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# Numerical Differentiation

Extracts from this document...

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

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= xand 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

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. RANGStrt: 0End: 10Ptch: 1EXE 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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