Find $1/\sqrt{a}$ with the Newton's method

Question

I have sequence defined by :

$$x_{n+1} = x_n - \frac{x_n^{-2} - a} {\frac {-2}{x_n^3}}$$

I need to find the sign of $x_{i+1} - x_i$ for $x_i ∈ ]0, \frac {1}{\sqrt{a}}[ $ and for now i only find this :

$ x_{i+1} - x_i $ $ = $ $ x_i - \frac{x_i^{-2} - a} {\frac {-2}{x_i^3}} - x_i $ $ = $ $ - \frac{x_i^{-2} - a} {\frac {-2}{x_i^3}} $ $ = $ $ \frac{x_i^{-2} - a} {\frac {2}{x_i^3}} $ $ = $ $ \frac{x_i - a{x_i^3}} {2} $ $ = $ $ \frac{1} {2} (x_i - a{x_i^3}) $

i don't know if i have to set $a$ to a value and if i can separate the interval.

my questions are, should i set $a$ ? and am i on the right way ?

Answer 1

Hint:

Notice that the iteration can be written

$$x_{n+1}=\frac12x_n(3-ax_n^2)$$

or

$$\sqrt a x_{n+1}=\frac12\sqrt ax_n(3-(\sqrt ax_n)^2)$$

and with $y:=\sqrt a x_n$,

$$y_{n+1}=\frac12y_n(3-y_n^2).$$

Hence it is enough to consider $a=1$.

Answer 2

In the recursive process

$x_{n+1}=x_{n}-\frac{\frac{1}{x_{n}^{2}}-a} {\frac{-2}{x_{n}^{3}}}$,

starting with $x_{0}=\frac{0.1}{\sqrt{a}}$,

the difference between two successive iterations is always positive and tends to zero;

after a few iterations we arrive at the solution

$x=\frac{1}{\sqrt{a}}$.