Sum of Cosines: Find the Infinite Series

[jacobi1]

Find \(\displaystyle \sum_{n=0}^\infty \frac{\cos(nx)}{2^n}\).

 

[M R]

Find \(\displaystyle \sum_{n=0}^\infty \frac{\cos(nx)}{2^n}\).

[sp]
By considering the real part of \(\displaystyle \left(\frac{e^{ix}}{2}\right)^n\) and summing a GP I get \(\displaystyle \frac{4-2\cos(x)}{5-4\cos(x)}\)...I think. :p
[/sp]

 

[chisigma]

Find \(\displaystyle \sum_{n=0}^\infty \frac{\cos(nx)}{2^n}\).

Applying the Euler's identity...

$\displaystyle \cos (n x) = \frac{e^{i n x} + e^{- i n x}}{2}\ (1)$

... You obtain...

$\displaystyle \sum_{n =0}^{\infty} \frac{\cos (n x)}{2^{n}} = \frac{1}{2}\ \sum_{n =0}^{\infty} (\frac{e^{i x}}{2})^{n} + \frac{1}{2}\ \sum_{n =0}^{\infty} (\frac{e^{- i x}}{2})^{n} =\frac{1}{2}\ \frac{1}{1 - \frac{e^{i x}}{2}} + \frac{1}{2}\ \frac{1} {1 - \frac{e^{- i x}}{2}}\ (2)$

The task of 'suppression' of the imaginary terms from the (2) is tedious but not very difficult and is left to You...

Kind regards

$\chi$ $\sigma$