Showing a limit of the recurrence relation

[Mr Davis 97]

Homework Statement

Let $x_1=1$ and $\displaystyle x_{n+1} = 3 x_n^2$ for $n \ge 1$.
a) Show if $a = \lim x_n$, then $a = 1/3$ or $a = 0$.
b) Does $\lim x_n$ actually exist?

Homework Equations

The Attempt at a Solution

I have proven before that, in general, $\lim s_{n+1} = \lim s_n$. Hence to answer a), we simply take the limit of both sides and solve the quadratic equation. Easy. But clearly, since $x_n$ is not bounded, the limit doesn't actually exist. Why am I getting values $1/3$ or $0$ when the limit doesn't actually exist?

 

[Orodruin]

What happens if $x_1=1/3$? What happens if $x_1<1/3$?

 

[fresh_42]

I found the $a=0$ case, but not the $a=\frac{1}{3}$. Anyway, if it this editor here allowed me to write letters upside down, I will win in the lottery tomorrow. That's 100% certain and true. What changed between a) and b) and have you used all information given for a) same as for b)?

 

[Mr Davis 97]

I think I see... In part a), are we saying that if $x_n$ converges, one of those would be the resulting value. In part b), I guess the question of whether the sequence actually converges is dependent on the initial value, as if $|x_1| < 1/3$ it converges to 0, $|x_1| = 1/3$ it converges to 1/3, and if $|x_1| > 1/3$ it diverges.

 

[fresh_42]

I think I see... In part a), are we saying that if $x_n$ converges, one of those would be the resulting value. In part b), I guess the question of whether the sequence actually converges is dependent on the initial value, as if $|x_1| < 1/3$ it converges to 0, $|x_1| = 1/3$ it converges to 1/3, and if $|x_1| > 1/3$ it diverges.

No, in b) you used $x_1=1$ to see that it does not converge. In a) you reasoned by $\lim s_n = \lim s_{n+1}$. Where's the initial value here? It is the other way around. And in a) we have if then, but if is never true.

 

[Ray Vickson]

Homework Statement

Let $x_1=1$ and $\displaystyle x_{n+1} = 3 x_n^2$ for $n \ge 1$.
a) Show if $a = \lim x_n$, then $a = 1/3$ or $a = 0$.
b) Does $\lim x_n$ actually exist?

Homework Equations

The Attempt at a Solution

I have proven before that, in general, $\lim s_{n+1} = \lim s_n$. Hence to answer a), we simply take the limit of both sides and solve the quadratic equation. Easy. But clearly, since $x_n$ is not bounded, the limit doesn't actually exist. Why am I getting values $1/3$ or $0$ when the limit doesn't actually exist?

If the limit exists (call it $a$) we have $a = 3 a^2$, so $a = 0$ or $a = 1/3$.

However, what if the limit does not exist? In that case, a third solution to $a = 3 a^2$ "exists" within the extended real-number system, namely: $a = +\infty$.

What do you think is the correct value of $a$ in this problem?