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Let $f$ be function defined by
$$f(x)=\frac{x^3+1}{3}$$ has 3 fixed points say $\alpha ,\beta ,\gamma$ where $-2 < \alpha < -1$,$0 < \beta < 1$,$1 <\gamma <2$

I want to show the following :

for arbitray choice of $x_1$, define $x_n$ as $x_{n+1}=f(x_n)$

a) If $x_1$ < $\alpha$ then $x_n \to -\infty$ as $n\to \infty$

b) If $\alpha < x_1 < \gamma$ then $x_n \to \beta $ as $n\to \infty$

c) If $x_1$ > $\gamma$ then $x_n \to \infty$ as $n\to \infty$

My Attempt:

$$g(x)=f(x)-x$$

I can show using Intermediate value property there exist 3 fixed as per given condition. And $f'(x)>0$ therefore function is increasing on $\mathbb{R}$. I am not able to proceed further. Any help will be appreciated

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    $\begingroup$ Let me do one case, all others is the same thing. Assume that $x_1<\alpha$. Then $x_2=(x_1^3+1)/2<x_1$. Derivatives are not really needed to see this inequality, since you already know that $(x^3+1)/2-x=(x-\alpha)(x-\beta)(x-\gamma)$ with $\alpha<\beta<\gamma$. Assume that $x_{n}<\alpha$, then $x_{n+1}=(x_n^3+1)/2=x_n$. If the sequence were bounded, then it would have a limit, but that limit can only be a root of $(x^3+1)/2=x$, but $x_n<x_1<\alpha$. Therefore, the sequence diverges. For (b) just remember to split into two cases, $\alpha<x_1\leq\beta$ and $\beta<x_1<\gamma$. $\endgroup$
    – user561348
    May 24, 2018 at 11:53
  • $\begingroup$ Thanks A lot Sir Providing Great Insights to Look at Problem.I understand a and c. In case of b for $\alpha <x_1 < \beta$ I get $\alpha <x_1<x_n < \beta$ and similar for $\beta < x_1 < \gamma$ I get $\beta < x_n< x_1 < \gamma$ .As they are monotonic and bounded by $\beta $ Is it enough to show that both sequences are converging to $\beta$ $\endgroup$
    – Curious student
    May 24, 2018 at 12:41
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    $\begingroup$ Yes, since they are monotonic and bounded they must converge. The only possible limits are solutions of $(x^3+1)/2=x$, equation that follows from taking limits of $x_{n+1}=(x_n^3+1)/2$ under the assumption that $x_n\to x$. But in the first case $x_n$ satisfies $\alpha<x_1<x_n\leq\beta<\gamma$. Therefore, the only possible limit would be $\beta$. Similarly one can argue for the second case, since $\alpha<\beta\leq x_n<x_1<\gamma$. $\endgroup$
    – user561348
    May 24, 2018 at 12:45

2 Answers 2

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As regards (c), note that for $t>\gamma$, $f'(t)=t^2>\gamma^2>1$. Hence if $x_n>\gamma$ then by the Mean Value Theorem, $$ x_{n+1}-\gamma=f(x_n)-f(\gamma)=f'(t)(x_n-\gamma)>\gamma^2(x_n-\gamma)$$ which implies that if $x_1>\gamma$ then the sequence $x_n$ is strictly increasing and it goes to infinity: $$x_{n+1}>\gamma+\gamma^{2n}(x_1-\gamma)\to +\infty.$$

What about (a) and (b)?

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Notations

Let

$a_1=x_1$

$a_2=f\left(a_1\right)$

$a_3=f\left(a_2\right)$

$a_4=f\left(a_3\right)$

.

.

.

. In general

$a_n=f\left(a_{n-1}\right)$ This is a sequence of numbers $a_1,a_2,a_3,a_4....a_n......$.If it converges to, say $x$.then because we have $a_n=\frac{a_{n-1}^3+1}{3}$

applying limits $lim_{\to \infty}$ both sides we get $x=\frac{x^3+1}{3}$ and since you assumed that $\alpha,\beta$ and $\gamma$ are fixed points.Then the values of x is nothing but $\alpha,\beta$ and $\gamma$ only

You can use induction to show that it is bounded. To be clear if $a_n<\alpha$ then $a_{n+1}<\alpha$ etcetra.

Now you are ready to use induction to prove that the sequence is increasing or decreasing, according to wether $x_1<\alpha$ or $\alpha <x_1<\beta$ or $\beta <x_1<\gamma$ or $\gamma <x_1$.

Once you had shown that it is monotonic in the intervals, use induction to show that it is bounded above if the sequence is increasing or it is bounded below if the sequence is decreasing.

In the cases when it diverges you need to show that it is not bounded.

I am giving you idea how to show its monotonic. In the same way you can show that it is bounded. Hint

$a_{n+1}-a_n=\frac{a_{n}^3+1}{3}-a_n=\frac{a_{n}^3-3a_n+1}{3}=\frac{\left(x-\alpha\right)\left(x-\beta\right)\left(x-\gamma\right)}{3}$

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