• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
Page
  1. 1
    1
  2. 2
    2
  3. 3
    3
  4. 4
    4
  5. 5
    5
  6. 6
    6
  7. 7
    7
  8. 8
    8
  9. 9
    9
  10. 10
    10
  11. 11
    11
  12. 12
    12

Methods of Advanced Mathematics (C3) Coursework.

Extracts from this document...

Introduction

Methods of Advanced Mathematics (C3) Coursework. Task: Candidates will investigate the solution of equations using the following three methods: * Systematic search for change of sign using one of three methods: decimal search, bisection or linear interpolation. * Fixed point iteration using the Newton raphson method. * Fixed point iteration after rearranging the equation f(x) = 0 into the form x = g(x) Change of Sign This method is based on how the function has a change of sign either +ve to -ve or -ve to +ve on either side of a route when crossing the x-axis. This method relies on this fact to find the points between the positive and the negative value where there is another change from positive to negative. This can be done until a useful number of decimal places are found. To investigate this method I plan to use the function f (x) = x^3-5x+1. When I plot this it shows: From the graph we can see that routes lie on the x-axis between the values -3 and -2, 0 and 1 and 1 and 2. If I consider the function between -3 and -2 I can see I change of sign on the function from -ve to +ve. ...read more.

Middle

All of them showing a change of signs meaning that the error bounds have been established to a suitable degree of accuracy. Like change of sign method it is also possible for Newton-Raphson to fail. To illustrate this I am going to use the equation f(x)=2x^3-5x^2+2. -3 -97 -2 -34 -1 -5 0 2 1 -1 2 -2 3 11 Again I started the method by looking for a change of sign. I decided to try and get a solution for the route between 0 and 1, so I placed my first guess at 0.1. I then put this through the iterative formula. xn f(x) f'(x) (xn)+1 0.1 1.952 -0.94 2.176596 2.176596 -1.0643 6.659457 2.336414 2.336414 0.21402 9.388835 2.313618 2.313618 0.004662 8.980798 2.313099 2.313099 2.39E-06 8.971577 2.313099 2.313099 6.32E-13 8.971573 2.313099 2.313099 0 8.971573 2.313099 2.313099 0 8.971573 2.313099 However this seems to have picked up a different solution. From the change of sign we can see there is one between 2 and 3 so it therefore must be this. It can also be seen on the graph below. It happens because the tangent is diverged away from the suspected route on the first iteration but then homes in on another route as shown below. ...read more.

Conclusion

One disadvantage of the method is that an approximate value of the route needs to be known, so a graph must first be plotted. The convergence time to the routes is relatively fast but it does require a great deal of work. If this method was to be done manually it would take a great deal of time more so than if it is done on excel due to the benefits of excel I have already mentioned. The final method of rearrangement was relatively easy to use and there were no major calculations that needed to be done apart from the rearrangement at the beginning that did not prove to be too difficult. I did find however that it seemed to tend to the strongest route and would ignore the weaker routes when they are required and still tend to the stronger ones. Also as we already know when a route has a gradient lager than 1 or less than -1 the formula did not work because it couldn't find the route and so failed. I found it the least successful of the methods although it could be used to find an accurate route quickly it often failed. The time in which it takes to use the method is greatly increased by the use of excel like both other methods, if it were to be used manually, again like both other methods, it would be very time consuming. ...read more.

The above preview is unformatted text

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

    C3 Coursework - different methods of solving equations.

    5 star(s)

    formula in to work out the new value of x and using it for the next iteration by simply dragging down the formula as many cells as you wish (each cell = an iteration). This makes it very easy to work out the repetitive part of the method and makes the speed of convergence much faster.

  2. Marked by a teacher

    The Gradient Function

    5 star(s)

    0.5 2 1.414214 0.353535 3 1.732051 0.2887 4 2 0.25 So far, I cannot conclude anything from these data since the numbers are not whole and therefore hard to work with. However, this evidence will help support the formula which I will attempt to create by using binomial expansion.

  1. I am going to solve equations by using three different numerical methods in this ...

    And also I am impressed how they work out the roots. However, there are differences in the speed of convergence and the ease of use. Therefore, I am going to compare three methods how they work. Afterwards, I am going to compare the speed of convergence and ease of use with available hardware and software.

  2. Solving Equations. Three numerical methods are discussed in this investigation. There are advantages and ...

    In this case it is -0.625 which means the root can be found and the sequence converges. Can other arrangement work? There are many ways to rearrange, and are only one way to rearrange this equation, but can other rearrangement find the same target root?

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

    (x-1) (x+4) - 4, which can be expressed as f(x) = x�-13x+8=0 From the graph shown above, the intervals lie in the [-4,-3], [0, 1], [3, 4] Graph 3.1 For example, f(x) = 0 can be arranged into: A. B. Here the b)

  2. OCR MEI C3 Coursework - Numerical Methods

    x f(x) -1.60 -1.08576 -1.59 -0.80215 -1.58 -0.52658 -1.57 -0.2589 -1.56 0.001042 Change of sign indicates root exists in interval [-1.57,-1.56] x=-1.565�0.005 x=-1.6 (1d.p.) x f(x) -1.570 -0.2589 -1.569 -0.23256 -1.568 -0.2063 -1.567 -0.18011 -1.566 -0.154 -1.565 -0.12797 -1.564 -0.10202 -1.563 -0.07614 -1.562 -0.05033 -1.561 -0.02461 -1.560 0.001042 Change of sign indicates root exists in interval [-1.561,-1.560] x=-1.5605�0.0005 x=-1.56 (2d.p.)

  1. Numerical Methods coursework

    to estimate the area. Dependent on the amount of rectangles (n) used the general formula is: 3. Simpson's Rule Simpson's Rule is a weighted average of and . The normal average is exactly in the middle of and . But in this rules the average is twice as close to as it is to .

  2. C3 COURSEWORK - comparing methods of solving functions

    0.9 8 0.7 -1.187 9 0.8 -0.568 10 0.81 -0.50026 11 0.82 -0.43143 12 0.83 -0.36151 13 0.84 -0.2905 14 0.85 -0.21838 The root lies between 0.87 and 0.88 15 0.86 -0.14514 16 0.87 -0.0708 17 0.871 -0.0633 18 0.872 -0.05579 19 0.873 -0.04827 20 0.874 -0.04074 21 0.875 -0.0332

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