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Prove that

$$ \lim_{N\to +\infty} \frac{\sum_{n=1}^{N} \left[\cos \left( n\Delta \right)\right]^2}{N} = \frac{1}{2}, {\quad \rm for\ } {m\pi} \neq \Delta \in \mathbb{R}, \quad m \in \mathbb{Z} $$

and that

$$ \lim_{N\to +\infty} \frac{\sum_{n=1}^{N} \left[\sin\left( n\Delta \right)\right]^2}{N} = \frac{1}{2}, {\quad \rm for\ } {m\pi}\neq\Delta \in \mathbb{R}, \quad m \in \mathbb{Z}. $$

PS: I have verified the above identities using MATLAB, but I fail to prove them completely with "proper" math skills.

gulugulu
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  • Minor point: It would be more appropriate to write "$0 \neq \Delta \in \Bbb R$" instead. – Aryaman Maithani Jun 29 '21 at 20:06
  • A similar result can be shown for the series where sine/cosine are not squared, and the typical proof of that (e.g. here) usually hinges on the facts that $$\begin{align} e^{i \theta} &= \cos \theta + i \sin \theta\ \Re(e^{i \theta}) &= \cos \theta\ \Im(e^{i \theta}) &= \sin \theta\ \Re(z_1 + z_2) &= \Re(z_1) + \Re(z_2) \ \Im(z_1 + z_2) &= \Im(z_1) + \Im(z_2) \end{align}$$ You usually start with the finite series using just $e^{i \theta}$, and play with the real/imaginary parts as necessary and get the end result. I imagine something similar can work here. – PrincessEev Jun 29 '21 at 20:07
  • Also, the second identity seems to be weird. Shouldn't you get $0$ whenever $\Delta$ is an integer multiple of $\pi$? (Similarly, the first identity should be $1$.) – Aryaman Maithani Jun 29 '21 at 20:08
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    To couple with @EeveeTrainer's hint: Recall that $$\cos^2\theta = \frac{1}{2}(1 + \cos(2\theta)),$$ $$\sin^2\theta = \frac{1}{2}(-1 + \cos(2\theta)).$$ – Aryaman Maithani Jun 29 '21 at 20:10
  • You can also note that it suffices to simply solve one of the two limits. If one is $L$, the other is $1 - L$. In particular, this shows that there's something weird about your conditions on $\Delta$. – Aryaman Maithani Jun 29 '21 at 20:12
  • @Aryaman Maithani Yes. $\Delta \neq i\pi, i = 0, 1, \cdots$, not only 0 – gulugulu Jun 29 '21 at 20:18
  • @Lázaro Albuquerque I have the same thinking in the very beginning, but I tried with MATLAB and it works for any $\Delta$ (rational or irrational) – gulugulu Jun 29 '21 at 20:20

2 Answers2

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This is not a full solution. But this compiles the ideas from the comments which will lead to the solution.

  1. Note that since $\sin^2 \theta + \cos^2 \theta = 1$ for all $\theta \in \Bbb R$, it suffices to solve just the first limit. If that is $L$, then the second limit is $1 - L$.
  2. Let us concentrate on the first limit. Note that $$\cos^2\theta = \frac{1}{2}(1 + \cos(2\theta)).$$
  3. The question is now simplified to solving $$\lim_{N \to \infty}\frac{1}{N}\sum_{n = 1}^N \cos(2n\Delta).$$
  4. Note that angles in the cosine in the sum form an arithmetic progression. Thus, you can evaluate the sum explicitly, see answers and formula here. (Be careful in applying the formula by considering when $\Delta$ is an integer multiple of $\pi$.)
    Actually, all you need from the formula is the fact that the summation is bounded with the bound only depending on $\Delta$ (and not $N$).

I'll point out that your result, as stated, is incorrect. It will fail precisely when $\Delta$ is an integer multiple of $\pi$. In that case, the first limit is $1$ (and hence, the second is $0$).

Aryaman Maithani
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Take the first one. With a few manipulations we have the sum is equal to

$$S = \lim_{N\to\infty}\frac{\sum_{n=1}^N 1 + \cos(2n\Delta)}{2N} = \frac{1}{2}+\lim_{N\to\infty}\frac{1}{N}\operatorname{Re}\left\{\sum_{n=1}^Ne^{2in\Delta}\right\} = \frac{1}{2}+\lim_{N\to\infty}\operatorname{Re}\left\{\frac{e^{i2\Delta}}{N}\frac{e^{2iN\Delta}-1}{e^{2i\Delta}-1}\right\}$$

$$= \frac{1}{2} + \lim_{N\to\infty} \frac{\cos(2N\Delta)}{2N}+\frac{\cot(\Delta)}{2N}+\frac{\cos(2\Delta)}{4N\sin^2(\Delta)}$$

which converges to $\frac{1}{2}$ precisely when the terms on the right are bounded i.e. $\sin(\Delta) \neq 0$. What will always be true regardless of the value of $\Delta$ is that

$$\lim_{N\to\infty}\frac{\sum_{n=1}^N \cos^2(n\Delta)}{N} + \frac{\sum_{n=1}^N \sin^2(n\Delta)}{N} = 1$$

so if you know one, you can find the other.

Ninad Munshi
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  • Using your method, I finally reach a different result $$ \frac{1}{2} -\lim_{N\to +\infty} \Big[ \frac{1}{4N} + \frac{\cos(2(N+1)\Delta)}{8N\sin^2\Delta} - \frac{\cos(2N\Delta)}{8N\sin^2\Delta} \Big]$$ – gulugulu Jun 29 '21 at 23:17