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CMC - Problem 10

Problem 10. (16 points) If

$\sum_{a=1}^\infty\sum_{b=1}^\infty\frac{1}{a^4+4b^4}=\frac{\pi^p}{q},$

then evaluate the integer value of $$p+q$$.

Announcement. This is the last CMC problem of the season. I expect to return in about two weeks.

Note by Cody Johnson
3 years, 8 months ago

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$$p = 4, q = 288 \Rightarrow p+q=292$$

I don't have a full answer because I couldn't prove one crucial step.

Because (I can't prove this) $$\displaystyle \sum_{a=1}^\infty \frac {8b^4}{a^4+4b^4} = b \pi \coth (b \pi) - 1$$

$\begin{eqnarray} \displaystyle \sum_{a=1}^\infty \sum_{b=1}^\infty \frac {1}{a^4+4b^4} & = & \sum_{b=1}^\infty \frac {1}{8b^4} (b \pi \coth (b \pi) - 1 ) \nonumber \\ & = & \frac {\pi}{8} \left ( \displaystyle \sum_{b=1}^\infty \frac { \coth (b \pi) }{b^3} \right ) - \frac {1}{8} \left ( \sum_{b=1}^\infty \frac {1}{b^4} \right ) \\ & = & \frac {\pi}{8} \left ( \displaystyle \sum_{b=1}^\infty \frac { \frac {e^{2b \pi} + 1 }{e^{2b \pi} - 1 } }{b^3} \right ) - \frac {1}{8} \left ( \sum_{b=1}^\infty \frac {1}{b^4} \right ) \\ & = & \frac {\pi}{8} \left ( \displaystyle \sum_{b=1}^\infty \frac {2 + e^{2b \pi} -1 }{b^3( e^{2b \pi} - 1 )} \right ) - \frac {1}{8} \left ( \sum_{b=1}^\infty \frac {1}{b^4} \right ) \\ & = & \frac {\pi}{8} \left ( \displaystyle \sum_{b=1}^\infty \frac {1}{b^3} + \sum_{b=1}^\infty \frac {2}{b^3( e^{2b \pi} - 1 )} \right ) - \frac {1}{8} \zeta (4) \\ & = & \frac {\pi}{8} \left ( \zeta (3) + \frac {7 \pi^3}{180} - \zeta (3) \right ) - \frac {1}{8} \zeta (4) \\ & = & \frac {\pi}{8} \left ( \frac {7 \pi^3}{180} - \frac {\pi^3}{90} \right ) = \frac {\pi^4}{288} \\ \end {eqnarray}$

Note: from this, $$\displaystyle \zeta (3) = \frac {7 \pi^3}{180} - 2 \sum_{n=1}^\infty \frac {1}{n^3( e^{2n \pi} - 1 )}$$, and $$\zeta (4) = \frac {\pi^4}{90}$$ · 3 years, 8 months ago

A completion of the starting step with the Poisson summation formula:

Fix $$b$$ and consider $f(a) = \frac{1}{a^4 + 4b^4}.$ First, we compute the (continuous) Fourier transform $\hat{f}(c) = \int_{-\infty}^{\infty} e^{-2i\pi ct}\cdot \frac{1}{t^4 + 4b^4}\,dt$ with a contour integral.

Denote the integrand as $$g(t)$$, i.e. $g(t) = e^{-2i\pi ct}\cdot \frac{1}{t^4 + 4b^4}.$

Suppose $$c \geq 0$$. Then if $$\Im(t) \leq 0$$, we have $$\Re(-2i\pi ct) \leq 0$$ and $$|e^{-2i\pi ct}| \leq 1$$. Consider the contour integral of $$g$$ around a large semicircle centered at the origin in the half-plane $$\Im(t) \leq 0$$ of the complex plane. Since $$\frac{1}{t^4 + 4b^4}$$ decays quickly enough if $$t$$ has large magnitude, if the semicircle's radius goes to infinity, the integral over the arc of the semicircle goes to 0, so that the contour integral converges to the integral over the real line, $\int_{\infty}^{-\infty} g(t)\,dt.$ (Note the orientation, since we need to go around the semicircle counterclockwise.)

At the same time, the integral can be evaluated with the residue formula. Let $$\omega = e^{i\pi/4}$$, a primitive eighth root of unity. $$g$$ has four simple poles at $$t = \sqrt{2}\omega^k b$$ for $$k = 1, 3, 5, 7$$; the relevant poles inside our contour are $$\sqrt{2}\omega^5b$$ and $$\sqrt{2}\omega^7b$$. The residues at those points can be evaluated in an L'Hôpital-esque manner to be:

\begin{align*} \operatorname{res}_{\sqrt{2}\omega^5b} g &= e^{-2i\pi c(\sqrt{2}\omega^5b)}\cdot \frac{1}{4(\sqrt{2}\omega^5b)^3} \\ &= e^{2\pi bc(-1+i)}\cdot \frac{1+i}{16b^3} \\ \operatorname{res}_{\sqrt{2}\omega^7b} g &= e^{-2i\pi c(\sqrt{2}\omega^7b)}\cdot \frac{1}{4(\sqrt{2}\omega^7b)^3} \\ &= e^{2\pi bc(-1-i)}\cdot \frac{-1+i}{16b^3} \\ \end{align*}

Additionally supposing $$c$$ is an integer, we then have $$e^{2\pi bci} = 1$$ and can further simplify to \begin{align*} \operatorname{res}_{\sqrt{2}\omega^5b} g &= e^{-2\pi bc}\cdot \frac{1+i}{16b^3} \\ \operatorname{res}_{\sqrt{2}\omega^7b} g &= e^{-2\pi bc}\cdot \frac{-1+i}{16b^3} \end{align*}

So, \begin{align*} \int_{\infty}^{-\infty} g(t)\,dt &= 2\pi i\left( e^{-2\pi bc}\cdot \frac{1+i}{16b^3} + e^{-2\pi bc}\cdot \frac{-1+i}{16b^3}\right) \\ &= -\frac{\pi}{4b^3} \cdot e^{-2\pi bc} \\ \hat{f}(c) &= \frac{\pi}{4b^3} \cdot e^{-2\pi bc} \\ \end{align*}

Note that since $$f$$ is even, $$\hat{f}(c) = \hat{f}(-c)$$. Now the Poisson summation formula can be evaluated as two geometric series.

$\sum_{a=-\infty}^\infty f(a) = \sum_{c=-\infty}^\infty \hat{f}(c) = \frac{\pi}{4b^3} \left(\frac{1}{1 - e^{-2\pi b}} + \frac{e^{-2\pi b}}{1 - e^{-2\pi b}}\right) = \frac{\pi \coth(b\pi)}{4b^3}$

Thus, again since $$f$$ is even, \begin{align*} \sum_{a=1}^\infty f(a) &= \frac{1}{2}\left(\sum_{a=-\infty}^\infty f(a) - f(0)\right) \\ &= \frac{\pi \coth(b\pi)}{8b^3} - \frac{1}{8b^4} \\ &= \frac{1}{8b^4}\left(b\pi \coth(b\pi) - 1\right), \end{align*} as desired.

I have no idea how the parent post managed to continue from here, or how Sophie-Germain can help, however. · 3 years, 8 months ago

mind = blown, 8 points for you · 3 years, 8 months ago

Now, how do you solve it using Sophie-Germain Identity? · 3 years, 8 months ago

I used the Sophie-Germain Identity to do partial fraction decomposition and used complex numbers to arrive there. · 3 years, 8 months ago

Really? That's possible? I got stuck halfway and can't continue.

If it's better than Brian Chen's answer, that is if you don't need to use Poisson summation formula, I would like to know... · 3 years, 8 months ago

Use partial fraction and reflection formula for poly gamma function to prove that. · 1 year, 1 month ago

Excellent progress, 7 points. Can you finish the proof using the hint? · 3 years, 8 months ago

Has anyone proven the first step? If not, I'd like to take a shot. · 3 years, 8 months ago

Do I really need the knowledge of Fourier to really understand or maybe Riemann Zeta Function? · 3 years, 8 months ago

how do you write the solution · 3 years, 8 months ago

Hint: this problem is related to this problem. · 3 years, 8 months ago

You should probably clear up any ambiguity relating to the fact that you used $$a, b$$ as dummy variables and also in the answer form. · 3 years, 8 months ago

Fixed. · 3 years, 8 months ago

Hint #2: Sophie-Germain Identity · 3 years, 8 months ago

I'm still waiting for an explanation of how this helps, and I bet I'm not the only one ;-) · 3 years, 8 months ago

See my comment under Pi Han Goh's comment. · 3 years, 8 months ago

cody i wanna ask you that how can i submit my solution .... using all the maths terms??? please help · 3 years, 8 months ago

Just post your solution to this thread. · 3 years, 8 months ago

what is this equation · 3 years, 8 months ago

I've always wondered this, what does the notation mean when you have two sigmas next to each other? Does this mean that they both start at 1, then both go to 2, then both go to 3? Or does it mean the sum of a = 1 and b = 1 to infinity, the sum of a = 2 and b = 1 to infinity, the sum of a = 3 and b =1 to infinity, etc..?

From this problem I think it's the latter or else it would just be stated as "$$\displaystyle \sum_{a = 1}^{\infty} \frac{1}{5a^4}$$. · 3 years, 8 months ago

$\displaystyle \sum_{a=1}^m \displaystyle\sum_{b=1}^n f(a, b)$ means $\displaystyle \sum_{a=1}^m \left(\displaystyle\sum_{b=1}^n f(a, b)\right),$ or $(f(1, 1) + f(1, 2) + \ldots + f(1, n)) \\+ (f(2, 1) + f(2, 2) + \ldots + f(2, n)) \\+ \ldots \\ + (f(m, 1) + f(m, 2) + \ldots + f(m, n)).$

In other words, we evaluate the inner sums first, taking outside variables as constant.

Note that (in this example at least) we could switch the order of the summation symbols, to get $\displaystyle \sum_{b=1}^n \displaystyle\sum_{a=1}^m f(a, b)$ or $(f(1, 1) + f(2, 1) + \ldots + f(m, 1)) \\+ (f(1, 2) + f(2, 2) + \ldots + f(m, 2)) \\+ \ldots \\ + (f(1, n) + f(2, n) + \ldots + f(m, n)).$

See how these two are actually the same? · 3 years, 8 months ago

Be careful with limits to infinity. You cannot just interchange the order of summation, as that can affect the sum itself. Analysis deals with this, and one of the results is that if the sequence converges absolutely to a finite value, then we can rearrange the terms and it will still converge to the same (finite) value.

For fun, take the sequence $$- \frac{1}{i}$$, and rearrange terms to get get it to converge to any value that you wish. The absolute value of this sequence is the harmonic sequence $$\frac{1}{i}$$, which sums to infinity. Staff · 3 years, 8 months ago

And if you're a computer-sciency type of person, the first way to write it is the same as

int S=0;
for (int a=1; a<=m; a++)
for (int b=1; b<=n; b++)
S += f(a, b);
return S;


and the second way is the same as

int S=0;
for (int b=1; b<=n; b++)
for (int a=1; a<=m; a++)
S += f(a, b);
return S;


which both return the same number. · 3 years, 8 months ago

$$p + q = 292$$ · 3 years, 8 months ago