A property of the fourier series

Question

Show that any periodic function $f(x)$ with period $2\pi$ which is both odd and satisfies $f(\pi-x)=f(x)$ has $b_{n}=0$ for $n$ even and so has a fourier series of the form $$f(x) = \sum^{\infty}_{m=0} c_{m}\sin{(2m+1)x}.$$

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So we know $f$ is odd and periodic with period $2\pi$ thus it has a fourier series of the form $$f(x) = \sum_{n=1}^{\infty}b_{n}\sin{(nx)}\quad \text{where}\quad b_{n}=\frac{2}{\pi}\int^{\pi}_{0}f(x)\sin{(nx)} \ {\rm d}x.$$

Integrating $b_{n}$ by parts yields

\begin{align}b_{n} &= \frac{2}{\pi}\left[-\frac{1}{n}\Big[f(x)\cos{(nx)}\Big]^{\pi}_{0}+\frac{1}{n}\int^{\pi}_{0}f'(x)\cos{(nx)} \ {\rm d}x\right]\\ &= -\frac{2}{n\pi}\Big[f(\pi)\cos{(n\pi)}-f(0)\Big]+\frac{2}{n\pi}\int^{\pi}_{0}f'(x)\cos{(nx)} \ {\rm d}x\end{align}

Notice however $f(x) = f(\pi-x)$ thus $f(\pi) = f(0)$ so $$b_{n} = -\frac{2f(0)}{n\pi}\Big[(-1)^{n}-1\Big] + \frac{2}{n\pi}\int^{\pi}_{0}f'(x)\cos{(nx)} \ {\rm d}x$$

Now from this point onwards I'm not sure what to do with the second integral since we don't know $f(x)$ it will never cancel. Is what I have done going on the right track and what would be the next steps?

Answer 1

$$f(\pi-x) = \sum_{n=1}^{\infty} b_n \sin{n(\pi-x)} = \sum_{n=1}^{\infty} (-1)^{n+1} b_n \sin{n x}$$

The only way $f(\pi-x)=f(x)$ is when $b_n=0$ when $n$ is even.

Answer 2

Use the fact that $f(\pi-x)=f(x)$ to show that for $n$ even,

$$\int_0^\pi f'(x)\cos nx\,dx=-\int_0^\pi f'(x)\cos nx\,dx.$$