• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# 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 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.

## 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

# Related AS and A Level Core & Pure Mathematics essays

1.  ## C3 Coursework - different methods of solving equations.

5 star(s)

If that g'(x) < -1 or g'(x) > 1, then we know that the function will not converge to the root but will diverge away. e.g. Let us take a the last g(x) function possible from my f(x) function. x= x5 + 6x2 + 4 --> g(x)

2.  5 star(s)

+ an(n-1) (xn-2)h� + ... + ahn - axn) 2 h I can now multiply this through by h to get - anxn-1 + an(n-1) (xn-2)h + ... ahn-1. 2 Like all previous binomial expansion, every single value contains an h term, and continually limits to 0. The only term left over is naxn-1 My investigation is now at end, and perhaps the coursework is now complete.

1. ## Numerical solutions of equations

0=x5+4x2-2 f'(x) = 5x4+8x The iterative formula for the Newton-Raphson method is: xn+1=xn- f(xn) f'(xn) So my iterative formula is: xn+1=xn- xn5+4xn2-2 5xn4+8xn I think that a good starting value for x1 would be 1.2 x1 = 1.2 x2 = 0.887083 x3 = 0.720596 x4 = 0.682450 x5 = 0.680772

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

I am going to show one root graphically which lies between [-2,-1]. Afterwards, I will combine my equation with Now I am using the Excel to help me solve it accurately. Xr f(Xr) f'(Xr) X(r+1) -5 -463 298 -3.446308725 -3.446308725 -132.249178 137.4171434 -2.483916588 -2.483916588 -35.6643303 67.00593259 -1.95166028 -1.95166028 -8.12437758 37.61105474

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

key call 'bisection iteration' then we can enter upper and lower bound and this program will be able to help us to find the root automatically. However this method is not always working properly. There are two situations which lead to the failure of the method 1)

2. ## C3 COURSEWORK - comparing methods of solving functions

the root in the interval of [0, 1] is near to –0.19609 of 5 significant figures. n xn 1 1.00000 0.78571 2 0.78571 0.72753 3 0.72753 0.72300 4 0.72300 0.72297 5 0.72297 0.72297 6 0.72297 0.72297 Graphical Interpretation of the Newton Raphson method Example: y=2x y=0.5x³+1.5x²–x–0.25 Graph of y=f(x)

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. ## Evaluating Three Methods of Solving Equations.

in which every new value of x is getting closer to the root, until we eventually get to it Let me now illustrate the point by taking an example. My equation for this method is: f(x)=x3+8x2-13x-23 From the graph of this function I find that its root must lie between • Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to 