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I'm struggling to find the sum of $S_n=\cos{x}+2\cos{2x}+3\cos{3x}...+n\cos{nx}$

I know that for $z=e^{ix}$, $2\cos{nx}=z^n+\frac{1}{z^n}$.
So I've tried $2S_n=(z+2z^2+3z^3+...+nz^n)+(\frac{1}{z}+\frac{2}{z^2}+\frac{3}{z^3}+...+\frac{n}{z^n})$ but I don't know how to find the sum of $\sum_{r=1}^nrz^r$ or $\sum_{r=1}^nrz^{-r}$

Is there a way to find that sum? Is the way I'm doing it going in the right direction?

Edit for the answer using Thomas Andrews' comment
$S_n=\frac{d}{dx}\frac{\sin{\frac{n+1}{2}x}\sin{\frac{n}{2}x}}{\sin{\frac{x}{2}}}$
$=\frac{\sin{\frac{x}{2}}(a\sin{bx}\cos{ax}+b\sin{ax}\cos{bx})-\sin{ax}\sin{bx}\cos{\frac{x}{2}}}{\sin^{2}{\frac{x}{2}}}$ where $a=\frac{n+1}{2}$ and $b=\frac{n}{2}$

Pen and Paper
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    Use $\cos z = \Re e^{i z}$ instead. To find the corresponding sum, differentiate the (finite) geometric series. – Gary Oct 12 '21 at 00:21
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    An alternate start: It’s the derivative of $\sin x+\sin 2x+\cdots +\sin nx.$ – Thomas Andrews Oct 12 '21 at 00:30
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    You could continue simplifying the result to obtain $$\sum_{k=1}^n k, \cos(kx)=\frac{1}{4} \csc ^2\left(\frac{x}{2}\right) ((n+1) \cos (n x)-n \cos ((n+1) x)-1)$$ – Claude Leibovici Oct 12 '21 at 02:02
  • @ClaudeLeibovici Could you show me how to get that result, I've been trying to simplify for ages and I can't match it with your result, I get $\frac{1}{4}\csc^2{\frac{x}{2}}(1+\cos{(n+1)x}-n\sin{(n+\frac12)x}\sin{\frac{x}{2}})$ – Pen and Paper Oct 12 '21 at 07:06
  • HINT Your idea is fine, let use now $$f(z)=\sum_{0}^n z^r \implies f’(z)=\frac1z\sum_{0}^n rz^r$$ – user Oct 12 '21 at 00:23

1 Answers1

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$S=e^{ix}+2e^{2ix}+3e^{3ix}+\ldots+ne^{inx} \quad \quad \quad \quad \quad \quad \quad \quad (1)$

$e^{ix}S=\quad\quad e^{2ix}+2e^{3ix}+\ldots+(n-1)e^{inx}+ne^{(n+1)ix} \quad (2)$

$(1) - (2) \implies (1-e^{ix})S=e^{ix}\frac{1-e^{inx}}{1-e^{ix}}-ne^{i(n+1)x}$

$\implies S_n = Re(S)=Re\left(\frac{e^{ix}-(n+1)e^{i(n+1)x}+ne^{i(n+2)x}}{(1-e^{ix})^2}\right)$

Sandipan Dey
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