[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. 
   
SOLUTION. 
   
If 
    A( +sqrt( A(x^2*F(x)^2) ) ) = x 
then 
    A( -sqrt( A(x^2*F(x)^2) ) ) = G(-x) 
where 
    G(x) = B(x*F(-x)) 
such that 
    B(x*F(x)) = x. 
  
FYI, 
     Paul  
  
 
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
>


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