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Re: Simplify 0/0 to 1?
*To*: mathgroup at smc.vnet.net
*Subject*: [mg77514] Re: Simplify 0/0 to 1?
*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>
*Date*: Mon, 11 Jun 2007 04:24:55 -0400 (EDT)
*References*: <934147.40360.qm@web43135.mail.sp1.yahoo.com> <2DFED10A-AAB0-4649-A747-93E238C30B75@mimuw.edu.pl>
*Reply-to*: Andrzej Kozlowski <akoz at mimuw.edu.pl>
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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