• 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...


Numerical Methods Coursework

Numerical Integration

The Problem

Integration means finding the area underneath a particular region of a function. At my current knowledge of maths I am not able to integrate various functions. Therefore I am going to use knowledge of numerical methods to produce an approximation to an area which does not have an analytic solution.

In my coursework I will integrate the function:


The graph below shows the graph of the function.


The red arrow determines the region between 0.25 and 1.25, which then leads to the integral:


I can not solve this problem using the knowledge of C1 and C2, because I am not able to integrate cos(x) yet. Due to this I suggest that this problem will be appropriate for numerical solution.

The Approximation Rules

To solve this problem, I am going to use knowledge of numerical integration studied in the “Numerical Methods” textbook. The approximate methods of definite integrals may be determined by numerical integration using:

1. The Trapezium Rule:

The Trapezium Rule divides the area underneath the curve into trapeziums. We can then use the formula image45.png (where a, b are the bases and h is the height of the trapezium) to estimate the area.


...read more.








Error Analysis

Because all three rules are just estimates, there is always an error involved. In this section I will give an estimate of how much error there is in my solution of the problem.

Firstly I want to refer to my graph of my function on page 1. The section of the curve is concave that means that the Trapezium Rule gives an overestimate and the Midpoint Rule gives an underestimate. For that reason the estimates of the Trapezium Rule are decreasing and they will reach the solution from above. On the other side the Midpoint Rule estimates are increasing and they will reach the solution from below.

Error in the Trapezium Rule

In the Trapezium Rule the error is proportional to h2. We add a constant k so that:


In general, if image41.png is the absolute error in image33.png then there is a constant (k) so that:


Where h is the strip width corresponding to n strips.

So the Trapezium Rule with 2n strips has a strip width of image43.png, such that:


That shows that halving h or doubling n will reduce the error by a factor of 0.25.

Therefore the “error multiplier” is 0.25.

...read more.


This can be proved by looking on the diagram below.




The distance between image50.png and image51.png is 4 times smaller than the distance between image52.png and image53.png, where X represents the actual solution. To get the next distance between image51.png and image55.png we multiply the “multiplier” 0.25 with the distance between image50.png and image51.png.

This development leads to improved solution by extrapolation, as stated in the “Formula Application” section.

Error in my solution

The error gets smaller the more strips are used as shown above. Therefore:



T2, M2


T4, M4


T8, M8


T4096, M4096











 My solution to the integral image63.png is: 0.885909(6 decimal places)

The solution refers to:        T4096=        0.885909120684554

M4096=        0.885909049759261

S256=        0.885909075135303

The fact that the Trapezium Rule and the Midpoint Rule give the same solution till the sixth decimal place guaranties that my solution is valid. Simpson’s Rule, the weighted average, proves the solution as well.

Therefore: My solution is proved by 3 different rules.

I can say with guarantee that there is an error involved, because all three rules are just approximation rules. But I can say that my solution is not far away from the actual solution as the error for the rules with 4096 stripes is very small (actual value stated in the “Error Analysis” section).

...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. Methods of Advanced Mathematics (C3) Coursework.

    0.0000000005 as shown in red above. Rearranging method Iteration 1 - Working to find root [-2, -3] y = (5x - 2)^0(1/3) x y 5 2.802039 4 2.571282 3 2.289428 2 1.912931 1 1.259921 0 -1.44225 -1 -2 -2 -2.35133 -3 -2.62074 -4 -2.84387 -5 -3.03659 -1 -2 -2 -2.35133

  2. C3 Coursework: Numerical Methods

    This is illustrated by the graph below. This graph shows the equation of the line y=x�+10x�+4.8x+0.576 The roots appear to be between -1 and 0. By taking increments of 0.1 between -1 and 0 it will be possible to use decimal search to attempt to look for a change in sign.

  1. MEI numerical Methods

    and g(x) and drag it down to 9 significant figures. A spreadsheet can be made so this method can once again be used to find the root/s of the equation between 0 and ?/2. Secant method: Earlier I claimed that using the Newton raphson method can't be used because differentiation

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

    The trapezium rule makes use of trapeziums instead of rectangles. Like the mid-point rule it is easier to describe by using a diagram: Figure 1.1 shows 4 trapezia, each one of length one (h), approximating the area under the curve, between 1 and 5.

  1. 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.

  2. Mathematics Coursework - OCR A Level

    One then looks for where this new formula (y=g(x)) crosses the line of y=x as it will cross at the same points as where the original formula crosses the x-axis thus giving me the roots of the original equation. Original equation I found an equation, the graph of which is shown below, and rearranged into from the form f(x)=0 to x=g(x).

  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