Numerical Differentiation and Integration

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NUMERICAL DIFFERENTIATION AND INTEGRATION

ENGR 351 Numerical Methods for Engineers

Southern Illinois University Carbondale

College of EngineeringDr. L.R. ChevalierDr. B.A. DeVantier

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Numerical Differentiation and Integration

• Calculus is the mathematics of change.• Engineers must continuously deal with

systems and processes that change, making calculus an essential tool of our profession.

• At the heart of calculus are the related mathematical concepts of differentiation and integration.

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Differentiation

• Dictionary definition of differentiate - “to mark off by differences, distinguish; ..to perceive the difference in or between”

• Mathematical definition of derivative - rate of change of a dependent variable with respect to an independent variable

x

xfxxf

xy ii

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y

x

f x x f x

xi i

f xi

f x xi

y

x

f(x)

x

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Integration• The inverse process of differentiation• Dictionary definition of integrate - “to

bring together, as parts, into a whole; to unite; to indicate the total amount”

• Mathematically, it is the total value or summation of f(x)dx over a range of x. In fact the integration symbol is actually a stylized capital S intended to signify the connection between integration and summation.

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f(x)

x

I f x dxa

b

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Mathematical Backgroundd

dxx

d

dxe

d

dxx

d

dxa

d

dxx

d

dxx

d

dxx

if u and v are functions of x

d

dxu

d

dxuv

x

x

n

n

sin ? ?

cos ? ?

tan ? ?

ln ?

? ( ) ?

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udv

u du

a dx

dx

x

e dx

n

bx

ax

?

?

?

?

?

Mathematical Background

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Overview

• Newton-Cotes Integration FormulasTrapezoidal ruleSimpson’s RulesUnequal SegmentsOpen Integration

• Integration of EquationsRomberg Integration Gauss QuadratureImproper Integrals

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Overview

• Numerical DifferentiationHigh accuracy formulasRichardson’s extrapolationUnequal spaced dataUncertain data

• Applied problems

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Specific Study Objectives• Understand the derivation of the

Newton-Cotes formulas• Recognize that the trapezoidal and

Simpson’s 1/3 and 3/8 rules represent the areas of 1st, 2nd, and 3rd order polynomials

• Be able to choose the “best” among these formulas for any particular problem

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Specific Study Objectives

• Recognize the difference between open and closed integration formulas

• Understand the theoretical basis of Richardson extrapolation and how it is applied in the Romberg integration algorithm and for numerical differentiation

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Specific Study Objectives

• Recognize why both Romberg integration and Gauss quadrature have utility when integrating equations (as opposed to tabular or discrete data).

• Understand the application of high-accuracy numerical-differentiation.

• Recognize data error on the processes of integration and differentiation.

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Newton-Cotes Integration

• Common numerical integration scheme• Based on the strategy of replacing a

complicated function or tabulated data with some approximating function that is easy to integrate

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I f x dx f x dx

f x a a x a x

a

b

n

a

b

n nn

0 1 ....

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I f x dx f x dx

f x a a x a x

a

b

n

a

b

n nn

0 1 .... fn(x) is an nth orderpolynomial

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The approximation of an integral by the area under- a first order polynomial- a second order polynomial

We can also approximated the integral by using a series of polynomials applied piece wise.

0

1

2

3

4

5

0 5 10

x

f(x)

0

1

2

3

4

5

0 5 10

x

f(x)

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0

1

2

3

4

5

0 1 2 3 4 5 6 7 8 9 10

x

f(x)

An approximation of an integral by the area under straight line segments.

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Newton-Cotes Formulas

• Closed form - data is at the beginning and end of the limits of integration

• Open form - integration limits extend beyond the range of data.

0

1

2

3

4

5

0 1 2 3 4 5 6 7 8 9 10

x

f(x)

0

1

2

3

4

5

0 1 2 3 4 5 6 7 8 9 10

x

f(x)

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Trapezoidal Rule

• First of the Newton-Cotes closed integration formulas

• Corresponds to the case where the polynomial is a first order

I f x dx f x dx

f x a a xa

b

a

b

n

1

0 1

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I f x dx f x dx

f x a a xa

b

a

b

n

1

0 1

A straight line can be represented as:

f x f af b f a

b ax a1

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I f x dx f x dx

f af b f a

b ax a dx

a

b

a

b

a

b

1

Integrate this equation. Results in the trapezoidal rule.

I b a

f a f b

2

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I b a

f a f b

2

The concept is the same but the trapezoid is on its side.

heig

ht

base

base

widthheig

ht

heig

ht

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Error of the Trapezoidal Rule

E f b a

where a b

t

1

123

' '

This indicates that is the function being integrated islinear, the trapezoidal rule will be exact.

Otherwise, for section with second and higher order derivatives (that is with curvature) error can occur.

A reasonable estimate of x is the average value of b and a

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Multiple Application of the Trapezoidal Rule

• Improve the accuracy by dividing the integration interval into a number of smaller segments

• Apply the method to each segment• Resulting equations are called

multiple-application or composite integration formulas

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I f x dx f x dx f x dx

I hf x f x

hf x f x

hf x f x

x

x

x

x

x

x

n n

n

n

( ) ( ) ( )0

1

1

2

1

0 1 1 2 1

2 2 2

Multiple Application of the Trapezoidal Rule

where there are n+1 equally spaced base points.

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I hf x f x

hf x f x

hf x f x

I b af x f x f x

n

x

x

x

x

x

x

n n

i ni

n

n

n

( ) ( ) ( )0

1

1

2

1

0 1 1 2 1

01

1

2 2 2

2

2

We can group terms to express a general form

} }width average height

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I b af x f x f x

n

i ni

n

01

1

2

2} }

width average height

The average height represents a weighted averageof the function values

Note that the interior points are given twice the weightof the two end points

Eb a

nfa

3

212' '

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EXAMPLEEvaluate the following integral using the trapezoidal rule and h = 0.1

I b af x f x f x

n

i ni

n

01

1

2

2

I e dxx2

1

1 6. Example Problem

0

10

20

30

40

50

60

0 0.5 1 1.5 2

xf(

x)

hb a

n

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Simpson’s 1/3 Rule

• Corresponds to the case where the function is a second order polynomial

I f x dx f x dx

f x a a x a x

a

b

a

b

n

2

0 1 22

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• Designate a and b as x0 and x2, and estimate f2(x) as a second order Lagrange polynomial

I f x dx f x dx

x x x x

x x x xf x dx

a

b

a

b

x

x

2

1 2

0 1 0 20

0

2

.......

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• After integration and algebraic manipulation, we get the following equations

Ih

f x f x f x

b af x f x f x

34

4

6

0 1 2

0 1 2} }width average height

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Error

E f b a

where a b

t

1

123

' '

Single application of Trapezoidal Rule.

Single application of Simpson’s 1/3 Rule

E f b at 1

28804 5( )

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Multiple Application of Simpson’s 1/3 Rule


I b a

f x f x f x f x

n

Eb a

nf

x

x

x

x

x

x

i j nj

n

i

n

a

n

n

( ) ( ) ( )

, , .., , ..

0

1

1

2

1

02 4 6

2

1 3 5

1

5

44

4 2

3

180

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I b a

f x f x f x f x

n

i j nj

n

i

n

02 4 6

2

1 3 5

1

4 2

3, , .., , ..

The odd points represent the middle term for each application. Hence carry the weight 4The even points are common to adjacent applications and are counted twice.

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• Corresponds to the case where the function is a third order polynomial

I f x dx f x dx

f x a a x a x a x

Ih

f x f x f x f x

a

b

a

b

n

3

0 1 22

33

0 1 2 3

38

3 3

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Integration of Unequal Segments

• Experimental and field study data is often unevenly spaced

• In previous equations we grouped the term (i.e. hi) which represented segment width.

I b af x f x f x

n

I hf x f x

hf x f x

hf x f x

i ni

n

n n

01

1

0 1 1 2 1

2

2

2 2 2

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Integration of Unequal Segments

• We should also consider alternately using higher order equations if we can find data in consecutively even segments

trapezoidalrule

1/3rule

3/8rule

trapezoidalrule

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EXAMPLEIntegrate the following using the trapezoidal rule, Simpson’s 1/3 Rule and a multiple application ofthe trapezoidal rule with n=2. Compare results withthe analytical solution.

xe dxx2

0

4

0

5000

10000

15000

20000

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5

x

f(x)

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0

5000

10000

15000

20000

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5

x

f(x)


f(2) = 109.196

Ih

f x f x f x

b af x f x f x

t

34

4

6

4 00 4 109 196 1192383

68240 41

5216 93 8240 41

5216 93100 57 96%

0 1 2

0 1 2

. ..

. .

..

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I b af x f x f x

n

i ni

n

t

01

1

2

2

4 00 2 109 196 1192383

2 212142 22

5216 93 12142 22

5216 93100 133%

. ..

. .

.

Multiple Application ofthe Trapezoidal Rule

We are obviously not doing very well on our estimates.Lets consider a scheme where we “weight” the estimates....end of example

0

5000

10000

15000

20000

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5

x

f(x)

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Integration of Equations

• Integration of analytical as opposed to tabular functions

• Romberg Integration Richardson’s ExtrapolationRomberg Integration Algorithm

• Gauss Quadrature• Improper Integrals

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Richardson’s Extrapolation• Use two estimates of an integral to compute a

third more accurate approximation• The estimate and error associated with a

multiple application trapezoidal rule can be represented generally as:I = I(h) + E(h)where I is the exact value of the integralI(h) is the approximation from an n-segment

applicationE(h) is the truncation errorh is the step size (b-a)/n

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Make two separate estimates using step sizesof h1 and h2 .

I(h1) + E(h1) = I(h2) + E(h2)

Recall the error of the multiple-application of the trapezoidalrule

Eb a

h f12

2 ' '

Assume that is constant regardless of the step sizef ' '

E h

E h

h

h1

2

12

22

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E h

E h

h

h

E h E hh

h

1

2

12

22

1 21

2

2

Substitute into previous equation:

I(h1) + E(h1) = I(h2) + E(h2)

E h

I h I h

h

h

21 2

1

2

2

1

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Thus we have developed an estimate of the truncation error in terms of the integral estimates and their step sizes. This estimate can then be substituted into:

I = I(h2) + E(h2)

to yield an improved estimate of the integral:

I I hh

h

I h I h

2

1

2

2 2 1

1

1

E h

I h I h

h

h

21 2

1

2

2

1

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I I hh

h

I h I h

2

1

2

2 2 1

1

1

What is the equation for the special case where the interval is halved?

i.e. h2 = h1 / 2

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hh

hh

I I h I h I h

collecting terms

I I h I h

1

2

2

2

2 2 2 1

2 1

2 2

1

2 1

4

3

1

3

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EXAMPLE

Use Richardson’s extrapolation to evaluate:

xe dxx2

0

4

0

5000

10000

15000

20000

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5

x

f(x)

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I I h I h

I I h I h

I I h I h

II I

m l

m l

j k

kj k j k

k

4

3

1

316

15

1

1564

63

1

63

4

4 1

2 1

11 1 1

1,, ,

We can continue to improve the estimate by successive halving of the step size to yield a general formula:

k = 2; j = 1

ROMBERG INTEGRATION

Note:the subscriptsm and l refer tomore and lessaccurate estimates

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Error orders for j values i = 1 i = 2 i = 3 i = 4j O(h2) O(h4) O(h6) O(h8)

1 h I1,1 I1,2 I1,3 I1,4

2 h/2 I2,1 I2,2 I2,3

3 h/4 I3,1 I3,2

4 h/8 I4,1

Trapezoidal Simpson’s 1/3 Simpson’s 3/8 Boole’s

Rule Rule Rule Rule

hI63

1hI

63

64I m

Following a similar pattern to Newton divided differences, Romberg’s Table can be produced

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Gauss Quadrature

f(x)

x

f(x)

x

Extend the areaunder the straightline

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Method of Undetermined Coefficients

Recall the trapezoidal rule

I b a

f a f b

2

This can also be expressed as

I c f a c f b 0 1

where the c’s are constant

Before analyzingthis method,answer this question.What are twofunctions thatshould be evaluated exactlyby the trapezoidalrule?

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The two cases that should be evaluated exactlyby the trapezoidal rule: 1) y = constant 2) a straight line

f(x)

x

y = 1

(b-a)/2-(b-a)/2

f(x)

x

y = x

(b-a)/2

-(b-a)/2

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Thus, the following equalities should hold.

I c f a c f b

c c dx

cb a

cb a

xdx

b a

b a

b a

b a

0 1

0 12

2

0 12

2

1

2 2

FOR y=1since f(a) = f(b) =1

FOR y =xsince f(a) = x =-(b-a)/2andf(b) = x =(b-a)/2

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Evaluating both integrals

c c b a

cb a

cb

0 1

0 12

1

20

For y = 1

For y = x

Now we have two equations and two unknowns, c0 and c1.

Solving simultaneously, we get :

c0 = c1 = (b-a)/2

Substitute this back into: I c f a c f b 0 1

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I b a

f a f b

2

We get the equivalent of the trapezoidal rule.

DERIVATION OF THE TWO-POINT GAUSS-LEGENDRE FORMULA

I c f x c f x 0 0 1 1

Lets raise the level of sophistication by:- considering two points between -1 and 1- i.e. “open integration”

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f(x)

x-1 x0 x1 1

Previously ,we assumed that the equation fit the integrals of a constant and linear function.

Extend the reasoning by assuming that it also fits the integral of a parabolic and a cubic function.

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c f x c f x dx c c

c f x c f x xdx c x c x

c f x c f x x dx c x c x

c f x c f x x dx c x c x

0 0 1 1

1

1

0 1

0 0 1 1

1

1

0 0 1 1

0 0 1 12

1

1

0 02

1 12

0 0 1 13

1

1

0 03

1 13

1 2 1 1 2

0 0

2 3 2 3

0 0

`

/ /

Solve these equations simultaneously

f(xi) is either 1, xi, xi2 or xi

3

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c c

x

x

I f f

0 1

0

1

1

1

31

31

3

1

3

This results in the following

The interesting result is that the simple additionof the function values at 1

3

1

3and

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However, we have set the limit of integration at -1 and 1.

This was done to simplify the mathematics. A simplechange in variables can be use to translate other limits.

Assume that the new variable xd is related to theoriginal variable x in a linear fashion.

x = a0 + a1xd

Let the lower limit x = a correspond to xd = -1 and the upperlimit x=b correspond to xd=1

a = a0 + a1(-1) b = a0 + a1(1)

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a = a0 + a1(-1) b = a0 + a1(1)

SOLVE THESE EQUATIONSSIMULTANEOUSLY

ab a

ab a

0 12 2

x a a x

b a b a xd

d

0 1 2

substitute

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x a a x

b a b a x

dxb a

dx

dd

d

0 1 2

2

These equations are substituted for x and dx respectively.

Let’s do an example to appreciate the theorybehind this numerical method.

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EXAMPLEEstimate the following using two-point Gauss Legendre:

xe dxx2

0

4

0

5000

10000

15000

20000

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5

x

f(x)

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Higher-Point Formulas

I c f x c f x c f xn n 0 0 1 1 1 1

For two point, we determined that c0 =c1 = 1

For three point:

c0 = 0.556 x0=-0.775c1= 0.889 x1=0.0c2= 0.556 x2=0.775

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Higher-Point Formulas

Your text goes on to provide additional weightingfactors (ci’s) and function arguments (xi’s)in Table 22.1 p. 623.

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Improper Integrals

f x dxb

How do we deal withintegrals that do nothave finite limitsand boundedintegrands?

Our answer will focuson improper integralswhere one limit (upper orlower) is infinity.

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f x dx f x dx f x dxb A

A

b

Choose -A so that is is sufficiently large toapproach zero asymptotically.

Once chosen, evaluate the second part using aNewton-Cotes formula.

Evaluate the first part with the following identity.

f x dxt

ft

dt for aba

b

b

a

1 1

021

1

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Numerical Differentiation• Forward finite divided difference• Backward finite divided difference• Center finite divided difference• All based on the Taylor Series

.........

!2

''' 2

1 hxf

hxfxfxf iiii

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Forward Finite Difference

21

1

21

2

'''

'

.........!2

'''

hOhxf

h

xfxfxf

hOh

xfxfxf

hxf

hxfxfxf

iiii

iii

iiii

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Forward Divided Difference

f xf x f x

x xO x x

f

hO hi

i i

i ii i

i'

1

11

f(x)

x

(xi, yi)

(x i+1,y i+1)

estimateac

tual

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f xf

hO hi

i'

first forward divided difference

Error is proportional tothe step size

O(h2) error is proportional to the square of the step size

O(h3) error is proportional to the cube of the step size

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f(x)

x

estimate

actual

(xi,yi)

(xi-1,yi-1)

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f x f x f x hf x

h

f x f x f x hf x

h

f xf x f x

h

f

h

i i ii

i i ii

ii i i

12

12

1

2

2

'' '

!......

'' '

!.....

'

Backward Difference Approximation of theFirst Derivative

Expand the Taylor series backwards

The error is still O(h)

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Centered Difference Approximation of theFirst Derivative

Subtract backward difference approximationfrom forward Taylor series expansion

211

211

1

1

21

2'

6

''''2

'

....!2

'''

hOh

xfxfxf

hxf

hxfxfxf

xx

xfxfxf

hxf

hxfxfxf

iii

iiii

ii

iii

iiii

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f(x)

x

estimate

actual

(xi,yi)

(xi-1,yi-1)

(xi+1,yi+1)

211

2' hO

hxfxf

xf iii

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f(x)

xf(x)

x

f(x)

xf(x)

x

true derivative forwardfinite divideddifference approx.

backwardfinite divideddifference approx.

centeredfinite divideddifference approx.

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Numerical Differentiation

• Forward finite divided differences Fig. 23.1

• Backward finite divided differences Fig. 23.2

• Centered finite divided differences Fig. 23.3

• First - Fourth derivative

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Richardson Extrapolation

• Two ways to improve derivative estimates¤ decrease step size¤ use a higher order formula that employs

more points

• Third approach, based on Richardson extrapolation, uses two derivatives estimates to compute a third, more accurate approximation

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Richardson Extrapolation

I I hh

h

I h I h

Special case where h h

I I h I h

In a similar fashion

D D h D h

2

1

2

2 2 1

21

2 1

2 1

1

1

24

3

1

3

4

3

1

3

For a centered differenceapproximation withO(h2) the application ofthis formula will yielda new derivative estimateof O(h4)

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EXAMPLEGiven the following function, use Richardson’s extrapolation to determine the derivative at 0.5.

f(x) = -0.1x4 - 0.15x3 - 0.5x2 - 0.25x +1.2

Note:

f(0) = 1.2f(0.25) =1.1035f(0.75) = 0.636f(1) = 0.2

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Derivatives of Unequally Spaced Data

• Common in data from experiments or field studies• Fit a second order Lagrange interpolating

polynomial to each set of three adjacent points, since this polynomial does not require that the points be equispaced

• Differentiate analytically

f x f xx x x

x x x xf x

x x x

x x x x

f xx x x

x x x x

ii i

i i i ii

i i

i i i i

ii i

i i i i

'

11

1 1 1

1 1

1 1

11

1 1 1

2 2

2

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Derivative and Integral Estimates for Data with

Errors• In addition to unequal spacing, the other

problem related to differentiating empirical data is measurement error

• Differentiation amplifies error• Integration tends to be more forgiving• Primary approach for determining derivatives

of imprecise data is to use least squares regression to fit a smooth, differentiable function to the data

• In absence of other information, a lower order polynomial regression is a good first choice

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0

100

200

0 10 20

t

y0

102030

0 5 10 15

t

dy/d

t

0

50

100

150

200

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Numerical Differentiation and Integration

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Transcript of Numerical Differentiation and Integration