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MathGroup Archive 2006

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Re: Simplify UnitStep expressions

  • To: mathgroup at smc.vnet.net
  • Subject: [mg69361] Re: Simplify UnitStep expressions
  • From: p-valko at tamu.edu
  • Date: Sat, 9 Sep 2006 03:26:37 -0400 (EDT)
  • References: <edm1qa$cps$1@smc.vnet.net><edolp2$hs6$1@smc.vnet.net>

Daniel Lichtblau pointed out to me that
"Assuming is meant to be used in tandem use of Refine, Simplify, or
related. Alone it does but little.
In[27]:= Assuming[a > 2, Refine[Not[a > 2]]] Out[27]= False
"
He is right. I should have not posted this silly example.
Peter

p-valko at tamu.edu wrote:
> With all due respect I am afraid the phenomenon has little to do with
> the actual algorithms used in Simplify[] and FullSimplify[], but rather
> with the way how assumptions are treated in general.
>
> To illustrate my statement, I show an example without any Simplify[] or
> FullSimplify[]:
>
> In:
> Assuming[a > 2, Not[a > 2] ]
>
> I would like to get an answer False, but Mathematica gives
>
> Out:
> a <= 2
>
> Pretty surprising result ! ! !
>
> Regards
> Peter
>
> Adam Strzebonski wrote:
> > Andrzej Kozlowski wrote:
> > >
> > > On 5 Sep 2006, at 16:20, Adam Strzebonski wrote:
> > >
> > >> Andrzej Kozlowski wrote:
> > >>
> > >>> On 1 Sep 2006, at 11:41, L. Dwynn Lafleur wrote:
> > >>>
> > >>>> The following is transcribed from a Mathematica 5.2 notebook in
> > >>>> Windows XP:
> > >>>>
> > >>>> In[1]:= Simplify[UnitStep[a-x/b], a-x/b > 0]
> > >>>> Out[1]= 1
> > >>>>
> > >>>> In[2]:= Simplify[UnitStep[a-Pi/b], a-Pi/b > 0]
> > >>>> Out[2]= UnitStep[a-Pi/b]
> > >>>>
> > >>>> Why does the second output different from the first?  I know it has
> > >>>> something to do with the fact that Pi is internally defined in
> > >>>> Mathematica
> > >>>> because a similar result occurs Pi is replaced with E, but what
> > >>>> logic is
> > >>>> being followed?
> > >>>>
> > >>>> --
> > >>>> ======================================
> > >>>>  L. Dwynn Lafleur
> > >>>>  Professor of Physics
> > >>>>  University of Louisiana at Lafayette
> > >>>>  lafleur at louisiana.edu
> > >>>> ======================================
> > >>>>
> > >>> Curiously, if you use FullSimplify rather then Simplify you will get:
> > >>> FullSimplify[UnitStep[a-Pi/b], a-Pi/b > 0]
> > >>> 1
> > >>> The same holds if Pi is replaces by E, or indeed by explicit
> > >>> functions of E or Pi such as Pi^2, E^Pi etc. In all such cases
> > >>> FullSimplify works but Simplify does not work. Strange.
> > >>> Andrzej Kozlowski
> > >>
> > >>
> > >> The cylindrical algebraic decomposition (CAD) algorithm used by  Simplify
> > >> to prove inference requires polynomial inequalities with rational  number
> > >> coefficients. a-x/b > 0 is equivalent to a polynomial inequality
> > >> -(a*b^2) + b*x < 0 which has rational number coefficients.
> > >> a-Pi/b > 0 is equivalent to a polynomial inequality -(a*b^2) + b*Pi  < 0
> > >> which has a numeric coefficient Pi which is not a rational number.
> > >>
> > >> Mathematica has two ways of dealing with nonrational numeric
> > >> coefficients in CAD. One is to replace each nonrational coefficient
> > >> with a new variable. This method always allows to decide inference
> > >> (modulo the ability to zero-test the exact numeric constants), but
> > >> it is potentially very expensive - CAD has a doubly exponential
> > >> complexity in the number of variables and we add a new variable for
> > >> each nonrational coefficient. The second method replaces nonrational
> > >> numeric coefficients with their approximations. This is much less
> > >> expensive, but in some cases it fails to decide inference.
> > >> Simplify uses the second method which in this case is insufficient.
> > >>
> > >> FullSimplify uses more transformations, and one of the additional
> > >> transformations succeeds.
> > >>
> > >> Best Regards,
> > >>
> > >> Adam Strzebonski
> > >> Wolfram Research
> > >>
> > >>
> > >>
> > >
> > > Thanks for the explanation. I feel I should have guessed it, but  there
> > > is one thing that still puzzles me. What exactly makes the  rational
> > > approximation fail for Pi or E, since it seems to work fine  for
> > > algebraic numbers such as Sqrt[2] or 3^(1/3)? It certainly can't  be
> > > anything to do with Pi or E not being algebraic, so presumably it  is
> > > something to do with the way the rational approximation is chosen?  This
> > > sounds very interesting; could you explain the exact reason why  the
> > > rational approximation in this case doe snot work and in what  other
> > > cases will it not work in general? It sounds like the reason  might be
> > > mathematically interesting (?).
> > >
> > > Andrzej Kozlowski
> > >
> >
> > For Pi or E the assumption mechanism uses inexact (bignum)
> > approximations and constructs CAD with inexact sample point
> > coordinates. If we have an inequality f(X)<0 and we do not
> > find a cell with a sample point P for which f(P)<0, but we
> > do find a cell with a sample point P for which f(P) is
> > a bignum zero (for instance 0``20) then we cannot tell
> > whether f(X)<0 has any solutions or not.
> >
> > For algebraic numbers the assumption mechanism does not use
> > approximations. It replaces the algebraic numbers with new
> > variables, because in this case it does not contribute that
> > much to the complexity. We do not need to compute projections
> > wrt. the new variables. Instead we make the variables last in
> > the projection ordering, and in the lifting phase we only lift
> > the one-point cell which corresponds to the new variables being
> > equal to the corresponding algebraic numbers.
> > 
> > Best Regards,
> > 
> > Adam Strzebonski
> > Wolfram Research


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