Newton iteration

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Let $\{a_n\}$ be defined by $a_1=2$ , and

$a_{n+1}=\frac {a_n}{2}+\frac {1}{a_n}$

Prove that

$\sqrt{2}<a_n<\sqrt{2}+\frac {1}{n}$

The left side is trivial, the right side requires more work.

**Posted:** Mon Sep 19, 2005 6:51 pm

Matnomi · Hodge Conjecture Offline Joined: 16 Aug 2005 Posts: 64

Left side is trivial indeed, and to prove right side we can use induction and left side.

... so we have

$a_n+1=\frac{a_n}{2}+\frac{1}{a_n}<\frac{1}{\sqrt{2}}+\frac{1}{2n}+\frac{1}{\sqrt{2}}<\sqrt{2}+\frac{1}{n+1}$

**Posted:** Mon Sep 19, 2005 9:36 pm