I need to find the fixed points (i.e. when $f(x) = x$) of the following function $f(x) := exp(x - 2)$. I understood that the fixed points should be the intersecation points between $f(x)$ and a function $g(x) = x$, which has infinitely many fixed points (if I am not wrong).
So, to find the intersecation points I decided to do: $$f(x) = g(x) \\ f(x) - g(x) = 0 \\ f(x) - x = 0 \\ exp(x - 2) - x = 0$$.
Plotting $f(x)$ and $g(x)$ in the same graph (in a range $-5$ to $5$), it seems that $f(x)$ intersects twice $g(x)$, so it might have 2 fixed points:
I have also discover that the fixed points of $f(x)$ are the roots of the following function: $$h := f(x) - x$$
So, I can simply find the roots for the function $h$, and I am done.
Apparently the following algorithm (which I don't know how is it called. Can you tell me?), should find the roots of a function:
Let $f: [a, b] \rightarrow \mathbb{R}$ be continuous function with $f(a) < 0 < f(b)$. Then we can find a $c$ with $f(c) = 0$ with any desired accuracy $\epsilon > 0$ as follows:
Let $a_0 := a$, $b_0 := b$ and $ k := 0$.
Let $c_k := (a_k + b_k)/2$ and:
$a_{k+1} := a_k$ and $b_{k+1} := c_k$, if $f(c_k) > 0$
$a_{k+1} := c_k$ and $b_{k+1} := b_k$, if $f(c_k) \leq 0$, and incremenet $k$ by $1$.
If $|b_k - a_k| < \epsilon$, then return $c^{'} := (a_n + b_n)/2$, otherwise continue with step $2$.
This algorithm recursively defines the terms of the sequences $a_k$ and $b_k$ in such a way that $a_k$ is increasing and $b_k$ is decreasing. As the the sequences are clearly bounded, they both converge.
This algorithm should find the roots of function, but I am not seeing why. Could you please explain me why this algorithm should find the roots?
Also, could you explaing me why the fixed points of $f(x)$ are the the roots of $h(x)$?
Basically, I need to find the fixed points of f(x) with that algorithm. I need to implement it using Maxima, but before I need to understand it.
To mention another possible solution: You can use the Lambert W function $W$ which is the inverse of the function $x \mapsto xe^x$. So you have:
$$\begin{align} e^{x-2} & = x \\ -e^{-2} & = -x e^{-x} \\ W(-e^{-2}) & = x \end{align}$$
Note, that there a two branches of $W$ in the range $[-\tfrac 1e, 0[$. Thus the above equation will give two solutions...