10/27/2015 1 Differentiation-Continuous Functions Computer Engineering Majors Authors: Autar Kaw, Sri Harsha Garapati.

04/20/23http://

numericalmethods.eng.usf.edu 1

Differentiation-Continuous Functions

Computer Engineering Majors

Authors: Autar Kaw, Sri Harsha Garapati

http://numericalmethods.eng.usf.eduTransforming Numerical Methods Education for STEM

Undergraduates

http://numericalmethods.eng.usf.edu/

Differentiation – Continuous Functions

http://numericalmethods.eng.usf.edu


http://numericalmethods.eng.usf.edu3

Forward Difference Approximation

x

xfxxf

xxf

Δ

Δ

0Δ

lim

For a finite

'Δ' x

x

xfxxfxf


x x+Δx

f(x)

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

Graphical Representation Of Forward Difference

Approximation


5

Example 1There is strong evidence that the first level of processing what we see is done in the retina. It involves detecting something called edges or positions of transitions from dark to bright or bright to dark points in images. These points usually coincide with boundaries of objects. To model the edges, derivatives of functions such as

0,1

0,1)(

xe

xexf

ax

ax

need to be found.

a)Use forward divided difference approximation of the first derivative of to calculate its derivative at for . Use a step size of . Also calculate the absolute relative true error.

b)Repeat the procedure from part (a) with the same data except choose . Does the estimate of the derivative increase or decrease? Also calculate the relative true error.

05.0x xf 24.0a1.0x

12.0a


6

Example 1 Cont.

x

xfxfxf iii

1'

1.0ix

15.005.01.0

1

xxxi

023714.01)1.0( )1.024.0(

ef

035360.01)15.0( )15.024.0(

ef

Solution:

24.0a

05.0x


7

Example 1 Cont.

05.0

1.015.01.0' ff

f

05.0

023714.00.035360

0.23291

The exact value of 1.0'f can be calculated by differentiating

0,1 xexf ax

as

xfdx

dxf '


8

Knowing that

axax aeedx

d

x

ax

ax

eae

edx

dxf

24.0

'

24.0

)1(

23431.0

)(24.0(1.0 )1.024.0('

ef

Example 1 Cont.

we find


9

The absolute relative true error is

%59761.0

1000.23431

0.232910.23431

100Value True

Value eApproximat-Value True

t

Example 1 Cont.


10

Example 1 Cont.

b)

011928.0

11.0 )1.012.0(

ef

017839.01)15.0( )15.012.0(

ef

12.0a

05.0

1.015.01.0' ff

f

05.0

011928.00.017838

0.11821


11

)1(' axedx

dxf

axae

)1.012.0(' )(12.0(1.0 ef

0.11856

Example 1 Cont.

xe 12.012.0


12

100Value True

Value eApproximat-Value Truet


1000.11857

0.118210.11857

0.29940%

The estimate of the derivative decreased.

Example 1 Cont.


Backward Difference Approximation of the

First Derivative

We know

x

xfxxf

xxf

Δ

Δ

0Δ

lim

For a finite 'Δ' x ,

x

xfxxfxf

If 'Δ' x is chosen as a negative number,

x

xfxxfxf

x

xxfxf

Δ

Δ


Backward Difference Approximation of the First Derivative Cont.

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

anotherpoint 'Δ' x behind as 1 ixx . This gives

x

xfxfxf iii

1

1

1

ii

ii

xx

xfxf

where

1Δ ii xxx


xx-Δx

x

f(x)

Figure 2 Graphical Representation of backward difference approximation of first derivative

Backward Difference Approximation of the First Derivative Cont.


16


0,1

0,1)(

xe

xexf

ax

ax

need to be found.

a)Use backward divided difference approximation of the first derivative of to calculate its derivative at for . Use a step size of . Also calculate the absolute relative true error.


05.0x xf 24.0a1.0x

12.0a


17

Example 2 Cont.Solution

a) x

xfxfxf iii

1

1.0ix05.0x

05.005.01.0

1

xxx ii

24.0a


18

)1.024.0(11.0 ef

0.023714

)05.024.0(1)05.0( ef

0.011928

05.0

05.01.01.0' ff

f

05.0

011928.0023714.0

0.23572

Example 2 Cont.


19


100Value True


1000.23431

0.235720.23431

%60241.0

Example 2 Cont.


20

b)

11893.005.0

0059820.0011928.005.0

05.01.01.0'

fff

)1.012.0(11.0 ef

0.011928)05.012.0(1)05.0( ef

0.0059820

Example 2 Cont.

12.0a


21


100Value True


1000.11857

0.118930.11857

%0.30060The estimate of the derivative decreased.

Example 2 Cont.


Derive the forward difference approximation from Taylor series

Taylor’s theorem says that if you know the value of a function '' f at a point

ixand all its derivatives at that point, provided the derivatives are

continuous between ix and 1ix , then

2

111 !2 iii

iiiii xxxf

xxxfxfxf

Substituting for convenience ii xxx 1Δ

2

1 Δ!2

Δ xxf

xxfxfxf iiii

xxf

x

xfxfxf iiii !2

1

xx

xfxfxf iii

01


Derive the forward difference approximation from Taylor series Cont.

The x0term shows that the error in the approximation is of the order

of xΔ 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 x0

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

the first derivative is called the Central difference approximation of

the first derivative.


Derive the forward difference approximation from Taylor series Cont.

From Taylor series

32

1 Δ!3

Δ!2

Δ xxf

xxf

xxfxfxf iiiii

32

1 Δ!3

Δ!2

Δ xxf

xxf

xxfxfxf iiiii

Subtracting equation (2) from equation (1)

3

11 Δ!3

2Δ2 x

xfxxfxfxf i

iii

211

!32x

xf

x

xfxfxf iiii

211 02

xx

xfxfxf iii


Central Divided Difference

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

error is of the order of . 2Δ0 x

x

f(x)

x-Δx x x+Δx

Figure 3 Graphical Representation of central difference approximation of first derivative

http://numericalmethods.eng.usf.e

du

26


0,1

0,1)(

xe

xexf

ax

ax

need to be found.

a)Use central divided difference approximation of the first derivative of to calculate its derivative at for . Use a step size of . Also calculate the absolute relative true error.


05.0x xf 24.0a1.0x

12.0a


du

27

Example 3 cont.

Solution

a) x

tftfxf xxi

211'

1.0ix

xxx ii 1

05.01.0 15.0

xxx ii 1

05.01.0

05.0

24.0a

05.0x


du

28

Example 3 cont.

05.02

05.015.01.0' ff

f

1.0

05.015.0 ff

)15.024.0(115.0 ef

0.035360

)05.024.0(105.0 ef

0.011928


du

29

Example 3 cont.

1.0

05.015.01.0' ff

f

1.0

0.0119280.035360

0.23431


100Value True


1000.23431

0.234310.23431

%0.0024000


du

30

Example 3 cont.

1.0

05.015.01.0' ff

f

)15.012.0(115.0 ef

0.017839

)05.012.0(105.0 ef

0.0059820

b) 12.0a

1.0

0.00598200.017839

0.11857


du

31

Example 3 cont.


100Value True

Value eApproximat-Value Truext

1000.11857

0.118570.11857x

%106.0000 4

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


du

32

Comparision

Type of DifferenceApproximation

ForwardBackwardCentral

0.23291 0.23572 0.23431

0.597610.602410.0024000

0.118210.118930.11857

0.29940 0.30060 6.0000

24.0),1.0(' af 24.0%, at 12.0),1.0(' af 12.0%, at

Table 1 Summary of using different divided difference approximations

410

1.0f


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 tv

tt

t 8.921001014

1014ln2000

4

4

at using the backward divided difference scheme. 16t

210.50.250.125

28.91529.28929.48029.57729.625

1.27920.647870.326040.16355


Finding the value of the derivative within a prespecified tolerance Cont.

Given in Table 2 are the values obtained using the backward difference approximation method and the corresponding absolute relative approximate errors.

t tv %a

Table 2 First derivative approximations and relative errors for different Δt values of backward difference scheme


Finding the value of the derivative 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 125.0t

the absolute relative approximate error is 0.16355%, meaning that

at least 2 significant digits are correct in the answer.


Finite Difference Approximation of Higher Derivatives

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

For example, to approximate xf , the Taylor series for

32

2 Δ2!3

Δ2!2

Δ2 xxf

xxf

xxfxfxf iiiii

where

xxx ii Δ22

321 !3!2

xxf

xxf

xxfxfxf iiiii

where

xxx ii Δ1


Finite Difference Approximation of Higher Derivatives Cont.

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

3212 ΔΔ2 xxfxxfxfxfxf iiiii

xxfx

xfxfxfxf i

iiii Δ

Δ

22

12

xx

xfxfxfxf iiii Δ0

Δ

22

12

(5)


Example 4

The velocity of a rocket is given by

300,8.921001014

1014ln2000

4

4

tt

tt

Use forward difference approximation of the second derivative of to calculate the jerk at . Use a step

size of .

tνst 16 st 2Δ


Example 4 Cont.

Solution

2

12 2

t

ttttj iiii

16it

18216

1

ttt ii

202216

22

ttt ii

22

161822016

j

2t


Example 4 Cont.

208.92021001014

1014ln200020

4

4

m/s35.517

188.91821001014

1014ln200018

4

4

sm /02.453

168.91621001014

1014ln200016

4

4

m/s07.392


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

twice as

tνdt

dta and ta

dt

dtj


Example 4 Cont.

Knowing that

t

tdt

d 1ln and

2

11

ttdt

d

8.921001014

1014

1014

210010142000

4

4

4

4

tdt

dtta

t

t

3200

4.294040

8.9210021001014

10141

1014

210010142000

24

4

4

4

t

t


Example 4 Cont.

2)3200(

18000

t

tadt

dtj

3

2

m/s77909.0 )]16(3200[

1800016

j


10077909.0

84515.077909.0

t

% 4797.8

Similarly it can be shown that


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 xΔ . 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

where

xxx ii Δ1

(6)


Higher order accuracy of higher order derivatives Cont.

4321 !4!3!2

xxf

xxf

xxf

xxfxfxf iiiiii

where

xxx ii Δ1

(7)

Adding equations (6) and (7), gives

12

ΔΔ2

42

11

xxfxxfxfxfxf iiiii

12

Δ

Δ

2 2

211 xxf

x

xfxfxfxf iiiii

2

211 Δ0

Δ

2x

x

xfxfxfxf iiii


Example 5

The velocity of a rocket is given by

300,8.921001014

1014ln2000

4

4

tt

tt

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

tνst 16 st 2Δ


Example 5 Cont.

Solution

2

11 2

t

tttta iiii

16it

18216

1

ttt ii

14216

1

ttt ii

22

141621816

j

2t


Example 5 Cont.

188.91821001014

1014ln200018

4

4

m/s02.453

168.91621001014

1014ln200016

4

4

m/s07.392

148.91421001014

1014ln200014

4

4

m/s24.334


Example 5 Cont.

22

141621816

j

4

24.33407.392202.453


10077908.0

78.077908.0

t

3m/s77969.0

%077992.0

Additional Resources

For all resources on this topic such as digital audiovisual lectures, primers, textbook chapters, multiple-choice tests, worksheets in MATLAB, MATHEMATICA, MathCad and MAPLE, blogs, related physical problems, please visit

http://numericalmethods.eng.usf.edu/topics/continuous_02dif.html



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10/27/2015 1 Differentiation-Continuous Functions Computer Engineering Majors Authors: Autar Kaw, Sri Harsha Garapati.

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