[Date Index]
[Thread Index]
[Author Index]
How to simplify an expression in version 5
*To*: mathgroup at smc.vnet.net
*Subject*: [mg59022] How to simplify an expression in version 5
*From*: "passwd9" <david at carter-hitchin.clara.co.uk>
*Date*: Wed, 27 Jul 2005 01:24:43 -0400 (EDT)
*Sender*: owner-wri-mathgroup at wolfram.com
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.
Prev by Date:
**Re: Why won't Horner[] work?**
Next by Date:
**Re: Why won't Hornerwork?**
Previous by thread:
**Re: Operating with binary numbers**
Next by thread:
**Re: How to simplify an expression in version 5**
| |