3

I was playing around with the sum $\sum_{k = 1}^{n} \sin k$, and using very loose rigour I arrived at the following:

Proposition. Let $n \equiv n_0 \pmod {44}$ and $n_0 \equiv n_1 \pmod {6}$. Then $$\sum_{k = 1}^{n} \sin k \sim \frac {1} {2} \left ( 3 - |3 - n_1| \right ).$$

Does heuristics support this claim?

Note: I can provide my "proof" on request.

  • I don't know about the claim, but there is an explicit formula for this sum that can be found using Euler's formula and the geometric series partial sums. – Jair Taylor Nov 26 '15 at 15:26
  • @JairTaylor I didn't say this result is unique, I asked whether I have deviated from the equality too much when I did some small errors in each step of the calculation. –  Nov 26 '15 at 15:29
  • What does that symbol $\sim$ mean? Are you considering $n \to \infty$? – Crostul Nov 26 '15 at 15:42
  • 1
    This is a consequence of the fact that $44$ is very nearly a multiple of $2 \pi$; indeed, $44 = 7 \cdot 2 \pi + \epsilon$, where $|\epsilon| < \frac{1}{300}$. In particular this should be a good approximation when $n \epsilon \ll 1$, but it will not be a good approximation when this is not true. – Travis Willse Nov 26 '15 at 15:48
  • @Travis Yes, it indeed is. I am not asking for the mechanics behind this near-equality, I was actually playing with those mechanics and they lead me to this. What I am asking is the total error that I've done is too large to actually dominate the terms or is it OK? What do you think, can one checks it against the values of $\frac {\cos 1/2 - \cos (n + 1/2)} {2 \sin 1/2}$ for large $n$? –  Nov 26 '15 at 15:56
  • 1
    No, like I suggested, the approximation is not in general good for large $n$. For example, if $n$ is a multiple of $44$, then $n_1 = 0$ and hence your formula gives an approximate value of $0$, but already for $n = 66 \cdot 44$ the value of the sum is $> 1$, whereas the sum is bounded above in magnitude (for all $n$) by $\frac{1 + \cos \frac{1}{2}}{2 \sin \frac{1}{2}} < 2$. – Travis Willse Nov 26 '15 at 16:14
  • One probably ought to be more precise about what one means by $\sim$, since the approximation only gives half-integer values but the sums on the left-hand side are probably dense in some interval. One could ask, for example, for the largest $n$ for which the right-hand side is the nearest half-integer to the sum on the left-hand side. – Travis Willse Nov 26 '15 at 16:16
  • You're absolutely correct, you seem to know all what I did in order to arrive at the result above. As a remark I should say that I didn't know what to use to mean approximately equals so I used $\sim$ but apparently, I've used it wrong. –  Nov 26 '15 at 16:46

1 Answers1

0

I think it appropriate to present my rigour here now that it created sufficient curiosity.

Consider $$f (n) = \sum_{k = 1}^{n} \sin k.$$ Let $a_k = \left [ \frac {2k} {\pi} \right ]$ for $1 \leqslant k \leqslant n$ and $r_1, r_2, \cdots, r_n$ be real numbers in the interval $[0, 1]$. Then, denote $$\sin k = \begin {cases} r_k, & \text {if } a_k \equiv 0 \pmod {4} \\ 1 - r_k, & \text {if } a_k \equiv 1 \pmod {4} \\ -r_k, & \text {if } a_k \equiv 2 \pmod {4} \\ r_k - 1, & \text {if } a_k \equiv 3 \pmod {4} \\ \end {cases}$$ Here $r_{\theta}$ is the side of the right triangle just in front of the angle $\theta$. By Weyl's criterion, we see that $r_k$ is equidistributed modulo $1$, so the average of real numbers $r_k$ is $\frac {1} {2}$. Hence, $$\sin k \approx \begin {cases} \frac {1} {2}, & \text {if } a_k \equiv 0 \pmod {4} \\ \frac {1} {2}, & \text {if } a_k \equiv 1 \pmod {4} \\ - \frac {1} {2}, & \text {if } a_k \equiv 2 \pmod {4} \\ - \frac {1} {2}, & \text {if } a_k \equiv 3 \pmod {4} \\ \end {cases}$$ Since $\frac {2mk} {\pi}$ is close to an integer when $m = 11 t$ and the period of $\sin k$ with respect to $a_k$ is $4$, it is safe to assume $f (44t + n_0) \approx f (n_0)$ for any integer $t$. Let $n = 44 t + n_1$. Also, since $a_{6j + i} \equiv a_i \pmod {4}$ for any integers $i, j$, we have by setting $i = n_1$ the desired result.

This is a slightly simplified and fast version of what I'd done to arrive at the ridiculous expression that my question was about.