Showing the sequence is monotone, bounded, and finding the limit

Question

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

J.W. Perry, thank you this is useful I am now attempting to compute a simular problem to ensure i am understanding — Dom, Sep 29 '13 at 23:33

score 3 · Answer 1 · answered Sep 29 '13 at 23:22

3

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 $x = 1$.

answered Sep 29 '13 at 23:22

This is an interesting approach to the problem, @brom. May i ask how you "knew" to start with your first line – Dom Sep 29 '13 at 23:35
@Dom That's a good question. These kinds of problems where $x_n = f(x_{n-1})$ are examples of very simple dynamical systems. In that case it is always important to understand the function $f$ and things you want to prove about ${x_n}$ translate into things you want to prove about $f$. So then you start thinking about what kind of statements imply that ${x_n}$ is bounded and clearly "once $x_n$ enters a certain region, it never leaves that region" does. Translating this means f(region) is still in region and then you think about what the region could be... – Sep 29 '13 at 23:49
...but long story short: it was a gut feeling and something that becomes easier to spot once you've had some practice. – Sep 29 '13 at 23:49

score 0 · Answer 2 · answered Sep 29 '13 at 23:11

0

If your second equation is supposed to be $x_{n+1} = x_n^2-2x_n+2$, it might be useful to note that this is equivalent to $x_{n+1}-1 = (x_n-1)^2$. Thus, you might want to study the sequence $y_n = x_n-1$.

answered Sep 29 '13 at 23:11

Abel

7,312

Thank you, This agrees with some computations that were made, the double check is much appreciated – Dom Sep 29 '13 at 23:34

J. W. Perry · Answer 3 · 2013-09-30T01:06:35.183

My turn... request check for typos:

We first rewrite the sequence as (Abel's bright idea: complete the square), $$x_{n+1}={x_n}^2-2x_n+2=(x_n-1)^2+1.$$

$\textbf{Boundedness:}$ We prove that $x_n$ is bounded on the half open interval $1 \leq x_n<2$ using proof by induction on $n$.

For a basis, let $n=1$. Then $x_1=\frac{3}{2}$, and since $1 \leq \frac{3}{2}<2$, the basis holds.

For the inductive hypothesis we assume that the sequence is bounded for $n=k$, which is to say that $$1 \leq x_k<2.$$ Our burden is to show that the sequence holds for $n=k+1$. \begin{align*} & 1 \leq x_k<2 \\ \Rightarrow & 0 \leq x_k-1<1 \\ \Rightarrow & 0 \leq (x_k-1)^2 <1 \\ \Rightarrow & 1 \leq (x_k-1)^2+1 <2 \\ \Rightarrow & 1 \leq x_{k+1} <2. \end{align*} Thus we have shown that the sequence is bounded for all $n \in \mathbb{N}$.

$\textbf{Monotonicity:}$ We now prove that the sequence is monotone decreasing on the half open interval $[1,2)$ by showing that for all $n$, $x_{n+1}-x_n \leq 0$. \begin{align*} x_{n+1}-x_n &= (x_n-1)^2+1-x_n \\ &= {x_n}^2 -3x_n+2 \\ &=(x_n-2)(x_n-1). \end{align*} Now since $1 \leq x_n <2$, $(x_n-2)(x_n-1) \leq 0$, and we conclude that the sequence is monotone decreasing.

$\textbf{Limit:}$ We can now safely assume that the limit exists.

Suppose that $$\lim_{n \rightarrow \infty} (x_n)= L.$$ Then \begin{align*} & x_{n+1} = {x_n}^2-2x_n+2 \\ \Rightarrow & \lim_{n \rightarrow \infty}x_{n+1} = \lim_{n \rightarrow \infty}\left({x_n}^2-2x_n+2 \right) \\ \Rightarrow &L=L^2-2L+2 \\ \Rightarrow &L^2-3L+2=0 \\ \Rightarrow & (L-1)(L-2)=0 \\ \Rightarrow & L=1 \vee L=2. \end{align*}

Now since the sequence is monotone decreasing on the interval, $L \neq 2$, and so we conclude that $$\lim_{n \rightarrow \infty}(x_n) = 1.$$

Showing the sequence is monotone, bounded, and finding the limit

3 Answers3