Q:

Proving the locus of a Fourier series is a system of perpendicular lines

From "Fourier's series and integrals" by H.S. Carslaw, there is the following question:

Prove the zero locus of $\displaystyle \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^2} \sin(n x) \sin(n y) = 0$ is represented by  two systems of lines at right angles dividing the $(x,y)$-plane into squares of area $\pi^2$.

Really I have no idea how to prove it. First of all, the $(-1)^{n-1}$ means the sign in front of the sines is changing from positive to negative and back, yes? I don't understand how if $n$ is not changing the sum can be zero? What if all of the terms are positive, or all of the terms are negative? I think so - am I wrong? Also how to prove the claim in question would be interesting. Please help.

A:

My answer is not really straightforward anyway... Let's rewrite your equation as $\displaystyle \sum_{n=1}^\infty \frac{(-1)^{n-1}}{2 n^2} \left(\cos(n (x-y)) - \cos(n (x+y))\right) = 0$.
This is equivalent to (convergence of the $f(t)$ series being clear) : 
$$f(x-y)=f(x+y)\ \ \mathrm{for}\ f(t)=\sum_{n=1}^\infty \frac{(-1)^{n-1}\cos(n t)}{n^2}$$
But $\displaystyle f(t)=-\sum_{n=1}^\infty \frac{\cos(n (t+\pi))}{n^2}$ and this last sum is well known (or may be obtained by integration of the classical 'Sawtooth Wave' $\sum_{n=1}^\infty \frac{\sin(n (u))}{n}=\frac{\pi-u}{2}$ ) : $$ \sum_{n=1}^\infty \frac{\cos(n u)}{n^2}=\frac{(\pi-u)^2}{4}-\frac{\pi^2}{12}\ \ \mathrm{for}\ u \in (0,2\pi)$$
So that $f(t)=\frac{\pi^2}{12}-\frac{t^2}{4}$ for $t \in (-\pi,\pi)$ and $f(t+2k\pi)=f(t)$ (of course $f$ is even).
At this point $f(x-y)=f(x+y)$ is possible only for $x-y=x+y \mod (2 \pi)$ or $y-x=x+y \mod (2 \pi)$ and I'll let you conclude (and reverify all this of course! :-)).

