Hi all,
I'm trying to find:
$\displaystyle \sum_{k=0}^{n} (e^{i\Theta})^{k}$
I manipulate this the following way:
$\displaystyle = 1 + \sum_{k=1}^{n} (e^{i\Theta})^{k} = 1 + \frac{1-e^{in\Theta}}{1-e^{i\Theta}}$
Is this right? Thanks
Hi all,
I'm trying to find:
$\displaystyle \sum_{k=0}^{n} (e^{i\Theta})^{k}$
I manipulate this the following way:
$\displaystyle = 1 + \sum_{k=1}^{n} (e^{i\Theta})^{k} = 1 + \frac{1-e^{in\Theta}}{1-e^{i\Theta}}$
Is this right? Thanks
I see, thanks. Now, If I want to obtain the sum from just the real part of $\displaystyle (e^{i\Theta})^{k}$,
that is, $\displaystyle \sum_{k=0}^{n} cos(k\Theta)$
Could I start from:
$\displaystyle \sum_{k=0}^{n} (e^{i\Theta})^{k} = \sum_{k=0}^{n} cos(k\Theta) + i\sum_{k=0}^{n} sin(k\Theta)$
and then continue this way:
$\displaystyle \sum_{k=0}^{n} cos(k\Theta) = \sum_{k=0}^{n} (e^{i\Theta})^{k} - i\sum_{k=0}^{n} sin(k\Theta) $
Or is there an easier method?