I'm asked to find the Fourier cosine and sine series for $x^2$. I know the steps. However, I'm conceptually confused because we learnt that even functions have only cosine terms, and odd functions have only sine terms. Since $x^2$ is an even function, how/why does it have a sine series?
[Math] Sine series for $f(x) = x^2$
fourier analysisfourier series
Best Answer
That's not the case. Usually $f$ is a periodic function, so the expression is only given on a certain interval. For example, we are given $f(x) = x^2$ on $[0,\pi]$. But this is the only information you knew: $f$ is periodic and $f|_{[0,\pi]} \colon x \mapsto x^2$. To get what is $f$ on the whole axis, we could make the $f$ become an odd function: $$ f(x ) := \begin{cases} x^2, & [0,\pi],\\ - x^2, & [-\pi, 0] \end{cases} $$ then extend it by periodicity $2\pi$, or or make it become an even function: $$ f(x ) := x^2, \quad [-\pi, \pi] $$ and extend it by periodicity $2\pi$, or a function that is neither even nor odd: extend $f (x) = x^2, x\in[0, \pi]$ by periodicity $\pi$. All of these are possible if no more additional requirements. Naturally the sine series of $f$ exists: take the 1st case above and compute its Fourier expansion.