Math Challenge - January 2021

[fresh_42]

Summary: Linear Programming, Trigonometry, Calculus, PDE, Differential Matrix Equation, Function Theory, Linear Algebra, Irrationality, Group Theory, Ring Theory.1. (solved by @suremarc , 1 other solutions possible) Let $A\in \mathbb{M}_{m,n}(\mathbb{R})$ and $b\in \mathbb{R}^m$. Then exactly one of the following two statements is true:

• $Ax=b\, , \,x\geq 0 \,,$ is solvable for a $x\in \mathbb{R}^n.$
• $A^\tau y\leq 0\, , \,b^\tau y> 0\,,$ is solvable for some $y\in \mathbb{R}^m.$

The ordering is meant componentwise.2. (solved by @Antarres ) Prove $\pi =\displaystyle{\lim_{n \to \infty}2^n\underbrace{\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\cdots +\sqrt{2}}}}}}_{n\text{ square roots }}}.$3. (solved by @julian ) Let $z(t)$ be a non-negative continuous real function on the interval $[a,b]$ and $t_0\in [a,b].$
Prove that if
$$
z(t) \leq C+L\left|\int_{t_0}^t z(s)\,ds\right|\quad (*)
$$
for all $t\in [a,b]$ with any constants $C,L\geq 0,$ then
$$
z(t)\leq Ce^{L|t-t_0|} \quad (**)
$$
for all $t\in [a,b].$4. (solved by @Fred Wright ) Solve the partial differential equation
\begin{align*}
u:D\longrightarrow \mathbb{R}\, &, \, D\subseteq \mathbb{R}^3\\[10pt]
xu_x+yu_y+(x^2+y^2)u_z&=0\\[10pt]
u(1,0,0)&=0\\u_x(1,0,1)&=0\\
u_y(-1,1,(\pi+2)/2)&=1\\u_z(-1,1,(\pi+2)/2)&=-1
\end{align*}5. (solved by @suremarc ) Let $A\in \mathbb{M}(n,\mathbb{R})$ be a real square matrix. Show that there is a parameterized path $x\, : \,\mathbb{R}\longrightarrow \mathbb{R}^n$ as solution of the differential equation $\dot x(t)=Ax(t)$ which is unique for any given initial condition $x(t_0)=x_0.$6. (solved by @etotheipi , and @julian ) Calculate

1. $\displaystyle{\int_{-\infty}^\infty \dfrac{x^2}{x^4+2x^2+1}\,dx}$
2. $\displaystyle{\int_{0}^{\frac{\pi}{2}} \dfrac{1}{1+\sin^2 t}\,dt}$

7. (solved by @nuuskur and @StoneTemplePython ) Prove the following well known theorem by using topological and analytical tools only.

For every real symmetric matrix $A$ there is a real orthogonal matrix $Q$ such that $Q^\tau AQ$ is diagonal.

Hint: 'Topological and analytical tools only' forbids the words 'characteristic' and 'eigen'. You can use Heine-Borel instead.8. (solved by @julian ) We define $e=\displaystyle{\sum_{k=0}^\infty \dfrac{1}{k!}}.$ Prove that $e^2$ is irrational.9. (solved by @fishturtle1 ) Let $p<q$ be two primes, $b\in \mathbb{N},$ and $G$ a group with $p^2q^b$ elements. Show that:

1. If there is no normal $q-$Sylow subgroup in $G$, then $(p,q)=(2,3),$ and there is a non trivial homomorphism from $G$ to $S_4.$
2. $G$ is always solvable.

10. (solved by @suremarc ) Let $f(x)=2x^5-6x+6\in \mathbb{Z}[x]$. In which of the following rings is $f$ irreducible and why?
(a) $\mathbb{Z}[x]$
(b) $(S^{-1}\mathbb{Z})[x]$ with $S=\{2^n\,|\,n\in \mathbb{N}_0\}$
(c) $\mathbb{Q}[x]$
(d) $\mathbb{R}[x]$
(e) $\mathbb{C}[x]$An extra question from @julian as all others have been solved:
https://www.physicsforums.com/threads/math-challenge-january-2021.997970/page-4#post-6447092

16. Let $r/s$ be any positive or negative rational number, prove that $e^{r/s}$ is an irrational number.
Hint: make use of the polynomial:
\begin{align*}
P_n (x) = \frac{x^n (1-x)^n}{n!} .
\end{align*}

High Schoolers only11. Given a set $A$ of $32$ pairwise distinct, positive integers less than $112$. Decide right or wrong:

1. (solved by @Not anonymous ) There is a number that occurs at least five times among the differences between two numbers of $A$.
2. (solved by @Not anonymous ) There is a number that occurs at least six times among the differences between two numbers of $A$.

Hint: A difference in this context is always positive, and only counted once between any two numbers of $A$.12. The harmonic numbers are
$$
H_n := \sum_{k=1}^n \dfrac{1}{k} = 1+ \dfrac{1}{2}+\dfrac{1}{3}+\ldots +\dfrac{1}{n}\; , \;(n\in \mathbb{N})
$$
We define
$$
T_n := \sum_{k=1}^n \dfrac{1}{k\cdot H_k^2} = \dfrac{1}{H_1^2}+ \dfrac{1}{2\cdot H_2^2}+\dfrac{1}{3\cdot H_3^2}+\ldots +\dfrac{1}{n\cdot H_n^2}\; , \;(n\in \mathbb{N})
$$
Show that $T_n<2$ for all $n\in \mathbb{N}.$13. We have a four sided pyramid with summit $S$ and a quadratic base $A,B,C,D.$ Let $A',B',C',D'$ be four points on the edges $AS,BS,CS,DS,$ resp. with positive distances $a,b,c,d$ from $S,$ resp. Show that $A',B',C',D'$ are coplanar if and only if
$$
\dfrac{1}{a}+\dfrac{1}{c}=\dfrac{1}{b}+\dfrac{1}{d}
$$

14. (solved by @Not anonymous ) Let $f(n)=\left[\,2\sqrt{n}\,\right]-\left[\,\sqrt{n-1}+\sqrt{n+1}\;\right]$ for $n\in \mathbb{N}.$ Determine all values of $n$ such that $f(n)=1$ and all $n$ such that $f(n)=0.$
Hint: If $r\in \mathbb{R}$ with $s\leq r < s+1$ then $\left[r\right]=\lfloor r \rfloor =s.$15. Determine all pairs of non negative integers $(m,n)$ such that $2^m-5^n=7.$

 

[etotheipi]

Spoiler: 6.1

$$
\begin{align*}
\int_{-\infty}^{\infty} \frac{x^2}{x^4 + 2x^2 + 1} &= \int_{-\infty}^{\infty} \frac{dx}{1+x^2} - \int_{-\infty}^{\infty} \frac{dx}{(1+x^2)^2} \\ \\

&= \left[ \arctan{x} \right]_{-\infty}^{\infty} - \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sec^2{\theta}}{\sec^4{\theta}} d\theta = \pi - \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2{\theta} d\theta\\

&= \pi - \left[\frac{1}{4} \sin{2\theta} + \frac{1}{2} \theta \right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \pi - \frac{\pi}{2} = \frac{\pi}{2}

\end{align*}
$$

Spoiler: 6.2

$$
\begin{align*}
\int_{0}^{\frac{\pi}{2}} \frac{dt}{1+\sin^2{t}} &= \int_{0}^{\frac{\pi}{2}} \frac{dt}{1+\cos^2{t}\tan^2{t}} = \int_{0}^{\frac{\pi}{2}} \frac{\sec^2 t}{\sec^2 t + \tan^2 t} dt \\

&= \int_0^\infty \frac{du}{1+2u^2} = \left[\frac{1}{\sqrt{2}} \arctan{\sqrt{2} u} \right]_0^{\infty} \\

&= \frac{\pi}{2\sqrt{2}}
\end{align*}
$$

 

[fresh_42]

Spoiler: 6.1

$$
\begin{align*}
\int_{-\infty}^{\infty} \frac{x^2}{x^4 + 2x^2 + 1} &= \int_{-\infty}^{\infty} \frac{1}{1+x^2} dx - \int_{-\infty}^{\infty} \frac{1}{(1+x^2)^2} dx \\ \\

&= \left[ \arctan{x} \right]_{-\infty}^{\infty} - \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sec^2{\theta}}{\sec^4{\theta}} d\theta = \pi - \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2{\theta} d\theta\\

&= \pi - \left[\frac{1}{4} \sin{2\theta} + \frac{1}{2} \theta \right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \pi - \frac{\pi}{2} = \frac{\pi}{2}

\end{align*}
$$

Spoiler: 6.2

$$
\begin{align*}
\int_{0}^{\frac{\pi}{2}} \frac{1}{1+\sin^2{t}} dt &= \int_{0}^{\frac{\pi}{2}} \frac{1}{1+\cos^2{t}\tan^2{t}} dt = \int_{0}^{\frac{\pi}{2}} \frac{\sec^2 t}{\sec^2 t + \tan^2 t} dt \\

&= \int_0^\infty \frac{du}{1+2u^2} = \left[\frac{1}{\sqrt{2}} \arctan{\sqrt{2} u} \right]_0^{\infty} \\

&= \frac{\pi}{2\sqrt{2}}
\end{align*}
$$

You could had elaborated it a bit more (the substitutions, how to get the secans etc.) for the readers who are less familiar with the formulas, but o.k.

Another proof is possible by the residue theorem.

 

[S.G. Janssens]

5. Let $A\in \mathbb{M}(n,\mathbb{R})$ be a real square matrix and $x\, : \,\mathbb{R}\longrightarrow \mathbb{R}^2$ a parameterized path. Prove that there exists a unique solution of the differential equation $\dot x(t)=Ax(t)$ for any initial condition $x(t_0)=x_0.$

While I think I know what you want to ask here, I do not understand the present formulation in a strict sense. (It is not my intention to participate in the challenge.)

 

[fresh_42]

While I think I know what you want to ask here, I do not understand the present formulation in a strict sense. (It is not my intention to participate in the challenge.)

Imagine a thread with this topic placed in our Topology and Analysis forum rather than in the Linear And Abstract Algebra forum where it perfectly fits without being moved. The first sentences could be:
The orthogonal real matrices build a compact subset of the Euclidean real space, because ... Now we define a continuous function ...

 

[Lament]

2. Prove $\pi =\displaystyle{\lim_{n \to \infty}2^n\underbrace{\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\cdots +\sqrt{2}}}}}}_{n\text{ square roots }}}.$

Spoiler: Solution

It is easy to check
$\sin\frac {\pi} {4} =\frac {\sqrt{2}} {2}$, $\sin\frac {\pi} {4*2} =\frac {\sqrt{2-\sqrt{2}}} {2}$, $\sin\frac {\pi} {4*2^2} =\frac {\sqrt{2-\sqrt{2+\sqrt{2}}}} {2}$ and so on.

Hence, it can be proved by induction that $$\sin\frac {\pi} {4*2^n} =\frac {\sqrt{2-\underbrace{\sqrt{2+\sqrt{2+\sqrt{2+\cdots+\sqrt{2}}}}}}_{\text{n square roots}}} {2}$$

Consider an arc in a unit circle which subtends an angle $\frac {\pi} {4}$ at the centre.
We can approximate the length of the arc by bisecting the angle successively to get
\begin {array} {lll}
\text {Length of the arc }
&\approx 2 \sin\frac {\pi} {4*2} \text {bisecting once}\\
&\approx 2^2 \sin\frac {\pi} {4*2^2} \text {bisecting twice}\\
&\approx 2^3 \sin\frac {\pi} {4*2^3} \text {bisecting thrice}\\
&\vdots\\
&\approx 2^n \sin\frac {\pi} {4*2^n} \text {bisecting n times}\\
i.e. \text { Length of the arc } &= \displaystyle{\lim_{n \to \infty}2^n \sin\frac {\pi} {4*2^n}}\\
or, \text { Length of the arc } &= \displaystyle{\lim_{n \to \infty}2^n \frac {\sqrt{2-\underbrace{\sqrt{2+\sqrt{2+\sqrt{2+\cdots+\sqrt{2}}}}}}_{\text{n square roots}}} {2}}\\
\end {array}

But, Length of the arc $=\frac {\pi} {4}$

Hence,
\begin {array} {lll}
\frac {\pi} {4} &= \displaystyle{\lim_{n \to \infty}2^n \frac {\sqrt{2-\underbrace{\sqrt{2+\sqrt{2+\sqrt{2+\cdots+\sqrt{2}}}}}}_{\text{n square roots}}} {2}}\\
or, \pi &= \displaystyle{\lim_{n \to \infty}2^{n+1} \sqrt{2-\underbrace{\sqrt{2+\sqrt{2+\sqrt{2+\cdots+\sqrt{2}}}}}}_{\text{n square roots}}}\\
or, \pi &= \displaystyle{\lim_{n \to \infty}2^{n+1} \underbrace{\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\cdots+\sqrt{2}}}}}}_{\text{n+1 square roots}}}\\
\end {array}

QED

 

[AndreasC]

8 is eaaaasy. It is well known that e is transcendental. Suppose e^2 is rational. Then e is a root to the equation x^2=e^2 which can be made into one with integer coefficients by multiplying by the denominator of the rational form of e^2, contradiction ;)

 

[AndreasC]

I'm not home right now and I won't be for some time, but I'm looking at 4 and thinking

Spoiler

The solution u is stationary along the direction (x, y, sqrt(x^2+y^2)). These vectors are perpendicular to (x, y, - 1). It probably shouldn't be too hard to eliminate one of the variables starting from here by finding the characteristics, right?

Is that correct as a first step?

 

[julian]

Problem 6: (Edited version)

Spoiler

Integral 1. Write

\begin{align*}
I (\beta) = \beta \int_{-\infty}^\infty \frac{1}{x^2 + 1} dx + \int_{-\infty}^\infty \frac{1}{x^2 + \beta} dx .
\end{align*}

Then

\begin{align*}
\frac{\partial}{\partial \beta} I (\beta) |_{\beta = 1} = \int_{-\infty}^\infty \frac{x^2}{(x^2 + 1)^2} dx .
\end{align*}

We evaluate $I (\beta)$ the following way:

\begin{align*}
I (\beta) & = \beta \int_{-\infty}^\infty \left( \int_0^\infty e^{- \alpha (x^2+1)} d \alpha \right) dx + \int_{-\infty}^\infty \left( \int_0^\infty e^{- \alpha (x^2+\beta)} d \alpha \right) dx \\
& = \beta \int_0^\infty \left( \int_{-\infty}^\infty e^{- \alpha x^2} dx \right) e^{- \alpha} d \alpha + \int_0^\infty \left( \int_{-\infty}^\infty e^{- \alpha x^2} dx \right) e^{- \alpha \beta} d \alpha \\
& = \beta \int_0^\infty \sqrt{\frac{\pi}{\alpha}} e^{-\alpha} d \alpha + \int_0^\infty \sqrt{\frac{\pi}{\alpha}} e^{-\alpha \beta} d \alpha \\
& = \sqrt{\pi} \left( \beta + \frac{1}{\sqrt{\beta}} \right) \int_0^\infty \frac{e^{-\alpha}}{\sqrt{\alpha}} d \alpha
\end{align*}

Putting $u^2 = \alpha$, we obtain

\begin{align*}
I (\beta) & = \sqrt{\pi} \left( \beta + \frac{1}{\sqrt{\beta}} \right) 2 \int_0^\infty e^{-u^2} d u \\
& = \sqrt{\pi} \left( \beta + \frac{1}{\sqrt{\beta}} \right) \int_{- \infty}^\infty e^{-u^2} d u \\
& = \pi \left( \beta + \frac{1}{\sqrt{\beta}} \right)
\end{align*}

So finally

\begin{align*}
I = \frac{\partial}{\partial \beta} I (\beta) |_{\beta = 1} = \frac{\pi}{2} .
\end{align*}

Alternative evaluation of integral 1.

Evaluation via complex analysis. We write

\begin{align*}
I & = \int_{-\infty}^\infty \frac{x^2}{(x^2 + 1)^2} dx \\
& = \oint \frac{z^2}{(z + i)^2 (z - i)^2} dz \\
\end{align*}

Where we have completed the contour integral into the upper half plane. The pole in the upper half plane is $z_0 = i$. So the integral is by the residue theorem:

\begin{align*}
I & = 2 \pi i \lim_{z \rightarrow i} \left( \frac{d}{dz} \left[ \frac{(z - i)^2 z^2}{(z + i)^2 (z - i)^2} \right] \right) \\
& = 2 \pi i \lim_{z \rightarrow i} \frac{2 z (z+i)^2 - z^2 2 (z+i)}{(z+i)^4} \\
& = 2 \pi i \frac{2i (-4) - (-1) 2 (2i)}{2^4} \\
& = \frac{\pi}{2} .
\end{align*}Integral 2.

\begin{align*}
I = \int_0^{\pi / 2} \frac{1}{1 + \sin^2 t} dt = \frac{1}{4} \int_0^{2 \pi} \frac{1}{1 + \sin^2 t} dt
\end{align*}

This can be converted into a contour integral around the unit circle via $z = e^{i t}$, $dz = e^{i t}i d t$, $\sin^2 t = - \frac{1}{4} \left( z - \frac{1}{z} \right)^2$

\begin{align*}
I & = \frac{1}{4} \oint \frac{dz}{iz} \frac{1}{1 - (z - 1/z)^2/4} \\
& = \oint \frac{dz}{iz} \frac{z^2}{4 z^2 - (z^2 - 1)^2} \\
& = - \oint \frac{dz}{i} \frac{z}{z^4 - 6 z^2 + 1} \\
& = \frac{1}{4 \sqrt{2}} \oint \frac{dz}{i} \left( \frac{1}{z^2 + 2 \sqrt{2} z + 1} - \frac{1}{z^2 - 2 \sqrt{2} z + 1} \right)
\end{align*}

In the first factor has the poles are at $z_1 = - \sqrt{2} + 1$ and $z_2 = - \sqrt{2} - 1$, but only the pole $z_1$ lies inside the unit circle. In the second factor has the poles are at $z_3 = \sqrt{2} + 1$ and $z_4 = \sqrt{2} - 1$, but only the pole $z_4$ lies inside the unit circle. By the residue theorem the integral is

\begin{align*}
I & = \frac{\pi}{2 \sqrt{2}} \left( \text{Res}_{z_1} [ \frac{1}{z^2 + 2 \sqrt{2} z + 1} ] - \text{Res}_{z_4} [ \frac{1}{z^2 - 2 \sqrt{2} z + 1} ] \right) \\
& = \frac{\pi}{2 \sqrt{2}} \left( \lim_{z \rightarrow - \sqrt{2} + 1} \frac{1}{z + \sqrt{2} + 1} -
\lim_{z \rightarrow \sqrt{2} - 1} \frac{1}{z - \sqrt{2} - 1}
\right) \\
& = \frac{\pi}{2 \sqrt{2}} .
\end{align*}

 

[fresh_42]

It is easy to check
sin⁡π4=22, sin⁡π4∗2=2−22, sin⁡π4∗22=2−2+22 and so on.

If it is so easy, why didn't you wrote it down?

Hence, it can be proved by induction that sin⁡π4∗2n=2−2+2+2+⋯+2⏟n square roots2

Consider an arc in a unit circle which subtends an angle π4 at the centre.
We can approximate the length of the arc by bisecting the angle successively to get
\begin {array} {lll}
\text {Length of the arc }
&\approx 2 \sin\frac {\pi} {4*2} \text {bisecting once}\
&\approx 2^2 \sin\frac {\pi} {4*2^2} \text {bisecting twice}\
&\approx 2^3 \sin\frac {\pi} {4*2^3} \text {bisecting thrice}\
&\vdots\
&\approx 2^n \sin\frac {\pi} {4*2^n} \text {bisecting n times}\
i.e. \text { Length of the arc } &= \displaystyle{\lim_{n \to \infty}2^n \sin\frac {\pi} {4*2^n}}\
or, \text { Length of the arc } &= \displaystyle{\lim_{n \to \infty}2^n \frac {\sqrt{2-\underbrace{\sqrt{2+\sqrt{2+\sqrt{2+\cdots+\sqrt{2}}}}}}_{\text{n square roots}}} {2}}\
\end {array}

Approximations without closer considerations are problematic. Firstly, they are no equations, and secondly, your "proof" is similar to mine which "proves" π=2:

The sum of the arc lengths is always π and they "converge" to a line of length 2.

My suggestion is to use a recursion and proper equations aka limits.

 

[fresh_42]

8 is eaaaasy. It is well known that e is transcendental. Suppose e^2 is rational. Then e is a root to the equation x^2=e^2 which can be made into one with integer coefficients by multiplying by the denominator of the rational form of e^2, contradiction ;)

In this case you will have to prove irrationality of $e$ instead. The problem requires a proof that the number $c^2$ is irrational, where $c$ is defined as $c=\sum n!^{-1}$.

 

[AndreasC]

In this case you will have to prove irrationality of e instead

I'm guessing you meant transcedentality, irrationality is not enough. But yeah, I figured, I was just joking.

 

[fresh_42]

I'm not home right now and I won't be for some time, but I'm looking at 4 and thinking

Spoiler

The solution u is stationary along the direction (x, y, sqrt(x^2+y^2)). These vectors are perpendicular to (x, y, - 1). It probably shouldn't be too hard to eliminate one of the variables starting from here by finding the characteristics, right?

Is that correct as a first step?

My solution doesn't use elimination, so I think the answer is no. I use the so called characteristic system, or characteristic flows.

 

[AndreasC]

My solution doesn't use elimination, so I think the answer is no. I use the so called characteristic system, or characteristic flows.

I think we mean the same thing, calling it eliminating variables was kinda bad phrasing. I'm not sure how to solve it to the end because I don't know much about PDEs and I've never tried to solve an equation with 3 independent variables with characteristics, but I figured it must be something like that...

 

[fresh_42]

Problem 6:

Integral 1. Write

\begin{align*}
I (\beta) = \int_{-\infty}^\infty \frac{1}{x^2 + 1} dx + \int_{-\infty}^\infty \frac{1}{x^2 + \beta} dx .
\end{align*}

Then

\begin{align*}
\frac{\partial}{\partial \beta} I (\beta) |_{\beta = 1} = \int_{-\infty}^\infty \frac{x^2}{(x^2 + 1)^2} dx .
\end{align*}

How? Is that a typo? Where does the $x^2$ in the denominator comes from if we differentiate by $\beta$?

We evaluate $I (\beta)$ the following way:

\begin{align*}
I (\beta) & = \int_{-\infty}^\infty \left( \int_0^\infty e^{- \alpha (x^2+1)} d \alpha \right) dx + \int_{-\infty}^\infty \left( \int_0^\infty e^{- \alpha (x^2+\beta)} d \alpha \right) dx \\
& = \int_0^\infty \left( \int_{-\infty}^\infty e^{- \alpha x^2} dx \right) e^{- \alpha} d \alpha + \int_0^\infty \left( \int_{-\infty}^\infty e^{- \alpha x^2} dx \right) e^{- \alpha \beta} d \alpha \\
& = \int_0^\infty \sqrt{\frac{\pi}{\alpha}} e^{-\alpha} d \alpha + \int_0^\infty \sqrt{\frac{\pi}{\alpha}} e^{-\alpha \beta} d \alpha \\
& = \sqrt{\pi} \left( 1 + \frac{1}{\sqrt{\beta}} \right) \int_0^\infty \frac{e^{-\alpha}}{\sqrt{\alpha}} d \alpha
\end{align*}

Can you elaborate the transformation to the exponential function?

Putting $u^2 = \alpha$, we obtain

\begin{align*}
I (\beta) & = \sqrt{\pi} \left( 1 + \frac{1}{\sqrt{\beta}} \right) 2 \int_0^\infty e^{-u^2} d u \\
& = \sqrt{\pi} \left( 1 + \frac{1}{\sqrt{\beta}} \right) \int_{- \infty}^\infty e^{-u^2} d u \\
& = \pi \left( 1 + \frac{1}{\sqrt{\beta}} \right)
\end{align*}

So finally

\begin{align*}
I = \frac{\partial}{\partial \beta} I (\beta) |_{\beta = 1} = \frac{\pi}{2} .
\end{align*}

Alternative evaluation of integral 1.

Evaluation via complex analysis. We write

\begin{align*}
I & = \int_{-\infty}^\infty \frac{x^2}{(x^2 + 1)^2} dx \\
& = \oint \frac{z^2}{(z + i)^2 (z - i)^2} dz \\
\end{align*}

Where we have completed the contour integral into the upper half plane. The pole in the upper half plane is $z_0 = i$. So the integral is by the residue theorem:

\begin{align*}
I & = 2 \pi i \lim_{z \rightarrow i} \left( \frac{d}{dz} \left[ \frac{(z - i)^2 z^2}{(z + i)^2 (z - i)^2} \right] \right) \\
& = 2 \pi i \lim_{z \rightarrow i} \frac{2 z (z+i)^2 - z^2 2 (z+i)}{(z+i)^4} \\
& = 2 \pi i \frac{2i (-4) - (-1) 2 (2i)}{2^4} \\
& = \frac{\pi}{2} .
\end{align*}Integral 2.

\begin{align*}
I = \int_0^{\pi / 2} \frac{1}{1 + \sin^2 t} dt = \frac{1}{4} \int_0^{2 \pi} \frac{1}{1 + \sin^2 t} dt
\end{align*}

This can be converted into a contour integral around the unit circle via $z = e^{i t}$, $dz = e^{i t}i d t$, $\sin^2 t = - \frac{1}{4} \left( z - \frac{1}{z} \right)^2$

\begin{align*}
I & = \frac{1}{4} \oint \frac{dz}{iz} \frac{1}{1 - (z - 1/z)^2/4} \\
& = \oint \frac{dz}{iz} \frac{z^2}{4 z^2 - (z^2 - 1)^2} \\
& = - \oint \frac{dz}{i} \frac{z}{z^4 - 6 z^2 + 1} \\
& = \frac{1}{4 \sqrt{2}} \oint \frac{dz}{i} \left( \frac{1}{z^2 + 2 \sqrt{2} z + 1} - \frac{1}{z^2 - 2 \sqrt{2} z + 1} \right)
\end{align*}

In the first factor has the poles are at $z_1 = - \sqrt{2} + 1$ and $z_2 = - \sqrt{2} - 1$, but only the pole $z_1$ lies inside the unit circle. In the second factor has the poles are at $z_3 = \sqrt{2} + 1$ and $z_4 = \sqrt{2} - 1$, but only the pole $z_4$ lies inside the unit circle. By the residue theorem the integral is

\begin{align*}
I & = \frac{\pi}{2 \sqrt{2}} \left( \text{Res}_{z_1} [ \frac{1}{z^2 + 2 \sqrt{2} z + 1} ] - \text{Res}_{z_4} [ \frac{1}{z^2 - 2 \sqrt{2} z + 1} ] \right) \\
& = \frac{\pi}{2 \sqrt{2}} \left( \lim_{z \rightarrow - \sqrt{2} + 1} \frac{1}{z + \sqrt{2} + 1} -
\lim_{z \rightarrow \sqrt{2} - 1} \frac{1}{z - \sqrt{2} - 1}
\right) \\
& = \frac{\pi}{2 \sqrt{2}} .
\end{align*}

 

[wrobel]

Problem 4.

Spoiler

By the characteristics method, the PDE has 2 solutions
$$F=y^2\Big(\frac{1}{2}-\frac{z}{x^2+y^2}\Big),\quad H=y/x.$$
I believe that the solution to the problem can be found as $c_1H+c_2F$.
Actually any solution to this PDE is presented as follows $u(x,y,z)=w(F,H)$

 

[julian]

How? Is that a typo? Where does the $x^2$ in the denominator comes from if we differentiate by $\beta$?

Can you elaborate the transformation to the exponential function?

Sorry, I should have written:

\begin{align*}
I (\beta) = \beta \int_{-\infty}^\infty \frac{1}{x^2 + 1} dx + \int_{-\infty}^\infty \frac{1}{x^2 + \beta} dx .
\end{align*}

I missed off a $\beta$ in front of the first integral, it doesn't alter the end result. Then I used that

\begin{align*}
\frac{\partial}{\partial \beta} I (\beta) |_{\beta = 1} & = \int_{-\infty}^\infty \frac{1}{x^2 + 1} dx + \int_{-\infty}^\infty \frac{\partial}{\partial \beta} \frac{1}{x^2 + \beta} |_{\beta = 1} \\
& = \int_{-\infty}^\infty \frac{1}{x^2 + 1} dx - \int_{-\infty}^\infty \frac{1}{(x^2 + 1)^2} dx \\
& = \int_{-\infty}^\infty \frac{x^2}{(x^2 + 1)^2} dx .
\end{align*}I'm then writing

\begin{align*}
\frac{1}{(x^2 + 1)} = \int_0^\infty e^{- \alpha (x^2+1)} d \alpha
\end{align*}

and inserting it into the $x$ integral. I then perform the integration over $x$; it is a simple Gaussian integral. The final integral (over $\alpha$) is then easily performed by substitution.

 

[julian]

Thanks @fresh_42. I'll just go back and edit my post now!

 

[Antarres]

Guess I'll take my attempt at problem 2. though the previous post hinted the idea which I was satisfied to discover earlier, so it just has to get rigorous.

Spoiler: Problem #2

We will use the following notation:
$$\Delta_n = \underbrace{\sqrt{2-\sqrt{2+\sqrt{2+\dots \sqrt{2}}}}}_{\text{n roots}}$$
$$a_n = \underbrace{\sqrt{2+\sqrt{2+\dots \sqrt{2}}}}_{\text{n roots}}$$
So we have the relation:
$$\Delta_{n+1} = \sqrt{2-a_n}$$
for $n\geq 1$.
We want to establish first a recursion between $\Delta_n$ and $\Delta_{n+1}$ for $n \geq 2$. We have an obvious recursion for $a_n$:
$$a_{n+1} = \sqrt{2+a_n}$$
Hence:
$$\Delta_{n+1} = \sqrt{2-a_n} = \sqrt{2-\sqrt{2+a_{n-1}}} = \sqrt{2-\sqrt{2+2-\Delta_n^2}} = \sqrt{2-\sqrt{4-\Delta_n^2}}$$
Now, it is easy to check that as $n \rightarrow \infty$, $a_n \rightarrow 2$ and $\Delta_n \rightarrow 0$. We won't prove this, but this inspires us to claim:
$$\sin \frac{\pi}{2^{n+1}} = \frac{\Delta_n}{2}$$
for $n \geq 2$.
We will prove this by induction. The base of induction is proven using half-angle formula:
$$\sin \frac{\pi}{8} = \sqrt{\frac{1-\cos\tfrac{\pi}{4}}{2}} = \sqrt{\frac{1-\tfrac{\sqrt{2}}{2}}{2}} = \frac{\sqrt{2-\sqrt{2}}}{2} = \frac{\Delta_2}{2}$$
In the induction step, we assume that the claim holds for some $n \geq 2$, and we prove it for $n+1$:
$$\sin \frac{\pi}{2^{n+2}} = \sqrt{\frac{1- \cos \tfrac{\pi}{2^{n+1}}}{2}} = \sqrt{\frac{1- \sqrt{1 - \sin^2 \tfrac{\pi}{2^{n+1}}}}{2}}$$
We use the induction hypothesis to simplify the previous expression, along with the recursion we established earlier:
$$\sin \frac{\pi}{2^{n+2}} = \sqrt{\frac{1- \sqrt{1 - \tfrac{\Delta_n^2}{4}}}{2}} = \sqrt{\frac{1- \tfrac{\sqrt{4 - \Delta_n^2}}{2}}{2}} = \frac{\sqrt{2-\sqrt{4-\Delta_n^2}}}{2} = \frac{\Delta_{n+1}}{2}$$
Hence by induction, our claim holds. We substitute it into the limit:
$$\lim_{n\rightarrow\infty} 2^n\Delta_{n} = \lim_{n\rightarrow\infty} 2^n \cdot 2\sin \frac{\pi}{2^{n+1}} = \lim_{n\rightarrow\infty} \pi \frac{\sin \tfrac{\pi}{2^{n+1}}}{\tfrac{\pi}{2^{n+1}}} = \pi$$
where we have used the known relation that $\tfrac{\sin x}{x} \rightarrow 1$ as $x \rightarrow 0$(we have used that function of sequence approaches the function of its limit, but I think that is standard knowledge, we're not dealing with very wild functions in this example). That concludes the proof.

 

[fishturtle1]

Question 9

Spoiler

Lemma: Let $G$ be a group of order $12 = 4\cdot 3$. Then $G$ is solvable.

Proof: Let $n_p$ denote the number of Sylow $p$ subgroups of $G$. By Sylow's theorems, $n_3 = 1 \text{(mod 3)}$ and $n_3 \vert 4$. So, $n_3 = 1$ or $4$. If $n_3 = 1$, then there is a unique, normal Sylow $3$ subgroup $P$ of $G$. We see $G \rhd P \rhd 1$ is a solvable series. On the other hand, if $n_3 = 4$, then there's $8$ distinct elements of order $3$. There's $4$ unaccounted for elements, and so $n_2 = 1$. So, there is a unique, normal Sylow $2$ subgroup $Q$ of $G$. We see $G \rhd Q \rhd 1$ is a solvable series. []

Now we prove 9:

1) Proof: By Sylow's theorem, $n_q = 1 \text{(mod q)}$ and $n_q \vert p^2$. There is no normal Sylow $q$ subgroup, so $n_q \neq 1$. If $n_q = p$, then $p = 1 \text{(mod q)}$ which contradicts our assumption that $p < q$. So, $n_q = p^2$. In particular, $p^2 = 1 \text{(mod q)}$. Equivalently, $q \vert (p^2 - 1)$. Factoring, we have $q \vert (p-1)$ or $q \vert (p+1)$. Since $p < q$, we must have $q \vert (p+1)$ and in particular $q = p + 1$. Since $p, q$ are primes, we must have $(p,q) = (2,3)$.

So, we have $\vert G \vert = 2^2 \cdot 3^b$. Let $Q$ be a Sylow $3$ subgroup of $G$. Then $[G:Q] = 4$. So, there is a homomorphism $f: G \rightarrow S_4$ with $\ker f \le Q$. Since $Q \neq G$, we have $\ker f \neq G$ which implies $f$ is not trivial. Moreover, $\vert G/\ker f \vert \vert 4!$. It follows $\vert \ker f\vert \ge 3^{b-1}$. But $\vert \ker f \vert \le 3^{b-1}$, otherwise $Q$ is normal. So, $\vert \ker f \vert = 3^{b-1}$. In particular, $G/\ker f$ has order $12$. So, $G/\ker f$ is solvable and $\ker f$ is solvable. It follows that $G$ is solvable. []

Proof: 2) On the other hand, if $Q$ is normal, then $G/Q$ is solvable, and $Q$ is solvable, and so $G$ is solvable. []
[\Spoiler]

 

[fresh_42]

Question 9

Lemma: Let $G$ be a group of order $12 = 4\cdot 3$. Then $G$ is solvable.

Proof: Let $n_p$ denote the number of Sylow $p$ subgroups of $G$. By Sylow's theorems, $n_3 = 1 \text{(mod 3)}$ and $n_3 \vert 4$. So, $n_3 = 1$ or $4$. If $n_3 = 1$, then there is a unique, normal Sylow $3$ subgroup $P$ of $G$. We see $G \rhd P \rhd 1$ is a solvable series. On the other hand, if $n_3 = 4$, then there's $8$ distinct elements of order $3$. There's $4$ unaccounted for elements, and so $n_2 = 1$. So, there is a unique, normal Sylow $2$ subgroup $Q$ of $G$. We see $G \rhd Q \rhd 1$ is a solvable series. []

Now we prove 9:

1) Proof: By Sylow's theorem, $n_q = 1 \text{(mod q)}$ and $n_q \vert p^2$. There is no normal Sylow $q$ subgroup, so $n_q \neq 1$. If $n_q = p$, then $p = 1 \text{(mod q)}$ which contradicts our assumption that $p < q$. So, $n_q = p^2$. In particular, $p^2 = 1 \text{(mod q)}$. Equivalently, $q \vert (p^2 - 1)$. Factoring, we have $q \vert (p-1)$ or $q \vert (p+1)$. Since $p < q$, we must have $q \vert (p+1)$ and in particular $q = p + 1$. Since $p, q$ are primes, we must have $(p,q) = (2,3)$.

So, we have $\vert G \vert = 2^2 \cdot 3^b$. Let $Q$ be a Sylow $3$ subgroup of $G$. Then $[G:Q] = 4$. So, there is a homomorphism $f: G \rightarrow S_4$ with $\ker f \le Q$.

Here is where you lost me. How do you define $f$, and why is its kernel in $Q$?

Since $Q \neq G$, we have $\ker f \neq G$ which implies $f$ is not trivial. Moreover, $\vert G/\ker f \vert \vert 4!$. It follows $\vert \ker f\vert \ge 3^{b-1}$.

While I could answer the former question on my own, it's here where I gave up. Say $k:=\dim \operatorname{ker} f$. Then $\dfrac{4\cdot 3^b}{k}\vert 24$ or $3^{b-1}\cdot a= 2k$ for some $a$. Oh, back on the track! Since $2$ is prime, it must divide $a,$ so $k=3^{b-1}\cdot a'\geq 3^{b-1}$.

But $\vert \ker f \vert \le 3^{b-1}$, otherwise $Q$ is normal. So, $\vert \ker f \vert = 3^{b-1}$. In particular, $G/\ker f$ has order $12$. So, $G/\ker f$ is solvable and $\ker f$ is solvable. It follows that $G$ is solvable. []

Proof: 2) On the other hand, if $Q$ is normal, then $G/Q$ is solvable, and $Q$ is solvable, and so $G$ is solvable. []
[\Spoiler][/Spoiler]

Ok, got it, except that I do not know whether your $f$ is the same I have in mind, since you did not define it. Only one remark: the orbit-stabilizer formula for the cardinality of the orbits - if your $f$ was mine - would have been much quicker to conclude (without your Lemma, and proving $f \neq 1$ en passant).

Oh, another remark: I am not sure whether it is my age, lack of practice, or simply the time of day night time, or all of it, but please be a bit more detailed, so that I do not have to scribble so much. It would also help other readers to follow the proof, especially the algebra proofs, since they are not that famous on PF.

 

[fishturtle1]

Thank you for your time and feedback; and sorry for confusion!

Here are two things I would add to make post #20 clearer:

Lemma: Let $G$ be a group and $H$ a subgroup. If $[G: H] = n$, then there exists a homomorphism $f : G \rightarrow S_n$ with $\ker f \le H$.

Proof: Let $X$ be the set of cosets of $H$ in $G$. For each $g \in G$, define $\sigma_g : X \rightarrow X$ by $aX \mapsto (ga)H$. We see $(\sigma_g \circ \sigma_{g^{-1}})(aH) = aH = (\sigma_{g^{-1}} \circ \sigma_g)(aH)$ for all $a \in G$. So, $\sigma_g$ is invertible. Equivalently, $\sigma_g$ is a permutation on the set $X$.

Define a function $f : G \rightarrow S_X$ by $g \mapsto \sigma_g$. For any $g, h, a \in G$, we have $\sigma_{gh}(aH) = (gha)(H) = g(ha)H = (\sigma_g \circ \sigma_h)(aH)$. So, $f$ is a homomorphism. Let $k \in \ker f$. Then $\sigma_k(H) = kH = H$. This implies $k \in H$. So, $\ker f \le H$. []
Lemma: Let $G$ be a group and $H$ be a subgroup. If $G/H$ and $H$ are solvable, then $G$ is solvable.

Proof: By assumption, $G/H$ has a solvable series
$$G/H = G_1/H \rhd G_2/H \rhd \dots \rhd G_k/H = H/H$$
Then we have
$$G = G_1 \rhd G_2 \rhd \dots \rhd G_k = H$$
For all $i$, $G_i/G_{i+1} \cong (G_i/H)/(G_{i+1}/H)$ is abelian. By assumption $H$ is solvable, and also has a solvable series
$$H = H_1 \rhd H_2 \rhd \dots \rhd H_m = 1$$
Combining series we have
$$G = G_1 \rhd G_2 \rhd \dots \rhd G_k = H = H_1 \rhd H_2 \rhd \dots \rhd H_m = 1$$
which is a solvable series for $G$. We may conclude $G$ is solvable. []

The first lemma show's why the $f$ in post 20 exists, and the second lemma is for the proof of part 2.

 

[fresh_42]

Thank you for your time and feedback; and sorry for confusion!

No problem. My homomorphism was indeed another one: The group conjugation operates transitive on the four 3-Sylow subgroups: number them and we have a homomorphism.

 

[wrobel]

Problem 8: may I use the fact that $e$ is a transcendental number? :)

 

[Lament]

Approximations without closer considerations are problematic. Firstly, they are no equations, and secondly, your "proof" is similar to mine which "proves" π=2:

The sum of the arc lengths is always π and they "converge" to a line of length 2.

This is the first time I have seen this. Can you give me some idea behind why such an erroneous result happens while we use approximations @fresh_42?

Is it because the derivative at those crossings of the arc with the diameter is not continuous as $n\to\infty$?

My suggestion is to use a recursion and proper equations aka limits.

I think @Antarres already did so I will leave it at that.

 

[fresh_42]

Problem 8: may I use the fact that $e$ is a transcendental number? :)

See post #11 and its correction in post #12.

 

[fresh_42]

This is the first time I have seen this. Can you give me some idea behind why such an erroneous result happens while we use approximations @fresh_42?

It is because the closer the curves get, the more errors we add up. Both cancel each other so that the initial difference remains the same. You can do the same thing with diagonals in a square to "prove $\sqrt{2}=2$".

The problem is the addition of errors. I'm not saying your solution did this, I only wanted to demonstrate that one has to be careful with approximations.

 

[wrobel]

See post #11 and its correction in post #12.

for what purpose? The solution of the problem is a trivial consequence of cited above feature of $e$. Assume that $e^2$ is rational that is $e^2=m/n,\quad m,n\in\mathbb{N}$. Thus $e$ is a root of polynomial $x^2=m/n$. Contradiction.

 

[fresh_42]

for what purpose? The solution of the problem is a trivial consequence of cited above feature of $e$. Assume that $e^2$ is rational that is $e^2=m/n,\quad m,n\in\mathbb{N}$. Thus $e$ is a root of polynomial $x^2=m/n$. Contradiction.

Sure, and that's already been discussed.

 

[wrobel]

Sure, and that's already been discussed.

ok

 

[AndreasC]

If I wanted to pretend I knew how to solve problem 8, I would look up a proof that e is transcendental and repackage it so that it only seems to cover the special case. But that would be lame...

 

[wrobel]

I also do not know how to solve #8 directly
If anybody could recover any theorem in five minutes then math would not be needed. There are classical facts which are easy to prove such as Gronwall inequality (Probl 3) and there are nontrivial facts. For example I can not imagine that anybody will solve #5 ex nihilo. But it easily follows from the standard ODE theorysorry for offtop

 

[StoneTemplePython]

7. Prove the following well known theorem by using topological and analytical tools only.
For every real symmetric matrix $A$ there is a real orthogonal matrix $Q$ such that $Q^\tau AQ$ is diagonal.

my read is compactness plus well chosen inequalities are fine. Is it fair game to use that if $A= \begin{bmatrix} a_{1,1}& a_{1,2}\\ a_{1,2} & a_{2,2}\\ \end{bmatrix}$
with positive diagonals, then $a_{1,1}\cdot a_{2,2} -a_{1,2}^2\leq a_{1,1}\cdot a_{2,2}$ with equality iff $A$ is diagonal? I ask because this is a determinant inequality (Hadarmard's, though I only want the obvious 2x2 case) and perhaps 2x2 determinants are out of bounds.

 

[fresh_42]

my read is compactness plus well chosen inequalities are fine. Is it fair game to use that if $A= \begin{bmatrix} a_{1,1}& a_{1,2}\\ a_{1,2} & a_{2,2}\\ \end{bmatrix}$
with positive diagonals, then $a_{1,1}\cdot a_{2,2} -a_{1,2}^2\leq a_{1,1}\cdot a_{2,2}$ with equality iff $A$ is diagonal? I ask because this is a determinant inequality (Hadarmard's, though I only want the obvious 2x2 case) and perhaps 2x2 determinants are out of bounds.

I don't see how you get along with $n=2$, i.e. I fear I'll have to read a nasty induction ...
Nevertheless, coordinates will be necessary and the off diagonal elements are the key.

 

[StoneTemplePython]

I don't see how you get along with $n=2$, i.e. I fear I'll have to read a nasty induction ...
Nevertheless, coordinates will be necessary and the off diagonal elements are the key.

I am aware of the compactness proof of Wilf but I find all the indices involved to be unpleasant... so I am trying to figure out a slicker approach that yes hides a lot of ugliness via a (nice?) induction.