05/22/2014, 10:16 PM
(This post was last modified: 05/23/2014, 04:24 AM by sheldonison.)

(05/22/2014, 08:31 AM)tommy1729 Wrote: ....Hey Tommy,

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.

....

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).

...

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

Much to learn. I bought Conway's graduate complex analysis book, which includes the Weierstrass factorization theorem. I will add it to my large and growing collection of graduate math books, to supplement my somewhat weak formal math education (BSEE).

For the truncated product equation, where z_n are the zeros of the Asymptotic half exponential, is this correct?

update I tried Tommy's factor, which works surprisingly well. So I guess this is the Weierstrass factorization of the asymptotic half exponential.

Here are the first 10 zeros of the asymptotic half exponential. I also have an approximation for the zeros of the asymptotic function, in terms of Kneser's half exponential.

Code:

`1 -0.71176762728441566602682009931906`

2 -4.2615192715738731444168590500003

3 -15.214306922947794235707847543680

4 -43.768867332590888558785594556450

5 -109.77901963743164514158613984269

6 -229.50542893029538607241128473700

7 -458.89117411149970236796071783155

8 -861.01099146084982202350824423190

9 -1534.9231313380922901088132019252

10 -2623.2160464901874760015847745797