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MathGroup Archive 2000

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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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