- Level: AS and A Level
- Subject: Maths
- Word count: 2716
Pure 2 coursework - Decimal Search Method
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Introduction
Pure 2 coursework
Decimal Search Method
In order to find the root of the function f(x)=3x3-7x2-11x+17, the decimal search method was used. This is the process whereby a table of values is constructed that show whether the value of f(x) with certain values of x is positive or negative. By zooming in on the places where the value of f(x) switches from positive to negative (or vice versa), values for the root of this function can be found.
x | f(x) |
-4 | -243 |
-3 | -94 |
-2 | -13 |
-1 | 18 |
0 | 17 |
1 | 2 |
2 | -9 |
3 | 2 |
4 | 53 |
5 | 162 |
From this table, it is clear that there are roots where f(x)=0 when values of x lie between -2 and -1, 1 and 2, and 2 and 3. By concentrating on the interval of [1,2], the rot can be found to a greater level of accuracy.
x | f(x) |
1 | 2 |
1.1 | 0.423 |
1.2 | -1.096 |
1.3 | -2.539 |
1.4 | -3.888 |
1.5 | -5.125 |
1.6 | -6.232 |
1.7 | -7.191 |
1.8 | -7.984 |
1.9 | -8.593 |
2 | -9 |
It is now clear that the root lies somewhere inside the interval of [1.1, 1.2]
x | f(x) |
1.1 | 0.423 |
1.11 | 0.268193 |
1.12 | 0.113984 |
1.13 | -0.03961 |
1.14 | -0.19257 |
1.15 | -0.34488 |
1.16 | -0.49651 |
1.17 | -0.64746 |
1.18 | -0.7977 |
1.19 | -0.94722 |
1.2 | -1.096 |
The root is in the interval [1.12, 1.13]
x | f(x) |
1.12 | 0.113984 |
1.121 | 0.098597 |
1.122 | 0.083216 |
1.123 | 0.067841 |
1.124 | 0.052472 |
1.125 | 0.037109 |
1.126 | 0.021753 |
1.127 | 0.006403 |
1.128 | -0.00894 |
1.129 | -0.02428 |
1.13 | -0.03961 |
The root is somewhere inside the interval [1.127, 1.128]. It is required that the root be found to 3 decimal places. Therefore it is necessary to know the value of f(x) when x=1.1275
x | f(x) |
1.1275 | -0.00127 |
Middle
1.44528
1.44528
1.44528
1.44528
1.44528
1.44528
1.44528
1.44528
1.44528
1.44528
1.44528
1.44528
1.44528
1.44528
1.44528
1.44528
1.44528
This table shows that the function has a root of 1.44528 ± 0.000005
x | xn+1 |
5.00000 | 6.07407 |
6.07407 | 5.76048 |
5.76048 | 5.72064 |
5.72064 | 5.72001 |
5.72001 | 5.72001 |
5.72001 | 5.72001 |
5.72001 | 5.72001 |
5.72001 | 5.72001 |
5.72001 | 5.72001 |
5.72001 | 5.72001 |
5.72001 | 5.72001 |
This table shows that the function has a root of 5.72001 ± 0.000005
Failure of Newton-Raphson Method
The Newton-Raphson method can fail to find the root of a function, as in the following example. The function is f(x)=20-((x-3)2+0.05)-1
It is clear from simply looking at the function that one root is going to be x=3. However, if the Newton-Raphson method is used, the following data is obtained
x | xn+1 |
3.4000 | 2.5600 |
2.5600 | 3.6320 |
3.6320 | 0.7936 |
0.7936 | 109.3000 |
As can be seen on the diagram below, these figures are clearly divergent, which means that xn+1 is further from the root than xn, and so are no use in finding the root of the function.
Fixed Point Iteration
This method to find the roots of a function is done by rearranging the function f(x) into the form x=g(x). This can be explained with an example. If the expression xn+1=0.8+2.6/xn is used to create a large table of values for xn+1, it is found that after x50 and x51 are both 2.061324773. This means that xn+1 converges on this figure. If we let the limit L=2.061324773, then L satisfies the equation L=0.8+2.6/L and therefore L2-0.8L-2.6=0 and therefore L is a root of f(x). The function that I will be finding the roots of is f(x)=x3-4x2-7x+11. In order to find the roots, the equation f(x)=0 must be rearranged into the form x=g(x). Therefore, g(x) = (x3-4x2+11)/7
Now, xn+1=xn3-4xn2+11 is used to generate a table of values for xn, the results
7
are as follows:
x | xn+1 |
2 | 0.428571 |
0.428571 | 1.477718 |
1.477718 | 0.784603 |
0.784603 | 1.288657 |
1.288657 | 0.928207 |
0.928207 | 1.193349 |
1.193349 | 1.000443 |
1.000443 | 1.142541 |
1.142541 | 1.038553 |
1.038553 | 1.115115 |
1.115115 | 1.058957 |
1.058957 | 1.100278 |
1.100278 | 1.069938 |
1.069938 | 1.092252 |
1.092252 | 1.075859 |
1.075859 | 1.087912 |
1.087912 | 1.079056 |
1.079056 | 1.085566 |
1.085566 | 1.080782 |
1.080782 | 1.084299 |
1.084299 | 1.081714 |
1.081714 | 1.083614 |
1.083614 | 1.082218 |
1.082218 | 1.083244 |
1.083244 | 1.08249 |
1.08249 | 1.083044 |
1.083044 | 1.082637 |
1.082637 | 1.082936 |
1.082936 | 1.082716 |
1.082716 | 1.082878 |
1.082878 | 1.082759 |
1.082759 | 1.082846 |
1.082846 | 1.082782 |
1.082782 | 1.082829 |
1.082829 | 1.082795 |
1.082795 | 1.08282 |
1.08282 | 1.082801 |
1.082801 | 1.082815 |
1.082815 | 1.082805 |
1.082805 | 1.082812 |
1.082812 | 1.082807 |
1.082807 | 1.082811 |
1.082811 | 1.082808 |
1.082808 | 1.08281 |
1.08281 | 1.082809 |
1.082809 | 1.08281 |
1.08281 | 1.082809 |
1.082809 | 1.082809 |
1.082809 | 1.082809 |
Conclusion
Of the three methods, it is clear that the method that is the slowest to converge on the root is the decimal search method. The quickest to converge is the Newton-Raphson method, closely followed by the fixed point iteration. However, although both iteration methods can find the root correct to several decimal places very quickly, there is more initial work in order to start using these methods than there is for the decimal search. Formulae must be rearranged or new formulae constructed in order to begin the search. Also, a lot of the speed of the two iteration methods is due to the ability of a computer to quickly do repeat calculations many times over. They are well suited to the use of computers whereas the decimal search method is relatively slow even with the use of computers, as it would take a human to spot the change in sign that determines the next set of calculations.
Finally, there is no reason why the decimal search method could fail when used in conjunction with an graph plotted accurately by calculator or computer to determine the rough approximations of the roots, whereas with the other two methods there are functions where it is impossible to obtain an estimate of the roots.
Paul McKay Page 09 May 2007
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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