- Level: AS and A Level
- Subject: Maths
- Word count: 2551
Methods of Advanced Mathematics (C3) Coursework.
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. ...read more.
Middle
All of them showing a change of signs meaning that the error bounds have been established to a suitable degree of accuracy. Like change of sign method it is also possible for Newton-Raphson to fail. To illustrate this I am going to use the equation f(x)=2x^3-5x^2+2. -3 -97 -2 -34 -1 -5 0 2 1 -1 2 -2 3 11 Again I started the method by looking for a change of sign. I decided to try and get a solution for the route between 0 and 1, so I placed my first guess at 0.1. I then put this through the iterative formula. xn f(x) f'(x) (xn)+1 0.1 1.952 -0.94 2.176596 2.176596 -1.0643 6.659457 2.336414 2.336414 0.21402 9.388835 2.313618 2.313618 0.004662 8.980798 2.313099 2.313099 2.39E-06 8.971577 2.313099 2.313099 6.32E-13 8.971573 2.313099 2.313099 0 8.971573 2.313099 2.313099 0 8.971573 2.313099 However this seems to have picked up a different solution. From the change of sign we can see there is one between 2 and 3 so it therefore must be this. It can also be seen on the graph below. It happens because the tangent is diverged away from the suspected route on the first iteration but then homes in on another route as shown below. ...read more.
Conclusion
One disadvantage of the method is that an approximate value of the route needs to be known, so a graph must first be plotted. The convergence time to the routes is relatively fast but it does require a great deal of work. If this method was to be done manually it would take a great deal of time more so than if it is done on excel due to the benefits of excel I have already mentioned. 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. ...read more.
This student written piece of work is one of many that can be found in our AS and A Level Core & Pure Mathematics section.
Found what you're looking for?
- Start learning 29% faster today
- 150,000+ documents available
- Just £6.99 a month