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OCR MEI C3 Coursework - Numerical Methods
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Introduction
C3 Coursework Raoul Harris
C3 Coursework
Change of sign (decimal search)
Finding a root of an equation with graphical illustration
f(x)=x3-7x2+2x+1
This is graph of y=x3-7x2+2x+1
The graph shows that roots of f(x)=0 exist in the intervals [-1,0]; [0,1]; [6,7]
We shall test for a root in the interval [6,7]
x | f(x) |
6.0 | -23 |
6.1 | -20.289 |
6.2 | -17.352 |
6.3 | -14.183 |
6.4 | -10.776 |
6.5 | -7.125 |
6.6 | -3.224 |
6.7 | 0.933 |
Change of sign indicates root exists in interval [6.6,6.7]
This means that x=6.65±0.05
x=7 (0d.p.)
x | f(x) |
6.60 | -3.224 |
6.61 | -2.81992 |
6.62 | -2.41327 |
6.63 | -2.00405 |
6.64 | -1.59226 |
6.65 | -1.17787 |
6.66 | -0.7609 |
6.67 | -0.34134 |
6.68 | 0.080832 |
Change of sign indicates root in interval [6.67,6.68]
This means that x=6.675±0.005
x=6.7 (1d.p.)
x | f(x) |
6.670 | -0.34134 |
6.671 | -0.29924 |
6.672 | -0.25711 |
6.673 | -0.21496 |
6.674 | -0.17278 |
6.675 | -0.13058 |
6.676 | -0.08835 |
6.677 | -0.04609 |
6.678 | -0.00381 |
6.679 | 0.038498 |
Change of sign indicates root in interval [6.678,6.679]
This means that x=6.6785±0.0005
x=6.68 (2d.p.)
X | f(x) |
6.6780 | -0.00381 |
6.6781 | 0.000419 |
Change of sign indicates root in interval [6.6780,6.6781]
This means that x=6.67805±0.00005
x = 6.678 (3d.p.)
Failure of the decimal search
Two of the roots of f(x)=0 where f(x)=5x3-20x2+2x-0.05 could not be found with this method. This is because they are so close together that there is no sign change between f(0) and f(0.1). The graph on the next page shows the graph of y=f(x)
Middle
Finding the remaining root
The other root can be found using x1=-2.
xn | f(xn) | f'(xn) | |
x1 | -2 | -2.12500 | 30.27344 |
x2 | -1.92981 | -0.87732 | 10.43071 |
x3 | -1.84570 | -0.29110 | 4.65311 |
x4 | -1.78314 | -0.05792 | 2.98408 |
x5 | -1.76373 | -0.00339 | 2.64527 |
x6 | -1.76245 | -0.00001 | 2.62496 |
x7 | -1.76244 | 0.00000 | 2.62488 |
x8 | -1.76244 | 0.00000 | 2.62488 |
x=2.62488 (5d.p.)
Failure of the Newton-Raphson method
The method fails for the same function if x1=0
xn | f(xn) | f'(xn) | |
x1 | 0 | 0.60317 | -0.14172 |
x2 | 4.25600 | 1.05148 | -0.02495 |
x3 | 46.40573 | 1.00046 | -0.00002 |
x4 | Value could not be found | 1.00000 | 0.00000 |
The value diverges, despite the starting value being close to the root. This is because the tangent crosses the asymptote, as shown below:
Fixed point iteration with x=g(x)
Finding a root
f(x)=x5-4x+3
The graph of y=f(x):
We shall rearrange the equation into the form x=g(x) to find roots of f(x)=0.
0= x5-4x+3
x5=4x-3
x=(4x-3)1/5
xn+1=(4xn-3)(1/5)
We shall take a starting value of x1=-1.
x1 | -1 |
x2 | -1.47577 |
x3 | -1.54849 |
x4 | -1.55848 |
x5 | -1.55983 |
x6 | -1.56001 |
x7 | -1.56004 |
x8 | -1.56004 |
x=-1.56004 (5d.p.)
Graphical illustration
The graph below shows the lines y=x and y=g(x). The lines intersect at the roots of f(x)=0.
The importance of the magnitude of g’(x)
For a root of f(x)
Conclusion
The Newton-Raphson method is more difficult to automate because f'(x) must be found. Some computer programs, such as Autograph, can carry out the iterations for you, which, if they are available to you, can make it easier to use than a decimal search. Care must be taken when choosing a starting value and asymptotes can cause the method to fail. In some cases you may not be able to differentiate f(x).
Fixed point iteration with x=g(x) can be carried using programs like Autograph. It must still be rearranged manually though, and a large proportion of rearrangements fail. Because of this it is the most difficult method to use, especially if you do not have software to automate the iterations.
Overall the change of sign method is the easiest as it is purely numerical calculation. Newton-Raphson is the second simplest because it fails far less often than x=g(x).
This student written piece of work is one of many that can be found in our AS and A Level Core & Pure Mathematics section.
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