Re: Demonstrate that 1==-1
- To: mathgroup at smc.vnet.net
- Subject: [mg23200] Re: [mg23171] Demonstrate that 1==-1
- From: Adam Strzebonski <adams at wolfram.com>
- Date: Mon, 24 Apr 2000 01:12:07 -0400 (EDT)
- Organization: Wolfram Reasearch, Inc
- References: <B52619D5.624D%andrzej@tuins.ac.jp>
- Sender: owner-wri-mathgroup at wolfram.com
Andrzej Kozlowski wrote:
>
> I do not think this has anything to do with Limit. In fact both answers are
> incorrect. The correct answer depends on whether Abs[a]>1 or Abs[a]<1. The
> case Abs[a]==1 is special, since clearly for a=1 the integral does not
> converge at all! So you should expect to get into trouble with an example
> like this. Mathematica can actually give the correct answer, but it takes a
> bit of work and knowing what you are doing (pretty common situation). Let's
> start by assuming that a is real. Mathematica deals with this case pretty
> well.
>
> Define two functions, say g and h which for a given u are the real and
> imaginary parts of your expression, i. e.
>
> In[1]:=
> f[u_] := (a/E^(I*u) + 1)/(a/E^(I*u) - 1)
>
> In[2]:=
> g[u_] := ComplexExpand[Re[f[u]], TargetFunctions -> {Re, Im}]
>
> In[3]:=
> h[u_] := ComplexExpand[Im[f[u]], TargetFunctions -> {Re, Im}]
>
> In[4]:=
> Integrate[h[u], {u, 0, 2Pi}] // Simplify
>
> Out[4]=
> 0
>
> In[5]:=
> Simplify[Integrate[g[u], {u, 0, 2Pi}], a > 1]
>
> Out[5]=
> 2 Pi
>
> In[6]:=
> Simplify[Integrate[g[u], {u, 0, 2Pi}], a < -1]
>
> Out[6]=
> 2 Pi
>
> In[7]:=
> Simplify[Integrate[g[u], {u, 0, 2Pi}], -1 < a < 1]
>
> Out[7]=
> -2 Pi
>
> (To Adam Strzebonski: unfortunately the following does to work
>
> In[8]:=
> Simplify[(2*Sqrt[(-1 + a)^2/(1 + a)^2]*(1 + a)*Pi)/(-1 + a),
> Element[a ,Reals] && Abs[a] < 1]
>
> Out[8]=
> 2
> (-1 + a)
> 2 Sqrt[---------] (1 + a) Pi
> 2
> (1 + a)
> ----------------------------
> -1 + a
It has been fixed in the later releases of Mathematica 4.
In[1]:= Simplify[(2*Sqrt[(-1 + a)^2/(1 + a)^2]*(1 + a)*Pi)/(-1 + a),
Element[a ,Reals] && Abs[a] < 1]
Out[1]= -2 Pi
>
> In[9]:=
> Simplify[(2*Sqrt[(-1 + a)^2/(1 + a)^2]*(1 + a)*Pi)/(-1 + a),
> Element[a,Reals] && Abs[a] > 1]
>
> Out[9]=
> 2 (1 + a) Pi Abs[-1 + a]
> ------------------------
> (-1 + a) Abs[1 + a])
>
This is more tricky. Abs[-1 + a] and Abs[1 + a], when taken separately,
cannot be further simplified, because -1 + a and 1 + a do not have
constant signs on the set of a satisfying the assumptions. Only the
ratio Abs[-1 + a]/Abs[1 + a] can be simplified. So Simplify would need
to use a rule like
In[1]:= t[Abs[f_]^n_. Abs[g_]^m_. h_., assum_] := f^n g^m h /;
OddQ[m] && OddQ[n] && Experimental`ImpliesQ[assum, f*g>0]
In[2]:= t[other_, assum_] := other
In[3]:= mySimplify[expr_, assum_] := Simplify[expr, assum,
TransformationFunctions->{Automatic, t[#, assum]&}]
In[4]:= mySimplify[(2*Sqrt[(-1 + a)^2/(1 + a)^2]*(1 + a)*Pi)/(-1 + a),
Element[a,Reals] && Abs[a] > 1]
Out[4]= 2
Pi
(and similar rules for f*g<0, and Sign instead of Abs etc.)
The rules have a rather bad complexity (searching for pairs of elements
in every Times expression), so probably FullSimplify would be a better
place to put them in.
> Should not Simplify be able to manage these, at least when a is real? For
> the complex case see below.)
>
> If a is not real one can still show that we get the answer 2Pi for Abs[a]<1
> and -2Pi for Abs[a]<1. However, in this case Mathematica needs more help. We
> again define, f,g, and h
>
> In[1]:=
> Clear[f, g, h]
>
> In[2]:=
> f[u_] := (a/E^(I*u) + 1)/(a/E^(I*u) - 1)
>
> In[3]:=
> g[u_] := ComplexExpand[Re[f[u]], {a}, TargetFunctions -> {Re, Im}]
>
> In[4]:=
> h[u_] := ComplexExpand[Im[f[u]], {a}, TargetFunctions -> {Re, Im}]
>
> Note that now we told Mathematica not to assume that a is real. Integrating
> the imaginary part still gives 0:
>
> In[5]:=
> Integrate[h[u], {u, 0, 2Pi}] // Simplify
>
> Out[5]=
> 0
>
> If we integrate the real parts we get identical looking answers:
>
> In[6]:=
> Simplify[Integrate[g[u], {u, 0, 2Pi}], Abs[a] > 1]
>
> Out[6]=
> 2 2
> 2 Pi (-1 + Im[a] + Re[a] )
> ---------------------------
> 2 2
> Abs[-1 + Im[a] + Re[a] ]
>
> In[7]:=
> Simplify[Integrate[g[u], {u, 0, 2Pi}], Abs[a] < 1]
>
> Out[7]=
> 2 2
> 2 Pi (-1 + Im[a] + Re[a] )
> ---------------------------
> 2 2
> Abs[-1 + Im[a] + Re[a] ]
>
> Although the answers look identical they are in fact not! In the case
> Abs[a]>1 the numerator is positive while in the case Abs[a]<1 it is
> negative. So in the former case the answer is 2Pi while in the latter -2Pi.
> Perhaps Mathematica also ought to be able to get this right without human
> help?
FullSimplify can do this (it has more rules relating Re, Im, and Abs).
In[5]:= f[u_] := (a/E^(I*u) + 1)/(a/E^(I*u) - 1)
In[6]:= g[u_] := ComplexExpand[Re[f[u]], {a}, TargetFunctions -> {Re,
Im}]
In[7]:= FullSimplify[Integrate[g[u], {u, 0, 2Pi}], Abs[a] > 1]
Out[7]= 2 Pi
In[8]:= FullSimplify[Integrate[g[u], {u, 0, 2Pi}], Abs[a] < 1]
Out[8]= -2
Pi
Best Regards,
Adam Strzebonski
Wolfram Research
>
> Andrzej Kozlowski
>
> on 00.4.21 12:48 PM, Alberto Verga at verga at marius.univ-mrs.fr wrote:
>
> > Compute
> > Integrate[(1 + a/E^(I*u))/(-1 + a/E^(I*u)), {u, 0, 2*Pi}]
> >
> > Mathematica gives -2Pi
> >
> > Now multiply the numerator and the denominator by -1
> >
> > Integrate[(-1 - a/E^(I*u))/(1 - a/E^(I*u)), {u, 0, 2*Pi}]
> >
> > Mathematica gets 2*Pi
> >
> > This is only possible if 1==-1
> >
> > Is this another bug in Limit?
> >
> > Alberto Verga
> > irphe - Marseille
> >
> >
> >
>
> --
> Andrzej Kozlowski
> Toyama International University
> Toyama, Japan
> http://sigma.tuins.ac.jp/
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