Re: Zero testing

*To*: mathgroup at smc.vnet.net*Subject*: [mg53135] Re: [mg53106] Zero testing*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Fri, 24 Dec 2004 05:59:43 -0500 (EST)*References*: <200412231259.HAA21206@smc.vnet.net> <42577F00-556A-11D9-8E89-000A95B4967A@mimuw.edu.pl>*Sender*: owner-wri-mathgroup at wolfram.com

A correction. The failure of FullSimplify[ComplexExpand[Unevaluated[ ((1 + I*(-167594143/78256779 + Pi))^4)^(1/2)]]] -(((167594143 + 78256779*I) - 78256779*Pi)^2/ 6124123459454841) to give the right answer was due to my thoughtless placing of Unevaluated. With correct placing we get the right answer: FullSimplify[ComplexExpand[ ((1 + I*Unevaluated[-167594143/78256779 + Pi])^4)^ (1/2)]] ((167594143 + 78256779*I) - 78256779*Pi)^2/ 6124123459454841 Andrzej On 24 Dec 2004, at 14:11, Andrzej Kozlowski wrote: > > On 23 Dec 2004, at 21:59, Maxim wrote: > >> By 'zero testing' I mean the problem of deciding whether a given >> numerical >> quantity is exactly equal to zero. Often Mathematica doesn't make >> necessary checks at all: >> >> In[1]:= >> a = Pi/4; b = I*Log[-(-1)^(3/4)]; >> Limit[(a - b)/x, x -> 0] >> Integrate[1/((x - a)*(x - b)), x] >> Integrate[1/(x - b), {x, 0, 1}] >> >> Out[2]= >> DirectedInfinity[Pi - (4*I)*Log[-(-1)^(3/4)]] >> >> Out[3]= >> (-4*((-2*I)*ArcTan[Log[-(-1)^(3/4)]/x] - 2*Log[Pi - 4*x] + Log[16*(x^2 >> + Log[-(-1)^(3/4)]^2)]))/(2*Pi - (8*I)*Log[-(-1)^(3/4)]) >> >> Out[4]= >> ((-I)*Pi - (2*I)*ArcTan[Log[-(-1)^(3/4)]] - Log[Log[-(-1)^(3/4)]^2] >> + Log[1 + Log[-(-1)^(3/4)]^2])/2 >> >> The constants a and b are equal to each other. We can see that Out[2] >> is >> incorrect because it is equal to ComplexInfinity and the limit of 0/x >> is >> 0; Out[3] is also equal to ComplexInfinity (the denominator is zero) >> and >> therefore wrong as well; finally, Out[4] is incorrect because the >> integral >> doesn't converge. > > This problem amounts to the difference between Simplify and > FullSimplify. FullSimplify is needed to determine that a and b are > equal, just Simplify cannot achieve this: > > In[1]:= > a = Pi/4; b = I*Log[-(-1)^(3/4)]; > > In[2]:= > Simplify[a - b] > > Out[2]= > (1/4)*(Pi - 4*I*Log[-(-1)^(3/4)]) > > In[3]:= > FullSimplify[a - b] > > Out[3]= > 0 > > Functions like integrate use Simplify automatically but not > FullSimplify. Why not? The answer you provided yourself (well, almost) > in an earlier posting: FullSimplify generally takes far too long to be > used automatically. Even using Simplify leads to problems, as you have > noted. Are you then arguing that FullSimplify ought to be used > automatically? I know it ould b enice if one were able to eat the cake > and still have it ... > > If you apply FullSimplify to the argument in the above examples you > will get the answer you stated as correct. In all cases, that is, > except in one when it is you and not Mathematica that blundered. The > case in question is: > > >> Integrate[1/((x - a)*(x - b)), x] > > > Are you seriously claiming that > > Integrate[1/((x - Pi/4)^2), x] > > is ComplexInfinity? At least Mathematica knows better than that: > > > Integrate[FullSimplify[1/((x - a)*(x - b))], x] > > 4/(Pi - 4*x) > >> >> A variation of the same problem is when we're numerically evaluating a >> discontinuous function for an argument close to a discontinuity >> point. In >> those cases Mathematica does perform necessary checks but handles more >> complicated cases incorrectly, especially when the branch cuts are >> concerned: >> >> In[5]:= >> Sign[Im[Sqrt[-1 - 10^-18*I]]] >> >> Out[5]= >> 1 > > > The problem here is that Mathematica has great difficulty in > determining Im[Sqrt[-1 - 10^-18*I] is -1. Algebraically. You ned ot > use > > > Sign[FullSimplify[ComplexExpand[Im[Sqrt[-1-10^-18*I]]]]] > > -1 > > > This I agree to be a problem that ought to be fixed. It seems to me > that numerical methods should be used since: > > > Sign[N[Im[Sqrt[-1-10^-18*I]]]] > > -1 > > works easily and reliably. On the other hand it could be argued that > this sort of thing should be approached numerically anyway. >> >> In[6]:= >> ((1 + I*(-167594143/78256779 + Pi))^4)^(1/2) >> >> Out[6]= >> (1 + I*(-167594143/78256779 + Pi))^2 >> >> In fact In[5] is equal to -1 and In[6] is equal to -(1 >> + I*(-167594143/78256779 + Pi))^2. In the last case we do not even >> apply >> any functions, the error is in the automatic simplifications. >> Sometimes >> setting a high value for $MinPrecision helps, but here it doesn't >> have any >> effect either. >> > > This one is indeed a problem. Even this > > FullSimplify[ComplexExpand[ > (1 + I*(-167594143/ > 78256779 + Pi))^4]]^ > (1/2) > > > -(((167594143 + 78256779*I) - > 78256779*Pi)^2/ > 6124123459454841) > > gives the wrong answer ( the sign is wrong). To get the right one you > need to use high precision arithmetic and Unevaluated: > > > N[Unevaluated[ > ((1 + I*(-167594143/ > 78256779 + Pi))^4)^ > (1/2)], 20] > > > 1.5635873239815087691232`4.043\ > 607143112669*^-16 - > 2.00000000000000015635873239\ > 81508708003`20.15051499783199*I > > The only way that I have found of getting the correct exact answer is: > > > FullSimplify[ComplexExpand[((1 + I*(-u + Pi))^4)^ > (1/2)] /. u -> 167594143/78256779] > > ((167594143 + 78256779*I) - 78256779*Pi)^2/ > 6124123459454841 > > In[54]:= > N[%, 20] > > Out[54]= > 1.5635873239815087691362`4.043607143112669*^-16 - > 2.00000000000000015635873239815087005589`20.15051499783\ > 1987*I > > It is curious that this method works while the one using Unevaluated > does not: > > > FullSimplify[ComplexExpand[((1 + I*(-u + Pi))^4)^ > (1/2)] /. u -> 167594143/78256779] > FullSimplify[ComplexExpand[Unevaluated[ > ((1 + I*(-167594143/78256779 + Pi))^4)^(1/2)]]] > > > ((167594143 + 78256779*I) - 78256779*Pi)^2/ > 6124123459454841 > > > -(((167594143 + 78256779*I) - 78256779*Pi)^2/ > 6124123459454841) > > > > While I think this Mathematica's dealing wiht this situation could be > improved (particularly the last example) I find it difficult to see > how such problems could be avoided in general. The problem seems to be > very subtle. On the one hand, one can't expect the Mathematica > evaluator to automatically apply powerful transformation s such as > ComplexExpand or FullSimplify. On the other hand, certain > transformations have to be automatically performed by the evaluator > otherwise Mathematica would become unbearably tedious to use (one > could leave every expression unevaluated until Simplify or > FullSimplify or possibly a weaker function were applied to it but it > would still require the user to judge when to apply which). > > As far as I can tell all this means that the user has to remain > vigilant at all times and always think carefully about the answers > returned by the program. > > > Andrzej Kozlowski > Chiba, Japan > http://www.akikoz.net/~andrzej/ > http://www.mimuw.edu.pl/~akoz/ >

**References**:**Zero testing***From:*Maxim <ab_def@prontomail.com>