let $f:A \to \mathbb{R}$ be a function that $A \subseteq \mathbb{R}$ . let $x_0$ arbitrary,

Question

let $f:A \to \mathbb{R}$ be a function that $A \subseteq \mathbb{R}$ . let $x_0$ arbitrary, $$x_{n+1}=f(x_n) \ \ \ :n \in \mathbb{N}$$

now my question : when $\left( x_n \right)_{n\in \mathbb{R}}$ is converge ? Do have to Always $A=\mathbb{R}$ and $f$ be a function which is differentiable on $\mathbb{R}$ ?

Answer 1

At least, you should take $A$ such that $f(A)\subset A$. Then, if $x_1\in A$, $(x_n)_{n\in\mathbb N}$ will be a sequence of elements of $A$.

Even if $f$ is differentiable, you cannot be sure that $(x_n)_{n\in\mathbb N}$ converges. For instance, take $A=[-1,1]$, $f(x)=-x$ and $x_1=-1$. Then $(\forall n\in\mathbb{N}):x_n=(-1)^n$ and the sequence $\bigl((-1)^n\bigr)_{n\in\mathbb N}$ does not converge.

Answer 2

That almost never happens, even for very good functions. E.g. take $f(x)=x+1$. Then $x_n=x_0+n$ which obviously does not converg. Note that our $f$ is very good: continuous, differntiable, smooth, polynomial, affine, etc. etc.

However there is an important class of functions for which the thesis holds: contraction maps, i.e. functions

$$f:\mathbb{R}\to\mathbb{R}$$

such that for some $0\leq k<1$ the inequality

$$|f(x)-f(y)|<k|x-y|$$ holds for all $x,y\in\mathbb{R}$. An example would be $f(x)=\frac{1}{2}x$. For such functions Banach fixed point theorem applies and whatever $x_0$ you pick the sequence will always converg and to the exact same, unique fixed point of $f$.

Side note: $f$ does not even have to be continous to have convergent $x_n$ sequence. Take

$$f(x)=\begin{cases} \frac{1}{2}x&x\neq 1 \\ 1&x=1 \end{cases}$$

The sequence $x_n$ converges for all $x_0$. If $x_0\neq 2^k$ then $x_n\to 0$. Otherwise $x_n\to 1$.

For a continuous but not differentiable example consider $f(x)=\frac{1}{2}|x|$.