Is Calculating Only the Fourier Sine Series Sufficient for f(x) = x^3 on [-1,1]?

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Homework Help Overview

The discussion revolves around determining the Fourier series representation for the function f(x) = x^3 on the interval [-1, 1]. Participants explore whether calculating only the Fourier sine series is sufficient due to the odd nature of the function.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Some participants question the necessity of calculating all Fourier coefficients (a0, an, bn) given that x^3 is an odd function. They consider whether starting directly with the Fourier sine series would suffice.

Discussion Status

Participants have noted that it is possible to restrict the calculation to the sine series. Some suggest that calculating all terms initially would confirm that the coefficients for even terms are zero, reinforcing the approach of using the sine series alone.

Contextual Notes

There is an acknowledgment of the properties of odd functions and how they relate to the Fourier series representation, as well as a reference to specific integral evaluations to support the reasoning.

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Homework Statement



Determine a general Fourier series representation for f(x) = x^3 -1<x<1

Homework Equations


The Attempt at a Solution



May seem like a stupid Q, but would i have to calculate a0, an, bn or since i know that x^3 is an odd function, could jump straight into calculating the Fourier sine series for odd functions. Would that give me a general representation?
 
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looking at it one can directly jump to Fourier sine series
if you calculate all the terms initially assuming the complete Fourier representation you will find that coefficients associated with the even terms will go to zero.
so its more like using a known result.
 
Yes, you can restrict yourself to sines.

If you want you can explicitly check it, with an argument along the lines of
[tex]\int_{-1}^1 \cos(x) x^3 dx = \int_{-1}^0 \cos(x) x^3 dx + \int_0^1 \cos(x) x^3 dx = \int_0^1 \cos(-x) (-x)^3 dx + \int_0^1 \cos(x) x^3 dx = 0[/tex]
because cos(-x) (-x)3 = - cos(x) x3
 
thanks.
 

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