# In this investigation I am going to investigate three methods of finding the roots to equations and then compare them. I will be using those techniques fully. The three methods that I am going to examine are:

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Introduction

Zeshan Amir 13C (MEI Mathematics)

Mei Mathematics Coursework

Aim:

In this investigation I am going to investigate three methods of finding the roots to equations and then compare them. I will be using those techniques fully. The three methods that I am going to examine are:

-Decimal Search/Change of Sign

-The Newton Raphson Method

-The Rearrangement Method

Initial Exploration of Methods:

At the beginning of this project i did not have a deep understanding of the methods I was using however In with the help of trial and error with the methods i was able to velop equations and got an idea of there shape, making it easier for me to create equations to my specifications.

Decimal Search/Change of Sign Method:

In the piece of course work we are asked to find as well as not find the roots of equations that we have chosen. Roots are values that can be given to x so that f(x) is equal to zero. This can be displayed graphically by plotting the values of x against values of f(x). This will produce a line that will be continuous. If the function has roots (not all functions have roots e.g. f(x)=x2+3)

Middle

-0.7164

-0.7164

-0.66686

-0.66686

-0.6653

-0.6653

-0.66529

-0.66529

-0.66529

-0.66529

-0.66529

-0.66529

-0.66529

-0.66529

-0.66529

-0.66529

-0.66529

The root is therefore found to be -0.66529 ± 0.000005, as the calculations are only accurate as required to 5 decimal places.

The finding of the roots has been illustrated in the diagram below:

Further roots can also be found, by adapting the same process but with a different initial xn.

AS i have been asked to find all the roots all the other roots have been presented below:

x | xn+1 |

1.00000 | 1.53846 |

1.53846 | 1.44711 |

1.44711 | 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 the 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 there is no use in finding the root of the function.

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 i will be explaining 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 |

Conclusion

Conclusion:

In Conclusion, there is no reason why the decimal search method could fail when used in conjunction with a graph plotted accurately using a calculator or computer to determine the rough approximations of the roots before actually obtaining the root, while with the other two methods some functions will be present where it will be impossible to obtain even an estimate of the roots.

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