• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
  1. 1
  2. 2
  3. 3
  4. 4
  5. 5
  6. 6
  7. 7
  8. 8
  9. 9
  10. 10
  11. 11
  12. 12
  13. 13
  14. 14
  15. 15
  16. 16

C3 Coursework: Numerical Methods

Extracts from this document...


C3 Coursework: Sapphire Mason-Brown C3 Coursework Numerical Methods The place in which the graph of a line crosses the x axis is known as the root of the equation. It is not always possible to find the solution of an equation by algebraic or analytical methods such as factorising. This applies to equations such as y=3x3-11x+7. To solve equations such as these, numerical methods such as change of sign, x=g(x) and Newton-Raphson can be used to give estimates of the roots. Change of Sign Method The Change of sign method is a method used to look for when a sequence of numbers in the boundary of a root change from negative the positive or vice versa. This change means that the root of the equation is somewhere between the interval where there is a change of sign. This is the graph of the equation y=3x3-11x+7 There are 3 roots to the equation y=3x3-11x+7, this is illustrated by the three intersections with the x axis. There appears to be a root between 0 and 1. By taking increments of 0.1 between 0 and 1 it will be possible to use decimal search to look for a change in sign. This will make it possible to find an approximation for to the root between 0 and 1. This table shows the results of the numbers in increments of 0.1 between 0 and 1. ...read more.


The x co-ordinates for the intersection between and y=x give a root of the equation y=x5+2x�-10x+6. This is the first iteration to find the root of the equation. The point x=1 is the starting point and it gives the estimate x=0.9. After 15 iterations, they eventually converge on the point x=0.675839. This means that the final estimate of the root for the curve, y=x5+2x�-10x+6, is 0.67584. I shall now attempt to use the Change of Sign method to establish error bounds. However, the x=g(x) method does not work for all equations. Some equations cause the iterations to diverge instead of converge. This is the graph of the equation of the curve, y=5x5+6x3+9x2-15x-2 I shall use the x=g(x) method to attempt to find root between 0.5 and 1. In order to find this root, it is necessary to find x=g(x). For this equation x=g(x) is. To find this root I shall use 1 as a starting point. When using the x=g(x) method, it is not possible to find the root of the equation y=2x5+6x3=15x+2. This is because the x=g(x) method uses the gradient of the points to find the equation of the root. As the gradient of the curve at the points used to find the root is greater than 1 the iterations diverge as opposed to converge. This means that the x=g(x) ...read more.


To conclude, I believe the method that was easiest to use with the software was the x=g(x). This is because, it required the least work and using the method was not as tedious. It was easy to notice any typing errors in my formula as the formula was not greatly complex. Although an element of manual work was required to use the x=g(x) method (re-arranging the equation) it was partially eliminated by the simplicity of typing in the x=g(x) equation and replicating it. The second easiest method to use with the software was the Decimal search method. Although this required slightly more work than the other two methods, the software made this additional work less time-consuming. As the formulae and intervals between the numbers could easily be replicated the decimal search method was fairly simple to use in conjunction with the software I used. The method that was the least easy to use with the software at hand was the Newton Raphson method. This is mainly because of the complexity of the formula. As a result of this complexity it was more difficult to notice any typing error made when inputting the formulae used. This method was also not as simple because the software does not calculate the whole formula. An element of manual work is still required to differentiate the formula and substitute the different areas of the equation into the Newton Raphson Formula. ?? ?? ?? ?? C3 Coursework: Numerical Methods Sapphire Mason-Brown 1 ...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)

    Despite this, it is a very effective method as failure chances are relatively very low and the root can be found to many decimal places if the right software is used. With the use of excel, it can be really easy to work out the values of y when you sub in the x values.

  2. Pure 2 coursework - Decimal Search Method

    1.4455 -0.00395 As the root is in the interval [1.445, 1.4455] and all the numbers within that range, when rounded to 3 decimal places, equals 1.445, then one of the roots of that function is 1.445 correct to 3 decimal places.

  1. Change of Sign Method.

    If y=f(x)=x3+2x2-4x-4.58, the roots of the equation can be found where f(x)=0 and so where x3+2x2-4x-4.58=0. Having illustrated the equation graphically, it is evident that one of the roots of the equation lies in the interval [-1,0]. To check that there is a sign change in this interval: x -1 0 f(x)

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

    As the figures show blow: 2 5.2 0 7.2 5.2 72.0672 1 38.688 72.0672 152833.4368 2 6318.898 152833.4368 1.42797E+15 3 2.8E+10 1.42797E+15 1.16471E+45 4 2.45E+30 1.16471E+45 6.32E+134 5 1.63E+90 6.32E+134 #NUM! 6 4.8E+269 #NUM! #NUM! 7 #NUM! #NUM! #NUM! 8 #NUM!

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

    5.48615 -0.139332173 -2.492246 3 -2.492246 -0.092318832 3x2+6x+0.5 4.180401 -0.022083725 -2.47016 4 -2.47016 -0.002172495 3x2+6x+0.5 3.984138 -0.000545286 -2.469617 5 -2.469617 -1.31124E-06 3x2+6x+0.5 3.979329 -3.29513E-07 -2.469617 6 -2.469617 -4.77618E-13 3x2+6x+0.5 3.979326 -1.20025E-13 -2.469617 7 -2.469617 0

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

    Graph 2.1--- In this case, I choose [0, 1] as the root presented graphically as follows. First of all, I am going to show the way to find the root in the interval [0, 1] Graph 2.2---Newton-Raphson for By using the help of autograph, in interval [0, 1], I get a root of 3 decimal places 0.908.

  1. Numerical integration coursework

    Ratio of differences The ?error multiplier? mentioned above is the same as the ratio of differences as shown by the image of my data below, the ratio of differences is tending towards 1/4 for the mid-point and trapezium rule and 1/16 for Simpson?s rule.

  2. Fractals. In order to create a fractal, you will need to be acquainted ...

    = 0 z3 = 02 + (-1) = -1 When the absolute value is applied to the resulting numbers, we see that that |z1|, |z2|, and |z3| are equal to 1, 0, and 1, respectively, and it stays within the boundary of 2, thus being in the Mandelbrot Set.

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