• 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. Marked by a teacher

    The Gradient Function

    5 star(s)

    80 3 81 405 4 256 1280 Points Gradient 1,1 5 2,32 80 3,243 405 4,(1024) 1280 The general pattern here between the two values of x and the gradient, the value of the gradient function here is 5x4, as seen when comparing the two smaller tables above.

  1. The open box problem

    I will now draw table sand graphs to find out the volume and the value for x. X 0.5 1 1.5 2 2.5 V 18 24 21 12 0 We can see here that the maximum volume is between 0.5 and 1.5 so I will draw another table focusing in on the data.

  2. Numerical integration can be described as set of algorithms for calculating the numerical value ...

    The above theory was vital in my investigation. Finding the value of M524288 under the curve is practically impossible, unless extrapolated estimates are used. This can be explained using table 1.0 below: Mn Values of M Difference Ratio of Differences M1 0.9582363067372210 0.0187590424355787 0.2634168677131520 M2 0.9394772643016420 0.0049414481996782 0.2550173797225570 M4 0.9345358161019640 0.0012601551719167 0.2514868156098100 M8 0.9332756609300470 0.0003169124113596 0.2503932714168220 M16 0.9329587485186880

  1. Solving Equations Using Numerical Methods

    The last number inputted was the same as the last number outputted which means that this is a root. F(x) = x3 -3x+0.5 The change of sign here confirms F(1.6415) = -0.00144 the accuracy of the root. F(1.6425) = 0.00365 The root is 1.6425 (to 3dp)

  2. Numerical Method (Maths Investigation)

    However, with or without the use of computer, Newton-Raphson Method seems to be the best method among the three method I have been used to compare here. It has an iteration formula: . I can find the gradient of the tangent to the curve at certain point by differentiate the equation of the curve and substitute the Xn into it.

  1. C3 Mei - Numerical Methods to solve equations

    0.3 -0.07327 0.31 -0.03943 0.32 -0.00519 0.33 0.029491 0.34 0.064634 0.35 0.100271 0.36 0.136435 0.37 0.173159 0.38 0.210478 0.39 0.248426 0.4 0.28704 [0,1] [0.3,0.4] [0.32,0.33] x f(x) 0.321 -0.00174 0.3211 -0.00139 0.3212 -0.00105 0.3213 -0.0007 0.3214 -0.00036 0.3215 -1.4E-05 0.3216 0.000331 0.3217 0.000677 0.3218 0.001022 0.3219 0.001367 0.322 0.001713 x f(x)

  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