Is there any symmetry I can use to find this Fourier sine series?

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SUMMARY

The discussion focuses on finding the Fourier sine series (FSS) for the function f(x) = x(π² - x²). The key equation used is a_n = (2/L) ∫₀^L f(x)sin(nπx/L)dx, specifically evaluated for L = π. The function exhibits odd symmetry, allowing the use of half-range sine expansion, which eliminates cosine terms. The integration by parts (IBP) technique is emphasized, with a recommendation to combine the integrals into a single expression for efficiency.

PREREQUISITES
  • Understanding of Fourier series, specifically Fourier sine series.
  • Proficiency in integration techniques, particularly integration by parts (IBP).
  • Knowledge of function symmetry, including odd and even functions.
  • Familiarity with the properties of trigonometric functions in calculus.
NEXT STEPS
  • Study the properties of odd and even functions in relation to Fourier series.
  • Practice integration by parts with various functions to enhance efficiency.
  • Explore half-range sine expansions and their applications in Fourier analysis.
  • Review examples of Fourier sine series derivations for different functions.
USEFUL FOR

Students preparing for exams in calculus or mathematical analysis, particularly those focusing on Fourier series and integration techniques.

richyw
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Homework Statement



I am going over a practice exam, and I need to find the FSS of [tex]f(x)=x(\pi^2-x^2)[/tex]

Homework Equations


[tex]f(x) \sim \sum^\infty_{n=1}a_n sin\left(\frac{n \pi x}{L}\right)[/tex]
[tex]a_n=\frac{2}{L}\int^L_0 f(x)sin\left(\frac{n\pi x}{L}\right)dx[/tex]

The Attempt at a Solution


I think I need to integrate[tex]\frac{2}{\pi}\int^\pi_0 x(\pi^2-x^2)sin(n\pi)dx[/tex]which is two integrals, the first one would need me to use IBP once, and the second one would need me to use IBP three times. This is on a practice exam (and my exam is in an hour), so I am guessing that this integral is easier if I can find some symmetry in it. Is this true?
 
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You can indeed use symmetry. Graph f(x) on your calc: what do you notice? remember if x(t)=x(-t) then there is even symmetry, x(t)=-x(-t) means it has odd symmetry.
 
richyw said:

Homework Statement



I am going over a practice exam, and I need to find the FSS of [tex]f(x)=x(\pi^2-x^2)[/tex]

Homework Equations


[tex]f(x) \sim \sum^\infty_{n=1}a_n sin\left(\frac{n \pi x}{L}\right)[/tex]
[tex]a_n=\frac{2}{L}\int^L_0 f(x)sin\left(\frac{n\pi x}{L}\right)dx[/tex]

The Attempt at a Solution


I think I need to integrate[tex]\frac{2}{\pi}\int^\pi_0 x(\pi^2-x^2)sin(n\pi)dx[/tex]which is two integrals, the first one would need me to use IBP once, and the second one would need me to use IBP three times. This is on a practice exam (and my exam is in an hour), so I am guessing that this integral is easier if I can find some symmetry in it. Is this true?

The fact that the function is odd is the reason you can use the half range sine expansion in the first place. That is why there are no cosine terms. You just need to bite the bullet and do the integration by parts. Don't do two separate integrals though. Write your integrand as$$
(\pi^2x-x^3)sin(n\color{red}x)$$and do it all at once. Note your typo correction.
 

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