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

Solution of equations by Numerical Methods.

Extracts from this document...


Ashley Hilsdon Div 3

Pure 2 Coursework

Solution of equations by Numerical Methods

Method 1: The Change of Sign method

The simplest method for solving an f(x) function is to use a change of sign method; these include the methods of bisection, decimal search and linear interpolations.

Unfortunately, as well as being the simplest methods, they are also relatively cumbersome. The bisection was employed below for the function of f(x)=x5+3x2+x+2. This was also displayed graphically in the graph below.


The actual calculations for this method are summarised below in this table. I placed these with relevant diagrams

The error bounds for this result are conveniently provided by the use of this method, and are -1.485<x<-1.475. The maximum error is therefore ± 0.005.

However, this method is clearly not particularly easy to work with. In addition, as with many methods there are some functions for x, which do not work. An example is the function f(x)=x3-2x2-x. This is shown below.


...read more.


Using an estimate for (a.) as 0, the Newton-Raphson iterative formula was then used.

This next diagram gives some indication as to how the Newton- Raphson method actually works. What it basically does is to ‘slide’ the tangent across and is indeed sometimes known as ‘tangent sliding’.


For this method, the root can be calculated much more quickly then using the Change of Sign method. In accordance with the question the root was calculated to five significant figures, this means that the real value of x exists between 0.16655 and 0.16665, which give an error value of ± 0.00005.

This method, despite its speed in calculation a root, does not work for every function; this is illustrated below.

Method 3: Fixed Point Estimation- Rearranging the function

The third and final method for numerically solving a function of x is by rearranging an existing function of x and then using fixed-point iteration. This is illustrated for the function f(x)=x3+x-3 below.

Diagram of f(x)=x3+x-3

As is displayed graphically, the function with regards to (g)

...read more.


Rearranging the function is another useful technique, especially when a ‘cobweb’ effect is produced, as this gives natural error bounds for the root. However, it relies on an important rearrangement of the function, in conjunction with the plotting of at least two graphs. Once this has been done, the iterative formula is used, though within my calculations did not prove as fast as the Newton-Raphson method.

As I have already mentioned, the rearrangement method, relies more heavily on suitable graphical generating software or hardware; this is therefore a clear disadvantage that it faces. However, the change of sign method requires an interval for the root and the Newton-Raphson method requires an approximation to the result. Though, as this investigation has proved clear, regardless of the method, it has been necessary to have some idea of the function before attempting calculations. This is particularly important in the case of turning points and roots close to each other.

...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

    C3 Coursework - different methods of solving equations.

    5 star(s)

    is -1.961705 � 0.00005. Newton Raphson Method Finally, I am going to use Newton Raphson method to work out the root. I am going to use the inbuilt function in AutoGraph to work it out initially and then I will use Excel to work the root out to more decimal places.

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

    First of all, for one root of f(x) = 0. We then can draw a tangent to the curve y=f(x) at the point (x1, f(x1)). The point which the tangent cuts the x axis gives the next approximation for the root.

  1. The open box problem

    Using this equation I will create tables and graphs to find out the maximum volume and value of x, and to see if the pattern relates to the ratio 2:1. X 0.5 1 1.5 V 8 7 0 This table shows that the maximum volume lies between 0.5 and 1.5

  2. C3 Coursework: Numerical Methods

    This means that the x=g(x) method fails. Comparison of Methods 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.

  1. This coursework is about finding the roots of equations by numerical methods.

    If f(X)= x�-5x�+4.67x-0.418 then f'(x)= 3 x�-10x+4.67 Xn+1= Xn- (Xn �-5 Xn �+4.67 Xn -0.418) / (3 Xn �-10 Xn +4.67 x y 1.0000 1.9665 1.9665 0.5902 0.5902 13.2493 13.2493 15.1964 15.1964 17.1002 X=g(x) method Example: y=-3x�-x�+x+2 The solution is near to 1 Take x1 = 1 Roots are given by -3x�-x�+x+2=0 Rearranges to: 3x�= -x�+x+2 x� = 1/3(-x�+x+2)

  2. MEI numerical Methods

    the formula described above we would have an approximation of the root closer to the real root X(3), then I would use x(2) and x(3). The secant method always uses the two most recent approximations of the root. I will produce an answer to 9 significant figures by the use

  1. Experimentally calculating the wavelength of an He-Ne laser by means of diffraction gratings

    Switch on the laser. 5. Measure the distance between the central bright beam and the first order fringe to the right of it. Record. 6. Repeat step 5 for the next two orders of fringes to the right of the central bright beam.

  2. C3 COURSEWORK - comparing methods of solving functions

    0.8xn3+2 xn2-1 1 0 -1 2 -1 0.2 3 0.2 -0.9136 4 -0.9136 0.05929 5 0.05929 -0.9928 6 -0.9928 0.188464 7 0.188464 -0.92361 From the table above, we can see that the iteration is diverging away from the root in interval [0, -1] This is illustrated graphically below: X3 y= 0.8x³+2x²-x–1 X1 y= x X2 We can conclude that g'(x)

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