# C3 Coursework

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• Created on: 19-03-12 10:25
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### Page 1

C3 Coursework

I have chosen the equation 2x3+2x2-3.6x+1=0. The function is f(x) =2x3+2x2-3.6x+1, illustrated as a
graph above. I shall attempt to solve this equation and thus find the roots of the function.

Change of sign

To find a root, I shall use the Decimal Search Method to find a…

### Page 2

There is a change of sign between x=-3 and x=-2. This is noted as [-3,-2] and suggests a root being
present between these values. Remember that a root will always have the value y=0, which is
between a positive value and a negative value.

I shall now continue using the…

### Page 3

The solution to this is to use smaller steps. This finds a change of sign in the region [0, 0.5]

However, on closer inspection you can see that there are not one, but two roots in this region.

Without the graph for analysis you would fail to see one of…

### Page 4

The Newton-Raphson MethodTM relies on the general iterative formula:

For this function, it becomes:

Starting at 2, the method quickly converges to the root

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### Page 5

The root is found to be between 2.0557 and 2.0556. This x value can be written as:
2.0556

The Newton-RaphsonTM method will not work if we start at x=0 as it encounters a division by 0 in its
process, when xn is…

### Page 6

Rearrangement of f(x) =0 to give x=g(x)

I will now carry out the Rearrangement of f(x)=0 to give x=g(x) on a new function and equation.

New Function:

F(x) =0.2x3+1.4x2+2.5x+2.5

New Equation:

0.2x3+1.4x2+2.5x+2.5=0

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### Page 7

3 2
Rearranging f(x) gives x = -2.5-0.2x -1.4x
2.5

This now provides our iterative formula:

-2.5-0.2(xn)3-1.4(xn)2
xn+1 = 2.5

This is called g(x) and is displayed as the blue curve:

The function y=x is also added to the graph. Starting at x=0, a line is drawn vertically to meet…

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### Page 9

After 17 iterations the method converges to -5.00005±0.000005.

However, if we choose a different rearrangement of f(x), we can get:

3
f(x)= -2.5-0.2x2-1.4x2
0.2

This graph is then compared with y=x and the method is carried out from y=0:

(Green line represents method working)

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### Page 10

Clearly, the method will continue to infinity, not providing a solution, even though the lines
y=x and y=g(x) cross.

The magnitude of g'(x) (the gradient of g(x)) must be less than 1 (the gradient of y=x), otherwise the
series will be divergent and never reach a solution.

Comparison of Methods…