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How to find Fourier sine series of $f(x)=x(1-x), 0\lt x \lt 1$?

This is not an odd functions, so how to proceed?

User
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1 Answers1

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Define $g(x)=f(x)$ where $0<x<1$ and $g(x)=-f(-x)$ for $-1<x<0$. Then $g$ is an odd function. So you have to expand $g$ and that's the same as expanding $f$ in a sin series.

Gregory Grant
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  • Why expanding g will be same as expanding f ? – User Feb 16 '16 at 16:28
  • Because in $(0,1)$ which is the domain of $f$ the two functions are equal. – Gregory Grant Feb 16 '16 at 16:29
  • And since $g$ is an odd function you know expanding it will automatically produce a sin series. – Gregory Grant Feb 16 '16 at 16:30
  • Okie. And for calculating $b_n$ , coefficient of series, we need to integrate from 0 to 1 , right ? – User Feb 16 '16 at 16:31
  • I believe you need to integrate from $-1$ to $1$. – Gregory Grant Feb 16 '16 at 16:34
  • But also since $g$ is odd, $g(x)\sin(nx)$ is even and so the integral from $-1$ to $1$ is just $2$ times the integral from $0$ to $1$. So in effect you do only need to integrate from $0$ to $1$, just make sure to multiply by two. – Gregory Grant Feb 16 '16 at 16:43
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    @Member FYI it's customary to up-vote and/or hit the check mark when somebody gives you a proper answer to your question. – Gregory Grant Feb 16 '16 at 16:52
  • Sorry sir for the delay, I will be upvoting your answer for sure. I am in hurry to submit the assignment so, I thought to complete that first. – User Feb 16 '16 at 17:02
  • @Member Thanks wasn't trying to rush you. Just saw that you are relatively new here so just wanted to make sure you knew the drill. Good luck with the assignment. – Gregory Grant Feb 16 '16 at 17:17
  • Also one more thing, it's probably easier to calculate the series for $x$ and $x^2$ separately, and then just subtract, than to do $x-x^2$ in one fell swoop. – Gregory Grant Feb 16 '16 at 17:18
  • I have done it sir, Thanks for your help. – User Feb 16 '16 at 17:34