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Re: Re: Fourier expansion of (Cos[t]+A Cos[3 t])^1/3
*To*: mathgroup at smc.vnet.net
*Subject*: [mg65377] Re: [mg65095] Re: Fourier expansion of (Cos[t]+A Cos[3 t])^1/3
*From*: "'Kale Oyedeji" <koyedeji at morehouse.edu>
*Date*: Wed, 29 Mar 2006 06:34:16 -0500 (EST)
*Reply-to*: <kaleoyedeji at comcast.net>
*Sender*: owner-wri-mathgroup at wolfram.com
Thanks once again for your insight answer to my question. I tried
Simplify[FourierTrigSeries[(Cos[t]+ a Cos[3 t])^(1/3), t, 3,
FourierParameters->{0,1/(2 pi)}]] and some complex response after an unusual
long time which goes to support your assertion that it is not a trivial
problem.
'Kale
-----Original Message-----
From: Paul Abbott [mailto:paul at physics.uwa.edu.au]
To: mathgroup at smc.vnet.net
Subject: [mg65377] Re: [mg65095] Re: Fourier expansion of (Cos[t]+A Cos[3 t])^1/3
>Paul, thanks for your suggestion. Can you take a look at what I have
>done and why I am getting the result in the attached? I seem to be
>getting complex results despite the RealRoot specification.
You are getting complex results is due to the machine precision algorithms
used by NIntegrate. If you change the WorkingPrecision, you obtain
explicitly real results.
RealCubeRoot[f_]:= Sign[f] Abs[f]^(1/3)
f[a_][t_] = Cos[t]+a Cos[3 t];
trigseries[n_Integer][a_?NumericQ]:=
NIntegrate[Cos[n t] RealCubeRoot[f[a] [t]], {t,0, 2 Pi},
WorkingPrecision->25] / Pi /; n >=0
trigseries[1][1.0]
0.67222228384695859
Similar behaviour arises when you expand out polynomials.
Product[(x - N[i]), {i, 25}]
Solve[Expand[%] == 0, x]
Clearly, the answers should all be real.
Cheers,
Paul
>-----Original Message-----
>From: Paul Abbott [mailto:paul at physics.uwa.edu.au]
To: mathgroup at smc.vnet.net
>Subject: [mg65377] [mg65095] 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 morehouse.edu]
To: mathgroup at smc.vnet.net
>>Subject: [mg65377] [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}]
>
> pl[1/2]
>
>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.
>
>Cheers,
>Paul
>
>Attachment converted: Fock:2006 RealCubeRootP.nb ( / ) (002DD781)
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