Numerical solutions of equations
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A2 Mathematics Coursework C3 Year 12 Numerical solutions of equations Solving 0 = x5+x-5 using the "Change Of Sign" Method The method I will use to solve 0 = x5+x-5 is the Change of Sign Method involving the Decimal Search method. I have drawn this graph using the Autograph Software, and the print screen of this is below: From my graph above, I can see that the root of this equation is between x =1 and x = 1.5. The table of x values and f(x) values is shown below. I can work out the f(x) values by substituting the x-values into the equation. x 1 1.1 1.2 1.3 1.4 1.5 f(x) -3 -2.28949 -1.31168 0.01293 1.77824 4.09375 From my table of values above, it is clear that the change of sign from negative to positive occurs between x = 1.2 and x = 1.3. So, I can narrow these values down further to find another change of sign. x f(x) 1.21 -1.19626 1.22 -1.07729 1.23 -0.95469 1.24 -0.82837 1.25 -0.69824 1.26 -0.56420 1.27 -0.42616 1.28 -0.28403 1.29 -0.13769 1.30 0.01293 I can see that the change of sign is between x = 1.29 and x = 1.30. x f(x) 1.291 -0.12283 1.292 -0.10792 1.293 -0.09296 1.294 -0.07797 1.295 -0.06293 1.296 -0.04784 1.297 -0.03271 1.298 -0.01754 1.299 -0.00233 1.300 0.01293 The change of sign is in the interval [1.299, 1.300] x f(x)
Therefore, the Newton-Raphson method has failed to find that particular root despite taking a starting value close to it. This is shown graphically in Figure 6b. Solving 0= x5+4x2-2 using the "Rearranging method" The graph of this is shown below (or see Figure 5): I will rearrange f(x)=0 in the form of x = g(x). As my equation is 0= x5+4x2-2, there are possible rearrangements of this. I will only pick out two possible rearrangements. Rearrangement 1: 0= x5+4x2-2 -x5= 4x2-2 x5= -4x2+2 x= (-4x2+2)1/5 On the Autograph software, I will draw the equation y= g(x)= (-4x2+2)1/5 and the line y = x. This is shown below (or see Figure 6): Rearrangement 2: 0= x5+4x2-2 -4x2= x5-2 4x2= -x5+2 x2= (-x5+2)/4 x= ((-x5+2)/4))1/2 On Autograph software, I will draw the equation y = g(x) = ((-x5+2)/4))1/2 and the line y = x. This is what it looks like below (or see Figure 7): I can see from the diagram directly above for Rearrangement 2 that g'(x) is between 1 and -1, so there should be a convergence. Therefore, my chosen rearrangement is Rearrangement 2, which is: x = ((-x5+2)/4))1/2 So my iterative formula for this rearrangement is: xn+1= ((-xn5+2)/4))1/2 I can see from my graph above that the point of intersections of the two lines is between x = 0 and x = 1.
Both the Newton-Raphson method and the Rearrangement method were fixed point estimates and involved an iterative process. Therefore, these methods were very similar. However, these two methods differ because there is a specific formula for the Newton-Raphson method. Although the Newton-Raphson method was more complicated than the Rearrangement method, it gave a much more rapid rate of convergence. In fact, the Newton-Raphson method gave the most rapid rate of convergence to 6 decimal places, whereas, for the Decimal Search method, it took me very many calculations to converge to 5 decimal places. So, the Decimal Search method gave the slowest rate of convergence. These two methods were not as time-consuming as the Decimal Search method because it did not involve having to find a change of sign in many particular intervals; there was already a convergence towards the particular root I wanted. However, it was really easy to make mistakes on the calculator due the order of the terms in the iterative formulae. I used the Autograph software to draw these graphs. This was so much quicker than drawing the graphs by hand, and Excel does not have the facilities to draw such advanced graphs. The Autograph software was very easy to use and it also helped me to find out which rearrangement of a function were suitable (when carrying out the Rearrangement method) and in which interval the root was in (when using the Decimal Search method). ?? ?? ?? ?? Kosturi Ash 12/3 1
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