Complex power series to calculate Fourier series

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SUMMARY

The discussion centers on calculating the Fourier series for the function [\cos(\sin \theta)]e^{\cos \theta}, concluding that it can be expressed as \sum^{\infty}_{n=0} \frac{\cos n \theta}{n!}. The key method involves substituting z=re^{i \theta} into the power series for e^z, which allows for the separation of real and imaginary components. The suggestion to use r = 1 simplifies the problem, aiding in the derivation of the Fourier series representation.

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terhorst
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I never took complex analysis in undergrad and always regretted it, so I'm working through the book Visual Complex Analysis on my own. Really enjoying it so far.

Homework Statement



Actually you can view the problem http://books.google.de/books?id=ogz5FjmiqlQC&printsec=frontcover&hl=en#PPA117,M1", it's #21 at the top of the page.

Show that the Fourier series for [tex][\cos(\sin \theta)]e^{\cos \theta}[/tex] is [tex]\sum^{\infty}_{n=0} \frac{\cos n \theta}{n!}[/tex].

Homework Equations



The author suggests substituting [tex]z=re^{i \theta}[/tex] into the power series for [tex]e^z[/tex] and then isolating the real and imaginary parts.

The Attempt at a Solution



I don't really understand why the substitution. So then you end up with [tex]\sum \frac{r^n [\cos(n \theta) + i \sin (n \theta)]}{n !}[/tex]. But I don't see how this gets me any closer to equating [tex][\cos(\sin \theta)]e^{\cos \theta}[/tex] and [tex]\sum \frac{r^n \cos(n \theta)}{n!}[/tex].

A nudge in the right direction would be appreciated. Thanks for looking.
 
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terhorst said:
Show that the Fourier series for [tex][\cos(\sin \theta)]e^{\cos \theta}[/tex] is [tex]\sum^{\infty}_{n=0} \frac{\cos n \theta}{n!}[/tex].

The author suggests substituting [tex]z=re^{i \theta}[/tex] into the power series for [tex]e^z[/tex] and then isolating the real and imaginary parts.

Hi terhorst! :smile:

(have a theta: θ :wink:)

Hints:

i] use r = 1

ii] what is cos(eisinθ) ? :smile:
 

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