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I was playing around with Desmos, and I put in the following equation:

$$y=\sum_{n=1}^{100x}\sin n$$

I'm not quite sure what I was expecting, but I noticed that the seemingly random dots it produced were never greater than y=2 (and never less than y=-0.25).

I'm wondering if there is any proof and/or explanation that that is the case.

Also, what branch of math would this problem fall under? It really interests me and I would like to know.

Thanks.

Polygon
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    You might be interested in seeing the derivation of the closed form for the sum; basically, it boils down to seeing this as a geometric series in the complex plane. Some of the answers here discuss that or other derivations, and this pdf also discusses the proof through complex numbers. – Milo Brandt Jul 30 '16 at 03:24
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    Note that what you call “random dots” have some nice patterns to them! Here’s the graph of your function for $x=1,2,\dots500$: http://i.imgur.com/uezESYx.gif – Steve Kass Jul 31 '16 at 20:53

3 Answers3

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Note that

$$\begin{align} \sum_{n=1}^N\sin(n)&=\text{Im}\left(\sum_{n=1}^Ne^{in}\right)\\\\ &=\text{Im}\left(\frac{e^i-e^{i(N+1)}}{1-e^i}\right)\\\\ &=\text{Im}\left(e^{i(N+1)/2}\frac{\sin(N/2)}{\sin(1/2)}\right)\\\\ &=\frac{\sin\left(\frac{N+1}{2}\right)\sin(N/2)}{\sin(1/2)}\\\\ &\le \csc(1/2)\\\\ &\approx 2.08582964293349 \end{align}$$

As pointed out by ClementC., we can write

$$\sin\left(\frac{N+1}{2}\right)\sin(N/2)=\frac12\left(\cos(1/2)-\cos(N+1/2)\right)$$

to reveal that

$$\begin{align} \sum_{n=1}^N\sin(n)&=\frac{\cos(1/2)-\cos(N+1/2)}{2\sin(1/2)}\\\\ &\le \frac{\cos(1/2)+1}{2\sin(1/2)}\\\\ &\approx 1.95815868232297 \end{align}$$

Moreover, we find that

$$\begin{align} \sum_{n=1}^N\sin(n)&=\frac{\cos(1/2)-\cos(N+1/2)}{2\sin(1/2)}\\\\ &\ge \frac{\cos(1/2)-1}{2\sin(1/2)}\\\\ &\approx -0.127670960610518 \end{align}$$

Mark Viola
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    As a side note: one can slightly improve the upper bound to get the bound of $2$ conjectured by the OP. Namely, instead of upper bounding the numerator by $1$, one can observe that $$ 2\sin x\sin(x+\frac{1}{2}) = \cos\frac{1}{2}-\cos(2x+\frac{1}{2}) \leq \cos\frac{1}{2} + 1$$ from which $$\sin\frac{N+1}{2}\sin\frac{N}{2} \leq \frac{\cos\frac{1}{2} + 1}{2}.$$ This leads to the bound $$ \frac{\cos\frac{1}{2} + 1}{2\sin\frac{1}{2}} \simeq 1.9582$$ – Clement C. Jul 30 '16 at 03:49
  • @ClementC. Yes, the addition angle formula $\cos(x\pm y)=\cos(x)\cos(y)\mp \sin(x)\sin(y)$ leads to $\sin(N/2)\sin((N+1)/2)=\frac12(\cos(1/2)-\cos(N+1/2))$ from which a tighter upper bound can be obtained. – Mark Viola Jul 30 '16 at 03:54
  • (I think it would not be too hard to show that the upper bound above cannot be improved, as $\cos( N+\frac{1}{2})$ (sorry, I missed a parenthesis in the above comment) gets arbitrarily close to $-1$ when $N$ ranges over the positive integers. (I would bet it's dense in $[-1,1]$ and not hard to prove so.) – Clement C. Jul 30 '16 at 03:58
  • Yes, I would agree. As for the lower bound on absolute value, well it seems that $0$ is the best we can do, I believe. – Mark Viola Jul 30 '16 at 04:02
  • Yes, it looks like it... – Clement C. Jul 30 '16 at 04:02
  • Not really -- it can get close to 0. This is a lower bound ($-0,128$) on the actual quantity, not the absolute value -- right? (Also, if you are interested in adding it to you answer, I have a discrete plot for the first 100 terms or so) – Clement C. Jul 30 '16 at 04:07
  • https://postimg.org/image/picvdgmiv/ – Clement C. Jul 30 '16 at 04:08
  • Thank you! This answer is great! However, I'm not very familiar with working with complex exponents and the "Im" function. Can you please explain how you got from $\text{Im}\left(\frac{e^i-e^{i(N+1)}}{1-e^i}\right)$ to $\text{Im}\left(e^{i(N+1)/2}\frac{\sin(N/2)}{\sin(1/2)}\right)$? – Polygon Jul 30 '16 at 04:14
  • @Polygon: first, factor the denominator by $e^{i/2}$ and the numerator by $e^{i(N+1)/2}$. Then, use the identity $\sin x = \frac{e^{i x}-e^{-ix}}{2i}$. (The $2i$ will cancel between numerator and denominator.) – Clement C. Jul 30 '16 at 04:16
  • You're welcome! My pleasure. – Mark Viola Jul 30 '16 at 04:17
  • @ClementC. Thanks a ton! – Polygon Jul 30 '16 at 04:18
  • @ClementC. Clement, I want to thank you for all of your valuable input to improve the quality of this answer! -Mark – Mark Viola Jul 30 '16 at 04:18
  • Literally, for nothing. That's a great way to spend time :) – Clement C. Jul 30 '16 at 04:19
  • Well, I'll up vote some of your recent posts if you like. ;-)) – Mark Viola Jul 30 '16 at 04:19
  • (Oh, again no need. I'm also unclear on how that would be perceived by the moderators.) – Clement C. Jul 30 '16 at 04:21
  • I should be free to choose which answers are useful to me without justifying. But, I'll refrain if you're uncomfortable. In any case, I've up voted several of your answers in the past based purely on their high quality and ingenuity. -Mark – Mark Viola Jul 30 '16 at 04:25
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    I for one am not a fan of this kind of proof, when the formula for $\cos A - \cos B$ suffices (telescoping sum). I would use complex numbers in one of these situations: If it makes a long proof shorter, or if a short elementary proof does not exist. What was the need for complex numbers here? –  Jul 30 '16 at 11:34
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    @mathguy There are more than one way forward for most problems. This one is quite straightforward and is a robust methodology for evaluating a broad class of series. It's unfortunate that your attitude is myopic. Thank you for your comment. – Mark Viola Jul 30 '16 at 13:45
  • @ClementC.I used my superpowers to edit your 1st comment and added the missing parens. Please check that it's ok now. – Jyrki Lahtonen Aug 08 '16 at 08:16
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$\sin x + \sin 2x + \cdots + \sin nx = (\cos \frac x 2 - \cos (n + \frac 1 2) x)/ 2 \sin \frac x 2$. Now put $x = 1$. It's trigonometry.

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Sum of $sines $ series when angles are in Arithmetical progression

$S=\sin \alpha +\sin (\alpha+\beta)+\sin (\alpha+2\beta) +\cdots n$ terms

We know that

$$2\sin A.\sin B=\cos(A-B) -\cos(A+B) $$ $$\implies 2\sin\alpha.\sin\frac{\beta}{2}=\cos\left(\alpha-\frac{\beta}{2}\right) -\cos(\alpha+\frac{\beta}{2})$$ $$\implies 2\sin(\alpha + \beta).\sin\frac{\beta}{2}=\cos\left(\alpha +\frac{\beta}{2}\right) -\cos\left(\alpha+\frac{3\beta}{2}\right)$$ $$\implies 2\sin\left(\alpha + 2\beta\right).\sin\frac{\beta}{2}=\cos\left(\alpha +\frac{3\beta}{2}\right) -\cos\left(\alpha+\frac{5\beta}{2}\right)$$ $\cdots$ $\cdots$ $$\implies 2\sin\left(\alpha + (n-1)\beta\right).\sin\frac{\beta}{2}=\cos\left(\alpha +( n-1)\beta -\frac{\beta}{2}\right) -\cos\left(\alpha+(n-1)\beta+\frac{\beta}{2}\right)$$

By adding $$2\sin\frac{\beta}{2}\left[\sin \alpha +\sin (\alpha+\beta)+\sin (\alpha+2\beta) +\cdots +\sin(\alpha+(n-1)\beta) \right]=\cos\left(\frac{\alpha-\beta}{2}\right) - \cos\left(\alpha+\frac{(n-1)\beta}{2}\right) $$ $$\implies 2\sin\frac{\beta}{2}.S=2\sin\left(\alpha+\frac{(n-1)\beta}{2}\right).\sin\frac{n\beta}{2}$$ $$=\frac{\sin{\frac{n\beta}{2}}}{\sin{\frac{\beta}{2}}}{\sin\left[ {\alpha + \frac{\beta}{2}{(n-1)}}\right]}$$

Aakash Kumar
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