[seqfan] Re: Constant Arising from a Certain Functional Eq

Paul D Hanna pauldhanna at juno.com
Tue Aug 9 01:53:00 CEST 2011

Andy (and Seqfans), 
       Thanks for your interest and insightful comments. 
The differential equation came to mind as I considered a few examples in the OEIS. 
EXAMPLE 1: A112934 - 
O.g.f. satisfies: A(x) = 1+x + 2*x^2*[d/dx A(x)]/A(x). 
Yields series:  
O.g.f.: A(x) = 1 + x + 2*x^2 + 6*x^3 + 26*x^4 + 158*x^5 + 1282*x^6 +... 
1/A(x) = 1 - x - x^2 - 3*x^3 - 15*x^4 - 105*x^5 -... -A001147(n)*x^(n+1)-... 
where A001147 is the double factorial numbers: (2*n-1)!!.
EXAMPLE 2: A112936  -  
O.g.f. satisfies: A(x) = 1+x + 3*x^2*[d/dx A(x)]/A(x)
Yields series:  
O.g.f.: A(x) = 1 + x + 3*x^2 + 15*x^3 + 111*x^4 + 1131*x^5 + 14943*x^6 +... 
1/A(x) = 1 - x - 2*x^2 - 10*x^3 - 80*x^4 - 880*x^5 -...-A008544(n)*x^(n+1)-...  
where A008544 is the triple factorial numbers: product(k=0..n-1, 3*k+2 ).
I wanted to explore some interesting variations of these cases, 
and that's when the question involving non-negative coefficients came to mind. 
A simple case is this diff.eq.: 
  F(x,t) = (1+x)^2 + t*x^2*F'(x,t)/F(x,t). 
The least t such that F(x,t) consists entirely of positive coefficients is 
t = 2.044417702215325822734054118424444123239275900404337673654769113888960412061213453842830399116759613...
As a final example, the solution F(x,t) to the diff.eq. 
  F(x,t) = 1+x^2 + t*x^2*F'(x,t)/F(x,t) 
has negative coefficients for t less than 
It would be nice if these constants had some nice closed form or predictable CF expression. 
---------- Original Message ----------
From: Andrew N W Hone <A.N.W.Hone at kent.ac.uk>
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
Subject: [seqfan] Re: Constant Arising from a Certain Functional Eq
Date: Mon, 8 Aug 2011 23:02:34 +0100

Hi - 

The equation for F is a Riccati equation in x. It can be solved by standard methods (see e.g. the book by Ince). 
It is possible that the series solution is only an asymptotic series, but I would need to look at this more closely. 

In general, the best thing to do with a Riccati equation is to transform it to a linear homogeneous equation of second order. 

However, this is a special Riccati equation: it is equivalent to an inhomogeneous linear 1st order ODE for 1/F, viz:

1 = exp(x) / F +  t*x^2*F'/F^2 

which is 

- t*x^2*G' + exp(x) * G = 1 ---(*)

upon setting G=1/F. 

Now this ODE can be solved (in principle) by the integrating factor method. The integrating factor 
is exp(-I(x)) where I is the integral of exp(x)/(t*x^2).  So I(x) looks like an incomplete gamma function (up to multiplying by 1/t).
The solution for G will then involve integrating minus the exponential of an incomplete gamma times -1/(t*x^2). I doubt this can 
be given in closed form.

It would be worth computing an asymptotic expansion for the solution of (*) around x=0.

I'll try to play around with this if I have any time next week.

All the best,

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