What is the Fourier series for a periodic function over [-\pi,\pi]?

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Homework Help Overview

The discussion revolves around finding the Fourier series for a periodic function defined over the interval [-π, π]. The original poster presents a specific function and attempts to derive its Fourier coefficients.

Discussion Character

  • Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to calculate the Fourier coefficients a₀, aₙ, and bₙ but expresses confusion regarding the variable n and its implications in the calculations. Other participants provide insights into the behavior of sine and cosine functions at specific values of n.

Discussion Status

Participants are actively engaging with the mathematical expressions and exploring the properties of sine and cosine functions. There is a focus on clarifying the calculations for the Fourier coefficients, but no consensus has been reached on the next steps or the interpretation of results.

Contextual Notes

The original poster's equations were initially misformatted, leading to some confusion. There is an ongoing discussion about the implications of specific values of n in the context of the Fourier series.

TFM
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Homework Statement



Find the Fourier series corresponding to the following functions that are periodic over the
interval [tex][-\pi,\pi][/tex]

[tex]f(x) = 1, -\pi/2 < x< \pi/2; f(x)[/tex] otherwise.

Homework Equations



Fourier Series:

[tex]f(x) = \frac{1}{2}a_0 + \sum^\infty_{n=1}a_n cos\frac{2*\pi*n*x}{l} + \sum^\infty_{n=1} b_n sin\frac{2*\pi*n*x}{l}[/tex]

[tex]\frac{1}{l}\int^{l/2}_{-l/2}f(x) dx[/tex]

[tex]a_n = \frac{1}{l}\int^{l/2}_{-l/2}f(x) cos frac{2*\pi*n*x}{l}dx[/tex]

[tex]a_n = \frac{1}{l}\int^{l/2}_{-l/2}f(x) sin frac{2*\pi*n*x}{l}dx[/tex]

The Attempt at a Solution



So far I have:

[tex]a_0 = 1[/tex]

[tex]a_n = \frac{1}{\pi n}[sin(nx)]^{\pi/2}_{-\pi/2}[/tex]

[tex]b_n = -\frac{1}{\pi n}[cos(nx)]^{\pi}_{-\pi}[/tex]

But I am not sure what to do now. I seem to be mainly confused about the n's

TFM
 
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My equations didn't come out quite right...Sorry. Should be:


[tex] a_0 = \frac{1}{l}\int^{l/2}_{-l/2}f(x) dx [/tex]

[tex] a_n = \frac{1}{l}\int^{l/2}_{-l/2}f(x) cos \frac{2*\pi*n*x}{l} dx [/tex]

[tex] b_n = \frac{1}{l}\int^{l/2}_{-l/2}f(x) sin \frac{2*\pi*n*x}{l}dx [/tex]

TFM
 
Note that:

[tex]\sin( n \cdot \pi /2)= 1[/tex] if n = 1,3,5...
[tex]\sin( n \cdot \pi /2) = 0[/tex] if n = 2,4,6,...and

[tex]\cos( n \cdot \pi ) = (-1)^{n}[/tex]
 
So for [tex]a_n[/tex]

[tex] a_n = \frac{1}{\pi n}[sin(nx)]^{\pi/2}_{-\pi/2} [/tex]

n = 1,3,5...

so

[tex]a_n = \frac{1}{\pi 1}[sin(1x)]^{\pi/2}_{-\pi/2},[/tex]

[tex]\frac{1}{\pi * 3}[sin(3x)]^{\pi/2}_{-\pi/2},[/tex]

[tex]\frac{1}{\pi * 5}[sin(5x)]^{\pi/2}_{-\pi/2} ...[/tex]


and

[tex] \cos( n \cdot \pi ) = (-1)^{n} [/tex]

cos of n pi always = -1? (as long as pi is a whole number)

TFM
 

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