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# Solving Equations using Numerical Methods.

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Introduction

A2 Pure Maths Coursework

Solving Equations using Numerical Methods

There are three methods with which you can solve an equation.

1. Change of Sign Method
1. Newton-Raphson Method
1. Rearranging  the equation f(x)=0 in the form x= g(x)

Hardware and Software Used

For the coursework I have used a computer for attaining more accurate results and to avoid errors.

The software I have used are

• Microsoft Word
• GraphCalc (A software used for drawing graphs)

Change of Sign Method

x4-4x +2 We see that the roots lie in between [0, 1] and [1, 2]

We confirm this by looking for a change of sign

 x 0 1 2 f(x) 2 -1 10

Middle

0.00275

0.0024

0.00206

 x 0.5175 0.5176 0.5177 0.5178 0.5179 0.518 f(x) 0.00172 0.00137 0.00103 0.00068 0.00034 -2e-06

Root is in between [0.5179, 0.5180]

Failure Case

There are some cases where this method cannot be used to solve the equation.

Newton-Raphson Method f `(x) = ∫f(x)

= ∫ (x5 – 4x +1)

= 5x4 -4

Roots are in between [0,1] and [1, 2].

We look for the roots in [0,1]

xn+1 =        xn – f (xn)

f ` (xn)

xn+1=    xn - (xn5 – 4xn+1)

(5xn4 -4)

When x1= 0

x2 =   0 – ((0)5 - 4(0) + 1)

5 (0)4 - 4

=     0 - 1

-4

=  0.25

x3  =  0.25 - ((0.25)5- 4(0.25) + 1)

5(0.25) 4

= 0.25024533856722

x4  =  0.20524 - ((0.20524)5- 4(0.20524) + 1)

5(0.20524) 4

=   0.25024534093233

Root will be in between 0.250235  and 0.250245

Failure Case

Rearranging the Equation in the form  x=g(x)

We start this method by putting f(x) = 0.

x4

Conclusion

1/2

x

xn+1 = (4xn - 2)1/2

xn

x1 = 1 x1 = 2

x2  = 1.414x2 = 1.2247

x3  = 1.352x3 = 1.3901

x4 = 1.3654x4 = 1.3573

x5 = 1.3626x5 = 1.3643

x6= 1.3632x6 = 1.3628

x7 = 1.3630x7 = 1.3631

x8 = 1.3631                                x8 = 1.3631

x9 = 1.3631

x = x4 + 2

4

is equivalent to solving

y = x

y =   x4 + 2       is y = g(x)

4

The solution is where the line crosses the graph Rearrangement will work if its gradient g’(x) is less than the gradient of y = x otherwise it fails. The gradient of y = x is 1.

| g’(x) |  < 1  method works

| g’(x) |  > 1  method fails

g(x) = x4  +  2

4        4

g’(x) =  4x3

4

g’(x)        = x3

In [0, 1] , say  x= 0.5

g’(x)  = (0.5)3

= 0.625 < 1 . The method works

In [1, 2] , say x= 1.5

g’(x)  = (1.5) 3

= 5.0625 > 1. The method fails.

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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