Re: " f^ [n](x) = exp( n/ phi' (x) D ) o x "

Dear G. A. Edgar ,

How may it work ?
Let us try a known simple phi(x) =1/x ;phi'(x) = -1/x^2
and exp(n/phi'(x) D ) = exp(-n*x^2 D) , D for d/dx
Development :
Id + (- n*x^2 D)/1! + (-n*x^2 D)^2 /2! + ...
exp(-n*x^2D) o x = Id o x -(n*x^2 D)/1! o x + . ,
with a recurrence relation between terms:
ti(x) = - n*x^2/ i *d/dx (t[i-1](x) ) , we obtain :
x -n*x^2 +n^2*x^3 - ... = x*( 1 -nx +(nx)^2 - (nx)^3 ... )
or x / (1+n*x) = (x /(1 +x) )^ [n ] , [n ] iterated
à suivre ,

Alain

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