# Hero's Method/Proof 2

## Theorem

Let $a \in \R$ be a real number such that $a > 0$.

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

Let $\sequence {x_n}$ be the sequence in $\R$ defined recursively by:

$\forall n \in \N_{>0}: x_{n + 1} = \dfrac {x_n + \dfrac a {x_n} } 2$

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

## Proof

First we have the following lemmata:

### Lemma 1

$\forall n \in \N_{>0}: x_n > 0$

$\Box$

### Lemma 2

$\forall n \ge 2: x_n \ge \sqrt a$

$\Box$

Let $a > 0$.

We make no statement about $x_1$.

We specify that:

$x_{n + 1} = \dfrac {x_n + \dfrac a {x_n} } 2$

Now:

 $\ds x_{n + 1} - \sqrt a$ $=$ $\ds \frac {x_n + \dfrac a {x_n} } 2 - \sqrt a$ $\ds$ $=$ $\ds \frac 1 {2 x_n} \paren {x_n^2 - 2 x_n \sqrt a + a}$ $\ds$ $=$ $\ds \frac 1 {2 x_n} \paren {x_n - \sqrt a}^2$ $\ds$ $=$ $\ds \frac 1 {2 x_n} \paren {\dfrac {\paren {x_{n - 1} - \sqrt a}^2} {2 x_{n - 1} } }^2$ $\ds$ $=$ $\ds \frac 1 {2 x_n} \frac 1 {\paren {2 x_{n - 1} }^2} \paren {x_{n - 1} - \sqrt a}^4$ $\ds$ $=$ $\ds \frac 1 {2 x_n} \frac 1 {\paren {2 x_{n - 1} }^2} \frac 1 {\paren {2 x_{n - 2} }^4} \paren {x_{n - 2} - \sqrt a}^8$ $\ds$ $=$ $\ds \frac 1 {2 x_n} \frac 1 {\paren {2 x_{n - 1} }^2} \cdots \frac 1 {\paren {2 x_1}^{2 n - 1} } \paren {x_1 - \sqrt a}^{2 n}$

If we now assume that $x_1 \ge \sqrt a$, then it follows from Lemma 2 that $x_n \ge \sqrt a$.

So:

 $\ds \size {x_{n + 1} - \sqrt a}$ $\le$ $\ds \paren {\frac 1 {2 \sqrt a} }^{1 + 2 + 2^2 + \cdots + 2^{n - 1} } \paren {x_1 - \sqrt a}^{2 n}$ $\ds$ $=$ $\ds \paren {\frac 1 {2 \sqrt a} }^{\dfrac {2^n - 1} {2 - 1} } \paren {x_1 - \sqrt a}^{2 n}$ $\ds$ $=$ $\ds 2 \sqrt a \paren {\frac {x_1 - \sqrt a} {2 \sqrt a} }^{2 n}$

If $\size y < 1$, then $y^n \to 0$ as $n \to \infty$ from Sequence of Powers of Number less than One.

$y^{2^n} \to 0$ as $n \to \infty$

Thus we see that:

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

provided that:

$\dfrac {x_1 - \sqrt a} {2 \sqrt a} < 1$

that is, that:

$\sqrt a \le x_1 < 3 \sqrt a$

We 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 $\sqrt a$.

$\blacksquare$