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Re: Demonstrate that 1==-1
*To*: mathgroup at smc.vnet.net
*Subject*: [mg23183] Re: [mg23171] Demonstrate that 1==-1
*From*: Andrzej Kozlowski <andrzej at tuins.ac.jp>
*Date*: Mon, 24 Apr 2000 01:11:55 -0400 (EDT)
*Sender*: owner-wri-mathgroup at wolfram.com
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
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])
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?
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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