1. ## Proove this identity

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

~Kalyan.

2. Originally Posted by topsquark
Well, you know that $\displaystyle e^{i \theta} = cos(\theta) + i~sin(\theta)$ is bounded, right? At most the sum of the cosine series is N because $\displaystyle | cos(\theta) | < 1$ 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 $\displaystyle 1$.

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

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

RonL

3. Originally Posted by kalyanram
Pls proove this
$\displaystyle y(p,q)=\lim_{N \to \infty}\frac{1}{N} \sum^{N-1}_{l=0} e^{i(pl+ql^{2})}=0$
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 $\displaystyle e^{i(pl+ql^{2})} = e^{i\pi l(l+1)} = 1$, so the limit will be 1 in that case.