[seqfan] Iteration and Matrix Log

Paul D Hanna pauldhanna at juno.com
Thu Jul 22 22:55:30 CEST 2010

Hello SeqFans, 
   Consider the remarkable functional equation: 
(*) L(x) = G(x)/(x*G'(x)) * L(G(x)) where G(0)=0, G'(0)=1. 
Amazingly, the solution set of G(x) in (*) for a fixed L(x) 
consists of all iterations of any function in that set. 
To restate, suppose G(x) is a solution to (*) for a given L(x), 
and let G_n(x) denote the n-th iteration of G(x) with G_0(x)=x, 
then the following statement holds for all n:  
(**) L(x) = G_n(x)/(x*G_n'(x)) * L(G_n(x)).
This can be shown using induction and the chain rule of differentiation. 
As an application, the solution to L(x) in (*) for a fixed G(x) completely 
describes the columns of the MATRIX LOG of the Riordan array (G(x)/x, G(x)). 
I recorded my results in entries A179320, A179330, A179199, 
(cf. http://www.research.att.com/~njas/sequences/A179330) 
and restate the main concepts below. 
Are the numbered points below found in the literature? 
Statement (1) may be obvious to the eye, but (2) thru (5) are not so easily seen. 
Let Log(T) denote the matrix logarithm of a triangular matrix T; 
this can be defined by the series Log(T) = Sum_{n>=1} -(I - T)^n/n. 
Given triangular matrix R = Riordan array (G(x)/x, G(x)) where G(0)=0, G'(0)=1, 
  L(x) = g.f. of column 0 in Log(R) 
(1) (k+1)*L(x) = g.f. of column k in Log(R) 
and L(x) satisfies: 
(2) L(x) =  G(x)/(x*G'(x)) * L(G(x)).
Further, if G(x) satisfies (2), then: 
(3) L(x) = G_n(x)/(x*G_n'(x)) * L(G_n(x)) for all n
where G_n(x) denotes the n-th iteration of G(x). 
If G(x) satisfies (2) for some fixed L(x), 
then a solution G(x) can be obtained from L=L(x) by the series: 
(4) G(x)/x = 1 + L + L*Dx(L)/2! + L*Dx(L*Dx(L))/3! + L*Dx(L*Dx(L*Dx(L)))/4! +... 
Further, the n-th iteration of this G(x) is given by the series: 
(5) G_n(x)/x = 1 + n*L + n^2*L*Dx(L)/2! + n^3*L*Dx(L*Dx(L))/3! + n^4*L*Dx(L*Dx(L*Dx(L)))/4! +... 
where Dx(F) = d/dx(x*F) is the theta operator. 
The inversion formulae (4) and (5) are a consequence of (1) and 
the exponential series 
  R^n = I + n*Log(R) + n^2*Log(R)^2/2! + n^3*Log(R)^3/3! +... 
which yields R^n = Riordan array (G_n(x)/x, G_n(x)) 
where G_n(x) is the n-th iteration of G(x) (which is well-known). 
Also from (5) we can see that column 0 in the matrix Log(R)^m 
is given by the "nested derivative product": 
in which L appears m times, and Dx(F) = d/dx(x*F). 
And since column 0 of R^n equals G_n(x)/x, then it follows that 
  G_n(x) = Sum_{m>=0} n^m*[column 0 of Log(R)^m]/m! 
which is essentially what (5) is stating. 
OEIS entry A179420 describes the fascinating eigenfunction case: 
   H(H(x)) = x*H'(x) 
which has some beautiful properties such as: 
   H(x) = H_{n+1}(x) * H_n(x)/[x*d/dx H_n(x)] 
which holds for all n where H_n(x) = n-th iteration of H(x). 
See more at: http://www.research.att.com/~njas/sequences/A179420 

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