MathGroup Archive 2011

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: Full simplify problem

  • To: mathgroup at smc.vnet.net
  • Subject: [mg122277] Re: Full simplify problem
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Sun, 23 Oct 2011 06:23:50 -0400 (EDT)
  • Delivered-to: l-mathgroup@mail-archive0.wolfram.com
  • References: <D3C63570-91CA-489E-B8B7-661844B401BC@mimuw.edu.pl> <C9A65308-E4DD-4DD6-967D-FFDF15F6E7DC@mimuw.edu.pl> <1319312994.3872.YahooMailNeo@web43140.mail.sp1.yahoo.com>

I think it is "considered" normal, in the sense that Adam Strzebonski  
has written several times (on this forum) that FullSimplify is not 
"optimised" for assumptions involving Equal. I think this means that 
such assumptions will work for "simple" functions, such as polynomials 
but not necessarily for more complicated ones, unless the normal 
simplification process (which uses ComplexityFunction) reduces the 
problem to something that involves polynomials.

To see what I mean, take a polynomial

f[x_] := x^4 - 3 x^2 + 5 x - 1;

Note that now:

In[23]:= Assuming[x == y + z, FullSimplify[f[x] - f[(y + z)]]]

Out[23]= 0

without any need to play around with ComplexityFunction. On the other 
hand if you replace f by something non-polynomial, eg. Log or Sin, you 
won't get 0 unless you use a suitable ComplexityFunction.

Andrzej Kozlowski





On 22 Oct 2011, at 21:49, dimitris anagnostou wrote:

> Hello again.
>
> Thank you very much for your time to reply to my messages.
>
> As regard now the current post do you think it is a normal behavior 
for FullSimplify?
> What about if somebody is not experienced user of FullSimplify's 
options?
>
> Dimitris
>
> From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
> To: dimitris anagnostou <dimmechan at yahoo.com>
> Sent: Saturday, October 22, 2011 10:45 PM
> Subject: Fwd: Full simplify problem
>
> Hello Dimitris
>
> I also sent a response to that message. It is quite similar to yours.
>
> Best regards
>
> Andrzej
>
> Begin forwarded message:
>
> > From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
> > Date: 22 October 2011 17:02:05 CEST
> > To: A. Lapraitis <ffcitatos at gmail.com>
> > Cc: mathgroup at smc.vnet.net
> > Subject: Re: Full simplify problem
> >
> > Because FullSimplify would have to increase the default complexity 
to obtain the cancellation. But this works:
> >
> > Assuming[x == y + z,
> > FullSimplify[E^x - E^(y + z),
> >  ComplexityFunction -> (Count[#, x, {1, Infinity}] &)]]
> >
> > 0
> >
> > Note that the reason why this works is that:
> >
> > Assuming[x == y + z,
> > FullSimplify[x, ComplexityFunction -> (Count[#, x, {1, Infinity}] 
&)]]
> >
> > y+z
> >
> >
> > Andrzej Kozlowski
> >
> >
> > On 22 Oct 2011, at 12:09, A. Lapraitis wrote:
> >
> >> Hello,
> >>
> >> Could anyone explain why the following does not give zero?
> >>
> >> In[72]:= Assuming[
> >> x == y + z,
> >> FullSimplify[
> >> E^x - E^(y + z)
> >> ]
> >> ]
> >>
> >> Out[72]= E^x - E^(y + z)
> >>
> >> Thanks!
> >>
> >
>
>
>




  • Prev by Date: Difficulty with patterns
  • Next by Date: Re: a bug in Integrate (2nd message)
  • Previous by thread: Re: Full simplify problem
  • Next by thread: Re: Full simplify problem