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# Numerical Methods coursework

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Introduction

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 (where a, b are the bases and h is the height of the trapezium) to estimate the area.   Middle                   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 is the absolute error in 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 , 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.

Conclusion

This can be proved by looking on the diagram below.         The distance between and is 4 times smaller than the distance between and , where X represents the actual solution. To get the next distance between and we multiply the “multiplier” 0.25 with the distance between and .

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:

 Rule Error T2, M2 T4, M4 T8, M8 … … T4096, M4096 --- --- S2 S4 … … S256 Interpretation

My solution to the integral 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).

This student written piece of work is one of many that can be found in our AS and A Level Core & Pure Mathematics section.

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