**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.

**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.

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TopNewest\(p = 4, q = 288 \Rightarrow p+q=292\)

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 \( \zeta (4) = \frac {\pi^4}{90} \) – Pi Han Goh · 3 years, 8 months ago

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Poisson summation formula:

A completion of the starting step with theFix \(b\) 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. – Brian Chen · 3 years, 8 months ago

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– Cody Johnson · 3 years, 8 months ago

mind = blown, 8 points for youLog in to reply

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

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– Cody Johnson · 3 years, 8 months ago

I used the Sophie-Germain Identity to do partial fraction decomposition and used complex numbers to arrive there.Log in to reply

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... – Pi Han Goh · 3 years, 8 months ago

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– Ishan Singh · 1 year, 1 month ago

Use partial fraction and reflection formula for poly gamma function to prove that.Log in to reply

– Cody Johnson · 3 years, 8 months ago

Excellent progress, 7 points. Can you finish the proof using the hint?Log in to reply

– Jacob Erickson · 3 years, 8 months ago

Has anyone proven the first step? If not, I'd like to take a shot.Log in to reply

– John Ashley Capellan · 3 years, 8 months ago

Do I really need the knowledge of Fourier to really understand or maybe Riemann Zeta Function?Log in to reply

– Gopal Chpidhary · 3 years, 8 months ago

how do you write the solutionLog in to reply

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

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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. – Michael Tang · 3 years, 8 months ago

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– Cody Johnson · 3 years, 8 months ago

Fixed.Log in to reply

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

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– Brian Chen · 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 ;-)Log in to reply

– Cody Johnson · 3 years, 8 months ago

See my comment under Pi Han Goh's comment.Log in to reply

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

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– Cody Johnson · 3 years, 8 months ago

Just post your solution to this thread.Log in to reply

what is this equation – Shankar Lal UttWani · 3 years, 8 months ago

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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 "\( \displaystyle \sum_{a = 1}^{\infty} \frac{1}{5a^4}\). – Michael Tong · 3 years, 8 months ago

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

Note that (in this example at least) we could

switchthe 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? – Michael Tang · 3 years, 8 months ago

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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. – Calvin Lin Staff · 3 years, 8 months ago

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and the second way is the same as

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

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\(p + q = 292\) – Michael Lee · 3 years, 8 months ago

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– Cody Johnson · 3 years, 8 months ago

I'll award 2 points for this answer, considering the magnitude of this problem. But the solution's where it's at.Log in to reply