Methods of Advanced Mathematics (C3) Coursework.

Task: Candidates will investigate the solution of equations using the following three methods:

        

  • Systematic search for change of sign using one of three methods: decimal search, bisection or linear interpolation.
  • Fixed point iteration using the Newton raphson method.
  • Fixed point iteration after rearranging the equation f(x) = 0 into the form x = g(x)

Change of Sign

This method is based on how the function has a change of sign either +ve to -ve or -ve to +ve on either side of a route when crossing the x-axis. This method relies on this fact to find the points between the positive and the negative value where there is another change from positive to negative. This can be done until a useful number of decimal places are found.

To investigate this method I plan to use the function f (x) = x^3-5x+1. When I plot this it shows:

From the graph we can see that routes lie on the x-axis between the values -3 and -2, 0 and 1 and 1 and 2. If I consider the function between –3 and -2 I can see I change of sign on the function from –ve to +ve.

To home in on a more accurate answer I need to investigate the decimal places between -3 and -2

This method is repeated until a sufficient number of decimal places are achieved. In this investigation I think 5 decimal places will be enough

                                

From this I can see that the route lies between -2.491 and -2.490±0.0005 and is found at 0.011751 ± 0.0000005.

This method works for most functions of x but can still cause anomalous results in the occasion of certain instances where the method will not work as there being two routes within an interval.

When I plot a graph with this characteristic both routes are clearly distinguishable visually, however when I find the solutions numerically it only shows one change of sign being the first one and the second is over looked.

The routes are visible on the graph but the table below shows only the one change of sign in red.

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As you can see I have found only the first route and without a graph I wouldn’t be aware of the two routes actually present between this interval. The reason for the failure is the fact of two routes being in one interval. If the intervals were smaller then it would have worked.

Newton Raphson Method

In this method an estimate of the route is taken a line equal to x is taken up until it hits the curve and then a tangent is drawn down to the x-axis. This new position on the x-axis is treated ...

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