Jay Fox's Linear approximation for tetration and slog

[sheldonison]

After reading about the new wikipedia page for the superlogarithm, I started thinking about my linear approximation for tetration.

I suppose I ought to write up something more formal. My linear approximation is the same as the "standard" linear approximation for base e, but otherwise it's different. And it has a couple big advantages over the "standard" linear approximation.

First of all, it's \( C^1 \) continuous, meaning it's continuous and once differentiable for all real x. The "standard" linear approximation is only \( C^0 \) continuous for bases other than e.

More importantly, it helps highlight the fact that there would be an inflection point on the critical interval that I define, which helps expose more information about the slog (and tetration). I've always been partial to simple formulae that expose additional insights.

Anyway, I first discussed my linear approximation on Google groups, which is where I first met Gottfried Helms incidentally. (Which reminds me, Gottfried had put forth a guess as to where the inflection point would be located. I still haven't tried to determine the precise location of the infleciton point and whether it's consistent from base to base.)
....

I discovered the same critical section, which must contain the inflection point, for base n, which follows the general definition that t(x) = n^(t(x-1)). The critical section is between x-1 and x, with the following equations. t refers to the tetration estimation function for base n.

t(x-1) = -ln(ln(n))/ln(n)
t(x) = 1/ln(n)

Note, t(x+1) = e, for all base n values. Between x and x+1, the value of t(x) may be estimated by n^t(x-1), where t(x-1) is a line. For base e, the critical section is between t(-1) and t(0), where a linear estimate at x=0 produces

t(-1) = -ln(ln(e))/ln(e) = 0
t(0) = 1/ln(e) = 1

And t(1)=e. We know it contains the inflection point because one can show (with a bit of algebra), that the first derivative at t'(x) is equal to the first derivative at t'(x-1). Start from the definition of tetration, and take the derivative.

t(x) = n^t(x-1)
t'(x) = n^(t(x-1))*ln(n)*t'(x-1)

Next, we are interested specifically in t(x)=1/ln(n). we can substitute this in and get
t'(x) = t'(x-1), if t(x)=1/ln(n)

There are many equations that have the same first derivative at t'(x) and t'(x-1), but the simplest is a line, which has the same first derivative on either side. Other solutions for curves from t(x-1) to t(x) can be found that have smooth second order derivatives at t(x-1) and t(x), but this requires a 3rd order polynomial equation instead of a linear equation. I got as far as deriving the general solution for a fourth order equation that has smooth 1st derivatives, 2nd derivatives, and 3rd derivatives, for a base(n) estimate over the critical region. The coefficients require solving a quadratic equation. For a 5th order estimate that is 4th order smooth, the non-linear equations get hopelessly complex, though I may attempt a numerical estimate.

Andrew Robbins has done much better by solving the estimation for the hyper log function, which is linear, as opposed to the tetration function, which is non-linear.

This is not formally worded, but all sorts of questions arise. Where is the inflection point? The inflection point is the second derivative. In general, one can hypothesize that all even derivatives (2nd, 4th, 6th etc), will probably have an inflection point, and will go from -infinity at t(-2) to a critical section at t(-1) to t(0) containing the inflection point. to t(1), t(2), t(3).... then going off to +infinity.

Continuing on the hypothesis, the odd derivative ought to have minimum values in the -1 to 0 range, going off to infinity on either side.

My estimates give results close to Andrew's solutions. One reason why I solved the "3rd order" smooth derivative versions of these equations, is that the smooth 1st/2nd order derivative equations are mirror image symmatrical about the critical section, with the inflection point exactly in the middle of the critical section. A 4th order equation with a smooth 3rd derivative is no longer symmetrical, so it give more insight into the true nature of the curve. I developed the general base=n equations, as an attempt to understand the more general patterns of the tetration curve.

3rd order estimates for e^^0.5
1.646354229 from Andrew Robbin's tetration paper
1.644772458 -0.09% base n=e 3rd derivative smooth estimate using equations below
1.648721271 +0.14% baes n=e 1st/2nd derivative smooth estimate

Here are my base=e equations, for a tetration estimate, t(y) = a0 + a1*(x+0.5) + a2*(x+0.5)^2 + a3*(x+0.5)^3 + a4*(x+0.5)^4, where x is in the range (-1 to 0), where t(1)=e, t(0)=1, and t(-1)=0

a0=0.49760
a1=0.94949
a2=0.01918
a3=0.20204
a4=-0.03837

I can also post the exact equations used to generate these coefficients if anyone is interested.

For my 3rd derivative smooth estimate for base 10, I get 10^^1.38748975 ~= 100, versus Andrew's result of 1.39233. The error is 0.35% which seems a little higher than I would've expected.

For base n=10. t(y) = a0 + a1*(x+0.96997) + a2*(x+0.96997)^2 + a3*(x+0.96997)^3 + a4*(x+0.96997)^4. T(y) is a tetration estimate in the range (-1.46997 to -0.46997), that has smooth first, second, and third derivatives. If you plug -1 into this estimate, you get t(-1)=0, which is how I figured out the 0.96997 normalization value.

t(0.46997+1)=e,
t(-0.46997)=-1/ln(10) = 0.43429
t(-1.46997)=-ln(ln(10))/ln(10) = -0.362216

a0=0.02105
a1=0.70413
a2=0.11992
a3=0.36953
a4=-0.23985

A 3rd order equation with a smooth second derivative has a2=0 and a4=0. Interestingly, the a1 and a3 values are not changed between the 2nd derivative smooth results and the 3rd derivative smooth results. This is true for all values of n. A second order smooth equation, centered on the critical section is always of the form
t(n) = a0 +a1x +a3*x^3, with the same a1 and a3 values for the 3rd order smooth result. a0 does change, and the normalization range changes by a little as well.