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# The aim of this coursework is to compare three different numerical methods of solving equations. This will allow us to determine which one is the most efficient, quickest and easiest method to use.

Extracts from this document...

Introduction

## Pure 2 Coursework

Solution of equations by numerical methods

## Introduction

The aim of this coursework is to compare three different numerical methods of solving equations. This will allow us to determine which one is the most efficient, quickest and easiest method to use.

The three methods I will use are:

• Decimal Search method
• Newton-Raphson method
• Fixed Point Iteration method

## Software used

For all of the three methods I will be using Autograph, which draws accurate graphs, and shows clearly the different roots. I will use it especially for the Newton-Raphson method and Fixed Point Iteration method to find the roots.

For the Decimal Search method I will be using Excel, because it makes the various calculations easier and faster.

Decimal Search

The decimal search is named as it employs the tactic of splitting the current interval of x values into 10 equal intervals of equal size and looking for a change of sign.

This process is then repeated, again splitting the current interval into 10 equal intervals of equal size and this can be continued until the root has been found to the required degree of accuracy.

The equation that I have chosen to solve is x5 – 2.7x + 1.8 = 0

The graph illustrates y = x5 – 2.7x + 1.8

Zoom of y = x5 – 2.7x + 1.8 between x = -2 and x = 0

We can notice from the graph that y = x5 – 2.7x + 1.8 crosses the x-axis between        x = -2 and x = -1.

To find the smallest root, I will take x = -4 as my starting value, and see the number of iterations required to find the root.

Middle

Newton-Raphson

This method involves fixed-point estimation, whereby a tangent to the curve from an initial value of x is drawn then it is calculated where the tangent intercepts the x-axis… this gives a better approximation to the root.  Repeating the process gives more and more accurate values for the root until the desired accuracy is reached.

The equation that I have chosen to solve is x5 – 2.1x + 0.6 = 0

Newton-Raphson formula is:

The iterative formula is:

The graph below illustrates y = x5 – 2.1x + 0.6.

The graph on the following page illustrates y = x5 – 2.1x + 0.6 and it shows the effect of drawing successive tangents from a starting value of x0 = -1.5.

This illustration demonstrates how we acquire the first root of the equation. The tangent is found at a point and then the point where this tangent crosses the x-axis is found.  In the table below, x is the approximation to the root and |Δ| is the modulus of the error.

x|Δ|

-1.33441057                0.16559

-1.274151543                0.060259

-1.266706903                0.0074445

-1.266601067                0.000105795

-1.266601067                2.11252E-08

From the various calculations, x0, x1, x2, … converge on the root –1.266601067 in 5 iterations.  The other 2 roots were found in the same way, repeating the same operations.

The graph below illustrates y = x5 – 2.1x + 0.6 and it shows the effect of drawing successive tangents from a starting value of x0 = 0, in the scope of showing how to find the root near x = 0.5.

 x |Δ| 0.2857142857 0.285714 0.2866355552 0.000921269 0.2866356508 9.61131E-08 0.2866356508 0

From the various calculations, x0, x1, x2, x3 … converge on the root 0.2866356508 in 4 iterations.

The graph below illustrates y = x5 – 2.1x + 0.

Conclusion

Newton-Raphson Method = 0.105587 in 4 iterationsFixed Point Iteration Method = 0.105588 in 8 iterations

From the 3 results shown above, I can clearly conclude that the Newton-Raphson method is the quickest one, because it gives the fastest convergence.  It does reply on being able to differentiate the f(x) though.

The Decimal Search method is very tedious because it takes a lot of time to prepare the spreadsheet on Excel: especially the various formulas for the different calculations. And as shown above it takes many iterations to find the root.

The Fixed Point Iteration method takes a lot of time to find the right rearrangement of an equation, because not all g(x) graphs will allow convergence to the required root. I had also problems using the software Autograph, because when I had to load the method, it didn’t work for the first few times. As shown above it takes 8 iterations. This time the number of iterations required was very small as it usually takes many more (and depends on the value of the gradient of g(x) near the root) – the closer this is to 1, the faster is the convergence.

The Newton-Raphson method requires much less time for the preparation than all the other methods, and is definitely much simpler. With just a click on the graph it shows exactly the number of iterations and it shows on the graph the various tangents. It’s the fastest method, especially because it required only 4 iterations to find the middle root of the equations.

To conclude, the Newton-Raphson method is the best between the three, because it’s quick and very easy to use.

P2 Coursework        by Francesco Egro

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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