Challenge Math Challenge - July 2020

[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|<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.

 

[Infrared]

@lavinia Manifolds are CW complexes by Morse theory, and a CW complex must be finite to be compact (otherwise picking an interior point from each cell gives an infinite discrete subspace), so I think that part of the argument is fine. The asphericity assumption looks like the main issue to me.

Your second argument has the same idea as my argument in post 46. You can take a look there to see one way to sharpen it.

 

[lavinia]

@lavinia Manifolds are CW complexes by Morse theory, and a CW complex must be finite to be compact (otherwise picking an interior point from each cell gives an infinite discrete subspace), so I think that part of the argument is fine. The asphericity assumption looks like the main issue to me.

Your second argument has the same idea as my argument in post 46. You can take a look there to see one way to sharpen it.

The assumption of aspherical is weaker than the problem demanded., It is used for getting a contradiction. The argument is correct.

Morse theory I think works for smooth manifolds but I don't know what works for other manifolds.

 

[Infrared]

I see, I didn't read carefully. It looks fine then.

 

[zinq]

"Morse theory I think works for smooth manifolds but I don't know what works for other manifolds."

Yup. PL manifolds are immediately CW complexes. I'm not sure about those non-PL and not even triangulable manifolds they now know exist in all dimensions ≥ 4. (I just read somewhere that an argument of Milnor shows that all topological manifolds have the homotopy type of a CW-complex.)

 

[Isaac0427]

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

(solved by @Isaac0427 with complex numbers, completely real solution still possible)

I'll leave that for someone else to answer-- I thought of the complex solution first for some reason, but I see that there's a much easier way to go about it.

 

[fresh_42]

I'll leave that for someone else to answer-- I thought of the complex solution first for some reason, but I see that there's a much easier way to go about it.

The solution I have in mind is basically only a real version of your solution, but it uses a function which helps here. That's why I wrote that remark. Something left to learn.

 

[Isaac0427]

The solution I have in mind is basically only a real version of your solution, but it uses a function which helps here. That's why I wrote that remark. Something left to learn.

Spoiler: Don't want to give away a hint

The other one I had in mind is an inductive proof.

 

[fresh_42]

Spoiler: Don't want to give away a hint

The other one I had in mind is an inductive proof.

I think formally it is always an induction. Even your pairing is formally an induction since commutativity is only defined for two factors. My solution works with matrices, but of course it is only another formalism than yours.

 

[Adesh]

Another method for Question 15.

Spoiler

By Stirling’s approximation we have $n! =\sqrt{2\pi n} \left( \frac{n}{e}\right)^n$.
Therefore,
$$
(n!)^2= 2\pi n \left(\frac{n}{e}\right)^{2n}
$$
And
$$
(2n)!= 2 \sqrt{\pi n} 4^n\left( \frac{n}{e}\right)^{2n}$$
$$\frac{(2n)!}{(n!)^2}= \frac{4^n}{\sqrt{\pi n}}
$$
So, we got to compare $\frac{4^n}{\sqrt{\pi n}}$ and $\frac{4^n}{n+1}$.
My claim: $\sqrt{\pi n} \lt n+1$,
$$
\pi n \lt n^2 + 1 + 2n
$$ above statement is true for n=2 and n=3 by trial and error method. For $n \gt 3$ we have
$$
\pi n \lt n^2 \\
\therefore
\pi n \lt n^2 +1 + 2n
$$
Hence, $\sqrt{\pi n} \lt (n+1)$, therefore
$$
\frac{4^n}{\sqrt {\pi n}} \gt \frac{4^n}{n+1}
$$

 

[fresh_42]

Another method for Question 15.

Spoiler

By Stirling’s approximation we have $n! =\sqrt{2\pi n} \left( \frac{n}{e}\right)^n$.
Therefore,
$$
(n!)^2= 2\pi n \left(\frac{n}{e}\right)^{2n}
$$
And
$$
(2n)!= 2 \sqrt{\pi n} 4^n\left( \frac{n}{e}\right)^{2n}$$
$$\frac{(2n)!}{(n!)^2}= \frac{4^n}{\sqrt{\pi n}}
$$
So, we got to compare $\frac{4^n}{\sqrt{\pi n}}$ and $\frac{4^n}{n+1}$.
My claim: $\sqrt{\pi n} \lt n+1$,
$$
\pi n \lt n^2 + 1 + 2n
$$ above statement is true for n=2 and n=3 by trial and error method. For $n \gt 3$ we have
$$
\pi n \lt n^2 \\
\therefore
\pi n \lt n^2 +1 + 2n
$$
Hence, $\sqrt{\pi n} \lt (n+1)$, therefore
$$
\frac{4^n}{\sqrt {\pi n}} \gt \frac{4^n}{n+1}
$$

Nice idea! The problem with Stirling is, that your equalities are only approximations. So in order for proof to count, you have to take the error margins into account. Especially the division needs watching, since a small error could theoretically turn into something large if you divide it. So you need to use expressions with upper and lower bounds.