Jay Fox's Linear approximation for tetration and slog

[sheldonison]

Thats indeed an interesting approach to extend this linear approximation on a unit interval to a polynomial approximation there.
It would think that this approach is different from the approach of approximating the super exponential by polynomials in a point (instead of the unit interval) such that the first some derivations are smooth.
Earlier I started a thread about this topic:
http://math.eretrandre.org/tetrationforu...php?tid=34 however didnt really come to conclusions.

I have also another question: Do your computations lead to the same interval as for the linear approximation, or is it different? And yes I would like to see some more equations

Thanks for your reply. I've been browsing the forum's more here, and there's a lot of interesting posts! This is my first attempt at using tex.

My approximation equations use sexp instead of slog.

The resulting equations are non-linear, instead of linear equations. Therefore, the approximation is limited to a fourth order equation, instead of an "nth" order equation.

The approximation equation is centered on the critical section. For base(e) the approximation is centered at x= -0.5 instead of at x=0 or at x=-1. For the general base(n), the approximation is centered about the critical unit section where first derivatives are equal on either side of the unit section.

Equations to solve for a0..a4, so that the 1st, 2nd, and 3rd derivatives are smooth. First, solve for the "odd" terms, a1 and a3

\( f(x) = a0 + a1*x + a2*x^2 + a3*x^3 + a4*x^4 \)

Starting with the definition of sexp for base n
\( f(x)= n^{f(x-1)} \)
\( f'(x)= n^{f(x-1)}*\ln(n)*f'(x-1) \)

The approximation is centered around x=0, for the range x[-0.5 to +0.5]. Later, the normalization coefficent will be estimated so that in the critical section, the following approximation holds.
\( \text{sexp}(x+\text{norm}) \) ~= \( f(x) \)

Next, continue by taking the first derivative.
\( f'(0.5) = n^{f(-0.5)}*\ln(n)*f'(-0.5) \)
if we set \( n^{f(-0.5)} \) equal to \( 1/\ln(n) \), then we have \( f'(0.5)=f'(-0.5) \)
\( f(-0.5) = -\ln(\ln(n))/\ln(n) \)
\( f(+0.5) = 1/\ln(n) \)
\( f(+1.5) = e \)

Continue by taking the 2nd and 3rd derivatives. The f''' third derivative equations are used to calculate a2 and a4.
\( f''(x) = n^{f(x-1)}*\ln(n)*f''(x-1) + n^{f(x-1)}*\ln(n)*\ln(n)*(f'(x-1))^{2} \)
\( f''(0.5) = f''(-0.5) + \ln(n)*(f'(-0.5))^2 \)
\( f'''(x) = n^{f(x-1)}*\ln(n)*f'''(x-1) + n^{f(x-1)}*\ln(n)*\ln(n)*\ln(n)*(f'(x-1))^{3} + n^{f(x-1)}*\ln(n)*\ln(n)*3*f'(x-1)*f''(x-1) \)
\( f'''(0.5) = f'''(-0.5) + \ln(n)*3*f'(-0.5)*f''(-0.5) + \ln(n)*\ln(n)*(f'(-0.5))^{3} \)

Now calculate the difference between f(+0.5) and f(-0.5). This difference is assigned to the shorthand constant "m", which is equal to the difference between f(0.5) and f(-0.5).
\( m = (1 + \ln(\ln(n))) / \ln(n) \)
\( f(+0.5) = a0 + a1/2 + a2/4 + a3/8 + a4/16 \)
\( f(-0.5) = a0 - a1/2 + a2/4 - a3/8 + a4/16 \)
\( f(+0.5) - f(-0.5) = m = a1 + a3/4 \)
\( a3 = 4*(m-a1) \)

Next, average f'(+0.5) and f'(-0.5), noting that the two are equal, and using the shorthand value "p" for the first derivative of f(x) at +/- 1/2.
\( f'(-0.5) = a1 - a2 + 3a3/4 -a4/2 \)
\( f'(+0.5) = a1 + a2 + 3a3/4 + a4/2 \)
\( (f'(+0.5) + f'(-0.5))/2 = f'(+0.5) \)
\( (f'(+0.5) + f'(-0.5))/2 = a1 + (3*a3)/4 \)
\( p = f'(-0.5) = f'(+0.5) = a1 + (3*a3)/4 \)

Now we calculate the difference between f''(+0.5) and f''(-0.5), and then substitute in for p from the equation just derived.
\( f''(+0.5) = 2*a2 + 3*a3 + 3*a4 \)
\( f''(-0.5) = 2*a2 - 3*a3 + 3*a4 \)
\( f''(+0.5) - f''(-0.5) = 6*a3 \)
\( f''(+0.5) - f''(-0.5) = 6*a3 = \ln(n)*p*p \)
\( 6*a3 = \ln(n)*( a1 + 3a3/4)^{2} \)

Now, substitute in from a3 = 4*(m-a1), and then with a little algebra, generate the quadratic equation for a1. Solve the quadratic equation for a1 in terms of the coefficients, a, b, and c. Then solve for a3 in terms of a1.
\( 24(m-a1) = \ln(n)*(a1 + 3(m-a1))^{2} \)
\( a = 4*\ln(n) \)
\( b = (24 - 12*\ln(n)*m) \)
\( c = 9*m*m*\ln(n)-24m \)
\( a1 = -(b + sqrt{b^{2} - 4ac})/2a \)
\( a3 = 4*(m-a1) \)

The interesting thing about this result is that the value for a1 and a3 are the same for a 3rd order approximation, as they are for a 4th order approximation. I will post the equations for a0, a2, and a4 next.

Moderators Note: Changed ln to \ln in tex code, this lets ln appear non-italic as it should be. Looks much nicer