Estimate $\sum_{n = 0}^N \cos (\alpha n^2)$

Question

I want to estimate this sum $$\sum_{n = 0}^N \cos (\alpha n^2)$$ where $\alpha$ is a constant less than $1$ and $N$ is an integer. One of the things that I tried was using Taylor expansion for cosine and then using Stirling's approximation for the factorial in it but summing over the powers of integers involves Bernoulli numbers and it gets tricky quite quickly.

Is there a way to convert this sum into an integral (with an appropriate error term maybe)? If not, is there any other way to estimate this sum? If you could point me to the relevant literature, that would be useful as well.

Answer 1

(Incomplete answer)

a) $\alpha\ll N^{-1}$

$$\begin{aligned}S&=\sum_{n=0}^N\cos(\alpha n^2)\\ &=\int_{-1/2}^{N-1/2}\cos(\alpha x^2)dx+\sum_{n=0}^{N-1}\int_{n-1/2}^{n+1/2}(\cos(\alpha x^2)-\cos(\alpha n^2))dx\\ &=\sqrt{\frac{\pi }{2\alpha}} C\left( (N-1/2) \sqrt{\frac{2}{\pi \alpha}}\right)-\sqrt{\frac{\pi }{2\alpha}} C\left( -1/2 \sqrt{\frac{2}{\pi \alpha}}\right)+r_\alpha(N) \end{aligned}$$where $C$ denotes the Fresnel C function.
Now, we estimate $r_\alpha(N)$.
$$\begin{aligned}\left|\int_{n-1/2}^{n+1/2}(\cos(\alpha x^2)-\cos(\alpha n^2))dx\right|&=\left|\int_{n-1/2}^{n+1/2}2\sin\frac{\alpha(x+n)(x-n)}{2}\sin\frac{\alpha(x^2+n^2)}2dx\right|\\ &\le\left|\int_{n-1/2}^{n+1/2}\alpha(x+n)(x-n)dx\right|\\ &=\alpha/12\end{aligned}$$ Sum them together, we get $|r_\alpha(N)|\le\alpha N/12\ll 1$.

b) $\alpha\not\ll N^{-1}$

I strongly believe that there is no good estimation as $\cos(\alpha x^2)$ starts to oscillate extremely quick when $\alpha x^2\gg 1$ for almost all $\alpha\in\mathbb R$. $\cos(\alpha n^2)$ starts to have "randomness". This MO link gives more information.

Answer 2

Integral is a nice idea. With the inverse Euler's formula you have: $$\cos t = \frac{e^{it} + e^{-it}}{2}$$

I'm not entirely sure but your sum might be: $$\sum_{n = 0}^N \cos (\alpha n^2) = \int_{0}^{N} \frac{e^{i\alpha n^2} + e^{-i\alpha n^2}}{2} dn = \frac{e^{i\alpha n^2}}{4\alpha n} - \frac{e^{-i\alpha n^2}}{4\alpha n} + C\Biggr|_{0}^{N}$$

Hope that helps. My first day at stackoverflow :)

Edit: This would make the sum following: $$\frac{e^{i\alpha N^2}}{4\alpha N} - \frac{e^{-i\alpha N^2}}{4\alpha N} - \frac{1}{0} + \frac{1}{0}$$ so either my equation is wrong or you can just start from 1 to N + cos(0)