4
$\begingroup$

I know that this can be proven inductively. However, I can't get passed the trig. I am pretty sure trig identities can show that the expression above is true for $n=0$, and that if the expression holds for $n=k$ it holds for $n=k+1$. But alas, I am getting lost in a sea of trig. Hopefully someone can shed some light on this.

$\endgroup$
6
  • 1
    $\begingroup$ Hint : $\sum_{k=1}^n cos(nx)=Re(\sum_{k=1}^n e^{inx})$ $\endgroup$ – Bertrand R Oct 9 '13 at 22:25
  • $\begingroup$ Looks like you can derive this result using Fourier series $\endgroup$ – peterwhy Oct 9 '13 at 22:26
  • $\begingroup$ @peterwhy That would be odd, since this is an important inequality to derive some identities concerning the Dirichlet kernel. $\endgroup$ – Pedro Tamaroff Oct 9 '13 at 22:33
  • $\begingroup$ @PedroTamaroff well, not exactly using a lot of Fourier series, but deriving using an idea from frequency domain is possible. $\endgroup$ – peterwhy Oct 9 '13 at 22:36
  • $\begingroup$ @peterwhy (I wouldn't say your answer is using Fourier series, but it is a nice one. +1) $\endgroup$ – Pedro Tamaroff Oct 9 '13 at 22:39
5
$\begingroup$

Hint: $$\frac{1}{2} + \sum_{k=1}^n \cos(kx) = \frac{1}{2}\sum_{k=-n}^n e^{ikx}$$

$\endgroup$
4
$\begingroup$

Hint: $$ 2\cos(kx)\,\sin(\frac{x}{2})=\sin\left(kx+\frac{x}{2}\right)- \sin\left(kx-\frac{x}{2}\right) $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.