The problem I am having is figuring out the way show the following sequence is monotone:
let $x_1 = \frac{3}{2}$ and $x_{n+1} = {x_n}^2-2x_n+2$, show that the sequence $x_n$ is monotone and bounded and find the limit.
I have found the first three terms, and found that the sequence is decreasing, I have followed an example in my text that is the opposite however my text is vague and I'm not sure how they found where the sequence is bounded I am led to believe by math that it is bounded by 1.
any hints or suggestions on how to approach the problem would bennefit me greatly
thanks
Best Answer
Let $f(x) = x^2 - 2x + 2$. Then it is easy to show that $f([1,2]) \subset [1,2]$. Since $x_1 \in [1,2]$, this implies that $$x_n = f(x_{n-1}) = (f \circ f)(x_{n-2}) = ... = (f\circ ... \circ f)(x_1) \in [1,2]$$ so $\{x_n\}$ stays inside $[1,2]$ and, in particular, bounded. Now when $1 \le x \le 2$ we have $$f(x) - x = x^2 - 3x + 1 \le 2x - 3x + 1 \le 0$$ so that $$x_{n+1} = f(x_n) \le x_n$$ and $\{x_n\}$ is decreasing. Hence $$x = \lim_{n \to \infty} x_n$$ exists. Since $f$ is a continuous function, $$x = \lim_{n \to \infty} x_{n+1} = \lim_{n \to \infty} f(x_n) = f \left( \lim_{n \to \infty} x_n \right) = f(x)$$ or, in other words, $$ 0 = f(x) - x = x^2 - 3x + 1 = (x-1)(x-2).$$ Therefore, $x \in \{1,2\}$. But since we showed $\{x_n\}$ is decreasing we must have $x \le x_1 = \frac{3}{2}$ and we conclude $x = 1$.