Question –
Consider the sequence $\langle x_n \rangle , \: n \geq 0$ defined by the recurrence relation $x_{n+1} = c . x^2_n -2$, where $c > 0$.
Suppose there exists a non-empty, open interval $(a, b)$ such that for all $x_0$ satisfying $a < x_0 < b$, the sequence converges to a limit. The sequence converges to the value?
$A. \frac{1+\sqrt{1+8c}}{2c}$
$B. \frac{1-\sqrt{1+8c}}{2c}$
$C. 2$
$D. \frac{2}{2c-1}$
Since it converges, we can write: $x=cx^2 -2$
or
$cx^2 -x -2 = 0$
Solving it gives-
$x=\frac{1\pm \sqrt{1+8c}}{2c}$
So both options (A) and (B) satisfy this.
The answer is (B). I am not able to proceed further. Please help
Best Answer
Let $f(x)=cx^2-2$. Indeed, $f(x)=x\iff x=\frac{1\pm\sqrt{1+8c}}2$. But$$f'\left(\frac{1+\sqrt{1+8c}}2\right)=1+\sqrt{1+8c}>0$$and therefore if $x_0$ is slightly larger than $f(x_0)$, then $f(x_0)>x_0$ and, in fact, the sequence $(x_n)_{n\in\mathbb N}$ is strictly increasing. Therefore, we don't have $\lim_{n\to\infty}x_n=x_0$. So, A. is not an option.