Re: Re: Re: Re: How to simplify to a result that is real
- To: mathgroup at smc.vnet.net
- Subject: [mg50802] Re: Re: [mg50788] Re: [mg50735] Re: How to simplify to a result that is real
- From: Bob Hanlon <hanlonr at cox.net>
- Date: Wed, 22 Sep 2004 00:11:01 -0400 (EDT)
- Reply-to: hanlonr at cox.net
- Sender: owner-wri-mathgroup at wolfram.com
I overfocused on getting rid of the second form of ArcTan. In the last step
use FullSimplify rather than just Simplify.
Bob Hanlon
>
> From: Bob Hanlon <hanlonr at cox.net>
To: mathgroup at smc.vnet.net
> Date: 2004/09/21 Tue AM 07:23:53 EDT
> To: Richard Chen <richard at doubleprime.com>,
<mathgroup at smc.vnet.net>
> Subject: [mg50802] Re: [mg50788] Re: [mg50735] Re: How to simplify to a result that
is real
>
> ArcTan[x, y] and ArcTan[y/x] are in general not the same.
>
> Plot3D[ArcTan[y/x],{x,-2,2},{y,-2,2}];
>
> Plot3D[ArcTan[x,y],{x,-2,2},{y,-2,2}];
>
> ch=Integrate[1/(a+b Cos[t]),{t,0,c},Assumptions->{a>b>0,0<c<Pi}];
>
> ch = Simplify[ComplexExpand[ch,
> TargetFunctions->{Re,Im}],{a>b>0,0<c<Pi}];
>
> ch = TrigToExp[ch];
>
> ch = Simplify[ComplexExpand[ch,
> TargetFunctions->{Re,Im}],{a>b>0,0<c<Pi}]
>
> (ArcTan[Sqrt[(a - b)/(a + b)]*Tan[c/2]] +
> ArcTan[((a - b)*Tan[c/2])/
> Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2]
>
>
> Bob Hanlon
>
> >
> > From: Richard Chen <richard at doubleprime.com>
To: mathgroup at smc.vnet.net
> > Date: 2004/09/21 Tue AM 03:49:05 EDT
> > To: mathgroup at smc.vnet.net
> > Subject: [mg50802] [mg50788] Re: [mg50735] Re: How to simplify to a result that is
> real
> >
> > 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
> > >
> >
> >
>