Re: Re: Weird result in Mathematica 6

*To*: mathgroup at smc.vnet.net*Subject*: [mg76620] Re: [mg76432] Re: [mg76393] Weird result in Mathematica 6*From*: Adam Strzebonski <adams at wolfram.com>*Date*: Thu, 24 May 2007 06:27:04 -0400 (EDT)*References*: <200705211001.GAA10071@smc.vnet.net> <EB6D3224-597F-4DD6-B05D-08B9F6A05D2D@mimuw.edu.pl> <200705220648.CAA19836@smc.vnet.net> <63B2BBD7-455D-42F6-AFB2-63F7D37D62D3@mimuw.edu.pl>*Reply-to*: adams at wolfram.com

Andrzej Kozlowski wrote: > *This message was transferred with a trial version of CommuniGate(tm) Pro* > > On 22 May 2007, at 15:48, Adam Strzebonski wrote: > >> Andrzej Kozlowski wrote: >>> *This message was transferred with a trial version of CommuniGate(tm) >>> Pro* >>> >>> On 21 May 2007, at 19:01, Sebastian Meznaric wrote: >>> >>>> I was playing around with Mathematica 6 a bit and ran this command to >>>> solve for the inverse of the Moebius transformation >>>> >>>> FullSimplify[ >>>> Reduce[(z - a)/(1 - a\[Conjugate] z) == w && a a\[Conjugate] < 1 && >>>> w w\[Conjugate] < 1, z]] >>>> >>>> This is what I got as a result: >>>> -1 < w < 1 && -1 < a < 1 && z == (a + w)/(1 + w Conjugate[a]) >>>> >>>> Why is Mathematica assuming a and w are real? The Moebius >>>> transformation is invertible in the unit disc regardless of whether a >>>> and w are real or not. Any thoughts? >>>> >>>> >>> >>> >>> Reduce and FullSimplify will usually deduce form the presence of >>> inequalities in an expression like the above that the variables >>> involved in the inequalites are real. In your case it "sees" >>> a*Conjugate[a]<1 and "deduces" that you wanted a to be real. This was >>> of coruse not your intention but you can get the correct behaviour by >>> using: >>> >>> FullSimplify[ >>> Reduce[(z - a)/(1 - Conjugate[a]*z) == w && Abs[a]^2 < 1 && Abs[w] ^2 < >>> 1, z]] >>> >>> >>> -1 < Re[w] < 1 && -Sqrt[1 - Re[w]^2] < Im[w] < Sqrt[1 - Re[w]^2] && >>> -1 < >>> Re[a] < 1 && >>> -Sqrt[1 - Re[a]^2] < Im[a] < Sqrt[1 - Re[a]^2] && >>> z == (a + w)/(w*Conjugate[a] + 1) >>> >>> Mathematica knows that the fact that an inequality involves Abs[a] does >>> not imply that a is real but it does not "know" the same thing about >>> a*Conjugate[a]. This is clearly dictated by considerations of >>> performance than a straight forward bug. >>> Andrzej Kozlowski >>> >> >> By default, Reduce assumes that all algebraic level variables appearing >> in inequalities are real. You can specify domain Complexes, to make >> Reduce assume that all variables are complex and inequalities >> >> expr1 < expr2 >> >> should be interpretted as >> >> Im[expr1]==0 && Im[expr2]==0 && Re[expr1]<Re[expr2] >> >> For more info look at >> >> http://reference.wolfram.com/mathematica/ref/Reduce.html >> http://reference.wolfram.com/mathematica/tutorial/RealReduce.html >> http://reference.wolfram.com/mathematica/tutorial/ComplexPolynomialSystems.html >> >> >> In your example we get >> >> In[2]:= Reduce[(z - a)/(1 - a\[Conjugate] z) == w && a a\[Conjugate] < >> 1 && >> w w\[Conjugate] < 1, z, Complexes] >> >> 2 2 >> Out[2]= -1 < Re[w] < 1 && -Sqrt[1 - Re[w] ] < Im[w] < Sqrt[1 - Re[w] ] && >> >> 2 2 >>> -1 < Re[a] < 1 && -Sqrt[1 - Re[a] ] < Im[a] < Sqrt[1 - Re[a] ] && >> >> a + w >>> z == ------------------ >> 1 + w Conjugate[a] >> >> >> Evaluate >> >> Reduce[x^2+y^2<=1, {x, y}, Complexes] >> >> to see why I think that assuming that variables appearing >> in inequalities are real is a reasonable default behaviour. >> >> Best Regards, >> >> Adam Strzebonski >> Wolfram Research >> > > Still, it seems to me that there is a certain problem with this, not > very important but still, a "logical difficulty". It concerns not > Reduce, where you can specify the domain to be Reals or Complexes etc, > but Simplify, where you can't. So for example: > > Simplify[Re[x], x*Conjugate[x] > 1] > x > > folowing the principle also used by reduce, Simplify assumed that x is > real. On the other hand: > > Simplify[Re[x], Abs[x] > 1] > Re(x) > > which also agrees with the principle, sicne Abs in non-algebraic. But, > unlike in the case of Reduce, there seems to be no way to make Simplify > treat the first assumption as taking place over the Complexes as in the > Reduce example: > > Simplify[Re[x], x*Conjugate[x] > 1 && Elment[x, Complexes]] > x Adding Element[x, Complexes] doesn't change anything, since Element[x, Reals] && Element[x, Complexes] <=> Element[x, Reals] The way to specify that variables used in an inequality are complex is to use Im[expr]==0 && Re[expr]>0 instead of expr>0. In[1]:= Simplify[Re[x], Im[x*Conjugate[x]]==0 && Re[x*Conjugate[x]]>1] Out[1]= Re[x] Best Regards, Adam Strzebonski Wolfram Research > > Simplify[Re[x] && Element[x, Complexes], x*Conjugate[x] > 1] > x > > In other words, it seems that when using Simplify one really needs to > use Abs in inequalities, if one does not want to force the assumption > that a variable is real. (?) > > Andrzej Kozlowski >

**References**:**Weird result in Mathematica 6***From:*Sebastian Meznaric <meznaric@gmail.com>

**Re: Weird result in Mathematica 6***From:*Adam Strzebonski <adams@wolfram.com>

**Re: Re: Weird result in Mathematica 6**

**Re: Mathematica 6.0 easier for me ... (small review)**

**Re: Re: Weird result in Mathematica 6**

**Re: Re: Weird result in Mathematica 6**