lecture on numerical differentiation

8/6/2019 Lecture on Numerical Differentiation

1/102

6/3/2011 1

Differentiation-Continuous

Functions


2/102

2

Forward Difference Approximation

x

xfxxf

x

xf

0

lim

p

!d

For a finite '' x

x

xfxxfxf(

(}d

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3

x x+x

f(x)

Figure 1 Graphical Representation of forward difference approximation of first derivative.

Graphical Representation OfForward Difference

Approximation

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5

Example 1 Cont.

Solution

t

ttta iii

(

}

RR 1

16!it

2 !t

18

2161

!

!(! ttt ii

2

161816

RR }a

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6

Example 1 Cont.

188.91821001014

1014ln200018

4

4

v

v!R

m/s02.453!

168.91621001014

1014ln200016

4

4

v

v!R

m/s07.392!

Hence

2

161816

RR }a

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7

Example 1 Cont.

2

07.39202.453 }

2m/s474.30}

The exact value of 16a can be calculated by differentiating

tt

t 8.921001014

1014ln2000

4

4

v

v!R

as

? Atdt

dta !

b)

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8

Example 1 Cont.

Knowing that

? At

t

dt

d 1ln ! and 2

11

ttdt

d!

8.921001014

1014

1014

210010142000

4

4

4

4

v

v

v

v!

tdt

dtta

8.9210021001014

101411014

21001014200024

4

4

4

vv

vv!

t

t

t

t

3200

4.294040

!

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9

Example 1 Cont.

163200164.294040

16

!a

2

m/s674.29!

The absolute relative true error is

100

ValueTrue

ValueeApproximat-ValueTruext !

100674.29

474.30674.29x

!

%6967.2!6/3/2011


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10

Backward Difference Approximation of the

First Derivative

We know

x

xfxxf

xxf

0

lim

p

!d

For a finite '' x ,

x

xfxxfxf

(

(}d

If '' x is chosen as a negative number,

x

xfxxfxf

(

(}d

x

xxfxf

!

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11


First Derivative Cont.

This is a backward difference approximation as you are taking a point backward from x. To find the value of xfd at ixx ! , we may choose anotherpoint '' x behind as

1

!i

xx . This gives

x

xfxfxf iii

(

}d 1

1

1

!

ii

ii

xx

xfxf

where

1 ! ii xxx

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12

xx-x

x

f(x)

Figure 2 Graphical Representation of backward differenceapproximation of first derivative


First Derivative Cont.

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13

Example 2

The velocity of a rocket is given by

300,8.9210010141014

ln2000 4

4

ee

v

v

! ttttR

where '' is given in m/s and ''t is given in seconds.

a) Use backward difference approximation of the first derivative ofto calculate the acceleration at . Use a step size of .

b) Find the absolute relative true error for part (a).

ts16!t st 2 !

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14

Example 2 Cont.Solution

t

ttta ii

(

} 1

RR

16!it

2 !t

14216

1

!

!(! ttt ii

2

141616

RR }a

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15

Example 2 Cont.

168.91621001014

1014ln200016

4

4

v

v!R

m/s07.392!

148.91421001014

1014ln200014

4

4

v

v!R

m/s24.334!

2

141616

RR }a

2

24.33407.392 !

2

m/s915.28}6/3/2011


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16

Example 2 Cont.


100674.29

915.28674.29xt

!

%5584.2!

The exact value of the acceleration at from Example 1 is

2m/s674.2916 !a

s16!t

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17

Derive the forward difference approximation

from Taylor series

Taylors theorem says that if you know the value of a function '' f at a point

ix and all its derivatives at that point, provided the derivatives are

continuous between ix and 1ix , then

-dd

d! 2

111!2

iii

iiiii xxxf

xxxfxfxf

Substituting for convenienceii xxx ! 1

-ddd!2

1 !2

xxfxxfxfxf iiii

-(dd

(

!d x

xf

x

xfxfxf iiii

!2

1

xx

xfxfxf iii (

(

!d 01

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18


from Taylor series Cont.

The x(0 term shows that the error in the approximation is of the order

of x Can you now derive from Taylor series the formula for backward

divided difference approximation of the first derivative?

As shown above, both forward and backward divided difference

approximation of the first derivative are accurate on the order of x(0

Can we get better approximations? Yes, another method to approximate

the first derivative is called the Central difference approximationof

the first derivative.

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from Taylor series Cont.

From Taylor series

-ddd

dd

d!32

1 !3

!2

xxf

xxf

xxfxfxf iiiii

-ddd

dd

d!32

1 !3

!2

xxf

xxf

xxfxfxf iiiii

Subtracting equation (2) from equation (1)

-ddd

d! 3

11 !3

22 xxfxxfxfxf iiii

-(ddd

(

!d

211

!32x

xf

x

xfxfxf iiii

211 0

2

x

x

xfxfxf iii (

(

!d

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20

Central Divided Difference

Hence showing that we have obtained a more accurate formula as the

error is of the order of . 20 x

x

f(x)

x-x x x+x

Figure 3 Graphical Representation of central difference approximation offirst derivative6/3/2011


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21

Example 3


300,8.9210010141014

ln2000 4

4

ee

v

v!

ttttR

where '' is given in m/s and ''t is given in seconds.

(a) Use central divided difference approximation of the first derivative ofto calculate the acceleration at . Use a step size of .

(b) Find the absolute relative true error for part (a).

tst 16! st 2 !

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22

Example 3 cont.

Solution

t

ttta iii

(

}

2

11 RR

16!it

18216

1

!!

(! ttt ii

142161

!!(! ttt

ii

221418

16RR

}a

4

1418 RR }

2!(t

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Example 3 cont.

188.91821001014

1014ln200018

4

4

v

v!R

m/s02.453!

148.91421001014

1014ln200014

4

4

v

v!R

m/s24.334!

4

141816 RR }a

4

24.33402.453 }

2m/s694.29}6/3/2011


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24

Example 3 cont.


100674.29

694.29674.29v

!t

%069157.0!

The exact value of the acceleration at from Example 1 is

2m/s674.2916 !as16!t

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25

Comparision of FDD, BDD, CDD

The results from the three difference approximations are given in Table 1.

Type of Difference

Approximation

ForwardBackward

Central

30.47528.915

29.695

2.69672.5584

0.069157

Table 1 Summary of a (16) using different divided difference approximations

16a

2/sm%t

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26

Finding the value of the derivative

within a prespecified tolerance

In real life, one would not know the exact value of the derivative so how

would one know how accurately they have found the value of the derivative.

A simple way would be to start with a step size and keep on halving the step

size and keep on halving the step size until the absolute relative approximate

error is within a pre-specified tolerance.

Take the example of finding for tvd t

tt 8.9

21001014

1014ln2000

4

4

v

v!R

at using the backward divided difference scheme.16!t6/3/2011


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2

1

0.5

0.25

0.125

28.91529.289

29.480

29.577

29.625

1.2792

0.64787

0.32604

0.16355

27


within a prespecified tolerance Cont.

Given in Table 2 are the values obtained using the backward differenceapproximation method and the corresponding absolute relativeapproximate errors.

t( tvd %a

Table 2 First derivative approximations and relative errors fordifferentt values of backward difference scheme

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within a prespecified tolerance Cont.

From the above table, one can see that the absolute relative

approximate error decreases as the step size is reduced. At 125.0!(t

the absolute relative approximate error is0.16355%

, meaning thatat least2 significant digits are correct in the answer.

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29

Finite Difference Approximation of

Higher Derivatives

One can use Taylor series to approximate a higher order derivative.

For example, to approximate xfdd , the Taylor series for

-ddd

dd

d!32

2 2!3

2!2

2 xxf

xxf

xxfxfxf iiiii

where

xxx ii 22 !

-321!3!2

xxf

xxf

xxfxfxf iiiii (ddd(dd(d!

where

xxx ii 1 !6/3/2011


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Finite Difference Approximation of

Higher Derivatives Cont.

Subtracting 2 times equation (4) from equation (3) gives

-3212 2 xxfxxfxfxfxf iiiii ddddd!

-ddd

!dd xxf

x

xfxfxfxf i

iii

i

22

12

x

x

xfxfxfxf iiii 0

22

12

$dd (5)

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31

Example 4


300,8.9210010141014

ln2000 4

4

ee

v

v

! ttttR

Use forward difference approximation of the second derivative

of to calculate the jerk at . Use a step size of .

tst 16! st 2 !

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33

Example 4 Cont.

208.92021001014

1014ln200020

4

4

v

v!R

m/s35.517!

188.91821001014

1014ln200018

4

4

v

v!R

s/02.453!

168.91621001014

1014ln200016

4

4

v

v!R

m/s07.392!6/3/2011


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34

Example 4 Cont.

4

07.39202.453235.51716

}j

3m/s84515.0}

The exact value of 16j can be calculated by differentiating

tt

t 8.921001014

1014ln2000

4

4

v

v!R

twice as

? Atdt

dta ! and ? Ata

dt

dtj !

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35

Example 4 Cont.

Knowing that

? At

t

dt

d 1ln ! and

2

11

ttdt

d!

8.921001014

1014

1014

210010142000

4

4

4

4

v

v

v

v!

tdt

dtta

t

t

3200

4.294040

!

8.92100210010141014

11014

210010142000 24

4

4

4

v

v

v

v

! t

t

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36

Example 4 Cont.

? A

2)3200(

18000t

tadt

dtj

!

!

3

2

m/s77909.0

)]16(3200[

1800016

!

!j


10077909.0

84515.077909.0v

!t

%4797.8!

Similarly it can be shown that

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37

Higher order accuracy of higher

order derivatives

The formula given by equation (5) is a forward difference approximation of

the second derivative and has the error of the order of x . Can we get

a formula that has a better accuracy? We can get the central difference

approximation of the second derivative.

The Taylor series for

-4321!4!3!2

xxf

xxf

xxf

xxfxfxf iiiiii (dddd

(ddd

(dd

(d!

where

xxx ii 1 !

(6)

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Higher order accuracy of higher

order derivatives Cont.

-4321!4!3!2

xxf

xxf

xxf

xxfxfxf iiiiii (dddd

(ddd

(dd

(d!

where

xxx ii 1 !

(7)

Adding equations (6) and (7), gives

12

2

42

11

xxfxxfxfxfxf iiiii ddddd!

12

22

2

11 xxf

x

xfxfxfxf iiiii

dddd

!dd

22

11 0

2x

x

xfxfxfxf iiii

!dd

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39

Example 5


300,8.921001014

1014ln2000

4

4

ee

v

v! tt

ttR

Use central difference approximation of second derivative of tocalculate the jerk at . Use a step size of .

tst 16! st 2 !

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41

Example 5 Cont.

188.91821001014

1014ln200018

4

4

v

v!R

m/s02.453!

168.91621001014

1014ln200016

4

4

v

v!R

m/s07.392!

148.91421001014

1014ln200014

4

4

v

v!R

m/s24.334!

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42

Example 5 Cont.

221416218

16RRR

}j

424.33407.392202.453

}


10077908.0

78.077908.0 v!t

3m/s77969.0}

%077992.0!

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44

Forward Difference Approximation

x

xfxxf

x

xf

0

lim

p

!d

For a finite '' x

x

xfxxfxf(

(}d

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45

x x+x

f(x)

Figure 1 Graphical Representation of forward difference approximation of first derivative.

Graphical Representation OfForward Difference

Approximation

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46

Example 1The upward velocity of a rocket is given as a function of time inTable 1.

Using forward divided difference, find the acceleration of the

rocket at .

t v(t)

s m/s

0 0

10 227.04

15 362.78

20 517.3522.5 602.97

30 901.67

Table 1 elocity as a function of time

s16!t6/3/2011


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47

Example 1 Cont.

t

ttta iii

(

}

RR 1

15!it

5

15201

!!!( ii ttt

To find the acceleration at , we need to choose thetwo values closest to , that also bracket to

evaluate it. The two points are and .

s16!ts16!t

s20!ts15!t

201 !it

s16!t

Solution

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48

Example 1 Cont.

2m/s914.30

5

78.36235.5175

152016

}

}

}

RRa

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49

Direct Fit Polynomials

'1' n nn

yxyxyxyx ,,,,,,,, 221100 -

thn

nn

n

nnxaxaxaaxP !

1

110 --

12121 12)(

!!dn

n

n

n

n

nxnaxanxaa

dx

xdPxP --

In this method, given data points

one can fit a order polynomial given by

To find the first derivative,

Similarly other derivatives can be found.

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50

Example 2-Direct Fit Polynomials

The upward velocity of a rocket is given as a function of time inTable 2.

Using the third order polynomial interpolant for velocity,find the acceleration of the rocket at .

t v(t)

s m/s

0 0

10 227.04

15 362.78

20 517.3522.5 602.97

30 901.67


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51

Example 2-Direct Fit Polynomials cont.

For the third order polynomial (also called cubic interpolation), we choose the velocity given by

332

210 tatataatv !

Since we want to find the velocity at , and we are using third order polynomial, we needto choose the four points closest to and that also bracket to evaluate it.

The four points are

04.227,10 !! oo tvt

78.362,15 11 !! tvt

35.517,20 22 !! tvt

97.602,5.22 33 !! tvt

Solution

s16!ts16!t s16!t

.5.22and,20,15,10 321 !!!! tttto

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52


such that

Writing the four equations in matrix form, we have

332

210 10101004.22710 aaaav !!

33

2

210

15151578.36215 aaaav !!

332

210 20202035.51720 aaaav !!

332

210 5.225.225.2297.6025.22 aaaav !!

!

97.602

35.517

78.362

04.227

1139125.5065.221

8000400201

3375225151

1000100101

3

2

1

0

a

a

a

a

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53


Solving the above four equations gives

3810.40 !a

289.211 !a

13065.02 !a

0054606.03 !aHence

5.2210,0054606.013065.0289.213810.4 32

3

3

2

210

ee!

!

tttt

tatataatv

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Figure 1 Graph of upward velocity of the rocket vs. time.6/3/2011


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56

Lagrange Polynomial

nn

yxyx ,,,, 11 - th

n 1In this method, given , one can fit a order Lagrangian polynomialgiven by

!

!n

i

iin xfxLxf0

)()()(

where n in )(xfn stands for the thn order polynomial that approximates the function

)(xfy ! given at )1( n data points as nnnn yxyxyxyx ,,,,......,,,, 111100 , and

{!

!

n

ijj ji

j

i

xx

xxx

0

)(

)(xi a weighting function that includes a product of )1( n terms with terms of

ij ! omitted.

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57

Then to find the first derivative, one can differentiate xfn

for other derivatives.

For example, the second order Lagrange polynomial passing through

221100 ,,,,, yxyxyx is

21202

10

12101

20

02010

21

2

xfxxxx

xxxxxf

xxxx

xxxxxf

xxxx

xxxxxf

!

Differentiating equation (2) gives

once, and so on

Lagrange Polynomial Cont.

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58

21202

1

2101

0

2010

2

222xf

xxxxxf

xxxxxf

xxxxxf

!dd

2

1202

101

2101

200

2010

212

222xf

xxxx

xxxxf

xxxx

xxxxf

xxxx

xxxxf

!

d

Differentiating again would give the second derivative as

Lagrange Polynomial Cont.

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59

Example 3

Determine the value of the acceleration at usingthe second order Lagrangian polynomial interpolation for

velocity.

t v(t)

s m/s

0 0

10 227.04

15 362.78

20 517.3522.5 602.97

30 901.67


s16!t

The upward velocity of a rocket is given as a function of time inTable 3.

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60

Solution

Example 3 Cont.

)()()()( 212

1

02

0

1

21

2

01

0

0

20

2

10

1 tvtt

tt

tt

tttv

tt

tt

tt

tttv

tt

tt

tt

tttv

!

0

2010

212 ttttt

tttta R

!

12101

202 ttttt

tttR

2

1202

102 ttttt

ttt

04.22720101510

201516216

!a

78.36220151015

2010162

35.517

15201020

1510162

35.51714.078.36208.004.22706.0 !

2m/s784.29!

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Backward DividedDifference

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62

Definition

ix

ix

.

x

xxfxf

x

xf

(

(

p(!d

0

lim

Slope at

f(x)

y

x

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63

Backward Divided Difference

xx (xx

x

xxfxfxf

(

($d

x

xxfxfxf iii

(

($d

)(xf

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64

Example

Example:


300,8.9210010141014

ln2000 4

4

ee

v

v

! ttttR

whereR given in m/s and t is given in seconds. Use backward difference approximation

Of the first derivative of t to calculate the acceleration at .16st! Use a step size of.2st!(

t

ttta iii

(

$ 1

RRSolution:

16!it6/3/2011


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65

2 !t

ttt ii (!1 14216 !!

2

141616

RR !a

168.91621001014

1014ln200016

4

4

v

v!R s/07.392!

148.91421001014

1014ln200014

4

4

v

v!R s/24.334!

Example (contd.)

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66

Hence


tt

t 8.921001014

1014ln2000

4

4

v

v!R

as

Example (contd.)

2/915.2824.33407.3922

)14()16(16 sma !!

!

RR

? A

2/674.29)16(

3200

4.294040)()(

sa

t

tt

dt

dta

!

!! R

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67


Example (contd.)

100v

!TrueValue

eValueApproximatTrueValuetI

%557.2

100674.29

915.28674.29

!

v

!

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68

Effect Of Step Size

xexf 49)( !

Value of )2.0('f Using backward Divided difference method.

h )2.0('

f aE aI % S g f digits

tE

tI %

0.05 72.61598 7.50 49 9. 65 77

0.025 76.24 76 .627777 4.758129 1 .87571 4.8 7418

0.0125 78.14946 1.905697 2.4 8529 1 1.97002 2.458849

0.00625 79.12627 0.976817 1.2 4504 1 0.99 20 1.2 9648

0.00 125 79.62081 0.4945 0.62111 1 0.49867 0.622404

0.00156 79.86962 0.248814 0. 11525 2 0.24985 0. 1185

0.000781 79.99442 0.124796 0.156006 2 0.12506 0.156087

0.000 91 80.05691 0.062496 0.078064 2 0.06256 0.078084

0.000195 80.08818 0.0 1272 0.0 9047 0.0 129 0.0 9052

9.77E-05 80.10 8 0.015642 0.019527 0.01565 0.019529

4.88E-05 80.11165 0.00782 0.009765 0.00782 0.0097656/3/2011


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69

Effect of Step Size in Backward

Divided Difference Method

8

9

1 3 9 11

Number of times the step size is halved, n

f'(0.

2

Initial step size=0.05

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Effect of Step Size on Approximate

Error

0

4

5 7

Number of steps involved, n

Ea


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Effect of Step Size on Least


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72


Number of Significant Digits

Correct

2

2.

.

2 7


Least

numberofsignif

ican

digitscorrect


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73

Effect of Step Size on True Error

2

3

7

2 6 2


Et


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Central Divided Difference


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76

Definition

ix

.

x

xxfxf

x

xf(

(

p(!d

0

lim

f(x)

Slope at

x

y

ix6/3/2011


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78

Example

Example:


300,8.9210010141014

ln2000 4

4

ee

v

v!

ttttR

where R given in m/s and t is given in seconds. Use central difference approximation of

the first derivative of t to calculate the acceleration at .16st! Use a step size of.2st!(

t

ttta iii

(

$

2

11 RR

Solution:

16!it6/3/2011


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79

2 !t

ttt ii 1 ! 18216 !!

4

1418

)2(2

141816

RRRR !

!a

188.91821001014

1014ln2000184

4

v

v!R s/02.453!

148.91421001014

1014ln200014

4

4

v

v!R sm /24.334!

Example (contd.)

ttt ii (!1 14216 !!

6/3/2011


80/102

80

Hence


tt

t 8.921001014

1014ln2000

4

4

v

v!R

as

Example (contd.)

2/695.2924.33402.453

4

)14()18(16 sma !!

!

RR

? A

2/674.29)16(

3200

4.294040)()(

sa

t

tt

dt

dta

!

!! R

6/3/2011


81/102

81


Example (contd.)

100v

!True alue

e aluepproxi atTrue aluetI

%070769.0

100674.29

695.29674.29

!

v

!

6/3/2011


82/102

82

Effect Of Step Size

xexf 49)( !

Value of )2.0('f Using Central Divided Difference difference method.

h )(

f aE aI % Sig ifica tdigits

tE

tI %

0.05 80.65467 -0.5 520 0.668001

0.025 80.25 07 -0.4016 0.500417 1 -0.1 60 0.16675

0.0125 80.15286 -0.100212 0.125026 2 -0.0 9 0.041672

0.00625 80.12782 -0.025041 0.0 1252 -0.008 5 0.010417

0.00 125 80.12156 -0.00626 0.00781 -0.00209 0.0026040.00156 80.12000 -0.001565 0.00195 4 -0.00052 0.000651

0.000781 80.11960 -0.000 91 0.000488 5 -0.0001 0.00016

0.000 91 80.11951 -9.78 -05 0.000122 5 -0.0000 4.07 -05

0.000195 80.11948 -2.45 -05 .05 -05 6 -0.00001 1.02 -05

9.77 -05 80.11948 -6.11 -06 7.6 -06 6 0.00000 2.54 -06

4.88 -05 80.11947 -1.5 -06 1.91 -06 7 0.00000 6. 6 -076/3/2011


83/102



84/102

84


Error


E(a)


6/3/2011

Effect of Step Size on Absolute


85/102

85


Relative Approximate Error

5

5 8

Number eps involved, n

|E(a)|,


6/3/2011



86/102

86



Correct

6

8

6 8

Number of s teps involved, n

Least

numberofsignificantdigits

correct


6/3/2011


87/102

87


8


E(t)


6/3/2011


88/102


89/102

6/3/2011 89

Forward Divided Difference


90/102

90

Definition

ix

.

x

xfxxf

xxf

0

lim

p!d

f(x)

Slope at

x

y

ix6/3/2011


91/102

91

Forward Divided Difference

xx ( xx

x

xfxxfxf

(

($d

x

xfxxfxf iii

(

($d

)(xf

6/3/2011


92/102

92

Example

Example:


300,8.921001014

1014ln2000

4

4

ee

v

v! tt

ttR

whereR given in m/s and t is given in seconds. Use forward difference approximation of

the first derivative of t to calculate the acceleration at .16st! Use a step size of.2st!(

t

ttta iii

1 $ Solution:

16!it6/3/2011


93/102

93

2 !t

ttt ii 1 ! 18216 !!

2

161816 RR !a

188.91821001014

1014ln200018

4

4

v

v!R s/02.453!

168.91621001014

1014ln200016

4

4

vv!R s/07.392!

Example (contd.)

6/3/2011


94/102

94

Hence


tt

t 8.921001014

1014ln2000

4

4

v

v!R

as

Example (contd.)

2/475.3007.39202.453

2

)16()18(16 sma !!

!

RR

? A

2/674.29)16(

3200

4.294040)()(

sa

t

tt

dt

dta

!

!! R

6/3/2011


95/102

95


Example (contd.)

100v

!TrueValue

eValueApproximatTrueValuetI

%6993.2

100674.29

475.30674.29

!

v

!

6/3/2011


96/102

96

Effect Of Step Size

xexf 49)( !

Value of )2.0('f Using forward difference method.

h )(

f aE aI % Sig ifica tdigits

tE

tI %

0.05 88.69 6 -8.57 89 10.701 8

0.025 84.262 9 -4.4 0976 5.258546 0 -4.14291 5.170918

0.0125 82.15626 -2.106121 2.56 555 1 -2.0 679 2.54219

0.00625 81.129 7 -1.0269 1.265756 1 -1.00989 1.260482

0.00 125 80.622 1 -0.507052 0.62892 1 -0.50284 0.627612

0.00156 80. 70 7 -0.251944 0. 1 479 2 -0.25090 0. 1 152

0.000781 80.24479 -0.125579 0.156494 2 -0.125 2 0.15641

0.000 91 80.18210 -0.062691 0.078186 2 -0.0626 0.078166

0.000195 80.15078 -0.0 1 21 0.0 9078 -0.0 1 0 0.0 907

9.77 -05 80.1 512 -0.015654 0.0195 5 -0.01565 0.0195 4

4.88 -05 80.127 0 -0.007826 0.009767 -0.00782 0.0097666/3/2011

Effect of Step Size in Forward


97/102

97

Effect of Step Size in Forward

Divided Difference Method

7

9

7 9

Number of times s tep size halved, n

f'(0.

2)


6/3/2011



98/102

98


Error

8

Number of times step size halved, n

Ea


6/3/2011



99/102

99


Relative Approximate Error

1

2

3

4

5

6

0 1 2 3 4 5 6 7 8 9 10 11 12

Nu be f t es step ze a e n

|Ea|%


6/3/2011



100/102

100


Correct

0

1

1 10 11

Number of times s tep size halved, n

Least

numberofsignific

antdigits

correct


6/3/2011


101/102

101



00 10 1

Number mes step size halved, n

Et


6/3/2011



102/102


Relative True Error

0

2

1 0

1 2

1 2 10 1 1

Number oft mes step size halved, n

|Et|

%


lecture on numerical differentiation

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