- Level: AS and A Level
- Subject: Maths
- Word count: 2551
Der Handschuh.
Extracts from this document...
Introduction
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.
-6 | -183 |
-5 | -97 |
-4 | -41 |
-3 | -9 |
-2 | 5 |
-1 | 7 |
0 | 3 |
1 | -1 |
To home in on a more accurate answer I need to investigate the decimal places between -3 and -2
x | f(x) |
-3 | -9 |
-2.9 | -6.889 |
-2.8 | -4.952 |
-2.7 | -3.183 |
-2.6 | -1.576 |
-2.5 | -0.125 |
-2.4 | 1.176 |
-2.3 | 2.333 |
-2.2 | 3.352 |
-2.1 | 4.239 |
-2 | 5 |
This method is repeated until a sufficient number of decimal places are achieved. In this investigation I think 5 decimal places will be enough
-2.5 | -0.125 |
-2.49 | 0.011751 |
-2.48 | 0.147008 |
-2.47 | 0.280777 |
-2.46 | 0.413064 |
-2.45 | 0.543875 |
-2.44 | 0.673216 |
-2.43 | 0.801093 |
-2.42 | 0.927512 |
-2.41 | 1.052479 |
-2.4 | 1.176 |
-2.5 | -0.125 |
-2.499 | -0.11126 |
-2.498 | -0.09753 |
-2.497 | -0.08382 |
-2.496 | -0.07012 |
-2.495 | -0.05644 |
-2.494 | -0.04277 |
-2.493 | -0.02912 |
-2.492 | -0.01548 |
-2.491 | -0.00186 |
-2.49 | 0.011751 |
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.
x | f(x) | x | f(x) | |
-4 | -301 | -4 | -301 | |
-3.5 | -197.875 | -3 | -121 | |
-3 | -121 | -2 | -31 | |
-2.5 | -66.625 | -1 | -1 | |
-2 | -31 | 0 | -1 | |
-1.5 | -10.375 | 1 | -1 | |
-1 | -1 | 2 | 29 | |
-0.5 | 0.875 | 3 | 119 | |
0 | -1 | 4 | 299 | |
0.5 | -2.875 | |||
1 | -1 | |||
1.5 | 8.375 | |||
2 | 29 | |||
2.5 | 64.625 | |||
3 | 119 | |||
3.5 | 195.875 | |||
4 | 299 |
Middle
This was the beginning stages of this method and shows the change of signs I had to investigate. There were three routes as shown by the highlighted cells.
-5 | -5624 |
-4 | -1727 |
-3 | -350 |
-2 | -23 |
-1 | 4 |
0 | 1 |
1 | -2 |
2 | 25 |
3 | 352 |
4 | 1729 |
5 | 5626 |
After taking an estimate of the roots to be –1.5, 0.5 and 1.5 I started to use the iterative formula. On the first route this idea becomes clear as we can see by the highlighted red cells that for the first value of the root drawn out by the equation becomes the starting point for the next set of figures to go through the iterative equation
Route 1 | |||
xn | f(xn) | f'(xn) | xn+1 |
-1.5 | 2.6875 | 16.875 | -1.659259259 |
-1.65925926 | -1.312787509 | 34.50075025 | -1.621208275 |
-1.62120827 | -0.093523598 | 29.65572012 | -1.61805463 |
-1.61805463 | -0.000604217 | 29.2730055 | -1.61803399 |
-1.61803399 | -2.57554E-08 | 29.27050994 | -1.618033989 |
Route 2 | |||
0.5 | 0.4375 | -3.125 | 0.64 |
0.64000000 | -0.095971635 | -4.4662784 | 0.618511945 |
0.61851195 | -0.002042156 | -4.274856611 | 0.618034232 |
0.61803423 | -1.03759E-06 | -4.270512042 | 0.618033989 |
0.61803399 | -2.68452E-13 | -4.270509831 | 0.618033989 |
Route 3 | |||
1.5 | -0.6875 | 16.875 | 1.540740741 |
1.54074074 | 0.077437779 | 20.74492848 | 1.537007887 |
1.53700789 | 0.000695055 | 20.37311978 | 1.536973771 |
1.53697377 | 5.76884E-08 | 20.36973796 | 1.536973768 |
1.53697377 | 0 | 20.36973768 | 1.536973768 |
The three routes I have found using this method are all highlighted in green. They can also be seen on the below graph
In this method I also looked at the error bounds of my solutions. I took the final figure up by 0.0000000005 and down by the same.
Conclusion
The final method of rearrangement was relatively easy to use and there were no major calculations that needed to be done apart from the rearrangement at the beginning that did not prove to be too difficult. I did find however that it seemed to tend to the strongest route and would ignore the weaker routes when they are required and still tend to the stronger ones. Also as we already know when a route has a gradient lager than 1 or less than –1 the formula did not work because it couldn’t find the route and so failed. I found it the least successful of the methods although it could be used to find an accurate route quickly it often failed. The time in which it takes to use the method is greatly increased by the use of excel like both other methods, if it were to be used manually, again like both other methods, it would be very time consuming.
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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