Sum to $n$ terms and also to infinity of the following series:$$\cos \theta+ 2\cos 2\theta+ \cdots + n\cos n\theta$$the solution provided by the book is $$S_n=\frac{(n+1)\cos n\theta-n\cos(n+1)\theta-1}{2(1-\cos\theta)}$$Can anyone help me to explain how to get $S_n$.

Thanks in advance.

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    $\begingroup$ $$\sum_{r=1}^nr\cos(rt)=$$ real part of $$\sum_{r=1}^nr(e^{it})^r$$ $\endgroup$ – lab bhattacharjee Aug 27 '18 at 9:01
  • $\begingroup$ Relevant (uses complex exponentials and differentiation and gives the answers for sin and cos): youtu.be/X9J3Cq2_5hA $\endgroup$ – aidangallagher4 Aug 27 '18 at 9:06
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    $\begingroup$ alternatively, look at $\sum_{r=1}^{r=n} \sin (r \theta)$ and then take $\frac{d}{d \theta}$ $\endgroup$ – jim Aug 27 '18 at 9:09
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    $\begingroup$ If you're given the answer, you could also prove it by induction. $\endgroup$ – J.G. Aug 27 '18 at 9:36

To find closed form for series of $\cos$ and $\sin$ functions, approach using complex numbers helps very much. Consider, $z=e^{i\theta}$, where, $i=\sqrt{-1}$. We need to construct a function whose real part or imaginary part(otherwise we will have $i$ in the sum, which we don't have in the sum we are interested to find out) will give the required sum. Now, if we take $f(z)=1+z+z^2+z^3+\cdots +z^n$, $$z\cdot f'(z)=z+2z^2+3z^3+\cdots+nz^{n}\\=\cos{\theta}+2\cos{2\theta}+\cdots+\cos{n\theta}+i\sum_{k=1}^{n}k\cdot\sin{(k\theta)} $$ So, notice that $\Re{\{z\cdot f'(z)\}}$ is the required sum. On the other hand, $f(z)=\frac{z^{n+1}-1}{z-1} $ gives, $$\require{cancel}z\cdot f'(z)=z\cdot \frac{(z-1)\cdot (n+1)z^n-z^{n+1}+1}{(z-1)^2}=z\cdot \frac{(n+\cancel{1})z^{n+1}-(n+1)z^n-\cancel{z^{n+1}}+1}{(z-1)^2}\\= z^{n+1}\cdot \frac{n(z-1)-1}{(z-1)^2} $$

Now, $\sum_{k=1}^{n}k\cdot \cos{(k\theta)}=\Re{\{z\cdot f'(z)\}}=\Re{\{z^{n+1}\cdot \frac{n(z-1)-1}{(z-1)^2}\}}$. After some calculation you will get the result.

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    $\begingroup$ tarit goswami, OP is clearly in precalculus. what is OP to do with this complex analysis answer? :| $\endgroup$ – BCLC Sep 1 '18 at 6:22
  • $\begingroup$ @BCLC Some introductory knowledge$\big(e^{ix}=\cos{x}+i\sin{x}\big)$ of Complex numbers will suffice to understand this :D . Though to be sure that differentiation works for complex polynomial will need complex analysis, OP can take it for granted now. This way of using $e^{ix}$ makes calculation of closed forms of lots of complicated series easier. $\endgroup$ – tarit goswami Sep 1 '18 at 6:41

Hint: Take integral you get:

$$ (\sin x+\sin 2x+\sin 3x+\dots+\sin nx)=\frac{\cos x/2 - \cos (n+ 1/2)x}{2 \sin x/2} $$

Now take the derivative.

  • $\begingroup$ $sin\theta +sin2\theta +sin3\theta... +sin n\theta=\frac{sin\frac{ n\theta}{2}}{sin\frac{\theta}{2}}.sin(n+1)\frac{\theta}{2}$ I couldn't understand how did you get $\frac{cos\frac{x}{2}-cos(n+1/2)x}{2sin\frac{x}{2}}$.could you explain please.And thank you for your Hint.@sirous $\endgroup$ – emonhossain Aug 27 '18 at 10:43
  • $\begingroup$ I think you are right. $\endgroup$ – Narasimham Aug 27 '18 at 12:05
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    $\begingroup$ md emon, it has a rather long procedure. Use expansion of $e^{i X}$ for $X= x, 2x, 3x ...$. The sum of LHS gives a geometric progression its result must be found. The RHS gives the sum of $ \cos nx$ and $\sin nx$. Writing LHS $e^{S}i=\cos s + i sin s$ gives the result where $\cos s$ is for the sum of $\cos ns$ and $\sin s$ gives the sum of $\sin nx$. $\endgroup$ – sirous Aug 27 '18 at 13:40
  • $\begingroup$ There is also a prof in this site. just write the formula and search $\endgroup$ – sirous Aug 27 '18 at 14:03
  • $\begingroup$ @sirous Yeah, I have done exactly in that way, it seems very complicated. $\endgroup$ – tarit goswami Sep 1 '18 at 6:47

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