3.6 Obtain Fourier Series approximation

3.6.1 Example 1

Obtain Fourier Series approximation of $f(x)=e^{-|x|}$ for $-1<x<1$

3.6.1.1 Maple

restart; 
f:=x->exp(-abs(x)); 
f_approx:=OrthogonalExpansions:-FourierSeries(f(x),x=-1..1,infinity ): 
f_approx:=subs(i=n,f_approx);
 

$$1-{\mathrm{e}}^{-1}+\sum _{n=1}^{\mathrm{\infty }}(-\frac{2({(-1)}^n{\mathrm{e}}^{-1}-1)\cos \left(\pi nx\right)}{{\pi }^2{n}^2+1})$$

3.6.1.2 Mathematica

Mathematica does not have a buildin function to give general series expression as the above with Maple. There is a user written package and answer here https://mathematica.stackexchange.com/questions/149468/a-more-convenient-fourier-series which provides this.

In Mathematica it is possible to obtain the terms using the command FourierSeries . For example the terms $n=0,n=-1,n=1$ can be obtained using

expr = Exp[-Abs[x]]; 
FourierSeries[expr, x, 1, FourierParameters -> {1, Pi}]
 

$$\frac{(1+e)e^{-i\pi x}}{e+e{\pi }^2}+\frac{(1+e)e^{i\pi x}}{e+e{\pi }^2}+\frac{e-1}{e}$$

see https://reference.wolfram.com/language/ref/FourierSeries.html for definitions of FourierParameters used above.

3.6.2 Example 2

Obtain Fourier Series approximation of

$$f\left(x\right)=\{\begin{array}{ccc}\frac{2xh}{L}& & 0\le x\le \frac{L}{2}\\ \frac{2h(L-x)}{L}& & \frac{L}{2}\le x\le L\end{array}$$

For $0<x<L$

3.6.2.1 Maple

restart; 
f:=x->piecewise(0<x and x<L/2,2*x*h/L, L/2<x and x<L, 2*h*(L-x)/L); 
f_approx:=OrthogonalExpansions:-FourierSeries(f(x),x=0..L,infinity ): 
f_approx:=subs(i=n,f_approx): 
simplify(%) assuming L>0
 

$$2\left(\sum _{n=1}^{\mathrm{\infty }}\frac{h({(-1)}^n-1)\cos \left(\frac{2\pi nx}{L}\right)}{{\pi }^2{n}^2}\right)+\frac{h}{2}$$

3.6.3 Example 3

Obtain Fourier Series approximation of $\cosh x$ for $-1<x<1$.

3.6.3.1 Maple

restart; 
f:=x->cosh(x); 
f_approx:=OrthogonalExpansions:-FourierSeries(f(x),x=-1..1,infinity ): 
f_approx:=subs(i=n,f_approx); 
convert(%,trig);
 

$$\sinh \left(1\right)+\left(\sum _{n=1}^{\mathrm{\infty }}\frac{2\sinh \left(1\right){(-1)}^n\cos \left(\pi nx\right)}{{\pi }^2{n}^2+1}\right)$$