Find the Fourier Coefficients

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Homework Statement:
Find the Fourier Coefficient
Relevant Equations:
##f:[0,1]\rightarrow \mathbb{R}## given by
$$f(x)=x^2$$
Consider the function ##f:[0,1]\rightarrow \mathbb{R}## given by
$$f(x)=x^2$$

(1) The Fourier coefficients of ##f## are given by
$$\hat{f}(0)=\int^1_0x^2dx=\Big[\frac{x^3}{3}\Big]^1_0=\frac{1}{3}$$
$$\hat{f}(k)=\int^1_0x^2e^{-2\pi i k x}dx$$

Can this second integral be evaluated?

Staff Emeritus
Homework Helper
Why couldn't it be? I don't see what potential problem you're thinking of.

docnet
Homework Helper
Integrate by parts twice.

Delta2 and docnet
docnet
@vela @pasmith thank you. Using integration by parts makes perfect sense for the integral.

I am confused about something else.

so... I know how to find the Fourier series of ##f(x)=x^2## over ##[-\pi,\pi]##, and I want to change the domain to ##[0,1]##. Is it okay to do a change of variables using ##x=\pi(2y-1)## and compute the Fourier series in ##y## space?

Not having studied the Fourier series in detail, I am asking silly questions. Could one say that ##[0,1] \subset [-\pi,\pi]## so the series over ##[-\pi,\pi]## contains the Fourier series for the other?

edit: the change of variables using ##x=\pi(2y-1)## scales the domain.

Last edited:
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Solution attempt:

Determining the Fourier series of a function ##f(x):[0,1]\rightarrow \mathbb{R}##

(1) Theorem: For an even function ##f(x)## over the symmetric range ##[-\pi,\pi]##, the Fourier series is given by
$$f(x)\sim \frac{a_0}{2}+\sum_{n=1}^\infty a_k cos(kx)$$ where $$a_k=\frac{2}{\pi}\int^\pi_0f(x)cos(kx)dx$$
For ##f(x)=x^2## we compute the Fourier coefficients $$\int^\pi_0x^2dx=\Big[\frac{x^3}{3}\Big]^\pi_0=\boxed{a_0=\frac{\pi^2}{3}}$$
$$a_k=\int^\pi_0x^2cos(kx)dx=\frac{2}{\pi}\Big[\frac{2xcos(kx)}{k^2}+\Big(\frac{x^2}{k}-\frac{2}{k^3}\Big)sin(kx)\Big]^\pi_0$$
$$=\frac{2}{\pi}\Big[\frac{2xcos(kx)}{k^2}\Big]_0^\pi=\frac{4cos(kx)}{k^2}\Rightarrow \boxed{a_k=(-1)^k\frac{4}{k^2}}$$
(2) The Fourier series of ##x^2## is
$$\boxed{x^2\sim \frac{\pi^2}{3}+\sum^\infty_{n=1}(-1)^k\frac{4}{k^2}cos(kx)}$$

Delta2
Staff Emeritus
Homework Helper
You found the series for ##f: [-\pi, \pi] \to \mathbb{R}##.

Did the original problem ask you to find the Fourier cosine series, in which case you would extend ##f## so you have an even function, or the Fourier series?

docnet
docnet
You found the series for ##f: [-\pi, \pi] \to \mathbb{R}##.

Did the original problem ask you to find the Fourier cosine series, in which case you would extend ##f## so you have an even function, or the Fourier series?
Hi vela,

I'm sorry but you should to explain like I'm 5 years old what the difference is?

Thanks - newb

Staff Emeritus