06/08/2010, 01:59 PM
(This post was last modified: 06/09/2010, 12:18 AM by sheldonison.)

(06/07/2010, 12:34 AM)tommy1729 Wrote:The correcting factor is(06/02/2010, 10:56 PM)sheldonison Wrote: 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? ... The value I'm using for k is 0.067838366.

no , i use iterations of exp(x) !

you only use logaritms and the superfunction of 2sinh(x) ...

i use logaritms , superfunction of 2sinh(x) and exp(x).

numerically it might not be a big difference for large values ( since 2sinh(x) and exp(x) are close for large x ) , but for small x or properties it might be a huge difference ... maybe that is why you get 0.92715... ?

i dont know if i need a correcting 'k'.

and btw as for the diff-eq , i made a correction to that post (typo mainly but important ) , maybe it will help.

regards

tommy1729

k=

If you try your equations to calculate TommySexp(0.5) with x=0, y=0.5, and k=3

Quote: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)]

then g(0,0.5) = (for k=3)

log log log (3 times) [f( g(0,3), 0.5)]

g(0,3) is =3814279.104....

Calculating f( g(3,0) ,0.5) is a a somewhat complex operation, and I'd like to understand if you would calculate that expression the same way I would. My approach would be to use the inverse superfunction, then add 0.5, then calculate the superfunction.

Update or alternatively, iterate

The inverse superfunction gives 3.0678383... which is 3 plus the normalization constant, then adding 0.5 ... then calculate the superfunction of 3.5678383.... 8.10306E+77

Finally, take the logarithm three times, and you get (for Tommy's k=3)

TommySexp(0.5) =~ f(0,0.5) =~ 0.4987433...

Numerically, this is exactly what the limit equations I've posted give.

- Sheldon

- Sheldon