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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.

Iguana
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    I don't know that this help you. But exists the Chebyshev polynomials and these have the property of $T_n(\cos(x))=\cos(nx)$. – Boris Valderrama Nov 17 '19 at 07:23
  • Thanks for that! It's worth giving it a shot. – Iguana Nov 17 '19 at 07:31
  • This question appears to be a duplicate. There are also a lot of other questions that derive the same result

    $$\sum\limits_{k=0}^{n}\cos(kx)=\frac{1}{2}+\frac{\sin(\frac{2n+1}{2}x)}{2\sin(x/2)}$$

     – Axion004 Nov 17 '19 at 07:32
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    @Axion004 It is a related question but note that there is $k^2$ as opposed to $k$ in this question and that makes it quite different. However, maybe there are some tricks that I can use from that question. Thanks for that! – Iguana Nov 17 '19 at 07:34
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    @Axion004 this is absolutely not a duplicate of that question. That question is trivial compared to this (the point being that there is no closed form analogue of the geometric series when one puts squares in the exponent). – shalin Nov 17 '19 at 07:55
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    Some observations: If $\alpha = \dfrac{\pi}{4}$, then the sum is $\dfrac{\sqrt{2}}{4}N +O(1)$. For each of $\alpha = 0.1, 0.5, 0.8, 0.99$, the sum is less than $2 \times 10^4$ for all $N \le 10^8$. The sum appears to grow slower when $\alpha$ isn't a rational multiple of $\pi$. – JimmyK4542 Nov 17 '19 at 08:30
  • @JimmyK4542 That is interesting because, for my purpose, one can assume $\alpha$ to be a rational multiple of $\pi$. – Iguana Nov 17 '19 at 08:56
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    @Iguana well that changes everything. It seemed like you were trying to gauge the behavior of the partial sums for generic values of alpha, but this is the complete opposite... – shalin Nov 17 '19 at 16:42
  • If $\alpha = \dfrac{p}{q}\pi$, then $\cos(\alpha n^2)$ will be $2q$-periodic. So, you can approximate $S_N = \dfrac{S_{2q}}{2q}N + O(q)$, where $S_N$ is the sum of the first $N$ terms and $S_{2q}$ is the sum of the first $2q$ terms. – JimmyK4542 Nov 17 '19 at 18:41

2 Answers2

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(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.

Kemono Chen
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    It seems from Terry Tao’s answer in that MO post that the partial sums should be growing slightly faster that $\sqrt N$ as $N$ becomes large. Fascinating... – shalin Nov 17 '19 at 08:01
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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)

Can Metan
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    Your integration is wrong and in any case, you can't just convert a sum into an integral without any sort of justification. – Iguana Nov 17 '19 at 06:45
  • Hmm. Would be great if you could tell me what's wrong with it. I hope this at least gives you some idea. – Can Metan Nov 17 '19 at 06:47
  • $\int e^{x^2} dx \neq \frac{e^{x^2}}{2x}$ – Iguana Nov 17 '19 at 06:50
  • @CanMetan The primitive of $e^{i\alpha n^2}$ isn't $e^{i\alpha n^2}/(2\alpha n)$ There isn't a common primitive for this function, is similar to the gaussian – Boris Valderrama Nov 17 '19 at 06:50