How to calculate $ \sum_{k=1}^{n} \sin(2k-1)x $?

Question

How do i calculate $ \sum_{k=1}^{n} \sin(2k-1)x $, using $e^{i*x}=cos(x) + i*sin(x)$?

score 3 · Accepted Answer · answered Nov 21 '16 at 16:59

3

Hint: $\sum_{k=1}^n e^{(2k-1)ix}$ is a geometric series.

answered Nov 21 '16 at 16:59

Robert Israel

With formula for geometric series, i need to have exponent $k$, so i need to transform it somehow. My idea is, that it should look like this:
$e^{-ix} \sum_{k=1}^{n} e^{2ix^{k}}$
– Zvnoky Brown Nov 21 '16 at 20:15

1 Answers1