Hero's Method/Lemma 2

Lemma for Hero's Method

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:

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

Proof

Lemma 1

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

$\Box$

We have:

\(\ds x_{n + 1}\)

\(=\)

\(\ds \frac {x_n + \dfrac a {x_n} } 2\)

\(\ds \leadstoandfrom \ \ \)

\(\ds 2 x_n x_{n + 1}\)

\(=\)

\(\ds x_n^2 + a\)

\(\ds \leadstoandfrom \ \ \)

\(\ds x_n^2 - 2 x_n x_{n + 1} + a\)

\(=\)

\(\ds 0\)

This is a quadratic equation in $x_n$.

We know that this equation must have a real solution with respect to $x_n$, because $x_n$ has been explicitly constructed by the iterative process.

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

$a = 1$

$b = -2 x_{n + 1}$

$c = a$

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

From Lemma 1:

$x_{n + 1} > 0$

It follows that:

$\forall n \ge 1: x_{n + 1} \ge \sqrt a$ for $n \ge 1$

Thus:

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

$\blacksquare$

Sources

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• 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 5$: Subsequences: $\S 5.5$: Example