1
$\begingroup$

QUESTION: The sequence $(x_n)$ is given by $x_{n+1}=x_n(2-x_n)$ and $0<x_1<1$. Show that the sequence is convergent.

MY ATTEMPT: As per the question, we have $x_{n+1}=x_n(2-x_n)$. But I can do nothing with this recursion. Is there any way that I can transform the recursion to a relation for $x_n$ in terms of $n$? Or is there some other way to prove the convergence?

P.S. Proof by monotone convergence theorem is preferred.

$\endgroup$
9
$\begingroup$

We have $$x_{n+1} = x_n(2-x_n) = 2x_n-x_n^2$$ This gives us $$1-x_{n+1} = 1-2x_n + x_n^2 = (1-x_n)^2$$ Setting $a_n=1-x_n$, we have the sequence $$a_{n+1} = a_n^2$$ where $a_1 \in (0,1)$. Now prove that $a_n<1$ and that $a_n$ is monotone decreasing and bounded below by $0$.

Now conclude that $\lim_{n \to \infty} a_n$ exists and compute the limit. Therefore compute $\lim_{n \to \infty} x_n$.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Let $f(x)=x(2-x)$. $[0,1]$ is stable by $f$, $f$ is strictly increasing on $[0,1]$ and over $y=x$, hence the sequence converges to $f(1)=1$

| cite | improve this answer | |
$\endgroup$

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.