The iteration
$x_{n+1} = \frac{1}{2} (x_{n} + \frac{2}{x_{n}}) , n \ge 0$
for a given $x_{0} \ne 0$ is an instance of
fixed point iteration for $f(x) = x^2 - 2$.
Newton's method for $f(x) = x^2 - 2$.
fixed point iteration for $f(x) = \frac{x^2 + 2}{2x}$.
Newton's method for $f(x) = x^2 + 2$.
It is well known that this is Newton's method for $f(x) = x^2 - 2$.
Also solving
$x = \frac{1}{2} (x + \frac{2}{x})$ [Equivalent of writing $f(x) = 0$ as $x= g(x)$]
we get
$x^2 - 2 = 0$
which shows the given iteration is a fixed point iteration for $f(x) = x^2 - 2$.
Thus the correct options are 2 & 1 but the correct options in the key are given to be 2 & 3.
This question appeared in CSIR Dec 2015. Please help!
Thanks in advance!
2) A newton method iteration scheme is $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$ , therefore taking $f(x) = x^2-2$ , we will get option 2 is correct and hence 1 & 4 are false.
3) A fixed point iteration scheme is given by $x_{n+1} = f(x_n)$, therefore taking $f(x) = \frac{x^2+2}{2x}$ we get option 3 is correct.