tetration base conversion, and sexp/slog limit equations

[sheldonison]

Consider what happens as b approaches e^(1/e) in the equation \( slog_b(x) \). The curve becomes more and more linear, and there are fewer and fewer degrees of freedom for how to extend the sexp function to real numbers, and still have an increasing "well behaved" function. It must be possible to describe this rigorously in terms of limits.

But it never gets completely linear, doesnt it? Otherwise it would not be analytic.

again, consider the region from \( \text{slog}_b(e) \) to \( \text{slog}_b(\text{slog}_b(e)) \).
Let m=\( \text{slog}_b(e)-\text{slog}_b(\text{slog}_b(e)) \)
and let average =\( 1/2(\text{slog}_b(e)+\text{slog}_b(\text{slog}_b(e))) \)

If we translate the region so it is centered over -0.5 to +0.5, then the linear approximation would be:
\( \text{sexp}_b(x) = a0 + a1*x + a2*x^2 + a3*x^3 + a4*x^4 ... \)
with, \( a0=\text{average, and } a1=m \text{, } a2=0 \text{, } a3=0 \text{, } a4=0 \)

As b approaches \( e^{1/e} \), m will approach zero, and average will approach e. The endpoints of this critical section of the curve have been chosen so that
\( \text{sexp}_b^'(-0.5) = \text{sexp}_b^'(+0.5) \)

Thus a linear approximation has a continuous first derivative. However the second derivative is not continuous. It takes some algebra to show that the second derivative can be made continuous with an a3 term that is a solution of this quadratic equation.
\( a3*a3 + b*a3 + c = 0 \text{ with }
b=(4*m-24*\text{slog}_b(e)) \text{ and }c=4*m^2 \)

As m gets smaller, the result for a3 approximates accurately as:
\( a3 \text{=~} m*m*ln(b)/6 \text{ and, } a1=m-a3/4 \)
The a3 maximum contribution =~ \( m^2*(1/(72*e))*sqrt(1/3)\text{ at } x=sqrt(1/12) \)

This 3rd order equation (with a2=0 and a4 and higher zero), approximates the critical section much better than the linear estimate. But the maximum contribution of the a3 term to the sexp equation is proportional to m squared, or proporional to m when divided by m, the delta of the critical section. This shows that the a3 contribution becomes smaller and smaller and the linear term for a1=m can be as accurate as desired as m goes to zero, as b approaches \( e^{1/e} \).

I have also done the algebra for a continuous 3rd derivative, which effects the a0, a2 and a4 terms, with a1 and a3 unchanged. This contribution is proportional to m cubed, so it is much smaller than the a3 contribution, and is proportional to m squared when divided by m, the delta of the critical section.

So I think a linear approximation does converge as closely as one would like to the actual function, as the base approaches \( e^{1/e} \). And then this approximation can be used to convert to any other arbitrary base, as described. The data posted earlier for base e and base 10 sexp results used a 3rd order polynomial to model the critical section for base b=1.485, and should be accurate to nearly six significant digits.