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Analyse the use of three methods which are called the: change of sign, Newton-Raphson and the rearrangement method and use them to find roots of different equations.

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MATHS P2 COURSEWORK There are many different kind of methods which can be used to find the roots of equations which can not be sold algebraically. In this coursework we are going to analyse the use of three of these methods which are called the: change of sign, Newton-Raphson and the rearrangement method and are going to use them to find roots of different equations. Change of sign method A root of an equation (where the graph crosses the x-axis) can be detected by finding where the solution of a formula changes sign from positive to negative. Where we find a change of sign on a graph using omnigraph we take the range of the numbers it is in and divide it by ten to find where it now changes sign. This procedure is then repeated to the required level of accuracy. Here is a graphical representation of the systematic decimal search: X 1 X 10-1 X10-2 Using change of sign method on excel First you look up between 2 values (e.g. 1 and 2, or 4 and 5) where the graph crosses the x-axis. Then you take that as your range and divide it up into 10 equal segments. For example, if between 1 and 2.. you use 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, etc. you work out the formula for each of these and where the change of sign occurs you repeat the process between that range. A B C D E E 1 X1 f(A1) A2+0.01 f(C1) C1+0.001 f(C1) 2 A1+0.1 f(A2) C1+0.01 f(C2) E1+0.001 f(C2) 3 A2+0.1 f(A3) C2+0.01 f(C3) E2+0.001 f(C3) 4 A3+0.1 f(A4) C3+0.01 f(C4) E3+0.001 f(C4) 5 This process carries on to the required accuracy and this is how we try to find the root of an equation, using omnigraph and excel. Example I'm going to use the change of sign method to investigate when the method works and when it doesn't. ...read more.


From this failure it is apparent that in order for the iteration to converge to a root, the gradient at the root generally needs to be between -1 and 1. Newton raphson A root of an equation can also be found by the following method: A (Xn,f(Xn)) B Xn y=f(X) * Take tangent at point (A) on y=f(x) * Find out when this tangent cuts the x axis (i.e. y=0) (B) * Use that value as the next x value * Repeat this procedure Equation of AB slope of AB = f'(xn) y=f'(xn)x + c f(xn)=f'(xn)xn + c c=f(xn) - f'(xn)xn y=f'(xn)x + f(xn) - f'(xn)xn At B, y=0 0=f'(xn)x + f(xn) - f'(xn)xn x=f'(xn)xn -f(xn) f'(xn) xn+1 = xn f(xn) f'(xn) * Method on Microsoft Excel First a value of x is typed into A1. Then in the cell below (B1), the A 1 X 2 A1-f(A1)/f'(A1) 3 A2-f(A2)/f'(A2) 4 A3-f(A3)/f'(A3) 5 A4-f(A4)/f'(A4) 6 A5-f(A5)/f'(A5) 7 A6-f(A6)/f'(A6) 8 A7-f(A7)/f'(A7) 9 A8-f(A8)/f'(A8) Drag down until two Consecutive values are identical This method will find different roots depending on the nature of the slope, and the original value of x that you start with. I am going to use the above method to find the root of f(x)= 0.5x3-2.5x+0.06 As we can see, the graph has 3 roots. To get to these roots, we need different starting points for each root. These starting points are going to be -2, -0.5 and 3. All of the roots found below are (to 8or9 D.P). f(x)= 0.5x3-2.5x+0.06 f'(x)=1.5x2-2.5 A B C x1 -2.00000000 -0.5 3.000000000 x2 -2.30285714 0.087058824 2.449090909 x3 -2.24986517 0.023844498 2.251747622 x4 -2.24797511 0.024002765 2.224478330 x5 -2.24797274 0.024002766 2.223970147 x6 -2.24797274 0.024002766 2.223969972 x7 -2.24797274 0.024002766 2.223969972 Therefore, we can see that the roots of the equation f(x)= 0.5x3-2.5x+0.06 are -2.24797, 0.02400 and 2.22397 (to 5DP). Below is the graphical demonstration of how the root between 2 and 3 was found. As evident in the graph, it only takes a few iterations to find the root. ...read more.


This is a greater number of calculations to perform than the other two methods, and the calculations are much more complex. Also, this method doesn't always converge to the required root. Different starting values of x may have to be used to obtain the required root. Rearrangement method This method isn't as mathematically advanced as the Newton-Raphson method, but it requires more mathematical ability than the change of sign method. The only calculation that you need to do is to rearrange the formula f(x)=0 into the form x=g(x), which can be done in a variety of ways. However, not all of the rearrangements work, and so many different rearrangements and starting values often have to be used. For the rearrangement to work, the gradient of g'(x) at the root needs to be between -1 and 1. CONCLUSION I think that the rearrangement method is the best method to use, because even though the convergence of the rearrangement and Newton-raphson methods are the same, the ease of use of the rearrangement method is much better. However, when I used the rearrangement before, I did not converge to a value until I repeated the process 32 times, this shows even thought the method can be fast, it can also be unreliable. However, when using this method you must be carefull in choosing the right rearrangement of f(x) because not all rearrangments work. In this sense the Newton raphson method is better because it is more likely to find a root, and only fails when there is a break in the magnitude of the graph. The slowest, but easiest method is the change of sign method. However, this method can take a very long time to do as it requires many repetitions of the same process. Due to the unreliability of the rearrangement method I think the Newton raphson method is the best method to use, because it shows reliability, accuracy and has fast convergences. ?? ?? ?? ?? ...read more.

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