Find Fourier Series/Coefficients with Mathematica

Question

Hello I want to find the Fourier series and/or the coefficients for a function like the following:

or

or

For the first one I did the following:

FourierTrigSeries[
Piecewise[{{0, -Pi <= x <= -Pi/2}, {Cos[x], -Pi/2 <= x <= Pi/2}, {0, 
Pi/2 < x < Pi}}], x, 5 ]

Which seems to be correct. How can I convert this to summation form?
For Taylor Series I'm using something like this:

series[expr_, x_, x0_] := 
Defer[expr = Sum[#, {n, 0, ∞}]] &[
FullSimplify@
SeriesCoefficient[expr, {x, x0, n}, 
 Assumptions -> n ⩾ { 0}] (x - x0)^n] 

series[1/2*(x^2 - 2 x + 5)/(x^2 - 6 x + 9), x, 1]

The goal would be to get something like this: Which would be the solution for the first one.

Answer 1

As mentioned in the comment above, you can try the functions in A more convenient Fourier series. To be specific, easyFourierTrigSeries therein is your friend:

expr = Piecewise[{{0, -Pi <= x <= -Pi/2}, {Cos[x], -Pi/2 <= x <= Pi/2}, {0, 
     Pi/2 < x < Pi}}];

expansion = easyFourierTrigSeries[expr, {x, -Pi, Pi}, n]

The form of the result is different from the one given in your question, but it's easy to show they're equivalent. Let's extract the summand:

summand = expansion[[2, 1, 1]]

By setting \[FormalK] to 1 we obtain the 2nd term in your result:

secondterm = summand /. \[FormalK] -> 1

When \[FormalK] is odd, the summand is 0:

oddterm = FullSimplify[summand /. \[FormalK] -> 2 k + 1, {k > 0, k ∈ Integers}]
(* 0 *)

When \[FormalK] is even and larger than 1, the summand is equivalent to:

eventerm = FullSimplify[summand /. \[FormalK] -> 2 k, {k > 0, k ∈ Integers}]

So expansion can be rewritten as

firstterm = expansion[[1]]

firstterm + secondterm + HoldForm[Sum[#, {k, Infinity}]] &@eventerm

which is (almost) the same as yours.

Answer 2

You may have to do this the old fashion way. Generally if there is more than one form of a correct answer, you probably will not get the exact one you are looking for without some manipulation.

Clear["Global`*"]
f[x_] = Cos[x]*(UnitStep[x + Pi/2] - UnitStep[x - Pi/2])

Your Piecewise works also

$Assumptions = n \[Element] Integers

Set f[x] to be the form

eqn1 = f[x] == an Cos[n  x] + bn Sin[n  x]

to get an

Simplify[Integrate[eqn1[[1]]*Cos[n*x], {x, -Pi, Pi}] == Integrate[eqn1[[2]]*Cos[n*x], {x, -Pi, Pi}]];

an = an /. Solve[%, an][[1]] // FullSimplify;
(*(2*Cos[(Pi*n)/2])/(Pi - Pi*n^2)*)

Except for Pi in the denominator, these are the same coefficients that bills computed in his answer. For n = 0, a0 is 1/2 the general n coefficient.

a0 = 1/2 an /. n -> 0
(*1/Pi*)

For n = 1, the denominator of an is 0, so take the limit.

a1 = Limit[an, n -> 1]
(* 1/2 *)

solve for bn

Simplify[Integrate[eqn1[[1]]*Sin[n*x], {x, -Pi, Pi}] == Integrate[eqn1[[2]]*Sin[n*x], 
    {x, -Pi, Pi}]]
(*bn==0*)

bn = 0;

The Sin terms had better go away for an even function. Look at the first few terms of the series for n > 1.

Table[an Cos[n x], {n, 2, 10}]
(*{(2*Cos[2*x])/(3*Pi), 0, -((2*Cos[4*x])/(15*Pi)), 0,(2*Cos[6*x])/(35*Pi)}*)

Terms for odd values of n are 0, so simplify by changing n to 2n and starting the series at n = 1.

an = an /. n -> 2 n // Simplify
(*(2*(-1)^n)/(Pi - 4*Pi*n^2)*)

and your Fourier series becomes:

series = a0 + a1*Cos[x] + Sum[an*Cos[2*n*x], {n, 1, Infinity}]
(*Sum[(2*(-1)^n*Cos[2*n*x])/(Pi - 4*Pi*n^2), {n, 1, Infinity}] + Cos[x]/2 + 1/Pi*)

which looks like the answer you are looking for.

Answer 3

If you want the general term of the Fourier expansion, you might be better off using FourierTransform instead of FourierTrigSeries. For example, define a function that is the FT

ft[w_] = FourierTransform[Piecewise[{{0, -Pi <= x <= -Pi/2}, 
       {Cos[x], -Pi/2 <= x <= Pi/2}, {0,Pi/2 < x < Pi}}], 
       x, w, FourierParameters -> {1, -1}];

 ft[w]
 -((2 Cos[(\[Pi] w)/2])/(-1 + w^2))

To recapture the first 10 (say) coefficients:

ft[#] & /@ Range[2, 10]
{2/3, 0, -(2/15), 0, 2/35, 0, -(2/63), 0, 2/99}

which match the coefficients of the Cos terms in the FourierTrigExpansion.