• 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...

Introduction

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.

Problem

image05.png.image06.png

image15.png

image00.png

image22.png

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.

Strategy

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.

Middle

= 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.

Conclusion

 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. The open box problem

    X 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 6.5 7.0 7.5 V 98 169 216 242 250 243 224 196 162 125 88 54 26 7 0 The table shows that the maximum volume lies between 2.0 and 3.0 so I will construct another table to give a closer up view of this.

  2. 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.

  1. 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.

  2. 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.

  1. Best shape for gutter and further alegbra - using Excel to solve some mathematical ...

    By using algebra I can find ? at which the optimum capacity exists. Angle to give Optimum capacity; =w2 sin ? = 0 Cos ? = 0 ? = 90� Semi-Circular cross-section l = ? x r l = ?

  2. Mathematics Coursework - OCR A Level

    the rearrangement method does not work for this rearrangement of the equation as it cannot calculate the y-value for x=0.332207 and beyond. The above is the graph of y=(3x5-x+0.31)1/2 (blue line) and y=x (red line). The purple line shows the method diverging and then it stops when it cannot work

  1. C3 COURSEWORK - comparing methods of solving functions

    22 0.876 -0.02565 23 0.877 -0.01809 The root lies between 0.879 and 0.880 24 0.878 -0.01051 25 0.879 -0.00293 26 0.8791 -0.00217 27 0.8792 -0.00141 28 0.8793 -0.00065 29 0.87931 -0.00057 30 0.87932 -0.0005 31 0.87933 -0.00042 The root lies between 0.8793 and 0.8794 32 0.87934 -0.00034 33 0.87935 -0.00027

  2. Numerical integration coursework

    You then multiply this by the height, which you can get by doing the function of the mid-point of each rectangle, and finally add up the area found of them. Hence you get this formula: Mn= h (f(mâ)+f(mâ)+…+f(mn)) I used this formula and many algorithms when working on Excel to

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