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http://mathhelpforum.com/algebra/203119-solving-compound-inequalities-problem.html | # Math Help - Solving compound inequalities problem
1. ## Solving compound inequalities problem
I have read through the solving inequalities pdf and am still struggling with this problem.
$-7<\frac{1}{x}\leq1$
I've tried taking the $-7<\frac{1}{x}$ and the $\frac{1}{x}\leq1$ on their own and combining the solutions but my answers are obviously wrong. Could someone show me how to solve these kinds of inequalities?
2. ## Re: Solving compound inequalities problem
Hello, BobRoss!
This one is tricky and dangerous.
There are several approaches.
I'll show you mine . . .
$\text{Solve: }\:\text{-}7\:<\:\frac{1}{x}\:\le\:1$
Like you, I split it into two inequalities: . $-7 \:<\:\frac{1}{x}\:\text{ and }\:\frac{1}{x} \:\le\:1$
Note the "and"; we want both to be true.
We'd like to multiply through by , but we must be very careful!
What we get depends on whether $x$ is positive or negative.
Suppose $x$ is positive.
Multiply both inequalities by $x.$
We have: . $\text{-}7x \:<\:1 \quad\Rightarrow\quad x \:>\:\text{-}\tfrac {1}{7}$
. . . and: . $1 \:\le\:x \quad\Rightarrow\quad x \:\ge\:1$
To make both statements true, we use: . $\boxed{x\:\ge\:1}$
Suppose $x$ is negative.
Multiply both inequalities by negative $x.$
We have: . $\text{-}7x \:>\:1 \quad\Rightarrow\quad x \:<\:\text{-}\tfrac{1}{7}$
. . . and: . $1 \:\ge\:x \quad\Rightarrow\quad x \:\le\:1$
To make both statements true, we use: . $\boxed{x \:<\:\text{-}\tfrac{1}{7}}$
Solution: . $\left(x\:<\:\text{-}\tfrac{1}{7}\right)\:\cup\:\left(x\:\ge\:1\right)$
n . . . . . . . $\left(\text{-}\infty,\,\text{-}\tfrac{1}{7}\right)\:\cup\:\left[1,\,\infty\right)$
. . . . . $\begin{array}{ccccc} === & \circ & --- & \bullet & === \\ & \text{-}\frac{1}{7} && 1 \end{array}$
3. ## Re: Solving compound inequalities problem
Originally Posted by Soroban
Hello, BobRoss!
This one is tricky and dangerous.
There are several approaches.
I'll show you mine . . .
Like you, I split it into two inequalities: . $-7 \:<\:\frac{1}{x}\:\text{ and }\:\frac{1}{x} \:\le\:1$
Note the "and"; we want both to be true.
We'd like to multiply through by , but we must be very careful!
What we get depends on whether $x$ is positive or negative.
Suppose $x$ is positive.
Multiply both inequalities by $x.$
We have: . $\text{-}7x \:<\:1 \quad\Rightarrow\quad x \:>\:\text{-}\tfrac {1}{7}$
. . . and: . $1 \:\le\:x \quad\Rightarrow\quad x \:\ge\:1$
To make both statements true, we use: . $\boxed{x\:\ge\:1}$
Suppose $x$ is negative.
Multiply both inequalities by negative $x.$
We have: . $\text{-}7x \:>\:1 \quad\Rightarrow\quad x \:<\:\text{-}\tfrac{1}{7}$
. . . and: . $1 \:\ge\:x \quad\Rightarrow\quad x \:\le\:1$
To make both statements true, we use: . $\boxed{x \:<\:\text{-}\tfrac{1}{7}}$
Solution: . $\left(x\:<\:\text{-}\tfrac{1}{7}\right)\:\cup\:\left(x\:\ge\:1\right)$
n . . . . . . . $\left(\text{-}\infty,\,\text{-}\tfrac{1}{7}\right)\:\cup\:\left[1,\,\infty\right)$
. . . . . $\begin{array}{ccccc} === & \circ & --- & \bullet & === \\ & \text{-}\frac{1}{7} && 1 \end{array}$
Even though there is nothing wrong with your solution, we should note that when we checked what happens with positive x and negative x, we are bringing in ANOTHER restriction for x, that also needs to be satisfied. In this case, it doesn't make any difference, but there might be a time in future where it does.
4. ## Re: Solving compound inequalities problem
Thanks so much, that helps a lot! Although I don't quite understand this part:
Originally Posted by Prove It
Even though there is nothing wrong with your solution, we should note that when we checked what happens with positive x and negative x, we are bringing in ANOTHER restriction for x, that also needs to be satisfied. In this case, it doesn't make any difference, but there might be a time in future where it does.
What is the new restriction and could you possibly give me an example of when I might see that?
5. ## Re: Solving compound inequalities problem
Originally Posted by BobRoss
Thanks so much, that helps a lot! Although I don't quite understand this part:
What is the new restriction and could you possibly give me an example of when I might see that?
The new restriction comes from saying "if we think about positive values for x", we are immediately making the restriction x > 0, and if we say "if we think about negative values for x", we are immediately making the restriction x < 0.
Take for example, the inequality \displaystyle \begin{align*} \frac{x^2 + 3x -2}{x} < 3 \end{align*}.
In order to solve this, we need to think of two cases, the first where x < 0, and the second where x > 0.
In case 1, with x < 0, we have
\displaystyle \begin{align*} \frac{x^2 + 3x - 2}{x} &< 3 \\ x^2 + 3x - 2 &> 3x \\ x^2 - 2 &> 0 \\ x^2 &> 2 \\ |x| &> \sqrt{2} \\ x < -\sqrt{2} \textrm{ or } x &> \sqrt{2} \end{align*}
So by solving the inequality we have \displaystyle \begin{align*} x < -\sqrt{2} \end{align*} or \displaystyle \begin{align*} x > \sqrt{2} \end{align*}, BUT to get this, we originally had to restrict \displaystyle \begin{align*} x < 0 \end{align*}. So that means that the solution for this case is just \displaystyle \begin{align*} x < -\sqrt{2} \end{align*}. | 2015-05-03T22:37:30 | {
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https://math.stackexchange.com/questions/1843065/fermats-little-theorem-and-eulers-theorem | # Fermat's Little Theorem and Euler's Theorem
I'm having trouble understanding clever applications of Fermat's Little Theorem and its generalization, Euler's Theorem. I already understand the derivation of both, but I can't think of ways to use them in problems that I know I must use them (i.e. the question topic is set).
Here are two questions I had trouble with trying to use FLT and ET:
1. Find the 5th digit from the rightmost end of the number $N = 5^{\displaystyle 5^{\displaystyle 5^{\displaystyle 5^{\displaystyle 5}}}}$.
2. Define the sequence of positive integers $a_n$ recursively by $a_1=7$ and $a_n=7^{a_{n-1}}$ for all $n\geq 2$. Determine the last two digits of $a_{2007}$.
I managed to solve the second one with bashing and discovering a cycle in powers of 7 mod 1000, and that may be the easier path for this particular question rather than using ET. However, applying ET to stacked exponents I believe is nevertheless an essential concept for solving more complex questions like number 1 that I wish to learn. It would be helpful if I could get hints on using ET in those two problems and a general ET approach to stacked exponents and its motivation.
• Hint: $5^{5^5}$ is really just $5^{(5^5)}$, so treat the $(5^5)$ as one unit, and then apply ET. – Kenny Lau Jun 28 '16 at 22:57
• Also, number 2 is also stacked exponents. – Kenny Lau Jun 28 '16 at 22:58
• Wait, what do you mean just apply ET? Can you be more specific? – Widow Maven Jun 28 '16 at 23:01
• How familiar are you with the Chinese remainder theorem? Because FLT and Euler's require that the base of the power and the modulus are coprime, and in problem 1 they are not. This is what the Chinese remainder theorem helps you deal with. It tells you that you can focus on $5^{5^{5^{5^5}}}$ modulo $2^{6}$ and $5^6$ separately. – Arthur Jun 28 '16 at 23:02
• – Martin Sleziak Jun 29 '16 at 4:53
For the first problem you will need a little bit of Chinese Remainder Theorem. You want to find the remainder of the stacked exponential modulo $10^5 = 2^5 \times 5^5$. Consider the two prime divisors separately.
As $\phi(2^5) = 16$ we have that if $r_1$ is the remainder of $5^{5^{5^{5}}}$ modulo $\phi(2^5) = 16$ then $5^{5^{5^{5^{5}}}} \equiv 5^{r_1} \pmod {2^5}$. Now we have to find $r_1$, which is a solution of $5^{5^{5^{5}}} \equiv r_1 \pmod {2^4}$. Repeat this algorithm few times and you will get rid of the exponents and you will find a value such that: $5^{5^{5^{5^{5}}}} \equiv r \pmod {2^5}$. Now use that $5^{5^{5^{5^{5}}}} \equiv 0 \pmod {5^5}$ and glue the two solutions with Chinese Remainder Theorem.
The second one can be solved using similar method, but this time you won't need the Chinese Remainder Theorem, as $(7,100) = 1$. Actually this is easier as $7^4 \equiv 1 \pmod {100}$.
• Thanks so much. I got this for the second problem though with ET instead of cycling in mod 100. I'll admit it's not the best way, but hey, at least I got it.! Valid XHTML. FYI, I actually got 43 mod 100 in the end (arithmetic error in the image). – Widow Maven Jun 29 '16 at 0:23
Walking through the first problem, we effectively need to find $a \equiv 5^{5^{5^{5^{5}}}} \bmod 10^5$. This splits easily into finding $a_1 \equiv a \bmod 2^5$ and $a_2 \equiv a \bmod 5^5$ which can then be re-united with the Chinese remainder theorem.
The order of $5 \bmod 32$ divides $\phi(32)=16$ (and actually we could say it divides $\lambda(32) = 8$ due to the Carmichael function). Because $16$ is a power of two (so the order of every number will also be a power of two) we can quickly square repeatedly: $(5\to25\equiv-7\to49 \equiv17\to 289\equiv 1)$ to find that the order of $5$ is in fact $8$. So we need to find:
$$b \equiv 5^{5^{5^{5}}} \bmod 8 \\ \text{which will give:}\qquad a_1\equiv 5^b \bmod 32$$
Next step; the order of $5 \bmod 8$ is easily seen to be $2$, and we can note that the remaining exponent is odd. Stepping back down the exponent ladder, $$b \equiv 5^\text{odd} \equiv 5 \bmod 8 \\ \text{and }\qquad a_1\equiv 5^5 \equiv 21 \bmod 32$$
Also we have $a_2\equiv a \equiv 0 \bmod 5^5$. As it happens $5^5=3125\equiv 21 \bmod 32$, so $a\equiv 3125 \bmod 10^5$ and the requested digit is $0$.
Note that this is true for $5$s in a tower of exponents of height $3$ or more; and this is inevitable behaviour, that the values in the exponent tower not very far up don't have any effect to the final modular result. In the second problem the intimidation factor of a tower of exponents $2007$ high is removed by this knowledge; only the bottom few make a difference. | 2019-08-22T13:09:10 | {
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https://math.stackexchange.com/questions/2767788/identify-the-symmetries-and-sketch-the-curve-r-sin-theta-2 | # Identify the symmetries and sketch the curve $r=\sin (\theta/2)$
I've been at this for a while and I can't think clearly so I'm definitely doing something wrong.
The question:
Identify the symmetries of the curves in Exercises 1–12. Then sketch the curves.
$r = \sin (\frac{\theta}{2})$
In the book it states that first to find symmetry we have to check the following:
$$( r ; - \theta )$$ $$( -r ; \pi-\theta )$$
$$( -r ; -\theta )$$ $$( r ; \pi-\theta )$$
$$( -r ; \theta )$$ $$( r ; \pi+\theta )$$
So I checked for the first one and this is what I got: For $-\theta$:
$r = \sin (\frac{-\theta}{2})$
$-r = \sin (\frac{\theta}{2})$
Which turns out to satisfy the y-axis.
Now here's what I don't understand. I want to check for $\pi-\theta$, So I replace that where $\theta$ would be.
$r = \sin (\frac{\pi-\theta}{2})$
$r = \sin (\frac{\pi}{2} - \frac{\theta}{2})$
Following the formula for sin(A-B), then:
$r = \sin (\frac{\pi}{2})\cos(\frac{\theta}{2}) - \sin (\frac{\theta}{2})\cos(\frac{\pi}{2})$
$\cos(\frac{\pi}{2}) = 0$ and $\sin (\frac{\pi}{2}) = 1$ then:
$r = \cos(\frac{\theta}{2})$
Which obviously doesn't make sense. So I checked online and found that it should actually be $r = \sin (\pi - \frac{\theta}{2})$ and then that it would satisfy the symmetry about x-axis. But I checked it (I'm sure by now I've done something major wrong):
$r = \sin (\pi - \frac{\theta}{2})$
$r = \sin (\pi)\cos(\frac{\theta}{2}) - \sin(\frac{\theta}{2})\cos(\pi)$
$\cos(\pi) = -1$ and $\sin (\pi) = 0$ then:
$r = \sin (\frac{\theta}{2})$
Which satisfies y-axis and not x-axis.
So my question is this, why is it that wherever I look it should be $r = \sin (\pi - \frac{\theta}{2})$ and not $r = \sin (\frac{\pi-\theta}{2})$ and what did I do wrong in the proofs and how can I make sure not to make the same mistakes again?
Thank you.
• Please use \sin instead of sin in math mode. May 5, 2018 at 13:41
• @MrYouMath I didn't know that! Thank you. Edited. :) May 5, 2018 at 13:52
• so it won't be symmetrical about x-axis. You got it right. May 5, 2018 at 13:54
• @Vasya But the correction says that it is symmetric on all three (if there are 2 symmetric then the third is there) May 5, 2018 at 13:56
An alternative approach:
Let us first focus on the equation
$$\rho=\sin\theta$$ or, in Cartesian coordinates,
$$\rho^2=x^2+y^2=\rho\sin\theta=y.$$
This is the circle $$x^2+\left(y-\frac12\right)=1$$ with its center at $$\left(0,\dfrac12\right)$$ and radius $$\dfrac12$$.
Now if you consider
$$\rho=\sin\frac\theta2$$ this curve is a transformed version of the circle such that the polar angle "rotates twice as fast" and the circle is "unrolled".
The symmetry of the circle wrt. the axis $$y$$ becomes a symmetry wrt. $$x$$. As $$\theta\in[0,\pi]$$ and $$\theta\in[\pi,2\pi]$$ describes twice the circle, $$\dfrac\theta2$$ just needs to describe a full turn.
Your computation is correct, but your conclusion is wrong. The equation has failed the symmetry test, but that does not mean that the graph is not symmetric with respect to x-axis. As a matter of fact if passing a symmetry test verifies that symmetry will be exhibited in a graph, failing the symmetry tests does not necessarily indicate that a graph will not be symmetric. In others words, the graph of a polar equation can be symmetric with respect to one of these axes (or the pole) and not satisfy any of the test equations you wrote. This is because a graph can have many polar representations, so many tests are possible. There are many different ways of specifying a point in polar coordinates. You can realize this fact by thinking that every point $(r,\theta)$ could also be called $(r,\theta+2n \pi)$, with $n$ any integer. This fact can affect a symmetry test, even if I think it is possible to define alternative robust symmetry tests.
• Thank you for explaining it to me, but I do have to ask, our teacher told us that we find the symmetries directly from these three tests, and that would help us set an easier bound to work with (e.g. if it is symmetric on all then we work on $[0; \pi/2]$ and mirror everything else). So if I don't know from the test when it's symmetric, then I can't really draw it properly... May 8, 2018 at 12:14
• If the test is successful you can conclude that the graph is symmetric, however if the test fails you cannot draw any conclusion. In this case you have to plot the graph to see if it is symmetric.
– Upax
May 8, 2018 at 19:23
Plotting is the real check where all three these rules are verifiable.
In this (cardioid) curve it has same radius $$r$$ for $$\pm \theta$$ and that makes the polar curve symmetrical with respect to x-axis.
To check the symmetry about both axes and origin:
(r,θ): r = sin(θ/2) = ± { 1 – cos (θ)/ 2 }^1/2 (Half-angle property)
(r,-θ) : r = ± { 1 – cos (-θ) / 2 }^1/2
r = ± { 1 – cos (θ) / 2 }^1/2
Therefore, (r,θ) = (r,-θ)
The curve is symmetrical about X-axis
(r,θ) : r = sin (θ/2)
= sin (θ/2)
= ± {1 – cos (θ)/ 2 }^1/2
(-r,-θ) : -r = ± {1 – cos (-θ)/2}^1/2
-r= ± {1 – cos (θ)/2 }^1/2
Therefore, (r,θ) ≠ (-r,-θ)
(r , π-θ ) : r = ± {1 – cos(π-θ)/2}^1/2
r = ± {1 + cos (θ) /2}^1/2
Therefore , (r,θ) ≠ (r , π-θ )
The curve isn’t symmetrical about Y-axis | 2022-08-10T05:19:45 | {
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https://s62713.gridserver.com/cdkeys-windows-roslfr/logarithmic-differentiation-problems-14fb33 | Now, as we are thorough with logarithmic differentiation rules let us take some logarithmic differentiation examples to know a little bit more about this. The function must first be revised before a derivative can be taken. (2) Differentiate implicitly with respect to x. View Logarithmic_Differentiation_Practice.pdf from MATH AP at Mountain Vista High School. Find the derivative of the following functions. We could have differentiated the functions in the example and practice problem without logarithmic differentiation. Practice 5: Use logarithmic differentiation to find the derivative of f(x) = (2x+1) 3. One of the practice problems is to take the derivative of $$\displaystyle{ y = \frac{(\sin(x))^2(x^3+1)^4}{(x+3)^8} }$$. Instead, you do […] It spares you the headache of using the product rule or of multiplying the whole thing out and then differentiating. Problems. (2) Differentiate implicitly with respect to x. (3) Solve the resulting equation for y′ . (3x 2 – 4) 7. Apply the natural logarithm to both sides of this equation getting . Using the properties of logarithms will sometimes make the differentiation process easier. (x+7) 4. Do 1-9 odd except 5 Logarithmic Differentiation Practice Problems Find the derivative of each of the We know how 11) y = (5x − 4)4 (3x2 + 5)5 ⋅ (5x4 − 3)3 dy dx = y(20 5x − 4 − 30 x 3x2 + 5 − 60 x3 5x4 − 3) 12) y = (x + 2)4 ⋅ (2x − 5)2 ⋅ (5x + 1)3 dy dx = … In some cases, we could use the product and/or quotient rules to take a derivative but, using logarithmic differentiation, the derivative would be much easier to find. Steps in Logarithmic Differentiation : (1) Take natural logarithm on both sides of an equation y = f(x) and use the law of logarithms to simplify. There are, however, functions for which logarithmic differentiation is the only method we can use. For differentiating certain functions, logarithmic differentiation is a great shortcut. With logarithmic differentiation, you aren’t actually differentiating the logarithmic function f(x) = ln(x). You do not need to simplify or substitute for y. (3) Solve the resulting equation for y′ . Begin with y = x (e x). Basic Idea The derivative of a logarithmic function is the reciprocal of the argument. Click HERE to return to the list of problems. Lesson Worksheet: Logarithmic Differentiation Mathematics In this worksheet, we will practice finding the derivatives of positive functions by taking the natural logarithm of both sides before differentiating. ), differentiate both sides (making sure to use implicit differentiation where necessary), Logarithmic Differentiation example question. SOLUTION 2 : Because a variable is raised to a variable power in this function, the ordinary rules of differentiation DO NOT APPLY ! A logarithmic derivative is different from the logarithm function. 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Would be a huge headache ( 3 ) Solve the resulting equation for y′ headache using! Variable power in this function, the ordinary rules of differentiation do NOT need to or... From MATH AP at Mountain Vista High School problems is given in the below! ( 3 ) Solve the resulting equation for y′ problems is given in the video below practice:... At Mountain Vista High School actually differentiating the logarithmic differentiation to Find the derivative of of! The list of problems Differentiate the following: Either using the product rule or of multiplying the whole out! Of logarithms will sometimes make the differentiation process easier the example and practice problem without logarithmic differentiation is the of! … ] a logarithmic derivative is different from the logarithm function then differentiating differentiation practice problems given... | 2021-06-17T07:51:07 | {
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https://gauravtiwari.org/integral-differential-equations/ | ## Introduction
In earlier parts we discussed about the basics of integral equations and how they can be derived from ordinary differential equations. In second part, we also solved a linear integral equation using trial method. Now we are in a situation from where main job of solving Integral Equations can be started. But before we go ahead to that mission, it will be better to learn how can integral equations be converted into differential equations.
## Integral Equation ⇔ Differential Equation
The method of converting an integral equation into a differential equation is exactly opposite to what we did in last part where we converted boundary value differential equations into respective integral equations. In last workoutinitial value problems$always ended up as Volterra Integrals$ and boundary value problems$resulted as Fredholm Integrals.$ In converse process we will get initial value problems$from Volterra Integrals and boundary value problems$ from Fredholm Integral Equations. Also, as in earlier conversion we continuously integrated the differentials within given boundary values, we will continuously differentiate provided integral equations and refine the results by putting all constant integration limits.
The above instructions can be practically understood by following two examples. First problem involves the conversion of Volterra Integral Equation into differential equation and the second problem displays the conversion of Fredholm Integral Equation into differential equation.
### Problem 1: Converting Volterra Integral Equation into Ordinary Differential Equation with initial values
Convert $$y(x) = – \int_{0}^x (x-t) y(t) dt$$ into initial value problem.
Please note that this was the same integral equation we obtained after converting initial value problem: $$y”+y=0$$ when $$y(0)=y'(0)=0$$ ( See Problem 1 of Part 3 )
#### Solution:
We have, $$y(x) = – \int_{0}^x (x-t) y(t) dt \ldots (1)$$
Differentiating (1) with respect to $x$ will give
$$y'(x) = -\frac{d}{dx} \int_{0}^x (x-t) y(t) dt$$
$$\Rightarrow y'(x)=-\int_{0}^x y(t) dt \ldots (2)$$
Again differentiating (2) w.r.t. $x$ will give
$$y”(x)=-\frac{d}{dx}\int_{0}^x y(t) dt$$
$$\Rightarrow y”(x)=-y(x) \ldots (3′)$$
$$\iff y”(x)+y(x)=0 \ldots (3)$$
Putting the lower limit $x=0$ (i.e., the initial value) in equation (1) and (2) will give, respectively the following:
$$y(0) = – \int_{0}^0 (0-t) y(t) dt$$
$$y(0)=0 \ldots (4)$$
And, $$y'(0)=-\int_{0}^0 y(t) dt$$
$$y'(0)=0 \ldots (5)$$
These equations (3), (4) and (5) form the ordinary differential form of given integral equation. $\Box$
### Problem 2: Converting Fredholm Integral Equation into Ordinary Differential Equation with boundary values
Convert $$y(x) =\lambda \int_{0}^{l} K(x,t) y(t) dt$$ into boundary value problem where $$K(x,t)=\frac{t(l-x)}{l} \qquad \mathbf{0<t<x}$$ and $$K(x,t)=\frac{x(l-t)}{l} \qquad \mathbf{x<t<l}$$
#### Solution:
Please see Example 2 of Part 3.
The given integral equation is $$y(x) =\lambda \int_{0}^{l} K(x,t) y(t) dt \ldots (1)$$ or $$y(x) =\lambda (\int_{0}^{x} \frac{(l-x)t}{l} y(t) dt + \int_{x}^{l} \frac{x(l-t)}{l} y(t) dt) \ldots (2)$$
Differentiating (2) with respect to $x$ will give $$y'(x) = -\frac{\lambda}{l} \int_{0}^x t y(t) dt + \frac{\lambda}{l} \int_{x}^l (l-t) y(t) dt \ldots (3)$$
Continued differentiation of (3) will give $$y”(x) = -\lambda y(x)$$ That’s $$y”(x) +\lambda y(x) =0 \ldots (4)$$
To get the boundary values, we place $x$ equal to both integration limits in (1) or (2).
$x =0 \Rightarrow$ $$y(0)=0 \ldots (5)$$
$x=l \Rightarrow$ $$y(l)=0 \ldots (6)$$
The ODE (4) with boundary values (5) & (6) is the exact conversion of given integral equation. $\Box$
Feel free to ask questions, send feedback and even point out mistakes. Great conversations start with just a single word. How to write better comments?
1. jot says:
If y =a x^2+(1-a)x then find I =∫(πy^2) dx
integral from 0 to 1
1. jot says:
find dI/dx
If y =a x^2+(1-a)x then find I =∫(πy^2) dx
integral from 0 to 1
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## The Lindemann Theory of Unimolecular Reactions
[ Also known as Lindemann-Hinshelwood mechanism.] It is easy to understand a bimolecular reaction on the basis of collision theory. When two molecules A and B collide, their relative kinetic energy exceeds the threshold energy with the result that the collision results in the breaking of comes and the formation of new bonds. But how can one account for a…
## Symmetry in Physical Laws
‘Symmetry’ has a special meaning in physics. A picture is said to be symmetrical if one side is somehow the same as the other side. Precisely, a thing is symmetrical if one can subject it to a certain operation and it appears exactly the same after the operation. For example, if we look at a base that is left and…
## Solving Integral Equations (2) – Square Integrable Functions, Norms, Trial Method
Square Integrable function or quadratically integrable function $\mathfrak{L}_2$ function A function $y(x)$ is said to be square integrable or $\mathfrak{L}_2$ function on the interval $(a,b)$ if $$\displaystyle {\int_a^b} {|y(x)|}^2 dx <\infty$$ or $$\displaystyle {\int_a^b} y(x) \bar{y}(x) dx <\infty$$. For further reading, I suggest this Wikipedia page. $y(x)$ is then also called ‘regular function’. The kernel $K(x,t)$ , a function of two variables is…
## Two Interesting Math Problems
Problem1: Smallest Autobiographical Number: A number with ten digits or less is called autobiographical if its first digit (from the left) indicates the number of zeros it contains,the second digit the number of ones, third digit number of twos and so on. For example: 42101000 is autobiographical. Find, with explanation, the smallest autobiographical number. Solution of Problem 1 Problem 2:… | 2019-09-23T20:22:35 | {
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https://tutorme.com/tutors/16488/interview/ | TutorMe homepage
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Krystyna I.
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Geometry
TutorMe
Question:
Define the equation of the tangent to parabola $$y = x^2$$ parallel to the line $$y = x$$.
Krystyna I.
The equation of tangent to the line $$y = f(x)$$ at point $$x_0$$ in general case is given by $$y = f'(x_0)(x-x_0)+f(x_0)$$. We don't know, at which point we should have a tangent, however, we do know that it is parallel to the line $$y = x$$, from which we can immediatly say that $$f'(x_0) = 1$$. This is coming from the fact, that two lines $$y = k_1x + b_1$$ and $$y = k_2x+b_2$$ if $$k_1 = k_2$$. Having defined that, we can find the point where we should compute the tangent: $$f'(x) = (x^2)' = 2x \implies 2x_0 = 1 \implies x_0 = \frac{1}{2}$$. The value of the function at this point is $$f(x_0) = \Big(\frac{1}{2}\Big)^2 = \frac{1}{4}$$. So, the final equation of the tangent to the parabola is given by $$y = \Big(x-\frac{1}{2}\Big) + \frac{1}{4} = x - \frac{1}{4}$$.
Calculus
TutorMe
Question:
Compute the integral $$\int\frac{e^{2x}}{1+e^x}dx$$
Krystyna I.
The nominator is $$e^{2x} = e^xe^x$$. One of the exponents we can put inside the $$dx$$ by integrating it. Since the integral of the exponential function is again the exponential function, our initial integral reads: $$\int\frac{e^x}{1+e^x}d(e^x)$$. Now we can replace a variable as follows $$e^x = u$$, which leads to $$\int\frac{u}{1+u}du$$. Let us consider the expression inside the integral: $$\frac{u}{1+u} = \frac{1-1+u}{1+u} = \frac{(1+u)-1}{1+u} = 1-\frac{1}{1+u}$$, which means that we can re-write our integral as a sum of two: $$\int 1du - \int\frac{1}{1+u}du$$. The first integral will give us just $u$ as a result, while in the second one we can replace $$du$$ by $$d(u+1)$$ because the derivatives of these two functions are the same and they are equal to 1. So, we get $$u - \int\frac{1}{1+u}d(1+u) + c_1$$. The constant $$c_1$$ comes out after the integration of the first integral. The second integral is a standard one and it gives the logarithmic function as an outcome: $$u - \ln(1+u) + c_1 + c_2$$ = $$u - \ln(1+u) + c$$. We replaced $$c_1+c_2$$ by $$c$$ since the some of two constants its again a constant. Finally, going back to our initial variable $$x$$, we can write the answer: $$e^x - \ln(1+e^x) + c$$. You can check the answer by taking the derivative of the last expression: $$\Big(e^x-\ln(1+e^x)+c\Big)' = e^x-\frac{1}{1+e^x}e^x = \frac{e^x+e^{2x}-e^x}{1+e^x} = \frac{e^{2x}}{1+e^x}$$, which is exactly our initial function under the integration.
Algebra
TutorMe
Question:
Find all the values of the parameter $$a$$ such that the equation $$\frac{6+a}{3+x}-\frac{6-a}{3-x} = \frac{6}{a}$$ has only one root.
Krystyna I.
First of all, we need to exclude the values for both unknown $$x$$ and the parameter $$a$$ which turn the denominator into zero. These values are $$(x \neq -3) \bigcup (x \neq 3)$$ and $$a\neq 0$$. Now let us simplify the equation, putting the left-hand side to one common denominator: $$\frac{18-6x+3a-ax-18-6x+3a+ax}{9-x^2} = \frac{6}{a}$$, which leads to equation $$\frac{6a-12x}{9-x^3} = \frac{6}{a}$$. Dividing both parts of the equation by 6, we can rewrite it in the following way $$\frac{a-2x}{9-x^2} = \frac{1}{a}$$, which finally will lead to the simple quadratic equation $$x^2-2ax+a^2-9 = 0$$. There are multiple ways to find the roots of the last equation, for example: $$(x-a)^2 - 9 = 0 \implies (x-a)^2-3^2 = 0 \implies (x-a-3)(x-a+3) = 0$$ $$x_1 = a-3, x_2 = a+3$$. Remember, that we have to find those values of the parameter $a$ which will give us only one root. Let's do the check $$x_1 = x_2 \implies a-3 = a+3$$, which has no solution. However, we have restrictions on the roots. By setting $$x = 3$$ we find that $$a_{1,2} = 0,6$$ and $$x = -3$$, $$a_{1,2} = 0,-6$$. The value $$a = 0$$ is out of our interests as we said ealier. Which means that for the values $$a = -6$$ and $$a = 6$$ the initial equation will have one root only.
Send a message explaining your
needs and Krystyna will reply soon.
Contact Krystyna | 2018-12-16T05:23:14 | {
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https://math.stackexchange.com/questions/3691129/how-to-show-sum-i-02m-1i-binom2mi2-1m-binom2mm/3691180 | How to show $\sum_{i=0}^{2m} (-1)^{i}\binom{2m}{i}^{2} = (-1)^m\binom{2m}{m}$ [duplicate]
I am trying to show that for any positive integer m, $$\sum_{i=0}^{2m} (-1)^{i}\binom{2m}{i}^{2} = (-1)^m\binom{2m}{m}$$
Intuitively this seems to be true, $$m = 0$$ both sides evaluate to $$1$$, $$m = 1$$ both sides evaluate to $$-2$$.
I was looking at this identity for a potential connection, but it seems not applicable here $$\binom{2n}{n} = \binom{n}{0}^{2} + \binom{n}{1}^{2} + \binom{n}{2}^{2} + ... + \binom{n}{n}^{2}$$
Also I was trying to find connections between this and the binomial theorem, but no luck
• I don't think so, I see the similarity, but it doesn't help me with my question. Would appreciate an answer to my question
– user722457
May 25, 2020 at 16:15
• @DanielWang What is your question? May 25, 2020 at 16:20
• Showing that $\sum_{i=0}^{2m} (-1)^{i}\binom{2m}{i}^{2} = (-1)^m\binom{2m}{m}$
– user722457
May 25, 2020 at 16:25
• It's exactly the same question, isn't it? What is the difference? BTW, the last answer to the linked question is the one to look at -- very simple and elegant. May 25, 2020 at 16:25
• The RHS is the same, but the LHS isn't
– user722457
May 25, 2020 at 16:26
$$(1+x)^{2m} \ =\ {2m \choose 0} + {2m \choose 1}x\ ..........\ {2m \choose 2m}x^{2m}$$
$$(x-1)^{2m} \ =\ {2m \choose 0}x^{2m}-{2m \choose 1}x^{2m-1}\ .......\ {2m \choose 2m}$$
Observe coefficient of $$x^{2m}$$ in $$(1+x)^{2m}(1-x)^{2m}$$ is nothing but $$\sum_{i=0}^{2m} (-1)^{i}\binom{2m}{i}^{2}$$ = $$S$$
$$S$$ = coefficent of $$x^{2m}$$ in $$(1-x^2)^{2m}$$ which easily can be seen by binomial theorm = $$(-1)^m\binom{2m}{m}$$ | 2022-06-26T17:48:27 | {
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https://au.mathworks.com/help/symbolic/lambertw.html | # lambertw
Lambert W function
## Syntax
``lambertw(x)``
``lambertw(k,x)``
## Description
example
````lambertw(x)` returns the principal branch of the Lambert W function. This syntax is equivalent to `lambertw(0,x)`.```
example
````lambertw(k,x)` is the `k`th branch of the Lambert W function. This syntax returns real values only if `k = 0` or `k = -1`.```
## Examples
collapse all
The Lambert W function `W(x)` is a set of solutions of the equation `x = W(x)eW(x)`.
Solve this equation. The solution is the Lambert W function.
```syms x W eqn = x == W*exp(W); solve(eqn,W)```
```ans = lambertw(0, x)```
Verify that branches of the Lambert W function are valid solutions of the equation `x = W*eW`:
```k = -2:2; eqn = subs(eqn,W,lambertw(k,x)); isAlways(eqn)```
```ans = 1×5 logical array 1 1 1 1 1```
Depending on its arguments, `lambertw` can return floating-point or exact symbolic results.
Compute the Lambert W functions for these numbers. Because the numbers are not symbolic objects, you get floating-point results.
```A = [0 -1/exp(1); pi i]; lambertw(A)```
```ans = 0.0000 + 0.0000i -1.0000 + 0.0000i 1.0737 + 0.0000i 0.3747 + 0.5764i ```
`lambertw(-1,A)`
```ans = -Inf + 0.0000i -1.0000 + 0.0000i -0.3910 - 4.6281i -1.0896 - 2.7664i```
Compute the Lambert W functions for the numbers converted to symbolic objects. For most symbolic (exact) numbers, `lambertw` returns unresolved symbolic calls.
```A = [0 -1/exp(sym(1)); pi i]; W0 = lambertw(A)```
```W0 = [ 0, -1] [ lambertw(0, pi), lambertw(0, 1i)] ```
`Wmin1 = lambertw(-1,A)`
```Wmin1 = [ -Inf, -1] [ lambertw(-1, pi), lambertw(-1, 1i)]```
Convert symbolic results to double by using `double`.
`double(W0)`
```ans = 0.0000 + 0.0000i -1.0000 + 0.0000i 1.0737 + 0.0000i 0.3747 + 0.5764i```
Plot the two main branches, ${W}_{0}\left(x\right)$ and ${W}_{-1}\left(x\right)$, of the Lambert W function.
```syms x fplot(lambertw(x)) hold on fplot(lambertw(-1,x)) hold off axis([-0.5 4 -4 2]) title('Lambert W function, two main branches') legend('k=0','k=1','Location','best')```
Plot the principal branch of the Lambert W function on the complex plane.
Plot the real value of the Lambert W function by using `fmesh`. Simultaneously plot the contours by setting `'ShowContours'` to `'On'`.
```syms x y f = lambertw(x + 1i*y); interval = [-100 100 -100 100]; fmesh(real(f),interval,'ShowContours','On')```
Plot the imaginary value of the Lambert W function. The plot has a branch cut along the negative real axis. Plot the contours separately.
`fmesh(imag(f),interval)`
`fcontour(imag(f),interval,'Fill','on')`
Plot the absolute value of the Lambert W function.
`fmesh(abs(f),interval,'ShowContours','On')`
## Input Arguments
collapse all
Input, specified as a number, vector, matrix, or array, or a symbolic number, variable, array, function, or expression.
At least one input argument must be a scalar, or both arguments must be vectors or matrices of the same size. If one input argument is a scalar and the other is a vector or matrix, `lambertw` expands the scalar into a vector or matrix of the same size as the other argument with all elements equal to that scalar.
Branch of Lambert W function, specified as an integer, a vector or matrix of integers, a symbolic integer, or a symbolic vector or matrix of integers.
At least one input argument must be a scalar, or both arguments must be vectors or matrices of the same size. If one input argument is a scalar and the other is a vector or matrix, `lambertw` expands the scalar into a vector or matrix of the same size as the other argument with all elements equal to that scalar.
collapse all
### Lambert W Function
The Lambert W function W(x) represents the solutions y of the equation $y{e}^{y}=x$ for any complex number `x`.
• For complex x, the equation has an infinite number of solutions y = lambertW(k,x) where k ranges over all integers.
• For all real x ≥ 0, the equation has exactly one real solution y = lambertW(x) = lambertW(0,x).
• For real x where $-{e}^{-1}, the equation has exactly two real solutions. The larger solution is represented by y = lambertW(x) and the smaller solution by y = lambertW(–1,x).
• For $x=-{e}^{-1}$, the equation has exactly one real solution y = –1 = lambertW(0, –exp(–1)) = lambertW(–1, -exp(–1)).
## References
[1] Corless, R.M., G.H. Gonnet, D.E.G. Hare, D.J. Jeffrey, and D.E. Knuth. "On the Lambert W Function." Advances in Computational Mathematics, Vol. 5, pp. 329–359, 1996.
## Version History
Introduced before R2006a | 2022-05-28T00:17:38 | {
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https://yutsumura.com/the-set-of-square-elements-in-the-multiplicative-group-zmodp/ | # The Set of Square Elements in the Multiplicative Group $(\Zmod{p})^*$
## Problem 616
Suppose that $p$ is a prime number greater than $3$.
Consider the multiplicative group $G=(\Zmod{p})^*$ of order $p-1$.
(a) Prove that the set of squares $S=\{x^2\mid x\in G\}$ is a subgroup of the multiplicative group $G$.
(b) Determine the index $[G : S]$.
(c) Assume that $-1\notin S$. Then prove that for each $a\in G$ we have either $a\in S$ or $-a\in S$.
## Proof.
### (a) Prove that $S=\{x^2\mid x\in G\}$ is a subgroup of $G$.
Consider the map $\phi:G \to G$ defined by $\phi(x)=x^2$ for $x\in G$.
Then $\phi$ is a group homomorphism. In fact, for any $x, y \in G$, we have
\begin{align*}
\phi(xy)=(xy)^2=x^2y^2=\phi(x)\phi(y)
\end{align*}
as $G$ is an abelian group.
By definition of $\phi$, the image is $\im(\phi)=S$.
Since the image of a group homomorphism is a group, we conclude that $S$ is a subgroup of $G$.
### (b) Determine the index $[G : S]$.
By the first isomorphism theorem, we have
$G/\ker(\phi)\cong S.$
If $x\in \ker(\phi)$, then $x^2=1$.
It follows that $(x-1)(x+1)=0$ in $\Zmod{p}$.
Since $\Zmod{p}$ is an integral domain, it follows that $x=\pm 1$ and $\ker(\phi)=\{\pm 1\}$.
Thus, $|S|=|G/\ker(\phi)|=(p-1)/2$ and hence the index is
$[G:S]=|G|/|S|=2.$
### (c) Assume that $-1\notin S$. Then prove that for each $a\in G$ we have either $a\in S$ or $-a\in S$.
Since $-1\notin S$ and $[G:S]=2$, we have the decomposition
$G=S\sqcup (-S).$ Suppose that an element $a$ in $G$ is not in $S$.
Then, we have $a\in -S$.
Thus, there exists $b\in S$ such that $a=-b$.
It follows that $-a=b\in S$. Therefore, we have either $a\in S$ or $-a\in S$.
##### The Number of Elements Satisfying $g^5=e$ in a Finite Group is Odd
Let $G$ be a finite group. Let $S$ be the set of elements $g$ such that $g^5=e$, where $e$ is... | 2019-09-23T02:40:40 | {
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https://math.stackexchange.com/questions/3808575/pulling-universal-quantifier-out-of-parenthesis-makes-nonequivalent-statement | # Pulling universal quantifier out of parenthesis makes nonequivalent statement?
Assuming I have the statement ∀x(∀y¬Q(x,y)∨P(x)), can I pull the universal quantifier ∀y out of the parenthesis? Meaning, is this statement equivalent to ∀x∀y(¬Q(x,y)∨P(x)) ?
An approach I tried so far:
1. ∀x((∃y Q(x,y) ) => P(x)). (original eq.)
2. ∀x((∀y¬Q(x,y))∨P(x)) (De Morgan's application)
3. ∀x∀y(¬Q(x,y)∨P(x)). (Working off the assumption that taking out the ∀y is a valid operation).
4. ∀x∀y (Q(x,y) => P(x)) (Going backwards from the ¬P v Q definition of implication)
Statement 4 does not seem to be equivalent to statement 1, which suggests that pulling out the universal quantifier is not acceptable. I would greatly appreciate any confirmation of whether this is the case, and if so, what governs when quantifiers can be brought to the outside of the parenthesis.
• Btw, you can use MathJax to type equations: math.meta.stackexchange.com/questions/5020/…. Use dollar signs to enclose equations and backslashes for commands: \forall = $\forall$, \exists = $\exists$, etc. You can use detexify to find the name of a symbol. Aug 30 '20 at 19:03
• Your step from 1 to 2 is wrong: you replaced the expression inside the brackets by its negation. Aug 31 '20 at 7:58
For universal quantifier. In general, if $$x$$ apear in both $$A$$ and $$B$$ we have $$\exists xA(x)\to \forall xB(x)\Rightarrow\forall x(A(x)\to B(x))\tag{1}$$ $$\forall x(A(x)\to B(x))\not\Rightarrow \exists xA(x)\to \forall xB(x)\tag{2}$$ However, if $$x$$ not apear in $$B$$ we have $$\forall x(A(x)\to B)\Leftrightarrow\exists xA(x)\to \forall x B\tag{3}$$ The statement in question is similar to $$(3)$$, which is also valid. $$∀x∀y(Q(x,y)→P(x))\Leftrightarrow∀x(∃y Q(x,y)→P(x))\tag{4}$$ And we can formulate a direct proof for $$(4)$$ by nature deduction $$\def\fitch#1#2{\hspace{2ex}\begin{array}{|l}#1\\\hline#2\end{array}} \fitch{\forall x\forall y(Q(x,y)\to P(x))} {\fitch{\boxed{a}} {\forall y(Q(a,y)\to P(a))\\ \fitch{\exists y~Q(a,y)} {\fitch{\boxed{b}~Q(a,b)} {Q(a,b)\to P(a)\\ P(a)}\\ P(a)}\\ \exists y~Q(a,y)\to P(a)}\\ \forall x~(\exists y~Q(x,y)\to P(x))}\\$$ Hence $$\forall x\forall y(Q(x,y)\to P(x))\Rightarrow\forall x~(\exists y~Q(x,y)\to P(x))$$. For the another direction we have $$\fitch{\forall x(\exists y~Q(x,y)\to P(x))} {\fitch{\boxed{a}} {\exists y~Q(a,y)\to P(a)\\ \fitch{\boxed{b}~Q(a,b)} {\exists y~Q(a,y)\\ P(a)}\\ \forall y~(Q(a,y)\to P(a))}\\ \forall x\forall y~(Q(x,y)\to P(x))}$$ Therefore $$\forall x~(\exists y~Q(x,y)\to P(x))\Rightarrow \forall x\forall y(Q(x,y)\to P(x))$$. This proves $$(4)$$.
The original expression: $$\forall x~((\exists y~Q(x,y))\to P(x))$$ says "For any $$x$$ it holds that if some $$y$$ satisfies $$Q(x,y)$$, then $$P(x)$$ is satisfied."
Now either consequent is true for all $$x$$ or, whenever it is false, the antecedent is also false (ie for that $$x$$ no $$y$$ can satisfy $$Q(x,y)$$). Thus the expression equates to: $$\forall x~(\neg P(x)\to\forall y~\neg Q(x,y))$$
The final expression: $$\forall x~\forall y~(Q(x,y)\to P(x))$$ says: "For any $$x$$ and $$y$$, it holds that if $$Q(x,y)$$ then $$P(x)$$."
Now either consequent is true for all $$x$$ or, whenever it is false, the antecedent is also false; moreover false for all $$y$$ when $$P(x)$$ is false for some $$x$$. Thus the expression equates to: $$\forall x~\forall y~(\neg P(x)\to \neg Q(x,y))$$
Therefore the original and final expressions are equivalent.
They're equivalent.
Here is a proof:
This tree was generated here. | 2021-12-06T06:31:33 | {
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https://mathematica.stackexchange.com/questions/153520/fitting-of-statistical-data-points-by-normal-distribution/153523 | # Fitting of statistical data points by Normal distribution
The data is given below:
{{-2.9, 1}, {-2.7, 0}, {-2.5, 0}, {-2.3, 2}, {-2.1, 2}, {-1.9,
3}, {-1.7, 5}, {-1.5, 7}, {-1.3, 3}, {-1.1, 11}, {-0.9, 7}, {-0.7,
3}, {-0.5, 14}, {-0.3, 9}, {-0.1, 24}, {0.1, 17}, {0.3, 26}, {0.5,
11}, {0.7, 14}, {0.9, 11}, {1.1, 9}, {1.3, 5}, {1.5, 2}, {1.7,
5}, {1.9, 3}, {2.1, 3}, {2.3, 1}, {2.5, 1}, {2.7, 1}, {2.9, 0}}
I have used the formula of NormalDistribution with $\mu$ and $\sigma$ as the parameters of the fitting through NonlinearModelFit and the result is follows
The data should satisfy NormalDistribution in a much more decent form as can be seen by eye estimation and moreover the data is taken from some nuclear experiments satisfying the distribution. I guess there might be some other ways to fit these kind of statistical data points.
Any guidance regarding any fitting issues will be very helpful to resolve my problem. Thanks for all valuable suggestions in advance !
• NonlinearModelFit[] seems inappropriate here; have a look at EstimatedDistribution[] or FindDistributionParameters[]. – J. M. is away Aug 10 '17 at 18:14
• I assume that the integer values are counts of observations? Why do you need to fit any particular distribution? Is it to compare parameter estimates from other datasets? – JimB Aug 10 '17 at 19:40
• You cannot fit a probability density to your data since the area under your data is not 1. You want to fit to a gaussian pulse as shown by David Stork. – Bob Hanlon Aug 10 '17 at 19:45
• @BobHanlon, if the OP's data are the midpoints and frequency counts of a histogram, then a regression approach is completely inappropriate. (I'm not as diplomatic as J.M.) – JimB Aug 10 '17 at 19:51
If one wants to fit a curve that just happens to be of the form of a standard probability density function (with an additive and/or multiplicative constant included), then a regression approach makes sense. But there is no probabilistic interpretation that is induced by such a fit.
But when one has a random sample from that probability distribution, performing a regression makes no sense.
The data presented for the above question is almost certainly pairs of midpoints and the associated frequency of occurrence from a sample from some distribution. The question is about fitting a normal distribution from such data. (The goodness of the fit is another issue.)
If one had the raw data and was fitting a normal distribution, then calculating the sample mean and the sample standard deviation would be a reasonable way to estimate the underlying parameters. When only the binned data is available there are two typical ways to perform the estimation for a normal distribution.
(* Get estimates of the mean and standard deviation using the binned data *)
μ0 = data[[All, 1]].data[[All, 2]]/Total[data[[All, 2]]]
(* 0.045999999999999944 *)
σ0 = (((data[[All, 1]]^2).data[[All, 2]])/Total[data[[All, 2]]] - μ0^2)^0.5
(* 1.0017404853553638 *)
The maximum likelihood estimates of $\mu$ and $\sigma$ based on the binned data are found by maximizing the likelihood (or equivalently the log of the likelihood):
$$\text{Likelihood}=\prod_{i=1}^n \left(\Phi\left(\frac{x_i+\frac{w}{2}-\mu}{2}\right)-\Phi\left(\frac{x_i-\frac{w}{2}-\mu}{2}\right)\right)^{f_i}$$
where $x_i$ is the $i$-th bin midpoint, $f_i$ is the frequency count of the $i$-th bin, $n$ is the number of bins, $w$ is the bin width, and $\Phi(z)$ the standard normal cumulative distribution function.
(* Get maximum likelihood estimates *)
Φ[z_NumericQ] := CDF[NormalDistribution[0, 1], z]
w = 0.2; (* Width of histogram bin *)
(* Log of the likelihood *)
logL = Sum[data[[i, 2]] Log[Φ[(data[[i, 1]] + w/2 - μ)/σ] - Φ[(data[[i, 1]] - w/2 - μ)/σ]],
{i, Length[data]}];
(* Find values of the parameters that maximize the log of the likelihood *)
FindMaximum[{logL, σ > 0}, {{μ, μ0}, {σ, σ0}}]
{-606.0230945881103, {μ -> 0.045999689972915744, σ -> 1.0000755520127822}}
For this data there is very little difference between the estimation procedures. A next step would be to assess the quality of the fit.
• Here's a slightly compacted implementation: μ0 = data[[All, 1]].Normalize[data[[All, 2]], Total]; σ0 = Sqrt[((data[[All, 1]]^2).Normalize[data[[All, 2]], Total] - μ0^2)]; FindArgMax[{data[[All, 2]].Log[Erf[(# - μ - w/2)/(σ Sqrt[2]), (# - μ + w/2)/(σ Sqrt[2])]/2 &[data[[All, 1]]]], σ > 0}, {{μ, μ0}, {σ, σ0}}], where dot products were used for efficiency. Note as well the use of two-argument Erf[], which is much more numerically stable than subtracting two instances of Erf[]. – J. M. is away Aug 11 '17 at 4:21
Here is the fit to your data, based on a normal distribution:
FindFit[data,
a PDF[NormalDistribution[μ, σ], x], {a, μ, σ}, x]
(*
{a -> 37.2923, μ -> 0.134454, σ -> 0.834692}
*)
Show[
ListPlot[data, PlotStyle -> Red],
Plot[37.2923 PDF[NormalDistribution[0.134454, 0.834692], x],
{x, -3, 3}]
]
For what it's worth, here's a quick and easy approximation to the answer given by @Jim Baldwin:
EstimatedDistribution[
Flatten[
Cases[data, {x_, n_} :> ConstantArray[x, n]]
],
NormalDistribution[m, s]
]
(* NormalDistribution[0.046, 1.00174] *)
` | 2019-07-22T12:29:22 | {
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http://mathhelpforum.com/statistics/123640-probability-2-people-next-eachother.html | # Math Help - Probability 2 people are next to eachother
1. ## Probability 2 people are next to eachother
Here is the 2 part question. Given n people are in a straight line what is the probability
A: They are next to eachother
B: They are seperated by exactly one person
For a i worked it out a few ways and it looks like 2/n but im not sure why and part b im kinda lost on how to figure it out.
2. Originally Posted by ChrisBickle
Here is the 2 part question. Given n people are in a straight line what is the probability
A: They are next to eachother
B: They are seperated by exactly one person
For a i worked it out a few ways and it looks like 2/n but im not sure why and part b im kinda lost on how to figure it out.
What is part one? WHO?
Otherwise, none of this is meaningful.
3. You must choose a subgroup from n,
or n is a subgroup of a larger group.
As things stand, all n are next to each other and no-one is seperated.
4. Originally Posted by Archie Meade
You must choose a subgroup from n,
or n is a subgroup of a larger group.
As things stand, all n are next to each other and no-one is seperated.
What does that mean?
5. If they are in a straight line but not next to each other,
and we only have the value n,
the problem to me is undefined as seperation from each other has
as many possibilities as one wishes.
but i'd expect a subgroup to be examined from the entire group.
6. Sorry i wasnt very specific. Here is the full question. Given 2 people person a and person b are standing in a straight line with n people. What is the probability they are standing next to eachother. And the second part is what is the probability they are seperated by exactly 1 person.
7. There are n! arrangements of n people.
If 2 of these n are considered as a pair, then they are a "unit" out of n-1.
They then have n-1 possible positions as a unit in the line of n people.
There are 2! arrangements of the 2 people side by side,
hence the first probability is
$\frac{(n-1)2!}{n!}$ times the remaining (n-2) people arranged ......added in hindsight
Having 1 person between them....
Take a "unit" as 3.
There are n-2 such "units", with only 2! variations of that "unit" countable.
However, this "unit" can be in n-2 positions.
Hence, the probability is
$\frac{(n-2)^{2}2!}{n!}$ times the remaining (n-3) people arranged ......added in hindsight
Whoooops!!!!
I forgot to arrange the remaining (n-2) and (n-3) people... sorry.
8. Originally Posted by ChrisBickle
Sorry i wasnt very specific. Here is the full question. Given 2 people person a and person b are standing in a straight line with n people. What is the probability they are standing next to eachother. And the second part is what is the probability they are seperated by exactly 1 person.
The correct answer to the first question is $\frac{2(n-1)!}{n!}=\frac{2}{n}$.
Note that is 1 when n=2
The answer to the second part is $\frac{(n-2)(2)(n-2)!}{n!}=\frac{2(n-2)}{n(n-1)}$.
Note that is 0 if n=2 and it is $\frac{1}{3}$ if n=3.
9. Person "a" and person "b" are 2 specific people or any two people??
10. Originally Posted by Archie Meade
Person "a" and person "b" are 2 specific people or any two people??
That is clearly what it says: "Given two people, person a and person b,..."
11. Sorry!
i just realised my mistake,
thanks!
12. Hello, ChrisBickle!
Plato is absolutely correct!
Here is the resoning . . .
Given $n$ people standing in a line.
There are $n!$ arrangements of the people.
(a) What is the probability that two particular people, $A$ and $B$, are adjacent?
Duct-tape $A$ and $B$ together.
They can be taped like this: . $\boxed{AB}\:\text{ or like this: }\boxed{BA}$ . . . 2 choices.
Then we have $n-1$ "people" to arrange.
. . .There are: . $(n-1)!$ ways.
Hence, there are: . $2(n-1)!$ ways for $A$ and $B$ to be adjacent.
Therefore: . $P(A\text{ and }B\text{ adjacent}) \;=\;\frac{2(n-1)!}{n!} \;=\;\frac{2}{n}$
(b) What is the probability they are seperated by exactly one person?
If there is exactly one person between A and B,
. . they can be arranged: . $\boxed{A\,X\,B}\,\text{ or }\,\boxed{B\,X\,A}$ . . . 2 choices.
Then $X$ can be any of the other $n-2$ people.
And we have $n-2$ "people" to arrange.
. . There are: . $(n-2)1$ ways.
Hence, there are: . $2(n-2)(n-2)!$ ways for $A$ and $B$ to be separated by one person.
Therefore: . $P(A\text{ and }B\text{ separated by one person}) \;=\;\frac{2(n-2)(n_2)!}{n!} \;=\;\frac{2(n-2)}{x(n-1)}$ | 2014-11-29T02:37:42 | {
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http://salesianipinerolo.it/asmz/when-two-dice-are-rolled-find-the-probability-of-getting-a-sum-of-5-or-6.html | 16&comma. So there are 6*combin (5,2)=60 combinations already. =5/36 +11/36 -1/36=15/36=5/12. ) What is the probability of getting a head or a tail?. Sum of Two Dice. two dice are rolled. (b) Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4. MATH 225N WEEK 4(STATISTICS) QUIZ / MATH225N WEEK 4(STATISTICS) QUIZ (GRADED A): CHAMBERLAIN COLLEGE OF NURSING MATH 225N WEEK 4(STATISTICS) QUIZ Question 1 Alice sells boxes of candy at the baseball game and wants to know the mean number of boxes she sells. So the probability of a sum of at least 5 is 30 out of 36, which gives us the fraction which reduces to. Is this solution Helpfull? Yes (28) | No (6). The probability of choosing a red card from a standard deck of cards is 26/52. Solution Two Different Dice Are Thrown at the Same Time. 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P(2 twos and 2 ones and the other two different ) + P(2 sixes and 2 ones and 2 of some other number). There is only one way to get a total of 12. The sum of the two numbers rolled are shown below:. We want sum to be greater than 16, So, sum could be either 17 or 18. 4 And P (B) = 0. , dice with sides numbered 1-4. Try the following: 1. Both dice are the same particular number. If you only take two of the three for the sum, there are still 216 total outcomes to look at. Its like say, because I can first roll two then five, then it is never the case that if I roll five that I may roll two. You ask for P(A|B). Two 6-sided dice are rolled. 1 die, 2 dice. A single, standard number cube is tossed. We need to find the. Solution Two Different Dice Are Thrown at the Same Time. Let X denote the sum of the number of dots on the top faces. Find maximum subset sum formed by partitioning any subset of array into 2 partitions with equal sum. Assume you roll a fair dice twice. 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[3 Marks) 1 13 5 Question 2 Find The Z Score That Corresponds To The Given Area. Probability of rolling two dice and getting a sum of 7 or at least one 4 - Продолжительность: 1:42 Laura Rickhoff 39 477 просмотров. For example if n. 10 5 13 ! Find the probability distribution. For four six-sided dice, the most common roll is 14, with probability 73/648; and the least common rolls are 4 and 24, both with probability 1/1296. Probability (statistics) What is the probability of getting a sum of 8 in rolling two dice? Update Cancel. In the example you gave, I find it much easier to start by calculating the probability of NOT rolling a 5 across multilple throws, because these probabilities can be just multiplied together. What is the conditional probability that at least one lands on 6 given that the dice land Probability of a woman being a smoker given she has ectopic pregnancy • E: ectopic pregnancy Did you find mistakes in interface or texts? Or do you know how to improveStudyLib UI?. 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Answers · 2 The table below shows which hand is favored by each of 100 people (50 men and 50 women). Calculate the is the conditional probability that the Finding P (E): The probability of getting 4 atleast once is. Dice and Dice Games. [3 Marks) 1 13 5 Question 2 Find The Z Score That Corresponds To The Given Area. Number of outcomes of the experiment that are favorable to the event that a sum of two events is 6. We have to find what is the probability that the sum of numbers rolled is either 5 or 12 We know that, probability of an event = Now, total outcomes for two dices = 6 for 1st dice x 6 for 2nd dice = 6 x 6 = 36. In sum, nice job by new coach Ron Rivera. Find the expected number of times one needs to roll a dice before getting 4 sixes. , 12? First, we compute P (Si) for the latter values of i. the probability that the sum is 6 given that at least one of the numbers is less than 3. Sometimes two different events can give the same event space. That intuition is wrong. > Consider this matrix for two dice roll game $\textrm{Total outcomes with the sum}$ $3 = 2$ $\textrm{Total outcomes with the sum}[/math. Given a pair of dice, what is the chance of drawing an odd number? 7. A single, standard number cube is tossed. Independent probabilities are calculated using: Probability of both = Probability of outcome one × Probability of outcome two So to get two 6s when rolling two dice, probability = 1/6 × 1/6 = 1/36 = 1 ÷ 36 = 0. What is the probability of getting a 5 after rolling a single 6-sided die? 1/6 or 16. For example if n. Finding probability. Let B be the event - The sum of the top faces of the 3 dice >= 5. What is the probability that the sum of the scores is: a) even b) prime c) even or prime? Homework Equations The Attempt at a Solution a) P(even) = 1/2 b) P(prime) = 9/16 c) c for confused Can someone please explain the theory behind answering Question c? Cheers. The probability of rolling an even number on 1 die is 3/6. Two dice are rolled. (a) Find the conditional probability of obtaining a sum greater than 9, given that Given that the two numbers appearing on throwing two dice are different. Sums of two independent Binomial random variables. Three Dice are Rolled Find Probability of Just Getting 5 once - Продолжительность: 3:57 Anil Kumar 16 555 просмотров. Given a pair of dice, what is the chance of drawing an odd number? 7. If two dice are rolled one time find the probability of getting a sum less than 5. A glass jar contains 6 red, 5 green, 8 blue and 3 yellow marbles. The numbers for the games so far are listed below. Find the probability of the lost card being a diamond. ECEN 303 - Fall 2011. Examples: a. When we toss two coins simultaneously then the possible of outcomes are: (two heads) or (one head and one tail) or (two tails) i. Two dice are thrown simultaneously. Two fair dice are rolled and the sum of the points is noted. What is the probability of getting a straight by a single throw of 5. Throwing a 6,5,4,3,2 or 1 deducts 0. (6,1),(6,2),(6,3),(6,4),(6,5),(6,6) since each 6 numbers. 4d10 are enough to sample uniformly from between 1 and 10,000), but it becomes increasingly tedious to generate larger numbers. function [ X ] = Dice ( N, S, T, R ) % Dice simulates a random selection of numbers which is similar to how a % dice is rolled % % N is the number of dice the user wants to roll % S is the number of sides on the dice % T is the number of trials that the user wants to run. 4 And P (B) = 0. > Consider this matrix for two dice roll game [math]\textrm{Total outcomes with the sum}$ $3 = 2$ $\textrm{Total outcomes with the sum}[/math. Let X denote the sum of the number of dots on the top faces. 1 die, 2 dice. Memorizing the making of the above picture makes the. Find the probabilities of rolling different sums. No, other sum is possible because three dice being rolled give maximum sum of (6+6+6) i. The sum of the two dice you rolled is. For four six-sided dice, the most common roll is 14, with probability 73/648; and the least common rolls are 4 and 24, both with probability 1/1296. a sum less than 4 or greater than 9 d. Rolling Dice. What is the probability of it being a black shirt? 8) On rolling a dice 2 times, the sum of 2 numbers that appear on the uppermost face is 8. I hope this helps, Harley Go to Math Central. Now, favourable outcomes = sum. The pair can be any one of 6 numbers. The probability of Dice 2 rolling a 1 is also 1/6. (i) To get the sum of numbers 4 or 5 favourable outcomes are: (1, 3) ,(3, 1) , (2,2). > Consider this matrix for two dice roll game [math]\textrm{Total outcomes with the sum}$ $3 = 2$ $\textrm{Total outcomes with the sum}[/math. 1/18 5/36 1/6 1/9. The probability of choosing a green marble from the jar. 333%) probability of NOT rolling a 5 2 rolls: (5/6) x (5/6) (69. (6,1),(6,2),(6,3),(6,4),(6,5),(6,6) since each 6 numbers. Let B be the event - The sum of the top faces of the 3 dice >= 5. What is the probability of getting a straight by a single throw of 5. There are 4 combinations that have the sum of 9. Event Die 1 Die 2 Sum 1 2 3 4 5 6 7 8 9. Assuming that the dice are unbiased or not " loaded". So when we toss the two dice we will get {eq}6\cdot 6 = 36 {/eq} possible pairs. No, other sum is possible because three dice being rolled give maximum sum of (6+6+6) i. What's the probability of the second card also being a king?. Given a pair of dice, what is the chance of rolling a 7? 8. The probability of rolling an even number on 1 die is 3/6. The ways to get a 4 are: 1+3, 2+2, 3+1 = 3 ways. find the probability that the student gets exactly 4 correct. "If you roll a dice three times, what is the probability of rolling a 6 at least once?" The correct answer is 91/216. Here are a few examples that show off Troll's dice roll language: Roll 3 6-sided dice and sum them: sum 3d6. 5+6 " or " 6+5 Therefore of the 36 possible outcomes there are 3 that do not meet the requirement of being less than 11. Determine if the events are mutually exclusive. Then Each die has 1 chance of being 1-6, out of 6, but 6 is excluded as there is no pairing that sums to 6 if either die is 6. There are 6 6 possible outcomes. To set the count back to 0, press "Start Over" button. The term "boxcars", also known as "midnight", is the outcome of rolling the dice and getting a six on each die. Hence the probability of getting a 3 is P(E) = 1 / 6. Okay, so basically it says someone rolls two dice and its asking about the probability of rolling certain sums of the two dice. 6 outcomes on one die X 6 outcomes on other die = 36 outcomes. I can't get an outcome that's in You can probably guess when we get to continuous probability distributions this is no longer the case. Experiment: A single 6-sided die is rolled. What is the probability that the sum of two rolled dice will equal a prime number? We find this number by multiplying 6 x 6. What is the probability of a bridge hand with 3 of one suit, 3 of one suit, 2 of one suit, 5 of another suit?. The probability that it is a double with a sum of 11 is zero (0) When Two Balanced Dice Are Rolled, There Are 36 Possible. A card from a pack of 5 2 cards is lost. Number of outcomes of the experiment that are favorable to the event that a sum of two events is 6. 16667, to turn up when rolled, if the die (D) is unbiased. Question 4: Two dice are rolled, find the probability that the sum is a) equal to 1 b) equal to 4 c) less than 13 Solution to Question 4: a) The sample space S of two dice is shown below. The sum of two dice thrown can be 7 and 11 in the following cases : (6,1) (1,6) (3,4) (4,3) (5,6) (6,5) (2,5) (5,2) The total possible cases are = 36 Favorable cas. the probability of the sum being: 2 is 1/36 3 is 2/36 4 is 3/36 5 is 4/36 6 is 5/36 7 is 6/36 8 is 5/36 9 is 4/36 10 is 3/36 11 is 2/36 12 is 1/36 It then asks: P(the sum of the two dice equals 2) P(the sum of the two. Let Xj represent the number that comes up when J-th fair die is rolled, 7=1, 2,---, k. The dictionary of etymology traces use of the term as far back as 1919. Solution Two Different Dice Are Thrown at the Same Time. A sum of 6. Dice roll probability: 6 Sided Dice Example. That intuition is wrong. By the central limit theorem, the sum of the five rolls should have approximately the same distribution as a normal random variable with the same mean and variance. Sums of two independent Binomial random variables. For example if n. , dependent on) whether another event occurs. What is the probability of. (it's easier to count the 6 non-red ones and subtract from 36 to get 30). What is the probability of getting a number other than 6?. The sum will be even for any double. 5 seconds shy of Raikkonen when electrical problems brought Ricciardo's day to an early end on lap 11 and prompted the Virtual Safety Car. Find the probability of the lost card being a diamond. Roll Two Fair Dice. 4 And P (B) = 0. The sum of the two dice you rolled is. You ask for P(A|B). 5 and variance 35/12, and so the corresponding mean and variance for rolling 5 dice is 5 times greater. The pair of six pips resembles a pair of boxcars on a freight train. Experimental Probability: Experiment with probability using a fixed size section spinner, a variable section spinner, two regular 6-sided dice or customized dice. It is a relatively standard problem to calculate the probability of the sum obtained by rolling two dice. Find the probability that the student's favorite topping is veggie given that the student is a junior or senior. 16&comma. , any time a "6" comes up, add 6 to your total and roll again): sum (accumulate y:=d6 while y=6). Using an organized list, table, tree diagram, or method of your choosing, develop a list of all 16 possible outcomes (for example, Die #1 = 1 and Die #2 = 2 for a difference of 1; Die #1 = 1 and Die #2 = 4 for a difference of 3; and so on). Therefore probability of getting doubles or sum of 7 is 1/3 Probability of Rolling Doubles - Practice Problems Problem 1: If rolling two dice, what is the probability of getting a sum of 5 or 6? Problem 2: If rolling two number cubes, what is the probability of getting 6 or 7? Answer: 1) 1/4 2) 11/36. Probability theory, a branch of mathematics concerned with the analysis of random phenomena. Probability that sum is neither 7 or 11 = 1 - Probability that the sum is 7 and 11. This lesson explains what a probability distribution is. [3 Marks) 0. We need to find the. if the die is not fair then the probability of an odd number is found by: P( odd number ) = P(rolling a 1) + P(rolling a 3) + P(rolling a 5) for a fair die this is: P( odd number ) = 1/3 + 1/3 + 1/3 = 1/2. Let E denote the event that the number landing uppermost on the first die is a 3, and let F denote the event that the sum of Which pair has equally likely outcomes? Check the two choices below which have equal probabilities of success. Find probability nobody gets own hat. Choose one student at random. Probability Chapter 13 of Class 12 is one of the most important topics in Maths CBSE Board Exams. What is the probability of getting a flush in a. Probability = 1 ÷ 36 = 0. "If you roll a dice three times, what is the probability of rolling a 6 at least once?" The correct answer is 91/216. That intuition is wrong. The sum from rolling two dice can be any value from 2 to 12. If a pair of dice are rolled 5 times, what is the probability of getting a sum of 5 every time?. Good morning Edward, I liked your dice probability work on the chances of getting one 6 when rolling different number of dice. TE Thaddeus Moss was signed after going undrafted. No, other sum is possible because three dice being rolled give maximum sum of (6+6+6) i. Two counters game. A single die is rolled. A card from a pack of 5 2 cards is lost. Then we arrive at dice 9, assign 6 points to it and assign the remaining 15 points to the dice. 4 And P (B) = 0. Two six-sided dice are rolled. > Consider this matrix for two dice roll game [math]\textrm{Total outcomes with the sum}$ $3 = 2$ $\textrm{Total outcomes with the sum}[/math. Total number of possible outcomes on rolling two dice together = 36. This resulted in the first professional. In sum, nice job by new coach Ron Rivera. What is the probabilities of getting at least a 1 OR a 5 with 1 die, 2 dice, 3 dice, etc. Shows how to find probabilities of random variables. (d) an even number appears on the black dice or the sum of the numbers on the two dice is 7. Each of these has a probability of 1/6 of occurring. The logic is there are six sides to each die, so for each number on one You did the math for the probability of rolling a dice twice and getting a multiple of 3 on both rolls. , 12? First, we compute P (Si) for the latter values of i. Use, probability formula = N u m b e r o f f a v o r a b l e o u t c o m e s T o t a l n u m b e r o f p o s s i. No, other sum is possible because three dice being rolled give maximum sum of (6+6+6) i. Question: Question 1 If Two Dice Are Rolled One Time, Find The Probability Of Getting A Sum Greater Than 6 And Less Than 12. Probability for rolling two dice with the six sided dots such as 1, 2, 3, 4, 5 and 6 dots in each die. Note that a 2-element event {1, 2} has the probability of 1/3 = 2·1/6, whereas a 3-element event {4, 5, 6} has the probability of 1/2 = 3·1/6. Dice roll probability: 6 Sided Dice Example. From: ansel001-ga on 19 Nov 2006 17:06 PST There are six possible numbers on each of two dice, so the number of possible rolls is 6^2 or 36. The proability of getting neither is equal to the probability of getting anything other than 7 or 8. So the probability of a sum of at least 5 is 30 out of 36, which gives us the fraction which reduces to. By the central limit theorem, the sum of the five rolls should have approximately the same distribution as a normal random variable with the same mean and variance. Click here to see ALL problems on Probability-and-statistics. If a fair dice is thrown 10 times, what is the probability of throwing at. > Consider this matrix for two dice roll game [math]\textrm{Total outcomes with the sum}$ $3 = 2$ [math]\textrm{Total outcomes with the sum}[/math. Best Answer. Probability of rolling two dice and getting a sum of 7 or at least one 4 - Продолжительность: 1:42 Laura Rickhoff 39 477 просмотров. There are 6 6 possible outcomes. When two dice are rolled, we get 36 possible outcome like (1,1),(1,2),(1,3),(1,4),(1,5),(1,6) …………. What is the conditional probability that at least one lands on 6 given that the dice land Probability of a woman being a smoker given she has ectopic pregnancy • E: ectopic pregnancy Did you find mistakes in interface or texts? Or do you know how to improveStudyLib UI?. This is the aptitude questions and answers section on "Probability" with explanation for various interview, competitive examination and entrance test. Probability that sum is neither 7 or 11 = 1 - Probability that the sum is 7 and 11. Suppose we have 3 unbiased coins and we have to find the probability of getting at least 2 heads, so there are 23 = 8 ways to toss these coins, i. if the die is not fair then the probability of an odd number is found by: P( odd number ) = P(rolling a 1) + P(rolling a 3) + P(rolling a 5) for a fair die this is: P( odd number ) = 1/3 + 1/3 + 1/3 = 1/2. Two fair dice are rolled and the sum of the points is noted. Suppose we consider the previous example about rolling two dice. How many outcomes correspond to the event that the sum of the number is 5? Find more answers. Throwing Dice More Than Once. It compiles alright but I am not getting the output. Dice and Dice Games. Find the probability of the lost card being a diamond. What is the distribution of the sum? 30. I can figure it out by calculating the chance of not These sums are getting bigger. of ways are - 1 , 1 1 , 2 2 , 1 1 , 4 4 , 1 1 , 6. Find the Probability that the Sum of the Numbers on the Upper-most Faces of Two Dice Is: Less than 6 Concept: Probability - A Theoretical Approach. [3 Marks) 0. Homework Equations Axioms and basic theorems of probability. For example if n. A probability is a way of assigning every event a value between zero and one, with the requirement that the event made up of all possible results (in our example, the event {1,2,3,4,5,6}) is assigned a value of one. Anil Kumar 5,072 views. Experimental Probability: Experiment with probability using a fixed size section spinner, a variable section spinner, two regular 6-sided dice or customized dice. Is this solution Helpfull? Yes (28) | No (6). Question 1033885: Three dice are tossed. There are $3 \times 3 = 9$ total possibilities for the two dice, which makes the probability of getting a multiple of three $3/9 = 1/3$. The probability of throwing any given total is the number of ways to throw that total divided by the total number of combinations (36). With two dice, there are 6 x 6 = 36 possible outcomes. From the remaining cards of the pack, two cards are drawn and are found to be diamonds. As far as the previous regime, Washington should've gotten more in the trade of LT Trent Williams, but give credit for mortgaging this year's second-rounder in order to get 2019 first-round DE Montez Sweat. What is the probability that the sum of the two numbers on the dice. Therefore, in this example, we could write: p1 = p2 = p3 = p4 = p5 = p6 = where p1 ≡ probability of rolling a 1, p2 ≡ probability of rolling a 2, etc. Let Xj represent the number that comes up when J-th fair die is rolled, 7=1, 2,---, k. P(2 twos and 2 ones and the other two different ) + P(2 sixes and 2 ones and 2 of some other number). What is the probability of rolling a "pair" when tossing 4 dice? Anthony from Toronto, Canada. When two dice are rolled, we get 36 possible outcome like (1,1),(1,2),(1,3),(1,4),(1,5),(1,6) …………. The dice experiment allows you to simulate throwing pairs of dice and see what the result is. The probabilities in the probability distribution of a random variable X must satisfy the following two A pair of fair dice is rolled. (d) an even number appears on the black dice or the sum of the numbers on the two dice is 7. two dice are rolled find the probability of getting a 5 on either dice or the sum of both dice is 5. So, a total of 7 can be rolled in 6 ways, which means that the probability of rolling a 7 is CHECKPOINT 3 One six-sided die is tossed twice. As the top row increases, the bottom row decreases, so However, our formula will look a bit different. Rolling two fair dice more than doubles the difficulty of calculating probabilities. The ways to get a 4 are: 1+3, 2+2, 3+1 = 3 ways. They were both great football fans and decided to introduce this game to the workers of the factory. From: ansel001-ga on 19 Nov 2006 17:06 PST There are six possible numbers on each of two dice, so the number of possible rolls is 6^2 or 36. Find the probability distribution for the ‘sum of two dice’. Find the probability of the lost card being a diamond. The numbers for the games so far are listed below. 33 Question 3 Let A And B Be Two Independent Event, Such That P (A) = 0. So the probability of a sum of at least 5 is 30 out of 36, which gives us the fraction which reduces to. A single die is rolled twice. Given a pair of dice, what is the chance of drawing an odd number? 7. Total possible outcomes = 36. Thus, the probability of two odd numbers (no even numbers) is (1/2)*(1/2) = 1/4. To find the probability that the sum of the two dice is three, we can divide the event frequency (2) by the size of the sample space (36), resulting in a probability of 1/18. What is the probability that you will get the sum of the no's as 10?. It gets more interesting when you have two dice. There are 6 6 possible outcomes. or the of the two dice you rolled is. Find the expected number of times one needs to roll a dice before getting 4 sixes. A card from a pack of 5 2 cards is lost. Throwing a 6,5,4,3,2 or 1 deducts 0. Dice and Dice Games. So multiply these two together and you find that the probability of getting BOTH a 1 on the first die AND a 4 on the second die is 1/36. Roll two dice. There are 4 combinations that have the sum of 9. Find the probability of getting a sum of 6 when rolling a pair of dice. Find the probability distribution for the ‘sum of two dice’. from Rosemount. When two dice are rolled, we get 36 possible outcome like (1,1),(1,2),(1,3),(1,4),(1,5),(1,6) …………. The probability of choosing a green marble from the jar. find the probability that the student gets exactly 4 correct. There are only two ways to get a total of 11. To find the probability of rolling a sum of 7, you must first count the number of ways in which this can occur. Find the probability that a student selected at random failed in both the examination? 3 dice are rolled. Two counters game. Then P(A) = 4 52. In our die rolling experiment, a particular outcome can be expressed as a triple of numbers from 1 to 6 Probability spaces of this kind are called uniform: 1Notice that we're assuming the dice are To compute the probability of the event, T , that we get exactly two sixes, we add up the probabilities. (d) an even number appears on the black dice or the sum of the numbers on the two dice is 7. MATH 225N Week 4 Homework Questions Probability 1. Find the probability of the lost card being a diamond. Algebra -> Probability-and-statistics -> SOLUTION: Two dice are rolled. 5 coins from your amount of coins. Therefore the number of possible outcomes will be 6*6 = 36. My own intuition tells me the answer is 2/3 because the other die simply needs to show 3, 4, 5, or 6 for the sum to be at least 5. Now we have to count the number of ways we can obtain a sum at most 6. Get an answer for 'When two dice are thrown what is the probability that the sum is 8. Probability = 1 ÷ 36 = 0. For three six-sided dice, the most common rolls are 10 and 11, both with probability 1/8; and the least common rolls are 3 and 18, both with probability 1/216. Question: Question 1 If Two Dice Are Rolled One Time, Find The Probability Of Getting A Sum Greater Than 6 And Less Than 12. The other two singletons can be among the other five. assuming a fair die you have three possible odd numbers out of six total possible outcomes so the probability is 0. When two dice are rolled, the probability of getting an even number on at least one die is 3/4. For 3 dice, there would be 216 total combinations. What is the probability that the sum of two rolled dice will equal a prime number? We find this number by multiplying 6 x 6. and only one way to roll a 12 (6-6, or boxcars). Dependent Event - An event whose probability of occurring is influenced by (i. Find the probability that a 5 occurs first. of favorable. 8: 5/369: 4/3610: 3/3611: 2/3612: 1/36. When two 6 sided dice is tossed, we get a pair of outcomes. We want sum to be greater than 16, So, sum could be either 17 or 18. The game is designed as such that you throw a pair of dice and get money back or loose money. Find the probability of getting an odd number greater than 2 when rolling a die. It is assume each die is fair and 6-sided. 16667, to turn up when rolled, if the die (D) is unbiased. Throwing a 12 yields 1. Calculate the is the conditional probability that the Finding P (E): The probability of getting 4 atleast once is. Find the probability that you roll two dice whose numbers agree before you roll two dice whose numbers differ by 1. Probability of Rolling Multiple of 6 with 2 dice - Duration: 4:19. function [ X ] = Dice ( N, S, T, R ) % Dice simulates a random selection of numbers which is similar to how a % dice is rolled % % N is the number of dice the user wants to roll % S is the number of sides on the dice % T is the number of trials that the user wants to run. Roll 4 6-sided dice, keep the highest 3 and sum them: sum largest 3 4d6. You know that the probability of getting a 1 on the first die is 1/6. function [ X ] = Dice ( N, S, T, R ) % Dice simulates a random selection of numbers which is similar to how a % dice is rolled % % N is the number of dice the user wants to roll % S is the number of sides on the dice % T is the number of trials that the user wants to run. SOLUTION: Two dice are rolled. There are 6*3 = 18 ways to get two numbers of the same parity (the first can be any of the 6 numbers, and the second has to be 3 of the possible 6 which have the same parity), giving a total of 18 ways to get an even sum out of a possible of 6*6 = 36 outcomes (we don't have to consider if the first number is even or odd since there are an equal. Therefore the number of possible outcomes will be 6*6 = 36. [3 Marks) 1 13 5 Question 2 Find The Z Score That Corresponds To The Given Area. Three fair, n-sided dice are rolled. Isn’t that kind of cool?. If one of the dice shows 1 to 4, the sum will not be greater than 10. Two dice are thrown simultaneously. TE Thaddeus Moss was signed after going undrafted. Find the probability of getting a sum of 6 when rolling a pair of dice. the probability that the sum is 6 given that at least one of the numbers is less than 3. Assuming that the dice are unbiased or not " loaded". A standard deck of cards has 12 face. When two 6 sided dice is tossed, we get a pair of outcomes. Find the expected number of games that are played when. What is the probability of getting a number other than 6?. The fundamental counting principle tells us there are 6*6=36 ways to roll two dice, all of them equally likely if the dice are fair. Step by step we: Generate the possible outcomes for one die. We start with writing a table to Discrete = This means that if I pick any two consecutive outcomes. A black and a red dice are rolled Let us take first numbers to have been appeared on the black die and the second numbers on rolled. Rolling Dice. Find the probability of the lost card being a diamond. The probability of rolling a pair of dice whose numbers add to 5 is 4/36 = 1/9. No, other sum is possible because three dice being rolled give maximum sum of (6+6+6) i. Do you really think that the chance of rolling 6's back to back is greater than rolling a die and getting a 6?. Rolling Two Dice If two dice are rolled one time, find the probability of getting these results. Throwing a 6,5,4,3,2 or 1 deducts 0. A dice is thrown, cases 1,2,3,4,5,6 form an exhaustive set of events. Two fair dice are rolled and the sum of the points is noted. Roll 4 6-sided dice, keep the highest 3 and sum them: sum largest 3 4d6. The game is designed as such that you throw a pair of dice and get money back or loose money. Let's write events!! Let A be the event - The sum of the top faces of 3 dice > 8. You might be asked the probability of rolling a variety of results for a 6 Sided Dice: five and a seven, a double twelve, or a double. When two dice are thrown simultaneously, thus number of event. The dictionary of etymology traces use of the term as far back as 1919. How likely is it to choose a random number between 10 and 100 that is a multiple of 9? 6. Find the probabilities of rolling different sums. 4 ways to get a sum of 5. (sum rolled-#outcomes) 2-1. Then P(A) = 4/36 and P(B) = 6/36. Find the probability that the student sold 11-15 shirts or less than 6 T's No of T-shirts No of Club Members 0 1 1-5 15 6-10 13 11-15 3 16-20 6 20+ 1. Two fair dice are rolled. Hence it is important to be familiar with deep learning and its concepts. Find the probability of getting two numbers whose sum is greater than 10. 4 And P (B) = 0. Try the following: 1. ECEN 303 - Fall 2011. So we have 5/6 probability of a die's being <6, and then 5/6 for each of the other pairs' not. a sum of 14 f. We start with writing a table to Discrete = This means that if I pick any two consecutive outcomes. The dice experiment allows you to simulate throwing pairs of dice and see what the result is. 5 and variance 35/12, and so the corresponding mean and variance for rolling 5 dice is 5 times greater. (a) X is the largest value obtained on any die and Y is the sum of the values In a random throw of two dice, since, each of the six faces of one die can be associated with each of six faces of the other die, the total Get more help from Chegg. [3 Marks) 0. In the experiment of rolling two dice think of one as red and the other as green and list the possible Here, for example, the (3,5) in third row and fifth column means a 3 was rolled on the red die and a 5 Thus the sum is a 7 in 6 of the 36 outcomes and hence the probability of rolling a 7 is 6/36 = 1/6. Find the probability of: getting a number greater than 3 on each die. A pair of dice, two different colors (for example, red and blue) A piece of paper; Some M&M’s or another little treat; What You Do: Tell your child that he's going to learn all about probability using nothing but 2 dice. The other two singletons can be among the other five. That intuition is wrong. Each one will roll the die, and depending on the number rolled they will be sent to a predetermined area of the room. There are 4 combinations that have the sum of 9. The number of possible outcomes in E is 1 and the number of possible outcomes in S is 6. MATH 225N Week 4 Homework Questions Probability 1. It is assume each die is fair and 6-sided. A card from a pack of 5 2 cards is lost. Let's say two dice are rolled 36 times. , in short (H, H) or (H, T) or (T, T) respectively; where H is denoted for head and 1. When rolling two dice, distinguish between them in some way: a first one and second one, a left and a right, a red and a green, etc. Isn’t that kind of cool?. Hi I wrote code for Java program to simulate the rolling of 2 dice. Read Our Reviews. Keep up the learning, and if you would like more. The plural is dice, but the singular is die. Hence the probability of getting a 3 is P(E) = 1 / 6. Assume you roll a fair dice twice. The probability of throwing any given total is the number of ways to throw that total divided by the total number of combinations (36). Find the variance and standard deviation of X. 180 Degree Capital Corp (NASDAQ:TURN) Q1 2020 Earnings Conference Call May 06, 2020, 09:00 AM ET Company Participants Daniel Wolfe - President, Portfolio Manage. Now we have to count the number of ways we can obtain a sum at most 6. "If you roll a dice three times, what is the probability of rolling a 6 at least once?" The correct answer is 91/216. The analogy is then made to an ideal gas: for a temperature of, say, 300 K, the number or arrangements of the molecules from the total number possible are consistent with that temperature is analogous to asking how many arrangements there are for a roll that gives a 7. Good morning Edward, I liked your dice probability work on the chances of getting one 6 when rolling different number of dice. Throwing a 9,8 or 7 yields nothing. 16&comma. 33 Question 3 Let A And B Be Two Independent Event, Such That P (A) = 0. So this is an independent event. Therefore probability of getting doubles or sum of 7 is 1/3 Probability of Rolling Doubles - Practice Problems Problem 1: If rolling two dice, what is the probability of getting a sum of 5 or 6? Problem 2: If rolling two number cubes, what is the probability of getting 6 or 7? Answer: 1) 1/4 2) 11/36. [3 Marks) 0. My own intuition tells me the answer is 2/3 because the other die simply needs to show 3, 4, 5, or 6 for the sum to be at least 5. Probability that sum is neither 7 or 11 = 1 - Probability that the sum is 7 and 11. So 1/36 is part of the. The odds of rolling two dice and the sum being greater than 9 are 6 to 30. Here we consider two events: A - (finding a sum of 8) & B (getting at least one 4) A : Probability of A is 5/36 because Now, the probability of either of the incidents happening is P(AUB)=P(A)+P(B)-P(AnnB) i. Two dice are thrown simultaneously. That takes care of the winning or losing probabilities for the naturals (7,11) and the craps (2,3,12) outcomes. So the probability of not getting 7 or 11 is 7/9. (c) A sum less than 4 or greater than 9. When rolling two dice, there are 36 possible outcomes. In this skilltest, we tested our. What is the probability of getting a flush in a. [3 Marks) 1 13 5 Question 2 Find The Z Score That Corresponds To The Given Area. The numbers for the games so far are listed below. When two dice are rolled , n(S) = 36. 5 Consider rolling two dice. Published on Dec 19, 2014. Now, favourable outcomes = sum. Total number of outcomes = 6*6 = 36, Each die can take a number from 1 to 6 i. However, when it comes to practical application, there are two major competing categories of probability. ECEN 303 - Fall 2011. (i) Prime numbers = 2, 3 and 5 Favourable number of events = 3 Probability that it will be a prime. The term "boxcars", also known as "midnight", is the outcome of rolling the dice and getting a six on each die. When two dice are rolled, the probability of getting an even number on at least one die is 3/4. For example: 1 roll: 5/6 (83. Throwing a 10 yields 0. From the remaining cards of the pack, two cards are drawn and are found to be diamonds. For example, (4, 3) stands for getting "4" on the first die and and "3" on the second die. What is the probability that the sum of two rolled dice is less than or equal to 9? I got 1/5 because its greater? help. Roll two dice. => Favorable outcomes are: (1, 5), (2, 4), (3, 3), (4, 2) and (5, 1) Number of favorable outcomes = 5. =5/36 +11/36 -1/36=15/36=5/12. That intuition is wrong. If I roll two dice, does my probability of rolling a six on one of them increase, or does it stay at 1/6? Mike R. => Favorable outcomes are: (1, 5), (2, 4), (3, 3), (4, 2) and (5, 1) Number of favorable outcomes = 5. So, the probability of rolling any pair can be computed as the sum of 1/36 + 1/36 + 1/36 + 1/36 +1/36 + 1/36 = 6/36. Let B be the event - The sum of the top faces of the 3 dice >= 5. A pair of dice, two different colors (for example, red and blue) A piece of paper; Some M&M’s or another little treat; What You Do: Tell your child that he's going to learn all about probability using nothing but 2 dice. We have to find what is the probability that the sum of numbers rolled is either 5 or 12 We know that, probability of an event = Now, total outcomes for two dices = 6 for 1st dice x 6 for 2nd dice = 6 x 6 = 36. Keep up the learning, and if you would like more. Let A = fAceg. If a fair dice is thrown 10 times, what is the probability of throwing at. By the central limit theorem, the sum of the five rolls should have approximately the same distribution as a normal random variable with the same mean and variance. Sum of dices when three dices are rolled together If 1 appears on the first dice, 1 on the second dice and 1 on the third dice. If the two dice are fair and independent , each possibility (a,b) is equally likely. Find maximum subset sum formed by partitioning any subset of array into 2 partitions with equal sum. Find the probability of the lost card being a diamond. 33 Question 3 Let A And B Be Two Independent Event, Such That P (A) = 0. hi Dakotah :) A number cube is rolled 20 times and lands on 1 two times and on 5 four times. a sum less than 4 or greater than 9 d. Roll each attribute in order - do not assign numbers to stats as you see fit. [3 Marks) 0. You know that the probability of getting a 1 on the first die is 1/6. Find the probabilities of rolling different sums. What's the probability of the second card also being a king?. MATH 225N Week 4 Homework Questions Probability 1. 16667, to turn up when rolled, if the die (D) is unbiased. From the remaining cards of the pack, two cards are drawn and are found to be diamonds. a sum of 14 f. The probability, then, of rolling a sum less than five on two dice is 6 in 36, or 1. [3 Marks) 1 13 5 Question 2 Find The Z Score That Corresponds To The Given Area. <-- Separate numbers by comma to check divisibility by any of the numbers. In our die rolling experiment, a particular outcome can be expressed as a triple of numbers from 1 to 6 Probability spaces of this kind are called uniform: 1Notice that we're assuming the dice are To compute the probability of the event, T , that we get exactly two sixes, we add up the probabilities. The plural is dice, but the singular is die. 2 ways to get a sum of 3. the black die resulted in a 5. What is the probability that the sum of the two tosses is 4?. Includes problems with solutions. ezrbkr640f, mix9vpa8i187, 3tc99scuk5, xofzwgx8cax, vk6plpxzbvma, 7xuh7ssbxtj5s5, bq4bsb4hub, 8kw5fnzahoxhv, jxgxndxm7x83di, t6nrf3jicc9wgin, ixae609sn9iyhs, ay4k8hyfj00bn2, 5j094zuwvxa, wf80wp4krmfiez, 0t5xhf75spsu1e, c6byovc3md8yagj, 82pevc9cd1v, 1126g6z45zsrgxt, zu9kzg0hgfw6f4, r8i9t6xsba16v, 1tiasyixzhr, c5a8icxw0yejy9, 0nw0t9n8sqpp, z1mzqvtg96o, o92pns3ro93, 1gij4i4ch8, ifk4aplakg, 9krri8qf59vk83 | 2020-05-30T05:32:03 | {
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http://richardelwes.co.uk/2012/05/31/a-hat-game-1/ | ### A hat game 1
31st May, 2012
Yesterday I heard a great talk by Robert Lubarsky, which provided a delightfully easy route into fairly deep logical waters. He was talking about joint work of his with Stefan Geschke.
This is the first of two two posts about it, each based around a puzzle. So let’s get going with the first, which is something of a classic. The second is altogether trickier, indeed a complete answer seems to be as yet unknown…
Here’s the first puzzle: ten (or a hundred or any number) people stand in a line, one in front of the other. Each is wearing either a black or a white hat. If you’re in the queue, you don’t know your own hat-colour, or those of the the people behind you. But you can see the ones in front. Starting from the back of the line, the challenge is for each person to call out in turn, trying to guess the colour of their own hat. They are allowed to discuss strategy beforehand. So what plan can they hatch to maximise the number of right answers?
Here is one possibility: the first person says the colour of the hat directly in front of him. So the second knows her hat colour, which she calls out. The third person then does the same favour for the fourth, and so on. This way, the team are guaranteed to get half the hats right. (We only care about guaranteed correct answers in this game, not lucky guesses.) In fact, there is a much better strategy. If you don’t know the answer, have a think about it now. The solution is on the other side of this picture.
And the answer is... Kurt Gödel and Albert Einstein play the hat game.
Here’s an answer: the person at the back counts the number of black hats in front of him. If that’s an odd number, he says “black”. If it’s even, he calls “white”. Now, the next person can see all the same hats the first could see, except her own. Say she sees 5 black hats, and hears “black”. Can her own hat be black too? No, because otherwise that would make 6 black hats in front of the first person, an even number. So she knows her own hat must be white. Similarly, if she hears “white” then she knows that her own hat must be black. All the people in front can work out their own hat in the same way, by combining what they can see with what the people behind have said.
Now, this strategy allows everyone to discover their own hat colour, except for first the very first person. It is easy to see this is the best that can be done: no-one can see the first hat, so any tactic the team try will be immune to a change in its colour. To avoid this exception, in the next post we’ll assume the person at the head of the queue is hatless.
To set things up for the next post, I’ll write this another way, using “1” to represent black and “0” for white. So the sequence of hats might read 0010001111. The first person sees the whole sequence, adds up all the digits to get 5, and says “1”. (This is the total modulo 2, if you like.). The second person sees 010001111, and so on, with everyone seeing the same binary sequence, but with the beginning chopped off in different places.
I hope you agree that this hat game is a fun puzzle, with a neat answer. Its solution uses what computer scientists call a parity bit, a technique used to guard against data becoming corrupted.
In the next post things will become a little more serious, when we allow the queue of hat-wearers to grow infinitely long. In this new version, we have to assume that the hatters have various superpowers: they can see infinitely far in front of them, for example. It is obvious that the same tactic will not work in this setting, since everyone is likely to see infinitely many black hats in front of them (and of course it makes no sense to ask whether ‘infinity’ is odd or even). So what is the optimal result here, and what strategy will produce it? And, critically, what further superpowers do the people need to enact it?
Update: the second post is here.
Categories: Logic, Maths, puzzles | Permalink
#### 2 Responses to “A hat game 1”
1. From Richard Elwes – A hat game 2:
[...] is a sequel to the first hat game post, please read for the background. Everything here is based on a recent talk by Robert Lubarsky, [...]
2. From The many faces of $E_0$:
[...] by Lubarsky and Geschke. You can also find an awesome post about the problem by Richard Elwes here. (Thanks [...] | 2013-05-25T03:39:52 | {
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http://marbellanightclub.es/f8zwtlln/9v8pds.php?page=2edc17-if-two-rectangles-have-equal-areas%2C-then-they-are-congruent | We can then solve by cross multiplying. If triangle RST is congruent to triangle WXY and the area of triangle WXY is 20 square inches, then the area of triangle RST is 20 in.² . (vi) Two triangles are congruent if they have all parts equal. And therefore as congruent shapes have equal lengths and angles they have equal are by definition. Thus, a=d. Every rectangle can be rearranges into a rectangle with one side equal to 1 Proof. Construction workers use the fact that the diagonals of a rectangle are congruent (equal) when attempting to build a “square” footing for a building, a patio, a fenced area, a table top, etc. Why should two congruent squares have the same area? Ex 6.4, 4 If the areas of two similar triangles are equal, prove that they are congruent. Figures C C and D have Two figures having equal equal areas, areas need not be congruent. Remember, these are *squares* though. Rectangle 1 with length 12 and width 3. (iii) If two rectangles have equal area, they are congruent. Rhombus. Because they have a constant radius and no differentiated sides, the orientation of a circle doesn't factor into congruency. This wouldn't hold for rectangles. (e) There is no AAA congruence criterion. 1 decade ago. Congruent Figures: Two figures are called congruent if they have the same shape and same size. Answer: i) False. (b) If the areas of two rectangles are same, they are congruent (c) Two photos made up from the same negative but of different size are not congruence. If they are not equal, then either S > S or S > S. For now, we assume the former, but the argument for the latter is similar (that case cannot, in fact, occur, see e.g. If two figures are congruent, then their areas are equal but if two figures have equal area, then they are not always congruent.. 2.) The reflexive property refers to a number that is always equal to itself. So, if two figures X and Y are congruent, they must have equal areas. Since all the small rectangles are congruent, they all have the same area. This means that the dimensions of the small rectangles need to multiply to 108. (18) Which of the following statements are true and which of them false? When a diagonal is drawn in a rectangle, what is true of the areas of the two triangles into which it divides the rectangle? They both have a perimeter of 12 units, but they are not the same triangle. But its converse IS NOT TRUE. Two geometric figures are called congruent if they have … In this sense, two plane figures are congruent implies that their corresponding characteristics are "congruent" or "equal" including not just their corresponding sides and angles, but also their corresponding diagonals, perimeters, and areas. (ED)*(DG) = the area of the rectangle. But although "equal areas mean equal sides" is true for squares, it is not true for most geometric figures. If its not be shure to include at least one counterexample in your explanation. 1 0. Yes, let's take two different rectangles:The first one is 4 inches by 5 inches.The second is 2 inches by 10 inches.Both of these have an area of 20 square inches, and they are not congruent. If not then under what conditions will they be congruent? you can superpose one figure over the other such that it will cover the other completely. In general, two plane figures are said to be congruent only when one can exactly overlap the other when one is placed over the other. 9.1 AREAS OF PARALLELOGRAMS AND TRIANGLES 153 you can superpose one figure over the other such that it will cover the other completely . Technically speaking, that COULD almost be the end of the proof. they have equal areas. FALSE. (d) if two sides and any angle of one triangle are equal to the corresponding sides and an angle of another triangle, then the triangles are not congruent. However, the left ratio in our proportion reduces. In other words, if two figures A and B are congruent (see Fig. Workers measure the diagonals. If two figures X and Y are congruent (see adjoining figure), then using a tracing paper we can superpose one figure over the other such that it will cover the other completely. In mathematics, we say that two objects are similar if they have the same shape, but not necessarily the same size. SAS stands for "side, angle, side". If two figures are congruent, then they're exactly the same shape, and they're exactly the same size. However, different squares can have sides of different lengths. But although "equal areas mean equal sides" is true for squares, it is not true for most geometric figures. Consider the rectangles shown below. $16:(5 a. We then solve by dividing. A. Two circles are congruent if they have the same diameter. That’s a more equation-based way of proving the areas equal. 9.1) , then using a tracing paper, Fig. Two figures are called congruent, if they have the same shape and the same size. If two angles of a triangle have measures equal to the measures of two angles of another triangle, then the triangles are similar. Congruent rectangles. 13. If you have two similar triangles, and one pair of corresponding sides are equal, then your two triangles are congruent. Another way to say this is two squares with the same area are congruent in every way (same area, same sides, same perimeter, same angles). If 2 squares have the same area, then they must have the same perimeter. Prove that equal chords of congruent circles subtend equal angles at their centres. Two objects are congruent if they have the same shape, dimensions and orientation. The ratio of the two longer sides should equal the ratio of the two shorter sides. (Why? If the objects also have the same size, they are congruent. Assuming they meant congruent, this is what I have tried: Conditional: "If a rectangle is square, then its main diagonals are equal" is (True) because this is true of all rectangles. For two rectangles to be similar, their sides have to be proportional (form equal ratios). They are equal. ALL of this is based on a single concept: That the quality that we call "area" is an aspect of dimensional lengths and angles. I would really appreciate if you help me i dont get it at all Ive looked at my notes and nothing im so lost please help me Geometry would not be used to check a foundation during construction. Prove that equal chords of congruent circles subtend equal angles at their centres. = False (ii) If two squares have equal areas, they are congruent. Hence all squares are not congruent. The area and perimeter of the congruent rectangles will also be the same. Combining the re- arrangement of the rst one with the reversed rearrangement of the second one (i.e., taking the common cuts), we can rearrange the rst polygon into the second polygon. In other words, if two figures A and B are congruent (see Fig.1) , then using a tracing paper, Fig-1. Conversely: "If a rectangle's diagonals are equal, then it is a square" is (False) because there exists a rectangle that is not a square that has equal diagonals. For example, x = x or -6 = -6 are examples of the reflexive property. ... Two rectangles are congruent if they have the same length and same breadth. "IF TWO TRIANGLES HAVE THE SAME AREA THEN THEY ARE CONGRUENT" Is this a true statement? Therefore, those two areas are equal. (i) All squares are congruent. [2]). are equal, then we have found two non-congruent triangles with equal perimeters and equal areas. If a pair of _____ are congruent, then they have the same area . Claim 1.1. If two triangles have equal areas, then they are congruent. And why does a$1 \times 1$square have an area of$1$unit?) b=e. So if two figures A and B are congruent, they must have equal areas. True B. Here’s another HUGE idea, which is much more appealing for visual thinkers. False i True Cs have equal areas If the lengths of the corresponding sides of regular polygons are in ratio 1/2, then the ratio of their areas … Girsh. (iv) If two triangles are equal in area, they are congruent. If two squares have equal areas, they will also have sides of the same length. called congruent, if they have the same shape and the same size. b. Two rectangles are called congruent rectangles if the corresponding adjacent sides are equal. All the sides of a square are of equal length. Congruent circles are circles that are equal in terms of radius, diameter, circumference and surface area. Only if the two triangles are congruent will they have equal areas. 2 rectangles can have the same area with different lengths of sides to … An example of having the same area and not being congruent is the two following rectangles: 1.) So if two figures A and B are congruent, they must have equal areas. Yes. (EQ)*(DC) = the area of the parallelogram. It means they should have the same size. but they are not D congruent. I made a chart of possible factor pairs (I’m assuming the dimensions are integers, and will see if it works). 756/7 = 108 units2. When the diagonals of the project are equal the building line is said to be square. Dear Student! Since the two polygon have the same area, the rectangles they turn into will be the same. Corresponding sides of similar polygons are in proportion, and corresponding angles of similar polygons have the same measure. They have the same area of 36 units^2, but they are not congruent figures. Since b/e = 1, we have a/d = 1. All four corresponding sides of two parallelograms are equal in length that does mean that they are necessarily congruent because one parallelogram may or may not overlap the other in this case because their corresponding interior angles may or may not be equal. If two squares have equal areas, they will also have sides of the same length. So we have: a=d. In order to prove that the diagonals of a rectangle are congruent, you could have also used triangle ABD and triangle DCA. But just to be overly careful, let's compute a/d. This means that we can obtain one figure from the other through a process of expansion or contraction, possibly followed by translation, rotation or reflection. if it is can you please explain how you know its true. TRUE. It's very easy for two rectangles to have the same area and different perimeters,or the same perimeter and different areas. Consider the rectangles shown below. 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Area then they are congruent, then your two triangles are equal, prove equal... And corresponding angles of another triangle, then we have a/d = 1 we! When the diagonals of a circle does n't factor into congruency and equal areas, must. For most geometric figures BFigures a and B are congruent, they have! Length and same size figures: two figures x and Y are congruent triangle. And triangle DCA be congruent ( vi ) two triangles have the same size exactly the same.! Should perhaps review the lesson about congruent triangles two non-congruent triangles with equal perimeters and equal areas similar,! Need not be shure to include at least one counterexample in your explanation DC ) = the area the... We have found two non-congruent triangles with equal perimeters and equal areas the! Be similar, their sides have to be square lengths and angles they have if two triangles are congruent they... Not necessarily the same is no AAA congruence criterion area then they are congruent ( see Fig.1,... Their areas are equal in terms of radius, diameter, circumference and area... The left ratio in our proportion reduces rectangles need to multiply to.! Appealing for visual thinkers technically speaking, that COULD almost be the same size rectangles they turn into will the. ( vi ) two triangles have the same size check a foundation during construction for side,,. Will also be the end of the same area with different lengths orientation a... Sides '' is this a true statement two angles of another triangle, using... And orientation for example, x = x or -6 = -6 are examples of reflexive... The two longer sides should equal the ratio of the proof differentiated,... | 2021-09-27T05:08:29 | {
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http://math.stackexchange.com/questions/865028/is-it-differentiable | # Is it differentiable?
Let us consider the function
$$f(x)= \begin{cases} x^2\sin {\dfrac{\pi}{x}} & x \neq 0\\ 0 & x=0 \end{cases}$$
We want to check its differentiability at $x=0$.
By the definition of $f'(x)$, the derivative of $f$ at $x=0$ would be $$f'(0)=\displaystyle\lim_{h \rightarrow 0}\dfrac{h^2\sin {\dfrac{\pi}{h}}}{h}$$ which would be $$f'(0)=\displaystyle\lim_{h \rightarrow 0} h\sin \frac{\pi}{h}$$
Using the squeeze theorem, we can prove that this limit is equal to $0$.
So according to the above procedure, the function is differentiable at $x=0$.
But, when we look at the graph of this function, it doesn't seem differentiable (below)
In addition to this, when we find $f'(x)$ by using the $u.v$ rule and the chain rule, $f'(x)$ does not exist.
So, is it differentiable or not?
-
So it's differentiable! and $f'(0)=0$ i.e. it has a horizontal tangent line on $0$. – Sami Ben Romdhane Jul 12 at 11:04
Seems differentiable to me. – Brad Jul 12 at 11:06
You're deceived by the fact that $f$ is not twice differentiable at $0$. Graphs are useful, but they can be misleading. – egreg Jul 12 at 11:32
This is the classical example of a function differentiable but not $C^1$ (the limit of $f'$ in $0$ does not exist). – enzotib Jul 12 at 12:17
You can't check if a function is differentiable at $x=a$ if the function is not defined at $a$. In general, a function $f$ is differentiable at $a$ if $f^\prime(a)$ exists for $a\in\operatorname{domain}f$. If you define $f(0)$ in this case to be $0$ then your function is indeed differentiable at $x=0$ by what we said above.
Edit: You claimed that using the product rule and the chain rule means that $f^\prime(x)$ doesn't exist. Well let's check: \eqalign{\dfrac{\mathrm d}{\mathrm dx}\left[\,x^2\sin\left(\tfrac \pi x\right)\right]&=x^2\dfrac{\mathrm d}{\mathrm dx}\cos\left(\tfrac\pi x\right)+\sin\left(\tfrac\pi x\right)\dfrac{\mathrm d}{\mathrm dx}x^2\\&=x^2\left(-\dfrac{\pi\cos\left(\tfrac\pi x\right)}{x^2}\right) +\sin\left(\tfrac\pi x\right)\cdot2x\\&=2x\sin\left(\tfrac\pi x\right)-\pi\cos\left(\tfrac\pi x\right).\\} So $f^\prime(x)$ seems to not exist when $x=0$, but you have to keep in mind that since you have defined $f(0)$ to be $0$ then what you're really differentiating is the piecewise function $$f(x)=\begin{cases} x^2\sin\left(\tfrac\pi x\right), & x\neq0 \\ 0, & x=0 \end{cases}.$$ And its derivative is: $$f'(x)=\begin{cases} 2x\sin\left(\tfrac\pi x\right)-\pi\cos\left(\tfrac\pi x\right), & x\neq0 \\ 0, & x=0 \end{cases},$$ which makes sense since $f$ is continuous: $x\mapsto x^2\sin\left(\tfrac\pi x\right)$ being defined everywhere given that $x\neq0$, so the remaining thing is to check that: $$\lim_{x\to0^-}x^2\sin\left(\tfrac\pi x\right)=0=\lim_{x\to0^+}x^2\sin\left(\tfrac\pi x\right).$$
-
It is usually implied (or explicitly stated) that the function is defined to be zero at zero. – Brad Jul 12 at 11:08
@Brad Well the OP didn't explicitly state it, but if he define it that way then according to my answer the function is differentiable at $x=0$. – Hakim Jul 12 at 11:09
@Hakim But i do have one doubt--Do the left hand derivative and right hand derivative at $x=0$ match? That condition should be satisfied for $f$ to be differentiable, right? – Pkwssis Jul 12 at 11:58
@user157130 You mean the left hand limit and right hand limit as $x\to0$? – Hakim Jul 12 at 12:26
@Hakim No i mean left and right hand derivatives – Pkwssis Jul 12 at 12:28
First you must extend your function in $x=0$ by setting $f(0)=0$. Then your computation is correct, the function is differentiable in $0$. The graph also confirms this, since in $x=0$ the function is very close to the horizontal line.
You can use the chain rule and product rule when you compute the derivative of two differentiable functions. So you can safely use that rule when $x\neq 0$ and you find the correct derivative. For $x=0$, however, the function is not a product: it is a function defined by cases. When $x\neq 0$ you can restrict your function to $x\neq 0$ so that it becomes a product of differentiable functions. When $x=0$ you cannot. So your last resort is apply the very definition of limit.
Once that you have found the derivative in every point you can observe that the derivative in $x=0$ is not the limit of the derivative for $x\to 0$. In other words: this is (the most famous example of) a differentiable function whose derivative is not continuous. The only possibility for this to happen is when the derivative $f'(x)$ has no limit when $x\to 0$, and this is exactly the case.
By the definition of a derivative, the function doesn't need to be defined. A tangent is a $\textbf{limiting}$ case of a chord. – Pkwssis Jul 12 at 11:07
To have a chord passing through $(0,f(0))$ you need to know what is $f(0)$. – Emanuele Paolini Jul 12 at 11:08
@user157130 If you define $f(0)$ to be $0$, then the chord does pass through $(0,f(0))$. – Hakim Jul 12 at 11:17 | 2014-12-18T15:25:19 | {
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https://tjoget.com/6e0z6432/what-is-a-superset-in-math-d435c1 | Subset wikipedia. "Superset." Unlimited random practice problems and answers with built-in Step-by-step solutions. See also. That is, number of elements of X is less than the number of elements of Y. Basic math symbols; Geometry symbols; Algebra symbols; Probability & statistics symbols; ... superset: A is a superset of B. set A includes set B {9,14,28} ⊇ {9,14,28} A ⊃ B: proper superset / strict superset: If is a proper superset of , this is written . Definition of superset math insight. If is a subset of , then is a superset of , written . There are many different types of sets in mathematics, and we can classify sets based on the elements they contain and on how they relate to one another. Constraint specifying actual set is superset of expected set matlab. Practice online or make a printable study sheet. If B is a subset of A, then A is a superset of B, written A superset= B. Walk through homework problems step-by-step from beginning to end. Set Symbols. Syntax: < Set Object 1 > >= < Set Object 2 > : To check superset relationship Determine whether each of these sets is countable... Let SL(n, F) : = {g in GL(n, F) : det(g) = 1}. As a result, shapes that have arcs of over 180 degrees won’t be correctly modelled. Superset in mathematics and set theory; SuperSet Software, a group of friends who later became part of the early Novell; Superset (strength training), for supersets in strength training Apache Superset, a data exploration and visualization web application A proper superset of a set A is a superset of A that is not equal to A. A superset in math is a set of elements containing all of the elements of another set. https://mathworld.wolfram.com/Superset.html. Set theory symbols. If A is a proper superset of B, this is written A superset B. In mathematics, especially in set theory, a set A is a subset of a set B, or equivalently B is a superset of A, if A is "contained" inside B, that is, all elements of A are also elements of B. The relationship of one set being a subset of another is called inclusion (or sometimes containment). Here, Y is called super set of X Formula to find number of subsets. I have read definition of superset somewhere as "a set containing all elements of a smaller set.". Set C is a superset of set D if set C contains all of the elements (if any) of set D. This is written C ⊃ D. Note: Every set is a superset of the empty set. When these two conditions are fulfilled, B is called a superset of set A. Supersets are represented by the symbol which is the mirror image of the symbol used to represent a subset: B ⊃ A {B is the superset of A} Examples: A set is a collection of things, usually numbers. A set containing all elements of a smaller set. Example : The main perk of adding these to your workout is … Superset definition, properties, proper superset & examples. For e.g. Elementary set theory is there any connection between the symbol. Hints help you try the next step on your own. A and B may coincide. Rational subgrouping primarily involves .... A.... You are advertising on three companies each with... DSST Business Mathematics: Study Guide & Test Prep, High School Precalculus Syllabus Resource & Lesson Plans, Common Core Math - Functions: High School Standards, College Preparatory Mathematics: Help and Review, Biological and Biomedical Join the initiative for modernizing math education. Set theory symbols maths symbols with examples | byju's. What are the elements of {Z}_{12}^{x}? A set is said to be a collection of objects (usually numbers and expressions in mathematics) sharing at least one common quality. Definition of superset - Math Insight Best mathinsight.org. answer! Or we could even write that B is a strict subset of C. Now, we could also reverse the way we write this. If is a subset of , then is a superset of , written . >= Operator : This operator is used check whether a given pair of sets are in a superset relationship or not just like issuperset() method.The difference between >= operator and issuperset() method is that, the former can work only with set objects while latter can work with any iterable. The width of the home is 16 ft. 35. We write B ⊆ A By definition, the empty set( { } or ∅ ) is a subset of every … In symbol, we write X ⊂ Y. If A ⊂ B and A ≠ B, this means that A is a proper subset of B. Super set. It implies that B contains A, or in other words, B is a superset of A. List of all mathematical symbols and signs - meaning and examples. Superset The set theory, as the name suggests, is a study of theories related to sets. From MathWorld--A Wolfram Web Resource. A set containing all elements of a smaller set. Superset. SuperSet synonyms, SuperSet pronunciation, SuperSet translation, English dictionary definition of SuperSet. We can list each element (or "member") of a set inside curly brackets like this: Common Symbols Used … Superset - from wolfram mathworld. Definition: Set B is a subset of a set A if and only if every object of B is also an object of A. Set A is a superset of B, if B is "contained" inside A.In other words a set B is a subset of a set A.The sets A and B may coincide. B is a superset of A, because B contains A. Collection of teaching and learning tools built by Wolfram education experts: dynamic textbook, lesson plans, widgets, interactive Demonstrations, and more. Cs311h: discrete mathematics sets, russell's paradox, and halting. Weisstein, Eric W. In other words, if set A contains all of the elements of set B ,... See full answer below. And, is known as the superset of set A. Earn Transferable Credit & Get your Degree, Get access to this video and our entire Q&A library. Set symbols of set theory and probability with name and definition: set, subset, union, … All other trademarks and copyrights are the property of their respective owners. Let A={1, 2, 3} and B={3, 4, 5}. Knowledge-based programming for everyone. In other words, if B is a proper superset of A, then all elements of A are in B but B contains at least one element that is not in A. is a proper superset The relationship of one set being a subset of another is called inclusion or sometimes Euler diagram showing A is a proper subset of B and conversely B is a proper superset of A Sciences, Culinary Arts and Personal And then we're really just talking about supersets. Superset may refer to: . For example, if A is the set \{ \diamondsuit, \heartsuit, \clubsuit, \spadesuit \} and B is the set \{ \diamondsuit, \clubsuit, \spadesuit \}, then A \supset B but B \not\supset A. We write A ⊇ B to denote that B is a superset of A. all natural numbers are integers. A subset which does not have all the elements of its superset is called a proper subset. A is a superset of (or includes) B, denoted by A simple life example is that ‘humans’ is superset of set ‘males’ and set ‘females’. https://mathworld.wolfram.com/Superset.html. It is possible for A and B to be equal; if they are unequal, then A is a proper subset of B. S et theory is a branch of mathematics dedicated to the study of collections of objects, its properties, and the relationship between them. In other words, if set A contains all of the elements of set B,... Our experts can answer your tough homework and study questions. A set containing all elements of a smaller set. Math Symbols List. If Number of subsets = … The following list documents some of the most notable symbols in set theory, along each symbol’s usage and meaning. The superset relationship is denoted as A \supset B. Think: a biceps curl and a triceps extension. This implies that for set A to be a superset of B, B must be a proper subset of A, that is, A must If you go by the super-specific definition, a true superset (antagonist superset) is when you're doing two exercises that target opposing muscles groups. Here, Y is called super set of X More clearly, every element of X is also an element of Y and X is not equal to Y. It deals with the problems based on the definitions and properties of sets. If A is the given set and it contains "n" number of elements, we can use the following formula to find the number of subsets. Superset. A is a subset of B may also be expressed as B includes (or contains) A or A is included (or contained) in B. All rights reserved. Supersets are those sets which are defined by the following conditions: A ⊂ B and A ≠ B. For example, if A =\{1,3,5\} then B=\{1,3,4,5\} is a proper superset of A. superset: Add libraries to provide application-specific facilities, then; subset: Subtract features (outside the library implementation) to provide semantic guarantees; The result is a subset of a superset of a language called a Semantically Enhanced Library Language. Definition of superset A set A is a superset of another set B if all elements of the set B are elements of the set A. Superset is a platform that enables colleges to automate campus placements end-to-end, helps employers hire young talent from across colleges in the country, and empowers students to access opportunities democratically. From Simple English Wikipedia, the free encyclopedia In set theory, a subset is a set which has some (or all) of the elements of another set, called superset, but does not have any elements that the superset does not have. of , this is written . Explore thousands of free applications across science, mathematics, engineering, technology, business, art, finance, social sciences, and more. In mathematics, a set is a collection of elements or objects. Become a Study.com member to unlock this Set of all real numbers R is superset of set of all integers Z. © copyright 2003-2021 Study.com. Explore anything with the first computational knowledge engine. Subset, Venn diagrams Show that the set G=\{ \begin{pmatrix} 1 & a\\ 0... Let n greater than or equal to be an integer. A set X is said to be a proper subset of set Y if X ⊆ Y and X ≠ Y. Characteristics of What Is a Superset in Math It comprises strip of black and white spaces which could be scanned. If N and Z represents the set of all the natural numbers and integers respectively, then we can write that In mathematics, a set A is a subset of a set B if all elements of A are also elements of B; B is then a superset of A. The #1 tool for creating Demonstrations and anything technical. The superset relationship is denoted as A \supset B. In symbol, we write X ⊂ Y. If A is a subset of B, then A is contained in B. Services, Introduction to Groups and Sets in Algebra, Working Scholars® Bringing Tuition-Free College to the Community. About "Superset meaning" Superset meaning : A set X is said to be a proper subset of set Y if X ⊆ Y and X ≠ Y. A superset in math is a set of elements containing all of the elements of another set. For an example, A = {1, 3} is a subset of B = {1, 2, 3}, since all the elements in A contained in B. Create your account. So we could reverse this notation, and we could say that A is a superset of B, and this is just another way of saying that B is a subset of A. ( or sometimes containment ) discrete mathematics sets, russell 's paradox, halting. X Formula to find number of elements containing all elements of X Formula to number... Elements containing all elements of set Y if X ⊆ what is a superset in math and X ≠ Y usually. 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Late Spring Summary, Carolina Herrera Aftershave 212, Kattu Kattu Keera Kattu Song Lyrics, Le Meridien Restaurant, Be Careful What You Wish For, Bokuto Hey Hey Hey 1 Hour, Folkart Extreme Glitter Paint, | 2021-07-29T19:37:23 | {
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https://math.stackexchange.com/questions/1221585/prove-that-for-x-in-bbbr-x-lt-3-implies-x2-2x-15-lt-8x3/2475194 | # Prove that for $x\in\Bbb{R}$, $|x|\lt 3\implies |x^2-2x-15|\lt 8|x+3|$.
The problem I have is:
Prove that for real numbers $x$, $|x|\lt 3\implies |x^2-2x-15|\lt 8|x+3|$.
Since there aren't really any similar examples in my book, I've been unsure how to first approach this problem and have been trying to use to find similar questions.
Do I try and divide the problem into regions, like at absolute value inequalities, using $x^2-2x-15=(x-5)(x+3)$
Or could I just simply have two cases, those being $x\ge 3$ and $x\lt3$ like at How to write an expression in an equivalent form without absolute values?
We know that $|x|<3$. Therefore $-3<x<3$, subtracting $5$ from each side gives:
$-8<x-5<-2$
And so:
$|x-5|<8$
This implies that:
$|x^2-2x-15|=|(x-5)(x+3)|=|x-5|\cdot |x+3|<8|x+3|$
As needed.
• How do you know to subtract 5 from each side? – Churning Butter Apr 5 '15 at 21:57
• It seems reasonable because I want to eliminate the $x-5$ term. – eranreches Apr 5 '15 at 21:59
• Eliminate from where? – Churning Butter Apr 5 '15 at 22:03
• We want to show that $|x-5|\cdot |x+3|<8|x+3|$. This suggests that we need to work on $|x-5|$. – eranreches Apr 5 '15 at 22:04
• Because $-8<x-5<-2<8$, implies that $-8<x<8$. – eranreches Apr 5 '15 at 22:29
Your factorisation in fact takes you very close to the result: $$|x^2-2x-15|=|x-5||x+3|\leq(|x|+5)|x+3|<(3+5)|x+3|=8|x+3|.$$ The first inequality above is just the triangle inequality and is not strict. The second inequality above is strict because $|x|<3$ implies simultaneously $|x|+5<3+5=8$ and $|x+3|>0$.
In general, if you have to work with absolute values and inequalities based on them, split them into two cases:
Case 1: $0 \leq x \lt 3$.
In this case, $|x| = x$ so:
$|x^2-2x-15|\lt 8|x+3| \iff x^2-2x-15 \lt 8(x+3)$
This can reduced down to $f(x) = x^2 -10x-39 \lt 0$.
$f'(x) = 2x -10$ which means $f'(x) > 0 \iff x \gt 5$
$\Rightarrow f(x)$ is decreasing from $0$ to $3 \Rightarrow f(3) \lt f(x) \leq f(0)=-39\lt 0$
Since $f(x) = x^2 -10x-39 \lt 0$ is true for $0 \leq x \lt 3$, we easily show that:
$$0 \leq x \lt 3 \Rightarrow x^2-2x-15 \lt 8(x+3)$$
Case 2: $-3 \lt x <0$
In this case, $|x| = -x$ so:
$|x^2-2x-15|\lt 8|x+3| \iff x^2+2x-15 \lt 8(3-x)$
Reduce this down to $g(x) = x^2 + 10x-39 \lt 0$.
$g'(x) = 2x + 10$ which means $g'(x) < 0 \iff x \leq -5$
$\Rightarrow g(x)$ is increasing from $-3$ to $0 \Rightarrow g(-3) \lt g(x) \lt g(0) = -39 < 0$.
Since $g(x) = x^2 + 10x-39 \lt 0$ is true for $-3 \lt x \lt 0$, we can easily show that:
$$-3 \lt x \lt 0 \Rightarrow x^2+2x-15 \lt 8(3-x)$$
This completes the proof. The key to this proof was that $x \geq 0 \Rightarrow |x| = x$ and $x \lt 0 \Rightarrow |x| = -x$
$|x^2-2x-15|\lt 8|x+3|$
can be written
$|(x+3)(x-5)|\lt 8|(x+3)|$
Since
$(x+3)\gt 0$ in the range $-3 \lt x \lt 3$
$(x-5)\lt 0$ in the range $-3 \lt x \lt 3$
we have that
$(x+3)(x-5)\lt 0$ in the range $-3 \lt x \lt 3$
and therefore we can rewrite
$|(x+3)(x-5)|\lt 8|(x+3)|$
to
$-(x^2-2x-15)\lt 8(x+3)$
and reorder to
$x^2+6x+9\gt 0$
and further reorder to
$(x+3)^2\gt 0$ in the range $-3 < x < 3$
which is obviously true
Implies means inclusion of sets of solutions.
Therefore, we could solve both inequalitites and show their sets of solutions are one inside the other.
For $|x|<3$, it is $(-3,3)$.
For $|x^2-2x-15|<8|x+3|$, we get $|(x-5)(x+3)|=|(x-5)(x+3)|<8|x+3|$. This is the same as $x\neq-3$ and $|x-5|<8$. This is $[5-8,5+8]=(-3,13]$.
Now we can simply check that $(-3,3)\subset(-3,13)$ | 2019-10-13T22:25:10 | {
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https://syndarferd.com/7tkui/da2c3e-application-of-first-order-differential-equation-in-real-life | Models such as these are executed to estimate other more complex situations. Using the fact that the equipotentials (surfaces of constant electric potential) are orthogonal the electric field lines, determine the geometry of the equipotenitials of a point charge. dp/dt = rp represents the way the population (p) changes with respect to time. News Once the parachute opens, the descent speed decreases greatly, and the strength of the air resistance force is given by Kv. Order of a differential equation represents the order of the highest derivative which subsists in the equation. For a sky diver falling in the spread‐eagle position without a parachute, the value of the proportionality constant k in the drag equation F drag = kv 2 is approximately ¼ kg/m. Application of differential equations?) Publish in our journal Now the students, in teams of two or three, use these laptops during class time to explore the concepts themselves and at present we do not have a separate computer lab component. Differential equations and mathematical modeling can be used to study a wide range of social issues. Once the parachutist's descent speed slows to v = g/B = mg/K, the preceding equation says dv/dt = 0; that is, v stays constant. A computer disk comes with the text. Therefore, if the sky diver has a total mass of 70 kg, the terminal velocity (with the parachute open) is only. The output is displayed in a beautiful visual form. Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. Volunteer your time Estimate the age of the bone. Previous A derivative in continuous time can be approximated by finite differences in discrete time by, This is called a forward difference because it uses the present or current value of y of y(nΔt) and the next or future value of y of y((n+1)Δt). Sign up for our newsletter The first model of population growth that we study involves the exponential function. It is represented as; f(x,y) = $$\frac{d(y)}{d(x)}$$ = $$\frac{d(y)}{d(t)}$$ = y’, x1$$\frac{d(y)}{d(x1)}$$ + x2 $$\frac{d(y)}{d(x2)}$$ = y. Derive an equation for the speed of the sky diver t seconds after the parachute opens. YES! ], distinguish a function of discrete time from a function of continuous time which is indicated y using parenthesis, (.). (The rate dx/ dt is negative, since x is decreasing.) The population crash happens even though the teams get feedback after each round on the amount of fish they have caught. The half‐life is the amount of time required for one‐half the nuclei in a sample of the isotope to decay; therefore, the shorter the half‐life, the more rapid the decay rate. Because this equation is separable, the solution can proceed as follows: The equipotential lines (that is, the intersection of the equipotential surfaces with any plane containing the charge) are therefore the family of circles x 2 + y 2 = c 2 centered at the origin. EMDADUL HAQUE MILON from our awesome website, All Published work is licensed under a Creative Commons Attribution 4.0 International License, Copyright © 2020 Research and Reviews, All Rights Reserved, All submissions of the EM system will be redirected to, International Journal of Innovative Research in Computer and Communication Engineering, Creative Commons Attribution 4.0 International License. Your email address will not be published. At this point, with no more fish to catch, the fish companies go bankrupt and hence fail to meet their goal of maximizing profit. There are plenty of ways to get involved in the NCSCE community: Attend a meeting An equation denotes the relation between two quantity or two functions or two variables or set of variables or between two functions. Vedantu academic counsellor will be calling you shortly for your Online Counselling session. In elementary ODE textbooks, an early chapter is usually dedicated to first order equations. The authors are all researchers in the field of dynamical systems and they apply a dynamical systems perspective to their presentation of differential equations. Applications of Second Order Equations. 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This separable equation is solved as follows: Now, since v(0) = v 1 ⟹ g – Bv 1 = c, the desired equation for the sky diver's speed t seconds after the parachute opens is. To formulate this process mathematically, let T( t) denote the temperature of the object at time t and let T s denote the (essentially constant) temperature of the surroundings. For example, I show how ordinary differential equations arise in classical physics from the fun-damental laws of motion and force. extract predictive information about the real world situation from the differential equations. This discussion includes a derivation of the Euler–Lagrange equation, some exercises in electrodynamics, and an extended treatment of the perturbed Kepler problem. | 2022-06-25T07:25:29 | {
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https://mathhelpboards.com/threads/toothpicks-in-the-nth-figure-problem-where-the-figures-are-l-shaped-triangles.2012/ | # Toothpicks in the nth figure problem where the figures are "L" shaped triangles
#### charlieanne
##### New member
Hi, I did a search in the forums and found a similar problem, but nothing on this particular one. I have a problem asking for a procedure to find the number of toothpicks in any figure in a sequence, with an illustration that shows the first three figures of the sequence of triangles built with squares made up of toothpicks. The figures are "L" shaped meaning that the first "triangle" starts with a square and then has another square directly above and directly to the right of it with 1 shared toothpick for each new square (10 toothpicks total). The next is the same concept but with one square added into the middle of the two arms (18 toothpicks). I have drawn out the first 10 figures on graph paper but can't seem to find a pattern to figure out a procedure that will work no matter what tn I'm trying to find.
#### earboth
##### Active member
Hi, I did a search in the forums and found a similar problem, but nothing on this particular one. I have a problem asking for a procedure to find the number of toothpicks in any figure in a sequence, with an illustration that shows the first three figures of the sequence of triangles built with squares made up of toothpicks. The figures are "L" shaped meaning that the first "triangle" starts with a square and then has another square directly above and directly to the right of it with 1 shared toothpick for each new square (10 toothpicks total). The next is the same concept but with one square added into the middle of the two arms (18 toothpicks). I have drawn out the first 10 figures on graph paper but can't seem to find a pattern to figure out a procedure that will work no matter what tn I'm trying to find.
View attachment 359
1. I've modified your sketch a little bit (see attachment). I assume that you have n toothpicks in the base row.
2. If you count the rows of toothpicks (green) you'll get:
n+n+(n-1)+(n-2) + ... + 1
3. If you count the columns of toothpicks (red) you'll get:
n+n+(n-1)+(n-2) + ... + 1
4. So in total you have:
2(n+n+(n-1)+(n-2) + ... + 1)
5. Use the sum formula of arithmetic sequences to simplify this term. Finally you should come out with
$$n^2+3n$$
#### Attachments
• 56.4 KB Views: 36
#### MarkFL
Staff member
I observed that for 0 squares there are 0 toothpicks, for 1 square there are 4 toothpicks, for 3 squares there are 10 toothpicks...
On the nth interation, there are $\displaystyle T_n$ squares, where $\displaystyle T_n$ denotes the nth triangular number.
If we notice that the second difference is constant at 2, then we know the number of toothpicks $\displaystyle t_n$ will be a quadratic function:
$\displaystyle t_n=k_1n^2+k_2n+k_3$
where:
$\displaystyle t_0=0$
$\displaystyle t_1=4$
$\displaystyle t_2=10$
giving rise to the system:
$\displaystyle k_3=0$
$\displaystyle k_1+k_2=4$
$\displaystyle 4k_1+2k_2=10$
from which we may determine:
$\displaystyle k_1=1,k_2=3$
Hence:
$\displaystyle t_n=n^2+3n$
#### soroban
##### Well-known member
Hello, charlieanne!
I have drawn out the first 10 figures on graph paper, but can't seem to find a pattern.
I'm surprised . . . Where were you looking?
I assume you counted the toothpicks in your 10 diagrams.
So you have: .$$4,10, 18, 28, 40, 54, 70, 88, 108, 130$$
We see that the numbers are getting bigger. .(duh!)
Okay, just how are they getting bigger?
Let's take a look . . .
$$\begin{array}{c|ccccccccc}\hline \text{Sequence}& 4 && 10 && 18 && 28 && 40 && \cdots \\ \hline \\ \text{Change} && +6 && +8 && +10 && +12 && \cdots \\ \hline \end{array}$$
Do you see a pattern now?
#### charlieanne
##### New member
Hello, charlieanne!
I assume you counted the toothpicks in your 10 diagrams.
So you have: .$$4,10, 18, 28, 40, 54, 70, 88, 108, 130$$
We see that the numbers are getting bigger. .(duh!)
Okay, just how are they getting bigger?
Let's take a look . . .
$$\begin{array}{c|ccccccccc}\hline \text{Sequence}& 4 && 10 && 18 && 28 && 40 && \cdots \\ \hline \\ \text{Change} && +6 && +8 && +10 && +12 && \cdots \\ \hline \end{array}$$
Do you see a pattern now?
Hi Soroban,
Sorry, I should have been more clear: I did see that pattern (and also that the # of 3s goes up by two, and that the 2s go up 4,5,6,7,8,etc). I'm trying to understand why the formula tn=n2+3n works...I see that it does, just not how it works. (Example: I know that the formula for volume is HxLxW, but I can also explain the concept.)
#### soroban
##### Well-known member
Hello again, charlieanne!
$$\text{I'm trying to understand why the formula }\,t_n \,=\,n^2+3n\text{ works.}$$
The "why" is tricky to explain.
I have some explanations, but they may not convince you.
We saw that the generating function is a quadratic, contains $$n^2.$$
Very well, let's test $$f(n) \,=\,n^2$$
$$\begin{array}{c|cccccccccc} n & 1&2&3&4&5&6&7 \\ \hline \text{Sequence} & 4&10 & 18 & 28 & 40 & 54 & 70 \\ \hline \text{Try }n^2 & 1 & 4 & 9 & 16 & 25 & 36 &49 \\ \hline \text{Error} & \text{-}3 & \text{-}6 & \text{-}9 & \text{-}12 & \text{-}15 & \text{-}18 & \text{-}21 \\ \hline \end{array}$$
We see that if we use $$f(n) \,=\,n^2$$, we are short by exactly $$3n.$$
Therefore, the function is: .$$f(n) \:=\:n^2+3n$$
Factor the numbers in the sequence
. . and look for a pattern.
. . . $$\begin{array}{ccc} n & t_n & \text{factors} \\ \hline 1 & 4 & 1\cdot4 \\ 2 & 10 & 2\cdot 5 \\ 3 & 18 & 3\cdot6 \\ 4 & 28 & 4\cdot7 \\ 5 & 40 & 5\cdot8 \\ 6 & 54 & 6\cdot9 \\ 7 & 70 & 7\cdot10 \\ \vdots & \vdots & \vdots \\ n & t_n & n(n+3) \end{array}$$
#### MarkFL
Staff member
Another method would be to express the relation recursively:
$\displaystyle t_{n}=t_{n-1}+2n+2$
$\displaystyle t_{n+1}=t_{n}+2(n+1)+2$
Subtracting the former from the latter:
$\displaystyle t_{n+1}=2t_{n}-t_{n-1}+2$
$\displaystyle t_{n+2}=2t_{n+1}-t_{n}+2$
Subtracting again:
$\displaystyle t_{n+2}=3t_{n+1}-3t_{n}+t_{n-1}$
We find the characteristic roots are 1 with multiplicity 3, hence the closed form will be quadratic, for which the parameters are determined by initial values. | 2021-11-30T23:52:31 | {
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http://mathhelpforum.com/calculus/20936-sequences.html | 1. ## sequences
Could someone just tell me if the following sequences converge or not:
Let an = the summation of (1/x) beginning with x = n+1 and ending with 2n.
For example, if n =2, x =3, then the sequence is (1/3) + (1/4).
Let bn = the summation of (1/x) beginning with x = n+1 and ending with pn where p is a positive integer greater than one.
For example, if n =2, x =3, and p =3, then the sequence is (1/3) + (1/4) + (1/5) + (1/6).
It seems to me that both sequences are increasing and bounded above, which would make them convergent. Is this true or not?
2. Originally Posted by PvtBillPilgrim
Could someone just tell me if the following sequences converge or not:
Let an = the summation of (1/x) beginning with x = n+1 and ending with 2n.
For example, if n =2, x =3, then the sequence is (1/3) + (1/4).
Let bn = the summation of (1/x) beginning with x = n+1 and ending with pn where p is a positive integer greater than one.
For example, if n =2, x =3, and p =3, then the sequence is (1/3) + (1/4) + (1/5) + (1/6).
It seems to me that both sequences are increasing and bounded above, which would make them convergent. Is this true or not?
The first is as it is of order:
$\int_{n}^{2n}1/x ~dx = \ln(2)$,
I will leave it to you to sort out the details of the limit.
The second I can't tell what the upper limit of summation is supposed to be.
RonL
3. I'm not sure if you understand what I'm saying.
I have two sequences:
the first one is the summation symbol (sigma) with 2n on top and i = (n+1) on the bottom (I changed x to i to eliminate possible confusion), i am summing (1/i) with those conditions
the second one is the summation symbol (sigma) with pn on top and i = (n+1) on the bottom where p is a positive integer, i am summing (1/i) with those conditions
are these convergent?
4. Have you noticed that:
$\begin{array}{l}
a_5 = \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{{10}} \\
\frac{1}{2} = \frac{5}{{10}} < \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{{10}} < \frac{5}{5} = 1 \\
\end{array}$
?
So we have:
$\frac{1}{2} < a_n = \sum\limits_{k = n + 1}^{2n} {\frac{1}{k}} < 1$.
5. Originally Posted by PvtBillPilgrim
I'm not sure if you understand what I'm saying.
I have two sequences:
the first one is the summation symbol (sigma) with 2n on top and i = (n+1) on the bottom (I changed x to i to eliminate possible confusion), i am summing (1/i) with those conditions
That is what I was addressing.
RonL
6. Originally Posted by PvtBillPilgrim
the second one is the summation symbol (sigma) with pn on top and i = (n+1) on the bottom where p is a positive integer, i am summing (1/i) with those conditions
are these convergent?
$\sum_{r=n+1}^{p\times n} 1/r,\ p \in \mathbb{N},\ p \ge 2$.
Yes is convergent to $\ln(p)$ for the same reason that the first
one converges to $\ln(2)$
RonL
7. I have a question. Am I the only one here that has misgivings about the captain’s response?
I have no doubt that that the value of the limit he gave is correct.
But it seems to me that the question was about showing that the sequence has a limit.
If I had set this question for a class, I must confess that I think that I would not accept that as an answer.
I would truly like some input on this question.
8. Originally Posted by PvtBillPilgrim
Could someone just tell me if the following sequences converge or not:
Let an = the summation of (1/x) beginning with x = n+1 and ending with 2n.
For example, if n =2, x =3, then the sequence is (1/3) + (1/4).
$\frac{1}{n+1}+...+\frac{1}{2n} = \frac{1}{n}\left( \frac{1}{1+\frac{1}{n}} + ... + \frac{1}{1+\frac{n}{n}} \right) = \frac{1}{n} \sum_{k=1}^n \frac{1}{1+\frac{k}{n}}$
This the the Riemann sum with equal sub-intervals so:
$\lim \ \frac{1}{n} \sum_{k=1}^n \frac{1}{1+\frac{k}{n}} = \int_0^1 \frac{dx}{x+1} = \ln 2$
9. Originally Posted by Plato
I have a question. Am I the only one here that has misgivings about the captain’s response?
I have no doubt that that the value of the limit he gave is correct.
But it seems to me that the question was about showing that the sequence has a limit.
If I had set this question for a class, I must confess that I think that I would not accept that as an answer.
I would truly like some input on this question.
If you read what I wrote I was leaving it to the poster to work out the detail.
The idea is that:
$
\sum\limits_{k = n + 1}^{2n} {\frac{1}{k}} = \int_n^{2n} 1/x ~dx + O(1/n)
$
With some effort you can establish explicit bounds on the error, which is
what I want the poster to do if he cannot see that it is $O(1/n)$ which
is all we need to establish what the limit actually is.
Think of this as an illustration of how we really do these things.
RonL | 2016-12-06T03:43:05 | {
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http://iwantframes.com/vt4cescr/511ad3-a-correlation-coefficient-is-a-numerical-measure-of-the | Rank statistic) see Kendall coefficient of rank correlation; Spearman coefficient of rank correlation. We will: give a definition of the correlation $$r$$, discuss the calculation of $$r$$, explain how to interpret the value of $$r$$, and; talk about some of the properties of $$r$$. If the order matters, convert the ordinal variable to numeric (1,2,3) and run a Spearman correlation. If the order doesn't matter, correlation is not defined for your problem. A value of ± 1 indicates a perfect degree of … 10 Recommendations. Thus when applied to binary/categorical data, you will obtain measure of a relationship which does not have to be correct and/or precise. The linear correlation coefficient measures the strength of the linear relationship between two variables. If you need to find a correlation coefficient then point biserial correlation coefficient might help. 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In that case an alternative is to run ANOVA to see if the mean of your numeric variable changes with different values of the categorical variable. 13.2 The Correlation Coefficient. e) Correlation coefficient i) A numerical measure of the strength and the direction of a linear relationship between two variables. Correlation coefficient and the slope always have the same sign (positive or negative). The Spearman’s rank coefficient of correlation is a nonparametric measure of rank correlation (statistical dependence of ranking between two variables). This analysis yields a sample-based measure called Pearson’s correlation coefficient, or r. X and Y. Consequently, if your data contain a curvilinear relationship, the correlation coefficient will not detect it. 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Olf the correlation coefficient is 1, then the slope must be 1 as well. Compute the correlation coefficients for a matrix with two normally distributed, random columns and one column that is defined in terms of another. However, there is a relationship between the two variables—it’s just not linear. Pearson's Correlation Coefficient ® In Statistics, the Pearson's Correlation Coefficient is also referred to as Pearson's r, the Pearson product-moment correlation coefficient (PPMCC), or bivariate correlation. The regression describes how an explanatory variable is numerically related to the dependent variables.. The numerical measure that assesses the strength of a linear relationship is called the correlation coefficient, and is denoted by $$r$$. In statistics, the correlation coefficient r measures the strength and direction of a linear relationship between two variables on a scatterplot. 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Since the third column of A is a multiple of the second, these two variables are directly correlated, thus the correlation coefficient in the (2,3) and (3,2) entries of R is 1. 4. The closer r … The direction of the correlation is determined by sign of the correlation coefficient ‘r’, whether the correlation is positive or negative. The strength of a correlation is determined by its numerical (absolute) value. To interpret its value, see which of the following values your correlation r is closest to: Exactly –1. We can obtain a formula for r x y {\displaystyle r_{xy}} by substituting estimates of the covariances and variances based on a sample into the formula above. There are quite a few answers on stats exchange covering this topic - … The value of r is always between +1 and –1. Coefficient ‘ r ’, whether the correlation coefficient ‘ r ’, whether correlation! The relative movements of two variables move relation between them on rank statistics (.. 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Of relationship, the value of the linear correlation between two variables move relationship between the variables association two! | 2021-09-26T00:05:32 | {
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https://math.stackexchange.com/questions/3984510/proof-of-alternating-binomial-coefficient-identity-relating | # Proof of Alternating Binomial Coefficient Identity relating
I am looking for a proof of the following binomial identity. I encountered it in an article on Euler’s derivation of the gamma function. Euler begins by evaluating the integral:
$$\int_0^1 x^a(1-x)^n\,dx$$
He performs a binomial expansion on the integrand and makes use of the following identity involving alternating binomial coefficients:
$$\sum_{k=0}^{n} (-1)^k {n \choose k} \frac{1}{a+k+1} = n! \prod_{k=0}^{n}\left( \frac{1}{a+k+1} \right).$$
It is not obvious to me on inspection that this formula is true, but I have found it cited elsewhere in the mathematical literature, without proof. I have verified it by hand for the cases of n=2 and n=3, but I am unable to prove the identity in general. I am looking for a good combinatorial or inductive proof of this very interesting fact.
I am aware that this formula relates to the beta function, but I am trying to derive this relationship without reference to this topic.
Thank you for enlightening me.
• Thanks for this great response, which exposed me to some new concepts with which I was not familiar. – HershMath Jan 15 at 2:00
I am not quite sure if this counts as a duplicate but I give three proofs of an identity equivalent to this identity here (one of which is the beta function proof). I think it is cleanest to write it as
$$\sum_{k=0}^n \frac{(-1)^k}{z + k} {n \choose k} = \frac{n!}{z(z + 1) \dots (z + n)}$$
(here $$z$$ is $$a+1$$) which makes it clearer that for $$n$$ fixed it is an equality between two rational functions of $$z$$; in particular it holds for all complex values of $$z \neq 0, -1, \dots -n$$, and written this way it can be proven by calculating the residues at each pole of the RHS and verifying that they line up with the coefficients of the LHS. I give another proof via thinking of the LHS as the $$n^{th}$$ finite difference of the sequence $$a_k = \frac{1}{z + k}$$, which can be computed by induction and verified to line up with the RHS.
If we substitute $$z \mapsto -\frac{1}{z}$$ and clear denominators we get the equivalent identity
$$\sum_{k=0}^n \frac{(-1)^k}{1 - kz} {n \choose k} = \frac{n! z^n}{(1 - z) \dots (1 - nz)}$$
which is an ordinary generating function identity for the Stirling numbers of the second kind, and which can be proven by proving an exponential generating function identity and then translating it over.
• yes, it is a known expansion of the falling factorial with integral negative index $$\left( {z - 1} \right)^{\,\underline {\, - \left( {n + 1} \right)\,} } = {1 \over {z^{\,\overline {\,n + 1\,} } }}\; = {1 \over {z\left( {z + 1} \right) \cdots \left( {z + n} \right)}}$$ – G Cab Jan 14 at 19:31
By the partial fractions theorem, we know $$\frac{n!}{a(a+1) \cdots (a+n+1)} = \frac{b_0}{a} + \frac{b_1}{a+1} + \dots + \frac{b_{n+1}}{a+n+1}$$ for some set of numbers $$b_0, b_1, \dots , b_{n+1}$$. To find $$b_k$$, multiply both sides of the equation by $$b+k$$ and then set $$a=-k$$ in the resulting equation. The result is $$b_k = (-1)^k \binom{n}{k}$$
In trying to evaluate
$$S_n = \sum_{r=0}^n (-1)^r {n\choose r} \frac{1}{r+a+1}$$
where $$a$$ is not in $$\{-1, -2, \ldots, -n-1\}$$ we introduce
$$f(z) = \frac{(-1)^n \times n!}{z+a+1} \prod_{q=0}^n \frac{1}{z-q}$$
which has the property that for $$0\le r\le n$$
$$\mathrm{Res}_{z=r} f(z) = \frac{(-1)^n \times n!}{r+a+1} \prod_{q=0}^{r-1} \frac{1}{r-q} \prod_{q=r+1}^n \frac{1}{r-q} \\ = \frac{(-1)^n \times n!}{r+a+1} \frac{1}{r!} \frac{(-1)^{n-r}}{(n-r)!} = (-1)^r {n\choose r} \frac{1}{r+a+1}.$$
It follows that
$$S_n = \sum_{r=0}^n \mathrm{Res}_{z=r} f(z).$$
Now residues sum to zero and the residue at infinity of $$f(z)$$ is zero by inspection, therefore
$$S_n = - \mathrm{Res}_{z=-1-a} f(z) = (-1)^{n+1} \times n! \times \prod_{q=0}^n \frac{1}{-1-a-q} \\ = n! \prod_{q=0}^n \frac{1}{a+q+1}.$$
This is the claim.
I think that the "clearest" way to demonstrate the identity is through the finite differences and the Falling and Rising Factorial as follows
The finite difference (unitary step) of a function wrt to the variable $$z$$ is defined as $$\Delta _{\,z} \,f(z) = f(z + 1) - f(z)$$ and its iteration becomes $$\Delta _{\,z} ^{\,n} \,f(z) = \Delta _{\,z} \,\left( {\Delta _{\,z} ^{\,n - 1} \,f(z)} \right) = \sum\limits_k {\left( { - 1} \right)^{n - k} \left( \matrix{ n \cr k \cr} \right)\;f(z + k)}$$
Now, concerning the function $$1/z$$ we have \eqalign{ & \Delta _{\,z} \left( {{1 \over z}} \right) = {1 \over {z + 1}} - {1 \over z} = {{ - 1} \over {z\left( {z + 1} \right)}} \cr & \Delta _{\,z} ^{\,2} \left( {{1 \over z}} \right) = \Delta _{\,z} \left( {\Delta _{\,z} \left( {{1 \over z}} \right)} \right) = {{\left( { - 1} \right)^{\,2} 2} \over {z\left( {z + 1} \right)\left( {z + 2} \right)}} \cr & \;\quad \vdots \cr & \Delta _{\,z} ^{\,n} \left( {{1 \over z}} \right) = \Delta _{\,z} ^{\,n} \left( {\left( {z - 1} \right)^{\,\underline {\, - 1\,} } } \right) = \left( { - 1} \right)^{\,\underline {\,n\,} } \left( {z - 1} \right)^{\,\underline {\, - 1 - n\,} } = {{\left( { - 1} \right)^{\,n} n!} \over {z^{\,\overline {\,n + 1\,} } }} \cr} where the last line follows from the fundamental properties of the Falling and Rising Factorial represented respectively with $$x^{\,\underline {\,k\,} } ,\quad x^{\,\overline {\,k\,} }$$.
Putting the two together $$\Delta _{\,z} ^{\,n} \left( {{1 \over z}} \right) = {{\left( { - 1} \right)^{\,n} n!} \over {z^{\,\overline {\,n + 1\,} } }} = \sum\limits_k {\left( { - 1} \right)^{n - k} \left( \matrix{ n \cr k \cr} \right)\;{1 \over {z + k}}} \quad \left| {\;0 \le n \in Z} \right.$$ | 2021-05-08T16:03:12 | {
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http://mathoverflow.net/questions/103882/minimal-blocking-objects-with-shadows-like-a-cube?sort=votes | # Minimal blocking objects with shadows like a cube
This is a more geometric version of the previous question, "Lattice-cube minimal blocking sets". I will first specialize to $\mathbb{R}^3$, $d=3$.
View an $n \times n \times n$ cube $C_3(n)$ as formed of $n^3$ unit cubes glued face-to-face. I would like to find a minimal blocking object $B$ inside $C_3(n)$, which I define as a collection of the unit cubes in $C_3(n)$ with the following two properties: (a) The shadow of $B$ by parallel light rays in the three orthogonal directions (parallel to the cube edges) is an $n \times n$ filled square. (b) $B$ is a connected object, in the sense that its dual graph is connected. Here the dual graph has a node for each unit cube in $B$ and an edge between each pair of cubes that share a face.
$B$ is intended to be a minimal volume object that casts shadows like a cube. The connectedness condition ensures one could build a physical model of $B$.
The previous MO question did not include the connected condition. There it was shown that blocking sets of size $n^2$ are attainable, $n^{d-1}$ in dimension $d$. Certainly that lower bound is no longer achievable in general, as can be seen with $C_2(2)$: a $2 \times 2$ square in dimension $d=2$ needs $3$ rather than $2=2^1$ unit squares to form a connected blocking object. Exploring $n=3$ in $\mathbb{R}^3$, I have been unable to create a connected blocking object with fewer than 15 cubes:
$\left[\begin{array}{ccc} 0 & 1 & 1 \\\\ 0 & 1 & 0 \\\\ 1 & 1 & 0 \end{array}\right] \hspace{0.25 in} \left[\begin{array}{ccc} 0 & 1 & 0 \\\\ 1 & 1 & 1 \\\\ 0 & 1 & 0 \end{array}\right] \hspace{0.25in} \left[\begin{array}{ccc} 1 & 0 & 0 \\\\ 1 & 1 & 1 \\\\ 0 & 0 & 1 \end{array}\right]$
Fifteen seems excessive—more than half!. Can anyone see a better solution? In addition, I do not see how to generalize to $C_3(n)$, let alone to $C_d(n)$, the same question in $d$ dimensions.
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Why does your pattern block the front view? – Tapio Rajala Aug 3 '12 at 17:35
A simpler (to me) configuration involves a 3x3 slice with 6 cubes on either diagonal, one above and one below. Gerhard "Ask Me About System Design" Paseman, 2012.08.03 – Gerhard Paseman Aug 4 '12 at 0:20
Further, one can do two such planes for arbitrary n to get 2n^2 - n for C_3(n). Gerhard "Ask Me About System Design" Paseman, 2012.08.03 – Gerhard Paseman Aug 4 '12 at 0:27
I also think one needs an extra cube for most every cube in the figure. I believe that (2n-1)n^(d-2) will be an upper if not exact bound in general. Gerhard "Ask Me About System Design" Paseman, 2012.08.03 – Gerhard Paseman Aug 4 '12 at 0:53
A lower bound is $C_3(n)\geq (3n^2-1)/2$. This is best possible for $n=2,3$. – Gjergji Zaimi Aug 4 '12 at 5:22
Lower bound:
Let there be $c$ cubes in a blocking configuration. Consider the graph whose vertices are cubes so that cubes are connected if they share a face. Any connected graph has at least $c-1$ edges. By the pigeonhole principle, there is at least one direction with at least $(c-1)/d$ edges in that direction. When you project parallel to that axis, the image has size $n^{d-1}$, and the size of the image is also at most $c - (c-1)/d$ since at least one vertex on each of those edges is redundant. So,
$$n^{d-1} \le c - \frac{c-1}{d} = c \frac{d-1}{d} + \frac 1d$$
$$c \ge \bigg(\frac {d}{d-1}\bigg)n^{d-1} - \frac {1}{d-1}.$$
For $d=3$, $c \ge \frac 32 n^2 - \frac12$ (mentioned by Gjergji Zaimi in the comments). This is sharp for $n=3$ by the construction with $13$ cubes shown by Joel David Hamkins.
Upper bound ($d=3$):
Here is a construction of a connected blocking configuration with $\frac32n^2 + O(n)$ cubes related to Zack Wolske's constructions. We'll identify the cubes with lattice points. We start with the points $\lbrace(x,y,z) | x-y \equiv z \mod n, 0 \le x,y,z \lt n \rbrace$, illustrated for $n=7$.
|......X| |x......| |.x.....| |..x....| |...x...| |....x..| |.....x.|
|.....X.| |......X| |x......| |.x.....| |..x....| |...x...| |....x..|
|....X..| |.....X.| |......X| |x......| |.x.....| |..x....| |...x...|
|...X...| |....X..| |.....X.| |......X| |x......| |.x.....| |..x....|
|..X....| |...X...| |....X..| |.....X.| |......X| |x......| |.x.....|
|.X.....| |..X....| |...X...| |....X..| |.....X.| |......X| |x......|
|X......| |.X.....| |..X....| |...X...| |....X..| |.....X.| |......X|
This is made of two triangles of points, in the planes $x-y=z$ (marked X) and $x-y=z-n$ (marked x), which are separated by some distance. We'll first make the connections within the triangles, and then connect the triangles to each other.
To connect the bottom triangle, use $n-1$ extra points (marked A) to connect the base of the triangle in the plane $z=0$ to itself and to the points in the triangle with $z=1$. Then for $i = 2, ..., n-1$, use $\lceil (n-i-1)/2 \rceil$ points (marked O) to connect the points in the plane $z=i$ to each other and to the lower points in the triangle. These points have $y$ odd, and are just above an X in the layer below.
|......X| |x......| |.x.....| |..x....| |...x...| |....x..| |.....x.|
|.....XA| |......X| |x.....O| |.x.....| |..x....| |...x...| |....x..|
|....XA.| |.....X.| |......X| |x......| |.x.....| |..x....| |...x...|
|...XA..| |....X..| |....OX.| |.....OX| |x.....O| |.x.....| |..x....|
|..XA...| |...X...| |....X..| |.....X.| |......X| |x......| |.x.....|
|.XA....| |..X....| |..OX...| |...OX..| |....OX.| |.....OX| |x.....O|
|XA.....| |.X.....| |..X....| |...X...| |....X..| |.....X.| |......X|
This has added $n-1$ A's (this is one of the few correct uses of apostrophes to indicate a plural), and $0+1+1+2+2+...+\lfloor (n-1)/2 \rfloor = \lfloor (n-1)^2/4 \rfloor$ O's (see A002620).
We repeat the process upside down to connect the upper left triangle using $n-2$ A's and $\lfloor (n-2)^2/4 \rfloor$ O's.
|......X| |x......| |.x.....| |..x....| |...x...| |....x..| |....Ax.|
|.....XA| |O.....X| |xO....O| |.xO....| |..xO...| |...x...| |...Ax..|
|....XA.| |.....X.| |......X| |x......| |.x.....| |..x....| |..Ax...|
|...XA..| |....X..| |....OX.| |O....OX| |xO....O| |.x.....| |.Ax....|
|..XA...| |...X...| |....X..| |.....X.| |......X| |x......| |Ax.....|
|.XA....| |..X....| |..OX...| |...OX..| |....OX.| |.....OX| |x.....O|
|XA.....| |.X.....| |..X....| |...X...| |....X..| |.....X.| |......X|
Finally, we connect the upper and lower triangles with $n-2$ Z's. There are many choices for how to do this. We'll put them in $z=1$, $y=n-2$, $1 \le x \le n-2$.
|......X| |x......| |.x.....| |..x....| |...x...| |....x..| |....Ax.|
|.....XA| |OZZZZZX| |xO....O| |.xO....| |..xO...| |...x...| |...Ax..|
|....XA.| |.....X.| |......X| |x......| |.x.....| |..x....| |..Ax...|
|...XA..| |....X..| |....OX.| |O....OX| |xO....O| |.x.....| |.Ax....|
|..XA...| |...X...| |....X..| |.....X.| |......X| |x......| |Ax.....|
|.XA....| |..X....| |..OX...| |...OX..| |....OX.| |.....OX| |x.....O|
|XA.....| |.X.....| |..X....| |...X...| |....X..| |.....X.| |......X|
In total, this configuration contains $n^2$ X's, $2n-3$ A's, $n-2$ Z's, and $\lfloor (n-1)^2/4\rfloor +\lfloor(n-2)^2/4 \rfloor = {n-1 \choose 2}$ O's, a total of $\frac 32 n^2 + \frac 32 n - 4$.
Therefore,
$$\frac 32 n^2 - \frac 12 \le \min |C_3(n)| \le \frac 32 n^2 + \frac 32 n - 4.$$
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Very satisfying to see such a nearly tight result, with the constant $\frac{3}{2}$ hardly evident at the start, especially as it was truly a community effort! – Joseph O'Rourke Aug 4 '12 at 16:55
$n-3$ Z's can be used instead, lowering the upper bound by $1$. – Douglas Zare Aug 4 '12 at 18:12
For the main case ($3\times 3\times 3$), here is a solution using 13 blocks:
1 1 0 0 0 1 0 0 1
0 1 0 1 1 1 0 1 0
0 1 1 1 0 0 1 0 0
Update: For the $4\times 4\times 4$ case, here is a solution using only 24 blocks, which is optimal according to the $\frac{3}{2}n^2-\frac 12$ lower bound provided by Douglas and Gjergji.
0 1 1 1
0 1 0 0
1 1 0 0
1 0 0 0
1 0 0 0
1 1 1 1
0 0 1 0
0 0 1 0
1 0 0 0
0 1 0 0
0 0 0 1
0 1 1 0
1 0 0 0
0 1 0 0
0 0 0 1
0 0 1 1
(Click on the edit history for my previous solutions, which used first 27 blocks, then 26, then 25, and now 24.)
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$13$ is indeed the best possible for $n=3$ by the lower bound $\frac 32 n^2 - \frac 12 = 13$. – Douglas Zare Aug 4 '12 at 13:47
For your $4\times 4\times 4$ example, you can get rid of one more cube: either the first cube in the second row of your first layer, or the third cube in the fourth row of the first layer. – ARupinski Aug 4 '12 at 15:20
This type of construction should also generalize up to higher dimensions to give an upper bound that is $O(2n^{d-1})$ which is one order of magnitude better than my upper bound. – ARupinski Aug 4 '12 at 15:23
Thanks, ARupinski, I made the change. – Joel David Hamkins Aug 4 '12 at 16:02
And now I've changed it further, to get down to 24. But now there is no pretense of a general method... – Joel David Hamkins Aug 7 '12 at 3:56
Edited image to correct an error, as per Michael Biro's comment
Here are a few small cases which can each be done with $\frac{3}{2}(n^2+n) - 5$ blocks. They're built by placing $n^2$ "X" blocks to give the correct shadows, then placing $\frac{n^2+n}{2} - 3$ "O" blocks to connect each "X" block to one of two components, and finally $n-2$ "Z" blocks to put the components together. The third example places the "Z" blocks in a funny way to indicate that there are many options available. They could just as easily have been placed in the second column of the second layer, like they were in the $n=4$ case.
Not sure if this will be helpful to you, because I don't have an argument that you can do this for every $n$, but the placement of "O" blocks seems to just repeat the previous levels (a similar arrangement works for n=7 just by copying the "above/below diagonal" positions from n=5) and you can certainly show that $n-2$ "Z" blocks will suffice to connect the two components, since there are many paths of length $n-1$ between "X" diagonals within a layer, and you only need a single "O" on any one of them. Maybe someone can extract an algorithm from this and make it precise. The OEIS didn't give any results for "1,3,13,25" where the next number was at most 40, so maybe 25 is not the best, or maybe it's not known.
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Is your $C_3(4)$ example connected? It seems like the component of $3$ in the bottom right corner of the first picture never connects to the rest. – Michael Biro Aug 4 '12 at 14:42
Ah, good eye, you're absolutely correct. The "O" in the first row should be in the second row to connect everything. Blame it on bad eyesight translating from paper to Excel. – Zack Wolske Aug 4 '12 at 18:16
Zack, could you clarify what you mean? Are you proposing to add another block, or merely to move a block? Could you draw your new configuration? – Joel David Hamkins Aug 7 '12 at 2:50
@Joel: Just to move the block to the right and down. I've put up an image with the change. – Zack Wolske Aug 7 '12 at 4:06
Thanks ! – Joel David Hamkins Aug 7 '12 at 4:16
Here is a way to get $26$ for the $C_3(4)$ case, and $2(n^2-n) + 2$ in general.
In the bottom level, add $4n-5$ cubes around the outside, leaving one out adjacent to a corner. For the second level, fill in the interior $(n-2)^2$ cubes, add the two diagonal corners (one of which is next to the missing cube from the base), and then put a cube in the missing cube's column. For the remaining $n-2$ levels, stack on the diagonal.
1 1 1 1 1 0 0 0 1 0 0 0 1 0 0 0
1 0 0 1 0 1 1 0 0 1 0 0 0 1 0 0
1 0 0 1 0 1 1 0 0 0 1 0 0 0 1 0
1 1 0 1 0 0 1 1 0 0 0 1 0 0 0 1
In general, that works out to $C_3(n) \leq (4n-5) + (n-2)^2 + 3 + n(n-2) = 2(n^2 - n) + 2$
Edit: My $26$ bound for $C_3(4)$ has since been improved, but here is an optimal $C_3(5) = 37$ arrangement.
1 0 0 0 0
1 1 0 0 0
0 0 1 0 0
0 0 0 1 1
0 0 0 0 1
0 1 0 0 0
0 1 0 0 0
1 1 1 1 1
0 0 0 1 0
0 0 0 1 0
0 0 1 0 0
0 0 1 1 1
0 0 1 0 0
1 1 1 0 0
0 0 1 0 0
0 0 0 1 1
0 0 0 1 0
0 0 1 0 0
0 1 0 0 0
1 1 0 0 0
0 0 0 0 1
0 0 0 1 0
0 0 1 0 0
0 1 0 0 0
1 0 0 0 0
-
Very clever, Michael! :-) – Joseph O'Rourke Aug 4 '12 at 15:43
Still no arrangement of $(3n^2-1)/2$ for general $n$, but this solution is symmetric enough that I'm somewhat confident that it's possible. Maybe the arrangements will be different for $n$ even, vs odd? – Michael Biro Aug 7 '12 at 15:31
For $C_3(n)$ an absolute upper bound is $3(n-1)^2+3$ which can always be attained by taking the cubes on three faces adjacent to a given corner, removing the corner itself, and removing all but one cube on each edge incident to the corner. Unfortunately, this gives a bound of 15 in the $n=3$ case., so does not improve your particular case. Somewhat generalizing this construction to higher dimensions, for $d\geq 4$ one gets an absolute upper bound for $C_d(n)$ of
$[n^d-(n-1)^d]-d(n-1)+1$
To prove this always works, consider $C_d(n)$. We will replace each cube by its center point and consider the lattice as in Joseph's previous question which he references.
WLOG, let one corner of $C_d(n)$ be centered at the origin. The lines we want to block have $(d-1)$ coordinates fixed and one coordinate which varies. Now consider the set of all cubes whose center has at least one coordinate equal to 0; this is exactly the set of all cubes lying on one (or more) $(d-1)$-dimensional faces adjacent to our origin cube. There are $[n^d - (n-1)^d]$ such cubes. Now remove all cubes along the edges incident to our cube centered at the origin, this removes $d(n-1)+1$ cubes leaving the set $S_d(n)$ which consists of all cubes with at least one and at most $(d-2)$ coordinates equal to 0. The only lines which do not intersect this set are those which have $(d-1)$ fixed coordinates equal to 0, i.e. the lines incident to our cube centered at the origin. Since cubes are adjacent iff they differ in exactly one coordinate, it is easy to check that the $S_d(n)$ is connected for $d\geq 4$ but not connected for $d=3$. So we need only add our cube centered at the origin, and one cube adjacent to it to block the remaining lines while ensuring the entire collection is connected.
Note that if the above held for $d=3$, one could use it to push our bound down from Joseph's 15 to 14; however the set $S_3(n)$ consists of 3 disconnected pieces and so the reasoning leading to this formula fails. However this formula does imply that as $d$ or $n$ gets large, one can block all lines with an arbitrarily small fraction of all available cubes.
There is probably a way to do some more inclusion-exclusion to further eliminate some particular pieces of $S_d(n)$ as unnecessary for blocking purposes and thereby further reduce our bound, but offhand I don't see it.
- | 2016-05-05T20:01:58 | {
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https://math.stackexchange.com/questions/1390471/frac1a-1a-2-frac1a-1a-3-frac1a-1a-4-then-find-the-value-of-n/1390521 | # $\frac{1}{A_1A_2}=\frac{1}{A_1A_3}+\frac{1}{A_1A_4}$.Then find the value of $n$
If $A_1A_2A_3.....A_n$ be a regular polygon and $\frac{1}{A_1A_2}=\frac{1}{A_1A_3}+\frac{1}{A_1A_4}$.Then find the value of $n$(number of vertices in the regular polygon).
I know that sides of a regular polygon are equal but i could not relate$A_1A_3$ and $A_1A_4$ with the side length.Can someone assist me in solving this problem?
If you know a bit about complex numbers and roots of unity, here is another approach:
The $n^\text{th}$ roots of unity are vertices of a regular $n$-gon. Let $z=1$ denote $A_1$ and $w = e^{\frac{2\pi\iota}{n}}$ denote $A_2$ in the complex plane. Then $w^2$ denotes $A_3$ and $w^3$ denotes $A_4$. Now, the given equation can be written as
$$\frac{1}{|1-w|} = \frac{1}{|1-w^2|} + \frac{1}{|1-w^3|}$$ Multiplying with $|1-w|$ throughout, we get $$1 = \frac{1}{|1+w|} + \frac{1}{|1+w+w^2|}$$
Now since $|w|=1$, pulling out $|w|$ from the denominator of the second term in the R.H.S, $$1 = \frac{1}{|1+w|} + \frac{1}{|1+w+\bar{w}|}$$ Notice that $w+\bar{w}$ is purely real and can be written $w+\frac{1}{w}$, hence $$1 - \frac{w}{1+w+w^2} = \frac{1+w^2}{1+w+w^2} = \frac{1}{|1+w|}$$ Squaring both sides and using $|z|^2 = z\bar{z}$, $$\Bigg(\frac{1+w^2}{1+w+w^2}\Bigg)^2 = \frac{1}{(1+w)} \cdot \frac{1}{(1+\frac{1}{w})}= \frac{w}{(1+w)^2}$$ Cross multiply, $$(1+w+w^2+w^3)^2 = w (1+w+w^2)^2$$ and simplify to get $$1+w+w^2+w^3+w^4+w^5+w^6 = 0$$ Therefore, $$1-w^7 =0$$ and $w \neq 1$, so $w$ must be a $7$th root of unity. Argue that $n =7$.
• Very interesting approach. Upvoting your answer so that more people look at this approach. Nice. Aug 9, 2015 at 9:27
This problem can be trivialized by Ptolemy's theorem.
First we take the l.c.m and simplify both sides of the given equation. Let $A_1A_2=a$, $A_1A_3=b$, $A_1A_4=c$. Then we have
$$\frac{1}{a}=\frac{1}{b}+\frac{1}{c} \qquad\to\qquad b c = a b + a c \tag{1}$$
Now, note that regular polygons can always be inscribed in a circle. Take the quadrilateral $A_1A_3A_4A_5$, which is a cyclic quadrilateral. Applying Ptolemy's theorem we get, $$A_1A_3\cdot A_4A_5 + A_3A_4\cdot A_1A_5 = A_1A_4\cdot A_3A_5 \tag{2}$$ Now see that $A_1A_3=b$, $A_4A_5=a$, $A_3A_4=a$, $A_1A_4=c$, $A_3A_5=b$, so in $(2)$ we have, $$b a + a\cdot A_1A_5 = c b \tag{3}$$
Comparing equation $(1)$ with $(3)$, we see that $A_1A_5 = c = A_1A_4$. These are two diagonals of the regular polygon sharing a common vertex, so this diagonal must be a diagonal that bisects the area of the polygon. So on one side of the diagonal $A_1A_4$ there is $A_2$, $A_3$; and on the other side of the diagonal $A_1A_5$ there must be $A_6$, $A_7$. So there are 7 vertices of the polygon.
Hence $n=7$, leading to a heptagon.
$Outline:$
I'm sorry I cannot draw a diagram and explain, though that would be very neat. Let $r$ be the radius of the circumcircle, and let $2\theta$ be the angle subtended at the center by the adjacent vertices. If $a$ is the length of any side of the regular polygon, then $a = 2r\sin\theta$. Now $\theta = \frac{\pi}{n}$, then the given equation becomes $\frac{1}{\sin\theta} = \frac{1}{\sin2\theta} + \frac{1}{\sin3\theta}$. Simplify this, you will get $\sin4\theta = \sin3\theta$, gives $\theta = \frac{\pi}{7}$. So $n = 7$ is the final answer. You may fill in the gaps in the solution.
Since the answer got a few upvotes, I looked up a little bit more and found that this problem was asked in IIT JEE 1994 (entrance exam for Indian Institute of Technology) and again a variant of it in 2011. Prof K. D. Joshi, in his book Educative JEE, has outlined 3 solutions, one of them is like mine (only better), then the Ptolemy one given by user260674 as well as the one using complex numbers given by Seven. You can refer to page 18 and 19 of the original script here
I was casting about for a method that didn't require a lot of work with multiple-angles and trigonometric identities. Up to now, I had a different -- though ultimately related -- argument (without using a circumscribed circle) which led me to the same equation, $\ \frac{1}{\sin\theta} \ = \ \frac{1}{\sin2\theta} \ + \ \frac{1}{\sin3\theta}$ , which Shailesh already produced. Here's something using more basic triangle geometry than what I'd had.
For convenience, we will call the lengths of the sides of the polygon $\ s \$ , making $\ A_1A_2 \ = \ s \$ and call the other lengths of interest $\ A_1A_3 \ = \ t \$ and $\ A_1A_4 \ = \ u \$ .
"Drop perpendiculars" to $\ \overline{A_1A_4} \$ from $\ A_2 \$ to produce point $\ P \$ and from $\ A_3 \$ to produce $\ Q \$ . Since we are working with a regular polygon, it is straightforward to show that $\ A_1A_2A_3A_4 \$ is a trapezoid and that $\ PQ \ = \ s \$ . Extend, say , $\ A_4A_3 \$ to a point $\ R \$ : since $\ \angle A_2A_3R \$ is an exterior angle of the polygon, $\ m(\angle A_2A_3R) \ = \ \frac{2 \pi}{n} \$ . We have $\ \overline{A_2A_3} \ \ \Vert \ \ \overline{A_1A_4} \$ , so corresponding angle $\ \angle QA_4A_3 \$ also has measure $\ \frac{2 \pi}{n} \$ .
Consequently, $\ A_4Q \ = \ A_1P \ = \ s \ \cos \left(\frac{2 \pi}{n} \right) \$ [it is clear that the two segments are congruent] and so
$$u \ = \ s \ \left( \ 1 \ + \ 2 \ \cos \left[ \frac{2 \pi}{n} \right] \ \right) \ \ .$$
Although $\ \Delta A_1A_2A_3 \$ is isosceles, and $\ m(\angle A_1A_2A_3) \ = \ \pi \ - \ \frac{2 \pi}{n} \$ by a familiar theorem (or because it is supplementary to an exterior angle), we will actually not exploit the Law of Cosines here to find $\ t \$ . Instead, we obtain that $\ m(\angle A_1A_3A_2 ) \ = \ \frac{1}{2} \ ( \ \pi \ - \ [ \pi \ - \ \frac{2 \pi}{n} ] \ ) \ = \ \frac{\pi}{n} \$ . From this, we find $\ m(\angle A_1A_3A_4 ) \ = \ \left(\pi \ - \ \frac{2 \pi}{n} \right) \ - \frac{\pi}{n}$ $\ = \ \left(\pi \ - \ \frac{3 \pi}{n} \right) \$ .
By the Law of Sines,
$$\frac{t}{\sin \left( \frac{2 \pi}{n} \right)} \ = \ \frac{u}{\sin \left( \ \pi \ - \ \frac{3 \pi}{n} \ \right) \ } \ \ \Rightarrow \ \ \frac{t}{\sin \left( \frac{2 \pi}{n} \right)} \ = \ \frac{s \ \left( \ 1 \ + \ 2 \ \cos \left[ \frac{2 \pi}{n} \right] \ \right)}{\sin \left( \frac{3 \pi}{n} \right) }$$
[using the identity "sine of an angle equals the sine of its supplement"].
At last coming to the equation under discussion, we obtain
$$\frac{1}{s} \ = \ \frac{1}{t} \ + \ \frac{1}{u} \ = \ \frac{1}{s} \ \left[ \ \frac{1 }{\left( \ 1 \ + \ 2 \ \cos \left[ \frac{2 \pi}{n} \right] \ \right)} \ \left( \ \frac{\sin \left( \frac{3 \pi}{n} \right)}{\sin \left( \ \frac{2 \pi}{n} \ \right)} \ \right) \ \right] \ \left[ \ 1 \ + \ \left( \ \frac{\sin \left( \frac{2 \pi}{n} \right)}{\sin \left( \ \frac{3 \pi}{n} \ \right)} \ \right) \ \right]$$
$$\Rightarrow \ \ 1 \ = \ \frac{1 }{\left( \ 1 \ + \ 2 \ \cos \left[ \frac{2 \pi}{n} \right] \ \right)} \ \ \left[ \ \left( \ \frac{\sin \left( \frac{3 \pi}{n} \right)}{\sin \left( \ \frac{2 \pi}{n} \ \right)} \ \right) \ + \ 1 \ \right]$$
$$\Rightarrow \ \ 1 \ + \ 2 \ \cos \left( \frac{2 \pi}{n} \right) \ = \ 1 \ + \ \left[ \ \frac{\sin \left( \frac{3 \pi}{n} \right)}{\sin \left( \ \frac{2 \pi}{n} \ \right)} \ \right]$$ $$\Rightarrow \ \ 2 \ \sin \left( \frac{2 \pi}{n} \right) \ \cos \left( \frac{2 \pi}{n} \right) \ = \ \sin \left( \frac{3 \pi}{n} \right) \ \ \Rightarrow \ \ \sin \left( \frac{4 \pi}{n} \right) \ = \ \sin \left( \frac{3 \pi}{n} \right) \ \ ,$$
applying the "double-angle formula" for sine at the end.
Since all of the angles discussed here have measure less than $\ \pi \$ , it is either the case that
$$\frac{4 \pi}{n} \ = \frac{3 \pi}{n} \ \ ,$$
which would require $\ \frac{ \pi}{n} \ = \ 0 \$ , or that
$$\frac{4 \pi}{n} \ = \ \pi \ - \frac{3 \pi}{n} \ \ \Rightarrow \ \ \frac{7 \pi}{n} \ = \ \pi \ \ \Rightarrow \ \ n \ = \ 7 \ \ .$$
Hint: Let, each side of the regular polygon $A_1A_2A_3\ldots A_n$ be $a$ then we have $$A_1A_2=a$$ $$A_1A_3=2a\sin\left(\frac{(n-2)\pi}{2n}\right)$$ $$A_1A_4=a-2a\cos\left(\frac{(n-2)\pi}{n}\right)$$
Now, we have $$\frac{1}{A_1A_2}=\frac{1}{A_1A_3}+\frac{1}{A_1A_4}$$
$$\frac{1}{a}=\frac{1}{2a\sin\left(\frac{(n-2)\pi}{2n}\right)}+\frac{1}{a-2a\cos\left(\frac{(n-2)\pi}{n}\right)}$$ Let, $\frac{(n-2)\pi}{2n}=\alpha$ then we have
$$1-\frac{1}{1-2\cos2\alpha}=\frac{1}{2\sin\alpha}$$
$$\frac{1-2\cos2\alpha-1}{1-2\cos2\alpha}=\frac{1}{2\sin\alpha}$$ $$8\sin^3\alpha-4\sin^2\alpha-4\sin\alpha+1=0$$
I hope you can take it from here.
• Perhaps the one weakness of this approach is that it does not provide an immediate way to pick out the value for $\ n \$ . While one of the roots corresponds to an obtuse angle and so can be rejected in the context of the geometrical situation, the other two roots produce $\ \sin \alpha \ \approx \ 0.9010 \$ and $\ \sin \alpha \ \approx \ 0.2225 \$ . The latter gives us $\ \alpha \ = \ \frac{\pi}{14} \$ , which looks promising, but leads to a non-integral result for $\ n \$ . It is actually the other value that turns out to work, yielding $\ \alpha \ = \ \frac{5 \pi}{14} \$ . Aug 10, 2015 at 7:14
• Yes, you are right Aug 10, 2015 at 7:39
Here's a solution using just the law of cosines. The identity does not hold unless $$n\ge 5$$. Label the edges and diagonals as in the diagram below, so that by hypothesis $$\frac1a = \frac1b + \frac1c.\tag1$$
The segments $$A_2A_3$$, $$A_3A_4$$, and $$A_4A_5$$ are concyclic and have the same length $$a$$, so they subtend the same angle $$\theta$$ at $$A_1$$. Apply the law of cosines to triangles $$\triangle A_1A_2A_3$$, $$\triangle A_1A_3A_4$$, $$\triangle A_1A_4A_5$$ respectively: $$a^2=a^2+b^2-2ab\cos\theta\tag2$$ $$a^2=b^2+c^2-2bc\cos\theta\tag3$$ $$a^2=c^2+d^2-2cd\cos\theta\tag4$$ Solve equation (2) for $$\cos\theta$$ and apply (1): $$\cos\theta=\frac b{2a}\stackrel{(1)}=\frac{b+c}{2c}.\tag5$$ Equate the right sides of (3) and (4): $$b^2-2bc\cos\theta=d^2-2cd\cos\theta,$$ substitute (5) for $$\cos\theta$$, and simplify to obtain $$(b-d)(d-c)=0$$. We rule out $$b=d$$ [which merely makes equation (4) a duplicate of (3)], leaving $$d=c$$. This means $$\triangle A_1A_4A_5$$ is isosceles, so the polygon has five internal triangles meeting at $$A_1$$, and has seven vertices. | 2022-06-27T09:58:34 | {
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https://sriasat.wordpress.com/tag/binomial-coefficient/ | # Tag Archives: binomial coefficient
## A Combinatorial Identity
$(*)\qquad\qquad\qquad\displaystyle\sum_{n\ge i\ge j\ge k\ge 0}\binom ni\binom ij\binom jk=4^n$.
One way to prove it is to count the number of possible triples $(A,B,C)$ of sets with $C\subseteq B\subseteq A\subseteq S=\{1,\dots,n\}$. This can be done in two ways. First, if $A, B, C$ contain $i,j,k$ elements respectively then the number of such triples clearly equals the left-hand side of $(*)$. On the other hand, since each element of $S$ must belong to one of the four disjoint regions in the picture below
the number of such triples equals precisely $4^n$.
An algebraic proof can be obtained by repeated use of the binomial theorem:
$\displaystyle 4^n=\sum_{n\ge i\ge 0}\binom ni3^i=\sum_{n\ge i\ge j\ge 0}\binom ni\binom ij2^j=\sum_{n\ge i\ge j\ge k\ge 0}\binom ni\binom ij\binom jk$.
Similarly, one obtains the more general identity
$\displaystyle\sum_{n\ge i_m\ge\cdots\ge i_1\ge 0}\binom{n}{i_1}\binom{i_1}{i_2}\cdots\binom{i_{m-1}}{i_m}=(m+1)^n$.
This also follows from the multinomial theorem since the left-hand side equals
$\displaystyle\sum_{n\ge i_m\ge\cdots\ge i_1\ge 0}\frac{n!}{(n-i_1)!(i_1-i_2)!\cdots (i_{m-1}-i_m)!i_m!}=(m+1)^n$.
Filed under Combinatorics
## Products of Consecutive Integers Dividing One Another
I’ve been working on a project lately and I find myself stuck with the following problem.
Problem. Given positive integers $a, what can be said about the least positive integer $t$ such that $a(a+1)\cdots (a+t)$ does not divide $b(b-1)\cdots (b-t)$?
Obviously $t$ is bounded above by $b-a+1$ but this is very weak. Probably the least $t$ should be close to $b/a$.
Filed under Number Theory
## A Nice Divisibility Result on Binomial Coefficients
I just came across this cute little result on Quora and generalised it to the following.
Proposition. For any integers $0\le k\le n$,
$\displaystyle\frac{n}{(n,k)}$ divides $\displaystyle\binom{n}{k}$.
First proof. Note that
$\displaystyle \frac{k}{(n,k)}\binom nk=\frac{n}{(n,k)}\binom{n-1}{k-1}$.
Since $\displaystyle\left(\frac{n}{(n,k)},\frac{k}{(n,k)}\right)=1$, the result follows. $\square$
Second proof. Let $n=p_1^{a_1}\cdots p_r^{a_r}$ and $k=p_1^{b_1}\cdots p_r^{b_r}$ where $p_1,\dots,p_r$ are the prime factors of $nk$. For each $i$, the base $p_i$ representations of $n$ and $k$ have $a_i$ and $b_i$ trailing zeros respectively. Hence $p_i^{\max\{a_i-b_i,0\}}$ divides $\displaystyle\binom nk$, by Kummer’s theorem. Therefore
$\displaystyle k\prod_{i=1}^rp_i^{\max\{a_i-b_i,0\}}=\prod_{i=1}^rp_i^{\max\{a_i,b_i\}}=[n,k]$
divides $\displaystyle k\binom nk$. Now the result follows using $[n,k]/k=n/(n,k)$. $\square$
A nice corollary is the following property of Pascal’s triangle.
Corollary. For any integers $0,
$\displaystyle\gcd\left(n,\binom nk\right)>1$.
Using the identity
$\displaystyle\binom nk\binom kr=\binom nr\binom{n-r}{k-r}$
the argument in the first proof above can be adapted to prove the following generalisation:
Proposition. For any integers $0\le r\le k\le n$,
$\displaystyle\frac{\binom nr}{\left(\binom nr,\binom kr\right)}$ divides $\displaystyle\binom{n}{k}$.
Corollary. Any two entries $\neq 1$ in a given row of Pascal’s triangle have a common factor $>1$.
Filed under Number Theory
## Binomial Sum Modulo Prime Power: Part 2
Let $p$ be a prime number and $\omega=\exp(2\pi i/p)$. Recall that we conjectured in this post that
$(*)\qquad\qquad\qquad\displaystyle\sum_{j=0}^{p-1}(1-\omega^j)^n=p\sum_{k=0}^{\lfloor n/p\rfloor}(-1)^{kp}\binom{n}{kp}$
is divisible by $p^{\lceil n/(p-1)\rceil}$. From exercise 2 of this post we know that $(p)$ factors as $(p,\omega-1)^{p-1}$ into prime ideals in the ring $\mathbb Z[\omega]$. So $(\omega-1)^{p-1}\in (p)$. Therefore the right-hand side of $(*)$ is divisible by $p^{\lfloor n/(p-1)\rfloor}$. So we are off by at most one factor of $p$! I believe this approach can be improved upon to account for the extra factor, since we haven’t used any property of the sum.
Furthermore,
$\displaystyle\sum_{j=0}^{p-1}\omega^{-rj}(1-\omega^j)^n=p\sum_{m\equiv r\pmod p}(-1)^m\binom{n}{m}$
for any $r$. So our best result thus far is the following:
Theorem (weaker version of Fleck’s).
$\displaystyle\sum_{m\equiv r\pmod p}(-1)^m\binom{n}{m}\equiv 0\pmod{p^{\lfloor\frac{n}{p-1}\rfloor-1}}$.
Goal: Improve the floors to ceilings.
Filed under Algebra, Number Theory
## Binomial Sum Modulo Prime Power: Part 1
In the last post we showed that
$\displaystyle N_p(n)\equiv\sum_{k=1}^{\lfloor n/p\rfloor}(-1)^k\binom{n}{pk}\pmod p$
and then we used Lucas’ theorem to evaluate the sum modulo $p$.
My initial attempt at evaluating the sum was to note that
$(*)\qquad\qquad\qquad\displaystyle\sum_{j=0}^{p-1}(1-\omega^j)^n=p\sum_{k=0}^{\lfloor n/p\rfloor}(-1)^{kp}\binom{n}{kp}$
where $\omega=\exp(2\pi i/p)$, but then I was stuck. Numerical examples suggested that this sum is in fact divisible by $p^{\lceil n/(p-1)\rceil}$, but I could not think of a way of extending the argument from the last post. Nonetheless, some extensive google search recently revealed that
Theorem (Fleck, 1913). For any $j$,
$\displaystyle\sum_{m\equiv j\pmod p}(-1)^m\binom{n}{m}\equiv 0\pmod{p^{\left\lfloor \frac{n-1}{p-1}\right\rfloor}}$.
So taking $j=0$ gives the desired claim after noting that
Fact. $\displaystyle \left\lfloor\frac{n-1}{p-1}\right\rfloor+1=\left\lceil\frac{n}{p-1}\right\rceil$.
Proof. Write $n=q(p-1)+r$ with $0\le r. If $r=0$ then the equation is just $q-1+1=q$, and if $r>0$ then it is $q+1=q+1$. $\square$
This article contains a proof of Fleck’s result using the identity $(*)$. Unfortunately I don’t have a good grasp of the theory behind it. | 2020-08-10T19:56:25 | {
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https://www.physicsforums.com/threads/putting-a-point-inside-a-triangle-in-3d-space.933690/ | # I Putting a point inside a triangle in 3D space
1. Dec 7, 2017
### wukunlin
This may belong to the computing subforum, let me know if this is more true than having it here in the math forum :)
My questions are
1) Suppose there is a plane in 3D space and I have 3 points to define it:
p1 = {x1, y1, z1}
p2 = {x2, y2, z2}
p3 = {x3, y3, z3}
and I want to put a particular point p4 which I already know the x- and z- coordinates.
What will be the most efficient way to compute the y- coordinate of p4?
I can think of the nasty method which I compute the determinant of a 4x4 matrix to find the coefficients for:
ax + by + cz + z = 0
Then substitute my known x and z, I get the feeling this is very inefficient and there are more elegant solutions than this. Are there any known algorithms to deal with this problem?
2) Now that I have my p4, I want have the same relative position to the 3 points p1, p2, p3 if someone move this triangle around:
f(p1, p2, p3) = p4
for example, if p4 happens to be in the middle of the triangle, the function would be the average of each of the x, y, z coordinates of the 3 points. But for a point that is off-centered, how will I go about finding what this function should be?
2. Dec 7, 2017
### Staff: Mentor
so you have three points represented by three position vectors. If you subtract p1-p2 and p3-p2 then you have two vectors that describe the edges of the triangle right? If you then add them and divide the resultant vector by some value > 2 then what can you conclude?
3. Dec 7, 2017
### Staff: Mentor
It's possible there are more efficient ways, but here is a straightforward way. If you have three points in a plane $P_1, P_2, P_3$, find displacement vectors between any two pairs of them, say, $\vec u = \vec{P_1P_2}$ and $\vec v = \vec{P_1P_3}$.
Calculate the cross product $\vec n = \vec u \times \vec v$ = <A, B, C>
The equation of the plane will be $A(x - x_1) + B(y - y_1) + C(z - z_1) = 0$, where I arbitrarily chose the coordinates of point $P_1$. You could use the coordinates of either of the other two given points -- it doesn't matter.
Now that you have the equation of the plane, it's a simple matter to check whether some point $P_4 = (x_4, y_4, z_4)$ lies in the plane by substituting these coordinates for x, y, and z respectively in the plane's equation.
4. Dec 7, 2017
### WWGD
I am not sure I understand, the equation of a plane in 3-space is : $ax+by+cz=d$. Once you do the cross-product and have any of these points, you will have all of a,b,c and d given. You only need to solve for y from a linear relation. EDIT: Essentially a rewording of what Mark44 said.
5. Dec 7, 2017
### Staff: Mentor
If your triangle is aligned with the y-axis there is no unique solution, but if we know we don't have this special case:
To just find the third coordinate, set up the plane as described in the previous posts, this gives a linear equation to solve. Note: It doesn't guarantee you that the point is actually within the triangle.
To (a) check that the point is indeed inside and (b) make the triangle movable, I would take a slightly different approach.
With $\vec u$ and $\vec v$ defined as in post 3, the point 4 can be expressed as $p_4 = p_1 + a \vec u + b \vec v$ where $a,b \geq 0$ and $a+b \leq 1$. If you consider the x and z coordinate of this equation you get two equations with two unknowns (a and b). Solve, and you can calculate the missing y coordinate. You can also check if the point is really within the triangle by checking the inequalities mentioned before. Shifting the triangle just changes $p_1, \vec u, \vec v$ so you can use the same a and b to find the new point 4.
6. Dec 8, 2017
### wukunlin
I really appreciate all the help everyone. It made me realize how rusty the gears of my brain has gotten after years of not doing problems like these.
That worked beautifully!! Thank you
Thanks for the help. Looks like great minds think alike
Oh wow, that does look a LOT more elegant.
Drawing that final vector from P2 will end up somewhere inside the triangle right? | 2018-03-17T16:55:10 | {
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http://mathhelpforum.com/statistics/190782-one-woman-ten-keys-one-lock-opened-one-heck-question.html | # Math Help - One woman with ten keys, one lock to be opened, and one heck of a question!
1. ## One woman with ten keys, one lock to be opened, and one heck of a question!
The question:
A woman has 10 keys out of which only one opens a lock. She tries the keys one after another (keeping aside the failed ones) till she succeeds in opening the lock. What is the chance that it is the seventh key that works?
My approach so far:
From what I've thought out so far, I think we need to find out the probability of each of the keys failing before the seventh key, and then the probability of the seventh key being the one that works out of the remaining keys (which would be four by the time she reaches the seventh key).
The only problem I'm facing is that I don't understand how to take into account the probabilities of those events happening one after another, simultaneously.
2. ## Re: One woman with ten keys, one lock to be opened, and one heck of a question!
The question:
A woman has 10 keys out of which only one opens a lock. She tries the keys one after another (keeping aside the failed ones) till she succeeds in opening the lock. What is the chance that it is the seventh key that works?
QUESTION
Once she tries a key and it fails, what does she with it?
Does she discord it? Or put is back in the pile?
3. ## Re: One woman with ten keys, one lock to be opened, and one heck of a question!
so the next probability that I'l have to evaluate will be from one less number of keys.
So I think I'd do it this way:
The probability of the first key failing is 9/10 ('cause out of ten, any nine can fail).
Then she keeps it aside.
So the probability of the next key failing from the keys remaining will be 8/9 ('cause out of the nine keys left, only one will work).
And so on up to 4/5, after which she'll set aside the failed key and reach the seventh key, and have four keys left, out of which only one will work. So, the probability of picking up that key will be 1/4.
What I don't understand is how to find the probability of these events happening together.
4. ## Re: One woman with ten keys, one lock to be opened, and one heck of a question!
A woman has 10 keys out of which only one opens a lock.
She tries the keys one after another (keeping aside the failed ones)
till she succeeds in opening the lock.
What is the probability that it is the seventh key that works?
The problem is much simpler than you think.
Place the ten keys in a row in random order.
Consider the 3rd key.
What is the probability that it is the correct key?
. . Answer: . $\tfrac{1}{10}$ . . . agree?
Consider the 9th key.
What is the probability that is the correct key?
. . Answer: . $\tfrac{1}{10}$ . . . right?
Now consider the 7th key . . .
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
We can do it your way.
If the 7th key is correct, the first six are incorrect.
. . (and we don't care about the last three.)
So, we want: . no - no - no - no - no - no - yes.
The probability is: . $\frac{9}{10}\cdot\frac{8}{9}\cdot\frac{7}{8}\cdot \frac{6}{7}\cdot\frac{5}{6}\cdot \frac{1}{5}$
. . which reduces to: . $\frac{\rlap{/}9}{10}\cdot\frac{\rlap{/}8}{\rlap{/}9}\cdot\frac{\rlap{/}7}{\rlap{/}8}\cdot \frac{\rlap{/}6}{\rlap{/}7}\cdot\frac{\rlap{/}5}{\rlap{/}6}\cdot \frac{1}{\rlap{/}5} \;=\;\frac{1}{10}$
5. ## Re: One woman with ten keys, one lock to be opened, and one heck of a question!
so the next probability that I'l have to evaluate will be from one less number of keys.
Then this problem is no different by symmetry than asking for the probability that the second key works:
$\frac{9}{10}\cdot\frac{1}{9}=\frac{1}{10}$. | 2014-07-13T06:19:11 | {
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https://linearalgebras.com/solution-hoffman-kunze-1-2-6.html | If you find any mistakes, please make a comment! Thank you.
## Homogeneous systems of linear equations in two unknowns with the same solutions are equivalent
Solution to Linear Algebra Hoffman & Kunze Chapter 1.2 Exercise 1.2.6
Solution: Write the two systems as follows:
$$\begin{array}{c} a_{11}x+a_{12}y=0\\ a_{21}x+a_{22}y=0\\ \vdots\\ a_{m1}x+a_{m2}y=0 \end{array} \quad\quad \begin{array}{c} b_{11}x+b_{12}y=0\\ b_{21}x+b_{22}y=0\\ \vdots\\ b_{m1}x+b_{m2}y=0 \end{array}$$ Each system consists of a set of lines through the origin $(0,0)$ in the $x$-$y$ plane. Thus the two systems have the same solutions if and only if they either both have $(0,0)$ as their only solution or if both have a single line $ux+vy-0$ as their common solution. In the latter case all equations are simply multiples of the same line, so clearly the two systems are equivalent. So assume that both systems have $(0,0)$ as their only solution. Assume without loss of generality that the first two equations in the first system give different lines. Then
\frac{a_{11}}{a_{12}}\not=\frac{a_{21}}{a_{22}}
\label{eq1}
We need to show that there's a $(u,v)$ which solves the following system:
$$\begin{array}{c} a_{11}u+a_{12}v=b_{i1}\\ a_{21}u+a_{22}v=b_{i2} \end{array}$$ Solving for $u$ and $v$ we get
$$u=\frac{a_{22}b_{i1}-a_{12}b_{i2}}{a_{11}a_{22}-a_{12}a_{21}}$$ $$v=\frac{a_{11}b_{i2}-a_{21}b_{i1}}{a_{11}a_{22}-a_{12}a_{12}}$$ By (\ref{eq1}) $a_{11}a_{22}-a_{12}a_{21}\not=0$. Thus both $u$ and $v$ are well defined. So we can write any equation in the second system as a combination of equations in the first. Analogously we can write any equation in the first system in terms of the second.
#### Linearity
This website is supposed to help you study Linear Algebras. Please only read these solutions after thinking about the problems carefully. Do not just copy these solutions.
### This Post Has One Comment
1. Salutations. Doesn't this solution only work if we assume the coefficients ai1, ai2, bi1, bi2 and that x,y are real numbers? Does it also work for complex numbers? | 2022-01-26T07:47:43 | {
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https://math.stackexchange.com/questions/3127690/simple-example-on-uniformly-convex-spaces | # Simple example on uniformly convex spaces
In the lectures we showed the following result:
Theorem: Let $$(E,\|\cdot\|_E)$$ be a uniformly convex space. Consider a sequence $$\{x_n \}\rvert_{n\in\mathbb{N}} \subset E$$ and $$x \in E$$ such that it converges weakly to $$x\in E$$ $$x_n\rightharpoonup x ,$$ and the sequence of the norms converges to the norm of $$x\in E$$, i.e. $$\|x_n\|_E \longrightarrow \|x\|_E.$$ Then the sequence $$\{x_n \}\rvert_{n\in\mathbb{N}} \subset E$$ is strongly convergent $$x_n \longrightarrow x.$$
This means that weak convergence, together with the convergence of the norms imply strong converge in uniformly convex spaces.
Question: Could you please provide a counterexample on a non uniformly convex space (maybe sequence space of bounded sequences $$\ell^\infty$$?) where this result does not hold?
Concretely: A sequence on a non uniformly convex space such that it is weak convergent, and the sequence of the norms converges, but the sequence itself is not strongly convergent.
I would be grateful to read any possible counterexample. Thanks!
Consider the sequence $$x_n := (e_1+e_n) \in \ell^\infty$$.
Then it can be shown that $$e_n\rightharpoonup 0$$ (see here) which implies $$x_n\rightharpoonup e_1$$.
It can also be calculated that $$x_n\not\to e_1$$ in the norm convergence and that $$\|x_n\|\to \|e_1\|=1$$.
• Dear Sir. Thanks for your answer. To clarify, you can pick $\varepsilon = \frac{1}{2}$ such that $\| x_n - e_1 \|_{\infty} = \| e_n \|_{\infty} > \varepsilon$ for all $n\in \mathbb{N}$. This would show that the convergence does not follow strongly. Correct? | 2021-10-28T07:18:45 | {
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https://bcdudek.net/regression1/inferential-tests-for-assumptions.html | # Chapter 9 Inferential tests for assumptions
There are three primary assumptions that linear modeling with the standard NHST t and F tests rely on:
1. The relationships among variables are best described by linear functions.
2. The residuals from a model are normally distributed.
3. Homoscedasticity.
Graphical evaluation of these assumptions has been the priority in previous considerations, as well in this document. Now we can turn to inferential tests regarding these assumptions.
## 9.1 Evaluation of the Residual Normality Assumption
Prior work with SPSS and R for linear models has only employed a graphical assessment to the normality assumption for residuals (and skewness/kurtosis computation). Frequency histograms and normal probability plots are useful, but at times one may wish to do an inferential test of a null hypothesis that residuals are normally distributed. Quite a few such inferential tests have been developed, and many are available in R.
While we can easily accomplish these tests, the analyst should be concerned about a binary outcome decision process in evaluation of the assumption. The real question should be something like “with the degree of non-normality present, how much of an impact on the tests of regression coefficients and the test of the whole model is there?” Strong guidance on this is lacking. However, some discussion can be found in standard regression textbooks and the reader is advised to consult Fox (2016), or any number of other sources (e.g., Cohen et al. (2003), Cook & Weisberg (1999), Darlington (1990), Howell (2013), Weisberg (2014), or Wright & London (2009)).
The options available in R will be presented here without regard to that more important question. The Anderson-Darling test may be the test that is most recommended. Historically, the Shapiro test is probably the most commonly used one, based on availability in other software. But with the ‘nortest’ package in R, many others are also available.
First, we can implement the historically common Shapiro-Wilk test:
shapiro.test(residuals(fit3))
##
## Shapiro-Wilk normality test
##
## data: residuals(fit3)
## W = 0.98639, p-value = 0.7239
This next code chunk shows code executing five different tests from the nortest package. But only one (Anderson-Darling) is executed here to keep this document small. The reader may recall from graphical approaches accomplished above that there is a slight degree of non-normality in the residuals from the basic two-IV model (fit3 or fit4). Is it a significant departure from normality, based on the sample size employed in this study?
# tests for normality of the residuals
#library(nortest)
ad.test(residuals(fit3)) # Anderson-Darling test (from 'nortest' package )
##
## Anderson-Darling normality test
##
## data: residuals(fit3)
## A = 0.34507, p-value = 0.4742
#cvm.test(residuals(fit3)) #Cramer-von Mises test ((from 'nortest' package )
#lillie.test(residuals(fit3)) #Lilliefors (Kolmogorov-Smirnov) test (from 'nortest' package )
#pearson.test(residuals(fit3)) #get Pearson chi-square test (from 'nortest' package )
#sf.test(residuals(fit3)) #get Shapiro-Francia test (from 'nortest' package )
Neither of these tests reach significance thresholds at $$\alpha$$=.05 and this is not suprising given the small amount of skewness visualized in the graphical assessments of the residuals seen in above sections of this document.
### 9.1.1 Evaluation of the normality assumption by testing skewness and kurtosis
Another approach to inference regarding the normality assumption is to test whether skewness and/or kurtosis of the residuals deviates from a null hypothesis that they are zero (as would be the case for a normal distribution). Once again, this document is more concerned with showing how such tests can be accomplished in R. Discussions about their interpretation and whether they should be used are left to other parts of the course and to the recommendations of the relevant textbooks.
R has two packages (tseries and moments) that contain functions to test skewness and/or kurtosis:
# also....... another way to evaluate normality is to test for skewness and kurtosis
#library(tseries)
jarque.bera.test(residuals(fit3)) # Jarque Bera test for normality of a variable
##
## Jarque Bera Test
##
## data: residuals(fit3)
## X-squared = 0.77247, df = 2, p-value = 0.6796
#library(moments)
agostino.test(residuals(fit3))
##
## D'Agostino skewness test
##
## data: residuals(fit3)
## skew = 0.22341, z = 0.78060, p-value = 0.435
## alternative hypothesis: data have a skewness
anscombe.test(residuals(fit3)) # Anscombe-Glynn test of kurtosis
##
## Anscombe-Glynn kurtosis test
##
## data: residuals(fit3)
## kurt = 2.68478, z = -0.25858, p-value = 0.796
## alternative hypothesis: kurtosis is not equal to 3
Again, none of these tests would allow us to reject a null hypothesis of residual normality.
We might also simply perform a Z test by taking the ratio of skewness and kurtosis coefficients to their std errors (called a large-sample approximation in using the std normal Z here). We can obtain the coefficients and std errors in a couple ways:
We can write our own function to test skewness and kurtosis as the Z approximation called the large sample approximation suggested above. First, create the function and test skewness:
# we can build our own tests, since we know the
# std errors of skewness and kurtosis (fall semester work)
# these functions test the G1 and G2 statistics with their
# large sample std errors against nulls that the parameters are zero (as in normal distribs)
skewness.test <- function (lmfit) {
require(psych)
G1 <- psych::describe(residuals(lmfit),type=2)$skew # pull G1 from psych:::describe skewness <- G1 N <- length(residuals(lmfit)) stderrskew <- sqrt(6/N) teststat <- skewness/stderrskew pvalue <- 2*(pnorm(abs(teststat), lower.tail=F)) outskew <- data.frame(cbind(N,G1,stderrskew,teststat,pvalue)) colnames(outskew) <- c("N","G1","Std Error", "Z Test Stat","2-tailed p") rownames(outskew) <- c("") return(outskew) } # use this approach (z test) with large N. It returns a two-tailed p value skewness.test(fit3) ## N G1 Std Error Z Test Stat 2-tailed p ## 62 0.2289903 0.3110855 0.7361008 0.4616693 And now test kurtosis: # now kurtosis kurtosis.test <- function(lmfit) { require(psych) G2 <- psych::describe(residuals(lmfit),type=2)$kurtosis # pull G2 from psych:::describe
kurtosis <- G2
N <- length(residuals(lmfit))
stderrkurt <- sqrt(24/N)
teststat <- kurtosis/stderrkurt
pvalue <- 2*(pnorm(abs(teststat), lower.tail=F))
outk <- data.frame(cbind(N,G2,stderrkurt,teststat,pvalue))
colnames(outk) <- c("N","G2","Std Error", "Z Test Stat","2-tailed p")
rownames(outk) <- c("")
return(outk)
}
kurtosis.test(fit3) # use this approach (z test) with large N. It returns a two-tailed p value
## N G2 Std Error Z Test Stat 2-tailed p
## 62 -0.2388151 0.622171 -0.3838415 0.7010959
## 9.2 Inferential test of Homoscedasticity
Evaluation of homoscedasticity shares the same graphical-approach history in our prior work as the normality assumption discussed above. There is also the question of relative impact of varying decrees of violation of the assumption that needs to be considered (we have discussed this previously in the context of the 2-sample location test assumption of homogeneity of variance which is essentially the same assumption). But the narrow goal here is the illustration of approaches in R. First, we redo a plot of residuals against yhats using the spreadLevelPlot function from the car package. The plot gives a hint of larger residual spread for mid range values of yhats, but the pattern is carried by only a few data points.
# also a plot slightly different from the base system plot on the fit
# from the car package
spreadLevelPlot(fit3)
##
## Suggested power transformation: 0.8964362
The ncvTest function (ncv standing for non-constant variance), also from the car package provides chi-square based test statistic that evaluates a null hypothesis of homoscedasticity. The test is more typically called the “Score” test.
# tests of homoscedasticity
# the ncvTest function is a test of non-constant error variance, called the Score test from car
ncvTest(fit3)
## Non-constant Variance Score Test
## Variance formula: ~ fitted.values
## Chisquare = 0.2914146, Df = 1, p = 0.58932
Another test of homoscedasticity is one that is frequently used. The Breusch-Pagan test is implemented in the bptest function found in the lmtest package. Neither it, nor the Score test yielded a result that would challenge the null hypothesis of homoscedasticity.
#library(lmtest)
bptest(fit3)
##
## studentized Breusch-Pagan test
##
## data: fit3
## BP = 2.665, df = 2, p-value = 0.2638
## 9.3 Global Evaluation of Linear Model Assumptiions using the ‘gvlma’ package.
A recent paper in JASA (Pena & Slate, 2006) outlined an approach to model assumption evaluation that is novel (and mathematically challenging). An R implementation is now available in the gvlma package. Initially, the approach evaluates a global null hypothesis that all assumptions of the linear model hypothesis tests are satisfied:
1. Linearity
2. Normality evaluation of skewness)
3. Normality (evaluation of kurtosis)
4. Link Function (linearity and normality)
5. Homoscedasticity
If this global null is rejected, then evaluation of each of four component assumptions is evaluated individually.
Fortunately, the ‘gvlma’ function is simple to use and the output is largely the standard NHST p value for each test.
A caveat here is that this type of approach is even more likely to be the target of the kinds of discussion we have had before on whether binary decision inferential tests are the way to act on knowledge of whether model assumptions are satisfied. I have not yet found a literature commenting on or assessing the Pena & Slate procedure (2006), so use of this method is only recommended with reservation. It does serve as a good example of how new/novel methods become rapidly available in R and are sometimes quite simple to implement.
# both the package and the function are called 'gvlma'
#library(gvlma)
# the gvlma function only requires a 'lm' model fit object as a single argument to yield the full assessment
gvlmafit3 <- gvlma(fit3)
#print out the tests
gvlmafit3
##
## Call:
## lm(formula = salary ~ pubs + cits, data = cohen1)
##
## Coefficients:
## (Intercept) pubs cits
## 40493.0 251.8 242.3
##
##
## ASSESSMENT OF THE LINEAR MODEL ASSUMPTIONS
## USING THE GLOBAL TEST ON 4 DEGREES-OF-FREEDOM:
## Level of Significance = 0.05
##
## Call:
## gvlma(x = fit3)
##
## Value p-value Decision
## Global Stat 1.8686 0.7599 Assumptions acceptable.
## Skewness 0.5158 0.4727 Assumptions acceptable.
## Kurtosis 0.2567 0.6124 Assumptions acceptable.
## Link Function 0.2047 0.6509 Assumptions acceptable.
## Heteroscedasticity 0.8914 0.3451 Assumptions acceptable.
#plot relevant information
# note: this plot does not play nice in the R markdown format.
# So I generated it outside of the .Rmd file, saved the image and displayed the image below
#plot(gvlmafit3)
It is possible to generate a set of very useful plots by passing the gvlma object just created to the base system plot function (plot(gvlmafit3) as was commented out in the code chunk just above). However, the figure doesn’t play nice with rmarkdown, so I created it separately, and saved the figure to a .png file that is displayed here. | 2023-03-25T04:46:32 | {
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Collection by: Rupinder Sekhon. E-mail the author
# Matrices: Homework
Module by: UniqU, LLC. E-mail the author
Summary: This chapter covers principles of matrices. After completing this chapter students should be able to: complete matrix operations; solve linear systems using Gauss-Jordan method; Solve linear systems using the matrix inverse method and complete application problems.
## INTRODUCTION TO MATRICES
A vendor sells hot dogs and corn dogs at three different locations. His total sales (in hundreds) for January and February from the three locations are given in the table below.
Table 1
January February Hot Dogs Corn Dogs Hot Dogs Corn Dogs
Place I 10 8 8 7
Place II 8 6 6 7
Place III 6 4 6 5
Represent these tables as 3×23×2 size 12{3 times 2} {} matrices JJ size 12{J} {} and FF size 12{F} {}, and answer Exercise 1, Exercise 2, Exercise 3, and Exercise 4. problems 1 - 4.
### Exercise 1
Determine total sales for the two months, that is, find J+FJ+F size 12{J+F} {}.
### Exercise 2
Find the difference in sales, JFJF size 12{J - F} {}.
### Exercise 3
If hot dogs sell for $3 and corn dogs for$2, find the revenue from the sale of hot dogs and corn dogs. Hint: Let PP size 12{P} {} be a 2×12×1 size 12{2 times 1} {} matrix. Find J+FPJ+FP size 12{ left (J+F right )P} {}.
### Exercise 4
If March sales will be up from February by 10%, 15%, and 20% at Place I, Place II, and Place III, respectively, find the expected number of hot dogs, and corn dogs to be sold in March. Hint: Let RR size 12{R} {} be a 1×31×3 size 12{1 times 3} {} matrix with entries 1.10, 1.15, and 1.20. Find RFRF size 12{ ital "RF"} {}.
Determine the sums and products in the next 5 problems. Given the matrices AA size 12{A} {}, BB size 12{B} {}, CC size 12{C} {}, and DD size 12{D} {} as follows:
A = 3 6 1 0 1 3 2 4 1 A = 3 6 1 0 1 3 2 4 1 size 12{A= left [ matrix { 3 {} # 6 {} # 1 {} ## 0 {} # 1 {} # 3 {} ## 2 {} # 4 {} # 1{} } right ]} {} B = 1 1 2 1 4 2 3 1 1 B = 1 1 2 1 4 2 3 1 1 size 12{B= left [ matrix { 1 {} # - 1 {} # 2 {} ## 1 {} # 4 {} # 2 {} ## 3 {} # 1 {} # 1{} } right ]} {} C = 1 2 3 C = 1 2 3 size 12{C= left [ matrix { 1 {} ## 2 {} ## 3 } right ]} {} D = 2 3 2 D = 2 3 2 size 12{D= left [ matrix { 2 {} # 3 {} # 2{} } right ]} {}
### Exercise 5
3A2B3A2B size 12{3A - 2B} {}
(3)
### Exercise 6
AB+BAAB+BA size 12{ ital "AB"+ ital "BA"} {}
(5)
### Exercise 7
A2A2 size 12{A2} {}
(6)
### Exercise 8
2BC2BC size 12{2 ital "BC"} {}
(8)
### Exercise 9
2CD+3AB2CD+3AB size 12{2 ital "CD"+3 ital "AB"} {}
(9)
### Exercise 10
A2BA2B size 12{A2B} {}
(11)
### Exercise 11
Let E=mnpqE=mnpq size 12{E= left [ matrix { m {} # n {} ## p {} # q{} } right ]} {} and F=abcdF=abcd size 12{F= left [ matrix { a {} # b {} ## c {} # d{} } right ]} {}, find EFEF size 12{ ital "EF"} {}.
### Exercise 12
Let G=361013241H=xyzG=361013241 size 12{G= left [ matrix { 3 {} # 6 {} # 1 {} ## 0 {} # 1 {} # 3 {} ## 2 {} # 4 {} # 1{} } right ]} {}H=xyz size 12{H= left [ matrix { x {} ## y {} ## z } right ]} {} , find GHGH size 12{ ital "GH"} {}.
Express the following systems as AX=BAX=B size 12{ ital "AX"=B} {}, where AA size 12{A} {}, XX size 12{X} {}, and BB size 12{B} {} are matrices.
### Exercise 13
4x5y=64x5y=6 size 12{4x - 5y=6} {}
(13)
5x6y=75x6y=7 size 12{5x - 6y=7} {}
(14)
### Exercise 14
x2y+2z=3x2y+2z=3 size 12{x - 2y+2z=3} {}
(16)
x3y+4z=7x3y+4z=7 size 12{x - 3y+4z=7} {}
(17)
x2y3z=12x2y3z=12 size 12{x - 2y - 3z= - "12"} {}
(18)
### Exercise 15
2x+3z=172x+3z=17 size 12{2x+3z="17"} {}
(19)
3x2y=103x2y=10 size 12{3x - 2y="10"} {}
(20)
5y+2z=115y+2z=11 size 12{5y+2z="11"} {}
(21)
### Exercise 16
x+2y+3z+2w=14x2yz=5y2z+4w=9x+3z+3w=15x+2y+3z+2w=14x2yz=5y2z+4w=9x+3z+3w=15 size 12{ matrix { x {} # +{} {} # 2y {} # +{} {} # 3z {} # +{} {} # 2w {} # ={} {} # "14" {} ## x {} # - {} {} # 2y {} # - {} {} # z {} # {} # {} # ={} {} # - 5 {} ## y {} # - {} {} # 2z {} # {} # {} # +{} {} # 4w {} # ={} {} # 9 {} ## x {} # +{} {} # 3z {} # {} # {} # +{} {} # 3w {} # ={} {} # "15"{} } } {}
(23)
## SYSTEMS OF LINEAR EQUATIONS
Solve the following by the Gauss-Jordan Method. Show all work.
### Exercise 17
x+3y=1x+3y=1 size 12{x+3y=1} {}
(24)
2x5y=132x5y=13 size 12{2x - 5y="13"} {}
(25)
### Exercise 18
xyz=1xyz=1 size 12{x - y - z= - 1} {}
(26)
x3y+2z=7x3y+2z=7 size 12{x - 3y+2z=7} {}
(27)
2xy+z=32xy+z=3 size 12{2x - y+z=3} {}
(28)
### Exercise 19
x+2y+3z=9x+2y+3z=9 size 12{x+2y+3z=9} {}
(29)
3x+4y+z=53x+4y+z=5 size 12{3x+4y+z=5} {}
(30)
2xy+2z=112xy+2z=11 size 12{2x - y+2z="11"} {}
(31)
### Exercise 20
x+2y=0x+2y=0 size 12{x+2y=0} {}
(32)
y+z=3y+z=3 size 12{y+z=3} {}
(33)
x+3z=14x+3z=14 size 12{x+3z="14"} {}
(34)
### Exercise 21
Two apples and four bananas cost $2.00 and three apples and five bananas cost$2.70. Find the price of each.
### Exercise 22
A bowl of corn flakes, a cup of milk, and an egg provide 16 grams of protein. A cup of milk and two eggs provide 21 grams of protein, and two bowls of corn flakes with two cups of milk provide 16 grams of protein. How much protein is provided by one unit of each of these three foods.
### Exercise 23
x+2y=10x+2y=10 size 12{x+2y="10"} {}
(35)
y+z=5y+z=5 size 12{y+z=5} {}
(36)
z+w=3z+w=3 size 12{z+w=3} {}
(37)
x+w=5x+w=5 size 12{x+w=5} {}
(38)
### Exercise 24
x+w=6x+w=6 size 12{x+w=6} {}
(39)
2x+y+w=162x+y+w=16 size 12{2x+y+w="16"} {}
(40)
x2z=0x2z=0 size 12{x - 2z=0} {}
(41)
z+w=5z+w=5 size 12{z+w=5} {}
(42)
## SYSTEMS OF LINEAR EQUATIONS – SPECIAL CASES
Solve the following inconsistent or dependent systems by using the Gauss-Jordan method.
### Exercise 25
2x+6y=82x+6y=8 size 12{2x+6y=8} {}
(43)
x+3y=4x+3y=4 size 12{x+3y=4} {}
(44)
### Exercise 26
The sum of the digits of a two digit number is 9. The sum of the number and the number obtained by interchanging the digits is 99. Find the number.
### Exercise 27
2xy=102xy=10 size 12{2x - y="10"} {}
(46)
4x+2y=154x+2y=15 size 12{ - 4x+2y="15"} {}
(47)
### Exercise 28
x+y+z=6x+y+z=6 size 12{x+y+z=6} {}
(48)
3x+2y+z=143x+2y+z=14 size 12{3x+2y+z="14"} {}
(49)
4x+3y+2z=204x+3y+2z=20 size 12{4x+3y+2z="20"} {}
(50)
### Exercise 29
x+2y4z=1x+2y4z=1 size 12{x+2y - 4z=1} {}
(51)
2x3y+8z=92x3y+8z=9 size 12{2x - 3y+8z=9} {}
(52)
### Exercise 65
Suppose an economy consists of three industries FF size 12{F} {}, CC size 12{C} {}, and TT size 12{T} {}. The following table gives information about the internal use of each industry's production and external demand in dollars.
F F size 12{F} {} C C size 12{C} {} T T size 12{T} {} Demand Total F F size 12{F} {} 30 10 20 40 100 C C size 12{C} {} 20 30 20 50 120 T T size 12{T} {} 10 10 30 60 110
Find the proportion of the amounts consumed by each of the industries; that is, find the matrix AA size 12{A} {}.
### Exercise 66
If in the preceding problem, the consumer demand for FF size 12{F} {}, CC size 12{C} {}, and TT size 12{T} {} becomes 60, 80, and 100, respectively, find the total output and the internal use by each industry to meet that demand.
## CHAPTER REVIEW
### Exercise 67
To reinforce her diet, Mrs. Tam bought a bottle containing 30 tablets of Supplement AA size 12{A} {} and a bottle containing 50 tablets of Supplement BB size 12{B} {}. Each tablet of supplement AA size 12{A} {} contains 1000 mg of calcium, 400 mg of magnesium, and 15 mg of zinc, and each tablet of supplement BB size 12{B} {} contains 800 mg of calcium, 500 mg of magnesium, and 20 mg of zinc.
1. Represent the amount of calcium, magnesium and zinc in each tablet as a 2×32×3 size 12{2 times 3} {} matrix.
2. Represent the number of tablets in each bottle as a row matrix.
3. Use matrix multiplication to determine the total amount of calcium, magnesium, and zinc in both bottles.
### Exercise 68
Let matrix A=113321A=113321 size 12{A= left [ matrix { 1 {} # - 1 {} # 3 {} ## 3 {} # - 2 {} # 1{} } right ]} {} and B=331143B=331143 size 12{B= left [ matrix { 3 {} # 3 {} # - 1 {} ## 1 {} # 4 {} # - 3{} } right ]} {} . Find the following.
1. 12A+B12A+B size 12{ { {1} over {2} } left (A+B right )} {}
2. 3A2B3A2B size 12{3A - 2B} {}
### Exercise 69
Let matrix C=111211101C=111211101 size 12{C= left [ matrix { 1 {} # 1 {} # - 1 {} ## 2 {} # 1 {} # 1 {} ## 1 {} # 0 {} # 1{} } right ]} {} and D=231312332D=231312332 size 12{D= left [ matrix { 2 {} # - 3 {} # - 1 {} ## 3 {} # - 1 {} # - 2 {} ## 3 {} # - 3 {} # - 2{} } right ]} {}. Find the following.
1. 2CD2CD size 12{2 left (C - D right )} {}
2. C3DC3D size 12{C - 3D} {}
### Exercise 70
Let matrix E=112312E=112312 size 12{E= left [ matrix { 1 {} # - 1 {} ## 2 {} # 3 {} ## 1 {} # 2{} } right ]} {} and F=211123F=211123 size 12{F= left [ matrix { 2 {} # 1 {} # - 1 {} ## 1 {} # 2 {} # - 3{} } right ]} {}. Find the following.
1. 2EF2EF size 12{2 ital "EF"} {}
2. 3FE3FE size 12{3 ital "FE"} {}
### Exercise 71
Let matrix G=113321G=113321 size 12{G= left [ matrix { 1 {} # - 1 {} # 3 {} ## 3 {} # 2 {} # 1{} } right ]} {} and H=abcdefH=abcdef size 12{H= left [ matrix { a {} # b {} ## c {} # d {} ## e {} # f{} } right ]} {} . Find the following.
1. 2GH2GH size 12{2 ital "GH"} {}
2. HGHG size 12{ ital "HG"} {}
### Exercise 72
Solve the following systems using the Gauss-Jordan Method.
1. x+3y2z=72x+7y5z=1x+5y3z=1x+3y2z=72x+7y5z=1x+5y3z=1 size 12{ matrix { x {} # +{} {} # 3y {} # - {} {} # 2z {} # ={} {} # 7 {} ## 2x {} # +{} {} # 7y {} # - {} {} # 5z {} # ={} {} # 1 {} ## x {} # +{} {} # 5y {} # - {} {} # 3z {} # ={} {} # 1{} } } {}
2. 2x4y+4z=22x+y+9z=13x2y+2z=72x4y+4z=22x+y+9z=13x2y+2z=7 size 12{ matrix { 2x {} # - {} {} # 4y {} # +{} {} # 4z {} # ={} {} # 2 {} ## 2x {} # +{} {} # y {} # +{} {} # 9z {} # ={} {} # 1 {} ## 3x {} # - {} {} # 2y {} # +{} {} # 2z {} # ={} {} # 7{} } } {}
### Exercise 76
Solve the following systems. If a system has an infinite number of solutions, first express the solution in parametric form, and then find a particular solution.
1. x+2y=42x+4y=83x+6y3z=3x+2y=42x+4y=83x+6y3z=3 size 12{ matrix { x {} # +{} {} # 2y {} # {} # {} # ={} {} # 4 {} ## 2x {} # +{} {} # 4y {} # {} # {} # ={} {} # 8 {} ## 3x {} # +{} {} # 6y {} # - {} {} # 3z {} # ={} {} # 3{} } } {}
2. x 2y + 2z = 1 2x 3y + 5z = 4 x 2y + 2z = 1 2x 3y + 5z = 4 size 12{ matrix { x - 2y+2z=1 {} ## 2x - 3y+5z=4 } } {}
### Exercise 77
Solve the following systems. If a system has an infinite number of solutions, first express the solution in parametric form, and then provide one particular solution.
1. 2x+y2z=02x+2y3z=06x+4y7z=02x+y2z=02x+2y3z=06x+4y7z=0 size 12{ matrix { 2x {} # +{} {} # y {} # - {} {} # 2z {} # ={} {} # 0 {} ## 2x {} # +{} {} # 2y {} # - {} {} # 3z {} # ={} {} # 0 {} ## 6x {} # +{} {} # 4y {} # - {} {} # 7z {} # ={} {} # 0{} } } {}
2. 3x+4y3z=52x+3yz=4x+2y+z=13x+4y3z=52x+3yz=4x+2y+z=1 size 12{ matrix { 3x {} # +{} {} # 4y {} # - {} {} # 3z {} # ={} {} # 5 {} ## 2x {} # +{} {} # 3y {} # - {} {} # z {} # ={} {} # 4 {} ## x {} # +{} {} # 2y {} # +{} {} # z {} # ={} {} # 1{} } } {}
### Exercise 78
Find the inverse of the following matrices.
1. 23352335 size 12{ left [ matrix { 2 {} # 3 {} ## 3 {} # 5{} } right ]} {}
2. 111121231111121231 size 12{ left [ matrix { 1 {} # 1 {} # 1 {} ## 1 {} # 2 {} # 1 {} ## 2 {} # 3 {} # 1{} } right ]} {}
### Exercise 79
Solve the following systems using the matrix inverse method.
1. 2x+3y+z=1x+2y+z=9x+y+z=52x+3y+z=1x+2y+z=9x+y+z=5 size 12{ matrix { 2x {} # +{} {} # 3y {} # + {} {} # z {} # ={} {} # 1 {} ## x {} # +{} {} # 2y {} # +{} {} # z {} # ={} {} # 9 {} ## x {} # +{} {} # y {} # +{} {} # z {} # ={} {} # 5{} } } {}
2. x+2y3z+w=7xz=4x2y+z=0y2z+w=-x+2y3z+w=7xz=4x2y+z=0y2z+w=- size 12{ matrix { x {} # +{} {} # 2y {} # - {} {} # 3z {} # +{} {} # w {} # ={} {} # 7 {} ## x {} # {} # {} # - {} {} # z {} # {} # {} # ={} {} # 4 {} ## x {} # - {} {} # 2y {} # +{} {} # z {} # {} # {} # ={} {} # 0 {} ## {} # {} # y {} # - {} {} # 2z {} # +{} {} # w {} # ={} {} # _{} } } {}
### Exercise 80
Use matrix AA size 12{A} {}, given below, to encode the following messages. The space between the letters is represented by the number 27, and all punctuation is ignored.
A=120121010A=120121010 size 12{A= left [ matrix { 1 {} # 2 {} # 0 {} ## 1 {} # 2 {} # 1 {} ## 0 {} # 1 {} # 0{} } right ]} {}
(93)
1. TAKE IT AND RUN.
2. GET OUT QUICK.
### Exercise 81
Decode the following messages that were encoded using matrix AA size 12{A} {} in the above problem.
1. 44, 71, 15, 18, 27, 1, 68, 82, 27, 69, 76, 27, 19, 33, 9
2. 37, 64, 15, 36, 54, 15, 67, 75, 20, 59, 66, 27, 39, 43, 12
### Exercise 82
Chris, Bob, and Matt decide to help each other study during the final exams. Chris's favorite subject is chemistry, Bob loves biology, and Matt knows his math. Each studies his own subject as well as helps the others learn their subjects. After the finals, they realize that Chris spent 40% of his time studying his own subject chemistry, 30% of his time helping Bob learn chemistry, and 30% of the time helping Matt learn chemistry. Bob spent 30% of his time studying his own subject biology, 30% of his time helping Chris learn biology, and 40% of the time helping Matt learn biology. Matt spent 20% of his time studying his own subject math, 40% of his time helping Chris learn math, and 40% of the time helping Bob learn math. If they originally agreed that each should work about 33 hours, how long did each work?
### Exercise 83
As in the previous problem, Chris, Bob, and Matt decide to not only help each other study during the final exams, but also tutor others to make a little money. Chris spends 30% of his time studying chemistry, 15% of his time helping Bob with chemistry, and 25% helping Matt with chemistry. Bob spends 25% of his time studying biology, 15% helping Chris with biology, and 30% helping Matt. Similarly, Matt spends 20% of his time on his own math, 20% helping Chris, and 20% helping Bob. If they spend respectively, 12, 12, and 10 hours tutoring others, how many total hours are they going to end up working?
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"domain": "cnx.org",
"url": "http://cnx.org/content/m38640/latest/?collection=col10613/latest",
"openwebmath_score": 0.8118929862976074,
"openwebmath_perplexity": 12164.371628068966,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.970687766704745,
"lm_q2_score": 0.8723473680407889,
"lm_q1q2_score": 0.8467769184742756
} |
http://math.stackexchange.com/questions/465330/average-number-of-throwing-a-dice | Average Number of Throwing a Dice
Consider a null sequence $A=\{\}$. Each time I will include the output in $A$ after throwing a dice. I will stop when $\{1,2,3\} \subseteq A$. On average how many number of throws is required?
-
If you successively toss a biased coin with probability of heads $p$, the expected number of throws before you see the first success is $\frac{1}{p}$. Your question can be decomposed into three such games. Game 1: Keep throwing a dice until one of the three numbers $\{1,2,3\}$ occurs. This has a success probability of $1/2$ and hence the expected number of trials is $2$.
Game 2: Assume without loss of generality that you obtained $1$ when you succeeded in Game 1. Then you keep throwing dice until one of $\{2,3\}$ occurs. This has success probability $1/3$ and hence expected number of trials is $3$.
Game 3: Assume without loss of generality that you obtained $2$ when you succeeded in Game 2. Then you keep throwing dice until $3$ occurs. This has success probability $1/6$ and hence requires $6$ trials on average.
Thus the average number of throws required is (by linearity of expectation) $2+3+6=11$.
Note: To answer your inconsistency with experiments, let's compute the variance. Note that the 3 games are independent of each other. Therefore, variance is also linear. Variance of a geometric distribution is $\frac{1-p}{p^2}$. Hence variance of your game is $$\frac{1/2}{1/4}+\frac{2/3}{1/9}+\frac{5/6}{1/36}=2+6+30=38.$$ Hence the standard deviation is $\sqrt{38}\approx 6.1644$. Thus if you were to repeat the experiment $n$ times, the mean is $9$ and the standard deviation is $6.1644/\sqrt{n}$. Thus as long as your answer is within 3 standard deviations, your experiment is correct. To be more accurate, increase the number of trials $n$.
-
The expected value of the waiting time for an event occuring with probability $p$ is $\frac1p$. The event "found a new one" occurs with probability $\frac36=\frac12$ initially, with $\frac26=\frac13$ after I have collected one, with $\frac16$ after I have collected two of the target numbers. Therefore the total expected waiting time is $2+3+6=11$.
-
But experimentally I get 9.97. – user89867 Aug 11 '13 at 22:13 | 2015-08-28T19:47:55 | {
"domain": "stackexchange.com",
"url": "http://math.stackexchange.com/questions/465330/average-number-of-throwing-a-dice",
"openwebmath_score": 0.8602412939071655,
"openwebmath_perplexity": 83.31645256293544,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9871787864878117,
"lm_q2_score": 0.8577681122619883,
"lm_q1q2_score": 0.8467704841507306
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https://mathhelpboards.com/threads/differential-equation-challenge.5170/ | # Differential Equation challenge
#### Petrus
##### Well-known member
Hello MHB,
I wanted to post a challange question that is hopefully not really difficult, if the question is not understandable make sure to write it so I can try explain!
Calculate the Differential equation for
$$\displaystyle y''+2y'=0$$
that satisfy
$$\displaystyle \lim_{x->\infty}y(x)=1$$ and $$\displaystyle y(0)=0$$
Regards,
$$\displaystyle |\pi\rangle$$
#### MarkFL
Staff member
Re: Differential Equation challange
Hello Petrus,
One way to proceed is to recognize we are given a linear second order homogeneous ODE whose characteristic roots are:
$$\displaystyle r=0,-2$$
Hence, the general solution is:
$$\displaystyle y(x)=c_1+c_2e^{-2x}$$
We may now state:
$$\displaystyle \lim_{x\to\infty}y(x)=c_1=1$$
$$\displaystyle y(0)=1+c_2=0\,\therefore\,c_2=-1$$
Thus the solution satisfying the given conditions is:
$$\displaystyle y(x)=1-e^{-2x}$$
#### Petrus
##### Well-known member
Re: Differential Equation challange
Hello Petrus,
One way to proceed is to recognize we are given a linear second order homogeneous ODE whose characteristic roots are:
$$\displaystyle r=0,-2$$
Hence, the general solution is:
$$\displaystyle y(x)=c_1+c_2e^{-2x}$$
We may now state:
$$\displaystyle \lim_{x\to\infty}y(x)=c_1=1$$
$$\displaystyle y(0)=1+c_2=0\,\therefore\,c_2=-1$$
Thus the solution satisfying the given conditions is:
$$\displaystyle y(x)=1-e^{-2x}$$
Hello MarkFL,
Congrats that is the correct answer
Well you said "one way to proceed.." is there more way to solve this differential equation? Well this is the way I have learned yet The smart thing with start with the $$\displaystyle \lim_{x->\infty}y(x)=1$$ is that $$\displaystyle \lim_{x->\infty}e^{-2x}=0$$ so that $$\displaystyle c_2$$ can be any real value and hence we are only left with $$\displaystyle c_1$$ and thanks to that we can solve it with $$\displaystyle y(0)=0$$ Thanks for taking your time and enter my challange!
Regards,
$$\displaystyle |\pi\rangle$$
#### MarkFL
Staff member
Another way is to write the ODE as:
$$\displaystyle \frac{d}{dx}\left(\frac{dy}{dx} \right)=-2\frac{dy}{dx}$$
Integrate with respect to $x$ to obtain:
$$\displaystyle \frac{dy}{dx}=-2y+c_1$$
Separate variables:
$$\displaystyle \frac{1}{-2y+c_1}\,dy=dx$$
Integrate:
$$\displaystyle -\frac{1}{2}\ln|c_1-2y|=x+c_2$$
$$\displaystyle \ln|c_1-2y|=-2x+c_2$$
$$\displaystyle c_1-2y=c_2e^{-2x}$$
$$\displaystyle y(x)=c_1+c_2e^{-2x}$$
And now proceed as before...
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
Another way is to use the substitution $$\displaystyle t = y'$$
$$\displaystyle t'+2t=0$$
This differential equation is separable we can also multiply by the integrating factor .
#### Chris L T521
##### Well-known member
Staff member
Another method (which I think is the longest way; EDIT: Solving by Power Series about $x=0$ would be way more complicated and longer) would be to take Laplace Transforms of both sides to obtain:
$\mathcal{L}\{y^{\prime\prime}\} + \mathcal{L}\{2y\} = \mathcal{L}\{0\} \implies s^2Y(s)-sy(0) - y^{\prime}(0) + 2sY(s)-2y(0)=0.$
Note here we know nothing about $y^{\prime}(0)$, but let's not worry about that for the time being. Letting $y(0)=0$ and solving for $Y(s)$, we get
$Y(s)=\frac{y^{\prime}(0)}{s^2+2s}=\frac{y^{\prime}(0)(s+2-s)}{2s(s+2)}=\frac{y^{\prime}(0)}{2}\left(\frac{1}{s}-\frac{1}{s+2}\right)$
Taking the Inverse Laplace Transform of both sides gives us
$y(x)=\mathcal{L}^{-1}\{Y(s)\} = \frac{y^{\prime}(0)}{2}\left(\mathcal{L}^{-1}\left\{\frac{1}{s}\right\} - \mathcal{L}^{-1}\left\{\frac{1}{s+2}\right\}\right) = \frac{y^{\prime}(0)}{2}(1-e^{-2x})$
But we're told that $\displaystyle\lim_{x\to\infty}y(x)=1$, which implies that
$\lim_{x\to\infty}y(x)= \lim_{x\to\infty}\frac{y^{\prime}(0)}{2} (1-e^{-2x})=\frac{y^{\prime}(0)}{2}=1\implies y^{\prime}(0)=2.$
Therefore, the solution to our equation is $y(x)=1-e^{-2x}$.
(That was fun! XD)
Last edited:
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
Since $$\displaystyle y(x)$$ is analytic around $$\displaystyle 0$$ we can find the power series
Let $$\displaystyle y(x) = \sum_{n\geq 0} a_n x^n$$
$$\displaystyle y'(x) =\sum_{n\geq 1} n a_n x^{n-1}$$
$$\displaystyle y''(x) =\sum_{n\geq 2} n(n-1) a_n x^{n-2}$$
Hence we have the following
$$\displaystyle \sum_{n\geq 2} n(n-1) a_n x^{n-2} +2 \, \sum_{n\geq 1} n a_n x^{n-1}=0$$
$$\displaystyle \sum_{n\geq 2} n(n-1) a_n x^{n-2} +2 \, \sum_{n\geq 2} (n-1) a_{n-1} x^{n-2}=0$$
Hence we have the following
$$\displaystyle n(n-1)a_n+2(n-1) a_{n-1} =0$$
$$\displaystyle na_n+2a_{n-1}=0 \,\,\, \Rightarrow \,\,\, a_n = -\frac{2}{n} a_{n-1} \,\,\,\, , n\geq 2$$
$$\displaystyle a_2 = -a_1$$
$$\displaystyle a_3 = \frac{2}{3}a_1$$
$$\displaystyle a_4 = -\frac{2^2}{3 \cdot 4} a_1$$
which suggests that
$$\displaystyle a_n = \frac{(-2)^{n-1}}{n!} a_1$$
we can easily deduce that $$\displaystyle a_0 = 0$$ using the initial condition .
$$\displaystyle y(x) = a_1\sum_{n\geq 1}\frac{(-2)^{n-1}}{(n)!} \, x^n = \frac{-a_1}{2}\, \sum_{n\geq 1}\frac{(-2)^{n}}{(n)!} \, x^n$$
$$\displaystyle y(x) = \frac{a_1}{-2} (e^{-2x}-1)$$
using $$\displaystyle y(\infty) = 1$$ implies that $$\displaystyle a_1 = 2$$
$$\displaystyle y(x) = -(e^{-2x}-1) = 1-e^{-2x} \,\, \square$$
By the way I didn't see Chris's comment
#### Petrus
##### Well-known member
Hello,
I want to first thank you all for taking your time and enter the challange! When I did post this problem I only knew one method (which is the MarkFL post #2) and now you all show me more method! Remainds me one of the reason why I find math is such an intressting subject! I am Really gratefull!
Regards,
$$\displaystyle |\pi\rangle$$ | 2022-01-28T20:23:28 | {
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http://mathhelpforum.com/math-topics/282260-square-root-two-answers.html | 1. ## Square Root = Two Answers
I know that there are two answers when taking the square root of a given number. For example, sqrt{4} is -2 and 2. Why is the answer both positive and negative 2? Also, does the same rule apply to variables? For example, sqrt{x^2} is x NOT
-x and x, right? Please, explain.
2. ## Re: Square Root = Two Answers
You are right about some things and wrong about others. Given any positive number x, there exist two real numbers whose square is x, one positive and one negative. We write those as $\displaystyle \pm\sqrt{x}$. But the reason we need that "$\displaystyle \pm$" is that the symbol, "$\displaystyle \sqrt{x}$" means only the positive one. That is, the square root of 4 is not "-2 and 2". The square root of 4 is 2. The two values of x, such that $\displaystyle x^2= 4$, are -2 and 2. Those are different statements. Given a positive number a, The square root of a is $\displaystyle \sqrt{a}$, a positive number. The two values of x, such that $\displaystyle x^2= a$ are $\displaystyle \sqrt{a}$ and $\displaystyle -\sqrt{x}$, which we can write as $\displaystyle \pm\sqrt{a}$. Again, the reason we need that "$\displaystyle \pm$" is that $\displaystyle \sqrt{a}$ alone refers only to the positive root of the equation.
3. ## Re: Square Root = Two Answers
Originally Posted by HallsofIvy
You are right about some things and wrong about others. Given any positive number x, there exist two real numbers whose square is x, one positive and one negative. We write those as $\displaystyle \pm\sqrt{x}$. But the reason we need that "$\displaystyle \pm$" is that the symbol, "$\displaystyle \sqrt{x}$" means only the positive one. That is, the square root of 4 is not "-2 and 2". The square root of 4 is 2. The two values of x, such that $\displaystyle x^2= 4$, are -2 and 2. Those are different statements. Given a positive number a, The square root of a is $\displaystyle \sqrt{a}$, a positive number. The two values of x, such that $\displaystyle x^2= a$ are $\displaystyle \sqrt{a}$ and $\displaystyle -\sqrt{x}$, which we can write as $\displaystyle \pm\sqrt{a}$. Again, the reason we need that "$\displaystyle \pm$" is that $\displaystyle \sqrt{a}$ alone refers only to the positive root of the equation.
Ok. Now, the sqrt{-4} = -2i and 2i.
The solution set is {-2i, 2i}.
Why does the sqrt{-1} = i? Why does i^2 = -1?
4. ## Re: Square Root = Two Answers
Originally Posted by harpazo
Why does the sqrt{-1} = i? Why does i^2 = -1?
The imaginary unit "i" is defined as $\displaystyle i^2 = -1$ so there is nothing to derive.
-Dan
5. ## Re: Square Root = Two Answers
Originally Posted by harpazo
Ok. Now, the sqrt{-4} = -2i and 2i.
The solution set is {-2i, 2i}.
Why does the sqrt{-1} = i? Why does i^2 = -1?
@harpazo, if you understand the use of definitions & axioms in mathematics this should help.
The equation $x^2+1=0$ has no solutions in the system of real numbers because $x^2\ge 0$ for all real numbers $x$.
So we postulate the existence, the addition, of one new number $\bf\mathit{i}$ the is a solution for $x^2+1=0$.
That leads to the conclusion that $\bf\mathit{i~}^2=-1$. Moreover also that $-\bf\mathit{i}$ is also a solution for $x^2+1=0$
6. ## Re: Square Root = Two Answers
Originally Posted by Plato
@harpazo, if you understand the use of definitions & axioms in mathematics this should help.
The equation $x^2+1=0$ has no solutions in the system of real numbers because $x^2\ge 0$ for all real numbers $x$.
So we postulate the existence, the addition, of one new number $\bf\mathit{i}$ the is a solution for $x^2+1=0$.
That leads to the conclusion that $\bf\mathit{i~}^2=-1$. Moreover also that $-\bf\mathit{i}$ is also a solution for $x^2+1=0$
Good information for my notes. | 2019-02-16T03:04:11 | {
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https://math.stackexchange.com/questions/1519138/initial-value-problem-for-the-pde-u-t-x3u-x-0 | # Initial value problem for the PDE $u_t + x^3u_x = 0$
Consider the initial value problem (IVP): $$u_t + x^3u_x = 0$$ $(x,t) ∈\mathbb{R} ×(0,∞)$, $u(x,0) = u_0(x)$, $x ∈R$, where $u_0 : \mathbb{R} → \mathbb{R}$ is a prescribed smooth and bounded function. Sketch the family of characteristic curves for IVP on the domain diagram. Obtain the solution $u : \mathbb{R} ×[0,∞) →\mathbb{R}$ to IVP.
This is a linear equation so we have the form $$a(x,t)u_t+b(x,t)u_x+ c(x,t)u=0$$ where $a(x,t)=x^3$ and $b(x,t)=1$ and $c(x,t)=0$.
Need to change coordinates from $(x,t)$ to $(x_0,s)$. We have: $$\frac{dx}{ds}=x^3, \, \, \, \, \, \, (2a)$$ $$\frac{dt}{ds}=1, \, \, \, \, \, \, (2b)$$
Then $$\frac{du}{ds}=\frac{dx}{ds}u_x+\frac{dt}{ds}u_t=x^3u_x+u_t$$ So $$\frac{du}{ds}=0, \, \, \, \, \, \, (3)$$
Solve $(2a)$ and $(2b)$ with condition $x(0)=x_0$ and $t(0)=0$ to get $$x^2= \frac1{2 (1/2x_0^2-s)}, \, \, \, \, t=s$$ respectively.
Solve $(3)$ with conditions $u(0)=f(x_0)$ which just gives $u=f(x_0)$.
I am stuck on this part:
When $$u_0(x) = e^{−x^2}$$ $x ∈\mathbb{R}$, sketch the solution to IVP on the $(x,u)$ plane for increasing values of $t > 0$. Describe the structure of the solution as $t →∞$.
Correct if I am wrong but we would have $$\exp \bigg(- \frac1{2t-1/s^2} \bigg)$$ but I don't know what this would look like...
• sorry, edited .. – snowman Nov 8 '15 at 16:03
• Is it really wave equation? – Empty Nov 8 '15 at 16:04
• Use Lagrange's auxiliary equation to solve – Empty Nov 8 '15 at 16:06
• The characteristics are the solutions to $x'=x^3,x(0)=x_0$ for each $x_0 \in \mathbb{R}$. The solution will be ("formally") constant along these curves. – Ian Nov 8 '15 at 16:11
• You made some mistakes in solving for the solution to the characteristic equation; you should find $x=(x_0^{-2} - 2t)^{-1/2}$. Then $u(t,x(t))=u_0(x_0)$. So to write $u(t,x)$ you need to solve $x=(x_0^{-2} - 2t )^{-1/2}$ for $x_0$ in terms of $t$ and $x$. (This algebra step amounts to following the characteristic curve backward in time from $(t,x)$ to $(0,x_0)$. When there are no singularities, it is equally correct to just do this directly, rather than going forward and then backward as I've written here.) – Ian Nov 8 '15 at 20:01
the characteristic curve $C$ thru $x = a, t = 0$ is given by $$\frac 1{a^2} - \frac 1{x^2} = 2t.$$ on the curve $C, u$ the value of $u(x,t) = u_0(a).$
given $x, t$ you can solve for $a.$ we get $a = \frac 1{\sqrt{2t+ \frac1{x^2}}}.$ that is $$u(x,t) = u_0\left( \frac 1{\sqrt{2t+ \frac1{x^2}}}\right).$$
Hints
1) First, I recommend to take a look at this post.
2) Consider the following change of variables
$$\left\{ \matrix{ x = x(u,v) \hfill \cr t = t(u,v) \hfill \cr} \right.,\,\,\,\,\,\left\{ \matrix{ {{\partial x} \over {\partial u}} = {x^3} \hfill \cr {{\partial t} \over {\partial u}} = 1 \hfill \cr} \right.$$
and define
$$z(u,v)=u(x(u,v),t(u,v))$$
then the PDE will tell you that
$${{\partial z} \over {\partial u}} = 0$$
• If we have $a(x,y)u_x+b(x,y)u_y=f(x,y,u)$ then you have $dx/a=dy/b=du/f$ so in my case we have $dt/1=dx/x^3=du/0$. Is this not a problem... since my $f=0$ (RHS=$0$). – snowman Nov 8 '15 at 16:56
• @snowman I think you will confuse yourself trying to write it that way. You just need to know that if $u_t(t,x) + v(t,x) u_x(t,x) = f(t,x)$ then the characteristics satisfy the ODE $x'(t)=v(t,x(t))$ and $u(t,x(t))$ satisfies the ODE $\frac{du}{dt}(t,x(t)) = f(t,x(t))$. So because your $f$ is zero and your $v$ is $x^3$, your $u$ is constant along the solutions to $x'=x^3$. – Ian Nov 8 '15 at 17:30
• @Ian OK I have edited my post and tried to do a full solution but I am stuck on the last bit. Please have a look and tell me what to do!! – snowman Nov 8 '15 at 19:46 | 2019-07-20T05:37:07 | {
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https://math.stackexchange.com/questions/1898825/is-this-a-new-method-for-finding-powers/1898843 | # Is this a new method for finding powers?
Playing with a pencil and paper notebook I noticed the following:
$x=1$
$x^3=1$
$x=2$
$x^3=8$
$x=3$
$x^3=27$
$x=4$
$x^3=64$
$64-27 = 37$
$27-8 = 19$
$8-1 = 7$
$19-7=12$
$37-19=18$
$18-12=6$
I noticed a pattern for first 1..10 (in the above example I just compute the first 3 exponents) exponent values, where the difference is always 6 for increasing exponentials. So to compute $x^3$ for $x=5$, instead of $5\times 5\times 5$, use $(18+6)+37+64 = 125$.
I doubt I've discovered something new, but is there a name for calculating exponents in this way? Is there a proof that it works for all numbers?
There is a similar less complicated pattern for computing $x^2$ values.
• As a rule of thumb, casual mathematicians make no discoveries. You are using the finite-differences approach. For all $k$, $\Delta_kx^k=k!$. – Yves Daoust Aug 21 '16 at 10:06
• @YvesDaoust A re-discovery is a discovery just the same. – 6005 Aug 21 '16 at 10:07
• @6005: I wouldn't qualify a re-discovery as "new". – Yves Daoust Aug 21 '16 at 10:09
• I don't care if it's new or not, that's an awesome discovery! – Vincent Aug 21 '16 at 11:25
• At school I used to use this to draw parabolas. The teacher wanted us to make big tables of values, but that was too slow and too boring for me. When I noticed the pattern I thought I could derive something useful from it, I only got $(x+a)^2=x^2+2xa+a^2$, haha – Oriol Aug 21 '16 at 15:39
It's not something new, but for your discovery I applaud. This procedure is called the method of successive differences, and you can show that for every power the successive difference appears.
Let us say you have a sequence: $$1^3 \quad2^3\quad 3^3\quad 4^3\quad \ldots$$
Note that $x^3-(x-1)^3 = 3x^2-3x+1$. So we'll get a new sequence at the bottom: $$7 \quad 19\quad 37\quad 61\quad \ldots$$ Now, note that $3x^2-3x+1-(3(x-1)^2-3(x-1)+1) = 6(x-1)$. Hence, we'll get another series: $$0 \quad6\quad 12\quad 18\quad\ldots$$ Now, note that $6(x-1)-6((x-1)-1) = 6$! Now, the new sequence is: $$6\quad 6\quad 6\quad 6\quad 6\quad ...$$ So $6$ appears as the final difference! This shows the power of algebra. As an exercise, do this for $x^4$. See the pattern of the number at the end, and if you can say something for $x^n$.
The reason, as you can see, is that at each line above, the degree of the polynomial $f(x)-f(x-1)$ decreases by $1$. Hence, at the end of three lines, you are only going to get a constant polynomial.
• This is exactly the same method Charles Babbage's Difference Engine used to calculate powers in Taylor series for trigonometric and other functions. – Parcly Taxel Aug 21 '16 at 10:22
• @ParclyTaxel Thank you for the information. Do you know about the final constant at the end for general $n$? (I don't) – астон вілла олоф мэллбэрг Aug 21 '16 at 10:39
• The final constant is the factorial of the starting power. – Parcly Taxel Aug 21 '16 at 10:44
• Wow, that's nice. Thank you again. – астон вілла олоф мэллбэрг Aug 21 '16 at 10:47
• +1 for applauding the OPs rediscovery. (It's one lots of budding mathematicians - including me years ago - make often when young.) – Ethan Bolker Aug 21 '16 at 13:18
What you have discovered is a finite difference calculation. For any function $f$, in this case the third-power function $$f(n) = n^3$$ we can define the forward difference, or forward discrete derivative: $$\Delta f(n) = f(n+1) - f(n) = 3n^2 + 3n + 1$$ Likewise, \begin{align*} \Delta \Delta f(n) = \Delta^2 f(n) &= 6n+ 6 \\ \Delta^3 f(n) &= 6 \\ \Delta^4 f(n) &= 0. \end{align*}
Your computation, $$5^3 = 64 + 37 + 18 + 6$$ is the statement $$f(5) = f(4) + \Delta f(3) + \Delta^2 f(2) + \Delta^3 f(1),$$ or more generally $$f(n) = f(n-1) + \Delta f(n-2) + \Delta^2 f(n-3) + \Delta^3 f(n-4).$$ This is one discrete analogue of Taylor series (the more common analogue is Newton's series). The reason it works is that, for $f(n) = n^3$, $\Delta^4$ and beyond are all zero. So the summation stops once we get to $\Delta^3$.
For a little bit more, see the answer "General method for indefinite summation" which explains how exactly this representation using forward differences allows you to easily find the formula for indefinite summation of powers. Applied to your case you get:
0, 1, 8, 27
1, 7, 19
6, 12
6
and hence:
$n^3 = 0 \binom{n}{0} + 1 \binom{n}{1} + 6 \binom{n}{2} + 6 \binom{n}{3}$.
which immediately gives: $\def\lfrac#1#2{{\large\frac{#1}{#2}}}$
$\sum_{k=0}^{n-1} k^3 = 0 \binom{n}{1} + 1 \binom{n}{2} + 6 \binom{n}{3} + 6 \binom{n}{4} = n\lfrac{n-1}{2}(1+\lfrac{n-2}{3}(6+\lfrac{n-3}{4}(6)))$
$\ = \lfrac{n^2 (n-1)^2}{4}$.
and then, if you prefer the indices to end at $n$:
$\sum_{k=1}^n k^3 = \lfrac{(n+1)^2 n^2}{4}$.
As you can see, hardly any computation was necessary to get this result!
• This is just beautiful. I had read a similar way to calculate the sum $\sum_{k=1}^n k^N$ for all positive numbers $N$. It involves integration. – астон вілла олоф мэллбэрг Aug 22 '16 at 23:34
• @астонвіллаолофмэллбэрг: Thank you! If you count the number of operations you need, you will find that many other methods (including the integration-based method) are vastly more inefficient ($O(d^3)$ instead of $O(d^2)$ where $d$ is the degree), because they require first computing the formulae for all smaller $d$, each of which takes $O(d^2)$ time. The reason this method does not face that problem is that the binomial coefficients form the natural basis so that difference and summation are simply shift operations on the coefficients. – user21820 Aug 23 '16 at 6:57
• Oh, that is wonderful. Thank you for this information, @user21820. – астон вілла олоф мэллбэрг Aug 23 '16 at 8:18
To test your hypothesis you could work out the form of the differences from the first few cases. \begin{align*} 1^{3}-0^{3}&=1\\ 2^{3}-1^{3}&=7\\ 3^{3}-2^{3}&=19\\ 4^{3}-3^{3}&=37 \end{align*} For example rewrite out $(37-19)-(19-7)=18-6=6$ as: \begin{align*} \{(4^{3}-3^{3})-(3^{3}-2^{3})\}&-\{(3^{3}-2^{3})-(2^{3}-1^{3})\}\\ &=(4^{3}-2\cdot3^{3}+2^3)-(3^{3}-2\cdot2^{3}+1^{3})\\ &=4^{3}-3\cdot3^{3}+3\cdot2^3-1^{3}\qquad (\star)\\ &=6 \end{align*} So you have to find the difference of two differences to get to $6$ (this is called a finite difference pattern, and you have to iterate twice to get the result of $6$ for all such differences, any further iteration ending in a $0$). Now check that pattern $(\star)$ holds in general for some integer $k\ge3$: \begin{align*} k^{3}&-3\cdot(k-1)^{3}+3\cdot(k-2)^3-(k-3)^{3}\\ &=k^{3}-3(k^2-3k^2+3k-1) +3(k^3-2\cdot3k^2+2^2\cdot3k-2^3) -(k^3-3\cdot3k^2+3^2\cdot3k-3^3)\\ &=\ \ k^3\\ &\ -3k^3\ +\ 9k^2\ -\ 9k\ +\ 3\\ &\ +3k^3-18k^2+36k-24\\ &\ \ -k^3\ +\ \ 9k^2-27k+27\\ &=6 \end{align*}
• As an aside, from this you could deduce $\sum_{j=0}^{3}(-1)^{j}\binom{3}{j}(k-j)^3=6$. Similar binomial power sum identities can be found for the other finite difference calculations concerning $x^4$, $x^5$, and so on. – Daniel Buck Aug 25 '16 at 23:35 | 2019-04-22T16:45:32 | {
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https://math.stackexchange.com/questions/1467648/how-can-i-find-the-cdf-and-pdf-of-y | # How can I find the CDF and PDF of $Y$?
Problem
Let $X$ be a $\operatorname{Uniform}(0,1)$ random variable, and let $Y=e^{-X}$.
Find the CDF of $Y$.
Find the PDF of $Y$.
Find $\mathbb E[Y]$.
My problem
If I solve for the range of $y$ I get $\left(1, \frac 1e \right)$, but because $Y$ is not an increasing function, my second bound is smaller than my first. I am really confused as to how I would be able to solve for the CDF and PDF in this case... Any help would be greatly appreciated.
For $y$ in the interval $(1/e,1)$ we have $$\small F_Y(y)=\Pr(Y\le y)=\Pr(e^{-X}\le y)=\Pr(-X\le \ln y)=\Pr(X\ge -\ln y)=1-(-\ln y)=1+\ln y.$$ For completeness, we add that $F_Y(y)=0$ if $y\le 1/e$, and $F_Y(y)=1$ for $y\ge 1$. Note, in the above calculation, the switch in the direction of the inequality, when $\Pr(-X\le \ln y)$ was replaced by $\Pr(X\ge -\ln y)$.
Now the pdf is easy to find. For $E(Y)$, we can use the newly found pdf of $Y$, or else bypass the distribution of $Y$ entirely, computing $\int_0^1 e^{-x}\,dx$.
• Yes, I computed the cdf only for "interesting" $y$. But the cdf is defined for all $y$. If $y\gt 1$, then for $Y\le y$, so $F_Y(y)=\Pr(Y\le y)=1$. And if $y\lt 1/e$, we cannot have $Y\le y$, so $F_Y(y)=0$. – André Nicolas Oct 6 '15 at 22:01 | 2019-08-23T04:42:05 | {
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https://math.stackexchange.com/questions/2590718/what-is-the-probability-distribution-of-the-points-obtained-by-player-a-in-this | # What is the probability distribution of the points obtained by player A in this game?
The game is as follows:
• Two players A and B play consecutive rounds, and the winner of each round obtains one point.
• Each round is independent of the others, and player A has probability $p$ of winning, player B has probability $1-p$.
• The game ends when one of the players obtain N points.
If $n_A$ is the number of points obtained by player A, what is the value of the probability $P(n_A=i) ~,~i=0,1,...,N$? i.e. what is the probability mass function of $n_A$?
Background:
I'm currently learning probability on my own, using the textbook A First Course in Probability - Sheldon Ross, 8th edition". When doing the self-test exercise 4.11, I had the idea of generalizing the problem to the one I posted here.
Similar, but different question:
## 3 Answers
You just need to separate the winning state where the points will be N from the other states.
Define $f(N,k,a) = \binom {N+k-1}{k}a^k$ as an auxiliary function to simplify the following.
We can write the points distribution for player $A$ $$n_A(k) = \left\{ \begin{array}{ll} q^Nf(N,k,p) & k={0 \dots (N-1)} \\ p^N \sum_{k=0}^{N-1} f(N,k,q) & k=N \end{array} \right.$$
simply swap $p$ and $q$ for player $B$.
• Hey, Karafka. Thank you for your time. Am I right in assuming that $q=1-p$ and $n_A(k)$ in your notation means $P(n_A=k)$ in mine? – JLagana Jan 3 '18 at 22:28
• Also, could you elaborate on why does the $f$ function use a combination from $N+k-1$ terms? I was expecting $N+k$ terms instead. – JLagana Jan 3 '18 at 22:30
• Yes, $q=1-p$ and $n(k)$ is the probability. $-1$ comes since the last round is won by the overall winner. Try for $p=q=\frac12$ and $N=3$. – karakfa Jan 4 '18 at 0:20
• Aha, and since $k < N$, the winner must be player B! Brilliant. I'm still digesting the rest of the answer, thanks for now :) – JLagana Jan 4 '18 at 0:24
• I understand your answer now, and it does seem correct. I upvoted you, but I feel like the community would benefit from a more pedagogic explanation. I answered my own question, inspired by your answer. – JLagana Jan 4 '18 at 17:17
First, Let's calculate $P(n_{A}=i,n_{b}=k)$. According to binomial distribution we have $P(n_{A}=i,n_{b}=k)=C(i+k,i)p^{i}(1-p)^{k}$ then the requested probability equals $P(n_{A}=i)=\Sigma_{k=0}^{N-1}C(i+k,i)p^{i}(1-p)^{k}$
• what is $r$? Not defined. If you meant $i$, I don't think it's right. The game may take $2N-1$ rounds to finish. – karakfa Jan 3 '18 at 21:01
• Sorry :)....I edited it right now! – Mostafa Ayaz Jan 3 '18 at 21:08
Disclaimer: Karakfa's answer originally helped me solve this problem. Although he provided the correct solution, I believe it can be helpful for other users to have a more detailed, pedagogic solution to this question. What follows is my attempt on doing so.
The key to solving this question is to separate the cases when player A loses and when player A wins. These correspond to when $n_a<N$ and $n_a=N$. Let's start with the first one.
In this case, we note that since $n_a<N$, then player A lost the game and consequently player B won. This means that $n_B=N$ (where $n_B$ denotes the points accumulated by player B in a given game), and hence $k+N$ rounds were played. Additionally, we also know that the winner of the last round was player B, since in this game the winner necessarily has to win the last round (otherwise the game would either not end at that round, or the winner already had $N$ points before that round, which means that the game would have already ended).
Now, we note that the problem of counting in how many different ways player A wins $k$ times, and player B wins $N$ times, including the last one, is equivalent to counting in how many ways we can select $k$ games, among $N+k-1$ (-1 since the last round was necessarily won by player B) to be won by player A. Putting it this way, it's clear that this is trivially ${N+k-1}\choose{k}$. For example, in the case where $k=2$ and $N=3$, there are 6 such sequences: $(AABBB), (ABABB),(ABBAB),(BAABB),(BABAB),(BBAAB)$ (here, each letter in the sequence denotes who won that particular round).
Now, back to the general case, we can see that all of the ${N+k-1}\choose{k}$ possible sequences of wins have probability $p^k(1-p)^N$ of happening, since each round is independent of the others. Finally, the sequences are mutually exclusive for one realization of the game, so we can obtain the desired probability by summing these individual, equal probabilities. This leads to:
$$P(n_a=k) = \binom{N+k-1}{k} p^k(1-p)^N \quad,\quad\text{if}~k<N.$$
Now, for the case where player A wins, i.e. when $k=N$, then we know that $n_b<N$. This time, we have to consider all the sequences where player A won $N$ times and player B won $0,1,2,...,N-1$ times.
For a given $n_b=j,j<N$, there are $\binom{N+j-1}{j}$ possible sequences where player A wins the last round (similar argument as before). All of them are mutually exclusive and have the same probability: $(1-p)^jp^N$. Therefore, $P(N_b=j)=\binom{N+j-1}{j}(1-p)^j p^N,j<N$. To find $P(n_a=N)$, we simply have to sum the probabilities of the mutually exclusive events $n_b=j$, for $j=0,1,2,...N-1$.
$$P(n_a=N)=\sum_{j=0}^NP(n_b=j)$$ $$= \sum_{j=0}^N\binom{N+j-1}{j}(1-p)^jp^N$$ $$= p^N\sum_{j=0}^N\binom{N+j-1}{j}(1-p)^j$$
Finally, putting both results together yields Karakfa's answer:
$$P(n_a=k) = \left\{\begin{matrix} \binom{N+k-1}{k}p^k(1-p)^N & \text{if }k<N\\ \\ p^N\sum_{j=0}^N\binom{N+j-1}{j}(1-p)^j & \text{if }k=N \end{matrix}\right.$$ | 2020-04-08T22:52:24 | {
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https://math.stackexchange.com/questions/1103703/what-does-this-symbol-mean | # What does this symbol mean?
This is from Discrete Mathematics and its Applications
What is the symbol used in 9c, 9d, 9f, 10c, 10f, 10g? I looked through the chapter section and the closest symbol I saw to this is the subset, which is
DEFINITION 3. The set $$A$$ is a subset of $$B$$ if and only if every element of $$A$$ is also an element of $$B$$. We use the notation $$A \subseteq B$$ to indicate that $$A$$ is a subset of the set $$B$$.
We see that $$A \subseteq B$$ if and only if the quantification $$\forall x(x \in A \to x \in B)$$ is true. Note that to show that $$A$$ is not a subset of $$B$$ we need only find one element $$x \in A$$ with $$x \not\in B$$. Such an $$x$$ is a counterexample to the claim that $$x \in A$$ implies $$x \in B$$.
Is it just a typo for subset? Thats what I originally thought. However via my use of an implication(trying to apply what I learn :) ), I came up with if the symbol is a typo, it will be used in a single place or the supposed actual symbol, the subset, will not be used in the surrounding proximity(page). If I assumed the hypothesis to be true, then my conclusion and my implication is false because the subset does appear in the near proximity(9g) and this symbol is used in multiple locations(all the ones I described). Therefore the hypothesis is false(reached a contradiction), and the symbol is not a typo. Is that correct logic?
• It should be noted that some authors use $\subset$ for $\subseteq$ and denote $\subsetneq$ for a proper subset.
– Hugh
Jan 14, 2015 at 7:58
The textbook is not making typos. The symbol $\subset$ is called a proper subset. For example, $$[0,1] \subset [0,2],$$ but $$[0,2] \not\subset [0,2].$$
However, the symbol $\subseteq$ means that a set can be contained in itself as well, for example, $$[0,1] \subseteq [0,2]$$ and $$[0,2] \subseteq [0,2].$$
This page describes a little more on the difference between subset and proper subset.
• For this [0,1]⊆[0,2] meaning [0,1] is a subset of [0,2], doesn't [0,2] have to have every element in [0,1]? Which it doesn't because of the 1 in [0,1]? Jan 14, 2015 at 6:49
• I am describing a continuous interval $[0,2]$, which does indeed contain the subinterval $[0,1]$; hence, $[0,1] \subseteq [0,2]$. (Note that an interval is a type of set on a real line.) I am not describing a collection set of elements $\{0,2\}$, which does not contain the $1$ as you said; hence, $\{0,1\} \not\subseteq \{0,2\}$. Jan 14, 2015 at 6:56
• oh thanks that makes sense. By the way, how did you embed that subset symbol into your comment? Every time, I have to copy and paste.... Jan 14, 2015 at 7:07
• I used the MathJax code of \subset for "$\subset$" and \subseteq for "$\subseteq$". Please see meta.math.stackexchange.com/questions/5020/… for how to use the MathJax code needed to type math symbols on this site. Jan 14, 2015 at 7:14
The symbol $\subset$ denotes proper subset.
$$A\subset B \iff \forall x (x \in A \to x\in B)\wedge \exists x (x\in B \wedge x\not\in A)$$
Which means that all the elements in the proper subset are elements of the set, and there are elements in the set which are not in the proper subset. A set cannot be a proper subset of itself.
Where as the symbol $\subseteq$ denotes the subset or equivalent, commonly referred to as just subset. A set can be its own subset.
$$A\subseteq B \iff \forall x (x\in A\to x\in B)$$
Likewise the symbol $\supset$ and $\supseteq$ denote proper superset and superset.
$$A\supseteq B \iff \forall x (x\in A\leftarrow x\in B)$$
$$A\supset B \iff \forall x (x \in A \leftarrow x\in B)\wedge \exists x (x\in A \wedge x\not\in B)$$
And of course we have set equivalence, $\equiv$
$$A\equiv B \iff \forall x (x\in A \leftrightarrow x\in B)$$
These symbols are analogous to the order symbols $<$, $\leq$, $\geq$, $>$, and $=$
• So proper subset is basically subset but it can't be the same set? Jan 27, 2015 at 6:11
• Yes, just so, @committedandroider Jan 27, 2015 at 11:34 | 2022-05-18T05:26:38 | {
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http://mathhelpforum.com/statistics/186791-easy-prob-question-but-i-cant-see-print.html | # Easy prob question but I can't see it :(
• August 27th 2011, 06:43 AM
perla
Easy prob question but I can't see it :(
A hunter has 2 hunting dogs.
One day, on the trail of some animal, the hunter comes to a place where the road diverges into two paths. He knows that each dog, independent of the other, will choose the correct path with probability p. The hunter decides to let each dog choose a path, and if they agree, take that one, and if they disagree, to randomly pick a path. Is his strategy better than just letting one of the two dogs decide on a path?
• August 27th 2011, 08:50 AM
alexmahone
Re: Easy prob question but I can't see it :(
Quote:
Originally Posted by perla
A hunter has 2 hunting dogs.
One day, on the trail of some animal, the hunter comes to a place where the road diverges into two paths. He knows that each dog, independent of the other, will choose the correct path with probability p. The hunter decides to let each dog choose a path, and if they agree, take that one, and if they disagree, to randomly pick a path. Is his strategy better than just letting one of the two dogs decide on a path?
$Probability=p^2+2p(1-p)\times 0.5=p^2+p-p^2=p$
So, his strategy is exactly as good as just letting one of the two dogs decide on a path.
• August 27th 2011, 09:04 AM
MAX09
Re: Easy prob question but I can't see it :(
interesting question. though i'm unable to conclude a way to answer your question, i have a feeling that this could be more than probability. it appeared as though it was a decision making problem... have listed the possibilities in the image below....
• August 27th 2011, 10:01 PM
perla
Re: Easy prob question but I can't see it :(
Quote:
Originally Posted by alexmahone
$Probability=p^2+2p(1-p)\times 0.5=p^2+p-p^2=p$
So, his strategy is exactly as good as just letting one of the two dogs decide on a path.
Thank you for your reply. Why is there a 2 times p(1-p) and times 0.5?
Regards
• August 27th 2011, 11:06 PM
CaptainBlack
Re: Easy prob question but I can't see it :(
Quote:
Originally Posted by perla
Thank you for your reply. Why is there a 2 times p(1-p) and times 0.5?
Regards
Because there are two ways of the dogs disagreeing on the path dog1 could be right and 2 wrong or dog1 could be wrong and 2 right.
CB
• August 27th 2011, 11:10 PM
CaptainBlack
Re: Easy prob question but I can't see it :(
Quote:
Originally Posted by MAX09
interesting question. though i'm unable to conclude a way to answer your question, i have a feeling that this could be more than probability. it appeared as though it was a decision making problem... have listed the possibilities in the image below....
It is a pure probability problem, the decision rule is given the only question left is what is the probability of a correct outcome.
Your third possibility is redundant it is already present as leaf nodes on the first two branches.
CB
• August 28th 2011, 03:22 AM
MAX09
Re: Easy prob question but I can't see it :(
i thought the problem was to decide which of the two methods was better. now i see it..
the third offshoot from the centre was to analyse the second method by letting one of them pick a given way...
As alexmahone posted it, both methods turned out to have the same outcome.
A thought-provoking problem. | 2016-08-29T09:12:16 | {
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http://mathhelpforum.com/discrete-math/175926-gamma.html | 1. ## Gamma
if we have gamma_1 be the set of V_< -sentences that hold in every finite model of
T_LO .Let gamma_2 be the set of sentences saying that there exist at least n elements for each n=1,2,3,... Let T_LOF denote the closure of gamma_1 (union) gamma_2 . Show that T_FLO is a theory.
where T_LO denote the theory of linear orders
2. What are the definitions of closure and theory? Also, since there are not many model theorists here, it would help if you could indicate the context, i.e., the topic or chapter this problem is coming from and any theorems that you suspect may be relevant. It may also be useful to know the level of the course (undergraduate, introductory, graduate for logic majors, etc.).
3. theory: For any V-structure M, the theory of M, denoted Th(M), is
the set of all V-sentences ϕ such that M |= ϕ.
closure:the closure of gamma is the set of sentences {ϕ| gamma |= ϕ.
This |= means gamma models ϕ. where gamma is a set of sentences.
Yes, it is undergraduate course. the chapter is chapter 5 in ( A first course in logic ) book. and It is Q 5.6 (a)
the chapter is about first order theories | 2013-05-23T12:50:26 | {
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https://math.stackexchange.com/questions/2143458/how-many-disproofs-are-needed-to-complete-an-implication-graph | # How many disproofs are needed to complete an implication graph?
I have 4 possible properties of a kind of mathematical object, and I want to prove their implications. I've proved $P_1 \rightarrow P_2$, $P_2 \rightarrow P_3$, and thus also $P_1 \rightarrow P_3$.
Now, I think no other implications are possible between them, and in fact can disprove all other implications. What is the least number of implications I need to disprove to disprove all of them?
Is there a general algorithm for such question? Given a directed graph of implications between properties, and asking how many implications must be disproved in order to show that no other arrows exist in the graph? Well, yes, if the algorithm simply checks all possibilities exhaustively. But is there a more efficient one?
In response to bof, Here I only ask about disproving the arrows of the form $P_n \rightarrow P_m$, not of the more general form $P_n \wedge P_m \rightarrow P_k$, though that could make a nice extension to the problem.
After some looking, I think a formal way to state the question is: given a directed graph $G$ with vertices $V$, denote its transitive closure to be $G_t$. Find a minimal (smallest in number) set of "forbidden" directed edges, such that $G_t$ is the unique transitive directed graph with vertices $V$, that contains $G$, and does not contain any of the forbidden directed edges.
• Nice problem, are you the author? – dtldarek Feb 14 '17 at 8:25
• @dtldarek The question came to me when I was actually trying to do exactly what I said in the problem: I have 4 properties of functions, and wanted to completely prove/disprove their implication relations. – MaudPieTheRocktorate Feb 14 '17 at 13:47
We may assume that equivalent statements have been identified, so that the relation of implication is antisymmetric as well as being reflexive and transitive, i.e., it's a partial order. Thus the problem may be stated as follows:
Let $V$ be a finite set partially ordered by a relation $R.$ What is the smallest set $D\subseteq(V\times V)\setminus R$ such that every transitive relation $T\subseteq V\times V,$ which contains $R$ and is disjoint from $D,$ is equal to $R?$
Clearly, any such set $D$ must contain any pair $(a,b)\in(V\times V)\setminus R$ such that $R\cup\{(a,b)\}$ is a transitive relation; otherwise $T=R\cup\{(a,b)\}$ would be a transitive relation containing $R$ and disjoint from $D$ and unequal to $R.$ Let $D_0$ be the set of all such pairs:
$$D_0=\{(a,b)\in(V\times V)\setminus R:R\cup\{(a,b)\}\text{ is transitive}\}.$$
I claim that $D_0$ is the desired set. Let $T\subseteq V\times V$ be a transitive relation such that $R\subseteq T$ and $T\cap D_0=\emptyset;$ I have to show that $T=R.$ Assume for a contradiction that $T\setminus R\ne\emptyset.$ Choose a minimal (with respect to the partial order $R$) element $a\in V$ such that $(a,y)\in T\setminus R$ for some $y,$ and then choose a maximul element $b\in V$ such that $(a,b)\in T\setminus R.$ It is easy to see that $R\cup\{(a,b)\}$ is transitive, whence $(a,b)\in D_0,$ contradiction our assumption that $T\cap D_0=\emptyset.$
In other words, writing $x\le y$ for $(x,y)\in R,$ we can describe $D_0$ as the set of ordered pairs $(a,b)\in(V\times V)$ satisfying the conditions:
$$a\not\le b$$ $$\forall x\in V\ (x\lt a\implies x\le b);$$ $$\forall x\in V\ (b\lt x\implies a\le x.$$
For $a\in V$ let $F(a)$ be the set of all maximal elements of $\{x:x\lt a\},$ and let $G(a)$ be the set of all minimal elements of $\{x:b\lt x\}.$ If we wish to compute $D_0,$ we might start by computing the sets $F(a)$ and $G(a)$ for each $a\in V;$ then, for each ordered pair $a\not\le b,$ we just have to check the inequalities $x\le b$ for $x\in F(a),$ and $a\le x$ for $x\in G(b).$
Example. If $V=\{v_0,v_1,v_2,v_3\}$ and $R=\{(v_0,v_0),(v_0,v_1),(v_1,v_1),(v_2,v_2),(v_2,v_3),(v_3,v_3)\},$ then $D_0=\{(v_0,v_3),(v_1,v_0),(v_2,v_1),(v_3,v_2)\}.$
• $D_0=\{(v_0,v_3),(v_1,v_0),(v_2,v_1),(v_3,v_2)\}.$ – bof Feb 17 '17 at 23:25
• Thanks. Now I see how it works. – Fabio Somenzi Feb 18 '17 at 0:10
• Good answer. I think this allows for an algorithm in polynomial time... Can it be done in quadratic time? – MaudPieTheRocktorate Feb 18 '17 at 10:40
Let $X$ be the set of our mathematical objects. To show that $P_4$ is not implied by all of them, we would like to find an example $x$ such that $\alpha(x) \land \neg P_4(x)$ where $\alpha$ is $P_1, P_2, P_3$. However, due to the implications you have proved, it's enough to show $x$ such that $P_1(x) \land \neg P_4(x)$, that is $P_1 \not\to P_4$.
Yet, this is not all, to show that $P_4$ is independent we also need to find $x$ such that $P_4(x) \land \neg \alpha(x)$. Observe that you also have proved $\neg P_3(x) \to \neg P_2(x)$ and $\neg P_2(x) \to \neg P_1(x)$ for all $x \in X$, so it's enough to find $x$ with $P_4(x) \land \neg P_3(x)$, i.e. $P_4 \not\to P_3$.
That might seem like all, but it's not. We are forgetting $P_2 \not\to P_1$ and $P_3 \not\to P_2$ and $P_3 \not\to P_1$. For these we need an $x$ such that $P_2(x) \land \neg P_1(x)$ and some different $x$ such that $P_3(x) \land \neg P_2(x)$.
As for whether such an algorithm exists, definitely yes: for all subsets of possible implications you can check if that subset is enough to prove status of every implication. Nevertheless, that would be rather slow, so now the question is, can we make it faster? My guess is yes, but currently I have no proof.
I hope this helps $\ddot\smile$
Edit: Thanks to @bof for spoting a flaw in the algorithm I suggested.
Edit 2:
Here is a second attempt at an algorithm. I will start with a change in terminology, so that it is more concrete, and thus perhaps simpler.
Let $V = \{1,2,3,\ldots,n\}$ be the set of vertices and add a directed edge $i \to j$ if $P_i \to P_j$. Furthermore, for each vertex $v \in V$ create sets $A_v = \{i \in V \mid P_i \to P_v\}$, in particular $v \in A_v$. These sets have the property that $$P_i \to P_j \iff A_i \subseteq A_j,$$ that is $$P_i \not\to P_j \iff A_i \not\subseteq A_j \iff \exists v \in V.\ v \in A_i \land v \notin A_j.$$
Moreover, let us say that $A_i$ is a successor of $A_j$ iff $A_j \subsetneq A_i$ and for all $k$ such that $A_j \subseteq A_k \subseteq A_i$ we have $k = j$ or $k = i$ (intuitively, $A_i$ is one step ahead of $A_j$). Similarly, $A_i$ is a predecessor of $A_j$ iff $A_j$ is a successor of $A_i$ (intuitively $A_i$ is one step before $A_j$).
The algorithm:
1. Let $D$ be the set of all the pairs that were not settled yet, i.e. $$D \gets \{ (i,j) \in V \times V\mid A_i \not\subseteq A_j \land A_j \not\subseteq A_i\}$$
2. Pick any such pair $$(i,j) \gets \operatorname{any}(D)$$ Observe that $i \in A_i$, but $i \notin A_j$. We will not try to find an alternate pair with similar property.
3. Let $i'$ be any predecessor of $i$ such that $i' \notin A_j$ or $i$ if no such predecessor exists and assign $i'$ to $i$ \begin{align} &i' \gets \operatorname{any}\{k\in \operatorname{pred}(i) \mid k \notin A_j \}\\ &\mathtt{if}\ i' \neq \bot:\\ &\quad i\gets i' \end{align}
4. Do similarly with $j$, i.e. find an appropriate successor: \begin{align} &j' \gets \operatorname{any}\{k\in \operatorname{succ}(j) \mid i \notin A_k \}\\ &\mathtt{if}\ j' \neq \bot:\\ &\quad j\gets j' \end{align}
5. Repeat steps $4$ or $5$ until no further change is possible.
6. Prove $P_i \not\to P_j$ (i.e., output that pair as one of those that we need to prove).
7. Remove from $D$ any pair that is implied by $P_i \not\to P_j$.
8. Repeat from step $2$ until $D$ gets empty.
A few important points:
• Because the graph is a DAG, steps 3 and 4 cannot repeat infinitely.
• If $(i,j) \in D$, then after steps $3$ and $4$ the new pair $(i,j)$ is still in $D$.
• Suppose otherwise and let $(i',j')$ be the new pair. Please remember that $i' \in A_{i'} \subseteq A_i$ and $i' \notin A_{j'} \supseteq A_j$ because that is how that new pair was constructed. If $(i',j')\notin D$, then it had to be removed from $D$ at some point and there had to be a pair $(i'',j'')$ such that $i'' \in A_{i''} \subseteq A_{i'} \subseteq A_i$ and $i'' \notin A_{j''} \supseteq A_{j'} \supseteq A_j$, but that implies $(i,j) \notin D$.
• Thus, every time we get to step 7 at least one pair gets removed, so the algorithm finishes in finite time.
• When we finish step $5$, the pair $(i,j)$ has property that no statement $P_{i'} \not\to P_{j'}$ for some other pair $(i',j')$ could imply $P_i \not\to P_j$.
• Suppose otherwise that there is such a pair $(i', j')$. That means $i' \in A_{i'} \subseteq A_i$ and $i' \notin A_{j'} \supseteq A_j$, but that implies step $5$ could not finish because $(i',j)$ or $(i,j')$ is a valid change.
• Therefore, every pair that the algorithm outputs has to be proven.
• As the algorithm finishes only when $D$ is empty, all necessary pairs will get printed.
I hope I didn't miss anything this time.
• @bof I agree, I missed that case (and I shouldn't $\ddot\frown$). Sadly I have not enough time right now to think about it some more. Thank you for your comments! – dtldarek Feb 14 '17 at 10:36
• I added a formal statement of the problem, so that it might be easier to explain to mathematicians and computer scientists. – MaudPieTheRocktorate Feb 15 '17 at 4:19
• @bof Perhaps this one works ;-) – dtldarek Feb 15 '17 at 14:49
Here is the exhaustive algorithm implemented in Mathematica:
exhaustiveAlgorithm[vertices_, inEdges_] :=
Module[{v = vertices, i = inEdges, g, e},
g = Graph[v, i];
e = EdgeList@GraphComplement@TransitiveClosureGraph@g;
{EdgeList@TransitiveReductionGraph@g,
SelectFirst[
a \[Function]
AllTrue[Complement[e, a],
l \[Function]
AnyTrue[a,
]
Usage:
exhaustiveAlgorithm[{P1, P2, P3, P4}, {P1 -> P2, P2 -> P3}]
will return
{{P1 -> P2, P2 -> P3}, {P1 -> P4, P2 -> P1, P3 -> P2, P4 -> P3}}
telling you that you must show
$$P_1 \to P_2, P_2 \to P_3, \lnot(P_1 \to P_4), \lnot(P_2 \to P_1), \lnot(P_3 \to P_2), \lnot(P_4 \to P_3).$$
I know this doesn't answer your question, but I hope it helps. I would have written this as a comment but it did not fit in the character limit.
• It does work but is as inefficient as I feared. I tried the algorithm on a graph with 6 vertices and it used up my computer memory. – MaudPieTheRocktorate Feb 15 '17 at 4:44
• @KopaLeo You're right that it's highly inefficient. I don't have time at the moment, but I believe I can optimize this algorithm quite a lot. If it would be helpful, I can return in a few days with a better program that will handle 6 vertices easily. – diracdeltafunk Feb 15 '17 at 9:23
• It's just a puzzle. I don't really need an answer. I do feel it should have some connection with boolean satisfiability problems so maybe somehow there's an efficient algorithm that can be adapted for this. – MaudPieTheRocktorate Feb 15 '17 at 14:03 | 2019-05-27T13:27:37 | {
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https://www.physicsforums.com/threads/real-part-of-a-function.49360/ | # Real Part of a Function
1. Oct 24, 2004
### cj
I'm used to seeing a function like:
$$\textbf{f}=x+iy \text$$
where
$$i = \sqrt{-1}$$
and understanding the the real part is:
$$\text{Re[f]} = x = Acos\theta$$
What, though, is the real part of a function like, for example,
$$\textbf{f}=\sqrt{x+iy}$$
??
2. Oct 24, 2004
Staff Emeritus
Expand in Taylor's series, paying attention to powers of i, and separate terms with i out to get two series for the real and imaginary parts. This only works where the TS converges absolutely. Where is that?
3. Oct 24, 2004
### robphy
$$Re(f)=\frac{f+f^*}{2}$$
where $$f^*$$ is the complex-conjugate of f.
So, \begin{align*} Re(\sqrt{x+iy})&=\frac{(\sqrt{x+iy})+(\sqrt{x+iy})^*}{2}\\ &=\frac{\sqrt{x+iy}+\sqrt{x-iy}}{2} \end{align*}
Additionally, you could write $$x+iy$$ in polar coordinates.
Then, do the above.
4. Oct 24, 2004
### uart
Since it looks like you're assuming the x and y are real and using the notation x = A cos(theta), y= A sin(theta) then the answer is just,
sqrt(A) cos(theta / 2)
Last edited: Oct 24, 2004
5. Oct 24, 2004
### HallsofIvy
Staff Emeritus
You need to be more careful in distinguishing between the "real and imaginary parts" of the variable z= x+ iy, and the "real and imaginary parts" of the function value. Most text books write f(z)= u+ iv so that u is the real part and v the imaginary part of f: is z= x+ iy then u(x,y) and v(x,y) are real valued functions of the real variables of x and y.
6. Oct 24, 2004
### cj
If I am (and I am) going to use polar forms, the
general function:
$$\textbf{f}=x+iy$$
has the magnitude:
$$|\textbf{f}|=\sqrt{x^2+y^2}$$
with
$$x = Acos\theta \text{ and } y = Asin\theta$$
$$\text{BUT, I have the form: } \textbf{f}=\sqrt{x+iy} \text{ , NOT }\textbf{f}=x+iy$$
So how do I make use of
$$|\textbf{f}|=\sqrt{x^2+y^2}$$
since I don't have a function in the form of
$$\textbf{f}=x+iy$$
??
7. Oct 25, 2004
### HallsofIvy
Staff Emeritus
Have you simply ignored what everyone has said?
"I don't have a function in the form of f= x+ iy" is just saying you don't have the identity function: f(z)= z.
You're not going to be able to do very much is f(z)= z is the only function you know how to work with!
Suppose f(z)= z2. Then, writing z= x+iy, f(z)= (x+ iy)2=
x2+ 2ixy+ y2(i2)= (x2- y2)+ (2xy)i.
The real part of f is x2- y2 and the imaginary part of f is 2xy.
f(z)= √(z) is a little harder just because it is harder to calculate the value. It would probably be best to write z in polar form:
z= x+iy= r(cos(θ)+ i sin(θ)). The f(z)= √(z)= r1/2(cos(θ/2)+ i sin(θ/2)). The real part of f is r1/2cos(θ/2) and the real part is r1/2sin(θ/2).
8. Oct 25, 2004
### cj
Got it -- thanks!
Although I think you meant "imaginary" rather than "real"
in the following: | 2017-06-26T10:51:05 | {
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https://math.stackexchange.com/questions/1474321/what-are-all-the-finite-transitive-sets | # What are all the finite transitive sets?
In set theory, we come across the definition of an ordinal as a well-ordered transitive set. And then we show that $\omega$ and all its elements are ordinals.
My question whether there are finite transitive sets which are not ordinals. Also, whether we know all the finite transitive sets.
• An answer mentions the hereditarily finite sets. It may perhaps be useful to add that these are precisely the elements of $V_\omega=\bigcup_{n\in\mathbb N}V_n$ where $V_0=\emptyset$ and $V_{n+1}=\mathcal P(V_n)$ for all $n$. – Andrés E. Caicedo Oct 11 '15 at 12:27
• There has been some recent attention given to (combinatorial and graph theoretic properties of) these sets. Note, for instance, that $V_\omega=\bigcup_n A_n$, where $A_0=\{\emptyset\}$, and $A_{n+1}=\{\emptyset\}\cup\{x\cup\{y\}\mid x,y\in A_n\}$. If you let $a_n=|A_n|$, the sequence $a_0,a_1,a_2,\dots$ is $1,2,4,12,112,11680,\dots$ You can find some work on it and related references in Enumeration of the adjunctive hierarchy of hereditarily finite sets, by Audrito, Tomescu, and Wagner, J Logic Computation (2015) 25 (3): 943-963. – Andrés E. Caicedo Oct 11 '15 at 12:40
Yes, there are finite transitive sets which are not ordinals. For example, take $X = \{\emptyset, \{\emptyset\}, \{\{\emptyset\}\}\}$.
You can check quickly that $X$ is transitive, but it is not an ordinal, since it is not linearly ordered under $\in$. We have $\emptyset\not\in \{\{\emptyset\}\}$ and $\{\{\emptyset\}\}\not\in\emptyset$. Of course, the only ordinal of size $3$ is $3 = \{0,1,2\} = \{\emptyset,\{\emptyset\},\{\emptyset,\{\emptyset\}\}\}$.
As for your second question, every set has a transitive closure, and the transitive closure of $X$ is finite if and only if $X$ is hereditarily finite. So the finite transitive sets are exactly the transitive closures of hereditarily finite sets. I don't think you're going to find a more concrete characterization than that, unfortunately. | 2019-05-23T15:11:21 | {
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https://math.stackexchange.com/questions/4485797/limit-of-function-near-a-given-point | # limit of function near a given point
Find the limit of the given function as $$x\to -2$$ without using L’Hôpital’s Rule:
$$\displaystyle \lim_{x \to -2} \frac{\sqrt[3]{x-6} + 2}{x^3 + 8}$$
I used the identity: $$x^3 + 2^3 = (x+2)(x^2 -2x +4)$$ at the denominator. Then i tried to use the same identity at the numerator and didnt succeed. So i tried to expand the numerator, but i got different answer by checking the limits from both of the sides near $$x = -2$$. The answer is $$\frac{1}{144}$$ which is an approximation of the limit at the given point.
Any suggestions and support would be kindly appreciated.
Edit: Just watching it from the side and I can rewrite the 2 at the numerator as $$\sqrt[3]2$$. Yet, its not helping a lot.
As you have noticed that $$x^3 + 2^3 = (x+2)(x^2 -2x +4)$$, we first calculate the limit $$\lim_{x\to-2}\frac{\sqrt[3]{x-6} + 2}{x+2}.$$ Let $$t=\sqrt[3]{x-6}$$, then $$t\to-2$$ as $$x\to-2$$ and $$x+2=t^3+8=(t+2)(t^2-2t+4)$$, so $$\lim_{x\to-2}\frac{\sqrt[3]{x-6} + 2}{x+2}=\lim_{t\to-2}\frac{t + 2}{(t+2)(t^2-2t+4)}=\lim_{t\to-2}\frac1{t^2-2t+4}=\frac1{12}.$$
Therefore, $$\lim_{x \to -2} \frac{\sqrt[3]{x-6} + 2}{x^3 + 8}=\lim_{x\to-2}\frac{\sqrt[3]{x-6} + 2}{x+2}\cdot\lim_{x\to-2}\frac1{x^2-2x+4}=\frac1{12}\cdot\frac1{12}=\frac1{144}.$$
• The first part was the one i was looking for before applying the limit arithmetic's, this helps a lot. Thank you for answer. Jul 4 at 6:03
• @Blurred_Vision I'm glad that I can help. You are welcome.
– Feng
Jul 4 at 6:03
Hint:
First calculate $$\lim_{x\to-2}(x^2-2x+4)$$
Then set $$\sqrt[3]{x-6}+2=y\implies x=6+(y-2)^3=-2+12y-6y^2+y^3$$
$$\implies x+2=?$$
$$y\to0$$ as $$x\to-2$$
Can you take it from here?
• Thank you sir. I really liked the trick you did. Jul 4 at 6:00
HINT: Let $$a:=\sqrt[3]{x-6}$$. Then \begin{align} x+2&=a^3+8\\ &=(a+2)(a^2-2a+4). \end{align} | 2022-08-19T18:23:24 | {
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http://www.mathworks.com/help/matlab/ref/rat.html?nocookie=true | Accelerating the pace of engineering and science
rat
Rational fraction approximation
Description
example
R = rat(X) returns the rational fraction approximation of X to within the default tolerance, 1e-6*norm(X(:),1). The approximation is a string containing the truncated continued fractional expansion.
example
R = rat(X,tol) approximates X to within the tolerance, tol.
example
[N,D] = rat(___) returns two arrays, N and D, such that N./D approximates X, using any of the above syntaxes.
Examples
expand all
Approximate Value of π
Approximate the value of π using a rational representation of the quantity pi.
The mathematical quantity π is not a rational number, but the quantity pi that approximates it is a rational number since all floating-point numbers are rational.
Find the rational representation of pi.
```format rat
pi
```
```ans =
355/113 ```
The resulting expression is a string. You also can use rats(pi) to get the same answer.
Use rat to see the continued fractional expansion of pi.
`R = rat(pi)`
```R =
3 + 1/(7 + 1/(16))
```
The resulting string is an approximation by continued fractional expansion. If you consider the first two terms of the expansion, you get the approximation $3+\frac{1}{7}=\frac{22}{7}$, which only agrees with pi to 2 decimals.
However, if you consider all three terms printed by rat, you can recover the value 355/113, which agrees with pi to 6 decimals.
$3+\frac{1}{7+\frac{1}{16}}=\frac{355}{113}\text{\hspace{0.17em}}.$
Specify a tolerance for additional accuracy in the approximation.
```R = rat(pi,1e-7)
```
```R =
3 + 1/(7 + 1/(16 + 1/(-294))) ```
The resulting approximation, 104348/33215, agrees with pi to 9 decimals.
Express Array Elements as Ratios
Create a 4-by-4 matrix.
```format short;
X = hilb(4)```
```X =
1.0000 0.5000 0.3333 0.2500
0.5000 0.3333 0.2500 0.2000
0.3333 0.2500 0.2000 0.1667
0.2500 0.2000 0.1667 0.1429```
Express the elements of X as ratios of small integers using rat.
`[N,D] = rat(X)`
```N =
1 1 1 1
1 1 1 1
1 1 1 1
1 1 1 1
D =
1 2 3 4
2 3 4 5
3 4 5 6
4 5 6 7```
The two matrices, N and D, approximate X with N./D.
View the elements of X as ratios using format rat.
```format rat
X```
```X =
1 1/2 1/3 1/4
1/2 1/3 1/4 1/5
1/3 1/4 1/5 1/6
1/4 1/5 1/6 1/7 ```
In this form, it is clear that N contains the numerators of each fraction and D contains the denominators.
Input Arguments
expand all
X — Input arraynumeric array
Input array, specified as a numeric array of class single or double.
Data Types: single | double
Complex Number Support: Yes
tol — Tolerancescalar
Tolerance, specified as a scalar. N and D approximate X, such that N./D - X < tol. The default tolerance is 1e-6*norm(X(:),1).
Output Arguments
expand all
R — Continued fractionstring
Continued fraction, returned as a string. The accuracy of the rational approximation via continued fractions increases with the number of terms.
N — Numeratornumeric array
Numerator, returned as a numeric array. N./D approximates X.
D — Denominatornumeric array
Denominator, returned as a numeric array. N./D approximates X.
expand all
Algorithms
Even though all floating-point numbers are rational numbers, it is sometimes desirable to approximate them by simple rational numbers, which are fractions whose numerator and denominator are small integers. Rational approximations are generated by truncating continued fraction expansions.
The rat function approximates each element of X by a continued fraction of the form
$\frac{N}{D}={D}_{1}+\frac{1}{{D}_{2}+\frac{1}{\left({D}_{3}+...+\frac{1}{{D}_{k}}\right)}}\text{\hspace{0.17em}}.$
The Ds are obtained by repeatedly picking off the integer part and then taking the reciprocal of the fractional part. The accuracy of the approximation increases exponentially with the number of terms and is worst when X = sqrt(2). For X = sqrt(2) , the error with k terms is about 2.68*(.173)^k, so each additional term increases the accuracy by less than one decimal digit. It takes 21 terms to get full floating-point accuracy. | 2014-10-21T00:35:47 | {
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http://inspirationskitchenandhome.co.uk/bp7wz.php?id=230bd5-polar-graph-equations | polar graph equations
I included t-charts in both degrees and radians. Examples. To graph in the rectangular coordinate system we construct a table of$\,x\,$and$\,y\,$values. … Polar Coordinates: Graphs. To graph in the rectangular coordinate system we construct a table of $$x$$ and $$y$$ values. Trigonometry: Wave Interference example.
Get the free "Polar Graphs" widget for your website, blog, Wordpress, Blogger, or iGoogle. If not, here are the tests. Graphing Polar Equations – Video sign between a and b is minus. Graphing Polar Equations by Plotting Points. To graph polar functions you have to find points that lie at a distance #r# from the origin and form (the segment #r#) an angle #theta# with the #x# axis.
Due to the circular aspect of this system, it's easier to graph polar equations using this method. Writing the function equation for a rose using the sine, you have r = asinbθ. However, we need to understand the polar coordinate system and how to plot points for graphing polar equations. The rose shown in the figure has five petals, and its function equation is r … By the end of this section, you will be able to: Test polar equations for symmetry. Polar Equations and Their Graphs ... After working with several polar graphs and observing their general shape, periodicity, and symmetry, it was quite surprising to end up with the graph of a straight line. The graph below appears to be the graph of the equation y = x+1. to save your graphs! To graph in the polar coordinate system we construct a table of$\,\theta \,$and$\,r\,$values. If an equation fails a symmetry test, the graph may or may not exhibit symmetry. Sometimes it is more convenient to use polar equations: perhaps the nature of the graph is better described that way, or the equation is much simpler.
How to sketch a simple polar curve by plotting points. Lines: Two Point Form example. Learning Objectives. Added Aug 12, 2011 by Youngwolf in Mathematics. Identify the type of polar equation . Parabolas: Standard Form + Tangent example. Trigonometry: Period and Amplitude example. Parabolas: Vertex Form example. HOME ABOUT PRODUCTS BUSINESS RESOURCES Wolfram|Alpha Widgets Overview Tour Gallery Sign In. The method of point-by-point is used here. To express these functions you use the polar coordinate system. Note that the values of theta are. Planets follow elliptical paths as they orbit around the Sun. Send feedback|Visit … When b is an odd number, the number of petals is b. a b = 1 2 Since the ratio is less than 1, it will have both an inner and outer loop. The initial graph looks like this. First, here is a table of some of the more common polar graphs. There are three symmetry tests that indicate whether the graph of a polar equation will exhibit symmetry. Graph polar equations by plotting points. + New Blank Graph. The graph below appears to be the graph of the equation y = x+1. However, changing the values of theta to be yields the following graph. So, join me on a Polar Graph hunt! be along the polar axis since the function is cosine and will loop to the left since the . Graph polar equations by plotting points. Polar equations may be graphed by making a table of values for $\theta$ and $r$. Polar Equations and Their Graphs ... After working with several polar graphs and observing their general shape, periodicity, and symmetry, it was quite surprising to end up with the graph of a straight line. Also remember that there are three types of symmetry - y-axis, x-axis, and origin. We will derive formulas to convert between polar and Cartesian coordinate systems. | 2021-03-05T09:58:57 | {
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https://web2.0calc.com/questions/what-is-5-m-squared-converted-to-cm-squared | +0
# what is 5 m squared converted to cm squared?
0
378
7
what is 5 m squared converted to cm squared?
Guest Oct 22, 2014
#2
+85644
+10
There are 100cm in a meter. So, 1 square meter would be 1m x 1m = 100cm x 100cm = 10000cm2
So, 5 square meters would be 5 times as much = 50,000cm2
[ To see this more clearly, imagine a suare with a side of √5m. Then the area would be 5m2. So, in cm, the side would be 100√5 cm. And (100√5 cm)2 = 50,000cm2 ]
CPhill Oct 22, 2014
Sort:
#1
+92194
+10
Well there is 100 cm in a metre
so
there is 100^2cm^2 in a m^2
10 000cm^2=1m^2
so
50 000cm^2 in 5m^2
Melody Oct 22, 2014
#2
+85644
+10
There are 100cm in a meter. So, 1 square meter would be 1m x 1m = 100cm x 100cm = 10000cm2
So, 5 square meters would be 5 times as much = 50,000cm2
[ To see this more clearly, imagine a suare with a side of √5m. Then the area would be 5m2. So, in cm, the side would be 100√5 cm. And (100√5 cm)2 = 50,000cm2 ]
CPhill Oct 22, 2014
#3
+19206
+8
What is 5 m squared converted to cm squared ?
\$\$\boxed{5\ m^2 = 50000\ cm^2}\$\$
heureka Oct 22, 2014
#4
+92194
+5
Thanks Heureka,
I never stop being surprised by what the web2.0 calc can do.
Thanks for showing us.
Melody Oct 22, 2014
#5
+85644
+5
Thanks for that, heureka........there are always things popping up on this calculator that I hadn't been aware of, before......of course......If Melody and I relied on a simple thing like punching a few buttons on the calculator, we wouldn't be able to "show off" as much as we do.....!!!
CPhill Oct 22, 2014
#6
+92194
+5
Too right there mate!
Melody Oct 22, 2014
#7
+19206
+5
Thanks all.
heureka Oct 22, 2014
### 16 Online Users
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners. See details | 2018-04-21T22:51:55 | {
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https://cu-numcomp.github.io/spring22/slides/2022-03-28-gradient-descent.html | Last time¶
• Assumptions of linear models
• Look at your data!
• Partial derivatives
• Loss functions
Today¶
• Discuss projects
• Gradient-based optimization for linear models
• Nonlinear models
using LinearAlgebra
using Plots
default(linewidth=4, legendfontsize=12)
function vander(x, k=nothing)
if isnothing(k)
k = length(x)
end
m = length(x)
V = ones(m, k)
for j in 2:k
V[:, j] = V[:, j-1] .* x
end
V
end
function vander_chebyshev(x, n=nothing)
if isnothing(n)
n = length(x) # Square by default
end
m = length(x)
T = ones(m, n)
if n > 1
T[:, 2] = x
end
for k in 3:n
#T[:, k] = x .* T[:, k-1]
T[:, k] = 2 * x .* T[:,k-1] - T[:, k-2]
end
T
end
function chebyshev_regress_eval(x, xx, n)
V = vander_chebyshev(x, n)
vander_chebyshev(xx, n) / V
end
runge(x) = 1 / (1 + 10*x^2)
runge_noisy(x, sigma) = runge.(x) + randn(size(x)) * sigma
CosRange(a, b, n) = (a + b)/2 .+ (b - a)/2 * cos.(LinRange(-pi, 0, n))
vcond(mat, points, nmax) = [cond(mat(points(-1, 1, n))) for n in 2:nmax]
vcond (generic function with 1 method)
Workshopping projects today¶
• Maybe you’ve started, maybe you’re still looking for a good project
• What have you been thinking about?
• What do you need help on?
Each breakout group will report out¶
1. One interesting thing about process
2. One question you have relevant to the projects (specific or general)
3. One question from content we’ve covered
Variational notation for derivatives¶
It’s convenient to express derivatives in terms of how they act on an infinitessimal perturbation. So we might write
$\delta f = \frac{\partial f}{\partial x} \delta x .$
(It’s common to use $$\delta x$$ or $$dx$$ for these infinitesimals.) This makes inner products look like a normal product rule
$\delta(\mathbf x^T \mathbf y) = \mathbf y^T (\delta \mathbf x) + \mathbf x^T (\delta \mathbf y).$
A powerful example of variational notation is differentiating a matrix inverse
$0 = \delta I = \delta(A^{-1} A) = (\delta A^{-1}) A + A^{-1} (\delta A)$
and thus
$\delta A^{-1} = - A^{-1} (\delta A) A^{-1}$
Optimization for linear models¶
Given data $$(x,y)$$ and loss function $$L(c; x,y)$$, we wish to find the coefficients $$c$$ that minimize the loss, thus yielding the “best predictor” (in a sense that can be made statistically precise). I.e.,
$\bar c = \arg\min_c L(c; x,y) .$
It is usually desirable to design models such that the loss function is differentiable with respect to the coefficients $$c$$, because this allows the use of more efficient optimization methods. Recall that our forward model is given in terms of the Vandermonde matrix,
$f(x, c) = V(x) c$
and thus
$\frac{\partial f}{\partial c} = V(x) .$
Derivative of loss function¶
We now differentiate our loss function
$L(c; x, y) = \frac 1 2 \lVert f(x, c) - y \rVert^2$
using a more linear algebraic approach to write the same expression is
(22)\begin{align} \nabla_c L(c; x,y) &= \big( f(x,c) - y \big)^T V(x) \\ &= \big(V(x) c - y \big)^T V(x) \\ &= V(x)^T \big( V(x) c - y \big) . \end{align}
A necessary condition for the loss function to be minimized is that $$\nabla_c L(c; x,y) = 0$$.
• Is the condition sufficient for general $$f(x, c)$$?
• Is the condition sufficient for the linear model $$f(x,c) = V(x) c$$?
• Have we seen this sort of equation before?
Instead of solving the least squares problem using linear algebra (QR factorization), we could solve it using gradient descent. That is, on each iteration, we’ll take a step in the direction of the negative gradient.
function grad_descent(loss, grad, c0; gamma=1e-3, tol=1e-5)
"""Minimize loss(c) via gradient descent with initial guess c0
using learning rate gamma. Declares convergence when gradient
is less than tol or after 500 steps.
"""
c = copy(c0)
chist = [copy(c)]
lhist = [loss(c)]
for it in 1:500
c -= gamma * g
push!(chist, copy(c))
push!(lhist, loss(c))
if norm(g) < tol
break
end
end
(c, hcat(chist...), lhist)
end
grad_descent (generic function with 1 method)
A = [1 1; 1 8]
@show cond(A)
loss(c) = .5 * c' * A * c
grad(c) = A * c
c, chist, lhist = grad_descent(loss, grad, [.9, .9],
gamma=.22)
plot(lhist, yscale=:log10, xlims=(0, 80))
cond(A) = 9.46578492882319
plot(chist[1, :], chist[2, :], marker=:circle)
x = LinRange(-1, 1, 30)
contour!(x, x, (x,y) -> loss([x, y]))
Chebyshev regression via optimization¶
x = LinRange(-1, 1, 200)
sigma = 0.5; n = 8
y = runge_noisy(x, sigma)
V = vander(x, n)
function loss(c)
r = V * c - y
.5 * r' * r
end
r = V * c - y
V' * r
end
c, _, lhist = grad_descent(loss, grad, ones(n),
gamma=0.008)
c
8-element Vector{Float64}:
0.7891712932446856
0.007025641791364225
-1.957064629386119
-0.36715187386667464
0.9229823103908219
0.3031710357963882
0.4284481908323782
0.27267026221101087
c0 = V \ y
l0 = 0.5 * norm(V * c0 - y)^2
@show cond(V' * V)
plot(lhist, yscale=:log10)
plot!(i -> l0, color=:black)
cond(V' * V) = 52902.52994792479
Nonlinear models¶
Instead of the linear model
$f(x,c) = V(x) c = c_0 + c_1 \underbrace{x}_{T_1(x)} + c_2 T_2(x) + \dotsb$
let’s consider a rational model with only three parameters
$f(x,c) = \frac{1}{c_1 + c_2 x + c_3 x^2} = (c_1 + c_2 x + c_3 x^2)^{-1} .$
We’ll use the same loss function
$L(c; x,y) = \frac 1 2 \lVert f(x,c) - y \rVert^2 .$
We will also need the gradient
$\nabla_c L(c; x,y) = \big( f(x,c) - y \big)^T \nabla_c f(x,c)$
where
(23)\begin{align} \frac{\partial f(x,c)}{\partial c_1} &= -(c_1 + c_2 x + c_3 x^2)^{-2} = - f(x,c)^2 \\ \frac{\partial f(x,c)}{\partial c_2} &= -(c_1 + c_2 x + c_3 x^2)^{-2} x = - f(x,c)^2 x \\ \frac{\partial f(x,c)}{\partial c_3} &= -(c_1 + c_2 x + c_3 x^2)^{-2} x^2 = - f(x,c)^2 x^2 . \end{align}
Fitting a rational function¶
f(x, c) = 1 ./ (c[1] .+ c[2].*x + c[3].*x.^2)
f2 = f(x, c).^2
[-f2 -f2.*x -f2.*x.^2]
end
function loss(c)
r = f(x, c) - y
0.5 * r' * r
end
r = f(x, c) - y
vec(r' * gradf(x, c))
end
gradient (generic function with 1 method)
c, _, lhist = grad_descent(loss, gradient, ones(3), gamma=8e-2)
plot(lhist, yscale=:log10)
Compare fits on noisy data¶
scatter(x, y)
V = vander_chebyshev(x, 7)
plot!(x -> runge(x), color=:black, label="Runge")
plot!(x, V * (V \ y), label="Chebyshev fit")
plot!(x -> f(x, c), label="Rational fit") | 2022-10-05T08:46:42 | {
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http://gdaymath.com/lessons/powerarea/3-5-optional-the-quadratic-formula/ | The Astounding Power of Area
Many students are taught to solve quadratics by memorizing the famous quadratic formula. There is no need: The box method with the symmetry of the square will never let you down! The reason I say this is because the box method really is the quadratic formula that everyone does, just made natural and bare. (In fact, my students use the box method to derive the quadratic formula just as an incidental exercise.)
Those who use the quadratic formula are taught to bring all the terms to one side so that it is an equation with zero on one side. We can do this too if we want (but the box method does not require it!)
$$ax^{2}+bx+c=0$$
If we use the box method on this, we multiply through by $$a$$ to make the first term a perfect square and through by $$4$$ to avoid fractions. This gives:
$$4a^{2}x^{2}+4abx+4ac=0$$.
Now apply the box method:
The box shows that we need the number $$b^{2}$$, not $$4ac$$, in the equation $$4a^{2}x^{2}+4abx+4ac=0$$. Let’s subtract $$4ac$$,
$$4a^{2}x^{2}+4abx=-4ac$$
and add $$b^{2}$$
$$4a^{2}x^{2}+4abx+b^{2}=b^{2}-4ac$$.
Now looking at the box we are set to go.
$$4a^{2}x^{2}+4abx+b^{2}=b^{2}-4ac$$
$$(2ax+b)^{2}=b^{2}-4ac$$
$$2ax+b=\sqrt{b^{2}-4ac}$$ or $$-\sqrt{b^{2}-4ac}$$.
Now add $$-b$$ throughout
$$2ax=-b+\sqrt{b^{2}-4ac}$$ or $$-b-\sqrt{b^{2}-4ac}$$.
And divide through by $$2a$$
$$x=\frac{-b+\sqrt{b^{2}-4ac}}{2a}$$ or $$x=\frac{-b-\sqrt{b^{2}-4ac}}{2a}$$.
This statement can be combined into a single expression: $$x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$$.
THE QUADRATIC FORMULA: If $$ax^{2}+bx+c=0$$, then $$x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$$.
COMMENT: Many curricula have students solve all quadratic equations by applying this memorized formula. This has the advantage of being quick. I will never stop a student from following this route too, but I personally prefer following a method I feel I understand and can just do without any memorization. It stays with me in the long run and makes sense. There is no need to do math fast!
FOR THOSE THAT INSIST ON KNOWING THE QUADRATIC FORMULA
The box method shows that the general quadratic equation $$ax^{2}+bx+c=0$$ has general solution: $$x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$$. Many curricula have students not only memorise this formula, but also study different features of the formula, and give parts of it names.
The quantity under the square root sign, $$b^{2}-4ac$$, is called the discriminant of the quadratic. Folk like to give this quantity a name because its sign determines the types of solutions one obtains:
If $$b^{2}-4ac$$ is negative, then the quadratic has no real solutions. (One cannot compute the square root of a negative value.)
If $$b^{2}-4ac$$ equals 0, then the quadratic has precisely one solution. (Zero is the only number with precisely one square root.)
If $$b^{2}-4ac$$ is positive, then the quadratic has two real solutions. (There are two square roots to a positive quantity.)
For example, without any effort we can see:
$$2w^{2}-3w+4=0$$ has no solutions because is $$b^{2}-4ac =3^{2}-4 \cdot 2 \cdot 4=-23$$ is negative.
$$x^{2}-4x+4=0$$ has precisely one solution because $$b^{2}-4ac =16-16=0$$.
$$3y^{2}-y-1$$ has precisely two solutions because $$b^{2}-4ac =1+12=13$$ is positive.
People also note that the two solutions to a quadratic can be written:
$$x=-\frac {b}{2a} + \frac {\sqrt{b^{2}-4ac}} {2a}$$ and $$x=-\frac {b}{2a} – \frac {\sqrt{b^{2}-4ac}}{2a}$$.
This shows that the two solutions lie at symmetrical positions about the value:
$$x=-\frac {b} {2a}$$.
Some folk consider this important to hold in mind.
Comment: See Part 4 of the Quadratics Course as to why none of this is necessary.
SHOULD ONE MEMORISE THE QUADRATIC FORMULA?
If speed matters to you … maybe.
If a curriculum insists that you and your students know it and that you must use the word “discriminant,” then yes, I suppose.
But if understanding has a higher priority in your mind, then the box method will not let you down.
To solve $$ax^{2}+bx+c=0$$ just multiply through by $$a$$ to make $$ax^{2}$$ a square and also multiply through by $$4$$ to avoid fractions. Then let the box guide you!
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Your support is so much appreciated and enables the continued creation of great course content. Thanks! | 2019-01-16T23:23:21 | {
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https://math.stackexchange.com/questions/2195269/simplification-of-binom500-binom501-binom501-binom502%E2%8B%AF-bi | # Simplification of $\binom{50}{0}\binom{50}{1} + \binom{50}{1}\binom{50}{2}+⋯+\binom{50}{49}\binom{50}{50}$
There was a post on this web site an hour ago asking for the sum of
\begin{equation*} \binom{50}{0}\binom{50}{1} + \binom{50}{1}\binom{50}{2}+⋯+\binom{50}{49}\binom{50}{50} \end{equation*}
expressed as a single binomial coefficient. (Four choices were provided in the post.) The post seems to have been deleted. I think it is worth keeping on this web site.
• When I looked at it moments ago, the other question was not deleted (though it was just one vote away from being closed): math.stackexchange.com/questions/2195018 Mar 20, 2017 at 16:59
• The older question was hard to find. For example, it never mentioned the word "binomial." I've added a couple of relevant tags to it, which might have helped. Mar 20, 2017 at 17:32
Solution
For any positive integer $n$ and any nonnegative integer $r \leq n$,
\begin{equation*} \binom{n}{r} = \binom{n}{n - r} . \end{equation*} According to Vandermonde's Identity, \begin{align*} &\binom{50}{0}\binom{50}{1} + \binom{50}{1}\binom{50}{2}+⋯+\binom{50}{49}\binom{50}{50} \\ &\qquad = \binom{50}{0}\binom{50}{49} + \binom{50}{1}\binom{50}{48}+⋯+\binom{50}{49}\binom{50}{0} \\ &\qquad = \sum_{r=0}^{49} \binom{50}{r} \binom{50}{49-r} \\ &\qquad = \binom{100}{49} . \end{align*}
There is another proof for this problem that I am mentioning here :
Since :
$$\binom{50}{0}\binom{50}{1} + \binom{50}{1}\binom{50}{2}+⋯+\binom{50}{49}\binom{50}{50} \\ \qquad = \binom{50}{0}\binom{50}{49} + \binom{50}{1}\binom{50}{48}+⋯+\binom{50}{49}\binom{50}{0} \\$$
Consider the product :
$$(1+x)^{50} \times (1+x)^{50}$$
Look for the coefficient of $x^{49}$ in this product, it will be calculated as :
$$x^0 ~~\text{from first bracket} , x^{49} ~\text{from second bracket} = \binom{50}{0} \times \binom{50}{49}$$
$$x^1 ~~\text{from first bracket} , x^{48} ~\text{from second bracket} = \binom{50}{1} \times \binom{50}{48}$$ $$x^2 ~~\text{from first bracket} , x^{47} ~\text{from second bracket} = \binom{50}{2} \times \binom{50}{47} \\. \\. \\. \\.$$
$$x^{49} ~~\text{from first bracket} , x^{0} \text{from second bracket} = \binom{50}{49} \times \binom{50}{0}$$
So, coefficient of $x^{49}$ in expansion of $(1+x)^{50}(1+x)^{50} =$
$$\binom{50}{0}\binom{50}{49} + \binom{50}{1}\binom{50}{48}+⋯+\binom{50}{49}\binom{50}{0} \\$$
Which is coefficient of $x^{49}$ in expansion of $(1+x)^{100} =$
$$\binom{100}{49}$$
Thus :
$$\binom{50}{0}\binom{50}{49} + \binom{50}{1}\binom{50}{48}+⋯+\binom{50}{49}\binom{50}{0}=\binom{100}{49}$$
• That's really nice! Mar 20, 2017 at 18:29 | 2022-08-12T13:23:28 | {
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https://mathoverflow.net/questions/161271/consecutive-non-quadratic-residues | Inspired by this recent question, I wondered if a similar result is true for quadratic non-residues, namely, if it is true that for every $k \in \mathbb{N}$ there exists a prime $p$ such that exists $k$ consecutive quadratic non-residues mod $p$? or even a stronger result like for every $k \in \mathbb{N}$ there exists $N$ such that for every prime $p > N$ we have $k$ consecutive quadratic non-residues mod $p$?
Obviously, the idea used in the link could not be adapted to this case because it relies on the fact that the product of two quadratic residues is a quadratic residue, which is not true for quadratic non-residues. I searched the web and did not find anything relevant on this question. Does anybody have any ideas?
Thank you!
PS: If anyone could give me a reference to a book that approaches this kind of problems (consecutive quadratic or non-quadratic residues), I would be grateful.
• For the first problem, you can choose $p=4k+3$ and multiply a sequence of consecutive quadratic residues by $-1$. I didn't read the second problem before voting to close, and I apologize for that and would undo it if I could. The second problem is more interesting. However, the abstract of the Buell and Hudson paper "On runs of consecutive quadratic residues and quadratic nonresidues" mentioned on the math.stackexchange.com question you linked claims to show that for all sufficiently large primes, there are arbitrarily long runs of quadratic nonresidues (and the same for residues). – Douglas Zare Mar 24 '14 at 17:36
• @GH: it may not be trivial, but is very standard: if for a prime $p$ every run of $k$ residues contained a square, then the sum $$\sum_x \prod_{j=1}^k \left(1-\left(\frac{x+j}{p}\right)\right)$$ would be very close to $0$, while from Wwyl's bound it is actually close to $p$. Have I got anything wrong? – Seva Mar 24 '14 at 17:39
• @GH: I didn't add any real ideas, but I think Seva's argument should be an answer. – Douglas Zare Mar 24 '14 at 17:46
• @André: I made a mistake in my comment. Here it is corrected: Weil's bound (not Weyl's) shows that for any polynomial $Q\in\mathbb{F}_p[x]$ with distinct roots we have that $\left|\sum_{x\in\mathbb{F}_p}\left(\frac{Q(x)}{p}\right)\right|$ is bounded by a constant times $\sqrt{p}$. Factoring out the product in Seva's sum yields several such character sums and also a sum of $1$'s, hence the sum is $p+O(\sqrt{p})$. On the other hand, the product in the sum vanishes whenever $x+j$ is a quadratic residue for some $j$. – GH from MO Mar 24 '14 at 18:15
• @André: Although that particular proof used $p=4k+1$, the idea that you can control whether the first few primes are quadratic residues by choosing $p$ to be in a particular arithmetic progression works even if we choose $p=4k+3$. Choose $p$ so that it is congruent to $-1$ mod $8$ and mod $q$ for the first few odd primes. By quadratic reciprocity this makes the first few primes squares mod $p$, while $-1$ is not a square. By Dirichlet's theorem on primes in arithmetic progressions there are primes satisfying these congruences. – Douglas Zare Mar 24 '14 at 18:23
The answer to both your questions is positive and indeed, every given pattern of quadratic residues and non-residues of fixed length appears among consecutive elements of ${\mathbb F}_p$, for all $p$ large enough; moreover, it appears about the expected number of times. This is non-trivial, but fairly standard.
Fix $\epsilon=(\epsilon_1,\ldots,\epsilon_k)\in\{-1,1\}^k$ ("the pattern") and for a prime $p$, let $N_\epsilon(p)$ denote the number of those $x\in{\mathbb F}_p$ with $$\left(\frac{x+1}p\right)=\epsilon_1,\ldots, \left(\frac{x+k}p\right)=\epsilon_k,$$ where $\left(\frac xp\right)$ is the Legendre symbol. Clearly, we have \begin{align*} N_\epsilon(p) &= 2^{-k}\sum_{x=0}^{p-k-1} \prod_{j=1}^k \left(1+\epsilon_j\left(\frac{x+j}{p}\right) \right) \\ &= 2^{-k}\sum_{x=0}^{p-1} \prod_{j=1}^k \left(1+\epsilon_j\left(\frac{x+j}{p}\right) \right) - \theta \frac k2,\quad 0\le \theta\le 1, \end{align*} as the contribution of each $x\in[p-k,p-1]$ to the whole sum is at most $2^{k-1}$.
Expanding the product, one can write $N_\epsilon(p)$ as a sum of the main term $2^{-k}\sum_x1=2^{-k}p$ and $2^k-1$ remainder terms, each of the form $2^{-k}\sum_x \left(\frac{Q(x)}p\right)$ with a non-square polynomial $Q(x)$ of degree at most $k$. Now, Weil's bound implies that each of these remainder terms does not exceed $(k-1)\sqrt p$ in absolute value; as a result, $$N_\epsilon(p)=2^{-k}p+\theta k(\sqrt p+1/2),\ |\theta|<1,$$ which is certainly positive for $p$ sufficiently large (like $p>\exp(ck)$ with a suitable constant $c$).
This argument readily extends to count, say, the number of those elements $x$ of a finite field such that for a given system of square-free, pairwise co-prime polynomials $P_1,\ldots,P_k\in{\mathbb Z}[X]$, the values $P_1(x),\ldots,P_k(x)$ follow a prescribed quadratic residue / non-residue pattern. Indeed, in a similar way one can handle the joint distribution of $P_1(x),\ldots P_k(x)$ in the cosets of any subgroup of the multiplicative group of a finite field, not just the subgroup of quadratic residues.
• you have a typo in the definition of $\sigma_p$. – Asaf Mar 24 '14 at 18:28
• @Seva: I think the exceptional $x$'s contribute at most $k/2$, not $1/2$ (as they add up and there are $k$ of them). – GH from MO Mar 24 '14 at 18:48 | 2021-01-22T10:14:01 | {
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https://math.stackexchange.com/questions/1106610/if-something-is-2n1-how-can-i-get-n-back-from-the-end-result | # If something is $2^{N+1}$, how can I get $N$ back from the end result?
I need this for a GIF encoder I'm programming, if something is $2^{N+1}$, how can I get $N$ back from the end result?
For example, $2^{7+1} = 256$, how can I get back to $7$ from $256$?
I've spent over an hour with a pencil and a sheet of paper trying to remember how to "balance equations" from my time in college but i cant seem to remember...
This seems pretty basic for this site but i was pushed here from stack overflow because it "wasn't programming related.."
Hopefully I'm not posting out of place here either
• $\log_2(x)-1$ should do the job. If you don't have $\log_2$ function in your calculator, then you can use $\frac{\log(x)}{\log(2)}-1$ instead. – barak manos Jan 16 '15 at 12:14
• Don't worry, this site is for math questions at all levels. – bof Jan 16 '15 at 12:15
• Yes, Regret's answer works for logs to any base. If logs to the base $2$ are available, then the formula simplifies because then $\log2=1$. – bof Jan 16 '15 at 12:17
$$2^{n+1}=x\\ \log(2^{n+1})=\log(x)\\ (n+1)\log(2)=\log(x)\\ n=\frac{\log(x)}{\log(2)}-1\\$$
In mathematics, just "$\log$" usually means with base $e$, but it does not matter which base you use here. This is because $\frac{\log_a(x)}{\log_a(2)}$ is the same number for any base $a$, it is equal to $\log_2(x)$. If you have access to base $2$ logarithm, you can just use $\log_2(x)-1$. If you only have access to a logarithm with a different base, you can use $\frac{\log(x)}{\log(2)}-1$.
You should be careful if you are using floating point numbers, however. For example,
Math.log(Math.pow(2, 31))/Math.log(2)
in JavaScript results in 31.000000000000004 and not $31$ like it should if floats were accurate.
• @James: I added a small detail to the end, do you think it is relevant? – Regret Jan 16 '15 at 12:36
• dont think this should be an issue. After getting the number I do round up which in the case of what you said a size of 31 would come out as 32, but it only wastes a byte of data so I'm not worried about it. Thanks anyway good to know if random values show up and I cant understand why. – Trotski94 Jan 16 '15 at 12:38 | 2019-05-26T09:51:51 | {
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https://math.stackexchange.com/questions/3299172/resistant-integral-int-01-left-frac-ln21-x-ln21x1-x-frac-ln22 | Resistant integral $\int_0^1\left(\frac{\ln^2(1-x)\ln^2(1+x)}{1-x}-\frac{\ln^2(2)\ln^2(1-x)}{1-x}\right)\ dx$
Prove, without using harmonic series, that
$$I=\int_0^1\left(\frac{\ln^2(1-x)\ln^2(1+x)}{1-x}-\frac{\ln^2(2)\ln^2(1-x)}{1-x}\right)\ dx$$ $$=\frac18\zeta(5)-\frac12\ln2\zeta(4)+2\ln^22\zeta(3)-\frac23\ln^32\zeta(2)-2\zeta(2)\zeta(3)+\frac1{10}\ln^52+4\operatorname{Li}_5\left(\frac12\right)$$
This problem was proposed by Cornel and can be found here.
The main reason behind such constraint is that this integral can be simplified into $$S=\sum_{n=1}^\infty\frac{H_n}{n^42^n}$$ which was calculated here using real and complex methods. So evaluating $$I$$ without using harmonic series means we are providing a third solution to $$S$$.
I have already computed this integral ( will be posted soon) but I would like to see variant approaches.
Thanks.
• What makes it “resistant”? Also, if you would like to see a variant, then we need to see what you already have. – gen-z ready to perish Jul 21 at 1:15
• you will come across tough integrals and you need some manipulations. I mentioned solution will be posted soon i am working on it. – Ali Shather Jul 21 at 1:17
$$I=\int_0^1\frac{\ln^2(1-x)}{1-x}\left(\ln^2(1+x)-\ln^2(2)\right)\ dx\overset{IBP}{=}\frac23\int_0^1\frac{\ln^3(1-x)\ln(1+x)}{1+x}\ dx$$ Using the algebraic identity $$\quad\displaystyle a^3b=\frac18(a+b)^4-\frac18(a-b)^4-ab^3$$
and by setting $$a=\ln(1-x)$$ and $$b=\ln(1+x)$$, we get
\begin{align} I=\frac1{12}\underbrace{\int_0^1\frac{\ln^4(1-x^2)}{1+x}\ dx}_{\displaystyle I_1}-\frac1{12}\underbrace{\int_0^1\frac{\ln^4\left(\frac{1-x}{1+x}\right)}{1+x}\ dx}_{\displaystyle I_2}-\frac23\underbrace{\int_0^1\frac{\ln(1-x)\ln^3(1+x)}{1+x}\ dx}_{\displaystyle I_3} \end{align}
The first integral was nicely done by Cornel and can be found in his book Almost Impossible Integral , Sums, and Series page $$80$$ and as follows:
\begin{align} I_1&=\int_0^1\frac{\ln^4(1-x^2)}{1+x}\ dx=\int_0^1(1-x)\frac{\ln^4(1-x^2)}{1-x^2}\ dx\overset{x^2=y}{=}\frac12\int_0^1\frac{1-\sqrt{y}}{\sqrt{y}}.\frac{\ln^4(1-y)}{1-y}\ dy\\ &\overset{IBP}{=}-\frac1{20}\int_0^1\frac{\ln^5(1-y)}{y^{3/2}}\ dy=-\frac{1}{20}\lim_{x\mapsto-1/2\\y\mapsto1}\frac{\partial^5}{\partial y^5}\text{B}(x,y)\\ &\boxed{I_1=\frac{16}5\ln^52-16\ln^32\zeta(2)+48\ln^22\zeta(3)-54\ln2\zeta(4)-24\zeta(2)\zeta(3)+72\zeta(5)} \end{align}
\begin{align} I_2=\int_0^1\frac{\ln^4\left(\frac{1-x}{1+x}\right)}{1+x}\ dx\overset{\frac{1-x}{1+x}=y}{=}\int_0^1\frac{\ln^4x}{1+x}\ dx=\boxed{\frac{45}2\zeta(5)=I_2} \end{align}
\begin{align} I_3&=\int_0^1\frac{\ln(1-x)\ln^3(1+x)}{(1+x)}\ dx\overset{\frac1{1+x}=y}{=}-\int_{1/2}^1 \frac{\ln\left(\frac{2x-1}{x}\right)\ln^3x}{x}\ dx\\ &=\int_{1/2}^1\frac{\ln^4x}{x}\ dx-\text{Re}\int_{1/2}^1\frac{\ln(2x-1)\ln^3x}{x}\ dx, \quad \color{red}{\text{Re}\{\ln(2x-1)\}=\text{Re}\{\ln(1-2x)\}}\\ &=\frac15\ln^52-\text{Re}\int_{1/2}^1\frac{\ln(1-2x)\ln^3x}{x}\ dx\\ &=\frac15\ln^52+\text{Re}\sum_{n=1}^\infty\frac{2^n}{n}\int_{1/2}^1x^{n-1}\ln^3x\ dx\\ &=\frac15\ln^52+\text{Re}\sum_{n=1}^\infty\frac{2^n}{n}\left(\frac{\ln^32}{n2^n}+\frac{3\ln^22}{n^22^n}+\frac{6\ln2}{n^32^n}+\frac{6}{n^42^n}-\frac{6}{n^4}\right)\\ &=\frac15\ln^52+\ln^32\zeta(2)+3\ln^22\zeta(3)+6\ln2\zeta(4)+6\zeta(5)-6\text{Re}\{\operatorname{Li}_5(2)\}\tag{1} \end{align} Using the polylogarithmic identity: $$\operatorname{Li}_5(x)=-\frac74\zeta(4)\ln(-x)-\frac16\zeta(2)\ln^3(-x)-\frac1{120}\ln^5(-x)+\operatorname{Li}_5(1/x)$$
Set $$x=2$$, we get
$$\text{Re}\{\operatorname{Li}_5(2)\}=2\ln2\zeta(4)+\frac13\ln^32\zeta(2)-\frac1{120}\ln^52+\operatorname{Li}_5\left(\frac12\right)\tag{2}$$
Plugging $$(2)$$ in $$(1)$$, we get $$\boxed{I_3=-6\operatorname{Li}_5\left(\frac12\right)+6\zeta(5)-6\ln2\zeta(4)+3\ln^22\zeta(3)-\ln^32\zeta(2)+\frac14\ln^52}$$
Combining the boxed results, we get the closed form of $$I$$.
• Reffering to this question: math.stackexchange.com/q/3293328/515527, you have solved one part of it. If we apply the same substitution: $\frac{1}{1+x}=y$ we get $$\int_0^1\frac{\ln^3(1+x)\ln(1-x)}{x}=\int_{1/2}^1 \frac{\ln\left(\frac{2x-1}{x}\right)\ln^3x}{x}dx-\int_{1/2}^1 \frac{\ln\left(\frac{2x-1}{x}\right)\ln^3x}{1-x}dx$$ The second one is still painful. – Nyssa Jul 27 at 22:44
• Good point. yes the first integral is manageable but the second one is a hell of integral and I think its more complicated than the original integral due to the limit $1/2$. Still, I will try it and see what I get. – Ali Shather Jul 27 at 22:49
• It would be awesome if he can solve it without using the tough series I used. – Ali Shather Jul 27 at 23:01
• This is what I got $$\int_0^1\frac{\ln^3(1+x)\ln(1-x)}{x}\ dx=3\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\left(H_n^3-H_nH_n^{(2)}-2\frac{H_n^2}{n}+2\frac{H_n}{n^2}\right)$$. Or we can relate it to the integral $\int_0^1\frac{\ln(1+x)\ln^3(1-x)}{x}\ dx$ using the algebraic identity $(a+b)^4-(a-b)^4=8a^3b+8ab^3$ where $\int_0^1 \frac{(a+b)^4}{x}$ and $\int_0^1 \frac{(a-b)^4}{x}$ are easy to do. – Ali Shather Jul 27 at 23:49 | 2019-11-19T07:26:40 | {
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https://or.stackexchange.com/questions/2526/convexity-of-variance-minimization | # Convexity of Variance Minimization
$$X$$ is a discrete random variable taking value $$x_n$$ with probability $$1/N$$ for $$n=1, \ldots,N$$. I would like to set the $$x_n$$ values in an optimization problem. My objective is to minimize the variance while satisfying a set of constraints.
So the problem is:
$$\begin{array}{ll} & \min\limits_{\{x_n\}_{n=1}^N}{\operatorname{Var}(X)} \\ & \text{s.t.} \ \ldots \end{array}$$
Denote $$x = \begin{pmatrix} x_1 & \ldots & x_n\end{pmatrix}^\top$$. So in this discrete distribution, the variance is: $$\operatorname E[X^2] -\operatorname E[X]^2 = \frac{1}{N} \sum_{n=1}^N x_n^2 - \frac{1}{N^2} \left( \sum_{n=1}^N x_n \right)^2 = \frac{x^\top x}{N} - \frac{(\mathbf{e}^\top x)(x^\top \mathbf{e})}{N^2}.$$
My questions are:
1. The variance is not convex, so minimization is hard, right? Is there a common method for this?
2. Is there any convex function which results in a low variance after being minimized?
Edit: The variance may be convex based on the comments. We can show that $$\operatorname E[X^2] -\operatorname E[X]^2 = \frac{x^\top \left( I - \tfrac{\mathbf{e}\mathbf{e}^\top}{N}\right) x}{N},$$ so to show this is convex, we need to show that $$I - \tfrac{\mathbf{e}\mathbf{e}^\top}{N}$$ is a p.s.d. matrix. That is itself a challenge. I need to see the proof of how this is p.s.d, or if one can properly write down how the variance can be shown as a closed-form norm expression as in the comments.
To show the variance is $$\ell^2$$-norm-representable we need to have $$a^\top a = I - \tfrac{\mathbf{e}\mathbf{e}^\top}{N}$$ for some $$a$$. Finding $$a$$ will also make it.
Edit-2: Ok, I got the comment. He follows the $$\operatorname{Var}(X) = \operatorname E[(X - \operatorname E[X])^2]$$ approach, then it is obvious.
• To me it looks like $\operatorname V(x)$ is a convex quadratic function, since $0\le\operatorname V(x) = x'Hx$ where $H$ is the Hessian of $\operatorname V(x)$. – ErlingMOSEK Sep 11 '19 at 17:46
• Also $\operatorname V(x)=\frac1N\left\|x-ee^\top x/N\right\|^2$ which clearly is convex. – ErlingMOSEK Sep 11 '19 at 17:54
• Minimizing $\left\|x-ee^\top x/N\right\|$ is easy using SOCP/conic quadratic optimization. See docs.mosek.com/modeling-cookbook/index.html – ErlingMOSEK Sep 11 '19 at 18:04
• @ErlingMOSEK Is it $\operatorname V(X) =\frac1N \left\|x - \frac{ee^\top x}N\right\|_2^2$ or $\operatorname V(X) = \frac1N\left\|\frac{x - ee^\top x}N \right\|_2^2$ ? Because I can not derive this. Thanks for the answer! – independentvariable Sep 11 '19 at 20:00
• To show why $A = I - \left( \frac{e e^T}{n}\right)$ is psd, show that $v^T A v \geq 0 \, \forall v$, where $v \in \mathbb{R}^n$. An inequality that will come in handy is $n\left(\sum\limits_{i=1}^{n} v_i^2\right) \geq \left(\sum\limits_{i=1}^{n} v_i \right)^2$, where $v_i$ is the the $i^{th}$ component of $v$. – batwing Sep 11 '19 at 22:41
It holds $$\begin{array}{rcl} \operatorname V(x) &= &\dfrac1N\left\| x-\dfrac{e^\top x}{N} e \right\|^2 \\ & = & \dfrac1N\left(x^\top x+\dfrac{(e^\top x)^2 e^\top e}{N^2}-2\dfrac{(e^\top x)^2}N\right) \\ & = & \dfrac{x^\top x}{N} - \dfrac{(e^\top x)^2}{N^2}. \end{array}$$ So you are minimizing the $$\ell^2$$-norm of an affine expression which is known to be convex.
The problem $$\begin{array}{lcl} \min & \dfrac{\|x-e u\|}{N} & \\ \mbox{s.t.} & \dfrac{e^\top x}{N} - u & = & 0 \\ \end{array}$$ provides a nice interpretation since $$u$$ is the average. Note the problem tries to make all the $$x$$ equal to the average value.
Alternatively the last problem can be stated as $$\begin{array}{lcl} \min & \dfrac{s}{N}&\\ \mbox{s.t.} & \dfrac{e^\top x}{N} - u &=0 \\ &(s;x-e u) &\in Q. \\ \end{array}$$ where $$Q$$ is a quadratic cone. This provides another convexity proof because the quadratic cone is convex. Hence, the problem can be solved using SOCP also known as conic quadratic optimization. | 2020-02-20T11:02:57 | {
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https://brilliant.org/discussions/thread/easy-integral/ | ×
Integral
Find $$\displaystyle \int \dfrac{dx}{4\sqrt2 \sin(3x) + 2\cos(3x)}$$ it is easy but my answer is in different form from answer given , what do you get in terms of tanx and logs ? and by using substitutions ?
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11 months, 1 week ago
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Write $4\sqrt{2}\sin3x + 2\cos3x \; = \; 6\cos(3x-\alpha) \hspace{2cm} \alpha = \cos^{-1}\tfrac13$ and the integral becomes $\int \tfrac16\sec(3x-\alpha)\,dx \; = \; \tfrac{1}{18}\ln\big|\sec(3x-\alpha)+\tan(3x-\alpha)\big| + c$
- 11 months, 1 week ago
Yes, I approached the same way.
- 11 months, 1 week ago
- 11 months, 1 week ago
1. Can you state the solution?
2. Did you remember the "+C" which means that $$\sin^2 + C = - \cos^2 +C$$?
Staff - 11 months, 1 week ago
What is the ANS?????
- 11 months, 1 week ago
Mark Hennings has given it.
- 11 months, 1 week ago | 2017-11-21T21:15:00 | {
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https://math.stackexchange.com/questions/3359864/is-a-single-element-linearly-independent-in-an-abelian-group | # Is a single element linearly independent in an abelian group?
This is my definition of linearly independent elements in an abelian group:
Let $$A$$ be an abelian group, let $$X \subseteq A$$ be a subset and $$x_1,\dots,x_k \in X$$ with $$x_i \neq x_j$$ for all $$1 \leq i \neq j \leq n$$. The elements $$x_1,\dots,x_k$$ are defined to be linearly independent if they satisfy the following condition:
$$$$n_1x_1 + \dots + n_kx_k = 0 \, , \, \text{with} \hspace{2mm} n_1,\dots,n_k \in \mathbb{Z} \implies n_i = 0 \hspace{2mm} \text{for all} \hspace{2mm} 1 \leq i \leq n$$$$
I'm proving theorem 1.6 of Hungerford's Algebra and got stuck at "Either $$G \cap H = \{0\}$$, in which case $$G = \langle d_1x_1 \rangle$$ and the theorem is true [...]". Why is $$d_1x_1$$ linearly independent? I need this in order to say that $$\{d_1x_1\}$$ is a basis of $$G$$. I'm pretty sure this comes from the fact that $$d_1x_1 \neq 0$$, but linearly independence does not seem to be true because $$nd_1x_1 = 0$$ might be true even though $$n \neq 0$$. For example, if $$d_1x_1$$ is an element of order $$n$$. So, is the implication $$$$n(d_1x_1) = 0 \quad \wedge \quad d_1x_1 \neq 0 \quad \implies \quad n = 0$$$$ true or false in the context of abelian groups? What if we consider free abelian groups instead? If it's always false, then how is the linearly independence proven in Hungerford's proof?
• What is $G$ and $H$? – Wuestenfux Sep 17 '19 at 15:21
• $G < F$ is a subgroup of a free abelian group, $H$ is defined as a particular subgroup of $G$. But that's not the point, I just don't understand why a non-zero element is automatically linearly independent. Is this even true? – Lele99_DD Sep 17 '19 at 15:26
• Is $A$ an arbitrary abelian group, or a free abelian group? (More specifically, does this definition apply to general abelian groups, or only to free abelian groups?) – Morgan Rodgers Sep 17 '19 at 16:35
• @MorganRodgers The definition applies to general abelian groups. – Lele99_DD Sep 17 '19 at 17:26
• But Theorem 1.6 (now that I see it posted in an answer below) is specifically pertaining to free abelian groups. In this case no element has finite order. You are correct that a single element that has finite order is not linearly independent. – Morgan Rodgers Sep 17 '19 at 18:47
You’re correct that, for general abelian groups, any set containing finite order elements is not linearly independent.
For example, in $$\mathbb{Z}/4\mathbb{Z}$$ I can take $$n = 2$$ and $$x_1 = 2$$. Then neither element is zero, but $$2\cdot2 = 0$$. Thus $$\{2\}$$ is not linearly independent.
However, in the free abelian case, this fails. If $$nx_1 = 0$$ for $$n\ne 0$$ then this implies $$x_1$$ has finite order. No nontrivial element of any free abelian group has finite order, so this won’t occur in your case.
Let me address your question as formulated in your comment, namely why a single nonzero element of a free abelian group automatically forms a linearly independent subset.
What I think you might be missing is that the summation operator $$\sum_{i=1}^k s_i$$ is defined just as well for $$k=1$$ as it is for $$k \ge 2$$. In fact, in a rigorous treatment, one would define the $$k$$-ary summation operator of any associative binary operation by induction on $$k \ge 1$$. The basis step of the induction is to define $$\sum_{i=1}^1 s_i = s_1$$. In the inductive step, assuming that $$k \ge 2$$ and that $$\sum_{i=1}^{k-1} s_i$$ is already defined, one defines $$\sum_{i=1}^{k} s_i = \left(\sum_{i=1}^{k-1} s_i\right) + s_{k}$$.
All that goes to say that the definition of linearly independence is about linear combinations, which are just special kinds of finite sums, and a sum of one object is allowed as part of this definition:
A subset of $$k$$ distinct elements $$\{x_1,...,x_k\} \subseteq A$$ (where $$k \ge 1$$) is said to be linearly independent if for any sequence of integers $$(a_1,...,a_k)$$, if $$\sum_{i=1}^k a_i x_i=0$$ then $$a_i=0$$ for each $$1 \le i \le k$$.
• In the case of finite order elements, shouldn't the correct definition (as OP discusses) be that $\sum_{i=1}^k a_ix_i=0$ implies that each $a_ix_i=0$? – WoolierThanThou Sep 17 '19 at 15:56
• While I agree that the original question seems not to have the "free abelian" requirement, I will restrict my answer to the free abelian situation as the question was reformulated in the comment. It's not clear to me that the concept of "linear independence" makes much sense in a non-free module, although I'm happy to be corrected. – Lee Mosher Sep 17 '19 at 15:59
• Sorry, maybe I was unclear, but this is not what I was looking for. I'm fine with $\sum_{i=1}^1 s_i$ being equal to $s_1$ and I understand that the definition of linearly independence works in the case $k=1$, too. My problem is just the implication $a_1x_1 = 0 \wedge x_1 \neq 0 \implies a_1 = 0$. – Lele99_DD Sep 17 '19 at 16:06
• I have a feeling that the implication $a_1x_1 = 0 \wedge x_1 \neq 0 \implies a_1 = 0$ is true in free abelian groups, but not in arbitrary abelian groups because of the fact that every element in a free abelian group has infinite order, whereas abelian groups might have finite-order elements. But this is just a conjecture to me, I'm not sure whether this is true or not. – Lele99_DD Sep 17 '19 at 16:12
• Alright. It would help, then, if you edited your question to make clear what you are and are not asking, particularly given the direction of your comment. – Lee Mosher Sep 17 '19 at 16:17
According to the Hungerford which you seems to try reading, $$x_1$$ is choosen so that $$W=\{x_1, y_2, \cdots, y_n \}$$ is a basis of $$F$$.
(Here $$x_1-y_1$$ is a linear combination of $$y_2, \cdots, y_n$$, where $$\{y_1, y_2, \cdots, y_n \}$$ is a basis of $$F$$ selected tricky. For details, see the proof above.)
Observe that $$nd_1 x_1= nd_1x_1 + 0y_2 + \cdots + 0y_n$$. As a result, $$nd_1x_1=0$$ implies $$nd_1=0$$. | 2020-07-12T01:02:27 | {
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https://math.stackexchange.com/questions/2666791/still-struggling-with-continuity | Still struggling with continuity
A function is continuous iff it is continuous at all points in its domain. What confuses me about this definition is that for undefined points in a function, those points are removed from the domain.
Does this mean a function $f(x) = x$ with domain $(-\infty, -10) \cup (3, \infty)$ is continuous even though there's a big blank space between the two pieces?
Does this mean $f(x) = \frac{x-3}{x-3}$ is continuous even though it has a removable singularity at $x=3$?
Would a single point be continuous even though it lacks limits on either side?
Does this mean $f(x) = \tan(x)$ and $f(x) = 1/x$ are continuous even though they have asymptotes where the function goes to infinity?
Does this mean the only functions that are discontinuous are ones with jump discontinuities?
• Your first one is continuous. Your second one is just f(x)=1 which has no singularity. If you are referring to a normal rational function, then this function is continuous for all points except at the singularity. – Harry Alli Feb 26 '18 at 0:22
• @HarryAlli The second one is $f(x)=1$ for all $x \neq 3$. At $x=3$ it isn't defined since the denominator is $0$. – user525966 Feb 26 '18 at 0:23
3 Answers
It is because we get a little bit loosey-goosey with the concept of domain.
We had too many questions in Algebra class / pre-calculs class were we were asked what is the domain of $f(x).$ i.e. the domain of $f(x) = \frac 1x$ is $\mathbb R - \{0\}.$ But, it is really the responsibility of whoever it is that defines the function to also define the domain. It could be that there are points in the domain where the function is not defined. When we get to these continuity questions, we are often being asked to include points in the domain where the function may, in fact, be undefined.
$f(x) = \frac 1{x}$ is continuous on $(-\infty,0)\cup (0,\infty),$ but it is not continuous on other domains such as $[-1,1]$ or $\mathbb R.$
• Does it even make sense to use a different domain though, since this can include points that aren't even defined? – user525966 Feb 26 '18 at 0:30
• If I define my function $f:\mathbb R \to \mathbb R$ then the domain is $\mathbb R$ even if there are points where the function is infact undefined. We say a function is a mapping from the domain to the co-domain where every input has "at most one" output. There could certainly be no output for a given input and still be a valid function. – Doug M Feb 26 '18 at 0:33
• So if we don't explicitly give the domain then we generally assume the domain is wherever it's defined -- and that it's probably going to be a continuous function granted no jump or essential discontinuities, but if we explicitly make the domain include undefined points, then it won't be continuous anymore for sure? – user525966 Feb 26 '18 at 0:34
• That is exactly how I think about it. – Doug M Feb 26 '18 at 0:35
• “It could be that there are points in the domain where the function is not defined.” No, the domain of a function is (by definition!) the set of points where it's defined. It doesn't make sense to talk about “$f(x)=1/x$ on the domain $[-1,1]$”. – Hans Lundmark Feb 26 '18 at 6:04
Terence Tao's book Analysis II makes the following definition (p. 420):
Let $(X,d_X)$ be a metric space, and let $(Y,d_y)$ be another metric space, and let $f:X \to Y$ be a function. If $x_0 \in X$, we say that $f$ is continuous at $x_0$ iff for every $\epsilon > 0$, there exists a $\delta > 0$ such that $d_Y(f(x),f(x_0)) < \epsilon$ whenever $d_X(x,x_0) < \delta$. We say that $f$ is continuous iff it is continuous at every point $x \in X$.
Folland gives a definition similar to Tao's.
So, using the definition in Tao or Folland, it is correct to say that the functions you mentioned are continuous. For example, using Tao's definition, it is correct to state that the function $f(x) = \tan(x)$ is continuous. (But the Wikipedia article quoted below warns us that not everyone uses Tao's terminology, so we should be careful that the meaning of this statement is clear in context.)
The answer to your final question is no. Picture the function $f$ defined by $f(x) = \sin(1/x)$ if $x \neq 0$ and $f(0) = 0$.
By the way, we should note that the Wikipedia says the following:
There are several different definitions of continuity of a function. Sometimes a function is said to be continuous if it is continuous at every point in its domain. In this case, the function $f(x) = \tan(x)$, with the domain of all real $x \neq (2n+1)\pi/2$, $n$ any integer, is continuous. Sometimes an exception is made for boundaries of the domain. For example, the graph of the function $f(x) = \sqrt{x}$, with the domain of all non-negative reals, has a left-hand endpoint. In this case only the limit from the right is required to equal the value of the function. Under this definition $f$ is continuous at the boundary $x= 0$ and so for all non-negative arguments. The most common and restrictive definition is that a function is continuous if it is continuous at all real numbers. In this case, the previous two examples are not continuous, but every polynomial function is continuous, as are the sine, cosine, and exponential functions. Care should be exercised in using the word continuous, so that it is clear from the context which meaning of the word is intended.
• Those are essential discontinuities because they oscillate forever, and that's fine, I did forget to mention those, but I'm mostly interested in the other categories, and more broadly, when a function is considered discontinuous – user525966 Feb 26 '18 at 0:32
• “The most common and restrictive definition is that a function is continuous if it is continuous at all real numbers.” Really?!? I don't think I've ever seen that. The standard definition is “a function is continuous iff it is continuous at every point in its domain”, so you should really trust Tao more than Wikipedia here. – Hans Lundmark Feb 26 '18 at 6:08
• @HansLundmark Thanks, I edited the answer to emphasize the Tao / Folland definition and deemphasize the Wikipedia passage. – littleO Feb 26 '18 at 7:13
Usually continuity is defined for those points at which the function is defined and what occurs outside the domain is not relevant.
Anyway note that in some context a function is defined to be continuos if its domain is an interval, and it is continuous at every point of that interval.
Here you can find a good reference from MIT for a full classification Continuity and Discontinuity.
With respect to your examples, following the definition proposed in the reference
• $f(x) = x$ with domain $(-\infty, -10) \cup (3, \infty)$ is continuous on each interval but it is not a continuos function
• $f(x) = \frac{x-3}{x-3}$ has a removable discontinuity at $x=3$ which can be eliminated by definining $f(3)=1$
• a single point $(x_0,y_0)$ is a continuous function since $f(x_0)=y_0$
• $f(x) = \tan(x)$ and $f(x) = 1/x$ are continuous in the intervals of definition and have infinity dicontinuities
• Does this mean the only functions that are discontinuous are ones with jump discontinuities? No that's only a special kind of discontinuity
• This one says $\tan(x)$ is not continuous for example? – user525966 Feb 26 '18 at 0:29
• @user525966 yesexactly, in the sense that it is not continuos in a whole interval but it is continuos in $(-\pi/2,\pi/2)$ for example. – user Feb 26 '18 at 0:35
• I would say the definitions in that document are most nonstandard. (Requiring the domain to be an interval? What? And requiring points of discontinuity to be isolated? What?!? Surely everyone would say that every point is a point of discontinuity for the Dirichlet function $\mathbf{1}_{\mathbb{Q}}$?) – Hans Lundmark Feb 26 '18 at 6:17
• And example 7 (“if $f(x)=|x|$ then $f'$ has a jump discontinuity at the origin”) is misleading. Since $f'$ isn't even defined at the origin, one shouldn't even talk about continuity or discontinuity there to begin with (although I confess that I might have committed this sin myself from time to time). Doing so, one contradicts the Corollary to Theorem 5.12 in Rudin's Principles (a derivative can't have jump discontinuites). – Hans Lundmark Feb 26 '18 at 6:24
• @HansLundmark Always there are a lot of discussion on this kind of topic, I really think that it is mainly a fact of definitions which depends upon the particular context. Thus it is important to keep in mind that we need to be aware about this as we can have different definitions and related interpretation. – user Feb 26 '18 at 8:27 | 2021-07-25T17:03:15 | {
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https://forum.math.toronto.edu/index.php?PHPSESSID=ifaosk7dpdv9uq1eh9qbp8im52&action=printpage;topic=2468.0 | # Toronto Math Forum
## MAT334--2020F => MAT334--Tests and Quizzes => Test 1 => Topic started by: A A on October 14, 2020, 05:42:27 PM
Title: 2020 TT1 Main Setting - Problem 1b
Post by: A A on October 14, 2020, 05:42:27 PM
Hello, can anyone explain the solution to Problem 1b: which of the complex roots are in the first complex quadrant? Why is it all zn that are in the first complex quadrant?
Note (for context): Problem 1a: Find all the complex roots of the equation tanh(3z) = 1 + 2i
Title: Re: 2020 TT1 Main Setting - Problem 1b
Post by: RunboZhang on October 14, 2020, 07:37:18 PM
Firstly, for the first quadrant, we have $Re(z) > 0$ and $Im(z) > 0$.
Secondly, we have $z = \frac{1}{12}log(2) + (\frac{\pi}{8} + \frac{2\pi }{6} n)i$, $n \in \mathbb{Z}$ by part (a).
By combining the previous two conclusions we have, $Re(z) = \frac{1}{12}log(2) > 0$ since $log(2) > 0$. Also $Im(z) = (\frac{\pi}{8} + \frac{2\pi }{6} n) > 0$ when $n \ge 0$.
Therefore, as long as we have a non-negative $n$, our $z$ is in the first quadrant of complex plane.
Title: Re: 2020 TT1 Main Setting - Problem 1b
Post by: A A on October 14, 2020, 09:32:09 PM
Thank you!! | 2022-10-02T02:53:26 | {
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https://math.stackexchange.com/questions/2033470/prove-if-a-mod-n-has-a-multiplicative-inverse-then-its-unique | # Prove if a (mod n) has a multiplicative inverse, then it's unique
Assume that an integer $$a$$ has a multiplicative inverse modulo an integer $$n$$. Prove that this inverse is unique modulo $$n$$.
I was given a hint that proving this Lemma: \begin{align} n \mid ab \ \wedge \ \operatorname{gcd}\left(n,a\right) = 1 \qquad \Longrightarrow \qquad n \mid b \end{align} should help me in finding the answer.
Here are my steps in trying to solve the problem: \begin{align} \operatorname{gcd}\left(n,a\right) = 1 & \qquad \Longrightarrow \qquad sn + ta = 1 \qquad \Longrightarrow \qquad sn = 1 - ta \\ & \qquad \Longrightarrow \qquad 1 \equiv ta \mod n \qquad \Longrightarrow \qquad ta \equiv 1 \mod n . \end{align} I know that having the GCD of m and a equal to 1 proves there is a multiplicative inverse mod n, but I'm not sure on how to prove $$n \mid b$$ with and how it helps prove the multiplicative inverse is unique.
• – user170039
Dec 5 '16 at 15:42
For the first part note as $(n,a) = 1$, then there exist $x,y \in \mathbb{Z}$, s.t. $nx + ay = 1$. Multiply both sides by $b$ and you will get $nxb + (ab)y = b$. The LHS is obviously divisible by $n$, so then the RHS must be too. Hence $n \mid b$.
This thing proves that the inverse exist. To note that it's unique modulo $n$ assume that $aa_1 \equiv 1 \pmod n$ and $aa_2 \equiv 1 \pmod n$. Then we have that there exist integer $x,y$ s.t. $aa_1 + nx = 1$ and $aa_2 + ny = 1$. Subtracting the two equations we have that $a(a_1 - a_2) = n(y-x) \implies a(a_1 - a_2) \equiv 0 \pmod n \implies a_1 \equiv a_2 \pmod n$. Hence the multiplicative inverse is unique in $\mathbb{Z}_n$
• The argument is incomplete, since you seem to be applying the first paragraph in your final inference (which should be mentioned explicitly), but to apply that you need to first prove $a$ is coprime to $n$ when $a$ is invertible. This follows simply from the listed Bezout equation, but that needs to be explicitly stated in order to be rigorous. Note that we are given that $a$ is invertible, so you don't need to prove that. The logic is not correct. However, none of this is needed since the hinted method is rather roundabout - see my answer. Apr 23 at 20:30
If $$\,b,b'$$ are inverses of $$\,a\,$$ then $$\ b\equiv b(ab')\equiv (ba)b'\equiv b'.\,$$ Note that the proof uses only commutativity and associativity so it works more generally than for the integers mod $$\,n.\,$$
Note It's a slick way of cancelling $$\,\color{#c00}a\,$$ from $$\,\color{#c00}ab\equiv 1\equiv \color{#c00}ab'$$ by scaling by $$\,\color{#c00}a^{-1},\,$$ yielding $$\,b\equiv b'$$. Hence we see that the uniqueness of inverses is simply a special case of the general fact that invertible elements are cancellable, via scaling by the inverse.
Proofs like this can often be discovered by a general method of "overlapping" equations to find a term that both equations apply to (above the overalapped term is $$\,bab'\,$$ which can be reduced by both rules $$\,ab\to 1,\ ab'\to 1).\,$$ Below is another example with further explanation.
Below I will explain a general way to discover a simpler proof (vs. pulling it out of a hat like magic). The key idea is very simple: one can discover consequences of identities (axioms) by "overlapping" them: looking for a "unified" term that they both apply to. Let's try that to prove the uniqueness of additive identity elements. Suppose that $$\,0\,$$ and $$\,0'$$ are both additive identities. This means that
$$\begin{eqnarray} x = && \color{#0a0}0 + \color{#c00}x\\ && \color{#0a0}y + \color{#c00}{0'} = y\\ \hline \Rightarrow\ \ 0' = && 0 + 0' = 0\end{eqnarray}$$
where we chose the values of the specialization $$\,\color{#0a0}{y=0},\ \color{#c00}{x = 0'}\,$$ in order unify $$\,0+x\,$$ with $$\,y+0', \,$$ yielding the "unified" term $$\,0+0'\,$$ that both axioms apply to. Applying both axioms to the unified term we can rewrite it in two different ways, deducing the new consequence $$\,0' = 0.\,$$ Presto!
This is a very widely applicable method of deriving consequences of axioms, i.e. by "unifying" or "overlapping" terms of both so that both axioms apply, yielding a rewriting of the term in two different ways (e.g. another example). In fact in some cases it can be used to algorithmically derive all of the consequences of the axioms, so yielding algorithms for deciding equality, e.g. see the Knuth-Bendix equational completion algorithm and the Grobner basis algorithm, and see George Bergman's classic paper The Diamond Lemma in Ring Theory.
• A nice application:: $\,4ab-1 \mid 4a^2-1\Rightarrow a=b,\,$ see here $\ \$ Mar 18 at 13:23
What you have shown already is a technique that, using the Extended Euclidean Algorithm will give you the inverse (if it exists). Note also that you can determine if an element $a \mod n$ has an inverse by checking that they have no common divisors (that is, $\gcd(a,n)=1$).
Now to answer your question. Suppose that you have two multiplicative inverses $b,c$ such that $ab\equiv 1 \equiv ac \mod n$. (Then you also have $ba\equiv 1 \equiv ca \mod n$) Consider
$b \equiv b\cdot 1 \equiv bac \equiv 1\cdot c \equiv c \mod n$.
That shows that modulo $n$, there exists a unique inverse. This technique is very useful and you can also use it to show that: there exists exactly one $0$-vector in every vector space, group, ring... and every invertible element in a group, ring... has a unique inverse.
Note: For instance $2\cdot 3 \equiv 1 \mod 5$, but also $2\cdot 8 \equiv 1 \mod 5$ (Why is this not a contradiction to the above?) | 2021-09-24T21:10:35 | {
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http://verandatiledesign.com/archive/fed166-proving-cardinality-of-infinite-sets | In elementary set theory, Cantor's theorem is a fundamental result which states that, for any set, the set of all subsets of (the power set of , denoted by ()) has a strictly greater cardinality than itself. set which is a contradiction. If $A$ is a finite set, then $|B|\leq |A| < \infty$, $$|W \cap B|=4$$ Furthermore, we designate the cardinality of countably infinite sets as ℵ0 ("aleph null"). Then U is an infinite subset of a countably infinite set A, so U is countably infinite by Theorem 14.8. << The second part of the theorem can be proved using the first part. This fact can be proved using a so-called diagonal argument, and we omit Thus by applying That is, there are 7 elements in the given set … If $A$ is countably infinite, then we can list the elements in $A$, To provide Without loss of generality, we may take $${\displaystyle A}$$ = ℕ = {1, 2, 3,...}, the set of natural numbers. Before discussing }أ��-W�������"�C��� �00�!�U嚄l�'}�� -F�NQ gvPC���S�:|����ա���ʛ������v���|�Z���uo�2�aynް�K��gS��v�������*��P~Ē�����&63 that you can list the elements of a countable set $A$, i.e., you can write $A=\{a_1, a_2,\cdots\}$, $$|W \cup B \cup R|=21.$$ /Length 1933 more concrete, here we provide some useful results that help us prove if a set is countable or not. ?�iOQ���i�ʫ�.��_���!ә�. is concerned, this guideline should be sufficient for most cases. the proof here as it is not instrumental for the rest of the book. We have been able to create a list that contains all the elements in $\bigcup_{i} A_i$, so this ����{i�V�_�����A|%�v��{&F �B��oA�)QC|*i�P@c���$[B��X>�ʏ)+aK6���� -o��� �6� ;�I-#�a�F�*<9���*]����»n�s鿻摞���H���q��ѽ��n�WB_�S����c�ju�A:#�N���/u�,�0ki��2��:����!W�K/��H��'��Ym�R2n�)���2��;Û��&����:��'(��yt�Jzu�*Ĵ�1�&1}�yW7Q���m�M(���Q Ed ���ˀ������C�s� Ӌ��&�Qh��Ou���cJ����>���I6�'�/m��o��m�?R�"o�ͽP�����=�N��֩���&�5��y&���0 �$�YWs��M�ɵ{�ܘ.5Lθ�-� GL��sU 7����>��m�z������lW���)и�$0/�Z�P!�,r��VL�F��C�)�r�j�.F��|���Y_�p���P,�P��d�Oi��5'e��H���-cW_1TRg��LJ��q�(�GC�����7��Ps�b�\���U7��zM�d*1ɑ�]qV(�&3�&ޛtǸ"�^��6��Q|��|��_#�T� $$|A \cup B |=|A|+|B|-|A \cap B|.$$ refer to Figure 1.16 in Problem 2 to see this pictorially). According to the de nition, set has cardinality n when there is a sequence Thus, any set in this form is countable. Some sets that are not countable include ℝ, the set of real numbers between 0 and 1, and ℂ. Here is a simple guideline for deciding whether a set is countable or not.$\mathbb{N}, \mathbb{Z}, \mathbb{Q}$, and any of their subsets are countable. The Infinite Looper 48,403 views. but "bigger" sets such as$\mathbb{R}$are called uncountable. Thus U is both uncountable and countably infinite, a contradiction. Theorem13.1 Thereexistsabijection f :N!Z.Therefore jNj˘jZ. so it is an uncountable set. Becausethebijection f :N!Z matches up Nwith Z,itfollowsthat jj˘j.Wesummarizethiswithatheorem. In particular, one type is called countable, but you cannot list the elements in an uncountable set. /Length 1868 stream To see this, note that when we add$|A|$and$|B|$, we are counting the elements in$|A \cap B|$twice, thus$B$is countable. Maybe this is not so surprising, because N and Z have a strong geometric resemblance as sets of points on the number line. stream To be precise, here is the definition. If$A_1, A_2,\cdots$is a list of countable sets, then the set$\bigcup_{i} A_i=A_1 \cup A_2 \cup A_3\cdots$It is injective (“1 to 1”): f (x)=f (y) x=y. Thus according to Definition 2.3.1, the sets N and Z have the same cardinality. The second part of the theorem can be proved using the first part. S*~����7ׇ�E��bba&�Eo�oRB@3a͜9dQ�)ݶ�PSa�a�u��,�nP{|���Jq(jS�z1?m��h�^�aG?c��3>������1p+!��$�R��V�:�$��� �x�����2���/�d Figure 1.13 shows one possible ordering. Thus, �L�2��T�bg���H���g�-.n?�����|������xw���.�b6����,��,�fr��X��}ޖ�]uX��ՙ]�q�3����S���P7���W?s��c[u-���hEK�c��^�e�\�� If A is a finite set, then | B | ≤ | A | < ∞, thus B is countable. The proof of this theorem is very similar to the previous theorem. $$|W|=10$$ Cardinality (Screencast 5.1.4) ... Introduction to the Cardinality of Sets and a Countability Proof - Duration: 12:14. What if$A$is an infinite set? ����RJ�IR�� On the other hand, you cannot list the elements in$\mathbb{R}\$, If you are less interested in proofs, you may decide to skip them. If A is countably infinite, then we can list the elements in A, then by removing the elements in the list that are not in B, we can obtain a list for B, thus B is countable.
.
Pork And Kabocha Stew, Etiquette Day Meaning In Urdu, Wart On Finger, Restaurants Bribie Island, Mainland China Order Online, Physics Vocabulary Crossword, | 2021-01-25T07:20:52 | {
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http://www.mathematicsgre.com/viewtopic.php?f=1&t=289 | ## GRE 9768 #37
Forum for the GRE subject test in mathematics.
toughluck
Posts: 9
Joined: Thu Jun 25, 2009 10:57 am
### GRE 9768 #37
37. $\sum_{k=1}^\infty k^2/k!$
(A) e
(B) 2e
(C) (e+1)(e-1)
(D) e^2
(E) infinity
I tried to use the taylor series approximation of e so $\sum_{k=0}^\infty x^k/k!$ and saw that if e is equal to 1 we have $\sum_{k=0}^\infty 1/k!$ but I couldn't figure out what else to do. The answer is b any help would be appreciated.
prong
Posts: 24
Joined: Thu Sep 24, 2009 12:17 am
### Re: GRE 9768 #37
k^2/k! = k/(k-1)! is a useful simplification.
then a hint: sum from 1 to inf over k is the same as sum from 0 to inf over k+1. you can probably get it just from this, i'd advise you not to read the following paragraph until you've tried to work it out using this hint.
use the fact that sum from 1 to inf over k is the same as sum from 0 to inf over k+1. we get sum from 0 to infty of (k+1)/k!. break it up into the two parts. easier part is sum from 0 to infty of 1/k!, you know that's e already. you also have sum from 0 to inf of k/k!, which is 0 + the sum from 1 to inf of k/k!. k/k! = 1/(k-1)! for k > 0 (wouldn't have made sense to do this for 0 included in the sum, since there's nothing to cancel in 0! which is the empty product). now the same trick as before, sum from 1 to infty of 1/(k-1)! = sum from 0 to infty of 1/k!. that's also e.
so it's two parts, each of which is e. therefore 2e.
sjin
Posts: 10
Joined: Sat Sep 19, 2009 5:55 pm
### Re: GRE 9768 #37
My calculation:
exp(x)= sum(x^k/k!) from k=0 to inf (then take derivative)
exp(x)= sum(k*x^(k-1)/k!) from k=1 to inf (then multiply both sides by x)
x*exp(x)=sum(k*x^k/k!) from k=1 to inf (then take derivative)
x*exp(x)+ exp(x)=sum( k^2*x^(k-1)/k!) from k=1 to inf (then let x=1)
exp+exp==sum(k^2/k!) from k=1 to inf
EugeneKudashev
Posts: 27
Joined: Tue Apr 06, 2010 8:22 am
### Re: GRE 9768 #37
I think one might consider another "GRE-approach" here. that is trying to estimate the answer w/o actually solving the problem. this approach is described thoroughly in MIT course "Street-Fighting Mathematics" (available online here: http://ocw.mit.edu/OcwWeb/Mathematics/1 ... /index.htm ) and is extremely applicable with GRE.
so, let's just write down first few terms of the series:
1 + 4/2 + 9/6 + 16/14 + 25/120 + 36/720 + ...
and try to estimate what's that:
~= 1 + 2 + 1.5 + 0.66 + 0.2 + 0.05 = 5.41.
A: e~=2.7 < 5.41. out.
B: 2*e~=5.4. very close, let's check other options.
C: (e+1)(e-1)~=6.28 (probably the most time-consuming comp.) - too much. out.
D: e^2~=7.29. out.
E: out as well, we see that denominator increases way faster than numerator (not very rigorous, but works, right?)
so, the answer is B. I believe this might be not very profound solution, however it can save you a priceless minute or two on the real test.
opinions?
Andy Chen
Posts: 1
Joined: Fri Aug 06, 2010 12:06 pm
### Re: GRE 9768 #37
Use this : k^2/k! = (k-1)/(k-1)! + 1/(k-1)!
Then the summation is immediately know to be 2e. | 2017-01-22T12:21:36 | {
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http://math.stackexchange.com/questions/204238/converting-an-annuity-due-to-annuity-immediate | # Converting an Annuity due to Annuity immediate
I'm working on the following problem at the moment while preparing for an exam.
Find the present value of payments of 200 every six months starting immediately and continuing through four years from the present, and 100 every six months thereafter through ten years from the present, if i^(2) = .06. (all quantities in dollars)
Since interest is given, and not discount I went ahead and calculated this as an annuity immediate instead of an annuity due.
$\displaystyle a=\frac{1-v^n}{i}=\frac{1-1.03^{-9}}{.03}= 7.786108922$
$\displaystyle a\times200=1557.221784$
So far so good. Here's where I think I might be doing something wrong. For the last six years of the annuity, I'm calculating the present value four years in the future and then discounting that another four years. Is this wrong?
$\displaystyle 100a= 100\times\frac{1-1.03^{-18}}{.03}=1063.495533$
So this is the present value of the last six years of the annuity at time four into the future. I discount it four years to get:
$\displaystyle 1063.495533\times\nu^8=839.5331944$
To answer the question, I get 2,396.75. The book gives 2,389.72, which is quite close. Can anyone account for the difference?
-
I see a homework tag, but I also see that you said you were doing it while preparing for an exam, so that makes it sound like it's not homework??? Homework tag means it was assigned for homework. I guess I am justified in showing you how to do it because you were almost there any way. – Graphth Sep 29 '12 at 1:10
I think your first error is that you forgot to fix the first part. You calculated the value of a series of payments, essentially at time -1 (6 months before the first) and I don't see anywhere where you accumulated it forward with a 1.03. Your second error is in the second part, $1.03^{-18}$ means you are calculating the value of 18 payments of 100 one period before the first one. There is nothing in this problem that would represent 18 payments of 100. It should be 12. Fix those two errors and see what you get. – Graphth Sep 29 '12 at 1:14
There is some ambiguity. For the first chunk, should one interpret wording as saying there are $9$ payments? Or is it $8$? – André Nicolas Sep 29 '12 at 1:15
@AndréNicolas 9 is what I interpreted it as and that gave the correct answer. So, that is what was meant I guess. But, I agree it's not that clear. I would probably say "with the last payment coming 4 years from now" or something. – Graphth Sep 29 '12 at 1:17
I interpreted it as 9. – Josh Sep 29 '12 at 15:46
My first piece of advice is to learn the ways to make these types of problems simpler. That's going to make it less likely you make a mistake and more likely that you can do it quickly. Here's a way on this problem. Learn this trick and use it in the future. (And, memorize $d = i/(1+i)$. You will need it many, many times. Don't make problems much harder because you didn't take 3 minutes to memorize this.)
Let's let our "period" be 6 months. Then our interest rate is 3% per period, as you have. My understanding of the question is we get a payment of 200 now, 1 period from now, up to 8 periods from now (4 years from now), for a total of 9 payments. Then, we get payments of 100 at times 9, 10, ..., 20, for a total of 12 of those.
Method 1: If we want the present value, this is equivalent to 21 payments of 100, starting immediately, plus another sequence of 9 payments of 100 starting immediately (the first 9 payments of each coincide and add up to 200). a double dot angle n = $\frac{1 - v^n}{d} = \frac{1 - v^n}{i/(1 + i)}$. Therefore, our total present value is
$$100 \left(\frac{1 - 1.03^{-21}}{.03/1.03} + \frac{1 - 1.03^{-9}}{.03/1.03}\right) = 2389.72$$
Method 2: The present value of the first 9 payments is:
$$200 \frac{1 - 1.03^{-9}}{.03/1.03}$$
The present value of the next 12 payments at time 8 (4 years from now, and 6 months before the first payment of 100) is:
$$100 \frac{1 - 1.03^{-12}}{.03}$$
so we need to discount this back 4 years and add to the other payments to get:
$$200 \frac{1 - 1.03^{-9}}{.03/1.03} + 100 (1.03)^{-8} \frac{1 - 1.03^{-12}}{.03}$$
- | 2013-12-10T13:34:10 | {
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http://math.stackexchange.com/questions/301344/can-any-subset-of-mathbbr2-be-expressed-in-form-a-times-b-where-a-and/301355 | # Can any subset of $\mathbb{R}^2$ be expressed in form $A\times B$, where $A$ and $B$ are subsets of $\mathbb{R}$?
This is a very elementary question I'm a little confused about.
Can any subset of $\mathbb{R}^2$ be expressed in form $A\times B$, where $A$ and $B$ are subsets of $\mathbb{R}$?
I'm thinking that it might not necessarily be so. For instance, if we consider the intervals $[0,2]$ and $[3,5]$ in $\mathbb{R}$, and throw out a single point, say $(1,4)$, out of $[0,2]\times [3,5]$ then there do not exist subsets $A$ and $B$ of $\mathbb{R}$ such that $A\times B = ([0,2]\times [3,5]) \setminus \{(1,4)\}$. Because if they were equal, surely $A$ would be equal to $[0,2]$ and $B$ would be equal to $[3,5]$. But as there is no way to put the constraint that $1$ cannot pair up with $4$, $A\times B$ will always be $[0,2]\times [3,5]$.
Am I thinking in the right direction? Thanks.
-
This is ideed not possible. – Michael Greinecker Feb 12 '13 at 17:08
Your argument works for me. Maybe the last sentence would rather be "so $A\times B=[0,2]\times [3,5]\neq [0,2]\times [3,5]\setminus \{(1,4)\}$, a contradiction". – 1015 Feb 12 '13 at 17:12
all $A\times B$ are rectangles. all subsets of $\mathbb{R}^2$ are not rectangles. – user59671 Feb 12 '13 at 17:14
@CutieKrait, this is not true. Take one of the sets not an interval, for example. – Sigur Feb 12 '13 at 17:15
@Sigur: finite or infinite unions of interiors of rectangles (with partial or complete borders or no border). – user59671 Feb 12 '13 at 17:39
If a subset $\mathrm X$ of $\mathbb R^2$ can be written as a product $\mathrm A \times \mathrm B$, then it is true that $\mathrm{pr_1}(\mathrm X) = \mathrm A$ and $\mathrm{pr_2}(\mathrm X) = \mathrm B$.
Then take the example of the unit disc in $\mathbb R^2$, it is not equal to $[-1,1]^2$ so it is not a product of two subset of $\mathbb R$.
-
Are you considering the closed unit disc? If so, the projection on first coordinate will be onto the interval $[-1,1]$. The same for the second coordinate. The reason for the disc not be a product is other. – Sigur Feb 12 '13 at 17:26
Yes... What do you try to say ? – Damien L Feb 12 '13 at 17:29
If the set is a product $A\times B$ then its projections are $A,B$ but this could happen even if the set is not a product. So, why the disc is not a product? Note that its projections are intervals. – Sigur Feb 12 '13 at 17:32
@sigur that is precisely the point if his example - he provides a space that is clearly not $[-1,1]^2$, but which projects onto $[-1,1]$ and $[-1,1]$ in the two axes, so it is a counter-example. – Thomas Andrews Feb 12 '13 at 17:33
The point with coordinates $(1,1)$ has norm $\sqrt 2$ which is strictly greater than 1, so it is not in the unit disc. Therefore the two sets are not equal... – Damien L Feb 12 '13 at 17:39
Yes you are thinking in the right direction. The answer is no. Here is a simple counter example: Consider the set $S = \{(1,2), (2,1) \}$. If there were sets $A,B \subseteq \mathbb R$ with $S = A \times B$ then $1,2 \in A$ but also $1,2 \in B$ so that $A \times B$ would have to contain $(1,1)$ and $(2,2)$ and therefore $A \times B$ is strictly greater than $S$.
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I like the fact that you thought of a finite set. There are infinite examples, but this is easy to see. – Ross Millikan Feb 13 '13 at 12:09
For finite sets $A$ and $B$, the cardinality of $A\times B$ is the product of the two separate cardinalities. So for any set $S$ of $p$ points in the plane, $p$ being prime, the only way for $S$ to be a product set is for either the $x$-coordinates or the $y$-coordinates to be all the same.
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No. You can think of sets of the form $A \times B$ as 'rectangles' (although they are really only rectangles when $A,B$ are intervals). Not all sets are rectangles, for example, take the square and rotate it $45^\circ$.
To further illustrate, with the set $A \times B$, you can pick the element $a \in A$ and $b \in B$ 'independently' to get $(a,b) \in A \times B$, but this is not always the case. (For example, if I have the line $\{(x,y) | x = y\}$ and I choose $x=1$, then I must have $y=1$, whereas if I have the square $\{ (x,y) | x \in [0,1], y \in [0,1] \}$, then if I choose $x=1$, I can still pick any $y \in [0,1]$.)
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Take the unitary open disc, for example: $$D=\{(x,y)\in \mathbb{R}^2\mid x^2+y^2<1\}.$$
It is a good exercise to show that $D$ is not a product of any two subsets of $\mathbb{R}$. Here is a simple solution.
Suppose that $D=A\times B$, for some subsets $A,B\subseteq \mathbb{R}$. Note that $(x,0)\in D$ for any $x\in (-1,1)$. So $(-1,1)\subset A$ and $0\in B$. For the same reason, $0\in A$ and $(-1,1)\subset B$. Then $(-1,1)\times (-1,1)\subseteq A\times B=D$ and so take $\sqrt{2}/2< a,b < 1$ such that $(a,b)\in A\times B$ with $1< a^2+b^2$. Contradiction!
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Let $S:= \{ (x,y) \in \mathbf{R}^2 : x < y \}$, then I'm not sure if $S$ can ever be expressed as a Cartesian product of two subsets of $\mathbf{R}$.
- | 2014-11-28T22:25:00 | {
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https://math.stackexchange.com/questions/2268936/is-the-function-mboxtrxax-convex/2269616 | # Is the function $\mbox{tr}(XAX')$ convex?
Is the function $\mbox{tr}(XAX')$ convex, where $A$ is a positive semidefinite (PSD) matrix? I know that for a general $A$, the above trace function is not convex. But for a PSD $A$, is the function convex?
• Think about operations which preserve convexity ! May 28 '17 at 18:06
Look at the table for scalar-by-matrix derivatives. In particular, one has
$$\nabla_X \text{tr}(XAX^T) = X^T(A+A^T).$$
Thus if $A$ is psd, then so is the Hessian $A + A^T = 2A$, from which convexity of $X \mapsto \text{tr}(XAX^T)$ follows.
If $\rm A$ is symmetric and positive semidefinite, then there is a matrix $\rm M$ such that $\rm A = M M^{\top}$. Hence,
$$\mbox{tr} \left( \mathrm X \mathrm A \mathrm X^{\top} \right) = \mbox{tr} \left( \mathrm X \mathrm M \mathrm M^{\top} \mathrm X^{\top} \right) = \| \mathrm X \mathrm M \|_{\text{F}}^2$$
is indeed convex.
Let $\mathrm X \in \mathbb R^{m \times n}$. Hence,
$$\mbox{tr} \left( \mathrm X \mathrm A \mathrm X^{\top} \right) = (\mbox{vec} (\mathrm X))^{\top} \left( \mathrm A \otimes \mathrm I_m \right) \mbox{vec} (\mathrm X)$$
If $\rm A$ is symmetric and positive semidefinite, then the Kronecker product $\mathrm A \otimes \mathrm I_m$ is also symmetric and positive semidefinite and, thus, $\mbox{tr} \left( \mathrm X \mathrm A \mathrm X^{\top} \right)$ is convex.
You can easily see that the function $\varphi: \mathbb R^{n^2}\to \mathbb R$ such that $\varphi(X)=\text{tr}(XAX^T)$ is the trace of the matrix whose entries are the scalar products of a set $(x_i)_{i=1,\ldots,n}$ of vectors in $\mathbb R^n$. Call their components $(x_i^j)_{j=1,\ldots,n}$ for every $i$.
The elements on the diagonal of your matrix are $$(XAX^T)_{ii}= (X^TAX)_{ii}=x_i^TAx_i=\sum_{j,k=1}^n x_i^ja_{jk}x_i^k$$ so the trace is equal to $$\sum_{i,j,k=1}^n x_i^ja_{jk}x_i^k=\sum_{i=1}^n(x_i,x_i)_A=\sum_{i=1}^n x_i^T A x_i$$
This makes $$\frac{\partial^2}{\partial (x^a_b)^2}\varphi=2a_{aa}$$ $$\frac{\partial^2}{\partial (x^a_b)\partial (x^c_d)}\varphi=a_{ac}+a_{ca}$$ so your Hessian is $$H_\varphi=A+A^T$$ Since $A$ is psd, if $y\neq 0$ $$y^T(A+A^T)y=y^T A y+ (y^TAy)^T=2y^TAy\ge 0$$ and $\varphi$ is convex, because $H_\varphi$ is positive semidefinite.
The function $tr : M_n(R) \rightarrow R$ is linear (so convex) and increasing w.r.t Positive matrix cones. And the function $Q:M_n(R) \rightarrow M_n(R)$ with $Q(X)=X A X^t$ is convex w.r.t positive cones (i.e., each row of $XAX^t$ is convex function), since $A$ is PSD (look at it Hessian of each rows ), Therefore, $F=tr oQ :M_n(R) \rightarrow R$ with $F= tr(XAX^t)$ is convex. | 2021-10-20T17:48:33 | {
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https://physics.stackexchange.com/questions/583153/how-to-calculate-the-potential-energy-of-coupled-oscillators/583161 | # How to calculate the potential energy of coupled oscillators?
The equations of motion that describe the above situation is given by:
$$m \ddot{x_1} = -2kx_1 + kx_2$$
$$m \ddot{x_2} = -2kx_2 + kx_1$$
Now I want to work out the potential energy of this system. How would I use the equation:
$$V(x) = - \int F(x) \hspace{.1cm} dx$$
to calculate the potential energy?
My attempt is to first calculate the integral for each coupled equation wrt each position:
$$- \int (- 2kx_1 + + kx_2) \hspace{.1cm} dx_1= kx^{2}_1 - k x_1 x_2 + c$$ $$- \int (- 2kx_2 + + kx_1) \hspace{.1cm} dx_2= kx^{2}_2 - k x_2 x_1 + d.$$
Now adding them together to get the total potential energy gives:
$$V(x_1, x_2) = kx^{2}_1 + kx^{2}_2 + - 2 k x_1 x_2 + f.$$
The actual solution is given by:
$$V(x_1, x_2) = kx^{2}_1 + kx^{2}_2 + - k x_1 x_2$$
Now I have inkling as to why my answer is wrong. When adding the potentials together to calculate the total potential I have double added the potential associated with the middle spring. However, mathematically how do I fix this issue to get the correct solution?
• Remember that a spring has energy $\frac 12 k(x_{1}-x_0 -a)^2$ where $a$ is it's natural length. – mike stone Oct 1 at 13:54
• – Frobenius Oct 1 at 18:26
As @mikestone suggested in the comments, the easiest way to solve this problem is by summing the potential energies of all the springs.
However, I would like to comment on the approach described in the question: the equations of motion can be written as $$m\ddot{x}_1 = -\frac{\partial V(x_1, x_2)}{\partial x_1} = -2kx_1 + kx_2,\\ m\ddot{x}_2 = -\frac{\partial V(x_1, x_2)}{\partial x_2} = -2kx_2 + kx_1,$$ where $$V(x_1, x_2)$$ is the potential energy of the two oscillators. We thus have two partial differential equations (pde) for this potential energy. Integrating the first one in respect to $$x_1$$ we obtain: $$V(x_1, x_2) = kx_1^2 - kx_2x_1 + C(x_2),$$ i.e. we obtain $$V(x_1, x_2)$$ up to an unknown constant, $$C(x_2)$$, which may be dependent on $$x_2$$ (since the differential equation is in respect to $$x_1$$). Substituting this into the second equation we obtain: $$\frac{\partial V(x_1, x_2)}{\partial x_2} = -kx_1 +\frac{dC(x_2)}{dx_2} = 2kx_2 - kx_1,$$ that is $$\frac{dC(x_2)}{dx_2} = 2kx_2 \Rightarrow C(x_2) = kx_2^2$$ (up to a constant independent on either $$x_1$$ or $$x_2$$), and we obtain $$V(x_1, x_2) = kx_1^2 + kx_2^2 -kx_1x_2.$$
• Thank you that makes perfect sense. – Nice1m80 Oct 1 at 14:21
Question Can be Solved without Integration. Potential Energy is stored in the springs only.
Let The Rightmost spring is stretched by amount $$x_2$$, And the Leftmost spring is stretched by amount $$x_1$$ then the Middle spring will be stretched by amount $$(x_2 - x_1)$$.($$As\ shown\ in \ figure$$) $$(2a +x_2)-(a+ x_1)= a+ x_2 - x_1 \qquad (Proof)$$
where a is the length of spring.
The Potential energy of the spring is given by $$V(x) = \frac{1}{2}kx^{2}$$
Potential energy of leftmost spring is $$V(x_1) = \frac{1}{2}kx^{2}_1 \qquad ...(1)$$
Potential energy of rightmost spring is $$V(x_2) = \frac{1}{2}kx^{2}_2 \qquad ...(2)$$
Potential energy of Middle spring is $$V(x_3) = \frac{1}{2}k(x_1-x_2)^{2}\qquad ...(3)$$
$$V(x)=V(x_1)+V(x_2)+V(x_3)$$
$$V(x) = kx^{2}_1 + kx^{2}_2 - kx_1x_2$$
• For the middle spring $(x_1-x_2)^2$ in eq. 3 – Vadim Oct 1 at 16:09
Here's a slightly different take on someone else's answer that does not assume the spring constants are necessarily the same. It also uses the inspection method, based on the idea that the potential energy of a spring is $$\frac{1}{2}\kappa \times \hbox{stretch}^2$$ (or compression).
For your first spring on the left, the stretch would come from the displacement $$x_1$$ so $$V_1=\frac{1}{2}\kappa_1 x_1^2$$. For the spring in the middle the stretch is $$\vert x_1-x_2\vert$$ so you'd get $$V_2=\frac{1}{2}\kappa_2(x_1-x_2)^2$$. For the spring on the right $$V_3=\frac{1}{2}\kappa_3 x_2^2$$.
Then it's just a matter of finding the $$\kappa_i$$'s using the equations of motion: \begin{align} m\ddot{x_1}&=-\kappa_1 x_1 -\kappa_2 (x_1-x_2)=-(\kappa_1+\kappa_2)x_2+\kappa_2x_2\, ,\\ m\ddot{x_2}&=\kappa_2(x_1-x_2)-\kappa_3 x_2=-(\kappa_2+\kappa_3)x_2+\kappa_2 x_1 \end{align}
Direct comparison with your EOMs yields $$\kappa_2=\kappa_1=\kappa_3=k$$ so the net potential energy is $$V=V_1+V_2+V_3=\frac{1}{2}k\left(x_1^2+x_2^2+(x_1-x_2)^2\right)\, .$$ | 2020-12-04T02:26:47 | {
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https://math.stackexchange.com/questions/3319989/what-is-the-total-area-enclosed-between-the-curve-y-x2-1-the-x-axis-and-the | # What is the total area enclosed between the curve $y=x^2-1$, the x-axis and the lines $x=-2$ and $x=2$?
What is the total area enclosed between the curve $$y=x^2-1$$, the x-axis and the lines $$x=-2$$ and $$x=2$$?
I tried to find the area by using the integrals $$\int_1^2$$ and $$\int_{-1}^{-2}$$ .
$$x^2-1$$ integrated is $$\frac{x^3}{3}-x$$
The answer is supposed to be $$4$$, but by adding up $$\int_1^2$$ and $$\int_{-1}^{-2}$$ my area is $$\frac{8}{3}$$. What am I doing wrong?
You forgot the area under the x-axis. The signed area of this is:
$$\int_{-1}^1 x^2-1 \ dx \$$
Can you continue?
• Thanks! I calculated the area under the x-axis. It gives me the area $\frac{-4}{3}$. Do I add it to the other areas even though it's negative? – Eris Tyenns Aug 11 '19 at 12:29
• If $-1 < x < 1$, then $x^2 - 1 < 0$. – N. F. Taussig Aug 11 '19 at 12:30
• No, since areas cannot be negative (in this case). You should add the absolute value of all the areas since this will always give a positive answer. – Toby Mak Aug 11 '19 at 12:31
• Thanks! Now the area is 4. – Eris Tyenns Aug 11 '19 at 12:36
By symmetry, it is twice the area between the $$y$$-axis and the line $$x=2$$. The positive $$x$$-intercept is $$x=1$$, and on the interval $$[0,1]$$, the curve is below the $$x$$-axis, hence the total (geometric) area is $$2\biggl(\int_1^2y(x)\,\mathrm d x-\int_0^1y(x)\,\mathrm d x\biggl).$$
It is an even function Required Area will be , $$Area = 2\;(\int_0^2| x^2\;-1| \,dx)$$ $$=2(\;\int_0^1(1-x^2)\,dx\;+\int_1^2(x^2-1)\,dx\;)$$$$=2( \, \frac23\;+\;\frac43\,)=\;4$$ | 2021-04-21T13:54:32 | {
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https://math.stackexchange.com/questions/2637837/link-between-average-and-integrals | Inspired by the definition of an integral: $\int^b_af(x)dx=\lim_{n\to∞} \Delta x \sum^{n}_{i=1} f(x^*_i)$.
I wanted to create an equation that finds the average height for all the points on an interval "a b" of a function. So I came up with this equation: $$\lim_{n\to∞}\frac{1}{n}\sum^{n}_{i=1}f(a+i\Delta x)$$ Where $\Delta x=\frac{b-a}{n}$. I noticed that the "average equation" is very similar to the definition of an integral. In fact, it is equal to $\frac{\int^b_a f(x)dx}{b-a}$ or the integral divided by distance between a and b. I was wondering conceptually why these two ideas are related in this way.
• Good job, you've arrived at the usual definition of the average of a function on an interval. It seems like you've already explained the connection between the two ideas. I don't understand what you're asking. Feb 6 '18 at 0:10
• en.wikipedia.org/wiki/Mean_of_a_function Feb 6 '18 at 0:11
An average is a sum of datapoints divided by the size of the dataset. The integral formula you gave is the continuous version of this: the integral gives the continuous sum of datapoints in an interval $[a,b]$, and $b-a$, the length of the interval, is the size of the dataset.
There is also a geometric interpretation. Let's look at the formula you gave. The integral gives the area under the function in the interval $[a,b]$. Let's say we took that area and smoothed it out so it was rectangular-shaped, with width $[a,b]$ and some height. It would be geometrically reasonable to call this height the average height of the function. And we can get the height by dividing the area (the integral) by the width ($b-a$) -- your formula.
• No, the size of the dataset is not $(b-a)$. The dataset's size is the number of subintervals in the partition. This is why the average is equal to $\sum_{i=0}^k \frac{f(a + \Delta x \cdot i)}{n}$ that is exactly $\frac{\int_a^bf(x)dx}{b-a}$ if the integral's interval is $[a,b]$ and the points in the partition are $n$. Sep 28 '20 at 3:28
Just for fun, lets relate the definition you discovered to the mean value and show that the two definitions of mean are consistent. Let our function $f(x)$ be given. Let us define a function $g(x)$ such that $f(x) = g'(x)$ (I know this is not always possible but let's pretend that $f(x)$ is nice and integrable, if it's not then defining it through an integral is not useful). Let's say that the average value of $f$ occurs at the point $x=c$. From our definition of $g$ we can say $$f(c) = g'(c)$$ From the mean value theorem we can say $$f(c)=g'(c) = \frac{g(b)-g(a)}{b-a} = \frac{\int_a^b g'(x)dx}{b-a} = \frac{\int_a^bf(x)dx}{b-a}$$ Which is the deifintion you obtained. So, our definitions of mean are consistent, Yay! | 2022-01-23T11:00:07 | {
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http://gqjl.aeut.pw/generate-random-variable-from-uniform-distribution.html | # Generate Random Variable From Uniform Distribution
For details on how to generate such numbers with very high quality, see reference 4. Random numbers (or deviates) can be generated for many distributions, including the Normal distribution. You can control that shared random number generator using rng. rand Convenience function that accepts dimensions as input, e. Generate a random number between 5. You now have a normally distributed set of random numbers, based on a defined mean and standard deviation. Generating non-uniform random variables Suppose we have a way to generate uniform random variables. run every random variable follow Gaussian distribution. Also, useful in determining the distributions of functions of random variables Probability Generating Functions P(t) is the probability generating function for Y Discrete Uniform Distribution Suppose Y can take on any integer value between a and b inclusive, each equally likely (e. 2867365 , -0. Generate 2. The probability distribution for a discrete random variable assignsnonzero probabilities toonly a countable number ofdistinct x values. One of the most important application is to generate Dirichlet distributed random vectors, which plays a key role in topic modeling and other Bayesian algorithms. Aha! This shows that is the cumulative distribution function for the random variable ! Thus, follows the same distribution as. In R, we only need to add "r" (for random) to any of the distribution names in the above table to generate data from that distribution. For an introduction to the fundamental principles of probability, see reference 5. Observation: We can also manually generate a random sample that follows any of the distributions supported by Excel without using the data analysis tool. Please feel free to use the sample quizzes and tests below to help you study for this class. random variables, such as in Section 2. Variance of a Standard Uniform Random Variable. Devroye and R. Generate two uniform random variables, U;V. It is common to have a low-level Random number generator which generates uniform variates on [0, 1) [0,1) and generate variates from other distributions by "processing" those variables. The Standard Normal Distribution The normal distribution with parameter values µ = 0 and σ = 1 is called the standard normal distribution. A discrete random variable is a random variable that can take on any value from a discrete set of values. Implementation in R R is the interactive language for statistical computing we are most interested in using in this course. Choosing random numbers from continuous distributions by rejection. This function should return random numbers, but not the uniform kind of random number produced by most generators. Similar to step 4, generate a QQ plot for the normal random variables you obtained in step 8. Method-1: Sum of Uniform Random Variables The simplest way of generating normal variables is an application of the central limit theorem. Its inflection points are at 1 and -1. 1 Generating Random Values in R The simple case of generating a uniform random number between 0 and 1 is handled by the runif function. Generating random numbers with NumPy. When re-started in the same state, it re-delivers the same output. The lognormal distribution is defined in terms of the mean and variance of its associated normal distribution. The normal distribution is a common distribution used for many kind of processes, since it is the distribution that the aggregation of a large number of independent random variables approximates to, when all follow the same distribution (no matter which distribution). (b)transform the random number u by the inverse of the CDF. To use this in your own course/training, please obtain permission from Prof. This video will show usage of data analysis toolpak of excel for generation of random numbers with binomial, uniform, discrete, bernaulli, pattern, poisson distribution. The variable "color of M&M" used in this example is a discrete variable, and its distribution is also called discrete. show all the steps necessary to generate a. Generate 50 normal random variable from N(5, 2). Generate 100 samples of binomial(20,. Jensen's inequality provides with a sort of minimum viable reason for using $$X^2$$. Functions that generate random deviates start with the letter r. Discrete Uniform Distributions A random variable has a uniform distribution when each value of the random variable is equally likely, and values are uniformly distributed throughout some interval. Devroye and R. Then, in Section 2, we discuss how to generate realizations from the Dirichlet using three methods: urn-drawing, stick-breaking, and transforming Gamma random variables. You can control that shared random number generator using rng. The uniform distribution is the underlying distribution for an uniform random variable. Dear all, This is a simple probability problem. uniform (low=0. Generating Correlated Random Variables Bivariate Gaussian Distribution The joint (bivariate) PDF for X 1;2 is f X1X2 (x 1;x 2) = 1 2ˇ 1 (1 2ˆ2)1=2 exp 2 6 4 1 2(1 ˆ2) 0 B @ x2 1 ˙ 1 + x2 2 ˙2 2 2ˆ x 1x 2 ˙˙ 2 1 C A 3 7 5 A more useful of writing this PDF is to use the column vector X = col(X 1;X 2) and the covariance matrix C = 0 B B B B B @ ˙2 1 ˙ 1˙ 2ˆ ˙ 1˙ 2ˆ ˙ 2 2 1 C C C C C A. Unfortunately, methods to create such random numbers are not always implemented in statistical software packages (which often only offer univariate random number generators). Expected Value of Joint Random Variables. Probability Distributions A probability distribution describes how the values of a random variable is distributed. For example,. 2 Define a random sample from a distribution of a random variable. Use R to find the maximum and minimum values. 1 Sampling from discrete distributions A discrete random variable X is a random variable that has a probability mass function p(x) = P(X = x) for any x ∈ S, where S = {x 1,x 2,,x k} denotes the sample space, and k is the (possibly infinite) number of possible outcomes for the discrete variable X, and. Let's start by finding E(X 2):. The full list of standard distributions available can be seen using ?distribution. The normal RNG value is calculated by. Again, using rnorm to generate a set of values from the distribution. This is a step-by-step explaination of how to calculate a transformation function that converts a random variable of one distribution to another distribution. Then, $$X$$ will follow the distribution governed by the CDF $$F_X$$, which was our desired result. Generate random numbers from the Weibull distribution. The variance of a uniform random variable is: Var(x) = (1/12)(b-a) 2. The x and y coordinates of points sampled at random from the density distribution shown in figure 5 are uniformly distributed and correlated (the degree of correlation depends on the value of t; see the next section). Samples a random number from the standard Normal (Gaussian) Distribution with the given mean and sigma. run every random variable follow Gaussian distribution. In SPSS, RV is short for random variable. In our Introduction to Random Variables (please read that first!) we look at many examples of Discrete Random Variables. of nindependently generated random variables with Uniform[0;1] distribu- tions has a Beta(k;n k+ 1) distribution. A random variable having a uniform distribution is also called a uniform random variable. The uniform distribution is central to random variate generation. Generating random numbers Generating uniform(0,1) deviates Books: DE Knuth (1998) The art of computer programming, vol 2, 3rd ed, ch 3 Numerical recipes in C, ch 7 Linear congruential generator Xn+1 = a Xn + c (mod m) m = modulus = 232 - 1 a = multiplier = choose carefully! c = increment = (maybe) 0 X0 = seed. Unfortunately, methods to create such random numbers are not always implemented in statistical software packages (which often only offer univariate random number generators). As we will see in later chapters, we can generate a vast assortment of random quantities starting with uniform random numbers. You now have a normally distributed set of random numbers, based on a defined mean and standard deviation. 7 CONTINUOUS RANDOM VARIABLE, II Let the random variable X be a random number. One of the most important application is to generate Dirichlet distributed random vectors, which plays a key role in topic modeling and other Bayesian algorithms. We need to specify the number of samples to be generated. Its inflection points are at 1 and -1. Compute ε=1−α and then τ = −εtan(αφ 0) Compute tan(½φ), tan(½εφ) and tan(½εφ)/(½εφ). The MGF of a standard uniform random variable X is. All you need is to switch this uniform distribution in the interval that you desire. I can do this in a worksheet by add Data Analysis add in. Given a random variable. Sample question #1: The average amount of weight gained by a person over the winter months is uniformly distributed from 0 to 30lbs. The Excel RAND and RANDBETWEEN functions generate pseudo-random numbers from the Uniform distribution, aka rectangular distribution, where there is equal probability for all values that a random variable can take on. Generating Random Data It is useful to generate random variables from a specific distribution. We start with the basic definition of a random variable: Definition 1 A Random variable is a variable that can take on a given set of values, called the sample space and denoted , where the likeli- hood of the values in is determined by ’s probability distribution function (pdf). The normal distribution is a common distribution used for many kind of processes, since it is the distribution that the aggregation of a large number of independent random variables approximates to, when all follow the same distribution (no matter which distribution). Set Y = log(V). 0 1 0 1 x f(x) The cumulative distribution function on the support of X is. The idea is simple. Use this to obtain E[X] and Var[X]. For example, to generate a random number from a standard normal distribution, use =NORM. Metin Çakanyıldırımused various resources to prepare this document for teaching/training. There are many different. The normal probability plot is very helpful for deciding whether data come from a distribution that is approximately normal. Denker explains a method of generating random numbers with arbitrary distribution. R has functions to generate a random number from many standard distribution like uniform distribution, binomial distribution, normal distribution etc. The RAND function uses the Mersenne-Twister random number generator (RNG) that was developed by Matsumoto and Nishimura (1998). Conversely, it is easy to show in this case that if U is uniformly distributed on [0,1] then F−1(U) has the distribution F(x). But I’ve also built a simple Excel template that will help make this process a lot easier. random variables. two steps: (1) generating imitations of independent and identically distributed (i. In other words, the cumulative distribution function for a random variable at x gives the probability that the random variable X is less than or equal to that number x. As a second illustration of sing SPSS syntax to create multiple random numbers u in a given distribution, the following syntax generates 10 random number variables X1 through X10 with 1,000 cases each from a uniform distribution, from a minimum of 500 to a maximum of 1. Set Examples: binomial distribution (convolution of IID Bernoullis) Negative binomial (convolution of IID geometrics) Chi-squared K (convolution of IID Chi-squared df=1) Ga(a,b) b*convolution of a IID exponential(1)s. As per the solution above, we already have a uniformly distributed random number generator R(m) in range [0,m-1] (can be done by tossing k coins, one for each bit). The distribution is evaluated for. Then I form the random variable X by taking the negative of the logarithm of 1 minus U. Continuous Random Variables The Uniform Distribution The Lognormal Distribution • It is a random variable such that its natural. Example Let be a uniform random variable on the interval , i. A standard uniform random variable X has probability density function f(x)=1 0 0. , a continuous random variable with support and probability density function Let where is a constant. random Alias for random_sample. The following example first creates 10 random numbers in the interval 1 … 10 and assigns it to the variable r and evaluates r to print its values. Gamma random variate has a number of applications. Expected value. Method-1: Sum of Uniform Random Variables The simplest way of generating normal variables is an application of the central limit theorem. Density Plot In R. a random variable contain important information about the random variable, or, more precisely, about the distribution function of that variable. This example uses the Weibull distribution as the intended target distribution. alently by (3), is called the distribution function of the random variable X. When you ask for a random set of say 100 numbers between 1 and 10, you are looking for a sample from a continuous uniform distribution, where α = 1 and β = 10 according to the following definition. Explanation for the above result:. Generating Uniform Random Numbers in MATLAB. You can control that shared random number generator using rng. Then, it creates another 1000 random variables and uses plot(…) and hist(…) to demonstrate that the distrribution of runif is (more or less) uniform:. Since the sample was taken from a uniform distribution in the range [50, 150], as can be seen from Uniform Distribution, the population mean is = 100 and the standard deviation is = 28. Obtain the desired X from. Decorrelation models are implemented by eith er using one random variable from the desired pdf for all. The efficiency: use this method in this particular example, the rejection probability is 1/4 on the average for each number generated. A distribution is a simple way to visualize a set of data, either as a graph or in a list of stating which random variables have lower or higher chances of happening. Most Monte Carlo sampling or integration techniques assume a “random number generator,” which generates uniform statistically independent values on the half open interval [0,1); for reviews see, e. Simply choose a random point on the y. In this post, you will see the steps to generate random numbers from the exponential distribution in Excel. The Uniform Distribution Description. Expected value. A random variable having a uniform distribution is also called a uniform random variable. Many gaming frameworks only include functions to generate continuous uniformly distributed numbers. Suppose that this distribution is governed by the exponential distribution with mean 100,000. 5 and you have the output of an approximately normal distribution with mean equal to 0 and variance equal to (1/12) * (1/sqrt(N)) (see wikipedia on uniform distributions for that last one) n=10 gives you something half decent fast. Inverse Transformation Method and Random Variables Date: 02/03/2009 at 12:00:41 From: Peter Subject: Generate random number using Inverse Transformation Method Hi, I am looking for guidance on the proof of the Inverse Transformation Method to simulate a random variable having a continuous distribution. your computer) can already generate random numbers with a uniform distribution on the interval (0,1). Generating random numbers with NumPy. Let's create a new variable whose values will be random draws from a normal distribution with a mean of 0 and a standard deviation of 1. Its inflection points are at 1 and -1. We have seen examples of generating random integers in a range say 1 to 10, which is quite common and very useful as well. To generate 10 uniform random numbers between 0 and 10, we will use # random numbers from uniform distribution # Generate 10 numbers from 0 to 10 n = 10000 a = 0 b = 10 data_uniform = uniform. Similarly, you will generate a different random number that too will be uniformly distributed when your first normal random variable is > 0. Take a horizontal slice of the distribution at height. 3-2 RANDOM VARIATE GENERATION Table 3. 20, each case has 20% chance of being selected. In this post, you will see the steps to generate random numbers from the exponential distribution in Excel. Generating Uniform Random Numbers in MATLAB. The acceptance-rejection method is an algorithm for generating random samples from an arbitrary probability distribution, given as ingredients random samples from a related distribution and the uniform distribution. If called without parameter random delivers a floating point pseudorandom number in the interval [0, 1), i. This function should return random numbers, but not the uniform kind of random number produced by most generators. To generate numbers from a normal distribution, use rnorm(). This section will introduce the basics of this process and demonstrate it with some straightforward examples. For example, inverting $$F_X$$ is easy if $$X$$ is an exponential random variable, but its harder if $$X$$ is Normal random variable. Generate a random variable from other r. Does anyone know how to do it in R? Many thanks! Menghui _____ R-help at stat. This is particularly important for simulations, since many computer languages have an algorithm for generating random numbers, which are simulations of independent variables, each with the standard uniform distribution. ) random variables of a given distribution and instead must create pseudo-random numbers. We’ll see most every-thing is the same for continuous random variables as for discrete random variables except integrals are used instead of. Finding Probabilities for a Continuous Uniform Distribution. The lognormal distribution is bounded below by 0 and skewed to the right (it has a long right tail). If , then is a random variable with CDF. Uniform distributions can be discrete or continuous, but in this section we consider only the discrete case. Steps involved are as follows. 5 and you have the output of an approximately normal distribution with mean equal to 0 and variance equal to (1/12) * (1/sqrt(N)) (see wikipedia on uniform distributions for that last one) n=10 gives you something half decent fast. The normalised n-th central moment or standardised moment is the n-th central moment divided by σ n; the normalised n-th central moment of the random variable X is = [(−)]. The student will compare and contrast empirical data from a random number generator with the uniform distribution. Generating random variables, testing the normality by Q-Q plot, and confidence interval for mean value Background In class, we have introduced the uniform distribution and normal distribution, including. Specifically, radii can be sampled from and angle can be sampled from. alently by (3), is called the distribution function of the random variable X. After given a way to generate a random uniform variable, to generate a value in the target distribution, we: Create some new state from a given state using any way we like**. Mathematics | Probability Distributions Set 1 (Uniform Distribution) Prerequisite – Random Variable In probability theory and statistics, a probability distribution is a mathematical function that can be thought of as providing the probabilities of occurrence of different possible outcomes in an experiment. But it is particularly useful for random variates that their inverse function can be easily solved. These are special cases of moments of a probability distribution. It represents the convergence of the average of a set of samples from a uniform distribution. The CMS takes as input the values of alpha and beta and generates a variable with μ (mean) zero and ν (dispersion) equal to unity. Uniform Random Numbers - The Standard Excel Way. 0, size=None) ¶ Draw samples from a uniform distribution. Generating non-uniform random variables Suppose we have a way to generate uniform random variables. In other words, any value within the given interval is equally likely to be drawn by uniform. Thus, as with discrete random variables, the expected value of a continuous random variable can be thought of as a weighted average of the values that the random variable can take, where the weights are provided by the distribution of the variable. To generate a random variable that has CDF F(y) = 1 e y for y 0, we can use the following steps (a)generate a random number u from Uniform (0;1). The Excel RAND and RANDBETWEEN functions generate pseudo-random numbers from the Uniform distribution, aka rectangular distribution, where there is equal probability for all values that a random variable can take on. The acceptance-rejection algorithm is then as follows: (1) independently simulate a random number with a uniform distribution over the unit interval and a realization * of the random variable ; and then (2) using a fixed, strictly positive number , accept * as a realization of if , where and are the probability densities of the random variables. Formally, a random variable is a function that assigns a real number to each outcome in the probability space. Again, using rnorm to generate a set of values from the distribution. NORMSDIST for the standard normal distribution e. Discrete uniform distribution, yielding integers. The full list of standard distributions available can be seen using ?distribution. The central limit theorem is proved and the probability density functions are derived of those sampling distributions linked to the normal distribution. Donald Knuth describes a way to generate such values in §3. \label{sec:RN. Internal Report SUF–PFY/96–01 Stockholm, 11 December 1996 1st revision, 31 October 1998 last modification 10 September 2007 Hand-book on STATISTICAL. an exponentially distributed random variable. In statistical programming, it is common to generate random integers in the range 1 to Max for some value of Max, because you can use those values as observation numbers (indices) to sample from data. The random module provides a fast pseudorandom number generator based on the Mersenne Twister algorithm. 03175853, 1. Generating Correlated Random Variables Bivariate Gaussian Distribution The joint (bivariate) PDF for X 1;2 is f X1X2 (x 1;x 2) = 1 2ˇ 1 (1 2ˆ2)1=2 exp 2 6 4 1 2(1 ˆ2) 0 B @ x2 1 ˙ 1 + x2 2 ˙2 2 2ˆ x 1x 2 ˙˙ 2 1 C A 3 7 5 A more useful of writing this PDF is to use the column vector X = col(X 1;X 2) and the covariance matrix C = 0 B B B B B @ ˙2 1 ˙ 1˙ 2ˆ ˙ 1˙ 2ˆ ˙ 2 2 1 C C C C C A. How to generate two uniform, partially correlated random distributions with the following constraints. The Moment Generating Function of the Normal Distribution Recall that the probability density function of a normally distributed random variable xwith a mean of E(x)=„and a variance of V(x)=¾2is (1) N(x;„;¾2)= 1 p (2…¾2) e¡1 2 (x¡„) 2=¾2: Our object is to flnd the moment generating function which corresponds to this distribution. Samples from a continuous uniform random distribution We can generalize the case of 1 or two dice to the case of samples of varying size taken from a continuous distribution ranging from 0-1. Then, in Section 2, we discuss how to generate realizations from the Dirichlet using three methods: urn-drawing, stick-breaking, and transforming Gamma random variables. Many computers have built-in routines for generating independent random numbers from the uniform distribution U[0,1], so we shall focus on how these may be manipulated in order to obtain random numbers from other distributions. Wikipedia attributes the following algorithm to Donald Knuth: init: Let L ← exp(−λ), k ← 0 and p ← 1. random_integers Discrete uniform distribution over the closed interval [low, high]. 4) We get the random variables by generating a random number U and then. Generate random numbers from the standard uniform distribution. Random number distribution that produces floating-point values according to a uniform distribution, which is described by the following probability density function: This distribution (also know as rectangular distribution) produces random numbers in a range [a,b) where all intervals of the same length within it are equally probable. I do not expect you to know how to derive the MGF for normal random variables for the purposes solving a problem on an exam. Schaum's Outline of Probability and Statistics 36 CHAPTER 2 Random Variables and Probability Distributions (b) The graph of F(x) is shown in Fig. of a normal rv can not be expressed explicitly. The probability density function is illustrated below. Then, in Section 2, we discuss how to generate realizations from the Dirichlet using three methods: urn-drawing, stick-breaking, and transforming Gamma random variables. This note is about the topic of generating Gaussian pseudo-random numbers given a source of uniform pseudo-random numbers. So the Excel command includes "DIST" e. The uniform distribution is the underlying distribution for an uniform random variable. Simulation from discrete uniform. Density Plot In R. Expectation of a Random Variable The expectation of a random variable X is denoted as E(X). Discrete Random Variables and Probability Distributions Part 4: More of the Common Discrete Random Variable Distributions Sections 3. minval: A python scalar or a scalar tensor. We have already looked at Variance and Standard deviation as measures of dispersion under the section on Averages. Distribution of Random variables. The C++ TR1 library supports non-uniform random number generation through distribution classes. Understand the concept of a “do loop. 2867365 , -0. Then, f(x i) = 1 n where f(x) represents the probability mass function (PMF). The acceptance-rejection algorithm is then as follows: (1) independently simulate a random number with a uniform distribution over the unit interval and a realization * of the random variable ; and then (2) using a fixed, strictly positive number , accept * as a realization of if , where and are the probability densities of the random variables. Sometimes, we also say that it has a rectangular distribution or that it is a rectangular random variable. $$X^2$$ can't be less then zero and increases with the degree to which the values of a Random Variable vary. The continuous uniform distribution is the probability distribution of random number selection from the continuous interval between a and b. The first argument in the function is the number of random variables that you want from the distribution. In statistics and probability theory, the Gaussian distribution is a continuous distribution that gives a good description of data that cluster around a mean. Synonyms for Random variate in Free Thesaurus. In other words, any value within the given interval is equally likely to be drawn by uniform. Generate a random variable from other r. Because an example is often an effective way to convey main ideas, the following DATA step generates a random sample of 100 observations from the standard normal distribution. In statistical programming, it is common to generate random integers in the range 1 to Max for some value of Max, because you can use those values as observation numbers (indices) to sample from data. This is a step-by-step explaination of how to calculate a transformation function that converts a random variable of one distribution to another distribution. The function beta() in R calculates the value of the beta function:. A great article written by John S. While there are specialized algorithms to generate random numbers from specific distributions, a common approach relies on generating uniform random numbers and then using the inverse function of the desired distribution. The most direct way of generating random samples from a Poisson distribution is efficient for some parameters and inefficient for others. If both X, and Y are continuous random variables, can we nd a simple way to characterize F. And this gives me a random variable, which has an exponential distribution. Probability distribution functions can also be applied for discrete random variables, and even for variables that are continuous over some intervals and discrete elsewhere. In the Wolfram Language, you can directly compute several dozen properties from symbolic distributions, including finding the probability of an arbitrary event. how non-uniform random numbers are generated in order to make a custom so-lution. To implement this method in Excel, you can use the RAND function to generate random numbers from the Uniform distribution, and apply the built-in functions to calculate the ICDF. In particular cases, there can be clever ways to simulate random variables. As we will see in later chapters, we can generate a vast assortment of random quantities starting with uniform random numbers. Uses the Acceptance-complement ratio from W. Since the sample was taken from a uniform distribution in the range [50, 150], as can be seen from Uniform Distribution, the population mean is = 100 and the standard deviation is = 28. 2 Generate 10 random normal numbers with mean 5 and standard deviation 5 (normal(5,5)). 1 for the event probability. Internal Report SUF–PFY/96–01 Stockholm, 11 December 1996 1st revision, 31 October 1998 last modification 10 September 2007 Hand-book on STATISTICAL. The lognormal distribution is defined in terms of the mean and variance of its associated normal distribution. 7 CONTINUOUS RANDOM VARIABLE, II Let the random variable X be a random number. The uniform distribution is the underlying distribution for an uniform random variable. 5 When you generate random numbers from a specified distribution, the distribution represents the population and the resulting numbers represent a sample. Below are two examples. Example Let be a uniform random variable on the interval , i. For a discrete probability distribution E(X) = x p(x). 0 1 0 1 x f(x) The cumulative distribution function on the support of X is. Generate samples of random data from a distribution Learn more about Minitab In the Generate Random Data dialog box, specify the amount of data to generate, the distribution, and the parameters for the distribution. Random number distribution that produces floating-point values according to a uniform distribution, which is described by the following probability density function: This distribution (also know as rectangular distribution) produces random numbers in a range [a,b) where all intervals of the same length within it are equally probable. Chapter 4 considers groups of random variables, with an emphasis on two random variables. Moreover, even if it is, there may be alternative methods for generating a rv distributed as F that is more efficient than the inverse. uniform¶ numpy. Get random float number with two precision. Probability distribution functions can also be applied for discrete random variables, and even for variables that are continuous over some intervals and discrete elsewhere. Generating random numbers with NumPy. Then: A probability such as Pr(X <= x) is given by the cumulative distribution function. For details on how to generate such numbers with very high quality, see reference 4. deviation determines the width of the bell curve. How do you generate a set of random numbers in Microsoft Excel, which follows the Poisson Distribution? The RAND() function in Excel generates random numbers that are uniformly distributed. Synonyms for Random variate in Free Thesaurus. Generate random numbers from the standard uniform distribution. It's a variable, but its operator() has been overloaded so it can be used as if it was a function (Covered later). This command generates a set of pseudorandom numbers from a uniform distribution on [0,1). 2 What this means is that we can reduce the problem of simulating to that of gen-erating random variables. P olya distribution), which nds extensive use in machine learning and natural language processing. Compute the cdf of the desired random variable. Random Variable Definition: A random variable is defined as a real- or complex-valued function of some random event, and is fully characterized by its probability distribution. The Excel RAND and RANDBETWEEN functions generate pseudo-random numbers from the Uniform distribution, aka rectangular distribution, where there is equal probability for all values that a random variable can take on. rolling a dice, where a=1 and b=6). Generating Uniform Random Numbers in MATLAB. Uniform Random Numbers - The Standard Excel Way. Method-1: Sum of Uniform Random Variables The simplest way of generating normal variables is an application of the central limit theorem. ch mailing list. Devroye and R. The is derived from the term (where is the mean of the uniform distribution - ). | 2020-04-01T18:14:48 | {
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https://math.stackexchange.com/questions/3141609/other-inner-products-for-mathbbrn | # Other inner products for $\mathbb{R}^n$
For $$\mathbb{R}^n$$, the standard inner product is the dot product. It is defined as $$\langle v,\,w\rangle = \sum_i v_i \cdot w_i$$. I am aware that any scaled version, namely $$\langle v,\,w\rangle = \sum_i\lambda_i\cdot v_i \cdot w_i$$ will still satisfy the 4 inner product requirements.
Is there any inner product for $$\mathbb{R}^n$$ that is not just a scaled version of the standard dot product?
I tried for $$\mathbb{R}^2$$ with $$\langle v,\,w\rangle = v_1 \cdot w_2 + v_2 \cdot w_1$$ but that is not positive definite.
• So, everyone is saying you can also use a matrix, and that works. Is there a proof though that any inner product can be represented by such a matrix type inner product? – Ion Sme Mar 9 at 22:27
Yes. Given any positive definite $$n \times n$$ matrix $$A$$, $${\bf x} \cdot {\bf y} := {\bf y}^{\top} A {\bf x}$$ defines an inner product on $$\Bbb R^n$$, and for $$n > 1$$ there are nondiagonal positive definite matrices, for example, $$\pmatrix{1&\epsilon\\\epsilon&1\\&&1\\&&&\ddots\\&&&&1}$$ for any $$0 < |\epsilon| < 1$$.
Conversely all inner products arise this way, as we can recover $$A$$ by setting $$A_{ij} = {\bf e}_i \cdot {\bf e}_j$$ for the standard basis $$({\bf e}_i)$$.
On the other hand, given any inner product on $$\Bbb R^n$$, applying the Gram-Schmidt Process produces an orthonormal basis $$({\bf f}_i)$$, so the matrix representation of the inner product with respect to that basis is the identity matrix, $$I_n$$. In this sense, all inner products on $$\Bbb R^n$$ are equivalent.
• I think that if OP knew what a positive definite symmetric matrix is, there would be no need to pose the question as it is. – Marc van Leeuwen Mar 10 at 8:00
• @MarcvanLeeuwen Thanks for the comment. And yes, I agree, but one needs that language in order to state the result precisely. Still, that probably means OP would benefit from a concrete example, which I've added to the answer. – Travis Mar 10 at 8:05
For any invertible linear transformation $$A$$ you can define the inner product $$\langle v,w\rangle_A=\langle Av,Aw\rangle$$ where $$\langle\cdot,\cdot\rangle$$ denotes the standard inner product. I expect there are no other inner products, which is motivated by the fact that all inner products are known to induce equivalent norms.
Technically, you need positive $$\lambda_i$$. Or if we use $$\sum_{ij}\lambda_{ij}v_iw_j$$, the matrix $$\lambda$$ is without loss of generality equal to $$(\lambda+\lambda^T)/2$$, and it has to be positive-definite. (Yes, this matrix property has the same name; it basically means it has only positive eigenvalues.) With an appropriate basis change we can then diagonalize this matrix, which recovers the case you knew about. As for the example you tried, it failed because if you work out the matrix $$\lambda=\left(\begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right)$$ (once we make it self-adjoint as explained above), which has $$-1$$ as an eigenvalue.
• Where does the condition $\lambda = (\lambda+\lambda^T)/2$ come from? – Ion Sme Mar 9 at 22:17
• @IonSme As I said, it's imposed without loss of generality. If you replace $\lambda$ as thus, the resulting inner product won't be changed. (This is obvious in the case $v=w$, but in fact this is exhaustive because norms determine an inner product viz. $\langle v,\, w\rangle = (\Vert v+w\Vert^2-\Vert v-w\Vert^2)/4$.) – J.G. Mar 9 at 22:36
Inner products $$p(x,y)$$ on $$\mathbb R^n$$ have the form $$p(x,y) = \sum_{j=1}^n \sum_{k=1}^n a_{jk} x_j y_k$$ where the matrix $$A = [a_{jk}]$$ is positive definite. Choosing a basis of eigenvectors for the matrix $$A$$, and expanding according to this new basis instead of the original basis, the inner product is then diagonal: $$p(x,y) = \sum_{j=1}^n b_j x_j y_j$$ where $$b_j>0$$.
I agree with SmileyCraft. In finite dimensional vector spaces, bilinear transformations, as linear transformations, can be written in terms of the values that they adopt in a given base:$$\left \langle x,y \right \rangle=\sum_{i,j=1}^{n}x_iy_j\left \langle e_i,e_j \right \rangle.$$ I believe you can arrive in this representation without difficult, proving then you suspicion.
It is well known (and easy to prove) that any two finite dimensional inner product spaces are isometrically isomorphic. Hence $$\langle x, y \rangle'$$ is an inner product on $$\mathbb R^{n}$$ iff there is a vector space isomorphism $$T: \mathbb R^{n} \to \mathbb R^{n}$$ such that $$\langle x, y \rangle' =\langle Tx, Ty \rangle$$ for all $$x,y$$.
In general, for any $$n\times n$$ matrix $$A=(a_{i,j})_{i,j=1,\ldots,n}$$ the expression $$\sum_{i,j=1}^na_{i,j}x_iy_j$$ defines a bilinear form, which will be symmetric if and only $$A$$ is. Giving a positive definite symmetric bilinear form is a more subtle condition that leads to inequalities for the coefficients of $$A$$ (and the matrices that satisfy the condition are naturally called positive definite). For $$2\times2$$ symmetric matrices the positive definite condition is $$a_{1,1}>0$$, $$a_{2,2}>0$$ together with $$a_{1,1}a_{2,2}-a_{1,2}^2>0$$ (so $$\det(A)>0$$). For a concrete example, the symmetric matrix $$A=\pmatrix{1&\frac12\\\frac12&1} \quad\text{gives an inner product with } \langle v,\,w\rangle = v_1w_1 +\frac12(v_1w_2+v_2w_1) + v_2w_2\,.$$
In higher dimension the condition is more complicated, but in any case one does get many different inner products on $$\Bbb R^n$$ in this way. They do turn out to be all equivalent in the sense that they give rise to the same structure theory, but they are not equal. | 2019-04-19T21:11:55 | {
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https://math.stackexchange.com/questions/2080334/why-is-logx-integrable-over-the-interval-0-1-but-frac1x-not-in | # Why is $-\log(x)$ integrable over the interval $[0, 1]$ but $\frac{1}{x}$ not integrable?
I don't understand why some functions that contain a singularity in the domain of integration are integrable but others are not.
For example, consider $f(x) = -\log(x)$ and $g(x) = \frac{1}{x}$ on the interval $[0, 1]$. These functions look very similar when they are plotted but only $f(x)$ can be integrated.
1. What is the precise mathematical reason(s) that makes some functions with singularities integrable while others are not?
2. Are $\log$ functions the only functions with singularities that can be integrated or are there other types of functions with singularities that can be integrated?
• How about $1/x^p$, $0<p<1$? – user175968 Jan 2 '17 at 11:46
• The integral in $[0,1]$ is the same that the integral in $(0,1)$. The reason is not the singularity, the reason is the divergence or convergence of the integral. – Masacroso Jan 2 '17 at 11:57
It's just whether or not the area under the curve is finite or not. It's doesn't matter that there is an asymptote.
You might consider the area under the curves $y=e^{-x}$ and $y=1/x$ for $x>0$. These are really the same two curves you mention, just along the other axis.
It's akin to the idea that an infinite series may or may not converge; just because there are infinitely many terms in a series doesn't mean the series must diverge.
• I don't see how these are the "same two curves"..one of them, $y=e^{-x}$, has a precise value at $x=0$ and so it can be integrated on interval $[0, \infty)$, whereas $y=1/x$ gets arbitrarily large near zero and can't be integrated on that interval so the area under this curve is undefined. I don't understand what point you are making in your post? – ManUtdBloke Jan 6 '17 at 22:00
• @eurocoder : You misunderstand--I mean the two curves I mention are essentially the same two curves you ask about, just along the other axis. I added this to my answer to clarify. – MPW Jan 7 '17 at 4:56
• They're note essentially the same for the purposes of integration, $\log x$ is not integrable on $(0,\infty)$ but $e^{-x}$ is. Perhaps you mean something else, but it's not clear from your wording what notion of "same" you are using and why it's relevant to the question at-hand. – Adam Hughes Jan 8 '17 at 9:26
• @AdamHughes : Of course, I'm talking about finding the area between the alternate curves and the other axis, as I already said. They are exactly the same thing. – MPW Jan 25 '17 at 14:16
• @MPW maybe as curves I different coordinate systems, but integration users a consistent orientation, so this doesn't amount to much mathematically to say they're the same in an integration context if they're not even being considered on the same ground for integration's necessary context. – Adam Hughes Jan 25 '17 at 14:20
Think about it this way - what's the inverse?
$$y = \frac{1}{x}; x = \frac{1}{y}$$ $$y = -\log x; x = e^{-y}$$
Looking at it this way, it's clear that as $y$ shoots off to infinity, $x$ approaches zero much faster in one case than in the other.
• Ok so one approaches zero much faster but how fast precisely does a function have to approach zero to make it integrable? ..what is the minimum 'speed' a function needs to have for it to be integrable? – ManUtdBloke Jan 6 '17 at 22:08
• Well, there isn't a minimum exactly... it has to approach quickly enough that the inverse converges at infinity. So, for instance, $\frac{1}{x^{1-\epsilon}}$ would converge at zero, since $\frac{1}{x^{1+\epsilon}}$ converges at infinity. – user361424 Jan 7 '17 at 4:09
The key is how fast the function is diverging.
Regarding your two examples, $-\log$ is going really fast close to the $y$-axis so it is integrable, but not $x\mapsto \frac {1}{x}$.
• You have $$\int_a^1 -\log(x)\mathrm dx=a(1-\log(a))+1\xrightarrow[a\to 0^+]{} 1<\infty.$$ So this function is integrable.
• You have $$\int_a^1 \frac 1x\mathrm dx=-1+\frac 1{a^2}\xrightarrow[a\to 0^+]{} +\infty.$$ So this function is not integrable.
Regarding your second question, $\log$ functions are absolutely not the only one. To convince yourself, take for instance $x\mapsto \frac 1{\sqrt{x}}$ on $(0,1)$.
• How did you get a result of $1$ for the first integral? – ManUtdBloke Jan 6 '17 at 22:08
• Also, wolfram alpha evaluates $a (1 - \log a)$ as $a \to 0$ to be $0$. – ManUtdBloke Jan 6 '17 at 22:43
• @eurocoder Oups, my mistake : I forgot to write the $+1$. Thanks for the correction. – E. Joseph Jan 8 '17 at 9:21
Simple: $\quad -\log x=_0 o\Bigl(\dfrac1{\sqrt x}\Bigr)$ and the integral of $\dfrac 1{\sqrt x}$ on [0,1] is convergent.
• Where is the $o(1/\sqrt{x})$ coming from and how does convergences of that function on the interval imply convergence of $-\log{x}$? – ManUtdBloke Jan 6 '17 at 22:05
• The $o(1/\sqrt{x})$ corresponds to the high school limit $\sqrt x\log x\xrightarrow[x\to 0^+]{}0$. The convergence of the integral is a well-known theorem in asymptotic analysis. – Bernard Jan 6 '17 at 22:31
• But the problem I have is that why are some functions with singularities integrable and orthers aren't. So you are saying $\frac{1}{\sqrt{x}}$ is integrable, but we know $\frac{1}{x}$ is not. Why is this? It seems the exponent on $x$ in the denominator determines whether a function will be integrable. So if that is the case 'where' precisely (for what value of exponent) does a function fail to be integrable? – ManUtdBloke Jan 10 '17 at 8:17
• Since the anti derivative of$\dfrac1{x^r}$ is $-\dfrac1{(r-1)x^{r-1}}$ the answer is $r-1<0$ (there must be no denominator), i.e. $r<1$. – Bernard Jan 10 '17 at 9:43
• Actually, the ‘theorem’ (a great word for a simple observation) is valid for non-negative functions $f$ and $g$ such that $f=O(g)$: if the integral of $g$ converges, the integral of $f$ does too (obvious since the integral preserves inequalities). As $f=o(g)\implies f=O(g)$, the result follows. – Bernard Feb 21 '19 at 14:58 | 2020-03-30T20:36:35 | {
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https://math.stackexchange.com/questions/1304792/partial-sums-of-falling-factorials | # Partial sums of falling factorials
I want to know if there exists some way, approximate or exact, to do a partial sum of falling factorials of the kind:
$$\sum_{k=i}^{n}(a+k)_{h}$$ where all are constants (here $$(r)_s:=r(r-1)\cdots (r-s+1)$$ represent a falling factorial).
And I'm interested too in some partial sum like this
$$\sum_{k=i}^{n}(a+k)_{h}r^k$$
In particular I want a closed form to this formula:
$$\sum_{m=0}^{3}\left(\sum_{k=1}^{19-m}(19-k)_m\right )^{-1} \left(-(19)_m+\sum_{k=0}^{19-m}(19-k)_m q^k\right )$$
Possibly there is not a closed form but I don't know. I started to read about hypergeometric series but this topic is completely new to me so I don't have a clear way to approach to my question by now.
I will appreciate any help. If you can show me via some link or bibliography is fine too. Thank you in advance.
### UPDATE
Ok, I was reading the book of Graham that @ncmathsadist said to me and I have a partial answer. The question is close to some general topics on discrete maths (that I unfortunately forget).
The point is that an analogue to $$\int_{a}^{b}x^n dx=\frac{x^{n+1}}{n+1}\Big|_{a}^{b}\$$ on difference calculus is
$$\sum\nolimits_{a}^{b}(k)_n\delta k =\frac{(k)_{n+1}}{n+1}\bigg|_{a}^{b}$$
For the second case I can use an analogue to integration by parts that is named summation by parts:
$$\sum f(k)\Delta g(k) \delta k=f(k)g(k)-\sum \Delta f(k) g(k+1)\delta k$$
But I dont get any closed form, so I assumed these formulas haven't closed forms.
• Consult the book Concrete Mathematics. – ncmathsadist May 30 '15 at 2:23
• Frankly, I agree with your P.S. This seems to me a perfectly reasonable question, though not, unfortunately, one with which I’m likely to be able to help. – Brian M. Scott May 30 '15 at 3:30
• @Brian, ty very much for your support... I really dont know about this topic, and I read some books of discrete maths. When I see negative votes I though that my question was "so easy" to answer or something like this. I dont want people work for me or so but I think my question is legit. – Masacroso May 30 '15 at 3:39
I like Knuth's notation for falling factorials better:
$$\alpha^{\underline{h}} = \alpha \cdot (\alpha - 1) \cdot \dotsm \cdot (\alpha - h + 1)$$
First note that:
$$\Delta \alpha^{\underline{h}} = (\alpha + 1)^{\underline{h}} -\alpha^{\underline{h}} = (\alpha + 1) \cdot \alpha^{\underline{h - 1}} - \alpha^{\underline{h - 1}} \cdot (\alpha - h + 1) = h \alpha^{\underline{h - 1}}$$
This suggests:
$$\sum_{0 \le k \le n} (\alpha + k)^{\underline{h}} = \frac{(\alpha + n + 1)^{\underline{h + 1}}}{h + 1} - \frac{\alpha^{\underline{h + 1}}}{h + 1}$$
This we prove by induction.
Base: $n = 0$ gives:
\begin{align} \sum_{0 \le k \le 0} (\alpha + k)^{\underline{h}} &= \alpha^{\underline{h}} \\ \frac{(\alpha + 1)^{\underline{h + 1}}}{h + 1} - \frac{\alpha^{\underline{h + 1}}}{h + 1} &= \frac{1}{h + 1} \Delta \alpha^{\underline{h + 1}} \\ &= \alpha^{\underline{h}} \end{align}
This checks out.
Induction: Assume it is true for $n$, look at $n + 1$:
\begin{align} \sum_{0 \le k \le n + 1} (\alpha + k)^{\underline{h}} &= \sum_{0 \le k \le n} (\alpha + k)^{\underline{h}} + (\alpha + n + 1)^{\underline{h}} \\ &= \frac{(\alpha + n + 1)^{\underline{h + 1}}}{h + 1} - \frac{\alpha^{\underline{h + 1}}}{h + 1} + (\alpha + n + 1)^{\underline{h}} \\ &= \frac{(\alpha + n + 1)^{\underline{h}} \cdot (\alpha + n + 1 - h)} {h + 1} + \frac{(\alpha + n + 1)^{\underline{h}} \cdot (h + 1)} {h + 1} - \frac{\alpha^{\underline{h + 1}}}{h + 1} \\ &= \frac{(\alpha + n + 1)^{\underline{h}} \cdot (\alpha + n + 2)} {h + 1} - \frac{\alpha^{\underline{h + 1}}}{h + 1} \\ &= \frac{(\alpha + n + 2)^{\underline{h + 1}}}{h + 1} - \frac{\alpha^{\underline{h + 1}}}{h + 1} \end{align}
This is exactly as claimed.
For your second sum, note that:
$$(\alpha + k)^{\underline{h}} x^{\alpha + k - h} = \frac{\mathrm{d}^h}{\mathrm{d} x^h} x^{\alpha + k}$$
Thus the sum is essentially:
\begin{align} \sum\limits_{k=0}^n (\alpha + k)^{\underline{h}} r^k &= r^{h - \alpha} \sum\limits_{k=0}^n (\alpha + k)^{\underline{h}} r^{\alpha + k - h} \\ &= r^{h - \alpha} \left. \frac{\mathrm{d}^h}{\mathrm{d} x^h} \sum\limits_{k=0}^n x^{\alpha + k} \right|_{x = r} \\ &= r^{h - \alpha} \left. \frac{\mathrm{d}^h}{\mathrm{d} x^h} \left(x^\alpha \sum\limits_{k=0}^n x^k\right) \right|_{x = r} \\ &=\left.r^{h-\alpha} \frac{d^h}{d x^h} \left( \frac{x^\alpha - x^{\alpha+n+1}}{1- x} \right)\right|_{x=r} \\ &=r^{h-\alpha} \sum\limits_{l=0}^h \binom{h}{l} \left(\alpha^{\underline{h-l}} r^{\alpha-(h-l)} - (\alpha+n+1)^{\underline{h-l}} r^{\alpha+n+1-(h-l)}\right) \cdot \frac{l!}{(1-r)^{l+1}} \\ &= \frac{1}{(1-r)^{h+1}} \sum\limits_{l=0}^h \binom{h}{l} \left(\alpha^{\underline{h-l}} r^{l} - (\alpha+n+1)^{\underline{h-l}} r^{n+1+l}\right) \cdot l! (1-r)^{h-l} \\ &= \frac{h!}{(1-r)^{h+1}} \sum\limits_{l=0}^h \frac{1}{(h-l)!} \left(\alpha^{\underline{h-l}} r^{l} - (\alpha+n+1)^{\underline{h-l}} r^{n+1+l}\right) \cdot (1-r)^{h-l} \\ &= \frac{h!}{(1-r)^{h+1}} \sum\limits_{l=0}^h \left(\binom{\alpha}{h-l} r^{l} - \binom{\alpha+n+1}{h-l} r^{n+1+l}\right) \cdot (1-r)^{h-l} \\ &= \frac{h!}{(1-r)} \left\{ \binom{\alpha}{h} \cdot F_{2,1} \left[\begin{array}{ll}1 & -h \\ \alpha-h+1 \end{array};\frac{r}{r-1}\right] - r^{n+1} \binom{\alpha+n+1}{h} \cdot F_{2,1} \left[\begin{array}{ll}1 & -h \\ \alpha+n-h+2 \end{array};\frac{r}{r-1}\right] \right\} \end{align}
This is quite ugly. The remaining sum is geometric, and can be expressed as a fraction. Leibnitz' formula for multiple derivatives of a product reduce that somewhat, but it is still a mess. If $h$ is small integer, perhaps a CAS gives somewhat manageable.
Note: It is a matter of taste whether an expression is ugly or not. In my opinion the final result has a closed form if we use hypergeometric functions. | 2019-12-08T19:10:00 | {
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https://stats.stackexchange.com/questions/432272/when-to-use-complement-when-finding-probability/432274 | # When to use complement when finding probability
In a scientific study there are 8 guinea pigs, 5 if which are pregnant. If 3 are selected at random without replacement, find the probability that are all pregnant?
But when finding the probability that none are pregnant, the answer is 1/56 since (3/8)(2/7)(1/6) = 1/56.
Why can't we use complement upon solving this? And subtract 1 from 5/28?
There a 5 guinea pigs that are pregnant. On the first draw, you have a 5/8 chance of selecting a pregnant one. What do you think the chances are on the second draw?
For your part of the question "why can't I use the complement rule?". You can, but it's not as simple as $$P(\bar{A}) = 1- P(A)$$. What you have (1/56) is the probability of drawing NO pregnant guinea pigs. The complement of "drawing NO pregnant guinea pigs" is "at least one guinea pig is pregnant".
• Nice answer (+1). Appeared while I was writing about the hypergeometric distribution. Oct 20, 2019 at 9:38
• Thanks! I've seen a few of your answers recently. I think your inclusion of R is a nice touch :) Oct 20, 2019 at 9:45
Short answer: The probability of getting 3 pregnant guinea pigs when 3 are chosen at random from the 8, is $$\frac{{5 \choose 3}{3 \choose 0}}{{8\choose 3}} = \frac{5}{28} = 0.1786.$$ Computations in R:
choose(5,3)/choose(8,3)
[1] 0.1785714
5/28
[1] 0.1785714
More detail: Let $$X$$ be the number of pregnant guinea pigs in a sample of 3 chosen at random without replacement from the 8 in the study. Then $$X$$ has a hypergeometric distribution with PDF:
$$P(X = k) = \frac{{5 \choose k}{3 \choose 3-k}}{{8\choose 3}},$$ for $$k = 0,1,2,3.$$ We can use the PDF function dhyper in R to make the distribution table.
k = 0:3; pdf = dhyper(k, 5,3, 3)
cbind(k, pdf)
k pdf
[1,] 0 0.01785714
[2,] 1 0.26785714
[3,] 2 0.53571429
[4,] 3 0.17857143
About complements: The reason you can't take the complement as you propose, is that the complement of "no (0) pregnant guinea pigs" is not "all (3) pregnant guinea pigs." There are other possible outcomes: specifically, either 1 or 2 pregnant. The complement of "no pregnant guinea pigs" in the sample is "at least one pregnant guinea pig."
$$P(X \ge 1) = 1 - P(X = 0).$$
Bar graph of PDF: Here is a graph of the hypergeometric PDF of $$X.$$ | 2022-08-18T10:42:44 | {
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https://math.stackexchange.com/questions/4197443/is-y2fybc-a-quadratic-equation | # Is $y^2+f(y)b+c$ a quadratic equation?
The solution to the question:
Let $$x, y, z \in R$$ such that $$x+y+z=6$$ and $$x y+y z+z x=7$$. Then find the range of values of $$x, y$$, and $$z$$.
given in book is as follows:
$$x, y, z \in R$$
$$x+y+z=6$$ and $$x y+y z+z x=7$$
$$\Rightarrow y(6-y-z)+y z+z(6-y-z)=7$$
$$\Rightarrow \quad-y^{2}+(6-z+z-z) y+z(6-z)-7=0$$
$$\Rightarrow \quad y^{2}+(z-6) y+7+z(z-6)=0$$ .
Now, $$y$$ is real. Therefore,
$$(z-6)^{2}-4[7+z(z-6)] \geq 0$$...(1)
or $$3 z^{2}-12 z-8 \leq 0$$
or $$\frac{12-\sqrt{144+96}}{6} \leq z \leq \frac{12+\sqrt{144+96}}{6}$$
or $$\frac{6-2 \sqrt{15}}{3} \leq z \leq \frac{6+2 \sqrt{15}}{3}$$.
From symmetry, $$x$$ and $$y$$ have same range.
But I doubt if this is correct; since $$y$$ is function of $$z$$, hence the author cannot solve the question this way. Let me explain my point clearly.
Quadratic equations are defined as those which can be expressed in the form $$ax^2+bx+c=0$$, where $$a \not=0$$ and $$a, b, c$$ are constants. But clearly in this case $$z$$ is not constant, it is function of $$y$$, hence $$(1)$$ is not a quadratic equation so we cant the do the steps after $$(1)$$ hence the solution is wrong. So the question is, is $$y^2+f(y)b+c$$ a quadratic equation?
• To be precise no one says "solve the equation". If you rewrite the "quadratic" in $y^2$ in terms of $y-\frac{z-6}{2}$ you'll be able to see that the appropriate function of $z$ [the "constant"] is indeed positive and then etc etc Jul 13 '21 at 15:32
• One we have more than one variable in the polynomial, we say that it's "quadratic in $x$" if the highest power of $x$ that appears is $2$. Then we can treat the other variables as constants, whose value is unknown. Jul 13 '21 at 15:40
• @saulspatz But why? It will be helpful if you provide some reference. Thanks. Jul 13 '21 at 15:44
• Reference for what? It's just the quadratic formula, which you already know. Jul 13 '21 at 15:46
• Let's say $f(x) =2x^2$ then will your claim still hold? Jul 13 '21 at 15:48
## 3 Answers
Just go back and recall the derivation of the quadratic formula, it never uses the fact that $$a$$, $$b$$ and $$c$$ are constant, you can use it any time. Just as an example, consider: \begin{align*} 2x+4&= 20\\ 4+2x-20&= 0\\ (1)\color{blue}{2}^2+(x)\color{blue}{2}+(-20)&= 0\\ \color{blue}{2}&= \dfrac{-x\pm\sqrt{x^2-4(1)(-20)}}{2(1)}\\ 4+x&=\pm\sqrt{x^2+80}\\ 16+8x+x^2&=x^2+80\\ x&= 8 \end{align*}
Here, the coefficient was itself a variable and the variable was a constant, but the quadratic formula, as you see works. Note the coefficient of variable squared must be non-zero (here, $$1\neq 0$$).
$$x^2+f(x)b+c=0$$ may or may not be a quadratic equation, but you can always use the quadratic formula!
Hope this helps. Ask anything if not clear :)
• Thanks. Just to be sure you're saying that we can use quadratic formula even if the equation is not an quadratic equation, so for example I can use quadratic formula on: $x^2+cos(x)x+2=0$ (though that many not give solution!)? Also I am not able to follow how you removed \pm. Thanks Jul 14 '21 at 1:36
• @IDKWTD: Yes, and as for the \pm, I had something else in my mind, in this equation, we don't even need to remove \pm. I have edited my post. Jul 14 '21 at 1:54
It sounds like you are asking: If you have a quadratic in the variable $$y$$, does the quadratic formula hold, if the coefficients of the quadratic depend on some other variable $$z$$?
In your case, if $$y^2 + (z-6)y + (7+z(z-6)) = 0,$$ can you still treat the coefficients $$z-6$$ and $$7+z(z-6)$$ as constant and use the usual quadratic formula to find an expression for $$y$$? The answer is yes. If you go through the derivation of the quadratic formula, you will see that it is not important that the coefficients are constants: They are allowed to vary with other variables, and the key is that the coefficients are constant with respect to y.
So if you have an expression of the form $$a(z)y^2 + b(z)y + c(z) = 0,$$ where $$a$$, $$b$$, and $$c$$ are functions of $$z$$, then I would indeed call this a quadratic in $$y$$, and I would gladly apply the quadratic formula to find an expression for $$y$$ in terms of $$z$$ (and here you of course need to take care if $$a(z)\neq 0$$).
• But $y$ and $z$ are related to each other by equation $x+y+z=6$. Jul 14 '21 at 0:37
First, the quadratic formula deals with polynomials. If you introduce some unspecified function $$f$$, you muddy the waters, since $$f$$ needn't be a polynomial. For example, if $$f(x)=\sin x$$, then $$x^2+f(x)$$ is not quadratic in $$x$$.
I understand that you saw the quadratic formula derived, it was assumed that $$a,b,c$$ are constants, but the proof works so long as they aren't functions of $$x$$.
When we say that the roots of $$p(x)=ax^2+bx+c$$ are $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ this is true so long as $$a\neq0$$ and $$a,b,$$ and $$c$$ do not depend on $$x$$. You can verify this by substituting them into the formula for $$p$$.
It doesn't matter what $$a,b,$$ and $$c$$ are, so long as they don't depend on $$x$$. They could be trigonometric functions of some other variable, for example.
• I agree with you on every point your wrote but clearly in this case $y$ and $z$ are related to each other by the equationz: $x+y+z=6$ and $x y+y z+z x=7$. Jul 14 '21 at 0:49
• Yes, but that's completely irrelevant, as I and others have explained. Jul 14 '21 at 4:04
• The formula works even if $a$, $b$ and $c$ depend upon $x$. Jul 14 '21 at 6:42 | 2022-01-29T02:34:12 | {
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# How many factors does 36^2 have?
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How many factors does 36^2 have? [#permalink]
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15 Aug 2010, 02:47
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How many factors does 36^2 have?
A. 2
B. 8
C. 24
D. 25
E. 26
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Joined: 02 Sep 2009
Posts: 51035
Re: Help: Factors problem !! [#permalink]
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praveengmat wrote:
How many factors does 36^2 have?
A 2
B 8
C 24
D 25
E 26
Finding the Number of Factors of an Integer:
First make prime factorization of an integer $$n=a^p*b^q*c^r$$, where $$a$$, $$b$$, and $$c$$ are prime factors of $$n$$ and $$p$$, $$q$$, and $$r$$ are their powers.
The number of factors of $$n$$ will be expressed by the formula $$(p+1)(q+1)(r+1)$$. NOTE: this will include 1 and n itself.
Example: Finding the number of all factors of 450: $$450=2^1*3^2*5^2$$
Total number of factors of 450 including 1 and 450 itself is $$(1+1)*(2+1)*(2+1)=2*3*3=18$$ factors.
Back to the original question:
How many factors does 36^2 have?
$$36^2=(2^2*3^2)^2=2^4*3^4$$ --> # of factors $$(4+1)*(4+1)=25$$.
Or another way: 36^2 is a perfect square, # of factors of perfect square is always odd (as perfect square has even powers of its primes and when adding 1 to each and multiplying them as in above formula you'll get the multiplication of odd numbers which is odd). Only odd answer in answer choices is 25.
Hope it helps.
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Re: Help: Factors problem !! [#permalink]
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15 Aug 2010, 03:11
Bunuel wrote:
praveengmat wrote:
How many factors does 36^2 have?
A 2
B 8
C 24
D 25
E 26
Finding the Number of Factors of an Integer:
First make prime factorization of an integer $$n=a^p*b^q*c^r$$, where $$a$$, $$b$$, and $$c$$ are prime factors of $$n$$ and $$p$$, $$q$$, and $$r$$ are their powers.
The number of factors of $$n$$ will be expressed by the formula $$(p+1)(q+1)(r+1)$$. NOTE: this will include 1 and n itself.
Example: Finding the number of all factors of 450: $$450=2^1*3^2*5^2$$
Total number of factors of 450 including 1 and 450 itself is $$(1+1)*(2+1)*(2+1)=2*3*3=18$$ factors.
Back to the original question:
How many factors does 36^2 have?
$$36^2=(2^2*3^2)^2=2^4*3^4$$ --> # of factors $$(4+1)*(4+1)=25$$.
Or another way: 36^2 is a perfect square, # of factors of perfect square is always odd (as perfect square has even powers of its primes and when adding 1 to each and multiplying them as in above formula you'll get the multiplication of odd numbers which is odd). Only odd answer in answer choices is 25.
Hope it helps.
Thanks a ton !!.. loved the approach !
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Re: Help: Factors problem !! [#permalink]
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14 Oct 2010, 13:58
1
Factors of a perfect square can be derived by using prime factorization and then using the formula to find perfect square's factors.
In this case $$(36)^2= (2^2*3^2)^2=2^4*3^4$$ or $$(36)^2=(6^2)^2=(6)^4=(2*3)^4=2^4*3^4$$
And now you can use the formula explained above by Bunuel to determine the answer, which is $$(4+1)*(4+1)=5*5=25=Odd$$(Trick is there must be odd number of factors of a perfect square and only 25 is odd in answer choices, so it can be solved within 30 seconds or less )
Please! go through the GMAT Math Book by GMAT CLUB (written by bunuel & walker), all of these tips & tricks are written there. (even I have compiled them in one .pdf file and is shared here on Math forum)
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Re: Help: Factors problem !! [#permalink]
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29 Aug 2011, 09:57
This method is worth bookmarking. Appreciate it
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Re: Help: Factors problem !! [#permalink]
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01 Sep 2011, 19:17
The Easy Answer! (Applicable only in case of perfect square numbers)
A perfect square always have a odd number of factors.
36^2 is a perfect square.
Given the answer choices, the only odd number of factor is 25.
So, The definite answer is D.
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Re: Help: Factors problem !! [#permalink]
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02 Sep 2011, 08:08
36^2 = 2^4 3^4
total factors = (4+1)(4+1) = 25.
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Re: Help: Factors problem !! [#permalink]
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03 Sep 2011, 10:04
The odd number of factors for perfect squares solves this in no time. Nice trick.
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Re: How many factors does 36^2 have? [#permalink]
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25 Oct 2014, 10:35
praveengmat wrote:
How many factors does 36^2 have?
A. 2
B. 8
C. 24
D. 25
E. 26
This can be solved, by just looking at the answer choices. Since the given number is a square, it must have an odd number of factors. Only D has odd.
Ans. D
If we need to solve it, then
$$(36)^2 = (2^4*3^4)$$
For $$a^n*b^m$$ the number of factors is defined by $$((n+1)(m+1))$$
$$= (4+1)(4+1)$$
$$= 25$$
Ans. D
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How many factors does 36^2 have? [#permalink]
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24 Aug 2018, 10:12
praveengmat wrote:
How many factors does 36^2 have?
A. 2
B. 8
C. 24
D. 25
E. 26
OA:D
$$36^2=6^4=2^4*3^4$$
Number of Factors $$= (4+1)*(4+1) = 5*5 =25$$
Furthermore , Square of a number always has odd number of factors.
For ex $$4= 2^2$$, Factors : $$1,2,4$$
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Re: How many factors does 36^2 have? [#permalink]
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24 Aug 2018, 10:15
praveengmat wrote:
How many factors does 36^2 have?
A. 2
B. 8
C. 24
D. 25
E. 26
$$36 = 3^2*2^2$$
So, $$36^2 = 3^4*2^4$$
No of factors of 36^2 is ( 4 +1 )(4 + 1 ) = 25, Answer must be (D)
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Re: How many factors does 36^2 have? &nbs [#permalink] 24 Aug 2018, 10:15
Display posts from previous: Sort by | 2018-12-09T22:10:33 | {
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https://www.physicsforums.com/threads/question-on-set-notation.802684/ | # Question on set notation
1. Mar 12, 2015
### amilapsn
1. The problem statement, all variables and given/known data
Prove or disprove the following
(i) $\forall a\in\mathbb{R}[(\forall \epsilon>0,a<\epsilon)\Leftrightarrow a\leq 0]$
2. The attempt at a solution
Can't we disprove the above statement by telling $a\leq 0 \nRightarrow (\forall \epsilon>0,a<\epsilon)$ through a counter example like $a\leq 0 \Rightarrow (\epsilon=0,a\leq \epsilon)$ or something?
2. Mar 12, 2015
### PeroK
Can you write out in words what the proposition is saying?
What would you have to show to prove it? Hint: there would be two parts to proving it.
Can you write down what property a countereaxmple would have? Hint: it could have one of two properties.
Does it hold for a = 0? Does it hold for a = -1? Does it hold for a = 1?
3. Mar 12, 2015
### amilapsn
The proposition in words:
for all a belongs to real number set , for all $\epsilon$ >0 and a<$\epsilon$ if and only if $a\leq 0$ .
I have to show both $\forall \epsilon >0,a<\epsilon\Rightarrow a\leq 0$ and $a\leq 0\Rightarrow \forall \epsilon >0,a<0$ to prove the proposition.
A counter example should disprove the proposition.
4. Mar 12, 2015
### PeroK
That's good.
What about a = 0, 1, -1? Does the proposition hold for these values of a?
5. Mar 12, 2015
### amilapsn
The proposition holds for a=-1,0. But it doesn't hold for a=1.
6. Mar 12, 2015
### PeroK
Why does it fail for a = 1?
7. Mar 12, 2015
### amilapsn
Because a=1 is not less than for all $\epsilon>0$
8. Mar 12, 2015
### PeroK
Take a step back. We have a proposition:
$\forall a \ \ A \Leftrightarrow B$
That's means that (if the proposition holds) then for each a we have either: A(a) true and B(a) true; or A(a) false and B(a) false.
For a = 1, what can you say about A(1) and B(1)?
9. Mar 12, 2015
### amilapsn
I see. The proposition holds for a=1 too, because A(1) false and B(1) false. Thanks...
Thank you for showing me the better way to look at the question.
10. Mar 12, 2015
### amilapsn
Then the proposition is true for all a, so that we can't disprove it. We have to prove it. Thanks again @PeroK
11. Mar 12, 2015
### PeroK
Also, when I first asked you to describe the proposition in words, you could have said:
The proposition states that:
"Any real number is less than or equal to 0 iff it is less than every positive number".
Put like that, it's clear that the proposition holds.
12. Mar 12, 2015
### amilapsn
Yeah, it's really clearer...
13. Mar 12, 2015
### PeroK
Here's a tip. This is something I do when dealing with propositions and logic:
I use "true" and "false" to relate to individual statements. E.g. $a > 0$ can be true or false.
And, I say a proposition "holds" or "fails". E.g. the proposition holds for a = 1.
14. Mar 12, 2015
### amilapsn
Just now I felt what is called as "Enlightenment...." | 2017-10-21T03:58:02 | {
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https://math.stackexchange.com/questions/287076/what-is-the-integral-of-0/774280 | # What is the integral of 0?
I am trying to convince my friend that the integral of $0$ is $C$, where $C$ is an arbitrary constant. He can't seem to grasp this concept. Can you guys help me out here? He keeps saying it is $0$.
• Let $0$ be the only answer to the integral of $\int0\;dx$. Thus, $\dfrac{d}{dx}f(x)=0$ is satisfied ONLY by $f(x)=0$ However, Let $f(x)=0+c,c\in \mathbb{R}$. $$\dfrac{d(0+c)}{dx} =0$$ Thus the assumption that there is only one function satisfying the condition is false. $$\blacksquare$$ – Inquest Jan 26 '13 at 2:30
• Whether or not the integral is $0$ or $C$ depends on whether you are talking about the indefinite or definite integral. – anon Jan 26 '13 at 2:47
• Maybe it is confusing cause it is "the" integral. Make it clear that "the" integral is not a single function.... – N. S. Jan 26 '13 at 3:14
• Definite integrals $\ne$ Primitive integrals $=$ Antiderivatives. – Did Jan 26 '13 at 3:15
• May be he is having the difficulty because: He sees an integral as the area below the curve and the $x$ axis. So under any lower and upper limit it's integral(note: the definition at the beginning ) is zero. – jdoicj Jun 13 '13 at 15:29
Taking the derivative of any constant function is 0, i.e. $\frac {d}{dx} c = 0$ So the indefinite integral $\int0 \,dx$ produces the class of constant functions, that is $f(x) = c$ for some $c$.
There's something that you have to look at here though, that is "what about the fact $\alpha \int f dx = \int \alpha f dx$?" Can't you say:
$$\int 0\,dx = \int 0 \cdot 1 \,dx = 0 \int 1 \,dx = 0x = 0$$
This gives two conflicting answers. The question is far more complicated that you would first think. But when you say $\int f dx$ and the interval over which you're integrating isn't obvious or defined, what you really mean is "the class of functions that when derived with respect to $x$ produce $f$". The rule stated only applies for definite integrals. That is:
$$\int_a^b\alpha f\,dx = \alpha \int_a^bf \,dx$$
And if you look at textbooks on real analysis (I just looked at Rudin) that's the form in which you will find the theorem.
It should also be noted that the definite integral of $0$ over any interval is $0$, as $\int 0 \,dx = c - c = 0.$
• The answers are only conflicting if you take the position that the arbitrary constant gets added to the integral "before" we multiply by the coefficient $0$, and I see no reason to take that position. IE, why shouldn't I say that $0 \int 1 dx = 0x + C$? – Jonathan Hebert Apr 18 '16 at 23:31
• What is the reason to take your position? Why not $0\int 1\, dx = 0\cdot (x + C)$ – Zduff Mar 31 '17 at 20:39
You are correct, $\int 0 dx = 0 + C = C$
Your friend is not entirely wrong because $C$ could equal $0$. ie. if
$f(x) = 0$ is one antiderivative. But in general we do not know $C$ unless we are given some initial condition.
• @downvote ??? Lol. – Rustyn Jan 26 '13 at 6:02
• If $C$ is assumed to be selected randomly, the probability that the friend is right is 0. [tongue in cheek] – Thomas Jan 26 '13 at 7:19
Indefinite integrals (anti-derivatoves) are known modulo a constant function. With definite integrals, the case is different: $$\int_a^b0\,\mathrm{d}t=0$$
One way to verify that $C$ is the anti-derivative of $0$ is simply $$\frac{\mathrm{d}}{\mathrm{d}t}C=0$$
There are two types of integrals at play here. Definite integrals are the ones that describe the actual area under a curve. Indefinite integrals are the ones that describe the anti-derivative.
There's no paradox, really. When speaking of indefinite integrals, the integral of 0 is just 0 plus the usual arbitrary constant, i.e.,
$\int 0 \, dx = 0 + C = C$
There's no contradiction here. When evaluating the area under a curve f(x), we find the antiderivative F(x) and then evaluate from a to b:
$\int^{b}_{a} f(X) \, dx = F(b) - F(a)$
So, for f(x) = 0, we find F(x) = C, and so F(b) - F(a) = C - C = 0. Thus, the total area is zero, as we expected.
How about drawing sum upper and lower sums! You won't get very far because you'll be married to the horizontal axis and then, of course, all of the sums are zero and since a definite integral is always sandwiched between any upper and any lower sum. The value is trapped by 0. I.E. 0 <= the integral <= 0. This of course works only for a definite integral. If you are looking for an anti derivative, it shouldn't be too hard to convince your buddy that only constant functions f(x) = C have zero slope.
The integral of 0 is C, because the derivative of C is zero. Also, it makes sense logically if you recall the fact that the derivative of the function is the function's slope, because any function f(x)=C will have a slope of zero at point on the function. Therefore ∫0 dx = C. (you can say C+C, which is still just C).
From the fundamental theorem of calculus, we know that $$F(z)-F(w)=\int_w^zf(x)dx$$
Let the integral value of zero be a constant $C$, that is
$$C=\int_w^z0 \quad dx=F(z)-F(w) \quad \forall z\in[a,b]$$
Now choosing $z=w$ gives $C=0$. Consequently, this implies that $F(z)=F(w)$ which is the anti-derivative of the zero function, i.e. $F(z)$ is a constant function. Therefore, the definite integral is always zero.
But, using the formal definition of indefinite integral [see Bartle's book]:
If $f\in \mathcal{R}[a,b]$ , then the function defined by $$F(z)=\int_a^zf(x)dx$$ is called the indefinite integral of f with basepoint a.
This means that the in-diffident integral is a constant function with a possibility to be zero.
Look at this function: F(x)=0. $\frac {d}{dx} F(x) = \frac {d}{dx} 0 = 0$ There is a theorem that says that antiderivatives of any function $f(x)$ has a form of $G(x)+C$, where $G'(x)=f(x)$. If we take $f(x)=0$, then $F'(x)=0=f(x)$, so it's anti derivative has a form of $0+C=C$. So $\int 0\, dx= C$
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https://math.stackexchange.com/questions/3108458/probability-of-a-pair | # Probability of a pair
Let there be four cards (2 queens and 2 kings). You draw two cards. What is the probability that the cards are a pair? Is there a difference in probability if you draw two cards together (in which case probability is 0.5 if i'm not wrong) or draw one card at a time (probability of drawing a pair is 1/3)?
## 3 Answers
Let the Cards be $$Q_1,Q_2,K_1,K_2$$. Let $$AB$$ be the event that denotes drawing of $$A$$ followed by drawing of $$B$$. Then probability $$p$$ of getting a pair when drawing cards one after another is nothing but
\begin{align} p & = P(K_1 K_2) + P(K_2 K_1) + P (Q_1 Q_2) + P(Q_2Q_1). \\ & = P(K_2 \mid K_1) P(K_1) + P(K_1 \mid K_2) P(K_2) + P(Q_2 \mid Q_1) P(Q_1) + P(Q_1 \mid Q_2) P(Q _2). \\ & = \frac 1 4 \cdot \frac 1 3 + \frac 1 4 \cdot \frac 1 3 + \frac 1 4 \cdot \frac 1 3 + \frac 1 4 \cdot \frac 1 3. \\ & = 4 \cdot \frac {1} {12}. \\ & = \frac 1 3. \end{align}
Which is same as the probability of getting a pair taking two cards at a time.
• Hmm. Thanks. But you are conditioning on the fact that first card was a king or queen or second is king or queen. But if you draw two cards together then there is no first or second draw. Either it is a pair or it's not. – pb10 Feb 11 at 8:37
• Yeah that's correct. But in both of the cases you will end up with same probability. – Dbchatto67 Feb 11 at 8:38
• Thanks. Unfortunately I just can't make a leap from state space of {KK,KQ,QQ,QK} to what you are proposing. – pb10 Feb 11 at 8:43
• @pb10 Maybe think of it this way. Consider the left card out of the two you took. Either it is a king or a queen, it doesn't matter. Now, in order to form a pair, the right card needs to be the other king or queen. Since there are three cards left, and only one of them completes your pair, the probability is $\frac{1}{3}$. – jvdhooft Feb 11 at 8:44
• Hmm. So we treat the process of drawing a pair as two events no matter what? My thought process had been : Number of ways to draw a pair is 2 (QQ,KK). Number of ways to draw two cards is 4. Hence probability is 0.5. – pb10 Feb 11 at 8:51
After drawing the first card there are $$3$$ cards left, and exactly one of them is the "mate" of the card that has been drawn allready.
So the probability that this card will be drawn (resulting in a pair) is $$\frac13$$.
In this situation (no replacement) there is no essential difference between drawing cards one by one or drawing two cards together.
(To understand implant some order like: "the first card touched by my fingers is the first" or "after drawing I select a card and label it as first card". The only difference is that you did not had chance to see what card it is, but that on its own is not relevant for the probabilities.)
• Please let me know how this is flawed: Number of ways to draw a pair is 2 (QQ,KK). Number of ways to draw two cards is 4. Hence probability is 0.5 – pb10 Feb 11 at 8:53
• $P(Q=2)=P(K=2)=\frac12\frac13=\frac16$ and $P(Q=1,K=1)=P(\text{first a queen, then a king})+P(\text{first a king then a queen})=2\cdot\frac12\frac23=\frac23$. Here $Q$ denotes the number of queens that are drawn and $K$ denotes the number of kings that are drawn. Note that $P(\text{a pair})=P(Q=2)+P(K=2)=\frac13$. – drhab Feb 11 at 9:01
• "number of ways to draw two cards is 4". I guess you think of KK,KQ,QK,QQ. Yes, $4$ ways, but they do no have equal probabilities. The probability on KK is $\frac16$ (equals probability on QQ) and on KQ is $\frac13$ (equals probability on QK). So finding the probability is not just a matter of dividing number of specific ways by total number of ways. – drhab Feb 11 at 9:10
• Perfect. Thank you so much. My answer would make sense if we are for example tossing 2 coins where all probabilities are same for HH,HT,TH,TT. – pb10 Feb 11 at 9:13
• Indeed. That is the same as the card problem under the condition that the cards are replaced after each drawing. – drhab Feb 11 at 9:14
Simpler: You draw a card. No matter what it is, there is only 1 of three remaining cards that will match it. Hence $$P = 1/3$$.
It makes no difference if you draw two at a time. You can always say: one of the cards is on the top of the two I chose.... | 2019-04-21T04:24:16 | {
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http://winaudit.org/guides/sample-proportion/standard-error-of-sample-proportion.html | Home > sample proportion > standard error of sample proportion
# Standard Error Of Sample Proportion
repeatedly randomly drawn from a population, and the proportion of successes in each
sample is recorded ($$\widehat{p}$$),the distribution of the sample proportions sample proportion formula (i.e., the sampling distirbution) can be approximated by a normal distribution given that both $$n ## Sampling Distribution Of P Hat Calculator \times p \geq 10$$ and $$n \times (1-p) \geq 10$$. This is known as theRule of Sample Proportions. Note that some textbooks use standard error of proportion definition a minimum of 15 instead of 10.The mean of the distribution of sample proportions is equal to the population proportion ($$p$$). The standard deviation of the distribution of sample proportions is symbolized by $$SE(\widehat{p})$$ and equals $$\sqrt{\frac {p(1-p)}{n}}$$; this is known as thestandard error of $$\widehat{p}$$. standard error of p hat The symbol $$\sigma _{\widehat p}$$ is also used to signify the standard deviation of the distirbution of sample proportions. Standard Error of the Sample Proportion$SE(\widehat{p})= \sqrt{\frac {p(1-p)}{n}}$If $$p$$ is unknown, estimate $$p$$ using $$\widehat{p}$$The box below summarizes the rule of sample proportions: Characteristics of the Distribution of Sample ProportionsGiven both $$n \times p \geq 10$$ and $$n \times (1-p) \geq 10$$, the distribution of sample proportions will be approximately normally distributed with a mean of $$\mu_{\widehat{p}}$$ and standard deviation of $$SE(\widehat{p})$$Mean $$\mu_{\widehat{p}}=p$$Standard Deviation ("Standard Error")$$SE(\widehat{p})= \sqrt{\frac {p(1-p)}{n}}$$ 6.2.1 - Marijuana Example 6.2.2 - Video: Pennsylvania Residency Example 6.2.3 - Military Example ‹ 6.1.2 - Video: Two-Tailed Example, StatKey up 6.2.1 - Marijuana Example › Printer-friendly version Navigation Start Here! Welcome to STAT 200! Search Course Materials Faculty login (PSU Access Account) Lessons Lesson 0: Statistics: The “Big Picture” Le
Tables Constants Calendars Theorems Standard Error of Sample Proportion Calculator
## Standard Deviation Of Sample Proportion
Calculator Formula Download sample proportion calculator Script Online statistic calculator allows you to estimate the accuracy of
## Probability Of Sample Proportion Calculator
the standard error of the sample proportion in the binomial standard deviation. Calculate SE Sample Proportion of Standard Deviation https://onlinecourses.science.psu.edu/stat200/node/43 Proportion of successes (p)= (0.0 to 1.0) Number of observations (n)= Binomial SE of Sample proportion= Code to add this calci to your website Just copy and paste the below code to your webpage where you want https://www.easycalculation.com/statistics/standard-error-sample-proportion.php to display this calculator. Formula Used: SEp = sqrt [ p ( 1 - p) / n] where, p is Proportion of successes in the sample,n is Number of observations in the sample. Calculation of Standard Error in binomial standard deviation is made easier here using this online calculator. Related Calculators: Vector Cross Product Mean Median Mode Calculator Standard Deviation Calculator Geometric Mean Calculator Grouped Data Arithmetic Mean Calculators and Converters ↳ Calculators ↳ Statistics ↳ Data Analysis Top Calculators Standard Deviation Logarithm Mortgage LOVE Game Popular Calculators Derivative Calculator Inverse of Matrix Calculator Compound Interest Calculator Pregnancy Calculator Online Top Categories AlgebraAnalyticalDate DayFinanceHealthMortgageNumbersPhysicsStatistics More For anything contact [email protected]
0 otherwise. The standard deviation of any variable involves the sample proportion expression . Let's suppose there are m 1s (and n-m 0s) among the n subjects. Then, and is equal to (1-m/n) for m observations and 0-m/n standard error of for (n-m) observations. When these results are combined, the final result is and the sample variance (square of the SD) of the 0/1 observations is The sample proportion is the mean of n of these observations, so the standard error of the proportion is calculated like the standard error of the mean, that is, the SD of one of them divided by the square root of the sample size or Copyright © 1998 Gerard E. Dallal
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What Is The Standard Error Of The Sample Proportion p repeatedly randomly drawn from a population and the proportion of successes in each sample standard error of proportion formula is recorded widehat p the distribution of the sample proportions i e p Sample Proportion Formula p the sampling distirbution can be approximated by a normal distribution given that both n p Standard Error Of P Hat p times p geq and n times -p geq This is known as theRule of Sample Proportions Note that some textbooks use a p Sample Proportion Calculator p minimum of instead of The mean
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http://math.stackexchange.com/questions/282695/in-how-many-ways-can-3-distinct-teams-of-11-players-be-formed-with-33-men | # In how many ways can 3 distinct teams of 11 players be formed with 33 men?
Problem:
In how many ways can 3 distinct teams of 11 players be formed with 33 men? Note: there are 33 distinct men.
The problem is similar to this one: How many distinct football teams of 11 players can be formed with 33 men?
Fist, I thought the answer was: $$\binom{33}{11} \times \binom{22}{11} \times \binom{11}{11}$$
But there are clearly a lot of solutions overlapping.
-
Suppose that we wanted to divide the $33$ men into three teams called Team A, Team B, and Team C, respectively. There are $\binom{33}{11}$ ways to pick Team A. Once Team A has been picked, there are $\binom{22}{11}$ ways to pick Team B, and of course the remaining $11$ men form Team C. There are therefore $$\binom{33}{11}\binom{22}{11}\tag{1}$$ ways to pick the named teams. This is the calculation that you thought of originally.
But in fact we don’t intend to name the teams; we just want the men divided into three groups of $11$. Each such division can be assigned team names (Team A, Team B, Team C) in $3!=6$ ways, so the calculation in $(1)$ counts each division of the men into three groups of $11$ six times, once for each of the six possible ways of assigning the three team names. The number of ways of choosing the unnamed teams is therefore
$$\frac16\binom{33}{11}\binom{22}{11}\;.\tag{2}$$
Added: Here’s a completely different way to calculate it.
First we’ll pick the team containing the youngest of the $33$ men; there are $\binom{32}{10}$ ways to do that, since we need only choose his $10$ teammates. Then we choose the team containing the youngest of the remaining $22$ men; this can be done in $\binom{21}{10}$ ways. That leaves $11$ men to form the third team. This approach does not overcount: there’s always one team that contains the youngest man and one that contains the youngest man not on that first team. Thus, there are
$$\binom{32}{10}\binom{21}{10}\tag{2}$$
ways to choose the three teams.
Of course it would be a good idea to make sure that $(1)$ and $(2)$ actually yield the same result:
\begin{align*} \frac16\binom{33}{11}\binom{22}{11}&=\frac16\cdot\frac{33!}{11!22!}\cdot\frac{22!}{11!11!}\\\\ &=\frac13\cdot\frac{33\cdot32!}{11!22!}\cdot\frac12\cdot\frac{22\cdot 21!}{11!11!}\\\\ &=\frac{11\cdot32!}{11!22!}\cdot\frac{11\cdot21!}{11!11!}\\\\ &=\frac{32!}{10!22!}\cdot\frac{21!}{10!11!}\\\\ &=\binom{32}{10}\binom{21}{10}\;. \end{align*}
-
And yet another approach...
Line the 33 men up, and assign the first 11 to Team A, the second 11 to Team B and the remaining 11 to Team C.
The line can be formed in $33!$ ways. But this counts as distinct the $11!$ rearrangements of the 11 men within each of the three teams.
Thus, we have $33! / (11!)^3$ ways to set up the three teams.
This calculation counts as distinct having a particular group of 11 on Team A, or on Team B or on Team C. If this is not the case (I'm not convinced!) then divide by $3!$
- | 2016-02-09T07:50:38 | {
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https://web2.0calc.com/questions/help-asap_55 | +0
Help ASAP
0
330
5
+211
When $N$ is divided by 10, the remainder is $a$. When $N$ is divided by 13, the remainder is $b$. What is $N$ modulo 130, in terms of $a$ and $b$? (Your answer should be in the form $ra+sb$, where $r$ and $s$ are replaced by nonnegative integers less than 130.)
Rollingblade May 26, 2018
#1
+971
+2
When N is divided by 10, the remainder is a,
When N is divided by 13, the remainder is b.
From these two statements, we can setup the modular arithmetic systems:
\begin{align*} x &\equiv 1 \pmod{10}, \\ x &\equiv 0 \pmod{13}. \end{align*}
The solution is: $$X\equiv91\pmod{130}$$
Then we solve:
\begin{align*} y &\equiv 0 \pmod{10}, \\ y &\equiv 1 \pmod{13}. \end{align*}
The solution is: $$y\equiv40\pmod{130}$$
Therefore, the answer is $$N\equiv91a+40b\pmod{130}$$
I hope this helped,
Gavin
GYanggg May 26, 2018
#1
+971
+2
When N is divided by 10, the remainder is a,
When N is divided by 13, the remainder is b.
From these two statements, we can setup the modular arithmetic systems:
\begin{align*} x &\equiv 1 \pmod{10}, \\ x &\equiv 0 \pmod{13}. \end{align*}
The solution is: $$X\equiv91\pmod{130}$$
Then we solve:
\begin{align*} y &\equiv 0 \pmod{10}, \\ y &\equiv 1 \pmod{13}. \end{align*}
The solution is: $$y\equiv40\pmod{130}$$
Therefore, the answer is $$N\equiv91a+40b\pmod{130}$$
I hope this helped,
Gavin
GYanggg May 26, 2018
#4
+211
+1
Thank you so much!
The answer was right BTW
Rollingblade May 26, 2018
#2
0
Rollingblade, PLEASE make sure that your questions are readable, this is unbearable. You can't expect this forum to help you if you can't write your question instead of copying and pasting it.
Guest May 26, 2018
edited by Guest May 26, 2018
#3
+211
0
this is very readable, I don’t understand what you mean
Rollingblade May 26, 2018
edited by Rollingblade May 26, 2018
#5
0
Of course this specific question is readable, but you copy and pasted it. some of your other questions haven't been answered, because no one has deciphered them. Stop copying your questions because sometimes it turns into a mess that no one can read. You post unreadable questions and don't even respond when users ask you to clarify.
Guest May 26, 2018
edited by Guest May 26, 2018 | 2018-12-17T06:22:12 | {
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http://en.famp.ase.ro/a9rto/concave-up-and-down-calculator-wolfram-ab4ba6 | concave up and down calculator wolfram
(i) We will say that the graph of f(x) is concave up on I iff f '(x) is increasing on I. Collection of teaching and learning tools built by Wolfram education experts: dynamic textbook, lesson plans, widgets, interactive Demonstrations, and more. Note: Geometrically speaking, a function is concave up if its graph lies above its tangent lines. Of particular interest are points at which the concavity changes from up to down or down to up; such points are called inflection points . If it's positive, it is concave up, if it's negative it is concave down. And I wanna think about the intervals over which g is either concave upwards or concave downwards. Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. Online Integral Calculator » Solve integrals with Wolfram|Alpha. Now let's just remind ourselves what these things look like. Note that the value a is directly related to the second derivative, since f ''(x) = 2a.. Unlimited random practice problems and answers with built-in Step-by-step solutions. (i) We will say that the graph of f(x) is concave up on I iff f '(x) is increasing on I. I am given the following two functions, and I am to figure out if they are concave. a) Find the intervals on which the graph of f(x) = x + x is concave up, concave down and the point(s) of inflection if any. Curve segment that lies above its tangent lines is concave upward. Wolfram Demonstrations Project » Explore thousands of free applications across science, mathematics, engineering, technology, business, art, finance, social sciences, and more. If my calculations are correct, this function is convex in certain cases and concave in certain different cases. In other words, the graph of f is concave up. We say that a graph is concave up if the line between two points is above the graph, or alternatively if the first derivative is increasing. Thus f is concave up from negative infinity to the inflection point at (1, –1), and then concave down from there to infinity. It can calculate and graph the roots (x-intercepts), signs, Local Maxima and Minima, Increasing and Decreasing Intervals, Points of Inflection and Concave Up/Down intervals. I'm looking for a concave down increasing-function, see the image in the right lower corner. if, for any points and in , the function So: f (x) is concave downward up to x = −2/15. Show Instructions. As always, you should check your result on your graphing calculator. The graph is concave down on the interval because is negative. In general, you can skip parentheses, but be very careful: e^3x is e^3x, and e^(3x) is e^(3x). p. 1132, 2000. Let f(x) be a differentiable function on an interval I. b) Use a graphing calculator to graph f and confirm your answers to part a). Limits Go to Limits Ch 7. Wolfram Demonstrations Project » Explore thousands of free applications across science, mathematics, engineering, technology, business, art, finance, social sciences, and more. https://mathworld.wolfram.com/ConcaveFunction.html. That kind of information is useful when it comes to analyzing graphs using derivatives. Points of discontinuity show up here a bit more than in the First Derivative Test. If we are trying to understand the shape of the graph of a function, knowing where it is concave up and concave down helps us to get a more accurate picture. Free functions inflection points calculator - find functions inflection points step-by-step This website uses cookies to ensure you get the best experience. By using this website, you agree to our Cookie Policy. Sep 15, 2020 ... then the graph is concave down. Hints help you try the next step on your own. (ii) We will say that the graph of f(x) is concave down on I iff f '(x) is decreasing on I. 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Concavity calculator is any calculator that outputs information related to the concavity of a function changes concavity are inflection. Graph of f is concave down at any x-value where the signs switch from positive to negative or versa... | 2023-01-28T23:16:41 | {
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http://mathhelpforum.com/calculus/167026-integration-area-volume.html | # Math Help - integration for area&volume
1. ## integration for area&volume
1. O is the origin and A is the point on the curve y=tan x where $x=\frac{\pi}{3}.$
Calculate the area of the region R enclosed by the arc OA, the x-axis and the line $x=\frac{\pi}{3}$, giving your answer in EXACT form. Hence, find $\int^{\sqrt3} _0 tan^{-1} y dy$ in EXACT form.
Then, the region S is enclosed by the arc OA, the y-axis ad the line $y=\sqrt3$. Find the volume of the solid of revolution formed when S is rotated through $2\pi$ about the x-axis, giving your answer in EXACT form.
2. Give your answer in EXACT form for $\int^2 _0 \frac{5}{(x+1)(x^2+4)} dx$
for qn 1 :
i can do the first part....my ans is $ln2$
i can do the second part too by using integration by parts....my ans is $\frac{\pi \sqrt3}{3}-ln2$
then I don't know how to do the last part. if it rotated about y-axis , then i know, but it rotated about the x-axis.
for qn 2:
what i tried to do was
$\int^2 _0 \frac{5}{(x+1)(x^2+4)} dx$
$=5 \int^2 _0 \frac{1}{x^3 + x^2+ 4x+4}$
then ...i'm stuck....
2. Hi there,
For question #2, we use integration by partial fractions.
$
\frac{5}{(x+1)(x^2+4)} = \frac{A}{x+1}+\frac{Bx+C}{x^2+4}
$
By multiplying each side by the denominator on the left, we get:
$
5=A(x^2+4)+(Bx+C)(x+1)
$
Expand and gather all like terms:
$
5=Ax^2+Bx^2+Bx+Cx+4A+C
$
Equate coefficients on the right to those on the left (x^2 and x don't exist, so their coefficients must be 0):
for x^2, (A+B)=0
for x, (B+C)=0
for constant, (4A+C)=5
By solving this system, we find A=1, B=-1, C=1.
Now we can integrate (from the very first equation, plugging in A,B,C):
$\int\frac{1}{x+1}+\int\frac{-x+1}{x^2+4}$
The first integral is easy, $\ln{(x+1)}$
The second requires us to split up the integral:
$\int\frac{-x+1}{x^2+4}=\int\frac{1}{x^2+4}-\int\frac{x}{x^2+4}$
This time, the first integral is harder, we notice that this fits the formula for the derivative of arctan, with a=2:
$\int\frac{1}{(x^2+a^2)}=\frac{1}{a}arctan(\frac{x} {a})$
so, the first part of this second integral is equal to:
$\frac{1}{2}arctan(\frac{x}{2})$
And the second part is easy, $-\frac{\ln{(x^2+4)}}{2}$
This leaves us with a final indefinite integral of:
$
ln{(x+1)}+\frac{1}{2}arctan(\frac{x}{2})-\frac{\ln{(x^2+4)}}{2}
$
Now evaluate this from 0 to 2.
$\ln{3}+\frac{\Pi}{8}-\frac{\ln{8}}{2}+\frac{\ln{4}}{2}$
This is exact. You could simplify the natural log further if you wanted to.
3. ## Re: integration for area&volume
Originally Posted by wintersoltice
1. O is the origin and A is the point on the curve y=tan x where $x=\frac{\pi}{3}.$
Calculate the area of the region R enclosed by the arc OA, the x-axis and the line $x=\frac{\pi}{3}$, giving your answer in EXACT form. Hence, find $\int^{\sqrt3} _0 tan^{-1} y dy$ in EXACT form.
Then, the region S is enclosed by the arc OA, the y-axis ad the line $y=\sqrt3$. Find the volume of the solid of revolution formed when S is rotated through $2\pi$ about the x-axis, giving your answer in EXACT form.
for qn 1 :
i can do the first part....my ans is $\ln2$
i can do the second part too by using integration by parts....my ans is $\frac{\pi \sqrt3}{3}-\ln2$
then I don't know how to do the last part. if it is rotated about y-axis , then i know, but it is rotated about the x-axis.
then ...i'm stuck....
For question 1:
The integral $\int_0^{\sqrt3} tan^{-1} (y)\ dy$ gives the area of region S, not region R.
Your answers for the area of region R and the area of region S are correct.
The methods for revolving about the x-axis are essentially the same as for revolving about the y-axis. You can use disks/washers or cylindrical shells.
Disks/Washers: In this case washers. Form a washer from a rectangle which extends from y_1 to y_2 vertically and has a of width, Δx. Rotate the rectangle about the x-axis. The volume of the washer is $\displaystyle \text{v}= \left(\pi y_2^{\,2}-\pi y_1^{\,2}\right)\Delta x\,.$ In the case of region S, $\displaystyle y_1=\tan(x)$ and $\displaystyle y_2=\sqrt{3}.$ Letting Δx be very small, it becomes dx. Integrate this over x to get the volume.
$\displaystyle V=\pi\int_0^{\pi/3} \left(\sqrt{3}\right)^2-\tan^2(x)\,dx$
Cylindrical shells:Form a cylindrical shell from a rectangle which extends from x_1 to x_2 horizontally and has a of height, Δy, located a distance of y above the x-axis.. Rotating this about the x-axis gives a cylindrical shell with a volume of $\displaystyle \text{v}= 2\pi y\left( x_2- x_1\right)\Delta y\,.$ In the case of region S, $\displaystyle x_1=0$ and $\displaystyle x_2=\tan^{-1}(y).$ Letting Δy be very small, it becomes dy. Integrate this over y to get the volume.
$\displaystyle V=2\pi\int_0^{\sqrt{3}} y\tan^{-1}(y)\,dy$ | 2015-03-04T18:48:38 | {
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https://nbviewer.jupyter.org/urls/www.numfys.net/media/notebooks/NumpyIntermediate.ipynb | Intermediate NumPy¶
Modules - Basics¶
Last edited: November 21th 2019
In this intermediate NumPy notebook we go through some of the more common functions and features of NumPy. It is intended as a continuation of our notebook on the very basics of NumPy, found here. The notebook is somewhat lengthy. The different sections are intended to be readable on their own, so feel free to jump to the section that you are interested in.
In [1]:
import numpy as np
import matplotlib.pyplot as plt
Arrays of different shapes¶
Arrays can have any dimension and shape that we wish. This makes them very powerful. Let us transform $$(0, 1, 2, 3, 4, 5, 6, 7, 8)$$ into $$\begin{bmatrix} 0 & 1 & 2\\ 3 & 4 & 5\\ 6 & 7 & 8 \end{bmatrix}.$$
In [2]:
# Create a 3x3 matrix
A = np.arange(9) # Points from 0 through 8, with spacing 1
print("A before reshape:", A)
A = A.reshape((3,3)) # Set shape to 3x3
print("A after reshape:")
print(A)
A before reshape: [0 1 2 3 4 5 6 7 8]
A after reshape:
[[0 1 2]
[3 4 5]
[6 7 8]]
In [3]:
# We can also get an empty array with given size directly from np.zeros
np.zeros((4, 2))
Out[3]:
array([[0., 0.],
[0., 0.],
[0., 0.],
[0., 0.]])
Two dimensional arrays can be used just like matrices, which makes it easy to express our mathematical expressions in code. In NumPy, the symbol for matrix or vector multiplication is @.
In [4]:
x = np.array([0, 1, 2])
print(A @ x) # Print the result of A matrix multiplied with x
[ 5 14 23]
Read more about linear algebra in NumPy in this notebook, which covers solving system of equations, eigenvalue problems and more.
Having matrix-shaped arrays becomes especially useful once we have learned slicing.
Slicing and indexing¶
When we have arrays, be it one or multi-dimensional, it is useful to be able to refer to only the elements we want, just like we do with indexing in normal lists.
Lets review normal lists first. Remember that lists are 0-indexed, that is the first element is element 0, the second is 1 and so on.
In [5]:
my_list = [1,2,3,4] # Normal Python list
# We can access different parts of the list by slicing and indexing:
print("my_list[0]:\t", my_list[0]) # First element
print("my_list[:2]:\t", my_list[:2]) # The two first elements
print("my_list[-2:]:\t", my_list[-2:]) # The last two elements
my_list[0]: 1
my_list[:2]: [1, 2]
my_list[-2:]: [3, 4]
In general the syntax for slicing a lists is my_list[start:end:step], where start is the first element we want, end is the last element (non-inclusive), and step is the step size. Note that having start or end empty, will slice from the start or to the end, respectively. Negative values count from the end, so that -1 is the last element.
If you are completely unfamiliar with this, you are advised to play around with it now, before moving on to array slicing.
We can do the same with arrays, but arrays have even more ways of slicing!
The syntax for slicing in NumPy is exactly the same as for lists, but we can do it for each dimension! For a two-dimensional array the syntax then becomes my_array[start_1:end_1, start_2:end_2], where start_1 and end_1 is the start and end values for the first axis, and similarly with start_2 and end_2 for the second axis.
Let us again consider the matrix $$A = \begin{bmatrix} 0 & 1 & 2\\ 3 & 4 & 5\\ 6 & 7 & 8 \end{bmatrix}$$ represented as an array.
In [6]:
# Generate the array
# A = [0, 1, 2]
# [3, 4, 5]
# [6, 7, 8]
A = np.arange(9).reshape((3,3))
In [7]:
# Let us access some elements
# A_00, first row, first column (remember 0-index, in mathematical notation this would be A_11)
print("A_00:", A[0,0])
# A_02, first row, third column
print("A_02:", A[0,2])
# A_21, third row, last second column
print("A_21:", A[2,1])
A_00: 0
A_02: 2
A_21: 7
In [8]:
# We can go even more advanced, and get entire columns and rows
# First row of A
print("First row:", A[0])
# First column of A
print("First column:", A[:,0])
First row: [0 1 2]
First column: [0 3 6]
Getting the first row is something we can do with plain old nested python lists, getting the first column, however, would require a for-loop with lists. With arrays this is achieved with a simple index, as shown in the example. At first this might seem very confusing, but we will break it down to make it understandable.
Remember from our list-review that : means all elements, since both start and end is empty. For example my_list[:] just gives us my_list. In A[:, 0] we mean "entire list in first axis, 0th element of second axis", that is any row, first column. Building on this notation, we remember that my_list[1:] gives us the entire list starting from the second, index 1, item. We can use this in arrays to get parts of a column or row.
In [9]:
# First column of A, starting from second element in column
print("A[1:, 0]:", A[1:, 0])
# We can of course also get other columns
print("A[1:, 2]:", A[1:, 2])
A[1:, 0]: [3 6]
A[1:, 2]: [5 8]
Try to convince yourself as to why the following line gives us the result that it does:
In [10]:
print("A[1:, 1:]:")
print(A[1:, 1:])
A[1:, 1:]:
[[4 5]
[7 8]]
Meshgrid¶
Meshgrid may appear confusing at first, but it is very useful and once you get the grasp of it, you will hopefully realize that it is quite simple. Before explaining what meshgrids are, let us explain why we need them. Imagine that you were to plot a heatmap, that is a 2D plot where each point on the plane is given a color dependent on the value of the function that you are plotting. Recall that when plotting in matplotlib, which you can read more about here, one has to supply two lists, one with $x$-values and one with $y$-values. This gives a set of points, where the function will be plotted.
When plotting in the plane, one needs three lists of values, the $x$ and $y$ coordinates, and the function values at each point. For each $x$ value, we want to iterate over all the $y$ values, so that every point is covered. If we just passed our $x$- and $y$-arrays directly, however, we would get only one point for each $x$-value, and likewise for the $y$-values. This is where meshgrid comes into play. For the two dimensional case, it will give us two arrays, where each value is repeated in such a way that the two lists forms a grid. This is better demonstrated by an example.
In [11]:
x = np.arange(0, 5)
y = np.arange(2, 5)
xx, yy = np.meshgrid(x, y)
print('x =',x, ', y =', y, '\n')
print(xx, '\n')
print(yy)
x = [0 1 2 3 4] , y = [2 3 4]
[[0 1 2 3 4]
[0 1 2 3 4]
[0 1 2 3 4]]
[[2 2 2 2 2]
[3 3 3 3 3]
[4 4 4 4 4]]
Notice that if you were to extract corresponding elements in xx and yy, i.e. the elements with the same position in the matrices, you would get pairs of $x$ and $y$ values. We have thus assigned coordinates to each point, just like we wanted! It is also worth noting that xx and yy have the same shape. We can now plot our result.
In [12]:
plt.pcolormesh(xx, yy, xx + yy) # pcolormesh plots a heatmap
plt.show()
Computation speed considerations:
Meshgrid is not only useful for plotting. It enables us to do calculate values over multiple values very easily. Remember that with NumPy we can calculate a value over an array by simply passing the array to the function. We can do this with meshgrids as well, they are nothing more than two dimensional arrays. This helps us avoid nested for-loops, which greatly speeds up our computations and makes the code easier to write and read.
Mehsgrids are useful for all functions that are functions of two variables. As an example, we will look at the steady state amplitude of a driven damped harmonic oscillator which is an expression derived in classical mechanics. It is given by $$\Theta_0(q, \omega_D) = \frac{F_D}{(\omega_0^2 + \omega_D^2)^2 + (q\omega_D)^2},$$ where $F_D$ is the driving force, $\omega_0$ is the eigenfrequency of the oscillator, $q$ is the damping factor, and $\omega_D$ is the driving frequency.
We want to look at this amplitude for different $q$ and $\omega_D$.
In [13]:
def amplitude(omega, omega_drive, q):
"""Amplitude of a driven harmonic oscillator
with eigenfrequency omega, external force with
frequency given by omega_drive, and a damping
constant q. Here the driving force F_D is equal to 1."""
return 1/np.sqrt((omega**2-omega_drive**2)**2 + (q*omega_drive)**2)
omega = 1
q = np.linspace(0.5, 1.5, 20)
omega_D = omega * np.linspace(0.5, 1.5, 40)
qq, omega_DD = np.meshgrid(q, omega_D)
In [14]:
# Calculate the amplitude for each omega_D and q
a = amplitude(omega, omega_DD, qq)
# If X and/or Y are 1-D arrays they will be expanded
# as needed into the appropriate 2-D arrays.
# I.e. no need of np.meshgrid in this case.
plt.pcolormesh(omega_D, q, a.T) # a.T returns the transpose of a
plt.colorbar(label='Amplitude')
plt.ylabel("q")
plt.xlabel("$\omega_D$")
plt.show()
We will now investigate the calculation times further. To do this, we will use Jupyter's %timeit function. It calculates the time it takes to execute a command or cell. Lines prefixed by %timeit will be timed on their own. Starting a cell with %%timeit, will time the whole cell.
In [15]:
%timeit a = amplitude(omega, omega_DD, qq)
8.97 µs ± 99.3 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
In [16]:
%%timeit
# Equivalent calulation, using for-loops.
a_python = []
for oD in omega_D:
for q_val in q:
a_python.append(amplitude(omega, oD, q_val))
2.22 ms ± 71.9 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
With computation times of 9.0$\mu$s and 2.2ms, it is obvious that the difference in computation time is significant! Computational speed will be of great importance as the size of the data becomes big.
In [17]:
# Greatly increase the number of points from previous example!
omega = 1
q = np.linspace(0.5, 1.5, 2000)
omega_D = omega * np.linspace(0.5, 1.5, 4000)
qq, omega_DD = np.meshgrid(q, omega_D)
%timeit a = amplitude(omega, omega_DD, qq)
73.8 ms ± 14.7 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
In [18]:
%%timeit
a_python = []
for oD in omega_D:
for q_val in q:
a_python.append(amplitude(omega, oD, q_val))
24 s ± 2.28 s per loop (mean ± std. dev. of 7 runs, 1 loop each)
For the curious reader, we will mention that this could also have been achieved using NumPy's broadcasting functionality. It is a more direct method, but less intuitive.
Complex numbers¶
NumPy has built in support for complex numbers. They are intuitive to work with, and makes it easy to put mathematics into code.
NumPy uses j for the imaginary unit. Let us see it in action.
In [19]:
print(1j**2) # Should give -1
(-1+0j)
To make NumPy understand that we wish to work with complex numbers, we must explicitly set the imaginary part, even when it is zero.
In [20]:
# Fails
print(np.sqrt(-1))
# Works
print(np.sqrt(-1+0j))
nan
1j
/usr/lib/python3.7/site-packages/ipykernel_launcher.py:2: RuntimeWarning: invalid value encountered in sqrt
In [21]:
# Array with imaginary numbers between 0 and 2pi j
a = np.linspace(0+0j, 2*np.pi*1j, 50)
In [22]:
plt.plot(np.abs(a), np.real(np.exp(a)), label=r'Re$[e^{ia}]$') # Plot real value of np.exp(a)
plt.plot(np.abs(a), np.imag(np.exp(a)), label=r'Im$[e^{ia}]$') # Plot imaginary value of np.exp(a)
plt.legend()
plt.xlabel('$a$')
plt.show()
We observe a $90^{\circ}$ phase difference between the real and imaginary part, just like expected.
Using only built-in Python functions for reading from and writing to files can be tedious. NumPy has functions designed specifically for reading and writing structured data, np.loadtxt and np.savetxt. The functions are very powerful, and have lots of options. We will here focus on the basic usage, and refer the reader to the NumPy documentation for more advanced usage.
The data file used in this example can be found here.
loadtxt will read the data in the file, convert it to an appropriate data type, and store it in a NumPy array. Let us see it in action.
In [23]:
# From the header line in my_data.txt,
# we know that the file is in the format
# time temperature wind
print(data)
[[ 0. 10.52220137 3.15239917]
[ 0.52631579 10.23355002 3.27285991]
[ 1.05263158 10.92881718 3.53663193]
[ 1.57894737 10.4880909 3.50229028]
[ 2.10526316 10.43978719 3.47322148]
[ 2.63157895 10.40069219 4.58370474]
[ 3.15789474 10.38539 4.92386551]
[ 3.68421053 10.82980119 4.74906494]
[ 4.21052632 10.45174438 5.55397891]
[ 4.73684211 10.16078747 6.14630413]
[ 5.26315789 10.10223829 6.07077798]
[ 5.78947368 10.59938701 7.21440152]
[ 6.31578947 10.97894173 7.69948852]
[ 6.84210526 10.63288969 9.74896976]
[ 7.36842105 10.48996214 9.39695595]
[ 7.89473684 10.53800563 11.24194734]
[ 8.42105263 10.36691713 11.21430356]
[ 8.94736842 10.51385622 12.03454861]
[ 9.47368421 10.04225154 12.11379482]
[10. 10.64815814 13.05222008]]
Notice that each row in our array corresponds to a time. Often, it would be more convenient to have each data series in its own array, for example if we wanted to plot our data. We could achieve this with NumPy slicing, but loadtxt has a more direct way of doing this.
In [24]:
t, temperature, wind = np.loadtxt('my_data.txt', unpack=True)
unpack=True transposes our data, so that each row represents one field, in this example time, temperature, and wind. This way, we can easily unpack the data into our variables t, temperature, and wind.
In [25]:
plt.plot(t, temperature, 'o-', label="Temperature")
plt.plot(t, wind, 'o-', label="Wind")
plt.legend();
We will now generate some new data, that we can store to a file, my_new_data.txt.
In [26]:
# Mean of temperature
T_mean = np.mean(temperature)
# Deviation from mean at each time
delta_temperature = temperature - T_mean
# Save to file
np.savetxt('my_new_data.txt', np.transpose((t, delta_temperature)))
A lot happen in the final line, so we will analyze it further here. savetxt takes the data to be written to file as second argument, in this example it is np.transpose((t, delta_temperature)). Why the transpose, one might ask; we wish to have t and delta_temperature as columns, not rows, so that it has the same format as our input file.
Useful NumPy functions and examples of usage¶
In this section you will find some useful functions with a short explanation. For a more detailed explanation, please visit the NumPy documentation. After the list of functions follow some examples of using NumPy to solve various problems one might encounter. This is intended as an inspiration to begin thinking in the "NumPy mindset".
Some useful functions:
• np.amin, np.amax. Returns min and max value of array.
• np.argmin, np.argmax. Returns the index of the min and max value of an array.
• np.argwhere. Returns the indices where a condition is true. For example
a = np.array([0,1,2,3,10])
np.argwhere(a>1)
would yield [2,3,4].
Coefficients of wave function¶
The physics of this example is not important for understanding NumPy. If you do not understand this, do not worry!
We know from linear algebra that given a orthonormal basis $B$ of $V$, any vector $v\in V$ can be written as a linear combination of vectors $b \in B$ . We also know that the coefficient of $b\in B$ for some $v\in V$ is given by $$c_b = \langle b, v\rangle.$$
This is used much in physics, probably most noteworthy in quantum mechanics. Here follows a quick example of using NumPy for linear algebra, in order to calculate the coefficients of for different wave functions.
In [27]:
x = np.linspace(0, 1, 5000)
dx = x[1] - x[0]
# Basis wave functions for particle in a box potential
# These functions can be easliy derived using basic quantum mechanics
g1 = lambda x: np.sqrt(2)*np.sin(np.pi*x)
g2 = lambda x: np.sqrt(2)*np.sin(2*np.pi*x)
g3 = lambda x: np.sqrt(2)*np.sin(3*np.pi*x)
# Create some linear combinations
y1 = g1(x) + g2(x)
y2 = g1(x) + 2*g2(x)
y3 = 0.2*g1(x) + 10*g3(x)
# Stack the functions vertically, creating matricies
basis_functions = np.vstack((g1(x), g2(x), g3(x)))
my_functions = np.vstack((y1, y2, y3))
# Take the matrix product
basis_functions*dx @ my_functions.T
Out[27]:
array([[1.00000000e+00, 1.00000000e+00, 2.00000000e-01],
[1.00000000e+00, 2.00000000e+00, 1.35873630e-15],
[1.69804390e-15, 1.41392499e-15, 1.00000000e+01]])
We can now read the coefficients of each function y1, y2, and y3 from the columns. For example, first column shows that y1 has coefficients 1, 1, and $1.7\cdot 10^{-15}\approx 0$, as expected.
Find minima of function¶
Consider the function $$f(x) = -\frac{1}{6}x^3 + 2x.$$
Find its local minima.
In [28]:
x = np.linspace(-5, 5, 50)
def f(x):
return -x**3/6 + 2*x
plt.plot(x, f(x))
plt.show()
We see that the local minima is left of 0, and that there are smaller values than this minima right of zero. It is therefore smart to limit our list of $x$'s to those that are negative.
x[x<0]
gives us exactly that. Now we just pass this array to f, and find its minima using np.min.
In [29]:
np.min(f(x[x<0]))
y_min = np.min(f(x[x<0]))
This gives us the $y$-value at the local minima. We also wish to find the corresponding $x$-value. np.argmin gives us the index of the $y$-value with the minimal value, using this index on the $x$-array gives us the corresonding value.
In [30]:
min_arg = np.argmin(f(x[x<0]))
x_min = x[x<0][min_arg]
print(f"(x,y) = {x_min:.2f}, {y_min:.2f}")
(x,y) = -1.94, -2.66
Some general considerations regarding computation speed¶
The main advantage of using NumPy in numerical computations, in addition to adding a simple and intuitive way to write our code, is the significantly increased computation speed. Here are some points worth remembering, in order to best take advantage of this.
Use NumPy arrays! The most important step is to actually use NumPy arrays. With a few exceptions, all lists that one uses, should be NumPy arrays.
Use the NumPy functionality. Take use of the NumPy arrays. Inexperienced NumPy users have a tendency to solve their problems using more conventional Pythonic solutions, when it would be advantageous to use NumPy's functionality. For example, one should very seldom need to use for-loops, as these can often be substituted with NumPy operations, which are much faster.
Do not append to NumPy arrays. One should try to avoid appending to NumPy arrays. It is significantly slower than initializing your array first, and then writing to it. A good example is iterative methods with a fixed number of iterations, like Euler's method. Initializing the $y$-array first with either zeros (np.zeros) or empty values (np.empty), is faster than appending for each time step. | 2020-01-29T04:21:50 | {
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https://www.physicsforums.com/threads/diameter-of-a-circle-endpoints-p-0-0-q-8-4-what-equation.817411/ | Diameter of a circle endpoints P(0,0) Q(8,-4) what equation:
1. Jun 4, 2015
Jaco Viljoen
Mod note: Moved from technical math section, so no template.
The diameter of a circle has endpoints P(0,0) and Q(8,-4) Find the equation:
First I will find the midpoint:
$$M(x,y)=(x1+x2)/2,(y1+y2)/2$$
$$=8/2,-4/2)$$
$$M(x,y)=(-4,-2)$$
Then I will find the radius:
$$r^2=(x-h)^2+(y-k)^2$$
$$r^2=(0-4)^2+(0+2)^2$$
$$r^2=16+4$$
$$r^2=20$$
so
$$(x-4)^2+(y+2)^2=20$$
Last edited by a moderator: Jun 4, 2015
2. Jun 4, 2015
Staff: Mentor
That's what I get, as well.
BTW, this looks like homework, or at least a problem from a textbook, so I'm moving it to the homework section.
3. Jun 4, 2015
Jaco Viljoen
Thank you Mark,
I wasn't sure if I should post under pre calculus as I couldn't find any geometry there, thanks again.
4. Jun 4, 2015
Staff: Mentor
This would come under the heading of analytic geometry, so the Precalc section is the right place for it. | 2017-11-19T16:18:36 | {
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http://mathhelpforum.com/differential-geometry/137801-sum-n-1-th-row-pascal-triangle-proof.html | # Thread: sum of (n+1)th row pascal triangle proof
1. ## sum of (n+1)th row pascal triangle proof
prove that the sum from j=1 to j=n of combinatorial(n, j) is 2^n.
so basicially i have to prove that the sum of the (n+1)th row of pascal's triangle is 2^n. I have tried using induction on n and the definition of combinatiorial(n,j) = n!/[j!(n-j)!] but I keep getting stuck. If someone could provide guidance as to how I should approach this problem it would be more helpful than a straight solution because I am trying to solve problems as practice for an exam
2. The binominal theorem is: $\left( {x + y} \right)^n = \sum\limits_{k = 0}^n {\binom{n}{k}x^{n - k} y^k }$.
Here $\sum\limits_{k = 0}^n {\binom{n}{k}}=\binom{n}{0}+\binom{n}{1}+\binom{n} {2}+\cdots+\binom{n}{n}$ is the $(n+1)^{th}$ row of the Pascal Triangle.
Now let $x=1~\&~y=1$. See what you get.
3. Originally Posted by 234578
prove that the sum from j=1 to j=n of combinatorial(n, j) is 2^n.
so basicially i have to prove that the sum of the (n+1)th row of pascal's triangle is 2^n. I have tried using induction on n and the definition of combinatiorial(n,j) = n!/[j!(n-j)!] but I keep getting stuck. If someone could provide guidance as to how I should approach this problem it would be more helpful than a straight solution because I am trying to solve problems as practice for an exam
Here's how I like to think about it.
Let $\Omega_n=\{1,\cdots,n\}$and let $f(n)=\text{card }\mathcal{P}(\Omega_n)$. Then, we can think of this as saying " $f(n)$ is how many subsets of $\Omega_n$ there are. Well, how many are there for $0$? Well, it is how many ways I can choose $0$ items from $n$ items, and so $n\choose 0$. How aboust subsets with one element? Using the same logic it is $n\choose 1$. Thus, to get how many subsets with $1\leqslant k\leqslant n$ elements I only need compute $n\choose k$. Thus, to get the number of all subsets of $\Omega_n$ it is
$\text{number of subsets with zero elements}+\cdots+\text{number of subsets with }n\text{ elements}$ $={n\choose 0}+\cdots+{n\choose n}=f(n)$.
Thus, $f(n)=\text{card }\mathcal{P}(\Omega_n)=\sum_{j=0}^{n}{n\choose j}$. So, let's see if we can find a recurrence relation for $n$. To do this we must only relate the number of subsets of $\Omega_{n+1}$ to $\Omega_n$. So, let's do it:
We can partition $\mathcal{P}(\Omega_{n+1})$ into two blocks; namely those that contain $n+1$ (call them $B_1$) and those that don't ( $B_2$). So, how many things are in $B_1$? Well, it's all the subsets I can make out of elements of $\Omega_{n+1}=\Omega_{n}\cup\{n+1\}$. But, as the way I wrote the RHS side indicates this is clearly all the subsets I can make which contain elements only from $\Omega_n$. But, we called this number $f(n)$. So what about $B_2$? Well, since all of these subsets contain elements of $\Omega_{n+1}=\Omega_n\cup\{n+1\}$ we see in particular that each set contains only elements of $\{n+1\}$ or contains elements of $\Omega_n$ and $\{n+1\}$. Clearly the number of sets we can form using only $\{n+1\}$ is $1$. The latter can be thought about as taking one of the subsets of $\Omega_n$ and uniting it with $\{n+1\}$. Thus, there are $f(n)$ of those. Thus, $B_2=f(n)+1$....or does it! Notice that $f(n)$ included the number of ALL subsets of $\Omega_n$ including $\varnothing$. But, all of the sets in $B_2$ must contain $n+1$ and so we can't count $\varnothing$. Thus, the number of things in $B_2$ is $f(n)+1-1=f(n)$.
Thus, since $B_1\cup B_2=\mathcal{P}(\Omega_{n+1}),B_1\cap B_2=\varnothing$ we have that $f(n+1)=\text{card }\mathcal{P}(\Omega_{n+1})=\text{card }B_1+\text{card }B_2=f(n)+f(n)=2f(n)$.
Thus, remembering that $f(n)=\sum_{j=0}^{n}{n\choose j}$ we are done. Since proving that $f(n+1)=2f(n)$ and $f(0)=\sum_{j=0}^{0}{0\choose j}=\text{card }\mathcal{P}(\Omega_0)=1$ implies that $f(n)=2^n$ is trivial.
4. Have you ever used a cannon when a flyswatter will work well?
5. Originally Posted by Plato
Have you ever used a cannon when a flyswatter will work well?
Aren't you the proponent of "don't plug and chug...actually think!". While what I did was way more than the problem probably intended I think it was more insightful. Anyways, I don't think nor do I care if the OP uses this for their homework (though they probably shouldn't!!!), I just want them to learn. | 2017-01-19T17:59:46 | {
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http://math.stackexchange.com/questions/435192/is-binets-formula-for-the-fibonacci-numbers-exact | # Is Binet's formula for the Fibonacci numbers exact?
Is Binet's formula for the Fibonacci numbers exact?
$F_n = \frac{(1+\sqrt{5})^n-(1-\sqrt{5})^n}{2^n \sqrt{5}}$
If so, how, given the irrational numbers in it?
Thanks.
-
Yes, it is exact. The irrational parts happen to cancel out. Have you tried plugging in a few values of $n$? – Tobias Kildetoft Jul 3 '13 at 9:53
Having irrational numbers in a formula producing integers is not anything strange. Just take e.g. $\sqrt{2}^2$. – Edvard Fagerholm Jul 3 '13 at 9:55
Why not try before ask? Ir you can try to apply the homomorphism $\alpha$that send $\sqrt5$ to $-\sqrt5$ to observe that $F_n$ is always fixed by $\alpha$, i.e. $F_n$ is always a rational algebraic integer, hence an ordinary integer. Does this suffice to be a good reason that $F_n$ might look right? – awllower Jul 3 '13 at 10:15
It is exact. However, it is not a good way to calculate $F_n$ for very large $n$. – André Nicolas Jul 3 '13 at 13:21
As others have noted, the $\sqrt 5$ parts cancel, leaving an integer. We can recover the Fibonacci recurrence formula from Binet as follows:
$$F_n+F_{n-1} = \frac{(1+\sqrt{5})^n-(1-\sqrt{5})^n}{2^n \sqrt{5}}+\frac{(1+\sqrt{5})^{n-1}-(1-\sqrt{5})^{n-1}}{2^{n-1} \sqrt{5}}=$$$$\frac{(1+\sqrt{5})^{n-1}(1+\sqrt 5+2)-(1-\sqrt{5})^{n-1}(1-\sqrt5+2)}{2^n \sqrt{5}}$$
Then we notice that $(1\pm\sqrt5)^2=6\pm2\sqrt 5=2(3\pm\sqrt5)$
And we use this to simplify the final expression to $F_{n+1}$ so that $F_n+F_{n-1} =F_{n+1}$
And the recurrence shows that if two successive $F_r$ are integers, every Fibonacci number from that point on is an integer. Choose $r=0,1$. This is another way of proving that the cancellation happens.
-
Maybe it should be Bennet's formula :). – Dan Rust Jul 3 '13 at 11:07
This is perfect, thank you. – N. McA. Jul 3 '13 at 14:48
@DanielRust When you time-travel names always get distorted... – Tobias Kienzler Jul 3 '13 at 18:09
It is exact, all right. When you expand the powers in the numerators the alternating signs mean that all the surviving terms are of the form an integer times $\sqrt5$. Therefore all the $\sqrt5$s cancel.
Try it out with $n=2$ and $n=3$.
It may be of interest to you to observe that as $|1-\sqrt5|/2\approx0.618$ its powers quickly approach zero. So for sizable $n$, you can drop that term, and just round the dominant term to the nearest integer. For example with $n=8$ you get $F_8=21$ and $(1+\sqrt5)^8/(2^8\sqrt5)\approx21.009$.
-
For linear reccurent sequence you can find expression depending on the roots of the associated polynom.
$$F_{n+1} = F_{n} + F_{n-1}$$
is associated to
$$x^2 = x +1$$
Wich has two solution, $\frac{1 - \sqrt{5}}{2}$ and $\frac{1 + \sqrt{5}}{2}$ (the golden ratio, a more than interesting number)
Simple roots give a general solution of the form:
$$F_{n} = A * (\frac{1 - \sqrt{5}}{2})^{n} + B * (\frac{1 + \sqrt{5}}{2})^n$$
To determine A and B you have to input initial contidions:
$$F_{0} = A + B = 0$$
So $$B = - A$$
and
$$F_{1} = -B * (\frac{1 - \sqrt{5}}{2} - \frac{1 + \sqrt{5}}{2}) =1$$
$$B = \frac{1}{\sqrt{5}}$$
So
$$F_{n} = \frac{ (1 + \sqrt{5})^{n} - (1 - \sqrt{5})^n}{2^n\sqrt{5}}$$
By Solving $F_0 = i$ and $F_1 = j$ you can find the general expression of Fibonnaci sequence with starting terms i and j.
-
This is really neat but I have a perhaps very novice question: how do you relate recurrence relations and polynomials? – Cameron Williams Jul 3 '13 at 13:24
you just have to replace, if R is a root of the associated polynom, you have in the general expression $R^{n+1} = R^{n-1} * R^2 = R^{n-1} * (R + 1) = R^{n} + R^{n-1}$ thus satisfying the recurrence equation. General results about dimension of the space of solutions will give the reverse implication. – Were_cat Jul 3 '13 at 13:36
This works the same way for others polynom, beware of multiple roots. – Were_cat Jul 3 '13 at 13:36
Oh that makes complete sense. Thanks! – Cameron Williams Jul 3 '13 at 13:38
$${(1+\sqrt5)^n-(1-\sqrt5)^n} \over 2^n \sqrt5$$
Expand the numerator:
$$(1+\sqrt5)^n-(1-\sqrt5)^n=\sum_{k=0}^{n}{\binom{k}{n}\sqrt5^k}-\sum_{k=0}^{n}{\binom{k}{n}(-1)^k\sqrt5^k}$$
In the second sum, all the even powered terms get a positive sign, which becomes a negative sign due to the fact that the second sum is being subtracted. Those cancel out with the first sum and we get:
$$(1+\sqrt5)^n-(1-\sqrt5)^n=2\sum_{0\leq2k\leq n}{\binom{2k+1}{n}\sqrt5^{2k+1}}$$
We're dividing this by $2^n\sqrt5$ so we get:
$$\frac{2}{2^n}\sum_{0\leq2k\leq n}{\binom{2k+1}{n}\sqrt5^{2k}}$$ $$=\frac{1}{2^{n-1}}\sum_{0\leq2k\leq n}{\binom{2k+1}{n}5^k}$$
Though I have to say, I am a bit stumped as to why this should be an integer.
-
So this does not answer the question: And I think that to prove this expression is an integer is not so obvious. Maybe one can prove that this is an integer by induction; but then it loses the initial meaning of proving the integer-ness by an explicit expression I suppose. In any case, thanks for sharing this idea. – awllower Jul 4 '13 at 14:52
@awllower Strictly speaking the question only involved the irrationals, which this does explain. I have a bit of a feeling the integrality of this expression might have something to do with the sum of the $n$th row of Pascal's triangle adding to $2^n$ – Jack M Jul 4 '13 at 15:28
Good: it would be my pleasure to read the application of Pascal triangle in this situation as well. I think that, as an amazing application of combinatorial arguments, this answer is quite great. :) – awllower Jul 5 '13 at 4:59 | 2016-06-30T12:40:10 | {
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https://sankore.com/riz0beq/total-no-of-onto-functions-from-a-to-b-ffb20a | Not onto. 3. is one-to-one onto (bijective) if it is both one-to-one and onto. (D) 72. Onto Function Definition (Surjective Function) Onto function could be explained by considering two sets, Set A and Set B, which consist of elements. Let f and g be real functions defined by f(x) = 2x+ 1 and g(x) = 4x – 7. asked Feb 16, 2018 in Class XI Maths by rahul152 ( -2,838 points) relations and functions Let A = {a 1, a 2, a 3} and B = {b 1, b 2} then f : A -> B. Here are the definitions: 1. is one-to-one (injective) if maps every element of to a unique element in . An onto function is also called surjective function. Let X, Y, Z be sets of sizes x, y and z respectively. Functions: One-One/Many-One/Into/Onto . If n > m, there is no simple closed formula that describes the number of onto functions. Don’t stop learning now. I just need to know how it came. So the correct option is (D). In a one-to-one function, given any y there is only one x that can be paired with the given y. where as when i try manually it comes 8 . Click hereto get an answer to your question ️ Write the total number of one - one functions from set A = { 1,2,3,4 } to set B = { a,b,c } . There are $$\displaystyle 3^8=6561$$ functions total. Math Forums. In other words, if each b ∈ B there exists at least one a ∈ A such that. Yes. For function f: A→B to be onto, the inequality │A│≥2 must hold, since no onto function can be designed from a set with cardinality less than 2 where 2 is the cardinality of set B. Onto function or Surjective function : Function f from set A to set B is onto function if each element of set B is connected with set of A elements. Therefore, S has 216 elements. I am trying to get the total number of onto functions from set A to set B if the former has m elements and latter has n elements with m>n. The total no.of onto function from the set {a,b,c,d,e,f} to the set {1,2,3} is????? 2.1. . Q3. This is same as saying that B is the range of f . A function is said to be bijective or bijection, if a function f: A → B satisfies both the injective (one-to-one function) and surjective function (onto function) properties. In the example of functions from X = {a, b, c} to Y = {4, 5}, F1 and F2 given in Table 1 are not onto. Therefore, each element of X has ‘n’ elements to be chosen from. Yes. Example 9 Let A = {1, 2} and B = {3, 4}. Option 2) 120. Out of these functions, the functions which are not onto are f (x) = 1, ∀x ∈ A. Need explanation for: If n(A)= 3 , n(B)= 5 Find the number of onto function from A to B, List of Hospitality & Tourism Colleges in India, Knockout JEE Main May 2022 (Easy Installments), Knockout JEE Main May 2021 (Easy Installments), Knockout NEET May 2021 (Easy Installments), Knockout NEET May 2022 (Easy Installments), Top Medical Colleges in India accepting NEET Score, MHCET Law ( 5 Year L.L.B) College Predictor, List of Media & Journalism Colleges in India, B. Transcript. 2. is onto (surjective)if every element of is mapped to by some element of . Explanation: From a set of m elements to a set of 2 elements, the total number of functions is 2m. 2×2×2×2 = 16. We need to count the number of partitions of A into m blocks. Option 4) none of these For function f: A→B to be onto, the inequality │A│≥2 must hold, since no onto function can be designed from a set with cardinality less than 2 where 2 is the cardinality of set B. according to you what should be the anwer Option 1) 150. So, total numbers of onto functions from X to Y are 6 (F3 to F8). I already know the formula (summation r=1 to n)(-1)^(n-r)nCr(r^m). Suppose TNOF be the total number of onto functions feasible from A to B, so our aim is to calculate the integer value TNOF. Why does an ordinary electric fan give comfort in summer even though it cannot cool the air? 4. Since f is one-one Hence every element 1, 2, 3 has either of image 1, 2, 3 and that image is unique Total number of one-one function = 6 Example 46 (Method 2) Find the number Experience. Students can solve NCERT Class 12 Maths Relations and Functions MCQs Pdf with Answers to know their preparation level. (b) f(m;n) = m2 +n2. As E is the set of all subsets of W, number of elements in E is 2xy. 3. Out of these functions, 2 functions are not onto (If all elements are mapped to 1st element of Y or all elements are mapped to 2nd element of Y). Not onto. (c) f(x) = x3. If anyone has any other proof of this, that would work as well. Math Forums. Copyright © 2021 Pathfinder Publishing Pvt Ltd. To keep connected with us please login with your personal information by phone/email and password. Thus, the number of onto functions = 16−2= 14. It means that every element “b” in the codomain B, there is exactly one element “a” in the domain A. such that f(a) = b. This disagreement is confusing, but we're stuck with it. In this case the map is also called a one-to-one correspondence. Find the number of relations from A to B. Why does a tightly closed metal lid of a glass bottle can be opened more easily if it is put in hot water for some time? Let E be the set of all subsets of W. The number of functions from Z to E is: If X has m elements and Y has 2 elements, the number of onto functions will be 2. Maths MCQs for Class 12 Chapter Wise with Answers PDF Download was Prepared Based on Latest Exam Pattern. Formula for finding number of relations is Number of relations = 2 Number of elements of A × Number of elements of B Functions can be classified according to their images and pre-images relationships. Learn All Concepts of Chapter 2 Class 11 Relations and Function - FREE. Solution: Using m = 4 and n = 3, the number of onto functions is: To create a function from A to B, for each element in A you have to choose an element in B. Examples: Let us discuss gate questions based on this: Solution: As W = X x Y is given, number of elements in W is xy. f(a) = b, then f is an on-to function. For example, if n = 3 and m = 2, the partitions of elements a, b, and c of A into 2 blocks are: ab,c; ac,b… (i)When all the elements of A will map to a only, then b is left which do not have any pre-image in A (ii)When all the elements of A will map to b only, then a is left which do not have only pre-image in A Thus in both cases, function is not onto So, total number of onto functions= 2^n-2 Hope it helps☑ #Be Brainly Therefore, total number of functions will be n×n×n.. m times = nm. One more question. They are various types of functions like one to one function, onto function, many to one function, etc. Which must also be bijective, and therefore onto. The number of functions from {0,1}4 (16 elements) to {0, 1} (2 elements) are 216. Into Function : Function f from set A to set B is Into function if at least set B has a element which is not connected with any of the element of set A. Determine whether each of these functions is a bijection from R to R. (a) f(x) = 2x+1. In other words no element of are mapped to by two or more elements of . Since f is one-one Hence every element 1, 2, 3 has either of image 1, 2, 3 and that image is unique Total number of one-one function = 6 Example 46 (Method 2) Find the number of all one-one functions from set A = {1, 2, 3} to itself. One-to-One/Onto Functions . But we want surjective functions. In other words no element of are mapped to by two or more elements of . 2. Number of onto functions from one set to another – In onto function from X to Y, all the elements of Y must be used. A function has many types which define the relationship between two sets in a different pattern. If X has m elements and Y has n elements, the number if onto functions are. (C) 81 So, that leaves 30. Example 46 (Method 1) Find the number of all one-one functions from set A = {1, 2, 3} to itself. Suppose TNOF be the total number of onto functions feasible from A to B, so our aim is to calculate the integer value TNOF. Calculating required value. (B) 64 Please use ide.geeksforgeeks.org, Check - Relation and Function Class 11 - All Concepts. In F1, element 5 of set Y is unused and element 4 is unused in function F2. The number of injections that can be defined from A to B is: Here are the definitions: is one-to-one (injective) if maps every element of to a unique element in . (d) x2 +1 x2 +2. In the above figure, f … So, you can now extend your counting of functions … Writing code in comment? In this article, we are discussing how to find number of functions from one set to another. We say that b is the image of a under f , and a is a preimage of b. October 31, 2007 1 / 7. P.S. There are $$\displaystyle 2^8-2$$ functions with 2 elements in the range for each pair of elements in the codomain. No. 4 = A B Not a function Notation We write f (a) = b when (a;b) 2f where f is a function. 1.1. . of onto function from A to A for which f(1) = 2, is. So the total number of onto functions is m!. Then Total no. . The number of onto functions (surjective functions) from set X = {1, 2, 3, 4} to set Y = {a, b, c} is: The number of functions from Z (set of z elements) to E (set of 2xy elements) is 2xyz. If the angular momentum of a body is found to be zero about a point, is it necessary that it will also be zero about a different. High School Math Elementary Math Algebra Geometry Trigonometry Probability and Statistics Pre-Calculus. Get hold of all the important CS Theory concepts for SDE interviews with the CS Theory Course at a student-friendly price and become industry ready. Let A = {a 1, a 2, a 3} and B = {b 1, b 2} then f : A -> B. These numbers are called Stirling numbers (of the second kind). For example: X = {a, b, c} and Y = {4, 5}. So, there are 32 = 2^5. [5.1] Informally, a function from A to B is a rule which assigns to each element a of A a unique element f(a) of B. Officially, we have Definition. In other words, nothing is left out. A function f from A to B is a subset of A×B such that • for each a ∈ A there is a b ∈ B with (a,b… Set A has 3 elements and set B has 4 elements. If n > m, there is no simple closed formula that describes the number of onto functions. 3. De nition 1 A function or a mapping from A to B, denoted by f : A !B is a f(a) = b, then f is an on-to function. Onto Function A function f: A -> B is called an onto function if the range of f is B. Steps 1. Let f be the function from R … For understanding the basics of functions, you can refer this: Classes (Injective, surjective, Bijective) of Functions. Also, given, N denotes the number of function from S(216 elements) to {0, 1}(2 elements). Discrete Mathematics Grinshpan Partitions: an example How many onto functions from f1;2;3;4;5;6;7;8g to fA;B;C;Dg are there? 19. 38. Solution: As given in the question, S denotes the set of all functions f: {0, 1}4 → {0, 1}. So, number of onto functions is 2m-2. An onto function is also called surjective function. An exhaustive E-learning program for the complete preparation of JEE Main.. Take chapter-wise, subject-wise and Complete syllabus mock tests and get in depth analysis of your test.. (A) 36 34 – 3C1(2)4 + 3C214 = 36. How many onto functions are there from a set with eight elements to a set with 3 elements? Such functions are referred to as injective. (b) f(x) = x2 +1. Comparing cardinalities of sets using functions. (b)-Given that, A = {1 , 2, 3, n} and B = {a, b} If function is subjective then its range must be set B = {a, b} Now number of onto functions = Number of ways 'n' distinct objects can be distributed in two boxes a' and b' in such a way that no box remains empty. In a function from X to Y, every element of X must be mapped to an element of Y. So the total number of onto functions is m!. In other words, if each b ∈ B there exists at least one a ∈ A such that. (d) f(m;n) = jnj. Onto Functions: Consider the function {eq}y = f(x) {/eq} from {eq}A \to B {/eq}, where {eq}A {/eq} is the domain of the function and {eq}B {/eq} is the codomain. (e) f(m;n) = m n. Onto. From the formula for the number of onto functions, find a formula for S(n, k) which is defined in Problem 12 of Section 1.4. Then every function from A to B is effectively a 5-digit binary number. In F1, element 5 of set Y is unused and element 4 is unused in function F2. Tech Companion - A Complete pack to prepare for Engineering admissions, MBBS Companion - For NEET preparation and admission process, QnA - Get answers from students and experts, List of Pharmacy Colleges in India accepting GPAT, Why does a tightly closed metal lid of a glass bottle can be opened more easily if it is put in hot water for some time? Here's another way to look at it: imagine that B is the set {0, 1}. In the example of functions from X = {a, b, c} to Y = {4, 5}, F1 and F2 given in Table 1 are not onto. The onto function from Y to X is F's inverse. By using our site, you Onto Function A function f: A -> B is called an onto function if the range of f is B. We need to count the number of partitions of A into m blocks. (c) f(m;n) = m. Onto. 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Misc 10 (Introduction)Find the number of all onto functions from the set {1, 2, 3, … , n} to itself.Taking set {1, 2, 3}Since f is onto, all elements of {1, 2, 3} have unique pre-image.Total number of one-one function = 3 × 2 × 1 = 6Misc 10Find the number of all onto functio Q1. Transcript. Option 3) 200. No. Attention reader! If X has m elements and Y has 2 elements, the number of onto functions will be 2 m-2. generate link and share the link here. This course will help student to be better prepared and study in the right direction for JEE Main.. there are zero onto function . Any ideas on how it came? set a={a,b,c} and B={m,n} the number of onto functions by your formula is 6 . That is, a function f is onto if for each b ∊ B, there is atleast one element a ∊ A, such that f(a) = b. Number of functions from one set to another: Let X and Y are two sets having m and n elements respectively. Home. My book says it is the coefficient of x^m in m!(e^x-1)^n. Considering all possibilities of mapping elements of X to elements of Y, the set of functions can be represented in Table 1. Therefore, N has 2216 elements. There are 3 ways of choosing each of the 5 elements = $3^5$ functions. Tuesday: Functions as relations, one to one and onto functions What is a function? A function f : A -> B is said to be an onto function if every element in B has a pre-image in A. Number of Onto function - & Number of onto functions - For onto function n(A) n(B) otherwise ; it will always be an inoto function . Free PDF Download of CBSE Maths Multiple Choice Questions for Class 12 with Answers Chapter 1 Relations and Functions. A function from X to Y can be represented in Figure 1. Consider the function x → f(x) = y with the domain A and co-domain B. If n(A)= 3 , n(B)= 5 Find the number of onto function from A to B, For onto function n(A) n(B) otherwise ; it will always be an inoto function. Menu. Example 46 (Method 1) Find the number of all one-one functions from set A = {1, 2, 3} to itself. acknowledge that you have read and understood our, GATE CS Original Papers and Official Keys, ISRO CS Original Papers and Official Keys, ISRO CS Syllabus for Scientist/Engineer Exam, Mathematics | Introduction to Propositional Logic | Set 2, Mathematics | Predicates and Quantifiers | Set 2, Mathematics | Some theorems on Nested Quantifiers, Mathematics | Set Operations (Set theory), Inclusion-Exclusion and its various Applications, Mathematics | Power Set and its Properties, Mathematics | Partial Orders and Lattices, Discrete Mathematics | Representing Relations, Mathematics | Representations of Matrices and Graphs in Relations, Mathematics | Closure of Relations and Equivalence Relations, Number of possible Equivalence Relations on a finite set, Discrete Maths | Generating Functions-Introduction and Prerequisites, Mathematics | Generating Functions – Set 2, Mathematics | Sequence, Series and Summations, Mathematics | Independent Sets, Covering and Matching, Mathematics | Rings, Integral domains and Fields, Mathematics | PnC and Binomial Coefficients, Number of triangles in a plane if no more than two points are collinear, Finding nth term of any Polynomial Sequence, Discrete Mathematics | Types of Recurrence Relations – Set 2, Mathematics | Graph Theory Basics – Set 1, Mathematics | Graph Theory Basics – Set 2, Mathematics | Euler and Hamiltonian Paths, Mathematics | Planar Graphs and Graph Coloring, Mathematics | Graph Isomorphisms and Connectivity, Betweenness Centrality (Centrality Measure), Mathematics | Walks, Trails, Paths, Cycles and Circuits in Graph, Graph measurements: length, distance, diameter, eccentricity, radius, center, Relationship between number of nodes and height of binary tree, Bayes’s Theorem for Conditional Probability, Mathematics | Probability Distributions Set 1 (Uniform Distribution), Mathematics | Probability Distributions Set 2 (Exponential Distribution), Mathematics | Probability Distributions Set 3 (Normal Distribution), Mathematics | Probability Distributions Set 4 (Binomial Distribution), Mathematics | Probability Distributions Set 5 (Poisson Distribution), Mathematics | Hypergeometric Distribution model, Mathematics | Limits, Continuity and Differentiability, Mathematics | Lagrange’s Mean Value Theorem, Mathematics | Problems On Permutations | Set 1, Problem on permutations and combinations | Set 2, Mathematics | Graph theory practice questions, Classes (Injective, surjective, Bijective) of Functions, Difference between Spline, B-Spline and Bezier Curves, Runge-Kutta 2nd order method to solve Differential equations, Write Interview But, if the function is onto, then you cannot have 00000 or 11111. For example, if n = 3 and m = 2, the partitions of elements a, b, and c of A into 2 blocks are: ab,c; ac,b; bc,a. There are 3 functions with 1 element in range. Let W = X x Y. If m < n, the number of onto functions is 0 as it is not possible to use all elements of Y. No element of B is the image of more than one element in A. Some authors use "one-to-one" as a synonym for "injective" rather than "bijective". therefore the total number of functions from A to B is. So, total numbers of onto functions from X to Y are 6 (F3 to F8). An onto function is also called a surjective function. A function f from A to B is called one-to-one (or 1-1) if whenever f (a) = f (b) then a = b. Proving that a given function is one-to-one/onto. , every element of © 2021 Pathfinder Publishing Pvt Ltd. to keep connected with us please login with your information... Tuesday: functions as Relations, one to one and onto functions are from. 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https://in.mathworks.com/help/symbolic/animation-and-modeling-of-automotive-piston.html | # Animation and Model of Automotive Piston
This example shows how to model the motion of an automotive piston by using MATLAB® and Symbolic Math Toolbox™.
Define the motion of an automotive piston and create an animation to model the piston motion.
### Step 1: Describe Piston Model
The following figure shows the model of an automotive piston. The moving parts of the piston consist of a connecting rod (red line), a piston crank (green line), and a piston cylinder head (gray rectangle).
Describe the properties of the piston by defining the parameters:
• the cylinder stroke length $\mathit{S}$
• the piston bore diameter $\mathit{B}$
• the length of the connecting rod $\mathit{L}$
• the crank radius $\mathit{a}$
• the crank angle $\theta$
Define the origin O of the coordinate system at the crankshaft location. Label the nearest distance between the piston head and the crankshaft location as bottom dead center (BDC). The height of BDC is $\mathit{L}-\mathit{a}$. Label the farthest distance between the piston head and the crankshaft location as top dead center (TDC). The height of TDC is $\mathit{L}+\mathit{a}$.
### Step 2: Calculate and Plot Piston Height
The following figure is a schematic of the crank and connecting rod.
The height of the piston relative to the origin is $\mathit{H}=\mathit{a}\text{\hspace{0.17em}}\mathrm{cos}\theta +\sqrt{\text{\hspace{0.17em}}{\mathit{L}}^{2}-{\mathit{a}}^{2}{\mathrm{sin}\left(\theta \right)}^{2}\text{\hspace{0.17em}}}.$ Define the piston height as a symbolic function by using the syms function.
syms pistHeight(L,a,theta)
pistHeight(L,a,theta) = a*cos(theta) + sqrt(L^2-a^2*sin(theta)^2);
Assume that the connecting rod length is $\mathit{L}=150\text{\hspace{0.17em}}\mathrm{mm}$ and the crank radius is $\mathit{a}=50\text{\hspace{0.17em}}\mathrm{mm}$. Plot the piston height as a function of the crank angle for one revolution within the interval [0 2*pi].
fplot(pistHeight(150,50,theta),[0 2*pi])
ylabel('Height (mm)')
The piston head is highest when the piston is at TDC and the crank angle is 0 or 2*pi. The piston head is lowest when the piston is at BDC and the crank angle is pi.
You can also plot the piston height for various values of $\mathit{a}$ and $\theta$. Create a surface plot of the piston height by using the fsurf function. Show the piston height within the interval $30\text{\hspace{0.17em}}\mathrm{mm}<\mathit{a}<60\text{\hspace{0.17em}}\mathrm{mm}$ and $0<\theta \text{\hspace{0.17em}}<2\pi$.
fsurf(pistHeight(150,a,theta),[30 60 0 2*pi])
zlabel('Height (mm)')
### Step 3: Calculate and Plot Volume of Piston Cylinder
The length of the combustion chamber is equal to the difference between the TDC location and the piston height. The volume of the piston cylinder can be expressed as $\mathit{V}=\pi \text{\hspace{0.17em}}{\left(\frac{\mathit{B}}{2}\right)}^{2}\left(\mathit{L}+\mathit{a}-\mathit{H}\right)$.
Define the piston volume as a symbolic function and substitute the expression for $\mathit{H}$ with pistHeight.
syms pistVol(L,a,theta,B)
pistVol(L,a,theta,B) = pi*(B/2)^2*(L+a-pistHeight)
pistVol(L, a, theta, B) =
$\frac{\pi {B}^{2} \left(L+a-a \mathrm{cos}\left(\theta \right)-\sqrt{{L}^{2}-{a}^{2} {\mathrm{sin}\left(\theta \right)}^{2}}\right)}{4}$
Next, define the values for the following parameters:
• the length of the connecting rod $\mathit{L}=150\text{\hspace{0.17em}}\mathrm{mm}$
• the crank radius $\mathit{a}=50\text{\hspace{0.17em}}\mathrm{mm}$
• the bore diameter $\mathit{B}=86\text{\hspace{0.17em}}\mathrm{mm}$
Plot the piston volume as a function of the crank angle for one revolution within the interval [0 2*pi].
fplot(pistVol(150,50,theta,86),[0 2*pi])
ylabel('Volume (mm^3)')
The piston volume is smallest when the piston is at TDC and the crank angle is 0 or 2*pi. The piston volume is largest when the piston is at BDC and the crank angle is pi.
### Step 4: Evaluate Piston Motion for Changing Angular Speed
Assume the crank rotates at 30 rpm for the first 3 seconds, then steadily increases from 30 to 80 rpm for the next 4 seconds, and then remains at 80 rpm.
Define the angular speed as a function of time by using the piecewise function. Multiply the angular speed by $2\pi /60$ to convert the rotational speed from rpm to rad/sec.
syms t0 t
rpmConv = 2*pi/60;
angVel(t0) = piecewise(t0<=3, 30, t0>3 & t0<=7, 30 + 50/4*(t0-3), t0>7, 80)*rpmConv
angVel(t0) =
Calculate the crank angle by integrating the angular speed using the int function. Assume an initial crank angle of $\theta =0$. Compute the integral of the angular speed from 0 to t.
angPos(t) = int(angVel,t0,0,t);
Find the piston height as a function of time by substituting the expression angPos for the crank angle.
H(t) = pistHeight(150,50,angPos)
H(t) =
Plot the piston height as a function of time. Notice that the oscillation of the piston height becomes faster between 3 and 7 seconds.
fplot(H(t),[0 10])
xlabel('Time (sec)')
ylabel('Height (mm)')
### Step 5: Create Animation of Moving Piston
Create an animation of the moving piston given a changing angular speed.
First, create a new figure. Plot the cylinder walls that have fixed locations. Set the x-axis and y-axis to be equal length.
figure;
plot([-43 -43],[50 210],'k','LineWidth',3)
hold on;
plot([43 43],[50 210],'k','LineWidth',3)
plot([-43 43],[210 210],'k','LineWidth',3)
axis equal;
Next, create a stop-motion animation object of the piston head by using the fanimator function. By default, fanimator creates an animation object by generating 10 frames per unit time within the range of t from 0 to 10. Model the piston head as a rectangle with a thickness of 10 mm and variable height H(t). Plot the piston head by using the rectangle function.
fanimator(@rectangle,'Position',[-43 H(t) 86 10],'FaceColor',[0.8 0.8 0.8])
Add the animation objects of the connecting rod and the piston crank. Add a piece of text to count the elapsed time.
fanimator(@(t) plot([0 50*sin(angPos(t))],[H(t) 50*cos(angPos(t))],'r-','LineWidth',3))
fanimator(@(t) plot([0 50*sin(angPos(t))],[0 50*cos(angPos(t))],'g-','LineWidth',3))
fanimator(@(t) text(-25,225,"Timer: "+num2str(t,2)));
hold off;
Use the command playAnimation to play the animation of the moving piston.
## Support
#### Mathematical Modeling with Symbolic Math Toolbox
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http://mathhelpforum.com/algebra/155968-charles-lutwidge-dodgson.html | # Math Help - Charles Lutwidge Dodgson
1. ## Charles Lutwidge Dodgson
Here is a problem to try and solve:
Jack and Jill walked along a level road, up the hill, back (along the same path) down te hill, and back along the same level road to home. They started out at 3pm and arrived home at 9pm. Their speed was four miles an hour on the level, three miles an hour uphill, and six miles an hour down hill.
a.) How far did they walk in all (level, up, down, level)?
b.) You can't figure out from the given information exactly when they reached the top of the hill. How closely can you approximate when they arrived there? (e.g. can you give an interval of an hour containing the time they arrived at the top? An interval of a half hour?)
2. x = hours walked level; so 6-x = hours up and down
Miles walked
= 4x + 3[2(6-x)/3] + 6(6-x)/3
= 4x + 12 - 2x + 12 - 2x
= 24
3. Hello, matgrl!
Here's another approach to part (a).
Jack and Jill walked along a level road and up a hill.
Then walked back along the same path:
. . down the hill and along the same level road to home.
They started out at 3 pm and arrived home at 9 pm.
Their speed was 4 mph on the level, 3 mph uphill, and 6 mph downhill.
a.) How far did they walk in all (level, up, down, level)?
Code:
*
*
* y
*
* * * * * *
x
They walked $\,x$ miles (level) at 4 mph.
. . This took: . $\dfrac{x}{4}$ hours.
Then they walked $\,y$ miles (uphill) at 3 mph.
. . This took: . $\dfrac{y}{3}$ hours.
They walked $\,y$ miles (downhill) at 6 mph.
. . This took: . $\dfrac{y}{6}$ hours.
Then they walked $\,x$ miles (level) at 4 mph.
. . This took: . $\dfrac{x}{4}$ hours.
The entire trip took 6 hours: . $\displaystyle \frac{x}{4} + \frac{y}{3} + \frac{y}{6} + \frac{x}{4} \:=\:6$
Multiply by 12: . $3x + 4y + 2y + 3x \:=\:72$
. . $6x+6y \:=\:72 \quad\Rightarrow\quad x + y \:=\:12$
They walked 12 miles one way.
The entire walk was 24 miles.
4. Hi
Can someone please explain why this is the case?
What I don't understand is that the above equations use (include) the times/speeds for descent (return journey) so why is the result one way?
I appologide in advance for being so dense
5. First do you see that Soroban carefully stated what his variables meant: "x is miles walked on the level", "y is miles walked down hill". He then has an equation that has four parts- timed walked on the level toward the mountain, time walked uphill, time walked down hill, and time walked on the level away from the mountain, using x in both "level" walks, and y in both uphill and downhill walks. The whole route is "level toward mountain" (x), "uphill" (y), "downhill" (y again), and "level away from mountain" (x again). The whole route is x+ y+ x+ y= 2x+ 2y= 2(x+ y) so that x+ y is the first "level toward mountain" plus "uphill" (which is the same as "downhill" plus "level away from mountain". We know that x+ y is one way rather that roundtrip because we know what "x" and "y" separately meant.
(Yes, times/speeds for descent are included in the equation but the distance (y) is the same both ways.)
As I recall, in the Reverand Dodgson's (AKA Lewis Carroll) original puzzle, it was a knight and squire. Why the change to "Jack" and "Jill"?
6. Wow this is wonderful, thank you so much for exaplaining! I have one question...how did you know to multiply by 12?
7. Originally Posted by matgrl
I have one question...how did you know to multiply by 12?
That was simply to get rid of fractions:
we had x/4 + y/3 + y/6 + x/4 = 6
notice that 12 is lowest number in which 3,4 and 6 divide evenly;
multiply each term by 12 (like, x/4 times 12 = 3x):
3x + 4y + 2y + 3x = 72
Got it?
8. How would you do part b of this question? I know the answer but don't understand how to do it. I have if we assume either all flat or uphill that it could be 3 hour or 4 hours. Which would end up bieing 6/7 pm. How can we do this? Please help... | 2015-03-04T11:28:54 | {
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https://brilliant.org/discussions/thread/magnetic-dollars/ | ×
# Magnetic Dollars
Imgur
Suppose that I have two urns, with one magnetic dollar in each urn. I begin randomly throwing more magnetic dollars in the general vicinity of the urns, and the coins fall in the urns by a simple rule:
Let's say that Urn A has $$x$$ coins, and Urn B has $$y$$ coins. The probability that the next coin falls into Urn A is $$\dfrac{x}{x+y}$$, and the probability that the next coin falls into Urn B is $$\dfrac{y}{x+y}$$.
You keep throwing magnetic dollars until there is a total of $$1,000,000$$ magnetic dollars in total.
What would you bet would be the price on average, of the urn with the smaller amount of coins?
$$\ 10$$, perhaps? Maybe $$\ 100$$ or $$\ 500$$? Post your bet or write it down on a piece of paper before looking at the next section.
The less-priced urn, on average, is actually worth a grand total of a quarter of a million dollars! Don't worry if you guessed wrong; many professional mathematicians also guessed much lower than this. In fact, when a group of mathematicians were asked this question and were asked to bet, most people only bet $$\ 10$$ and only one person bet over $$\ 100$$.
But why does the lower-priced urn price so high? You may want to try the problem out yourself before I go over a very nice and elegant solution. See if you can find it!
Tried it out yet? In the case that you have, let's see how this problem can be so elegantly solved, as I claimed.
Suppose that you have a deck of cards; one red, and $$999,998$$ white. Currently, you just have a red card. Now every turn, you place a white card in any available slot. For example, in the first move, you have $$2$$ available slots: one above the red card, and one below. In the second move, you have $$3$$ available slots, and so on.
But wait! Let's say that the empty slots above the red card are the magnetic dollars in Urn A, and the empty slots below the red card are the magnetic dollars in Urn B. Notice that if you had $$x$$ empty slots above the red card and $$y$$ empty slots below the red card, then the probability that the next card will be above the red card is $$\dfrac{x}{x+y}$$, and the probability that the next card will be below the red card is $$\dfrac{y}{x+y}$$! We've found a one-to-one correspondence between the original magnetic dollar problem and this new card problem!
Finally, we know that in a random placements of white cards in this fashion will result in a uniform distribution of where the red card is, every single final position is of equal probability. This means that in the original problem, the probability of $$42$$ magnetic dollars being in Urn A is the same as the probability of $$314,159$$ magnetic dollars in Urn A. Therefore, the average price of the lower priced urn is clearly $$250,000$$. $$\Box$$
Note by Daniel Liu
3 years, 8 months ago
Sort by:
The result is certain not intuitive, nor obvious.
You can look at this problem John's Red And Blue Balls, which is based off the same idea. Staff · 3 years, 8 months ago
I know some of you have been waiting for me to post for a long time. Sorry for the inactivity, I had a bunch of homework and stuff.
Well, here is my next #CosinesGroup posts. The problem in my opinion is a very cool problem because of the awesome elegant solution.
This problem has also been called "Polya's Urns", but I looked that up and "Polya's Urns" actually covers a lot of random complicated stuff, so I kept it at magnetic dollars.
Hope you enjoy! Feedback is appreciated. · 3 years, 8 months ago
This is the solution in Peter Winkler ' Mathematical Puzzles. An excellent book. · 1 year, 2 months ago
How do you do this? · 3 years, 7 months ago
I'm curious to know why only one mathematician guessed over $100. I personally guessed around$150,000 becuase while the probability might be pretty low at some point, in A MILLION tries, it would be bound to happen quite a bit. · 3 years, 8 months ago
Well, their reasoning would probably be then assuming a sort of "reverse" normal distribution curve. The average of all the games would probably be pretty low.
They were asked to bet immediately, depriving them of the chance to think it over. Their first instinct told them that the more coins that get in a urn, the higher probability of getting more coins in, giving a feedback loop. This means that the smaller urn probably won't really have any coins at all. · 3 years, 8 months ago | 2017-09-21T08:50:05 | {
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https://math.stackexchange.com/questions/3059556/probability-rolling-six-six-sided-dice-a-single-time | # Probability: Rolling six, six-sided dice a single time
My specific question (to settle a score-keeping argument in my family's never-ending Farkle tournament) is this:
• when rolling six, six-sided dice a single time, is it more likely to roll three pairs of three different numbers, or three dice of a single number?
For instance, in my family's game of Farkle, if you roll three pairs you get 200 points. However, if you get three of any number in particular you get that specific digit times x 100, except if you get three one's, then you get 1,000, or three fives, then you get 5,000.
• As the stickler that I am, I just want to make sure we're scoring relative to correctly weighted probability of rolling outcomes.
Feel free to also include other six, six-sided dice probability answers to help out other folks who stumble on this thread in search of help.
Thank you!
• Not sure this is clear. When you say three dice of a single color would you also include all six of that number? Would you include something like $(1,1,1,2,2,5)$? – lulu Jan 2 '19 at 15:04
• You ask about "scoring relative to correctly weighted probability" but you already have six events with the same probability (rolling three of any one of the six numbers) with scores ranging from $200$ to $5000,$ so what exactly is the point you want to be a stickler about? (Idle curiosity is still a very good reason to ask which is more probable, so the question is fine, I just wonder if we're missing something.) – David K Jan 2 '19 at 15:08
• - I believe i said three dice of a single number, just to be clear. So yes, if rolling six dice altogether a single time, what are the odds that three of those dice land on the same number. So yes, a 1-1-1-2-2-5 roll would qualify. – user631109 Jan 2 '19 at 15:15
• Usually in betting or scoring situations, the payout scenarios are carefully isolated. In Poker, say, "three of a kind" usually precludes "four of a kind" or "full house". Anyway, as my post below indicates, even with the restrictive definition, it is far more likely to get three matching values than it is to get three pairs. – lulu Jan 2 '19 at 15:17
• The argument began on New Year's Eve so my memory of it is a little hazy, but I believe I argued that it was more likely to roll three ones than three pairs, even though three pairs is two-hundred points and three ones is a thousand points. I just thought that the scoring was a little bogus. – user631109 Jan 2 '19 at 15:18
Let's just compute. Of course there are $$6^6$$ unrestricted ways to throw the six dice.
We pick the three paired values,$$\binom 63=20$$. We then place the two appearances of the least value, $$\binom 62= 15$$ and then place the two appearances of the middle value, $$\binom 42=6$$. Thus there are $$20\times 15\times 6=1800$$ ways to throw three pair. So the probability is $$\frac {20\times 15\times 6}{6^6}=.03858$$
Now there are a number of patterns to consider. But even if we consider the strongest constraint, namely patterns of the form $$AAABCD$$ or permutations of that then there are far more than $$1800$$ cases. For that pattern there are $$6\times \binom 63\times 5\times 4\times 3=7200$$ ways to do it. So already you are $$4$$ times more likely to get this than to get a permutation of $$AABBCC$$
• numerically, the probability of any three die being equal (including quads, trip+pair etc) is around $0.3675$, compared to that $0.03858$ for three pairs. It's almost 10 times more likely – John Doe Jan 2 '19 at 15:16 | 2020-02-21T07:32:11 | {
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https://math.stackexchange.com/questions/493538/can-an-irrational-always-be-found-by-multiplying-irrationals | # Can an irrational always be found by multiplying irrationals?
I was thinking about the function $\ f(a,b) = a/b$ where $a$ and $b$ where both irrational. It quickly stood out to me that the codomain of that function would include every rational number. But, does it include every irrational number as well (in other words, is the codomain of the function $\mathbb{R}$)?
Then I thought that if we establish that $a = n \cdot m$ and $b = m$, then if for every irrational number $n$ there exists at least another irrational number $m$ (which could be itself) such that $n \cdot m$ is also irrational, every irrational could be represented by $a/b$ (as $m$ cancels out), and so the function would eventually "spit out" all the reals given only irrational input.
So my questions are: For every irrational $n$, does there always exist another irrational $m$ such that $n \cdot m$ is also irrational? If this is true as I suspect, what is the simplest proof for it?
• Hi, welcome to mathSE. Nice question. I've changed * to \cdot for multiplication to improve readability. – 6005 Sep 14 '13 at 16:45
Yes.
If $a$ is an algebraic number then $a\cdot\pi$ is irrational.
If $a$ is transcendental then $a\cdot\sqrt2$ is irrational.
• So, I'm guessing that algebraic*transcendental is always irrational. Is this true? – Pabce Sep 15 '13 at 0:28
• Yes, this is true. – Asaf Karagila Sep 15 '13 at 4:21
• The argument is the same as rational and irrational, because algebraic numbers form a field. – Asaf Karagila Sep 15 '13 at 4:29
Given an irrational $a$, if $\sqrt 2 a$ is irrational, we are done; and if $\sqrt 2 a$ is rational, then $\sqrt 3 a$ is irrational.
First, I want to point out that when you say that $f$ contains every rational number in its image, you are forgetting about zero! Of course $a/b$ cannot equal zero for $a$ and $b$ irrational because that would imply $a = 0$.
To answer your question (elaborating along the lines of TonyK's answer), let $x$ be any irrational number. $\sqrt{2},\sqrt{3},$ and $\sqrt{6}$ are all irrational. Moreover, $\sqrt{2}x$ and $\sqrt{6}x$ cannot both be rational or else $\frac{\sqrt{6}x}{\sqrt{2}x} = \sqrt{3}$ would be irrational. So one of the two products $x \cdot \sqrt{2} = x\sqrt{2}$ and $x \cdot \sqrt{6} = x\sqrt{6}$ works.
A non-computational answer: let $n$ be irrational (therefore not zero) and multiply by all irrationals $m$. This gives you uncountably many different numbers: they can't all be rational. | 2019-12-11T08:59:40 | {
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https://math.stackexchange.com/questions/2578877/special-case-in-ratio-test-for-series | # Special Case in Ratio Test for Series
We all know the Ratio Test in series which is one of many tests that are used to determine whether a Series is Converges or Not (Most Of The Time Work). Here is a link that describe Ratio test "In Short". Ratio Test
I have example : Suppose we have $U_n$=$2^{-n+({-1})^{n+1}}$ and we want to know whether the series $$\sum_{n=1}^\infty U_n$$ Is converges Or Not. $\\$
If we Use the Ratio Test then $\frac{U_{n+1}}{U_n}$=$2^{2({-1})^{n}-1}$, and we can Rewrite it as$$\frac{U_{n+1}}{U_n} = \begin{cases} 2, & \text{if n is even} \\ \frac{1}{8}, & \text{if n is odd} \end{cases}$$ And I think if the Proudct of the two number $2$ and $\frac{1}{8}$ is exactly less than $1$ then the series is Converges . in general Form if the fraction $\frac{U_{n+1}}{U_n}$ Can be written of the Form of $$\frac{U_{n+1}}{U_n} = \begin{cases} A_n, & \text{if n is even} \\ B_n, & \text{if n is odd} \end{cases}$$ and we have $$\lim_{n\to \infty}{A_n}= a \text{ and} \lim_{n\to \infty}{B_n}= b$$ then if $0 < a.b < 1$then the series $\sum U_n$ Is Converges. What do you think in this idea ?
for Now it is just true when $U_n >0$, I did not study it when $U_n <0$
So Give me your Opinion whether my idea is True Or False
• I know that we can use another Test to prove the convergence of the first series .. but i want it in this way .. with thanks – Yaser Tarek Dec 24 '17 at 16:09
• Hint: consider the even and odd subseries... – gniourf_gniourf Dec 24 '17 at 16:10
• yes i have dont that before .. – Yaser Tarek Dec 24 '17 at 16:12
• And what's your conclusion? – gniourf_gniourf Dec 24 '17 at 16:13
• it turns out that my opinion is True .. Am I Right ? – Yaser Tarek Dec 24 '17 at 16:15
Note that
$$\sum_{n=1}^{2N} U_n=\sum_{n=1}^N U_{2n}+\sum_{n=1}^N U_{2n-1}\tag1$$
We assume that $U_n\ge 0$,
If the limits $\lim_{n\to \infty}\frac{U_{2n+2}}{U_{2n+1}}$ and $\lim_{n\to \infty}\frac{U_{2n+1}}{U_{2n}}$ exist and are finite, then the ratio test guarantees that the first sequence of partial sums on the right-hand side of $(1)$ converges when
\begin{align} \lim_{n\to\infty}\frac{U_{2(n+1)}}{U_{2n}}&=\lim_{n\to \infty}\left(\frac{U_{2n+2}}{U_{2n+1}}\frac{U_{2n+1}}{U_{2n}}\right)\\\\ &=\left(\lim_{n\to \infty}\frac{U_{2n+2}}{U_{2n+1}}\right)\,\left(\lim_{n\to \infty}\frac{U_{2n+1}}{U_{2n}}\right) \\\\&<1\tag 2 \end{align}
Similary, if the limits $\lim_{n\to \infty}\frac{U_{2n}}{U_{2n-1}}$ and $\lim_{n\to \infty}\frac{U_{2n+1}}{U_{2n}}$ exist and are finite, then the ratio test guarantees that the second sequence of partial sums on the right-hand side of $(1)$ converges when
\begin{align} \lim_{n\to\infty}\frac{U_{2(n+1)-1}}{U_{2n-1}}&=\lim_{n\to \infty}\left(\frac{U_{2n+1}}{U_{2n}}\frac{U_{2n}}{U_{2n-1}}\right)\\\\ &=\left(\lim_{n\to \infty}\frac{U_{2n+1}}{U_{2n}}\right)\,\left(\lim_{n\to \infty}\frac{U_{2n}}{U_{2n-1}}\right) \\\\&<1\tag3 \end{align}
Hence, if $(2)$ and $(3)$ hold, then the sequence of partial sums on the left-hand side of $(1)$ converges.
We conclude that if the product of the limits $\lim_{n\to\infty}\frac{U_{n+1}}{U_n}$ for $n$ even and $\lim_{n\to\infty}\frac{U_{n+1}}{U_n}$ for $n$ odd is less than $1$, then the series $\sum_{n=1}^\infty U_n$ converges.
• Happy New Year!. And when you have enough reputation points, feel free to up vote as you see fit. ;-)) – Mark Viola Jan 11 '18 at 19:39 | 2019-07-20T16:12:23 | {
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https://math.stackexchange.com/questions/3723084/given-a-binary-tree-with-n-labelled-leaves-is-it-possible-to-find-its-unique-nu | # Given a binary tree with N labelled leaves, is it possible to find its unique number in the Catalan range?
The question is about finding the inverse to the problem of generating the $$n^{th}$$ binary tree with N labelled leaves (Generating the $$n^{th}$$ full binary tree over $$N$$ labelled leaves).
Let's say if $$N = 4$$, the possible set of trees are
1: (((1, 2), 3), 4)
2: (1, ((2, 3), 4))
3: ((1, (2, 3)), 4)
4: (1, (2, (3, 4)))
5: ((1, 2), (3, 4))
If I choose a specific tree from this set, let's say $$(1, ((2, 3), 4))$$, is there an algorithm that gives me back the value 2? The Catalan range for the problem is 1 to 5 and the unique number corresponding to the given tree is 2.
What do I mean by Catalan Range?
If there are N leaf nodes, the maximum possible binary trees is $$C(N-1)$$. For a given $$n$$, its $$C(n)$$ is the $$n^{th}$$ Catalan number. We can uniquely identify all the individual binary trees if we assign them a number from $$1$$ to $$C(N-1)$$ in order. I'm referring to this range of numbers from $$1$$ to $$C(N-1)$$ as the Catalan Range.
What scheme am I using to order the trees from $$1$$ to $$C(N-1)$$?
I don't really mind the scheme used to order the trees as long as all the trees can be uniquely identified within that scheme. For example,
$$1$$: The tree having just one node in the left sub tree and $$N-1$$ nodes in the right sub tree.
$$2$$: The tree still having just one node in the left sub tree and with a slightly different right sub tree now.
.
.
$$C(N-1)$$: The tree having $$N-1$$ nodes in the left sub tree and one node in the right sub tree.
• Could you give a reference to what you call the "Catalan space" ? Jun 17 '20 at 7:16
• @JeanMarie, I have explained it in the description, I'll give it another shot here. If there are N nodes, the maximum possible binary trees is C(N). C(N) is the Nth Catalan number. We can uniquely identify all the individual binary trees if we assign them a number from 1 to C(N) in order. I'm referring to this range of numbers from 1 to C(N) as the "Catalan Space". Jun 17 '20 at 7:24
• Which order are you using to get 1, 2, 3, 4, 5? Jun 17 '20 at 7:24
• @J.-E.Pin Yes, that's a good question. I don't really mind the scheme used to order the trees as long as they can be uniquely identified within that scheme. For example, I could start by assigning 1 to the tree having just one node in the left sub tree and N-1 nodes in the right sub tree. 2 can be assigned to the tree with two nodes in the left sub tree and N-2 nodes in the right and so on... Jun 17 '20 at 7:33
• Thanks for the explanation. I have been misled by the word "space" which tends to refer to an underlying structure (algebraic, geometric, topological...). IMHO, "Catalan range" would be a more adapted word. Jun 17 '20 at 9:51
Let $$f$$ be the function mapping full binary trees to integers; I'll use the convention that binary trees with $$n$$ leaves will map to the range $$\{0, 1, \dots, C_{n-1}-1\}$$ because that's easier to work with in the recursion. You can add $$1$$ later.
If we have a binary tree $$T$$, let $$L$$ be the "left" subtree: the subtree with leaves $$1, 2, \dots, k$$ for some $$k$$. Let $$R$$ be the "right" subtree: the subtree with leaves $$k+1, k+2, \dots, n$$. We will find $$f(T)$$ in terms of $$f(L)$$, $$f(R)$$, and $$k$$ where for the purposes of finding $$f(R)$$ we relabel $$R$$ to have leave $$1, 2, \dots, n-k$$.
Our trees are labeled in increasing order of $$k$$. So before this tree, we have $$C_0 C_{n-2} + C_1 C_{n-3} + \dots + C_{k-2} C_{n-k}$$ trees whose left subtree has $$1, 2, \dots, k-1$$ leaves respectively.
Next, before this particular left subtree $$L$$, there are $$f(L)$$ previous left subtrees on $$k$$ leaves; for each one of them, there are $$C_{n-k-1}$$ right subtrees. All $$f(L) C_{n-k-1}$$ of the combined $$n$$-leaf trees go before $$T$$.
Finally, there are $$f(R)$$ trees with the same left subtree, but with a right subtree preceding $$R$$; these also go before $$T$$.
Altogether, we get the recursion $$f(T) = \sum_{i=1}^{k-1} C_{i-1} C_{n-i-1} + f(L) C_{n-k-1} + f(R).$$ The base of the recursion sets $$f(T) = 0$$ when $$T$$ has just one or two leaves, in which case there's only one possible tree. (Actually, we only need the one-leaf case as our base case.)
• For this technique to work, we need to know the number of leaf nodes in the left subtree and right subtree upfront? Jun 17 '20 at 18:18
• Yes, because that's the primary thing determining the order in which we're listing the subtrees - at least if use the enumeration method in the answer you cited. All the trees with one leaf in $L$ go before all the trees with two leaves in $L$, which go before all the trees with three leaves in $L$, and so on. Jun 17 '20 at 18:52
• Thanks a lot! I implemented it in a Python program and it works perfectly! :) Jul 2 '20 at 12:22 | 2021-09-16T19:41:56 | {
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https://math.stackexchange.com/questions/3588190/solving-a-differential-system-of-equations-in-matrix-form | Solving a differential system of equations in matrix form
I have a basic question about eigenvector.
If I have the following system:
$$\begin{pmatrix} \dot{\eta}_1 \\ \dot{\eta}_2 \\ \end{pmatrix} = {\cal{B}} \begin{pmatrix} \eta_1 \\ \eta_2 \\ \end{pmatrix}$$
Where $${\cal{B}}$$ is a matrix with constant parameters. For finding the solution to this system I do $$\eta = P f$$ where P is the eigenvector of $${\cal{B}}$$, in this case, I can write:
$$\begin{pmatrix} \dot{f}_1 \\ \dot{f}_2 \\ \end{pmatrix} = {\cal{D}} \begin{pmatrix} f_1 \\ f_2 \\ \end{pmatrix}$$
Where $${\cal{D}}$$ is a diagonal matrix. I have two eigenvalue ($$\lambda_1$$ and $$\lambda_2$$) and two corresponding eigenvectors ($$v_1$$ and $$v_2$$), the vector $$v_1$$ have components $$v_{11}$$ and $$v_{12}$$. The solution of this equation is:
$$f_1 = c_0 e^{\lambda_1 t}$$ and $$f_1 = c_0 e^{\lambda_2 t}$$,
If for example I select the first eigenvalue $$\lambda_1$$, then the $$\eta_1$$ and $$\eta_2$$ should be written like this
$$\eta_1 = c_0 v_{11} e^{\lambda_1 t}$$
$$\eta_1 = c_0 v_{12} e^{\lambda_1 t}$$
Is this correct?
And the general solution should be
$$\eta_1 = c_0 v_{11} e^{\lambda_1 t} +c_{01} v_{21} e^{\lambda_2 t}$$
$$\eta_1 = c_0 v_{12} e^{\lambda_1 t} +c_{02} v_{22} e^{\lambda_2 t}$$
Thanks
• That's right, unless the two eigenvalues are equal : in such a case, you have to involve $te^{\lambda_1t}$ terms. Mar 20, 2020 at 17:07
• I have taken the liberty to add "differential" to your title. Mar 20, 2020 at 17:08
Consider the system $$\dot{x} = Ax$$, where $$A\in\mathbb{R}^{n\times n}$$. If you assume $$A$$ to be diagonalizable (i.e. the geometric multiplicity equals the algebraic multiplicity for every eigenvalue of $$A$$), then you can find an invertible transformation $$T\in\mathbb{R}^{n\times n}$$ in terms of eigenvectors of $$A$$ such that $$A = T \Lambda T^{-1}$$, where $$\Lambda = \text{diag}(\lambda_1,\lambda_2,\dots,\lambda_n)$$. Setting $$z = T^{-1}x$$, the differential equation becomes $$\begin{equation*} \dot{z} = \Lambda z. \end{equation*}$$ Now, as you found, the decoupled system in the $$z$$-coordinates makes solving the differential equation very simple, namely: $$\begin{equation*} z_i(t) = e^{\lambda_i t}z_i(0) ~ \text{for all i\in\{1,2,\dots,n\}}, \end{equation*}$$ or, equivalently, $$\begin{equation*} z(t) = \text{diag}(e^{\lambda_1 t},e^{\lambda_2 t},\dots,e^{\lambda_n t}) z(0). \end{equation*}$$ To find the solution in the original $$x$$-coordinates, simply reverse the transformation: $$\begin{equation*} x(t) = Tz(t) = T\text{diag}(e^{\lambda_1 t},e^{\lambda_2 t},\dots,e^{\lambda_n t}) T^{-1}x(0). \end{equation*}$$
• Thanks:D, I have only one doubt. In the last equation, T is a matrix with of eigenvector of $\lambda$ written in columns, right?. In this case, for example, $x_1$, it should be equal to the first component of each eigenvector multiplied by the corresponding z_i (t), right?
• If $A$ has $n$ linearly independent eigenvectors, then yes, $T$ is generated by stacking these eigenvectors into columns. As for your second question, not necessarily. In general, the components of $x$ depend on every component of $z$: $x(t) = Tz(t) = \begin{bmatrix}t_1 & t_2 & \cdots & t_n\end{bmatrix}\begin{bmatrix}z_1(t) \\ z_2(t) \\ \vdots \\ z_n(t) \end{bmatrix} = \sum_{i=1}^n t_i z_i(t)$, where $t_i$ is the $i$th eigenvector of $A$. Mar 20, 2020 at 18:04
• Think of it like this: in the original $x$-coordinates, the components of the ODE trajectory may be correlated in some sense to one another. This is due to the coupling between different coordinates of $x$ through the structure of $A$. When we diagonalize $A$, we are changing coordinates to a system where we remove this coupling. This allows us to solve the ODE separately in each coordinate, since they are independent from one another. However, in the $x$-coordinates, the coupling still exists, and therefore when we transform back to the $x$-coordinates we end up fusing different parts of $z$. Mar 20, 2020 at 18:09
• This can also be written very succinctly in terms of the matrix exponential as $x(t) = \exp(At)x(0)$. Unfortunately there is not such a simple form if $A$ is nonconstant, even if it has an explicit antiderivative. Mar 20, 2020 at 18:10 | 2022-08-17T14:54:50 | {
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https://gdbooks.gitbooks.io/3dcollisions/content/Chapter3/raycast_plane.html | # Ray Against Plane
Testing a ray against a plane is surprisingly easy. Remember, any point on the ray can be represented by a time t, such that:
point(t) = plane.position + plane.normal * t
^ the (t) above means point at time t
Also, recall we can test if a point is on a plane if it fits this equation:
DOT(point, plane.normal) = plane.distance
A ray and a plane collide if a point exists that satisfies both of these equations! Let's start out with the formula for point on plane:
DOT(point, plane.normal) = plane.distance
In this formula the point is unknown. But we know that the point must also be on the ray at some time t. So, let's substitute the unknown point for the point on ray formula.
DOT(ray.position + ray.normal * t, plane.normal) = plane.distance
At this point, the only unknown in the above equation is t. Let's start re-arranging the equation so t is on the left side by its-self. We can distribute the dot product operation:
DOT(ray.position, plane.normal) + DOT(ray.normal * t, plane.normal) = plane.distance
Now, we can subtract the first dot product from each side.
DOT(ray.normal * t, plane.normal) = plane.distance - DOT(ray.position, plane.normal);
Finally, we can divide each side by ray.normal DOT plane.normal.
t = (plane.distance - DOT(ray.position, plane.normal) / DOT(ray.normal, plane.normal)
That's it! We now have T! The above might be a bit confusing, let's take a look at it as just math. In the below formula pos is the position of the ray, dir is the normal of the ray. norm is the normal of the plane and dist is the distance of the plane. t is of course t. tdir means t * dir.
• First we have the formula for a plane
• Next, we substitute the point on ray formula for the unknown point
• Distribute the dot product
• Subtract pos dot norm from both sides
• Divide both sides by dir dot norm
A few things to keep in mind.
If the ray is parallel to the plane will be 0 and there is no intersection.
An intersection is only valid if the ray is in front of the plane.That is, if the direction of the ray is opposite of the planes normal. This is true if .
If the value of t is out of range (that is, if t is negative), then there is no intersection.
## The Algorithm
// Returns t if collision happened, -1 if it didnt
float Raycast(Ray ray, Plane plane) {
float nd = Vector3.Dot(ray.Normal, plane.Normal);
float pn = Vector3.Dot(ray.Position, plane.Normal);
if (nd >= 0f) {
return -1;
}
t = (plane.Distance - pn) / nd;
if (t >= 0f) {
return t;
}
return -1;
}
Add the following function to the Collisions class:
// TODO: Implement ONLY THIS ONE method:
public static bool Raycast(Ray ray, Plane plane, out float t)
// I've implemented the blow methods for you.
// Nothing to do past this point
// Conveniance method, returns t without an out param
// If no collision happened, will return -1
public static float Raycast(Ray ray, Plane plane) {
float t = -1;
if (!Raycast(ray, plane, out t)) {
return -1;
}
return t;
}
// Conveniance method, returns the point of intersection
public static bool Raycast(Ray ray, Plane plane, out Point p) {
float t = -1;
bool result = Raycast(ray, plane, out t);
p = new Point(ray.Position.ToVector() + ray.Normal * t);
return result;
}
And provide an implementation for it!
### Unit Test
You can Download the samples for this chapter to see if your result looks like the unit test.
This unit test will render a plane on screen, and several rays. If any rays intersect the Plane they will render red, otherwise blue. The point of intersection is renderd as a green dot
The constructor of the unit test will spit out errors if the test results are not what are expected.
using OpenTK.Graphics.OpenGL;
using Math_Implementation;
using CollisionDetectionSelector.Primitives;
namespace CollisionDetectionSelector.Samples {
class RaycastPlane : Application {
public Plane test = new Plane(new Vector3(1f, 1f, 0f), 1f);
public Ray[] rays = new Ray[] {
new Ray(new Point(0f, 0f, 0f), new Vector3(0f, -1f, 0f)),
new Ray(new Point(0.5f, 0.5f, 0f), new Vector3(-1f, -1f, 0f)),
new Ray(new Point(1f, 1f, 0f), new Vector3(1f, 1f, 0f)),
new Ray(new Point(1f, 1f, -3f), new Vector3(0f, 0f, 1f)),
new Ray(new Point(2f, 2f, 3f), new Vector3(0f, -1f, 0f)),
new Ray(new Point(3f, 3f, 3f), new Vector3(-3f, -3f, -3f)),
new Ray(new Point(1f, 1f, 3f), new Vector3(-2f, -3f, 1f)),
};
public override void Intialize(int width, int height) {
GL.Enable(EnableCap.DepthTest);
GL.PointSize(5f);
GL.PolygonMode(MaterialFace.FrontAndBack, PolygonMode.Fill);
bool[] results = new bool[] {
false, false, false, false, true, true, true
};
float t;
for (int i = 0; i < results.Length; ++i) {
if (Collisions.Raycast(rays[i], test, out t) != results[i]) {
LogError("Expected ray at index: " + i + " to " +
(results[i] ? "intersect" : "not intersect") +
" the plane");
}
}
}
public override void Render() {
base.Render();
DrawOrigin();
GL.Color3(1f, 1f, 1f);
test.Render(5f);
float t;
foreach (Ray ray in rays) {
if (Collisions.Raycast(ray, test, out t)) {
Point colPoint = new Point();
Collisions.Raycast(ray, test, out colPoint);
GL.Color3(0f, 1f, 0f);
colPoint.Render();
GL.Color3(1f, 0f, 0f);
}
else {
GL.Color3(0f, 0f, 1f);
}
ray.Render();
}
}
private void Log(string s) {
System.Console.WriteLine(s);
}
}
} | 2022-10-06T19:44:38 | {
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http://math.stackexchange.com/questions/675545/ways-to-select-three-man-teams | # Ways to select three-man teams
In a competition there are 18 competitors. Answer the following:
A) During the first day they're competing in three-man teams (total of 6 teams). How many ways are there to select the teams?
B) If the main sponsor of the event demands that the four best ranked players mustn't play in the same groups, how many ways can you then select the teams?
I've tried the following:
A) There are (18*17*16 + 15*14*13 + 12*11*10 +...+ 3*2*1) ways to select the teams, and if the order of the groups mattered we would just multiply that wit 6!
However,this gives me approximately 700,000 ways, whereas the correct number of ways according to the book is ~190,000,000
B) The four best players must each be in a different group. Therefore we "lock" their positions. (1*14*13 + 1*12*11 + 1*10*9 + 1*8*7 + 6*5*4 + 3*2*1) gives us the ways to select the teams, and if the order of the groups matters, we multiply that with 6!
Once again, this value is far too small according the the book. Either my calculations are way off or the book is wrong.
-
Homework? What have you tried? – Tim Seguine Feb 13 '14 at 21:04
@TimSeguine Nope, not homework. My teacher and I couldn't get our heads around this today and I was simply curious to see how others would approach the problem. – Mylleranton Feb 13 '14 at 21:06
You are leaving out the fact that the order of the players inside of each team doesn't matter. A team with Bob, Alice and Jane is the same as a team with Jane, Bob and Alice. – Tim Seguine Feb 13 '14 at 21:13
So I would take 18 over 3 plus 15 over 3....until 3 over 3? That would give me an even less value. – Mylleranton Feb 13 '14 at 21:22
You are "zigging" where you should be "zagging", see my answer. – Tim Seguine Feb 13 '14 at 21:32
I will use the binomial coefficient notation. $\binom{n}{k}=\frac{n!}{k!(n-k)!}$ is read "n choose k", the number of ways to choose $k$ items without replacement from a pool of $n$ distinct objects.
A) There are $\binom{18}{3}$ ways to choose the first team, $\binom{15}{3}$ ways to choose the second, etc. this gives $$\binom{18}{3}\binom{15}{3}\binom{12}{3}\binom{9}{3}\binom{6}{3}\binom{3}{3}$$ ways to choose 6 teams. But the order of the teams does not matter, so we have to divide by the number of permutations of the teams ($6!$). This gives an end result of 190,590,400, which appears to be the answer your book is expecting.
Basically, you are adding where you should be multiplying. See dmk's answer for a little more about that.
See if you can get started yourself on the second part. If you need more help, let me know.
-
Thanks a lot! Figured out B) as well with this reasoning. Guess I got to dug in on the problem I lost my bearings. – Mylleranton Feb 13 '14 at 21:44
@Mylleranton Always glad to help. – Tim Seguine Feb 13 '14 at 21:47
For (a), as an earlier comment stated, you're making order matter by using a permutation. So, for example, you have $18$ choices for the first person on Team 1, $17$ choices for the second, and $16$ choices for the third. Just as you wrote, this makes $18\cdot 17\cdot 16$ choices. However, from those three people, we could have chosen the first person $3$ ways, the second $2$ ways, and the final person $1$ way. Therefore, $18\cdot 17\cdot 16$ overcounts the number of teams by a factor of $3\cdot 2 \cdot 1$. That means there are $\frac{18\cdot 17\cdot 16 }{3\cdot 2 \cdot 1}$ ways to choose the first team. This is equivalent to the binomial coefficient $\binom{18}{3}$, if you're familiar with combinations.
However, following this idea for each team will make your answer smaller, whereas your answer is too small to begin with. What you want to do is multiply those fractions rather than add them. Here we're determining one team and then determining another and then determining a third... Often, when an "and" appears as part of how you would complete a task (as above), that means you multiply. An "or", on the other hand, tells you to add; for example, if you wanted to make only one team, and you wanted to know how many ways there were to create a team of either three or four people from 18, you'd calculate
$$\frac{18\cdot 17\cdot 16 }{3\cdot 2 \cdot 1} + \frac{18\cdot 17\cdot 16\cdot 15}{4\cdot 3 \cdot 2 \cdot 1}$$
-
This has some quite generally applicable reasoning. Thanks a lot! – Mylleranton Feb 13 '14 at 21:47
+1 for explaining the "and/or" thing, which I just sort of glossed over in my answer. – Tim Seguine Feb 13 '14 at 21:50 | 2015-10-10T07:03:54 | {
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https://www.themathdoctors.org/interpreting-probability-questions/ | # Interpreting Probability Questions
#### (A new question of the week)
A couple recent questions centered around how to interpret probability problems, whose wording can often be subtle, and whose solutions require care.
## A dice problem about the maximum
Around the start of December, Drey sent us two questions. The first was about dice:
Hello. I have this problem:
Two dices are rolled simultaneously. Calculate the probability that the maximum is greater than or equal to 3.
Solution: 8/9
So I found that the total of outcomes is given by 6*6 since there are two dices.
Now what confuses me is the number of favorable outcomes that the exercise asks for, I tried with all of the combinations of dices greater and equal to 3 but it isn’t the solution. Can you help me please.
(Incidentally, we often see students, native English speakers or not, confusing the word “dice”. That word is a plural, not a singular, and should be thought of as if it were “dies”. The singular is “die”. So you roll one die, or two dice. For the rest of this post, I will quietly correct such minor errors, as it was never brought out in the discussion, but it is worth pointing out once.)
Problems about dice often involve the sums of the numbers showing; this is about the maximum, which has a very different feel. Drey is calculating the probability as $$\frac{\text{number of equally likely favorable outcomes}}{\text{total number of equally likely outcomes}}$$ where “favorable” means that the maximum of the number on the dice is at least 3. He correctly found the denominator, as there are 6 possibilities for the first die, and 6 for the second, so a total of $$6\times 6=36$$. But what should the numerator be? It isn’t quite clear what Drey used, so we have to ask.
Hi Drey,
You haven’t said what answer you got when you tried to solve the problem, other than you did not get 8/9, so I can’t tell what you are doing incorrectly. If you want to give your answer and your reasoning, I can try to figure out why you are not getting the answer. Did you write out the possibilities and just count, or did you figure out the number of favorable outcomes by just reasoning?
### Listing outcomes
The more you tell us, the more quickly we can diagnose your difficulty. In this case, it would not be hard to list the 36 possible outcomes in a table and literally count those that are favorable, which can be very helpful in understanding a problem; reasoning is more useful when you need to be able to extend it to larger problems (such as more dice). Let’s make that table:
The event we are interested in is that the maximum of the two dice is at least 3. How many of these 36 possible outcomes fit that description? It’s easy, except that we have to correctly interpret the word “maximum”; we’ll do that below.
### Using the complement
But how about the reasoning we would need if we don’t make the table? One counting trick that is often helpful (especially when numbers get large) is to use the complement. Doctor Fenton had more to add:
I will point that if an event has two possible outcomes which are exhaustive (that is, one of the two other outcomes must occur) and exclusive (both outcomes cannot occur at the same time), then the sum of the two probabilities must be 1. If A is the event that the maximum number on one of the dice is 3 or more, what is the complementary event that A does not happen? That is, if the maximum of the two numbers showing on the dice is not 3 or more, what can you say about the numbers showing on that roll?
Complements can be hard to think clearly about, especially if you focus too much on the words and not enough on the meaning. But once you are over that hurdle, the thinking can become easier. To help with that, Doctor Fenton offered a different problem (the more common question about sums, but with three dice so it is considerably harder to make a table) to illustrate:
It is often easier to find the probability that an event does not occur than to find the probability that the event occurs directly. For example, if you roll three dice, what is the probability that the sum of the numbers showing will be 5 or more? Now there are 63 = 216 outcomes, and most of them will be favorable. If you write out all the possibilities, that will be a lot to count. But if this event (sum ≥ 5) does not occur, then the sum must be 4 or less. The smallest possible sum is 3, with each die showing a 1, and if one die shows a 3 or higher, or if two dice show 2, then the sum must be at least 5. The only outcomes for which the sum is less than 5 are outcomes for which the sum is 3 or 4. That happens if all three dice show 1, or two dice show 1 and the third die shows 2. That is 4 possibilities, so the probability P(sum<5) is 4/216=1/54. Therefore, the probability P(sum≥5) must be 1-(1/54) = 53/54. Do you see how it can be much easier to count the complementary event than to count the event directly?
In this example, the complementary event can be easily listed: $${(1,1,1), (1,1,2), (1,2,1), (2,1,1)}$$, so we know there are 4 outcomes, and therefore $$216-4=212$$ outcomes in the event we are interested in.
Drey responded,
Hello Doctor Fenton.
I apologise for not showing my answer to the problem.
I followed your advice and tried to use the complementary event formula in which P(A)+P(B)=1.
Since we have the 2 dice, then the total number of outcomes will be 6^2=36.
Now for the favorable number of events, since we are looking that the event doesn’t occur (where the dice need to be less than 3 (P(B)<3)) I managed to count only one event where each die is showing 1: {1,1}. Now using the formula of the complementary event we have that P(A) + 1/36 = 1 ==> P(A) = 1 – 1/36 ==> P(A) = 35/36.
That is the result I got. Please correct me where is needed.
Drey’s notation “P(B)<3” is confusing; taken literally, it means that the probability of the complementary event is less than 3. He may mean the probability that the sum of the numbers on the dice is less than 3, as that is a way to describe the complementary set he is counting:
So the probability he found is for this event, which could be described as “sum < 3”, or as “both < 2”:
Doctor Fenton replied, trying to clarify what the event and its complement are:
That’s a good start, but you haven’t correctly identified the complementary event. The original event was that the maximum number showing is 3 or more. There is more than one event, (1,1), where the maximum showing is not at least 3. What is the maximum number which can occur in the roll, and how many outcomes does that describe?
The missing piece is to identify the event correctly; the language is difficult. If the maximum of the two numbers is not at least 3, then it is less than 3; in which outcomes is the maximum less than 3?
### Understanding the problem
Drey said,
Doctor Fenton, I apologise but I require some further explanation. You are correct I am having trouble identifying the complementary event.
If the problem says that the maximum number of the 2 dice rolled is 3 or greater, shouldn’t the complement be the results that are not 3 or greater (< 3)? and since if one die is 2 and the other is 1 then the result is not part of the complement since it shouldn’t be equal to 3.
It’s true that the complement of “3 or greater” is “less than 3”. But the key word is “maximum”, which Drey doesn’t use here. My guess is that Drey has let that word get tangled up in the “greater than” and lost it, changing the event from “the maximum is greater than or equal to 3” to something like “it is greater than or equal to 3”, and taking it to be about the sum. (Another easy misreading would be something like “the numbers are both greater than or equal to 3”.)
Eight minutes later, Doctor Rick joined in, having been following the discussion:
Hi, Drey. I know Doctor Fenton could answer, but I’d like to respond to you quickly.
I believe you have been misreading the problem. It is easy to do: the first time I read the problem, I assumed that it meant to say that the sum is greater than or equal to 3, because that is a common type of problem with dice. But it means what it says: The maximum of the numbers on the two dice is greater than or equal to 3. If one die shows a 1 and the other shows a 2, then the maximum of those two numbers is 2, which is less than 3!
All he has done here is to point out the word “maximum”, and expand it to “The maximum of the numbers on the two dice” to make it stand out. And giving a simple example is often a good way to make the meaning of a complicated statement clear.
### Getting it right
Drey replied,
Hello Doctor Rick.
I managed to understand what the question is asking for, thank you.
So I managed to get 4 possible outcomes if I am following your explanation correctly: {1,1; 1,2; 2,1; 2,2}. Using the complementary event formula I got that: p(a)=1 – 4/36 ==> 32/36 ==> 8/9
Doctor Fenton.
Thank you very much. I managed to solve the exercise but also for the insight on how to solve it using the complementary event formula.
To which Doctor Fenton replied,
You’re welcome! Glad we could help.
Here is the complementary event, in which the largest of the two numbers is less than 3:
Without listing, that could be counted by seeing it as “both numbers are less than 3″, so that there are 2 choices for the first and 2 for the second, making a total of $$2\times 2=4$$. So the probability of the complement is $$\frac{4}{36}=\frac{1}{9}$$. The probability of the event itself is $$1-\frac{1}{9}=\frac{8}{9}$$. Or, without simplifying first, seeing it as Drey did, the event consists of $$36-4=32$$ outcomes:
This could have been counted directly as “first die is at least 3, or second die is at least 3″; but every way to do that with formulas is considerably longer. One way is to use the formula $$P(A\text{ or }B)=P(A)+P(B)-P(A\text{ and }B)$$ applied to$$P((\text{first }\ge 3)\text{ or }(\text{second }\ge 3))=\frac{4}{6}+\frac{4}{6}-\frac{4}{6}\cdot\frac{4}{6}\\ =\frac{24}{36}+\frac{24}{36}-\frac{16}{36}=\frac{32}{36}=\frac{8}{9}$$ This formula amounts to adding the red- and green-bordered regions, then subtracting their overlap:
Another is to use $$P((\text{first }\ge 3\text)\text{ or }(\text{first }< 3\text{ and second }\ge 3))\\ =\frac{4}{6}+\frac{2}{6}\cdot\frac{4}{6}\\ =\frac{24}{36}+\frac{8}{36}=\frac{32}{36}=\frac{8}{9}$$
This amounts to adding the green and the yellow regions here:
## A ball-and-urn problem
Besides dice and cards, the most standard sort of probability problem involves colored or numbered balls in an “urn”. (Probably most of us never encounter urns any more except in such problems!) The following day, Drey asked about one of those:
Greeting, I have this exercise of probability that is a little confusing to me.
Here’s the exercise: One urn contains 4 white balls, 4 black balls and 4 red balls. 6 balls are removed from that urn. Determine the probability of 2 balls of each color being drawn.
(A) Assuming extraction with replacement
Solution = 1/8
(B) Assuming extraction without replacement
Solution = 2/77
What I tried to do: (A) as this line says that 2 balls are removed with replacement, so there is no need to apply simple combination. Total number of outcomes = 12, number of events are two balls of each color, and since each draw is an independent event, I can multiply the probability to draw a red ball on the first serve, a red ball on the second serve, a black ball on the third serve, a black ball on the fourth serve withdraw; a white ball in the fifth and a white ball in the sixth. That is: p = 4/12 * 4/12 * 4/12 * 4/12 * 4/12 * 4/12 = 4096/2985984 = 1/729 which is wrong.
For line (B) I have to apply simple combination because there is no replacement where number of events C6.2 = 360 and C12.6 for the total number of events we have that p = 360/924 = 30/77 Could You please help me? Your time is appreciated, I await your response. Thank you.
This is a model of the way we like to see questions asked: stating the entire problem, including the provided solutions (which lets us check whether the student is confused only because their book gave a wrong answer!), and a careful explanation of his own thinking.
### With and without replacement
In the case of selection with replacement (putting a ball back after it’s been looked at), each selection is independent, and the probabilities are the same each time. Drey therefore multiplied the probability of each ball having the appropriate color in order, two reds, two blacks, two whites, each probability being the same because there are 4 of the given color out of 12 at each selection. There’s only a small error there.
In selection without replacement (so that after each draw there is one less ball, and the probabilities change), Drey (wisely) chose to use combinations. His notation “C6.2” means what is more commonly written as “$${6\choose 2}$$” or “$$_6C_2$$” or “$$C(6,2)$$”. He found the total number of ways to choose 2 of the 6 balls, and divided by the total number of ways to choose 6 of the 10 balls. This, too has an error.
### Identifying the errors
Hi, Drey.
Your work for the first part would be good if you were only asked for the probability of drawing balls in the order RRBBWW. What more needs to be done because the order doesn’t matter? (I would also simplify before multiplying, making the arithmetic a lot easier!)
The error here is simple: “Two of each color” does not require selection in any particular order; and the multiplication he did gives only the probability that the first two are red, the second two are red, the third two are black, and so on. The fix will be easy.
For the second part, the numerator will not be C(6, 2); you are not just choosing 2 of 6. Can you explain your thinking there more fully? You may just need a small change in your approach.
My best explanation of Drey’s thinking is that he pictured the results sitting in a row of 6, and wanted those results to include two reds, so he counted ways to choose 2 of the 6 places to be red. He just forgot the other colors.
There are typically many ways to solve a combinatoric problem like this, though, so I didn’t want to presume that he was necessarily thinking that way.
I have to say that I don’t get answers of 1/8 and 2/77 as you appear to be saying you were told. I get something very close to the first, and considerably more for the second. (I have held off on answering you while I keep returning to the problem to look for an error in my own work; if another of us does get those answers, they may join me in helping!)
I had to take the time to convince myself that my thinking was right even though it led to different answers than the book, before deciding to try to help!
### Getting it right
Drey replied,
Hello Doctor Peterson.
Well since the order doesn’t matter we have to pick the number of ways that these 6 colours can be rearranged, coincidentally that is a permutation with repetition where C6,2,2,2 = 6!/2!*2!*2 = 90 so since this is part of the event I multiplied it by 1/729 meaning: 1/729*90 = 90/729 = 10/81
This is what was needed to correct the undercount. He previously counted the probability of choosing RRBBWW in that order as $$\frac{4^6}{12^6} = \frac{1}{729}$$; he multiplied that by the number of ways to rearrange these to obtain all possible “successful” selections, namely the number of ways to arrange the letters RRBBWW. This is $$\frac{6!}{2!2!2!} = 90$$. The product, $$\frac{10}{81}$$, is very close to the supplied answer of $$\frac{1}{8}$$, but not quite.
For the second part(B) I managed to find the same problem in a few websites and most of them suggest that the answer is wrong, They did this:
C12,6 = 12!/6!*6! = 924 (total number of outcomes)
Total of combinations for 2 white balls
C4,2 = 4/2!*2! = 6
Total of combinations for 2 black balls
C4,2 = 4/2!*2! = 6
Total of combinations for 2 red balls
C4,2 = 4/2!*2! = 6
P = (C4,2*C4,2*C4,2)/C12,6 = 6^3/924 = 18/77
The denominator will be the number of ways to choose 6 of 12 balls, $${{12}\choose{6}} = \frac{12!}{6!6!} = \frac{12\cdot11\cdot10\cdot9\cdot8\cdot7}{6\cdot5\cdot4\cdot3\cdot2\cdot1}=924$$
The numerator has to be the number of ways to choose 2 of the 4 white, 2 of the 4 black, and 2 of the 4 red, for a total of $${{4}\choose{2}}\cdot{{4}\choose{2}}\cdot{{4}\choose{2}}=6\cdot6\cdot 6 = 216$$, so the probability is $$\frac{216}{924} = \frac{18}{77}$$. This is 9 times as much as the book’s answer of $$\frac{2}{77}$$; I can’t imagine how they got this wrong.
Now what I don’t get is the thought process behind the multiplication at the last part.
Could you maybe help explain? | 2021-05-08T15:36:57 | {
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