• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

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:

Extracts from this document...

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.

| Page

This student written piece of work is one of many that can be found in our AS and A Level Probability & Statistics 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

Related AS and A Level Probability & Statistics essays

1.  C3 Coursework - different methods of solving equations.

5 star(s)

-2.0 -2 -1.9 2.79901 -1.8 6.34432 -1.7 8.84143 -1.6 10.47424 -1.5 11.40625 -1.4 11.78176 -1.3 11.72707 -1.2 11.35168 -1.1 10.74949 -1.0 10 X f(x) -1.970 -0.41553 -1.969 -0.36493 -1.968 -0.31448 -1.967 -0.26416 -1.966 -0.21399 -1.965 -0.16395 -1.964 -0.11405 -1.963 -0.0643 -1.962 -0.01468 -1.961 0.034796 -1.960 0.084135 x f(x)

2.  5 star(s)

9 729 2916 General proof - (x+h)^4 - x^4 = x^4 + h^4 +4hx� + 6x�h� + 4xh� - x^4 = x + h - x h h (h� + 4x� +6x�h +4xh�) = h� + 4x� +6x�h +4xh� h H tends to 0 again here, and every term contains an h except for (4x�).

1. MEI numerical Methods

False position method: If we apply the first order of convergence principles to this method we get the following: Once again the last value isn't remotely close to a K, granted that the false position method's first order convergence has a smaller range of K than secant method did.

2. Sequences and series investigation

We can simplify this equation to: 6a + b = 6 My next calculation is below: N =3 _27a + 9b + 3c = 26 12a + 6b + 3c = 12 15a + 3b =14 (15a + 3b = 14)

1. Numerical solutions of equations

0.68071 -3.816012x10-4 0.68072 -3.164083x10-4 0.68073 -2.512140x10-4 0.68074 -1.860182x10-4 0.68075 -1.208210x10-4 0.68076 -5.562232x10-5 0.68077 9.577756x10-6 I can see that the change of sign is between x = 0.68076 and x = 0.68077 As there is a change of sign, I can finally say that the root of this function is in the interval [0.68076, 0.68077].

2. C3 COURSEWORK - comparing methods of solving functions

0.904741021 0.867836107 4 0.867836107 0.884382712 5 0.884382712 0.877172596 6 0.877172596 0.880355145 7 0.880355145 0.878958208 8 0.878958208 0.879572894 9 0.879572894 0.879302711 10 0.879302711 0.879421526 11 0.879421526 0.879369287 12 0.879369287 0.879392257 13 0.879392257 0.879382157 14 0.879382157 0.879386598 15 0.879386598 0.879384645 16 0.879384645 0.879385504 17 0.879385504 0.879385126 18 0.879385126 0.879385292 f(0.879385)

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

- 1303.22i 1358.8 z8 = -1.5504*106 + 1.0026*106 1.8463*106 As you can see, the fourth iteration does not stay inside the circle. Here is a better illustration of why this point is not in the Julia set: Here is another example of a point that is in the Julia set

2. Evaluating Three Methods of Solving Equations.

I will now try to find the same root by applying the other two methods. Decimal Search Method: From the graph of the function I know the root of the equation lies between 5<x<6. So I carried out the decimal search between 5 and 6, hoping to get the same root as before. • Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to 