For what values $\alpha$ does the iteration of $e^{-\alpha x}$ converge?

Question

For what values of $\alpha >0$ will the fixed point iteration $x_{k+1} = g(x_k)$ converge for $g(x)=e^{-\alpha x}$ assuming that $x_0= [0,1]$.

Would it suffice to find the derivative of say $h(x) = g(x) - x$ and if that is less than $1$ it is good?

Answer 1

You are iterating a special case of the exponential map:

$$g_\lambda(x)=e^{\lambda\cdot x}$$

for $\lambda=-a$, so let's assume that $\lambda\neq 0$. The fixed points of $g_\lambda$ are found using the Lambert $W$ function as:

$$ \begin{align} g_\lambda(x)&=x\Rightarrow\\ e^{\lambda\cdot x}&=x\Rightarrow\\ x\cdot e^{-\lambda\cdot x}&=1\Rightarrow\\ -\lambda\cdot x\cdot e^{-\lambda\cdot x}&=-\lambda\Rightarrow\\ -\lambda\cdot x&=W(-\lambda)\Rightarrow\\ x&=\frac{W(-\lambda)}{-\lambda} \end{align} $$

Checking the derivative at these fixed points:

$$ \begin{align} \lvert g_\lambda'(x)\rvert&=\lvert\lambda\cdot e^{\lambda\cdot \frac{W(-\lambda)}{-\lambda}}\rvert\\ &=\lvert\lambda\cdot e^{-W(-\lambda)}\rvert\\ &=\lvert\lambda\cdot\frac{W(-\lambda)}{-\lambda}\rvert\\ &=\vert-W(-\lambda)\rvert\\ \end{align} $$

By L. Euler's Theorem (here) the iterated exponential of base $e^{\lambda}$ converges, provided:

$$ \begin{align} e^{-e}&\le e^\lambda\le e^{\frac{1}{e}}\Leftrightarrow\\ -e&\le \lambda\le \frac{1}{e}\\ \end{align} $$

Then (excluding the end points, which need to be checked separately),

$$ \begin{align} -\frac{1}{e}&\lt -\lambda\lt e\Leftrightarrow\,\,\,\text{($W$ monotone in range)}\\ W\left(-\frac{1}{e}\right)&\lt W(-\lambda)\lt W(e)\Leftrightarrow\\ -1&\lt W(-\lambda)\lt 1\,\,\,\text{(special values of $W$)}\Leftrightarrow\\ \lvert -W(-\lambda)\rvert&\lt 1 \end{align} $$

Therefore, the iterates of $g_\lambda$ will converge by fixed point iteration, provided:

$$-\frac{1}{e}\lt -\lambda= \alpha\lt e$$

and provided the seed value $x_0$ similarly satisfies:

$$e^{-e}\le x_0\le e^{\frac{1}{e}}$$

The case $\alpha=0$ gives the trivial iterates of base 1, so it is omitted.

The case $\alpha=e$ implies $\lambda=-\alpha=-e$ and then $g_\lambda$ will iterate the exponential of base $e^{-e}$, which is known to converge (by Euler, above).

The case $\alpha=-\frac{1}{e}$ implies $\lambda=-\alpha=\frac{1}{e}$ and then $g_\lambda$ will iterate the exponential of base $e^{\frac{1}{e}}$, which is also known to converge (again by Euler).

Concluding, the iterates of $g_{-\alpha}(x)=e^{-\alpha\cdot x}$ will converge, iff $\alpha\in [-\frac{1}{e},e]$ and $x\in[e^{-e},e^{1/e}]$.

You can deduce the answer now by looking at your question's restrictions for $\alpha$ and $x$.