Find Fourier Series of the given function.

Question

How can I find the Fourier Series of the function f(x)=(1-x)(1+x) on the interval [-1/2, 1/2].

Answer 1

The Fourier coefficients on an interval $[-L/2,L/2]$ are given by $$ A_n=\frac{2}{L}\int_{-L/2}^{L/2}f(x)\cos\frac{2\,\pi\,n\,x}{L}\,dx,\quad B_n=\frac{2}{L}\int_{-L/2}^{L/2}f(x)\sin\frac{2\,\pi\,n\,x}{L}\,dx. $$ In your case $L=1$ and, integrating by parts, $$ A_n=2\int_{-1/2}^{1/2}(1+x)(1-x)\,\cos(2\,\pi\,n\,x)\,dx=2\int_{-1/2}^{1/2}(1-x^2)\,\cos(2\,\pi\,n\,x)\,dx=\frac{(-1)^{n+1}}{\pi^2\,n^2}. $$ Since $f$ is even, $B_n=0$.

Answer 2

In your reply to my comment, you were asking about the difference in changing the period from $[-1,1]$ to $[-1/2,1/2]$.
Let us fix this point first of all. If $$ f(x/T)\quad \to \quad a_{\,n} \cos \left( {\frac{{2\pi n}} {T}x} \right) $$ then $$ g(x/T') = f(x/T)\quad \to \quad a_{\,n} \cos \left( {\frac{{2\pi n}} {{T'}}x} \right) $$

That is:
if you change the $x$ scale and consequently the period, keeping the ordinate scale, then the Fourier coefficients will be the same, but will apply to frequencies which are in the inverse relation with the periods.

In your case: $$ x^{\,2} = \left( {\frac{x} {1}} \right)^{\,2} \quad \to \quad a_{\,n} \cos \left( {\frac{{2\pi n}} {1}x} \right)\quad \Rightarrow \quad x^{\,2} = \left( {\frac{1} {2}} \right)^{\,2} \left( {\frac{x} {{1/2}}} \right)^{\,2} \quad \to \quad \frac{{a_{\,n} }} {4}\cos \left( {\frac{{2\pi n}} {{1/2}}x} \right) $$ Which means, that the coefficients of the series for: $$ x^{\,2} \quad \left| {\;x \in \left[ { - 1/2,\;1/2} \right]} \right. $$ will be $1/4$ of those corresponding to: $$ x^{\,2} \quad \left| {\;x \in \left[ { - 1,\;1} \right]} \right. $$ and that of course they will apply to harmonics with double frequencies.

That's because you have a scaling either in $x$ and in $y$.