Convergence of fixed point iteration for Omega constant

Question

In wikipedia, it is introduced that Omega constant can be computed by fixed-point iteration solving $\Omega_n = e^{-\Omega_{n-1}}$. Yet, does it converges with any starting value $\Omega_1$?

While I was confirming that it converges to omega by actually trying it with starting point around the true $\Omega$, I found that it converges even with negatively large value such as -100. However, the fixed-point iteration is not guaranteed to converge if $|f'(x)| > 1$. At -100, it has much larger gradient than $1$ so I'm confused at which starting point, it converges and where it doesn't.

Answer 1

Set $f(x)=\exp(-x)$ and $g(x)=f(f(x))=\exp(-\exp(-x))$ $(x \in \mathbb{R})$. Start iterating $f$ for any $x_0 \in \mathbb{R}$, that is $x_{n+1}=f(x_n)$ $(n \ge 0)$. Then $x_{n+2}=g(x_n)$ $(n \ge 0)$. Since $g$ is a contraction $(0\le g'(x) \le 1/e)$ the sequence $(x_{2n})$ is convergent. Its limit $y$, say, is the unique fixed point of $g$. Moreover $f(f(y))=y$ implies $g(f(y))=f(y)$. Since the fixed point of $g$ is unique $f(y)=y$. Finally $x_{2n+1}=f(x_{2n}) \to f(y)=y$ $(n \to \infty)$, so $x_n \to y$ $(n \to \infty)$.

Answer 2

Consider $$f(x)=e^{-x}-x \qquad\qquad f'(x)=-e^{-x}-1<0\qquad\qquad f''(x)=e^{-x} >0$$ $f(x)$ is continuous and its first derivatives do not change sign anywhere. So, Newton will converge to the solution for any $x_0$.

However, if we start at $x_0$ such that $f(x_0) >0$, by Darboux theorem, there will not be any overshoot of the solution during Newton iterations since $f''(x_0)>0$. For illustration, start with $x_0=-10$; you will have $20$ exact figures after $15$ iterations.