Hero's Method/Proof 2

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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.

So, by Limit of Subsequence equals Limit of Real Sequence:

$y^{2^n} \to 0$ 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 $\sqrt a$.

$\blacksquare$


Sources

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