# Quod Erat Demonstrandum

## 2009/07/03

### 計到即存在?

Filed under: HKALE,Pure Mathematics — johnmayhk @ 8:41 上午
Tags: ,

$\sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \dots}}}}$ = ?

$x = \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \dots}}}}$

$x^2 = 2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \dots}}}$

$x^2 = 2 + x$

$\sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \dots}}}} = 2$

$\sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \dots}}}}$

$\left \{ \begin{array}{ll} a_1 = 2\\a_{n+1} = \sqrt{2 + a_{n}} (n \in \mathbb{N})\end{array}\right.$

$\sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \dots}}}} = \lim_{n \rightarrow \infty}a_n$

$\left \{ \begin{array}{ll} b_1 = 1\\b_{n+1} = \sqrt{1 - b_n} (n \in \mathbb{N})\end{array}\right.$

$\sqrt{1 - \sqrt{1 - \sqrt{1 - \dots}}} = \lim_{n \rightarrow \infty}b_n$

$y = \sqrt{1 - \sqrt{1 - \sqrt{1 - \dots}}}$

$y^2 = 1 - \sqrt{1 - \sqrt{1 - \dots}}$

$y^2 = 1 - y$

「計出」 $y = \frac{-1 \pm \sqrt{5}}{2}$

$b_1 = 1$
$b_2 = \sqrt{1 - 1} = 0$
$b_3 = \sqrt{1 - 0} = 1$
$\dots$

## 1 則迴響 »

1. Hi,

An interesting analysis. Thanks!

Cheers,
Wen Shih

迴響 由 Wen Shih — 2009/07/04 @ 5:47 上午 | 回應