Convergence of $x_{n+1} = \frac{(x_n)^3 + 3ax_n}{3(x_n)^2+a} $

Question

Let $a \geq 0$ and let $(x_n)_n$ be a sequence such that $$x_{n+1} = \frac{(x_n)^3 + 3ax_n}{3(x_n)^2+a} $$ with $x_0 \geq 0$. Does this sequence converge to $\sqrt{a}$ for every $x_0 \geq 0$?

If $x_0 > \sqrt{a}$, then we can easily prove that the derivative of the function $ \displaystyle x \mapsto \frac{x^3+3ax}{3x^2+a} $ is bounded above by $\frac{1}{3}$ and below by $0$, so we get that $x_n \to \sqrt{a}$ as $n \to \infty$.

However, I don't know how to approach this problem if $x_0 < \sqrt{a}$. I think that $x_n \not\to \sqrt{a}$ as $n \to \infty$ if $x_0 < \sqrt{a}$. Using the same idea as before, we can get that $x_n < \sqrt{a}, \forall n \in \mathbb{Z}_+$. Then I tried to prove that the sequence is decreasing, so it will be convergent, but with limit smaller than $\sqrt{a}$. However, I have not been able to do that. How should I proceed?

Answer 1

The map $\displaystyle\;\varphi: x \mapsto \frac{x^3 + 3ax}{3x^2 + a}$ has $3$ fixed points: $0, \pm \sqrt{a}$.

Instead of the sequence $x_{n+1} = \varphi(x_n)$, one can study auxiliary sequence of the form $z_n = f(x_n)$ where $f(x)$ is a rational function over $x$ depending on the fixed points of $\varphi$.

If one is lucky, with suitable choice of $f(x)$, the recurrence sequence of $z_n$ will be significantly simpler than that of $x_n$. This will allow us to extract the asymptotic behavior of $x_n$ more easily.

In this case, if one define two auxiliary sequences $y_n, z_n$ by $$x_n = \sqrt{a} y_n\quad\text{ and }\quad z_n = \frac{1-y_n}{1+y_n} \iff y_n = \frac{1-z_n}{1+z_n}$$ One can verify $z_n$ satisfies a recurrence relation $$z_{n+1} = z_n^3, \forall n \quad\implies\quad z_n = z_0^{3^n}$$

For any $x_0 > 0$, $|z_0| < 1$ and hence

$$\lim_{n\to\infty} z_n = \lim_{n\to\infty} z_0^{3^n} = 0 \implies \lim_{n\to\infty} y_n = 1 \implies \lim_{n\to\infty} x_n = \sqrt{a}$$

When $x_0 = 0$, $z_0 = 1$ and hence all $z_n = 1 \implies y_n = 0 \implies x_n = 0$.
In this case, the limit is $0$ instead.

Answer 2

Suppose $(x_n)$ converges, say $\lim x_n = l \in \mathbb{R}$. Since the function $f : x \mapsto \frac{x^3 + 3 a x}{3 x^2 + a}$ is continuous, we have $\lim f(x_n) = f(\lim x_n) = f(l)$, but we also have $\lim f(x_n) = \lim x_{n+1} = \lim x_n = l$. Hence, we must have $l = f(l)$.

Possible limits
The equation $f(x) = x$ has three solutions : $-\sqrt{a}$, $0$ and $\sqrt{a}$.

In particular, the answer to the problem is no. Take eg $x_0 = 0$, then $x_n = 0$ for all $n$ hence $\lim x_n = 0$.

limit for a given $x_0$
For this, we need to solve the inequation $f(x)>x$, hence to establish the variation chart of the function $g$ defined by $g(x) = f(x) - x$. You'll get that :

$f(x)> x \Longleftrightarrow x \in ]- \infty;-\sqrt{a}[\cup ]0;\sqrt{a}[$ and $f(x)< x \Longleftrightarrow x \in ]-\sqrt{a};0[\cup ]\sqrt{a};+\infty[$

To sum it up :

• If $x_0 < -\sqrt{a}$, then for all $n$, $x_n < f(x_n) < -\sqrt{a}$ hence the sequence $(x_n)$ converges since it is increasing and majored. We get $\lim x_n = -\sqrt{a}$ in this case.
• If $-\sqrt{a} < x_0 < 0$, then for all $n$, $ -\sqrt{a} < f(x_n) < x_n$. With a similar argument, we get $\lim x_n = -\sqrt{a}$ in this case.
• Using the symetry ($f$ is odd), one gets that $\lim x_n = \sqrt{a}$ whenever $x_0>0$.