(05/22/2014, 07:08 AM)sheldonison Wrote:(05/22/2014, 12:16 AM)JmsNxn Wrote:Thanks James,(05/18/2014, 06:14 PM)sheldonison Wrote: So, anyway, lets say we have these zeros of the asymptotic half iterate in an infinite list. Can we recover the asymptotic half iterate with the Weiestrass factorization theorem?

- Sheldon

If you have some more to add on top of the theorem. All it will give you is the half iterate upto multiplication by an entire never zero function. These are usually solved using additional properties of the function. I'm sure there must be a way.

It looks like the "multiplication by an entire never zero function" is exp(g(z)), where g(z) is any entire function. For our purposes, g(z)=k might work, if that is a legal choice. That would just be multiplication by a constant. Multiplication by exp of anything else would probably grow faster than the half iterate of exponentiation, which by definition grows slower than exp(z).

- Sheldon

"probably" is not so logical as you might think.

Remember the fake log for instance.

And there are many other " counter-intuitive " functions.

Im carefull with the word counter-intuitive because i do not " believe " in that word ; everybody has a different intuition.

Another " counter-intuitive " (entire) function is this one :

this entire f(z) is bounded outside the strip .

( this can be shown with contour integration or substitution )

So its almost everywhere constant.

And one could take the fake log of this f(z) once or twice too. Or f(fake log(z)).

The fact that a function has no 0 does not limit exp(g(z)) much.

So exp(g(z)) can be very different from exp(polynomial).

Therefore the "probably " idea is not very solid.

On the other hand : if the truncated Taylor series can be approximated well with a truncated product (1+b_n x) then its likely that g(z) is indeed a constant.

This is also likely because of the simplification that all derivates are positive and it might be possible to find a recursion to solve for b_n ...

So I think truncation and induction will show us the way.

Thinking ...

regards

tommy1729