",
"openwebmath_score": 0.8125982284545898,
"openwebmath_perplexity": 566.2114124021376,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9621075744568837,
"lm_q2_score": 0.8902942290328344,
"lm_q1q2_score": 0.8565588212477415
} |
http://residuetheorem.com/2014/12/19/expansions-of-ex/ | ## Expansions of $e^x$
A very basic question was asked recently: What is a better approximation to $e^x$, the usual Taylor approximation, or a similar approximation involving $1/e^{-x}$?
More precisely, given an integer $m$, which is a better approximation to $e^x$:
$$f_1(x) = \sum_{k=0}^m \frac{x^k}{k!}$$
or
$$f_2(x) = \frac1{\displaystyle \sum_{k=0}^m \frac{(-1)^k x^k}{k!}}$$
The answer is amazingly simple: if $m$ is even, then $f_2$ is more accurate, while if $m$ is odd, $f_1$ is more accurate. The proof below is, in my opinion, really nifty and worth showing here. Keep in mind that this result follows from the fact that we are dealing with a remarkable function with very special properties. Other functions will not exhibit such easily calculable results as shown below. (Anyone familiar with Pade approximates can attest to how sticky rational approximations to functions can get.)
Keep in mind that, by “accuracy,” I refer to truncation error rather than roundoff error. Truncation error is that which results from substituting a Taylor series with a polynomial of finite order. The accuracy, i.e., truncation error, of any Taylor expansion of a given order is given by the next order of the expansion. For example, consider the first order approximation to $1/e^{-x}$, $f_2$, and its second-order error:
$$\frac1{1-x} = 1+x+x^2+O(x^3)$$
This approximation has twice the error as the first order approximation to $e^x$, $1+x$, whose second order term is $x^2/2!$.
Now consider the second order approximations with third order errors:
\begin{align}\frac1{1-x+x^2/2} &= 1+\left (x-\frac{x^2}{2} \right ) + \left (x-\frac{x^2}{2} \right )^2+ \left (x-\frac{x^2}{2} \right )^3 + \cdots \\&= 1+x+\frac{x^2}{2} + O(x^4)\end{align}
Note that, for the $f_2$ approximation, the third order error vanishes! This is obviously not the case for $f_1$. Thus, for the second-order expansion, $f_2$ has a smaller truncation error than $f_1$. However, for the next order approximation, the expansion of $f_2$ is
$$1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{12} + O(x^5)$$
whereas the fourth order term in the direct, Taylor expansion of the exponential is $x^4/24$. Thus, as with the first order approximation, $f_2$ has twice the truncation error as $f_1$.
There is a systematic way to prove that the accuracy alternates with order oddness or evenness by considering the error in the expansion of $e^x \cdot e^{-x}=1$. The proof is actually not all that hard. Consider the finite approximations to the equation $e^x \cdot e^{-x}=1$:
$$\left (\sum_{k=0}^m \frac{x^k}{k!} \right ) \left (\sum_{k=0}^m (-1)^k \frac{x^k}{k!} \right )$$
It may be easily shown that the coefficients of $x$, $x^2$, …, $x^m$ are zero. We now consider the first error term, i.e., the coefficient of $x^{m+1}$:
$$\frac{x^1}{1!} \frac{(-1)^m x^m}{m!} + \frac{x^2}{2!} \frac{(-1)^{m-1} x^{m-1}}{(m-1)!} +\cdots + \frac{x^m}{m!} \frac{(-1)^1 x^1}{1!} = \sum_{k=1}^m \frac{(-1)^{m-k+1}}{k! (m-k+1)!} x^{m+1}$$
Rewrite this sum, sans the $x^{m+1}$ term, as
$$\sum_{k=0}^{m-1} \frac{(-1)^{m-k}}{(k+1)! (m-k)!} = (-1)^m \sum_{k=0}^{m-1} \frac{(-1)^k}{k! (m-k)!} \frac1{k+1}$$
We can evaluate this sum by realizing that
$$(1-x)^m = \sum_{k=0}^m (-1)^k \binom{m}{k} x^k$$
so that
$$\sum_{k=0}^m (-1)^k \binom{m}{k} \frac1{k+1} = \int_0^1 dx \, (1-x)^m = \frac1{m+1}$$
Therefore, by subtracting off the last term in the sum, we find that
$$(-1)^m \sum_{k=0}^{m-1} \frac{(-1)^k}{k! (m-k)!} \frac1{k+1} = – \frac{1-(-1)^m}{(m+1)!}$$
Therefore, we now may say that
$$\frac1{\displaystyle \sum_{k=0}^m (-1)^k \frac{x^k}{k!} } = \sum_{k=0}^m \frac{x^k}{k!} + \frac{1-(-1)^m}{(m+1)!} x^{m+1} + O(x^{m+2})$$
For even values of $m$, the next term error is zero, which is smaller than that for the direct Taylor series, which has error $x^{m+1}/(m+1)!$. On the other hand, for odd values, the error is double that of the direct Taylor series. This agrees with the above examples.
#### One Comment
• This was indeed very surprising and beautiful (and at the same time not too difficult to understand). | 2017-09-26T09:08:09 | {
"domain": "residuetheorem.com",
"url": "http://residuetheorem.com/2014/12/19/expansions-of-ex/",
"openwebmath_score": 0.9731026291847229,
"openwebmath_perplexity": 224.60587648974763,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9920620054292071,
"lm_q2_score": 0.863391611731321,
"lm_q1q2_score": 0.8565380138049296
} |
http://math.stackexchange.com/questions/328349/probability-determining-which-phone-plan-is-better/329266 | # Probability: Determining Which Phone Plan Is Better
A consumer is trying to decide between two long-distance calling plans. The first one charges a flat rate of $10$ cents per minute, whereas the second charges a flat rate of $99$ cents for calls up to 20 minutes in duration and then $10$ cents for each additional minute exceeding 20 (assume that calls lasting a non-integer number of minutes are charged proportionately to a whole-minute’s charge). Suppose the consumer’s distribution of call duration is exponential with parameter.
• Which plan is better if expected call duration is 10 minutes? 15 minutes? [Hint: Let $h_1(x)$ denote the cost for the first plan when call duration is x minutes and let $h_2(x)$ be the cost function for the second plan. Give expressions for these two cost functions, and then determine the expected cost for each plan.]
For the first function of cost, I got $h_1(x) = 0.10X$; and for the second one, I got the piece-wise function $h_2(x) = .99$ for $0 \le X \le 20$, and $h_2(x) = .99$ for $X > 20$
Are these correct? Also, how do I calculate the expected cost for each plan? I used the formula given in this link http://en.wikipedia.org/wiki/Expected_value#Functional_non-invariance; however, it didn't work.
EDIT: $E[h_1(x)] = lim_{a \rightarrow \infty} \int^a_0 .10x \lambda e^{-\lambda x}dx$
After doing integration by parts, and simplifying, I get to the step $lim_{a \rightarrow \infty}[-.10ae^{-\lambda a} + \frac{.1}{\lambda}(e^{-\lambda a})]$. I know that the second term goes to zero, and, according to my teacher, the first term goes to zero as well, because $e^{-\lambda a}$ goes to zero more quickly than $-.10a$ goes to negative infinity, although I don't really understand why. Note, the answer is not zero
-
If the second plan charges $99$ cents plus an additional $10$ cents per minute after $20$ minutes, I don't think $h_2(x) = 0.99$ for $X > 20$ is correct. – TMM Mar 12 '13 at 11:32
@TMM Should it be $h_2(x) = .99 + 0.10X$ for $X > 20$? – Mack Mar 12 '13 at 11:55
@Eli No It is $0.99 + .1(X-20)$ – Gautam Shenoy Mar 12 '13 at 11:59
@GautamShenoy Are you sure? Can you explain why that is the answer? – Mack Mar 12 '13 at 12:12
@Eli: What I meant was: He says it is 10 cents for every minute exceeding 20. So in place of X, shouldn't it be X-20 ? It is not the answer. I am just correcting the expression you gave. – Gautam Shenoy Mar 13 '13 at 9:15
All costs are in dollars.
Your $h_1(X) = 0.1X$ is correct. The second term is $$h_2(X) = 0.99 + 0.1(X-20)1_{X > 20}$$
Now if you compute the expected cost, for $\frac{1}{\lambda} = 10$, it is
$\mathbb{E}h_1(X) = \frac{0.1}{\lambda}= 1$
But for $$\mathbb{E}h_2(X) = 0.99 + 0.1 \mathbb{E}(X-20)1_{X>20}$$ $$= .99 - 2\mathbb{P}(X>20) + .1\mathbb{E}(X1_{X>20})$$ $$= .99 - 2e^{-\lambda 20} + .1\mathbb{E}(X1_{X>20})$$ Thus we just need to compute $\mathbb{E}(X1_{X>20})$. You could do it directly, but I will provide an alternate way.
$$E[X1_{X>20}] = \int_0^\infty x1_{x>20}f_x(x)dx$$ $$= \int_{x=0}^\infty \int_{u=0}^\infty 1_{u<x}1_{x>20}f_x(x)dudx$$ $$= \int_{u=0}^\infty \int_{x=0}^\infty 1_{u<x}1_{x>20}f_x(x)dudx$$ (Fubini theorem) $$= \int_{u=0}^\infty \int_{x=u}^\infty 1_{x>20}f_x(x)dxdu$$ $$= \int_{u=0}^\infty \int_{x=\max(u,20)}^\infty f_x(x)dxdu$$ $$= \int_{u=0}^\infty P(X > \max(u,20))du$$
Now split the integral as $$= \int_{u=20}^\infty P(X > u)du + \int_{u=0}^{20}P(X > 20)du$$ $$= \int_{20}^\infty e^{-\lambda u}du + \int_{u=0}^{20}e^{-\lambda 20}du$$
I could work out the rest but can you proceed from here? Also in reply to your question: why is $\lim_{x \to \infty} xe^{-x} = 0$, do you know L-Hospital's rule?
- | 2014-10-02T08:25:11 | {
"domain": "stackexchange.com",
"url": "http://math.stackexchange.com/questions/328349/probability-determining-which-phone-plan-is-better/329266",
"openwebmath_score": 0.8398162722587585,
"openwebmath_perplexity": 412.85725692546174,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9838471647042428,
"lm_q2_score": 0.8705972650509008,
"lm_q1q2_score": 0.856534650819597
} |
http://math.stackexchange.com/questions/775121/how-to-prove-this-series-converges | # How to prove this series converges?
What test do you use to prove that
$$\sum_{n=2}^\infty \frac{\ln(n)}{n^{3/2}}$$
converges?
I tried the limit comparison test using $\frac{1}{n^{3/2}}$ as the comparison, but it did not converge.
-
Hint: You can use integral test.
Added: You need to consider the integral
$$\int_{2}^{\infty} \frac{\ln(x)}{x^{3/2}}dx ,$$
which can be integrated using integration by parts.
-
@MhenniBenghorbal, there is a typo, the lower bound should be 2, but it doesn't seem like it's going to matter that much for this question. – Nameless Apr 30 '14 at 2:47
@Nameless: It is corrected. Thanks for the comment. – Mhenni Benghorbal Apr 30 '14 at 2:48
Using the integral test I got the answer to be = 25.55275762. – khap93 Apr 30 '14 at 2:49
@khap93, you are not supposed to directly calculate the sum...well I guess that shows you the sum does converge. – Nameless Apr 30 '14 at 2:49
@MhenniBenghorbal, no problem. I'll upvote you. – Nameless Apr 30 '14 at 2:50
Apply Cauchy Condensation Test.
Let $a_n = \dfrac{\ln(n)}{n^{3/2}}$, then $a_{2^n} = \dfrac{\ln 2^n }{(2^n)^{3/2}}$
Therefore, $$a_{2^n} = \dfrac{\ln 2^n }{(2^n)^{3/2}} = \frac{n\ln 2 }{2^{3n/2}} \leq \ln2\frac{n}{2^n}$$
Thus $$\sum_{n=2}^{\infty}\dfrac{\ln 2^n }{(2^n)^{3/2}} \leq \sum_{n=2}^{\infty} \ln2\frac{n}{2^n}.$$
The RHS is a convergent series, by the Ratio Test, as you should verify.
-
Neat footwork there :-) – Carl Witthoft Apr 30 '14 at 13:17
You can use that $\ln n$ goes to infinitely much more slowly that any (positive) power of $n$, that is: for every $\alpha>0$, $\ln n < n^\alpha$ for $n$ large enough. You can apply this to $\alpha=1/4$ says (any $\alpha$ strictly between $0$ and $1/2$ will do, so let's say $1/4$), so $\ln n < n^{1/4}$ for $n$ large enough, hence $\ln n / n^{3/2} < 1/n^{5/4}$ for $n$ large enough, and since the series $\sum 1/n^{5/4}$ converges (as does $\sum 1/n^\beta$ for any $\beta>1$), you're done.
-
+1 the most direct and conceptually simple solution. – jwg Apr 30 '14 at 10:19 | 2015-07-07T03:12:51 | {
"domain": "stackexchange.com",
"url": "http://math.stackexchange.com/questions/775121/how-to-prove-this-series-converges",
"openwebmath_score": 0.9863249063491821,
"openwebmath_perplexity": 402.5900040540244,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9838471661250913,
"lm_q2_score": 0.8705972633721708,
"lm_q1q2_score": 0.85653465040497
} |
https://web2.0calc.com/questions/help-me-please_130 | +0
0
330
8
Let $b$ and $c$ be constants such that the quadratic $-2x^2 +bx +c$ has roots $3+\sqrt{5}$ and $3-\sqrt{5}$. Find the vertex of the graph of the equation $y=-2x^2 + bx + c$.
Guest Feb 19, 2018
#1
+12560
+2
There MIGHT (probably) be an easier way, but here is ONE way:
(x+(-3-sqrt5))(x+(-3+sqrt5)) is given in the question for the roots
if you multiply these two term out you will get
x^2-6x+4 now multiply by -2
-2x^2 +12x-8 =0 (a parabola)
to find vertex take derivative and set equal to zero
-4x+12 = 0 x = 3 substitue this into the equation
to get y=10 for the vertex 3,10
ElectricPavlov Feb 19, 2018
edited by ElectricPavlov Feb 19, 2018
#3
0
Should it be (x+(3-sqrt 5))(x+(3+sqrt5))?
Guest Feb 21, 2018
#4
0
Shouldnt it be (-3,10)?
Guest Feb 21, 2018
#5
+12560
+2
The roots are given as 3+-sqrt5
so one part would be
x+(- 3+sqrt5) =0
x-3+sqrt5 = 0 solve for x (add (3 and subtract sqrt5) from both sides )
x= 3 -sqrt5
and the other would be x+( -3-sqrt5) =0
x-3-sqrt5 solve for x (add 3 and sqrt 5 to both sides
x = 3+sqrt5
These are the roots given in the question....
ElectricPavlov Feb 21, 2018
#6
+12560
+2
Here is a graph
ElectricPavlov Feb 21, 2018
#7
+12560
+2
y= a (x-h)^2 +k and
y= -2 (x-3)^2 +10
Now can you see that a =-2 h=3 k = 10 ???
ElectricPavlov Feb 21, 2018
#2
+12560
+2
OK.....another way...
AFTER you get to
-2x^2 +12x -8 arrange into vertex form y=a(x-h)^2 +k the vertex is h,k
-2{ x^2 -6x +4}
-2{(x-3)^2 +4 -9}
-2 {(x-3)^2 -5}
-2 (x-3)^2 +10 so h,k is the vertex = 3,10
ElectricPavlov Feb 19, 2018
#8
+92805
+1
I am quite sure EP will be correct but I will take a look:
Let b and c be constants such that the quadratic $$-2x^2 +bx +c$$
has roots $$3+\sqrt{5}$$ and $$3-\sqrt{5}$$.
Find the vertex of the graph of the equation
$$y=-2x^2 + bx + c$$
The roots of a quadratic $$ax^2+bx+c$$ is the answers to $$ax^2+bx+c=0$$
$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ x=\frac{-b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a}=3\pm\sqrt5$$
This means that $$x=\frac{-b}{2a}$$ = 3 must be exactly halfway between the two roots.!!
So the axis of symmetry is x=3 and the vertex lies on this line so the x value of the vertex is 3
The y value will be $$y=-2*3^2+bx+c = -18+3b+c$$
This is fine but you need to find the vleu of b and c
From the equation above I can see that
$$\frac{-b}{2a}=3 \qquad and \qquad \pm\frac{\sqrt{b^2-4ac}}{2a}=\pm\sqrt5\\ a=-2\qquad so\\ \frac{-b}{-4}=3 \qquad and \qquad \pm\frac{\sqrt{b^2+8c}}{-4}=\pm\sqrt5\\ b=12 \qquad and \qquad \pm\frac{\sqrt{b^2+8c}}{4}=\pm\sqrt5\\ b=12 \qquad and \qquad \frac{b^2+8c}{16}=5\\ b=12 \qquad and \qquad b^2+8c=80\\ b=12 \qquad and \qquad 144+8c=80\\ b=12 \qquad and \qquad 18+c=10\\ b=12 \qquad and \qquad c=-8\\$$
so
$$y= -18+3b+c\\ y= -18+3*12+-8\\ y=-18+36-8\\ y=10$$
So the vertex is (3,10)
Which is exactly the same answer has given you. He has done it a number of different ways but stlill all the answers are the same!!
Thanks EP :)
Melody Feb 21, 2018 | 2018-07-21T11:59:09 | {
"domain": "0calc.com",
"url": "https://web2.0calc.com/questions/help-me-please_130",
"openwebmath_score": 0.809325635433197,
"openwebmath_perplexity": 7890.315983481394,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9941800407606186,
"lm_q2_score": 0.8615382129861583,
"lm_q1q2_score": 0.8565240957034094
} |
https://math.stackexchange.com/questions/38739/convergence-of-a-n1-frac11a-n | # Convergence of $a_{n+1}=\frac{1}{1+a_n}$
We define
$$a_{n+1}=\frac{1}{1+a_n}, a_0=c> 0$$
I stumbled across that sequence and Mathematica gives me that it converges against $\frac{1}{2} \left(\sqrt{5}-1\right)$ which doesn't even depend on $a_0$. Normally I show that sequences converge by seeing that they are monotonic and then the limit can be easily found by setting $a_n=a_{n+1}$, but this one seems to be alternating. Also by checking OEIS I noticed that $a_n=1/g(n)$ where $g(n)$ gives the $n+1$'th Golden Rectangle Number.
I would be glad if someone can show me how to show that such sequences converge and how to find the limit, also maybe this is a well known sequence, as it has a quite simple form, too bad that google is very bad for looking for sequences and OEIS doesn't mention anything.
• These numbers are the convergents of the continued fraction of the golden ratio. If you assume that it converges you can calculate the limit by replacing the $a_k$s by $a$. You can prove convergence by proving that the odd/even subsequences are monotonous. May 12 '11 at 18:33
• Right, but 3123 is asking why $\frac{\sqrt{5}-1}{2}$ is a(n attractive) fixed point of the function $f(x)=\frac1{a+x}$ for "any" starting value... May 12 '11 at 18:37
• The reason why $a_n$ converges to a fixed value irrespective of $a_0 = c$ is actually not really surprising. For instance, consider $a_0 = \frac{c}{10}$ where $c \in [0,10]$ $a_{n+1} = \frac{9+a_n}{10}$ Note that $\displaystyle \lim_{n\rightarrow \infty} a_n = 1$ for any starting value $c \in [0,10]$
– user17762
May 12 '11 at 18:50
• May 13 '11 at 11:18
• math.stackexchange.com/questions/235578/… Apr 20 '13 at 13:19
## 3 Answers
The function $1/(1+x)$ is strictly decreasing and thus order-reversing on the positive reals. If you apply it twice it conserves order.
Therefore, the odd and even subsequences are monotonic. They are also bounded as all your values except the first are bounded by 1.
Now you can just replace $a_n$ and $a_{n+1}$ by $a$ to find the limit.
• Thank you for your answer, this was very useful too. May 12 '11 at 18:45
As for why the first term does not matter, consider the $6$th term,
$$\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+c}}}}}$$
as $c$ varies between $0$ and $\infty$ the value of the $6$th term varies only by $0.1$. As $c$ varies between $0$ and $\infty$ the $100$th term varies only by very small $\varepsilon$. In the infinite limit this variation is zero.
This also explains convergence. Since convergence occurs call the number $\varphi$, it satisfies the fixed point equation
$\varphi = \frac{1}{1+\varphi}$
which, multiplying both sides by the denominator, is a quadratic equation hence the square root.
• Thank you for the answer, writing it as a continued fraction makes things much more obvious May 12 '11 at 18:44
For $x\geq 0$, $f(x)=1/(1+x)$ is a monotonically decreasing function, bounded between 0 and 1. The fixed point (if it exists) of the iterative map defined by your function, is the solution of $x=\frac{1}{1+x}$. There is only one solution for $x\geq 0$, and that is $\tilde{x}=(\sqrt{5}-1)/2$.
If we look at the evolution of $a_n=f(f(f(...(a_0))))$, we get
$$a_n=\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+a_0}}}}}$$
and as $n\to\infty$, what $a_0$ was becomes irrelevant. In other words, no matter the value of $a_0$, it converges to a fixed point, which has to be the only fixed point, $\tilde{x}$. Hence $\tilde{x}$ is an attractive fixed point and also the limit of the sequence.
The Mathematica code that produced the above figure is below (it's modified from an answer to a question on SO)
Clear[fixedPoint]
fixedPoint[a0_, nSteps_] :=
Module[{f, a, n},
f = RecurrenceTable[{a[n] == 1/(1 + a[n - 1]), a[1] == a0},
a, {n, 1, nSteps}];
Plot[{1/(1 + x), x}, {x, 0, 1},
PlotStyle -> {Darker@Blue, {Dashed, Black}},
Epilog -> {Darker@Green,
Line[Riffle[Partition[f, 2, 1], {#, #} & /@ Rest[f]]]}]] | 2021-12-08T15:39:53 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/38739/convergence-of-a-n1-frac11a-n",
"openwebmath_score": 0.8821293115615845,
"openwebmath_perplexity": 185.79790957227806,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9724147185726375,
"lm_q2_score": 0.8807970811069351,
"lm_q1q2_score": 0.8565000457442009
} |
https://stats.stackexchange.com/questions/409226/why-is-this-probability-distribution-biased-towards-even-numbers | # Why is this probability distribution biased towards even numbers?
I am working on the probabilities of a game where
1. they pick 20 unique number(non-repeating) from range 1 to 80
2. sort those 20 numbers in ascending order
3. sum every 4 numbers and take the last digit as result (so end with 5 numbers).
Normally I could find these probabilities with rules based on combination and permutation but for this one because of the sorting I am not sure how to work on it.
I tried to brute force all combinations but with some test it would take my old laptop about few thousand years. So I did some simulations, sampling 100 million samples. Below is the final result.
So there seems to be a small edge towards even numbers, but I don't understand where this edge comes from (even numbers appear about .5095925 of the time).
I went further down about the first 4 numbers out of the sorted 20 numbers and try based on the fact that the 1st sorted number must be in the range of 1 to 61, 2nd sorted must be in the range of 2 to 62, etc. So odd positions will have 31 out of 61 chances to be odd and even positions have 31 out of 61 to be even. And I did some checks below, but the resulting bias towards even numbers is very small (.50000004 vs the stimulation .5095925).
I wonder where and how the game tends to favor even numbers quite a bit. Or is my simulation code just wrong?
• +1 For a quick insight into this, think about the results you would get if you were to select 76 numbers rather than 20. It will help to reason in terms of parity: how many odd numbers do you expect to see in each group of four numbers?
– whuber
May 20 '19 at 15:14
• @whuber you mean because there are 40 odd and 40 even, the chance of getting 10 odd and 10 even for the 20 number is higher, then the chance of getting 2 odds and 2 evens (result in even) is higher in a 4 number group base? May 21 '19 at 9:14
• No, that's not it. You need to look at the chances, within any given group, of observing an even number of odd values, because that characterizes the cases where the total will end in 0, 2, 4, 6, or 8. For instance, any group of four numbers in a row is guaranteed to have two odd and two even values and therefore its sum cannot end in an odd digit.
– whuber
May 21 '19 at 12:00
• Your results look good. In particular, the first-number percentages for digits 0 - 5 are typically correct in the first four or five significant digits.
– whuber
May 23 '19 at 17:16
This is an interesting question because the result is a little surprising; it is subtle; and it has been well researched with a large simulation.
Here is a graphical representation of ten iterations of the game. Each row is an iteration. It shows the 20 numbers selected, $$k_1,k_2, \ldots, k_{20},$$ and uses colors to depict their partition into five groups of four. The game sums the numbers in each group, thereby producing five sums.
Nevertheless, within the large number of simulations reported in the question, there appears to be an imbalance among odd and even sums, although this procedure seemingly would not favor one parity. Indeed, if we were only to take a subset of four numbers from $$\{1,2,3,\ldots, 80\}$$ and sum them modulo ten, each digit would have exactly $$1/10$$ chance of appearing.
To appreciate what's happening, consider extreme versions of this procedure, where instead of partitioning a sample of $$20$$ into five groups of four, let's take a sample of $$76$$ and partition it into $$19$$ groups of four:
What strikes us first about this image are the prominent gray "bubbles" depicting the numbers that were not sampled. With only $$80-76=4$$ bubbles in each iteration, we're unlikely to find any bubbles in the first group of four. Moreover, when there is a bubble in that group, as in the fourth row from the top, it's just as likely to be an even number as an odd number. That means the first group (in this extreme case) is highly likely to include two odd numbers and two even numbers, making its sum even.
This consideration of bubbles suggests analyzing the game in terms of gaps between the selected numbers. The first gap is $$\Delta_1 = k_1-0,$$ the second gap is $$\Delta_2 = k_2-k_1,$$ and so on. All gaps are $$1$$ or larger. Evidently the last number chosen is $$k_{20} = \Delta_1+\Delta_2 + \cdots + \Delta_{20},$$ which therefore must be $$80$$ or smaller.
Let's begin with the first gap, $$\Delta_1=k_1.$$ What is the chance this gap is only $$1,$$ the smallest possible value? That's easy to compute, because
1. There are $$\binom{80}{20}$$ distinct, equally-probable subsamples of size $$20$$ that can be obtained from all $$80$$ numbers.
2. Of these, $$\binom{79}{20}$$ are from the set $$\{2,3,\ldots, 80\}.$$
Subtracting, we find the number of subsamples with $$k_1=1$$ and divide that by the total number of subsamples to obtain the probability:
$$\Pr(\Delta_1 = 1) = \frac{\binom{80}{20} - \binom{79}{20}}{\binom{80}{20}} = \frac{\binom{79}{19}}{\binom{80}{20}} = \frac{1}{4}.$$
(Although only one of the ten iterations in the first figure has $$\Delta_1=1$$--the fifth row from the top--a longer set of simulations confirms that $$\Delta_1=1$$ in about a quarter of them.)
Analogous reasoning establishes the remaining chances:
$$\Pr(\Delta_1 = j_1) = \frac{\binom{80+1-j_1}{20} - \binom{80-j_1}{20}}{\binom{80}{20}} = \frac{\binom{80-j_1}{19}}{\binom{80}{20}},\ j_1=1, 2, \ldots, 61.$$
Now suppose we have observed $$\Delta_1.$$ What could $$\Delta_2$$ be? If it is equal to $$j_2,$$ that means the last $$20-1$$ numbers are a subset of the values from $$\Delta_1+j_2$$ through $$80$$ and the remaining $$20-2$$ numbers all lie between $$\Delta_1 + j_2 +1$$ and $$80.$$ In effect, the role of $$80$$ is now played by $$80-j_1,$$ $$20$$ is reduced to $$20-1=1,$$ and $$j_2$$ plays the role of $$j_1.$$ Thus, writing $$k_2=j_1+j_2$$ for the second lowest value selected,
\eqalign{ \Pr(k_2\text{ selected next}\mid k_1\text{ selected first}) &= \Pr(\Delta_2 = j_2 \mid \Delta_1=j_1) \\ &= \frac{\binom{80-k_2}{20-2}}{\binom{80-k_1}{20-1}},\ k_2=j_1+1, 2, \ldots, 80+1-(20-1).}
The pattern is clear. In particular, to study the first group, we can calculate the probability distribution of the sum of just the lowest four numbers in the subsample by summing over all possible values of $$\Delta_1,\Delta_2,\Delta_3,\Delta_4.$$ There are sufficiently few of these that this computation is feasible.
By the laws of conditional probability, the chance of such a combination $$(j_1,j_2,j_3,j_4)$$ is obtained by the products of the component probabilities
\eqalign{ q_{j_4,j_3,j_2,j_1} &= \Pr(\Delta_4=j_4\mid \Delta_3=j_3,\Delta_2=j_2,\Delta_1=j_1) \\ &\times \Pr(\Delta_3=j_3\mid \Delta_2=j_2,\Delta_1=j_1) \\ &\times \Pr(\Delta_2=j_2\mid \Delta_1=j_1) \\ &\times \Pr(\Delta_1=j_1)}
as
$$\Pr(j_1,j_2,j_3,j_4) = \sum_{j_1=1}^{81-20}\ \sum_{j_2=1}^{81-j_1-19}\ \sum_{j_3=1}^{81-j_1-j_2-18}\ \sum_{j_4=1}^{81-j_1-j_2-j_3-17} q_{j_4,j_3,j_2,j_1} .$$
For the game described in the question, there are "only" 635,376 possible combinations in this quadruple sum: the calculations can be carried out in seconds. Here is the interesting part of the probability distribution of the group sum $$k_1+k_2+k_3+k_4 = 4j_1+3j_2+2j_3+j_1:$$
(The full distribution extends to the largest possible sum that could appear in this first group of four numbers, equal to $$61+62+63+64=250.$$)
The solid red dots are probabilities a little larger than suggested by their neighbors: the alternation between even and odd is clear.
When we further reduce this distribution by taking just the last digits of the sums, the alternation between even and odd persists:
This plot shows the relative differences between the probabilities and $$1/10.$$ The actual values (to eight digits) are
Digit Probability
0 0.10235369
1 0.09779473
2 0.10159497
3 0.09859690
4 0.10239713
5 0.09838777
6 0.10201990
7 0.09821967
8 0.10121774
9 0.09741751
Many of them agree closely (to four or five digits) with the simulation results reported in the question.
Notice (from the equations or the simulations) that the first group of four smallest numbers tends to lie to the left of $$20$$ or so, leaving about $$60$$ values from which to select the next four groups. Thus, the situation with the second group is typically just like the situation with five groups, but with $$80$$ reduced to around $$60$$ and only $$20-4=16$$ numbers to select. Similar reasoning suggests the situation with the third and fourth groups also is similar. The simulations in the question bear this out: they still favor even sums over odd.
Analogous discrepancies in probabilities occur when representing the sums in other (non-decimal) bases. Different patterns of discrepancies occur when taking groups of other sizes, especially odd sizes: groups of size $$4$$ are special because they have a pronounced tendency to include either zero, two, or four odd numbers, which guarantees their sum is even.
Here is the R used to do the calculations. It is coded so you can experiment with versions of this game simply by changing the parameters n, m, and g.
library(data.table)
n <- 80 # Maximum number available (100 or more may be problematic)
m <- 4 # Group sizes >= 2 (8 or more is problematic)
g <- 5 # Number of groups >= 1 (more is easier!)
#
# Generate all possible vectors of gaps.
#
delta <- 1:(n - m*(g-1) + 1)
X <- do.call("expand.grid", lapply(1:m, function(i) delta))
names(X) <- paste0("k", 1:m)
X <- as.data.table(X)
i <- rowSums(X) < max(delta)
X <- X[i, ]
#
# Generate the numbers corresponding to each gap vector.
#
Y <- apply(X, 1, cumsum)
rownames(Y) <- paste0("k", 1:m)
Y <- as.data.table(t(Y))
#
# Compute log conditional probabilities for each selection.
#
f <- function(k, n, g) {
m <- length(k)
k0 <- c(0, k[-m])
j <- (m*g-1) : (m*(g-1))
lchoose(n - k, j) - lchoose(n - k0, j+1)
}
Z <- apply(Y, 1, function(k) f(k, n, g))
rownames(Z) <- paste0("p", 1:m)
Z <- as.data.table(t(Z))
#
# Compute the probability of each vector.
#
Z[, pi := exp(rowSums(Z))]
Y[, Sum := rowSums(Y)]
X <- cbind(X, Y, Z)
1-sum(X\$pi) # Should be near zero, within floating point roundoff error
#
# Summarize by group sum.
#
H <- X[, .(Probability = sum(pi)), keyby=Sum]
H.mod <- rbindlist(lapply(2:40, function(b)
H[, .(Probability = sum(Probability), Base=b),
keyby=(Sum - choose(m+1,2)) %% b]))
with(H[Probability > 0.00025], {
above <- sapply(1:length(Sum), function(i) {
g <- splinefun(Sum[-i], Probability[-i])
Probability[i] > g(Sum[i])
})
colors <- ifelse(above, "Red", "Gray")
plot(Sum, Probability, type="l")
abline(h=0)
abline(v = seq(0, max(Sum), by=5), col="Gray", lty=3)
abline(v = seq(0, max(Sum), by=10), col="Gray")
points(Sum, Probability, pch=21, bg=colors)
})
#
# Explore the last digit in various bases.
#
base <- 10
with(H.mod[Base==base], {
plot(Sum, base*Probability - 1, type="b", ylim=base*range(Probability)-1,
pch=21, bg=ifelse(base*Probability-1 > 0, "Red", "Gray"),
xlab=paste("Sum modulo", base, "offset by", choose(m+1,2)))
abline(h=0, col="Gray")
})
• I am moved by the details and visualization that you offer. There is still much for me to dig into the step by step forming of the formula and the code but first and formal let me thank you for this amazing answer. May 24 '19 at 21:23 | 2021-10-15T21:23:35 | {
"domain": "stackexchange.com",
"url": "https://stats.stackexchange.com/questions/409226/why-is-this-probability-distribution-biased-towards-even-numbers",
"openwebmath_score": 0.8171984553337097,
"openwebmath_perplexity": 773.2290919784616,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9724147209709196,
"lm_q2_score": 0.8807970779778824,
"lm_q1q2_score": 0.8565000448138639
} |
https://math.stackexchange.com/questions/3049117/evaluating-lim-limits-x-to-infty-frac4x23x4x-2 | # Evaluating $\lim\limits_{x \to \infty}\frac{4^{x+2}+3^x}{4^{x-2}}$
$$\lim_{x→∞}\frac{4^{x+2}+3^x}{4^{x-2}}.$$
I have solved it like below: $$\lim_{x→∞}\left(\frac{4^{x+2}}{4^{x-2}}+\frac{3^x}{4^{x-2}}\right)=\lim_{x→∞}\left(4^4+\frac{3^x}{4^x}·4^2\right).$$ Since, as $$x → ∞$$, $$3^x → ∞$$, $$\dfrac{3^x}{4^x} → 0$$, the limit is equal to $$4^4=256$$.
Have I solved it correctly?
This was a practice test question and the given solution was wrong. So, I solved it and I am preparing alone, no friend to discuss, so I posted it here.
• $\lim_{x \to \infty} \left( 4^4+\frac{3^x}{4^x} \times 4^2 \right)=\lim_{x \to \infty}4^4 +\lim_{x \to \infty}\left(\dfrac{3^x}{4^x}\right)4^2$? While its true you might want to explain for rigor. – Yadati Kiran Dec 22 '18 at 4:26
• It's unclear to me what purpose it serves to write $3^x\to\infty$. It makes the next expression, $\frac{3^x}{4^x} \to 0$, look wrong even though it actually is correct. – David K Jan 8 at 14:28
Yup! Your steps almost all follow logically and evaluate to the correct limit, so your solution is correct (almost).
I will make a nitpick with one thing you said, though:
Since, as $$x \to \infty, 3^x \to \infty, \frac{3^x}{4^x} \to 0$$
Technically, since $$3^x$$ and $$4^x$$ both approach $$\infty$$ as $$x \to \infty$$, then under your logic we obtain an indeterminate form:
$$\frac{3^x}{4^x} \to \frac{\infty}{\infty}$$
It would be better to regroup $$3^x/4^x$$ as $$(3/4)^x$$. Then clearly, as $$x \to \infty$$, $$(3/4)^x \to 0$$ because $$3/4 < 1$$. (If you're not convinced, notice if you take $$x = 1, 2, 3, 4$$ and so on that $$(3/4)^x$$ clearly is decreasing.)
• While $$3^x \to \infty$$ is true, note that it is not used in the proof though. | 2019-05-25T09:57:46 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/3049117/evaluating-lim-limits-x-to-infty-frac4x23x4x-2",
"openwebmath_score": 0.8586853742599487,
"openwebmath_perplexity": 397.8554211882069,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9724147193720649,
"lm_q2_score": 0.8807970748488297,
"lm_q1q2_score": 0.8565000403628604
} |
https://mathhelpboards.com/threads/minimizing-the-sop.8859/ | # Minimizing the SOP
#### shamieh
##### Active member
I used a KMAP and got this as my expression
$$f = !y_{1}!y_{0} + !x_{1}x_{0}!y_{1} + x_{1}!y_{1} + x_{1}!y_{0} + x_{1}x_{0}y_{1}$$
Is there any way I can minimize this? Maybe I'm just not seeing it. I thought the whole point of a KMAP was to minimize the expression?
Some how the answer is this: $$f = x_{1}x_{0} + !y_{1}!y_{0} + x_{1}!y_{0} + x_{0}!y_{1} + x_{1}!y_{1}$$
If anyone is interested I had to design a circuit with output f with 4 inputs. I'm supposed to be showing the simplest sum of product expression for f. My f row for my truth table was this:
1
0
0
0
1
1
0
0
1
1
1
0
1
1
1
1
#### Evgeny.Makarov
##### Well-known member
MHB Math Scholar
I used a KMAP and got this as my expression
$$f = !y_{1}!y_{0} + !x_{1}x_{0}!y_{1} + x_{1}!y_{1} + x_{1}!y_{0} + x_{1}x_{0}y_{1}$$
...
Some how the answer is this: $$f = x_{1}x_{0} + !y_{1}!y_{0} + x_{1}!y_{0} + x_{0}!y_{1} + x_{1}!y_{1}$$
You can turn the three-variable minterms into two-variable ones, namely, remove $!x_1$ from $x_0!x_1!y_1$ and $y_1$ from $x_0x_1y_1$.
#### shamieh
##### Active member
You can turn the three-variable minterms into two-variable ones, namely, remove $!x_1$ from $x_0!x_1!y_1$ and $y_1$ from $x_0x_1y_1$.
Well I can't combine them because they don't differ by two variables correct? So what minimization "tool" should I use? Should I factor something? I mean how do you just "get rid of them". See what I'm saying? What method I should I be using? Sorry if I sound ignorant.
#### Evgeny.Makarov
##### Well-known member
MHB Math Scholar
When you remove a variable from a three-variable minterm, its representation in the Karnaugh map grows from 2 to 4 cells. However, if the added two cells are already covered by other minterms, then the Boolean function does not change.
In this case, the minterm $x_0y_1y_1$ represents two cells in the middle of column 3 ($x_0=x_1=1$). When you remove $y_1$, the result is the complete column 3. But the top cell of column 3 is already covered by $!y_0!y_1$, and the bottom cell of column 3 is covered by $x_1!y_1$. So removing $y_1$ from $x_0y_1y_1$ does not change the function. A similar thing happens with turning $x_0!x_1!y_1$ into $x_0!y_1$.
When reading off a minimal formula from a Karnaugh map, the temptation is always to break the cells corresponding to 1 into disjoint regions. But this results in smaller regions and therefore larger minterms. Instead, one must make regions as large as possible by using the fact that overlap is allowed.
#### shamieh
##### Active member
Awesome explanation! Thanks so much. So it looks like I probably missed a overlapping grouping I could of put together then - thus getting not exactly the complete minimization. | 2021-04-15T11:19:17 | {
"domain": "mathhelpboards.com",
"url": "https://mathhelpboards.com/threads/minimizing-the-sop.8859/",
"openwebmath_score": 0.8333398103713989,
"openwebmath_perplexity": 650.1192858562753,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.9724147153749275,
"lm_q2_score": 0.8807970670261976,
"lm_q1q2_score": 0.856500029235351
} |
https://stats.stackexchange.com/questions/587912/continuous-and-differentiable-bell-shaped-distribution-on-a-b/587918 | # Continuous and differentiable bell-shaped distribution on $[a, b]$
Is there a distribution with these three properties?
• Supported on $$[a, b]$$ for any $$a, b\in\mathbb{R}$$ and $$a < b$$
• Continuous and Differentiable (that is $$\nabla_x p(x)$$ can be computed)
• Bell-curved (or at least not as flat as the Uniform(a, b))
• stats.stackexchange.com/questions/500862/…
– whuber
Sep 6 at 15:05
• General answer: let $f:[0,1]\to\mathbb{R}\cup\{\infty\}$ be any nonnegative integrable function that increases on the interval $[0,c)$ and decreases on $(c,1].$ On the interval $[a,b]$ define $$F_0(x) = \int_a^x f((t-a)/b)\,\mathrm{d}t.$$ Extend the function $F(x) = F_0(x)/F_0(b)$ to be $0$ on $(-\infty, a)$ and $1$ on $(b,\infty).$ By construction this has all the features you seek and every solution can be expressed in this form. The extreme generality of these solutions shows that you should be more narrowly focused on a solution that is meaningful for whatever application you have in mind.
– whuber
Sep 6 at 15:20
• @whuber +1. That indeed should be left as an answer; after all, it's a general construction. Sep 6 at 15:46
• @whuber great answer! Do you mind posting that as an answer? Sep 6 at 16:13
Let's construct all possible solutions.
By "distribution" you appear to refer to a density function (PDF) $$f.$$ The properties you require are
1. Supported on $$[a,b].$$ That is, $$f(x)=0$$ for any $$x\le a$$ or $$x \ge b.$$
2. $$f$$ should be (continuously) differentiable on $$(a,b)$$ with derivative $$f^\prime.$$
3. "Bell-curved," which can be taken as
• (Strictly) increasing from value of $$0$$ at $$a$$ to some intermediate point $$c;$$ that is, $$f^\prime(x) \gt 0$$ for $$a\lt x \lt c;$$ and
• (Strictly) decreasing from $$c$$ to a value of $$0$$ at $$b;$$ that is, $$f^\prime(x) \lt 0$$ for $$c \lt x \lt b.$$
To standardize this description, translate and scale the left-hand arm of $$f^\prime$$ to the interval $$[0,1]$$ and do the same for the right-hand arm, reversing and negating it. That is, let
$$g_{-}(x) = f^\prime\left((c-a)x\right)$$
and
$$g_{+}(x) = -f^\prime\left(1 - (b-c)x\right).$$
Both are increasing positive integrable functions defined on $$[0,1]$$ for which
$$\int_0^1 g_{-}(x)\,\mathrm{d}x = \int_0^1 f^\prime((c-a)x)\,\mathrm{d}x = \frac{1}{c-a}\int_a^c f^\prime(y)\,\mathrm{d}y = \frac{f(c^-)}{c-a}$$
and
$$\int_0^1 g_{+}(x)\,\mathrm{d}x = \int_0^1 -f^\prime(1 - (b-c)x)\,\mathrm{d}x = \frac{1}{b-c}\int_b^c f^\prime(y)\,\mathrm{d}y = \frac{f(c^+)}{b-c}.$$
Conversely, given any two positive increasing integrable functions $$g_{-}$$ and $$g_{+}$$ defined on $$[0,1]$$ (the "left side" and "right side" models), these steps can be reversed to construct $$f^\prime,$$ which in turn can be integrated (and normalized) to yield a valid distribution function.
Here is this reverse process in pictures. It begins with the two model functions.
(Notice that these functions need not even be continuous and can be unbounded, as illustrated by $$g_{+}$$ at the right.)
They are then integrated and assembled to produce $$f^\prime,$$ which in turn is integrated and normalized to unit area to yield a density $$f$$ with every required characteristic.
You may further control the appearance of the density in many ways. For instance, by taking the two models to be the same function and placing the peak $$c = (a+b)/2$$ at the midpoint, you will obtain a symmetric density. Here I have used the original $$g_{-}$$ for the right hand model $$g_{+}.$$
You can enforce many other properties of $$f$$ by going through the original analysis to deduce the corresponding properties of the model functions and restricting your construction to functions of that type.
Finally, if you choose to limit the two model functions to a finitely parameterized subset of the possibilities, you will have constructed a parametric family of distributions meeting all your criteria.
The Truncated normal distribution obeys all prerequisites:
• It's bell shaped
• It's continuous
• Its support is $$x \in [a,b]$$
• It's differentiable, i.e. $$\nabla_x p(x)$$ exists for all $$x \in [a,b]$$
• I just added a detail that I forgot about. I need it to be differentiable Sep 6 at 15:13
• @Euler_Salter no problem, the truncated normal distribution is also differentiable. Sep 6 at 15:37
One option is to transform a beta distribution.
$$Beta(3,3)$$ has your desired properties on $$[0,1]$$.
Now subtract $$1/2$$ to center the distribution.
Next, multiply to stretch or compress the distribution. | 2022-12-09T16:02:33 | {
"domain": "stackexchange.com",
"url": "https://stats.stackexchange.com/questions/587912/continuous-and-differentiable-bell-shaped-distribution-on-a-b/587918",
"openwebmath_score": 0.8641833066940308,
"openwebmath_perplexity": 387.9795726937566,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9653811611608242,
"lm_q2_score": 0.8872045892435128,
"lm_q1q2_score": 0.8564905965511144
} |
http://mathhelpforum.com/discrete-math/31673-couting-questions.html | # Math Help - Couting Questions
1. ## Couting Questions
Hi all,
I have a couple of questions regarding counting, so it will be fantastic if you can help me out. Thanks in advance!
Qns 1:
Consider strings of length n over the set {a,b,c,d}. How many such strings contain at least one pair of adjacent characters that are the same?
By adjacent do they mean (a,b) in length n, etc? Is the combination (b,a) allowed or equivalent to (a,b)?
Qns 2:
How many integers from 1 through 999999 contain each of the digits 1,2 and 3 at least once? (Hint: For each i let $A_{i}$ be the set of integers from 1 through 999999 that do not contain the digit i)
The solution asked me to use the inclusion and exclusion method, so how do i know when to use this?
Qns 3:
If $n=p_{1}p_{2}p_{3}p_{4}p_{5}$ where $p_{i}$ are distinct primes, in how many ways can n be expressed as a product of 2 positive integers? (n=ab and n=ba are considered the same)
What is the answer if $n=p_{1}p_{2}...p_{k}$?
From the solution, they mentioned the number of ways to choose a is $2^{5}$ which i understand. It is stated that "it is a duplication by a factor of 2 since $p_{1}p_{2}$ and $p_{3}p_{4}p_{5}$ both appear as subsets but they give the same factorisation". I do not understand where the duplication is, can somehow show me an example?
2. Hello, shaoen01!
3) If $n=p_1p_2p_3p_4p_5$ where $p_i$ are distinct primes,
In how many ways can n be expressed as a product of 2 positive integers?
( $n=ab\text{ and }n=ba$ are considered the same)
What is the answer if $n=p_1p_2\hdots p_k$?
From the solution, they mentioned the number of ways to choose a is $2^{5}$ which i understand.
It is stated that "it is a duplication by a factor of 2 since $p_{1}p_{2}$ and $p_{3}p_{4}p_{5}$ both appear
as subsets but they give the same factorisation".
I do not understand where the duplication is.
Can somehow show me an example?
Suppose $n \:=\:30 \:=\:2\cdot3\cdot5$
With three factors to choose from, there are: $2^3 = {\bf8}$ possible subsets.
They are: . $\{\;\},\;\{2\},\;\{3\},\;\{5\},\;\{2,3\},\;\{3,5\} ,\;\{2,5\},\;\{2,3,5\}$
But the question is: how many two-set partitions are there?
There are four: . $\begin{array}{cc}\{\;\} &\{2,3,5\} \\ \{2\} & \{3,5\} \\ \{3\} & \{2,5\} \\ \{5\} & \{2,3\} \end{array}\quad \text{ The products are: }\begin{Bmatrix}1\cdot30 \\ 2\cdot15 \\ 3\cdot10 \\ 5\cdot6\end{Bmatrix}$
The partition $\{2\}\;\;\{3,5\}$ is the same as $\{3,5\}\;\;\{2\}$
. . because $2\cdot15\text{ and }15\cdot2$ are the same factoring.
3. Originally Posted by shaoen01
Qns 1:
Consider strings of length n over the set {a,b,c,d}. How many such strings contain at least one pair of adjacent characters that are the same?
By adjacent do they mean (a,b) in length n, etc? Is the combination (b,a) allowed or equivalent to (a,b)?
Qns 2:
How many integers from 1 through 999999 contain each of the digits 1,2 and 3 at least once? (Hint: For each i let $A_{i}$ be the set of integers from 1 through 999999 that do not contain the digit i
I have no gotten a clear model for question #1. But I think that it is clear that you would allow neither ‘ab’ nor ‘ba’. I am pretty sure that one counts these with recursion.
Question #2 is lengthy but straightforward.
If T=999999 is the total number then we need to remove $
\left( {A_1 \cup A_2 \cup A_3 } \right)$
, that is removing any number that fails to contain a 1, 2, or 3.
You know that:
$\# \left( {A_1 \cup A_2 \cup A_3 } \right) =$ $\# \left( {A_1 } \right) + \# \left( {A_2 } \right) + \# \left( {A_3 } \right) - \# \left( {A_1 \cap A_2 } \right) - \# \left( {A_1 \cap A_3 } \right) - \# \left( {A_3 \cap A_2 } \right) + \# \left( {A_1 \cap A_2 \cap A_3 } \right)$.
I will help with one of those: $\# \left( {A_1 \cap A_2 } \right) = 8^6 - 1$.
4. Originally Posted by Soroban
Hello, shaoen01!
Suppose $n \:=\:30 \:=\:2\cdot3\cdot5$
With three factors to choose from, there are: $2^3 = {\bf8}$ possible subsets.
They are: . $\{\;\},\;\{2\},\;\{3\},\;\{5\},\;\{2,3\},\;\{3,5\} ,\;\{2,5\},\;\{2,3,5\}$
But the question is: how many two-set partitions are there?
There are four: . $\begin{array}{cc}\{\;\} &\{2,3,5\} \\ \{2\} & \{3,5\} \\ \{3\} & \{2,5\} \\ \{5\} & \{2,3\} \end{array}\quad \text{ The products are: }\begin{Bmatrix}1\cdot30 \\ 2\cdot15 \\ 3\cdot10 \\ 5\cdot6\end{Bmatrix}$
The partition $\{2\}\;\;\{3,5\}$ is the same as $\{3,5\}\;\;\{2\}$
. . because $2\cdot15\text{ and }15\cdot2$ are the same factoring.
Hi Soroban:
Thank you for your reply. So for example, 3x10 is also equivalent to 10x3 right? So this kind of repeated value is not allowed. If i use your example, i can't seem to get the answer of $2^{4}=16$. And must $n=p_{1}p_{2}p_{3}p_{4}p_{5}$ be consecutive values? For example, 1 to 5?
5. Originally Posted by Plato
I have no gotten a clear model for question #1. But I think that it is clear that you would allow neither ‘ab’ nor ‘ba’. I am pretty sure that one counts these with recursion.
Question #2 is lengthy but straightforward.
If T=999999 is the total number then we need to remove $
\left( {A_1 \cup A_2 \cup A_3 } \right)$
, that is removing any number that fails to contain a 1, 2, or 3.
You know that:
$\# \left( {A_1 \cup A_2 \cup A_3 } \right) =$ $\# \left( {A_1 } \right) + \# \left( {A_2 } \right) + \# \left( {A_3 } \right) - \# \left( {A_1 \cap A_2 } \right) - \# \left( {A_1 \cap A_3 } \right) - \# \left( {A_3 \cap A_2 } \right) + \# \left( {A_1 \cap A_2 \cap A_3 } \right)$.
I will help with one of those: $\# \left( {A_1 \cap A_2 } \right) = 8^6 - 1$.
Hi Plato:
Qns 1:
Should i use the unordered and no repetition method to calculate this? Do you have any clues or hints to give me?
Qns 2:
But i am curious to know how did you get the value $
\left( {A_1 \cap A_2 } \right) = 8^6 - 1
$
. I do know the inclusion/exclusion rule you written above, the thing is i do not know how to get the value like how you did.
6. Originally Posted by shaoen01
I do not know how to get the value like how you did.
The set $A_1$ is the set of all integers 1 to 999999 that do not contain the digit 1. Because we can only use $\left\{ {0,2,3,4,5,6,7,8,9} \right\}$ there $9^6-1$ numbers in $A_1$, we subtract the 1 to account for 000000000.
7. Hello, shaoen01!
Plato's approach to #2 is correct . . .
I would further assume that an integer does not begin with zero.
2) How many integers from 1 through 999999
contain each of the digits 1,2 and 3 at least once?
The solution asked me to use the inclusion-exclusion method,
so how do i know when to use this?
It is used when counting the items which are not in the set
is easier than counting the items that are in the set.
There are 999,999 integers.
How many do not contain at least {1,2,3} ?
Let $n(i')$ = number of integers that do not contain an $i.$
We want:
$n(1' \,\cup \,2' \,\cup \,3') \;=\;n(1') \,+\, n(2') \,+ \,n(3') \,-\, n(1' \,\cap \,2') \,- \,n(2' \,\cap \,3') \,- \,n(1' \,\cap \,3') \,+$ $n(1' \cap 2' \cap 3')$
Consider $n(1')$
1-digit numbers: 8 choices . (It must not be 0 or 1.)
2-digit numbers: $8\cdot9$ choices.
3-digit numbers: $8\cdot9^2$ choices.
4-digit numbers: $8\cdot9^3$ choices.
5-digit numbers: $8\cdot9^4$ choices.
6-digit numbers: $8\cdot9^5$ choices.
There are: . $8(1 + 9 + 9^2 + \hdots + 9^5) \;=\;8\,\frac{9^5-1}{9-1} \;=\;59,048$ such numbers.
The same reasoning holds for $n(2')\text{ and }n(3')$
. . Hence: . $n(1') \:=\:n(2') \:=\:n(3') \:=\:59,048$
Consider $n(1' \cap 2')$
1-digit numbers: 7 choices. .(It must not be 0, 1, or 2.)
2-digit numbers: $7\cdot8$ choices.
3-digit numbers: $7\cdot8^2$ choices.
4-digit numbers: $7\cdot8^3$ choices.
5-digit numbers: $7\cdot8^4$ choices.
6-digit numbers: $7\cdot8^5$ choices.
There are: . $7(1 + 8 + 8^2 +\hdots+8^5) \:=\:7\,\frac{8^5-1}{8-1} \:=\:32,767$ such numbers.
The same reasoning holds for $n(2'\cap3')\text{ and }n(1'\cap3')$
. . Hence: . $n(1'\cap2') \:=\:n(2' \cap 3') \:=\:n(1' \cap 3') \:=\:32,767$
Consider $n(1' \cap 2' \cap 3')$
1-digit numbers: 6 choices. .(It must not be 0, 1, 2, or 3.)
2-digit numbers: $6\cdot7$ choices.
3-digit numbers: $6\cdot7^2$ choices.
4-digit numbers: $6\cdot7^3$ choices.
5-digit numbers: $6\cdot7^4$ choices.
6-digit numbers: $6\cdot7^5$ choices.
There are: . $6(1 + 7 + 7^2 + \hdots + 7^5) \:=\:6\,\frac{7^5-1}{7-1} \:=\:16,806$ such numbers.
. . Hence: . $n(1' \cap 2' \cap 3') \:=\:16,806$
Then: . $n(1' \cup 2' \cup 3') \:=\:3(59,048) - 3(32,767) + 16,806 \:=\:95,649$
Therefore: . $n(1 \cap 2\cap 3) \;=\;999,999 - 95,649 \;=\;\boxed{904,350}$
But check my work and my reasoning . . . please!
.
8. Originally Posted by Soroban
I would further assume that an integer does not begin with zero. But check my work and my reasoning . . . please!
But in the case "assume that an integer does not begin with zero", how would account for 234?
That contains no 1's but has the form 000234.
9. Originally Posted by shaoen01
Qns 1:
Consider strings of length n over the set {a,b,c,d}. How many such strings contain at least one pair of adjacent characters that are the same?
If that interpretation of the question is correct, then the answer is quite easy. There are $4^n$ possible strings altogether, but $4\times3^{n-1}$ of these are not allowed. (For a string not to be allowed, its first letter can be any one of the four, but each subsequent letter must be different from its predecessor, so there are only three choices for it.) Thus the number of allowable strings is $4(4^{n-1} - 3^{n-1})$.
There are: . $8(1 + 9 + 9^2 + \hdots + 9^5) \;=\;8\,\frac{9^5-1}{9-1} \;=\;59,048$ such numbers.
Bit of slippage here, I think! It should be $8\,\frac{9^{\textstyle\mathbf6}-1}{9-1} \;=\;531440$ (and similarly for the other two summations). | 2014-11-01T03:23:34 | {
"domain": "mathhelpforum.com",
"url": "http://mathhelpforum.com/discrete-math/31673-couting-questions.html",
"openwebmath_score": 0.8626995086669922,
"openwebmath_perplexity": 567.1806657916333,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9653811571768047,
"lm_q2_score": 0.8872045907347108,
"lm_q1q2_score": 0.8564905944560486
} |
http://wildjusticemusic.com/american-satan-mkkicgb/biconditional-statement-truth-table-ad83f6 | biconditional statement truth table
You can enter logical operators in several different formats. • Identify logically equivalent forms of a conditional. a. Worksheets that get students ready for Truth Tables for Biconditionals skills. Edit. Solution: Yes. Notice that in the first and last rows, both P ⇒ Q and Q ⇒ P are true (according to the truth table for ⇒), so (P ⇒ Q) ∧ (Q ⇒ P) is true, and hence P ⇔ Q is true. Biconditional: Truth Table Truth table for Biconditional: Let P and Q be statements. The biconditional uses a double arrow because it is really saying “p implies q” and also “q implies p”. The biconditional, p iff q, is true whenever the two statements have the same truth value. Compound Propositions and Logical Equivalence Edit. first condition. A biconditional is true only when p and q have the same truth value. Conditional Statements (If-Then Statements) The truth table for P → Q is shown below. In writing truth tables, you may choose to omit such columns if you are confident about your work.) The following is truth table for ↔ (also written as ≡, =, or P EQ Q): To show that equivalence exists between two statements, we use the biconditional if and only if. Sign up or log in. biconditional Definitions. Let's put in the possible values for p and q. Also how to do it without using a Truth-Table! Let p and q are two statements then "if p then q" is a compound statement, denoted by p→ q and referred as a conditional statement, or implication. If a is even then the two statements on either side of $$\Rightarrow$$ are true, so according to the table R is true. Is there XNOR (Logical biconditional) operator in C#? BNAT; Classes. text/html 8/17/2008 5:10:46 PM bigamee 0. Two formulas A 1 and A 2 are said to be duals of each other if either one can be obtained from the other by replacing ∧ (AND) by ∨ (OR) by ∧ (AND). 3. • Construct truth tables for conditional statements. SOLUTION a. It is denoted as p ↔ q. How to find the truth value of a biconditional statement: definition, truth value, 4 examples, and their solutions. In Example 3, we will place the truth values of these two equivalent statements side by side in the same truth table. In this implication, p is called the hypothesis (or antecedent) and q is called the conclusion (or consequent). Whenever the two statements have the same truth value, the biconditional is true. Now you will be introduced to the concepts of logical equivalence and compound propositions. en.wiktionary.org. Give a real-life example of two statements or events P and Q such that P<=>Q is always true. P Q P Q T T T T F F F T F F F T 50 Examples: 51 I get wet it is raining x 2 = 1 ( x = 1 x = -1) False (ii) True (i) Write down the truth value of the following statements. When P is true and Q is true, then the biconditional, P if and only if Q is going to be true. The connectives ⊤ … "A triangle is isosceles if and only if it has two congruent (equal) sides.". Now I know that one can disprove via a counter-example. A biconditional statement is really a combination of a conditional statement and its converse. All birds have feathers. Two line segments are congruent if and only if they are of equal length. Also, when one is false, the other must also be false. About Us | Contact Us | Advertise With Us | Facebook | Recommend This Page. Name. Examples. The compound statement (pq)(qp) is a conjunction of two conditional statements. Just about every theorem in mathematics takes on the form “if, then” (the conditional) or “iff” (short for if and only if – the biconditional). The truth table for the biconditional is . Writing Conditional Statements Rewriting a Statement in If-Then Form Use red to identify the hypothesis and blue to identify the conclusion. In each of the following examples, we will determine whether or not the given statement is biconditional using this method. (a) A quadrilateral is a rectangle if and only if it has four right angles. So we can state the truth table for the truth functional connective which is the biconditional as follows. Definitions are usually biconditionals. It is helpful to think of the biconditional as a conditional statement that is true in both directions. 2. Compare the statement R: (a is even) $$\Rightarrow$$ (a is divisible by 2) with this truth table. Summary: A biconditional statement is defined to be true whenever both parts have the same truth value. A biconditional statement is often used in defining a notation or a mathematical concept. Mathematicians abbreviate "if and only if" with "iff." Watch Queue Queue. In Example 3, we will place the truth values of these two equivalent statements side by side in the same truth table. You are in Texas if you are in Houston. The conditional operator is represented by a double-headed arrow ↔. The correct answer is: One In order for a biconditional to be true, a conditional proposition must have the same truth value as Given the truth table, which of the following correctly fills in the far right column? In the truth table above, when p and q have the same truth values, the compound statement (pq)(qp) is true. Notice that the truth table shows all of these possibilities. Is this statement biconditional? Make a truth table for ~(~P ^ Q) and also one for PV~Q. If I get money, then I will purchase a computer. 1. The biconditional connects, any two propositions, let's call them P and Q, it doesn't matter what they are. In this section we will analyze the other two types If-Then and If and only if. We can use the properties of logical equivalence to show that this compound statement is logically equivalent to $$T$$. Biconditional statement? text/html 8/18/2008 11:29:32 AM Mattias Sjögren 0. A double implication (also known as a biconditional statement) is a type of compound statement that is formed by joining two simple statements with the biconditional operator. When proving the statement p iff q, it is equivalent to proving both of the statements "if p, then q" and "if q, then p." (In fact, this is exactly what we did in Example 1.) So to do this, I'm going to need a column for the truth values of p, another column for q, and a third column for 'if p then q.' Biconditional Statements (If-and-only-If Statements) The truth table for P ↔ Q is shown below. Construct a truth table for ~p ↔ q Construct a truth table for (q↔p)→q Construct a truth table for p↔(q∨p) A self-contradiction is a compound statement that is always false. Mathematics normally uses a two-valued logic: every statement is either true or false. Sign up to get occasional emails (once every couple or three weeks) letting you know what's new! When we combine two conditional statements this way, we have a biconditional. Otherwise it is false. • Use alternative wording to write conditionals. Unit 3 - Truth Tables for Conditional & Biconditional and Equivalent Statements & De Morgan's Laws. A biconditional statement will be considered as truth when both the parts will have a similar truth value. The statement qp is also false by the same definition. Use a truth table to determine the possible truth values of the statement P ↔ Q. Therefore, a value of "false" is returned. Ask Question Asked 9 years, 4 months ago. Definition. The biconditional statement $$p\Leftrightarrow q$$ is true when both $$p$$ and $$q$$ have the same truth value, and is false otherwise. Logical equivalence means that the truth tables of two statements are the same. Determine the truth values of this statement: (p. A polygon is a triangle if and only if it has exactly 3 sides. Table truth table for ( p↔q ) ∧ ( p↔~q ), is this a self-contradiction p >. A: it is always biconditional statement truth table functional connective which is the biconditional an! Implies q, is true but the back is false ; otherwise, it is always true up, can! For better understanding, you can think of the following sentences using iff next lesson ; Last. Two inputs, say a and b: we will not play of determining truth values of statements! Shown below form can be useful when writing proof or when showing logical biconditional statement truth table that this statement! P\Right ) \ ) table with 8 rows to cover all possible scenarios ( logical biconditional or implication... Making statements based on opinion ; back them up with references or personal experience biconditional. Side in the next section ( T\ ) represents the sentence, passed. And problem packs statement will be introduced to the concepts of logical biconditional or double.. Helpful to think of the biconditional to an equivalence statement first order of following. And binary operations along with their truth-tables at BYJU 's if they are when showing logical equivalencies state! So, the truth of biconditional statement truth table ( or 'if ' ) statements their! ( T\ ) false by the definition of a biconditional is true and q such that p < >. I know that one can disprove via a counter-example figure out the way it does in table. Is equivalent to p q, since these statements have the same truth for! For each truth table for p ↔ q implies that p < = q! That p and q have the same truth value the table below, have. Answer is provided in the first row naturally follows this definition a two-way arrow ( ) and q shown. And compound propositions involve the assembly of multiple statements, we can focus on the truth table for --. Polygon is a compound statement that is always false higher. equal length is either true or.. To show that ~q p is a quadrilateral, then the polygon has only four sides, then the has! That ~q p is true as well the former statement is one of the rows that! You use truth tables of two conditional statements this way, we can use the biconditional is true definition! ( p↔q ) ∧ ( p↔~q ), is false, the conditional, equivalence compound!: definition, truth value logical equivalence and compound propositions between two statements, multiple... The connectives ⊤ … we still have several conditional geometry statements and their.... 4 using this abbreviation and compound propositions involve the assembly of multiple statements we. How to do a truth table for p and q what they are logically equivalent to \ ( )! Biconditional connects, any two propositions, let 's call them p and q is a quadrilateral, then quadrilateral! That get students ready for truth tables for conditional & biconditional and equivalent statements side side. I will purchase a computer the following: 1 it comes out the way it does ∧! Weeks ) letting you know what 's new a compound statement been defined, we will look at a table! ] this is often used in defining a notation or a mathematical.. If q is true regardless of the biconditional operator is denoted by a double-headed arrow.! Defined, we will place the truth table information and to our privacy policy is saying if. To mathematical Thinking Making statements based on opinion ; back them up references... Each sentence from examples 1 through 4 using this abbreviation T. F. F. F. T. Note is... According to when p and q such that p < = > q is below... Provided in the possible truth values also be false ) ( qp ) is a hypothesis and q be.. Their converses from above a polygon is a truth table for p↔ ( )!, since these statements have the same truth value of q to biconditional statements occur frequently in mathematics every! Information and to our privacy policy '' operator two types If-Then and if and only if q is ;! B are true, conditional, p iff q, its inverse, converse, and problem packs parts... A diadic operator in natural language and code a declarative sentence which has one and only ''! Confident about your work. will place the truth value ( qp ) is hypothesis. you passed the exam iff you scored 65 % or higher..! ⊤ … we still have several conditional geometry statements and their converses from above equivalence and biconditional T.! The form ' p if and only if I get money, then x + 7 = 11 x! This form can be useful when writing proof or when showing logical.. This guide, we can use the properties of logical equivalence and compound propositions involve the of... Boolean algebra, which is a rectangle if and only if y, ” where x is a rectangle and... A triangle iff it has exactly 3 sides. true only p! Examples, we have two propositions, let 's look at a truth table all. \Rightarrow p\right ) \ ) mathematical concept both in natural language and code logic: every is... Choose a different button 4 examples, and contrapositive the same truth table truth table for the truth to. Are of equal length on opinion ; back them up with references or experience! Abbreviate if and only if '' with iff '' logical operators in several different.... There XNOR ( logical biconditional or double implication worksheets that get students for... These possibilities a computer and problem packs or double implication using a Truth-Table 65 % higher! Or three weeks ) letting you know what 's new statement is either or... Lesson ; your Last operator of Example 1 and thus be true whenever both parts have the same hence! Every couple or three weeks ) letting you know what 's new operator in c?! First row naturally follows this definition mistake, choose a different button p is true well! Triangle iff it has exactly 3 sides. will determine whether or the... Byju 's for truth tables for these statements weeks ) letting you know what 's new to examples! Not biconditional feedback to your answer is provided in the possible values called truth values of two! And their converses from above immediately follow and thus be true whenever both parts have same. The different types of unary and binary operations along with their truth-tables at BYJU.! Use the biconditional pq 's put in the first row naturally follows this definition Rewriting a is! Be correct combination of a biconditional statement is a conclusion 4 using this abbreviation + 2 = 7 and... Recommend this Page your work. signing up, you may choose omit... Types If-Then and if and only if. algebra, which involves only true or false answer! Truth-Tables at BYJU 's a statement is [... if and only if q ' statement is logically equivalent biconditional! Always false or ⇔ themselves that must be correct can be useful when proof... Purchase a computer if q ' ∧ ( p↔~q ), is true whenever the two have. 11 iff x = 5, both a and b: we will not.... Rs is true in both directions this out is the biconditional, which involves only true false! And compound propositions involve the assembly of multiple statements, using multiple operators iff your.! Antecedent ) and q have the same truth value isosceles if and only if am... A counter-example self-contradiction is a hypothesis and blue to identify the hypothesis and blue identify! Tables above show that equivalence exists between two statements have the same truth value then examine biconditional! Statements occur frequently in mathematics the given statement is really a combination a. Iff consequent ) back them up with references or personal experience bi-conditionals are represented a... Letting you know what 's new are listed in the same truth value of true is returned this guide we! Q will immediately follow and thus biconditional statement truth table true whenever both parts have same. Then examine the biconditional x→y denotes “ x if and only if x = 5, both a b. Lesson, 2 practice sheets, homework sheet, and a biconditional statement is...! Two equivalent statements side by side in the same definition implies p ” logical equivalencies four right.! Rows to cover all possible scenarios exact truth values for p ↔ q is shown.... Or a mathematical concept ) the truth or falsity of its components side... Antecedent ) and a biconditional is true regardless of the statement p ↔ q is a conclusion sometimes!, Solution: rs represents, I am breathing if and if... Raining and b are true ) a self-contradiction multiple operators iff it has two congruent ( )... Written as iff , let 's call them p and q such that p < = q... Can have a biconditional statement is one of the form if and only if make... B = c, then a = c. 2 back is false when one true! Arrow because it is equivalent to: \ ( biconditional statement truth table m \wedge \sim )! Other must also be false now you will be considered as truth when both components are or... P and q and b are true ( qp ) is a.! | 2021-05-08T20:26:29 | {
"domain": "wildjusticemusic.com",
"url": "http://wildjusticemusic.com/american-satan-mkkicgb/biconditional-statement-truth-table-ad83f6",
"openwebmath_score": 0.5925559997558594,
"openwebmath_perplexity": 734.4940030100662,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.952574129515172,
"lm_q2_score": 0.899121375242593,
"lm_q1q2_score": 0.8564797613501972
} |
https://mathematica.stackexchange.com/questions/159614/building-a-noisy-function | # Building a Noisy Function
Here's a nice function with a nice plot:
f[t_] := Exp[-Abs[t]] Sin[t];
Plot[f[t], {t, -10, 10}, PlotRange -> All]
What I would like to do is to make it noisy... to add something to f[t] so that it returns a noisy version. A roundabout way to accomplish this is to discretize/sample f[t], and then add noise to the samples:
range = Range[-10, 10, 0.1];
ListLinePlot[f[#] & /@ range +
RandomReal[{-0.03, 0.03},Length[range]], PlotRange -> All]
Is there a more direct way? It seems like there are lots of functions that generate random processes, but they seem to invariably end up with a TemporalData object or a list of values. Is there any way to add noise directly to the function?
Update: I'm sorry I can't accept all the answers! David's has the advantage of simplicity (and of showing me a new way that the Random functions can work). A.G.'s answer provides clear flexibility in choosing the correlation of the random process. J.M.'s is probably the slickest, building on his earlier Perlin noise function.
• Do you want the whole function or is a plot enough? – A.G. Nov 10 '17 at 1:12
• I was hoping for a function that could be used elsewhere. – bill s Nov 10 '17 at 1:54
• @bills I think that your question deserves more thinking. Noise is always characterised, in a first approximation, by a mean, a variance, and an amplitude. Therefore, I think that in the proposal some modelling of your additive noise seems to be worth considered. – José Antonio Díaz Navas Nov 11 '17 at 16:37
• I wasn't too picky about the character of the noise, but the answers do provide a variety of possibilities -- for example, J.M.s suggestion of RandomVariate provides just about any independent distribution, while other answers address how to incorporate correlation over time. – bill s Nov 11 '17 at 21:49
Plot[
f[t] + Sum[.01 RandomReal[n] Cos[n t/10], {n, 10}],
{t, -10, 10},
PlotRange -> All]
• Even simpler: Plot[f[t] + RandomReal[1], {t, -10, 10}]. I had no idea RandomReal could be used like this. Thanks! – bill s Nov 10 '17 at 0:08
• @bill, you can use RandomVariate[] too if you want: Plot[f[t] + RandomVariate[NormalDistribution[0, 1/50]], {t, -10, 10}, PlotRange -> All] – J. M.'s ennui Nov 10 '17 at 4:51
• An incredibly minor quibble: this new function will always have the same derivative as the original function at $t = 0$. This probably doesn't matter for most purposes, but it'd be easy enough to get around by adding in a random phase to the cosine: Cos [n t/10 + RandomReal[{0, 2 Pi}]. – Michael Seifert Nov 10 '17 at 15:31
If you need your function to look "ragged", but still be "continuous" and "reproducible", you can use one-dimensional Perlin noise (previously used here):
fBm = With[{permutations = Apply[Join, ConstantArray[RandomSample[Range[0, 255]], 2]]},
Compile[{{x, _Real}},
Module[{xf = Floor[x], xi, xa, u, i, j},
xi = Mod[xf, 16] + 1;
xa = x - xf; u = xa*xa*xa*(10. + xa*(xa*6. - 15.));
i = permutations[[permutations[[xi]] + 1]];
j = permutations[[permutations[[xi + 1]] + 1]];
(2 Boole[OddQ[i]] - 1)*xa*(1. - u) +
(2 Boole[OddQ[j]] - 1)*(xa - 1.)*u],
RuntimeAttributes -> {Listable},
RuntimeOptions -> {"EvaluateSymbolically" -> False}]];
and then do something like
Plot[f[t] + Sum[fBm[5 2^k t]/2^k, {k, 0, 2}]/20, {t, -10, 10},
PlotPoints -> 55, PlotRange -> All]
If you want something like the example you present, you'll likely need to have the errors serially correlated as the number of evaluated data points gets large. Otherwise you'll get fuzzy caterpillars when the errors are independent.
Update: Hopefully a more justified presentation than before.
Suppose that we add noise to f[t] such that the distribution of the random noise is normally distributed with mean zero and variance $\sigma^2$. But rather than having independent random noise at times $t_1$ and $t_2$, we have $\operatorname{Correlation}(e_{t_1}, e_{t_2}) = \rho^{|t_1 - t_2|}$ with $\rho\ge 0$.
So that defines a random function with a continuous time index. Nothing has been said so far that makes this a discrete time model.
But when we want a particular realization of this function with random noise, we need to choose specific times to produce a display of the function. This doesn't make the function discrete in $t$. It might be a semantics issue but I wouldn't call this a sampled/discretized version of the function despite the way the values of the realization of this function are produced in code.
The following code uses values of $t$ that are equally spaced however the functions involved will take any unequally-spaced values of $t$. The parameter $\rho$ is the correlation of two errors one unit apart.
Manipulate[
(* Values of t to consider *)
t = Table[-10 + 20 i/n, {i, 0, n}];
(* Differences in sucessive values of t *)
d = Differences[t];
(* Autoregressive errors of order 1 *)
e = ConstantArray[0, n + 1];
e[[1]] = RandomVariate[NormalDistribution[0, σ], 1][[1]];
Do[e[[i]] = ρ^d[[i - 1]] e[[i - 1]] +
Sqrt[1 - ρ^(2 d[[i - 1]])] RandomVariate[NormalDistribution[0, σ], 1][[1]],
{i, 2, n + 1}];
(* Values of underlying function *)
fTrue = f[#] & /@ t;
(* Function plus errors *)
y = fTrue + e;
(* Plot true function and contaminated function *)
ListPlot[{Transpose[{t, fTrue}], Transpose[{t, y}]},
PlotRange -> All, Joined -> True,
PlotStyle -> {{Thickness[0.01]}, {Red}}],
(* Sliders *)
{{σ, 0.05}, 0.01, 0.2, Appearance -> "Labeled"},
{{ρ, 0.5}, 0, 0.999, Appearance -> "Labeled"},
{{n, 100}, 10, 2000, 1, Appearance -> "Labeled"},
TrackedSymbols :> {σ, ρ, n},
Initialization :> (
f[t_] := Exp[-Abs[t]] Sin[t];
)]
When there is a large number of points close together and independent errors are used ($\rho=0$), then the result looks like a fuzzy caterpillar and probably unrealistic for any physical process. (Would an observation a very short distance away from an observation at $t$ be consistently wildly different?)
• Like in my example, this operates on a sampled version of the function. My hope was to be able to define a noisy function that could be used elsewhere. – bill s Nov 10 '17 at 1:56
In order to get a function it is possible to add a noise function. That noise will be more complex if the function is to be used on a large interval (unless it is periodic, which is not so good for a "random" noise).
Here is an example that produces a smooth function while avoiding periodical noise.
SeedRandom[1]; (* change for different noise *)
n = 200; (* increase for more randomness *)
a = 10; (* use function on interval [-a, a] *)
w = 1/500; (* weight of random noise *)
c = Table[RandomVariate[NormalDistribution[]], n]; (* coefficients *)
\[Epsilon][x_] := w Sum[ c[[i]] Cos[ \[Pi] i (x - a)/(2 a)],
{i, 1, n}]; (* noise *)
Plot[\[Epsilon][t], {t, -10, 10}, PlotRange -> All]
f[t_] := Exp[-Abs[t]] Sin[t] + \[Epsilon][t];
Plot[f[t], {t, -a, a}, PlotRange -> All]
Noise:
Function:
• Instead of Table[RandomVariate[NormalDistribution[]], n], you could do RandomVariate[NormalDistribution[], n]. Then, you can do ε[x_] = w c.Cos[π Range[n] (x - a)/(2 a)]. – J. M.'s ennui Nov 10 '17 at 4:47 | 2021-03-01T10:31:36 | {
"domain": "stackexchange.com",
"url": "https://mathematica.stackexchange.com/questions/159614/building-a-noisy-function",
"openwebmath_score": 0.38297948241233826,
"openwebmath_perplexity": 2139.226606922408,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.952574129515172,
"lm_q2_score": 0.8991213745668094,
"lm_q1q2_score": 0.8564797607064634
} |
https://mathhelpboards.com/threads/compute-a-square-root-of-a-sum-of-two-numbers.8337/ | # Compute a square root of a sum of two numbers
#### anemone
##### MHB POTW Director
Staff member
Compute $\sqrt{2000(2007)(2008)(2015)+784}$ without the help of calculator.
#### MarkFL
Staff member
My solution:
$$\displaystyle 2000\cdot2015=4030028-28$$
$$\displaystyle 2007\cdot2008=4030028+28$$
$$\displaystyle 784=28^2$$
Hence:
$$\displaystyle 2000\cdot2007\cdot2008\cdot2015+784=4030028^2-28^2+28^2=4030028^2$$
And so:
$$\displaystyle \sqrt{2000\cdot2007\cdot2008\cdot2015+784}=\sqrt{4030028^2}=4030028$$
#### anemone
##### MHB POTW Director
Staff member
This is probably the most genius way to collect the four numbers $2000$, $2007$, $2008$ and $2015$ in such a manner so that their product takes the form $(a+b)(a-b)$ and what's more, $b^2=784$!
Bravo, MarkFL!
##### Well-known member
Letting 2000 = a
we have 2000 * 2007 *2008 * 2015 + 784
a(a+7)(a+8)(a+15) + 784
= a(a+15) (a+7) (a+8) + 784
= (a^2+15a) (a^2 + 15a + 56) + 784
= $(a^2 + 15 a + 28 - 28 ) (a^2 + 15a + 28 + 28) + 28^2$
= $(a^2 + 15 a + 28)^2 - 28^2 + 28^2$
= $(a^2 + 15 a + 28)^2$
hence square root is $a^2 + 15a + 28$ or 4000000 + 30000 + 28
#### anemone
##### MHB POTW Director
Staff member
Letting 2000 = a
we have 2000 * 2007 *2008 * 2015 + 784
a(a+7)(a+8)(a+15) + 784
= a(a+15) (a+7) (a+8) + 784
= (a^2+15a) (a^2 + 15a + 56) + 784
= $(a^2 + 15 a + 28 - 28 ) (a^2 + 15a + 28 + 28) + 28^2$
= $(a^2 + 15 a + 28)^2 - 28^2 + 28^2$
= $(a^2 + 15 a + 28)^2$
hence square root is $a^2 + 15a + 28$ or 4000000 + 30000 + 28
Hey kaliprasad, thanks for participating and your method is good and I'm particularly very happy to see you finally picking up on LaTeX!
#### agentmulder
##### Active member
Almost identical to MarkFL's , i just factored 16 to make the multiplications a bit smaller.
$\sqrt{(2000)(2007)(2008)(2015)+ 784} \ =$
$\sqrt{16 [ (\frac{2000}{4})(2007) ( \frac{2008}{4} ) (2015) + \frac{16 \cdot 7^2}{16} ]} \ =$
$4 \sqrt{(500)(2015)(502)(2007) \ + \ 7^2} \ =$
$4 \sqrt{(1007507 -7)(1007507 + 7) + 7^2} \ =$
$4 \sqrt{(1007507)^2} \ = \ 4030028$
Admittedly , had i not seen MarkFL's method i probably would not have discovered it on my own.
Last edited:
#### anemone
##### MHB POTW Director
Staff member
Almost identical to MarkFL's , i just factored 16 to make the multiplications a bit smaller.
$\sqrt{(2000)(2007)(2008)(2015)+ 784} \ =$
$\sqrt{16 [ (\frac{2000}{4})(2007) ( \frac{2008}{4} ) (2015) + \frac{16 \cdot 7^2}{16} ]} \ =$
$4 \sqrt{(500)(2015)(502)(2007) \ + \ 7^2} \ =$
$4 \sqrt{(1007507 -7)(1007507 + 7) + 7^2} \ =$
$4 \sqrt{(1007507)^2} \ = \ 4030028$
Admittedly , had i not seen MarkFL's method i probably would not have discovered it on my own.
Nice one, agentmulder! | 2021-09-26T21:35:42 | {
"domain": "mathhelpboards.com",
"url": "https://mathhelpboards.com/threads/compute-a-square-root-of-a-sum-of-two-numbers.8337/",
"openwebmath_score": 0.6741387844085693,
"openwebmath_perplexity": 5247.750325238607,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9919380075184121,
"lm_q2_score": 0.8633916099737806,
"lm_q1q2_score": 0.8564309533055059
} |
https://math.stackexchange.com/questions/1527353/are-there-infinitely-many-rational-pairs-a-b-which-satisfy-given-equation | # Are there infinitely many rational pairs $(a,b)$ which satisfy given equation?
I saw on some facebook page this concrete example:
$1.2^2+0.6^2=1.2+0.6$
The question that immediately arises is:
Are there infinitely many pairs $(a,b)$ of rational numbers such that we have $a^2+b^2=a+b$?
• HINT: Your equation represents a circle. It has infinitely many rational points on it. – user167045 Nov 13 '15 at 15:49
• Great. Please post this as an answer. – Shailesh Nov 13 '15 at 15:50
• $$a=\frac{t(t+k)}{t^2+k^2}$$ $$b=\frac{k(t+k)}{t^2+k^2}$$ – individ Nov 13 '15 at 17:20
• @individ Would you like to post this as an answer? – Farewell Nov 13 '15 at 17:25
• Why? The equation is simple. – individ Nov 13 '15 at 17:30
Hint: $$a^2+b^2=a+b\implies{}a^2-a+b^2-b=0\implies{}2\left(a-\frac{1}{2}\right)^2+2\left(b-\frac{1}{2}\right)^2=1.$$
What can you say about this equation with respect to the cartesian plane? What is the parametrization of this figure?
• @AntePaladin: A circle with a rational centre will always have infinitely many rational points while a circle with an irrational centre will have at most two rational points. See here fore the proof: books.google.co.in/… – Jack Frost Nov 13 '15 at 16:16
• @Jack Frost a circle with rational centre and rational radius has infinitely many rational points. A circle with rational centre and transcendental radius has no rational points. – Robert Israel Nov 13 '15 at 16:41
• In this case the radius is the square root of a rational: that's more delicate. There wouldn't be any if the radius was $1/\sqrt{3}$. – Robert Israel Nov 13 '15 at 16:46
• @AntePaladin Consider the map $(x,y)\rightarrow\left(\frac{x+y}2,\frac{x-y}2\right)$. It takes the circle of radius $1$ to a circle of radius $\frac{1}{\sqrt{2}}$ and preserves rational points. – Milo Brandt Nov 13 '15 at 16:48
• Yes, in fact a parametrization of this circle is $$a = \dfrac{t+1}{t^2+1}, \ b = \dfrac{t^2 + t}{t^2 + 1}$$ Any rational value of $t$ gives you a rational point $(a,b)$. – Robert Israel Nov 13 '15 at 17:30
Real solutions:
The equation $$a^2 + b^2 = a + b$$ can be transformed: $$a^2 + b^2 = a + b \iff \\ a^2 - a + b^2 -b = 0 \iff \\ (a - 1/2)^2 + (b - 1/2)^2 = 1/4 + 1/4 = 1/2 = (1/\sqrt{2})^2 \quad (*)$$ The solutions form a circle with origin $(1/2, 1/2)$ and radius $1/\sqrt{2}$ in $\mathbb{R}^2$. They are $$(a, b) = (a, b(a)) = \left(a, (1/2) \pm\sqrt{(1/\sqrt{2})^2 - (a - 1/2)^2}\right)$$ for $a \in [1/2 - 1/\sqrt{2}, 1/2 + 1/\sqrt{2}] = [a_-, a_+]$.
Rational solutions:
Among all those infinite many real solutions $(a, b)$ we expect infinite many rational solutions, but my topology is too bad to give a good argument here. If we insert arbitrary rational $a$ not all $b = b(a)$ are rational.
I tried to find all rational $a$ for which $b = b(a)$ is rational, but ran against a wall. However an infinite subset of such rational $a$ is still infinite, so we try to pick an easy one.
An infinite rational subset of solutions:
The subset I came up with consists of those $a$ which have the form $$a_n = \frac{1}{2} + \frac{1}{2n} < a_+ \quad (n \in \mathbb{N})$$ which gives \begin{align} b_n &= \frac{1}{2} \pm \sqrt{\frac{1}{2} - \frac{1}{4n^2}} \\ &= \frac{1}{2} \pm \frac{\sqrt{2n^2 - 1}}{2n} \\ &= \frac{n \pm \sqrt{2n^2 - 1}}{2n} \end{align}
Then $b_n$ is rational, if $$2n^2 -1 = m^2$$ for some $m \in \mathbb{N} \cup \{ 0 \}$.
We can rewrite this as $$m^2 - 2 n^2 = -1 \quad (**)$$ which is a Diophantine equation related to the Pell equation with constant $D = 2$, see negative Pell equation.
Solution of the negative Pell equation:
One solution to $(**)$ is $m = 1$ and $n = 1$, from which we can generate infinite many more solutions:
For odd $k \in \mathbb{N}$ we have: $$-1 = (1^2 - 2\cdot 1^2)^k = (m^2 - 2n^2) \Rightarrow \\ -1 = (1 + \sqrt{2})^k (1-\sqrt{2})^k = (m + \sqrt{2} n)(m - \sqrt{2} n)$$ which because of $$(1 + \sqrt{2})^k = \sum_{i=0}^k \binom{k}{i} 1^i (\sqrt{2})^{n-i} = c_1 \cdot 1 + c_2 \sqrt{2} \quad (c_1, c_2 \in \mathbb{N}) \\ (1 - \sqrt{2})^k = \sum_{i=0}^k \binom{k}{i} 1^i (-\sqrt{2})^{n-i} = d_1 \cdot 1 - d_2 \sqrt{2} \quad (d_1, d_2 \in \mathbb{N})$$ gives $$m + \sqrt{2} n = (1 + \sqrt{2})^k \\ m - \sqrt{2} n = (1 - \sqrt{2})^k \\$$ and $$m = \frac{(1 + \sqrt{2})^k + (1 - \sqrt{2})^k}{2} \\ n = \frac{(1 + \sqrt{2})^k - (1 - \sqrt{2})^k}{2 \sqrt{2}}$$
Here are the first ten solutions:
k = 1, m = 1, n = 1
k = 3, m = 7, n = 5
k = 5, m = 41, n = 29
k = 7, m = 239, n = 169
k = 9, m = 1393, n = 985
k = 11, m = 8119, n = 5741
k = 13, m = 47321, n = 33461
k = 15, m = 275807, n = 195025
k = 17, m = 1607521, n = 1136689
k = 19, m = 9369319, n = 6625109
Plugging the Pell solutions into the circle solution:
We have \begin{align} a_k &= \frac{1}{2} + \frac{1}{2n} \\ &= \frac{1}{2} + \frac{\sqrt{2}}{(1 + \sqrt{2})^k - (1 - \sqrt{2})^k} \in \mathbb{Q} \end{align} and \begin{align} b_k &= \frac{1}{2} \pm \frac{m}{2n} \\ &= \frac{1}{2} \pm \frac{\sqrt{2}}{2} \frac{(1 + \sqrt{2})^k + (1 - \sqrt{2})^k} {(1 + \sqrt{2})^k - (1 - \sqrt{2})^k} \in \mathbb{Q} \end{align} for any odd $k \in \mathbb{N}$.
Here is a visualization:
Featured are the real solutions (blue circle) and a few of the rational solutions $Q_k^{(+)} = (a_k, b_k^{(+)})$ (green points), $Q_k^{(-)} = (a_k, b_k^{(-)})$ (red points) for $k \in \{ 1,3,5 \}$. The other ones pile up too much to be distinguishable in the plot.
• FYI: When your number of edits exceeded ten, a system flag was raised. It used to be a rule that at that point a post became Community-Wiki = free for all to edit, and no more rep gained from votes. This is because some users wanted to "bump" their post to the front page by frivolous edits. It is clear to me that you had no such intentions. Yet, the bumping may irritate other users. Even though sometimes the post needs a lot of polishing and such. – Jyrki Lahtonen Nov 14 '15 at 6:50
• (cont'd) To address those needs a sandbox was created in meta. So if you foresee the need to edit a post many times, my advice is to copy/paste its content to a vacated answer box of the sandbox, and do the editing there. When you are happy with it copy/paste it back here. No real harm done. I just wanted to make sure that you are aware of the downsides of plenty of edits, and the recommended solution. – Jyrki Lahtonen Nov 14 '15 at 6:52
• How do you know that from this $(1 + \sqrt{2})^k (1-\sqrt{2})^k = (m + \sqrt{2} n)(m - \sqrt{2} n)$ follows this $m + \sqrt{2} n = (1 + \sqrt{2})^k \\ m - \sqrt{2} n = (1 - \sqrt{2})^k \\$? – Farewell Nov 14 '15 at 13:01
• It looks obvious but... – Farewell Nov 14 '15 at 13:01
• @AntePaladin I added the expansion via binomial theorem. – mvw Nov 14 '15 at 23:19 | 2019-05-23T15:43:15 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/1527353/are-there-infinitely-many-rational-pairs-a-b-which-satisfy-given-equation",
"openwebmath_score": 0.9885804057121277,
"openwebmath_perplexity": 567.0416955906717,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9898303446461048,
"lm_q2_score": 0.865224073888819,
"lm_q1q2_score": 0.8564250432534765
} |
https://math.stackexchange.com/questions/2763504/convergence-of-a-n2-sqrta-n-sqrta-n1 | Convergence of $a_{n+2} = \sqrt{a_n} + \sqrt{a_{n+1}}$ [duplicate]
Let $a_1$ and $a_2$ be positive numbers and suppose that the sequence {$a_n$} is defined recursively by $a_{n+2} = √a_n + √a_{n+1}$. Show that this sequence is convergent.
So, I have been able to show the convergence taking three different cases namely,
Case 1: Both $a_1$ , $a_2$ <4, then I proved that the sequence will be monotonically increasing as well as bounded and so converging
Case 2: Both $a_1$, $a_2$ >4, in this case the sequence is monotonically decreasing and bounded and hence convergent.
Case 3: One of $a_1$ and $a_2$ is <4 & other >4, lets say, $a_1$<4<$a_2$ in this case sequence will alternatively increase and decrease i.e. $a_{2n-1}$ will be increasing and $a_{2n}$ will be decreasing and $a_{2n}-a_{2n-1}$ will converge to Zero.
My question is that is there any general method through which we don't have to take all these cases and can prove the convergence of series in more generality?
marked as duplicate by Gabriel Romon, Martin R, Sil, Siméon, Sangchul Lee real-analysis StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); May 2 '18 at 18:54
• The things you say you proved under Case 1 and Case 2 cannot both be true. (I don't see why any of the three is true, actually...) – David C. Ullrich May 2 '18 at 15:35
• @DavidC.Ullrich: He is splitting up the question into cases. In the first possible case, we have $a_1,a_2<1$. In a second possible case, we have $a_1,a_2>1$. In a third possible case, we have $a_1<1<a_2$ or vice versa. No matter the sequence $a_i$, $a_1$ and $a_2$ must be in one of these three cases. – Clayton May 2 '18 at 15:39
• If $a_n$ converges to $l$ then $l=2 \sqrt{l}$. So $l=0$ or $l=4$. I think this is why @DavidC.Ullrich say that both affirmation cannot be true, if Case $1$ is true then for $n$ large enough $a_n>1$ which is problematic with Case 2. – Delta-u May 2 '18 at 15:54
• @Clayton I understood he's splitting it into cases... that doesn't explain why his conclusions in those cases are correct. – David C. Ullrich May 2 '18 at 16:05
• Sorry, I made a mistake. I should have written 4 instead of 1. Edited it. – Lord KK May 2 '18 at 16:07
If we can make a guess of what the limit should be, then it is often easier to show the convergence by playing with the difference between $a_n$ and the limit candidate.
In our case, the limit value must solve the constraints $x = 2\sqrt{x}$ and $x\geq 0$, hence $x = 0$ or $4$. We claim that $4$ is the limit.
We first establish the boundedness of $(a_n)$ away from $0$ and $\infty$.
• $a_{n+2} \geq \sqrt{a_{n+1}}$ for all $n\geq1$. So $a_{n+2} \geq (a_2)^{1/2^n}$ and hence $\liminf_{n\to\infty} a_n \geq 1$.
• Let $M=\max\{4,a_1,a_2\}$. Then we inductively check that $a_n \leq M$ for all $n\geq1$.
Now define $\epsilon_n = \lvert a_n - 4 \rvert$ and $\bar{\epsilon} = \limsup_{n\to\infty} \epsilon_n$. Then $\bar{\epsilon} \in [0, \infty)$ and
$$\epsilon_{n+2} \leq \frac{\epsilon_n}{\sqrt{a_n} + 2} + \frac{\epsilon_{n+1}}{\sqrt{a_{n+1}} + 2}.$$
Now taking $\limsup_{n\to\infty}$ to both sides yields $\bar{\epsilon} \leq \frac{\bar{\epsilon}}{3} + \frac{\bar{\epsilon}}{3}$, which is enough to conclude that $\bar{\epsilon} = 0$ and hence $a_n \to 4$.
Well, if $a_n\to L$ then $L=\sqrt L+\sqrt L$, so $L=4$ or $L=0$.
So instead of trying to show the sequence converges, we try to show it converges to $4$ or $0$; that may be simpler.
Let's get rid of the square roots by defining $b_n=\sqrt{a_n}$. Now$$b_{n+2}^2=b_n+b_{n+1},$$and we want to show $b_n\to 2$. If you subtract $4$ from both sides, factor the left side and apply the triangle inequality you get $$|b_{n+2}-2|\le\frac{|b_n-2|+|b_{n+1}-2|}{b_{n+2}+2}.$$It seems likely to me you can use this to show $b_n\to2$ or $0$ (if it happens that $b_n\ge c>0$ for all $n$ then it follows that $b_n\to 2$). I have to go to class now, sorry...
Edit: How stupid of me - the other answer points out that it's more or less obvious that $b_n\ge c>0$. In case it's not clear why we care about that, it shows that $$|b_{n+2}-2|\le\frac2{2+c}\max(|b_n-2|,|b_{n+1}-2|);$$hence $b_n\to2$. since $2/(2+c)<1$.
• But how can you claim the boundedness on the right hand side? – Lord KK May 2 '18 at 16:13
• @AloknathFurr I'm not sure what you're referring to. If $b_n\ge c>0$ then the inequality shows that $|b_n-2|$ decreases geometrically. – David C. Ullrich May 2 '18 at 16:15
• ok.. ok.. I got it. Thanks – Lord KK May 2 '18 at 16:17 | 2019-09-20T20:40:23 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2763504/convergence-of-a-n2-sqrta-n-sqrta-n1",
"openwebmath_score": 0.8820814490318298,
"openwebmath_perplexity": 322.0400184138422,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9857180677531124,
"lm_q2_score": 0.868826777936422,
"lm_q1q2_score": 0.8564182527596524
} |
https://math.stackexchange.com/questions/2276402/limit-of-sequence-in-which-each-term-is-defined-by-the-average-of-preceding-two/2276434 | # Limit of sequence in which each term is defined by the average of preceding two terms
We have a sequence of numbers $x_n$ determined by the equality
$$x_n = \frac{x_{n-1} + x_{n-2}}{2}$$
The first and zeroth term are $x_1$ and $x_0$.The following limit must be expressed in terms of $x_0$ and $x_1$ $$\lim_{n\rightarrow\infty} x_n$$
The options are:
A)$\frac{x_0 + 2x_1}{3}$ B)$\frac{2x_0 + 2x_1}{3}$
C)$\frac{2x_0 + 3x_1}{3}$ D)$\frac{2x_0 - 3x_1}{3}$
Since it was a multiple choice exam I plugged $x_0=1$ and $x_1=1$. Which means that all terms of this sequence is $1$,i.e, $$x_n=1, n\in \mathbb{N}$$
From this I concluded that option A was correct.I could not find any way to solve this one hence I resorted to this trick. What is the actual method to find the sequence's limit?
• Your 'trick' was very smart, in the setting of a multiple choice exam. It makes it very easy to exclude options. But it is even smarter that you want to know the 'actual' method to solve this! – user193810 May 11 '17 at 15:32
• @Pakk thanks for the compliment! – Ananth Kamath May 12 '17 at 2:21
• @Pakk Hehe, me on those factoring problems back in ye olde days o' Algebra. Just plug $x=0$ in and you pretty much done... – Simply Beautiful Art May 27 '17 at 0:29
$$2x_n = x_{n-1} + x_{n-2}$$
$$2x_2 = x_{1} + x_{0}\\ 2x_3 = x_{2} + x_{1}\\ 2x_4 = x_{3} + x_{2}\\ 2x_5 = x_{4} + x_{3}\\ ...\\ 2x_n = x_{n-1} + x_{n-2}$$
Now sum every equation and get
$$2x_n+x_{n-1}=2x_1+x_0$$
Supposing that $x_n$ has a limit $L$ then making $n\to \infty$ we get:
$$2L+L=2x_1+x_0\to L=\frac{2x_1+x_0}{3}$$
• What did you sum to arrive at $2x_n+x_{n-1}=2x_1+x_0$? When I summed the numbered equations, I got a proportion of $x_0$ and $x_1$, but not $2:1$, and summing the indexed equations was quite boring. – user121330 May 12 '17 at 22:40
• @user121330: just those equations as written (equivalently, you can rewrite them bringing the RHS terms to LHS and negating them). Note that each (non-terminal) $x_i$ occurs once as $2x_i$ on the LHS, and twice as $x_i$ on the RHS. So they all cancel out except for the terminal ones. – smci May 12 '17 at 23:23
• I fondly remember this as exercise 3.5.10 of Bartle's and Sherbert's Introduction to Real Analysis, how my teacher solved it by finding an invariant as you did, and how I had wondered how she came up with it (she just said to 'play with it'). The nostalgia. – Vandermonde May 14 '17 at 22:27
• No you're not summing to infinity. You're summing for LHS's from $x_2$ to $x_n$. That's finite. After you get that sum, we then apply the condition that if ${x_n}$ has a limit L as $n \to \infty$ then both $x_n, x_n+1 \to L$, and that sets up a useful equality involving $x_1, x_0$. – smci May 16 '17 at 6:40
• But one cannot assume that as n→∞ $x_n$ = $x_n-1$ when it is not given that the sequence converges. – Ahmad Lamaa Sep 24 '18 at 21:09
Yet another trick: you can write the recurrence in matrix form: $$\left(\begin{array}{c} x_n\\ x_{n-1} \end{array} \right) = \left(\begin{array}{cc} 1/2 & 1/2\\ 1 & 0\\ \end{array} \right) \left(\begin{array}{c} x_{n-1}\\ x_{n-2} \end{array} \right).$$ Then, $$\left(\begin{array}{c} x_n\\ x_{n-1} \end{array} \right) = \left(\begin{array}{cc} 1/2 & 1/2\\ 1 & 0\\ \end{array} \right)^{n-1} \left(\begin{array}{c} x_1\\ x_0 \end{array} \right).$$ Diagonalizing/converting the matrix to the Jordan form, you can find a closed form for the sequence.
• I wouldn't call it a trick, I think it's the most natural approach to these problems. And it justifies the method in Arnaud's answer. – Sasho Nikolov May 12 '17 at 19:11
First, notice that you can rewrite the recurrence relation as $$2x_{n+2}-x_{n+1}-x_n=0,\quad n\geq 0.$$ Now the key point is that this recurrence relation is linear, and thus if $(y_n)_{n\geq 0}$ and $(z_n)_{n\geq 0}$ satisfy this relation, then for any $\alpha,\beta\in \Bbb R$, $(\alpha y_n+\beta z_n)_{n\geq 0}$ will also satisfy it. So we can try to express $(x_n)$ as a linear combination of simpler sequences $(y_n)$ and $(z_n)$. An example of simple sequence would be a geometric sequence $y_n=r^n$ for some $r$. Can we find a sequence of this form satisfying the recurrence relation? $r$ would have to be such that $$2r^{n+2}-r^{n+1}-r^n=r^{n}(2r^2-r-1)=0,\quad n\geq 0,$$thus it is enough that $$2r^2-r-1=0.$$ This will hold if and only if $r\in \left\{1,\frac{-1}{2}\right\}$.
Thus we know that any sequence of the form
$$\alpha +\beta \left(\frac{-1}{2}\right)^n$$ satisfies our recurrence relation. Then it suffices to check that the first two term are the same, and all the others will follow. Thus you need to find $\alpha,\beta$ such that $$\left\{ \begin{array}{}\alpha+\beta & = & x_0\\ \alpha-\frac{\beta}{2} & = & x_1.\end{array} \right.$$
This is a simple linear system, whose solution is given by $\alpha=\frac{x_0+2x_1}{3}$ and $\beta=\frac{2x_0-2x_1}{3}$.
Thus the $n$-th term must be given by $$x_n=\frac{x_0+2x_1}{3}+\frac{(2x_0-2x_1)(-1)^{n}}{3\cdot 2^n}$$ and thus $\lim_{n\to \infty }x_n = \frac{x_0+2x_1}{3}$.
This method works for a large variety of cases; in fact, it can be applied to give a formula for $x_n$ for any linear recurrence relations (there are some difficulties if the corresponding polynomial equation has multiple roots, because you don't get enough geometric sequences, but you can find other solutions in those case). For example, you can apply the same method to the Fibonacci sequence, and it gives you the Binet formula (you can find more details in the answers to this question).
• How can we come to the conclusion that $x_n = r^n$ ? Thanks for answering by the way. – Ananth Kamath May 11 '17 at 14:19
• We can't. It's more "wishful thinking". Maybe I can explain that better. – Arnaud D. May 11 '17 at 14:22
• It does makes sense, but I believe such an approach is only good for this particular question? – Ananth Kamath May 11 '17 at 14:30
• @AnanthKamath No, it's a general method for linear recurrence relations. You can find more information for example at en.wikipedia.org/wiki/… – Arnaud D. May 11 '17 at 14:44
• I never knew that was possible. Thanks for sharing info. – Ananth Kamath May 11 '17 at 14:58
Here's yet another way to look at the problem, that might not lead directly to a proof, but might give intuitive insight into the sequence.
Note that from $x_0$, we go $x_1-x_0$ up to $x_1$, then $\frac12(x_1-x_0)$ down to $x_2$, then $\frac14(x_1-x_0)$ up to $x_3$, etc. You might be able to prove this by induction, but might not want to. If $x_0=0$ and $x_1=1$, one can put this in a picture: ($A=(0,x_0)$, $B=(1,x_1)$, ...)
We conclude that, since the jump from $x_0$ to $x_1$ can really be seen a a jump of $1=(-1/2)^0$ relative to $x_0$:
$$x_n = x_0 + (x_1-x_0)\sum_{i=0}^{n-1} \left(-\frac12\right)^i$$
Then the limit for $n \rightarrow \infty$ is:
$$\lim_{n\rightarrow\infty} x_n = x_0 + (x_1-x_0)\sum_{i=0}^\infty \left(-\frac12\right)^i = x_0 + (x_1-x_0)\frac1{1+\frac12}$$
by the geometric series (since $\lvert-\frac12\rvert < 1$). Then this is equal to:
\begin{align*} x_0 + (x_1-x_0)\frac1{1+\frac12} &= x_0 + \frac23(x_1-x_0) = \frac{3x_0+2x_1-2x_0}3 \\ &= \frac{x_0 + 2x_1}3. \end{align*}
• Can't we upvote an answer twice? :) – Kartik May 12 '17 at 10:26
• This is what Soumik did in his answer. Unfortunately he did not feel like working it out. – Carsten S May 12 '17 at 12:39
• @CarstenS Oh right I see; luckily I wasn't the only one to see this :) – tomsmeding May 12 '17 at 16:29
• Yes, +1 for the illustration. – Carsten S May 12 '17 at 16:31
$$x_n-x_{n-1}=-\frac{1}{2}(x_{n-1}-x_{n-2})$$ use repeatedly to get $$x_n-x_{n-1}=(-\frac{1}{2})^{n-1}(x_{1}-x_{0})$$ Now it is fairly straightforward to compute
• Could you show a little more work on how you got to your first step? – AlgorithmsX May 11 '17 at 19:04
• subtract $x_{n-1}$ from both sides of the equation and iterate. – user379195 May 11 '17 at 19:05
• I know how you got there, but I think it's better for you to show some less obvious intermediate steps. It's still a great post. – AlgorithmsX May 11 '17 at 19:08
Here is a way to find the general solution of the recurrence using generating functions and then find the limit. We wish to solve the recurrence $$a_n = \frac{a_{n-1}}{2} + \frac{a_{n-2}}{2}\quad (n\geq 2)\tag{1}$$ where $a_0=x_0$ and $a_1=x_1$. We first make the recurrence valid for all $n\geq 0$. To this end, set $a_n=0$ for $n<0$ and note that the recurrence $$a_n = \frac{a_{n-1}}{2} + \frac{a_{n-2}}{2}+x_0\delta_{n,0}+(x_1-x_0/2)\delta_{n,1}\tag{2}$$ where $\delta_{ij}$ is the Kronecker Delta does the trick. Let $A(x)= \sum_{n=0}^\infty a_nx^n$ be the generating function and multiply by $x^n$ and sum on $n$ in (2). Then we get $$A(x)\left(1-\frac{1}{2}x-\frac{1}{2}x^2\right)=x_0+x\left(x_1-\frac{x_0}{2}\right)\tag{3}$$ and hence $$A(x) =\frac{x_0+(x_1-x_0/2)x}{1-\frac{1}{2}x-\frac{1}{2}x^2} =\frac{x_0+(x_1-x_0/2)x}{(1-x)(1+x/2)} =\frac{2x_1+x_0}{3(1-x)}+\frac{2x_0-2x_1}{3(1+x/2)}\tag{4}$$ by partial fractions. By using the geometric series we get that \begin{align} A(x) &=\sum_{n=0}^\infty \left[\frac{2x_1+x_0}{3}\right]x^n+ \sum_{n=0}^\infty \left[\frac{2x_0-2x_1}{3}\right]\left(\frac{-x}{2}\right)^n\\ &=\sum_{n=0}^\infty \left[\frac{(2x_1+x_0)}{3}+\frac{(2x_0-2x_1)(-1)^n}{3(2^n)}\right]x^n\tag{5} \end{align} Hence
$$a_n=\frac{x_0+2x_1}{3}+\frac{(2x_0-2x_1)(-1)^{n}}{3\cdot 2^n}\tag{6}$$ and thus $$\lim_{n\to \infty }a_n = \color{blue}{\frac{x_0+2x_1}{3}.}\tag{7}$$
Here is a formal derivation of your result. The sequence you have found is a generalization of the Fibonacci sequence.
There have been many extensions of the sequence with adjustable (integer) coefficients and different (integer) initial conditions, e.g., $f_n=af_{n-1}+bf_{n-2}$. (You can look up Pell, Jacobsthal, Lucas, Pell-Lucas, and Jacobsthal-Lucas sequences.) Maynard has extended the analysis to $a,b\in\mathbb{R}$, (Ref: Maynard, P. (2008), “Generalised Binet Formulae,” $Applied \ Probability \ Trust$; available at http://ms.appliedprobability.org/data/files/Articles%2040/40-3-2.pdf.)
We have extended Maynard's analysis to include arbitrary $f_0,f_1\in\mathbb{R}$. It is relatively straightforward to show that
$$f_n=\left(f_1-\frac{af_0}{2}\right) \frac{\alpha^n-\beta^n}{\alpha-\beta}+\frac{f_0}{2} (\alpha^n+\beta^n)$$
where $\alpha,\beta=(a\pm\sqrt{a^2+4b})/2$.
The result is written in this form to underscore that it is the sum of a Fibonacci-type and Lucas-type Binet-like terms. It will also reduce to the standard Fibonacci and Lucas sequences for $a=b=1 \ \text{and} \ f_0=0, f_1=1$. Notice that there is nothing in this derivation that limits the results to integer values for the initial conditions, scaling factors, or the sequence itself.
So, specializing to your case, we can say
$$x_n=\frac{x_{n-1}}{2} +\frac{x_{n-2}}{2}$$
and
$$\alpha,\beta=\frac{\frac{1}{2}\pm\sqrt{\frac{1}{4}+2}}{2}=1,\frac{1}{2}$$
For the limit as $n\to\infty$, we note that the $\alpha$-term dominates, and is in fact, always unity, ergo
$$\lim_{n\to\infty} x_n=\left(x_1-\frac{x_0}{4}\right)\frac{\alpha^n}{\alpha-\beta}+\frac{x_0}{2}\alpha^n=\left(x_1-\frac{x_0}{4}\right)\frac{2}{3}+\frac{x_0}{2}=\frac{2x_1+x_0}{3}$$
as was shown previously.
This proves the OP's assertion. The advantage of this method is that it will apply to all such problems.
$$x_n = \frac{x_{n-1}+x_{n-2}}{2}$$
Add $\frac{x_{n-1}}{2}$ to both sides:
$$\frac{2x_{n}+x_{n-1}}{2} = \frac{2x_{n-1}+x_{n-2}}{2}$$
Let $y_{n} = \frac{2x_{n}+x_{n-1}}{2}$:
$$y_{n} = y_{n-1} = \ldots = y_{1}$$
As $n \rightarrow \infty, x_{n-1} \rightarrow x_{n}$, so:
$$\frac{3x_{n}}{2} = \frac{2x_{1}+x_{0}}{2}$$
Note that this is the "same" approach as some of the other answers, I just added it because it is most clear to me.
My solution:
Map $[x_0,x_1]$ to $[0,1]$ by setting $x = x_0 + (x_1-x_0)y$ so that $y_0 = 0$ and $y_1 = 1$.
Now the sequence runs $0, 1, \frac{1}{2}, \frac{3}{4}, \frac{5}{8}, \frac{11}{16}$, etc
Take the difference between each successive term i.e. $y_k-y_{k-1}$
You get $+1, -\frac{1}{2}, +\frac{1}{4}, -\frac{1}{8}, +\frac{1}{16}$, etc
This looks like the expansion of $\frac{1}{1-z}$ for $|z|<1$ i.e. $1+z+z^2+z^3+...$
In our case, $z = -\frac{1}{2}$, so the sum is $\frac{1}{1-(-1/2)} = \frac{2}{3}$
So the limit of $y_k$ is $\frac{2}{3}$, given $y_0 = 0$ and $y_1 = 1$. So now map back to $x$
$x = x_0 + (x_1-x_0)\frac{2}{3} = \frac{x_0 + 2x_1}{3}$
• Use this link to know how to write mathematics on this site. – StubbornAtom May 14 '17 at 10:25
• Apologies, have used the maths editor on the site before but this was from my phone, so it's not particularly easy to do. Appreciate the edit, meant to follow up this evening with the same. Best, james – James Spencer-Lavan May 14 '17 at 10:44
Unless $x_0=x_1$ which is clear, consider the tranformation $y_n:=\frac{x_n-x_0}{x_1-x_0}$. Then $y_0=0$, $y_1=1$, and $y_{n+1}$ is the average of $y_n$ and $y_{n-1}$. It is clear that $\lim y_n$ is $$\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}-\frac{1}{32}+\cdots=\frac{1}{2}\left(1+\frac{1}{4}+\frac{1}{16}+\cdots\right)=\frac{2}{3}.$$ Going back, you have $\lim x_n=\frac{1}{3}(x_0+2x_1)$.
• This was my instant reaction to the problem, comparison with the binary number given by 0.101010101010... which is 2/3. – richard1941 May 16 '17 at 18:10
Since the recurrence relation is linear, the sequence's limit is, if it exists, of the form $ax_0+bx_1$. If you consider the sequence $x'_n=x_{n+1}$, it clearly has the same limit therefore $$ax_0+bx_1=ax'_0+bx'_1=ax_1+bx_2=ax_1+\frac b2(x_0+x_1)=\frac{b}2x_0+\left(a+\frac b2\right)x_1.$$ Since this is true for any $x_0$, $x_1$, we have $a=\frac b2$ (and also $b=a+\frac b2$ but this is equivalent). This is enough to answer the question, but not to find $a$ and $b$. To find $a$ and $b$, you can simply take the same trick you used by considering the constant sequence $1$ and you get $a+b=1$, which solves the problem.
I came to this page because I wanted to justify the formula I came up with for the series with xn = pxn-1+(1-p)xn-2. This is more general than the original problem, though not as general as the linear combination problem. Based on what I read here, I can show the solution and justify it using just a little insight and basic algebra. The first insight is to assume the solution has the form rx1+(1-r)x0. Let's simplify things by using x0=0 and x1=1. Then x2=p. We must also have rx2+(1-r)x1 as the solution. Taking rx2+(1-r)x1 = rx1+(1-r)x0 and substituting gives p=(1-r)+rp. Solving for r gives r =1/(2-p) and (1-r) = (1-p)/(2-p). Having found the solution to the problem, we need to justify it. It is straightforward, though a little tedious, to show that, for the value of r that we found, that rxn + (1-r)xn-1 = rxn-1 + (1-r)xn-2. Just substitute for xn and r and combine terms. By induction, this means that rxn + (1-r)xn-1 = rx1 + (1-r)x0. As was done in a previous post, we can take xn-1 on the left side to equal xn, which proves our assertion provided we have convergence. To find where we have convergence, use the expression for xn to show that xn - xn-1 = -(1-p)(xn-1 - xn-2), giving convergence when |1-p| < 1.
• There is a simpler approach for finding r. Find a and b such that ax<sub>n</sub>+bx<sub>n-1</sub>=ax<sub>n-1</sub>+bx<sub>n-2</sub>. The constraints for x<sub>n-1</sub> are the same as for x<sub>n-2</sub>, giving b=a(1-p). What we need to realize is that if a and b satisfy the equation, so do ta and tb. We need the additional constraint of a+b=1. Then things work out very simply. – user1153980 Nov 12 '17 at 10:10
First, the sequence $(x_n)$ has clearly a limit $\ell$ because $$\left|x_{n+1}-x_{n}\right|=\frac{|x_1-x_0|}{2^n}.$$ Now, sum the $n-1$ equations $2x_{i+2}=x_{i+1}+x_i$ for $i=0,1,\ldots,n-2$ and we obtain $$2x_n+x_{n-1}=2x_1+x_0.$$ Passing to the limit we reach $$\ell=\frac{2x_1+x_0}{3}.$$ | 2020-07-13T04:56:41 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2276402/limit-of-sequence-in-which-each-term-is-defined-by-the-average-of-preceding-two/2276434",
"openwebmath_score": 0.9318219423294067,
"openwebmath_perplexity": 254.52874176837292,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9857180677531124,
"lm_q2_score": 0.868826771143471,
"lm_q1q2_score": 0.8564182460637179
} |
http://mathhelpforum.com/calculus/164757-integrating-enclosed-area.html | 1. ## Integrating enclosed area
Calculate the area enclosed by x = 9-y^2 and x = 5 in two ways: as an integral along the y-axis and as an integral along the x-axis.
Please tell me if my figure is correct, and also if the shaded area is also the are I am looking for.
Also is there any formulas I need to use for this problem ? How do I do this both ways ?
Thank You
2. Yes, your graph is fine.
To integrate over the x-axis, you need
$\displaystyle\int_{5}^9f(x)dx$
so $x=9-y^2\Rightarrow\ y^2=9-x\Rightarrow\ y=\pm\sqrt{9-x}=f(x)$
Take the positive one and double your result as the x-axis is an axis of symmetry.
$2\displaystyle\int_{5}^9\left(9-x\right)^{0.5}dx$
To integrate over the y-axis, you could integrate $f(y)$ from $y=-3$ to $y=3.$
If you like, again use the x-axis as an axis of symmetry and double the integral from $y=0$ to $y=3.$
This integral includes the unshaded part against the y-axis, so you have a few ways to cope with that.
The easiest way is to subtract 5 from $x=f(y)$
$x-5=f(y)-5=4-y^2$
The new function is $x=4-y^2$
so the limits of integration become $\pm2$
You can then calculate
$2\displaystyle\int_{0}^2\left(4-y^2\right)}dy$
3. Thank You very much. I'll give it a shot.
4. The first integral can be evaluated by making a substitution,
while the 2nd one doesn't require any substitution.
5. Thank you for all you help.
6. Ok so I solved the first one I get .5. I let u = 9-x , du = -dx so I multiplied by a negative inside the integral and a negative outside. then I get, -2(.5(9-x)^-.5). Then using the fundemental theorem of calculus I used the limts 5 to 9 and my final answer was .5. is this correct ?
7. Originally Posted by wair
Ok so I solved the first one I get .5. I let u = 9-x , du = -dx so I multiplied by a negative inside the integral and a negative outside. then I get, -2(.5(9-x)^-.5). Then using the fundemental theorem of calculus I used the limts 5 to 9 and my final answer was .5. is this correct ?
Not quite,
you differentiate to introduce the substitution,
but you are also differentiating when you should be integrating!
$u=9-x\Rightarrow\ du=-dx\rightarrow\ dx=-du$
$x=5\Rightarrow\ u=4$
$x=9\Rightarrow\ u=0$
the integral becomes
$\displaystyle\ -2\int_{4}^0u^{0.5}}du=(-2)\left[-\int_{0}^4u^{0.5}}du\right]=2\int_{0}^4u^{0.5}}du$
$=2\displaystyle\left[\frac{u^{\frac{3}{2}}}{\frac{3}{2}}\right]$ from u=0 to u=4
8. so I just had to chnage my limits ? and then invert them to get rid of the what was it called "signed area" ? But I don't understand how you got the last step. wouldn't it be .5(u)^-.5 ?
9. Originally Posted by wair
so I just had to chnage my limits ? and then invert them to get rid of the what was it called "signed area" ? But I don't understand how you got the last step. wouldn't it be .5(u)^-.5 ?
No, that's what you get when you differentiate.
$\displaystyle\frac{d}{du}u^{0.5}=0.5u^{-0.5}$
But
$\displaystyle\int{u^{0.5}}du=\frac{u^{1.5}}{1.5}+c$
because
$\displaystyle\frac{d}{du}\left[\frac{u^{1.5}}{1.5}+c\right]=\frac{1.5}{1.5}u^{0.5}$
To integrate, apply differentiation in reverse.
10. Oh right right. I understand now thank you .
11. Ok so for the first one I get -21.1. (1/1.5)(9-x)^1.5. 2(7.45)-2(18)=-21.1. Is that correct ?
12. I get 5.3 for the second one. shouldn't I get the same answer for both ?
13. Originally Posted by wair
Ok so for the first one I get -21.1. (1/1.5)(9-x)^1.5. 2(7.45)-2(18)=-21.1. Is that correct ?
If you work with 9-x instead of u, then there is no need to change the limits of integration.
$2\displaystyle\int_{x=5}^{x=9}{(9-x)^{0.5}}dx=-2\int_{x=5}^{x=9}{(9-x)^{0.5}}d(9-x)$
$=2\displaystyle\int_{x=9}^5{(9-x)^{0.5}}d(9-x)=2\frac{(9-x)^{1.5}}{1.5}$ from x=9 to 5 (start calculations at x=5)
$=2\displaystyle\frac{(9-5)^{1.5}-(5-5)^{1.5}}{1.5}=2\frac{4^{1.5}}{1.5}=2\frac{8}{1.5} =2\frac{16}{3}=\frac{32}{3}$
You must get a positive value for area, so as our graph is symmetrical about the x-axis, we integrate the part above the x-axis and double it.
Using the u-substitution, we get
$\displaystyle\ 2\int_{0}^4{u^{0.5}}du=2\left[\frac{u^{1.5}}{1.5}\right]$ from u=0 to u=4 (start calculations at u=4)
$\displaystyle\ =2\frac{4^{1.5}}{1.5}-0=2\frac{8}{1.5}=\frac{16}{1.5}=\frac{32}{3}$
For the 2nd integral..
$\displaystyle\ 2\int_{0}^2{\left(4-y^2\right)}dy=2\int_{0}^2{4}dy-2\int_{0}^2{y^2}dy=2\left[4y-\frac{y^3}{3}\right]$ evaluated from y=0 to y=2
(start calculations at y=2)
comes out as the same value.
Review your calculations for both integrals.
14. Thank you very much. So if I want to use the new limits I don't need to replacex u with what u equals. And both values must be the same. OK Thank you
15. Originally Posted by wair
Thank you very much. So if I want to use the new limits I don't need to replacex u with what u equals. And both values must be the same. OK Thank you
I think you mean.... "don't need to replace u with 9-x".
Yes, it is simplest to work with "u", having made the substitution, then integrate f(u)du using the "u" limits.
This area you are now calculating is on a different graph but evaluating the new "u" integral will give the same area as the shaded region on the original graph.
If you prefer not to change the limits, here's how to do it....
$\displaystyle\ -2\int_{x=5}^{x=9}{u^{\frac{1}{2}}}du=-2\left[\frac{u^{\frac{3}{2}}}{\frac{3}{2}}\right]$ from x=5 to x=9 (start calculations at x=9)
$=-\displaystyle\frac{4}{3} (9-x)^{\frac{3}{2}}$ from x=5 to x=9
$=-\displaystyle\frac{4}{3}\left[ (9-9)^{1.5}-(9-5)^{1.5}\right]=\frac{32}{3}$
You've got to practice. Keep going until you can reproduce the solution for both integrals. | 2017-11-24T00:56:30 | {
"domain": "mathhelpforum.com",
"url": "http://mathhelpforum.com/calculus/164757-integrating-enclosed-area.html",
"openwebmath_score": 0.9478228688240051,
"openwebmath_perplexity": 643.7519994399651,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9857180681726683,
"lm_q2_score": 0.8688267660487572,
"lm_q1q2_score": 0.8564182414062878
} |
http://mathhelpforum.com/advanced-algebra/87241-method-finding-minimal-polynomial.html | Math Help - Method for finding a Minimal Polynomial?
1. Method for finding a Minimal Polynomial?
Hello, I cannot seem to find in my notes nor in any text a method of finding minimal polynomials. Most minimal polynomials I've encountered are somewhat trivial, but I came across one that I may be tested on that I am unable to solve using a brute-force method:
$m_{\sqrt 2+i}$ over $\mathbb Q$
The reason I ask about a method of finding the min. poly. is because I am taking an oral test, and I don't want to be floundering about if I am asked something different than this one in particular.
Anyway, I would appreciate any help. Thanks!
P.s.: That's m_{\sqrt 2 +i}; sorry that the "i" is so tiny.
2. Originally Posted by Dark Sun
Hello, I cannot seem to find in my notes nor in any text a method of finding minimal polynomials. Most minimal polynomials I've encountered are somewhat trivial, but I came across one that I may be tested on that I am unable to solve using a brute-force method:
$m_{\sqrt 2+i}$ over $\mathbb Q$
The reason I ask about a method of finding the min. poly. is because I am taking an oral test, and I don't want to be floundering about if I am asked something different than this one in particular.
Anyway, I would appreciate any help. Thanks!
P.s.: That's m_{\sqrt 2 +i}; sorry that the "i" is so tiny.
$x=\sqrt{2}+i$
$x^2=(\sqrt{2}+i)^2 \iff x^2=2+2i\sqrt{2}-1 \iff x^2-1=2i\sqrt{2}$
$(x^2-1)^2=(2i\sqrt{2})^2$
$x^4-2x^2+1=-8$
$x^4-2x^2+9=0$
3. Originally Posted by TheEmptySet
$x=\sqrt{2}+i$
$x^2=(\sqrt{2}+i)^2 \iff x^2=2+2i\sqrt{2}-1 \iff x^2-1=2i\sqrt{2}$
$(x^2-1)^2=(2i\sqrt{2})^2$
$x^4-2x^2+1=-8$
$x^4-2x^2+9=0$
you also need to prove that $p(x)=x^4 - 2x^2 + 9$ is irreducible over $\mathbb{Q}$: we only need to show that it cannot be factored into two quadratic polynomials over $\mathbb{Z}.$ why?
suppose $p(x)=(x^2+ax+b)(x^2+cx+d),$ for some $a,b,c,d \in \mathbb{Z}.$ then we'll have $bd=9$ and $b+d+2=a^2,$ which is easily seen to have no solutions in $\mathbb{Z}.$
4. Because irreducible over $\mathbb Z$ implies irreducible over $\mathbb Q$.
$x^4-2x^2+9$ clearly does not have any degree 1 factors. Also, a degree 3 factor would imply a degree 1 factor, so if a factor exists, then it will be degree 2, which you have just shown is a contradiction.
Thank you EmptySet and NonCommAlg for presenting me with this truly powerful method! | 2015-03-05T01:16:04 | {
"domain": "mathhelpforum.com",
"url": "http://mathhelpforum.com/advanced-algebra/87241-method-finding-minimal-polynomial.html",
"openwebmath_score": 0.8886801600456238,
"openwebmath_perplexity": 290.746664530634,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9857180681726682,
"lm_q2_score": 0.8688267660487572,
"lm_q1q2_score": 0.8564182414062876
} |
https://www.physicsforums.com/threads/find-formula-for-series.762785/ | # Find formula for series
1. Jul 23, 2014
### Maxo
1. The problem statement, all variables and given/known data
By using the general equation for an arithmetic series, find a formula for calculating the series
$$1+3+5+...+(2n-1)$$
2. Relevant equations
General equation for an arithmetic series:
$$\sum ^{n}_{k=1}k=\frac{n(n+1)}{2}$$
3. The attempt at a solution
Using the general equation we have
$$\sum ^{n}_{k=1}(2k-1)$$
but how can we go from there? What is the next step?
I tried replacing n with (2k-1) in the general equation but that gives the wrong answer. So how can we do it?
Last edited: Jul 23, 2014
2. Jul 23, 2014
### Mentallic
$$\sum (a+b) = \sum a + \sum b$$
and
$$\sum (ab) = a\sum b$$
if a is a constant that is independent of the summation index (k for example).
So using these formulae, you should be able to solve your problem.
edit:
And keep in mind that
$$\sum_{k=1}^{n}1 = 1+1+1+...+1 = n$$
because you go from k=1 to n while summing 1 each time (hence n times).
3. Jul 23, 2014
### Maxo
Allright, so we have
$$\sum ^{n}_{k=1}(2k-1)=\sum ^{n}_{1}2k-\sum ^{n}_{1}1$$
The second term is easy since we see that we have 1 n times so 1*n = n as you already wrote. Can it be also calculated using the general equation? When I try I get (1(1+1))/2=2/2=1 which is not correct. Also if we just put in n it gets n(n+1)/2 which does not equal n! I guess I am using the general equation wrong? How should it be used then?
Then the first term, how is that calculated? Using the general equation I get 2k(2k+1)/2 = k(2k+1) which is again wrong. Please tell help me how to use the general equation.
4. Jul 23, 2014
### Mentallic
Notice the similarity?
$$\sum^n_1 2k =2\cdot 1+2\cdot 2+2\cdot 3+...+2\cdot n \\ = 2(1+2+3+...+n) \\ = 2\sum^n_1 k$$
Do you know how to prove the general summation formula? If you do, you'll also see why you can't use it in the instances that you've tried to. If not, begin with
$$S=1+2+3+...+n$$
and then consider
$$S=n+...+3+2+1$$
and add both of these values together, term by term (add the values in the same column in pairs)
$$2S=(1+n)+(2+(n-1))+(3+(n-2))+...+((n-2)+3)+((n-1)+2)+(n+1)$$
and simplify the right side, then solve for S.
Now, once you've learned how to prove the sum of the first n integers, now use a similar technique to find the general summation formula
$$S=a+(a+d)+(a+2d)+...+(a+(n-1)d)$$
where a is your starting point, and d is the difference between each value.
5. Jul 23, 2014
### Maxo
Thank you, you explain very well! I understand everything better now.
So if we write this again the opposite direction and then add column wise, we get that each term is equal to 2a+(n-1)d and we have n such terms, so 2S = n(2a+(n-1)d) = 2an + nd(n-1), so S = an + nd(n-1)/2.
In other words, the general formula for an arithmetic sum is:
$$Sn=an+\frac{n^2d-nd}{2}$$
Correct?
That's interesting, I have never seen this more "more general"(?) summation formula before. Does it have some particular name? In my math book, which is supposed to be for university level math, it's not even mentioned.
Btw if you have some suggestions where I can learn more and where it's explained as well as you explain here, please feel free to share it.
Last edited: Jul 23, 2014
6. Jul 23, 2014
### Mentallic
You're welcome! Yes, that's correct! Although, the factored form is more generally used.
$$S_n = \frac{n}{2}\left(2a+(n-1)d\right)$$
and now, if we let d=1 and a=1, then this says we are taking the sum of a linear series that starts with 1 (a=1) and increases by a value of 1 each time (d=1) which would give you your sum of positive integers. If you want to find
$$\sum_{k=1}^n 1$$
then this is like letting a=1 and d=0, which would give you S=n as expected.
Also,
$$\sum_{k=1}^n 2k = 2+4+6+...+2n$$
Which as we can see from the expanded form, we would let a=2 and d=2 since it starts at the value of 2 and increases by a value of 2 between each term. The formula, as well as the method I showed you earlier that moves the 2 outside of the sum would of course give you the same result of S=n(n+1)
It's called an arithmetic progression. Just google that and you'll find plenty on it
7. Jul 24, 2014
### Staff: Mentor
Isn't the formula for the sum of an arithmetic series n(a+l)/2, where a is the first term, l is the last term, and n is the number of terms?
Chet
8. Jul 25, 2014
### HallsofIvy
Yes, it is. Another way of looking at it is that the average value in an arithmetic sequence is the same as the average of the first and last term.
By the way, here's another way of summing that original series, $$\sum_{ix= 1}^n 2n-1$$. That is, of course, the sum of odd integers from 1 to 2n-1. It is the same as the sum of all integers from 1 to 2n minus the sum of all even numbers from 2 to 2n.
The sum of all numbers from 1 to 2n is, using that formula, $2n(1+ 2n)/2= n(2n+ 1)$. The sum of all even numbers is $2+ 4+ \cdot\cdot\cdot+ 2n= 2(1+ 2+ 3+ \cdot\cdot\cdot+ n)= 2(n(n+1)/2)= n(n+ 1)$.
9. Jul 31, 2014
### Maxo
Is there any way to write this general sum using the sigma notation? If so, how could it be written with Sigma notation?
10. Jul 31, 2014
### Mentallic
Well you aren't going to be writing the formula above using sigma notation, because sigma is used as a shorthand form to express summations that follow an expressible pattern. But the arithmetic sum itself can be expressed in sigma notation of course.
$$a+(a+d)+(a+2d)+...+(a+(n-1)d)$$
$$= \sum_{k=0}^{n-1}a+k\hspace{1 mm}d$$
This is a valid summation, but we can do more with it. Notice the constant a is added n times, hence to remove it from the sum, we exchange 'a' in the summation for 'an' outside of the summation. Also the constant d can be factored out of the sum. It's for the same reason that
$$1d+2d+3d+...+nd = d(1+2+3+...+n)$$
$$= an + d\sum_{k=0}^{n-1}k$$
Now at this point, k=0 clearly doesn't give us anything. We only needed k=0 in the first place for when a was within the sum (else if we started with k=1 then we'd only be adding the value of a (n-1) times.). So we can remove k=0 and start at k=1.
$$=an + d\sum_{k=1}^{n-1} k$$
And that would be your summation for a general arithmetic series.
11. Aug 1, 2014
### Maxo
Thank you, I would pretty much call this a perfect answer. :) | 2018-03-22T18:02:26 | {
"domain": "physicsforums.com",
"url": "https://www.physicsforums.com/threads/find-formula-for-series.762785/",
"openwebmath_score": 0.8787103891372681,
"openwebmath_perplexity": 423.95371650000294,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9857180652357771,
"lm_q2_score": 0.8688267677469951,
"lm_q1q2_score": 0.8564182405286218
} |
https://math.stackexchange.com/questions/3465091/how-often-is-the-product-of-the-distinct-prime-factors-of-a-number-greater-than | # How often is the product of the distinct prime factors of a number greater than that of the the next number?
The radical $$\mathrm{rad}\; n$$ of a positive integer $$n$$ is the product of $$n$$'s distinct prime factors. For example $$\mathrm{rad}\; 12=2\cdot 3=6$$.
Denote the sum of $$n$$'s distinct prime factors by $$\mathrm{spf}\; n$$.
Claim: Given any $$k \ge 1$$, for nearly half of positive integers $$n$$, $$\mathrm{spf}\; n > \mathrm{spf}\; (n+k)$$, but for only one-third of positive integers $$n$$, $$\mathrm{rad}\; n > \mathrm{rad}\; (n+k)$$.
Detailed explanation:
For any $$k \ge 1$$, there will be $$n$$ such that $$\mathrm{rad}(n) > \mathrm{rad}(n+k)$$ e.g. $$n=15, k = 1$$. We define
$$a_{n,k} = \begin{cases} 1 & \frac{\mathrm{rad}(n)}{\mathrm{rad}(n+k)} > 1 \\ 0 &\text{ otherwise} \end{cases}$$
Question: What is limiting value
$$\lim_{x \to \infty}\dfrac{1}{x}\sum_{n \le x}a_{n,k}$$
Instead of product when I used sum, the limiting value approached $$0.5$$ which means that it is equally likely the sum of distinct prime factors of a natural number is greater or less than that of the next natural number which makes intuitive sense.
Similarly I was expecting that product of the distinct prime factors of a number is equally likely to be greater than or less than that of the the next number but this was not the case. I was surprised to see that for $$x = 1.5 \times 10^8$$
$$\dfrac{1}{x}\sum_{n \le x}a_{n,k} \approx 0.3386$$ and was decreasing with $$x$$ hence the eventual limit is expected be be slightly less than this.
Note: The initial version of this question was with $$k=1$$ but the result doesn't change with $$k$$, hence I have updated the question.
• $p_n$ for primes $p(n)$ for partitions to be precise :) – Nilotpal Kanti Sinha Dec 6 '19 at 4:30
• Well the Euler product is written $\prod_p \frac1{1-p^{-s}}$. $p$ is reserved for primes in this context, use $\Pi$ or anything. – reuns Dec 6 '19 at 4:35
Each of these products will be $$\frac{n}{p}$$ and $$\frac{n+1}{q}$$ for some $$p,q\in\mathbb{Z}^+$$, where $$p$$ and $$q$$ are the "leftover" primes in the factorizations of $$n$$ and $$n+1$$ when one copy of each distinct prime has been removed.
If $$p\neq q$$, $$\frac{n}{p}>\frac{n+1}{q}\iff p. Which of $$p$$ and $$q$$ is larger will be effectively "random", in that the behavior will jump around a lot and one will be larger than the other with a limiting ratio of $$0.5$$. (A formal proof might prove difficult, but you can certainly gather some numerical evidence to support this - I think it makes as much intuitive sense as the sum case you mentioned.)
When $$p=q$$, the comparison will always go in favor of $$n+1$$. But since $$n$$ and $$n+1$$ are relatively prime, this only happens when $$p=q=1$$ - i.e., when $$n$$ and $$n+1$$ are each squarefree. So we want to find out what fraction of consecutive integer pairs are squarefree, and hence what limiting ratio to exclude from the even split mentioned above.
Fortunately, we can express this as an infinite product. For a number to be squarefree is just for it to not be divisible by $$4, 9, 25, \ldots, p^2,\ldots$$. The limiting fraction of $$(n,n+1)$$ pairs which do not contain a multiple of 4 is $$2/4$$; likewise, for neither to be a multiple of 9 is $$7/9$$, and so on.
However, congruence modulo the first $$N$$ primes is independent over large ranges by the Chinese remainder theorem. So our probability that a "random" pair are both squarefree is
$$\prod_p\frac{p^2-2}{p^2}$$
which looks to be around $$0.322634...$$, or a touch under one third. This means that your limiting value will be
$$\frac12\left(1-\prod_p\frac{p^2-2}{p^2}\right) \approx 0.33868295...$$
If $$n+1$$ is squarefree then $$p(n+1)=n+1\implies a_n=0$$ thus $$\lim\sup_{n\to \infty} \dfrac{1}{x}\sum_{n \le x}a_n \le 1-\frac{6}{\pi^2}\approx 0.393$$
• You can quickly verify with some numerical calculations that 0.39 is much too high. Try n from 1 to 10,000. – RavenclawPrefect Dec 6 '19 at 4:20
• Sure this is just a rigorous upper bound, your approach is a conjectural estimate. It seems hard to make it rigorous, you should show us a plot of how well does it fit. – reuns Dec 6 '19 at 4:23
• For $x = 1.65 \times 10^8$, the value is $0.3386851$ so I guess there is scope to tighten the upper bound. – Nilotpal Kanti Sinha Dec 6 '19 at 4:28
• If your value is correct the convergence should be in $O(x^{-1/2+\epsilon})$ (that's what predicts the random model for the primes) – reuns Dec 6 '19 at 4:29
• Ah, I missed the ≤. Apologies. A graph of the ratios from n=1000 to 100,000 is at imgur.com/3LOWorl, with the blue line the constant estimated to around 9 digits. – RavenclawPrefect Dec 6 '19 at 4:41 | 2020-01-26T03:07:48 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/3465091/how-often-is-the-product-of-the-distinct-prime-factors-of-a-number-greater-than",
"openwebmath_score": 0.9285973310470581,
"openwebmath_perplexity": 237.79102174026804,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9817357189762611,
"lm_q2_score": 0.8723473879530492,
"lm_q1q2_score": 0.8564145901091501
} |
https://math.stackexchange.com/questions/464692/distance-between-k-bit-numbers-in-random-sequence-of-n-bit-numbers | # Distance between k-bit numbers in random sequence of n-bit numbers.
In a random sequence of n-bit numbers, what is the average distance apart of each k-bit number and what is the average distance apart of each odd k-bit number.
Numbers are all positive integers.
Definition: An n-bit number is a number consisting of n bits - set or clear.
Definition: A k-bit number is a number of just k set bits and therefore n-k clear bits.
E.g. in an n-bit number with n=8 there are 8 k-bit numbers where k = 1 and 28 k-bit numbers where k = 2.
Let me clarify with an example.
If I generate a sequence of all 256 8-bit numbers at random I will generate k-bit numbers as 1 0-bit number, 8 1-bit numbers, 28 2-bit numbers etc. Note that the list 1, 8, 28, 56, 70, 56, 28, 8, 1 is a row in Pascals triangle.
Taking only the odd values we get 1, 7, 21, 35, 35, 21, 7, 1 - the next lower row of the triangle.
I would like to get an estimate of the mean distance between these numbers.
In one run of my example I get the positions of each odd 2-bit number as [45, 90, 112, 121, 168, 229, 242] giving gaps of [45, 22, 9, 47, 61, 13] and an average gap of 32.83.
I need to predict this average gap for any n and k. This is beyond my schoolboy maths so I hope someone here can help.
In this example a(8,2) = 32.83. What, for example would a(96,13) be, i.e. the average gap between numbers with 13 bits set in a random sequence of 96-bit numbers. And what would o(96,13) be, i.e. the average gap between odd numbers with 13 bits set in a random sequence of 96-bit numbers.
Please assume good randomness and as many trials as needed to achieve stable averages.
Happy with a value derived from a row of Pascals triangle.
Approximate results from trial runs:
o(10,2)=84.75
o(10,3)=26.28
o(10,4)=11.86
o(10,5)=8.0
• I don't understand: how is $90$ an odd 2-bit number? How is it odd? How is it a 2-bit number, under either your n-bit definition or your k-bit definition? Aug 11 '13 at 1:00
• @Henry - The first odd 2-bit number in my trial run occurred at position 45 (it could have been 3 or 5 or 9 or 17 or 33 or 65 or 129), the second odd 2-bit number occurred at position 90 in the sequence. Does that help? Aug 11 '13 at 1:04
• It still seems strange, but I starting to understand what you are saying. Aug 11 '13 at 1:11
• Perhaps if you ignore the odd requirement it will begin to make sense. I am sure I will be able to extend any ideas you have to only work with odd numbers - it looks like all I need to do is step up one row in Pascals Triangle. Aug 11 '13 at 1:16
If you have $p$ people arranged in a row at random and $s$ of them are special, then the expected average gap between special people is $\frac{p+1}{s+1}$. Imagine having an additional special person, and arranging all $p+1$ people round a circular table with $s+1$ gaps between the special people: the gaps are identically distributed so each have the same expectation, whether or not you ignore the gaps next to the additional person.
So all you need to do is find $s$ and $p$ for your particular question.
There are $2^n$ numbers with $n$ bits. ${n \choose k}$ of them have $k$ bits set, and ${n-1 \choose k-1}$ have $k$ bits set including the last one: these are the numbers in Pascal's triangle you have found.
So the answer to your not-necessarily-odd question is $\dfrac{2^n+1}{{n \choose k}+1}$, while the answer to your odd question is $\dfrac{2^n+1}{{n-1 \choose k-1}+1}$.
If $n=8$ and $k=2$ then this last expression is $\dfrac{257}{8}=32.125$, so your random example of $32.83$ was not too far away. For $n=10$ and $k=2,3,4,5$ it gives about $102.5, 27.7, 12.1, 8.1$, and if I were you I would check the first of these.
• That all looks spot on. Could you expand a little on ${n \choose k}$ please? My schoolboy maths is a little rusty I'm afraid. I'm happy with the discrepancy between my 84.75 against your 102.5, after all this was just one sample run. Aug 11 '13 at 17:35
• ${n \choose k}$ is the number of ways of choosing $k$ different items from $n$ possibilities. Its value appears in Pascal's triangle in the $n$th row, $k$th item along (both counts starting at $0$). Using factorials ${n \choose k}=\frac{n!}{k! (n-k)!}$. For example ${5 \choose 2} = 10$ as there are $10$ was of choosing two items from five. Aug 11 '13 at 18:00
• Works a treat - perfect! Aug 12 '13 at 19:22 | 2021-09-17T12:57:43 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/464692/distance-between-k-bit-numbers-in-random-sequence-of-n-bit-numbers",
"openwebmath_score": 0.6845873594284058,
"openwebmath_perplexity": 226.76213042236165,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9817357173731317,
"lm_q2_score": 0.8723473846343394,
"lm_q1q2_score": 0.8564145854525684
} |
https://math.stackexchange.com/questions/1750285/find-the-product-of-fx-and-gx-given-one-of-them | # Find the product of $f(x)$ and $g(x)$ given one of them
I'm given the final answer which is $$(g \cdot f)(x) = \frac{1}{x^2+4}\;.$$
Also, i'm given $f(x) = x^2+1$.
I've solved this using the composition, however the second part of the question asks me to find the $g(x)$ which would make this multiplication true. How would I do this? Do I divide the final answer by $f(x)$?
• is that $$(f\cdot g)(x)=\frac1{x^2+4}$$ or $$(f\cdot g)(x)=\frac1{x^2}+4\;?$$ Either way, the answer to your question is yes. – Brian M. Scott Apr 19 '16 at 21:34
• the first one, and how would i go about doing that? I know once I divide i take the reciprocal of g(x) and multiply using the FOIL method, however my answer makes no sense – Saad Siddiqui Apr 19 '16 at 21:35
• thank you for the edit, the question is now formatted correctly. My math is however not giving me the correct answer, could someone guide me through it? – Saad Siddiqui Apr 19 '16 at 21:35
• Your best bet is to learn to use basic MathJax; there’s a tutorial here. Until then, be sure to use enough parentheses to make your expressions completely unambiguous. – Brian M. Scott Apr 19 '16 at 21:37
• For the division, just do it: $$\frac{\frac1{x^2+4}}{x^2+1}=\frac1{x^2+4}\cdot\frac1{x^2+1}=\ldots$$ – Brian M. Scott Apr 19 '16 at 21:38
Yes, becasue by definition $(g\cdot f)(x)=g(x)\cdot f(x)$. Hence $g(x)=\frac1{(x^4+1)(x^2+1)}$.
If we had $(g\circ f)(x)=\frac1{x^4+1}$ instead, one possible $g$ would be $g(x)=\frac1{(x-1)^2+4}$.
• when taking the reciprocal and multiplying, I get a quartic function which makes no sense – Saad Siddiqui Apr 19 '16 at 21:39
• @SaadSiddiqui $\frac1{(x^4+4)(x^2+1)}\cdot(x^2+1)=\frac1{x^4+4}$ – Hagen von Eitzen Apr 19 '16 at 21:40
• Why does a quartic function make no sense?... or really a ratio of 1/ a quartic function. – Doug M Apr 19 '16 at 21:40
• well when multiplying out the denominators, I get an answer of x^4 + 5x^2 + 5. I'm struggling to understand how this, when multiplied by f(x) which is x^2+1 gives me the final answer. – Saad Siddiqui Apr 19 '16 at 21:44
So you are told that $(g \times f)(x) = \dfrac{1}{x^{2} + 4}$, and also told that $f(x) = x^{2} + 1$.
$(g \times f)(x)$ is just the name we give for the product of the two functions, i.e., $(g \times f)(x)$ really means $g(x)f(x)$.
So we know what this product is. It is $g(x)f(x) = \dfrac{1}{x^{2} + 4}$. We also know that $f(x) = x^{2} + 1$. So that means:
$$g(x)(x^{2} + 1) = \dfrac{1}{x^{2} + 4}$$
and to solve for $g(x)$, just divide both sides by $x^{2} + 1$ to get:
$$g(x) = \dfrac{\left (\frac{1}{x^{2} + 4} \right )}{x^{2} + 1}$$
Now, how do we simplify this? Well, $x^{2} + 1$ is the same as $\dfrac{x^{2} + 1}{1}$, so the fraction is really $$\dfrac{\left (\frac{1}{x^{2} + 4} \right )}{\left (\frac{x^{2} + 1}{1}\right)}$$ and when we divide two fractions, we invert the bottom one and multiply, so we get:
$$\dfrac{1}{x^{2} + 4} \cdot \dfrac{1}{x^{2} + 1}$$
And this is just $\dfrac{1}{(x^{2} + 4)(x^{2} + 1)}$, which is your final answer (unless you want to multiply the denominator out using the FOIL method).
• I got to this last part, but it's the multiplying out which has stumped me – Saad Siddiqui Apr 19 '16 at 21:41
• @SaadSiddiqui Ok, so $(x^2 + 4)(x^{2} + 1) = x^{4} + x^{2} + 4x^{2} + 4 = x^{4} + 5x^{2} + 4$, so your final answer should be $\dfrac{1}{x^{4} + 5x^{2} + 4}$. – layman Apr 19 '16 at 21:42
• I got the same answer, but how does this (when multiplied with the original f(x) ) give me my final answer? – Saad Siddiqui Apr 19 '16 at 21:44
• @SaadSiddiqui Well, first you need to understand that $x^{4} + 5x^{2} + 4 = (x^{2} + 4)(x^{2} + 1)$, which we know since we multiplied the right hand side out to get the left hand side. So $\dfrac{1}{x^{4} + 5x^{2} + 4} = \dfrac{1}{(x^{2} + 4)(x^{2} + 1)}$. So multiplying the left hand side fraction by $x^{2} + 1$ should give the same thing as multiplying the right hand side fraction by $x^{2} + 1$, since the fractions are equal. But what happens when we multiply the right hand side fraction by $x^{2} + 1$? We get $(x^{2} + 1) \cdot \dfrac{1}{(x^{2} + 4)(x^{2} + 1)}$, which is... – layman Apr 19 '16 at 21:47
• Yes completely, by not multiplying out the denominator, you can simply cancel out in the next step. Thank you so much! – Saad Siddiqui Apr 19 '16 at 21:54 | 2020-10-24T01:22:31 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/1750285/find-the-product-of-fx-and-gx-given-one-of-them",
"openwebmath_score": 0.8249678611755371,
"openwebmath_perplexity": 246.3432266685997,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9817357237856482,
"lm_q2_score": 0.8723473730188543,
"lm_q1q2_score": 0.8564145796431738
} |
http://math.stackexchange.com/questions/1600227/theoretical-question-about-the-rank-and-existence-of-an-inverse-of-a-block-matri | # Theoretical question about the rank and existence of an inverse of a block matrix,
Let A and B be two $n \times n$ square matrices with complex coefficients, and consider the $2n \times 2n$ matrix $M$ given by
$$M = \begin{bmatrix} A & A \\ A & B \\ \end{bmatrix}$$
1. Determine the rank of $M$ in terms of $A$ and $B$.
2. What is the condition for $M$ to have an inverse $M^{-1}$? Compute $M^{-1}$ when it exists.
Any ideas on how to get started on this problem are welcome.
Thanks,
-
If $A$ is rank deficient, then $M$ is also rank deficient. However in the case that $A$ is full rank, there's not a lot we can say except that the rank of $M$ is at least the rank of $A$ (this is so whatever the rank of $A$ is). – hardmath Jan 4 at 23:18
You can do Gaussian elimination to compute the inverse, except instead of working with individual entries, instead you work with entire blocks. The only difference is that you need to keep track of the order of the blocks, as they won't commute. – Nick Alger Jan 4 at 23:18
Hi @hardmath, thanks so much, yes, I agree with you. I've had this answer for awhile and am thinking, "there has to be many more cases to consider". It looks like you also think that there is not much more to say...hmmm... – User001 Jan 4 at 23:21
Hi @NickAlger, would it also be correct, if I augmented this $2nx2n$ matrix with a $2nx2n$ identity matrix I, and then row-reduce until I get a $2nx2n$ identity matrix on the left side? And, are you saying to do something different, like this: get $nxn$ identity blocks, so, 4 in total, on the left side? So, start with 4 $nxn$ identity blocks on the right-side of the augmented matrix? Thanks so much, – User001 Jan 4 at 23:25
@User001 If you want to go the augmented route, then you can augment with the 2n-by-2n identity. In the meantime it looks like some people have answered explaining this in more detail. If you are unsure of which is right, you can always multiply out the matrix times it's inverse and see whether you get the identity back again. – Nick Alger Jan 4 at 23:49
## 4 Answers
\begin{align} [ M \mid I ] &= \left[ \begin{array}{cc|cc} A & A & I & 0 \\ A & B & 0 & I \end{array} \right] \\ &\to \left[ \begin{array}{cc|cc} A & A & I & 0 \\ 0 & B-A & -I & I \end{array} \right] \\ & \to \left[ \begin{array}{cc|cc} I & I & A^{-1} & 0 \\ 0 & I & -(B-A)^{-1} & (B-A)^{-1} \end{array} \right] \\ & \to \left[ \begin{array}{cc|cc} I & 0 & A^{-1} +(B-A)^{-1} & -(B-A)^{-1} \\ 0 & I & -(B-A)^{-1} & (B-A)^{-1} \end{array} \right] = [ I \mid M^{-1} ] \end{align}
-
@mvw....wow...this is too cool...! :-) Thanks so much! – User001 Jan 4 at 23:56
The derivation above uses that the block matrices are like numbers themselves to a certain degree, matrices form a ring, so that some algorithms from linear algebra can be applied to them. – mvw Jan 5 at 0:08
Hi @mvw, in your second row, after just one row operation, we already can determine the rank of M, which is just rank(A) + rank (B-A), since the the matrix is block upper-triangular (and also row operations are rank-preserving). Also, the determinant of $M$ is just det(A)*det(B-A), because of the block-diagonal structure of the matrix, which shows that $M^{-1}$ exists if and only if det(A)*det(B-A) is non-zero (also, your first row operation is a type that preserves the determinant, too). – User001 Jan 5 at 0:33
And finally, computation of the inverse is what you have shown, with block row-operations on an augmented matrix [M | I]. Do I have it all correct? Thanks so much @mvw, – User001 Jan 5 at 0:33
I just wanted to get the inverse. But the intermediate forms seem to be already useful for your other issues. – mvw Jan 5 at 0:48
Note that $$M= \underbrace{\begin{bmatrix}I&0\\I&I\end{bmatrix}}_{=P} \underbrace{\begin{bmatrix}A&0\\0&B-A\end{bmatrix}}_{=M^\prime} \underbrace{\begin{bmatrix}I&I\\0&I\end{bmatrix}}_{=Q}$$ But $P$ and $Q$ are invertible so $\DeclareMathOperator{rank}{rank}\rank M=\rank M^\prime$. Since $M^\prime$ is block-diagonal we have $$\rank M=\rank A+\rank(B-A)$$
To see why $\rank M=\rank M^\prime$, note the two facts:
Fact 1. Let $A$ be an $m\times n$ matrix. If $B$ is an $n\times k$ matrix of rank $n$, then $\rank(AB)=\rank(A)$.
In our case take $A=M$ and $B=Q^{-1}$. This implies that $\rank(MQ^{-1})=\rank(M)$.
Fact 2. Let $A$ be an $m\times n$ matrix. If $C$ is an $l\times m$ matrix of rank $m$, then $\rank(CA)=\rank(A)$.
In our case take $A=M^\prime$ and $C=P$. This implies that $\rank(PM^\prime)=\rank(M^\prime)$.
Since $MQ^{-1}=PM^\prime$ we see that $$\rank(M)=\rank(MQ^{-1})=\rank(PM^\prime)=\rank(M^\prime)$$ For more about rank, see wikipedia. Also, try to prove these facts yourself!
-
Hi @BrianFitzpatrick, really cool factorization :-). Can I ask you a naive follow-up question? Since $P$ and $Q$ are both invertible, why does determining the rank of $M$ just reduce to checking the rank of $M'$? The only rule / theorem that I know about rank is that rank (AB) = rank (BA), and then some rank inequalities proved very early in the standard textbooks...thanks, – User001 Jan 4 at 23:39
...that's why I feel that when I go to solve a theoretical question regarding rank, inverse or trace, I feel that I have so little tools at my disposal to work with...unlike advanced calculus and complex analysis questions, where I usually immediately see at least two or three ways to approach the problem...@BrianFitzpatrick, thanks, – User001 Jan 4 at 23:41
@User001 See the properties of rank here: en.wikipedia.org/wiki/Rank_(linear_algebra)#Properties – Brian Fitzpatrick Jan 4 at 23:42
Hi @BrianFitzpatrick, with the matrix $M$, and following all of the comments and answers I've gotten so far, I block-row-reduced, but stopped after just one operation, as I think it may be enough: I knocked out the lower left block matrix A. The entries left are A,A on the first row and B in the lower-right corner. With this matrix, which is block-upper-triangular, we can conclude that the rank of $M$ is the rank of A + the rank of (B-A), by standard (non-block) row-reduction techniques. Is my thinking correct? – User001 Jan 5 at 0:13
Also, from this reduced matrix, the determinant of $M$ is just det(A)*det(B-A) (I believe it works because of the zero block matrix below or above the diagonal), as the row operation of adding a scalar multiple of a column / row to another column / row does not change the determinant. Also, my reason for reading off the rank of A and (B-A) to conclude the rank of $M$ is the same reason: row operations are rank-perserving. What do you think? Thanks, @BrianFitzpatrick, – User001 Jan 5 at 0:13
In this case we have rank($M$) = rank($A$) + rank($B-A$). So we can say a lot about the rank of $M$, just not in terms of rank($A$) and rank($B$).
-
Hi @hardmath, why rank($B-A$)? Thanks, – User001 Jan 4 at 23:33
Block row elimination: subtract top row blocks from bottom row blocks (or the equivalent block column elimination). – hardmath Jan 4 at 23:36
First note that $$\begin{bmatrix} A & A \\ A & B \\ \end{bmatrix} = \begin{bmatrix} A & 0 \\ 0 & I \\ \end{bmatrix} \begin{bmatrix} I & 0 \\ A & I \\ \end{bmatrix} \begin{bmatrix} I & 0 \\ 0 & (B-A) \\ \end{bmatrix} \begin{bmatrix} I & I \\ 0 & I \\ \end{bmatrix}$$ (You can derive this by doing the steps in Gaussian elimination and then writing them as multiplications by "primitive" operation matrices.)
So you can immediately see that $M$ is invertible if and only if both $A$ and $B-A$ are invertible.
Now in that product, the second and fourth matrices are of full rank. Thus the product of the three rightmost matrices will have the same rank deficiency as that of $B-A$. And then the rank deficiency of $M$ is always somewhere between the sum of the deficiencies of $A$ and $B-A$ as an upper bound, and the minimum of those two deficiencies as a lower bound.
- | 2016-06-30T16:26:09 | {
"domain": "stackexchange.com",
"url": "http://math.stackexchange.com/questions/1600227/theoretical-question-about-the-rank-and-existence-of-an-inverse-of-a-block-matri",
"openwebmath_score": 0.9768921732902527,
"openwebmath_perplexity": 338.30605092368046,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9817357216481429,
"lm_q2_score": 0.8723473614033683,
"lm_q1q2_score": 0.8564145663751892
} |
http://vanmauchonloc.vn/tulsa-social-rknq/d31aad-find-function-from-points-calculator | Wed, 12 / 2020 6:16 am |
If you know two points that fall on a particular exponential curve, you can define the curve by solving the general exponential function using those points. Instructions: Use this step-by-step Logarithmic Function Calculator, to find the logarithmic function that passes through two given points in the plane XY. Can you find a 4th order polynomial? To derive the equation of a function from a table of values (or a curve), there are several mathematical methods. For example, the points (0,0), (1.40,10), (2.41,20), and (4.24,40) would yield the cubic function y=-0.18455x^3+1.84759x^2+4.191795+0. Linear equation with intercepts. Function point = FP = UFP x VAF. no data, script or API access will be for free, same for Function Equation Finder download for offline use on PC, tablet, iPhone or Android ! Also I would define it in single line as "A Method of quantifying the size and c… How to reconstruct a function? Indeed, by dividing both sides of the equations: In order to solve for $$A_0$$ we notice from the first equation that: It is not always growth. In practice, the type of function is determined by visually comparing the table points to graphs of known functions. The idea of this calculator is to estimate the parameters $$A_0$$ and $$k$$ for the function $$f(t)$$ defined as: so that this function passes through the given points $$(t_1, y_1)$$ and $$(t_2, y_2)$$. Find Equation Of Exponential Function Given Two Points Calculator Tessshlo. BYJU’S online function calculator tool makes the calculations faster, and it displays the graph of the function by calculating the x and y-intercept values, slope values in a fraction of seconds. The parameter $$k$$ will be zero only if $$y_1 = y_2$$ (the two points have the same height). In L1, enter the x-coordinates given. Can you find a line that goes through them? Curve sketching means you got a function and are looking for roots, turning and inflection points. Tool to find the equation of a function from its points, its coordinates x, y=f(x) according to some interpolation methods and equation finder algorithms. the 5 parameters provided in the question, VAF = Value added Factor i.e. Sure, lot of them - an infinite number of them. We need to find a function with a known type (linear, quadratic, etc.) This calculator will find either the equation of the parabola from the given parameters or the axis of symmetry, eccentricity, latus rectum, length of the latus rectum, focus, vertex, directrix, focal parameter, x-intercepts, y-intercepts of the entered parabola. Line through two points Instructions: Use this step-by-step Exponential Function Calculator, to find the function that describe the exponential function that passes through two given points in the plane XY. New coordinates by rotation of points. UFP = Sum of all the complexities i.e. Inflection Point Calculator is a free online tool that displays the inflection point for the given function. Press [STAT]. The simple function point method can be used on any piece of software to be developed, however the number of function points estimated for engineering projects may lack precision. 2009/06/10 09:07 Male/30 level/An office worker/Very/ Purpose of use Initially, for a watch manual - but finding these computations if fantastic! This means: You calculate the difference of the y-coordinates and divide it by the difference of the x-coordinates. How to calculate the equation of a linear function from two given points? In case you have any suggestion, or if you would like to report a broken solver/calculator, please do not hesitate to contact us. Trending Posts. a bug ? Also, explore hundreds of other calculators addressing math, finance, health, fitness, and more. You need to provide the points $$(t_1, y_1)$$ and $$(t_2, y_2)$$, and this calculator will estimate the appropriate exponential function and will provide its graph. Yes. Linear equation with intercepts. dCode tries to propose the most simplified solutions possible, based on affine function or polynomial of low degree (degree 2 or 3). An online curve-fitting solution making it easy to quickly perform a curve fit using various fit methods, make predictions, export results to Excel,PDF,Word and PowerPoint, perform a custom fit through a user defined equation and share results online. The shortest distance between two points on the surface of a sphere is an arc, not a line. find power function from two points calculator, Power in physics is the amount of work done divided by the time it takes, or the rate of work.Here’s what that looks like in equation form: Assume you have two speedboats of equal mass, and you want to know which one will … For an introduction to what are Function Points please read my earlier article here. Second calculator finds the line equation in parametric form, that is, . 0.65 + (0.01 * TDI), TDI = Total Degree of Influence of the 14 General System Characteristics. 3. Make use of the below calculator to find the vertical asymptote points and the graph. Enter the point and slope that you want to find the equation for into the editor. How to Use the Calculator. Roots at and Further point on the Graph: It also shows plots of the function and illustrates the domain and range on a number line to enhance your mathematical intuition. How to find an equation from a set of points. Clear any existing entries in columns L1 or L2. To graph a parabola, visit the parabola grapher (choose the "Implicit" option). To find the equation of sine waves given the graph: Find the amplitude which is half the distance between the maximum and minimum. First calculator finds the line equation in slope-intercept form, that is, . Cartesian to Polar coordinates. Indeed, if the parameter $$k$$ is positive, then we have exponential growth, but if the parameter $$k$$ is negative, then we have exponential decay. New coordinates by rotation of axes. Definition: A stationary point (or critical point) is a point on a curve (function) where the gradient is zero (the derivative is équal to 0). Exponential functions, constant functions and polynomials are also supported. When 3 points are input, this calculator will generate a second degree equation. Write The Equation For Photosynthesis. But "Why re-invent the wheel?" Help. absolute extreme points f ( x) = ln ( x − 5) $absolute\:extreme\:points\:f\left (x\right)=\frac {1} {x^2}$. Mathepower finds the function. What we do here is the opposite: Your got some roots, inflection points, turning points etc. If you are familiar with graphing algebraic equations, then you are familiar with the concepts of the horizontal X-Axis and the Vertical Y-Axis. Learn how to use the Stat plot feature of the TI-84+ Calculator to find the equation of those points. absolute extreme points f ( x) = 1 x2. The calculator will find the domain, range, x-intercepts, y-intercepts, derivative, integral, asymptotes, intervals of increase and decrease, critical points, extrema (minimum and maximum, local, absolute, and global) points, intervals of concavity, inflection points, limit, Taylor polynomial, and graph of the single variable function. The method used to calculate function point is knows as FPA (Function Point Analysis). My question is - Can I find the function using points? As a result we should get a formula y=F(x), named the empirical formula (regression equation, function approximation), which allows us to … It accepts inputs of two known points, or one known point and the slope. How about an ellipse or hyperbola? In practice, this means substituting the points for y and x in the equation y = ab x. Instructions: Use this step-by-step Exponential Function Calculator, to find the function that describe the exponential function that passes through two given points in the plane XY. dCode is free and its tools are a valuable help in games, maths, geocaching, puzzles and problems to solve every day!A suggestion ? The vertical graph occurs where the rational function for value x, for which the denominator should be 0 and numerator should not be equal to zero. The method used to calculate function point is knows as FPA (Function Point Analysis). Are there any convenient websites out there with this functionality? Method 2: use a interpolation function, more complicated, this method requires the use of mathematical algorithms that can find polynomials passing through any points. A Function Calculator is a free online tool that displays the graph of the given function. What about a circle that touches the two points? It was easy example, but my graphic is not a liner function. A stationary point is therefore either a local maximum, a local minimum or an inflection point.. an idea ? Area of a triangle with three points. Primarily, you have to find … and are looking for a function having those. Let's take a simple case: two points. Example: a function has for points (couples $(x,y)$) the coordinates: $(1,2) (2,4), (3,6), (4,8)$, the ordinates increase by 2 while the abscissas increase by 1, the solution is trivial: $f (x) = 2x$. equation,coordinate,curve,point,interpolation,table, Source : https://www.dcode.fr/function-equation-finder. Thank you! Yes. Function Point Calculator: Main Description Details Uses: Calculator. These online calculators find the equation of a line from 2 points. Also I would define it in single line as "A Method of quantifying the size and complexity of a software system in terms of the functions that the system delivers to the user". NB: for a given set of points there is an infinity of solutions because there are infinite functions passing through certain points. You need to provide the points $$(t_1, y_1)$$ and $$(t_2, y_2)$$, and this calculator will estimate the appropriate exponential function and will provide its graph. Of course? There can be various methods to calculate function points; you can define your custom too based on your specific requirements. First, we have to calculate the slope m by inserting the x- and y- coordinates of the points into the formula . What about a circle that touches the two points? y=F(x), those values should be as close as possible to the table values at the same points. You need to provide the points $$(t_1, y_1)$$ and $$(t_2, y_2)$$, and this calculator will estimate the appropriate exponential function and will provide its graph. Area of a triangle with three points. Functions: What They Are and How to Deal with Them, Normal Probability Calculator for Sampling Distributions, Exponential Function Calculator from Two Points, exponential function calculator given points. You will be asked to provide simple estimates of the software you plan to develop. Intersection of two lines. absolute extreme points y = x x2 − 6x + 8. find power function from two points calculator, The terminal coordinates program may be used to find the coordinates on the Earth at some distance, given an azimuth and the starting coordinates. (Try this with a string on a globe.) By using this website, you agree to our Cookie Policy. Casio is a great company, providing durable products. This free slope calculator solves for multiple parameters involving slope and the equation of a line. Method 1: detect remarkable solutions, like remarkable identities, it is sometimes easy to find the equation by analyzing the values (by comparing two successive values or by identifying certain precise values). Two point form calculator This online calculator can find and plot the equation of a straight line passing through the two points. Cartesian to Polar coordinates. The blue line is my function. What about a cubic? New coordinates by rotation of axes. Wolfram|Alpha is a great tool for finding the domain and range of a function. Polar to Cartesian coordinates Of course you can. Thus function points can be calculated as: Polar to Cartesian coordinates Visualize the exponential function that passes through two points, which may be dragged within the x-y plane. Given the 3 points you entered of (14, 4), (13, 16), and (10, 18), calculate the quadratic equation formed by those 3 pointsCalculate Letters a,b,c,d from Point 1 (14, 4): b represents our x-coordinate of 14 a is our x-coordinate squared → 14 2 = 196 c is always equal to 1 Intersection of two lines. In calculus we know that we can figure out the curve of a cubic function by simply knowing the location of four points. Linear equation given two points. To derive the equation of a function from a table of values (or a curve), there are several mathematical methods.. If the period is more than 2π then B is a fraction; use the formula period = 2π/B to find … Press [STAT] again. Distance between the asymptote and graph becomes zero as the graph gets close to the line. 1 - Enter the x and y coordinates of three points A, B and C and press "enter". I … a feedback ? Linear equation given two points. Analyze the critical points of a function and determine its critical points (maxima/minima, inflection points, saddle points) symmetry, poles, limits, periodicity, roots and y-intercept. Two equations are displayed: an exact one (top one) where the coefficients are in fractional forms an the second with approximated coefficients whose number of … Technically, in order to find the parameters you need to solve the following system of equations: Solving this system for $$A_0$$ and $$k$$ will lead to a unique solution, provided that $$t_1 = \not t_2$$. Enter any number (even decimals and fractions) and our calculator will calculate the the slope intercept form (y=mx+b), point slope (y-y1)= m(x-x1) and the standard form (ax+by=c). Have a tried and tested method given by IFPUG by their experiences and case study the. Watch manual - but finding These computations if fantastic passing through the two points additional point the. 'Re ok with this functionality if I know only points and tested method given by by. This, but you can opt-out if you wish: points\: f\left ( x\right ) =\sqrt { }... Amplitude which is the horizontal X-Axis and the slope ) =\sqrt { x+3 } $the following points!, interpolation, Newtonian interpolation and Neville interpolation, visit the parabola grapher ( choose ... Description Details Uses: calculator mean find formula/function f ( x ), there are an infinite of! 1 x2 known points, or one known point and the slope the equation of exponential function, a... And press Enter '' and use the arrow buttons on your specific requirements 09:07! Male/30 level/An office worker/Very/ Purpose of use Initially, for a watch manual - but finding These if... The Vertical Y-Axis a globe. the Logarithmic function calculator, to find the Vertical.! Points ( with integer x- and y-values ) an infinite number of them point, interpolation table. Function, use a graphing calculator to find … exponential functions, constant functions and polynomials are supported! We 'll assume you 're ok with this, but my graphic calculator, an... Of solutions because there are infinite functions passing through the two points to calculate the difference of the to. The type of function is determined by visually comparing the find function from points calculator values at same... This website, you agree to our Cookie Policy - an infinite number of functions can! A sphere is an infinity of solutions because there are infinite functions passing through the two points calculator.! You have to find … exponential functions, constant functions and polynomials are also.! 'Re ok with this, but you can opt-out if you wish if the for. There are several mathematical methods =\sqrt { x+3 }$ } $0-127 right! Lazy to find my graphic calculator, calculating an equation from a table values!, B and C and press Enter '' intercept parameters and displays on. It displays the inflection point of function is determined by visually comparing the points. For finding the domain and range on a graph, we have to find the Logarithmic function passes! Wolfram|Alpha is a free online tool that displays the inflection point in a fraction of seconds online equation... Y = ab x Neville interpolation it by the difference of the calculator... Values at the same points degree of Influence of the points will snap to the points. Or an inflection point calculator is a great company, providing durable products graph a parabola, visit the grapher! Computer, you agree to our Cookie Policy already have a tried and tested method by. That there are several mathematical methods also has the ability to provide step by step solutions estimates of the into. B and C and press Enter '' have the following data points, points. That displays the inflection point for the function which is the opposite: got... Use Initially, for a given set of points there is an arc, a! Use Initially, for a watch manual - but finding These computations fantastic... The following data points, turning points etc. can you find a line estimates of the given.. Of the 14 General System Characteristics the domain and range of a line the! Byju ’ S online inflection point this website, you agree to our Cookie Policy you ok. A local minimum or an inflection point for the given function values at the same points, has... I have the following data points, turning points etc. great tool for finding the and... Lazy to find the period of the points will snap to the grid points with... A curve ), TDI = Total degree of Influence of the points is 0, which means the and. Second calculator finds find function from points calculator line equation in slope-intercept form, that is, software costs! To develop my question is - can I find the equation of a is... Roots, inflection points, or one known point and slope that you want to find the equation of points! For help find function from points calculator the calculation faster, and it displays the inflection point calculator makes... { x } { x^2-6x+8 }$ are infinite functions passing through the two points on the surface a..., point, interpolation, Newtonian interpolation and Neville interpolation number of them Discord for requests... And Neville interpolation two given points in the question, VAF = Value added Factor i.e:! A function from a table of values ( or a curve ), there are an number... Best 'Function equation Finder ' tool, so feel free to write of functions that can go through set! Details Uses: calculator a 6th, a 6th, a 7th polynomial! Function to repeat we do here is the horizontal X-Axis and the of. Great company, providing durable products also supported local maximum, a 7th order polynomial Vertical Y-Axis, my... Step-By-Step explanation on how to find the amplitude which is half the between... } { x^2-6x+8 } $find from two points calculator this online calculator can find and plot the of. Parabola, visit the parabola grapher ( choose the Implicit '' option.! The distance between two points are there any convenient websites out there with this functionality with graphing algebraic equations then. Period of the points will snap to the table points to graphs of known functions infinity! Ok with this functionality, so feel free to write by the difference of y-coordinates... I find the function using points introduction to what are function points please read my earlier article here {! Of a line from 2 points find from two on curve function find function from points calculator you castle learning reference writing that through! Try this with a string on a number line to enhance your mathematical.... Point on the graph the curve of an exponential function, use a calculator! Between the maximum and minimum a watch manual - but finding These computations if fantastic x } { x^2-6x+8$... In practice, this means substituting the points for y and x in the question, VAF Value... Use this step-by-step Logarithmic function that passes through two given points in the question, VAF = Value added i.e... Be as close as possible to the grid points ( with integer x- and y-values ) x- and coordinates! Given 2 points find from two on curve function finding you castle learning reference writing passes., interpolation, Newtonian interpolation and Neville interpolation, VAF = Value added Factor i.e slope by. Of a line Value added Factor i.e company, providing durable products a explanation..., B and C and press Enter '' the following data points, ( left column... + 8 These computations if fantastic function which is half the distance between two on...
find function from points calculator
Rate this post | 2021-04-12T01:07:49 | {
"domain": "vanmauchonloc.vn",
"url": "http://vanmauchonloc.vn/tulsa-social-rknq/d31aad-find-function-from-points-calculator",
"openwebmath_score": 0.6647066473960876,
"openwebmath_perplexity": 607.7967962292461,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_q1_score": 0.9817357195106374,
"lm_q2_score": 0.8723473614033683,
"lm_q1q2_score": 0.8564145645105419
} |
http://math.stackexchange.com/questions/271167/what-is-the-definition-of-first-last-element-in-a-poset?answertab=oldest | # What is the definition of first/last element in a poset?
I have read the terms first element/last elements in the context of a basic course in set theory.
When I learned about posets I didn't encounter those terms. I tried looking up the definitions but I didn't find them.
Can someone please write down the definitions for first/last element in a poset ?
-
The first element of a poset $\langle P,\le\rangle$ is simply the unique minimum element of $P$, if there is one: $p_0$ is the first element of $P$ if $p_0\le p$ for all $p\in P$. Similarly, the last element of $P$ is the unique maximum element of $P$, if there is one: $p_1$ is the last element of $P$ if $p\le p_1$ for all $p\in P$.
A poset need not have a first or last element.
-
Thanks Brian! I will accept when the system lets me (3 minutes) – Belgi Jan 5 '13 at 22:09
@Belgi: As always, you’re welcome! – Brian M. Scott Jan 5 '13 at 22:09
If by "first" element you mean "an element preceding all others", then there need not be one (similarly for "last" element).
Consider the sets $\{1\}$, $\{2\}$, $\{3\}$, $\{1,2\}$, and $\{1,3\}$ ordered by inclusion. There is no first or last element, but there are three minimal elements (an element with no predecessor) and two maximal elements (an element with no successor).
In the case where there is an element preceding all others, it is generally called a minimum element. Similarly, the element that is preceded by all others is called a maximum element.
-
So $a$ is first if $a\leq b$ for all other elements $b$ ? – Belgi Jan 5 '13 at 22:05
@Belgi That is a reasonable definition for "first", but I think "minimum" is a more common term for it. – Austin Mohr Jan 5 '13 at 22:06
Thanks for the answer Austin, I upvoted it – Belgi Jan 5 '13 at 22:08
A poset need not have a first or last element.
• An element $a_1$ is first (the unique minimum) in a poset $P$ if $a_1 \le a\;\;\forall a \in P$.
• An element $a_n$ is last (the unique maximum) in a poset P if $a \le a_n \;\;\forall a \in P$.
(1) There may not be a "minimum" (first) nor "maximum" (last) element in a poset.
(2) You might have a unique minimum (first) element, but no maximum element (last) in a poset.
(3) Likewise, a poset may not have a minimum "first" element, but have a maximum ("last") element.
(4) And if there is a unique minimum and a unique maximum element in the poset, then those are the first and last elements in the poset, respectively.
Exercise: try to find an "example" poset that models/represents each of these four possibilities) (one per possibility)!
-
But I like yours – Babak S. Mar 17 '13 at 9:31 | 2015-07-29T22:36:50 | {
"domain": "stackexchange.com",
"url": "http://math.stackexchange.com/questions/271167/what-is-the-definition-of-first-last-element-in-a-poset?answertab=oldest",
"openwebmath_score": 0.7926149964332581,
"openwebmath_perplexity": 452.4268497006021,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.975946446405965,
"lm_q2_score": 0.8774767874818409,
"lm_q1q2_score": 0.8563703525466247
} |
http://lxlj.emporiumgalorium.de/alternating-series-test.html | # Alternating Series Test
Share a link to this widget: More. Infinite series whose terms alternate in sign are called alternating series. 6 Alternating Series and Conditional Convergence Page 1 Theorem 15 - The Alternating Series Test The series 1 12 3 4 1 (1)n n n uuu uu ∞ + = ∑− =−+−+" converges if the following conditions are satisfied: 1. 1 2 1 4 1 16 1 256. Express the series in a sum End,tD÷ = 1-÷t±-¥tF ' Et. A retrospective record review of performance on the three-step Luria test was conducted on 383 participants from a university-based dementia clinic. The sequence of (positive) terms b n eventually decreases. for all of you pump and level control needs including alternating relays and isolated switches. The terms alternate, and the computation above shows that the terms decrease in absolute value. our series will diverge. This is a very useful lecture in Calculus. The series is an absolutely convergent series; The series is a conditionally convergent series; The series diverges to or ; The partial sums of the series have differing values of limit superior and limit inferior. No, it does not establish the divergence of an alternating series unless it fails the test by violating the condition lim_{n to infty}b_n=0, which is essentially the Divergence Test; therefore, it established the divergence in this case. By the ratio test, the power series converges. Let’slookateachpartmoreclosely: (n+1)!: Prettyself-explanatory. However, given the same series Ʃ((-1)^n)(a_{n}), if I apply condition 1 of the alternating series test, which is the Nth term test on just the (a_{n}) portion, the test is inconclusive if the limit is anything other than zero. Alternating Series Remainder. So here is a good way of testing a given alternating series: if you. Proof Example of a divergent series. Suppose that: 1. Alternating series test - Series, Calculus, Mathematics Summary and Exercise are very important for perfect preparation. Series Convergence Tests Math 122 Calculus III D Joyce, Fall 2012 Some series converge, some diverge. Alternating Series and P-series "convergence" I couldn't resist trying out a pun. Overview of Ratio Test; 4 Examples; Root Test. Trust Ram Meter Inc. then the series is convergent. " From MathWorld--A Wolfram Web Resource. There are two things we have to verify: we need the sequence fa ng= n+1 n2 to be decreasing, and we need this sequence to have limit zero. Alternating Series Test. Personality defines a unique, recognizable individual and is developed as a result of the interaction of the inherited elements and the life-time environment. Correct! This is the correct answer. In order to use this test, we first need to know what a converging series and a diverging series is. 15 hours ago · The year before 107k Penn State fans watched a 42-13 stomping; the year before that Michigan hammered PSU 49-10. Y —l) n an Odd terms are neg. Blood, Part 1 - True Blood: Crash Course A&P #29 - Duration: 10:00. At , the series is. Teach yourself calculus. 15 kV Class 3/C termination samples were built using 3M™ 3/C Phase Rejacketing System RJS. Electricity flows in two ways: either in an alternating current (AC) or in a direct current (DC). : GB 1984-2014: Status: valid remind me the status change. What test to use? When you're looking at a positive series, what's the best way to determine whether it converges or diverges? This is more of an art than a science, that is, sometimes you have to try several things in order to nd the answer. Alternating Series Test The last two tests that we looked at for series convergence have required that all the terms in the series be positive. Topics for Test 2 Convergence tests for series. Fingerpicking for Ukulele - Alternating Thumb Style, 2nd Edition is an new expanded edition which focuses on the alternating thumb fingerpicking style through a series of graduated lessons-chapters incorporating your index and middle fingers with the alternating thumb. Use a known series to find a power series in x that has the given function as its sum: (a) xsin(x3) Recall the Maclaurin series for sinu = X∞ n=0 (−1)n u2n+1 (2n+1)! Therefore, sin(x3) = X∞ n=0 (−1)n (x3)2n+1 (2n+1)! =. The alternating series test can only tell you that an alternating series itself converges. An alternating seriesalternates because it has a factor of -1. Looking for abbreviations of AST? It is Alternating series test. Alternating Series (6. Converges by. an infinite series whose terms are alternately positive and negative: u 1 - u 2 + u 3 - u 4 + … + (-I) n-1 u n + …. Alternating Series and P-series "convergence" I couldn't resist trying out a pun. Hosts Julia Collin Davison and Bridget Lancaster and the Test Kitchen cooks prepare America's favorite recipes, passing along valuable tips as they go. Since the terms of an alternating series change sign, the partial sums for any alternating series will jump back and forth over some line. For problems with multiple parts you can view the solution to each part by clicking the Show Solution link after the problem statement for that part or you can view the solutions to all parts by clicking the Show All Solutions link near the top of the solution. In fact, when checking for absolute convergence the term 'alternating series' is meaningless. 0 nF, R = 100Ω, and the source voltage is 220 V. The applet shows the series called the alternating harmonic series because its terms alternate sign: The harmonic series diverges, but maybe the minus signs change the behavior in this case. Line jumping is the idea behind our first convergence test, the alternating series test. Free online storage and sharing with Screencast. CONVERGENCE TESTS FOR SERIES: COMMENTS AND PROOFS PART IV: THE ALTERNATING SERIES TEST Math 112 The convergence tests for series have nice intuitive reasons why they work, and these are fairly easy to turn into rigorous proofs. Alternating Series Test Recall that the Integral Test, Direct Comparison Test, and Limit Comparison Test all require that the terms of the series are positive. It’s important to rely on the de nition of an in nite series when trying to telescope a series. We note that S 2 n + 1 - S 2 n = a 2 n + 1. com is online education portal for providing cost effective entrance examination practice to students for Railway (RRB), SSC, Banking (IBPS) Clear / PO, BPSC, UPSC, JPSC, Medical, Engineering, MCA, MBA, etc. Here is a set of practice problems to accompany the Alternating Series Test section of the Series & Sequences chapter of the notes for Paul Dawkins Calculus II course at Lamar University. For the series above, the root test determines that the series converges for and divergesk kB " # for. The Alternating Series Test If the alternating series satisfies for k = 2, 3, 4, 5, , and. All together, the series converges for , and diverges for and for. Course Description Calculus emphasizes a multi-representational approach, with concepts, results, and problems being expressed graphically, numerically, analytically, and verbally. Using this method, it was recently shown that increasing alpha (10 Hz) oscillations improved creative ideation with figural material and that increasing gamma (40 Hz. The total resistance of the circuit is found by simply adding up the resistance values of the individual resistors:. With no loss of generality we can assume that the series begins at n =1. The participants. Notice that the series in question is alternating, and we can verify that the hypotheses of the alternating series test apply: (1)To show that the (absolute value) of the terms of the series are decreasing, we’ll compute a derivative. CLP-2 Integral Calculus. To answer that question, you must investigate the positive series with a different test. Solved examples with detailed answer description, explanation are given and it would be easy to understand. 6 Alternating Series and Conditional Convergence Page 1 Theorem 15 - The Alternating Series Test The series 1 12 3 4 1 (1)n n n uuu uu ∞ + = ∑− =−+−+" converges if the following conditions are satisfied: 1. A score is given for each subtest, and then it is averaged into an overall Full Scale IQ. Sequences and Series Review Video (PatrickJMT) Assignment #7: Series Flow. For : The first and second conditions are satisfied since the terms are positive and are decreasing after each term. One test that is specifically designed to handle series whose terms alternate positive and negative is the Alternating Series Test. For example. MATH 1920 Alternating Series Test. Using this method, it was recently shown that increasing alpha (10 Hz) oscillations improved creative ideation with figural material and that increasing gamma (40 Hz. Q1: The alternating series test does not apply to the series ∞ ( − 1 ) 𝑛 𝑛 + 1. In fact, when checking for absolute convergence the term 'alternating series' is meaningless. Rather than oscillating back and forth, DC provides a constant voltage or current. But the alternating series approximation theorem quickly shows that. then is this the same as the divergence test, and is it safe to say that the series diverges. Answer to: Using the Alternating Series Test, determine whether \sum_{k=1}^{\infty} ( (-1)^{k} - ( (k + 2)/(4^{k}) ) converges. The book guides students through the core concepts of calculus and helps them understand how those concepts apply to their lives and the world around them. The Alternating Series Test. Use the Alternating Series Test to determine the convergence status of the following series. Tell the patient that you are going to show them a series of hand movements. Some very interesting and helpful examples are included. Search streaming video, audio, and text content for academic, public, and K-12 institutions. Free-Response Questions 1. Determine whether the alternating series $$\sum\limits_{n = 2}^\infty {\large\frac{{{{\left( { - 1} \right)}^{n + 1}}\sqrt n }}{{\ln n}}\normalsize. The following is a Taylor Series evaluated a particular value of x, find the sum of the series. The Alternating Series Test Math. Alternating Series Test 1 Alternating Series Test If the terms of the alternating series ( 1)n 1b n b1 b2 b3 n 1 where bn 0 satisfy (1) bn 1 bn for all n 1 (bn is decreasing) (2) lim n bn 0 then the series is convergent. The table below will help show you how the scores derived from the various subtests is used within the various index scores. Battaly 2017 2 April 21, 2017 Calculus Home Page Class Notes: Prof. Note: AST doesn't apply when either of the conditions is not met, and so never is a test for divergence. Free online storage and sharing with Screencast. an infinite series whose terms are alternately positive and negative: for uk > 0. So in other words, an alternating series will converge if it passes the n-th term test and the absolute value of the terms decrease. Applications of integration including finding areas and volumes. Alphabetical Listing of Convergence Tests. The terms alternate, and the computation above shows that the terms decrease in absolute value. Handout on the Alternating Series Test - Part I (Maynooth University) Handout on the Alternating Series Test - Part II (Maynooth University) Handout on the Alternating Series Test - Part III (Maynooth University) Video on Alternating Series (Patrick JMT). a mathematical series in which consecutive terms are alternatively positive and negative…. Therefore, we will have to look at the alternating series to determine if it converges or not. Alternating series, absolute and conditional convergence You have to know the de nition of what it means for a series to be alternating and con-vergent. A quantity that measures how accurately the nth partial sum of an alternating series estimates the sum of the series. Therefore, the sums converge to the same limit if and only if a n → 0 as n → ∞. Alternating Series & AS Test Objectives: Be able to describe the convergence of alternating series. However, it can be remedied by choosing any sequence which goes to zero fast enough, and putting it in the place of the zero terms, as you do. an infinite series whose terms are alternately positive and negative: u 1 - u 2 + u 3 - u 4 + … + (-I) n-1 u n + …. After defining alternating series, we introduce the alternating series test to determine whether such a series converges. series diverges by Limit Comparison, and the original series does not converge absolutely. A retrospective record review of performance on the three-step Luria test was conducted on 383 participants from a university-based dementia clinic. Infinite series whose terms alternate in sign are called alternating series. (This can usually b e done b y insp ection). Proof: Look at the. So this is a geometric series with common ratio r = –2. If you need to review this test, please refer to the supplemental notes 23. Infinite series whose terms alternate in sign are called alternating series. Then P∞ k=1 (−1) ka k converges. A divergent alternating series whose terms go to zero. Course Description Calculus emphasizes a multi-representational approach, with concepts, results, and problems being expressed graphically, numerically, analytically, and verbally. (f) Prove that the alternating harmonic series X1 n=1 ( 1)n n converges. Here is how one can find the derivative of arctan x: The above is a modern proof, Gregory used the derivative of arctan from the work of others. 5 kV through 765 kV and IEEE Standard 82, IEEE Standard Test Procedure for Impulse Voltage Tests on Insulated Conductors. ) Summarizing the above work, we know that 4 is not included, but 6 is. An alternating series is a series whose terms are al-ternately positive and negative. 88 11 Note: and ; therefore, 89 1 b) lim lim 0 8 Therefore. It is one of the most commonly used tests for determining the convergence or divergence of series. It is not obvious that the sequence b n decreases monotonically to 0. Overview of Alternating Series Test; 3 Examples; Conditional and Absolute Convergence for Alternating Series; 2 Examples; Ratio Test. In mathematical analysis, the alternating series test is a method used to prove that an alternating series with terms that decrease in absolute value is a convergent series. Converges by ratio test. This test does not prove absolute convergence. toit toshowthatthe series converges. Chroma's 63800 Series AC & DC Electronic Loads include built-in 16-bits precision measurement circuits to measure the steady-state and transient responses for true RMS voltage, true RMS current, true power(P), apparent power(S), reactive power(Q), crest factor, power factor, THDv and peak repetitive current. Run multiple programs, render videos and more — the Yoga 920 is designed to multitask with ease. The Alternating Series Test If k 1 ak is an Alternating Series and lim 0 k k a and ak eventually becomes strictly decreasing. 01 Single Variable Calculus, Fall 2005 Prof. However, the third condition is not valid since and instead approaches infinity. Overview of Root Test; 3 Examples; Sequences. Since this is an alternating series whose terms decrease to zero, we know that the series converges. The alternating series. In mathematical analysis, the alternating series test is a method used to prove that an alternating series with terms that decrease in absolute value is a convergent series. Free series convergence calculator - test infinite series for convergence step-by-step. Here is how one can find the derivative of arctan x: The above is a modern proof, Gregory used the derivative of arctan from the work of others. Thus, the even and odd series both converge. In this worksheet, we will practice determining whether an alternating series is convergent or divergent using the alternating series test. Series Convergence Flowchart doesa n! 0? Isa n > 0? Diverges by Divergence Test Is it alternating in sign and ja n decreasing? Are there any easy comparisons? Does it feel likea n `looks like' someb n? Try Ratio Test: lim a n+1 a n = c if 0 c < 1 then P a n converges if c > 1 then P a n diverges if c= 1 then test is inconclusive Try Integral. Converges by ratio test. The total resistance of the circuit is found by simply adding up the resistance values of the individual resistors:. One test that is specifically designed to handle series whose terms alternate positive and negative is the Alternating Series Test. Several convergence testing methods, such as the alternating series test, ratio test and root test, will be presented. 0 < a n+1 <= a n), and approaching zero, then the alternating series. EX 4 Show converges absolutely. 1 Examples 2 Alternating series test. Alternating series definition is - a mathematical series in which consecutive terms are alternatively positive and negative. Bhagwan Singh Vishwakarma 129,314 views. (b) Prove the Alternating Series Test using the Nested Interval Property (Theorem 1. This is a test which we'll use to show lots of alternating series converge. Here are a bunch of Single Variable Calculus applets. Overview of Alternating Series Test; 3 Examples; Conditional and Absolute Convergence for Alternating Series; 2 Examples; Ratio Test. (LTspice is also called SwitcherCAD by its manufacturer, since they use it primarily for the design of switch mode power supplies (SMPS). Use a known series to find a power series in x that has the given function as its sum: (a) xsin(x3) Recall the Maclaurin series for sinu = X∞ n=0 (−1)n u2n+1 (2n+1)! Therefore, sin(x3) = X∞ n=0 (−1)n (x3)2n+1 (2n+1)! =. ) The following test says that if the terms of an alternating series decrease toward 0 in absolute value, then the series converges. 5 – Notes Page 2 of 3 Ex 3. 이 감소수열이므로 라고 하면. (g) State the Alternating Series Estimation Theorem. For those that diverge, say which hypotheses of the alternating series test do not. Sequences and Series: Alternating Series Test (MathsCasts). (i) The series (−1)n is an alternating. The alternating series test requires that the a n alternate sign, get smaller and approach zero as n approaches infinity, which is true in this case. If s= P ( 1)n 1b. If a k → 0, then X (−1)ka k converges. If, as our intuition tells us should be true, the rearrangement does not change the sum, then we have just seen that. AC versus DC. When a series alternates (plus, minus, plus, minus,) there's a fairly simple way to determine whether it converges or diverges: see if the terms of the series approach 0. 8 = angle Pod sino g 3! as 3! a 7 " Functions can often be represented by an infinite series. Lecture 27 :Alternating Series The integral test and the comparison test given in previous lectures, apply only to series with positive terms. • Be sure that you apply the alternating series test only to alternating series. But the alternating series approximation theorem quickly shows that. Part (a) asked students to use the ratio test to determine the interval of convergence for the given Maclaurin series. The integral test, which is my favorite test in general, tends to be awkward with alternating series. 7 Alternating Series, Absolute Convergence notes by Tim Pilachowski So far, we have pretty much limited our attention to series which are positive. The alternating series test is a simple test we can use to find out whether or not an alternating series converges (settles on a certain number). (You probably figured out that with this naked summation. The Alternating Series Test (Leibniz's Theorem) This test is the sufficient convergence test. The idea of hopping back and forth to a limit is basically. Rearrangements. What test to use? When you're looking at a positive series, what's the best way to determine whether it converges or diverges? This is more of an art than a science, that is, sometimes you have to try several things in order to nd the answer. Alternating Series Test 1 Alternating Series Test If the terms of the alternating series ( 1)n 1b n b1 b2 b3 n 1 where bn 0 satisfy (1) bn 1 bn for all n 1 (bn is decreasing) (2) lim n bn 0 then the series is convergent. We look at a couple of examples. 7 Alternating Series, Absolute Convergence notes by Tim Pilachowski So far, we have pretty much limited our attention to series which are positive. We're learning alternating series test and the whole class is confused on why the series needs to be decreasing to pass the test. EXPECTED SKILLS: Determine if an alternating series converges using the Alternating Series Test. The un 's are all positive. If it does, then try applying the Ratio Test i. The alternating series simply tells us that the absolute value of each of the terms decreases monotonically, i. Instead, I'll give a more elementary proof of the Alternating series test which does not use Abel's formula (but we will see Abel's formula later). What does alternating personality mean in law?. Just better. An alternating sequence will have numbers that switch back and forth between positive and negative signs. Electricity or "current" is nothing but the movement of electrons through a conductor, like a wire. It should be noted that Theorem 1. CLP-2 Integral Calculus. Here are a bunch of Single Variable Calculus applets. 7: Series, alternating series test - solutions Apply the AST (possibly in combination with other tests) and state your conclusion about convergence. Embed this widget ». Finally, by L'Hôpital's Rule, By the Alternating Series Test, the series converges. of series with positive and negative terms and whether or not they converge. So, given the series look at the limit of the non-alternating part: So, this series converges. Jason Starr. The well-known Leibniz Criterion or alternating series test of convergence of alternating series is generalized for the case when the absolute value of terms of series are “not absolutely monotonously” convergent to zero. Alexander Street is an imprint of ProQuest that promotes teaching, research, and learning across music, counseling, history, anthropology, drama, film, and more. With the Alternating Series Test, all we need to know to determine convergence of the series is whether the limit of b[n] is zero as n goes to infinity. Alternating series test for convergence. Find more Mathematics widgets in Wolfram|Alpha. ) Note: Some of this was written using SwitcherCad III, and some was written using LTspice IV. 5 kV through 765 kV and IEEE Standard 82, IEEE Standard Test Procedure for Impulse Voltage Tests on Insulated Conductors. There are two main types of current used in most electronic circuits today. The table below will help show you how the scores derived from the various subtests is used within the various index scores. Alternating Series Test If for all n, a n is positive, non-increasing (i. [email protected] This is the logical reasoning questions and answers section on "Number Series Type 2" with explanation for various interview, competitive examination and entrance test. A series in which successive terms have opposite signs is called an alternating series. I Absolute convergence test. The Alternating Series Test (Leibniz's Theorem) This test is the sufficient convergence test. b n+1 ≤ b n for all n > N. Theorem 4 : (Comparison test ) Suppose 0 • an • bn for n ‚ k for some k: Then. You can see some Alternating Series, Absolute and Conditional Convergence - Notes, Engineering, Semester sample questions with examples at the bottom of this page. I believe we sometimes overemphasize the importance of this test because we want to make clear the distinction between absolute convergence and convergence. When a sum does this, we say it ‘telescopes’. The Absolute-Convergence Test (ACT) For series that have infinitely many negative terms and infinitely many positive terms and that aren't alternating, the tests discussed in previous sections and the AST can't be applied. Definition of alternating personality in the Legal Dictionary - by Free online English dictionary and encyclopedia. The part of the summation makes up the series alternator. Calculus 141, section 9. Suppose that {a i} is a sequence of positive numbers such that a i > a i+1 for all i. Teach yourself calculus. Course Material Related to This Topic:. The lower beds of the Peuquenes ridge, and of the several great lines to the westward of it, are composed of a vast pile, many thousand feet in thickness, of porphyries which have flowed as submarine lavas, alternating with angular and rounded fragments of the same rocks, thrown out of the submarine craters. iii) if ρ = 1, then the test is inconclusive. Line jumping is the idea behind our first convergence test, the alternating series test. The common ratio of a geometric series may be negative, resulting in an alternating sequence. Geometric series. Continuing with post on sequences and series New Series from Old 1 Rewriting using substitution New Series from Old 2 Finding series by differentiating and integrating New Series from Old 3 Rewriting rational expressions as geometric series Geometric Series – Far Out A look at doing a question the right way and the “wrong” way?…. Hence, the interval of convergence is: (−8,10] and the radius convergence is: R = 10. Therefore, the sums converge to the same limit if and only if a n → 0 as n → ∞. Alternating Series Test. Absolute Ratio Test Let be a series of nonzero terms and suppose. In a alternating series every term will have a sign different than the term before it. The alternating series simply tells us that the absolute value of each of the terms decreases monotonically, i. (One is the harmonic series; the other can be proved divergent by comparison with the harmonic series. An alternating series is a series whose terms are al-ternately positive and negative. An alternating sequence will have numbers that switch back and forth between positive and negative signs. The short circuit test is maintained by default. Alternating series test listed as AST Alternating series. Then the series converges if both of the following conditions hold. If property 3 is respected but property 1 and/or property 2 do not hold, then the alternating series test is inconclusive. More Alternating Series Examples - Finding whether a given alternating series converges or diverges. Alternating Series (6. The Alternating Series Test. So, the series converges by the alternating series test. −1 3 2 4 −3 5 4 6 −5 7. The test that we are going to look into in this section will be a test for alternating series. The series above is thus an example of an alternating series, and is called the alternating harmonic series. Alternating Series and Leibniz’s Test Let a 1;a 2;a 3;::: be a sequence of positive numbers. In order to use this test, we first need to know what a converging series and a diverging series is. The integral test, which is my favorite test in general, tends to be awkward with alternating series. Overview of Root Test; 3 Examples; Sequences. jxjn+1: Rememberthatjxjrepresentsthedistancebetweenx and0. このコンテンツの表示には、Adobe Flash Playerの最新バージョンが必要です。 http://www. Warm Up: Find the sum of the infinite series. Alternating Series Test. Does ∑ (−1)𝑛3𝑛 4𝑛−1 ∞ 𝑛=1 converge or diverge? For an alternating series, how close is 𝑠𝑛 to the sum of the infinite number of terms?. Mathispower4u 70,297 views. Chroma's 63800 Series AC & DC Electronic Loads include built-in 16-bits precision measurement circuits to measure the steady-state and transient responses for true RMS voltage, true RMS current, true power(P), apparent power(S), reactive power(Q), crest factor, power factor, THDv and peak repetitive current. Alternating Current Circuits 5 Open-Ended Problems 57. Topics for Test 2 Convergence tests for series. SERIES CONVERGENCE/ DIVERGENCE FLOW CHART nVergeS TEST FOR DIVERGENCE GEOMETRIC SERIES ALTERNATING SERIES TELESCOPING SERIES ? May to etc. Ratio and Root Tests. 1997 views. Holmes May 1, 2008 The exam will cover sections 8. In this section we introduce alternating series—those series whose terms alternate in sign. Proof: Look at the. The series above is thus an example of an alternating series, and is called the alternating harmonic series. alternating series test. It is important that you verify the conditions of the Alternating Series Test are met; otherwise some-one might not believe your conclusion is valid. An alternating series is any series, , for which the series terms can be written in one of the following two forms. Alternating series convergence: a visual proof Richard H. The Alternating Series Test. alternating with phrase. Incorrect! Remember the conditions of the Alternating Series Test. For : The first and second conditions are satisfied since the terms are positive and are decreasing after each term. The test was used by Gottfried Leibniz and is sometimes known as Leibniz's test , Leibniz's rule , or the Leibniz criterion. The alternating series test can tell us if it's safe to open that box. Fourier series are used in the analysis of periodic functions. Solution: Let I1 be the closed interval [0, s1]. It is important that you verify the conditions of the Alternating Series Test are met; otherwise some-one might not believe your conclusion is valid. 1, which asks for a proof of the Alter-nating Series Test using the Cauchy Criterion for series (Theorem 2. The alternating series test says that if the absolute value of each successive term decreases and \lim_{n\to\infty}a_n=0, then the series converges. I Absolute and conditional convergence. LTspice Tutorial Introduction While LTspice is a Windows program, it runs on Linux under Wine as well. 1 Is there a short name for series which satisfy the hypothesis of the alternating series test?. This is a very useful lecture in Calculus. Printing the sum of an alternating series [Basic Python] I've been stuck on this little problem for the best part of today, and it's driving me insane now. Hence, the interval of convergence is: (−8,10] and the radius convergence is: R = 10. Write the three rules that are used to satisfy convergence in an alternating series test. Assignment #4: Alternating Series. Therefore, we will have to look at the alternating series to determine if it converges or not. Correct! This is the correct answer. ( 1) Use Alternating Series Test to show con verges. What is alternating personality? Meaning of alternating personality as a legal term. To see how this works, let \(S$$ be the sum of a convergent alternating series, so. The alternating series test requires that the a n alternate sign, get smaller and approach zero as n approaches infinity, which is true in this case. Theorem (Alternating series test) If the terms of the series ∑ n = 1 ∞ (-1) n an have the property thatall ofthe an terms are positive and an+1 < an forall n, thenthe series converges. Alternating Series (6. I have a bachelors in Mathematics Education from Slippery Rock University, and a Masters in Administration and Supervision from The College of Notre Dame. Alternating Series Test 1 Alternating Series Test If the terms of the alternating series ( 1)n 1b n b1 b2 b3 n 1 where bn 0 satisfy (1) bn 1 bn for all n 1 (bn is decreasing) (2) lim n bn 0 then the series is convergent. Q1: The alternating series test does not apply to the series ∞ ( − 1 ) 𝑛 𝑛 + 1. It's also known as the Leibniz's Theorem for alternating series. (Warning: Do not use a multimeter to measure the wall outlets in your home. In this video, Krista King from integralCALC Academy talks about the Alternating Series Test (Calculus problem example). Transcranial alternating current stimulation (tACS) is a non-invasive brain stimulation method that allows to directly modulate brain oscillations of a given frequency. Drag up for fullscreen M M. 01 Single Variable Calculus, Fall 2005 Prof. The test was used by Gottfried Leibniz and is sometimes known as Leibniz's test , Leibniz's rule , or the Leibniz criterion. Looking for abbreviations of AST? It is Alternating series test. Mathematics Assignment Help, Steps for alternating series test, Steps for Alternating Series Test Suppose that we have a series ∑a n and either a n = (-1) n b n or a n = (-1) n+1 b n where b n > 0 for all n. The common ratio of a geometric series may be negative, resulting in an alternating sequence. We can therefore. n satis es the requirements for the alternating series test. However, it is not enough to have having a limit of zero, you also need decreasing, as the following example shows. Solved examples with detailed answer description, explanation are given and it would be easy to understand. You’ll do that one for homework. Alternating Current Circuits 5 Open-Ended Problems 57. We note that S 2 n + 1 - S 2 n = a 2 n + 1. Many, but not all, of the problems will have. Alternating series test for convergence. Alternating Series Estimation Theorem. The fact that sums, products, integrals, antiderivatives of Taylor series are also Taylor series is in 8. 6 Alternating Series and Conditional Convergence Page 1 Theorem 15 - The Alternating Series Test The series 1 12 3 4 1 (1)n n n uuu uu ∞ + = ∑− =−+−+" converges if the following conditions are satisfied: 1. According to the alternating series test, we know that this series converges to some number. Click on the name of the test to get more information on the test. Many series such as 8œ" 8œ" 8œ" ∞ ∞ ∞ # sin 8 " 88 8 8 8 8" ß " ß "and do not have all positive terms and thus cannot be investigated using the above mentioned tests. A t-test is one of the most frequently used procedures in statistics. Alternating series have the simplest of sign patterns. The terms alternate. This is a correct reasoning to show the divergence of the above series. | 2019-12-10T08:01:54 | {
"domain": "emporiumgalorium.de",
"url": "http://lxlj.emporiumgalorium.de/alternating-series-test.html",
"openwebmath_score": 0.7549763321876526,
"openwebmath_perplexity": 830.88619511814,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.97594644290792,
"lm_q2_score": 0.8774767858797979,
"lm_q1q2_score": 0.8563703479136634
} |
https://math.stackexchange.com/questions/352163/show-that-theres-a-unique-minimum-spanning-tree-if-all-edges-have-different-cos/352212 | # Show that there's a unique minimum spanning tree if all edges have different costs
Show that there's a unique minimum spanning tree (MST) in case the edges' weights are pairwise different $(w(e)\neq w(f) \text{ for } e\neq f)$.
I thought that the proof can be done for example by contradiction, saying that we have $2$ different MST meaning that somewhere was possible to pick from more edges, so $w(e) = w(f)$ for $e\neq f$, contradiction. Apparently this is not correct.
How would you show that a graph has a unique MST if all edges have distinct weights?
If $T_1$ and $T_2$ are distinct minimum spanning trees, then consider the edge of minimum weight among all the edges that are contained in exactly one of $T_1$ or $T_2$. Without loss of generality, this edge appears only in $T_1$, and we can call it $e_1$.
Then $T_2 \cup \{ e_1 \}$ must contain a cycle, and one of the edges of this cycle, call it $e_2$, is not in $T_1$.
Since $e_2$ is a edge different from $e_1$ and is contained in exactly one of $T_1$ or $T_2$, it must be that $w ( e_1 ) < w ( e_2 )$. Note that $T = T_2 \cup \{ e_1 \} \setminus \{ e_2 \}$ is a spanning tree. The total weight of $T$ is smaller than the total weight of $T_2$, but this is a contradiction, since we have supposed that $T_2$ is a minimum spanning tree.
• How do we know that $w ( e_1 ) < w ( e_2 )$? Mar 4, 2017 at 23:57
• @Alexander Because $e_1$ is the edge of least weight that is contained in exactly one of the MSTs, and $e_2$ is a different edge that is contained in exactly one of the MSTs. Mar 5, 2017 at 3:49
• Is $e_1$ the edge of least weight out of the edges of $T_1 \cup T_2$, which is in either $T_1$ or $T_2$? Or is $e_1$ the edge of least weight out of the edges of $T_1 \bigtriangleup T_2$? Mar 6, 2017 at 1:43
• I suspect it's the latter of two, otherwise I don't understand how we know this edge isn't in both $T_1$ and $T_2$. Mar 6, 2017 at 1:47
• @Alexander As I said, $e_1$ is the edge of least weight that is contained in exactly one of the MSTs. If it was in both $T_1$ and $T_2$ it would not be contained in exactly one of the MSTs. Mar 6, 2017 at 3:41
A proof using cycle property:
Let $G=(V,E)$ be the original graph.
Suppose there are two distinct MSTs $T_1=(V,E_1)$ and $T_2=(V,E_2)$. Since $T_1$ and $T_2$ are distinct, the sets $E_1-E_2$ and $E_2-E_1$ are not empty, so $\exists e\in E_1-E_2$.
Since $e\notin E_2$, adding it to $T_2$ creates a cycle. By cycle property the most expensive edge of this cycle (call it $e'$) does not belong to any MST. But
If $e'=e$ then $e'\in E_1$ (because $e\in E_1-E_2$)
If $e'\neq e$ then $e'\in E_2$
Both cases are contradicting with the fact that $e'$ is not in any MST.
The best explanation I found was on wikipedia, Proof:
1. Assume the contrary, that there are two different MSTs $A$ and $B$.
2. Since $A$ and $B$ differ despite containing the same nodes, there is at least one edge that belongs to one but not the other. Among such edges, let e1 be the one with least weight; this choice is unique because the edge weights are all distinct. Without loss of generality, assume $e1$ is in $A$.
3. As $B$ is a MST, $\{e1\}\cup B$ must contain a cycle $C$.
4. As a tree, $A$ contains no cycles, therefore $C$ must have an edge $e2$ that is not in $A$.
5. Since $e1$ was chosen as the unique lowest-weight edge among those belonging to exactly one of $A$ and $B$, the weight of $e2$ must be greater than the weight of $e1$.
6. Replacing $e2$ with $e1$ in $B$ therefore yields a spanning tree with a smaller weight.
7. This contradicts the assumption that $B$ is a MST.
More generally, if the edge weights are not all distinct then only the (multi-)set of weights in minimum spanning trees is certain to be unique; it is the same for all minimum spanning trees [1].
This is a slightly different (though more lengthy) proof than the other's in that it is an algorithmic proof. We can use Prim's algorithm and demonstrate the proof by studying the spanning tree it builds in such a graph $$G$$. Here is Prim's algorithm:
• $$V_T \leftarrow \{ v_0 \}$$
• $$T_0 \leftarrow \emptyset$$
• for $$i \leftarrow 1$$ to $$|V| - 1$$ do:
• find an edge $$e_i=(v_i,u_i)$$ of minimum weight such that $$v_i$$ is in $$V_T$$ and $$u_i$$ is not.
• $$V_T \leftarrow V_T \cup \{ u_i \}$$
• $$T_i \leftarrow T_i \cup \{ e_i \}$$
We demonstrate the following : At iteration $$i$$, the choice $$e_i$$ that Prim's algorithm makes belongs to every minimum spanning tree. We can show this by using the inductive hypothesis:
Suppose $$T_{i-1}$$ is a subset of every minimum spanning tree. Then, $$T_i = T_{i-1} \cup \{ e_i \}$$ is a subset of every minimum spanning tree.
• Base case: $$T_1 = e_1$$ where $$e_1$$ is incident to $$v_0$$. Suppose that there is some minimum spanning tree $$T$$ that does not contain $$e_1$$. Then, $$T$$ contains another edge, $$\bar{e}_1$$ incident to $$v_0$$. By the greedy choice of Prim's and the distinctness of weights, $$w(e_1) < w(\bar{e}_1)$$ and we can replace $$\bar{e}_1$$ with $$e_1$$ to decrease the weight of $$T$$, which is a contradiction.
• Inductive step: Suppose, to the contrary, that there exists some minimum spanning tree $$T$$ which contains the edges in $$T_{i-1}$$ but does not contain $$e_i$$. Consider the set of edges $$T \setminus T_{i-1}$$, which must be a forest composed of trees $$\bar{T}^1_{i-1}, \bar{T}^2_{i-1}, \ldots, \bar{T}^k_{i-1},$$ where each tree $$\bar{T}^t_{i-1}$$ is connected to $$T_{i-1}$$ in $$T$$ by an edge $$e^t_{i-1}$$.
However, note that one of those trees must be incident to the vertex $$u_i$$. Suppose that tree is $$\bar{T}^x_{i-1}$$. Then, by Prim's greedy choice and since the weights of edges are distinct, we know that $$w(e_i) < w(e^x_{i-1})$$. Which means that we replace $$e^x_{i-1}$$ with $$e_i$$ and decrease the weight of $$T$$. Hence, $$T$$ is not a minimum spanning tree. Then, $$e_i$$ must be in every minimum spanning tree and $$T_i$$ must be a subset of every minimum spanning tree.
Thus, the minimum spanning tree produced by Prim's algorithm is the only minimum spanning tree of $$G$$.
– fhy
Dec 29, 2019 at 12:30
This is a (slightly) more detailed version of the currently accepted answer
(For the sake of contradiction) Let $$T_1 = (V, E_1)$$ and $$T_2 = (V, E_2)$$ be two distinct MSTs of the graph $$G = (V, E)$$
Note that both have the same vertex set $$V$$ since both are spanning trees of $$G$$
Consider the set $$E_{\Delta} = E_1 \triangle E_2$$
Let $$e = (u, v)$$ be the edge in $$E_{\Delta}$$ having the least cost (or weight)
Note that since all costs are unique, and $$E_{\Delta}$$ is non-empty, $$e$$ must be unique.
Without loss of generality, assume $$e \in E_1$$
Now, there must be a path $$P$$, with $$e \notin P$$, in $$T_2$$ connecting $$u$$ and $$v$$, since trees are by definition, connected.
Note that at least one edge (say $$e'$$) that occurs in $$P$$ must not be in $$E_1$$, thus, $$e' \in E_{\Delta}$$
This is because, if $$P \subset E_1$$, $$T_1$$ will contain a cycle formed by the path $$P$$ and the edge $$e$$, this leads to a contradiction, since trees by definition are acyclic.
Note that by definition of $$e$$ and the fact that all costs are distinct, $$\text{cost}(e) < \text{cost}(e')$$
Now, consider $$T_2' = (V, E_2')$$, where $$E_2' = (E_2 \backslash\{e'\})\cup\{e\}$$, this has a strictly lesser total cost than $$T_2$$
As $$T_2$$ was a MST, this leads to a contradiction. | 2023-03-21T07:42:29 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/352163/show-that-theres-a-unique-minimum-spanning-tree-if-all-edges-have-different-cos/352212",
"openwebmath_score": 0.8433116674423218,
"openwebmath_perplexity": 125.79959717816376,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9759464485047916,
"lm_q2_score": 0.8774767762675405,
"lm_q1q2_score": 0.8563703434437397
} |
https://thinairmedia.org/at45qi/zat0b.php?931dcd=diagonal-of-parallelogram | The diagonals of a parallelogram bisect each other. Diagonal of Parallelogram Formula The formula of parallelogram diagonal in terms of sides and cosine β (cosine theorem) if x =d 1 and y = d 2 are the diagonals of a parallelogram and a and b are the two sides. Show that it is a rhombus. The shape has the rotational symmetry of the order two. The pair of opposite sides are equal and they are equal in length. DOWNLOAD PDF / PRINT . The diagonals bisect the angles. If they diagonals do indeed bisect the angles which they meet, could you please, in layman's terms, show your proof? Then, substitute 4.8 for in each labeled segment to get a total of 11.2 for the diagonal … This is the currently selected item. Consecutive angles are supplementary. Diagonals of rectangles and general parallelograms, however, do not. Type your answer here… Related Topics. The diagonals of a parallelogram bisect each other. Construction of a parallelogram given the length of two diagonals and intersecting angles between them - example Construct a parallelogram whose diagonals are 4cm and 5cm and the angle between them is … Some of the properties of a parallelogram are that its opposite sides are equal, its opposite angles are equal and its diagonals bisect each other. Parallelogram definition, a quadrilateral having both pairs of opposite sides parallel to each other. That is, each diagonal cuts the other into two equal parts. 1 answer. These parallelograms have different areas. Learn more about Diagonal of Parallelogram & Diagonal of Parallelogram Formula at Vedantu.com The diagonals of a parallelogram. Area of the parallelogram using Trignometry: $$\text{ab}$$$$sin(x)$$ where $$\text{a}$$ and $$\text{b}$$ are the length of the parallel sides and $$x$$ is the angle between the given sides of the parallelogram. The rectangle has the following properties: All the properties of a parallelogram apply (the ones that matter here are parallel sides, opposite sides are congruent, and diagonals bisect each other). where is the two-dimensional cross product and is the determinant.. As shown by Euclid, if lines parallel to the sides are drawn through any point on a diagonal of a parallelogram, then the parallelograms not containing segments of that diagonal are equal in area (and conversely), so in the above figure, (Johnson 1929).. Vice versa, if the diagonals of a parallelogram are perpendicular, then this parallelogram is a rhombus. There are several rules involving: the angles of a parallelogram ; the sides of a parallelogram ; the diagonals of a parallelogram Calculate certain variables of a parallelogram depending on the inputs provided. You get the equation = . Test the conjecture with the diagonals of a rectangle. Because the parallelogram has adjacent angles as acute and obtuse, the diagonals split the figure into 2 pairs of congruent triangles. A parallelogram is a quadrilateral in which both pairs of opposite sides are parallel. The area of the parallelogram represented by the vectors A = 4 i + 3 j and vector B = 2 i + 4 j as adjacent side is. In the figure below diagonals AC and BD bisect each other. A parallelogram where all angles are right angles is a rectangle! Area of the parallelogram when the diagonals are known: $$\frac{1}{2} \times d_{1} \times d_{2} sin (y)$$ where $$y$$ is the angle at the intersection of the diagonals. Apply the formula from the Theorem. MCQ in Plane Geometry. These properties concern its sides, angles, and diagonals. If you just look […] Diagonals divide the parallelogram into two congruent triangles; Diagonals bisect each other; There are three special types of parallelogram, they are: Rectangle; Rhombus; Square; Let us discuss these special parallelograms one by one. You can use the calculator for each formula. In a parallelogram, the sides are 8 cm and 6 cm long. For instance, please refer to the link, does $\overline{AC}$ bisect $\angle BAD$ and $\angle DCB$? A parallelogram is a quadrilateral made from two pairs of intersecting parallel lines. Proof: Diagonals of a parallelogram. The two bimedians in a quadrilateral and the line segment joining the midpoints of the diagonals in that quadrilateral are concurrent and are all bisected by their point of intersection. If you make the diagonals almost parallel to one another - you will have a parallelogram with height close to zero, and thus an area close to zero. Notice the behavior of the two diagonals. You can rotate the two diagonals around this joint, and form different parallelogram (by connecting the diagonals's end points). See more. Opposite sides are congruent. If ∠Boc = 90° and ∠Bdc = 50°, Then ∠Oab = - Mathematics If ∠Boc = 90° and ∠Bdc = 50°, Then ∠Oab = - Mathematics Question By … General Quadrilateral; Kite; Rectangle; Rhombus; Square; Discover Resources. Make a conjecture about the diagonals of a parallelogram. The diagonals are perpendicular bisectors of each other. person_outlineTimurschedule 2011-03-28 14:49:28. In a parallelogram, the diagonals bisect each other, so you can set the labeled segments equal to one another and then solve for . There are three cases when a parallelogram is also another type of quadrilateral. The adjacent angles of the parallelogram are supplementary. The length of the shorter diagonal of a parallelogram is 10.73 . Solution Let x be the length of the second diagonal of the parallelogram. The diagonal of the parallelogram will divide the shape into two similar congruent triangles. The Diagonals of a Parallelogram Abcd Intersect at O. So we have a parallelogram right over here. Check the picture. Perimeter of a Parallelogram. Area of a Parallelogram : The Area is the base times the height: Area = b × h (h is at right angles to b) Example: A parallelogram has a base of 6 m and is 3 m high, what is its Area? Solution (1) AC=24 //Given Type your answer here… Can you now draw a rectangle ? Properties concern its sides, angles, diagonals, height, perimeter and area diagonal of parallelogram parallelograms type of &! Which both pairs of opposite sides parallel do bisect the angles which the meet and cm! Being said, I was wondering if within parallelogram the diagonals of a parallelogram are,. Parallelogram in which measure of each interior angle is \ ( 90^\circ \ ) properties concern sides. Cm long your answer here… Can you now draw a rectangle and rhombus! \ ) and they are equal in length, in layman 's terms, show your?! Certain variables of a diagonal of parallelogram is also another type of quadrilateral and students a day preparation. Thus, the diagonals ( lines linking opposite corners ) bisect each other definition of quadrilateral (. Educates thousands of reviewers and students a day in preparation for their board … the diagonals of parallelogram. Lengths, corner angles, diagonals, height, perimeter and area of parallelograms do indeed bisect angles. Diagonals around this joint, and form different parallelogram ( by connecting the split! Square may be considered as rectangle which has equal adjacent sides, angles, diagonals,,. Intersecting parallel lines do bisect the angles which the meet figure into pairs... 6 m × 3 m = 18 m 2 ), or a rhombus this Drag the orange dots each... Lines linking opposite corners ) bisect each other shorter diagonal of the parallelogram has the following properties: opposite are. Within parallelogram the diagonals bisect diagonal of parallelogram angles are all … the diagonals of a parallelogram is also another of! Is “ maybe. ” diagonals of a parallelogram diagonal of parallelogram Intersect at O 6 ×! Not equal ( lines linking opposite corners ) bisect each other you Can rotate the two around... Has the following properties: opposite sides are parallel Can rotate the two diagonals around this joint, and.! Quadrilateral having both pairs of intersecting parallel lines: the diagonals of a and... Abcd Intersect at O each vertex to reshape the parallelogram has adjacent angles acute! The conjecture with the help of law of cosines special quadrilaterals: rectangle is a quadrilateral from. & special quadrilaterals: diagonal of parallelogram, square,... all Questions Ask Doubt Solving in Plane Geometry 1 AC=24! Parallel lines the order two triangles, both acute and obtuse, the diagonals bisect the angles the., each diagonal cuts the other into two equal parts opposite sides parallel to each other diagonal of parallelogram as! And a rhombus square,... all Questions Ask Doubt Ask Doubt are perpendicular = m... Two diagonals around this joint, and diagonals in length with opposite are... Obtuse are congruent you now draw a rectangle Kite are perpendicular, then this parallelogram is another! ; square ; Discover Resources sides are 8 cm and 6 cm long Questions... Rectangle is a quadrilateral with opposite sides are 8 cm and 6 cm long students... Test the conjecture with the help of law of cosines the triangles, both and! Into two equal parts now draw a rectangle a right angle adjancent from! Properties concern its sides, angles, and diagonals 3 m = 18 m 2 a day preparation! Which both pairs of opposite sides parallel to each other angle between sides are true about.... And area of a parallelogram is a rhombus that being said, I was wondering if within parallelogram the of!, show your proof now draw a rectangle not equal two equal parts try this Drag orange... Angles are right angles is a rectangle you now draw a rectangle calculate the between. The angle between diagonals of rectangles and general parallelograms, however, do not 04:15: PM each angle. Of parallelograms “ maybe. ” diagonals of a parallelogram and adjancent angles from lengths... Rectangle which has equal adjacent sides, angles, and form different parallelogram by. Parallelogram and adjancent angles from side lengths and angle square ; Discover Resources diagonals do indeed bisect angles! The parallelogram are not equal quadrilateral made from two pairs of intersecting parallel lines 6 ×! Parallelogram bisect each other other into two equal parts done with the help of law of cosines and rhombus. In the figure below diagonals AC and BD bisect each other, as above rectangles and parallelograms. Split the figure into 2 pairs of opposite sides parallel in any parallelogram, the sides are parallel by.. Dots on each vertex to reshape the parallelogram \ ( 90^\circ \ ) side lengths angle! Parallelogram and adjancent angles from side lengths and angle between sides diagonals the... X diagonal /2 ), or 24x10/2=120, as above the rotational symmetry of the order two which of... Parallelograms, however, do not has equal adjacent sides, angles diagonals. And adjancent angles from side lengths and angle parallel to each other, which are parallelograms do. Both acute and obtuse are congruent joint, and form different parallelogram ( by connecting diagonals! Said, I was wondering if within parallelogram the diagonals split the figure into 2 pairs of opposite sides equal... Thus, the diagonals of a parallelogram if given 1.Sides and diagonal 2.Sides and area of.... Is \ ( 90^\circ \ ) that are true about it a parallelogram angles... Inoybix educates thousands of reviewers and students a day in preparation for their board … diagonals! 16Th Aug, 2017, 04:15: PM two of the shorter sides of the parallelogram a... Are parallel by definition, show your proof diagonals around this joint, and different. Versa, if the diagonals bisect the angles maybe. ” diagonals of a is. Below diagonals AC and BD bisect each other those things that are true about it special. And students a day in preparation for their board … the diagonals bisect the angles which they,! Angles, diagonals, height, perimeter and area of a parallelogram depending on the inputs provided not equal to! Both acute and obtuse, the diagonals of a parallelogram and adjancent from. The figure into 2 pairs of intersecting parallel lines quadrilateral having both pairs of opposite sides parallel to other!, then this parallelogram is 10.73 has equal adjacent sides, angles, and diagonals each other another type quadrilateral..., height, perimeter and area of parallelograms adjacent sides, angles and... To reshape the parallelogram is a rhombus this joint, and form different parallelogram by! Diagonals around this joint, and diagonals of a parallelogram are simply those things that are true about it,! Diagonals bisect the angles which the meet … the diagonals of a parallelogram are simply those things that are about. Any parallelogram, the diagonals bisect the angles as above: Latest Problem Solving in Plane Geometry perpendicular... Pair of opposite sides parallel to each other parallelogram is a quadrilateral with opposite are. To each other AC=24 //Given a diagonal of parallelogram shape into two similar congruent triangles congruent triangles into similar... Proof: the diagonals bisect the angles which the meet your proof are true it... Proof: diagonal of parallelogram diagonals ( lines linking opposite corners ) bisect each other of parallelograms measure each. 1 ) AC=24 //Given a parallelogram is also another type of quadrilateral each diagonal of parallelogram reshape...,... all Questions Ask Doubt p inoyBIX educates thousands of reviewers students... Ac=24 //Given a parallelogram is a quadrilateral with opposite sides are parallel by.! That is, each diagonal cuts the other into two equal parts 90^\circ. Side lengths and angle it is done with the help of law of cosines the second diagonal of triangles... Diagonals 's end points ) do bisect the angles which the meet not equal x be the of... General quadrilateral ; Kite ; rectangle ; rhombus ; square ; Discover Resources rectangle. Side lengths and angle between sides they meet, could you please, in layman terms... Parallelogram if given 1.Sides and diagonal 2.Sides and area of a parallelogram is ( diagonal x diagonal )! Triangles, both acute and obtuse, two of the second diagonal of the parallelogram diagonals,,! Both acute and obtuse are congruent end points ) also another type of.... Two diagonals around this joint, and form different parallelogram ( by connecting the diagonals the. Cuts the other into two equal parts ( diagonal x diagonal /2 ), or a rhombus with right! In the figure below diagonals AC and BD bisect each other parallelogram whose are... By | 16th Aug, 2017, 04:15: PM a Kite are.... Type your answer here… Can you now draw a rectangle properties: opposite sides are parallel by.! Of rectangles and general parallelograms, do bisect the angles be considered as rectangle which equal. Which measure of each interior angle is \ ( 90^\circ \ ) is “ maybe. ” diagonals a. And a rhombus with a right angle variables of a rectangle was wondering if within parallelogram the diagonals of parallelogram!: opposite sides are equal in length rectangle and a rhombus with a right angle the conjecture the. Solving in Plane Geometry things that are true about it parallelograms, do not parallelogram,. Ask Doubt, angles, diagonals, height, perimeter and area of parallelograms draw a.. Was wondering if within parallelogram the diagonals of a parallelogram whose angles are right angles is a rectangle the provided. The length of the parallelogram will divide the shape into two similar congruent triangles 18 2... The help of law of cosines opposite sides parallel to each other area of parallelograms diagonal! Has all the properties of the second diagonal of a parallelogram is a rectangle and a rhombus with a angle! Angles as acute and obtuse, the diagonals bisect the angles which they,...
Uncategorized | 2021-05-09T13:31:05 | {
"domain": "thinairmedia.org",
"url": "https://thinairmedia.org/at45qi/zat0b.php?931dcd=diagonal-of-parallelogram",
"openwebmath_score": 0.7166275382041931,
"openwebmath_perplexity": 854.741629216889,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.9669140235181257,
"lm_q2_score": 0.8856314632529871,
"lm_q1q2_score": 0.8563294814881909
} |
https://math.stackexchange.com/questions/2729586/monotonically-and-strictly-increasing-functions | # Monotonically and strictly increasing functions
This is a question on terminology.
What is the difference between a (i) strictly increasing function, and a (ii) monotonically increasing function? Is it that a monotonically increasing function may also include functions that are constant in some intervals, while strictly increasing function must always have a positive derivative where it is defined?
If so, is it correct to say, that
Strictly increasing functions $\implies$ monotonically increasing, while the converse is not true? And a strictly increasing function is equivalent to a 'strictly monotonically increasing' function?
Thanks.
• Yes. Given $x>y$, "monotonically increasing" means that $f(x)≥f(y)$ while "strictly increasing" means $f(x)>f(y)$. This means, for example, that a constant function is both monotonically increasing and montonically decreasing. – lulu Apr 9 '18 at 16:41
• And a strictly increasing function is equivalent to a 'strictly monotonically increasing' function? How do you define "strictly monotonically increasing"? I've never heard that. As lulu said, yes for everything else. – anderstood Apr 9 '18 at 16:42
• To avoid ambiguity, functions satisfying $x\le y\implies f(x)\le f(y)$ are sometimes called non-decreasing. – Julián Aguirre Apr 9 '18 at 16:42
• Thanks for your clarifications. @anderstood Perhaps I could change that to 'strictly monotone functions', which would refer to strictly increasing or decreasing functions? – T J. Kim Apr 9 '18 at 16:46
• I would avoid non-standard usage. We already have the phrase "strictly increasing", why introduce new terminology for the same thing? – lulu Apr 9 '18 at 16:48
You almost have it right. The condition is better stated without referring to derivatives. A function $f(x)$ is strictly increasing if for all $(x,y)$ such that $y>x$,
$$f(y) > f(x)$$
and is monotonic increasing if for all $(x,y)$ such that $y>x$, $$f(y) \geq f(x)$$
Your definition involving derivatives would say that the sawtooth $$g(x) = x - \lfloor x \rfloor$$ is strictly monotonic (since the derivative is not defined at integer $x$), but it is not monotonic at all.
Your last sentence is completely correct.
• I see, thanks for your answer. Would the definition involving derivatives hold if $f$ is defined to be differentiable for all x in the domain, in which case the sawtooth function could not be considered? – T J. Kim Apr 9 '18 at 16:58
A strictly increasing function: let $$P=\{x_1,...,x_n; x_i<x_{i+1}\}$$ then $f(x_i)<f(x_{i+1})$ for all $x_i,x_{i+1} \in P$. A monotonic increasing function: let $$P=\{x_1,...,x_n; x_i<x_{i+1}\}$$ then $f(x_i)\leq f(x_{i+1})$ for all $x_i,x_{i+1} \in P$. So a monotonic function can be constant for some interval $(x_k, x_l)$ or can be increasing on that interval too, a strictly increasing funcion has always a greater image provided x increases. | 2019-10-16T09:17:38 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2729586/monotonically-and-strictly-increasing-functions",
"openwebmath_score": 0.8859436511993408,
"openwebmath_perplexity": 489.9478911180118,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9740426420486938,
"lm_q2_score": 0.8791467785920306,
"lm_q1q2_score": 0.8563264509683796
} |
http://mathhelpforum.com/advanced-statistics/38179-product-two-normal-distributions.html | # Math Help - Product of Two Normal Distributions
1. ## Product of Two Normal Distributions
Hello,
I am trying to find the distribution of the product of two normal densities (different means and standard deviations) . Does the product follow normal distribution also? If so, what is the mean and standard deviation of the resultant distribution? Thanks a lot.
2. Originally Posted by vioravis
Hello,
I am trying to find the distribution of the product of two normal densities (different means and standard deviations) . Does the product follow normal distribution also? If so, what is the mean and standard deviation of the resultant distribution? Thanks a lot.
No.
If the two random variables X and Y are independent, then the pdf of Z = XY is probably (I haven't done the calculation) a Bessel function. See 3. of properties at Normal distribution - Wikipedia, the free encyclopedia.
3. Originally Posted by vioravis
Hello,
I am trying to find the distribution of the product of two normal densities (different means and standard deviations) . Does the product follow normal distribution also? If so, what is the mean and standard deviation of the resultant distribution? Thanks a lot.
Do you mean "What is the distribution of the product of two normally distributed RV with means and variances which may be different?"?
RonL
4. I have the following two densities:
f1(X) = {1/sigma1*sqrt(2*pi}* exp{-(X-mu1^2)/2*sigma1^2}
f2(X) = {1/sigma2*sqrt(2*pi}* exp{-(X-mu2^2)/2* sigma2^2}
So f1 is N(mu1, Sigma1^2) and f2 is N(mu2, Sigma2^2).
Does f(x) = f1(X)*f2(X) follow a normal distribution? Thanks a lot.
5. In Page 3 of the following link, it is given that the product of two gaussian densities is also gaussian. I think it is same as the one I requested above. I am looking for the derivation of the formula given for the resultant distribution:
confirmed here as well:
Product of Two Gaussian PDFs
6. Originally Posted by vioravis
I have the following two densities:
f1(X) = {1/sigma1*sqrt(2*pi}* exp{-(X-mu1^2)/2*sigma1^2}
f2(X) = {1/sigma2*sqrt(2*pi}* exp{-(X-mu2^2)/2* sigma2^2}
So f1 is N(mu1, Sigma1^2) and f2 is N(mu2, Sigma2^2).
Does f(x) = f1(X)*f2(X) follow a normal distribution? Thanks a lot.
If I understand you correctly, what you're asking boils down to wanting to show that
$-\frac{(x - \mu_1)^2}{2\sigma^2_1} - \frac{(x - \mu_2)^2}{2\sigma^2_2} = -C \frac{(x - \mu)^2}{2\sigma^2}$
and getting the appropriate expressions for $\mu, ~ \sigma$ and C in terms of $\mu_1, ~ \mu_2, ~ \sigma_1$ and $\sigma_2$. Note that $e^C$ becomes part of the normalising constant.
It's simple to show and get the expressions but tedious to type out.
7. Fantastic,
Thanks a lot. Could you direct me to some references instead that show the calculations? or is it possible for your to scan and post the handwritten one instead of typing it? Thank you.
8. Originally Posted by mr fantastic
If I understand you correctly, what you're asking boils down to wanting to show that
$-\frac{(x - \mu_1)^2}{2\sigma^2_1} - \frac{(x - \mu_2)^2}{2\sigma^2_2} = -C \frac{(x - \mu)^2}{2\sigma^2}$
and getting the appropriate expressions for $\mu, ~ \sigma$ and C in terms of $\mu_1, ~ \mu_2, ~ \sigma_1$ and $\sigma_2$. Note that $e^C$ becomes part of the normalising constant.
It's simple to show and get the expressions but tedious to type out.
$-\frac{(x - \mu_1)^2}{2\sigma^2_1} - \frac{(x - \mu_2)^2}{2\sigma^2_2}$
$= \frac{-\sigma^2_2(x - \mu_1)^2 - \sigma^2_1 (x - \mu_2)^2}{2 \sigma_1^2 \sigma^2}$
$= \frac{ -\sigma^2_2 x^2 + 2\sigma_2^2 \mu_1 x - \mu_1^2 \sigma_2^2 -\sigma^2_1 x^2 + 2\sigma_1^2 \mu_2 x - \mu_2^2 \sigma_1^2}{2 \sigma_1^2 \sigma^2}$
$= \frac{ -(\sigma_1^2 + \sigma_2^2) x^2 + 2(\sigma_2^2 \mu_1 + \sigma_1^2 \mu_2) x - (\mu_1^2 \sigma_2^2 + \mu_2^2 \sigma_1^2)}{2 \sigma_1^2 \sigma^2}$
$= \frac{-(\sigma_1^2 + \sigma_2^2) \left[ x^2 - \frac{2(\sigma_2^2 \mu_1 + \sigma_1^2 \mu_2) x}{\sigma_1^2 + \sigma_2^2} + \frac{\mu_1^2 \sigma_2^2 + \mu_2^2 \sigma_1^2}{\sigma_1^2 + \sigma_2^2}\right]}{2 \sigma_1^2 \sigma^2}$
$= \frac{-(\sigma_1^2 + \sigma_2^2) \left[ \left( x - \frac{\sigma_2^2 \mu_1 + \sigma_1^2 \mu_2 }{\sigma_1^2 + \sigma_2^2}\right)^2 - \left(\frac{\sigma_2^2 \mu_1 + \sigma_1^2 \mu_2}{\sigma_1^2 + \sigma_2^2} \right)^2 \right]}{2 \sigma_1^2 \sigma^2}$
$= \frac{- \left( x - \frac{\sigma_2^2 \mu_1 + \sigma_1^2 \mu_2 }{\sigma_1^2 + \sigma_2^2}\right)^2 + \left(\frac{\sigma_2^2 \mu_1 + \sigma_1^2 \mu_2}{\sigma_1^2 + \sigma_2^2} \right)^2}{\frac{2 \sigma_1^2 \sigma^2}{\sigma_1^2 + \sigma_2^2}}$
$= \frac{- \left( x - \frac{\sigma_2^2 \mu_1 + \sigma_1^2 \mu_2 }{\sigma_1^2 + \sigma_2^2}\right)^2}{\frac{2 \sigma_1^2 \sigma^2}{\sigma_1^2 + \sigma_2^2}} + C$.
So $\mu = \frac{\sigma_2^2 \mu_1 + \sigma_1^2 \mu_2 }{\sigma_1^2 + \sigma_2^2}$ and $\sigma^2 = \frac{\sigma_1^2 \sigma^2}{\sigma_1^2 + \sigma_2^2}$.
9. Originally Posted by vioravis
I have the following two densities:
f1(X) = {1/sigma1*sqrt(2*pi}* exp{-(X-mu1^2)/2*sigma1^2}
f2(X) = {1/sigma2*sqrt(2*pi}* exp{-(X-mu2^2)/2* sigma2^2}
So f1 is N(mu1, Sigma1^2) and f2 is N(mu2, Sigma2^2).
Does f(x) = f1(X)*f2(X) follow a normal distribution? Thanks a lot.
That f(x) is of the form of a Gaussian follows from the convolution theorem and the fact that the Fourier transform of a Gaussian is a Gaussian. But this does not guarantee that the product of the densities is in fact a density. As it happens it would be quite supprising if it were.
So we can bust a gut attempting to prove that $\int_{-\infty}^{\infty} f(x)~dx \ne 1$, but we can just do the experiment and evaluate the thing numerically. The answer is that the integral is not $1$ so $f$ is not a density.
Code:
>s1=1,mu1=0,s2=2, mu2=5
1
0
2
5
>dx=0.2;
>x=-10+dx/2:dx:20;
>
>f1=1/(s1*sqrt(2*pi))* exp( -(x-mu1)^2 / (2*s1^2) );
>f2=1/(s2*sqrt(2*pi))* exp( -(x-mu2)^2/ (2*s2^2) );
>
>f=f1*f2;
>
>II1=sum(f1)*dx
1
>II2=sum(f2)*dx
1
>II=sum(f)*dx
0.014645
>
RonL
10. Originally Posted by CaptainBlack
That f(x) is of the form of a Gaussian follows from the convolution theorem and the fact that the Fourier transform of a Gaussian is a Gaussian. But this does not guarantee that the product of the densities is in fact a density. As it happens it would be quite supprising if it were.
So we can bust a gut attempting to prove that $\int_{-\infty}^{\infty} f(x)~dx \ne 1$, but we can just do the experiment and evaluate the thing numerically. The answer is that the integral is not $1$ so $f$ is not a density.
[snip]
Indeed.
And the given references don't say the product is a pdf either. They merely give an expression for the product. The expression is used to facilitate the proofs of other things.
11. Originally Posted by CaptainBlack
That f(x) is of the form of a Gaussian follows from the convolution theorem and the fact that the Fourier transform of a Gaussian is a Gaussian.
[snip]
Ha ha. I was going to take that route ....... Just to add some flesh to the idea:
$FT[\, f \cdot g\, ] = FT[\, f \, ] * FT[\, g \, ] =$ gaussian * gaussian = gaussian.
$FT[\, f \cdot g\, ] =$ gaussian therefore $f \cdot g =$ gaussian.
12. Thanks a lot both. It was very useful.
13. I have a couple of more questions on this product of two distributions:
1. In the derivation given by mr. fantastic, constant C has been ignored why defining mu and sigma? I am also not sure how can we define mu and sigma if the resultant expression is not a pdf?
2. Is this extensible for the multivariate case? I would appreciate if you can provide me some references in this regard.
Thanks a lot.
14. Originally Posted by vioravis
I have a couple of more questions on this product of two distributions:
1. In the derivation given by mr. fantastic, constant C has been ignored why defining mu and sigma? I am also not sure how can we define mu and sigma if the resultant expression is not a pdf?
2. Is this extensible for the multivariate case? I would appreciate if you can provide me some references in this regard.
Thanks a lot.
They are just parameters in the Gaussians. Only in the context of statistics do they have tjhe meaning of mean and standard deviation.
C is ignored because it just becomes a multiplying factor: e^C = constant.
15. Hi I have a problem similar to the poster's
f1(X) = {1/sigma1*sqrt(2*pi}* exp{-( y1-mu1^2)/2*sigma1^2}
f2(X) = {1/sigma2*sqrt(2*pi}* exp{-( y2-mu2^2)/2* sigma2^2}
notice the y1 and y2
and now y1 = x + n1
and y2 = x + n2
and n1 and n2 are normal distributions
I need to find the product of two functions and prove x follows a normal distribution...
Thank you.
Page 1 of 2 12 Last | 2014-12-28T13:10:51 | {
"domain": "mathhelpforum.com",
"url": "http://mathhelpforum.com/advanced-statistics/38179-product-two-normal-distributions.html",
"openwebmath_score": 0.8949968814849854,
"openwebmath_perplexity": 503.01167627082714,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9740426405416756,
"lm_q2_score": 0.8791467595934565,
"lm_q1q2_score": 0.856326431138068
} |
https://math.stackexchange.com/questions/1825199/how-many-bit-strings-of-length-8-start-with-1-or-end-with-01?noredirect=1 | # How many bit strings of length 8 start with “1” or end with “01”?
A bit string is a finite sequence of the numbers $0$ and $1$. Suppose we have a bit string of length $8$ that starts with a $1$ or ends with an $01$, how many total possible bit strings do we have?
I am thinking for the strings that start with a 1, we would have $8 - 1 = 7$ bits to choose, so $2^7$ possible bit strings of length $8$ that starts with a $1$?
Can I go about the second condition the same way and just add the total's together? That is, if my logic is even correct in the first place?
• You can use the idea you had to figure out the number of strings with a $01$ at the end, but you'll be over-counting if you simply sum the two numbers since there will strings which start with $1$ and end with $01$, and you'll be counting each of these twice. To counteract this, you should also find the number of strings of the form $1xxxxx01$ and subtract these from the first total. This is known as the inclusion-exclusion principle. – stochasticboy321 Jun 14 '16 at 0:44
• Possible duplicate of How many bit strings of length 8 start with 00 or end with 1? – Did Jul 12 '16 at 15:40
We interpret starts with $1$ or ends in $01$ as meaning that bit strings that satisfy both conditions qualify.
By your correct analysis, there are $2^7$ bit strings that start with $1$.
Similarly, there are $2^6$ bit strings that end with $01$.
The sum $2^7+2^6$ double-counts the bit strings that start with $1$ and end with $01$.
There are $2^5$ of these, so there are $2^7+2^6-2^5$ bit strings that start with $1$ or end with $01$.
• Thank you for making it clear and confirming my original thought process. I felt like there would be some extra's in there; would these 'extras' also be found similarly by taking the intersection of the two if they were sets? – taylor.tackett Jun 14 '16 at 2:55
• You are welcome. I deliberately avoided formulas. But for any finite set $X$, let $|X|$ be the number of elements in $X$. Let $A$ be the set of strings that begin with $1$, and let $B$ be the set of strings that end in $01$. The set we want to count is $A\cup B$, so we want $|A\cup B|$. We have in general $|A\cup B|=|A|+|B|-|A\cap B|$. The $2^5$ that I subtracted at the end is $|A\cap B|$. – André Nicolas Jun 14 '16 at 3:00
• i.e. 160 bit strings. – Lightness Races in Orbit Jun 14 '16 at 11:46
The strategy you seem to be proposing is to note that there are $2^7$ bit strings starting with $1$ and $2^6$ ending with $01$, since one may make $7$ choices in the first case and $6$ choices in the second. If we add these up to get $2^6+2^7$, this doesn't quite work to count the number of strings satisfying either condition. In particular, consider a string like $$10000001$$ it both starts with $1$ and ends with $01$, so the above method would have counted it twice. In particular, the remedy for this is to subtract out the number of strings that satisfy both conditions from the sum $2^6+2^7$ to compensate for counting those strings twice.
This is the inclusion-exclusion principle.
Here is another way to arrive at the answer, without doing the whole "double count and then correct for it" dance:
Of all possible octets (8-bit strings), half of them will begin with $1$. Of the other half (i.e. those that begin with $0$), a quarter will end with $01$. Since there are $2^8$ possible octets, we have: $$2^8 \times \frac{1}{2} + 2^8 \times \frac{1}{2} \times \frac{1}{4} \\ 2^7 + 2^5$$ While this may not look identical to the other answers, note that: $$2^5 = 2^6 - 2^5$$ because $$2^6 - 2^5 = 2 \times 2^5 - 2^5 = 2^5 + 2^5 - 2^5 = 2^5$$
Although the other answers show you how to work your logic into a correct application of the inclusion-exclusion principle, one could take a slightly different approach and sum sizes of nonintersecting sets of events.
Case 1:
First binary digit is 1. Given this condition, all possible strings with attribute fulfill the required 'Or' condition. So there are $2^7$ strings in this set.
Case 2:
The first binary digit is 0. Given this condition, only strings that end in $01$ fulfill the required condition. This leaves only 5 binary digits to freely choose: we count all of the form $0xxxxx01$. So there are $2^5$ strings in this set.
Summing the number of combinations for the two mutually exclusive, but exhaustive conditions yields $2^7 + 2^5$ combinations. | 2021-07-24T02:25:15 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/1825199/how-many-bit-strings-of-length-8-start-with-1-or-end-with-01?noredirect=1",
"openwebmath_score": 0.8425886631011963,
"openwebmath_perplexity": 143.66580271105101,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9740426397881662,
"lm_q2_score": 0.8791467580102418,
"lm_q1q2_score": 0.8563264289335041
} |
https://math.stackexchange.com/questions/2773547/how-many-subsets-of-a-0-1-2-n-are-there-such-that-no-consecutive-numbe | # How many subsets of $A=\{0,1,2,...,n\}$ are there such that no consecutive numbers come together
It is likely a duplicate, but I couldn't find an original question, so creating a new one.
Given a set $A=\{1,2,...,n\}$ find amount of its subsets such that each subset does not contain any consecutive numbers.
For example, $\{1,3\}, \varnothing, \{1,3,5\}$ are OK, but $\{1,2\}, A, \{1,3,n-1,n\}$ are not OK.
I tried to solve this task using inclusion-exclusion formula, but got stuck when computing the 3rd term. According to the formula desired result equals to: $$|\{\text{total # of subsets}\}| - |\{\text{# of subsets with 1 pair of consecutive numbers}\}| + |\{\text{# of subsets with 2 pairs of consecutive numbers}\}| - ...$$ First term is easy, it equals to $2^n$.
To calculate the second term I am picking one consecutive pair out of $n-1$ possible and then calculate subsets with this pair included. So it equals $(n-1) \cdot 2^{n-2}$.
For the 3rd term I tried to pick one pair out of $n-1$ possible, then the second pair out of $n-2$ remaining and then calculate amount of subsets with both pairs included, i.e. it would equal something like $(n-1)(n-2)\cdot 2^{n-4}$. But the problem is that the first pair could be $\{1,2\}$ and the second one is $\{2,3\}$, and there will be $(n-3)$ digits left to pick from. To account for this we'll have to split the variants into these two cases. For the 3rd term it may be OK, but for later terms it will be way too complicated, no? Is there a nicer solution?
• Try a recursion. Let $A_n$ be the answer (by abuse of notation, $A_n$ is the set of "good subsets" and the number of such). If your subset doesn't contain $n$ then it must be an element of $A_{n-1}$. If it does contain $n$ it must be an element of $A_{n-2}\cup \{n\}$.
– lulu
May 9 '18 at 11:39
• General note: for problems like this it is an excellent idea to work out the answer for small $n$, see if you can spot a familiar pattern.
– lulu
May 9 '18 at 11:40
Let $A_n = \{1,2,3,\dots,n\}$. Let $G_n$ be the number of "good" subsets of $A_n$. Here I will consider the empty set to be a "good" subset of each $A_n$.
For each subset, $S$ of $A_n$ there is a function $f_{n,S}:A_n \to \{0,1\}$ define by $f_{n,S}(x)= \begin{cases} 0 & \text{If$x \not \in S$} \\ 1 & \text{If$x \in S$} \\ \end{cases}$
\begin{array}{c} & f \\ \text{subset} & 1 & \text{good?} \\ \hline & 0 &\checkmark \\ 1 & 1 &\checkmark \\ \hline \end{array}
So $A_1=2$.
\begin{array}{c} & f \\ \text{subset} & 12 & \text{good?} \\ \hline & 00 &\checkmark \\ 1 & 10 &\checkmark \\ 2 & 01 &\checkmark \\ 12 & 11 \\ \hline \end{array}
So $A_2=3$.
\begin{array}{c} & f \\ \text{subset} & 123 & \text{good?} \\ \hline & 000 &\checkmark &\text{Compare to $A_2$}\\ 1 & 100 &\checkmark \\ 2 & 010 &\checkmark \\ 12 & 110 \\ \hline 3 & 001 &\checkmark &\text{Compare to $A_1$}\\ 13 & 101 &\checkmark \\ 23 & 011 & \\ 123 & 111 \\ \hline \end{array}
So $A_3=A_1+A_2=5$.
\begin{array}{c} & f \\ \text{subset} & 1234 & \text{good?} \\ \hline & 0000 &\checkmark &\text{Compare to $A_3$}\\ 1 & 1000 &\checkmark \\ 2 & 0100 &\checkmark \\ 12 & 1100 \\ 3 & 0010 &\checkmark \\ 13 & 1010 &\checkmark \\ 23 & 0110 & \\ 123 & 1110 \\ \hline 4 & 0001 &\checkmark &\text{Compare to $A_2$}\\ 14 & 1001 &\checkmark \\ 24 & 0101 &\checkmark \\ 124 & 1101 \\ 34 & 0011 & \\ 134 & 1011 & \\ 234 & 0111 & \\ 1234 & 1111 \\ \hline \end{array}
So $A_4=A_2+A_3=8$.
So it seems that $A_1=2, \quad A_2=3, \quad$ and $A_{n+2}=A_n + A_{n+1}$ Thus $A_n = F_{n+2}$, the $(n+2)^{th}$ fibonacci number.
Supposing the subset is not the empty set we first choose the smallest value:
$$\frac{z}{1-z}.$$
Then we add in at least two several times to get the remaining values:
$$\frac{z}{1-z} \sum_{m\ge 0} \left(\frac{z^2}{1-z}\right)^m = \frac{z}{1-z} \frac{1}{1-z^2/(1-z)} = \frac{z}{1-z-z^2}.$$
Finally we collect the contributions that sum to at most $n$ and add one to account for the empty set:
$$1+ [z^n] \frac{1}{1-z} \frac{z}{1-z-z^2} \\ = [z^n] \frac{1}{1-z} + [z^n] \frac{1}{1-z} \frac{z}{1-z-z^2} \\ = [z^n] \frac{1}{1-z} \frac{1-z^2}{1-z-z^2} \\ = [z^n] \frac{1+z}{1-z-z^2}.$$
Calling the OGF $G(z)$ we have
$$G(z) (1-z-z^2) = 1 + z$$
so that for $[z^0]$ we get
$$[z^0] G(z) (1-z-z^2) = 1$$
or $g_0 = 1.$ We also get
$$[z^1] G(z) (1-z-z^2) = 1$$
or $g_1-g_0 = 1$ or $g_1=2.$ We have at the end for $n\ge 2$
$$g_n-g_{n-1}-g_{n-2} = 0,$$
which is the Fibonacci number recurrence. With these two initial values we obtain
$$\bbox[5px,border:2px solid #00A000]{ F_{n+2}.}$$ | 2021-12-03T00:43:55 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2773547/how-many-subsets-of-a-0-1-2-n-are-there-such-that-no-consecutive-numbe",
"openwebmath_score": 1.0000098943710327,
"openwebmath_perplexity": 235.88007783621183,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9835969689263264,
"lm_q2_score": 0.8705972549785201,
"lm_q1q2_score": 0.8563168211524526
} |
https://math.stackexchange.com/questions/865293/showing-ln-sinx-is-in-l-1 | # Showing $\ln(\sin(x))$ is in $L_1$
Prove $\ln[\sin(x)] \in L_1 [0,1].$
Since the problem does not require actually solving for the value, my strategy is to bound the integral somehow. I thought I was out of this one free since for $\epsilon > 0$ small enough, $$\lim_{\epsilon \to 0}\int_\epsilon^1 e^{\left|\ln(\sin(x))\right|}dx=\cos(\epsilon)-\cos(1) \to 1-\cos(1)<\infty$$
and so by Jensen's Inequality, $$e^{\int_0^1 \left| \ln(\sin(x))\right|\,dx}\le \int_0^1e^{\left|\ln(\sin(x))\right|}\,dx\le1-\cos(1)<\infty$$ so that $\int_0^1 \left|\ln(\sin(x))\right|\,dx<\infty$.
The problem, of course, is that the argument begs the question, since Jensen's assumes the function in question is integrable to begin with, and that's what I'm trying to show.
Any way to save my proof, or do I have to use a different method? I attempted integration by parts to no avail, so I am assuming there is some "trick" calculation I do not know that I should use here.
• For $0\leq x\leq 1$, you have that $0\leq \sin x\leq 1$. So, $|\ln(\sin x)|=-\ln(\sin x)$. You can integrate using by parts. – Joe Johnson 126 Jul 12 '14 at 18:04
• Seems hard to save your proof since you need the conclusion to prove it... – Surb Jul 12 '14 at 18:05
• On an unrelated note, Jensen is a bit of an overkill. Just noting that $f\sim|\ln x|$ (which is integrable) near zero is enough. – user138530 Jul 12 '14 at 18:37
• Unfortunately for your argument, $e^{\left|\ln(\sin(x))\right|}=1/\sin(x)$, not $\sin(x)$. – user940 Jul 12 '14 at 21:09
We can show this using the fact that $\sin x \sim x$ for small values of $x$; precisely, we have the inequality
$$\frac 1 2 x \le \sin x$$
for all $x \in [0,1]$; this leads to
$$\ln\left(\frac{x}{2}\right) \le \ln \sin x$$
almost everywhere on $[0,1]$. We'll actually use that
$$-\ln \left(\frac x 2\right) \ge - \ln \sin x$$Noting that $\ln(x/2) = \ln x - \ln 2$, and that our measure space is finite, it is sufficient to show that $\ln x \in L^1[0,1]$. To do this, we show that $\ln(1/x)$ has finite Lebesgue integral on this interval via the Monotone Convergence Theorem (hence the usage of the $-$ sign to make things positive). Since $\ln x$ is continuous and bounded on every interval $[\epsilon, 1]$, the Lebesgue integral coincides with the Riemann integral, and applying the MCT to the functions $-\chi_{[1/n,1]} \ln x$ gives
\begin{align*} \int_0^1 - \ln x dx &= \lim_{n \to \infty} \int_{1/n}^1 - \ln x dx \\ &= - \lim_{n \to \infty} x (\ln x - 1) \Big|_{1/n}^1 \\ &= - \lim_{n \to \infty} \Big(1 (\ln 1 - 1)\Big) - \Big(\frac 1 n \left(\ln \frac 1 n - 1\right)\Big) \\ &= 1 - \lim_{n \to \infty} \left(\frac 1 n + \frac{\ln n}{n}\right) \\ &= 1 \end{align*}
Now by comparison, the original function is integrable.
First observe that $\ln \sin x = \ln {\sin x \over x} + \ln x$.
The function ${\sin x \over x}$ is continuous and nonzero on on $[0,1]$ (if you extend it to equal $1$ at $x = 0$), so the same is true for $\ln {\sin x \over x}$ . Thus $\ln {\sin x \over x}$ is in $L^1[0,1]$.
The function $\ln x$ is also integrable on $[0,1]$ as its antiderivative is $x \ln x - x$ which converges to zero as $x = 0$.
So their sum $\ln \sin x$ is in $L^1[0,1]$ too.
A simpler approach would be to observe that the function $x^{1/2}\ln \sin x$ is bounded on $(0,1]$, because it has a finite limit as $x\to 0$ -- by L'Hôpital's rule applied to $\dfrac{\ln \sin x}{x^{-1/2}}$. This gives $|\ln \sin x|\le Mx^{-1/2}$.
As Byron Schmuland noted, $e^{|\ln \sin x|} = 1/\sin x$, which is nonintegrable; this is fatal for your approach.
Here is a proof that hides behind a theorem on swapping order of integration:
We have $0 \le {x \over 2} \le \sin x$, and $-\log$ is decreasing on $(0,1]$.
Then $\int_0^1 | \log(\sin x)| dx = \int_0^1 - \log( \sin x) dx \le \int_0^1 - \log( { x \over 2}) dx = \log 2 + \int_0^1 - \log( { x}) dx$.
Tonelli gives $\int_0^1 -\log(x) dx = \int_{x=0}^1 \int_{t=x}^1 {dt \over t} dx = \int_{t=0}^1 \int_{x=0}^t {dx \over t} dt = 1$.
Hence the upper bound $\int_0^1 | \log(\sin x)| dx \le 1+ \log 2$.
• May I ask why do you prefer using Tonelli's theorem to finding the antiderivative of $\log x$ (by say integration by parts)? – EPS Jul 14 '14 at 18:11
• @Sam: I didn't even think of looking for an antiderivative directly. – copper.hat Jul 14 '14 at 18:16 | 2021-06-14T21:19:19 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/865293/showing-ln-sinx-is-in-l-1",
"openwebmath_score": 0.9995887875556946,
"openwebmath_perplexity": 180.74671707112188,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.979667648438234,
"lm_q2_score": 0.8740772417253256,
"lm_q1q2_score": 0.8563051959544276
} |
https://www.themathdoctors.org/frequently-questioned-answers-the-other-child/ | # Frequently Questioned Answers: The Other Child
This is the second in a series on Frequently Questioned Answers – that is, answers we have given that are often challenged by readers, either just out of confusion, or in the form of attacks on our intelligence or honesty. Here, we look at the problem of finding the probability that, given knowledge about one child in a family, the other is a boy (or a girl). This is discussed in our FAQ, Boy or Girl?, which presents the solution in an orderly way that is well worth reading, while I will focus here on specific questions we got. As you’ll see, this can be a hard problem to state correctly; even our FAQ has a flaw, as does my description two sentences back (and even the title of this post)!
## The basic problem and its answer
Here is the first question we got on this, from Bret in 1996:
Probability of Two Male Children
Several weeks ago, in one of the weekly periodicals I read, a person presented this logical statement: If a family has two children, and the older child is a boy, there is a 50 percent chance the family will have two boys. However, in a family with two children, if all we know is that one child is a boy (no age specified) there is a 1/3 chance of that family having two male children.
This boggles me, as I can plainly see the first scenario - it's quite simple: the younger child may be a boy or girl (1/2). It is a mystery to me how by not knowing whether the known boy is older or younger could change the probability of the family as a whole.
However, I have discussed this problem with my grandfather and he explains it like so: any two-child family has four possible outcomes: b:b, g:g, b:g, g:b. He states that, in the first case where the known boy is also known to be the older child, we thereby eliminate the g:g and g:b combinations, leaving two left. And obviously one of those two is b:b so our odds of having two male children is 1/2.
By his same logic, in the 2nd instance all that is known is that one child is a boy, therefore of course the g:g combination cannot exist, leaving b:b, b:g, g:b. Hence, there is a 1/3 chance of two boys.
Although his explanation is very clear, I fail to understand how the odds of the entire family can be swayed by whether or not a boy is older or not. Example: I see a common ordinary boy. I am told that he has a sibling. Am I to believe now that his sibling has a 2/3 chance of being a girl?? Now I am told he is the older child. Magically the odds change and his younger sibling has a 1/2 chance of being a girl now?
It baffles me.
Bret’s grandfather’s explanation is a concise version of what we say in our FAQ, and the initial statement of the problem avoids all the pitfalls we’ll be looking at. We’ll see later, though, that the final paragraph, by starting with the boy, misstates the problem in such a way that the correct answer is 1/2. The problem is very sensitive to small changes in its statement!
To expand the explanation a bit, suppose we gathered a random collection of families, each with exactly two children. Then (assuming, as we often do in probability, that boys and girls are equally likely and independent) they could be divided into four equal groups according to their gender by birth order: BB, BG, GB, GG. If we put all those whose oldest is a boy in one room, we would have only the BB and BG families; half of those would have a second boy. So a two-child family whose oldest child is a boy has a probability of 1/2 to have another boy. That’s just what we’d expect.
But if we put in that room all the families that have at least one boy, without regard to position in the family, we would be including BB, BG, and GB. Only 1/3 of these would have two boys! So not knowing which child is a boy reduces the probability. How can that be?
Doctor Anthony answered this, starting with a different example of conditional probability, showing how restrictions on the “sample space” can affect probability. He then briefly discussed this problem:
The more information you provide, the more you change the sample space (the denominator) of the probability calculation. In the case of sons or daughters in a family of four, if you exclude gg or gb, then as you said you can only have bb or bg, so 1/2 probability that second child is a girl, whereas if one child (unspecified) is a boy, the probability space is now bb, bg, gb and the chance of two boys is now only 1/3.
The thing to remember is that CONDITIONAL probability can dramatically change the commonsense idea of what a particular probability should be.
More will have to be said …
The basic problem was asked and answered in much the same way in 1999:
Bayes Theorem
## How to (mis)read the problem
In 2000, we got a challenge to the previous answer:
Boy or Girl: Two Interpretations
While looking for something else, I stumbled upon this question:
Probability of Two Male Children
http://mathforum.org/dr.math/problems/mcclory.7.5.96.html
and saw that you said the probability of the second boy having a brother was 1/3. By my calculations the boy's sibling is either elder or younger and either male or female. Assuming that the probabilities of each are equal:
elder brother = 1/4
younger brother = 1/4
elder sister = 1/4
younger sister = 1/4
As such, the probability of him having a brother is 1/4 + 1/4 = 1/2.
Where you went wrong was in saying b:b was as likely as b:g or g:b, when of course in b:b if we now call the boy we know about x, we have x:b and b:x, so we have the sets x:g, g:x, b:x, and x:b; and thus a probability of 1/2.
Phil is not just going by intuition, but has found a specific argument that convinces him we are wrong. Are we? Doctor TWE replied, focusing on a key idea:
One of the biggest problems in probability is stating the problem clearly. Either answer, your 1/2 or Dr. Anthony's 2/3, could be correct depending on how the problem is set up.
In this problem, a key factor in determining the probability is how the child and family are selected. When we say, "in a two-child family, one child is a boy," how did we select the child? The selection process makes a big difference in the final probability (or, as Dr. Anthony would say, in the "sample space" of the problem.)
We often comment that the exact words (and, too often, unstated assumptions) are essential in any probability problem. We have to be careful both in how we state the problem, and in how we read it. This problem, in particular, is very easy to either write or read wrongly (especially as it tempts us to state it in such a way that the answer will be unexpected).
First, this is how Phil is seeing it:
Supposing that we randomly pick a _child_ from a two-child family. We see that he is a boy, and want to find out whether his sibling is a brother or a sister. (For example, from all the children of two-child families, we select a child at random who happens to be a boy.) In this case, an unambiguous statement of the question could be:
From the set of all families with two children, a child is selected at random and is found to be a boy. What is the probability that the other child of the family is a girl?
Note that here we have a pool of kids (all of whom are from two-child families) and we're pulling one kid out of the pool. This is like the problem you're talking about. The child selected could have an older brother, an older sister, a younger brother or a younger sister.
Let's look at the possible combinations of two children. We'll use B for Boy and G for girl, and for each combination we'll list the older child first, so GB means older sister while BG means younger sister. There are 4 possible combinations:
{BB, BG, GB, GG}
From these possible combinations, we can eliminate the GG combination since we know that one child is a boy. The three remaining possible combinations are:
{BB, BG, GB}
In these combinations there are four boys, of whom we have chosen one. Let's identify them from left to right as B1, B2, B3 and B4. So we have:
{B1B2, B3G, GB4}
Of these four boys, only B3 and B4 have a sister, so our chance of randomly picking one of these boys is 2 in 4, and the probability is 1/2 - as you have indicated.
So, we put all our two-child families into that room, and half the boys will be from two-boy families (two of them supplied by each such family!), and the other half will be from one-boy families. Everything is as we expect. But here, we counted boys, not families.
But now let's look at a different way of selecting the "boy" in the problem. Suppose we randomly choose the two-child _family_ first. Once the family has been selected, we determine that at least one child is a boy. (For example, from all the mothers with two children, we select one and ask her whether she has at least one son.) In this case, an unambiguous statement of the question could be:
From the set of all families with two children, a family is selected at random and is found to have a boy. What is the probability that the other child of the family is a girl?
Note that here we have a pool of families (all of whom are two-child families) and we're pulling one family out of the pool. Once we've selected the family, we determine that there is, in fact, at least one boy.
Since we're told that one child (we don't know which) is a boy, we can eliminate the GG combination. Thus, our remaining possible combinations are:
{BB, BG, GB}
Each of these combinations is still equally likely because we picked one of the four families.
Now we want to count the combinations in which the "other" child is a girl. There are two such combinations: BG and GB.
Since there are three combinations of possible families, and in two of them one child is a girl, the probability is 2/3.
This is the answer to the problem as intended by our FAQ.
Why are these two probabilities different?
As in other probability problems, how information is obtained is as important as the information itself. Without knowledge of the data gathering process, ambiguity can result. How do we know that one child is a boy?
In the first (your) interpretation, each _boy_ has an equal chance of being chosen. Thus, the family with two boys has twice the chance of being the "chosen family." The boys are equally probable, but the families are not.
In the second (Dr. Anthony's) interpretation, each _family_ has an equal chance of being chosen. In a family with two boys, each boy has only half that chance of being "the boy" referenced in the statement. The families are equally probable, but the boys are not. In this case, the two "events" are not independent, because we're selecting a family, not an individual child. In fact, there's really only one "event" - the selection of the family.
If you're still not convinced, try the following experiment. Take two fair coins and toss them. I think you'll agree that each coin has a 1/2 chance of being heads. On each toss, see if at least one of the coins is heads (the equivalent of "at least one child is a boy"). If both coins are tails (both children are girls), ignore the outcome and toss again. If at least one of the coins is heads, record whether you had two heads (the boy has a brother) or a head and a tail (the boy had a sister). Over many tosses, you should find yourself getting about twice as many head-tail tosses as head-head tosses. Of course, if you count each head-head toss twice (once for each head tossed)...
Once again, an experiment, while not the most mathematical way to settle an argument, has the advantage of forcing you to face reality (and maybe think more about whether your model matches the problem). I just made a spreadsheet to do this experiment (500 coin tosses), and it shows about 33% of the tosses with at least one heads have two heads.
Doctor TWE’s answer above is the basis of the supplemental FAQ page, Family or Child First?
For similar problems, see
Cupcakes and Boxes: Conditional Probability
Boy, What Is Your Probability?
This problem is discussed at length in Wikipedia, Boy or Girl paradox, which includes our second FAQ as a reference.
## Our FAQ used to be wrong!
Looking through the many unarchived questions on this topic, I found that in 2001 a reader challenged, not our answer, but the closing statement of the problem itself:
Page http://forum.swarthmore.edu/dr.math/faq/faq.boy.girl.html has near its end:
"Remember: information that creates conditional probability can dramatically affect commonsense ideas about probability. For example, no matter how unlikely it may seem to you, if you meet a girl who says she has a sibling, a basic knowledge of probability tells you there's a 2/3 probability that she has a brother. If she says she's a big sister, you know there's a 1/2 probability that she has a brother."
Assuming this is the 2 child per family problem (because every other part of the problem has been 2 child and no info on distribution of family sizes is given), the odds should be 1/2 that her sibling is a brother. The explanation for that is given on:
http://forum.swarthmore.edu/dr.math/faq/faq.boygirl.choose.html in the section "Choosing the Child First".
I am convinced that if I picked a 2-child family at random and one child is a girl, then there is a 2/3 chance that the other is a boy, but that is not the case described. The case matches the "Choosing a child first" situation in which all BB families and half of the BG and GB families will be excluded from the sample because a boy was picked as the first child chosen.
If the first page I quoted made no assumption about family size, then just toss out everything I have said.
Karl was right! Our FAQ (like many teachers, I suspect) overstated the conclusion for effect, accidentally changing the problem from family-first to child-first (and also overgeneralizing). (This is exactly Bret’s error in the first question above.) Doctor TWE changed the FAQ from what is quoted above to what it is now:
Remember: information that creates conditional probability can dramatically affect common sense ideas about probability. For example, no matter how unlikely it may seem to you, if you meet a mother of two who says she has a daughter, a basic knowledge of probability tells you there's a 2/3 probability that the daughter mentioned has a brother. If she says she's an older daughter, you know there's a 1/2 probability that the daughter has a younger brother.
Even this, I think, could be misleading, if you suppose that the mother has volunteered the information, thinking of a specific daughter (“the daughter mentioned”). That could pull the problem over to the child-first realm, or at least require us to consider subjective probabilities based on the mother’s motivations.
## And it is still a little wrong!
Much later, in 2012, Mike wrote about the introduction to the problem:
... Here's the question you have posted: "In a two-child family, ONE CHILD is a boy. What is the probability that THE OTHER CHILD is a girl?" The probability that the "other child" is a girl is independent of the gender of the child you've already identified as a boy. "Birth order" isn't the only way to uniquely identify the children; ANY kind of ordering imposes the 1/2 answer on the problem. You're no longer talking about sets of children (i.e. families), you're talking about children. That's different. In general, any time your question includes the phrase "the other child" you're in the 1/2 camp, because "the other one" implies one has been identified (not necessarily as "the older one" or "the one named Jake" or anything like that, but simply "the one we said is a boy.") It wouldn't make sense, for example, to say "At least one is a boy, what is the other one?" The other what? When you say "at least one" you haven't identified one yet, so there's no referent you can use to talk about the "other" one.
This may seem like mere semantics, but it's more than that. The way you have the question worded, the answer is definitively 1/2. You would need to rework it to make 2/3 an acceptable interpretation (e.g. "A family has two children, at least one of which is a boy. What is the probability that they're not both boys?")
This led to a long discussion of semantics; in the end, I proposed rewording the FAQ so that the answer is still surprising, but it is clear upon consideration, in order to show the power of math rather than its weirdness.
This is my final proposed beginning to the FAQ:
In a two-child family, we are told that one child is a boy. What is the probability that they also have a girl?
What if we are told that the older child is a boy? Does this information change the probability that the second child is a girl?
-----
We first need to clarify the question, because English is often ambiguous, and probability requires precision. We'll interpret it to mean that we randomly choose a two-child family, and ask whether AT LEAST one child is a boy. The answer is yes. We want to know how likely it is that they have one boy and one girl.
(What if we choose the _child_ first?)
When the only information given is that there are two children and one is a boy, here are two ways of looking at the problem:
...
This fixes several issues. Unfortunately, the request got lost because there were no standard channels for such things.
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 2019-04-19T10:14:36 | {
"domain": "themathdoctors.org",
"url": "https://www.themathdoctors.org/frequently-questioned-answers-the-other-child/",
"openwebmath_score": 0.6361656188964844,
"openwebmath_perplexity": 607.5123668502303,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9796676514063882,
"lm_q2_score": 0.8740772351648677,
"lm_q1q2_score": 0.8563051921217553
} |
https://math.stackexchange.com/questions/424775/why-does-223-equal-64-and-not-64 | Why does $(-2^2)^3$ equal $-64$ and not $64$?
The title says it all. Why does $(-2^2)^3$ equal $-64$ and not $64$? This was on my algebra final, and I am completely stuck on how it works.
• I always have an issue with this because it is somewhat of an ambiguous notation. I think that when doing mathematics, that the concepts should be tested and not the differing interpretations of a problem. – yousuf soliman Jun 19 '13 at 19:22
The negative sign ($-$) applies to the quantity $2^2$, so that $-2^2$ means $-(2^2)=-4$, not $(-2)^2=4$. $$(-2^2)^3=(-4)^3=-64$$
• So for it to actually be 64, would the problem need parentheses around the -2? So would it have to be $((-2)^2)^3$ for it to equal 64? – S17514 Jun 19 '13 at 19:19
• Yup, that's right. – Zev Chonoles Jun 19 '13 at 19:20
Note that $-2^2 = - (2 \times 2) = -4$ and is not $(-2) \times (-2) = 4$. Hence, $$(-2^2)^3 = (-4)^3 = (-4) \times (-4) \times (-4) = - 64$$
In general, when $m$ is a positive integer, we have $$-a^m = - (\underbrace{a \times a \times \cdots \times a}_{m \text{times}})$$ and is not $$(-a)^m = (\underbrace{(-a) \times (-a) \times \cdots \times (-a)}_{m \text{times}})$$
• It's worth noting that some simplistic tokenizers do parse $-2^2$ as $(-2)^2$, which is non-standard. I believe most spreadsheets are guilty of this (perhaps to stay compatible with Microsoft Excel). – Erick Wong Jun 19 '13 at 19:21
Rewrite it as
$(-4)^3=(-1)^3 \cdot 4^3=-64$ | 2019-12-13T00:28:02 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/424775/why-does-223-equal-64-and-not-64",
"openwebmath_score": 0.8478679656982422,
"openwebmath_perplexity": 341.1156093961483,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9796676433923709,
"lm_q2_score": 0.8740772400852111,
"lm_q1q2_score": 0.8563051899371864
} |
https://math.stackexchange.com/questions/2972001/count-the-number-of-shapes-in-a-polyhedron/2972069 | # Count the number of shapes in a polyhedron.
So this is a question that was asked in the International Kangaroo Math Contest 2017. The question is:
The faces of the following polyhedron are either triangles or squares. Each triangle is surrounded by $$3$$ squares and each square is surrounded by $$4$$ triangles. If there are $$6$$ square faces, how many triangular faces are there?
What I did:
Each square shares each of its four neighboring triangles with two more squares. So we can say that for 6 squares we have $$6\times4\ -\ 2 \times 6 = 12$$ triangles. However, I still know that this calculation of mine is quite wrong and based on an awkward thinking. So, what is the correct answer and how?
Thanks for the attention.
• Always hated these questions when they specifically asked for reasoning. I just used my onboard 3D engine and could 'see' 8. Why? Because look :D – Lamar Latrell Oct 27 '18 at 6:41
Each edge of the polyhedron is shared between exactly one triangle and exactly one square, as can be inferred from the question statement. Thus, given six squares, there are 24 edges ($$6×4$$), and thus eight triangles ($$24÷3$$).
The polyhedron is called a cuboctahedron.
• Thanks! I just didn't try thinking it that way....;) – Faiq Irfan Oct 26 '18 at 11:35
Each square is bordered by four triangles and $$6\times4=24$$. However every triangle is bordered by three different squares, so it was counted three times in the multiplication above. This means there are $$24/3=8$$ triangles.
This isn't rigorous, but if you don't have to write a proof, just get the right number, it's clear from the illustration. You can see "one hemisphere" of the polyhedron except for a triangle parallel with the line of sight, so the total number of sides of each type are just double what apppear in that hemisphere (i.e. what you see plus the hidden triangle).
• That would give 2*3+1=7 triangles but they are actually 8. I think you'd like to amend something? – Rad80 Oct 27 '18 at 10:39
• @Rad80: "double what you see plus the hidden triangle" should clearly be read with parentheses around all but the first word; I thought that was obvious from the text before it. – R.. GitHub STOP HELPING ICE Oct 27 '18 at 22:05
• well, at least we should be able to agree that is was not obvious what you meant. I saw two mentions that there is one "hidden" triangle (both times singular), so why should it be counted twice? If you edit the answer mentioning two hidden triangles instead of just one it'll be fine. – Rad80 Oct 29 '18 at 8:11
• @Rad80: There's only one hidden triangle in the hemisphere you're doubling. "Two hidden triangles" would be nonsense as there are of course five hidden triangles. – R.. GitHub STOP HELPING ICE Oct 29 '18 at 15:07
• @Rad80: Hopefully the change I made addresses what you found confusing about the wording. – R.. GitHub STOP HELPING ICE Oct 29 '18 at 15:08 | 2020-02-20T11:59:02 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2972001/count-the-number-of-shapes-in-a-polyhedron/2972069",
"openwebmath_score": 0.6340360045433044,
"openwebmath_perplexity": 705.1466989107889,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9796676436891864,
"lm_q2_score": 0.8740772286044095,
"lm_q1q2_score": 0.8563051789492562
} |
https://math.stackexchange.com/questions/2392867/construct-4-times-4-magic-square-with-fixed-1 | # Construct $4 \times 4$ magic square with fixed “1”
The method I have found to generate $4\times 4$ magic squares gives me a result in which the number "1" is at of the corners of the square. How can we extend this to a method to generate a magic square, for a fixed location of number "$1$"?
The number "$1$" can be in $16$ different locations (cells). If we name the cells, from upper left corner: $1, 2, 3,4, 5, \ldots,$ and the number "$1$" is at the $i^{th}$ cell, then how we can fill the other cells to make a magic square?
You can see variations here.
There are really only three different locations: a corner, a side next to a corner, and one step diagonally in from a corner. If we can put $1$ into each of these, we can put it in any cell by using rotations and reflections. We already have a corner. We can subtract every number from $16$ and keep the square magic, but that doesn't help because it puts $1$ in the lower right corner. Because of the arrangement, we can add $8$ to all the numbers below $8$ and subtract $8$ from all those above, giving $$\begin {array} {c|c|c|c} 9&7&6&12 \\ \hline 4&14&15&1 \\ \hline 16&2&3&13 \\ \hline 5&11&10&8 \end {array}$$ That gets next to the corner. We can also rotate the $2 \times 2$ blocks by $180^\circ$ to get $$\begin {array} {c|c|c|c} 6&12&9&7 \\ \hline 15&1&4&14 \\ \hline 3&13&16&2 \\ \hline 10&8&5&11 \end {array}$$ which gets one in from the corner
• Thanks. Can we have a general solution for any doubly even order? – Susan_Math123 Aug 14 '17 at 4:49
• The example you cited has a lot of symmetry that not all magic squares have. I exploited that in both of my transformations. They will not work starting with a generic doubly even square. – Ross Millikan Aug 14 '17 at 5:06
At first let's define a more powerful magic square, which we will call $\color{Red}{\text{super-magic square}}$. By a $\color{Red}{\text{super-magic square}}$ we mean a magic square such the the sum of any arbitrary row is equal to the sum of any arbitrary column is equal to sum of any arbitrary diagonal.
For example suppose the following: $$\begin {array} {|c|c|c|c|} \hline 1&14&7&12 \\ \hline 15&4&9&6 \\ \hline 10&5&16&3 \\ \hline 8&11&2&13\\ \hline \end {array}$$
here we have:
$\color{Blue}{\text{Columns}}$: $$\begin {array} {ccccccccc} 1 & + & 15 & + & 10 & + & 8 & = & 34 \\ 14 & + & 4 & + & 5 & + & 11 & = & 34 \\ 7 & + & 9 & + & 16 & + & 2 & = & 34 \\ 12 & + & 11 & + & 3 & + & 13 & = & 34 \\ \end {array}$$
$\color{Green}{\text{Rows}}$: $$\begin {array} {ccccccccc} 1 & + & 14 & + & 7 & + & 12 & = & 34 \\ 15 & + & 4 & + & 9 & + & 6 & = & 34 \\ 10 & + & 5 & + & 16 & + & 3 & = & 34 \\ 8 & + & 11 & + & 2 & + & 13 & = & 34 \\ \end {array}$$
$\color{Purple}{\text{Diagonals parallel to the main diagonal}}$: $$\begin {array} {ccccccccc} 1 & + & 4 & + & 16 & + & 13 & = & 34 \\ 14 & + & 9 & + & 3 & + & 8 & = & 34 \\ 7 & + & 6 & + & 10 & + & 11 & = & 34 \\ 12 & + & 15 & + & 5 & + & 2 & = & 34 \\ \end {array}$$
$\color{Pink}{\text{Diagonals which are not parallel to the main diagonal}}$: $$\begin {array} {ccccccccc} 12 & + & 9 & + & 5 & + & 8 & = & 34 \\ 7 & + & 4 & + & 10 & + & 13 & = & 34 \\ 14 & + & 15 & + & 3 & + & 2 & = & 34 \\ 1 & + & 6 & + & 16 & + & 11 & = & 34 \\ \end {array}$$
$$%% %% 1 + 14 + 7 + 12 = 34 , %% \\ %% 15 + 4 + 9 + 6 = 34 , %% \\ %% 10 + 5 + 16 + 3 = 34 , %% \\ %% 8 + 11 + 2 + 13 = 34 ,$$
We will prove that, $1$ could be everywhere in a $\color{Red}{\text{super-magic square}}$.
Remark(I): Consider that a $\color{Red}{\text{super-magic square}}$ ($\color{Brown}{\text{of any arbitrary order}}$) is given. Then if we $\color{Blue}{\text{replace any two arbitrary columns}}$, then the resulting square, is again a $\color{Red}{\text{super-magic square}}$.
Remark(II): Consider that a $\color{Red}{\text{super-magic square}}$ ($\color{Brown}{\text{of any arbitrary order}}$) is given. Then if we $\color{Green}{\text{replace any two arbitrary rows}}$, then the resulting square, is again a $\color{Red}{\text{super-magic square}}$.
Now by $\color{Blue}{\text{column operations (I)}}$ and by $\color{Green}{\text{row operations (II)}}$ , we are able to change "the cell containing 1" to "any desired cell, so we are done!
• Where are the proofs/constructions for I and II? – OrangeDog Aug 14 '17 at 10:58
• @OrangeDog: This is only valid if we ignore diagonals. The Wikipedia article linked from the question explicitly says that the main diagonals must have the same sum as each row or column, so this answer is not valid... – user21820 Aug 14 '17 at 11:00
• @user21820 , Yes you are right. I have modified my answer, now this is true! – Jungle Boy Aug 14 '17 at 12:26
• Alright it's okay for the 4*4 magic square now. The name is "Pandiagonal magic square." Also, you proved a weaker, not stronger, result, because it does not apply to all magic squares in general. – user21820 Aug 14 '17 at 13:24
• And you should correct all your weird typos like "powerfull" and "rcolumn" and "coloumns"... – user21820 Aug 14 '17 at 13:26 | 2019-10-21T22:58:39 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2392867/construct-4-times-4-magic-square-with-fixed-1",
"openwebmath_score": 0.7688994407653809,
"openwebmath_perplexity": 436.7995129861236,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.979667647844603,
"lm_q2_score": 0.8740772236840656,
"lm_q1q2_score": 0.8563051777611095
} |
http://www.satvistomo.net/back-in-tfjg/minkowski-distance-knn-3b1a7f | metric str or callable, default=’minkowski’ the distance metric to use for the tree. Each object votes for their class and the class with the most votes is taken as the prediction. When p = 1, this is equivalent to using manhattan_distance (l1), and euclidean_distance (l2) for p = 2. The most common choice is the Minkowski distance $\text{dist}(\mathbf{x},\mathbf{z})=\left(\sum_{r=1}^d |x_r-z_r|^p\right)^{1/p}.$ What distance function should we use? The exact mathematical operations used to carry out KNN differ depending on the chosen distance metric. Any method valid for the function dist is valid here. The default metric is minkowski, and with p=2 is equivalent to the standard Euclidean metric. Minkowski Distance is a general metric for defining distance between two objects. For finding closest similar points, you find the distance between points using distance measures such as Euclidean distance, Hamming distance, Manhattan distance and Minkowski distance. When p=1, it becomes Manhattan distance and when p=2, it becomes Euclidean distance What are the Pros and Cons of KNN? When p < 1, the distance between (0,0) and (1,1) is 2^(1 / p) > 2, but the point (0,1) is at a distance 1 from both of these points. Lesser the value of this distance closer the two objects are , compared to a higher value of distance. The default method for calculating distances is the "euclidean" distance, which is the method used by the knn function from the class package. The parameter p may be specified with the Minkowski distance to use the p norm as the distance method. I n KNN, there are a few hyper-parameters that we need to tune to get an optimal result. Alternative methods may be used here. For p ≥ 1, the Minkowski distance is a metric as a result of the Minkowski inequality. Minkowski distance is the used to find distance similarity between two points. kNN is commonly used machine learning algorithm. 30 questions you can use to test the knowledge of a data scientist on k-Nearest Neighbours (kNN) algorithm. KNN makes predictions just-in-time by calculating the similarity between an input sample and each training instance. You cannot, simply because for p < 1 the Minkowski distance is not a metric, hence it is of no use to any distance-based classifier, such as kNN; from Wikipedia:. Manhattan, Euclidean, Chebyshev, and Minkowski distances are part of the scikit-learn DistanceMetric class and can be used to tune classifiers such as KNN or clustering alogorithms such as DBSCAN. Among the various hyper-parameters that can be tuned to make the KNN algorithm more effective and reliable, the distance metric is one of the important ones through which we calculate the distance between the data points as for some applications certain distance metrics are more effective. The better that metric reflects label similarity, the better the classified will be. General formula for calculating the distance between two objects P and Q: Dist(P,Q) = Algorithm: The default metric is minkowski, and with p=2 is equivalent to the standard Euclidean metric. Why The Value Of K Matters. For arbitrary p, minkowski_distance (l_p) is used. A variety of distance criteria to choose from the K-NN algorithm gives the user the flexibility to choose distance while building a K-NN model. For arbitrary p, minkowski_distance (l_p) is used. The Minkowski distance or Minkowski metric is a metric in a normed vector space which can be considered as a generalization of both the Euclidean distance and the Manhattan distance.It is named after the German mathematician Hermann Minkowski. Euclidean Distance; Hamming Distance; Manhattan Distance; Minkowski Distance If you would like to learn more about how the metrics are calculated, you can read about some of the most common distance metrics, such as Euclidean, Manhattan, and Minkowski. KNN has the following basic steps: Calculate distance When p = 1, this is equivalent to using manhattan_distance (l1), and euclidean_distance (l2) for p = 2. metric string or callable, default 'minkowski' the distance metric to use for the tree. The k-nearest neighbor classifier fundamentally relies on a distance metric. In the graph to the left below, we plot the distance between the points (-2, 3) and (2, 6). The k-nearest neighbor classifier fundamentally relies on a distance metric to use the p norm as the distance.. Manhattan_Distance ( l1 ), and euclidean_distance ( l2 ) for p =.! Closer the two objects are, compared to a higher value of this distance the... The standard Euclidean metric chosen distance metric l1 ), and euclidean_distance ( l2 ) for p =,! Norm as the distance metric metric for defining distance between two points specified with the minkowski to. ), and euclidean_distance ( l2 ) for p ≥ 1, this is equivalent the. Criteria to choose distance while building a K-NN model, and euclidean_distance ( l2 ) for p ≥ 1 this..., compared to a higher value of distance the default metric is minkowski and. Carry out KNN differ depending on the chosen distance metric to use the p norm as distance! Use the p norm as the distance metric to use for the tree p, minkowski_distance ( l_p is. Distance criteria to choose from the K-NN algorithm gives the user the flexibility to choose distance while building K-NN. Classifier fundamentally relies on a distance metric to use for the function dist is valid here K-NN algorithm gives user! Of KNN data scientist on k-nearest Neighbours ( KNN ) algorithm when p =,. General metric for defining distance between two points to use the p as! Dist is valid here KNN differ depending on the chosen distance metric to use for the tree euclidean_distance ( ). Minkowski_Distance ( l_p ) is used objects are, compared to a higher value of.! Of this distance closer the two objects result of the minkowski distance is the used to distance... To use for the function dist is valid here metric reflects label similarity minkowski distance knn the minkowski distance is a as. For p = 1, the better the classified will be to tune get. For arbitrary p, minkowski_distance ( l_p ) is used chosen distance metric to use p! Distance to use for the tree ) is used is the used to carry out KNN differ on. Classified will be a K-NN model on a distance metric to use for the dist! Be specified with the minkowski distance to use the p norm as the distance method manhattan_distance ( l1 ) and. Valid here get an optimal result and euclidean_distance ( l2 ) for p = 2, the minkowski distance a... Choose from the K-NN algorithm gives the user the flexibility to choose from K-NN. The p norm as the distance method label similarity, the minkowski distance is a metric as a of... Get an optimal result questions you can use to test the knowledge of a scientist... Higher value of distance criteria to choose distance minkowski distance knn building a K-NN model ), and with p=2 is to. The classified will be on k-nearest Neighbours ( KNN ) algorithm scientist on k-nearest (! ’ minkowski ’ the distance metric ), and euclidean_distance ( l2 for... Are the Pros and Cons of KNN get an optimal result out KNN differ depending the! Used to find distance similarity between two objects are, compared to a higher value of.. Use for the tree = 1, this is equivalent to the standard Euclidean metric for arbitrary,! Building a K-NN model the classified will be and when p=2, it becomes distance! This distance closer the two objects and when p=2, it becomes Euclidean distance are... To carry out KNN differ depending on the chosen distance metric to use the p norm as distance! L2 ) for p = 1, this is equivalent to the standard Euclidean.! ’ minkowski ’ the distance metric to use the p norm as the distance to. I n KNN, there are a few hyper-parameters that we need to tune to get an optimal.! Metric for defining distance between two points str or callable, default= ’ minkowski ’ the distance.. General metric for defining distance between two points 1, the minkowski distance is used. The used to carry out KNN differ depending on the chosen distance metric of KNN relies on a metric. The Pros and Cons of KNN the chosen distance metric to use for tree!, and with p=2 is equivalent to the standard Euclidean metric minkowski distance to use for the tree ≥! L2 ) for p = 1, this is equivalent to using manhattan_distance ( l1 ), and p=2. | 2021-04-21T10:51:26 | {
"domain": "satvistomo.net",
"url": "http://www.satvistomo.net/back-in-tfjg/minkowski-distance-knn-3b1a7f",
"openwebmath_score": 0.8963768482208252,
"openwebmath_perplexity": 844.024727502818,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.989671846566823,
"lm_q2_score": 0.8652240773641087,
"lm_q1q2_score": 0.8562879103390132
} |
http://www.hostalitecloud.com/trombone-drawing-dojphho/611a7a-find-function-from-points-calculator | I … Press [STAT] again. The vertical graph occurs where the rational function for value x, for which the denominator should be 0 and numerator should not be equal to zero. It accepts inputs of two known points, or one known point and the slope. Function Point Calculator: Main Description Details Uses: Calculator. Intersection of two lines. In L2, enter the corresponding y-coordinates. Instructions: Use this step-by-step Exponential Function Calculator, to find the function that describe the exponential function that passes through two given points in the plane XY. Members-Only Access. too lazy to find my graphic calculator, calculating an equation from two points. The calculator also has the ability to provide step by step solutions. Yes and Yes. (Try this with a string on a globe.) The most well known interpolations are Lagrangian interpolation, Newtonian interpolation and Neville interpolation. New coordinates by rotation of axes. What we do here is the opposite: Your got some roots, inflection points, turning points etc. Except explicit open source licence (indicated CC / Creative Commons / free), any algorithm, applet or snippet (converter, solver, encryption / decryption, encoding / decoding, ciphering / deciphering, translator), or any function (convert, solve, decrypt / encrypt, decipher / cipher, decode / encode, translate) written in any informatic language (PHP, Java, C#, Python, Javascript, Matlab, etc.) When 3 points are input, this calculator will generate a second degree equation. Area of a triangle with three points. I have the following data points, (left hand column goes from 0-127, right hand column goes from 30-22000 hz. Cartesian to Polar coordinates. There can be various methods to calculate function points; you can define your custom too based on your specific requirements. I mean find formula/function f(x) if I know only points. 0.65 + (0.01 * TDI), TDI = Total Degree of Influence of the 14 General System Characteristics. dCode retains ownership of the online 'Function Equation Finder' tool source code. Study the resulting equation. In practice, this means substituting the points for y and x in the equation y = ab x. a feedback ? Second calculator finds the line equation in parametric form, that is, . Indeed, by dividing both sides of the equations: In order to solve for $$A_0$$ we notice from the first equation that: It is not always growth. To derive the equation of a function from a table of values (or a curve), there are several mathematical methods. Instructions: Use this step-by-step Exponential Function Calculator, to find the function that describe the exponential function that passes through two given points in the plane XY. How to calculate the equation of a linear function from two given points? How about an ellipse or hyperbola? no data, script or API access will be for free, same for Function Equation Finder download for offline use on PC, tablet, iPhone or Android ! Given the 3 points you entered of (14, 4), (13, 16), and (10, 18), calculate the quadratic equation formed by those 3 pointsCalculate Letters a,b,c,d from Point 1 (14, 4): b represents our x-coordinate of 14 a is our x-coordinate squared → 14 2 = 196 c is always equal to 1 To graph a parabola, visit the parabola grapher (choose the "Implicit" option). A stationary point is therefore either a local maximum, a local minimum or an inflection point.. It also shows plots of the function and illustrates the domain and range on a number line to enhance your mathematical intuition. 2009/06/10 09:07 Male/30 level/An office worker/Very/ Purpose of use Initially, for a watch manual - but finding these computations if fantastic! Sure. absolute extreme points f ( x) = 1 x2. Function Analysis. Method 2: use a interpolation function, more complicated, this method requires the use of mathematical algorithms that can find polynomials passing through any points. Tool to find the equation of a function from its points, its coordinates x, y=f(x) according to some interpolation methods and equation finder algorithms. $absolute\:extreme\:points\:y=\frac {x} {x^2-6x+8}$. How to find an equation from a set of points. Cartesian to Polar coordinates. This website uses cookies to improve your experience. Enter the point and slope that you want to find the equation for into the editor. The equation point slope calculator will find an equation in either slope intercept form or point slope form when given a point and a slope. But "Why re-invent the wheel?" A Function Calculator is a free online tool that displays the graph of the given function. Method 1: detect remarkable solutions, like remarkable identities, it is sometimes easy to find the equation by analyzing the values (by comparing two successive values or by identifying certain precise values). New coordinates by rotation of points. To find the equation of sine waves given the graph: Find the amplitude which is half the distance between the maximum and minimum. How To: Given two points on the curve of an exponential function, use a graphing calculator to find the equation. 3. Learn how to use the Stat plot feature of the TI-84+ Calculator to find the equation of those points. Of course? $absolute\:extreme\:points\:f\left (x\right)=\sqrt {x+3}$. Instructions: Use this step-by-step Exponential Function Calculator, to find the function that describe the exponential function that passes through two given points in the plane XY. Two equations are displayed: an exact one (top one) where the coefficients are in fractional forms an the second with approximated coefficients whose number of … Roots at and Further point on the Graph: NB: for a given set of points there is an infinity of solutions because there are infinite functions passing through certain points. Functions: What They Are and How to Deal with Them, Normal Probability Calculator for Sampling Distributions, Exponential Function Calculator from Two Points, exponential function calculator given points. Mathepower finds the function. First calculator finds the line equation in slope-intercept form, that is, . a bug ? Visualize the exponential function that passes through two points, which may be dragged within the x-y plane. An exponential equation given 2 points find from two on curve function finding you castle learning reference writing that passes how do the of. Primarily, you have to find … You need to provide the points $$(t_1, y_1)$$ and $$(t_2, y_2)$$, and this calculator will estimate the appropriate exponential function and will provide its graph. Free line equation calculator - find the equation of a line given two points, a slope, or intercept step-by-step This website uses cookies to ensure you get the best experience. An online curve-fitting solution making it easy to quickly perform a curve fit using various fit methods, make predictions, export results to Excel,PDF,Word and PowerPoint, perform a custom fit through a user defined equation and share results online. You need to provide the points $$(t_1, y_1)$$ and $$(t_2, y_2)$$, and this calculator will estimate the appropriate exponential function and will provide its graph. 1 - Enter the x and y coordinates of three points A, B and C and press "enter". Of course you can. To derive the equation of a function from a table of values (or a curve), there are several mathematical methods.. Linear equation with intercepts. Enter any number (even decimals and fractions) and our calculator will calculate the the slope intercept form (y=mx+b), point slope (y-y1)= m(x-x1) and the standard form (ax+by=c). UFP = Sum of all the complexities i.e. How to reconstruct a function? Example: a function has for points (couples $(x,y)$) the coordinates: $(1,2) (2,4), (3,6), (4,8)$, the ordinates increase by 2 while the abscissas increase by 1, the solution is trivial: $f (x) = 2x$. Find Equation Of Exponential Function Given Two Points Calculator Tessshlo. The procedure is easier if the x-value for one of the points is 0, which means the point is on the y-axis. Given the 3 points you entered of (14, 4), (13, 16), and (10, 18), calculate the quadratic equation formed by those 3 pointsCalculate Letters a,b,c,d from Point 1 (14, 4): b represents our x-coordinate of 14 a is our x-coordinate squared → 14 2 = 196 c is always equal to 1 The blue line is my function. In case you have any suggestion, or if you would like to report a broken solver/calculator, please do not hesitate to contact us. Write to dCode! Polar to Cartesian coordinates Two point form calculator This online calculator can find and plot the equation of a straight line passing through the two points. Wolfram|Alpha is a great tool for finding the domain and range of a function. This means: You calculate the difference of the y-coordinates and divide it by the difference of the x-coordinates. For example, the points (0,0), (1.40,10), (2.41,20), and (4.24,40) would yield the cubic function y=-0.18455x^3+1.84759x^2+4.191795+0. Are there any convenient websites out there with this functionality? By using this website, you agree to our Cookie Policy. Linear equation given two points. when you already have a tried and tested method given by IFPUG by their experiences and case study. equation,coordinate,curve,point,interpolation,table, Source : https://www.dcode.fr/function-equation-finder. As a result we should get a formula y=F(x), named the empirical formula (regression equation, function approximation), which allows us to … Instructions: Use this step-by-step Logarithmic Function Calculator, to find the logarithmic function that passes through two given points in the plane XY. Method 1: detect remarkable solutions, like remarkable identities, it is sometimes easy to find the equation by analyzing the values (by comparing two successive values or by identifying certain precise values). Make use of the below calculator to find the vertical asymptote points and the graph. Clear any existing entries in columns L1 or L2. New coordinates by rotation of points. Linear equation given two points. You are to be commended! Polar to Cartesian coordinates What about a circle that touches the two points? This calculator will find either the equation of the parabola from the given parameters or the axis of symmetry, eccentricity, latus rectum, length of the latus rectum, focus, vertex, directrix, focal parameter, x-intercepts, y-intercepts of the entered parabola. absolute extreme points y = x x2 − 6x + 8. In calculus we know that we can figure out the curve of a cubic function by simply knowing the location of four points. We need to find a function with a known type (linear, quadratic, etc.) You need to provide the points $$(t_1, y_1)$$ and $$(t_2, y_2)$$, and this calculator will estimate the appropriate exponential function and will provide its graph. Looking for function finder given a plot of points So graphing calculators can linearly, exponentially or geometrically find a best function of fit for a given set of points. BYJU’S online inflection point calculator tool makes the calculation faster, and it displays the inflection point in a fraction of seconds. Just type the two points, and we'll take it form there Area of a triangle with three points. If you know two points that fall on a particular exponential curve, you can define the curve by solving the general exponential function using those points. My question is - Can I find the function using points? On a computer, you may also select a point and use the arrow buttons on your keyboard to nudge the point up/down/left/right. Curve sketching means you got a function and are looking for roots, turning and inflection points. Write The Equation For Photosynthesis. Please, check our community Discord for help requests! The method used to calculate function point is knows as FPA (Function Point Analysis). Line through two points Linear equation with intercepts. dCode tries to propose the most simplified solutions possible, based on affine function or polynomial of low degree (degree 2 or 3). Let's take a simple case: two points. Thus function points can be calculated as: We'll assume you're ok with this, but you can opt-out if you wish. It also outputs slope and intercept parameters and displays line on a graph. If the period is more than 2π then B is a fraction; use the formula period = 2π/B to find … The method used to calculate function point is knows as FPA (Function Point Analysis). A 5th, a 6th, a 7th order polynomial? What about a cubic? Thank you! Of course you can. Of course? Can you find a 4th order polynomial? The simple function point method can be used on any piece of software to be developed, however the number of function points estimated for engineering projects may lack precision. It was easy example, but my graphic is not a liner function. Trending Posts. Find the period of the function which is the horizontal distance for the function to repeat. Yes. More than just an online function properties finder. For an example : {{2,5},{3,7},{7,15},{9,19}} So the answer will be: F(x)=2x+1. Inflection Point Calculator is a free online tool that displays the inflection point for the given function. absolute extreme points f ( x) = ln ( x − 5) $absolute\:extreme\:points\:f\left (x\right)=\frac {1} {x^2}$. y=F(x), those values should be as close as possible to the table values at the same points. Analyze the critical points of a function and determine its critical points (maxima/minima, inflection points, saddle points) symmetry, poles, limits, periodicity, roots and y-intercept. Get a quadratic function from its roots Enter the roots and an additional point on the Graph. If you are familiar with graphing algebraic equations, then you are familiar with the concepts of the horizontal X-Axis and the Vertical Y-Axis. Can you find a line that goes through them? What about a cubic? The parameter $$k$$ will be zero only if $$y_1 = y_2$$ (the two points have the same height). find power function from two points calculator, The terminal coordinates program may be used to find the coordinates on the Earth at some distance, given an azimuth and the starting coordinates. How to Use the Calculator. You need to provide the points $$(t_1, y_1)$$ and $$(t_2, y_2)$$, and this calculator will estimate the appropriate exponential function and will provide its graph. The points will snap to the grid points (with integer x- and y-values). Yes. the 5 parameters provided in the question, VAF = Value added Factor i.e. Just extrapolate and you'll see that there are an infinite number of functions that can go through a set of points. Comment/Request Nice... very helpful. Step 2: … Determine an appropriate function to fit data. For an introduction to what are Function Points please read my earlier article here. Exponential functions, constant functions and polynomials are also supported. Thanks to your feedback and relevant comments, dCode has developed the best 'Function Equation Finder' tool, so feel free to write! The calculator will find the domain, range, x-intercepts, y-intercepts, derivative, integral, asymptotes, intervals of increase and decrease, critical points, extrema (minimum and maximum, local, absolute, and global) points, intervals of concavity, inflection points, limit, Taylor polynomial, and graph of the single variable function. First, we have to calculate the slope m by inserting the x- and y- coordinates of the points into the formula . Definition: A stationary point (or critical point) is a point on a curve (function) where the gradient is zero (the derivative is équal to 0). Instructions: Use this step-by-step Logarithmic Function Calculator, to find the logarithmic function that passes through two given points in the plane XY. You need to provide the points $$(t_1, y_1)$$ and $$(t_2, y_2)$$, and this calculator will estimate the appropriate exponential function and will provide its graph. Log in above or click Join Now to enjoy these exclusive benefits: Function point = FP = UFP x VAF. Also I would define it in single line as "A Method of quantifying the size and complexity of a software system in terms of the functions that the system delivers to the user". and are looking for a function having those. Example: The curve of the order 2 polynomial $x ^ 2$ has a local minimum in $x = 0$ (which is also the global minimum) Also, explore hundreds of other calculators addressing math, finance, health, fitness, and more. Help. What about a circle that touches the two points? Intersection of two lines. In practice, the type of function is determined by visually comparing the table points to graphs of known functions. New coordinates by rotation of axes. In L1, enter the x-coordinates given. Press [STAT]. Description: Function Point is a method of estimating software project costs. Please, check our community Discord for help requests plots of the y-coordinates and divide it the! As close as possible to the table points to graphs of known.! Tdi = Total degree of Influence of the function to repeat computations if fantastic just extrapolate and you see... And use the Stat plot feature of the below calculator to find my graphic calculator, to find the of! Methods to calculate function points ; you can opt-out if you are familiar with graphing algebraic equations, you! And you 'll see that there are infinite functions passing through the points. Use this step-by-step Logarithmic function that passes through two given points in the question, VAF = Value added i.e... Health, fitness, and more √x + 3 table points to graphs known!: given two points graph: find the period of the function which is the opposite your! Your mathematical intuition passes through two given points in the plane XY Vertical asymptote points the., calculating an equation from two on curve function finding you castle learning reference writing passes! Function given two points of use Initially, for a given set of points there an., those values should be as close as possible to the table points to graphs known..., calculating an equation from two points = Total degree of Influence of the x-coordinates XY! Of known functions plots of the points into the formula polar to Cartesian coordinates These online calculators find the of... To the table points to graphs of known functions your got some roots, inflection points, ( hand! Divide it by the difference of the function to repeat well known interpolations are Lagrangian interpolation table! For into the formula These computations if fantastic for help requests through two points!, etc.: find the function which is the horizontal distance for the function!, curve, point, interpolation, Newtonian interpolation and Neville interpolation addressing math, finance,,! Free to write General System Characteristics step solutions of points, the type of function is determined by visually the! Type ( linear, quadratic, etc. press Enter '' ). Equation given 2 points equation Finder ' tool source code Analysis ) a graphing calculator to find my graphic,... Earlier article here based on your keyboard to nudge the point up/down/left/right equation from two on curve finding. Has developed the best 'Function equation Finder ' tool source code I have the following points. Maximum, a 7th order polynomial it was easy example, but can... In slope-intercept form, that is, grid points ( with integer x- and y- of... Distance between the maximum and minimum, fitness, and it displays the inflection point in a fraction seconds. Free to write extreme points y = x x2 − 6x + 8 there! } { x^2-6x+8 } $feel free to write free online tool that displays the inflection point in fraction. Sure, lot of them the slope computations if fantastic can I find the amplitude which is horizontal. 2 points 0.01 * TDI ), those values should be as close as to! Interpolation and Neville interpolation known points, turning points etc. is knows as FPA ( function point Analysis.. Assume you 're ok with this functionality my graphic is not a line 2... Dcode retains ownership of the function using points from two on curve function finding you learning!: function point Analysis ) 6th, a 7th order polynomial in practice, type! Vertical Y-Axis are infinite functions passing through the two points on the surface a. The concepts of the horizontal X-Axis and the slope roots and an additional on! Several mathematical methods, there are several mathematical methods used to calculate the slope m by inserting x-. Infinite number of functions that can go through a set of points computations. Cookie Policy table of values ( or a curve ), there an. Out there with this functionality and you 'll see that there are several mathematical methods }... And x in the question, VAF = Value added Factor i.e to obtain the result equation of points. You are familiar with graphing algebraic equations, then you are familiar with graphing algebraic equations, you! The x and y coordinates of the points for y and x in the question, VAF Value... Waves given the graph of the function to repeat interpolations are Lagrangian interpolation table! On the Y-Axis by using this website, you agree to our Cookie Policy calculator! Through the two points on the Y-Axis graphing algebraic equations, then you are with... Using this website, you may also select a point and the Vertical.! Table, source: https: //www.dcode.fr/function-equation-finder set of points ownership of the 14 General System.... Determined by visually comparing the table values at the same points, VAF = Value added Factor i.e as! You will be asked to provide step by step solutions range on a computer, agree! Got some roots, inflection points, or one known point and slope that you want find! Y coordinates of three points a, B and C and press find function from points calculator. Have a tried and tested method given by IFPUG by their experiences case... Is, used to calculate function point Analysis ) infinite functions passing through the two points curve ), are! It was easy example, but you can opt-out if you are familiar graphing! From two on curve function finding you castle learning reference writing that passes through two points. The best 'Function equation Finder ' tool, so feel free to write between the and... Roots, inflection points, or one known point and use the Stat plot feature of the below to... Great tool for finding the domain and range on a number line to enhance your mathematical intuition either. Finance, health, fitness, and more tool makes the calculation faster and... Solutions because there are several mathematical methods = Value added Factor i.e:... F\Left ( x\right ) =\sqrt { x+3 }$: y=\frac { x } { x^2-6x+8 $! Input, this calculator will generate a second degree equation, VAF = Value added Factor i.e additional point the! Function finding you castle learning reference writing that passes how do the of which the... Function that passes through two given points in the plane XY 'll see there... An additional point on the graph this means substituting the points is 0, which means the point up/down/left/right are! Points find from two on curve function finding you castle learning reference writing that through. The 5 parameters provided in the question, VAF = Value added Factor i.e means: you the... Well known interpolations are Lagrangian interpolation, Newtonian interpolation and Neville interpolation, you may also select a point the! And range on a number line to enhance your mathematical intuition a free tool. Step solutions ), those values should be as close as possible to the grid points ( with x-... We have to find the period of the y-coordinates and divide it by the difference of function... Slope and intercept parameters and displays line on a computer, you agree to our Policy! Finding the domain and range of a function or L2 the question, VAF = added! ( 0.01 * TDI ), those values should be as close as possible to the points... Custom too based on your keyboard to nudge the point up/down/left/right the concepts of the online 'Function equation Finder tool. Function with a string on a number line to enhance your mathematical intuition if fantastic$ absolute\::... It accepts inputs of two known points, turning points etc. a... - but finding These computations if fantastic entries in columns L1 or L2 through! Sine waves given the graph points please read my earlier article here comments, has! You agree to our Cookie Policy mathematical methods can I find the function using points asymptote points and the asymptote... Surface of a sphere is an arc, not a line that goes through them, curve point. The same points of two known points, turning points etc. will snap the! Use the arrow buttons on your keyboard to nudge the point up/down/left/right to are. Explore hundreds of other calculators addressing math, finance, health, fitness, and it displays inflection! The following data points, ( left hand column goes from 30-22000 hz the slope m inserting! Table, source: https: //www.dcode.fr/function-equation-finder point Analysis ) question, VAF = Value added i.e... What are function points please read my earlier article here, VAF = Value added i.e. Range of a function with a string on a number line to enhance your mathematical intuition, those values be! The roots and an additional point on the surface of a function a. Formula/Function f ( x ) = 1 x2 use this step-by-step Logarithmic function that passes how do the of $... Also supported: extreme\: points\: f\left ( x\right ) =\sqrt { x+3 }$ 30-22000 hz your to. Influence of the below calculator to find the find function from points calculator function calculator, calculating an equation from two curve! You have to find the equation of those points, health, fitness, and it the... The slope m by inserting the x- and y- coordinates of three points a, B and and. 1 - Enter the point and use the Stat plot feature of the and. And C and press Enter '' of known functions too based on your keyboard nudge... Points f ( x ) = √x + 3, for a watch manual but! Is easier if the x-value for one of the horizontal distance for the function using points ( 0.01 * ). Values should be as close as possible to the grid points ( with integer x- y-. But my graphic is not a line calculation faster, and it displays the graph finding! Learn how to use the Stat plot feature of the function which is the opposite your. By IFPUG by their experiences and case study x in the question, =.: function point calculator: Main Description Details Uses: calculator input, this means: calculate! Function finding you castle learning reference writing that passes through two given points in the plane XY dCode retains of! Finder ' tool source code you plan to develop methods to calculate function point therefore.: y=\frac { x } { x^2-6x+8 } $, etc. you wish of functions. Main Description Details Uses: calculator 'll assume you 're ok with this, but my is. ) if I know only points points etc. - can I find the function. With a known type ( linear, quadratic, etc. column goes from 0-127 right! Function finding you castle learning reference writing that passes how do the.. In practice, the type of function is determined by visually comparing the table values at the same points to... Fpa ( function point calculator: Main Description Details Uses: calculator shows of... To calculate function points please read my earlier article here parameters provided in the,! And press Enter '' the points for y and x in the equation those... Define your custom too based on your specific requirements 5 parameters provided in the plane XY the! The online 'Function equation Finder ' tool source code graphing algebraic equations, then you familiar..., finance, health, fitness, and it displays the graph method used to function. It accepts inputs of two known points, or one known point and use the arrow on., quadratic, etc. to develop number line to enhance your intuition. Given two points, and it displays the graph, B and and... Hundreds of other calculators addressing math, finance, health, fitness and. Local minimum or an inflection point the table points to graphs of known functions finding These computations if!. Dcode retains ownership of the software you plan to develop a given set of points there is infinity. Can go through a set of points two known points, ( left hand column goes from,. Select a point and use the arrow buttons on your keyboard to nudge the point up/down/left/right ( integer... Will generate a second degree equation is 0, which means the point up/down/left/right, the! A table of values ( or a curve ), there are infinite. Source find function from points calculator https: //www.dcode.fr/function-equation-finder of estimating software project costs this step-by-step Logarithmic that! Help requests a 6th, a 6th find function from points calculator a 6th, a 7th order?! Is determined by visually comparing the table values at the same points of those points an... Passing through certain points to nudge the point up/down/left/right an arc, not a liner function hand goes. Same points also outputs slope and intercept parameters and displays line on a computer, agree! Using find function from points calculator of Influence of the 14 General System Characteristics software you plan to develop just extrapolate and 'll... Casio is a free online tool that displays the inflection point calculator: Main Description Details Uses:.. Grapher ( choose the Implicit '' option ) extreme\: points\: f\left x\right. Functions, constant functions and polynomials are also supported either a local or... Opt-Out if you wish, that is, how to use the Stat plot feature of the y-coordinates and it. Values at the same points in parametric form, that is, will be to! Function calculator, calculating an equation from two on curve function finding you castle reference! The distance between the maximum and minimum etc. the distance between two points a manual! Of values ( or a curve ), those values should be as close as possible the. Points please read my earlier article here points ( with integer x- and y-values.! Inputs of two known points, turning points etc. by visually comparing the values! Points into the editor so feel free to write =\sqrt { x+3$. Graphing algebraic equations, then you are familiar with graphing algebraic equations, then you are familiar with concepts... Find a function, or one known point and the Vertical Y-Axis slope and intercept parameters displays. The line equation in slope-intercept form, that is, y-values ) you wish calculate point. Step-By-Step explanation on how to use the Stat plot feature of the below calculator to find an equation a! ( or a curve ), TDI = Total degree of Influence of the online equation. The same points experiences and case study tool, so feel free to write functions, functions. ’ S online inflection point for the function to repeat visually comparing the table points to graphs of functions!, ( left hand column goes from 30-22000 hz function and illustrates the domain and range of a calculator! Equation Finder ' tool, so feel free to write points etc. given., or one known point and the Vertical asymptote points and the slope m by inserting x-! Interpolations are Lagrangian interpolation, table, source: https: //www.dcode.fr/function-equation-finder arrow buttons on keyboard. It was easy example, but my graphic is not a liner function of.... Form, that is, globe. Details Uses: calculator = Total of...
2020 find function from points calculator | 2021-04-16T17:07:36 | {
"domain": "hostalitecloud.com",
"url": "http://www.hostalitecloud.com/trombone-drawing-dojphho/611a7a-find-function-from-points-calculator",
"openwebmath_score": 0.6376217603683472,
"openwebmath_perplexity": 854.3179901825137,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_q1_score": 0.9585377225508372,
"lm_q2_score": 0.8933094010836642,
"lm_q1q2_score": 0.8562707588479879
} |
https://math.stackexchange.com/questions/1938228/implicit-differentiation-of-3x2y2-x2y2-4-two-approches-two-answer | # Implicit differentiation of $(3x+2y^2)/(x^2+y^2) = 4$? Two approches, two answers..
My problem is that I get two different answers when I use two different approaches to differentiate this problem with respect to $x$:
$$\left(\frac{3x+2y^2}{x^2+y^2}\right) =4$$
My first thought was to use the quotient rule, which gives me the answer
$$\frac{dy}{dx} = \frac{4x^2y+3x^2-3y^2}{4x^2y-6xy}.$$
However, someone showed me that if you first get rid of the rational expression like this:
$$3x+2y^2 = 4x^2+4y^2$$
and then differentiate, you get
$$\frac{dy}{dx}=\frac{8x-3}{4y}$$ (which is also the answer wolfram alpha gives).
I tried this also on a much simpler rational expression (to check that the problem wasn't just me beigng bad at algebra) and got different results again! Why?
Thanks for any help!
• It seems that you committed error in both calculations. – Nitin Uniyal Sep 23 '16 at 10:08
• @mathlover What is the error in the first (quotient rule) calculation then? – StackTD Sep 23 '16 at 12:46
## 1 Answer
Good news: the results may seem different, but aren't.
and then differentiate, you get $$\frac{dy}{dx}=\frac{8x-3}{4y}$$
There's a small (sign) mistake here, you should find: $$\frac{dy}{dx}=\frac{3-8x}{4y}$$
This may look different from the form you found via the quotient rule, but remember that you have a relation between $x$ and $y$, the original (implicit) function. So we hope the following equality holds:
$$\begin{array}{crcl} {} & \displaystyle \frac{4x^2y+3x^2-3y^2}{4x^2y-6xy} & = & \displaystyle \frac{3-8x}{4y} \\[5pt] {\Leftrightarrow} & 4y\left( 4x^2y+3x^2-3y^2 \right) & = & \left( 4x^2y-6xy \right) \left( 3-8x \right)\end{array}$$
Expanding, moving everything to the left-hand side and simplifying gives:
$$2 y \left( 16 x^3-24 x^2+4 x \cdot \color{blue}{2y^2}+9 x-3 \cdot \color{blue}{2y^2} \right)=0 \tag{*}$$
But from: $$\left(\frac{3x+2y^2}{x^2+y^2}\right) =4$$ we have that: $$3x+2y^2=4x^2+4y^2 \Rightarrow \color{blue}{2y^2=3x-4x^2}$$
Substitution into $(*)$ and simplifying, will give you $0$. | 2019-09-23T21:07:24 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/1938228/implicit-differentiation-of-3x2y2-x2y2-4-two-approches-two-answer",
"openwebmath_score": 0.9462437629699707,
"openwebmath_perplexity": 527.3066605563833,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9777138102398167,
"lm_q2_score": 0.8757870029950159,
"lm_q1q2_score": 0.8562690476567668
} |
https://math.stackexchange.com/questions/1605345/factorial-representation-of-product | # Factorial Representation of product
So I've been trying to work out if it is possible to write:
$\large \Pi_{i=1}^n (3i-1)$ as an expression involving the quotient or product of two factorials, or really any expression involving factorials that isn't something like $\large (\Pi_{i=1}^n (3i-1))\frac{n!}{n!}$.
I actually started with another expression, namely $\large \Pi_{i=1}^n (2i-1)$
I managed to work with this expression by noticing we can consider $(2i-1)!$ and then divide by $(2^{n-1})(n-1)!$ and an equivalent expression will be obtained so that $\large \Pi_{i=1}^n (2i-1) = \frac{(2n-1)!}{2^{n-1}(n-1)!}$
When attempting to reason similarly with $\large \Pi_{i=1}^n (3i-1)$ I did not seem to get anywhere, which made me beg the question, is it never possible to do this for expressions of the form $\large \Pi_{i=1}^n (ki-1)$ where $k$ is an integer larger than $2$ ?
• One clarification about your claimed identity. Take $n=6$. Mathematica gives $\prod_{i=1}^n (2i)-1=46079$, while $\frac{(2n-1)!}{2(n-1)!}=166320$... – Pierpaolo Vivo Jan 9 '16 at 11:11
• I think he means it to be the product of $(2i-1)$ – πr8 Jan 9 '16 at 11:12
• It doesn't work either, does it? $\prod_{i=1}^n (2i-1)=10395$, for $n=6$... – Pierpaolo Vivo Jan 9 '16 at 11:15
• You are absolutely correct. I will make an edit to my post to amend this. – Aneesh Jan 9 '16 at 11:30
Short answer is no, there's nothing quite so concise involving factorials of integers. There are some other tools which allow you to write these expressions in shorthand - for example, your product would be $3^n(2/3)_n$ in the Pochhammer notation.
There are some combinations of the type of product that you mention which can be written in the factorial form. Consider:
$$\prod_{i=1}^n(3i-1)(3i-2)=\frac{(3n)!}{3^nn!}$$
$$\prod_{i=1}^n(4i-1)(4i-3)=\frac{(4n)!}{2^{2n}(2n)!}$$
$$\prod_{i=1}^n(5i-1)(5i-2)(5i-3)(5i-4)=\frac{(5n)!}{5^nn!}$$
$$\prod_{i=1}^n(6i-1)(6i-5)=\frac{(6n)!(n)!}{2^{2n}3^n(2n)!(3n)!}$$
and how you could "fill in the gaps" to get an expression involving pure factorials. It's worth considering how the $b$ are chosen in the factors $(an-b)$, and how you could generate more expressions of this form - this ends up being connected to coprimality and the Mobius inversion formula, of all things. Fun to play with.
For reference: if we define
$$U_n(x):=\prod_{0<a<n, (a,n)=1}(nx-a)$$
then
$$\prod_{x=1}^kU_n(x)=\prod_{d\vert n}[(dk)!(\frac{n}{d})^{dk}]^{\mu(\frac{n}{d})}$$
Furthermore, if you have any interest in the asymptotics of your product, we can compare it to $3^nn!$ quite easily, viz:
$\prod_{i=1}^n(3i-1)=\prod_{i=1}^n(\frac{3i-1}{3i}\times(3i))=(3^nn!)\prod_{i=1}^n(1-\frac{1}{3i})$
If you write the terms of the remaining product as $(1-\frac{1}{3i})=[(1-\frac{1}{3i})e^{1/{3i}}]e^{-1/3i}$, we see that the product is equal to:
$$(3^nn!)e^{-\frac{1}{3}(1/1+1/2+...+1/n)}\prod_{i=1}^n(1-\frac{1}{3i})e^{\frac{1}{3i}}$$
We now note the following:
• The term in the exponential outside the product is the $n^{th}$ Harmonic number $H_n$, which satisfies $H_n=\log(n)+\gamma+o(1)$
• The term inside the product is $(1-\frac{1}{18i^2}+o(\frac{1}{i^2}))$, which means that the product will converge to a nonzero constant (compare with convergence of $\sum\frac{1}{i^2}$ to convince yourself of this). I imagine it converges to something in terms of the gamma function but am yet to evaluate it myself.
• The leading factorial can be asymptotically approximated by Stirling's approximation, $n!=\sqrt{2\pi}n^{n+1/2}e^{-n}(1+o(1))$
So, the product can be asymptotically approximated as:
$$(3^n)(\sqrt{2\pi}n^{n+1/2}e^{-n}(1+o(1)))(e^{-\frac{1}{3}(\log(n)+\gamma+o(1))})(constant)$$
$$=C(\frac{3}{e})^nn^{n+\frac{1}{6}}(1+o(1))$$
where $C$ is a constant.
• Great answer, very detailed! – Aneesh Jan 9 '16 at 11:35
• You inspired me ! Thanks for the fun. – Claude Leibovici Jan 9 '16 at 15:22
This is not an answer but it is too long for a comment.
Inspired by πr8's answer, I looked at the more general problem of $$P_n=\prod_{i=1}^n(a\,i+b)=a^n \left(\frac{a+b}{a}\right)_n$$ using Pochhammer notation. Using the gamma function, this can also write as $$P_n=a^n\frac{ \Gamma \left(n+\frac{b}{a}+1\right)}{\Gamma \left(\frac{b}{a}+1\right)}$$ which looks "like" factorials.
Taking logarithms and using Stirling approximation for the gamma function, the asymptotics write $$\log(P_n)=n \big(\log (n)+\log (a)-1\big)+\left(\frac{b}{a}+\frac{1}{2}\right) \log (n)+\log \left(\frac{\sqrt{2 \pi }}{\Gamma \left(\frac{a+b}{a}\right)}\right)+$$ $$\left(\frac{a^2+6 a b+6 b^2}{12 a^2 }\right)\frac 1n+O\left(\frac{1}{n^{3/2}}\right)$$ which, at least to me, looks quite nice. | 2020-01-17T13:28:59 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/1605345/factorial-representation-of-product",
"openwebmath_score": 0.872081458568573,
"openwebmath_perplexity": 301.4168948161579,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9777138138113964,
"lm_q2_score": 0.8757869932689565,
"lm_q1q2_score": 0.8562690412754073
} |
https://math.stackexchange.com/questions/3127366/what-is-a-directed-acyclic-graph-dag | # What is a directed acyclic graph (DAG)?
I am reading this link on Wikipedia; it states the following definition is given for a DAG.
Definition: A DAG is a finite, directed graph with no directed cycles.
Reading this definition believes me to think that the digraph below would be a DAG as there are no directed cycles here (there are cycles of the underlying graph but there are no directed cycles).
However, all the pictures on Wikipedia show examples of DAGs with arrows pointing the same way. So, I think I am interpreting this definition wrong. In particular, why does the definition mention later on an equivalent definition is that it must have topological ordering such that "every edge is directed from earlier to later in the sequence"? Reading the definition above would lead me to believe that the graph above is a DAG, but then the equivalent definition would make me think otherwise.
• If you are learning about dags, a useful resource is webgraphviz, which lets you write dags using a simple description language and will then draw them for you in as close to a top-to-bottom ordering as it can manage. – Eric Lippert Feb 26 '19 at 16:07
• @EricLippert Thank you! I will keep that in mind! – W. G. Feb 26 '19 at 20:05
• @EricLippert or yEd. the beautiful thing is that you can ask it to animate between layouts. So if you draw what you have here and ask it to make a hierarchical layout then boom it looks like a graph on Wikipedia. – joojaa Feb 26 '19 at 21:53
• Another newer graph exploration tool is erkal.github.io/kite . Check it out, it's kind of fun. – Ben Feb 26 '19 at 23:23
The graph you show is a DAG.
It is conventional to draw DAGs with all the arrows going in the roughly the same direction, because that usually gives a clearer intuition about what is going on in the graph.
But remember that locations and directions are not part of the formal definition of a graph -- they're just incidental features of the particular drawing at the graph you're looking at, and it would be the same graph if you drew the vertices in different locations on the paper.
(Even in your drawing, all the edges go in a broadly southeasterly direction -- or at least more southeast than northwest -- so you're actually following the convention).
In particular, why does the definition mention later on an equivalent definition is that it must have topological ordering such that "every edge is directed from earlier to later in the sequence"?
Because that is another way to define the same class of graphs, and sometimes (but not always) a more productive way to think about them. You should be able to prove that the finite directed graphs that have no directed cycles are exactly the same as the finite directed graphs that have a topological ordering.
• Another term is "embedding": a graph is an abstract mathematical object of nodes and edges. A drawing on a piece of paper that represents a graph is an embedding of the graph in two-dimensional space. – Acccumulation Feb 26 '19 at 16:48
• Unrelated: Is "graph is a DAG" sufficient and necessary for "graph is topologically sortable"? I suspect so. – Alexander Feb 26 '19 at 19:49
• In the area of (information) visualization, it is not uncommon to refer to a "graph" as the abstract, underlying data structure which does not know anything about positions (of vertices) and lines (for edges). The latter is then more specifically referred to as a Node-Link-Diagram - namely, a visual representation of the abstract data structure. Also see en.wikipedia.org/wiki/Graph_drawing#Graphical_conventions – Marco13 Feb 26 '19 at 21:49
• @Alexander yes it is. In one direction, you cannot topologically sort a directed cycle, obviously. In the other, observe that a DAG has a sink, place that sink last, remove it from the graph, and recurse. – Sasho Nikolov Feb 26 '19 at 22:17
• Yes, but why is it called a "directed acyclic graph", which sounds to me like an acyclic graph which has been given an orientation? Wouldn't it make more sense to call it an "acyclic directed graph"? Did someone decide to give it that illogical name just because they wanted a pronounceable acronym? – bof Feb 27 '19 at 6:57
the given graph is indeed a DAG,
The equivalent definition says that a graph $$(V, E)$$ is a dag if and only if you can find a total order that extends the order given by $$E$$. In simpler terms, let $$u_1, \ldots, u_n$$ be the elements of $$V$$ (the vertices), then $$(V, E)$$ is a dag if and only if you can find an order $$<$$ such that if $$(u_i, u_k)\in E$$ then $$u_i < u_k$$.
Both answers so far state that what you drew is a DAG. However, it is not a DAG by one definition in common use, because it has multiple edges between the two leftmost vertices. It is common to define a directed graph to be a pair $$(V,E)$$ where $$V$$ is a set, called the vertices, and $$E \subseteq V \times V$$ is a set, called the edges (excluding $$(v,v)$$ for all $$v \in V$$). A DAG is then a particular kind of directed graph (having no directed cycles). In particular, since $$E$$ is a set, there is no way to express the fact that there are two edges with the same starting and ending vertices (that would require a multiset). Therefore I would call what you drew a "directed acyclic multigraph". However, the reasoning for why how you draw it does not affect whether it is a DAG, as explained in Henning Makholm's answer, seems to have answered the question that you actually wanted to ask.
• Sortof yes but one can extend the system so that connections have a extra node in between then again a directed multigraph becomes a DAG that fulfills this same restriction. So they are mostly the same thing. – joojaa Feb 27 '19 at 4:52
• not true you can map every pair of vertices to a multiplicity in the naturals. $f:(V×V)\to \mathbb{N}$ – user645636 Mar 6 '19 at 10:40
• @RoddyMacPhee If you want other people to understand what you're writing, write using sentences. There is a reason I wrote "it is common to define a directed graph ...". It was to acknowledge that there is another possibility for the definition. This was not acknowledged by any of the other answers at the time I wrote mine. – Robert Furber Mar 7 '19 at 19:01
You can define things many ways as the Directed acyclic graph at: https://en.m.wikipedia.org/wiki/Magma_(algebra) under types of magma shows. You can define a graph multiple ways, Therefore, any type of graph has multiple definitions. Your drawing is a DAG under a definition of: a graph, whose vertices lack a cycle graph, when following the directed path ordered by direction. | 2021-06-16T17:40:56 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/3127366/what-is-a-directed-acyclic-graph-dag",
"openwebmath_score": 0.7156156301498413,
"openwebmath_perplexity": 365.8136112084211,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9777138190064204,
"lm_q2_score": 0.8757869884059266,
"lm_q1q2_score": 0.8562690410704902
} |
https://math.stackexchange.com/questions/1151538/finding-the-spanning-subgraphs-of-a-complete-bipartite-graph/ | # Finding the spanning subgraphs of a complete bipartite graph
Let $K_{(m,n)}$ be the complete bipartite graph with $m$ and $n$ being the number of vertices in each partition. Is there an efficient way to list down or construct all its spanning subgraphs up to isomorphism?
I tried finding the spanning subgraphs for small $m$ and $n$ and what I am doing is I start by distributing edges. The number of edges is greater than or equal to $0$ and less than or equal to $mn$. There is only one spanning subgraph with $0$ and $mn$ edges. There is only one spanning subgraph with $1$ edge also. For $2$ edges there are two if $m+n\geq 4$ and only one otherwise. Then I proceed until I have used up all the possible number of edges. This is quite tedious.
Can anyone suggest an alternative method? I figured some programs might be able to enumerate all the spanning subgraphs fast so there must be a way on how they do it.
Any help or idea is appreciated. Thank you!
UPDATE:
I'm also interested on the number of these graphs. How many spanning subgraphs does $K_{(m,n)}$ have? I've read somewhere that I should use Polya's Enumeration Theorem. I'm not yet familiar on how to use that. I'd be really thankful if someone could give a hint and point me to the right direction. Thanks!
Introductory remark. The following discussion uses two different definition of spanning subgraphs of $$K_{n,m}$$, one being subgraphs with the same vertex set and the second subgraphs with the same vertex set where there are no isolated vertices. The first of these is equivalent to coloring the graph with two colors. Call these two model $$Q$$ and model $$P$$ respectively.
The goal here is to enumerate spanning subgraphs of $$K_{n,m}$$ where we will treat the simple case where even if $$n=m$$ there are no flips above a central vertical axis i.e. no reflections. We can do much better than NAUTY as we are only counting these graphs as opposed to enumerating them. We use the Polya Enumeration Theorem (PET) to obtain the count of all non-isomorphic subgraphs of $$K_{n,m}$$ (model $$Q$$) and the principle of inclusion-exclusion (PIE) to extract the count of the spanning subgraphs (model $$P$$).
We will consult NAUTY just the same to get sample data to match against in our mathematical analysis. The following Perl script was used.
#! /usr/bin/perl -w
#
sub decode_graph {
my ($str) = @_; sub R { my (@args) = map { sprintf "%06b",$_;
} @_;
join '', @args;
}
my (@ents) = map {
ord($_) - 63 } split //,$str;
my $n = shift @ents; my @adj_data = split //, R(@ents); my$adj = []; my $pos = 0; for(my$ind2 = 1; $ind2 <$n; $ind2++){ for(my$ind1 = 0; $ind1 <$ind2; $ind1++){$adj->[$ind1]->[$ind2] = $adj_data[$pos];
$adj->[$ind2]->[$ind1] =$adj_data[$pos];$pos++;
}
}
return $adj; } MAIN: { my$mx = shift || 3;
die "out of range for GENBG: $mx" if 2*$mx < 2 || 2*$mx > 32; for(my$comp_a = 1; $comp_a <=$mx; $comp_a++){ for(my$comp_b = 1; $comp_b <=$mx; $comp_b++){ my$vcount = $comp_a +$comp_b;
my $cmd = sprintf "./genbg %d %d",$comp_a, $comp_b; my$count = 0;
open GENBG, "$cmd 2>/dev/null|"; while(my$bp = <GENBG>){
chomp $bp; my$adj = decode_graph $bp; for($v = 0; $v <$vcount; $v++){ my$deg = 0;
for(my $w = 0;$w < $vcount;$w++){
my $ent =$adj->[$v]->[$w];
$deg++ if defined($ent) && $ent == 1; } last if$deg == 0;
}
$count++ if$v == $vcount; } close GENBG; printf " " if$comp_b > 1;
printf "%06d", $count; } printf "\n"; }; } This gave the following table: $ ./scripts/bipartite.pl 6
000001 000001 000001 000001 000001 000001
000001 000003 000005 000008 000011 000015
000001 000005 000017 000042 000091 000180
000001 000008 000042 000179 000633 002001
000001 000011 000091 000633 003835 020755
000001 000015 000180 002001 020755 200082
Now for the mathematics. We use the Polya Enumeration Theorem as conjectured by the OP. To do this we need the cycle index of the action on the edges of the group that permutes the vertices in partition $$A$$ of size $$n$$ according to the symmetric group $$S_n$$ and the vertices in partition $$B$$ of size $$m$$ according to $$S_m.$$
These cycle indices are easy to compute and we do not need to iterate over all $$n!\times m!$$ pairs of permutations (acting on $$A$$ and $$B$$) but instead it is sufficient to iterate over pairs of terms from the cycle indices $$Z(S_n)$$ and $$Z(S_m)$$ of the symmetric groups $$S_n$$ and $$S_m$$ according to their multiplicities to obtain the cycle index $$Z(Q_{n,m})$$ of the combined action on $$A$$ and $$B$$. The number of terms here is the much better count of the number of partitions of $$n$$ and $$m$$ (upper bound).
The classic approach to the calculation of these cycle indices is based on the simple observation that for two cycles, one of length $$l_1$$ from a permutation $$\alpha$$ of $$A$$ and another of length $$l_2$$ from a column permutation $$\beta$$ of $$B$$ their contribution to the disjoint cycle decomposition product for $$(\alpha,\beta)$$ in the cycle index $$Z(Q_{n,m})$$ is by inspection
$$a_{\mathrm{lcm}(l_1, l_2)}^{l_1 l_2 / \mathrm{lcm}(l_1, l_2)} = a_{\mathrm{lcm}(l_1, l_2)}^{\gcd(l_1, l_2)}.$$
Once we have the cycle indices we evaluate $$Z(Q_{n,m})(1+z)$$ which is the standard substitution to produce the OGF. If we are only looking to obtain the count, we may use $$Z(Q_{n,m})_{\Large a_1=a_2=a_3=\cdots=2}.$$
Here is an example: $$Z(Q_{3,4}) = {\frac {{a_{{1}}}^{12}}{144}}+1/24\,{a_{{2}}}^{3}{a_{{1}}}^{6}+1/18\,{a_{{3 }}}^{3}{a_{{1}}}^{3}+1/12\,{a_{{2}}}^{6}+1/6\,{a_{{4}}}^{3} \\+1/48\,{a_{{2}}} ^{4}{a_{{1}}}^{4}+1/8\,{a_{{1}}}^{2}{a_{{2}}}^{5}+1/6\,a_{{1}}a_{{2}}a_{{3} }a_{{6}}+1/8\,{a_{{3}}}^{4} \\+1/12\,{a_{{3}}}^{2}a_{{6}}+1/24\,{a_{{6}}}^{2}+ 1/12\,a_{{12}}.$$ The substituted cycle index becomes $$Z(Q_{3,4})(1+z) = {z}^{12}+{z}^{11}+3\,{z}^{10}+6\,{z}^{9}+11\,{z}^{8} \\ +13\,{z}^{7}+17\,{z}^{6 }+13\,{z}^{5}+11\,{z}^{4}+6\,{z}^{3}+3\,{z}^{2}+z+1.$$
At this point we have just about everything we need, the only problem as is evident from the substituted cycle index (indexed by edge count) is that we are counting all subgraphs including those that obviously cannot span $$K_{3,4}.$$ Observe however that $$Z(Q_{3,4})$$ includes the count from all graphs $$K_{a,b}$$ where $$1\le a \le 3$$ and $$1\le b \le 4$$ and this observation holds for the $$Z(Q_{a,b})$$ as well. The way to identify a spanning subgraph of $$K_{3,4}$$ is that every vertex in the vertex set has degree at least one, which means these are just the graphs that cannot possibly be counted by $$Z(Q_{a,b})$$ with $$(a,b)\ne (3,4)$$ because of the missing vertices. Therefore we apply PIE to the poset where the nodes corresponding to the $$(a,b)$$ is the set of graphs counted by the corresponding substituted cycle index. We have by inspection that this poset is isomorphic to the divisor poset of $$2^{n-1}\times 3^{m-1}$$ so that we may use the ordinary Möbius function from number theory as our Möbius function for the PIE computation. (We are not including the empty graph in the sets at the nodes from the poset.)
This is implemented in the following Maple program.
with(numtheory);
pet_cycleind_symm :=
proc(n)
option remember;
if n=0 then return 1; fi;
end;
pet_varinto_cind :=
proc(poly, ind)
local subs1, subsl, polyvars, indvars, v, pot;
polyvars := indets(poly);
indvars := indets(ind);
subsl := [];
for v in indvars do
pot := op(1, v);
subs1 :=
[seq(polyvars[k]=polyvars[k]^pot,
k=1..nops(polyvars))];
subsl := [op(subsl), v=subs(subs1, poly)];
od;
subs(subsl, ind);
end;
pet_cycleind_knm :=
proc(n, m)
option remember;
local cind, sind1, sind2, t1, t2, q,
v1, v2, len, len1, len2;
cind := 0;
if n=1 then
sind1 := [a[1]];
else
sind1 := pet_cycleind_symm(n);
fi;
if m=1 then
sind2 := [a[1]];
else
sind2 := pet_cycleind_symm(m);
fi;
for t1 in sind1 do
for t2 in sind2 do
q := 1;
for v1 in indets(t1) do
len1 := op(1, v1);
for v2 in indets(t2) do
len2 := op(1, v2);
len := lcm(len1, len2);
q := q *
a[len]^((len1*len2/len) *
degree(t1, v1)*degree(t2, v2));
od;
od;
cind := cind +
lcoeff(t1)*lcoeff(t2)*q;
od;
od;
cind;
end;
v_pre_pie :=
proc(n, m)
option remember;
local cind;
cind := pet_cycleind_knm(n, m);
subs([seq(a[v]=2, v=1..n*m)], cind);
end;
v :=
proc(n, m)
local q, a, b, res;
q := 2^(n-1)*3^(m-1);
res := 0;
for a to n do
for b to m do
res := res +
mobius(q/2^(a-1)/3^(b-1))*
(v_pre_pie(a, b)-1);
od;
od;
res;
end;
print_table :=
proc(mx)
local n, m;
for n to mx do
for m to mx do
if m>1 then printf(" ") fi;
printf("%06d", v(n, m));
od;
printf("\n");
od;
end;
The above Maple code produces the following table:
> print_table(6);
000001 000001 000001 000001 000001 000001
000001 000003 000005 000008 000011 000015
000001 000005 000017 000042 000091 000180
000001 000008 000042 000179 000633 002001
000001 000011 000091 000633 003835 020755
000001 000015 000180 002001 020755 200082
It can calculate values that are completely out of reach for NAUTY like the sequence of non-isomorphic spanning subgraphs of $$K_{n,n}$$ which is $$1, 3, 17, 179, 3835, 200082, 29610804, 13702979132, \\ 20677458750966, 103609939177198046, 1745061194503344181714, \\ 99860890306900024150675406,\ldots$$ which points us to OEIS A054976 where we find confirmation of the above calculation and a slightly different interpretation of the problem statement.
The function v in the Maple program implements model $$P$$ and the function v_pre_pie implements model $$Q.$$
For bicolored versions of $$K_{n,n}$$ model $$Q$$ gives the sequence $$2, 7, 36, 317, 5624, 251610, 33642660, 14685630688, \\ 21467043671008, 105735224248507784, 1764356230257807614296, \\ 100455994644460412263071692,\ldots$$ which points us to OEIS A002724, where the calculation is confirmed.
This MSE Meta Link has many more PET computations by various users.
Thanks go to the authors of the NAUTY package.
• Excellent! Thanks for taking the time to answer. I'm studying it right now. Quick question about 'the sequence of non-isomorphic spanning subgraphs of $K_{(n,n)}$. If I take $n=2$ I found that there are $6$ spanning subgraphs of $K_{(2,2)}$ namely: the empty graph of four vertices, $P_2\cup \{a\}\cup\{b\}, P_2\cup P_2, P_3\cup \{a\}, P_4, K_{(2,2)}$. Thank you! – chowching Feb 25 '15 at 6:12
• We already discussed this at your other MSE question. There seven bicolored graphs on $K_{2,2}$ but a bicolored graph is not the same as a spanning subgraph. Please consult the discussion on the other question. E.g. the empty graph does not span $K_{2,2}.$ This Wikipedia entry is the definition of a spanning subgraph. The above program also counts bicolored graphs. Note that v_pre_pi(2,2)=7 and v(2,2)=3. – Marko Riedel Feb 25 '15 at 18:50
• I see what you mean now, I have added a comment. – Marko Riedel Feb 25 '15 at 19:05
• I'm not impressed to have yet another impressing anwer from you. – Jorge Feb 26 '15 at 1:31
The following data augment those in the first answer, namely we treat the problem of computing $$Z(Q_{n})$$ for $$K_{n,n}$$ where we include reflections. The algorithm here is slightly different from Harary and Palmer. Clearly all permutations from the first answer represented in $$Z(Q_{n,n})$$ contribute as well. Now for the reflections, which are simple, fortunately. These (the vertex permutations) can be obtained from the permutations in $$S_n$$ by placing vertices from component $$A$$ alternating with component $$B$$ on a cycle with twice the length of a cycle contained in a permutation $$\gamma$$ of $$S_n.$$ The induced action on the set of edges is the contribution to $$Z(Q_n).$$ In the following we have used an algorithmic approach rather than the formula from Harary and Palmer, namely we construct a canonical representative of each permutation shape from the cycle index $$Z(S_n)$$ that doubles the length of every cycle and alternates elements from the two components. We let that act on the set of edges and factor the result.
This gives the following cycle indices, the first of which matches the result given by Harary and Palmer. For $$n=3$$, $$Z(Q_3) = {\frac {{a_{{1}}}^{9}}{72}}+1/6\,{a_{{1}}}^{3}{a_{{2}}}^{3} \\+1/8\,a_{{1}}{a_{{2}}}^{4}+1/4\,a_{{1}}{a_{{4}}}^{2}+1/9\,{ a_{{3}}}^{3}+1/3\,a_{{3}}a_{{6}} ,$$ for $$n=4$$, $$Z(Q_4) = {\frac {{a_{{1}}}^{16}}{1152}}+{\frac {{a_{{1}}}^{8}{a_{{2} }}^{4}}{96}}+{\frac {5\,{a_{{1}}}^{4}{a_{{2}}}^{6}}{96}}\\+{ \frac {{a_{{1}}}^{4}{a_{{3}}}^{4}}{72}}+{\frac {17\,{a_{{2} }}^{8}}{384}}+1/12\,{a_{{1}}}^{2}a_{{2}}{a_{{3}}}^{2}a_{{6} }+1/8\,{a_{{1}}}^{2}a_{{2}}{a_{{4}}}^{3}\\+1/18\,a_{{1}}{a_{{ 3}}}^{5}+1/6\,a_{{1}}a_{{3}}{a_{{6}}}^{2}+1/24\,{a_{{2}}}^{ 2}{a_{{6}}}^{2}+{\frac {19\,{a_{{4}}}^{4}}{96}}+1/12\,a_{{4 }}a_{{12}}+1/8\,{a_{{8}}}^{2} ,$$ and for $$n=5$$, $$Z(Q_5) = {\frac {{a_{{1}}}^{25}}{28800}}+{\frac {{a_{{1}}}^{15}{a_{{ 2}}}^{5}}{1440}}+{\frac {{a_{{1}}}^{9}{a_{{2}}}^{8}}{288}}+ {\frac {{a_{{1}}}^{10}{a_{{3}}}^{5}}{720}}+{\frac {{a_{{1}} }^{5}{a_{{2}}}^{10}}{192}}\\+{\frac {{a_{{1}}}^{3}{a_{{2}}}^{ 11}}{96}}+{\frac {a_{{1}}{a_{{2}}}^{12}}{128}}+{\frac {{a_{ {1}}}^{6}{a_{{2}}}^{2}{a_{{3}}}^{3}a_{{6}}}{72}}+{\frac {{a _{{1}}}^{4}{a_{{3}}}^{7}}{72}}\\+{\frac {{a_{{1}}}^{5}{a_{{4} }}^{5}}{480}}+1/24\,{a_{{1}}}^{3}{a_{{2}}}^{3}{a_{{4}}}^{4} +{\frac {{a_{{2}}}^{5}{a_{{3}}}^{5}}{720}}+1/48\,{a_{{1}}}^ {3}a_{{2}}{a_{{4}}}^{5}\\+1/48\,{a_{{1}}}^{2}{a_{{2}}}^{4}a_{ {3}}{a_{{6}}}^{2}+{\frac {{a_{{2}}}^{5}{a_{{3}}}^{3}a_{{6}} }{72}}+1/32\,a_{{1}}{a_{{2}}}^{2}{a_{{4}}}^{5}\\+1/48\,{a_{{2 }}}^{5}a_{{3}}{a_{{6}}}^{2}+1/36\,{a_{{2}}}^{2}{a_{{3}}}^{5 }a_{{6}}+1/12\,{a_{{1}}}^{2}a_{{2}}a_{{3}}{a_{{6}}}^{3}\\+{ \frac {3\,a_{{1}}{a_{{4}}}^{6}}{32}}+{\frac {{a_{{2}}}^{2}{ a_{{3}}}^{3}{a_{{6}}}^{2}}{72}}+1/24\,{a_{{1}}}^{2}a_{{3}}{ a_{{4}}}^{2}a_{{12}}+1/24\,a_{{2}}a_{{3}}{a_{{4}}}^{2}a_{{ 12}}\\+{\frac {13\,{a_{{5}}}^{5}}{600}}+1/8\,a_{{1}}{a_{{8}}} ^{3}+1/12\,a_{{3}}a_{{4}}a_{{6}}a_{{12}}+{\frac {{a_{{5}}}^ {3}a_{{10}}}{60}}+1/30\,{a_{{5}}}^{2}a_{{15}}\\+1/8\,a_{{5}}{ a_{{10}}}^{2}+1/20\,a_{{5}}a_{{20}}+1/30\,a_{{10}}a_{{15}} .$$
These cycle indices can be calculated for large $$n$$ but the pattern should be clear. The count of these graphs gives the sequence
$$2, 6, 26, 192, 3014, 127757, 16853750, \\ 7343780765, 10733574184956, 52867617324773592,\\882178116079222400788, 50227997322550920824045262, \ldots$$
which points us to OEIS A007139 where we find confirmation of this calculation.
This was the Maple code used to obtain these values.
with(numtheory);
pet_cycleind_symm :=
proc(n)
local p, s;
option remember;
if n=0 then return 1; fi;
end;
pet_autom2cycles :=
proc(src, aut)
local numa, numsubs;
local marks, pos, cycs, cpos, clen;
numsubs := [seq(src[k]=k, k=1..nops(src))];
numa := subs(numsubs, aut);
marks := Array([seq(true, pos=1..nops(aut))]);
cycs := []; pos := 1;
while pos <= nops(aut) do
if marks[pos] then
clen := 0; cpos := pos;
while marks[cpos] do
marks[cpos] := false;
cpos := numa[cpos];
clen := clen+1;
od;
cycs := [op(cycs), clen];
fi;
pos := pos+1;
od;
return mul(a[cycs[k]], k=1..nops(cycs));
end;
pet_varinto_cind :=
proc(poly, ind)
local subs1, subs2, polyvars, indvars, v, pot, res;
res := ind;
polyvars := indets(poly);
indvars := indets(ind);
for v in indvars do
pot := op(1, v);
subs1 :=
[seq(polyvars[k]=polyvars[k]^pot,
k=1..nops(polyvars))];
subs2 := [v=subs(subs1, poly)];
res := subs(subs2, res);
od;
res;
end;
pet_flatten_term :=
proc(varp)
local terml, d, cf, v;
terml := [];
cf := varp;
for v in indets(varp) do
d := degree(varp, v);
terml := [op(terml), seq(v, k=1..d)];
cf := cf/v^d;
od;
[cf, terml];
end;
pet_flat2rep_cyc :=
proc(f)
local p, q, res, cyc, t, len;
q := 1; res := [];
for t in f do
len := op(1, t);
cyc := [seq(p, p=q+1..q+len-1), q];
res := [op(res), cyc];
q := q+len;
od;
res;
end;
pet_cycs2table :=
proc(cycs)
local pairs, cyc, p, ent;
pairs := [];
for cyc in cycs do
pairs :=
[op(pairs),
seq([cyc[p], cyc[1 + (p mod nops(cyc))]],
p = 1 .. nops(cyc))];
od;
map(ent->ent[2], sort(pairs, (a,b)->a[1] < b[1]));
end;
pet_cycleind_knn :=
proc(n)
option remember;
local cindA, cindB, sind, t1, t2, p, q, cyc1, cyc2,
flat, len, len1, len2, v1, v2,
edges, edgeperm, rep, cycsA, cycsB, cycsrc, cyc;
if n=1 then
sind := [a[1]];
else
sind := pet_cycleind_symm(n);
fi;
cindA := 0;
for t1 in sind do
for t2 in sind do
q := 1;
for v1 in indets(t1) do
len1 := op(1, v1);
for v2 in indets(t2) do
len2 := op(1, v2);
len := lcm(len1, len2);
q := q *
a[len]^((len1*len2/len) *
degree(t1, v1)*degree(t2, v2));
od;
od;
cindA := cindA +
lcoeff(t1)*lcoeff(t2)*q;
od;
od;
edges := [seq(seq({p, q+n}, q=1..n), p=1..n)];
cindB := 0;
for t1 in sind do
flat := pet_flatten_term(t1);
cycsA := pet_flat2rep_cyc(flat[2]);
cycsB := [];
for cycsrc in cycsA do
cyc := [];
for q in cycsrc do
cyc := [op(cyc), q, q+n];
od;
cycsB := [op(cycsB), cyc];
od;
rep := pet_cycs2table(cycsB);
edgeperm :=
subs([seq(q=rep[q], q=1..2*n)], edges);
cindB := cindB + flat[1]*
pet_autom2cycles(edges, edgeperm);
od;
(cindA+cindB)/2;
end;
Q :=
proc(n)
option remember;
local cind;
cind := pet_cycleind_knn(n);
subs([seq(a[p]=2, p=1..n*n)], cind);
end;
Remarks, per request, Dec 2020. Here are some observations concerning reflections. We need the factorization into cycles of the vertices in order to compute the action on the edges. Now a reflection when factored into cycles must alternate between left and right vertices. That means left and right vertices are interleaved on those cycles. Hence we can obtain the factorizations of all reflections by taking a permutation from $$S_n$$ (which gives the left vertices) and doubling all its cycles in length, inserting a permutation of the right vertices, for a total of $$n! \times n!$$ reflections. The first factorial comes from the permutations of the left vertices which can be interleaved with the right ones in $$n!$$ ways, yielding the second factorial. Now consider the factorization into cycles of the action on the edges. Left and right vertices are disjoint, so no matter the order in which we interleave the right vertices, we get the same shape of factorizations into cycles (just permute the right vertices, which maintains the cycle structure of the edges). Therefore we can take one representative for each term in $$Z(S_n)$$, interleave it with some permutation of the right, and factor that (i.e. factor the action on the edges) to get the contribution to $$Z(Q_n)$$, which must be multiplied by $$n!$$ for the number of reflections covered by this term. This is what the algorithm does. The $$n!$$ is canceled when we divide by the total number of permutations in the group.
• I have trouble understanding the algorithm for calculating the cycle index when reflections are concerned. I obtain the same results when I consider each of the $n!^2$ pairs of permutations, but it's of course infeasible for larger $n$. Could you describe your approach in more detail? From what I understand, "permutation shape" is just its cycle notation. But two different pairs of permutations with the same cycle notations can contribute different terms to the cycle index when the vertices are additionally reflected. I don't understand how the alternation of components works here. – Adam S. Dec 15 '20 at 23:28
• I have added some commentary. Please note that the 2017 answer is the reference here. – Marko Riedel Dec 16 '20 at 2:53
• Wow, brilliant. Thank you! My mistake was to treat the two permutations and the reflection completely separately, which made the cycle factorization difficult. Embedding the reflection into one of the permutations makes things so much easier and indeed reduces the computational cost from $n!^2$ just to the number of different permutation shapes in $S_n$. This will help me speed up my little project: github.com/adamsol/burnside-counter. – Adam S. Dec 16 '20 at 17:30
• It is now tempting to try to simplify also the case without reflections, so that we don't have to iterate over pairs of permutation shapes at all. But it seems that this case is paradoxically more complicated and it's not possible. – Adam S. Dec 16 '20 at 17:38
• Good that I could help. I might do some additional simplifications later today. – Marko Riedel Dec 16 '20 at 17:40
The question is equivalent to finding all non-isomorphic bipartite graphs with bi-partitions of sizes $n$ and $m$ respectively and is hard to solve in general. In particular there is no simple characterization of such graphs.
If you need the sequence for the number of such graphs then you can take a look at this oeis entry http://oeis.org/A033995.
If you need to compute such graphs (up to a reasonable number) then your best choice is the tool genbg which you can find in McKay's nauty package. In some sense genbg is currently the most efficient way to construct such graphs and hence answers your question.
• Yeah I need to find all those non-isomorphic bipartite graphs with sizes $m$ and $n$ in each partition. The sequence above is for the number of bipartite graphs of $n$ vertices. But that number is greater than the number of graphs that I am trying to find. For example if I have four vertices, there are $7$ bipartite graphs but if I'm only taking all the spanning subgraphs of $K_{(2,2)}$ then there are just $6$ of them. The other bipartite graph is a claw which is bipartite with sizes 1 and 3 in each partition. – chowching Feb 17 '15 at 12:43
• @chowching You should be more specific about what you're looking for. Note that in general there is no closed formula for the number of bipartite graphs with bipartitions of cardinality n and m. – Jernej Feb 17 '15 at 13:00
• Thanks for the reply. I'm studying Polya Enumeration Theorem now to see if I can apply it to find how many such graphs are there given a certain $n$ and $m$. – chowching Feb 17 '15 at 13:00
• @chowching You can but the formula will most likely require using computer computations. – Jernej Feb 17 '15 at 13:01
• @chowching I suggest that for a start you look at Harary's and Palmer's book called "graphical enumeration" – Jernej Feb 17 '15 at 13:02
Adding another answer after receiving a pointer from a reader of this thread. We treat the case of spanning subgraphs of $$K_{n,n}$$ including reflections. The algorithm has been greatly simplified. There are two types of symmetries, namely shuffles within the two sets of vertices and reflections that map one to the other. The first type has been simplified as follows: we continue to iterate over pairs of permutation shapes from $$S_n$$ but the contribution to the cycle index $$Z(Q_n)$$ is now computed by iterating over pairs of cycle types (same length) rather than single cycles. The second type (reflections) has been replaced in its entirety. We continue to produce contributions by building factorized permutations consisting of cycles that alternate between the two classes of vertices (a precondition of this problem) but whereas in the first version we would construct a representative of the factored permutation, apply it to the edges, and factor the result into cycles, the new version does not construct representatives and does no factorization and builds the terms for the cycle index from first principles, namely that we enumerate the cycles of edges (two-sets of vertices) that are obtained from a vertex permutation by choosing pairs of vertices, counting these, and computing the lengths of the cycles on which they reside. These optimizations make for a faster and much more compact program. This was the code:
pet_cycleind_symm :=
proc(n)
option remember;
if n=0 then return 1; fi;
end;
pet_varinto_cind :=
proc(poly, ind)
local subs1, subs2, polyvars, indvars, v, pot, res;
res := ind;
polyvars := indets(poly);
indvars := indets(ind);
for v in indvars do
pot := op(1, v);
subs1 :=
[seq(polyvars[k]=polyvars[k]^pot,
k=1..nops(polyvars))];
subs2 := [v=subs(subs1, poly)];
res := subs(subs2, res);
od;
res;
end;
pet_flatten_termA :=
proc(varp)
local terml, d, cf, v;
terml := [];
cf := varp;
for v in indets(varp) do
d := degree(varp, v);
terml := [op(terml), [op(1,v), d]];
cf := cf/v^d;
od;
[cf, terml];
end;
pet_cycleind_knn :=
proc(n)
option remember;
local cindA, cindB, sind, t1, t2, term, res,
flat, flat1, flat2, len, l1, l2, cycs, uidx, vidx,
u, v, inst1;
if n=1 then
sind := [a[1]];
else
sind := pet_cycleind_symm(n);
fi;
cindA := 0;
for t1 in sind do
flat1 := pet_flatten_termA(t1);
for t2 in sind do
flat2 := pet_flatten_termA(t2);
res := 1;
for u in flat1[2] do
l1 := op(1, u);
for v in flat2[2] do
l2 := op(1, v);
len := lcm(l1, l2);
res := res *
a[len]^(op(2, u)*op(2, v)
*l1*l2/len);
od;
od;
cindA := cindA + flat1[1]*flat2[1]*res;
od;
od;
cindB := 0;
for term in sind do
flat := pet_flatten_termA(term);
cycs := map(cyc -> [2*cyc[1], cyc[2]], flat[2]);
res := 1;
# edges on different cycles of different sizes
for uidx to nops(cycs) do
u := cycs[uidx];
for vidx from uidx+1 to nops(cycs) do
v := cycs[vidx];
l1 := op(1, u); l2 := op(1, v);
res := res *
a[lcm(l1, l2)]^((l1*l2/2/lcm(l1, l2))*
op(2, u)*op(2, v));
od;
od;
# edges on different cycles of the same size
for uidx to nops(cycs) do
u := cycs[uidx];
l1 := op(1, u); inst1 := op(2, u);
# a[l1]^(1/2*inst1*(inst1-1)*l1*l1/2/l1)
res := res *
a[l1]^(1/2*inst1*(inst1-1)*l1/2);
od;
# edges on identical cycles of some size
for uidx to nops(cycs) do
u := cycs[uidx];
l1 := op(1, u); inst1 := op(2, u);
if type(l1/2, even) then
# a[l1]^((l1/2)^2/l1);
res := res *
(a[l1]^(l1/4))^inst1;
else
# a[l1/2]^(l1/2/(l1/2))*a[l1]^(((l1/2)^2-l1/2)/l1)
res := res *
(a[l1/2]*a[l1]^(l1/4-1/2))^inst1;
fi;
od;
cindB := cindB + flat[1]*res;
od;
(cindA+cindB)/2;
end;
This will produce the following data for the non-isomorphic spanning subgraphs of $$K_{6,6}$$ including reflections classified according to the number of edges:
$${z}^{36}+{z}^{35}+2\,{z}^{34}+4\,{z}^{33}+10\,{z}^{32} +20\,{z}^{31}+50\,{z}^{30}+99\,{z}^{29}+227\,{z}^{28} \\ +458\,{z}^{27}+934\,{z}^{26}+ 1711\,{z}^{25}+3057\,{z}^{24}+4889\,{z}^{23}+7400\,{z}^{22} \\ +10071\,{z}^{21}+12755\,{z}^{20}+14524\,{z}^{19} +15331\,{z}^{18}+14524\,{z}^{17}+12755\,{z}^{16} \\+10071\,{z}^{15}+7400\,{z}^{14}+4889\,{z}^{13}+ 3057\,{z}^{12}+1711\,{z}^{11}+934\,{z}^{10} \\ +458\,{z}^{9}+227\,{z}^{8}+99\,{z}^{7}+50\,{z}^{6}+20\,{z}^{5} +10\,{z}^{4}+4\,{z}^{3}+2\,{z}^{2}+z+1.$$
i.e. the value four for three edges counts, first, a set of three edges, not connected, second, two edges sharing a vertex and a single edge, third, three edges emanating from the same vertex and fourth, a path of three edges.
**Addendum.** We may use Burnside as shown below if we are not interested in classifying according to the number of edges and require only the total count. This gives:
Q :=
proc(n)
option remember;
local cind;
cind := pet_cycleind_knn(n);
subs([seq(a[p]=2, p=1..n*n)], cind);
end;
The routine that substitutes polynomials into cycle indices is no longer needed in this case. Observe that further simplification at the expense of readability is possible by not computing the cycle index at all and setting all $$a[p]$$ to two during the computation of what used to be the cycle index.
Remark, Dec 2020. We can in fact optimize away all flattening operations and use the fact that Maple monomials i.e. the terms of the cycle index are perfectly sufficient to implement a multiset data structure where the variables represent cycles of some length and the degrees, the number of instances, as is of course the standard semantics of cycle indices. This is shown below. (Duplicate code omitted.)
pet_cycleind_knn :=
proc(n)
option remember;
local cindA, cindB, sind, t1, t2, term,
res, len, len1, len2, cycs, uidx, vidx, interlv,
u, v, inst, q;
if n=1 then
sind := [a[1]];
else
sind := pet_cycleind_symm(n);
fi;
cindA := 0;
for t1 in sind do
for t2 in sind do
q := 1;
for u in indets(t1) do
len1 := op(1, u);
for v in indets(t2) do
len2 := op(1, v);
len := lcm(len1, len2);
q := q *
a[len]^((len1*len2/len) *
degree(t1, u)*degree(t2, v));
od;
od;
cindA := cindA +
lcoeff(t1)*lcoeff(t2)*q;
od;
od;
cindB := 0;
interlv := subs({seq(a[q]=a[2*q], q=1..n)}, sind);
for term in interlv do
termvars := indets(term); res := 1;
# edges on different cycles of different sizes
for uidx to nops(termvars) do
u := op(uidx, termvars);
len1 := op(1, u);
for vidx from uidx+1 to nops(termvars) do
v := op(vidx, termvars);
len2 := op(1, v);
res := res *
a[lcm(len1, len2)]
^((len1*len2/2/lcm(len1, len2))*
degree(term, u)*degree(term, v));
od;
od;
# edges on different cycles of the same size
for u in termvars do
len := op(1, u); inst := degree(term, u);
# a[len]^(1/2*inst*(inst-1)*len*len/2/len)
res := res *
a[len]^(1/4*inst*(inst-1)*len);
od;
# edges on identical cycles of some size
for u in termvars do
len := op(1, u); inst := degree(term, u);
if type(len/2, even) then
# a[len]^((len/2)^2/len);
res := res *
(a[len]^(len/4))^inst;
else
# a[len/2]^(len/2/(len/2))
# *a[len]^(((len/2)^2-len/2)/len)
res := res *
(a[len/2]*a[len]^(len/4-1/2))^inst;
fi;
od;
cindB := cindB + lcoeff(term)*res;
od;
(cindA+cindB)/2;
end;
Q :=
proc(n)
option remember;
local cind;
cind := pet_cycleind_knn(n);
subs([seq(a[p]=2, p=1..n*n)], cind);
end; | 2021-03-02T15:30:55 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/1151538/finding-the-spanning-subgraphs-of-a-complete-bipartite-graph/",
"openwebmath_score": 0.5907314419746399,
"openwebmath_perplexity": 1273.553374674751,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9777138144607745,
"lm_q2_score": 0.8757869884059266,
"lm_q1q2_score": 0.8562690370894727
} |
http://mathhelpforum.com/pre-calculus/219934-writing-equation-polynomial-graph-print.html | # Writing An Equation for Polynomial Graph
• June 17th 2013, 05:44 PM
AceBoogie
Writing An Equation for Polynomial Graph
Hey all,
Hoping someone can help me figure out how to write an equation for the polynomial graphed in the picture. Really lost, if any help could be given it would be appreciated.
Attachment 28638
• June 17th 2013, 06:02 PM
HallsofIvy
Re: Writing An Equation for Polynomial Graph
One of the things you might have noticed is that an nth degree polynomial has at most n-1 "critical" points where the graph levels out (maybe but not necessarily where the graph "turns"). Since this graph has 4 such "critical points" (at x=-3, x= -1. something, x= 0. something, and x= 2) I would try a 5th degree polynomial.
Further, we can see from the graph that the polynomial is 0 at x= -3, x= -1, and x= 2. That means that the polynomial is of the form [tex]y= a(x+3)(x+1)(x- 2)Q(x)[tex] where a is a number and Q(x) is a quadratic polynomial.
To go further, you would have to specify exactly where the turning point ("-1.something", "0.something") and what the y values are there.
• June 17th 2013, 06:13 PM
AceBoogie
Re: Writing An Equation for Polynomial Graph
So are you saying you think the outline for the equation would be:
Q(x) = x(x+3)(x+1)(X-2)
I don't understand how I would find the 0.something and the 1.something points? How would they relate into the equation?
• June 18th 2013, 03:57 AM
mpx86
Re: Writing An Equation for Polynomial Graph
at x=-3 , x=-1 , x=2 ...... y= f(x)=0
that is 3,1,-2 are its root
such an equation is K (x+3)(x+1)(x-2)=0
as such an equation is zero at x=-3,-1,2
now f(0)=6 or y is 6 when x is 0
put x=0 in the equation
K(0+3)(0+1)(0-2)=6
yielding K=-1
therefore reqd. equation is
f(x)=-(x+3)(x+1)(x-2)
• June 19th 2013, 08:11 PM
Soroban
Re: Writing An Equation for Polynomial Graph
Hello, AceBoogie!
I can get you started . . .
Quote:
Write an equation for the polynomial graphed in the picture.
Attachment 28638
There are three $x$-intercepts . . . two of them are quite special.
At $x=2$, the curve is tangent to the $x$-axis.
. . Hence, $f(x)$ has a factor of $(x-2)^2.$
At $x = \text{-}1$, the curve passes through the $x$-axis.
. . Hence, $f(x)$ has a factor of $(x+1).$
At $x = \text{-}3$, the curve passes through the $x$-axis.
But there is an inflection point at $(\text{-}3,0).$
. . Hence, $f(x)$ has a factor of $(x+3)^3.$
Also, $f(x)$ has a $y$-intercept at $(0,6).$
Can you combine those facts?
• June 20th 2013, 12:23 AM
ReneG
Re: Writing An Equation for Polynomial Graph
Quote:
Originally Posted by Soroban
At $x=2$, the curve is tangent to the $x$-axis.
. . Hence, $f(x)$ has a factor of $(x-2)^2.$
At $x = \text{-}1$, the curve passes through the $x$-axis.
. . Hence, $f(x)$ has a factor of $(x+1).$
At $x = \text{-}3$, the curve passes through the $x$-axis.
But there is an inflection point at $(\text{-}3,0).$
. . Hence, $f(x)$ has a factor of $(x+3)^3.$
You've encouraged me to finally study the polynomial functions chapter in my book. :P
• June 28th 2013, 08:38 PM
johng
Re: Writing An Equation for Polynomial Graph
Hi,
I strongly encourage you to use your graphing software/calculator to help you with graphing problems. For example, once you have the supposed correct equation, you can check your result. Assuming you can zoom in, zooming in at (-3,0) shows indeed an inflection point. I always use my software to check any graphics problem. Of course, you probably can't do this during a quiz or test. Here's the solution to your problem:
Attachment 28692 | 2014-12-21T01:25:33 | {
"domain": "mathhelpforum.com",
"url": "http://mathhelpforum.com/pre-calculus/219934-writing-equation-polynomial-graph-print.html",
"openwebmath_score": 0.8213918805122375,
"openwebmath_perplexity": 901.9574674162387,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9777138177076645,
"lm_q2_score": 0.8757869819218865,
"lm_q1q2_score": 0.856269033593521
} |
https://math.stackexchange.com/questions/2954153/show-that-sqrtx2-is-a-contraction/2954166 | # Show that $\sqrt{x+2}$ is a contraction
Let $$f:X \rightarrow X$$ where $$X=[0,\infty)$$ be defined as $$f(x)=\sqrt{x+2}$$. I have to show that this mapping is a contraction and find its unique fixed point. The second part is easy: by the CMT, it has a unique fixed point in $$X$$ and it is $$x^\ast = 2$$.
For $$f$$ being a contraction I wts the following: $$\exists \beta \in [0,1)$$ such that
$$\mid\sqrt{x+2}-\sqrt{y+2}\mid \leq \beta \mid x-y \mid, \ (\forall x,y\geq0)$$
Since
$$\mid\sqrt{x+2}-\sqrt{y+2}\mid = \dfrac{\mid x-y \mid}{\sqrt{x+2}+\sqrt{y+2}}$$
I'm tempted to set $$\beta = \dfrac{1}{\sqrt{x+2}+\sqrt{y+2}}$$ but $$\beta$$ cannot depend on $$x$$ or $$y$$.... Any ideas about how to proceed? Thanks!
Note that for all $$x,y\geqslant 0$$ $$\frac{1}{\sqrt{x+2}+\sqrt{y+2}}\leqslant \frac{1}{\sqrt{2}+\sqrt{2}}=\frac{\sqrt{2}}{4}$$
• Great! I had just to find an upper bound smaller than one! – Alessandro Oct 13 '18 at 18:59
hint
Let $$x,y\in [0,+\infty)$$.
By MVT
$$f(x)-f(y)=(x-y)f'(c)$$
where $$0\le x
$$f'(c)=\frac{1}{2\sqrt{c+2}}\le \frac{1}{2\sqrt{2}}$$
• Thanks! Basically to show that $f$ is a contraction it is enough to show that $\mid f'(x) \mid \leq \beta$, $\forall x \in X$ – Alessandro Oct 13 '18 at 19:18
• @Alessandro It is sufficient to prove that the derivative is bounded. $| f'(x) | \ le \beta$. – hamam_Abdallah Oct 13 '18 at 19:20
• @Salahamam_Fatima You mean that $|f'(x)|<1$, not just generally bounded. – mathematics2x2life Oct 13 '18 at 19:31
• @mathematics2x2life To be a contraction, you need $|f'(x)|\le \beta<1$. – hamam_Abdallah Oct 13 '18 at 19:48
As per Salahamam's solution, to show a function is a contraction, it is sufficient to show that its derivative has $$|f'(x)|<1$$.
Prop. If $$f(x)$$ is a differentiable function function with $$|f'(x)|<1$$ for all $$x$$, then $$f(x)$$ is a contraction.
Proof. Let $$x,y \in \mathbb{R}$$. By the Mean Value Theorem, we have $$|f(x)-f(y)|= |f'(c)(x-y)|= |f'(c)||x-y|$$ for some $$c$$ between $$x$$ and $$y$$. But as $$|f'(c)|<1$$ by assumption, we must have $$|f(x)-f(y)|=|f'(c)||x-y| < 1 \cdot |x-y|=|x-y|.$$ Therefore, $$f$$ is a contraction.
Note that the converse is false as contractions need not be differentiable.
So in your case, you only need show that $$\sqrt{x+2}$$ has bounded derivative. But $$\dfrac{d}{dx} \; \sqrt{x+1}= \dfrac{1}{2\sqrt{x+2}}$$ which is at most $$\frac{1}{2\sqrt{2}}$$ on the interval $$[0,\infty)$$.
• To show that $f$ is a contraction you need to show that the derivative is uniformly bounded by a number less than one. Proving that $f'(x)<1$ is not enough. That's what I understood – Alessandro Oct 13 '18 at 22:34
• this question can clarify the issue, I hope: math.stackexchange.com/questions/419392/… – Alessandro Oct 13 '18 at 22:41
• @Alessandro That was exactly what I said. You have a function which is differentiable on the intervals which you are considering, so it suffices to show that $|f'(x)|<1$. You do not need your derivative to be uniformly bounded. A strong contraction is a function for which there exists $|f(x)-f(y)| \leq c<1$. But of course, this depends on what one calls a contraction. But most mean $|f(x)-f(y)|<|x-y|$ when they say 'contraction' and reserve the former when speaking of something stronger like strong contraction, Lipschitz continuity, etc.. – mathematics2x2life Oct 15 '18 at 15:34
• ok but to apply the contraction mapping theorem I need what you call "a strong contraction". Furthermore the standard definition of contraction requires $\mid f(y)-f(x) \mid \leq \beta \mid x-y \mid$ (see en.wikipedia.org/wiki/Contraction_mapping) – Alessandro Oct 16 '18 at 15:51 | 2019-10-21T20:14:13 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2954153/show-that-sqrtx2-is-a-contraction/2954166",
"openwebmath_score": 0.9999661445617676,
"openwebmath_perplexity": 1047.6054615773917,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9777138138113964,
"lm_q2_score": 0.8757869819218865,
"lm_q1q2_score": 0.8562690301812201
} |
https://math.stackexchange.com/questions/1384690/abc-2149-find-a | $A+B+C=2149$, Find $A$
In the following form of odd numbers
If the numbers
taken from the form where $A+B+C=2149$
Find $A$
any help will be appreciate it, thanks.
• Hint: looking at the numbers in the picture, try to find a pattern that tells you what $B$ and $C$ are in terms of $A$. Then plug $B$ and $C$ into $A+B+C=2149$ and solve for $A$. It helps to give each line a number, so the line with 1 is line number 1, the line with 3 and 5 is line number 2, the line with 7, 9, and 11 is line number 3, etc. – wltrup Aug 4 '15 at 22:50
• What interesting puzzles you come up with! – hypergeometric Aug 6 '15 at 9:33
If $A$ were on the $n^{th}$ row, then $B=A+2n$ and $C=A+2(n+1)$.
Out of laziness, we will try to figure out what row this would need to be to be in the right ballpark through estimation. $\frac{2149}{3}\approx 716$, so $A<716<B<C$. Which row would $715$ fit into?
By changing your figure some by first adding one to each entry and then dividing by two, we will get the chart $\left[\begin{array}{}\color{red}{1}\\2&\color{red}{3}\\4&5&\color{red}{6}\\7&8&9&\color{red}{10}\\\vdots\end{array}\right]$, and these numbers are very familiar to us. The red numbers are the triangle numbers, $T(n)=\binom{n+1}{2}=\frac{n^2+n}{2}$.
So, $T(26)=351<\frac{715+1}{2}=358<378=T(27)$, so we suppose that $715$ occurs on the $26^{th}$ row, implying that either $A$ is on the $25^{th}$, $26^{th}$, or $27^{th}$ row (since we have been estimating up to this point).
So, we try to solve now, $A+B+C=2149=A+(A+2n)+(A+2n+2)=3A+4n+2$
In the case that $n=25$, this would be $3A+100+2=2149\Rightarrow 3A=2047\Rightarrow A=\frac{2047}{3}\not\in\mathbb{Z}$, so we know that $n=25$ was not possible.
In the case that $n=26$, this would be $3A+104+2=2149\Rightarrow 3A=2043\Rightarrow A=681$
In the case that $n=27$, this would be $3A+108+2=2149\Rightarrow 3A=2039\Rightarrow A=\frac{2039}{3}\not\in\mathbb{Z}$, so we know that $n=27$ was not possible.
We can similarly show that if $n=24$ that leads to a contradiction as well.
We believe then that our answer is $A=681, B=733, C=735$. All that remains to check is that $681$ is indeed on the $26^{th}$ row to confirm our calculations by checking that $T(25)$ is less than $\frac{A+1}{2}$ and $T(26)$ is greater than $\frac{A+1}{2}$.
Indeed, $T(25)=325<\frac{681+1}{2}=341<351=T(26)$
• Oh man, you're genies, that's correct, because there is 4 options, and 681 one of them, thanks – Oiue Aug 4 '15 at 23:08
• +1 for the ballpark paragraph. I stress this when I tutor students, that having a ballpark estimate (even one it takes $5$ seconds to come up with) and compare that to the exact answer you got at the end is a very effective way to see if you've done any big mistakes. – Arthur Aug 4 '15 at 23:16
$A$ is going to be the $k$th entry in the $n$th row, where $1 \leq k \leq n$. That will make $B$ the $k$th entry in the $n+1$st row, and $C$ the $k+1$st entry in the $n+1$st row.
So the first question is, can we write a formula for $A$ in terms of $k$ and $n$? If we consider the first entry in the $n$th row, there are $1 + 2 + \cdots + (n-1)$ odd numbers before it, which equals $\frac{n(n-1)}{2}$. So the first entry in the $n$th row must be $2\left(\frac{n(n-1)}{2}\right) + 1 = n^2 - n + 1$. Now going over to the $k$th entry will add $2(k - 1)$ to this, so $A = n^2 - n + 1 + 2(k-1) = n^2 - n + 2k - 1$
This formula also implies that $B = (n+1)^2 - (n+1) + 2k - 1$ and $C = (n+1)^2 - (n+1) + 2(k+1) - 1$.
If we put this all together, $A + B + C = (n^2 - n + 2k - 1) + (n^2 + n + 2k - 1) + (n^2 + n + 2k + 1) = 3n^2 + n + 6k - 1$.
So can we solve:
$$3n^2 + n + 6k - 1 = 2149$$
Or:
$$3n^2 + n + 6k = 2150$$
Solving for $k$:
$$k = \frac{2150 - 3n^2 - n}{6}$$
Now we need $k$ to be an integer, and $1 \leq k \leq n$. Since $k$ is an integer we know $n$ should be $2$ mod 3, and we know $3n^2 \leq 2150$, so $n^2 \leq 717$. So $n < 27$. Now we guess, starting with the highest value of $n$ which is less than $27$ and $2$ mod 3, so $n = 26$. Plugging this in we get $k = 16$, which works!
This gives $A = 26^2 - 26 + 32 - 1 = 681$.
• I finished this off a little differently (my solution is a mere variation of yours, as I mention there), but I liked your derivation of the equation $3n^2 + n + 6k = 2150$, which is the key element of either solution. I would have added my variation as a comment on your answer but it didn't look like it would fit the allotted space. – David K Aug 5 '15 at 5:30
I like this one.
Look, let's fill the blanks in the middle column with pair numbers. You will find that it forms the sequence 1, 4, 9, 16, 25, 36.. that is evidently x².
All the numbers in a row are sequential, so we can establish a variable to this position i.e. n. And establish that in the A position we have n, so in the B position we have (n - 1) and in the C position we have (n + 1).
Given all this we can say:
A = x² + n
B = (x + 1)² + (n - 1)
C = (x + 1)² + (n + 1)
So we can put this in the original equation and we will have an equation with two values to solve (x, n). Nevertheless we also know that n can not be greater than x, ensuring the selected values are in the triangle.
So
(x² + n) + (x + 1)² + (n - 1) + (x + 1)² + (n + 1) = 2149
3x² + 4x + (3n) + 2 = 2149
3x² + 4x + (3n - 2147) = 0
With this equation we can find:
x = 26
n = 5
And so:
A = 681 B = 733 C = 735
Hope it helps!
• If you "fill in" the table so that, for example, row $4$ is 13 14 15 16 17 18 19, then the middle A is still $x^2$ and $A+B+C = 3x^2+4x+2$. Solving $3x^2+4x+2=2149$ for $x$, we get $x=26.094$. So we are in row $26$ and our A is just to the right of $A=26^2$, where $A+B+C=2134$. In the filled in table, each shift to the right adds $1$ to $A, B,$ and $C$ and adds $3$ to $A+B+C$. $(2149-2134)/3=5$. So $A = 26^2 + 5 =681$ – steven gregory Aug 5 '15 at 14:56
Let's write the numbers as $a_k:=2k-1$, starting at $k=1$. Then, $a_1=1$, $a_2=3$, $a_3=5$, and so on. When $A=a_k$ is in row $n$ (the first row is row $0$), then $B=a_{k+n+1}$ and $C=a_{k+n+2}=B+2$. Hence, we have
\begin{align*} && 2149 &= a_k + 2a_{2k+n+1} +2 \\ &\Rightarrow& 2147 &= 2k-1 + 2(2(k+n+1)-1) = 6k + 4n + 1 \end{align*}
Note that the $n$-th row (starting at $n=0$) of the triangle starts with $a_{k_n}$, where $$k_n = 1+\cdots+n = \frac{n(n+1)}2.$$ Therefore, $k=k_n+q$ where $0\le q\le n$. Substituting, we get \begin{align*} && 2147 &= 3n(n+1) + 10q + 4n + 1 \\ &\Rightarrow & 0 &= 3 n^2 + 7 n + 10 q -2146 \end{align*} Observe that the linear term $10q$ does not affect the position of the zeros of this quadratic polynomial much. Hence, I just plugged in $q=n/2$ and this polynomial has a zero around $25$. So, $n=25$ seems a good guess.
Indeed, plugging in $n=25$ into our earlier equation gives
\begin{align*} && 2147 &= 6*k + 4*25 + 1 \\ &\Rightarrow & 2046 &= 6*k \\ &\Rightarrow & 341 &= k \end{align*}
So we have $A=a_{341}=681$ in row $n=25$ and $B=2*(341+25+1)-1=733$, $C=735$.
Looking at the first element in each row, we see
row A B C A+B+C
1 1 3 5 9
2 3 7 9 19
3 7 13 15 35
4 13 21 23 57
5 21 31 33 85
6 31 43 45 139
Using finite differences, we find that
\begin{align} A &= n^2 - n + 1\\ B &= n^2 + n + 1 = A+2n\\ C &= n^2 + n + 3 = A+2n+2\\ A+B+C &= 3n^2+n+5 \end{align} where A is the first element in row $n$.
For each shift of A to the right in the same row, A,B, and C increase by 2 and A+B+C increases by $6$.
Solving $3n^2+n+5 = 2149$ for n, we get $n = 26.567$ So we are in row $26$.
For the first element in row $26, A = 651,\; B = 703,\; C = 705,$ and $A+B+C = 2059$.
Since $\dfrac{2149-2059}6 = 15$, we need to increase $A, B,$ and $C$ by $2\cdot15 = 30$.
So
\begin{align} A &= 681\\ B &= 733\\ C &= 735 \end{align}
• Ok, you just should edit the first row $a=1, b=3, c=5, a+b+c=9$ – Oiue Aug 5 '15 at 0:08
• @Oiue. Yep. It's nine in my notes too. Thanks. – steven gregory Aug 5 '15 at 1:18
Let $r$ be the row number, and $U$ be the unshown number between $B$ and $C$ and directly below $A$, i.e.
\begin{align} &A &&\leftarrow\text {row}\; r\\ B \quad [&U]\quad C &&\leftarrow\text {row}\; r+1 \end{align}
Note that the difference between two vertically adjacent numbers (both shown and unshown) is $(r+1)^2-r^2=2r+1$, i.e. $U-A=2r+1$. Hence \begin{align} 2149&=A+B+C&&\text{(given)}\\ &=A+2U&&\text{(as}\; B=U-1, C=U+1\text{)}\\ &=3A+4r+2&&\text{(using } U=A+2r+1\text{)}\\ 3A+4r-2147&=0&& && .....(1)\\ \end{align}
Note also that the centre column (including unshown numbers) corresponds to $r^2$. As an initial approximation, assume that $A$ is the centre column, i.e. $A=r^2$. Substituting in $(1)$ gives
\begin{align}3r^2+4r-2147&=0\\ \color{blue}{r}&\color{blue}{\approx 26} \; (r>0)\qquad \qquad && && && &&& .....(2)\end{align}
Putting $(2)$ in $(1)$ gives
$$\color{red}{A=681}\qquad\blacksquare\qquad \Rightarrow \color{blue}{B=733, C=735}$$
Check: $A+B+C=681+733+735=2149$
NB: The number "$A$" is in row $26$ but is not in the centre column; the element in the centre column is $26^2=676$ which is blank as it is an even number.
• every solutions her always adds something new, thanks :) – Oiue Aug 5 '15 at 18:41
This is a slight variation on the nice answer by @Alex Zorn.
In that answer we find that if the $1$ at the top of the triangle of numbers is considered entry number $1$ (counting from the left) in row number $1$ (counting from the top), then entry number $k$ in row $n$ is $n^2−n+2k−1$, and $A$ is entry $k$ in row $n$ where $n,k$ are a solution to
$$3n^2 + n + 6k = 2150.$$
We have $1 \leq k \leq n$ by the way the entries in a row are counted, so for all $n > 0$ we have $$3n^2 + n = 2144 \leq 3n^2 + n + 6k = 2150 \leq 3n^2 + 7n = 2150. \tag 1$$
Solving the two equations
$$3n^2 + n = 2144, \tag 2$$ $$3n^2 + 7n = 2150, \tag 3$$
we see that they each have one positive root. The positive root of Equation $(2)$ is $r_1 \approx 26.642$ and the positive root of Equation $(3)$ is $r_2 \approx 25.629$.
But because of Inequality $(1)$, whatever the value of $k$ is (provided that $1 \leq k \leq n$) there is a positive root of $3n^2 + n + 6k = 2150$ that is not less than $r_2$ and not greater than $r_1$ (draw a graph if you have trouble seeing this). The only integer that satisfies both conditions is $26$, so if $3n^2 + n + 6k = 2150$ has a solution for integer $n$ then the only possibility is $n = 26$.
Setting $n = 26$, we have $3(26^2) + 26 + 6k = 2150$, which is a simple linear equation in $k$ with the solution $k = 16$. We then plug $n$ and $k$ into the formula for $A$ to get the answer. | 2019-06-27T04:37:41 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/1384690/abc-2149-find-a",
"openwebmath_score": 0.9880356192588806,
"openwebmath_perplexity": 232.46094820076627,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9875683487809279,
"lm_q2_score": 0.8670357563664174,
"lm_q1q2_score": 0.8562570702488057
} |
https://math.stackexchange.com/questions/3021679/values-of-x-satisfying-sin-x-cdot-cos3-x-sin3x-cdot-cos-x | Values of $x$ satisfying $\sin x\cdot\cos^3 x>\sin^3x\cdot\cos x$
For what values of $$x$$ between $$0$$ and $$\pi$$ does the inequality $$\sin x\cdot\cos^3 x>\sin^3x\cdot\cos x$$ hold?
My Attempt $$\sin x\cos x\cdot(\cos^2x-\sin^2x)=\frac{1}{2}\cdot\sin2x\cdot\cos2x=\frac{1}{4}\cdot\sin4x>0\implies\sin4x>0\\ x\in(0,\pi)\implies4x\in(0,4\pi)\\ 4x\in(0,\pi)\cup(2\pi,3\pi)\implies x\in\Big(0,\frac{\pi}{4}\Big)\cup\Big(\frac{\pi}{2},\frac{3\pi}{4}\Big)$$ But, my reference gives the solution, $$x\in\Big(0,\dfrac{\pi}{4}\Big)\cup\Big(\dfrac{3\pi}{4},\pi\Big)$$, where am I going wrong with my attempt?
The given solution is wrong; you are correct. At $$x=\frac{7\pi}8\in\left(\frac{3\pi}4,\pi\right)$$, we have that $$\frac14\sin4x=\frac14\sin\frac{7\pi}2=-\frac14<0$$ which is a contradiction.
Finishing what you did
$$\sin(4x)>0\iff$$
$$2k\pi <4x<(2k+1)\pi \iff$$
$$\frac{k\pi}{2}
$$k=0$$ gives $$0
and $$k=1$$ gives $$\frac{\pi}{2}.
As an alternative for a full solution we can consider two cases
• $$\sin x \cos x >0$$ that is $$x\in(0,\pi/2)\cup(\pi,3\pi/2)$$
$$\sin x\cdot\cos^3 x>\sin^3x\cdot\cos x \iff\cos^2x>\sin^2x \iff2\sin^2 x<1$$
$$-\frac{\sqrt 2}2<\sin x<0 \,\land\, 0<\sin x<\frac{\sqrt 2}2 \iff \color{red}{x\in(0,\pi/4)}\cup(\pi,5\pi/4)$$
• $$\sin x \cos x <0$$ that is $$x\in(\pi/2,\pi)\cup(3\pi/2,2\pi)$$
$$\sin x\cdot\cos^3 x>\sin^3x\cdot\cos x \iff\cos^2x<\sin^2x \iff2\sin^2 x>1$$
$$-1<\sin x<-\frac{\sqrt 2}2\,\land\, \frac{\sqrt 2}2<\sin x <1 \iff \color{red}{x\in(\pi/2,3\pi/4)}\cup(3\pi/2,7\pi/4)$$
and then your solution is correct. | 2019-05-25T06:56:33 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/3021679/values-of-x-satisfying-sin-x-cdot-cos3-x-sin3x-cdot-cos-x",
"openwebmath_score": 0.9619792103767395,
"openwebmath_perplexity": 3921.7755985756467,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9854964211605606,
"lm_q2_score": 0.8688267796346599,
"lm_q1q2_score": 0.8562256819384124
} |
http://jeonjucj.kr/mexican-chili-octfec/90jrx.php?tag=inverse-trigonometric-functions-derivatives-44b03e | The inverse of six important trigonometric functions are: 1. What are the derivatives of the inverse trigonometric functions? For example, the sine function $$x = \varphi \left( y \right)$$ $$= \sin y$$ is the inverse function for $$y = f\left( x \right)$$ $$= \arcsin x.$$ Then the derivative of $$y = \arcsin x$$ is given by, ${{\left( {\arcsin x} \right)^\prime } = f’\left( x \right) = \frac{1}{{\varphi’\left( y \right)}} }= {\frac{1}{{{{\left( {\sin y} \right)}^\prime }}} }= {\frac{1}{{\cos y}} }= {\frac{1}{{\sqrt {1 – {\sin^2}y} }} }= {\frac{1}{{\sqrt {1 – {\sin^2}\left( {\arcsin x} \right)} }} }= {\frac{1}{{\sqrt {1 – {x^2}} }}\;\;}\kern-0.3pt{\left( { – 1 < x < 1} \right).}$. Derivatives of Inverse Trig Functions. ${y^\prime = \left( {\arctan \frac{1}{x}} \right)^\prime }={ \frac{1}{{1 + {{\left( {\frac{1}{x}} \right)}^2}}} \cdot \left( {\frac{1}{x}} \right)^\prime }={ \frac{1}{{1 + \frac{1}{{{x^2}}}}} \cdot \left( { – \frac{1}{{{x^2}}}} \right) }={ – \frac{{{x^2}}}{{\left( {{x^2} + 1} \right){x^2}}} }={ – \frac{1}{{1 + {x^2}}}. Derivatives of the Inverse Trigonometric Functions. In Table 2.7.14 we show the restrictions of the domains of the standard trigonometric functions that allow them to be invertible. Suppose \textrm{arccot } x = \theta. We then apply the same technique used to prove Theorem 3.3, “The Derivative Rule for Inverses,” to differentiate each inverse trigonometric function. The Inverse Cosine Function. }$, ${y^\prime = \left( {\frac{1}{a}\arctan \frac{x}{a}} \right)^\prime }={ \frac{1}{a} \cdot \frac{1}{{1 + {{\left( {\frac{x}{a}} \right)}^2}}} \cdot \left( {\frac{x}{a}} \right)^\prime }={ \frac{1}{a} \cdot \frac{1}{{1 + \frac{{{x^2}}}{{{a^2}}}}} \cdot \frac{1}{a} }={ \frac{1}{{{a^2}}} \cdot \frac{{{a^2}}}{{{a^2} + {x^2}}} }={ \frac{1}{{{a^2} + {x^2}}}. The derivatives of $$6$$ inverse trigonometric functions considered above are consolidated in the following table: In the examples below, find the derivative of the given function. The usual approach is to pick out some collection of angles that produce all possible values exactly once. Table 2.7.14. Quick summary with Stories. Dividing both sides by \cos \theta immediately leads to a formula for the derivative. The Inverse Tangent Function. 1. which implies the following, upon realizing that \cot \theta = x and the identity \cot^2 \theta + 1 = \csc^2 \theta requires \csc^2 \theta = 1 + x^2, Inverse Trigonometric Functions - Derivatives - Harder Example. Arcsine 2. The inverse trigonometric functions actually perform the opposite operation of the trigonometric functions such as sine, cosine, tangent, cosecant, secant, and cotangent. Inverse Trigonometry Functions and Their Derivatives. The process for finding the derivative of \arctan x is slightly different, but the same overall strategy is used: Suppose \arctan x = \theta. These cookies will be stored in your browser only with your consent. 3 Definition notation EX 1 Evaluate these without a calculator. Practice your math skills and learn step by step with our math solver. Section 3-7 : Derivatives of Inverse Trig Functions. If f(x) is a one-to-one function (i.e. Related Questions to study. Inverse trigonometric functions provide anti derivatives for a variety of functions that arise in engineering. Because each of the above-listed functions is one-to-one, each has an inverse function. Presuming that the range of the secant function is given by (0, \pi), we note that \theta must be either in quadrant I or II. x = \varphi \left ( y \right) x = φ ( y) = \sin y = sin y. is the inverse function for. \frac{d}{dx}(\textrm{arccot } x) = \frac{-1}{1+x^2}, Finding the Derivative of the Inverse Secant Function, \displaystyle{\frac{d}{dx} (\textrm{arcsec } x)}. -csc^2 \theta \cdot \frac{d\theta}{dx} = 1 Derivative of Inverse Trigonometric Function as Implicit Function. Inverse Trigonometric Functions Note. Inverse trigonometric functions have various application in engineering, geometry, navigation etc. Here we will develop the derivatives of inverse sine or arcsine, , 1 and inverse tangent or arctangent, . They are cosecant (cscx), secant (secx), cotangent (cotx), tangent (tanx), cosine (cosx), and sine (sinx). Email. Here, for the first time, we see that the derivative of a function need not be of the same type as the … Necessary cookies are absolutely essential for the website to function properly. Review the derivatives of the inverse trigonometric functions: arcsin (x), arccos (x), and arctan (x). Derivatives of Inverse Trigonometric Functions To find the derivatives of the inverse trigonometric functions, we must use implicit differentiation. One example does not require the chain rule and one example requires the chain rule. }$, $\require{cancel}{y^\prime = \left( {\arcsin \left( {x – 1} \right)} \right)^\prime }={ \frac{1}{{\sqrt {1 – {{\left( {x – 1} \right)}^2}} }} }={ \frac{1}{{\sqrt {1 – \left( {{x^2} – 2x + 1} \right)} }} }={ \frac{1}{{\sqrt {\cancel{1} – {x^2} + 2x – \cancel{1}} }} }={ \frac{1}{{\sqrt {2x – {x^2}} }}. \frac{d\theta}{dx} = \frac{-1}{\csc^2 \theta} = \frac{-1}{1+x^2} This video covers the derivative rules for inverse trigonometric functions like, inverse sine, inverse cosine, and inverse tangent. View Lesson 9-Differentiation of Inverse Trigonometric Functions.pdf from MATH 146 at Mapúa Institute of Technology. 2 The graph of y = sin x does not pass the horizontal line test, so it has no inverse. We'll assume you're ok with this, but you can opt-out if you wish. Derivatives of inverse trigonometric functions. Lesson 9 Differentiation of Inverse Trigonometric Functions OBJECTIVES • to The derivatives of the inverse trigonometric functions can be obtained using the inverse function theorem. Derivatives of Inverse Trigonometric Functions using First Principle. It is mandatory to procure user consent prior to running these cookies on your website. Arccosecant Let us discuss all the six important types of inverse trigonometric functions along with its definition, formulas, graphs, properties and solved examples. The corresponding inverse functions are for ; for ; for ; arc for , except ; arc for , except y = 0 arc for . In modern mathematics, there are six basic trigonometric functions: sine, cosine, tangent, secant, cosecant, and cotangent. }$, ${y’\left( x \right) }={ {\left( {\arctan \frac{{x + 1}}{{x – 1}}} \right)^\prime } }= {\frac{1}{{1 + {{\left( {\frac{{x + 1}}{{x – 1}}} \right)}^2}}} \cdot {\left( {\frac{{x + 1}}{{x – 1}}} \right)^\prime } }= {\frac{{1 \cdot \left( {x – 1} \right) – \left( {x + 1} \right) \cdot 1}}{{{{\left( {x – 1} \right)}^2} + {{\left( {x + 1} \right)}^2}}} }= {\frac{{\cancel{\color{blue}{x}} – \color{red}{1} – \cancel{\color{blue}{x}} – \color{red}{1}}}{{\color{maroon}{x^2} – \cancel{\color{green}{2x}} + \color{DarkViolet}{1} + \color{maroon}{x^2} + \cancel{\color{green}{2x}} + \color{DarkViolet}{1}}} }= {\frac{{ – \color{red}{2}}}{{\color{maroon}{2{x^2}} + \color{DarkViolet}{2}}} }= { – \frac{1}{{1 + {x^2}}}. We know that trig functions are especially applicable to the right angle triangle. Derivative of Inverse Trigonometric functions The Inverse Trigonometric functions are also called as arcus functions, cyclometric functions or anti-trigonometric functions. Derivatives of Inverse Trigonometric Functions Learning objectives: To find the deriatives of inverse trigonometric functions. The derivatives of the above-mentioned inverse trigonometric functions follow from trigonometry … This category only includes cookies that ensures basic functionalities and security features of the website. In the following discussion and solutions the derivative of a function h(x) will be denoted by or h'(x) . To be a useful formula for the derivative of \arctan x however, we would prefer that \displaystyle{\frac{d\theta}{dx} = \frac{d}{dx} (\arctan x)} be expressed in terms of x, not \theta. The derivatives of the inverse trigonometric functions are given below. And To solve the related problems. Inverse Trigonometric Functions: •The domains of the trigonometric functions are restricted so that they become one-to-one and their inverse can be determined. For example, the domain for $$\arcsin x$$ is from $$-1$$ to $$1.$$ The range, or output for $$\arcsin x$$ is all angles from $$– \large{\frac{\pi }{2}}\normalsize$$ to $$\large{\frac{\pi }{2}}\normalsize$$ radians. Upon considering how to then replace the above \sin \theta with some expression in x, recall the pythagorean identity \cos^2 \theta + \sin^2 \theta = 1 and what this identity implies given that \cos \theta = x: So we know either \sin \theta is then either the positive or negative square root of the right side of the above equation. The inverse of these functions is inverse sine, inverse cosine, inverse tangent, inverse secant, inverse cosecant, and inverse cotangent. Then it must be the case that. Derivative of Inverse Trigonometric Functions using Chain Rule. 11 mins. This website uses cookies to improve your experience. Dividing both sides by -\sin \theta immediately leads to a formula for the derivative. Arcsecant 6. Arctangent 4. We can use implicit differentiation to find the formulas for the derivatives of the inverse trigonometric functions, as the following examples suggest: Finding the Derivative of Inverse Sine Function, \displaystyle{\frac{d}{dx} (\arcsin x)}, Suppose \arcsin x = \theta. AP.CALC: FUN‑3 (EU), FUN‑3.E (LO), FUN‑3.E.2 (EK) Google Classroom Facebook Twitter. Since \theta must be in the range of \arccos x (i.e., [0,\pi]), we know \sin \theta must be positive. If we restrict the domain (to half a period), then we can talk about an inverse function. Examples: Find the derivatives of each given function. Derivatives of Exponential, Logarithmic and Trigonometric Functions Derivative of the inverse function. Thus, Finally, plugging this into our formula for the derivative of \arcsin x, we find, Finding the Derivative of Inverse Cosine Function, \displaystyle{\frac{d}{dx} (\arccos x)}. Differentiation of Inverse Trigonometric Functions Each of the six basic trigonometric functions have corresponding inverse functions when appropriate restrictions are placed on the domain of the original functions. Arccotangent 5. Inverse trigonometric functions are literally the inverses of the trigonometric functions. But opting out of some of these cookies may affect your browsing experience. Example: Find the derivatives of y = sin-1 (cos x/(1+sinx)) Show Video Lesson. In this section we are going to look at the derivatives of the inverse trig functions. As such. Implicitly differentiating with respect to x yields 1 du Out of these, the cookies that are categorized as necessary are stored on your browser as they are essential for the working of basic functionalities of the website. Derivatives of Inverse Trigonometric Functions We can use implicit differentiation to find the formulas for the derivatives of the inverse trigonometric functions, as the following examples suggest: Finding the Derivative of Inverse Sine Function, d d x (arcsin These functions are used to obtain angle for a given trigonometric value. The differentiation of trigonometric functions is the mathematical process of finding the derivative of a trigonometric function, or its rate of change with respect to a variable. •Since the definition of an inverse function says that -f 1(x)=y => f(y)=x We have the inverse sine function, -sin 1x=y - π=> sin y=x and π/ 2 <=y<= / 2 Of course |\sec \theta| = |x|, and we can use \tan^2 \theta + 1 = \sec^2 \theta to establish |\tan \theta| = \sqrt{x^2 - 1}. Upon considering how to then replace the above \cos \theta with some expression in x, recall the pythagorean identity \cos^2 \theta + \sin^2 \theta = 1 and what this identity implies given that \sin \theta = x: So we know either \cos \theta is then either the positive or negative square root of the right side of the above equation. Previously, derivatives of algebraic functions have proven to be algebraic functions and derivatives of trigonometric functions have been shown to be trigonometric functions. Using similar techniques, we can find the derivatives of all the inverse trigonometric functions. Domains and ranges of the trigonometric and inverse trigonometric functions Coming to the question of what are trigonometric derivatives and what are they, the derivatives of trigonometric functions involve six numbers. Using this technique, we can find the derivatives of the other inverse trigonometric functions: \[{{\left( {\arccos x} \right)^\prime } }={ \frac{1}{{{{\left( {\cos y} \right)}^\prime }}} }= {\frac{1}{{\left( { – \sin y} \right)}} }= {- \frac{1}{{\sqrt {1 – {{\cos }^2}y} }} }= {- \frac{1}{{\sqrt {1 – {{\cos }^2}\left( {\arccos x} \right)} }} }= {- \frac{1}{{\sqrt {1 – {x^2}} }}\;\;}\kern-0.3pt{\left( { – 1 < x < 1} \right),}\qquad$, ${{\left( {\arctan x} \right)^\prime } }={ \frac{1}{{{{\left( {\tan y} \right)}^\prime }}} }= {\frac{1}{{\frac{1}{{{{\cos }^2}y}}}} }= {\frac{1}{{1 + {{\tan }^2}y}} }= {\frac{1}{{1 + {{\tan }^2}\left( {\arctan x} \right)}} }= {\frac{1}{{1 + {x^2}}},}$, ${\left( {\text{arccot }x} \right)^\prime } = {\frac{1}{{{{\left( {\cot y} \right)}^\prime }}}}= \frac{1}{{\left( { – \frac{1}{{{\sin^2}y}}} \right)}}= – \frac{1}{{1 + {{\cot }^2}y}}= – \frac{1}{{1 + {{\cot }^2}\left( {\text{arccot }x} \right)}}= – \frac{1}{{1 + {x^2}}},$, ${{\left( {\text{arcsec }x} \right)^\prime } = {\frac{1}{{{{\left( {\sec y} \right)}^\prime }}} }}= {\frac{1}{{\tan y\sec y}} }= {\frac{1}{{\sec y\sqrt {{{\sec }^2}y – 1} }} }= {\frac{1}{{\left| x \right|\sqrt {{x^2} – 1} }}.}$. }\], ${y^\prime = \left( {\text{arccot}\frac{1}{{{x^2}}}} \right)^\prime }={ – \frac{1}{{1 + {{\left( {\frac{1}{{{x^2}}}} \right)}^2}}} \cdot \left( {\frac{1}{{{x^2}}}} \right)^\prime }={ – \frac{1}{{1 + \frac{1}{{{x^4}}}}} \cdot \left( { – 2{x^{ – 3}}} \right) }={ \frac{{2{x^4}}}{{\left( {{x^4} + 1} \right){x^3}}} }={ \frac{{2x}}{{1 + {x^4}}}.}$. Important Sets of Results and their Applications Then it must be the case that. 3 mins read . The sine function (red) and inverse sine function (blue). f(x) = 3sin-1 (x) g(x) = 4cos-1 (3x 2) Show Video Lesson. The domains of the other trigonometric functions are restricted appropriately, so that they become one-to-one functions and their inverse can be determined. Dividing both sides by $\sec^2 \theta$ immediately leads to a formula for the derivative. The derivatives of inverse trigonometric functions are quite surprising in that their derivatives are actually algebraic functions. Click or tap a problem to see the solution. In both, the product of $\sec \theta \tan \theta$ must be positive. Sec 3.8 Derivatives of Inverse Functions and Inverse Trigonometric Functions Ex 1 Let f x( )= x5 + 2x −1. One way to do this that is particularly helpful in understanding how these derivatives are obtained is to use a combination of implicit differentiation and right triangles. For example, the derivative of the sine function is written sin′ = cos, meaning that the rate of change of sin at a particular angle x = a is given by the cosine of that angle. Trigonometric Functions (With Restricted Domains) and Their Inverses. Check out all of our online calculators here! Formula for the Derivative of Inverse Cosecant Function. All the inverse trigonometric functions have derivatives, which are summarized as follows: Proofs of the formulas of the derivatives of inverse trigonometric functions are presented along with several other examples involving sums, products and quotients of functions. Definition of the Inverse Cotangent Function. }\], ${y^\prime = \left( {\text{arccot}\,{x^2}} \right)^\prime }={ – \frac{1}{{1 + {{\left( {{x^2}} \right)}^2}}} \cdot \left( {{x^2}} \right)^\prime }={ – \frac{{2x}}{{1 + {x^4}}}. Arccosine 3. Thus, To be a useful formula for the derivative of \arcsin x however, we would prefer that \displaystyle{\frac{d\theta}{dx} = \frac{d}{dx} (\arcsin x)} be expressed in terms of x, not \theta. However, since trigonometric functions are not one-to-one, meaning there are are infinitely many angles with , it is impossible to find a true inverse function for . Upon considering how to then replace the above \sec^2 \theta with some expression in x, recall the other pythagorean identity \tan^2 \theta + 1 = \sec^2 \theta and what this identity implies given that \tan \theta = x: Not having to worry about the sign, as we did in the previous two arguments, we simply plug this into our formula for the derivative of \arccos x, to find, Finding the Derivative of the Inverse Cotangent Function, \displaystyle{\frac{d}{dx} (\textrm{arccot } x)}, The derivative of \textrm{arccot } x can be found similarly. Another method to find the derivative of inverse functions is also included and may be used. The basic trigonometric functions include the following $$6$$ functions: sine $$\left(\sin x\right),$$ cosine $$\left(\cos x\right),$$ tangent $$\left(\tan x\right),$$ cotangent $$\left(\cot x\right),$$ secant $$\left(\sec x\right)$$ and cosecant $$\left(\csc x\right).$$ All these functions are continuous and differentiable in their domains. Problem. Derivatives of inverse trigonometric functions Calculator Get detailed solutions to your math problems with our Derivatives of inverse trigonometric functions step-by-step calculator. These six important functions are used to find the angle measure in a right triangle when two sides of the triangle measures are known. 7 mins. The inverse sine function (Arcsin), y = arcsin x, is the inverse of the sine function. Inverse Sine Function. Since \theta must be in the range of \arcsin x (i.e., [-\pi/2,\pi/2]), we know \cos \theta must be positive. Here, we suppose \textrm{arcsec } x = \theta, which means sec \theta = x. Thus, Finally, plugging this into our formula for the derivative of \arccos x, we find, Finding the Derivative of the Inverse Tangent Function, \displaystyle{\frac{d}{dx} (\arctan x)}. To be a useful formula for the derivative of \arccos x however, we would prefer that \displaystyle{\frac{d\theta}{dx} = \frac{d}{dx} (\arccos x)} be expressed in terms of x, not \theta. You can think of them as opposites; In a way, the two functions “undo” each other. VIEW MORE. Like before, we differentiate this implicitly with respect to x to find, Solving for d\theta/dx in terms of \theta we quickly get, This is where we need to be careful. We also use third-party cookies that help us analyze and understand how you use this website. In this section we review the definitions of the inverse trigonometric func-tions from Section 1.6. If $$f\left( x \right)$$ and $$g\left( x \right)$$ are inverse functions then, This lessons explains how to find the derivatives of inverse trigonometric functions. In the previous topic, we have learned the derivatives of six basic trigonometric functions: \[{\color{blue}{\sin x,\;}}\kern0pt\color{red}{\cos x,\;}\kern0pt\color{darkgreen}{\tan x,\;}\kern0pt\color{magenta}{\cot x,\;}\kern0pt\color{chocolate}{\sec x,\;}\kern0pt\color{maroon}{\csc x.\;}$, In this section, we are going to look at the derivatives of the inverse trigonometric functions, which are respectively denoted as, ${\color{blue}{\arcsin x,\;}}\kern0pt \color{red}{\arccos x,\;}\kern0pt\color{darkgreen}{\arctan x,\;}\kern0pt\color{magenta}{\text{arccot }x,\;}\kern0pt\color{chocolate}{\text{arcsec }x,\;}\kern0pt\color{maroon}{\text{arccsc }x.\;}$. a) c) b) d) 4 y = tan x y = sec x Definition [ ] 5 EX 2 Evaluate without a calculator. Then $\cot \theta = x$. These cookies do not store any personal information. Similarly, we can obtain an expression for the derivative of the inverse cosecant function: ${{\left( {\text{arccsc }x} \right)^\prime } = {\frac{1}{{{{\left( {\csc y} \right)}^\prime }}} }}= {-\frac{1}{{\cot y\csc y}} }= {-\frac{1}{{\csc y\sqrt {{{\csc }^2}y – 1} }} }= {-\frac{1}{{\left| x \right|\sqrt {{x^2} – 1} }}.}$. The process for finding the derivative of $\arccos x$ is almost identical to that used for $\arcsin x$: Suppose $\arccos x = \theta$. Now let's determine the derivatives of the inverse trigonometric functions, y = arcsinx, y = arccosx, y = arctanx, y = arccotx, y = arcsecx, and y = arccscx. g ( x) = arccos ( 2 x) g (x)=\arccos\!\left (2x\right) g(x)= arccos(2x) g, left parenthesis, x, right parenthesis, … {\displaystyle {\begin{aligned}{\frac {d}{dz}}\arcsin(z)&{}={\frac {1}{\sqrt {1-z^{2}}}}\;;&z&{}\neq -1,+1\\{\frac {d}{dz}}\arccos(z)&{}=-{\frac {1}{\sqrt {1-z^{2}}}}\;;&z&{}\neq -1,+1\\{\frac {d}{dz}}\arctan(z)&{}={\frac {1}{1+z^{2}}}\;;&z&{}\neq -i,+i\\{\frac {d}{dz}}\operatorname {arccot}(z)&{}=-{\frac {1}{1+z^{2}}}\;;&z&{}\neq -i,+i\\{\frac {d}{dz}}\operatorname {arcsec}(z)&{}={\frac {1}{z^{2… Derivatives of Inverse Trigonometric Functions. This implies. The inverse functions exist when appropriate restrictions are placed on the domain of the original functions. Inverse Functions and Logarithms. Formula for the Derivative of Inverse Secant Function. \dfrac {d} {dx}\arcsin (x)=\dfrac {1} {\sqrt {1-x^2}} dxd arcsin(x) = 1 − x2 , we must use implicit differentiation is also included and may be used function (.! Will be stored in your browser only with your consent we can talk about an inverse function of. Functions follow from trigonometry … derivatives of the inverse trig functions are literally the Inverses of the trigonometric functions used... Arctangent, this, but you can think of them as opposites ; in a right triangle two. You can opt-out if you wish LO ), arccos ( x ) = 4cos-1 ( 2! } x = \theta $immediately leads to a formula for the derivative use implicit differentiation their Inverses inverse. That ensures basic functionalities and security features of the standard trigonometric functions you also have the option to opt-out these. ; in a way, the two functions “ undo ” each other restrictions are placed on the of. User consent prior to running these cookies pick out some collection of that! Inverse functions exist when appropriate restrictions are placed on the domain ( to half period. Rule and one example does not require the chain rule trig functions f x ( ) = 4cos-1 3x! Arcsin ), and arctan ( x ) = 3sin-1 ( x ), (. Inverse cotangent of examples and worked-out practice problems half a period ), FUN‑3.E.2 ( EK ) Google Facebook... Obtain angle for a given trigonometric value rule and one example does not require the chain rule and example. We can talk about an inverse function one-to-one function ( red ) and their inverse can be obtained the! Restricted so that they become one-to-one and their inverse can be determined differentiation of inverse trigonometric:. A calculator$ \textrm { arccot } x = \theta $immediately leads to formula... With this, but you can opt-out if you wish chain rule and one example does not require chain... Be determined is mandatory to procure user consent prior to running these may. Trigonometry … derivatives of inverse functions is also included and may be used opposites. = arcsin x, is the inverse trigonometric functions derivative of the inverse functions derivatives... By$ \cos \theta $must be positive the triangle measures are known the product$! Sec \theta = x $yields examples: find the derivatives of y = sin x does not pass horizontal...$ \cos \theta $, which means$ sec \theta = x $inverse trigonometric functions derivatives opt-out of these will. Domains ) and inverse tangent using the inverse trigonometric functions that allow them to be invertible basic. Usual approach is to pick out some collection of angles that produce all possible values exactly once:. Be the cases that, Implicitly differentiating the above with respect to$ x $yields functions is one-to-one each... To obtain angle for a given trigonometric value our math solver be used angle measure in a way the! Opt-Out of these cookies will be stored in your browser only with your consent basic! The domain ( to half a period ), FUN‑3.E.2 ( EK ) Google Classroom Facebook Twitter = x yields... Ek ) Google Classroom Facebook Twitter are placed on the domain of inverse! While you navigate through the website derivatives for a given trigonometric value various application in engineering may be used think. Basic functionalities and security features of the inverse function = x$ yields may used. Functions OBJECTIVES • to there are particularly six inverse trig functions are used to find the angle measure a! The restrictions of the inverse of the inverse of the trigonometric functions ( i.e chain rule and one requires. You also have the option to opt-out of these cookies will be stored your! If you wish in your browser only with your consent included and be! Inverse secant, inverse cosine, tangent, secant, inverse sine function ( red ) and tangent. Right triangle when two sides of the original functions not require the chain rule and example! To be trigonometric functions: sine, cosine, inverse cosine, inverse secant, cosecant, inverse trigonometric functions derivatives trigonometric...: 1 if you wish functions are especially applicable to the right angle triangle a function... Let f x ( ) = 3sin-1 ( x ) 3sin-1 ( x ) = (... Of $\sec \theta \tan \theta$ must be positive here, we suppose ${. Of examples and worked-out practice problems that trig functions are especially applicable to the right angle triangle the with! Product of$ \sec \theta \tan \theta $immediately leads to a formula for the derivative trigonometry ratio in mathematics... To$ x $yields step with our derivatives of inverse sine function,... Trigonometric func-tions from section 1.6 functions to find the derivative we are going to look at derivatives. We review the derivatives of inverse sine or arcsine,, 1 and inverse sine arcsine.$ \sec \theta \tan \theta $immediately leads to a formula for the website to function properly of... Facebook Twitter, secant, inverse secant, inverse cosine, tangent, inverse cosine, sine. No inverse method to find the derivatives of the inverse of these functions is also included and may be.! Stored in your browser inverse trigonometric functions derivatives with your consent Definition notation EX 1 Let x. Horizontal line test, so that they become one-to-one functions and derivatives of the above-mentioned inverse trigonometric.... On your website restricted appropriately, so that they become one-to-one functions and derivatives trigonometric! Is inverse sine or arcsine,, 1 and inverse cotangent this, you. The deriatives of inverse trigonometric functions like, inverse cosine, and cotangent functions OBJECTIVES • to are... And worked-out practice problems functions follow from trigonometry … derivatives of the measures! Third-Party cookies that help us analyze and understand how you use this website uses cookies to improve experience. Exponential, Logarithmic and trigonometric functions Learning OBJECTIVES: to find the angle measure in a right triangle two... That produce all possible values exactly once some collection of angles that produce all values... = x5 + 2x −1 placed on the domain ( to half a period ), arctan... Classroom Facebook Twitter are placed on the domain of the inverse trigonometric functions be used is the inverse and! Functions calculator inverse trigonometric functions derivatives detailed solutions to your math skills and learn step by step with our math solver,! •The domains of the original functions trigonometric functions have proven to be functions.: arcsin ( x ) = x5 + 2x −1 step with our math solver however imperfect to! The sine function ( arcsin ), y = sin x does not require the chain and! To opt-out of these functions is inverse sine, inverse cosecant, and cotangent then we can talk an... The domain ( to half a period ), y = sin x does not the. Suppose$ \textrm { arcsec } x = \theta $immediately leads to a for. With this, but you can opt-out if you wish rules for inverse trigonometric functions can be determined all. To procure user consent prior to running these cookies to look at the derivatives of the trigonometric derivative... Algebraic functions and their inverse can be obtained using the inverse trigonometric functions are restricted so they. The website to function properly especially applicable to the right angle triangle to function.. Deriatives of inverse trigonometric functions Learning OBJECTIVES: to find the derivative rules for trigonometric... Use implicit differentiation given below security features of the trigonometric functions cookies on your website to obtain angle for given! ( 1+sinx ) ) Show Video Lesson not require the chain rule and one example does not the... The Inverses of the original functions may affect your browsing experience half a period ) FUN‑3.E... Have proven to be algebraic functions and derivatives of the inverse function has plenty of examples and practice! We review the derivatives of inverse trigonometric functions are: 1 we can talk about inverse. X ( ) = x5 + 2x −1 sides by$ \cos \theta immediately. From trigonometry … derivatives of the triangle measures are known, there are particularly six inverse functions... Ek ) Google Classroom Facebook Twitter functions and their inverse can be obtained using the inverse trigonometric functions anti. To these functions is inverse sine, inverse cosine, and inverse tangent, inverse cosine, tangent secant! Without a calculator to procure user consent prior to running these cookies will be stored in your only!, cosecant, and arctan ( x ) is a one-to-one function ( arcsin ), we! Cookies may affect your browsing experience by $\cos \theta$ immediately leads a. Important trigonometric functions to find the derivatives of algebraic functions have various application in engineering, geometry navigation... The domains of the above-listed functions is inverse sine function red ) and inverse sine function ( red and! Definitions of the inverse trigonometric functions step-by-step calculator inverse secant, cosecant, and inverse tangent arctangent..., derivatives of inverse trigonometric functions can be determined math skills and learn step step... Have something like an inverse to these functions is one-to-one, each has an inverse..: arcsin ( x ) g ( x ), arccos ( x,! Lo ), arccos ( x ) is a one-to-one function ( i.e section 1.6 Facebook.. To opt-out of these cookies = sin x does not require the chain.. Covers the derivative we restrict the domain of the inverse of the inverse functions... Six important trigonometric functions have proven to be invertible $\textrm { arccot } x = \theta$ leads. Plenty of examples and worked-out practice problems triangle when two sides of the trigonometric functions provide anti for. Basic functionalities and security features of the above-mentioned inverse trigonometric functions that arise in engineering leads. Absolutely essential for the derivative x $that, Implicitly differentiating the above with respect to$ \$. X ) affect your browsing experience examples: find the derivatives of trigonometric are!
Best Glue For Fresh Flowers, Mystical White Hair, Essay On A Holiday I Enjoyed, Which Of These Sound Effects Describe The Lg Soundbar Sn4?, Small Stakes Hold Em Pdf, Senior Vice President Of Sales Salary, Sofa Cloth Price In Pakistan, Cordless Belt Sander Reviews, | 2021-04-13T00:55:18 | {
"domain": "jeonjucj.kr",
"url": "http://jeonjucj.kr/mexican-chili-octfec/90jrx.php?tag=inverse-trigonometric-functions-derivatives-44b03e",
"openwebmath_score": 0.8942282795906067,
"openwebmath_perplexity": 1072.3048836770301,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.985496421586532,
"lm_q2_score": 0.8688267694452331,
"lm_q1q2_score": 0.856225672266864
} |
http://artofproblemsolving.com/wiki/index.php?title=1984_AIME_Problems/Problem_14 | # 1984 AIME Problems/Problem 14
## Problem
What is the largest even integer that cannot be written as the sum of two odd composite numbers?
## Solution 1
Take an even positive integer $x$. $x$ is either $0 \bmod{6}$, $2 \bmod{6}$, or $4 \bmod{6}$. Notice that the numbers $9$, $15$, $21$, ... , and in general $9 + 6n$ for nonnegative $n$ are odd composites. We now have 3 cases:
If $x \ge 18$ and is $0 \bmod{6}$, $x$ can be expressed as $9 + (9+6n)$ for some nonnegative $n$. Note that $9$ and $9+6n$ are both odd composites.
If $x\ge 44$ and is $2 \bmod{6}$, $x$ can be expressed as $35 + (9+6n)$ for some nonnegative $n$. Note that $35$ and $9+6n$ are both odd composites.
If $x\ge 34$ and is $4 \bmod{6}$, $x$ can be expressed as $25 + (9+6n)$ for some nonnegative $n$. Note that $25$ and $9+6n$ are both odd composites.
Clearly, if $x \ge 44$, it can be expressed as a sum of 2 odd composites. However, if $x = 42$, it can also be expressed using case 1, and if $x = 40$, using case 3. $38$ is the largest even integer that our cases do not cover. If we examine the possible ways of splitting $38$ into two addends, we see that no pair of odd composites add to $38$. Therefore, $\boxed{038}$ is the largest possible number that is not expressible as the sum of two odd composite numbers.
## Solution 2
Let $n$ be an integer that cannot be written as the sum of two odd composite numbers. If $n>33$, then $n-9,n-15,n-21,n-25,n-27,$ and $n-33$ must all be prime (or $n-33=1$, which yields $n=34=9+25$ which does not work). Thus $n-9,n-15,n-21,n-27,$ and $n-33$ form a prime quintuplet. However, only one prime quintuplet exists as exactly one of those 5 numbers must be divisible by 5.This prime quintuplet is $5,11,17,23,$ and $29$, yielding a maximal answer of 38. Since $38-25=13$, which is prime, the answer is $\boxed{038}$.
1984 AIME (Problems • Answer Key • Resources) Preceded byProblem 13 Followed byProblem 15 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions
ACS WASC
ACCREDITED
SCHOOL
Our Team Our History Jobs | 2017-08-17T07:38:50 | {
"domain": "artofproblemsolving.com",
"url": "http://artofproblemsolving.com/wiki/index.php?title=1984_AIME_Problems/Problem_14",
"openwebmath_score": 0.8392579555511475,
"openwebmath_perplexity": 173.72552905808487,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9916842198021455,
"lm_q2_score": 0.863391602943619,
"lm_q1q2_score": 0.8562118281488666
} |
https://nhigham.com/2021/01/05/what-is-the-log-sum-exp-function/ | # What Is the Log-Sum-Exp Function?
The log-sum-exp function takes as input a real $n$-vector $x$ and returns the scalar
$\notag \mathrm{lse}(x) = \log \displaystyle\sum_{i=1}^n \mathrm{e}^{x_i},$
where $\log$ is the natural logarithm. It provides an approximation to the largest element of $x$, which is given by the $\max$ function, $\max(x) = \max_i x_i$. Indeed,
$\notag \mathrm{e}^{\max(x)} \le \displaystyle\sum_{i=1}^n \mathrm{e}^{x_i} \le n \mskip1mu \mathrm{e}^{\max(x)},$
and on taking logs we obtain
$\notag \qquad\qquad \max(x) \le \mathrm{lse}(x) \le \max(x) + \log n. \qquad\qquad (*)$
The log-sum-exp function can be thought of as a smoothed version of the max function, because whereas the max function is not differentiable at points where the maximum is achieved in two different components, the log-sum-exp function is infinitely differentiable everywhere. The following plots of $\mathrm{lse}(x)$ and $\max(x)$ for $n = 2$ show this connection.
The log-sum-exp function appears in a variety of settings, including statistics, optimization, and machine learning.
For the special case where $x = [0~t]^T$, we obtain the function $f(t) = \log(1+\mathrm{e}^t)$, which is known as the softplus function in machine learning. The softplus function approximates the ReLU (rectified linear unit) activation function $\max(t,0)$ and satisfies, by $(*)$,
$\notag \max(t,0) \le f(t) \le \max(t,0) + \log 2.$
Two points are worth noting.
• While $\log(x_1 + x_2) \ne \log x_1 + \log x_2$, in general, we do (trivially) have $\log(x_1 + x_2) = \mathrm{lse}(\log x_1,\log x_2)$, and more generally $\log(x_1 + x_2 + \cdots + x_n) = \mathrm{lse}(\log x_1,\log x_2,\dots,\log x_n)$.
• The log-sum-exp function is not to be confused with the exp-sum-log function: $\exp \sum_{i=1}^n \log x_i = x_1x_2\dots x_n$.
Here are some examples:
>> format long e
>> logsumexp([1 2 3])
ans =
3.407605964444380e+00
>> logsumexp([1 2 30])
ans =
3.000000000000095e+01
>> logsumexp([1 2 -3])
ans =
2.318175429247454e+00
The MATLAB function logsumexp used here is available at https://github.com/higham/logsumexp-softmax.
Straightforward evaluation of log-sum-exp from its definition is not recommended, because of the possibility of overflow. Indeed, $\exp(x)$ overflows for $x = 12$, $x = 89$, and $x = 710$ in IEEE half, single, and double precision arithmetic, respectively. Overflow can be avoided by writing
\notag \begin{aligned} \mathrm{lse}(x) &= \log \sum_{i=1}^n \mathrm{e}^{x_i} = \log \sum_{i=1}^n \mathrm{e}^a \mathrm{e}^{x_i - a} = \log \left(\mathrm{e}^a\sum_{i=1}^n \mathrm{e}^{x_i - a}\right), \end{aligned}
which gives
$\notag \mathrm{lse}(x) = a + \log\displaystyle\sum_{i=1}^n \mathrm{e}^{x_i - a}.$
We take $a = \max(x)$, so that all exponentiations are of nonpositive numbers and therefore overflow is avoided. Any underflows are harmless. A refinement is to write
$\notag \qquad\qquad \mathrm{lse}(x) = \max(x) + \mathrm{log1p}\Biggl( \displaystyle\sum_{i=1 \atop i\ne k}^n \mathrm{e}^{x_i - \max(x)}\Biggr), \qquad\qquad (\#)$
where $x_k = \max(x)$ (if there is more than one such $k$, we can take any of them). Here, $\mathrm{log1p}(x) = \log(1+x)$ is a function provided in MATLAB and various other languages that accurately evaluates $\log(1+x)$ even when $x$ is small, in which case $1+x$ would suffer a loss of precision if it was explicitly computed.
Whereas the original formula involves the logarithm of a sum of nonnegative quantities, when $\max(x) < 0$ the shifted formula $(\#)$ computes $\mathrm{lse}(x)$ as the sum of two terms of opposite sign, so could potentially suffer from numerical cancellation. It can be shown by rounding error analysis, however, that computing log-sum-exp via $(\#)$ is numerically reliable.
## References
This is a minimal set of references, which contain further useful references within.
## Related Blog Posts
1. Nice post! It’s nice to know the sign issue doesn’t cause problem, and to formalize the pulling-the-max-out trick. How is the scipy.special.logsumexp implementation?
(1) if we want to approximate $\|x\|_\infty$, then we can use $f(x) = \log( \sum_i exp^{x_i} + exp^{-x_i} )$. In this case, we’d pull out $max |x_i|$ instead of $max x_i$, but I’m not sure we’d want to do log1p
(2) as for the logsumexp and its relationship to softmax (its derivative), many times each $x_i$ is parameterized by a vector $\theta$ and we want the gradient with respect to $\theta$, so then we have to modify the softmax formula to include the gradients of the $x_i$ terms. I’m thinking the naive implementation is not stable, but there ought to be similar tricks. | 2021-04-20T22:42:46 | {
"domain": "nhigham.com",
"url": "https://nhigham.com/2021/01/05/what-is-the-log-sum-exp-function/",
"openwebmath_score": 0.9465411901473999,
"openwebmath_perplexity": 479.7915498684911,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9814534338604444,
"lm_q2_score": 0.8723473846343393,
"lm_q1q2_score": 0.8561683361685501
} |
https://math.stackexchange.com/questions/3011414/wrong-inflection-point?noredirect=1 | # Wrong Inflection point
I have the next function: $$f(x)=x^3-3x^2+4$$.
I need to find an inflection point so I had done the following steps:
I found the first derivative: $$f'(x)=3x^2-6x$$
Then the second one: $$f''(x)=6x-6$$
Comparative to zero: $$6x-6=0$$
$$6x=6\rightarrow x=1$$
I checked values before and after the point in the second derivative:
$$f''(0)=-6<0$$
$$f''(2)=12>0$$
So the point (1,2) is inflection point but when I chaked the graph, there were not any:
Can someone explain why? Thank You.
• The first derivative gives you critical points not the second derivative. The point $(1,2)$ is inflection point if in addition to $f''(1)=0$, it is the root of the first derivative as well. – Yadati Kiran Nov 24 '18 at 10:56
• Yes, I know. But I need an inflection point and not critical points. – violettagold Nov 24 '18 at 10:59
• So there are none in this case. – Yadati Kiran Nov 24 '18 at 11:00
• What do you mean by root? – violettagold Nov 24 '18 at 11:02
• 1,2 is inflection point – Akash Roy Nov 24 '18 at 11:26
Draw the graph together with the tangent at $$(1,f(1))$$: $$1$$ is where the second derivative vanishes and $$f(1)=2$$; since $$f'(1)=-3$$, the tangent has equation $$y-2=-3(x-1)$$ or $$y=-3x+5$$.
As you see, the tangent “crosses” the graph, because the curve is concave for $$x<1$$ and convex for $$x>1$$.
So at $$x=1$$ there's indeed an inflection point.
• Thank you very much for the help! – violettagold Nov 24 '18 at 12:03
In order to find out critical points first find the points where derivative equals to zero and check the concavity of graph about that point.
For example if $$0$$ is an inflection point,
Then $$f^{''}(0^{+})>0$$ and $$f^{''}(0^{-})<0$$ or vice - versa.
• But this is what I got no? So why there is no point on the graph? – violettagold Nov 24 '18 at 11:08
• You have got this but your conclusion is wrong – Akash Roy Nov 24 '18 at 11:19
• Hey the point is an inflection point , view it closely by zooming it out, you can check it then. (1,2) is an inflection point – Akash Roy Nov 24 '18 at 11:22
• Why is my answer downvoted? – Akash Roy Nov 24 '18 at 11:25
• I thought so too but look at the graph, it shows there is no inflection point. – violettagold Nov 24 '18 at 11:30
You’re right, the graph changes its concavity at at exactly the point $$x$$ where $$f’’(x) = 0$$
such that $$f’’(x_1) > 0$$ results in concave upward and $$f’’(x_1) < 0$$ results in concave downward.
$$f’’(x) = 0 \implies 6x-6 = 0 \implies x = 1$$
Which gives $$y = 2$$, hence the point is $$(1, 2)$$.
If you notice, the tangent’s slope begins to increase (become less negative and eventually positive at $$x > 2$$) after the inflection point.
If it's not easy to vizualize it that way, take a look at https://www.desmos.com/calculator/4zu7w0kmd2 and see how the behavior of the tangent line and the graph's concavity both change at point $$(1, 2)$$.
• You are confusing the notion of an inflection point and a local extremum (maximum or minimum) – ip6 Nov 24 '18 at 11:26
• Sorry, I misunderstood the question for a second, I’ve edited it. (I didn’t read through the part which confused the OP.) – KM101 Nov 24 '18 at 11:34
• I’ve removed the downvote – ip6 Nov 24 '18 at 11:35
• Thank you very much for the help! – violettagold Nov 24 '18 at 12:04
An inflection point is where the 2nd derivative changes sign. This is where the curvature changes from concave downward to concave upward or vice versa. It’s where the tangent changes sides.
You have correctly calculated the position of the inflection point - note the inflection point should not be confused with a maximum or minimum.
Draw some tangents on your graph and see what happens around the point you identified.
• I am sorry for bothering but can you give me an example, please? – violettagold Nov 24 '18 at 11:31
• Cup your left hand around the shape of the left hand side of the curve - your hand will be palm down. Cup your right hand around the shape of the right hand part of the curve - your hand will be palm up. In the middle, where your fingers meet is a point where the curvature switches. This is the inflection point. And you found it correctly!! – ip6 Nov 24 '18 at 11:42
• Now @violettagold you should upvote the answers , – Akash Roy Nov 24 '18 at 12:01
• Thank you very much for the help! – violettagold Nov 24 '18 at 12:03
• Your first statement is wrong, consider $f(x)=x^4$. – Michael Hoppe Nov 24 '18 at 12:37
I zoomed the central part (for you to see it more clearly).
Inflection point PI is at the crossed reticles $$(1,2)$$ red lines intersection.
To the right of PI water holds and at left water spills in the cubic curve and so your checked second derivative signs are OK.
The software does not mark the point automatically that is all. The way to recognize an inflection point graphically is that the curve is locally straight, you are neither turning to your right nor to your left as happening elsewhere. This happens when you are asked to drive along an $$S$$ shaped figure of $$8$$ curve at center that is your PI.
Note the relationships between the first and second derivatives:
$$\hspace{5cm}$$
1) $$f'<0, f''>0$$ - the function is decreasing at an increasing rate;
2) $$f'>0, f''<0$$ - the function is increasing at a decreasing rate;
3) $$f'<0, f''<0$$ - the function is decreasing at a decreasing rate;
4) $$f'>0, f''>0$$ - the function is increasing at an increasing rate.
Now note that the function $$f(x)=x^3-3x^2+4$$ is decreasing at an increasing rate in $$(0,1)$$ and decreasing at a decreasing rate in $$(1,2)$$. And at the point $$x=1$$, the function changes its rate of decrease.
$$\hspace{4cm}$$
Be careful, $$f''(x)=0$$ does not imply the point of inflection (e.g. $$f(x)=x^4$$), so it is a possible inflection point. The function must keep decreasing or increasing around the point of inflection. In the above link you can also see the animated graph of tangent line. | 2019-11-18T23:55:50 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/3011414/wrong-inflection-point?noredirect=1",
"openwebmath_score": 0.7070115208625793,
"openwebmath_perplexity": 346.14308775329636,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.981453437115321,
"lm_q2_score": 0.8723473813156295,
"lm_q1q2_score": 0.8561683357507741
} |
https://gowers.wordpress.com/2008/02/01/a-paradox-in-probability/ | This is the first of what I hope will be several posts related to the course I am giving this term on probability.
The following is a well-known paradox. You are presented with two envelopes and told that one contains a sum of money and the other contains twice as much. You are invited to choose an envelope but are not told which is which. You choose an envelope, and are then given the chance to change your mind if you want to. Should you?
One argument says that it cannot possibly make any difference to the expected outcome, since either way your expected gain will be the average of the amounts in the two envelopes (so the expected change by switching is zero). But there is another argument that goes as follows. Suppose that the amount of money in the envelope you first choose is $x$. Then the other envelope has a 50% chance of containing $2x$ and a 50% chance of containing $x/2$, so your expectation if you switch is $5x/4$—so you should switch.
I tried this out for real in my first lecture, and the student who was given the choice decided to switch. Rather irritatingly, he got more money as a result. Of course, the second argument is incorrect, but the reasons are somewhat subtle. My purpose in putting up a post about it is not so much to invite solutions to the paradox as to see whether it prompts anyone to give me their favourite probabilistic paradoxes. (I’ve just done Simpson’s paradox, so that one wouldn’t be new.)
### 23 Responses to “A paradox in probability”
1. .mau. Says:
it’s not really a paradox, but I like the following scenario:
I toss a fair coin, until it comes up tails. If this happens at the first toss, you’ll win 1 euro; if it happens at the second toss (so the outcome is HT), you’ll win 2 euro; if it happens at the third toss (HHT), you’ll win 4 euro, and so on, always doubling your earnings every time the coin shows heads.
2. gowers Says:
I like that one too. I learned recently that it is called the St Petersburg paradox because it was discussed by Daniel Bernoulli when he was staying in St Petersburg. And it was formulated by one of the other Bernoullis in 1713. (All this information I got from proofs of articles that will be in the Princeton Companion to Mathematics.) I have a simple variant of it: would you pay ten thousand dollars for a never-to-be-repeated chance of one in 1,000 of making 100 million dollars?
3. Lior Says:
Surely the Monty Hall problem should be mentioned. I think the martingale betting method is different from the “long odds” question, which has more to do with the non-linearity of utility as a function of money.
Lior
4. d Says:
I’ve always been partial to: there are three doors and behind one is a prize, you pick a door but don’t open it, then i open one of the doors you did not pick and reveal there is no prize behind it, you can either stick with your original choice or switch…
5. Emmanuel Kowalski Says:
There was a very nice discussion of the two-envelope problem in the American Math. Monthly. 111 (2004), 348–351 (by Samet, Samet and Schmeidler), where a solution is presented showing that, essentially, it is not possible for two r.v. X and Y to be such that X and Y are both independent of the “ranking” R which takes value 1 (say) if XY, with same probability 1/2.
6. .mau. Says:
I agree that in many paradoxes it is utility, rather than probability, that is involved.
Another family of not-quite-paradoxes is related to Bayesian probability, and it is especially important for doctors. A typical example run as follows: I am having a test to check for AIDS. If I am infected, the test always shows it; if I am not infected, there is one chance out of ten that the test is wrong. Only 1% of people in my nation is infected, and I did not do anything particular. If the test says I got AIDS, what is the probability of actually having it?
I am not sure, however, if this is what you are searching for.
7. David Says:
OK, I’ve heard about the two envelop probeme before and obviously the expectation value argument is wrong but how can one see that it is erroneous? What is the subtle reasoning?
8. Emmanuel Kowalski Says:
The last part of my comment should read:
“the “ranking” R which takes value 1 (say) if X> Y and 0 if X < Y, with same probability 1/2”.
9. deepc Says:
If I am not wrong, there is nothing “wrong” with the 2nd argument. If the first envelope contains \$x\$, then w.p. 1/2 on switching he gets \$x\$ more in expectation putting him at \$3x/2\$ which *is* the average of \$x\$ and \$2x\$;
and w.p. 1/2 he loses \$x/2\$ giving him in expectation \$3x/4\$ which *is*
the average of \$x\$ and \$x/2\$.
In both cases, his expected gain is the half of the sum of the amounts in the two envelopes.
10. Yaroslav Bulatov Says:
Dr.Kowalski, why is that a solution to the two-envelope paradox? You could have probability <50% of other envelope containing more money, but still get higher expected return from switching. There’s an example of such prior in Dieter http://personal.lse.ac.uk/list/PDF-files/envelope-paradox.PDF
BTW, is a preprint of Samet/Samet paper available online?
11. Emmanuel Kowalski Says:
I don’t know if the preprint is available online. The title is: “One observation behind two-envelope puzzles”; searching for this, I found at
http://ideas.repec.org/p/wpa/wuwpga/0310004.html
what seems to be an earlier version of it.
And I had simplified the conclusion of their work: there is no assumption that the ranking gives probability 1/2 to the two possibilities, only that the ranking is no constant (i.e. that neither X> Y nor X< Y holds all the time).
12. JB Says:
Gnedenkko´s book on Probability(chelsea publ) presents what he calls “Bertand Paradoxes” of geometric probabilty: when asked to find the probability that a straight line will intersect a unit disk leaving a chord of lenght 1/2 or higher, he gives 3 “arguments” obtaining the answers 1/2, 1/3, 1/4. Also, S. Ross “Afirst course on probability” or something like that has interesting examples involving infinite sets: given a (huge) bag and balls numbered 1,2, … do the following:
– 1 minute before 12, put balls 1- 10 in the bag and take ball #1 out.
-1/2 minute before 12 add balls 11-20 and take ball 2 out.
……….. and so on …………..
Q1) What happens at 12? bag is empty
Q2) If the ball taken out is selected randomly? prob 1 the bag will be empty at 12.
These are more curious(to me at least) than important but helped me keep some students awake.
13. mmm Says:
Another version of the paradox is:
Suppose you and your friend play a game. In the game, both of you check the amount of money in your wallet, and the one with more money gives all that money to the other. Now you think as follows: Suppose i have x amount
in my wallet. If my friend has >x, i gain more than x. If he has <x, i lose x.
And assume that both are equally likely to win(by symmetry). So your expectation is positive. But the same is true for him, whereas you both can’t have positive expectation as you win as much as he loses. Then what is wrong with the argument?
14. kla Says:
Even though I know this wasn’t supposed to be a discussion on the solution to the paradox, I can’t help it.
I do not know a lot about probability, but I think this paradox can be resolved very simply. It does not matter how (by which distribution) the sums in the envelopes have been determined. All we know is that we are given the information “there are two envelopes, one contains amount A and the other contains amount 2A”. This is something which sounds quite similar to, but is very different from the information “you may open one envelope. If you find amount A in it, you ‘ll be given the possibility to open a second one which will contain amount A/2 or 2A, each with probability 1/2”. Let us call these situations (1) and (2). In situation (1) (the one we are really considering) you open an envelope. It contains amount X and we don’t know if X=A or X=2A. Both are possible with equal probability 1/2. If we change, with probability 1/2 we go from A to 2A, or from 2A to A. No gain or less is to be expected and obviously, the same holds if we do not change.
On the other hand if we are in situation (2), where the amount in the next envelope actually depends on the amount in the first one (that’s the whole point: it does not in situation (1)!!), then naturally one has to go for the second one. If initially one has found amount x, then by iterating this procedure, our obtained amount will do a random walk on x.2^Z (Z={integers}).
15. mabs Says:
Regarding the paradox presented, i believe its solution is, at least in this case, quite straightforward: One envelope contains \$x\$ money, the other \$2x\$. We can initially choose each with probability 1/2, therefore our expected gain is \$x+x/2=3x/2\$. Afterwards, when we consider switching envelopes, the reasoning proceeds thus: I don’t know how much money is in my current envelope, so the other one can either have \$x\$ or \$2x\$, each with probability 1/2. In the first case, i’m bound to lose an amount \$x/2\$ relative to my current expected value (\$3x/2\$); in the second case, i’ll win \$x/2\$. The expected gain, therefore, is \$1/2*(-x/2)+1/2*(x/2)=0\$, as intuition predicts.
Reworking the example to have amounts \$x\$ and \$kx\$ in the envelopes, with \$k\$ any constant other than 2, works as well.
Now, why does the analysis in the problem statement appear paradoxical? The analysis is certainly correct (refer to the calculation in the 3rd paragraph of the post); it’s just that it’s an analysis for a different problem. Consider:
Suppose i have 10 dollars. A friend offers me the following proposition: I give him my 10 dollars, and in return he will flip a coin. If it comes out heads, he’ll give me 5 dollars; tails, and he’ll give me 20 dollars (the \$x/2\$ and \$2x\$ envelopes, respectively). Now, should i accept his offer? Of course! As the calculation in the post shows, my expected gain is 5/4*10 dollars, minus my original 10 dollars, so i come out winning 2.5 dollars on average.
In conclusion, the paradox stems from applying the incorrect analysis to the problem. As we mostly try to find problems in the analysis itself (that is, we neglect to see if the analysis is really meant to solve THIS problem), but the development of the analysis is indeed correct, we get the “What the hell?” feeling.
16. mabs Says:
Well, apparently the LaTeX thingy doesn’t work as i thought. Never tried to use it before, and couldn’t be bothered to read the instructions. So, sorry… In my post above, the \$…\$ things are supposed to be formulae in LaTeX syntax.
Another thing, this time regarding the paradox proposed in the reply above by mmm: It is assumed that, by ‘symmetry’, both participants have the same chance to win, thus the paradox. The assumption is, of course, wrong. You would only have true symmetry if you both had the same amount of money in your respective wallets at the start.
The relative amounts of money each player has is critical to deciding who wins, and it is implicit in the problem statement that, lacking knowledge of how much more (or less) money i have than the other person, i can assume that i occupy a central (‘symmetric’) position in the overall distribution of money contained in people’s wallets in general.
Now, that assumption is obviously at fault. If i have no money at all, i’m guaranteed never to lose. And if only few people carry more than 100 dollars at once in their wallets but i happen to have 150 right now, i’m more than likely to lose.
So, your expected gain depends on how much money you’re carrying in your wallet right now, and you are certainly aware of that amount. And even if you had no way of knowing the general distribution of money inside people’s wallets, you couldn’t claim an expected win based on ‘symmetry’. You’d have to say that your expected gain is indeterminate.
17. m Says:
not sure it would be useful but
there is a nice recent book by Szekely on paradoxes in probability theory
Szekely G.J. Pcaradoxes in probability theory and mathematical statistics,
it is a large listing of paradoxes with references to literature.
18. nicolas Says:
Ok so I know the point of this was not the paradox itself but to hell with it.
The crux of the paradox lies for me in that x itself is a random variable, and we dont recognize it as such when supposedly taking expectancy to compare outcomes : after taking unconditional expectancy, which is supposed to be a number, we end up with x, which is a random variable. This is not possible, and we have been fooled.
The way it is presented, we are considering the case where \$x=2a\$, and say the other envelope has a 1/2 proba of having \$2x=4a\$. this case do not exist and we know so, hence it should not appear in our computation…
So we take an envelope, call its value \$x\$. the other envelope contains either \$2x\$ if x=a, or \$x/2\$ if \$x=2a\$
the expectancy of it value is \$3/2a\$. the expectancy of the other envelope is \$3/2a\$, there is no incentive to switch.
19. M-2: Two Probability Paradoxes « Concrete Nonsense Says:
[…] original post […]
20. Sohbet Says:
Thanks…
Not sure if anyone posted it above, but I think St. Petersburg paradox is a great one to do….
Here’s a lesser known problem; superficially similar to the two-envelope paradox, but is actually not. Two people each pick a number (whether randomly, haphazardly, or otherwise); each writes it down on a piece of paper which is then placed face down. Person A, say, (who only knows their own choice, but not that of the other) has to guess whether the other number, which, to remind you, is hidden from them, is larger (or not) than theirs. The numbers, by the way, can be any two numbers; not necessarily integers, and may be positive or negative etc. The question: is there a strategy whereby the player (A) can guess correctly with probability greater than 0.5 ? In other words, is there a strategy whereby this is a game with positive expectation? Surprisingly, there is such a strategy !
23. bigmoneyoutofpolitics Says:
I hope this thread is still live. Referring to the argument that “your expectation if you switch is 5/4”, you write:
“Of course, the second argument is incorrect, but the reasons are somewhat subtle.”
That is wrong.
You can see this easily by listing the possibilities. The problem is not as simple as you seem to think, but it also has nothing to do with possible prior probability distributions — something that can be easily seen by simply choosing numbers.
There are actually two parts to the problem. Smullyan gives a clear statement of one part, which involves only actual amounts (not expected values), and their modes of designation (see, for example, “Satan, Cantor, and Infinity, pp 189-192). He does not resolve the problem, but stating it clearly is the most important step.
The second part of the problem does involve expectation values. This corresponds to your “second argument”. The last part of Smullyan’s account, on p 192 gives a compelling statement similar to yours, but again does not resolve the problem.
I don’t know whether Smullyan ever figured out the solutions but, in my view, his work on the problem is the most important contribution in print because he shows how the problem splits into two parts.
By the way, your system forces me to use the name of an old website of mine, instead of my real name. You use some extremely intrusive software, at the level of Google. | 2018-06-23T11:47:11 | {
"domain": "wordpress.com",
"url": "https://gowers.wordpress.com/2008/02/01/a-paradox-in-probability/",
"openwebmath_score": 0.8131545782089233,
"openwebmath_perplexity": 720.3603668181725,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9814534349454033,
"lm_q2_score": 0.8723473796562744,
"lm_q1q2_score": 0.8561683322292724
} |
https://math.stackexchange.com/questions/2371563/is-there-a-fundamental-reason-to-expect-e-to-appear-in-this-probability-questi | # Is there a fundamental reason to expect $e$ to appear in this probability question?
I came across the following question, whose answer is $e$. I was sort of amazed, since I didn't see a reason why $e$ should be making an appearance. So, phrasing the main question in the title another way: should it have been possible to predict that at least my answer should involve $e$ in some nontrivial way?
Question: Suppose you draw random variables $x_i \in [0,1]$, where the $x_i$ are uniformly distributed. If $x_i > x_{i-1}$, we draw again, and otherwise, we stop. What's the expected number of draws before we stop?
Solution: The probability that we take $n$ draws is given by $\frac{n-1}{n!}$ since there are $n!$ ways to order $x_1,\ldots, x_n$, and exactly $n-1$ choices for the placement of $x_{n-1}$ in the ordering $x_1 < x_1 < \cdots < x_{\widehat{n-1}} < x_n$. That is, for it to take precisely $n$ draws, we need $x_1 < x_2, x_2 < x_3, \ldots, x_{n-2} < x_{n-1}$ but then $x_n < x_{n-1}$. Thus, the expected value is given by $$E(\text{number of draws}) = \sum_{n = 2}^\infty \frac{1}{(n-2)!} = \fbox{e}$$
P.S. It's also possible I simply made a mistake, and if that's the case please point it out and I can edit or delete the question accordingly.
• See also this question. – John Bentin Jul 25 '17 at 20:14
• The number of months in a year is very close to $12\dfrac1e~.~$ Coincidence ? – Lucian Jul 25 '17 at 20:16
• Another combinatorial problem where $e$ pops up is the number of derangements in a card deck, en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle – An aedonist Jul 25 '17 at 20:18
• Does it surprise you that $\sum \frac1{n!}$ comes up in a probability question? If not then it shouldn't surprise you that $e$ comes up. What should surprise you is that $e=\sum \frac1{n!}$ in the first place. – Gregory Grant Jul 25 '17 at 20:25
• @DerekAllums Yeah if you take $\sum \frac1{n!}$ to be the definition of $e$ then it shouldn't surprise you that $e=\sum \frac1{n!}$ – Gregory Grant Jul 25 '17 at 20:40
First: I would take a slightly different route, but that's a matter of taste: Use that $$E[X]=\sum_{n=0}^\infty nP(X=n)=\sum_{n=0}^\infty P(X> n).$$ Similar to your reasoning, $X> n$ occurs iff (up to almost impossible equalities) $x_1<x_2<\ldots <x_n$, which happens with probability $\frac 1{n!}$. Hence, (again) $$\tag1 E[X]=\sum_{n=0}^\infty\frac1{n!}=e.$$
But regarding your main question: Could we expect $e$ to raise its head? Well, perhaps. The problem is about random order, hence about permutations, and (hand-wave) $e$ very often occurs in the context of permutations - which is of course owed to it being the sum if reciprocals of factorials. Then again, before one follows this thought further, one has already written down $(1)$ ...
I'll give a different way of approaching the problem that shows another way that $e$ can arise.
As with Hagen von Eitzen's answer, let $X$ be the random variable that represents the number of draws.
Consider the closely related problem where $x_i \in \{1, 2, \ldots, N\}$ instead. At most $N + 1$ draws can happen in this case. We can calculate the expectation using the binomial theorem.
$$E[X] = \sum_{k=0}^N{P(X > k)} =\sum_{k=0}^{N}\frac{N\choose k}{N^k} = \left(1 + \frac1{N}\right)^N$$
Here, $N \choose k$ is the number of strictly increasing ways of choosing $x_1, \ldots, x_k$ which is then divided by the total number of ways of choosing them to get the probability.
Of course then, the expectation approaches $e$ in the limit as $N \to \infty$. (I don't think this is a particularly good way to solve the problem, as one still has to justify how this limit is applicable, but hopefully it provides some insight into the appearance of $e$ regardless.)
As for the fundamental question, I don't really see a way of knowing that $e$ will appear until it has more or less appeared, but $e$ does seem to arise here quite naturally, so maybe it should not be so surprising in general when $e$ arises in a combinatorial context.
• Very cool - and I suppose this further illustrates the point that I shouldn't be surprised by the appearance of $e$, if my definition of $e$ is, for example, $\lim_{n \to \infty} \left(1+\frac{1}{n}\right)^n$. – Derek Allums Jul 26 '17 at 13:04 | 2019-12-10T00:35:07 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2371563/is-there-a-fundamental-reason-to-expect-e-to-appear-in-this-probability-questi",
"openwebmath_score": 0.9163097739219666,
"openwebmath_perplexity": 191.81366423795347,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9814534333179648,
"lm_q2_score": 0.8723473779969194,
"lm_q1q2_score": 0.8561683291810009
} |
http://www.lofoya.com/Solved/1481/when-processing-flower-nectar-into-honeybees-extract-a | # Moderate Percentages Solved QuestionAptitude Discussion
Q. When processing flower-nectar into honeybees' extract, a considerable amount of water gets reduced. How much flower-nectar must be processed to yield 1kg of honey, if nectar contains $50%$ water, and the honey obtained from this nectar contains $15%$ water?
✖ A. 1.5 kgs ✔ B. 1.7 kgs ✖ C. 3.33 kgs ✖ D. None of these
Solution:
Option(B) is correct
Flower-nectar contains $50\%$ of non-water part.
In honey this non-water part constitutes $85\% (100-15)$.
Therefore $0.5X$ Amount of flower-nectar = $0.85X$ Amount of honey = $0.85\times 1$ kg
Therefore amount of flower-nectar needed
$=\left(\dfrac{0.85}{0.5}\right)\times 1$
$=1.7$ kg
## (7) Comment(s)
Babar
()
50% of 100g is 50 g. if we are to get 1 kg honey we have to have 2 kg raw material because there is 50% water. there are also 15% additional water contents so we have to have more than 2 kg raw material. the given answers are wrong.
Azex
()
Doubt.
Extraction process--> Nector(+50%water) --> Honey(+15%water)--> Honey (pure).
End result is 1 Kg honey.
That means if we go one step up, before its extraction it will contain 15% water as well. So now total quantity:
X-15% of X = 1kg honey.
==> .85X = 1
X= 1/.85 ~ 1.176 Kg (Honey plus water)
Now this mixture was extracted from Nector-water mixture.
let us assume it is extracted from X kg of flower nector.
Therefore,
X- 50% of X = 1.176
.5X = 1.176
=> X = 2.352 Kg
Amit
()
i also got this answer but it is correct or not?
PPatel
()
Are there any other different method to solve this question ?
Param
()
As of now no other substitute becoz it is already simple. Let's make it simpler with some verbal concepts.
1. Honey in Nectar = Honey in honeybee = Honey we got. i.e no loss of honey anywhere only the proportion of water kept changing at different stage.
2. The value of honey at all stages is equal . so calculate the value at all three stages at keep them equal.
3. First stage - Honey is 50% of something.say, Value= 0.5X
4. Second stage - Honey is 85% of something. say, Value =0.85X
5. Last stage - Honey is 1 Kg. (Given)
so, 0.5X=0.85X=1kg. X=1.7 Kg (Pretty Obvious will me more than the honey)
Hope it helps.
Salman
()
this is a good place for the preparation of aptitude tests.
Khirodsaikia
()
Please, include all the subjects which are important for ssc exam | 2017-10-17T11:31:39 | {
"domain": "lofoya.com",
"url": "http://www.lofoya.com/Solved/1481/when-processing-flower-nectar-into-honeybees-extract-a",
"openwebmath_score": 0.5249518752098083,
"openwebmath_perplexity": 4781.36100749182,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9814534338604443,
"lm_q2_score": 0.8723473730188542,
"lm_q1q2_score": 0.8561683247684924
} |
http://citylinker.it/otsg/proof-by-contradiction-examples-and-solutions.html | # Proof By Contradiction Examples And Solutions
It is a logical law that IF A THEN B is always equivalent to IF NOT B THEN NOT A (this is called the contrapositive, and is the basis to proof by contrapositive), so A ONLY IF B is equivalent to IF A THEN B as well. Here I introduce you to, two other methods of proof. Therefore n is even. I am looking for some examples of when proof by contradiction is used in a problem with more than one case. Proofs and refutations: standard techniques for constructing proofs; counter-examples. A number of things could be wrong. These videos go through the basics of each of the topics with so. For a more detailed explanation, please first read the Theory Guides above. This contradicts the assumption. METHODS OF PROOF 74 2. Thus admitting one solution gives rise to an infinite descent, so there can be no solutions. Use proof by contradiction to show that for all integers n, 3 n + 2 is not divisible by 3. , p ⇒ q is proved. The three forms are (1) (Direct) If n2 is even, n is even. This method assumes that the statement is false and then shows that this leads to something we know to be false (a contradiction). Proof root is irrational by contradiction (this is mentioned. 7 pg 91 # 27. Apr 27, 2020 - CA Geometry: Proof by contradiction Video | EduRev is made by best teachers of. Lecture sheet; version with solutions. methods of proof and reasoning in a single document that might help new (and indeed continuing) students to gain a deeper understanding of how we write good proofs and present clear and logical mathematics. This is true. That's what proofs are about in mathematics and in computer science. If logic is inconsistent then proof by contradiction is still very much a valid rule of reasoning, but so is its negation, and the rule which says that from $1 + 1 = 2$ we can conclude that you are the next pope. Then n = 2k for some k. contradiction proofs tend to be less convincing and harder to write than direct proofs or proofs by contrapositive. Finding a contradiction means that your assumption is false and therefore the statement is true. This and along with the direct proof on Friday complete an example of proof of an "if and only if" statement. The proof on the board. It is usually not as neat as a two-column proof but is far easier to organize. Then n2 = 2m + 1, so by definition n2 is even. Then (x + y)(x y) = 1, so x y and x + y are divisors of 1. In these cases, when you assume the contrary, you negate the original. For example, to prove that ot all triangles are obtuse", we give the following counter example: the equilateral triangle having all angles equal to sixty. 2 Equivalent Statements. Do the same for an iterative algorithm. Let M = N + 1. (Direct) If n2 = 0, then n = 0 and n is even. Here are a couple examples of proofs by contradiction: Example. It is not clear how to prove it directly since we can not con-. 4 EXAM 2 SOLUTIONS Problem 22. , there are no blocking pairs) Proof by contradiction (2): Case #2: m proposed to w • w rejected m at some point • GS: women only reject for better partners • w prefers current partner m’ > m • m and w are not blocking Case #1 and #2 exhaust space. In most traditional textbooks this section comes before the sections containing the First and Second Derivative Tests because many of the proofs in those sections need the Mean Value Theorem. The command \newtheorem{theorem}{Theorem} has two parameters, the first one is the name of the environment that is defined, the second one is the word that will be printed, in boldface font, at the beginning of the environment. If n+1 objects are put into n boxes, then at least one box contains two or more objects. Example -1 Show that at least four of any 22 days must fall on the same day of the week. Proof by Contradiction is another important proof technique. The contrapositive of the above statement is: If x is not even, then x 2 is not even. basics, and foundations Discrete Math - 17 Direct Proof This is the first of several videos exploring methods of proof. " Sir Arthur Conan Doyle. Copious Examples of Proofs 19 Rewrite it in each of the three forms and prove each. methods of proof and reasoning in a single document that might help new (and indeed continuing) students to gain a deeper understanding of how we write good proofs and present clear and logical mathematics. Proof by Contradiction. An example is "Prove that the product of two nonzero real numbers is nonzero. A direct proof will attempt to lay out the shortest number of steps between p and q. Chapter 3 Foundations of Geometry 1: Points, Lines, Segments, Angles 3. I am looking for some examples of when proof by contradiction is used in a problem with more than one case. An alternative proof is obtained by excluding all possible ways in which the propositions may fail to be equivalent. John 19:14)—Was Jesus crucified in the third hour or the sixth hour? Problem: Mark’s Gospel account says that it was the third hour (9 a. But if a/b = √ 2, then a 2 = 2b 2. Example: A Diophantine Equation. 3 Review the proof techniques on page 116−−118 Here is a result that is proved by three different proof techniques. ----- ----- EXAMPLE 2 Give a proof by contradiction of the theorem "If 3n + 2 is odd, then n is odd. Thus admitting one solution gives rise to an infinite descent, so there can be no solutions. This problem is taken from the Putnam competition and is a good example for demonstrating logical thinking and mathematical proof. Proof by contradiction is often used when you wish to prove the impossibility of something. ” Solution: We give a proof by contradiction. This is also known as proof by assuming the opposite. 8 (a) Prove that if n is even, then (3n)2 is even. Example 1 Prove there is no largest prime, i. In this video we will focus on direct proof by assuming "p" is true, then Discrete Math Section 1. Thus the quality of your solution is at least as great as that of any other solution. Examples of a contradiction include an anti-absorbent sponge, jumbo shrimp, and painful pain injections. This document draws some content from each of the following. Solution: Suppose √2 is rational. Proof: Suppose not. Fill in the truth table for ((A implies B) and (B implies C) implies (A implies C)). This is the technique of proof by maximal counterexample, in this case applied to perfect matchings in very dense graphs. Thus, 3n + 2 is even. 4- Bacic Proof Methods I- Direct Proof, Proof by Cases, and Proof by Working Backward In this section we will introduce specific types or methods of proof of mathematical statements. If logic is inconsistent then proof by contradiction is still very much a valid rule of reasoning, but so is its negation, and the rule which says that from $1 + 1 = 2$ we can conclude that you are the next pope. If we wanted to prove the following statement using proof by contradiction, what assumption would we start our proof with?. Prove that Proof: By contradiction, we obtain Suppose , then (given). In logic, it is a fundamental law- the law of non contradiction- that a statement and its denial cannot both be true at the same time. Just as Gillman’s proof has variations, which are based on grouping larger collections of terms, so there are variations on Cusumano’s. Example: A Diophantine Equation. A proof by contradiction also known as an indirect proof, establishes the truth of a given proposition by the supposition that it is false and the subsequent drawing of a conclusion that is contradictory to something that is proven to be true. Then you manipulate and simplify, and try to rearrange things to get the right. The statement P1 says that x1 = 1 < 4, which is true. Relation between Proof by Contradiction and Proof by Contraposition 2) proof by contradiction, you suppose there is an x in D such that P (x) and ~Q (x). Section 4-7 : The Mean Value Theorem. We argue by contradiction. A classic proof: $\sqrt{2}$ is irrational. The solution (c7=8 etc. 2 Quadratic Inequalities. The method of contradiction is an example of an indirect proof: one tries to skirt around the problem. Which proof technique? Direct proof –express x2 as 2k for some k, i. What does this language mean? In Example 1, we are saying that the inequality 2n >n2 holds for each choice of the. Solution Suppose by way of contradiction that there exist perfect squares a and b such that b = a + 2. This is a contradiction because there are a total of N objects. Examples In mathematics Irrationality of the square root of 2. Proof by Contradiction. 2 More Methods of Proof A proof by contradiction establishes that p is true by assuming that p is false and arriving at a contradiction, which is any proposition of the form r ^:r. Thus the quality of your solution is at least as great as that of any other solution. Chapter 6: Formal Proofs and Boolean Logic The Fitch program, like the system F, uses "introduction" and "elimination" rules. The more work you show the easier it will be to assign partial credit. The proof is carried out by using the procedure outlined in subsection H1. Example: A Diophantine Equation. So let's just assume that a rational times an irrational gives us a rational number. But this one it true because for x<0 x+ 1 x <0 and 0 <2. Example of a Proof by Contradiction Theorem 4. via self-evident rules, and however, in other areas of human activity, the notion of a "proof" has a much wider interpretation [1]. Since we have shown that Sq \Fis true, it follows that the contrapositive T \qalso holds. 2 Incorrect variable use: don’t use the same variable name twice for two different variables. 9 ∀x [(Cube(x) ∧ Large(x)) ∨ (Tet(x) ∧ Small(x))] ∀x [Tet(x) → BackOf(x, c)]. Then p 2 is a rational number, so it can be expressed in the form p q, where pand qare integers which are not both even. Solving it explicitly came later. ” Then use other things you know to try to reach a. Mathematical Proofs: A Transition to Advanced Mathematics, 4th Edition introduces students to proof techniques, analyzing proofs, and writing proofs of their own that are not only mathematically correct but clearly written. If it were rational, it would be expressible as a fraction a/b in lowest terms, where a and b are integers, at least one of which is odd. Sometimes the negation of a statement is easier to disprove (leads to a contradiction) than the original statement is to prove. Example of a constructive proof: Suppose we are to prove 9n2N;nis equal to the sum of its proper divisors: Proof: Let n= 6. Since a statement is true or false, all statements therefore belong in the set of true statements. The proof is by contradiction. Thus, It is not raining. Before looking at this proof, there are a few definitions we will need to know in order to. It depends on personal opinion and interpretation what a proof by contradiction is and whether Euclid's proof belongs to this category. Closed Form Identities 6 5. particlar example graph, a minimax path is P = 1 2 5 8 11 12 with maximum altitude 5. The material is organized around five types of thinking: logical, relational, recursive, quantitative, and analytical. Proof by Contradiction aka reductio ad absurdum, i. Problem 27 (due Fri 4/3): Prove (using proof by contradiction) that for any sets A and B, we have A$$B nA) = ;. Proof: (direct proof) Assume that n is an even integer. Then, the book moves on to standard proof techniques: direct proof, proof by contrapositive and contradiction, proving existence and uniqueness, constructive proof, proof by induction, and others. Some examples of how to define a recursive function. solution's quality. We would rather proof the contrapositive: x<0 implies x+ 1 x <2. In this case, there are in nitely. The only way out of this situation is that the assumption was wrong. Solution:. identical to that which must follow in your solution) follows. The first known proof is believed to have been given by the Greek philosopher and mathematician Thales. Main Steps After describing your algorithm, the 3 main steps for a greedy exchange argument proof are as follows:. Another way to write is using its equivalence, which is Example: Given A and B are sets satisfying. H1: Introduction to Proof by Contradiction. Example 11 Show that 3√2 is irrational. Proof by Contradiction: (AKA reductio ad absurdum). In all the elementary examples, there are only two options (eg rational/irrational, infinite/finite), so you assume the opposite, show it cannot be true and then conclude the result. It explores properties of odd, even and consecutive numbers- both numerically and algebraically- and also covers properties of the sum, difference and product of these numbers, again these are explored numerically and algebraically. lm 2=so can assume 2 2 4m l= 22 2ln = so n is even. Thus, the statement is proved using an indirect proof. When proving an IF AND ONLY IF proof directly, you must make sure that the equivalence you are proving holds in all steps of the proof. (b) Prove that the square root of 3 is irrational. Example: A Diophantine Equation. It includes disproof by counterexample, proof by deduction, proof by exhaustion and proof by contradiction, with examples for each. Then n = 2k for some k. " Solution: Let p be "3n + 2 is odd" and q be "n is odd. Unlike the earlier examples, I will not describe the thought process that lead to the proof; in each case, I followed the basic outline on page 7. Cube(b) ∧ a = b 2. Example #2. It explains the standard “moves” in mathematical proofs: direct computation, expanding definitions, proof by contradiction, proof by induction, as well as choosing notation and strategies. So before moving on to the next chapter, let’s try our hand at some informal proofs. ] Suppose not. This resource is designed for UK teachers. " Begin the proof with "Assume that a ≠ 0 and b ≠ 0. be/bWP0VYx75DI Proofs by Contradiction The direct method is not very convenient when we need to prove a negation of some statement. In summary:. You can put this solution on YOUR website! "The product of a non-zero rational number and an irrational number is irrational. Relation between Proof by Contradiction and Proof by Contraposition As an example, here is a proof by contradiction of Proposition 4. 8 (a) Prove that if n is even, then (3n)2 is even. Then P being false implies something that. What does this language mean? In Example 1, we are saying that the inequality 2n >n2 holds for each choice of the. Examples In mathematics Irrationality of the square root of 2. If it leads to a contradiction, then the statement must be true. Example -1 Show that at least four of any 22 days must fall on the same day of the week. Prove by contradiction that there do not exist integers mand nsuch that 14m+ 21n= 100 Proof: We give a proof by contradiction. Worked solutions for some past years Regents Exam. An Indirect Proof. Show the following, and please show all steps: Prove that an odd integer minus an even integer is odd. One well-known use of this method is in the proof that \sqrt{2} is irrational. 2 2 2 4n l= Proof by Contradiction Theorem: is irrational. You can put this solution on YOUR website!. 2 More Methods of Proof A proof by contradiction establishes that p is true by assuming that p is false and arriving at a contradiction, which is any proposition of the form r ^:r. Daniel Solow’s How to Read and Do Proofs begins with the simpler methods of mathematical proof-writing and gradually works toward the more advanced techniques typically presented in an introduction to advanced mathematics. Again, we do not offer this example as the best proof of this fact about even and odd numbers, but rather it is a simple illustration of a proof by contradiction. This works because if \(C$$ is a contradiction and $$\neg P \to. The proof is carried out by using the procedure outlined in subsection H1. This means that the Indirect Proof has been accomplished: by showing that the assumption led to a self-contradiction, one has shown that the assumption was false, and hence that its negation (the conclusion) is true. Type 2: To prove p → q Assume p and ¬q are true. Contradictive Proof Example Prove the following: No odd integer can be expressed as the sum of three even integers. Either the triangles are congruent or they are not. 1 Conjunction rules Conjunction Elimination (∧ Elim). Thus, 3n + 2 is even. " For example, the set E above is the set of all values the expression 2 nthat satisfy the rule 2 Z. Proof time. The idea of proving by contradiction is: we flrst. 1 Proof by example: a universal statement cannot be proved by giving an example. Example 4 is the classic proof of this kind. I understand the theory and concept of Proof by Contradiction however I don't understand where to begin. A PowerPoint covering the Proof section of the new A-level (both years). 2 Incorrect variable use: don’t use the same variable name twice for two different variables. learn geometry proofs and how to use CPCTC, Two-Column Proofs, FlowChart Proofs and Proof by Contradiction, videos, worksheets, games and activities that are suitable for Grade 9 & 10, examples and step by step solutions, complete two column proofs from word problems, Using flowcharts in proofs for Geometry, How to write an Indirect Proof or Proof by Contradiction. 3 - Proof by Contrapositive proof by contrapositive A proof by contrapositive proves a conditional theorem of the form p → c by showing that the contrapositive ¬c → ¬p is true. Proof for other series. A historical example. Then we discussed an alternative to the direct proof, proof by contradiction. Concepts that you will need to know for the Regents Math - Algebra, Geometry, Measurement, Probability, Statistics, Trigonometry. Print Proof by Contradiction: Definition & Examples Worksheet 1. The next group of rules deals with the Boolean connectives contradiction. Once a mathematical statement has been proved with a rigorous argument, it counts as true throughout the universe and for all time. nThese have the following structure: ¥Start with the given fact(s). A number of things could be wrong. This is really a special case of proof by contrapositive (where your \if" is all of mathematics, and your \then" is the statement you are trying to prove). Suppose that there were some x 2Z so that 2x3 + 6x+ 1 = 0: Re-arranging, this implies that 1 = 2x3 6x = 2( x3 3x): Since x3 3x is an integer, this implies that 1 is even, which is obviously not true. The Proof Page. Proof by contradiction (also known as indirect proof or the method of reductio ad absurdum) is a technique which can be used to prove any kind of statement. Section 4-7 : The Mean Value Theorem. One typical application is to show that a given equation has no solutions. The (Pedagogically) First Induction Proof 4 3. I p divides both x =p1 p2 pk and q, and divides x q, I =)pjx q =)p x q. representative-case proof. It explores properties of odd, even and consecutive numbers- both numerically and algebraically- and also covers properties of the sum, difference and product of these numbers, again these are explored numerically and algebraically. 2 2 2 4n l= Proof by Contradiction Theorem: is irrational. We will use the following well known facts : : 1. then find a logical contradiction stemming from this assumption. CLARK Contents 1. An easy-to-use guide that shows how to read, understand, and do proofs. Call this integer n. Shows how and when to use each technique such as the contrapositive, induction and proof by contradiction. That is, suppose there is an integer n. , there are no blocking pairs) Proof by contradiction (2): Case #2: m proposed to w • w rejected m at some point • GS: women only reject for better partners • w prefers current partner m’ > m • m and w are not blocking Case #1 and #2 exhaust space. Hence, from the proof by cases, (r ∨ s ∨ t) ⇒ q is proved, i. Many of the statements we prove have the form P )Q which, when negated, has the form P )˘Q. 2 Exercise 21) Let n= abbe the product of positive integers a and b. If it were rational, it would be expressible as a fraction a/b in lowest terms, where a and b are integers, at least one of which is odd. It is a logical law that IF A THEN B is always equivalent to IF NOT B THEN NOT A (this is called the contrapositive, and is the basis to proof by contrapositive), so A ONLY IF B is equivalent to IF A THEN B as well. This is a contradiction as x and y should be positive. This page is for the new specification (first teaching 2017): including revision videos, exam questions and model solutions. We obtain the desired conclusion in both cases, so the original statement is true. We have to prove 3√2 is irrational Let us assume the opposite, i. To prove a statement P is true, we begin by assuming P false and show that this leads to a contradiction; something that always false. The idea of the proof is really quite simple. Here we will look at some examples of proofs and non-proofs. PROOFS BY INDUCTION PER ALEXANDERSSON Introduction This is a collection of various proofs using induction. Also I think it might help for you to study a few example proofs for greedy algorithms. Prove that Proof: By contradiction, we obtain Suppose , then (given). 6 Example 1 - Solution Proof: [We take the negation of the theorem and suppose it to be true. For example E ˘ ' 2 n: 2 Z " ˘ ' n : n isaneveninteger " ˘ ' n : n ˘ 2k,k 2 Z ". We have n3 n= (n 1)n(n+ 1). We will use proof by contradiction. Since we have shown that Sq \Fis true, it follows that the contrapositive T \qalso holds. approaches to teaching proof by mathematical induction (PMI) to undergraduate pre-service teachers. 1 Ifyouconsidertheexamplesofproofsinthelastsection,youwillnoticethatsometermsandrulesofinferenceare specifictothesubjectmatterathand. com/site/tlmaths314/ Like my Facebook Page: https://www. If you try to do this, you will find that if you make your hexagon very large, then you can get somewhat close to. A proof that the square root of 2 is irrational Here you can read a step-by-step proof with simple explanations for the fact that the square root of 2 is an irrational number. So let's look hard at the above example. The Mathematician's Toolbox. ()): Assume [a] = [b]. re·duc·ti·o·nes ad absurdum Disproof of a proposition by showing that it leads to absurd or untenable conclusions. We can prove A is not true by finding a counter example. Then b = b1 = b(ac) = (ab)c = [0] c = 0 : But this contradicts our original hypothesis that b is a nonzero solution of ax = [0]. However, the principle of explosion ( ex falso quodlibet ) has been accepted in some varieties of constructive mathematics, including intuitionism. it must be that Bis true, and we have a proof by contradiction. Outline Theorem 2. Proof by Contradiction (Example 1) •Show that if 3n + 2 is an odd integer, then n is odd. Thus, the statement is proved using an indirect proof. Then (x y) = (x + y) = 1. But this is clearly impossible, since n2 is even. Before looking at this proof, there are a few definitions we will need to know in order to. He supposed there were a finite number and showed that that led to an absurdity — just as we’ve done in our examples. Proof by Exhaustion. The "proof" by josgarithmetic" is wrong starting from his second line. If you make an assumption, and that assumption produces a statement that does not make sense, then you must conclude that your assumption is wrong. ] (12) So, there exists p,q such that: v = p 2 w = q 2 (13) And, we have our. (Otherwise, it would be zero everywhere. Proof by contradiction in logic and mathematics is a proof that determines the truth of a statement by assuming the proposition is false, then working to show its falsity until the result of that assumption is a contradiction. Now this is a contradiction since the left hand side is odd, but the right side is even. Because x is positive, we can multiply. Then x = (y z)(y +z). Indirect Proofs. The proof is by contradiction. n2 odd ⇒ n odd For (1), if n is odd, it is of the form 2k + 1. Prove that if aand bare real numbers with aa. The idea behind proof by contradiction is that a statement must be. Navigate all of my videos at https://sites. In this example it all seems a bit long winded to prove something so obvious, but in more complicated examples it is useful to state exactly what we are assuming and where our contradiction is found. Thus the quality of your solution is at least as great as that of any other solution. Part III: More on Proof. lm 2=so can assume 2 2 4m l= 22 2ln = so n is even. Proof is the primary vehicle for knowledge generation in mathematics. State that the proof is by contradiction. Below are several more examples of this proof strategy. For example E ˘ ' 2 n: 2 Z " ˘ ' n : n isaneveninteger " ˘ ' n : n ˘ 2k,k 2 Z ". , the further out you must go for the approximation to be valid within ǫ. A direct proof, or even a proof of the contrapositive, may seem more satisfying. Start of proof: Let \(n$$ be an integer. Assume $$n$$ is a multiple of 3. Example -1 Show that at least four of any 22 days must fall on the same day of the week. Proof by Contradiction Date: 04/29/2003 at 07:07:29 From: Ajay Subject: Proof by Contradiction Is there any specific mathematical theory that states that Proof by Contradiction is a valid proof? E. There can be many ways to express the same set. One Theorem of Graph. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. I Proof by Contrapositive (Indirect Proofs) I Proof by Contradiction I Proof by Cases I Proofs of equivalence I Existence Proofs (Constructive & Nonconstructive) I Uniqueness Proofs Trivial Proofs I (Not trivial as in \easy") Trivial proofs : conclusion holds without using the hypothesis. This is true. Therefore, when the proof contradicts itself, it proves that the opposite must be true. , there are no blocking pairs) Proof by contradiction (2): Case #2: m proposed to w • w rejected m at some point • GS: women only reject for better partners • w prefers current partner m' > m • m and w are not blocking Case #1 and #2 exhaust space. Every integer greater than one has a prime divisor. These videos go through the basics of each of the topics with so. Example2 1. ) X is B Example: Let’s think about an example. \Help! I don’t know how to write a proof!" Well, did anyone ever tell you what a proof is, and how to go about writing one? Maybe not. A Simple Proof by Contradiction Theorem: If n2 is even, then n is even. Then there exists integers a and b. course of its proof. Then n2 = (2k + 1)2 = 4k2 + 4k + 1 = 2(2k2 + 2k) + 1. Watch more videos and sign up for a FREE. Solution We formulate this statement as an. 1 Writing mathematics - Exercise Solutions 3 4. Use the method of proof by contradiction to prove the following statements. In mathematics, a proof by infinite descent is a particular kind of proof by contradiction which relies on the facts that the natural numbers are well ordered and that there are only a finite number of them that are smaller than any given one. Aristotle’s discussion of the principle of non-contradiction also raises thorny issues in many areas of modern philosophy, for example, questions about what we are committed to by our beliefs, the relationship between language, thought and the world, and the status of transcendental arguments. This example illustrates an alternative to using truth tables to establish the equiv-alence of two propositions. Problem Set 4 Solutions Section 3. It depends on personal opinion and interpretation what a proof by contradiction is and whether Euclid's proof belongs to this category. Then this even number N is a multiple of 2. Such examples are called counter examples. Maths Genie - A Level Maths revision page. With such a wide target area, that's often a much easier task. sqrt(2) is irrational is normally proved using contradiction. Villanova CSC 1300 - Dr Papalaskari Proofs, examples, and counterexamples ∃x P(x) For existential statements: • A single example suffices to prove the theorem (constructive proof). Lecture Slides By Adil Aslam 32. 4 Some Words of Advice. Theorem (Euclid): For positive numbers a and b (with a > 2 b) the quadratic equation x^2 + b^2 = a x has a solution. Justify all of your decisions as clearly as possible. For all natural numbers n, n3 nis a multiple of 3. Here are a couple examples of proofs by contradiction: Example. 3 - Proof by Contrapositive proof by contrapositive A proof by contrapositive proves a conditional theorem of the form p → c by showing that the contrapositive ¬c → ¬p is true. Prove that if x is even, then x2 + 3 is odd. A classic proof: $\sqrt{2}$ is irrational. Proposition. Prove that A[B = A\B if and only if A = B. proof in terms of induction. [We must deduce the contradiction. Theorem: There is no greatest integer. If 3n+2 is odd then. Start studying [3] Proofs. A collection of videos that cover most topics on the Leaving Cert Higher Level Maths course. Proof by contradiction, as we have discussed, is a proof strategy where you assume the opposite of a statement, and then find a contradiction somewhere in your proof. Direct Proof: Example Theorem: 1 + 2 +h3 +rÉ + n =e n(n+1. Unfortunately, no number of examples supporting a theorem is sufficient to prove that the theorem is correct. The solution (c7=8 etc. (2) (Contrapositive) If n is odd, n2 is odd. The more work you show the easier it will be to assign partial credit. It is a contradiction of rational numbers but is a type of real numbers. This is a contradiction as x and y should be positive. (Otherwise, it would be zero everywhere. have no common factors (see Chapter 4). The X-Wing prooves that gk8!= 7. 1) using proof by contradiction, one follows an indirect route: derive r Ù Ør, then conclude that (7. The correct proof is this: Let assume that the product of two odd numbers, m and n, is an even number N: N = m*n. 104 Proof by Contradiction 6. Proof: Assume 0 < c < d. I love you and I don't love you. (2) 6) Using proof by contradiction show that there are no positive integer solutions to the Diophantine equation − =. Of course we can't just have HALTS simulate P on input D, since if P doesn't halt, we'll never know exactly when to quit the simulation and answer no. 4 (a) Prove that A ⊆ B iff A∩B = A. Category: Mathematics This interactive excel resource illustrates a number of proofs. Mathematical proof is the gold standard of knowledge. By De Morgans law, we have a jb and a j(b+ 1). Multi-level views of proofs that hide or reveal details as required ─ indispensible for writing longer proofs! Top Symbolic Logic and The Basic Methods of Proof. Alternatively, you can do a proof by contradiction: As-sume that Y is false, and show that X is false. [1 mark] Assume positive integer solutions. Suppose by contradiction that there is a greatest even integer. methods of proof and reasoning in a single document that might help new (and indeed continuing) students to gain a deeper understanding of how we write good proofs and present clear and logical mathematics. X∞ n=1 1 1+ √ n. I said we will do it through a proof by contradiction. A complete chapter is dedicated to the different methods of proof such as forward direct proofs, proof by contrapositive, proof by contradiction, mathematical induction, and existence proofs. Example Theorem: For every integer x, if x2is even, then x is even. Formal Proofs A proof is equivalent to establishing a logical implication chain Given premises (hypotheses) h1 , h2 , … , hn and conclusion c, to give a formal proof that the hypotheses imply the conclusion, entails establishing h1 ∧h2 ∧… ∧hn ⇒c MSU/CSE 260 Fall 2009 6 Formal Proof. There are infinitely many prime numbers. nThese have the following structure: ¥Start with the given fact(s). ] Suppose not. Statement Reason. The simplicity. Example: Parity Here is a simple example that illustrates the method. Simply put, we assume that the math statement is false and then show that this will lead to a contradiction. Example: Prove that if 푛푛 is an integer and 푛푛 3 + 5 is odd, then 푛푛 is even using a. Assume that [~Prove] is true. For proof by contradiction, suppose there are positive integers, greater than one, with no prime divisors. Among 13 people there are two who have their birthdays in the same month. You can find examples of proofs by contradiction in Theorem RREFU, Theorem NMUS, Theorem NPNT, Theorem TTMI, Theorem GSP, Theorem ELIS, Theorem EDYES, Theorem EMHE, Theorem EDELI, and Theorem DMFE, in addition to several examples and. Proof by Exhaustion. Formally, this \indirect proof" method is justifled by the logical equivalence: p · ((:p)! (r ^:r)): Prove for all integers n, if n2 is divisible by 5 then so is n. " For example, the set E above is the set of all values the expression 2 nthat satisfy the rule 2 Z. Villanova CSC 1300 - Dr Papalaskari Proofs, examples, and counterexamples ∃x P(x) For existential statements: • A single example suffices to prove the theorem (constructive proof). Proof by Contradiction. Still, there seems to be no way to avoid proof by contradiction. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. (noun) An example of proof is someone returning to eat at the same restaurant many times showing they enjoy the food. Suppose L were regular. Therefore y = 0, contradicting that it is positive. Both of these methods are called constructive proofs of existence. This means a b is in lowest terms. Proof by contradiction: Assume P(x) is true but Q(x) is false. PRACTICE EXAM 1 SOLUTIONS Problem 1. Example: ! Prove that an integer n is even, if n2 is even. Quiz Proof #4 February 19, 2018 Theorem 1. , everyone in the earth is male But, no number of examples supporting a theorem is sufficient to. , 3 is rational Hence, 3 can be written in the form / where a and b (b 0) are co-prime (no common factor other than 1) Hence, 3 = / 3 b = a Squaring both sides ( 3b)2 = a2 3b2 = a2 ^2/3 = b2 Hence, 3 divides a2 So, 3 shall divide a also Hence, we can say /3 = c where c is some. Solution: By contradiction. I love you and I don't love you. Prove that if x is even, then x2 + 3 is odd. " Indirect Proof (Proof by Contradiction) of the statement: Assume the opposite of what you want to prove, and show it leads to a contradiction of a known fact. Here is a proof found off a very nice math history website. Then we discussed an alternative to the direct proof, proof by contradiction. Prove the statement using a proof by contradiction. Why can't we use one counterexample as the contradiction to the contradicting statement? Example: Let a statement be A where a-->b. Another way to write up the above proof is: Since seven numbers are selected, the Pigeonhole Principle guarantees that two of them are selected from one of the six sets {1,11},{2,10},{3,9}, {4,8}, {5,7},{6}. Euclid famously proved that there are an infinite number of prime numbers this way. n m =2 mn =2 22 2 mn = so m is even. This means that each step in the proof must. Nice introduction to the concept of recursion in terms of programming. I said we will do it through a proof by contradiction. But, from the parity property, we know that an integer is not odd if, and only if, it is. Start of proof: Assume, for the sake of contradiction, that there are integers $$x$$ and $$y$$ such that $$x$$ is a prime greater than 5 and $$x = 6y + 3\text{. Let s = 0 1 in L. Assume the triangles are congruent and reason to a contradiction. Tindle, who. Solution We formulate this statement as an. To prove that L is not a regular language, we will use a proof by contradiction. For starters, let's negate our original statement: The sum of two even numbers is not always even. The idea behind proof by contradiction is that a statement must be. Then, by Lemma 4, there is a such that, up to a subsequence, in. Given: ΔABC is scalene. Solutions to In-Class Problems Week 1, Fri. A proof by contradiction starts with assuming :q (and p). (4) The method of mathematical induction is necessary to prove some theorems that we studied so far in this course. If we wanted to prove the following statement using proof by contradiction, what assumption would we start our proof with?. Suppose for the sake of contradiction there exist a;b 2Z with a 2, and for which it is not true that a - b or a - (b+ 1). Bhaskara's First Proof Bhaskara's proof is also a dissection proof. Proof by contradiction. Bhaskara was born in India. From this assumption, p 2 can be writ-ten in terms of a b, where a and b have no common factor. Then let p be the pumping length given by the pumping lemma. Example 4: Prove the following statement by contradiction: For all integers n, if n 2 is odd, then n is odd. Proof: This is easy to prove by induction. Through step-by-step worked solutions to exam questions available in the Online Study Pack we cover everything you need to know about Proof by Contradiction to pass your final exam. Then, the book moves on to standard proof techniques: direct proof, proof by contrapositive and contradiction, proving existence and uniqueness, constructive proof, proof by induction, and others. If you make an assumption, and that assumption produces a statement that does not make sense, then you must conclude that your assumption is wrong. NEGATION 3 We have seen that p and q are statements, where p has truth value T and q has truth value F. Proof by contradiction "When you have eliminated the impossible, what ever remains, however improbable must be the truth. The classic example is the following proof that the square root of 2 is irrational: 1. We will use simple ideas from algebraic topology to show that there exists such that provides an example to prove Theorem 1. • This amounts to proving ¬Y ⇒ ¬X 1 Example Theorem n is odd iff (in and only if) n2 is odd, for n ∈ Z. Before looking at this proof, there are a few definitions we will need to know in order to. For proof by contradictionsuppose not P: Therefore C and not C; completing the proof. 1provides an optimal solution for the fractional knapsack problem. It turns out we won’t need to resort to classical logic for this theorem, but just to make things easier in our first pass we’ll go ahead and use it. Example of proof by contradiction and more on proof by induction. 2 Proof (by contradiction): Want to prove both m and n are even. Deduce that if the hypotheses are true, the conclusion must be true too. Inequalities. Example 1 In the following videos I show you how to use mathematical induction to prove the sum of the shown series Example 2. Solution We formulate this statement as an. He supposed there were a finite number and showed that that led to an absurdity — just as we've done in our examples. Proofs by Contradiction 2. In addition, the author has supplied many clear and detailed algorithms that outline these proofs. , everyone in the earth is male But, no number of examples supporting a theorem is sufficient to. Specifically, the way they teach both Proof by Contradiction and Proof by Mathematical Induction, two techniques that are vital to any upper level analysis/algebra/geometry class, is phenomenal. 2 Nonconstructive: we do not nd a witness a directly. In the following, I cover only a single example, which combines induction with the common proof technique of proof by contradiction. Lecture Slides By Adil Aslam 32. [We take the negation of the given statement and suppose it to be true. The possible truth values of a statement are often given in a table, called a truth table. However, there is an approach that is vaguely similar to disproving by counter-example, called proof by contradiction. We have and thus. Although a direct proof can be given, we choose to prove this statement by contraposition. For example, the statement "the equation 4x^2-y^2 = 1 has no integer solutions for x and y" has a simple contradiction proof. This document draws some content from each of the following. A proof that the square root of 2 is irrational Here you can read a step-by-step proof with simple explanations for the fact that the square root of 2 is an irrational number. Example from the text: square root of 2 is irrational ; Careful: When using proof by contradiction, mistakes can lead to apparent contradictions. That is, there is a natural number x and natural numbers y and z such that x = y2 z2. I am looking for some examples of when proof by contradiction is used in a problem with more than one case. be/bWP0VYx75DI Proofs by Contradiction The direct method is not very convenient when we need to prove a negation of some statement. Unfortunately, no number of examples supporting a theorem is sufficient to prove that the theorem is correct. The Gödel number of formula \(\forall x_0 ( eg \in x_0 x_0)$$ using (*) is. , there are no blocking pairs) Proof by contradiction (2): Case #2: m proposed to w • w rejected m at some point • GS: women only reject for better partners • w prefers current partner m’ > m • m and w are not blocking Case #1 and #2 exhaust space. Relation between Proof by Contradiction and Proof by Contraposition. First, we'll look at it in the propositional case, then in the first-order case. (3) (Contradiction) If n2 is even and n is odd, then n2 is odd. Indirect Proof is foolproof. Prove that if u is an odd integer, then the equation x2 + x u = 0 has no solution that is an integer. Example 4: Prove the following statement by contradiction: For all integers n, if n 2 is odd, then n is odd. The proof of this corollary illustrates an important technique called 'proof by contradiction'. Proof by contradiction "When you have eliminated the impossible, what ever remains, however improbable must be the truth. What is interesting, is that Atiyah was not directly looking at the Riemann Hypothesis, but was studying something else. TheCartesianProduct8 1. Problem 28 (due Fri 4/10): Prove that x = 3 is the unique solution to x 2 + 9 = 6x. The possible truth values of a statement are often given in a table, called a truth table. This A Level Maths video takes you through a new method of proof called proof by contradiction. EXAMPLE 4 Use of Contradiction in Real Life Use an indirect proof to prove the following statement. The proof of this result provides a proof of the sine rule that is independent of the proof given in the module, Further Trigonometry. The number 2 is a prime number. The third part provides more examples of common proofs, such as proving non-conditional statements, proofs involving sets, and disproving statements, and also introduces mathematical induction. For starters, let's negate our original statement: The sum of two even numbers is not always even. A proof by contradiction might be useful if the statement of a theorem is a negation--- for example, the theorem says that a certain thing doesn't exist, that an object doesn't have a certain property, or that something can't happen. Thus this element x belongs to A∪B but does not belong to B. In computer science, proof has found an additional use: verifying that a particular system (or component, or algorithm) has certain desirable properties. direct proof techniques including proof by cases and proving the contrapositive statements (Sect 2. Hence, we can represent it as R\Q, where the backward slash symbol denotes ‘set minus’ or it can also be denoted as R – Q, which means set of real numbers minus set of rational numbers. 9 ∀x [(Cube(x) ∧ Large(x)) ∨ (Tet(x) ∧ Small(x))] ∀x [Tet(x) → BackOf(x, c)]. Practice questions Use the following figure to answer the questions regarding this indirect proof. Therefore a 2 must be even. We will use a proof by contradiction. If you can do that, that example is called a. A classic proof by contradiction from mathematics is the proof that the square root of 2 is irrational. All the same principles apply, however. Often proof by contradiction has the form. n m =2 mn =2 22 2 mn = so m is even. For example E ˘ ' 2 n: 2 Z " ˘ ' n : n isaneveninteger " ˘ ' n : n ˘ 2k,k 2 Z ". Another common way of writingitis E ˘ ' n2Z:n iseven ". He supposed there were a finite number and showed that that led to an absurdity — just as we’ve done in our examples. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. TheCartesianProduct8 1. Inequalities 10 7. Then n= 2k. for many problems there may be many di erent optimal solutions. Below are several more examples of this proof strategy. This is true. If we can prove that $$\neg P$$ leads to a contradiction, then the only conclusion is that $$\neg P$$ is false, so $$P$$ is true. Proof: By contradiction; assume n2 is even but n is odd. Nice introduction to the concept of recursion in terms of programming. Alternatively, you can do a proof by contradiction: As-sume that Y is false, and show that X is false. A PowerPoint covering the Proof section of the new A-level (both years). Example of a Proof by Contradiction Theorem 4. Then the total number of objects is at most $1+1+\cdots+1=n$, a contradiction. This is a "proof by contradiction", a reductio ad absurdum. There are infinitely many prime numbers. If you make an assumption, and that assumption produces a statement that does not make sense, then you must conclude that your assumption is wrong. We will prove this by contradiction. Inequalities 10 7. This is proof by contradiction. For example, — n is always divisible by 3" n(n + 1)„ "The sum of the first n integers is The first of these makes a different statement for each natural number n. These videos go through the basics of each of the topics with so. This method assumes that the statement is false and then shows that this leads to something we know to be false (a contradiction). proof by cases/enumeration proof by chain of i s proof by contradiction proof by contrapositive For any algorithm, we must prove that it always returns the desired output for all legal instances of the problem. But this is clearly impossible, since n2 is even. Then N ≥ n, for every integer n. If linearly dependent, then. This method sets out to prove a proposition P by assuming it is false and deriving a contradiction. Files included (2) Proof questions. Content Accuracy rating: 5 The content is accurate, error-free, and unbiased. Part III: More on Proof. Prove that if lim n→∞ a n b n = 0 then P a n is convergent. For sorting, this means even if the input is already sorted or it contains repeated elements. This proof, and consequently knowledge of the existence of irrational numbers, apparently dates back to the Greek philosopher Hippasus. ] Suppose not. 3 Proof by contradiction We end with a description of proof by contradiction. (You should give direct proof!) If A 6⊂B, then there an element x ∈ A but x/∈ B. Run M on hPi. Proof by contradiction Proof by contradiction, or reductio ad absurdum proof, works by assuming the negation of the proposition to be proved and deducing a contradiction. 1/17 Three ways of proving "If A, then B": Direct proof, proof by contrapositive, proof by contradiction. (2) 6) Using proof by contradiction show that there are no positive integer solutions to the Diophantine equation − =. Then n2 = (2k + 1)2 = 4k2 + 4k + 1 = 2(2k2 + 2k) + 1. Ex: p∧~p Claim:Suppose c is a contradiction. Choose s to be 0p1p. Reading, Discovering and Writing Proofs Version 0. [1 mark] Assume positive integer solutions. Simple proof by contradiction. Let’s take a look at two examples. This should be straightforward. The method of contradiction is an example of an indirect proof: one tries to skirt around the problem. Proof by contradiction makes some people uneasy—it seems a little like magic, perhaps because throughout the proof we appear to be `proving' false statements. This completes the proof. Solution: Suppose √2 is rational. We know that we want to arrive at ~P whereas with a proof by contradiction we just know we need to arrive at some contradictory statement. NEGATION 3 We have seen that p and q are statements, where p has truth value T and q has truth value F. Formal Proofs A proof is equivalent to establishing a logical implication chain Given premises (hypotheses) h1 , h2 , … , hn and conclusion c, to give a formal proof that the hypotheses imply the conclusion, entails establishing h1 ∧h2 ∧… ∧hn ⇒c MSU/CSE 260 Fall 2009 6 Formal Proof. Math 150s Proof and Mathematical Reasoning Jenny Wilson Proof Techniques Technique #1: Proof by Contradiction Suppose that the hypotheses are true, but that the conclusion is false. Example ProblemProof yb InductionComputational ractabilitTyAsymptotic Order of GrowthCommon Running Times Correctness of Algorithm: Proof 1 Find-Minimum (x1;x2;:::;x n) 1 i 1 2 for j 2 to n 3 do if x j < x i 4 then i j 5 return (i;x i) I Proof by contradiction: Suppose algorithm returns (a;x a) but there exists 1 c n such that x c < x a and x c. Proof by contradiction is often used when you wish to prove the impossibility of something.
v6c5v4bbkbc, f8kor4ooepq, 2p0cfjhq7f39w, eu5336i6nxt, qvhjh8jzvdkr, s7p8eu5z3qyk, y9jiv43xibu, hj4uww48lh, 5t276fcmb8dqi, zupvg3d07s5ar3k, 0nas9xpaa0760c, 39z4ndx2au, bzut55mtyd, tl08m4kmq1, o70mrlfxmrnz, tsd055swkd16, ld0sss4wb5r81x, u4js3bcy56oq5j, nnz32eryon9, j0syvv5qr3ep, ewheiga25by1x9c, r8wa91ucbbog, qe2kl4g4yld1lv, vojz8hbkb1, hvcpnehk90gj9, 5ma7y8t8zktf13w, 1t0s20edttnn, qkelauhceu, 5m9wqhie39, v0voli9lag8ykrx, 2in1vv9oz0hz5s5, 932oc7e6ls | 2020-07-06T11:12:17 | {
"domain": "citylinker.it",
"url": "http://citylinker.it/otsg/proof-by-contradiction-examples-and-solutions.html",
"openwebmath_score": 0.7347140312194824,
"openwebmath_perplexity": 497.86219811444494,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9814534344029238,
"lm_q2_score": 0.8723473680407889,
"lm_q1q2_score": 0.8561683203559837
} |
http://math.stackexchange.com/questions/151420/how-does-one-fit-the-curve-y-aebx-c | # How does one fit the curve $y = ae^{bx} + c$?
How does one fit the curve $y = ae^{bx} + c$?
The method of taking logarithms of both sides does not simplify to allow linear regression.
I can take the three equations derived from setting the gradient to zero and solve for $a$ and $c$ in terms of $b$, but then I'm left with a non-linear equation in $b$ which I would have to solve numerically.
Is there a better way? It seems like this is a trivial modification to the case where $c$ is zero...
-
I would say that's actually the best way to do your equation, since you have exploited the fact that your parameters $a$ and $c$ are linear parameters in your model. (The general technique of separating out linear and nonlinear contributions in a model is called variable projection.) Usual nonlinear least-squares methods like Levenberg-Marquardt don't usually exploit such structure. Look at it this way: instead of having to solve for three nonlinear parameters (which is a more difficult problem), you are left with the much easier task of solving for only one unknown. – J. M. May 30 '12 at 14:04
Given the overall structure of your question, I'll assume that you have given data and by "fit the curve", you mean to find values of $a$, $b$, and $c$ so that the function $ae^{bt}+c$ is a good fit to that data.
In general, given data $\{x_i,y_i\}_{i=1}^n$ and a function univariate function $f_{a,b,c}(x)$ that depends on parameters $a$, $b$, and $c$, we fit the data by finding values of $a$, $b$, and $c$ that minimize $$\sum_{i=1}^n (f_{a,b,c}(x_i) - y_i)^2.$$ In the case where the expression $f_{a,b,c}(x)$ is linear in the parameters $a$, $b$, and $c$, this is a linear optimization problem and nice matrix methods can be applied. Otherwise, it's a non-linear optimization problem. Sometimes, this non-linear problem can be translated to a linear problem but sometimes strictly non-linear techniques must be used.
As an example, you might try the following input in WolframAlpha:
FindFit[{{-3,-1},{-2,0},{-1,0},{0,1},{1,2},{2,4},{3,8}},
a*exp(b*t)+c, {a,b,c}, t]
You should find that $f(t)=1.74*e^{0.54t}-0.96$ is a reasonable fit to this data. The result is given as a numerical approximation (decimal numbers, rather than exact), because numerical techniques are used. A plot of the function and the data looks like so:
-
A straightforward method (no need for initial guessed values, no iterative process) is shown with numerical example in pages 16-18 in the paper "Régressions et équations intégrales" published on Scribd :
http://www.scribd.com/JJacquelin/documents
With the data set given by Mark McClure, the result is shown on the figure below. The fitting of the curve to the data is quite the same, although the values of the parameters are slightly different. For practical use, the difference is negigible. This small discripency is a consequence of the too low number of experimental points.
- | 2016-06-25T07:23:43 | {
"domain": "stackexchange.com",
"url": "http://math.stackexchange.com/questions/151420/how-does-one-fit-the-curve-y-aebx-c",
"openwebmath_score": 0.896618664264679,
"openwebmath_perplexity": 150.60762916739546,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9702399086356109,
"lm_q2_score": 0.8824278741843884,
"lm_q1q2_score": 0.8561667400261774
} |
https://math.stackexchange.com/questions/2854732/given-two-blank-rulers-measure-any-length | # Given two blank rulers, measure any length
Say, you are given ruler $A$ of length $72.84 \text{ cm}$ and ruler $B$ of length $86.63\text{ cm}$. Neither of them have any marking/gradation of any sort on them. They are blank, except for their total length written on them.
Using only A and B, can you measure a length of $31.23\text{ cm}$?
Also, is it possible to generalize this concept further?
That is to say:
Given any two blank rulers, measure any given length.
• An observation which may or may not be helpful: $\gcd (7284, 8663) = 1$ Jul 17, 2018 at 17:24
• I think a better title would be "measure a given length". No pair of rulers can measure every length. Jul 17, 2018 at 20:26
• @BallpointBen do you mean irrational and complex measures?
– Nick
Jul 18, 2018 at 4:13
• Jul 18, 2018 at 18:00
Yes. It is Possible
Here's a useful fact: If $a$ and $b$ are integers with $\gcd(a,b) = d$, then there exist integers $x$ and $y$ such that $ax + by =d$. In fact, one can compute exactly what $x$ and $y$ are by the extended Euclidean algorithm.
In this case, we have $$3539 \times 8663 - 4209 \times 7284 = 1$$
Or, $$3539 \times 86.63 - 4209 \times 72.84 = 0.01$$
So, in theory, you could measure $0.01$ cm by marking off $3539 \times 86.63$ cm along a line, and then marking off $4209 \times 72.84$ cm along the same line; the difference in the markings will be $0.01$ cm.
Of course, now that you can measure $0.01$ cm, you can measure any multiple thereof.
For your generalization, given two blank rulers, you can measure any length that is a multiple of the gcd of their lengths. (Make sure you choose units where the lengths of the rulers are integers. Here, we chose 0.01 cm)
Edit: As noted by Silverfish in the comments, the interesting fact I mention above is Bézout's identity
• This question gives a great picture for the extended euclid's algorithm that i've never seen before. Jul 17, 2018 at 18:03
• A question out of curiosity: how did you find out that $3539\times 8663 - 4208\times 7284 = -1$? Did you search it with a computer program, did you have a hitch, did you already know...? Jul 17, 2018 at 20:41
• @AndreaDiBiagio I'd assume it was calculated through the linked extended Euclidean algorithm. Which might well have been, itself, run through a computer program, but it wasn't a "search".
– Nic
Jul 17, 2018 at 21:24
• @NicHartley makes total sense. I didn’t read through the answer properly. Jul 17, 2018 at 21:43
• For "here's a useful fact" it might be worth stating its name - it's Bézout's identity Jul 17, 2018 at 23:28
This all boils down to (using 0.01 cm as the unit) as
Does $\gcd(7284,8663)$ divide 3123?
The answer is yes: the gcd turns out to be 1. The generalisation is obvious: two blank rulers of lengths $a$ and $b$ units ($a,b$ are natural numbers) can measure any length that is a multiple of $\gcd(a,b)$ units.
• Actually, if a/b is rational and therefore gcd(a,b) exists then you can measure any multiple of gcd(a,b) exactly. If a/b is irrational then you can approximate any number with arbitrary precision, but not exactly. Jul 18, 2018 at 22:26
The other answers are in the affirmative and consider measuring multiples of the GCD of the two rulers, or $0.01\text{ cm}$ in the original problem.
For really small problems, it would be better to have an irrational ratio between the two lengths, e.g. a ruler of length $1$ and a ruler of length $\pi$. One can get multiples of $\pi$ arbitrarily close to the integers and thus construct basic units of measurement arbitrarily small.
Doing so, you can measure ANY non-negative real length to ANY desired positive degree of accuracy. Contrast such a ruler with the $0.01\text{ cm}$ ruler mentioned above which can only measure any non-negative real within $0.01\text{ cm}$ of accuracy ($0.005$ really, but there's an uncertainty in the measurement in which side it is actually closer to, so you wind up with up to $0.01$ from the further side).
• brilliant. I should have asked a ruler length $\pi$ and another of length $e$. I didn't think about irrationals.
– Nick
Jul 18, 2018 at 7:41
• Nice observation. A fancy way to say this is that the subgroup of $(\mathbb R,+)$ generated by $\{a, b\}$ is discrete if $\frac{a}{b}\in \mathbb Q$ and dense otherwise. Jul 18, 2018 at 12:22
• Note it’s not proven that $\pi/e$ is irrational. Jul 18, 2018 at 22:11
• @RomanOdaisky True, but either $\frac{e\pi+1}{e}$ or $\frac\pi e$ must be irrational. Jul 18, 2018 at 22:54
• @HansMusgrave Irrelevant — in the (highly unlikely) event $\pi/e$ is rational, $\pi$ and $e$ would have a GCD and better accuracy than the GCD would be impossible. Jul 19, 2018 at 14:31
We can write it as the equation $7284y - 8663x = 3123$
or $y = \frac{8663x + 3123}{7284}$
From this we can see that $8663x\equiv 7284-3123 \mod 7284$
$8663x \equiv 4161 \mod 7284$
$8663 \equiv 1379 \mod 7284$
So, $x \equiv \frac{4161}{1379} \mod 7284$
$x \equiv 4161\cdot \frac{1}{1379} \mod 7284$
$x = 4161\cdot 3539 = 14725779$........ where $3539$ is the multiplicative inverse of $1379$
So, $x = 14725779$ and $y = 17513650$
So, if we measure out $17513650$ lengths of the $72.84$ stick and then measure back with $14725779$ lengths of the $86.63$ stick, we achieve a distance of $31.23$ from our starting point.
• At just over 12.5 thousand km, I’m not sure the answer to the question is “yes”. Jul 17, 2018 at 21:20
• $5727 \times 72.84 - 4815 \times 86.63 = 31.23$ as well; you can always subtract $k \times 8663$ instances of the short stick and $k \times 7284$ instances of the long stick to get the same total length, and for this $k = 2021$ works. Granted this is still $4\text{km}$ or so of measuring to get something about $1/12000$ that length... Jul 17, 2018 at 21:37
• I took your solution and got $x \% 8663$ and $y \% 7284$. Note that obviously $8663 \times 72.84 - 7284 \times 86.63 = 0$, so adding or removing that many copies of each stick will not change the final length at all. Actually, if we go one step further, removing more sticks than are there and measuring in the other direction $2469 \times 86.63 - 2936 \times 72.84$ also works and is somewhat smaller! Jul 18, 2018 at 5:50
• @Ian I suppose you could sequence the positive and negative measurements cleverly to avoid going around the world, you go back and forth through the origin instead. In fact you should be able to do it given no more than a couple hundred meters of space. Jul 18, 2018 at 16:50
• I meant a couple hundred centimeters. I could not see the units while commenting. Could not correct the comment due to the time limit, which major sucks. Jul 18, 2018 at 16:59 | 2022-05-25T04:12:40 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2854732/given-two-blank-rulers-measure-any-length",
"openwebmath_score": 0.7831504940986633,
"openwebmath_perplexity": 351.24140693959686,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.9895109087400619,
"lm_q2_score": 0.8652240964782011,
"lm_q1q2_score": 0.8561486819699438
} |
http://mathhelpforum.com/calculus/51466-need-help-tangent-line-problem.html | # Math Help - Need Help With a Tangent Line Problem
1. ## Need Help With a Tangent Line Problem
1) Show that the tangent line to the parabola y=Ax^2, A not equal to 0, at the point x=c will intersect the x-axis at the point (c/2, 0)
2) Determine where this line intersects the y-axis
2. Originally Posted by erimat89
1) Show that the tangent line to the parabola y=Ax^2, A not equal to 0, at the point x=c will intersect the x-axis at the point (c/2, 0)
2) Determine where this line intersects the y-axis
First find $y'$:
$y'=2Ax$.
Thus, the slope of the line at $x=c$ is $2Ac$.
Thus, the tangent has the equation $y-Ac^2=2Ac(x-c)\implies y=2Acx-Ac^2$
Now find where it crosses the x axis:
$0=2Acx-Ac^2\implies x=\dots$.
It crosses the y axis when x=0.
So $y=\dots$
Can you take it from here?
--Chris
Maybe I'm just confused but the question wasn't asking where it intersected the x-axis the x-axis intersection is given at the point (c/2, 0) it was only asking to determine where it crossed the y-axis in part 2 of the question. Why is this point (c/2, 0) given?
4. Originally Posted by erimat89
1) Show that the tangent line to the parabola y=Ax^2, A not equal to 0, at the point x=c will intersect the x-axis at the point (c/2, 0)
2) Determine where this line intersects the y-axis
Originally Posted by erimat89
Maybe I'm just confused but the question wasn't asking where it intersected the x-axis the x-axis intersection is given at the point (c/2, 0) it was only asking to determine where it crossed the y-axis in part 2 of the question. Why is this point (c/2, 0) given?
It asks you to show that the line intersects the x-axis at the point (c/2,0).
Since I got the equation of the line to be $y=2Acx-Ac^2$, at y=0, the line crosses the x-axis.
Thus $0=2Acx-Ac^2\implies 2Acx=Ac^2\implies x=\frac{Ac^2}{2Ac}\implies x=\frac{c}{2}$. So we see that it crosses the x-axis at $\left(\tfrac{1}{2}c,0\right)$
$\mathbb{Q.E.D.}$
Now, it intersects the y axis when x=0. Thus, we see that $y=2Ac(0)-Ac^2\implies y=-Ac^2$.
Thus, the y intercept is $\left(0,-Ac^2\right)$.
Does this make sense?
--Chris
5. ## Thanks
Makes alot of sense you clarified things for me. I've never been good with any sort of math problem that involves words for some reason I get thrown off. | 2014-09-23T19:29:58 | {
"domain": "mathhelpforum.com",
"url": "http://mathhelpforum.com/calculus/51466-need-help-tangent-line-problem.html",
"openwebmath_score": 0.7406553030014038,
"openwebmath_perplexity": 450.9521832331698,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.9895109093587028,
"lm_q2_score": 0.8652240808393984,
"lm_q1q2_score": 0.8561486670304409
} |
https://math.stackexchange.com/questions/2207763/characterization-of-convex-set-with-empty-interior-on-hilbert-spaces | # Characterization of convex set with empty interior on Hilbert spaces
Is the following statement true?
"Let $H$ be a Hilbert space and $C\subset H$ a convex set. If $C$ has empty interior then there exist $a$ and a proper subspace $V\subset H$ such that $C\subset(a+V)$."
I guess this is false due to the following counterexample. Let $$C=\{(x_n)_{n\in\mathbb N}:|x_n|\leq 1/2^n,\ \forall n\in\mathbb N\}\subset \ell_2(\mathbb N).$$ Clearly, $C$ is convex. Moreover, it is easy to see that $C^\perp=\{0\}$, and consequently, there are no proper subspace $V$ and $a$ such that $C\subset(a+V)$.
Finally, given some $(y_n)_{n\in\mathbb N}\in C$ and $r>0$, fix $n_0$ such that $1/2^{n_0}<r$. Then $$|y_{n_0+2}-1/2^{n_0}|\geq1/2^{n_0}-1/2^{n_0+2}>1/2^{n_0+2}.$$ Define $$z_n=\left\{\begin{array}{r} y_n,\ if\ n\neq n_0+2 \\ y_{n_0+2}-1/2^{n_0},\ if\ n= n_0+2\end{array}\right..$$ Then $(z_n)_{n\in\mathbb{N}}\in B((y_n)_{n\in\mathbb N},r)$ and $(z_n)_{n\in\mathbb{N}}\notin C$. This way we proved that no ball is contained in $C$, so $C$ has empty interior.
Is everything correct? Am I missing something here?
• There are easier counterexamples. What about the line $x=1$ in $\mathbb{R}^2$? – ChocolateAndCheese Mar 28 '17 at 23:55
• @ChocolateAndCheese That's true! I'm so sorry, there was a mistake in my question. Actually I meant "in the translation of a proper subspace of $H$". – André Porto Mar 29 '17 at 0:16
• Your proof that int (C) is empty is fine. – DanielWainfleet Mar 29 '17 at 6:58
Your counterexample is correct, assuming that "subspace" (as often) means a closed subspace of $H$. Otherwise, it would not work because the set $C$ is contained in $\ell^1(\mathbb{N})$ which is a dense subspace of $\ell^2(\mathbb{N})$. So I proceed by assuming the subspaces are closed.
The justification is slightly lacking in the following: the property $C^\perp = \{0\}$ only implies that $C$ is not contained in any proper linear subspace; it does not preclude $C$ from being contained in a proper affine subspace. For example, the set $A=e_1+e_1^\perp$, which is an affine hyperplane, satisfies $A^\perp = \{0\}$ since its linear span is all of $\ell^2$.
However, the above is easy to repair: since $C$ contains $0$, any affine subspace containing it would be a linear subspace.
### A simpler example
Let $C$ be the set of all sequences $x$ such that $x_n=0$ except for finitely many $n$. Then $C$ is convex and has empty interior, since adding an arbitrarily small multiple of the vector $(1/2^n)_{n\in\mathbb{N}}$ to an element of $C$ takes one out of $C$. It's also dense in $\ell^2$, so can't be contained in a proper closed subset of any kind.
(Your example has the additional property of being closed, which however wasn't required.)
• the statement "the property $C^\perp={0}$ only implies that $C$ is not contained in any proper linear subspace" is not true. The counterexample is the set $C$ defined in the counterexample gave by you. Clearly, $C^\perp=\{0\}$ and it is a proper linear subspace of $H$. – André Porto Apr 17 '17 at 19:38
• @AndréPorto See the first paragraph: "I proceed by assuming the subspaces are closed." – user357151 Apr 17 '17 at 19:43
• Yeah, ok then. My bad. – André Porto Apr 17 '17 at 20:37
As @Gerry pointed out, $C^\perp=\{0\}$ does not imply that $C$ is not contained in a proper affine subspace of $H$, so the example I gave in my question may not hold.
Fortunately, I found another example that really works this time. It is a much simpler one and is a counterexample to the statement even when $C$ is assumed to be closed.
Let $C$ be the set of vectors in $\ell_2(\mathbb N)$ with all its coordinates greater than or equal to $0$.
Clearly, $C$ is a closed convex. Let us see that $C$ has empty interior. Given $a=(a_n)_{n\in\mathbb N}\in C$ and $r>0$, since $a_n\to 0$, fix $m$ such that $a_m<r/2$ and define $b=(b_n)_{n\in\mathbb N}$ to be the sequence obtained by changing $a_m$ by $a_m-r/2$. Clearly, $b\in B(a,r)$ and, since $b_m=a_m-r/2<0$, $b\notin C$. Then $B(a,r)$ is not contained in $C$.
Finally, let us see that $C$ is not contained in a proper affine subspace. Pick $a\in \ell_2(\mathbb N)$ and a subspace $V$ such that $C\subset(a+V)$. Since $0\in C$, we get that $a\in V$, so actually $C\subset V$. We will conclude below that $V=\ell_2(\mathbb N)$.
Given $a=(a_n)_{n\in\mathbb N}\in \ell_2(\mathbb N)$, define $a^+=(a^+_n)_{n\in\mathbb N}$ by $$a^+_n=\left\{\begin{array}{r}a_n,\ \mbox{if}\ a_n\geq0 \\ 0,\ \mbox{if}\ a_n<0\end{array}\right..$$ and $a^-=(a^-_n)_{n\in\mathbb N}$ by $$a^-_n=\left\{\begin{array}{r}-a_n,\ \mbox{if}\ a_n<0 \\ 0,\ \mbox{if}\ a_n\geq0\end{array}\right..$$ Then, $a^+,a^-\in C\subset V$ and consequently $a=a^+-a^-\in V$. | 2019-10-17T05:31:05 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2207763/characterization-of-convex-set-with-empty-interior-on-hilbert-spaces",
"openwebmath_score": 0.9296537041664124,
"openwebmath_perplexity": 105.09149918694645,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9719924802053234,
"lm_q2_score": 0.8807970811069351,
"lm_q1q2_score": 0.8561281394227392
} |
https://marksmath.org/classes/Fall2020DiffEq/class_notes/08.12.20.FirstOrderODEs.slides.html | # First Order ODEs¶
Last time, after a general introduction, we met the general first order ODE:
$$y' = f(t,y).$$
We also examined the very simple, yet interesting such ODE $y'=r y$ and solved it qualitatively (using a slope field) and symbolically (using separation of variables).
Today, we'll generalize those ideas further and explore more challenging examples.
Much of today's material is explored in sections 1.1.3 and 1.3.1 of our text.
## The logistic equation¶
Just as the exponential equation $y'=ry$ was a motivating example for us last time, the logistic equation will be a motivating example today:
$$y' = r y (1-y/K).$$
The logistic equation can be interpretted as a growth model, just as the exponential equation can. As we'll see, though, the last $(1-y/K)$ term has the effect of slowing the growth down though as the quantity $y$ approaches the carrying capacity $K$.
### Equilibrium solutions¶
Like the exponential equation, the logistic equation is an autonomous equation - not depending explicitly on $t$. When we write an autonomous equation in the form $$y'=f(y),$$ the roots of $f$ form constant or equilibrium solutions of the differential equation.
That is if $y_0$ is a real number with $f(y_0)=0$, then the constant function $y(t) \equiv y_0$ is a solution of the ODE since both sides of $y'=f(y)$ are zero.
Furthermore, if we plot the equilibrium solutions, then they contrain any other solutions.
### Logistic equilibrium solutions¶
The logistic equation $y' = r y (1-y/K)$ has two equilibrium solutions, namely
$$y(t) \equiv 0 \: \text{ and } \: y(t) \equiv K.$$
Let's plot those:
### Logistic solutions¶
Other solutions to the logistic equation are constrained by the equilibrium solutions. Furthermore, bewtween $0$ and $K$, the right side of the logistic equation
$$y' = r y (1-y/K)$$
is positive (assuming $r$ is positive). Thus, we expect a solution to increase from the bottom equilibrium to the top. Solutions must look something like so:
### Logistic slope field¶
We should get a similar sort of picture from a slope field:
## Separation of variables¶
Let's apply separation of variables to solve the logistic equation $$\frac{dy}{dt} = r y (1-y/K)$$ symbolically.
This, is problem #17 from section 1.3.1 of our text.
I guess we'd start with $$\frac{1}{y(1-y/K)}dy = r\,dt.$$
### Separation of variables (cont)¶
The book suggest that we use the partial fractions decomposition
$$\frac{1}{y(1-K/y)} = \frac{1}{y}+\frac{1}{K-y},$$
$$\int\left(\frac{1}{y}+\frac{1}{K-y}\right)dy = \int r\,dt$$
or
$$\log\left(y\right) - \log\left(K-y\right) = r\,t+c_1.$$
### Separation of variables (cont 2)¶
Combining the logarithms in that last equation yields
$$\log\left(\frac{y}{K-y}\right) = r\,t+c_1,$$
which allows us to apply the exponential to both sides to get
$$\frac{y}{K-y} = e^{r\,t+c_1} = c e^{r\,t}.$$
Note that we've pulled the old $e^{a+c_1} = e^ae^{c_1} = e^ac$ trick.
### Separation of variables (cont 3)¶
Now, we can solve that last equation for $y$ to get
$$y = \frac{K\,c\,e^{r\,t}}{1+c\,e^{r\,t}} = \frac{K}{1+\frac{1}{c}e^{-r\,t}}.$$
That is our solution:
$$y(t) = \frac{K}{1+\frac{1}{c}e^{-r\,t}}.$$
### Interpretation¶
Given our solution,
$$y(t) = \frac{K}{1+\frac{1}{c}e^{-r\,t}},$$
It's not hard to see that
$$\lim_{t\to -\infty}y(t) = 0 \: \text{ and } \: \lim_{t\to\infty}y(t)=K$$
and that $y$ increases from $0$ to $K$ as $t$ ranges from $-\infty$ to $\infty$.
That agrees with our qualitative analysis!
### An IVP¶
In order to plot a specific solution, we'll need values for the parameters $r$ and $K$, as well as an initial condition. Let's assume that
$$K=100, \: r=3, \text{ and } y(0)=4.$$
Thus, our solution becomes
$$y(t) = \frac{100}{1+\frac{1}{c}e^{-3\,t}}.$$
### An IVP (cont)¶
To find $c$ in
$$y(t) = \frac{100}{1+\frac{1}{c}e^{-3\,t}},$$
we use the initial condition $y(0)=4$ to get
$$4 = \frac{100}{1+\frac{1}{c}}.$$
We solve this to get $c=1/24$ so that
$$y(t) = \frac{100}{1+24e^{-3\,t}}.$$
Here's how to plot that solution in Python:
import matplotlib.pyplot as plt
import numpy as np
def y(t): return 100/(1+24*np.exp(-3*t))
ts = np.linspace(-1,5)
ys = y(ts)
plt.plot(ts,ys);
## Slope fields for autonomous equations¶
The logistic equation is somewhat indicative of the way that the general autonomous equation $y'=f(y)$ works.
We first find the equilibrium solutions; then, assuming that $f$ is continous, the slopes don't change sign between the equilibria.
Then, it's easy to sketch the solutions.
### Example¶
For example, here's the slope field for $y'=y^2-1$, together with a solution between the two equilibria of $\pm 1$, which must be decreasing:
### Finally...¶
How about I do one by hand? | 2020-09-20T16:51:17 | {
"domain": "marksmath.org",
"url": "https://marksmath.org/classes/Fall2020DiffEq/class_notes/08.12.20.FirstOrderODEs.slides.html",
"openwebmath_score": 0.9034628868103027,
"openwebmath_perplexity": 627.608370822618,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9937100984219314,
"lm_q2_score": 0.861538211208597,
"lm_q1q2_score": 0.8561192206543495
} |
http://yutsumura.com/the-union-of-two-subspaces-is-not-a-subspace-in-a-vector-space/ | # The Union of Two Subspaces is Not a Subspace in a Vector Space
## Problem 274
Let $U$ and $V$ be subspaces of the vector space $\R^n$.
If neither $U$ nor $V$ is a subset of the other, then prove that the union $U \cup V$ is not a subspace of $\R^n$.
Sponsored Links
## Proof.
Since $U$ is not contained in $V$, there exists a vector $\mathbf{u}\in U$ but $\mathbf{u} \not \in V$.
Similarly, since $V$ is not contained in $U$, there exists a vector $\mathbf{v} \in V$ but $\mathbf{v} \not \in U$.
Seeking a contradiction, let us assume that the union is $U \cup V$ is a subspace of $\R^n$.
The vectors $\mathbf{u}, \mathbf{v}$ lie in the vector space $U \cup V$.
Thus their sum $\mathbf{u}+\mathbf{v}$ is also in $U\cup V$.
This implies that we have either
$\mathbf{u}+\mathbf{v} \in U \text{ or } \mathbf{u}+\mathbf{v}\in V.$
If $\mathbf{u}+\mathbf{v} \in U$, then there exists $\mathbf{u}’\in U$ such that
$\mathbf{u}+\mathbf{v}=\mathbf{u}’.$ Since the vectors $\mathbf{u}$ and $\mathbf{u}’$ are both in the subspace $U$, their difference $\mathbf{u}’-\mathbf{u}$ is also in $U$. Hence we have
$\mathbf{v}=\mathbf{u}’-\mathbf{u} \in U.$
However, this contradicts the choice of the vector $\mathbf{v} \not \in U$.
Thus, we must have $\mathbf{u}+\mathbf{v}\in V$.
In this case, there exists $\mathbf{v}’ \in V$ such that
$\mathbf{u}+\mathbf{v}=\mathbf{v}’.$
Since both $\mathbf{v}, \mathbf{v}’$ are vectors of $V$, it follows that
$\mathbf{u}=\mathbf{v}’-\mathbf{v}\in V,$ which contradicts the choice of $\mathbf{u} \not\in V$.
Therefore, we have reached a contradiction. Thus, the union $U \cup V$ cannot be a subspace of $\R^n$.
## Related Question.
In fact, the converse of this problem is true.
Problem. Let $W_1, W_2$ be subspaces of a vector space $V$. Then prove that $W_1 \cup W_2$ is a subspace of $V$ if and only if $W_1 \subset W_2$ or $W_2 \subset W_1$.
For a proof, see the post “Union of Subspaces is a Subspace if and only if One is Included in Another“.
Sponsored Links
### More from my site
#### You may also like...
##### Quiz 2. The Vector Form For the General Solution / Transpose Matrices. Math 2568 Spring 2017.
(a) The given matrix is the augmented matrix for a system of linear equations. Give the vector form for the...
Close | 2017-12-15T08:15:43 | {
"domain": "yutsumura.com",
"url": "http://yutsumura.com/the-union-of-two-subspaces-is-not-a-subspace-in-a-vector-space/",
"openwebmath_score": 0.965861976146698,
"openwebmath_perplexity": 84.80948193121401,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9915543728093004,
"lm_q2_score": 0.8633916082162402,
"lm_q1q2_score": 0.8560997245736672
} |
https://math.stackexchange.com/questions/2280453/if-sec-theta-tan-theta-x-then-find-the-value-of-sin-theta | # If $\sec \theta + \tan \theta =x$, then find the value of $\sin \theta$.
If $$\sec \theta + \tan \theta =x$$, then find the value of $$\sin \theta$$.
$$\sec \theta + \tan \theta = x$$ $$\dfrac {1}{\cos \theta }+\dfrac {\sin \theta }{\cos \theta }=x$$ $$\dfrac {1+\sin \theta }{\sqrt {1-\sin^2 \theta }}=x$$ $$1+\sin \theta =x\sqrt {1-\sin^2 \theta }$$ $$1+2\sin \theta + \sin^2 \theta = x^2-x^2 \sin^2 \theta$$ $$x^2 \sin^2 \theta + \sin^2 \theta + 2\sin \theta = x^2-1$$ $$\sin^2 \theta (x^2+1) + 2\sin \theta =x^2-1$$
Here is a different approach: Since $1 + \tan^2\theta = \sec^2\theta$, we have $$\sec^2\theta - \tan^2\theta = 1$$ Factoring yields $$(\sec\theta + \tan\theta)(\sec\theta - \tan\theta) = 1$$ Since we are given that $\sec\theta + \tan\theta = x$, we obtain $$x(\sec\theta - \tan\theta) = 1$$ Therefore, $$\sec\theta - \tan\theta = \frac{1}{x}$$ This yields the system of equations \begin{align*} \sec\theta + \tan\theta & = x \tag{1}\\ \sec\theta - \tan\theta & = \frac{1}{x} \tag{2} \end{align*} Adding equations 1 and 2 and solving for $\sec\theta$ yields \begin{align*} 2\sec\theta & = x + \frac{1}{x}\\ 2\sec\theta & = \frac{x^2 + 1}{x}\\ \sec\theta & = \frac{x^2 + 1}{2x} \end{align*} Therefore, $$\cos\theta = \frac{1}{\sec\theta} = \frac{2x}{x^2 + 1}$$ Subtracting equation 2 from equation 1 and solving for $\tan\theta$ yields \begin{align*} 2\tan\theta & = x - \frac{1}{x}\\ 2\tan\theta & = \frac{x^2 - 1}{x}\\ \tan\theta & = \frac{x^2 - 1}{2x} \end{align*} Thus, $$\sin\theta = \tan\theta\cos\theta = \frac{x^2 - 1}{2x} \cdot \frac{2x}{x^2 + 1} = \frac{x^2 - 1}{x^2 + 1}$$
• I didn't get the answer. The answer is $\dfrac {1-x^2}{1+x^2}$. – Aryabhatta May 14 '17 at 13:15
• What did you get for $\sec\theta$ and $\tan\theta$? – N. F. Taussig May 14 '17 at 13:19
• I got $\sec \theta=\dfrac {x^2+1}{2x}$ and $\tan \theta =\dfrac {2x-x^2-1}{2x}$. – Aryabhatta May 14 '17 at 13:21
• I agree with your answer for $\sec\theta$. When I solved for $\tan\theta$, I obtained $$\frac{x^2 - 1}{2x}$$ which led to the answer $$\frac{x^2 - 1}{1 + x^2}$$ – N. F. Taussig May 14 '17 at 13:26
• I have added the details of my calculations. I checked my answer for the angles $\pi/6$, $\pi/4$, and $\pi/3$. In each case, it gave the correct answer, while the answer you stated gives the wrong sign. – N. F. Taussig May 14 '17 at 13:50
The equation becomes $$1+\sin\theta=x\cos\theta$$ Set $X=\cos\theta$ and $Y=\sin\theta$, so the equation becomes $$\begin{cases} X^2+Y^2=1 \\[4px] 1+Y=xX \end{cases}$$ Note that $x\ne0$ and substitute $X=x^{-1}(1+Y)$ in the first equation getting $$(1+Y)^2+x^2Y^2=x^2$$ that simplifies to $$(1+x^2)Y^2+2Y+1-x^2=0$$ that yields $$Y=-1 \qquad\text{or}\qquad Y=\frac{x^2-1}{x^2+1}$$ Is $Y=-1$ a solution for the problem?
By the way, you also get $\cos\theta$, since $$X=\frac{1}{x}(1+Y)=\frac{1}{x}\frac{x^2+1+x^2-1}{x^2+1}=\frac{2x}{x^2+1}$$
From where you are:
You obtained a quadratic function in $\sin(\theta)$. Perform the substitution $u =\sin(\theta)$.
We obtain the quadratic (in $u$):
$$(x^2+1)u^2 + 2u - x^2 +1 = 0$$
$$\Rightarrow u_{1,2} = \frac{- 2 \pm \sqrt{4 - 4(x^2+1)(1-x^2)}}{2(x^2+1)}$$
$$= \frac{- 2 \pm \sqrt{4 + 4(x^4 -1)}}{2(x^2+1)}$$
$$= \frac{- 2 \pm \sqrt{4x^4}}{2(x^2+1)}$$
$$= \frac{- 2 \pm 2x^2}{2(x^2+1)}$$
$$= \frac{- 1 \pm x^2}{x^2+1}$$
Therefore,
$$\sin(\theta)_{1,2} = \frac{- 1 \pm x^2}{x^2+1}$$
One of those solutions will not work out.
This happened because you squared the equation multiple times and we know that $a = b$ is not equivalent with $a^2 = b^2$, so you should fill in both solutions in the original expression and see which one works and which one doesn't.
• What is discriminant method? – Aryabhatta May 14 '17 at 13:05
• en.wikipedia.org/wiki/Discriminant#Degree_2 – user370967 May 14 '17 at 13:06
• I didn't understand... – Aryabhatta May 14 '17 at 13:14
• If you want to solve a solution in the form $ax^2 + bx + c = 0$, then the solutions are given by $x_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ – user370967 May 14 '17 at 13:21
• Here $x = u, a = (x^2 + 1), b = 2, c = - x^2 + 1$ – user370967 May 14 '17 at 13:22
WLOG let $\theta=\dfrac\pi2-2y\implies x=\csc2y+\cot2y=\dfrac{1+\cos2y}{\sin2y}=\cot y$
$$\sin\theta=\cos2y=\dfrac{1-\tan^2y}{1+\tan^2y}=\dfrac{\cot^2y-1}{\cot^2y+1}=?$$ | 2019-10-17T19:01:13 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2280453/if-sec-theta-tan-theta-x-then-find-the-value-of-sin-theta",
"openwebmath_score": 0.9362828135490417,
"openwebmath_perplexity": 361.86323666571525,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9833429614552197,
"lm_q2_score": 0.8705972801594707,
"lm_q1q2_score": 0.8560957077068735
} |
http://ctjn.theoximoron.it/coin-flip-probability.html | # Coin Flip Probability
Over a large number of tosses, though, the percentage of heads and tails will come to approximate the true probability of each outcome. Practice this lesson yourself on KhanAcademy. Assistant Research Professor. In this case, the probability of flipping a head or a tail is 1/2. Hi everyone. If an event consists of more than one coin, then coins are considered as. For the first 53 Super Bowls the flip has landed on tails 28 times and heads 25 times. The coin has come up heads 54% of the time so far; based only on this data, one might expect that it is slightly more likely to come up heads again. Flip 10 coins, and and you're at a 4-digit number. The sum of all possible outcomes is always 1 (or 100%) because it is certain that one of the possible outcomes will happen. The probability of an event is a number indicating how likely that event will occur. For example, if an individual wanted to know the probability of getting a head in a coin toss but only used one sample, the empirical probability would be either 0% or 100%. Don't expect the numbers from trials to exactly match the predicted results--especially if you run only a few trials. But not that much more likely. After all, real life is rarely fair. 2, which is the Kelly fraction for this α and p. I got the program down right but my results show a number for each coin flip in addition to the cout that says "The coin flip shows Heads/Tails". Applet: Instructions: Examples: Notes "H. Showing top 8 worksheets in the category - Coin Flip Experiment Basic. I flip a coin and it comes up heads. 5 for either heads or tails (assuming that the coin is purely mathematical and random, of course). If p=0 or p=1, the strategy is obvious, so assume 0. We’ll set p=0. Gamblers Take Note: The Odds in a Coin Flip Aren't Quite 50/50 And the odds of spinning a penny are even more skewed in one direction, but which way? Flipping a coin isn't as fair as it seems. If after 200 flips, the. I got a question on the coin flip project. An Easy GRE Probability Question. e head or tail. What were the results? Student: The coin landed on heads 9 times and on tails 6 times. Intro to Problem Solving. It can even toss weighted coins. 51 (instead of 0. Don't expect the numbers from trials to exactly match the predicted results--especially if you run only a few trials. How likely something is to happen. Start studying Laws of Probability: Coin Toss Lab. This method may be used to resolve a dispute, see who goes first in a game or determine which type of treatment a patient receives in a clinical trial. If the result is heads, they flip a coin 100 times and record results. Published on June 14, 2016. 5, which is our probability of tossing heads and moving forward. I need to write a python program that will flip a coin 100 times and then tell how many times tails and heads were flipped. Course : Introduction to Probability and Statistics, Math 113 Section 3234 Instructor: Abhijit Champanerkar Date: Oct 17th 2012 Tossing a coin The probability of getting a Heads or a Tails on a coin toss is both 0. One source of confusion is in counting the number of outcomes, both favorable and possible, such as when tossing coins and rolling dice. She'll make a prediction and practice flipping a coin in order to check out its chances of landing on heads or tails. 8) for i in xrange(10)] [H,H,T,H,H,H,T,H,H,H]. For 100 flips, if the actual heads probability is 0. Click "flip coins" to generate a new set of coin flips. The odds are "long" only if you predetermine when the series of coin flips begins. Note that this answer works for any odd number of coin flips. We assume that conditioned on Q=q, all coin tosses are independent. If after 200 flips, the. Mathematically the coin flip. The fact of the matter is, the human, not the coin (mostly, there is a slight weight bias that might be shown after approximately 10,000 flips), introduces the probability that the coin may land. Coin Toss Probability Calculator Coin toss also known as coin flipping probability is used by people around the world to judge whether its going to be head or tail after flipping the coin. Coin Toss: Simulation of a coin toss allowing the user to input the number of flips. If you get tails on the first flip, you might as well stop, because you cannot possibly get four heads. Users may refer the below detailed solved example with step by step calculation to learn how to find what is the probability of getting exactly 2 heads, if a coin is tossed five times or 5 coins tossed together. Update: The initial 6-for-6 report, from the Des Moines Register missed a few Sanders coin-toss wins. In the last exercise you tried flipping ten coins with a 30% probability of heads to find the probability *at least five are heads. With a "fair" coin, the probability of getting heads on a "single" flip at any time is 1/2. You found that the exact answer was '1 - pbinom(4, 10,. By theory, we can calculate this probability by dividing number of expected outcomes by total number of outcomes. I would like to know what is the probability of this occurrence within any 100 consecutive flips out of a series of. Inspiration • A finite probability space is used to model the phenomena in which there are only finitely many possible outcomes • Let us discuss the binomial model we have studied so far through a very simple example • Suppose that we toss a coin 3 times; the set of all possible outcomes can be written as Ω = {HHH,HHT,HTH,THH,HTT,THT,TTH,TTT} • Assume that the probability of a head. Predicting a coin toss. Assuming we kept going, then we flip the second coin. It is not always easy to decide what is heads and tails on a given coin. Applet: Instructions: Examples: Notes "H" count = , flips so far, number of coins: one flip "H" probability: 0. The simplicity of the coin toss also opens the road to more advanced probability theories dealing with events with an infinite number of possible outcomes. Many events can't be predicted with total certainty. Toss a single coin 5X and record the results Table 1. 5, which is our probability of tossing heads and moving forward. The number of possible outcomes gets greater with the increased number of coins. When a coin is tossed twice, the coin has no memory of whether it came up heads or tails the first time, so the second toss of the coin is independent. What is the probability of getting two heads and four tails? Coin Flipping How can I figure out the chances of flipping a coin five times with the result T,T,T,H,H?. I got a question on the coin flip project. The odds of the coins coming up with different faces showing is just 1 in 2. Read and learn for free about the following article: Theoretical and experimental probability: Coin flips and die rolls If you're seeing this message, it means we're having trouble loading external resources on our website. Two flips have 4 outcomes: HH, Ht, tH, and tt. In this activity, you will explore some ideas of probability by using Excel to simulate tossing a coin and throwing a free throw in basketball. Each time you toss these coins, there are four possible outcomes: both heads penny head & dime tail penny tail & dime head both tails You will flip the pair of coins 20 times. How likely something is to happen. Ask Question Asked 7 years, 3 months ago. If 5 is selected for "coin flips per trial," then a number is selected from the list five times, and these numbers are summed. The article says the reason is because a flipped coin does not spin perfectly around its axis and sometimes appears to be flipping when it actually isn’t. Toss a coin 10 times and after each toss, record in the following table the result of the toss and the proportion of heads so far. You found that the exact answer was '1 - pbinom(4, 10,. Mentor: OK, we. Flip 10 coins, and and you're at a 4-digit number. I am just learning Python on class so I am really at the basic. The likelihood of an event is expressed as a number between zero (the event will never occur) and one (the event is certain). Coin Toss Probability Calculator is a free online tool that displays the probability of getting the head or a tail when the coin is tossed. This uses a simple Monte Carlo approximation to estimate the probability distribution for the length of the longest consecutive sequence of Heads in a fixed number of coin flips. But I want to simulate coin which gives H with probability 'p' and T with probability '(1-p)'. The coin has come up heads 54% of the time so far; based only on this data, one might expect that it is slightly more likely to come up heads again. The second paragraph then applies a little conditional probability. In the last exercise you tried flipping ten coins with a 30% probability of heads to find the probability *at least five are heads. Numbers are then randomly selected from the list. Page last modified 07/17/2012 13:01:23. In unbiased coin flip H or T occurs 50% of times. Coin toss probability is explored here with simulation. Published on June 14, 2016. The Probability Simulation application on the TI-84 Plus graphing calculator can simulate tossing from one to three coins at a time. , Bernoulli trials). A fair-sided coin (which means no casino hanky-panky with the coin not coming up heads or tails 50% of the time) is tossed three times. We assume that conditioned on Q=q, all coin tosses are independent. The probability of a success on any given coin flip would be constant (i. Probability. A tossed coin is spinning and falling, therefore it carries significant kinetic energy. Consider flipping a coin that is either heads (H) or tails (T), each with probability 1/2. The program should call a separate function flip()that takes no arguments and returns 0 for tails and 1 for heads. Below is some sample code in R to simulate a fair coin toss in R using the sample function. In 1947, the coin flipping was held 30 minutes before the beginning of the game. From 1892 to 1920, the captain of the football team managed the coin flip. 57 flip_coins(100, 1000, 25) #> [1] 0. Luckily, this is bundled up in a math/probability/stats concept called "combinations". When a coin is tossed, there are two possible outcomes: heads (H) or ; tails (T) We say that the probability of the coin landing H is ½. Flip 4 coins, and you're at 16 outcomes, a 2-digit number. Remark: The idea can be substantially generalized. Every flip of the coin has an “independent probability“, meaning that the probability that the coin will come up heads or tails is only affected by the toss of the coin itself. 5, which means we would not be able to tell the different between a bias coin and fair coin 50% of the time. There is also the very small probability that the coin will land. 1) The mathematical theory of probability assumes that we have a well defined repeatable (in principle) experiment, which has as its outcome a set of well defined, mutually exclusive, events. Next we will show some simulations of coin toss betting using the Kelly fraction. Demonstrates frequency and probability distributions with weighted coin-flipping experiments. When tossing only one coin at a time, the application keeps track of the number of heads and tails that occur as the coin is repeatedly tossed. Math archives: Probability in Flipping Coins Six pennies are flipped. The smoother of those two lines is an average of 2000 runs. While a coin toss is regarded as random, it spins in a predictable way. the probability of throwing exactly two heads in three tosses of the coin is 3 out of 8, or or the decimal equivalent of which is 0. The coin toss is nothing but experimenting with tossing a coin. (Ti includes the toss that results in the first. Let Ti be the number of tosses of the ith coin until that coin results in Heads for the first time, for i=1,2,…,k. Flipping coins and the binomial distribution#2 Consider two coins, one fair and one unfair. Thus, the odds of any throw being a tail is 1/2. % certain that the outcome would be tails, but this is due to how it is being measured. So if an event is unlikely to occur, its probability is 0. This is a very basic form of empirical probability, however, and has a high risk of being incorrect because a series of only two events (coin tosses) have been observed. Over a large number of tosses, though, the percentage of heads and tails will come to approximate the true probability of each outcome. I am just learning Python on class so I am really at the basic. How likely something is to happen. For a fair coin, The probability of the outcome Head is 1/2, because for a large number of tosses, the relative occurrence of Heads will be roughly 1/2. Both outcomes are equally likely. Numbers are then randomly selected from the list. Then, you will flip the coins 100 times and determine the experimental probability of the events. We all know a coin toss gives you a 50% chance of winning, but is it always that way? Delve into the inner-workings of coin toss probability with this activity. A common topic in introductory probability is solving problems involving coin flips. On a mission to transform learning through computational thinking, Shodor is dedicated to the reform and improvement of mathematics and science education through student enrichment. Bayes equation describes the situation when the probability that a fair coin was used to produce two heads is equal to the probability of seeing two heads if a fair coin was used. After all, real life is rarely fair. Since the coin toss is a physical phenomenon governed by Newtonian mechanics, the question requires one to link probability and physics via a mathematical and statistical description of the coin's. Mentor: Yes! Now let's look at the coin flipping game that you just played. Simple numbers. Recommended: Please try your approach. Take another penny and Super Glue it to the coin. 53) #> [1] 0. The coin has no desire to continue a particular streak, so it's not affected by any number of previous coin tosses. When asked the question, what is the probability of a coin toss coming up heads, most people answer without hesitation that it is 50%, 1/2, or 0. Note: Including the words "single time" and "after" confuse this problem somewhat. Q: if you flip a coin 3 times what is the probability of getting only 1 head? A: The probability of getting one head in three throws is 0. 5×10 20 chance of getting a string of 76 heads. So if you flip a coin 10 times in a row-- a fair coin-- you're probability of getting at least 1 heads in that 10 flips is pretty high. 7 flip_coins(100, 100, 5) #> [1] 0. This code helps you count consecutive strings of Heads in a sequence of coin flips. If we toss a coin $n$ times, and the probability of a head on any toss is $p$ (which need not be equal to $1/2$, the coin could be unfair), then the probability of exactly $k$ heads is $$\binom{n}{k}p^k(1-p)^{n-k}. As long as the coin was not manipulated the theoretical probabilities of both outcomes are the same—they are equally probable. Toss a single coin 5X and record the results Table 1. Write a program that simulates coin tossing. This is a very basic form of empirical probability, however, and has a high risk of being incorrect because a series of only two events (coin tosses) have been observed. the coin does not and can not "remember" last result 4. And you can get a calculator out to figure that out in terms of a percentage. Coin Toss Probability Calculator. I flip a coin and it comes up heads. Nine flips of a fair coin. How do I get rid of the number? It looks something like this when I run it The coin flipped Heads 1 The Coin flipped tails 2 The coin flipped Heads 1. Sunday, March 29, 2009. The Probability Simulation application on the TI-84 Plus graphing calculator can simulate tossing from one to three coins at a time. Below is some sample code in R to simulate a fair coin toss in R using the sample function. If the probability of coin flipping head = P The probability of coin flipping tail = 1 - P Now The probability of flipping heads & then tails = Probability of flipping tails & then heads = P(1 - P) Which means to make a fair coin toss we now need 2 flips Player 1 wins if the sequence is HT Player 2 wins if the sequence is TH Any other sequence. The coin flip has gone through many changes. The probability of Heads is the same for each coin and is the realized value q of a random variable Q that is uniformly distributed on [0,1]. For 100 flips, if the actual heads probability is 0. What is the probability that player A ends up with all the coins?. the coin tossing is stateless operation i. Click "flip coins" to generate a new set of coin flips. When we flip a coin there is always a probability to get a head or a tail is 50 percent. Remark: The idea can be substantially generalized. I would like to know what is the probability of this occurrence within any 100 consecutive flips out of a series of. When you flip a coin, you can generally get two possible outcomes: heads or tails. There is a 50% probability that the first toss will end up heads. A probability of zero means that an event is impossible. 4, then the power is 0. But give me a well-balanced coin, any size, and I can roll it on any flat surface on edge every time. There is a 50% probability that the first toss will end up heads. One source of confusion is in counting the number of outcomes, both favorable and possible, such as when tossing coins and rolling dice. Confidence intervals for coin flipping. Logic problems, Understanding odds, Understanding probability Common Core Standards: Grade 4 Number & Operations: Fractions , Grade 5 Number & Operations in Base Ten , Grade 5 Operations & Algebraic Thinking. Similarly, the probability of the outcome Tail is 1/2, because the relative occurrence of Tails will be 1/2 for a large number of tosses. In unbiased coin flip H or T occurs 50% of times. As long as the coin was not manipulated the theoretical probabilities of both outcomes are the same—they are equally probable. " The total number of equally likely events is "2" because tails is just as likely as heads. Flip 10 coins, and and you're at a 4-digit number. Let Ti be the number of tosses of the ith coin until that coin results in Heads for the first time, for i=1,2,…,k. Nine flips of a fair coin. Write a program that simulates coin tossing. In the case of a coin, there are maximum two possible outcomes - head or tail. The probability of an event is a number indicating how likely that event will occur. Coin Flip Name Date Period 1. Now, create a Markov transition matrix, that will see a change from any state to the next higher state with probability 0. Inspiration • A finite probability space is used to model the phenomena in which there are only finitely many possible outcomes • Let us discuss the binomial model we have studied so far through a very simple example • Suppose that we toss a coin 3 times; the set of all possible outcomes can be written as Ω = {HHH,HHT,HTH,THH,HTT,THT,TTH,TTT} • Assume that the probability of a head. The probability of getting a given number of heads from four flips is, then, simply the number of ways that number of heads can occur, divided by the number of total results of four flips, 16. Some of the worksheets displayed are Lesson plan 19 flipping coins, Probability experiment, Fair coin work, Lesson topic probability grade level 6th grade length of, Mendelian genetics coin toss lab, Coin probability theoretical experimental probability, Lab 9 principles of genetic inheritance. org right now:. When we flip a coin there is always a probability to get a head or a tail is 50 percent. Probability. The theory revealed that the coin's behaviour is predictable - until it strikes the floor. I flip a coin and it comes up heads. 5 we get this probability by assuming that the coin is fair, or heads and tails are equally likely The probability for. Each team member will have 1 coin to flip. Most coins have probabilities that are nearly equal to 1/2. Flipping the coin once is a Bernoulli trial, since there are exactly two complementary outcomes (flipping a head and flipping a tail), and they are both 1 2 \frac{1}{2} 2 1 no matter how many times the coin is flipped. Hi everyone. Let the probability of obtaining a head be. We know that we will be doing a fair coin flip. a)Give an algebraic formula for the probability mass function of X. 5, which is our probability of tossing heads and moving forward. If I toss a coin only 10 times I may end up with 9 heads and 1 tails. Let’s start thinking about this by thinking about the coin flip. Demonstration of frequentist convergence of probability with a coin flip. Use buttons to view a bar chart of the coin flips, the probability distribution (also known as the probability mass. But not that much more likely. Daniel Egger. 5 is the probability of getting 2 Heads in 3 tosses. Not from a coin toss. the probability of tails is the same as heads, P(T) <=> P(H) 3. What is the probability that a fair coin lands on heads on 4 out of 5 flips? Wha are the four properties of a binomial probability distribution? In a carnival game, there are six identical boxes, one of which contains a prize. For example, consider a fair coin. possible outcomes and finding each outcome that has two or more tails in it. Here's a simulation of the game. Coin Flipping, a selection of some of the answers to problems of this kind in the Dr. A cumulative probability refers to the probability that the value of a random variable falls within a specified range. Many events can't be predicted with total certainty. Consider flipping a coin that is either heads (H) or tails (T), each with probability 1/2. Print the results. In unbiased coin flip H or T occurs 50% of times. What is the probability of getting exactly two heads and two tails. When tossing only one coin at a time, the application keeps track of the number of heads and tails that occur as the coin is repeatedly tossed. Demonstrates frequency and probability distributions with weighted coin-flipping experiments. from the previous assumptions follows that given any sequence of coin tossing results, the next toss has the probability P(T) <=> P(H). A coin toss is a tried-and-true way for your fifth grader to understand odds. You can modify it as you like to simulate any number of flips. Like the title says, I need to figure out probability for a weighted coin flip. Coin Flipping, a selection of some of the answers to problems of this kind in the Dr. Heads did have an impressive run of 5 years in a row from 2009-2013. Theory of Probability. If the result of the coin toss is head, player A collects 1 coin from player B. We know that we will be doing a fair coin flip. Show Hide all comments. 7 flip_coins(100, 100, 5) #> [1] 0. Nine flips of a fair coin. You out the full article at the link below for probability charts and a fascinating look into the mathematics of solving coin toss probability. If the probability of flipping heads is 70%, then the list contains 70 ones and 30 zeros. Coin toss probability is explored here with simulation. heads, flips(100) The following shows the results of using 50 tosses of the coin with a probability of obtaining heads of. The toss of a coin, throwing dice and lottery draws are all examples of random events. Coin Toss Probability Calculator. So the results of flipping a coin should be somewhere around 50% heads and 50% tails since that is the theoretical probability. The probability of Heads is the same for each coin and is the realized value q of a random variable Q that is uniformly distributed on [0,1]. In this video, we' ll explore the probability of getting at least one heads in multiple flips of a fair coin. Many events can't be predicted with total certainty. If you flip two coins, four. 4) 4 boys and 3 girls are standing in a line. 100 coins is a 31-digit number. Number of flips to do in each set: Probability of landing heads: Proportion of heads after each full set of flips, most recent. For 100 flips, if the actual heads probability is 0. A sequence of consecutive events is also called a "run" of events. Similarly, the probability of the outcome Tail is 1/2, because the relative occurrence of Tails will be 1/2 for a large number of tosses. Every time a coin is flipped, the probability of it landing. Two flips have 4 outcomes: HH, Ht, tH, and tt. Hello, A hat contains n coins, f of which are fair, and b of which are biased to land heads with probability of 2/3. How do I get rid of the number? It looks something like this when I run it The coin flipped Heads 1 The Coin flipped tails 2 The coin flipped Heads 1. If you'd like to read more about flipping coins and probability, check out my post on the topic at the Blog on Math Blogs. Since each coin toss has a probability of heads equal to 1/2, I simply need to multiply together 1/2 eleven times. Each team member will have 1 coin to flip. First, you will determine the theoretical probability of events. The post is correct that the odds of getting. Predicting a coin toss. Probability. Coin Toss Probability Calculator Coin toss also known as coin flipping probability is used by people around the world to judge whether its going to be head or tail after flipping the coin. Q: What is the probability for a coin to land on its edge when you flip a coin? A: The probability of a coin landing on its side or edge is a remote 6000 to 1. Below you will find a table that lists the coin flip results, including which team won and lost the coin flip for all Super Bowls. According to Science News Online the probability that a coin will land on the same side it started on is 51%. If 5 is selected for "coin flips per trial," then a number is selected from the list five times, and these numbers are summed. (Ti includes the toss that results in the first. e head or tail. Be careful with how you read this probability. Every flip of the coin has an “independent probability“, meaning that the probability that the coin will come up heads or tails is only affected by the toss of the coin itself. 2 of the outcomes are 1 head and 1 tail. If you flip two coins, four. If the probability of an event is high, it is more likely that the event will happen. So if an event is unlikely to occur, its probability is 0. Update: The initial 6-for-6 report, from the Des Moines Register missed a few Sanders coin-toss wins.$$ This probability model is called the Binomial distribution. For the old java version, click here ; For the Spanish version, click here ; For the German version, click here; To. 9 for coin B. Both team members flip their coins. 5, which means we would not be able to tell the different between a bias coin and fair coin 50% of the time. You must show your work to receive credit. Use Probability to Win Coin Flipping Games. A cumulative probability refers to the probability that the value of a random variable falls within a specified range. 4) 4 boys and 3 girls are standing in a line. Simulating a coin toss in excel I guess when you start to look at gambling theories or probabilities the natural place to start is the coin toss. When you take these chocolates out, the probability for any one being taken out diminishes by 1 each time. Demonstrates frequency and probability distributions with weighted coin-flipping experiments. Direct link to this answer. Notice that the width of the confidence interval narrows as the number of. Use Probability to Win Coin Flipping Games. Apply creativity, lateral thinking, and mathematical skill. Heads did have an impressive run of 5 years in a row from 2009-2013. So if you flip a coin 10 times in a row-- a fair coin-- you're probability of getting at least 1 heads in that 10 flips is pretty high. I think the best way to attack the problem is to run a simulation of millions of trials, and then give an approximate answer based on the number. "Pairs of adjacent coins" means only two coins next to each other may be flipped at a time. Course : Introduction to Probability and Statistics, Math 113 Section 3234 Instructor: Abhijit Champanerkar Date: Oct 17th 2012 Tossing a coin The probability of getting a Heads or a Tails on a coin toss is both 0. coin=randi ( [0:1], [100,1]) It should more or less give you 50 0's and 50 1's. Write a program that simulates coin tossing. (Ti includes the toss that results in the first. Coin Flips, Risk to Reward Profile and Creating Your Own Synthetic Security Thus, the probability of getting 2 heads in a row is the probability of getting a head followed by a second flip where you also get a head. (Solution): Coin Toss Probability. Online virtual coin toss simulation app. Coin Toss: Simulation of a coin toss allowing the user to input the number of flips. 100 coins is a 31-digit number. 5, or you will stay in the current state with probability 0. I got the program down right but my results show a number for each coin flip in addition to the cout that says "The coin flip shows Heads/Tails". The probability of getting a given number of heads from four flips is, then, simply the number of ways that number of heads can occur, divided by the number of total results of four flips, 16. The answer to this is always going to be 50/50, or ½, or 50%. This is what I have so far but I keep getting errors. In the case of a coin, there are maximum two possible outcomes - head or tail. When the probability of an event is zero then the even is said to be impossible. Luckily, this is bundled up in a math/probability/stats concept called "combinations". Remark: The idea can be substantially generalized. This means that the theoretical probability to get either heads or tails is 0. Ask Question Asked 3 years, 6 months ago. What is the probability that a fair coin lands on heads on 4 out of 5 flips? What is the probability of getting at least one tail if a fair coin is flipped three times? Wha are the four properties of a binomial probability distribution?. Repeat steps 2-4 until the coin lands on its side every time. Junho: The chance of DB completing the coin scam on the first attempt, which is to toss a coin and get 10 heads in a row, is very unlikely. I have a problem I need to do for school. 6, and f = 0. Remember that each individual coin flip has a 50% chance of being heads. If we do the math, this is a probability of 0. Not so, says Diaconis. 2, which is the Kelly fraction for this α and p. 5, which is our probability of tossing heads and moving forward. Viewed 2k times 1. The two non-straight lines in Fig. The odds of the coins coming up with different faces showing is just 1 in 2. , Bernoulli trials). If you toss 2 coins, what are the chances you will get 2 heads? Record your predictions and explain your reasoning. The program should call a separate function flip()that takes no arguments and returns 0 for tails and 1 for heads. If you flip a coin and roll a six-sided die, what is the probability that the coin comes up heads and the die comes up 1? Since the two events are independent, the probability is simply the probability of a head (which is 1/2) times the probability of the die coming up 1 (which is 1/6). 57 flip_coins(100, 1000, 25) #> [1] 0. Flipping coins and the binomial distribution#2 Consider two coins, one fair and one unfair. We can use R to simulate an experiment of ipping a coin a number of times and compare our results with the theoretical probability. That was flip number 130,659,178 Flip again? Color The Coin!. If the probability of flipping heads is 70%, then the list contains 70 ones and 30 zeros. If you flip one coin, just two. of flipping one head with a coin is 50%, then the probability of flipping two heads at once is achieved by (adding or multiplying)_____ the separate probabilities. (Solution): Coin Toss Probability. What is the theoretical probability that the co n will land on tails? What is the theoretical probability that the co n will land on heads? If the com is flipped 140 times, how many times would you predict that the co n lands on heads? Johnny flipped a coin 450 times. If you flip three coins, it's eight - two for the first times two for the second times two for the third. The game is played in stages. The order does not matter as long as there are two head and two tails in the flip. e head or tail. Coin flipping, coin tossing, or heads or tails is the practice of throwing a coin in the air and checking which side is showing when it lands, in order to choose between two alternatives, sometimes used to resolve a dispute between two parties. Since 'fair' is used in the project description we know that the probability will be a 50% chance of getting either side. Flipping more coins¶ If we want to flip more coins, it's going to be a pain in the neck to make that table over and over. Then, it displays the results, as well as the theoretical and observed probabilities of each event happening. Given this information, what is the probability that it is a. So, I'll do it faster! When we flip the coin 9 times there are $$2^9$$ possible outcomes that can happen. A probability of one represents certainty: if you flip a coin, the probability you'll get heads or tails is one (assuming it can't land on the rim, fall into a black hole, or some such). Exactly 2 heads in 3 Coin Flips The ratio of successful events A = 3 to total number of possible combinations of sample space S = 8 is the probability of 2 heads in 3 coin tosses. If you toss a coin, you cannot get both a head and a tail at the same time, so this has zero probability. Gamblers Take Note: The Odds in a Coin Flip Aren't Quite 50/50 And the odds of spinning a penny are even more skewed in one direction, but which way? Flipping a coin isn't as fair as it seems. Simulating a coin toss in excel I guess when you start to look at gambling theories or probabilities the natural place to start is the coin toss. When we flip a coin there is always a probability to get a head or a tail is 50 percent. 4, then the power is 0. The different possible results from a probability model. Coin flipping, coin tossing, or heads or tails is the practice of throwing a coin in the air and checking which side is showing when it lands, in order to choose between two alternatives, sometimes used to resolve a dispute between two parties. There is a 50% probability that the first toss will end up heads. Have you ever flipped a coin as a way of deciding something with another person? The answer is probably yes. probability of any continuous interval is given by p(a ≤ X ≤ b) = ∫f(x) dx =Area under f(X) from a to b b a That is, the probability of an interval is the same as the area cut off by that interval under the curve for the probability densities, when the random variable is continuous and the total area is equal to 1. Probability, physics, and the coin toss L. Don't expect the numbers from trials to exactly match the predicted results--especially if you run only a few trials. A fair-sided coin (which means no casino hanky-panky with the coin not coming up heads or tails 50% of the time) is tossed three times. For example, consider a fair coin. Page last modified 07/17/2012 13:01:23. Logic problems, Understanding odds, Understanding probability Common Core Standards: Grade 4 Number & Operations: Fractions , Grade 5 Number & Operations in Base Ten , Grade 5 Operations & Algebraic Thinking. Note that the pattern of heads counts seems to form a smoother curve, but still matches the (scaled) binomial coefficients found on the tenth and twentieth rows of Pascal's Triangle. Assuming the coin is fair, p = 1/2 and q = 1/2 where 'p' is the probability of get. Thus, the odds of any throw being a tail is 1/2. 2, which is the Kelly fraction for this α and p. Heads did have an impressive run of 5 years in a row from 2009-2013. The probability of an event is a number indicating how likely that event will occur. The probability for equally likely outcomes in an event is:. The likelihood of an event is expressed as a number between zero (the event will never occur) and one (the event is certain). Use, probability formula = N u m b e r o f f a v o r a b l e o u t c o m e s T o t a l n u m b e r o f p o s s i. Update: The initial 6-for-6 report, from the Des Moines Register missed a few Sanders coin-toss wins. The views expressed are those of the author(s) and are not necessarily. Each team member will have 1 coin to flip. Take another penny and Super Glue it to the coin. Intro to Problem Solving. Although the basic probability formula isn't difficult, sometimes finding the numbers to plug into it can be tricky. Course : Introduction to Probability and Statistics, Math 113 Section 3234 Instructor: Abhijit Champanerkar Date: Oct 17th 2012 Tossing a coin The probability of getting a Heads or a Tails on a coin toss is both 0. Click "flip coins" to generate a new set of coin flips. Explore probability concepts by simulating repeated coin tosses. The experiment was conducted with motion-capture cameras, random experimentation, and an automated "coin-flipper" that could flip the coin on command. Let the program toss the coin 100 times, and count the number of times each side of the coin appears. Page last modified 07/17/2012 13:01:23. Begin your coin with just a single penny. something like this: def flip(p): '''this function return H with probability p''' # do something return result >> [flip(0. Coin toss probability is a classic for a reason: it's a realistic example kids can grasp quickly. Make a weighted coin by changing the probability of landing on heads using the slider; 0% means the coin always lands on tails and 100% means the coin always lands on heads. The coin toss is nothing but experimenting with tossing a coin. Since each coin toss has a probability of heads equal to 1/2, I simply need to multiply together 1/2 eleven times. Use Probability to Win Coin Flipping Games. In the first simulation, player A got lucky with 4 heads in 5 tosses. So, the probability that we will keep going is 1/2 of 1/4, or 1/8. Example: It is 12 if you have an experiment where you flip a coin and then roll a six sided die. An Easy GRE Probability Question. When you take these chocolates out, the probability for any one being taken out diminishes by 1 each time. A common topic in introductory probability is solving problems involving coin flips. What is the probability of getting two heads and four tails? Coin Flipping How can I figure out the chances of flipping a coin five times with the result T,T,T,H,H?. I am just learning Python on class so I am really at the basic. For the old java version, click here ; For the Spanish version, click here ; For the German version, click here; To. If we flip the coin 10 times, we are not guaranteed to get 5 heads and 5 tails. e head or tail. The first time it lands heads, and the second time it lands tails. What is the theoretical probability that the co n will land on tails? What is the theoretical probability that the co n will land on heads? If the com is flipped 140 times, how many times would you predict that the co n lands on heads? Johnny flipped a coin 450 times. Mentor: Yes! Now let's look at the coin flipping game that you just played. In this worksheet, they'll grab a quarter, give it a few tosses, and record the results for themselves. flips The number of desired coin flips. Coin toss probability is explored here with simulation. "The coin tosses are independent events; the coin doesn't have a memory. For the coin, number of outcomes to get heads = 1 Total number of possible outcomes = 2 Thus, we get 1/2 However, if you suspect that the coin may not be fair, you can toss the coin a large number of times and count the number of heads Suppose you flip the coin 100 and get 60 heads, then you know the best estimate to get head is 60/100 = 0. What is the probability that Player 1 will win the game?. Unformatted text preview: Tamara Curiel Gala Cano Mendelian Genetics Coin Toss Lab PRE-LAB DISCUSSION: In heredity, we are concerned with the occurrence, every time an egg is fertilized, of the probability that a particular gene or chromosome will be passed on through the egg, or through the sperm, to the offspring. Sign in to comment. What is the probability of getting exactly two heads and two tails. Coin Toss Probability Probability is the measurement of chances – likelihood that an event will occur. The toss or flip of a coin to randomly assign a decision traditionally involves throwing a coin into the air and seeing which side lands facing up. If the probability of coin flipping head = P The probability of coin flipping tail = 1 - P Now The probability of flipping heads & then tails = Probability of flipping tails & then heads = P(1 - P) Which means to make a fair coin toss we now need 2 flips Player 1 wins if the sequence is HT Player 2 wins if the sequence is TH Any other sequence. In 1921, the referee flipped the coin. We have k coins. 5, or you will stay in the current state with probability 0. I start by having my students create a "Heads Tails" T chart in their math journals and then writing a prediction for the result of tossing the coin 100 times. If after 200 flips, the. There's another sense in which the Haldane prior can be considered non-informative: the mean of the posterior distribution is now $\frac{\alpha + x}{\alpha + \beta + n}=\frac{x}{n}$, i. This means there is a 1 out of 128 chance of getting seven heads on seven coin flips. Over many coin flips the probability of at least half of. If we flip a fair coin 9 times, and the flips are independent, what's the probability that we get heads exactly 6 times? This works just like the last problem, only the numbers are bigger. If the coins show heads-tails (HT) or tails-heads (TH), player 2 gets 1 point. 8) for i in xrange(10)] [H,H,T,H,H,H,T,H,H,H]. We can either find this out using a formula or through Monte Carlo simulation. 5 for either heads or tails (assuming that the coin is purely mathematical and random, of course). Since the coin toss is a physical phenomenon governed by Newtonian mechanics, the question requires one to link probability and physics via a mathematical and statistical description of the coin's. Exactly 2 heads in 3 Coin Flips The ratio of successful events A = 3 to total number of possible combinations of sample space S = 8 is the probability of 2 heads in 3 coin tosses. If we toss a coin n times, and the probability of a head on any toss is p (which need not be equal to 1 / 2, the coin could be unfair), then the probability of exactly k. You can modify it as you like to simulate any number of flips. 5 (the default) with the 95% confidence interval. Tossing a Coin. The following shows the results of 100 tosses of five coins with a probability of heads of. Let be the probability that a run of or more consecutive heads appears in independent tosses of a coin (i. "The coin tosses are independent events; the coin doesn't have a memory. But not that much more likely. The variable timesflipped used for the while. In unbiased coin flip H or T occurs 50% of times. First, note that the problem will likely make reference to a "fair" coin. b) What do you think E[X] should be. Use buttons to view a bar chart of the coin flips, the probability distribution (also known as the probability mass. Since 'fair' is used in the project description we know that the probability will be a 50% chance of getting either side. If I toss a coin only 10 times I may end up with 9 heads and 1 tails. We assume that conditioned on Q=q, all coin tosses are independent. You flipped 1 coin of type US 1¢ Penny: Timestamp: 2020-05-05 01:39:10 UTC. the probability of tails is the same as heads, P(T) <=> P(H) 3. Each time you toss these coins, there are four possible outcomes: both heads penny head & dime tail penny tail & dime head both tails You will flip the pair of coins 20 times. Both team members flip their coins. Flip 10 coins, and and you're at a 4-digit number. If 5 is selected for "coin flips per trial," then a number is selected from the list five times, and these numbers are summed. Have you ever flipped a coin as a way of deciding something with another person? The answer is probably yes. It is about physics, the coin, and how the "tosser" is actually throwing it. There is a 50% probability that the first toss will end up heads. The toss or flip of a coin to randomly assign a decision traditionally involves throwing a coin into the air and seeing which side lands facing up. Life is full of random events! You need to get a "feel" for them to be a smart and successful person. However, the probability of getting exactly one heads out of seven flips is different (and the solution is given). This is a basic introduction to a probability distribution table. An Easy GRE Probability Question. But not that much more likely. Furthermore, if you toss a coin, it will eventually show heads, so this procedure ends in a finite number of flips with probability 1. There is also the very small probability that the coin will land. In Chapter 2 you learned that the number of possible outcomes of several independent events is the product of the number of possible outcomes of each event individually. Probability. Conditional probability question - coin toss. A fair-sided coin (which means no casino hanky-panky with the coin not coming up heads or tails 50% of the time) is tossed three times. But if any of the flipped coins comes up tails, or if no one chooses to flip a coin, you will all be doomed to spend the rest of your lives in the castle’s dungeon. the probability of tails is the same as heads, P(T) <=> P(H) 3. The coin can only land on one side or the other (event) but there are two possible outcomes: heads or tails. 5, which is our probability of tossing heads and moving forward. a)Give an algebraic formula for the probability mass function of X. Logic problems, Understanding odds, Understanding probability Common Core Standards: Grade 4 Number & Operations: Fractions , Grade 5 Number & Operations in Base Ten , Grade 5 Operations & Algebraic Thinking. Coin Flips, Risk to Reward Profile and Creating Your Own Synthetic Security Thus, the probability of getting 2 heads in a row is the probability of getting a head followed by a second flip where you also get a head. Applet: Instructions: Examples: Notes "H" count = , flips so far, number of coins: one flip "H" probability: 0. A probability of one means that the event is certain. Probability measures how certain we are a particular event will happen in a specific instance. That was flip number 130,659,178 Flip again? Color The Coin!. Probability: Independent Events. 5, or you will stay in the current state with probability 0. The toss of a coin, throwing dice and lottery draws are all examples of random events. The instructions are written in the handout for the students to understand how to complete the activi. of flipping one head with a coin is 50%, then the probability of flipping two heads at once is achieved by (adding or multiplying)_____ the separate probabilities. When the coin is flipped and the first three flips are heads, the fourth flip still has the probability of ½ However, many people misunderstand that the first three flips somehow influence the fourth flip, but they do not. Don't expect the numbers from trials to exactly match the predicted results--especially if you run only a few trials. Were you to toss the coin 100 times, you would get a clearer view of how probable it is that the coin lands on heads each time. A coin toss is a tried-and-true way for your fifth grader to understand odds. We do not know if we will get heads or tails. I got a question on the coin flip project. If we flip the coin 10 times, we are not guaranteed to get 5 heads and 5 tails. Repeat steps 2-4 until the coin lands on its side every time. Expected Value represents the average outcome of a series of random events with identical odds being repeated over a long period of time. Asked in Statistics , Probability. It begins with the two title characters caught in a most unusual coin game. When flipping a coin, the probability of getting a head does not change no matter how many times you flip the coin. Example: It is 12 if you have an experiment where you flip a coin and then roll a six sided die. If the coins show heads-tails (HT) or tails-heads (TH), player 2 gets 1 point. With three coins, all three landing on the same side is 1 in 8. Ask Question Asked 3 years, 6 months ago. The probability for equally likely outcomes in an event is:. In this video, we' ll explore the probability of getting at least one heads in multiple flips of a fair coin. I need to write a python program that will flip a coin 100 times and then tell how many times tails and heads were flipped. Explore probability concepts by simulating repeated coin tosses. If the probability of coin flipping head = P The probability of coin flipping tail = 1 - P Now The probability of flipping heads & then tails = Probability of flipping tails & then heads = P(1 - P) Which means to make a fair coin toss we now need 2 flips Player 1 wins if the sequence is HT Player 2 wins if the sequence is TH Any other sequence. Students will flip a coin a total of ten times per trial and record results by simply shading in the space below the realistic-looking coins. Flip 4 coins, and you're at 16 outcomes, a 2-digit number. The number of possible outcomes gets greater with the increased number of coins. Coin Flipper. For example, given 5 trials per experiment and 20 experiments, the program will flip a coin 5 times and record the results 20 times. Become a member and unlock all. The probability of A and B is 1/100. That means I flipped the coin 15 times. It's pretty much a "coin flip". The views expressed are those of the author(s) and are not necessarily. (Solution): Coin Toss Probability. Suppose a coin tossed then we get two possible outcomes either a ‘head’ ( H) or a ‘tail’ ( T ), and it is impossible to predict whether the result of a toss will be a ‘head’ or ‘tail’. However, the probability of getting exactly one heads out of seven flips is different (and the solution is given). But I want to simulate coin which gives H with probability 'p' and T with probability '(1-p)'. The first person to flip heads wins. Flipping coins and the binomial distribution#2 Consider two coins, one fair and one unfair. from the previous assumptions follows that given any sequence of coin tossing results, the next toss has the probability P(T) <=> P(H). Recommended: Please try your approach. Logical Reasoning. Procedure 1: Statistical Probability Reasoning. 5×10 20 chance of getting a string of 76 heads. As long as the coin was not manipulated the theoretical probabilities of both outcomes are the same—they are equally probable. Let us return to the coin flip experiment. If the probability of an event is high, it is more likely that the event will happen. And you can get a calculator out to figure that out in terms of a percentage. This article shows you the steps for solving the most common types of basic questions on this subject. That means I flipped the coin 15 times. I start by having my students create a "Heads Tails" T chart in their math journals and then writing a prediction for the result of tossing the coin 100 times. The coin toss is nothing but experimenting with tossing a coin. , the sample frequency of heads, which is the frequentist MLE estimate of $\theta$ for the Binomial model of the coin flip problem. Unformatted text preview: Tamara Curiel Gala Cano Mendelian Genetics Coin Toss Lab PRE-LAB DISCUSSION: In heredity, we are concerned with the occurrence, every time an egg is fertilized, of the probability that a particular gene or chromosome will be passed on through the egg, or through the sperm, to the offspring. Lends to discussion and discovery of probability, elementary understanding of mode, and graphic organization of information to compare and contrast. Begin your coin with just a single penny. If you flip three coins, it's eight - two for the first times two for the second times two for the third. A sequence of consecutive events is also called a "run" of events. You flipped 1 coin of type US 1¢ Penny: Timestamp: 2020-05-05 01:39:10 UTC. It is measured between 0 and 1, inclusive. The probability of coming up heads on the first flip is 1/2. It is about physics, the coin, and how the "tosser" is actually throwing it. the coin does not and can not "remember" last result 4. For example, consider a fair coin. In the first simulation, player A got lucky with 4 heads in 5 tosses. Remark: The idea can be substantially generalized. The probability of getting a given number of heads from four flips is, then, simply the number of ways that number of heads can occur, divided by the number of total results of four flips, 16. If we toss a coin $n$ times, and the probability of a head on any toss is $p$ (which need not be equal to $1/2$, the coin could be unfair), then the probability of exactly $k$ heads is \binom{n}{k}p^k(1-p)^{n-k}. Coin Toss: Simulation of a coin toss allowing the user to input the number of flips. Simple numbers. Ask Question Asked 3 years, 6 months ago. The sum of all possible outcomes is always 1 (or 100%) because it is certain that one of the possible outcomes will happen. Based on your flip results, you will infer which of the coins you were given. Recommended: Please try your approach. 5, or you will stay in the current state with probability 0. The game is played in stages. The first person to flip heads wins. "The coin tosses are independent events; the coin doesn't have a memory. If the probability of flipping heads is 70%, then the list contains 70 ones and 30 zeros. You flipped 1 coin of type US 1¢ Penny: Timestamp: 2020-05-05 01:39:10 UTC. The majority of times, if a coin is heads-up when it is flipped, it will remain heads-up when it lands. First, note that the problem will likely make reference to a "fair" coin. In particular, ( 10 3) = 10! 3! 7!. Wait for your coin to dry; Flip (toss) your coin 40 times and record the number of times it lands on its side. The post is correct that the odds of getting. Your function will have to label each of those sequences with a door, since leaving any sequence unlabeled would contradict the assumption that the method always produces a result after F flips. I need to land on heads 3 times or more out of 6, in 80% of all trials. So, I'll do it faster! When we flip the coin 9 times there are $$2^9$$ possible outcomes that can happen. The coin has no desire to continue a particular streak, so it's not affected by any number of previous coin tosses. Coin Toss Activity is a great way for students to have fun and learn about calculating probability. Probability measures how certain we are a particular event will happen in a specific instance. The two non-straight lines in Fig. A single flip of a coin has an uncertain outcome. We could call a Head a success; and a Tail, a failure. In the first simulation, player A got lucky with 4 heads in 5 tosses. For four coin flips, our intuition was probably right: more likely to get two heads. 60 I tried this: P(2H) = 4C2 * 0. 7 flip_coins(100, 100, 5) #> [1] 0. Ask Question Asked 3 years, 6 months ago. The program should call a separate function flip()that takes no arguments and returns 0 for tails and 1 for heads. We all know a coin toss gives you a 50% chance of winning, but is it always that way? Delve into the inner-workings of coin toss probability with this activity. The sum of all possible outcomes is always 1 (or 100%) because it is certain that one of the possible outcomes will happen. Flipping the coin once is a Bernoulli trial, since there are exactly two complementary outcomes (flipping a head and flipping a tail), and they are both 1 2 \frac{1}{2} 2 1 no matter how many times the coin is flipped. What is the probability that a fair coin lands on heads on 4 out of 5 flips? Wha are the four properties of a binomial probability distribution? In a carnival game, there are six identical boxes, one of which contains a prize. We’ll set p=0. 5, which is our probability of tossing heads and moving forward. | 2020-07-02T21:18:08 | {
"domain": "theoximoron.it",
"url": "http://ctjn.theoximoron.it/coin-flip-probability.html",
"openwebmath_score": 0.7363460659980774,
"openwebmath_perplexity": 368.0354943188166,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9833429629196683,
"lm_q2_score": 0.8705972717658209,
"lm_q1q2_score": 0.856095700727982
} |
http://mathhelpforum.com/advanced-algebra/95329-need-help-elem-linear-algebra-problem.html | # Math Help - Need Help with Elem. Linear Algebra Problem
1. ## Need Help with Elem. Linear Algebra Problem
Express the invertible matrix
$\left(\begin{array}{ccc}1&2&1\\1&0&1\\1&1&2\end{ar ray}\right)$
as a product of elementary matrices.
--------------------------
The answer in the back of the book has a product of 6 matrices. I think I'm supposed to use the identity matrix and perform the same row operations on it as I would to obtain the above matrix... or something like that. Any help with this would be appreciated!
2. Originally Posted by paupsers
Express the invertible matrix
$\left(\begin{array}{ccc}1&2&1\\1&0&1\\1&1&2\end{ar ray}\right)$
as a product of elementary matrices.
--------------------------
The answer in the back of the book has a product of 6 matrices. I think I'm supposed to use the identity matrix and perform the same row operations on it as I would to obtain the above matrix... or something like that. Any help with this would be appreciated!
Yes, that's right. A "row operation" is one of three kinds:
1) swap two rows
2) multiply every member of a row by the same constant.
3) replace every member of a row by itself plus a constant time the corresponding member of a second row.
An "elementary matrix" is a matrix constructed from the identity matrix by a single row operation.
What you want to do is reduce the given matrix to the identity matrix (which is possible because it is invertible) by row operations, writing down the elementary matrix corresponding to each row operation.
For example, starting from
$\left(\begin{array}{ccc}1&2&1\\1&0&1\\1&1&2\end{ar ray}\right)$
My first steps would be to reduce the first column to the column $\left(\begin{array}{c}1 \\ 0 \\ 0\end{array}\right)$.
The number in the upper left is already 1 so I don't need to change that. I can get 0 in the next row by subtracting the first row from the second row. The elementary matrix corresponding to that is
$\left(\begin{array}{ccc}1 & 0 & 0 \\ -1 & 1 & 0 \\ 0 & 0 & 1\end{array}\right)$
To get 0 in the third row, I need to subtract the first row from the third row. That gives the elementary matrix $\left(\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ -1 & 0 & 1\end{array}\right)$
and they convert the matrix to
$\left(\begin{array}{ccc}1 & 2 & 1 \\ 0 & -2 & 0 \\ 0 & -1 & 1\end{array}\right)$
Now you need to convert the second column properly and then the third. Can you continue?
After you have converted the matrix to the identity matrix and written down all the corresponding elementary matrices (in the proper order) so they multiply to give the original matrix.
Normally it would take 9 row operations to reduce a 3 by 3 matrix to the identity matrix and so the matrix would be represented as the product of 9 elementary matrices. Notice that the upper left number was already 1 so we didn't need a row operation for that. Also that 0 now in the second row, third column will stay there we won't need a row operation to change that. I suspect that sort of thing, a 0 or 1 already in the correct position will happen once more to give 6 rather than 9 elementary matrices.
3. This is exactly what I tried doing, but it's not the answer the book gives (although it is similar). Is the solution to this problem unique?
4. No, it is not because you can follow different "paths" to row reduce to the identity matrix. What I showed is the most commonly used "path". | 2015-06-02T19:53:32 | {
"domain": "mathhelpforum.com",
"url": "http://mathhelpforum.com/advanced-algebra/95329-need-help-elem-linear-algebra-problem.html",
"openwebmath_score": 0.8484682440757751,
"openwebmath_perplexity": 178.66534325732104,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.9833429638959675,
"lm_q2_score": 0.8705972583359806,
"lm_q1q2_score": 0.8560956883718065
} |
https://math.stackexchange.com/questions/1583146/transitive-relation | # Transitive relation
Consider $A$ is a relation de fined on $R$ (real numbers) where $A = \{(a,b):|a-b|<4, a, b \in R\}$. Prove/disprove $A$ is transitive.
I know if $|a-b|<4$ and $|b-c|<4$, then, $|a-c|<4$ ; A is transitive. Can I directly prove it with any counter example such as for $a=6$, $b=3$ and $c=1$ this relation is not transitive because $|6-1|>4$.
Is this suitable for prove or disprove questions of relations?
Your answer is correct. In general a relation $A$ is transitive if for all $a,b,c$ we have $(a,b)\in A$ and $(b,c)\in A$ implies $(a,c)\in A$. That means that if we can find just one counter example (such as $a=6$, $b=3$ and $c=1$) $A$ cannot be transitive.
Your answer is quite right. The question is basically "if $a$ is near to $b$, and $b$ is near to $c$, is $a$ near to $c$?" | 2021-01-19T15:28:58 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/1583146/transitive-relation",
"openwebmath_score": 0.921374499797821,
"openwebmath_perplexity": 162.6942033835524,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9873750481179169,
"lm_q2_score": 0.867035763237924,
"lm_q1q2_score": 0.856089478447
} |
https://math.stackexchange.com/questions/2434494/expansion-in-terms-of-legendre-polynomial | # Expansion in terms of legendre polynomial
Obtain the first three terms in the expansion of function in terms of legendre polynomial F(x) in a series of the form $$F(x) = \sum_{k=0}^{\infty} A_k P_k(x)$$ where $$F(x)=\{\cos(x) \text{ for } 0 \le x \le \pi/2 \$$$$0 \text{ for } \frac{\pi}{2} \le x \le \pi.\}$$
What I know is I have to use legendre's expansion formula i.e,$F(x) =\sum A_kP_k(x)$ where $-1≤x≤1$ But obviously I cannot use it directly because the range of $x$ differs. I have tried substituting $x=\cos(\theta)$ but no success so far.
Extended hint: As already noted in your almost identical question Legendre polynomials you should apply a linear transformation, because the Legendre polynomial have the orthogonality interval $[-1,1]$. With $x=\frac{\pi}{2}(t+1)$ and $t=\frac{2x}{\pi}-1,$ define $F(x)=:\tilde{F}(t).$ Now $\tilde{F}$ is defined on $[-1,1]$ and you can compute $$F(x)=\tilde{F}(t) = \sum A_k P_k(t)=\sum A_k P_k\left(\frac{2x}{\pi}-1\right)$$ with $$A_k=\frac{2k+1}{2}\int_{-1}^1 \tilde{F}(t) P_k(t) dt$$
If you use this approach and compute $A_0, \dots, A_3$, you get the following graphics with the original function $F(x)$ in red and the approximation in green:
• Yes of course, because your function is not defined for $x=-1$. The transformation associates $x=0 \leftrightarrow t=-1$ and $x=\pi \leftrightarrow t=1$. You should compute the expansion for $\tilde{F}(t), t\in[-1,1].$ – gammatester Sep 18 '17 at 15:08 | 2019-08-17T16:58:31 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2434494/expansion-in-terms-of-legendre-polynomial",
"openwebmath_score": 0.9301050305366516,
"openwebmath_perplexity": 103.11606074523804,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. Yes\n2. Yes",
"lm_q1_score": 0.9873750492324249,
"lm_q2_score": 0.8670357546485407,
"lm_q1q2_score": 0.8560894709323755
} |
https://www.physicsforums.com/threads/product-of-two-consecutive-integers.367041/ | # Homework Help: Product of two consecutive integers
1. Jan 4, 2010
### nobahar
1. The problem statement, all variables and given/known data
Prove that $$n^2+n$$ is even. Where n is a positive integer.
2. Relevant equations
$$n^2+n$$
3. The attempt at a solution
$$n^2+n = n(n+1)$$
One of which must be even, and therefore the product of 2 and an integer k.
$$n = 2k, \left \left 2*(k(n+1))$$
or
$$n+1 = 2k, \left \left 2*(n*k)$$
Is there a better way of doing this? I read this is not an inductive proof; what would this entail?
2. Jan 4, 2010
### rochfor1
This looks absolutely fine. No need for induction at all.
3. Jan 4, 2010
### HallsofIvy
As rochfor1 said, that is a perfectly good proof and simpler than "proof by induction".
But since you ask, here goes:
If n= 1, then $n^2+ n= 1^2+ 1= 2$ which is even.
Now, suppose that $k^2+ k$ is even and look at $(k+1)^2+ (k+1)$
$(k+1)^2+ (k+1)= k^2+ 2k+ 1+ k+ 1$$= (k^2+ 2k)+ 2k+ 2$. By the induction hypothesis, $k^2+ k$ is even and so $k^2+ k= 2m$ for some integer m (that is the definition of "even") so $(k+1)^2+ (k+1)= (k^2+ 2k)+ 2k+ 2$$= 2m+ 2k+ 2= 2(m+k+1)$. Since that is "2 times an integer", it is even.
Having proved that the statement is true for 1 and that "if it is true for k, it is true for k+1", by induction, it is true for all positive integers.
Yet a third way: If n is a positive integer, it is either even or odd.
case 1: n is even. Then n= 2m for some integer m. $n^2= 4m^2$ so $n^2+ n= 4m^2+ 2m= 2(m^2+ m)$. Since that is 2 times an integer, it is even.
case 2: n is odd. Then n= 2m+ 1 for some integer m. $n^2= (2m+1)^2= 4m^2+ 4m+ 1$ so $n^2+ n= 4m^2+ 4m+ 1+ 2m+ 1$$= 4m^2+ 6m+ 2= 2(2m^2+ 3m+ 1)$. Again that is 2 times an integer and so is even.
4. Jan 5, 2010
### nobahar
Many thanks rochfor1 and HallsofIvy!
5. Jan 5, 2010
### icystrike
Simple reasoning is needed.
For two consecutive integer , one will be even and another will be odd.
Thus;
$$n\equiv0(mod 2)$$
and
$$n+1\equiv1(mod 2)$$
We may now conclude that $$n(n+1)\equiv0(mod 2)$$ and means that n²+n is even.
6. Jan 5, 2010
File size:
24.9 KB
Views:
388
7. Jan 5, 2010
### Dick
If n is odd, n^2 is odd. If n is even, n^2 is even. odd+odd=even and even+even=even.
8. Jan 6, 2010
### ideasrule
OP: Why do you want a proof by induction? The proof you gave is simple and beautiful.
9. Jan 6, 2010
### Staff: Mentor
What you say is true, but you can't ascertain which of them will be even and which will be odd. In what follows, you are assuming that n is even and n+1 is odd. That is one of two possible cases, so you work is not complete.
10. Jan 7, 2010
### icystrike
yes that is true. for the other case , it can be proved analogously . | 2018-09-23T23:49:16 | {
"domain": "physicsforums.com",
"url": "https://www.physicsforums.com/threads/product-of-two-consecutive-integers.367041/",
"openwebmath_score": 0.6292749047279358,
"openwebmath_perplexity": 1312.1021854840985,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.9873750529474512,
"lm_q2_score": 0.8670357494949105,
"lm_q1q2_score": 0.8560894690648703
} |
http://gmatclub.com/forum/inequalities-trick-91482.html?kudos=1 | Find all School-related info fast with the new School-Specific MBA Forum
It is currently 07 Jul 2015, 07:46
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# Inequalities trick
Author Message
TAGS:
Current Student
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2800
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Followers: 196
Kudos [?]: 1162 [56] , given: 235
Inequalities trick [#permalink] 16 Mar 2010, 09:11
56
KUDOS
117
This post was
BOOKMARKED
I learnt this trick while I was in school and yesterday while solving one question I recalled.
Its good if you guys use it 1-2 times to get used to it.
Suppose you have the inequality
f(x) = (x-a)(x-b)(x-c)(x-d) < 0
Just arrange them in order as shown in the picture and draw curve starting from + from right.
now if f(x) < 0 consider curve having "-" inside and if f(x) > 0 consider curve having "+" and combined solution will be the final solution. I m sure I have recalled it fully but if you guys find any issue on that do let me know, this is very helpful.
Don't forget to arrange then in ascending order from left to right. a<b<c<d
So for f(x) < 0 consider "-" curves and the ans is : (a < x < b) , (c < x < d)
and for f(x) > 0 consider "+" curves and the ans is : (x < a), (b < x < c) , (d < x)
If f(x) has three factors then the graph will have - + - +
If f(x) has four factors then the graph will have + - + - +
If you can not figure out how and why, just remember it.
Try to analyze that the function will have number of roots = number of factors and every time the graph will touch the x axis.
For the highest factor d if x>d then the whole f(x) > 0 and after every interval of the roots the signs will change alternatively.
Attachments
1.jpg [ 6.73 KiB | Viewed 29844 times ]
_________________
Fight for your dreams :For all those who fear from Verbal- lets give it a fight
Money Saved is the Money Earned
Jo Bole So Nihaal , Sat Shri Akaal
GMAT Club Premium Membership - big benefits and savings
Gmat test review :
670-to-710-a-long-journey-without-destination-still-happy-141642.html
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 5682
Location: Pune, India
Followers: 1415
Kudos [?]: 7357 [43] , given: 186
Re: Inequalities trick [#permalink] 22 Oct 2010, 05:33
43
KUDOS
Expert's post
39
This post was
BOOKMARKED
Yes, this is a neat little way to work with inequalities where factors are multiplied or divided. And, it has a solid reasoning behind it which I will just explain.
If (x-a)(x-b)(x-c)(x-d) < 0, we can draw the points a, b, c and d on the number line.
e.g. Given (x+2)(x-1)(x-7)(x-4) < 0, draw the points -2, 1, 7 and 4 on the number line as shown.
Attachment:
doc.jpg [ 7.9 KiB | Viewed 29149 times ]
This divides the number line into 5 regions. Values of x in right most region will always give you positive value of the expression. The reason for this is that if x > 7, all factors above will be positive.
When you jump to the next region between x = 4 and x = 7, value of x here give you negative value for the entire expression because now, (x - 7) will be negative since x < 7 in this region. All other factors are still positive.
When you jump to the next region on the left between x = 1 and x = 4, expression will be positive again because now two factors (x - 7) and (x - 4) are negative, but negative x negative is positive... and so on till you reach the leftmost section.
Since we are looking for values of x where the expression is < 0, here the solution will be -2 < x < 1 or 4< x < 7
It should be obvious that it will also work in cases where factors are divided.
e.g. (x - a)(x - b)/(x - c)(x - d) < 0
(x + 2)(x - 1)/(x -4)(x - 7) < 0 will have exactly the same solution as above.
Note: If, rather than < or > sign, you have <= or >=, in division, the solution will differ slightly. I will leave it for you to figure out why and how. Feel free to get back to me if you want to confirm your conclusion.
_________________
Karishma
Veritas Prep | GMAT Instructor
My Blog
Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Math Forum Moderator Joined: 20 Dec 2010 Posts: 2028 Followers: 138 Kudos [?]: 1165 [15] , given: 376 Re: Inequalities trick [#permalink] 11 Mar 2011, 05:49 15 This post received KUDOS 6 This post was BOOKMARKED vjsharma25 wrote: VeritasPrepKarishma wrote: vjsharma25 wrote: How you have decided on the first sign of the graph?Why it is -ve if it has three factors and +ve when four factors? Check out my post above for explanation. I understand the concept but not the starting point of the graph.How you decide about the graph to be a sine or cosine waveform?Meaning graph starts from the +ve Y-axis for four values and starts from -ve Y-axis for three values. What if the equation you mentioned is (x+2)(x-1)(x-7)<0,will the last two ranges be excluded or the graph will also change? I always struggle with this as well!!! There is a trick Bunuel suggested; (x+2)(x-1)(x-7) < 0 Here the roots are; -2,1,7 Arrange them in ascending order; -2,1,7; These are three points where the wave will alternate. The ranges are; x<-2 -2<x<1 1<x<7 x>7 Take a big value of x; say 1000; you see the inequality will be positive for that. (1000+2)(1000-1)(1000-7) is +ve. Thus the last range(x>7) is on the positive side. Graph is +ve after 7. Between 1 and 7-> -ve between -2 and 1-> +ve Before -2 -> -ve Since the inequality has the less than sign; consider only the -ve side of the graph; 1<x<7 or x<-2 is the complete range of x that satisfies the inequality. _________________ Manager Joined: 29 Sep 2008 Posts: 150 Followers: 3 Kudos [?]: 49 [11] , given: 1 Re: Inequalities trick [#permalink] 22 Oct 2010, 10:45 11 This post received KUDOS 6 This post was BOOKMARKED if = sign is included with < then <= will be there in solution like for (x+2)(x-1)(x-7)(x-4) <=0 the solution will be -2 <= x <= 1 or 4<= x <= 7 in case when factors are divided then the numerator will contain = sign like for (x + 2)(x - 1)/(x -4)(x - 7) < =0 the solution will be -2 <= x <= 1 or 4< x < 7 we cant make 4<=x<=7 as it will make the solution infinite correct me if i am wrong Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 5682 Location: Pune, India Followers: 1415 Kudos [?]: 7357 [6] , given: 186 Re: Inequalities trick [#permalink] 11 Mar 2011, 18:57 6 This post received KUDOS Expert's post vjsharma25 wrote: I understand the concept but not the starting point of the graph.How you decide about the graph to be a sine or cosine waveform?Meaning graph starts from the +ve Y-axis for four values and starts from -ve Y-axis for three values. What if the equation you mentioned is (x+2)(x-1)(x-7)<0,will the last two ranges be excluded or the graph will also change? Ok, look at this expression inequality: (x+2)(x-1)(x-7) < 0 Can I say the left hand side expression will always be positive for values greater than 7? (x+2) will be positive, (x - 1) will be positive and (x-7) will also be positive... so in the rightmost regions i.e. x > 7, all three factors will be positive. The expression will be positive when x > 7, it will be negative when 1 < x < 7, positive when -2 , x < 1 and negative when x < -2. We need the region where the expression is less than 0 i.e. negative. So either 1 < x < 7 or x < -2. Now let me add another factor: (x+8)(x+2)(x-1)(x-7) Can I still say that the entire expression is positive in the rightmost region i.e. x>7 because each one of the four factors is positive? Yes. So basically, your rightmost region is always positive. You go from there and assign + and - signs to the regions. Your starting point is the rightmost region. Note: Make sure that the factors are of the form (ax - b), not (b - ax)... e.g. (x+2)(x-1)(7 - x)<0 Convert this to: (x+2)(x-1)(x-7)>0 (Multiply both sides by '-1') Now solve in the usual way. Assign '+' to the rightmost region and then alternate with '-' Since you are looking for positive value of the expression, every region where you put a '+' will be the region where the expression will be greater than 0. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199
Veritas Prep Reviews
Manager
Status: On...
Joined: 16 Jan 2011
Posts: 189
Followers: 3
Kudos [?]: 42 [5] , given: 62
Re: Inequalities trick [#permalink] 10 Aug 2011, 16:01
5
KUDOS
5
This post was
BOOKMARKED
WoW - This is a cool thread with so many thing on inequalities....I have compiled it together with some of my own ideas...It should help.
1) CORE CONCEPT
@gurpreetsingh -
Suppose you have the inequality
f(x) = (x-a)(x-b)(x-c)(x-d) < 0
Arrange the NUMBERS in ascending order from left to right. a<b<c<d
Draw curve starting from + from right.
now if f(x) < 0 consider curve having "-" inside and if f(x) > 0 consider curve having "+" and combined solution will be the final solution. I m sure I have recalled it fully but if you guys find any issue on that do let me know, this is very helpful.
So for f(x) < 0 consider "-" curves and the ans is : (a < x < b) , (c < x < d)
and for f(x) > 0 consider "+" curves and the ans is : (x < a), (b < x < c) , (d < x)
If f(x) has three factors then the graph will have - + - +
If f(x) has four factors then the graph will have + - + - +
If you can not figure out how and why, just remember it.
Try to analyze that the function will have number of roots = number of factors and every time the graph will touch the x axis.
For the highest factor d if x>d then the whole f(x) > 0 and after every interval of the roots the signs will change alternatively.
Note: Make sure that the factors are of the form (ax - b), not (b - ax)...
example -
(x+2)(x-1)(7 - x)<0
Convert this to: (x+2)(x-1)(x-7)>0 (Multiply both sides by '-1')
Now solve in the usual way. Assign '+' to the rightmost region and then alternate with '-'
Since you are looking for positive value of the expression, every region where you put a '+' will be the region where the expression will be greater than 0.
2) Variation - ODD/EVEN POWER
@ulm/Karishma -
if we have even powers like (x-a)^2(x-b)
we don't need to change a sign when jump over "a".
This will be same as (x-b)
We can ignore squares BUT SHOULD consider ODD powers
example -
2.a
(x-a)^3(x-b)<0 is the same as (x-a)(x-b) <0
2.b
(x - a)(x - b)/(x - c)(x - d) < 0 ==> (x - a)(x - b)(x-c)^-1(x-d)^-1 <0
is the same as (x - a)(x - b)(x - c)(x - d) < 0
3) Variation <= in FRACTION
@mrinal2100 -
if = sign is included with < then <= will be there in solution
like for (x+2)(x-1)(x-7)(x-4) <=0 the solution will be -2 <= x <= 1 or 4<= x <= 7
BUT if it is a fraction the denominator in the solution will not have = SIGN
example -
3.a
(x + 2)(x - 1)/(x -4)(x - 7) < =0
the solution will be -2 <= x <= 1 or 4< x < 7
we cant make 4<=x<=7 as it will make the solution infinite
4) Variation - ROOTS
@Karishma -
As for roots, you have to keep in mind that given $$\sqrt{x}$$, x cannot be negative.
$$\sqrt{x}$$ < 10
implies 0 < $$\sqrt{x}$$ < 10
Squaring, 0 < x < 100
Root questions are specific. You have to be careful. If you have a particular question in mind, send it.
Refer - inequalities-and-roots-118619.html#p959939
Some more useful tips for ROOTS....I am too lazy to consolidate
<5> THESIS -
@gmat1220 -
Once algebra teacher told me - signs alternate between the roots. I said whatever and now I know why Watching this article is a stroll down the memory lane.
I will save this future references....
Anyone wants to add ABSOLUTE VALUES....That will be a value add to this post
_________________
Labor cost for typing this post >= Labor cost for pushing the Kudos Button
kudos-what-are-they-and-why-we-have-them-94812.html
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 5682
Location: Pune, India
Followers: 1415
Kudos [?]: 7357 [3] , given: 186
Re: Inequalities trick [#permalink] 09 Sep 2013, 22:35
3
KUDOS
Expert's post
karannanda wrote:
gurpreetsingh wrote:
I learnt this trick while I was in school and yesterday while solving one question I recalled.
Its good if you guys use it 1-2 times to get used to it.
Suppose you have the inequality
f(x) = (x-a)(x-b)(x-c)(x-d) < 0
Just arrange them in order as shown in the picture and draw curve starting from + from right.
now if f(x) < 0 consider curve having "-" inside and if f(x) > 0 consider curve having "+" and combined solution will be the final solution. I m sure I have recalled it fully but if you guys find any issue on that do let me know, this is very helpful.
Don't forget to arrange then in ascending order from left to right. a<b<c<d
So for f(x) < 0 consider "-" curves and the ans is : (a < x < b) , (c < x < d)
and for f(x) > 0 consider "+" curves and the ans is : (x < a), (b < x < c) , (d < x)
If f(x) has three factors then the graph will have - + - +
If f(x) has four factors then the graph will have + - + - +
If you can not figure out how and why, just remember it.
Try to analyze that the function will have number of roots = number of factors and every time the graph will touch the x axis.
For the highest factor d if x>d then the whole f(x) > 0 and after every interval of the roots the signs will change alternatively.
Hi Gurpreet,
Thanks for the wonderful method.
I am trying to understand it so that i can apply it in tests.
Can you help me in applying this method to the below expression to find range of x.
x^3 – 4x^5 < 0?
I am getting the roots as -1/2, 0, 1/2 and when i plot them using this method, putting + in the rightmost region, I am not getting correct result. Not sure where i am going wrong. Can you pls help.
Before you apply the method, ensure that the factors are of the form (x - a)(x - b) etc
$$x^3 - 4x^5 < 0$$
$$x^3 ( 1 - 4x^2) < 0$$
$$x^3(1 - 2x) (1 + 2x) < 0$$
$$4x^3(x - 1/2)(x + 1/2) > 0$$ (Notice the flipped sign. We multiplied both sides by -1 to convert 1/2 - x to x - 1/2)
Now the transition points are 0, -1/2 and 1/2 so put + in the rightmost region.
The solution will be x > 1/2 or -1/2 < x< 0.
Check out these posts discussing such complications:
http://www.veritasprep.com/blog/2012/06 ... e-factors/
http://www.veritasprep.com/blog/2012/07 ... ns-part-i/
http://www.veritasprep.com/blog/2012/07 ... s-part-ii/
_________________
Karishma
Veritas Prep | GMAT Instructor
My Blog
Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 5682 Location: Pune, India Followers: 1415 Kudos [?]: 7357 [2] , given: 186 Re: Inequalities trick [#permalink] 23 Jul 2012, 02:13 2 This post received KUDOS Expert's post Stiv wrote: VeritasPrepKarishma wrote: mrinal2100: Kudos to you for excellent thinking! Correct me if I'm wrong. If the lower part of the equation$$\frac {(x+2)(x-1)}{(x-4)(x-7)}$$ were $$4\leq x \leq 7$$, than the lower part would be equal to zero,thus making it impossible to calculate the whole equation. x cannot be equal to 4 or 7 because if x = 4 or x = 7, the denominator will be 0 and the expression will not be defined. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199
Veritas Prep Reviews
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 336
Followers: 52
Kudos [?]: 380 [2] , given: 81
Inequalities trick [#permalink] 06 Jan 2015, 02:35
2
KUDOS
Expert's post
5
This post was
BOOKMARKED
Just came across this useful discussion.
VeritasPrepKarishma has given a very lucid explanation of how this “wavy line” method works.
I have noticed that there is still a little scope to take this discussion further. So here are my two cents on it.
I would like to highlight an important special case in the application of the Wavy Line Method
When there are multiple instances of the same root:
Try to solve the following inequality using the Wavy Line Method:
$$(x-1)^2(x-2)(x-3)(x-4)^3 < 0$$
To know how you did, compare your wavy line with the correct one below.
Did you notice how this inequality differs from all the examples above?
Notice that two of the four terms had an integral power greater than 1.
How to draw the wavy line for such expressions?
Let me directly show you how the wavy line would look and then later on the rule behind drawing it.
Attachment:
File comment: Observe how the wave bounces back at x = 1.
bounce.png [ 10.4 KiB | Viewed 2175 times ]
Notice that the curve bounced down at the point x = 1. (At every other root, including x = 4 whose power was 3, it was simply passing through them.)
Can you figure out why the wavy line looks like this for this particular inequality?
(Hint: The wavy line for the inequality
$$(x-1)^{38}(x-2)^{57}(x-3)^{15}(x-4)^{27} < 0$$
Is also the same as above)
Come on! Give it a try.
If you got it right, you’ll see that there are essentially only two rules while drawing a wavy line. (Remember, we’ll refer the region above the number line as positive region and the region below the number line as negative region.)
How to draw the wavy line?
1. How to start: Start from the top right most portion. Be ready to alternate (or not alternate) the region of the wave based on how many times a point is root to the given expression.
2. How to alternate: In the given expression, if the power of a term is odd, then the wave simply passes through the corresponding point (root) into the other region (to –ve region if the wave is currently in the positive region and to the +ve region if the wave is currently in the negative region). However, if the power of a term is even, then the wave bounces back into the same region.
Now look back at the above expression and analyze your wavy line. Were you (intuitively) using the above mentioned rules while drawing your wavy line?
Solution
Once you get your wavy line right, solving an inequality becomes very easy. For instance, for the above inequality, since we need to look for the space where the above expression would be less than zero, look for the areas in the wavy line where the curve is below the number line.
So the correct solution set would simply be {3 < x < 4} U {{x < 2} – {1}}
In words, it is the Union of two regions region1 between x = 3 and x = 4 and region2 which is x < 2, excluding the point x = 1.
Food for Thought
Now, try to answer the following questions:
1. Why did we exclude the point x = 1 from the solution set of the last example? (Easy Question)
2. Why do the above mentioned rules (especially rule #2) work? What is/are the principle(s) working behind the curtains?
Foot Note: Although the post is meant to deal with inequality expressions containing multiple roots, the above rules to draw the wavy line are generic and are applicable in all cases.
- Krishna
_________________
Last edited by EgmatQuantExpert on 10 Jan 2015, 20:41, edited 1 time in total.
Current Student
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2800
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Followers: 196
Kudos [?]: 1162 [1] , given: 235
Re: Inequalities trick [#permalink] 19 Mar 2010, 11:59
1
KUDOS
ttks10 wrote:
Can u plz explainn the backgoround of this & then the explanation.
Thanks
i m sorry i dont have any background for it, you just re-read it again and try to implement whenever you get such question and I will help you out in any issue.
sidhu4u wrote:
I have applied this trick and it seemed to be quite useful.
Nice to hear this....good luck.
_________________
Fight for your dreams :For all those who fear from Verbal- lets give it a fight
Money Saved is the Money Earned
Jo Bole So Nihaal , Sat Shri Akaal
GMAT Club Premium Membership - big benefits and savings
Gmat test review :
670-to-710-a-long-journey-without-destination-still-happy-141642.html
Math Forum Moderator
Joined: 20 Dec 2010
Posts: 2028
Followers: 138
Kudos [?]: 1165 [1] , given: 376
Re: Inequalities trick [#permalink] 26 May 2011, 19:55
1
KUDOS
chethanjs wrote:
mrinal2100 wrote:
if = sign is included with < then <= will be there in solution
like for (x+2)(x-1)(x-7)(x-4) <=0 the solution will be -2 <= x <= 1 or 4<= x <= 7
in case when factors are divided then the numerator will contain = sign
like for (x + 2)(x - 1)/(x -4)(x - 7) < =0
the solution will be -2 <= x <= 1 or 4< x < 7
we cant make 4<=x<=7 as it will make the solution infinite
correct me if i am wrong
Can you please tell me why the solution gets infinite for 4<=x<=7 ?
Thanks.
(x -4)(x - 7) is in denominator.
Making x=4 or 7 would make the denominator 0 and the entire function undefined.
Thus, the range of x can't be either 4 or 7.
4<=x<=7 would be wrong.
4<x<7 is correct because now we removed "=" sign.
_________________
Senior Manager
Joined: 16 Feb 2012
Posts: 257
Concentration: Finance, Economics
Followers: 4
Kudos [?]: 123 [1] , given: 121
Re: Inequalities trick [#permalink] 22 Jul 2012, 02:03
1
KUDOS
VeritasPrepKarishma wrote:
mrinal2100: Kudos to you for excellent thinking!
Correct me if I'm wrong.
If the lower part of the equation$$\frac {(x+2)(x-1)}{(x-4)(x-7)}$$ were $$4\leq x \leq 7$$, than the lower part would be equal to zero,thus making it impossible to calculate the whole equation.
_________________
Kudos if you like the post!
Failing to plan is planning to fail.
Current Student
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2800
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Followers: 196
Kudos [?]: 1162 [1] , given: 235
Re: Inequalities trick [#permalink] 18 Oct 2012, 04:17
1
KUDOS
GMATBaumgartner wrote:
gurpreetsingh wrote:
ulm wrote:
if we have smth like (x-a)^2(x-b)
we don't need to change a sign when jump over "a".
yes even powers wont contribute to the inequality sign. But be wary of the root value of x=a
Hi Gurpreet,
Could you elaborate what exactly you meant here in highlighted text ?
Even I have a doubt as to how this can be applied for powers of the same term . like the example mentioned in the post above.
If the powers are even then the inequality won't be affected.
eg if u have to find the range of values of x satisfying (x-a)^2 *(x-b)(x-c) >0
just use (x-b)*(x-c) >0 because x-a raised to the power 2 will not affect the inequality sign. But just make sure x=a is taken care off , as it would make the inequality zero.
_________________
Fight for your dreams :For all those who fear from Verbal- lets give it a fight
Money Saved is the Money Earned
Jo Bole So Nihaal , Sat Shri Akaal
GMAT Club Premium Membership - big benefits and savings
Gmat test review :
670-to-710-a-long-journey-without-destination-still-happy-141642.html
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 5682
Location: Pune, India
Followers: 1415
Kudos [?]: 7357 [1] , given: 186
Re: Inequalities trick [#permalink] 18 Oct 2012, 09:27
1
KUDOS
Expert's post
GMATBaumgartner wrote:
Hi Gurpreet,
Could you elaborate what exactly you meant here in highlighted text ?
Even I have a doubt as to how this can be applied for powers of the same term . like the example mentioned in the post above.
In addition, you can check out this post:
http://www.veritasprep.com/blog/2012/07 ... s-part-ii/
I have discussed how to handle powers in it.
_________________
Karishma
Veritas Prep | GMAT Instructor
My Blog
Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Math Expert Joined: 02 Sep 2009 Posts: 28348 Followers: 4485 Kudos [?]: 45444 [1] , given: 6761 Re: Inequalities trick [#permalink] 02 Dec 2012, 06:25 1 This post received KUDOS Expert's post GMATGURU1 wrote: Going ahead in this inequalities area of GMAT, can some one has the problem numbers of inequalities in OG 11 and OG 12? Cheers, Danny Search for hundreds of question with solutions by tags: viewforumtags.php DS questions on inequalities: search.php?search_id=tag&tag_id=184 PS questions on inequalities: search.php?search_id=tag&tag_id=189 Hardest DS inequality questions with detailed solutions: inequality-and-absolute-value-questions-from-my-collection-86939.html Hope it helps. _________________ Current Student Status: Nothing comes easy: neither do I want. Joined: 12 Oct 2009 Posts: 2800 Location: Malaysia Concentration: Technology, Entrepreneurship Schools: ISB '15 (M) GMAT 1: 670 Q49 V31 GMAT 2: 710 Q50 V35 Followers: 196 Kudos [?]: 1162 [1] , given: 235 Re: Inequalities trick [#permalink] 20 Dec 2012, 20:04 1 This post received KUDOS VeritasPrepKarishma wrote: The entire concept is based on positive/negative factors which means <0 or >0 is a must. If the question is not in this format, you need to bring it to this format by taking the constant to the left hand side. e.g. (x + 2)(x + 3) < 2 x^2 + 5x + 6 - 2 < 0 x^2 + 5x + 4 < 0 (x+4)(x+1) < 0 Now use the concept. Yes this is probable but it might not be possible always to group them. So in case you are unsure just follow the number plugging approach. But most of the times this trick would be very handy. _________________ Fight for your dreams :For all those who fear from Verbal- lets give it a fight Money Saved is the Money Earned Jo Bole So Nihaal , Sat Shri Akaal Support GMAT Club by putting a GMAT Club badge on your blog/Facebook GMAT Club Premium Membership - big benefits and savings Gmat test review : 670-to-710-a-long-journey-without-destination-still-happy-141642.html Math Expert Joined: 02 Sep 2009 Posts: 28348 Followers: 4485 Kudos [?]: 45444 [1] , given: 6761 Re: Inequalities trick [#permalink] 11 Nov 2013, 06:51 1 This post received KUDOS Expert's post anujpadia wrote: Can you please explain the above mentioned concept in relation to the following question? Is a > O? (1) a^3 - 0 < 0 (2) 1- a^2 > 0 Can you please explain the scenario when (x-a)(x-B)(x-C)(x-d)>0? Sorry, but finding it difficult to understand. Check alternative solutions here: is-a-0-1-a-3-a-0-2-1-a-86749.html Hope this helps. _________________ Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 5682 Location: Pune, India Followers: 1415 Kudos [?]: 7357 [1] , given: 186 Re: Inequalities trick [#permalink] 23 Apr 2014, 03:43 1 This post received KUDOS Expert's post 1 This post was BOOKMARKED PathFinder007 wrote: I have a query. I have following question x^3 - 4x^5 < 0 I can define this as (1+2x).x^3(1-2x). now I have roots -1/2, 0, 1/2. so in case of >1/2 I will always get inequality value as <0 and in case of -1/2 and 0 I will get value as 0. So How I will define them in graph and what range I will consider for this inequality. Thanks The factors must be of the form (x - a)(x - b) .... etc x^3 - 4x^5 < 0 x^3 * (1 - 4x^2) < 0 x^3 * (1 - 2x) * (1 + 2x) < 0 x^3 * (2x - 1) * (2x + 1) > 0 (Note the sign flip because 1-2x was changed to 2x - 1) x^3 * 2(x - 1/2) *2(x + 1/2) > 0 So transition points are 0, 1/2 and -1/2. ____________ - 1/2 _____ 0 ______1/2 _________ This is what it looks like on the number line. The rightmost region is positive. We want the positive regions in the inequality. So the desired range of x is given by x > 1/2 or -1/2 < x< 0 For more on this method, check these posts: http://www.veritasprep.com/blog/2012/06 ... e-factors/ http://www.veritasprep.com/blog/2012/07 ... ns-part-i/ http://www.veritasprep.com/blog/2012/07 ... s-part-ii/ The links will give you the theory behind this method in detail. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199
Veritas Prep Reviews
Intern
Joined: 24 Sep 2013
Posts: 12
Followers: 0
Kudos [?]: 2 [1] , given: 56
Re: Inequalities trick [#permalink] 11 Nov 2014, 08:52
1
KUDOS
For statement I
(1) (1-2x)(1+x)<0
Once you get this into the req form
2 (x-1/2) (x+1) > 0
sol +++++ -1 ------ 1/2 +++++
As Karishma pointed out ponce in the req form, the rightmost will always be positive and the alternating will happen from there.
So sol for this x> 1/2 and x<-1
Integers greater than 1/2 and less than -1 , thus |x| may be >= 1. Unsure
Thus Insufficient.
For Statement II
(2) (1-x)(1+2x)<0
Once you get this into the req form
2(x+1/2) (x-1) > 0
sol +++++ -1/2 ------ 1 +++++
So sol for this x>1 and x< -1/2
Integers greater than 1 and less than -1/2 , thus |x| may be >= 1. Unsure
Thus Insufficient.
Combining Both the statements
x<-1 and x>1
Thus Integers for this range will give |x| > 1
Thus Sufficient.
Hope this helps. I am not very confident of my solution though. Its my first solution gmatclub
mayankpant wrote:
gurpreetsingh wrote:
I learnt this trick while I was in school and yesterday while solving one question I recalled.
Its good if you guys use it 1-2 times to get used to it.
Suppose you have the inequality
f(x) = (x-a)(x-b)(x-c)(x-d) < 0
Just arrange them in order as shown in the picture and draw curve starting from + from right.
now if f(x) < 0 consider curve having "-" inside and if f(x) > 0 consider curve having "+" and combined solution will be the final solution. I m sure I have recalled it fully but if you guys find any issue on that do let me know, this is very helpful.
Don't forget to arrange then in ascending order from left to right. a<b<c<d
So for f(x) < 0 consider "-" curves and the ans is : (a < x < b) , (c < x < d)
and for f(x) > 0 consider "+" curves and the ans is : (x < a), (b < x < c) , (d < x)
If f(x) has three factors then the graph will have - + - +
If f(x) has four factors then the graph will have + - + - +
If you can not figure out how and why, just remember it.
Try to analyze that the function will have number of roots = number of factors and every time the graph will touch the x axis.
For the highest factor d if x>d then the whole f(x) > 0 and after every interval of the roots the signs will change alternatively.
Hi
Can you please explain this question to me using the graph. I am missing the point when graph is being used here?
If x is an integer, is |x|>1?
(1) (1-2x)(1+x)<0
(2) (1-x)(1+2x)<0
For me ,the first equations roots are -1 and 1/2. Now I am struggling to get to the correct sign using the graph method here.
Same for second equation: roots are 1 and -1/2 but struggling for the sign.
THanks
Intern
Joined: 08 Mar 2010
Posts: 6
Followers: 0
Kudos [?]: 0 [0], given: 4
Re: Inequalities trick [#permalink] 16 Mar 2010, 09:46
Can u plz explainn the backgoround of this & then the explanation.
Thanks
Re: Inequalities trick [#permalink] 16 Mar 2010, 09:46
Go to page 1 2 3 4 5 6 Next [ 105 posts ]
Similar topics Replies Last post
Similar
Topics:
127 Tips and Tricks: Inequalities 27 14 Apr 2013, 08:20
inequality, 11 09 Sep 2007, 12:18
Inequalities 3 09 Dec 2006, 04:28
inequalities 1 06 Aug 2006, 07:23
inequalities 3 05 Dec 2005, 11:18
Display posts from previous: Sort by | 2015-07-07T15:46:01 | {
"domain": "gmatclub.com",
"url": "http://gmatclub.com/forum/inequalities-trick-91482.html?kudos=1",
"openwebmath_score": 0.6233364343643188,
"openwebmath_perplexity": 1919.4069545627315,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9632305381464927,
"lm_q2_score": 0.8887587920192298,
"lm_q1q2_score": 0.8560796095191096
} |
https://math.stackexchange.com/questions/2243813/why-do-we-care-about-the-fact-that-a-series-converges-but-not-what-it-converges | # Why do we care about the fact that a series converges, but not what it converges to?
In the 10 tests of convergence/divergent (that I know), them being,
by defn, integral test, div test, comparison test, limit comparison test, gs test, alternating series test, p-series test, root test, ratio test
the only ones that can tell me what the series converges to is by defn and the gs test.
The other ones can tell me if the series converges or diverges.
But I've always wondered, why do we care whether it converges, if we cannot exactly figure out what it converges to? Why is it helpful to know only the behaviour and not the actual value?
• Of course we care about both, but in practice it's generally a lot easier to decide convergence than to actually compute the sum. A good convergence result often comes with information about the speed of convergence...so that we can use it to compute the series numerically, which is often the best we can manage. – lulu Apr 20 '17 at 16:58
• There are ways (in specific cases) to prove that if a sequence converges, then it must converge to $L$. For example, some recurrences, if they converge, converge to a fixed point. Also, for many of the tests, you can get estimates on the remaining error. – Michael Burr Apr 20 '17 at 16:59
• You can numerically compute the sum and see it goes to a certain approximate value, but what if it diverges slowly? The most dangerous and famous example is the harmonic series, which diverges but really slowly. So, knowing if it converges in the first place is essential. – AspiringMathematician Apr 20 '17 at 17:10
• Well, a professor of mine once said "Before you go digging around in the mud, you want to know that there's a potato in there". In context it depends on what you want to do. If the goal is not to actually solve an equation but just to understand a behavior--- say, you want to know if a function is bounded or not. If it is bounded/unbounded you can say such and such will happen. We just want to know the behavior, we don't care what the bound itself is nescessarily. Except when we do. – fleablood Apr 20 '17 at 17:15
Here are some ideas:
• Tests like the integral test actually include bounds on the error. For example, if the integral test applies to $\sum a_n$, then you could compute $\sum_{n=1}^Na_n$ and use the integral test on the remainder to bound the remainder.
• The alternating series test also comes with a bound on the error, so you know how close your partial sum is to the right answer (if you were working on a computer, for example).
• The ratio test essentially tells you that the the terms (eventually) act like a geometric series. Therefore, if you can get a handle on how close something is to being a geometric series, you can use the geometric series computation to bound the remainder.
• If you have a recurrence like $x_n=ax_{n-1}+b$, then you can prove that if the sequence converges, then it must converge to a fixed point, i.e., a point that satisfies $L=aL+b$, $L=\frac{b}{1-a}$. However, to use this, one must first justify why the sequence converges at all.
• :o I didn't know about this "bounds on error" thing... thanks! – K Split X Apr 20 '17 at 17:08
If the series $\sum_{n=0}^\infty a_n$ converges, then I already know what it converges to: the actual value is $\sum_{n=0}^\infty a_n$.
There are lots of ways to represent numeric values, and "the sum of this series" is one of them. Often, it is a very useful way to write the actual value. Sometimes, it's even the best way we have to do so.
I don't want to replace any of the answers given so far and rather stress just one point: To sometimes consider the questions of convergence independently from the computation of the limit is an important educational step, in order to build up a more abstract point of view on convergence.
• I guess, there are many reasons to discuss convergence of series as one of the first topics in an analysis course. Analysis of series is a bit more abstract, than analysis of sequences, but simple enough to present elementary proofs for the important theorems. Students see how exchange of limits can lead to wrong results. (This can be also archived with sequences, but harmonic series as counterexample is simpler to remember.) They can to practice how to estimate and handle an abstract term like a sum with unknown entries.
Why is it important to consider convergence and computation of limit points separate ?
• Some spaces are constructed to be the completion of certain smaller spaces. For example the real numbers can be constructed as the set of all limit points of rational Cauchy-convergent sequences (with identifying some limit points as being equal). Or sometimes you build function spaces in this way, for example the set of limits of smooth functions with respect to some Sobolev norm. When working in these spaces, it is often important to handle convergence of sequences without using a closed form for the limit point.
• Many numerical simulations use some kind of series or sequences to approximate complicated terms, like solutions of differential equations. In these situations the computer will compute sufficiently many terms to estimate the exact limit. It is of great interest to prove theoretically, that these series converge to a unique limit. In many cases the limit of the sequence exist only as an abstract object, being a solution of a certain differential equation. It is common the consider the general convergence to some point (~ Stability) independently from conditions to ensure that an existing unique limit is indeed the correct solution (~ Consistency). Therefore I consider it to be an important lesson to see how convergence and computing the limit point can be treated somehow independently from each. It can happen surprisingly easy, that numerical simulations converge to a false solutions if one of both condition is not satisfied, see for example the plots one the first slides here: Slides 3-5. | 2019-06-19T09:23:24 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2243813/why-do-we-care-about-the-fact-that-a-series-converges-but-not-what-it-converges",
"openwebmath_score": 0.8936986327171326,
"openwebmath_perplexity": 236.96948826206506,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.975576912786245,
"lm_q2_score": 0.8774767906859264,
"lm_q1q2_score": 0.8560460984989582
} |
https://s108528.gridserver.com/opposite-of-sgretr/difference-between-rhombus-and-square-e6d981 | A rhombus is a quadrilateral with all sides of equal length. Next: Rectangle→ Chapter 3 Class 8 Understanding Quadrilaterals; Concept wise; Rhombus, Rectangle, Square. Square. The main difference between Rhomboid and Rhombus is that the Rhomboid is a parallelogram in which adjacent sides are of unequal lengths and angles are oblique and Rhombus is a quadrilateral in which all sides have the same length. Another interesting thing is that the diagonals (dashed lines) meet in the middle at a right angle. What is the difference between Rhombus and Rectangle? Yes, because a square is just a rhombus where the angles are all right angles. Main Difference . The difference between a kite and a rhombus is that a kite does not always have four equal sides or two pairs of parallel sides like a rhombus. Is Rhombus a Square? Keep in mind that the question "Is a rhombus a square?" 1 Descriptions; 2 Rhombus vs Parallelogram; 3 Comparison Chart; Descriptions A rhombus . (c) All angles are equal to 90 degrees. Venn diagrams are schematic diagrams used in logic and in the branch of mathematics known as … Rectangle Square Example 7 Example 8 Ex 3.4, 5 Ex 3.4, 6 … The three special parallelograms — rhombus, rectangle, and square — are so-called because they’re special cases of the parallelogram. that's the main difference!! Here are the differences between a square and a rhombus. As nouns the difference between rhombus and square is that rhombus is while square is any simple object with four nearly straight and nearly equal sides meeting at nearly right angles. Square… ADVERTISEMENT. Rhomboid. Trigonometry. Think about what happens when you "tilt" a square into a rhombus, keeping the horizontal lines of the square horizontal, but rotating the vertical lines a little bit clockwise. The other opposite sides need not be parallel. Please find the attached file for the answer. In plane Euclidean geometry, a rhombus (plural rhombi or rhombuses) is a quadrilateral whose four sides all have the same length. On the other hand, if you bisect the diagonals of a rectangle, they bisect each other at equal length but … To calculate the perimeter, you have 4 x length of one side or 4A, where A is the length of the side. If there are 32.6 million Americans in this age group, find the tot … al U.S. … Among these, many people get confused with rhombus and parallelograms and wonder if they are similar or if the terms are used interchangeably. Therefore, the difference between a square and a rhombus is that {eq}\displaystyle{ \rm \boxed{ \text{1) A square has all sides equal and all angles equal, but a rhombus … The table below offers a clear distinction between them. The plural is rhombi or rhombuses, and, rarely, rhombbi or rhombbuses (with a double b). On a square, both of the diagonals equal each other, but on a rhombus … I seems clear that the orientation of the figure has a lot to do with the name applied to it. • Considering the diagonals; – The diagonals of the rhombus bisect each other at right angles, and the triangles formed are equilateral. Also opposite sides are parallel and opposite angles are equal. Trapezoid A quadrangle which has only one of the two opposite sides as parallel is called a Trapezoid. WHAT IS DIFFERENCE BETWEEN SQUARE AND RHOMBUS WHAT IS DIFFERENCE BETWEEN SQUARE AND RHOMBUS JUSTIFY YOUR ANSWER pls ans both u will get 28 pt I(t) = 27, 992 * (1.017) ^ t covert to log Americans who are 65 years of age or older make up 13.1% of the total population. To distinguish a rectangle from rhombus following property should be kept in mind: 2. All sides of the rectangle are at right angle while this is not the case with rhombus shape. In other words … 3. While it may be a little bit confusing since a square is also a rectangle, there is one difference that makes it easy to distinguish one from the other. Thus a rhombus is not a square unless the angles are all right angles. Compare the new area to the original area. Its properties are (a) All sides are equal. The rhombus has a square as a special case, and is a special case of a kite and parallelogram. What’s the difference between a square and a rectangle? rhombus . The most common types of quadrilaterals are a square, rectangle, rhombus, parallelogram, trapezium, and kite. Rhombus Rhombus and Kite You are here. Rhombuses, rectangles, and squares 2. If both pair of opposite sides of a … All the angles of a square are equal … As nouns the difference between rhomboid and rhombus is that rhomboid is a parallelogram which is neither a rhombus nor a rectangle while rhombus is . 3. – The … English (fish) (wikipedia rhombus) Noun … The top right one, the original “diamond” is actually a square. 10 thoughts on “ Rhombus vs Diamond ” howardat58 February 11, 2016 at 2:57 am. As with all quadrilaterals, the sum … in the above quadrilateral family tree works just like. In this article, let us discuss the difference between rhombus and parallelogram in detail. Square A square is a parallelogram with right angles and equal sides. Rhombus and square have all sides equal, to distinguish a rhombus from square following property should be kept in Each angle of square has to be 90 degrees unlike rhombus. A square must have 4 right angles. (d) The diagonals are equal. Difference Between Rhombus and Parallelogram. (b) Opposite sides are equal and parallel. Unlike rhombus square needs to have all angles equal to 90 degree. A rhombus, also known as “rhom” or “diamond,” is an equilateral quadrilateral, a term that refers to a figure with four parallel sides (the lines will never intersect even if they … Solved Problems : Home >> Parallelogram >> Difference & Similarity between Square & Parallelogram >> Difference & Similarity between Square & Parallelogram. The Rhombus. To distinguish a rectangle from rhombus following property should be kept in mind: 2. Unlike rhombus only opposite sides of rectangle are equal. Rhombus vs … Harlon Moss. Area of rhombus = (1/2) x (d₁ x d₂) Here d₁ and d₂ are diagonals. A square is a quadrilateral with all sides equal in length and all interior angles right angles. This is true for all squares, so ALL SQUARES ARE RHOMBI (the plural of rhombus). but square is a quadilateral having all 4 sides equal but having all angles of its 4 sides as 90 degree . (In addition, the square is a special case or type of both the rectangle and the rhombus.) means Is every rhombus also always a square? As a adjective square is shaped like a (the polygon). Trigonometry Ratios. A kite is one type of four sided figure known as a quadrilateral. A kite is a four-sided shape that has two sets of adjacent sides that have equal lengths. It is more common to call this shape a rhombus, but some people call it a rhomb or even a diamond. Each angle of rectangle has to be 90 degrees unlike rhombus. Views: 2. A rhombus, on the other hand, does not have any rules about its angles, so there are many many, examples of a rhombus that are not also squares. In contrast, a rhombus is a shape having four equal sides but in a diamond shape. A rhombus is a four-sided shape where all sides have equal length (marked "s"). All rhombuses and squares are also kites. A rhombus is a quadrilateral with all sides equal in length. To distinguish a rectangle from rhombus following property should be kept in mind: 2. Same is the case with a rhombus … In general, a rhombus has the following special properties • All four sides are equal in length. Other Names. The rhombus is often called a diamond, after the … Each angle of square has to be 90 degrees unlike rhombus. Updated: February 27, 2019. As you can see in diagram … The shaded region represents the result of the operation. The different quadrilaterals are square, rectangle, rhombus, parallelogram, kite, trapezium which have some similar characteristics. As a verb square is to adjust so as to align with or place at a right angle to something else. Rhomboid vs. Rhombus. Rhombus and square have all sides equal, to distinguish a rhombus from square following property should be kept in Each angle of square has to be 90 degrees unlike rhombus. Sum of interior angles will be 360 degree Difference between square and rhombus: A square also fits the definition of a rectangle (all angles are 90°), and a rhombus (all sides are equal length). Each angle of rectangle has to be 90 degrees unlike rhombus. There are many shapes which give the impression of being associated to 1 one other, nevertheless you probably have a take a look at it, there are only some variations between them. 5 Min Read. As a adjective rhomboid is resembling, or shaped like a rhombus or rhomboid. Similarly, the area is A = b x h. Main Difference Between Parallelogram vs Rhombus . • Area of any can be calculated using the formula base ×height. The square can be considered as a special case of the rhombus, where the internal angles are right angles. Each angle of rectangle has to be 90 degrees unlike rhombus. Unlike rhombus only opposite sides of rectangle are equal. A square, however, has four equal interior angles. Rhombus and Parallelograms are different although they both have four sides and four vertices and look almost similar. (AB=DC=AD=BC) • The diagonals of the rhombus bisect each other at right angles; diagonals are perpendicular to each other, in addition to the following properties of a … Area of Parallelogram … Note that a square is also a rhombus because it has 4 equal sides as well. Rhombus, Rectangle, Square; Rhombus and Kite. Opposite sides of a rhombus are parallel. Rectangle and rhombus are special cases of the parallelograms. rhombus is a quadilateral having all 4 sides equal but not having all angles 90 degree. Learn difference between square & rhombus topic of maths in details explained by subject experts on vedantu.com. The difference is a rhombus is made up of 2 acute and 2 obtuse angles and a square is made up of 4 right angles. Another name is equilateral quadrilateral, since equilateral means that all of its sides are equal in length. Rhombus Square Three dimensional object Trapezium Triangle Quadrilateral. Data-Handling. Answer: Rhombus is a four-sided quadrilateral with all its four sides equal in lengths.Rhombus is also referred to as a slanting square. You can come back to reference it in the future if you have any … Published: 14 Jan, 2021. • Rhombus and rectangle are quadrilaterals. CONTINUE READING BELOW. ADVERTISEMENT. Arithmetic Mean Frequency Distribution Table Graphs Median Mode Range . The major difference between rectangle and rhombus is that, the opposite sides of the rectangle are equal. Contents. The properties of square and rhombus are similar, the only distinguishing property is that square has all the angles equal to 90 0, and rhombus does not.. Rhombus is also called an equilateral quadrilateral. This article will explore the difference between a rhombus and a parallelogram. Square… Opposite angles are congruent in a rhombus. A square however is a rhombus … Diagram 2. 3. A Rhombus is a quadrilateral: A Square is also a quadrilateral: Opposite sides of Rhombus are parallel to each other : Opposite sides of Square are also parallel to each other: All sides of Rhombus are of equal length: All side of Square are also of equal length: Adjacent Angles of Rhombus are supplementary: Adjacent angles of Square are also supplementary : Diagonals of Rhombus … Rhomboid is a related term of rhombus. … Both a square and a rectangle may have their four angles at 90°, have parallel opposite sides with the same length, and have their diagonals equal in length, but only a square … Answer: No, a rhombus is not a square. ! Last updated at Sept. 25, 2018 by Teachoo Subscribe to our Youtube Channel - https://you.tube/teachoo. But these shapes are different from each other because of their few properties. The three-level hierarchy you see with. The name "rhombus" comes from the Greek word rhombos: a piece of wood whirled on a string to … Properties of rhombus: It has four equal sides ; Both diagonals are intersecting each other at right angle. 3 difference between Rhombus & square 2 See answers sufiyaanwar200451 sufiyaanwar200451 Answer: Here is four diffence b/w rhombus and square hope it help u. Step-by-step explanation: Purplehidie Purplehidie Answer: The sides of a square are perpendicular to each other whereas the sides of a rhombus are not perpendicular to each other. Unlike rhombus only opposite sides of rectangle are equal. Main Difference. Parallelograms are different although they both have four sides all have the length... H. Main difference between square and a parallelogram with right angles and equal sides ; both are... Rectangle and rhombus are special cases of the figure has a lot to do with the name to... A shape having four equal sides each angle of square has to be 90 unlike... Or place at a right angle almost similar special properties • all four sides and vertices! Represents the result of the rectangle and rhombus is a quadrilateral with all four. Rectangle→ Chapter 3 Class 8 Understanding Quadrilaterals ; Concept wise ; rhombus rectangle... Sides that have equal lengths referred to as a special case or type four. Angle to something else its sides are equal in lengths.Rhombus is also referred to as a adjective is. Object Trapezium Triangle quadrilateral equal sides to calculate the perimeter, you have x. Diagonals of the rectangle and the triangles formed are equilateral 1/2 ) x ( d₁ d₂! It has four equal sides two opposite sides of the side Here d₁ and d₂ are.. Square… answer: No, a rhombus. square unless the angles are equal Quadrilaterals ; wise... S '' ) shape a rhombus is a four-sided shape where all sides of equal length and kite however! The difference between rhombus and parallelogram in detail not having all angles 90 degree the table below a! In general, a rhombus and parallelogram in detail ( d₁ x d₂ ) d₁! Be 360 degree difference between a rhombus and a rectangle from rhombus following property be! One of the rectangle are equal a slanting square sets of adjacent that. Angles, and the rhombus is a rhombus ( plural rhombi or rhombuses ) a. Chapter 3 Class 8 Understanding Quadrilaterals ; Concept wise ; rhombus, but some call. The question is a quadrilateral with all sides are equal called a.. To call this shape a rhombus. the Area is a quadilateral having all 4 sides in... Equal sides 1/2 ) x ( d₁ x d₂ ) Here d₁ and are. Have four sides equal in lengths.Rhombus is also referred to as a slanting square parallelogram ; Comparison... Between rhombus and parallelograms and wonder if they are similar or if the terms are used interchangeably in. Right angles of interior angles diamond shape below offers a clear distinction between.! - https: //you.tube/teachoo and all interior angles right angles rectangle from rhombus following property be. Article will explore the difference between square & rhombus topic of maths in explained. Are parallel and opposite angles are equal … a rhombus is not a square is a special case type! To have all angles of its sides are equal in length d₁ and d₂ are diagonals you see. Or if the terms are used interchangeably the question is a quadrilateral the middle a... Its four sides equal but not having all 4 sides equal but not having all 4 equal! Call this shape a rhombus and a parallelogram the branch of mathematics known as difference! Dimensional object Trapezium Triangle quadrilateral shape having four equal interior angles two sets of adjacent sides that have equal.... Not having all angles of a square? Here d₁ and d₂ are diagonals all of... To align with or place at a right angle to something else rhombuses and! 1 Descriptions ; 2 rhombus vs parallelogram ; 3 Comparison Chart ; Descriptions a rhombus where the internal are! Verb square is just a rhombus where the angles are equal the angles are right.... Thus a rhombus is a four-sided quadrilateral with all sides have equal lengths square is a special case the! Rhombus. to do with the name applied to it “ diamond ” is a! Addition, the sum … each angle of rectangle are equal to degree! A related term of rhombus: it has four equal interior angles right angles Considering the diagonals ; the. Clear distinction between them since equilateral means that all of its 4 sides as 90 degree tree works just.! Case or type of both the rectangle are equal angles, and the triangles formed equilateral. Rhombus a square and rhombus: it has four equal interior angles are similar if... Where a is the length of the two opposite sides of the rhombus is that the diagonals ; the. Or type of four sided figure known as … difference between square and rhombus: it has four equal angles... … a rhombus and parallelograms are different from each other at right angle to something else ;. The differences between a square however is a four-sided quadrilateral with all sides of the figure has a lot do... Rhombus has the following special properties • all four sides and four vertices and look almost similar is... And look almost similar diagrams are schematic diagrams used in logic and in the middle at right! 4 sides as 90 degree square and a parallelogram 4A, where the angles... Rectangle has to be 90 degrees or if the terms are used interchangeably call! The major difference between rectangle and rhombus are special cases of the parallelograms square to! This article will explore the difference between a square and a rectangle from rhombus following property should be kept mind... But in a diamond shape details explained by subject experts on vedantu.com their few properties have the same length discuss! As with all sides have equal lengths two sets of adjacent sides that have equal length or 4A, a! Are parallel and opposite angles are right angles is also referred to as a adjective square is just a is!, where a is the length of one side or 4A, where the angles! Any can be considered as a slanting square has only one of the operation the has! And, rarely, rhombbi or rhombbuses ( with a double b ) opposite sides are …... H. Main difference between a square place at a right angle interesting thing is that the question is... People get confused with rhombus shape interesting thing is that the diagonals ( dashed lines ) meet in the at..., rectangle, square ; rhombus and kite, rarely, rhombbi rhombbuses! Used interchangeably often called a trapezoid they both have four sides all have the same.. And equal sides but in a diamond 1/2 ) x ( d₁ x d₂ ) Here d₁ and d₂ diagonals. In lengths.Rhombus is also referred to as a adjective square is a quadilateral having all angles degree! To as a slanting square rhombi ( the polygon ) thus a rhombus a. Four sided figure known as … difference between a square? “ diamond ” is actually a is. Four equal sides base ×height distinguish a rectangle of one side or 4A, where the are... Triangles formed are equilateral equal sides are special cases of the figure has a to! Used in logic and in the middle at a right angle to else. ( 1/2 ) x ( d₁ x d₂ ) Here d₁ and d₂ are diagonals quadrilateral, since equilateral that... Updated at Sept. 25, 2018 by Teachoo Subscribe to our Youtube Channel - https: //you.tube/teachoo this,. Mind: 2 the original “ diamond ” is actually a square is to adjust as... A rhombus is not a square and a rhombus is that, the original “ diamond ” is a. A quadrilateral the middle at a right angle to something else as 90 degree or,... Quadilateral having all angles 90 degree to our Youtube Channel - https: //you.tube/teachoo of maths details... Rhombus topic of maths in details explained by subject experts on vedantu.com 2 rhombus parallelogram. ; Concept wise ; rhombus and parallelograms and wonder if they are or. Quadrilateral, since equilateral means that all of its sides are equal ” is actually a square are in! And the triangles formed are equilateral a ) all sides equal in and! All angles are all right angles and equal sides but in a diamond shape a term! Is that the question is a parallelogram difference between rhombus and square a right angle while this is for. Angles of a square however is a four-sided shape that has two sets of adjacent that... And in the middle at a right angle = ( 1/2 ) x ( d₁ x d₂ ) Here and! Shaped like a ( the polygon ) us discuss the difference between square and rhombus special... Not a square are equal other at right angles and kite to adjust so as to align or. The orientation of the rectangle and rhombus are special cases of the rectangle are at right angles equal! Shapes are different from each other at right angle to something else square rhombus! Rarely, rhombbi or rhombbuses ( with a double b ): Rectangle→ Chapter 3 8. Explained by subject experts on vedantu.com in plane Euclidean geometry, a rhombus and parallelograms are different each... Where the internal angles are equal but square is just a rhombus is a parallelogram maths in explained. Square… answer: rhombus is not the case with rhombus and kite plural rhombus... Since equilateral means that all of its 4 sides equal but having all angles 90 degree these, people! Comparison Chart ; Descriptions a rhombus or rhomboid a quadilateral having all 4 equal. Have four sides all have the same length after the … Learn difference between parallelogram vs rhombus. its are! The following special properties • all four sides all have the same.! Rectangle are equal and d₂ are diagonals the rectangle are equal having all angles are all right angles so. For all squares, so all squares, so all squares are (...
difference between rhombus and square 2021 | 2022-05-21T05:17:58 | {
"domain": "gridserver.com",
"url": "https://s108528.gridserver.com/opposite-of-sgretr/difference-between-rhombus-and-square-e6d981",
"openwebmath_score": 0.592010498046875,
"openwebmath_perplexity": 901.8399469256652,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. Yes\n2. Yes",
"lm_q1_score": 0.9755769099458929,
"lm_q2_score": 0.8774767810736693,
"lm_q1q2_score": 0.8560460866291191
} |
http://mathhelpforum.com/number-theory/132423-jacobi-symbol.html | # Math Help - Jacobi Symbol
1. ## Jacobi Symbol
Evaluate the Jacobi symbol ((n−1)(n+1)/n) for any odd natural number n.
Trying out some numbers, I THINK it alternates between 1 and -1, but how can we PROVE it formally?
Any help is appreciated!
[also under discussion in math link forum]
2. $\left(\frac{(n-1)(n+1)}{n}\right) = \left(\frac{n-1}{n}\right)\left(\frac{n+1}{n}\right) = \left(\frac{-1}{n}\right)\left(\frac{1}{n}\right) = \left(\frac{-1}{n}\right)$
For $n$ odd, $\left(\frac{-1}{n}\right) = (-1)^\frac{n-1}{2} = \begin{cases} \;\;\,1 & \text{if }n \equiv 1 \pmod 4\\ -1 &\text{if }n \equiv 3 \pmod 4\end{cases}$.
For $n$ even, take $n=2k$, then $\left(\frac{-1}{n}\right) = \left(\frac{-1}{2k}\right) = \left(\frac{-1}{2}\right)\left(\frac{-1}{k}\right) = \left(\frac{-1}{k}\right) = \begin{cases} \;\;\,1 & \text{if }k \equiv 1 \pmod 4\\ -1 &\text{if }k \equiv 3 \pmod 4\end{cases}$.
3. Originally Posted by chiph588@
$\left(\frac{(n-1)(n+1)}{n}\right) = \left(\frac{n-1}{n}\right)\left(\frac{n+1}{n}\right) = \left(\frac{-1}{n}\right)\left(\frac{1}{n}\right) = \left(\frac{-1}{n}\right)$
For $n$ odd, $\left(\frac{-1}{n}\right) = (-1)^\frac{n-1}{2} = \begin{cases} \;\;\,1 & \text{if }n \equiv 1 \pmod 4\\ -1 &\text{if }n \equiv 3 \pmod 4\end{cases}$.
For $n$ even, take $n=2k$, then $\left(\frac{-1}{n}\right) = \left(\frac{-1}{2k}\right) = \left(\frac{-1}{2}\right)\left(\frac{-1}{k}\right) = \left(\frac{-1}{k}\right) = \begin{cases} \;\;\,1 & \text{if }k \equiv 1 \pmod 4\\ -1 &\text{if }k \equiv 3 \pmod 4\end{cases}$.
Thank you!
But I thought the Jacobi symbol is defined only for ODD positive integers at the bottom. In your proof, why is there a case where "n" is even??? How is this possible??
4. Originally Posted by kingwinner
Thank you!
But I thought the Jacobi symbol is defined only for ODD positive integers at the bottom. In your proof, why is there a case where "n" is even??? How is this possible??
Oops! You're right!
5. But I'm concerned with one special case. For the case n=1, ((n−1)(n+1)/n)=(0/1)
Is (0/1)=1 or (0/1)=0 ?? Which one is the correct answer and why?
Thanks!
6. Originally Posted by kingwinner
But I'm concerned with one special case. For the case n=1, ((n−1)(n+1)/n)=(0/1)
Is (0/1)=1 or (0/1)=0 ?? Which one is the correct answer and why?
Thanks!
$\left(\frac{a}{n}\right)
= \begin{cases}
\;\;\,0\mbox{ if } \gcd(a,n) \ne 1
\\\pm1\mbox{ if } \gcd(a,n) = 1\end{cases}$
So sub in $0$ and $1$ and see what you get.
7. Originally Posted by chiph588@
$\left(\frac{a}{n}\right)
= \begin{cases}
\;\;\,0\mbox{ if } \gcd(a,n) \ne 1
\\\pm1\mbox{ if } \gcd(a,n) = 1\end{cases}$
So sub in $0$ and $1$ and see what you get.
gcd(0,1)=1, so that rule says that (0/1)=+1 OR -1, but how do we know whether it is +1 or -1?
8. Jacobi symbol
Apparently the answer is $0$. I haven't had too much exposure to the Jacobi symbol so I can't really tell you why. Check out the site for yourself though.
9. $\left( \frac{0}{1} \right)=1$ because the set of prime factors of $1$ is empty. if $n > 1$ is an odd integer, then $\left( \frac{0}{n} \right)=0.$
10. Originally Posted by NonCommAlg
$\left( \frac{0}{1} \right)=1$ because the set of prime factors of $1$ is empty. if $n > 1$ is an odd integer, then $\left( \frac{0}{n} \right)=0.$
Is there any reason why (0/1)=1?? Is this simply becuase by convention, we define it to be that way?
1 has no prime factorization, so the product is empty. Is it conventional to define the "empty" product to be equal to +1??
Also, is it true that, by definition, (a/1)=1 for any integer a?
Can someone clarify this? Thank you!
11. Originally Posted by kingwinner
Is there any reason why (0/1)=1?? Is this simply becuase by convention, we define it to be that way?
1 has no prime factorization, so the product is empty. Is it conventional to define the "empty" product to be equal to +1??
Also, is it true that, by definition, (a/1)=1 for any integer a?
Can someone clarify this? Thank you!
12. Originally Posted by NonCommAlg
Is there any real reason why we define (a/1)=1 for any integer a?
Why not define it to be 0? why not -1?
13. Originally Posted by kingwinner
Is there any real reason why we define (a/1)=1 for any integer a?
Why not define it to be 0? why not -1? | 2016-02-13T14:03:00 | {
"domain": "mathhelpforum.com",
"url": "http://mathhelpforum.com/number-theory/132423-jacobi-symbol.html",
"openwebmath_score": 0.9109496474266052,
"openwebmath_perplexity": 609.76899833776,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9755769078156283,
"lm_q2_score": 0.8774767794716264,
"lm_q1q2_score": 0.8560460831969453
} |
http://math.stackexchange.com/questions/158602/number-of-elements-vs-cardinality-vs-size/158635 | # Number of elements vs cardinality vs size
I have been wondered the definition of cardinality and number of elements. One mathematician told me that one can't said that the cardinality or size of the set $\{1\}$ is one, it should be said that the number of elements in the set is one. I guess that his opinion is that there is no term size in mathematics. Is these true? On the other hand, if we have an infinite set like $\mathbb Z$, can we say that the number of elements of $\mathbb Z$ is infinite or is the term "number of elements" used only in finite sets?
-
I have no idea what the mathematician was thinking. The cardinality of $\{1\}$ is indeed $1$. Of course it’s also true that $1$ is the number of elements in the set. I would probably not use the expression number of elements when the set is infinite, but I’d not complain if someone said ‘$\Bbb R$ has $2^\omega$ elements’ instead of ‘the cardinality of $\Bbb R$ is $2^\omega$’. – Brian M. Scott Jun 15 '12 at 10:51
For finite sets there is absolutely no problem; a set has cardinality or size $n$ if it is in bijection with $\{ 1, 2, ... n \}$, and it has cardinality or size $0$ if it is empty. (The empty set is unique!) It is silly to prevent yourself from being able to say this. – Qiaochu Yuan Jun 15 '12 at 12:39
Converting my comment to an answer:
I have no idea what the mathematician was thinking: the cardinality of the set $\{1\}$ is indeed $1$, and that’s a perfectly normal way to express the fact. Of course it’s also true that $1$ is the number of elements in the set.
I would probably not myself use the expression number of elements when speaking of an infinite set, but ‘$\Bbb R$ has $2^\omega$ elements’ is a perfectly fine synonym of ‘the cardinality of $\Bbb R$ is $2^ω$’.
-
Okay. I guess the mathematician wanted that I won't be his graduate student as he asked me to modify my bachelor thesis so heavily. That was one point where I made a "mistake" and got a bad grade. And is the word "size" a synonym of the cardinality? – mathenthusiast Jun 15 '12 at 15:52
@mathenthusiast: I wouldn’t say so: it’s more general, as it can also refer to length, area, volume, etc. – Brian M. Scott Jun 15 '12 at 21:38
The question you need to ask yourself is "what is the meaning of the notion number?"
Natural numbers can be used to measure size, length, etc. while rational numbers measure ratio between two integers, real numbers measure length... what do complex numbers measure?
In mathematics we allow ourselves to define new ways of measuring things, and usually we refer to the values of these measures as numbers. Indeed there is no real difference between the length of a $100$ meters running track, and a $100$ meters long hot dog...
When measuring the size of a set we came up with a clever way to discuss infinite sets. This way is "cardinality", and the cardinal of a set represents in a good sense the "number of elements in the set". So it has a perfectly good meaning to say that a set has $\aleph_0$ many elements, or that a set is of size $\aleph_1$.
- | 2014-12-22T23:34:46 | {
"domain": "stackexchange.com",
"url": "http://math.stackexchange.com/questions/158602/number-of-elements-vs-cardinality-vs-size/158635",
"openwebmath_score": 0.87849360704422,
"openwebmath_perplexity": 186.55819648630023,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.9755769092358048,
"lm_q2_score": 0.8774767778695834,
"lm_q1q2_score": 0.856046082880201
} |
https://math.stackexchange.com/questions/3210927/integration-by-partial-fractions-int-sec3xdx | # Integration by Partial Fractions $\int \sec^3xdx$
I have looked at multiple ways to do partial fractions to integrate $$\sec^3x$$, but there is a part where I keep getting stuck when I see the partial fractions get split up.
$$\int \frac{\cos x}{\cos^4 x} = \int \frac{1}{(1-y^2)^2}$$
The next step I have seen is shown like this on Wikipedia:
$$\frac{1}{(1-u^2)^2}=\frac{1/4}{1-u}+\frac{1/4}{(1-u)^2}+\frac{1/4}{1+u}+\frac{1/4}{(1+u)^2}.$$
Where does the $$\frac{1}{4}$$ come from in the numerator?
And for the denominator, why are there two additional positive fraction decompositions $$1+u$$ and $$(1+u)^2$$?
• See this page for the general theorem in use. – Bernard May 2 at 12:35
• – cgiovanardi May 2 at 14:54
And for the denominator, why are there two additional positive fraction decompositions
Because $$1-u^2 = (1-u)(1+u).$$ It's the "difference of squares" factorization. And so, squaring both sides, we also have $$(1-u^2)^2 = (1-u)^2(1+u)^2$$.
In partial fractions decomposition, all irreducible factors must appear as denominators, to the highest power that appears, as well as to all lower powers. So that means we need a denominator for $$(1-u)$$ and $$(1-u)^2,$$ but also for $$(1+u)$$ and $$(1+u)^2.$$
Where does the $$\frac{1}{4}$$ come from in the numerator?
For partial fraction decomposition, the numerators must be chosen so that if you were to give every term common denominator and combine the numerators, you would get the starting fraction back. It takes a bit of algebra to find what numerators do the job, but that is where the $$\dfrac{1}{4}$$ comes from.
But let's do the algebra. The normal way to find these numerators is to introduce variables for them and solve. Like so:
$$\dfrac{1}{(1-u^2)^2} = \dfrac{1}{(1+u)^2(1-u)^2} = \dfrac{A}{1+u} + \dfrac{B}{(1+u)^2} + \dfrac{C}{1-u} + \dfrac{D}{(1-u)^2}$$
here the "variable" numerators are called $$A, B, C,$$ and $$D.$$
We solve for them by clearing denominators:
$$1 = A(1-u)^2(1+u) + B(1-u)^2 + C(1-u)(1+u)^2 + D(1+u)^2$$
Then in general you may have to expand all those binomials and match like terms of the resulting polynomials in $$u$$. But a shortcut can be to sub in the values of the roots in $$u$$. For example, at $$u=1,$$ the above equation becomes
$$1 = A(1-1)^2(1+1) + B(1-1)^2 + C(1-1)(1+1)^2+ D(1+1)^2 = D(1+1)^2.$$
So we have $$D=1/4.$$
Similarly, by next subbing $$u=-1,$$
$$1 = B(1-(-1))^2$$
So we have $$B=1/4.$$
To figure out that $$A$$ and $$C$$ are also $$1/4$$, we can either sub in some other, non-root values of $$u$$, or bite the bullet and do the algebra. (But if this were a partial fractions problem with no powers of irreducibles, we'd be done).
I'll do the algebra, I guess. Expand, and gather like terms in $$u$$, treating the coefficients $$A, B, C,$$ and $$D$$ as variables.
$$1 = A(1-u-u^2+u^3) + B(1-2u+u^2) + C(1+u-u^2-u^3) + D(1+2u+u^2) \\ = (A+B+C+D) + (-A-2B+C+2D)u + (-A+B-C+D)u^2 + (A-C)u^3.$$
Now for two polynomials in $$u$$ to be equal for all values of $$u$$, every coefficient must be equal. On the left-hand side we have the polynomial $$1$$, which has a constant coefficient of $$1$$ and all higher coefficients are $$0$$.
So this gives us the following equations:
$$1 = A+B+C+D\\ 0 = -A-2B+C+2D\\ 0 = -A+B-C+D\\ 0 = A-C.$$
We could solve this system of four linear equations in four unknowns, using Gaussian elimination or substitution or matrix methods. But first let's remember that we already know that $$B=D=1/4.$$ So we don't have four unknowns any more, our only unknowns left are $$A$$ and $$C$$. By the fourth equation, $$A=C$$, so our only unknown is really $$A$$. The first equation becomes
$$1 = A+B+C+D = A + \dfrac{1}{4} + A + \dfrac{1}{4}.$$
So $$2A=1-\dfrac{2}{4} = \dfrac{1}{2}.$$ And so $$A=B=C=D=\dfrac{1}{4}.$$
• Oh, seeing the way you wrote that makes sense (going from $1-u^2$ to $(1-u^2)^2)$, just adding ^2 to the factors – Evan Kim May 2 at 12:38
This is actually a problem of partial fractions.
When dividing one fraction to partial fractions, all possible cases of denominator should be considered. In this case, the denominator can be factorised to $$(1-u)^2(1+u)^2$$, so there are actually in total 8 possible fractions, with denominators being $$1-u$$, $$(1-u)^2$$, etc. Setting all the numerators to A, B, C, D, etc. for different denominator, layout the expression and compare coefficients, the result should then be that 1 fraction equal to 4 partial fractions.
Note that $$(1-y^2)^2=(y-1)^2(y+1)^2$$. On the other hand, every rational function of the form$$\frac{p(x)}{(x-r_1)^{m_1}\times\cdots\times(x-r_k)^{m_k}},$$where $$\deg p(x), can be written as a linear combination of rational functions of the type $$\frac1{(x-r_j)^N}$$, with $$N\in\{1,2,\ldots,m_j\}$$. The rest is a matter of computations. | 2019-09-21T17:38:26 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/3210927/integration-by-partial-fractions-int-sec3xdx",
"openwebmath_score": 0.9416100382804871,
"openwebmath_perplexity": 242.41495961374093,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9755769078156283,
"lm_q2_score": 0.8774767778695834,
"lm_q1q2_score": 0.8560460816340292
} |
https://www.khanacademy.org/math/calculus-1/cs1-derivatives-chain-rule-and-other-advanced-topics/cs1-differentiation-using-multiple-rules/v/applying-the-chain-rule-and-product-rule | If you're seeing this message, it means we're having trouble loading external resources on our website.
If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.
Main content
# Applying the chain rule and product rule
AP Calc: FUN‑3 (EU)
## Video transcript
- [Instructor] What we're going to do in this video is try to find the derivative with respect to X of X squared sin of X. All of that to the third power. And what's going to be interesting is that there's multiple ways to tackle it. And I encourage you to pause the video and see if you could work through it on your own. So there's actually multiple techniques. One path is to do the chain rule first. So I'll just say CR for chain rule first. And so I have, I'm taking the derivative with respect to X of something to the third power. So, if I take the derivative it would be the derivative with respect to that something. So that would be three times that something squared times the derivative with respect to X of that something. Where the something, in this case, is X squared sin of X. X squared sin of X. This is just an application of the chain rule. Now, the second part, what would that be? The second part here do this in another color. In orange. Well here, I would apply the product rule. I have a product of two expressions. So I would take the derivative of, let me write this down. So this is gonna be the product rule. Product rule. I would take the derivative of the first expression. So, X, derivative of X squared is two X. Let me write a little bit to the right. This is gonna be two X times the second expression sin of X. Plus the first expression X squared times the derivative of the second one. Cosin of X. That's just the product rule as applied to this part right over here. And all of that, of course, is being multiplied by this up front. Which actually, let me just rewrite that. So all of this I could rewrite as let's see, this would be three times if I have the product raised to the second power I could take each of them to the second power and then take their products. So X squared squared is X to the fourth. And then sin of X squared is sin squared of X. And then all of that is being multiplied by that. And, if we want, we can algebraically simplify. We can distribute everything out. In which case, what would we get? Well let's see. Three times two is six. X to the fourth times X is X to the fifth. Sin squared of X times sin of X is sin of X to the third power. And then, let's see, three so plus three, X to the fourth times X squared is X to the sixth power. And then I have sin squared of X sin squared of X cosin of X. So there you have it, that's one strategy chain rule first, and then product rule. What would be another strategy pause the video and try to think of it. Well, we could just algebraically use our exponent properties first. In which case, this is going to be equal to the derivative with respect to X of if I'm taking X squared times sin of X to the third power instead I could say X to the third power which is going to be X to the sixth. And then sin of X to the third power. Sin of X to the third power. I'm using the same exponent property that we used right over here to simplify this. If I have if I'm doing the product things to some exponent, well that's the same thing if each of them raised to the exponent and then the product of the two. Now how would we tackle this? Well, I here, I would do the product rule first. So let's do that. So let's do the product rule. So, we're gonna take the derivative of the first expression. So, derivative of X to the sixth is six X to the fifth. Times the second expression. Sin to the third of X. Or, sin of X to the third power. Plus the first X to the sixth times the derivative of the second and I'm just gonna write that D DX of sin of X to the third power. Now, to evaluate this right over here it does definitely make sense to use the chain rule. Use the chain rule. And so what is this going to be? Well I have the derivative of something to the third power. So that's going to be three times that something squared times the derivative of that something. So in this case, the something is sin of X. And the derivative of sin of X is cosin of X. Then I have all of this business over here. I have six X to the fifth. Sin to the third or sin of X to the third power. Plus X to the sixth. And if I were to just simplify this a little bit in fact, you see it very clearly. These two things are equivalent. This this term is exactly equivalent to this term the way it's written. And then this is exactly, if you multiply three times X to the sixth sin of X squared cosin of X. So, the nice thing about math if we're doing things that make logical sense we should get to the same endpoint. But the point here is that there's multiple strategies. You could use a chain rule first and then the product rule. Or you could use a product rule first, and then the chain rule. In this case, you could debate which one is faster. It looks like the one on the right might be a little bit faster. But sometimes these two are pretty close. But sometimes it'll be more clear than not which one is preferable. You really want to minimize the amount of hairiness, the amount the number of steps the chances for careless mistakes you might have. | 2020-08-08T04:12:21 | {
"domain": "khanacademy.org",
"url": "https://www.khanacademy.org/math/calculus-1/cs1-derivatives-chain-rule-and-other-advanced-topics/cs1-differentiation-using-multiple-rules/v/applying-the-chain-rule-and-product-rule",
"openwebmath_score": 0.8288611769676208,
"openwebmath_perplexity": 240.07129107804414,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9936116789011081,
"lm_q2_score": 0.8615382094310355,
"lm_q1q2_score": 0.8560344267102258
} |
https://math.stackexchange.com/questions/2819906/number-of-rectangles-with-odd-area | # Number of rectangles with odd area.
We have a $10\times 10$ square.
How many rectangles with odd area are on the picture?
I say lets choose a vertex first, there are $11\cdot11=121$ possibilities.
Now, choose odd width and side (left or right) and odd length and side (up or down). There are $5$ possibilities to each, so in total we have $\dfrac{121\cdot 25}{4}$ rectangles. We divide by $4$ because every rectangle is counted $4$ times, one time for each vertex of the rectangle. But, the result is not a whole number. Where am I wrong? Thanks.
The width must be odd and the height must be odd. There are $10$ choices of a pair of $x$-coordinates to have width $1$, $8$ choices for width $3$, $6$ choices for with $5$, etc. So we have $10+8+6+4+2=30$ ways to choose the two $x$ coordinates and likewise $30$ ways to choose $y$-coordinates. This gives us a total of $30\cdot 30=900$ odd area rectangles.
What you did was to pick one vertex and than assume that each possible odd width and height could be realized with this vertex.
• But where was I wrong? – Omer Jun 14 '18 at 19:29
• @Omer: In counting the number of possible starting vertices (121). Not all of them are valid for all of the sub-cases you count. So, you go wrong at the very first step. Compare your thinking to how Hagen von Eitzen counted the cases, to correct the error. – Nominal Animal Jun 14 '18 at 19:55
• "and than assume that each possible odd width and height could be realized with this vertex." - no, s/he just assumed that 5 different odd x pairs and 5 different odd y pairs included the x and y of this vertex. If you start at (0,0) then in the x direction you can go to (1,0), (3,0), (5,0), (7,0), (9,0) but if you start at (5,0) you can go to (0,0), (2,0), (4,0), (6,0), (8,0), (10,0), that's 6 choices! – immibis Jun 15 '18 at 3:55
The reasoning is ok, except for this:
There are 5 possibilities to each
Sadly, no. If the chosen vertex is at $(1,1)$ (grid starting at $(0,0)$) you have 6 possibilities for each direction.
In general, you have 6 possibilities if the coordinate is odd, 5 if it's even.
Fix: Because there $36$ all-even-coordinates vertices, $25$ all-odd, and $121-36-25=60$ mixed vertices, the correct counting is
$$36 \times 5^2+ 25 \times 6^2 + 60 \times5 \times 6=3600$$
Dividing by $4$ you get the $900$ rectangles.
I assume you're looking for rectangles with sides parallel to those of the square and vertices that are lattice points. A rectangle with integer sides has odd area iff the sides are both odd. For each pair of odd integers $x, y$ with $1 \le x,y \le 9$, there are $(11-x)(11-y)$ possible positions for a rectangle of size $x \times y$. Thus the number of rectangles is $$\sum_{x \in \{1,3,5,7,9\}} \sum_{y \in \{1,3,5,7,9\}} (11-x)(11-y) = 900$$ | 2019-09-16T20:29:29 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2819906/number-of-rectangles-with-odd-area",
"openwebmath_score": 0.7377194762229919,
"openwebmath_perplexity": 283.9377867944591,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9793540698633748,
"lm_q2_score": 0.8740772433654401,
"lm_q1q2_score": 0.8560311056649034
} |
http://math.stackexchange.com/questions/286315/fundamental-group-of-this-space | Fundamental group of this space
Based on this question: What is the homology groups of the torus with a sphere inside?
I'm trying to find the fundamental group of this space using the Seifert–van Kampen theorem. If $U$ is the torus and $V$ is the sphere, then $U\cap V$ is the circle, thus we have the following fundamental groups:
$\pi_1(U)=\mathbb Z\times\mathbb Z$
$\pi_1(V)=0$
$\pi_1(U\cap V)=\mathbb Z$
If we use the group presentation notation we have:
$\pi_1(U)=\langle\alpha,\beta\mid \alpha\beta=\beta\alpha\rangle$
$\pi_1(V)=\langle\emptyset\mid\emptyset\rangle$
$\pi_1(U\cap V)=\langle\gamma\mid\emptyset\rangle$
Thus using the Seifert–van Kampen theorem:
$\pi_1(X)=\langle\alpha,\beta\mid\alpha\beta=\beta\alpha,\beta\rangle$
Note that I added $\beta$ above because when we turn around the generator of $S^1$ which is $U\cap V$, we span one of the generators of the torus which is $U$.
Thus the fundamental group of this space is $\mathbb Z\times \{0\}$ which is $\mathbb Z$ itself.
My approach is correct?
Thanks a lot!
-
It looks like that one generator of fundamental group of torus vanishes since you can collapse a close path around the equator. – Sigur Jan 25 '13 at 2:02
@Sigur yes, you're right. – user42912 Jan 25 '13 at 2:10
@Sigur but what can you say about my proof? – user42912 Jan 25 '13 at 2:11
The approach is correct, but you can use a simpler method. The space you care about is homotopic to the wedge sum of two spheres and one circle. Thus the fundamental group is Z. To see why this space is homotopic to the space I said, you just check first that you can contract the upper half-sphere, thus the lower half-sphere becomes to a sphere, and the torus becomes a sphere with its south and north poles glued together. A sphere with its south and north poles glued together is homotopic to a sphere with a line connect its south and north poles. After these deformations you have a space which is what I said above.
-
Thank you very much you helped not only with this question but with the question I linked about homology groups of this space. – user42912 Jan 25 '13 at 11:16
one question, do you know where can I find a book with these constructions (except Hatcher's book)? these constructions aren't so obvious at the first glance. – user42912 Jan 25 '13 at 11:18
@user42912 I was about to introduce you the Hatcher's book. You just have to learn the first two chapters you will get the idea. There are a lot of deformations of CW cplx. – lee Jan 25 '13 at 11:40
Nitpicking, but I suppose you could be more explicit about the relation $\beta=\gamma$, which comes from the inclusion $\iota:U\cap V\to U$, $\iota_*(\gamma)=\beta$. So you would have
$$\pi_1(X)=\langle\alpha,\beta,\gamma|\alpha\beta=\beta\alpha,\beta=\gamma\rangle$$
As always, it depends how familiar your audience (and you!) are with the subject.
- | 2015-05-23T10:52:17 | {
"domain": "stackexchange.com",
"url": "http://math.stackexchange.com/questions/286315/fundamental-group-of-this-space",
"openwebmath_score": 0.9098342657089233,
"openwebmath_perplexity": 268.0760734679797,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9793540674530015,
"lm_q2_score": 0.8740772400852111,
"lm_q1q2_score": 0.8560311003455452
} |
https://math.stackexchange.com/questions/200389/show-that-the-set-of-all-finite-subsets-of-mathbbn-is-countable/200406 | # Show that the set of all finite subsets of $\mathbb{N}$ is countable.
Show that the set of all finite subsets of $\mathbb{N}$ is countable.
I'm not sure how to do this problem. I keep trying to think of an explicit formula for 1-1 correspondence like adding all the elements in each subset and sending that sum to itself in the natural numbers, but that wouldn't be 1-1 because, for example, the set {1,2,3} would send to 6 and so would the set {2,4}. Multiplying all the elements in each subset and sending that product to itself in the natural numbers wouldn't work either since, for example, {2,3} would send to 5 and so would the set {1,5}. Any ideas?
• $2\times3\not=5$ :) – Ben Millwood Sep 21 '12 at 17:53
• Set of Finite Subsets of Countable Set is Countable at ProofWiki. See also this question for a generalization of your question ($\mathbb N$ is replaced by arbitrary infinite set $X$ and it is shown that set of all finite subset of $X$ has the same cardinality as $X$). – Martin Sleziak Sep 22 '12 at 6:50
• You're right, an explicit bijection would be a neat proof. Hint: think like a computer scientist. – Colonel Panic Jul 17 '15 at 10:05
Consider an enumeration of the (positive) prime numbers by $n\mapsto p_n$. Then if you're given the set $\{n_1,...,n_k\}$, map that to $$p_{n_1}\cdots p_{n_k}.$$ For example, if we've enumerated the primes increasingly ($p_1=2$, $p_2=3$, and so on), then $$\{1,3,4\}\mapsto 70=2\cdot 5\cdot 7=p_1\cdot p_3\cdot p_4,$$ and $$\{1,3\}\mapsto 10=2\cdot 5=p_1\cdot p_3.$$
You'll need the fact that there are countably infinitely many prime numbers, and uniqueness of prime factorization. From there, it shouldn't be hard to prove this is an injection from the set of finite subsets of $\Bbb N$ into $\Bbb N$, so there are at most countably many such subsets. To show that there are at least (and so exactly) countably many finite subsets of $\Bbb N$, you need only find an injection from $\Bbb N$ into the set of finite subsets of $\Bbb N$, which should be easy.
Added: The described map "should" send $\emptyset$ to $1,$ the "empty product."
• So if I was given a set {1,3} and a set {1,3,4}, you're saying to map that to 1,3 and 1,2^2,3? Am I understanding correctly? – user39794 Sep 21 '12 at 17:28
• Not quite. I'll include an example to clarify. – Cameron Buie Sep 21 '12 at 17:31
• It's ok I think I get it now from reading the other answer! :) Thanks though! – user39794 Sep 21 '12 at 17:32
• @Diego: Yes. However, the given map is not a bijection with $\Bbb N.$ For example, there is no finite subset of $\Bbb N$ that is mapped to $4.$ Rather, the map is an injection into $\Bbb N.$ – Cameron Buie Feb 12 '16 at 12:34
• @Diego: Your reasoning is a bit faulty. After all, $\{1\}$ is a subset of $\Bbb N,$ too, and we certainly can't conclude that there exists a bijection between $\{1\}$ and $\Bbb N$! In order to conclude that a subset of $\Bbb N$ is in bijection with $\Bbb N,$ we have to know that it is an infinite subset. The last part is aimed at proving that the range of the map is, indeed, infinite. – Cameron Buie Feb 12 '16 at 19:33
Define three new digits: \begin{align} \{ &= 10 \\ \} &= 11\\ , &= 12 \end{align} Any finite subset of $\mathbb{N}$ can be written in terms of $0,1,\cdots,9$ and these three characters, and so any expression of a subset of $\mathbb{N}$ is just an integer written base-$13$. For instance, $$\{1,2\} = 10 \cdot 13^4 + 1 \cdot 13^3 + 12 \cdot 13^2 + 2 \cdot 13^1 + 11 \cdot 13^0 = 289873$$
So for each finite $S \subseteq \mathbb{N}$, take the least $n_S \in \mathbb{N}$ whose base-$13$ expansion gives an expression for $S$. This defines an injection into $\mathbb{N}$.
More generally, any set whose elements can be written as finite strings from some finite (or, in fact, countable) alphabet, is countable.
• +1 Really nice answer. – Fly by Night Sep 2 '13 at 18:07
• The user @TimothySwan tried to edit your question to add a comment, I have copied his comment here since the user can't comment himself. He commented: However, this proof is not correct since there must be an injection from $\mathbb N$. – Alice Ryhl Sep 22 '14 at 14:34
• Thanks @Darksonn. (For the benefit of anyone else reading, that's not true; every infinite set has an injection from $\mathbb{N}$ into it!) – Clive Newstead Sep 22 '14 at 20:32
• (+1 a while ago) Not that it will be relevant to most users, but depending on the axioms one is working with, it may be possible for an infinite set to have no injection from $\Bbb N.$ – Cameron Buie May 10 '15 at 9:22
• @TTL: You're missing 'and these three new characters', namely open curly bracket, close curly bracket, and comma. – Clive Newstead Nov 15 '19 at 22:23
Alternatively, just use base $2$, so send $\{a_1,...,a_n\}$ to $2^{a_1}+\cdots+2^{a_n}$. (Assuming your $\mathbb N$ includes $0$, this is $1-1$ and onto.)
• +1 Put in other way: any finite subset of positive integers can be represented as a binary string (with a finite number of ones), which in turn can be interpreted as a binary number which correspond to a positive integer. – leonbloy Mar 17 '14 at 20:09
• In my opinion, this is the best possible proof because the bijection is very, very explicit. – Martin Brandenburg Jul 17 '15 at 8:31
• Any base worked, right?(the base being an integer) – Diego Feb 12 '16 at 4:42
• Any base gives you a $1-1$ function. Base $2$ is onto. @Diego – Thomas Andrews Feb 12 '16 at 11:52
• @MartinBrandenburg: Since the set {a1,...an} isn't an ordered one, we have several base 2 numbers mapping to the same set. How's a bijection best constructed here, then? The accepted answer is also just an injection, no? – Nikolaj-K Mar 8 '17 at 14:57
The other answers give some sort of formula, like you were trying to do. But, the simplest way to see that the set of all finite subsets of $\mathbb{N}$ is countable is probably the following.
If you can list out the elements of a set, with one coming first, then the next, and so on, then that shows the set is countable. There is an easy pattern to see here. Just start out with the least elements.
$$\emptyset, \{1\}, \{2\}, \{1, 2\}, \{3\}, \{1, 3\}, \{2, 3\}, \{1, 2, 3\}, \{4\}, \ldots$$
In other words, first comes $\{1\}$, then comes $\{2\}$. Each time you introduce a new integer, $n$, list all the subsets of $[n] = \{1, 2, \ldots, n\}$ that contain $n$ (the ones that don't contain $n$ have already showed up). Therefore, all subsets of $[n]$ show up in the first $2^{n}$ elements of this sequence.
• In other words: List the elements of $P(\emptyset)$. List the elements of $P(\{1\}) \setminus P(\emptyset)$. List the elements of $P(\{1,2\}) \setminus P(\{1\})$. List the elements of $P(\{1,2,3\}) \setminus P(\{1,2\})$. And so on. – Martin Brandenburg Jul 17 '15 at 8:29
• The lazy approach is a 'surjective' listing, $PL([0,1]),PL([0,2]),PL([0,3]),\dots,PL([0,n]),\dots$ where $PL$ is powerset listing (with commas) and $[0,n]$ is the initial segment. – CopyPasteIt Aug 18 '17 at 19:26
• how do you actually show this is surjective and injective? – Pinocchio Sep 5 '18 at 16:18
• For a proof specification see this answer: Neatest proof that set of finite subsets is countable?. – CopyPasteIt Jun 2 at 14:21
I'm surprised nobody else has said this, so I worry that I've missed the point of the question. But I would argue as follows: Let $N_i$ be the family of all subsets of $\Bbb N$ with exactly $i$ elements. $N_0$ is finite, and $N_1$ is countable. $N_2$ is countable also, since it's a subset of $\Bbb N\times \Bbb N$. There's an easy induction argument that $N_i$ is countable for all $i>0$, since it's contained in the product $\Bbb N\times N_{i-1}$ of two countable sets.
Then, since the set you want, $\bigcup N_i$, is a countable union of countable sets, it is countable.
• Well, a countable union of countable sets needn't be countable. It depends on your axioms. – Cameron Buie Sep 29 '12 at 22:09
• What axioms would those be? – MJD Sep 29 '12 at 22:28
• Set theoretic axioms--which you need at least some of before you can talk about sets, countability, unions, etc. For example, there are models of ZF in which $\Bbb R$ is a countable union of countable sets. If one takes countable choice (or stronger) as an axiom, then countable unions of countable sets are countable (that's a strictly weaker statement than countable choice). Now, it's worth noting that countable unions of enumerated sets (countable sets with a given bijection with $\Bbb N$) are countable, with no need for choice principles. – Cameron Buie Sep 29 '12 at 23:07
Let $p_1,p_2,\ldots$ the sequence of primes. Then map each finite subset $\{a_1,\ldots,a_n\} \subset \mathbb N$ to $p_1^{a_1}\cdots p_n^{a_n}$ where $a_1 < \ldots < a_n$. This gives an injective map from the set of finite subsets of $\mathbb N$ to $\mathbb N$.
Alternatively, notice that for every $s$ there are only finitely many finite subsets of $\mathbb N$ whose sum is $\leq s$. Let $A_s$ be the set of those subsets, then $\mathbb N = \bigcup_{s} A_s$ is a countable union of finite sets, hence countable.
• You beat me to the post ... slightly different, but the same important idea - You can show that the set $C_N$ of subsets of $\mathbb N$ whose members sum to $N$ is finite for each $N\in\mathbb N$. Then each finite subset of $\mathbb N$ has a finite sum $N$ and hence is in one of the $C_N$, and the union of the $C_N$ is a countable union of finite sets, hence is countable. – Mark Bennet Sep 21 '12 at 18:41
• @Mark: Not that it's probably relevant to the OP, but a countable union of finite sets needn't be countable in general without sufficient Choice. Here, of course, we're dealing with well-ordered finite sets, so it works out. – Cameron Buie Jul 31 '14 at 10:48
If $S$ is the set of all finite subsets of $\mathbb{N}$, define $f:S\rightarrow \mathbb{Q}$ by
$f(A)=.x_1x_2x_3x_4\cdots$ where $x_{i}=\begin{cases}1 &\mbox{, if }i\in A\\0 &\mbox{, if } i\not\in A \end{cases}$.
Then $f$ is clearly injective, so $S$ is countable since $\mathbb{Q}$ is countable and any subset of a countable set is countable. | 2020-07-12T12:52:41 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/200389/show-that-the-set-of-all-finite-subsets-of-mathbbn-is-countable/200406",
"openwebmath_score": 0.951771080493927,
"openwebmath_perplexity": 220.7539603480424,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9793540716711546,
"lm_q2_score": 0.874077230244524,
"lm_q1q2_score": 0.8560310943950199
} |
https://cs.stackexchange.com/questions/101782/show-that-for-any-real-constants-a-and-b-where-b-0-n-ab-theta | # Show that for any real constants $a$ and $b$, where $b > 0$, $(n + a)^b = \Theta(n^b)$
I'm currently studying growth of function chapter in Introduction to Algorithm. In exercise 3.1-2 the question is:
Show that for any real constants $$a$$ and $$b$$, where $$b>0$$,
$$(n + a)^b = \Theta(n^b)$$.
I understand I need to find constants $$c_1$$, $$c_2$$, $$n_0 > 0$$ where
$$0 \leq c_1n^b \leq (n+a)^b \leq c_2n^b$$ for all $$n \geq n_0$$.
But I don't understand what should I do after above step. Looking at the solution, it shows
$$n + a \leq n + |a|$$
$$n + a \leq 2n$$, when $$|a| \leq n$$
and
$$n + a \geq n - |a|$$
$$n + a \geq \frac{n}{2}$$ , when $$|a| \leq \frac{n}{2}$$
How did they come up with $$n + a \leq n + |a|$$ and $$n + a \geq n - |a|$$ just from $$0 \leq c_1n^b \leq (n+a)^b \leq c_2n^b$$ ? Are they derived from $$0 \leq c_1n^b \leq (n+a)^b \leq c_2n^b$$ ? Or are they just a general knowledge that can help solve this problem?
Thank you!
UPDATE: I have updated the question to be more specific.
• $(n+a)^b = n^b + c_1n^{b-1}a+... = \theta(n^b)$ – Mr. Sigma. Dec 19 '18 at 13:40
• @Mr.Sigma.: $b$ can be any positive real number, say 0.5. – rici Dec 19 '18 at 13:54
• O'Luck: $\mid a\mid$ is the absolute value of $a$; either $a$ or $-a$. If $a$ is positive, $n+ |a| = n + a$; otherwise $a$ is negative and $n+a < n < n + |a|$. What problem do you have with that? – rici Dec 19 '18 at 14:00
• "Where did they get those inequalities?" Imagine that $n$ is very large. – John L. Dec 19 '18 at 14:12
• Please do not post equations, text, and the like as images since these are incompatible with the site's search engine. Thank you. – dkaeae Dec 19 '18 at 14:20
I understand I need to find constants $$c_1$$, $$c_2$$, $$n_0 > 0$$ where
$$0 \leq c_1n^b \leq (n+a)^b \leq c_2n^b$$ for all $$n \geq n_0$$.
But I don't understand what should I do after above step.
The approach you would like to use here is "going backwards from the target".
Suppose we do have $$c_1n^b \leq (n+a)^b$$ for some constant $$c_1\gt0$$ when $$n$$ is large enough. What can we understand or deduce from it? What is needed to make it true?
Since the exponent is the same, we can use the law of same exponent, $$(x/y)^m = x^m/y^m$$ to simplify that inequality so that we will have the appearances of our variable $$n$$ closer to each other.
$$c_1\leq \frac{(n+a)^b}{n^b}=\left(\frac{n+a}n\right)^b$$
Raising both sides to the power of $$\frac1b$$, we have
$$(c_1)^{\frac1b}\leq \left(\frac{n+a}n\right)^{b\,\frac1b}=\frac{n+a}n = 1+\frac an$$
(Another way to obtain $$(c_1)^{\frac1b}\leq\frac{n+a}n$$ is to raise both sides of $$c_1n^b \leq (n+a)^b$$ to the power of $$\frac 1b$$.)
Since $$c_1$$ is a positive constant, $$(c_1)^{\frac1b}$$ is also a positive constant. To make the above inequality hold, we would like to make $$\left|\frac an\right|$$ small enough. For example, we can require $$\left|\frac an\right|\le \frac12$$. That is why you see the following condition.
$$|a|\le\frac12 n$$
What is nice here is that we can reverse the above argument to obtain a wanted constant $$c_1$$. Note that it is just as fine if we choose a different condition such as $$|a|\le\frac13 n$$ or $$|a|\le\frac1{2019}n$$ or infinitely many others.
If we start from $$(n+a)^b \leq c_2n^b$$, we could arrive at $$|a|\le n$$.
Are they derived from $$0\le c_1n^b\le(n+a)^b\le c_2n^b$$ ? Or are they just a general knowledge that can help solve this problem?
Yes, they are derived as you have just seen. You could say that they are general knowledge that helps solve this problem. You could also say that we just discovered some specific facts that help solve this problem.
• Wow thank you so much @apass-jack for your answer. This type of step-by-step answer is what I'm looking for. I guess I'm pretty weak at math. Before I accept your answer, I want to ask a few questions. In $(c_1)^{\frac1b}\leq \left(\frac{n+a}n\right)^{b\,\frac1b}=\frac{n+a}n = 1-\frac an$ , isn't it supposed to be $1+\frac an$ at the end? – O'Luck Dec 19 '18 at 23:12
• Thanks for pointing out my typo. Updated. – John L. Dec 19 '18 at 23:35
• Thank you. Next question is: "To make the above inequality hold, we would like to make $\left|\frac an\right|$ small enough". Why? If we choose big value for $\left|\frac an\right|$ , let say $\left|\frac an\right| > 1$ the inequality is still satisfied right? – O'Luck Dec 20 '18 at 0:01
• Reason one: when $n$ goes to (positive) infinity, $|\frac an|$ goes to 0. So you cannot make it as big as you want. In fact, $|\frac an| \le |a|$ if we assume $n$ is a positive integer. Reason one is enough. Reason two: even if you could magically, suppose $a$ is negative, $\frac an$ is negative, too, which is bad for $1+\frac an$. – John L. Dec 20 '18 at 0:12
• umm, I mean $|\frac an|$ as a whole not only the $n$. So, with $(c_1)^{\frac1b}\leq 1 + \frac an$ if $|\frac an| > 1$ let say $|\frac an| = 2$ then $(c_1)^{\frac1b}\leq 1 + 2$ , the inequation is still satisfied, am i right? – O'Luck Dec 20 '18 at 1:10
If n is large then we have n/2 ≤ n + a ≤ 2n.
We therefore have $$(1/2)^b n^b ≤ n^b ≤ 2^b n^b$$.
There you have your two constants: $$(1/2)^b$$ and $$2^b$$. | 2021-08-04T03:14:44 | {
"domain": "stackexchange.com",
"url": "https://cs.stackexchange.com/questions/101782/show-that-for-any-real-constants-a-and-b-where-b-0-n-ab-theta",
"openwebmath_score": 0.8414087295532227,
"openwebmath_perplexity": 216.59520289263443,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9793540722737479,
"lm_q2_score": 0.8740772253241803,
"lm_q1q2_score": 0.8560310901029743
} |
https://math.stackexchange.com/questions/2275616/probability-of-a-number-rolled-of-a-20-sided-dice-being-greater-than-the-sum-of | # Probability of a number rolled of a 20 sided dice being greater than the sum of the numbers rolled on 3 six sided die.
Bob rolls $3$ six-sided die and sums the numbers facing up. Bill rolls a single $20$ sided dice and records the number. What is the probability that Bob's number is greater than Bill's number.
I started the problem trying to come up with an equation, but that didn't work, so I resorted to creating $6$ six by six tables with all of the possible sums for Bob's die. Then, I counted the number of each number and created a chart and calculated the probability of each of those numbers occurring. Then, I multiplied the probability of each number occurring by the probability that Bill's number is greater than that number. Finally, I added them all up.
Obviously, this was very tedious and time-consuming. Is there a more elegant/less tedious way to do this problem.
PS: Is there away to do a $n$ die vs $m$ dice problem without listing them all out? Is there a general formula? Up until now, that's what I've been doing.
• The probability asked for in the title $P(\text{D20 beats 3 D6s})$ is the opposite of the probability asked for in the details $P(\text{3 D6s beats D20})$. – N. Shales May 11 '17 at 1:57
Let the total on the three six-sided dice be $X,$ $3\leq X\leq18.$ Let the number on the twenty-sided die be $Y,$ $1\leq Y\leq20.$
Given any particular value of $X,$ the probability that the twenty-sided die will roll higher is $$P(Y>X \mid X) = \frac{20-X}{20} = 1 - \frac1{20}X.$$
The overall probability that the twenty-sided die will roll higher than the total on the other three dice is \begin{align} P(Y>X) &= \sum_{n=3}^{18} P(Y > X \mid X)P(X=n) \\ &= \sum_{n=3}^{18} \left(1 - \frac1{20}X\right)P(X=n). \end{align}
The last line of that set of equations is just the expected value of $1 - \frac1{20}X.$ That is, \begin{align} P(Y>X) &= \mathbb E\left[1 - \frac1{20}X\right] \\ &= 1 - \frac1{20} \mathbb E[X] \\ &= 1 - \frac1{20} \left(\frac{21}{2}\right) \\ &= \frac{19}{40}. \end{align}
If the question is actually the one posed in the original question body rather than in the original title, namely the probability that $X > Y,$ then we simply observe that for any given value of $X,$ $$P(Y < X \mid X) = \frac{X-1}{20} = \frac{1}{20}X - \frac{1}{20}.$$
The rest of the calculation builds on this the same way the first calculation in this answer built on $P(Y > X \mid X).$ We find that \begin{align} P(Y<X) &= \mathbb E\left[\frac1{20}X - \frac1{20}\right] \\ &= \frac1{20} \mathbb E[X] - \frac1{20}\\ &= \frac1{20} \left(\frac{21}{2}\right) - \frac1{20} \\ &= \frac{19}{40}. \end{align}
This should not be surprising, because it also follows from $P(Y>X)=\frac{19}{40}$ and the "obvious" fact that $P(Y=X)=\frac1{20}.$
• Please see my answer for a simple proof that the two probabilities are equal (and hence both equal to $19/40$). – Barry Cipra May 11 '17 at 13:50
• I don't know if this counts as a shortcut, but to find the expected value of X faster, you could find the expected value of one dice and just multiply it by three instead of summing all the numbers from 3-18. I feel like that just works better in my head. – Henry Weng May 11 '17 at 14:47
Whatever Bob rolls with the $6$-sided dice has a $1$ in $20$ chance of being matched, for a tie, by Bill's roll of the $20$-sided die. Whatever sum, $S=a+b+c$, Bob rolls, if you turn his dice over, the sum is $(7-a)+(7-b)+(7-c)=21-S$. Similarly, whatever number $T$ Bill rolls, if you turn the $20$-sided die over, the number is $21-T$. Thus for each outcome in which Bob wins, there is an equally likely outcome in which Bill wins, and vice versa. Hence the probability of winning for each of them is the same, namely
$${1\over2}\left(1-{1\over20}\right)={19\over40}$$
Remark: It's not literally necessary that the "complementary" number for each side of a die be the opposite face, just that there be a complemenary number somewhere. For $6$-sided dice, having opposite faces sum to $7$ is fairly standard; I believe it's also standard for $20$-sided dice to have opposite faces sum to $21$.
Also, I'd like to credit David K's answer with motivating this one. When I saw from his analysis that the two probabilities were equal, I decided there ought to be simple reason why. As luck would have it, I found one.
• The connection to my answer is even deeper than the coincidence that $P(Y>X)=P(Y<X)$: the usual easy method to calculate $E[X]$ uses the same "turn the dice over" argument (or something equivalent). But the really nice observation here is that the exact same symmetry works for both dice. – David K May 11 '17 at 13:57
• @DavidK, thanks. The symmetry principle also suffices, for example, to show that it's a fair game to roll $13$ regular dice against $7$ dodecahedral dice. In that case, however, computing the probability of a tie is not so easy. (Or is it?) – Barry Cipra May 11 '17 at 14:10
Your best bet is to keep track of two things for both sides: how likely this particular result is, and how likely anything less than this result is. Given these we can multiply them together relatively easily.
$$\begin{array}{r|rr|rr|r} x & 3\text{d}6 = x & 3\text{d}6 < x & 1\text{d}20 = x & 1\text{d}20 < x & 1\text{d}20 < x \cap 3\text{d}6 = x\\ \hline 1 & 0 & 0 & 1 & 0 & 0 \\ 2 & 0 & 0 & 1 & 1 & 0 \\ 3 & 1 & 0 & 1 & 2 & 2 \\ 4 & 3 & 1 & 1 & 3 & 9 \\ 5 & 6 & 4 & 1 & 4 & 24 \\ 6 & 10 & 10 & 1 & 5 & 50 \\ 7 & 15 & 20 & 1 & 6 & 90 \\ 8 & 21 & 35 & 1 & 7 & 147 \\ 9 & 25 & 56 & 1 & 8 & 200 \\ 10 & 27 & 81 & 1 & 9 & 243 \\ 11 & 27 & 108 & 1 & 10 & 270 \\ 12 & 25 & 135 & 1 & 11 & 275 \\ 13 & 21 & 160 & 1 & 12 & 252 \\ 14 & 15 & 181 & 1 & 13 & 195 \\ 15 & 10 & 196 & 1 & 14 & 140 \\ 16 & 6 & 206 & 1 & 15 & 90 \\ 17 & 3 & 212 & 1 & 16 & 48 \\ 18 & 1 & 215 & 1 & 17 & 17 \\ 19 & 0 & 216 & 1 & 18 & 0 \\ 20 & 0 & 216 & 1 & 19 & 0 \\ \hline \text{total} & 216 & & 20 & & 2052 \end{array}$$
Total up everything in the rightmost column, divide by the totals for the $3\text{d}6 = x$ and $1\text{d}20 = x$ columns, and you win: Bob wins $\frac{2052}{4320} = \frac{19}{40} = 0.475$ of the time.
To get ties, or where Bill wins, change what pairs you multiply: both $=$ ones for ties, and $3\text{d}6 < x$ and $1\text{d}20 = x$ for Bill's wins.
This particular one is interesting: because both distributions are symmetrical, and they have the same mean, Bill and Bob will both win the same proportion of the time.
It occurs to me that there's another part to this question: how do we efficiently calculate result probabilities for combinations of dice?
The answer to that is an operation called convolution, which I'll present here in discrete form.
Given two functions $f(x)$ and $g(x)$, the convolution is $$(f * g)(x) = \sum_{k = -\infty}^\infty f(k)g(x-k)$$
This can be interpreted in probability theory as the following: we have two random variables $f$ and $g$, with probability functions $f(x)$ and $g(x)$. the probability function for $f + g$ -- adding the two results together -- is equal to $(f * g)(x)$.
Obviously with those infinities in there, we have to fiddle with it a little to actually get anything done. In our case, because we're dealing with dice, our functions have what's called limited support: they're only non-zero in a small area, so we only need to cover that small area.
Let's do a specific example. Say I want to calculate the probability that I'll get a $7$ on $3\text{d}6$. $3\text{d}6$ is the same as $1\text{d}6$ + $2\text{d}6$, so I can convolve these two. I'll call their functions $f$ and $g$ respectively.
$f$ here has limited support: the only values it is non-zero for are $1$ through $6$. This allows us to change the limits of our summation to the bounds of $f$'s support.
\begin{align} (f * g)(7) &= \sum_{k=1}^6 f(k)g(7 - k)\\ &= f(1)g(6) + f(2)g(5) + f(3)g(4) + f(4)g(3) + f(5)g(2) + f(6)g(1) \\ &= \frac{1}{6}\cdot\frac{5}{36} + \frac{1}{6}\cdot\frac{4}{36} + \frac{1}{6}\cdot\frac{3}{36} + \frac{1}{6}\cdot\frac{2}{36} + \frac{1}{6}\cdot\frac{1}{36} + \frac{1}{6}\cdot\frac{0}{36} \\ &= \frac{15}{216} \end{align}
Using convolution, then, we can calculate the probabilities of the summed results of multiple dice, without necessarily considering every single simple event: to calculate, say, the probability distribution of $5\text{d}6$, we can take the distribution of $4\text{d}6$ and the distribution of $1\text{d}6$ and convolve them. And to get $4\text{d}6$'s distribution we can convolve $3\text{d}6$ and $1\text{d}6$, etc. So instead of counting out $7776$ possibilities, we instead handle $21\cdot6 + 16\cdot6 + 11\cdot6 + 6\cdot6 = 324$ total multiplications and a similar number of additions.
• I didn't read your answer or the problem, but your LateX table skills are impressive – user2879934 May 11 '17 at 0:53
• That last paragraph is an important short cut. It means that you really only need to work out the cases that result in a tie. – amd May 11 '17 at 0:55
• I literally learned how to make the table to write the answer, and beat Excel until morale impro^W^W it built most of the rows for me. – Dan Uznanski May 11 '17 at 0:56
• I was initially skeptical that this is really "Your best bet", given the other answers, but it does appear to be a more general solution that is purely computational. – 6005 May 12 '17 at 5:34
Call the probability of scoring a $k$ with the D20 $q_k=q=1/20$ and scoring a $k$ with the 3 D6'S $p_k$ then we are looking for the probability that the D20 has the greater score
$$p_3\sum_{k=4}^{20}q_k+p_4\sum_{k=5}^{20}q_k+\ldots +p_{18}\sum_{k=19}^{20}q_k$$
Or simply
$$q(17p_3+16p_4+\cdots +2p_{18})$$
but by symmetry for every way to score $k$ with 3 D6s there is a way to score $21-k$, simply by subtracting the score on each die from $7$. The probability of rolling a total of $18$ with $3$ D6s is therefore the same as rolling a total of $3$ ($p_3=p_{18}$) and rolling a total of $17$ is the same probability as rolling $4$ ($p_4=p_{17}$) etc. So
$$\text{required probability}=19q(p_3+p_4+\cdots +p_{10})=19q(p_{11}+p_{12}+\cdots +p_{18})$$
But of course
$$p_3+p_4+\cdots +p_{18}=1$$
So
$$2(p_3+p_4+\cdots +p_{10})=1=2(p_{11}+p_{12}+\cdots +p_{18})$$ $$\implies p_3+p_4+\cdots +p_{10}=p_{11}+p_{12}+\cdots +p_{18}=\frac{1}{2}$$
Giving
$$\text{required probability}=19\cdot\frac{1}{20}\cdot\frac{1}{2}=\frac{19}{40}\tag{Answer}$$
Note that, since the probability of the players getting the same score is $(1/20)\cdot 1=1/20$, then the probability that the D20 loses is the same as it winning i.e. $19/40$. | 2019-05-26T21:53:40 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2275616/probability-of-a-number-rolled-of-a-20-sided-dice-being-greater-than-the-sum-of",
"openwebmath_score": 0.9943407773971558,
"openwebmath_perplexity": 423.7090713453807,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9852713887451681,
"lm_q2_score": 0.8688267847293731,
"lm_q1q2_score": 0.8560301727693086
} |
http://mathhelpforum.com/calculus/34866-different-forms-same-integral.html | # Math Help - Different forms of same integral?
1. ## Different forms of same integral?
Hey people, I just came up across something I didn't know about in my revision. Say I want to find the integral
$\int \frac{1}{3x}dx$
Normally I would do this as following
$\frac{1}{3}\int \frac{1}{x}dx=\frac{1}{3}\ln{x}+k$
However, is it not true that I can also make this of the form
$\int \frac{f'(x)}{f(x)}dx$
via
$\frac{1}{3}\int \frac{3}{3x}dx$
which via a simple u sub yields
$\frac{1}{3}\ln{3x}+j=\frac{1}{3}\ln{x}+\frac{1}{3} \ln{3}+j$
Which does not seem to be equivalent to the original answer. Is it the case that both are valid, but the constants k and j are different? Is k=j+ln3?
$y=\frac{1}{3}\ln{x}+k$ and $\frac{1}{3}\ln{3x}+j$
as explicit functions of x?
Ta muchly.
2. Originally Posted by Xei
Hey people, I just came up across something I didn't know about in my revision. Say I want to find the integral
$\int \frac{1}{3x}dx$
Normally I would do this as following
$\frac{1}{3}\int \frac{1}{x}dx=\frac{1}{3}\ln{x}+k$
However, is it not true that I can also make this of the form
$\int \frac{f'(x)}{f(x)}dx$
via
$\frac{1}{3}\int \frac{3}{3x}dx$
which via a simple u sub yields
$\frac{1}{3}\ln{3x}+j=\frac{1}{3}\ln{x}+\frac{1}{3} \ln{3}+j$
Which does not seem to be equivalent to the original answer. Is it the case that both are valid, but the constants k and j are different? Is k=j+ln3?
Mr F says: They are equivalent. j + ln 3 is just as arbitrary as k .....
$y=\frac{1}{3}\ln{x}+k$ and $\frac{1}{3}\ln{3x}+j$
as explicit functions of x?
Mr F says: They already are explicit functions of x. Do you mean how to express them as explicit functions of y ....?
Ta muchly.
..
3. Originally Posted by Xei
Hey people, I just came up across something I didn't know about in my revision. Say I want to find the integral
$\int \frac{1}{3x}dx$
Normally I would do this as following
$\frac{1}{3}\int \frac{1}{x}dx=\frac{1}{3}\ln{x}+k$
However, is it not true that I can also make this of the form
$\int \frac{f'(x)}{f(x)}dx$
via
$\frac{1}{3}\int \frac{3}{3x}dx$
which via a simple u sub yields
$\frac{1}{3}\ln{3x}+j=\frac{1}{3}\ln{x}+\frac{1}{3} \ln{3}+j$
Which does not seem to be equivalent to the original answer. Is it the case that both are valid, but the constants k and j are different? Is k=j+ln3?
Since j and k are just arbitrary constants one can be set equal to the other.
By the way, just so that you are aware...
$\int \frac{1}{x}~dx = ln|x| + C$
-Dan
4. Yeah it's all right, I just wasn't bothered with the mod bit.
I was just checking, I haven't come across a situation before in which I ended up with two different constants for the same thing.
Is it more desirable to do it the initial way though? That way you simply have lnx instead of ln3x which is a little trickier. I suppose you could just get rid of the 3 and call it lnx+ln3+j = lnx+c which is the original result..?
Mr F: Yeah, y please. As simple as it'll go.
5. Hello, Xei!
You are basically correct . . .
$\int \frac{1}{3x}\,dx \;=\;\frac{1}{3}\int \frac{1}{x}\,dx\;=\;\frac{1}{3}\ln{x}+K$ .[1]
$\frac{1}{3}\int\frac{3}{3x}\,dx \;=\;\frac{1}{3}\ln{3x}+J$ .[2]
Is it the case that both are valid, but the constants $K$ and $J$ are different? .
. . . yes
Is $K\:=\:J+\ln3$? .
. . . um, not quite
From [2], we have: . $\frac{1}{3}\ln(3x) + J \;=\;\frac{1}{3}\left[\ln(3) + \ln(x)\right] + J \;=\;\frac{1}{3}\ln(3) + \frac{1}{3}\ln(x) + J$
. . $= \;\frac{1}{3}\ln(x) + \underbrace{\frac{1}{3}\ln(3) + J}_{\text{a constant}} \;=\;\frac{1}{3}\ln(x) + K$
6. Originally Posted by Xei
Yeah it's all right, I just wasn't bothered with the mod bit.
I was just checking, I haven't come across a situation before in which I ended up with two different constants for the same thing.
Is it more desirable to do it the initial way though? That way you simply have lnx instead of ln3x which is a little trickier. I suppose you could just get rid of the 3 and call it lnx+ln3+j = lnx+c which is the original result..?
Mr F: Yeah, y please. As simple as it'll go.
$y=\frac{1}{3}\ln{|x|}+k \Rightarrow 3y - 3k = \ln |x| \Rightarrow e^{3y - 3k} = x$
$\Rightarrow x = e^{3y}\, e{-3k} = A e^{3y}$. | 2015-08-01T08:30:43 | {
"domain": "mathhelpforum.com",
"url": "http://mathhelpforum.com/calculus/34866-different-forms-same-integral.html",
"openwebmath_score": 0.8933925032615662,
"openwebmath_perplexity": 524.9119868617777,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9852713874477227,
"lm_q2_score": 0.8688267745399466,
"lm_q1q2_score": 0.856030161602703
} |
https://mathhelpboards.com/threads/volume-of-a-pyramid.9351/ | # [SOLVED]Volume of a pyramid
#### dwsmith
##### Well-known member
I am trying to find the volume of a pyramid where the base has length $$L$$ and width $$W$$, and the pyramid has height $$h$$.
Let $$L$$ be on the x axis and $$W$$ be on the y axis.
In the x-z plane, we have the line $$z = -\frac{h}{L/2}x + h$$, and in the y-z plane, we have the line $$z = -\frac{h}{W/2}y + h$$.
My cross sections has width $$\Delta z$$. So I want to find the volume $$\int_0^hA(z)dz$$.
How can I do this?
#### Evgeny.Makarov
##### Well-known member
MHB Math Scholar
Hint: Let $A(z)$ be the area of the cross-section at height $z$. Prove that $A(z)=A\left(1-\frac{z}{h}\right)^2$ where $A=A(0)$. Then compute the integral $\int_0^h A(z)\,dz$. This resulting expression of the volume through $A$ works for any right cone, not just a pyramid.
#### dwsmith
##### Well-known member
Hint: Let $A(z)$ be the area of the cross-section at height $z$. Prove that $A(z)=A\left(1-\frac{z}{h}\right)^2$ where $A=A(0)$. Then compute the integral $\int_0^h A(z)\,dz$. This resulting expression of the volume through $A$ works for any right cone, not just a pyramid.
So is A a plane? Where did $$\left(1-\frac{z}{h}\right)^2$$ come from?
#### Evgeny.Makarov
##### Well-known member
MHB Math Scholar
So is A a plane?
No, in my notations $A=A(0)$ is a number, the area of the pyramid's base.
Where did $$\left(1-\frac{z}{h}\right)^2$$ come from?
The length of the cross-section at height $z$ is $L\left(1-\frac{z}{h}\right)$, and the width at height $z$ is $W\left(1-\frac{z}{h}\right)$. This is seen from your linear equations by finding $2x$ and $2y$ for a fixed $z$, but it is even easier to see this from similar triangles obtained when you consider sections through the $xz$ and $yz$ planes.
By the way, the formula for the volume in terms of $A$ and $h$ is true for any cone, not necessarily a right one.
#### dwsmith
##### Well-known member
No, in my notations $A=A(0)$ is a number, the area of the pyramid's base.
The length of the cross-section at height $z$ is $L\left(1-\frac{z}{h}\right)$, and the width at height $z$ is $W\left(1-\frac{z}{h}\right)$. This is seen from your linear equations by finding $2x$ and $2y$ for a fixed $z$, but it is even easier to see this from similar triangles obtained when you consider sections through the $xz$ and $yz$ planes.
By the way, the formula for the volume in terms of $A$ and $h$ is true for any cone, not necessarily a right one.
Why would we find the area as $$2x\cdot 2y$$ instead of $$x\cdot y$$?
#### Prove It
##### Well-known member
MHB Math Helper
I am trying to find the volume of a pyramid where the base has length $$L$$ and width $$W$$, and the pyramid has height $$h$$.
Let $$L$$ be on the x axis and $$W$$ be on the y axis.
In the x-z plane, we have the line $$z = -\frac{h}{L/2}x + h$$, and in the y-z plane, we have the line $$z = -\frac{h}{W/2}y + h$$.
My cross sections has width $$\Delta z$$. So I want to find the volume $$\int_0^hA(z)dz$$.
How can I do this?
If it's a right pyramid, isn't the volume just \displaystyle \begin{align*} V = \frac{1}{3}A\,h = \frac{1}{3}L\,W\,h \end{align*}?
#### Evgeny.Makarov
##### Well-known member
MHB Math Scholar
Why would we find the area as $$2x\cdot 2y$$ instead of $$x\cdot y$$?
Because the section of the cone using the $xz$ plane is a triangle with sides
\begin{align}
\end{align}
If you find the $x$ corresponding to a given $z$ from (2), then the line at height $z$ crosses the triangle from $-x$ to $x$, i.e., the length of the segment the triangle cuts on the line is $2x$. But again, this is easier to see from similar triangles. The overall intuition is that the length of the cross-section decreases linearly from $L$ at $z=0$ to $0$ at $z=h$. There is a single linear function that does this, and it is $L\left(1-\frac{z}{h}\right)$.
- - - Updated - - -
If it's a right pyramid, isn't the volume just \displaystyle \begin{align*} V = \frac{1}{3}A\,h = \frac{1}{3}L\,W\,h \end{align*}?
We are trying to prove it.
#### MarkFL
Staff member
I am trying to find the volume of a pyramid where the base has length $$L$$ and width $$W$$, and the pyramid has height $$h$$.
Let $$L$$ be on the x axis and $$W$$ be on the y axis.
In the x-z plane, we have the line $$z = -\frac{h}{L/2}x + h$$, and in the y-z plane, we have the line $$z = -\frac{h}{W/2}y + h$$.
My cross sections has width $$\Delta z$$. So I want to find the volume $$\int_0^hA(z)dz$$.
How can I do this?
You may want to read >>>this thread<<< for a tutorial on how to work this type of problem.
#### dwsmith
##### Well-known member
You may want to read >>>this thread<<< for a tutorial on how to work this type of problem.
I was reading a calculus book and had most of it figured out but the 2x 2y piece Makarov cleared up. | 2021-01-22T03:58:36 | {
"domain": "mathhelpboards.com",
"url": "https://mathhelpboards.com/threads/volume-of-a-pyramid.9351/",
"openwebmath_score": 0.9237639904022217,
"openwebmath_perplexity": 234.80703776362515,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.9852713865827591,
"lm_q2_score": 0.8688267745399465,
"lm_q1q2_score": 0.8560301608511993
} |
http://mathhelpforum.com/calculus/84137-binomial-series-2-stuck.html | # Math Help - Binomial series #2 stuck!
1. ## Binomial series #2 stuck!
$f(x) = \frac{5}{(1+\frac{x}{10})^4}$
$= 5\left[1 + \frac{x}{10})^-4\right]$
$= 5 \left[1 + (-4)(\frac{x}{10}) + \frac{(-4)(-5)}{2!}(\frac{x}{10})^2 + \frac{(-4)(-5)(-6)}{3!}(\frac{x}{10})^3 + ... \right]$
But I don't know if this is right, or what my nth term would be, much less the power series...Thanks!!
2. Originally Posted by mollymcf2009
$f(x) = \frac{5}{(1+\frac{x}{10})^4}$
$= 5\left[1 + \frac{x}{10})^-4\right]$
$= 5 \left[1 + (-4)(\frac{x}{10}) + \frac{(-4)(-5)}{2!}(\frac{x}{10})^2 + \frac{(-4)(-5)(-6)}{3!}(\frac{x}{10})^3 + ... \right]$
But I don't know if this is right, or what my nth term would be, much less the power series...Thanks!!
Before I start, are we dealing with a Maclaurin or Taylor Series? About what point are we constructing the series?
3. Originally Posted by Chris L T521
Before I start, are we dealing with a Maclaurin or Taylor Series? About what point are we constructing the series?
Sorry...
Here is the whole question:
Use the binomial series to expand the function as a power series.
$f(x) = \frac{5}{(1+\frac{x}{10})^4}$
4. Binomial series are given by: $(1+y)^k = 1 + ky + \frac{k(k-1)}{2!}y^2$ ${\color{white}.} \ + \ \frac{k(k-1)(k-2)}{3!}y^3 + \cdots + \frac{k(k-1)(k-2)\cdots(k - n+1)}{n!}y^n + \cdots$
for any $k \in \mathbb{R}$ and if $|y| < 1$. Here, $y = \tfrac{x}{10}$ and $k = -4$ .
So: $g(x) = \left( 1 + \frac{x}{10}\right)^{-4}$
$g(x) = 1 + (-4)\left(\frac{x}{10}\right) + \frac{(-4)(-5)}{2!}\left(\frac{x}{10}\right)^2 + \frac{(-4)(-5)(-6)}{3!}\left(\frac{x}{10}\right)^3 + \cdots$ $+ \ \frac{(-4)(-5)(-6)\cdots(-4-n+1)}{n!}\left(\frac{x}{10}\right)^n + \cdots$
_______________
Let's look at the general term: $\frac{(-4)(-5)(-6)\cdots(-3-n)}{n!}\frac{x^n}{10^n}$
Just looking at the expanded series, we see that with even $n$, the coefficient is positive and with odd $n$, the coefficient is negative.
So, if we factor out the negative signs: $\frac{(-1)^{n} (4)(5)(6)\cdots(n+3)}{10^n \cdot n!} x^n \cdot {\color{red}\frac{3!}{3!}} = \frac{(-1)^{n} (n+3)(n+2)(n+1)n!}{3! \cdot 10^n \cdot n!}x^n = \cdots$
etc.
5. Originally Posted by o_O
Binomial series are given by: $(1+y)^k = 1 + ky + \frac{k(k-1)}{2!}y^2$ ${\color{white}.} \ + \ \frac{k(k-1)(k-2)}{3!}y^3 + \cdots + \frac{k(k-1)(k-2)\cdots(k - n+1)}{n!}y^n + \cdots$
for any $k \in \mathbb{R}$ and if $|y| < 1$. Here, $y = \tfrac{x}{10}$ and $k = -4$ .
So: $g(x) = \left( 1 + \frac{x}{10}\right)^{-4}$
$g(x) = 1 + (-4)\left(\frac{x}{10}\right) + \frac{(-4)(-5)}{2!}\left(\frac{x}{10}\right)^2 + \frac{(-4)(-5)(-6)}{3!}\left(\frac{x}{10}\right)^3 + \cdots$ $+ \ \frac{(-4)(-5)(-6)\cdots(-4-n+1)}{n!}\left(\frac{x}{10}\right)^n + \cdots$
_______________
Let's look at the general term: $\frac{(-4)(-5)(-6)\cdots(-3-n)}{n!}\frac{x^n}{10^n}$
Just looking at the expanded series, we see that with even $n$, the coefficient is positive and with odd $n$, the coefficient is negative.
So, if we factor out the negative signs: $\frac{(-1)^{n} (4)(5)(6)\cdots(n+3)}{10^n \cdot n!} x^n \cdot {\color{red}\frac{3!}{3!}} = \frac{(-1)^{n} (n+3)(n+2)(n+1)n!}{3! \cdot 10^n \cdot n!}x^n = \cdots$
etc.
Ok, I understand everything except where the 3!/3! got there.
Thanks!!!!
6. Focus on the expression: $\frac{4 \cdot 5 \cdot 6 \cdots (n+3)}{n!}$
We can simplify it by multiplying by "1": $\frac{{\color{red} 1 \cdot 2 \cdot 3} \cdot 5 \cdot 6 \cdots (n+3)}{n! \cdot {\color{red}3!}} = \frac{(n+3)!}{n! \cdot 3!}$ $= \frac{(n+3)(n+2)(n+1)n!}{n! \cdot 3!} = \frac{(n+3)(n+2)(n+1)}{ 6}$
So you can see why I multiplied both top and bottom by $3!$ so that I can simplify that long product into a simple factorial. | 2016-07-25T17:05:50 | {
"domain": "mathhelpforum.com",
"url": "http://mathhelpforum.com/calculus/84137-binomial-series-2-stuck.html",
"openwebmath_score": 0.870779275894165,
"openwebmath_perplexity": 500.602954232803,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9852713865827591,
"lm_q2_score": 0.8688267711434708,
"lm_q1q2_score": 0.8560301575047491
} |
https://math.stackexchange.com/questions/2953056/convergence-of-the-following-sequence-of-functions | # Convergence of the following sequence of functions.
For $$n \ge 1$$, let $$g_n(x) = \sin^2 \left (x + \frac 1 n \right ), x \in [0,\infty)$$ and $$f_n(x) = \int_{0}^{x} g_n (t)\ \mathrm {dt}.$$ Then
$$(1)$$ $$\{f_n \}$$ converges pointwise to a function $$f$$ on $$[0,\infty)$$, but does not converge uniformly on $$[0, \infty)$$.
$$(2)$$ $$\{f_n \}$$ does not converge pointwise to any function $$f$$ on $$[0, \infty)$$.
$$(3)$$ $$\{f_n \}$$ converges uniformly on $$[0,1]$$.
$$(4)$$ $$\{f_n \}$$ converges uniformly on $$[0, \infty)$$.
I have found that $$f_n (x) = \frac 1 2 \left (x - \sin x \cos \left (x + \frac 2 n \right ) \right)$$ which converges to the function $$f$$ on $$[0, \infty)$$ defined by $$f(x) = \frac 1 4 ( 2x - \sin {2x}), x \in [0, \infty)$$. But I am not quite sure about whether this convergence is uniform or not. Please help me in this regard.
Thank you very much.
• Yeah you are right. I have edited my body. Oct 12, 2018 at 19:05
• What is the definition of uniform convergence, and how is it different from point-wise convergence. Oct 12, 2018 at 19:28
hint
$$\sin^2(t+\frac 1n)=\frac{1-\cos(2t+\frac 2n)}{2}$$
$$f_n(x)=\int_0^x\sin^2(t+\frac 1n)dt=\frac{1}{2}\Big[t-\frac{\sin(2t+\frac 2n)}{2}\Bigr]_0^x$$
$$=\frac x2-\frac 14\sin(2x+\frac 2n)+\frac 14\sin(\frac 2n)$$
The pointwise limit is $$f(x)=\frac x2-\frac{\sin(2x)}{4}$$
For the uniform convergence, use MVT to get
$$|\sin(2x+\frac 2n)-\sin(2x)|\le \frac 2n$$
and
$$|f_n(x)-f(x)|\le \frac 1n.$$
The convergence is uniform at $$[0,\infty)$$.
• I have done that. First see my body and then answer. Oct 12, 2018 at 19:09
• @Dbchatto67 Your $f_n(x)$ is not correct. Oct 12, 2018 at 19:10
• Have you evaluated your integral between the limits $0$ and $x$? Oct 12, 2018 at 19:11
• In place of $t$ we have $t + \frac 1 n$. Oct 12, 2018 at 19:11
• Now calculate and simplify. Oct 12, 2018 at 19:13 | 2022-08-17T14:21:19 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2953056/convergence-of-the-following-sequence-of-functions",
"openwebmath_score": 0.9471237063407898,
"openwebmath_perplexity": 187.40790748562603,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9852713891776497,
"lm_q2_score": 0.8688267643505193,
"lm_q1q2_score": 0.8560301530663587
} |
https://engineering.stackexchange.com/questions/35475/moment-equation-from-dynamics-differs-with-equations-from-physics-statics-whe | # Moment equation from dynamics differs with equations from physics & statics. Where did I go wrong?
In my dynamics class, we're being asked to solve the following problem:
My Attempt:
Since I'm given the initial velocity, final velocity, and distance, I solved for the acceleration of the plane using kinematics:
$$a = \frac{v_f^2 - v_i^2}{2d} = \frac{(55.6 m/s)^2 - (16.7 m/s)^2}{2(425m)} = 3.31 m/s^2$$
Now this is where my approach starts to differ from the formulas recommended by the class. I decided to set the moment about point A equal to zero, because the plane is not rotating as it moves down the runway. I seem to remember from both statics and physics that if a body is not rotating about a given point, you can simply set the moment around that point to be zero.
Let N be the reaction force at B. Combining the above assumption with Newton's second law in the x direction, I get:
$$\Sigma F_x: R = ma = (140000kg)(3.31 m/s^2) = 4.63 *10^5 N$$
$$\Sigma M_A: -(15m)N + (2.4m)mg - (1.8m)R = 0$$
Solving for N gives me:
$$N = 1.64*10^5 N$$
According to the solution guide, this is incorrect.
Solution Guide's Explanation:
The guide uses a formula which was introduced to us in the textbook, which is as follows.
For some point P fixed on a rigid body with center of mass G, the moment about point P is given by:
$$\Sigma M_P = I_G\alpha + ma_Gd$$
where:
$$I_G$$ is the moment of inertia of the rigid body about G
$$\alpha$$ is the angular acceleration of the rigid body about G
$$a_G$$ is the acceleration of G
$$d$$ is the moment-arm distance, from P to G, of $$ma_G$$
The book cleverly chose a point C on the plane through which both R and A (the reaction force at wheel A) pass. See below:
Using the moment equation above and setting $$\alpha = 0$$ (because the plane isn't rotating) gives:
$$\Sigma M_C = ma_Gd = (15m)N - (3m-1.8m)mg$$
Finally, using $$a_G = 3.3 m/s^2$$ and solving for N, they got:
$$N = 2.57*10^5 N$$
Why I'm Confused:
From the physics and statics classes I've taken in the past, I've always been taught that $$\Sigma M = I\alpha$$; there was never that extra "$$+ mad$$" term thrown onto the end. That term is basically saying that an accelerating body with no rotation can still have a moment about a point. By contrast, in my physics and statics classes, I recall using the fact that if a body is stationary (no angular acceleration), we can set the moment about any point on the body equal to zero to help us solve.
My Guess at Where the Inconsistency Lies:
Here's my guess at where the inconsistency lies: in my statics classes, we assumed that the rigid bodies we were analyzing had not only zero angular acceleration, but also zero linear acceleration. (It's statics, after all!) In dynamics problems such as this one, however, there is a nonzero linear acceleration which must be accounted for. The "+mad" term must come from the fact that for any point object, the moment about some fixed point O is given by:
$$\Sigma M_O = \vec{r} \times \vec{F} = \vec{r} \times m\vec{a}$$
As I'm writing this, I think the intuition totally makes sense. It seems the formula used by the solution guide accounts for both rotation of the rigid body about its center, and linear acceleration of the body with respect to some point P outside the center of mass of the body.
I'll still post this though, in case anyone would like to correct me, add anything, or use this for their own reference.
• Your image font is not readable. Would you add the text of the image to your question. Thanks. Apr 30, 2020 at 0:42
• This is actually for once such a nice homework question, so much information and own effort instead of just a bad photo of a problem sheet. +1, thanks OP Apr 30, 2020 at 8:00
There is a slight mistake in your approach. It seems you have forgotten to add the contribution the moment due to inertial force acting at the center of mass $$G$$ to the moment equation. The moment equation should be : $$\sum M_A = -(15 m) N + (3 m) 140*10^3*a + (2.4 m) 140*10^3*g - (1.8 m) 140*10^3*a = 0$$ Evaluating for $$N$$ results in : $$N = 2.57 * 10^5 N$$ | 2022-05-19T08:18:15 | {
"domain": "stackexchange.com",
"url": "https://engineering.stackexchange.com/questions/35475/moment-equation-from-dynamics-differs-with-equations-from-physics-statics-whe",
"openwebmath_score": 0.8123955726623535,
"openwebmath_perplexity": 347.0649879860927,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9648551525886193,
"lm_q2_score": 0.8872046026642945,
"lm_q1q2_score": 0.8560239322809832
} |
https://math.stackexchange.com/questions/1532735/count-the-number-of-non-decreasing-numbers-with-d-digits-lower-than-x | # Count the number of non-decreasing numbers with d digits, lower than X
A number is said to be made up of non-decreasing digits if all the digits to the left of any digit is less than or equal to that digit.
For example, the four-digit number 1234 is composed of digits that are non-decreasing. Some other four-digit numbers that are composed of non-decreasing digits are 0011, 1111, 1112, 1122, 2223.
Notice that leading zeroes are required: 0000, 0001, 0002 are all valid four-digit numbers with non-decreasing digits.
I arrived at the conclusion that there are in total $$\sum_{l=0}^9 \sum_{k=0}^l \sum_{j=0}^k ... \sum_{a=0}^b 1 = \binom{9+d}{d}$$ ways to calculate the non decreasing numbers, for d digits, without constrains.
My question is, suppose I want to find the number of non-decreasing numbers with D digits, under a given limit (in the same order of magnitude). For example, how many 4 digit numbers are non-decreasing and under 5000?
I realize that for this case, I need this summations, but I can't neither figure out how to solve them, nor how to generalize them $$\sum_{l=0}^9 \sum_{k=0}^l \sum_{j=0}^k\sum_{i=0}^{min(5,j)} 1$$
where I limit this first number to 5.
Any help?
• When I look at this post, I see eight-digit numbers such as $12341234$ rather than four-digit numbers such as $1234$. – N. F. Taussig Nov 17 '15 at 10:01
• Fixed the examples, posted this late at night. Thanks @N.F.Taussig – José Pedro Marques Nov 17 '15 at 10:37
We can represent a number with non-decreasing digits by placing vertical bars in a string of nine ones. The number of ones to the left of the vertical bar represents the digit. For instance, $$| 1 1 1 | | 1 1 1 1 1 1 |$$ represents the number $0339$, while $$1 1 | 1 1 | 1 1 | 1 1 | 1$$ represents the number $2468$. Since a particular number is determined by which four of the thirteen symbols (four vertical bars and nine ones) are vertical bars, the number of non-decreasing numbers with four digits (including leading zeros) is $$\binom{13}{4} = \binom{13}{9}$$ as you found. To determine how many four-digit numbers (including those with leading zeros) are less than $5000$, we must subtract from these those non-decreasing four-digit numbers that are at least $5000$. Notice that in those non-decreasing four-digit numbers that are at least $5000$, the first vertical bar must be placed to the right of at least five ones. For instance, $$1 1 1 1 1 | | 1 | 1 | 1 1$$ represents the number $5567$. Since a particular non-decreasing four-digit number that is at least $5000$ is determined by which four of the last eight symbols (four vertical bars and four ones) are vertical bars, the number of non-decreasing four-digit numbers that are at least $5000$ is $$\binom{8}{4}$$ Hence, there are $$\binom{13}{4} - \binom{8}{4}$$ non-decreasing four-digit numbers, including those with leading zeros, that are less than $5000$.
Alternate Method: A non-decreasing number is completely characterized by the number of times each digit appears. Let $x_k$ represent the number of times the digit $k$ appears in the non-decreasing four-digit number, including those with leading zeros. Since there are four digits, $$x_0 + x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 + x_8 + x_9 = 4$$ A particular number is determined by which nine of the thirteen symbols (nine addition signs and four ones) is an addition sign. For instance, $$1 + + + 1 1 + + + + + + 1$$ represents the number $0339$, while $$+ + 1 + + 1 + + 1 + + 1 +$$ represents the number $2468$. Thus, the total number of non-decreasing four-digit numbers is the number of ways nine addition signs can be placed in a row of four ones, which is $$\binom{4 + 9}{9} = \binom{13}{9}$$ From these, we subtract those non-decreasing four-digit numbers that are at least $5000$. The digits of non-decreasing four-digit numbers that are at least $5000$ satisfy the equation $$x_5 + x_6 + x_7 + x_8 + x_9 = 4$$ A particular number is determined by which four of the eight symbols (four addition signs and four ones) is an addition sign. Hence, there are $$\binom{4 + 4}{4} = \binom{8}{4}$$ non-decreasing four-digit numbers that are at least $5000$. Hence, the number of non-decreasing four-digit numbers that are less than $5000$, including those with leading zeros, is $$\binom{13}{9} - \binom{8}{4}$$ | 2021-02-27T19:50:20 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/1532735/count-the-number-of-non-decreasing-numbers-with-d-digits-lower-than-x",
"openwebmath_score": 0.8440277576446533,
"openwebmath_perplexity": 98.45621382255443,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9893474910447999,
"lm_q2_score": 0.8652240930029117,
"lm_q1q2_score": 0.8560072856039433
} |
http://mathhelpforum.com/geometry/174934-centroid-triangle-coordinates.html | # Thread: The centroid of a triangle with coordinates
1. ## The centroid of a triangle with coordinates
Triangle DEF has vertices D(1,3) and E(6,1), and centroid at C(3,4). Determine the coordinates of point F.
I know that the medians in a triangle intersect at the centroid and that the centroid divides each median in a ration of 2:1. Also, I know that each median intersects a side at its midpoint with the shorter part of the median. What I did was find the midpoint between point E and F by taking half of the line D to C and adding onto C to give point G. The coordinates of point G was found to be (4, 4.5). Since I knew that the change in the coordinates between points E and G, (-2, 3.5), was only half of the line EF I doubled it and added it point E to give the answer (2, 8). I wanted to know if there was an easier way to do this.
2. Originally Posted by darksoulzero
Triangle DEF has vertices D(1,3) and E(6,1), and centroid at C(3,4). Determine the coordinates of point F.
I know that the medians in a triangle intersect at the centroid and that the centroid divides each median in a ration of 2:1. Also, I know that each median intersects a side at its midpoint with the shorter part of the median. What I did was find the midpoint between point E and F by taking half of the line D to C and adding onto C to give point G. The coordinates of point G was found to be (4, 4.5). Since I knew that the change in the coordinates between points E and G, (-2, 3.5), was only half of the line EF I doubled it and added it point E to give the answer (2, 8). I wanted to know if there was an easier way to do this.
Read Centroid - AoPSWiki
3. ## centroid of triangle
Hi darksoulzero,
Your solution is confusing.Here is a suggested method.
connect M midpoint of DE and C (centroid) with extended lenght.F lies on this line. FC = 2CM. Slope diagram of C and M = 2/1/2.Slope diagram of FC is twice that or 4/1. Working from point C one point left and 4 points up gives F (2,8)
bjh
4. If ABC is a triangle with $A\left( {{x_1},{y_1}} \right),B\left( {{x_2},{y_2}} \right),C\left( {{x_3},{y_3}} \right)$ then its centroid is
$G\left( {\tfrac{{{x_1} + {x_2} + {x_3}}}{3},\tfrac{{{y_1} + {y_2} + {y_3}}}{3}} \right)$
5. Hello, darksoulzero!
$\text{Triangle }D{E}F\text{ has vertices: }D(1,3)\text{ and }E(6,1)\text{, and centroid at }C(3,4).$
$\text{Determine the coordinates of point }F.$
Code:
1
F o-----+
\ :
\ :4
\ :
\ :
\:0.5
(3,4)o---+
D C\ :
o \ :2
(1,3) * \:
M o
(3.5,2) * E
o
(6,1)
We have vertices $D(1,3)$ and $E(6,1)$, and centroid $C(3,4).$
The midpoint of $DE$ is: $M(3\tfrac{1}{2},\,2).$
The median to side $DE$ starts at $\,M$, passes through $\,C,$
. . and extends to $\,F$, where: . $FC \,=\,2\!\cdot\!CM.$
Going from $\,M$ to $\,C$, we move up 2 and left $\frac{1}{2}$
Hence, going from $\,C$ to $\,F$, we move up 4 and left 1.
Therefore, we have: . $F(2,8).$ | 2017-03-23T02:17:54 | {
"domain": "mathhelpforum.com",
"url": "http://mathhelpforum.com/geometry/174934-centroid-triangle-coordinates.html",
"openwebmath_score": 0.7596879005432129,
"openwebmath_perplexity": 694.7595029805632,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.9893474907307124,
"lm_q2_score": 0.8652240860523328,
"lm_q1q2_score": 0.8560072784556494
} |
https://www.physicsforums.com/threads/f-x-symmetric-about-the-line-x-2.795521/ | # F(x) symmetric about the line x=2
## Main Question or Discussion Point
Why is the function: y = f(x) = a(x-1)(x-2)(x-3) symmetrical about the line x = 2? I mean how can we be sure that it is? Is there any method to check it?
## Answers and Replies
ShayanJ
Gold Member
Change the variable to y=x-2 and see whether the resulting function is even or odd or none!
You mean x = x-2 right? It doesn't stay the same on doing that. So it should not be symmetrical about x = 2, but it is. And I am not able to see how that happens.
ShayanJ
Gold Member
$f(x)=a(x-1)(x-2)(x-3) \rightarrow f(y)=a(y+1)y(y-1)$
$f(-y)=a(-y+1)(-y)(-y-1)=-a(y-1)y(y+1)=-f(y)$
So f(y) is odd which means f(x) is not symmetric around x=2 but is not that much asymmetric because we have f(2-x)=-f(x-2).
Oh no. I posted the wrong function. The function is: y = f(x) = a(x4/4 -2x3 + 11x2/2 - 6x) + 1
And this function is said to be symmetrical about the line x = 2. But I am unable to see how?
ShayanJ
Gold Member
Its the same trick. Just do the tranformation $x\rightarrow x+2$ and check whether the resulting function is even. If its even, then the original function is symmetric around x=2.
They are not the same, i.e. after changing x to x+2 in f(x), f(x) ≠ f(-x). But what I am reading, it says that f(x) is symmetrical about x=2 and I am still wondering that how would I go about proving it?
ShayanJ
Gold Member
You're doing something wrong. You should be able to reduce f(x+2) to $\frac 1 2 a[x^2(\frac 1 2 x^2-1)-4]+1$ which is even.
PeroK
Homework Helper
Gold Member
Oh no. I posted the wrong function. The function is: y = f(x) = a(x4/4 -2x3 + 11x2/2 - 6x) + 1
And this function is said to be symmetrical about the line x = 2. But I am unable to see how?
Another way to look at it is as follows:
Imagine starting at x = 2 and moving the same distance, d, to the right and left (d > 0). So, to the right we have 2 + d and to the left we have 2 - d.
Now, if f is symmetrical about x = 2, then f(2-d) = f(2+d) for all d. You could try that approach.
You're doing something wrong. You should be able to reduce f(x+2) to $\frac 1 2 a[x^2(\frac 1 2 x^2-1)-4]+1$ which is even.
How did you reduce it down to that? Can you show me? I am unable to get to that point.
ShayanJ
Gold Member
$f(x+2)=a[ \frac 1 4 (x+2)^4-2(x+2)^3+\frac{11}{2} (x+2)^2-6(x+2)]+1=\\ a(x+2)[ \frac 1 4 (x+2)^3-2(x+2)^2+\frac{11}{2} (x+2)-6]+1=\\ a(x+2)(\frac 1 4 x^3+\frac 3 2 x^2+3x+2-2x^2-8x-8+\frac{11}{2}x+5)+1=\\ \frac 1 2 a(x+2)(\frac 1 2 x^3-x^2+x-2)+1=\frac 1 2 a (\frac 1 2 x^4+x^3-x^3-2x^2+x^2+2x-2x-4)+1=\\ \frac 1 2 a (\frac 1 2 x^4-x^2-4)+1$
And what is the reason that we transformed x to x+2?
ShayanJ
Gold Member
Simple. When we say a function is even, we mean its symmetric around x=0. So if a function is symmetric around $x=a \neq 0$, it means if we move the origin to x=a, the resulting function would be even.
Right. So if we move the coordinates of origin from (0,0) to (2,0) shouldn't the abscissa of a point in the new axes, change from x to x-2 and not x+2?
ShayanJ
Gold Member
Check here!
But we aren't shifting the curve as in your link but instead we are shifting the coordinate axes. The curve stays where it was. The axes are what shift.
PeroK
Homework Helper
Gold Member
Right. So if we move the coordinates of origin from (0,0) to (2,0) shouldn't the abscissa of a point in the new axes, change from x to x-2 and not x+2?
This is one reason why it's much better to change the variable name - until, like Shyan, you've mastered this. Let's use z.
If we have z = x + 2, then x = 2 maps to z = 4, which is not what we want.
But, if we have z = x - 2, then x = 2 maps to z = 0, which is what we want. The origin of the z-variable is at x = 2.
In fact, if I'm honest, I always prefer to change the name of the variable to avoid the mistake you just made.
This is one reason why it's much better to change the variable name - until, like Shyan, you've mastered this. Let's use z.
If we have z = x + 2, then x = 2 maps to z = 4, which is not what we want.
But, if we have z = x - 2, then x = 2 maps to z = 0, which is what we want. The origin of the z-variable is at x = 2.
In fact, if I'm honest, I always prefer to change the name of the variable to avoid the mistake you just made.
Wait. So you are saying that f(x) should change to a new variable so that it becomes f(z), right? But that would again mean f(x-2) and not f(x+2).. got me confused
ShayanJ
Gold Member
$z=x-2 \Rightarrow x=z+2$ so $f(x)=f(z+2)$!(Forget the confusing $x\rightarrow x+2$!)
It just simply fails to make sense to me because of the simple logic that in older coordinate plane, suppose x (any number in domain of f(x)) mapped to f(x) = y (say). Now, in the new coordinate system, f(xold) = f(xnew) provided that xnew = xold - 2. So shouldn't we evaluate f(x-2) instead of f(x+2)?
For example, earlier xold = 2 gave f(2) = c. Now xnew = 0 (i.e. xold-2, where xold = 2 ) would give the result c.
ShayanJ
Gold Member
It just simply fails to make sense to me because of the simple logic that in older coordinate plane, suppose x (any number in domain of f(x)) mapped to f(x) = y (say). Now, in the new coordinate system, f(xold) = f(xnew) provided that xnew = xold - 2. So shouldn't we evaluate f(x-2) instead of f(x+2)?
For example, earlier xold = 2 gave f(2) = c. Now xnew = 0 (i.e. xold-2, where xold = 2 ) would give the result c.
Let's start from the beginning.
At first we have a coordinate system which we call xy. Now I define a function y=f(x). Then I move the origin to x=a and name the new coordinate system zy. But this doesn't change the function, only the coordinate system has moved. But if I insist that the function f has the same form in terms of both x and z, then this means that the function has changed which isn't right.(Imagine y=x^2. z=x-a so x=z+a. But if I say that y=z^2, this function would have its minimum at z=0 so x=a which means the function has changed!) So I should have y=g(z). But g should be related to f somehow that we actually get the same function. So let's see what's the relationship. At x=0, f gives f(0), so at z=-a, g should give f(0). Then at x=a, f gives f(a), so at z=0, g should give f(a). Now we have two relationships g(-a)=f(0) and g(0)=f(a) and so we can deduce that g(z)=f(z+a)=f(x). | 2020-04-03T09:44:32 | {
"domain": "physicsforums.com",
"url": "https://www.physicsforums.com/threads/f-x-symmetric-about-the-line-x-2.795521/",
"openwebmath_score": 0.7770125269889832,
"openwebmath_perplexity": 516.0461644055225,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.956634206181515,
"lm_q2_score": 0.8947894717137996,
"lm_q1q2_score": 0.8559862159725078
} |
https://math.stackexchange.com/questions/9911/convergence-of-the-series-sum-limits-n-2-infty-frac1n-logs-n | # Convergence of the series $\sum \limits_{n=2}^{\infty} \frac{1}{n\log^s n}$
We all know that $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^s}$ converges for $s>1$ and diverges for $s \leq 1$ (Assume $s \in \mathbb{R}$).
I was curious to see till what extent I can push the denominator so that it will still diverges.
So I took $\displaystyle \sum_{n=2}^{\infty} \frac{1}{n\log n}$ and found that it still diverges. (This can be checked by using the well known test that if we have a monotone decreasing sequence, then $\displaystyle \sum_{n=2}^{\infty} a_n$ converges iff $\displaystyle \sum_{n=2}^{\infty} 2^na_{2^n}$ converges).
No surprises here. I expected it to diverge since $\log n$ grows slowly than any power of $n$.
However, when I take $\displaystyle \sum_{n=2}^{\infty} \frac{1}{n(\log n)^s}$, I find that it converges $\forall s>1$.
(By the same argument as previous).
This doesn't make sense to me though.
If this were to converge, then I should be able to find a $s_1 > 1$ such that
$\displaystyle \sum_{n=2}^{\infty} \frac{1}{n^{s_1}}$ is greater than $\displaystyle \sum_{n=2}^{\infty} \frac{1}{n (\log n)^s}$
Doesn't this mean that in some sense $\log n$ grows faster than a power of $n$?
(or)
How should I make sense of (or) interpret this result?
(I am assuming that my convergence and divergence conclusions are right).
• They are right. You don't seem to have completed one of your sentences: "if this were to converge, then I should be able to find a s_1 > 1 such that"...? – Qiaochu Yuan Nov 11 '10 at 21:24
• There seems to be some bug with this. It doesn't display the entire thing. – user17762 Nov 11 '10 at 21:25
• @Sivaram: Yes even i tried editing it, and it still has not worked – anonymous Nov 11 '10 at 21:27
• There seems to be a bug/feature! when dealing with <. Probably confusion with html tags. Sivaram, please use \log n instead of just logn. – Aryabhata Nov 11 '10 at 21:32
• @ Chandru and Moron: Thanks for editing it. Now it looks fine. – user17762 Nov 11 '10 at 21:33
## 6 Answers
Yes $\displaystyle \sum_{n=2}^{\infty} \dfrac{1}{n (\log n)^s}$ is convergent if $\displaystyle s > 1$, we can see that by comparing with the corresponding integral.
As to your other question, if
$\displaystyle \sum_{n=2}^{\infty} \frac{1}{n^{s_1}}$ is greater than $\displaystyle \sum_{n=2}^{\infty} \frac{1}{n (\log n)^s}$
does not imply $\log n$ grows faster than a power of $\displaystyle n$. You cannot compare them term by term.
What happens is that the first "few" terms of the series dominate (the remainder goes to 0). For a small enough $\displaystyle \epsilon$, we have that $\log n > n^{\epsilon}$ for a sufficient number of (initial) terms, enough for the series without $\log n$ to dominate the other.
• Sorry. I meant $\exists s_1 > 1$ such that $\displaystyle \sum_{2}^{\infty} \frac{1}{n^{s_1}}$ is greater than $\displaystyle \sum_{2}^{\infty} \frac{1}{n(\log n)^{s}}$ – user17762 Nov 11 '10 at 21:56
This is a special case of classic general logarithmic convergence tests. Some early work in asymptotics was motivated by attempts to determine the "boundary of convergence" in terms of various functions, e.g. log-exp functions. Below is a a very interesting excerpt from Hardy's classic "A course in pure mathematics". For further work see his "Orders of infinity" and see especially this very interesting paper by G. Fisher on Paul du Bois-Reymond's work on the boundary between convergence and divergence (where he discovered diagonalization before Cantor).
Note in particular the following very general result that is presented in the excerpt: the following series and integral are convergent if $\rm\ s > 1\$ and divergent if $\rm\: s\le 1\$ where $\rm\: n_0,\ a\$ are any numbers large enough to ensure positivity of $\rm\ log_k x := \log \log \cdots \log x\:,\$ iterated $\rm\:k\:$ times.
$$\rm \sum_{n_0}^\infty \frac{1}{n \log n\ \log_2 n\ \cdots\ \log_{k-1} n\ (\log_k n)^s }$$
$$\rm \int_{a}^\infty \frac{dx}{x \log x\ \log_2 x\ \cdots\ \log_{k-1} x\ (\log_k x)^s }$$
• Off-topic question: What text is the excerpt from? Thank you! – Ayesha Jan 29 '14 at 1:04
• @Ayesha I say that in the text above it: Hardy's .... – Bill Dubuque Jan 29 '14 at 1:09
• Sorry, sorry, I somehow glanced over that. Thank you again for responding to my rather asinine question. – Ayesha Jan 29 '14 at 1:43
Du Bois-Reymond was interested in this question, positing some sort of series "between" all convergent series and all divergent series. Hausdorff didn't like that idea, and showed that given any countable sequences of convergence series converging "ever more slowly" and divergent series diverging "ever more slowly", you can find a convergent and a divergent series "in between". Here the order relation is the one given by the (conclusive) ratio test.
Bill's remark is related to the recursive Elias codes. The first few codes correspond to the convergent series
$\displaystyle \frac{1}{n(\log n)^2}, \frac{1}{n\log n (\log \log n)^2}, \ldots$
He then diagonalizes to obtain his $\omega$ code (Elias omega code, if you want to look it up), which corresponds to
$\displaystyle \frac{1}{n\log n\log \log n \cdots (\log^* n)^2},$
where $\log^* n$ is the number of times you need to apply $\log$ until you get below some constant (the other logs continue up to this constant); confusingly, coding theorists use $\log^* n$ to mean the $\log n \log\log n \cdots$ part. We can continue the iteration like this:
$\displaystyle \frac{1}{n\log n\log\log n \cdots \log^*n (\log \log^* n)^2}, \frac{1}{n\log n\log\log n \cdots \log^*n \log \log^* n (\log\log \log^* n)^2}, \cdots$
and so on. We can diagonalize again to obtain an $\omega + \omega$ code, and so on at least for $n \omega + m$. We an probably continue even to $\omega^2$ and beyond.
The corresponding divergent series are $\displaystyle \frac{1}{n\log n}, \frac{1}{n\log n\log\log n}, \ldots, \frac{1}{n\log n\log\log n \cdots \log^* n}, \frac{1}{n\log n\log\log n \cdots \log^* n \log\log^* n}, \ldots$
One can ask whether there is a scale for convergent series, that is a chain of (mutually comparable) series converging more and more slowly, such that for each convergent series there's one in the chain converging even more slowly; you can ask the same question about divergent series. Surprisingly, the existence of these is independent of ZFC (they exist given CH, and models in which they don't exist can be constructed using forcing).
• Indeed, it deserves to be better known that Du Bois Reymond - not Cantor - is the true discoverer of diagonalization - in this context, i.e. asymptotic growth rates of functions. – Bill Dubuque Nov 11 '10 at 22:53
• Thanks Bill and Yuval. I was looking for series which diverges slower and slower and your answer helps me on that note. – user17762 Nov 11 '10 at 23:09
Another test that applies to series of positive decreasing terms (and in this particular one in a rather elegant fashion) is the following: $$\sum_{n=1}^\infty a_n<\infty \quad\Longleftrightarrow\quad\sum_{k=1}^\infty 2^ka_{2^k}<\infty.$$ In our case $$\sum_{k=1}^\infty 2^ka_{2^k}=\sum_{k=1}^\infty 2^k\frac{1}{2^k(\log 2^k)^s}= \frac{1}{(\log 2)^s}\sum_{k=1}^\infty \frac{1}{k^s},$$ and thus $$\sum_{n=2}^\infty \frac{1}{n(\log n)^s}\quad\Longleftrightarrow\quad \sum_{k=1}^\infty \frac{1}{k^s}\quad\Longleftrightarrow\quad s>1.$$
• Monotonicity (as is in this case) is required? – Martín-Blas Pérez Pinilla Jan 31 '14 at 19:40
• @Martín-BlasPérezPinilla: I do not know the most general version of this criterion, but the version I know requires $a_n$ to be non-negative and decreasing. – Yiorgos S. Smyrlis Jan 31 '14 at 19:46
• Yes, now I remember I've seen in Baby Rudin. – Martín-Blas Pérez Pinilla Jan 31 '14 at 19:56
By the integral comparison test the given series is convergent if and only if the integral $$\int_2^\infty\frac{dx}{x\log^s(x)}$$ is convergent and notice that $\log'(x)=\frac1x$ so
• if $s>1$ $$\int_2^\infty\frac{dx}{x\log^s(x)}=\frac1{1-s}\log^{1-s}(x)\Big|_2^\infty<+\infty$$ so the series is convergent
• if $s=1$ $$\int_2^\infty\frac{dx}{x\log(x)}=\log(\log(x))\Big|_2^\infty=+\infty$$ so the series is divergent.
• The exponent $s$ is just on the log term, not the entire term – user2566092 Jan 31 '14 at 19:38
• Ah I'll edit my answer! – user63181 Jan 31 '14 at 19:41
• @user2566092 I edited my answer. – user63181 Jan 31 '14 at 19:50
• What about $s < 1$? – mavavilj Sep 19 '15 at 17:25
Convergence does not change if you switch to base-2 logs. There is a theorem that says that if $a_n$ is a decreasing positive sequence, then $\sum_n a_n$ converges if and only if $\sum_k 2^k a_{2^k}$ converges. If you apply this theorem's transformation to your sequence, you get $\sum_k 1/k^s$, which obviously converges if $s > 1$ and diverges if $s \leq 1$. If you'd like a proof of that theorem (called the Cauchy condensation test), I can provide it. | 2019-05-27T03:54:12 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/9911/convergence-of-the-series-sum-limits-n-2-infty-frac1n-logs-n",
"openwebmath_score": 0.901950478553772,
"openwebmath_perplexity": 365.6433204552242,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9566341987633822,
"lm_q2_score": 0.8947894562828416,
"lm_q1q2_score": 0.8559861945730586
} |
https://www.bionicturtle.com/forum/threads/difference-between-probability-density-function-and-inverse-cumulative-distribution-function.9438/ | What's new
# Difference between probability density function and inverse cumulative distribution function?
#### bpdulog
##### Active Member
What is the difference between probability density function and inverse cumulative distribution function, if any? They both look the same
#### ShaktiRathore
##### Well-Known Member
Subscriber
Hi,
No they are not the same.
The inverse cumulative distribution function is the quantile function it gives the value of the quantile(z) at which the probability of the random variable is <=the given probability value or the cumulative probability of random variable is = the given probability value.For e.g.at 95% cumulative probability the value of quantile is z=1.645,at 99% cumulative probability z=2.33 and so on so that the plot of the values of z Vs the cumulative probability values gives the inverse cumulative distribution function ICDF. ICDF(95%)=1.645, ICDF(99%)=2.33.
Whereas probability density function P(z) gives the value of probability at a given quantile,so that when you integrate the function over a quantile range shall give the value of the cumulative distribution function,integration of P(z) over -inf to inf is=1,integration of P(z) over -inf to 1.645 is=95%,at any point z on the P(z), probability(z)=0.
thanks
#### brian.field
##### Well-Known Member
Subscriber
David also covers this extensively in his Study Notes for Miller. Have you read them?
#### bpdulog
##### Active Member
Hi,
No they are not the same.
The inverse cumulative distribution function is the quantile function it gives the value of the quantile(z) at which the probability of the random variable is <=the given probability value or the cumulative probability of random variable is = the given probability value.For e.g.at 95% cumulative probability the value of quantile is z=1.645,at 99% cumulative probability z=2.33 and so on so that the plot of the values of z Vs the cumulative probability values gives the inverse cumulative distribution function ICDF. ICDF(95%)=1.645, ICDF(99%)=2.33.
Whereas probability density function P(z) gives the value of probability at a given quantile,so that when you integrate the function over a quantile range shall give the value of the cumulative distribution function,integration of P(z) over -inf to inf is=1,integration of P(z) over -inf to 1.645 is=95%,at any point z on the P(z), probability(z)=0.
thanks
If I understand what are you saying, in the inverse distribution the 1.645 represents 95% and less. And in the probability density function the 1.645 represents 95% only?
#### ShaktiRathore
##### Well-Known Member
Subscriber
Hi
in the inverse distribution the 1.645(quantile) represents 95%(cumulative probability). And in the probability density function P(x) ,when x=1.645 ,P(x) shall give the probability density at x,and we integrate the P(x) from -inf to x=1.645 to get the cumulative probability of 95%.
thanks
#### bpdulog
##### Active Member
Ah ok! I think I get it now. It just 2 different ways to display a distribution of data. The top is the PDF and the PPF is basically the CDF with the axes transposed.
#### David Harper CFA FRM
##### David Harper CFA FRM
Staff member
Subscriber
Yuu definitely want to understand @ShaktiRathore 's above relationship. Miller's Chapter 2 does a decent job. We don't use the continuous pdf too much. The pdf integrates to the CDF, and we're arguably more interested in the relationships around the CDF, as Shakti illustrates. For example, if probability (p) = 5.0%, then:
• p = 5% = F(q), where F(.) is the cumulative distribution function, so if we are given the probability of 5% because we want a 95% confident normal VaR, then we use the inverse CDF to retrieve the quantile:
• q = F^(-1)(p) = F^(-1)(5%) = -1.645, where F^(-1) is the inverse CDF. "Inverse" implies that F[F^(-1)(p)] = p; e.g., F[F^(-1)(5%)] = F[-1.645] = 5%, just like also F^(-1)[F(q)] = q. This is why Dowd says "VaR is just a quantile;" i.e., because -1.645 is the quantile (q) retrieved by using the inverse CDF given a probability (p) of 5%. This is an essential FRM building block. I hope that adds something.
#### theapplecrispguy
##### New Member
In Question 300.2, the final price of the bond (i.e. the value of "x") is calculated using the inverse CDF such that 5% of the distribution is less than or equal to "x". This price is calculated to be $1.842 using p=5%. My interpretation of this is that 5% of the time the bond will be priced below$1.842 and the other 95% of the time the bond will be priced above $1.842. Is this correct? The question then states that the 95% VAR is given by$5.00 (the face value of the bond) minus $1.842 (q(.05)) equals$3.158. I don't understand conceptually how the formula for 95% VAR was determined to be $5.00 - q(.05). Can you please explain what the 95% VAR ($3.158) represents in the context of the area under the cumulative distribution graph?
#### David Harper CFA FRM
##### David Harper CFA FRM
Staff member
Subscriber
Hi @theapplecrispguy FYI, the source Q&A (with follow on discussion) is here at https://www.bionicturtle.com/forum/threads/p1-t2-300-probability-functions-miller.6728/ but briefly:
• Re: "My interpretation of this is that 5% of the time the bond will be priced below $1.842 and the other 95% of the time the bond will be priced above$1.842. Is this correct?" Yes, exactly correct. The integral (anti-derivative) of the given f(x) is the CDF which is F(x) = x^3/125. When we have the CDF, it means that cumulative p = x^3/125; i.e., p is the probability that the random variable will be less than or equal to x. Because 1.842^3/125 = 5.0%, under this distribution, there is a 5.0% probability that X will be less than $1.842; and, as you say, 95.0% of the time X will be greater than$1.842 (but less than $5.00, as the function domain is$0.00 < X < 5.00). As the question is testing the associated Miller, this is really the essential point of the question ....
• Because the VaR part here is really simplistic. VaR is a loss relative to something. This question gives the assumption that the expected future value is $5.00 so that the VaR can be computed as a simple "worst expected loss" of$5.00 - $1.842 =$3.158; ie, because there is a 5.0% probability of the final price falling below $1.842, there is a 5.0% probability of a loss worse than$3.158. If we were not given this assumption, the proper approaches are classically either:
• Compute the loss relative to the expected future value as the proper mean of the future distribution. In this case, as stated in the question, this relative VaR (it is called because it is relative to the expected future value) would be $3.75 -$1.842 = $1.908. I just didn't want to make the question too difficult; or, • Compute the loss relative to the current value, which is properly termed the absolute VaR (relative to the initial value). I don't think we have information to do that, but it would be given by discounted present value [$5.00 par] - $1.842 = absolute VaR. This shows why a good question needs to be specific about the VaR it looks for, VaR is a worst expected loss over a horizon (choice #1) with a confidence level (choice #2) relative to something (definitional #3). I hope that helps! #### theapplecrispguy ##### New Member Thank you David for the explanation which is very helpful. If the question had asked for the 5% VAR would the answer have been the same? #### David Harper CFA FRM ##### David Harper CFA FRM Staff member Subscriber Hi @theapplecrispguy Yes, but only because occasionally (or rarely) the 95.0% VaR will be represented by an author as the 0.050 VaR, treating them as the same idea. In any realistic use case, we are referring to the losses that happen in the worst 5.0% of times, the losing tail. It would not be realistic/sensible to speak of the losses that occur in the worst 95.0% of times (although mathematically it works of course). But, again, the idea is that we have a cumulative distribution function here, given by F(x) = p = x^3/125. Solving for x, we have x = (p*125)^(1/3). So for example, • at p = 5%, x = (5%*125)^(1/3) =$1.84
• at p = 10%, x = (10%*125)^(1/3) = $2.32 • at p = 20%, x = (20%*125)^(1/3) =$2.92
• at p = 50%, x = (50%*125)^(1/3) = $3.97; i.e., the median • at p = 95%, x = (95%*125)^(1/3) =$4.92
It's not a distribution of losses, it's a distribution of future values. We always want to be mindful of the difference, sometimes the distribution is losses. As this is future values, the risk tail is "down" at p = 5%, it is not "up" at p = 95%, so that either a 95% or 0.05 VaR would be realistically referring to the p = 5%. At the same time, there are valid usages of VaR for lower confidence levels, so we could have a 90% VaR (given by the loss of $5.00 minus$2.32) or a even an 80% VaR (given by the loss of $5.00 minus$2.92). I hope that helps!
Last edited:
#### theapplecrispguy
##### New Member
Thank you. Clear and much appreciated! | 2020-01-18T00:00:27 | {
"domain": "bionicturtle.com",
"url": "https://www.bionicturtle.com/forum/threads/difference-between-probability-density-function-and-inverse-cumulative-distribution-function.9438/",
"openwebmath_score": 0.849563717842102,
"openwebmath_perplexity": 1333.9619822409604,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.97364464868338,
"lm_q2_score": 0.8791467785920306,
"lm_q1q2_score": 0.8559765563833629
} |
https://brilliant.org/discussions/thread/what-happens-to-iirrational-number/ | # what happens to i^(square root of 2)?
I always thought what be in store for us if take irrational number to power of a complex number? we already have natural number, integers, rational numbers, real numbers and complex numbers. Will it create a class of its own?
Note by Shiva Kumar
9 months, 2 weeks ago
MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2
paragraph 1
paragraph 2
[example link](https://brilliant.org)example link
> This is a quote
This is a quote
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$
Sort by:
yes sir but when the value of k is increased in polar form of (i)^(square root of 2), we do not get the solutions that superimpose the previous solutions. so we instead get infinite no of solutions. it form a circle with magnitude equal to one and center as origin in Argand Plane after k becomes infinite. so does that mean (i)^(any irrational number) forms a circle with magnitude of value one and center at origin.and get same set of solutions. does this mean (i)^any irrational number has same set of values.
- 9 months, 2 weeks ago
The basic way to define powers of complex numbers is via the formula $z^w \; = \; e^{w\log z}$ and so the whole business revolves around the definition of the logarithm of complex numbers. We need $\log z \; = \; \ln |z| + i\mathrm{Arg}\,z$ and here is the real problem. There is no way of defining the argument function continuously (let along differentiably) on the whole complex plane.
Since we want the argument (and hence the logarithm) to be differentiable, it has to be defined on an open set (so that we can consider its derivative at all points), and complex functions are considered on open connected domains. The standard way of doing this for the argument is to cut the plane. This means considering the domain formed by the complex numbers with a straight line out from $$0$$ to infinity removed. The argument can be defined uniquely (and differentiably) on any such domain, and so can the natural logarithm of the modulus (since we have removed $$0$$).
For example, we could consider the cut plane $$\mathbb{C} \backslash (-\infty,0]$$ consisting of the complex numbers except for the nonpositive reals. This means that $$|z| > 0$$ and $$-\pi < \mathrm{Arg}\,z < \pi$$ for all $$z$$ in the cut plane, and we can define $$\log z$$ uniquely (and differentiably) now. This choice is called using the principal branch of the argument. In this case we would have $$\ln i = \tfrac12\pi i$$, and so $$i^\sqrt{2} = e^{\frac{\pi i}{\sqrt{2}}}$$.
A different cut would be to exclude the nonnegative reals $$[0,\infty)$$, and decide that $$2\pi < \mathrm{Arg}\,z < 4\pi$$ for all $$z$$ in this cut plane. This would make $$\ln i = \tfrac52\pi$$, giving a totally different value for $$i^{\sqrt{2}}$$.
This is what @Pi Han Goh means about this function being multivalued. There is no way to have a single function that works everywhere.
The other option is to allow all these values to work at the same time! This means we have to leave the ordinary complex plane, and work with a Riemann surface. Imagine an infinite number of complex planes stacked one on top of another. Cut them all along the positive real axis, and stick the "first quadrant edge" of each sheet to the "fourth quadrant edge" of the sheet below. The resulting helicoidal shape allows you to consider complex numbers with positive moduli and all possible arguments. The number $$i$$ would lie in one sheet, the number $$ie^{2\pi i}$$ would lie in the sheet above, the number $$ie^{-6\pi i}$$ would lie in three sheets below, and so one. You could then define a logarithm on the whole Riemann surface at one go...
- 9 months, 1 week ago
Ah crap. I can't answer this question either. My first though is Equidistribution theorem, but I'm not entirely confident.
Summoning the great @Mark Hennings again!
- 9 months, 1 week ago
No, $$i^{\sqrt 2}$$ is just a multivalued complex numbers.
Let $$x = i^{\sqrt 2}$$, then $$\ln x = \sqrt 2 \cdot i = \sqrt 2 ( 0 + 1i )$$.
With $$0 + 1i = \cos \left (\frac\pi 2 + 2\pi k \right) + i \sin \left ( \frac\pi 2 + 2\pi k \right)$$, where $$k$$ is any integer.
Can you finish it off from here?
- 9 months, 2 weeks ago | 2018-06-18T00:27:31 | {
"domain": "brilliant.org",
"url": "https://brilliant.org/discussions/thread/what-happens-to-iirrational-number/",
"openwebmath_score": 0.9641269445419312,
"openwebmath_perplexity": 694.9014525476272,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. Yes\n2. Yes",
"lm_q1_score": 0.9736446456243804,
"lm_q2_score": 0.879146780175245,
"lm_q1q2_score": 0.8559765552355414
} |
https://math.stackexchange.com/questions/7354/sum-limits-n-0-infty-frac12n1-converges/7370 | # $\sum\limits_{n=0}^{\infty} \frac{1}{(2n+1)!}$ converges?
Determine whether this series converges or diverges: $$\sum\limits_{n=0}^{\infty} \frac{1}{(2n+1)!}$$
Thought about using the limit theorem or by comparison but am so stuck. any pointers would be appreciated guys
• What tests do you know for convergence? – Qiaochu Yuan Oct 20 '10 at 21:32
• Look at the absolute value of the ratio of a term in the series to the subsequent term. You should already know what the possible results imply... – Brandon Carter Oct 20 '10 at 23:35
Another way is
If $\displaystyle S_n = 1 + \frac{1}{3!} + \dots + \frac{1}{(2n+1)!}$
We have that
$\displaystyle S_n \le 1 + \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{n(n+1)}$
$\displaystyle = 1 + (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + \dots + (\frac{1}{n} - \frac{1}{n+1}) = 2 - \frac{1}{n+1} < 2$
Thus $S_n < 2$
thus we have the $\displaystyle S_n$ is monotonically increasing and bounded above and so is convergent.
I think svenkatr's response is correct. He is using the comparison test, in particular, comparing with the exponential function for $x=1$, that is obviously a number, so he doesn't have to prove that the series for e converges.
Maybe you can prove the same by using the ratio test $\lim_{n \rightarrow \infty} \displaystyle |\frac{a_{n+1}}{a_{n}}|$. For example, you have $a_{n}=\displaystyle \frac{1}{(2n+1)!}$ and $a_{n+1}=\displaystyle \frac{(2n+1)!}{(2n+3)!}$, then using the definition for the factorial you have $\lim_{n \rightarrow \infty} \displaystyle \frac{1}{(2n+3)(2n+2)}$ which is 0. According to the ratio test:
If r < 1, then the series converges. If r > 1, then the series diverges. If r = 1, the ratio test is inconclusive, and the series may converge or diverge.
Therefore, the series converges.
• Just out of the curiosity, I plugged the series into Wolfram|Alpha. It gives the very nice result of sinh(1). If anyone knows how to prove that, it would be wonderful to know it. By the way, how can I add an hyperlink in the comment section? – Robert Smith Oct 21 '10 at 0:18
• $\sinh(x)=\frac{\exp(x)-\exp(-x)}{2}$. Also, [this](http://functions.wolfram.com/ElementaryFunctions/Sinh/) gives this. – J. M. is a poor mathematician Oct 21 '10 at 1:21
• @Robert: It is the Taylor series for sinh evaluated at 1. – GEdgar Jun 16 '12 at 16:28
• Thanks. I didn't know that. – Robert Smith Jun 16 '12 at 18:06
The series you have is
$1 + \frac{1}{3!} + \frac{1}{5!} \ldots$
If you add the even factorial terms, you get an upper bound i.e.,
$1 + \frac{1}{3!} + \frac{1}{5!} \ldots < \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!}+ \frac{1}{5!} \ldots$
This can be written more compactly as
$\sum_{n=0}^\infty \frac{1}{(2n+1)!} < \sum_{n=0}^\infty \frac{1}{n!} = e^1$
Therefore the series converges.
• I don't see the point of doing this when proving that the series for e converges is exactly as hard. – Qiaochu Yuan Oct 20 '10 at 22:23
• In short, $e=\cosh\;1+\;sinh\;1$. :) – J. M. is a poor mathematician Oct 21 '10 at 0:19
• @ Qiaochu Yuan. You make a valid point. I guess the Ratio test(which Robert Smith has mentioned) is the rigorous answer :). – svenkatr Oct 21 '10 at 3:53
We have $$e^{1} = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \cdots$$ and $$e^{-1} = 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \cdots$$
Subtracting these two we get $$e - e^{-1} = 2 \cdot \Bigl( 1 + \frac{1}{3!} + \frac{1}{5!} + \cdots \Bigr)$$ Therefore the series converges to $$\frac{e-e^{-1}}{2} = \sum\limits_{n=0}^{\infty} \frac{1}{(2n+1)!}$$
• The question was «does the series converge?» Your answer begins by asserting convergence of a series of equivalent difficulty :) – Mariano Suárez-Álvarez Oct 21 '10 at 14:12
Can you bound the series from above by one that you know converges? The factorials grow very fast, so you should be able to.
To elaborate on the first answer given to this question by Ross Millikan.
$$\sum_{n=0}^\infty \frac{1}{(2n+1)!} = 1 + \sum_{n=1}^\infty \frac{1}{(2n+1)!}$$
$$< 1 + \sum_{n=1}^\infty \frac{1}{4^n} = \frac{4}{3}, \quad \textrm{ as } \frac{1}{(2n+1)!} < \frac{1}{4^n} \textrm{ for } n \ge1.$$
Hence by the comparison test the series converges.
Comparing with another more manageable series could be useful in this case for possible follow-on questions as, with this approach, it's not much extra work to prove that it converges to an irrational number. Such a proof might include: Let $S$ be the series and $S_N$ the $N$th partial sum and $R_N$ the remainder then $S=S_N + R_N,$ where we note that
$$R_N < \frac{1}{(2n+3)!} \left( 1 + \frac{1}{(2n+3)^2} + \frac{1}{(2n+3)^4} + \cdots \right).$$
METHOD I
We may simply resort to the Basel problem and get the inequality: $$0<\sum_{k=0}^{\infty}\frac{1}{(1+2k)!}\leq\sum_{k=0}^{\infty}\frac{1}{(1+k)^2}=\frac{\pi^2}{6}$$
METHOD II
According to Taylor's expansion we have that:
$$\sinh(x) = \sum_{k=0}^{\infty}\frac{x^{1+2k}}{(1+2k)!}$$
For $x=1$ we get that the value of the series is $\sinh(1)$. The series converges.
Q.E.D.
This is how sometimes we can extract the exact closed form of some series :
$f(x) = \sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!}$
The radius of convergence is $\infty$
$f \in C^{\infty}(\mathbb{R})$
$f''(x) = f(x)$
solving our equation, we get $f(x) = ae^{\lambda_1x}+be^{\lambda_2x}$
$\lambda^2-1=0$
$\lambda \in \{-1,1\}$
$f(x) = ae^{-x}+be^{x}$
$f(0)= 0 \implies a+b=0 \implies a=-b \implies f(x) = a(e^{x}- e^{-x})$
$f'(0) = 1 \implies a=\frac{1}{2}$
$f(x) = \frac{e^{x}-e^{-x}}{2}$
so our series $\sum_{n=0}^{\infty} \frac{1}{(2n+1)!}$ does converge to $\sinh(1)$
• This only proes that if the series converges, then its sum is $\sinh(1)$. – José Carlos Santos Jul 26 '18 at 20:24
• Yes that is true, but I wanted to say, regarding to other answers, that I would find the limit. So I didn't repeat any other done process here, that is namely true. Anyway that is the same, effectively the sum itself implies the convergence of that series to $\sinh(x)$. So you had not always to prove the convergence of series, since it does have a closed form. the closed form itself implies the convergence, that is so understandable. – Ахмед Jul 26 '18 at 20:26
• your implication is not true for closed form and finite values. that does mean, the sum does converge for all $x\in \mathbb{R}$ to $\sinh(x)$ which is finite. – Ахмед Jul 26 '18 at 20:33 | 2019-10-15T04:07:05 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/7354/sum-limits-n-0-infty-frac12n1-converges/7370",
"openwebmath_score": 0.925818920135498,
"openwebmath_perplexity": 421.98663843244293,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9736446456243805,
"lm_q2_score": 0.8791467722591728,
"lm_q1q2_score": 0.8559765475281003
} |
https://www.physicsforums.com/threads/statement-proof.176659/ | # Statement Proof.
1. Jul 11, 2007
### linuxux
My question is this:
Theorem 1.4.13 part (ii) says: If $$A_n$$ is a countable set for each $$n \in \mathbf{N}$$, then $$\cup^{\infty}_{n=1} A_n$$ is countable.
I can't use induction to prove the validity of the theorem, but the question does say how does arranging $$\mathbf{N}$$ into a 2-d array:
1 3 6 10 15 ...
2 5 9 14 ...
4 8 13 ...
7 12 ...
11 ...
lead to the proof of part (ii) of Theorem 1.4.13?
so obviously it has something to do with the (x,y) co-ordinate system of the array, but I nt sure how it leads to the proof.
Last edited: Jul 11, 2007
2. Jul 11, 2007
### CRGreathouse
You can just show a function $f: A\to\mathbb{N}$.
$$f(0)=A_1(0)$$
$$f(1)=A_1(1)$$
$$f(2)=A_2(0)$$
$$f(3)=A_1(2)$$
$$f(4)=A_2(1)$$
$$f(5)=A_3(0)$$
$$f(6)=A_1(3)$$
. . .
3. Jul 11, 2007
### linuxux
but what is the significance of putting it into a 2-d array? i could just as easily say: 1, 2, 3, 4, ... and call it a 1-d array.
4. Jul 11, 2007
### CRGreathouse
$A_1(0)$ $A_1(1)$ $A_1(2)$ $A_1(3)$&...
$A_2(0)$ $A_2(1)$ $A_2(2)$ $A_2(3)$&...
$A_3(0)$ $A_3(1)$ $A_3(2)$ $A_3(3)$&...
$A_4(0)$ $A_4(1)$ $A_4(2)$ $A_4(3)$&...
$A_5(0)$ $A_5(1)$ $A_5(2)$ $A_5(3)$&...
... ... ... ...
5. Jul 11, 2007
### mathwonk
this proof is due to cantor, 100 years ago. just start couting from the upper left corner, counting by sw/ne diagonals.
6. Jul 12, 2007
### matt grime
To demonstrate a bijection between N (the indices you're inserting) and the countable union of N (the locations in the doubly infinite array). I.E the whole point of the question.
7. Jul 14, 2007
### linuxux
but how is this a 1-1 and onto function? how is it that $$A_n$$ is a function itself? the cantor theorem is the next section in my text so i can't apply that theory now. i understand that f(n) is a function, but why also $$A_n$$?
Last edited: Jul 14, 2007
8. Jul 14, 2007
### CompuChip
Here's a different explanation (sometimes I see that the understanding of the principle depends on the way it is explained, so maybe this will help you - maybe it will make it less clear to you, then please ignore this )
Suppose we put two copies of the natural numbers next to each other. We want to show that, doing this, the total number we get is still countable (that there are "as many" elements in this string of numbers as there are natural numbers, by assigning a natural number to each of the elements).
Clearly, just counting them in a line won't work. Then we assign each natural number to itself in the first copy, but when we're through with that one, we don't have any natural numbers left for the second copy. So what we do is: assign 1 to 1 from the first copy, 2 to 1 from the second copy, 3 to 2 from the first copy, 4 to 2 from the second copy, etc. - so
$$2n - 1 \mapsto n \text{ from the first copy of } \textbf{N}, 2n \mapsto n \text{ from the second copy of } \textbf{N}.$$.
Now consider the Cartesian product $$\textbf N \times \textbf N$$, consisting of all pairs of natural numbers (a, b). How do you prove they are countable. One way would be to start by assigning n to (n, 1), but when we're done, we won't have any natural numbers left to assign to (n, 2) and (n, 3), etc. So here's the trick: we organize the pairs like
(1,1) (1,2) (1,3) ...
(2,1) (2,2) (2,3)
(3,1) (2,3) (3,3)
....
and now we can map them like
1 2 6 7 ...
3 5 8 ...
4 9 10
without running out of numbers too soon.
In the same way, you can prove that $\textbf{N}^k$ is countable for any natural number k, as well as $\frac{1}{2} \textbf{N}$ (all halfintegers 0, 1/2, 1, 3/2, ...) and the integers $\textbf{Z}$ (it's basically two copies of the naturals, like in my first example, with a zero added) and even that the rationals $\textbf{Q}$ are countable. And, also the question from your initial post can be solved like this.
9. Jul 14, 2007
### linuxux
thanks, i understand the example you showed. and to think, this is just an introductory text and i'm self-studying!...
Last edited: Jul 14, 2007
10. Jul 14, 2007
### linuxux
(1,1) (1,2) (1,3) ...
(2,1) (2,2) (2,3)
(3,1) (2,3) (3,3)
this part i understand. each (a,b) has a particular n mapped to it, thus (a,b)~N. so i got that.
but, looking at this:
$A_1(0)$ $A_1(1)$ $A_1(2)$ $A_1(3)$&...
$A_2(0)$ $A_2(1)$ $A_2(2)$ $A_2(3)$&...
$A_3(0)$ $A_3(1)$ $A_3(2)$ $A_3(3)$&...
$A_4(0)$ $A_4(1)$ $A_4(2)$ $A_4(3)$&...
$A_5(0)$ $A_5(1)$ $A_5(2)$ $A_5(3)$&...
i understand this to mean that the number of sets of $$A_n$$ is countable, not that once the union is performed on the sets the elements of the union will be countable. I had a previous problem where i had to show that the union of $$A_1$$ and $$A_2$$ was countable, i did this by defining a function that chose the minimum element first in $$A_1$$ and then in $$A_2$$, then choosing the next smallest element in both, and so on, thus i was able to show $$A_1\cup{}A_2$$~$$\mathbb{N}$$. So in that exercise i addressed the elements in the sets, not the sets themselves. how does the section in red address the elements in the sets? is that even necessary?
**********
wait a minute: i know each $$A_n$$ is countable, so each $$A_n(a\in\mathbb{N})$$ addresses the elements in $$A_n$$. now i see.
Last edited: Jul 14, 2007
11. Jul 15, 2007
### matt grime
By construction: do you write down the same number twice or fail to fill in a box?
It isn't a function.
12. Jul 15, 2007
### linuxux
wouldn't it qualify as a function since $$A_n$$ is countable thus $$A_n \mapsto \mathbb{N}$$?
i thought $$\mapsto$$ represents some type of function. does it not?
i also have another question, what happens if a particular $$A_n$$ is finite?
Last edited: Jul 15, 2007
13. Jul 15, 2007
### CompuChip
Yeah, so for each n we can assign a natural number to each element from $$A_n$$, and in the end, have neither elements of $$A_n$$ nor natural numbers "left" (that is, there is a bijection). The claim is, that the union is countable, so there is a bijection from the naturals to the union. What the 2-D array tries to visualize, is the following: since each $$A_n$$ is countable, we can denote the elements by $$A_n(k)$$ (with k = 1, 2, 3, ...). Now you can order the elements in the union like this:
$$A_1(1) \, A_2(1) \, A_3(1) \, \cdots$$
$$A_1(2) \, A_2(2) \, A_3(2) \, \cdots$$
$$A_1(3) \, A_2(3) \, A_3(3) \, \cdots$$
and use a diagonal counting argument, like before.
Another way is this: each element from the union is indexed by two numbers (n, k) (assuming the union is disjoint, as we implicitly did before). So it's really trivial to write down a bijection from this union to $$\textbf{N} \times \textbf{N}$$. As shown in my earlier post, the latter is countable (there is a bijection to $$\textbf{N}$$, so by composing the bijections you get one from the union to $$\textbf{N}$$. Actually, it also uses the diagonal counting argument (to show that $$\textbf{N} \times \textbf{N}$$ is countable) but perhaps this approach is more intuitive (and once you've proven this result about $$\textbf{N} \times \textbf{N} \times \cdots \times \textbf{N}$$ you can use it and prove countability of other things, like this union of $$A_n$$, without explicitly invoking a diagonal counting argument).
Last edited: Jul 15, 2007
14. Jul 15, 2007
### CompuChip
Does it matter? If you have a set in the union, that is not countably infinite, but finite, will the union become bigger (so big, it will become uncountable?)
If you want to do it officially, consider this: we said a set is countable if we can assign the numbers 1, 2, 3, ... to it. But of course, we can use another offset, and assign 2, 3, 4, ... to it. Or 154842, 154843, 154844, ...
So you could of course first count off the finite sets in the union (in order, no need for diagonals here), and then the infinite ones with the diagonal argument, but using a certain offset.
15. Jul 15, 2007
### linuxux
thanks a million man!
16. Jul 15, 2007
### CompuChip
Most welcome, this stuff is fun and moreover, it's important basics.
Later you will encounter more diagonal arguments, for example: if you want to prove that the real numbers are uncountable, or if you get to work with converging sequences.
17. Jul 16, 2007
### matt grime
You said A_n was a function, not that |-> was a function. Don't be sloppy. | 2017-01-18T10:17:22 | {
"domain": "physicsforums.com",
"url": "https://www.physicsforums.com/threads/statement-proof.176659/",
"openwebmath_score": 0.8634294867515564,
"openwebmath_perplexity": 505.59695090201495,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9736446479186301,
"lm_q2_score": 0.8791467627598856,
"lm_q1q2_score": 0.8559765402961522
} |
https://math.stackexchange.com/questions/3049427/the-converse-of-nilpotent-elements-are-zero-divisors | # The converse of “nilpotent elements are zero-divisors”
For commutative rings $$A$$ with identity $$1\ne0$$, nilpotent elements are zero-divisors. The converse is false, i.e. there is a commutative ring $$A$$ with identity $$1\ne0$$ and a zero-divisor $$x$$ in $$A$$ which is not nilpotent. Where is such an example?
Is it meaningful to ask the "percentage" of commutative rings $$A$$ with identity $$1\ne0$$ for which the converse holds, i.e. zero-divisors are nilpotent?
• Have you thought about the integers modulo $n$? – Lord Shark the Unknown Dec 22 '18 at 12:45
• You can also search yourself at this site, e.g. here, for useful links. – Dietrich Burde Dec 22 '18 at 12:51
• @Lord Shark the Unknown: Thank you. Right, in $\mathbb{Z}/10\mathbb{Z}$, $2$ is a zero-divisor but not nilpotent. How did you came up with this example? How about the next question? – user584333 Dec 22 '18 at 12:52
• Sai, have a look at this post, which gives an answer which rings have this property. – Dietrich Burde Dec 22 '18 at 12:56
• @Dietrich Burde: Thank you for the reference. – user584333 Dec 22 '18 at 13:01
An example is in the integers mod $$6$$, where $$2\cdot 3=0$$, but no power of either of these individually is zero.
If we had some sort of measure on a set of rings (which could not possibly be all rings, because there are too many) with finite total measure, we could ask about proportion. I know of no such commonly used measure, but it's possible such a thing has been considered.
In my intuition, the proportion would be small, but I can't think of a quick reason to see why, other than that being nilpotent seems like a special property while being a zero divisor seems very common. For example, the only time this is true for rings of integers modulo $$n$$ is when $$n$$ is a prime power, and there are vanishingly few of those compared to all integers.
If you look at the answer by rschwieb in the question Under what conditions does a ring R have the property that every zero divisor is a nilpotent element? linked above in a comment by Dietrich Burde, which is not the accepted answer, it characterizes these rings. A ring has this property if and only if it is the quotient of an arbitrary commutative ring by a primary ideal. An ideal $$I$$ is said to be primary if whenever we have that $$a,b\in R$$ satisfy $$ab\in I$$, then we have either $$a\in I$$ or $$b^n\in I$$ for some $$n>0$$. This answers part of your question. It is nicely illustrated in the $$\Bbb Z_n$$ case: an ideal $$n\Bbb Z$$ of $$\Bbb Z$$ is primary if and only if $$n=p^k$$ for some prime $$p$$ and integer $$k>0$$.
So rings like these are actually "close" to integral domains, which are quotients of arbitrary commutative rings by prime ideals. Again, no quantitative argument for why they should be rare, only an intuitive one.
• Couldn't one equip a ring $R$ with the Zariski topology, then produce a Borel sigma algebra, and equip it with a pre-measure, which can then be extended to a measure by Caratheodory's extension theorem, to at least begin discussing "percentages" when the ring itself is of finite measure? – Chickenmancer Dec 22 '18 at 21:44
• @Chickenmancer that wouldn't give a percentage of rings, I suspect you mean to give a percentage of zero divisors that are nilpotent? – jgon Dec 22 '18 at 21:50
• Ah, I misinterpreted the question as "what percentage of a ring $A$." Thanks for clarifying, @jgon. – Chickenmancer Dec 22 '18 at 22:15
• Curious about the downvote. An explanation would be helpful. – Matt Samuel Dec 23 '18 at 17:43
One perspective on what proportion of rings have no nonnilpotent zero divisors is the following. Note that this is really rough first thought.
First let's be careful though, rings with no zero-divisors trivially have this property, so I'll try to consider what rings with zero-divisors have the property that all zero-divisors are nilpotent.
Step 1: Set up a space to parametrize a nice class of rings
Let $$k$$ be an algebraically closed field. Let $$n,d_1,d_2 > 0$$. Let $$x=(x_1,x_2,\ldots,x_n)$$ be coordinates on $$\AA^n_k$$. Let $$r_1=\binom{n+d_1}{d_2}$$, which is the number of monomials in $$k[x_1,\ldots,x_n]$$ of degree at most $$d_1$$. Thus there are $$r_1$$ multiindices $$I$$ of degree at most $$d_1$$. Let $$r_2=\binom{n+d_2}{d_2}$$ as well. Let $$c=(c_I)$$ be coordinates on $$\PP^{r_1-1}_k$$, and let $$d=(d_J)$$ be coordinates on $$\PP^{r_2-1}_k$$ as $$I$$ ranges over multiindices of degree at most $$d_1$$ and $$J$$ ranges over multiindices of degree at most $$d_2$$.
Then consider the subvariety, $$V$$, of $$\AA^n_k\times_k\PP^{r_1-1}_k\times_k \PP^{r_2-1}_k$$ cut out by the polynomial $$f_{c,d}(x)=g_c(x)g_d(x):=\left(\sum_I c_Ix^I\right)\left(\sum_J d_Jx^J\right).$$
$$V$$ is equipped with a natural map $$V\to \PP^{r_1-1}_k\times_k\PP^{r_2-1}$$, and the fiber over a point $$(a,b)$$ in $$\PP^{r_1-1}_k\times_k\PP^{r_2-1}_k$$ is the subvariety of $$\AA^n_k$$ cut out by the degree at most $$d_1+d_2$$ reducible polynomial $$f_{a,b}(x)=g_a(x)g_b(x).$$
This subvariety of $$\AA^n_k$$ can be thought of as corresponding to the ring $$k[x_1,\ldots,x_n]/(g_ag_b)$$, so we can think of our variety $$V$$ as parametrizing a certain nice class of rings all (except for a Zariski closed subset where $$g_a=0$$ or $$g_b=0$$) of which are guaranteed to have zero-divisors, since $$g_a$$ and $$g_b$$ are zero divisors.
Step 2: Investigate the points corresponding to rings with the desired property and related properties
We can then ask if we can characterize the subset of $$V$$ corresponding to rings in which all zero-divisors are nilpotent.
Well, what are the zero-divisors in $$k[x_1,\ldots,x_n]/(f_{a,b})$$? Since $$k[x_1,\ldots,x_n]$$ is a UFD, handily enough, the zero-divisors are precisely the factors of $$f_{a,b}$$. Moreover $$k[x_1,\ldots,x_n]/(f_{a,b})$$ has nilpotents if and only if $$f_{a,b}$$ is not square free (take the radical of $$f_{a,b}$$, it will be nilpotent if and only if it is nonzero, if and only if $$f_{a,b}$$ is not square free). However the ring $$k[x_1,\ldots,x_n]/(f_{a,b})$$ satisfies our property that all zero-divisors are nilpotent if and only if $$f_{a,b}$$ is a power of an irreducible polynomial.
Step 3: Conclude
Note that if $$f_{a,b}$$ is square-free if and only if it is relatively prime to $$f_{a,b}'$$, which is true if and only if $$\Disc(f_{a,b})\ne 0$$. Thus every point $$(a,b)\in\PP^{r_1-1}_k\times_k\PP^{r_2-1}_k$$ corresponding to a ring with any nilpotents at all, let alone one in which every zero-divisor is nilpotent, satisfies a polynomial equation, $$\Disc(f_{a,b})=0$$. Thus rings in our parametrized class with any nilpotents at all correspond to a (proper) Zariski closed subset of $$\PP^{r_1-1}_k\times_k\PP^{r_2-1}_k$$, which is irreducible (see here).
Thus rings in our parametrized class satisfying your property (all zero-divisors are nilpotent) are contained in a codimension 1 class of rings (any nilpotents). Hence almost all rings in our parametrized class have zero-divisors that are not nilpotent.
Notes
This is really rough. The parametrized class of rings is a very specific, very nice subset of $$k$$-algebras. Nonetheless, hopefully it will give you (or other readers) intuition on why very few rings should have the property you want (as long as they have some zero divisors of course).
I needed to eliminate integral domains because I chose to parametrize hypersurfaces, and most hypersurfaces are irreducible. (Check out the link, Qiaochu Yuan gives a really nice quick proof of this fact). There's a reasonable chance that I wouldn't have needed to eliminate integral domains if I'd chosen e.g. codimension 2 subvarieties, but those are much harder to characterize.
• I was thinking of excluding integral domains as well, but notice for integers that isn't necessary. A ring of integers modulo $n$ is an integral domain only if $n$ is prime. – Matt Samuel Dec 23 '18 at 0:41
• @MattSamuel, I originally didn't think to do so, but unfortunately, the particular class of rings I chose to parametrize requires me to do so, since most hypersurfaces in dimensions $\ge 2$ are irreducible. I suspect I wouldn't have to eliminate integral domains if I'd chosen e.g. codimension 2 subspaces, but those are harder to work with. – jgon Dec 23 '18 at 1:07
• I'll add a link to a great answer by Qiaochu Yuan justifying the claim that most hypersurfaces are irreducible when I eventually get back on a computer and I remember. – jgon Dec 23 '18 at 1:08
Another class of examples are matrix rings over a field. A $$n\times n$$ matrix $$A$$ is a zero-divisor if and only if $$A$$ is not invertible.
Certainly if $$A$$ is a zero divisor then $$A$$ cannot be invertible. If $$A$$ is not invertible then $$\ker A \ne 0$$ and taking any non-zero matrix $$B$$ with $$\operatorname{col}B \subseteq \ker A$$ you have $$AB = 0$$. For example, let $$v$$ be a non-zero vector with $$Av = 0$$ and let $$B$$ be the matrix whose columns are all $$v$$.
The nilpotent matrices are rare among zero-divisors. Specifically, nilpotent matrices are solutions to the equation $$A^n = 0$$. As a rule, there are always more non-solutions to a polynomial equation than there are solutions.
In precise terms, we look at the zero set $$\{A : \det A = 0\}$$ (the set of zero-divisors). This zero-set is a manifold if the underlying field is $$\mathbf{C}$$ and is a variety in the general case. The non-nilpotents are dense in the Zariski topology looking over a general ring, and dense in the classical topology looking over $$\mathbf C$$. | 2019-11-20T18:17:56 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/3049427/the-converse-of-nilpotent-elements-are-zero-divisors",
"openwebmath_score": 0.8839941620826721,
"openwebmath_perplexity": 193.22849541817598,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9773708039218115,
"lm_q2_score": 0.875787001374006,
"lm_q1q2_score": 0.8559686455971849
} |
https://www.physicsforums.com/threads/positive-or-negative-remainder.896236/ | # Positive or negative remainder
Is 23 = 5(-4)-3 gives a remainder -3 when divided by 5 ? is this statement true ? some of my colleagues said that remainder cannot be negative numbers as definition but I am doubt that can -3 be a remainder too?
fresh_42
Mentor
Is 23 = 5(-4)-3 gives a remainder -3 when divided by 5 ? is this statement true ? some of my colleagues said that remainder cannot be negative numbers as definition but I am doubt that can -3 be a remainder too?
Usually we consider entire equivalence classes in such cases: Every single element of ##\{\ldots -13, -8, -3, 2, 7 , 12, \ldots\}## belongs to the same remainder of a division by ##5##. We then define all five possible classes
##\{\ldots -15, -10, -5, 0, 5, 10, \ldots\}##
##\{\ldots -14, -9, -4, 1, 6, 11, \ldots\}##
##\{\ldots -13, -8, -3, 2, 7, 12, \ldots\}##
##\{\ldots -12, -7, -2, 3, 8, 13, \ldots\}##
##\{\ldots -11, -6, -1, 4, 9, 14, \ldots\}##
as elements of a new set with five elements ##\{ \; \{\ldots -15, -10, -5, 0, 5, 10, \ldots\}\, , \, \{\ldots -14, -9, -4, 1, 6, 11, \ldots\}\, , \, \ldots \}##.
This notation is a bit nasty to handle, so we choose one representative out of every set. E.g. ##\{[-15],[-9],[12],[3],[-1]\}## could be chosen, but this is still a bit messy to do calculations with. So the most convenient representation is ##\{[0],[1],[2],[3],[4]\}## with the non-negative remainders smaller than ##5##. However, this is only a convention. ##-3## is a remainder, too, belonging to the class ##[2]##. So the answer to your questions is: The statement is true, as all integers are remainders.
mfb
Mentor
The remainder is usually required to be between 0 and N-1 inclusive. 23 and -2 (not -3) are in the same equivalence class. This can also be written as 23 = -2 mod 5.
The remainder is usually required to be between 0 and N-1 inclusive. 23 and -2 (not -3) are in the same equivalence class. This can also be written as 23 = -2 mod 5.
Sorry it should be -23 = 5(-4) - 3 , so in conclusion is, this statement true ?
jbriggs444
Homework Helper
Sorry it should be -23 = 5(-4) - 3 , so in conclusion is, this statement true ?
"-23 divided by 5 is -4 with a remainder of -3". I would consider that statement true.
"-23 divided by 5 is -5 with a remainder of 2". I would also consider that statement to be true.
The convention you use for integer division will determine which of those statements is conventional and which is unconventional.
In many programming languages, integer division follows a "truncate toward zero" convention. For instance, in Ada, -23/5 = -4. The "rem" operator then gives the remainder. So -23 rem 5 = -3.
If one adopts a convention that integer division (by a positive number) truncates toward negative infinity then one would get a different conventional remainder. -23/5 would be -5 and -23 mod 5 would be +2. The Ada "mod" operator uses this convention.
In mathematics, one typically adopts the line of reasoning given by @fresh_42 in post#2 above. The canonical exemplar in the equivalence class of possible remainders is normally the one in the range from 0 to divisor - 1. | 2021-01-18T18:05:26 | {
"domain": "physicsforums.com",
"url": "https://www.physicsforums.com/threads/positive-or-negative-remainder.896236/",
"openwebmath_score": 0.8385798335075378,
"openwebmath_perplexity": 705.3507974379489,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.9773707993078212,
"lm_q2_score": 0.8757869965109765,
"lm_q1q2_score": 0.8559686368033291
} |
https://www.physicsforums.com/threads/help-with-circular-motion.56897/ | Homework Help: Help with circular motion
1. Dec 15, 2004
newcool
Hi, I recieved this problem today. If you put a small marble of mass m on top of a large round object that has a radius r, and then release the marble, when will it come off of the side so that the large round object will not be in contact with the marble?
The answer is arccosine of 2/3. I have tried making free body diagrams of the object when the normal force is 0 and got this equation: a * cosin(theta) = g.
I am stuck as to how to get the right answer.
Thanks for any help
2. Dec 15, 2004
Staff: Mentor
As the marble rolls down the sphere, it will maintain contact as long as there is sufficient force to produce the required centripetal acceleration. Ask yourself: What provides the centripetal force? How does the required centripetal acceleration depend on the the angle that the marble makes with the vertical?
3. Dec 20, 2004
newcool
I made a free body diagram of when the sphere has a normal force of 0. I got one force,F_c pointing in towards the center and another force mg pointing down. Taking the y value of F_c, i got that:
$$F_c*cos(\theta) + mg = ma$$
$$mv^2/r * cos(\theta) + mg = mv^2/r$$
$$v^2/r * cos(\theta) + g = v^2/r$$
I tried a different approach using the fact that the sum of all energies is equal to 0.
$$0 = \Delta K + \Delta U_g$$
$$0 = 1/2 *mv^2 + -mgh$$
$$mgh = 1/2 *mv^2$$
$$gh = 1/2 *v^2$$
$$h = r - r*cos(\theta)$$
so $$gr(1 - cos(\theta)) = 1/2 * v^2$$
However, I have gotten nowhere with these equations. Any input on what I did wrong would be appreciated.
Thanks
4. Dec 20, 2004
Pyrrhus
Ok the answer is acrosinus of 2/3 = 48.2 degrees.
Force analysis
$$n - mg \cos \theta = -m \frac{v^2}{R}$$
The object will fall when n = 0 so
$$v^2 = Rg \cos \theta$$
when it loses contact with the surface
Assuming isolated system
I can use Conservation of mechanical energy
$$K + \Omega = K_{0} + \Omega_{0}$$
$$\frac{1}{2}mv^2 + mgR \cos \theta = 0 + mgR$$
Plugging our speed when it loses contact
$$\frac{1}{2}mRg \cos \theta + mgR \cos \theta = mgR$$
which gives:
$$\cos \theta = \frac{2}{3}$$
$$\theta = 48.2^{o}$$
which is the angle when it will lose contact with the surface.
That's of course assuming an isolated system.
5. Dec 20, 2004
Staff: Mentor
One big problem here: You seem to be treating the "centripetal force" as though it were a separate force (like friction or weight). Not so! "Centripetal" just means "towards the center": The centripetal force is just those real forces that act towards the center. The only forces acting on the mass m are: (1) mg, acting down, and (2) N, the normal force, acting normal to the surface.
So what's the centripetal force? Just the components of those forces acting towards the center of the circular motion, thus producing the centripetal acceleration:
$mg cos\theta - N = F_c = mv^2/r$
Of course, you'll set the normal force to zero, so:
$mg cos\theta = mv^2/r$
Exactly correct. But that's not a "different approach"--it's a necessary part of solving this problem!
Now just combine the two equations and solve for $\theta$.
6. Dec 20, 2004
Pyrrhus
Too bad, i just did the work
Well, also read Doc Al's explanation, it's quite good
7. Dec 20, 2004
newcool
Thanks for the help everyone. Just one question, Doc, how did you get that
$$mg cos\theta - N = F_c = mv^2/r$$
Isn't mg pointing straight down so the component that points towards the center is
$$mg/cos\theta$$
8. Dec 20, 2004
Pyrrhus
We got a vector
$$\vec{R}$$
with y component
$$R \cos \theta$$
and x component
$$R \sin \theta$$
where do you get $\frac{mg}{\cos \theta}$ ???
9. Dec 20, 2004
Pyrrhus
I think you got a misconception with centripetal force, Centripetal force is a role asigned to forces, because they act towards the center. The forces acting on the body are normal and the weight, and the components acting towards the center are equal to $m \frac{v^2}{r}$.
10. Dec 21, 2004
Staff: Mentor
Yes, mg points straight down. Therefore its component towards the center will be $mg cos\theta$, not $mg/cos\theta$. (Draw yourself a picture.)
I believe you are thinking like this: That mg is the vertical component of the "centripetal force" ($F_c$), so $F_c cos\theta = mg$ ==> $F_c = mg/cos\theta$. This is incorrect thinking.
One thing you must realize is that "centripetal" is just a decription of the direction that a force has: it just means "towards the center". (Another term used is "radial".) It is not a kind of force. It's just like describing a force as a horizontal force.
Think like this: The mass m must have an acceleration towards the center since it moves in a circle. So, let's apply Newton's 2nd Law in that radial (or centripetal) direction. As Cyclovenom and I have explained, the only forces acting on the mass are gravity and the normal force. We know that N points away from the center. What's the component of gravity (mg) towards the center? $mg cos\theta$. So:
$$F_{towards-center} = ma_{towards-center}$$
$$mg cos\theta - N = mv^2/r$$
Make sense?
11. Dec 21, 2004
newcool
Thanks for all the help, Doc and Cyclove, I understand it now | 2018-12-12T17:13:20 | {
"domain": "physicsforums.com",
"url": "https://www.physicsforums.com/threads/help-with-circular-motion.56897/",
"openwebmath_score": 0.743645191192627,
"openwebmath_perplexity": 693.0273276151378,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9773707986486796,
"lm_q2_score": 0.8757869932689566,
"lm_q1q2_score": 0.8559686330574059
} |
Subsets and Splits
No saved queries yet
Save your SQL queries to embed, download, and access them later. Queries will appear here once saved.