Kurt Foster on Mon, 23 Jul 2018 14:50:34 +0200

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Re: generating function solution to poly

 To: Kevin Ryde <user42_kevin@yahoo.com.au>, Pari Users <pariusers@pari.math.ubordeaux.fr>
 Subject: Re: generating function solution to poly
 From: Kurt Foster <drsardonicus@earthlink.net>
 Date: Mon, 23 Jul 2018 06:50:29 0600
 Deliverydate: Mon, 23 Jul 2018 14:50:34 +0200
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 References: <87fu0dgf09.fsf@blah.blah>
On Jul 21, 2018, at 2:38 AM, Kevin Ryde wrote:
I have a generating function (and more terms too)
g = x^2 + x^3 + x^4 + 3*x^5 + 6*x^6 + 12*x^7 + 29*x^8 + 67*x^9 +
O(x^10);
which satisfies a cubic
(1+x)*g^3  2*g^2 + (1x+2*x^2)*g  x^2 == 0
Is there an easy or good way to have gp solve that for series g?
I know how to work upwards to get g term by term (after deciding
lowest
should be 0), but maybe gp already has it. I wondered only for
interest
or generality though, since this one comes from recurrences which are
easy to calculate. I attempted serreverse() without joy (change
variables to solve in x, but I may have confused myself).
On Jul 22, 2018, at 2:15 AM, Peter Pein wrote:
Hi Kevin,
I'm just curious: are you sure the coefficients in g are correct?
starting Bill Allombert's suggestion to iterate via Newton but
starting
with just X^2+O(X^3) gives another sequence for g and a difference
for
coeffs of x^7 and higher powers
? P=substvec((1+x)*g^32*g^2+(1+x+2*x^2)*gx^2,[x,g],[X,y])
%1 = (X + 1)*y^3  2*y^2 + (2*X^2 + X + 1)*y  X^2
Different problem, different solution. Compare coefficients of y, 1x
+2*x^2 and 2*X^2 + X + 1. Never mind the change of variable name or
order of terms. A minus sign has ben changed to a plus sign.
Take it from a holder of a Black Belt of transfinite degree in making
typos  copypaste is the way to go in transcribing algebraic
expressions.