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

[tommy1729]

I'm pretty sure that this function can't be analytic at the fix points of exp.
If it was, then if c is a fix point we could directly evaluate the derivatives of f at c, which are actually polynomials of c multiplied by f(c)
\( f \(c\) = \alpha \)
\( f' \(c\) = f\(e^c \) = f\(c\) = \alpha \)
\( f'' \(c\) = e^c f'\(e^c )\) = c \alpha \)
\( f^{[3]} \(c\) = c^3 \alpha + c \alpha \)
\( f^{[4]} \(c\) = c^6 \alpha + c^4 \alpha + 3*c^2 \alpha + c \alpha \)
\( f^{[5]} \(c\) = c^10 \alpha + c^8 \alpha + 9*c^6 \alpha + c^5 \alpha + + 6c^4 \alpha + 7*c^3 \alpha + c \alpha \)

and in general the degree of the polynomial appears to be \( \frac{n(n+1)}{2} \) which obviously grows too fast to converge.

Sorry Guys but my 3rd derivative in the op was wrong as clearly shown here.

I guess these are named polynomials , not ?

Regards

Tommy1729