0
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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 $f(x) = x^2(x-3)/4$.

enter image description here

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 $\ell = 0$.

$\endgroup$
1
$\begingroup$

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?

$\endgroup$
4
  • $\begingroup$ Thanks. My best guess would be $0$ since it's decreasing. I'm thinking of finding the formula for this recurrence, but I'm not sure will it work out. $\endgroup$
    – Chan
    Oct 24 '12 at 1:22
  • $\begingroup$ If $b=1/2$, then $a_0=-1/2$: can a decreasing sequence that starts at $-1/2$ converge to zero? But, maybe it isn't decreasing.... $\endgroup$ Oct 24 '12 at 1:31
  • $\begingroup$ Thanks once again. I finally got it solved ;) $\endgroup$
    – Chan
    Oct 25 '12 at 23:51
  • $\begingroup$ Good! Then you can post your solution as an answer. Later, you can accept your solution. $\endgroup$ Oct 26 '12 at 3:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.