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

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 image00.png, 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.

image01.jpg

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:

image07.png

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.

...read more.

Middle

image10.jpg

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 image11.png, and the upper value from image12.png).

The derivative of image09.png is image13.png.

image14.png

In the spreadsheet above, the first root for image15.png is found, to 4 significant figures, so image16.png. I will now try to find the second root for image15.png

image18.png

My initial value of image03.png 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].

image19.png

This third and final root can be expressed as image20.png.

image21.jpg

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.

image22.jpg

The graph shown above is image23.png. It can be seen that there is a root in the interval [2, 3]. However, if image24.png

...read more.

Conclusion

image52.png, image53.png and image54.png.

Comparison of Methods

To compare the methods, I am going to use the equation that I used for the fixed point iteration method, image30.png. 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.

image55.png

image56.png

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.

...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)

    I have noticed that the gradient is always equal to 9x�. I predict that when x = 5, the gradient will be equal to 225. 5 125 375 225 My theory was correct. Binomially expanding this, yet again, will allow me to get to the root of the equation, and see whether it is correct using algebra.

  2. Investigation of circumference ratio - finding the value of pi.

    24-sided: The triangle ACD in 24-sided is one twelfth of the . Known: Segment BC=1 Angle ABC=90 degree Angle ACB=7.5 degree Angle ACD=15 degree Segment AB = Segment BD So Segment BC=the radius of circle So Then we use the area formula of triangle: As we known, the area of triangle ACD is one twenty-fourth of area of equilateral polygon.

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

    -0.32 1.753E-06 6.10352E-05 15 -0.320251 1.75303E-06 -0.32019 -6.13511E-07 -0.32 5.6981E-07 3.05176E-05 16 -0.320221 5.69814E-07 -0.32019 -6.13511E-07 -0.32 -2.183E-08 1.52588E-05 17 -0.320221 5.69814E-07 -0.32021 -2.18343E-08 -0.32 2.7399E-07 7.62939E-06 18 -0.320213 2.73993E-07 -0.32021 -2.18343E-08 -0.32 1.2608E-07 3.8147E-06 19 -0.32021 1.2608E-07 -0.32021 -2.18343E-08 -0.32 5.2123E-08 1.90735E-06 20 -0.32021 5.21233E-08 -0.32021 -2.18343E-08 -0.32 1.5145E-08 9.53674E-07 Spreadsheet 1.8 Graph1.9--- bisection of y = (x+0.3)

  2. Different methods of solving equations compared. From the Excel tables of each method, we ...

    Use Autograph that follows the same steps as Equation 1 we can only get one of the three roots, which is shown below graphically. Using Excel, if we start the interval with x=0 and x=1 and follow the same steps as done in Equation 1, we end up with only one root.

  1. The method I am going to use to solve x&amp;amp;#8722;3x-1=0 is the Change ...

    0.4364621 x6 = V[(1-3(x5)^5)/5] = 0.4364590 x7 = V[(1-3(x6)^5)/5] = 0.4364594 x8 = V[(1-3(x7)^5)/5] = 0.4364594 I can see convergence from x4 , and there is no change between x7 and x8 for this number of decimal places. I know that this method has worked successfully to find a root.

  2. Finding the root of an equation

    The next part of the investigation is establishing where, between -0.2606 and -0.2605, a change of sign occurs. X Y 0.2606 0.000355557 -0.26059 0.000310821 -0.26058 0.000266086 -0.26057 0.000221351 -0.26056 0.000176616 -0.26055 0.000131882 -0.26054 0.000087148 -0.26053 0.000042414 -0.26052 -0.000002319558 -0.26051 -0.000047053 -0.2605 -0.000091785 There is a change of sign and therefore a root between -0.26053 and -0.26052.

  1. The Gradient Fraction

    128.96 - 128 = 0.96 4.01 - 4 = 0.01 = 96 'y=x4' solved by the 'Increment' method x -3 -2 -1 0 1 2 3 x4 81 16 1 0 1 16 81 y 81 16 1 0 1 16 81 Results x Gradient 1 4.06 2 32.34 3

  2. maths pure

    = (x-1)(x+2)2 but since there is no change of sign, it will not be detected by the decimal search. 2. Newton-Raphson Method The Newton-Raphson Method states that if a is a first approximation to a root of f(x) = 0, a better approximation is, in general,???

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