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OCR MEI C3 Coursework - Numerical Methods

Extracts from this document...

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

image00.png

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]

image01.png

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±0.05image12.png

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.)

image13.png

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.)image14.png

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)

...read more.

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:

image04.png


Fixed point iteration with x=g(x)

Finding a root

f(x)=x5-4x+3

The graph of y=f(x):

image05.png

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.

image06.pngimage07.png

image08.png

The importance of the magnitude of g’(x)

For a root of f(x)

...read more.

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).

...read more.

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