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

[JmsNxn]

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).
\( \exp_\eta^{\alpha t}(a) = \text{cheta}(\text{cheta}^{-1}(a) + t) \)

Logarithmic semi-operators contain infinite rings and infinite abelian groups. In so far as {t} and {t-1} always form a ring and {t-1} is always an abelian group (therefore any operator greater than {1} is not commutative and is not abelian). There is an identity function S(t), however its values occur below e and are therefore still unknown for operators less than {1} (except at negative integers where it is a variant of infinity (therefore difficult to play with) and at 0 where it is 0). Greater than {1} operators have identity 1.

The difficulty is, if we use the lower superfunction of \( \exp_\eta(x) \) to define values less than e we get a hump in the middle of our transformation from \( a\,\{t\}\,b \). Therefore we have difficulty in defining an inverse for rational exponentiation. however, we still have a piecewise formula:

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

therefore rational roots, the inverse of rational exponentiation is defined so long as \( a > e \) and \( \frac{1}{b} \exp_\eta^{\alpha 1-t}(a) > e \).
rational division and rational subtraction is possible if \( a,b > e \) and \( \exp_\eta^{\alpha-t}(a) - \exp_\eta^{\alpha -t}(b) > e \).

Here are some graphs, I'm sorry about their poor quality but I'm rather new to pari-gp so I don't know how to draw graphs using it. I'm stuck using python right now. Nonetheless here are the graphs.

the window for these ones is xmin = -1, xmax = 2, ymin = 0, ymax = 100




If there's any transformation someone would like to see specifically, please just ask me. I wanted to do the transformation of \( f(x) = x\, \{t\}\, 3 \) as we slowly raise t, but the graph doesn't look too good since x > e.

Some numerical values:
\( 4\,\{1.5\} 3 = 4\,\{0.5\}\,4\,\{0.5\}\, 4= 21.01351835\\
3\, \{1.25\}\, 2 = 3\, \{0.25\}\, 3 = 6.46495791\\
5\, \{1.89\}\, 3 = 5\, \{0.89\}\,5\,\{0.89\}\, 5 = 81.307046337\\ \)
(I know I'm not supposed to be able to calculate the second one, but that's the power of recursion)

I'm very excited by this, I wonder if anyone has any questions comments?

for more on rational operators in general, see the identities they follow on this thread http://math.eretrandre.org/tetrationforu...hp?tid=546

thanks, James

PS: thanks go to Sheldon for the taylor series approximations of cheta and its inverse which allowed for the calculations.