Finding \sum_{n=0}^{N-1} Cos (nx)

[ognik]

Hi - going in circles about a step in an example, to find $ \sum_{n=0}^{N-1} Cos (nx) $

The book says we find \(\displaystyle \sum_{n=0}^{N-1} \left( e^{ix} \right)^n = \frac{1-e^{iNx}}{1-e^{ix}}\) , then take the real part.

But I see $ \sum_{n=0}^{N-1} \left( e^{ix} \right)^n $ as a geometric series with a=1, $ r= e^{ix}$ and n=N-1 and \(\displaystyle S_n = \frac{1-r^n}{1-r} = \frac{1-e^{i(N-1)x}}{1-e^{ix}}\)

If I take the sum from n=1 to N I still don't get their result, 'cos then the 1st term would be $e^{ix}$ instead of 1

What am I missing with that N-1 slight of hand please?

 

[Deveno]

Try $n = N$, the formula for a geometric partial sum is:

$S_n = \sum\limits_{k=0}^{n-1} ar^k$

there's no "index sleight-of-hand".

 

[ognik]

Hi - I know that if I just use n=N it works, but the series is only summed to N-1? This would add an extra term...

 

[Prove It]

Hi - going in circles about a step in an example, to find $ \sum_{n=0}^{N-1} Cos (nx) $

The book says we find \(\displaystyle \sum_{n=0}^{N-1} \left( e^{ix} \right)^n = \frac{1-e^{iNx}}{1-e^{ix}}\) , then take the real part.

But I see $ \sum_{n=0}^{N-1} \left( e^{ix} \right)^n $ as a geometric series with a=1, $ r= e^{ix}$ and n=N-1 and \(\displaystyle S_n = \frac{1-r^n}{1-r} = \frac{1-e^{i(N-1)x}}{1-e^{ix}}\)

If I take the sum from n=1 to N I still don't get their result, 'cos then the 1st term would be $e^{ix}$ instead of 1

What am I missing with that N-1 slight of hand please?

Remember that $\displaystyle \begin{align*} \mathrm{e}^{\mathrm{i}\,n\,\theta} \equiv \cos{ \left( n\,\theta \right) } + \mathrm{i} \sin{ \left( n\,\theta \right) } \end{align*}$, so that means

$\displaystyle \begin{align*} \sum_{n = 0}^{N - 1}{ \left( \mathrm{e}^{\mathrm{i}\,x} \right) ^n } &= \sum_{n = 0}^{N-1}{ \left[ \cos{ \left( n\,x \right) } + \mathrm{i}\sin{ \left( n\,x \right) } \right] } \\ \frac{1 - \left( \mathrm{e}^{i\,x} \right) ^N }{ 1 - \mathrm{e}^{\mathrm{i}\,x} } &= \sum_{n = 0}^{N-1}{ \cos{ \left( n \,x \right) }} + \mathrm{i} \sum_{n = 0}^{\infty}{ \sin{ \left( n \,x \right) } } \\ \frac{1 - \left[ \cos{ \left( N \,x \right) } + \mathrm{i}\sin{ \left( N\,x \right) } \right] }{1 - \left[ \cos{(x)} + \mathrm{i}\sin{(x)} \right] } &= \sum_{n = 0}^{N-1}{ \cos{ \left( n\,x \right) } } + \mathrm{i} \sum_{n = 0}^{\infty}{ \sin{ \left( n\,x \right) } } \\ \frac{\left[ 1 - \cos{( \left( N\,x \right) } - \mathrm{i}\sin{ \left( N \, x \right) } \right] \left[ 1 - \cos{(x)} + \mathrm{i}\sin{(x)} \right]}{\left[ 1 - \cos{(x)} \right] ^2 + \left[ \sin{(x)} \right] ^2} &= \sum_{n = 0}^{N-1}{ \cos{ \left( n\,x \right) } } + \mathrm{i} \sum_{n = 0}^{\infty}{ \sin{ \left( n\,x \right) } } \end{align*}$

Cleaning up the left hand side so that you can read off its real and imaginary parts will enable you to answer the question.