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/