# Fixed point iteration: Rearrangement method explained

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Introduction

Maths C3 coursework

Fixed point iteration: Rearrangement method

Another numerical method is called the rearrangement method. As the name states it requires you to rearrange the original function. From that point you can tabulate and calculate your results. Rearranging the original function is key to this otherwise you cannot repeat the number of iterations you do.

The function I will be using is: f (x) = x3+2x2-9x-11:

This image (left) shows that there are 3 roots to this equation. This can be deduced because it has 3 points of intersection with the ‘x’ axis. To find one of these roots I will use the rearrangement method. This I need to rearrange f(x) into a form so that x = g(x). To find a root with this new function I need to also have the values for y = x.

g(x) = 3√ (9x+11-2x2). The steps I followed to derive this were:

f(x) = x3 +2x2 – 9x – 11

Middle

-0.0613

8

-3.72855

-3.69288

-0.0357

9

-3.69288

-3.67197

-0.0209

10

-3.67197

-3.65967

-0.0123

11

-3.65967

-3.65241

-0.0073

12

-3.65241

-3.64811

-0.0043

13

-3.64811

-3.64558

-0.0025

14

-3.64558

-3.64407

-0.0015

15

-3.64407

-3.64318

-0.0009

16

-3.64318

-3.64266

-0.0005

17

-3.64266

-3.64234

-0.0003

18

-3.64234

-3.64216

-0.0002

19

-3.64216

-3.64205

-0.0001

20

-3.64205

-3.64199

-0.0001

21

Conclusion

Again the generic formula I will use is: xn+1 = g (xn)

Iterations | x | g(x) | ∆x |

1 | -5 | -9.555555556 | 4.5556 |

2 | -9.55556 | -77.87639079 | 68.3208 |

3 | -77.8764 | -51131.22153 | ######## |

4 | -51131.2 | -1.48525E+13 | ######## |

5 | -1.5E+13 | -3.64045E+38 | ######## |

6 | -3.6E+38 | -5.3607E+114 | ######## |

7 | -5E+114 | #NUM! | #NUM! |

From this table I can see that ∆x iscontinually increasing. This shows that the difference between x and g(x) is getting larger. Hence the values are diverging. From this I can deduce that the root cannot be obtained with this new form of g(x). Again the magnitude of g’(x) is key. If I look at the graph I can see why there is no convergence: This shows that at the point of intersection the gradient of g(x) is greater than the gradient of y = x. since the gradient of y = x is always 1 it means that the gradient of g(x) at the point of intersection is greater than 1. This means that if I was to continue iterating with my calculations I would get diverging values:

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