[seqfan] Re: A Remarkable Eigenfunction - A179497

[Paul D Hanna]

SeqFans, 
      Here is a related series expression for A(x) using nested derivatives. 
Using abbreviations A=A(x), A_n = n-th iteration of A(x), and operator D(F)=d/dx F, 
then A satisfies the series: 
  
   A = x + A^2 + A^2*D(A^2)/2! + A^2*D(A^2*D(A^2))/3! + A^2*D(A^2*D(A^2*D(A^2)))/4! +... 
 
Likewise, the n-th iteration of A may be expressed in a similar fashion using n as an argument:  
  
(**)   A_n = x + n*A^2 + n^2*A^2*D(A^2)/2! + n^3*A^2*D(A^2*D(A^2))/3! + n^4*A^2*D(A^2*D(A^2*D(A^2)))/4! +...
 
which holds for real n as well as intergers.  I find this appealing. 
  
Here is some rudimentary PARI code for (**):
     A_(k,A)=local(D=x);x+sum(n=1,#A,k^n*(D=A^2*deriv(D))/n!)  
where A_(k,A) returns the k-th iteration of A.
  
 
>From the above series (**), is it any easier now to see why the relation 
    [A_{n+1}(x)]^2 = A(x)^2 * A_n'(x)
holds?   
 
We may write: 
     A^2*A_n' = A^2 + n*A^2*D(A^2) + n^2*A^2*D(A^2*D(A^2))/2! + n^3*A^2*D(A^2*D(A^2*D(A^2)))/3! + n^4*A^2*D(A^2*D(A^2*D(A^2*D(A^2))))/4! +...
 
but it is still not readily apparent to me that this sum should equal [A_{n+1}(x)]^2.
 
 
At any rate, does the series (**) lead to any nice formulas? 
 
Thanks, 
-- Paul 
  
---------- Original Message ----------
From: "Paul D Hanna" <pauldhanna at juno.com>
To: seqfan at list.seqfan.eu
Subject: [seqfan]  A Remarkable Eigenfunction - A179497
Date: Sun, 1 Aug 2010 13:58:48 GMT

SeqFans, 
    
Consider the eigenfunction described by:

  E.g.f. satisfies: A(A(x))^2 = A(x)^2 * A'(x) 
 
  [http://www2.research.att.com/~njas/sequences/A179497].
 
Explicitly, the series begins: 

  A(x) = x + 2*x^2/2! + 18*x^3/3! + 312*x^4/4! + 8240*x^5/5! + 297000*x^6/6! + 13705776*x^7/7! + 776778688*x^8/8! + 52511234688*x^9/9! + 4143702216960*x^10/10! +..


Let A_n(x) denote the n-th iteration of A(x) where A_{n+1}(x) = A_n(A(x)) with A_0(x)=x;

then the pattern continues: 

  [A_3(x)]^2 = A(x)^2 * A_2'(x), 
 
  [A_4(x)]^2 = A(x)^2 * A_3'(x),

  [A_5(x)]^2 = A(x)^2 * A_4'(x), ...

so that the iterations of A(x) satisfy: 

(*) [A_{n+1}(x)]^2 = A(x)^2 * A_n'(x)   for all n. 


I find formula (*) to be remarkable, even though the proof for n>0 
is an easy exercise in induction and the chain rule. 


Surely A(x) must have some other nice properties. 

What else can be said about A(x)? 

  Paul 

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