Hero's Method

Theorem
Let $$a \in \mathbb{R}$$ be a real number such that $$a > 0$$.

Let $$x_1 \in \mathbb{R}$$ be a real number such that $$x_1 > 0$$.

Let $$\left \langle {x_n} \right \rangle$$ be the sequence in $\mathbb{R}$ defined inductively by:

$$\forall n \in \mathbb{N}^*: x_{n+1} = \frac {x_n + \frac a {x_n}} 2$$

Then $$x_n \to \sqrt a$$ as $$n \to \infty$$.

Proof
It is clear that $$x_n > 0$$ (if necessary, this can be proved by induction on $$n$$).

Also:

$$ $$ $$

This is a Quadratic Equation in $$x_n$$.

We know that this equation must have a real solution, because $$x_n$$ originally comes from the iterative process defined above.

Thus its discriminant $$b^2 - 4 a c \ge 0$$, where:


 * $$a = 1$$
 * $$b = -2 x_{n+1}$$
 * $$c = a$$

Thus $$x_{n+1}^2 \ge a$$.

Since $$x_{n+1}$$ it follows that $$x_{n+1} \ge \sqrt a$$ for $$n \ge 1$$.

Thus $$x_n \ge \sqrt a$$ for $$n \ge 2$$.

Now, consider $$x_n - x_{n+1}$$.

$$ $$ $$

So, providing we ignore the first term (about which we can state nothing), the sequence $$\left \langle {x_n} \right \rangle$$ is decreasing and bounded below by $$\sqrt a$$.

Thus by Monotone Bounded Sequence Converges, $$x_n \to l$$ as $$n \to \infty$$, where $$l \ge \sqrt a$$.

Now we want to find exactly what that value of $$l$$ actually is.

By Limit of a Subsequence we also have $$x_{n+1} \to l$$ as $$n \to \infty$$.

But $$x_{n+1} = \frac {x_n + \frac a {x_n}} 2$$.

Because $$l \ge \sqrt a$$ it follows that $$l \ne 0$$.

So by the Combination Theorem for Sequences, $$x_{n+1} = \frac {x_n + \frac a {x_n}} 2 \to \frac {l + \frac a l} 2$$ as $$n \to \infty$$.

Since a Sequence has One Limit at Most, that means $$l = \frac {l + \frac a l} 2$$ and so $$l^2 = a$$.

Thus $$l = \pm \sqrt a$$ and as $$l \ge +\sqrt a$$ it follows that $$l = +\sqrt a$$.

Hence the result.

Further Investigation
This time, we suppose $$a > 0$$ but make no statement about $$x_1$$.

Again, we specify $$x_{n+1} = \frac {x_n + \frac a {x_n}} 2$$.

Now:

$$ $$ $$ $$ $$ $$ $$

If we now assume that $$x_1 \ge \sqrt a$$, then it follows (as above) that $$x_n \ge \sqrt a$$.

So:

$$ $$ $$

If $$\left|{y}\right| < 1$$, then $$y^n \to 0$$ as $$n \to \infty$$ from ... (work in progress).

So, by Limit of a Subsequence, $$y^{2^n} \to 0$$ as $$n \to \infty$$.

Thus we see that $$x_n \to \sqrt a$$ as $$n \to \infty$$ provided that $$\frac {x_1 - \sqrt a} {2 \sqrt a} < 1$$, that is, that $$\sqrt a \le x_1 < 3 \sqrt a$$.

We have assumed (above) that $$x_1 \ge \sqrt a$$.

Now we have shown that $$x_n \to \sqrt a$$ as $$n \to \infty$$ provided that $$\sqrt a \le x_1 < 3 \sqrt a$$.

However, we have already shown that $$x_n \to \sqrt a$$ as long as $$x_1 \ge 0$$.

The advantage to this analysis is that this gives us an opportunity to determine how close $$x_n$$ gets to $$\sqrt a$$.