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

Andrew N W Hone A.N.W.Hone at kent.ac.uk
Tue Aug 9 00:02:34 CEST 2011

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,
Andy

________________________________________
From: seqfan-bounces at list.seqfan.eu [seqfan-bounces at list.seqfan.eu] On Behalf Of Benoît Jubin [benoit.jubin at gmail.com]
Sent: 01 August 2011 07:54
To: Sequence Fanatics Discussion list
Subject: [seqfan] Re: Constant Arising from a Certain Functional Eq

Hi Paul,

How did you find this differential equation?

Your coefficients c(n) = C(n,.) are given by
c(0) = c(1) = 1
c(n) = 1 + t*n*(n-1)*c(n-1) + \sum_{m=1}^{n-2}
\choose(n,m)*c(m)*(1-c(n-m))   for n>=2
(of course, one can also find recursions for the coefficients of these
polynomials).

Do you know if there actually is a nonzero solution through x=0, that

Here is a closed form solution away from x=0, say x>0.  Not sure if it
helps.  I rewrite the equation as
t*x^2*y' - y^2 + y*e^x = 0
and assume t \neq 0 (in the case t=0, there are exactly two continuous
solutions, 0 and exp).

If we do the substitution y = e^(f(x))*z, we find that a good choice
of f is such that t*x^2*f' + e^x = 0, for instance f(x) = -1/t
\int_1^x e^u/u^2 du.  Then we can separate variables (since a nonzero
solution vanishes nowhere, from Picard-Lindelof) and we find, modulo
mistakes,
y(x) = e^(f(x)) * t / (A - \int_1^x e^(f(v))/v^2 dv)
It seems that this function is equivalent to t*x^2 at x=0, which is
not the case of the power series solution... (caution: I didn't check
my computations)

Also, what kind of variants of this equation are you thinking of?

Regards,
Benoit

On Fri, Jul 29, 2011 at 11:07 AM, Paul D Hanna <pauldhanna at juno.com> wrote:
> SeqFans,
>    Consider the functional equation:
>
> (1) F(x,t) = exp(x) + t*x^2*F'(x,t)/F(x,t)
>
> where the logarithmic derivative is taken wrt x (not t).
>
> Clearly, a solution F(x,t) to (1) as a series in x is determined by the parameter t.
> For example,
> F(x,1) = 1 + x + 3*x^2/2! + 13*x^3/3! + 73*x^4/4! + 481*x^5/5! + 3001*x^6/6! - 8819*x^7/7! - 1883951*x^8/8! +...
>
> F(x,2) = 1 + x + 5*x^2/2! + 49*x^3/3! + 865*x^4/4! + 25921*x^5/5! + 1236001*x^6/6! + 87338161*x^7/7! +...
>
> Observe that in F(x,1) negative coefficients appear,
> while F(x,2) consists of only positive coefficients.
>
> In fact, some coefficients in F(x,t) are negative for any t<=1,
> and F(x,t) consists entirely of positive coefficients when t>=2.
>
>
> So then, what is the least t such that F(x,t) consists entirely of positive coefficients?
>
>
> It is interesting that such a threshhold value of t exists, and equals (approx):
>
> 1.029360728932960666742865617647360272428697824009903266405204289055749580312045961621266246011201846...
>
> Can this constant be expressed in terms of other known constants?
>
>
> This constant is obtained by regarding the coefficients C(n,t) of x^n in F(x,t) as polynomials in t:
>
>   F(x,t) = Sum_{n>=0} C(n,t)*x^n/n!
>
> and then finding the limit of the roots of the polynomials that are found in a suitable interval.
> That is, let r(n) be within the interval (1,2) and satisfy
>   C(n,r(n)) = 0,
> then the above constant equals limit r(n).
> (Here we determined the interval (1,2) from our observation of negative and positive coefficients;
> more specific conditions on the roots are needed for the general case.)
>
> The coefficient polynomials begin:
> C(0,t) = 1;
> C(1,t) = 1;
> C(2,t) = 2*t + 1;
> C(3,t) = 12*t^2 + 1;
> C(4,t) = 144*t^3 - 72*t^2 + 1;
> C(5,t) = 2880*t^4 - 2640*t^3 + 240*t^2 + 1;
> C(6,t) = 86400*t^5 - 108000*t^4 + 25200*t^3 - 600*t^2 + 1;
> C(7,t) = 3628800*t^6 - 5503680*t^5 + 2036160*t^4 - 171360*t^3 + 1260*t^2 + 1;
> C(8,t) = 203212800*t^7 - 352235520*t^6 + 172730880*t^5 - 26530560*t^4 + 940800*t^3 - 2352*t^2 + 1;
> C(9,t) = 14631321600*t^8 - 27991111680*t^7 + 16634419200*t^6 - 3674522880*t^5 + 272160000*t^4 - 4451328*t^3 + 4032*t^2 + 1;
> C(10,t) = 1316818944000*t^9 - 2719858176000*t^8 + 1860093849600*t^7 - 525486528000*t^6 + 60495724800*t^5 - 2356361280*t^4 + 18869760*t^3 - 6480*t^2 + 1;
> ...
> From this we see that the real roots in the interval (1,2) are already converging:
> r(7) = 1.004549380068308349645605642...
> r(8) = 1.022716018771956066492600608...
> r(9) = 1.027860824033290897499412851...
> r(10) = 1.029071227426111538733060520...
> ...
>
> It may be interesting to study variants of (1) to arrive at other constants.
>
> Thanks,
>   Paul
>
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>
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>

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