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Re: Substitute expressions with FullSimplify

• To: mathgroup at smc.vnet.net
• Subject: [mg108696] Re: Substitute expressions with FullSimplify
• From: Guido Walter Pettinari <coccoinomane at gmail.com>
• Date: Sun, 28 Mar 2010 04:06:31 -0500 (EST)
• References: <hoi2n9$qj7$1@smc.vnet.net> <hoklk6$snd$1@smc.vnet.net>

Thank you all for the answers!

@Andrzej
Thank you for pointing me to replacementFunction! It works well in
most cases, however it seems to fail when the analyzed expression
contains lists.

For example, this works fine:

replacementFunction[HypergeometricPFQ [ a + b, d, x ], a + b - c, {a}]
-> HypergeometricPFQ[c, d, x]

however:

replacementFunction[HypergeometricPFQ[ {a + b},  {d, e},  x ],  a + b
- c, {a}]

returns this error:
Part::partw: Part 2 of {{{1},c}} does not exist. >>
HypergeometricPFQ::hdiv: HypergeometricPFQ[{{{1},c}}[[2]],{{0},e},x]
does not exist. Arguments are not consistent. >>

I cannot avoid using a list as an argument since I need to compute
1F2. Moreover, I am afraid I have not the expertise to fix
replacementFunction by myself.

As a side note, the code for replacementFunction was messed up in you
post: the = symbol was always repeated twice. This is the corrected
version I am using:

replacementFunction[expr_, rep_, vars_] :=
 Module[{num = Numerator[expr], den = Denominator[expr],
   hed = Head[expr], base, expon},
  If[PolynomialQ[num, vars] &&
    PolynomialQ[den, vars] && ! NumberQ[den],
   replacementFunction[num, rep, vars]/
    replacementFunction[den, rep, vars],
   If[hed === Power && Length[expr] == 2,
    base = replacementFunction[expr[[1]], rep, vars];
    expon = replacementFunction[expr[[2]], rep, vars];
    PolynomialReduce[base^expon, rep, vars][[2]],
    If[Head[Evaluate[hed]] === Symbol &&
      MemberQ[Attributes[Evaluate[hed]], NumericFunction],
     Map[replacementFunction[#, rep, vars] &, expr],
     PolynomialReduce[expr, rep, vars][[2]]]]]]

Example:
expr = x/6 + y/6 - 1/2*(3 + x + y);
replacementFunction[expr, x + y - r, {x, y}] // Simplify
-> 1/6 (-9 - 2 r)


@Nasser
My example was probably too simple, as David Park noted. Let me
consider two more complex expressions:

expr1 = x^2/6 + y^2/6 - 1/2*(3 + x^2 + y^2)
expr2 = x/6 + y/6 - 1/2*(3 + x + y)

The simple use of ReplaceAll on both of them:

expr1 /. x^2 + y^2 -> r
expr2 /. x + y -> r

produces an unmodified output because, as far as I know, ReplaceAll
works synctatically i.e. it only looks for the exact expression to
substitute and will ignore something simply related to that like x^2/6
+ y^2/6.

On the other hand, the use of FullSimplify:

FullSimplify[expr1, x^2 + y^2 == r]
FullSimplify[expr2, x + y == r]

works only in the first case, where I succesfully obtain as an output
the expression

1/6 (-9 - 2 r) ,

while it is useless in the x+y == r case.

Using replacementFunction, I achieve what I am looking for:

replacementFunction[expr, x + y - r, {x, y}] // Simplify
-> 1/6 (-9 - 2 r)

Cheers,

Guido




On Mar 27, 10:11 am, "Nasser M. Abbasi" <n... at 12000.org> wrote:
> "Guido Walter Pettinari" <coccoinom... at gmail.com> wrote in messagenews:ho=
i2n9$qj7$1 at smc.vnet.net...
>
>
>
> > Hello world!
>
> > This is my first post in this group, but it has been a while since I
> > started reading it. I always found it quite useful, therefore I wish
> > to thank everibody for their contributions!
>
> > Here is my problem. Let's say I have an expression. I would like to
> > substitute all the occurences of a given subexpression in this
> > expression with a symbol. I want to do it in an intelligent way, i.e.
> > by using FullSimplify instead of ReplaceAll.
>
> > If my expression is:
>
> > x^2 + y^2
>
> > I know that:
>
> > FullSimplify [ x^2 + y^2,   x^2 + y^2 == r ]
>
> > will produce 'r' as a result, which is what I want.
>
> > However, if my expression is
>
> > x + y ,
>
> > then
>
> > FullSimplify [ x + y,   x + y == r ]
>
> > produces 'x + y' and not 'r' ! I tried to use
>
> > FullSimplify [ x + y,   x + y == r, ComplexityFunction -> LeafCou=
nt ]
>
> > but I still get 'x+y' as a result.
>
> > Do you have any idea on how to substitute x+y with r in an expression?
>
> Good question. I do not know myself, may be a Mathematica expert can tell
> us. But I am just curious, why do you consider
>
> Simplify [ x + y,   x + y == r ]
>
> more "intelligent" than
>
> In[39]:= x + y /. {x + y -> r}
> Out[39]= r
>
> ?
>
> --Nasser

• Follow-Ups:
  • Re: Re: Substitute expressions with FullSimplify
    • From: Andrzej Kozlowski <akoz@mimuw.edu.pl>