- Level: AS and A Level
- Subject: Maths
- Word count: 1858
'Change of Sign Method'
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Introduction
Mathematics Coursework
Numerical Methods can be used to solve those equations which cannot be solved analytically.
One such method is the ‘Change of Sign Method’, focusing on Decimal Search.
Decimal search is based on the principal that f(x) changes sign as a curve passes through the x axis at a root.
Above the x axis, f(x) is positive and below the x axis f(x) is negative.
Using this principal, values of f(x) are initially found and consecutive values are found in increments of 0.1 until a change in sign is found. If further accuracy is required, consecutives values could then be found in increments of 0.01 and 0.001, until a change of accuracy is found, therefore giving one root of the equation.
Evidence
Equation: y=0.6938x3 – 0.9157x2 – 1.421x + 1.671
Graph:
Table of results
First Interval
x | f(x) |
1 | 0.0281 |
1.1 | -0.07665 |
1.2 | -0.1539 |
1.3 | -0.1996 |
1.4 | -0.2094 |
1.5 | -0.1792 |
1.6 | -0.105 |
1.7 | 0.01757 |
1.8 | 0.1926 |
1.9 | 0.4242 |
The table shows that the sign changes within the interval 1.6 to 1.7. This tell us that the roots lie somewhere within this interval;
Therefore the root is 1.65 ± 0.05
Second Interval
X | f(x) |
1.6 | -0.105 |
1.61 | -0.09497 |
1.62 | -0.08447 |
1.63 | -0.07348 |
1.64 | -0.06199 |
1.65 | -0.05001 |
1.66 | -0.03752 |
1.67 | -0.02452 |
1.68 | -0.01101 |
1.69 | 0.003019 |
The table of values shows that the root lies in the interval [1.68, 1.69]. Therefore the root of the equation is 1.685 ± 0.05
Third Interval
x | f(x) |
1.68 | -0.01101 |
1.681 | -0.009628 |
1.682 | -0.008243 |
1.683 | -0.006854 |
1.684 | -0.005459 |
1.685 | -0.004059 |
1.686 | -0.002654 |
1.687 | -0.001243 |
1.688 | 0.0001725 |
1.689 | 0.001593 |
1.69 | 0.003019 |
Middle
Using this method, a first approximation, x1 is made. The intercept of the tangent at (x1,y1) with the x axis is found as the next approximation, x2. This process is repeated until the desired accuracy is achieved. This method is also known as the ‘Tangent Sliding’ method.
This method can be demonstrated manually, using the equation: y= 20x3 + 15x2 – 15x – 12
F (x) = 20x3 + 15x2 – 15x – 12
F’(x) = 60x2 + 30x – 15
First estimate x1 will be 1
First iteration
F (1) = 20(13) + 15(12) -15(1) – 12 = 8
F’(1) = 60(12) + 30(1) – 15 = 75
X2 = 1 – 8/75
X2 = 0.8933
Second Iteration
F (0.8933) = 20(0.89333) + 15(0.89332) – 15(0.8933) – 12 = 0.82906074
F’(0.8933) = 60(0.89332) + 30(0.8933) – 15 = 59.68266667
X3 = 0.8933 – 0.829/59.68266667
X3 = 0.879442165
X3 = 0.8794
Third iteration
F (0.8794) = 20(0.87943) + 15(0.87942) -15(0.8794) – 12 = 0.01318348
F’(0.8794) = 60(0.87942) + 30(0.8794) – 15 = 57.78837624
X4 = 0.8794 – 0.013/57.79
X4 = 0.879214031
X4 = 0.8792
This can also be demonstrated graphically using the computer programme autograph:
y= 20x3 + 15x2 – 15x – 12
First iteration
X = 0.8933
Second iteration
X = 0.8794
Third Iteration
X = 0.8792
The results have converged to 0.8792
To confirm that this is a root by looking for a change of sign
F (x) = 20x3 + 15x2 – 15x – 12
F (0.8791) = -6.5827 x 10-3 = negative
F (0.8792) = positive
Therefore the root is 0.87925 ±0.00005
Where the method cannot work
If the initial estimate is too far from the required root, the method can fail.
This can be shown using the equation y=20x3 + 15x2 – 8x – 19
Take the initial estimate as x1 = -1
First Iteration
x = - 0.2727
Second Iteration
x = -1.647
Third Iteration
x=-1.13
Evidently, this method fails in this instance.
Conclusion
Whilst the Newton Raphson method allows quick convergence, and is easy to perform on autograph, the method could be criticised, as it requires calculus, unlike decimal search. It can also be subject to failure, for example, if the initial estimate is made too far from the required root. Furthermore, to find an error bound for increased accuracy, decimal search needs to be carried out, again adding to the amount of work required. This disadvantage can also be applied to the graphical convergence method. In this method, there is also a high failure rate, as it can diverge, oscillate or converge to the wrong root, as shown in the calculations. Also, you may not be able to rearrange f(x) =0, in which case this method could not be used.
Therefore, in conclusion, whilst each method can be used to find roots, a d=sketch of the graph should be drawn, before deciding upon which method will work the best, and the level of accuracy that is desired.
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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