[seqfan] Re: Functional Equation Challenge
Paul D Hanna
pauldhanna at juno.com
Fri Jun 17 18:51:51 CEST 2016
SeqFans, [resending prior email since it was garbled]
The answer is easier than I expected.
A( +sqrt( A(x^2*F(x)^2) ) ) = x
A( -sqrt( A(x^2*F(x)^2) ) ) = G(-x)
G(x) = B(x*F(-x))
B(x*F(x)) = x.
On Wed, Jun 8, 2016 at 5:20 AM, Paul D Hanna <pauldhanna at juno.com> wrote:
> 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.
> 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).
> For the same A(x), F(x), and G(x), we have
> A( +sqrt( A(x^2 * F(x)) ) ) = x
> 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?
More information about the SeqFan