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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?

KSmarts
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Oleg
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    You want to show that what is even or odd? The Fourier sine series? Cosine series? Full Fourier series? Because the function you have given is neither even nor odd (with respect to $x=0$). – JohnD Dec 03 '14 at 18:08
  • The concept of odd/even function requires the domain of the function to be symmetric (i.e., $\forall x\in \text{dom}(f)(-x\in \text{dom}(f))$). It doesn't make sense to ask whether a function is even or odd if its domain isn't symmetric. What's being asked of you is to look at the $\pi$-periodic extension of $f$ to $\mathbb R$. A partial reiteration of the present comment can be read here. – Git Gud Dec 03 '14 at 18:08
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    Even with respect to what? Reflection across $x=\pi/2$? You need to check whether $f(x)=f(\pi - x)$ (even) or $f(x)=-f(\pi - x)$ (odd), or neither. – mjqxxxx Dec 03 '14 at 18:08
  • @mjqxxxx I have no idea to respect to what , the professor just said that identifying whether a function is odd or even will simplify the process of computing the fourier series. so my answer would be to check if $f(x)=f(\pi-x)$ because of my x can only be between 0 and pi ? – Oleg Dec 03 '14 at 18:14
  • Yes, that's what I'd do. Since $f(\pi - x)=-(\pi - x)^2 + (\pi - x)\pi = \pi x - x^2 = f(x)$, the function is even with respect to $x=\pi/2$, and so you only need terms with that symmetry (i.e., the $\sin$ terms). – mjqxxxx Dec 03 '14 at 19:04

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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.

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 $f$).

JohnD
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