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I'm trying to use Parseval's identity to evaluate the values of the series $$\sum_{k=0}^{\infty}\frac{1}{(2k+1)^6}$$ using a Fourier series that I have derived earlier as $f(x)=x(\pi-|x|) = \sum_{k=0}^\infty \frac{8}{\pi (2k+1)^3}\sin((2k+1)x)$. What I know is a special case of the Parseval identity, namely:$${\frac {1}{\pi }}\int _{-\pi }^{\pi }f(x)^{2}\,\mathrm {d} x={\frac {a_{0}^{2}}{2}}+\sum _{k=1}^{\infty }(a_{k}^{2}+b_{k}^{2}).$$ I was able to use this to evaluate $\sum_{n=1}^\infty \frac{1}{n^4}$ given $g(x)=x^2=\frac{pi^2}{3}+\sum_{n=1}^{\infty}\frac{4(-1)^n}{n^2}\cos(nx)$ which was straightforward but I'm struggling with $f(x)$ since the series has a different form. Am I supposed to express the part with $\sin((2k+1)x)$ in a different way so that I can apply the Parseval identity?

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  • $\begingroup$ The sum of every squared Fourier coefficient is just$$\sum_{k=1}^\infty b_k^2=\sum_{k=0}^\infty\frac{64}{\pi^2(2k+1)^6}$$ $\endgroup$ Jun 22, 2020 at 13:07
  • $\begingroup$ @PeterForeman So you are saying I can just take the coefficient even though my Fourier series is not of the form $f(x)\sim {\frac {a_{0}}{2}}+\sum _{k=1}^{\infty }\left(a_{k}\,\cos(kx)+b_{k}\,\sin(kx)\right)$? $\endgroup$
    – MJP
    Jun 22, 2020 at 13:25
  • $\begingroup$ If you think about it, what Peter Foreman said is exactly the same as what you have there. In both cases, we are summing the squares of the coefficients. In your expression, you're simply adding a bunch of zeros (coming from the odd terms) $\endgroup$ Jun 22, 2020 at 13:29
  • $\begingroup$ @PeterForeman Just to clarify, I was asking if the above mentioned special case (which I could only find in the German wikipedia-article about the Parceval identity: de.wikipedia.org/wiki/Parsevalsche_Gleichung) is applicable in cases where the sine or cosine-function have an argument different than simply kx. $\endgroup$
    – MJP
    Jun 22, 2020 at 19:27
  • $\begingroup$ @MJP Yes as Benjamin says above we just need the sum of the squares of every coefficient of $\sin{(kx)}$ and $\cos{(kx)}$. In your case we have every other coefficient equal to zero so we can just add the odd indexed ones to get your result. $\endgroup$ Jun 22, 2020 at 19:42

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But if you know the sum of $\;1/n^6\;$ then you're done:

$$\frac{\pi^6}{945}=\sum_{n=1}^\infty\frac1{n^6}=\sum_{n=1}^\infty\frac1{(2n)^6}+\sum_{n=1}^\infty\frac1{(2n-1)^6}=\frac1{64}\frac{\pi^6}{945}+\sum_{n=1}^\infty\frac1{(2n-1)^6}\implies$$

$$\sum_{n=1}^\infty\frac1{(2n-1)^6}=\left(1-\frac1{64}\right)\frac{\pi^6}{945}=\ldots$$

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