Sum of Cosines: Find the Infinite Series

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SUMMARY

The infinite series $$\sum_{n=0}^\infty \frac{\cos(nx)}{2^n}$$ can be evaluated using Euler's identity. By expressing the cosine function in terms of exponential functions, the series simplifies to $$\frac{1}{2} \left( \frac{1}{1 - \frac{e^{ix}}{2}} + \frac{1}{1 - \frac{e^{-ix}}{2}} \right)$$. This results in the final expression $$\frac{4 - 2\cos(x)}{5 - 4\cos(x)}$$, confirming the convergence of the series for real values of x.

PREREQUISITES
  • Understanding of infinite series and convergence
  • Familiarity with Euler's identity and complex numbers
  • Knowledge of geometric series summation techniques
  • Basic trigonometric identities and their applications
NEXT STEPS
  • Study the derivation of geometric series and its applications in complex analysis
  • Explore advanced topics in Fourier series and their convergence properties
  • Learn about the implications of Euler's formula in signal processing
  • Investigate the relationship between trigonometric functions and complex exponentials
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Mathematicians, physics students, and anyone interested in series convergence, complex analysis, and the application of Euler's identity in solving infinite series problems.

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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