# Sum of product of Fourier series

I want to compute the following product

$$\frac{1}{N}\sum_{t=1}^{N}\left(\sum_{s=-\infty}^{\infty}a_{s}\exp(2\pi is\frac{t}{N}\right)\left(\sum_{z=-\infty}^{\infty}a_{z}\exp(-2\pi iz\frac{t}{N})\right)$$ If $N\rightarrow\infty$ we can approximate it with integral: $$\int_{0}^{1}\left(\sum_{s=-\infty}^{\infty}a_{s}\exp(2\pi isx\right)\left(\sum_{z=-\infty}^{\infty}a_{z}\exp(2\pi izx)\right)dx=\sum_{s=-\infty}^{\infty}a_{s}\sum_{z=-\infty}^{\infty}a_{z}\int_{0}^{1}\exp(2\pi i(s-z))dx$$ Hence $s=z$ and we have $$\sum_{s=-\infty}^{\infty}a_{s}^{2}$$

On the other hand this product is equal to: $$\frac{1}{N}\left(\sum_{s=-\infty}^{\infty}a_{s}\right)\left(\sum_{z=-\infty}^{\infty}a_{z}\right)\left(\sum_{t=1}^{N}\exp(2\pi i(s-z)\frac{t}{N})\right)=\sum_{w=-\infty}^{\infty}\sum_{s=-\infty}^{\infty}a_{s}a_{s+Nw}$$

We have $$\sum_{s=-\infty}^{\infty}a_{s}^{2}+\sum_{w=-\infty,w\neq0}^{\infty}\sum_{s=-\infty}^{\infty}a_{s}a_{s+Nw}$$

Now I don't see why the second term will disappear.

Which is correct?

• There is an error somewhere in your penultimate formula. Could you please re-edit? Also, is this an isolated calculation or exercise? if not, could you give us a bit more background context (for instance, why you're interested)? – Yemon Choi Apr 10 '10 at 19:37
• You are taking limits as $N\to\infty$. In the sum $\sum_{w\ne0}a_{s+Nw}$ the terms thin out as $N$ increases. Assuming $a_n\to0$ quickly enough as $n\to\pm\infty$ then $\sum_{w\ne0}a_{s+Nw}$ will tend to $0$. – Robin Chapman Apr 10 '10 at 20:02
• Following Robin's comment, you need some hypotheses on the decay of your coefficients in order for the Fourier series to converge. – S. Carnahan Apr 10 '10 at 20:11
• Yemon Choi, thanks, I have re-edited. I have obtained this expression, while solving some least square problem. And I use Fourier series of basis functions. – WBT Apr 10 '10 at 20:17
• Robin Chapman, Scott Carnahan In my case Fourier coefficients $a_s=sinc(s)^p$ function. But still, I am confused about the limits. Because $s->\infty$, hence $s+Nw$ can be small, depending in which order I'll take limits. – WBT Apr 10 '10 at 20:25

In details: $$\sum_{w\neq0}\sum_{s} a_s a_{s+Nw}$$ is better written $$\sum_{s}\sum_{z} b_{s,z}^N$$ where $$b_{s,z}^N = a_{s}a_{z} \mbox{ when } z=s \mbox{ mod } N,\, N\neq0 \mbox{ and } b_{s,z}^N=0 \mbox{ otherwise.}$$ Now, $|b_{s,z}^N|\leq |a_{s}||a_{z}|$ for all $N$, and $$\sum\sum|a_{s}||a_{z}|=(\sum |a_{s}|)^2<\infty.$$ On the other hand, for any fixed $z, s$ $$\lim_{N\to \infty} b_{s,z}^N = 0 \,(\mbox{ since } a_{s+Nw} \to 0 \mbox{ with }N)$$ Now apply the DCT to conclude.