In Part I, we see that Iterative Substitution doesn't converge. The reason is that in each iteration step, \(x_{new}\) is not reduced; therefore, the subsequent \(x\) is simply the \(x\) before the last \(x_{new}\). So, what can we do to reduce \(x_{new}\)? How about we take a average of \(x_{new}\) and \(x_{old}\)?
$$\ x_{int} = \frac{b}{x_{old}}$$
$$\ x_{new} = \frac{1}{2}(x_{old}+x_{int})$$
$$\ x_{new} = \frac{1}{2}(x_{old}+x_{int})$$
where \(x_{int}\) is the intermediate solution. Combining these two into one equation, and using the iteration count, we come to the following equation.
$$\ x_{n+1} = \frac{1}{2}(x_{n}+\frac{b}{x_{n}})$$
This is also called the Babylonian method.
Example
Let's plug in \(b=9\) and \(x_{0}=2\) again into this equation, we see that \(x_{0}=2\), \(x_{1}=3.25\), \(x_{2}=3.009615\), \(x_{3}=3.000015\), and by the fifth iteration, \(x_{4}=3.000000\).
| Xn | |
|---|---|
| 0 | 2.000000 |
| 1 | 3.250000 |
| 2 | 3.009615 |
| 3 | 3.000015 |
| 4 | 3.000000 |
| 5 | 3.000000 |
See Next: Part III: Fixed Point Iteration
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