- #1

libelec

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## Homework Statement

Using the Fourier trigonometrical series for [tex]f(x) = {x^2},{\rm{ }}0 \le x < 2\pi [/tex], prove that [tex]\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}} = \frac{{{\pi ^2}}}{6}[/tex]

## The Attempt at a Solution

This is more of a "what am I doing wrong question". First, because I'm not in the period defined by the trigonometrical coefficients, I have to change the limits using that:

[tex]0 \le x < 2\pi [/tex]

[tex] - \pi \le x - \pi < \pi [/tex]

[tex]\begin{array}{l}

- \pi \le t < \pi , \\

t = x - \pi , \\

f(t) = f(x - \pi ) \\

\end{array}[/tex]

Then I find the Fourier coefficients with this in hand:

[tex]{a_0} = \frac{1}{\pi }\int\limits_{ - \pi }^\pi {f(t).dt = } \frac{1}{\pi }\int\limits_0^{2\pi } {f(x).dx = } \frac{1}{\pi }\int\limits_0^{2\pi } {{x^2}.dx = } \frac{8}{3}{\pi ^2}[/tex]

[tex]{a_n} = \frac{1}{\pi }\int\limits_{ - \pi }^\pi {f(t).\cos (nt)dt = } \frac{1}{\pi }\int\limits_0^{2\pi } {f(x).\cos (n(x - \pi ))dx = } \frac{1}{\pi }\int\limits_0^{2\pi } {{x^2}.\left[ {\cos (nx)\cos (n\pi ) - \sin (nx)\sin (n\pi )} \right].dx = }

[/tex]

[tex]{a_n} = \frac{1}{\pi }\int\limits_0^{2\pi } {{x^2}.\cos (nx).{{( - 1)}^n}dx = } \frac{{{{( - 1)}^n}}}{\pi }\left( {\left. {\frac{{2x.\cos (nx)}}{{{n^2}}}} \right|_0^{2\pi }} \right) = \frac{{2{{( - 1)}^n}}}{{{n^2}}}[/tex]

[tex]{b_n} = \frac{1}{\pi }\int\limits_{ - \pi }^\pi {f(t).\sin (nt).dx = } \frac{1}{\pi }\int\limits_0^{2\pi } {f(x).\sin (n(x - \pi )).dx = } \frac{1}{\pi }\int\limits_0^{2\pi } {{x^2}.\left[ {\sin (nx)\cos (n\pi ) + \sin (n\pi )\cos (nx)} \right].dx = } [/tex]

[tex]{b_n} = \frac{{{{\left( { - 1} \right)}^n}}}{\pi }\int\limits_0^{2\pi } {{x^2}.\sin (nx).dx = } \frac{{{{\left( { - 1} \right)}^n}}}{\pi }\left( {\left. {(\frac{2}{{{n^3}}} - \frac{{{x^2}}}{n})\cos (nx)} \right|_0^{2\pi }} \right) = \frac{{4{{\left( { - 1} \right)}^{n + 1}}}}{n}[/tex]

Then, using Parseval's identity:

[tex]{\left\| {f(x)} \right\|^2} = \frac{{{a_0}^2}}{2} + \sum\limits_{n = 1}^\infty {{a_n}^2 + {b_n}^2} [/tex]

[tex]\frac{1}{\pi }\int\limits_0^{2\pi } {{{({x^2})}^2}.dx} = \frac{{{{\left( {\frac{8}{3}{\pi ^2}} \right)}^2}}}{2} + \sum\limits_{n = 1}^\infty {{{\left( {\frac{{2{{( - 1)}^n}}}{{{n^2}}}} \right)}^2} + {{\left( {\frac{{4{{( - 1)}^{n + 1}}}}{n}} \right)}^2}} [/tex]

[tex]\frac{{32}}{5}{\pi ^4} = \frac{{32{\pi ^4}}}{9} + \sum\limits_{n = 1}^\infty {\frac{4}{{{n^4}}} + \frac{{16}}{{{n^2}}}} [/tex]

[tex]\frac{{112}}{9}{\pi ^4} = \sum\limits_{n = 1}^\infty {\frac{{4 + 16{n^2}}}{{{n^4}}}} [/tex]

And here I reach a deadpoint. What do I do?