[seqfan] Re: Functional Inverse
Andrew N W Hone
A.N.W.Hone at kent.ac.uk
Wed Apr 21 13:13:47 CEST 2010
As far as I know there is no closed form expression for the inverse of a hypergeometric function. However, you can quickly obtain a power series for it by using
reversion of the series. Maple has the
command which does this.
Given y = f(x) = x + c_2 x^2 + c_3 x^3 + ..., the reverted series x=g(y) is obtained recursively by successive approximations,
starting from x=y, then set x=y(1 + \delta y) where \delta = O(1) as y -> 0, and substitute back in y=f(x)
to get 0 = \delta y^2 +c_2 y^2 + O(y^3), which gives \delta = -c_2 + O(y), so replace \delta by -c_2 + \delta ' y to
get the correction at next order, and so on.
For the hypergeometric series f(x) = 1+ (ab/c) x + (a(a+1)b(b+1)/(2c(c+1))) x^2 + ... the series reversion will give a
power series in z = y-1.
Maybe this will be useful for you.
From: seqfan-bounces at list.seqfan.eu [seqfan-bounces at list.seqfan.eu] On Behalf Of Karol [penson at lptl.jussieu.fr]
Sent: 21 April 2010 10:22
To: Sequence Fanatics Discussion list; penson at lptl.jussieu.fr
Subject: [seqfan] Functional Inverse
Does anybody know how to obtain quickly in closed form the functional
denoted by Inv(a, b, c ; x) , of the Gauss hypergeometric
function ( in Maple notation)
hypergeom( [a, b], [c], x), satisfying
hypergeom( [a, b], [c], Inv(a, b, c ; x) ) = x ?
Thanks in advance, Karol A. Penson
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