## Solving Equations By Iteration

Sometimes we cannot solve an equation exactly. If we cannot solve an equation exactly we may be able to solve it approximately using iteration.
Example: Solve the equation

$x^2-3x+1=0$

We can rearrange this equation.
$x^2-3x+1=0 \rightarrow x^2=3x-1 \rightarrow x=\sqrt{3x-1}$

From this we can get the iteration rule
$x_{n+1}= \sqrt{3x_n-1}$

If we take
$x_0=1$
then
$x_1=\sqrt{3 \times -1}=1.414$
to 3 decimal places. Continuing in this way, we have the table below.

 $n$ $x_n$ $x_{n+1}=\sqrt{3x_n-1}$ 0 1 1.414 1 1.414 1.801 2 1.801 2.098 3 2.098 2.301 4 2.301 2.430 5 2.430 2.508 6 2.508 2.594 7 2.594 2.581 8 2.581 2.597

We could go on. The iterates seem to be getting closer together. In fact because the original equation is a quadratic we can solve it exactly.

$x= \frac{3 \pm \sqrt{(-3)^2-4 \times 1 \times 1}}{2 \times 1} = \frac{3 \pm \sqrt{5}}{2} = 2.618$
or
$0.382$

The iteration is converging to the first of this. An iteration can only converge to one solution at a time, and iteration is not guaranteed. The solution that the the iterations converge to may be different if the original equation is rearranged to give a different iteration formula.