Find the limit of a sequence II

[lfdahl]

Let the sequence ${x_n}$ be defined by $x_0=2$ and $x_n=\frac{x_{n-1}}{2}+\frac{1}{x_{n-1}}$ for $n \ge 1$.
Find the limit.

 

[chisigma]

Let the sequence ${x_n}$ be defined by $x_0=2$ and $x_n=\frac{x_{n-1}}{2}+\frac{1}{x_{n-1}}$ for $n \ge 1$.
Find the limit.

[sp]The solving procedure for this type of problems is illustrated in...

http://mathhelpboards.com/discrete-mathematics-set-theory-logic-15/difference-equation-tutorial-draft-part-i-426.html#post2492

Writing the difference equation in the form...

$\displaystyle \Delta_{n} = a_{n+1} - a_{n} = \frac{1}{a_{n}} - \frac{a_{n}}{2} = f(a_{n})\ (1)$

... we discover that f(x) has two attractive fixed points in $x= \sqrt{2}$ and $x=- \sqrt{2}$. For both the fixed points the criteria for convergence are satisfied, so that for any $a_{0}< 0$ the sequence converges to $- \sqrt{2}$ and for any $a_{0}>0$ the sequence converges to $\sqrt{2}$...[/sp]

Kind regards

$\chi$ $\sigma$

 

[I like Serena]

Let the sequence ${x_n}$ be defined by $x_0=2$ and $x_n=\frac{x_{n-1}}{2}+\frac{1}{x_{n-1}}$ for $n \ge 1$.
Find the limit.

I recognize this one. ;)

Spoiler

Let $f(x)=x^2-2$.
Then the root is approximated using the Newton-Raphson method with:
$$x_n = x_{n-1} - \frac{f(x_{n-1})}{f'(x_{n-1})}
= x_{n-1} - \frac{x_{n-1}^2-2}{2x_{n-1}}
= \frac{x_{n-1}}{2}+\frac{1}{x_{n-1}}$$
In other words, the limit is $\sqrt 2$.

 

[chisigma]

I recognize this one. ;)

Spoiler

Let $f(x)=x^2-2$.
Then the root is approximated using the Newton-Raphson method with:
$$x_n = x_{n-1} - \frac{f(x_{n-1})}{f'(x_{n-1})}
= x_{n-1} - \frac{x_{n-1}^2-2}{2x_{n-1}}
= \frac{x_{n-1}}{2}+\frac{1}{x_{n-1}}$$
In other words, the limit is $\sqrt 2$.

In fact the equation $f(x)=x^{2} - 2 = 0$ has two solutions, $x=\sqrt{2}$ and $x=-\sqrt{2}$, and the Newton-Raphson method converges to the positive or negative root according to the initial value $x_{0}$...

Kind regards

$\chi$ $\sigma$

 

[lfdahl]

Good job, well done chisigma and I like Serena! Thankyou very much for two different and very clever solutions indeed!

Here is an alternative approach by other:

Spoiler

We assume that the limit \[L=\lim_{n \to \infty }\left \{ x_n \right \}\] exists. Then
$x_n=\frac{x_{n-1}}{2}+\frac{1}{x_{n-1}}$ for $n \ge 1 \Rightarrow $

\[x_nx_{n-1}=\frac{x_{n-1}^{2}}{2}+1 \: \: \: for \: \: \: n\geq 1 \Rightarrow \lim_{n \to \infty }\left \{ x_nx_{n-1} \right \}=\lim_{n \to \infty }\frac{x_{n-1}^{2}}{2}+1 \Rightarrow L^2 = \frac{L^2}{2}+1 \Rightarrow L^2 = 2\]Therefore $L = \sqrt{2}$ or $L = -\sqrt{2}$.$x_0 = 2 > 0$ and if $x_n > 0$ then $x_{n+1}= \frac{x_n}{2}+\frac{1}{x_n} > 0 + 0 = 0$. Hence, by induction $x_n > 0$ for all $n \ge 0$. It follows that $L=\lim_{n \to \infty }\left \{ x_n \right \} \geq 0$. We conclude that the limit is $\sqrt{2}$.