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Numerical solutions of equations
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Introduction
Kosturi Ash 12/3
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) |
1.2991 | -0.000805 |
1.2992 | 0.000720 |
1.2993 | 0.002244 |
The root of this equation lies in the interval [1.2991, 1.2992]. This means that I can take the root to be x = 1.29915(5 decimal places).
Middle
3+6x2+3
...read more.
The iterative formula is: xn+1 = xn- f(xn)
f’(xn)
So the iterative formula for this equation is: xn+1 = xn- xn4+2xn3+3xn-4
4xn3+6xn2+3
On my diagram of this function, I have taken my x1 value to be about –1.6
x1 = -1.6
x2 = 3.682591
x3 = 2.658976
x4 = 1.899865
x5 = 1.353797
x6 = 1.003365
x7 = 0.847116
x8 = 0.819172
x9 = 0.818390
x10 = 0.818389
x11 = 0.818389
From that, I can quite clearly see that there is a convergence beginning to show towards the positive root. However, my aim was to find the negative root, and I could not find it. Therefore, the Newton-Raphson method has failed to find that particular root, even though the starting value was close to it.
Another example where the Newton-Raphson would not work is the function (x+21/2)1/2 ln (x+21/2) = 0. The graph and calculations of this is below:
The function is: (x+21/2)1/2 ln (x+21/2) = 0
f’(x) = ((x+21/2)1/2X 1 ) + (ln (x+21/2) X ½ (x+21/2) -1/2)
x + 2 1/2
(x+21/2)1/2 + ln (x+21/2) (x+21/2) -1/2
(x+21/2)1 2
1 + ln(x+21/2)
(x+21/2)1/2 2(x+21/2)1/2
= 2+ ln((x+21/2)
2(x+21/2)1/2
The iterative formula for the Newton-Raphson method is:
xn+1 = xn = f’(xn)
f’(xn)
x n+1 = xn - 2(x+21/2)ln(x+21/2)
2 + ln(x+21/2)
I choose my starting value of x (x1) to beabout –1.3.
x1 = -1.3
x2 = -4.22079
x3 = not defined
x4 =
Another example where the Newton-Raphson method would fail is with the function x3-5x+0.1=0.
Conclusion
Decimal Search method.
...read more.
Both the Newton-Raphson method and the Rearrangementmethod 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).
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