show that x_n converges to \sqrt{a}

**wopashui** · September 20th 2011, 06:54 PM

Fix a>0 and let $x_1>\sqrt{a}$ . For n >=1, define

$x_{n+1}$ = $1/2(x_n +a/x_n)$

Show that $(x_n)$ converges and that lim n--> infinity $x_n = \sqrt{a}$ .

Here is what I have got so far, $x_{n+1}$ = $1/2(x_n +a/x_n)$ $>= \sqrt{a}$ since $\sqrt{ab}<=(a+b)/2$

**mr fantastic** · September 20th 2011, 07:11 PM

Originally Posted by wopashui

Fix a>0 and let $x_1>\sqrt{a}$ . For n >=1, define

$x_{n+1}$ = $1/2(x_n +a/x_n)$

Show that $(x_n)$ converges and that lim n--> infinity $x_n = \sqrt{a}$ .

Here is what I have got so far, $x_{n+1}$ = $1/2(x_n +a/x_n)$ $>= \sqrt{a}$ since $\sqrt{ab}<=(a+b)/2$

1. Use induction to show that $x_n < x_{n=1}$ .

2. Let the limit be l. Solve $l = \frac{1}{2} \left(l + \frac{a}{l}\right)$ .

**wopashui** · September 21st 2011, 03:40 PM

Originally Posted by mr fantastic

1. Use induction to show that $x_n < x_{n=1}$ .

2. Let the limit be l. Solve $l = \frac{1}{2} \left(l + \frac{a}{l}\right)$ .

you meant to show $x_n > x_{n+1}$ , since the sequence has to be decreasing to get to $\sqrt{a}$

**mr fantastic** · September 21st 2011, 05:59 PM

Originally Posted by wopashui

you meant to show $x_n > x_{n+1}$ , since the sequence has to be decreasing to get to $\sqrt{a}$

Yes, typo on my part.

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