06/08/2011, 11:47 PM

(06/08/2011, 08:32 PM)sheldonison Wrote: [ -> ]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

Woah, I wonder what consequences this will have on the algebra.

I guess \( \log_f^{\circ q}(a\,\,\bigtriangleup_{1+q}\,\,b) = b\log_f^{\circ q}(a) \)

\( \log_f^{\circ q}(a\,\,\bigtriangleup_{q}\,\,b) = \log_f^{\circ q}(a) + b \)

which I guess isn't too drastic.

But ,the question is, of course, does the following still hold \( 0 \le q\le1 \):

\( a\,\,\bigtriangleup_{1+q}\,\,2 = a\,\,\bigtriangleup_q\,\,a \)

\( \exp_{2^{\frac{1}{2}}}^{\circ 1+q}(\exp_{2^{\frac{1}{2}}}^{\circ -q-1}(a) + 2) \neq \exp_{a^{\frac{1}{a}}}^{\circ q}(a + a) \), so no it doesn't. That's not good, we want operators to be recursive.

And I'm unsure if the inverse is still well defined, so I think we lose:

\( (a\,\,\bigtriangleup_{1+q}\,\, -1)\,\,\bigtriangleup_q\,\,a = S(q) \)

where S(q) is the identity function. We may even lose the identity function altogether, this is really bad.

We also lose:

\( (a\,\,\bigtriangleup_{1+q}\,\,b)\,\,\bigtriangleup_{1+q}\,\, c = a\,\,\bigtriangleup_{1+q}\,\,bc \)

\( (a\,\,\bigtriangleup_{1+q}\,\,b)\,\,\bigtriangleup_q\,\,(a\,\,\bigtriangleup_{1 + q} \,\,c) = a\,\,\bigtriangleup_{1+q}\,\,b+c \)

These are all too many valuable qualities that are lost when redefining semi-operators the way that you do. Sure it's analytic over \( (-\infty, 2] \), but it loses all its traits which make it an operator in the first place. I'm going to have to stick with the original definition of \( \vartheta \) that isn't fully analytic.

However, I am willing to concede the idea of changing from base eta to base root 2.

That is to say if we define:

\( \vartheta(a,b,\sigma) = \exp_{2^{\frac{1}{2}}}^{\circ \sigma}(\exp_{2^{\frac{1}{2}}}^{\circ -\sigma}(a) + h_b(\sigma))\\\\

[tex]h_b(\sigma)=\left{\begin{array}{c l}

\exp_{2^{\frac{1}{2}}}^{\circ -\sigma}(b) & \sigma \le 1\\

\exp_{2^{\frac{1}{2}}}^{\circ -1}(b) & \sigma \in [1,2]

\end{array}\right. \)

This will give the time honoured result, and aesthetic necessity in my point of view, of:

\( \vartheta(2, 2, \sigma) = 2\,\,\bigtriangleup_\sigma\,\, 2 = 4 \) for all \( \sigma \)

I like this also because it makes \( \vartheta(a, 2, \sigma) \) and \( \vartheta(a, 4, \sigma) \) potentially analytic over \( (-\infty, 2] \) since 2 and 4 are fix points.

I also propose writing

\( a\,\,\bigtriangle_\sigma^f\,\,b = \exp_f^{\circ \sigma}(\exp_f^{\circ -\sigma}(a) + h_b(\sigma))\\\\

[tex]h_b(\sigma)=\left{\begin{array}{c l}

\exp_f^{\circ -\sigma}(b) & \sigma \le 1\\

\exp_f^{\circ -1}(b) & \sigma \in [1,2]

\end{array}\right. \)

Sheldon's analytic function is then:

\( g(\sigma) = a\,\,\bigtriangle_\sigma^{b^{\frac{1}{b}}}\,\,b \)

that's still very pretty though, that \( g \) isn't piecewise over \( (-\infty, 2] \) and potentially analytic.