Micromass' big October challenge

[micromass]

Awesome, it appears (a) and (c) are solved then.

 

[micromass]

And I saw that (b) was solved too!

 

[TeethWhitener]

Ok, I'll try advanced problem number 7. First, we note that $X=b\tan{\theta}$. To get the expected value of $X$, we use the law of the unconcscious statistician:
$$E[X(\theta)] = \int^{\frac{\pi}{2}}_{-\frac{\pi}{2}} X(\theta) f(\theta)d\theta$$
where $f(\theta)$ is the probability density function of $\theta$. We know that $\theta$ is uniformly distributed over $(-\pi/2,\pi/2)$, so $f(\theta) = 1/\pi$. Thus the expected value of $X$ is
$$E[X(\theta)] =\frac{b}{\pi} \int^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \tan{\theta} d\theta$$
Since $\tan \theta$ is an odd function and the integration is symmetric about $x=0$, the expected value is $E[X] = 0$.

For the variance, we have:
$$\sigma^2 = E([X^2])-(E[X])^2$$
so we need to evaluate $E([X^2])$. We use the unconscious statistician again, which gives us the integral:
$$E[X^2] = \frac{b^2}{\pi} \int^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \tan^2{\theta} d\theta$$
This integral diverges to infinity as we take the bounds of integration out to $\pm \pi/2$. This would imply that the variance is infinite $(\sigma^2 = \infty)$. I'm not sure if this is right, but my intuition says this makes sense, since there's a finite probability of $|X|$ being arbitrarily large. But I don't know that for certain, and I'm not sure how to make it more mathy.

 

[micromass]

Ok, I'll try advanced problem number 7. First, we note that $X=b\tan{\theta}$. To get the expected value of $X$, we use the law of the unconcscious statistician:
$$E[X(\theta)] = \int^{\frac{\pi}{2}}_{-\frac{\pi}{2}} X(\theta) f(\theta)d\theta$$
where $f(\theta)$ is the probability density function of $\theta$. We know that $\theta$ is uniformly distributed over $(-\pi/2,\pi/2)$, so $f(\theta) = 1/\pi$. Thus the expected value of $X$ is
$$E[X(\theta)] =\frac{b}{\pi} \int^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \tan{\theta} d\theta$$
Since $\tan \theta$ is an odd function and the integration is symmetric about $x=0$, the expected value is $E[X] = 0$.

For the variance, we have:
$$\sigma^2 = E([X^2])-(E[X])^2$$
so we need to evaluate $E([X^2])$. We use the unconscious statistician again, which gives us the integral:
$$E[X^2] = \frac{b^2}{\pi} \int^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \tan^2{\theta} d\theta$$
This integral diverges to infinity as we take the bounds of integration out to $\pm \pi/2$. This would imply that the variance is infinite $(\sigma^2 = \infty)$. I'm not sure if this is right, but my intuition says this makes sense, since there's a finite probability of $|X|$ being arbitrarily large. But I don't know that for certain, and I'm not sure how to make it more mathy.

Your expected value is wrong.

 

[TeethWhitener]

Your expected value is wrong.

Did I cut too many corners? Upon closer inspection, I get
$$\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \tan{\theta} d \theta = \infty - \infty$$
so it's an indeterminate form. I have no idea how to handle this if not by the symmetry of the function.

 

[Erland]

Your expected value is wrong.

Really? As far as I can see, the symmetry of the problem implies that the expectation is $0$. If we take two infinitesimal intervals $[-x-dx, x]$ and $[x,x+dx]$ on the wall, corresponding to the angle intervals $[-\theta-d\theta, \theta]$ and $[\theta,\theta+d\theta]$, respectively, both having probability $d\theta/\pi$, their contributions to the expectation are $-xd\theta/\pi$ and $xd\theta/\pi$, respectively, so they cancel each other out. Every infinitesimal interval has such a "mirror" interval on the other side of $a$, so the expectation must be $0$.

How could it be in any other way?

 

[micromass]

Really? As far as I can see, the symmetry of the problem implies that the expectation is $0$. If we take two infinitesimal intervals $[-x-dx, x]$ and $[x,x+dx]$ on the wall, corresponding to the angle intervals $[-\theta-d\theta, \theta]$ and $[\theta,\theta+d\theta]$, respectively, both having probability $d\theta/\pi$, their contributions to the expectation are $-xd\theta/\pi$ and $xd\theta/\pi$, respectively, so they cancel each other out. Every infinitesimal interval has such a "mirror" interval on the other side of $a$, so the expectation must be $0$.

How could it be in any other way?

Then by that same reasoning, would you say that $\int_{-\infty}^{+\infty} xdx = 0$ too?

 

[TeethWhitener]

Then by that same reasoning, would you say that $\int_{-\infty}^{+\infty} xdx = 0$ too?

So is it just an indeterminate form, like in my post #75 above?

 

[micromass]

So is it just an indeterminate form, like in my post #75 above?

Yes, the expectation value doesn't exist.

You can do the test here: http://www.math.uah.edu/stat/apps/CauchyExperiment.html Run it for 1000 or 10000 turns. You'll see it getting close to 0, but then suddenly it'll jump away from 0 again.

 

[Erland]

Yes, the expectation value doesn't exist.

Yes, Ok, I agree. The integral $\int_{-\pi/2}^{\pi/2}\tan\theta d\theta$ diverges. Sorry...

But then, the variance is undefined also, since it is defined in terms of the expectation: $V(X)=E((X-E(X))^2)$.

So the entire problem was a poser...

 

[micromass]

So the entire problem was a poser...

Haha, yes indeed!

 

[TeethWhitener]

But then, the variance is undefined also

Good point. I was assuming that $E[X]=0$. Call it the "renormalized" answer

 

[TeethWhitener]

Ok, I'll try the unsolved advanced number 2. Last month, I was somewhat unadvisedly trying to show that $dM_p/dp$ was positive everywhere, but the hint this month helps out a whole bunch. We want to know the nature of the relationship:
$$\left(\sum_i {\frac{x_i^p}{n}}\right)^\frac{1}{p} \sim \left(\sum_i {\frac{x_i^q}{n}}\right)^\frac{1}{q}$$
For the moment, let us consider both terms raised to the power of $p$. We have:
$$\sum_i {\frac{x_i^p}{n}}$$
and
$$\left(\sum_i {\frac{x_i^q}{n}}\right)^\frac{p}{q}$$
We notice that
$$\sum_i {\frac{\left(x_i^q\right)^{\frac{p}{q}}}{n}} = \sum_i {\frac{x_i^p}{n}}$$
which allows us to apply Jensen's inequality:
$$\left(\sum_i {\frac{x_i^q}{n}}\right)^\frac{p}{q} \leq \sum_i {\frac{\left(x_i^q\right)^{\frac{p}{q}}}{n}} = \sum_i {\frac{x_i^p}{n}}$$
for $x^{p/q}$ convex (2nd derivative $> 0$) and
$$\left(\sum_i {\frac{x_i^q}{n}}\right)^\frac{p}{q} \geq \sum_i {\frac{x_i^p}{n}}$$
for $x^{p/q}$ concave (2nd derivative $< 0$). For positive $x$, $x^{p/q}$ is convex when $p/q > 1$ and $p/q < 0$ and concave otherwise. (This is true because the second derivative of $x^a$ is $a(a-1)x^{a-2}$ and the function $a(a-1)$ is a parabola which is only negative between 0 and 1).

We want to know what happens for $p>q$. This condition breaks the problem into 3 cases: $p$ and $q$ both positive (convex), $p$ positive and $q$ negative (convex), and $p$ and $q$ both negative (concave). For the convex case, we have
$$(M_p)^p \geq (M_q)^p$$
Since $p$ is positive whenever $x^{p/q}$ is convex, this result implies that $M_p \geq M_q$. For the concave case, we have
$$(M_p)^p \leq (M_q)^p$$
Since $p$ is negative whenever $x^{p/q}$ is concave, this result also implies that $M_p \geq M_q$, which completes the proof.

 

[person_random_normal]

https://mail.google.com/mail/u/0/?u...68608010&rm=15794b3f1927dbe0&zw&sz=w1366-h662

https://mail.google.com/mail/u/0/?ui=2&ik=8afd5aa52a&view=fimg&th=15794b3f1927dbe0&attid=0.1&disp=inline&realattid=1547350341991792640-local0&safe=1&attbid=ANGjdJ9AL_gRE5sXmUYaevAhKSEVWeCDGMrM2iIlWdWoI_HQXuUtcdRj5gRNTsNVhx1fD7uUlFXVMOekFVnhgzPXCNVO8-rWTr1U5G2r1pmTcTPjii6g4adAZZBM6Ic&ats=1475668608010&rm=15794b3f1927dbe0&zw&sz=w1366-h662
Is it right ??

free adult image hosting

 

[micromass]

free adult image hosting

Those integrals are both incorrect.

 

[Erland]

Attempting Advanced Problem 6:

$X$ and $Y$ are independent stochastic variables which both have uniform distrubution on $[0,1]$. This means that for every Lebesgue mesaurable set $A\subseteq \mathbb R^2$, $P((X,Y)\in A)=m(A\cap[0,1]^2)$, where $m$ is the Lebesgue measure on $\mathbb R^2$. Also, for every Lebesgue measurable set $B\subseteq \mathbb R$, $P(X^Y\in B)=m(\{(x,y)\in[0,1]^2\,|\,x^y\in B\})$.

The distribution of $X^Y$ is determined by its cumulative distribution function $F:\mathbb R \to[0,1]$, given by $F(t)=P(X^Y\le t)=m(\{(x,y)\in[0,1]^2\,|\,x^y\le t\})$, for all $t\in\mathbb R$. To find the distribution, it should suffice to find $F$.

The function $f(x,y)=x^y$, defined on $[0,1]^2$ (for definiteness, we define $f(0,0)=0^0=1$, but this does not really matter) is increasing in $x$ and decreasing in $y$. $f$ is continuous on all $[0,1]^2$ except at $(0,0)$, so it is Lebesgue measurable. We have $f(x,0)=1$ and $f(x,1)=x$, for all $x\in[0,1]$, and $f(0,y)=0$ and $f(1,y)=1$ for all $y\in\,]0,1]$, so the range of $f$ is $[0,1]$.

Hence, $F(t)=0$ for $t<0$ and $F(t)=1$ for $t>1$.
For $t\in[0,1]$ and $(x,y)\in[0,1]\times\,]0,1]$, we have $x^y\le t\Leftrightarrow x\le t^{\frac1y}(\le 1)$, since $\frac1y\ge 1$. For each $\epsilon\in\,]0,1[\,$, we then have $m(\{(x,y)\in[0,1]\times[\epsilon,1]\,|\,x^y\le t\})=\int_\epsilon^1 t^{\frac1y}dy$. This tends to both $F(t)$ (since $m(\{(x,0)\,|\,0\le x\le 1\})=0)$ and $\int_0^1t^{\frac1y}$ (since this improper integral converges, because the integrand is continuous and bounded on $]0,1]\,$) as $\epsilon\to 0$. Hence $F(t)=\int_0^1 t^{\frac1y} dy$.
I don't know of any closed expression for this integral, and I doubt that such an expression exists or is known.

So my answer is that the desired distribution is given by the cumulative distribution function:

$$F(t)=\begin{cases} 0 & t<0,\\ \int_0^1 t^{\frac1y}dy,& 0\le t \le 1, \\1,&t>1.\end{cases}$$

 

[jostpuur]

I just posted one answer, and only after replying I noticed that it said this:

CHALLENGES FOR HIGH SCHOOL AND FIRST YEAR UNIVERSITY:

before that problem. Whoops... Did that mean that I should not have posted that answer?

 

[Nugatory]

Whoops... Did that mean that I should not have posted that answer?

Yes, and we've removed it.

 

[Chestermiller]

My solution to this problem for x vs y, expressed parametrically in terms of $\theta$ is as follows:
$$x=x_0\left[1-\frac{1}{(\sec{\theta}+\tan{\theta})^{V/v}}\right]$$
$$y=x_0\left[\int_0^{\theta}{\frac{\sec^2{\theta '}d\theta '}{(\sec{\theta '}+\tan{\theta '})^{V/v}}}-\frac{\tan {\theta}}{(\sec{\theta}+\tan{\theta})^{V/v}}\right]$$where $\theta '$ is a dummy variable of integration.

Charles, your analytic solution to this problem should match mine, and should thus somehow provide the result of correctly integrating of my "mystery integral" in the equation for y. Could you please see if you can back out the integral evaluation? Thanks.

Chet

I finally got the integral in the parameteric equation for y. @fresh_42 submitted the integrand to Wolfram Alpha, and it provided the closed-form result. Thanks you so much @fresh_42. Here is the desired result:
$$y=\frac{x_0(V/v)}{\left(\frac{V}{v}\right)^2-1}\left[1-\frac{(\sec{\theta}+(V/v)\tan{\theta})}{(\sec{\theta}+\tan{\theta})^{V/v}}\right]$$
I hope I did the "arithmetic" correctly and that this result agrees with Charles Links' results.

 

[MAGNIBORO]

problem 4 highshool:
x coordinate = $-1+\frac{1}{2}-\frac{1}{4}... = -1 + \frac{1}{2}(1-\frac{1}{2}+\frac{1}{3}...) = -1 + \frac{log(2)}{2}$
y coordinate = $1-\frac{1}{3}+\frac{1}{5}..=\frac{\pi }{4}$
converges to the point $\left ( -1+\frac{log(2)}{2},\frac{\pi }{4} \right )$

 

[micromass]

problem 4 highshool:
x coordinate = $-1+\frac{1}{2}-\frac{1}{4}... = -1 + \frac{1}{2}(1-\frac{1}{2}+\frac{1}{3}...) = -1 + \frac{log(2)}{2}$
y coordinate = $1-\frac{1}{3}+\frac{1}{5}..=\frac{\pi }{4}$
converges to the point $\left ( -1+\frac{log(2)}{2},\frac{\pi }{4} \right )$

Read the problem more carefully. The lengths are defined recursive.

 

[MAGNIBORO]

Read the problem more carefully. The lengths are defined recursive.

my mistake,
x coordinate: series of cos is $1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}...$ so $-1+\frac{1}{2!}-\frac{1}{4!}+\frac{1}{6!}...=-cos(1)$
y coordinate: series of sin is $x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}...$ so $1-\frac{1}{3!}+\frac{1}{5!}-\frac{1}{7!}...=sin(1)$

for problem 9:
$I_{n}= \int_{0}^{1}x^n\sqrt{1-x}\, dx = \beta (n+1,\frac{3}{2})=\frac{n! \, \frac{\sqrt{\pi }}{2}}{(n+\frac{3}{2})!}$
and $I_{n-1}= \frac{(n-1)! \, \frac{\sqrt{\pi }}{2}}{(n+\frac{1}{2})!}$
so
$I_{n-1} \: \frac{2n}{2n+3}= \frac{(n-1)! \, \frac{\sqrt{\pi }}{2}}{(n+\frac{1}{2})!} \: \frac{n}{n+\frac{3} {2}} = \frac{n! \, \frac{\sqrt{\pi }}{2}}{(n+\frac{3}{2})!} = I_{n}$

 

[MAGNIBORO]

for high school problem 5 (very entertaining) :
$A(1,n) = A(0,A(1,n-1)) = 1 + A(1,n-1)$
$1 + A(1,n-1) = 2 + A(1,n-2)$
$...$
$A(1,n) = n + A(1,0) = n+2$

$A(2,0) = A(1,1) = 1 + 2 = 3$
$A(2,n) = A(1,A(2,n-1)) = A(2,n-1) + 2$
$2 + A(2,n-1) = 4 + A(2,n-2)$
$...$
$A(2,n) = 2n + 3$

$A(3,0) = A(2,1) = 5$
$A(3,1) = A(2,A(3,0)) = 2 A(3,0) + 3 = 13$
$A(3,2) = 2 A(3,1) +3 = 29$
$A(3,3) = 2 A(3,2) + 3 = 61$
now note that
$A(3,1) - A(3,0) = 8$
$A(3,2) - A(3,1) = 16$
$A(3,3) - A(3,2) = 32$
so
$A(3,n+1) = 2 A(3,n) + 3 = A(3,n) + 2^{n+3}$
$A(3,n) = 2^{n+3} - 3$

 

[MAGNIBORO]

Problem 10:
$$\int_{-1}^{1}\frac{dx}{x^2+1}=\frac{\pi }{2}$$
$$\int_{-1}^{1}\frac{e^{x}+1}{(x^2+1)(e^{x}+1)}=\frac{\pi }{2}$$
$$\int_{-1}^{1}\frac{e^{x}}{(x^2+1)(e^{x}+1)}+\int_{-1}^{1}\frac{1}{(x^2+1)(e^{x}+1)}=\frac{\pi }{2}$$
we can assume that the 2 integrals are equal, and this is true if
$$\int_{-1}^{1}\frac{1-e^{x}}{(x^2+1)(e^{-x}+1)}=0$$
That would be true if $\frac{1-e^{x}}{(x^2+1)(e^{-x}+1)}$ is a odd function
$$f(-x)=-f(x)$$
$$\frac{1-e^{-x}}{(x^2+1)(e^{-x}+1)}=-\frac{1-e^{x}}{(x^2+1)(e^{x}+1)}$$
$$\frac{-1+e^{x}}{(x^2+1)(e^{x}+1)}=-\frac{1-e^{x}}{(x^2+1)(e^{x}+1)}$$
so
$$\int_{-1}^{1}\frac{e^{x}}{(x^2+1)(e^{x}+1)}=\int_{-1}^{1}\frac{1}{(x^2+1)(e^{x}+1)}$$
$$\int_{-1}^{1}\frac{1}{(x^2+1)(e^{x}+1)}=\frac{\pi }{4}$$
this problem is very good, and It has a very beautiful relationship:
$$\int_{-1}^{1}\frac{1}{(x^2+1)(e^{-x}+1)}+\int_{-1}^{1}\frac{1}{(x^2+1)(e^{x}+1)}=\frac{\pi }{2}$$Problem 6:
$$I=\int_{0}^{\frac{\pi }{2}}\frac{sin(x)cos(x)}{x+1}=\frac{1}{2}\left ( \left[ \frac{sin^2(x)}{x+1} \right]_{0}^{\frac{\pi }{2}}+\int_{0}^{\frac{\pi }{2}}\frac{sin^2(x)}{(x+1)^2}
\right )=\frac{1}{2}\left ( \frac{2}{\pi+2 }+ \frac{1}{2}\left ( \int_{0}^{\frac{\pi }{2}}\frac{1}{(x+1)^2}-\int_{0}^{\frac{\pi }{2}}\frac{cos(2x)}{(x+1)^2} \right )\right )$$
$$=\frac{1}{\pi +2}+\frac{\pi}{4\pi+8}-\frac{1}{4}\int_{0}^{\frac{\pi}{2}}\frac{cos(2x)}{(x+1)^2}$$
$$=\frac{1}{\pi +2}+\frac{\pi}{4\pi+8}-\frac{1}{4}\int_{1}^{1+\frac{\pi}{2}}\frac{cos(2u-2)}{u^2}$$
$$=\frac{1}{\pi +2}+\frac{\pi}{4\pi+8}-\frac{1}{4}\left (cos(2)\int_{1}^{1+\frac{\pi}{2}}\frac{cos(2u)}{u^2}+sin(2)\int_{1}^{1+\frac{\pi}{2}}\frac{sin(2u)}{u^2} \right )$$
$$I=\frac{1}{\pi +2}+\frac{\pi}{4\pi+8}-\frac{1}{2}\left (cos(2)\int_{2}^{2+\pi}\frac{cos(u)}{u^2}+sin(2)\int_{2}^{2+\pi}\frac{sin(u)}{u^2} \right )$$
introduce M
$$I=\frac{1}{\pi +2}+\frac{\pi}{4\pi+8}-\frac{M}{2}$$

now the other integral.

$$J=\int_{0}^{\pi }\frac{cos(x)}{(x+2)^2}=\int_{2}^{\pi +2}\frac{cos(u-2)}{u^2}=cos(2)\int_{2}^{2+\pi}\frac{cos(u)}{u^2}+sin(2)\int_{2}^{2+\pi}\frac{sin(u)}{u^2}$$
$$J=M$$
$$I=\frac{1}{\pi +2}+\frac{\pi}{4\pi+8}-\frac{J}{2}$$

 

[zinq]

In Problem 2) ("Let p≠0 be a real number ...") it is interesting to look at what happens as p → 0. The problem is still meaningful in that case!

 

[Stella.Physics]

so what about the answer to the birds on wire problem? Really eager to find out what the answer and proof is :)

 

[Erland]

A clue to advanced problem 3 would also be appreciated...

 

[parshyaa]

CHALLENGES FOR HIGH SCHOOL AND FIRST YEAR UNIVERSITY:

1) let A,B,C,D be a complex numbers with length 1. Prove that if A+B+C+D=0, then these four numbers form a rectangle.

2) On an arbitrary triangle, we produce on each side an equilateral triangle. Prove that the centroids of these three triangles forms an equilateral triangle

• Are they unsolved

 

[parshyaa]

CHALLENGES FOR HIGH SCHOOL AND FIRST YEAR UNIVERSITY:

1) let A,B,C,D be a complex numbers with length 1. Prove that if A+B+C+D=0, then these four numbers form a rectangle.

2) On an arbitrary triangle, we produce on each side an equilateral triangle. Prove that the centroids of these three triangles forms an equilateral triangle

• Are they unsolved

Hey i got the answer to the second of these questions, please tell me it is solved or unsloved

 

[parshyaa]

 

[mfb]

If they are not marked as solved (and if the last 2-3 posts don't cover them), then no one posted a solution yet.

 

[parshyaa]

If they are not marked as solved (and if the last 2-3 posts don't cover them), then no one posted a solution yet.

It means they are unsloved, thanks

 

[parshyaa]

@micromass

 

[parshyaa]

Where is micromass

 

[mfb]

Currently too busy to maintain the challenges threads.