Let me summarize, we have a polynomial \( p \) in some interval \( [x_0-1,x_0) \). (For for being unambiguous I will use the variable \( b \) for the base instead of \( n \) which is usually reserved for natural numbers.)
\( p(x) = a_0 + a_1*x + a_2*x^2 + a_3*x^3 + a_4*x^4 \)
There are 5 constants and \( x_0 \) to be determined. To do that we consider 4 equations \( f^{(n)}(x_0)=(b^{f(x)})^{(n)}(x_0-1) \), \( n=0,\dots,3 \) plus \( f'(x_0)=f'(x_0-1) \). These are (for simplicity and that I can reuse your text I set \( x=x_0 \) and \( f=p \)):
(0) \( f(x)= b^{f(x-1)} \)
(1) \( f'(x)= b^{f(x-1)}*\ln(b)*f'(x-1)=f(x)\ln(b)f'(x-1) \)
(*) \( f'(x)=f'(x-1) \)
(2) \( f''(x) = b^{f(x-1)}*\ln(b)*f''(x-1) + b^{f(x-1)}*\ln(b)*\ln(b)*(f'(x-1))^{2} \)
(3) \( f'''(x) = b^{f(x-1)}*\ln(b)*f'''(x-1) + b^{f(x-1)}*\ln(b)*\ln(b)*\ln(b)*(f'(x-1))^{3} + b^{f(x-1)}*\ln(b)*\ln(b)*3*f'(x-1)*f''(x-1) \)
From (0), (1) and (*) we get
\( f(x) = 1/\ln(b) = \log_b(e) \)
As we have 6 values to be determined (5 constants and \( x_0 \)) but only have 5 equations, we need another equation.
But as we know the values \( p(x_0-1)=\log_b(\log_b(e)) \) and \( p(x_0)=\log_b(e) \) independently from \( x_0 \) we just search for the \( n \) such that \( f(n)=\exp_b^n(1) \) is inside the interval \( [p(x_0-1),p(x_0)) \), there can only be on such \( n \). And then the additional equation
\( p(n)=\exp_b^n(1) \)
now completely determines \( a_0,\dots,a_4,x_0 \).
This is the approach that I would generalize from JayD's linear approximation considering your explanations. Is it how you calculate your values? Particularly I dont see why you need the special values -0.5 and 0.5.
And writing this, I wonder whether it would be more straight forward, to take the first 4 derivations (instead of now 3) into account, instead of introducing equation (*)? But then perhaps \( p(x_0) \) would depend on \( x_0 \) and we could not find an appropriate \( n \), who knows.
\( p(x) = a_0 + a_1*x + a_2*x^2 + a_3*x^3 + a_4*x^4 \)
There are 5 constants and \( x_0 \) to be determined. To do that we consider 4 equations \( f^{(n)}(x_0)=(b^{f(x)})^{(n)}(x_0-1) \), \( n=0,\dots,3 \) plus \( f'(x_0)=f'(x_0-1) \). These are (for simplicity and that I can reuse your text I set \( x=x_0 \) and \( f=p \)):
(0) \( f(x)= b^{f(x-1)} \)
(1) \( f'(x)= b^{f(x-1)}*\ln(b)*f'(x-1)=f(x)\ln(b)f'(x-1) \)
(*) \( f'(x)=f'(x-1) \)
(2) \( f''(x) = b^{f(x-1)}*\ln(b)*f''(x-1) + b^{f(x-1)}*\ln(b)*\ln(b)*(f'(x-1))^{2} \)
(3) \( f'''(x) = b^{f(x-1)}*\ln(b)*f'''(x-1) + b^{f(x-1)}*\ln(b)*\ln(b)*\ln(b)*(f'(x-1))^{3} + b^{f(x-1)}*\ln(b)*\ln(b)*3*f'(x-1)*f''(x-1) \)
From (0), (1) and (*) we get
\( f(x) = 1/\ln(b) = \log_b(e) \)
As we have 6 values to be determined (5 constants and \( x_0 \)) but only have 5 equations, we need another equation.
But as we know the values \( p(x_0-1)=\log_b(\log_b(e)) \) and \( p(x_0)=\log_b(e) \) independently from \( x_0 \) we just search for the \( n \) such that \( f(n)=\exp_b^n(1) \) is inside the interval \( [p(x_0-1),p(x_0)) \), there can only be on such \( n \). And then the additional equation
\( p(n)=\exp_b^n(1) \)
now completely determines \( a_0,\dots,a_4,x_0 \).
This is the approach that I would generalize from JayD's linear approximation considering your explanations. Is it how you calculate your values? Particularly I dont see why you need the special values -0.5 and 0.5.
And writing this, I wonder whether it would be more straight forward, to take the first 4 derivations (instead of now 3) into account, instead of introducing equation (*)? But then perhaps \( p(x_0) \) would depend on \( x_0 \) and we could not find an appropriate \( n \), who knows.
