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

[JmsNxn]

But James, this is not analytic at \( t=1 \), if we reformulate:
\(
f(t) = a\, \{t\}\, b =
\left\{
\begin{array}{c l}
\exp_\eta^{\circ t}(\exp_\eta^{\circ-t}(a) + \exp_\eta^{\circ -t}(b)) & t \in (-\infty,1]\\
\exp_\eta^{\circ t}(\exp_\eta^{\circ -t}(a)+\exp_\eta^{\circ -1}(b)) & t \in [1,2]
\end{array}
\right.
\)
We can say:
\( a\, \{t\}\, b = \exp_\eta^{\circ t}(\exp_\eta^{\circ -t}(a) + h_b(t)) \)
where
\( h_b(t)=\left{\begin{array}{c l}
\exp_\eta^{\circ -t}(b) & t\le 1\\
\exp_\eta^{\circ -1}(b) & t\in [1,2]
\end{array}\right. \)

\( f \) is addition and composition of analytic functions, except this one function \( h_b \). The whole function \( f(t) \) can not be analytic. I wonder why it looks so smooth.

I like your definition better--it seems sleeker . I was sort of aware that there was no way I was gonna produce an analytic function over the whole complex domain, I'm happy with analytic in a few regions.

But I see you gracefully avoided that problem by just defining it for a,b > e

well hopefully I'll be having to tackle that problem soon.

PS:
1. \( g(t) = a\, \} t \{ \, b \), This notation is ambiguous, compare \( \{ a \} t \{ b \} + c \). Please invent a better one!

Alright, from henceforth I shall refer to logarithmic semi operators with the following notation:
\( a\, \bigtriangleup_t\, b = a \,\{t\}\,b \)
And the inverse is given by:
\( a\, \bigtriangledown_t\, b = a \,\}t\{\, b \)

therefore:
\( a \bigtriangleup_0 b = a + b\\
a \bigtriangleup_1 b = a * b\\
a \bigtriangledown_0 b = a - b \) etc etc..

2. \( \exp_\eta^{\alpha t} \), not \alpha but \circ belongs in the exponent: \( \exp_\eta^{\circ t} \). This notation is derived from the symbol for function composition \( f\circ g \).

I knew there was something off about my equations. lol

Hey James, try my code snippet, which I updated while you were posting. It will work for values of a and b<e, seamlessly.
- Shel

I'm wary about using \( \text{sexp}_\eta(x) \) for defining bases less than e. My complaints are explained by the following points:

fatb(e+0.0001, 2, pi*I) = -0.999999 - 0.00115551*I
fatb(e+0.0001, 1.8, pi*I) = -1.883265702 - 0.00194696*I
fatb(e+0.0001, 1.5, pi*I) = -5.707515375 - 0.011242371*I
fatb(e+0.0001, 1.3, pi*I) = -4.091499848 - 8.531525563*I
fatb(e+0.0001, 1.1, pi*I) = -1.002757644 - 8.536029475*I
fatb(e+0.0001, 1, pi*I) = -8.53659263001*I

Ignoring the drastic jumps in values, observe the hump that occurs in the real transformation. For no reason the values just spike to -5 randomly. This happens with all regular superfunctions of the logarithm. That's what makes the cheta function unique.

But I think, is it possible to create an upper superfunction for \( \exp_{2^{\frac{1}{2}}}(x) \)?, perhaps it will give similar smooth results. Except it will be defined for a,b > 2... at least I think so.