Re: Re: How to simplify to a result that is real

• To: mathgroup at smc.vnet.net
• Subject: [mg50788] Re: [mg50735] Re: How to simplify to a result that is real
• From: Richard Chen <richard at doubleprime.com>
• Date: Tue, 21 Sep 2004 03:49:05 -0400 (EDT)
• References: <cidt38\$brv\$1@smc.vnet.net> <200409180948.FAA00572@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```I have seen quite a few responses here. I'll just Thank all of you
with one response.

I still feel that at this stage, simplifying expressions
in mathematica is still a kind of art and requires intimate knowledge
of how esoteric mathematica options work. For example,
I just tried this technique on a problem which is essentially
the same:

ch = Integrate[1/(a + b Cos[t]), {t, 0, c},
Assumptions -> {a > b > 0, 0 < c < Pi}]

FullSimplify[ComplexExpand[ch, TargetFunctions -> {Re, Im}], {
a > b > 0, 0 < c < Pi}]

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. 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
>

```

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