• 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...


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.


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.

...read more.



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.


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


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].


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


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

...read more.


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.



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)

    First I shall investigate fractional powers - I will try to investigate 2 or 3 of these to find a general rule. y = x1/2 x y second point x second point y gradient 3 1.73205080756888 3.1 1.760681686 0.286309 3 1.73205080756888 3.01 1.734935157 0.288435 3 1.73205080756888 3.001 1.732339459 0.288651 4

  2. Investigate the number of winning lines in the game of connect 4.

    3 10 17 24 first layer 7 7 7 second layer Like before the first to equations do not follow the pattern so they are excluded. Also like before the connects has 2 layers so we use this original equation. C=aH+b We use the same method as before. 3=a3+b (1)

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

    3 13.6838346 4 197.712422 5 594509.728 6 1.6163E+16 7 3.2483E+47 8 2.637E+141 9 #NUM! In spite of the effectiveness of this method, it fails occasionally. Besides the root we have worked out, there is two other interception points. To test it, I use 4 as an approximation in applying the

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

    Segment AC and Segment BC are radius r of circle. Angle ACD dependent by the n, n decide how many equilateral triangle in circle. So we use n to divide 360 degree (the degree of central angle degree), then we will got the degree of angle ACD.

  1. Analyse the use of three methods which are called the: change of sign, Newton-Raphson ...

    Here is an example of an equation that cannot be found using the systematic decimal search: f(x)=x3+5.283x2-0.2144x-22.56371908 As shown there is a definite root between 1 and 2 and a possible root between -3 and -4. Using systematic decimal search it is not clear whether a root lies between -3 and -4.

  2. Change of Sign Method.

    Fixed Point Iteration Method Let y=f(x)=x3+2x2-4x-4.58. The graph is shown below: The roots of the equation can be found where f(x)=0. From the graph, it is evident that the roots of the equation lie in the intervals [-3,-2],[-1,0] and [1,2].

  1. Although everyone who gambles at all probably tries to make a quick mental marginal ...

    8100 Total = 8.07333 Due to the inaccuracy, the first estimate was obviously wrong. Adjusting the numbers is necessary: P = + Table 3:Second Estimate for Case 1 Reciprocal of odds Actual Payout Estimated Payout Difference Squared 10068347520 $2 000 000.00 $2 057 737.48 3333616596.75 228826080 $1721.80 $1833.71 12523.85 5085024

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

    The table below shows part of the formulas used in the Excel. A f(a) b f(b) c f( c) 0 =0.2*(A2-4)*(A2+2)*(2*A2-1)+1 1 =0.2*(C2-4)*(C2+2)*(2*C2-1)+1 =(A2+C2)/2 =0.2*(E2-4)*(E2+2)*(2*E2-1)+1 =IF(F2>0,E2,A2) =0.2*(A3-4)*(A3+2)*(2*A3-1)+1 =IF(F2<0,E2,C2) =0.2*(C3-4)*(C3+2)*(2*C3-1)+1 =(A3+C3)/2 =0.2*(E3-4)*(E3+2)*(2*E3-1)+1 =IF(F3>0,E3,A3) =0.2*(A4-4)*(A4+2)*(2*A4-1)+1 =IF(F3<0,E3,C3) =0.2*(C4-4)*(C4+2)*(2*C4-1)+1 =(A4+C4)/2 =0.2*(E4-4)*(E4+2)*(2*E4-1)+1 Failure in this method This method cannot always be applied to every equation successfully.

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