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Problem of the Week #83 - October 28th, 2013

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Chris L T521

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Jan 26, 2012
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Thanks again to those who participated in last week's POTW! Here's this week's problem!

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Problem: Given the Fourier series $\displaystyle t=2\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\sin(nt)$ ($-\pi < t <\pi$), show that $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{90}$.

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Hint:
Integrate the Fourier series an appropriate number of times, and then evaluate the result at an appropriate value for $t$.


Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Chris L T521

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Jan 26, 2012
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No one answered this week's problem. You can find my solution below.

We first note that if $f$ is a piecewise continuous periodic function of period $2L$ and
\[f(t) = \frac{a_0}{2}+\sum_{n=1}^{\infty} \left[a_n\cos\left(\frac{n\pi t}{L}\right) + b_n\sin\left(\frac{n\pi t}{L}\right)\right]\]
is it's Fourier series, then the series that results when we integrate term by term is given as follows:
\[\begin{aligned} F(t) = \int_0^t f(s)\,ds &= \int_0^t\frac{a_0}{2}\,ds+\sum_{n=1}^{\infty} \int_0^t\left[a_n\cos\left(\frac{n\pi s}{L}\right) + b_n\sin\left(\frac{n\pi s}{L}\right)\right]\,ds \\ &= \frac{a_0 t}{2} + \sum_{n=1}^{\infty}\frac{L}{n\pi}\left[a_n\sin\left(\frac{n\pi t}{L}\right) - b_n\left(\cos\left(\frac{n\pi t}{L}\right) - 1\right)\right]\end{aligned}.\]
This new series is actually convergent for all $t$ (proof omitted); furthermore, if $a_0\neq 0$, the series for $F(t)$ is not a Fourier series due to the linear term $\dfrac{a_0t}{2}$.


In this problem, we're given that the Fourier sine series for $f(t)=t$ over the interval $-\pi < t < \pi$ is
\[t=\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\sin(nt).\]

Our objective is to show that $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^4}= \frac{\pi^4}{90}$, so we need to get some variant of $\dfrac{1}{n^4}$ to appear in the summation; this is easily attained by integrating the given series term by term three times. We see that after one integration, we have
\[\frac{t^2}{2} = 2\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}\left(1-\cos(nt)\right)\]
After integrating for the second time, we have
\[\frac{t^3}{6} = 2\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}\left( t-\frac{\sin(nt)}{n}\right)\]
After integrating for the third and final time, we have
\[\begin{aligned}\frac{t^4}{24} &= 2\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}\left(\frac{t^2}{2} +\frac{\cos(nt)}{n^2} - \frac{1}{n^2}\right) \\ &= \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2}t^2 - 2\sum_{n=1}^{\infty}\frac{(-1)^n}{n^4} \cos(nt) - 2\sum_{n=1}^{\infty}\frac{(-1)^{n+1} }{n^4}\end{aligned}\]

We now note that
\[\begin{aligned} \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2} &= 1-\frac{1}{2^2}+\frac{1}{3^2}-\frac{1}{4^2}+\ldots\\ &= \left(1+\frac{1}{2^2}+\frac{1}{3^3}+\frac{1}{4^2}+ \ldots\right) - 2\left(\frac{1}{2^2}+\frac{1}{4^2} +\ldots\right)\\ &= \left(1+\frac{1}{2^2} + \frac{1}{3^2}+\ldots\right) - \frac{2}{2^2}\left(1 +\frac{1}{2^2}+\frac{1}{3^2}+\ldots\right) \\ &= \sum_{n=1}^{\infty}\frac{1}{n^2} - \frac{1}{2} \sum_{n=1}^{\infty}\frac{1}{n^2}\\ &= \frac{1}{2}\sum_{n=1}^{\infty} \frac{1}{n^2} \\ &= \frac{\pi^2}{12} \end{aligned}\]
Likewise, we see that
\[\begin{aligned} \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^4} &= 1 - \frac{1}{2^4} + \frac{1}{3^4} - \frac{1}{4^4} + \ldots\\ &= \left(1 +\frac{1}{2^4} + \frac{1}{3^4} + \frac{1}{4^4} + \ldots\right) - 2\left(\frac{1}{2^4} + \frac{1}{4^4} + \ldots\right)\\ &= \left(1+\frac{1}{2^4}+\frac{1}{3^4}\ldots\right) - \frac{2}{2^4}\left( 1 +\frac{1}{2^4} + \frac{1}{3^4} + \ldots\right) \\ &= \sum_{n=1}^{\infty}\frac{1}{n^4} - \frac{1}{8}\sum_{n=1}^{\infty} \frac{1}{n^4}\\ &= \frac{7}{8} \sum_{n=1}^{\infty} \frac{1}{n^4}.\end{aligned}\]

Thus, we now see that

\[\frac{t^4}{24} = \frac{\pi^2t^2}{12} - 2\sum_{n=1}^{\infty} \frac{(-1)^n}{n^4}\cos(nt) - \frac{7}{4}\sum_{n=1}^{\infty} \frac{1}{n^4}\]

Letting $t=\pi$ gives us

\[\begin{aligned} \frac{\pi^4}{24} = \frac{\pi^4}{12}- 2\sum_{n=1}^{\infty}\frac{(-1)^n}{n^4}\cos(n\pi) - \frac{7}{4}\sum_{n=1}^{\infty} \frac{1}{n^4} &\implies -\frac{\pi^4}{24} = -2\sum_{n=1}^{\infty} \frac{(-1)^n}{n^4}\cdot (-1)^n - \frac{7}{4}\sum_{n=1}^{\infty} \frac{1}{n^4} \\ &\implies -\frac{\pi^4}{24} = -2\sum_{n=1}^{\infty} \frac{1}{n^4}- \frac{7}{4}\sum_{n=1}^{\infty} \frac{1}{n^4}\\ &\implies -\frac{\pi^4}{24} = -\frac{15}{4}\sum_{n=1}^{\infty}\frac{1}{n^4} \\ &\implies \sum_{n=1}^{\infty}\frac{1}{n^4} = -\frac{4}{15}\left(-\frac{\pi^4}{24}\right) = \frac{\pi^4}{90} \end{aligned}\]

This now completes the justification that $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{90}$.$\hspace{.25in}\blacksquare$
 
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