Challenge Math Challenge - June 2020

[fresh_42]

#8) a)

Let $I(\alpha ) := \int_0^\pi \log (1-2\alpha \cos(x)+\alpha^2)\,dx\quad (\alpha \geq 1)$, then

$$\begin{gathered} \tfrac{dI}{d\alpha } = \int_0^\pi \tfrac{\partial }{\partial \alpha} \log (1-2\alpha \cos(x)+\alpha ^2)\,dx = 2 \int_0^\pi \tfrac{\alpha -\cos (x) }{1-2\alpha \cos (x)+\alpha ^2} \,dx \\ = \tfrac{1}{\alpha} \int_0^\pi\left(1- \tfrac{1-\alpha ^2}{1-2\alpha \cos(x)+\alpha^2} \right) \,dx \\ \end{gathered}$$

I used a table of integrals here and got

$$\tfrac{dI}{d\alpha } =\tfrac{\pi}{\alpha }-\tfrac{2}{\alpha}\left[ \arctan \left(\tfrac{1+\alpha}{1-\alpha}\tan \left( \tfrac{x}{2}\right)\right)\right| _{x=0}^{\pi} =\tfrac{2\pi}{\alpha}$$

hence

$$I(\alpha )=2\pi \int\tfrac{d\alpha}{\alpha}= 2\pi\log | \alpha | + C$$

I'll leave the rest of the solution (determining the constant) to somebody else. Good luck!

Note: I've read this problem somewhere before.

The Feynman trick is the correct idea, but I'd like to see the complete integration, since this is a bit tricky to do. And what about $|\alpha|=1$?

 

[fresh_42]

Although, it is clearly written “n different integers”, but then also I want to ask if all of them are different or not all of them are equal? I mean is there a possibility that two integers could be same?

We consider $n$ different integers. Otherwise we would have $n-1$ many. It is the starting situation, not a resulting property.

 

[fresh_42]

A few words: I suspect the step when I said "let's add $q$ to our set and we will get $k+1$ integers" because I have to make sure that $q$ doesn't occur in the set $\{a_1, a_2, \cdots a_k\}$ (because we're asked in the question to prove for "$n$ different integers").

This is not the point. You cannot "add" an integer you like, since then you could always add $n$ and the statement becomes trivial. We have given any $n$ different integers.

Induction is the wrong idea, sum and division, however, was a good idea. Try different sums.

 

[fresh_42]

$$i^i=\{e^{-\pi/2+2\pi n}\mid n\in\mathbb{Z}\}$$

You can do this with every complex number: $z=re^{i\varphi}=re^{i(\varphi + 2n\pi)}.$ We usually assume $0\leq \varphi < 2\pi$ as representation.

 

[wrobel]

You can do this with every complex number: z=reiφ=rei(φ+2nπ).z=re^{i\varphi}=re^{i(\varphi + 2n\pi)}.

$e^{i(\varphi+2n\pi)}$ is one number while $i^i$ is the infinite set of numbers

 

[benorin]

Ignore this post, it contains an arithmetic error, for the correct solution, read post #40.

4. Let $S:=\{\,(x,y,z)\in \mathbb{R}^3\,|\,x^2+y^2=(2-z)^2,\,0\leq z\leq 2\,\}$ be the surface of a cone $C$ with a circular cross section and a peak at $(0,0,2)$. The orientation of $S$ be such, that the normal vectors point outwards. Calculate the flux through $S$ of the vector field (FR)
$$
F\, : \,\mathbb{R}^3 \longrightarrow \mathbb{R}^3\, , \,F(x,y,z)=\begin{pmatrix}xy^2\\x^2y\\(x^2+y^2)(1-z)\end{pmatrix}.$$

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

$$\begin{gathered} \iint_{S_1}\vec{F}\cdot d\vec{S} = \iint_{\text{Projection of } S_1\text{ in the }xy\text{-plane}}\left(-Pg_x-Qg_y+R\right)\, dA \\ = \iint_{x^2+y^2\leq 2^2}\left[\tfrac{2x^2y^2}{\sqrt{x^2+y^2}}+(x^2+y^2)\left(-1+\sqrt{x^2+y^2}\right)\right] \, dA\\= \int_{0}^{2\pi}\int_0^2\left(\tfrac{r^4\sin ^2\theta\cos ^2\theta}{r}+r^2(r-1)\right) r\, dr d\theta\\ =\int_{0}^{2\pi}\sin ^2\theta\cos ^2\theta\, d\theta\int_0^2r^4\, dr+\int_{0}^{2\pi}d\theta \int_0^2 (r^4-r^3)\, dr=\boxed{\tfrac{32\pi}{5}}\\ \end{gathered} $$

Let $S_2$ denote the circular disk $z=h(x,y)=0,\quad (0\leq x^2+y^2\leq 2^2 )$ with downward pointing normal. Then

$$\begin{gathered} \iint_{S_2}\vec{F}\cdot d\vec{S}=-\iint_{x^2+y^2\leq 2^2}(x^2+y^2)\, dA\\ = -\int_0^{2\pi}\, d\theta \int_0^2 r^3\, dr=\boxed{-8\pi}\\ \end{gathered}$$

Hence, adding the fluxes, we get $\boxed{ \iint_{S}\vec{F}\cdot d\vec{S}=-\tfrac{8\pi}{5}}$.

 

[fresh_42]

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

$$\begin{gathered} \iint_{S_1}\vec{F}\cdot d\vec{S} = \iint_{\text{Projection of } S_1\text{ in the }xy\text{-plane}}\left(-Pg_x-Qg_y+R\right)\, dA \\ = \iint_{x^2+y^2\leq 2^2}\left[\tfrac{2x^2y^2}{\sqrt{x^2+y^2}}+(x^2+y^2)\left(-1+\sqrt{x^2+y^2}\right)\right] \, dA\\= \int_{0}^{2\pi}\int_0^2\left(\tfrac{r^4\sin ^2\theta\cos ^2\theta}{r}+r^2(r-1)\right) r\, dr d\theta\\ =\int_{0}^{2\pi}\sin ^2\theta\cos ^2\theta\, d\theta\int_0^2r^4\, dr+\int_{0}^{2\pi}d\theta \int_0^2 (r^4-r^3)\, dr=\boxed{\tfrac{32\pi}{5}}\\ \end{gathered} $$

Let $S_2$ denote the circular disk $z=h(x,y)=0,\quad (0\leq x^2+y^2\leq 2^2 )$ with downward pointing normal. Then

$$\begin{gathered} \iint_{S_2}\vec{F}\cdot d\vec{S}=-\iint_{x^2+y^2\leq 2^2}(x^2+y^2)\, dA\\ = -\int_0^{2\pi}\, d\theta \int_0^2 r^3\, dr=\boxed{-8\pi}\\ \end{gathered}$$

Hence, adding the fluxes, we get $\boxed{ \iint_{S}\vec{F}\cdot d\vec{S}=-\tfrac{8\pi}{5}}$.

This is the wrong answer. The mistake is in the first part of your calculation, which I do not really understand what you did. Cover the cone with a disc $D$ and calculate $\int_{D\cup S}F\vec{n}dS$.

 

[benorin]

I used a formula out of my old Stuart Calculus book, 4th ed., pg 1100: If $\vec{F}=\left< P,Q,R\right>$ and $S:=\left\{(x,y,z) | z=g(x,y)\forall (x,y)\in D\right\}$ then (with upward directed normal)
$$\iint_{S}\vec{F}\cdot d\vec{S}=\iint_{D}\left( -P \tfrac{\partial g}{\partial x}-Q \tfrac{\partial g}{\partial y}+R\right) \, dA$$

where the instance of $z$ in any of $P,Q,R$ is replaced by $z=g(x,y)$. This is how I learned to calcullate flux through surfaces. Is this incorrect? I checked my integration and arithmetic with WolframAlpha.

 

[benorin]

Oh I see that I "lost" a factor of 2 on a single term along the way. 1 sec.

 

[benorin]

Corrected arithmetic error from previous post, $\boxed{\boxed{\text{corrections are double boxed}}}$.

4. Let $S:=\{\,(x,y,z)\in \mathbb{R}^3\,|\,x^2+y^2=(2-z)^2,\,0\leq z\leq 2\,\}$ be the surface of a cone $C$ with a circular cross section and a peak at $(0,0,2)$. The orientation of $S$ be such, that the normal vectors point outwards. Calculate the flux through $S$ of the vector field (FR)
$$
F\, : \,\mathbb{R}^3 \longrightarrow \mathbb{R}^3\, , \,F(x,y,z)=\begin{pmatrix}xy^2\\x^2y\\(x^2+y^2)(1-z)\end{pmatrix}.$$

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

$$\begin{gathered} \iint_{S_1}\vec{F}\cdot d\vec{S} = \iint_{\text{Projection of } S_1\text{ in the }xy\text{-plane}}\left(-Pg_x-Qg_y+R\right)\, dA \\ = \iint_{x^2+y^2\leq 2^2}\left[\tfrac{2x^2y^2}{\sqrt{x^2+y^2}}+(x^2+y^2)\left(-1+\sqrt{x^2+y^2}\right)\right] \, dA\\ = \int_{0}^{2\pi}\int_0^2\left(\tfrac{\boxed{\boxed{-2}}r^4\sin ^2\theta\cos ^2\theta}{r}+r^2(r-1)\right) r\, dr d\theta\\ =\boxed{\boxed{-2}}\int_{0}^{2\pi}\sin ^2\theta\cos ^2\theta\, d\theta\int_0^2r^4\, dr+\int_{0}^{2\pi}d\theta \int_0^2 (r^4-r^3)\, dr=\boxed{\boxed{\tfrac{8\pi}{5}}}\\ \end{gathered} $$

Let $S_2$ denote the circular disk $z=h(x,y)=0,\quad (0\leq x^2+y^2\leq 2^2 )$ with downward pointing normal. Then

$$\begin{gathered} \iint_{S_2}\vec{F}\cdot d\vec{S}=-\iint_{x^2+y^2\leq 2^2}(x^2+y^2)\, dA\\ = -\int_0^{2\pi}\, d\theta \int_0^2 r^3\, dr=\boxed{-8\pi}\\ \end{gathered}$$

Hence, adding the fluxes, we get $\boxed{\boxed{ \iint_{S}\vec{F}\cdot d\vec{S}=-\tfrac{32\pi}{5}}}$.

 

[fresh_42]

Corrected arithmetic error from previous post, .
I assume the cone has a base because it's oriented. Let denote the upper cone-shaped surface given by . Also denote the and components of by and , respectively. ThenLet denote the circular disk with downward pointing normal. ThenHence, adding the fluxes, we get .

Divergence is zero, flux is not. And consider the correct sign! The orientation is given in the problem statement.

 

[benorin]

Ahh, yes. Seems it was a negative two that I fudged. I've edited the corrected post #40, please refresh the page @fresh_42 .

 

[benorin]

The 3b1b (3 blue 1 brown) guy said you never really stop making stupid little mistakes, or something close to that. I see some evidence of this claim. Lol.

 

[fresh_42]

The 3b1b (3 blue 1 brown) guy said you never really stop making stupid little mistakes, or something close to that. I see some evidence of this claim. Lol.

There is no fifth in the solution. It is quite simple if you use Gauß' theorem.

 

[Halc]

I need clarification on problem 11: How to get a rope to fall if either nail is removed.
From what I can tell, the trick is to not have the rope fall, which I suppose could be done by making a loop of it. Maybe the problem is suggesting a loop (the phrase 'hang the rope around' sort of suggests one).

Spoiler

Suppose a 3m rope draped symmetrically between a pair of nails separated by a meter. The rope hands straight down on either side and kind of a curve between. Remove either nail and the rope is over twice as long on one side of the remaining nail as the other, so it just pulls the short side up and falls.

If the rope is a loop, then just reduce a 6 meter loop to a 3 meter rope by doubling it up.

This is way too easy of an answer, and so I presume there's something unstated in the problem that I'm not getting.

 

[benorin]

I will circle back to the flux thing after this post.

The Feynman trick is the correct idea, but I'd like to see the complete integration, since this is a bit tricky to do. And what about $|\alpha|=1$?

Edit: The unabridged version.

#8) a) Let $I(\alpha ) := \int_0^\pi \log (1-2\alpha \cos(x)+\alpha^2)\,dx\quad ( | \alpha | > 1)$, then

$$\begin{gathered} \tfrac{dI}{d\alpha } = \int_0^\pi \tfrac{\partial }{\partial \alpha} \log (1-2\alpha \cos(x)+\alpha ^2)\,dx = 2 \int_0^\pi \tfrac{\alpha -\cos (x) }{1-2\alpha \cos (x)+\alpha ^2} \,dx \\ = \tfrac{1}{\alpha} \int_0^\pi\left(1- \tfrac{1-\alpha ^2}{1-2\alpha \cos(x)+\alpha^2} \right) \,dx \\ \end{gathered}$$

now do Karl Weierstass' trick and apply the change the change of variables $t=\tan \left(\tfrac{x}{2}\right) , 0\leq t\leq \infty\implies \cos (x)= \tfrac{1-t^2}{1+t^2}\text{ and } dx=\tfrac{2\, dt}{1+t^2}$ to get

$$\begin{gathered} \tfrac{dI}{d\alpha } = 2\int_0^\infty \tfrac{\alpha -\tfrac{1-t^2}{1+t^2}}{1-2\alpha \left(\tfrac{1-t^2}{1+t^2}\right)+\alpha ^2}\cdot \tfrac{2\, dt}{1+t^2} = 4\int_0^\infty \tfrac{(\alpha+1)t^2+(\alpha -1)}{\left[ (\alpha +1)^2t^2+(\alpha -1) ^2\right] (1+t^2)}\, dt\\ \end{gathered}$$

Aside: partial faction decomposition of the integrand. Let $A,B, C,$ and $D$ be rational functions of $\alpha$. Then set

$$\begin{gathered}\tfrac{(\alpha+1)t^2+(\alpha -1)}{\left[ (\alpha +1)^2 t^2+(\alpha -1) ^2\right] (1+t^2)}=\tfrac{At+B}{1+t^2}+\tfrac{Ct+D}{ (\alpha +1)^2 t^2+ (\alpha -1) ^2}\\ \implies (\alpha +1) t^2 + (\alpha -1)= (At+B) \left[ (\alpha +1)^2 t^2 + (\alpha -1) ^2 \right] + (Ct+D)(1+t^2)\\ \end{gathered}$$

solving this system of equations is quickest by plugging in enough values of $t$, namely,

$$\begin{gathered} \boxed{t=0} : \alpha -1 = B(\alpha -1)^2+D\implies D=(\alpha -1)\left[ 1-B(\alpha -1)\right] \quad (eqn\, 1) \\ \boxed{t=i} : -2 = (Ai+B)\left[ (\alpha -1)^2+(\alpha +1)^2\right]\implies A=0 \quad (eqn \, 2)\text{ and } B=\tfrac{1}{2\alpha}\quad (eqn \, 3)\\ \end{gathered} $$

Now $(eqn \, 2)$ implies $C=0$ and substituting $(eqn \, 3)$ into $(eqn \, 1)$ gives $D=\tfrac{1}{2}\left(\alpha - \tfrac{1}{\alpha}\right)$. End aside.

Hence we have

$$\begin{gathered} \tfrac{dI}{d\alpha } = 2\int_0^\infty\left( \tfrac{\tfrac{1}{\alpha}}{1+t^2}+\tfrac{\alpha -\tfrac{1}{\alpha}}{ (\alpha +1)^2 t^2+(\alpha -1) ^2}\right) \, dt \\ = \tfrac{2}{\alpha} \left[ \tan ^{-1} (t) \right| _{t=0}^{\infty}+2\tfrac{\alpha -\tfrac{1}{\alpha}}{(\alpha +1) ^2}\int_0^\infty \tfrac{dt}{ t^2+\left( \tfrac{\alpha -1}{\alpha +1}\right) ^2} \\= \tfrac{\pi}{\alpha} +2\tfrac{\alpha -\tfrac{1}{\alpha}}{(\alpha +1) ^2}\cdot \left( \tfrac{\alpha +1}{\alpha -1}\right)\left[ \tan ^{-1} \left( t\cdot \tfrac{\alpha +1}{\alpha -1}\right) \right| _{t=0}^{\infty} \\ = \boxed{\tfrac{2\pi}{\alpha}=\tfrac{dI}{d\alpha}}\\ \end{gathered}$$

hence

$$I(\alpha )=2\pi \int\tfrac{d\alpha}{\alpha}= 2\pi\log | \alpha | + C\quad (\text{for } | \alpha | > 1) $$

As for the case of $\alpha =1$, I know that $I(1)=0$ but infuriatingly enough I cannot prove it with any clever trick thus far. I think there's going to be something on par with the integral of a odd function over a symmetric interval vanishes--but I did trig, log, and basic integral identities for a half hour: no joy! I think I'm just tired. I hope it will come to me later...

Note: I've read this problem somewhere before.

Edit 2: Did I say after this post? Right. Well not right after...

 

[fresh_42]

I hope it will come to me later...

You're closing in. Your answer is almost right and the two cases $|\alpha|=1$ are missing.

 

[fresh_42]

I will circle back to the flux thing after this post.

Note my hint in post #37.

 

[Infrared]

@benorin Your answer is correct, but you can simplify: What is $2\lambda(2)$?

@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.

 

[StoneTemplePython]

Thanks for the the catch @StoneTemplePython Condition 1 should be "The $\textbf{off-diagonal}$ entries of $A$ are all positive." I should have checked more carefully.

so the 'right answer' is to apply Gerschgorin Discs / strict diagonal dominance and the result falls out. Here's a different approach:

Spoiler

the problem implies all diagonal entries are negative
$A = -D +B$
where $B$ is strictly positive except the diagonal is zero'd out, and $D$ is a diagonal matrix with positive diagonal entries. Multiplication by an invertible matrix doesn't change rank and it is convenient to homogenize the diagonal, so it suffices to check the invertibility of

$D^{-1}A = -I + D^{-1}B = -I + C$
where $C$ is strictly positive except its diagonal is zero

the problem tells us
$A\mathbf 1 \leq -\alpha \mathbf 1$
for some $\alpha \gt 0$
where all inequalities are interpreted component-wise

re-scaling by positive numbers:
$\big(-I + C\big)\mathbf 1 = D^{-1}A\mathbf 1 \leq -\alpha D^{-1}\mathbf 1 \leq -\sigma \mathbf 1$
where $\sigma := \alpha \cdot \min_i (D^{-1}_{i,i})$ which implies $\sigma$ is positive. If for some reason $\sigma \gt 1$ then we can use an even looser upper bound and select $\sigma := \frac{1}{2}$. Thus we have $\sigma \in (0,1)$

re-arranging terms
$C\mathbf 1 \leq (1-\sigma)\mathbf 1$
and using the non-negativity in $C$
$C^2\mathbf 1 \leq (1-\sigma)C\mathbf 1 \leq (1-\sigma)^2\mathbf 1$
$\longrightarrow C^k\mathbf 1 \leq (1-\sigma) C^{k-1}\mathbf 1\leq ... \leq (1-\sigma)^k\mathbf 1$
- - - - -
to help with interpretation:
$\mathbf x := C^{k-1}\mathbf 1$
$\mathbf x = C^{k-1}\mathbf 1 \leq (1-\sigma)^{k-1} \mathbf 1$
so we know $0 \leq x_j \leq (1-\sigma)^{k-1}$ and by non-negativity of $C$, we can specialize to its jth column giving us
$x_j \mathbf c_j \leq (1-\sigma)^{k-1}\mathbf c_j$
summing over the bound:
$C^{k}\mathbf 1 = C \mathbf x = \sum_{j=1}^n x_j \mathbf c_j\leq \sum_{j=1}^n (1-\sigma)^{k-1} \mathbf c_j =(1-\sigma)^{k-1} \cdot \sum_{j=1}^n \mathbf c_j = (1-\sigma)^{k-1} C\mathbf 1 \leq (1-\sigma)^k \mathbf 1$
- - - - -

if $C$ had an eigenvalue with modulus $\geq 1$, we'd have

$1$
$\leq \big \Vert C^k\big \Vert_F$
$= \Big(\sum_{i=1}^n\sum_{j=1}^n \vert c^{(k)}_{i,j}\vert^2 \Big)^\frac{1}{2}$
$\leq \Big(\sum_{i=1}^n\sum_{j=1}^n \vert c^{(k)}_{i,j}\vert \Big) \text{ (triangle inequality)}$
$=\sum_{i=1}^n\sum_{j=1}^n c^{(k)}_{i,j}$
$= \mathbf 1^T C^k \mathbf 1$
$\leq (1-\sigma)^k \mathbf 1^T \mathbf 1$
$=(1-\sigma)^k \cdot n$

but the upper bound may be arbitrarily small for large enough k thus the spectral radius of $C$ is $\lt 1$, and in particular $C$ can't have an eigenvalue of 1 telling us
$\det\big(-I+C\big ) \neq 0\longrightarrow \det\big(A\big) \neq 0$

 

[benorin]

@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 :)

 

[Fred Wright]

# 8

Spoiler

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 $2\sum_{k=1}^{\infty}\frac{\sin(kx)}{k^2\alpha^k}|_0^{\pi}=0$.

 

[Halc]

@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.

 

[Fred Wright]

#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]

@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.

 

[Infrared]

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.

 

[Halc]

#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.

Spoiler

 

[Infrared]

@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]

@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.

Spoiler