• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
Page
  1. 1
    1
  2. 2
    2
  3. 3
    3
  4. 4
    4
  5. 5
    5
  6. 6
    6
  7. 7
    7
  8. 8
    8
  9. 9
    9
  10. 10
    10
  11. 11
    11
  12. 12
    12

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.

This student written piece of work is one of many that can be found in our AS and A Level Core & Pure Mathematics section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related AS and A Level Core & Pure Mathematics essays

  1. Marked by a teacher

    C3 Coursework - different methods of solving equations.

    5 star(s)

    graph is wrong. The iterations are also wrong as they diverge away from the point as well. If you differentiate the y = g(x) function and substitute x for the approximation of the root. If that g'(x) < -1 or g'(x)

  2. C3 Coursework: Numerical Methods

    When differentiated, y=log(x+3)-x is This does not work as an equation so it is impossible to find the root of this equation using the Newton Raphson method. X=g(x) Method The x=g(x) method is another fixed point iteration method. This means that an estimate of the root is need as a starting point.

  1. MEI numerical Methods

    This means for the false position method to work, the two approximations of the root used in the formula, their functions would have to be opposites. Hence unlike the secant method, we would use the most recent approximation and the most recent opposite signed approximation's function.

  2. Arctic Research (Maths Coursework)

    order for the researchers to arrive at observation site C, their plane must leave the base camp at a bearing of 90o and must expect 0.16 hrs (0.16 hrs � 60 = 9.68 minutes) for the journey in which they will travel with a R.V.

  1. I am going to solve equations by using three different numerical methods in this ...

    =2*B4^3-3*B4^2-8*B4+23 =2*C4^3-3*C4^2-8*C4+23 =(B4+C4)/18 =2*F4^3-3*F4^2-8*F4+23 =IF(G4>0,B4,F20) =IF(G4>0,F4,C20) =2*B4^3-3*B4^2-8*B4+24 =2*C4^3-3*C4^2-8*C4+24 =(B4+C4)/19 =2*F4^3-3*F4^2-8*F4+24 =IF(G4>0,B4,F21) =IF(G4>0,F4,C21) =2*B4^3-3*B4^2-8*B4+25 =2*C4^3-3*C4^2-8*C4+25 =(B4+C4)/20 =2*F4^3-3*F4^2-8*F4+25 =IF(G4>0,B4,F22) =IF(G4>0,F4,C22) =2*B4^3-3*B4^2-8*B4+26 =2*C4^3-3*C4^2-8*C4+26 =(B4+C4)/21 =2*F4^3-3*F4^2-8*F4+26 =IF(G4>0,B4,F23) =IF(G4>0,F4,C23) =2*B4^3-3*B4^2-8*B4+27 =2*C4^3-3*C4^2-8*C4+27 =(B4+C4)/22 =2*F4^3-3*F4^2-8*F4+27 Bisection failure I am going to show the failure of bisection method.

  2. Solving Equations. Three numerical methods are discussed in this investigation. There are advantages and ...

    =B3^3-5*B3^2+4*B3+2 =IF(G2>0,D2,F2) =D3^3-5*D3^2+4*D3+2 3 =IF(G3>0,F3,B3) =B4^3-5*B4^2+4*B4+2 =IF(G3>0,D3,F3) =D4^3-5*D4^2+4*D4+2 4 =IF(G4>0,F4,B4) =B5^3-5*B5^2+4*B5+2 =IF(G4>0,D4,F4) =D5^3-5*D5^2+4*D5+2 5 =IF(G5>0,F5,B5) =B6^3-5*B6^2+4*B6+2 =IF(G5>0,D5,F5) =D6^3-5*D6^2+4*D6+2 c f(c) Error =(B2+D2)/2 =F2^3-5*F2^2+4*F2+2 =(B2+D2)/2 =(B3+D3)/2 =F3^3-5*F3^2+4*F3+2 =H2/2 =(B4+D4)/2 =F4^3-5*F4^2+4*F4+2 =H3/2 =(B5+D5)/2 =F5^3-5*F5^2+4*F5+2 =H4/2 =(B6+D6)/2 =F6^3-5*F6^2+4*F6+2 =H5/2 This is a sample I select form my actual table to show how a spread sheet can help.

  1. Solving Equations Using Numerical Methods

    Method 2: The Newton-Raphson Method Root ? Root ? For the second method, the Newton-Raphson Method, I will be using a different function. The equation I will be using is x3-7x2+x+3=0. This means that I will be drawing the function of f(x)=x3-7x2+x+3.Root ? Firstly, I started by entering the equation into Autograph to get a sketch of the graph.

  2. C3 COURSEWORK - comparing methods of solving functions

    However, if we do the integer search, we can only get 1 change of sign. The search will find the first root 0.61074687 in this interval [0, 1]. In order to find all the change of signs, we better draw a graph.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work