• 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

Der Handschuh.

Extracts from this document...

Introduction

Methods of Advanced Mathematics (C3) Coursework.

Task: Candidates will investigate the solution of equations using the following three methods:

  • Systematic search for change of sign using one of three methods: decimal search, bisection or linear interpolation.
  • Fixed point iteration using the Newton raphson method.
  • Fixed point iteration after rearranging the equation f(x) = 0 into the form x = g(x)

Change of Sign

This method is based on how the function has a change of sign either +ve to -ve or -ve to +ve on either side of a route when crossing the x-axis. This method relies on this fact to find the points between the positive and the negative value where there is another change from positive to negative. This can be done until a useful number of decimal places are found.

To investigate this method I plan to use the function f (x) = x^3-5x+1. When I plot this it shows:

image02.png

From the graph we can see that routes lie on the x-axis between the values -3 and -2, 0 and 1 and 1 and 2. If I consider the function between –3 and -2 I can see I change of sign on the function from –ve to +ve.

-6

-183

-5

-97

-4

-41

-3

-9

-2

5

-1

7

0

3

1

-1

To home in on a more accurate answer I need to investigate the decimal places between -3 and -2

x

f(x)

 -3

-9

-2.9

-6.889

-2.8

-4.952

-2.7

-3.183

-2.6

-1.576

-2.5

-0.125

-2.4

1.176

-2.3

2.333

-2.2

3.352

-2.1

4.239

-2

5

This method is repeated until a sufficient number of decimal places are achieved. In this investigation I think 5 decimal places will be enough

-2.5

-0.125

-2.49

0.011751

-2.48

0.147008

-2.47

0.280777

-2.46

0.413064

-2.45

0.543875

-2.44

0.673216

-2.43

0.801093

-2.42

0.927512

-2.41

1.052479

-2.4

1.176

-2.5

-0.125

-2.499

-0.11126

-2.498

-0.09753

-2.497

-0.08382

-2.496

-0.07012

-2.495

-0.05644

-2.494

-0.04277

-2.493

-0.02912

-2.492

-0.01548

-2.491

-0.00186

-2.49

0.011751

From this I can see that the route lies between -2.491 and -2.490±0.0005 and is found at 0.011751 ± 0.0000005.

This method works for most functions of x but can still cause anomalous results in the occasion of certain instances where the method will not work as there being two routes within an interval.

When I plot a graph with this characteristic both routes are clearly distinguishable visually, however when I find the solutions numerically it only shows one change of sign being the first one and the second is over looked.

image03.png

The routes are visible on the graph but the table below shows only the one change of sign in red.

x

f(x)

x

f(x)

-4

-301

-4

-301

-3.5

-197.875

-3

-121

-3

-121

-2

-31

-2.5

-66.625

-1

-1

-2

-31

0

-1

-1.5

-10.375

1

-1

-1

-1

2

29

-0.5

0.875

3

119

0

-1

4

299

0.5

-2.875

1

-1

1.5

8.375

2

29

2.5

64.625

3

119

3.5

195.875

4

299

...read more.

Middle

This was the beginning stages of this method and shows the change of signs I had to investigate. There were three routes as shown by the highlighted cells.

-5

-5624

-4

-1727

-3

-350

-2

-23

-1

4

0

1

1

-2

2

25

3

352

4

1729

5

5626

After taking an estimate of the roots to be –1.5, 0.5 and 1.5 I started to use the iterative formula. On the first route this idea becomes clear as we can see by the highlighted red cells that for the first value of the root drawn out by the equation becomes the starting point for the next set of figures to go through the iterative equation

Route 1

xn

f(xn)

f'(xn)

xn+1

-1.5

2.6875

16.875

-1.659259259

-1.65925926

-1.312787509

34.50075025

-1.621208275

-1.62120827

-0.093523598

29.65572012

-1.61805463

-1.61805463

-0.000604217

29.2730055

-1.61803399

-1.61803399

-2.57554E-08

29.27050994

-1.618033989

Route 2

0.5

0.4375

-3.125

0.64

0.64000000

-0.095971635

-4.4662784

0.618511945

0.61851195

-0.002042156

-4.274856611

0.618034232

0.61803423

-1.03759E-06

-4.270512042

0.618033989

0.61803399

-2.68452E-13

-4.270509831

0.618033989

Route 3

1.5

-0.6875

16.875

1.540740741

1.54074074

0.077437779

20.74492848

1.537007887

1.53700789

0.000695055

20.37311978

1.536973771

1.53697377

5.76884E-08

20.36973796

1.536973768

1.53697377

0

20.36973768

1.536973768

The three routes I have found using this method are all highlighted in green. They can also be seen on the below graphimage04.png

In this method I also looked at the error bounds of my solutions. I took the final figure up by 0.0000000005 and down by the same.

...read more.

Conclusion

The final method of rearrangement was relatively easy to use and there were no major calculations that needed to be done apart from the rearrangement at the beginning that did not prove to be too difficult. I did find however that it seemed to tend to the strongest route and would ignore the weaker routes when they are required and still tend to the stronger ones. Also as we already know when a route has a gradient lager than 1 or less than –1 the formula did not work because it couldn’t find the route and so failed. I found it the least successful of the methods although it could be used to find an accurate route quickly it often failed. The time in which it takes to use the method is greatly increased by the use of excel like both other methods, if it were to be used manually, again like both other methods, it would be very time consuming.

...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

    The Gradient Function

    5 star(s)

    By trying to find a general trend from this, I can conclude that when a, in axn, is constantly 1, and the values of n are whole numbers, the gradient function for this in general is: nxn-1 However, I will need to prove this using binomial expansion.

  2. Marked by a teacher

    Estimate a consumption function for the UK economy explaining the economic theory and statistical ...

    3 star(s)

    became bigger and the scale in Figure1 (b) followed the same. However, a fundamental assumption behind OLS is that the behavior of the residuals is random. The Figure 1(b) shows that the residuals are related to one another, therefore the absolute consumption equation not fits data well as the graph is pattern.

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

    = f(x1) [x-x1] This tangent cuts the x-axis at (x2, 0), so 0- f(x1) = f(x1) [x2-x1] Rearranging this gives The Newton-Raphson iterative formula is then formed On viewing the graph 2.1, the intervals in which the roots lie are [-3,-2], [-1, 0], [0, 1] respectively.

  2. The Gradient Fraction

    the -4 has been multiplied by +4. These results are not exactly accurate due to the graph scales, but they show an approximate of how close they are. The gradients obtained are all multiples of 4. The Increment Method The increment method is another method which calculates the gradient of a line.

  1. Methods for Advanced Mathematic

    x f(x) -3 -31 -2 -8 -1 5 0 2 1 1 2 44 Newton Raphson Method Newton Raphson Method is an iterative process. It is used for finding approximations to the roots of a real valued equation. Here is the procedure to perform the Newton Raphson Method: 1.

  2. Math Portfolio Type II - Applications of Sinusoidal Functions

    daylight and a smaller range between the smallest number of hours of daylight and greatest number of hours of daylight as you go south towards the equator. This relationship is illustrated by the differences in the parameters in the two equations, since a, which is the amplitude of the function, is higher in Toronto (3.255)

  1. Mathematical Investigation

    Continuation from Part I-III: For the sake of a clear and easier understanding of the sine functions, the reflections of the sine functions will be explained here. Sine functions of y=-sin(x) and y=sin (-x) are investigated: Figure #6: Graphs of the sine functions of y=sin (-x)

  2. Estimate a consumption function for the UK economy explaining the economic theory and statistical ...

    0.001826?t It is visibly obvious now that the equation is far more complex than the original Keynesian consumption function however whether it is a better prediction of consumption remains to be seen. Statistically and graphically this equation is the most accurate one with an R� value of 0.998, meaning that

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