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Re: Fourier expansion of (Cos[t]+A Cos[3 t])^1/3

>Below is the question I sent last week to the Mathematical Group and 
>later forwarded a copy to you. I shall appreciate answer.

As far as I can see your posting has not appeared on MathGroup, nor 
did I receive any copy from you?

>From: 'Kale Oyedeji [mailto:koyedeji at]
To: mathgroup at
>Sent: Wednesday, March 08, 2006 11:58 AM
>To: 'MathGroup at'
>Subject: [mg65095] Verifying A Result
>My problem is to find the Fourier expression for (Cos[t]+A Cos[3 t])^1/3.

Do you mean (Cos[t] + a Cos[3 t])^(1/3) or (Cos[t] + a Cos[3 t])/3 
(which is what you have typed here)? The latter is trivial, the 
former not. How does this combination arise in a practical situation? 
How should you handle the situation when Cos[t]+a Cos[3 t] < 0 ? I 
assume that you want "real" cube roots.

>I used FourierTrigSeries[(Cos[t]+A Cos[3 t])^1/3,t,k] and obtained 
>even functions Cos[2ntPi].

Which agrees with the documentation for FourierTrigSeries:

FourierTrigSeries[expr, t, k] gives the kth-order Fourier 
trigonometric series approximation to the periodic function of t that 
is equal to expr for -1/2 <= t <= 1/2, and has a period of 1.

FourierTrigSeries[expr, t, k, FourierParameters -> {a, b}] gives the 
kth-order Fourier trigonometric series approximation to the periodic 
function of t that is equal to expr for -1/(2 Abs[b]) <= t <= 1/(2 
Abs[b]), and has a period of 1/Abs[b].

>My colleague thinks that this is wrong and the result should be in 
>terns of odd functions Cos[(2n+1)tPi]. Is my colleague right and 
>Mathematica wrong?

Mathematica is not wrong.

   FourierTrigSeries[(Cos[t] + a Cos[3 t]), t, 3]

is a series in Cos[2 n t Pi], However, if you enter

   Simplify[FourierTrigSeries[(Cos[t] + a Cos[3 t]), t, 3,
      FourierParameters -> {0, 1/(2 Pi)}]]

you will get odd functions. In fact you will just get your input as 
it is the fourier trig series that you are trying to compute.

Assuming you want the fourier trig series of (Cos[t]+a Cos[3 
t])^(1/3), here is one way to proceed (it is a hack). First defined

   RealCubeRoot[f_] := Sign[f] Abs[f]^(1/3)

A plot shows what your function looks like:

   f[a_][t_] = Cos[t] + a Cos[3 t];

   pl[a_] := Plot[RealCubeRoot[f[a][t]], {t, 0, 2Pi}]


I doubt that the fourier trig series you are after can be computed in 
closed form. Instead, use the definition:

  trigseries[n_Integer?Positive][a_?NumericQ] :=
    NIntegrate[Cos[n t] RealCubeRoot[f[a][t]], {t, 0, 2 Pi}] / Pi

Then compute trigseries[n][a] for n = 0, 1, 2, 3, ... and numeric a. 
You will see that only terms with odd n are non-zero.


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