Locally quadratic convergence

Question

Show that the iteration method $$x_{n+1}=\frac{1}{2}x_n+\frac{a}{2x_n}, \ \ a>0$$ $n=1, 2, \ldots $ converges locally quadratic to $\sqrt{a}$.

Could you explain to me how we could show that?

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EDIT:

We have that $|x_{n+1}-\sqrt a|=\frac1{2x_n}|x_n-\sqrt a|^2$.

To get local convergence we want that the starting point is near the root. Then this should hold for all approximations. So it must hold that $\left |x_n-\sqrt{a}\right |<\epsilon$, with $\epsilon>0$. Then we get: $$\left |x_n-\sqrt{a}\right |<\epsilon \Rightarrow \frac{1}{2(\sqrt{a}+\epsilon)}<\frac{1}{2x_n}<\frac{1}{2(\sqrt{a}-\epsilon)}$$ So we get $$|x_{n+1}-\sqrt a| = \frac1{2x_n}|x_n-\sqrt a|^2 < \frac{1}{2(\sqrt{a}-\epsilon)}|x_n-\sqrt a|^2$$ If we choose for example $\epsilon=\frac{\sqrt{a}}{2}$ we get $$|x_{n+1}-\sqrt a| < \frac{1}{\sqrt{a}}|x_n-\sqrt a|^2 \text{ with } \frac{1}{\sqrt{a}}>0$$ So the method converges locally and quadratic to $\sqrt{a}$.

Is that correct?

Answer 1

One needs to assume $a\ne 0$, otherwise the convergence is linear ($\frac{x_{n+1}}{x_n}=\frac12$). From the iteration procedure $$ x_{n+1}-\sqrt a = \frac1{2x_n}(x_n^2 + a - 2\sqrt a x_n) = \frac1{2x_n}(x_n-\sqrt a)^2 $$ This is quadratic convergence if $(x_k)$ converges to $\sqrt a$.

Now, if the iteration is started at $x_1$ close to $\sqrt a$, i.e., $x_1 \in (\sqrt a-\delta,\sqrt a+\delta)$, then $$ \frac1{2x_1}|x_1-\sqrt a| \le \frac\delta{2(\sqrt a-\delta)} $$ This factor is equal to $\frac12$ if $\delta = \frac12\sqrt a$. Per induction one can show that $x_k \in (\sqrt a-\delta,\sqrt a+\delta)$ for all $k$ and $x_k \to \sqrt a$. This is local convergence.

This iteration is nothing else than Newtons method applied to the equation $x^2=a$.