# Math Challenge - June 2020

• Challenge
• Featured
benorin
Homework Helper
@benorin Your answer is correct, but you can simplify: What is ##2\lambda(2)##?
##2\lambda (2) =\zeta (2) +\hat{\zeta} (2) = \tfrac{\pi ^2}{6}+\tfrac{\pi ^2}{12} = \tfrac{\pi ^2}{4}##
where ##\hat{\zeta} ## is the alternating Zeta function, but I'm not going to prove the value of those sums. They are known :)

• dextercioby
# 8
Part a.)
Factor the logarithm:
$$\log(1-2\alpha \cos(x) +\alpha^2)= \log[(\alpha - e^{ix})(\alpha - e^{-ix})]\\ =2\log(\alpha) + \log(1-\frac{ e^{ix}}{\alpha}) + \log(1-\frac{ e^{-ix}}{\alpha})\\$$
The Integral becomes:
$$I=\int_0^{\pi}\log(1-2\alpha \cos(x) +\alpha^2)dx= I_1+I_2+I_3\\ I_1=2\log(\alpha)\int_0^{\pi}dx=2\pi \log(\alpha)\\ I_2=\int_0^{\pi} \log(1-\frac{ e^{ix}}{\alpha})dx\\$$
Make a substitution in ##I_2##:
$$u=\frac{ e^{ix}}{\alpha}\\ du=iudx\\ I_2=i\int_{\frac{-1}{\alpha}}^{\frac{1}{\alpha}}\frac{\log(1-u)}{u}du\\$$
Make a substitution in ##I_3##:$$I_3=\int_0^{\pi} \log(1-\frac{ e^{-ix}}{\alpha})dx\\ u=\frac{ e^{-ix}}{\alpha} du=-iudx\\$$
and find,$$I_3=-I_2\\$$
and thus,for ##\alpha \geq 1##,$$I=2\pi \log(\alpha)\\$$
for ##\alpha \leq -1## take the principal branch of the logarithm,$$I=2\pi(\log(|\alpha |)+ i\pi)\\$$
Part b.)
Extend the the Taylor series for ##\log(1-x)## to the complex plane,$$\log(1-x)=\sum_{k=1}^{\infty}(-1)^{k+1}\frac{(-x)^k}{k}=-\sum_{k=1}^{\infty}\frac{x^k}{k}\\ \log(1-z)=-\sum_{k=1}^{\infty}\frac{z^k}{k}\\$$
This series converges for ##|z| \leq 1##.
From part a.),$$I=2\pi \log(\alpha) + \int_0^{\pi} \log(1-\frac{ e^{ix}}{\alpha})dx + \int_0^{\pi} \log(1-\frac{ e^{-ix}}{\alpha})dx\\ =2\pi \log(\alpha) -\sum_{k=1}^{\infty}\ \int_0^{\pi}\frac{e^{ikx}+ e^{-ikx} }{k\alpha^k}=2\pi \log(\alpha)-2\sum_{k=1}^{\infty} \int_0^{\pi} \frac{\cos(kx)}{k\alpha^k}dx\\ =2\pi \log(\alpha)-2\sum_{k=1}^{\infty}\frac{\sin(kx)}{k^2\alpha^k}|_0^{\pi}=2\pi \log(\alpha)\\$$
We recover the result of Part a.) because ##2\sum_{k=1}^{\infty}\frac{\sin(kx)}{k^2\alpha^k}|_0^{\pi}=0##.

• benorin and dextercioby
Halc
Gold Member
@Halc The condition "will fall down" means that the rope is not supported by a nail. You could take the rope to be a loop, if that's helpful.
My solution (the latter part) works in that case.

etotheipi
@Halc The condition "will fall down" means that the rope is not supported by a nail. You could take the rope to be a loop, if that's helpful.

The question sort of reminds me of the Borromean rings... I'm sure the answer is some clever figure-of-eight type arrangement of the loop around the pins.

• Halc
#7
If all the off diagonal elements are positive and the sum of the entries in each row is negative implies that all the diagonal entries are negative. This implies that no column is a multiple of another column and therefore all the columns are linearly independent. The invertible matrix theorem states that if all the columns are linearly independent then the matrix is invertible.

Infrared
Gold Member
@StoneTemplePython I actually hadn't heard of Gerschgorin Discs and instead had a method in mind with just row reduction, but your solution looks good too.

@benorin Yes, you've solved it fully now. In case you weren't sure how to get those values, from ##\zeta(2)=\sum_{k=1}^\infty\frac{1}{k^2}=\pi^2/6##, you get ##\sum_{k=1}^\infty \frac{1}{(2k)^2}=\frac{1}{4}\zeta(2)=\pi^2/24##, and then subtracting these two sums gives the sum of the reciprocals of the odd squares.

@Halc I think you're misunderstanding the problem. If you're treating the rope as a loop, then just letting it rest on top of the nails doesn't count as a valid arrangement, since it could just fall off. To rephrase it with this language: find a way of arranging a loop of rope around two nails in a wall such that if any part of the rope is pulled down, the rope will stay on the wall, but this will no longer be the case if either nail is removed.

@etotheipi Yes, and there's a general solution for ##n## nails (removing any nail will cause the rope to fall). This was just a fun puzzle a friend showed me in undergrad that I thought was fun enough to share here.

• benorin
Infrared
Gold Member
This implies that no column is a multiple of another column and therefore all the columns are linearly independent

This isn't true. For example, none of the vectors ##\begin{bmatrix}1\\0\\0\end{bmatrix},\begin{bmatrix}0\\1\\0\end{bmatrix},\begin{bmatrix}1\\1\\0\end{bmatrix}## are a multiple of another, and yet they are still linearly dependent. Also for ##n=2##, having the diagonal terms negative and off-diagonal terms positive is consistent with the columns being linearly dependent.

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Halc
Gold Member
#11

To rephrase it with this language: find a way of arranging a loop of rope around two nails in a wall such that if any part of the rope is pulled down, the rope will stay on the wall, but this will no longer be the case if either nail is removed.
OK, I figured it was something like that.
Cheat answer: Bend the two nails downward but with the heads touching, forming a closed loop trapping the rope.

The question sort of reminds me of the Borromean rings... I'm sure the answer is some clever figure-of-eight type arrangement of the loop around the pins.
Good description. You should get credit for it then, or at least my like. Infrared
Gold Member
@Halc That picture can work, but could you re-draw it so that it's clear which strands are over/under in all of the crossings?

And, if anyone has a solution for the general problem with ##n## nails, feel free to post!

Halc
Gold Member
@Halc That picture can work, but could you re-draw it so that it's clear which strands are over/under in all of the crossings?
Also, apologies for the free-hand drawing using paint, crude, but adequate I guess. • Infrared
Delta2
Homework Helper
Gold Member
Corrected arithmetic error from previous post, ##\boxed{\boxed{\text{corrections are double boxed}}}##.

I assume the cone has a base because it's oriented. Let ##S_1## denote the upper cone-shaped surface given by ##z=g(x,y):=2-\sqrt{x^2+y^2},\quad ( 0\leq x^2+y^2\leq 2^2)##. Also denote the ##x,y,## and ##z## components of ##\vec{F}## by ##P,Q,## and ##R##, respectively. Then

$$\begin{gathered} \iint_{S_1}\vec{F}\cdot d\vec{S} = \iint_{\text{Projection of } S_1\text{ in the }xy\text{-plane}}\left(-Pg_x-Qg_y+R\right)\, dA \\ = \iint_{x^2+y^2\leq 2^2}\left[\tfrac{2x^2y^2}{\sqrt{x^2+y^2}}+(x^2+y^2)\left(-1+\sqrt{x^2+y^2}\right)\right] \, dA\\ = \int_{0}^{2\pi}\int_0^2\left(\tfrac{\boxed{\boxed{-2}}r^4\sin ^2\theta\cos ^2\theta}{r}+r^2(r-1)\right) r\, dr d\theta\\ =\boxed{\boxed{-2}}\int_{0}^{2\pi}\sin ^2\theta\cos ^2\theta\, d\theta\int_0^2r^4\, dr+\int_{0}^{2\pi}d\theta \int_0^2 (r^4-r^3)\, dr=\boxed{\boxed{\tfrac{8\pi}{5}}}\\ \end{gathered}$$
I don't understand why u have -2 inside the box while it should have been +2. Then after calculating the integrals as in the last line(where -2 I put +2) above I get $$2\frac{\pi}{4}\frac{32}{5}+2\pi\frac{12}{5}=40\frac{\pi}{5}=8\pi$$

Another way to verify this result:The divergence of the vector field F can be easily found and its 0. Hence by the divergence theorem, the flux through any closed surface is zero.

I believe you calculated correctly the flux through the circular disk at the base of the cone as##-8\pi##, this means that the flux through the surface ##S_1## must be ##8\pi## so they add together and give result 0, because the surface ##S_1## and the circular disk at the base form a closed surface.

Mentor
# 8

for ##\alpha \leq -1## take the principal branch of the logarithm,$$I=2\pi(\log(|\alpha |)+ i\pi)$$
It is a real integral with a real value! If you use complex analysis, you should eliminate it again at the end.
Part b.)
Extend the the Taylor series for ##\log(1-x)## to the complex plane,$$\log(1-x)=\sum_{k=1}^{\infty}(-1)^{k+1}\frac{(-x)^k}{k}=-\sum_{k=1}^{\infty}\frac{x^k}{k}\\ \log(1-z)=-\sum_{k=1}^{\infty}\frac{z^k}{k}\\$$
This series converges for ##|z| \leq 1##.
...
Yes, but it is ##|\alpha|## in the solution to be exact.

Btw., at all engineers, physicists and all who deal with real problems out there:
is a must have!

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• benorin
benorin
Homework Helper
I don't understand why u have -2 inside the box while it should have been +2. Then after calculating the integrals as in the last line(where -2 I put +2) above I get $$2\frac{\pi}{4}\frac{32}{5}+2\pi\frac{12}{5}=40\frac{\pi}{5}=8\pi$$

Another way to verify this result:The divergence of the vector field F can be easily found and its 0. Hence by the divergence theorem, the flux through any closed surface is zero.

I believe you calculated correctly the flux through the circular disk at the base of the cone as##-8\pi##, this means that the flux through the surface ##S_1## must be ##8\pi## so they add together and give result 0, because the surface ##S_1## and the circular disk at the base form a closed surface.
You’re correct of course, I must have missed either the minus from the formula or from the power rule yesterday. Thanks. It’s amazing what sleep does. Lol

• fresh_42 and Delta2
benorin
Homework Helper
@benorin Yes, you've solved it fully now. In case you weren't sure how to get those values, from ##\zeta(2)=\sum_{k=1}^\infty\frac{1}{k^2}=\pi^2/6##, you get ##\sum_{k=1}^\infty \frac{1}{(2k)^2}=\frac{1}{4}\zeta(2)=\pi^2/24##, and then subtracting these two sums gives the sum of the reciprocals of the odd squares.
Iirc I did these in Fourier analysis way back when... but that is easier way. I was thinking the trick to analytically continue the zeta fcn to Re[z]>0 when I did that. Thanks.

benorin
Homework Helper
It can be done with real analysis and elementary methods. However, part one is rather tricky, whereas part two only requires to find the series expansion which fits best. Here is a hint for part a):
$$\displaystyle{\int_0^{\pi/2} \log(\sin(x))\,dx=-\frac{\pi}{2}\log(2)}$$
#8) The case of ##| \alpha |=1##. Recall ##I(\alpha )## is define to be the integral in question. Then
$$I(+1)= \int_{0}^\pi \log (2-2\cos (x)) \, dx$$
Make the substitution ##u=\pi -x## (I’m skipping some elementary details here) to get
$$I(+1)= \int_{0}^\pi \log (2+2\cos (x)) \, dx=I(-1)$$
Let ##\beta := I(\pm 1)## so that

$$\begin{gathered} 2\beta = \int_0^\pi \log\left[ (2-2\cos (x))(2+2\cos (x))\right]\, dx \\ = \int_0^\pi \log (4\sin ^2 (x))\,dx \\ = 2\pi\log 2+4\left( -\tfrac{\pi}{2}\log 2\right) =0\\ \end{gathered}$$

Where I have used some trig identities and log properties and symmetry over ##\left[ 0,\pi \right]## of the absolute value of ##\sin x## in the last equality. Apologies for skipping steps but I’m on my phone typing this and it’s a pain but I’m sure fresh_42 can follow me well enough. If not, I will type it up in full on my Mac later.

• Delta2 and fresh_42
Question 3.
Let B = {en | n ∈ Z+} be an orthonormal basis for the separable infinite-dimensional Hilbert space H
of square-summable sequences of real numbers.

Let O = CC({±en | n ∈ Z+}) be the closed convex hull of ±B. Then O may be described as the Hilbert orthoplex (the generalization to Hilbert space of the regular octahedron). The set O has empty interior.* Now let

X = RO = {t⋅y | t ∈ R+, y ∈ O }

Then X is a convex subset of H that cannot be all of H because it contains no interior. And X is contained in no half-space because it contains all the axes R⋅en, n ∈ Z+.
_____
* For, the distance of the origin to the set {y ∈ H | Σn yn = 0} ⊂ O, where y = (yn), is zero. For instance, consider the sequence {zn} where zn = e1/n - (e2 + ... + en2+1)/n2, for which ||zn|| = √2/n.

In question 13 does “ ##n\in \mathbb{N_0}##” mean that ##n=0## is allowed ?

Infrared
Gold Member
@Adesh Yes, but ##n=0## does not satisfy the condition "there are no integers ##a,b\in\mathbb{Z}## with ##3a^3+b^3=n^3.##

@Adesh Yes, but ##n=0## does not satisfy the condition "there are no integers ##a,b\in\mathbb{Z}## with ##3a^3+b^3=n^3.##
For ##n=0## both ##a## and ##b## cannot be an integer to satisfy the equation. One of them can be an integer but not both.

Infrared
Gold Member
Why not ##a=b=0##?

• Why not ##a=b=0##?
Oh yeah! I didn’t take that into account, thanks.

wrobel
In my opinion the solution of probl. 3 by zinq is not completely formal but is it completely correct and nice. I would consider probl. 3 solved by zinq
I also think that the Baire category theorem is needed to turn zinq's argument into a formal proof

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• jim mcnamara
strangerep
2. Let ##A## and ##B## be complex ##n\times n## matrices such that ##AB-BA## is a linear combination of ##A## and ##B##. Show that ##A## and ##B## must have a common eigenvector. (IR)

Thank you to @Infrared for posing this question. It made me discover some holes in my understanding of even the simpler case where ##AB=BA##. (E.g., certain conditions on the intersection of the nullspaces of ##A## and ##B## must be met.)

Anyway, I'll start with a non-general partial solution to get things moving...

For this initial simple case, assume we are given $$AB - BA = \alpha A + \beta B ~,~~~~~ (1)$$ where the scalar coefficients ##\alpha,\beta## are non-zero.

Let ##|a\rangle## be an eigenvector of ##A## with eigenvalue ##a##, i.e., ##A |a\rangle = a |a\rangle##. Then ##|a\rangle## is also an eigenvector of ##(A+k)##, with eigenvalue ##a+k##.

Now, (1) implies $$0 ~=~ AB|a\rangle - BA|a\rangle - \alpha A |a\rangle - \beta B|a\rangle ~=~ AB|a\rangle - aB |a\rangle - a \alpha|a\rangle - \beta B|a\rangle~.~~~~~ (2)$$We need to recast (2) in the form: $$0 ~=~ A(B+v) |a\rangle ~-~ w (B+v) |a\rangle ~=~ AB|a\rangle + va|a\rangle - w (B+v) |a\rangle ~,~~~~~ (3)$$ because that means ##(B+v)|a\rangle## is an eigenvector of ##A##, with eigenvalue ##w##.

Rearranging (3) and comparing with (2), we find that we need ##w=a+\beta## and ##a\alpha = v (w-a)##, which implies ##v = a\alpha/\beta##.

Therefore, ##(B+ a\alpha/\beta) |a\rangle## is an eigenvector of ##A## with eigenvalue ##(a+\beta)##.

[Edit: I have struck out the following paragraph. Anyone following the proof above should now go straight to the later post in this thread by PeroK, and later by me, which complete the answer.]
The proof then continues (in the simple case of distinct ##A##-eigenvectors) by the standard technique of realizing that the vector ##(B+ a\alpha/\beta) |a\rangle## must be a multiple of an ##A##-eigenvector, resulting (after a shift of the eigenvalue) in an eigenvector of ##B##.

A more general solution would require a condition on the intersection of the nullspaces of ##A## and ##B##, among other things (IIUC). As an illustration of what can go wrong in the simpler case of ##AB=BA##, consider $$A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} ~~~\mbox{and}~~ B = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} ~,$$ for which ##\begin{pmatrix} 1 \\ 0 \end{pmatrix}## is an eigenvector of ##B\;## but not ##A## (since that vector lies in the nullspace of ##A##).

I'll leave it at that for now, and ask the question: how far did you [ @Infrared ] envisage that the answer should go? I'm guessing you'll require total completeness to qualify as a proper answer? Last edited:
• I want to elaborate on wrobel's point: If A = r e (r > 0) and B = x+iy are two complex numbers, then the definition of AB is the set

AB = {exp(B log(A))} as log(A) ranges over all its values. In other words

AB = {exp((ln(r) + i(θ+2nπ)) ⋅ (x + iy) | n ∈ Z), i.e.,

AB = {exp(x ln(r) - y (θ+2nπ) + i (y ln(r) + x (θ+2nπ))) | n ∈ Z}

where ln(r) = ##\int_{1}^{r} t^{-1} ~dt## as usual.

By linearity of $T$, it suffices to show $T$ is continuous at $0$. Take $(x_n) \subseteq H_1$ such that $x_n \xrightarrow[n\to\infty]{}0\in H_1$. Let $z\in H_2$, then
$$\langle z,Tx_n \rangle = \langle Sz, x_n \rangle \xrightarrow[n\to\infty]{}0 \in \mathbb K.$$
As $z$ is arbitrary it implies $Tx_n\xrightarrow[n\to\infty]{}0 \in H_2$.

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• • wrobel, member 587159 and Delta2