CMC - Problem 10

Problem 10. (16 points) If

a=1b=11a4+4b4=πpq,\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+qp+q.

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

Note by Cody Johnson
5 years, 6 months ago

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11 votes

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p=4,q=288p+q=292p = 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) a=18b4a4+4b4=bπcoth(bπ)1 \displaystyle \sum_{a=1}^\infty \frac {8b^4}{a^4+4b^4} = b \pi \coth (b \pi) - 1

\begin{aligned} \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, ζ(3)=7π31802n=11n3(e2nπ1) \displaystyle \zeta (3) = \frac {7 \pi^3}{180} - 2 \sum_{n=1}^\infty \frac {1}{n^3( e^{2n \pi} - 1 )} , and ζ(4)=π490 \zeta (4) = \frac {\pi^4}{90}

Pi Han Goh - 5 years, 6 months ago

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

Fix bb and consider f(a)=1a4+4b4.f(a) = \frac{1}{a^4 + 4b^4}. First, we compute the (continuous) Fourier transform f^(c)=e2iπct1t4+4b4dt \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)g(t), i.e. g(t)=e2iπct1t4+4b4.g(t) = e^{-2i\pi ct}\cdot \frac{1}{t^4 + 4b^4}.

Suppose c0c \geq 0. Then if (t)0\Im(t) \leq 0, we have (2iπct)0\Re(-2i\pi ct) \leq 0 and e2iπct1|e^{-2i\pi ct}| \leq 1. Consider the contour integral of gg around a large semicircle centered at the origin in the half-plane (t)0\Im(t) \leq 0 of the complex plane. Since 1t4+4b4\frac{1}{t^4 + 4b^4} decays quickly enough if tt 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, g(t)dt. \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 ω=eiπ/4\omega = e^{i\pi/4}, a primitive eighth root of unity. gg has four simple poles at t=2ωkbt = \sqrt{2}\omega^k b for k=1,3,5,7k = 1, 3, 5, 7; the relevant poles inside our contour are 2ω5b\sqrt{2}\omega^5b and 2ω7b\sqrt{2}\omega^7b. The residues at those points can be evaluated in an L'Hôpital-esque manner to be:

res2ω5bg=e2iπc(2ω5b)14(2ω5b)3=e2πbc(1+i)1+i16b3res2ω7bg=e2iπc(2ω7b)14(2ω7b)3=e2πbc(1i)1+i16b3 \begin{aligned} \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{aligned}

Additionally supposing cc is an integer, we then have e2πbci=1e^{2\pi bci} = 1 and can further simplify to res2ω5bg=e2πbc1+i16b3res2ω7bg=e2πbc1+i16b3 \begin{aligned} \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{aligned}

So, g(t)dt=2πi(e2πbc1+i16b3+e2πbc1+i16b3)=π4b3e2πbcf^(c)=π4b3e2πbc \begin{aligned} \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{aligned}

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

a=f(a)=c=f^(c)=π4b3(11e2πb+e2πb1e2πb)=πcoth(bπ)4b3 \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 ff is even, a=1f(a)=12(a=f(a)f(0))=πcoth(bπ)8b318b4=18b4(bπcoth(bπ)1), \begin{aligned} \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{aligned} as desired.

I have no idea how the parent post managed to continue from here, or how Sophie-Germain can help, however.

Brian Chen - 5 years, 6 months ago

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mind = blown, 8 points for you

Cody Johnson - 5 years, 6 months ago

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@Cody Johnson What are you talking about? This answer easily deserve 1000 points!

Now, how do you solve it using Sophie-Germain Identity?

Pi Han Goh - 5 years, 6 months ago

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@Pi Han Goh I used the Sophie-Germain Identity to do partial fraction decomposition and used complex numbers to arrive there.

Cody Johnson - 5 years, 6 months ago

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@Cody Johnson 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...

Pi Han Goh - 5 years, 6 months ago

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@Pi Han Goh Use partial fraction and reflection formula for poly gamma function to prove that.

Ishan Singh - 2 years, 11 months ago

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Excellent progress, 7 points. Can you finish the proof using the hint?

Cody Johnson - 5 years, 6 months ago

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

Jacob Erickson - 5 years, 6 months ago

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how do you write the solution

gopal chpidhary - 5 years, 6 months ago

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

John Ashley Capellan - 5 years, 6 months ago

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You should probably clear up any ambiguity relating to the fact that you used a,ba, b as dummy variables and also in the answer form.

Michael Tang - 5 years, 6 months ago

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Fixed.

Cody Johnson - 5 years, 6 months ago

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Hint: this problem is related to this problem.

Cody Johnson - 5 years, 6 months ago

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cody i wanna ask you that how can i submit my solution .... using all the maths terms??? please help

gopal chpidhary - 5 years, 6 months ago

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Just post your solution to this thread.

Cody Johnson - 5 years, 6 months ago

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Hint #2: Sophie-Germain Identity

Cody Johnson - 5 years, 6 months ago

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I'm still waiting for an explanation of how this helps, and I bet I'm not the only one ;-)

Brian Chen - 5 years, 6 months ago

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See my comment under Pi Han Goh's comment.

Cody Johnson - 5 years, 6 months ago

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p+q=292p + q = 292

Michael Lee - 5 years, 6 months ago

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

Cody Johnson - 5 years, 6 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 "a=115a4 \displaystyle \sum_{a = 1}^{\infty} \frac{1}{5a^4}.

Michael Tong - 5 years, 6 months ago

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a=1mb=1nf(a,b)\displaystyle \sum_{a=1}^m \displaystyle\sum_{b=1}^n f(a, b) means a=1m(b=1nf(a,b)),\displaystyle \sum_{a=1}^m \left(\displaystyle\sum_{b=1}^n f(a, b)\right), or (f(1,1)+f(1,2)++f(1,n))+(f(2,1)+f(2,2)++f(2,n))++(f(m,1)+f(m,2)++f(m,n)).(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 b=1na=1mf(a,b)\displaystyle \sum_{b=1}^n \displaystyle\sum_{a=1}^m f(a, b) or (f(1,1)+f(2,1)++f(m,1))+(f(1,2)+f(2,2)++f(m,2))++(f(1,n)+f(2,n)++f(m,n)).(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 - 5 years, 6 months ago

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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 1i - \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 1i \frac{1}{i}, which sums to infinity.

Calvin Lin Staff - 5 years, 6 months ago

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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.

Michael Tang - 5 years, 6 months ago

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what is this equation

ShanKar LAl UttWani - 5 years, 6 months ago

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