# Proving the locus of a Fourier series is a system of perpendicular lines [closed]

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.

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## closed as not a real question by Andrés E. Caicedo, William, Rudy the Reindeer, Grigory M, J. M.Sep 20 '12 at 17:08

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Please try to make this more understandable. – Alex Becker Jan 11 '12 at 20:05
no no @Alex Becker,i have added also from the book,but somehow it haven't shown it,it is just browser problem maybe ,i have added it – dato datuashvili Jan 11 '12 at 20:06
I have no issue with the question from the book except for the fact that you didn't use LaTeX. But the rest of what you've written, i.e. "proof it", "n is not changing and how is sum of these items zero", "i have added also from the book,but somehow it haven't shown it,it is just browser problem maybe ,i have added it", etc. makes very little sense. – Alex Becker Jan 11 '12 at 20:10
aa last ones are just my opinions,they are not part of question – dato datuashvili Jan 11 '12 at 20:19

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! :-)).

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At a glance, I'd start by thinking about what happens to the expression inside the summation when $x=k\pi, k\in\mathbb{Z}$, or when $y=k\pi, k\in\mathbb{Z}$.

The $\dfrac{(-1)^{n-1}}{n^2}$ part does alternate between positive and negative, but without knowing the signs of the sines, it's not clear to me that the terms being summed alternate between positive and negative.

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so it means that sin(nx)*sin(ny) is equal zero,so because sin(kpi) is zero,it means that yes x=kpi or y=m*pi,but does it divide plane into square are of pi^2? – dato datuashvili Jan 11 '12 at 20:32
@dato: those two sets of lines do divide the plane into squares with sides of length π. But that's only showing that those lines are included in the locus; you'd still have to show that nothing else is in the locus (that is, that when $x\neq k\pi$ and $y\neq k\pi$, the sum of the infinite series is not 0). – Isaac Jan 11 '12 at 20:33
ok @Isaac thanks for helping,just i am surprising who is downvoting – dato datuashvili Jan 11 '12 at 20:35
when x!=kpi and y!=kpi, then it is clear that their sum can't be equal to zero ,i think so – dato datuashvili Jan 11 '12 at 20:39
@dato: Right, the product of the sines is not zero when neither $x$ nor $y$ is an integer multiple of π, but that doesn't mean the sum of the infinite series is nonzero. It's quite likely that there are some positive terms and some negative terms... – Isaac Jan 11 '12 at 20:52