Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have this question:

enter image description here

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!

share|cite|improve this question
Go to…. 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…, and…. You will get what you want. – xpaul Apr 28 '13 at 13:38
up vote 1 down vote accepted

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.

share|cite|improve this answer
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
Okay I think I see, thank you. – Mike Miller Apr 27 '13 at 13:36

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.