Demonstrating the convergence of $x_{n+1} = 1/2(x_n + c/x_n)$

Question

I'm curious how to demonstrate the convergence of such a sequence. It is given as: Let $c > 0$. Define $g(x) = \dfrac{1}{2}\left(x+\dfrac{c}{x}\right)$ when $x \ne 0$. Let $x_0 > 0$, $(x_0)^2 > c$ and define a sequence recursively that $x_{n+1} = g(x_n)$. I am asked to show that $x_n \to \sqrt{c}$.

Answer 1

$x_n=\frac{1}{2}(x_{n-1}+\frac{c}{x_{n-1}})\geq\sqrt{c}$, so $x_{n+1}=\frac{1}{2}(x_n+\frac{c}{x_n})\leq x_n$, $\{x_n\}$ decreases and have a lower bound, so $x_n$ convergence, let $n\rightarrow\infty$ in $x_{n+1}=\frac{1}{2}(x_n+\frac{c}{x_n})$ leads that $\lim_{n\rightarrow\infty}x_n=\sqrt{c}$.

Answer 2

Observe that $g'(x) = \dfrac{1}{2}\left(1 - \dfrac{c}{x^2}\right)$. We see that $g'(x) > 0$ when $1 > c/x^2 \iff x > \sqrt{c}$, which means that $x > \sqrt{c}$ implies $g(x) > f(\sqrt{c}) = \sqrt{c}$. Hence $\sqrt{c}$ is a lower bound for $\{x_n\}$.

Suppose $x_n > \sqrt{c}$, i.e., $x_n^2 > c$ or $x_n > c/x_n$. Then $x_{n+1} = \dfrac{1}{2} \left(x_n + \dfrac{c}{x_n}\right) \le x_n$. Since $x_0 > \sqrt{c}$, all of $\{x_n\}$ is monotonically decreasing. By the Monotone Convergence Theorem, $\{x_n\}$ converges.

The limit as $n \to \infty$ must satisfy $L = \dfrac{1}{2} \left(L + \dfrac{c}{L}\right)$ and reside in the interval $[\sqrt{c},x_0)$. It is straightforward to see that $L = \sqrt{c}$.

Answer 3

We have $x_0 = \sqrt c + u_0$ for some $u_0 \gt 0$.

Also for any $u \gt 0$ we can write the identify

$$\tag 1 0.5 \bigr(\,(\sqrt c + u) + \frac{c}{\sqrt c + u}\,\bigr) = \sqrt c + \frac{0.5 u^2}{\sqrt c + u} $$

It is easily shown using $\text{(1)}$ that

$\tag 2 s \gt \sqrt{c} \implies g(s) \gt \sqrt{c}$

and

$\tag 3 s,t \gt \sqrt{c} \land s \lt t \implies g(s) \lt g(t)$

We'll now prove using induction that for all $n \ge 1$,

$\tag{C} x_n \gt \sqrt{c} \land x_n - \sqrt{c} \lt (.5)^n u_0$

For the base case $n = 1$, we get $x_1 \gt \sqrt{c}$ using $\text{(2)}$. Also using $\text{(1)}$

$ x_1 - \sqrt{c} = g(x_0) - \sqrt{c} = \frac{0.5 u_0^2}{\sqrt c + u_0} \lt (.5)^1 u_0$

and so $\text{(C)}$ is true for $n = 1$.

For the inductive step we assume that $\text{(C)}$ is true for some integer $n \ge 1$. Then there must exists $u \gt 0$ such that

$ x_n = \sqrt{c} + u \;\text{and } u \lt (.5)^n u_0$

We get $x_{n+1} \gt \sqrt{c}$ using $\text{(2)}$.

Moreover, by $\text{(3)}$

$$ x_{n+1} - \sqrt{c} \lt g(\sqrt{c} + (.5)^n u_0) - \sqrt{c}$$

and using $\text{(1)}$

$ g(\sqrt{c} + (.5)^n u_0) = \sqrt{c} + \frac{0.5 [(.5)^n u_0]^2}{\sqrt{c} + (.5)^n u_0} \lt \sqrt{c} + (.5)^{n+1} u_0$

and so

$$x_{n+1} - \sqrt{c} \lt (.5)^{n+1} u_0$$

and the proof by induction is complete.

Since for $n \ge 1$

$$ \sqrt{c} \lt x_n \lt \sqrt{c} + \frac{u_0}{2^n} $$

we can use the squeeze theorem to deduce that the sequence $(x_n)$ converges to $\sqrt c$.