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Expand $f(x)=x(\pi-x)$ into a Fourier cosine series on $(0;\pi)$

My idea is to expand the function into an even function on $[-\pi;\pi]$ so that the $b_n=0$ and we'll get a Fourier cosine series. But I don't know how. Any one help me please?

anvo
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    Follow the recipe for a Fourier cosine series on $[0,L]$: $f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty a_n \cos\left(\frac{2\pi n x}{L}\right)$ with $$a_n = \frac{2}{L}\int_0^{L} f(x)\cos\left(\frac{2\pi nx}{L}\right),{\rm d}x$$ For your case $L = \pi$ and $f(x) = x(\pi-x)$. Perform the integral to determine $a_n$. – Winther Nov 03 '16 at 16:10
  • Thank you for your answer. But we don't needto expand $f(x)$ into an even function? – anvo Nov 03 '16 at 16:14
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    By assuming $f(x)$ only has $\cos$-terms and not $\sin$-terms we are implicitly assuming that $f$ (when evaluated outside of $[0,L]$) is continued as an even function. Note that what you suggest would also work and give the same result, i.e. if you try to compute the Fourier series of an even continued $f(x)$ on $[-\pi,\pi]$ by using the standard formulas for this case. – Winther Nov 03 '16 at 16:17
  • Oh I see, thank you so much. – anvo Nov 03 '16 at 16:21

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