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.

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