On 9 fév, 06:32, Matt <matt271829-n...@yahoo.co.uk> wrote: > It's well known and straightforward that if f(x) = phi^-1(1 + phi(x)) > for some function phi, then f^n(x) = phi^-1(n + phi(x)), where "f^n" > denotes iteration of f. But say we have an expression for f^n that we > found some other way and we know is a valid continuous function > iteration, then how to recover phi? As an alternative to guessing the > correct form followed by some trial-and-error, I found > > phi(x) = Integral dx/g(0,x) > > where g(n,x) = d/dn f^n(x). > > Probably nothing new, but kind of cute I thought.
With smooth functions from phi(f^[r](x)=phi(x)+r , r real we've got d/dr f^[r](x) = phi'(x)* d/dx f^[r](x). There are other relations: f^[r](x) = exp(1/phi'(x)* d/dx ) O x phi^-1(L)= f^[L-phi(x)](x) = c constant, Example f(x) = 2*x +1 phi^-1(L)= (2x+1)^[L-phi(x)] = c ,constant 2^(L-phi(x))*(x+1)-1 = c all done phi(x) = ln(x+1)/ln(x) + k ,any ct.
ISOLATING POWER r in g(r,x) : Ex. g(r,x) = 5^r*x/(1+(5^r-1)*x) 5^r*x/(x-1) = gr/(gr-1) and a little later : ln(gr/(gr-1))/ln(5) = ln(x/(x-1))/ln(5) + r So phi(x) = ln(x/(x-1))/ln(5) counts iterations of f(x) = g(1,x)=5x/(1+4*x)