# Parseval's Theorem Q

I have this question:

I know Parseval's theorem is given by $2a_0^2 + \sum_1^{\infty} (a_n^2 + b_n^2) = \frac {2}{T} \int_{-T/2}^{T/2} f(x)^2 dx$, where T is the period.

$f(x)$ is even, so I know I only need the $a_0$, $a_n$ coefficients. I seem to differ from the solution by my calculation of $a_0$. I've said:

$$a_0 = \frac {4}{T} \int_0^{T/2} f(x) dx = \frac {2}{\pi} \int_0^{\pi} x^2 dx = \frac {2\pi^2}{3}$$

However, the solution has this result as $\frac {\pi^2}{3}$. Could anyone suggest where I've gone wrong? Thanks!

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Go to math.stackexchange.com/questions/8337/…. Maybe you will get what you want. –  xpaul Apr 26 '13 at 15:44
I understand the question and the solution, the only thing that differs in my solution is my calculation of $a_0$. I can't see where I went wrong. –  Mike Miller Apr 26 '13 at 15:46
I think your identity is wrong. It should be $\frac{1}{2}a_0^2$. –  xpaul Apr 26 '13 at 15:52
Go to these two links math.stackexchange.com/questions/8337/…, and math.stackexchange.com/questions/28329/…. You will get what you want. –  xpaul Apr 28 '13 at 13:38

The formula for $a_0$ is not the one you have written. Namely, $$a_0=\frac{1}{T}\int_{0}^{T}f(x)dx,$$ or, for even function $f$, $$a_0=\frac{2}{T}\int_{0}^{T/2}f(x)dx.$$ This explains the difference.
P.S. I think many people were confused by $\frac{2\pi}{3}$ instead of $\frac{2\pi^2}{3}$ (which should in fact be $\frac{\pi^2}{3}$) in your last formula.
Sorry, I did mean $2\frac {\pi^2}{3}$. My formula for $a_0$ is $\frac {2}{T} \int_{-\frac{T}{2}}^{\frac {T}{2}} f(x) dx$ or for an even function $\frac {4}{T} \int_{0}^{\frac {T}{2}} f(x) dx$. Is this wrong? Edit: I think you are saying it is wrong and what you are saying is right. In that case I've just written it down wrong in my notes and it clears everything up for me! Thank you. –  Mike Miller Apr 26 '13 at 23:34
In fact, the definition I am given is $a_0 =\frac{2}{T} \int_{-\frac{T}{2}}^{\frac{T}{2}} f(x) dx = \frac{4}{T} \int_{0}^{\frac{T}{2}} f(x) dx$ for an even function. The period is $2\pi$ so I can't see where I've gone wrong. –  Mike Miller Apr 26 '13 at 23:50
You can well use your definition, but then in the expansion of $f$ you wil have $a_0/2$ instead of $a_0$, and $2a_0^2$ in your version of Parceval's theorem by $a_0^2/2$ (as was fairly noticed by xpaul). Otherwise Parceval's theorem will be incorrect (try it for $f(x)=1$, where all Fourier coefficients except $a_0$ are zero). –  Start wearing purple Apr 27 '13 at 7:33