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Let $f(x)=\sin(\pi x)$ and let $0<x<1$

Find the Fourier series of the above function.

Now at first we should rescale $x$ to a symmetric interval : Let $s=x- 1/2 $

Thus $-1/2<s<1/2$

Now the task is to find the Fourier series of $f(s)=\sin(\pi (s+1/2))$

Which is very long to do but at the end I hopefully got the right answer (using the rescaled inner product to find the coefficients)

$FS(\sin(\pi x))= 2/\pi + \sum_{k=1}^\infty \frac{8}{\pi(1-4k^2)} \cos2k\pi x$

Now my first question is , is this right, and for the second, was there another way to compute its Fourier series because this method took 2 pages.

msm
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Mike Harb
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  • At what point do you want to evaluate the Fourier series next? – msm Oct 16 '16 at 08:25
  • I don't want to evaluate my Fourier series, my task is just to find it, and I'm asking if there is some less complicated method to do so – Mike Harb Oct 16 '16 at 08:30
  • You have one period only. It is up to you how to extend the function to make it periodic (since the FS applies to periodic functions). Usually it is extended in a way that the point at which the Fourier series wants to be evaluated be smooth. If there is no restriction, then extend it either way. – msm Oct 16 '16 at 08:50
  • I make an odd extension or an even one by my choice? – Mike Harb Oct 16 '16 at 08:53
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    It seems to be the case. Note that shifting the $\sin$ is not required. Just integrate over one period ($[0,1]$ if it is even). – msm Oct 16 '16 at 08:54

1 Answers1

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It is not necessary to rescale the whole setup. Extending your $f$ to a periodic function of period $1$ gives $$f(t)=\bigl|\sin(\pi t)\bigr|\qquad(-\infty<t<\infty)\ ,$$ which is an even function. Its Fourier series is therefore of the form $$f(t)\rightsquigarrow{a_0\over2}+\sum_{k=1}^\infty a_k\,\cos(2k\pi t)\ ,$$ whereby the $a_k$ are given by $$a_k=2\int_0^1 f(t)\,\cos(2k\pi t)\>dt=2\int_0^1 \sin(\pi t)\,\cos(2k\pi t)\>dt\qquad(k\geq0)\ .$$ Now use the formula $$2\sin\alpha\cos\beta=\sin(\alpha+\beta)+\sin(\alpha-\beta)$$ to "linearize" the integrand on the RHS.

Christian Blatter
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