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The recurrence is defined as follows: $ a_{n+1} = \dfrac{a_n^2(a_n - 3)}{4} \text{ where } 0 < b < 3$ and $a_0 = b - 1$.
My approach is to show that is $(a_n)$ is bounded and decreasing, then by setting: $l = \dfrac{l^2(l - 3)}{4}$, I could solve for the limit $l$. However, I got 3 solutions for this equation $\{0, -1, 4\}$ which doesn't help much. So could anyone give me a hint how to solve this problem?
In fact, possibly except for the first term, we always have
$$-1 \leq a_2 \leq a_3 \leq a_4 \leq \cdots \leq 0.$$
To prove this, let $f(x) = x^2(x-3)/4$.
Then on $(0, 2)$,
$$f'(x) = \frac{3x^2 - 6x}{4} = \frac{3x(x-2)}{4} < 0 $$
and $f$ is strictly decreasing on $[0, 2]$. Since $f(0) = 0$ and $f(2) = -1$, we must have $-1 \leq f(x) \leq 0$ whenever $0 \leq x \leq 2$.
Now we divide into cases:
Case 1) Assume $a_2 = -1$. Then $f(-1) = -1$ implies that $a_n = -1$ for all $n \geq 2$. Thus we must have $\lim a_n = -1$.
Case 2) Assume $a_2 = 0$. Then $f(0) = 0$ and by the same arguments as in Case 1, we have $\lim a_n = 0$.
Case 3) Assume $a_2 \in (-1, 0)$. Then for $x \in (-1, 0)$,
$$f(x) - x = \frac{x^3 - 3x^2 - 4x}{4} = \frac{x(x-4)(x+1)}{4} > 0.$$
Thus $f(x) > x$ on $(-1, 0)$. Also, it is straightforward to check that $f'(x) > 0$ on $(0, 1)$. Thus we must have $-1 < x < f(x) < 0$ on $(-1, 0)$. Then it follows from by induction that
$$-1 < a_n < a_{n+1} < 0$$
for all $n \geq 2$. Indeed, assume $a_n \in (-1, 0)$. Then $a_n < f(a_n) = a_{n+1} \in (-1, 0)$ and the claim follows. This implies that $a_n$ is bounded and monotone increasing, hence it converges to some limit $\ell$. Clearly we have $-1 < \ell \leq 0$, and therefore the only possible value for $\ell$ is $\ell = 0$.
You have $-1\lt a_0\lt2$. If the sequence really is decreasing, could it converge to 4? If it starts below 0, could it converge to 0?
