(05/11/2014, 04:29 PM)JmsNxn Wrote: If \( F \) is holomorphic for \( \Re(z) < b \) and \( |F(z)| < C e^{\alpha |\im(z)| + \rho|\Re(z)|} \) then:
\( \beta(x) = \sum_{n=0}^\infty \frac{(-x)^n}{n!}F(-n) \)
When F has singularities we see we pull on a second balancing function \( \psi \) such that:
\( \beta(x) = \psi(x) + \sum_{n=0}^\infty \frac{(-x)^n}{n!}F(-n) \)
For simple functions like \( F(z) = 1/(z^2 - 1) \) then \( \psi \) is very easy to calculate. For more complicated functions like tetration, which no doubt has essential singularities instead of poles, it becomes much more complicated. But the general result on essential singularities is just applying cauchy's residue formula around all the poles of \( F \) on the function \( \G(z)x^{-z} \)
Well, I guess then these formulas aren't of much use for continuum-summing tetration, since it is most definitely not bounded with the bound
\( |F(z)| < C e^{\alpha |\Im(z)| + \rho |\Re(z)|} \)
. In fact it is unbounded on the right half-plane (where it behaves chaotically) and has branch point singularities (which are neither poles nor essential singularities) on the left half-plane (these are logarithmic, double-logarithmic, triple-logarithmic, etc. in that order at \( z = -2 \), \( z = -3 \), \( z = -4 \), ...).
I'm curious: how did you get that first formula? Is it possible to get a similar formula for
\( \mathrm{semi}\beta_U(x) = \frac{1}{2\pi i} \int_{\sigma}^{\sigma + i\infty} \Gamma(z) F(z) x^{-z} dz \)
and
\( \mathrm{semi}\beta_L(x) = \frac{1}{2\pi i} \int_{\sigma - i\infty}^{\sigma} \Gamma(z) F(z) x^{-z} dz \)
and \( F \) satisfying the given bound? As then I might have something, perhaps. I'll have to see, though.
