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solving f(g(x)) = f(x) converging to f(exp(x)) = f(x)
#1
Let and be truncated taylor taylors series of resp up to orders .

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
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#2
I strongly believe in this.
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#3
Guess I never saw this thread.

I absolutely believe this will converge pointwise at least. But can we be sure that the Taylor series is convergent?

Sounds like Andrew's slog, but I like this more, seems more tangible.

I do not think it'll be easy to show the radius of convergence is non zero though. And we need it to be \(|L|\). Sounds really hard to be honest.
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