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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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