• 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

Pure 2 coursework - Decimal Search Method

Extracts from this document...

Introduction

Pure 2 coursework

Decimal Search Method

In order to find the root of the function f(x)=3x3-7x2-11x+17, the decimal search method was used. This is the process whereby a table of values is constructed that show whether the value of f(x) with certain values of x is positive or negative. By zooming in on the places where the value of f(x) switches from positive to negative (or vice versa), values for the root of this function can be found.

x

f(x)

-4

-243

-3

-94

-2

-13

-1

18

0

17

1

2

2

-9

3

2

4

53

5

162

From this table, it is clear that there are roots where f(x)=0 when values of x lie between -2 and -1, 1 and 2, and 2 and 3. By concentrating on the interval of [1,2], the rot can be found to  a greater level of accuracy.

x

f(x)

1

2

1.1

0.423

1.2

-1.096

1.3

-2.539

1.4

-3.888

1.5

-5.125

1.6

-6.232

1.7

-7.191

1.8

-7.984

1.9

-8.593

2

-9

It is now clear that the root lies somewhere inside the interval of [1.1, 1.2]

x

f(x)

1.1

0.423

1.11

0.268193

1.12

0.113984

1.13

-0.03961

1.14

-0.19257

1.15

-0.34488

1.16

-0.49651

1.17

-0.64746

1.18

-0.7977

1.19

-0.94722

1.2

-1.096

The root is in the interval [1.12, 1.13]

x

f(x)

1.12

0.113984

1.121

0.098597

1.122

0.083216

1.123

0.067841

1.124

0.052472

1.125

0.037109

1.126

0.021753

1.127

0.006403

1.128

-0.00894

1.129

-0.02428

1.13

-0.03961

The root is somewhere inside the interval [1.127, 1.128]. It is required that the root be found to 3 decimal places. Therefore it is necessary to know the value of f(x) when x=1.1275

x

f(x)

1.1275

-0.00127

...read more.

Middle

1.44528

1.44528

1.44528

1.44528

1.44528

1.44528

1.44528

1.44528

1.44528

1.44528

1.44528

1.44528

1.44528

1.44528

1.44528

1.44528

1.44528

This table shows that the function has a root of 1.44528 ± 0.000005

x

xn+1

5.00000

6.07407

6.07407

5.76048

5.76048

5.72064

5.72064

5.72001

5.72001

5.72001

5.72001

5.72001

5.72001

5.72001

5.72001

5.72001

5.72001

5.72001

5.72001

5.72001

5.72001

5.72001

This table shows that the function has a root of 5.72001 ± 0.000005

Failure of Newton-Raphson Method

The Newton-Raphson method can fail to find the root of a function, as in the following example. The function is f(x)=20-((x-3)2+0.05)-1

It is clear from simply looking at the function that one root is going to be x=3. However, if the Newton-Raphson method is used, the following data is obtained

x

xn+1

3.4000

2.5600

2.5600

3.6320

3.6320

0.7936

0.7936

109.3000

As can be seen on the diagram below, these figures are clearly divergent, which means that xn+1 is further from the root than xn, and so are no use in finding the root of the function.

image00.png

Fixed Point Iteration

This method to find the roots of a function is done by rearranging the function f(x) into the form x=g(x). This can be explained with an example. If the expression xn+1=0.8+2.6/xn is used to create a large table of values for xn+1, it is found that after x50 and x51 are both 2.061324773. This means that xn+1 converges on this figure. If we let the limit L=2.061324773, then L satisfies the equation L=0.8+2.6/L and therefore L2-0.8L-2.6=0 and therefore L is a root of f(x). The function that I will be finding the roots of is f(x)=x3-4x2-7x+11. In order to find the roots, the equation f(x)=0 must be rearranged into the form x=g(x). Therefore, g(x) = (x3-4x2+11)/7

Now, xn+1=xn3-4xn2+11  is used to generate a table of values for xn, the results

  7

are as follows:

x

xn+1

2

0.428571

0.428571

1.477718

1.477718

0.784603

0.784603

1.288657

1.288657

0.928207

0.928207

1.193349

1.193349

1.000443

1.000443

1.142541

1.142541

1.038553

1.038553

1.115115

1.115115

1.058957

1.058957

1.100278

1.100278

1.069938

1.069938

1.092252

1.092252

1.075859

1.075859

1.087912

1.087912

1.079056

1.079056

1.085566

1.085566

1.080782

1.080782

1.084299

1.084299

1.081714

1.081714

1.083614

1.083614

1.082218

1.082218

1.083244

1.083244

1.08249

1.08249

1.083044

1.083044

1.082637

1.082637

1.082936

1.082936

1.082716

1.082716

1.082878

1.082878

1.082759

1.082759

1.082846

1.082846

1.082782

1.082782

1.082829

1.082829

1.082795

1.082795

1.08282

1.08282

1.082801

1.082801

1.082815

1.082815

1.082805

1.082805

1.082812

1.082812

1.082807

1.082807

1.082811

1.082811

1.082808

1.082808

1.08281

1.08281

1.082809

1.082809

1.08281

1.08281

1.082809

1.082809

1.082809

1.082809

1.082809

...read more.

Conclusion

Of the three methods, it is clear that the method that is the slowest to converge on the root is the decimal search method. The quickest to converge is the Newton-Raphson method, closely followed by the fixed point iteration. However, although both iteration methods can find the root correct to several decimal places very quickly, there is more initial work in order to start using these methods than there is for the decimal search. Formulae must be rearranged or new formulae constructed in order to begin the search. Also, a lot of the speed of the two iteration methods is due to the ability of a computer to quickly do repeat calculations many times over. They are well suited to the use of computers whereas the decimal search method is relatively slow even with the use of computers, as it would take a human to spot the change in sign that determines the next set of calculations.

Finally, there is no reason why the decimal search method could fail when used in conjunction with an graph plotted accurately by calculator or computer to determine the rough approximations of the roots, whereas with the other two methods there are functions where it is impossible to obtain an estimate of the roots.

Paul McKay        Page         09 May 2007

...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)

    of the chord PQ also becomes closer to the actual gradient of PQ. Here, he tends to 0, so therefore 2x + h = 2x + 0 = 2x. Therefore the gradient is 2x. According to the previous values used in the increment method, x� is equal to 2x, so

  2. Marked by a teacher

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

    3 star(s)

    graph are relate to another, as they display a recurrent cyclical pattern. Even the Figure 2(a) shows that the predict value fits the actual value well. Different way to measure permanent income will lead to different equation. If the permanent income has been treated as the lagged income, the following

  1. Methods of Advanced Mathematics (C3) Coursework.

    The change of sign method was the simplest available to me and was very quick and easy to use. Unlike the two other methods there were no formula required. Each time I repeated the method the solution became more accurate and the route's accuracy increased to another significant figure was found.

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

    -1.328125 0.625 0.1191406 0.125 4 0.625 0.11914063 0.75 -1.328125 0.6875 -0.612549 0.0625 5 0.625 0.11914063 0.6875 -0.6125488 0.65625 -0.248627 0.03125 6 0.625 0.11914063 0.65625 -0.2486267 0.640625 -0.065212 0.015625 7 0.625 0.11914063 0.640625 -0.0652122 0.6328125 0.0268483 0.0078125 8 0.6328125 0.02684832 0.640625 -0.0652122 0.63671875 -0.019211 0.00390625 9 0.6328125 0.02684832 0.6367188 -0.0192111 0.634765625

  1. Arctic Research (Maths Coursework)

    In my investigation I will also calculate the resultant velocity, journey time and bearing of travel, for each observation site A - H. I will make separate calculations for the departing and returning journeys, as this will allow me to compare how they are affected by the wind.

  2. Experimentally calculating the wavelength of an He-Ne laser by means of diffraction gratings

    = (dx)/(mL). To calculate whether this hypothesis is reached, it is necessary to calculate the percent error for the calculated wavelengths for each fringe in each diffraction grating: 600 lines/mm 1st Fringe Percent Error = abs((theoretical value - experimental value)/theoretical value)100% Percent Error = abs((632.8 - 788)/632.8)100% Percent Error �

  1. Decimal Search.

    0 24.7 0.1 18.709 0.2 12.844 0.3 7.429 0.4 3.004 0.5 0.325 0.6 0.364 0.7 4.309 0.8 13.564 0.9 29.749 1 54.7 This would indicate that the graph does not cross the axis and therefore has no roots, but if the graph is sketched: This method would therefore appear ineffective.

  2. C3 COURSEWORK - comparing methods of solving functions

    roots and therefore, Newton Raphson is the best method to solve the equation. If you only had a graphic calculator By using graphic calculator, we can draw the graph out first, which may let us see if there is going to be a clear change of sign.

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