I'm trying to sketch several periods of the periodic function $f(x)$ below and expand it in an apporpriate Fourier series. I know there's a formula for expanding fourier series, but when I try I always get an error
$$f(x) = x^2,\,\, 0 < x < 10$$
I'm trying to sketch several periods of the periodic function $f(x)$ below and expand it in an apporpriate Fourier series. I know there's a formula for expanding fourier series, but when I try I always get an error
$$f(x) = x^2,\,\, 0 < x < 10$$
Working purely formally, so not worrying about convergence and justifying the interchange of the summation and integration, suppose we have
$$x^{2}=\sum_{n=-\infty}^{\infty}{c_{n}e^{inx}}.$$
Then, recalling that $\frac{1}{\pi}\int_{-\pi}^{\pi}{e^{i(n-m)x}\ dx}=0$ if $n\neq m$ and $2$ otherwise, we can multiply each side of the expression by $e^{imx}$ and integrate, getting
$$\int_{-\pi}^{\pi}{x^{2}e^{imx}\ dx}=\sum_{n=-\infty}^{\infty}{\int_{-\pi}^{\pi}c_{-n}e^{i(m-n)x}\ dx}=2\pi c_{-m},$$
and so
$$c_{n}=\frac{1}{2\pi}\int_{-\pi}^{\pi}{x^{2}e^{-inx}\ dx}.$$
You can calculate this using integration by parts, and you'll find that
$$c_{n}=\frac{2((\pi^{2}n^{2}-2)\sin(\pi n)+2\pi n\cos(\pi n))}{n^{3}}.$$
At this point, notice that, as $n$ is always an integer, $\sin(\pi n)=0$ for each $n$, and $\cos(\pi n)=1$ if $n$ is even, $-1$ if $n$ is odd, i.e. $\cos(\pi n)=(-1)^{n}$. Then we have
$$c_{n}=\frac{4\pi n(-1)^{n}}{n^{3}}=\frac{4\pi(-1)^{n}}{n^{2}},$$
and so the Fourier series is
$$x^{2}=\sum_{n=-\infty}^{\infty}{\frac{4\pi(-1)^{n}}{n^{2}}e^{inx}}.$$
The only thing left to worry about now is convergence, which I suspect you aren't too bothered about. But just for the sake of giving a complete answer, $x^{2}$ is square-integrable on $[-\pi,\pi]$, and so Carleson's Theorem tells us that the Fourier series converges pointwise to $x^{2}$ for almost all $x$.
(In fact, as $\sin$ is an odd function, you'll be able to see that the series is actually just a sum of cosines)