[seqfan] Re: Functional Inverse

Andrew N W Hone A.N.W.Hone at kent.ac.uk
Wed Apr 21 13:13:47 CEST 2010

Dear Karol,

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.

Best wishes,
Andy Hone
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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