solving f(g(x)) = f(x) converging to f(exp(x)) = f(x) tommy1729 Ultimate Fellow Posts: 1,853 Threads: 400 Joined: Feb 2009 02/22/2022, 12:44 AM (This post was last modified: 02/22/2022, 12:49 AM by tommy1729.) 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 tommy1729 Ultimate Fellow Posts: 1,853 Threads: 400 Joined: Feb 2009 05/26/2022, 10:59 PM I strongly believe in this. JmsNxn Ultimate Fellow Posts: 1,176 Threads: 123 Joined: Dec 2010 05/26/2022, 11:07 PM (This post was last modified: 05/26/2022, 11:08 PM by JmsNxn.) 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. « Next Oldest | Next Newest »

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