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#### jacobi

##### Active member

- May 22, 2013

- 58

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

- Thread starter jacobi
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- Thread starter
- #1

- May 22, 2013

- 58

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

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

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.

- Feb 13, 2012

- 1,704

Applying the Euler's identity...Find \(\displaystyle \sum_{n=0}^\infty \frac{\cos(nx)}{2^n}\).

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

... 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

$\chi$ $\sigma$