Let )
and )
be truncated taylor taylors series of resp ,g(x))
up to orders 
.
Consider the equation
)=f(x))
More specific = 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 = 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
 = 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
 = 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
,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(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
More specific
We add the condition
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
and
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
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
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
