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On the function $f(x)$ = $x^3$ on $(-1,1)$ find Fourier coefficients for this function and then use Parseval's Theorem to evaluate:

$$\sum_{n=1}^{\infty}\frac{1}{n^6}$$

Current work:

I have used Mathematica to find the sum, which results in $\frac{\pi^6}{945}$

Since $a_0$ and $a_n = 0$ as it is an odd function, my problem is calculating $b_n$. I am using Mathematica and resulting in:

$$\int_a^b x^3 \sin{(n x \pi)} \,dx= \frac17$$ in which $a$ is $-1$ and $b$ is $1$, the right hand side is calculated from the average value formula, by squaring Integral$[f(x)dx]$.

This is the problem, as my result for the $b_n$ coefficients, is resulting in a nasty rational:

$$-2(-6+n^2\pi^2)cos(n\pi) / n^3\pi^3$$

where I can't seem to factor out a $\frac{1}{n^6}$ term to evaluate the sum using the theorem.

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  • $\begingroup$ are you sure $\int_{-1}^1 x^3 \sin(nx\pi)dx = \frac{1}{7}$, it contradicts Riemann-Lesbegue Lemma. $\endgroup$
    – Yimin
    Commented Feb 15, 2013 at 19:36
  • $\begingroup$ sorry, what i meant from that was that, there are two sides of the theorem's equation. the left hand side which computes $b_n$ which is equal to the average value of just $f(x)^2$, which computes to 1/7. The left hand side of the equation, the integral, is just set up in the equation form, unless I am doing something wrong.. $\endgroup$
    – julesverne
    Commented Feb 15, 2013 at 19:41

1 Answer 1

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When you integrate (using integration by parts or a table) \begin{equation*} b_n=\int_{-1}^{1}x^3\sin (n \pi x) \, dx = 2\int_{0}^{1}x^3 \sin (n \pi x) \, dx \end{equation*} you get \begin{equation*} -\frac{2(-1)^n}{n\pi}\left( 1-\frac{6}{n^2\pi^2} \right). \end{equation*} Squaring this, we obtain \begin{equation*} \frac{4}{n^2\pi^2}-\frac{48}{n^4\pi^4}+\frac{144}{n^6\pi^6}. \end{equation*} If we use the fact that \begin{equation*} \sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}, \quad \sum_{n=1}^{\infty}\frac{1}{n^4}=\frac{\pi^4}{90} \end{equation*} together with Parseval's identity \begin{equation*} \frac{1}{1}\int_{-1}^{1}|x^3|^2 \, dx = \sum_{n=1}^{\infty}|b_n|^2\quad (\text{since } a_n=0) \end{equation*} the result follows.

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  • $\begingroup$ superb.. thank you so much. I was stuck after the term expansion, not knowing that using the $1/n^2$ identity to help reduce would essentially cancel out all the unnecessary terms. $\endgroup$
    – julesverne
    Commented Feb 15, 2013 at 21:09

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