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

[sheldonison]

I can't believe I overlooked this! This is huge!
....
therefore:
\( \vartheta(a, e, \sigma) = \exp_\eta^{\circ \sigma}(\exp_\eta^{\circ -\sigma}(a) + e) \)

And therefore \( \vartheta \) is potentially analytic, and isn't piecewise, for all a, and b = e, \( \sigma \in (-\infty, 2] \)


This is crazy.

So we have a definition for t=0..2 (addition, multiplication, exponentiation), for all values of a. Nice!
\( a\{t\}e = \exp_\eta^{\circ t}(\exp_\eta^{\circ -t}(a)+e) \)

So, now, I'm going to define the function f(b), which returns the base which has the fixed point of "b".
\( f(e)=\eta \), since e is the fixed point of b=\( \eta \)
\( f(2)=\sqrt{2} \), since 2 is the lower fixed point of b=sqrt(2)
\( f(3)=1.442249570 \), since 3 is the upper fixed point of this base
\( f(4)=\sqrt{2} \), since 4 is the upper fixed point of b=sqrt(2)
\( f(5)=1.379729661 \), since 5 is the upper fixed point of this base
I don't know how to calculate the base from the fixed point, but that's the function we need, and we would like the function to be analytic! This also explains why the approximation of using base eta pretty well, since the base we're going to use isn't going to be much smaller than eta, as b gets bigger or smaller than e.

Now, we use this new function in place of eta, in James's equation. Here, f=f(b).
\( a\{t\}b = \exp_f^{\circ t}(\exp_f^{\circ-t}(a)+b) \)
- Sheldon