Challenge Math Challenge - July 2020

[fresh_42]

1. (solved by @nuuskur ) Let $V$ be an infinite dimensional topological vector space. Show that the weak topology on $V$ is not induced by a norm. (MQ)

2. The matrix groups $U(n)$ and $SL_n(\mathbb{C})$ are submanifolds of $\mathbb{C}^{n^2}=\mathbb{R}^{2n^2}$. Do they intersect transversely? (IR)

3. (solved by @zinq ) Let $f:\mathbb{R}\to \mathbb{R},\quad f>0$ be a continuous and $1-$periodic function. Show that
$$\int_0^1\frac{f(x+a)}{f(x)}dx\ge 1$$ for any $a\in\mathbb{R}$. (WR)

4. (solved by @nuuskur ) Prove that $C[0,1]$ is not dual to a Banach space. (WR)

5. (solved by @julian , @mathwonk ) Let $(M,g)$ be a Riemannian manifold. Let $f:M\to\mathbb{R}$ be a smooth function such that $|\nabla f|=1$ everywhere on $M$. Show that all integral curves of $\nabla f$ are geodesics. (IR)

6. (solved by @Incand ) Let $f: (a,b) \to \mathbb{R}$ be a continuous function that is midpoint convex, i.e. $f(\frac{1}{2}x + \frac{1}{2}y) \leq \frac{1}{2}f(x) + \frac{1}{2}f(y)$ for all $x,y \in (a,b)$ . Show that $f$ is convex, i.e. $f(tx + (1-t)y) \leq tf(x) + (1-t)f(y)$ for all $x,y \in (a,b)$ and $0 \leq t \leq 1$. (MQ)

7. (solved by @nuuskur ) Consider $D:=\{f: \mathbb{R} \to \mathbb{R}\mid \mathrm{\ f \ is \ not \ continuous}\}$. What is the cardinality of $D$? (MQ)

8. (solved by @zinq ) Let $X$ be a compact manifold such that $\pi_1(X)$ (the fundamental group of $X$) is finite and nontrivial. Show that $\pi_k(X)$ is also non-trivial for some $k\geq 2.$ (IR)

9. (solved by @mathwonk ) Let $(X,d)$ be a compact metric space and $f:X\to X$ be a mapping onto. Assume that $d(f(x),f(y))\le d(x,y),\quad \forall x,y\in X.$ Show that $d(f(x),f(y))= d(x,y),\quad \forall x,y\in X.$ (WR)

10. (solved by @etotheipi ) Calculate the electrostatic potential $U(a)$ of a surface $S=\{\,(x,y,z)\in \mathbb{R}^3\,|\,x^2+y^2=z^2,\,0\leq z\leq 1\,\}$ charged with a field of homogeneous density $\rho$ at the point $a=(0,0,1)$. (FR)

High Schoolers only11. (solved by @Isaac0427 , @ItsukaKitto ) Prove that the product of a finite number of sums of two integers squares is again a sum of two integers squared.
$$
(a_1^2+b_1^2)\cdot (a_2^2+b_2^2)\cdot \ldots \cdot (a_n^2+b_n^2)=a^2+b^2
$$

12. (solved by @etotheipi ) Given a positive integer in decimal representation without zeros. We build a new integer by concatenation of the number of even digits, the number of odd digits, and the number of all digits (the sum of the former two). Then we proceed with that number.
Determine whether this algorithm always comes to a halt. What is or should be the criterion to stop?

13. (solved by @Lament ) List all real functions $f\, : \,\mathbb{R}\longrightarrow \mathbb{R}$ with the following properties:
\begin{align*}
f(xy)&=f(x)f(y)-f(x)-f(y)+2\\
f(x+y)&=f(x)+f(y)+2xy-1\\
f(1)&=2
\end{align*}

14. (solved by @ItsukaKitto ) Find all real solutions $(x,y)$ such that
$$
\sin^4x = y^4+x^2y^2-4y^2+4\, , \,\cos^4x=x^4+x^2y^2-4x^2+1
$$

15. (solved by @etotheipi , @Adesh) Prove
$$
\dfrac{(2n)!}{(n!)^2}>\dfrac{4^n}{n+1}
$$
for all natural numbers $n>1.$

 

[etotheipi]

Spoiler: Problem 15

The statement holds for $n=2$, now assume for $n=k$ that$$\dfrac{(2k)!}{(k!)^2}>\dfrac{4^k}{k+1}$$Then for $n=k+1$,$$\dfrac{(2(k+1))!}{((k+1)!)^2} = \frac{2(2k+1)}{k+1} \dfrac{(2k)!}{(k!)^2} > \frac{2(2k+1)}{k+1} \dfrac{4^k}{k+1} = \frac{(2\times 4^k)(2k+1)}{(k+1)^2}$$Now consider the expression $$\frac{(2\times 4^k)(2k+1)}{(k+1)^2} - \frac{4^{k+1}}{k+2} = 2 \times 4^k \left (\frac{2k+1}{(k+1)^2} - \frac{2}{k+2} \right) = 2 \times 4^k \left( \frac{k}{(k+1)^2 (k+2)} \right) > 0$$ since $k>1$. That means that $$\dfrac{(2(k+1))!}{((k+1)!)^2} > \frac{4^{k+1}}{(k+1)+1}$$so if the statement is true for $n=k$ it also holds for $n=k+1$. Hence the statement is true for all $n \in \mathbb{N}$

 

[Incand]

Spoiler: Question 6

We study the inequality
\begin{equation}
f(tx+(1-t)y) \le tf(x)+(1-t)f(y).
\end{equation}
Take points $x,y \in (a,b)$. We know the inequality is satisfied for $t=1/2$.

By using the midpoint inequality for the points $x,y,(x+y)/2$ we also get that the inequality is true for $t=1/4,3/4$. Proceeding iteratively this way we get that the inequality is true for any $t$ of the form
\begin{equation}
t = \frac{m}{2^n},
\end{equation}
for $m,n\in \mathbb{Z}^+$ with $m\le 2^n$. (Cf. binary search.)

Next we show that the fractions of the above form are dense in $[0,1]$. For any $r\in [0,1]$ and $n\in \mathbb{Z}^+$ we have with an appropriate choice of $m$ that
\begin{equation}
|r-\frac{m}{2^n}| \le \frac{1}{2^n},
\end{equation}
which we can make arbitrary small by increasing $n$.

Since $\mathbb{R}$ is Hausdorff, $f$ is determined by its valued on a dense subset. This forces the inequality for all $t\in [0,1]$.

 

[member 587159]

Spoiler: Question 6

We study the inequality
\begin{equation}
f(tx+(1-t)y) \le tf(x)+(1-t)f(y).
\end{equation}
Take points $x,y \in (a,b)$. We know the inequality is satisfied for $t=1/2$.

By using the midpoint inequality for the points $x,y,(x+y)/2$ we also get that the inequality is true for $t=1/4,3/4$. Proceeding iteratively this way we get that the inequality is true for any $t$ of the form
\begin{equation}
t = \frac{m}{2^n},
\end{equation}
for $m,n\in \mathbb{Z}^+$ with $m\le 2^n$. (Cf. binary search.)

Next we show that the fractions of the above form are dense in $[0,1]$. For any $r\in [0,1]$ and $n\in \mathbb{Z}^+$ we have with an appropriate choice of $m$ that
\begin{equation}
|r-\frac{m}{2^n}| \le \frac{1}{2^n},
\end{equation}
which we can make arbitrary small by increasing $n$.

Since $\mathbb{R}$ is Hausdorff, $f$ is determined by its valued on a dense subset. This forces the inequality for all $t\in [0,1]$.

Yes! Well done! The key is indeed to realize that the so called dyadic rationals
$$D:= \left\{\frac{m}{2^n}\mid 0 \leq n; 0 \leq m \leq 2^n\right\}$$ are dense in $[0,1]$.

 

[Isaac0427]

Summary:: Functional Analysis, Topology, Differential Geometry, Analysis, Physics
Authors: Math_QED (MQ), Infrared (IR), Wrobel (WR), fresh_42 (FR).

High Schoolers only

I just graduated high school...do I count?

 

[Delta2]

I find the formulation of 10. a bit strange. @fresh_42 Does it mean to find the potential U(a) at point a, created by the given surface that is charged with a uniform surface charge density $\rho$?
Also for 2. what's the definition of 1-periodic.

 

[Adesh]

Also for 2. what's the definition of 1-periodic.

I confirmed it with some people and they said it means a function with period as $1$, that is $f(x+1)=f(x)$.

 

[member 587159]

I just graduated high school...do I count?

Yes. Definitely.

 

[etotheipi]

Spoiler: Problem 10

I'm pretty sure this is massively incorrect, but in any case I'll write it up.

A ring-shaped surface element on the cone at a height $z$ has charge $\rho dA = 2\sqrt{2} \pi \rho z dz$, and the distance $d$ from this element to the point $a$ is $d = \sqrt{2z^2 -2z + 1}$. That means $$dU = \frac{2\sqrt{2} \pi \rho z}{4\pi \epsilon_0 \sqrt{2z^2 -2z + 1}}dz$$ $$U = \frac{\rho}{2 \epsilon_0} \int_0^1 \frac{z}{\sqrt{z^2 -z+\frac{1}{2}}} dz$$Then we let $u = z-\frac{1}{2}$, so that$$\begin{align*}

U = \frac{\rho}{2 \epsilon_0} \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{u}{\sqrt{u^2 + \frac{1}{4}}} + \frac{\frac{1}{2}}{\sqrt{u^2 + \frac{1}{4}}} \, du &= \frac{\rho}{2\epsilon_0} \left[\text{arsinh}({2u}) \right]_{0}^{\frac{1}{2}} \\

&= \frac{\rho}{2\epsilon_0} \text{arsinh}(1)

\end{align*}
$$

 

[fresh_42]

Spoiler: Problem 10

I'm pretty sure this is massively incorrect, but in any case I'll write it up.

A ring-shaped surface element on the cone at a height $z$ has charge $\rho dA = 2\sqrt{2} \pi \rho z dz$, and the distance $d$ from this element to the point $a$ is $d = \sqrt{2z^2 -2z + 1}$. That means $$dU = \frac{2\sqrt{2} \pi \rho z}{4\pi \epsilon_0 \sqrt{2z^2 -2z + 1}}dz$$ $$U = \frac{\rho}{2 \epsilon_0} \int_0^1 \frac{z}{\sqrt{z^2 -z+\frac{1}{2}}} dz$$Then we let $u = z-\frac{1}{2}$, so that$$\begin{align*}

U = \frac{\rho}{2 \epsilon_0} \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{u}{\sqrt{u^2 + \frac{1}{4}}} + \frac{\frac{1}{2}}{\sqrt{u^2 + \frac{1}{4}}} \, du &= \frac{\rho}{2\epsilon_0} \left[\text{arsinh}({2u}) \right]_{0}^{\frac{1}{2}} \\

&= \frac{\rho}{2\epsilon_0} \text{arsinh}(1)

\end{align*}
$$

I consider the density includes permittivity, but this isn't important. What is your result?

 

[etotheipi]

I consider the density includes permittivity, but this isn't important. What is your result?

If the density includes the permittivity, then I get about $U = \frac{1}{2}\text{arsinh}(1)\rho \approx 0.441 \rho$...

 

[fresh_42]

If the density includes the permittivity, then I get about $U = \frac{1}{2}\text{arsinh}(1)\rho \approx 0.441 \rho$...

I meant the exact real number. Your potential integral looks ok, modulo constant factors, but what is $x$ in $U(a)=x\cdot \rho\;$?

$0.4...$ is wrong, but as you didn't show anything I cannot tell where or whether you were mistaken anywhere. It looks as if you failed on the lower bound of the integral.

 

[etotheipi]

I meant the exact real number. Your potential integral looks ok, modulo constant factors, but what is $x$ in $U(a)=x\cdot \rho\;$?

$0.4...$ is wrong, but as you didn't show anything I cannot tell where or whether you were mistaken anywhere.

I guess in that case I would have $x = \frac{\ln{(1+\sqrt{2}})}{2}$... it's very likely I've messed up so I'll have another look. Numerically that's the same as what I get if I calculate the first integral involving $z$

 

[fresh_42]

I guess in that case I would have $x = \frac{\ln{(1+\sqrt{2}})}{2}$... it's very likely I've messed up so I'll have another look.

Close ... but wrong denominator and missing $\pi$.

I hope I didn't make the mistake. But where did you get the zero from in the lower bound?

 

[etotheipi]

and missing $\pi$.

That is strange... for me the $\pi$'s cancel

I hope I didn't make the mistake. But where did you get the zero from in the lower bound?

For that I just used the fact that the second term in the integrand is an even function.

 

[fresh_42]

That is strange... for me the $\pi$'s cancel
For that I just used the fact that the second term in the integrand is an even function.

The $\pi$ comes in by the parametrization of the surface. O.k. it may cancel depending on the definition of $\rho$. I used the formula
$$
U(a)=\int\int_S \dfrac{\rho}{|x-a|}\,dO
$$
$\operatorname{arsinh}$ isn't even and I haven't your second term at all. But as you didn't show us what you have done, we discuss it here instead.

 

[etotheipi]

$\operatorname{arsinh}$ isn't even.

I meant$$\begin{align*}
U &= \frac{\rho}{2 \epsilon_0} \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{u}{\sqrt{u^2 + \frac{1}{4}}} + \frac{\frac{1}{2}}{\sqrt{u^2 + \frac{1}{4}}} \, du \\

&= \frac{\rho}{2 \epsilon_0} \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{\frac{1}{2}}{\sqrt{u^2 + \frac{1}{4}}} \, du \\

&= \frac{\rho}{2 \epsilon_0} \int_{0}^{\frac{1}{2}} \frac{1}{\sqrt{u^2 + \frac{1}{4}}} \, du \\

&= \frac{\rho}{2\epsilon_0} \left[\text{arsinh}({2u}) \right]_{0}^{\frac{1}{2}}

\end{align*}
$$because $\frac{\frac{1}{2}}{\sqrt{u^2 + \frac{1}{4}}}$ is even and $\frac{u}{\sqrt{u^2 + \frac{1}{4}}}$ is odd.

 

[fresh_42]

I have $U(a)=2\,\pi\rho\,\int_0^1 \dfrac{t}{\sqrt{t^2-t+\frac{1}{2}}}\,dt \,,$ too, but then we differ. Maybe I used a different split than you. I have no idea what you did between the above integral and the $u-$substitution.

 

[fresh_42]

I meant$$\begin{align*}
U &= \frac{\rho}{2 \epsilon_0} \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{u}{\sqrt{u^2 + \frac{1}{4}}} + \frac{\frac{1}{2}}{\sqrt{u^2 + \frac{1}{4}}} \, du \\

&= \frac{\rho}{2 \epsilon_0} \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{\frac{1}{2}}{\sqrt{u^2 + \frac{1}{4}}} \, du \\

&= \frac{\rho}{2 \epsilon_0} \int_{0}^{\frac{1}{2}} \frac{1}{\sqrt{u^2 + \frac{1}{4}}} \, du \\

&= \frac{\rho}{2\epsilon_0} \left[\text{arsinh}({2u}) \right]_{0}^{\frac{1}{2}}

\end{align*}
$$because $\frac{\frac{1}{2}}{\sqrt{u^2 + \frac{1}{4}}}$ is even and $\frac{u}{\sqrt{u^2 + \frac{1}{4}}}$ is odd.

As you basically solved it and we are only discussing the integration, would you agree if I gave you the credits and post my calculation instead, so that people can read the solution in one step instead of spread all over the place?

 

[etotheipi]

As you basically solved it and we are only discussing the integration, would you agree if I gave you the credits and post my calculation instead, so that people can read the solution in one step instead of spread all over the place?

I don't mind , but I think we agree on the integration of $\int_0^1 \dfrac{t}{\sqrt{t^2-t+\frac{1}{2}}}\,dt$, it's just that your leading factor was $2\pi \rho$ and mine was $\frac{\rho}{2}$ (or with the $\epsilon_0$ added in, $\frac{\rho}{2\epsilon_0}$).

 

[fresh_42]

The general formula (Coulomb) for the potential is
$$
U(a)=\int\int_S \dfrac{\rho}{|x-a|}\,dO
$$
With the parameterization $\Phi(t,\varphi)=(t\cos\varphi,t\sin\varphi,t)$ we get $\Phi_t=(\cos \varphi,\sin \varphi ,1)$ and $\Phi_\varphi=(-t\sin \varphi,t\cos\varphi,0)$. The fundamental quantities are $$E=\Phi_t\cdot \Phi_t\, , \,F=\Phi_t\cdot \Phi_\varphi\, , \,G=\Phi_\varphi \cdot \Phi_\varphi$$
These are in our case $E=2,\,F=0,\,G=t^2$ and the scalar surface element is $dO=\sqrt{EG-F^2}\,dt\,d\varphi= \sqrt{2}t\,dt\,d\varphi$ since $t\geq 0.$ Thus
\begin{align*}
U(a)&=\int_0^1 \int_0^{2\pi} \dfrac{\rho}{\sqrt{(t\cos \varphi - 0)^2+(t\sin\varphi-0)^2+(t-1)^2}} \sqrt{2}t\,dt\,d\varphi \\
&=\int_0^1 dt\int_0^{2\pi}d\varphi\, \,\dfrac{\sqrt{2}\rho\, t}{\sqrt{t^2+(t-1)^2}}=2\,\pi\rho\,\int_0^1 \dfrac{t}{\sqrt{t^2-t+\frac{1}{2}}}\,dt\\
&=2\,\pi\rho\,\int_0^1 \dfrac{1}{2}\left(\dfrac{2t-1}{\sqrt{t^2-t+\frac{1}{2}}} + \dfrac{1}{2\sqrt{t^2-t+\frac{1}{2}}}\right)\\
&=\pi\rho \left[\sqrt{t^2-t+\frac{1}{2}}\right]_0^1 +\pi\rho \int_0^1\dfrac{dt}{\sqrt{t^2-t+\frac{1}{2}}}\\
&\stackrel{\tau=t-1/2}{=}\pi\rho \int_{-1/2}^{1/2}\dfrac{d\tau}{\sqrt{\tau^2+\frac{1}{4}}}=\pi\rho \int_{-1/2}^{1/2}\dfrac{d(2\tau)}{\sqrt{(2\tau)^2+1}}\\
&=\pi\rho \left[\operatorname{arsinh}(2\tau)\right]_{-1/2}^{1/2}=\pi\rho \left[\log\left(2\tau + \sqrt{(2\tau)^2+1}\right)\right]_{-1/2}^{1/2}\\
&=\pi\rho \log\left(\dfrac{1+\sqrt{2}}{-1+\sqrt{2}}\right)=(\log(3+2\sqrt{2}))\pi\rho
\end{align*}

 

[etotheipi]

The general formula (Coulomb) for the potential is
$$
U(a)=\int\int_S \dfrac{\rho}{|x-a|}\,dO
$$

I think it is because I am using the form of Coulomb's law equivalent to $$U(a) = \int \int_S \frac{\rho}{4 \pi \epsilon_0 |x - a|} dO$$That would explain the factor of $4$ between our answers, in addition to the un-cancelled $\pi$ in yours.

 

[fresh_42]

I think it is because I am using the form of Coulomb's law equivalent to $$U(a) = \int \int_S \frac{\rho}{4 \pi \epsilon_0 |x - a|} dO$$That would explain the factor of $4$ between our answers, in addition to the un-cancelled $\pi$ in yours.

Sure, but I think you've made an integration error anyway. You skipped what was my lines 3 and 4 in the above calculation and that's where the problems with your solution started. As a result we had different lower limits in the final term.

 

[etotheipi]

Sure, but I think you've made an integration error anyway. You skipped what was my lines 3 and 4 in the above calculation and that's where the problems with your solution started. As a result we had different lower limits in the final term.

When I do $$I = \int_0^1 \frac{z}{\sqrt{z^2 -z+\frac{1}{2}}} dz$$ on my calculator (which you have in line 2), I get $I = 0.881$, which is the same as $\text{arsinh}(1)$ in my answer.

And the ratio of our answers is $$\frac{\ln{(3+2\sqrt{2})}\pi \rho}{\frac{\rho}{2\epsilon_0} \text{arsinh}(1)} = 4 \pi \epsilon_0$$So I think everything just comes down to the form of Coulomb's law.

 

[fresh_42]

When I do $$I = \int_0^1 \frac{z}{\sqrt{z^2 -z+\frac{1}{2}}} dz$$ on my calculator (which you have in line 2), I get $I = 0.881$, which is the same as $\text{arsinh}(1)$ in my answer.

And the ratio of our answers is $$\frac{\ln{(3+2\sqrt{2})}\pi \rho}{\frac{\rho}{2\epsilon_0} \text{arsinh}(1)} = 4 \pi \epsilon_0$$So I think everything just comes down to the form of Coulomb's law.

I concentrated on the difference between my $\operatorname{arsinh}(1)-\operatorname{arsinh}(-1)$ and your $\operatorname{arsinh}(1)-0$. But as $\operatorname{arsinh}$ is an odd function, this difference could well be hidden in a factor $2$ and we are debating about literally nothing. A typical confusion online versus us in front of a blackboard with a piece of chalk in hand each.

 

[etotheipi]

Yes, I probably should have shown more steps given that I had quite a lot of $2$'s flying around. Effectively I just changed $\text{arsinh}(1) - \text{arsinh}(-1) \rightarrow 2\text{arsinh}(1)$.

and we are debating about literally nothing.

It's beginning to look to me too that this is the case

 

[PeroK]

Re question 10. If we use SI units first. For a ring on the surface at neight $z$ we have:
$$dq = 2\pi R \rho dl = 2\pi z \rho (\sqrt 2 dz)$$
Where $R = z$ is the radius of a ring and $dl = \sqrt 2 dz$ is the length of the ring along the surface. Therefore, we have:
$$V = \int \frac{dq}{4\pi \epsilon_0 r} = \int_0^1 \frac{2\pi z \rho (\sqrt 2 dz)}{4\pi \epsilon_0 \sqrt{z^2 + (1-z)^2}} = \frac{\rho}{2\epsilon_0}\int_0^1 \frac{z dz}{\sqrt{(z - \frac 1 2)^2 + \frac 1 4}} $$
Using $u = z - \frac 1 2$, we get:
$$V = \frac{\rho}{2\epsilon_0}\int_{- \frac 1 2}^{\frac 1 2} \frac{(u + \frac 1 2) du}{\sqrt{u^2 + \frac 1 4}} = \frac{\rho}{2\epsilon_0}\int_{0}^{\frac 1 2} \frac{du}{\sqrt{u^2 + \frac 1 4}} $$
The first term is an odd function, so the integral vanished and the second is even. This gives:
$$V = \frac{\rho}{2\epsilon_0}[\ln(\frac 1 2 + \sqrt{\frac 1 2}) - \ln(\frac 1 2)] = \frac{\rho}{2\epsilon_0}[\ln(1 + \sqrt 2)]$$
The answer in SI and cgs units should be:
$$V_{SI} = \frac{\rho}{2\epsilon_0}[\ln(1 + \sqrt 2)]$$
$$V_{cgs} = 2\pi \rho[\ln(1 + \sqrt 2)]$$

 

[Isaac0427]

This may or may not be the intended solution, but it's the first thing that came to mind when I saw the problem. Let's factor every term on the LHS:
$$(a_1^2+b_1^2)(a_2^2+b_2^2)(...)(a_n^2+b_n^2)=(a_1+b_1i)(a_1-b_1i)(a_2+b_2i)(a_2-b_2i)(...)(a_n+b_ni)(a_n-b_ni).$$
We can now regroup these terms in the following way:
$$(a_1+b_1i)(a_2+b_2i)(...)(a_n+b_ni)*(a_1-b_1i)(a_2-b_2i)(...)(a_n-b_ni).$$
The group of terms on the left is going to be some complex number with integer real and imaginary parts (assuming all $a_n$ and $b_n$ are integers). We will call this new complex number a+bi. But, if the group of terms on the left is a+bi, then the group of terms on the right will be a-bi, due to the property of complex conjugates
$$\left(z_1*z_2*...*z_n\right)^*=z_1^**z_2^**...*z_n^*.$$
Thus, our expression reduces to (a+bi)(a-bi) which is equal to $a^2+b^2$.

 

[nuuskur]

Uhh, been a while since I did cardinal arithmetic.. here goes..

Spoiler: Problem 7

Firstly, |\mathbb R^{\mathbb R}| = |\mathcal P(\mathbb R)|. Indeed, since |\mathbb R| = |\mathbb R^2| we have |\mathcal P(\mathbb R)| = |\mathcal P(\mathbb R^2)|. Given f:\mathbb R\to\mathbb R, it is, as a relation, a subset
<br /> \{(x,f(x)) \mid x\in\mathbb R\} \subseteq \mathbb R^2.<br />
On the other hand, given F\subseteq \mathbb R, we can map it to its characteristic function \chi _F, so we have a bijection and thus |\mathcal P(\mathbb R)| = |2^{\mathbb R}|. Certainly the \chi _F\in\mathbb R^{\mathbb R}, so we have
<br /> |\mathcal P(\mathbb R)|=|2^{\mathbb R}|\leq |\mathbb R^{\mathbb R}| \leq \mathcal P(\mathbb R).<br />
By Cantor-Bernstein-Schroeder, we have |\mathbb R^{\mathbb R}| = |\mathcal P(\mathbb R)|.

Let C be the subset of continuous maps. On the one hand |\mathbb R| \leq |C| since mapping to constant functions is injective. On the other hand, if two continuous maps coincide on \mathbb Q, then they coincide on \mathbb R, thus mapping f\mapsto f\vert_{\mathbb Q} is injective. Thus, |C| \leq |\mathbb R^{\mathbb Q}| = |\mathbb R|. Again, by CBS, |C| = |\mathbb R|.
By the axiom of choice, we have for infinite cardinals \kappa + \lambda = \max\{\kappa,\lambda\}, thus
<br /> |\mathcal P(\mathbb R)| = |\mathbb R^{\mathbb R} \setminus C|+|C| = \max\{|\mathbb R^{\mathbb R} \setminus C|,|C|\}.<br />
By Cantor's theorem |\mathbb R| &lt; |\mathcal P(\mathbb R)|, so it must be |D|=|\mathbb R^{\mathbb R} \setminus C| = |\mathcal P(\mathbb R)|.

 

[member 587159]

Uhh, been a while since I did cardinal arithmetic.. here goes..

Spoiler: Problem 7

Firstly, |\mathbb R^{\mathbb R}| = |\mathcal P(\mathbb R)|. Indeed, since |\mathbb R| = |\mathbb R^2| we have |\mathcal P(\mathbb R)| = |\mathcal P(\mathbb R^2)|. Given f:\mathbb R\to\mathbb R, it is, as a relation, a subset
<br /> \{(x,f(x)) \mid x\in\mathbb R\} \subseteq \mathbb R^2.<br />
On the other hand, given F\subseteq \mathbb R, we can map it to its characteristic function \chi _F, so we have a bijection and thus |\mathcal P(\mathbb R)| = |2^{\mathbb R}|. Certainly the \chi _F\in\mathbb R^{\mathbb R}, so we have
<br /> |\mathcal P(\mathbb R)|=|2^{\mathbb R}|\leq |\mathbb R^{\mathbb R}| \leq \mathcal P(\mathbb R).<br />
By Cantor-Bernstein-Schroeder, we have |\mathbb R^{\mathbb R}| = |\mathcal P(\mathbb R)|.

Let C be the subset of continuous maps. On the one hand |\mathbb R| \leq |C| since mapping to constant functions is injective. On the other hand, if two continuous maps coincide on \mathbb Q, then they coincide on \mathbb R, thus mapping f\mapsto f\vert_{\mathbb Q} is injective. Thus, |C| \leq |\mathbb R^{\mathbb Q}| = |\mathbb R|. Again, by CBS, |C| = |\mathbb R|.
By the axiom of choice, we have for infinite cardinals \kappa + \lambda = \max\{\kappa,\lambda\}, thus
<br /> |\mathcal P(\mathbb R)| = |\mathbb R^{\mathbb R} \setminus C|+|C| = \max\{|\mathbb R^{\mathbb R} \setminus C|,|C|\}.<br />
By Cantor's theorem |\mathbb R| &lt; |\mathcal P(\mathbb R)|, so it must be |D|=|\mathbb R^{\mathbb R} \setminus C| = |\mathcal P(\mathbb R)|.

Yes, well done. To see that $|\mathcal{P}(\mathbb{R})| = |\mathbb{R}^\mathbb{R}|$, one can also argue that
$$|\mathbb{R}^\mathbb{R}| = (2^{\aleph_0})^{2^{\aleph_0}} = 2^{\aleph_0. 2^{\aleph_0}} = 2^{2^{\aleph_0}}= |\mathcal{P}(\mathbb{R})|$$
where it was used that $2^{\aleph_0} = |\mathbb{R}|$.

 

[Adesh]

@wrobel Сэр, я получаю равенство, а не неравенство, here is my attempt.

Spoiler: Question 3

$f(x+a)$ has period 1, $f(x)$ has also period 1, therefore $g(x) = \frac{f(x+a)}{f(x)}$ will have a period 1.

Trigonometric Fourier Series for $g(x) = \frac{f(x+a)}{f(x)}$ is
$$
\frac{f(x+a)}{f(x)} = a_0 \sum_{n=1}^{\infty} \left[ a_n \cos (2\pi n x) + b_n \sin (2\pi nx)\right]
$$
$$
\int_{0}^{1} \frac{f(x+a)}{f(x)} = \int_{0}^{1} a_0 dx + 0 $$
$$\int_{0}^{1} \frac{f(x+a)}{f(x)} = a_0
$$
Fourier exponential series for $f(x)$ and $f(x+a)$
$$
f(x+a) = \sum_{n=-\infty}^{\infty} c_n e^{i 2\pi n (x+a) } = \sum_{n=-\infty}^{\infty} c_n e^{i 2\pi n x} e^{i2\pi na}
$$
$$
f(x) = \sum_{n=-\infty}^{\infty} c'_n e^{i2\pi nx}
$$
$$
g(x) = \frac{f(x+a)}{f(x)} = \frac{\sum_{n=-\infty}^{\infty} c_n e^{i2\pi nx} e^{i2\pi na} }
{ \sum_{n=-\infty}^{\infty} c'_n e^{i2\pi nx} }
$$
$$
a_0 = \frac{c_0}{c'_0}
$$
$c_0$ and $c'_0$ are the average values of $f(x+a)$ and $f(x)$.
$$
c_0 = \int_{0}^{1} f(x+a) dx = \int_{a}^{1+a} f(u) du $$
$u = x+a$.
$$
c'_0 = \int_{0}^{1} f(x) dx
$$
Therefore, $c_0 = c'_0$. Hence, $a_0 = \frac{c_0}{c'_0} = 1$. So, finally we have
$$
\int_{0}^{1} \frac{f(x+a)}{f(x)} = 1
$$

I don't know where the mistakes lies because of which I'm not getting an inequality.

 

[wrobel]

here is a mistake:
$$a_0=\frac{c_0}{c_0'}$$

 

[Adesh]

here is a mistake:
$$a_0=\frac{c_0}{c_0'}$$

Okay, but what’s the mistake in that?

 

[member 587159]

@wrobel Сэр, я получаю равенство, а не неравенство, here is my attempt.

Spoiler: Question 3

$f(x+a)$ has period 1, $f(x)$ has also period 1, therefore $g(x) = \frac{f(x+a)}{f(x)}$ will have a period 1.

Trigonometric Fourier Series for $g(x) = \frac{f(x+a)}{f(x)}$ is
$$
\frac{f(x+a)}{f(x)} = a_0 \sum_{n=1}^{\infty} \left[ a_n \cos (2\pi n x) + b_n \sin (2\pi nx)\right]
$$
$$
\int_{0}^{1} \frac{f(x+a)}{f(x)} = \int_{0}^{1} a_0 dx + 0 $$
$$\int_{0}^{1} \frac{f(x+a)}{f(x)} = a_0
$$
Fourier exponential series for $f(x)$ and $f(x+a)$
$$
f(x+a) = \sum_{n=-\infty}^{\infty} c_n e^{i 2\pi n (x+a) } = \sum_{n=-\infty}^{\infty} c_n e^{i 2\pi n x} e^{i2\pi na}
$$
$$
f(x) = \sum_{n=-\infty}^{\infty} c'_n e^{i2\pi nx}
$$
$$
g(x) = \frac{f(x+a)}{f(x)} = \frac{\sum_{n=-\infty}^{\infty} c_n e^{i2\pi nx} e^{i2\pi na} }
{ \sum_{n=-\infty}^{\infty} c'_n e^{i2\pi nx} }
$$
$$
a_0 = \frac{c_0}{c'_0}
$$
$c_0$ and $c'_0$ are the average values of $f(x+a)$ and $f(x)$.
$$
c_0 = \int_{0}^{1} f(x+a) dx = \int_{a}^{1+a} f(u) du $$
$u = x+a$.
$$
c'_0 = \int_{0}^{1} f(x) dx
$$
Therefore, $c_0 = c'_0$. Hence, $a_0 = \frac{c_0}{c'_0} = 1$. So, finally we have
$$
\int_{0}^{1} \frac{f(x+a)}{f(x)} = 1
$$

I don't know where the mistakes lies because of which I'm not getting an inequality.

A more serious flaw is that not every continuous function has a (pointwise convergent) Fourier series so your argument does not work.

 

[wrobel]

Okay, but what’s the mistake in that?

take a function $f(x)=2+\cos(2\pi x)$ and try to prove this equality

 

[Adesh]

A more serious flaw is that not every continuous function has a (pointwise convergent) Fourier series so your argument does not work.

It is given that the function is bounded and periodic therefore it does have a convergent Fourier series.

 

[Adesh]

take a function $f(x)=2+\cos(2\pi x)$ and try to prove this equality

Yes I got $a_0 \gt \frac{c_0}{c’_0}$.

 

[member 587159]

It is given that the function is bounded and periodic therefore it does have a convergent Fourier series.

This is false. As a corollary of the uniform boundedness principle you can prove that there are Fourier series that do not converge pointwise to the function itself. See Papa Rudin for a proof.

You always have $L_2$-convergence but that's not enough for what you did.

 

[lavinia]

I make a rule not to answer Math Challenges but if anyone wants I can throw out some thoughts for problem 8.

 

[mathwonk]

I am interested in your thoughts on prob. #8. The only solution I know is via non trivial properties of Eilenberg Maclane spaces, i.e. if a K(G,1) has a finite dimensional CW structure, then G is torsion free. Might be enough to know the cohomology of (infinite) lens space.

 

[Infrared]

@lavinia I don't mind you giving your thoughts (unless you've seen an equivalent problem before). I proposed it knowing that it might be less accessible than most of the other problems.

@mathwonk There is a solution only using standard tools in a first algebraic topology class, but it might be a little tricky to find (hint: by Whitehead, an equivalent problem is to show that the universal cover of $X$ is not contractible)

 

[mathwonk]

Nice hint. (That seems to do it assuming your compact manifolds have no boundary.) But I confess that theorem of Whitehead was not in my alg top course (although postnikov towers were). To be fair, his presumed parent theorem that a map of CW complexes induces a homotopy equivalence if it induces isoms on homotopy groups was. Of course Whitehead's 1978 book, in which I have seen the theorem cited, came out after my graduate school experience. And I am probably confusing JHC with GW.

 

[Infrared]

To be fair, his presumed parent theorem that a map of CW complexes induces a homotopy equivalence if it induces isoms on homotopy groups was.

This was the theorem I was referring to. Did you have something else in mind?

 

[mathwonk]

No but I was wondering where the map comes from. In case all groups vanish, do you just map in a one point space, and deduce that the space is homotopy equivalent to a point? That's a nice example of a case where just knowing the groups does all the work! ( I have always internalized the warning about that theorem that just knowing two spaces have the same homotopy groups does not (usually) mean they have the same homotopy type, unless those group isomorphisms are induced by a map of the spaces.)

 

[zinq]

Spoiler: Problem 8.

8. Let be a compact manifold M such that π₁(M) is finite and nontrivial. Show that π_n(M) is also non-trivial for some n ≥ 2.

Let G = π₁(M). Suppose that for all n ≥ 2, π_n(M) is trivial. Then M is a K(G,1) (Eilenberg-MacLane) space. Let g ∈ G be any nontrivial element and let H denote the finite subgroup of G generated by g. H is a nontrivial finite cyclic group, say H = Z_p. Let M' → M be the covering space of M corresponding to the subgroup H of G = π₁(M). That is, π₁(M') = Z_p.

Then by standard covering space theory, π_n(M') = π_n(M) is trivial by assumption. Hence M' is a K(Z_p, 1).

Since the group H determines the homotopy type of any K(H, 1), this implies that each K(Z_p, 1) space is homotopy equivalent to the lens space L_p = S^∞ / Z_p, where S^∞ is the infinite ascending union of S¹ ⊂ S³ ⊂ ... ⊂ S^2n-1 ⊂ ..., where

S^2n-1 = {(z₁, ..., z_n) ∈ ℂⁿ | |z₁|² + ... + |z_n|² = 1},

and Z_p acts on S^∞ via multiplication by exp(2πi/p) on each component.

However, we know (Hatcher, Algebraic Topology (2001), Example 3.41, p. 251) that each even-dimensional cohomology group H^2k(L_p; Z_p) is nonzero.

This shows that any K(Z_p, 1) — and in particular the covering space M' — cannot be a finite cell complex. But since every compact manifold is a finite cell complex, and every finite covering space of a finite cell complex is also a finite cell complex, this is contradicts our assumption that M is a K(G,1). Hence π_n(M) is also non-trivial for some n ≥ 2, QED.

 

[Infrared]

@mathwonk Right, this is the one case where it is possible! Of course the other pitfall is that you have to verify your space is in fact a cell complex.

@zinq Yes, that's right, nice work. I think this was the argument @mathwonk had in mind in post 40.

I'll just point out a short/elementary but tricky solution: Suppose for contradiction the universal cover $\tilde{X}$ were contractible. The Lefschetz fixed point theorem applies, and so in particular any covering transformation of $\tilde{X}\to X$ has a fixed point and is thus the identity. This contradicts $\pi_1(X)\neq 1$. Hence $\tilde{X}$ has a nontrivial homotopy group, necessarily in degree at least $2$, and so does $X$.

 

[Adesh]

I tried something few things more, I worked on finding the minimum sufficient codition for $\int_{0}^{1} \frac{f(x+a)}{f(x)}\geq 1$ but couldn’t anything better than $\frac{f(x+a)}{f(x)} \geq$.

If $a$ is an integer then $\frac{f(x+a)}{f(x)}$ will be 1, if $a$ is not an integer then let $\{a\}$ denote the fractional part of $a$, so we have
$$
\int_{0}^{1}\frac{f(x+\{a\})}{f(x)} \geq 1$$
But again not much useful.

Can I get a little hint about this problem? (If a little hint can solve the whole problem then, no issues, I shall wait for the solution).

Thanks @wrobel for this question.

 

[nuuskur]

I understand we can assume without loss |a|&lt;1. I tried using Lagrange's MVT (integrand is continuous), that means the integral is equal to \frac{f(u+a)}{f(u)} for some u\in (0,1). No brilliant ideas, though.

 

[zinq]

Spoiler: Problem 8.

I'll just point out a short/elementary but tricky solution: Suppose for contradiction the universal cover were contractible. The Lefschetz fixed point theorem applies, and so in particular any covering transformation of has a fixed point and is thus the identity. This contradicts . Hence has a nontrivial homotopy group, necessarily in degree at least , and so does .

Wow, that is really elegant!

P.S. Before posting, I had not seen any prior posts on Problem 8. except the statement of the problem.

 

[lavinia]

Spoiler: Problem 8.

A couple of ideas: Maybe all wrong

Assume that the manifold is aspherical in the sense that all of its homotopy groups of dimension 2 through n are zero where n is its dimension.

Proof 1:By the exact homotopy sequence of the fibration the universal covering manifold is also aspherical and since it is also simply connected Hurewicz's Theorem says that its integer homology is zero in dimensions 1 to n. So it has Euler characteristic 1. The covering projection divides the Euler characteristic by the order, p, of the fundamental group of the base manifold so the base manifold has Euler characteristic 1/p. This can't happen because Euler characteristics are integers. ( This Euler characteristic argument I think uses that the base manifold is a finite CW complex so I am not sure if all compact manifolds are accounted for. )

Proof 2: If a finite p-group acts on a compact topological space and that space's homology mod p is acyclic, then the action has a fixed point. In this case the universal cover's homology is mod p acyclic for all p. So any p subgroup of the group of covering transformation must have a fixed point.

This theorem of p group actions is due to P. A. Smith.