Prove that there exists $n\in \mathbb{N}$ s.t. $x_n=\frac12$

Question

Let $(x_n)_n$ a sequence given by $2x_{n+1}=2x_n^2-5x_n+3$ with $x_1\in \mathbb{Q}$. I know that the sequence is convergent. I know that the limit of the sequence should be $\dfrac{1}{2}$ or $3$.

I want to prove that there exists $n\in \mathbb{N}$ s.t. $x_n=\dfrac{1}{2}$ if the sequence goes to $\dfrac{1}{2}$. Similar if the sequence goes to $3$.

I tried by definition with $\epsilon$ but I didn’t succeed.

Answer 1

Let $f(x)=x^2-\frac{5}{2}x+\frac{3}{2}$, so that $x_{n+1}=f(x_n)$. Suppose that $x_n$ converges to $l$, where $l\in\lbrace \frac{1}{2},3\rbrace$. Then

$$ y_n=\frac{x_{n+1}-l}{x_n-l}=\frac{f(x_{n})-f(l)}{x_n-l} \to f'(l) \textrm{ when } n \to \infty \tag{1} $$

Note that $f'(\frac{1}{2})=-\frac{3}{2}$ and $f'(3)=\frac{7}{2}$. So $|f'(l)| \geq \frac{3}{2}$ in both cases, and hence $|f'(l)| \gt \frac{5}{4}$ in both cases. It follows that there is a $n_0$ such that $|y_n|\gt \frac{5}{4}$ for all $n\geq n_0$. Then

$$|x_{n+1}-l| \geq \frac{5}{4} |x_n-l| \textrm{ for all } n\geq n_0 \tag{2}$$. By induction, we deduce $$|x_n-l| \geq \big(\frac{5}{4}\big)^{n-n_0}|x_{n_0}-l| \textrm{ for all } n\geq n_0 \tag{3}$$

If $x_{n_0}\neq l$, we would deduce $\lim_{n\to\infty}{|x_n-l|}=\infty$, which is impossible. So $x_{n_0}=l$, which finishes the proof.

Answer 2

As pointed out in Ewan Delanoy's answer, you're basically looking at the iterations of the function $f(x)=x^2-\frac{5}{2}x+\frac{3}{2}$. It might help to get some intuition about the behavior if you actually draw this function and use the graph to get a few iterations, starting from different points.

You can see one example below. You can find out more in the Wikipedia articles: Cobweb plot and Fixed-point iteration.

Answer 3

Let $\overline{x} = \lim_n x_n \in \{\frac12,3\}$. Consider the series $y_n = x_n - \overline{x}.$

$$ \begin{align} y_{n+1} &= x_{n+1} - \overline{x}\\ &= x_n^2 - \frac52x_n+\frac32 - \overline{x}\\ &= \big(y_n+\overline{x}\big)^2 - \frac52\big(y_n+\overline{x}\big)+\frac32 -\overline{x}\\ &= y_n\Big(y_n+2\overline{x}-\frac52\Big)+\underbrace{\overline{x}^2-\frac72\overline{x}+\frac32}_{=0 \text{ for } \overline{x} \in\{\frac12,3\}} \end{align} $$

By definition of $\overline{x}$, the series $(y_n)_n$ converges to $0$.

Consider the case $\overline{x} = \frac12$. Then $$ y_{n+1} = y_n \Big(y_n - \frac32\Big). $$ Let $n_0$ be such that $|y_n| < \frac14$ for all $n \ge n_0$. This implies $|y_n-\frac32| > \frac54$ for all $n \ge n_0$. By induction (as in the answer by Ewan Delanoy), we get: $$ |y_{n}| > \frac54^{n-n_0} |y_{n_0}| \qquad \text{for all } n > n_0. $$ For large enough $n$, this contradicts $|y_n| < \frac14$, unless $|y_{n_0}| = 0.$ Conclude that there is an $n_0$ such that $y_{n_0} = 0$ and thus $x_{n_0} = \frac12.$

For $\overline{x}=3$, the same ideas apply.