06/04/2010, 04:48 AM
(This post was last modified: 06/04/2010, 04:52 AM by sheldonison.)

I think I know what Tommy may have in mind, based on these posts, and his other posts on half iterates.

For small values of x, 2sinh(x) = 2x, (the derivative at x=0), so the "half iterate" of 2sinh(x) would be . Using this "half iterate" of 2sinh(x), along with Tommy's equations for converting back and forth between f and g (the same as the base change equations, and the same as Levy's equations!), we can calculate TommySexp(0.5) and, yes, the results are the same as what I posted earlier. Perhaps Tommy can verify that this is what he means when he says he does it differently.

The idea of using the 2sinh(x) half iterate limit of is attractive due to its simplicity. That the TommySexp(0.5) is different than the Knesser sexp(0.5) by a small amount is another example of the base change function almost working, but not quite. It would be nice to graph the 1-cyclic wobble between TommySexp(x) and Kouznetsov's published sexp. Anyway, for large enough numbers, if we iterate exponents enough times, the inverse superfunctions (for any function with exponential growth, for any base) all agree as exactly as desired on what a full iterate means, but they all seem to disagree on what a fractional iterate means!

I see Henryk has posted something on base changes for bases between 0 and eta, and now I will go read that, and get some sleep!

- Shel

tommy1729 Wrote:... if we take the half-iterate of 2sinh(x) by using taylor series , we get a good approximation of the half-iterate of exp(x) for x large.

if this good approximation is analytic in say [e^e,e^e^e^e] we could use that interval and take logs or exp of it to compute the half-iterate for [-oo,+oo] up to a relatively high precision.

(06/02/2010, 10:13 PM)tommy1729 Wrote: ...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.

then g(x,y) = lim k -> oo

log log log ... (k times ) [f( g(x,k) ,y)]

For small values of x, 2sinh(x) = 2x, (the derivative at x=0), so the "half iterate" of 2sinh(x) would be . Using this "half iterate" of 2sinh(x), along with Tommy's equations for converting back and forth between f and g (the same as the base change equations, and the same as Levy's equations!), we can calculate TommySexp(0.5) and, yes, the results are the same as what I posted earlier. Perhaps Tommy can verify that this is what he means when he says he does it differently.

The idea of using the 2sinh(x) half iterate limit of is attractive due to its simplicity. That the TommySexp(0.5) is different than the Knesser sexp(0.5) by a small amount is another example of the base change function almost working, but not quite. It would be nice to graph the 1-cyclic wobble between TommySexp(x) and Kouznetsov's published sexp. Anyway, for large enough numbers, if we iterate exponents enough times, the inverse superfunctions (for any function with exponential growth, for any base) all agree as exactly as desired on what a full iterate means, but they all seem to disagree on what a fractional iterate means!

I see Henryk has posted something on base changes for bases between 0 and eta, and now I will go read that, and get some sleep!

- Shel