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!


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 $2 \pi x$.

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

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