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

Solution of Equations by Numerical Methods

Extracts from this document...

Introduction

Damian Chandler 13MP        66555 0474        Maths Coursework

Solution of Equations by Numerical Methods

Change of Sign Method – Numerical Search

By using the computer program ‘Omnigraph’, I have produced this graph of the equation y = x^3-9x+7.

image00.png

The roots of this equation are shown where the line crosses the X-axis. I can now see that the three roots of this equation lie between (-4,-3), (0,1) and (2,3). Due to the fact that I have only declared the roots to be within wide margins, I will need to use a different method to acquire a more accurate answer. Instead of simply saying “the root is about -3.3”, I can use a method called decimal search to find an answer to any degree of accuracy that I decide.

Decimal search works by finding two points on the x-axis of a graph, where between, the sign of Y changes, i.e. where the line crosses the x-axis. Without the graph I would be able to work out that the sign changes three times, meaning there are three roots.

This table shows the values of Y when X is between -5 and 4:

X

-5

-4

-3

-2

-1

0

1

2

3

4

Y

- 73

-21

7

17

15

7

-1

-3

7

35

From this I can now work out a more accurate answer to where the roots lie. I now know that a root lies between (-4,-3).

...read more.

Middle

2

3

4

Y

-43.26

-10.58

4.096

6.776

3.456

0.136

2.816

17.496

It appears from the table that there is only a single root. However the curve of the equation looks like this:

image01.png

Rearranging into the Form x = g(x)

The Equation y = x^3-14x-4, can be written in many different ways. By rearranging the equation I could write it as: ‘14x = x^3-4’ or ‘x = (x^3-4)/14.

The graph of the equation y = x^3-14x-4 looks like this:

image02.png

If I now use the rearranged equation x = (x^3-4)/14 and draw the lines of both sides of the equation on a graph i.e. y = x and y = (x^3-4)/14, then the points where the two lines cross are the roots from the original equation.

This shows the graph of the two lines y=x and y = (x^3-4)/14.

image03.png

x

-4

-3

-2

-1

0

1

2

3

4

f(x)

-12

11

16

9

-4

-17

-24

-19

4

The roots lie between (-4,-3), (-1,0) and (3,4).

Comparing this to my graph, I can see that it looks correct.

For my equation y = x^3-14x-4, to find the root in the interval (-1,0), I start at X1 = 1.

Xn^3 – 4       = Xn+1

                                               14

X2 = (1^3 – 4)/14 = -0.21429

X3 = (-0.21429^3 – 4)/14 = -0.28642

X4 = (-0.28642^3 – 4)/14 = -0.28738

X5 = (-0.28738^3 – 4)/14 = -0.28741

X6 = (-0.28741^3 – 4)/14 = -0.28741

Root is -0.28741 to 5 d.p. This means the root is x = 0.28741 ± 0.000005

Failure of rearranging into the Form x = g(x)

...read more.

Conclusion

The rearranging into the form g(x) is a very easy method as it simply requires just putting the new answer into another simple formula. This is still a tedious method as it requires a lot of repetition.

Although the Newton-Raphson method can be slightly confusing and a mistake can be made easily, it is by far the quickest. The main problem occurring, by an estimate being made near a turning point that doesn’t occur close to the x-axis can be eliminated by using the decimal search method to find the rough points at which the roots cross the x-axis. The other problem that can cause this method to take a long time is by the original estimate being made not close to the root. This will result in it taking a long time to reach the root. This can however be eliminated by combining the decimal search method with Newton-Raphson. This will allow decimal search to find the rough points at which the roots occur, then allowing Newton-Raphson to close in on the root without suffering the repetition of decimal search.

Page  of

...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)

    Despite this, it is a very effective method as failure chances are relatively very low and the root can be found to many decimal places if the right software is used. With the use of excel, it can be really easy to work out the values of y when you sub in the x values.

  2. The open box problem

    So x =1/6L or x=L/6, and V=x(l-2x)^2. Method 2: Using Algebra Although I have already solved the squares problem, I will now attempt to solve it using algebra. To do this I will use the equation to find the volume and merge this with some other sort of expression to

  1. This coursework is about finding the roots of equations by numerical methods.

    If f(X)= x�-5x�+4.67x-0.418 then f'(x)= 3 x�-10x+4.67 Xn+1= Xn- (Xn �-5 Xn �+4.67 Xn -0.418) / (3 Xn �-10 Xn +4.67 x y 1.0000 1.9665 1.9665 0.5902 0.5902 13.2493 13.2493 15.1964 15.1964 17.1002 X=g(x) method Example: y=-3x�-x�+x+2 The solution is near to 1 Take x1 = 1 Roots are given by -3x�-x�+x+2=0 Rearranges to: 3x�= -x�+x+2 x� = 1/3(-x�+x+2)

  2. C3 Coursework: Numerical Methods

    y=3x3-11x+7, when differentiated, is. This means that the formula to solve this equation is: . I will now use this formula to try to find the root between 0 and 1. As with the Change of Sign method, I will use x=0 as my starting point.

  1. Numerical solutions of equations

    I choose my starting value of x (x1) to be about -1.3. x1 = -1.3 x2 = -4.22079 x3 = not defined x4 = Another example where the Newton-Raphson method would fail is with the function x3-5x+0.1=0. The graph of this, along with the calculations is shown below: f(x)

  2. C3 Numerical Solutions to Equations

    = -1.47*10^-6. f(- 2.72115744) = 8.29*10^-7. Therefore there is a root as the function is continuous. f(x)=0 can be rearranged into the form x=g(x)=(x4-2x�+5x�-7x-2)0.2 The x=g(x) iterations are applied to give the results: Therefore x = -2.72115745 � 0.000000005 f(- 2.72115746) = -1.47*10^-6. f(- 2.72115744)

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

    bound is 2.8 and the function of the upper bound is 1.74() and they have no sign change due to it has an asymptote between 1 and 2 therefore there is no sign change and this method will fail when it is not continuous function.

  2. Numerical solution of equations, Interval bisection---change of sign methods, Fixed point iteration ---the Newton-Raphson ...

    9 0.63508886 x=(x�+8)/13 n X 1 -2 =A3+1 =(B3^3+8)/13 =A4+1 =(B4^3+8)/13 =A5+1 =(B5^3+8)/13 =A6+1 =(B6^3+8)/13 =A7+1 =(B7^3+8)/13 =A8+1 =(B8^3+8)/13 =A9+1 =(B9^3+8)/13 =A10+1 =(B10^3+8)/13 Spreadsheet 3.4 Spreadsheet 3.5 Therefore we get the solution of 0.63509 with error bounds �0.000005, giving solution bounds of 0.635085 < x < 0.635095.

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