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MathGroup Archive 2004

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

  • To: mathgroup at smc.vnet.net
  • Subject: [mg50801] 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 00:10:59 -0400 (EDT)
  • Reply-to: hanlonr at cox.net
  • Sender: owner-wri-mathgroup at wolfram.com

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: [mg50801] [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
> > 
> 
> 


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