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MathGroup Archive 2005

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Re: How to simplify an expression in version 5

  • To: mathgroup at smc.vnet.net
  • Subject: [mg59071] Re: How to simplify an expression in version 5
  • From: Peter Pein <petsie at dordos.net>
  • Date: Thu, 28 Jul 2005 02:26:26 -0400 (EDT)
  • References: <dc76b4$jpk$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

passwd9 schrieb:
> 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,

don't know if Marily Monroe has been capable of doing it...

When you are doing Fourierseries with Mathematica, I'm sure you are an
advanced user, who noticed the existence of the basic Functions
Expand[], Simplify[] etc. before.

-- 
Peter Pein
Berlin


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