- #1
AppleFritters
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Homework Statement
I'm given the following function
[tex]f(x) = \begin{cases} x &-2<x<2\\ f(x+4) &\mbox{otherwise} \end{cases} [/tex]
And I'm asked to find the Fourier sine series. Then I'm supposed to use Parseval's theorem to obtain a certain sum.
Homework Equations
Since I have a sine Fourier series, Parseval's Theorem for this says
[tex] \frac{1}{b-a} \int_a^b |f(x)|^2 dx = \sum_{n=1}^\infty b_n^2 [/tex]
The Attempt at a Solution
So I worked through and got the Fourier sine series for this function which is
[tex] f(x) = \frac{4}{\pi} \left[ \sin \left( \frac{\pi x}{2} \right) - \frac{1}{2} \sin \left( \frac{2 \pi x}{2} \right) + \frac{1}{3} \sin \left( \frac{3 \pi x}{2} \right) - \frac{1}{4} \sin \left( \frac{4 \pi x}{2} \right) + \ldots \right] [/tex]
Now I apply Parseval's Theorem:
[tex] \frac{1}{b-a} \int_a^b |f(x)|^2 dx = \sum_{n=1}^\infty b_n^2 \\
\frac{1}{2-(-2)} \int_{-2}^2 x^2 dx = \sum_{n=1}^\infty b_n^2 \\
[/tex]
On the left hand side:
[tex]\frac{1}{4} \int _{-2}^{2} x^2 dx = \frac{16}{12} [/tex]
On the right hand side:
[tex] \left(\frac{4}{\pi} \right)^2 \sum_{n=1}^\infty \frac{1}{n^2} = \frac{16}{\pi ^2} \sum_{n=1}^\infty \frac{1}{n^2} [/tex]
Now equating the left and right hand sides:
[tex] \frac{16}{12} = \frac{16}{\pi ^2} \sum_{n=1}^\infty \frac{1}{n^2}\\ [/tex]
[tex]\frac{\pi ^2}{12} \sum_{n=1}^\infty \frac{1}{n^2} [/tex]
The answer I'm supposed to be getting is
[tex]\frac{\pi ^2}{6} \sum_{n=1}^\infty \frac{1}{n^2} [/tex]
So I'm off by a factor of a half somewhere but I can't figure it out. Some help would be appreciated. Thank you.