03/06/2016, 03:16 PM
(This post was last modified: 03/07/2016, 05:32 AM by sheldonison.)

(03/05/2016, 11:38 PM)tommy1729 Wrote: So ... It is analytic !

Here is a sketch of the proof.

Lets consider the half-exp for simplicity, the general case follows by analogue.

Proof sketch

1) 2sinh^[1/2] (x) = g(x) is analytic in [0,e] so it is analytic for x>= 0.

2) therefore f = ln(g(exp(x)) is analytic in [1,oo].

X* = x - eps , X** = x + eps.

Let f(x,n) = ln^[n](f( exp^[n](x) ). + continuation ( like ln(exp) = id ).

Clearly for finite n (integer >0)

F(x,n) is analytic in x for x > 1.

3) we can choose small eps > 0 , independant of n such that

For the points w* in the radius eps around x,

And points w** around exp(x) with radius exp(x**) - exp(x),

( Abs(f(w*,n )) - Abs(ln( f(w**,n ) )^2 < (Abs(f(x*,n)) - Abs(ln( f(exp(x**) ,n)) ))^2 { follows from squeeze theorem and analiticity } < 10/n^3 { follows from fast convergeance of the n th step for real x > 1 }.

Obviously I don't agree with Tommy's proof. Here is a graph of the logarithmic singularities of f(z,1) = ln(2sinh^[0.5](exp(z))). The singularities occur approximately where . Now consider f(z,n). For n=1, the radius of convergence is reasonable, for n=3 at z=1,

the radius of convergence is about 0.21; where the 41st singularity is the nearest singularity to exp(exp(1)) at 11.8293629488482 + 7.81997442041279i.

For n=4, at z=1, the radius is about 0.0012; I don't expect to convince Tommy, which is perhaps unfortunate. The conjecture is that all of the derivatives of f(z,n) do converge as n gets arbitrarily large, even though the function is nowhere analytic.

- Sheldon