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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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