Re: Simplify 0/0 to 1?

*To*: mathgroup at smc.vnet.net*Subject*: [mg77516] Re: Simplify 0/0 to 1?*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Mon, 11 Jun 2007 04:25:58 -0400 (EDT)*References*: <934147.40360.qm@web43135.mail.sp1.yahoo.com> <2DFED10A-AAB0-4649-A747-93E238C30B75@mimuw.edu.pl> <B338F0C9-E8DF-405A-802C-C1820700EB83@mimuw.edu.pl>

I need to correct somehtin gin my post below. The three argument usage of If in the examples below will lead to serious trouble; the correct way is to use four arguments: $Pre = (If[Simplify[Denominator[#]] == 0, Print["Zero denominator!"], #, #] &); and $Pre = If[N[Denominator[#1], 20] == 0, Print["Probably zero denominator!"], #1, #] &; Andrzej Kozlowski On 11 Jun 2007, at 10:31, Andrzej Kozlowski wrote: > In fact, if you are really concerned with this, you could do > something like: > > $Pre = (If[Simplify[Denominator[#]] == 0, Print["Zero > denominator!"], #] &); > > You will then get: > > o = (Log[2]*Cos[Pi/12] - Log[2]*Sin[Pi/12] - 2*Cos[Pi/12] + 2*Sin > [Pi/12] + > Sqrt[2] + 2*Log[Cos[Pi/12] - Sin[Pi/12]]*Cos[Pi/12] - > 2*Log[Cos[Pi/12] - Sin[Pi/12]]*Sin[Pi/12])/(Log[2]*Cos[Pi/12] - > Log[2]*Sin[Pi/12] + 2*Log[Cos[Pi/12] - Sin[Pi/12]]*Cos[Pi/12] - > 2*Log[Cos[Pi/12] - Sin[Pi/12]]*Sin[Pi/12]) > > "Zero denominator!" > > before you even apply Simplify. Similarly: > > (1 - Cos[Pi/23]^2 - Sin[Pi/23]^2)^2/(1 - 2*Cos[Pi/23]^2 + > Cos[Pi/23]^4 - 2*Sin[Pi/23]^2 + 2*Cos[Pi/23]^2*Sin[Pi/23]^2 + > Sin[Pi/23]^4) > "Zero denominator!" > > Unfortunately this will produce dramatic slowdowns in any slighlty > complex algebraic computations (I haven't tried it but I am pretty > sure). It might be better to replace this with a numerical check, > but you will have to use extended precision arithemtic, since > machine precision will often fail to detect zeros. So you could > instead use: > > $Pre =. > $Pre = If[N[Denominator[#1], 20] == 0, > Print["Probably zero denominator!"], #1] & ; > o > N::meprec:Internal precision limit $MaxExtraPrecision = 50. reached > while evaluating -((-1+) log(2))/(2 )+((1+) log(2))/(2 )-((-1+) log > ((1+)/(2 )-<<1>>/(2 <<1>>)))/+((1+) log((1+)/(2 )-(-1+)/(2 )))/. >> > "Probably zero denominator!" > > Peronally I think it is much better to leave thigs as they are, > keep this in mind and be a little careful. > > Andrzej Kozlowski > > > On 10 Jun 2007, at 22:20, Andrzej Kozlowski wrote: > >> *This message was transferred with a trial version of CommuniGate >> (tm) Pro* >> I would not call this bad performance. In fact this is inevitable, >> not only in Mathematica but in every CAS that exists today >> (although not necssarily in the same example). ALl you need to do >> is to choose numerator and denominator equal to zero but >> sufficiently complicated and such that the CAS will see that they >> are equal before it can check that they are both 0. Every CAS will >> then simplify the expression to 1, because for reason of >> performance they cancel common factors as soon as they see them. >> Here is a very trivial example: >> >> (1 - Cos[Pi/23]^2 - Sin[Pi/23]^2)/(1 - Cos[Pi/23]^2 - Sin[Pi/23]^2) >> 1 >> >> This was turned into 1 by the evaluator withot any Simplify even >> though both, the numerator and the denominator are clearly zero. >> You can avoid this by mapping Sinmplify on the numerator and the >> denominator but you have to use Unevaluated first: >> >> Simplify /@ >> Unevaluated[(1 - Cos[Pi/23]^2 - Sin[Pi/23]^2)/(1 - Cos[Pi/23]^2 - >> Sin[Pi/23]^2)] >> The same approach will work in your example: >> >> Simplify/@o >> >> I am convinced that this phenomenon is general will be observed in >> every CAS, although not necessarily in the same examples. But in >> every CAS you can construct expressions equal to zero that will be >> sufficiently complicated for the CAS not to be able to see that >> they are zero before seeing that they are equal. Here is another >> example, which is just again based on the Sin^2[a]+Cos^2[a]==1 >> identity, but just made a bit more complicated, so that ti doe not >> reduce to 1 immediately: >> >> >> Simplify[(1 - Cos[Pi/23]^2 - Sin[Pi/23]^2)^2/(1 - 2*Cos[Pi/23]^2 + >> Cos[Pi/23]^4 - >> 2*Sin[Pi/23]^2 + 2*Cos[Pi/23]^2*Sin[Pi/23]^2 + Sin[Pi/23]^4)] >> >> 1 >> >> As you can see again, the numerator and the denominator are both >> zero but Simplify sees that they are equal before it can see that >> they are both zero and at that point the Evaluator performs the >> cancelation. I think for reason of performance all CAS have to >> work like this in principle. >> >> Andrzej Kozlowski >> >> >> >> >> >> On 10 Jun 2007, at 20:57, dimitris anagnostou wrote: >> >>> >>> Hi. >>> This appeared in another forum as part of a question >>> what another CAS does. >>> Just of curiosity I check Mathematica's performance (5.2). >>> The result was poor! >>> >>> Here is the expression >>> >>> >>> In[16]:= >>> o = (Log[2]*Cos[Pi/12] - Log[2]*Sin[Pi/12] - 2*Cos[Pi/12] + 2*Sin >>> [Pi/ >>> 12] + Sqrt[2] + >>> 2*Log[Cos[Pi/12] - Sin[Pi/12]]*Cos[Pi/12] - 2*Log[Cos[Pi/12] - >>> Sin[Pi/12]]*Sin[Pi/12])/ >>> (Log[2]*Cos[Pi/12] - Log[2]*Sin[Pi/12] + 2*Log[Cos[Pi/12] - >>> Sin[Pi/ >>> 12]]*Cos[Pi/12] - >>> 2*Log[Cos[Pi/12] - Sin[Pi/12]]*Sin[Pi/12]) >>> >>> >>> Out[16]= >>> (Sqrt[2] + (-1 + Sqrt[3])/Sqrt[2] - (1 + Sqrt[3])/Sqrt[2] - ((-1 + >>> Sqrt[3])*Log[2])/(2*Sqrt[2]) + >>> ((1 + Sqrt[3])*Log[2])/(2*Sqrt[2]) - ((-1 + Sqrt[3])*Log[-((-1 + >>> Sqrt[3])/(2*Sqrt[2])) + (1 + Sqrt[3])/(2*Sqrt[2])])/ >>> Sqrt[2] + ((1 + Sqrt[3])*Log[-((-1 + Sqrt[3])/(2*Sqrt[2])) + >>> (1 + >>> Sqrt[3])/(2*Sqrt[2])])/Sqrt[2])/ >>> (-(((-1 + Sqrt[3])*Log[2])/(2*Sqrt[2])) + ((1 + Sqrt[3])*Log[2])/ >>> (2*Sqrt[2]) - >>> ((-1 + Sqrt[3])*Log[-((-1 + Sqrt[3])/(2*Sqrt[2])) + (1 + Sqrt >>> [3])/ >>> (2*Sqrt[2])])/Sqrt[2] + >>> ((1 + Sqrt[3])*Log[-((-1 + Sqrt[3])/(2*Sqrt[2])) + (1 + Sqrt[3])/ >>> (2*Sqrt[2])])/Sqrt[2]) >>> >>> Watch now a really bad performance! >>> >>> In[17]:= >>> (Simplify[#1[o]] & ) /@ {Numerator, Denominator} >>> >>> Out[17]= >>> {0, 0} >>> >>> That is Mathematica simplifies succesfully both the numerator >>> and denominator to zero. So, you wonder what goes wrong? >>> >>> Try now to simplify the whole expression! >>> >>> In[19]:= >>> Simplify[o] >>> >>> Out[19]= >>> 1 >>> >>> A very weird result to my opinion! >>> Simplification of 0/0 to 1? >>> I think no simplification or some >>> warning messages would be much better >>> than 1! >>> >>> Note also that >>> >>> In[20]:= >>> RootReduce[o] >>> >>> Out[20]= >>> 1 >>> >>> Dimitris >>> >>> Take the Internet to Go: Yahoo!Go puts the Internet in your >>> pocket: mail, news, photos & more. >> >

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