• 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

    Estimate a consumption function for the UK economy explaining the economic theory and statistical ...

    3 star(s)

    0.000 0.9966 sigma 0.0246121 RSS 0.0321049445 R^2 0.996608 F(1,53) = 1.557e+004 [0.000]** log-likelihood 126.726 DW 0.348 no. of observations 55 no.of parameters 2 mean(LC) 12.6508 var(LC) 0.172083 LC = + 0.7504 + 0.9383*LY (SE) (0.0954) (0.00752) Figure 2 Table 3 EQ( 3)

  2. OCR MEI C3 Coursework - Numerical Methods

    Finding the root using Newton-Raphson f'(x)=5x4-4 xn F(xn) f'(xn) x1 -2 -21 76 x2 -1.72368 -5.32084 40.1368 x3 -1.59112 -0.83341 28.0463 x4 -1.56140 -0.03491 25.71855 x5 -1.56004 -7E-05 25.61534 x6 -1.56004 -2.8E-10 25.61513 -1.56004 (5d.p.) is therefore a root of f(x)=0.

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

    To conclude the results I have found, Newton-Raphson Method is the one has the fastest speed of convergence, which can be explained as in this method, finding a root by the intercepts of tangents to the curve with x-axis seems to be more efficient, by comparing to the Interval Bisection Method and the method of Rearranging the equation f(x)

  2. Numerical Methods coursework

    * There is a connection between the Trapezium Rule and the Midpoint Rule which can be used to shorten calculations: * Quick calculation of using and * All three rules simplify the working out on the spreadsheet due to less difficult formulae Formula Application I used the programme "Microsoft Excel 2003" to produce the spreadsheet.

  1. MEI numerical Methods

    and F=(K+2)) C= F/K Thus C = 0.678176364 Assuming c = common factor, then ? at k =4 x C will equal ? when K = 5. 0.197900255 x 0.678176364 = 0.134211275 0.134211275 = 0.165395943, therefore this isn't a geometric progression.

  2. Arctic Research (Maths Coursework)

    (resultant velocity) 2 R.V.2 = 115600 - 900 R.V.2 = 114700 R.V. = V114700 R.V. = 338.67 km/h Trigonometry to find the angle ? and direction Sin ? = opposite Hypotenuse Sin ? = 30 340 Sin ? = 0.088 ?

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

    0.001858218 1.27002 0.000648492 1.269531 1.27002 -0.000566192 0.000648492 1.269775 4.17701E-05 1.269531 1.269775 -0.000566192 4.17701E-05 1.269653 -0.000262056 1.269653 1.269775 -0.000262056 4.17701E-05 1.269714 -0.000110104 1.269714 1.269775 -0.000110104 4.17701E-05 1.269745 -3.41573E-05 1.269745 1.269775 -3.41573E-05 4.17701E-05 1.26976 3.80883E-06 1.269745 1.26976 -3.41573E-05 3.80883E-06 1.269753 -1.51736E-05 1.269753 1.26976 -1.51736E-05 3.80883E-06 1.269756 -5.68226E-06 1.269756 1.26976 -5.68226E-06 3.80883E-06 1.269758

  2. C3 COURSEWORK - comparing methods of solving functions

    This is because of the steep gradient where the graph crosses the x-axis, meaning that the resulting tangent is directed further away from the root.

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