The purpose of Chapter 5 is to develop the basic principles of numerical integration

Usefule Words

integrate, integral积分（的） , integration积分（法） ,

quadrature求积分 , integrand 被积函数 definite integral定积分 , indefinite integral不定积分 , tra

pezoidal梯形（的） ,

composite trapezoidal quadrature复合求积公式algebraic accuracy 代数精度

The solution can not be found directly.

x dxx

dx

dy

22 )()( dydxds

2022 )()( dydxs

dxy 2021 )(

dxx 2021 sin

A Real ProblemFind the length of the curve from to .xy cos 0x

2

x

8961.1

2

2

4

2

4

2

0

2 sin1sin1()sin1sin1(2

4

xxxx

xxxx

)sin1sin1(2

2 2

2

0

2

xx

xx

9100.1

Definition Suppose that . A formula

with the property that

is called a numerical integration or a quadrature formula.

bxxxa M 10

M

jjj xffQ

0)(][

][][)( fEfQdxxfba

ba

)(xfy This definition implies that

a quadrature formula is dependent on the choice of and .jx j

Question arises:

(1) How to choose and ? jx j

(2) How to evaluate a quadrature formula?

5.1 Introduction to quadrature

Degree of Precision ( accuracy of formula)

Definition The degree of precision of a quadratrue formula is apositive integer such that for all polynomials of degree but for some polynomials ofdegree .

0][ iPE01 ][ nPEni

)(xPi

)(xPn 11n

n

][][)( fEfQdxxfba

If the degree of precision of a quadratrue formula is n, then

(1) The formula produces exact results for all with .

)(xPi ni

(2) The error term is of the form where k is a suitable constant and c is dependent on x.

][ fE )(][ )( cfkfE n 1

Example Prove that the degree of precision of the formula

)()()( bfafabdxxfba

2is only one.

Example Find constants and so that the formula)()()()( hfAhfAfAdxxfh 20 210

30

,0A 2A1A

has the highest possible degree of precision. Determine the the degree n.

(See Ex. 5.4 (P209))

49,0,

4 210h

AAh

A

Suppose that and are in [a,b], and and

Two useful theorems

Weighted Mean value Theorem for Integrals Suppose

],,[ baCf

the Riemann integral of g exists on [a,b], and g(x) does not

change sign on [a,b]. Then there exists a number c in (a,b) with

.)()()()( b

a

b

adxxgcfdxxgxf

1x 2x 1c 2c

are positive constants. Then a number exists between and 1x 2x

with .)()()(21

2211

ccxfcxfc

f

Derivation of Quadrature formulas

],,[ baCf

Let are M+1 equidistant nodes on the interval

and is the Mth Lagrange interpolating polynomial for .

Mxxx ,,, 10 ],[ ba

with truncation error

M

k

ba kMk

ba M

ba dxxLxfdxxPdxxf

0)()()()( ,

ba

M

ii

Mba M dxxxxf

MdxxPxffE

0

1

11 )())((

!)()()(][ )(

.)()()( ,

M

kkMkM xLxfxP

0

)(xf)(xPM

)]()([

)]()([)(

bfafab

xfxfhdxxfba

2

2 10

M=1 Trapezoidal Rule

)()()( fab

fE

12

3

M=2 Simpson’s rule

)(][ )( 45

90fhfE

)()(

)()()()(

bfba

fafab

xfxfxfhdxxfba

24

6

43 210

That is Then

See Th.5.1 and Cor.5.1, page 204

(Newton--Cotes formulas)

),(12

][2

)(3

10 cfh

ffh

dxxfb

a

M=1 Trapezoidal Rule

abh

）（ 01

10

01

01

10

101 )()()()( xx

hf

xxh

fxxxx

xfxxxx

xfxP

))((2

)()(

))((2

)()()(

101

01

10

101

1

xxxxcf

xxhf

xxh

f

xxxxcf

xPxf

）（

1

0

1

0

1

0

))((2

)()()( 101

01

10 x

x

x

x

x

xdxxxxx

cfdxxx

hf

xxh

fdxxf ）（

03

10 12)()(

2h

cfff

h

M=2 Simpson’s rule

),(90

)()(4)(3

)( )4(5

210 fh

xfxfxfh

dxxfb

a 2

abh

))(())(()(

))(())(()(

))(())(()()(

1202

102

2101

201

2010

2102 xxxx

xxxxxf

xxxxxxxx

xfxxxx

xxxxxfxP

))()((!3

)()( 2101

2 xxxxxxcf

xE

Integrating P2(x) on [x0, x2] directly provides only an error term involving f (3). By approaching the problem in another way, a higher-order term involving f (4) can be derived.

Expand f(x) about x1 in the third Taylor polynomial.

41

1)4(

21

121

1111 )(

!4)()(

!3)()(

!2)())(()()( xx

cfxx

xfxx

xfxxxfxfxf

Integrating f(x) on [x0, x2] gives

2

0

2

0

21

121

1111 )(

!3)()(

!2)())(()()(

x

x

x

xdxxx

xfxx

xfxxxfxfdxxf

,)()(4)(3

),( 210 xfxfxfh

hfT

)(12

)()(2)()( 2)4(

2

2210

1 cfh

hxfxfxf

xf

).(90

][ )4(5

cfh

fE

Then (Simpson’s rule)

))((2

)()()( 101

01

10 xxxx

cfxx

hf

xxh

fxf

）（

Differentiating f(x) and evaluating f(x0) give the forward-difference formulah

cfh

ffxf

2)()( 01

0

Interpolating f(x) with nodes x0 and x1

Expand f(x) in the 3rd Taylor polynomial about x1, evaluate f(x0) and f(x2), and then add them.

.)(!4

)(2

0

41

1)4(

x

xdxxx

cf

and

Example Use quadrature formulas (5.4) to (5.7) to approximate the integral .))sin((

10 41 dxxe x

Solution ).sin()( xexf x 41 Let The result of using the trapezoidal rule with h=1 is

..))()(()()( 8607901021

2 1010 ffffhdxxf

Using Simpson’s rule with h=1/2, we have ..))().()(()()( 32128115040

614

3 21010 ffffffhdxxf

For Simpson’s 3/8 rule with h=1/3, we obtain

..))()/()/()(( /

)()(

3144011323313081

3383

321010

ffff

ffffhdxxf

The result of using the Boole’s rule with h=1/4 is

..))()/()/()/()(( /

)()(

3085911743322112413207901

73212327452

4321010

fffff

fffffhdxxf

30825060461.

1

0 41 dxxe x ))sin((30825060461.

860790.T

321281.S

2062.1)1()5.0(2)0(25.0

2 fffT

2836.1)1()75.0(2)5.0(2)25.0(2)0(225.0

4 fffffT

Composite

Quadratrue Formulas

))1()875.0(2)75.0(2)625.0(2

)5.0(2)375.0(2)25.0(2)125.0(2)0((2125.0

8

ffff

fffffT

5.2 Composite Trapezoidal and Simpson’s Rule

. ,and ,, , Where. with],[

lssubinterva tequidistan into devided is],[ that Suppose

bxa xMkM

abhxx

Mba

Mkk

01 10

)(),( and )()()( ) ,( fhabhfEbfxfafhhfT TM

jj

21

1 122

2

Then the composite trapezoidal rule is expressed as follows:

. ,and 2 ,, , Where. with],[

lssubinterva tequidistan 2 into devided is],[ that Suppose

bxa xMkM

abhxx

Mba

Mkk

201 102

)(),( and

)()( )()(),(

)( 44

112

1

12

180

423

fhabhfE

bfxfxfafhhfS

S

M

kk

M

kk

Then the composite Simpson’s rule is expressed as follows:

Example2 Approximate using the given entries .

a. Use the Composite Trapezoidal rule .

b. Use the Composite Simpson’s rule .

94569080210281

8187

1

10 .)()()(sin

,,.

k

ixfffdxx

xhMa

dxx

x

10

sin

0.841470910.87719257/80.90885163/40.93615565/80.95885101/20.97672673/80.98961581/40.99739781/81.00000000f(x)=sinx/xx

94608320

4210381

81824

112

3

12

10

.

)()()()(sin,,.

i

ii

i xfxfffdxx

xhMb

Solution

Example3 Determine the values M and h such that the

truncation error Es(f, h) is less than when the

composite Simpson’s rule is used to approximate .dxx

72

1

9105

Solution

)(),( )( 44

180fhabhfES

.)(x

xf 1Let

.)( and )(,)( ,)( )(5

4432

24621x

xfx

xfx

xfx

xf

The first four derivatives of f are

),(),( )( 44

180fhabhfES

According to we have

.),( 94

54

54 105

48224

1802724

18027

hhx

hhfES

So h< 0.0221336 and M=(b-a)/2h> 112.95.In consideration of M being a positive integer, the result is M= 113 and h=1/2M.

5.3 Recursive Rules and Romberg IntegrationRecursive Quadrature RulesRecursive Trapezoidal Rules

J

abh

2

a b

.2

Where.2

)0( 010ab

abhffh

T

Let

.2

where,2

)0( 2

2 2

)1( 11120210ab

hhfT

hfffh

fffh

T

)222(2

)( 212210 MM fffffh

JT

)()22(2

221

123122220 MMM fffhffffh

M

kkfh

JT1

122)1(

Then,

,2

)1(

2 2

222 2

)2(

31

3142043210

ffhT

ffhfffh

fffffh

T

.2

where 2

abh

Example1 Approximate using 51

1 dxx . (3) and )(, )(),( TTTT 210

ln 5≈1.6094379124

.4000002 ])5()1([2

)0(,15,0 .ffhThJ

.866666.1 )3(2

)0()1(,2

15,1 1

fhT

ThJ

.683333.1 )]4()2([2

)1()2(,12

15,2 2

ffhT

ThJ

.628968.1 )]5.4()5.3()5.2()5.0([2

)2()3(

,5.02

15,3 3

ffffhT

T

hJ 1 5

Recursive Simpson Rules

MMM ffffffffh

JTJT

2224212310 )(2)(43

3)1()(4

)(JS

Recursive Boole Rules

15)1()(16)(

JSJS

JB

)222(2

)1( 222420 MM fffffh

JT

)222(2

)( 212210 MM fffffh

JT

Richardson’s Improvement

. ])()([)(, 4000002512

015 .ffhTh

..1 )()()(, 866666320

12 fhT

Th

.. )]()([)()(, 68333314221

21 ffhT

Th

.. )].().().().([)()(,. 62896815453525022

350 ffffhT

Th

1

Example1 Approximate using 51

1 dxx . (3) and )(, )(),( TTTT 210

5

ln 5≈1.6094379124

The results are as follows:

3)1()(4)(

JTJT

JS

15)1()(16)(

JSJS

JB

k T(J) S(J) B(J)

0 2.4

1 1.866666 1.688888

2 1.683333 1.622222 1.617778

3 1.628968 1.610846 1.610088

(1))()( ),()( 263

42

21 hOJThahahahfTdxxfQ

b

a

Example

then (2))()1( 64164)2,( 263

42

21 hOJThahahahfTQ

If

Subtract (2) from 4×(1) (i.e 4(1)-2):

).()2,(),(4 6012),(4)2,(3 463

42 hOhfThfThahahfThfTQ

Thus,).(

3)2,(),(4 4hO

hfThfTQ

Recursive Simpson Rule

)()()(3

)1()(4 44 hOJShOJTJT

Q

That is,

Similarly, ),()( ),( 483

62

41 hOJShbhbhbhfSQ if

we obtain )()()(15

)1()(16 66 hOJBhOJSJS

Q

Recursive Boole Rule

(Richardson’s Improvement)

for all ,NTheorem If ],[ baCf N then

. ),()( 6

34

22

1 hahahahfTdxxfb

a

)()( 2hOJT

15)1()()(

JSJS

JB16

Define ,, all for ),()( 21 JhfSJS

3)1()()(

JTJT

JS4

Theorem .,, all for ),()( 32 JhfBJB Then

and

and

3,2,3

)1()(4)(

JJTJT

JS

4,3,15

)1()(16)(

JJSJS

JB

,5,4,63

)1()(64)(

JJBJB

dxxfb

a

.,2,1 allfor )( 2

)1()(1

12

JxfhJT

JTM

kk

First improvement

Second improvement

Third improvement

Give preliminary approximations using the Composite Trapezoidal rule firstly and then improve them by Richardson’s extrapolation process

(2) ),( 4223

2212

21 44412 KKKKKK hchchcKhRQ

Richardson’s Improvement

Jabh 2/)(

a b122 Jabh /)(

Suppose that . (1) ),( 423

222

211 KKK hchchcKhRQ

. )(),( KhOKhRQ 21 ThenThat is

Subtract (2) from 4×(1). Then

)(),(),( 22

141214

K

K

KhO

KhRKhRQ

K=1

),( 0hR),( 02hR

30204 ),(),( hRhR

K=2),( 12hR),( 1hR 15

12116 ),(),( hRhR

K=3),( 22hR

),( 2hR 6322264 ),(),( hRhR

)( 4hO )( 8hO)( 6hO

)0,(JR )1,(JR )2,(JR )3,(JR),( 00R),( 01R),( 02R),( 03R

),( 11R),( 12R),( 13R

),( 22R),( 22R )3,3(R

)(0T)(1T)(2T)(3T

)(1S)(2S)(3S

)(2B)(3B

t)Improvemen Third(63

)2()3(64 BB

Romberg Integration Tabuleau

)(14

)1,2()1,(4 22

K

K

K

hOKhRKhR

Q

14)1,1()1,(4

K

K KJRKJR

Example2 Approximate using Romberg integration.51

1 dxx

2.400000

1.866667 1.688888

1.683334 1.622222 1.617778

1.628968 1.610846 1.610088 1.609490

4.2511

24)0()0,0(

TR

3)0,1()0,(4)1,(

JRJR

JR

866667.1)3(22

)0()1()0,1( fT

TR

683334.1)4()2(12

)1()2()0,2( ffT

TR

,628969.1)5.4()5.3()5.2()5.1(5.02

)2()3()0,3( ffffT

TR

15)1,1()1,(16)2,(

JRJR

JR

63)2,1()2,(64)3,(

JRJR

JR

R(J,0) R(J,1) R(J,2) R(J,3)

Solution We use J=3. The Romberg table is given as follows:

Where and

ln 5≈1.6094379124

Example2 Approximate using R(5,5).

0

sin dxx

Solution Compute R(J,k) where J=0,1,2,3,4,5 and k=0,1,2,3,4,5. The results are listed in the following table.

0 0

1 1.57079633 2.09439511

2 1.89611890 2.00455976 1.99857073

3 1.97423160 2.00026917 1.99998313 2.00000555

4 1.99357034 2.00001659 1.99999975 2.00000001 1.99999999

5 1.99839336 2.00000103 2.00000000 2.00000000 2.00000000 2.00000000

J R(J,0) R(J,1) R(J,2) R(J,3) R(J,4) R(J,5)

.00000000.2)5,5(sin0

Rdxx

Therefore,

ba

)(xfy

5.4 Gauss-Legendre IntegrationGaussian quadrature choose the points for evaluation in an optimal , rather than equally spaced way .

Definition Choose the parameters of the quadrature formula

properly such that it is likely of degree of

precision 2N-1 . This formula is called Gaussian formula .

)()(1

N

iii

b

axfcdxxf

)()(1

1

1N

iii xfcdxxf

1

1

1

1

22)(

22)

22()( dttg

abdt

abbat

abfdxxf

batabxb

a

322211

2

1

1

1,,,1)(,)()()()(,2 xxxxfxfcxfcxfcdxxfN

iii

,21,1)( 21

11 ccdxxf

,0,)( 221111 xcxcdxxxxf

,32,)( 2

22211

11

22 xcxcdxxxxf

,0,)( 322

311

11

33 xcxcdxxxxf

33,

33

,1,1

21

21

xx

cc

)33()

33()()(

2

1

11 ffxfcdxxf

iii

xxfxfcdxxfN ,1)(,)()(,1 11

1

1

0,021,2)(

,21,1)(

1111

111

xxdxxf

cdxxf)0(2)(1

1 fdxxf

Gaussian nodes

is likely of degree of precision 2N-1.

Nxxx ,,, 21 are the zeros of the Nth Legendre polynomial.

31

23)()(1)( 2

210 xxPxxPxP

,2,11!2

1)(1)( 20

Nx

dxd

NxPxP N

N

N

NN

353

76

835)(,

53

25)( 24

42

3 xxxPxxxP

Theorem Suppose that are the roots of the NthLegendre polynomial . For each , the numbers are defined by

Nxxx ,,, 21 )(xPN Ni ,,2,1

iw .1

11

dx

xxxx

wN

ijj ji

ji

If is any polynomial of degree less than 2N, then)(xP

N

iii xPwdxxP

1

1

1)()(

)0()()( 111

1

1fcxfcdxxf

has the degree of precision 1.

Midpoint Rule

two-point Gaussian formula

)515(

95)0(

98)

515(

95)(1

1 fffdxxf

Three-point Gaussian formula

.2,11

11 dxwN

.,

,21

1

1

112

12

21

21

dxxxxx

wdxxxxx

w

N

)33()

33()(1

1 ffdxxf has the degree of precision 3.

31

23)( 2

2 xxP

xxP )(1

53

25)( 2

3 xxxP

,))((

))((1

13121

321

dx

xxxxxxxx

w ,))((

))((1

13212

312

dx

xxxxxxxx

w .))((

))((1

12313

213

dx

xxxxxxxx

w

)()()( )2(

1

1

1N

N

iii kfxfwdxxf

Truncation errors for Gaussian formulas

?),()0(2)( 11

1

1

kfkfdxxf

),()33()

33()( )4(

2

1

1fkffdxxf

Example

Let .)( 2xxf We have .31

1 k

For

.135

12 k See Table 5.8, P240Let .)( 4xxf We have

Change of variables

1

1

1

1)(

22)

22()( dttg

abdt

abbat

abfdxxf

b

a

22ba

tab

x

Example Approximate the integral using Gaussian

quadrature with n=2 and n=3.

5.11

2

dxe x

1

125.125.0

25.2

25.0

5.11

22

25.0 dtedxe ttx

x

094003.125.0

,2 222 25.15773502692.025.025.15773502692.025.05.1

1

eedxe

nx

)33()

33()(1

1 ffdxxf

093642.1]5555555556.0

8888888889.0

5555555556.0[25.0

,3

2

2

22

25.15773502692.025.0

25.1025.0

25.15773502692.025.05.11

e

e

edxe

nx

)515(

95)0(

98)

515(

95)(1

1 fffdxxf