0
$\begingroup$

Find the Fourier series for $f(x)=x^2$ in terms of $\{1,\cos(2\pi\theta),\sin(2\pi\theta),\cos(4\pi\theta),\sin(4\pi\theta),\dots\}$


So I know how to find the Fourier expansion in terms of the standard orthonormal basis of trig functions:

$f(x)=\frac{a_0}{2}+\sum_{n=1}^{+\infty} a_n \ \cos(nx) \ + \ b_n \ \sin(nx)$

$a_0=\frac{1}{\pi} \int_{-\pi}^{\pi} x^2 \ dx=\frac{2}{3} \pi^2$

$a_n=\frac{1}{\pi} \int_{-\pi}^{\pi} x^2 \ \cos(nx) \ dx=\frac{1}{\pi} \ \frac{2(\pi^2 n^2-2) \sin(n \pi)+4 \pi n \cos(n \pi)}{n^3}=\frac{4 (-1)^n}{n^2}$

$b_n=0 \qquad \forall n\ge 1$

and then since $f$ is even:

$f(x)=\frac{\pi^2}{3}+4 \ \sum_{n=1}^{+\infty} \frac{(-1)^n}{n^2} \ \cos(nx)$


If somebody could help walk me through it with the new basis I'd appreciate it... I'm so bad at this stuff! Thanks!

$\endgroup$
1
$\begingroup$

Hint: If you expand $\frac 1 {4\pi^{2}} x^{2}$ as $\sum a_n \sin nx+\sum b_n \cos nx$ on $[-\pi, \pi]$ the you get the required expansion of $x^{2}$ on $|x| \leq \frac 1 2$ by just replacing $x$ by $2 \pi x$.

$\endgroup$
1
  • $\begingroup$ So that's the only difference? all $x$ becomes $2 \pi x$? Amazing!! $\endgroup$
    – Num Toez
    Apr 10 '20 at 0:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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