CHILD ABUSE, CHILDHOOD & HISTORY
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- Level: AS and A Level
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Numerical Differentiation
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Introduction
Numerical Differentiation
Introduction
When finding the roots of an equation, the first thing you would try to do is factorise it. However sometimes factorizing does not find the root, therefore there are Numerical Methods that can be used to approximate the root, and the three different methods that I am using in this coursework are: Decimal search, Fixed Point Iteration, and Newton-Raphson.
These are the three main curves that I am using:
Change of Sign
Change of sign works by taking increments between two values, then substituting the incremented values into the equation to get the y values. The intersection of the axis is found by the change of sign, because the line must have gone through the x-axis because it has changed from positive to negative or vice versa. The positive and negative values are then used as the bounds for the increments, and so the process is continued as you refine your search to the required degree of accuracy.
Finding the root of the equation between x=2 and x=3.
x | y | ||||||
2 | 3 | ||||||
3 | -3.5 | 2 | < x < | 3 | |||
x | y | ||||||
2.0 | 3 | ||||||
2.1 | 2.206 | ||||||
2.2 | 1.428 | ||||||
2.3 | 0.672 | ||||||
2.4 | -0.056 | 2.3 | < x < | 2.4 | |||
x | y | ||||||
2.30 | 0.672 | ||||||
2.31 | 0.597841 | ||||||
2.32 | 0.523968 | ||||||
2.33 | 0.450387 | ||||||
2.34 | 0.377104 | ||||||
2.35 | 0.304125 | ||||||
2.36 | 0.231456 | ||||||
2.37 | 0.159103 | ||||||
2.38 | 0.087072 | ||||||
2.39 | 0.015369 | ||||||
2.40 | -0.056 | 2.39 | < x < | 2.40 | |||
x | y | ||||||
2.390 | 0.015369 | ||||||
2.391 | 0.008216971 | ||||||
2.392 | 0.001068288 | ||||||
2.393 | -0.006077043 | 2.392 | < x < | 2.393 | |||
x | y | ||||||
2.3920 | 0.001068288 | ||||||
2.3921 | 0.000353604 | ||||||
2.3922 | -0.000361047 | 2.3921 | < x < | 2.3922 | |||
x | y | ||||||
2.39210 | 0.000353604 | ||||||
2.39211 | 0.000282137 | ||||||
2.39212 | 0.000210671 | ||||||
2.39213 | 0.000139205 | ||||||
2.39214 | 6.77397E-05 | ||||||
2.39215 | -3.72549E-06 | 2.39214 | < x < | 2.39215 | |||
x | y | ||||||
2.392140 | 6.77397E-05 | ||||||
2.392141 | 6.05932E-05 | ||||||
2.392142 | 5.34467E-05 | ||||||
2.392143 | 4.63001E-05 | ||||||
2.392144 | 3.91536E-05 | ||||||
2.392145 | 3.20071E-05 | ||||||
2.392146 | 2.48606E-05 | ||||||
2.392147 | 1.7714E-05 | ||||||
2.392148 | 1.05675E-05 | ||||||
2.392149 | 3.42102E-06 | ||||||
2.392150 | -3.72549E-06 | 2.392149 | < x < | 2.392150 | x= 2.39215 (6s.f) | ||
Error Bounds
Middle
0
...read more.
5.29
1
1.69
2
0.09
3
0.49
4
2.89
To show another limitation of The Decimal search method I have used the equation:
x |
(x+3.9)(x–3.9) |
As you can see from the table below, there must be a sign change between x=3 and x=4, however looking at the graph shows that there is actually a discontinuity.
x | y |
0 | 0 |
1 | -0.070373 |
2 | -0.1784121 |
3 | -0.4830918 |
4 | 5.0632911 |
5 | 0.5107252 |
Newton-Raphson Method
xr+1 = xr – f (xr)
f ’(xr)
y= 4ln|x| - x + 2
dy= -1 + 4
dx
x0 | 1 | x0 | 6 | |
x1 | 0.666666667 | x1 | 15.50111363 | |
x2 | 0.724372086 | x2 | 12.08113601 | |
x3 | 0.727506177 | x3 | 11.90987844 | |
x4 | 0.727514477 | x4 | 11.90926753 | |
x5 | 0.727514477 | x5 | 11.90926752 | |
α= 0.727514 (6s.f) | β= 11.9093 (6s.f) | |||
Error Bounds
x= 0.727514 ± 0.0000005
f(0.7275145) = 4xln0.7275145-0.7275145+2
= 1.04545x10-07 >0
f(0.7275135) = 4xln0.7275135-0.7275135+2
= -4.39363 x10-06 <0
When the bounds are put into the formula, they show a change of sign which means that there is a root between the two intervals, therefore confirming that:
α= 0.727514 ± 0.0000005.
Therefore:
α= 0.727514 (6s.f.)
1. |
| The first stage is to find the tangent to the curve at the point x0. This is done by differentiating the function. Then with the equation of the tangent the intersection with the x-axis is found. |
2. |
| The point of intersection x value is then used as the new point for the tangent of the line, and again the intersection of the axis with the tangent is found. |
3. |
| This process is repeated, and it converges very quickly to the value. |
When Newton-Raphson does not work
y= 20x5– 3x2 + 0.1
dy= 100x4-6x
dx
x0= 0
x1= 0-(20x05-3x02+0.1)
100x04-6x0
x1= 0-0.1
0
Cannot divide by 0.
Fixed Point Iteration
My Function is:
This can be arranged to: y= x and either | ||
g(x)= ln|-7+10x| | g(x)= 0.1ex+ 0.7 | |
0 = 0.2ex–2x+1.4 -1.4+2x = 0.2ex -7+10x = ex ln|-7+10x|= x | or | 0 = 0.2ex–2x+ 1.4 2x = 0.2ex+ 1.4 x= 0.1ex+ 0.7 |
|
| |
Using this rearrangement α cannot be found because its gradient is approximately 3.8. However β has a gradient between 1 and -1, therefore it can converge. | Conversely with this rearrangement α is possible to work out because the g(x) has a magnitude between 1 and -1, however β has a root with a magnitude greater than 1. | |
Therefore both functions must be used to find all the roots of the equation. | ||
x0 | 0 | |
x1 | 0.8 | |
x2 | 0.922554093 | |
x3 | 0.951570754 | |
x4 | 0.958977437 | |
x5 | 0.960902721 | |
x6 | 0.961405517 | |
x7 | 0.961536984 | |
x8 | 0.96157137 | |
x9 | 0.961580364 | |
x10 | 0.961582717 | |
x11 | 0.961583332 | |
x12 | 0.961583493 | |
x13 | 0.961583535 | |
x14 | 0.961583546 | α= 0.961584 (6s.f.) |
x0 | 3 | |
x1 | 3.135494216 | |
x2 | 3.192734792 | |
x3 | 3.215965511 | |
x4 | 3.225241724 | |
x5 | 3.228921883 | |
x6 | 3.230378172 | |
x7 | 3.23095386 |
|
x8 | 3.231181344 | |
x9 | 3.231271221 | |
x10 | 3.231306729 | |
x11 | 3.231320756 | |
x12 | 3.231326297 | |
x13 | 3.231328487 | β= 3.32133 (6s.f.) |
Conclusion
x
...read more.
EXE
Then pressing execute again repeats the calculation so you can work it out very quickly.
Decimal Search
Decimal search takes the longest to converge with 41 iterations, however the method contains the easiest calculations. Decimal search is a bit more difficult to set up on calculators than the other two methods. Graphical calculators with a table function are probably the easiest to set up. If a calculator with a table function is not available the values would need to be written down to keep track of the calculations. With a good knowledge of a spreadsheet program such as Microsoft Excel and using IF and THEN statements, would take the human error out, and speed up the process. The main problems with Decimal search is that the process stops when one sign change is found, therefore other roots are missed, and when the root is repeated or just goes through the axis but comes back out before the next increment.
On my calculator these are the steps used to set it up:
0.2eX-2x+1.4 | This is inputting the function into the calculator. |
RANG Strt: 0 End: 10 Ptch: 1 EXE | Pressing the RANG button allows you to edit the range of the table. I set it to go from 0 to 10 in intervals of 1. |
EXE | This runs the table and gives you the x and f(x) values. Then all that is needed is to change the range of the table to the bounds from the sign change and change the intervals. |
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