Sequence $x_{n}=\frac{1}{2}(x_{n-1}+\frac{a}{x_{n-1}}).$

Question

Let $a$ and $x_{0}$ be positive numbers, and define the sequence $\{x_{n}\}$ recursively $$x_{n}=\frac{1}{2}(x_{n-1}+\frac{a}{x_{n-1}}).$$

How to prove that the sequence converges and how to find its limit ? Actually i am thinking to prove that the sequence is monotone and bounded then limit can be find by solving the equation $x^{2}-x-a=0.$ But monotone part is dependent on the real number $a.$ Please give me simplest way to handle the problem. Thanks a lot.

It is not monotone, but it turns around its limit, which is $\sqrt a$. You can show the subsequences of terms of odd rank and of terms of even rank are adjacent. You should study the defining function and use the mean value inequality to show convergence. — Bernard, Sep 17 '16 at 15:56
See here: http://math.stackexchange.com/questions/1541378/recursive-monotone-decreasing-sequence-proof-x-k-frac12-leftx-k-1?rq=1 — carmichael561, Sep 17 '16 at 15:56
@carmichael561 i looked at this solution i am trying different and simplest way... — neelkanth, Sep 17 '16 at 15:57
@Bernard defining function is $f(x)=\frac{1}{2}(x+\frac{a}{x})$ — neelkanth, Sep 17 '16 at 15:58
This does not mean a recursive sequence defined with the function is increasing. — Bernard, Sep 17 '16 at 17:02
A recursive sequence defined by an increasing function must be monotone... — neelkanth, Sep 18 '16 at 04:12

score 1 · Answer 1 · edited Sep 17 '16 at 20:55

1

$\displaystyle\frac12(x_n+\frac a{x_n})\geq\sqrt a$ by AM-GM

$x_n-x_{n-1}\displaystyle=\frac12(\frac{a-{x_{n-1}}^2}{x_{n-1}})<0$

Therefore $x_n$ monotonically decreases to $\sqrt a$.

edited Sep 17 '16 at 20:55

user84413

27,211

answered Sep 17 '16 at 19:06

Takahiro Waki

2,310

(Notice that it is possible to have $x_0<x_1$.) – user84413 Sep 17 '16 at 20:55
Provided you prove that there exists $N$ with $x_{N}^2>a$ as the base of the induction. If $x_0^2<a$ you're in trouble, aren't you? – egreg Sep 17 '16 at 21:07

Sequence $x_{n}=\frac{1}{2}(x_{n-1}+\frac{a}{x_{n-1}}).$

1 Answers1