Jay Fox's Linear approximation for tetration and slog

[sheldonison]

.... And yes I would like to see some more equations

There are three more unknown coefficients to calculate, based on three equations. First, the average of f(+0.5,-0.5) is calculated in terms of the coefficients, and in terms of the boundary definition. Second, the difference between f'(+0.5,-0.5) is calculated in terms of the coefficients, and is also know to be zero due to the boundary definition. Third, the difference between f'''(+0.5,-0.5) is calculated in terms of the coefficients. That difference can also be calculated in terms of the other derivatives and boundary conditions.

There are also important values used in this section that have already been defined, such as f(0.5), f(-0.5), a1, a3, and p from my previous post.
\( f(+0.5) = 1/\ln(n) \)
\( f(-0.5) = -\ln(\ln(n))/\ln(n) \)
\( m = (1 + \ln(\ln(n))) / \ln(n) \)
\( ave = (f(-0.5) + f(0.5))/2 = 1/2(-\ln(\ln(n)) / \ln(n)) + 1/2(1/\ln(n)) \)
\( ave = (1-\ln(\ln(n))/(2*\ln(n)) \)
\( ave = a0 + a2/4 + a4/16 \)
\( f'(+0.5) - f'(-0.5) = 0 = 2a2 + a4 \)
\( a4 = -2*a2 \)
\( ave = a0 - a2/8 \)
\( f''(-0.5) = (-3*a3 + 2*a4) \)

At this point, it is worth noting that if a third order estimate were used, there would not be an a4 term, and therefore a4=0 and a2=0. For that third order approximation, it is then trivial to determine that a0 is equal to "ave", which was just calculated. Continuing on, calculate the difference between f'''(0.5) and f'''(-0.5), in terms of the a0..a4 coefficients, and then substitute in the known values for f'(x)=p, and substitute in the equation for a4 in terms of a2, and substitute in the equation for f''(-0.5) in terms of a3, a4.
\( f'''(+0.5) - f'''(-0.5) = 24*a4 = \ln(n)*3*p*f''(-0.5) + \ln(n)*\ln(n)*p^{3} \)
\( 24*a4 = \ln(n)*3*p*f''(-0.5) + \ln(n)*\ln(n)*p^3 \)
\( 24*a4 = \ln(n)*3*p*(-3*a3 + 2*a4) + \ln(n)*\ln(n)*p^{3} \)
\( 24*a4 - \ln(n)*3*p*2*a4 = \ln(n)*3*p*(-3*a3) + \ln(n)*\ln(n)*p^{3} \)
\( 24*a4 - \ln(n)*p*6*a4 = -\ln(n)*9*p*a3 + \ln(n)*\ln(n)*p^{3} \)
\( a4*(24 - \ln(n)*p*6) = -\ln(n)*9*p*a3 + \ln(n)*\ln(n)*p^{3} \)
\( a4 = (-\ln(n)*9*p*a3 + \ln(n)*\ln(n)*p^{3}) / (24 - \ln(n)*6*p \)

Next, use the value calculated for a4 to generate a2, and a0, using the value of ave in terms of the ln(n), from above.
\( a2 = -a4/2 \)
\( a0 = ave - (a2/ \)

Finally, the normalization value must be approximated by iteration to determine sexp(x+norm) ~= f(x). Find the value of x where f(x) has a known defined value of sexp, either a value 0, or a value of 1, corresponding to sexp(-1) or sexp(0).

I calculated the normalization values for 2,e,3, and 10 in the following spread sheet for 3rd and 4th order approximations.