can fixed point iteration be used on a quadratic equation ?
i was trying to follow this link ... but the examples are a bit complicated ...
http://www.ce.utexas.edu/prof/mckinney/ce311k/Overheads/14-NonlinearEquations_1_FixedPoint.pdf
It depends on the quadratic equation and how you rearrange it. For example suppose we take the equation
x
2 - 21 x + 20 = 0
We can re-arrange this to
x = \(\displaystyle \frac{x^2\, +\, 20}{21}\)
which will good for the root at x=1 but horrible for the root at x=20. For x=1, start with an initial guess of x=5 and you get
| 5.00000 |
| 2.14286 |
| 1.17104 |
| 1.01768 |
| 1.00170 |
| 1.00016 |
| 1.00002 |
| 1.00000 |
| 1.00000 |
but start with x=20.01 for the root at x=20 and get
| 20.01000 |
| 20.01905 |
| 20.03631 |
| 20.06922 |
| 20.13208 |
| 20.25240 |
| 20.48380 |
| 20.93267 |
| 21.81795 |
Doing a fixed point interation
x = g(x)
works only if \(\displaystyle |g'(x)|\, \lt\, 1\) in the vicinity of the root and you start close enough. If we instead rearrange the above equation as
x = \(\displaystyle 21\, -\, \frac{20}{x}\)
then this would word well for the root at x=20
| 100.00000 |
| 20.80000 |
| 20.03846 |
| 20.00192 |
| 20.00010 |
| 20.00000 |
| 20.00000 |
| 20.00000 |
| 20.00000 |
but not well at all for the root at x=1.
