OCR MEI C3 Coursework - Numerical Methods

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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]

Change of sign indicates root exists in interval [6.6,6.7]

This means that x=6.65±0.05

x=7 (0d.p.)


Change of sign indicates root in interval [6.67,6.68]

This means that x=6.675±0.005

x=6.7 (1d.p.)

Change of sign indicates root in interval [6.678,6.679]

This means that x=6.6785±0.0005

x=6.68 (2d.p.)

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), which shows that the roots cannot be found using a decimal search.

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Newton-Raphson

Finding a root

 

The graph of y=f(x):

Roots of f(x)=0 exist in the intervals [-2,-1] and [0,1].

f(x)=(x2+0.9x-2.52)-1+1

f’(x)=-(2x+0.9)(x2+0.9x-2.52)-2

We shall find a root by taking x1=1

f(0.862435)=0.000014

f(0.862445)=-0.000012

The change of sign indicates that the root is 0.86244±0.000005.

Graphical illustration

The following is a graphical illustration of the process for finding this root with x1=1:

Finding the remaining root

The other root can be found using x1=-2.

x=2.62488 (5d.p.)

Failure of ...

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