# Identifying the iteration Scheme

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

1. fixed point iteration for $f(x) = x^2 - 2$.

2. Newton's method for $f(x) = x^2 - 2$.

3. fixed point iteration for $f(x) = \frac{x^2 + 2}{2x}$.

4. 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.

• The key is correct. Why do you think 1 is correct? A fixed point iteration for 1 gives $x_{n+1} = x_n^2 -2$ which looks nothing like the above... – copper.hat May 31 '16 at 17:34
• Is it so? Actually I have never studied these iteration schemes but now when I am referring articles on fixed point iteration schemes everywhere they are asking to write the original equation $f(x) = 0$ in the form $x=g(x)$ for some function $g$ . By this definition it is clearly the iteration scheme for 1. Am I wrongly interpreting the definition? – Shivani Goel May 31 '16 at 17:38
• A fixed point iteration for $f$ is $x_{n+1} = f(x_n)$. A Newton iteration is $x_{n+1} = x_n - {f(x_n) \over f'(x_n)}$. Just evaluate these with the given $f$s and compare. – copper.hat May 31 '16 at 17:45
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.