Rational operators (a {t} b); a,b > e solved

[sheldonison]

Well alas, logarithmic semi operators have paid off and have given a beautiful smooth curve over domain \( (-\infty, 2] \). This solution for rational operators is given by \( t \in (-\infty, 2] \) \( \{a, b \in \R| a, b > e\} \):

\(
f(t) = a\, \{t\}\, b =
\left\{
\begin{array}{c l}
\exp_\eta^{\alpha t}(\exp_\eta^{\alpha-t}(a) + \exp_\eta^{\alpha -t}(b)) & t \in (-\infty,1]\\
\exp_\eta^{\alpha t-1}(b * \exp_\eta^{\alpha 1-t}(a)) & t \in [1,2]\\
\end{array}
\right.
\)

Which extends the ackerman function to domain real (given the restrictions provided).
the upper superfunction of \( \exp_\eta(x) \) is used (i.e: the cheta function).
\( \exp_\eta^{\alpha t}(a) = \text{cheta}(\text{cheta}^{-1}(a) + t) \) .....

Hey James,

Sounds very exciting. fyi, I admit I don't yet understand your new functions, but I made some graphs, and it looks very promising. I thought I'd post the following snippet of pari-gp code, which implements the rational operator function you posted, which can be used with mylatest kneser.gp code, which includes \( \text{sexp}_\eta(z) \) support. This would also make it easier for other to try your new function. If a<e, then I used sexpeta(invsexpeta(a)+t), instead of cheta(invcheta(a)+t). I think this is legal since the two are in many ways the same function when they are being to generate fractional iterations of eta. This expansive definition eliminates your restrictions, and covers all of the reals, working seamlessly from -infinity to infinity. It also matches your second example exactly.

Code:

expeta(t,a) = {
  if (real(a)<exp(1), sexpeta(invsexpeta(a)+t), cheta(invcheta(a)+t));
}
fatb(a,t,b) = {
  if (t<1,  return (expeta(t,expeta(-t,a)+expeta(-t,b))));
  if (t>=1, return (expeta(t-1,b*expeta(1-t,a))));
}

I've gotten as far as quickly verifying that for t=0, we have addition, t=1, we have multiplication, and t=2 is exponentiation. The existing code has problems for invcheta(z) or invsexpeta(z), where z is too close to e, and invcheta(z)<=-1000, or invsexpeta(z)>1000. Other than that, it seems to work great. For example,
fatb(3,0,4)=7, which is 3+4
fatb(3,1,4)=12, which is 3x4
fatb(3,2,4)=81, which is 3^4
I wonder what it means that fatb(3,-1,4)=5.429897..?
Also, is there a smooth continuation to a function for t=3, which would be tetration?
- Sheldon