• 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. Marked by a teacher

    The Gradient Function

    5 star(s)

    I will need to try this, however, with 2 more cases, to find a respective pattern between all of them. With the next method, I shall use solely the table/increment method to find respective gradients. It is perhaps not a preferred method to algebraic proof.

  1. MEI numerical Methods

    Hence the equation, x +ktanx = 1, turns into, x + ktanx - 1 = 0, this is our function (f). As explained earlier the method of bisection requires an interval which supports the sign change argument. This means that the f(a) must equal negative and f(b) must be positive.

  2. Numerical solutions of equations

    = x3-5x+0.1 f'(x) = 3x2-5 The iterative formula for the Newton-Raphson method is: xn+1 = xn- f(xn) f'(xn) My iterative formula is: xn+1 = xn- x3-5x+0.1 3x2-5 I will take my starting value of x (x1) to be 1.1. x1 = 1.1 x2 = -1.870073 x3 = -2.400053 x4 =

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

    -0.32749 1.662 -0.04103 1.6662 -0.00419 1.2 0.112 1.63 -0.27216 1.663 -0.0324 1.6663 -0.00329 1.3 0.088 1.64 -0.20902 1.664 -0.02368 1.6664 -0.0024 1.4 -0.224 1.65 -0.13775 1.665 -0.01488 1.6665 -0.0015 1.5 -0.5 1.66 -0.05802 1.666 -0.00598 1.6666 -0.0006 1.6 -0.416 1.67 0.030502 1.667 0.003005 1.6667 0.0003 1.7 0.352 There appears to be only one root in the interval (1, 2)

  2. C3 Numerical Solutions to Equations

    f(x)=x5+x4-2x�+5x�-7x-2=0 and finding the root between -3 and -2 found above using the change of sign method Using the Newton-Raphson method with the starting point -3 gives 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 ...

    15 1.529297 8.43E-05 1.529419 -0.00044 1.529358 -0.00018 6.1E-05 16 1.529297 8.43E-05 1.529358 -0.00018 1.529327 -4.6E-05 3.05E-05 17 1.529297 8.43E-05 1.529327 -4.6E-05 1.529312 1.9E-05 1.53E-05 18 1.529312 1.9E-05 1.529327 -4.6E-05 1.52932 -1.4E-05 7.63E-06 19 1.529312 1.9E-05 1.52932 -1.4E-05 1.529316 2.7E-06 3.81E-06 20 1.529316 2.7E-06 1.52932 -1.4E-05 1.529318 -5.5E-06 1.91E-06 After 20th terms and approximation of the root to is found.

  2. Fractals. In order to create a fractal, you will need to be acquainted ...

    and as we defined before, the Mandelbrot set consists of the points that are connected in the Julia set. This simple recursive formula provides the creation of these types of complex fractals: Created at www.easyfractalgenerator.com with constant -0.80102 - 0.10772i The SierpiÅksi Triangle is a preeminent type of fractal that

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