MHB Sum of Cosines: Find the Infinite Series

AI Thread Summary
The discussion focuses on finding the infinite series sum $$\sum_{n=0}^\infty \frac{\cos(nx)}{2^n}$$ using Euler's identity. By expressing cosine in terms of exponential functions, the series can be rewritten and simplified into two geometric series. The resulting expression is $$\frac{1}{2}\left(\frac{1}{1 - \frac{e^{ix}}{2}} + \frac{1}{1 - \frac{e^{-ix}}{2}}\right)$$. Participants note that while the suppression of imaginary terms in the final expression is tedious, it is manageable. The thread emphasizes the importance of careful manipulation of series and identities in deriving the solution.
jacobi1
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Find $$\sum_{n=0}^\infty \frac{\cos(nx)}{2^n}$$.
 
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jacobi said:
Find $$\sum_{n=0}^\infty \frac{\cos(nx)}{2^n}$$.
[sp]
By considering the real part of $$\left(\frac{e^{ix}}{2}\right)^n$$ and summing a GP I get $$\frac{4-2\cos(x)}{5-4\cos(x)}$$...I think. :p
[/sp]
 
jacobi said:
Find $$\sum_{n=0}^\infty \frac{\cos(nx)}{2^n}$$.

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

$\chi$ $\sigma$
 
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