       Re: Re: Re: Cannot Factor an expression

```Andrzej,

There was a syntax problem with your PolynomialReduce  function
below ... (complete inputs) should have been:-

q = Sum[(Subscript[z, i] - 1)*(=CE=BC - Subscript[x, i])^2, {i, 1, 3}]

p = Expand[q]

Factor[First[
PolynomialReduce[
p, {Subscript[z, 1] - 1, Subscript[z, 2] - 1,
Subscript[z, 3] - 1}, {Subscript[x, 1], Subscript[x, 2],
Subscript[x,
3]}]]].{Subscript[z, 1] - 1, Subscript[z, 2] - 1, Subscript[z,
3] - 1}

There is a missing ", {Subscript[x, 1], Subscript[x, 2], Subscript[x, =

3]}]"

It seems your postings are sometimes missing lines because I am sure
you would have tested your function yet it clearly fails 6.0.2's
syntax checking.

HTH Syd

Syd Geraghty B.Sc, M.Sc.

sydgeraghty at mac.com

My System

Mathematica 6.0.1 for Mac OS X x86 (64 - bit) (June 19, 2007)
MacOS X V 10.5 .20
MacBook Pro 2.33 Ghz Intel Core 2 Duo  2GB RAM

On Apr 15, 2008, at 2:55 AM, Andrzej Kozlowski wrote:

> 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:=
> 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
>>>
>>
>>
>
>

```

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