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I am given $f(x) = x$ for $0 \le x \le 2$. The question wants me to evaluate $\sum_{n=1}^\infty \frac{1}{(2n+1)^4}$ by first evaluating the Fourier sine of $f(x)$ by extending it outside the interval.

I get $f(x) = 2 + \sum_{n=1}^\infty \frac{-4}{\pi}\sin(\frac{n\pi x}{2})$. And I know that I probably have to make use of Parseval's theorem. Yet, I am not sure how to proceed as I don't know how to get an expression to the power of $4$ from the Fourier sine expression of $f(x)$.

Thanks!

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  • $\begingroup$ your fourier coefficient looks wrong, you have a constant $-\frac{4}{\pi}$ which is independent of $n$, you should get sth that scales like $\frac{1}{n}$ $\endgroup$ – felasfa Oct 11 '16 at 5:32
  • $\begingroup$ This would require a good deal of differentiation (or integration, if you start the other way around). Do you know how the same was done with squares, to begin with? $\endgroup$ – Ivan Neretin Oct 11 '16 at 6:18
  • $\begingroup$ Similar question, to show you how it looks like: math.stackexchange.com/questions/845036/… $\endgroup$ – Jean-Claude Arbaut Oct 11 '16 at 6:48
  • $\begingroup$ The Fourier series may be written, on $]0,2[$, $$f(x)=1-\frac2{\pi}\sum_{n=1}^{\infty} \frac{\sin (n\pi x)}{n}$$ $\endgroup$ – Jean-Claude Arbaut Oct 11 '16 at 7:45
  • $\begingroup$ Have a look at this similar question: math.stackexchange.com/questions/1948206/… $\endgroup$ – Jack D'Aurizio Oct 11 '16 at 10:28
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Once you manage to get that $$ g_1(x)=\frac{4}{\pi}\sum_{n\geq 1}\frac{\sin((2n-1)x)}{2n-1} \tag{1}$$ is the Fourier series of a rectangle wave that equals $1$ over $(0,\pi)$ and $-1$ over $(\pi,2\pi)$, by termwise integration $$ g_2(x) = \frac{4}{\pi}\sum_{n\geq 1}\frac{1-\cos((2n-1)x)}{(2n-1)^2}=C+\frac{4}{\pi}\sum_{n\geq 1}\frac{\cos((2n-1)x)}{(2n-1)^2}\tag{2}$$ is the Fourier series of a triangle wave that equals $x$ over $(0,\pi)$ and $2\pi-x$ over $(\pi,2\pi)$.
In particular $$ C=\frac{1}{2\pi}\left(\int_{0}^{\pi}x\,dx+\int_{\pi}^{2\pi}(2\pi-x)\,dx\right) = \frac{\pi}{2}\tag{3}$$ in order to ensure that $g_2(x)-C$ has mean zero over $(0,2\pi)$. By Parseval's identity $$ \int_{-\pi}^{\pi}(g_2(x)-C)^2\,dx = \frac{16}{\pi}\sum_{n\geq 1}\frac{1}{(2n-1)^4}\tag{4} $$ and $$ \eta(4)=\sum_{n\geq 1}\frac{1}{(2n-1)^4}=\frac{\pi}{16}\int_{-\pi}^{\pi}\left(|x|-\frac{\pi}{2}\right)^2\,dx =\color{red}{\frac{\pi^4}{96}}.\tag{5} $$

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Using your function, $f(x)=x$ on $[0,2]$ then extended by periodicity (that is, the period is $T=2$), you get Fourier coefficients

$$a_0=\frac1T\int_0^T f(x)\mathrm{d}x=1$$

And for all $n>0$ $$a_n=\frac2T\int_0^T f(x)\cos \frac{2n\pi x}{T}\mathrm{d}x=\int_0^2 x\cos (n\pi x)\mathrm{d}x=\left[x\frac{\sin(n\pi x)}{n\pi}\right]_0^2-\int_0^2 \frac{\sin(n\pi x)}{n\pi}\mathrm{d}x=0$$ $$b_n=\frac2T\int_0^T f(x)\sin \frac{2n\pi x}{T}\mathrm{d}x=\int_0^2 x\sin (n\pi x)\mathrm{d}x=\left[-x\frac{\cos(n\pi x)}{n\pi}\right]_0^2+\int_0^2 \frac{\cos(n\pi x)}{n\pi}\mathrm{d}x\\=-\frac{2}{n\pi}$$

Thus, since the function $f$ is piecewise continuous, the Fourier series converges to $\hat f$, and on $]0,2[$, where it's continuous, you have

$$x=1-\frac2\pi\sum_{n=1}^\infty \frac{\sin(n\pi x)}n$$

You can integrate the Fourier series, and you get

$$\frac{x^2}2-x+C=\frac2{\pi^2}\sum_{n=1}^\infty \frac{\cos(n\pi x)}{n^2}$$

The mean of the RHS is zero on $[0,2]$, and to get zero mean on the LHS, you need $C=\frac13$.

You may also write

$$\frac{x^2}2-x=-\frac13+\frac2{\pi^2}\sum_{n=1}^\infty \frac{\cos(n\pi x)}{n^2}$$

Then, Parseval's theorem yields

$$\frac12\int_0^2 \left(\frac{x^2}2-x\right)^2\mathrm{d}x=\frac19+\frac{2}{\pi^4}\sum_{n=1}^\infty \frac1{n^4}$$

Then

$$\frac12\int_0^2 \left(\frac{x^2}2-x\right)^2\mathrm{d}x=\frac12\left[\frac{x^5}{10}+\frac{x^3}{3}-\frac{x^4}{4}\right]_0^2=\frac{2}{15}$$

And $\dfrac{2}{15}-\dfrac{1}{9}=\dfrac{1}{45}$, hence

$$\sum_{n=1}^\infty \frac1{n^4}=\frac{\pi^4}{90}$$

Now,

$$\sum_{n=1}^\infty \frac1{(2n)^4}=\frac{1}{16}\sum_{n=1}^\infty \frac1{n^4}=\frac{\pi^4}{90\times16}$$

And finally

$$\sum_{n=0}^\infty \frac1{(2n+1)^4}=\sum_{n=1}^\infty \frac1{n^4}-\sum_{n=1}^\infty \frac1{(2n)^4}=\frac{\pi^4}{90}\left(1-\frac1{16}\right)=\frac{\pi^4}{96}$$

But!

Your sum starts from $n=1$, so your sum is

$$\sum_{n=1}^\infty \frac1{(2n+1)^4}=\frac{\pi^4}{96}-1$$

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