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 But if it holds in the complex plane, that would really be something, since g and f are such different functions. f is the super function of 2sinh, and g is the iterated natural logarithm (n times) of f(z+n). Do you think g(x) is analytic (as n=>oo)?

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.