|Kevin Ryde on Mon, 23 Jul 2018 09:37:35 +0200|
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|Re: generating function solution to poly|
Karim Belabas <Karim.Belabas@math.u-bordeaux.fr> writes: > > ? seralgdep(g,3,2) > %2 = (y + 1)*x^3 - 2*x^2 + (2*y^2 - y + 1)*x - y^2 I didn't know that one. Though actually I wanted the other way, have the cubic, want the series :). Bill Allombert <Bill.Allombert@math.u-bordeaux.fr> writes: > > ? g=truncate(g)-subst(P,y,truncate(g))/subst(P',y,g) > %3 = > %X^2+X^3+X^4+3*X^5+6*X^6+12*X^7+29*X^8+67*X^9+157*X^10 + ... Thanks, and starting from say just 3 terms of series too. It comes out all terms correct in each doubled series length is it? Nice. Peter Pein <firstname.lastname@example.org> writes: > > ? P=substvec((1+x)*g^3-2*g^2+(1+x+2*x^2)*g-x^2,[x,g],[X,y]) ^^^^^^^^^^^ Cut and paste gone astray, should be 1-x+2*x^2. :) > I'm just curious: are you sure the coefficients in g are correct? I think correct. They're counts (of some trees) so are to be integers and >=0 in any case. There might be other series "solutions" to the cubic, starting series low terms 1*x^0 -1*x^1 and above that not integers by my reckoning, so not what I had sought.