Searching for an asymptotic to exp[0.5]

[JmsNxn]

I'm not sure if this will help, but I know some about taylor series and can do some fractional calculus (Where can't we do fractional calculus)


If \( \phi \) is holo on \( \Re(z) < 1 \) and satisfies some fast growth at imaginary infinity and negative infinity. And if \( \phi(z) \neq \mathbb{N} \) in the half plane. Then fix \( 1>\tau > 0 \):

\( \frac{1}{2\pi i}\int_{\tau-i\infty}^{\tau+i\infty}\frac{\pi}{\sin(\pi z)\G(\phi(z))} w^{-z}\,dz = \sum_{n=0}^\infty \frac{w^n}{(\phi(-n))!} \)


Maybe that might help some of you? The unfortunate part is as \( w \to \infty \) we're going to get decay to zero. I'm not sure about the iterates. We can also note this is a modified fourier transform and so we can apply some of Paley Wiener's theorems on bounding fourier transforms from the original functions. I.e: We can bound the taylor series by the function in the integral. Therefore maybe if we get very fast decay to zero we can talk about asymptotics of \( 1/\exp^{0.5}(x) \).

Hey James,

Actually, the half iterate is really well behaved in the complex plane, if you follow the well behaved branch for real(z)>0, especially as compared to Tetration, so something like that should work. There are singularities at L, and at z+pi i, I need to remember where the singularities are; Mike has done a nice complex plane plot where if you put the branches at L, The singularity at L, L* are the only two singularities you see.
- Sheldon

Welllll!!! If that's so I have a different proposal for you.

Take \( \phi(z)=\frac{1}{\exp^{0.5}(z+1)} \). If it is holo in \( \Re(z) > -1 \) (i.e. the half iterate of the exponential has no zeroes) and If we can show \( |\phi(z)| < C e^{\alpha|\Im(z)| + \rho|\Re(z)|} \) for \( 0< \alpha < \pi/2 \) and \( \rho \ge 0 \) THEN!

\( \vartheta(w) = \sum_{n=0}^\infty \frac{w^n}{n!\exp^{0.5}(n+1)} \)

we have in \( 0 < \Re(z) < 1 \)

\( \phi(-z) = \frac{1}{\G(z)} \int_0^\infty \vartheta(-w)w^{z-1}\,dw \)

Transforming the mellin transform into a fourier transform is easily done by looking at the inverse mellin transform. Then we can talk about some bounded conditions. I suggest if you want to know what I'm talking about and where this comes from, to read the paper I posted in the other thread on tetration. It's all rigorously shown, and is quite simple and elegant, but remarkably powerful so far.


Tell me what you think, maybe we can do this with \( \exp^{\ell}(z+1) \) as well, with \( \ell \in (0,1) \)?