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Re: Re: how to have a blind factorization of a polinomial?
*To*: mathgroup at smc.vnet.net
*Subject*: [mg54151] Re: [mg54123] Re: how to have a blind factorization of a polinomial?
*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>
*Date*: Sat, 12 Feb 2005 01:56:59 -0500 (EST)
*References*: <cud789$2t3$1@smc.vnet.net> <cuf453$gfb$1@smc.vnet.net> <200502110833.DAA09170@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
On 11 Feb 2005, at 08:33, foice wrote:
> On Thu, 10 Feb 2005 07:57:23 +0000 (UTC), "Jens-Peer Kuska"
> <kuska at informatik.uni-leipzig.de> wrote:
>
>> Sqrt[x + y] /. a_ + x :> x*(a/x + 1)
> Thanks for your help. Probably is a viable method, but is a little
> less than you expect
> for a WorldWideFamous software as Mathematica.
> In this way you can chose the form of the output, you can also do
> mistaques typing
> something like
> Sqrt[x^3 + y^2] /. a_ + x :> x^3*(a/x + 1)
> and mathematica will not prevent you from the error giving an output
> in the form
> Sqrt[x^3*(y^2/x+1)]
>
> Is mathematica so feature poor to allow only a "requested form"
> factorization?
> I can't belive it.
> Why there isn't somethig as Collect but that can use also negatieve
> powers to collect?
>
> i.e. Collect[x+y,x,Blind] giving x(1+y/x)
>
> Or better, at least for my task, a Mathematica command converting a
> function f(x,y) into a
> function f(x,y/x) (if possible)
>
> Thanks
>
Such a function does not exist because there are infinitely many
similar "factorizations" that might be useful to some users but not to
99% of others. Besides, it is easy to define it yourself:
CollectBlind[expr_, {x_, y_}] := Block[{
u}, MapAll[Factor, (expr /. y -> u x)] /. u -> y/x]
Then
CollectBlind[x + y, {x, y}]
x*(y/x + 1)
and also, for example,
CollectBlind[x^2 + y, {x, y}]
x*(x + y/x)
CollectBlind[x^2 + y, {x^2, y}]
x^2*(y/x^2 + 1)
Andrzej Kozlowski
Chiba, Japan
http://www.akikoz.net/~andrzej/
http://www.mimuw.edu.pl/~akoz/
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