A Fourier Cosine Series 

>

restart:

 

>	f:=abs(x);

 

(1.1)

 

>	plot(f, x=-Pi..Pi);

 

>	an:=int(f*cos(n*x),x=-Pi..Pi)/Pi;

 

(1.2)

 

>	assume(n,integer);#Letting MAPLE know n is an integer simplifies the calculation an:=int(f*cos(n*x),x=-Pi..Pi)/(Pi);

 

(1.3)

 

>	A:=n->int(f*cos(n*x),x=-Pi..Pi)/(Pi);

 

(1.4)

 

>	seq(A(n),n=1..10);

 

(1.5)

 

>	g:= M->A(0)/2 +sum(A(n)*cos(n*x),n=1..M);

 

(1.6)

 

>

h:=g(3);

 

(1.7)

 

>	plot({f,h},x=-Pi..Pi);

 

>	s:=seq(g(2*M+1),M=0..4);

 

(1.8)

 

>	plot({s,f},x=-3*Pi..3*Pi);

 

>	plot({s,f},x=0..0.5,y=0..0.5);

 

Bessel's Inequality and Parseval's Theorem 

Let's compute the (norm)^2 of F(x) and the n-term approximation of F 

>	norm2:=h->int(h^2,x=-Pi..Pi);

 

(2.1)

 

Compute the L^2 norm of f 

>	f2:=norm2(f);

 

(2.2)

 

And now compare it to the n-term approximation for n=1 to 20. Note the difference is always positive (as a result of Bessel's inequality) and converges to zero (as a result of Parseval's Theorem). 

>	E2:=seq(evalf(f2-norm2(g(N))),N=1..20);

 

(2.3)

 

Parseval's Identity tells us that  

‖f‖²=π[a₀² + ∑(a₀² /2+ b₀² ) 

which yields for this problem that 

>	Sum (1/(2*n-1)^4,n=1..infinity)=sum (1/(2*n-1)^4,n=1..infinity);

 

(2.4)

 

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