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

Numerical Methods Coursework

Extracts from this document...


Paul Koleoso        30083490

For this coursework, I am going to use knowledge of numerical methods to produce an approximation to an area which does not have an analytic solution.






I would be finding the approximation of the integral on the graph between the shaded portion of the graph above which is from 0 to 1 using numerical integration.

This problem was chosen because it cannot be integrated by any analytical method, therefore approximation method would be used. Due to this I suggest that this problem will be appropriate for numerical solution.


To solve this problem, I am going to use knowledge of numerical integration studied in the Numerical Methods textbook.

Numerical integration is a method used to approximate an area under the graph. According to syllabus on NM module, the approximate methods of definite integrals may be determined by numerical integration using;

  1. Mid – point rule
  2. Trapezium rule
  3. Simpson’s rule

Since there are lots mathematical functions which can not be integrated in real life, an alternative approach to these problems are to sub-divide the area under the graph into strips or shapes such as rectangles which approximately covers the area.

...read more.


= kimage08.png²   = image09.pngimage10.pngimage11.png

This means that halving h, or equivalently doubling n will reduce the error by a factor of image12.png= 0.25. image13.pngSince the absolute error is proportional to h². It is a second order method.

Also trapezium rule is similar to the mid- point rule. Error is also proportional to h². Therefore this also means halving h, or equivalently doubling n will reduce the error by a factor of image12.png= 0.25. image13.pngSince the absolute error is proportional to h². It is a second order method.

Viewing error in terms of differences and ratio differences

When the values for the number of strips double, the ratio difference between successive estimates is the same.

...read more.


 The problem specifiedimage05.png.image06.png can also be written as  image23.png ³.                            Microsoft Excel might find it complicated to solve square roots. So I suggest error might have occurred, therefore affecting the validity of my result.

For example;

image24.png)² - image25.pngimage26.pngimage25.png

If I had more time I could have increased the accuracy of my result by finding the real answer and not using extrapolated values to find “M16”, “T32” ,“M32”and “T64”, as this could have improved the validity of my result and therefore making my answer to the solution more accurate by producing the solution to a higher degree of significant figures. Also to improve the validity of my result Simpson’s rule could have been used. This is because in Simpson’s rule error is proportional to image27.png, which means it is a fourth order method  and also when you halve the width of the strip or you equivalently doubled the number of strips, this would reduced the absolute error with a scale factor of 0.0625.                    

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

    f(??? ?f(??) > the area of the underneath the curve in figure 1.0 is thus h f(??) + h f(??????h f(??) + h f(??? ???h f(??) > this can be generalised to Mn= h [f(??) + f(?????(??) + f(???????????f(?n)] Note that Mn is dependent on the number of rectangles.

  2. MEI numerical Methods

    To use fix point iteration we do rearrange f(x) = 0, to G(x) = x. We then do G(x1), and so on; this is called a recurrence relation. By doing this, most of the time the sequence gets closer to the final answer thus it converges.

  1. C3 Coursework: Numerical Methods

    The able shows that there is change of sign from 0.752 to 0.753. This means that the root of the equation must lie between 0.752 and 0.753. Our estimate of the root is with a maximum error of � 0.0005.

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

    * Distance between laser and diffraction grating Dependent: * Angle at which laser is diffracted at the first order fringe ? * Distance between the central bright beam and the first fringe x * Calculated wavelength of the He-Ne laser ?

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

    =B3^3-5*B3^2+4*B3+2 =IF(G2>0,D2,F2) =D3^3-5*D3^2+4*D3+2 3 =IF(G3>0,F3,B3) =B4^3-5*B4^2+4*B4+2 =IF(G3>0,D3,F3) =D4^3-5*D4^2+4*D4+2 4 =IF(G4>0,F4,B4) =B5^3-5*B5^2+4*B5+2 =IF(G4>0,D4,F4) =D5^3-5*D5^2+4*D5+2 5 =IF(G5>0,F5,B5) =B6^3-5*B6^2+4*B6+2 =IF(G5>0,D5,F5) =D6^3-5*D6^2+4*D6+2 c f(c) Error =(B2+D2)/2 =F2^3-5*F2^2+4*F2+2 =(B2+D2)/2 =(B3+D3)/2 =F3^3-5*F3^2+4*F3+2 =H2/2 =(B4+D4)/2 =F4^3-5*F4^2+4*F4+2 =H3/2 =(B5+D5)/2 =F5^3-5*F5^2+4*F5+2 =H4/2 =(B6+D6)/2 =F6^3-5*F6^2+4*F6+2 =H5/2 This is a sample I select form my actual table to show how a spread sheet can help.

  2. Mathematics Coursework - OCR A Level

    x-value x-y 0.2509 0.0008579 0.2492 0.001614 0.2523 0.00303 0.2466 0.005714 0.2572 0.01068 0.2369 0.02031 0.2744 0.03749 0.2006 0.07381 0.3322 0.1316 Overflow Overflow - it reached overflow because there are certain x values where there is no point of the graph present.

  1. C3 COURSEWORK - comparing methods of solving functions

    34 0.87936 -0.00019 The root lies between 0.87938 and 0.87939 So to 4 d.p. the root has value 0.8794. 35 0.87937 -0.00012 36 0.87938 -4.00E-05 Secondly, I will use Newton Raphson method to find the same root in the interval [0, 1]: First of all, I need to rearrange the equation x³+3x²–3=0 to the formulae of Newton Raphson ( )

  2. Numerical integration coursework

    make an approximation from Mâ up to Mââ, these algorithms made it much easier and quicker to come to a final answer thanks to the ability to just simply pull down the box with the algorithm in to transfer it to other boxes.

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