04/17/2015, 02:15 AM
(This post was last modified: 04/17/2015, 01:13 PM by fivexthethird.)
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.
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.