# Fixed point iteration

1. Jan 24, 2013

### Huumah

1. The problem statement, all variables and given/known data
Apply fixed point iteration to find the solution of each equation to eight correct decimal places

x3=2x+2

3. The attempt at a solution
I have tried to rewrite the equation for in every possible way to solve for one x and pluggin in my guess( have tried -2,-1,0,1,2,3,4)

and finding x1 and then x2 and plugging them all inn seperatly.

But my answer switches from positive values to negative values and never seems to be converging to the answer which is 1.76929235

I can understand the sample problem but i'm stuck on this problem.

2. Jan 24, 2013

### Staff: Mentor

The formula you show above is incorrect. It should be
$$x_{n+1} = \frac{2(x_n + 1)}{x_n^2}$$
If you start with x0 = 1, what are the next three numbers you get?
If you start with x0 = 2, what are the next three numbers you get?

3. Jan 25, 2013

### jfgobin

Don't forget that there are conditions for a function to have a fixed point.

The expression I used is:

$x_{n+1}=\sqrt { \frac { 2\left ( x_n + 1 \right )}{x_n} }$

Try with $x_0 = 1$ and $x_0 = 2$ and let me know.

J.