Convergence of the sequence $a_{n+1}=\frac{a_n^2(a_n-3)}{4}, a_0=-\frac12$

Question

I approached the problem as follows.
First, note that if $a_n<0$ then $a_{n+1}<0$. Since $a_0<0$, it follows that the sequence is bounded above by zero. Now, consider the difference $a_{n+1}-a_n=\frac{a_n}{4}(a_n+1)(a_n-4)$. We know already that $a_n<0$, so there are two possibilities two consider: 1) $a_n\in (-1,0)\implies a_{n+1}>a_n$ and 2) $a_n\leqslant -1\implies a_{n+1}\leqslant a_n$.
In the first case we have a bounded monotone sequence, hence it converges to $\lim_{n\to\infty}a_n=a^*$, where $-\frac12 \leqslant a^*\leqslant 0$. Taking the limit as $n\to \infty$ of $a_{n+1}-a_n$ we obtain $\frac{a^*}{4}(a^*+1)(a^*-4)=0$. It follows that $a^*\in\{-1,0,4\}\cap\left[-\frac12,0\right]$, so in this case $a^*=0$.
To eliminate the second possibility we use $a_0=-\frac12$ to prove by induction that $\left|a_n\right|\leqslant\frac12$ implies $\left|a_{n+1}\right|\leqslant\frac{1}{4\cdot4}\left(3+\frac12\right)=\frac{7}{32}\leqslant\frac12$ and hence $\left|a^*\right|\leqslant \frac12$.
I have the following questions: is this solution correct? Are there any cookbook style methods for solving problems of this kind? My approach seems a bit ad hoc to me, so I am looking for a more general or straightforward one. Thanks.

Answer 1

Your solution is correct. But you don't have to eliminate the second possibility: As $a_0=-{1\over2}$ and $a_n<a_{n+1}<0$ when $a_n\in(-1,0)$ the second possibility can never arise.

Given a recursion $a_{n+1}=f(a_n)$ it is always good practice to determine the fixed points of $f$, i.e., the solutions of $f(x)=x$. In your case the equation $$x=f(x):={x^2(x-3)\over4}$$ has the solutions $-1$, $0$, and $4$. As $f'(-1)={9\over4}>1$, $f'(0)=0$, and $f'(4)=6>1$ only the point $0$ can be attractive. In order to prove that in fact $\lim_{n\to\infty}a_n=0$ I'd more rely on estimates than on monotonicity arguments, because in many examples the $a_n$ oscillate around the limit $a^*$. Now from the recursion equation we get $$|a_{n+1}|\leq |a_n|^2{3+|a_n|\over 4}<|a_n|^2$$ if $|a_n|<1$, and this implies that the $a_n$ converge even "quadratically" to $0$ if $|a_0|<1$.

Answer 2

The systematic way is to consider this as a fixed-point iteration. One general result is that the sequence $a_{n+1}=g(a_n)$ converges to a fixed point $x^*$ of $g$ when $|g'(x^*)|<1$ and $a_0$ is close enough to $x^*$.

The sequence is a fixed-point iteration of $g(x)=\frac{x^2(x-3)}{4}$. The fixed points of $g$ are $-1, 0, 4$. Of these, only $|g'(0)|<1$. This means that the fixed-point iteration of $g$ converges to $x^*=0$ if $a_0$ is in the region $|g'(x)|<1$, which is approximately $-0.527<x<2.527$. This is the case for $a_0=-\frac12$.