Matrixexponential and Normaldistribution/Gauss' function (II)

[Paul D. Hanna]

Sorry ... when I wrote:  
>      Sum_{n>=0} e(n)/n!*x^n = Sum_{n>=1} d(n)*x^n/n! 
>          = exp( d(1)*x + d(2)*x^2/2! + d(3)*x^3/3! +...)
> Thus, each k-th diagonal of T has its contribution simply as 
> d(k)*x^k/k! in the e.g.f. of column 0. 
  
I should have included the exp() of the sum on the right: 
 
>      Sum_{n>=0} e(n)/n!*x^n = exp( Sum_{n>=1} d(n)*x^n/n! )
>          = exp( d(1)*x + d(2)*x^2/2! + d(3)*x^3/3! +...)
> Thus, each k-th diagonal of T has its contribution simply as 
> d(k)*x^k/k! in the e.g.f. of column 0. 
 
This simple relation holds true only because each diagonal k of T is a 
fixed multiple, d(k), of binomial coefficients {C(n+k,k)|n=0,1,2,3,..}. 
 
Paul