Fourier Series Proof where $f(x)$ satisfies the symmetry condition $f(L-x)=f(x)$

Question

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.

Answer 1

(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.