[seqfan] Re: Functional Equation Challenge

[Olivier Gerard]

On Wed, Jun 8, 2016 at 5:20 AM, Paul D Hanna <pauldhanna at juno.com> wrote:

> SeqFans,
>        Here is a challenge for you, the solution being as yet unknown to
> me, and yet does not seem to be intractable.
> And there are sequences of coefficients for certain A(x) that may be
> appropriate for submission to the OEIS.
>     OBJECTIVE.
> Given F(x) with F(0)=1, suppose A(x) satisfies
>     A( +A(x)^2 * F(A(x)) ) = x^2
> then find G(x) such that
>     A( -A(x)^2 * F(A(x)) ) = G(-x^2).
>
> OBSERVATION:
> For the same A(x), F(x), and G(x), we have
>     A( +sqrt( A(x^2 * F(x)) ) ) = x
> and
>     A( -sqrt( A(x^2 * F(x)) ) ) = G(-x).
>
> Below I give 6 simple examples.
>
> Note that we may generate A(x) for a given F(x) by iterating the relation
>    A(x) = Series_Reversion( sqrt( A(x^2 * F(x)) ) ).
>
>
> How does one determine G(x) from F(x)?
>
> Solution, anyone?
>       Paul
>
>
Dear Paul,

If what you want is to have the integer coefficients of the whole trio,
you may better do things the other way, starting from A
and using a simple polynomial equation solver.



The series for F is uniquely determined by the series for A.

If I take

A(x)= x + a[2] x^2 + a[3] x^3 + ...

I got

F(x) = 1 + (-2 a[2]) x   +   (a[2] (-1 + 5 a[2]) -  2 a[3]) x^2 +  -2 (a[2]
(a[2] (-2 + 7 a[2]) - 6 a[3]) + a[4])  x^3 + ...

or

G(x) = - x^2 +  2 a[2] x^4    -  4 a[2]^2  x^6 +  (10 a[2]^3 - 4 a[2] a[3]
+ 2 a[4]) x^8   +   (- 28 a[2]^4 + 24 a[2]^2 a[3] - 12 a[2] a[4] ) x^10 +
...


The coefficients of F(x) are straightforward in terms of A(x), you only
need up to a[i+1] to have the i-th derivative of F,
G(x) is lacunary and you only need up to a[i] to have the  2i-th and
(2i+2)-th coefficient of G

To have A from F, you just identify each expression in terms of a[i]  for a
given n-th derivative of F at 0 with
the actual value. It gives you a polynomial system which is easy to solve.
Then you plug the value into the expression for G.

To have A from G, same thing, but you will need many more coefficients of G
for a given number of coefficient of A.


A(x) =   x  - (f[1]/2) x^2   +   1/8 (2 f[1] + 5 f[1]^2 - 4 f[2])  x^3  +
1/4 (-f[1]^2 -  4 f[1]^3 + 6 f[1] f[2] - 2 f[3]) x^4  + ...


If I were you I would be very interested in F(x) =   (3x -1)/(x+1) = 1 - 2
x - 2 x^2 - 2 x^3 - ...

because it gives an integral  A(x) which is not yet in the OEIS

{0,1, 1, 3, 14, 66, 338, 1799, 9917 , ...)



Olivier