Proove this identity

**kalyanram** · June 22nd 2008, 11:37 PM

Pls proove this
$y(p,q)=\lim_{N \to \infty}\frac{1}{N} \sum^{N-1}_{l=0} e^{i(pl+ql^{2})}=0$
for p≠0, 2π, 4π, ... and q≠0, 2π, 4π, ...

~Kalyan.

**CaptainBlack** · June 23rd 2008, 05:02 AM

Originally Posted by topsquark

Well, you know that $e^{i \theta} = cos(\theta) + i~sin(\theta)$ is bounded, right? At most the sum of the cosine series is N because $| cos(\theta) | < 1$ given your constraints. So what does that mean for the limit of the cosine series? You can do the same with the sine series.

I'll let you fill in the details.

-Dan

All you are going to get from this approach is that the limit is less than or equal to $1$ .

A hand-waving argument would go that the complex exponentials are of unit magitude and random phase so as $N$ becomes large the sum goes as $\sqrt{N}$ , but we divide by $N$ so $y$ goes to $0$ .

Unfortunatly I see no way to make this rigourous so some more thought is needed.

RonL

**Opalg** · June 23rd 2008, 08:37 AM

Originally Posted by kalyanram

Pls proove this
$y(p,q)=\lim_{N \to \infty}\frac{1}{N} \sum^{N-1}_{l=0} e^{i(pl+ql^{2})}=0$
for p≠0, 2π, 4π, ... and q≠0, 2π, 4π, ...

~Kalyan.

You will need a stronger restriction on p and q than that. If p=q=π then $e^{i(pl+ql^{2})} = e^{i\pi l(l+1)} = 1$ , so the limit will be 1 in that case.

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