1
$\begingroup$

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:

enter image description here

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:

  1. Let $a_0 := a$, $b_0 := b$ and $ k := 0$.

  2. 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 $1$.

  3. 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.

$\endgroup$
0
$\begingroup$

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...

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.