07/28/2014, 12:17 PM

Let f(z) be the entire fake half-iterate of exp(z).

Now we do not have a sexp and hence also no theta(z) function.

However at least at the positive real line there must other entire functions that satisfy the functional equation very well.

Let u(z),v(z),w(z) give these other solutions f_i(z) :

f_1(z) = f(z+u(z)) = f(z)(1+v(z)) = f(z) + w(z).

(Here I assumed u,v,w are Always related , that is another subject)

Now it seems logical to assume u,v,w also grow much slower then exp , likely like exp^[1/2].

This gives then a similar hadamard product for these functions.

Let m be an integer > -1.

Now U := u,v,w must satisfy U(f^[m](0)) = 0.

Which btw is consistant with carlson and hadamard.

So U is completely determined by the other zero's apart from U(f^[m](0)) = 0 (and a constant C of course).

And that determines how well the functional equation is satisfied off the real line !!

It would be amazing if one of u,v,w is somehow optimal WITHOUT other zero's then U(f^[m](0)) = 0.

Of particular intrest is the cases where u,v,w have no zero's in the region where our Original f(z) satisfies the functional equation well.

Because if it has a zero there (point "Q") then it must also have a zero at point f(Q) which complicates matters.

regards

tommy1729

Now we do not have a sexp and hence also no theta(z) function.

However at least at the positive real line there must other entire functions that satisfy the functional equation very well.

Let u(z),v(z),w(z) give these other solutions f_i(z) :

f_1(z) = f(z+u(z)) = f(z)(1+v(z)) = f(z) + w(z).

(Here I assumed u,v,w are Always related , that is another subject)

Now it seems logical to assume u,v,w also grow much slower then exp , likely like exp^[1/2].

This gives then a similar hadamard product for these functions.

Let m be an integer > -1.

Now U := u,v,w must satisfy U(f^[m](0)) = 0.

Which btw is consistant with carlson and hadamard.

So U is completely determined by the other zero's apart from U(f^[m](0)) = 0 (and a constant C of course).

And that determines how well the functional equation is satisfied off the real line !!

It would be amazing if one of u,v,w is somehow optimal WITHOUT other zero's then U(f^[m](0)) = 0.

Of particular intrest is the cases where u,v,w have no zero's in the region where our Original f(z) satisfies the functional equation well.

Because if it has a zero there (point "Q") then it must also have a zero at point f(Q) which complicates matters.

regards

tommy1729