Re: Re: Cannot Factor an expression
- To: mathgroup at smc.vnet.net
- Subject: [mg87683] Re: [mg87618] Re: [mg87563] Cannot Factor an expression
- From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
- Date: Tue, 15 Apr 2008 05:55:25 -0400 (EDT)
- References: <200804130731.DAA11432@smc.vnet.net> <200804140946.FAA08308@smc.vnet.net>
I got my p's and q's mixed up in addition to having some garbled
input. So here we go again:
q = Sum[(Subscript[z, i] - 1)*
(\[Mu] - Subscript[x, i])^2, {i, 1, 3}]
(Subscript[z, 1] - 1)*(\[Mu] - Subscript[x, 1])^2 +
(\[Mu] - Subscript[x, 2])^2*(Subscript[z, 2] - 1) +
(\[Mu] - Subscript[x, 3])^2*(Subscript[z, 3] - 1)
In[2]:=
p = Expand[q];
Factor[First[PolynomialReduce[p,
{Subscript[z, 1] - 1, Subscript[z, 2] - 1,
Subscript[z, 3] - 1}]]] .
{Subscript[z, 1] - 1, Subscript[z, 2] - 1,
Subscript[z, 3] - 1}
(Subscript[z, 1] - 1)*(\[Mu] - Subscript[x, 1])^2 +
(\[Mu] - Subscript[x, 2])^2*(Subscript[z, 2] - 1) +
(\[Mu] - Subscript[x, 3])^2*(Subscript[z, 3] - 1)
All free and no need to load any Add On packages ;-)
Andrzej Kozlowski
On 14 Apr 2008, at 18:46, Andrzej Kozlowski wrote:
> Factor is not "effectively the reverse of Expand", except in the case
> of when you expand an expression that has been factored in *the usual
> sense*, in other words, into multiplicative factors (don't forget that
> a "factor" is the same thing as a "divisor"). Other kinds of "pseudo-
> factorizations", like the one you want, are not unique, and
> Mathematica can't read your mind to find out which one you like best.
> For example:
>
> q = Sum[(Subscript[z, i] - 1)*(=CE=BC - Subscript[x, i])^2, {i, 1, 3}]
>
> (Subscript[z, 1] - 1)*(=CE=BC - Subscript[x, 1])^2 +
> (=CE=BC - Subscript[x, 2])^2*(Subscript[z, 2] - 1) +
> (=CE=BC - Subscript[x, 3])^2*(Subscript[z, 3] - 1)
>
> p = Expand[q];
>
> FullSimplify[p]
>
> (Subscript[z, 1] + Subscript[z, 2] + Subscript[z, 3] - 3)*\[Mu]^2 -
> 2*Subscript[x, 1]*(Subscript[z, 1] - 1)*\[Mu] -
> 2*Subscript[x, 2]*(Subscript[z, 2] - 1)*\[Mu] -
> 2*Subscript[x, 3]*(Subscript[z, 3] - 1)*\[Mu] +
> Subscript[x, 1]^2*(Subscript[z, 1] - 1) +
> Subscript[x, 2]^2*(Subscript[z, 2] - 1) +
> Subscript[x, 3]^2*(Subscript[z, 3] - 1)
>
> in by many criteria not worse then the one you want.
>
> However, there is, in fact, a way to get the form q by starting from
> p, as long as you know that that is what you want:
>
>
> Factor[First[PolynomialReduce[q, {Subscript[z, 1] - 1,
> Subscript[z, 2] - 1, Subscript[z, 3] - 1}]]] .
> {Subscript[z, 1] - 1, Subscript[z, 2] - 1, Subscript[z, 3] - 1}
>
> (Subscript[z, 1] - 1)*(\[Mu] - Subscript[x, 1])^2 +
> (\[Mu] - Subscript[x, 2])^2*(Subscript[z, 2] - 1) +
> (\[Mu] - Subscript[x, 3])^2*(Subscript[z, 3] - 1)
>
> Andrzej Kozlowski
>
>
> On 13 Apr 2008, at 16:31, Francogrex wrote:
>> If Factor is effectively the reverse of expand, then how come factor
>> cannot find back the expression below?
>>
>> Expand[Sum[(Subscript[z, i] - 1)*(\[Mu] - Subscript[x, i])^2, {i, 1,
>> 3}]]
>>
>> Factor[-3*\[Mu]^2 + 2*\[Mu]*Subscript[x, 1] - Subscript[x, 1]^2 +
>> 2*\[Mu]*Subscript[x, 2] - Subscript[x, 2]^2 + 2*\[Mu]*Subscript[x,
>> 3] -
>> Subscript[x, 3]^2 + \[Mu]^2*Subscript[z, 1] -
>> 2*\[Mu]*Subscript[x, 1]*Subscript[z, 1] + Subscript[x, 1]^2*
>> Subscript[z, 1] + \[Mu]^2*Subscript[z, 2] -
>> 2*\[Mu]*Subscript[x, 2]*Subscript[z, 2] + Subscript[x, 2]^2*
>> Subscript[z, 2] + \[Mu]^2*Subscript[z, 3] -
>> 2*\[Mu]*Subscript[x, 3]*Subscript[z, 3] + Subscript[x, 3]^2*
>> Subscript[z, 3]]
>>
>> Is there some additional arguments to the function factor so that it
>> can find back the original:
>> Sum[(Subscript[z, i] - 1)*(\[Mu] - Subscript[x, i])^2, {i, 1, 3}]
>>
>> Thanks for help with this
>>
>
>
- Follow-Ups:
- Re: Re: Re: Cannot Factor an expression
- From: Syd Geraghty <sydgeraghty@mac.com>
- Re: Re: Re: Cannot Factor an expression
- References:
- Cannot Factor an expression
- From: Francogrex <franco@grex.org>
- Re: Cannot Factor an expression
- From: Andrzej Kozlowski <akoz@mimuw.edu.pl>
- Cannot Factor an expression