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Solving Equations using Numerical Methods.

Extracts from this document...

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

image00.png

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

...read more.

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

-0.000002

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

image01.png

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

...read more.

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

image02.png

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.        

...read more.

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