Odd or Even for Fourier Series?

Question

I have the function $f(x) = -x^2 + x\pi$ and $0\le x\le \pi$ and without seeing the graph I want to show if it is odd or even, but of course $f(x) = f(-x)$ doesn't show that it is even because I can't take $-x$. So how can I show that it's even?

Answer 1

The question is not well-formed, but I will take a shot in an effort to help form it.

The answer depends on how you extend $f(x)=-x^2+\pi x$, $0< x< \pi$ to $-\pi< x< 0$. If you do an even extension, $$f_e(x):=\begin{cases}f(-x), &-\pi<x<0,\\ f(x), &0<x<\pi,\end{cases}$$ then the full Fourier series will end up being just a Fourier cosine series (all sine terms vanish) and thus the resulting Fourier series is even. On the other hand, if you do an odd extension, $$f_o(x):=\begin{cases}-f(-x), &-\pi<x<0,\\ f(x), &0<x<\pi,\end{cases}$$ then the full Fourier series will end up being just a Fourier sine series (all cosine terms vanish) and thus the resulting Fourier series is odd. Finally, you could extend $f$ in some arbitrary way and the result be neither even nor odd.

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The only other thing I can think that you might be asking is, if I just use $f(x)$ as defined on $0<x<\pi$, will its Fourier series end up being even or odd?

The answer there is (again), it depends on whether you use a Fourier sine or cosine series. If you do a Fourier cosine series for $f$ on $0<x<\pi$, the series will be even (and it will correspond to the even periodic extension of $f$), but if you do a Fourier sines series, the series will be odd (and the series will correspond to the odd periodic extension of $f$).