06/02/2010, 10:56 PM
(This post was last modified: 06/02/2010, 11:50 PM by sheldonison.)
(06/02/2010, 10:13 PM)tommy1729 Wrote: thanks for the reply.How do you calculate f(x,y)? I would drop the "x" term entirely... I calculate f(y) using the superfunction equation I gave. If f(z)=2^z is a small number, then it approximates the behavior of the superfunction for 2sinh. Do you have a different equation, and could you share it?
that is intresting.
however i dont think it is the same ,
i do it differently , although the end result may be the same ;
for real x and y , both >=0 :
let g(x,y) be the y'th iterate of exp(x) evaluated at x.
let f(x,y) be the y'th iterate of 2 sinh(x) evaluated at x.
After that, do you agree that converting from your f(y) to g(y) by iterating natural logarithms would appear to be exactly the same? Here, the superfunction <=> f(y), and TommySexp <=> g(y). The only reason I threw in the "k" constant was so that the resulting sexp_e(z) can be normalized so that TommySexp_e(0)=1. Otherwise, with the k constant=0, I get TommySexp(0)=0.92715... The value I'm using for k is 0.067838366.
Quote:then g(x,y) = lim k -> ooNope, certainly didn't, nor do I understand it
log log log ... (k times ) [f( g(x,k) ,y)]
regards
tommy1729
btw g(x,y) satisfies ( derivative with respect to x )
g'(x,y) / g'(exp(x),y) = exp(x)/F(exp(x)
but i guess you already knew that.
