Chapter 7.11, Problem 11P

Let $f(x)$ on $(0,2l)$ satisfy $f(2l-x)=f(x),$ that is, $f(x)$ is symmetric about $x=l.$ If you expand $f(x)$ on $(0,2l)$ in a sine series ${\displaystyle \sum b_n}\sin \frac{n\pi x}{2l},$ show that for even $n,b_n=0.$ Hint: Note that the period of the sines is 4 $l$.

Sketch an $f(x)$ which is symmetric about $x=l,$ and on the same axes sketch a few sines to see that the even ones are antisymmetric about $x=l$.

Alternatively, write the integral for $b_n$ as an integral from 0 to $l$ plus an integral from $l$ to $2\bar{l}$, and replace $x$ by $2l-x$ in the second integral.

(b) Similarly, show that if we define $f(2l-x)=-f(x),$ the cosine series has $a_n=0$ for even $n$.