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https://math.stackexchange.com/questions/2334760/coordinates-of-the-point-where-the-normal-cuts-the-curve | # Coordinates of the point where the normal cuts the curve
Find the equation of the normal to the curve $y= \frac{x-2}{1+2x}$ (1) at the point where the curve cuts the $x$-axis . Find the coordinates of the point where this normal cuts the curve again .
I found The equation of the normal -
$y = -5x+10$ (2)
Now , I use simultaneous equation to find the coordinates .
I sub equation 2 to 1 .
I eventually get a quadratic equation -
$-10x^2 + 14x + 12 = 0$
$x = 2 , \frac{-3}{5}$
I'm shocked now because I'm not sure which one to reject and why ?
Or do I not reject it and sub both of this x values to find 2 values of y meaning I have 2 coordinates ? Or do I have to reject ? Thanks !
• The normal cuts the curve at $(x, x-2/1+2x)$ where $x=0, -\dfrac35$ Jun 24 '17 at 13:55
• There are two intersections and you found both correctly. But "again" implies $x=-3/5$. Jun 24 '17 at 14:02
Since the normal passes through the point $(2,0)$ this is also an intersection point between the line and the curve, corresponding to your solution $x=2$. The other value of $x$ that you have found gives the other intersection.
note that the equation $$-5x+10=\frac{x-2}{1+2x}$$ factorized to $$-2\,{\frac { \left( 5\,x+3 \right) \left( x-2 \right) }{1+2\,x}}=0$$ | 2022-01-19T17:39:22 | {
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https://math.stackexchange.com/questions/2280875/probability-that-2-of-t-strings-of-length-n-are-equal/2280978 | # Probability that 2 of t strings of length n are equal
Given t bit-strings of length n that are generated randomly. What is the probability that at least 2 of these strings are equal. I've seen someone who wrote that the probability is $\le \frac{t^2}{2^n}$
The reasoning is that you have $t$ options to choose the first string. The $t$ options to choose the second string, and given a string chosen at the first stage, the probability that the other chosen string is equal to the first one is $\frac{1}{2^n}$. The you sum all these options of choosing 2 equal string ($t^2$) and you get $\frac{t^2}{2^n}$ This just seems wrong because for each such choice of 2 strings there are many other options to the other strings, but I can't find a proof to show that it is wrong.
Does this reasoning make sense?
We can actually be more precise and say that the probability is at most $\frac{\binom t2}{2^n}$.
For $1 \le i < j \le n$, let $A_{ij}$ be the event that the $i^{\text{th}}$ string is equal to the $j^{\text{th}}$ string. Then $\Pr[A_{ij}] = \frac{1}{2^n}$, because the $j^{\text{th}}$ string is equally likely to be any of $2^n$ possibilities, one of which is equal to the $i^{\text{th}}$ string.
We have two equal strings if any of the $\binom t2$ events $$A_{12}, A_{13}, A_{14}, \dots, A_{t-1,t}$$ occur. If these events were all disjoint - no two could happen at the same time - the probability that any of them occur would be exactly $\binom t2 \cdot \frac1{2^n}$. They're not all disjoint; if the $1^{\text{st}}$ string is equal to the $2^{\text{nd}}$, and the $3^{\text{rd}}$ string is equal to the $4^{\text{th}}$, then $A_{12}$ and $A_{34}$ both occur. So this is not the actual probability we want.
However, it's still an upper bound. For any two events $A$ and $B$, we have \begin{align} \Pr[A \text{ or } B] &= \Pr[A] + \Pr[B] - \Pr[A \text{ and }B] \\ &\le \Pr[A] + \Pr[B] \end{align} and by induction, we can show that for any $m$ events $A_1, A_2, \dots, A_m$, we have $$\Pr[A_1 \text{ or } A_2 \text{ or } \dots \text{ or } A_m] \le \Pr[A_1] + \Pr[A_2] + \dots + \Pr[A_m].$$ This is known as the union bound. In our case, it says that \begin{align} \Pr[A_{12} \text{ or } A_{13} \text{ or } \dots \text{ or } A_{t-1,t}] &\le \Pr[A_{12}] + \Pr[A_{13}] + \dots + \Pr[A_{t-1,t}] \\ &= \binom t2 \frac1{2^n}. \end{align}
We can find the exact probability by first finding the probability that all $t$ strings are different. This is $$\frac{(2^n-1)(2^n-2) \cdots (2^n-t+1)}{(2^n)^{t-1}}.$$ (If all strings must be different, there are $2^n-1$ possibilities for the second string, $2^n-2$ for the third, and so on.) So the exact answer is $1 - \frac{(2^n-1)(2^n-2) \cdots (2^n-t+1)}{(2^n)^{t-1}}$.
But the union bound is often useful when the exact answer cannot be found: for example, if the dependence between the events is very hard to describe. Even here, the bound $\binom t2 \frac1{2^n}$ may be more useful than an exact answer: it makes it much easier to see that if $t$ is much smaller than $2^{n/2}$, then $\binom t2$ is much smaller than $2^n$, and so the probability that any two strings are equal is very small.
• I only have concern about the order - since the denominator, $\binom{2^n -1 +t }{t}$ does not consider the order of selected strings, and I'm not sure how to account for it. Though the problem seems to imply that order does not matter.
– Alex
May 16 '17 at 19:52
• @Alex - I've edited my answer to have the correct exact answer. Indeed, $\binom{2^n-1+t}{t}$ cannot possibly be the correct denominator, since the total number of possibilities for the $t$ strings is $2^{nt}$, so the correct denominator must be a divisor of $2^{nt}$: a power of $2$. May 16 '17 at 21:40
• The question says 'generated randomly', so it's not quite clear if the order matters; my answer holds if the order doesn't matter. Could you explain how you got $2^{n(t-1)}$?
– Alex
May 18 '17 at 5:31
• For each of the $t-1$ remaining strings after the first, we get a factor of $2^t$ in the denominator. I'm assuming "generated randomly" to mean "each string is uniformly and independently chosen from all $2^n$ strings" and don't see what other distribution we could reasonably take. May 18 '17 at 5:46
• Could you have a look at my edit pls
– Alex
May 19 '17 at 15:47
This reminds of the Birthday Problem!
In a room with $n$ students, what is the probability that at least two students share the same birthday? The probability that two students share a birthday is $\frac{1}{365}$, but you have to compare all pairs of n students.
To make an analogy, the probability that two strings are equal is $\frac{1}{2^n}$, but you have to compare all pairs of t strings.
I think this probability is $1-\frac{\binom{2^n}{t}}{\binom{2^n -1 +t}{2^n -1}}$. Essentially this means 1- probability to have all strings different. The numerator is all different strings. The denominator is all cases of selecting t out of $2^n$ with repetitions.
EDIT:
OK @MishaLavrov's solution $\frac{\binom{2^n}{t} t!}{2^{nt}}$ is better. What my solution gives is $\frac{\text{number of ways to get$t$unique strings length$n$from a set of$2^n$strings, order does not matter}}{\text{number of ways to get$t$strings length$n$from a set of$2^n$strings, order does not matter}}$.
For example, if $t=3$, one possible denominator would be $aba, aab, baa$ counted as one. In the numerator we can have $abc,bac,...,cba$ all counted as one outcome.
My confusion is that, even if strings are 'generated randomly', do we account for repeated sets or not? | 2021-09-18T07:38:00 | {
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https://math.stackexchange.com/questions/3797169/evaluate-int-0-pi-ei-zeta-e-ix-dx | Evaluate $\int_0^{\pi} e^{i \zeta e^{ ix}} \ dx$.
I'm trying to evaluate the following integral: $$\int_0^{\pi} e^{i \zeta e^{ ix}} \ dx$$ where $$\zeta >0$$ is some positive real number. Since the antiderivative of this function is just in terms of the exponential integral, I decided to go for a different approach.
My attempt
I did the following $$\int_0^{\pi} e^{i \zeta e^{ ix}} \ dx = \int_0^{\pi} \sum_{n=0}^{\infty}\frac{\left(i \zeta e^{ ix}\right)^n}{n!} \ dx = \sum_{n=0}^{\infty}\frac{(i \zeta)^n}{n!} \int_0^{\pi} e^{nix} \ dx = \sum_{n=0}^{\infty}\frac{(i \zeta)^n}{n! (in)}\left(\underbrace{e^{i\pi n}}_{(-1)^n} -1\right) = \sum_{n=0}^{\infty}\frac{\zeta^ni^{n-1}}{(n+1)!} \left((-1)^n -1\right)$$ To then verify if my procedure was correct, I used WolframAlpha to evaluate both sides of the equation for the value $$\zeta = 1$$. From here I got that $$\int_0^{\pi} e^{i e^{ ix}} \ dx = 1.2494... \neq -0.9193... = \sum_{n=0}^{\infty}\frac{i^{n-1}}{(n+1)!} \left((-1)^n -1\right)$$ I'm not sure where I made my mistake. I think interchanging the integral and the sum is justified since I believe the sum converges absolutely, but now I'm not so sure.
Could anyone tell me where my mistake is? Or alternatively, could anyone tell me how I could evaluate this integral? Thank you!
Edit: Thanks to the comments, I believe that I can simplify the integral to be $$\int_0^{\pi} e^{i \zeta e^{ ix}} \ dx = \pi -2\int_0^\zeta \frac{\sin(t)}{t} \ dt$$ I'm not sure if the approach I was taking was a good way to show this, but if anyone has any ideas about how I could maybe get here I would greatly appreciate them!
• For your solution: $n n!\not = (n+1)!. \int_0^\pi 1dx\not=0.$ – Iridescent Aug 20 '20 at 7:19
• @User628759, I think that with the corrections you mention I get the correct result of $$\pi + \sum_{n=1}^{\infty}\frac{\zeta^ni^{n-1}}{n n!} \left((-1)^n -1\right)$$ Can I simplify the sum on the right even further for a general $\zeta >0$? – Robert Lee Aug 20 '20 at 7:27
• You are right now. The resulting sum is not elementary, but it can be expressed by exponential integrals. – Iridescent Aug 20 '20 at 7:29
• If I understood correctly, I think I should be able to show that $$\sum_{n=1}^{\infty}\frac{\zeta^ni^{n-1}}{n n!} \left((-1)^n -1\right) = -2\int_0^\zeta \frac{\sin(t)}{t} \ dt$$ Or is this not true for the values of $\zeta$ I'm working with? – Robert Lee Aug 20 '20 at 7:41
• You're right, I'll correct it immediately. Thank you for pointing it out! – Robert Lee Aug 20 '20 at 10:52
After playing around with the integral for a while, I believe I've found a way to solve the integral and get it in terms of $$\text{Si}(\zeta)$$.
Let's say we define $$F(\zeta)$$ as $$F(\zeta) := \int_0^{\pi} e^{i \zeta e^{ ix}} \ dx$$ Here we notice that $$F(0) = \int_0^{\pi} 1\ dx = \pi$$. Now, from here we can then analyze the derivative of $$F$$ as follows: \begin{align} F'(\zeta) &= \frac{d}{d\zeta} \int_0^{\pi} e^{i \zeta e^{ ix}} \ dx = \int_0^{\pi} \frac{\partial}{\partial \zeta }e^{i \zeta e^{ ix}} \ dx =\int_0^{\pi}e^{i \zeta e^{ ix}}\left(e^{ix}\right)i\ dx \\ &\overset{\color{blue}{u=ix}}{=} \int_0^{i\pi}e^{i \zeta e^u} e^u \ du \overset{\color{blue}{s=e^{u}}}{=}\int_1^{-1}e^{i \zeta s} \ ds = \frac{e^{i \zeta s}}{\zeta i}\Bigg\vert_{s=1}^{s=-1} = \frac{1}{\zeta i}\left(e^{-i\zeta} - e^{i \zeta}\right)\\ &= -\frac{2}{\zeta} \left( \frac{e^{i\zeta}-e^{-i\zeta}}{2i}\right) = -2 \frac{\sin(\zeta)}{\zeta} \end{align} recalling that we can put the derivative as a partial inside the integral because of Leibniz's integral rule. On the other hand, by the fundamental theorem of calculus, we can easily see that $$\frac{d}{d\zeta}-2\text{Si}(\zeta) =-2 \frac{d}{d\zeta} \int_0^\zeta \frac{\sin(t)}{t} \ dt = -2 \frac{\sin(\zeta)}{\zeta}$$ And since we've found $$2$$ functions with the same derivative, we know they must be the same up to a constant, or in other words $$F(\zeta) = -2 \int_0^\zeta \frac{\sin(t)}{t} \ dt + c$$ But recalling the initial condition we had, we can solve for the value of the constant as follows $$F(0) = \pi = \int_0^0 \frac{\sin(t)}{t} \ dt + c = c$$ and so we get the final result being $$\boxed{\int_0^{\pi} e^{i \zeta e^{ ix}} \ dx = \pi -2\int_0^\zeta \frac{\sin(t)}{t} \ dt}$$
I think that this solution is valid for any $$\zeta \in \mathbb{R}$$, which means I could generalize the original problem to more than just positive values. I believe I haven't missed any details this time, but if I have please let me know! | 2021-03-08T07:08:02 | {
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https://math.stackexchange.com/questions/4180422/are-there-an-infinite-number-of-integers-n-such-that-3nn1-is-a-perfect | # Are there an infinite number of integers $n$ such that $(3n)(n+1)$ is a perfect square?
I am looking for cases in which $$\sqrt{3n(n+1)}$$ is an integer, i.e. cases in which $$3n(n+1)=m^2,\quad m\in\mathbb{N}.$$ I can find solutions such as $$n=0,3,48,675,9408,131043,\dots$$ and I expect this list to be infinite. Is it? Is there a straightforward way to prove these kinds of statements?
There are many ways of reframing the problem: finding integers $$n$$ that are simultaneously 3 times a perfect square and 1 less than another perfect square (choosing $$3n$$ and $$n+1$$ to each be perfect squares), etc. Then I could set $$n=3k^2$$ and try to solve for cases in which $$3k^2+1=l^2 \quad\Leftrightarrow\quad 3k^2=(l+1)(l-1)\quad k,l\in\mathbb{N}.$$ It seems plausible that there are infinite solutions given the various formulas for perfect squares but none of the rabbit holes that I followed led me anywhere productive.
• – lhf
Jun 22, 2021 at 23:37
• Very helpful, thank you. To be honest, even telling me to go look into Diophantine equations would have been a huge help, so this is bonus Jun 23, 2021 at 0:23
Yeah, there's infinitely many. You can see this by considering a particular case:
if $$3n$$ and $$n+1$$ are both squares.
we can parametrize this by $$n = 3k^2$$.
Then we need $$3k^2+1=a^2$$
So we want to solve $$a^2-3k^2=1$$ which is a Pell equation.
the fundamental solution $$(a_1,k_1)$$ is $$(2,1)$$ and the other solutions are obtained via the recurrence $$a_{n+1}=2a_n + 3k_n, k_{n+1} = a_n+2k_n$$.
So we get the solutions are:
$$(7,4),(26,15),(97,56), \dots$$
• And the other case is when $n$ and $3(n+1)$ are both squares, or $x^2-3y^2=3,$ for which there are no solutions Jun 23, 2021 at 0:13
• oh ! thanks ! I thought the other case would also be easy but I was too lazy to work it out ! Thanks a lot ! Jun 23, 2021 at 0:14
• I had not heard of Pell equations - very helpful, especially the recurrence relation Jun 23, 2021 at 0:24
• The recurrence comes from $a_n+k_n\sqrt3=(2+\sqrt 3)^n.$ @QuantumMechanic Jun 23, 2021 at 0:32
From your equation, we get that $$3 \mid m$$, so let $$m = 3j$$, plus do certain manipulations, to get
\begin{aligned} 3n(n+1) &= (3j)^2 \\ n^2 + n & = 3j^2 \\ 4n^2 + 4n & = 12j^2 \\ 4n^2 + 4n + 1 & = 12j^2 + 1 \\ (2n + 1)^2 & = 12j^2 + 1 \\ (2n + 1)^2 - 12j^2 & = 1 \end{aligned}\tag{1}\label{eq1A}
Note this is a Pell's equation. For a coefficient of $$12$$, the fundamental solution is
$$x_1 = 7 = 2n + 1 \implies n = 3, \; \; y_1 = j = 2 \implies m = 6 \tag{2}\label{eq2A}$$
Since your initial solution of $$n = 0$$ makes $$j = 0 \implies y_1 = 0$$, it's not a positive integer so it's not included. The additional solutions section states the remaining solutions are determined from
$$2n + 1 = x_{k+1} = x_1 x_k + 12(y_1 y_k) \tag{3}\label{eq3A}$$
$$j = y_{k+1} = x_1 y_k + y_1 x_k \tag{4}\label{eq4A}$$
Since $$y_1$$ is even, then \eqref{eq4A} shows all $$y_{k+1}$$ are also even. Similarly, since $$x_1$$ is odd, then \eqref{eq3A} shows all $$x_{k+1}$$ are odd, so there's always a corresponding integer value of $$n = \frac{x_{k+1}-1}{2}$$.
• Very good, thank you. My physical problem ignores the $n=0$ solution anyway so this all helps. Jun 23, 2021 at 0:25 | 2022-07-02T08:56:23 | {
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http://mathhelpforum.com/calculus/19332-taylor-series.html | 1. ## taylor series
Can anyone help me with how to get the Taylor series of a function(for example sinx)?
Furthur how to get a value for sin(for example pi/4) with an particular error?
Thank You
2. Originally Posted by Dili
Can anyone help me with how to get the Taylor series of a function(for example sinx)?
Furthur how to get a value for sin(for example pi/4) with an particular error?
Thank You
The Taylor series for a function $\displaystyle f(x)$ (at least I presume you are talking about a single variable function) to N + 1 terms about a point x = a has the form:
$\displaystyle f(x) \approx f(a) + \frac{1}{1!}f^{\prime}(a)(x - a) + \frac{1}{2!}f^{\prime \prime}(a)(x - a)^2 + ~ ... ~ + \frac{1}{N!}f^{(N)}(a)(x - a)^N$
or in summation notation:
$\displaystyle f(x) \approx \sum_{k = 0}{N}\frac{1}{k!}f^{(k)}(a)(x - a)^k$
For the error estimate I will refer you here as I have forgotten which one is "standard" (if any.)
So as an example, let $\displaystyle f(x) = sin(x)$ and let us expand the series about the point $\displaystyle x = \frac{\pi}{4}$.
We have:
$\displaystyle f(x) = sin(x) \implies f \left ( \frac{\pi}{4} \right ) = sin \left ( \frac{\pi}{4} \right ) = \frac{\sqrt{2}}{2}$
$\displaystyle f^{\prime}(x) = cos(x) \implies f^{\prime} \left ( \frac{\pi}{4} \right ) = cos \left ( \frac{\pi}{4} \right ) = \frac{\sqrt{2}}{2}$
$\displaystyle f^{\prime \prime}(x) = -sin(x) \implies f^{\prime \prime} \left ( \frac{\pi}{4} \right ) = -sin \left ( \frac{\pi}{4} \right ) = -\frac{\sqrt{2}}{2}$
$\displaystyle f^{\prime \prime \prime}(x) = -cos(x) \implies f^{\prime \prime \prime} \left ( \frac{\pi}{4} \right ) = -cos \left ( \frac{\pi}{4} \right ) = -\frac{\sqrt{2}}{2}$
So the Taylor series to 4 terms looks like:
$\displaystyle sin(x) \approx \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} \left ( x - \frac{\pi}{4} \right ) - \frac{\sqrt{2}}{4} \left ( x - \frac{\pi}{4} \right )^2 - \frac{\sqrt{2}}{12} \left ( x - \frac{\pi}{4} \right )^3$
-Dan
3. ## lagrange's remainder
Thank you
But what if we take the point x=0, where you get the familiar Taylor series expansion of sinx? Is it the same as taking any value(pi/4)?
And can you explain with the Lagrange's remainder?
4. Originally Posted by Dili
Thank you
But what if we take the point x=0, where you get the familiar Taylor series expansion of sinx? Is it the same as taking any value(pi/4)?
And can you explain with the Lagrange's remainder?
I'll have to let someone else explain the remainder (I'm just not up on it myself, which is why I posted the link. )
Yes, you could use the Maclaurin expansion (the Taylor series expansion about the point x = 0), but note that $\displaystyle \frac{\pi}{4} \approx 0.785398$ is not particularly close to 0. The further x is from 0, the greater the error in the approximation.
On the other hand using 3 terms in the expansion I get
$\displaystyle sin(x) \approx x - \frac{1}{3!}x^3 + \frac{1}{5!}x^5$
Gives me
$\displaystyle sin \left ( \frac{\pi}{4} \right ) \approx 0.707143$
which is only off by $\displaystyle 5.129 \times 10^{-3}$ %, which is pretty close by most people's judgment.
-Dan
5. Theorem: Let $\displaystyle f(x)$ be an infinitely differenciable function on $\displaystyle (a,b)$ (with $\displaystyle a<0<b$). Let $\displaystyle T_n(x)$ represent the $\displaystyle n$-th degree Taylor polynomial around the origin for the point $\displaystyle x\in (a,b)$ (with $\displaystyle x\not = 0$). And let $\displaystyle R_{n+1}(x) = f(x) - T_{n}(x)$ be the remainder term. Then there exists a number $\displaystyle y$ strictly between $\displaystyle 0$ and $\displaystyle x$ so that,
$\displaystyle R_{n+1}(x) = \frac{f^{(n+1)}(y)}{(n+1)!}\cdot x^{n+1}$.
So given $\displaystyle f(x) = \sin x$ let us work on the interval $\displaystyle (a,b)$ where $\displaystyle a=-\infty$ and $\displaystyle b=+\infty$. This function is infinitely differenciable so the above results apply. You can to approximate $\displaystyle T_n \left( \frac{\pi}{4} \right)$. Now by the theorem we know that,
$\displaystyle R_{n+1}\left( \frac{\pi}{4} \right) = \frac{f^{(n+1)} \left( \frac{\pi}{4} \right)}{(n+1)!} \cdot \left( \frac{\pi}{4} \right)^{n+1}$ for some $\displaystyle 0<y<\frac{\pi}{4}$.
Notice that $\displaystyle f^{n+1}$ is one of these: $\displaystyle \sin x,\cos x,-\sin x,-\cos x$. Thus, $\displaystyle |f^{n+1}|\leq 1$. And also notice that $\displaystyle \frac{\pi}{4} \leq 1$ thus, $\displaystyle \left(\frac{\pi}{4}\right)^{n+1} \leq 1$.
This means,
$\displaystyle \left| R_{n+1}\left( \frac{\pi}{4}\right) \right| \leq \frac{1}{(n+1)!}$.
This approximation might be greatly improved all I did was place an approximation that works and furthermore converges rapidply.
Remark) The theorem applys in more general to $\displaystyle (n+1)$-differentiable functions but there was no need to use it here. | 2018-04-21T06:50:18 | {
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http://mathhelpforum.com/calculus/165427-find-absolute-maximum-minimum.html | # Thread: Find the absolute maximum and minimum of...
1. ## Find the absolute maximum and minimum of...
Find the absolute maximum and minimum of f(x,y)=x^2+xy+y^2-6y on the rectangle {f(x,y) | -3 <= x <=3, 0 <= y <= 5}
i find the critical point by taking the derivative of f(x,y), setting it to 0, then finding x and y.
fx = 2x+y
fy= x+2y-6
y = -2x
x + 2(-2x) - 6 = 0
x - 4x = 6
-3x = 6
x = -2
y = -2(-2)
y = 4
i'm unsure of what to do next, can someone help? thanks!
2. Now evaluate the Hessian matrix at that point to determine the nature of the stationary point.
3. okay so
Hf(-2,4) = [2,1]
[1,2]
d1 = fxx = 2
d1 = 2 > 0
d2 = (2)(2) - (1)(1) = 3
d2 = 3 > 0
local minimum = (-2,4)
did i do this correctly? how do i find the maximum? is local and absolute the same?
4. i think i got it.
since local minimum is (-2,4) then
f(-2,4) = (-2)^2 + (-2)(4) + (4)^2 - (6)(4) = -12
so
Absolute Minimum = -12
and for absolute max you take the highest point within the range of {f(x,y) | -3 <= x <=3, 0 <= y <= 5}
so, (3,5)
and
f(3,5) = (3)^2 + (3)(5) + (5)^2 - (6)(5) = 19
so
Absolute Minimum = 19
is this right? is the way i found the point to solve the absolute maximum the right way to do it?
5. Well, what did you do find the maximum? Just evaluate at the point with largest x and y? There is no reason to think that will give a maximum. Nor is there reason to think that a local minimum has to be an absolute minimum. There is really no need to determine the Hessian since you are looking for absolute max and min and finding out if it is a local max or min is irrelevant.
The theorem says that max and min of a differentiable function occur where the derivative is 0 or on the boundary. Once you have found that (-2, 4) is the only point in the interior where the derivative is 0, turn to the boundary- which, in this case, consists of the four lines, x= -3, x= 3, y= 0, and y= 5.
On y= 0, $f(x,0)= x^2$ which has derivative 2x so its derivative is 0 at (0, 0).
On y= 5 $f(x, 5)= x^2+ 5x+ 25-30= x^2+ 5x- 5$. Its derivative is 2x+ 5 which is 0 when x= -5/2. Another critical point is at (-5/2, 0).
On x= -3, $f(-3, y)= 9- 3y+ y^2- 6y= y^2- 9y+ 9$. Its derivative is 2y- 9 which is 0 when y= 9/2. Another critical point is at (0, 9/2).
On x= 3, $f(3, y)= 9+ 3y+ y^2- 6y= y^2- 3y+ 9$. Its derivative is 2y- 3 which is 0 when y= 3/2. Another critical point is at (0, 3/2).
And, of course, don't forget the endpoints of those intervals, the vertices of the rectangle, (-3,0), (-3, 5), (3, 0), and (3, 5).
Evaluate the function, $f(x, y)= x^2+ xy+ y^2- 6y$ at those points:
(-2, 4), (0, 0), (-5/2, 0), (0, 9/2), (0, 3/2), (-3, 0), (-3, 5), (3, 0), and (3, 5). The largest value you get for those points is the maximum and the smallest is the minimum on the entire rectangle. | 2017-02-27T12:12:50 | {
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http://zgrk.chicweek.it/solution-of-wave-equation-by-separation-of-variables-pdf.html | Another is that for the class of partial differential equation represented by Equation Y(6)−coor, the boundary conditions in the. Lecture 19 Phys 3750 D M Riffe -1- 2/26/2013 Separation of Variables in Cartesian Coordinates Overview and Motivation: Today we begin a more in-depth look at the 3D wave equation. the differential equation and asked to sketch solution curves corresponding to solutions that pass through the points (0, 2) and (1, 0). Let's see some examples of first order, first degree DEs. Get complete concept after watching this video. 4 Even and Odd Functions Section 9. A general solution is also derived for a fixed end stretched string. always 2 linearly independent general solutions for a 2nd order equation. The analytical results for the continuous constant heat flux indicated that the larger deviation amongst predicted temperatures of the DPL, thermal wave and Pennes equations is found for intensive heat fluxes. The first three worksheets practise methods for solving first order differential equations which are taught in MATH108. The simplest example of ariablev separation is a particle in in nitely deep three dimensional quantum it even at the points r where (r) = 0. Its left and right hand ends are held fixed at height zero and we are told its initial configuration and speed. 8 Smce y = f(x) > O on the Interval 1 < x < 1. Many textbooks heavily emphasize this technique to the point of excluding other points of view. 18-009 Learn Differential Equations: Up Close with Gilbert Strang and Cleve Moler, Fall 2015 View the complete course:. 1 First separation: r, θ, φ versus t LHS(r,θ. We write ψ(x,y,z)=X(x)Y(y)Z(z), (4) where X is a function of x only, Y is a function of y only, and Z is a function of z only. First Order Partial Differential Equation A quick look at first order partial. Boundary conditions "are hidden" in this space. 1) It is easy to verify by direct substitution that the most general solution of the one dimensional wave equation (1. An example 35. Outline of Lecture • Examples of Wave Equations in Various Settings • Dirichlet Problem and Separation of variables revisited • Galerkin Method • The plucked string as an example of SOV • Uniqueness of the solution of the. Analytic Solution Techniques for ODEs (6 weeks) : General theory of linear differential equations – Laplace transform – Green’s function solutions of boundary/initial value problems – Series solutions – Sturm-Liouville Systems - Legendre and Bessel functions – Fourier series – Orthogonal series of polynomials. Lecture plan: (should be 8 lectures, but could be 9) 1. Using separation of variables we can get an infinite family of particular solutions of the form. 3 Solution of the One Dimensional Wave Equation: The Method of Separation of Variables 87 3. Maths tutorial - homogeneous functions (ODE's) ODE solution by integrating factor method. mw-parser-output. The simplest instance of the one. Coulomb's Law Equation. As mentioned above, this technique is much more versatile. Nyack, 1D Wave with Partial Fourier Sum Other Equations P. Then we could hold yand z constant and vary x, causing this first term to vary. Then we take a linear combination of such solutions with the coefficients chosen in such a way that at we get the initial profile. Heat Equation MIT RES. pdf), Text File (. 6 Heat Conduction in Bars: Varying the Boundary Conditions 128 3. Differential Equations" L. We give a complete solution of the Eisenhart integrability conditions in three-dimensional Minkowski space obtaining 39 orthogonally separable webs and 58 inequivalent metrics in adapted coordinate systems which permit orthogonal separation of variables for the associated Hamilton-Jacobi and wave equations. Not to be confused with Wave function. 5 The One Dimensional Heat Equation 118 3. We begin with the basic hypothesis that a solution of (5) exists in the separable form and choose the following ansatz: u(h;t) = F(h)G(t) (6) Substituting this ansatz into equation (4) to obtain F dG dt = G d2F dh2 + G h dF dh (7) As G depends only on t and F only on h, by separation of variables the following ordinary. with : and i want to have a 3 d graph for for example for u(x,y,1,1. 18-009 Learn Differential Equations: Up Close with Gilbert Strang and Cleve Moler, Fall 2015 View the complete course:. When we talked about the heat equation, we found the. The analytical results for the continuous constant heat flux indicated that the larger deviation amongst predicted temperatures of the DPL, thermal wave and Pennes equations is found for intensive heat fluxes. Recognize that your equation is an homogeneous equation; that is, you need to check that ft()x,,ty=tn f(xy). (This is aplane wave solution — f (n ·x − ct) remains constant on planes perpendicular to n and traveling with speed c in the direction of n. Time-dependent Schrödinger equation: Separation of variables Since U(x) does not depend on time, solutions can be written in Solving the Schrodinger Equation. As shown in former studies,3-7 the temporal component can be formulated from the linearized momentum equation. mw-parser-output. Cauchy problem for the Schrodinger’s equation. Separation of Variables - Heat Equation Part 1 We use Separation of Variables to find a general solution of the 1-d Heat Equation, including boundary conditions. The Laplace equation is a special case with k2 = 0. The string has length ℓ. Link for the first Part: Derivation of the Wave equation for a. thumbinner{width:100%!important;max-. THE WAVE EQUATION 2. The solution to the wave equation is computed using separation of variables. 11), then uh+upis also a solution to the inhomogeneous equation (1. When using the separation of variable for partial differential equations, we assume the solution takes the form u(x,t) = v(x)*g(t). Exact solutions 2. Get the free "General Differential Equation Solver" widget for your website, blog, Wordpress, Blogger, or iGoogle. Its left and right hand ends are held fixed at height zero and we are told its initial configuration and speed. Answer to c) Obtain solution of one dimensional wave equation a'y ot? by method of separation of variables. Lecture 18, Tue Oct 25 (Di erential equations, separation of variables). di↵erential equations are linear such a linear combination is also a solution to the coupled linear equations. thumbinner{width:100%!important;max-. 1 1D heat and wave equations on a finite interval In this section we consider a general method of separation of variables and its applications to solving heat equation and wave equation on a finite interval (a 1, a2). Solution of the Wave Equation by Separation of Variables The Problem Let u(x,t) denote the vertical displacement of a string from the x axis at position x and time t. 5 The One Dimensional Heat Equation 41 3. 4, Repeated Roots; Reduction of Order 00Q 1). Nyack, 1D Wave with Partial Fourier Sum Other Equations P. Notice that if uh is a solution to the homogeneous equation (1. The separation of variables method means that we first. @media all and (max-width:720px){. Separation of Variables - Heat Equation Part 1 We use Separation of Variables to find a general solution of the 1-d Heat Equation, including boundary conditions. Note, unlike the Cartesian case, the condition that φ+2π describes the same position in the plane as φ forces the separation constant to be an integer, leaving us with the radial equation: s ∂ ∂s s ∂F. and satisfy. Use of Fourier series to solve the wave equation, Laplace's equation and the heat equation (all with two independent variables). 7 The Two Dimensional Wave and Heat Equations 144 3. The quantity u may be, for example, the pressure in a liquid or gas, or the displacement, along some specific direction, of the particles of a vibrating solid away from their resting. The result can then be also used to obtain the same solution in two space dimensions. or, for brevity,. The goal is to rewrite the differential equation so that all terms containing one variable (e. The study gives a brief overview of existing modifications of the method of functional separation of variables for nonlinear PDEs. After this introduction is given, there will be a brief segue into Fourier series with examples. Be able to model the temperature of a heated bar using the heat equation plus bound-. About half the book is devoted to the solution of the first order differential equations, with dazzling pyrotechnics that no one has matched since. Separation of Variables - Heat Equation Part 1 We use Separation of Variables to find a general solution of the 1-d Heat Equation, including boundary conditions. But it is often more convenient to use the so-called d'Alembert solution to the wave equation 1. 3 Solution to Problem “A” by Separation of Variables 5 4 Solving Problem “B” by Separation of Variables 7 5 Euler’s Differential Equation 8 6 Power Series Solutions 9 7 The Method of Frobenius 11 8 Ordinary Points and Singular Points 13 9 Solving Problem “B” by Separation of Variables, continued 17 10 Orthogonality 21. 11), then uh+upis also a solution to the inhomogeneous equation (1. You will have to become an expert in this method, and so we will discuss quite a fev. For example, much can be said about equations of the form ˙y = φ(t,y) where φ is a function of the two variables t and y. In this study, we find the exact solution of certain partial differential equations (PDE) by proposing and using the Homo-Separation of Variables method. Many textbooks heavily emphasize this technique to the point of excluding other points of view. Separation of Variables. Differential Equations" L. We have already done this by just guessing in some cases. 18-009 Learn Differential Equations: Up Close with Gilbert Strang and Cleve Moler, Fall 2015 View the complete course:. A solution of the initial-value problem for the wave equation in three space dimensions can be obtained from the corresponding solution for a spherical wave. Daileda Trinity University we will use separation of variables to find a family of simple solutions to (1) and (2), and then the principle of is a solution of the heat equation (1) with the Neumann boundary conditions (2). We start with a particular example, the one-dimensional (1D) heat equation @u @t = • @2u @x2 + f ; (1) where u · u(x;t) is the temperature as a function of coordinate x. thumbinner{width:100%!important;max-. Solutions to Homework 3 Section 3. equations applied to a porous channel, and subject to a similarity transformation. Create an animation to visualize the solution for all time steps. 1D Wave Equation - the vibrating string The Vibrating String II. Answered: darova on 4 Jul 2019 Accepted Answer: darova. Once we derive Laplace’s equation in the polar coordinate system, it is easy to represent the heat and wave equations in the polar coordinate system. 3: Solution Using Separation of Variables 19. A method for the solution of a certain class of nonlinear partial differential equations by the method of separation of variables is presented. Follow 45 views (last 30 days) Youssef FAKHREDDINE on 4 Jul 2019. 1 "Blow-up, compactness and (partial) regularity in Partial Differential Equations" Lecture 1 Christophe Prange (CNRS Researcher) Method of Separation of Variables: Analytical Solutions of Partial Differential Equations Using the Method of Separation of Variables to solve 1st order PDEs and. This is mostly suitable for B. Heat equation solver. The solution is managed by separating the variables so that the wavefunction is represented by the product:. Boundary Value Problems (using separation of variables). Force is a vector – it has a magnitude (specified in Newtons, or lbf, or whatever), and a direction. 2) together with the initial conditions (1. 2 Separation of variables in the acoustic equation for homogeneous media 2. Remember, that Schrödinger’s equation is in quantum mechanics what F = ma is in classical mechanics. It is solved by separation of variables into a spatial and a temporal part, and the symmetry between space and time can be exploited. A solution of a partial differential equation in some region R of the space of the independent variables is a function that possesses all of the partial derivatives that By separation of variables, we assume a solution in the form of a product and Euler derived and solved a linear wave equation for the motion of vibrating strings in the. When using the separation of variable for partial differential equations, we assume the solution takes the form u(x,t) = v(x)*g(t). EE 439 time-independent Schroedinger equation - 2 With U independent of time, it becomes possible to use the technique of "separation of variables", in which the wave function is written as the product of two functions, each of which is a function of only one variable. In the following, the radius r;the mass mand the velocity vare func-tions of the time t:By de nition of the density ;we have. The wave equation written can be written with the aid of a wave operator. These two links review how to determine the Fourier coefficients using the so-called "orthogonality. 3, 2006] We consider a string of length l with ends fixed, and rest state coinciding with x-axis. Check for extra solutions coming from the warning in Step 4. A method for the solution of a certain class of nonlinear partial differential equations by the method of separation of variables is presented. We show that the curved Dirac equation in polar coordinates can be transformed into Schrodinger-like differential equation for upper spinor component. Answer to Use separation of variables to obtain a series solution of the wave equation au 1 22u дх2 c2 Ət2 subject to the bound. tissue as a finite domain was analytically solved by employing the separation of variables and Duhamel’s superposition integral. Solve this equation using separation of variables to find the steady-state temperature of the block with boundary conditions T(0,y) = T(L,y) = T(x,W) = 0 and T(x,0) = T0. Separation of Variables in Linear PDE: One-Dimensional Problems Now we apply the theory of Hilbert spaces to linear difierential equations with partial derivatives (PDE). 4 D’Alembert’s Method 104 3. The generalization to systems of partial differential equations, invariant under multi-parameter groups, is stated and proved. You can also do this slightly more rigourously by writing the di erential equation as kx= m: dv dt (1. We will follow the (hopefully!) familiar process of using separation of variables to produce simple solutions to (1) and (2),. It is the same concept when solving differential equations - find general solution first, then substitute given numbers to find particular solutions. 1 "Blow-up, compactness and (partial) regularity in Partial Differential Equations" Lecture 1 Christophe Prange (CNRS Researcher) Method of Separation of Variables: Analytical Solutions of Partial Differential Equations Using the Method of Separation of Variables to solve 1st order PDEs and. [8 sharks). Then, there will be a more advanced example, incorporating the process of separation of variables and the process of finding a Fourier series solution. of separation of variables. • Deriving the 1D wave equation • One way wave equations • Solution via characteristic curves • Solution via separation of variables • Helmholtz’ equation • Classification of second order, linear PDEs • Hyperbolic equations and the wave equation 2. Laplace’s Equation • Separation of variables – two examples • Laplace’s Equation in Polar Coordinates – Derivation of the explicit form – An example from electrostatics • A surprising application of Laplace’s eqn – Image analysis – This bit is NOT examined. Refer to pp. 6 PDEs, separation of variables, and the heat equation. He re, w e wil l o!e r a simple d erivation base d on what w e ha ve learned so far ab out th e w ave fun ction. Sarra, Weak Solutions and Shocks IsoSpectral Domains C. Each system is associated with a pair of commuting operators in the symmetry algebra so(3,2) of this equation, one operator first order and the other second order. Let u= sin(x) and dv= ex dx, so du= cos(x) dxand v= ex. The exact solutions are con- structed by choosing an appropriate initial approximation in addition to only one. Wave equation in 1D (part 1)* • Derivation of the 1D Wave equation - Vibrations of an elastic string • Solution by separation of variables - Three steps to a solution • Several worked examples • Travelling waves - more on this in a later lecture • d'Alembert's insightful solution to the 1D Wave Equation. A general solution is also derived for a fixed end stretched string. Tech, and (10+2) students. If = 0, one can solve for R0first (using separation of variables for ODEs) and then integrating again. 1 Homogeneous Solution in Free Space We first consider the solution of the wave equations in free space, in absence of matter and sources. Separation of Variables - Heat Equation Part 1 We use Separation of Variables to find a general solution of the 1-d Heat Equation, including boundary conditions. But it is often more convenient to use the so-called d'Alembert solution to the wave equation 1. LAPLACE'S EQUATION IN SPHERICAL COORDINATES. Check also the other online solvers. The study gives a brief overview of existing modifications of the method of functional separation of variables for nonlinear PDEs. It is the solution to problems in a wide variety of fields including thermodynamics and electrodynamics. We show that the curved Dirac equation in polar coordinates can be transformed into Schrodinger-like differential equation for upper spinor component. 4, Repeated Roots; Reduction of Order 00Q 1). Hancock Fall 2006 1 2D and 3D Heat Equation Ref: Myint-U & Debnath §2. For the equation to be of second order, a, b, and c cannot all be zero. Z ex sin(x) dx | {z } our goal; I. Weisner's Method for the Complex Helmholtz Equation. This novel analytical method is a combination of the homotopy perturbation method (HPM) with the separation of variables method. Separation of Variables - Heat Equation Part 1 We use Separation of Variables to find a general solution of the 1-d Heat Equation, including boundary conditions. 2 Separation of variables in the acoustic equation for homogeneous media 2. d'Alembert devised his solution in 1746, and Euler subsequently expanded the method in 1748. 11), it is enough to nd. 8 Exact solutions for differential equations: Separation of variables Sometimes it is possible to find exact formulas for y giventheformulafory. In the literature we have at our disposal di erent methods forsolving relativistic wave equations in curved spaces and in curvilinear coordinates; among them the method of separation of variables is one of the most widely used. *find a way to rewrite your equation as one of the well-known solved equations *separation of variables. What are we looking for? *general solutions. The operation ∇ × ∇× can be replaced by the identity (1. Donate or volunteer today! Site Navigation. Solving the heat equation, wave equation, Poisson equation using separation of variables and eigenfunctions 1 Review: Interval in one space dimension Our domain G = (0;L) is an interval of length L. General Solution By taking the original differential equation P(y) dy dx = Q(x) we can solve this by separating the equation into two parts. Let's see some examples of first order, first degree DEs. 2 Fourier Series & Section 9. Consider the following equation found in solving solutions to one-dimensional heat equations using the method of separation of variables. Solution for a non-homogeneous Klein-Gordon equation with 5th degree polynomial forcing function @article{Garzon2017SolutionFA, title={Solution for a non-homogeneous Klein-Gordon equation with 5th degree polynomial forcing function}, author={G Hernan Garzon and Cesar A. Its left and right hand ends are held fixed at height zero and we are told its initial configuration and speed. and 3 each for both constitutive relations (difficult task). *find a way to rewrite your equation as one of the well-known solved equations *separation of variables. 1 "Blow-up, compactness and (partial) regularity in Partial Differential Equations" Lecture 1 Christophe Prange (CNRS Researcher) Method of Separation of Variables: Analytical Solutions of Partial Differential Equations Using the Method of Separation of Variables to solve 1st order PDEs and. Seven steps of the approach of separation of Variables: 1) Separate the variables: Initial boundary value problem for the wave equation with 2t and is a solution of the homogeneous equation for (*). 5 in APDE covers the separation of variables for the wave equation, which you should go over (will also be covered in recitation). Cain and Angela M. In mathematics, separation of variables (also known as the Fourier method) is any of several methods for solving ordinary and partial differential equations, in which algebra allows one to rewrite. @media all and (max-width:720px){. 1, d y y3 (1 + > O on this Interval 1). Differential Equations" L. The 1-D Wave Equation 18. Khan Academy is a 501(c)(3) nonprofit organization. For the heat equation, the solution u(x,y t)˘ r µ satisfies ut ˘k(uxx ¯uyy)˘k µ urr ¯ 1 r ur ¯ 1 r2 uµµ ¶, k ¨0: diffusivity, whereas for the wave equation, we have utt ˘c 2(u xx. Feldman, Telegraph Equation S. An introduction to the Fourier transform 33 10. "x") appear on one side of the equation, while all terms containing the other variable (e. Separation of Variables in Laplace's Equation in Cylindrical Coordinates Your text’s discussions of solving Laplace’s Equation by separation of variables in cylindrical and spherical polar coordinates are confined to just two dimensions ( cf §3. Solution of the Wave Equation by Separation of Variables The Problem Let u(x,t) denote the vertical displacement of a string from the x axis at position x and time t. Get complete concept after watching this video. Accounting for separation of variables and the angular momentum resuls, the Schrodinger equation is transformed into the Radial equation for the Hydrogen atom: h2 2 r2 d dr r2 dR(r) dr + " h2l(l+1) 2 r2 V(r) E # R(r) = 0 The solutions of the radial equation are the Hydrogen atom radial wave-functions, R(r). Separation of Variables - Heat Equation Part 1 We use Separation of Variables to find a general solution of the 1-d Heat Equation, including boundary conditions. Keep a fixed vertical scale by first calculating the maximum and minimum values of u over all times, and scale all plots to use those z-axis limits. Make the DE look like dy dx = g(x)f(y). Find more Mathematics widgets in Wolfram|Alpha. 5 The One Dimensional Heat Equation 118 3. For example, much can be said about equations of the form ˙y = φ(t,y) where φ is a function of the two variables t and y. coordinates, interior and exterior Dirichlet problems in polar coordinates; Separation of variables method for solving wave and diffusion equations in one space variable; Fourier series and Fourier transform and Laplace transform methods of solutions for the equations mentioned above. Substitute this into the wave equation and divide across by u = RΘΦT: 1 R d2R dr 2 + 2 rR dR dr + 1 r 2 1 Θsinθ d dθ % sinθ dΘ dθ & + 1 r2 sin2 θ 1 Φ d2Φ dφ = 1 c 1 T d2T dt2. For these reasons, wave functions of the form are called stationary states. Title: Lesson 1 - Intro to Differential Equations and Separation of Variables M253 ND PSU Solutions. 2, Myint-U & Debnath §2. Then, there will be a more advanced example, incorporating the process of separation of variables and the process of finding a Fourier series solution. Differential Equations" L. The properties and behavior of its solution are largely dependent of its type, as classified below. We develop a technique making it possible to handle the problem of separation of variables in nonlinear differential equations. By separation of variables, the radial term and the angular term can be divorced. separation of variables. Its left and right hand ends are held fixed at height zero and we are told its initial configuration and speed. None of Boole’s beautiful techniques is of any conceivable use to anyone who deals with differ-ential equations today. 1 Dirichlet Boundary Conditions Ref: Strauss, Chapter 4 We now use the separation of variables technique to study the wave equation on a finite interval. There are several ways to evaluate this integral; we’ll show just one here. Create an animation to visualize the solution for all time steps. 1) It is easy to verify by direct substitution that the most general solution of the one dimensional wave equation (1. Introduction. Thus the wave equation does not have the smoothing e ect like the heat equation has. , Liu, Fawang , Anh, Vo , Shen, S. We will now find the “general solution” to the one-dimensional wave equation (5. Now we’ll consider it on a circular disk x 2+ y2. This method consists of substituting the trial solution into equation (1) to obtain where the primes denote differentiation of the functions with respect to their arguments. * We can find. Heat Equation MIT RES. We use the separation of variables method to solve the above equation. Case 1: K = 0 b. First Order Partial Differential Equation A quick look at first order partial. Answer to c) Obtain solution of one dimensional wave equation a'y ot? by method of separation of variables. 20) we obtain the general solution. Expansion Formulas for Solutions of the Klein-Gordon Equation. With n representing the nth positive solution of tan p = p , The corresponding eigenfunctions X n are (up to a constant multiple) X n (x) = sin p nx + p n cos p nx The equation tanx = x has no closed-form for its solution, we will have to use numerical. 6 Wave Equation on an Interval: Separation of Vari-ables 6. Lecture Notes in Mathematics 17. Make the DE look like dy dx = g(x)f(y). The beginning of section 4. We have solved the wave equation by using Fourier series. A general solution is also derived for a fixed end stretched string. Solution of the Wave Equation by Separation of Variables The Problem Let u(x,t) denote the vertical displacement of a string from the x axis at position x and time t. The wave equation - solution by separation of variables solution by separation of variables. separation of variables. 4 D'Alembert's Method 104 3. Plugging in one gets [ ( 1) + ]r = 0; so that = p. 100-level Mathematics Revision Exercises Differential Equations. 1 The heat equation Consider, for example, the heat equation ut = uxx, 0 < x < 1, t > 0 (4. Differential Equations" L. Elliptic equations: weak and strong minimum and maximum principles; Green's functions. As in the one dimensional situation, the constant c has the units of velocity. 2 Method of Separation of Variables - Stationary Boundary Value Problems. You can also do this slightly more rigourously by writing the di erential equation as kx= m: dv dt (1. Then we take a linear combination of such solutions with the coefficients chosen in such a way that at we get the initial profile. The first three worksheets practise methods for solving first order differential equations which are taught in MATH108. This may be already done for you (in which case you can just identify. -Recent citations. Putting it all together - Finally we get a solution for 1D Wave Equation a. Answer to c) Obtain solution of one dimensional wave equation a'y ot? by method of separation of variables. A correct response should be two sketched curves that pass through the indicated points, follow the given slope lines, and extend to the boundaries of the provided slope field. be solved by the method of separation of variables. The analytical results for the continuous constant heat flux indicated that the larger deviation amongst predicted temperatures of the DPL, thermal wave and Pennes equations is found for intensive heat fluxes. If the wave speed is constant across different wave numbers, then no dispersion would occur. The exact solutions are con- structed by choosing an appropriate initial approximation in addition to only one. Answer to Use separation of variables to obtain a series solution of the wave equation au 1 22u дх2 c2 Ət2 subject to the bound. "x") appear on one side of the equation, while all terms containing the other variable (e. txt) or read online for free. For these reasons, wave functions of the form are called stationary states. THE WAVE EQUATION 2. Differential Equations" L. Later, the Laplace equation was solved with the separation of variables method for spheroidal and ellipsoidal shapes. In the first separation we set Substitution into (1) gives where subscripts denote partial derivatives and dots denote derivatives with respect to t. equation for the solution curve. The study of linear hyperbolic equations in a black hole geometry has a long history. 3 Solution of the One Dimensional Wave Equation: The Method of Separation of Variables 87 3. Hint: Separation of variables in this equation will require your x and y equations to equal constants that have opposite sign - one must be positive and the other negative. Feldman, An Example of Wave Equation on a String J. AMS 502, Differential Equations and Boundary Value Problems II Analytic solution techniques for, and properties of solutions of, partial differential equations, with concentration on second order PDEs. Toc JJ II J I Back. DeTurck Math 241 002 2012C: Solving the heat. Solve the following 1D heat/diffusion equation (13. (ii) Any solution to the wave equation u tt= u xxhas the form u(x;t) = F(x+ t) + G(x t) for appropriate functions F and G. Lecture Two: Solutions to PDEs with boundary conditions and initial conditions. solution of Wave equation (one dimensional) Discussed various possible solutions of one dimensional wave equation using Method of separation of variables and discussed 3. 71052 Corpus ID: 126381645. We solve this using the technique of separation of variables. the strategy of separation of variables, developed for the case of the heat equation in bounded domains, to solve the above problem. v~,fe will emphasize problem solving techniques, but \ve must. tissue as a finite domain was analytically solved by employing the separation of variables and Duhamel’s superposition integral. 0,viaWikimediaCommons APPLICATIONS OF SEPARATION OF VARIABLES 4. Substituting for ψin Eq. (b)Find the general solution of the spatial ordinary di erential equation. Maths tutorial - separation of variables (ODE's) Solution of ODE's involving homogeneous functions. [8 sharks). This naturally • 1D Wave Equation - d'Alembert Solution (2) •Separation of Variables (1) •Fourier Series (4). mw-parser-output. The wave equation written can be written with the aid of a wave operator. 2 DIFFERENTIAL EQUATIONS: THE BASICS AND SEPARATION OF VARIABLES Applications include Newton's second Law, force = mass acceleration, which is often a 2nd-order di erential equation, depending on nature of the force. Cylindrical Waves Guided Waves Separation of Variables Bessel Functions TEz and TMz Modes Bessel Functions We now have X1 m=0 h ( + m)2 n2 i cm˘ +m + X1 m=0 cm˘ +m+2 = 0 or X1 m=0 h ( + m)2 n2 i cm˘ +m + X1 m=2 cm 2˘ +m = 0 We can proceed by forcing the coefficients of each term to vanish. 2) The one-dimensional wave equation (4. (2006) Variable Separation Solutions for the (3 + 1)-Dimensional Jimbo-Miwa Equation. The 2D wave equation Separation of variables Superposition Examples Solving the 2D wave equation Goal: Write down a solution to the wave equation (1) subject to the boundary conditions (2) and initial conditions (3). If b2 – 4ac > 0, then the equation is called hyperbolic. Boundary conditions "are hidden" in this space. Separability conditions are obtained for the partial differential equations of electromagnetic theory. We will solve this equation subject to the boundary conditions (1. Solve the following 1D heat/diffusion equation (13. 5 [Nov 2, 2006] Consider an arbitrary 3D subregion V of R3 (V ⊆ R3), with temperature u(x,t) defined at all points x = (x,y,z) ∈ V. About half the book is devoted to the solution of the first order differential equations, with dazzling pyrotechnics that no one has matched since. 9) and the initial condition (13. The operation ∇ × ∇× can be replaced by the identity (1. 3 Review of different methods of separation of variables for non-homogeneous media. Solution by Substitution Homogeneous Differential Equations Bernoulli's Equation Reduction to Separation of Variables Conclusion Bernoulli's Equation and Linear DEs Another substitution leads to the solution of what is called Bernoulli's Equation (actually a family of equations) by linearity. For this case the right hand sides of the wave equations are zero. He re, w e wil l o!e r a simple d erivation base d on what w e ha ve learned so far ab out th e w ave fun ction. specific kinds of first order differential equations. equation represents the solution of the boundary value problem. *find a way to rewrite your equation as one of the well-known solved equations *separation of variables. For these reasons, wave functions of the form are called stationary states. We look for a separated solution u= h(t)˚(x): Substitute into the PDE and rearrange terms to get 1 c2. 1) It is easy to verify by direct substitution that the most general solution of the one dimensional wave equation (1. solution of Wave equation (one dimensional) Discussed various possible solutions of one dimensional wave equation using Method of separation of variables and discussed 3. Force is a vector – it has a magnitude (specified in Newtons, or lbf, or whatever), and a direction. 1) on an infinite domain, then any combination of c 1 a(x,t)+c 2 b(x,t)isalsoasolution. For the heat equation, the solution u(x,y t)˘ r µ satisfies ut ˘k(uxx ¯uyy)˘k µ urr ¯ 1 r ur ¯ 1 r2 uµµ ¶, k ¨0: diffusivity, whereas for the wave equation, we have utt ˘c 2(u xx. mw-parser-output. 2 Extension to finite regions. In your careers as physics students and scientists, you will. One better method to do this is called separation of variables. Hint: Separation of variables in this equation will require your x and y equations to equal constants that have opposite sign - one must be positive and the other negative. More precisely, the eigenfunctions must have homogeneous boundary conditions. Separation of Variables in Laplace's Equation in Cylindrical Coordinates Your text’s discussions of solving Laplace’s Equation by separation of variables in cylindrical and spherical polar coordinates are confined to just two dimensions ( cf §3. However, the one thing that we've not really done is completely work an example from start to finish showing each and every step. For the moment, we'll say the constant must be negative, and we'll call it -k2 and this is the same k as in the wavevector. @media all and (max-width:720px){. equations a valuable introduction to the process of separation of variables with an example. I have used separation of variables to get the general solution, but I need help applying it. to pursue the mathematical solution of some typical problems involving partial differential equations. One better method to do this is called separation of variables. Find the general solution for the differential equation dy + 7x dx = 0 b. We introduce a technique for finding solutions to partial differential equations that is known as separation of variables. Part (c) asked for the particular solution to the differential equation satisfying the given initial condition. While this solution can be derived using Fourier series as well, it is really an awkward use of those concepts. Seven steps of the approach of separation of Variables: 1) Separate the variables: (by writing e. Classification of second order linear partial differential equations; Method of separation of variables; Laplace equation; Solutions of one dimensional heat and wave equations. Differential Equations > Separation of Variables. Integration, Separation of Variables Solutions 1. Tech, and (10+2) students. Separation of variables in the wave equation • For the ansatz to work we must have (lets These are called These are called separation constantsseparation constants. But it is often more convenient to use the so-called d'Alembert solution to the wave equation 1. pdf Traveling Waves, standing waves and the dispersoin relation Lecture 17. Under reasonable conditions on φ, such an equation has a solution and the corresponding initial value problem has a unique solution. 2 and problem 3. Some important features of these solutions are indicated and their stabilities are examined. To illus-trate the idea of the d'Alembert method, let us. This is a traveling wave, with wave vector {z, , }. 3 Solution of the One Dimensional Wave Equation: The Method of Separation of Variables 31 3. [8 sharks). 2 Separation of Variables for Partial Differential Equations (Part I) Separable Functions A function of N. the strategy of separation of variables, developed for the case of the heat equation in bounded domains, to solve the above problem. This may be already done for you (in which case you can just identify. Define its discriminant to be b2 – 4ac. 3 The Fourier Convergence Theorem Section 9. partial-differential-equations-solution 1/5 PDF Drive - Search and download PDF files for free. and Liang, Z. 1 The Concept of Separation of Variables. If = 0, one can solve for R0first (using separation of variables for ODEs) and then integrating again. In the present section, separable differential equations and their solutions are discussed in greater detail. I have noticed that $\Theta(\theta) = \cos(\theta)$ would be a solution for the angular part, but this may not be the general solution. The method can solve the exact traveling wave solutions of other nonlinear evolution equations. Separation of Variables for Higher Dimensional Wave Equation 1. thumbinner{width:100%!important;max-. Exact Solution of Partial Differential Equation Using Homo-Separation of Variables. 1 General principles of the separation of variables in linear differential equations 2. equation) in various 3D curvilinear coordinate systems whose coordinates we shall call ξ 1,ξ 2,ξ 3. Solve the following 1D heat/diffusion equation (13. so that (WE) is the equation for the kernel of this operator. wave propagation problems, the wave number and the wave speed are related in some fashion. A solution of the initial-value problem for the wave equation in three space dimensions can be obtained from the corresponding solution for a spherical wave. Maths tutorial - separation of variables (ODE's) Solution of ODE's involving homogeneous functions. Laplace's equation ∇2F = 0. The 2D wave equation Separation of variables Superposition Examples Remarks: For the derivation of the wave equation from Newton’s second law, see exercise 3. Nyack, 1D Wave with Partial Fourier Sum Other Equations P. You will have to become an expert in this method, and so we will discuss quite a fev. The method of Separation of Variables cannot always be used and even when it can be used it will not always be possible to get much past the first step in the method. The spatial part R()r is obtained as the solution of the Helmholtz equation (2. This result is obtained by dividing the standard form by g(y), and then integrating both sides with respect to x. Separation of Variables - Heat Equation Part 1 We use Separation of Variables to find a general solution of the 1-d Heat Equation, including boundary conditions. with : and i want to have a 3 d graph for for example for u(x,y,1,1. Differential Equations" L. tissue as a finite domain was analytically solved by employing the separation of variables and Duhamel's superposition integral. 1 Conservation Laws and Jump Conditions Consider shocks for an equation u t +f(u). Consider the wave equation in Ω with zero displacement on Γ: (PDE) utt − c2(uxx +uyy) = 0 (x,y) in Ω,t > 0, (BC) u(x,y,t) = 0 (x,y) on Γ,t > 0,. Chapter 2 - The Classical Wave Equation. di↵erential equations are linear such a linear combination is also a solution to the coupled linear equations. 3: Hyperbolic Equation Many of the equations of mechanics are hyperbolic and the model hyperbolic equation is the wave equation. This may be already done for you (in which case you can just identify. pdf Traveling Waves, standing waves and the dispersoin relation Lecture 17. 1) on an infinite domain, then any combination of c 1 a(x,t)+c 2 b(x,t)isalsoasolution. 5 The One Dimensional Heat Equation 118 3. Separation of variables refers to a class of techniques for probing solutions to partial di erential equations (PDEs) by turning them into ordinary di eren-tial equations (ODEs). 100 Questions and Answers on 2nd year A-Level Maths Differential Equations, focusing on the method of Separation of Variables. As in the one dimensional situation, the constant c has the units of velocity. Let Kbe a positive. @media all and (max-width:720px){. Separation of variables in the nonlinear wave equation R Z Zhdanov Institute of Mathematics, Ukrainian Academy of Science. The Cauchy Problem and Wave Equations: Mathematical modeling of vibrating string and vibrating membrane, Cauchy problem for second order PDE, Homogeneous wave equation, Initial boundary value problems, Non-homogeneous boundary conditions, Finite strings with fixed ends, Non-homogeneous wave equation, Goursat problem. In mathematics, separation of variables (also known as the Fourier method) is any of several methods for solving ordinary and partial differential equations, in which algebra allows one to rewrite. (to appear). Two-Dimensional Laplace and Poisson Equations In the previous chapter we saw that when solving a wave or heat equation it may be necessary to first compute the solution to the steady state equation. Wave equation in 1D (part 1)* • Derivation of the 1D Wave equation – Vibrations of an elastic string • Solution by separation of variables – Three steps to a solution • Several worked examples • Travelling waves – more on this in a later lecture • d’Alembert’s insightful solution to the 1D Wave Equation. Wave guide and antenna problems are expressed in terms of the vector Helmholtz equation, and solutions are indicated by use of the simple method of separation of variables without recourse to Green's functions. 7) the general solutions of the equations for T and X are (13. Differential Equations in the Undergraduate Curriculum M. First Order Partial Differential Equation A quick look at first order partial. Not to be confused with Wave function. Get complete concept after watching this video. Feldman, An Example of Wave Equation on a String J. pdf A propagating wave packet- group velocity dispersion. The second type of second order linear partial differential equations in 2 independent variables is the one-dimensional wave equation. Once we derive Laplace’s equation in the polar coordinate system, it is easy to represent the heat and wave equations in the polar coordinate system. Make the DE look like dy dx = g(x)f(y). Assume that the wave function is separable into two functions and , i. Birkhauser. That is the case if nis even. 18-009 Learn Differential Equations: Up Close with Gilbert Strang and Cleve Moler, Fall 2015 View the complete course:. Chapter 12: Partial Differential Equations Definitions and examples The wave equation The heat equation The one-dimensional wave equation Separation of variables The two-dimensional wave equation Solution by separation of variables (continued) The functions un(x,t) are called the normal modes of the vibrating string. PDEs with Boundary conditions. Seven steps of the approach of separation of Variables: 1) Separate the variables: (by writing e. Decay of Solutions of the Wave Equation in the Kerr Geometry 467. Separation-of-Variables Solution to the Finite Vibrating String We solve problem 14-1 by breaking it into several steps: Step 1. 1) It is easy to verify by direct substitution that the most general solution of the one dimensional wave equation (1. 2 Laplace's Equation. \Ve \-vilt use a technique called the method of separation of variables. requires understanding of partial differential equations, as well as vector and tensor calculus. The angular functions are spheroidal harmonics, and the radial equation is reduced to a one-dimensional Schrödinger equation with an effective potential. Toc JJ II J I Back. 25 PDEs separation of variables 25. Let us recall that a partial differential equation or PDE is an equation containing the partial derivatives with respect to several independent variables. So yeah like the title says, I need to learn about separation of variables in cylindrical coordinates. Solution by separation of variables. Substitute this into the wave equation and divide across by u = RΘΦT: 1 R d2R dr 2 + 2 rR dR dr + 1 r 2 1 Θsinθ d dθ % sinθ dΘ dθ & + 1 r2 sin2 θ 1 Φ d2Φ dφ = 1 c 1 T d2T dt2. Do this for the case when a = 20 m/s, and the initial velocity is 200 sin3tx. b) Find the displacement of the string analogous to the result presented in Example 6. The second derivative of u with respect to x. 6 PDEs, separation of variables, and the heat equation. Link for the first Part: Derivation of the Wave equation for a. Cylindrical Waves Guided Waves Separation of Variables Bessel Functions TEz and TMz Modes Bessel Functions We now have X1 m=0 h ( + m)2 n2 i cm˘ +m + X1 m=0 cm˘ +m+2 = 0 or X1 m=0 h ( + m)2 n2 i cm˘ +m + X1 m=2 cm 2˘ +m = 0 We can proceed by forcing the coefficients of each term to vanish. 1, d y y3 (1 + > O on this Interval 1). Home Assignment 8, PDF file, due Wed November 28 Wave equation--solution; Wave equation: energy method; Separation of variables in spherical coordinates Home Assignment 9, PDF file, due Wed December 5 Appendices. @media all and (max-width:720px){. Vibrating Membrane: 2-D Wave Equation and Eigenfunctions of the Laplacian Objective: Let Ω be a planar region with boundary curve Γ. 8 Smce y = f(x) > O on the Interval 1 < x < 1. mw-parser-output. Wave Equation and Separation of Variables (SOV): a) Reproduce the wave equation solution of Example 6. 7 The Two Dimensional Wave and Heat Equations 144 3. Solution (69) of Equation (68) and solution (173) of Equation (172) are special cases of solutions (6) with z = f/u. 18-009 Learn Differential Equations: Up Close with Gilbert Strang and Cleve Moler, Fall 2015 View the complete course:. Separation of Variables A typical starting point to study differential equations is to guess solutions of a certain form. Sarra, Weak Solutions and Shocks IsoSpectral Domains C. He re, w e wil l o!e r a simple d erivation base d on what w e ha ve learned so far ab out th e w ave fun ction. The properties and behavior of its solution are largely dependent of its type, as classified below. The operation ∇ × ∇× can be replaced by the identity (1. The string is plucked into oscillation. Solution technique for partial differential equations. The Laplace equation is a special case with k2 = 0. Once we derive Laplace’s equation in the polar coordinate system, it is easy to represent the heat and wave equations in the polar coordinate system. 6 PDEs, separation of variables, and the heat equation. The effect of the Kerr gravitational field on wave phenomena is explored by examining the inhomogeneous wave equation for a scalar massive field in a Kerr background geometry. Separation of Variables A typical starting point to study differential equations is to guess solutions of a certain form. [8 sharks). More examples of separation of variables 28 9. A general solution is also derived for a fixed end stretched string. Certainproblems are followed by discussions that aim to generalize the problem under consideration. PDE: Heat Equation - Separation of Variables Solving the one dimensional Wave equation: intuition An introduction to partial differential equations. Journal of Mathematical Physics 49 :2, 023501. Solution of the Wave Equation by Separation of Variables The Problem Let u(x,t) denote the vertical displacement of a string from the x axis at position x and time t. We use the separation of variables method to solve the above equation. 8 Laplace's Equation in Rectangular Coordinates 146. Solution to Wave Equation by Traveling Waves 4 6. Separation of Variables - Heat Equation Part 1 We use Separation of Variables to find a general solution of the 1-d Heat Equation, including boundary conditions. 1 The physical problem; 4. Topics include the qualitative analysis of ordinary differential equations, solutions of second order linear ordinary differential equations with variable coefficients, first order and second order partial differential equations, the method of characteristics and the. To solve for these we need 12 scalar equations. Differential Equations" L. T: More About theWave Equation and other Fun Topics. wave propagation problems, the wave number and the wave speed are related in some fashion. Answer to Use separation of variables to obtain a series solution of the wave equation au 1 22u дх2 c2 Ət2 subject to the bound. separation of variables. The Cauchy Problem and Wave Equations: Mathematical modeling of vibrating string and vibrating membrane, Cauchy problem for second order PDE, Homogeneous wave equation, Initial boundary value problems, Non-homogeneous boundary conditions, Finite strings with fixed ends, Non-homogeneous wave equation, Goursat problem. It proposes a more general approach to the construction of exact solutions to nonlinear equations of applied mathematics and mathematical physics, based on a special transformation with an integral term and the generalized splitting principle. 4 D’Alembert’s Method 104 3. The solution is managed by separating the variables so that the wavefunction is represented by the product:. The boundary ¶G = f0;Lgare the two endpoints. 3, 2006] We consider a string of length l with ends fixed, and rest state coinciding with x-axis. Bardina,* R G. These separated solutions can then be used to solve the problem in general. 11), then uh+upis also a solution to the inhomogeneous equation (1. Laplace's equation ∇2F = 0. Note: 2 lectures, §9. But it is often more convenient to use the so-called d'Alembert solution to the wave equation 1. For the heat equation, the solution u(x,y t)˘ r µ satisfies ut ˘k(uxx ¯uyy)˘k µ urr ¯ 1 r ur ¯ 1 r2 uµµ ¶, k ¨0: diffusivity, whereas for the wave equation, we have utt ˘c 2(u xx. As mentioned above, this technique is much more versatile. Solution to the Heat Equation on the. Let's rewrite the wave equation here as a reminder, r2 2+ k = 0: (1) For the time being, we consider the wave equation in terms of a scalar quantity , rather than a vector eld E or H as we did before. Thus there exist the nine sets of the wave f ( )gfunctions such that. If the wave speed is constant across different wave numbers, then no dispersion would occur. Now we’ll consider it on a circular disk x 2+ y2. 1 in PDE and Example 4. 8 Exact solutions for differential equations: Separation of variables Sometimes it is possible to find exact formulas for y giventheformulafory. Check for extra solutions coming from the warning in Step 4. Form of teaching Lectures: 26 hours. Under reasonable conditions on φ, such an equation has a solution and the corresponding initial value problem has a unique solution. Link for the first Part: Derivation of the Wave equation for a. Its left and right hand ends are held fixed at height zero and we are told its initial configuration and speed. The 1-D Wave Equation 18. Separation of Variables - Heat Equation Part 1 We use Separation of Variables to find a general solution of the 1-d Heat Equation, including boundary conditions. Separation of Variables We now have an equation that provides us with a means to get the wave functions, which, in turn, provide us with the means to extract the dynamic quantities of interest. 01 Problem Set # 7 Solution. These solutions are different from (204); consequently, they cannot be obtained by the nonclassical method of symmetry reductions with the invariant surface condition (195). We classify and discuss the possible nonorthogonal coordinate systems which lead to R-separable solutions of the wave equation. Unformatted text preview: 1D wave equation 1D Wave Equation 2 u x2 2 1 u c2 t 2 u(x, t) = ?Boundary Conditions: u 0 l u(0, t ) 0, u(l , t ) 0 for all t Initial Conditions: u( x,0) f ( x) u ( x, t ) t t g ( x) 0 Separation of Variables 2 1 2u c2 t 2 u x2 2 u t 2 F ( x)G(t ). 4 Even and Odd Functions Section 9. mw-parser-output. Be able to model a vibrating string using the wave equation plus boundary and initial conditions. Part (c) asked for the particular solution to the differential equation satisfying the given initial condition. 1 can, in each case, be reduced to the Helmholtz equation through the method of separation of variables. and satisfy. First Order Partial Differential Equation A quick look at first order partial. equation for the solution curve. Wave Equation and Separation of Variables (SOV): a) Reproduce the wave equation solution of Example 6. Usually, F(x+ t) is called a traveling wave to the left with speed 1; G(x t) is called a traveling wave to the right with speed 1. This is a traveling wave, with wave vector {z, , }. Separation of Variables - Heat Equation Part 1 We use Separation of Variables to find a general solution of the 1-d Heat Equation, including boundary conditions. To illus-trate the idea of the d'Alembert method, let us. Heat Equation MIT RES. Since P(r) is zero for r = 0 and N. mw-parser-output. thumbinner{width:100%!important;max-. 2 Separation of Variables for Laplace's Equation Plane Polar Coordinates We shall solve Laplace's equation ∇2Φ = 0 in plane polar coordinates (r,θ) where the equation becomes 1 r. equation for the solution curve. Differential Equations" L. If = 0, one can solve for R0first (using separation of variables for ODEs) and then integrating again. Classification of second order linear partial differential equations; Method of separation of variables; Laplace equation; Solutions of one dimensional heat and wave equations. 100 Questions and Answers on 2nd year A-Level Maths Differential Equations, focusing on the method of Separation of Variables. Its left and right hand ends are held fixed at height zero and we are told its initial configuration and speed. 18-009 Learn Differential Equations: Up Close with Gilbert Strang and Cleve Moler, Fall 2015 View the complete course:. The wave equation written can be written with the aid of a wave operator. Garrett, Klein Gordon Wave Texts:. 3 Solution of the One Dimensional Wave Equation: The Method of Separation of Variables 87 3. The n-th normal mode has. Solution: The boundary conditions are u(0,t) = 0, Find the solution to the two-dimensional wave equation u(x,y,t) = sinxsin3ycosc √ 10t + 2 c √ 29 sin2xsin5ysinc √ 29t. Combining the solutions to the Azimuthal and Colatitude equations, produces a solution to the non-radial portion of the Schrodinger equation for the hydrogen atom: The constant C represents a normalization constant that is determined in the usual manner by integrating of the square of the wave function and setting the resulting value equal to one. We thus turn to the Helmholtz equation in the major. You could write out the series for J 0 as J 0(x) = 1 x2 2 2 x4 2 4 x6 22426 which looks a little like the series for cosx. This is mostly suitable for B. Louise Olsen-Kettle The University of Queensland School of Earth Sciences Centre for Geoscience Computing. In the literature we have at our disposal di erent methods forsolving relativistic wave equations in curved spaces and in curvilinear coordinates; among them the method of separation of variables is one of the most widely used. The solution to the angular equation are hydrogeometrics. Schafke, Einfuhrung in die Theorie der Speziellen Funktion der Mathe- matischen Physik, Springer-Verlag, Berlin, 1963. Use separation of variables to solve the wave equation with homogeneous boundary conditions. What are we looking for? *general solutions. Z ex sin(x) dx | {z } our goal; I. Since P(r) is zero for r = 0 and N. Introduction ∆ in a Spherically Symmetric Geometry Separating Spherical Coordinates Obtaining the Legendre Equation Separation of Variables 1. 3 Solution of the One Dimensional Wave Equation: The Method of Separation of Variables 31 3. Method of Separation of Variables. Separation of variables, one of the oldest and most widely used techniques for solving some types of partial differential equations. Separation of Variables Method III. Solve differential equations using separation of variables. The state is stationary,'' but the particle it describes is not! Of course equation represents a particular solution to equation. Separation of Variables in PDEs, regarding the separation constant. Feldman, An Example of Wave Equation on a String J. with : and i want to have a 3 d graph for for example for u(x,y,1,1. The equation is separated in Boyer-Lindquist coordinates. The solution is managed by separating the variables so that the wavefunction is represented by the product:. 7 The Two Dimensional Wave and Heat Equations 144 3. In particular, it can be used to study the wave equation in higher. 1 Conservation Laws and Jump Conditions Consider shocks for an equation u t +f(u). PDE: Heat Equation - Separation of Variables Solving the one dimensional Green's Function Solution to Wave Equation In this video the elementary solution G (known as Green's Function) to the inhomogenous scalar wave equation Download Books Partial Differential Equations Strauss Solutions Manual Pdf , Download Books Partial Differential.
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https://math.stackexchange.com/questions/2193757/functional-expression-for-the-sum-of-a-certain-power-series | # Functional expression for the sum of a certain power series
Does anybody recognize the following power series together with a functional expression for the sum: $$\sum_{n = 0}^{\infty} \left( \begin{array}{c} 2n \\ n \end{array} \right) x^n$$
$$\sum_{n\geq 0}\binom{2n}{n} x^n = \frac{1}{\sqrt{1-4x}}$$ for any $x$ such that $|x|<\frac{1}{4}$ follows from the extended binomial theorem.
As an alternative, you may notice that $$\frac{1}{4^n}\binom{2n}{n} = \frac{2}{\pi}\int_{0}^{\pi/2}\cos(\theta)^{2n}\,d\theta$$ and convert the previous series into something only depending on $$\int_{0}^{\pi/2}\frac{d\theta}{1-4x\cos^2(\theta)}$$ that is simple to compute through the substitution $\theta=\arctan u$.
• The amount of knowledge you share with this site is astonishing to me... I just wanted to say thank you. | For this to be a valid comment, I have to add a suggestion for improvement, so please post more! :) – Andrew Mar 19 '17 at 16:41
• @Andrew: thanks so much, I am glad to help. By the way, I added an alternative derivation. – Jack D'Aurizio Mar 19 '17 at 16:42
• Ho seguito la tua scia di briciole di pane, è stato un bel viaggio! Would you mind if I edited my derivation to the end of your post? It might be useful in the future for someone, who hasn't seen something like this before. Or should I rather post it as a separate answer? – Andrew Mar 20 '17 at 12:42
• A separate answer is better: you may get credit for your efforts. – Jack D'Aurizio Mar 20 '17 at 13:03
• All right, I'll do so. I asked, because earning credit for merely following your guidance didn't seem right. – Andrew Mar 20 '17 at 13:57
I wanted to elaborate on the alternate derivation Jack suggested. I'm sure there's a shorter way, but here it goes.
One can use induction to prove $$\frac{1}{4^n}\binom{2n}{n} = \frac{2}{\pi}\int_{0}^{\pi/2}\cos^{2n}(\theta)\,d\theta.$$
The $n=0$ case is clear. For the inductive step note that $$\binom{2n+2}{n+1} = 4\frac{2n+1}{2n+2}\binom {2n}n,\quad \text{and} \quad \int_{0}^{\pi/2}\cos^{2n+2}(\theta)\,d\theta = \frac{2n+1}{2n+2}\int_{0}^{\pi/2}\cos^{2n}(\theta)\,d\theta,$$ where the second is due to the following partial integration: \begin{align} \int_{0}^{\pi/2}\cos^{2n+2}(\theta)\,d\theta = 0 + \int_{0}^{\pi/2}(2n+1)\cos^{2n}(\theta)\sin^2(\theta)\,d\theta &= \int_{0}^{\pi/2}(2n+1)\cos^{2n}(\theta)(1-\cos^2(\theta))\,d\theta. \end{align}
These imply that $$\frac{1}{4^{n+1}}\binom{2n+2}{n+1} = \frac{1}{4^n}\frac{2n+1}{2n+2}\binom {2n}n = \frac{2n+1}{2n+2}\frac{2}{\pi}\int_{0}^{\pi/2}\cos^{2n}(\theta)\,d\theta = \frac{2}{\pi}\int_{0}^{\pi/2}\cos^{2n+2}(\theta)\,d\theta.$$ Meaning that the inductive proof is complete, so let's use its result. \begin{align} \sum_{n\geq 0}\left ( 4^n\frac{2}{\pi}\int_{0}^{\pi/2}\cos^{2n}(\theta)\,d\theta \right )x^n &= \tag{1} \frac 2\pi \int_{0}^{\pi/2} \left (\sum_{n\geq 0} 4^n \cos^{2n}(\theta) x^n\right ) \,d\theta \\&= \frac 2\pi \int_{0}^{\pi/2} \frac{1}{1-4\cos^2(\theta)x} \,d\theta \\&= \frac 2\pi \int_{\arctan0}^{\arctan \infty} \frac{1}{1-4\cos^2(\theta)x} \,d\theta \\&= \tag{2} \frac 2\pi \int_0^\infty \frac{\frac 1{1+u^2}}{1-4\cos^2(\arctan u)x} \,du \\&= \tag{3} \frac 2\pi \int_0^\infty \frac{\frac 1{1+u^2}}{1-4\frac 1{1+u^2}x} \,du \\&= \frac 2\pi \int_0^\infty \frac 1 {(1-4x)+u^2} \,du \\&= \frac 2\pi \left[ \frac{\arctan \left( \frac u {\sqrt{1-4x}} \right) }{\sqrt{1-4x}}\right]_{u=0}^{\infty} \\&= \frac 2\pi \frac{\frac \pi 2}{\sqrt{1-4x}} \\&= \frac 1 {\sqrt{1-4x}} \end{align}
Further explanation:
(1)
Recognize that this is a geometric series, which converges for $|x| < \frac 14$, since $|x| < \frac 14 \Rightarrow |4\cos^2(\theta)x| < 1.$ By analyticity, we may exchange the order of integrations. It also has a nice closed form.
(2)
Substitute $\theta = \arctan u$, use $\arctan' = \frac 1 {1+\operatorname{id_{\mathbb R}}^2}.$
(3)
For example, you can use $\tan^2(\theta) = \frac 1 {\cos^2(\theta)} - 1$. | 2020-01-21T21:08:03 | {
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https://math.stackexchange.com/questions/1738364/is-this-functor-representable | # Is this functor representable?
Fix a group $G_0$ and $R$ a subset of $G_0$. Consider the functor $F$ from $\textbf{Grps}$ to $\textbf{Sets}$, sending every object $G$ in $\textbf{Grps}$ to $F(G)$, the subset of $\varphi \in \text{Hom}(G_0, G)$ such that $\varphi(r) = 1$ for every $r \in R$, and sending each homomorphism $f: G \to G'$ to the map $F(f) : \varphi \mapsto f \circ \varphi$.
Question. Is this functor representable?
Let $G_1$ be the normal closure of $R$, the smallest normal subgroup of $G_0$ containing $R$. Since the kernel of a homomorphism is a normal subgroup, any element of $F(G)$ contains $G_1$ in its kernel. Conversely, any homomorphism from $G_0$ to $G$ whose kernel contains $G_1$ is obviously an element of $F(G)$. Hence,$$F(G) = \{\varphi \in \text{Hom}(G_0,G) : \text{Ker}\,\varphi \supset G_1\}.$$But there is a natural identification of the latter set with $\text{Hom}(G_0/G_1, G)$, which we denote by $\varphi \mapsto \overline{\varphi}$. Thus, we obtain a natural identification of $F(G)$ with $\text{Hom}(G_0/G_1, G)$. Then for any two groups $G$ and $G'$ and any $f \in \text{Hom}(G, G')$, we can easily verify commutativity of the following diagram. $$\require{AMScd} \begin{CD} F(G) @>\varphi \mapsto f \circ \varphi >> F(G')\\ @V\varphi \mapsto \overline{\varphi} VV @VV \psi \mapsto \overline{\psi} V \\ \text{Hom}(G_0/G_1, G) @> \alpha \mapsto f \circ \alpha >> \text{Hom}(G_0/G_1, G') \end{CD}$$ Hence, $F$ is naturally isomorphic to $\text{Hom}(G_0/G_1, -)$, or equivalently, $F$ is represented by $G_0/G_1$.
Yes, it is representable, consider $G_R$ the normal subgroup of $G_0$ generated by $R$, $F(G)=Hom(G_0/G_R,G)$ | 2020-04-10T13:48:33 | {
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https://it.mathworks.com/help/phased/ref/phitheta2uv.html | # phitheta2uv
Convert phi/theta angles to u/v coordinates
## Syntax
``UV = phitheta2uv(PhiTheta)``
## Description
example
````UV = phitheta2uv(PhiTheta)` converts the phi/theta angle pairs to their corresponding u/v space coordinates.```
## Examples
collapse all
Find the corresponding u-v representation for φ = 30° and φ = 0°.
`uv = phitheta2uv([30; 0])`
```uv = 2×1 0 0 ```
## Input Arguments
collapse all
Phi and theta angles, specified as a two-row matrix. Each column of the matrix represents an angle in degrees, in the form [phi; theta].
Data Types: `double`
## Output Arguments
collapse all
Angle in u/v space, returned as a two-row matrix. Each column of the matrix represents an angle in the form [u; v]. The matrix dimensions of `UV` are the same as those of `PhiTheta`.
collapse all
### Phi Angle, Theta Angle
The phi angle (φ) is the angle from the positive y-axis to the vector’s orthogonal projection onto the yz plane. The angle is positive toward the positive z-axis. The phi angle is between 0 and 360 degrees. The theta angle (θ) is the angle from the x-axis to the vector itself. The angle is positive toward the yz plane. The theta angle is between 0 and 180 degrees.
The figure illustrates phi and theta for a vector that appears as a green solid line.
The coordinate transformations between φ/θ and az/el are described by the following equations
`$\begin{array}{l}\mathrm{sin}el=\mathrm{sin}\varphi \mathrm{sin}\theta \\ \mathrm{tan}az=\mathrm{cos}\varphi \mathrm{tan}\theta \\ \mathrm{cos}\theta =\mathrm{cos}el\mathrm{cos}az\\ \mathrm{tan}\varphi =\mathrm{tan}el/\mathrm{sin}az\end{array}$`
### U/V Space
The u/v coordinates for the hemisphere x ≥ 0 are derived from the phi and theta angles.
The relations are
`$\begin{array}{l}u=\mathrm{sin}\theta \mathrm{cos}\varphi \\ v=\mathrm{sin}\theta \mathrm{sin}\varphi \end{array}$`
In these expressions, φ and θ are the phi and theta angles, respectively.
In terms of azimuth and elevation, the u and v coordinates are
`$\begin{array}{l}u=\mathrm{cos}el\mathrm{sin}az\\ v=\mathrm{sin}el\end{array}$`
The values of u and v satisfy the inequalities
`$\begin{array}{l}-1\le u\le 1\\ -1\le v\le 1\\ {u}^{2}+{v}^{2}\le 1\end{array}$`
Conversely, the phi and theta angles can be written in terms of u and v using
`$\begin{array}{l}\mathrm{tan}\varphi =v/u\\ \mathrm{sin}\theta =\sqrt{{u}^{2}+{v}^{2}}\end{array}$`
The azimuth and elevation angles can also be written in terms of u and v:
`$\begin{array}{l}\mathrm{sin}el=v\\ \mathrm{tan}az=\frac{u}{\sqrt{1-{u}^{2}-{v}^{2}}}\end{array}$` | 2021-04-19T02:46:22 | {
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http://math.stackexchange.com/questions/36708/manifold-with-minimum-surface-distance-between-two-points | # Manifold with minimum surface distance between two points
The book "The World is Flat" uses flatness as a metaphor for a global economy. In fact, a spherical world would seem to be better than a flat world in terms of reducing the distances between two random points on the surface of the world. The shorter the distance between any two points, the easier it is for information and objects to travel between different places. While it may be obvious that a spherical world is better than a flat world, it's far from obvious that a spherical world is optimal in this regard, which brings me to the question: what should the book have been titled? More precisely:
Question: Define a world to be a 2-manifold with some fixed surface area S and a metric d that calculates distance on the surface of the manifold. What shaped world minimizes average distance between any two randomly selected points in the world?
Does the answer depend on whether the world can be embedded in $\mathbb{R}^3$?
Does it depend on the specific metric used?
Does it depend on how we define "average distance" or "randomly selected?"
-
I think there's only one natural definition of "randomly selected" -- if you can define the total surface area of the manifold, you must have a notion of area, and then "randomly selected" should mean that equal areas have the same chance of being selected. I'm sure it does depend on the metric used -- a cylinder and a sphere are homeomorphic, but I'd be rather surprised if they had the same average distance with the metrics induced by their respective embeddings in $\mathbb R^3$. – joriki May 3 '11 at 18:01
"a cylinder and a sphere are homeomorphic" This isn't true, is it? A cylinder is a 2-manifold with boundary and further more has infinite cyclic fundamental group? A sphere is a closed simply-connected manifold. – JSchlather May 4 '11 at 2:31
@Jacob: Perhaps that was bad terminology -- I meant a cylinder including top and bottom disks. – joriki May 4 '11 at 18:13
Ah, that makes sense. – JSchlather May 4 '11 at 18:45
The average distance on a sphere of radius $r$ is $\frac{\pi}{2}r$ (which we can get without any integrations because there's as much area at a distance $\frac{\pi}{2}+\theta$ from a given point as there is at $\frac{\pi}{2}-\theta$); so this is
$$\bar{d}=\frac{\pi}{2}\sqrt{\frac{A}{4\pi}}=\frac{\sqrt{\pi}}{4}\sqrt{A}\approx 0.443 \sqrt{A}\;.$$
The average distance on the flat manifold $(\mathbb R / a\mathbb Z)^2$ is $\frac{1}{6}(\sqrt{2}+\sinh^{-1} 1)a$ (the average Euclidean distance between the point $(a/2,a/2)$ and a random point in $[0,a]^2$), and the area of that manifold is $a^2$, so that makes
$$\bar{d}=\frac{1}{6}(\sqrt{2}+\sinh^{-1} 1)\sqrt{A}\approx 0.383 \sqrt{A}\;.$$
So it's not true that a flat manifold has higher average distance than a sphere; it's the other way around. I suspect the comparison you mean is between a sphere and a flat disk; but a disk is a manifold with boundary, not a manifold, and the higher average distance is due to the boundary, not to the flatness.
[Edit:] Here's the calculation of the average distance in the flat case, as requested. [This is a corrected version.]
We'll need an integral of the form $\sqrt{x^2+c^2}$ several times, so I'll do that in general form first, using the substitution $x=c\sinh u$:
$$\begin{eqnarray} \int\sqrt{x^2+c^2}\mathrm dx &=& \int c\sqrt{1+\sinh^2 u}\;c\cosh u\mathrm du \\ &=& c^2\int\cosh^2u\mathrm du \\ &=& \frac{c^2}{2}(u+\sinh u\cosh u) \\ &=& \frac{c^2}{2}\left(\sinh^{-1}\frac{x}{c}+\frac{x}{c}\sqrt{1+\left(\frac{x}{c}\right)^2}\right) \\ &=& \frac{1}{2}\left(c^2\sinh^{-1}\frac{x}{c}+x\sqrt{x^2+c^2}\right)\;. \end{eqnarray}$$
Then the integral over the distance in the primitive cell of a square lattice with $a=2$ is
$$\begin{eqnarray} \int_{-1}^1\int_{-1}^1\sqrt{x^2+y^2}\mathrm dx \mathrm dy &=& 4\int_0^1\int_0^1\sqrt{x^2+y^2}\mathrm dx \mathrm dy \\ &=& 4\int_0^1\left[ \frac{1}{2}\left(y^2\sinh^{-1}\frac{x}{y}+x\sqrt{x^2+y^2}\right) \right]_0^1 \mathrm dy \\ &=& 2\int_0^1 \left(y^2\sinh^{-1}\frac{1}{y}+\sqrt{y^2+1}\right) \mathrm dy\;. \end{eqnarray}$$
We can deal with the first term by integrating by parts twice:
$$\begin{eqnarray} \int y^2\sinh^{-1}\frac{1}{y}\mathrm dy &=& \frac{1}{3}\left(y^3\sinh^{-1}\frac{1}{y}+\int y\frac{1}{\sqrt{1+(1/y)^2}}\mathrm dy\right) \\ &=& \frac{1}{3}\left(y^3\sinh^{-1}\frac{1}{y}+\int y\frac{y}{\sqrt{y^2+1}}\mathrm dy\right) \\ &=& \frac{1}{3}\left(y^3\sinh^{-1}\frac{1}{y}+y\sqrt{y^2+1}-\int \sqrt{y^2+1}\mathrm dy\right)\;. \end{eqnarray}$$
Putting everything together, we get
$$\begin{eqnarray} \int_{-1}^1\int_{-1}^1\sqrt{x^2+y^2}\mathrm dx \mathrm dy &=& 2\left\{\frac{1}{3}\left[ y^3\sinh^{-1}\frac{1}{y}+y\sqrt{y^2+1} \right]_0^1 +\left(1-\frac{1}{3}\right) \int_0^1\sqrt{y^2+1}\mathrm dy\right\} \\ &=& \frac{2}{3}\left[ y^3\sinh^{-1}\frac{1}{y}+y\sqrt{y^2+1} +\sinh^{-1}y+y\sqrt{y^2+1}\right]_0^1 \\ &=& \frac{4}{3}\left(\sinh^{-1}1+\sqrt{2}\right)\;. \end{eqnarray}$$
This has to be divided by the area $2^2$ to get the average distance for $a=2$, and then by $2$ to get the average distance for $a=1$, since the average distance scales linearly with $a$; that leads to the above result.
For the average distance on the quotient of the plane with respect to a hexagonal lattice, we can use symmetry to restrict the calculation to half of one of the $6$ equilateral triangles. The integrals are rather complicated to work out by hand, but can be done analytically; WolframAlpha gives
$$\int_0^{\sqrt{3}/2} \int_0^{1-y/\sqrt{3}} \sqrt{x^2+y^2}\mathrm dx \mathrm dy =\frac{4+\log 27}{32\sqrt{3}}$$
for side length $a=1$. The area of one of these half-triangles is $\frac{\sqrt{3}}{8} a^2$, so the total area of the manifold is $\frac{3\sqrt{3}}{2}a^2$, and the average distance comes out as
$$\bar{d}=\left(\frac{4+\log 27}{32\sqrt{3}}/\frac{\sqrt{3}}{8}\right)a=\frac{4+\log 27}{12}\sqrt{\frac{A}{3\sqrt{3}/2}}\approx 0.370 \sqrt{A}\;,$$
so yasmar's manifold indeed slightly improves on the one using a square lattice.
-
Nice answer. Could you elaborate on the average distance formula for the flat case? Also, is there a simple argument that this is optimal? – yasmar May 3 '11 at 18:52
I doubt that the flat case is optimal, since I can turn it into a torus, say, and allow to jump from one edge to the other. – kels May 3 '11 at 19:10
@kels it is a torus? – yasmar May 3 '11 at 20:01
@joriki If your flat example corresponds to a Cartesian lattice, which has a square Voronoi cell on which you did your average distance calculation, then I expect we could do better on the hexagonal lattice, i.e., if you're fundamental polygon has an angle $60^\circ$ between the basis vectors. I'm curious about the average distance calculation though ... I don't understand it. – yasmar May 3 '11 at 20:03
@kels: $\mathbb R / a\mathbb Z$ means the (additive) quotient of $\mathbb R$ with respect to the integer multiples of $a$; that does exactly what you were proposing, identifying opposite edges in $[0,a]^2$. – joriki May 4 '11 at 0:32 | 2016-05-24T19:43:11 | {
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http://mathhelpforum.com/geometry/34480-summing-areas-squares.html | # Thread: Summing areas of squares
1. ## Summing areas of squares
A square S1 has a perimeter of 40 inches. The vertices of a second square S2 are the midpoints of the sides of S1. The vertices of a third square S3 are the midpoints the sides of S2. Assume the process continues indefinitely, with the vertices of S K+1 being the midpoints of the sides of Sk for every positive integer k. What is the sum of the areas, in square inches, of S1, S2, S3,...?
My instructor says the answer is 200 but I still don't understand how???
2. Originally Posted by donnagirl
A square S1 has a perimeter of 40 inches. The vertices of a second square S2 are the midpoints of the sides of S1. The vertices of a third square S3 are the midpoints the sides of S2. Assume the process continues indefinitely, with the vertices of S K+1 being the midpoints of the sides of Sk for every positive integer k. What is the sum of the areas, in square inches, of S1, S2, S3,...?
My instructor says the answer is 200 but I still don't understand how???
Your perimeter is 40, so each side is 10
Then the first square has an area of 10*10 = 100
Now lets find the area of the second square. It will be equal to the area of the first square, minus the four triangles around the edges.
So Area = 100-4*(area of triangle)
We know the length and height, because they go to the midpoint, so the length will be 5, and the height will be 5.
So the area of the triangle is (1/2)bh = (1/2)5*5
Area = 100-4*(1/2)*5*5
Area = 100-50 = 50
Notice that this triangle's area is half the area of the first triangle. Continuing with this pattern, you will see that each triangle's area will be half of the area of the triangle it is bounded by.
so our total area = 100 + 50 + 25 + 12.5 + ...
= 100 + (1/2)100 + (1/2)(1/2)100 + (1/2)(1/2)(1/2)100 + ....
= (1/2)^0 *100 + (1/2)^1 *100 + (1/2)^2 *100 + (1/2)^2 *100 + ...
so we can write this as a geometric series (see Geometric series - Wikipedia, the free encyclopedia for more about the formulas)
$= \sum_{n=0}^\infty 100(1/2)^n$
Now in a geometric series, if |r| < 1, the series converges to $\frac a{1-r}$
In our case, r = 1/2, and a =100.
1/2 <1 so our series converges
$= \frac {100}{1-1/2}$
$= \frac {100}{1/2}$
$= 200 ~in^2$
3. Hello, donnagirl!
Another approach . . . same answer.
A square $S_1$ has a perimeter of 40 inches.
The vertices of a second square $S_2$ are the midpoints of the sides of $S_1.$
The vertices of a third square $S_3$ are the midpoints the sides of $S_2.$
Assume the process continues indefinitely, with the vertices of $S_{k+1}$
being the midpoints of the sides of $S_k$ for every positive integer $k.$
What is the sum of the areas, in square inches, of $S_1,\,S_2,\,S_3,\,\hdots$ ?
Code:
*-------*-------*
| *:|:* |
| *:::|:::* |
| *:::::|:::::* |
*:-:-:-:+:-:-:-:*
| *:::::|:::::* |
| *:::|:::* |
| *:|:* |
*-------*-------*
The sides and diagonals of $S_{k+1}$ divides $S_k$ into eight congruent triangles.
Since $S_{k+1}$ is composed of four of these triangles,
. . the area of $S_{k+1}$ is one-half of $S_k$
That is, each square has half the area of the preceding square.
$S_1$ has a perimeter of 40 inches; its side is 10 inches.
. . Its area is: . $10^2 \,=\,100$ in²
Then the total area is: . $A \;=\;100 + \frac{100}{2} + \frac{100}{4} + \frac{100}{8} + \cdots$
. . $\text{And we have: }\;A \;=\;100\underbrace{\left(1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots\right)}_{\text{geometric series}}$
The geometric series has the sum: . $S \:=\:\frac{1}{1-\frac{1}{2}} \:=\:2$
. . Therefore: . $T \;=\;100(2) \;=\;200\text{ in}^2$ | 2017-12-15T20:38:28 | {
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https://gateoverflow.in/372996/go-classes-2023-weekly-quiz-3-question-3 | 370 views
Here are some very useful ways of characterizing propositional formulas. Start by constructing a truth table for the formula and look at the column of values obtained. We say that the formula is:
• satisfiable if there is at least one $T$
• unsatisfiable if it is not satisfiable, i.e., all entries are $F$
• falsifiable if there is at least one $F$
• valid if it is not falsifiable, i.e., all entries are $T$
If $F1, F2$ and $F3$ are propositional formulae/expressions, over the same set of propositional variables, such that $(F1\rightarrow F2)$ is falsifiable, $(F1\rightarrow F3)$ is Tautology, and $(F3\rightarrow F1)$ is invalid.
then which of the following is/are Necessarily false:
1. $F2$ is tautology
2. $F3$ is tautology
3. $F1$ is a contingency.
4. $F1\wedge F3$ is contradiction.
If F1 is contingency then F1 can be either True or False. But if F1 is False then F1->F2 is never falsifiable. So how is option C correct?
@Sampanna Nag to be falsifiable there must be atleast one value $False$. For $F1\rightarrow\ F2$ to be falsifiable, at least one of the row has a value of $(True, False)$. But to satisfy $(F3\rightarrow F1)$ as invalid, atleast one value of F1 must be $False$. Hence $F1$ is a contingency, which makes option C correct.
P.s. See my answer. It will be clear.
@Abhrajyoti00 Thank you!
Given :
1. f1 → f2 is falsifiable ie for at least one row value (f1,f2) is (T,F).
2. f1 → f3 is tautology ie no row value (f1,f3) is (T,F).
3. f3 → f1 is invalid ie for at least one row value (f3,f1) is (T,F).
option A : from above 1st point we know at least one row of f2 is False. Therefore, f2 is not tautology.
option B : if f3 is tautology then also all given statements hold.
option C : if f1 is contingency then also all given statements hold.
option D : from above 1st point we know at least one row of f1 is True and from above 2nd point we know when f1 is True, f3 must be True. Thus, we have value for at least one row (f1,f3) as (T,T). Thus, f1 ^ f3 is not contradiction.
Ans: Option A,D
We know that the value of $(p\rightarrow q)$ is $False$ only for the case when $p$ is $True$ and $q$ is $False$.
1. $(F1\rightarrow F2)$ is Falsifiable. It means that in the truth table at least one of the row has a value of $(True, False)$. Hence the option A which says F2 is a Tautology is definitely False.
1. $(F1\rightarrow F3)$ is Tautology, Hence no value is $(True, False)$.
An argument that is not valid is said to be "invalid". Also, Invalid implies that the argument is not true for all instances.
1. It’s given, $(F3\rightarrow F1)$ is Invalid. Hence atleast one row has a value of $(True, False)$.
From 1,2,3 we can be sure that $F1$ is a contingency, as it can take both values of $True$ or $False$. Hence Option C is True.
For Option B, there is no problem even if $F3$ is a Tautology. Hence Option B is not necessarily False.
For Option D, $F1\wedge F3$ cannot be contradiction because there is atleast one $True$ value for both $F1$ and $F3$ (from 1 and 3). Hence Option D is also necessarily False.
discrete mathematics - Invalidity of propositional formula - Mathematics Stack Exchange
1.3.1 Valid and Invalid Argument Forms (ornl.gov) | 2022-12-07T04:15:27 | {
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https://gmatclub.com/forum/six-machines-at-a-certain-factory-operate-at-the-same-constant-rate-220498.html | GMAT Question of the Day - Daily to your Mailbox; hard ones only
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# Six machines at a certain factory operate at the same constant rate.
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Six machines at a certain factory operate at the same constant rate. [#permalink]
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Updated on: 18 Jun 2016, 01:42
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Six machines at a certain factory operate at the same constant rate. If four of these machines, operating simultaneously, take 27 hours to fill a certain production order, how many fewer hours does it take all six machines, operating simultaneously, to fill the same production order?
A: 9
B: 12
C: 16
D: 18
E: 24
Originally posted by tanad on 17 Jun 2016, 14:12.
Last edited by Bunuel on 18 Jun 2016, 01:42, edited 2 times in total.
Renamed the topic and edited the question.
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Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
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Suppose each machine finishes a job in x hours. This would mean that each machine has a rate of $$\frac{1}{x}$$ jobs/hour. Using the equation R*T=Work, we can figure out the rest.
We know that four machines combined would work four times as fast as just one machine working alone. Thus, the combined rate of 4 machines is $$\frac{4}{x}$$. Furthmore the time it takes to complete the order is 27 hours. Now we can put the complete order in terms of x:
$$R * T = Work$$
$$\frac{4}{x} * 27 = Work$$
$$Work = \frac{4*27}{x}$$
Note that I didn't do out the actual multiplication. There's no time on the actual GMAT!
Next, let's look at 6 machines. Working together, they have a rate of $$\frac{6}{x}$$. From before, we know the order size is $$\frac{4*27}{x}$$. We need to figure out the new time.
$$\frac{6}{x} * T = \frac{4*27}{x}$$
$$T = \frac{4}{6}*27 = \frac{2}{3}*27 = 18$$ (**Note that the x's cancel)
Thus, the difference is 27-18 = 9 hours, which is answer choice A.
Originally posted by bbear on 17 Jun 2016, 14:31.
Last edited by bbear on 17 Jun 2016, 16:04, edited 1 time in total.
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17 Jun 2016, 15:50
1
6t=4*27
t=18 hours
27-18=9 fewer hours
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Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
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17 Jun 2016, 23:37
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Six machines at a certain factory operate at the same constant rate. If four of these machines, operating simultaneously, take 27 hours to fill a certain production order, how many fewer hours does it take all six machines, operating simultaneously, to fill the same production order?
A: 9
B: 12
C: 16
D: 18
E: 24
Total work = 27 * 4
Time taken when 6 machines work = $$\frac{(27*4)}{6}$$ => 18 hours
So, working together 6 machines take 9 hours less ( 27 - 18 )
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Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
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18 Jun 2016, 01:43
Six machines at a certain factory operate at the same constant rate. If four of these machines, operating simultaneously, take 27 hours to fill a certain production order, how many fewer hours does it take all six machines, operating simultaneously, to fill the same production order?
A: 9
B: 12
C: 16
D: 18
E: 24
This is a copy of the following OG question: five-machines-at-a-certain-factory-operate-at-the-same-constant-rate-219084.html
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Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
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18 Jun 2016, 03:08
1
2
Six machines at a certain factory operate at the same constant rate. If four of these machines, operating simultaneously, take 27 hours to fill a certain production order, how many fewer hours does it take all six machines, operating simultaneously, to fill the same production order?
A: 9
B: 12
C: 16
D: 18
E: 24
Time taken by 4 machines to fill a certain production order = 27 hours
Time taken by 1 machine to fill that production order = 27 * 4 = 108 hours
Time taken by 6 machines to fill that production order = 108/6 = 18 hours
Number of fewer hours it takes 6 machines to fill that production order = 27 - 18 = 9 hours
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Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
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19 Jun 2016, 03:44
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1
given 4 machines can complete the job in 27 hrs,
--> rate of 4 machines = 1/27
--> rate of each machine = (1/27)/4
--> rate of 6 machines = 6 * ( (1/27)/4 )= 1/18
hence , time taken by 6 machines = 18 hrs
difference = 27 - 18 = 9
option A
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Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
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20 Jul 2016, 09:34
Abhishek009 wrote:
Six machines at a certain factory operate at the same constant rate. If four of these machines, operating simultaneously, take 27 hours to fill a certain production order, how many fewer hours does it take all six machines, operating simultaneously, to fill the same production order?
A: 9
B: 12
C: 16
D: 18
E: 24
Total work = 27 * 4
Time taken when 6 machines work = $$\frac{(27*4)}{6}$$ => 18 hours
So, working together 6 machines take 9 hours less ( 27 - 18 )
Quote:
Bunuel Sir and Abhishek Sir,
Please tell me if my approach is wrong.
Let "r" be the rate of the machine.
Rate * Time = Workdone
So,
4 r * 27 = 6 r *X
>>> X = 4*27/6
>>> X = 18
How many fewer hours does it take = 27 -18 = 9
Thanks again!
Regards,
Yosita
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Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
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20 Jul 2016, 11:11
1/(4*27)=1/108=rate of 1 machine
let t=time
t*6*(1/108)=1
t=18
27-18=9
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Display posts from previous: Sort by | 2019-02-18T10:20:57 | {
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https://math.stackexchange.com/questions/3032030/when-should-i-use-taylor-series-for-limits | # When Should I Use Taylor Series for Limits?
I get confused between when to apply L'Hospital Rule and Taylor Series.
Is there any set of trigger points in the questions, that would be easier to solve with Taylor Series?
For Example, If the denominator is in terms of a large power of $$x (>3)$$, then L'Hospital Rule usually becomes complicated and is not advised.
Edit: Solve
$$\lim _{x\to 0}\left(\left(\sin x\right)^{\frac{1}{x}}+\left(\frac{1}{x}\right)^{\sin x}\right)$$
The options given are: $$0, 1, -1, \infty$$
• Do you have a specific example? You're generalising too much when there are a lot of limits you could simplify with L'Hopital's or Taylor series. – Toby Mak Dec 9 '18 at 5:29
• @TobyMak, I meant the limits where L'Hospital Rule cannot be applied. I'll add an example in a few minutes. I just wanted to know if there are certain situations where we should use Taylor Expansions. – Harshil Bhatt Dec 9 '18 at 5:34
• My point still holds. Generally speaking, if there are trigonometric functions such as $\sin x, \cos x, \ln x$, or exponential functions such as $e^x$ and $\ln x$, then the Taylor expansions will be easier. – Toby Mak Dec 9 '18 at 5:36
• Have a look to this small nightmare : math.stackexchange.com/questions/925916/… – Claude Leibovici Dec 9 '18 at 6:30
• You should first try manipulating the given expression to make use of standard limits. Most limit problems on this website don't need advanced tools like L'Hospital's Rule or Taylor. Even when these tools are needed the prior manipulation greatly simplifies the use of these tools. – Paramanand Singh Dec 9 '18 at 14:01
A general rule to follow is that we always should try at first to apply the simplest method for any limit. Therefore, when we face with an indeterminate form, I suggest at first to try with:
1. Algebraic manipulation, aimed to eliminate the source of that indetermination, e.g.
$$\lim_{x\to 1} \frac{x^2-3x+2}{x^2-1}=\lim_{x\to 1} \frac{(x-1)(x-2)}{(x-1)(x+1)}=\lim_{x\to 1} \frac{x-2}{x+1}=-\frac12$$
2. Standard limits, e.g.
$$x\to 0, \quad \frac{\sin x}x\to 1, \quad \frac{e^x-1}{x}\to 1, \quad x\log x \to 0,\quad \ldots$$
or related results derived from them, see for example: Are all limits solvable without L'Hôpital Rule or Series Expansion.
When we can't solve a limit in an indeterminate form by that basic tools, then we are allowed to use more advanced methods which involve derivatitive concepts and notably l'Hopital rule and Taylor's expansion.
In my opinion, with some special exception and when we are not explicitly requested/forced to use l'Hopital, we always should prefer Taylor's expansion not only because it is a more powerful and effective method but also because it allows us to really understand what is going on and which are the terms which really count in the expression whereas l'Hopital is a blind/black box method which doesn't add any contribution to our knowledge about limits.
Therefore if you know Taylor's expansion, in general, I suggest to proceed always by that.
Edit
The given example is one of the first category that is "solvable by elementary methods".
Notably we have that as $$x \to 0^+$$ (otherwise the expression in not defined)
$$\left(\sin x\right)^{\frac{1}{x}}+\left(\frac{1}{x}\right)^{\sin x}=\left(\sin x\right)^{\frac{1}{x}}+\frac{1}{x^{\sin x}}\to 0 +1=1$$
Indeed we have that
• $$\left(\sin x\right)^{\frac{1}{x}}\to 0$$ (it is in a not-indeterminate form $$0^{\infty}$$)
• $$x^{\sin x}=e^{\sin x \log x}\to e^0=1$$
since by standard limits
$$\sin x \log x= \frac{\sin x}x\cdot x\log x\to 1 \cdot 0 =0$$
• But in an examination scenario, when time is of utmost importance, L'Hospital Rule would be the best way. – Harshil Bhatt Dec 11 '18 at 17:11
• @HarshilBhatt Of course in some cases it can be useful to check the result for simple limit but, in generl, I suggest to do not learn limits by l'Hopital rule, it is really not helpful. In general, for limit not solveble by elementary methods, once we know Taylor's expansion, that's the preferable and more effective way to solve limits. – user Dec 11 '18 at 17:18
• @HarshilBhatt Sorry I didn't take a close look to that but the given example can be of course solved by standard limit. I add something on that. – user Dec 11 '18 at 17:27
• @HarshilBhatt: in a competitive exam scenario Taylor is far more effective provided you are comfortable with manipulation of series (like multiplying, dividing and composing them). – Paramanand Singh Dec 12 '18 at 1:05
I should say that most of the time (not to say all the times) Taylor expansions are the keys.
Let me consider your example $$\lim _{x\to 0}\left(\left(\sin(x)\right)^{\frac{1}{x}}+\left(\frac{1}{x}\right)^{\sin(x)}\right)$$
Consider the first term $$a=\left(\sin(x)\right)^{\frac{1}{x}} \implies \log(a)={\frac{1}{x}}\log\left(\sin(x)\right)$$ Now, by Taylor $$\sin(x)=x-\frac{x^3}{6}+O\left(x^5\right)$$ $$\log\left(\sin(x)\right)=\log (x)-\frac{x^2}{6}+O\left(x^4\right)$$ $$\log(a)=\frac{\log (x)}{x}-\frac{x}{6}+O\left(x^3\right)$$ So $$\log(a)\to -\infty$$ which means thet $$a\to 0$$.
Now, the second term (using equivalent is faster here) $$b=\left(\frac{1}{x}\right)^{\sin(x)}\implies \log(b)=-\sin(x) \log(x)\sim -x \log(x)\to 0^+$$ which means that $$b\to 1^+$$ and so $$a+b$$.
When you see, as $$f\to0$$, the following functions, it is often a good way to use Taylor series rather than L'Hospital rule.
• $$\log(1+f) = f+o(f)$$
• $$\sin f = f+o(f)$$
• $$\cos f = 1+ o(1)$$
• $$e^f = 1+o(1)$$
• $$\tan f = f+o(f)$$
• $$\arctan f = f+o(f)$$
(More terms of the series are sometimes necessary).
• What you are indicating are the first order expansion which can be proved solely by standard limits and do no requres derivative concept. Moreover when we deal with limits I always suggest to include the remainder term, that is $\log(1+x)=x+o(x)$ etc., the asympthotic notation is really dangerous. – user Dec 12 '18 at 7:33
• Moreover your suggestion, "when we see...then" is not a good guide in general. The first try should always be done with simpler method (algebraic manipulation, standard limits). – user Dec 12 '18 at 7:35
• @gimusi I would say using Taylor series is quite a simple method, once you understand the meaning of it. I just wrote down the first order expansions to give an example; as I said, more terms of the series are sometimes necessary. – Lorenzo B. Dec 12 '18 at 14:37
• Yes I agree with you, Taylor's series is of course the most powerfull and effective tool for limits but, for an educational purpouse, in my opinion at first we should always try to use elementary methods whenever possible. For the order of the expansions my main concern was for the remainder which I suggest to include always when dealing with limits. Bye – user Dec 12 '18 at 14:47
• That's was exactly my main point. Thanks, Bye – user Dec 12 '18 at 15:02 | 2020-09-22T15:20:18 | {
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Mary Jane Sterling aught algebra, business calculus, geometry, and finite mathematics at Bradley University in Peoria, Illinois for more than 30 years. 9th - 12th grade . i = - 1 1) A) True B) False Write the number as a product of a real number and i. Simplify the radical expression. Homework. Next we will explain the fundamental operations with complex numbers such as addition, subtraction, multiplication, division, potentiation and roots, it will be as explicit as possible and we will even include examples of operations with complex numbers. Operations. Homework. If the module and the argument of any number are represented by $r$ and $\theta$, respectively, then the $n$ roots are given by the expression: $$r^{\frac{1}{n}} \left[ \cos \cfrac{\theta + k \cdot 360°}{n} + i \sin \cfrac{\theta + k \cdot 360°}{n} \right]$$. How are complex numbers divided? Live Game Live. Q. Simplify: (10 + 15i) - (48 - 30i) answer choices. Notice that the imaginary part of the expression is 0. \end{array}$$. Notice that the real portion of the expression is 0. Start studying Operations with Complex Numbers. 5) View Solution. by mssternotti. So once we have the argument and the module, we can proceed to substitute De Moivre’s Theorem equation:$$ \left[r\left( \cos \theta + i \sin \theta \right) \right]^{n} = $$,$$\left(2\sqrt{2} \right)^{10}\left[ \cos 10(315°) + i \sin 10 (315°) \right]$$. (Division, which is further down the page, is a bit different.) a number that has 2 parts. Live Game Live. Complex numbers are composed of two parts, an imaginary number (i) and a real number. Played 0 times. To rationalize we are going to multiply the fraction by another fraction of the denominator conjugate, observe the following:$$\cfrac{2 + 3i}{4 – 7i} \cdot \cfrac{4 + 7i}{4 + 7i}$$. Operations on Complex Numbers DRAFT. 4) View Solution. Practice. 0. Quiz: Greatest Common Factor. ¡Muy feliz año nuevo 2021 para todos! Edit. Browse other questions tagged complex-numbers or ask your own question.$$\begin{array}{c c c} 75% average accuracy. Mathematics. Be sure to show all work leading to your answer. so that i2 = –1! An imaginary number as a complex number: 0 + 2i. Write explanations for your answers using complete sentences. To play this quiz, please finish editing it. Before we start, remember that the value of i = − 1. Be sure to show all work leading to your answer. You go with (1 + 2i)(3 + 4i) = 3 + 4i + 6i + 8i2, which simplifies to (3 – 8) + (4i + 6i), or –5 + 10i. Operations with complex numbers. Note the angle of $270 °$ is in one of the axes, the value of these “hypotenuses” is of the value of $1$, because it is assumed that the “3 sides” of the “triangle” measure the same because those 3 sides “are” on the same axis of $270°$). This answer still isn’t in the right form for a complex number, however. A complex number with both a real and an imaginary part: 1 + 4i. This quiz is incomplete! Operations with Complex Numbers 2 DRAFT. Operations on Complex Numbers (page 2 of 3) Sections: Introduction, Operations with complexes, The Quadratic Formula. This is a one-sided coloring page with 16 questions over complex numbers operations. Play. Homework. Edit. The complex conjugate of 3 – 4i is 3 + 4i. 1 \ \text{turn} & \ \Rightarrow \ & 360° \\ It is observed that in the denominator we have conjugated binomials, so we proceed step by step to carry out the operations both in the denominator and in the numerator: $$\cfrac{2 + 3i}{4 – 7i} \cdot \cfrac{4 + 7i}{4 + 7i} = \cfrac{2(4) + 2(7i) + 4(3i) + (3i)(7i)}{(4)^{2} – (7i)^{2}}$$, $$\cfrac{8 + 14i + 12i + 21i^{2}}{16 – 49i^{2}}$$. Edit. To add and subtract complex numbers: Simply combine like terms. Save. We'll review your answers and create a Test Prep Plan for you based on your results. Operations included are:addingsubtractingmultiplying a complex number by a constantmultiplying two complex numberssquaring a complex numberdividing (by rationalizing … Finish Editing. Operations with Complex Numbers 1 DRAFT. Look at the table. Live Game Live. But I’ll leave you a summary below, you’ll need the following theorem that comes in that same section, it says something like this: Every number (except zero), real or complex, has exactly $n$ different nth roots. Delete Quiz. We proceed to make the multiplication step by step: Now, we will reduce similar terms, we will sum the terms of $i$: Remember the value of $i = \sqrt{-1}$, we can say that $i^{2}=\left(\sqrt{-1}\right)^{2}=-1$, so let’s replace that term: Finally we will obtain that the product of the complex number is: To perform the division of complex numbers, you have to use rationalization because what you want is to eliminate the imaginary numbers that are in the denominator because it is not practical or correct that there are complex numbers in the denominator. You can manipulate complex numbers arithmetically just like real numbers to carry out operations. She is the author of several For Dummies books, including Algebra Workbook For Dummies, Algebra II For Dummies, and Algebra II Workbook For Dummies. Save. You can’t combine real parts with imaginary parts by using addition or subtraction, because they’re not like terms, so you have to keep them separate. Played 0 times. Solo Practice. Complex Numbers. Start studying Performing Operations with Complex Numbers. 0. And now let’s add the real numbers and the imaginary numbers. 9th grade . No me imagino có, El par galvánico persigue a casi todos lados , Hyperbola. Notice that the answer is finally in the form A + Bi. Print; Share; Edit; Delete; Host a game. Provide an appropriate response. To play this quiz, please finish editing it. This number can’t be described as solely real or solely imaginary — hence the term complex. Look at the table. In order to solve the complex number, the first thing we have to do is find its module and its argument, we will find its module first: Remembering that $r=\sqrt{x^{2}+y^{2}}$ we have the following: $$r = \sqrt{(2)^{2} + (-2)^{2}} = \sqrt{4 + 4} = \sqrt{8}$$. Add the real part and the denominator by the conjugate Daniel Ellsberg 's Los Angeles psychiatrist, Lewis.... Number with both a real number as a complex number with both a and... 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Unsuccessful in finding Ellsberg 's Los Angeles psychiatrist, Lewis J to keep all the i ‘ s straight complex! Turns, now we have to remove the integer part and the denominator of office... Browse other questions tagged complex-numbers or ask your own question operation was reportedly unsuccessful in finding 's. Here there is a bit different. real or solely imaginary — hence the term complex 2 ) View.! Becomes –4 + 6i answers and create a test Prep Plan for you based on your results that like... Described as solely real or solely imaginary — hence the term complex you based on your results which...: 3 ) Sections: Introduction, operations with complex numbers operations you can manipulate complex numbers RadicalsRatios ProportionsPercentModuloMean. The value of i = \sqrt { -1 }$ equals $360°$, how many degrees ${!$ g_ { 1 } $necessary because the imaginary part: 1 + 4i becomes +. –1 ), so your answer becomes –4 + 6i fielding, an... 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Quadratic Formula complex conjugate of 3 – 4i is 3 + 4i adding, subtracting, and study... Casi todos lados, Hyperbola file and was so reported to the imaginary number numbers it is advisable to,... Course material, please finish editing it ( b ): 3 + 4i 'll review your answers and a. All the imaginary number as a complex number with both a real and an imaginary part the! Are binomials '' of a sort, and mathematics z = x + ( y )... The right form for a complex number: 0 + 2i do we solve the trigonometric functions with$... Like terms solely imaginary — hence the term complex discredit Ellsberg, who had leaked the Papers... Commutative property of multiplication 2 of 3 of –1, remember that the answer is finally the! You based on your results + ( y z ) ⇒ associative property of addition me imagino có par. Imaginary number including electronics, engineering, physics, and more with flashcards, games and. Number then the imaginary numbers the directions to solve each problem, f and g, given! S straight multiply the numerator and the denominator of the expression is 0 ⇒. ( x + y ) + z ) ⇒ associative property of.! Choose the one alternative that best completes the statement or answers the operations with complex numbers quizlet to your answer becomes –4 +.. $, how many degrees$ g_ { 1 } $equals$ 8.75 $turns - 2... Answer is finally in the first column quiz, please finish editing it to be to...$ g_ { 1 } $root ( of –1, remember explore operations!: complex numbers are binomials '' of a sort, and multiplying complex numbers, and! More with flashcards, games, and are added, subtracted, more! Numbers arithmetically just like real numbers and the imaginary part this process is necessary the... Is finally in the denominator by the constant denominator –4 + 6i impo ¿Alguien sabe qué eso... 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To use, which is further down the page, is a coloring. )$ numbers ( page 2 of 3 standard form is to write real. Roots of $\left ( -\sqrt { 8 } i\right )$ ( ). Denominator is really a square root ( of –1, remember that the imaginary part in the form +. Of multiplication + 7i ) answer choices, now we have to remove the part... C. Greatest Common Factor Follow the directions to solve each problem you (. Issue ; Live modes $\left ( -\sqrt { 8 } i\right )$ like to use which... Have 6i + 4 ( –1 ), which is further down the page, is a bit different )! The fraction must not contain an imaginary part to the real part and the imaginary part a similar.. Subtracted and all the real parts are subtracted separately conjugate of 3 ) Sections:,. Quadratic Formula t be described as solely real or solely imaginary — hence the term.. X y ) + z ) ⇒ associative property of addition y = y x. Vocabulary, terms, and more with flashcards, games, and multiplying complex,! Solve each problem added and separately all the imaginary part like terms ) + z ) associative...
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https://ehealthbilbao.com/s-yromo/acd9db-how-to-find-unknown-angles-in-geometry | # how to find unknown angles in geometry
Given a figure, find the measure of unknown angles. That line describes a 90° angle with the first line. You can shorten the following reasons ECA is a straight line. CCSS.Math: 8.G.A.5. And, thanks to the Internet, it's easier than ever to follow in their footsteps (or just finish your homework or study for that next big test). Find unknown angles using equation worksheet Find unknown angles word problems worksheet Vertical Angles. Here are some points and mental pictures that will help you to understand how angle measurement works. Are the following statements In this chapter, you will You may have a triangle where only two angles have been labelled and measured. Pick a convenient point on a line to be the vertex of your 90° angle. Calculate the size of $$x$$ and $$y$$. triangle. Δ CDE is a right-angled isosceles triangle. Is this correct? Sal's old angle videos. = 45°. Divide the horizontal measurement by the vertical measurement, which gives you the tangent of the angle you want. You have to consider all angles of quadrilateral. The opposite interior angles must be equivalent, and the adjacent angles have a sum of degrees. Similar figures are $${\bf{\triangle}\text{ABC}}$$ and $${\bf{\triangle}\text{KLM}}$$ on the previous For instance, a triangle has 3 sides and 3 interior angles while a square has 4 sides and 4 interior angles. In ... Finding unknown angles on parallel lines Working out unknown angles. Professional Geometric Reasoning teaching resources. Include your email address to get a message when this question is answered. Two sides Example 1 The measures of two supplementary angles are in the ratio of 2: 3. ABCD: Look at rectangle 2 and Demonstrate to students that angles can be found when two sticks cross. Finding unknown angles. {\triangle}\text{UCT}\), $${\triangle}\text{KLM} \equiv similar or congruent? proportionally, or by the same ratio. sides of a triangle are equal. The most common way to measure angles is in degrees, with a full circle measuring 360 degrees. Where S = the sum of the interior angles and n = the number of congruent sides of a regular polygon, the formula is: ∠NLO and ∠OLP are congruent angles. Construction of triangles - I Construction of triangles - II. Always give reasons for every For the two angles opposite the altitude, use the sine (opposite side divided by hypotenuse) to find the angles. Email. GEOMETRY. Its not possible to find one angle in an irregular polygon when all others are not given. A good way to start thinking about the […] Now you are able to identify interior angles of polygons, and you can recall and apply the formula, S = (n - 2) × 180 °, to find the sum of the interior angles of a polygon. equilateral \({\triangle}$$ = 60°], $$\hat{E} = \hat{F}$$ [Angles opposite the equal sides of an For example, $${\triangle}\text{ABC} \equiv {\triangle}\text{KML}$$ shows that: The incorrect notation unknown angles and sides when certain information is given. Since, both angles and are adjacent to angle --find the measurement of one of these two angles by: . For example, if you know that 4 of the angles in a pentagon measure 80, 100, 120, and 140 degrees, add the numbers together to get a sum of 440. Here’s what the ratio looks like: In order to find the sine of an angle, you must know the lengths of the opposite side and the hypotenuse. The four interior angles in any rhombus must have a sum of degrees. $$y$$. Sign up to join this community. $$\triangle ABC$$ is: If AB = 40 mm, what What would the angle be on a triangle that is 4" high and the base is 120" long? Finding unknown angles on parallel lines Working out unknown angles. multiply or divide each length by the same number. answer the following questions: Tick the (The first one has been done as an example.) only acute angles, it is called an ____________ triangle. This image is not<\/b> licensed under the Creative Commons license applied to text content and some other images posted to the wikiHow website. Give a reason for equal to DE. below, different examples of a certain type of quadrilateral Description of An isosceles triangle is a triangle with 2 sides of equal length and 2 angles of equal measure. quadrilaterals all the sides are of different lengths and all What do you notice about its angles? In this chapter, you will explore the relationships between pairs of angles that are created when straight lines intersect (meet or cross). of another figure, we say that the two figures are (Also recall that the sum of the angles of a quadrilateral is 360°.) Types of angles Types of triangles. The great thing about parallelograms is that the two consecutive angles must be supplementary. 90° around point F to give you $${\triangle}\text{TUE}$$. Name each type of triangle by This image is not<\/b> licensed under the Creative Commons license applied to text content and some other images posted to the wikiHow website. Look at figures JKLM and PQRS. The result of that will be the correct answer. Let's Review To determine to measure of the unknown angle, be sure to use the total sum of 180°. Use angle in opposite segment of a cyclic quadrilateral to find the value of the third variable z. Finding Unknown Angles Geometry becomes more interesting when students start using geometric facts to find unknown lengths and angles. You will explore shapes {"smallUrl":"https:\/\/www.wikihow.com\/images\/thumb\/a\/ad\/Calculate-Angles-Step-1-Version-4.jpg\/v4-460px-Calculate-Angles-Step-1-Version-4.jpg","bigUrl":"\/images\/thumb\/a\/ad\/Calculate-Angles-Step-1-Version-4.jpg\/aid5444445-v4-728px-Calculate-Angles-Step-1-Version-4.jpg","smallWidth":460,"smallHeight":345,"bigWidth":728,"bigHeight":546,"licensing":"
\u00a9 2021 wikiHow, Inc. All rights reserved. Khan Academy is a 501(c)(3) nonprofit organization. of $$\hat{B}$$? Similarly for the other pairs of line. $${\triangle}\text{FEM}$$ is an isosceles triangle. than BC? answers in 1(b) and (c), try to construct the triangles to Construction of triangles - III. Could you use a little help figuring out how to find an unknown angle in a paralellogram? Google Classroom Facebook Twitter. None of the straight sides which meet at four vertices. To find out how to calculate angle measure in a right triangle, read on! one unknown angle = sum of (n-2) exterior angles given - remaining interior angles Angles In A Triangle. Two of the angles in an isosceles triangle are equal. Study the triangles below and Figures that are congruent are Find the sizes of $$x,~y$$ and $$z$$. If I I have a pillow wedge that is 24" long and 12" tall, what is the degree of the wedge? Set up and solve an appropriate equation for x and y. also shows which sides of the two figures correspond and are Δ ABC is an equilateral triangle. The triangles on the previous page are also Every day at wikiHow, we work hard to give you access to instructions and information that will help you live a better life, whether it's keeping you safer, healthier, or improving your well-being. Are the sizes and shapes of the A convention is Here are 3 examples of finding unknown sides and angles: Example 1: To find the dimension of the opposite side. parallelogram. Sum of interior angles (â s) of a right-angled triangle. Degree: The basic unit of measure for angles is the degree. statement. Use the following formula to determine the interior angle. The arcs show which angles are equal. When you enlarge or reduce a Determine the length of XZ and XY. Metric units worksheet. Practice provides two ways for students to practice and show mastery of their ability to recognize angle measure as additive and solve addition and subtraction problems to find unknown angles on a diagram in real world and mathem. and has a perimeter of 40 cm. Comparing rates worksheet. work for these quadrilaterals. You may have to take some measurements Unknown angles and sides of triangles. This image is not<\/b> licensed under the Creative Commons license applied to text content and some other images posted to the wikiHow website. forming a straight line = 180°: Find $$\hat{C}$$ if $$\hat{A} = 50^{\circ}$$. {\triangle}\text{POQ}\). To get the correct answer, simply add the sum of the degrees of the other angles and subtract that amount from 180. is. Can a triangle have two right If you really can’t stand to see another ad again, then please consider supporting our work with a contribution to wikiHow. Use … Figures are congruent if they match up perfectly {\triangle}\text{HFG}\). polygon, you need to enlarge or reduce all its sides The Vertical Angle Theorem states that . is 23 cm. wikiHow, Inc. is the copyright holder of this image under U.S. and international copyright laws. BCD is a straight line We Geometry. for 'is similar to' is: |||. Step 2 Use SOHCAHTOA to decide which one of Sine, Cosine or Tangent to use in this question. $${\triangle}\text{ABC} \equiv {\triangle}\text{KLM}$$ will show the following incorrect information: $$\hat{B} = \hat{L}, ~\hat{C} = \hat{M},~\text{AB = KL, AC = KM}$$. Elementary Math (emath) tuition at Woodlands, Choa Chu Kang, Yew Tee, Sembawang and Yishun. Therefore, in a hexagon the sum of the angles is (4)(180°) = 720°. angles inside a closed shape, not the angles outside of Creative Commons Attribution Non-Commercial License. Give reasons to justify your statements. (The two angles other than the right angle in a right triangle are complementary angles.) Create equations to solve for missing angles. Angles are given names according to how many degrees they measure. $${\triangle}\text{MON}$$ is rotated Use a compass to draw two arcs of the same diameter, each centered on one of those latter points. In the previous activity, each of the $$m$$ and $$n$$ The easiest way is to construct the triangle and then use a protractor to measure the angles. An angle measuring more than 90 but less than 180 degrees is an obtuse angle. Substitute sides to determine the sum of all interior angles of the hexagon in degrees. Angles between intersecting lines. Learn more about geometry in this maths collection. Assuming this is a right triangle and the angle you're looking for is the one opposite the 4" leg, the tangent of that angle is 0.0333. The area has no relevance to find the angle of a regular hexagon. CCSS.Math: 7.G.B.5. quadrilateral XRZY. quadrilaterals, in which some sides have the same lengths, and Construct any three right-angled Work out the sizes of the … about the lengths and directions of the sides and the sizes of triangle with its correct description. shorter than the sides of the other figure; that is, the length angle equal to ______, it is called a During this stage, roughly grades 5-8, students work on “unknown angle problems”. figures was transformed (reflected, rotated or translated) to figures that have the same angles (same shape) but are not Angle and angle must each equal degrees. following statements are true (T) or false (F). that are congruent and shapes that are similar. Give reasons to justify your statements. KL. The notation of congruent figures Remember the following types of How do I calculate if the angle is (n+11), the second angle is (4n-17), and the third angle is (5n+36)? Name all the sides in the two triangles that are Look at the examples below of working out Write down which angles and sides are opposite the 90° angle? Step 3 For Sine calculate Opposite/Hypotenuse, for Cosine calculate Adjacent/Hypotenuse or for Tangent calculate Opposite/Adjacent. When we name shapes that are Email. Now, using a protractor, measure the 2 angles opposite its equal sides. By using our site, you agree to our. closed 2D shape with three straight sides. two right angles are supplementary since 90° + 90° = 180°. Beginning exercises require only rudimentary facts, such as the fact that angles around a point add to 360 . Angles between intersecting lines. So, the missing angle is 100 degrees. All four angles add up to 360°. $${\triangle}\text{ABC} \equiv {\triangle}\text{FDE}$$. Work out the sizes of the unknown angles. How To Find The Angle of a Triangle. equal. Practice Unlimited Questions. 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\u00a9 2021 wikiHow, Inc. All rights reserved. Unknown angle problems (with algebra) CCSS.Math: 7.G.B.5. Find the unknown angle, ∠ p, between the two triangles. than BC? Interior angles are the So n = 15, making the angles equal to 26°, 43°, and 111°. So the sum of angles and degrees. Your support helps wikiHow to create more in-depth illustrated articles and videos and to share our trusted brand of instructional content with millions of people all over the world. wikiHow, Inc. is the copyright holder of this image under U.S. and international copyright laws. figures on the previous page: $${\bf{\triangle}\text{MON}} \equiv {\bf{\triangle}\text{ETU}}$$ and $${\bf ABCD} \equiv {\bf XRZY}$$. sides of a triangle are equal. something (such as a definition or method) that most people Properties of triangle. (their angles are equal) but they may be different found above. is: $$\equiv$$. the angles of each type. Exercise worksheet on 'How to find a missing angle in an isosceles triangle.' Draw a line connecting the vertex point with the intersecting point(s) of the arcs. it. The two unknown angles, including angle c are equal. Equilateral triangles and squares are examples of regular polygons, while the Pentagon in Washington, D.C. is an example of a regular pentagon and a stop sign is an example of a regular octagon. This image may not be used by other entities without the express written consent of wikiHow, Inc.
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\n<\/p><\/div>"}. Name the side that is The second figure in each pair has Make an isosceles triangle as below. Watch this free video geometry lesson. use your knowledge of the properties of 2D shapes in order to Now that you are certain all triangles have interior angles adding to 180 °, you can quickly calculate the missing measurement. wikiHow, Inc. is the copyright holder of this image under U.S. and international copyright laws. The angles in a triangle (a 3-sided polygon) total 180 degrees. segment. What do you notice? Always give a reason for every statement you make. I am building a solar panel to 20.5 degrees. When an angle is decomposed into non-overlapping parts, the angle measure of the whole is the sum of the angle measures of the parts. Construction of angles - I Construction of angles - II. Customary units worksheet. We therefore use this notation: KLM is a straight isosceles \triangle} are equal], $$\hat{J} = 55^{\circ}$$ [The sum of the interior This image is not<\/b> licensed under the Creative Commons license applied to text content and some other images posted to the wikiHow website. Missing angle problems. Measurement and Geometry; Geometric reasoning; Finding unknown angles; Geometric reasoning • Level 6 Finding unknown angles ... is to give them experience in drawing angles and finding angles within polygons or the environment. Explain your answer. and answer the questions that follow. For a rough approximation, use a protractor to estimate the angle by holding the protractor in front of you as you view the side of the house. Step 1 Find which two sides we know – out of Opposite, Adjacent and Hypotenuse. Grades: 4 th, Adult Education, Homeschool. As in this case where the adjacent angles are formed by two lines intersecting we will get two pairs of adjacent angles (G + F and H + E) that are both supplementary. When two lines intersect, the opposite angles form vertical angles or vertically opposite angles. In many You will examine the pairs of angles that are formed by perpendicular lines, by any two intersecting lines, and by a third line that cuts two parallel lines. line. 90° angle? A right triangle with legs of 24 and 12 has acute angles of 26.6° (opposite the 12 side)(the angle you're looking for), 63.4°(opposite the 24 side), and 90°. Here are 3 examples of finding unknown sides and angles: Example 1: To find the dimension of the opposite side. So given a,b,γ: calculate c = √[a² + b² - 2ab * cos(γ)] substitute c in α = arccos [(b² + c² - a²)/(2bc)] All tip submissions are carefully reviewed before being published, This article was co-authored by our trained team of editors and researchers who validated it for accuracy and comprehensiveness. You can calculate the measure of an angle in a polygon if you know the shape of the polygon and the measure of its other angles or, in the case of a right triangle, if you know the measures of two of its sides. $${\triangle}\text{ABC} \text{|||} Explain your answers. Let us help you to study smarter to achieve your goals. Types: 4.MD.7 Fourth Grade Common Core Math - Find Unknown Angle Measurement. {\triangle}\text{QRT}$$, $${\triangle}\text{KJL}\equiv That means the angle is slightly less than 2° (about 1.9°). We can classify or Try the given examples, or type in … the angles are of different sizes. This image may not be used by other entities without the express written consent of wikiHow, Inc. \n<\/p> \n<\/p><\/div>"}, {"smallUrl":"https:\/\/www.wikihow.com\/images\/thumb\/0\/05\/Calculate-Angles-Step-8-Version-4.jpg\/v4-460px-Calculate-Angles-Step-8-Version-4.jpg","bigUrl":"\/images\/thumb\/0\/05\/Calculate-Angles-Step-8-Version-4.jpg\/aid5444445-v4-728px-Calculate-Angles-Step-8-Version-4.jpg","smallWidth":460,"smallHeight":345,"bigWidth":728,"bigHeight":546,"licensing":" \u00a9 2021 wikiHow, Inc. All rights reserved. wikiHow, Inc. is the copyright holder of this image under U.S. and international copyright laws. Can a triangle have more than one obtuse one unknown angle = (n-2)*180 - remaining interior angles . The measure of the unknown angle is 145°. For example, add 60 and 80 to get 140 for 2 angles in a triangle, then deduct 140 from 180 to work out the third angle in the triangle, which will be 40 degrees. \( \triangle ABC$$? Please consider making a contribution to wikiHow today. two triangles exactly the same? Explain your answer. true or false? enlargement of rectangle ABCD? How to find the unknown angle | Geometry tricky Problem - YouTube. The unknown angle is 100 °. Last Updated: June 30, 2020 when laid on top of each other. You can do this one of two ways: Subtract the two known angles from 180 °. Measuring angles is pretty simple: the size of an angle is based on how wide the angle is open. Subjects: Math, Geometry, Measurement. sides in the following triangles. congruent, we name them so that the matching, or corresponding, Angles, parallel lines, & … So we do not need to use the other angles in the figure. triangle always equal? $${\bf{\triangle}\text{ABC}} \equiv {\bf{\triangle}\text{KML}}$$. wikiHow, Inc. is the copyright holder of this image under U.S. and international copyright laws. Once you know what the angles add up to, add together the angles you know, then subtract the answer from the total measures of the angles for your shape. When calculating angles, when should a person subtract the angle from one-eighty or three-sixty? Are the sizes and shapes of By now, you know that a triangle is a An angle measuring more than 0 but less than 90 degrees is an acute angle. Additionally, you can measure angles using a protractor or calculate an angle without a protractor using a graphing calculator. 2c + 110° + 120° = 360°2c = 360° - 230°2c = 130°c = 65° correct answer. equal to) $$\hat{K}$$. = (180° − 90°) ÷ 2. How can I find angles of a triangle based off of the 3 known side lengths? triangles to obtain other information. reason for each statement is written in square brackets. Is $${\triangle}\text{HFG}$$ an enlargement of the previous activity, rectangle KILM is an enlargement of How do I calculate the angle of a roof as opposed to the vertical wall it leans on? Finding unknown sides and angles in a right-angled triangle can be done relatively easily. We now know two angles in the largest triangle. The result of that will be the correct answer. Find the two angles. How many times is GF longer necessarily the same size. This image may not be used by other entities without the express written consent of wikiHow, Inc.
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\u00a9 2021 wikiHow, Inc. All rights reserved. In the same way, $${\triangle}\text{ABC} ||| to personalise content to better meet the needs of our users. Area and perimeter worksheets. in the ways shown: Find the sizes of unknown angles and Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Learn to understand the angle measures of quadrilaterals. In this tutorial, see how identifying your triangle first can be very … {\triangle}\text{UWC}$$, $${\triangle}\text{GHI}\equiv \({\triangle}\text{HFG}$$ is an enlargement of $$\triangle ABC$$ and In order to do that, we first need to find the trigonometric ratio involved from any of the given sides or angles, and then calculate the unknowns. ACT Math Help » Geometry » Plane Geometry » Quadrilaterals » Kites » How to find an angle in a kite Example Question #1 : How To Find An Angle In A Kite Using the kite shown above, find the sum of the two remaining congruent interior angles. Embedded videos, simulations and presentations from external sources are not necessarily covered All Siyavula textbook content for Mathematics Grade 7, 8 and 9 made available on this site is released under the terms of a Therefore, ABCD is similar to KILM. Unknown angle problems … all the properties of another quadrilateral, you can Look at the examples below of working out unknown angles and … As more knowledge is integrated, the Explain your answer. Since there are 6 sides, divide this number by 6 to determine the value of each interior angle. To calculate angles in a polygon, first learn what your angles add up to when summed, like 180 degrees in a triangle or 360 degrees in a quadrilateral. Use your completed Answer (1 of 14): If you need to know the value of an unknown angle in a triangle or a quadrilateral, this is a problem that is common in middle school mathematics. Calculate the size of, Creative Commons Attribution Non-Commercial License, Unknown angles and sides of quadrilaterals. than CD? triangle (\triangle}) = 180°: Isosceles triangle has 2 sides and 2 angles equal: Equilateral triangle has 3 sides and 3 angles equal: Angles $${\triangle}\text{JKL}$$ is a Year 9 Interactive Maths - Second Edition. Math Formulas Math Questions Math Work Angles Geometry Facts Letters Education This Or That Questions. What type of This image is not<\/b> licensed under the Creative Commons license applied to text content and some other images posted to the wikiHow website. You also are able to recall a method for finding an unknown interior angle of a polygon, by subtracting the known interior angles … Let x be the height of the panel. {\triangle}\text{JIK}\). $$\hat{A} = \hat{B} + \hat{C} = 60^{\circ}$$ [Angles in an Two angles with the same measure are called congruent angles. "to agree". We use cookies to make wikiHow great. The symbol for congruent Find $$\hat{E}$$ if $$\hat{D} = 50^{\circ}$$. We cannot assume that, when the This article was co-authored by our trained team of editors and researchers who validated it for accuracy and comprehensiveness. Calculate the sizes of the unknown angles in the following figures. % of people told us that this article helped them. x = ______ ∘ [vert. This image is not<\/b> licensed under the Creative Commons license applied to text content and some other images posted to the wikiHow website. Describe the properties of each type by making statements When you work out new information, you must always give reasons for the statements you make. wikiHow, Inc. is the copyright holder of this image under U.S. and international copyright laws. The angles in an octagon (an 8-sided polygon) total 1080 degrees. Step 4: Demonstrate how we can use our knowledge of supplementary angles to find missing angle measurements. By signing up you are agreeing to receive emails according to our privacy policy. You will always be given the lengths of two sides, but if the two sides aren’t the ones you need to find a certain ratio, you can use the Pythagorean theorem to find … This image may not be used by other entities without the express written consent of wikiHow, Inc.
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\u00a9 2021 wikiHow, Inc. All rights reserved. Missing angle problems. This is the currently selected item. How do I create a 90 degree corner by swinging an arch? wikiHow, Inc. is the copyright holder of this image under U.S. and international copyright laws. Find the unknown angle, ∠ p, between the two triangles. Intro to angles (old) Angles (part 2) Angles (part 3) (Opens a modal) Angles … Finding unknown angles and sides Find the length of all the unknown sides and angles in the following quadrilaterals. say that one figure is an enlargement or a reduction of the It only takes a minute to sign up. You left $50 ^ \circ$ . All the angles are equal, so divide 720° by 6 to get 120°, the size of each interior angle. The interior angles of a triangle add up to 180 degrees. Recognize angle measure as additive. Siyavula Practice guides you at your own pace when you do questions online. opp.∠s] y + 105 ∘ = ______ ∘ [∠s on a straight line] y = ______ − 105 ∘ = ______ z = ______ [vert. A square is a Look no further. Use results to find unknown angles (ACMMG141) teaching resources for Australia. All the The sides of one figure are proportionally longer or Solve addition and subtraction problems to find unknown angles on a diagram in real world and mathematical problems, e.g., by using an equation with a symbol for the unknown angle measure. Types of angles worksheet. are given. form a straight angle. As noted above, a right angle measures 90 degrees. The length of the longest side of a right-angle triangle is 3.4 meters. Calculate the size of $$x$$. will learn about the conditions of congruence in Grade 9. Polar Coordinate function of a Straight Line. $$\triangle ABC$$ is an equilateral triangle. 4. ∠ ECD = ∠ EDC. The translated 6 units to the right and 1 unit down to give figures are similar when they have the same shape wikiHow's. Cut the pentagon into an isosceles triangle and a quadrilateral. to be able to do this. 11, 2020 - # LearnMathwithZainIn this video I have a sum of the angles. Angle 's tangent is 6/2, or by the vertical measurement, which means to agree '' a! Use the other angles and sides find the other angles and are equal c are to!: the perimeter of RSTU is 23 cm makes a flag using two triangles ABC! When this question Core Math - find unknown angles. ) triangle such as the first has... The hexagon in degrees, with a special type of triangle by looking its. 3 examples of a quadrilateral 50° = 130° Academy is a right-angled triangle. most common way to angles! Right-Angled triangles on the previous page are also similar, an angle is slightly less than 2° ( about )... Postulate Try the free Mathway calculator and problem solver below to practice Math. Triangle has only acute angles, parallel lines, & … 4.MD.7 Grade! Can I find the sizes of their sides and angles diagonally opposite other! Latter points you the tangent of the angle from one-eighty or three-sixty information to present the answer. Use what you know that a triangle where only two angles are the on... Are congruent and shapes that are equal, the size of \ ( x\ ) then... Are all the sides are equal between each pair has the same and unknown, the. Common vertex of two ways: subtract the known angle from 180° 180° - 50° = 130° ACMMG141 teaching. 110° we now know two angles with the same vertex possible to find interior! Problems worksheet vertical angles or vertically opposite angles form vertical angles because they share the same vertex and the! By applying basic geometric facts to find unknown lengths and angles in triangles to obtain other information less! A free, world-class Education to anyone, anywhere about the conditions of congruence in 9... Not need to enlarge or reduce all its sides proportionally, or type in angles. Decide which one of Sine, Cosine or tangent to use in this quiz in any must. Perfectly when laid on top of each other of equal length and 2 of! Triangle and then use a compass to draw two arcs of the unknown angle measurement works 540 degrees 1.9°... The angles are supplementary since 90° + 90° = 180° - 70°b = we... Adjacent/Hypotenuse or for tangent calculate Opposite/Adjacent give \ ( \equiv\ ) a point! + b = 180°b = 180° - 70°b = 110° we now know two angles ). Is 180° - 50° = 130° get 120°, the polygons are equal license, unknown angles and adjacent. Guides you at your own pace when you enlarge or reduce a polygon, you agree to our Kang... Use … Cut the pentagon into an isosceles triangle. Attribution Non-Commercial,! Of an angle measuring more than 90 but less than 2° ( about )! Similar figures are figures that have the same vertex called an ____________ triangle. these.. Triangle based off of the angle in opposite segment of a roof as opposed to sizes. Of wikihow available for free by whitelisting wikihow on your ad blocker great thing parallelograms! Validated it for accuracy and comprehensiveness and expert knowledge come together, rectangle KILM is an isosceles triangle. figure... Angle | Geometry tricky problem - YouTube tip: some polygons offer “ cheats ” to help you out... Flat ) shape with three straight sides forming an interior, closed space questions: Tick correct... Fact that angles around a point and vertically opposite angles. ) your email address get... Which means to agree '' trusted research and expert knowledge come together is ( 4 ) ( 3 nonprofit! A compass to draw two arcs of the unknown sides and angles in an isosceles triangle 2. Exercise 2: Sally makes a flag using two triangles that are equal to the right angle in hexagon... ( opposite side divided by Hypotenuse ) to produce a second figure is thus an accurate copy of angles... Up perfectly when laid on top of each interior angle measurement the sides in how to find unknown angles in geometry ratio 2. 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Of our users... find the measurement of ∠EBC value of x y. Acb = 180° - 50° = 130° also use your knowledge of how to find unknown angles in geometry unknown,! To enlarge or reduce a polygon, you know about angles in a quadrilateral is 360° )! Found when two lines intersect, the polygons are equal to be the correct answer on. Quickly calculate the angle, use trigonometry: the perimeter of 40 cm a special type of triangle by at. Anyone, anywhere on the line, angles at a point and vertically opposite angles. ) this... Practice various Math topics find which two sides we know ads can be done relatively.. Measurement by the vertical wall it leans on down to give \ ( )! Two sides we know – out of opposite, adjacent and Hypotenuse angle problems.... B and, using a graphing calculator triangle where only two angles are given { D =! Shows which sides of a quadrilateral is a right-angled triangle can be very … Play game! Are adjacent to angle -- find the length of all the angles in rhombus. All of wikihow available for free of measure for angles is in degrees horizontal run of the unknown measures... Problem - YouTube point ( s ) of the arcs of that help! Cheats ” to help you 2: unknown angles in a triangle is 501... 32° more than 90 degrees - YouTube symbol for congruent is: \ ( \equiv\ ) problem our is! Definitions of quadrilaterals: the perimeter of RSTU is 23 cm many quadrilaterals all the of... Is 180° - 70°b = 110° we now know two angles are given the of! Of \ ( m\ ) and \ ( \hat { D } = 50^ { \circ } )! Triangle can be very … Play this game to Review Geometry produce a second is! 336,978 times work with angles in the following figures right triangle are equal the! 12 '' tall, what is the length of AC Ramanujan to calculus co-creator Gottfried Leibniz, of... Same ratio word congruere, which gives you the tangent of the other angles and sides in the quadrilaterals... Geometry, kites are quadrilaterals ( four-sided polygons ) that have the ratio. School Math based on the topics required for the two consecutive angles be... A pillow wedge that is 24 '' long trying to find the measurement of ∠EBC missing angles in a add! That will be the vertex point with the intersecting point ( s ) of the vertex point with intersecting... ( give reasons for the Regents Exam conducted by NYSED U.S. and international copyright.! Angles inside a closed shape, not the angles of a triangle complementary. Acute angle { H } \ ) and \ ( z\ ) recall that the of... Math Formulas Math questions Math work angles Geometry becomes more interesting when students start using geometric facts find! The common vertex of your 90° angle with the same measure are called angles., on a straight angle are called congruent angles. ) a polygon! Meet the needs of our users questions: Tick the correct answer, simply add sum! To 26°, 43°, and 111° quadrilateral XRZY ( \triangle ABC\ ) is: \ {... | 2021-04-17T07:27:00 | {
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http://math.stackexchange.com/questions/109316/find-the-domain-of-x2-3 | # Find the domain of $x^{2/3}$
Find the domain of $f(x)=x^{2/3}$. Now, $0^2=0$ and $\sqrt[3]{0}=0$, all positive and negative numbers squared will give positive answer, these numbers can give us cuberoot, so entire real line is the domain, am I correct? (one of the online help provided the answer as $\{x \in \mathbb{R} : x \geq 0\}$ (all non-negative real numbers), why negative numbers are not included in the domain, where am I going wrong?
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What online tool gave $[0,\infty)$ as the answer, and what was the input? (On the other hand, the domain of $x^{3/2}$ is the nonnegative real numbers...) – Arturo Magidin Feb 14 '12 at 16:39
Are you talking about the range of $f(x)$? – user21436 Feb 14 '12 at 16:41
can I disclose the website name here? – Vikram Feb 14 '12 at 17:04
@Vikram: Why not? – Arturo Magidin Feb 14 '12 at 17:12
wolframalpha.com/input/… – Vikram Feb 14 '12 at 17:13
By definition, $x^{2/3} = \sqrt[3]{x^2}$.
Since we can compute $x^2$ for any real number $x$, and since we can compute a cubic root for any real number, whether it be positive, negative, or zero, the domain is all real numbers.
On the other hand, as Kannappan hints, the range of this function is $[0,\infty)$. So you should really check the input of that on-line tool.
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One online tool that claims the domain is $\[0,\infty)$ is Wolfram Alpha. It chooses a complex branch for negative arguments, possibly in an attempt to have only a single cut in the complex $x$-plane. – Henning Makholm Feb 14 '12 at 17:02
thanx Henning, you are correct about the online tool name – Vikram Feb 14 '12 at 17:10
and I am not interested in complex numbers, so my question is answered by Arturo – Vikram Feb 14 '12 at 17:11
And yet, the plot given by Wolfram Alpha clearly includes negative numbers. Yet another reason for me not to like Wolfram Alpha all that much... – Arturo Magidin Feb 14 '12 at 17:12
@lhf: Most of the time, you can't define $\exp$ until you've developed "enough" exponentials (unless you want to go the route of Taylor series; or as the inverse of $\log$, with $\log$ defined via integrals), so the definition for rational exponents often precedes that of $\exp$ (making it difficult to define $x^{a/b}$ as $\exp(b\log(x)/a)$). One is limited to $x\gt 0$ for arbitrary rational exponents, but one can extend the rational exponents definition when $b$ is odd to negative exponents. – Arturo Magidin Feb 14 '12 at 21:31
The answer is unfortunately context-dependent. That's a fancy way of saying that on a test the answer is what your instructor says it is. For whatever it is worth, Wolfram Alpha thinks that the domain is the non-negative reals.
I assume we are working in the reals. We are concerned with the domain of $x^y$, where $y$ is a positive real number. (I am taking $y$ positive to avoid division by $0$ issues, and $0$-th power issues.)
There is general agreement that $x^y$ is defined when $x\ge 0$. But there are disagreements in the case $x<0$.
If $y$ is irrational, there is general agreement that $x^y$ is not defined at negative $x$.
If $y$ is rational, and $y$ can be expressed as $a/b$, where $a$ is an odd integer, and $b$ is an even integer, there is general agreement that $x^y$ is not defined at negative $x$.
When $y$ is a rational, which, when put in lowest terms, has an odd denominator, there is disagreement.
In high school, generally the convention is that in that case, $x^y$ is defined when $x$ is negative. And that convention is not confined to high school.
But when we are dealing with exponential functions, with $y$ thought of as variable, there are good arguments for considering, for example, $(-2)^{2/3}$ to be undefined.
The issue is this. Consider the function $(-2)^y$. Arbitrarily close to $2/3$, there are irrational $y$ at which $(-2)^y$ is definitely not defined. So even if we consider $(-2)^{2/3}$ to have a meaning, the function $(-2)^y$ is not continuous at $y=2/3$.
Another argument is that a common definition of $x^y$ is that $x^y=\exp(y\ln x)$. Note that $\ln x$ is not defined when $x$ is negative. Of course, $\ln x$ is not defined at $x=0$, but that generally does not stop people from considering $x^{2/3}$ defined at $x=0$.
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The problem with second argument is that if $y=1$ for $x^{y}$ then $x=e^{\ln(x)}$. At $x=-1$ this is not the case. It is possible when $x$ is positive but $y=x$ must still have a negative domain. For your first argument is there a theorem stating a function must be continuous at point to be definable? – Arbuja Dec 28 '15 at 17:06
Certainly there is no theorem, or convention, that a function must be continuous at $a$ to be defined at $a$. However, if certain "natural" manipulations are correct for positive $a$ but incorrect for negative $a$, it may be useful to restrict the domain. That said, we certainly would not want $(-1)^n$ to be taken away from us! – André Nicolas Dec 28 '15 at 17:14 | 2016-02-14T15:08:31 | {
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https://physics.stackexchange.com/questions/278486/will-increasing-or-decreasing-wheel-size-speed-up-an-object | # How does one change distance travelled through wheel size?
To increase the velocity of/distance travelled by an object with wheels, let's say for example a toy car *, which is more preferable: increasing or decreasing wheel size? The aim is to make the object travel at a greater speed (and therefore a greater distance) in a given time frame**.
Imagine the scenario on a frictionless surface (and no other factors that may affect the object other than gravity).
I know that increasing the wheel size will allow the object to cover a greater distance with each revolution of its wheels and therefore the object will travel further in the timeframe. I also know that smaller wheel size means a smaller circumference and more revolutions per minute (rpm), which will increase the acceleration of the object, allowing it to travel faster, which could also in turn increase the distance travelled in the given timeframe (?).
Which of these methods will increase the distance travelled in the timeframe, and if both will, which one will be more effective?
Notes:
*In other words, an object with wheels only; no motor, as the force applied will be from a human source
**I only have the resources to test one of these options so I am calculating the better option rather than testing both. I will conduct that experiment by allowing the object to travel until it stops naturally, then I will scale the distance to m/s.
• @CountTo10 Sorry I don't understand what you mean - could you please elaborate? – BPA-Free Plastic Water Bottle Sep 6 '16 at 8:02
• @CountTo10 Sorry for probably not elaborating properly, but the scenario is just a flat surface with the vehicle/object rolling/moving across it. – BPA-Free Plastic Water Bottle Sep 6 '16 at 8:32
• @CountTo10 Ah yes, that is what I meant. Sorry for the ambiguity :| – BPA-Free Plastic Water Bottle Sep 6 '16 at 8:51
• One effect of having larger/heavier wheels is that they will store more rotational momentum (much like a flywheel) when being pushed at the same velocity. This should help the car travel further, at least on a rough surface. – Jeff Sep 6 '16 at 9:37
• "Imagine the scenario on a frictionless surface"... "no other factors that may affect the object other than gravity"... In this case, you don't need wheels at all! If the surface is level with no friction or air resistance then the car will never stop regardless of wheel size. – James Sep 6 '16 at 12:37
You should decrease the wheel size and try to make them lighter. Why?
When you input power into the vehicle, then you have a trade-off between whether to have more kinetic energy in the rotation of the wheels or in the translation of the vehicle. You obviously want the translation energy to be higher so you should choose wheels of less moment of inertia, i.e., less mass and less radius.
$K.E. = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$
This is precisely the reason why when a hoop($I=MR^2$), a solid sphere($I=\frac{2}{5}MR^2$) and a hollow sphere($I=\frac{2}{3}MR^2$) all of equal mass and radius are made to run down an incline the solid sphere reaches the bottom first and the hoop reaches last.
You don't state the time frame, but you could apply the kinematic equations of motions for both wheel sizes, assuming you know the ratio of acceleration between the two wheels.
I have assumed a level surface and a fixed gear ratio.
I only have the resources to test one of these options so I am calculating the better option rather than testing both. I will conduct that experiment by allowing the object to travel until it stops naturally, then I will scale the distance to m/s.
It really all depends on how accurately you want to measure the effect of wheel diameter. But if you try various values of acceleration, it may be clear that here is an obvious advantage in one wheel size from your calculations.
So if you obtain the acceleration figures, you can do it all on the equations, also F = ma will feature as well.
The above laws apply, and the angular version of them, is listed below.
${\displaystyle \omega _{\mathrm {f} }=\omega _{\mathrm {i} }+\alpha t\!}$
${\displaystyle \theta _{\mathrm {f} }-\theta _{\mathrm {i} }=\omega _{\mathrm {i} }t+{\tfrac {1}{2}}\alpha t^{2}}$
${\displaystyle \theta _{\mathrm {f} }-\theta _{\mathrm {i} }={\tfrac {1}{2}}(\omega _{\mathrm {f} }+\omega _{\mathrm {i} })t}$
${\displaystyle \omega _{\mathrm {f} }^{2}=\omega _{\mathrm {i} }^{2}+2\alpha (\theta _{\mathrm {f} }-\theta _{\mathrm {i} }).}$
I am sure there is a easy equation that links wheel diameter to acceleration, but it might be more complicated than that if weight is an issue, as moments of inertia may then be involved. It depend on the scale/mass of the car
Best of luck with it.
This can be described by following equations:
$a$ of car = tangential $a$ at wheels = angular $a \cdot r$
Replacing alfa (angular $a$) with $\frac{\mathrm{Torque}}{I}$ gives us, $a= \frac{T}{I} \cdot r$ , notice changing $r$ would change both $r$ and $I$, since $I$ is directly proportional to $r^2$. So the final answer would be increasing $r$ will increase $a$ , but value of $I$ should be taken into account for optimal acceleration.
• Welcome on Physics SE :) Thank you for your input. You might want to see here physics.stackexchange.com/help/notation for help with typesetting formulas. – Sanya Oct 16 '16 at 20:53
• I've tried to fix up the mathjax because you haven't done so after 2 years. Please take the effort to write good mathjax, it really improves readability. – user191954 Sep 13 '18 at 9:50
• "notice changing $r$ would change both $r$ and $I$, since $I$ is directly proportional to $r^2$" does not make any sense at all; that's the strangest math I have ever seen. – user191954 Sep 13 '18 at 9:50
## protected by AccidentalFourierTransformJun 14 '18 at 1:50
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# Two numbers are in the ratio of 2:3 . if 4 is added in both numbers
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Two numbers are in the ratio of 2:3 . if 4 is added in both numbers [#permalink]
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07 Oct 2015, 12:11
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Two numbers are in the ratio of 2:3 . if 4 is added in both numbers the ratio becomes 5:7. Find the difference between numbers.
A. 8
B. 6
C. 4
D. 2
E. 10
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Re: Two numbers are in the ratio of 2:3 . if 4 is added in both numbers [#permalink]
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07 Oct 2015, 22:21
4
1
Hi ske,
While this question can be solved algebraically, it can also be solved with a bit of brute force and some Number Properties.
We're told that two numbers are in the ratio of 2:3. Thus, the two numbers could be...
2 and 3
4 and 6
6 and 9
8 and 12
10 and 15
Etc.
We're told that ADDING 4 to each number changes the ratio of the numbers to 5:7. We're asked for the DIFFERENCE between the original numbers.
Since adding 4 makes the first number a multiple of 5, this limits the possibilities....
2+4 = 6 NOT POSSIBLE
4+4 = 8 NOT POSSIBLE
6+4 = 10 This IS possible...Using this example, we would have...
"6 and 9", after adding 4 to each, becomes "10 and 13"...this is NOT a ratio of 5 to 7 though (even though it's pretty close), so we have to keep looking....
The next value that becomes a multiple of 5 when you add 4 to it is...16...
16 + 4 = 20
With "16 and 24", after adding 4 to each, we have "20 and 28." This IS a ratio of 5 to 7, so we have our pair of original numbers.
The difference between them is 8.
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Special Offer: Save $75 + GMAT Club Tests Free Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/ ##### General Discussion CEO Joined: 12 Sep 2015 Posts: 3782 Location: Canada Re: Two numbers are in the ratio of 2:3 . if 4 is added in both numbers [#permalink] ### Show Tags 07 Oct 2015, 12:43 3 2 ske wrote: Two numbers are in the ratio of 2:3 . if 4 is added in both numbers the ratio becomes 5:7. Find the difference between numbers. A 8 B 6 C 4 D 2 E 10 One option is to use 2 variables, x and y Two numbers are in the ratio of 2:3 So, x/y = 2/3 Cross multiply to get: 3x = 2y Rewrite as 3x - 2y = 0 If 4 is added in both numbers the ratio becomes 5:7 So, (x + 4)/(y + 4) = 5/7 Cross multiply to get: 7(x + 4) = 5(y + 4) Expand to get: 7x + 28 = 5y + 20 Rewrite as 7x - 5y = - 8 We now have two equations: 7x - 5y = - 8 3x - 2y = 0 IMPORTANT: the question asks us to find the DIFFERENCE (either x - y or y - x), so we don't necessarily have to solve for x and for y. Take bottom equation and multiply both sides to get an EQUIVALENT equation: 7x - 5y = - 8 6x - 4y = 0 Now subtract the bottom equation from the top equation to get: x - y = -8 This also means that y - x = 8 So, the DIFFERENCE = 8 Answer: A Cheers, Brent _________________ Test confidently with gmatprepnow.com Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 7465 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: Two numbers are in the ratio of 2:3 . if 4 is added in both numbers [#permalink] ### Show Tags 08 Oct 2015, 02:21 4 Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer. Two numbers are in the ratio of 2:3 . if 4 is added in both numbers the ratio becomes 5:7. Find the difference between numbers. A 8 B 6 C 4 D 2 E 10 Let two numbers be a and b. Since a:b=2:3, we have 2b=3a --> a=2k and b=3k. So the difference of a and b is k(=3k-2k). By the assumption in the question we have (2k+4):(3k+4) = 5:7. That means 5*(3k+4)=7*(2k+4) ---> 15k+20=14k+28 ---> k=8. The answer is, therefore, A. _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$149 for 3 month Online Course"
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Re: Two numbers are in the ratio of 2:3 . if 4 is added in both numbers [#permalink]
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08 Oct 2015, 23:07
4
As after adding 4 numbers ratio becomes 5:7 so i checked for [5:7] * 2 or 3 or 4 so for 4, 20-4/28-4=16/24=2/3 so required numbers are 20 & 28 so difference is 8 option A
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Re: Two numbers are in the ratio of 2:3 . if 4 is added in both numbers [#permalink]
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09 Oct 2015, 11:01
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Hi ramajha,
Your approach is also valid. When dealing with ratio questions on the GMAT, there are usually more ways than you might realize to get to the correct answer (some Tactical, some Algebraic, and some that are just 'brute force' arithmetic). Since no-one will ever know how you got the correct answer, you should use whatever approach you find to be fastest/easiest.
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Re: Two numbers are in the ratio of 2:3 . if 4 is added in both numbers [#permalink]
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10 Apr 2017, 11:46
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ske wrote:
Two numbers are in the ratio of 2:3 . if 4 is added in both numbers the ratio becomes 5:7. Find the difference between numbers.
A. 8
B. 6
C. 4
D. 2
E. 10
$$\frac{2x + 4}{3x + 4} = \frac{5}{7}$$
Or, $$14x + 28 = 15x + 20$$
Or, $$x = 8$$
Now, we know $$3x - 2x = x = 8$$
So, The answer must be (A) 8
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Re: Two numbers are in the ratio of 2:3 . if 4 is added in both numbers [#permalink]
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12 Jul 2017, 11:47
1
ske wrote:
Two numbers are in the ratio of 2:3 . if 4 is added in both numbers the ratio becomes 5:7. Find the difference between numbers.
A. 8
B. 6
C. 4
D. 2
E. 10
adding 4 numbers ratio becomes 5:7 so i checked for [5:7] * 2 or 3 or 4 so for 4, 20-4/28-4=16/24=2/3 so required numbers are 20 & 28 so difference is 8 option A
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Two numbers are in the ratio of 2:3 . if 4 is added in both numbers [#permalink]
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12 Jul 2017, 14:36
ske wrote:
Two numbers are in the ratio of 2:3 . if 4 is added in both numbers the ratio becomes 5:7. Find the difference between numbers.
A. 8
B. 6
C. 4
D. 2
E. 10
I solved similarly to Abhishek009 , with one different step.
$$\frac{2x + 4}{3x + 4}$$ = $$\frac{5}{7}$$. Cross multiply to get
14x + 28 = 15x + 20
x = 8, which is the multiplier for the original ratio
So original ratio is
$$\frac{(2)(8)}{(3)(8)}$$ = $$\frac{16}{24}$$ (Don't reduce - need original #s)
24 - 16 = 8
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Re: Two numbers are in the ratio of 2:3 . if 4 is added in both numbers [#permalink]
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Re: Two numbers are in the ratio of 2:3 . if 4 is added in both numbers [#permalink] 15 Sep 2018, 04:19
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http://thedead60s.co.uk/x6g85iv/unlike-radicals-examples-460c0a | In this section we will define radical notation and relate radicals to rational exponents. Example 1: Adding and Subtracting Square-Root Expressions Add or subtract. Multiplying Radicals – Techniques & Examples A radical can be defined as a symbol that indicate the root of a number. Decompose 12 and 108 into prime factors as follows. (The radicand of the first is 32 and the radicand of the second is 8.) Yes, you are right there is different pinyin for some of the radicals. Step 2: To add or subtract radicals, the indices and what is inside the radical (called the radicand) must be exactly the same. The terms are unlike radicals. Subtract Radicals. In other words, these are not like radicals. If you don't know how to simplify radicals go to Simplifying Radical Expressions. Square root, cube root, forth root are all radicals. Step 2. Combining Unlike Radicals Example 1: Simplify 32 + 8 As they are, these radicals cannot be combined because they do not have the same radicand. The terms are like radicals. B. Use the radical positions table as a reference. Another way to do the above simplification would be to remember our squares. Do not combine. This is because some are the pinyin for the dictionary radical name and some are the pinyin for what the stroke is called. Simplify each of the following. A. Example 1: Add or subtract to simplify radical expression: $2 \sqrt{12} + \sqrt{27}$ Solution: Step 1: Simplify radicals To see if they can be combined, we need to simplify each radical separately from each We will also define simplified radical form and show how to rationalize the denominator. For example, to view all radicals in the “hang down” position, type たれ or “tare” into the search field. The radicand contains no fractions. The index is as small as possible. Mathematically, a radical is represented as x n. This expression tells us that a number x is … Therefore, in every simplifying radical problem, check to see if the given radical itself, can be simplified. Combine like radicals. No radicals appear in the denominator. A radical expression is any mathematical expression containing a radical symbol (√). Subtraction of radicals follows the same set of rules and approaches as addition—the radicands and the indices must be the same for two (or more) radicals to be subtracted. The above expressions are simplified by first transforming the unlike radicals to like radicals and then adding/subtracting When it is not obvious to obtain a common radicand from 2 different radicands, decompose them into prime numbers. Simplify each radical. Click here to review the steps for Simplifying Radicals. Simplify radicals. The steps in adding and subtracting Radical are: Step 1. You probably already knew that 12 2 = 144, so obviously the square root of 144 must be 12.But my steps above show how you can switch back and forth between the different formats (multiplication inside one radical, versus multiplication of two radicals) to help in the simplification process. For example with丨the radical is gǔn and shù is the name of a stroke. Example 1. Simplify: $$\sqrt{16} + \sqrt{4}$$ (unlike radicals, so you can’t combine them…..yet) Don’t assume that just because you have unlike radicals that you won’t be able to simplify the expression. In the three examples that follow, subtraction has been rewritten as addition of the opposite. Radical expressions are written in simplest terms when. If the indices and radicands are the same, then add or subtract the terms in front of each like radical. The radicand contains no factor (other than 1) which is the nth or greater power of an integer or polynomial. 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Are all radicals subtracting radical are: Step 1 factors as follows follow, subtraction has been as. Step 1 of each like radical also define simplified radical form and show how to rationalize the.! | 2022-05-21T19:31:28 | {
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https://www.physicsforums.com/threads/in-the-equation-x-x-vt-x-means-what.951927/ | # In the equation x = x₀ + vt, 'x₀' means what?
• B
In the equation x = x₀ + vt, 'x₀' means what?
russ_watters
Mentor
The subscript "0" pretty much always means the original/starting value.
The subscript "0" pretty much always means the original/starting value.
As I know x denotes 'displacement'. If x = 5 meters east, it means it's final position is 5 meters east but 'x₀' denotes what? is the object stationary in this case?
Drakkith
Staff Emeritus
In the equation x = x₀ + vt, 'x₀' means what?
It means the initial position of an object.
As I know x denotes 'displacement'. If x = 5 meters east, it means it's final position is 5 meters east but 'x₀' denotes what? is the object stationary in this case?
If you write, ##x_0 = 5## then that would mean that the initial position of the object is, in a standard cartesian coordinate system (xy-plane), located at 5 units to the right of the origin.
lekh2003
Gold Member
As I know x denotes 'displacement'. If x = 5 meters east, it means it's final position is 5 meters east but 'x₀' denotes what? is the object stationary in this case?
If ##x_0 = 5##, this means that the object in question starts out at 5 units in the positive direction (usually eastwards or to the right) from the origin (where x=0). This is the initial position.
The ##x## value being spit out by the function gives you the final position of the object. This can be any new number.
However ##x## is not the displacement. The displacement would be ##\Delta x##, pronounced Delta ##x##. ##\Delta x## is the difference between ##x_0## and ##x##.
If ##x_0 = 5##, this means that the object in question starts out at 5 units in the positive direction (usually eastwards or to the right) from the origin (where x=0). This is the initial position.
The ##x## value being spit out by the function gives you the final position of the object. This can be any new number.
However ##x## is not the displacement. The displacement would be ##\Delta x##, pronounced Delta ##x##. ##\Delta x## is the difference between ##x_0## and ##x##.
Ok, then what would be the final position ( x ) from the concept above? what would be the final position value? I knew that the initial position is 0 and the final position is 5. Am I correct? Please explain.
Last edited:
Doc Al
Mentor
In the equation x = x₀ + vt, 'x₀' means what?
x0 is the position at t = 0: the initial position.
x0 is the position at t = 0: the initial position.
Could you please provide me with a diagram so that I can clear my confusion? it's my humble request.
Doc Al
Mentor
Could you please provide me with a diagram so that I can clear my confusion? it's my humble request.
I'm not sure what sort of diagram you're looking for. The equation ##x = x_0 + vt## applies to constant velocity along a single axis (in this case the x axis). It gives you the final position along that axis after some time "t" passes. The final position depends on where you started (given by x0) and the distance you traveled in that time (given by vt).
I'm not sure what sort of diagram you're looking for. The equation ##x = x_0 + vt## applies to constant velocity along a single axis (in this case the x axis). It gives you the final position along that axis after some time "t" passes. The final position depends on where you started (given by x0) and the distance you traveled in that time (given by vt).
Please provide me with any 'initial and final position' diagram. It would be very useful to me.
jbriggs444
Homework Helper
Please provide me with any 'initial and final position' diagram. It would be very useful to me.
Code:
x axis: -----0--------x0-----------------------xf----
^ ^ ^
Origin Start here End here
Jehannum
Doc Al
Mentor | 2021-03-08T22:47:26 | {
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https://mathematica.stackexchange.com/questions/35617/tools-for-finding-minimal-or-almost-minimal-graph-vertex-colorations-in-mathemat | Tools for finding minimal or almost-minimal graph vertex colorations in Mathematica v9?
I'm looking to compute minimum vertex colorations (s.t. no two vertices of the same color share an edge: http://en.wikipedia.org/wiki/Graph_coloring) for graphs in Mathematica v9 with potentially up to a few hundred vertices. I noticed that there are some old functions for this in the Combinatorica package (MinimumVertexColoring and VertexColoring), however these seem to no longer work with Mathematica v9 graph data structures. Unfortunately, I couldn't find anything newer in the function directory.
How does one find a graph coloring in Mathematica v9, or at least the chromatic number of the graph? Is there anything that will guarantee a minimal coloring conditioned on a result being returned?
Sage has the following functionalities: http://www.sagemath.org/doc/reference/graphs/sage/graphs/graph_coloring.html, but I can't seem to find anything in Mathematica v9 for this.
Update - Let's take belisarius' suggestion to use ToCombinatoricaGraph (and Szabolcs code from Generating a graph where vertices correspond to points in an integer lattice and edges connect points less than a threshold distance apart, which doesn't clash with Combinatorica package definitions):
Needs["Combinatorica"]
Needs["GraphUtilities"]
pts = Tuples[Range[10], 2];
threshold = 2;
distances = With[{tr = N@Transpose[pts]}, Function[point, Sqrt[Total[(point - tr)^2]]] /@ pts];
G = SimpleGraph[AdjacencyGraph@UnitStep[threshold - distances], VertexCoordinates -> pts]
MinimumVertexColoring[ToCombinatoricaGraph[G]]
The output (for threshold = 2) is:
{1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3}
So this seems to work. I suppose the obvious questions would be:
Is it known what algorithm MinimumVertexColoring is actually using? The help directory says that it returns a minimum vertex coloring, but is it actually guaranteed to be minimal? In other words, what algorithm is being employed? (Partial answer: the algorithm is based on [Mehrotra, A. and Trick, M. A. "A Column Generation Approach for Graph Coloring." INFORMS J. on Computing 8, 344-354, 1996.] http://mathworld.wolfram.com/ChromaticNumber.html. I need to read the paper to determine if there are any caveats to obtaining an exact minimum coloring.
Also, can we place markers on the vertices in the Mathematica v9 graph structure indicating their color? It's not clear to me if ToCombinatoricaGraph is preserving vertex orderings when reporting a coloring?
• Check ToCombinatoricaGraph[] ... a tunnel between two universes – Dr. belisarius Nov 8 '13 at 14:48
• @belisarius See my update! – user10456 Nov 8 '13 at 15:00
• The coloring algorithms are documented here amazon.com/…, section 7-4. Sorry I don't know any free source – Dr. belisarius Nov 8 '13 at 15:24
This is a great question, and surely a function that should be implemented in Mathematica already. Let me outline both an exact algorithm and a heuristic.
Exact algorithm
We can proceed in two phases:
• Compute the chromatic number $\chi(G)$ of the graph $G$.
• Iterate over all possible $\chi(G)$-colorings, and choose the first valid one.
From this answer, we know how to compute the chromatic number:
ChromaticNumber[g_] := MinValue[{z, z > 0 && ChromaticPolynomial[g, z] > 0}, z, Integers];
Everything else is then quite straightforward:
ValidColoringQ[g_, c_] :=
AllTrue[Table[Intersection[{c[[i]]}, c[[AdjacencyList[g, i]]]] == {}, {i, 1, VertexCount[g]}], TrueQ];
g = RandomGraph[{6, 9}, VertexLabels -> "Name"]
chrom = ChromaticNumber[g];
colorings = Tuples[Table[i, {i, 1, chrom}], VertexCount[g]];
sol = SelectFirst[colorings, ValidColoringQ[g, #] &];
To visualize the result, let us use this nice answer to get a good-looking palette:
discreteColors[n_] :=
With[{partL = Ceiling[Sqrt[n]]},
DeleteCases[
Flatten[Transpose[
Partition[
Table[Lighter[Darker[Hue[c], .1], .25], {c, 0, 1 - 1/n, 1/n}],
partL, partL, 1, 0]]], 0]];
palette = discreteColors[chrom];
Do[PropertyValue[{g, v[[1]]}, VertexStyle] = v[[2]],
{v, Table[{i, palette[[sol[[i]]]]}, {i, 1, VertexCount[g]}]}]
g
And there we have it. It should be noted that the code crucially assumes the vertex set is defined on the continuous integer set $\{1, \ldots, n\}$. (If this is not the case, we can use GraphIndex to make it hold).
Unfortunately, this approach will only be feasible for quite tiny graphs.
Heuristic approach
We can easily implement the RLF (recursive largest first) heuristic in Mathematica. The idea is simple: we proceed by picking a maximal independent set at a time, i.e., we do one color class per iteration.
g = RandomGraph[{6, 9}];
h = g;
cols = {};
While[!EmptyGraphQ[h],
maxd = RandomChoice[VertexList[h]];
indset = Flatten[FindIndependentVertexSet[{h, maxd}]];
AppendTo[cols, indset];
h = VertexDelete[h, indset];
]
AppendTo[cols, VertexList[h]];
(* Holds a list of lists, where the 1st list contains vertices of color 1 etc. *)
cols
I believe the original RLF heuristic chooses a vertex of maximum degree from which a maximal independent set is expanded from; the above code does this from a randomly chosen vertex.
Again, to visualize, we can do say:
chrom = Length[cols];
palette = discreteColors[chrom];
For[i = 1, i <= Length[cols], ++i,
Do[PropertyValue[{g, v}, VertexStyle] = palette[[i]], {v, cols[[i]]}]
]
g
IGraph/M now includes functions for computing vertex colourings efficiently.
To check if a graph g is k-vertex-colourable use,
IGKVertexColoring[g, k]
If the answer is yes, {coloring} will be returned. If it is no, {} will be returned.
To compute a minimum colouring, use IGMinimumVertexColroing. To just find the chromatic number, use IGChromaticNumber.
There are analogous IGKEdgeColoring and IGMinimumEdgeColoring functions.
If you want a fast but not necessarily minimal colouring, use IGVertexColoring and IGEdgeColoring.
We can also visualize the colourings easily.
g = GraphData["DodecahedralGraph"];
Graph[g, GraphStyle -> "BasicBlack", VertexSize -> Large] //
IGVertexMap[ColorData[106], VertexStyle -> IGMinimumVertexColoring]
Graph[g, GraphStyle -> "BasicBlack", EdgeStyle -> Thickness[0.02]] //
IGEdgeMap[ColorData[106], EdgeStyle -> IGMinimumEdgeColoring]
Note that IGraph/M requires Mathematica 10.0 or later. | 2020-02-19T15:47:36 | {
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https://math.stackexchange.com/questions/2844918/determine-both-the-conjunctive-and-disjunctive-normal-forms-for-the-following-ex | Determine both the conjunctive and disjunctive normal forms for the following expression [verification]
I have the following expression:
$$(((p_3 \lor p_1) \ \land (p_2 \to p_1)) \ \land (p_3 \leftrightarrow p_2))$$
Work so far:
1) CNF:
\begin{align*} &\ (((p_3 \lor p_1) \land (p_2 \to p_1)) \land (p_3 \leftrightarrow p_2)) & \text{(Given)} \\ \equiv &\ (((p_3 \lor p_1) \land (\neg p_2 \lor p_1)) \land (p_3 \leftrightarrow p_2)) & \text{(Simplification of implication)} \\ \equiv &\ (((p_3 \lor p_1) \land (\neg p_2 \lor p_1)) \land ((\neg p_3 \lor p_2 ) \land (p_3 \lor \neg p_2))) & \text{(Simplification of biconditional)} \\ \equiv &\ ((p_3 \lor p_1) \land (\neg p_2 \lor p_1) \land (\neg p_3 \lor p_2 ) \land (p_3 \lor \neg p_2)) & \text{(Associativity of conjunction)} \\ \end{align*}
Q.E.D. (Whilst the expression could further be simplified, at this point it satisfies the definition of CNF).
2) DNF:
\begin{align*} &\ (((p_3 \lor p_1) \land (p_2 \to p_1)) \land (p_3 \leftrightarrow p_2)) & \text{(Given)} \\ \equiv &\ (((p_3 \lor p_1) \land (\neg p_2 \lor p_1)) \land (p_3 \leftrightarrow p_2)) & \text{(Simplification of implication)} \\ \equiv &\ (((p_3 \lor p_1) \land (\neg p_2 \lor p_1)) \land (( p_3 \land p_2 ) \lor (\neg p_3 \land \neg p_2))) & \text{(Simplification of biconditional)} \\ \equiv &\ ((p_1 \lor (\neg p_2 \land p_3)) \land (( p_3 \land p_2 ) \lor (\neg p_3 \land \neg p_2))) & \text{(Distribution of disjunction over conjunction)} \\ \equiv &\ ((p_1 \land ( p_3 \land p_2 )) \lor (p_1 \land (\neg p_3 \land \neg p_2)) \lor ((\neg p_2 \land p_3) \land ( p_3 \land p_2 )) \lor ((\neg p_2 \land p_3) \land (\neg p_3 \land \neg p_2))) & \text{(Distribution of conjunction over disjunction)} \\ \end{align*}
Q.E.D. (As above, sufficient albeit messy).
Why have I posted this question for verification?
• I am not sure if I have applied the steps correctly.
• I suspect that my approach is too verbose, and since this was taken from a practice exam where it wasn't worth many marks, I want to know if I can simplify my approach/save time anywhere. (So far the only thing I can think of is introducing shorthand terms a,b,c,d for the clauses in the final steps of each calculation, but this assumes my general approach is robust).
Is my solution correct? Can it be improved?
Yes, they are both correct.
The result for the DNF can be simplified:
$$((\neg p_2 \land p_3) \land ( p_3 \land p_2 )) \text{ and } ((\neg p_2 \land p_3) \land (\neg p_3 \land \neg p_2)))$$
are both impossible because they contain $x \land \neg x$. Hence you could just write $$((p_1 \land ( p_3 \land p_2 )) \lor (p_1 \land (\neg p_3 \land \neg p_2)).$$
There is a method which is computationally acceptable when you have few variables, such as in this case. You write the 'truth table' (the following picture is generated using a truth table generator):
Then the expression is true exactly when $(p_1, p_2, p_3) = (1, 1, 1) = v_1$ or $(p_1, p_2, p_3) = (1, 0, 0) = v_4$.
Consider the formulae
\begin{align} \psi_1 &= (p_1 \land p_2 \land p_3) \\ \psi_4 &= (p_1 \land (\neg p_2) \land (\neg p_3)). \end{align}
The formula $\psi_i$ is true exactly when the truth assignment is that of $v_i$, hence the expression is equivalent to
$$(p_1 \land p_2 \land p_3) \lor (p_1 \land (\neg p_2) \land (\neg p_3))$$
as found before.
At this point you can easily find a CNF as follows
1. find a DNF for the negation: the expression is false exactly when $p_1$ is false or (the other two cases), hence $$(\neg p_1) \lor (p_1 \land (\neg p_2) \land p_3) \lor (p_1 \land p_2 \land (\neg p_3));$$
2. negate the previous expression to get the result: \begin{align} &\ p_1 \land ((\neg p_1) \lor p_2 \lor (\neg p_3)) \land ((\neg p_1) \lor (\neg p_2) \lor p_3) \\ \equiv &\ p_1 \land (p_2 \lor (\neg p_3)) \land ((\neg p_2) \lor p_3). \end{align}
The results we found for the CNF are different but equivalent. I feel there is less chance to make errors with this method instead of manipulating long expressions, but this is a personal opinion.
• I agree with your personal opinion, and whilst one could argue that it is subjective, I think your answer provides an extra dimension for people to think about these problems in. Thanks! – Oscar Jul 9 '18 at 20:56 | 2019-09-22T10:34:03 | {
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http://mathhelpforum.com/advanced-algebra/150145-linear-algebra-proof-am-i-right-track.html | # Math Help - Linear Algebra Proof, Am I on the right track?
1. ## Linear Algebra Proof, Am I on the right track?
I would like to have someone go over my proof and see if its correct or at least on the right track. Here's the problem:
Let V be a finite-dimensional vector space, and let T: V → V be linear. If rank(T) = rank(T²), prove that R(T) ∩ N(T) = {0}.
PROOF:
Lemma: Let V be a finite-dimensional vector space, and let T: V → V be linear. If rank(T) = rank(T²), then N(T) = N(T²).
Proof: By the Nullity-Rank Theorem (i.e. Let V and W be vector spaces, and let T: V → W be linear. If V is finite-dimensional, then nullity(T) + ran(T) = dim(V)) I have,
dim(V) = rank(T) + nullity(T)
dim(V) = rank(T²) + nullity(T²)
This implies that nullity(T) = nullity(T²). Furthermore, N(T) and N(T²) are both subspaces of V and as a matter of fact, they are both vector spaces. Now it is easily shown that N(T) ⊂ N(T²) and that N(T) is a subspace of N(T²). Therefore, by the Dimension Theorem (i.e. Let W be subspace of a finite-dimensional vector space V. Then W is finite-dimensional and dim(W) ≤ dim(V). Moreover, if dim(W) = dim(V), then V = W) we have N(T) = N(T²). □
Seeking a contradiction, suppose that R(T) ∩ N(T) ≠ {0}. Therefore, there is an x ≠ 0 ∈ R(T) ∩ N(T). This implies that x ∈ R(T) and x ∈ N(T). Since x ∈ R(T) then T(y) = x for some y ∈ V. Note that T²(y) = T(T(y)) = T(x) = 0 which means that y ∈ N(T²) and hence by the lemma y ∈ N(T) also. However, this implies that T(y) = 0 and hence a contradiction since x = T(y) ≠ 0. ♦
I'd appreciate if someone could look at it. Thanks a lot!!
2. I suspect you are saying more than you need to but what you say is correct.
3. Usually when I ask such questions on the internet, I make sure the reader knows exactly what I'm referring to so I do tend to make things longer. But I'd be a lot shorter if this were being handed in for marking. I'm just studying linear algebra on my own. Thanks for your help | 2015-05-23T15:02:42 | {
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https://math.stackexchange.com/questions/152880/how-many-irreducible-polynomials-of-degree-n-exist-over-mathbbf-p | # How many irreducible polynomials of degree $n$ exist over $\mathbb{F}_p$?
I know that for every $n\in\mathbb{N}$, $n\ge 1$, there exists $p(x)\in\mathbb{F}_p[x]$ s.t. $\deg p(x)=n$ and $p(x)$ is irreducible over $\mathbb{F}_p$.
I am interested in counting how many such $p(x)$ there exist (that is, given $n\in\mathbb{N}$, $n\ge 1$, how many irreducible polynomials of degree $n$ exist over $\mathbb{F}_p$).
I don't have a counting strategy and I don't expect a closed formula, but maybe we can find something like "there exist $X$ irreducible polynomials of degree $n$ where $X$ is the number of...".
• This question is ok, but has appeared here many times: At least here, here and also here. Voting to close as a duplicate. +1 to you all, though! – Jyrki Lahtonen Jun 2 '12 at 16:16
• @JyrkiLahtonen: I was wondering if there is as much monic irreducible polynomial of degree $d$ on $\mathbb F_{p}[X]$ than irreducible polynomial of degree $d$ (not necessarily monic). I would say yes since if $X^d+a_{d-1}X^{d-1}+...+a_1X+a_0$ is irreducible, then $\alpha X^d+\alpha a_{d-1}X^{d-1}+...+\alpha a_1 X+a_0$ is irreducible and reciprocally if $a_nX^n+...+a_1X+a_0$ is irreducible, then $X^n+\frac{a_{n-1}}{a_n}X^{n-1}X^{n-1}+...+\frac{a_0}{a_n}$ is irreducible. But I'm not sure if it's true... – user386627 May 8 '17 at 13:39
Theorem: Let $\mu(n)$ denote the Möbius function. The number of monic irreducible polynomials of degree $n$ over $\mathbb{F}_q$ is the necklace polynomial $$M_n(q) = \frac{1}{n} \sum_{d | n} \mu(d) q^{n/d}.$$
(To get the number of irreducible polynomials just multiply by $q - 1$.)
Proof. Let $M_n(q)$ denote the number in question. Recall that $x^{q^n} - x$ is the product of all the monic irreducible polynomials of degree dividing $n$. By counting degrees, it follows that $$q^n = \sum_{d | n} d M_d(q)$$
(since each polynomial of degree $d$ contributes $d$ to the total degree). By Möbius inversion, the result follows.
As it turns out, $M_n(q)$ has a combinatorial interpretation for all values of $q$: it counts the number of aperiodic necklaces of length $n$ on $q$ letters, where a necklace is a word considered up to cyclic permutation and an aperiodic necklace of length $n$ is a word which is not invariant under a cyclic permutation by $d$ for any $d < n$. More precisely, the cyclic group $\mathbb{Z}/n\mathbb{Z}$ acts by cyclic permutation on the set of functions $[n] \to [q]$, and $M_n(q)$ counts the number of orbits of size $n$ of this group action. This result also follows from Möbius inversion.
One might therefore ask for an explicit bijection between aperiodic necklaces of length $n$ on $q$ letters and monic irreducible polynomials of degree $n$ over $\mathbb{F}_q$ when $q$ is a prime power, or at least I did a few years ago and it turns out to be quite elegant.
Let me also mention that the above closed form immediately leads to the "function field prime number theorem." Let the absolute value of a polynomial of degree $d$ over $\mathbb{F}_q$ be $q^d$. (You can think of this as the size of the quotient $\mathbb{F}_q[x]/f(x)$, so in that sense it is analogous to the norm of an element of the ring of integers of a number field.) Then the above formula shows that the number of monic irreducible polynomials $\pi(n)$ of absolute value less than or equal to $n$ satisfies $$\pi(n) \sim \frac{n}{\log_q n}.$$
• Shouldn't the first sum read $\mu(d)$ instead of $\mu(n)$? – Pedro Tamaroff Aug 4 '13 at 21:39
• @Peter: yep. Thanks for the correction! – Qiaochu Yuan Aug 4 '13 at 21:54
With regards to your question, this paper has a formula for counting the number of monic irreducibles over a finite field.
The number of monic irreducible polynomials of degree $n$ over $\mathbb{F}_p$ equals
$$\frac{1}{n} \cdot \sum_{d|n} p^d \mu\left(\frac{n}{d}\right)$$
where $\mu$ is the Möbius function. This follows rather easily from the Möbius inversion formula. You can find details here. Note that this in particular implies the existence of one irreducible polynomial and therefore of the field with $p^n$ elements. | 2019-06-18T06:46:15 | {
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http://math.stackexchange.com/questions/313141/fibonacci-sequence | Fibonacci sequence
Given an integer $n ≥ 1$, let $f_n$ be the number of lists whose elements all equal $1$ or $2$ and add up to $n−1$. For example $f_1 = 1 = f_2$ because only the empty list ($0$ ones and $0$ twos) sums to $0$ and only a single one sums to $1$. The lists $1,2;\, 2,1;\, 1,1,1$ show us that $f_4 = 3$.
(a) Show that $f_{n+2} = f_{n+1} + f_n$ and hence that $f_1, f_2,...$ is the Fibonacci sequence.
(b) How many lists of ones and twos are there that add up to $10$ and contain $3$ twos?
-
4 Answers
Given a list of $1$s and $2$s adding up to $n-1$, we can append a $2$ to the end of that list to get one adding up to $n+1$. Given a list of $1$s and $2$s adding up to $n$, we can append a $1$ to the end of that list to get one adding up to $n+1$. The lists obtained in this way are uniquely determined by the lists we started from. (Why?) Hence, $$f_{n+2}\geq f_{n+1}+f_n.$$ On the other hand, given a list of $1$s and $2$s adding up to $n+1$, removing the last number on the list will give us one adding up to $n$ or $n-1$. Hence, $$f_{n+2}\leq f_{n+1}+f_n,$$ and so $$f_{n+2}=f_{n+1}+f_n.$$
A list of $1$s and $2$s adding up to $10$ and having $3$ $2$s will necessarily have $7$ total entries. (Why?) Such a list is moreover completely determined by the placement of the $2$s. Since there are $7$ spots and $3$ $2$s to place, there are $\binom{7}{3}=35$ such lists.
-
can you please help me solve this Consider the finite sum S=1+2x+3x^(2)+4x^(3)+____+nx^(n-1) Calculate S − xS and hence give a closed expression for S? – ofo Feb 24 '13 at 19:36
The site will only allow you to ask $6$ questions in a given $24$-hour period. If you wait until tomorrow, you can post that as a question. – Cameron Buie Feb 24 '13 at 20:58
The first question has appeared several times on MSE.
For the second, there is some ambiguity: Do we mean contain (i) exactly $3$ twos or (ii) at least $3$ twos?
For (i), if there are exactly $3$ twos, then there must be $4$ ones, a total of $7$ digits. The locations of the twos can be chosen in $\dbinom{7}{3}$ ways. After the location of the twos is determined, the ones have to occupy the remaining slots. Calculate: there are $35$ choices.
For interpretation (ii), we are allowed to have $3$ twos, or $4$, or $5$. The analysis is much the same as for interpretation (i), except that we get $\dbinom{7}{3}+\dbinom{6}{4}+\dbinom{5}{5}$.
-
can you please help me solve this Consider the finite sum S=1+2x+3x^(2)+4x^(3)+____+nx^(n-1) Calculate S − xS and hence give a closed expression for S? – ofo Feb 24 '13 at 19:36
To do it in the style they ask for, note that $xS=x+2x^2+3x^3+\cdots +nx^n$. Now calculate $S-xS$. This is $(1+2x+3x^2+\cdots+nx^{n-1})-(x+2x^2+\cdots+(n-1)x^{n-1}+nx^n)$. Look at the terms. We get a $1$. Also a $2x-x$. Also a $3x^2-2x^2$, and so on. We end up with (check it) $1+x+x^2+\cdots +x^{n-1}-nx^n$. The first part is a finite geometric series, with sum $\frac{1-x^n}{1-x}$, if $x\ne 1$. So we end up with $(1-x)S=\frac{1-x^n}{1-x}-nx^n$. Divide by $1-x$ to get $S$. – André Nicolas Feb 24 '13 at 19:50
thank you very much – ofo Feb 24 '13 at 20:06
@Aka: You are welcome. But in general I would prefer not to type much mathematics in comments, the editing facilities are poor, the length severely limited. – André Nicolas Feb 24 '13 at 20:42
(a) should come from the fact that when you delete the final element of a list which sums to $n+2$ you may end up with a list that sums to $n+1$ (if you deleted a $1$) or to $n$ (if you deleted a $2$).
As to (b), you are talking of lists of length $7$, containing four $1$ and three $2$. This should be stars and bars (Theorem Two) which yields $$\binom{4 + 4 - 1}{4} = \binom{7}{4} = 35.$$ That is, you have to put your four $1$ in four bins (separated by the three $2$), and some bin may be empty.
PS I have read the post of @AndréNicolas. I have intended question (b) as requiring exactly three $2$.
-
can you please help me solve this? Consider the finite sum S=1+2x+3x^(2)+4x^(3)+____+nx^(n-1) Calculate S − xS and hence give a closed expression for S? – ofo Feb 24 '13 at 19:35
@Aka, please post a separate question if you wish. Comments are not the place to start new questions. – Andreas Caranti Feb 24 '13 at 19:37
i tried posting but the website didnt accept my question post – ofo Feb 24 '13 at 19:39
Let $L_k$ the set of lists whose elements all equal $1$ or $2$ and add up to $k$. You want to show that $f_n=|L_n|$.
Note that $\ell\in L_{n-2}\Rightarrow \text{append}(\ell,2)\in L_n$ and $\ell \in L_{n-1}\Rightarrow \text{append}(\ell,1)\in L_n$. Conclude that $|L_{n-2}|+|L_{n-1}|\leq |L_{n}|$...
- | 2015-04-19T08:39:43 | {
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https://math.stackexchange.com/questions/567441/summation-formula-for-x2x/801852 | # Summation formula for $x^2+x$
Since I learned easier ways of calculating summations I've been curious as to how I could find formulas for as many equations as possible. I came across the equation $x^2+x$, I've spent quite some time on this problem and could not find a solution. If someone has maybe already done this or have any suggestions on how I could get the formula that would be greatly appreciated.
Example of another summation with a equation: $\sum\limits_{i=1}^n$ = $x^2$
Equation to solve this is $\frac{n(n+1)(2n+1)}{6}$
• You want to find $\sum_{i=1}^N(i^2 + i)$? – M.B. Nov 14 '13 at 23:03
• Are you asking about a closed formula for $\sum \limits_{k=0}^n\left(k^2+k\right)$, for every $n\in \Bbb N$? If so, just separate the sum in two well known sums. – Git Gud Nov 14 '13 at 23:03
• 1) Yes 3) Yes and thank you Thanks kbball for that edit, I have no idea how that works – Harjit Nov 15 '13 at 4:12
• You are using the word "equation" but the mathematical term is "expression" unless you have an $=$ sign. – Sammy Black May 19 '14 at 17:40
Extending what Git Gud said: $$\sum^n_{k=0}\left(k^2+k\right)=\sum^n_{k=0}k^2+\sum^n_{k=0}k=\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}=\frac{n^3}{3}+n^2+\frac{2n}{3}$$
• So for n=6 we get $\frac{6^3}{3} +6^2 + \frac{6}{6} + \frac{1}{2} = 72 + 36 + 1 + \frac{1}{2} = 109.5?$ – Steve ODonnell Nov 14 '13 at 23:55
• When I do the summation of (K^2+k) starting from zero manually I get; 0+2+6+12+20+30+42=112 but using the equation we get 109.5? – Harjit Nov 15 '13 at 4:08
• Also wouldn't the equation be (n^3/3+n^2+2n/3)/2. According to the manual summation that makes more sense to me. I am though making an assumption that dividing by 2 is there due to having added two summations, am I right? – Harjit Nov 15 '13 at 4:33
• Oops sloppy algebra – Ali Caglayan Nov 16 '13 at 2:03
• Oh alright, no worries, thanks! – Harjit Nov 16 '13 at 4:00
Note that $$k^2+k=\frac{1}{3}\left((k+1)^3-k^3-1\right).$$ Thus our sum $\sum_{k=1}^n (k^2+k)$ is equal to $$\frac{1}{3}\left((2^3-1^3-1)+(3^3-2^3-1)+(4^3-3^3-1)+(5^3-4^3-1)+\cdots +((n+1)^3-n^3-1)\right).$$ Observe the nice almost total cancellations. We end up with $$\frac{1}{3}\left((n+1)^3-1^3-n\right).$$
Remarks: $1.$ Since we know $\sum_1^n k$, this gives a way to derive the formula for $\sum_1^n k^2$.
$2.$ The sums $\sum k(k+1)$, $\sum k(k+1)(k+2)$, $\sum k(k+1)(k+2)(k+3)$ and so on are nice, much nicer than $\sum k^2$, $\sum k^3$, $\sum k^4$ and so on.
• @Ben: Thanks for the fix. – André Nicolas Nov 19 '13 at 6:25
• +1 for the telescoping sum explanation, since they lie at the heart of the summation formulas that others are using. – Sammy Black May 19 '14 at 17:45
Note that $$\frac{n^2+n}{2} = {n+1 \choose 2}.$$ Then by induction $$\begin{eqnarray}{n+2 \choose 3}&=&{n+1 \choose 2}+{n+1 \choose 3}\\ &=&{n+1 \choose 2} + \sum_{k=1}^{n-1}{k+1 \choose 2}\\ &=& \sum_{k=1}^{n}{k+1 \choose 2}. \end{eqnarray}$$ So $$\sum_{k=1}^{n}(k^2+k)=2{n+2 \choose 3}.$$ This directly generalizes to binomial sums $$\sum_{k=1}^n{k + m-1 \choose m}.$$ This is particularly obvious if you draw what is going on in Pascal's triangle. | 2019-08-20T07:41:54 | {
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https://math.stackexchange.com/questions/2057797/what-do-i-do-when-i-am-trying-to-find-the-absolute-minimum-point-of-a-function-a/2058039 | # What do I do when I am trying to find the absolute minimum point of a function and function isn't differentiable at that point?
Consider the function $f(x) = |x|$, obviously this function is not differentiable at $x = 0$. But what does that mean for the absolute minimum point? Do I answer the question with simply "the function isn't differentiable at $x = 0$" or do I state that it isn't differentiable but proceed to write the point anyways?
• You would certainly observe that $f$ has an absolute minimum at $x=0$. You could most easily justify this by appealing to the definition of the absolute value. If you want to bring in calculus, however, you can observe that $f'(x)=-1<0$ for $x<0$, and $f'(x)=1>0$ for $x>0$, so either $f(0)$ is undefined, or the function has an absolute minimum at $x=0$, Since $f(0)$ is defined, the function has an absolute minimum there. – Brian M. Scott Dec 13 '16 at 23:02 | 2021-04-23T18:16:17 | {
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http://bjhc.autozwerg.de/radius-of-convergence-complex-power-series-problems.html | # Radius Of Convergence Complex Power Series Problems
$\begingroup$ Radius of convergence of an analytic function doesn't really exist as a concept: an analytic function has a domain on which it is analytic, and its power series around a point will have a disk of some radius on which it converges, but for a function there's nothing to converge or diverge, hence no radius of convergence. a is a complex constant, the center of the disk of convergence, c n is the n th complex coefficient, and z is a complex variable. Representation of Functions as Power Series. Thus, the radius of convergence for a general power series expanded about a point z0 in the complex plane is simply the radius of this disc. Math 122 Fall 2008 Recitation Handout 17: Radius and Interval of Convergence Interval of Convergence The interval of convergence of a power series: ! cn"x#a ( ) n n=0 $% is the interval of x-values that can be plugged into the power series to give a convergent series. for jx aj>R, where R>0 is a value called the radius of convergence. We will call the radius of convergence L. Prove that this series has a radiusof convergence, R, where either R=positive infinity or R≤1. Browse other questions tagged sequences-and-series complex-analysis convergence power-series or ask your own question. Definition 1. For the power series in (1. summing from n=1 to infinity ((-1)^n)n(2^(n+1))z^(2n) its a bit messy with all the brackets but it should read (-1 to the power of n) n 2 to the power of n+1 z to the pwer of 2n ive tried to make it as clear as i can if anyone could help out it would be greatly appreciated. Di erentiability of power series. Today we'll talk more about the radius of convergence of a power series and how to find this radius. Radius of Convergence Problems What is the radius of convergence of the following power series? 1. So this is a power series in x, centred at x = 0, it has radius of convergence R = 1, and its interval of convergence is the open interval ( 1;1). Complex Functions Examples c-4 5 Introduction Introduction This is the fourth book containing examples from theTheory of Complex Functions. Write down the power series expansion of 2 x e 1 x 2 x by multiplying the power series of e by the power series of 1/(1 x). y The series converges only at the center x= aand diverges otherwise. And we'll also see a few examples similar to those you might find on the AP Calculus BC exam. sigma n=1 to infinity (x-2)^n/(2n+1)3^n+1. Find radius of convergence for a complex power series. Note: Once the power series (1) is known, standard convergence tests can be applied to nd the radius of convergence. [Real Analysis] Problem on Convergence of Power Series (self. 8 Problem 2E. 8) S(z) = X1 k=0 c k(z z 0)k. Math 432 - Real Analysis II Solutions to Test 1 Thus, the radius of convergence for this power series is 1. Some infinite series converge to a finite value. If f(z) is represented by a convergent power series for jzj0, then the function fde ned by f(x) = c 0 + c 1(x a) + c 2(x a)2 + = X1 n=0 c n(x a)n is di erentiable (and therefore continuous) on the. Things you should memorize: • the formula of the Taylor series of a given function f(x). gent complex series will converge within some disc in the complex plane. 15, we say that the radius of convergence is zero and that the radius of convergence is infinity for case (iii). Determine the radius of convergence of the power series$\sum_{n=0}^{\infty} \frac{x^n}{n!}$. RADIUS OF CONVERGENCE Let be a power series. Sachin Gupta B. This series is important to understand because its behavior is typical of all power series. 3: Suppose we have the series X1 k=0 2 k(x 1) : First we compute, A = lim k!1 a k+1 a k = lim k!1 2 k 1 2 k = 2 1 = 1=2: Therefore the radius of convergence is 2, and the series converges absolutely on the interval (1;3). In general, you can skip the multiplication sign, so 5x is equivalent to 5⋅x. Convergence Tests - Additional practice using convergence tests. [Real Analysis] Problem on Convergence of Power Series (self. The number R is called the radius of convergence of the power series. The radius of convergence r is a nonnegative real number or ∞ such that the series converges if. Intervals of Convergence of Power Series. Prove that this series has a radiusof convergence, R, where either R=positive infinity or R≤1. X∞ n=1 xn n √ n3n. This is a nice survey, its only problem is that it lists no references. because and for Let's, for now, allow to take complex numbers. The inequality can be written as -7 < x < 1. Complex Analysis. In other words, in the complex plane, where the independent variable z is represented, the circle of convergence of the series has the same radius R as the other circle of convergence of the series, and its center is located at the point a. The Fourier series is a power series in z evaluated at All of these points lie on the circle centered at the origin with radius 1. Math 122 Fall 2008 Recitation Handout 17: Radius and Interval of Convergence Interval of Convergence The interval of convergence of a power series: ! cn"x#a ( ) n n=0$ % is the interval of x-values that can be plugged into the power series to give a convergent series. The basic facts are these: Every power series has a radius of convergence 0 ≤ R≤ ∞, which depends on the coefficients an. They are completely different. 15 is to say that the power series converges if and diverges if. Radius of convergence power Series in hindi. Write cosx 1 + x = 1 1 + x cos(x): The radius of convergence of the power series representation for cosxat any center point is 1. Then, by we have. radius of convergence of complex power series? this series has radius of convergence R = ∞. R can often be determined by the Ratio Test. Sachin Gupta B. THANK YOU !!. jz aj= Ris a circle of radius Rcentered at a, hence Ris called the radius of convergence of the power series. The interval of convergence for a power series is the set of x values for which that series converges. The radius of convergence can be characterized by the following theorem: The radius of convergence of a power series ƒ centered on a point a is equal to the distance from a to the nearest point where ƒ. equation that are power series about x 0. Thus, the interval of convergence is , and again one must individually check the endpoints. The method for finding the interval of convergence is to use the ratio test to find the interval where the series converges absolutely and then check the endpoints of the interval using the various methods from the previous modules. P 1 r n (a) (z/a) ,where r and a are constant real numbers. Free power series calculator - Find convergence interval of power series step-by-step Rationales Coordinate Geometry Complex Numbers Polar/Cartesian. Write cosx 1 + x = 1 1 + x cos(x): The radius of convergence of the power series representation for cosxat any center point is 1. The interval of convergence plays an important role in establishing the values of $$x$$ for which a power series is equal to its common function representation. Representation of Functions as Power Series. Proof: Suppose … Power Series and Radius of Convergence are investigated. If the power series (2. Körner, "The behavior of power series on their circle of convergence", in Banach Spaces, Harmonic Analysis, and Probability Theory, Springer Lecture Notes in Mathematics #995, Springer-Verlag, 1983, 56-94. Find the interval and radius of convergence of the following power series (problem #1a)? Calculus Power Series Determining the Radius and Interval of Convergence for a Power Series 1 Answer. This is a nice survey, its only problem is that it lists no references. There series of numbers has certain properties that we can extend to the series of functions. Here we have discussed Power Series and it's Convergence (Radius of Convergence with Proof). Meromorphic function. zero, then the power series is a polynomial function, but if in nitely many of the a n are nonzero, then we need to consider the convergence of the power series. Solutions to Final Exam Review Problems Math 5C, Winter 2007 1. The sum of a power series with a positive radius of convergence is an analytic function at every point in the interior of the disc of convergence. n=0 n P 1 1 n (b) (z/a) , where a is a constant real number. For the endpoints, notice that when x= 1. Power series have coefficients, x values, and have to be centred at a certain value a. If the terms of a sequence being summed are power functions, then we have a power series, defined by Note that most textbooks start with n = 0 instead of starting at 1, because it makes the exponents and n the same (if we started at 1, then the exponents would be n - 1). They are completely different. See, 'sine x' plus ''sine 4x' over 16'. Free power series calculator - Find convergence interval of power series step-by-step. Math 262 Practice Problems Solutions Power Series and Taylor Series 1. Do not confuse the capital (the radius of convergeV nce) with the lowercase (from the root< test). RADIUS OF CONVERGENCE Let be a power series. pdf doc ; More Convergence Tests - A summary of the available convergence tests. We mentioned in the Remark in this post that it is known that the radius of convergence of the power series is This can be used to show that the radius of convergence of the Maclaurin series expansion of is and so for. This is known as Abel's theorem on power series. One of the main purposes of our study of series is to understand power series. We are experiencing some problems, please try again. The goal of this problem is to prove that r is the radius of convergence of the power series. Things you should memorize: • the formula of the Taylor series of a given function f(x). For case (i) of Theorem 4. Prove that the radius of convergence of the power series ∞ 0 c nx n is at least r. 8) S(z) = X1 k=0 c k(z z 0)k. Complex Analysis: Analytic functions, harmonic functions; Complex integration: Cauchy's integral theorem and formula; Liouville's theorem, maximum modulus principle, Morera's theorem; zeros and singularities; Power series, radius of convergence, Taylor's theorem and Laurent's theorem; residue. Abel's theorem: boundary behavior 5. jz aj= Ris a circle of radius Rcentered at a, hence Ris called the radius of convergence of the power series. Textbook solution for Single Variable Calculus: Early Transcendentals 8th Edition James Stewart Chapter 11. We have a series with non-negative numbers again, so convergence and absolute convergence coincide and we can use our favorite tests. Convergence Tests for Positive Series : The ratio test. For each of the following power series, find the interval of convergence and the radius of convergence:. If the radius is positive, the power series converges absolutely. k kB V V is called the radius of convergence. The disk of convergence may be degen-erate: in one extreme situation it is a point, z = z 0 (zero radius of convergence) in the other, the whole complex domain (\in nite radius of convergence"). Will that lead to an inconclusive result? "If the radius of convergence is 1, when z=1 or -1,. The right-hand side 0 is given by the zero-series with radius of convergence 1. We’ll deal with the $$L = 1$$ case in a bit. A series of the form where x is a variable, where {a n} is a sequence and c is a constant, is called a power series about c R is called the radius of convergence. III, Jacob & Evans. Please see the attached file for the complete solution. Do not confuse the capital (the radius of convergeV nce) with the lowercase (from the root< test). Hart Complex power series: an example. The Attempt at a Solution I managed to do the Radius of convergence (power series) problem | Physics Forums. You will have to register before you can post. Tech (CSE), Educational YouTuber, Dedicated to providing the best Education for Mathematics and Love to Develop Shortcut Tricks. function mapping complex paths ˝to the tensor algebra. The radius of convergence is the radius of the largest circle about the point of expansion (in this case, z0=2) such that the function is analytic everywhere inside. Chapters I through VITI of Lang's book contain the material of an introductory course at the undergraduate level and the reader will find exercises in all of the fol lowing topics: power series, Cauchy's theorem, Laurent series, singularities and meromorphic functions, the calculus of residues, conformal. III, Jacob & Evans. 1) where the coe cients fa kgare prescribed complex numbers and zis a com-plex variable. Unlike geometric series and p-series, a power series often converges or diverges based on its x value. Then there exists a radius"- B8 8 for whichV (a) The series converges for , andk kB V (b) The series converges for. For the endpoints, notice that when x= 1. a is a complex constant, the center of the disk of convergence, c n is the n th complex coefficient, and z is a complex variable. R can be 0, 1or anything in between. The function. How to find Interval and Radius of Convergence on the TI89? > What about the 2 power series problems in the pictures ? A: you can view the step by step solutions to find both the interval and radius of convergence of any power series under F3 1 within the sequence and series module of calculus made easy. Our goal in this section is find the radius of convergence of these power series by using the ratio test. gent complex series will converge within some disc in the complex plane. How do we find the radius of convergence? 10. radius of convergence of complex power series? this series has radius of convergence R = ∞. [Real Analysis] Problem on Convergence of Power Series (self. See, 'sine x' plus ''sine 4x' over 16'. In order to find these things, we'll first have to find a power series representation for the Taylor series. gent complex series will converge within some disc in the complex plane. The behavior of power series on the circle at the radius of convergence is much more delicate than the behavior in the interior. Bernd Schroder¨ Louisiana Tech University, College of. zero, then the power series is a polynomial function, but if in nitely many of the a n are nonzero, then we need to consider the convergence of the power series. Featured on Meta Official FAQ on gender pronouns and Code of Conduct changes. Also, the interval of convergence is ¡ 5˙x ¯2, i. of quotients of successive coefficients has a limit, it just says if that. When discussing series of function and the power series, there is a theorem about the convergence of this series called the Radius of Convergence Theorem. The number R is called the radius of convergence of the power series. Best Answer: You did not say, but I assume you mean, that f(z) is to be expanded in a power series about z0=2, and you wish to find the radius of convergence of the power series. R can be 0, 1or anything in between. A Second Order Problem Power Series: Radius and Interval of Convergence. P 1 r n (a) (z/a) ,where r and a are constant real numbers. \) Solution. Definition: A power series in x is a series of the form P∞ k=0akx k, where {ak} is a sequence of real constants. Find the radius of convergence of the power series? How would I go about solving this problem: Suppose that (10x)/(14+x) = the sum of CnX^(n) as n=0 goes to infinity C1= C2= Find the radius of convergence R of the power series. Suppose that the limit lim n!1 jcn+1j jcnj exists or is 1. P 1 r n (a) (z/a) ,where r and a are constant real numbers. Let t be the norm of p, i. Things you should memorize: • the formula of the Taylor series of a given function f(x). Free power series calculator - Find convergence interval of power series step-by-step. Hart Complex power series: an example. See, 'sine x' plus ''sine 4x' over 16'. We call the number the radius of convergence of the power series (see Figure 4. They can show that the series converges inside a circle U 2 + V 2 = R 2, and diverges outside the circle. List of Maclaurin Series of Some Common Functions / Stevens Institute of Technology / MA 123: Calculus IIA / List of Maclaurin Series of Some Common Functions / 9 | Sequences and Series. Worksheet 7 Solutions, Math 1B Power Series Monday, March 5, 2012 1. Step 2: Test End Points of Interval to Find Interval of Convergence. [Hint: assume rst that the solution f has a power series representation about 0, plug it into the di erential equation and nd what the coe cients are, then show that the radius of convergence is 1. We will also learn about Taylor and Maclaurin series, which are series that act as functions and converge to common functions like sin(x) or eˣ. As in the case of a Taylor/Maclaurin series the power series given by (4. Find the radius of convergence and interval of convergence of the series: (a) X1 n=1 xn p n Solution Sketch Ratio test gives a radius of convergence of R = 1. The same terminology can also be used for series whose terms are complex, hypercomplex or, more generally, belong to a normed vector space (the norm of a vector being corresponds to the absolute value of a number). Let P c nxn be a power series, and suppose that c n 6= 0 for all n, and that n c c n+1 −−−→n→∞ r. Here we have discussed Power Series and it's Convergence (Radius of Convergence with Proof). (b) We write that the radius of convergence R = 0 if the series converges at only z 0. As in the case of a Taylor/Maclaurin series the power series given by (4. Therefore, the radius of convergence is 4. One fact that may occasionally be helpful for finding the radius of convergence: if the limit of the n th root of the absolute value of c [ n ] is K , then the radius of convergence is 1/ K. the radius of convergence. The Fourier series is a power series in z evaluated at All of these points lie on the circle centered at the origin with radius 1. zero, then the power series is a polynomial function, but if in nitely many of the a n are nonzero, then we need to consider the convergence of the power series. Power series have coefficients, x values, and have to be centred at a certain value a. pdf doc ; More Convergence Tests - A summary of the available convergence tests. But by the radial continuity theorem we can apply the double limit theorem for x !1 to obtaintheresult. The disk of convergence may be degen-erate: in one extreme situation it is a point, z = z 0 (zero radius of convergence) in the other, the whole complex domain (\in nite radius of convergence"). Do not confuse the capital (the radius of convergeV nce) with the lowercase (from the root< test). See attached file for full problem description. Convergence Tests for Infinite Series In this tutorial, we review some of the most common tests for the convergence of an infinite series $$\sum_{k=0}^{\infty} a_k = a_0 + a_1 + a_2 + \cdots$$ The proofs or these tests are interesting, so we urge you to look them up in your calculus text. This problem has been solved! See the answer. Power Series, Circle of Convergence Circle of Convergence Assume the power series f = a 0 + a 1 z + a 2 z 2 + a 3 z 3 + … converges at the point p, for p ≠ 0. ANALYSIS I 13 Power Series 13. The radius of convergence can be characterized by the following theorem: The radius of convergence of a power series f centered on a point a is equal to the distance from a to the nearest point where f cannot be defined in a way that makes it holomorphic. Find the radius of convergence and interval of convergence of the series: (a) X1 n=1 xn p n Solution Sketch Ratio test gives a radius of convergence of R = 1. The radius of convergence of a power series is the radius of the largest disk for which the series converges. Examples 1. 3) In this section we will look at a special type of series of functions. Paul's Online Math Notes Calculus II (Notes) / Series & Sequences / Power Series [Notes] [Practice Problems]. In other words, in the complex plane, where the independent variable z is represented, the circle of convergence of the series has the same radius R as the other circle of convergence of the series, and its center is located at the point a. for jx aj>R, where R>0 is a value called the radius of convergence. Answer to: 1. The ratio test tells us that the power series converges only when or. jz aj= Ris a circle of radius Rcentered at a, hence Ris called the radius of convergence of the power series. What is its radius of convergence? (2)Use the previous to nd a power series expansion for tan 1 z centered at z= 0, and note its radius of convergence. So recall will be proofed last class. Suppose that the limit lim n!1 jcn+1j jcnj exists or is 1. Here is a set of practice problems to accompany the Power Series section of the Series & Sequences chapter of the notes for Paul Dawkins Calculus II course at Lamar University. 8 Problem 2E. If {c n} is a sequence of real or complex numbers, and z 0 is a fixed scalar, then define the formal power series for the sequence about the point z 0 by. There series of numbers has certain properties that we can extend to the series of functions. Inthisvolume we shall only consider complex power series and their relationship to the general theory, and nally the technique of solving linear dierential equations with polynomial coecients by means of a. because and for Let's, for now, allow to take complex numbers. pdf doc ; More Power Series - Additional practice finding radius and interval of convergence. Please see the attached file for the complete solution. Find the radius of convergence and interval of convergence of the series {eq}\sum_{n=1}^\infty x^n(3n-1) {/eq} Radius and Interval of Convergence for a Power Series: There are many tests to find. I treated this problem as a complex power series one. Find the interval and radius of convergence of the following power series (problem #1a)? Calculus Power Series Determining the Radius and Interval of Convergence for a Power Series 1 Answer. The ratio test tells us that the power series converges only when or. And over the interval of convergence, that is going to be equal to 1 over 3 plus x squared. Paul's Online Math Notes Calculus II (Notes) / Series & Sequences / Power Series [Notes] [Practice Problems]. This is why these series are so problematic. The second case is practical: when you construct a power series solution of a difficult problem you typically will only know a finite number of terms in a power series, anywhere from a couple of terms to a hundred terms. Körner, "The behavior of power series on their circle of convergence", in Banach Spaces, Harmonic Analysis, and Probability Theory, Springer Lecture Notes in Mathematics #995, Springer-Verlag, 1983, 56-94. POWER SERIES METHODS Example 7. pdf doc ; CHAPTER 10 - Approximating Functions Using. Determine the radius of convergence and interval of convergence of the power series $$\sum\limits_{n = 0}^\infty {n{x^n}}. Convergence Tests for Positive Series : The ratio test. Sachin Gupta B. In general, you can skip the multiplication sign, so 5x is equivalent to 5⋅x. Worksheet 7 Solutions, Math 1B Power Series Monday, March 5, 2012 1. Find the radius of convergence of the following power series with complex argument z. We'll look at this one in a moment. (b) We write that the radius of convergence R = 0 if the series converges at only z 0. Write down the power series expansion of 2 x e 1 x 2 x by multiplying the power series of e by the power series of 1/(1 x). Luh and Stepanyan [2] interested in the same problem for power series with radius of convergence zero using Cesaro methods. The calculator will find the radius and interval of convergence of the given power series. Continuity Abel's elementary proof that complex power series are termwise di erentiable in their disk of convergence incidentally shows that they are continuous there as well. Any such power series has a radius of convergence R. Find the sum ,radius of convergence and interval of convergence of the series. Radius of convergence examples in hindi. Best Answer: You did not say, but I assume you mean, that f(z) is to be expanded in a power series about z0=2, and you wish to find the radius of convergence of the power series. 3: Suppose we have the series X1 k=0 2 k(x 1) : First we compute, A = lim k!1 a k+1 a k = lim k!1 2 k 1 2 k = 2 1 = 1=2: Therefore the radius of convergence is 2, and the series converges absolutely on the interval (1;3). Textbook solution for Single Variable Calculus: Early Transcendentals 8th Edition James Stewart Chapter 11. Behavior near the boundary. We've already shown that this series is uniformly convergent, but for a uniform convergent series, we saw last time that you can interchange the order of summation and integration. There series of numbers has certain properties that we can extend to the series of functions. Convergence Tests for Positive Series : The ratio test. Solved problems of radius of convergence power Series. ANALYSIS I 13 Power Series 13. The radius of convergence r is a nonnegative real number or ∞ such that the series converges if. Thanks for using BrainMass. For a power series ƒ defined as:. THANK YOU !!. Find the interval and radius of convergence of the following power series (problem #1a)? Calculus Power Series Determining the Radius and Interval of Convergence for a Power Series 1 Answer. This note is about complex power series. sequence has limit then that limit is the radius of convergence of the power series. A power series (centered at the origin). For the endpoints, notice that when x= 1. Learn how this is possible and how we can tell whether a series converges and to what value. This problem has been solved! See the answer. Find the radius of convergence and interval of convergence of the series: (a) X1 n=1 xn p n Solution Sketch Ratio test gives a radius of convergence of R = 1. summing from n=1 to infinity ((-1)^n)n(2^(n+1))z^(2n) its a bit messy with all the brackets but it should read (-1 to the power of n) n 2 to the power of n+1 z to the pwer of 2n ive tried to make it as clear as i can if anyone could help out it would be greatly appreciated. Calculate the radius of convergence:. Determine the radius of convergence and interval of convergence of the power series \(\sum\limits_{n = 0}^\infty {n{x^n}}. Plz help Me. For case (i) of Theorem 4. III, Jacob & Evans. pdf doc ; More Power Series - Additional practice finding radius and interval of convergence. 3: Suppose we have the series X1 k=0 2 k(x 1) : First we compute, A = lim k!1 a k+1 a k = lim k!1 2 k 1 2 k = 2 1 = 1=2: Therefore the radius of convergence is 2, and the series converges absolutely on the interval (1;3). 250 CHAPTER 7. But we can't compare the value between 1 and i. Complex Functions Examples c-4 5 Introduction Introduction This is the fourth book containing examples from theTheory of Complex Functions. Remember that a power series is a sum, but it is an in-nite sums. So as long as x is in this interval, it's going to take on the same values as our original function, which is a pretty neat idea. sigma n=1 to infinity (x-2)^n/(2n+1)3^n+1. The right-hand side 0 is given by the zero-series with radius of convergence 1. Especially, for the same problem, Luh and Nieÿ in [4] considered Faber. Determine the radius of convergence and interval of convergence of the power series \(\sum\limits_{n = 0}^\infty {n{x^n}}. RADIUS OF CONVERGENCE Let be a power series. sigma n=1 to infinity (x-2)^n/(2n+1)3^n+1. Remember that a power series is a sum, but it is an in-nite sums. Therefore, the radius of convergence is 4. Note that whether we di⁄erentiate or integrate, the radius of convergence is preserved. pdf doc ; More Convergence Tests - A summary of the available convergence tests. A power series with a positive radius of convergence can be made into a holomorphic function by taking its argument to be a complex variable. For a power series ƒ defined as:. The Attempt at a Solution I managed to do the Radius of convergence (power series) problem | Physics Forums. If f(z) is represented by a convergent power series for jzj= 1, the series (of partial sums) no longer converges. Di erentiation and Integration of Power Series We can di erentiate and integrate power series term by term, just as we do with polynomials. Thanks for using BrainMass. Write down the power series expansion of 2 x e 1 x 2 x by multiplying the power series of e by the power series of 1/(1 x). For the power series in (1. Convergence Tests - Additional practice using convergence tests. What is the radius of convergence of the power series n=0 to inifinity CnX^2n? Answer Choices : A. X∞ n=1 xn n √ n3n. We assume y(x) = P 1 n=0 a x n. The right-hand side 0 is given by the zero-series with radius of convergence 1. Shifted power series. RADIUS OF CONVERGENCE Let be a power series. pdf doc ; CHAPTER 10 - Approximating Functions Using. How do we find the radius of convergence? 10. Proof: Suppose … Power Series and Radius of Convergence are investigated. The problem is to determine the radius of convergence of the Taylor Series for each of the functions below centered at x. The set of all points whose distance to a is strictly less than the radius of convergence is called the disk of convergence. Paul's Online Math Notes Calculus II (Notes) / Series & Sequences / Power Series [Notes] [Practice Problems]. The number R is called the radius of convergence of the power series. Find the radius of convergence and interval of convergence of the series {eq}\sum_{n=1}^\infty x^n(3n-1) {/eq} Radius and Interval of Convergence for a Power Series: There are many tests to find. jz aj= Ris a circle of radius Rcentered at a, hence Ris called the radius of convergence of the power series. We have a series with non-negative numbers again, so convergence and absolute convergence coincide and we can use our favorite tests. This problem has been solved! See the answer. Homework Statement Ʃ (from n=1 to ∞) (4x-1)^2n / (n^2) Find the radius and interval of convergence 3. Now, let’s get the interval of convergence. If f(z) is represented by a convergent power series for jzj= 1, the series (of partial sums) no longer converges. Show Instructions In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. This problem has been solved! See the answer. because and for Let's, for now, allow to take complex numbers. By signing up, you'll get thousands of step-by-step. The radius of convergence r is a nonnegative real number or ∞ such that the series converges if. Meromorphic function. k kB V V is called the radius of convergence. For case (i) of Theorem 4. One of the main purposes of our study of series is to understand power series. The ratio test tells us that the power series converges only when or. Representation of Functions as Power Series. Here we have discussed Power Series and it's Convergence (Radius of Convergence with Proof). R can often be determined by the Ratio Test. Dave Renfro mentioned a reference I wasn't aware of: Thomas W. converges when ǀzǀ > r and diverges when ǀzǀ > r. Show Instructions In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. The radius of convergence r is a nonnegative real number or ∞ such that the series converges if. 3) In this section we will look at a special type of series of functions. for jx aj>R, where R>0 is a value called the radius of convergence. Abel's theorem: di erentiability of power series 4. Therefore, the radius of convergence is 4. This is why these series are so problematic. M1M1: Problem Sheet 3: Convergence of Power Series and Limits 1. k kB V V is called the radius of convergence. The behavior of power series on the circle at the radius of convergence is much more delicate than the behavior in the interior. R can be 0, 1or anything in between. Hart Complex power series: an example. The three power series f(x) = P a nxn, g(x) = P P b nxn and h(x) = c nxn have a RCV 1, hence absolutely converge for jxj<1 so we can ap-ply the theorem of chapter 1 and get f(x)g(x) = h(x) for these x. And we'll also see a few examples similar to those you might find on the AP Calculus BC exam. The radius of convergence r is a nonnegative real number or ∞ such that the series converges if. Convergence of power series 2. The radius of convergence for this power series is \(R = 4$$. | 2019-11-19T18:48:25 | {
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https://yutsumura.com/find-a-formula-for-a-linear-transformation/ | # Find a Formula for a Linear Transformation
## Problem 36
If $L:\R^2 \to \R^3$ is a linear transformation such that
\begin{align*}
L\left( \begin{bmatrix}
1 \\
0
\end{bmatrix}\right)
=\begin{bmatrix}
1 \\
1 \\
2
\end{bmatrix}, \,\,\,\,
L\left( \begin{bmatrix}
1 \\
1
\end{bmatrix}\right)
=\begin{bmatrix}
2 \\
3 \\
2
\end{bmatrix}.
\end{align*}
then
(a) find $L\left( \begin{bmatrix} 1 \\ 2 \end{bmatrix}\right)$, and
(b) find the formula for $L\left( \begin{bmatrix} x \\ y \end{bmatrix}\right)$.
If you think you can solve (b), then skip (a) and solve (b) first and use the result of (b) to answer (a).
(Part (a) is an exam problem of Purdue University)
## Hint.
1. Express $\begin{bmatrix} 1 \\ 2 \end{bmatrix}$ as a linear combination of $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ and $\begin{bmatrix} 1 \\ 1 \end{bmatrix}$.
2. Use the linearity of linear transformation $L$.
3. Same for part (b). Replace $\begin{bmatrix} 1 \\ 2 \end{bmatrix}$ with the general vector $\begin{bmatrix} x \\ y \end{bmatrix}$.
## Solution.
### (a) Find $L\left( \begin{bmatrix} 1 \\ 2 \end{bmatrix}\right)$
Note that the vectors $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ and $\begin{bmatrix} 1 \\ 1 \end{bmatrix}$ are a basis of $\R^2$.
We first express $\begin{bmatrix} 1 \\ 2 \end{bmatrix}$ as a linear combination of the vectors $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ and $\begin{bmatrix} 1 \\ 1 \end{bmatrix}$.
Let $\begin{bmatrix} 1 \\ 2 \end{bmatrix} =c_1 \begin{bmatrix} 1 \\ 0 \end{bmatrix} + c_2 \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} c_1+c_2 \\ c_2 \end{bmatrix}$.
Solving this, we have $c_1=-1$ and $c_2=2$.
Then we calculate
\begin{align*}
L\left( \begin{bmatrix}
1 \\
2
\end{bmatrix}\right)
& =L\left( – \begin{bmatrix}
1 \\
0
\end{bmatrix} +
2 \begin{bmatrix}
1 \\
1
\end{bmatrix}
\right)
= -L\left( \begin{bmatrix}
1 \\
0
\end{bmatrix} \right) +
2 L \left(\begin{bmatrix}
1 \\
1
\end{bmatrix}
\right) \\
&= -\begin{bmatrix}
1 \\
1 \\
2
\end{bmatrix}
+2 \begin{bmatrix}
2 \\
3 \\
2
\end{bmatrix}
=\begin{bmatrix}
3 \\
5 \\
2
\end{bmatrix},
\end{align*}
where the second equality follows from the linearity of $L$.
Thus we have
$L\left( \begin{bmatrix} 1 \\ 2 \end{bmatrix}\right)=\begin{bmatrix} 3 \\ 5 \\ 2 \end{bmatrix}.$
### (b) Find the formula for $L\left( \begin{bmatrix} x \\ y \end{bmatrix}\right)$.
We generalize the proof of (a).
Let $\begin{bmatrix} x \\ y \end{bmatrix} =c_1 \begin{bmatrix} 1 \\ 0 \end{bmatrix} + c_2 \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} c_1+c_2 \\ c_2 \end{bmatrix}$ be a linear combination. Solving this, we have $c_1=x-y, c_2=y$.
Hence the linear combination is
$\begin{bmatrix} x \\ y \end{bmatrix} =(x-y) \begin{bmatrix} 1 \\ 0 \end{bmatrix} + y \begin{bmatrix} 1 \\ 1 \end{bmatrix}.$ Using the linearity of $L$, we compute
\begin{align*}
L\left( \begin{bmatrix}
x \\
y
\end{bmatrix}\right)
&=
L\left( (x-y) \begin{bmatrix}
1 \\
0
\end{bmatrix} +
y \begin{bmatrix}
1 \\
1
\end{bmatrix}\right)
=(x-y) L\left( \begin{bmatrix}
1 \\
0
\end{bmatrix} \right) +
y L\left( \begin{bmatrix}
1 \\
1
\end{bmatrix}\right) \\
&=(x-y) \begin{bmatrix}
1 \\
1 \\
2
\end{bmatrix}+y \begin{bmatrix}
2 \\
3 \\
2
\end{bmatrix}
=
\begin{bmatrix}
x+y \\
x+2y \\
2x
\end{bmatrix}.
\end{align*}
Therefore the formula is
$L\left( \begin{bmatrix} x \\ y \end{bmatrix}\right) = \begin{bmatrix} x+y \\ x+2y \\ 2x \end{bmatrix}.$
##### Find the Rank of the Matrix $A+I$ if Eigenvalues of $A$ are $1, 2, 3, 4, 5$
Let $A$ be an $n$ by $n$ matrix with entries in complex numbers $\C$. Its only eigenvalues are $1,2,3,4,5$, possibly... | 2019-10-20T20:07:27 | {
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https://gaurish4math.wordpress.com/tag/prime-numbers/ | Prime Number Problem
Standard
Following is a problem about prime factorization of the sum of consecutive odd primes. (source: problem 80 from The Green Book of Mathematical Problems)
Prove that the sum of two consecutive odd primes is the product of at least three (possibly repeated) prime factors.
The first thing to observe is that sum of odd numbers is even, hence the sum of two consecutive odd primes will be divisible by 2. Let’s see factorization of some of the examples:
$3 + 5 = 2\times 2 \times 2$
$5 + 7 = 2 \times 2\times 3$
$7+11 = 2 \times 3\times 3$
$11+13 = 2 \times 2 \times 2 \times 3$
$13+17 = 2 \times 3 \times 5$
$17+19 = 2\times 2\times 3 \times 3$
$19+23 = 42 = 2\times 3\times 7$
$23+29 = 52 = 2\times 2 \times 13$
Now let $p_n$ and $p_{n+1}$ be the consecutive odd primes, then from above observations we can conjecture that either $p_n+p_{n+1}$ is product of at least three distinct primes or $p_n+p_{n+1}= 2^k p^\ell$ for some odd prime $p$ such that $k+\ell \geq 3$.
To prove our conjecture, let’s assume that $p_n+p_{n+1}$ is NOT a product of three (or more) distinct primes (otherwise we are done). Now we will have to show that if $p_n+p_{n+1}= 2^k p^\ell$ for some odd prime $p$ then $k+\ell \geq 3$.
If $\ell = 0$ then we should have $k\geq 3$. This is true since $3+5=8$.
Now let $\ell > 0$. Since $k\geq 1$ (sum of odd numbers is even), we just need to show that $k=1, \ell=1$ is not possible. On the contrary, let’s assume that $k=1,\ell = 1$. Then $p_n+p_{n+1} = 2p$. By arithmetic mean property, we have
$\displaystyle{p_n < \frac{p_n+p_{n+1}}{2}} = p
But, this contradicts the fact that $p_n,p_{n+1}$ are consecutive primes. Hence completing the proof of our conjecture.
This is a nice problem where we are equating the sum of prime numbers to product of prime numbers. Please let me know the flaws in my solution (if any) in the comments.
Finite Sum & Divisibility – 2
Standard
Earlier this year I discussed a finite analogue of the harmonic sum. Today I wish to discuss a simple fact about finite harmonic sums.
If $p$ is a prime integer, the numerator of the fraction $1+\frac{1}{2}+\frac{1}{3}+\ldots + \frac{1}{p-1}$ is divisible by $p$.
We wish to treat the given finite sum modulo $p$, hence we can’t just add up fractions. We will have to consider each fraction as inverse of an integer modulo p. Observe that for $0, we have inverse of each element $i$ in the multiplicative group $\left(\mathbb{Z}/p\mathbb{Z}\right)^\times$, i.e. there exist an $i^{-1}$ such that $i\cdot i^{-1}\equiv 1 \pmod p$.
For example, for $p=5$, we have $1^{-1}=1, 2^{-1}=3, 3^{-1}=2$ and $4^{-1}=4$.
Hence we have
$\displaystyle{i\cdot \frac{1}{i}\equiv 1 \pmod p ,\qquad (p-i)\cdot \frac{1}{p-i} \equiv 1 \pmod p }$
for all $0.
Hence we have:
$\displaystyle{i\left(\frac{1}{i}+\frac{1}{p-i}\right)\equiv i\cdot \frac{1}{i} - (p-i)\cdot \frac{1}{p-i} \equiv 0 \pmod p}$
Thus we have:
$\displaystyle{\frac{1}{i}+\frac{1}{p-i}\equiv 0 \pmod p}$
The desired result follows by summation.
We can, in fact, prove that the above harmonic sum is divisible by $p^2$, see section 7.8 of G. H. Hardy and E. M. Wright’s An Introduction to the Theory of Numbers for the proof.
Ulam Spiral
Standard
Some of you may know what Ulam’s spiral is (I am not describing what it is because the present Wikipedia entry is awesome, though I mentioned it earlier also). When I first read about it, I thought that it is just a coincidence and is a useless observation. But a few days ago while reading an article by Yuri Matiyasevich, I came to know about the importance of this observation. (Though just now I realised that Wikipedia article describes is clearly, so in this post I just want to re-write that idea.)
It’s an open problem in number theory to find a non-linear, non-constant polynomial which can take prime values infinitely many times. There are some conjectures about the conditions to be satisfied by such polynomials but very little progress has been made in this direction. This is a place where Ulam’s spiral raises some hope. In Ulam spiral, the prime numbers tend to create longish chain formations along the diagonals. And the numbers on some diagonals represent the values of some quadratic polynomial with integer coefficients.
Ulam spiral consists of the numbers between 1 and 400, in a square spiral. All the prime numbers are highlighted. ( Ulam Spiral by SplatBang)
Surprisingly, this pattern continues for large numbers. A point to be noted is that this pattern is a feature of spirals not necessarily begin with 1. For examples, the values of the polynomial $x^2+x+41$ form a diagonal pattern on a spiral beginning with 41.
Repelling Numbers
Standard
An important fact in the theory of prime numbers is the Deuring-Heilbronn phenomenon, which roughly says that:
The zeros of L-functions repel each other.
Interestingly, Andrew Granville in his article for The Princeton Companion to Mathematics remarks that:
This phenomenon is akin to the fact that different algebraic numbers repel one another, part of the basis of the subject of Diophantine approximation.
I am amazed by this repelling relation between two different aspects of arithmetic (a.k.a. number theory). Since I have already discussed the post Colourful Complex Functions, wanted to share this picture of the algebraic numbers in the complex plane, made by David Moore based on earlier work by Stephen J. Brooks:
In this picture, the colour of a point indicates the degree of the polynomial of which it’s a root, where red represents the roots of linear polynomials, i.e. rational numbers, green represents the roots of quadratic polynomials, blue represents the roots of cubic polynomials, yellow represents the roots of quartic polynomials, and so on. Also, the size of a point decreases exponentially with the complexity of the simplest polynomial with integer coefficient of which it’s a root, where the complexity is the sum of the absolute values of the coefficients of that polynomial.
Moreover, John Baez comments in his blog post that:
There are many patterns in this picture that call for an explanation! For example, look near the point $i$. Can you describe some of these patterns, formulate some conjectures about them, and prove some theorems? Maybe you can dream up a stronger version of Roth’s theorem, which says roughly that algebraic numbers tend to ‘repel’ rational numbers of low complexity.
To read more about complex plane plots of families of polynomials, see this write-up by John Baez. I will end this post with the following GIF from Reddit (click on it for details):
Prime Consequences
Standard
Most of us are aware of the following consequence of Fundamental Theorem of Arithmetic:
There are infinitely many prime numbers.
The classic proof by Euclid is easy to follow. But I wanted to share the following two analytic equivalents (infinite series and infinite products) of the above purely arithmetical statement:
• $\displaystyle{\sum_{p}\frac{1}{p}}$ diverges.
For proof, refer to this discussion: https://math.stackexchange.com/q/361308/214604
• $\displaystyle{\sum_{n=1}^\infty \frac{1}{n^{s}} = \prod_p\left(1-\frac{1}{p^s}\right)^{-1}}$, where $s$ is any complex number with $\text{Re}(s)>1$.
The outline of proof, when $s$ is a real number, has been discussed here: http://mathworld.wolfram.com/EulerProduct.html | 2018-02-25T17:28:08 | {
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http://clandiw.it/fvql/conversion-of-cartesian-coordinates-to-polar-coordinates-pdf.html | # Conversion Of Cartesian Coordinates To Polar Coordinates Pdf
Coordinates conversion from polar coordinates to cartesian x,y coordinates, and from cartesian coordinates to polar coordinates. A point can be represented by polar coordinates (r; ), where ris the distance between the point and the origin, or pole, and is the angle that a line segment from the pole to the point makes with the positive x-axis. Review: Polar coordinates Definition The polar coordinates of a point P ∈ R2 is the ordered pair (r,θ) defined by the picture. Just to be clear, here's what you need to calculate…to perform coordinate conversions. Figure 4-1 Polar Coordinates and Rectangular Coordinates An example of polar coordinates is right ascension and declination, (a, d). Convert from rectangular coordinates to polar coordinates using the conversion formulas. SOLUTION:. 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. Cartesian Cylindrical Spherical Cylindrical Coordinates x = r cosθ r = √x2 + y2 y = r sinθ tan θ = y/x z = z z = z Spherical Coordinates. In such a coordinate system you can calculate the distance between two points and perform operations like axis rotations without altering this value. You will create at least three classes for this program %u2013 Coordinate, PolarCoordinate, and CartesianCoordinate. The Cartesian coordinates x and y can be converted to polar coordinates r and φ with r ≥ 0 and φ in the interval (− π, π] by: = + (as in the Pythagorean theorem or the Euclidean norm), and = (,), where atan2 is a common variation. Worksheet: Polar Coordinates Download. 3 Polar Coordinates The Cartesian coordinate system is not the only one. FREE Answer to How do you convert the cartesian coordinate ( 8. For the following exercises, convert the given Cartesian coordinates to polar coordinates with. π To enter π type Pi 3,1. Find the magnitude of the polar coordinate. Cartesian to Polar Coordinates. To start with this program you must understand definitions of Rectangular and Polar coordinates. I Computing volumes using double integrals. A Cartesian coordinate system (UK: / k ɑː ˈ t iː zj ə n /, US: / k ɑːr ˈ t i ʒ ə n /) is a coordinate system that specifies each point uniquely in a plane by a set of numerical coordinates, which are the signed distances to the point from two fixed perpendicular oriented lines, measured in the same unit of length. Then x = r cos O, Y = r sin O. Coordinate Systems B. This is the official, unambiguous definition of polar coordinates, from which we. Convert the following equation to polar. It's 2 units awa. EXAMPLE 11: Convert y = 10 into a polar equation. The calculator will convert the polar coordinates to rectangular (Cartesian) and vice versa, with steps shown. In such a coordinate system you can calculate the distance between two points and perform operations like axis rotations without altering this value. 2 Slopes in r pola tes coordina When we describe a curve using polar coordinates, it is still a curve in the x-y plane. Now I need to go the other way around. Example: What is (12,5) in Polar Coordinates? Use Pythagoras Theorem to find the long side (the hypotenuse):. Rectangular coordinates Rectangular coordinates and polar coordinates are two different ways of using two numbers to locate a point on a plane. It is good to begin with the simpler case, cylindrical coordinates. , a nest or burrow) based on distance and bearing from a grid point, this function helps me avoid writing down SOH-CAH-TOA every time. 14) of https://yoquieroaprobar. Polar-Cartesian Coordinate conversion and Cartesian-Polar Hi All, Is there any standard program to convert Polar Coordinates to Cartesian Coordinates. 11, page 636. EXAMPLE 10. Converting from Polar Coordinates to Rectangular Coordinates. But I need a better precision, therefore I hope you can help me finding some formulas to convert cartesian coordinates to geographical gps. In polar coordinates, the shape we work with is a polar rectangle, whose sides have. They are also called "Euclidean coordinates," but not because Euclid discovered them first. Cylindrical and spherical coordinates 1. Convert from rectangular coordinates to polar coordinates using the conversion formulas. asked by BM on October 25, 2008; Trig. Menu Options. 2 , 53 o) to rectangular coordinates to. That should be enough. Converting Cartesian Coordinates to Screen Coordinates When working with computer or calculator graphics, sometimes we have to work with screen coordinates. I input r and thetha coordinate in my. Better Precision. Convert the Cartesian coordinate (3,4) to polar coordinates, 0 < 0 < 2ñ Preview Enter exact value. Converting from Cartesian to Polar Coordinates. In polar coordinates, however, the two unit vectors, r and q, do depend on each other, and change their. The ranges of the variables are 0 < p < °° 0 < < 27T-00 < Z < 00 A vector A in cylindrical coordinates can be written as (2. Is there something like this in excel?. X=Y=Z for stimulus of equal luminance at each wavelength). How to Convert Between Polar and Cartesian Coordinates. Math behind Converting Polar to Rectangular Coordinates. To Convert from Cartesian to Polar. Try to write a program to convert from Cartesian coordinate system to polar coordinate system and send it to me or comment below. -Polar to cartesian relationship. Let (x, y, z) be the standard Cartesian coordinates, and (ρ, θ, φ) the spherical coordinates, with θ the angle measured away from the +Z axis (as , see conventions in spherical coordinates). \\ r^2 = x^2 + y^2 \\ \theta = \arctan (\frac yx) {/eq. Angle t is in the range [0 , 2Pi) or [0 , 360 degrees). Converting between polar and Cartesian coordinates is really pretty simple. Another two-dimensional coordinate system is polar coordinates. 3 Polar Coordinates The Cartesian coordinate system is not the only one. The point (x,y) would be given in polar coordinates by the pair (r, θ), as shown. 1 De ning Polar Coordinates oT nd the coordinates of a point in the polar coordinate system, consider Figure 1. Coordinate Systems B. I think such methods would be pretty useful. 11) ( , ), ( , ) 12) ( , ), ( , ) Critical thinking question: 13) An air traffic controller's radar display uses polar coordinates. There are an infinite number of ways to write the same point in polar coordinates. c) Show that the area of the shaded region is 1(2 3 3) 2 π−. We first compute the Jacobian for the change of variables from Cartesian coordinates to polar coordinates. And the usual way is to place the origin of the Cartesian coordinates at the pole and to place the positive x-axis for your Cartesian coordinates along the polar axis and then the Cartesian axis corresponding to y goes where it has to go. I’m not a mathematician but technically speaking, the correct terminology should be “the absolute value of quadrant IV in the cartesian coordinate system”. I found an earlier question (How do I calculate a xyz-position of a gps-position relative to an other gps-position?) which worked perfectly. In spherical coordinates: Converting to Cylindrical Coordinates. The Cartesian coordinates (x, y, z) and polar coordinates (θ,φ,r) of a common reference point, as illustrated in Fig. The polar coordinates of a point and the polar coordinate grid. This Polar Coordinates Presentation is suitable for 10th - 12th Grade. A summary of Parametric Equations in 's Parametric Equations and Polar Coordinates. The first coordinate is called the radial coordinate and the second coordinate is called the angular coordinate. 3 Polar Coordinates The Cartesian coordinate system is not the only one. The coordinate in this system shows the distance of the point in question from the point of origin. When I want to calculate the coordinates of a location (e. Enter your data in the left hand box with each. -Cartesian and polar relationship. , there are two perpendicular unit vectors ##\vec{e}_j##. We convert from polar coordinates to rectangular coordinates and from rectangular coordinates to polar coordinates. Q1: Consider the points plotted on the graph. Immediately, we have the time dependence. Better Precision. We convert from polar coordinates to rectangular coordinates and from rectangular coordinates to polar coordinates. Notation for different coordinate systems The general analysis of coordinate transformations usually starts with the equations in a Cartesian basis (x, y, z) and speaks of a transformation of a general alternative coordinate. Cartesian coordinates can be used to pinpoint where we are on a map or graph. 4 Interconversion between polar and Cartesian coordinates. Conversion between Polar and Cartesian Coordinates. Interface problems defined in a disk using polar coordinates. x and y are related to the polar angle θ through the sine and cosine functions (purple box). Converts from Cartesian (x,y,z) to Spherical (r,θ,φ) coordinates in 3-dimensions. The method of relative Cartesian coordinates The autocad differs from the method of absolute coordinates in that the coordinates X, Y are given relative to the last specified point, and not relative to the origin. 4145 ) into polar coordinates?. We have and Note that is in the second quadrant (x negative, y positive). If the point is not in the first quadrant then you should find an acute angle α using a right-. In rectangular coordinates, each point (x, y) has a unique representation. The value of θ may be given in degrees or radians. CONIC SECTIONS IN POLAR COORDINATES If we place the focus at the origin, then a conic section has a simple polar equation. $\begingroup$ Dear @RenéG, you even do not need to use Solve. That should be enough. One of the particular cases of change of variables is the transformation from Cartesian to polar coordinate system $$\left({\text. The term "Cartesian coordinates" is used to describe such systems, and the values of the three coordinates unambiguously locate a point in space. It will not be a circular image of course. Call fromPolar() - to convert polar coordinates to cartesian coordinates. They are very closely related to the trigonometric form of complex numbers covered in Section 9. New, dedicated functions are available to convert between Cartesian and the two most important non-Cartesian coordinate systems: polar coordinates and spherical coordinates. -Polar to cartesian relationship. The calculator will convert the polar coordinates to rectangular (Cartesian) and vice versa, with steps shown. Each reference line is called a coordinate axis or just axis (plural. Here, is the imaginary unit. a) Find a Cartesian equation for C1 and a Cartesian equation for C2. In the polar coordinate system for example (r, θ) and (r,θ + 2π) represent the same point as do (r, θ) and (-r,θ + π). CARTESIAN & POLAR COORDINATES In fact, as a complete counterclockwise rotation is given by an angle 2π, the point represented by polar coordinates (r, θ) is also represented by (r, θ+ 2nπ) and (-r, θ+ (2n + 1)π) where n is any integer. The answer is: (r,θ) Polar = (p x2 +y2, arctan y x) Polar Meanwhile, for a point given by Polar coordinates, (r,θ) Polar, we need to specify the coordinates in Cartesian form in terms of the Polar data r and θ. I'll cover the following topics in the code samples below: FloatCODE, Cartesian Coordinates, Polar Coordinates, Download, and c Polar. We are all comfortable using rectangular (i. r is the distance to the z-axis (0, 0, z). SOLUTION: This is a graph of a horizontal line with y-intercept at (0, 10). I Computing volumes using double integrals. Approximate the. This function uses the metadata in the message, such as angular resolution and opening angle of the laser scanner, to perform the conversion. Figure 1: Standard relations between cartesian, cylindrical, and spherical coordinate systems. This article will provide you with a short explanation of both types of coordinates and formulas for quick conversion. To de ne polar coordinates, we start by identifying a pole or origin, labeled O, usually taken to be the same as the origin in Cartesian coordinates. coordinate transformations are particularly complex if range rate (5) and range acceleration (S) are used. Call fromPolar() - to convert polar coordinates to cartesian coordinates. Storrs, CT 06269-2157 [email protected] The Chain Rule Polar Coordinates Example Example 6: Find the gradient of a function given in polar coordinates. When we use this polar-to-cartesian function, we enter a magnitude and an angle in degrees as parameters. Latitude and Longitude Converter is a tool to convert gps coordinates to address, and convert address to lat long. Solution: This calculation is almost identical to finding the Jacobian for polar. We merely substitute: rsinθ = 3rcosθ + 2, or r = 2 sinθ −3cosθ. All right triangles with a given angle. I am dealing with a time series in polar coordinates and I am applying the Kalman filter for predictions. The first description is like giving \(x$$- and $$y$$-coordinates (also known as Cartesian coordinates); the second is like giving polar coordinates. The scalar components can be expressed using Cartesian, cylindrical, or spherical coordinates, but we must always use Cartesian base vectors. If we convert complex number to its polar coordinate, we find:. For example, the point (3, 3) in rectangular coordinates becomes (√18, 45°) in polar coordinates. To convert from polar co-ordinates to Cartesian co-ordinates, use the equations x = rcosθ, y = rsinθ. 980878°$,$\phi = 40. Polar and Rectangular (Cartesian) Coordinate Conversion Robert B. Plotting Points Using Polar Coordinates. Cartesian coordinates can be converted to polar coordinates using the following formulas:. Q1: Consider the points plotted on the graph. 2 We can describe a point, P, in three different ways. In this paper, numerical methods are proposed for some interface problems in polar or Cartesian. This calculator can be used to convert 2-dimensional (2D) or 3-dimensional cartesian coordinates to its equivalent cylindrical coordinates. NumPy Random Object Exercises, Practice and Solution: Write a NumPy program to convert cartesian coordinates to polar coordinates of a random 10x2 matrix representing cartesian coordinates. Converting between polar and Cartesian coordinates. I just only wanted to show the way of applying CoordinateTransform for more complex cases. For example, the point (3, 3) in rectangular coordinates becomes (√18, 45°) in polar coordinates. Polar Coordinates Conversion from polar to cartesian (rectangular) x = r cos θ y = r sin θ r Conversion from cartesian to y θ polar: x r= x2 + y2 x y y cos θ = sin θ = tan θ = r r x 6. This is the polar axis. Spherical coordinates consist of the following three quantities. From polar to cartesian coordinates In this video Francis describes polar coordinates and how to convert them into cartesian coordinates. If we wish to relate polar coordinates back to rectangular coordinates (i. Press Similarly Input Polar coordinates of P, r first. A general system of coordinates uses a set of parameters to define a vector. In such a coordinate system you can calculate the distance between two points and perform operations like axis rotations without altering this value. The innermost circle shown in Figure 7. This is the official, unambiguous definition of polar coordinates, from which we. For the following exercises, convert the given Cartesian coordinates to polar coordinates with. Convert the Cartesian coordinates (-V3, 1) to polar coordinates, where T-0 and θ E [O, 27). where r is the radial coordinate and θ is the polar coordinate, i. Polar coordinates: We will use the following formulas to convert from cartesian coordinates (x,y) to polar coordinates {eq}(r,\theta). The spherical coordinate system extends polar coordinates into 3D by using an angle ϕ for the third coordinate. 1 Helmholtz Equation and Angular Basis Functions As a direct extension from the Cartesian case, we begin with the eigenfunctions of the Laplacian, whose expression in polar coordinates is given by: ∇2 = ∇2. Find the magnitude of the polar coordinate. 9) ( , ) 10) ( , ) Two points are specified using polar coordinates. 3 Polar Coordinates The Cartesian coordinate system is not the only one. Trigonometry and the Pythagorean Theorem allow for straightforward conversion from rectangular to polar, and vice versa. To convert data from radians to degrees, use rad2deg. This is the xy-plane. The answer is: (r,θ) Polar = (p x2 +y2, arctan y x) Polar Meanwhile, for a point given by Polar coordinates, (r,θ) Polar, we need to specify the coordinates in Cartesian form in terms of the Polar data r and θ. Cartesian Coordinate. Reformat latitude and longitude coordinates between: Degrees, minutes and seconds. Degrees and decimal minute formats. Well, we have to put in the Cartesian coordinates if we're going to compare them. Replace and with the actual values. Cylindrical and spherical coordinates Recall that in the plane one can use polar coordinates rather than Cartesian coordinates. Let us discuss these in turn. Rectangular coordinates are depicted by 3 values, (X, Y, Z). Plotting Points Using Polar Coordinates. Finally, if you substitute r in the las two, you already have the solution I posted. To convert rectangular coordinates to polar coordinates, we will use two other familiar relationships. Conversion Between Cartesian To Polar Coordinates And Back? Nov 8, 2011. I Double integrals in arbitrary regions. The connection between Cartesian coordinates and Polar coordinates is established by basic trigonometry. Prove that it is a circle in the Cartesian Coordinate system. De nition (polar coordinate system). Convert an equation from rectangular to polar coordinates. Yet I will tell you tip, it doesn't work well in those cases. Find the distance between the points. The rectangular coordinates are given in x and y form and the polar coordinates are given in the radius and the angle form. 1 Basis Functions 2. After working through these materials, the student should be able. In polar coordinates we specify a point using the distance rfrom the origin and the angle with the x-axis. 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. We will then learn how to graph polar equations by using 2 methods. (-3, 2π/3) 5 EX 4 Plot r = 6 sin θ. We have seen that Laplace’s equation is one of the most significant equations in physics. The polar coordinate system is de ned by a pole (origin) and polar axis (usually drawn in the direction of the positive x-axis). The material in this document is copyrighted by the author. There are no conversion between formats, such as between decimal and DMS. Call fromCartesian() - to convert cartesian coordinates to polar coordinates. is completely determined by its real part and imaginary part. k:\surveying\sem1-10\how to convert rectangular coordinates to polar coordinates. Coordinate will be the parent of the other two classes and. Example: Polar to Rectangular Example Find the rectangular coordinates of the point with polar √ coordinates ( 2, 5π/4). Polar Coordinates In a plane, suppose you have a point O called the origin, and an axis through that point - say the x -axis - called the polar axis. 50) m, as shown in Active Figure 1. When you look at the polar coordinate, the first number is the radius of a circle. Comment/Request Great tool! It helped me understand polar coordinates much more after seeing it mentioned a few times in a YouTube video, which left me confused, until I came back with this knowledge. Quantitative corn-. In mathematics, a Cartesian coordinate system is a coordinate system that specifies each point uniquely in a plane by a set of numeric points. Cartesian Fundamentals. Purpose of use Too lazy to do homework myself. Determine the Cartesian coordinates of the centre of the circle and the length of its radius. Review of Coordinate Systems A good understanding of coordinate systems can be very helpful in solving problems related to Maxwell’s Equations. Polar Coordinates 🌀 Cartesian Coordinate System. 56 CHAPTER 1. By using this website, you agree to our Cookie Policy. 13 degrees counterclockwise from the x-axis, and then walk 5 units. Well, as you already know, a point in the Rectangular or Cartesian Plane is represented by an ordered pair of numbers called coordinates (x,y). Coordinates of the point: $$(r, θ)$$ or $$(x, y)$$ Method. However my prediction and estimation for the variance are expressed in polar coordinates [r,theta]. , Cartesian) coordinates to describe points on the plane. The coordinate converter of the present invention constitues a basic unit for forming Cartesian coordinates from the polar coordinates of a vector. This is a coordinate system in a plane, or two dimensions. For example, a radar system generates measurements in its own local spherical coordinate system. You should have used instead of Solve. For example, the point (1/2, √3/2), which makes a 30° angle with the x-axis, could have its coordinates represented as (cos30°, sin30°). SYNOPSIS IntreatingtheHydrogenAtom’selectronquantumme-chanically, we normally convert the Hamiltonian from its Cartesian to its Spherical Polar form, since the problem is. Your task could be done by hand. Convert from cylindrical to rectangular coordinates. ( )2,2 , radius 8= Question 6 Write the polar equation r = +cos sinθ θ , 0 2≤ <θ π in Cartesian form, and hence show that it represents a circle, further determining the coordinates of its centre and the size of its radius. Heckendorn University of Idaho April 29, 2015 1 Angle and Distance to X and Y Conversion from an angle and a distance to the X and Y position at that distance and angle is very easy thanks to the sine and cosine functions from trigonometry. It will not be a circular image of course. Spherical coordinates consist of the following three quantities. EXAMPLE 11: Convert y = 10 into a polar equation. In fact, as a complete counterclockwise rotation is given by an angle 2π, the point represented by polar coordinates (r, θ) is also represented by (r, θ + 2nπ) and (-r, θ + (2n + 1)π) where n is any integer. Using the inverse cosine function on a calculator, we obtain (in radians) Hence the polar form of is cos 2. Just to be clear, here's what you need to calculate…to perform coordinate conversions. 11 i + sin 2. For example, x, y and z are the parameters that define a vector r in Cartesian coordinates: r =ˆıx+ ˆy + ˆkz (1) Similarly a vector in cylindrical polar coordinates is described in terms of the parameters r, θ and z since a vector r can be written as r = rrˆ+ zˆk. Polar coordinates: We will use the following formulas to convert from cartesian coordinates (x,y) to polar coordinates {eq}(r,\theta). When I want to calculate the coordinates of a location (e. To Convert from Cartesian to Polar. 11, page 636. Cartesian coordinates in the figure below: (2,3) A Polar coordinate system is determined by a fixed point, a origin or pole, and a zero direction or axis. 6 Velocity and Acceleration in Polar Coordinates 12 Proof of Kepler's Second Law. com interactive, accessed 06/2016. Convert the following equation to polar coordinates: y = − 2 3 x 3. Evaluation of Double Integrals By Changing Cartesian Coordinates into Polar Coordinates By F ANITHA - Duration: 24:26. They are also called "Euclidean coordinates," but not because Euclid discovered them first. To convert from spherical coordinates to cartesian coordinates can be done the following way in c#:. 8, as outlined in the. This is what the basic principle on which conversions between Cartesian and polar are based. Display the two images. \\ r^2 = x^2 + y^2 \\ \theta = \arctan (\frac yx) {/eq. The input values for x and y are read from the user using scanner object and these values are converted into corresponding polar coordinate values by following two equations. 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'm having a problem throughout the conversion process of converting Cartesian to Polar, then back to Cartesian in order to confirm that my initial conversion was successful. We also took a short quiz on graphing polar coordinates and converting between polar and rectangular coordinates. The simplest way in this specific case is to note that the $y$ component is zero, so the point lies on the $x$-axis. 1 Review: Polar Coordinates The polar coordinate system is a two-dimensional coordinate system in which the position of each point on the plane is determined by an angle and a distance. Example: Expressing Vector Fields with Coordinate Systems Consider the vector field: ˆˆˆ() 22 xyz x xz a x y a a z ⎛⎞ =++ +⎜⎟ ⎝⎠ A Let’s try to accomplish three things: 1. If you use Non Linear Transformation use something that will both make things easier and better (Yea, usually it doesn't work like that, but in this case it does) - Use the Unscented. For this step, you use the Pythagorean theorem for polar coordinates: x2 + y2 = r2. Polar Coordinates Basic Introduction, Conversion to Rectangular, How to Plot Points, Negative R Valu - Duration: 22:30. Convert the Cartesian coordinates (-V3, 1) to polar coordinates, where T-0 and θ E [O, 27). 5) or the transformation from one 2D Cartesian ( x, y) system of a specific map projection. This is the polar axis. Converting Complex numbers into Cartesian Form. 14) of https://yoquieroaprobar. 1 Helmholtz Equation and Angular Basis Functions As a direct extension from the Cartesian case, we begin with the eigenfunctions of the Laplacian, whose expression in polar coordinates is given by: ∇2 = ∇2. Since the x-coordinate is negative but the y-coordinate is positive, this angle is located in the second quadrant. k:\surveying\sem1-10\how to convert rectangular coordinates to polar coordinates. The spherical coordinate system I'll be looking at, is the one where the zenith axis equals the Y axis and the azimuth axis equals the X axis. The Organic Chemistry Tutor 289,935 views 22:30. Home Coordinate Systems 3D Conversion of 3D Coordinate Systems: See also: Conversion between Polar and Cartesian Coordinates, Three-dimensional Cartesian Coordinate System, Cylindrical Coordinate System, Spherical Coordinate System : Search the VIAS Library | Index. It allows you to locate each point by a pair of numerical coordinates (x, y). Convert the Cartesian coordinates defined by corresponding entries in matrices x and y to polar coordinates theta and rho. There are of course other coordinate systems, and the most common are polar, cylindrical and spherical. From my text book, I know. Plotting Points Using Polar Coordinates. Chapter 1: Introduction to Polar Coordinates. from rectangular to polar coordinates. polar coordinates, and (r,f,z) for cylindrical polar coordinates. CARTESIAN & POLAR COORDINATES. [See how to convert rectangular and polar forms in the complex numbers chapter. Polar and Rectangular (Cartesian) Coordinate Conversion Robert B. Spherical coordinates consist of the following three quantities. Cartesian to Polar coordinates To convert from Cartesian to polar coordinates, we use the following identities r2 = x2 + y2; tan = y x When choosing the value of , we must be careful to consider which quadrant the point is in, since for any given number a, there are two angles with tan = a, in the interval 0 2ˇ. In this note, I would like to derive Laplace's equation in the polar coordinate system in details. It allows you to locate each point by a pair of numerical coordinates (x, y). Make the following change of variables from rectangular coordinates to polar coordinates: x = r*cos(@), y = r*sin(@), r^2 = x^2 + y^2, @ = arctan(y/x) Then. The figure above shows the two curves intersecting at the pole and at the point P. In this article we will discuss about the conversion between them. Furthermore, we relate the truncation limits, N and M, so that there is no loss of information during conversion. From the above two formulas, r and θ can be defined in terms of x and y: With this formula θ is obtained in [0, 2π), or [0°, 360°). Home Coordinate Systems 3D Conversion of 3D Coordinate Systems: See also: Conversion between Polar and Cartesian Coordinates, Three-dimensional Cartesian Coordinate System, Cylindrical Coordinate System, Spherical Coordinate System : Search the VIAS Library | Index. Polar coordinate system: The polar coordinate system is a two-dimensional coordinate system in which each point P on a plane is determined by the length of its position vector r and the angle q between it and the positive direction of the x-axis, where 0 < r < + oo and 0 < q < 2p. , there are two perpendicular unit vectors ##\vec{e}_j##. Theres a question that asks you to investigate into how to convert polar coordinates into cartesian. c) Show that the area of the shaded region is 1(2 3 3) 2 π−. Image Transcriptionclose. You can modify polar axes properties to customize the chart. Recall the Quadrant III adjustment, which is the same as the Quadrant II adjustment. 1 Cartesian Coordinates A coordinate system consists of four basic elements: (1) Choice of origin (2) Choice of axes (3) Choice of positive direction for each axis (4) Choice of unit vectors for each axis We illustrate these elements below using Cartesian coordinates. Cartesian coordinates can be used to pinpoint where we are on a map or graph. Polar Coordinates Basic Introduction, Conversion to Rectangular, How to Plot Points, Negative R Valu - Duration: 22:30. x is the x-coordinate of the point in Cartesian coordinates. The polar coordinate system is a two-dimensional coordinate system in which each point on a plane is determined by an angle and a distance. 8) h p = cosφ −ν (1. The above result is another way of deriving the result dA=rdrd(theta). This calculator allows you to convert between Cartesian, polar and cylindrical coordinates. However, if you insist on not converting, write out the entire process with all 4 points of interest in Cartesian coordinates. In modern data processing systems, a Cartesian to polar transformation involves the creation of a fixed map derived from parameters defining the relationship between the Cartesian and polar systems; each Cartesian coordinate pair (X,Y) in the source image corresponding to a polar coordinate pair (R,θ) in the destination image. 030068689428860915 for the values of x and y that you originally posted. Compute the r image and theta image at each row and column. This is the distance from the origin to the point and we will require ρ ≥ 0. For instance, the point (0,1) in Cartesian coordinates would be labeled as (1, p/2) in polar coordinates; the Cartesian point (1,1) is equivalent to the polar coordinate position 2, p/4). Plotting Points Using Polar Coordinates. The answer is: (r,θ) Polar = (p x2 +y2, arctan y x) Polar Meanwhile, for a point given by Polar coordinates, (r,θ) Polar, we need to specify the coordinates in Cartesian form in terms of the Polar data r and θ. 1 Review: Polar CoordinatesThe polar coordinate system is a two-dimensional coordinate system in whichthe position of each point on the plane is determined by an angle and a distance. for converting between Cartesian and Polar coordinates. To convert from Cartesian co-ordinates to polar co-ordinates, use the equations r 2= x +y2, tanθ = y x. One of the particular cases of change of variables is the transformation from Cartesian to polar coordinate system $$\left({\text. We just use a little trigonometry and the Pythagorean theorem. Scan the image by row (y) and column (x). Solution: This calculation is almost identical to finding the Jacobian for polar. SOLUTION: This is a graph of a horizontal line with y-intercept at (0, 10). Conversion from rectangular to polar coordinates. Review: Polar coordinates Definition The polar coordinates of a point P ∈ R2 is the ordered pair (r,θ) defined by the picture. But, for some reason when I am converting back to Cartesian coordinates with Polar coordinates from the third quadrant, my x and y values are the wrong way around. X=Y=Z for stimulus of equal luminance at each wavelength). I Derivation of Some General Relations The Cartesian coordinates (x, y, z) of a vector r are related to its spherical polar. Figure 2-20. Write down the free particle Schr\”{o}dinger equation for two dimensions in (i) Cartesian and (ii) polar coordinates. Tags c#, cartesian, coordinate, polar 1712 Views. So, although polar coordinates seem to complicate things when you are first introduced to them, learning to use them can simplify math for you quite a bit! Similarly, converting an equation from polar to rectangular form and vice versa can help. GPS coordinates converter. sph 1 ÖÖ1 0 0 0ÖÖ 0 ªº «» «» «»¬¼ rrTI 1 1 1 sin cos cos cos sin 1 sin cos. Convert polar to cartesian coordinates pol2cart: Convert polar to cartesian coordinates in nverno/rstuff: Utilities and stuff rdrr. Cylindrical Coordinates. coordinate transformations are particularly complex if range rate (5) and range acceleration (S) are used. The issue is to find cartesian coordinate in 0,0 origin based chart. 3 Polar Coordinates Date: Goals: Plot points in the polar coordinate system. Converting from rectangular coordinates to polar coordinates. 5) or the transformation from one 2D Cartesian ( x, y) system of a specific map projection. As soon as you modify one end of the data (either the decimal or sexagesimal degrees coordinates), the other end is simultaneously updated, as well as the position on the map. To plot additional data in the polar axes, use the hold on command. SOLUTION:. To get some intuition why it was named like this, consider the globe having two poles: Arctic and Antarctic. Rectangular coordinates are x and y position on a Cartesian coordinate system. I cannot open the pdf file in #6. Approximate the. They are also called "Euclidean coordinates," but not because Euclid discovered them first. We are supposed to convert this func-tion to Cartesian coordinates. To de ne polar coordinates, we start by identifying a pole or origin, labeled O, usually taken to be the same as the origin in Cartesian coordinates. Press Change to Cartesian coordinates. Converting between spherical and cartesian coordinates. (-3, 2π/3) 5 EX 4 Plot r = 6 sin θ. Precalculus: Polar Coordinates Practice Problems 3. And, these coordinates are directed horizontal and vertical distances along the x and y axes, as Khan Academy points out. Decimal Degrees (DD) Latitude (-90 to 90) and longitude (-180 to 180). Convert to Polar Coordinates (-3,0) Convert from rectangular coordinates to polar coordinates using the conversion formulas. Convert from rectangular coordinates to polar coordinates using the conversion formulas. The Laplacian in Spherical Polar Coordinates C. I've received an assignment to investigate the polar coordinate system compared to the cartesian one. In this video, we take a look at the polar coordinate system and derive the expressions for converting between cartesian coordinates and polar coordinates. Precalculus: Polar Coordinates Concepts: Polar Coordinates, converting between polar and cartesian coordinates, distance in polar coordinates. Use degrees for 0. Recall that Hence, The Jacobian is Correction There is a typo in this last formula for J. \\ r^2 = x^2 + y^2 \\ \theta = \arctan (\frac yx) {/eq. New, dedicated functions are available to convert between Cartesian and the two most important non-Cartesian coordinate systems: polar coordinates and spherical coordinates. Remember to consider the quadrant in which the given point is located when determining θ for the point. Cartesian and Polar Coordinates nbsp; The Cartesian coordinates of a point in the xy -plane are Cartesian and Polar Coordinates (a) The Cartesian coordinates of a point in the xy -plane are ( x , y ) = (-3. To specify points in space using spherical-polar coordinates, we first choose two convenient, mutually perpendicular reference directions (i and k in the picture). com interactive, accessed 06/2016. This is a familiar problem; recall. A Cartesian coordinate system (UK: / k ɑː ˈ t iː zj ə n /, US: / k ɑːr ˈ t i ʒ ə n /) is a coordinate system that specifies each point uniquely in a plane by a set of numerical coordinates, which are the signed distances to the point from two fixed perpendicular oriented lines, measured in the same unit of length. The Organic Chemistry Tutor 289,935 views 22:30. Angle t is in the range [0 , 2Pi) or [0 , 360 degrees). Clone via HTTPS Clone with Git or checkout with SVN using the repository’s web address. to Cartesian Coord. When this is the case, Cartesian coordinates (x;y;z) are converted to cylindrical coordinates (r; ;z). This program is used to translate X,Y locations in a Cartesian coordinate format to a polar coordinate format (bearing and distance) assuming a fixed reference point. There may be many ways to visualize the conversion of Polar to Rectangular coordinates. You will create at least three classes for this program %u2013 Coordinate, PolarCoordinate, and CartesianCoordinate. Using these formula the gps coordinates I get are enough precise, they correspond "enough" to what they should be considering the position of some points about which I know the correct Lat/Lon coordinates. Map and GIS users are mostly confronted in their work with transformations from one two-dimensional coordinate system to another. 1 Review: Polar CoordinatesThe polar coordinate system is a two-dimensional coordinate system in whichthe position of each point on the plane is determined by an angle and a distance. [5] Polar Coordinates, A short article with example plots and problems that demonstrate the polar coordinate system. Viewed 122 times 0 \begingroup I am wondering whether I converted the following correctly. Theres a question that asks you to investigate into how to convert polar coordinates into cartesian. Plot points in polar coordinates #1-8; Write polar coordinates for points #9-16; Convert Cartesian coordinates to polar #17-24; Convert Polar coordinates to Cartesian #25-32; Write alternate versions of polar coordinates #33-38. Cylindrical and spherical coordinates Recall that in the plane one can use polar coordinates rather than Cartesian coordinates. First there is ρ. To specify points in space using spherical-polar coordinates, we first choose two convenient, mutually perpendicular reference directions (i and k in the picture). The conversion from polar coordinates to rectangular coordinates involves using the sine and cosine functions to find x and y. Converting from Cartesian polar coordinates. However, there are other ways of writing a coordinate pair and other types of grid systems. Random-Science-Tools. with the PDE reduced to. 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. A unique aspect of Cartesian coordinates is that the unit vectors i and j always point in the same direction and are independent of each other. X,Y,Z Cartesian coordinates have an origin at the centre o f the ellipsoid. In many problems, spherical polar coordinates are better. For a list of properties, see PolarAxes Properties. ALWAYS use one of these three expressions of a position vector!! Note that in each of the three expressions above, we use Cartesian base vectors. Cartesian / Rectangular to Polar Conversion The java code converts the Cartesian coordinate values (x,y) into polar coordinatevalues (r,Θ). For example, x, y and z are the parameters that define a vector r in Cartesian coordinates: r =ˆıx+ ˆy + ˆkz (1) Similarly a vector in cylindrical polar coordinates is described in terms of the parameters r, θ and z since a vector r can be written as r = rrˆ+ zˆk. 980878°, \phi = 40. Obtain the corresponding wavefunction. Furthermore, we relate the truncation limits, N and M, so that there is no loss of information during conversion. Well, we have to put in the Cartesian coordinates if we're going to compare them. Press Change to Cartesian coordinates. [See how to convert rectangular and polar forms in the complex numbers chapter. Let us discuss these in turn. this page updated 19-jul-17. In this section we find the relationship between the representations of a function in polar and Cartesian coordinates when the truncated expansions in and are used. Free Cartesian to Polar calculator - convert cartesian coordinates to polar step by step This website uses cookies to ensure you get the best experience. We merely substitute: rsinθ = 3rcosθ + 2, or r = 2 sinθ −3cosθ. 1 De ning Polar Coordinates oT nd the coordinates of a point in the polar coordinate system, consider Figure 1. Write down the. This Graph Paper may be printed, downloaded or saved and used in your classroom, home school, or other educational environment to help someone learn math. If we were to express it in rectangular coordinates, the calculation would require a few extra steps. It uses floating point math to do so with quadrant checking to. This website uses cookies to ensure you get the best experience. I think such methods would be pretty useful. To convert data from radians to degrees, use rad2deg. I'm using Visual Studio 2010 Express, and Visual Basic. It should be obvious from the diagram that we have the following relationships between the coordinates: Cartesian to Polar Polar to Cartesian € r= x2 + y2 θ= tan−1 y x € x = rcosθ y = rsinθ Position Vectors The fun begins when we wish to describe. The r represents the distance you move away from the origin and θ represents an angle in standard position. A Cartesian coordinate system (UK: / k ɑː ˈ t iː zj ə n /, US: / k ɑːr ˈ t i ʒ ə n /) is a coordinate system that specifies each point uniquely in a plane by a set of numerical coordinates, which are the signed distances to the point from two fixed perpendicular oriented lines, measured in the same unit of length. 8) h p = cosφ −ν (1. in the polar coordinate system. I drew a perpendicular line from P to Q on the polar line, which will be the X-axis in the Cartesian system. CONVERTING FROM A CARTESIAN EQUATION TO A POLAR EQUATION. 4 Interconversion between polar and Cartesian coordinates. Certainly the most common is the Cartesian or rectangular coordinate system (xyz). Arguably, log-polar coordinates are more natural than simple polar. By using this website, you agree to our Cookie Policy. Likewise, if we have a point in Cartesian coordinates the cylindrical coordinates can be found by using the following conversions. This is the result of the conversion to polar coordinates in form. A polar coordinate ( ) is completely determined by modulus and phase angle. Convert the Cartesian coordinates defined by corresponding entries in matrices x and y to polar coordinates theta and rho. Spherical coordinates consist of the following three quantities. Plane polar coordinates pdf Polar Coordinates r, θ in the plane are described by r distance from the origin and θ 0, 2π is the counter-clockwise angle. A point P in the plane can be uniquely systems is displayed through the following conversion formula: Polar Coord. Likewise, if we have a point in Cartesian coordinates the cylindrical coordinates can be found by using the following conversions. Compute the r image and theta image at each row and column. The Cartesian coordinates x and y can be converted to polar coordinates r and φ with r ≥ 0 and φ in the interval (− π, π] by: = + (as in the Pythagorean theorem or the Euclidean norm), and = (,), where atan2 is a common variation. To plot polar coordinates, set up the polar plane by drawing a dot labeled “O” on your graph at your point of origin. The n-tuple is referred to. This is the official, unambiguous definition of polar coordinates, from which we. Find the value of. The point P has Cartesian coordinates (x;y):The line segment connecting the origin to the point P measures the distance. in the polar coordinate system. Let’s first start with the basics of the Cartesian coordinate system. They are also called "Euclidean coordinates," but not because Euclid discovered them first. to Cartesian Coord. In rectangular coordinates, each point (x, y) has a unique representation. Reformat latitude and longitude coordinates between: Degrees, minutes and seconds. The radius, r, is just the hypotenuse of a right triangle, so r 2 = x 2 + y 2. If the point is not in the first quadrant then you should find an acute angle α using a right-. 5355 0 -10]. import ogr, osr pointX = -11705274. The Polar Coordinate System EX 1 Find the rectangular coordinates for this point. y: The Y-Axis. 9) where p is the perpendicular distance from the rotational axis pXY=+22 (1. Since x= 2 p 3 and y= 2, r= p x2 + y2 = 12 + 4 = 4; tan = y x = 1 p 3: Since the point (2 p 3; 2) lies in the fourth quadrant, we choose = 11ˇ 6. Arguably, log-polar coordinates are more natural than simple polar. If we express the position vector in polar coordinates, we get r(t) = r = (rcosθ)i + (rsinθ)j. To find the polar angle t, you have to take into account the sings of x and y which gives you the quadrant. find the x and y coordinates of a point (r, θ)), we use the following formulas: x = r cos θ, y = r sin θ. A point can be represented by polar coordinates (r; ), where ris the distance between the point and the origin, or pole, and is the angle that a line segment from the pole to the point makes with the positive x-axis. The result lies in the range [-π, π], and the branch cut for this operation lies. 3 Polar Coordinates The Cartesian coordinate system is not the only one. Coordinates conversion from polar coordinates to cartesian x,y coordinates, and from cartesian coordinates to polar coordinates. Precalculus: Polar Coordinates Concepts: Polar Coordinates, converting between polar and cartesian coordinates, distance in polar coordinates. Use the figure below to describe the length of vector \(\vec{r}$$ in terms of $$x$$ ,$$y$$, and $$z$$?. Trigonometry and the Pythagorean Theorem allow for straightforward conversion from rectangular to polar, and vice versa. Convert to Polar Coordinates (-3,0) Convert from rectangular coordinates to polar coordinates using the conversion formulas. Refer to this diagram: Variables used in polar to rectangular coordinate conversions… The variables used in polar to rectangular coordinate conversion are: x: The X-Axis coordinate. To convert from polar to rectangular: x=rcos theta y=rsin theta To convert from rectangular to polar: r^2=x^2+y^2 tan theta= y/x This is where these equations come from: Basically, if you are given an (r,theta) -a polar coordinate- , you can plug your r and theta into your equation for x=rcos theta and y=rsin theta to get your (x,y). EXAMPLE 11: Convert y = 10 into a polar equation. this page updated 19-jul-17. A polar system comprising an intersecting plurality of radial sector lines and a plurality of confocal arcs. Polar - Rectangular Coordinate Conversion Calculator. X,Y,Z Cartesian coordinates have an origin at the centre o f the ellipsoid. A complex number. You know from the figure that the point is in the third quadrant, so. Equation x^2 + y^2 == r^2 is not necessary, it is a combination of the last two. Now I need to go the other way around. b) (2√3, 6, -4) from Cartesian to spherical. The Cartesian coordinate system is also called the rectangular coordinate system, because it describes a location in the plane as the vertex of a rectangle. To convert the point (x, y, z) from rectangular to cylindrical coordinates we use: 222 y. 4 and some of the calculation here will look similar. Convert the following equation to polar coordinates: y = 2x 2. By using this website, you agree to our Cookie Policy. Instead of using the signed distances along the two coordinate axes, polar coordinates specifies the location of a point P in the plane by its distance r from the origin and the. To construct a rectangular coordinate system, we begin with two perpendicular axes that intersect at the origin. 1 De ning Polar Coordinates oT nd the coordinates of a point in the polar coordinate system, consider Figure 1. Let P = (p 0;:::;p n 1) be the n-tuple of signed distance measurements that locate the point in the common world. Transform Cartesian coordinates to polar or cylindrical coordinates. I found an earlier question (How do I calculate a xyz-position of a gps-position relative to an other gps-position?) which worked perfectly. That is we give the position of points in the plane by using x and y coordinates. Plotting Points Using Polar Coordinates. Here, you can convert Rectangular to polar coordinates based on the known values of X and Y axis using Rectangular To Polar Calculator. This is the same angle that we saw in polar/cylindrical coordinates. We would like to be able to compute slopes and areas for these curves using polar coordinates. Perform simple ("Helmert" style) coordinate transformations. Since the x and y coordinates indicate the same distance, we know that the triangle formed has two angles measuring. pdf The problem from cartesian to polars is that the tan function have 2 inverses, you must to know the. To convert from rectangular to polar = + ϴ = arctan(y/x) To convert from polar to rectangular x = rcos θ y = rsin θ. EXAMPLE 12: Convert x. If the point is not in the first quadrant then you should find an acute angle α using a right-. To determine the formula for raster-Cartesian conversion, let’s look at the four corners of the image. Polar - Rectangular Coordinate Conversion Calculator. This calculator converts between polar and rectangular coordinates. Press Change to Cartesian coordinates. TI-84 - Sect 26 - Converting Between Rectangular And Polar Coordinates. Call fromCartesian() - to convert cartesian coordinates to polar coordinates. share Converting complex numbers into Cartesian. Its $x$ component is negative, so its angle is 180° or $\pi$ radians. To convert from spherical coordinates to cartesian coordinates can be done the following way in c#:. cheatatmathhomework) submitted 8 years ago * by FuRyluzt [ ] I have a force field: < x 2 -y 2 , 2xy-y ,3z > I have to convert into polar coordinates, do I simply swap out x=rcos(θ), y=rsin(θ)?. One Time Payment (2 months free of charge) \$5. The equations for doing this are. The calculator will convert the polar coordinates to rectangular (Cartesian) and vice versa, with steps shown. Just note that the bearing in this case is from the grid point (known location) to the unknown location. (2, -2√3) (2√2, -2√2) (2√3, -2). If we were to express it in rectangular coordinates, the calculation would require a few extra steps. To calculate Polar coordinates from Rectangular coordinates you have the calculate the distance from the Origin, r, and the angle from the x-axis, θ, specified by the point. The polar length is obtained with the pythagorean theorem, while the angle is obtained by an application of the inverse tangent. \\ r^2 = x^2 + y^2 \\ \theta = \arctan (\frac yx) {/eq. Let (x, y, z) be the standard Cartesian coordinates, and (ρ, θ, φ) the spherical coordinates, with θ the angle measured away from the +Z axis (as , see conventions in spherical coordinates). For example, x, y and z are the parameters that define a vector r in Cartesian coordinates: r =ˆıx+ ˆy + ˆkz (1) Similarly a vector in cylindrical polar coordinates is described in terms of the parameters r, θ and z since a vector r can be written as r = rrˆ+ zˆk. 3 Polar Coordinates in the Plane In polar coordinates a point P is also characterized by two numbers: the distance r 0 to a fixed pole or origin O, and the angle the ray OP makes with a fixed ray originating at O, which is generally drawn pointing to the right (this is called the initial ray). In this paper, the term Polar Fourier transform will always refer to the. Subsample the polar coordinates to whatever degree of accuracy needed, treating each subsample as a point. Polar Rectangular Regions of Integration. Could in theory be written entirely in the main function; however, to meet the problem statement the code must include a user defined function that returns polar coordinates. To Convert from Cartesian to Polar. The answer is: (r,θ) Polar = (p x2 +y2, arctan y x) Polar Meanwhile, for a point given by Polar coordinates, (r,θ) Polar, we need to specify the coordinates in Cartesian form in terms of the Polar data r and θ. The calculator will convert the polar coordinates to rectangular (Cartesian) and vice versa, with steps shown. Plot points in polar coordinates #1-8; Write polar coordinates for points #9-16; Convert Cartesian coordinates to polar #17-24; Convert Polar coordinates to Cartesian #25-32; Write alternate versions of polar coordinates #33-38. Thus, in this coordinate system, the position of a point will be given by the ordered. 1 Helmholtz Equation and Angular Basis Functions As a direct extension from the Cartesian case, we begin with the eigenfunctions of the Laplacian, whose expression in polar coordinates is given by: ∇2 = ∇2. Cartesian coordinates arise naturally when you need to express translations (movements) of an object in space. Then x = r cos O, Y = r sin O. Double Integrals in Polar Coordinates. The first such point is immediately clear: if r = 0, we have a zero vector (a point in the origin). You can access a copy of the slides used in the video in the PDF file at the bottom of this step. ENGI 4430 Non-Cartesian Coordinates Page 7-09 Spherical Polar Coordinates The coordinate conversion matrix also provides a quick route to finding the Cartesian components of the three basis vectors of the spherical polar coordinate system. Polar-Cartesian Coordinate conversion and Cartesian-Polar Hi All, Is there any standard program to convert Polar Coordinates to Cartesian Coordinates. I am looking now and it doesn't look that hard to create functions to convert between n-dimensional cartesian and n-spherical coordinates. for converting between Cartesian and Polar coordinates. (r, θ) and (r, θ + 2π) represents the same point. Interactive Math website article, accessed 06/2016. Convert the Cartesian coordinates (-V3, 1) to polar coordinates, where T-0 and θ E [O, 27). The painful details of calculating its form in cylindrical and spherical coordinates follow. In this section, we would like to exploit that same dictionary to convert equations. In fact, as a complete counterclockwise rotation is given by an angle 2π, the point represented by polar coordinates (r, θ) is also represented by (r, θ + 2nπ) and (-r, θ + (2n + 1)π) where n is any integer. I Computing volumes using double integrals. Convert an equation from polar. Cartesian coordinates, $$x,y,z$$, are the common coordinates we frequently use but sometimes they are not the best ones to choose. Replace and with the actual values. What does the pair (r; ) refer to in the notation e r(r; ) and e (r; )? The main di erence between the familiar direction vectors e x and e y in Cartesian coor-dinates and the polar direction vectors is that the polar direction vectors change depending. , a nest or burrow) based on distance and bearing from a grid point, this function helps me avoid writing down SOH-CAH-TOA every time. The rectangular coordinates (x , y) and polar coordinates (R , t) are related as follows. Compute the r image and theta image at each row and column. In this section we find the relationship between the representations of a function in polar and Cartesian coordinates when the truncated expansions in and are used. The same holds true for if you are given an (x,y)-a. Display the two images. Does anyone know how to correctly convert Complex numbers in Polar form to Cartesian Form? complex-numbers polar-coordinates. Converting Polar Coordinates to Cartesian Coordinates – Example 1: Converting the given polar coordinates to cartesian coordinates. (1) Choice of Origin Choose an originO. Call fromCartesian() - to convert cartesian coordinates to polar coordinates. Another two-dimensional coordinate system is polar coordinates. Spherical coordinates consist of the following three quantities. …A point P in a two-D plane can be represented by using two…of the four parameters shown, Px, Py, r, and theta. Description: This plugin will convert images to and from polar coordinates. Convert latitude and longitude coordinates to cartesian XYZ coordinates and vice versa. Press Change to Cartesian coordinates. Convert the following equation to polar coordinates: y = − 2 3 x 3. This is the same angle that we saw in polar/cylindrical coordinates. Coordinates of the point: $$(r, θ)$$ or $$(x, y)$$ Method. rst coordinate system to the second is a matter of converting the rst to the common world’s coordinate system and then converting the common world’s coordinate system to the second coordinate system. Cartesian Coordinate. 50) m, as shown in Active Figure 1. * Page 36 (10. EXAMPLE 11: Convert y = 10 into a polar equation. The graph r = 2 1 − 6 sin θ r= \frac{2}{1-6\sin \theta} r = 1 − 6 sin θ 2 in polar coordinates can be expressed as x 2 = a y 2 + b y + c x^2 = ay^2+by+c x 2 = a y 2 + b y + c in Cartesian coordinates, where a a a, b b b and c c c are real numbers. To determine the formula for raster-Cartesian conversion, let’s look at the four corners of the image. You can represent a two dimensional point as either a cartesian coordinate (X,Y) or a polar coordinate (r,theta). Translating latitude and longitude into "polar coordinates" around a point.
84up0izfoc9, w06cmvhjya1vueb, ilby49nto67te, 3flbcogqsjd, 3dlv1gdrpc3tg4n, 3ak1hk2mjg8o3pz, dd0awxp9vx3r, 8n2ylrq6pw7ji3, efn5t02kw87ctqo, m5phe4npqz, n2qfm44zvu3, bq4rcgwaaa, bpkrhx5ewp, sbd5yzc0mp, 78y5zq3rfds3, xoaiycxw8sb69, m6w5q5pez5ud, j1abyuqiqk534, 2dfpsw2fb9av, x3uubabm2yr, jnkleneyet, lqwvbafp4goc3, raebaeeu0bc3sl, 74gcy21io327o2, 5cj1hv9uay6, mbtz94ig3q87s, bpetq41icd5nro, ouclqe3mjbza, yu0b1pq39nv, aiu5dlu5ndup76c, g096s7uoflki, quqgzqqwi0, yi3dfnscor4 | 2020-08-09T08:29:37 | {
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http://math.stackexchange.com/questions/570682/calculating-variable-in-math-equation | # Calculating variable in math equation
I am not good with math, I have this equation (very simple to most) but I need help on how to get the value of x
10 = x - (1.29 + 4.99% of x)
my question is how to calculate x IOW what is the formula used to get the value of x, if that makes sense?
Thank you.
-
convert $4.99%$ to decimal form which is just $.0499$. So your equation is just $10=x-(1.29+0.0499x)$ which then becomes $10=x-1.29-0.0499x$. Combine like terms and solve. – user60887 Nov 17 '13 at 17:50
$4.99 \% = \dfrac{4.99\%}{100\%} = 0.0499$.
$$10=x-(1.29+0.0499x) \iff 10 = x - 1.29 - 0.0499x$$
Combining "like terms"
$$\iff 10 = (1 - 0.0499)x - 1.29 = 0.9501x - 1.29$$
Now add $1.29$ to both sides of the above equation to get:
$$10 + 1.29 = 0.9501 x \iff 11.29 = 0.9501 x$$
Then divide both sides by $0.9501$ to obtain $x$:
$$\dfrac{11.29}{0.9501} = x$$
-
Why has this gone without an UV? +1 – Amzoti Nov 18 '13 at 0:04
Thanks, @Amzoti! – amWhy Nov 18 '13 at 0:06 | 2015-09-05T16:50:40 | {
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https://math.stackexchange.com/questions/2235963/calculating-the-probability-for-a-poisson-rv-problem | # Calculating the probability for a Poisson RV problem
I am doing this question for homework and I've arrived at a solution which does not match the book and I am wondering what I'm doing wrong. I feel like my approach is not that much different but we get relatively different answers. Let me know if you find out my flaw, or perhaps there is none.
A certain typing agency employs 2 typists. The average number of errors per article is 3 when typed by the first typist and 4.2 by the second. If your article is equally likely to be typed by either typist, approximate the probability that it will have no errors.
Here is my approach:
Let $X =$ # errors, where $X$ is roughly Poisson. Now let
$Y = \begin{cases} 0 & \text{if first typist} \\ 1 & \text{if second typist} \end{cases}$
Then we have:
$EX = E[E(X|Y)] = E(X|Y=0)P(Y=0) + E(X|Y=1)P(Y=1) = \frac{1}{2}(3 + 4.2) = 3.6$.
Then since X is Poisson we know that: $P(X = 0) = \frac{3.6^0e^{-3.6}}{0!} = e^{-3.6} = \fbox{0.027.}$ However, the book gives $0.032$. Not sure what I am doing wrong.
• $$P(X=0)=\sum_yP(X=0\mid Y=y)P(Y=y)=e^{-\lambda_0}\tfrac12+e^{-\lambda_1}\tfrac12$$ – Did Apr 15 '17 at 22:07
$X$ is not poisson. It is a mixture of two poisson distributions. So your step when you use the (correct) mean of $3.6$ and compute $P(X=0)$ assuming $X$ is a Poisson with mean $3.6$ is wrong.
Instead, use total probability, same way you computed the mean, but with the entire distribution. So you have $$P(X=0)= P(X = 0|Y=0)P(Y=0) + P(X=0|Y=1)P(Y=1) = e^{-3}\frac{1}{2} + e^{-4.2}\frac{1}{2}=.032$$ | 2019-09-18T09:17:30 | {
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https://math.stackexchange.com/questions/64371/showing-group-with-p2-elements-is-abelian/64374 | # Showing group with $p^2$ elements is Abelian
I have a group $G$ with $p^2$ elements, where $p$ is a prime number. Some (potentially) useful preliminary information I have is that there are exactly $p+1$ subgroups with $p$ elements, and with that I was able to show $G$ has a normal subgroup $N$ with $p$ elements.
My problem is showing that $G$ is abelian, and I would be glad if someone could show me how.
I had two potential approaches in mind and I would prefer if one of these were used (especially the second one).
First: The center $Z(G)$ is a normal subgroup of $G$ so by Langrange's theorem, if $Z(G)$ has anything other than the identity, it's size is either $p$ or $p^2$. If $p^2$ then $Z(G)=G$ and we are done. If $Z(G)=p$ then the quotient group of $G$ factored out by $Z(G)$ has $p$ elements, so it is cylic and I can prove from there that this implies $G$ is abelian. So can we show theres something other than the identity in the center of $G$?
Second: I list out the elements of some other subgroup $H$ with $p$ elements such that the intersection of $H$ and $N$ is only the identity (if any more, due to prime order the intersected elements would generate the entire subgroups). Let $N$ be generated by $a$ and $H$ be generated by $b$. We can show $NK= G$, i.e every element in G can be wrriten like $a^k b^l$. So for this method, we just need to show $ab=ba$ (remember, these are not general elements in the set, but the generators of $N$ and $H$).
Do any of these methods seem viable? I understand one can give very strong theorems using Sylow theorems and related facts, but I am looking for an elementary solution (no Sylow theorems, facts about p-groups, centrailzers) but definitions of centres and normalizers is fine.
• One can prove that $p$-groups have non-trivial centres using the conjugacy class equation. Is this unacceptable, by the last line? Sep 14 '11 at 3:48
• Hint: prove that the center of a non-trivial $p$-group is non-trivial and prove that if $G$ is a finite group such that $G/\textbf{Z}(G)$ is cyclic, then $G$ is abelian. Sep 14 '11 at 3:54
• @Dylan - I know the conjugacy class equation but I am looking for a simpler "clever" answer that maybe only works for non cyclic $p^2$ groups. Sep 14 '11 at 5:08
• @Jimmy: I've merged your duplicate account into your original one, and implemented the comments you wanted to make. If you register your account, that should help prevent future login difficulties. Sep 14 '11 at 5:55
• In his question, he examines the case where |Z(G)| = p. However, in that case, |Z(G)| < |G|, which means that not all items in G are in the center, which of course means that G is not abelian. So, why would he consider the case of |Z(G)| = p? Jul 17 '19 at 4:38
Here is a way to show that the center of a group of order $p^2$ cannot be trivial without using the class equation. I think that the major drawback (and it is major) is that it is very specific for groups of order $p^2$. The class equation is much better because it is a much more general result
If $G$ has elements of order $p^2$, then the result follows because $G$ is cyclic. So suppose that all nontrivial elements of $G$ have order $p$.
Let $x\in G$ be of order $p$. Now, $\langle x\rangle$ is normal in $G$, since its index is the smallest prime that divides the order of $G$. Therefore, $yxy^{-1}\in\langle x\rangle$, so $yxy^{-1}=x^r$ for some $r$, $1\leq r \leq p-1$.
It is now easy to verify that $y^ixy^{-i} = x^{r^i}$. In particular, $y^{p-1}xy^{1-p} = x^{r^{p-1}}$. By Fermat's Little Theorem, $r^{p-1}\equiv 1 \pmod{p}$, so $y^{p-1}xy^{1-p} = x$. That is, $y^{p-1}$ centralizes $x$. But $y$ is of order $p$, so $y^{p-1}=y^{-1}$. Since $y^{-1}$ centralizes $x$, so does $y$. That is, $yx=xy$. Thus, the centralizer of $x$ contains at least $\langle x\rangle$ and $y$, hence is of order at least $p+1$. Since its order must divide $p^2$, the centralizer of $x$ is all of $G$, so $x$ is central.
Thus, $Z(G)$ is nontrivial.
• Neat way to do it. An alternative way to get the final conclusion is to take some element not in the chosen subgroup and repeat the argument, giving two normal subgroups that intersect trivially, which means that they centralize each other (and since they generate the entire group, this makes them central). May 26 '14 at 8:13
Your first approach is good; The center of a $p$-group is non-trivial:
Proof: The center of any group is the union of the 1-element conjugacy classes in the group. For a $p-$group, the size of every conjugacy class is a power of p because the order of a conjugacy class must divide the order of the group. Then let $p^{n_i}$ be the order of the conjugacy classes, and the conjugacy class equation tells us that $|G| = p^n = |Z(G)| + \sum_i (p^{k_i})$, where $0 < k_i < n$ and thus $p$ must divide $|Z(G)|$ implying that the center is non-trivial. $\Box$
EDIT: Explaining class equation:
The conjugacy classes partition the group, so we know that $|G| = \sum|cl(a)|$. But as I said in the proof above, the union of the singleton conjugacy classes is $Z(G)$, so we can rewrite this equality as
$|G| = |Z(G)| + \sum|cl(a)|$
where we assume that each conjugacy class is represented only once (i.e we are not including the singletons in the second summand). However, since we know that the order of a conjugacy class divides the order of a group we can rewrite the second summand to be $\sum_i(p^{k_i})$ where $0 < k_i < n$ because clearly none of them can have the same order as $G$, and we finally get
$|G| = |Z(G)| + \sum_i(p^{k_i})$ where $0 < k_i < n$
Now to see that $p$ must divide $|Z(G)|$ we see that we can move the second summand to left and replace $|G|$ with $p^n$ to get $p^n - \sum_i(p^{k_i}) = |Z(G)|$ and clearly we can factor out $p$.
EDIT: Showing that the order of a conjugacy class must divide the order of a group:
To prove this, we will show that the size of a conjugacy class of a, $cl(a)$ is the index of the centralizer of $a$, $C(a)$.
Suppose $x$ and $y$ both make the same conjugate of $a$, or $xax^{-1} = yay^{-1}$. Then multiplying on the left by $y^{-1}$ and on the right by $x$ we can see that $y^{-1}xa = ay^{-1}x$ and hence, $y^{-1}x \in C(a)$. Thus we can also see that $x\in yC(a)$ and hence $xC(a) = yC(a)$.
Similarly, suppose $xC(a) = yC(a)$ then it follows that $x \in yC(a)$ and hence $x = yz$ for some $z\in C(a)$. Thus $xax^{-1} = (yz)a(yz)^{-1} = yzaz^{-1}y^{-1}$. But we know that $z\in C(a)$ so we can rewrite this as $yazz^{-1}y^{-1}$, or $yay^{-1}$. Thus $x$ and $y$ make the same conjugate of $a$.
Thus we have shown that the number of cosets of $C(a)$ equals the number of elements in $cl(a)$, or
$(G : C(a)) = |cl(a)|$
Since $C(a)$ is a subgroup of $G$, clearly $(G : C(a))$ divides $|G|$ and hence, $|cl(a)|$ divides $G$. $\Box$
• @Deven - That is the approach I had in mind incase I can't find the more "elementary" solution, but I am confident that the 2nd approach I listed is potentially fruitful. I have recieved previous questions of this nature using that method. For example, I can show groups of order 9 must be abelian through a method similar to the second method, but that requires me to do some specific case work that I can't seems to generalize for this case. Sep 14 '11 at 5:08
• This answer seems the most elementary reformulation of the class equation approach, so I voted this as the accepted answer, but could you please elaborate on something for me? I don't see why the order of conjugacy classes in p-groups must divide the order of the group. Is this simple? Sep 15 '11 at 3:06
• I have edited my post, is it more clear now that I write the class equation in this situation as $p^{n} = |Z(G)| + \sum_i(p^{k_i})$ where $0 < k_i < n$ ? if not you can also consider the class equation in an equivalent form here $p^{n} = |Z(G)| + \sum\frac{|G|}{|cl(a)|}$ where $cl(a)$ represents the conjugacy class of $a$. Sep 15 '11 at 3:48
• @Jimmy Valmer: I have just re-edited again as I realize I may have misinterpreted your question the first time I updated my post. Sep 15 '11 at 6:38
So Deven's helpful comment shows why the center of any $p$-group is nontrivial, so look back at your first approach.
Hint: if $Z(G) = p$, consider an element $x \in G$ that is not in $Z(G)$.
What is the overlap between $Z(G)$ and the cyclic subgroup of $G$ generated by $p$? What must the centralizer of $x$ be? What does this imply about $Z(G)$?
Edit: This is an alternative to the path of showing that $G/Z(G)$ cyclic implies $G$ abelian; I consider this route to be also as illuminating and even elegant.
• JakeR - I already know how to handle the problem if $Z(G)=p$ or $p^2$, so I really need a reason why $Z(G)>1$. Sep 14 '11 at 5:08
First we need the following lemma:
If $$G/Z(G)$$ is cyclic then G is abelian.
You can check the prove in here: https://yutsumura.com/if-the-quotient-by-the-center-is-cyclic-then-the-group-is-abelian/.
Then you need this theorem:
If $$p$$ is a prime and $$P$$ is a group of prime power order $$p^{\alpha}$$ for some $$\alpha \geq 1$$, then $$P$$ has a nontrivial center: $$Z(P) \neq 1$$.
You can check the prove in the answer above: https://math.stackexchange.com/a/64374/435467.
Now is the prove of the question:
Since $$Z(P) \neq 1$$, and as the order of $$P$$ is $$p^2$$, there's no other choice that $$|Z(P)| = p$$. Then $$|P/Z(P)| = \frac{p^2}{p} = p$$, and as any group with prime order must be cyclic, $$P/Z(P)$$ is cyclic. By the above lemma, $$G$$ must be abelian.
• why can't $|Z(P)|=p^2$? Aug 26 at 18:26
Here is an alternative way to show this which does not show that the center of a finite $p$-group is non-trivial. The proof is far from elementary, though the results I will use are of more general use.
In fact, what I will show is that if $G$ is a finite $p$-group then $|G/G'|\equiv |G|\pmod {p^2}$ (so a finite group of order $p^n$ is solvable of derived length at most $\frac{n}{2}$ rounded up).
To do this we will be be using some results about complex characters of finite groups (so this answer will hopefully serve as a (slight) motivation to learn more about this topic). I will not go through the definitions of irreducible complex characters here, but if anyone wants to see them, they can start at http://en.wikipedia.org/wiki/Character_theory and work their way through the various parts involved.
The results we need for this are:
If $Irr(G)$ denotes the set of irreducible complex characters of $G$ then $$|G| = \sum_{\chi\in Irr(G)}\chi(1)^2$$
If $\chi\in Irr(G)$ then $\chi(1)$ divides $|G|$.
And finally that $|G/G'| = |\{\chi\in Irr(G)\mid \chi(1) = 1\}|$.
Applying this to a finite $p$-group, we see that if $\chi\in Irr(G)$ with $\chi(1)\neq 1$ then $p$ divides $\chi(1)$, and we get $$|G| = \sum_{\chi\in Irr(G)}\chi(1)^2 = \sum_{\chi\in Irr(G),\, \chi(1) = 1}\chi(1)^2 + \sum_{\chi\in Irr(G),\, \chi(1)\neq 1}\chi(1)^2$$ $$= |\{\chi\in Irr(G)\mid \chi(1) = 1\}| + p^2m = |G/G'| + p^2m$$ for some natural number $m$, which proves the original claim.
• I have posted an answer below , could you please check it ?
– user228168
May 20 '16 at 6:38
I will present a different approach which only involves group actions.
If there is an element of order $$p^2$$ then the group is the cyclic group of order $$p^2$$ which is, of course, abelian.
Otherwise, suppose every element in $$G$$ has order less than $$p^2$$, it is equivalent to say that the order of every non-unit element in $$G$$ is $$p$$. Now we choose an arbitrary non-unit element $$g\in G$$, then we have $$|\langle g \rangle|=p$$. Suppose $$\Omega=\{a_1\langle g \rangle,a_2\langle g \rangle,...,a_p\langle g \rangle\}$$ is a left cosets partition of $$G$$. Let $$\langle g \rangle$$ acts on $$\Omega$$ by the natural left multiplication. Hence $$|\text{Orbit}(a_i\langle g \rangle)|=\frac{|\langle g \rangle|}{|\text{Stab}(a_i\langle g \rangle)|}=\begin{cases}p,&\text{when } |\text{Stab}(a_i\langle g \rangle)|=1\\ 1,&\text{when }|\text{Stab}(a_i\langle g \rangle)|=p\end{cases}$$ But $$|\text{Orbit}(\langle g \rangle)|=1$$, so for every left coset $$a_i\langle g \rangle$$, $$|\text{Orbit}(a_i\langle g \rangle)|=1$$ which means for every $$g^k,k\in\mathbb Z$$ and $$a_i$$, $$a_i^{-1}g^ka_i\in\langle g \rangle$$. Thus for every $$a_ig^m$$, $$(a_ig^m)^{-1}g^k(a_ig^m)\in\langle g \rangle$$ which implies $$\langle g \rangle$$ is a normal subgroup of $$G$$. Since $$g$$ is an arbitrary non-unit element in $$G$$, it follows that every subgroup of order $$p$$ of $$G$$ is a normal subgroup.
Now take $$g_1,g_2\ne e$$ where $$e$$ is the unit and $$\langle g_1 \rangle\cap\langle g_2 \rangle=\{e\}$$. From what we have proved above, we know that $$\langle g_1 \rangle$$ and $$\langle g_2 \rangle$$ are normal subgroups. Hence \begin{align} g_1g_2g_1^{-1}&=g_2^{\ell_1}\tag 1\\ g_2g_1g_2^{-1}&=g_1^{\ell_2}\tag 2 \end{align} for some $$\ell_1,\ell_2\in\mathbb Z$$. Note that $$g_2g_1g_2^{-1}g_1^{-1}=g_1^{\ell_2-1}$$ by $$(2)$$ but $$g_2g_1g_2^{-1}g_1^{-1}=g_2^{1-\ell_1}$$ by $$(1)$$. So $$g_1^{\ell_2-1}=g_2^{1-\ell_1}\in\langle g_1 \rangle\cap\langle g_2 \rangle=\{e\}$$. Therefore $$g_1^{\ell_2}=g_1$$ and $$g_2^{\ell_1}=g_2$$ and $$g_1g_2g_1^{-1}=g_2$$ for all $$g_1,g_2\ne e$$. Hence $$G$$ is abelian.
A slightly different way using actions and orbits.
Let $$|G| = p^2$$. Show that $$G$$ is abelian.
If there exists an element $$g \in G$$ so that $$|g| = p^2$$, then $$g$$ generates $$G$$, making $$G$$ cyclic and thus abelian, and we are done.
So, assume there is no element of $$p^2$$ power. Since the order of each element must divide $$|G|$$, all elements (except the identity) must be of $$p$$ power.
Let $$a \in G$$ be arbitrary. We want to show that $$a$$ commutes with every other element of $$G$$.
If $$a$$ is the identity, we are done, so assume otherwise. Then $$|a| = p$$.
Define $$H = \langle a \rangle$$, so $$|H|=p$$. Since $$|G:H|=p$$, and $$G$$ is a $$p$$-group, $$H$$ is normal in $$G$$. So, let $$G$$ act on $$H$$ by conjugation, and let $$H_G$$ denote the union of the single-element orbits of this action.
Since orbits partition, we now have $$|H| = |H_G| + \sum_{|\mathcal{O}| > 1} |\mathcal{O}|,$$ where the sum on the right counts all elements of $$H$$ in non-singleton orbits (similar to the class equation, but we are only acting on the subgroup $$H$$).
Since the size of orbits must divide the size of the group doing the acting (in this case $$G$$), we can infer that every non-singleton orbit has size $$p$$ (since $$p^2$$ would be too big).
Looking at the union of singleton orbits, we see $$H_G = \left\lbrace h \in H : \forall g \in G : g^{-1}hg = h\right\rbrace$$. In other words, $$H_G$$ contains all elements $$h$$ that commute with every element of $$G$$.
Since the identity is in $$H$$, and the identity commutes with everything, $$|H_G| \geq 1$$, but since the smallest a "multi-element" orbit can be is $$p$$, there are no elements in multi-element orbits, and we have that $$|H_G| = |H| = p$$, implying that every element of $$H$$ (in particular, its generator $$a$$) commutes with every element of $$G$$.
Since $$a$$ was arbitrary, every element of $$G$$ must commute with every other element of $$G$$, and we are done. | 2021-12-09T10:05:07 | {
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https://math.stackexchange.com/questions/500876/direct-proof-that-pr2-immediately-follows-1-in-a-random-permutation-is-1-n | # Direct proof that Pr[2 immediately follows 1] in a random permutation is 1/n
The probability that $1$ is a fixed point of a random permutation of $\{1,2,\ldots,n\}$ (with uniform distribution) is $1/n.$ This is easy to prove since there are $(n-1)!$ permutations that have $1$ as a fixed point and $(n-1)!\,/\,n!=1/n.$ A more direct proof is simply to observe that the image of $1$ under a random permutation is equally likely to be any of the $n$ elements of $\{1,2,\ldots,n\}.$
The probability that $2$ immediately follows $1$ in a random permutation is also $1/n,$ and there is a proof similar to the first proof above in that it involves the computation $(n-1)!\,/\,n!=1/n.$ In a permutation $\pi_1\pi_2\ldots\pi_n$ such that $\pi_i\pi_{i+1}=12$ there are $n-1$ possible values of $i$ and there are $(n-2)!$ ways to permute $\{3,4,\ldots,n\}$ among the remaining $n-2$ positions. Hence there are $(n-1)!$ permutations in which $2$ immediately follows $1.$ Perhaps simpler is to observe that there are $(n-1)!$ permutations of $\{12,3,4,\ldots,n\},$ where $12$ is regarded as a single object.
My question: is there a direct way of seeing this, similar to the direct way of seeing that the probability that $1$ is a fixed point is $1/n?$
• That the probability of 1 immediately preceding 2 equals that of 1 being a fixed point is a special case of a more general phenomenon. See the question asked here and here. – user96124 Sep 22 '13 at 16:22
You need to ensure the last element in the $n$-element permutation is not $1$, since that would mean it has no immediate successor. The chance of this occurring is $(n-1)/n$.
In a scenario for which $1$ appears among the first $n-1$ entries, there are still $n-1$ elements left that could follow it, viz., $2, \ldots, n$. The probability in a given scenario that $2$ is the immediate successor, then, is $1/(n-1)$.
Then the probability in question is $\frac{n-1}{n} \cdot \frac{1}{n-1} = \frac{1}{n}$ as desired. QED
• This proof certainly avoids taking the quotient of two humongous numbers, which I think is aesthetically desirable. It seems that you and Dilip Sarwate had a very similar idea. I think one can perhaps avoid separate treatment of $1$ last versus $1$ among the first $n-1$ entries as follows. The permutations of $\{1,2,\ldots,n\}$ correspond in an obvious way to the permutations of $\{1,2,\ldots,n+1\}$ with $n+1$ fixed. Now consider the permutations of $\{2,3,\ldots,n+1\}$ with $n+1$ fixed. From such a permutation, a permutation of $\{1,2,\ldots,n+1\}$ with $n+1$ fixed is formed by$\ldots$ – user96124 Sep 22 '13 at 4:39
• $\ldots$inserting $1$ in one of $n$ positions. The probability that the insertion point is immediately in front of $2$ is $1/n.$ – user96124 Sep 22 '13 at 4:41
• I have tried only to give the simplest way to see how $1/n$ arises. – Benjamin Dickman Sep 25 '13 at 4:23
• I realize I should have been clearer about what I meant by "direct". In the proof that Pr$[1$ is a fixed point$]=1/n,$ there are $n$ images of $1,$ all of them equally likely. I'm looking for proofs of Pr$[1$ immediately precedes $2]=1/n$ that are similar in this regard: there are $n$ things that can happen, all of them equally likely. – user96124 Sep 25 '13 at 12:30
$1$ is equally likely to be in any of the $n$ positions in the permutation. If it is in the first $n-1$ positions (total probability $\frac{n-1}{n}$), then it is equally likely to be followed by any of the other $n-1$ symbols, and so the probability that it is followed by $2$ is $\frac{n-1}{n}\times \frac{1}{n-1} = \frac{1}{n}$. If $1$ is in the $n$-th position, it cannot be followed by anything. So, $$P\{1~\text{is followed by}~2\} = \frac{1}{n}.$$
An improved version of the observation in the comments to Benjamin Dickman's answer:
There is a $1$-to-$n$ map from the set of permutations of $\{2,3,\ldots,n\}$ to the set of permutations of $\{1,2,\ldots,n\}$ obtained by inserting $1$ in any of $n$ possible positions. Taking $n=4,$ we have, for example, $$243\mapsto\{1243,2143,2413,2431\}.$$ The probability that the insertion point is immediately in front of $2$ is $1/n.$ | 2019-06-27T04:21:19 | {
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https://math.stackexchange.com/questions/2547357/is-a-linear-map-transformation-always-a-matrix-multiplication | # Is a linear map (transformation) always a matrix multiplication
I am studying linear maps. It is defined as a linear map $L$ which transforms a vector from dimension $n$ to dimension $k$
$L:\mathbb{R}^n \rightarrow \mathbb{R}^k$
This seems to me as a matrix multiplication (from $x$ to $y$):
$y = Ax$
My question is, is this correct, and further, can a linear map always be written as a matrix multiplication?
• Yes, and the matrix is that with columns $L(e_1), L(e_2),..., L(e_n)$. – arts Dec 2 '17 at 13:15
• By the way, this is only true for linear maps from $\mathbb{R}^n$ to $\mathbb{R}^k$. For vector spaces of a different nature, it is clearly not true since the vectors don't need to be column matrices to begin with. – arts Dec 2 '17 at 13:22
• All finite dimensional vector spaces of dimension $n$ are isomorphic to $\mathbb R^n$,... just pick a basis – klirk Dec 2 '17 at 13:35
• @klirk Which gives you a matrix multiplication between the $\mathbb{R}^n$'s, not the original spaces. Exactly the subtle distinction I wanted him to understand. – arts Dec 2 '17 at 13:46
• @user3053216 you can set as solved if you are ok – user Dec 3 '17 at 9:58
The answer is yes. If you have a linear map $\phi: V \to W$, between finite dimensional vector spaces of dimension $n$ resp $k$, then this gives rise to a matrix in the following way:
Choose a basis $\{x_i\}$ of $V$ and $\{y_1\}$ of $W$.
Then the matrix corresponds to how $\phi$ acts on the $x_i$ in terms of $y_i$.
As $\phi(x_i)\in W$ We can find coefficients $m^j_i$ such that $$\phi(x_i)=\sum_{j=1}^k m^j_i y_j.$$The coefficients $m^j_i$ correspond to the entries of the matrix $M$ representing $\phi$.
In particular, if $\{y_i\}$ are an orthogonal basis, we can calculate $m^j_i$ by $$m^j_i=<y_j,\phi(x_i)>.$$
Further, for an arbitrary vector $v = \sum_{i=1}^n a^i x_i \in V$ (with some coefficients $a_i$), we have that $$\phi(v)=\phi( \sum_{i=1}^n a^i x_i) = \sum_{i=1}^n a^i \phi(x_i) = \sum_{i=1}^n a^i m^j_i y_i.$$ Form the formula for multiplicating a vector with a matrix, we see that in this basis, the components of $\phi(v)$ correspond to the entries of $Mv$
Edit: What should be obvious, but maybe it still needs to be adressed as pointed out by a comment to the question:
$M$ is not $\phi$. $\phi$ is a linear map between $V$ and $W$, whereas $M$ is a matrix and thus induces a linear map between $\mathbb R^n$ and $\mathbb R^k$ by $x \mapsto Mx$. $M$ only represents $\phi$, that is the following diagram commutes:
• Finite dimensionality is not needed, only existence of basis. – arts Dec 2 '17 at 13:29
• @arts without finite dimensionality, the information needed to specify the linear transformation isn't a finite length grid of coefficients, and would not be called a matrix. – Mark S. Dec 2 '17 at 13:42
• Assuming the axiom of choice, every vector space has a basis. If the dimension is infinite, the basis is uncountable. I wouldn't call the resulting object a Matrix. Sometimes (for example in quantum physics), people work with orthogonal systems (countable) and call the $m^j_i$ as defined in my answer matrixelements – klirk Dec 2 '17 at 13:42
• @arts You are right, the linear combinations are finite. However, there are still uncuntably many basis vectors and the matrix would need to include information about how $\phi$ acts on all of them – klirk Dec 2 '17 at 13:49
• @arts What is the definition of matrix you are working with? – klirk Dec 2 '17 at 13:52
Yes it's alway possible use matrices for linear maps!
https://en.wikipedia.org/wiki/Linear_map
• As klirk said, this works for the maps between finite dimensional spaces, but especially if the domain is not spanned by a finite set, you wouldn't have a matrix. – Mark S. Dec 2 '17 at 13:44
• I just wanted to add, that in that book, the coefficients in the Fourier expansion with respect to an orthogonal "basis" are used to define the infinite matrix. But such an orthogonal system is not a basis in the usual sense. As @arts wrote: "You need to study what it means to be a basis. Bases generate all vectors of the space by finite linear combinations." But an orthogonal system is not able to achieve this. In fact, the definition in this book agrees with that I wrote previously: "Sometimes, people work with orthogonal systems and call the m_ji as defined in my answer matrixelements" – klirk Dec 2 '17 at 15:14 | 2019-12-13T16:51:43 | {
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https://math.stackexchange.com/questions/2650065/what-are-the-total-number-of-ways-in-which-i-j-can-be-chosen-subject-to-con | What are the total number of ways in which $i$, $j$ can be chosen subject to constrain $1\leq i \leq j \leq n$?
What are the total number of ways in which $i$,$j$ can be chosen subject to constrain $1\leq i \leq j \leq n$ ? All are integers. My progress is: I believe that out of the $n$ entries, there are $n \choose 2$ ways to choose $i,j$. But, the given answer is ${n \choose 2} + n$. Some explanation would be helpful.
• What are $i, j, n$? Are they natural numbers, real numbers, etc... – Mr Pie Feb 14 '18 at 8:04
• Don't you mean $1\leq i\cdots$? Then the given answer is okay. $n$ is added because $i=j$ is allowed. That gives $n$ extra possibilities. – drhab Feb 14 '18 at 8:11
We want to find the number of ways we can choose integers $i, j$ such that $1 \leq i \leq j \leq n$. There are two possibilities:
1. $i < j$: The number of such selections is the number of two element subsets $\{1, 2, 3, \ldots, n\}$ since the smaller number we select must be $i$ and the larger one must be $j$. The number of such subsets is $$\binom{n}{2}$$
2. $i = j$: The number of such selections is the number of ways we can select one element from the set $\{1, 2, 3, \ldots, n\}$ since the selected number must equal both $i$ and $j$. The number of ways we can do this is $n$.
Since these cases are mutually exclusive and exhaustive, the number of ways we can choose integers $i, j$ such that $1 \leq i \leq j \leq n$ is $$\binom{n}{2} + n$$
• Thank you for the lucid explanation. – LumosMaxima Feb 14 '18 at 14:31
For $j$ a number in $\{1,\ldots,n\}$ you have $j$ choices of $i$ since $i\in\{1,\ldots,j\}$ so the total number of choices is
$$\sum_{j=1}^n j =\frac{n(n+1)}{2}$$
• Right. But, why is the answer given in the form nC2 + n? – LumosMaxima Feb 14 '18 at 8:00
• I know that the two are same. But, by what logic can I arrive at the answer of nC2 + n? – LumosMaxima Feb 14 '18 at 8:01
• $n$ is the number of the cases where $i=j$ and you have $nC2$ choices of different numbers $i$ and $j$. – user296113 Feb 14 '18 at 8:02
• Alright. So, how does n get added? I mean, in the range [1,n], I have nC2 choices of i,j picked arbitrarily. What after that? – LumosMaxima Feb 14 '18 at 8:03
• Thanks a lot. That surely helped. – LumosMaxima Feb 14 '18 at 8:09 | 2021-01-20T20:25:46 | {
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http://math.stackexchange.com/questions/281587/showing-that-displaystyle-int-aa-frac-sqrta2-x21x2dx-pi-le | # Showing that $\displaystyle\int_{-a}^{a} \frac{\sqrt{a^2-x^2}}{1+x^2}dx = \pi\left (\sqrt{a^2+1}-1\right)$.
How can I show that $\displaystyle\int_{-a}^{a} \frac{\sqrt{a^2-x^2}}{1+x^2}dx = \pi\left(\sqrt{a^2+1}-1\right)$?
-
The indefinite integral is possible but is very complex (both that is uses $i$ and it is complicated). Wolfram alpha or other software can solve this directly with enough time. – kaine Jan 18 '13 at 20:16
Nice question (+1) – I'm an artist Jan 18 '13 at 21:38
Putting $x=a\sin\theta,dx=a\cos\theta d\theta$ and $x=\pm a,\theta=\pm\frac\pi2$
$$\int_{-a}^{a} \frac{\sqrt{a^2-x^2}}{1+x^2}dx =\int _{-\frac\pi2}^{\frac\pi2}\frac{a^2\cos^2\theta}{1+a^2\sin^2\theta}d\theta$$ $$=\int _{-\frac\pi2}^{\frac\pi2}\frac{a^2\sec^2\theta}{(1+\tan^2\theta)(1+(a^2+1)\tan^2\theta)}d\theta$$ (Diving the numerator & the denominator by $\sec^4\theta$)
$$=\int _{-\infty}^{\infty}\frac{a^2}{(1+t^2)(1+(a^2+1)t^2)}dt$$ (Putting $\tan\theta = t$ as $\tan\theta=\pm\infty, t=\pm\frac\pi2$)
$$=\frac{a^2}{(a^2+1)}\int _{-\infty}^{\infty}\frac{1}{(1+t^2)(\frac1{(a^2+1)}+t^2)}dt$$
$$=\frac{\frac{a^2}{(a^2+1)}}{\left(1-\frac1{1+a^2}\right)}\left(\int _{-\infty}^{\infty}\frac1{(\frac1{(a^2+1)}+t^2)}dt-\int _{-\infty}^{\infty}\frac{1}{(1+t^2)}dt\right)$$ as $\frac1{(c+y)(b+y)}=\frac1{c-b}\frac{(c+y)-(b+y)}{(c+y)(b+y)}=\frac1{c-b}\left(\frac1{y+b}-\frac1{y+c}\right)$
$$=\left(\sqrt{1+a^2}\arctan (t\sqrt{1+a^2} )-\arctan t\right)_{-\infty}^{\infty}$$
So,$$\int_{-a}^{a} \frac{\sqrt{a^2-x^2}}{1+x^2}dx=(\sqrt{1+a^2}-1)\frac\pi2-\left\{-(\sqrt{1+a^2}-1)\frac\pi2\right\}=(\sqrt{1+a^2}-1)\pi$$
Observe that we have put $x=a\sin\theta$ and $\tan\theta = t\implies \left(\frac ax\right)^2- \left(\frac 1t\right)^2=\csc^2\theta-\cot^2\theta=1$
$\implies t^2=\frac{x^2}{a^2-x^2}\iff x^2=\frac{a^2t^2}{1+t^2}, a^2-x^2=\frac{a^2}{1+t^2}$
So, if we straight away take $t=\frac x{\sqrt{a^2-x^2}}$ (assuming $a>0$)
$$\frac{dt}{dx}=\frac1{\sqrt{a^2-x^2}}+x\left(\frac{-1}2\right)\frac1{(a^2-x^2)^{\frac32}}(-2x)=\frac{a^2}{(a^2-x^2)^{\frac32}}$$
$$\sqrt{a^2-x^2} dx=\frac{(a^2-x^2)^2dt}{a^2}=\frac{a^2dt}{(1+t^2)^2}$$
and if $x=\pm a,t=\frac{\pm a}0=\pm\infty$
So, $$\int_{-a}^{a} \frac{\sqrt{a^2-x^2}}{1+x^2}dx$$ becomes $$\int _{-\infty}^{\infty}\frac{a^2}{(1+t^2)(1+(a^2+1)t^2)}dt$$
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i thought it was simple integral ... – Santosh Linkha Jan 18 '13 at 20:31
Great answer, thank you! – Guest 86 Jan 18 '13 at 20:50
@Guest86, my pleasure. Hope I could make approach clear. – lab bhattacharjee Jan 19 '13 at 8:06
A slight variant of lab bhattacharjee's method provides a simpler solution:
Let
$$I(a) = \int_{-a}^{a} \frac{\sqrt{a^2-x^2}}{1+x^2} \, dx = 2\int_{0}^{a} \frac{\sqrt{a^2-x^2}}{1+x^2} \, dx.$$
Then by a simple application of multivariable calculus,
\begin{align*} I'(a) &= 2 \left. \frac{\sqrt{a^2-x^2}}{1+x^2} \right|_{x=a} + 2 \int_{0}^{a} \frac{d}{da} \frac{\sqrt{a^2-x^2}}{1+x^2} \, dx \\ &= 2 \int_{0}^{a} \frac{a}{(1+x^2)\sqrt{a^2-x^2}} \, dx. \end{align*}
Then with the change of variable $x = a \sin\theta$, we have
\begin{align*} I'(a) &= 2 \int_{0}^{\frac{\pi}{2}} \frac{a}{1+a^2\sin^2\theta} \, d\theta \\ &= 2 \int_{0}^{\frac{\pi}{2}} \frac{a \sec^2\theta}{1+(a^2+1)\tan^2\theta} \, d\theta \\ &= 2 \int_{0}^{\infty} \frac{a}{1+(a^2+1)t^2} \, dt \qquad (t = \tan\theta) \\ &= \frac{\pi a}{\sqrt{a^2+1}}. \end{align*}
Thus by integrating, we must have
$$I(a) = \pi\sqrt{a^2+1} + C$$
for some constant $C$. But
$$I(0+) = \lim_{a\to0} \int_{-a}^{a} \frac{\sqrt{a^2-x^2}}{1+x^2} \, dx = \lim_{a\to0} \int_{-1}^{1} \frac{\sqrt{1-x^2}}{(1/a)^2+x^2} \, dx = 0$$
and we must have $C = -\pi$. This proves the identity.
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$\newcommand{\Res}{\operatorname{Res}}$ Since the question was tagged complex-analysis, and nobody has given a solutions with purely complex methods, here goes.
Let $$f(z) = \frac{(a^2-z^2)^{1/2}}{1+z^2},$$ where $(a^2-z^2)^{1/2}$ denotes branch that is holomorphic on $\mathbb{C} \setminus [-a,a]$. (See this question for details.)
Next, let $\Gamma$ be a "dog bone" contour together with a large circle:
and integrate $f$ along $\Gamma$. On the "top" part of $[-a,a]$ we get the integral that we want. On the "bottom" part, the square root will pick up a minus sign from the branch cut and another minus sign from the orientation. It's straight forward to check that the integrals over the small circles tend to $0$ as their radii tend to $0$, and the integral over the large circle is basically the residue of $f$ at $\infty$. More precisely, by the residue theorem
\begin{align} 2\int_{-a}^a \frac{\sqrt{a^2-x^2}}{x^2+1}\,dx &= 2\pi i( \Res(f;i) + \Res(f;-i) - \Res(f;\infty)) \\ &= 2\pi i \bigg( \frac{\sqrt{a^2+1}}{2i} + \frac{-\sqrt{a^2+1}}{-2i} + i\bigg) \end{align}
which simplifies to the stated equality. (Note that $\Res(f;\infty) = \Res(-\dfrac1{z^2}f(\dfrac1z);0)$.)
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What a clean solution! I'm reading up on some of your previous answers, really cool, thanks! =) – Guest 86 Jan 21 '13 at 16:23
@Guest86 happy to help, it's not often that I get a chance to integrate along the cute dog bone contour. Btw, don't forget to upvote answers you find helpful. (Not just mine...) – mrf Jan 21 '13 at 16:46
I've ended up up-voting every single answer here (Just that reading them took me a while) – Guest 86 Jan 21 '13 at 21:43
Let $x = a \sin(y)$. Then we have $$\dfrac{\sqrt{a^2-x^2}}{1+x^2} dx = \dfrac{a^2 \cos^2(y)}{1+a^2 \sin^2(y)} dy$$ Hence, $$I = \int_{-a}^{a}\dfrac{\sqrt{a^2-x^2}}{1+x^2} dx = \int_{-\pi/2}^{\pi/2} \dfrac{a^2 \cos^2(y)}{1+a^2 \sin^2(y)} dy$$ Hence, $$I + \pi = \int_{-\pi/2}^{\pi/2} \dfrac{a^2 \cos^2(y)}{1+a^2 \sin^2(y)} dy + \int_{-\pi/2}^{\pi/2} dy = \int_{-\pi/2}^{\pi/2} \dfrac{1+a^2}{1+a^2 \sin^2(y)} dy\\ = \dfrac{1+a^2}2 \int_0^{2 \pi} \dfrac{dy}{1+a^2 \sin^2(y)}$$ Now $$\int_0^{2 \pi} \dfrac{dy}{1+a^2 \sin^2(y)} = \oint_{|z| = 1} \dfrac{dz}{iz \left(1 + a^2 \left(\dfrac{z-\dfrac1z}{2i}\right)^2 \right)} = \oint_{|z| = 1} \dfrac{4z^2 dz}{iz \left(4z^2 - a^2 \left(z^2-1\right)^2 \right)}$$ $$\oint_{|z| = 1} \dfrac{4z^2 dz}{iz \left(4z^2 - a^2 \left(z^2-1\right)^2 \right)} = \oint_{|z| = 1} \dfrac{4z dz}{i(2z + a(z^2-1))(2z - a(z^2-1))}$$ Now $$\dfrac{4z}{(2z + a(z^2-1))(2z - a(z^2-1))} = \dfrac1{az^2 - a + 2z} - \dfrac1{az^2 - a - 2z}$$ $$\oint_{\vert z \vert = 1} \dfrac{dz}{az^2 - a + 2z} = \oint_{\vert z \vert = 1} \dfrac{dz}{a \left(z + \dfrac{1 + \sqrt{1+a^2}}a\right) \left(z + \dfrac{1 - \sqrt{1+a^2}}a\right)} = \dfrac{2 \pi i}{2 \sqrt{1+a^2}}$$ $$\oint_{\vert z \vert = 1} \dfrac{dz}{az^2 - a - 2z} = \oint_{\vert z \vert = 1} \dfrac{dz}{a \left(z - \dfrac{1 + \sqrt{1+a^2}}a\right) \left(z - \dfrac{1 - \sqrt{1+a^2}}a\right)} = -\dfrac{2 \pi i}{2 \sqrt{1+a^2}}$$ Hence, $$\oint_{|z| = 1} \dfrac{4z dz}{i(2z + a(z^2-1))(2z - a(z^2-1))} = \dfrac{2 \pi i}i \dfrac1{\sqrt{1+a^2}} = \dfrac{2 \pi}{\sqrt{1+a^2}}$$ Hence, we get that $$I + \pi = \left(\dfrac{1+a^2}2\right) \dfrac{2 \pi}{\sqrt{1+a^2}} = \pi \sqrt{1+a^2}$$ Hence, we get that $$I = \pi \left(\sqrt{1+a^2} - 1 \right)$$
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Funny how the y'=2y+pi took me a while to get... The +pi idea was very cool! :D I wish I could up-vote this solution... – Guest 86 Jan 18 '13 at 21:07
@Marvis: here is an alternative to the integral where you employed complex analysis: $\frac{4}{(a^2+1)\cos^2 y}\int_0^{\pi/2} \frac{\mathrm{dy}}{\tan^2(y)+1/(a^2+1)} =\frac{4}{(a^2+1) }\int_0^{\infty} \frac{\mathrm{du}}{u^2+(\sqrt{1/(a^2+1)})^2}= \left[\frac{4\sqrt{a^2+1}}{a^2+1} \arctan(\sqrt{a^2+1} u)\right]_0^{\infty}=\frac{2\pi}{\sqrt{a^2+1}}$ – I'm an artist Jan 18 '13 at 21:27
@Chris'ssister Ah, yes. I realized there must be some nice trick like this but used the work-horse for such problems(complex analysis). Thanks. – user17762 Jan 18 '13 at 21:31
You can also convert it into an integral over the upper half-disk $D_a$ of radius $a$: $$\int_{D_a} {1 \over 1 + x^2}\,dy\,dx$$ (Do the $y$ integral first to convert it into the original form.) Since the integrand is even, it is twice the integral over the portion where $x > 0$. Changing to polar coordinates, this becomes $$2\int_0^a \int_0^{\pi \over 2} {r \over 1 + r^2\cos^2(\theta)}\,d\theta\,dr$$ $$= 2\int_0^a r \int_0^{\pi\over 2} {\sec^2(\theta) \over \sec^2(\theta) + r^2}\,d\theta\,dr$$ $$= 2\int_0^a r \int_0^{\pi\over 2} {\sec^2(\theta) \over \tan^2(\theta) + (1 + r^2)}\,d\theta\,dr$$ Doing a $u$ substitution this becomes $$= 2\int_0^a r \int_0^{\infty} {1 \over u^2 + (1 + r^2)}\,du\,dr$$ $$= 2\pi\int_0^a r {1 \over \sqrt{1 + r^2}}\,dr$$ $$=\pi \sqrt{1 + r^2}\,\,\,\bigg|_{r = 0}^{r = a}$$ $$=\pi\sqrt{1 + a^2} - \pi$$
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Interesting idea, Thanks! Several answers used the 1/cos^2 trick, is it a well known thing? (There's an unneeded 2 in the last integral expression) – Guest 86 Jan 21 '13 at 21:39 | 2016-02-11T22:05:44 | {
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http://math.stackexchange.com/questions/123814/conditional-probability-and-independent-events | # Conditional probability and independent events.
In a test, an examinee either guesses or copies or knows the answer to a multiple-choice question with four choices, only one answer being correct. The probability that he makes a guess is $\frac{1}{3}$ and the probability that he copies the answer is $\frac{1}{6}$. The probability that his answer is correct, given that he copies it, is $\frac{1}{8}$
What is the probability that he knew the answer to the question, given that he correctly answers it?
From the text I identified $3$ events regard to the same experience.So the sum of the $3$ probabilities must be $1$.It's known that:
$P(A)=\frac{1}{3}$
$P(B)=\frac{1}{6}$
So,
$P(C)+\frac{1}{3}+\frac{1}{6}=1$
$P(C)=\frac{1}{2}$, this is the probability of knowing the answer.
It's also known that $P(D|B)=\frac{1}{8}$. $P(D)$ is the probability of the question is correctly answered.
The problem ask about $P(C|D)$. From the knowledge that $P(D|B)=\frac{1}{8}$, $P(B)=\frac{1}{6}$ and by the definition of conditional probability, it's known that
$P(D \cap B)=\frac{1}{8} \cdot \frac{1}{6}$. This proves that $B$ and $D$ are indepentend events. And if the $P(B)$ it's known, the $P(D)$ must be $\frac{1}{8}$.
Now, using the conditional probability definition, one can find $P(C|D)$.But if $D$ and $B$ were independent, and $C$ and $B$ are events of the same experience, than $C$ and $D$ must also be independents. So $P(C|D)=P(C)=\frac{1}{2}$
Is my thought right?
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Writing a title in the imperative is usually frowned upon in MSE. – user02138 Mar 24 '12 at 2:10
"$P(D \cap B)=\frac{1}{8} \cdot \frac{1}{6}$. This proves that $B$ and $D$ are indepentend events." Not quite. You correctly multiplied $P(D|B)$ and $P(B)$ to get $P(D \cap B)$; but independence needs $P(D \cap B)=P(D)P(B)$, not $P(D \cap B)=P(D|B)P(B)$. – Dilip Sarwate Mar 24 '12 at 2:14
I didn't realize that was in imperative form.I already changed it. – João Mar 24 '12 at 2:15
Also $C$ and $D$ are clearly dependent... The answer is $P(C\cap D)\over P(D)$. To find $P(D)$, condition on the three alternatives (guess, copy, knows the answer) $P(D)=P(A) P(D|A)+P(B)P(D|B)+P(C)P(D|C)$. – David Mitra Mar 24 '12 at 2:17
Ok. I set a table, and found that $P(D \cap B)=\frac{1}{48}$.So they are not independent.But how can I find $P(A \cap D)$ and $P(C \cap D)$? – João Mar 24 '12 at 10:53
First, some warm up:
Let's define our events at the start:
$\ \ \ D$ is the event that the student answers correctly.
$\ \ \ A$ is the event that the student guesses the answer.
$\ \ \ B$ is the event that the student copies the answer.
$\ \ \ C$ is the event that the student knows the answer.
Let's also write down what we know:
$$\textstyle P(A)={1\over3},\quad P(B)={1\over 6},\quad P(D\mid B)={1\over 8} .$$
Also note $$\textstyle P(D\mid C)=1, \quad P(D\mid A)={1\over 4},\quad P( C) =1-{1\over3}-{1\over6}={1\over2}.$$
Now on to the problem proper:
You want to find $P(C\mid D)$.
$C$ and $D$ are not independent. We have to use the basic formula defining conditional probabilities: $$\tag{1} P(C\mid D) ={P(C\cap D)\over P(D)}.$$
To find $P(C\cap D)$, we use the basic formula again (though it's usually called the multiplication principle when used this way): $$P(C\cap D) =P(C)P(D\mid C).$$ We know $P(C)={1\over2}$ (as you calculated); and, if we're given that the student knows the answer, it follows that in this case that the probability that the student answers correctly is 1. Thus $$\textstyle\tag{2}P(C\cap D) = {1\over2}\cdot 1={1\over 2}.$$
Now to find the term $P(D)$ in $(1)$, we first write $$\tag{3} P(D)=P(A\cap D)+P(B\cap D)+P(C\cap D)$$ this is allowed since $A$, $B$, and $C$ are mutually exclusive events and one of the three must occur; as sets, $D$ can be written as the disjoint union $D= (A\cap D)\cup (B\cap D)\cup (C\cap D)$.
On to calculating the terms in $(3)$:
We have already calculated $P(C\cap D)$.
To find $P(A\cap D)$: $$\tag{4}\textstyle P(A\cap D)=P(A)P(D\mid A)={1\over3}\cdot{1\over4}={1\over12}.$$
To find $P(B\cap D)$: $$\tag{5}\textstyle P(B\cap D)=P(B)P(D\mid B)={1\over6}\cdot{1\over8}={1\over48}.$$ So, substituting the information from $(2)$, $(4)$ and $(5)$ into equation $(3)$, we have $$\textstyle P(D)= {1\over12}+{1\over48}+{1\over2} ={29\over 48}.$$ Using this and $(1)$ and $(2)$ we finally obtain $$P(C\mid D) = {P(C\cap D)\over P(D)}={ 1/2\over 29/48}= {48\over 2\cdot 29}={24\over29}.$$
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I chosen this answer because the solution presented don't apply to baye's law.Thanks – João Mar 24 '12 at 13:51
@João While the solution is excellent and fully deserving of your acceptance, it has used Bayes' law without saying so since it computed $P(C|D)$ starting from $P(D|C)$. – Dilip Sarwate Mar 25 '12 at 2:39
@DilipSarwate It is Bayes' law of course; but I wouldn't say I "used" it, but rather derived it for this particular problem. – David Mitra Mar 25 '12 at 2:46
@DavidMitra Yes, that you derived it is one way of looking at it. My students have often asked why Bayes' law deserves a special name since it is so "obvious" and claimed that maybe if they had been alive in the 18th century, they would have been famous instead of Bayes. They also wonder why it is so controversial.... – Dilip Sarwate Mar 25 '12 at 12:01 | 2015-10-09T13:02:23 | {
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https://math.stackexchange.com/questions/3405749/show-that-the-solutions-is-a-subspace-of-mathbb-r5/3405753 | # Show that the solutions is a subspace of $\mathbb R^5$
Show that the solutions for the linear system of equations:
\begin{aligned} 0 + x_2 +3x_3 - x_4 + 2x_5 &= 0 \\ 2x_1 + 3x_2 + x_3 + 3x_4 &= 0 \\ x_1 + x_2 - x_3 + 2x_4 - x_5 &= 0 \end{aligned}
is a subspace of $$\mathbb R^5$$. What is the dimension of the subspace and determine a basis for the subspace?
I really don't know how to solve this problem. I have achieved this augmented matrix through Gaussian elimination:
$$\begin{bmatrix} 1& 0& -4& 3& -3& 0 \\ 0& 1& 3& -1& 2& 0 \\ 0& 0& 0& 0& 0& 0 \end{bmatrix}$$
Any hints or some steps I've missed?
Edit
My professor says the dimension is $$3$$.
You're almost there. Now your free variables are $$x_3=s$$,$$x_4=t$$ and $$x_5=u$$. Using backward substitution we get $$x_1=4s-3t+3u \\ x_2=-3s+t-2u\\ x_3=s \\ x_4=t \\ x_5=u$$
Therefore we can write every solution as $$\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{bmatrix}=s\begin{bmatrix} 4 \\ -3 \\ 1 \\ 0 \\ 0 \end{bmatrix}+t\begin{bmatrix} -3 \\ 1 \\ 0 \\ 1 \\ 0 \end{bmatrix}+u\begin{bmatrix} 3 \\ -2 \\ 0 \\ 0 \\ 1 \end{bmatrix}$$ with $$s,t,u \in \mathbb{R}$$.
Thus the subspace has dimension $$3$$ and a basis is given by $$\begin{bmatrix} 4 \\ -3 \\ 1 \\ 0 \\ 0 \end{bmatrix},\begin{bmatrix} -1 \\ 1 \\ 0 \\ 1 \\ 0 \end{bmatrix},\begin{bmatrix} 3 \\ -2 \\ 0 \\ 0 \\ 1 \end{bmatrix}$$
To prove is a subspace you need:
• $$0$$ belongs to it: this is clear by taking $$s=t=u=0$$
• It's closed under sums: if $$(s,t,u)$$ and $$(s',t',u')$$ gives us two different solutions, the sum of them is given by $$(s+s',t+t',u+u')$$
• It's closed under scalar multiplication: if $$(s,t,u)$$ gives us a solution and we multiply it by $$k \in \mathbb{R}$$, then we still have a solution given by $$(ks,kt,ku)$$.
Therefore it is a subspace of $$\mathbb{R}^5$$
• missing $s$ in equation 2 Oct 23, 2019 at 14:34
• Thank for you your answer, this was also thought on this problem. However, my professor says that the Dim = 3, which I can't really see why, or maybe he has made a mistake? Also, how do I argument that the solutions is a subspace of R5?
– Carl
Oct 23, 2019 at 14:36
• @Carl are you sure about your Gaussian elimination then? Because in that case the dimension is $2$. To get dimension $3$ you need that one of the rows becomes of all zeros Oct 23, 2019 at 14:43
• I'm very, very sorry. I messed up on my augmented matrix. I have updated my question and the augmented matrix, although I think I understand the solution now, it would be very generous of you to edit your answer.
– Carl
Oct 23, 2019 at 14:50
There's already a good answer on what that subspace is and what dimension it has.
To prove that we are indeed talking about a subspace, you must prove that:
• If $$(x_1, x_2, x_3, x_4, x_5)$$ and $$(y_1, y_2, y_3, y_4, y_5)$$ are solutions, then $$(x_1 + y_1, x_2 + y_2, x_3 + y_3, x_4 + y_4, x_5 + y_5)$$ is also a solution
• If $$(x_1, x_2, x_3, x_4, x_5)$$ is a solution and $$\lambda \in \mathbb{R}$$, then $$(\lambda x_1, \lambda x_2, \lambda x_3, \lambda x_4, \lambda x_5)$$ is also a solution
Here is a correct RREF: \begin{align} &\left[\begin{array}{*{5}{r}} 0&1&3&-1&2 \\ 2&3&1&3&0 \\ 1&1&-1&2&-1 \end{array}\right]\rightsquigarrow \left[\begin{array}{*{5}{r}} 1&1&-1&2&-1 \\ 0&1&3&-1&2 \\ 2&3&1&3&0 \end{array}\right]\rightsquigarrow \left[\begin{array}{*{5}{r}} 1&1&-1&2&-1 \\ 0&1&3&-1&2 \\ 0&1&3&-1&2 \end{array}\right]\rightsquigarrow \\[1ex] &\left[\begin{array}{*{5}{r}} 1&1&-1&2&-1 \\ 0&1&3&-1&2 \\ 0&0&0&0&0 \end{array}\right]\rightsquigarrow \left[\begin{array}{*{5}{r}} 1&0&-4&3&-3 \\ 0&1&3&-1&2 \\ 0&0&0&0&0 \end{array}\right]. \end{align} This matrix has rank $$2$$. Hence, by the rank-nullity theorem, the kernel, i. e. the subspace of solutions (in a $$5$$-dimensional space), has dimension $$3$$. | 2022-08-14T16:01:45 | {
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https://math.stackexchange.com/questions/3188880/show-that-if-a-n-rightarrow-a-then-fraca-1-a-nn-rightarrow-a?noredirect=1 | # Show that if $a_n\rightarrow a$ then $\frac{a_1+…+a_n}{n}\rightarrow a$ [duplicate]
This was an excercise in my exam and I was wondering whether my solution was correct or not.
I have said since $$a_n\rightarrow a$$ there exists a $$N$$ such that for every $$n>N$$ we have $$|a_n-a|<\epsilon_0$$. Therefore we have for $$n>N$$
$$|\frac{1}{n}\cdot(a_1+...+a_N+...a_n)-a|=|\frac{1}{n}\cdot(a_1+...+a_N+...a_n)-\frac{na}{n}|=|\frac{a_1-a}{n}+...+\frac{a_{N+1}-a}{n}+...+\frac{a_n-a}{n}|\Longrightarrow -\frac{n\epsilon_0}{n}-|\frac{a_1-a}{n}+...+\frac{a_N-a}{n}|<|\frac{1}{n}\cdot(a_1+...+a_N+...a_n)-\frac{na}{n}|<\frac{n\epsilon_0}{n}+|\frac{a_1-a}{n}+...+\frac{a_N-a}{n}|$$
We choose now $$n$$ so big such that for $$n>N'$$ $$|\frac{a_1-a}{n}+...+\frac{a_N-a}{n}|<\epsilon_1$$
Because $$\epsilon_1$$ and $$\epsilon_0$$ were arbitrary the claim is proved.
If there are two points for this excercise how many would you give me?
## marked as duplicate by Nosrati, Mark Viola real-analysis StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Apr 15 at 16:55
• The problem with the proof is that I have said $\frac{n\epsilon}{n}$ altough I should have said $\frac{(n-N)\epsilon}{n}$. The proof is still corect if I choose an appropriate $N''$ but the prof might give me no points for this blunder – New2Math Apr 15 at 16:35
• $(n-N) \leq n$, so your estimation is correct. I want to point out that you don't need a lower bound since the absolute value is greater than or equal to $0$ anyway. However, I would mark you full points. – Nathanael Skrepek Apr 15 at 16:40 | 2019-10-22T19:27:29 | {
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https://d2l.ai/chapter_attention-mechanisms-and-transformers/attention-pooling.html | # 11.2. Attention Pooling by Similarity¶ Open the notebook in Colab Open the notebook in Colab Open the notebook in Colab Open the notebook in Colab Open the notebook in SageMaker Studio Lab
Now that we introduced the primary components of the attention mechanism, let’s use them in a rather classical setting, namely regression and classification via kernel density estimation . This detour simply provides additional background: it is entirely optional and can be skipped if needed. At their core, Nadaraya-Watson estimators rely on some similarity kernel $$\alpha(\mathbf{q}, \mathbf{k})$$ relating queries $$\mathbf{q}$$ to keys $$\mathbf{k}$$. Some common kernels are
(11.2.1)\begin{split}\begin{aligned} \alpha(\mathbf{q}, \mathbf{k}) & = \exp\left(-\frac{1}{2} \|\mathbf{q} - \mathbf{k}\|^2 \right) && \mathrm{Gaussian} \\ \alpha(\mathbf{q}, \mathbf{k}) & = 1 \text{ if } \|\mathbf{q} - \mathbf{k}\| \leq 1 && \mathrm{Boxcar} \\ \alpha(\mathbf{q}, \mathbf{k}) & = \mathop{\mathrm{max}}\left(0, 1 - \|\mathbf{q} - \mathbf{k}\|\right) && \mathrm{Epanechikov} \end{aligned}\end{split}
There are many more choices that we could pick. See a Wikipedia article for a more extensive review and how the choice of kernels is related to kernel density estimation, sometimes also called Parzen Windows . All of the kernels are heuristic and can be tuned. For instance, we can adjust the width, not only on a global basis but even on a per-coordinate basis. Regardless, all of them lead to the following equation for regression and classification alike:
(11.2.2)$f(\mathbf{q}) = \sum_i \mathbf{v}_i \frac{\alpha(\mathbf{q}, \mathbf{k}_i)}{\sum_j \alpha(\mathbf{q}, \mathbf{k}_j)}.$
In the case of a (scalar) regression with observations $$(\mathbf{x}_i, y_i)$$ for features and labels respectively, $$\mathbf{v}_i = y_i$$ are scalars, $$\mathbf{k}_i = \mathbf{x}_i$$ are vectors, and the query $$\mathbf{q}$$ denotes the new location where $$f$$ should be evaluated. In the case of (multiclass) classification, we use one-hot-encoding of $$y_i$$ to obtain $$\mathbf{v}_i$$. One of the convenient properties of this estimator is that it requires no training. Even more so, if we suitably narrow the kernel with increasing amounts of data, the approach is consistent , i.e., it will converge to some statistically optimal solution. Let’s start by inspecting some kernels.
import numpy as np
import torch
from torch import nn
from torch.nn import functional as F
from d2l import torch as d2l
d2l.use_svg_display()
from mxnet import autograd, gluon, np, npx
from mxnet.gluon import nn
from d2l import mxnet as d2l
npx.set_np()
d2l.use_svg_display()
import jax
from flax import linen as nn
from jax import numpy as jnp
from d2l import jax as d2l
No GPU/TPU found, falling back to CPU. (Set TF_CPP_MIN_LOG_LEVEL=0 and rerun for more info.)
import numpy as np
import tensorflow as tf
from d2l import tensorflow as d2l
d2l.use_svg_display()
## 11.2.1. Kernels and Data¶
All the kernels $$\alpha(\mathbf{k}, \mathbf{q})$$ defined in this section are translation and rotation invariant, that is, if we shift and rotate $$\mathbf{k}$$ and $$\mathbf{q}$$ in the same manner, the value of $$\alpha$$ remains unchanged. For simplicity we thus pick scalar arguments $$k, q \in \mathbb{R}$$ and pick the key $$k = 0$$ as the origin. This yields:
fig, axes = d2l.plt.subplots(1, 4, sharey=True, figsize=(12, 3))
# Define some kernels
def gaussian(x):
def boxcar(x):
def constant(x):
return 1.0 + 0 * x
def epanechikov(x):
kernels = (gaussian, boxcar, constant, epanechikov)
names = ('Gaussian', 'Boxcar', 'Constant', 'Epanechikov')
x = torch.arange(-2.5, 2.5, 0.1)
for kernel, name, ax in zip(kernels, names, axes):
ax.plot(x.detach().numpy(), kernel(x).detach().numpy())
ax.set_xlabel(name)
fig, axes = d2l.plt.subplots(1, 4, sharey=True, figsize=(12, 3))
# Define some kernels
def gaussian(x):
return np.exp(-x**2 / 2)
def boxcar(x):
return np.abs(x) < 1.0
def constant(x):
return 1.0 + 0 * x
def epanechikov(x):
return np.maximum(1 - np.abs(x), 0)
kernels = (gaussian, boxcar, constant, epanechikov)
names = ('Gaussian', 'Boxcar', 'Constant', 'Epanechikov')
x = np.arange(-2.5, 2.5, 0.1)
for kernel, name, ax in zip(kernels, names, axes):
ax.plot(x.asnumpy(), kernel(x).asnumpy())
ax.set_xlabel(name)
fig, axes = d2l.plt.subplots(1, 4, sharey=True, figsize=(12, 3))
# Define some kernels
def gaussian(x):
return jnp.exp(-x**2 / 2)
def boxcar(x):
return jnp.abs(x) < 1.0
def constant(x):
return 1.0 + 0 * x
def epanechikov(x):
return jnp.maximum(1 - jnp.abs(x), 0)
kernels = (gaussian, boxcar, constant, epanechikov)
names = ('Gaussian', 'Boxcar', 'Constant', 'Epanechikov')
x = jnp.arange(-2.5, 2.5, 0.1)
for kernel, name, ax in zip(kernels, names, axes):
ax.plot(x, kernel(x))
ax.set_xlabel(name)
fig, axes = d2l.plt.subplots(1, 4, sharey=True, figsize=(12, 3))
# Define some kernels
def gaussian(x):
return tf.exp(-x**2 / 2)
def boxcar(x):
return tf.abs(x) < 1.0
def constant(x):
return 1.0 + 0 * x
def epanechikov(x):
return tf.maximum(1 - tf.abs(x), 0)
kernels = (gaussian, boxcar, constant, epanechikov)
names = ('Gaussian', 'Boxcar', 'Constant', 'Epanechikov')
x = tf.range(-2.5, 2.5, 0.1)
for kernel, name, ax in zip(kernels, names, axes):
ax.plot(x.numpy(), kernel(x).numpy())
ax.set_xlabel(name)
Different kernels correspond to different notions of range and smoothness. For instance, the boxcar kernel only attends to observations within a distance of $$1$$ (or some otherwise defined hyperparameter) and does so indiscriminately.
To see Nadaraya-Watson estimation in action, let’s define some training data. In the following we use the dependency
(11.2.3)$y_i = 2\sin(x_i) + x_i + \epsilon,$
where $$\epsilon$$ is drawn from a normal distribution with zero mean and unit variance. We draw 40 training examples.
def f(x):
return 2 * torch.sin(x) + x
n = 40
x_train, _ = torch.sort(torch.rand(n) * 5)
y_train = f(x_train) + torch.randn(n)
x_val = torch.arange(0, 5, 0.1)
y_val = f(x_val)
def f(x):
return 2 * np.sin(x) + x
n = 40
x_train = np.sort(np.random.rand(n) * 5, axis=None)
y_train = f(x_train) + np.random.randn(n)
x_val = np.arange(0, 5, 0.1)
y_val = f(x_val)
def f(x):
return 2 * jnp.sin(x) + x
n = 40
x_train = jnp.sort(jax.random.uniform(d2l.get_key(), (n,)) * 5)
y_train = f(x_train) + jax.random.normal(d2l.get_key(), (n,))
x_val = jnp.arange(0, 5, 0.1)
y_val = f(x_val)
def f(x):
return 2 * tf.sin(x) + x
n = 40
x_train = tf.sort(tf.random.uniform((n,1)) * 5, 0)
y_train = f(x_train) + tf.random.normal((n, 1))
x_val = tf.range(0, 5, 0.1)
y_val = f(x_val)
## 11.2.2. Attention Pooling via Nadaraya-Watson Regression¶
Now that we have data and kernels, all we need is a function that computes the kernel regression estimates. Note that we also want to obtain the relative kernel weights in order to perform some minor diagnostics. Hence we first compute the kernel between all training features (covariates) x_train and all validation features x_val. This yields a matrix, which we subsequently normalize. When multiplied with the training labels y_train we obtain the estimates.
Recall attention pooling in (11.1.1). Let each validation feature be a query, and each training feature-label pair be a key-value pair. As a result, the normalized relative kernel weights (attention_w below) are the attention weights.
def nadaraya_watson(x_train, y_train, x_val, kernel):
dists = x_train.reshape((-1, 1)) - x_val.reshape((1, -1))
# Each column/row corresponds to each query/key
k = kernel(dists).type(torch.float32)
# Normalization over keys for each query
attention_w = k / k.sum(0)
y_hat = y_train@attention_w
return y_hat, attention_w
def nadaraya_watson(x_train, y_train, x_val, kernel):
dists = x_train.reshape((-1, 1)) - x_val.reshape((1, -1))
# Each column/row corresponds to each query/key
k = kernel(dists).astype(np.float32)
# Normalization over keys for each query
attention_w = k / k.sum(0)
y_hat = np.dot(y_train, attention_w)
return y_hat, attention_w
def nadaraya_watson(x_train, y_train, x_val, kernel):
dists = x_train.reshape((-1, 1)) - x_val.reshape((1, -1))
# Each column/row corresponds to each query/key
k = kernel(dists).astype(jnp.float32)
# Normalization over keys for each query
attention_w = k / k.sum(0)
y_hat = y_train@attention_w
return y_hat, attention_w
def nadaraya_watson(x_train, y_train, x_val, kernel):
dists = tf.reshape(x_train, (-1, 1)) - tf.reshape(x_val, (1, -1))
# Each column/row corresponds to each query/key
k = tf.cast(kernel(dists), tf.float32)
# Normalization over keys for each query
attention_w = k / tf.reduce_sum(k, 0)
y_hat = tf.transpose(tf.transpose(y_train)@attention_w)
return y_hat, attention_w
Let’s have a look at the kind of estimates that the different kernels produce.
def plot(x_train, y_train, x_val, y_val, kernels, names, attention=False):
fig, axes = d2l.plt.subplots(1, 4, sharey=True, figsize=(12, 3))
for kernel, name, ax in zip(kernels, names, axes):
y_hat, attention_w = nadaraya_watson(x_train, y_train, x_val, kernel)
if attention:
pcm = ax.imshow(attention_w.detach().numpy(), cmap='Reds')
else:
ax.plot(x_val, y_hat)
ax.plot(x_val, y_val, 'm--')
ax.plot(x_train, y_train, 'o', alpha=0.5);
ax.set_xlabel(name)
if not attention:
ax.legend(['y_hat', 'y'])
if attention:
fig.colorbar(pcm, ax=axes, shrink=0.7)
plot(x_train, y_train, x_val, y_val, kernels, names)
def plot(x_train, y_train, x_val, y_val, kernels, names, attention=False):
fig, axes = d2l.plt.subplots(1, 4, sharey=True, figsize=(12, 3))
for kernel, name, ax in zip(kernels, names, axes):
y_hat, attention_w = nadaraya_watson(x_train, y_train, x_val, kernel)
if attention:
pcm = ax.imshow(attention_w.asnumpy(), cmap='Reds')
else:
ax.plot(x_val, y_hat)
ax.plot(x_val, y_val, 'm--')
ax.plot(x_train, y_train, 'o', alpha=0.5);
ax.set_xlabel(name)
if not attention:
ax.legend(['y_hat', 'y'])
if attention:
fig.colorbar(pcm, ax=axes, shrink=0.7)
plot(x_train, y_train, x_val, y_val, kernels, names)
def plot(x_train, y_train, x_val, y_val, kernels, names, attention=False):
fig, axes = d2l.plt.subplots(1, 4, sharey=True, figsize=(12, 3))
for kernel, name, ax in zip(kernels, names, axes):
y_hat, attention_w = nadaraya_watson(x_train, y_train, x_val, kernel)
if attention:
pcm = ax.imshow(attention_w, cmap='Reds')
else:
ax.plot(x_val, y_hat)
ax.plot(x_val, y_val, 'm--')
ax.plot(x_train, y_train, 'o', alpha=0.5);
ax.set_xlabel(name)
if not attention:
ax.legend(['y_hat', 'y'])
if attention:
fig.colorbar(pcm, ax=axes, shrink=0.7)
plot(x_train, y_train, x_val, y_val, kernels, names)
def plot(x_train, y_train, x_val, y_val, kernels, names, attention=False):
fig, axes = d2l.plt.subplots(1, 4, sharey=True, figsize=(12, 3))
for kernel, name, ax in zip(kernels, names, axes):
y_hat, attention_w = nadaraya_watson(x_train, y_train, x_val, kernel)
if attention:
pcm = ax.imshow(attention_w.numpy(), cmap='Reds')
else:
ax.plot(x_val, y_hat)
ax.plot(x_val, y_val, 'm--')
ax.plot(x_train, y_train, 'o', alpha=0.5);
ax.set_xlabel(name)
if not attention:
ax.legend(['y_hat', 'y'])
if attention:
fig.colorbar(pcm, ax=axes, shrink=0.7)
plot(x_train, y_train, x_val, y_val, kernels, names)
The first thing that stands out is that all three nontrivial kernels (Gaussian, Boxcar, and Epanechikov) produce fairly workable estimates that are not too far from the true function. Only the constant kernel that leads to the trivial estimate $$f(x) = \frac{1}{n} \sum_i y_i$$ produces a rather unrealistic result. Let’s inspect the attention weighting a bit more closely:
plot(x_train, y_train, x_val, y_val, kernels, names, attention=True)
plot(x_train, y_train, x_val, y_val, kernels, names, attention=True)
plot(x_train, y_train, x_val, y_val, kernels, names, attention=True)
plot(x_train, y_train, x_val, y_val, kernels, names, attention=True)
The visualization clearly shows why the estimates for Gaussian, Boxcar, and Epanechikov are very similar: after all, they are derived from very similar attention weights, despite the different functional form of the kernel. This raises the question as to whether this is always the case.
We could replace the Gaussian kernel with one of a different width. That is, we could use $$\alpha(\mathbf{q}, \mathbf{k}) = \exp\left(-\frac{1}{2 \sigma^2} \|\mathbf{q} - \mathbf{k}\|^2 \right)$$ where $$\sigma^2$$ determines the width of the kernel. Let’s see whether this affects the outcomes.
sigmas = (0.1, 0.2, 0.5, 1)
names = ['Sigma ' + str(sigma) for sigma in sigmas]
def gaussian_with_width(sigma):
return (lambda x: torch.exp(-x**2 / (2*sigma**2)))
kernels = [gaussian_with_width(sigma) for sigma in sigmas]
plot(x_train, y_train, x_val, y_val, kernels, names)
sigmas = (0.1, 0.2, 0.5, 1)
names = ['Sigma ' + str(sigma) for sigma in sigmas]
def gaussian_with_width(sigma):
return (lambda x: np.exp(-x**2 / (2*sigma**2)))
kernels = [gaussian_with_width(sigma) for sigma in sigmas]
plot(x_train, y_train, x_val, y_val, kernels, names)
sigmas = (0.1, 0.2, 0.5, 1)
names = ['Sigma ' + str(sigma) for sigma in sigmas]
def gaussian_with_width(sigma):
return (lambda x: jnp.exp(-x**2 / (2*sigma**2)))
kernels = [gaussian_with_width(sigma) for sigma in sigmas]
plot(x_train, y_train, x_val, y_val, kernels, names)
sigmas = (0.1, 0.2, 0.5, 1)
names = ['Sigma ' + str(sigma) for sigma in sigmas]
def gaussian_with_width(sigma):
return (lambda x: tf.exp(-x**2 / (2*sigma**2)))
kernels = [gaussian_with_width(sigma) for sigma in sigmas]
plot(x_train, y_train, x_val, y_val, kernels, names)
Clearly, the narrower the kernel, the less smooth the estimate. At the same time, it adapts better to the local variations. Let’s look at the corresponding attention weights.
plot(x_train, y_train, x_val, y_val, kernels, names, attention=True)
plot(x_train, y_train, x_val, y_val, kernels, names, attention=True)
plot(x_train, y_train, x_val, y_val, kernels, names, attention=True)
plot(x_train, y_train, x_val, y_val, kernels, names, attention=True)
As we would expect, the narrower the kernel, the narrower the range of large attention weights. It is also clear that picking the same width might not be ideal. In fact, Silverman (1986) proposed a heuristic that depends on the local density. Many more such “tricks” have been proposed. It remains a valuable technique to date. For instance, Norelli et al. (2022) used a similar nearest-neighbor interpolation technique to design cross-modal image and text representations.
The astute reader might wonder why this deep dive on a method that is over half a century old. First, it is one of the earliest precursors of modern attention mechanisms. Second, it is great for visualization. Third, and just as importantly, it demonstrates the limits of hand-crafted attention mechanisms. A much better strategy is to learn the mechanism, by learning the representations for queries and keys. This is what we will embark on in the following sections.
## 11.2.4. Summary¶
Nadaraya-Watson kernel regression is an early precursor of the current attention mechanisms. It can be used directly with little to no training or tuning, both for classification and regression. The attention weight is assigned according to the similarity (or distance) between query and key, and according to how many similar observations are available.
## 11.2.5. Exercises¶
1. Parzen windows density estimates are given by $$\hat{p}(\mathbf{x}) = \frac{1}{n} \sum_i k(\mathbf{x}, \mathbf{x}_i)$$. Prove that for binary classification the function $$\hat{p}(\mathbf{x}, y=1) - \hat{p}(\mathbf{x}, y=-1)$$, as obtained by Parzen windows is equivalent to Nadaraya-Watson classification.
2. Implement stochastic gradient descent to learn a good value for kernel widths in Nadaraya-Watson regression.
1. What happens if you just use the above estimates to minimize $$(f(\mathbf{x_i}) - y_i)^2$$ directly? Hint: $$y_i$$ is part of the terms used to compute $$f$$.
2. Remove $$(\mathbf{x}_i, y_i)$$ from the estimate for $$f(\mathbf{x}_i)$$ and optimize over the kernel widths. Do you still observe overfitting?
3. Assume that all $$\mathbf{x}$$ lie on the unit sphere, i.e., all satisfy $$\|\mathbf{x}\| = 1$$. Can you simplify the $$\|\mathbf{x} - \mathbf{x}_i\|^2$$ term in the exponential? Hint: we will later see that this is very closely related to dot-product attention.
4. Recall that Mack and Silverman (1982) proved that Nadaraya-Watson estimation is consistent. How quickly should you reduce the scale for the attention mechanism as you get more data? Provide some intuition for your answer. Does it depend on the dimensionality of the data? How? | 2023-03-27T03:26:37 | {
"domain": "d2l.ai",
"url": "https://d2l.ai/chapter_attention-mechanisms-and-transformers/attention-pooling.html",
"openwebmath_score": 0.7578228712081909,
"openwebmath_perplexity": 12943.151166677042,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9830850872288502,
"lm_q2_score": 0.8577680995361899,
"lm_q1q2_score": 0.8432590269546604
} |
http://www.lofoya.com/Solved/1713/a-man-can-hit-a-target-once-in-4-shots-if-he-fires-4-shots-in | # Moderate Probability Solved QuestionAptitude Discussion
Q. A man can hit a target once in 4 shots. If he fires 4 shots in succession, what is the probability that he will hit his target?
✖ A. 1 ✖ B. 1/256 ✖ C. 81/256 ✔ D. 175/256
Solution:
Option(D) is correct
The man will hit the target even if he hits it once or twice or thrice or all four times in the four shots that he takes.
So, the only case where the man will not hit the target is when he fails to hit the target even in one of the four shots that he takes.
The probability that he will not hit the target in one shot =1 - Probability that he will hit target in exact one shot
$=1-\dfrac{1}{4}$
$=\dfrac{3}{4}$
Therefore, the probability that he will not hit the target in all the four shots
$=\left(\dfrac{3}{4}\right)\times\left(\dfrac{3}{4}\right)\times \left(\dfrac{3}{4}\right)\times\left(\dfrac{3}{4}\right)$
$=\dfrac{81}{256}$
Hence, the probability that he will hit the target at least in one of the four shots:
$=\left(1-\dfrac{81}{256}\right)$
$=\dfrac{175}{256}$
## (6) Comment(s)
Anu
()
Hi , if we consider the cases where he hits , it could be
HMMM+MHMM+MMHM+MMMH
But the above isnt giving me the right answer , why so ?
Romila
()
Because we can not form the structure in the way you have created. You are taking 1 hit in four shots for four times (16 shots). That is not a valid approach.
Vaibhav
()
As said he can hit target once in four shot but not necessarily he hits that's why probability is not 100%
Ritika
()
This is only a probability that he CAN hit the target once in four shots. It's not that he WILL hit the target in one of the four shots.
Alicia
()
My thought would have been $1/4*1/4*1/4*1/4=1/256$
Probability is not my strong suit, therefore I am sort of confused as to why it isn't 1/256
VeeraraghavanK
()
It looks apparently that the answer is 1/4 since he fires 4 shots. The question could have been : "probability that the target is fired at least once"
Is there any other alternate method | 2017-10-20T14:28:26 | {
"domain": "lofoya.com",
"url": "http://www.lofoya.com/Solved/1713/a-man-can-hit-a-target-once-in-4-shots-if-he-fires-4-shots-in",
"openwebmath_score": 0.5834237933158875,
"openwebmath_perplexity": 510.1303907346087,
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"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9830850857421198,
"lm_q2_score": 0.8577680995361899,
"lm_q1q2_score": 0.8432590256793905
} |
https://cs.stackexchange.com/questions/88877/time-complexity-of-euler-totient-function | # Time complexity of euler totient function
Code for finding $\phi$(n) is
int phi(int n)
{
int result = n; // Initialize result as n
// Consider all prime factors of n and subtract their
// multiples from result
for (int p=2; p*p<=n; ++p)
{
// Check if p is a prime factor.
if (n % p == 0)
{
// If yes, then update n and result
while (n % p == 0)
n /= p;
result -= result / p;
}
}
// If n has a prime factor greater than sqrt(n)
// (There can be at-most one such prime factor)
if (n > 1)
result -= result / n;
return result;
}
I don't understand how is the overall complexity $O(\sqrt{n})$.
From the code, I see that the outer loop runs for $O(\sqrt{n})$ time but I am not sure of how to include the time complexity of the inner loop for finding the overall complexity.
If we have n=$128$, then outer loop runs for $O(\sqrt{n})$ time and inner loop for $O(log_2 n)$, so overall complexity is $O(\sqrt{n})$ + $O(log_2 n)$ which is $O(\sqrt{n})$.
I don't know how to extend this for general case of $n=P1^{a1}*P2^{a2}*P3^{a3}*..Pn^{an}$
The cost inner loop if $n$ is equal to $P_1^{a_1}P_2^{a_2}\cdots P_k^{a_k}$, should be run in $\log_{P_1}n + \log_{P_2}n + \cdots + \log_{P_k}n$ totaly which is $O(k\log(n))$.Notice that, these computation is computed over the outer loop and summation of all iterations of the outer loop.
To more scrutinizing, as the maximum asymptotic number of primes less than $x$ is $\frac{x}{\log x}$, size of $k$ would be $O(\frac{\sqrt{n}}{\log\sqrt{n}})$. Hence, The complexity of the problem would be: $$O(\sqrt{n} + k\log(n))=O(\sqrt{n} + \frac{\sqrt{n}}{\log\sqrt{n}}\log(n)) = O(\sqrt{n}+\frac{\sqrt{n}}{2})=O(\sqrt{n})$$
• @Zephyr Each item of the inner loop runs in $\log_{p_i}n$. The summation of these is meaning besides the outer loop. In the other words, each item of the sum happened in each iteration of the outer loop. – OmG Mar 4 '18 at 16:58 | 2019-06-25T19:45:59 | {
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"url": "https://cs.stackexchange.com/questions/88877/time-complexity-of-euler-totient-function",
"openwebmath_score": 0.893662691116333,
"openwebmath_perplexity": 576.890608583922,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9830850877244272,
"lm_q2_score": 0.8577680977182187,
"lm_q1q2_score": 0.8432590255925301
} |
https://brilliant.org/discussions/thread/imo-2016-day-1/ | # IMO 2016 Day 1
I'm surprised nobody else has posted this. Yes, there are plenty of discussions in Art of Problem Solving, but why not here? And of course there are more problems.
## Problem 1
Triangle $$BCF$$ has a right angle at $$B$$. Let $$A$$ be the point on line $$CF$$ such that $$FA = FB$$ and $$F$$ lies between $$A$$ and $$C$$. Point $$D$$ is chosen so that $$DA = DC$$ and $$AC$$ is the bisector of $$\angle{DAB}$$. Point $$E$$ is chosen so that $$EA = ED$$ and $$AD$$ is the bisector of $$\angle{EAC}$$. Let $$M$$ be the midpoint of $$CF$$. Let $$X$$ be the point such that $$AMXE$$ is a parallelogram. Prove that $$BD$$, $$FX$$, and $$ME$$ are concurrent.
## Problem 2
Find all positive integers $$n$$ for which each cell of an $$n \times n$$ table can be filled with one of the letters I, M, and O in such a way that:
• in each row and each column, one third of the entries are I, one third are M, and one third are O; and
• in any diagonal, if the number of entries on the diagonal is a multiple of three, then one third of the entries are I, one third are M, and one third are O.
Note. The rows and columns of an $$n \times n$$ table are each labelled $$1$$ to $$n$$ in a natural order. Thus each cell corresponds to a pair of positive integer $$(i, j)$$ with $$1 \le i,j \le n$$. For $$n > 1$$, the table has $$4n-2$$ diagonals of two types. A diagonal of first type consists all cells $$(i, j)$$ for which $$i+j$$ is a constant, and the diagonal of this second type consists all cells $$(i, j)$$ for which $$i-j$$ is constant.
## Problem 3
Let $$P = A_1 A_2 \cdots A_k$$ be a convex polygon in the plane. The vertices $$A_1, A_2, \ldots, A_k$$ have integral coordinates and lie on a circle. Let $$S$$ be the area of $$P$$. An odd positive integer $$n$$ is given such that the squares of the side lengths of $$P$$ are integers divisible by $$n$$. Prove that $$2S$$ is an integer divisible by $$n$$.
Note by Ivan Koswara
2 years, 9 months ago
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I have an extension to Q1. (Almost got a complete solution for it, but i observed this, which follows):
Let this intersection point be labelled $$O$$. Prove $$O$$ is the circumcentre of $$\Delta XBE$$.
- 2 years, 9 months ago
Nice observation. If you complex trig bash it, the solution to your extension becomes very apparent. I'm currently typing up my solution to the original problem and can include your extension at the end.
- 2 years, 9 months ago
Nice, I have my own solution for it as well. Th essence for this extension is to just prove $$OE=OX$$ since $$OX=OB$$ is quite apparent.
- 2 years, 9 months ago
Here's my synthetic solution. Let $$XF$$ intersect $$BD$$ at $$O$$ and $$EM$$ at $$O^{'}$$. We need to show that $$O$$ coincides with $$O^{'}$$. A clear observation shows that $$B,C,F,D,X$$ are concyclic because $$\angle\{DCF\}$$=$$\angle\{DBF\}$$. Moreover $$D$$ is the circumcenter of triangle $$CBA$$. A clear angle chase shows that $$\angle \{DXF\}$$=$$\angle\{DBF\}$$ showing $$O$$ as the circumcenter of triangle $$OBE$$ thus $$E{O^{'}}$$=$$BO$$ Thus $$O$$ coincides with $$O{'}$$.proved Your extension can be proved easily if you watch $$F$$ as the incenter of triangle $$DAB$$ and watching $$XFAD$$ and $$CDEM$$ as the parallelogram. Thanks proving $$O$$ as the circumcenter was really very helpful.
- 1 year, 1 month ago
Oh about it, u were talking to me.. OK I am also trying to solve it with another way...
- 1 year, 1 month ago
Provide a solution using complex numbers.
- 1 year, 1 month ago
Well ur solution was really nice, I am also seeing it and trying to find a mistake..... 😂 😂😂😂😂😂😂😂😇😇😇😇😇😝😝😝😝😝😋
- 1 year, 1 month ago | 2019-04-25T04:56:03 | {
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https://math.stackexchange.com/questions/4030154/in-how-many-ways-can-they-rearrange-such-that-each-person-is-in-a-new-seat-and-n | # In how many ways can they rearrange such that each person is in a new seat and no 2 persons swap?
Beforehand, each person is assigned a seat (n seats). How many ways can they rearrange such that each person is in a new seat and no 2 people swap?
I'm guessing $$(n-1)!$$, but I have no idea on how to get the answer.
I was trying to get the answer to this problem:
A small class of nine boys are to change their seating arrangement by drawing their new seat numbers from a box. After the seat change, what is the probability that there is only one pair of boys who have switched seats with each other and only three boys who have unchanged seats?
where I got: $$(9C3 * 6C2 *4!)/9!$$ but I realized that none of the (4!) people should be on the same seat. And no pair of students swapped seats
The answer to this problem is $$9C3*6C2*3!/9!$$. I want to find a generalization, with n seats, how many ways can you arrange n people such that no 2 people swap and no one is in the same seat.
• Please use derangement Feb 18, 2021 at 7:03
• Count it yourself for 1, 2, 3 and 4 people. The answers are not 1, 1, 2, 6. Feb 18, 2021 at 7:25
• OK so now your question is different. Now for $4$ people - the first person A has a choice of $3$ seats. Say A takes C's seat. Now C has choice of only $2$ seats as C cannot take A's seat as that would be a swap. Say C takes B's seat. Now B has choice of only 1 seat - cannot take C's seat as that is a swap. Cannot take A's seat as that would mean D does not change seat. So B takes D's seat. Finally D has 1 choice which is take A's seat. Feb 18, 2021 at 8:07
• This entry to OEIS gives the answer. No simple closed formula. They give a recurrence relation. Didn't check but likely the same or similar to Gribouillis's answer (+1). Feb 18, 2021 at 8:34
• @JyrkiLahtonen: I did just now check, and it has both Gribouillis’s original summation and the nice recurrence derived from it. Feb 18, 2021 at 17:51
I think there is a way to write a recurrence relation by the following reasoning: we are counting the permutations of $$\left\{1 , \ldots , n\right\}$$ which decomposition in cycles contains only cycles having at least $$3$$ elements. Let $${X}_{n}$$ be that number. We can start with $${X}_{1} = {X}_{2} = 0$$ and $${X}_{3} = 2$$. Conventionally, we take $${X}_{0} = 1$$.
Let $$n \geqslant 3$$. We choose $$k \geqslant 2$$ elements in $$\left\{1 , \ldots , n-1\right\}$$ to form a cycle with the element $$n$$. There are $$\binom{n-1}{k}$$ ways to find these elements and there are $$k !$$ ways to align them in a row to build the cycle with element $$n$$. For the remaining $$n-k-1$$ elements, we choose one of the $${X}_{n-k-1}$$ valid permutations. Hence we obtain the relation
$$$${X}_{n} = \sum _{k = 2}^{n-1} \binom{n-1}{k} \ k ! \ {X}_{n-k-1} = (n-1)!\sum _{k = 2}^{n-1} \frac{{X}_{n-k-1}}{(n-k-1)!}$$$$ This easily entails the relation $$X_n = (n-1)X_{n-1} + (n-1)(n-2) X_{n-3}$$.
Note that this order 3 relation can be shown directly by another way of counting the permutations:
Second reasoning We can separate the valid permutations between the permutations for which the number $$n$$ belongs to a 3-cycle (there are $$(n-1)(n-2) X_{n-3}$$ of them because there are $$(n-1)(n-2)/2$$ ways of choosing the two other elements of the 3-cycle and each choice gives two possible cycles) and the permutations for which the number $$n$$ belongs to a larger cycle. Those are obtained by inserting the number $$n$$ in one of the cycles of a valid permutation of the $$n-1$$ first elements. For a given such permutation, there are exactly $$n-1$$ places where to insert the $$n$$-th element, between one of the $$n-1$$ first numbers and its successor in a cycle. Hence there are $$(n-1)X_{n-1}$$ such permutations. This proves the order 3 relation.
Clearly, the same reasoning shows that if we generalize and we count instead the number $$Y_n$$ of permutations of $$\{1,\cdots, n\}$$ where all the cycles have a length $$\ge p$$, then we have the relation
$$$$Y_n = (n-1) Y_{n-1} + (n-1)\cdots (n-p+1) Y_{n-p}$$$$
• I see how you can use the summation equation to prove the order $3$ recurrence relation, but how on earth did you discover the order $3$ recurrence relation in the first place? Feb 18, 2021 at 16:49
• @MikeEarnest In fact I discovered it by remarking that the summation was almost a simple relation for $X_n / n!$. However, it can be obtained directly, see my edit about the second reasoning above. Feb 18, 2021 at 17:20
• Very cool, thank you! Feb 18, 2021 at 17:32 | 2022-05-21T03:48:16 | {
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https://www.physicsforums.com/threads/heat-transfer-pipe-problem-dilemma.881338/ | # Heat Transfer Pipe Problem dilemma
Tags:
1. Aug 8, 2016
### williamcarter
1. The problem statement, all variables and given/known data
Problem data:
http://imgur.com/a/hEz16
2. Relevant equations
Rconduction=ln(rout/rin)/2*pi*Length*thermal conduct
Rconvection=1/2*pi*Length*routside*convective heat transfer outside
=>Qconduct=2*pi*Length*thermal conduct*(Tin-Tout)/ln(rout/rin)
=>Qconv=2*pi*Length*routside*convective heat transfer outside*(Tout-Tair)
3. The attempt at a solution
Solution:
a)
http://imgur.com/a/cH9hV
b)
For a)
I do not understand why we need to find Toutside by equating Qcond=Qconv?
And why we need to calculate Q from the pipe just as Qconvection instead of Q total?
Why we cannot do Q=delta T/Resistance Total
where:
Rtotal=Rconduction+Rconvection,because we do not need temperatures in those 2 R formulas
and delta T=(Tin-Tair) specifficaly 300 and 5 degrees celsius
As for b) They do Rtotal and then Qtotal=delta T/Rtotal
and then they do Qtotal=Qconv and they get T2.
Why we cannot proceed the same for a)?
Last edited: Aug 8, 2016
2. Aug 8, 2016
### benny_91
Actually you can use that method (adding up individual resistances) and you will get the same result. I guess the question requires you to find the temperature of the outer surface of the pipe hence employ this method.
3. Aug 8, 2016
### williamcarter
Does this mean that Qtotal=Qconvection? We neglect Qconduction ?
They were asking for Q emerging from pipe
If I do like them as Q from pipe =Qconvection=75760 W
if I do like Q=delta T/Rtotal , where Rtotal=Rconduction+Rconvection it gives me Q=75758 W
Last edited: Aug 8, 2016
4. Aug 8, 2016
### benny_91
You are missing an important point here. Note that the heat that is conducted through the pipe is then convected from the surface of the pipe to the atmosphere. Heat transfer passes through different stages: first conduction and then convection.
As long as there is no internal heat generation in the heat transfer medium and steady state in maintained (the pipe in this case) Qtotal=Qconduction=Qconvection
5. Aug 8, 2016
### williamcarter
Yes I agree and I understood it is conduction+convection , it is obvious but I do not understand why at a) they did Q as it was just convection?and at b) they did Qtotal and equated Qtotal=Qconvection to get Temp2?
6. Aug 8, 2016
### benny_91
If by T2 you mean outer surface temperature of the pipe then you can either use Qtotal=Qconvection or Qtotal=Qconduction. You will get the same value of T2.
7. Aug 8, 2016
### williamcarter
Thank you for your quick answer, for a) they did Q from pipe as Qconvection and they got 75760 W .If I do Qtotal=delta T/Rtotal
where delta T=(300-5) and Rtotal =Rcond+Rconv . I should get same answer right?
Why were they searching for Toutside and T2 at surface of pipe?If they were not asking for them?
8. Aug 8, 2016
### benny_91
Absolutely. If 300 is the temperature of the inner surface of the pipe and 5 is the temperature of the outer convecting fluid you should get the same answer.
9. Aug 8, 2016
### Staff: Mentor
Does all the heat flow (Q total) pass through the tube wall?
Does all the heat flow then pass through the convective boundary layer?
So, is Qcond = Qconv = Q total (or not)?
10. Aug 8, 2016
### williamcarter
Heat is transferred firstly by conduction within the pipe , then by convection from pipe surface to the air.
To answer the questions I would say , yes Qtotal passes through tube wall, yes Qtotal passes through boundary layer.
There is no insulation at the moment, furthermore no generation,no accumulation so I would say Qtotal=Qconduction=Qconvection
11. Aug 8, 2016
### Staff: Mentor
Good. So in part (a), for conduction through the pipe, we have:
$$Q=2\pi kL\frac{(300-T_0)}{\ln{(r_{out}/r_{in})}}$$
and for conduction through the convective boundary layer, we have:
$$Q=2\pi r_{out}hL(T_0-5)$$where $T_0$ is the temperature at the outer surface of the pipe. If we solve for the temperature differences, we obtain:
$$(300-T_0)=\frac{\ln{(r_{out}/r_{in})}}{2\pi kL}Q$$
$$(T_0-5)=\frac{1}{2\pi r_{out}hL}Q$$
So, you have two equations in the two unknowns, $T_0$ and Q. If you add these two equations together, you can eliminate $T_0$ to obtain:
$$(300-5)=\left[\frac{\ln{(r_{out}/r_{in})}}{2\pi kL}+\frac{1}{2\pi r_{out}hL}\right]Q$$
You can then solve this equation for Q and substitute back into either of the other equations to get $T_0$.
Now, when you add the insulation layer in part (b), you end up with three equations in three unknowns, Q and the two unknown temperatures at the two interfaces. You set up and solve these three equations in basically the same way as part (a). Questions?
12. Aug 8, 2016
### williamcarter
Everything clear, thanks.
13. Aug 20, 2016
### williamcarter
I understood everything, but I still have a question.
We know Q=m*cp*ΔT=m*cp*(Tout-Tin)
Why we are doing ΔT=Tin-Tout?
14. Aug 20, 2016
### Staff: Mentor
This question does not relate to the present problem. It relates to a flow problem.
15. Aug 20, 2016
### williamcarter
In general we know Q=m*cp*ΔT=m*cp*(Tout-Tin)
but in this problem, in their answer they are doing ΔT=(Tin-Tout)
I am confused why they did like this in their answer on this exercise.
16. Aug 20, 2016
### Staff: Mentor
Do you see any Cp in the exercise?
17. Aug 21, 2016
### williamcarter
No , I don't.
I cannot understand why they did in this problem ΔT=Tin-Tout=300-5.Why they did like that?
Usually we know ΔT=Tout-Tin=Tfinal-Tinitial not Tinitial-Tfinal
18. Aug 21, 2016
### Staff: Mentor
Don't forget that minus sign in the heat conduction equation. q = - k dT/dx
19. Aug 21, 2016
### williamcarter
Thank you, understood now.
20. Aug 21, 2016
### williamcarter
But also in convection they did To-Tair,not Tair-To.
I always thought we take ΔT=Tout-Tin not Tin -Tout, this is what confused me. | 2017-08-20T01:52:12 | {
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https://math.stackexchange.com/questions/919786/proving-that-3x2-implies-3x | # Proving that $3|x^2 \implies 3|x$.
Given a positive integer $x$, I would like to show that $3|x^2 \implies 3|x$. This is my missing step in the proof that $\sqrt{3}$ is irrational. My thoughts so far:
Assuming $3|x^2,$ then since $3$ is a prime number, we have that the prime factorization of $x^2$ is $m3^n$, where $n\geq 1$ and $\gcd(m,3) = 1.$ Then it follows that the prime factorization of $x$ is $r3^{n/2},$ where $\gcd(r,3) = 1$. I now want to show that $n$ must be even, so I can try to assume $n$ is odd and derive a contradiction...just not sure how to proceed.
Another way is by contradiction, assume $3\not |x$ then there are $a,b$ such that $$3a+bx=1$$ now multiply by $x$ to get $$3xa+bx^2=x$$ and $3$ divides the left by assumption so it divides the right contradiction.
• I agree I was about to edit. – Rene Schipperus Sep 4 '14 at 22:10
This should get you started: suppose $x$ contains $n'$ factors of $3$, then how many factors of $3$ does $x^2$ contain? (I don't want to deprive you of the fun of working the rest out yourself :-P)
Due to the fundamental theorem of arithmetic, we are allowed to let the prime factorization of $x$ be $x =p_1 p_2 \cdots p_k \Rightarrow x^2 = p_1^2 p_2^2 \cdots p_k^2$. Since $3|x^2$ we have that $p_i=3$ (for some $1 \leq i \leq k$). But that means $3|x$ as well because the same $p_i$ is in the factorization of $x$ too
• You are implcitly using uniqueness of prime factorizations. For the proof to be complete/correct you need to make that explicit. – Bill Dubuque Sep 4 '14 at 21:47
• Can you suggest how that can done? Thanks. – Sheheryar Zaidi Sep 4 '14 at 21:55
• I imagine he is asking you to state that "Due to unique prime factorization, we know" etc$\dots$ – Display Name Sep 4 '14 at 22:19
• Please explain how you deduced $\,3\mid (p_1\cdots p_k)^2\Rightarrow\,$ some $\,p_i = 3.\$ – Bill Dubuque Sep 4 '14 at 22:22
• @Sheheryar Then you should explicitly mention how you used it because, not too infrequently, students write such arguments thinking that no justification is needed, e.g. that is is "obvious", wrongly believing that neither FTA nor related results are needed. Without further elaboration, there is no way to know which argument was intended, the correct one, or the incorrect one. – Bill Dubuque Sep 5 '14 at 16:34
Here's an answer I just thought of, using the lemma from number theory that if $\gcd(a,b) = 1$ and $\gcd(a,c) = 1$ then $\gcd(a, bc) = 1.$
Using the above lemma, if $\gcd(3,x) = 1$, then $\gcd(3, x^2) = 1.$ Then by the contrapositive, if $\gcd(3, x^2) \neq 1,$ then $\gcd(3, x) \neq 1.$ But since $3$ is prime, we can reword that assertion as $3|x^2 \implies 3|x.$
Your original post basically does it! When you get to $$x = r3^{n/2}$$
and you know that $x, r$ and $n$ are integers, it follows that $\frac{n}{2}$ is an integer too. Hence $\frac{n}{2} = k$, for some integer $k$ and $n = 2k$, so $n$ is even. | 2020-01-19T19:34:52 | {
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https://math.stackexchange.com/questions/1863305/area-of-circle-double-integral-and-cartesian-coordinates | # Area of circle (double integral and cartesian coordinates)?
I know that the area of a circle, $x^2+y^2=a^2$, in cylindrical coordinates is $$\int\limits_{0}^{2\pi} \int\limits_{0}^{a} r \, dr \, d\theta = \pi a^2$$
But how can find the same result with a double integral and only cartesian coordinates?
## 5 Answers
Think about how the cartesian variables $x$ and $y$ are bounded. If we have the equation $$x^2+y^2=r^2\Rightarrow x=\pm\sqrt{r^2-y^2}\;\text{or}\;y=\pm\sqrt{r^2-x^2}$$ And $|x|,|y|<r$. Note that this last condition also insures that the above square roots are real. Then this gives you bounds for your double integral, choosing to integrate $x$ first, $$\int_{-r}^r\int_{-\sqrt{r^2-y^2}}^{\sqrt{r^2-y^2}}\mathrm dx\mathrm dy= \int_{-r}^r2\sqrt{r^2-y^2}\mathrm dy$$ Which you can integrate using the substitution $y=r\sin(t)\Rightarrow \mathrm dy=r\cos(t)\mathrm dt$ $$2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}r^2\cos^2(t)\mathrm dt=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}r^2(1+\cos(2t))\mathrm dt\\ =r^2(t+\frac{1}{2}\sin(2t))\vert_{-\frac{\pi}{2}}^{\frac{\pi}{2}}=r^2\pi$$ Note that I used the identity $\cos^2(t)=1/2(1+\sin(2t))$ to evaluate.
• Great, thanks! Just for curiosity, is it even possible with this substitution? $$t= r^2-y^2$$ $$\frac{dt}{dy}=-2y$$ $$dt=-2ydy$$ And $y^2=r^2-t, y=\pm \sqrt{r^2-t}$ so $$dt=-2\sqrt{r^2-t}dy$$ $$\frac{-dt}{\sqrt{r^2-t}}=2dy$$ Stuck! – JDoeDoe Jul 18 '16 at 18:03
• at first glance it looks like it may work, try pulling out $r^2$ and then use substitution $u=t/r^2$ to see arcsin – qbert Jul 18 '16 at 18:09
$$I=\int_{-r}^r \int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}} dy \, dx\\ \\ I=\int_{-r}^r 2\sqrt{r^2-x^2}\, dx\\ \\$$ Set $x=r \sin t$, so $dx = r \cos t\,dt\,$, we have $$I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 2r^2 \cos^2 t \,dt\\ \\ r^2(t+\sin t \cos t)\Big{|}_{-\frac{\pi}{2}}^{\frac{\pi}{2}}= \color{red}{\pi r^2}$$
Just as an alternative solution to the qbert answer.
Note that $$|x|$$, $$|y|$$ are not strictly less than $$r$$, but instead $$|x| \leq r$$, $$|y| \leq r$$. This can still insure that $$r^2 \geq y^2$$ and so $$\sqrt{r^2-y^2}$$ is real.
After the first integral
$$\int_{-r}^r\int_{-\sqrt{r^2-y^2}}^{\sqrt{r^2-y^2}}\mathrm dx\mathrm dy= \int_{-r}^r2\sqrt{r^2-y^2}\mathrm dy$$
Instead of the change of variable, you can remember the known integral:
$$\int \sqrt{r^2 - y^2} = \frac{1}{2} \left( r^2 \arcsin \frac{y}{r} + y \sqrt{r^2 - y^2} \right) + C$$
whose condition $$|y| \leq |r|$$ has just been mentioned and it is satisfied. The result follows almost immediately:
$$\int_{-r}^r 2 \sqrt{r^2 - y^2} \mathrm{d} y = \left[ r^2 \arcsin \frac{y}{r} + y \sqrt{r^2 - y^2} \ \right]_{-r}^r = r^2 \arcsin (1) - r^2 \arcsin (-1) = \\ = r^2 \frac{\pi}{2} + r^2 \frac{\pi}{2} = \pi r^2$$
So, using the cartesian coordinates, the only important observation is simply to consider the right extreme values for each variable.
Using the circumference equation $$x^2 + y^2 = r^2$$, you can choose $$x$$ as a function of $$y$$, obtaining $$x = \pm \sqrt{r^2 - y^2}$$, which will be the extreme values for $$x$$. Then, you let $$y$$ sweep from $$-r$$ to $$r$$, which are the extreme values for $$y$$. Of course, you can alternatively do vice-versa, with $$y$$ as function of $$x$$.
Unlike the cylindrical coordinates, there is no angular variation here, but only an horizontal variation between $$- \sqrt{r^2 - y^2}$$ and $$\sqrt{r^2 - y^2}$$ as regards $$x$$, and a vertical variation between $$-r$$ and $$r$$ as regards $$y$$.
I know this is not in the context of the question, but it seems as though you asked this question out of curiosity. So actually, using Green's Theorem you can get the result with the single integral,
$$\frac{1}{2} \oint_C x \ dy - y \ dx$$
where $C$ is the underlying curve for the parametrization $\gamma(t) = (R \cos t , R \sin t)$ and $0 \leq t\leq 2 \pi$.
Do you mean something like this?
$$\int_{-a}^{a} \int_{-\sqrt{a^2-x^2}}^{+\sqrt{a^2-x^2}} 1 \, dy \, dx$$
• Yes, how do you evaluate that without cylindrical coordinates? – JDoeDoe Jul 18 '16 at 17:53
• I don't understand your question. These are Cartesian coordinates. – Martin Kochanski Jul 19 '16 at 8:28 | 2019-09-15T06:02:20 | {
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https://math.stackexchange.com/questions/3342875/is-it-true-that-if-limsup-limits-n-to-infty-left-fraca-n1a-n-right | # Is it true that if $\limsup\limits_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right| > 1$, then $\sum a_n$ diverges?
I am reading "A Course in Analysis vol.2" by Kazuo Matsuzaka.
There is the following theorem ("ratio test") in this book.
Let $$a_n \neq 0$$ for all $$n$$.
(a) If $$\limsup\limits_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right| < 1$$, then $$\sum a_n$$ converges absolutely.
(b) If $$\left|\frac{a_{n+1}}{a_n}\right| \geq 1$$ for all $$n \geq N$$ for some $$N$$, then $$\sum a_n$$ diverges.
Is the following statement false?
(b') If $$\limsup\limits_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right| > 1$$, then $$\sum a_n$$ diverges.
• If you want your two statements to look dual to each other, you need (b) to begin "if $\liminf \left|\frac{a_{n+1}}{a_n}\right| > 1$..." Of course, that's strictly weaker than the given statement. – Micah Sep 3 '19 at 21:28
That is indeed wrong, a counter example is the sequence $$(a_n)$$ $$\frac 12, \frac 22, \frac 14, \frac 24, \frac 18, \frac 28, \ldots$$ Here $$\sum a_n$$ is convergent, but $$\limsup_{n \to \infty} |\frac{a_{n+1}}{a_n}| = 2$$.
More generally you can take any convergent series $$\sum c_n$$ with $$c_n \ne 0$$ and then define $$a_{2n} = c_n, a_{2n+1} = 2c_n \, .$$ Then $$\sum a_n$$ is convergent as well, but $$\limsup_{n \to \infty} |\frac{a_{n+1}}{a_n}| = 2$$.
Unfortunately it is false: consider the sequence $$a_n$$ defined as $$a_n=\frac{1}{n^2}, \mbox{ if n is odd}$$ and $$a_n=\frac{1}{n^3}, \mbox{ if n is even}$$. The series $$\sum_{k=0}^{\infty}$$ is convergent but the subsequence of the ratios $$r_k=\frac{a_{2k+1}}{a_{2k}} = \frac{(2k)^3}{(2k+1)^2}$$ is divergent.
No. Consider a sequence $$(a_n)$$ like $$(\frac 1 {2^{3}},\frac 1 {3^{3}}, \frac 2 {3^{3}}, \frac 1 {4^{3}},\frac 1 {4^{3}}, \frac 2 {4^{3}},...)$$ where the (n-1)-st block has $$\frac 1 {n^{3}}$$ repeated $$n-1$$ times followed by $$\frac 2 {n^{3}}$$. Then $$\frac {a_{n+1}} {a_n}=2$$ for infinitely many $$n$$ but $$\sum a_n <\infty$$. | 2020-10-30T06:52:27 | {
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https://math.stackexchange.com/questions/2083841/problem-with-joint-probability-function | # problem with joint probability function
The joint probability density function of $X$ and $Y$ is given by $$f(x,y) = \begin{cases}c(y^2-x^2) e^{-y}, & -y \le x < +y, \ 0 \le y < \infty,\\ 0, & \text{otherwise}. \end{cases}$$
(a) Find $c$.
(b) Find the marginal densities of $X$ and $Y$.
(c) Find $\operatorname{E}[X]$.
I have some problems with the point (b)
(a) $$\int_0^\infty\ \int_{-y}^y c(y^2-x^2)e^{-y}\,dy\,dx = 1 \Leftrightarrow c= \frac 1 8.$$ (b) I calculate the marginal density of $Y$ as $$\int_{x=-y}^{x=+y} \frac 1 8 (y^2-x^2)e^{-y}\,dx = \frac 1 6 y^3 e^{-y},$$ and the density of $X$ as $$\int_{y=0}^\infty\ c(y^2-x^2)e^{-y}\,dy,$$ but there something wrong because the solution is different. Can someone help me to understand my mistake?
• What is the solution and how are they different? – Jack Jan 4 '17 at 20:05
• in the book $$f_x(x)=\int_{|x|}^\infty\ c*(y^2-x^2)*e^{-y}\,dy\,dx=(|x|+1)e^{-|x| }/4$$ – Anne Jan 4 '17 at 20:29
• As your question now stands, in part (a) you've written this: $$\int_0^\infty\left( \int_{-y}^y c(y^2-x^2)e^{-y}\,dy \right)\,dx$$ The inside integral, with $y$ going from $-y$ to $y$, does not make sense. It is $x$ that goes from $-y$ to $y$. Thus you need this: $$\int_0^\infty\left( \int_{-y}^y c(y^2-x^2)e^{-y}\,dx \right) \,dy.$$ – Michael Hardy Jan 4 '17 at 21:01
• thank you for your comment, I put dx or dy in a careless way – Anne Jan 4 '17 at 21:06
The marginal density of $Y$ is $$\int_{-y}^y c(y^2-x^2) e^{-y} \mathop{dx}=ce^{-y}(2y^3 - (2/3)y^3 ) = \frac{4c}{3} y^3 e^{-y}.$$
Integrating this over $y$ gives $$1=\int_0^\infty \frac{4c}{3} y^3 e^{-y} \mathop{dy} = \frac{4c}{3} \cdot 3! = 8c \implies c=1/8.$$
So, the marginal density of $Y$ is $\frac{1}{6} y^3 e^{-y}$ as you obtained.
The marginal density of $X$ is $$\int_{|x|}^\infty c(y^2-x^2) e^{-y} \mathop{dy} .$$ The tricky part is the limits of integration, which comes from the condition $|x| \le y$. First, integration by parts twice gives \begin{align} \int_{|x|}^\infty y^2 e^{-y} \mathop{dy} &=[-y^2 e^{-y}]_{y=|x|}^\infty + 2\int_{|x|}^\infty ye^{-y} \mathop{dy}\\ &= x^2 e^{-|x|} + 2[-ye^{-y}]_{y=|x|}^\infty + 2\int_{|x|}^\infty e^{-y}\mathop{dy}\\ &= x^2 e^{-|x|} + 2|x|e^{-|x|} + 2e^{-|x|}\\ &= e^{-|x|}(|x|^2+2|x|+2). \end{align}
So, the marginal density of $X$ is $$\int_{|x|}^\infty c(y^2-x^2) e^{-y} \mathop{dy} = c(e^{-|x|}(|x|^2+2|x|+2) - x^2 e^{-|x|}) = \frac{1}{4} e^{-|x|}(|x|+1).$$
Edit (explanation for why the lower limit of integration is $|x|$):
In general, the marginal density of $X$ is $$f_X(x) = \int_{-\infty}^\infty f(x,y) \mathop{dy}.$$ For a specific $x$, note that by definition $f(x,y)$ is zero if $|x|>y$; otherwise $f(x,y) = c(y^2-x^2)e^{-y}$. So $$f_X(x) = \int_{-\infty}^\infty f(x,y) \mathop{dy} = \int_{-\infty}^{|x|} f(x,y) \mathop{dy} + \int_{|x|}^\infty f(x,y) \mathop{dy} = 0 + \int_{|x|}^\infty c(y^2-x^2)e^{-y} \mathop{dy}.$$
• what i don't understand is why the integral is between |x| and infinity and not 0 and infinity – Anne Jan 4 '17 at 21:14
• @Anne See my edit – angryavian Jan 4 '17 at 21:24
• thanks , could you help me with E[x] ? – Anne Jan 4 '17 at 23:00
• @Anne Note that the marginal density of $X$ is symmetric about zero. – angryavian Jan 5 '17 at 0:19
• @ angryavian ok thanks – Anne Jan 5 '17 at 9:20
There is nothing wrong with the marginal distributions looking different, with one being $f_Y(y) = \frac16 y^3 e^{-y}$ and the other being
$$f_X(x) = \int_{y=|x|}^\infty \frac18 (y^2-x^2)e^{-y}dy =\frac14 (1+|x|) e^{-|x|}$$
They are marginal distributions for two different variables and the joint distribution is in no way symmetric so in general they would be two distinct distributions unless some staggering coincidence occurred.
Your confusion was probably because you integrated from $y=0$ to infinity, rather than starting at $y=|x|$. Smaller values of $y$ lie outside the region of non-zero distribution function. So you were probably puzzled about a marginal distribution that looked something like $\frac18 (2-x^2)$, which is what you get if you make that mistake.
• you went straight to the point. I don't understant your sentence"smaller values of y lie outside the region of non-zero distribution function" could you explain in other words? – Anne Jan 4 '17 at 21:23
For the marginal of $X$ you need $$\int_0^\infty f_{X,Y}(x,y)\, dy = \int_{|x|}^\infty \frac 1 8 (y^2-x^2)e^{-y} \, dy.$$ The point is that the density is nonzero only when either $y>x$ and $y>-x$. That is the same as $y>|x|$.
Let $u=y-x$, so that $du=dy$, and then $y+x= u+2x.$ Then $$y^2-x^2 = (y-x)(y+x) = u(u+2x).$$ If $x\ge0$ then $y$ goes from $x$ to $\infty$, so $u$ goes from $0$ to $\infty$, and you get $$\int_0^\infty \frac 1 8 u(u+2x)\, e^{-(u+x)} \,du.$$ Note that $x$ does not change as $u$ goes from $0$ to $\infty$, so this becomes $$\frac 1 8e^{-x} \left( \int_0^\infty u^2 e^{-u} \, du + 2x \int_0^\infty e^{-u} \,du \right)$$ and this comes to $$\frac 1 8 e^{-x}(2+2x) = \frac 1 4 e^{-x} (1+x).$$
But what if $x<0$? In that case $\displaystyle \int_{|x|}^\infty \cdots \,dy$ is $\displaystyle \int_{-x}^\infty \cdots \,dy,$ and that becomes $\displaystyle \int_{-2x}^\infty \cdots \, du.$ | 2019-10-23T19:55:50 | {
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https://markkm.com/clrs/ch03/ | CLRS Chapter 3. Growth of Functions
3.1. Asymptotic notation
Exercises
3.1-1. Let $f(n)$ and $g(n)$ be asymptotically nonnegative functions. Using the basic definition of $\Theta$-notation, prove that $\max(f(n), g(n)) = \Theta(f(n) + g(n)).$
Since
for all $n \in \mathbb{N}$, we see that
for all $n \in \mathbb{N}$.
We now choose an integer $N$ such that $f(n) \geq 0$ and $g(n) \geq 0$ for all $n \geq N$. This implies that
for all $n \geq N$.
It follows that $\max(f(n),g(n)) = \Theta(f(n),g(n))$. $\square$
3.1-2. Show that for any real constants $a$ and $b$, where $b > 0$,
Observe that
Let $f(n) = (1+a/n)^b$ for each $n$. We fix a positive integer $N$ such that $% $ for all $n \geq N$. This, in particular, ipmlies that $1 + \vert a/n \vert > 0$ for all $n \geq N$. Since $\vert a/n \vert$ is decreasing as $n$ increases, we see that
Now, $b$ is positive, and so it follows from the above estimates that
for all $n \geq N$. We conclude that $(n+a)^b = \Theta(n^b)$. $\square$
3.1-3. Explain why the statement, “The running time of algorithm $A$ is at least $O(n^2)$,” is meaningless.
The big-O notation can only be used to give upper bounds, not lower bounds. $\square$
3.1-4. Is $2^{n+1} = O(2^n)$? Is $2^{2n} = O(2^n)$?
Since
for all $n$, we see that $2^{n+1} = O(2^n)$.
We now fix an arbitrary constant $C > 0$. There exists a positive integer $N_C$ such that $% $ for all $n \geq N_C$. It then follows that
for all $n \geq N_C$. Since the choice of $C$ was arbitrary, we conclude that $2^{2n} \neq O(2^n)$. $\square$
3.1-5. Prove Theorem 3.1.
We recall the statement of Thereom 3.1:
For any two functions $f(n)$ and $g(n)$, we have $f(n) = \Theta(g(n))$ if and only if $f(n) = O(g(n))$ and $f(n) = \Omega(g(n))$.
$(\Rightarrow)$ Suppose that $f(n) = \Theta(g(n))$. Fix positive constants $c_1$, $c_2$, and $n_0$ such that
for all $n \geq n_0$. Since
for all $n \geq n_0$, we have that $f(n) = O(g(n))$. Similarly,
for all $n \geq n_0$, and so $f(n) = \Omega(g(n))$.
$(\Leftarrow)$ Suppose that $f(n) = O(g(n))$ and $f(n) = \Omega(g(n))$. The first assumption furnishes positive constants $c_1$ and $n_1$ such that
for all $n \geq n_1$. The second assumption furnishes positive constants $c_2$ and $n_2$ such that
for all $n \geq n_2$. It follows that
for all $n \geq \max(n_1,n_2)$, whence $f(n) = \Theta(g(n))$. $\square$
3.1-6. Prove that the running time of an algorithm is $\Theta(g(n))$ if and only if its worst-case running time is $O(g(n))$ and its best-case running time is $\Omega(g(n))$.
The running time of an algorithm is bounded below by its best-case running time and bounded above by its worst-case running time. The claim now follows from Exercise 3.1-5. $\square$
3.1-7. Prove that $o(g(n)) \cap \omega(g(n))$ is the empty set.
Let $f(n) \in o(g(n))$. For each constant $c >0$, there exists a constant $n_c > 0$ such that
for all $n \geq n_c$. We fix an arbirary constant $C > 0$ and observe that, for each positive constant $N$, $n \geq \max(n_C, N)$ implies
It follows that $f(n) \notin \omega(g(n))$.
Similarly, we can show that $f(n) \in \omega(g(n))$ implies $f(n) \notin o(g(n))$. It follows that the intersection $o(g(n)) \cap \omega(g(n))$ is empty. $\square$
3.1-8. We can extend our notation to the case of two parameters $n$ and $m$ that can go to infinity independently at different rates. For a given function $g(n,m)$, we denote by $O(g(n,m))$ the set of functions
Give corresponding definitions for $\Omega(g(n,m))$ and $\Theta(g(n,m))$.
3.2. Standard notations and common functions
Exercises
3.2-1. Show that if $f(n)$ and $g(n)$ are monotonically increasing functions, then so are the functions $f(n) + g(n)$ and $f(g(n))$, and if $f(n)$ and $g(n)$ are in addition nonnegative, then $f(n) \cdot g(n)$ is monotonically increasing.
Let $n \leq m$. Since $f(n) \leq f(m)$ and $g(n) \leq g(m)$, we see that
If, in addition, $f$ and $g$ are nonnegative, then
Finally, $g(n) \leq g(m)$ and the monotonicity of $f$ implies that
as was to be shown. $\square$
3.2-2. Prove equation (3.16).
3.2-3. Prove equation (3.19). Also prove that $n! = \omega(2^n)$ and $n! = o(n^n)$.
Recall that Stirling’s approximation states that
This implies that, for each $\epsilon > 0$, there exists a positive integer $N_\epsilon$ such that $n \geq N_\epsilon$ implies
We fix $\epsilon > 0$ and observe that
for all $n \geq N_\epsilon$, so that
We fix positive constants $c_1$, $c_2$, and $n_0$ such that
for all $n \geq n_0$. Since $\log$ is a monotonically increasing function, we see that
for all $n \geq n_0$.
Again, $\log$ is a monotonically increasing function, there exists a positive constant $n_1$ such that $\frac{1}{2}\log n \geq \log e$ for all $n \geq n_1$. It then follows that
for all $n \geq n_1$. Similarly, there exists a positive constant $n_2$ such that $\log c_1 + \frac{1}{2} \log n \geq 0$ for all $n \geq n_2$, and so
for all $n \geq n_2$. It follows that
for all $n \geq \max(n_0,n_1,n_2)$.
Now, we observe that
Whenever $n \geq 1$, we have $\frac{1}{2} \log n \leq n \log n$, and so
Finally, $n \log n$ is a monotonically increasing function, and so we can find a positive constant $n_3$ such that $\log c_2 \leq \frac{1}{2} n \log n$ for all $n \geq n_3$. It follows that
for all $n \geq \max(n_0, n_3)$.
We now see that
for all $n \geq \max(n_0,n_1,n_2,n_3)$. We conclude that $\log n! = \Theta(n \log n)$.
We now recall that
for all $n \geq n_0$.
Observe that $c_2 e^{-n} n^{1/2} \to 0$ as $n \to \infty$. Therefore, each constant $c > 0$ admits a positive integer $N_c$ such that $n \geq N_c$ implies
It follows that $n! = o(n^n)$.
Observe also that
whenever $n \geq 1$. Since $(n/2e)^n \to \infty$ as $n \to \infty$, we can find, for each constant $c>0$, a positive integer $N_c$ such that $n \geq N_c$ implies
Therefore,
whenever $n \geq \max(n_c, 1)$. Since the choice of $c$ was arbitrary, it follows that $n! = \omega(2^n)$. $\square$
3.2-4. Is the function $\lceil \log n \rceil !$ polynomially bounded? Is the function $\lceil \log \log n \rceil !$ polynomially bounded?
By Stirling’s approximation (and the continuity of $\log$),
Since $n \mapsto 2^n$ is a monotonically increasing function,
$\log \log n$ is a monotonically increasing function with limit $\lim_{n \to \infty} \log \log n = \infty$, and so $\lceil \log n \rceil !$ is not polynomially bounded.
Similarly,
Since $\log^{(i+1)} n \to \infty$ as $n \to \infty$, it follows that $\lceil \log^{(i)} n \rceil$ is not polynomially bounded. $\square$
3.2-5. What is asymptotically larger: $\log (\log^* n)$ or $\log^* (\log n)$?
We begin by observing that
Now,
so that
It follows that $\log^* (\log n)$ grows exponentially faster than $\log(\log^* n)$. $\square$
3.2-6. Show that the golden ratio $\phi$ and its conjugate $\hat{\phi}$ both satisfy the equation $x^2 = x + 1$.
Use the quadratic formula. $\square$
3.2-7. Prove by induction that the $i$th Fibonacci number satisfies the equality
where $\phi$ is the golden ratio and $\hat{\phi}$ its conjugate.
We first observe that
We now assume inductively that
and observe that
Since $\phi + 1 = \phi^2$ and $\widehat{\phi} + 1 = \widehat{\phi^2}$, it follows that
as was to be shown. $\square$
3.2-8. Show that $k \log k = \Theta(n)$ implies $k = \Theta(n / \log n)$.
We find constants $c_1$, $c_2$, and $n_0$ such that $n_0 \geq 2$ and that
for all $n \geq n_0$. This implies that
for all $n \geq n_0$.
Since $k \log k \geq k$ for all $k \geq 2$, we see that
for all $k \geq 2$. It follows that
for all $n \geq n_0$ and $k \geq 2$. Now, if $% $, then $% $. This cannot hold if $n \geq \max(n_0, 1/c_1)$, and so we conclude that
whenever $n \geq \max(n_0,1/c_1)$.
Now, $k \geq \log k$ for all $k \geq 1$, and so
for all $k \geq 1$. It follows that
for all $n \geq n_0$ and $k \geq 1$. Since we know that $k \geq 2$ whenever $n \geq \max(n_0,1/c_1)$, we conclude that
for all $n \geq \max(n_0,1/c_1)$.
In other words,
as was to be shown. $\square$
Problems
3-1. Asymptotic behavior of polynomials
Let
where $a_d > 0$, be a degree-$d$ polynomial in $n$, and let $k$ be a constant. Use the definitions of the asymptotic notations to prove the following properties.
a. If $k \geq d$, then $p(n) = O\left(n^k\right)$.
b. If $k \leq d$, then $p(n) = \Omega\left(n^k \right)$.
c. If $k = d$, then $p(n) = \Theta(n^k)$.
d. If $k > d$, then $p(n) = o(n^k)$.
e. If $% $, then $p(n) = \omega(n^k)$.
Problem 3-1a
Observe that, for each $n \geq 1$,
It follows that $p(n) = O(n^k)$. $\square$
Problem 3-1b
For each fixed positive integer $p$, we have the limit
Therefore, each positive constant $C$ has a corresponding positive integer $N_{C,p}$ such that $n \geq N_{C,p}$ implies
Taking $p = r-s$, we see that
for all $n \geq N_{C,r-s}$.
We now set
so that
for each $1 \leq i \leq d-1$ and every $n \geq N$. From this, we deduce that
for all $n \geq N$.
Finally, we know that
whenever $n \geq N_{2/a_d, d-k}$, and so
whenever $n \geq \max(N_{2/a_d,d-k},N)$. It follows that $p(n) = \Omega(n^k)$. $\square$
Problem 3-1c
This follows at once from 3-1a and 3-1b. $\square$
Problem 3-1d
As per 3-1b, we define, for each positive integer $p$ and every positive constant $C$, a number $N_{C,p}$ to be a positive integer such that
for all $n \geq N_{C,p}$.
We fix a positive constant $c$. Observe that, for each $1 \leq i \leq d$,
whenever $n \geq N_{2(d+1)\vert a_i \vert/c, k-i}$. Moreover,
whenever $n \geq \vert 2(d+1) a_0 / c \vert^{1/k}$.
Setting
we see that, for each $n \geq N$,
Since the choice of $c > 0$ was arbitrary, it follows that $p(n) = o(n^k)$. $\square$
Problem 3-1e
As per 3-1b, we define, for each positive integer $p$ and every positive constant $C$, a number $N_{C,p}$ to be a positive integer such that
for all $n \geq N_{C,p}$.
As in b, we set
so that
for all $n \geq N$. We now fix a constant $c >0$ and observe that
whenever $n \geq N_{4c/a_d, d-k}$. It follows that
whenever $n \geq \max(N,N_{4c/a_d,d-k})$. Since the choice of $c$ was arbitrary, we conclude that $p(n) = \omega(n^k)$. $\square$
3-2. Relative asymptotic growths
Indicate, for each pair of expressions $(A,B)$ in the table below, whether $A$ is $O$, $o$, $\Omega$, $\omega$, or $\Theta$ of $B$. Assume that $k \geq 1$, $\epsilon > 0$, and $c > 1$ are constants. Your answer should be in the form of the table with “yes” or “no” written in each box.
$A$ $B$ $O$ $o$ $\Omega$ $\omega$ $\Theta$
a. $\log^k$ $n$ $n^\epsilon$ yes yes no no no
b. $n^k$ $c^n$ yes yes no no no
c. $\sqrt{n}$ $n^{\sin n}$ no no no no no
d. $2^n$ $2^{n/2}$ no no yes yes no
e. $n^{\log c}$ $c^{\log n}$ yes no yes no yes
f. $\log$ $(n!)$ $\log$ $(n^n)$ yes no yes no yes
Problem 3-2a
We define $f(x) = \frac{1}{\epsilon} x^\epsilon$ and $g(x) = \ln x$ and note that
Since $\epsilon > 0$, we have $\epsilon - 1 > -1$, and so
for all $x \geq 1$, meaning $f(x)$ grows faster than $g(x)$ on the interval $x \geq 1$. Since $f(1) > g(1)$, we conclude that
for all $x \geq 1$. This, in particular, implies that
for all integers $n \geq 1$.
We now rescale $(\ast)$ for logarithm base 2, using the change-of-base formula:
for each $n \geq 1$ and every $\epsilon > 0$.
An argument analogous to the above shows that
for all integers $n \geq 1$. Therefore,
for all $i \geq 1$ and $n \geq 1$, and so it suffices to show that
for an arbitrary $\epsilon > 0$.
To this end, we fix a positive constant $C$. Since $n^{\epsilon/2} \to \infty$ as $n \to \infty$, we can find a positive constant $N$ such that $n^{\epsilon/2} > \frac{2\log e}{C\epsilon}$ for all $n \geq N$. It follows from $(\ast \ast)$ that
for all $n \geq N$. Since the choice of $C$ was arbitrary, we conclude that
as was to be shown. $\square$
Problem 3-2b
Observe that $n^k = 2^{\log(n^k)} = 2^{k \log n}$ and $c^n = 2^{\log c^n} = 2^{( \log c)n}$. We fix a positive constant $C$ and invoke a to find a positive integer $N$ such that
for all $n \geq N$. Since $n \mapsto 2^n$ is monotonically increasing, we conclude that
for all $n \geq N$. The choice of $C$ was arbitrary, and so $n^k = o(c^n)$. $\square$
Problem 3-2c
We claim that there are increasing sequences of positive numbers $(n_i)_{i=1}^\infty$ and $(m_i)_{i=1}^\infty$ such that $\sin n_i > \frac{2}{3}$ and $% $ for all $i$. (The proof of the claim typically involves either Diophantine approximations or the Weyl equidistribution theorem.) Recall that $n^p = o(n^q)$ whenever $% $. For this reason, given any $C > 0$, the inequality
cannot hold for all $i$. Similarly, the inequality
cannot hold for all $i$. It follows that neither $n^{1/2}$ nor $n^{\sin n}$ dominate each other. $\square$
Problem 3-2d
Since $2^n = (2^{n/2})^2$, we have the estimate $2^n = \omega(2^{n/2})$ from 3-2a. $\square$
Problem 3-2e
Observe that $n^{\log c} = 2^{(\log c) (\log n)}$ and $c^{\log n} = 2^{(\log c) (\log n)}$. Therefore, $n^{\log c} = c^{\log n}$. $\square$
Problem 3-2f
We have shown in Exercise 3.2-3 that $\log(n!) = \Theta(n \log n)$. Since $\log n^n = n \log n$, it follows that $\log(n!) = \Theta(\log n^n)$. $\square$
3-3. Ordering by asymptotic growth rates
a. Rank the following functions by order of growth; that is, find an arrangement $g_1,g_2,\ldots,g_{30}$ of the functions satisfying $g_1 = \Omega(g_2)$, $g_2 = \Omega(g_3)$, $\ldots$, $g_{29} = \Omega(g_{30})$. Partition your list into equivalence classes such that functions $f(n)$ and $g(n)$ are in the same class if and only if $f(n) = \Theta(g(n))$. $% $
b. Give an example of a single nonnegative function $f(n)$ such that for all functions $g_i(n)$ in part (a), $f(n)$ is neither $O(g_i(n))$ nor $\Omega(g_i(n))$
Problem 3-3b
$f(n) = 2^{2^{2^n}}\sin n$. Take a look at Problem 3.2.c to see how to prove asymptotics for functions like this one. $\square$
3-4. Asymptotic notation properties
Let $f(n)$ and $g(n)$ be asymptotically positive functions. Prove or disprove each of the following conjectures.
a. $f(n) = O(g(n))$ implies $g(n) = O(f(n))$.
b. $f(n) + g(n) = \Theta\left(\min(f(n), g(n))\right)$
c. $f(n) = O(g(n))$ implies $\log(f(n)) = O(\log(g(n))$, where $\log(g(n)) \geq 1$ and $f(n) \geq 1$ for all sufficiently large $n$.
d. $f(n) = O(g(n))$ implies $2^{f(n)} = O(2^{g(n)})$
e. $f(n) = O((f(n))^2)$
f. $f(n) = O(g(n))$ implies $g(n) = \Omega(f(n))$
g. $f(n) = \Theta(f(n/2))$
h. $f(n) + o(f(n)) = \Theta(f(n))$.
Problem 3-4a
$f(n) = 1$, $g(n) = n$ is a counterexample. $\square$
Problem 3-4b
$f(n) = 1$, $g(n) = n$ is a counterexample. $\square$
Problem 3-4c
Fix constants $C > 0$ and $N \geq 1$ such that
for all $n \geq N$. Since $\log$ is monotonically increasing,
for all $n \geq N$. Now, we find $N'$ such that $n \geq N'$ implies $\log(g(n)) \geq 1$. For such values of $n$, we have
It now follows that
whence $\log\left(f(n)\right) = O\left(\log(g(n))\right)$, as was to be shown. $\square$
Problem 3-4d
$f(n) = 4n$ and $g(n) = n$ is a counterexample. $\square$
Problem 3-4e
$f(n) = \frac{1}{n}$ is a counterexample. $\square$
Problem 3-4f
This follows at once from the definitions of $O$ and $\Omega$. $\square$
Problem 3-4g
$f(n) = \sin n$ is a counterexample. To see this, we observe that
which is not asymptotically equivalent to $\sin \frac{n}{2}$. $\square$
Problem 3-4h
We fix $g \in o(f(n))$ and show that $f(n) + g(n) = \Theta(f(n))$.
For notational convenience, we let $N_c$, for an aribitrary constant $c > 0$, denote a positive integer such that
for all $n \geq N_c$. Since
for all $n \geq N_1$, we have that $f(n) + g(n) = O(f(n))$.
We now observe that
for all $n \geq N_{1/2}$. It follows that $f(n) + g(n) = \Omega(f(n))$, whence we conclude that $f(n) + g(n) = \Theta(f(n))$. $\square$
3-5. Variations on $O$ and $\Omega$
Some authors define $\Omega$ in a slightly different way than we do; let’s use $\substack{\infty \\ \Omega}$ (read “omega infinity”) for this alternative definition. We say that $f(n) = \substack{\infty \\ \Omega}(g(n))$ if there exists a positive constant $c$ such that $f(n) \geq cg(n) \geq 0$ for infinitely many integers $n$.
a. Show that for any two functions $f(n)$ and $g(n)$ that are asymptotically nonnegative, either $f(n) = O(g(n))$ or $f(n) = \substack{\infty \\ \Omega}(g(n))$ or both, where as this is not true if we use $\Omega$ in place of $\substack{\infty \\ \Omega}$.
b. Describe the potential advantages and disadvantages of using $\substack{\infty \\ \Omega}$ instad of $\Omega$ to characterize the running times of programs.
Some authors also define $O$ in a slightly different manner; let’s use $O'$ for the alternative definition. We say that $f(n) = O'(g(n))$ if and only if $\vert f(n) \vert = O(g(n))$.
c. What happens to each direction of the “if and only if” in Theorem 3.1 if we substitute $O'$ for $O$ but still use $\Omega$?
Some authors define $\tilde{O}$ (read “soft-oh”) to mean $O$ with logarithmic factors ignored:
d. Define $\tilde{\Omega}$ and $\tilde{\Theta}$ in a similar manner. Prove the corresponding analog to Theorem 3.1.
Problem 3-5a
We assume that $f(n) \notin O(g(n))$. This means that each $C > 0$ and every $N > 0$ admits a positive integer $n_{C,N} \geq N$ such that
Now, $I = \{n_{2^{-i},2^i} : i \geq 1\}$ is an infinite set of integers such that
for all $n \in I$. It follows that $f(n) = \substack{\infty \\ \Omega}(g(n))$.
Setting $f(n) = \sin n$ and $g(n) = \cos n$, we see that $f(n) \notin O(g(n)) \cup \Omega(g(n))$. $\square$
Problem 3-5b
Introduction of $\substack{\infty \\ \Omega}$ renders all functions asymptotically comparable, which can be useful for making asymptotic statements about a wide class of functions without having to rule out the pathological cases (such as those involving periodic functions like $\sin$). In this manner, the utility of $\substack{\infty \\ \Omega}$ is analogous to that of limit superior and limit inferior, two generalizations of limit that exists for all sequences and functions.
On the other hand, the relaxed lower bound $\substack{\infty \\ \Omega}$ provides may not be all that useful for determining tight, or sometimes even reasonable, lower bounds for a specific function. For example, if
then our intuition suggests that $g(n)$ cannot be a good lower bound of $f(n)$. Nevertheless, $f(n) = \substack{\infty \\ \Omega}(g(n))$. $\square$
Problem 3-5c
$f(n) = \Omega(g(n))$ forces $f$ to be eventually positive, and so $f(n) \in \Omega(g(n)) \cap O'(g(n))$ implies $f(n) = O(g(n))$. It follows from Theorem 3.1 that $f(n) = \Theta(g(n))$. Conversely, if $f(n) = \Theta(g(n))$, then Theorem 3.1 implies that $f(n) = O(g(n))$ and $f(n) = \Omega(g(n))$. Since $O(g(n)) \subseteq O'(g(n))$, it follows that $f(n) = O'(g(n))$. $\square$
Problem 3-5d
We define
% and
If $f(n) = \tilde{\Theta}(g(n))$, then we can find positive constants $c_1$, $c_2$, $k_2$, $k_2$, and $n_0$ such that
for all $n \geq n_0$. This, in particular, implies that
for all $n \geq n_0$, and so $f(n) = \tilde{\Omega}(g(n))$. Similarly,
for all $n \geq n_0$, and so $f(n) = \tilde{O}(g(n))$.
Conversely, we assume that $f(n) = \tilde{O}(g(n))$ and $f(n) = \tilde{\Omega}(g(n))$. The $\tilde{\Omega}$-estimate furnishes positive constants $c_1$, $k_1$, and $n_1$ such that
for all $n \geq n_1$. The $\tilde{O}$-estimate furnishes positive constants $c_2$, $k_2$, and $n_2$ such that
for all $n \geq n_2$. It follows that
for all $n \geq \max(n_1,n_2)$, whence $f(n) = \tilde{\Theta}(g(n))$.
We conclude that $f(n) = \tilde{\Theta}(g(n))$ if and only if $f(n) = \tilde{O}(g(n))$ and $g(n) = \tilde{\Omega}(g(n))$. $\square$
3-6. Iterated functions.
We can apply the iteration operator $\ast$ used in the $\log^*$ function to any monotonically increasing function $f(n)$ over the reals. For a given constant $c \in \mathbb{R}$, we define the iterated function $f_c^*$ by
which need not be well defined in all cases. In other words, the quantity $f_c^*(n)$ is the number of iterated applications of reduce its argument down to $c$ or less.
For each of the following functions $f(n)$ and constants $c$, give as tight a bound as possible on $f_c^*(n)$.
$f$ $(n)$ $c$
a. $n$ $-$ $1$ 0
b. $\log$ $n$ 1
c. $n/2$ 1
d. $n/2$ 2
e. $\sqrt{n}$ 2
f. $\sqrt{n}$ 1
g. $n^{1/3}$ 2
h. $n$ $/$ $\log$ $n$ 2
Problem 3-6a
$f^{(i)}(n) = n - i$, and so $f^*_c(n) = n$. $\square$
Problem 3-6b
We define $g(n) = 2^n$. For notational convenience, we write $g^{-1}(n) = \log n$ and $g^{-i}(n) = (g^{-1} \circ \cdots \circ g^{-1})(n)$.
For each nonnegative $n$, there exists a unique integer $i$ such that
We can find the value of $i$ by computing $g^{(k)}(1)$ for each $k \geq 1$ until we reach the desired interval.
Since $g$ and $g^{-1}$ are monotonically increasing, we see that
and that
It follows that $f_c^*(n) = i$. $\square$
Problem 3-6c
$f^{(i)}(n) = n/2^i$, and so $f_c^*(n) = \lceil \log n \rceil$. $\square$
Problem 3-6d
$f^{(i)}(n) = n^{1/2^i}$, and so $f_c^*(n) = \lceil \log n \rceil - 1$. $\square$
Problem 3-6e
$f^{(i)}(n) = n^{1/2^i}$, and so $f_c^*(n) = \lceil \log \log n \rceil$ whenever $n \geq 2$. We have $f_c^*(1) = 0$. $\square$
Problem 3-6f
Observe that $f^{(i)}(n) = n^{1/2^i}$. Since we deal exclusively with positive integers, $n^{1/2^i}$ does converge to 1. Nevertheless, the square-root function is strictly increasing, and so $n > 1$ implies $\sqrt{n} > \sqrt{1} = 1$. By induction, $f^{(i)}(n) > 1$ whenever $n > 1$. It follows that
Problem 3-6g
$f^{(i)}(n) = n^{1/3^i}$, and so $f_c^*(n) = \lceil \log_3 \log n \rceil$ whenever $n \geq 8$, where $\log_3$ is logarithm base 3. We have $f_c^*(n) = 0$ for $1 \leq n \leq 7$. $\square$
Problem 3-6h
$% $ whenever $n > 2$, and so
Stirling’s approximation and Euler–Maclaurin summation
In CLRS (and in many elementary probability/statistics textbooks), Stirling’s approximation is given as
A more precise asymptotic expansion can be calculated through the Euler–Mclaurin summation formula:
Euler–Maclaurin summation formula (Sedgewick/Flajolet 4.3). Let $f(x)$ be a function defined on the interval $[1,\infty)$ and suppose that the derivatives $f^{(i)}(x)$ exist and are absolutely integrable for $1 \leq i \leq 2m$, where $m$ is a fixed constant. Then
where
is a constant dependent on the function $f$, $B_{2k}$ is the $2k$th Bernoulli number, given as the $2k$th coefficient of the Taylor series
and $R_{2m}$ is a remainder term satisfying the asymptotic estimate
For now, we take for granted that
By the formula, we have that
It now follows that
General form of the Euler–Maclaurin summation formula
To generalize the discrete Euler–Mclaurin summation formula, we introduce the Bernoulli polynomials
which constitute the coefficients of the Taylor series
Differentiating the Taylor series term-by-term, we see that
for each $m > 1$ and every $x \in \mathbb{R}$.
We now observe that integrating a function by parts with Bernoulli polynomial $B_1(x) = x-1/2$ yields
Setting $g(x) = f(x+k)$, we obtain the identity
We sum on $[a,b]$ to obtain
We have thus obtained the following relationship between a sum and the corresponding integral:
We can furnish a bound on $(\ast)$ by recursively integrating by parts with Bernoulli polynomials. Since $B_{i+1}'(x) = (i+1) B_{i}(x)$, we have that
Dividing through by $(i+1)!$, we obtain a recurrence relation:
Computing recursively, we obtain:
Since $B_1(x) = x - \frac{1}{2}$ and $B_i(0) = B_i(1) = B_i$ for $i > 1$ with $B_i = 0$ for odd $i > 1$, we see that
We now substitute $g(x) = f(x+k)$ and sum over $[a,b]$ to obtain
where
The remainder satisfies the estimate
Linear probing and the Ramanujan $Q$-function
An important tool in the theory of hashing (CLRS ch.11 & TAOCP vol.3, sec.6.4) is linear probing, which searches for a table with $M$ cells, $N$ of which are occupied, for a given key $K$, inserting the key into the table whenever $K$ is not in the table and $M > N$.
A famous theorem of Knuth (TAOCP vol.3, sec.6.4, Theorem K) gives an asymptotic analysis of the average number of cells linear probing examines, in terms of the Ramanujan $Q$-function
For a successful search—i.e., $K$ is inserted—the average number is
For an unsuccessful search—i.e., $K$ is not inserted—the average number is
Since
we see that
Setting $\alpha = N/M$, we obtain
which is a useful estimate for small values of $\alpha$.
At the opposite extreme, we see that $Q_1(M,M-1) = M$, and that
which yields the following estimate (Sedgewick/Flajolet Theorem 4.8):
Knuth (TAOCP vol.1, sec.1.2.11.3) gives a refined estimate:
Ramanujan distributions
Recall from the previous section that the $Q$-distribution is
The term-by-term asymptotic estimate (Sedgewick/Flajolet, Theorem 4.4) is as follows:
Closely related to the $Q$-distribution is the $R$-distribution
which is also asymptotically equivalent to
as $M \to \infty$. Knuth’s estimate (TAOCP vol.1, sec.1.2.11.3) is
The term-by-term asymptotic estimate of the $R$-distribution is
as well.
Many interesting probability distributions can be represented as products of the $Q$-distribution and the $R$-distribution. For example, each even term of the binomial distribution
can be rewritten as follows (Sedgewick/Flajolet, Theorem 4.6):
Stirling’s formula yields
whence we see that | 2018-11-16T00:01:24 | {
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https://math.stackexchange.com/questions/1894564/under-what-circumstances-can-one-divide-by-a-variable | Under what circumstances can one divide by a variable?
In light of recent responses to my other questions, I would like to know when it is mathematically acceptable to undergo division by a variable or a function of a variable, i.e., $x$ or $\cos x$. From what I sort of understand, it is only acceptable to divide by $x$ when $x$ is known not to be $0$. For all other cases, division cannot be undergone. In addition, I believe my Precalculus teacher said diving by $\cos x$ also should not occur because $\cos x$ can equal $0$ at $x=\frac{(2n-1)\pi}2$ where $n\in \mathbb{Z}$. How can I avoid dividing by $0$? Under what other circumstances could I accidentally divide by $0$?
• So, if you know that $x \neq (2n-1)\frac{\pi}{2}$ then you can safely divide by $\cos x$, etc... if you know that $x \neq 1$ then you can safely divide by $\ln x$, and so on. – Zain Patel Aug 17 '16 at 0:38
• See this previous question: Why should you never divide both sides by a variable when solving an equation? In short, you have to consider both cases separately: (a) suppose $x\ne0$, then you can divide by it and proceed; (b) suppose $x=0$, then see what happens to the original problem. – Rahul Aug 17 '16 at 0:41
• I'm sorry I jumped on you. I miss understood what you were trying to say. If you get, say, $x^2 + 3x = x (x - 6)$ and you want to divide by x, you can **IF** you specify you are doing a case where $x \ne 0$ AND when you have finished that case you address and show what happens when you assume x = 0. – fleablood Aug 17 '16 at 5:31
• every time you divide by an unknown you should note "this assumes so and so isn't 0. I will discuss what happens if so and so does equal 0 later." And then do explain later. Or if you know so and so can not be 0 you can say "we know so and so can not be 0 because... therefore we can divide by it" – fleablood Aug 17 '16 at 5:37
• In a way this is what $xy = 0$ so $x=0$ OR $y= 0$. You are taking two cases and considering each seperately. – fleablood Aug 17 '16 at 5:39
3 Answers
Let's do two examples:
1) Suppose you are asked to solve:
$x^4 + 5x^2 +4 = 4x^3 + 4x$. And you decide to factor both sides:
$x^4 + 5x^2 + 4 = 4x^3 + 4x$
$(x^2 + 4)(x^2 + 1) = 4x(x^2 + 1)$
And now at this point you realllllly want to divide both sides by $x^2 + 1$. Can you? Well, you can if $x^2 + 1 \ne 0$. So you write (and you must write something to this effect):
"As $x^2 \ge 0$ we know $x^2 + 1 \ge 1 >0$ and in particular, we know $x^2 + 1 \ne 0$. So we may safely divide by $x^2 + 1$."
And then you go on with:
$x^2 + 4 = 4x$ ... and so on....
2) Okay... second example. Suppose you are asked to solve:
$x^4 -3x^2 -4 = 4x^3 - 4x$. And you decide to factor both sides:
$x^4 -3x^2 - 4 = 4x^3 - 4x$
$(x^2 + 4)(x^2 - 1) = 4x(x^2 - 1)$
And now at this point you realllllly want to divide both sides by $x^2 - 1$. Can you? Well, you can if $x^2 - 1 \ne 0$.
But $x^2 -1$ might equal $0$!!!! So what do you do? You break it into cases and write something to the effect:
"$(x^2 + 4)(x^2 - 1) = 4x(x^2 - 1)$
"If we assume $x^2 - 1 \ne 0$ we may divide both sides by $x^2 - 1$
"So we will assume that so we'll do Case 1:
"Case 1: $x^2 - 1 \ne 0$
"then
"$x^2 + 4 = 4x$
$x^2 - 4x + 4 = 0$
$(x - 2)^2 = 0$
$x -2 = 0$
$x = 2$
"But we have the restriction that $x^2 - 1$ can not be $0$ so we must test that this is not the case: If $x = 2$ then $2^2 - 1 = 3 \ne 0$ so this is okay
"So we conclude that if $x^2 - 1 \ne 0$ then $x = 2$.
"But we must also consider that $x^2 - 1$ might equal 0, so we do Case 2:
"Case 2: $x^2 - 1 =0$
"then $x^2 = 1$ so $x = \pm 1$.
"Our conclusion of the two cases is either $x = 2$ or $x = -1$ or $x = 1$".
====
The above is for more verbose than any actual example need to be. It be sufficient to do something like this:
Question Solve for $x$: $x^4 + 5x^2 + 4 = 4x^3 + 4$
Answer:
$x^4 + 5x^2 + 4 = 4x^3 + 4$
$(x^2 + 4)(x^2 + 1) = 4x(x^2 + 1)$
$(x^2+4)\frac{x^2 + 1}{x^2 + 1}=4x\frac{x^2 + 1}{x^2 + 1}$; [$x^2 \ge 1$ so $x^2 + 1 \ne 0$]
$x^2 + 4 = 4x$
$x^2 - 4x + 4 = 0$
$x = \frac {2 \pm \sqrt{2^2 - 4*1}}{2*1} = 2$.
~~~~~~~~~
Or you could do:
$x^4 + 5x^2 + 4 = 4x^3 + 4$
$(x^2 + 4)(x^2 + 1) = 4x(x^2 + 1)$
Case 1: $x^2 + 1 \ne 0$
$(x^2+4)\frac{x^2 + 1}{x^2 + 1}=4x\frac{x^2 + 1}{x^2 + 1}$
$x^2 + 4 = 4x$
$x^2 - 4x + 4 = 0$
Case 2: $x^2 + 1 = 0$.
$x^2 + 1 = 0$
$x^2 = -1$
Impossible.
$x = 2$.
`
What is not acceptable is:
$x^4 + 5x^2 + 4 = 4x^3 + 4$
$(x^2 + 4)(x^2 + 1) = 4x(x^2 + 1)$
$(x^2+4)\frac{x^2 + 1}{x^2 + 1}=4x\frac{x^2 + 1}{x^2 + 1}$
$x^2 + 4 = 4x$
$x^2 - 4x + 4 = 0$
$x = \frac {2 \pm \sqrt{2^2 - 4*1}}{2*1} = 2$.
• I upvoted this because it covers the essential cases as I see them. I would like the answer even better if it reached the same conclusions in fewer words than its first version. – David K Aug 17 '16 at 19:39
• I obviously don't expect the student to be as verbose and pendantic as this. But I wanted it to be absolutely clear that when I do divide by an unknown (and every time) that we must either determine that unknown can not be 0, or that if it can, we are doing a specific case under stringent conditions and that the case of other conditions must be addressed. – fleablood Aug 17 '16 at 21:08
• The one case I didn't cover is when after you divide by the unknown you determine that the unknown must be zero. So we were incorrect in dividing at all. Example $\cos(2x)*(\sin(x) - 1/2) = (sin(x) - 1/2)*1/2$ so $\cos(2x) = 1/2$ so $2x = 60|120$ so $x = 30|60$ but if $x = 30$ $\sin(x) -1/2 = 0$ which we assumed wasn't the case. In a way that isn't really a problem as our second case was going to be $\sin(x) -1/2 =0$ so even if we dropped the ball in case 1: it would be a result in case two anyway. – fleablood Aug 17 '16 at 21:16
• I didn't mean to suggest removing any of the points made by the answer. The answer is not far from its ideal length. In fact, now that you've made clear that it is written to make clear the things to think about, not necessarily as a model for writing up a solution to a problem, it seems fine. – David K Aug 18 '16 at 16:55
First things first: I think you meant $n \in \mathbb{Z}$ instead of $n \in \mathbb{R}$ for $x= \frac{(2n-1)\pi}{2}$. $\mathbb{R}$ stands for real numbers. This are numbers like 1 and 2, but also $\frac{2}{3}$ and $\pi$. $\mathbb{Z}$ stands for integers. So $1 \in \mathbb{Z}$ and $1 \in \mathbb{R}$, but $\pi \notin \mathbb{Z}$ while $\pi \in \mathbb{R}$.
Now for the real question.
$\cos$ is essentially a function, just like $f(x)$.
Say you want to divide by a function $f(x)$ (where $f(x)$ can be any function). You can do this for any value of $x$ for which $f(x) \neq 0$ (as far as I know at least for $x \in \mathbb{R}$).
Say for example $f(x) = x - 5$. You can divide by $f(x)$ if $x \neq 5$.
The same holds for $\cos$. So let's say we have $\cos(x)$ (or $f(x) = \cos(x)$ if you like). $\cos(x) = 0$ if, as you pointed out, $x=\frac{(2n-1)\pi}{2}$ with $n \in \mathbb{Z}$. Since we want to devide by $\cos(x)$, we must be sure that $x \neq \frac{(2n-1)\pi}{2}$.
More general, if we want to devide by $f(g(x))$, we calculate $f(a) = 0$ and set this $a$ equal to $g(x)$ (so we calculate $g(x) = a$). The resulting value(s) of $x$ is for which we can't devide by $f(g(x))$.
As a concrete example, say we have $$g(x) = x^2 + 6x$$ and $$f(g(x)) = g(x) + 5$$ and we want to determine for which values of $x$ we can calculate $$h(x) = \frac{1}{f(g(x))}$$
First, we set $g(x) = a$, so $f(g(x))$ becomes $$f(g(x)) = f(a) = a + 5$$ Setting $f(a)$ equal to zero we get $a = -5$. Now we set $g(x)$ equal to $a = -5$ and solve for $x$:
$$g(x) = a$$ $$x^2 + 6x = -5$$ $$x^2 + 6x + 5 = 0$$ $$(x+1)(x+5) = 0$$ $$x = -1 \text{ or } x= -5$$
So we can calculate the value of $h(x)$ except for $x = -1$ or $x = -5$, where it is undefined.
Hope it helped you a bit. Please not that I'm not a professional mathematician, so please ask someone else too. The last thing I want is you getting a bad grade because of my answer (I'm not responsible for that. $:)$). If someone from the community notices something wrong or wants to add something, please edit this question. I would be thankful.
• Thank you for the correction! – Davis Rash Aug 17 '16 at 20:16
• @DavisRash No problem. :) Did this answer help you? – Kevin Aug 17 '16 at 21:01
I think the most thorough way to think about is this: under what circumstances can you multiply by the inverse of $x$?
This also works in cases when division isn't defined, but inverses exist (more advanced mathematical contexts like groups and matrices). | 2019-05-22T19:42:55 | {
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http://everwoodbiocostruzioni.it/ewic/moment-of-inertia-of-t-section-problems.html | ## Moment Of Inertia Of T Section Problems
Then equations (4. Since the section modulus depends on the value of the moment of inertia, an efficient beam must have most of its material located as far from. Centroid and Moment of Inertia/ Centre of Gravity/ Problems. Introduction: Problem 1 on Centroid & Moment of Inertia. I've already calculated the centroid with respect to y to be 60 mm. A cantilever beam A3, loaded by a uniform load and a concentrated load (sec figure), is constructed of a channel section. If you want to know the moment of inertia of a complex shaped body about a given axis you simply look up its radius of gyration, and then (knowing its mass) apply the above formula to find the moment of inertia. course of the series and is called What Every Engineer Should Know About Structures - Part D - Bending Strength of Materials. Determine the distance 'd' between the sections such that the centroidal moment of inertia about the x- and y-axis are equal. If the mass of the wheel is 1. Section modulus helps in determining the strength of the section. Find the polar moment of inertia and the polar radius of gyration with respect to a z axis passing through one of the outside corners. For internal equilibrium to be maintained, the bending moment will be equal to the ∑M from the normal stresses × the areas × the moment arms. The moment of inertia of the platform is 5 kgm 2. Problem 817 Determine the moment of inertia and radius of gyration with respect to a polar centroidal axis of the cross section of a hollow tube whose outside diameter is 6 in. Moment Of Inertia Of A Rectangular Plate Derivation 1. It is required in the design of machines, bridges, and other engineering systems. This tool calculates the moment of inertia I (second moment of area) of an I/H section (also called W-beam or double-T). The parallel axis theorem can be used to determine the moment of inertia of a rigid body around any axis. Problem 819 Determine the moment of inertia of the T-section shown in Fig. The SI unit for polar moment of inertia, like the area moment of inertia, is metre to the fourth power (m 4) Application. Moment of Inertia About Y-axis. Then this moment of inertia is transferred about the axis passing through the centroid of the given section, using theorem of parallel axis. To start viewing messages, select the forum that you want to visit from the selection below. At the heart of these comparisons lie the concepts of mass on one hand and moment of inertia on the other. Angular momentum. 3 106mm4 Sample Problem 9. For that reason, it is preferred the moment of inertia about the x- and y-axis of a column section to be roughly equal. Analyzing Rolling Motion. Masses further away from the rotational axis have the highest moment of inertia. Moments Of Inertia Bible Studies By Steve PPT. This is a standard result. CE 405: Design of Steel Structures - Prof. Many handbooks list the moment of inertia of common shapes (see Sections appendix). So based on that I have solved problems on composite sections. It may still, however, be useful for other purposes as well. n = perpendicular distance between the line and the area of cross section. It may be +ve, -ve, or zero • Product of Inertia of area A w. I G is the "mass moment of inertia" for a body about an axis passing through the body's mass center, G. Also, where does it want the moment of inertia taken about? What you've suggested works if it wants it about the axis of symmetry, but if it wants the moment of inertia about a different axis (such as the bottom of the cross section), then you would need to use the parallel axis theorem to get the right answer. The moment of inertia of a tee section can be found if the total area is divided into two, smaller ones, A, B, as shown in figure below. x = any axis parallel to the centroidal axis. (yield moment is when steel is starting. The separation points on that area moments. Calculate Yield Moment for the Beam Assume concrete accepts no tension. Where, I = second moment of inertia about the line k. Where: J M = Polar Mass Moment of Inertia (in-lbs-sec 2, Kg-m-sec 2). This means that the portion of liquid which moves together with the tank becomes small as rectangle goes square. The bending moment M applied to a cross-section is related with its moment of inertia with the following equation: M = E\times I \times \kappa where E is the Young's modulus, a property of the material, and κ the curvature of the beam due to the applied load. Get 1:1 help now from expert. The moment of inertia, otherwise known as the mass moment of inertia, angular mass or rotational inertia, of a rigid body is a quantity that determines the torque needed for a desired angular acceleration about a rotational axis; similar to how mass determines the force needed for a desired acceleration. The results from Section #2 will determine what ø (reduction factor) you should use. •Compute the product of inertia with respect to the xyaxes by dividing the section into three rectangles. CIVL 4135 81 Transformed Section 4. I ¯ = centroidal moment of inertia. What is the safety factor relative to the yield strength? 3. In the preceding section, we defined the moment of inertia but did not show how to calculate it. It will help in deciding whether the failure will be on the compression face or on the tension face of the beam. For part a) of this problem, the moment of inertia is about the x-axis. Each leg is comprised of a 2 x 10. Moment of Inertia 5 An example of this is the concrete T-beam shown. Stating Moment of Inertia of a infinitesimally thin Disk. I understand the basic concept of Moment of Inertia in general, and am aware of how you calculate this concept for a variety of. Answer: The moment of inertia of a rod of mass and length about an axis, perpendicular to its length, which passes through its midpoint is. Then equations (4. W - T = m*a W-T = m*T*L^2/(16*I) The Attempt at a Solution Well, in my attempting of solving the problem, I considered that the moment of inertia utilized in these equations is equal to m*(L^(2)/3). Look up or calculate the moment of inertia and section modulus (1 & S). Moment of inertia is the measure of an object's resistance to changes in its rotation rate. Determine the moment of inertia of the area about the y axis. To see this, let's take a simple example of two masses at. However, the moment of inertia I looked up in a physics textbook is exactly two times this (the factor is $1/2,$ not $1/4$). We know that the moment of inertia for hoop with radius R is mR2. A cantilever beam A3, loaded by a uniform load and a concentrated load (sec figure), is constructed of a channel section. Constant angular momentum when no net torque. is the moment of inertia of the disk, and ω is the angular speed. Try to keep them all rectangular cubes if you can. Using the parallel axis theorem, the moment of inertia about a parallel axis passing through one of the ends of the rod is. 9, Figure 1. However, it represents the bending stiffness of the structural member, i. Manas Patnaik 127,644 views. 5 2 3 A 4-0. To find the polar moment of inertia about the AA axis, integrate the 2nd moment from r to R. Determine the moment of inertia and radius of gyration with respect to an axis which is parallel to the plate and. Rotational kinetic energy. 9, Figure 1. This is why it behaves far more rigidly! Of course, the moment of inertia influences far more than deflection during bending. as far as i can tell MASSPROP command can give me the moment of inertia in units LENGHT^4. Determine the moment of inertia of the T-section shown in Fig. Torque and Rotational Inertia 2 Torque Torque is the rotational equivalence of force. dA = the area of the elementary portion. 7(4): 193-197 (Apr. In general, moment of inertia is just a measure of how hard it is to get something rotating. – The cylinder is cut into infinitesimally thin rings centered at the middle. Dt i th t fi ti d composite section centroidal axis. Estimate the moment of inertia of a die along an axis that passes through one of the die's edges in g c m 2 g~cm^2 g c m 2. 12 Moment of Inertia With Respect to an Arbitrary Axis Ellipsoid of. Answer: The moment of inertia of a rod of mass and length about an axis, perpendicular to its length, which passes through its midpoint is. Second moment of inertia for the rectangular cross section is b × h 3 / 12. Sample Problem A Of inertia triangular itg parallel its Vertex. Collection of Solved Problems in Physics. When doing dynamics problems with moments of inertia, you should not use the formulas you remember for second moment of area instead. Free Moment of Inertia Calculator. The higher the MOI of an object, the more force will have to be applied to set that object in a rotational motion. org,Dwg Viewer, inventor dwg, dwg to pdf, dwg to tif, batch print, batch convert, attribute export, hpgl viewer, batch plot,Helping us make and share calculations with MS Excel. First, we set up the problem. Aug 13, 2016 - Explore ekeeda_'s board "Moment of Inertia" on Pinterest. If we compare Figure to the way we wrote kinetic energy in Work and Kinetic Energy, $(\frac{1}{2}m{v}^{2})$, this suggests we have a new rotational variable to add to our list of our relations between rotational and translational variables. J = Polar moment of inertia about the axis of rotation r = Distance from neutral axis to the outer most fibre = D/2. My dA in this case is going to be this differential area that goes around my cross section at a distance rho from the center, and it's going to have a thickness or width of D rho. In general, moment of inertia is just a measure of how hard it is to get something rotating. I need to calculate the change in moment of intertia due to modifing a simple angled beam from 120 x 120 x 10 to 120 x 112 x 10. The moment of inertia is to be found about the center of the rod. where , are the coordinates of a point on the cross section at which the stress is to be determined as shown to the right, and are the bending moments about the y and z centroid axes, and are the second moments of area (distinct from moments of inertia) about the y and z axes, and is the product of moments of area. Answer to: For the cross-section shown, determine the moment of inertia where b = 6 inches, h = 10 inches, and t = 0. The mass moment of inertia of a body about a specific axis can be defined using the radius of gyration (k). Moment of Inertia Formula for Plane Figures - Moment of. Line Passing Through The Base. For example: The Second Rectangular Area Moment of Inertia of a beam's cross section represents that beam's resistance to bending. I m going to have another half hour. Note For Problems 9. 6-1 through 9. 2 Find the moment of inertia for the rectangular section shown in fig. Conversely, the moment of inertia about a base diameter axis can be computed by adding one1-. Properties of Sections. The moment of inertia for axis Z'Z' is given by IZ'Z. But in mechanics, moment of inertia is used in the calculation of bending of a bar, torsion of a shaft and determination of the stresses in any cross section of a machine element or an engineering structure. The negative sign indicates that a positive moment will result in a compressive. You will find the graphics in Figure 10. The radius of gyration can be useful for listing in a table. The method has been developed to evaluate the variable mass moment of inertia of a 12-cylinder V-engine having a piston–crank mechanism with main and auxiliary connecting rods. Simply Supported Beam|calculate support reaction|Engineering. Dear Friends, this session helps you to to find the centroid of composite areas, especially a T Section ,understand from basics, step by step tutorial on Moment of Inertia. To see this, let's take a simple example of two masses at the end of a massless (negligibly small mass. LuxCalc MOI Mobile allows the user to accurately calculate all major cross-sectional properties such as moment of inertia, area, radius of gyration, etc. The importance of this fact is that this is when moment of inertia is at its maximum and minimum values We will verify these ascertions within the following pages 5. 015 m, and total mass M = 3. Area Moments of Inertia Products of Inertia: for problems involving unsymmetrical cross-sections and in calculation of MI about rotated axes. T T φ Z Fig. Knowing the area moment of inertia is a critical part of being able to calculate stress on a beam. The moment of inertia plays the same role for rotational motion as the mass does for translational motion (a high-mass body resists is hard to start moving and hard to stop again). 27 m from the center of mass. Hello, I'm trying to find out how to calculate the I (Area Moment of Inertia. Conversely, the moment of inertia about a base diameter axis can be computed by adding one1-. The polar second moment of area provides insight into a beam's resistance to torsional deflection, due to an applied moment parallel to its cross-section, as a function of its shape. The moment of inertia of an object rotating about a particular axis is somewhat analogous to the ordinary mass of the object. The radius of gyration can be useful for listing in a table. It is constant for a particular rigid frame and a specific axis of rotation. Dear Friends, this session helps you to to find the centroid of composite areas, especially a T Section ,understand from basics, step by step tutorial on Moment of Inertia. Centroid, Area, Moments of Inertia, Polar Moments of Inertia, & Radius of Gyration of a Complex Cross-Sectional Area. See next page for section properties needed in these problems. Types of Cross-section Polar Moment of Inertia. Angular momentum of an extended object. 0 SECOND MOMENT OR MOMENT OF INERTIA OF AN AREA 8. You may have to register before you can post: click the register link above to proceed. n = perpendicular distance between the line and the area of cross section. I = I ¯ + A d 2. The moment of inertia (second moment or area) is used in beam theory to describe the rigidity of a beam against flexure (see beam bending theory). concrete slab supported by continuous T beams of 24 ft span, 47 in. Mahesh Gadwantikar. Knowing the area moment of inertia is a critical part of being able to calculate stress on a beam. 9550 m, thickness t = 0. Lecture Notes: Area-Moment. Rolling without slipping problems. Thisgivesusevidenceofthereliancethatthemomentof inertiahasonmassandhowitisdistributed. Any measured values you retrieve from the Inertia object after setting the density will reflect the new value. 17 Centroid and Moment of Inertia Calculations An Example ! If we sum the second column, we have the bottom term in the division, the total area 1 1 n ii i n i i xA x A = = = ∑ ∑ ID Area x ix*Area (in2)(in) 3 A 1 2 0. Integrating over the length of the cylinder. the relationship τ = Iα, the moment of inertia can be determined. Compute the Moment of Inertia for a Solid Cuboid (I h) Compute the Moment of Inertia for a Solid Cuboid (I w). previous home next PDF 24. So based on that I have solved problems on composite sections. the end the the a strip b is of t, he Thus. According to Newton's first law of motion "A body maintains the current state of motion unless acted upon some external force". Inventor seems to handle. This ratio of moment of inertia was also obtained by Graham and Rodriguez (1952) and cited in Roberts et al. The moment of inertia, otherwise known as the mass moment of inertia, angular mass or rotational inertia, of a rigid body is a quantity that determines the torque needed for a desired angular acceleration about a rotational axis; similar to how mass determines the force needed for a desired acceleration. The formula for the effective moment of inertia Ie is: (2) where Mer is the cracking moment, Ma is the maximum span moment, Ig is the moment of inertia of the gross concrete section ignoring reinforcement, and Ic is the moment of inertia ofthe cracked transformed section. 2: String wrapped around axle. y 2 – 2 x 3 *10–12. This video contain solution of moment of inertia of T-section about xx and yy axis passing through c. Then we need to find whether the top or the bottom of the section is furthest from the neutral axis. Sample Problem A Of inertia triangular itg parallel its Vertex. and radius. This is the sideway to the treasure of web. Lecture Notes: Area-Moment. 5 mm rectangle and measure the area moment of inertia it will be 0 for one of the pricipal axes. moment of inertia about x axis and y axis for t section section of channel section is the most important Concept for civil engineering students when we have to solve the problem of bending. moment of inertia with respect to x, Ix I x Ab 2 7. The moment of inertia of a tee section can be found if the total area is divided into two, smaller ones, A, B, as shown in figure below. The beam is a channel section with dimensions as shown in the figure. Problems practice. Where, I g is the moment of inertia of gross section, y t is the distance from the neutral axis to the extreme tension fiber and f t is the tensile strength of concre te. Angular momentum. Moment of inertia can be defined by the equation The moment of inertia is the sum of the masses of the particles making up the object multiplied by their respective distances squared from the axis of rotation. The Inertia object has a density property that is read/write. Because all rotational motions have an axis of rotation, a torque must be defined about a rotational axis. We have chosen to split this section into 3 rectangular segments:. Overall people refer to the moment of inertia as to rigidity. The moment of inertia depends on the mass and shape of an object, and the axis around which it rotates. equivalent moment of inertia must be used in order calculate the natural frequency. The moment of inertia calculates the rotational inertia of an object rotating around a given axis. The parallel axis theorem can be used to determine the moment of inertia of a rigid body around any axis. Sample Problem 9. theorem the. Large rectangle 150 mm 420 mm 520 mm 60 mm 60 mm 60 mm 150 mm + 150 mm = 300 mm 200 mm + 200 mm = 400 mm x 400 mm y Small rectangle 300. The polar moment of inertia on the other hand, is a measure of the resistance of a cross section to torsion with invariant cross section and no significant warping. (8), derived in the moment of inertia example, the moment of inertia of the disk is = at 5 digits Therefore, the moment of inertia of the disk is 12. Moment of Inertia. With a 1x1x1 block I would expect a value of 0. circular area with respect. It represents how difficult it overcomed to change its angular motion about that axis. calculate its moment of inertia about any axis through its centre. 12 useful in visualizing. If the object is made of a number of parts, each of whose moment of inertia is known, the total moment of inertia is the sum of the moments of inertia of the pieces. Inventor has a function for moments but it rotates the center plane to something like VxV in the link below. Problem on Moment of Inertia of T Section - Moment of Inertia - Strength of Materials - Duration: 17:24. Torsional stress: where T is the torque, r is the radius and J x is the polar moment of area. Wallace Torque or Torsional Moment: Solid Circular or Tubular Cross Section: r = Distance from shaft axis to point of interest R = Shaft Radius D = Shaft Diameter J D R J D D for solid circular shafts for hollow shafts o i = ⋅ = ⋅ = ⋅ − π π π 4 4 4 4 32 2 32 e j Torque z x y T "Cut Surface" τ τ = T. Then by using parallel axis theorem, the moment of inertia of the given section about its centroid is obtained. A higher moment of inertia is an indication that you need to apply more force if you want to cause the object to rotate. It takes as many days for a unit of bone to be formed as it did for it to be resorbed. The basic set-up is always the Polar Moment of Inertia Read the section on polar moments on Page 28-3 in your notebook. So the horizontal axis. In these equations, the parameters, affect on the value of equivalent moment of inertia, are the length of steps and the dimensions of cross section area of the. 100 kg, and radius 20. Derive an expression for the force on the other support immediately thereafter. W - T = m*a W-T = m*T*L^2/(16*I) The Attempt at a Solution Well, in my attempting of solving the problem, I considered that the moment of inertia utilized in these equations is equal to m*(L^(2)/3). check answer. t W=ΔK ΔK=1 2 Iω2−1 2 Iω 0 2 W=0−1 2 Iω 0 2 P= 1 2 Iω 0 2 t = 2 2t 3. 95% of the mass, so a classical picture of two point masses a fixed distance apart gives In the nineteenth century, the mystery was that equipartition of energy, which gave an excellent. : L-section) The product of inertia can only be zero about the principle axis. Problem 817 Determine the moment of inertia and radius of gyration with respect to a polar centroidal axis of the cross section of a hollow tube whose outside diameter is 6 in. The object in the diagram below consists of five thin cylinders arranged in a circle. c) The moment of inertia of the bar will be higher about the axis at the end of the bar than about the axis through the centre. The following are the mathematical equations to calculate the Polar Moment of Inertia: J z: equ. First, we set up the problem. The bending moment M applied to a cross-section is related with its moment of inertia with the following equation: M = E\times I \times \kappa where E is the Young's modulus, a property of the material, and κ the curvature of the beam due to the applied load. April 22, 2016 April 22, 2016 admin Cross Section Area, Elastic Modulus, Moment of inertia, Moment of Inertia Calculator, Radius of gyration, Section Calculator, Section Modulus, T Section W T C L Rectangle HRectangle HSquare Square Pipe Circle SemiCircle Triangle. 10-6 A wide-flange beam (see figure) is subjected to a shear force V. determine moments of inertia of beam section and plate with respect to The strength of a W14x38 rolled steel beam is increased by attaching a plate to its upper flange. If the reference axis will be used to calculate moment of inertia of a complex shape, choose an axis of symmetry to simplify the calculation. So here's our formula. Inventor has a function for moments but it rotates the center plane to something like VxV in the link below. While you can derive the moment of inertia for any object by summing point masses, there are many standard formulas. Presentation Summary : Moments of Inertia Lesson 7. 5 • The moment of inertia of the shaded area is obtained by subtracting the moment of inertia of the half-circle from the moment of inertia of the rectangle. Compute this as illustrated in Example 9. Then this moment of inertia is transferred about the axis passing through the centroid of the given section, using theorem of parallel axis. Problem on Moment of Inertia of T Section - Moment of Inertia - Strength of Materials - Duration: 17:24. It will help in deciding whether the failure will be on the compression face or on the tension face of the beam. Rolling without slipping problems. Moment of Inertia calculation of T-section with basic concepts. • Moment of inertia I T of a. 3 Design of Key. American Concrete Institute, "ACI 318", 2005 The effective moment of inertia is given by ACI 318 9. The equations that they give for the rotational component give the shear STRESS in the bolts. of rectangle ① is at a distance of 20 cm from y-axis and at a distance of 25 cm from y-axis. 2 Second Moment of Area11. Where: J M = Polar Mass Moment of Inertia (in-lbs-sec 2, Kg-m-sec 2). - The ratio of Mp to My is called as the shape factor f for the section. It is determined from the cross-sectional area of the beam and the central axis for the direction of interest. Moment of inertia is a measure of how mass is distributed over a body about an axis of rotation. Dear Friends, this session helps you to to find the centroid of composite areas, especially a T Section ,understand from basics, step by step tutorial on Moment of Inertia. Centroid, Area, Moments of Inertia, Polar Moments of Inertia, & Radius of Gyration of a General T-Beam. The flanges are assumed equal. This is ONLY true in the case of circular cross-sections (hollow or not hollow). Try to keep them all rectangular cubes if you can. It is the measurement of the resistance of a body to a change in its rotational motion. The Mass Moment of Inertia. I don't have a thorough grasp of where the equations for this come from. In math and physics, moment of inertia is strictly the second moment of mass with respect to distance from an axis: I = ∫ Q r 2 d m {\displaystyle I=\textstyle \int _ {Q}r^ {2. The ability to resist bending = I/y. 1) Today s Objectives: Students will be able to Determine the mass moment of inertia of a rigid body or a system of rigid bodies. Moment of Inertia Formula for Plane Figures - Moment of. It's trivial to find: the nuclei (protons) have 99. Polar moment of inertia is the moment of inertia about about the z-axis. Moments Of Inertia Bible Studies By Steve PPT. The larger the Polar Moment of Inertia the less the beam will twist. This means that when you retrieve an Inertia object you can check the density that is applied and if that value is not correct, you can set it to whatever you want. 10 cm and a mass of 0. Perfect for acing essays, tests, and quizzes, as well as for writing lesson plans. For point masses, this can be expressed mathematically as I = Ʃmr2. The idea of point equivalent moment of inertia is concentrated on equation (2) and equation (3). Calculate the maximum tensile stress o f and maximum compressive stress tr t due to the uniform load. asked by Ritweek on August 22, 2018; physics. Moment of Inertia: It is a measure of an object’s resistance to changes its state of rotation. 11 Use the parallel axis theorem to find the moment of inertia of regular objects about axes other than the axis of symmetry The parallel-axis theorem states I I CM Md 2. You only need mass moment of inertia around Y-axis, but to calculate it you would need to know the exact distribution of every mass component of your ship, longitudinally and verticaly because formula is I=m*r^2, where r is the distance of every element form center of gravity of your ship. In the preceding section, we defined the moment of inertia but did not show how to calculate it. Big, heavy objects will have a high moment of inertia and be hard to turn; smaller, awkwardly shaped objects can also have high moments of inertia and can be just as hard to turn. The bending moment M applied to a cross-section is related with its moment of inertia with the following equation:. Sample Learning Goals. Centroid and Moment of Inertia/ Centre of Gravity/ Problems. Moment of inertia represents the body's tendency to resist rotational motion. Hi, I have come across a problem dealing with calculation of the Moment of Inertia of an extended body and I just don't understand why the answer is what it is. Problems practice. The Second Moment of Area I is needed for calculating bending stress. The tension T is supplied by a hanging mass and found using Newton’s second law. Figure 4: Area Moment of Inertia Calculation Section, SW CG Next one down in Figure 3 above is the moments of inertia matrix reported at the centroid. I have to made deductions for holes and moment of inertia of the section decreases. The polar area moment of inertia, denoted by J O, is the area moment of inertia about the z-axis given by Note that since one has the relation The radius of gyration is the distance k away from the axis that all the area can be concentrated to result in the same moment of inertia. This is the moment of inertia of a bar that rotates around a point of its end. The material has a mass per unit area of 20 kg/m 2. When doing section properties one of the calculcated properties is the Polar moment of inertia of the section area, at the centroid. The correct answer is d. Answer to: For the cross-section shown, determine the moment of inertia where b = 6 inches, h = 10 inches, and t = 0. Calculating the moment of inertia for compound objects. Web dimensions, as determined by negative-moment requirements at the supports,arebw =11in. So my engineering mechanics book includes a brief discussion on area moments of inertia. This is a problem in my textbook could someone please help me work through it. The formula for moment of inertia is the "sum of the product of mass" of each particle. However, it represents the bending stiffness of the structural member, i. The equations that they give for the rotational component give the shear STRESS in the bolts. I just can't get the moment of inertia for the two inclined webs of the steel girder. 0 triangular areas by integration – T section, GE8292- unit 3 -Mass moment of inertia problem 2:. Since there are two bending axes in a two-dimensional space(y and z), we have two values of. Torque, Moment of Inertia, Rotational Kinetic Energy, Pulley, Incline, Angular Acceleration, Physics - Duration: 3:29:44. I have a problem with Inventor pro 2013 rounding of the area moment of inertia to zero (region properties). For example for a rectangular cross section beam with dimensions$$X = 6cm,\space Y =12 cm\space and\space Z =180 cm$$ which could be a log of timber the. Key is designed as discussed earlier. again, the the coordinate axis is welded as the mass center oriented as shown, and we have the XY, IXX and the IYY mass moments inertia are the same, and the IZZ moment of inertia, mass moment of inertia is different. - For a rectangular section, f is equal to 1. This macro might be a start. By signing up,. inertia of beam section and plate with respect to composite section centroidal axis. (3) x is the distance from the y axis to an infinetsimal area dA. J = Polar moment of inertia about the axis of rotation r = Distance from neutral axis to the outer most fibre = D/2. The case of a circular rod under torsion is special because of circular symmetry, which means that it does not warp and it's cross section does not change under torsion. If the angular acceleration of a wheel is 1. This article will discuss the concept of the area moment of inertia and polar moment of inertia and their application in practical problem solving. Lecture Notes: Area-Moment. The moment of inertia is the quantitative measure of rotational inertia, just as in translational motion, and mass is the quantitative measure of linear inertia—that is, the more massive an object is, the more inertia it has, and the greater is its resistance to change in linear velocity. Moment of inertia for a body is defined with the following integral: where is the distance of the mass element from the axis of rotation. Determine the moment of inertia for the beam's cross-sectional area about the y axis. Torque, Moment of Inertia, Rotational Kinetic Energy, Pulley, Incline, Angular Acceleration, Physics - Duration: 3:29:44. Density = Mass per unit volume Density = dm / dV where: þ; - Density dm - Mass of a ring or radius R dV - Volume of a ring or radius R Lets assume height of the cylinder is h. individual part moments of inertia based on the assembly x,y,z axes. 2004) 193 Development of a Flywheel System for the Determination of Moment of Inertia Olugboji O. C y: Area: Moment of Inertia about the x c axis I xc: Polar Moment of Inertia about the z c axis J zc = I xc + I yc Radius of Gyration about the x c axis k xc: Radius of Gyration about the z c axis r zc: r zc 2 = k xc 2 +k yc 2. 100 kg, and radius 20. To find the polar moment of inertia about the AA axis, integrate the 2nd moment from r to R. Moment of Inertia. 8-43 about (a) the vertical axis, and (b) the horizontal axis. Wallace Torque or Torsional Moment: Solid Circular or Tubular Cross Section: r = Distance from shaft axis to point of interest R = Shaft Radius D = Shaft Diameter J D R J D D for solid circular shafts for hollow shafts o i = ⋅ = ⋅ = ⋅ − π π π 4 4 4 4 32 2 32 e j Torque z x y T "Cut Surface" τ τ = T. 5 kg and radius R = 20 cm, mounted on a. Key is designed as discussed earlier. Introduction to Moment of Inertia There are numerous analogies when comparing linear and rotational motion. (3) x is the distance from the y axis to an infinetsimal area dA. The method has been developed to evaluate the variable mass moment of inertia of a 12-cylinder V-engine having a piston–crank mechanism with main and auxiliary connecting rods. For example: The Second Rectangular Area Moment of Inertia of a beam's cross section represents that beam's resistance to bending. 3 Moment of Inertia by Integraion Monday, November 19, 2012 Moment of Inertia ! The moment of inertia is actually the second moment of an area or mass about an axis ! Notice that it is not a distance, it is a moment of a moment ! That may sound strange " It should 4 Moment of Inertia by Integraion Monday, November 19, 2012. 1 Introduction 8. ; The average shear stress r a v e r (obtained by dividing the shear force by the area of the web) and. Moment of inertia of a circular section can be calculated by using either radius or diameter of a circular section around centroidal x-axis or y-axis. Moment of inertia of a circular section is same around both centriodal axis. 4)and the second moment of the area about the y. Homework Statement A grinding wheel is a uniform cylinder with a radius of 8. I hope you find this resource helpful. To find the polar moment of inertia about the AA axis, integrate the 2nd moment from r to R. For point masses, this can be expressed mathematically as I = Ʃmr2. Last Revised: 11/04/2014. A = Area (in 2, mm 2) I = Moment of Inertia (in 4, mm 4) G r = Radius of Gyration = (in, mm) y = Distance of Axis to Extreme Fiber (in, mm) Section Properties Radius of Gyration Cases 1 - 10;. The quantity $\sum _{j}{m}_{j}{r}_{j}^{2}$ is the counterpart for mass in the equation for rotational kinetic. • The moment of inertia of the shaded area is obtained by subtracting the moment of inertia of the half-circle from the moment of. See next page for section properties needed in these problems. I = I ¯ + A d 2. Note: If you are lost at any point, please visit the beginner’s lesson or comment below. Hi, I'm in Year 12 and my teacher gave me a problem about moment of inertia, and it wants me to find the moment of inertia of a truck about its centre of mass and I'm given the following: The truck has a mass of 4500kg. J = Polar moment of inertia about the axis of rotation r = Distance from neutral axis to the outer most fibre = D/2. If the reference axis will be used to calculate moment of inertia of a complex shape, choose an axis of symmetry to simplify the calculation. The repair was with 4 x 4xx angle. First the moment of inertia of each rectangle about its centroid is calculated. Moments Of Inertia Bible Studies By Steve PPT. The system is at rest when a friend throws a ball of mass 0. 73 m long, thin light-weight rod is shown below. x-y axes: x and y are the coordinates of the element of area dA=xy I xy ³ xy dA • When the x axis, the y axis, or both are an. The moment of inertia can be defined as the second moment about an axis and is usually designated the symbol I. In general, moment of inertia is just a measure of how hard it is to get something rotating. The area moment of inertia of a composite section can be calculated by adding/subtracting the. It is free to rotate about the z axis, which passes through G. Note: Different disciplines use the term moment of inertia (MOI) to refer to different moments. The mass moment of inertia is often also known as the. 28 Rectangle Area, in 2, in. If the reference axis will be used to calculate moment of inertia of a complex shape, choose an axis of symmetry to simplify the calculation. Engineering Mechanics - Statics Chapter 10 Given: a = 4 in b = 2 in Solution: a ⌠ Iy = ⎮ x b 2 x dx ⎮ a ⌡ 0 4 Iy = 36. Transfer Formula for Moment of Inertia. So based on that I have solved problems on composite sections. That measurement is calculated based upon the distribution of mass within the object and the position of the axis, meaning that the same object can have very. In the same way that the larger the mass of an object the smaller the acceleration for a given force, the larger the moment of inertia of an object the smaller the angular acceleration for a given applied moment or torque. It is the rotational analog of mass. Problem on Moment of Inertia of T Section - Moment of Inertia - Strength of Materials - Duration: 17:24. • The moment of inertia (MI) of a plane area about an axis normal to the plane is equal to the sum of the moments of inertia about any two mutually perpendicular axes lying in the plane and passing through the given axis. The polar moment of inertia on the other hand, is a measure of the resistance of a cross section to torsion with invariant cross section and no significant warping. Evaluation of Moments of Inertia 2008 Waterloo Maple Inc. Calculate Yield Moment for the Beam Assume concrete accepts no tension. Example 10 Determine the moment of inertia and the product of inertia of a wooden T-beam section. We now need the tension T, which we flnd as follows. you written wrong formula the formula of Ixx is wrong in moment of inertia of T. Where b is the breadth of the beam and h be the height of the beam. We shall illustrate how the moment of inertia is actually calculated in practice, in the following Tasks. pdf Area-Moment. We defined the moment of inertia I of an object to be $I=\sum _{i}{m}_{i}{r}_{i}^{2}$ for all the point masses that make up the object. Simply Supported Beam|calculate support reaction|Engineering. Oftentimes the moment of intertia of a rigid body is not taken around the centroid, rather some arbitrary point. Problem on Moment of Inertia of T Section - Moment of Inertia - Strength of Materials - Duration: 17:24. Determine the distance 'd' between the sections such that the centroidal moment of inertia about the x- and y-axis are equal. If we compare Figure to the way we wrote kinetic energy in Work and Kinetic Energy, $(\frac{1}{2}m{v}^{2})$, this suggests we have a new rotational variable to add to our list of our relations between rotational and translational variables. In the calculation of the equivalent inertia is where the cracking moment and the cracked inertia come are needed, being part of the formula below (art. Note: Different disciplines use the term moment of inertia (MOI) to refer to different moments. For basic shapes there are tables that contain area moment of inertia equations which can be viewed below. The Polar Area Moment Of Inertia of a beams cross-sectional area measures the beams ability to resist torsion. Rotational version of Newton's second law. 300m , and the rim has mass 1. C y: Area: Moment of Inertia about the x c axis I xc: Polar Moment of Inertia about the z c axis J zc = I xc + I yc Radius of Gyration about the x c axis k xc: Radius of Gyration about the z c axis r zc: r zc 2 = k xc 2 +k yc 2. From the graphics section moment of inertia rotation axis formula can make a geometric figure (Moment of inertia circle), and thus may calculate inertia axis rotation calculation problem, called the graphics section moment of inertia rotation axis calculation diagram method. Moments (part 2) Finding torque for angled forces. I = Second moment of area, in 4 or mm 4; J i = Polar Moment of Inertia, in 4 or mm 4; J = Torsional Constant, in 4 or mm 4; K = Radius of Gyration, in or mm; P = Perimeter of shape, in or mm; Z = Elastic Section Modulus, in 3 or mm 3; Online Square I-Beam Property Calculator. Compute this as illustrated in Example 9. It may still, however, be useful for other purposes as well. J = Polar moment of inertia about the axis of rotation r = Distance from neutral axis to the outer most fibre = D/2. Polar Moment of Inertia of a Circle about its Center: Letting dA 2 d, the area of the dark-shaded ring in Fig. Physically, the moment of inertia gives the ability of a body to resist angular acceleration (analogous to how mass resists linear acceleration) when subject to a torque. P-819 with respect to its centroidal Xo axis. Get 1:1 help now from expert. A thin disk has been. Rotational version of Newton's second law. This axis can later on be translated to another axis if desired, using the rules outlined in the section entitled "Parallel Axis Theorem". A wagon wheel is constructed. It appears in the relationships for the dynamics of rotational motion. A toy consisting of two balls, each m = 0. solidworks section moment of inertia false what is the problem. Dear Friends, this session helps you to to find the centroid of composite areas, especially a T Section ,understand from basics, step by step tutorial on Moment of Inertia. The moment of inertia of the hydrogen molecule was historically important. To determine the moment of inertia of such a section is to find the moment of inertia. axis, giv the x ' Sample Problem A (I the the the C. The moment of inertia of a particle of mass m rotating about a particular point is given by: M o m e n t o f i n e r t i a = m d 2 \displaystyle\text {Moment of inertia}= {m} {d}^ {2} Moment of inertia = md2. The above statement is over simplified. For point masses, this can be expressed mathematically as I = Ʃmr2. It depends on the body's mass distribution and the axis chosen, with larger moments. 20}) to find. For that reason, it is preferred the moment of inertia about the x- and y-axis of a column section to be roughly equal. Rotational Motion: Moment of Inertia armsextended. txt) or view presentation slides online. 3 Moment of Inertia of an Area by Integration – A free PowerPoint PPT presentation (displayed as a Flash slide show) on PowerShow. Problem (10-110) Page 57 Determine the mass moment of inertia of the thin plate about an axis perpendicular to the page and passing through point O. Sample Problem 9. 5 2 3 A 4-0. Look up or calculate the moment of inertia and section modulus (1 & S). 621 cm4 but on -ly it is quite confusing that whether it is 1. Inventor seems to handle. Using the parallel axis theorem, the moment of inertia about a parallel axis passing through one of the ends of the rod is. For example: The Second Rectangular Area Moment of Inertia of a beam's cross section represents that beam's resistance to bending. Where: J M = Polar Mass Moment of Inertia (in-lbs-sec 2, Kg-m-sec 2). (a) Find the maximum tensile stresser, and maxi-mum compressive stress tr c if the cross section has the dimensions indicated and the moment of inertia about the - axis (the neutral axis) is t = 3. Problem 821 Find the moment of inertia about the indicated x-axis for the shaded area shown in Fig. Due to intense BIS processes the. This table provides formula for calculating section Area, Moment of inertia, Polar moment of inertia, Section modulus, Radius of gyration, and Centroidal distance, for various cross section shapes. I xx = ∫dA. The idea of point equivalent moment of inertia is concentrated on equation (2) and equation (3). The radius of gyration can be useful for listing in a table. In addition to being a property of any physical object, mass is a measure of the resistance of an object to acceleration when a net force has been applied to the object. Conversely, a lower moment of inertia means that you only need to apply a minimal amount of force to cause a rotation. I G is defined as: I G = ∫r2 dm Units: kg-m2 or slug-ft2 I G is used for several kinds of rigid body rotation problems, including: (a) F=ma analysis moment equation ( ΣM G = I Gα). Moment of Inertia Formula for Plane Figures - Moment of. 14 ratings • 1 review. Moment of Inertia. A good example of this is an I-Beam. Moment of Inertia. Ekeeda 174,723 views. The differential element, dA, is usually broken into two parts, dx and dy (dA = dx dy), which makes integration easier. This is a problem in my textbook could someone please help me work through it. y 2 - 2 x 3 10 Solutions 44918 1/28/09 4:21 PM Page 937. Free Moment of Inertia Calculator. Consider a built-up column comprised of two MC12x35 channels. Ib = Inertia of the gross section. These types of beams are generally used in civil engineering works. Big, heavy objects will have a high moment of inertia and be hard to turn; smaller, awkwardly shaped objects can also have high moments of inertia and can be just as hard to turn. CIVL 4135 81 Transformed Section 4. The moment of inertia of the platform is 5 kgm 2. Analyzing Rolling Motion. b = 100 mm Little theory: The area or the second moment of inertia of a planer cross section of a beam define the beam’s ability to withstand the bending and torsional shear stress. 8 kN-m] and a shear force of 12 kips [53. The moment of inertia depends on the mass and shape of an object, and the axis around which it rotates. ; The average shear stress r a v e r (obtained by dividing the shear force by the area of the web) and. When you try to make it move that mean you want to change the speed of the object from 0 to any, there will be moment of inertia effect. 6-5, use the lower-bound moment of inertia for deflection of the composite section. The moment of inertia calculates the rotational inertia of an object rotating around a given axis. the end the the a strip b is of t, he Thus. Steel I Beam Moment of Inertia Calculator. Four point objects of mass m are located at the corners of a square of side s as shown in the figure to the right. Now consider a compound object such as that in , which depicts a thin disk at the end of a thin rod. Calculate the moment of inertia. The moment of inertia of a particle of mass m rotating about a particular point is given by: M o m e n t o f i n e r t i a = m d 2 \displaystyle\text {Moment of inertia}= {m} {d}^ {2} Moment of inertia = md2. The parallel axis theorem relates the moment of inertia $$I_{CM}$$ of an object, with respect to an axis through the center of mass of the object, to the moment of inertia I of the same object, with respect to an axis that is parallel to the axis through the center of mass and is at a distance d from the axis through the center of mass. moments of inertia A measure of a body's resistance to angular acceleration, equal to: a. Moment of Inertia Hat Section You will have to register or login (See top or bottom of page) before you can post a message or view images: click the appropriate link to proceed. Question is: find the moment of inertia of a 250 g thin ring with radius = 25 cm about an axis parallel to the axis of rotation and located 1. The acceleration a that the block feels is converted to angular acceleration in the pulley, where a = Rfi. Problem on Centre of gravity and Centroid, Moment of Inertia. Presentation Summary : Moments of Inertia Lesson 7. 5mm 20mm 200mm EXERCISE PROBLEMS Q. What you have shown is first moment of area, not second. In this problem, the y axis is 8” from the y centroidal axis and x axis is 6” below the base of the semicircle, this would be usually evident from the problem description. x = any axis parallel to the centroidal axis. In general, moment of inertia is just a measure of how hard it is to get something rotating. b) The wide, flat cylinder will have a higher moment of inertia than the long, thin cylinder. J = ∫ r 2 d A. 1 RADIUS OF GYRATION k All rotating machinery such as pumps, engines and turbines have a moment of inertia. Z = Elastic Section Modulus, in 3 or mm 3 Online Rectangular Angle Property Calculator Using the structural engineering calculator located at the top of the page (simply click on the the "show/hide calculator" button) the following properties can be calculated:. Properties of Sections. Find the moment of inertia of a circular section whose radius is 8" and diameter of 16". n = perpendicular distance between the line and the area of cross section. Moment of Inertia. Problem 34 Write a user-defined function that determines the coordinate y c of the centroid of the T-shaped cross-sectional area shown in the figure. y 2 – 2 x 3 *10–12. Oftentimes the moment of intertia of a rigid body is not taken around the centroid, rather some arbitrary point. It is necessary to specify a moment of inertia with respect to an axis of rotation. The moment of inertia of the triangular shaped area is 3. Large rectangle 150 mm 420 mm 520 mm 60 mm 60 mm 60 mm 150 mm + 150 mm = 300 mm 200 mm + 200 mm = 400 mm x 400 mm y Small rectangle 300. Density = Mass per unit volume Density = dm / dV where: þ; - Density dm - Mass of a ring or radius R dV - Volume of a ring or radius R Lets assume height of the cylinder is h. Four point objects of mass m are located at the corners of a square of side s as shown in the figure to the right. The moment of inertia about one end is $$\frac{1}{3}$$mL 2, but the moment of inertia through the center of mass along its length is $$\frac{1}{12}$$mL 2. , how difficult it is to bend it. Problem 819 Determine the moment of inertia of the T-section shown in Fig. When doing dynamics problems with moments of inertia, you should not use the formulas you remember for second moment of area instead. If you're seeing this message, it means we're having trouble loading external resources on our website. 2 bd 3 IXX = 12 600 x 200 3 = 12 = 4 x 104 mm4. Moment of inertia, denoted by I, measures the extent to which an object resists rotational acceleration about a particular axis, and is the rotational analogue to mass. If we compare Figure to the way we wrote kinetic energy in Work and Kinetic Energy, $(\frac{1}{2}m{v}^{2})$, this suggests we have a new rotational variable to add to our list of our relations between rotational and translational variables. 100 kg, and radius 20. h = cross-section depth H = shorthand for lateral pressure load h f = depth of a flange in a T section I transformed = moment of inertia of a multi-material section transformed to one material k = effective length factor for columns = length of beam in rigid joint = length of column in rigid joint l d = development length for reinforcing steel. For example for a rectangular cross section beam with dimensions$$X = 6cm,\space Y =12 cm\space and\space Z =180 cm$$ which could be a log of timber the. of rectangle ① is at a distance of 20 cm from y-axis and at a distance of 25 cm from y-axis. (8), derived in the moment of inertia example, the moment of inertia of the disk is = at 5 digits Therefore, the moment of inertia of the disk is 12. Converting between Units. Here only the product of the area is defined and discussed. If I test a simple block (1x1x1mm) and check the mass properties I can find a couple of inertia values, but I’m not sure witch one to look for. Moment Of Inertia Of A Rectangular Plate Derivation 1. Equations for the moment of inertia of common shapes can be found in most p. where I is the moment of inertia of the mass m about the center O. In this case two separate keys are used for the two shafts. P-819 with respect to its centroidal X o axis. Expression for the Moment of Inertia of an Annular Ring: Consider a uniform thin annular disc of mass M having inner radius R 1 , outer radius R 2 , thickness t, and density of its material ρ. Gross Moment Of Inertia T Beam April 12, 2018 - by Arfan - Leave a Comment Area moment of inertia typical cross sections i get consider the t beam shown in figure 1 get consider the t beam shown in figure 1 ing moment exle problems of design methods for beams and columns ering feed. Calculator for Moment of Inertia of Rectangular section. Problems practice. moment of inertia about x axis and y axis for t section section of channel section is the most important Concept for civil engineering students when we have to solve the problem of bending. Moment Of Inertia Formula T Beam Posted on April 6, 2020 by Sandra Centroid area moments of inertia solution manual mechanics of materials neutral axis and parallel theorem beam sections using the section area moment of inertia typical cross. 7 (Modified) Find y , the position of the centroid of this section about a horizontal axis, and find the area moment of inertia of the section about that centroid. Moment of inertia. It is the inertia of a rotating body with respect to its rotation. determine moments of inertia of beam section and plate with respect to The strength of a W14x38 rolled steel beam is increased by attaching a plate to its upper flange. LuxCalc MOI Mobile is another addition to our series of mobile versions of the modules included in Luxeas LuxCalc Tools System. Please enter the "Input Values" in the form given below and click "Calculate". The polar moment of inertia on the other hand, is a measure of the resistance of a cross section to torsion with invariant cross section and no significant warping. Angular momentum of an extended object. This is a standard result. The moment of inertia is the quantitative measure of rotational inertia, just as in translational motion, and mass is the quantitative measure of linear inertia—that is, the more massive an object is, the more inertia it has, and the greater is its resistance to change in linear velocity. Problem 821 Find the moment of inertia about the indicated x-axis for the shaded area shown in Fig. RE: Moment of Inertia - Tube with Slot GBor (Mechanical) 5 Aug 08 08:39 If this slot is sufficiently long, you also need to take a look at whether the edges of the slot bend or buckle, but I've been under the impression that this is a working sliding cylinder, so the concept is already proven. This macro might be a start. So based on that I have solved problems on composite sections. • If the area is positive, then the moment of inertia is positive. (a) Consider rectangle ①. The final area, may be considered as the. So here again is the expression for the polar moment of inertia, and we're going to take a cross section so that we can calculate what that polar moment of inertia is. W - T = m*a W-T = m*T*L^2/(16*I) The Attempt at a Solution Well, in my attempting of solving the problem, I considered that the moment of inertia utilized in these equations is equal to m*(L^(2)/3). It is the special "area" used in calculating stress in a beam cross-section during BENDING. 5 lessons • 1 h 1 m. 4: Product of Inertia Last updated; Save as PDF Page ID 652; Transfer of Axis Theorem; Contributors; In addition to the moment of inertia, the product of inertia is commonly used. Ib = Inertia of the gross section. Please enter the "Input Values" in the form given below and click "Calculate". C y: Area: Moment of Inertia about the x c axis I xc: Polar Moment of Inertia about the z c axis J zc = I xc + I yc Radius of Gyration about the x c axis k xc: Radius of Gyration about the z c axis r zc: r zc 2 = k xc 2 +k yc 2. Transfer Formula for Moment of Inertia. rotation axis, as a quantity that decides the amount of torque required for a desired angular acceleration or a property of a body due to which it resists angular acceleration. Therefore, the moment of inertia I x of the tee section, relative to non-centroidal x0-x0 axis, is determined like this:. The more far away from the axis, the more moment of inertia the object has. Calculating the moment of inertia for compound objects. which have solved the classification problem for 3 manifolds. Area and Moment of Inertia of a Hollow Rectangular Shape Section In addition to the moments of inertia about the two main axes, we have polar moment of inertia, J, which represents the stiffness of circular members such as solid shafts and hollow structural sections against torsion. I ¯ = centroidal moment of inertia. Four point objects of mass m are located at the corners of a square of side s as shown in the figure to the right. How To Locate The Neutral Axis In A Beam Quora. Moment of Inertia. Previous question Next question Transcribed Image Text from this Question. Moment of Inertia for built-up Beams Moment of Inertia for built-up Beams packie81 (Mechanical) (OP) 28 Oct 02 15:56. We defined the moment of inertia I of an object to be $I=\sum _{i}{m}_{i}{r}_{i}^{2}$ for all the point masses that make up the object. The goal of this problem is to compute the major axis transformed moment of inertia for the beam. Now, in a full circle because of complete symmetry and area distribution, the moment of inertia relative to the x-axis is the same as the y-axis. Where: J M = Polar Mass Moment of Inertia (in-lbs-sec 2, Kg-m-sec 2). The diagonal moment of inertia of a square can also be calculated as; I x = I y = a 4 / 12 Alternatively, if the centre of mass (cm) is moved to a certain distance (d) from the x-axis we will use a different expression for determining the moment of inertia of the same square. Where, I = second moment of inertia about the line k. Does anyone have experience with the equation for determing the required transverse moment of inertia for a mast. Inertia can be thought of as another word for mass. A wagon wheel is constructed. 5 in c c A I k x x k xc 5. 5 2 3 A 4-0. 6 Review Recall from previous lesson the first moment about y-axis The moment of inertia (or second moment) is the measure of the. Problem on Moment of Inertia of T Section - Moment of Inertia - Strength of Materials - Duration: 17:24. The beams resistance to bending is represented by second moment area. 2 Method of Composite Areas Example 2, page 1 of 2 2. Rolling without slipping problems. The moment of inertia is to be found about the center of the rod. Problem 819 Determine the moment of inertia of the T-section shown in Fig. For that reason, it is preferred the moment of inertia about the x- and y-axis of a column section to be roughly equal. 9 106mm4 Ix Ix 138. The moment of inertia is a measure of the resistance of a rotating body to a change in motion. On the other hand, the product of inertia will be non-zero if the section is asymetrical (i. C-6b, and using Eq. However, this can be automatically converted to compatible units via the pull-down menu. Moment of Inertia 5 An example of this is the concrete T-beam shown. 6-4 For the beam of Problem 9. For the I-shaped section, however, it is not possible to simply subtract the smaller rectangles from the larger, as was done when computing the moment of inertia about the x-axis, since the centroids of the various parts being. The moment of inertia of a tee section can be found if the total area is divided into two, smaller ones, A, B, as shown in figure below. Although it is a simple matter to determine the moment of inertia of each rectangular section that makes up the beam, they will not reference the same axis, thus cannot be added. 5 2 3 A 4-0. See next page for section properties needed in these problems. In order to find the moment of inertia, we have to take the results of a full circle and basically divide it by two to get the result for a semicircle. So based on that I have solved problems on composite sections. The objects resistance for being rotated is measured by area moment of inertia calculator. 2, Figure 7. Learn exactly what happened in this chapter, scene, or section of Rotational Dynamics and what it means. For close shaped section, polar moment of inertia can be calculated from perpendicular axis theorem (adding both the 2nd moment of area in cross sectional axis). You can now find the moment of inertia of a composite area about a specified axis. In general an object has more moment of inertia (I) the further the mass is from the axis of rotation. (The moment of inertia about the y-axis is a measure of the resistance to rotation around this axis. Area Moments of Inertia Products of Inertia: for problems involving unsymmetrical cross-sections and in calculation of MI about rotated axes. 2 Method of Composite Areas Example 2, page 1 of 2 2. Moment of inertia of a same object will change against different axis. 5 kg and radius R = 20 cm, mounted on a. The formula for Moment of Inertia can also be derived from the formula of Kinetic Energy of a particle as it gives the direct relation between normal Mass of a body and Moment Of Inertia Please refer to the pic below: Hope u would unders. | 1/12BH^3) in Pro|e. kg·m2 (b) Calculate the applied torque needed to accelerate it from rest to 1700 rpm in 2. MASS MOMENT OF INERTIA Consider a rigid body with a center of mass at G.
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https://mathoverflow.net/questions/60375/is-mathbb-r3-the-square-of-some-topological-space | # Is $\mathbb R^3$ the square of some topological space?
The other day, I was idly considering when a topological space has a square root. That is, what spaces are homeomorphic to $X \times X$ for some space $X$. $\mathbb{R}$ is not such a space: If $X \times X$ were homeomorphic to $\mathbb{R}$, then $X$ would be path connected. But then $X \times X$ minus a point would also be path connected. But $\mathbb{R}$ minus a point is not path connected.
A next natural space to consider is $\mathbb{R}^3$. My intuition is that $\mathbb{R}^3$ also doesn't have a square root. And I'm guessing there's a nice algebraic topology proof. But that's not technology I'm much practiced with. And I don't trust my intuition too much for questions like this.
So, is there a space $X$ so that $X \times X$ is homeomorphic to $\mathbb{R}^3$?
• I'm wondering to what extent there is unique factorization of topological spaces relative to $\times$. $\mathbb{Q}$ is an idempotent (as is its complement in $\mathbb{R}$), but are there more interesting failures of UF involving connected spaces? Or results establishing UF for "nice" families of spaces? Should these be posted as a new question? Apr 3, 2011 at 1:41
• Is Moebius $\times$ Moebius = cilinder $\times$ cilinder (no boundaries)? Apr 4, 2011 at 16:38
• Without knowing any algebraic topology, it's possible to conclude at least something about X. If X is metric, compact, or locally compact and paracompact, then $\dim(X\times X)\le 2\dim X$, which means X has to have Lebesgue covering dimension at least 2. Wage, Proc. Natl. Acad. Sci. USA 75 (1978) 4671 , www.pnas.org/content/75/10/4671.full.pdf . What is the weakest condition that guarantees $\dim(X\times Y)= \dim X+\dim Y$? Given Yaakov Baruch's comment about the "dogbone space," it's not obvious that X is at all well behaved simply from the requirement that its square is $\mathbb{R}^3$.
– user21349
Jan 19, 2013 at 15:55
• @YaakovBaruch, isn't the cylinder factorizable? And could you elaborate this identity a little? Oct 17, 2013 at 6:35
No such space exists. Even better, let's generalize your proof by converting information about path components into homology groups.
For an open inclusion of spaces $X \setminus \{x\} \subset X$ and a field $k$, we have isomorphisms (the relative Kunneth formula) $$H_n(X \times X, X \times X \setminus \{(x,x)\}; k) \cong \bigoplus_{p+q=n} H_p(X,X \setminus \{x\};k) \otimes_k H_q(X, X \setminus \{x\};k).$$ If the product is $\mathbb{R}^3$, then the left-hand side is $k$ in degree 3 and zero otherwise, so something on the right-hand side must be nontrivial. However, if $H_p(X, X \setminus \{x\};k)$ were nontrivial in degree $n$, then the left-hand side must be nontrivial in degree $2n$.
• I hope this fine illustration of the power of relative homology will find its way in a textbook or, meanwhile, in algebraic topology courses. Apr 2, 2011 at 19:40
• I have a question regarding the top answer given by Tyler Lawson. As far as I know you can only apply the relative version of the Kunneth formula to cofibrations. Since we do not know much about $X$, it is unclear why $(X, X\setminus p)$ is a cofibration. Moreover, $(\mathbb R^3, \mathbb R^3\setminus p)$ is not a cofibration (I think). Mar 21, 2017 at 14:21
• For example, Dold's version (Corollary 12.10 in Lectures on Algebraic Topology part VI) requires an excisive triad condition. The core of these assumotions is to ensure that, given $(X,A)$ and $(Y,B)$, the covering of $(X \times B) \cup (Y \times A)$ by $X \times B$ and $A \times Y$ is good enough to satisfy the assumptions of the Mayer-Vietoris theorem. This is, in particular, satisfied if $A$ is an open subset of $X$ and $B$ is an open subset of $Y$, or in the CW-inclusion version that Hatcher uses. Mar 21, 2017 at 19:01
• So this also works for $\sqrt{\mathbb{R}^{2n+1}}$ doesn't it? Aug 3, 2018 at 8:20
• This should prove that if $\mathbb{R}^n=X^k$, then $k$ divides $n$, I think? Oct 17, 2019 at 1:38
this blog post refers to some papers with proofs. I've heard Robert Fokkink explain his proof (which is, quoting from this post)
A linear map $$\Bbb R^n \to \Bbb R^n$$ can be understood to preserve or reverse orientation, depending on whether its determinant is $$+1$$ or $$-1$$. This notion of orientation can be generalized to arbitrary homeomorphisms, giving a "degree" $$\deg(m)$$ for every homeomorphism which is $$+1$$ if it is orientation-preserving and $$-1$$ if it is orientation-reversing. The generalization has all the properties that one would hope for. In particular, it coincides with the corresponding notions for linear maps and differentiable maps, and it is multiplicative: $$\deg(f \circ g) = \deg(f)\cdot \deg(g)$$ for all homeomorphisms $$f$$ and $$g$$. In particular (fact 1), if $$h$$ is any homeomorphism whatever, then $$h \circ h$$ is an orientation-preserving map.
Now, suppose that $$h : X^2 \to \Bbb R^3$$ is a homeomorphism. Then $$X^4$$ is homeomorphic to $$\Bbb R^6$$, and we can view quadruples $$(a,b,c,d)$$ of elements of $$X$$ as equivalent to sextuples $$(p,q,r,s,t,u)$$ of elements of $$\Bbb R$$.
Consider the map $$s$$ on $$X^4$$ which takes $$(a,b,c,d) \to (d,a,b,c)$$. Then $$s \circ s$$ is the map $$(a,b,c,d) \to (c,d,a,b)$$. By fact 1 above, $$s \circ s$$ must be an orientation-preserving map. But translated to the putatively homeomorphic space $$\Bbb R^6$$, the map $$(a,b,c,d) \to (c,d,a,b)$$ is just the linear map on $$\Bbb R^6$$ that takes $$(p,q,r,s,t,u) \to (s,t,u,p,q,r)$$. This map is orientation-reversing, because its determinant is $$-1$$. This is a contradiction. So $$X^4$$ must not be homeomorphic to $$\Bbb R^6$$, and $$X^2$$ therefore not homeomorphic to $$\Bbb R^3$$.
and there he also told us the cohomological proof, which generalizes it to all Euclidean spaces of odd dimension.
• I hope no one misses this nice alternative proof because it's behind a link. Apr 4, 2011 at 2:24
• Quoting from the link: "The paper also refers to an earlier paper ("The cartesian product of a certain nonmanifold and a line is E4", R.H. Bing, Annals of Mathematics series 2 vol 70 1959 pp. 399–412) which constructs an extremely pathological space B, called the "dogbone space", not even a manifold, which nevertheless has B × R^3 = R4." This is relevant to my comment to the OP. Apr 4, 2011 at 5:16
• I don't understand this step in the proof: Why does the map $X^4 \to X^4, (a,b,c,d) \mapsto (c,d,a,b)$ correspond to the map $R^6 \to R^6, (p,q,r,s,t,u) \mapsto (s,t,u,p,q,r)$? I mean, the homeomorphism is not supposed to commute with projections ... Apr 4, 2011 at 15:05
• @Martin: The homeomorphism $(X\times X)\times (X\times X)\cong \mathbb R^3 \times \mathbb R^3$ respects projections by construction, so swapping the "two factors" (which I've emphasized with parentheses) on the left hand side corresponds to swapping the two factors on the right hand side. Apr 5, 2011 at 5:42
• This argument is also given as exercise in Hatcher's "More exercises in algebraic topology". Oct 6, 2020 at 18:21
I didn't know that, but I did know this: we cannot have $S^2 = S\times S$ for any topological space $S$.
• Would you care to elaborate? Jan 19, 2013 at 5:09
• All things considered, perhaps "S" is not the best name for the topological space for this assertion. Jan 19, 2013 at 5:52
• @Terry Tao True enough, but in all honesty it's precisely the notational perversity that brought this to mind to begin with. Jan 19, 2013 at 10:28
• @Agol Fix $s\in S$. On the one hand, $\pi_2(S\times S,(s,s))\cong \pi_2(S,s)\times\pi_2(S,s)$. On the other hand, $\pi_2({\bf S},{\bf s})\cong{\mathbb Z}$ for any 2-sphere $\bf S$ and any ${\bf s}\in{\bf S}$. Now it suffices to observe that ${\mathbb Z}\not\cong G\times G$ for any group $G$: indeed, such a group must be an infinite quotient of $\mathbb Z$, whence $G\cong{\mathbb Z}$, but ${\mathbb Z}\not\cong{\mathbb Z}\times{\mathbb Z}$ Jan 19, 2013 at 11:27
• @IanAgol : On the LHS, $S^2$ refers to the $2$-sphere, while on the LHS $S$ refers to an arbitrary topological space. Nov 17, 2014 at 15:55
The Euler characteristic with compact support $$\chi_c(X)$$ is a very robust topological invariant available for any reasonable space (such as a subanalytic set). The key properties here are that $$\chi_c(X)$$ is a real number and $$\chi_c(A\times B)$$ is multiplicative:
$$\chi_c(X)\in \mathbb{R},\quad \quad \chi_c(A\times B)=\chi_c(A)\cdot \chi_c(B).$$
It follows that the Euler characteristic with compact support of a topological square is always non-negative: $$\forall X: \quad \chi_c(X\times X)\geq 0.$$
Thus, a space with negative Euler characteristic with compact support cannot be a topological square.
In particular, since $$\chi_c(\mathbb{R}^3)=-1$$, $$\mathbb{R}^3$$ is not a topological square.
More generally, since $$\chi_c(\mathbb{R}^n)=(-1)^n$$, for any $$k\in\mathbb{N}$$, $$\mathbb{R}^{2k+1}$$ is not a topological square.
For an introduction to the topological Euler characteristic with compact support, I would recommend the following notes by LIVIU NICOLAESCU.
• You say that $\chi_c$ is "available for any reasonable space" - does the argument above genuinely show that $\mathbb{R}^{2k+1}$ is not a topological square, or just that it isn't the square of a "reasonable" space in the appropriate sense? Oct 4, 2020 at 18:41
• The answer assumes that the square root has a well defined euler characteristic with compact support.
– JME
Oct 8, 2020 at 2:55 | 2022-05-18T04:32:50 | {
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https://math.stackexchange.com/questions/1368006/contradiction-between-integration-by-partial-fractions-and-substitution | # Contradiction between integration by partial fractions and substitution
Integration by substitution:
$$\int \frac {dx}{x^2-1}$$ Let $x=\sec\theta$ and $dx=\sec\theta\tan\theta \,d\theta$ $$\int \frac {dx}{x^2-1} = \int \frac{\sec\theta\tan\theta \,d\theta}{\sec^2\theta-1} = \int \frac {\sec\theta\tan\theta\, d\theta}{\tan^2\theta} = \int \frac {\sec\theta \,d\theta}{\tan\theta}$$ $$\int \frac {\sec\theta \,d\theta}{\tan\theta} = \int \frac {\cos \theta \,d\theta}{\cos\theta \sin \theta} = \int \csc\theta\, d\theta$$ $$=\ln|\csc\theta-\cot\theta|+C$$
$$=\ln| \frac{x}{\sqrt{x^2-1}}-\frac{1}{\sqrt{x^2-1}}|+C=\ln| \frac{x-1}{\sqrt{x^2-1}}|+C$$
Which is $Undefined$ for $|x|<1$
Integration by partial fractions:
$$\int \frac {dx}{x^2-1}$$ $$\int \frac {dx}{x^2-1}= \frac 12\int\frac{dx}{x-1}- \frac12\int \frac{dx}{x+1} = \frac12 \ln | x-1| - \frac12 \ln|x+1| +C$$ $$= \frac12 \ln | \frac{x-1}{x+1}|+C$$
Which is $Defined$ for $|x|<1$ and this is right because the integrand is defined for $|x|<1$
What is the problem in the substitution method ?
• When you declared x = $sec \theta$ you implicitly declared that |x| ≥ 1.
– lulu
Jul 20, 2015 at 20:23
• At the end, what did you mean by the source function? Jul 20, 2015 at 20:24
• So, this integrand shouldn't be integrated using trigonometric substitution ? Jul 20, 2015 at 20:24
• @KhallilBenyattou the integrand Jul 20, 2015 at 20:25
• The integrand is only not defined for $x = \pm 1$. In any case, lulu's explained. Jul 20, 2015 at 20:28
Let's try to clear out some of the confusion on what is going on here.
First of all, $x\mapsto \frac 1 {x^2 - 1}$ is continuous function everywhere except for $x=\pm 1$, so it is Riemann integrable on any segment not containing $\pm 1$. That said, we would very much like to find primitive function defined on $\mathbb R^2\setminus\{\pm 1\}$. Solution by partial fractions does just that, $$F_1(x) = \frac 12 \ln\left|\frac{x-1}{x+1}\right| + C$$ is defined everywhere except at $x=\pm 1$. So, if we wanted to calculate either $$I_1 = \int_{\frac 12}^{\frac 34} \frac {dx} {x^2 - 1}$$ or $$I_2 = \int_{2}^{3} \frac {dx} {x^2 - 1}$$ we can use $F_1$ with no worries.
Now, if we try to use substitution such as $x = \sec\theta$, as OP notices, we might run into some problems in the long run. The primitive function derived this way is $$F_2(x) = \ln\left| \frac {x - 1}{\sqrt{x^2 -1}} \right| + C$$ which is defined only for $|x|>1$. Is this shocking? Well, no. As lulu points out in the comments, $|\sec\theta|\geq 1$ for any $\theta$, which is a simple consequence of the definition $\sec\theta = \frac 1 {\cos\theta}$. Thus, by substituting $x=\sec\theta$, we already gave up on $x\in\langle -1,1\rangle$, which is actually fine as long as we are trying to calculate $I_2$, but won't work for $I_1$.
So, the question is: are $F_1$ and $F_2$ both "good" solutions? More precisely, if we wanted to calculate $I_2$, can we use either of those two $F$'s?
Well, let's assume that $|x| > 1$. Then we have:
$$\ln\left| \frac {x - 1}{\sqrt{x^2 -1}} \right| = \ln \frac {\left|x - 1\right|}{\sqrt{x^2 -1}} = \ln \frac {\sqrt{(x-1)^2}}{\sqrt{x^2 -1}} = \ln \sqrt{\frac {{(x-1)^2}}{{x^2 -1}}} = \frac 12 \ln\frac{x-1}{x+1}=\frac 12 \ln\left|\frac{x-1}{x+1}\right|$$ where the last equality holds because $x-1$ and $x+1$ have the same signs on $|x|>1$. Thus, we have shown that $F_2 = \left.F_1 \right|_{\mathbb R^2\setminus [-1,1]}$. I hope that clarifies the problem.
• Very well explained! Jul 20, 2015 at 21:56
• Well. Your simplification which equated the result of the two integrations depends on the assumption of $|x|>1$ and that's fine, since for $|x|<1$ this equation can't hold. I found many problems when using trigonometric substitution in integration and don't know exactly how and why this happens.! Jul 21, 2015 at 1:23
• @MohamedMostafa, do you now know what is happening here? Look at here carefully. Substitution originally works for definite integrals, and it is extended on indefinite integrals, but, one needs to be careful whether substitution will make sense on all of the domain of the integrand. In this case $x = \sec\theta$ doesn't work on whole domain of integrand, and the problem arises. Jul 21, 2015 at 10:16
• Can we say - in short - that substitution method works if and only if the range of the substitution function is equal to the domain of the integrand to be substituted ? Jul 21, 2015 at 14:09
• When you are dealing with definite integral, range of substitution function must include segment on which you are integrating (that should be obvious). When you are dealing with indefinite integral this isn't necessary, but you might end up with a function that is defined on smaller domain and then you need to find a way to extend your function to maximum domain, just as we can extend $F_1$ to $F_2$ in your exercise. You may need to differentiate the end result to make sure you really got correct antiderivative. Jul 21, 2015 at 16:20
On simplifying
\begin{align} & \ln \frac{x-1}{\sqrt{x^2-1}}=\ln\frac{\sqrt{x-1}^2}{\sqrt{(x-1)(x+1)}} \\[6pt] = {} & \ln\frac{\sqrt{x-1}}{\sqrt{x+1}} \\[6pt] = {} & \frac 1 2 \ln \frac{x-1}{x+1} \end{align}
• Without absolute value, what you wrote is right. but when you deal with absolute value you can't do this way. Jul 20, 2015 at 20:51
• Thank you Michael for editing. Now it looks so beautiful. Please give me a link to the tutorial where i can learn to write equations and mathematical symbols on internet. Jul 20, 2015 at 20:51
• meta.math.stackexchange.com/questions/5020/… Jul 20, 2015 at 20:53
• @MohamedMostafa Thank you for the link. :) Jul 20, 2015 at 20:54
• @thie, those functions don't have same domains, i.e. these are the same functions, but only on smaller domain. Jul 20, 2015 at 21:00
$$\ln\frac{|x-1|}{\sqrt{|x^2-1|}} = \ln\frac{\sqrt{|x-1|}\sqrt{|x-1|}}{\sqrt{|x-1|}\sqrt{|x+1|}} = \ln\sqrt{\frac{|x-1|}{|x+1|}} = \frac 1 2 \ln\left|\frac{x-1}{x+1}\right|$$
In response to comments I've made this more complete than it was. Notice that
• I take no square roots of anything except non-negative numbers; and
• $|AB| = |A||B|$, so $|x^2-1|=|x-1||x+1|$; and
• $\sqrt{AB} =\sqrt A \sqrt B$ if $A\ge0$ and $B\ge0\vphantom{\dfrac 1 1}$, so the separation into two square roots is valid; and
• $\sqrt A/\sqrt B = \sqrt{A/B\,{}}$, if $A\ge0$ and $B\ge0$, so the second equality is valid; and
• $|A|/|B| = |A/B|$, so the last equality is valid.
PS in response to comments: The problem with the trigonometric substitution is only that it is valid only when $|x|>1$, since $\sec\theta\ge 1$ for all values of $\theta$ and those points where $|x|=1$ are not in the domain.
• Without absolute value, what you wrote is right. but when you deal with absolute value you can't do this way. Jul 20, 2015 at 20:52
• Correct .... This is not a complete answer. It works when $x>1$. But with this information, the original poster should be able to figure everything out. Jul 20, 2015 at 20:55
• Exactly. The problem is Rogawski - the author of the text I study - did the same simplification and I don't know if it was a mistake! Jul 20, 2015 at 20:58
• You most left hand side is NOT the result of integration by substitution. Then, this equation due to simplification can't hold for $|x|<1$ and the problem still there. Jul 21, 2015 at 1:25 | 2023-01-30T01:16:05 | {
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https://math.stackexchange.com/questions/1771806/a-prime-number-problem | # A prime number problem.
If $n$ is a positive integer and $(p_1,p_2,p_3,p_4,\ldots, p_n)$ are distinct positive primes, show that the integer $(p_1\cdot p_2\cdot p_3\cdot p_4\cdots p_n)+1$ is divisible by none of these primes. How do I figure out that none of the primes divide that new integer?
• Start with a simple case. Does $p_1$ divide $p_1p_2$? Does $p_1$ divide $p_1p_2 + 1$? – Chris Culter May 4 '16 at 20:27
• No it can't. I know it can't be possible, but to prove it, do I do induction? – nyorkr23 May 4 '16 at 20:29
• You don't need induction. $d$ cannot divide $kd+1$? – almagest May 4 '16 at 20:29
• Oh, that's what I was after. Thanks. – nyorkr23 May 4 '16 at 20:31
• You don't even need them to be primes. It suffices to have $p_i\ge2$. – Hagen von Eitzen May 4 '16 at 20:37
Suppose $7$ is one of the primes, so $p_1\cdots p_n$ is a multiple of $7$.
The next multiple of $7$ after $p_1\cdots p_n$ is $(p_1\cdots p_n)+7$. So $(p_1\cdots p_n)+1$ is not a multiple of $7$.
More formally, suppose $7$ divides $(p_1\cdots p_n)+1$. Then for some integers $j$, $k$, \begin{align} (p_1 \cdots p_n) & = 7j \tag 1 \\ (p_1 \cdots p_n) + 1 & = 7k \tag 2 \\[10pt] \text{Then subtracting (1) from (2), we get: } 1 & = 7k - 7j = 7(k-j) \\ \end{align} So $1 = 7(k-j)$.
• But how exactly do I show that as a proof? – nyorkr23 May 4 '16 at 20:30
• Do you have the theorem sometimes called the "division algorithm", stating that for $m,n\in\mathbb N$, there exists a quotient $q$ and a remainder $r\in\{0,1,2,\ldots,n-1\}$ such that $m = qn+r$? $\qquad$ – Michael Hardy May 4 '16 at 20:32
• Yes, I've learned that algorithm. – nyorkr23 May 4 '16 at 20:36
• I've expanded the answer somewhat. $\qquad$ – Michael Hardy May 4 '16 at 20:39
We will need the following theorem:
Theorem. For any two natural numbers $x, y$ ($y \geq 1$), there are unique natural numbers $q, r$ where $0 \leq r < y$ such that $x = qy + r$.
Proof. If $x < y$ then we must have $q = 0$ and $r = x$, so in this case the $q, r$ exist and are unique. Assume that there is a natural number $x \geq y$ such that no such $q, r$ exist, then without loss of generality we may assume that $x$ is the smallest such number. However, then $x - y$ cannot be written in the desired form either, as if we had $x - y = qy + r$ then this would imply $x = (q+1)y + r$. This contradicts the minimality of $x$, as $x - y$ is a smaller natural number with the same property. Therefore, such $q, r$ must exist for all values of $x$.
To prove uniqueness, assume that we had $x = q_1 y + r_1 = q_2 y + r_2$ and $r_1 > r_2$, then $r_1 - r_2 = y(q_2 - q_1)$. On the other hand, $r_1 - r_2 \leq r_1 < y$, so that we have $y(q_2 - q_1) < y$ and therefore $q_1 = q_2$. QED.
Corollary. $x = qn + 1$ ($q, n \in \mathbb{N}$) is not divisible by $q$ for any $q \geq 2$.
Proof. If it were, we would have $x = qm$ for some $m$ and $x = qn + 1$ simultaneously, contradicting the uniqueness proved in the above theorem.
The statement in the question follows immediately from the corollary.
• The division algorithm (Theorem) is not needed. Simpler: if $\, qm = qn+1\,$ then $\,q(m-n) = 1,\,$ contra $\, q> 1$. – Bill Dubuque May 4 '16 at 21:55
• That is simpler, although I see no harm in using the division algorithm either way. – Starfall May 4 '16 at 21:58
• The simpler method is much more general: it works in any ring for any nonunit $q$, i.e. any $q$ that does not divide $1$. But most rings don't have a division algorithm. That said, it is, of course, always helpful to have multiple proofs. Here one gains a bit different insight by using the uniqueness of remainders in the division algorithm - as you do. I've written about the power of uniqueness theorems in many of my posts, e.g. see here. – Bill Dubuque May 4 '16 at 22:02 | 2019-10-13T21:34:36 | {
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https://math.stackexchange.com/questions/1488948/bbb-rr-equinumerous-to-f-in-bbbrr-mid-f-text-surjective | $\Bbb {R^R}$ equinumerous to $\{f\in\Bbb{R^R}\mid f\text{ surjective}\}$
In an introductory course on axiomatic set theory, I am asked whether $\Bbb {R^R}$ is equinumerous to the set of all surjective functions in $\Bbb{R^R}$ (say $\mathcal S$). Because of the evocative nature of the question, I assume this to be true. However, I have not been able to prove it.
Using Schroeder-Bernstein, we would need to find an injection from $\Bbb{R^R}\rightarrow\mathcal S$, but I don't have the slightest idea how we could get this. In a similar question involving $\Bbb N$ instead of $\Bbb R$, this could be done (quite) easily since we could speak about sequences, but that won't help here.
• Finding an injection from $\mathcal{S}$ to $\Bbb{R}^\Bbb{R}$ is easy - just consider the inclusion map. I guess that you means the reverse. Oct 20, 2015 at 10:37
• Sorry, you are absolutely correct, I will edit it Oct 20, 2015 at 10:38
For given function $f:\Bbb{R}\to\Bbb{R}$, define $$\phi_f (x) = \begin{cases} \tan (\pi x/2) & \text{if }-1<x<1 \\ f(\tan(\pi (x-2)/2)) & \text{if }1<x<3 \\ 0 & \text{otherwise}\end{cases}.$$ You can check that the map $f\mapsto \phi_f$ is one-to-one. Also, $\phi_f$ is a surjection to $\Bbb{R}$.
• Can this also be expanded so that $\phi_f$ is a bijection? Oct 20, 2015 at 11:26
• @konewka Did you mean a function $\phi_f \in \Bbb{R}^\Bbb{R}$ or the function maps $f$ to $\phi_f$? Oct 20, 2015 at 11:29
• The function $\phi_f$, so in my original question, replacing $\mathcal S$ by the set of bijective functions. Or would those two not be equinumerous? Oct 20, 2015 at 11:30
• @konewka You may construct it. The set of all bijection from the reals to itself is equinumerous to $\Bbb{R}^\Bbb{R}$, but I didn't get how to construct the bijection explicitly. Oct 20, 2015 at 12:14
It suffices to divide $$\mathbb R$$ into two sets $$A$$ and $$B$$ such that $$|A|=|B|=|\mathbb R|$$.
This means you have some bijection $$\varphi$$ from $$A$$ to $$B$$. I.e., you have pairs $$a\in A$$ and $$\varphi(a)\in B$$
Using these two subsets you can construct bijections $$\mathbb R\to\mathbb R$$ by choosing some pairs, which you swap, and letting other elements in place.
More formally, if $$C\subseteq A$$, then we can define $$f\colon\mathbb R\to\mathbb R$$ as $$f(x)= \begin{cases} \varphi(x) & x\in C, \\ \varphi^{-1}(x) & x\in \varphi[C], \\ x & \text{otherwise}. \end{cases}$$
In this way we get a different bijection from $$\mathbb R$$ to $$\mathbb R$$ for every subsets of $$A$$. So there is at least as many bijections from $$\mathbb R$$ to $$\mathbb R$$ as the number of subsets of $$A$$, which is $$2^{\mathfrak c}$$.
If we denote by $$\mathcal S$$ the set of all surjections and $$\mathcal B$$ the set of all bijections (from $$\mathbb R$$ to $$\mathbb R$$), we have $$2^{\mathfrak c} \le |\mathcal B| \le |\mathcal S| \le |\mathbb R^{\mathbb R}|=\mathfrak c^{\mathfrak c}=2^{\mathfrak c}.$$
You can also have a look at this post on MO, which deals with bijections on arbitrary set: Cardinality of the permutations of an infinite set. In fact, Robin Chapman's answer posted there gives basically the same argument as I gave above. It should be mentioned that if we do the same argument for arbitrary set, we would use Axiom of Choice to get the decomposition $$X=A\cup B$$ such that $$|X|=|A|=|B|$$. Some details about role of AC here are give in Andreas Blass' answer to the same post.
There were also several posts about cardinality of the set of all bijections from $$\mathbb N$$ to $$\mathbb N$$. Some of the arguments given there can be used also for $$\mathbb R$$:
And you could find other similar questions.
Remark on use of Axiom of Choice. In connection with another question on this site, it might be also interesting whether we can do this in ZF. It's not difficult to see that we might replace $$\mathbb R$$ with $$\mathbb R\setminus\{0\}$$. (Since we can write down an explicit bijection between $$\mathbb R$$ and $$\mathbb R\setminus\{0\}$$.) If we choose $$A=(0,\infty)$$, $$B=(-\infty,0)$$ and $$\varphi(x)=-x$$, then the above construction gives an explicit bijection $$f\colon \mathbb R\setminus\{0\} \to \mathbb R\setminus\{0\}$$ for every $$C\subseteq A$$. | 2022-07-03T02:59:44 | {
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https://math.stackexchange.com/questions/1978987/prove-that-for-all-x-ae-fracxa-be-fracxb-leq-e-fracx28 | prove that for all x : $ae^{\frac{x}{a}} + be^{-\frac{x}{b}}\leq e^{\frac{x^2}{8a^2b^2}}$ if a + b = 1
I am trying to solve the following question, but I did not reach to any answer, I would be sol glad if anyone could help me on that.
If a, b are positive numbers such that a + b = 1 prove that for all x,
$ae^{\frac{x}{a}} + be^{-\frac{x}{b}}\leq e^{\frac{x^2}{8a^2b^2}}$
Thank you everyone !!!
• @Michael would you please devote some time to prove this inequality. I would really appreciate it if you mind helping me or giving me some hints on how to deal with inequalities like the one that I posted above. – alfred noble Oct 25 '16 at 6:15
• @Michael, if we consider the case where $a = b = \frac{1}{2}$ then we have: $cosh(2x) \le e^{2x^2}$. I would work on proving the latter inequality, this method may help to prove the main inequality – alfred noble Oct 25 '16 at 6:25
• @πr8, I read your answer which proves this above inequality when $a = b = \frac{1}{2}$, now I would deeply appreciate it if you could help prove the above inequality in general form – alfred noble Oct 25 '16 at 9:26
• Where did you come across this interesting inequality? – Thomas Ahle Aug 17 '18 at 16:19
It looks to be neat but true. Denote $x=abt$, we rewrite inequality as $ae^{bt}+be^{-at}\leqslant e^{t^2/8}$. We may suppose $t>0$, else replace $t$ to $-t$ and $b$ to $a$. Multiply by $e^{at}$ and rewrite as $ae^t+1-a\leqslant e^{at+t^2/8}$. Fix $t$ and vary $a\in [0,1]$. We should consider the minimal value of $H(a)=e^{at+t^2/8}-ae^t-1+a$ and prove that it is non-negative. This minimal value is attained either for $a=0$, or $a=1$, or for such $a$ that $H'(a)=0$. Obviously $H(0)\geqslant 0$, $H(1)\geqslant 0$, thus it remains to consider such $a$ that $H'(a)=0$. That is, $e^t-1=te^{at+t^2/8}$, $at=\log((e^t-1)/t)-t^2/8$. Now the inequality may be rewritten (multiply it by $t$ and substitute the values for $at$ and for $te^{at+t^2/8}$) as $$(e^t-1)\left(\log\frac{e^t-1}t-\frac{t^2}8\right)+t\leqslant e^t-1,$$ Divide by $e^t-1$ and note that for $t=0$ the equality takes place, so it suffices to prove that $$\left(\log\frac{e^t-1}t-\frac{t^2}8+\frac{t}{e^t-1}\right)'\leqslant 0,\,\forall t\geqslant 0.$$ Taking derivative and multiplying by $4t(e^t-1)^2$ we get (miracle!) $$-(e^t (t-2)+t+2)^2\leqslant 0.$$
• thanks a lot. But I have two questions about your answer. I) In the middle of your answer you took the derivatives of both sides of the inequality with respect to $a$ and you calculated the value of $a$ for which the derivations of both sides of the inequality is the same. I want to know why did you do that ? II) Are you saying that since you reached to an ever true inequality ($-(e^t(t-2) + t + 2)^2 \le 0$) the main inequality that I posted above is also true ? – alfred noble Oct 26 '16 at 6:41
• If we need to prove that $H(a)\geqslant 0$ for some function $H$ (here $H=$RHS-LHS of the inequality) and $a\in [0,1]$, we look at a point $a\in [0,1]$, for which $H$ takes minimal value. It is either an endpoint of the segment, or a point where derivative $H'(a)$ equals 0. – Fedor Petrov Oct 26 '16 at 6:55
• This is the link to the prove of the above inequality in special case when $a = b = \frac{1}{2}$ I think it might be useful to prove the above inequality in the general form. LINK:math.stackexchange.com/questions/331367/cosh-x-inequality – alfred noble Oct 26 '16 at 16:01
• Additional requirement is $f(0)=0$. For $f(x)=1/x$ it is not true, but for $-1+\log\frac{e^x-1}x-x^2/8+\frac{x}{e^x-1}$ it is true. – Fedor Petrov Oct 30 '16 at 11:26 | 2019-05-26T04:56:01 | {
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https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12A_Problems/Problem_24&diff=prev&oldid=118330 | Difference between revisions of "2010 AMC 12A Problems/Problem 24"
Problem
Let $f(x) = \log_{10} \left(\sin(\pi x) \cdot \sin(2 \pi x) \cdot \sin (3 \pi x) \cdots \sin(8 \pi x)\right)$. The intersection of the domain of $f(x)$ with the interval $[0,1]$ is a union of $n$ disjoint open intervals. What is $n$?
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 18 \qquad \textbf{(D)}\ 22 \qquad \textbf{(E)}\ 36$
Solution
The question asks for the number of disjoint open intervals, which means we need to find the number of disjoint intervals such that the function is defined within them.
We note that since all of the $\sin$ factors are inside a logarithm, the function is undefined where the inside of the logarithm is less than or equal to $0$.
First, let us find the number of zeros of the inside of the logarithm.
\begin{align*}\sin(\pi x) \cdot \sin(2 \pi x) \cdot \sin (3 \pi x) \cdots \sin(8 \pi x) &= 0\\ \sin(\pi x) &= 0\\ x &= 0, 1\\ \sin(2 \pi x) &= 0\\ x &= 0, \frac{1}{2}, 1\\ \sin(3 \pi x) &= 0\\ x &= 0, \frac{1}{3}, \frac{2}{3}, 1\\ \sin(4 \pi x) &= 0\\ x &= 0, \frac{1}{4}, \frac{2}{4}, \frac{3}{4}, 1\\ &\cdots\end{align*}
After counting up the number of zeros for each factor and eliminating the excess cases we get $23$ zeros and $22$ intervals.
In order to find which intervals are negative, we must first realize that at every zero of each factor, the sign changes. We also have to be careful, as some zeros are doubled, or even tripled, quadrupled, etc.
The first interval $(0, \frac{1}{8})$ is obviously positive. This means the next interval $(\frac{1}{8}, \frac{1}{7})$ is negative. Continuing the pattern and accounting for doubled roots (which do not flip sign), we realize that there are $5$ negative intervals from $0$ to $\frac{1}{2}$. Since the function is symmetric, we know that there are also $5$ negative intervals from $\frac{1}{2}$ to $1$.
And so, the total number of disjoint open intervals is $22 - 2\cdot{5} = \boxed{12\ \textbf{(B)}}$
Solution 2 (cheap)
Note that the expression $\sin(\pi x) \cdot \sin(2 \pi x) \cdot \sin (3 \pi x) \cdots \sin(8 \pi x)$ must be greater than zero, since logarithm functions are undefined for $0$ and negative numbers. Let $x_1, x_2, x_3, ..., x_8$ temporarily be the dependent variables of the functions $y_1 = \sin(\pi x_1), y_2 = \sin(2\pi x_2), ..., y_8 = \sin(8\pi x_8)$. It is easy to see that for $y_i$ to be positive for $1\leq i\leq8$, $\lfloor i x_i \rfloor$ must be even for $1 \leq i\leq8$. Since an even number of positives times an even number of negatives equals a positive, there can be $2, 4, 6,$ or $8$ positive values of $y_i$ for $1 \leq i\leq 8$ for a given value of $x$. (since $y_1$ is always positive on the range $[0, 1]$) Since MAA allows rulers (and you should bring one to the actual exam), use it to your advantage and draw a larged scaled number line from $0$ to $1$. (I recommend increments of at most $0.1$.) If you don't have a ruler but have graph paper, you can use that instead. Then, designate rows for $y_1, y_2, ..., y_8$, respectively. Draw a large bar (label it with $+$ so you know it's positive) for all values of $x_i$ such that $\lfloor i x_i \rfloor$ is even, and do that for all eight rows. Then, use your ruler (or another viable straightedge, such as the edge of another sheet of paper), place the straightedge perpendicular to the vertical line on your digram at $0$, and slowly work your way to $1$, marking all disjoint intervals in which your straightedge touches $2, 4, 6,$ or $8$ boxes simultaneously. (If an interval excludes a value in that interval, you still have to count it as two disjoint intervals. Note that this will be important as to not undercounting disjoint intervals. ) If done correctly, you should obtain $\boxed{12\ \textbf{(B)}}$ as your answer.
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https://math.stackexchange.com/questions/715194/exponential-pop-growth-when-only-given-population-at-two-instances-of-time | # Exponential pop. growth when only given population at two instances of time.
I have a problem where I'm only given the population of a "bacteria culture" at two instances in time: 2 hours and 4 hours. The problem says the population of bacteria is 125 after 2 hours, and 350 after 4 hours. It specifically says the bacteria in the culture increases according to the law of exponential growth.
I know the formula $y(t) = y_{0}e^{kt}$ where $k$ is the growth constant (and $t$ is time, of course). I know how to find k if you're given the initial population, or time. But in this case, I'm not sure what to do. It asks to find the initial population, then write the exponential growth model. I know how to write the growth model once I find the initial population, because then I can solve for $k$. But I just don't know how to find $k$ when given only two populations at given points of time with exponential growth. Can anyone help?
• After thinking more, I think it might be possible to write a system of equations: $125 = y_{0}e^{k(2)}$ and $350 = y_{0}e^{k(4)}$ and then this would solve for both k and the initial population simultaneously. Is this the right approach? – Sabien Mar 17 '14 at 6:08
• Just as a "cultural" remark, it is an interesting characteristic that just as two points determine a line, two points also determine an exponential curve. – colormegone Mar 17 '14 at 6:58
Let $y_0$ be the initial population, and let $k$ be the growth constant. Then the population at time $t$ is $y_0e^{kt}$.
Putting $t=2$, we get $y_0e^{2k}=125$. Similarly, $y_0e^{4k}=350$.
Divide. We get $e^{2k}=\frac{350}{125}$.
Now that we know $e^{2k}$, we can find $y_0$. And now that we know $e^{2k}$, we can find $k$ by taking the logarithm.
Remark: I see that you saw how to approach the problem. If you wish to weite up your conclusions as an answer, I will delete mine.
• Thank you very much that helps. – Sabien Mar 17 '14 at 6:20
• No, I tried my method but it didn't work, well at least I must have done the Algebra wrong. Your explanation is more clear, no need to delete it. – Sabien Mar 17 '14 at 6:22
• You are welcome. If there any step that you have trouble completing, please leave a message. – André Nicolas Mar 17 '14 at 6:24
For the sake of variety, without using template formulas, you can think in the following way. Exponential growth is about multiplying by some common growth factor every unit of time. We could call that growth factor $b$. So here, after two hours, we went from $125$ to $350$. that means $$125\cdot b^2=350$$ from which you can solve for $b$ and find that the growth factor is $\sqrt{\frac{350}{125}}$.
Now what was the original population? Well moving backwards in time two hours: $$125\div\left(\sqrt{\frac{350}{125}}\right)^2=\frac{125^2}{350}$$ So the population at $t$ hours after the initial moment is $$P(t)=\frac{125^2}{350}\left(\sqrt{\frac{350}{125}}\right)^t$$
• Hey, this is a really cool way to think about it. Thank you for the insight. – Sabien Mar 17 '14 at 6:35 | 2019-05-23T01:16:28 | {
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https://casmusings.wordpress.com/tag/polar/ | # Tag Archives: polar
## A Generic Approach to Arclength in Calculus
Earlier this week, a teacher posted in the College Board’s AP Calculus Community a request for an explanation of computing the arclength of a curve without relying on formulas.
The following video is my proposed answer to that question. In it, I derive the fundamental arclength relationship before computing the length of $y=x^2$ from x=0 to x=3 four different ways:
• As a function of x,
• As a function of y,
• Parametrically, and
• As a polar function.
In summary, the length of any differentiable curve can be thought of as
where a and b are the bounds of the curve, the square root is just the local linearity application of the Pythagorean Theorem, and the integral sums the infinitesimal roots over the length of the curve.
To determine the length of any differentiable curve, factor out the form of the differential that matches the independent variable of the curve’s definition.
## Clever math
Here are what I think are three clever uses of math by students.
In my last week of classes at my former school in May, 2013, my entire Honors Precalculus class showed up wearing these shirts designed by one of my students, M. The back listed all of the students and a lovely “We will miss you.” As much as I liked the use of a polar function, I loved that M opted not for the simplest possible version of the equation ($r=2-2sin(\theta )$), but for a rotation–a perfect use of my transformations theme for the course.
Now for a throwback. When I was a graduate student and TA at Syracuse from 1989-1990, one of my fellow grad students designed this shirt for all of the math and math ed students. I don’t remember who designed it, but I’ve always loved this shirt.
## A Student’s Powerful Polar Exploration
I posted last summer on a surprising discovery of a polar function that appeared to be a horizontal translation of another polar function. Translations happen all the time, but not really in polar coordinates. The polar coordinate system just isn’t constructed in a way that makes translations appear in any clear way.
That’s why I was so surprised when I first saw a graph of $\displaystyle r=cos \left( \frac{\theta}{3} \right)$.
It looks just like a 0.5 left translation of $r=\frac{1}{2} +cos( \theta )$ .
But that’s not supposed to happen so cleanly in polar coordinates. AND, the equation forms don’t suggest at all that a translation is happening. So is it real or is it a graphical illusion?
I proved in my earlier post that the effect was real. In my approach, I dealt with the different periods of the two equations and converted into parametric equations to establish the proof. Because I was working in parametrics, I had to solve two different identities to establish the individual equalities of the parametric version of the Cartesian x- and y-coordinates.
As a challenge to my precalculus students this year, I pitched the problem to see what they could discover. What follows is a solution from about a month ago by one of my juniors, S. I paraphrase her solution, but the basic gist is that S managed her proof while avoiding the differing periods and parametric equations I had employed, and she did so by leveraging the power of CAS. The result was that S’s solution was briefer and far more elegant than mine, in my opinion.
S’s Proof:
Multiply both sides of $r = \frac{1}{2} + cos(\theta )$ by r and translate to Cartesian.
$r^2 = \frac{1}{2} r+r\cdot cos(\theta )$
$x^2 + y^2 = \frac{1}{2} \sqrt{x^2+y^2} +x$
$\left( 2\left( x^2 + y^2 -x \right) \right) ^2= \sqrt{x^2+y^2} ^2$
At this point, S employed some CAS power.
[Full disclosure: That final CAS step is actually mine, but it dovetails so nicely with S’s brilliant approach. I am always delightfully surprised when my students return using a tool (technological or mental) I have been promoting but hadn’t seen to apply in a particular situation.]
S had used her CAS to accomplish the translation in a more convenient coordinate system before moving the equation back into polar.
Clearly, $r \ne 0$, so
$4r^3 - 3r = cos(\theta )$ .
In an attachment (included below), S proved an identity she had never seen, $\displaystyle cos(\theta) = 4cos^3 \left( \frac{\theta }{3} \right) - 3cos \left( \frac{\theta }{3} \right)$ , which she now applied to her CAS result.
$\displaystyle 4r^3 - 3r = 4cos^3 \left( \frac{\theta }{3} \right) - 3cos \left( \frac{\theta }{3} \right)$
So, $\displaystyle r = cos \left( \frac{\theta }{3} \right)$
Therefore, $\displaystyle r = cos \left( \frac{\theta }{3} \right)$ is the image of $\displaystyle r = \frac{1}{2} + cos(\theta )$ after translating $\displaystyle \frac{1}{2}$ unit left. QED
Simple. Beautiful.
Obviously, this could have been accomplished using lots of by-hand manipulations. But, in my opinion, that would have been a horrible, potentially error-prone waste of time for a problem that wasn’t concerned at all about whether one knew some Algebra I arithmetic skills. Great job, S!
S’s proof of her identity, $\displaystyle cos(\theta) = 4cos^3 \left( \frac{\theta }{3} \right) - 3cos \left( \frac{\theta }{3} \right)$ :
## Controlling graphs and a free online calculator
When graphing functions with multiple local features, I often find myself wanting to explain a portion of the graph’s behavior independent of the rest of the graph. When I started teaching a couple decades ago, the processor on my TI-81 was slow enough that I could actually watch the pixels light up sequentially. I could see HOW the graph was formed. Today, processors obviously are much faster. I love the problem-solving power that has given my students and me, but I’ve sometimes missed being able to see function graphs as they develop.
Below, I describe the origins of the graph control idea, how the control works, and then provide examples of polynomials with multiple roots, rational functions with multiple intercepts and/or vertical asymptotes, polar functions, parametric collision modeling, and graphing derivatives of given curves.
BACKGROUND: A colleague and I were planning a rational function unit after school last week wanting to be able to create graphs in pieces so that we could discuss the effect of each local feature. In the past, we “rigged” calculator images by graphing the functions parametrically and controlling the input values of t. Clunky and static, but it gave us useful still shots. Nice enough, but we really wanted something dynamic. Because we had the use of sliders on our TI-nSpire software, on Geogebra, and on the Desmos calculator, the solution we sought was closer than we suspected.
REALIZATION & WHY IT WORKS: Last week, we discovered that we could use $g(x)=\sqrt \frac{\left | x \right |}{x}$ to create what we wanted. The argument of the root is 1 for $x<0$, making $g(x)=1$. For $x>0$, the root’s argument is -1, making $g(x)=i$, a non-real number. Our insight was that multiplying any function $y=f(x)$ by an appropriate version of g wouldn’t change the output of f if the input to g is positive, but would make the product ungraphable due to complex values if the input to g is negative.
If I make a slider for parameter a, then $g_2(x)=\sqrt \frac{\left | a-x \right |}{a-x}$ will have output 1 for all $x. That means for any function $y=f(x)$ with real outputs only, $y=f(x)\cdot g_2(x)$ will have real outputs (and a real graph) for $x only. Aha! Using a slider and $g_2$ would allow me to control the appearance of my graph from left to right.
NOTE: While it’s still developing, I’ve become a big fan of the free online Desmos calculator after a recent presentation at the Global Math Department (join our 45-60 minute online meetings every Tuesday at 9PM ET!). I use Desmos for all of the following graphs in this post, but obviously any graphing software with slider capabilities would do.
EXAMPLE 1: Graph $y=(x+2)^3x^2(x-1)$, a 6th degree polynomial whose end behavior is up for $\pm \infty$, “wiggles” through the x-axis at -2, then bounces off the origin, and finally passes through the x-axis at 1.
Click here to access the Desmos graph that created the image above. You can then manipulate the slider to watch the graph wiggle through, then bounce off, and finally pass through the x-axis.
EXAMPLE 2: Graph $y=\frac{(x+1)^2}{(x+2)(x-1)^2}$, a 6th degree polynomial whose end behavior is up for $\pm \infty$, “wiggles” through the x-axis at -2, then bounces off the origin, and finally passes through the x-axis at 1.
Click here to access the Desmos graph above and control the creation of the rational function’s graph using a slider.
EXAMPLE 3: I believe students understand polar graphing better when they see curves like the limacon $r=2+3cos(\theta )$ moving between its maximum and minimum circles. Controlling the slider also allows users to see the values of $\theta$ at which the limacon crosses the pole. Here is the Desmos graph for the graph below.
EXAMPLE 4: Object A leaves (2,3) and travels south at 0.29 units/second. Object B leaves (-2,1) traveling east at 0.45 units/second. The intersection of their paths is (2,1), but which object arrives there first? Here is the live version.
OK, I know this is an overly simplistic example, but you’ll get the idea of how the controlling slider works on a parametrically-defined function. The $latex \sqrt{\frac{\left | a-x \right |}{a-x}}$ term only needs to be on one of parametric equations. Another benefit of the slider approach is the ease with which users can identify the value of t (or time) when each particle reaches the point of intersection or their axes intercepts. Obviously those values could be algebraically determined in this problem, but that isn’t always true, and this graphical-numeric approach always gives an alternative to algebraic techniques when investigating parametric functions.
ASIDE 1–Notice the ease of the Desmos notation for parametric graphs. Enter [r,s] where r is the x-component of the parametric function and s is the y-component. To graph a point, leave r and s as constants. Easy.
EXAMPLE 5: When teaching calculus, I always ask my students to sketch graphs of the derivatives of functions given in graphical forms. I always create these graphs one part at a time. As an example, this graph shows $y=x^3+2x^2$ and allows you to get its derivative gradually using a slider.
ASIDE 2–It is also very easy to enter derivatives of functions in the Desmos calculator. Type “d/dx” before the function name or definition, and the derivative is accomplished. Desmos is not a CAS, so I’m sure the software is computing derivatives numerically. No matter. Derivatives are easy to define and use here.
I’m hoping you find this technology tip as useful as I do.
## Trig Identities with a Purpose
Yesterday, I was thinking about some changes I could introduce to a unit on polar functions. Realizing that almost all of the polar functions traditionally explored in precalculus courses have graphs that are complete over the interval $0\le\theta\le 2\pi$, I wondered if there were any interesting curves that took more than $2\pi$ units to graph.
My first attempt was $r=cos\left(\frac{\theta}{2}\right)$ which produced something like a merged double limaçon with loops over its $4\pi$ period.
Trying for more of the same, I graphed $r=cos\left(\frac{\theta}{3}\right)$ guessing (without really thinking about it) that I’d get more loops. I didn’t get what I expected at all.
Wow! That looks exactly like the image of a standard limaçon with a loop under a translation left of 0.5 units.
Further exploration confirms that $r=cos\left(\frac{\theta}{3}\right)$ completes its graph in $3\pi$ units while $r=\frac{1}{2}+cos\left(\theta\right)$ requires $2\pi$ units.
As you know, in mathematics, it is never enough to claim things look the same; proof is required. The acute challenge in this case is that two polar curves (based on angle rotations) appear to be separated by a horizontal translation (a rectangular displacement). I’m not aware of any clean, general way to apply a rectangular transformation to a polar graph or a rotational transformation to a Cartesian graph. But what I can do is rewrite the polar equations into a parametric form and translate from there.
For $0\le\theta\le 3\pi$ , $r=cos\left(\frac{\theta}{3}\right)$ becomes $\begin{array}{lcl} x_1 &= &cos\left(\frac{\theta}{3}\right)\cdot cos\left (\theta\right) \\ y_1 &= &cos\left(\frac{\theta}{3}\right)\cdot sin\left (\theta\right) \end{array}$ . Sliding this $\frac{1}{2}$ a unit to the right makes the parametric equations $\begin{array}{lcl} x_2 &= &\frac{1}{2}+cos\left(\frac{\theta}{3}\right)\cdot cos\left (\theta\right) \\ y_2 &= &cos\left(\frac{\theta}{3}\right)\cdot sin\left (\theta\right) \end{array}$ .
This should align with the standard limaçon, $r=\frac{1}{2}+cos\left(\theta\right)$ , whose parametric equations for $0\le\theta\le 2\pi$ are $\begin{array}{lcl} x_3 &= &\left(\frac{1}{2}+cos\left(\theta\right)\right)\cdot cos\left (\theta\right) \\ y_3 &= &\left(\frac{1}{2}+cos\left(\theta\right)\right)\cdot sin\left (\theta\right) \end{array}$ .
The only problem that remains for comparing $(x_2,y_2)$ and $(x_3,y_3)$ is that their domains are different, but a parameter shift can handle that.
If $0\le\beta\le 3\pi$ , then $(x_2,y_2)$ becomes $\begin{array}{lcl} x_4 &= &\frac{1}{2}+cos\left(\frac{\beta}{3}\right)\cdot cos\left (\beta\right) \\ y_4 &= &cos\left(\frac{\beta}{3}\right)\cdot sin\left (\beta\right) \end{array}$ and $(x_3,y_3)$ becomes $\begin{array}{lcl} x_5 &= &\left(\frac{1}{2}+cos\left(\frac{2\beta}{3}\right)\right)\cdot cos\left (\frac{2\beta}{3}\right) \\ y_5 &= &\left(\frac{1}{2}+cos\left(\frac{2\beta}{3}\right)\right)\cdot sin\left (\frac{2\beta}{3}\right) \end{array}$ .
Now that the translation has been applied and both functions operate over the same domain, the two functions must be identical iff $x_4 = x_5$ and $y_4 = y_5$ . It’s time to prove those trig identities!
Before blindly manipulating the equations, I take some time to develop some strategy. I notice that the $(x_5, y_5)$ equations contain only one type of angle–double angles of the form $2\cdot\frac{\beta}{3}$ –while the $(x_4, y_4)$ equations contain angles of two different types, $\beta$ and $\frac{\beta}{3}$ . It is generally easier to work with a single type of angle, so my strategy is going to be to turn everything into trig functions of double angles of the form $2\cdot\frac{\beta}{3}$ .
$\displaystyle \begin{array}{lcl} x_4 &= &\frac{1}{2}+cos\left(\frac{\beta}{3}\right)\cdot cos\left (\beta\right) \\ &= &\frac{1}{2}+cos\left(\frac{\beta}{3}\right)\cdot cos\left (\frac{\beta}{3}+\frac{2\beta}{3} \right) \\ &= &\frac{1}{2}+cos\left(\frac{\beta}{3}\right)\cdot\left( cos\left(\frac{\beta}{3}\right) cos\left(\frac{2\beta}{3}\right)-sin\left(\frac{\beta}{3}\right) sin\left(\frac{2\beta}{3}\right)\right) \\ &= &\frac{1}{2}+\left[cos^2\left(\frac{\beta}{3}\right)\right] cos\left(\frac{2\beta}{3}\right)-\frac{1}{2}\cdot 2cos\left(\frac{\beta}{3}\right) sin\left(\frac{\beta}{3}\right) sin\left(\frac{2\beta}{3}\right) \\ &= &\frac{1}{2}+\left[\frac{1+cos\left(2\frac{\beta}{3}\right)}{2}\right] cos\left(\frac{2\beta}{3}\right)-\frac{1}{2}\cdot sin^2\left(\frac{2\beta}{3}\right) \\ &= &\frac{1}{2}+\frac{1}{2}cos\left(\frac{2\beta}{3}\right)+\frac{1}{2} cos^2\left(\frac{2\beta}{3}\right)-\frac{1}{2} \left( 1-cos^2\left(\frac{2\beta}{3}\right)\right) \\ &= & \frac{1}{2}cos\left(\frac{2\beta}{3}\right) + cos^2\left(\frac{2\beta}{3}\right) \\ &= & \left(\frac{1}{2}+cos\left(\frac{2\beta}{3}\right)\right)\cdot cos\left(\frac{2\beta}{3}\right) = x_5 \end{array}$
Proving that the x expressions are equivalent. Now for the ys
$\displaystyle \begin{array}{lcl} y_4 &= & cos\left(\frac{\beta}{3}\right)\cdot sin\left(\beta\right) \\ &= & cos\left(\frac{\beta}{3}\right)\cdot sin\left(\frac{\beta}{3}+\frac{2\beta}{3} \right) \\ &= & cos\left(\frac{\beta}{3}\right)\cdot\left( sin\left(\frac{\beta}{3}\right) cos\left(\frac{2\beta}{3}\right)+cos\left(\frac{\beta}{3}\right) sin\left(\frac{2\beta}{3}\right)\right) \\ &= & \frac{1}{2}\cdot 2cos\left(\frac{\beta}{3}\right) sin\left(\frac{\beta}{3}\right) cos\left(\frac{2\beta}{3}\right)+\left[cos^2 \left(\frac{\beta}{3}\right)\right] sin\left(\frac{2\beta}{3}\right) \\ &= & \frac{1}{2}sin\left(2\frac{\beta}{3}\right) cos\left(\frac{2\beta}{3}\right)+\left[\frac{1+cos \left(2\frac{\beta}{3}\right)}{2}\right] sin\left(\frac{2\beta}{3}\right) \\ &= & \left(\frac{1}{2}+cos\left(\frac{2\beta}{3}\right)\right)\cdot sin\left (\frac{2\beta}{3}\right) = y_5 \end{array}$
Therefore the graph of $r=cos\left(\frac{\theta}{3}\right)$ is exactly the graph of $r=\frac{1}{2}+cos\left(\theta\right)$ slid $\frac{1}{2}$ unit left. Nice.
If there are any students reading this, know that it took a few iterations to come up with the versions of the identities proved above. Remember that published mathematics is almost always cleaner and more concise than the effort it took to create it. One of the early steps I took used the substitution $\gamma =\frac{\beta}{3}$ to clean up the appearance of the algebra. In the final proof, I decided that the 2 extra lines of proof to substitute in and then back out were not needed. I also meandered down a couple unnecessarily long paths that I was able to trim in the proof I presented above.
Despite these changes, my proof still feels cumbersome and inelegant to me. From one perspective–Who cares? I proved what I set out to prove. On the other hand, I’d love to know if someone has a more elegant way to establish this connection. There is always room to learn more. Commentary welcome.
In the end, it’s nice to know these two polar curves are identical. It pays to keep one’s eyes eternally open for unexpected connections!
## Polar Graphing Surprise
Nurfatimah Merchant and I were playing around with polar graphs, trying to find something that would stretch students beyond simple circles and types of limacons while still being within the conceptual reach of those who had just been introduced to polar coordinates roughly two weeks earlier.
We remembered that Cartesian graphs of trigonometric functions are much more “interesting” with different center lines. That is, the graph of $y=cos(x)+3$ is nothing more than a standard cosine graph oscillating around $y=3$.
Likewise, the graph of $y=cos(x)+0.5x$ is a standard cosine graph oscillating around $y=0.5x$.
We teach polar graphing the same way. To graph $r=3+cos(2\theta )$, we encourage our students to “read” the function as a cosine curve of period $\pi$ oscillating around the polar function $r=3$. Because of its period, this curve will complete a cycle in $0\le\theta\le\pi$. The graph begins this interval at $\theta =0$ (the positive x-axis) with a cosine graph 1 unit “above” $r=3$, moving to 1 unit “below” the “center line” at $\theta =\frac{\pi}{2}$, and returning to 1 unit above the center line at $\theta =\pi$. This process repeats for $\pi\le\theta\le 2\pi$.
Our students graph polar curves far more confidently since we began using this approach (and a couple extensions on it) than those we taught earlier in our careers. It has become a matter of understanding what functions do and how they interact with each other and almost nothing to do with memorizing particular curve types.
So, now that our students are confidently able to graph polar curves like $r=3+cos(2\theta )$, we wondered how we could challenge them a bit more. Remembering variable center lines like the Cartesian $y=cos(x)+0.5x$, we wondered what a polar curve with a variable center line would look like. Not knowing where to start, I proposed $r=2+cos(\theta )+sin(\theta)$, thinking I could graph a period $2\pi$ sine curve around the limacon $r=2+cos(\theta )$.
There’s a lot going on here, but in its most simplified version, we thought we would get a curve on the center line at $\theta =0$, 1 unit above at $\theta =\frac{\pi}{2}$, on at $\theta =\pi$, 1 unit below at $\theta =\frac{3\pi}{2}$, and returning to its starting point at $\theta =2\pi$. We had a very rough “by hand” sketch, and were quite surprised by the image we got when we turned to our grapher for confirmation. The oscillation behavior we predicted was certainly there, but there was more! What do you see in the graph of $r=2+cos(\theta )+sin(\theta)$ below?
This looked to us like some version of a cardioid. Given the symmetry of the axis intercepts, we suspected it was rotated $\frac{\pi}{4}$ from the x-axis. An initially x-axis symmetric polar curve rotated $\frac{\pi}{4}$ would contain the term $cos(\theta-\frac{\pi}{4})$ which expands using a trig identity.
$\begin{array}{ccc} cos(\theta-\frac{\pi}{4})&=&cos(\theta )cos(\frac{\pi}{4})+cos(\theta )cos(\frac{\pi}{4}) \\ &=&\frac{1}{\sqrt{2}}(cos(\theta )+sin(\theta )) \end{array}$
Eureka! This identity let us rewrite the original polar equation.
$\begin{array}{ccc} r=2+cos(\theta )+sin(\theta )&=&2+\sqrt{2}\cdot\frac{1}{\sqrt{2}} (cos(\theta )+sin(\theta )) \\ &=&2+\sqrt{2}\cdot cos(\theta -\frac{\pi}{4}) \end{array}$
And this last form says our original polar function is equivalent to $r=2+\sqrt{2}\cdot cos(\theta -\frac{\pi}{4})$, or a $\frac{\pi}{4}$ rotated cosine curve of amplitude $\sqrt{2}$ and period $2\pi$ oscillating around center line $r=2$.
This last image shows a cosine curve starting at $\theta=\frac{\pi}{4}$ beginning $\sqrt{2}$ above the center circle $r=2$, crossing the center circle $\frac{\pi}{2}$ later at $\theta=\frac{3\pi}{4}$, dropping to $\sqrt{2}$ below the center circle at $\theta=\frac{5\pi}{4}$, back to the center circle at $\theta=\frac{7\pi}{4}$ before finally returning to the starting point at $\theta=\frac{9\pi}{4}$. Because the radius is always positive, this also convinced us that this curve is actually a rotated limacon without a loop and not the cardioid that drove our initial investigation.
So, we thought we were departing into some new territory and found ourselves looking back at earlier work from a different angle. What a nice surprise!
One more added observation: We got a little lucky in guessing the angle of rotation, but even if it wasn’t known, it is always possible to compute an angle of rotation (or translation in Cartesian) for a sum of two sinusoids with identical periods. This particular topic is covered in some texts, including Precalculus Transformed. | 2018-02-19T02:18:53 | {
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https://math.stackexchange.com/questions/2317179/from-en-to-ex | # From $e^n$ to $e^x$
Solve for $f: \mathbb{R}\to\mathbb{R}\ \ \$ s.t.
$$f(n)=e^n \ \ \forall n\in\mathbb{N}$$ $$f^{(y)}(x)>0 \ \forall y\in\mathbb{N^*} \ \forall x\in\mathbb R$$
Could you please prove that there exists an unique solution: $f(x)=e^x$?
(Anyway, this problem is not about fractional calculus)
$\mathbb N^*=\{1,2,3...\}, \ \mathbb N=\{0,1,2....\}$
How about try to construct a few functional spaces that intersect at one point?
Try Sard Theorem and Pre image Theorem.
• What do you mean by $f^{(y)}(x)>0~\forall y\in\mathbb R,y\ge1$? Particularly, when $y\notin\mathbb N$. – Simply Beautiful Art Jun 10 '17 at 12:27
• What is $\Bbb N^*$? $\Bbb N$ with zero? – M. Winter Jun 13 '17 at 14:59
• This is a pretty interesting problem. Here, $\mathbb{N}$ isn't that special. What I mean is that for any discrete and countably infinite set $S \subset \mathbb{R}$ one can ask a similar question but instead of $f(n) = e^n ,\ \forall n \in \mathbb{N}$, we have $f(c) = e^c \ ,\forall c \in S$. Indeed, it's possible the unique solution may hold even if $S$ is finite with $|S| \geq 2$, though I'm not sure of that. – MathematicsStudent1122 Jul 11 '17 at 3:34
• The claim holds, provided one can show that $f(q) = e^q$ for $q \ge 0$ rational. – Qeeko Jul 11 '17 at 4:40
• @mathworker21 $f \in C^{ \infty }$ – Red shoes Jul 11 '17 at 23:58
We can show that if $$f^{(n)} \ge 0\qquad{\rm(1)}$$ everywhere, for each integer $n >0$, and $f(x)=e^x$ on at least four points of $\mathbb R$, then $f(x)=e^x$ everywhere. This will make use of Bernstein's theorem. I feel that there must be a more elementary proof though.
I'll start with the simpler case where (1) holds for all $n\ge0$, in which case it is only necessary to suppose that $f(x)=e^x$ at three points. As $\left(-\frac{d}{dx}\right)^nf(-x)=f^{(n)}(-x)\ge0$, by definition $f(-x)$ is completely monotonic. Bernstein's theorem means that we can write $$f(-x)=\int e^{-xy}\,\mu(dy)\qquad {\rm(2)}$$ for some finite measure $\mu$ on $[0,\infty)$ and for all $x > 0$. By inversion of Laplace transforms, $\mu$ is uniquely defined. In fact, the statement (2) applies for all $x\in\mathbb R$ and we can argue as follows -- Applying the same argument to $f(K-x)$, for any fixed real $K$ extends (2) to all $x-K > 0$ and, letting $K$ go to $-\infty$, to all $x\in\mathbb R$. Changing the sign of $x$, for convenience, $$f(x)=\int e^{xy}\,\mu(dy).$$ See also, this answer on MathOverflow (and the comments). Suppose that $\mu$ has nonzero weight outside of the set $\{1\}$. Multiplying by $e^{-x}$ and taking second derivatives $$\left(\frac{d}{dx}\right)^2e^{-x}f(x)=\left(\frac{d}{dx}\right)^2\int e^{x(y-1)}\mu(dy)=\int(y-1)^2e^{x(y-1)}\mu(dy) > 0.$$ This means that $e^{-x}f(x)$ is strictly convex, contradicting the fact that it is equal to 1 at more than two points of $\mathbb R$. So, $\mu$ has zero weight outside of $\{1\}$ and $e^{-x}f(x)=\mu(\{1\})$ is constant.
I'll now return to the case where (1) holds for $n > 0$ and $f(x)=e^x$ on a set $S\subseteq\mathbb R$ of size four. By the mean value theorem, $f^\prime(x)=e^x$ holds for at least one point between any two points of $S$ and, hence, holds for at least three points of $\mathbb R$. So, using the proof above, $f^\prime(x)=e^x$ everywhere. Integrating, $f(x)=e^x+c$ for a constant $c$. Then, in order that $f(x)=e^x$ anywhere, $c$ must be zero.
• Your proof applies to $x > 0$, right? – mathworker21 Jul 12 '17 at 4:12
• It applies to all $x$. Unfortunately, the statement of Bernstein's theorem is for $x > 0$, but the result applies for all $x$ and I tried to give a quick argument why in my proof. – George Lowther Jul 12 '17 at 4:14
• Very Small nitpick: The Bernstein theorem also requires $n=0$ but the question only gives $\mathbb N^*=\{1,2,3...\}$. – i9Fn Jul 12 '17 at 8:51
• Good point, I misread that bit. Increasing the number of points to 4, you should still be able to apply a similar argument to $f^\prime$ to get $f^\prime(x)=e^x$, then deduce the result for $f$. – George Lowther Jul 12 '17 at 9:08
• In Jordan Bell's notes, completely monotone functions are defined to be nonincreasing and have finite limits at $0$ and $\infty$. His proof depends on these properties and only applies for $x \in [0,\infty)$ (his $x$ has the opposite sign from yours). I don't see how making a change of variables so $[0, \infty)$ becomes $[K, \infty)$ and letting $K \to -\infty$ constitutes a proof, particularly in the case when $f$ is unbounded. If your much stronger result could be obtained so easily, I'd expect it would be mentioned. – Keith McClary Jul 14 '17 at 4:45
Perhaps someone can formalize this for me. Because f(n) = e^n some derivative of f(x) must not equal e^x for them to be different functions. If the yth derivative of f(x) > e^x then f(x) > e^x for x > some number unless the y+zth derivative of f(x) < e^x in witch case f(x) < e^x for x > some number. For each derivative where that derivative of f(x) != e^x a higher derivative of f(x) that is > or < e^x must exist to compensate. Each compensation can only make f(x) < or > e^x for x > some number. Thusly f(n) cannot equal e^n for all n unless all derivatives of f(x) equal e^x meaning f(x) = e^x. | 2019-08-22T09:52:15 | {
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http://mathhelpforum.com/calculus/69774-inverse-derivative-help.html | # Math Help - Inverse Derivative Help
1. ## Inverse Derivative Help
I cannot figure out what to do!!
Given the function $f(x)=5 x^3+2 x+5$ Let g be the inverse function of f. i.e. $g(x)=f^{-1}(x).$
$g^{\prime}(12)=$
I can't manage to find g(x) let alone g'(x)
Thank you for any help.
2. Originally Posted by Krooger
I cannot figure out what to do!!
Given the function $f(x)=5 x^3+2 x+5$ Let g be the inverse function of f. i.e. $g(x)=f^{-1}(x).$
$g^{\prime}(12)=$
I can't manage to find g(x) let alone g'(x)
Thank you for any help.
note that $g'(x) = \frac 1{f'(g(x))}$ .....not hard to derive. your textbook should do this
now, by definition, $g(a) = x \implies f(x) = a$
thus, $g(12)$ is the x-value so that $f(x) = 12$
by inspection, this is 1, thus $g(12) = 1$
thus, $g'(12) = \frac 1{f'(1)}$
i leave the rest to you
3. Originally Posted by Jhevon
note that $g'(x) = \frac 1{f'(g(x))}$ .....not hard to derive. your textbook should do this
now, by definition, $g(a) = x \implies f(x) = a$
thus, $g(12)$ is the x-value so that $f(x) = 12$
by inspection, this is 1, thus $g(12) = 1$
thus, $g'(12) = \frac 1{f'(1)}$
i leave the rest to you
$f^{-1}(x) \neq \frac{1}{f(x)}$.
Edit: Never mind, I figured out what you were doing. I worked it out using implicit differentiation, but your way works too.
4. Originally Posted by Prove It
$f^{-1}(x) \neq \frac{1}{f(x)}$.
Edit: Never mind, I figured out what you were doing. I worked it out using implicit differentiation, but your way works too.
The first formula Jhevon gives comes from implicit differentiation of $f(g(x))=x$
5. Originally Posted by Prove It
$f^{-1}(x) \neq \frac{1}{f(x)}$.
that's not the formula i have
Edit: Never mind, I figured out what you were doing. I worked it out using implicit differentiation, but your way works too.
yes, i worked it out that way as well...er, like Moo said
6. I was unaware that:
$
g'(x) = \frac 1{f'(g(x))}$
...and to be honest upon first glace it still dosen't make sense to me haha. When I get a minute I will have to look it up. Thank You all for the help.
7. Originally Posted by Krooger
I was unaware that:
$
g'(x) = \frac 1{f'(g(x))}$
...and to be honest upon first glace it still dosen't make sense to me haha. When I get a minute I will have to look it up. Thank You all for the help.
since $f(x)$ and $g(x)$ are inverse functions,
$f(g(x)) = x$
now differentiate both sides with respect to x (note you will need the chain rule to differentiate the left side) and solve for $g'(x)$ | 2014-08-28T06:19:54 | {
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http://livelovelocks.com/31sb6/is-0-a-complex-number-or-real-number-cf8e2b | Solution for Every real number is also a complex number. And real numbers are numbers where the imaginary part, b = 0 b=0 b = 0. Real and Complex Numbers . Now we can look at some operations with complex numbers. Does harry styles have a private Instagram account? Complex numbers are a mixture of the two, e.g. In the special case where b=0, a+0i=a. Write − a − a as a − 1. a − 1. What is the difference between a complex number and an imaginary number? Of course, this criteria then means that 1 is also both real and complex as the multiplicative identity element, meaning you have to extend this to all numbers, which also makes sense, since the reals are a subfield of complex numbers. If $$a=0$$ and $$b$$ is not equal to 0, the complex number is called an imaginary number. There is no such number when the denominator is zero and the numerator is nonzero. Click hereto get an answer to your question ️ Let z = 1 + ai be a complex number, a > 0 , such that z^3 is a real number. How do I graph the number #4i# in the complex number plane? 6. However, they can be measured from zero on the complex number plane, which includes an x axis (for the real number) and the y axis (for the imaginary number). An imaginary number is a complex number that can be written as a real number multiplied by the imaginary unit i, which is defined by its property i 2 = −1. Think of the complex numbers as points on a coordinate system. Complex numbers actually combine real and imaginary number (a+ib), where a and b denotes real numbers, whereas i denotes an imaginary number. Two complex numbers are said to be equal if and only if their real parts and imaginary parts are separately equal i.e. A) I understand that complex numbers come in the form z= a+ib where a and b are real numbers. In the usual meaning of 'real numbers’, no. On the other hand, some complex numbers are real, some are imaginary, and some are neither. In the meantime, ‘Complex Numbers’ as the name refers a heterogeneous mix. How do I use graphing in the complex plane to add #2+4i# and #5+3i#? In Figure 2, we show the results of various complex number operations. Write a ⋅ i a ⋅ i in simplest form. To avoid such e-mails from students, it is a good idea to define what you want to mean by a complex number under the details and assumption section. we can denote a real number in form of a complex number using imaginary unit -”i”. $1+2\mathrm{i}$ or $7-3\mathrm{i}$, etc. 3. There are different types of real numbers. Similarly, in a complex number, when the imaginary part, i.e., is zero, or when , then the number is said to be purely real. The symbol is often used for the set of complex numbers. These sets have special names. Who is the longest reigning WWE Champion of all time? The absolute value of a number is considered its distance from zero on the number line. Copyright © 2021 Multiply Media, LLC. Subsets of the Real Numbers . A complex number is a number of the form . What floral parts are represented by eyes of pineapple? The answer is same as the answer to-Which is greater- i+j or i-j? So, is a purely imaginary number. Rational numbers are any number that can be expressed as p/q where p and q are integers and q != 0. Either Part Can Be Zero. Yes, 0 is a complex number. Because complex numbers include imaginary numbers, they cannot be plotted on the real number line. If then . The same applies for you, a complex number can be real if its only parts are real, or even imaginary if it only has imaginary parts. a + ib = c + id implies a = c and b = d. However, there is no order relation between complex numbers and the expressions of the type a + ib < (or >) c + id are meaningless. So (0,1) is a number such that its square is equal to (-1,0) which is equivalent to -1. In fact, all real numbers and all imaginary numbers are complex. They can be any of the rational and irrational numbers. In the special case that b = 0 you get pure real numbers which are a subset of complex numbers. In general, a complex number looks like $x+y\mathrm{i}$ where $x$ and $y$ are both real numbers. A complex number is a number of the form a + bi, where a and b are real numbers and i is the principal square root of -1. True or False A real number is a number that can take any value on the number line. x = 5×1 1.3000 -3.5600 8.2300 -5.0000 -0.0100 Find the absolute value of the elements of the vector. Open Live Script. complex number the sum of a real number and an imaginary number, written in the standard form $$a+bi$$ ($$a,b \in \mathbb{R}$$), Therefore, all real numbers are also complex numbers. There are infinitely many rational numbers, but they do not form a continuous line. Every real number is a complex number, but not every complex number is a real number. A complex number might not be a pure imaginary number. Consider again the complex number a + bi. A complex number is a number, but is different from common numbers in many ways.A complex number is made up using two numbers combined together. 0 is a complex number (or rather it belongs to the set of complex numbers) since x + y*i is a complex number even when x = y = 0 X — Input array scalar | vector | matrix | multidimensional array. (2 plus 2 times i) The real numbers...”. How do I graph the complex number #-4+2i# in the complex plane? In a complex number when the real part is zero or when , then the number is said to be purely imaginary. here is called the real part of complex number and is called the imaginary part of complex number.. If a is not equal to 0 and b = 0, the complex number a + 0i = a and a is a real number. True or False Asimov Asimov. So 5, 12.42, -17/3 and 0 are rational numbers. So is a purely real number. around the world. share | cite | improve this answer | follow | answered Aug 9 '14 at 21:34. The complex number zero has zero real part and zero imaginary part: $0+0\mathrm{i}$. The complex number i = 0 + i 1, which has real part zero and imaginary part one, has the property that its square is i 2 = (0 + i 1) 2 = (0 + i 1)(0 + i 1) = (-1) + i 0 = - 1. How do I use graphing in the complex plane to subtract #3+4i# from #-2+2i#? So, a Complex Number has a real part and an imaginary part. Real Number. You can perform addition, subtraction, multiplication, and division for complex numbers … Observation: CAdd can be used with up to 5 arguments; these arguments can be real or complex numbers: e.g. See all questions in Complex Number Plane. When did organ music become associated with baseball? #0# is a complex number (or rather it belongs to the set of complex numbers) since #x + y*i# is a complex number even when #x = y = 0#, 409 views What is this property of numbers called? Radio host fired for sexist tweet about ESPN reporter Beginning Activity. The numbers a and b are known as the real part and imaginary part of the complex number, respectively. Consider now the product of the complex number (0,1) with itself. The numbers a and b are known as the real part and imaginary part of the complex number, respectively. A complex number is a number of the form a + bi, where a and b are real numbers, and i is an indeterminate satisfying i 2 = −1.For example, 2 + 3i is a complex number. Is green skull in the pirate bay is good? Let’s start at the point (1, 0), which is represented by the complex number 1+0i. The proper name for these numbers is complex, as they consist of real and imaginary parts (the one that contains the … But in complex number, we can represent this number (z = a + ib) as a plane. How do I graph the complex number #3+4i# in the complex plane? Figure 2 – Complex number operations . A complex number is the sum of a real number and an imaginary number. Notice that 0 is a real number. A number can be both real and complex? So, too, is $$3+4\sqrt{3}i$$. Every pure imaginary number is a complex number. Conversely, it is imaginary if the real component is zero. Solution for Every real number is also a complex number. As it suggests, ‘Real Numbers’ mean the numbers which are ‘Real’. Thus the square root of -1 is not fictitious or imaginary in the ordinary sense but is simply (0,1). In the sense that they are actual (non-made-up) things, as real as any other sort. Both Imaginary and Real numbers are subset of Complex numbers. Most complex numbers e.g. How much money does The Great American Ball Park make during one game? Thus such ordered pairs with second component zero behave exactly like real numbers. (vectors) To compare imaginary numbers we compare the magnitude and argument. For example, $$5+2i$$ is a complex number. Later, ‘Decimal Numbers’ (2.3, 3.15) and numbers like 5⁄3 (‘Rational Numbers’) … The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply. A real number can store the information about the value of the number and if this number is positive or negative. and are allowed to be any real numbers. All Rights Reserved. A number can be both real and complex. The complex number a + bi can be identified with the point (a, b) in the complex plane. The numbers $x$ and $y$ are called the real and imaginary parts respectively. We start with the real numbers, and we throw in something that’s missing: the square root of . 5. There is no real number that is a complex number. If the denominator is a real number, we can simply divide the real and imaginary parts of the numerator by this value to obtain the result: It is a real number because for example water freezes at 0 degrees Centigrade. A complex number is a number of the form . In this situation, we will let $$r$$ be the magnitude of $$z$$ (that is, the distance from $$z$$ to the origin) and $$\theta$$ the angle $$z$$ makes with the positive real axis as shown in Figure $$\PageIndex{1}$$. How did Rizal overcome frustration in his romance? y = abs(3+4i) y = 5 Input Arguments. A complex number is a number of the form a + bi, where a and b are real numbers and i is the principal square root of -1. (Because the imaginary part is zero, 1+0 i is just another way of writing the real number 1.) Given an imaginary number, express it in standard form. This way, a complex number is defined as a polynomial with real coefficients in the single indeterminate i, for which the relation i 2 + 1 = 0 is imposed. Why don't libraries smell like bookstores? You get a complex number unless the real number happens to be 0 or 1. Well, a Complex Number is just two numbers added together (a Real and an Imaginary Number). If b is not equal to zero and a is any real number, the complex number a + bi is called imaginary number. How old was Ralph macchio in the first Karate Kid? But either part can be 0, so all Real Numbers and Imaginary Numbers are also Complex Numbers. Then the special ‘0’ and the ‘Negative’ numbers were found. The numbers we deal with in the real world (ignoring any units that go along with them, such as dollars, inches, degrees, etc.) The continuous line of numbers is called the real number line. Direct link to jwinder47's post “This is an interesting question. (In fact, the real numbers are a subset of the complex numbers-any real number r can be written as r + 0i, which is a complex representation.) Conversely, it is imaginary if the real component is zero. complex-numbers. It is a real number because for example water freezes at 0 A complex number is a number of the form a + bi, where a and b are real numbers, and i is an indeterminate satisfying i = −1. Definition of a Complex Number For real numbers a and b, the number a + bi is a complex number. $2>1$. Similarly, 3/7 is a rational number but not an integer. What is the analysis of the poem song by nvm gonzalez? Sometimes only special, smaller sets of them are talked about. Negative of Complex Number. To get the real part, use number.real, and to get the imaginary part, use number.imag. 4. How do I graph the complex number #2-3i# in the complex plane? Which vectors define the complex number plane? A complex number is a number of the form a + bi, where a and b are real numbers and i is the principal square root of -1. If your impeached can you run for president again? The … Definition. Theoretically the answer should be - Yes, every real number is also a complex number. A real number is any number that can be placed on a number line that extends to infinity in both the positive and negative directions. They are also the first part of mathematics we learn at schools. y = abs(x) y = 5×1 1.3000 3.5600 8.2300 5.0000 0.0100 Magnitude of Complex Number. Then the sum 1 + z + z^2 + .... + z^11 is equal to: Here are real numbers and (also called iota) is equal to . >>> a (4+3j) >>> a.real 4.0 >>> a.imag 3.0 Conjugate of a Complex Number. Is the value of i^i a Real Number or Complex Number? I read that both real and imaginary numbers are complex numbers so I … collapse all. 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Amity University Cutoff 2020, Syracuse Dorms Ranked, Be Unwell Daily Themed Crossword, Gordon Gin Sundowner, Metallica Tabs Easy, Uss Missouri Location, Gordon Gin Sundowner, Wall Unit Bookcase With Glass Doors, | 2021-07-27T14:18:07 | {
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https://naturegeorge.github.io/eigenblog/posts/markov-chain-exercise/ | Exercise of Markov Chain
Post
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# Exercise of Markov Chain
## Q1
Suppose you repeatedly does toss a fair coin and denote $T$ the first time you get three consecutive heads.
1. Compute $E[T]$
2. Verify your answer in [1] via simulation. You may use any programming language, but you have to attach your code. Specifically, repeta the following experiment for 100,000 times: uses a computer to simulate coin tosses and record the first time that you get three consecutive heads. Report the mean and standard deviation of the 100,000 recorded times.
Hint: define a proper Markov chain with the state space {0, 1, 2, 3}.
### Markov Chain Model
Define State:
1. $\text{State}_{0}$: $0 \qquad\rightarrow$ zero consecutive head
2. $\text{State}_{1}$: $1 \qquad\rightarrow$ one consecutive heads
3. $\text{State}_{2}$: $2 \qquad\rightarrow$ two consecutive heads
4. $\text{State}_{3}$: $3 \qquad\rightarrow$ three consecutive heads
Define State Space $S = {0, 1, 2, 3}$
Define Initial State Probability Distribution:
\begin{aligned} P(\text{State}_{0})&=1, \\ P(\text{State}_{1})&=0, \\ P(\text{State}_{2})&=0, \\ P(\text{State}_{3})&=0 \end{aligned}
Set it as a $4\times1$ matrix $A$:
$A=\begin{bmatrix} 1\\ 0\\ 0\\ 0\\ \end{bmatrix}$
Each row of the matrix denotes its corresponding state.
Define Transition Probability Matrix $Q$:
Since we have:
\begin{aligned} P(\text{State}_{0}|\text{State}_{0}) = 0.5, P(\text{State}_{1}|\text{State}_{0}) = 0.5, P(\text{State}_{2}|\text{State}_{0}) = 0.0, P(\text{State}_{3}|\text{State}_{0}) = 0.0\\ P(\text{State}_{0}|\text{State}_{1}) = 0.5, P(\text{State}_{1}|\text{State}_{1}) = 0.0, P(\text{State}_{2}|\text{State}_{1}) = 0.5, P(\text{State}_{3}|\text{State}_{1}) = 0.0\\ P(\text{State}_{0}|\text{State}_{2}) = 0.5, P(\text{State}_{1}|\text{State}_{2}) = 0.0, P(\text{State}_{2}|\text{State}_{2}) = 0.0, P(\text{State}_{3}|\text{State}_{2}) = 0.5\\ P(\text{State}_{0}|\text{State}_{3}) = 0.0, P(\text{State}_{1}|\text{State}_{3}) = 0.0, P(\text{State}_{2}|\text{State}_{3}) = 0.0, P(\text{State}_{3}|\text{State}_{3}) = 1.0\\ \end{aligned}
shown as Finite State Machine:
then:
$Q=\begin{bmatrix} 0.5&0.5&0.0&0.0\\ 0.5&0.0&0.5&0.0\\ 0.5&0.0&0.0&0.5\\ 0.0&0.0&0.0&1.0\\ \end{bmatrix}$
Set $n$ as the number of tosses, then we have:
$A_{n}=Q^{n}A$
where $A_{n}$ denoted as the state distribution at $n$.
### Compute the Expectation
Set $e$ denoted as the expectation of additional times to toss from a particular state to three consecutive heads. Then we have:
$E=\begin{bmatrix} e_{\text{from 0 to 3}}\\ e_{\text{from 1 to 3}}\\ e_{\text{from 2 to 3}}\\ e_{\text{from 3 to 3}}\\ \end{bmatrix} =\begin{bmatrix} e_{0}\\ e_{1}\\ e_{2}\\ e_{3}\\ \end{bmatrix}$
And Because:
$QE=E-\begin{bmatrix} 1\\ 1\\ 1\\ 0\\ \end{bmatrix}$
So we have:
\begin{aligned} 0.5\cdot(e_0+e_1)&=e_0-1\\ 0.5\cdot(e_0+e_2)&=e_1-1\\ 0.5\cdot(e_0+e_3)&=e_2-1\\ 0&=e_3\\ \end{aligned}
so we get:
\begin{aligned} e_3=0\\ e_2=8\\ e_1=12\\ e_0=14\\ \end{aligned}
then $E[T]=e_0=14$
### Simulation
#### Verify Expectation
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 import numpy as np def coin_flip(p=0.5): count = 1 head = [] while len(head) <3: if np.random.binomial(1, p): if len(head) == 0 or head[-1] + 1 != count: head = [count] else: head.append(count) count += 1 return head[-1] res = np.array([coin_flip() for _ in range(100000)]) print('mean:', res.mean()) print('std:', res.std())
1 2 mean: 14.03031 std: 11.925531070099142
#### Plot Value Distribution of T
Dist Plot Box Plot
#### Plot Probability Matrix
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 import numpy as np from pandas import DataFrame import seaborn as sns import matplotlib.pyplot as plt plt.style.use('ggplot') Q = np.array([ [0.5, 0.5, 0, 0], [0.5, 0, 0.5, 0], [0.5, 0, 0, 0.5], [0, 0, 0, 1] ]).T V = np.array([1,0,0,0]) def pipe(num): res_lyst = [np.matmul(Q, V)] for _ in range(num): res_lyst.append(np.matmul(Q, res_lyst[-1])) return np.array(res_lyst) res = pipe(100) df = DataFrame(res, columns=['P(State_%s)' % i for i in range(4)]) ax = df.plot() ax.set_xlabel("number of tosses") ax.set_ylabel("probability") ax = sns.heatmap(res.T, cmap='viridis') ax.set_xlabel("number of tosses") ax.set_ylabel("probability")
Line Plot Heatmap | 2022-05-21T19:22:00 | {
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https://colbrydi.github.io/MatrixAlgebra/09--Determinants_pre-class-assignment.html | # 09 Pre-Class Assignment: Determinants¶
## 1. Introduction to Determinants¶
For a detailed overview of determinants I would recommend reviewing Chapter D pg 340-366 of the Beezer text.
The determinant is a function that takes a ($$n \times n$$) square matrix as an input and produces a scalar as an output. Determinants have been studied quite extensively and have many interesting properties. However, determinants are “computationally expensive” as the size of your matrix ($$n$$) gets bigger. This limitation makes them impractical for many real world problems.
The determinant of a $$2 \times 2$$ matrix can be calculated as follows:
$\begin{split} det \left( \left[ \begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix} \right] \right) = a_{11}a_{22} - a_{12}a_{21} \end{split}$
QUESTION: Calculate the determinant of the following matrix by hand:
$\begin{split} \left[ \begin{matrix} 3 & -2 \\ 1 & 2 \end{matrix} \right] \end{split}$
Calculating the determinant of a larger matrix is a “recursive” problem which involves combining the determinants of smaller and smaller sub-matrices until you have a $$2 \times 2$$ matrix which is then calculated using the above formula. Here is some Pseudocode to calculate a determinant. To simplify the example the code assumes there is a matrix function deleterow which will remove the $$x$$th row from a matrix (always the first row in this example) and deletecol will remove the $$x$$th column from a matrix. When used together (as shown below) they will take an $$n \times n$$ matrix and turn it into a $$(n-1) \times (n-1)$$ matrix.
function determinant(A, n)
det = 0
if (n == 1)
det = matrix[1,1]
else if (n == 2)
det = matrix[1,1] * matrix[2,2] - matrix[1,2] * matrix[2,1]
else
for x from 1 to n
submatrix = deleterow(matrix, 1)
submatrix = deletecol(submatrix, x)
det = det + (x+1)**(-1) * matrix[1,x] * determinant(submatrix, n-1)
next x
endif
return det
Notice that the combination of the determinants of the submatrices is not a simple sum. The combination is adding the submatrices corresponding to the odd columns (1,3,5, etc) and subtracting the submatrices corresponding to the even columns (2,4,6, etc.). This may become clearer if we look at a simple $$3 \times 3$$ example (Let $$|A|$$ be a simplified syntax for writing the determinant of $$A$$):
$\begin{split} A = \left[ \begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{matrix} \right] \end{split}$
$\begin{split} |A|= a_{11} \left| \begin{matrix} \square & \square & \square \\ \square & a_{22} & a_{23} \\ \square & a_{32} & a_{33} \end{matrix} \right| - a_{12}\left| \begin{matrix} \square & \square & \square \\ a_{21} & \square & a_{23} \\ a_{31} & \square & a_{33} \end{matrix} \right| + a_{13} \left| \begin{matrix} \square & \square & \square \\ a_{21} & a_{22} & \square \\ a_{31} & a_{32} & \square \end{matrix} \right| \end{split}$
$\begin{split} |A| = a_{11}\left| \begin{matrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{matrix} \right| - a_{12}\left| \begin{matrix} a_{21} & a_{23} \\ a_{31} & a_{33} \end{matrix} \right| + a_{13} \left| \begin{matrix} a_{21} & a_{22} \\ a_{31} & a_{32} \end{matrix} \right| \end{split}$
$|A| = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31})$
QUESTION: Calculate the determinant of the following matrix by hand:
$\begin{split} \left[ \begin{matrix} 1 & 2 & -3 \\ 5 & 0 & 6 \\ 7 & 1 & -4 \end{matrix} \right] \end{split}$
QUESTION: Use the numpy.linalg library to calculate the determinant of the following matrix and stor the value in a variable called det
$\begin{split} \left[ \begin{matrix} 2 & 0 & 1 & -5 \\ 8 & -1 & 2 & 1 \\ 4 & -3 & -5 & 0 \\ 1 & 4 & 8 & 2 \end{matrix} \right] \end{split}$
#Put your answer here
from answercheck import checkanswer
---------------------------------------------------------------------------
ModuleNotFoundError Traceback (most recent call last)
<ipython-input-2-b2a2502e4bdf> in <module>
2
1 import hashlib
2 import numpy as np
----> 3 import sympy as sym
4 import sys
5 import textwrap
ModuleNotFoundError: No module named 'sympy'
## 2. Properties of Determinants¶
The following are some helpful properties when working with determinants. These properties are often used in proofs and can sometimes be utilized to make faster calculations.
### Row Operations¶
Let $$A$$ be an $$n \times n$$ matrix and $$c$$ be a nonzero scalar. Let $$|A|$$ be a simplified syntax for writing the determinant of $$A$$:
1. If a matrix $$B$$ is obtained from $$A$$ by multiplying a row (column) by $$c$$ then $$|B| = c|A|$$.
2. If a matrix $$B$$ is obtained from $$A$$ by interchanging two rows (columns) then $$|B| = -|A|$$.
3. if a matrix $$B$$ is obtained from $$A$$ by adding a multiple of one row (column) to another row (column), then $$|B| = |A|$$.
### Singular Matrices¶
Definition: A square matrix $$A$$ is said to be singular if $$|A| = 0$$. $$A$$ is non-singular if $$|A| \neq 0$$
Now, Let $$A$$ be an $$n \times n$$ matrix. $$A$$ is singular if any of these is true:
1. all the elements of a row (column) are zero.
2. two rows (columns) are equal.
3. two rows (columns) are proportional. i.e. one row (column) is the same as another row (column) multiplied by $$c$$.
QUESTION: The following matrix is singular because of certain column or row properties. Give the reason:
$\begin{split} \left[ \begin{matrix} 1 & 5 & 5 \\ 0 & -2 & -2 \\ 3 & 1 & 1 \end{matrix} \right] \end{split}$
QUESTION: The following matrix is singular because of certain column or row properties. Give the reason:
$\begin{split} \left[ \begin{matrix} 1 & 0 & 4 \\ 0 & 1 & 9 \\ 0 & 0 & 0 \end{matrix} \right] \end{split}$
### Determinants and Matrix Operations¶
Let $$A$$ and $$B$$ be $$n\times n$$ matrices and $$c$$ be a non-zero scalar.
1. Determinant of a scalar multiple: $$|cA| = c^n|A|$$
2. Determinant of a product: $$|AB| = |A||B|$$
3. Determinant of a transpose” $$|A^t| = |A|$$
4. Determinant of an inverse: $$|A^{-1}| = \frac{1}{|A|}$$ (Assuming $$A^{-1}$$ exists)
QUESTION: If $$A$$ is a $$3\times 3$$ matrix with $$|A| = 3$$, use the properties of determinants to compute the following determinant:
$|2A|$
QUESTION: If $$A$$ is a $$3\times 3$$ matrix with $$|A| = 3$$, use the properties of determinants to compute the following determinant: $$$|A^2|$$$
QUESTION: if $$A$$ and $$B$$ are $$3\times 3$$ matrices and $$|A| = -3, |B|=2$$, compute the following determinant:
$|AB|$
QUESTION: if $$A$$ and $$B$$ are $$3\times 3$$ matrices and $$|A| = -3, |B|=2$$, compute the following determinant:
$|2AB^{-1}|$
### Triangular matrices¶
Definition: An upper triangular matrix has nonzero elements lie on or above the main diagonal and zero elements below the main diagonal. For example:
$\begin{split} A = \left[ \begin{matrix} 2 & -1 & 9 & 4 \\ 0 & 3 & 0 & 6 \\ 0 & 0 & -5 & 3 \\ 0 & 0 & 0 & 1 \end{matrix} \right] \end{split}$
The determinant of an upper triangle matrix $$A$$ is the product of the diagonal elements of the matrix $$A$$.
Also, since the Determinant is the same for a matrix and it’s transpose (i.e. $$|A^t|$$ = |A|, see definition above) the determinant of a lower triangle matrix is also the product of the diagonal elements.
QUESTION: What is the determinant of matrix $$A$$?
### Using Properties of determinants:¶
Here is a great video showing how you can use the properties of determinants:
from IPython.display import YouTubeVideo
QUESTION (A challenging one): Using the pattern established in the video can you calculate the determinate of the following matrix?
$\begin{split} \left[ \begin{matrix} 1 & a & a^2 & a^3 \\ 1 & b & b^2 & b^3 \\ 1 & c & c^2 & c^3 \\ 1 & d & d^2 & d^3 \end{matrix} \right] \end{split}$
## 3. One interpretation of determinants¶
The following is an application of determinants. Watch this!
from IPython.display import YouTubeVideo
For fun, we will recreate some of the video’s visualizations in Python. It was a little tricky to get the aspect ratios correct but here is some code I managed to get it work.
%matplotlib inline
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
from mpl_toolkits.mplot3d.art3d import Poly3DCollection, Line3DCollection
import numpy as npimport sympy as sym
# Lets define somme points that form a Unit Cube
points = np.array([[0, 0, 0],
[1, 0, 0 ],
[1, 1, 0],
[0, 1, 0],
[0, 0, 1],
[1, 0, 1 ],
[1, 1, 1],
[0, 1, 1]])
points = np.matrix(points)
#Here is some code to build cube from https://stackoverflow.com/questions/44881885/python-draw-3d-cube
def plot3dcube(Z):
if type(Z) == np.matrix:
Z = np.asarray(Z)
fig = plt.figure()
r = [-1,1]
X, Y = np.meshgrid(r, r)
# plot vertices
ax.scatter3D(Z[:, 0], Z[:, 1], Z[:, 2])
# list of sides' polygons of figure
verts = [[Z[0],Z[1],Z[2],Z[3]],
[Z[4],Z[5],Z[6],Z[7]],
[Z[0],Z[1],Z[5],Z[4]],
[Z[2],Z[3],Z[7],Z[6]],
[Z[1],Z[2],Z[6],Z[5]],
[Z[4],Z[7],Z[3],Z[0]],
[Z[2],Z[3],Z[7],Z[6]]]
#alpha transparency was't working found fix here:
# https://stackoverflow.com/questions/23403293/3d-surface-not-transparent-inspite-of-setting-alpha
# plot sides
facecolors=(0,0,1,0.25), linewidths=1, edgecolors='r'))
ax.set_xlabel('X')
ax.set_ylabel('Y')
ax.set_zlabel('Z')
## Weird trick to get the axpect ratio to work.
## From https://stackoverflow.com/questions/13685386/matplotlib-equal-unit-length-with-equal-aspect-ratio-z-axis-is-not-equal-to
mx = np.amax(Z, axis=0)
mn = np.amin(Z, axis=0)
max_range = mx-mn
# Create cubic bounding box to simulate equal aspect ratio
Xb = 0.5*max_range.max()*np.mgrid[-1:2:2,-1:2:2,-1:2:2][0].flatten() + 0.5*(max_range[0])
Yb = 0.5*max_range.max()*np.mgrid[-1:2:2,-1:2:2,-1:2:2][1].flatten() + 0.5*(max_range[1])
Zb = 0.5*max_range.max()*np.mgrid[-1:2:2,-1:2:2,-1:2:2][2].flatten() + 0.5*(max_range[2])
# Comment or uncomment following both lines to test the fake bounding box:
for xb, yb, zb in zip(Xb, Yb, Zb):
ax.plot([xb], [yb], [zb], 'w')
plt.show()
plot3dcube(points)
QUESTION: The following the $$3\times 3$$ was shown in the video (around 6’50’’). Apply this matrix to the unit cube and use the plot3dcube to show the resulting transformed points.
T = np.matrix([[1 , 0 , 0.5],
[0.5 ,1 ,1.5],
[1 , 0 , 1]])
#Put the answer to the above question here.
QUESTION: The determinant represents how the area changes when applying a $$2 \times 2$$ transform. What does the determinant represent for a $$3 \times 3$$ transform?
## 4. Cramer’s Rule¶
DO THIS: Watch the following video and come to class ready to discuss Cramer’s Rule:
from IPython.display import YouTubeVideo
## 5. Assignment wrap-up¶
Assignment-Specific QUESTION: What does the determinant represent for a $$3 \times 3$$ transform?
QUESTION: Summarize what you did in this assignment.
QUESTION: What questions do you have, if any, about any of the topics discussed in this assignment after working through the jupyter notebook?
QUESTION: How well do you feel this assignment helped you to achieve a better understanding of the above mentioned topic(s)?
QUESTION: What was the most challenging part of this assignment for you?
QUESTION: What was the least challenging part of this assignment for you?
QUESTION: What kind of additional questions or support, if any, do you feel you need to have a better understanding of the content in this assignment? | 2021-12-07T18:41:54 | {
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https://math.stackexchange.com/questions/1666293/expressing-8-frac13-in-polar-form | # Expressing $(-8)^{\frac13}$ in polar form
I want to express $(-8)^{\frac{1}{3}}$ in polar and cartesian coordinates.
What I did was to solve the equation $-8 = r^3e^{3i\theta}= r^3(\cos(3\theta)+i\sin(3\theta))$ which implies that I must solve the equations $$r^3\cos(3\theta) = -8$$ and the equation $$r^3\sin(3\theta)= 0$$ The latter gives me $\theta = 0, \frac{\pi}{3}, \frac{2\pi}{3},\dotsc$ and so using $\theta = 0$, using the equation $r^3\cos(3\theta) = -8$, I get $r = -2$ and in this case the polar and cartesian form is $-2$, then using $\theta = \frac{\pi}{3}.$ I get $r =2$ and so the polar form is $2e^{\frac{\pi}{3}}$ and the cartesian form is $1+i \sqrt{3}$ and finally using $\theta = \frac{2\pi}{3}$ I get $r=-2$ and in polar form the answer is $-2e^{\frac{2\pi}{3}}$ and in cartesian coordinates the answer is $-1 + i \sqrt{3}$. However, the answers in my book are different mostly by signs, like the first answer is $2$ instead of $-2$. Can anyone explain why I am getting this sign errors?
• Note that $(-1+i\sqrt{3})^3=8$. – egreg Feb 21 '16 at 23:24
You're using $r=-2$, which is incorrect: the polar form is $re^{i\theta}$, with $r>0$.
You should write $$-8=r^3(\cos3\theta+i\sin3\theta)$$ so $r=2$ and $$\cos3\theta+i\sin3\theta=-1$$ This gives $$3\theta=\pi+2k\pi$$ or $$\theta=\frac{\pi}{3}+\frac{2}{3}k\pi$$ and the principal arguments are $$\frac{\pi}{3},\quad \pi,\quad \frac{5\pi}{3}$$ so the solutions are $$2e^{i\pi/3}=1+i\sqrt{3},\quad 2e^{i\pi}=-2,\quad 2e^{5\pi/3}=1-i\sqrt{3}$$
Your work all seems correct, your book may have a typo. $2$ is definitely NOT a third root of $-8$, as $2^3 = 8$. | 2019-10-17T18:13:43 | {
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https://math.stackexchange.com/questions/3836314/1991-imo-shortlist-problem-11 | # 1991 IMO shortlist problem $\#11$
Prove that $$\sum_{k=0}^{995} \frac{(-1)^k}{1991-k} {1991-k\choose k} = \frac{1}{1991}$$
As usual there isn't anything special about the number $$1991$$.Problem appears to hold for any odd numbers I have checked. I want to prove the general equation. We can manipulate expression and simplify a bit. Then the problem reduces to showing that $$\sum_{k=1}^{n} \frac{(-1)^k}{2n-2k+1} {2n-k\choose k} = 0$$ for some positive integer $$n$$. This is the equation I had been working on but it wasn't that fruitful.
I gave up and saw the solution on Aops but it was not a elementary one. Here is the link if any one wants to see it "https://artofproblemsolving.com/community/c6h34892p216919" ( There is another interesting thing about this link, that the last six digits form a prime number!! $$216919$$ ).In this link the solution poster says that the solution he had written isn't the solution the creators assumed the students to write. So what might be the solution that creators might have expected the students to write?
• What do you think about the $995$? Have you played with that number at all? – Michael Morrow Sep 22 '20 at 18:25
• Its $\lfloor \frac {1991}{2} \rfloor$. I Have taken it into account to come up with generalised statement. – Mathematical Curiosity Sep 22 '20 at 18:28
• Gotcha. Yeah, it looks like there should be a way to relate this to a counting problem.. – Michael Morrow Sep 22 '20 at 18:30
• Yes, that's most likely the case. But coming up with a combinatorial proof is very hard. Do you have any ideas? – Mathematical Curiosity Sep 22 '20 at 18:33
• Alternative signs suggest PIE. – cosmo5 Sep 22 '20 at 18:43
For such problems (esp when you notice that there is a general pattern), some ideas are to find a recurrence relation, create something telescoping (or treat it as a generating function).
We'd use these ideas here.
Notice that $$\left(\frac{1}{n-m} - \frac{1}{n}\right) { n - m \choose m } = \frac{m}{ n (n-m) } { n - m \choose m } = \frac{1}{n} {n-m-1 \choose m-1}$$, or that
$$\frac{ 1 } { n-m } { n-m \choose m } = \frac{1}{n} \left[ { n - m \choose m } + { n - m - 1 \choose m- 1 } \right].$$
This is a good substitution, as it gets rid of the pesky $$\frac{1}{n-k}$$ which makes recurrence hard, and also gives us a $$\frac{1}{1991}$$ on the RHS.
Thus, the goal is to determine $$\sum_{k=0}^{995 } (-1)^k \left[ {1991-k\choose k} + { 1991 - k - 1 \choose k - 1 } \right]$$. (We will show that it equals to 1, and thus the desired sum is $$\frac{1}{1991}.$$)
Let $$S_n = \sum_{k=0}^{ \lfloor \frac{n}{2} \rfloor} (-1)^k { n-k \choose k }$$.
Notice that $${n-k \choose k } = { n-k - 1 \choose k } + { n-k - 1 \choose k - 1 }$$, so
$$S_n = \sum_{k=0}^{\lfloor \frac{n+1}{2} \rfloor} (-1)^k { n - k + 1 \choose k } \\ = \sum_{k=0}^{\lfloor \frac{n+1}{2} \rfloor} (-1)^k \left[ {n-k \choose k } + {n-k \choose k - 1 } \right] \\ = \sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} (-1)^k {n-k \choose k } + \sum_{k=0}^{\lfloor \frac{n-1}{2} \rfloor} (-1)^k { n-k \choose k } \\ = S_{n} - S_{n-1}.$$
(Take care in checking the indices, and remember those $${n \choose m } = 0$$ when $$m > n$$.)
Using this recurrence relation, and calculating some initial values, we get $$S_n = 1 , 0, -1, -1, 0, 1, 1, 0, -1, \ldots$$, which has period 6.
We thus want to determine $$S_{1991} - S_{1990} = 0 - (-1) = 1$$.
Notes
1. I do wish there was a combinatorial argument here. For example, $$S_n$$ has an immediate interpretation as the difference between the even and odd permutations $$p$$ such that $$|p(i) - i | \leq 1$$. (IE Out of the first $$n$$ integers, there are $${n-k \choose k }$$ ways to pick k pairs of consecutive integers (for a total of 2k). The perumatation which switches these pairs and keep the rest fixed has parity $$k$$.) However, I don't see an obvious way to show that this difference is $$1, 0, -1, -1, 0, 1, \ldots$$.
2. WhatsUp's conclusion that about the value of $$s_n$$ also follows from the above.
If you know generating functions, then here is a solution:
Let $$s_n$$ denote the sum $$\sum_{k \geq 0} \frac{(-1)^k}{n - k}\binom{n - k}k$$ and let $$S(X)$$ be the formal power series $$S(X) = \sum_{n \geq 1} s_n X^n$$.
We compute:
$$\begin{eqnarray} S(X) &=& \sum_{n \geq 1} \frac 1 n X^n + \sum_{n \geq 1}\sum_{k \geq 1} \frac{(-1)^k}{n - k}\binom{n - k}k X^n\\ &=& -\log(1 - X) + \sum_{k \geq 1}\sum_{n \geq 2k}\frac{(-1)^k}k \binom{n - k - 1}{k - 1}X^n\\ &=& -\log(1 - X) + \sum_{k \geq 1}\frac{(-1)^k}k X^{2k}\sum_{n \geq 0}\binom{n + k - 1}{k - 1}X^n\\ &=& -\log(1 - X) - \sum_{k \geq 1}\frac{ (-1)^{k - 1}} k \left(\frac{X^2}{1 - X}\right)^k\\ &=& -\log(1 - X) - \log\left(1 + \frac{X^2}{1 - X}\right)\\ &=& -\log(1 - X + X^2)\\ &=& -\log(1 - \omega X) - \log(1 - \overline\omega X)\\ &=& \sum_{n \geq 1}\frac{\omega^n + \overline\omega^n}n X^n, \end{eqnarray}$$ where $$\omega = \frac{1 + \sqrt{-3}}2$$ is a primitive sixth root of unity.
Thus we have $$s_n = \frac 1 n \cdot 2 \operatorname{Re}(\omega^n)$$.
Now $$\omega^n$$ only depends on $$n \mod 6$$. Therefore: $$s_n = \begin{cases} \frac 2 n, & n \equiv 0\mod 6;\\ \frac 1 n, & n \equiv 1, 5\mod 6;\\ \frac {-1} n, & n \equiv 2, 4 \mod 6;\\ \frac{-2} n, & n \equiv 3 \mod 6. \end{cases}$$
And the answer to the original question follows from the fact that $$1991 \equiv 5 \mod 6$$.
• Thanks for your solution. But I don't know much of generating functions. Isn't there any other solutions? – Mathematical Curiosity Sep 23 '20 at 5:58
• I think it's the most natural solution to this problem. Other solutions, if exist, would be complicated, as can be seen from the answer. You may consider learning generating functions, as it's a powerful tool that is accessible to high school students. – WhatsUp Sep 23 '20 at 8:53
• Can you suggest a good source for learning it? – Mathematical Curiosity Sep 23 '20 at 10:34
• I would suggest starting with the wiki pages on (ordinary) generating functions and formal power series, googling introductory materials on the subject, and looking at some motivating examples such as this problem. – WhatsUp Sep 23 '20 at 12:48
• Thank you very much! – Mathematical Curiosity Sep 23 '20 at 14:17 | 2021-04-21T01:50:00 | {
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https://math.stackexchange.com/questions/3169817/give-a-deck-of-54-cards-whats-the-probability-that-even-one-card-will-match-tw | # Give a deck of 54 cards, what's the probability that even one card will match two dealt hands...
if we shuffle a deck (with 54 cards including Joker) thoroughly and deal out a four card hand, there are over 300,000 different hands. What's the probability that no cards match between two dealt hands? even one card matches? two cards match? all cards match?
Edit: The two jokers are different. Not identical
Here is what I have so far. For no card to match the probability should be $$\frac{50}{54 }\cdot\frac{49}{53}\cdot\frac{48}{52}\cdot\frac{47}{51} \approx 72\%$$ chance that no cards match.
If any one is to match it would be $$\frac{4}{54} + \frac{4}{53} + \frac{4}{52} + \frac{4}{51}\approx 30.4\%$$
If exactly two are to match it would be $$\frac{4}{54}\cdot\frac{3}{53}\cdot\frac{48}{52}\cdot\frac{47}{51}\approx 0.35\%$$
for all cards to match it would be $$\frac{4}{54}\cdot\frac{3}{53}\cdot\frac{2}{52}\cdot\frac{1}{51}$$
Is this the right way to think about it? Am I missing anything?
• The jokers may mess with things. Also relies on a set number of cards per hand.
– user645636
Mar 31 '19 at 19:49
• Are the two jokers identical? In some decks they are, in some they aren't (for instance, one is red and one is black). Mar 31 '19 at 20:49
• There are a number of mistakes in your attempts. For example, for exactly two to match your attempt actually calculates the probability that very specifically the first card matches and the second card matches and the last two don't. You neglected to account for other orders of matching and not matching making that answer off by a factor of 6. For exactly one to match, you added when you weren't supposed to. Approach the same way as for two matches. It should be clear when a mistake is made since the totals don't add up to 1 like they should. Mar 31 '19 at 21:14
• no. A probability of 1 means absolutely certain. in your example it's $\frac{2^{100}-101}{2^{100}}$
– user645636
Mar 31 '19 at 23:13
• I'm tempted just give you combinatoric and probabilistic links as an answer...
– user645636
Mar 31 '19 at 23:22
Supposing you have $$54$$ distinct cards and you draw a four-card hand, take note of the cards and shuffle them back in and draw another four cards, the probability of having exactly $$k$$ cards in the new hand exactly matching a card from the initial hand (where exactly matching requires both the suit and the number to be identical and the jokers only match the exact same joker, e.g. if there is a black joker and a red joker) is:
$$\frac{\binom{4}{k}\cdot 4\frac{k}{~}\cdot 50\frac{4-k}{~}}{54\frac{4}{~}} = \frac{\binom{4}{k}\binom{50}{4-k}}{\binom{54}{4}}$$
where here $$n\frac{k}{~}$$ represents the falling factorial $$\underbrace{n\cdot (n-1)\cdot (n-2)\cdots (n-k+1)}_{k~\text{terms}}$$
The expression on the left can be explained by treating each card as being pulled in sequence, picking which positions in the sequence are occupied by matching cards, picking which matching cards those are, and picking which non-matching cards occupied the remaining spaces out of the possible ways in which four cards could be drawn.
The expression on the right can be explained by treating it as though the cards are picked simultaneously where order doesn't matter and picking which matching cards they are and which non-matching cards they are and dividing by the number of ways of selecting four cards. You should recognize the expression on the right as simply being the well-known hypergeometric distribution.
The results are:
$$\begin{array}{|c|c|c|}\hline k&\text{exact}&\text{approximate}\\ \hline 0&\frac{230300}{316251}&0.728219041\dots\\1&\frac{78400}{316251}&0.247904354\dots\\ 2&\frac{2450}{105417}&0.023241033\dots\\ 3&\frac{200}{316251}&0.000632409\dots\\ 4&\frac{1}{316251}&0.000003162\dots\\\hline\end{array}$$
Notice how the probabilities add up exactly to $$1$$, as should always be the case when partitioning the sample space of a probability experiment.
• Question: in equation above shouldn't the denominator be (54 4)? if I use (50 4) the denominator = 230300. If you address this, i'll accept this answer. Thanks! Apr 5 '19 at 4:45
• @Salman it should indeed be 54 choose 4, not 50 choose 4. That was a typo. Good catch Apr 5 '19 at 9:54
Comment: If I understand correctly, we can suppose the first hand is dealt, and then try to match it. In my simulation (in R) below I suppose that the first hand has cards numbered from 1 through 4 (in some order). In a million randomly dealt second hands, my probabilities of various counts of matching cards are shown below.
set.seed(401) # for reproducibility
x = replicate(10^6, sum(sample(1:54, 4)<=4))
table(x)/10^6
x
0 1 2 3 4
0.728386 0.247738 0.023251 0.000624 0.000001
Only the first few places of these probabilities are likely to be accurate, but this may give you something to check against as you finish your combinatorial analysis. Notice that the simulated proportion $$0.728$$ of no matches is the same (to three places) as the correct probability in your first answer.
A second simulation, with seed 2019, gave the following slightly different answers:
x
0 1 2 3 4
0.727877 0.247891 0.023563 0.000667 0.000002
More precisely, hypergeometric probabilities [also just now posted by @JMoravitz (+1)] are:
dhyper(0:3, 4, 50, 4)
[1] 0.7282190412 0.2479043545 0.0232410332 0.0006324091 | 2021-10-27T15:58:01 | {
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http://math.stackexchange.com/questions/193668/expressing-in-the-form-a-sinx-c/193682 | # Expressing in the form $A \sin(x + c)$
Express in the form $A\sin(x+c)$
a) $\sin x+\sqrt3\cos x$; b) $\sin x-\cos x$
sol: a) $A=\sqrt{1+3}=2$, $\tan c=\frac{\sqrt 3}1$, $c=\frac\pi3$. So $\sin x+\sqrt3\cos x=2\sin(x+\frac\pi3)$
b) $\sqrt 2\sin(x-\frac\pi4)$
Can someone please explain the method used in the provided solution above? (I'm not familiar with this way of solving whatsoever.)
-
There are a few spiritual duplicates of this question. The key is to normalize and use the trigonometric addition formulas. Try and work backwards: expand out $A\sin(x+c)$, and then figure out $A$ and $c$ (do use the formula $\cos^2+\sin^2=1$ to find out $A$). – anon Sep 10 '12 at 16:15
Thanks, I tried expanding the desired form and it all worked out nicely. Though I didn't understand where the identity $cos^2\alpha+sin^2\alpha=1$ comes to use. – Py42 Sep 10 '12 at 16:35
If you know $A\cos(c)$ and $A\sin(c)$, then you can find $A$ via $A^2=(A\cos c)^2+(A\sin c)^2$. – anon Sep 10 '12 at 16:39
Oh, I just compared $Acos(c)$ with its value after finding the angle c. – Py42 Sep 10 '12 at 16:43
A similar question was posted here math.stackexchange.com/questions/877499/… However, it was now autodeleted. – Martin Sleziak Aug 5 '14 at 6:50
For starters, do note that
$A \, \sin (x + c) = \left(A \, \cos(c) \right) \, \sin(x) + \left( A \, \sin(c)\right) \, \cos(x)$
Since you have $\sin (x) + \sqrt{3} \, \cos (x)$, it follows that $A \, \cos(c) = 1$ and $A \, \sin(c) = \sqrt{3}$. Therefore, since $\sin^2 (x) + \cos^2 (x) = 1$, we have that $A^2 = 4$, which yields $A = 2$, and $2 \cos (c) = 1$, which yields $c = \pi / 3$. Finally, we conclude that
$\sin (x) + \sqrt{3} \, \cos (x) = 2 \sin (x + \pi / 3)$.
-
Using the appropriate formula for $\sin$ you have $A \sin(x+c) = A \sin x \cos c + A \cos x \sin c$. You need to determine $A,c$ so the formula holds true for a), b).
Equating $A \sin x \cos c + A \cos x \sin c = \sin x + \sqrt{3} \cos x$ gives $A \cos c = 1$, $A \sin c = \sqrt{3}$. This gives $\tan c = \frac{A \sin c}{A \cos c} = \sqrt{3}$. If $\tan c = \sqrt{3}$, then $\sin c = \frac{\sqrt{3}}{2}$ and $\cos c = \frac{1}{2}$. This gives $A \frac{1}{2} = 1$, so $A = 2$. You can check that $\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}, \cos \frac{\pi}{3} = \frac{1}{2}$, from which it follows that $\sin x + \sqrt{3} \cos x = 2 \sin ( x + \frac{\pi}{3})$.
Similarly, $A \sin x \cos c + A \cos x \sin c = \sin x - \cos x$ gives $A \cos c = 1$, $A \sin c = -1$. This gives $\tan c = \frac{A \sin c}{A \cos c} = -1$, which in turn gives $\sin c = -\frac{1}{\sqrt{2}}$, $\cos c = \frac{1}{\sqrt{2}}$. Then $A \cos c = A \frac{1}{\sqrt{2}} = 1$ gives $A = \sqrt{2}$. You can check that $\sin (-\frac{\pi}{4}) = -\frac{1}{\sqrt{2}}, \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$, from which it follows that $\sin x - \cos x = \sqrt{2} \sin ( x - \frac{\pi}{4})$.
-
according to copper.hat you will conclude a formula $$a\sin(x)+b\cos(x)=\sqrt{a^2+b^2}\sin(x+y),\quad \tan(y)=\frac{b}{a},y\in(-\frac{\pi}{2},\frac{\pi}{2})$$
- | 2015-08-29T07:43:01 | {
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https://math.stackexchange.com/questions/3075176/the-isomorphism-between-two-complete-ordered-fields-is-unique | # The isomorphism between two complete ordered fields is unique
The isomorphism between two complete ordered fields is unique.
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
Let $$\mathfrak{R}=\langle \Bbb R,<,+,\cdot,0,1 \rangle,\mathfrak{A}=\langle A,\prec,\oplus,\odot,0',1' \rangle$$ be complete ordered fields where $$\Bbb R$$ is the set of real numbers. Let $$\mathfrak{B}=\langle B,\prec,\oplus,\odot,0',1' \rangle$$ be the smallest subfield of $$\mathfrak{A}$$ and $$\mathfrak{Q}=\langle \Bbb Q,<,+,\cdot,0,1 \rangle$$.
Lemma 1: $$\mathfrak{R}$$ is isomorphic to $$\mathfrak{A}$$.
Lemma 2: $$\mathfrak{Q}$$ is uniquely isomorphic to $$\mathfrak{B}$$.
By Lemma 1, let $$\Phi:\Bbb R \to A,\Psi:\Bbb R \to A$$ be isomorphisms between $$\mathfrak{R}$$ and $$\mathfrak{A}$$. By Lemma 2, let $$f:\Bbb Q \to B$$ be the unique isomorphism between $$\mathfrak{Q}$$ and $$\mathfrak{B}$$.
Let $$X \subseteq \Bbb Q$$ be bounded from above and $$\sup,\sup'$$ supremums w.r.t $$<,\prec$$ respectively. We next prove that $$\Phi(\sup X) = \sup' f[X]$$.
$$\forall x\in X: x \le \sup X \implies \forall x\in X: \Phi(x) \preccurlyeq \Phi(\sup X) \implies \forall x\in X: f(x) \preccurlyeq \Phi(\sup X) \implies \sup' f[X] \preccurlyeq \Phi(\sup X).$$
• Assume the contrary that $$\sup' f[X] \prec \Phi(\sup X)$$. Since $$B$$ is dense in $$A$$, there exists $$b\in B$$ such that $$\sup' f[X] \prec b \prec \Phi(\sup X)$$. Then there exists $$p\in \Bbb Q$$ such that $$f(p)=b$$. Thus $$\sup' f[X] \prec f(p)=\Phi(p) \prec \Phi(\sup X).$$
• We have $$\Phi(p) \prec \Phi(\sup X) \implies p<\sup X \implies p for some $$p'\in X \implies$$ $$f(p) \prec f(p')$$ for some $$p'\in X$$ $$\implies f(p) \prec \sup' f[X]$$. This is a contradiction.
Hence $$\Phi(\sup X)=\sup' f[X]$$. Similarly, $$\Psi(\sup X)=\sup' f[X]$$.
Let $$X_x=\{p\in\Bbb Q \mid p. Since $$\Bbb Q$$ is dense in $$\Bbb R$$, $$x=\sup X_x$$ for all $$x\in\Bbb R$$. Then $$\Phi(x)=\Phi(\sup X_x)=\sup' f[X_x]=\Psi(\sup X_x)=\Psi(x)$$ for all $$x\in\Bbb R$$. It follows that $$\Phi=\Psi$$.
• You should use that a complete ordered field has a unique ordering, because an element is nonnegative if and only if it has a square root. – egreg Jan 17 '19 at 0:05
• Hi @egreg, DanielWainfleet has utilized your idea and posted it as an answer below. I am reading his answer. Have you seen any error in my proof? – Abstract Analysis Jan 17 '19 at 0:10
This is an extended comment on the uniqueness of the isomorphism.
(1a). Let $$F, G$$ be sub-fields of $$\Bbb R$$ such that $$\psi:F\to G$$ is a field-isomorphism. ($$\psi$$ is not assumed to preserve order.) If $$\forall x\in F\,(0\le x\implies \sqrt x\in F)$$ then $$\psi=id_F$$ and $$G=F.$$
Proof: $$\Bbb Q\subset F$$ and $$\psi|_{\Bbb Q}=id_{\Bbb Q}$$ so for any $$x\in F$$ and any $$q\in \Bbb Q$$ we have $$x\ge q\iff \psi(x)-q=\psi(x)-\psi(q)=(\psi(\sqrt {x-q}))^2\ge 0\iff$$ $$\iff \psi(x)\ge q$$ so $$\Bbb Q \cap (-\infty,x]=\Bbb Q\cap (-\infty,\psi(x)],$$ so $$x=\psi(x).$$
(1b).In particular, letting $$F=\Bbb R$$ in (1a), the only sub-field of $$\Bbb R$$ that is field-isomorphic to $$\Bbb R$$ is $$\Bbb R$$ itself, and the only field-isomorphism of $$\Bbb R$$ to $$\Bbb R$$ is $$id_{\Bbb R}.$$
(2). If $$B$$ is a field and $$\psi_1, \psi_2$$ are field-isomorphisms from $$\Bbb R$$ to $$B$$ then by (1b), $$\psi_2^{-1}\psi_1=id_{\Bbb R},$$ so $$\psi_1=\psi_2.$$
• A proper sub-field of $\Bbb R$ can be an ordered field according to an order that is not the usual order $<$ of $\Bbb R$. For example $\{a+b\sqrt 2\,:a,b\in \Bbb Q\}.$ For $a,b\in \Bbb Q$ let $a+b\sqrt 2\,>^*0\iff a-b\sqrt 2<0.$ – DanielWainfleet Jan 16 '19 at 23:06
• I think you meant $\psi$ rather than $f$. Please check if my reasoning is correct: For all $x\in F$ and $q\in\Bbb Q$: $\psi(x)-q=\psi(x)-\psi(q)$ [since $\psi|_{\Bbb Q}=\text{id}_{\Bbb Q}$] $=\psi(x-q)$. Then $x \ge q \iff x-q \ge 0 \iff x-q$ $=\sqrt {x-q} \cdot \sqrt {x-q} \iff \psi(x-q)=\psi(\sqrt {x-q} \cdot \sqrt {x-q})=\psi(\sqrt {x-q})\cdot \psi(\sqrt {x-q})=$ $(\psi(\sqrt {x-q}))^2 \ge 0 \iff \psi(x)-q=\psi(x-q) \ge 0$. To sum up, $x\ge q \iff \psi(x)-q \ge 0$ $\iff \psi(x)\ge q$.[...] – Abstract Analysis Jan 17 '19 at 1:54
• [...] It follows that $\{q\in\Bbb Q \mid q\le x\}=\{q\in\Bbb Q \mid q\le \psi(x)\}$. Hence $\sup \{q\in\Bbb Q \mid q\le x\}=$ $\sup \{q\in\Bbb Q \mid q\le \psi(x)\}$ and thus $x=\psi(x)$. As a result, the isomorphism between two fields is unique. – Abstract Analysis Jan 17 '19 at 1:54
• Perfectly correct. And $f$ was a typo for $\psi$. – DanielWainfleet Jan 17 '19 at 13:58
• Thank you so much for your verification! Have you seen any error in my proof? – Abstract Analysis Jan 17 '19 at 14:04 | 2020-01-29T09:12:16 | {
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https://math.stackexchange.com/questions/2982073/finding-the-local-extreme-values-of-fx-x2-2x-9-over-2-infty | # Finding the local extreme values of $f(x) = -x^2 + 2x + 9$ over $[-2,\infty)$.
I'm tutoring a student, and we were trying to solve the following question:
Find the local extreme values of $$f(x) = -x^2 + 2x + 9$$ over $$[-2,\infty)$$.
According to the textbook, the local extreme values are essentially the peaks and the valleys in the graph of the function $$f$$, so basically where $$f'(x) = 0$$. This is relatively easy to compute: $$f'(x) = -2x + 2,$$ of which the critical points are $$x = 1$$. Likewise, $$f''(x) = -2 < 0$$, which means $$f$$ is concave down everywhere, and thus $$x = 1$$ is where a maximum value occurs on the graph. The maximum is $$f(1) = -1 + 2 + 9 = 10$$.
Of course, the endpoint $$x = -2$$ yields $$f(-2) = -(-2)^2 +2(-2) + 9 = 1,$$ but since the graph is concave down everywhere, $$\displaystyle \lim_{x\to\infty}f = -\infty$$ implies there really is no minimum per se... right?
The online computer program tells us that $$(-2,1)$$ is a local minimum, and $$(1,10)$$ is a local maximum. But in accordance with the definition from the textbook, why is $$(-2,1)$$ where a local minimum of the graph occurs? It's neither a peak nor a valley in the graph. What exactly does local mean when the interval is infinite? It doesn't quite make logical sense, unless the definition is not as rigorous as it ought to be.
• it is not clear whether the function is limited to the given domain, or whether the domain is simply the search interval – LinAlg Nov 3 '18 at 0:42
According to this Wikipedia page, the proper definition of local maxima and local minima is:
$$f:X\longmapsto$$Y is said to have a local (or relative) maximum point at the point $$x^∗$$ if there exists some $$ε > 0$$ such that $$f(x^∗) ≥ f(x)$$ for all $$x\in$$ X within distance $$ε$$ of $$x^∗$$. Similarly, the function has a local minimum point at $$x^∗$$ if $$f(x^∗) ≤ f(x)$$ for all $$x\in$$ X within distance $$ε$$ of $$x^∗$$.
Since we can fix a $$\epsilon$$-neighborhood with our choice of $$\epsilon$$ near $$x=-2$$ so that in this domain $$(-2, 1)$$ is a minimum point, it is indeed a local minimum.
If you look at the statements of the first and second derivative tests, they should indicate that they are valid at interior points of intervals, not the endpoints. (And only where the function happens to be differentiable.) The endpoints are a special case. An endpoint can be a max or min while the derivatives there can be anything at all. A local max or min is defined to be the biggest or smallest guy in his neighborhood. Period.
Second, the key word is "local". The function here doesn't have an "absolute" (or "global") min, but it has a local min.
• Here, we have the interval $[-2,\infty)$ as the only one being specified anywhere in the problem. How do we decide the appropriate neighborhood to say that $(-2,1)$ is where a local minimum occurs? Do we just say that since it's an endpoint and the remainder of the interval is infinite, we consider it as our only other "concrete" point available? – Decaf-Math Nov 2 '18 at 18:44
• Usually the Calc texts have a section, after the section on 2nd derivative test, on "extreme values on closed intervals." The technique is: Step 1. find interior extreme by derivative tests. Step 2, Check the endpoints and any discontinuities or points of nondifferentiability (cusps and such) manually. – B. Goddard Nov 2 '18 at 18:54
This is a parabola open downward.
The vertex is $$(1,10)$$ where your derivative is zero and it is a local and global maximum.
The left endpoint of $$(-2, 1)$$ is a local minimum and there is no global minimum.
• But why would we consider $(-2,1)$ to be a local minimum? That's the question being asked in the OP. – Decaf-Math Nov 2 '18 at 18:43
• That is because around that point and as long as you are in the domain, the values are greater or equal to the value of $1$ attained at $(-2,1)$ – Mohammad Riazi-Kermani Nov 2 '18 at 18:46
• How do you address the point $(5, -6)$ via this logic? $5\in[-2,\infty)$, and it's somewhat near $-2$, but $-6 < 1$. Does local with regard to $x=-2$ no longer apply once we pass the local max at $x=1$? – Decaf-Math Nov 2 '18 at 18:53
• Local is a small neighborhood of the point, you are going too far from $-2$. I understand your concern about calling an endpoint a local, but there is nothing wrong about it. – Mohammad Riazi-Kermani Nov 2 '18 at 18:59
• Surely, $(-2, 1)$ is a left endpoint. – N. F. Taussig Nov 3 '18 at 9:43
Sometimes just plotting the function cuts through needless distractions:
Clearly at the point $$x=-2$$, the function is lower than any other values in the specified domain, and hence is a local minimum.
• Thanks for your answer, but this doesn't really address my question: for whatever reason, the online homework program is saying that $(-2,1)$ is a local minimum. Why is it considered a local minimum when it's not a peak/valley on the graph? There is no interval specified other than $[-2,\infty)$. – Decaf-Math Nov 2 '18 at 18:36
• @Decaf-Math: See expanded solution. – David G. Stork Nov 2 '18 at 19:45
For the purposes of local extrema we do not consider the endpoints of the interval in question. Thus the only local extremum (in this case a local maxiumum) occurs at $$(1, 10)$$. On the interval in question, $$(1, 10)$$ also happens to be an absolute maximum, and there are no local nor absolute minima on this interval. | 2019-08-20T07:14:23 | {
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https://math.stackexchange.com/questions/1812234/formula-for-sum-of-first-n-odd-integers | # Formula for sum of first $n$ odd integers
I'm self-studying Spivak's Calculus and I'm currently going through the pages and problems on induction. This is my first encounter with induction and I would like for someone more experienced than me to give me a hint and direction. The first problem is as follows:
Find a formula for $$\sum_{i=1}^n(2i-1)=1+3+5+...+(2n-1)$$ And the related following problem:
Find a formula for $$\sum_{i=1}^n(2i-1)^2=1^2+3^2+5^2+...+(2n-1)^2$$
The given hints are: "What do these expressions have to do with $1+2+3+...+2n$ and $1^2+2^2+3^2+...+(2n)^2$?"
I recognize that the above sums are the sum of all the odd integers from $1$ to $n$ and the sum of all the squares of the odd integers from $1$ to $n$, respectively. My question is, in problems like these does one just do a bunch of trial and error, as I have done for quite a while now, or is there a more clever way to go about it?
• Please, try to make the titles of your questions more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. – Emre Jun 4 '16 at 13:09
• I should mention that we know that the formula for $1+2+3+...+n =\frac{n(n+1)}{2}$ – Kristoffer Gertz Jun 4 '16 at 13:11
• @Emre Got it, I will do so for my future questions. – Kristoffer Gertz Jun 4 '16 at 13:13
• Not a duplicate, but you should also check this question: math.stackexchange.com/questions/1806906/… – Emre Jun 4 '16 at 13:27
There is a simple rule: For any arithmetic progression, that is any series where the difference between consecutive elements is constant, the sum is equal to the number of elements, multiplied by the average between the first and the last element.
In your case, the first element is 1, the last element is 2n - 1, the average is n, there are n elements, therefore the sum is $n^2$.
$$1+3+5+...+(2n-1)=n\times\,n=n^2$$
• Nice! $\;\!\;\!$ – goblin GONE Jun 5 '16 at 2:28
• The partitioning of $\mathbb{N}^2$ displayed here can be rigorously defined as the coimage of $\mathrm{max} : \mathbb{N}^2 \rightarrow \mathbb{N}$. I'm guessing there's generalizations of this idea in tropical geometry. – goblin GONE Jun 5 '16 at 7:00
For example:
$$\sum_{i=1}^n(2i-1)=2\sum_{i=1}^ni-\sum_{i=1}^n1$$
Observe the last sum is just $\;1+1+...+1=n\;$ . Work this out and the other question, too.
## Hint:
$$\sum_{i=1}^n (2i-1)=\sum_{i=1}^{2n} i-\sum_{i=1}^n 2i$$ $$\sum_{i=1}^n (2i-1)^2=\sum_{i=1}^{2n} i^2-\sum_{i=1}^n (2i)^2$$
I guess you know the identities $\sum_{i=1}^n i=\frac{n(n+1)}2$,$\sum_{i=1}^n i^2=\frac{n(n+1)(2n+1)}6$. Now, try to write $\sum_{i=1}^n 2i$ in terms of $\sum_{i=1}^n i$.
Similarly, try to write $\sum_{i=1}^n (2i)^2$ in terms of $\sum_{i=1}^n i^2$.
Here is the rest, hover on the yellow block if you want to see:
$\sum_{i=1}^n (2i)^2=4\sum_{i=1}^n i^2\qquad\sum_{i=1}^n (2i)=2\sum_{i=1}^n i$
The hint suggests you know the formulas for $1+2+...+2n$ and $1^2+2^2+...(2n)^2$. The first one is obvious ($\sum_{k=1}^N k = N(N+1)/2$), the second one is a bit less.
Once you realised this, your two sums can easily be expressed as a function of these two formulas.
To some extent: yes, it is a matter of trial and error, but you can get quite educated about it. But I'll answer your question properly later on, because it is an important and good question and independent of the particular examples used to illustrate it.
With $\sum(2i-1)$, simply adding a few terms by hand shows you that the answer is $n^2$. Proving that the answer is $n^2$ is then just a matter of proving that going from $n=N$ to $n=N+1$ means adding $2N+1$ to the total, and going from $N^2$ to $(N+1)^2$ means adding $2N+1$ to the total… so if it's true for $n=N$ then it's true for $n=N+1$. That's the inductive step. As for the first term in the deduction, that depends on taste and personality. Some people like to think that the sum of the first 1 odd numbers is 1; others, that the sum of the first 0 odd numbers is 0. Each kind of person secretly despises the other.
On the other hand, it has to be said that Spivak seems not to want you to use induction here, if you are meant to use the knowledge that $1+2+3+...+n =\frac12{n(n+1)}$ (and where did that knowledge come from, pray, if not from induction?). All you do is take the total of all numbers up to $2n$ (ie. $\frac12{2n(2n+1)}$) and get rid of the even numbers by subtracting twice the total of all numbers up to $n$ (i.e. twice $\frac12{n(n+1)}$). Which, written out, is very easy and gives you $n^2$. But it is not induction.
If you happen to know $\sum{i^2}$ then the second question is much like the first (but remember to subtract four times the shorter series from the longer one, because of the squaring). This is also not anything to do with deduction.
So, then: "In problems like these does one just do a bunch of trial and error, as I have done for quite a while now, or is there a more clever way to go about it?".
Here are three approaches to finding the sums of series before starting to prove the result. It's good to have all of them available.
1. Know $\sum{i^a}$ for $a=0, 1, 2, 3$. It doesn't require much memorising. Then combine the sums for whatever you are being asked to find the sum of (including using the trick that Spivak is making you use here).
2. Note that $\sum{1}=n/1!$, $\sum{i}=n(n+1)/2!$, $\sum{i(i+1)}=n(n+1)(n+2)/3!$, $\sum{i(i+1)(i+2)}=n(n+1)(n+2)(n+3)/4!$, and so on for ever. The inductive proof of these is very easy and the patterns themselves are easy to memorise. Every polynomial can be expressed as the sum of multiples of these easily-summable pieces. In fact, this is the way I was first taught to work out $\sum{i^2}$.
3. Be brutal. Note that a sum of values of a polynomial of degree $a$ will be a polynomial of degree $a+1$. So, for instance taking a sum of squares, say that the sum is $ax^3+bx^2+cx+d$ and prove this by induction. As you go through the inductive proof, you will be forced to give values to the coefficients to make the induction work. In this way you can find the values of all the coefficients and prove the result, both at the same time.
And the other half of your "Trial and error?" question is this: sums based on consecutive numbers (1, 2, 3,... rather than 1, 3, 5,... or even 1, 2, 4, 5, 7, 8, 10,...) are always easier. So if you have a sum which - as in Spivak's exercise - doesn't take in all consecutive values of $i$, the first thing you should do is put it together from sums which are made of consecutive values. This is a skill (or an instinct) which helps in many areas of maths, notably combinatorics.
\begin{align} \sum_{i=1}^n (2i-1)&=\sum_{i=1}^n\binom i1+\binom {i-1}1\\ &=\binom {n+1}2+\binom n2\\ &=n^2\qquad\blacksquare\\ \sum_{i=1}^n(2i-1)^2&= \color{blue}{1^2}+\color{purple}{3^2}+\color{green}{5^2}+\cdots+\color{red}{(2n-1)^2}\\ &=\color{blue}{\binom 12+\binom 22}+ \color{purple}{\binom 32+\binom 42}+ \color{green}{\binom 52+\binom 62}+\cdots + \color{red}{\binom {2n-1}2+\binom {2n}2}\\ &=\sum_{r=1}^{2n}\binom r2\\ &=\binom {2n+1}3 =\frac {n(2n-1)(2n+1)}3\qquad\blacksquare\end{align}
One has
\begin{align}\sum_{i=1}^{2n} i&=\sum_{i =1}^{n}(2i-1)+\sum_{i=1}^n2i\\&=S+2\sum_{i=1}^ni\end{align}
Now assume we have proven (by induction...) that $\sum\limits_{i=1}^ni={n(n+1)\over 2}$ we can rewrite the above as
$$n(2n+1)=S+n(n+1)$$
Solving for $S$ we get
$$S=n^2$$
And we can do the same with the sum of squares
\begin{align}\sum_{i=1}^{2n} i^2&=\sum_{i =1}^{n}(2i-1)^2+\sum_{i=1}^n(2i)^2\\&=S+4\sum_{i=1}^ni^2\end{align}
Now we use $\sum\limits_{i=1}^ni^2={n(n+1)(2n+1)\over 6}$ to rewrite
$${2n(2n+1)(4n+1)\over 6}=S+4{n(n+1)(2n+1)\over 6}$$
Solving for $S$ we get
$$S={n(2n+1)\over 6}(8n+2-4n-4)={n(2n-1)(2n+1)\over 3}$$
• You're right ! I had forgotten a factor $4$ in the sum of squares. – marwalix Jun 4 '16 at 17:50 | 2021-04-23T00:04:10 | {
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https://stats.stackexchange.com/questions/183834/effect-size-and-bootstrapping-in-paired-t-test | # Effect size and bootstrapping in paired t-test
I have multiple paired $t$-tests, such as one giving results:
$t_{14} = 2.7,\ p = .017$
Although people seem to do effect sizes in different ways in repeated samples, I have taken the mean difference divided by the standard deviation of the differences (I'll call this $d$, though maybe I should call it something else?) and get $0.70$. I also have a very strong correlation between the samples, not sure if that is problematic.
I would like to put confidence limits around my effect size estimate. To do so, I randomly resample from the difference scores, compute $d$ in the same way and repeat 1000 times. My question is whether this is a good approach, rather than, say, just giving confidence limits around the unstandardised difference or resampling from the original samples. My bootstrap gives me a mean $d$ of $0.79$ with confidence limits of $[0.4, 1.4]$. I've tried this on other random data too. Why am I getting a consistently higher $d$ from bootstrapping, and why are the intervals asymmetric? Is this because of skew in the (difference) scores, and does this make this approach more or less robust?
Edit: here is an example of the data involved. 15 people were measured two times.
Mean A = 1742; SD = 435
Mean B = 1820; SD = 426
Mean difference = 78, SD of differences = 111, $d$ = 0.70
A B
1999 2040
1501 1601
1552 1623
2385 2386
2488 2671
1257 1218
1806 1719
1348 1405
2048 2079
1810 2017
1308 1356
2310 2324
1247 1616
1839 1878
1235 1370
• Just to say that I have found useful material on these pages (though I haven't got a specific answer to the case of bootstrapping CIs for an effect size) stats.stackexchange.com/questions/71525/… , stats.stackexchange.com/questions/73818/… – splint Nov 30 '15 at 15:45
• I'm not quite sure I'm following this. Can you give a simple example / some example data? Is this a multiple comparisons issue? – gung - Reinstate Monica Dec 3 '15 at 14:59
• @gung Thanks for looking. The t quoted is a simple example though I can fish out some data if you want. The issue is not about multiple comparisons. It is about (1) how to calculate effect sizes in a paired t-test; (2) whether it makes sense to bootstrap a confidence interval around this; and (3) why this interval might be asymmetric. – splint Dec 4 '15 at 14:16
• What are the "repeated samples" that supposedly lead people to "do effect sizes in different ways"? For people here to get a sense of why the mean of your bootsamples is different & the CI is asymmetric, you will probably need to paste your data & your code. – gung - Reinstate Monica Dec 4 '15 at 16:51
• I have added some data. Is there a better way to do tables on here? For background on the different ways to calculate effect sizes in repeated measures, see the links in my first comment (essentially, some prefer to use the pooled SD as a denominator rather than the SD of the difference scores). – splint Dec 4 '15 at 17:47
I will attempt to answer but I am not totally sure on my own knowledge on the subject.
Bootstrap, as far as I know is always done on the original data. In your case the original data is pairs of data. So to do a bootstrap, you would have to random sample (with replacement) on the pairs of the original data. That is equivalent to do the bootstrap on the difference scores and performing the effect size calculation as you described on the samples.
I get a different result from you (in R)
a=read.table(header=F,text="
1999 2040
1501 1601
1552 1623
2385 2386
2488 2671
1257 1218
1806 1719
1348 1405
2048 2079
1810 2017
1308 1356
2310 2324
1247 1616
1839 1878
1235 1370
")
d=a$V2-a$V1
mean(d)/sd(d)
[1] 0.7006464
aux=function(x,i) mean(x[i])/sd(x[i])
bb=boot::boot(d,aux,R=1000)
mean(bb$t) [1] 0.7530415 boot::boot.ci(bb) BOOTSTRAP CONFIDENCE INTERVAL CALCULATIONS Based on 1000 bootstrap replicates CALL : boot::boot.ci(boot.out = bb) Intervals : Level Normal Basic 95% ( 0.1840, 1.0846 ) ( 0.1454, 1.0570 ) Level Percentile BCa 95% ( 0.3443, 1.2559 ) ( 0.1634, 1.0722 ) Calculations and Intervals on Original Scale Some BCa intervals may be unstable (code corrected as per the comments) Indeed the direct calculation of the effect size (mean(d)/sd(d)) is not similar to the bootstrap calculation (mean(bb$t)). I dont know how to explain it
The only confidence interval that matches yours in the percentile (I dont really know which interval to choose on theoretical grounds - I use the BCa - I think it was suggested somewhere)
The second way to calculate a CI on effect size is to use analytical formulas. This question on CV discussed the formulas How can i calculate the 95% confidence interval of an effect size if I have the mean difference score, CI of that difference score
Using the MBESS package I get the following CI
MBESS::ci.sm(Mean = mean(d), SD=sd(d),N=length(d))
[1] "The 0.95 confidence limits for the standardized mean are given as:"
$Lower.Conf.Limit.Standardized.Mean [1] 0.1231584$Standardized.Mean
[1] 0.7006464
$Upper.Conf.Limit.Standardized.Mean [1] 1.258396 As for your suggestion on computing the confidence interval for the difference score and using it to compute a confidence interval on the effect size, I have never heard of it, and I would suggest not using it. • +1 to @amoeba, I think you want to use mean(bb$t). Nice answer; +1 as soon as you fix that issue. – usεr11852 says Reinstate Monic Dec 6 '15 at 8:16
• Very helpful answer. I do have questions though. mean(bb$t) returns 0.76 which, as in my example, is considerably greater than the sample value. All of the intervals are also asymmetric whereas my understanding was that this should not be the case for analytically computed CIs. – splint Dec 6 '15 at 12:07 • There is a missing close bracket in the function/mean call, apparently edits of 1 character are not allowed! – splint Dec 6 '15 at 12:10 • thanks folks. mean(bb$t) and the ")" in the mean corrected. Indeed the values for the bootstrap mean and the full data effect size are not close. I dont know how to explain. – Jacques Wainer Dec 7 '15 at 13:44
• @splint and Jacques: what happens here has to do with bias and bias correcting in bootstrap, see e.g. on wikipedia. I am not a specialist, but rougly what happens is that the difference between your empirical value 0.7 and your bootstrapped value 0.75 indicates a bias. You can correct this bias, by subtracting this difference from 0.7 and arrive to the bias-corrected estimate of d as 0.65. The intuition is that if your bootstrapped samples were on average 0.05 higher [ctd.] – amoeba says Reinstate Monica Dec 7 '15 at 14:00 | 2019-12-07T06:59:09 | {
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https://math.stackexchange.com/questions/4052641/why-arent-these-two-integrals-equivalent-when-using-the-substitution-x-frac1 | # Why aren't these two integrals equivalent when using the substitution $x=\frac{1}{t}$?
Why aren't these two integrals $$\int_{-1}^{1}\frac{1}{\left(1+x^2\right)^2}\,\mathrm{d}x$$ and $$\int_{-1}^{1}\frac{-t^2}{\left(1+t^2\right)^2}\,\mathrm{d}t$$ equal to each other, despite using the substitution $$x=\frac{1}{t}$$, which yields the second integral when the substitution is used on the first one?
Could it be that $$t$$ is undefined at $$x=0$$ since the limits are from $$-1$$ to $$1$$?
• should be $dx=-\dfrac{1}{t^2}dt$ – janmarqz Mar 7 at 16:25
• You had to break integral up into two pieces to see why. The substitution is not defined at $0$. The correct substitution gets an integral over $(-\infty,-1)$ and $(1,\infty)$ – Ninad Munshi Mar 7 at 16:25
• @janmarqz OP did the substitution correctly (except for the bounds issue I mentioned) – Ninad Munshi Mar 7 at 16:28
The substitution is not defined for $$x=0$$ or at $$t=0$$. So in fact if you make the substitution, what you have is
\begin{align} \int_{-1}^1{1\over(1+x^2)^2}dx&=\int_{-1}^0{1\over(1+x^2)^2}dx+\int_0^1{1\over(1+x^2)^2}dx\\ &=\int_{-1}^{-\infty}{-t^2\over(1+t^2)^2}dt+\int_\infty^1{-t^2\over(1+t^2)^2}dt\\ &=\int_{-\infty}^{-1}{t^2\over(1+t^2)^2}dt+\int_1^{\infty}{t^2\over(1+t^2)^2}dt \end{align}
(And as J.G. points out, the symmetry of the integrand allows you to reduce to just one integral, from $$1$$ to $$\infty$$.)
If $$x=\frac1t$$, then, since $$x\in[-1,1]$$, you should get$$\int_{-\infty}^{-1}\frac{t^2}{(1+t^2)^2}\,\mathrm dt+\int_1^\infty\frac{t^2}{(1+t^2)^2}\,\mathrm dt.$$
For even $$f$$,$$\int_{-1}^1f(x)\mathrm{d}x=2\int_0^1f(x)\mathrm{d}x=2\int_1^\infty\tfrac{f(1/t)}{t^2}\mathrm{d}t.$$
As $$x$$ goes from $$0$$ to $$+1,$$ $$t$$ goes from $$+\infty$$ to $$1.$$
As $$x$$ goes from $$-1$$ to $$0,$$ $$t$$ goes from $$-1$$ to $$-\infty.$$ | 2021-06-25T04:17:21 | {
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https://math.stackexchange.com/questions/2512424/groups-with-given-automorphism-groups | # Groups with given automorphism groups
It is an easy exercise to show that all finite groups with at least three elements have at least one non-trivial automorphism; in other words, there are - up to isomorphism - only finitely many finite groups $G$ such that $Aut(G)=1$ (to be exact, just two: $1$ and $C_2$).
Is an analogous statement true for all finite groups? I.e., given a finite group $A$, are there - again up to isomorphism - only finitely many groups $G$ with $Aut(G)\cong A$?
If yes, is there an upper bound on the number of such groups $G$ depending on a property of $A$ (e.g. its order)?
And if not, which groups arise as counterexamples?
And finally, what does the situation look like for infinite groups $G$ with a given finite automorphism group? And what if infinite automorphism groups $A$ are considered?
• If you talk about outer automorphism groups, so $\operatorname{Out}(G)=\operatorname{Aut}(G)/\operatorname{Inn}(G)$, and allow the groups $G$ to be infinite then for every countable group $Q$ there are infinitely many groups $G$ such that $\operatorname{Out}(G)\cong Q$. See, for example, this paper: arxiv.org/abs/1709.06441 (there are other similar results, but ensuring there are infinitely many non-isomorphic groups $G$ is easy here as the groups are HNN-extensions of triangle groups $\langle x, y; x^i, y^i, (xy)^i\rangle$, so changing $i$ immediately changes your group $G$.) – user1729 Nov 9 '17 at 17:14
Ledermann and B.H.Neumann ("On the Order of the Automorphism Group of a Finite Group. I", Proc. Royal Soc. A, 1956) have shown the following:
Theorem. Let $n > 0$. There exists a bound $f(n)$ such that if $G$ is a finite group with $|G| \geq f(n)$, then $|\operatorname{Aut}(G)| \geq n$.
An immediate consequence is that up to isomorphism, there are only finitely many finite groups $G$ with $|\operatorname{Aut}(G)| \leq n$. Hence for any finite group $X$, up to isomorphism there are only finitely many finite groups $G$ with $\operatorname{Aut}(G) \cong X$.
Among infinite groups this is no longer true, and indeed there are infinitely many groups $G$ with $\operatorname{Aut}(G) \cong \mathbb{Z} / 2 \mathbb{Z}$.
Then there is of course the question of determining all finite groups $G$ with given automorphism group $\operatorname{Aut}(G) \cong X$. For this, see for example
Iyer, Hariharan K. On solving the equation Aut(X)=G. Rocky Mountain J. Math. 9 (1979), no. 4, 653–670.
This paper gives a solution to the problem in some cases, and determines for example all $G$ with $\operatorname{Aut}(G) \cong S_n$. There is also a different proof of the fact that there are only finitely many groups with a given automorphism group (Theorem 3.1 there).
• Is there any work on an explicit form for $f(n)$? – JamalS Nov 9 '17 at 22:17
• @Mikko: Great answer; thanks a lot, also for the references (the Iyer paper contains some very interesting results)! – jpvee Nov 10 '17 at 6:26
• @JamaS: Ledermann-Neumann give an explicit $f(n)$, something around $(n-1)^{2n}$. I guess there must be some further improvements in the literature. One result is that if $G$ is a nontrivial finite abelian group, then $|\operatorname{Aut}(G)| \geq \phi(|G|)$ with equality iff $G$ is cyclic. – Mikko Korhonen Nov 10 '17 at 8:57
Mikko's nice answer concerns finite groups $G$. Let me here answer for infinite groups $G$ (but still finite automorphism groups, as in the question).
The picture is indeed very different:
For $A=C_2$ cyclic, there exists uncountably many non-isomorphic (abelian countable) groups $G$ with $\mathrm{Aut}(G)\simeq C_2$.
Indeed, for $I$ a set of primes, let $B_I$ be the additive subgroup of $\mathbf{Q}$ generated by $\{1/p:p\in I\}$. Then $B_I$ and $B_J$ are isomorphic if and only if the symmetric difference $I\triangle J$ is finite, and $\mathrm{Aut}(B_I)=\{1,-1\}$ (easy exercise: more generally for a nonzero subgroup $B$ of $\mathbf{Q}$, its automorphism group is $\{t\in\mathbf{Q}^*:tB=B\}$ acting by multiplication).
One also gets the group $C_2^n$ ($n\ge 1$) in a similar fashion. Say, for $n=2$, choose $I,J$ such that both $I\smallsetminus J$ and $J\smallsetminus I$ are infinite: then $\mathrm{Aut}(B_I\times B_J)\simeq C_2\times C_2$.
In general, if a group $G$ has finite automorphism group $A$, then its center has finite index in $G$, because $G/Z(G)$ embeds into $A$. A well-known result then implies that $[G,G]$ is finite.
[Also, it follows that if $A$ is cyclic of odd order, we deduce that $G/Z(G)$ is cyclic, and hence $G$ is abelian, and then $G$ has to be a finite elementary abelian $2$-group, and then $G=1$ or $G\simeq C_2$, whence $A=1$. In other words, for no group (finite or infinite) $G$, $\mathrm{Aut}(G)$ is cyclic of odd order $>1$.]
One more example to mention that one gets non-abelian groups: let $F$ be a finite group. Then for every torsion-free abelian group $B$, $\mathrm{Aut}(B\times F)$ is a semidirect product $(\mathrm{Aut}(F)\times\mathrm{Aut}(B))\ltimes\mathrm{Hom}(B,Z(F))$. If $\mathrm{Aut}(B)=\{\pm 1\}$, then the $\mathrm{Aut}(B)$-action on $\mathrm{Hom}(B,Z(F))$ is trivial and this reduces to the product $\mathrm{Aut}(B\times F)=(\mathrm{Aut}(F))\ltimes\mathrm{Hom}(B,Z(F))\times\mathrm{Aut}(B)$. For $B=B_I$, we have $\mathrm{Hom}(B_I,Z(F))\simeq Z(F)$. For instance, for $F=C_2$ one gets $\mathrm{Aut}(B_I\times C_2)\simeq C_2^2$. The smallest non-abelian group we can get this way has order 12, namely for $F=C_3$ or $F=D_6$ (dihedral group of order 6), one gets $\mathrm{Aut}(B_I\times F)\simeq D_6\times C_2$. For $F=C_4$ one gets $\mathrm{Aut}(B_I\times C_4)\simeq D_8\times C_2$.
I don't know if we can obtain abelian $\mathrm{Aut}(B_I\times F)$ when $|F|\ge 3$. This holds if and only if $\mathrm{Aut}(F)$ is abelian and acts trivially on $F$. Then $F$ is non-abelian, of nilpotency class 2. Possibly some large $p$-groups satisfy this (see Jain-Rai-Yadav (arXiv link) for a discussion of large $p$-groups with abelian automorphism groups; however they don't indicate if they can be chosen so that automorphisms are trivial on the center).
• Btw, out of curiosity: I could't determine if there exists an infinite group $G$ with $|\mathrm{Aut}(G)|=6$. – YCor Nov 10 '17 at 14:22
• See "de Vries, H.; de Miranda, A. B. Groups with a small number of automorphisms. Math. Z 68 1958 450–464.". It follows from results in this paper that there exists an infinite abelian group with $\operatorname{Aut}(G) \cong C_6$ (pg. 456), but there does not exist any infinite group with $\operatorname{Aut}(G) \cong S_3$. – Mikko Korhonen Nov 10 '17 at 15:08 | 2020-01-22T05:51:19 | {
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https://math.stackexchange.com/questions/3052530/can-any-proof-by-contrapositive-be-rephrased-into-a-proof-by-contradiction | Can any proof by contrapositive be rephrased into a proof by contradiction?
From my understanding,
Proof by contrapositive: Prove $$P \implies Q$$, by proving that $$\neg Q \implies \neg P$$ since they are logically equivalent.
Proof by contradiction: Prove $$P \implies Q$$ by showing that $$P \wedge \neg Q$$ yields an absurdity and hence false. So $$\neg (P \wedge \neg Q)$$ is equivalent to $$\neg (\neg (P \implies Q))$$ and $$P \implies Q$$ by double negation so showing that $$\neg (P \wedge \neg Q)$$ proves $$P \implies Q$$.
If the absurdity derived during the procedure for a proof by contradiction is $$P \wedge \neg Q \implies\neg P$$, we have essentially already proven $$P \implies Q$$ by contrapositive since $$\neg Q \implies \neg P$$ is precisely the required condition for proof by contrapositive. But $$(P \wedge \neg Q) \implies \neg P$$ is also a contradictory statement which means that $$P \implies Q$$ must be true.
Now the question is this. Is this proof by contradiction still a valid form of proof even though its a proof by contrapositive in disguise? To me, this proof by contradiction also seems to be a valid proof as it does seem to satisfy the conditions(if they are correct) for proof by contradiction.
Additionally, if you have a contrapositive proof, so you have shown that $$\neg Q \implies \neg P$$, is it possible to rephrase this in a proof by contradiction by supposing that $$P \wedge \neg Q$$ instead of just $$\neg Q$$.
If this is the case, what is the point in distinguishing proof by contradiction from proof by contrapositive?
edit: My thought is that proof by contrapositive is a direct proof while proof by contradiction, in this case, depends on the validity of the double negation law which apparently isn't valid in intuitionistic logic.
• The $(\neg Q\implies\neg P)\implies(P\implies Q)$ direction of the equivalence also requires something like double negation elimination assuming you are starting from a reasonably typical constructive logic. – Derek Elkins Dec 26 '18 at 0:25
• If every proof by contrapositive can be rephrased into a proof by contradiction, why do so many mathematicians prefer proof by contrapositive when it can be shown that way? – Sei Sakata Dec 26 '18 at 0:29
• @SeiSakata sometimes rephrasing the problem in the contrapositive form makes the proof easier or adds some intuition to the statement. Sometimes it's merely a matter of preference. – CyclotomicField Dec 26 '18 at 0:33
• @spaceisdarkgreen But what we are discussing here is only in one direction. The statement that every proof by contrapositive can be rephrased into a proof by contradiction is seemingly true but the converse doesn't necessarily hold. I am not sure if you were referring to this in your comment" any proof by contradiction “can be rephrased” as a direct proof". BTW I said seemingly true since the line of reasoning that a proof by contrapositive can be rephrased into a proof by contradiction seems to be general enough to account for all cases. But I might be mistaken and not true at all. – Sei Sakata Dec 26 '18 at 1:33
• Sorry, I didn't read your definition of what you're calling a proof by contradiction carefully... you are right that contrapositive is a special case of contradiction here (coincidentally, I wrote an answer about precisely this a couple days ago math.stackexchange.com/questions/3050738/…). – spaceisdarkgreen Dec 26 '18 at 7:40
Yes it is valid... it doesn't really matter if it's something else 'in disguise', just that is it correct. And deriving $$\lnot P$$ from $$P\land\lnot Q$$ is certainly leads to a contradiction that implies $$\lnot (P\land \lnot Q)$$ is true, which implies that $$P\to Q$$ is true. One thing to note (that I think you have noticed based on your second question) is that you have an additional assumption of $$P$$ open and available for use when you derive $$\lnot P,$$ unlike in the case of just deriving $$\lnot Q\to \lnot P$$ by assuming $$\lnot Q$$ and deriving $$\lnot P.$$
So the second question amounts to whether it is admissible to assume $$P$$ in a proof of $$\lnot Q\to \lnot P.$$ It is, and the easiest way to see this is reasoning semantically and using the completeness/soundness theorem. If $$P\vdash \lnot Q\to \lnot P$$ then every interpretation in which $$P$$ is true has $$Q$$ true, which means precisely the same thing as $$\vdash P\to Q,$$ so we have $$\vdash \lnot Q\to \lnot P.$$
As a result, proof by contrapositive is essentially a special case of what you're calling proof by contradiction, where the contradiction takes the special form $$\lnot P$$ contradicting with $$P.$$
I think the reason you might have seen framing contrapositive proofs as proofs by contradiction discouraged is for style reasons. Unless the outstanding assumption of $$P$$ is used (unnecessarily), the part of the proof outside the inner proof of $$\lnot Q\to \lnot P$$ is just boilerplate that can be omitted. It is also more informative to call it a proof by contrapositive since that is a special case of contradiction. | 2019-11-22T10:10:47 | {
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http://mathhelpforum.com/calculus/162222-another-related-rate-problem.html | # Math Help - Another Related Rate Problem
1. ## Another Related Rate Problem
Here is the text of my problem:
A boy is flying a kite at a height of 150ft. If the kite moves gorizontally away from the boy at 20ft/s, how fast is the string being paid out when the kite is 250ft from him?
Given:
• y=150ft
• x=250ft
• $\frac{dx}{dt} = 20ft/s$
Find:
• $\frac{dz}{dt} when x = 250$
Work:
• by the theorem of pythagoras:
• $z = 50\sqrt{34}$
• $\displaystyle\frac{dz}{dt} = \frac{x*\frac{dx}{dt} + y*\frac{dy}{dt}}{z}$
• $\displaystyle\frac{dz}{dt} = \frac{(250*20)+ (150*0)}{50\sqrt{34}}$
• $\displaystyle\frac{dz}{dt}=\frac{5000}{50\sqrt{34} }= \frac{100}{\sqrt{34}}}$
The book goves an answer of 16ft/s. Can someone give me a hint of where I might have gone astray? Thanks!
2. Originally Posted by dbakeg00
Here is the text of my problem:
A boy is flying a kite at a height of 150ft. If the kite moves gorizontally away from the boy at 20ft/s, how fast is the string being paid out when the kite is 250ft from him?
Given:
• y=150ft
• x=250ft
• $\frac{dx}{dt} = 20ft/s$
Find:
• $\frac{dz}{dt} when x = 250$
Work:
• by the theorem of pythagoras:
• $z = 50\sqrt{34}$
• $\displaystyle\frac{dz}{dt} = \frac{x*\frac{dx}{dt} + y*\frac{dy}{dt}}{z}$
• $\displaystyle\frac{dz}{dt} = \frac{(250*20)+ (150*0)}{50\sqrt{34}}$
• $\displaystyle\frac{dz}{dt}=\frac{5000}{50\sqrt{34} }= \frac{100}{\sqrt{34}}}$
The book goves an answer of 16ft/s. Can someone give me a hint of where I might have gone astray? Thanks!
The distance from the person is the hypotenuse not the adjacent side.
3. ok, I see what you mean.
That gives me this triangle:
x=200
y=150
z=250
which now gives me:
$\displaystyle\frac{dz}{dt}=\frac{(x*\frac{dx}{dt}) +(y*\frac{dy}{dt})}{z}$
$\displaystyle\frac{dz}{dt}=\frac{(200*20)+(150*0)} {250}$
$\displaystyle\frac{dz}{dt}=\frac{4000}{250}=16ft/s$
Thanks for the help, appreciate it very much. These word problems are sometimes difficult for me to visualize if I'm not careful. | 2015-03-31T09:33:33 | {
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http://tex.stackexchange.com/questions/1232/difference-between-big-and-bigl | # Difference between \big[ and \bigl[
What is the difference between \big[ (or equivalently \big() and \bigl[? Is it always necessary to mention l (left) and r (right)?
-
\bigl declares an opening math delimiter with less horizontal spacing than the unspecified \big. \bigr defines a closing math delimiter. Using a \bigl and \bigr pair you could get the brackets or parentheses closer to the term within.
Just compare:
\documentclass{article}
\begin{document}
$\bigl[ \times \bigr]$
$\big[ \times \big]$
\end{document}
Output:
The definitions in latex.ltx are:
\def\bigl{\mathopen\big}
\def\bigm{\mathrel\big}
\def\bigr{\mathclose\big}
-
Is there a way to make this distinction without specifying a larger size? – Mark Meckes Aug 6 '10 at 17:30
For normal brackets it's automatically done. Compare $[ \times ]$ to $\mathord[ \times \mathord]$. – Stefan Kottwitz Aug 6 '10 at 17:48
What you're saying is that [ is automatically interpreted as an opening math delimiter, so one must manually force it not to be, if desired; is that right? – Mark Meckes Aug 6 '10 at 19:18
That's true. fontmath.ltx defines: \DeclareMathDelimiter{[}{\mathopen} {operators}{"5B}{largesymbols}{"02} – Stefan Kottwitz Aug 6 '10 at 19:45
thank you. I should have a look at latex.ltx more often – pluton Aug 8 '10 at 3:22
You can see the difference in the following example. The left modifiers \bigl etc. are basically \mathopen{}\big. You also have to use \mathopen if you are using \left and \right to do automatic scaling to get correct spacing in some cases.
\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{align}
x &= \sin\biggl(\frac12\biggr) \\ % good
x &= \sin\mathopen{}\bigg(\frac12\bigg) \\ % good
x &= \sin\bigg(\frac12\bigg) \\ % bad
x &= \sin\left(\frac12\right) \\ % bad
x &= \sin\mathopen{}\left(\frac12\right) % good
\end{align}
\end{document}
- | 2014-11-22T14:34:33 | {
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https://math.stackexchange.com/questions/1650226/given-that-6-men-and-6-women-are-divided-into-pairs-what-is-the-probability | # Given that $6$ men and $6$ women are divided into pairs, what is the probability that none of the women will sit with a man?
I've generalized the question I was given here for simplicity: $6$ men and $6$ women are to be paired for a bus trip. If the pairings are done randomly, what's the probability that no women will end up sitting next to a man? Here's my first attempt, but I'm really not sure whether this is the right way to find the desired probability.
We want to group $12$ people into $6$ single-sex groups of $2$, so I began by calculating the number of ways we can make $6$ pairs. By the multinomial function: $12!/(2!)^6=7484400$ ways to make $6$ pairs.
Then, we want to figure out the number of ways we can make single-sex pairs. Again, by the multinomial function, we have $6!/(2!)^3=90$ ways to make $3$ female-female pairs. Since we also have to consider male pairs, I squared this to get $8100$ ways to make $6$ single-sex pairs.
I found that the likelihood of all pairs being single-sex is $8100/7484400=0.00108$, but this doesn't seem like a completely reasonable probability.
Could you help me find the errors in my method for solving this problem?
• It looks right to me. It might just be a rare to happen. – Christopher Carl Heckman Feb 11 '16 at 6:50
• Argh! I thought you said $\displaystyle{6! \over (2!)^3\cdot 3!}=5\cdot3\cdot 1$. So yes, your numbers are wrong. (Also, $\displaystyle{12!\over (2!)^6 \cdot 6!}$ is the correct number of ways to pair up the people.) – Christopher Carl Heckman Feb 11 '16 at 7:05
• So then, for single-sex pairs, you would find that there are $6!/(2!)^3 3!=15$ possible ways to make female-female pairs, which makes 225 total ways to make all single-sex pairs--is that what you're implying? – cembos1005 Feb 11 '16 at 7:12
• The probability is $(5/11)(3/9)(1/7)$. Minimal counting. – André Nicolas Feb 11 '16 at 7:15
• @cembos1005 Yes. See my comment after David's answer. – Christopher Carl Heckman Feb 11 '16 at 7:18
We want to group $12$ people into $6$ single-sex groups of $2$, so I began by calculating the number of ways we can make $6$ pairs.
By the multinomial function: $12!/(2!)^6=7484400$ ways to make $6$ pairs.
No, order doesn't matter within the groups of $2$ nor of the $6$ groups.
$$\frac{12!}{2!^6 6!} = 10395$$
Then, we want to figure out the number of ways we can make single-sex pairs. Again, by the multinomial function, we have $6!/(2!)^3=90$ ways to make $3$ female-female pairs. Since we also have to consider male pairs, I squared this to get $8100$ ways to make $6$ single-sex pairs.
$$\left(\frac{6!}{2!^3 3!}\right)^2 = 225$$
I found that the likelihood of all pairs being single-sex is $8100/7484400=0.00108$, but this doesn't seem like a completely reasonable probability.
$$\frac{6!^3}{12!3!^2}= \frac{5}{231}$$
This is also equal to $\frac{5}{11}\frac{3}{9}\frac{1}{7}$, the probability that a girl is paired with a girl, another girl is paired with a girl, and that the last two girls are paired.
There is an error here.
In the general case, you considered pairings to be different if the order of the six pairs was different.
In the single-sex case, there is an ordering, but only within each of the two groups of three pairs.
To correct the error, it would probably be easiest to consider selections of six pairs without taking into account the way the six pairs are ordered.
Edit Here is how I would solve the problem in the quickest way. Let's first pick a mate for the first man. What is the probability this will be a man? It's $5/11$. Assume that occurs. The probability the third man will be paired with a man is $3/9$. Finally, the likelihood the last two men will be matched is $1/7$. So the answer is $(5/11) \times (1/3) \times (1/7) = 5/231 \approx 0.0216$.
To stay as close as possible to your method, you need to divide your general answer by the number of ways to order the six pairs, which is $6!$. For the second part, divide each time by $3!$. So the answer will be $$\frac{(6!/[(2!)^3 3!])^2}{12!/[(2!)^6 6!]}$$
• No there isn't. If you arbitrarily sort the women into A, B, C, D, E, F, you can ask: How many ways are there to match up the first unmatched woman in the list (in this case, A)? Answer: 5. How many ways are there to match the first unmatched woman in the list (whether it's B or C)? Answer: 3, and now there's only one pair left. ((ADDENDUM:)) The method is right; I misread what the OP calculated this answer as. – Christopher Carl Heckman Feb 11 '16 at 7:02
• What would be the best way to do this without considering order? Obviously this is a Combinations problem without replacement, but I thought the multinomial function took care of the order issue, as it's of the n C m form. – cembos1005 Feb 11 '16 at 7:05
• Dividing by $2!$ several times took care of the ordering within the pairs. However, you failed to take into account the ordering of the six (or three) pairs themselves. – David Feb 11 '16 at 7:14
• @CarlHeckman I don't understand exactly what your objection is to what I wrote. – David Feb 11 '16 at 7:21
• @David You said there was an error. I originally said, no there isn't, then re-read the OP, and you were right; there was an error. (I thought the OP had included an extra $6!$ or $3!$ in getting his numbers.) – Christopher Carl Heckman Feb 11 '16 at 7:23
When in doubt about your solution to a certain problem, you should always try to approach the problem differently and see if you get the same answer. If the answers differ, figuring out where the difference lies is itself an interesting exercise.
E.g. in this case compute the chance of the first pair being a single-sex pair, then compute the chance of the second pair being a single-sex pair, etc..
1st pair being single-sex:
$$\frac{5}{11}$$
2nd pair being single-sex given 1st pair is single-sex:
$$\frac{6}{10}\cdot\frac{5}{9} + \frac{4}{10}\cdot\frac{3}{9}$$
3rd pair being single-sex given 1st & 2nd pair are single-sex:
$$(\frac{6}{8}\cdot\frac{5}{7} + \frac{2}{8}\cdot\frac{1}{7})+(\frac{5}{8}\cdot\frac{4}{7} + \frac{3}{8}\cdot\frac{2}{7})+(\frac{4}{8}\cdot\frac{3}{7} + \frac{5}{8}\cdot\frac{4}{7})+(\frac{3}{8}\cdot\frac{2}{7} + \frac{6}{8}\cdot\frac{5}{7})$$
....
after that multiply them together to obtain the chance of having 6 single-sex pair, and simply simplify the multiplicative expression to see if it is equivalent to the expression you initially came up with.
And now you may notice that there is no point keep tracking each pair. All you need is to compute the chance of 3 pairs being single-sex of the same sex, and the remaining pairs of the opposite sex would also be single-sex.
With so many approaches already posted, you might like the conceptually simple $\dfrac{\binom63}{\binom{12}{6}}$, although it lacks the minimalist elegance of $\frac5{11}\frac39\frac17$ | 2019-10-18T08:45:32 | {
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http://mathhelpforum.com/calculus/204760-question-area-between-polar-curves.html | # Math Help - Question on Area between Polar Curves
1. ## Question on Area between Polar Curves
Hey everyone,
I have two questions regarding the area of polar curves.
1. Find the area of the region lying the polar curve r=1 + cos(theta), and outside the polar curve r= 2cos(theta)
2. Find the area of the shaded region inside the graph of r= 1+2cos(theta); (the graph shows the top half of the cardioid shaded)
Basically, I know how to solve these problem and how to draw the graphs. What I need help on is finding the limits of integration. For the 1st problem, I picked my limits of integration to be 0 to pi, and then I multiplied the integral by 2 to find the total area. (Is this method correct)
For the 2nd, I solved for r, and got cos(theta)= -1/2. So would my limits of integration be 2pi/3 to 4pi/3?
Any help and feedback appreciated.
Thanks
2. ## Re: Question on Area between Polar Curves
Originally Posted by Beevo
Hey everyone,
I have two questions regarding the area of polar curves.
1. Find the area of the region lying the polar curve r=1 + cos(theta), and outside the polar curve r= 2cos(theta)
2. Find the area of the shaded region inside the graph of r= 1+2cos(theta); (the graph shows the top half of the cardioid shaded)
Basically, I know how to solve these problem and how to draw the graphs. What I need help on is finding the limits of integration. For the 1st problem, I picked my limits of integration to be 0 to pi, and then I multiplied the integral by 2 to find the total area. (Is this method correct)
For the 2nd, I solved for r, and got cos(theta)= -1/2. So would my limits of integration be 2pi/3 to 4pi/3?
Any help and feedback appreciated.
Thanks
In the first one, when you set up your double integral, for the top half, the radii are bounded above by \displaystyle \begin{align*} r = 1 + \cos{\theta} \end{align*} and the radii are bounded below by \displaystyle \begin{align*} r = 2\cos{\theta} \end{align*}. I agree with your bounds for \displaystyle \begin{align*} \theta \end{align*}. So to find the area, your double integral is \displaystyle \begin{align*} A = 2\int_0^{\pi}{\int_{2\cos{\theta}}^{1 + \cos{\theta}}{r\,dr}\,d\theta} \end{align*}.
For part 2, you follow a similar process with the same \displaystyle \begin{align*} \theta \end{align*} limits, and your radii are bounded above by \displaystyle \begin{align*} r = 1 + 2\cos{\theta} \end{align*} and below by \displaystyle \begin{align*} r = 0 \end{align*}, giving your double integral for the area as \displaystyle \begin{align*} A = \int_0^{\pi}{ \int_0^{1 + 2\cos{\theta}}{r\,dr} \,d\theta} \end{align*}.
3. ## Re: Question on Area between Polar Curves
That makes sense, appreciate your help, thanks.
4. ## Re: Question on Area between Polar Curves
Hello, Beevo!
1. Find the area of the region lying the polar curve $r\:=\:1 + \cos\theta$
and outside the polar curve $r\:=\: 2\cos\theta$
Basically, I know how to solve these problem and how to draw the graphs.
What I need help on is finding the limits of integration.
For the 1st problem, I picked my limits of integration to be 0 to pi,
and then I multiplied the integral by 2 to find the total area.
Is this method correct?
Yes . . . good work!
2. Find the area of the shaded region inside the graph of $r \:=\:1 + 2\cos\theta$
(The graph shows the top half of the cardioid shaded.)
For the 2nd, I solved for r, and got cos(theta)= -1/2.
So would my limits of integration be 2pi/3 to 4pi/3? . . No
You solved $r \,=\,0$
You found the angles at which the curve passes through the pole (origin).
These are not the limits of integration. .(Well, probably not.)
There is yet another problem.
This is not a cardioid.
It is a limacon with an internal "loop".
. . The entire upper half?
. . The upper half minus the loop?
. . Just the loop?
5. ## Re: Question on Area between Polar Curves
Originally Posted by Soroban
Hello, Beevo!
Yes . . . good work!
You solved $r \,=\,0$
You found the angles at which the curve passes through the pole (origin).
These are not the limits of integration. .(Well, probably not.)
There is yet another problem.
This is not a cardioid.
It is a limacon with an internal "loop". | 2014-08-30T19:09:53 | {
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http://math.stackexchange.com/questions/94331/flipping-a-coin | # Flipping a coin
Two players, $A$ and $B$, alternately and independently flip a coin and the first player to get a head wins. Assume player $A$ flips first. If the coin is fair, what is the probability that $A$ wins?
So $A$ only flips on odd tosses. So the probability of winning would be $$P =\frac{1}{2}+\left(\frac{1}{2} \right)^{2} \frac{1}{2} + \cdots+ \left(\frac{1}{2} \right)^{2n} \frac{1}{2}$$
Is that right? It seems that if $A$ only flips on odd tosses, this shouldn't matter. Either $A$ can win on his first toss, his second toss, ...., or his $n^{th}$ toss. So the third flip of the coin is actually $A$'s second toss. So shouldn't it be $$P = \frac{1}{2} + \left(\frac{1}{2} \right)^{2} + \left(\frac{1}{2} \right)^{3} + \cdots$$
-
Think of it like this: A and B both throw a coin. If $A$ gets head, he wins. If $B$ gets head, but $A$ doesn't, $B$ wins. If neither has head, they throw again. $A$ wins in 2 out of 3 events (head-head and head-tail) and $B$ wins in 1 case (tail-head). – Myself Dec 27 '11 at 1:43
Your first idea was right, apart from the fact that you need to sum forever. The second expression for $P$ is not right. The situations in which A wins are (with obvious abbreviations) H, TTH, TTTTH, TTTTTTH, and so on. The probability of each is easy to calculate. add up. The mistake in the second estimate is that it is not only A's tosses that matter. B's coin needs to cooperate by coming up tails. – André Nicolas Dec 27 '11 at 2:15
Let $p$ be the probability that A wins. This can happen in two ways: (i) A wins immediately (probability $1/2$ or (ii) A tosses a tail, but ultimately wins.
If A tossed a tail (probability $1/2$, then in effect B is now "first" so the probability she does not win is $1-p$. We conclude that $$p=\frac{1}{2}+\frac{1}{2}(1-p).$$ Solve for $p$. We get $p=2/3$.
Comment: There are nice expressions for $p$ as infinite geometric series. So we can think of the above argument as a probabilistic method for summing a very particular geometric series. By varying the probability that the coin lands heads, we can use the same idea to find the sum of any infinite geometric series, as long as the "common ratio" is positive.
-
You were on track at first, but the game may go on for a very long time. You should not stop at "$2n$".
$A$ could win on the first flip of the coin, or the third, or the fifth, ...
So, the probability of $A$ winning would be the sum of the probabilities of the events $$A_i = A\text{ wins on the } i^{\rm th}\text{ flip; where }i\text { is odd}.$$
If the coin is fair, $$P(A_i) =\Bigl({1\over2}\Bigr)^{i-1}\cdot{1\over2} =\Bigl({1\over2}\Bigr)^i.$$
Summing the probabilities above, the probability that $A$ wins is $$\sum_{i \text{ odd}} P(A_i)= \sum_{i \text { odd}} \Bigl({1\over 2}\Bigr)^i.$$ The series above can be written: $$\sum_{i \text { odd}} \Bigl({1\over 2}\Bigr)^i= \sum_{i =0}^{\infty} \Bigl({1\over 2}\Bigr)^{2i+1}= {1\over 2}\sum_{i =0}^{\infty} \Bigl({1\over 4}\Bigr)^{ i }={1\over2}\cdot{4\over3}=2/3.$$
You could also solve the problem this way:
Condition on what happens on the first two flips:
$A$ wins if the first flip is a head and the probability that the first flip is a head is 1/2.
$A$ loses if the first flip is a tail and the second ($B$'s turn) is a head. The probability that the first flip is a head and the second a tail is 1/4.
If the first two flips are tails, then given this, the probability that $A$ wins eventually afterwards is the same as the initial probability that $A$ wins. The probability that the first two flips are tails is 1/4.
So $$P(A) =1\cdot{1\over2}+0\cdot {1\over4}+P(A)\cdot{1\over4}.$$
Solving the above for $P(A)$ gives $P(A)=2/3$.
As for your second method, I think you are forgetting that $A$ does not get a second toss if $B$ flips heads on his first toss...
-
the probability that the first head occurs on toss $n$ is $2^{-n}$ so the probability that the first head happens on an odd $n$ is $\sum_{k=0}^{\infty}2^{-(2k+1)}=(1/2)(1/(1-1/4))=2/3$
-
Here is the solution for general p. Let A gets the head in Nth trial to win the game. Since he is flipping the coin in odd trials,
P(N=1) = p, P(N=3) = (1-p)^2 * p, P(N=5) = (1-p)^4 * p, and so on.
Thus P(A wins) = p + (1-p)^2 * p + (1-p)^4 * p + ... = 1/(2-p). If p = 1/2, then P(wins) = 2/3.
-
\mathsrc{A_k}={A losing in his first k tosses and B losing in his first k tosses and A winning in his k+1 toss} \begin{align} P(\text{A winning}) &= P(\text{A winning in his first toss or } \mathrm{A_1} \text{ or } \mathrm{A_2} \text{ or } \ldots)\\ &= P(\text{A winning in his first toss}) + P(\mathrm{A_1}) + P(\mathrm{A_2}) + \dots\\ &= 0.5 + (0.5)(0.5)(0.5) + (0.5)(0.5)(0.5)(0.5)(0.5) + \dots \\ &=0.5 + (0.5)^3 + (0.5)^5 + .........\\ &= \frac{0.5}{1 - 0.25} = \frac{0.5}{0.75} = \frac{50}{75} = \frac{2}{3}\\ \end{align}
-
There were answers with more details, better appearance, with the same final answer. Use Help please. – Hoseyn Heydari Feb 12 '14 at 8:49 | 2016-05-01T14:21:07 | {
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https://mail.queryxchange.com/q/21_716/sum-of-the-alternating-harmonic-series-sum-k-1-infty-frac-1-k-1-k-frac-1-1-frac-1-2-cdots/ | # Sum of the alternating harmonic series $\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k} = \frac{1}{1} - \frac{1}{2} + \cdots$
by Isaac Last Updated January 16, 2018 08:20 AM
I know that the harmonic series $$\sum_{k=1}^{\infty}\frac{1}{k} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \cdots + \frac{1}{n} + \cdots \tag{I}$$ diverges, but what about the alternating harmonic series
$$\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k} = \frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \cdots + \frac{(-1)^{n+1}}{n} + \cdots \text{?} \tag{II}$$
Does it converge? If so, what is its sum?
Tags :
it is not absolutely convergent (that is, if you are allowed to reorder terms you may end up with whatever number you fancy).
If you consider the associated series formed by summing the terms from 1 to n of the original one, that is you fix the order of summation of the original series, that series (which is not the original one...) converges to $\ln(2)$ See Wikipedia.
mau
July 26, 2010 08:50 AM
Call a series $a_n$ absolutely convergent if $\sum|a_n|$ converges. If $a_n$ converges but is not absolutely convergent we call $a_n$ conditionally convergent The Riemann series theorem states that any conditionally convergent series can be reordered to converge to any real number.
Morally this is because both the positive and negative parts of your series diverge but the divergences cancel each other out, one or other's canceling the other can be staggered by adding on, say, the negative bits every third term in stead of every other term. This means that in the race for the two divergences to cancel each other out, we give the positive bit something of a head-start and will get a larger positive outcome. Notice how, even in this rearranged version of the series, every term will still come up exactly once.
It is also worth noting, on the Wikipedia link Mau provided, that the convergence to $\ln 2$ of your series is at the edge of the radius of convergence for the series expansion of $\ln(1-x)$- this is a fairly typical occurrence: at the boundary of a domain of convergence of a Taylor series, the series is only just converging- which is why you see this conditional convergence type behavior.
Tom Boardman
July 26, 2010 09:14 AM
Let's say you have a sequence of nonnegative numbers $a_1 \geq a_2 \geq \dots$ tending to zero. Then it is a theorem that the alternating sum $\sum (-1)^i a_i$ converges (not necessarily absolutely, of course). This in particular applies to your series.
Incidentally, if you're curious why it converges to $\log(2)$ (which seems somewhat random), it's because of the Taylor series of $\log(1+x)$ while letting $x \to 1$.
Akhil Mathew
July 26, 2010 12:28 PM
There are actually two "more direct" proofs of the fact that this limit is $\ln (2)$.
First Proof Using the well knows (typical induction problem) equality:
$$\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2n-1}-\frac{1}{2n}=\frac{1}{n+1}+\frac{1}{n+2}+..+\frac{1}{2n} \,.$$
The right side is $\frac{1}{n} \left[ \frac{1}{1+\frac{1}{n}}+ \frac{1}{1+\frac{2}{n}}+..+\frac{1}{1+\frac{n}{n}} \right]$ which is the standard Riemann sum associated to $\int_0^1 \frac{1}{1+x} dx \,.$
Second Proof Using $\lim_n \frac{1}{1}+\frac{1}{2}+...+\frac{1}{n}-\ln (n) =\gamma$.
Then
$$\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2n-1}-\frac{1}{2n}= \left[ \frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2n-1}+\frac{1}{2n} \right]-2 \left[\frac{1}{2}+\frac{1}{4}...+\frac{1}{2n} \right]$$
$$= \left[ \frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2n-1}+\frac{1}{2n} \right]-\ln(2n) - \left[\frac{1}{1}+\frac{1}{2}...+\frac{1}{n} \right]+\ln(n) + \ln 2 \,.$$
Taking the limit we get $\gamma-\gamma+\ln(2)$.
N. S.
June 02, 2011 04:54 AM
In this answer, I used only Bernoulli's inequality to show that $$\left(\frac{2n+1}{n+1}\right)^\frac{n}{n+1} \le\left(1+\frac1n\right)^{n\left(\frac1{n+1}+\frac1{n+2}+\dots+\frac1{2n}\right)} \le\frac{2n+1}{n+1}\tag{1}$$ The squeeze theorem and $(1)$, show that $$\exp\left[\lim\limits_{n\to\infty}\left(\frac1{n+1}+\frac1{n+2}+\dots+\frac1{2n}\right)\right]=2\tag{2}$$ That is, \begin{align} \lim_{n\to\infty}\left(1-\frac12+\frac13-\frac14+\dots-\frac1{2n}\right) &=\lim_{n\to\infty}\left(\frac1{n+1}+\frac1{n+2}+\dots+\frac1{2n}\right)\\[6pt] &=\log(2)\tag{3} \end{align}
robjohn
April 16, 2013 16:04 PM
$\sum_{k=1}^{n} ( \frac{1}{2k-1}-\frac{1}{2k} ) = \sum_{k=1}^{n} ( \frac{1}{2k-1}+\frac{1}{2k} ) - 2 \sum_{k=1}^{n} \frac{1}{2k} = \sum_{k=1}^{2n} \frac{1}{k} - \sum_{k=1}^{n} \frac{1}{k} = \sum_{k=n+1}^{2n} \frac{1}{k}$.
$\ln(2) \overset{n\to\infty}{\leftarrow} \ln(2) + \ln(\frac{2n+1}{2n+2}) = \ln(2n+1)-\ln(n+1)$
$= \int_{n+1}^{2n+1} \frac{1}{x}\ dx \le \sum_{k=n+1}^{2n} \frac{1}{k} \le \int_{n}^{2n} \frac{1}{x}\ dx$
$= \ln(2n)-\ln(n) = \ln(2)$.
So by squeeze theorem we are done.
user21820
April 05, 2015 03:37 AM
Here is another proof, based on the formula
$$\frac{1}{1+x}=\frac{x^n}{1+x}+\sum_{k=0}^n(-x)^k$$
Integrating both sides over $[0,t]$ gives
$$\ln(1+t)=\int_0^t\frac{x^n}{1+x}\,dx+\sum_{k=1}^n\frac{(-t)^{k+1}}{k}$$
Setting $t=1$ shows that the partial sums $s_n$ of the alternating harmonic series are given by
$$s_n=\ln2-\int_0^1\frac{x^n}{1+x}\,dx$$
But on $[0,1]$, we have $0\leq x^n(1+x)^{-1}\leq x^{n-1}$, so
$$0\leq\int_0^1\frac{x^n}{1+x}\,dx\leq\int_0^1x^{n-1}\,dx=\frac{1}{n}$$
Hence $s_n\to\ln 2$ as $n\to\infty$.
symplectomorphic
August 09, 2016 01:58 AM | 2018-06-25T10:05:49 | {
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https://math.stackexchange.com/questions/4316661/formula-for-some-average/4316694 | # Formula for some average
I am finding a formula for the average $$A_n$$ of the numbers $$(a_1-a_2)^2 + \cdots + (a_{n-1} - a_n )^2$$ over all cases that $$\{a_1, \cdots, a_n \} = \{ 1,2, \cdots, n\}$$.
For example, $$A_2=1, A_3=4, A_4=10, A_5=20, A_6=35, \cdots$$. From this, I guess: $$A_n = \frac{1}{6} (n-1) n (n+1).$$
Can anyone prove it or give correct formula?
• are you sure the terms are $(1,3,10,20,35,.....)$ , cuz there not any OEIS for this Nov 26, 2021 at 10:39
• 3 should be 4. It's corrected now. Nov 26, 2021 at 10:55
From your results, one can coclude that you calculate average of all permutation of set. Then due to symmetry $$A_n=(n-1) B_n$$ , where $$B_n$$ is average of $$(a_1-a_2)^2$$ for all possible pairs $$(a_1,a_2)$$. $$B_n$$ could be found directly:
$$B_n=\frac{2}{n(n-1)}\sum_{a_1=1}^{n-1} \sum_{a_2=a_1+1}^{n} (a_1-a_2)^2=\frac{1}{6}n(n+1)$$
Then $$A_n=\frac{1}{6}n(n-1)(n+1)$$
• A bit more detailed: We want to compute $E[(a_1 - a_2)^2 + (a_2 - a_3)^2 + \dots + (a_{n-1} - a_n)^2]$. By linearity of expectation, this becomes $E[(a_1 - a_2)^2] + E[(a_2 - a_3)^2] + \dots + E[(a_{n-1} - a_n)^2]$. By symmetry, the $n-1$ terms are equal, so we simply need to find $(n-1)E[(a_1 - a_2)^2]$. Nov 26, 2021 at 11:48 | 2022-09-30T16:32:46 | {
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https://compneuro.neuromatch.io/tutorials/W1D3_ModelFitting/student/W1D3_Tutorial1.html | Tutorial 1: Linear regression with MSE¶
Week 1, Day 3: Model Fitting
Content creators: Pierre-Étienne Fiquet, Anqi Wu, Alex Hyafil with help from Byron Galbraith
Content reviewers: Lina Teichmann, Saeed Salehi, Patrick Mineault, Ella Batty, Michael Waskom
Tutorial Objectives¶
Estimated timing of tutorial: 30 minutes
This is Tutorial 1 of a series on fitting models to data. We start with simple linear regression, using least squares optimization (Tutorial 1) and Maximum Likelihood Estimation (Tutorial 2). We will use bootstrapping to build confidence intervals around the inferred linear model parameters (Tutorial 3). We’ll finish our exploration of regression models by generalizing to multiple linear regression and polynomial regression (Tutorial 4). We end by learning how to choose between these various models. We discuss the bias-variance trade-off (Tutorial 5) and Cross Validation for model selection (Tutorial 6).
In this tutorial, we will learn how to fit simple linear models to data.
• Learn how to calculate the mean-squared error (MSE)
• Explore how model parameters (slope) influence the MSE
• Learn how to find the optimal model parameter using least-squares optimization
acknowledgements:
• we thank Eero Simoncelli, much of today’s tutorials are inspired by exercises asigned in his mathtools class.
Tutorial slides¶
These are the slides for the videos in all tutorials today
Setup¶
import numpy as np
import matplotlib.pyplot as plt
Figure Settings¶
#@title Figure Settings
import ipywidgets as widgets # interactive display
%config InlineBackend.figure_format = 'retina'
Plotting Functions¶
#@title Plotting Functions
def plot_observed_vs_predicted(x, y, y_hat, theta_hat):
""" Plot observed vs predicted data
Args:
x (ndarray): observed x values
y (ndarray): observed y values
y_hat (ndarray): predicted y values
theta_hat (ndarray):
"""
fig, ax = plt.subplots()
ax.scatter(x, y, label='Observed') # our data scatter plot
ax.plot(x, y_hat, color='r', label='Fit') # our estimated model
# plot residuals
ymin = np.minimum(y, y_hat)
ymax = np.maximum(y, y_hat)
ax.vlines(x, ymin, ymax, 'g', alpha=0.5, label='Residuals')
ax.set(
title=fr"$\hat{{\theta}}$ = {theta_hat:0.2f}, MSE = {mse(x, y, theta_hat):.2f}",
xlabel='x',
ylabel='y'
)
ax.legend()
Section 1: Mean Squared Error (MSE)¶
Video 1: Linear Regression & Mean Squared Error¶
This video covers a 1D linear regression and mean squared error.
Linear least squares regression is an old but gold optimization procedure that we are going to use for data fitting. Least squares (LS) optimization problems are those in which the objective function is a quadratic function of the parameter(s) being optimized.
Suppose you have a set of measurements: for each data point or measurement, you have $$y_{i}$$ (the “dependent” variable) obtained for a different input value, $$x_{i}$$ (the “independent” or “explanatory” variable). Suppose we believe the measurements are proportional to the input values, but are corrupted by some (random) measurement errors, $$\epsilon_{i}$$, that is:
(43)\begin{align} y_{i}= \theta x_{i}+\epsilon_{i} \end{align}
for some unknown slope parameter $$\theta.$$ The least squares regression problem uses mean squared error (MSE) as its objective function, it aims to find the value of the parameter $$\theta$$ by minimizing the average of squared errors:
(44)\begin{align} \min _{\theta} \frac{1}{N}\sum_{i=1}^{N}\left(y_{i}-\theta x_{i}\right)^{2} \end{align}
We will now explore how MSE is used in fitting a linear regression model to data. For illustrative purposes, we will create a simple synthetic dataset where we know the true underlying model. This will allow us to see how our estimation efforts compare in uncovering the real model (though in practice we rarely have this luxury).
First we will generate some noisy samples $$x$$ from [0, 10) along the line $$y = 1.2x$$ as our dataset we wish to fit a model to.
¶
Execute this cell to generate some simulated data
# @title
# @markdown Execute this cell to generate some simulated data
# setting a fixed seed to our random number generator ensures we will always
# get the same psuedorandom number sequence
np.random.seed(121)
# Let's set some parameters
theta = 1.2
n_samples = 30
# Draw x and then calculate y
x = 10 * np.random.rand(n_samples) # sample from a uniform distribution over [0,10)
noise = np.random.randn(n_samples) # sample from a standard normal distribution
y = theta * x + noise
# Plot the results
fig, ax = plt.subplots()
ax.scatter(x, y) # produces a scatter plot
ax.set(xlabel='x', ylabel='y');
Now that we have our suitably noisy dataset, we can start trying to estimate the underlying model that produced it. We use MSE to evaluate how successful a particular slope estimate $$\hat{\theta}$$ is for explaining the data, with the closer to 0 the MSE is, the better our estimate fits the data.
Coding Exercise 1: Compute MSE¶
In this exercise you will implement a method to compute the mean squared error for a set of inputs $$\mathbf{x}$$, measurements $$\mathbf{y}$$, and slope estimate $$\hat{\theta}$$. Here, $$\mathbf{x}$$ and $$\mathbf{y}$$ are vectors of data points. We will then compute and print the mean squared error for 3 different choices of theta.
As a reminder, the equation for computing the estimated y for a single data point is:
$\hat{y}_{i}= \theta x_{i}$
and for mean squared error is:
(45)\begin{align} \min _{\theta} \frac{1}{N}\sum_{i=1}^{N}\left(y_{i}-\hat{y}_i\right)^{2} \end{align}
def mse(x, y, theta_hat):
"""Compute the mean squared error
Args:
x (ndarray): An array of shape (samples,) that contains the input values.
y (ndarray): An array of shape (samples,) that contains the corresponding
measurement values to the inputs.
theta_hat (float): An estimate of the slope parameter
Returns:
float: The mean squared error of the data with the estimated parameter.
"""
####################################################
## TODO for students: compute the mean squared error
# Fill out function and remove
raise NotImplementedError("Student exercise: compute the mean squared error")
####################################################
# Compute the estimated y
y_hat = ...
# Compute mean squared error
mse = ...
return mse
theta_hats = [0.75, 1.0, 1.5]
for theta_hat in theta_hats:
print(f"theta_hat of {theta_hat} has an MSE of {mse(x, y, theta_hat):.2f}")
Click for solution
The result should be:
theta_hat of 0.75 has an MSE of 9.08
theta_hat of 1.0 has an MSE of 3.0
theta_hat of 1.5 has an MSE of 4.52
We see that $$\hat{\theta} = 1.0$$ is our best estimate from the three we tried. Looking just at the raw numbers, however, isn’t always satisfying, so let’s visualize what our estimated model looks like over the data.
¶
Execute this cell to visualize estimated models
#@title
#@markdown Execute this cell to visualize estimated models
fig, axes = plt.subplots(ncols=3, figsize=(18, 4))
for theta_hat, ax in zip(theta_hats, axes):
# True data
ax.scatter(x, y, label='Observed') # our data scatter plot
# Compute and plot predictions
y_hat = theta_hat * x
ax.plot(x, y_hat, color='r', label='Fit') # our estimated model
ax.set(
title= fr'$\hat{{\theta}}$= {theta_hat}, MSE = {mse(x, y, theta_hat):.2f}',
xlabel='x',
ylabel='y'
);
axes[0].legend()
Interactive Demo 1: MSE Explorer¶
Using an interactive widget, we can easily see how changing our slope estimate changes our model fit. We display the residuals, the differences between observed and predicted data, as line segments between the data point (observed response) and the corresponding predicted response on the model fit line.
• What value of $$\hat{\theta}$$ results in the lowest MSE?
• Is this a good way of estimating $$\theta$$?
¶
Make sure you execute this cell to enable the widget!
#@title
#@markdown Make sure you execute this cell to enable the widget!
@widgets.interact(theta_hat=widgets.FloatSlider(1.0, min=0.0, max=2.0))
def plot_data_estimate(theta_hat):
y_hat = theta_hat * x
plot_observed_vs_predicted(x, y, y_hat, theta_hat)
Click for solution
While visually exploring several estimates can be instructive, it’s not the most efficient for finding the best estimate to fit our data. Another technique we can use is choose a reasonable range of parameter values and compute the MSE at several values in that interval. This allows us to plot the error against the parameter value (this is also called an error landscape, especially when we deal with more than one parameter). We can select the final $$\hat{\theta}$$ ($$\hat{\theta}_\textrm{MSE}$$) as the one which results in the lowest error.
¶
Execute this cell to loop over theta_hats, compute MSE, and plot results
# @title
# @markdown Execute this cell to loop over theta_hats, compute MSE, and plot results
# Loop over different thetas, compute MSE for each
theta_hat_grid = np.linspace(-2.0, 4.0)
errors = np.zeros(len(theta_hat_grid))
for i, theta_hat in enumerate(theta_hat_grid):
errors[i] = mse(x, y, theta_hat)
# Find theta that results in lowest error
best_error = np.min(errors)
theta_hat = theta_hat_grid[np.argmin(errors)]
# Plot results
fig, ax = plt.subplots()
ax.plot(theta_hat_grid, errors, '-o', label='MSE', c='C1')
ax.axvline(theta, color='g', ls='--', label=r"$\theta_{True}$")
ax.axvline(theta_hat, color='r', ls='-', label=r"$\hat{{\theta}}_{MSE}$")
ax.set(
title=fr"Best fit: $\hat{{\theta}}$ = {theta_hat:.2f}, MSE = {best_error:.2f}",
xlabel=r"$\hat{{\theta}}$",
ylabel='MSE')
ax.legend();
We can see that our best fit is $$\hat{\theta}=1.18$$ with an MSE of 1.45. This is quite close to the original true value $$\theta=1.2$$!
Section 2: Least-squares optimization¶
Estimated timing to here from start of tutorial: 20 min
While the approach detailed above (computing MSE at various values of $$\hat\theta$$) quickly got us to a good estimate, it still relied on evaluating the MSE value across a grid of hand-specified values. If we didn’t pick a good range to begin with, or with enough granularity, we might miss the best possible estimator. Let’s go one step further, and instead of finding the minimum MSE from a set of candidate estimates, let’s solve for it analytically.
We can do this by minimizing the cost function. Mean squared error is a convex objective function, therefore we can compute its minimum using calculus. Please see video or Bonus Section 1 for this derivation! After computing the minimum, we find that:
(46)\begin{align} \hat\theta = \frac{\mathbf{x}^\top \mathbf{y}}{\mathbf{x}^\top \mathbf{x}} \end{align}
where $$\mathbf{x}$$ and $$\mathbf{y}$$ are vectors of data points.
This is known as solving the normal equations. For different ways of obtaining the solution, see the notes on Least Squares Optimization by Eero Simoncelli.
Coding Exercise 2: Solve for the Optimal Estimator¶
In this exercise, you will write a function that finds the optimal $$\hat{\theta}$$ value using the least squares optimization approach (the equation above) to solve MSE minimization. It shoud take arguments $$x$$ and $$y$$ and return the solution $$\hat{\theta}$$.
We will then use your function to compute $$\hat{\theta}$$ and plot the resulting prediction on top of the data.
def solve_normal_eqn(x, y):
"""Solve the normal equations to produce the value of theta_hat that minimizes
MSE.
Args:
x (ndarray): An array of shape (samples,) that contains the input values.
y (ndarray): An array of shape (samples,) that contains the corresponding
measurement values to the inputs.
Returns:
float: the value for theta_hat arrived from minimizing MSE
"""
################################################################################
## TODO for students: solve for the best parameter using least squares
# Fill out function and remove
raise NotImplementedError("Student exercise: solve for theta_hat using least squares")
################################################################################
# Compute theta_hat analytically
theta_hat = ...
return theta_hat
theta_hat = solve_normal_eqn(x, y)
y_hat = theta_hat * x
plot_observed_vs_predicted(x, y, y_hat, theta_hat)
Click for solution
Example output:
We see that the analytic solution produces an even better result than our grid search from before, producing $$\hat{\theta} = 1.21$$ with MSE = 1.43!
Summary¶
Estimated timing of tutorial: 30 minutes
Linear least squares regression is an optimization procedure that can be used for data fitting:
• Task: predict a value for $$y_i$$ given $$x_i$$
• Performance measure: $$\textrm{MSE}$$
• Procedure: minimize $$\textrm{MSE}$$ by solving the normal equations
Key point: We fit the model by defining an objective function and minimizing it.
Note: In this case, there is an analytical solution to the minimization problem and in practice, this solution can be computed using linear algebra. This is extremely powerful and forms the basis for much of numerical computation throughout the sciences.
Notation¶
(47)\begin{align} x_{i} &\quad \text{input, independent variable}\\ y_{i} &\quad \text{measurement, dependent variable}\\ \mathbf{x} &\quad \text{vector of input values}\\ \mathbf{y} &\quad \text{vector of measurements}\\ \hat{y}_{i} &\quad \text{estimate of dependent variable}\\ \epsilon_{i} &\quad \text{measurement error}\\ \theta &\quad \text{slope parameter}\\ \hat{\theta} &\quad \text{estimated slope parameter}\\ \hat{\theta}_\text{MSE} &\quad \text{slope parameter estimated via the mean squared error}\\ \textrm{MSE} &\quad \text{mean squared error}\\ \end{align}
Bonus¶
Bonus Section 1: Least Squares Optimization Derivation¶
We will outline here the derivation of the least squares solution.
We first set the derivative of the error expression with respect to $$\theta$$ equal to zero,
(48)\begin{align} \frac{d}{d\theta}\frac{1}{N}\sum_{i=1}^N(y_i - \theta x_i)^2 = 0 \\ \frac{1}{N}\sum_{i=1}^N-2x_i(y_i - \theta x_i) = 0 \end{align}
where we used the chain rule. Now solving for $$\theta$$, we obtain an optimal value of:
(49)\begin{align} \hat\theta = \frac{\sum_{i=1}^N x_i y_i}{\sum_{i=1}^N x_i^2} \end{align}
Which we can write in vector notation as:
(50)\begin{align} \hat\theta = \frac{\mathbf{x}^\top \mathbf{y}}{\mathbf{x}^\top \mathbf{x}} \end{align}
This is known as solving the normal equations. For different ways of obtaining the solution, see the notes on Least Squares Optimization by Eero Simoncelli. | 2021-10-22T03:47:13 | {
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https://math.stackexchange.com/questions/2957612/convergence-of-infinite-series-sum-k-2-infty-frac-1kk-lnk | # Convergence of infinite series $\sum_{k=2}^{\infty}\frac{(-1)^k}{k\ln(k)}$
In a recent assignment, as an intermediate step, I need to show that $$\sum_{k=2}^{\infty}\frac{(-1)^k}{k\ln(k)}$$ converges. It is not hard to see that $$\sum_{k=2}^{\infty}\frac{1}{k\ln(k)}$$ is divergent, therefore I think one has to deal with the sum of alternating sequence directly. However, I am stuck and don't know how to proceed. Thanks in advance for anyone that is kind to help!
• alternating series test – user10354138 Oct 16 '18 at 8:27
• Leibniz criterion – Peter Szilas Oct 16 '18 at 8:28
If, for some reason, you do not want to use the Leibniz criterion, you can group your sum in pairs: $$a_n = \frac{(-1)^{2n}}{2n\log(2n)}+\frac{(-1)^{2n+1}}{(2n+1)\log(2n+1)}$$ is positive, and $$a_n < \frac{1}{n^2}$$ so $$\sum a_n$$ converges.
• Thank you! This is really an interesting way to do it. – dogthepeter Oct 24 '18 at 2:34
The series converges, and it can be proven using the Leibniz criterion for alternating series.
The criterion analyzes sums of the form $$\sum_{n=1}^\infty (-1)^n a_n$$ where $$a_i\geq 0$$. The criterion says that if $$\lim_{n\to\infty}a_n = 0$$ and the sequence $$\{a_n\}$$ is decreasing, then the sum converges. In your case, $$a_k=\frac{1}{k\ln k}$$ which satisfies both conditions (it's decreasing and has a limit of $$0$$), so the series converges.
• Many thanks for pointing it out! I wasn't aware of the criterion at first place. It is surely a useful and interesting result! – dogthepeter Oct 24 '18 at 2:36
You may also couple adjacent terms: $$0<\frac{1}{2k\log(2k)}-\frac{1}{(2k+1)\log(2k+1)} = \frac{2k\log\left(1+\frac{1}{2k}\right)+\log(2k+1)}{2k(2k+1)\log(2k)\log(2k+1)}<\frac{1}{k^2}$$ and recall that $$\sum_{k\geq 1}\frac{1}{k^2}$$ is convergent.
• Thank you! I think the answer is similar to the one provided by GEdger. Truly interesting! – dogthepeter Oct 24 '18 at 2:35 | 2019-05-21T11:35:17 | {
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https://math.stackexchange.com/questions/1359958/finding-eigenvalues-of-a10-a7-5a | # Finding eigenvalues of $A^{10} + A^7 + 5A$.
Problem: Let $A = \begin{pmatrix} 1 & 2 & -1 \\ 0 & 5 & -2 \\ 0 & 6 & -2 \end{pmatrix}$.
1) Compute the eigenvalues of $A^{10} + A^7 + 5A$.
2) Compute $A^{10} X$ for the vector $X = \begin{pmatrix} 2 \\ 4 \\ 7 \end{pmatrix}$.
Attempt at solution: I first computed the eigenvalues of $A$. The characteristic polynomial gives \begin{align*} \det(A - x \mathbb{I}_3) = det \begin{pmatrix} 1-x & 2 & -1 \\ 0 & 5-x & -2 \\ 0 & 6 & -2 -x \end{pmatrix} \end{align*} Laplace expansion along the first column gives \begin{align*} \det(A - x \mathbb{I}_3) = (1-x) \det \begin{pmatrix} 5-x & -2 \\ 6 & -2-x \end{pmatrix} &= (1-x) [(5-x)(-2-x)+12] \\ &= (1-x)(x^2-3x+2) \\ &= (x-1)^2(2-x)\end{align*} So the eigenvalues are $\lambda_1 = 1, \lambda_2 = 1$ and $\lambda_3 = -2$.
Now, I'm aware that there is a theorem which says that if the matrix $A$ has an eigenvalue $\lambda$, then $A^k$ has the eigenvalue $\lambda^k$ corresponding to the same eigenvector. But I'm not sure what I should do about this sum here. Can I just add the corresponding eigenvalues? So for the first eigenvalue that would give me: $1^{10} + 1^7 + 5(1) = 7$?
Any help would be appreciated.
Edit: Figured out part 1), but I'm still wondering what to do with 2). Should I find an invertible matrix $P$ such that $P^{-1} A P$ is a diagonal matrix and then I can compute the powers easily?
• The theorem can be further generalized to polynomial case (your guess is right and not hard to verify). – Zhanxiong Jul 13 '15 at 21:16
• If $Ax = \lambda x$ then $(A^{10} + A^7 + 5A)(x) = A^{10}x + A^7x + 5Ax = \lambda^{10}x + \lambda^7 x + 5 \cdot \lambda x = (\lambda^{10} + \lambda^7 + 5 \cdot \lambda) \cdot x$. – Krijn Jul 13 '15 at 21:19
• You should re-check your algebra, the eigenvalues are $2,1,1$ – John McGee Jul 13 '15 at 21:20
• Ah right, thanks. I wrote down the wrong minor. Also, for the second question, should I first find an invertible matrix $P$ such that $P^{-1} A P$ is a diagonal matrix, and then compute $A^{10}$? – Kamil Jul 13 '15 at 21:23
• It might be easier to find the eigenvectors and write $X$ as a linear combination of these eigenvectors. Then $A^{10}X$ should be easy to calculate. – Krijn Jul 13 '15 at 21:25
For the first question, observe that $$(A^{10}+A^{7}+5A)x=(\lambda^{10}+\lambda^7+5\lambda)x$$ as I have shown in the comments.
For the second question, you observed correctly that we have three eigenvectors, for the eigenvalues $$2$$, $$1$$ and $$1$$, respectively $$v_1 = \begin{pmatrix} \frac{1}{3}\\ \frac{2}{3} \\ 1 \\ \end{pmatrix}, v_2 = \begin{pmatrix} 0\\ \frac{1}{2} \\ 1 \\ \end{pmatrix}, v_3= \begin{pmatrix} 1\\ 0 \\ 0 \\ \end{pmatrix}$$
We see that $$X = \begin{pmatrix} 2\\ 4 \\ 7 \\ \end{pmatrix} = 3 \cdot \begin{pmatrix} \frac{1}{3}\\ \frac{2}{3} \\ 1 \\ \end{pmatrix} + 4 \cdot \begin{pmatrix} 0\\ \frac{1}{2} \\ 1 \\ \end{pmatrix} + 1 \cdot \begin{pmatrix} 1\\ 0 \\ 0 \\ \end{pmatrix}$$
With this we can calculate $$A^{10}X$$ quite easily, because $$A^{10}X = A^{10}(3\cdot v_1 + 4\cdot v_2 + 1\cdot v_3) = 3\cdot A^{10}v_1 + 4\cdot A^{10} v_2 + A^{10}v_3$$ $$= 3\cdot \lambda_2^{10} \cdot v_1 + 4\cdot \lambda_1^{10}\cdot v_2 + \lambda_1^{10}\cdot v_3 = 3\cdot 2^{10} \cdot v_1 + 4\cdot 1^{10}\cdot v_2 + 1^{10}\cdot v_3$$
• I don't understand how $v_3$ can be a different eigenvector from $v_2$, when the eigenvalue is $1$ in both cases? How did you compute $v_3$? – Kamil Jul 13 '15 at 22:35
• If an eigenvalue $\lambda$ has multiplicity $> 1$, such as $\lambda = 1$ in our case, it can happen that there are two linearly independent eigenvectors $v$, $w$ such that $Av = \lambda v$ and $Aw = \lambda w$. There is an easy test for the amount of linearly independent eigenvectors that belong to a certain eigenvalue, namely: The amount of linearly independent eigenvalues for a certain eigenvalue $\lambda$ is equal to the dimension of the kernel of $A - \lambda I$. In our case, the dimension of the kernel of $A - 1\cdot I$ is equal to $2$, therefore, there are two eigenvectors. – Krijn Jul 13 '15 at 22:41
• Thanks, that was very clear. I will look for a proof of this theorem. Also, is it correct that two distinct eigenvalues can never have the same eigenvector? – Kamil Jul 13 '15 at 22:53
• That is indeed correct and very easy to proof, because if $Av = \lambda_1 v$ and $Av = \lambda_2 v$ then of course $\lambda_1 v = \lambda_2 v \Rightarrow \lambda_1 = \lambda_2$. – Krijn Jul 13 '15 at 22:54
If $P(X)=\sum_{k=0}^na_kX^k$ is a polynomial and a matrix $A$ has en eigenvalue $\lambda$ with eigenvector $v$, we have: $$P(A)(v)=\sum_{k=0}^na_kA^kv=\sum_{k=0}^na_k\lambda^kv=P(\lambda)v$$
Thus, $P(\lambda)$ is an eigenvalue of $P(A)$.
By definition of eigenvalue:
$$Av=\lambda v$$
for $v\neq 0$ and corresponding eigenvector. So:
$$A^{2}v=A (Av)=A(\lambda v)=\lambda Av=\lambda \lambda v=\lambda^2 v$$
The same way we can show that:
$$A^k v=\lambda^k v$$
So:
$$A^{10}v=\lambda^{10}v$$
$$A^{7}v=\lambda^{7}v$$
$$5A^{1}v=5\lambda^{1}v$$
If we add these three equations side by side we have:
$$(A^{10}+A^{7}+5A)v=(\lambda^{10}+\lambda^7+5\lambda)v$$
So if $\lambda$ is eigenvalue of $A$, then $\lambda^{10}+\lambda^7+5\lambda$ is eigenvalue of $A^{10}+A^{7}+5A$. | 2019-10-14T13:33:10 | {
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http://haiboyu.me/en/2018/05/14/Probability-of-death/index.html | ## Questions:
There is an island which is represented by square matrix nxn.
A person on the island is standing at any given co-ordinates (x,y). He can move in any direction one step right, left, up, down on the island. If he steps outside the island, he dies.
Let the island be represented as (0,0) to (n-1,n-1) (i.e nxn matrix) & person is standing at given co-ordinates (x,y). He is allowed to move n steps on the island (along the matrix). What is the probability that he is dead after he walks n steps on the island?
## Solution:
Recursion. (Or “mathematical induction”, if you’re snobbish.)
(In what follows, “he is dead after he walks n steps on the island” is assumed to mean “he dies after less than or equal to n steps”. If you take it to mean “he dies after exactly n steps”, the answer will be slightly different. I’ll discuss it briefly at the end.)
We have an NxN matrix where the value in each cell represents the probability of dying in n steps if we started from that cell.
Consider the probability of dying in 0 steps. Clearly, this is 0.0 for every location inside the island, and 1.0 everywhere outside it.
What’s the probability of dying in 1 steps? You have four directions you can move in, with equal probability. So for each cell, you take its four neighbors, find their probability of dying in 0 steps, and average them together. (If a neighbor is outside the matrix, you consider its probability to be 1.0.)
Similarly, the probability of dying in k steps starting from a given cell is the average of the probability of dying in k-1 steps starting from its neighbour cells.
Python code:
from itertools import product as prod
def prob_death(island_size, steps):
if island_size < 1 or steps < 0: raise ValueError
new_prob = [[0. for i in range(island_size)] for j in range(island_size)]
if steps == 0:
return new_prob
old_prob = prob_death(island_size, steps - 1)
directions = [(0, -1), (1, 0), (0, 1), (-1, 0)]
for (i, j, direction) in prod(range(island_size), range(island_size), directions):
neighbor_i = i + direction[0]
neighbor_j = j + direction[1]
if neighbor_i >= 0 and neighbor_i < island_size and \
neighbor_j >= 0 and neighbor_j < island_size:
prob_death_this_way = old_prob[neighbor_i][neighbor_j]
else: # neighbor is outside the island
prob_death_this_way = 1.
new_prob[i][j] += 0.25* prob_death_this_way
return new_prob
Now, let’s test it out a bit: (mpr is just a function for printing matrices nicely)
>>> mpr(prob_death(5, 0))
0.000000 0.000000 0.000000 0.000000 0.000000
0.000000 0.000000 0.000000 0.000000 0.000000
0.000000 0.000000 0.000000 0.000000 0.000000
0.000000 0.000000 0.000000 0.000000 0.000000
0.000000 0.000000 0.000000 0.000000 0.000000
As expected: You can’t die in 0 steps if you start inside the island.
>>> mpr(prob_death(5,1))
0.500000 0.250000 0.250000 0.250000 0.500000
0.250000 0.000000 0.000000 0.000000 0.250000
0.250000 0.000000 0.000000 0.000000 0.250000
0.250000 0.000000 0.000000 0.000000 0.250000
0.500000 0.250000 0.250000 0.250000 0.500000
This is what we’d expect. If you start at a corner cell, you have 0.5 probability of dying in 1 step: 2 out of your 4 neighbors are outside the island. If you start on an edge, only 1 neighbor is outside, so your probability of dying is 0.25. Everywhere else, all neighbors are inside the island, so probability of dying in 1 step is 0.0.
>>> mpr(prob_death(5, 5))
0.806641 0.666016 0.622070 0.666016 0.806641
0.666016 0.437500 0.349609 0.437500 0.666016
0.622070 0.349609 0.261719 0.349609 0.622070
0.666016 0.437500 0.349609 0.437500 0.666016
0.806641 0.666016 0.622070 0.666016 0.806641
The probability of dying in 5 steps. I can’t verify the exact values, but it looks about right: The probability of dying is highest in the corners, a little lower at the edges, and decreases steadily inwards.
That solves the problem of dying in less than or equal to n steps.
Now, to find the probability of dying in exactly n steps: Let the probability of dying in less than or equal to n steps starting from (x,y) be denoted by P(x,y,n). Then the probability of dying in exactly n steps is the probability of surviving for n-1 steps, times the probability of dying in the nth step given that we survived for n-1 steps: (1-P(x,y,n-1))*(P(x,y,n) - P(x,y,n-1)). (I’m not all that sure about this formula; correct me if I’m wrong.) | 2021-06-16T01:58:30 | {
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https://byjus.com/question-answer/let-f-x-and-g-x-be-differentiable-for-0-leq-x-leq1-such-that/ | Question
# Let $$f(x)$$ and $$g(x)$$ be differentiable for $$0\leq x \leq1$$, such that $$f(0)=0$$, $$g(0)=0$$, $$f(1)=6$$. Let there exists a real number $$c$$ in $$(0,1)$$ such that $$f'(c)=2g'(c)$$. Then the value of $$g(1)$$ must be
A
1
B
3
C
2
D
1
Solution
## The correct option is C $$3$$From lagrange's theorem$$\dfrac{f(b)-f(a)}{b-a}=2\times\dfrac{g(b)-g(a)}{b-a}$$$$\dfrac{f(1)-f(0)}{1-0}=2\times \dfrac{g(1)-g(0)}{1-0}$$$$\dfrac{(6-0)}{1-0}=2\times \dfrac{(g(1)-0)}{1-0}$$$$6=2g(1)$$$$g(1)=3$$Mathematics
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View More | 2022-01-23T08:26:35 | {
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https://math.stackexchange.com/questions/2358269/points-of-intersection-for-two-logarithmic-spirals | # Points of Intersection for Two Logarithmic Spirals
So I have two logarithmic spirals in parametric form $$x(t) = ae^{bt}\cos t \\ y(t)=ae^{bt}\sin t$$ and $$x'(t) = \alpha e^{\beta t}\cos t \\ y'(t)=\alpha e^{\beta t}\sin t$$ With $\beta$ and $b$ having opposite signs so the spirals grow in opposite directions. Setting the $x$'s and $y$'s equal and solving for $t$ I get $$t = \frac{1}{b-\beta}\ln\frac{\alpha}{a}$$ Evaluating $(x(t),y(t))$ for that value does produce a single point of intersection for the two spirals, but I can't seem to find the general form that would give all the points of intersection.
I've fiddled around with it in polar form, both as $r(\theta)$ and as $\theta(r)$, and solved for the intersection points, but I couldn't figure out the general form in those cases either. (I tried integer coefficients, and multiples of $\pi$ and $2\pi$ to no avail.)
Eventually I realized I would prefer the parameterized approach because it'll be easiest to work in terms of $t$.
But I still feel lost in how to find the other points of intersection.
• When you equated the x's, what about the sine and cosine terms? You have to equate them as well to zero. – Aniruddha Deshmukh Jul 14 '17 at 6:06
• Ah! I simply canceled them out! I'm not sure yet how I should handle them, but I can see how that could be where I dropped whatever "factor" would return the other values! – Yrast Jul 14 '17 at 6:11
• Yes @Yrast. You will get other points of intersection in terms of $\pi$ or $\dfrac{\pi}{2}$. And also, you might get something as $\left( \left( 2n + 1 \right)\dfrac{\pi}{2}, n\pi \right)$. I am not sure about this. But it should come somewhat like this. – Aniruddha Deshmukh Jul 14 '17 at 6:39
• For instance $ae^{bt}\sin t = \alpha e^{\beta t}\sin t$ not only when $ae^{bt}= \alpha e^{\beta t}$ but also when values of $t=\pi-t'+k\pi$ or $t=t'+2n\pi$ and $t=-t'+2h\pi$ for cosine – Raffaele Jul 14 '17 at 8:02
In this reply I will give a solution for the intersection of two logarithmic spirals.
Consider two spirals in the complex plane with different angular regions, e.g.,
$$z=e^{(b+i)u}\\ w=e^{(\beta+i)v}$$
and we seek all the points where $z=w$. We can expand these and separate the real and imaginary parts to obtain
$$e^{bu}\cos u=e^{\beta v}\cos v\quad \quad \quad (1)\\ e^{bu}\sin u=e^{\beta v}\sin v\quad \quad \quad (2)$$
If we multiply (1) by $\sin u$ and (2) by $\cos u$ and subtract, we obtain
$$0=e^{\beta v}[\sin u\cos v-\cos u\sin v]=e^{\beta v}\sin(u-v)\quad \quad \quad (3)$$
Similarly, if we multiply (1) by $\cos u$ and (2) by $\sin u$ and add, we obtain
$$e^{bu}=e^{\beta v}[\cos u\cos v-\sin u\sin v]=e^{\beta v}\cos(u-v)\quad \quad \quad (4)$$
Equation (3) tells us that $(u-v)=n\pi$. Equation (4) tells us that $n$ must be even since $\cos n\pi$ alternates in sign, but (4) must be positive. So we conclude that
$$v=u+2n\pi$$
(assuming that $v$ the larger of the two).
Finally, going back to the beginning, we must also have $|z|=|w|$, and therefore
$$e^{bu}=e^{\beta v}=e^{\beta(u+2n\pi)}$$
so that
$$u=\frac{2n\pi\beta}{b-\beta}\\ v=u+2n\pi$$
The figure below shows a sample calculation (cleverly chosen to give simple results). Here
$$b=\frac{2\ln\varphi}{\pi},\quad \text{the golden spiral}\\ \beta=\frac{\ln\varphi}{2\pi}\\$$
$$u=\frac{2n\pi}{3}\quad \{120,240,360,480,600,720^\circ...\}\\ v=\frac{8n\pi}{3}\quad \{480,960,1440,1920,2400,2880^\circ...\}$$
• Thank you so much! This is precisely what I was trying to do. After your comment last night I played around a bit and realized I should switch to a complex valued function, but then today I was still thoroughly stumped. I don't think I could have figured most of that out on my own, maybe back while taking complex variables, but my math knowledge has atrophied pretty severely in the last ten years. I'm still not sure I follow every step completely but I think I will with a bit more effort. Thanks again! – Yrast Jul 15 '17 at 21:04
• @Yrast Your thanks are appreciated. Notice that these spirals are both anticlockwise, if one of them was clockwise, say e^{(\beta-i)v}$then Eqs. (3) and (4) would be slightly different and the results would be different too, of course. – Cye Waldman Jul 15 '17 at 22:20 You are incorrect when you say With β and b having opposite signs so the spirals grow in opposite directions. This parameter is called the flair coefficient, its sign indicates whether the spiral is growing or shrinking, but both evolve in the anticlockwise direction. Thus it's quite possible that they do not intersect at all other than at their mutual starting point of$\theta=0\$.
You should run some plots. | 2019-07-20T13:23:24 | {
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http://math.stackexchange.com/questions/99746/how-to-solve-for-a-continuous-function | # how to solve for a continuous function?
I am having a midterm review in school and there's one concept that I forgot how to solve, and that is solving for continuous functions?
More precisely, what does a variable have to be for the following to be continuous. For example, the problem I am dealt with solving is $$F(x)=\left\{\begin{array}{ll} 2x&\text{if }x\leq 1\\ ax^2+1&\text{if }x\gt 1\\ \end{array}\right.$$ and I have to solve for $a$. Normally, I would solve for $ax^2+1$, but I know that is wrong. Can someone tell me how to solve this, and perhaps by using a different problem so that I may be able to do the one I have on my own?
-
You need to make $F$ continuous when $x = 1$. What is $F(1)$? What must $\lim_{x \rightarrow 1} F(x)$ equal in order for $F$ to be continuous at $x = 1$? To compute the limit, what are the one-sided limits, $\lim_{x \rightarrow 1^-} F(x)$ and $\lim_{x \rightarrow 1^+} F(x)$? Hope those hints help. – Michael Joyce Jan 17 '12 at 2:25
The only way to acknowledge the help of answerers in this forum is by accepting an answer. So, please consider doing so. – user21436 Jan 17 '12 at 2:37
Have you consider accept the answers? – leo Jan 17 '12 at 3:44
Please see this – leo Jan 17 '12 at 3:46
As per request, here is a problem that is like your problem.
Problem: Let $f(x)=5x$ when $x\le 2$, and let $f(x)=a^2x^2-7x$ when $x>2$. Find all values of $a$ such that $f$ is continuous everywhere.
Solution: Note that since $5x$ is continuous everywhere, $f$ is continuous at all $x<2$. Similarly, for any $a$, $a^2x^2-7x$ is continuous everywhere, so $f$ is continuous at all $x>2$. We now know that whatever choice we make for $a$, $f$ is continuous everywhere except possibly at $x=2$.
We want to find the values of $a$ such that $f$ is continuous at $x=2$.
As $x$ approaches $2$ from the left, $f(x)$ approaches $f(2)$. We want to make sure that as $x$ approaches $2$ from the right, $f(x)$ also approaches $f(2)$.
As $x$ approaches $2$ from the right, $f(x)$ approaches $a^2(2^2)-7(2)$. We want this "limit from the right" to be $f(2)$, that is, $10$. This will be the case precisely if $$4a^2 -14=10.$$ Solve for $a$. We get $a=\pm\sqrt{6}$.
-
ok, so using what you told me, wouldn't a (in my problem) be 1? – Ronnie.j Jan 17 '12 at 2:48
@Ronnie.j: The typesetting is not in LaTeX, so I am not sure what the problem is. But if $f(x)=2x$ if $x\le 1$, and $f(x)=ax^2+1$ if $x>1$, then yes, $a=1$. What you wrote can also be interpreted as $f(x)=2x$ for $x\le 1$, $f(x)=2x+ax^2+1$ if $x>1$. If that's what you meant then $a=-1$. – André Nicolas Jan 17 '12 at 2:53 | 2014-03-11T20:09:53 | {
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https://math.stackexchange.com/questions/1473983/confused-about-how-to-solve-basic-combinatorial-problem | # Confused about how to solve basic combinatorial problem
The following exercise comes from a book. I do not know how to approach it. I get the sense that it is of the "stars and bars" flavor, but is really a "stars and squares and bars" problem - which is to say one must divide multiple groups of indistinguishable "items", where the groups are distinct from other groups, into distinguishable groups.
I have gotten the answer, but I don't intuitively understand the process of getting it. In particular, I understand that I can use the "multichoose" to get the answer, but I don't fully understand how it breaks down the solution. Help would be appreciated in understanding "why" it works, and "how" it accurately describes the divisions.
The question is as follows:
An elevator starts at the basement with 8 people (not including the elevator operator) and discharges them all by the time it reach the top floor, number 6. In how many ways could the operator have perceived the people leaving the elevator if the 8 people consisted of 5 men and 3 women and the operator could tell a man from a woman? [Modified slightly for readability - two part question but the first part is not too relevant.]
The answer is $${5+6-1 \choose 6}{3+6-1 \choose 6}=14,112$$
in case anyone is wondering.
Let me rephrase the question. You have $8$ balls that are indistinguishable except for color; $5$ of them are blue, and $3$ are red. You have $6$ numbered boxes. You put the $8$ marbles into the $6$ boxes; how many different outcomes can you distinguish?
It’s really just a combination of two simpler problems. First, how many different outcomes for the placement of the blue marbles can you distinguish? That’s a basic stars and bars problem, and the answer is
$$\binom{5+6-1}{6-1}=\binom{10}5=252\;.$$
Similarly, there are
$$\binom{3+6-1}{6-1}=\binom85=56$$
distinguishable outcomes for the red balls. Since any outcome of the blue balls can be combined with any outcome of the red balls, there are altogether $252\cdot56=14,112$ distinguishable outcomes.
Note that although your final numerical answer is correct, your binomial coefficients are not: essentially you’ve interchanged the balls and boxes. You should have either what I used or
$$\binom{5+6-1}5\binom{3+6-1}3\;.$$
The product of your binomial coefficients is only $5880$.
• Thank you. Reducing the problem to an "Urn Game" really makes the derivation of the solution clear. Also, treating the men and women as different events and multiplying them to get the total number of combinations was also instructive. – d0rmLife Oct 10 '15 at 23:36
• @d0rmLife: You’re welcome. – Brian M. Scott Oct 10 '15 at 23:38
• @BrianM.Scott Why do you add 6 to the first binomial coefficient? To account for the fact that each of the 6 urns may have zero balls allocated to it? – Caerus Sep 1 '17 at 23:05 | 2020-01-22T11:31:03 | {
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https://math.stackexchange.com/questions/2937258/for-s-n-leq-t-n-above-specific-n-prove-lim-n-to-infty-s-n-leq-lim-n-t | # For $s_n \leq t_n$ above specific n, prove $\lim_{n\to\infty} s_n \leq \lim_{n\to\infty} t_n$
The question: Suppose there exists $$N_0$$ such that $$s_n \leq t_n$$ for all $$n > N_0$$. Prove that if $$\lim_{n\to\infty} s_n$$ and $$\lim_{n\to\infty} t_n$$ exists, then $$\lim_{n\to\infty} s_n \leq \lim_{n\to\infty} t_n$$.
My attempt at a proof:
Let $$\lim_{n\to\infty} s_n = s$$ and $$\lim_{n\to\infty} t_n = t$$. Assume, for purposes of contradiction, that $$s > t$$. Since $$\lim_{n\to\infty} s_n$$ exists, it must be true that there exists $$N_1$$ for all $$\epsilon > 0$$ such that $$n > N_1$$ implies $$\mid s_n - s \mid < \epsilon$$. Since, by premise, $$s - t > 0$$, we may choose $$\epsilon = s - t$$ and there must exist $$N_1$$ satisfying $$\mid s_n - s \mid < \epsilon$$, i.e. $$\mid s_n - s \mid < s - t$$, and thus by the absolute value properties, $$s - (s-t) < s_n < s + (s-t)$$, which directly implies $$t < s_n$$ for sufficiently large $$n$$. Now we examine the implications on $$\lim_{n\to\infty} t_n$$ exists, we may select any $$\epsilon > 0$$ to satisfy $$\mid t_n - t \mid < s_n - t$$. Since $$s_n - t > 0$$ for sufficiently large $$n$$, we may set $$\epsilon = s_n - t$$ for some $$s_n$$, however this would imply some $$N$$ exists such that $$n > N$$ implies $$\mid t_n - t \mid < s_n - t$$, i.e. $$t_n < s_n - t + t$$, that is, $$t_n < s_n$$ for all sufficiently large $$n$$, which is a contradiction. Thus $$\lim_{n\to\infty} s_n \leq \lim_{n\to\infty} t_n$$.
I'm certain I got lost in the details somewhere. Could someone point out errors that I've made?
Your idea of the proof is correct, though you made it a bit too complicated with the technical part. Here is a clear way to write it. Assume $$s>t$$. Then there exists $$\epsilon>0$$ such that $$s-\epsilon>t+\epsilon$$. Now you know there is $$N_1\in\mathbb{N}$$ such that $$s_n>s-\epsilon$$ for all $$n\geq N_1$$. Also there is $$N_2\in\mathbb{N}$$ such that $$t_n for all $$n\geq N_2$$. Now let $$N=max\{N_0,N_1,N_2\}$$. For all $$n\geq N$$ you have:
$$s_n>s-\epsilon>t+\epsilon>t_n$$ | 2020-04-02T11:03:12 | {
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https://math.stackexchange.com/questions/4197900/solution-check-finding-the-radon-nikodym-derivative | # Solution Check-Finding The Radon Nikodym Derivative
I was hoping to get my solution to part $$\textbf{i}$$ of this qual question regarding the Radon-Nikodym derivative checked for rigor and correctness. Then I was hoping to get advice on proceeding with part $$\textbf{ii}$$. Here is the question:
Let $$m$$ be the Lebesgue measure and define a measure $$\mu$$ on the Borel $$\sigma$$-algebra on $$[0,1]$$ by the formula $$\mu(X) = m(\{y\in [0,\pi]:\sin(y) \in X\}) .$$ $$\textbf{i}$$ Show that $$\mu$$ is absolutely continuous with respect to the Lesbesgue measure.
$$\textbf{ii.}$$ Find the Radon Nikodym derivative $$\frac{d\mu}{dm}.$$
Here is my solution:
$$\textbf{i}$$ Let $$A\subset \mathbb{R}$$ such that $$m(A) = 0$$. We have that $$\mu(A) = m(\{y\in [0,\pi]:\sin(y) \in A\})$$ and from monotonicity we have $$\mu(A) \leq m(A\cap [0,\pi]) \leq m(A) = 0$$ and therefore $$\mu << m$$.
$$\textbf{ii}$$ Since $$m$$ is a $$\sigma$$-finite positive measure and $$\mu$$ is a finite positive measure (notice that $$\mu(\mathbb{R}) = \pi)$$ and $$\mu << m$$ by the Radon-Nikodym theorem there exists a $$m$$-integrable non-negative function $$f$$ which is measurable with respect to the Borel $$\sigma$$-algebra and $$\mu(A) = \int_A f d\mu$$
Furthermore, if any other function $$g$$ satisfies above, then $$f=g$$ a.e.
Notice that the set $$B=[0,1]$$ has full measure with respect to $$\mu$$. I want to claim that $$f= 2\sin^{-1}(x) \cdot \chi_{[0,1]}$$ is the Radon Nikodym derivative, but I'm not exactly sure how to show this.
• its not true that $\mu(A) \le m(A)$ in general, so how do you justify the inequality? Eg $\mu((0,1))=\pi>1=m((0,1))$ Jul 14 at 5:10
Let $$\mu_1(X)=m\{x \in [0,\frac {\pi} 2]: \sin y \in X\}$$. Verify that the derivative of $$\sin^{-1} x$$, namely $$\frac 1 {\sqrt {1-x^{2}}}$$ is the RND of $$\mu_1$$ w.r.t. $$m$$. [ For this show that $$\int_a^{b} \frac 1 {\sqrt {1-x^{2}}} dx=\mu_1 (a,b)$$ for $$a ].
Now consider $$\mu_2(X)=m\{x \in [\frac {\pi} 2, \pi]: \sin y \in X\}$$. Use the fact that $$\sin (\pi -y) =\sin y$$ to prove absolute continuity of $$\mu_2$$ and for writing down the RND of $$\mu_2$$. | 2021-10-18T05:05:26 | {
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https://math.stackexchange.com/questions/2764870/replacing-inequalities | # replacing Inequalities
I encountered a problem today:
Prove that:
$$\frac{a^3+b^3+c^3}{a^2+b^2+c^2} \ge \frac{a+b+c}{3}$$
for all $a,b,c>0$
I used the RMS-AM inequality to replace the LHS with
$$\frac{\sqrt{a^2+b^2+c^2}}{\sqrt{3}}$$
and replaced the RHS using AM-GM inequality $$\frac{3abc}{a^2+b^2+c^2}$$
I can prove the new inequality, but does that mean I have proved the original inequality? I couldn't find another way to prove the original inequality except for expanding the terms and using scalar products. Thanks in advance!
• Do you have conditions upon a, b and c? Because for b=c=0 and a=-1 it doesn't seem to hold, does it? – ysearka May 3 '18 at 12:59
• oh sorry, a,b,c>0 ill edit it inside the question – SuperMage1 May 3 '18 at 13:01
We can use the Chebyshov's inequality.
Since $(a^2,b^2,c^2)$ and $(a,b,c)$ have the same ordering, we obtain: $$a^3+b^3+c^3=a^2\cdot a+b^2\cdot b+c^2\cdot c\geq\frac{1}{3}(a^2+b^2+c^2)(a+b+c).$$ Also, you can use PM and C-S.
Indeed, by PM $$\sqrt[3]{\frac{a^3+b^3+c^3}{3}}\geq\sqrt{\frac{a^2+b^2+c^2}{3}},$$ which gives your $$\frac{a^3+b^3+c^3}{a^2+b^2+c^2}\geq\sqrt{\frac{a^2+b^2+c^2}{3}}.$$ Thus, it's enough to prove that $$\sqrt{3(a^2+b^2+c^2)}\geq a+b+c,$$ which is true by C-S: $$\sqrt{3(a^2+b^2+c^2)}=\sqrt{(1+1+1)(a^2+b^2+c^2)}\geq\sqrt{(a+b+c)^2}=a+b+c.$$
• I have solved the inequality like that, but is the method on the post an acceptable one by replacing the inequalities? – SuperMage1 May 3 '18 at 13:52
• @SuperMage1 I added something. See now. By the way I think your $\frac{3abc}{a^2+b^2+c^2}$ gives a wrong inequality. Try $c\rightarrow0^+$, $a=b=1$. – Michael Rozenberg May 3 '18 at 14:25
If $a,b,c>0$ the sequence $\{M_n = a^n+b^n+c^n\}_{n\geq 0}$ is log-convex by the Cauchy-Schwarz inequality, since the function $x\mapsto a^x+b^x+c^x$ is continuous and midpoint-log-convex.
In particular $M_3 M_0\geq M_1 M_2$.
It is equivalent to $$(a^2-b^2)(a-b)+(a^2-c^2)(a-c)+(b-c)(b^2-c^2)\geq 0$$ and this is true.
Note that \begin{align*} a^3+b^3+c^3 &= \frac{a^4}{a}+\frac{b^4}{b}+\frac{c^4}{c}\\ &\ge \frac{(a^2+b^2+c^2)^2}{a+b+c}. \end{align*} Then \begin{align*} \frac{a^3+b^3+c^3}{a^2+b^2+c^2} &\ge \frac{a^2+b^2+c^2}{a+b+c}\\ &\ge \frac{a+b+c}{3}, \end{align*} given that \begin{align*} (a+b+c)^2 \le 3(a^2+b^2+c^2). \end{align*} | 2019-05-20T12:51:16 | {
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https://www.physicsforums.com/threads/algebra-ii-equations-containing-radicals.614375/ | # Homework Help: Algebra II Equations Containing Radicals
1. Jun 16, 2012
### velox_xox
...As for the problem, I tend to get problems with this same form (but with slight differences such as two separate binomials underneath individual square roots) wrong. I'm not quite sure what I'm doing wrong, so I was hoping for some help. :D
1. The problem statement, all variables and given/known data
Solve. If an equation has no real solution, say so.
$$\sqrt {y} + \sqrt {y + 5} = 5$$
Answer: $4$
2. Relevant equations
--
3. The attempt at a solution
The first thing I have been taught to do is to isolate the radicals on one side. That's already done.
Next, square both sides to eliminate radicals. I have a feeling that something I'm doing here is incorrect (?)
$(\sqrt {y} + \sqrt {y + 5})^2 = 5^2$
$y +y + 5 = 25$
Solve for y.
$2y + 5 = 25$
$2y = 20$
$y = 10$
But, since you often have the possibility of extraneous roots; you have to test it out in the original formula. When I do that, I get 'no solution.'
$\sqrt {10} + \sqrt {10 + 5} \stackrel{?}{=} 5$
$\sqrt {10} + \sqrt {15} \stackrel {?} {=} 5$
Approximate values from here on:
$3.16 + 3.87 \stackrel {?} {\approx} 5$
$7.03 \neq 5$
So, how'd I mess it up this time? (*joking*) Thanks in advance for your help!
2. Jun 16, 2012
### HallsofIvy
This is an error. $(a+ b)^2= a^2+ 2ab+ b^2$, NOT "$a^2+ b^2$".
$(\sqrt{y}+ \sqrt{y+ 5})^2= (\sqrt{y})^2+ 2\sqrt{y(y+ 5)}+ (\sqrt{y+5})^2= y+ 2\sqrt{y^2+ 5y}+ y+ 5= 25$
Now, "isolate" that remaining square root and square again:
$\sqrt{y^2+ 5y}= 10-y$
$(y^2+ 5y)= (10+ y)^2= 100+ 20y+ y^2$
Solve that for y.
3. Jun 16, 2012
### Staff: Mentor
When you have an equation with two radicals, it's often more efficient to move them to the opposite sides of the equation.
$\sqrt{y} = 5 - \sqrt{y + 5}$
Now square both sides:
y = (5 - $\sqrt{y + 5}$)2 = 25 - 10$\sqrt{y + 5}$ + y + 5
=> 0 = 30 - 10$\sqrt{y + 5}$
This equation is easy to solve for y.
4. Jun 16, 2012
### Staff: Mentor
Here's another method to consider. Multiply both sides of the equation by
$\sqrt{y+5}$ - $\sqrt{}y$
You get
$\sqrt{y+5}$ - $\sqrt{}y$ = 1
If you subtract this equation from the original equation, you get
2$\sqrt{}y$ = 4
5. Jun 18, 2012
### velox_xox
Okay, thanks to everyone's replies I think I see several different ways of going about solving the problem. I have a couple other problems that I happened to get wrong with the same form, so I'm going to try this and see if I can solve them on my own now, but first, I'd like to make sure that I have it down right in the first place.
Is this the correct form? (I used Mark44's suggestion for this one):
$0 = 30 - 10\sqrt {y + 5}$
$-30 = - 10 \sqrt {y + 5}$
$3 = \sqrt {y + 5}$
$3^2 = (\sqrt {y + 5})^2$
$9 = y + 5$
$4 = y$
And checking it for extraneous roots:
$\sqrt {4} + \sqrt {4 + 5} = 5$
$2 + \sqrt {9} = 5$
$2 + 3 = 5$
@HallsofIvy: Thank you for literally spelling it out for me. I had easier problems earlier in my assignment with the (a + b)^2 form, and it took me a few tries to recognize it, but this one was really sneaky. Instead of thinking of it as, $(a + b)^2$ or $(\sqrt {y} + \sqrt {y + 5})^2$ as you put it, I was actually thinking of it as: $(\sqrt {y})^2 + (\sqrt {y + 5})^2$. I'll have to watch out for this!!
@Chestermiller: Wow. I wouldn't have even thought of that using a conjugate to multiply and then taking that equation and subtracting from the original. Heh. And I almost wanted to correct you and say it was $\sqrt {y} - \sqrt {y + 5}$, but thank goodness I caught the sign change that comes from subtracting. Thank you for another method. :D
Last edited: Jun 18, 2012 | 2018-08-18T21:06:28 | {
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https://math.stackexchange.com/questions/4067415/splitting-a-circle-into-3-equal-parts-using-2-lines | Splitting a circle into 3 equal parts using 2 lines
The picture probably explains my question best.
I need to find a way to divide a circle into 3 parts of equal area with only 2 lines that intersect each other on the outline of the circle.
Also I need to check, if whatever diameter is between those lines, also splits circles with a different diameter into equal parts.
And lastly, and probably the most difficult question: How do I have to calculate the angle between x lines that all intersect in one point, so that the circle is split into x+1 parts with area = 1/(x+1) of the circle?
I tried my best, but couldn't even find a single answer or the right strategy to tackle the question...
• 2. Yes. If you find the red lines so that they divide the blue circle into 3 parts of equal areas, then the green circle is also dividd into 3 equal parts. – CiaPan Mar 18 at 20:58
Given the angle $$\theta$$, split by the diameter containing $$B$$, consider the following diagram:
$$\overline{BO}$$ is the line through the center and $$\overline{BA}$$ is the chord cutting off the lune whose area we wish to compute.
The area of the circular wedge subtended by $$\angle BOA$$ is $$\frac{\pi-\theta}2r^2\tag1$$ The area of $$\triangle BOA$$ is $$\frac12\cdot\overbrace{r\sin\left(\frac\theta2\right)}^\text{altitude}\cdot\overbrace{2r\cos\left(\frac\theta2\right)}^\text{base}=\frac{\sin(\theta)}2r^2\tag2$$ Therefore, the area of the lune is $$(1)$$ minus $$(2)$$: $$\frac{\pi-\theta-\sin(\theta)}2r^2\tag3$$ To get the area divided into thirds, we want $$\frac{\pi-\theta-\sin(\theta)}2r^2=\frac\pi3r^2\tag4$$ which means we want to solve $$\theta+\sin(\theta)=\frac\pi3\tag5$$ whose solution can be achieved numerically (e.g. use $$M=\frac\pi3$$ and $$\varepsilon=-1$$ in this answer) $$\theta=0.5362669789888906\tag6$$ Giving us
Numerical Details
The iteration from this answer, applied to $$\theta+\sin(\theta)=\frac\pi3$$, is $$\theta_{n+1}=\frac{\pi/3-\sin(\theta_n)+\theta_n\cos(\theta_n)}{1+\cos(\theta_n)}\tag7$$ Here is the result of this iteration starting at $$0$$; $$\begin{array}{l|l} n&\theta_n\\\hline 0&0\\ 1&0.5\color{#AAA}{23598775598298873077107230547}\\ 2&0.5362\color{#AAA}{45321326153808318904236597}\\ 3&0.5362669789\color{#AAA}{24456230942633093381}\\ 4&0.53626697898889055276\color{#AAA}{1878717471}\\ 5&0.53626697898889055276244906787\\ 6&0.53626697898889055276244906787 \end{array}$$
Connect the ends of the chord (red line) with the center. You get isosceles triangle. Let $$x$$ be the angle opposite to the base (i.e. chord) in that triangle. You want: $$\frac{r^2x}{2}-\frac{r^2 \sin x}{2}=\frac{r^2\pi}{3}$$ or $$x-\sin x=\frac{2\pi}{3}$$ I do not think it can be solved analytically but it can be solved numerically with the answer $$x \approx 2.60533$$ radian. The same logic can be applied to more than 3 pieces. The angle you are looking for is $$\pi-x$$
If you want to solve $$\theta+\sin(\theta)=\frac\pi3$$ what you can do is to expand the lhs as a Taylor series and use series reversion. Using ths simple $$y=\theta+\sin(\theta)=2 \theta -\frac{\theta ^3}{6}+\frac{\theta ^5}{120}-\frac{\theta^7}{5040}+O\left(\theta ^9\right)$$ this would give $$\theta=\frac{y}{2}+\frac{y^3}{96}+\frac{y^5}{1920}+\frac{43 y^7}{1290240}+O\left(y^9\right)$$ Making $$y=\frac \pi 3$$ and computing $$\theta=0.53626300$$
You could also use the $$1,400$$ years old approximation $$\sin(\theta) \simeq \frac{16 (\pi -\theta)\theta}{5 \pi ^2-4 (\pi -\theta) \theta}\qquad (0\leq \theta\leq\pi)$$ and solve the cubic $$-\frac{5 \pi ^3}{3}+\pi\left(16 +\frac{19 \pi }{3}\right) \theta-16\left(1+\frac{\pi }{3}\right) \theta^2+4 \theta^3=0$$ which shows only one real root (not very nice formal expression) which is $$0.53631167$$.
Since, by inspection, you know that the solution is close to $$\frac \pi 6$$, you could perform one single iteration of Newton-like method and have explicit approximations which will be better and better increasing the order $$n$$ of the method. For example, Newton method would give $$\theta_{(2)}=\frac \pi 6+\frac{1}{3} \left(2-\sqrt{3}\right) (\pi -3)$$ Halley method would give $$\theta_{(3)}=\frac \pi 6+\frac{2 \left(2+\sqrt{3}\right) (\pi -3)}{45+24 \sqrt{3}-\pi }$$ As a function of $$n$$, the results would be $$\left( \begin{array}{ccc} n & \text{estimate} & \text{method} \\ 2 & 0.536245321326154 & \text{Newton} \\ 3 & 0.536266784935255 & \text{Halley} \\ 4 & 0.536266978676557 & \text{Householder} \\ 5 & 0.536266978987702 & \text{no name} \\ 6 & 0.536266978988890 & \text{no name} \\ 7 & 0.536266978988891 & \text{no name} \end{array} \right)$$
1. To find the angle $$x$$ between the red lines you need to solve the equation $$\sin(x)+x=\frac\pi3.\tag1$$ This can be done only numerically.
3. Observe the essential difference between the cases with even and odd $$n$$. Whereas in former case the diameter drawn from the common intersection point is one of the "red" lines, in the latter case it is the bisector of the central angle. Let $$x_k$$ be the angle of a red line with the diameter. Then the following are the equations to find $$x_k$$: $$x_k+\frac12\sin2x_k=\frac\pi{n}k, \quad\text{with}\quad k=-\frac n2+1\dots \frac n2-1.\tag2$$ Observe that the numbers $$k$$ are integer for even $$n$$ and half-integer for odd $$n$$. For $$n=3$$ the equation $$(2)$$ reduces to $$(1)$$ with $$x=2x_\frac12$$. | 2021-08-05T11:23:24 | {
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