04/30/2014, 11:17 AM
(This post was last modified: 04/30/2014, 06:01 PM by sheldonison.)
(04/03/2014, 02:14 PM)JmsNxn Wrote: Hi everyone. I've been fiddling with this for a while and it looks like we should have some kind of convergence on an expression for tetration I've found. I'm not certain if i'm making mistakes or not but any help would be great.
....
\( \frac{1}{(^z e)}=\frac{1}{\Gamma(-z)} \sum_{n=0}^\infty \frac{(-1)^n}{n!(^n e)(n-z)} + \frac{1}{\Gamma(-z)} \int_1^\infty \vartheta(x)x^{-z-1}\,dx \)
...
There we are. Analytic tetration for \( \Re(z) > -1 \)
This will work on probably all real numbers....
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
