Let \( f(x,n) \) and \( g(x,m) \) be truncated taylor taylors series of resp \( f(x),g(x) \) up to orders \( n,m \).
Consider the equation
\( f(g(x))=f(x) \)
More specific \( f(x) = a_1+a_2 g(x)+a_3 g(x)^2+ a_4 g(x)^3+ a_5 g(x)^4+ ... \)
We add the condition \( f(x) = a_1 + a_2 g(x) + a_3 g(2x) + a_4 g(3x) + a_5 g(4x) + ... \)
By using the taylor coefficients of g(x) we can set up a system of equations ...
And by truncating and dropping terms , we can get exact solutions for the coefficients that are approximations for f and g.
For instance we solve
\( f(x,4) = a_1 + a_2 g(x,2)+ a_3 g(x,2)^2+ a_4 g(x,2)^3+ a_5 g(x,2)^4 \)
and
\( f(x,4) = a_1 + a_2 g(x,2) + a_3 g(2x,2) + a_4 g(3x,2) + a_5 g(4x,2) \)
The first equations has degree 4*2 = 8.
The second equation has degree 2.
dropping terms gives a solvable system and a relatively good approximation of orders (2,2).
This approximation then is an approximation to
\( f(g(x,2),2)=f(x,2) \)
Increasing n and m increases all precisions.
The conjecture is that since exp(c N x) = exp(c x)^N ( for integer N) we start to approximate
\( f(\exp*(x)) =f(x) \)
Where exp*(x) is a truncated taylor for exp(x) with order growing with n and m.
Then f(x,n) is clearly a truncated taylor solution for slog(x)/dx as n goes to +infinity.
Yes Im aware we have probably seen these equations before.
They probably converge to Andrews slog.
He took a different approach by taking derivatives.
But that is similar to using taylors since the nth derivative relates to the taylor coef.
regards
tommy1729
Consider the equation
\( f(g(x))=f(x) \)
More specific \( f(x) = a_1+a_2 g(x)+a_3 g(x)^2+ a_4 g(x)^3+ a_5 g(x)^4+ ... \)
We add the condition \( f(x) = a_1 + a_2 g(x) + a_3 g(2x) + a_4 g(3x) + a_5 g(4x) + ... \)
By using the taylor coefficients of g(x) we can set up a system of equations ...
And by truncating and dropping terms , we can get exact solutions for the coefficients that are approximations for f and g.
For instance we solve
\( f(x,4) = a_1 + a_2 g(x,2)+ a_3 g(x,2)^2+ a_4 g(x,2)^3+ a_5 g(x,2)^4 \)
and
\( f(x,4) = a_1 + a_2 g(x,2) + a_3 g(2x,2) + a_4 g(3x,2) + a_5 g(4x,2) \)
The first equations has degree 4*2 = 8.
The second equation has degree 2.
dropping terms gives a solvable system and a relatively good approximation of orders (2,2).
This approximation then is an approximation to
\( f(g(x,2),2)=f(x,2) \)
Increasing n and m increases all precisions.
The conjecture is that since exp(c N x) = exp(c x)^N ( for integer N) we start to approximate
\( f(\exp*(x)) =f(x) \)
Where exp*(x) is a truncated taylor for exp(x) with order growing with n and m.
Then f(x,n) is clearly a truncated taylor solution for slog(x)/dx as n goes to +infinity.
Yes Im aware we have probably seen these equations before.
They probably converge to Andrews slog.
He took a different approach by taking derivatives.
But that is similar to using taylors since the nth derivative relates to the taylor coef.
regards
tommy1729
