Extending f(x) as an Even Function: Obtain Cosine Fourier Series

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Homework Help Overview

The problem involves extending the function f(x) = sin(x) defined on the interval [0, π] as an even function and obtaining its cosine Fourier series representation. The context is within the study of Fourier series and function extension.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the method of extending sin(x) to an even function by defining it as -sin(x) for the interval [-π, 0]. Questions arise regarding the integration process for calculating the Fourier coefficients, particularly the a0 term.

Discussion Status

Some participants are verifying the integration steps and discussing the implications of the results on the a0 term. There is acknowledgment of a potential mistake in the integration process, and a participant has corrected their calculation, leading to a clearer understanding of the Fourier series coefficients.

Contextual Notes

Participants are navigating the constraints of the problem, including the requirement to extend the function and the specific intervals involved in the integration for Fourier coefficients.

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Homework Statement


f(x) = sin(x) for 0[tex]\leq[/tex]x<[tex]\pi[/tex]. Extend f(x) as an even function . Obtain a cosine Fourier series for f.


Homework Equations


[tex]a_{0}[/tex]/2 + [tex]\sum[/tex] [tex]a_{n}[/tex]cos(nx)


The Attempt at a Solution


So as far as I know, to extend sin(x) as an even function you have to make f(x)=-sin(x) for [tex]-\pi\leq[/tex]x<0 and then just use that to integrate for an but this gives a series without an a0 term which the question points to it having. What have I done wrong?
 
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Have you integrated correctly?

Did you remember that the integral over [-pi,pi] breaks to [-pi,0] where it's -sin(x) and [0,pi] where it's sin(x) (or alternatively, the integral over [-pi,pi] is double the integral over [0,pi])

Can you please show us your calculation?
 
sure.
As you've said, I've just used:
[tex]\frac{2}{\pi}[/tex][tex]\int^{\pi}_{0}[/tex]sin(x) dx
so integration gets
[-cos(x)[tex]]^{\pi}_{0}[/tex].
expanding gives [1 - 1] so I lose the [tex]a_{0}[/tex] term. Is that correct?

Oops, i see what I did. Forgot the - - so it should actually be 4/pi. Is that correct?
 
Yeah, that's right.
 

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