$\sum_{k=0}^{n} Re(\cos(kx))=\sum_{k=0}^{n} Re(\cos(x)^k)$?

Asked Aug 26 '17 at 14:55

Active Aug 26 '17 at 14:55

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I recently saw the following (Proving $\sum\limits_{k=0}^{n}\cos(kx)=\frac{1}{2}+\frac{\sin(\frac{2n+1}{2}x)}{2\sin(x/2)}$), and I saw that they said the following is true.

$$\sum_{k=0}^{n} Re(\cos(kx))=\sum_{k=0}^{n} Re(\cos(x)^k)$$

At the moment I believe this to be false, I tried a small case with, $n=2$ and $x=pi/4$. Which gives $$\sum_{k=0}^{2} Re(\cos(k\frac{\pi}{4}))=\sum_{k=0}^{2} Re(\cos(\frac{\pi}{4})^k),$$ and this is not true.

The most I could show was, $\cos(kx)=Re((\cos(x)+i\sin(x))^k)$.

Could any one show me why the person said the first equation is true?

asked Aug 26 '17 at 14:55

Andy Nguyen

2

Probably because they were confusing $\cos$ with $\exp$. Or maybe they momentarily believed ${\rm Re}((\cos+i\sin)^k)={\rm Re}(\cos^k)$ because ${\rm Re}(\cos+i\sin)=\cos$. – anon Aug 26 '17 at 15:03
Yea, I'm surprised nobody had caught that mistake then. – Andy Nguyen Aug 26 '17 at 15:16
Good point for you and $\to +1$ – Claude Leibovici Aug 26 '17 at 15:31

$\sum_{k=0}^{n} Re(\cos(kx))=\sum_{k=0}^{n} Re(\cos(x)^k)$?

0 Answers0