For what value of $x_1$ this simple iteration will converge.

Question

Question: The function f defined by

$f(x)=\frac{x^3+1}{3}$

has 3 fixed points $\alpha,\beta,\gamma$, where

$-2 <\alpha<-1,\;\;\;0 <\beta<1,\;\;\;1 <\gamma<2$.

For arbitrarily chosen $x_1$, define ${x_n}$ by setting $x_{n+1} = f(x_n)$

(a) If $x_1 <\alpha$ , prove that $x_n\rightarrow -\infty \;\;\;as\;\;\; n\rightarrow \infty$ .

(b) If $\alpha<x_1 <\gamma$ , prove that $x_n\rightarrow \beta \;\;\;as\;\;\; n\rightarrow \infty$ .

(c) If $\gamma<x_1$ , prove that $x_n\rightarrow \infty \;\;\;as\;\;\; n\rightarrow \infty$ .

# I know that for a fixed point $\xi$ if $|f'(\xi)|<0$ then it's stable and only for $\beta$ the condition is satisfied so for $x_1$ close to $\beta$ sequence will converge.

But I don't know how to show the above 3 conditions. So I need help in proving the above 3 conditions. Thankyou.

Answer 1

Note $g(x)=f(x) -x$. You have $g^\prime(x)= x^2-1$.

Hence:

1. $g$ is increasing from $(-\infty, -1)$ and vanishes at $\alpha <-1$.
2. is decreasing on $(-1,1)$ and vanishes for $\beta \in (-1,1)$
3. is increasing on $(1,\infty)$ and vanishes at $\gamma \in (1,\infty)$.

$(x_n)$ can only converge to a fixed point of $f$.

If $x_1 <\alpha$, you can prove based on 1. by induction that $(x_n)$ is decreasing. It can’t be bounded below as this would imply that the sequence would converge to a point less that $\alpha$ which is the lowest fixed point. Hence in that case $\lim\limits_{n \to \infty} x_n =-\infty$.

With a similar approach, you can prove the other required results.