• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
Page
  1. 1
    1
  2. 2
    2
  3. 3
    3
  4. 4
    4
  5. 5
    5
  6. 6
    6
  7. 7
    7
  8. 8
    8
  9. 9
    9
  10. 10
    10
  11. 11
    11
  12. 12
    12
  13. 13
    13

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.

image00.png

These are the three main curves that I am using:


image01.pngimage10.png

Change of Sign

image11.png

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

...read more.

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)

image14.png

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

image15.png

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.

image16.png

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.

image02.png

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.

image03.png

This process is repeated, and it converges very quickly to the value.

When Newton-Raphson does not work

image04.png

 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:  

image05.png

This can be arranged to:

y= x

and either

g(x)= ln|-7+10x|

g(x)= 0.1ex+ 0.7

0 = 0.2ex2x+1.4

           -1.4+2x = 0.2ex

-7+10x = ex

          ln|-7+10x|= x

or

         0 = 0.2ex2x+ 1.4

                    2x = 0.2ex+ 1.4

                       x= 0.1ex+ 0.7

image06.png

image07.png

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.

image08.png

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

image09.png

x8

3.231181344

x9

3.231271221

x10

3.231306729

x11

3.231320756

x12

3.231326297

x13

3.231328487

β= 3.32133 (6s.f.)

...read more.

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.

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.

        -  -

...read more.

This student written piece of work is one of many that can be found in our AS and A Level Core & Pure Mathematics section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related AS and A Level Core & Pure Mathematics essays

  1. Marked by a teacher

    The Gradient Function

    5 star(s)

    2.00000000000000 4.1 2.024845673 0.248457 4 2.00000000000000 4.01 2.002498439 0.249844 4 2.00000000000000 4.001 2.000249984 0.249984 2 1.41421356237310 2.1 1.449137675 0.349241 2 1.41421356237310 2.01 1.417744688 0.353113 2 1.41421356237310 2.001 1.414567072 0.353509 1 1.00000000000000 1.1 1.048808848 0.488088 1 1.00000000000000 1.01 1.004987562 0.498756 1 1.00000000000000 1.001 1.000499875 0.499875 x x 0.5 Gradient 1 1

  2. Solving Equations. Three numerical methods are discussed in this investigation. There are advantages and ...

    0.03125 6 0.6875 0.002319336 0.71875 -0.049697876 0.703125 -0.023702621 0.015625 7 0.6875 0.002319336 0.703125 -0.023702621 0.6953125 -0.010694265 0.0078125 8 0.6875 0.002319336 0.6953125 -0.010694265 0.69140625 -0.004188031 0.00390625 9 0.6875 0.002319336 0.69140625 -0.004188031 0.689453125 -0.000934478 0.001953125 10 0.6875 0.002319336 0.689453125 -0.000934478 0.688476563 0.000692398 0.000976563 11 0.688476563 0.000692398 0.689453125 -0.000934478 0.688964844 -0.000121048 0.000488281 12

  1. Numerical integration can be described as set of algorithms for calculating the numerical value ...

    Algebraically this can be donated as [f(?) + f(?+h)](h/2). The length of the next trapezium in figure 1.1 is [f(2) + f(3)](h/2); similarly in algebraic terms [f(?+h) + f(?????h?]??h???? As a result the total area in figure 1.1 can be given as follows: [f(?) + f(?+h)](h/2) + [f(?+h) + f(?????h?]?h??????[f(?+2h) + f(?????h?]?h??? + [f(?+3h) + f(?????h?]?h?????? [f(??????[f(??h)

  2. C3 Coursework: Numerical Methods

    The first root I shall attempt to find is the root at point C. The initial estimate (starting point) is x=2. The tangent to the point x=2 crosses the x axis at 1.63. This gives a second estimate of 1.63.

  1. The Gradient Fraction

    This method is much more accurate than the 'triangle method'. Here is an example of how to use the Increment Method: The first thing you need to do is to draw the tangent across the point you want to calculate.

  2. decimal search

    Newton-Raphson Method This method works using tangents to cut the x-axis from a fixed estimated point. The f(x) graph is sketched, and then an estimate value for x is taken using the sketch. Where the estimate point of x touches the f(x)

  1. Solving Equations Using Numerical Methods

    I will use the equation y=1.676x� + 3.018x� -1.851x + 0.2427. Here is the equation. I will try to find the middle root of the equation using the decimal search method. I will start with the integer immediately below the root (0)

  2. MEI numerical Methods

    Fixed point iteration: An approximation of the root can be calculated via this method. An equation can have more than one root, in order to find both roots we have to rearrange the equation so it equals x and use fixed point iteration for both equations.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work