07/28/2014, 10:30 PM

(07/28/2014, 12:17 PM)tommy1729 Wrote: 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

It turns out 1 + v(z) cannot both be entire and grow slower then O(exp(|z|/|a|) and still satisfy the functional equation well.

Hence there is no (nice) v(z) !!

I can prove that if you want, its trivial. It follows from considering the zero's of both 1+v(z) and f(z).

hence the attention goes to f_1(z) = f(z+u(z)).

It seems u(z) changes the function in the sense that the region where f_1^[2](z) - exp(z) << 1 is different from f(f(z)) - exp(z) << 1.

This can be explained by induction starting from the real line.

However that does assume the conjectures made about f(z).

It seems u(z) makes a " wobble " just like we saw with the theta functions.

Im a bit puzzled by u(z).

If u(z) has so many zero's this suggests (carlson , hadamard) u(z) grows at least like exp^[0.5](|z|/|a|) for z with negative real part.

But then |f(z+u(z))| grows like O(exp(|z|/|a|)) ?

This implies that f(f(z)) grows like exp(z) near the positive real line and like exp(exp(z)) near (but not on) the negative real line for sufficiently large z ?!?

That would be one weird function.

Like the half-iterate of both exp(z) and exp(exp(z)).

Thus the half-iterate of a twice fake/asymptotic function itself.

Fake function theory seems like a neverending story.

regards

tommy1729