Re: Re: Q: Factor with Polynominals?

*To*: mathgroup at smc.vnet.net*Subject*: [mg27048] Re: [mg26985] Re: Q: Factor with Polynominals?*From*: "Allan Hayes" <hay at haystack.demon.co.uk>*Date*: Thu, 1 Feb 2001 03:00:35 -0500 (EST)*References*: <B69E204A.A7EF%andrzej@tuins.ac.jp>*Sender*: owner-wri-mathgroup at wolfram.com

Andrzej, Following up your idea, I arrived at a kind of Horner transformation: he[f_, 0, _] = f; he[0, g_, _] = 0; he[f_, g_, 0] = f; he[f_, g_, n_:Infinity] := {he[#1, g, n - 1], #2} & @@ Flatten[PolynomialReduce[f, g]].{g, 1} Examples he[Expand[(x^2 + y^2)^4 + y (x^2 + y^2) + y], x^2 + y^2] y + (x^2 + y^2)*(y + (x^2 + y^2)^3) he[Expand[(x^2 + y^2)^4 + y (x^2 + y^2) + y], x^2 + y^2, 2] y + (x^2 + y^2)*(y + (x^2 + y^2)*(x^4 + 2*x^2*y^2 + y^4)) he[Expand[(x^2 + y^2)^4 + y (x^2 + y^2) + y], x^2 + y^2, 1] y + (x^2 + y^2)*(x^6 + y + 3*x^4*y^2 + 3*x^2*y^4 + y^6) Of course, if we always want to go as far as possible we can simplify the code: Clear[he] he[f_, 0] = f; he[0, g_] = 0; he[f_, g_] := {he[#1, g], #2} & @@ Flatten[PolynomialReduce[f, g]].{g, 1} Allan --------------------- Allan Hayes Mathematica Training and Consulting Leicester UK www.haystack.demon.co.uk hay at haystack.demon.co.uk Voice: +44 (0)116 271 4198 Fax: +44 (0)870 164 0565 ----- Original Message ----- From: "Andrzej Kozlowski" <andrzej at tuins.ac.jp> To: mathgroup at smc.vnet.net <hay at haystack.demon.co.uk> Subject: [mg27048] Re: [mg26985] Re: Q: Factor with Polynominals? > I have found a bit of time to consider your problem more carefully and I > concluded that while in principle what you want done is quite easy, there > seems to be an ambiguity in what you call "factoring a polynomial". > > Here is the function which will produce the output you wanted in the first > case, while in your second case it will essentially the same output: > > PolynomialFactor[f_, g_] := > If[PolynomialReduce[f, g][[1, 1]] === 0, PolynomialReduce[f, g][[2]], > PolynomialFactor[PolynomialReduce[f, g][[1, 1]], g]*g + > PolynomialReduce[f, g][[2]]] > > Here is what happens with your two examples: > > In[2]:= > PolynomialFactor[3 x^2 + 3 y^2 + x + y^3 + y x^2, x^2 + y^2] > > Out[2]= > x + (3 + y)*(x^2 + y^2) > > This is just what you wanted. > > In[3]:= > PolynomialFactor[Expand[(x^2 + y^2)^4 + y (x^2 + y^2) + y], x^2 + y^2] > > Out[3]= > 2 2 2 2 3 > y + (x + y ) (y + (x + y ) ) > > This looks somewhat different, but actually that's because it is factored > more deeply than you asked for. This sort of problem will occur in other > examples, e.g. > > In[15]:= > PolynomialFactor[(1 + x^2)^2*(x + 3) + (1 + x^2)*(x - 3), 1 + x^2] > > Out[15]= > 2 2 > (1 + x ) (-3 + x + (3 + x) (1 + x )) > > Again, this is factored further than is the case in your examples. One can > probably fix it with some further programming but quite frankly I do not > think it is worth the effort, having got to this point it should be easy to > re-arrange things by hand. > > -- > Andrzej Kozlowski > Toyama International University > JAPAN > > http://platon.c.u-tokyo.ac.jp/andrzej/ > http://sigma.tuins.ac.jp/ > > > on 01.1.31 1:22 PM, Robert at robert.schuerhuber at gmx.at wrote: > > > > > > > thanks again for your answers! > > unfortunately your method works with the (rather simple) examples, but it > > fails > > with longer, more complex poynomials. > > robert > > > > Allan Hayes wrote: > > > >> Here is a technique that works, with little thought, for both of the > >> examples given so far in this thread. > >> > >> poly = (x^2 + y^2)^4 + y (x^2 + y^2) + y; > >> poly2 = 3 x^2 + 3 y^2 + x + y^3 + y x^2; > >> > >> tf[u___ + a_ b_ + v___ + a_ c_ + w___] := u + a(b + c) + v + w; > >> tf[z_] := z; > >> > >> fs = FullSimplify[poly, TransformationFunctions -> {Automatic, tf}, > >> ComplexityFunction -> (Length[#] &) > >> ] > > > > > > > >