Calculating Fourier Series of f(x) = |x| - \pi

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Homework Help Overview

The discussion revolves around calculating the Fourier series for the function f(x) = |x| - π over the interval (-π, π]. Participants explore the properties of the function, particularly its evenness, and how this affects the Fourier series expansion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the nature of the function as even and its implications for calculating Fourier coefficients. There are questions about the convergence of the series and the application of Dirichlet conditions.

Discussion Status

Some participants have provided insights into the advantages of recognizing the function's evenness, suggesting that it simplifies the calculation of certain coefficients. There is an ongoing exploration of convergence issues and the behavior of the series at specific points.

Contextual Notes

Participants mention the concept of half-range expansions and the implications of the function being even, which may influence the approach to finding the Fourier series. There is also a reference to the Dirichlet conditions regarding convergence.

squenshl
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How do work out the Fourier series of f(x) = |x| - [tex]\pi[/tex] on ([tex]\pi[/tex],[tex]\pi[/tex]].
 
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squenshl said:
How do work out the Fourier series of f(x) = |x| - [tex]\pi[/tex] on ([tex]\pi[/tex],[tex]\pi[/tex]].

What have you tried? Have you noticed that is an even function?
 
I know that |x| = x, x [tex]\geq[/tex] 0 & -x, x < 0.
So does that mean |x| - [tex]\pi[/tex] = x - [tex]\pi[/tex], x [tex]\geq[/tex] 0 & -x - [tex]\pi[/tex], x < 0.
Then just find the Fourier series as a sum of these functions. If not I don't have a clue where to start.
 
Yes, that would work. But the fact that the function is even can reduce the work by half. Have you studied half-range expansions?
 
I see. Since f(x) is an even function, when it goes to finding bn you multiply an even function with sin which an odd function to get an odd function and the integral of an odd function is always zero
 
squenshl said:
I see. Since f(x) is an even function, when it goes to finding bn you multiply an even function with sin which an odd function to get an odd function and the integral of an odd function is always zero

Yes. And for the an you have even times even and you can use symmetry, which will help you with the absolute values. Again, look at "half range" expansions.
 
Cheers. Got it, that was easy after all.
 
Are there values of x at which this series fails to converge to f(x). To what values does it converge at these points.
 
Draw a couple of periods of f(x). Read what the Dirichlet conditions say about convergence of the FS and apply it to this f(x).
 

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