Fourier Cosine series of cos(x)

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SUMMARY

The discussion centers on the Fourier Cosine series of the function f(x) = cos(x). Participants confirm that the Fourier series for cos(x) on the interval [-π, π] results in a single non-zero coefficient for n=1, which corresponds to cos(x) itself. The confusion arises when considering the limit as α approaches π, leading to the realization that the series does not vanish but retains the fundamental frequency component. This highlights the importance of correctly identifying coefficients in Fourier series expansions.

PREREQUISITES
  • Understanding of Fourier series and their applications
  • Familiarity with periodic functions and their properties
  • Knowledge of trigonometric identities and transformations
  • Basic calculus, particularly limits and continuity
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  • Study the derivation of Fourier series for different types of functions
  • Explore the concept of convergence in Fourier series
  • Learn about the implications of periodicity in Fourier analysis
  • Investigate the role of coefficients in Fourier series expansions
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Mathematicians, physics students, and engineers interested in signal processing or harmonic analysis will benefit from this discussion on Fourier Cosine series.

sarahisme
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Hello peoples,

I think this is a trick question... well sort of :P

http://img133.imageshack.us/img133/472/picture8ox1.png

for part (a) i get that the cosine Fourier Series for f(x) = cos(x) to be:

http://img138.imageshack.us/img138/6114/picture9sq2.png

i hope that is ok, but its part (b) that is troubling me...

is all that happens as http://img138.imageshack.us/img138/7436/picture10gf7.png is that the cosine Fourier series of cos(x) goes to 0?

i am guessing i am missing some trick to this question?

Cheers! :D

Sarah
 
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When you let [itex]\alpha\to\pi[/itex] then the interval you're computing the Fourier series on becomes [itex][-\pi,\pi][/itex]. The original function you're given, [itex]\cos(x)[/itex] is exactly periodic on this interval, and so you should find that your Fourier series has only coefficient, for [itex]n=1[/itex], which corresponds to [itex]cos(x)[/itex].
 
jpr0 said:
When you let [itex]\alpha\to\pi[/itex] then the interval you're computing the Fourier series on becomes [itex][-\pi,\pi][/itex]. The original function you're given, [itex]\cos(x)[/itex] is exactly periodic on this interval, and so you should find that your Fourier series has only coefficient, for [itex]n=1[/itex], which corresponds to [itex]cos(x)[/itex].

ah yep i see now. i need to be more careful with the n = 1 term when alpha = pi

thanks for the help! :D

Sarah
 

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