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?

  • $\begingroup$ 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)? $\endgroup$ – Yemon Choi Apr 10 '10 at 19:37
  • $\begingroup$ 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 $0$. $\endgroup$ – Robin Chapman Apr 10 '10 at 20:02
  • $\begingroup$ Following Robin's comment, you need some hypotheses on the decay of your coefficients in order for the Fourier series to converge. $\endgroup$ – S. Carnahan Apr 10 '10 at 20:11
  • 1
    $\begingroup$ 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. $\endgroup$ – WBT Apr 10 '10 at 20:17
  • $\begingroup$ 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. $\endgroup$ – WBT Apr 10 '10 at 20:25

This is a consequence of the Dominated Convergence Theorem, see This question.

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.