MaplePrimes - answers and comments on Question, fourier series ..can someone explain this to me in more details ? thanks

someone help me ...

casperyc — Sat, 02 Feb 2008 01:13:13 Z

Fouriercoeff:=proc(f,n) > local k,acoeff,bcoeff,azero: > > acoeff:=seq(evalf(Int(f(x)*cos(k*x),x=-Pi..Pi))/evalf(Pi),k=1..n): > > bcoeff:=seq(evalf(Int(f(x)*sin(k*x),x=-Pi..Pi))/evalf(Pi),k=1..n): > > azero:=evalf(Int(f(x),x=-Pi..Pi))/evalf(2*Pi): > > print("a[n]=",acoeff \n); > print("b[n]=",bcoeff \n); > print("a[0]=",azero \n); > > end: and Fouriercoeff:=proc(f) > local k,acoeff,bcoeff,azero,n: > > assume(n::integer); > > acoeff:=evalf(Int(f*cos(n*x),x=-Pi..Pi))/evalf(Pi): > > bcoeff:=evalf(Int(f*sin(n*x),x=-Pi..Pi))/evalf(Pi): > > azero:=evalf(Int(f,x=-Pi..Pi))/evalf(2*Pi): > > print("a[n]=",acoeff \n); > print("b[n]=",bcoeff \n); > print("a[0]=",azero \n); > > end: is the best I can do to satisfy that question someone have any better idea to that questiones ( see exercise 4 and 5 on http://img20.imageshack.us/img20/3518/56351860db7.jpg) thanks

Fourier series

John Fredsted — Sat, 02 Feb 2008 18:58:52 Z

Concerning Exercise 4: You can define the following procedure:

Fouriercoeff := proc(f::algebraic,numTerms::posint)
   1/Pi*int(f,x=-Pi..Pi),
   seq([
      seq(1/Pi*int(f*F(k*x),x=-Pi..Pi),k=1..numTerms)
   ],F in [cos,sin])
end proc:

It is used as follows:

a0,a,b := Fouriercoeff(x^2,10);

Note that besides the function f, the procedure must be provided also a positive integer numTerms specifying the number of terms in the returned lists of Fourier coefficients.

Concerning Exercise 5: Using the above, define the expression:

expr := a0/2
	+ add(a[i]*cos(i*x),i=1..nops(a))
	+ add(b[i]*cos(i*x),i=1..nops(b));

From that expression it can be seen that the appropiate value is x = Pi, from which one can obtain for the famous Euler sum the value Pi^2/6.

some assumption on n?

djc — Sun, 03 Feb 2008 13:03:32 Z

I have changed this row in your code to get a simplified formula:

seq(1/Pi*int(f*F(n*x),x=-Pi..Pi),F in [cos,sin]) assuming n::posint

a0,an,bn := Fouriercoeff(x^2);

a0, an, bn := 2/3*Pi^2, 4/n^2*(-1)^n, 0

Nice

John Fredsted — Sun, 03 Feb 2008 13:43:05 Z

Thanks for that significant improvement.

thank you all very much

casperyc — Sun, 03 Feb 2008 19:15:17 Z

this looks much better! cheers chen

only one input is allowed

casperyc — Sun, 03 Feb 2008 00:03:04 Z

Fouriercoeff(x^2,10); it still has two inputs... according to the question.it says i have to assume n to be integer in order to simplify as far as possible.i have no idea how to do that