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Re: How to simplify an expression in version 5


passwd9 wrote:
> Hi,
> 
> I'm trying to simplify the expression that comes out of the
> FourierTrigSeries function, and I've tried 'Simplify', 'Collect' and
> 'Factor' but no luck.  My input is as follows:
> 
> << Calculus`FourierTransform`
> FourierTrigSeries[Cos[t], t, 5, FourierParameters -> {0, 1/Pi}]
> 
> And the output is:
> 
> \!\(\(2\/\@\[Pi] + \(4\ Cos[2\ t]\)\/\(3\ \@\[Pi]\) - \(4\ Cos[4\
> t]\)\/\(15\ \
> \@\[Pi]\) + \(4\ Cos[6\ t]\)\/\(35\ \@\[Pi]\) - \(4\ Cos[8\ t]\)\/\(63\
> \@\
> \[Pi]\) + \(4\ Cos[10\ t]\)\/\(99\ \@\[Pi]\)\)\/\@\[Pi]\)
> 
> That's rather horrid looking in ascii, so I'll try to make something
> more readable:
> 
> (2/Sqrt(Pi) + 4Cos(2t)/3Sqrt(Pi) - 4Cos(4t)/15Sqrt(Pi) + ... )
> ---------------------------------------------------------------
>                             Sqrt(Pi)
> 
> 
> Now the Sqrt(Pi) in the denominator could be 'brought up' into the
> terms in the numerator to give:
> 
> 
> (2/Pi + 4Cos(2t)/3Pi - 4Cos(4t)/15Pi + ... )
> 
> Much more elegant!  Anyone know if MM can do this?
> 
> Thanks.
> 
Hi David,

The command *ExpandAll* will do the trick:

In[1]:=
Needs["Calculus`FourierTransform`"]

In[2]:=
ExpandAll[FourierTrigSeries[Cos[t], t, 5, FourierParameters -> {0, 1/Pi}]]

Out[2]=
2/Pi + (4*Cos[2*t])/(3*Pi) - (4*Cos[4*t])/(15*Pi) + (4*Cos[6*t])/(35*Pi) -
   (4*Cos[8*t])/(63*Pi) + (4*Cos[10*t])/(99*Pi)

Regards,
/J.M.


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