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# Change of Sign Method

Extracts from this document...

Introduction

Numerical Methods Coursework

## To look at how this method I will find the roots of the equation x5-7x+5=0 by the change of sign method. Here is a graph of y=x5-7x+5 and an integer search showing where the roots of x5-7x+5=0:

 x -5 -4 -3 -2 -1 0 1 2 3 4 5 y -3085 -991 -217 -13 11 5 -1 23 227 1001 3095

## This equation has 3 roots as shown in the table as the y value changes sign 3 times in the table, one between x=-2 and x=-1 another between x=0 and x=1 and another between x=1 and x=2. This is also shown in the graph. A root can be more accurately found using a decimal search or zooming in on a graph and identifying when a change of sign occurs. I will do this to find the root of this equation between x=-2 and x=-1:

 x -2 -1.9 -1.8 -1.7 y -13 -6.461 -1.2957 2.7014

## As there is a change in sign between x=-1.8 and x=-1.7 there must be a root between x=-1.8 and x=-1.7 as shown on the graph above. Therefore a decimal search to two decimal places will be more accurate.

 x -1.8 -1.79 -1.78 -1.77 y -1.2957 -0.8466 -0.409 0.0173

## Therefore there is a root between x=-1.

Middle

1

1.25

1.27359

1.27389

1.27389

1.27389

1.27389

1.27389

1.27389

When x=1.27388 y=0.000034 and when x=1.27390 y=-0.000030. There is a change of sign between the two therefore:

x=1.2739 (5s.f.)

1.27388<x<1.27390

Now I will find the root between x=2 and x=3 and I will use x0=3 as this is close to the root and is most likely to find that root.

 3 2.58333 2.41243 2.37854 2.37720 2.37720 2.37720 2.37720 2.37720

When x=2.37719and when. There is a change of sign between the two therefore:

x=2.3772 (5s.f.)

2.37719<x<2.37721

### Failure of the Newton Raphson Method

When the gradient of the gradient of a curve is small the Newton Raphson method may give the root of the equation at a point you were not looking for. For example if I was looking for the root of the equation between and . The iterative formula for would be:

A sensibleto take would behowever the result would look like this.

 1 -1 -2.33333 -1.90831 -1.6162 -1.46103 -1.41736 -1.41423 -1.41421 -1.41421

This shows that the Newton Raphson method has failed to find the required root between x=0 and x=1 but the root at x=-1.4142 therefore it is a failure.

Rearranging f(x)=0 in the form x=g(x)

To show how this method works, I will show how this method can be used to find all of the roots of the equation x³-3x²-4x+7=0. Firstly I will require values for x0 and I also need to know how many roots I am looking for. This can be done with an integer search and a graph of y=x³-3x²-4x+7.

 x -5 -4 -3 -2 -1 0 1 2 3 4 5 y -173 -89 -35 -5 7 7 1 -5 -5 7 37

Conclusion

The Newton Raphson method was again fairly easy to perform and also gave reliable results although an accuracy check was required to ensure the root had been found but the method was not likely to fail. It was again also very easy to use with a spreadsheet program and graphing software although some time would have been spent in deriving the iterative formula. This method is slightly harder to perform with a graphical calculator as an iterative formula would need to be input, however this method is very fast.

The rearrangement to x=g(x) method was again easy to perform and gave reliable results although again an accuracy check would be required. However this method fails many times and more than one iterative formula was required to find all the roots of the equation. After calculating an iterative formula this method was again easy to perform with a spreadsheet program and graphing software. This method is similarly easy as the Newton raphson method however is more likely to take more iterations.

In conclusion this investigation has shown that the Newton Raphson method of solving equations has shown to be the better of the three methods. This is due to its speed, ease to use and likelihood not to fail.

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