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# Finding one root of an equation by a change of sign method, to 4 figures.

Extracts from this document...

Introduction

Halley Porkess        Pure 2 Coursework        08/05/2007

## Finding one root of an equation by a change of sign method, to 4 figures

Using the bisection method, I am going to find one root for the equation , the graph of which is shown below, where one division on the x-axis represents 1, and one division on the y-axis represents 14.

From the graph it can be seen that there is a root between 1 and 2, it is this root that I will try to find using the bisection method:

In the spreadsheet above, n is the iteration number. an is the lower bound, bn is the upper bound, and xn the bisection of the interval [1,2] and f(xn) is the value of y, for that particular value for x. If y > 0 then an remains the same, and the last value of xn becomes bn. If however, y < 0 then bn remains the same, and an becomes the value of xn in the previous iteration. f(an) and f(bn) are included to give an idea of how close to the root an and bn are.

Middle

It can be seen that there are roots in the intervals [-1, 0], and [1, 2] (twice, therefore, the lower value will be found by coming from , and the upper value from ).

The derivative of  is .

In the spreadsheet above, the first root for  is found, to 4 significant figures, so . I will now try to find the second root for

My initial value of  was 3, to avoid confusion. The second root in the interval [1, 2] is 1.9805 ± 0.0005.

I will now find the third root, which is in the interval [-1, 0].

This third and final root can be expressed as .

The graph above is the third root, zoomed in upon, to show the method. The green lines represents x0, the initial value, in this case –1. The purple line is the tangent to the curve at x0. Where the tangents meets the x-axis, this is x1. This happens again, from the blue line, to the pink line, and so on, until the root is reached.

The graph shown above is . It can be seen that there is a root in the interval [2, 3]. However, if

Conclusion

,  and .

Comparison of Methods

To compare the methods, I am going to use the equation that I used for the fixed point iteration method, . To find the root (to 4 figures) using the fixed point iteration method 25 iterations were required. Below are the spreadsheets of the bisection method and the Newton-Raphson method respectively, each finding the same root.

Out of these methods, the Newton- Raphson method converged on the root the fastest, with only 3 iterations required. The second fastest was the bisection method, with 14 iterations, and the slowest was fixed point iteration, which required 25 iterations.

With a computer spreadsheet, all of the methods are easy to apply, but they all have failings, that there are some equations that can not be solved with that particular method. All of the methods require a starting value that is close to the root, so a graph is required for each method, again with appropriate software this is easy.

Without a computer, the bisection method generally takes the most time, due to the calculations involved. The fixed point iteration may be slow, and the Newton- Raphson method requires the ability to differentiate.

This student written piece of work is one of many that can be found in our AS and A Level Core & Pure Mathematics section.

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1.1054687500 1.1035156250 -0.0275378227 0.0164708495 -0.0055966303 1.1035156250 1.1054687500 1.1044921875 -0.0055966303 0.0164708495 0.0054213097 1.1035156250 1.1044921875 1.1040039063 -0.0055966303 0.0054213097 -0.0000916085 1.1040039063 1.1044921875 1.1042480469 -0.0000916085 0.0054213097 0.0026638633 1.1040039063 1.1042480469 1.1041259766 -0.0000916085 0.0026638633 0.0012858806 1.1040039063 1.1041259766 1.1040649414 -0.0000916085 0.0012858806 0.0005970744 1.1040039063 1.1040649414 1.1040344238 -0.0000916085 0.0005970744 0.0002527175 1.1040039063 1.1040344238 1.1040191650 -0.0000916085 0.0002527175 0.0000805507 1.1040039063 1.1040191650 1.1040115356

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