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


  • $\begingroup$ See if this is not useful. $\endgroup$ – J. W. Perry Sep 29 '13 at 23:22
  • $\begingroup$ J.W. Perry, thank you this is useful I am now attempting to compute a simular problem to ensure i am understanding $\endgroup$ – Dom Sep 29 '13 at 23:33

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 $x = 1$.

  • $\begingroup$ This is an interesting approach to the problem, @brom. May i ask how you "knew" to start with your first line $\endgroup$ – Dom Sep 29 '13 at 23:35
  • $\begingroup$ @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... $\endgroup$ – user38355 Sep 29 '13 at 23:49
  • $\begingroup$ ...but long story short: it was a gut feeling and something that becomes easier to spot once you've had some practice. $\endgroup$ – user38355 Sep 29 '13 at 23:49

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 $y_n = x_n-1$.

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

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 $$\lim_{n \rightarrow \infty}(x_n) = 1.$$


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