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

[JmsNxn]

mylatest kneser.gp code, which includes \( \text{sexp}_\eta(z) \) support. This would also make it easier for other to try it.

Code:

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

Oh that's great! I'd love to see what happens for complex operator values, I'll just modify your code slightly to say real(t) under the if conditions.

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.

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
what does it mean that fatb(3,-1,4)=5.429897..?
Is there a smooth continuation to a function for t=3, which would be tetration? if a<e=2.718, then it appears that the non integer nodes have complex values, though perhaps the sexpeta/invsexpeta functions could be used in place of cheta/invcheta (it might even be a smooth transition).
- Sheldon

fatb(3, -1, 4) = 5.429897, is the value of 3 delta 4; where deltation is the operator which satisfies the following conditions:
\( a\, \bigtriangleup\, b = \log_\eta(e^{\frac{a}{e}} + e^{\frac{b}{e}}) \), it's commutative and addition is spread across it:
\( (a\, \bigtriangle\, b) + c = (a + c)\, \bigtriangleup\,(b+c) \), and finally, it's defined by the following recursion formula:
\( a_1\, \bigtriangle\, a_2\, \bigtriangle\, a_3 ... \bigtriangle\, a_n = a + \log_\eta(n) \)

Basically, the whole domain of operators from \( (-\infty,0) \) are designed to: continue the ring structure of {t-1} and {t} found across addition and multiplication, and multiplication and exponentiation; and are designed to provide a smooth extension of operators to the negative domain. It differs from zeration insofar as zeration is defined by proper recursion \( a + 2 = a \circ a \) and it's discontinuous. It's better than zeration insofar as the series of operators can be continued infinitely.

As of now, it's possible to compute \( a\, \{2 + q\}\, n \) for a>e, and \( n \in N \), which would be rational tetration. So actually yes, I can evaluate \( 4\, \{t\}\, 3 \) over domain \( t \in (-\infty, 3] \)
The formula is given by:
\( a\, \{2 + q\}\, n = a_1\, \{1+q\}\, (a_2 \,\{1+q\}\, (a_3\,\{1+q\}...a_n \)

And lastly, I tried using sexp/slog base \( \eta \) for a,b < e and the graph of \( a\, \{t\} \,b \) is not smooth and there are two humps at about {0.5} and {1.5} which is very unflattering. I was surprised when the cheta function worked, but it did. I have to give myself time to think about it first, but right now I'm sure the extension for a,b <= e is right infront of our eyes.

Edit:
I edited your code and tested if
fatb(3, i, 3) = fatb(3, i+1, 2), and it does. Ackerman function is entering the complex plane me thinks.

Edit again:

if a<e=2.718, then it appears that the non integer nodes have complex values

I was wondering if you could explain what you mean by this? Is there a way to calculate a,b < e using the cheta function?