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

[tommy1729]

Well, yes and no. I wanted:

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

Because this is a first order ODE; and is perfectly solvable. It's simply an equation:

\[
y'_{n+1} = g_n(x,y_{n+1})\\
\]

Which is really easy to solve for locally. Once you have \(f_n\), you get \(g_n\); this completely determines \(f_{n+1}(x)\) and ensures differentiability. It produces an exotic function yes; but it's approximating a half-exponential, which is the most exotic function I can think of.

It is more so a numerical procedure than a closed form expression, I'll give you that.

I started a new thread to get perhaps better initial guess by using sums like f_0(x) = exp(ln(x)^2) + exp(exp(ln(ln(x))^2))/10 + ... and such ideas.

see

https://math.eretrandre.org/tetrationfor...p?tid=1707



and then perhaps you could use half-plane integral representations or series representations
( meaning series or integrals valid on a half plane , like Laplace or Dirichlet etc ) to solve the recursively defined differential equation.


regards

tommy1729