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Re: How to simplify to a result that is real
*To*: mathgroup at smc.vnet.net
*Subject*: [mg50810] Re: How to simplify to a result that is real
*From*: Paul Abbott <paul at physics.uwa.edu.au>
*Date*: Wed, 22 Sep 2004 00:11:19 -0400 (EDT)
*Organization*: The University of Western Australia
*References*: <cidt38$brv$1@smc.vnet.net> <200409180948.FAA00572@smc.vnet.net> <cion2b$r9a$1@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
In article <cion2b$r9a$1 at smc.vnet.net>,
Richard Chen <richard at doubleprime.com> wrote:
> This time, mathematica comes back with an expression involving
> ArcTan[u,v] which is just ArcTan[v/u], by direct inspection.
> So the 2 terms involving different ArcTan are actually the same.
> But I cannot easily make Mathematica to recognize that they
> are the same.
I'm not sure if I follow you here, but ArcTan[u,v] is equal to
ArcTan[v/u] only if u > 0. For example, try
FullSimplify[ArcTan[u,v] == ArcTan[v/u], u > 0]
However, if u < 0 then
ArcTan[u,v] - ArcTan[v/u] = Sign[v] Pi
(though I cannot see how to get FullSimplify to deduce this fact).
Cheers,
Paul
> Even if I use the rule ArcTan[u_,v_]->ArcTan[v/u]
> it still does not think the 2 expressions are the same. It is easier
> to simply copy and paste an anwser than manipulate mathematica
> to get a simpler result.
>
> Perhaps future versions of mathematica will be smarter and does not
> require as much intervention from the user.
>
> Thanks
>
> Richard
>
> On Sat, Sep 18, 2004 at 05:48:55AM -0400, Peter Valko wrote:
> > Richard Chen <richard at doubleprime.com> wrote in message
> > news:<cidt38$brv$1 at smc.vnet.net>...
> > > The command:
> > >
> > > Integrate[1/(1 + e Cos[t]), {t, 0, a},
> > > Assumptions -> {-1 < e < 1, 0 < a < Pi}]
> > >
> > > leads to a complex valued result. I could not make
> > > mathematica to render the result in a form that is
> > > purely real. ComplexExpand, Refine all do not seem to work.
> > >
> > > Does anyone know how to make mathematica to simplify this
> > > result into a real form?
> > >
> > > Thanks for any info.
> > >
> > > Richard
> >
> >
> >
> > Richard,
> >
> > I think this will work:
> >
> >
> > ch = Integrate[1/(1 + e Cos[t]), {t, 0, a}, Assumptions -> {-1 < e <
> > 1, 0 < a < Pi}]
> >
> > FullSimplify[ComplexExpand[ch, TargetFunctions -> {Re, Im}], {-1 < e <
> > 1, 0 < a < Pi}]
> >
> >
> > The result is
> >
> > (-2*ArcTan[((-1 + e)*Tan[a/2])/Sqrt[1 - e^2]])/Sqrt[1 - e^2]
> >
> >
> > Peter
> >
>
--
Paul Abbott Phone: +61 8 6488 2734
School of Physics, M013 Fax: +61 8 6488 1014
The University of Western Australia (CRICOS Provider No 00126G)
35 Stirling Highway
Crawley WA 6009 mailto:paul at physics.uwa.edu.au
AUSTRALIA http://physics.uwa.edu.au/~paul
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