Let $f(x)$, (where $−L < x < L$), be an odd function that satisfies the symmetry condition $f (L − x) = f (x)$, for all $x$. Show that $A_n =0$, $\forall n$ while $Bn =0$ for all even $n$.
Some definitions:
Fourier Series: $A_0+\sum_{n=1}^{\infty}(A_ncos(\frac{n\pi x}{L})+B_nsin(\frac{n\pi x}{L}))$
where,
$A_0=\frac{1}{2L}\int_{-L}^{L}f(x)dx$
$A_n=\frac{1}{L}\int_{-L}^{L}f(x)cos(\frac{n\pi x}{L})dx$
$B_n=\frac{1}{L}\int_{-L}^{L}f(x)sin(\frac{n\pi x}{L})dx$.
Since $f(x)$ is odd it follows that $A_o=A_n=0$. However, I'm not sure how to use the symmetry condition to prove $Bn =0$ for all even $n$.Any comments are welcome.
(Hoping no calculation mistakes will happen)
For all $n$ you have
$$ B_{n} = \frac{2}{L} \int_{0}^{L} f(x) \sin \left( \frac{2 \pi x}{L} \right) dx $$
Specifically for $2n$ you have
$$ B_{2n} = \frac{2}{L} \int_{0}^{L} f(x) \sin \left( \frac{2 \pi n x}{L} \right) dx $$
Defining $g(x) = f\left(x - \frac{L}{2} \right)$ you have
$$ g(x) = f \left( x - \frac{L}{2} \right) = f \left(-L + x + \frac{L}{2} \right) = -f \left(L - x - \frac{L}{2} \right)$$ $$= \color{red}{-f \left(L - (x + \frac{L}{2}) \right)} = - f \left(x + \frac{L}{2} \right) = f \left(-x - \frac{L}{2} \right) = g(-x) $$
i.e. $g(x) = g(-x)$, $g$ is even
$$ B_{2n} = \frac{2}{L} \int_{-L/2}^{L/2} g(x) \sin \left( \frac{2 \pi n (x-L/2)}{L} \right) dx $$
We specifically have $$ \sin \left( \frac{2 \pi n (x-L/2)}{L} \right) = \sin \left( \frac{2 \pi n x }{L} \right) \cos \left( n \pi \right) - \cos \left( \frac{2 \pi n x }{L} \right) \sin \left( n \pi \right) = (-1)^n \sin \left( \frac{2 \pi n x }{L} \right) $$
Which allows to state
$$ B_{2n} = \frac{2}{L} \int_{-L/2}^{L/2} g(x) \sin \left( \frac{2 \pi n (x-L/2)}{L} \right) dx = (-1)^n \frac{2}{L} \int_{-L/2}^{L/2} g(x) \sin \left( \frac{2 \pi n x}{L} \right) dx = 0 $$
Because $g$ is even and the sin is odd in a symmetric interval.