(04/30/2014, 11:17 AM)sheldonison Wrote: That was my first reaction, for complex z, the inverse of the super-exponential is really badly behaved, and gets arbitrarily large near the real axis. For convergence testing, would it be better to focus on large real values of z, or small real values of z, sexp(0)=1, sexp(1)=e? I assume its the integral that causes problems?
Do we know what the values of any of these integrals are? If you have generated them, could you post them?
\( \vartheta(x) = \sum_{n=0}^\infty \frac{(-x)^n}{n!(^n e)} \)
\( \frac{1}{\Gamma(1)}\int_1^\infty \vartheta(x)\,dx \)
\( \frac{1}{\Gamma(0)}\int_1^\infty \vartheta(x)x^{-1}\,dx \)
\( \frac{1}{\Gamma(0.5)}\int_1^\infty \vartheta(x)x^{-3/2}\,dx \)
\( \frac{1}{\Gamma(-1)}\int_1^\infty \vartheta(x)x^{-2}\,dx \)
update
It looks like this equation is different than the one in the thread entire function close to sexp, which
has an additional exponential term in integral and may be entire, whereas I don't think the version above, without the exponential term converges. Here is the version that converges. Presumably, James is aware of the problems...
\( \int_1^\infty e^{-\lambda x}\vartheta(x)x^{-z-1}\,dx) \)
- Sheldon
Hey Sheldon.
Firstly \( \frac{1}{\Gamma(-k)} = 0 \) for \( k \in \mathbb{N} \cup \{0\} \). So it necessarily interpolates tetration. I haven't evaluated anything. I'm not very numerical. I'm more interested in just trying to show convergence. I haven't been able to evaluate anything yet. The problem at hand can be stated as such:
If \( \vartheta(w) = \sum_{n=0}^\infty \frac{w^n}{n!(^n e)} \)
Then if \( \int_0^\infty |\vartheta(-w)|w^{\sigma-1}\,dw < \infty \) for \( 0 < \sigma < 1 \) is condition 1. Then this is what we want: Condition 1 to be satisfied.
If this happens then for \( 0<\Re(z) < 1 \):
\( \frac{1}{(^{-z} e)} = \frac{1}{\Gamma(z)} \int_0^\infty \vartheta(-w)w^{z-1}\,dw \)
we will have a Tetration with the remarkable property of being well behaved as we move the imaginary argument. AND:
\( \frac{1}{2\pi i} \int_{\sigma - i\infty}^{\sigma + i\infty} \Gamma(z)\frac{w^{-z}}{(^{-z} e)}\,dz = \vartheta(-w) \)
Where this tetration is well behaved enough so that happens.
Once we have that condition 1 is satisfied I can extend the tetration for all \( \Re(z) > -1 \) by
\( \frac{1}{(^z e)} = \frac{1}{\Gamma(-z)} (\sum_{n=0}^\infty \frac{(-1)^n}{n!(^n e)(n-z)} + \int_1^\infty \vartheta(-w)w^{-z-1}\,dw) \)
Since the gamma function has simple zeroes at the non positive integers we see they cancel out the poles and all that's left is \( (^n e) \) and also the simple zero cancels out the integral on the right.
It's very difficult but what that's left to prove is that \( e^{-(^z e)} \) is holomorphic for \( \Re(z) > -1 \) then I can prove the convergence of
\( |\int_{\sigma - i\infty}^{\sigma + i\infty} \Gamma(z) \frac{w^{-z}}{e^{(^{-z} e)}}\,dz| < \infty \)
And then everything that's left is my theorems that I have in the paper I'm writing. My professor has looked over it and everything is right--I'm just working out the final knots of it. It shows a very competent way of iterating superfunctions, however it lacks theorems on convergence of mellin transforms. I.e. the convergence of the mellin transform of \( \vartheta \). This would give us Tetration.
I am wary on the convergence of the mellin transform though. I am not sure that it will converge. However. If you can find a tetration such that it satisfies: \( 1/(^{-z} e) \) is holomorphic on \( \Re(z) < 1 \) and \( |1/(^{-z} e)| < C e^{\alpha |\Im(z)| + \rho |\Re(z)|} \) for \( 0 < \alpha < \pi/2 \) and \( \rho \ge 0 \) then the mellin transform of \( \vartheta \) converges. This isn't hard to show but I'm a little too lazy to type it out. It just means the inversemellin transform converges and it equals vartheta and by this the mellin transform of vartheta necessarily converges and equals that tetration. This is mind bending when you think about uniqueness though. It should imply only one function satisfies these conditions. And that \( \alpha >0 \) so that if we take some entire periodic function \( f(z+1) = f(z) \) which NECESSARILY blows up \( f \sim e^{2\pi |\Im(z)|} \) so this implies \( |\frac{1}{(^{-z + f(z)} e)}| \not < C e^{\alpha |\Im(z)| \) and \( |\frac{1}{(^{-zf(z)} e)}| \not < C e^{\alpha |\Im(z)| \)
\( |\frac{f(z)}{(^{-z} e)}| \not < C e^{\alpha |\Im(z)| \)
So we have uniqueness!
