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Re: how to have a blind factorization of a polinomial?


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

-------------------------------------------------------------
tutto ciò che ho scritto è sempre In My Humble Opinion (IMHO)
probabilmente l'ho scritto di fretta, quindi scusate se sono stato sbrigativo.


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