1
$\begingroup$

For the question below:

Let $x_1 = \alpha$ and $x_{n+1} = 3x^2_n$ for $n ≥ 1$.

(b) Does $\lim x_n$ exist? Explain.

I proved in the first part that if the limit of $x_n =x$ then $x=0$ or $x=1/3$

In this part, I'm required to show if limit exists or not. I know that for $\alpha=1/3$, limit is 1/3. For $0<\alpha<1/3$, the sequence is decreasing and converges to $0$. And that for $\alpha>1/3$, the sequence diverges

Can anyone please help me formally prove the divergence for $\alpha>1/3$. Any suggestions/hints?

Thank you.

$\endgroup$
1
$\begingroup$

If $x_1 = \alpha, x_{n+1} = 3x_n^2$ then first you form the geometric picture in your mind, which you did correctly. Now, to see what exactly is happening, you must first inspect $x_{n+1} = 3x_n^2$ as a growth pattern, and then ram home the argument formally.

This lemma is our key tool.

If $0 < \alpha < \frac 13$, then $x_{n+1} < x_n$ for all $n$.

Proof : By induction. $x_1 = \alpha$, and $x_2= 3\alpha^2$, so $x_1 - x_2 = \alpha(1- 3\alpha) > 0$ since both the terms are greater than $0$.

For general $n$, start with $x_{n+1} = 3x_n^2$. By induction, $x_n < x_{n-1} ... < x_1 < \frac 13$. So, $x_n - x_{n+1} $ $= x_n(1-3x_n) > 0$ since $x_n > 0$ (since it's three times the square of something positive) and the other term is positive as $x_n < \frac 13$. Therefore, $x_n > x_{n+1}$ , completing the proof.

A bounded decreasing sequence converges.

This is easy to see. The candidate for the limit is the infimum of the sequence (as a set). Use $\epsilon - \delta$ and the definition of infimum.

Now, you showed that if there is a limit, it is $0$ or $\frac 13$. But the limit cannot be $\frac 13$ here : we saw that $x_n < \alpha$ for all $n$, so $|x_n - \frac 13| > |\alpha - \frac 13|$ for all $n$.

So, the sequence must go to $0$.


If $\alpha > \frac 13$, then a similar argument shows that $x_{n+1} > x_n$ for all $n$. However, here something different happens : if $x_n$ did converge, then it would go to either zero or $\frac 13$, but here neither is possible. Therefore, by my second lemma's contrapositive, $x_n$ must be divergent.


If $\alpha = \frac 13$ then the sequence is constant.


I leave you to see the nature of the sequence for $\alpha \leq 0$. In particular, you will notice that sign doesn't matter, since the square appears in $x_2$ itself, so this is only a question of the absolute value of $\alpha$.

$\endgroup$
  • $\begingroup$ You said: $ x_1 - x_2 = \alpha(1- 3\alpha^2) > 0$ since both the terms are greater than 0, what do you mean by both terms greater than 0? Is it because $\alpha<1/3$ And therefore $3\alpha <1$? $\endgroup$ – A.Asad Mar 14 '18 at 11:35
  • 1
    $\begingroup$ Oh, it's a mistake. I should correct it. Essentially, $1-3\alpha$ is a positive number and $\alpha$ is a positive number , so their product is positive. Yes, the edit is done. $\endgroup$ – астон вілла олоф мэллбэрг Mar 14 '18 at 11:37
  • 1
    $\begingroup$ Not only will you get it, you will never forget it as well. This is an important step in your pursuit of mathematics. Keep it up, and $+1$ for your question. $\endgroup$ – астон вілла олоф мэллбэрг Mar 14 '18 at 11:44
  • 1
    $\begingroup$ Speaking of which, if I have fully answered your question, then you may either accept my answer, or wait for a better one for a few days before making your decision. $\endgroup$ – астон вілла олоф мэллбэрг Mar 14 '18 at 11:49
  • 1
    $\begingroup$ Yes. But that is fine. Congratulations for being the first to avail this offer. In particular, I upvote your other question as well (it is not part of the terms and conditions, but the question is also good). $\endgroup$ – астон вілла олоф мэллбэрг Mar 15 '18 at 2:42

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.