06/05/2011, 08:47 PM
(06/05/2011, 06:24 PM)JmsNxn Wrote: Thank you very much for the series approximations sheldon, but sadly the humps still occur in base \( \eta \). I'm wondering now if there is a better base to work with or if it's smarter to dump the idea of logarithmic semi-operators altogether, as they seem to be a poor extension of ackerman function to domain real.Cheta(-1)=e*(log(2)+1).
The only interesting thing I have to report is that:
just like \( \lim_{p\to -\infty} \text{cheta}(p) = e \), \( \lim_{p\to -\infty} e\, \{p\}\, e = \ln^{\alpha -p}(2 \exp^{\alpha -p}(e)) = e \)
or that \( \lim_{p\to -\infty} \text{cheta}(p) = e\, \{p\}\, e = e \) where \( \{p\} \) is a logarithmic semi operator.
So my question was, what's the radius of convergence for the cheta series you gave me, and whats the recurrence relation so that I can produce the full \( \text{cheta}(x) \) function. I just want to test some values. for ex: if \( \text{cheta}(-1) = e + \ln(2) \) then I think we have something, but if it doesn't, oh well.
cheta(z-1)=\( \log_\eta(\text{cheta}(z))=e*\log(\text{cheta}(z)) \)
cheta(z+1)=\( \exp_\eta(\text{cheta}(z))=\exp(\text{cheta}(z)/e) \)
Cheta(z) is entire, but the series convergence for a finite number of terms is limited by how close we are to the region of superexponential growth, and how many terms are used. It probably has an effective radius of convergence of about 2 with the number of terms I posted. If you want more convergence as real(z) increases or decreases, use the recurrence relation below. If you want more convergence as imag(z) increases, one way to get that is to generate the series for cheta(z-100), centered at -100. For cheta/sexpeta, I actually use Newton Polynomial interpolation, centered around -100 for cheta, and +100 for sexpeta, with a 50 term polynomial, which has pretty good convergence out to a radius of 25 (~28 digits), and is accurate to 50 digits within a unit radius. I also recently included cheta/sexpeta support in my latest kneser.gp program.
- Sheldon
