I was toying around on Desmos, and I noticed that when successive sine functions were added $(\sin(x) + \sin(2x) + \sin(3x) + \ldots)$, it seemed to form a shape. Here is the graph of that shape.

It looks a little bit like a contorted tangent function or a logarithmic function, but I was wondering how I could find a function that exactly traced out

  1. the maximum values of the sine function,
  2. the minimum values of the sine function, and
  3. the average values of the peaks and valleys.

The function traced by the derivative is also quite interesting; it looks somewhat like a secant function. Any further information on this topic or directions in which to further my understanding on this topic would be greatly appreciated!

EDIT: On a second reread, I am looking more for a function that accomplishes this without the oscillations of the sine waves. In this graph the tangent function almost accomplishes this for the average case, but not quite.

  • $\begingroup$ A few more for you to $ $play with:\begin{align}\sum_{n=1}^\infty\frac{\sin(nx)}n&=\sin(x)+\frac{\sin(2x)}2+\frac{\sin(3x)}3+\dotsb\\ \sum_{n=1}^\infty\frac{\sin((2n-1)x)}{2n-1}&= \sin(x)+\frac{\sin(3x)}3+\frac{\sin(5x)}5+\dotsb\\ \sum_{n=1}^\infty\frac{\cos(nx)}{n^2}&=\cos(x)+\frac{\cos(2x)}4+ \frac{\cos(3x)}9+\dotsb\end{align} $\endgroup$ – Akiva Weinberger Jul 13 '17 at 2:09
  • $\begingroup$ (Also, see what happens if you make those alternating sums, say by putting a $(-1)^n$ after the $\sum$ sign) $\endgroup$ – Akiva Weinberger Jul 13 '17 at 2:18

Using Euler's identity, the value of your sum is $$ \sum_{k=1}^n \sin(kx)=\csc \frac{x}{2}\sin \frac{xn}{2}\sin\frac{x(n+1)}{2}. $$ I'll spare you the tedious proof of this, and perhaps you will want to try it yourself.

Does this help you confirm or reject any of your conjectures?

I'm sorry to say that the finding of maxima and minima will be rather tedious, because the derivative is less than beautiful:


In order to find the max/minima, you will need to solve the trigonometric equation $$(n+1)\cos(nx)-n\cos((n+1)x)-1=0$$ For $x$ and $n$, and find the general solution. I will probably edit later once I have done that, if you can't figure it out.

Best wishes, and cool question!

  • $\begingroup$ A bit... from the research that I have done into this series, I was able to find this condensed version of the series, but I couldn't find anything relating to a single function that traces out the maximum points in the function. What I am looking for is a single function that traces out the maximum points without the oscillations of the sine function (and perhaps a different one to trace out the minimum as well). I have edited the question to match this. $\endgroup$ – louie mcconnell Jul 12 '17 at 23:52
  • $\begingroup$ If you just want to connect the top of the waves...$\sin a\sin b = (\frac 12 (\cos (a-b) - \cos (a+b)) \implies csc \frac {x}{2}\sin\frac{xn}{2}\sin\frac {x(n+1)}{2} = \frac 12 (\csc \frac {x}{2}(\cos\frac {x}{2} -\cos\frac {x(2n+1)}{2})$ Why does that matter? For most $x$ there will exist an $n$ such that $\cos\frac {x(2n+1)}{2} = \pm 1$ giving us $\frac 12 (\cot \frac {x}{2} \pm \csc \frac {x}{2})$ for the max and the min. $\endgroup$ – Doug M Jul 13 '17 at 0:32
  • $\begingroup$ That $\frac12(\cot\frac x2\pm\csc\frac x2)$ should be in the answer. $\endgroup$ – Akiva Weinberger Jul 13 '17 at 2:12
  • $\begingroup$ Fascinating! This is exactly what I was looking for, Akiva/Doug. Thank you! I'm still al little spotty on your explanation though, Doug. $\endgroup$ – louie mcconnell Jul 13 '17 at 19:35
  • $\begingroup$ @DougM If you want, you can copy that out of the comments and write your own answer, and I'll delete mine for you. Great answer! $\endgroup$ – Frpzzd Jul 13 '17 at 20:39

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