exp(x) x f ' (exp(x)) = exp(f(x)) f(x)

[tommy1729]

When taking the half iterate \(f(f(x)) = \exp(x)\); we know that:

\[
f'(x) = \frac{\exp(x)}{f'(f(x))}\\
\]

Let's try to sequentially solve this equation for \(|x| < \delta\)--by writing:

\[
f_{n+1}'(x) = \frac{\exp(x)}{f_n'(f_{n+1}(x))}\\
\]

I think everyone here considered this at least once.

The problem here is you get even more complicated integrals and differential type equations at every step , especially if you want a closed form.
But even numerical.

Maybe you wanted to write 

 

\[
f_{n+1}'(x) = \frac{\exp(x)}{f_n'(f_{n-1}(x))}\\
\]

That seems easier because your method depends on earlier functions and not present or future ones.

On the other hand that might give worse solutions or worse convergeance.

Another thing I do not like is that those integrals are still asymptotics to closed form functions , so you are getting very slowly towards an exotic function.

I think I called it pade-tommy method or so in the past :

( inspired by pade approximations )

\[
f_{n+1}'(x) = \frac{2\exp(x)}{(f_n'(f_{n-1}(x)) + f_n'(f_{n-2}(x))} \mod x^{2n+4}\\
\]

IN mod x^(2n+4)

I guess you can see the pade analogue.

So we get increasing degree polynomials that should converge to the semi-exp(x).
We set the initial guess so that f_w(0) is between 0 and 1 ( because id(z) and exp(z) is between 0 and 1 ).


regards

tommy1729