Examining, analysing and comparing three different ways in which to find the roots to an equation.

Authors Avatar
Advanced Mathematics (C3) Coursework

Introduction:

For my investigation, I will be examining, analysing and comparing three different ways in which to find the roots to an equation. It will include the "Change of Sign Method", the "Newton-Raphson Method" and finally the method of rearranging "f(x) = 0" into the form "x = g(x)". I will be finding roots of equations using the methods, and hence compare the merits and flaws of the methods with each other. I will analyse which is the best in terms of factors such as a speed of convergence and ease of use with available software and hardware.

Change of Sign Method:

This method finds a root to an equation by looking at when values of f(x) change sign from positive to negative or vice versa. This works because when the value of f(x) changes its sign, it must have passed through the x axis, and thus the root is somewhere between the two values that changed. The technique of the method can be done in many ways such as by interval bisection or linear interpolation. However, I will be using the decimal search method.

With decimal search, you first draw the graph of the curve that is to be investigated, and look for where the curve passes through the x axis, as these points are where the roots are. But as you cannot guess exactly where the roots are, you take intervals: the two points surrounding where the curve passes through the x axis. These points are taken as whole numbers. The next step is then to investigate where exactly between this interval, the value of f(x) changes to positive, or negative. To do this you go one decimal point further between these values. For example if the root was between 3 and 4, you would now investigate every 0.1 value, such as 3.1, 3.2, 3.3 and so on.

To investigate these values you put them into the formula to receive a value of f(x), and you can thus see where the change of sign occurs. When this has been found, it will be again between two values, so again you adjust the decimal places by one, and you are now investigating to 2 decimal places, such as 3.22 and 3.23. You keep going further into more decimal places, in order to increase the accuracy of root. For my investigation I will be going to 3 decimal places.

To investigate this method I will be using this equation:

f(x) = x³-15x-11

I used Autograph in order to plot the graph, and this is shown on the following page.

As the graph shows, there are several roots. I have chosen to investigate the root in the interval of [4, 5]. I will then begin by using 1 decimal place, and I put the values back into the equation in order to get the values of f(x).

I calculated each of the f(x) values in order to see where the change in sign was, as shown in the results below:

4

4.1

4.2

4.3

4.4

4.5

4.6

4.7

4.8

4.9

5

-7.00

-3.58

0.09

4.01

8.18

2.63

7.34

22.32

27.59

33.15

39.00

As the graph and the table show, there is a change of sign between 4.1 and 4.2, and thus the new interval was [4.1, 4.2]. I then took this a stage further, and did the decimal search to 2 decimal places. The results were:

4.1

4.11

4.12

4.13

4.14

4.15

4.16

4.17

4.18

4.19

4.2

-3.58

-3.22

-2.87

-2.51

-2.14

-1.78

-1.41

-1.04

-0.67

-0.29

0.09

The result here was an interval between [4.19, 4.2], and so I then went further, to 3 decimal places.

4.19

4.191

4.192

4.193

4.194

4.195

4.196

4.197

4.198

4.199

4.2

-0.29

-0.25

-0.21

-0.18

-0.14

-0.10

-0.06

-0.03

0.01

0.05

0.09

Here the interval was between [4.197, 4.198] but in order to decide on a root with a 3 decimal place accuracy, I must go to 4 decimals and see towards which value the root lies closest to:
Join now!


4.197

4.1971

4.1972

4.1973

4.1974

4.1975

4.1976

4.1977

4.1978

4.1979

4.198

-0.0256

-0.0219

-0.0181

-0.0143

-0.0105

-0.0067

-0.0029

0.0009

0.0046

0.0084

0.012

As this shows, the root lies between the interval [4.1976, 4.1977] and this means that the root is closer towards 4.198 than 4.197, and so the solution of the answer is:

4.198 to 3 decimal places with an error bound of + or - 0.005

Failure Case for Change of Sign Method:
...

This is a preview of the whole essay