numerical integration roger crawfis. june 9, 2015osu/cse 5412 quadrature we talk in terms of...

Numerical Integration

Roger Crawfis

April 18, 2023 OSU/CSE 541 2

Quadrature

• We talk in terms of Quadrature Rules– 1. The process of making something square. 2.

Mathematics The process of constructing a square equal in area to a given surface. 3. Astronomy A configuration in which the position of one celestial body is 90° from another celestial body, as measured from a third.

– The American Heritage® Dictionary: Fourth Edition. 2000

April 18, 2023 OSU/CSE 541 3

Outline

• Definite Integrals• Lower and Upper Sums

– Reimann Integration or Reimann Sums

• Uniformly-spaced samples– Trapezoid Rules– Romberg Integration– Simpson’s Rules– Adaptive Simpson’s Scheme

• Non-uniformly spaced samples– Gaussian Quadrature Formulas

April 18, 2023 OSU/CSE 541 4

What does an integral represent?

Basic definition of an integral:

Motivation

b

adxxf area)(

d

c

b

adxdyxf volume)(

n

kk

b

a nxxfdxxf

1

)(lim)(

sum of height width

f(x)

x

n

abx

where

April 18, 2023 OSU/CSE 541 5

• Evaluate the integral, without doing the calculation analytically.

• Necessary when either:

– Integrand is too complicated to integrate analytically

– Integrand is not precisely defined by an equation, i.e., we are given a set of data (xi, yi), i = 1, 2, 3, …,n

Motivation

b

adxxfI )(

dxex

x x5.02

0 5.01

)1cos(2

April 18, 2023 OSU/CSE 541 6

• Integration is a summing process. Thus virtually all numerical approximations can be represented by

in which wi are the weights, xi are the sampling points, and Et is the truncation error

• Valid for any function that is continuous on the closed and bounded interval of integration.

t

b

a i

n

ii ExfwdxxfI

)()(1

Reimann Integral Theorem

April 18, 2023 OSU/CSE 541 7

• The most common numerical integration formula is based on equally spaced data points.

• Divide [x0 , xn] into n intervals (n1)

nx

xdxxf

0

)(

Partitioning the Integral

1 2

00 1 1

( ) ( ) ( ) ( )n

n

n

xx xx

xx x x

f x dx f x f x f x

April 18, 2023 OSU/CSE 541 8

Upper Sums

• Assume that f(x)>0 everywhere.

• If within each interval, we could determine the maximum value of the function, then we have:

• where

0

1

10

( )nx n

i i iix

f x M x x

1sup ( ) :i i iM f x x x x Supremum - least upper bound

April 18, 2023 OSU/CSE 541 9

Upper Sums

• Graphically:

x0 x1 x2 x4x3

April 18, 2023 OSU/CSE 541 10

Lower Sums

• Likewise, still assuming that f(x)>0 everywhere.

• If within each interval, we could determine the minimum value of the function, then we have:

• where

0

1

10

( )nx n

i i iix

f x m x x

1inf ( ) :i i im f x x x x Infimum - greatest

lower bound

April 18, 2023 OSU/CSE 541 11

Lower Sums

• Graphically

x0 x1 x2 x4x3

April 18, 2023 OSU/CSE 541 12

Finer Partitions

• We now have a bound on the integral of the function for some partition (x0,..,xn):

• As n, one would assume that the sum of the upper bounds and the sum of the lower bounds approach each other.

• This is the case for most functions, and we call these Riemann-integrable functions.

0

1 1

1 10 0

( )nxn n

i i i i i ii ix

m x x f x M x x

April 18, 2023 OSU/CSE 541 13

Bounding the Integral

• Graphically

x0 x1 x2 x4x3

April 18, 2023 OSU/CSE 541 14


• Halving each interval (much like Lab1):

x0 x3 x5 x9x7

April 18, 2023 OSU/CSE 541 15


• One more time:

x0 x5 x7 x11x9

April 18, 2023 OSU/CSE 541 16

Monotonic Functions

• Note that if a function is monotonically increasing (or decreasing), then the lower sum corresponds to the left partition values, and the upper sum corresponds to the right partition values.

x0 x3 x5 x9x7

April 18, 2023 OSU/CSE 541 17

Lab1 and Integration

• Thinking back to lab1, what were the limits or the integration?

• Is the sin function monotonic on this interval?

• Should the Reiman sum be an upper or lower sum?

April 18, 2023 OSU/CSE 541 18

Polynomial Approximation

• Rather than search for the maximum or minimum, we replace f(x) with a known and simple function.

• Within each interval we approximate f(x) by an mth order polynomial.

mmm xaxaxaaxp ...)( 2

210

April 18, 2023 OSU/CSE 541 19

• The m’s (order of the polynomials) may be the same or different.

• Different choices for m’s lead to different formulas:

1 1 2

1

0 0

1 20 0 0

( ) ( ) ( ) ... ( )n m m m n

nm n mn

x x x x

m m mx x x xf x dx p x dx p x dx p x dx

)(3/8 sSimpson'cubic3

)(1/3 sSimpson'quadratic2

)(Trapezoidlinear1

ErrorFormulaPolynomial

4

4

2

hO

hO

hO

m

Newton-Cotes Formulas

April 18, 2023 OSU/CSE 541 20

• Simplest way to approximate the area under a curve – using first order polynomial (a straight line)

• Using Newton’s form of the interpolating polynomial:

• Now, solve for the integral:

b

a

b

adxxpdxxfI 1

axab

afbfafxp

1

Trapezoid Rule

April 18, 2023 OSU/CSE 541 21

Trapezoid Rule

b

adxax

ab

afbfafI

bfafab

I

2

a b

f(a) f(b)

height average width I

Trapezoid Rule

April 18, 2023 OSU/CSE 541 22

Trapezoid Rule

• Improvement?

x0 x1 x2 x4x3

April 18, 2023 OSU/CSE 541 23

• The integration error is:

• Where h = b - a and is an unknown point where a < < b (intermediate value theorem)

• You get exact integration if the function, f, is linear (f = 0)

23

12

)(

12

1hf

abhfEt

)( 3hO

Trapezoid Rule Error

April 18, 2023 OSU/CSE 541 24

2

)( xexf Integrate from a = 0 to b = 2

0183.1)(1

)0()2(2

)02(

2

04

2

0

2

ee

ffbfafab

dxeI x

Use trapezoidal rule:

Example

April 18, 2023 OSU/CSE 541 25

Estimate error: 3

12

1hfEt

Where h = b - a and a < < b

Don’t know - use average value2

)42()( 2 xexxf 2564.0)2(

2)0(

f

f

58.0

2

20

12

23

ff

EE at

202 h

Example

April 18, 2023 OSU/CSE 541 26

More intervals, better result [error O(h2)]

0

1

2

3

4

5

6

7

3 5 7 9 11 13 15

n = 2

0

1

2

3

4

5

6

7

3 5 7 9 11 13 15

n = 3

0

1

2

3

4

5

6

7

3 5 7 9 11 13 15

n = 4

0

1

2

3

4

5

6

7

3 5 7 9 11 13 15

n = 8

April 18, 2023 OSU/CSE 541 27

• If we do multiple intervals, we can avoid duplicate function evaluations and operations:

• Use n+1 equally spaced points. • Each interval has:• Break up the limits of integration and expand.

n

abh

b

hb

ha

ha

ha

adxxfdxxfdxxfI ...

2

Composite Trapezoid Rule

April 18, 2023 OSU/CSE 541 28

• Substituting the trapezoid rule for each integral.

• Results in the Composite Trapezoid Formula:

2

22

2 2

....2

...a h a h b

a a h b h

a h a a h a hf a f a h f a h f a h

b b hf b h f b

I f x dx f x dx f x dx

bfihafafh

In

i

1

1

22


April 18, 2023 OSU/CSE 541 29

n

bfihafafab

bfihafafh

I

n

i

n

i

2

2

22

1

1

1

1

width

Average height


• Think of this as the width times the average height.

April 18, 2023 OSU/CSE 541 30

• The error can be estimated as:

• Where, is the average second derivative.

• If n is doubled, h h/2 and Ea Ea/4

• Note, that the error is dependent upon the width of the area being integrated.

f

n

abf

habEa

2

32

1212

f

)( 2hO

Error

April 18, 2023 OSU/CSE 541 31

0

5

10

15

20

25

30

35

40

0 0.2 0.4 0.6 0.8 1 1.2

Example

• Integrate:• from

a=0.2 to b=0.8

5432 200810730140203.0 xxxxxxf

April 18, 2023 OSU/CSE 541 32

26.112

81.322.342.08.0

2

bfafabI

Example

• A single application of the Trapezoid rule.

• Error: 3

12

1abfEt

April 18, 2023 OSU/CSE 541 33

Example

• We don’t know so approximate with average f

32 400097204380280 xxxxf

432 10003240219028020 xxxxxf

6.1312.08.0

)2.0()8.0(

2.08.0

8.0

2.0

ff

dxfxf

April 18, 2023 OSU/CSE 541 34

32

2

3

12 121

131.6 0.8 0.2 2.3712

t

b a h b aE f f

n

Example

• The error can thus be estimated as:

April 18, 2023 OSU/CSE 541 35

0

5

10

15

20

25

30

35

40

0 0.2 0.4 0.6 0.8 1 1.2

True value of integral is 12.82. Trapezoid rule is 11.26 - within approx error - Et is 12%

April 18, 2023 OSU/CSE 541 36

• Use intervals (0.2,0.4),(0.4,0.6),(0.6,0.8):– (n = 3, h = 0.2)

57.12 6

22.3416.3093.13231.36.0

)3)(2(

8.06.04.022.02.08.0

2

21

1

ffffn

bfihafafabI

n

i

True value of integral is 12.82

Using Three Intervals

April 18, 2023 OSU/CSE 541 37

0

5

10

15

20

25

30

35

40

0 0.2 0.4 0.6 0.8 1 1.2

Et is now 2%

April 18, 2023 OSU/CSE 541 38

76.12 12

22.3422.3516.3018.2293.1334.7231.36.0

)6)(2(

8.07.06.05.04.03.022.02.08.0

fffffffI

Using Six Intervals

• Use intervals (0.2,0.3),(0.3,0,4), etc. – (n = 6, h = 0.1)

True value of integral is 12.82

April 18, 2023 OSU/CSE 541 39

0

5

10

15

20

25

30

35

40

0 0.2 0.4 0.6 0.8 1 1.2

Et is now 0.5%

April 18, 2023 OSU/CSE 541 40

Multi-dimensional Integration

• Consider a two-dimensional case.

1 1 1

00 0 0

1

0 0

0 0

0 0

, ,

,

,

,

n

ii

n

ii

n n

i ji j

n n

i ji j

if x y dxdy A f y dy

n

iA f y dy

n

i jA A f

n n

i jA A f

n n

April 18, 2023 OSU/CSE 541 41


• For the Trapezoid Rule, this leads to weights in the following pattern:

1 2 2 2 2 2 1

2 4 4 4 4 4 2

2 4 4 4 4 4 2

2 4 4 4 4 4 2

2 4 4 4 4 4 2

2 4 4 4 4 4 2

1 2 2 2 2 2 1

2

1 0, 0,

2 1, , 2 1, 112 {1, 1} 1, , 14

4 1, , 2 1, , 2

ij

i n j n

i n j nA

i n j nn

i n j n

1

2

2

2

2

2

1

1 2 2 2 2 2 1

April 18, 2023 OSU/CSE 541 42


• If we use the weights from the Trapezoid rule, the error is still O(h2).

• However, there are now n2 function evaluations.– Equally-spaced samples on a square region.

April 18, 2023 OSU/CSE 541 43


• In general, given k dimensions, we have N= nk function evaluations:

• If the dimension is high, this leads to a significant amount of additional work in going from hh/2.– Remember this for Monte-Carlo Integration.

22

2 2 k kkO h O n O n O N

April 18, 2023 OSU/CSE 541 44

Reducing the Error

• To improve the estimate of the integral, we can either:– Add more intervals– Use a higher order polynomial– Use Richardson Extrapolation to examine the

limit as h0.• Called Romberg Integration

April 18, 2023 OSU/CSE 541 45

Adding More Intervals

• If we have an estimate for one value of h, do we need to recompute everything for a value of h/2?

1

1

22

n

hi

hI f a f a ih f b

April 18, 2023 OSU/CSE 541 46

Adding More Intervals

• This is called the Recursive Trapezoid Rule in the book.

• We have n 2n and hh/2.

2 1

12

1 1

1 0

1

0

( ) 2 ( )4 2

( ) 2 2 ( )4 2

22 4 2

n

hi

n n

i i

nh

i

h hI f a f a i f b

h hf a f a ih f a ih f b

I h hf a ih

April 18, 2023 OSU/CSE 541 47

• Given two numerical estimates obtained using different h’s, compute higher-order estimate

• Starting with a step size h1, the exact value is

• Suppose we reduce step size to h2

)()( 11nhOhAA

)()( 22nhOhAA

Recall Richardson Extrapolation

April 18, 2023 OSU/CSE 541 48

• Multiplying the 2nd eqn by (h1/h2)n and

subtracting from the 1st eqn:

• The new error term is generally O(h1n+1) or

O(h1n+2).

1

)()(

2

1

122

1

n

n

hh

hAhAhh

A

Richardson Extrapolation

April 18, 2023 OSU/CSE 541 49

• The true integral value can be written

• This is true for any iteration


hEhII

2211 hEhIhEhII

April 18, 2023 OSU/CSE 541 50


• For example: Using (n = 2)

• where c is a constant

• Therefore: 2

2

21

2

1

h

h

hE

hE

fchE 2[error = O(h2)]

order of error in trapezoidal rule

April 18, 2023 OSU/CSE 541 51

• This leads to:

• For integration, we have:

22

21

21h

hhEhE

2222

21

21 hEhIh

hhEhI


April 18, 2023 OSU/CSE 541 52


• Solving for E(h2):

• And plugging back into the estimated integral.

22

21

212

1h

h

hIhIhE

22 hEhII

April 18, 2023 OSU/CSE 541 53

• Leads to:

• Letting h2 = h1 /2


12221

21/

1hIhI

hhhII

12

1222

3

1

3

412

1

hIhII

hIhIhII

April 18, 2023 OSU/CSE 541 54

• We combined two O(h2) estimates to get an O(h4) estimate.

• Can also combine two O(h4) estimates to get an O(h6) estimate.

lm hIhII15

1

15

16

Romberg Integration

April 18, 2023 OSU/CSE 541 55

Romberg Integration

• Greater weight is placed on the more accurate estimate.

• Weighting coefficients sum to unity– i.e, (16-1)/15=1

• Can continue, by combining two O(h6) estimates to get an O(h8) estimate.

lm hIhII63

1

63

64

April 18, 2023 OSU/CSE 541 56

• General pattern is called Romberg Integration

– j : level of subdivision, j+1 has more intervals.– k : level of integration, k = 1 is original trapezoid

estimate [O(h2)], k = 2 is improved [O(h4)], etc.

, 1,, 1 , , 1,

4 1

4 1 4 1

kj k j k

j k j k j k j kk k

I II I I I

Romberg Integration

April 18, 2023 OSU/CSE 541 57

Romberg Integration

• For example, j = 1, k = 1 leads to

1,0 0,0 11,1 1

4 4 1

3 3 2 3

I I hI I I h

April 18, 2023 OSU/CSE 541 58

• Consider the function:

• Integrate from a = 0 to b = 0.8

• Using the trapezoidal rule yields the following results:

5432 400900675200252.0)( xxxxxxf

Example

April 18, 2023 OSU/CSE 541 59

intervals Integral

1 0.8 0.1728

2 0.4 1.0688

4 0.2 1.4848

h

Example

• Trapezoid Rules:

Exact integral is 1.64053334

3674667.1)1728.0(3

1)0688.1(

3

4I (j=1, k=1)

k

j

k = 0 k = 1

j = 0

j = 1

j = 2

April 18, 2023 OSU/CSE 541 60

2 4segments ( ) ( )

1 0.8 0.1728

2 0.4 1.0688 1.3674667

4 0.2 1.4848

h O h O h k

j

62346667.1)0688.1(3

1)4848.1(

3

4I (j=2, k=1)

k = 1k = 0


Example

April 18, 2023 OSU/CSE 541 61

Example

2 4segments ( ) ( )

1 0.8 0.1728

2 0.4 1.0688 1.3674667

4 0.2 1.4848 1.62346667

h O h O h k

j

(j=2, k=2)

k = 2k = 1


64053334.1)3674667.1(15

1)62346667.1(

15

16I

April 18, 2023 OSU/CSE 541 62

2 4 6segments ( ) ( ) ( )

1 0.8 0.1728

2 0.4 1.0688 1.3674667

4 0.2 1.

h O h O h O h

4848 1.62346667 1.64053334

k

jk = 3k = 2k = 1

Example

April 18, 2023 OSU/CSE 541 63

2 4 6 8segments ( ) ( ) ( ) ( )

1 0.8 0.1728

2 0.4 1.0688 1.3674667

4

h O h O h O h O h

0.2 1.4848 1.62346667 1.64053334

8 0.1 ?? ?? ?? ??

??)64053334.1(63

1(??)

63

64I (j=3, k=3)

k

j

k = 3

Example

• Better and better results can be obtained by continuing this

April 18, 2023 OSU/CSE 541 64

Romberg Integration

• Is this that significant?• Consider the cost of computing the

Trapezoid Rule for 1000 data points.– Refinement would lead to 2000 data points.

• Implies an additional 1003 operations using the Recursive Trapezoid Rule.

• Not to mention the 1000 (expensive) function evals.

– Romberg Integration cost:• Three additional operations – no function evals!!!

April 18, 2023 OSU/CSE 541 65

Higher-Order Polynomials

• Recall:

)(3/8 sSimpson'cubic3

)(1/3 sSimpson'quadratic2

)(Trapezoidlinear1

ErrorFormulaPolynomial

4

4

2

hO

hO

hO

m

1 1 2

1

0 0

1 20 0 0

( ) ( ) ( ) ... ( )n m m m n

nm n mn

x x x x

m m mx x x xf x dx p x dx p x dx p x dx

April 18, 2023 OSU/CSE 541 66

• If we use a 2nd order polynomial (need 3 points or 2 intervals):– Lagrange form.

dxxf

xxxx

xxxx

xfxxxx

xxxxxf

xxxx

xxxxI

x

x

21202

10

12101

200

2010

212

0

2

201

xxx

Simpson’s 1/3 Rule

April 18, 2023 OSU/CSE 541 67


• Requiring equally-spaced intervals:

2

0

0 0 0 00 1

0 02

2 2

2

2

x

x

x x h x x h x x x x hI f x f x

h h h h

x x x x hf x dx

h h

April 18, 2023 OSU/CSE 541 68

• Integrate and simplify:

0

2

4

6

8

10

12

3 5 7 9 11 13 15

210 43

xfxfxfh

I 2

abh


QuadraticPolynomial

April 18, 2023 OSU/CSE 541 69

• If we use a = x0 and b = x2, and x1 = (b+a)/2


6

4 210 xfxfxfabI

widthaverage height

April 18, 2023 OSU/CSE 541 70

• Error for Simpson’s 1/3 rule

Integrates a cubic exactly: 04 f

45

45

288090f

abf

hEt

2

abh

)( 5hO


April 18, 2023 OSU/CSE 541 71

Composite Simpson’s 1/3 Rule

• As with Trapezoidal rule, can use multiple applications of Simpson’s 1/3 rule.

• Need even number of intervals– An odd number of points are required.

April 18, 2023 OSU/CSE 541 72


• Example: 9 points, 4 intervals

-5

0

5

10

15

20

25

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5

April 18, 2023 OSU/CSE 541 73

• As in composite trapezoid, break integral up into n/2 sub-integrals:

• Substitute Simpson’s 1/3 rule for each integral and collect terms.


n

n

x

x

x

x

x

xdxxfdxxfdxxfI

2

4

2

2

0

...

1 2

01,3,5 2,4,6

4 2

3

n n

i j ni j

f x f x f x f x

I b an

n+1 data points, an odd number

April 18, 2023 OSU/CSE 541 74

• Odd coefficients receive a weight of 4, even receive a weight of 2.

• Doesn’t seem very fair, does it?

coefficients on numerator

14 1

1 41

14 1

i = 0

i = n


April 18, 2023 OSU/CSE 541 75

• The error can be estimated by:

• If n is doubled, h h/2 and Ea Ea/16

)4(4

)4(5

180180f

habf

nhEa

)4(f is the average 4th derivative

)( 4hO

Error Estimate

April 18, 2023 OSU/CSE 541 76

• Integrate from a = 0 to b = 2.• Use Simpson’s 1/3 rule:

22

0 1 20

0 1 4

1 4 ( )

31

0 4 (1) 23

1 ( 4 ) 0.82994

3

xI e dx h f x f x f x

f f f

e e e

2

)( xexf

2 12

0 12 210

bx

baxax

abh

Example

April 18, 2023 OSU/CSE 541 77

• Error estimate:

• Where h = b - a and a < < b• Don’t know

– use average value

45

90f

hEt

4 4 4

5 50 1 241 1

90 90 3t a

f x f x f xE E f

Example

April 18, 2023 OSU/CSE 541 78

• Let’s look at the polynomial again:

– From a = 0 to b = 0.8

5432 400900675200252.0)( xxxxxxf

8.0 4.02

0 4.02 210

bx

baxax

abh

36746667.1

8.0)4.0(403

)4.0(

)(43

1 )( 210

2

0

fff

xfxfxfhdxxfI


Another Example

April 18, 2023 OSU/CSE 541 79

• Actual Error: (using the known exact value)

• Estimate error: (if the exact value is not available)

• Where a < < b.

45

90f

hEt

27306666.036746667.1-1.64053334 E 16%

Error

April 18, 2023 OSU/CSE 541 80

• Compute the fourth-derivative

• Matches actual error pretty well.

Error

xxf 4800021600)()4(

27306667.04.090

4.0

90

4.0 45

14

5

fxfEE at

middle point

April 18, 2023 OSU/CSE 541 81

• If we use 4 segments instead of 1:– x = [0.0 0.2 0.4 0.6 0.8]

0.2b a

hn

232.0)8.0( 464.3)6.0(

456.2)4.0( 288.1)2.0( 2.0)0(

ff

fff

6234667.1 12

232.0)456.2(2)464.3288.1(42.08.0

)4)(3(

)8.0()6.0(4)4.0(2)2.0(4)0(08.0

3

241

5,3,1

2

6,4,20

fffffn

xfxfxfxf

abIn

n

i

n

jji


Example Continued

April 18, 2023 OSU/CSE 541 82

• Actual Error: (using the known exact value)

• Estimate error: (if the exact value is not available)

01706667.06234667.1-1.64053334 E 1%

0085.04.090

2.0

90

2.0 45

24

5

fxfEE at

middle point

Error

April 18, 2023 OSU/CSE 541 83

Error

• Actual is twice the estimated, why?

• Recall:xxf 4800021600)()4(

(4) (4)

0,0.8

(4)

max ( ) (0) 21600

(0.4) 2400

xf x f

f

April 18, 2023 OSU/CSE 541 84

Error

• Rather than estimate, we can bound the absolute value of the error:

• Five times the actual, but provides a safer error metric.

5 5

4 40.2 0.20 0.0768

90 90aE f f

April 18, 2023 OSU/CSE 541 85

Simpon’s 1/3 Rule

• Simpson’s 1/3 rule uses a 2nd order polynomial– need 3 points or 2 intervals

– This implies we need an even number of intervals.

• What if you don’t have an even number of intervals? Two choices:

1. Use Simpson’s 1/3 on all the segments except the last (or first) one, and use trapezoidal rule on the one left.– Pitfall - larger error on the segment using trapezoid

2. Use Simpson’s 3/8 rule.

April 18, 2023 OSU/CSE 541 86

• Simpson’s 3/8 rule uses a third order polynomial– need 3 intervals (4 data points)

33

22103 )()( xaxaxaaxpxf

3

0

3

03

x

x

x

xdxxpdxxfI


April 18, 2023 OSU/CSE 541 87


• Determine a’s with Lagrange polynomial• For evenly spaced points

3

abh

3210 338

3xfxfxxfhI

April 18, 2023 OSU/CSE 541 88

• Same order as 1/3 Rule.– More function evaluations.– Interval width, h, is smaller.

• Integrates a cubic exactly:

04 f

453

80tE h f )( 4hO

Error

April 18, 2023 OSU/CSE 541 89

Comparison

• Simpson’s 1/3 rule and Simpson’s 3/8 rule have the same order of error– O(h4)– trapezoidal rule has an error of O(h2)

• Simpson’s 1/3 rule requires even number of segments.

• Simpson’s 3/8 rule requires multiples of three segments.

• Both Simpson’s methods require evenly spaced data points

April 18, 2023 OSU/CSE 541 90

• n = 10 points 9 intervals– First 6 intervals - Simpson’s 1/3

– Last 3 intervals - Simpson’s 3/8

Simpson’s 1/3

Simpson’s 3/8

Mixing Techniques

April 18, 2023 OSU/CSE 541 91

• We can examine even higher-order polynomials.– Simpson’s 1/3 - 2nd order Lagrange (3 pts)

– Simpson’s 3/8 - 3rd order Lagrange (4 pts)

• Usually do not go higher.

• Use multiple segments.– But only where needed.

Newton-Cotes Formulas

April 18, 2023 OSU/CSE 541 92

Adaptive Simpson’s Scheme

• Recall Simpson’s 1/3 Rule:

• Where initially, we have a=x0 and b=x2.

• Subdividing the integral into two:

210 43

xfxfxfh

I

1 2 34 2 46

hI f a f x f x f x f b

April 18, 2023 OSU/CSE 541 93


• We want to keep subdividing, until we reach a desired error tolerance, .

• Mathematically:

1

1 2 3

43

4 2 46

b

a

b

a

hf x dx f a f x f b

hf x dx f a f x f x f x f b

April 18, 2023 OSU/CSE 541 94


• This will be satisfied if:

• The left and the right are within one-half of the error.

1 2

2 3

2

4 ,6 2

4 ,6 2

2

c

a

b

c

hf x dx f a f x f x and

hf x dx f x f x f b where

a bc x

April 18, 2023 OSU/CSE 541 95


• Okay, now we have two separate intervals to integrate.

• What if one can be solved accurately with an h=10-3, but the other requires many, many more intervals, h=10-6?

April 18, 2023 OSU/CSE 541 96


• Adaptive Simpson’s method provides a divide and conquer scheme until the appropriate error is satisfied everywhere.

• Very popular method in practice.

• Problem:– We do not know the exact value, and hence do

not know the error.

April 18, 2023 OSU/CSE 541 97


• How do we know whether to continue to subdivide or terminate?

5

4

, , ,

, 4 ,6 2

1,

90 2

b

a

I f x dx S a b E a b where

b a a bS a b f a f f b and

b aE a b f

April 18, 2023 OSU/CSE 541 98


• The first iteration can then be defined as:

• Subsequent subdivision can be defined as: 2 , ,S S a c S c b

1 1

1 1

,

, , ,

I S E where

S S a b E E a b

April 18, 2023 OSU/CSE 541 99


• Now, since

• We can solve for E(2) in terms of E(1).

5 52 4 4

54 1

4

1 / 2 1 / 2

90 2 90 2

1 1 1

2 90 2 16

h hE f f

hf E

2 , ,E E a c E c b

April 18, 2023 OSU/CSE 541 100


• Finally, using the identity:

• We have:

• Plugging into our definition:

1 1 2 2I S E S E

2 1 1 2 215S S E E E

2 2 2 2 11

15I S E S S S

April 18, 2023 OSU/CSE 541 101


• Our error criteria is thus:

• Simplifying leads to the termination formula:

2 2 11

15I S S S

2 1 15S S

April 18, 2023 OSU/CSE 541 102


• What happens graphically:

April 18, 2023 OSU/CSE 541 103

12 15S subd vS i ide

April 18, 2023 OSU/CSE 541 104

April 18, 2023 OSU/CSE 541 105

12 152

subdivideSS

April 18, 2023 OSU/CSE 541 106

2 12

1

15SI S S

12 154

S oS d ne

April 18, 2023 OSU/CSE 541 107

12 154

subdivideSS

April 18, 2023 OSU/CSE 541 108

2 12

1

15SI S S

April 18, 2023 OSU/CSE 541 109

2 12

1

15SI S S

April 18, 2023 OSU/CSE 541 110

I=Ileft + Iright

Iright =Ileft + Iright

April 18, 2023 OSU/CSE 541 111

April 18, 2023 OSU/CSE 541 112

April 18, 2023 OSU/CSE 541 113


• We gradually capture the difficult spots.

April 18, 2023 OSU/CSE 541 114

Adaptive Simpson’s Code

• Simple Recursive Programstatic const int m_nMaximum_Divisions = 1000;Real IntegrationSimpson( const Real (*f) (Real x), const Real start, const Real end, const Real

error_tolerance, int &level ){

level += 1;Real h = (end – start);Real midpoint = (start + end) / 2.0;Real f_start = f(start);Real f_end = f(end);Real f_mid = f(midpoint );oneLevel = h*( f_start + 4.0*f_mid + f_end) / 6.0;Real leftMidpoint = (start+ midpoint ) / 2.0;Real rightMidpoint = (end+ midpoint ) / 2.0Real f_midLeft = f(leftMidpoint );Real f_midRight = f(rightMidpoint );twoLevel = h*(f_start + 4.0* f_midLeft + 2.0* f_mid + 4.0* f_midRight + f_end) / 12.0;if( level >= m_nMax_Divisions ) // Terminate the process, converging too slow

return twoLevel;

April 18, 2023 OSU/CSE 541 115

Adaptive Simpson’s Code

if( absf( twoLevel – oneLevel) < 15.0*error_tolerance) // Desired solution reachedreturn twoLevel + (twoLevel-oneLevel) / 15.0;

//// Otherwise, split the interval in two and recursively evaluate each half.//leftIntegral = IntegrationSimpson( f, start, midpoint , error_tolerance/2.0, level );rightIntegral = IntegrationSimpson( f, midpoint , end, error_tolerance/2.0, level );return leftIntegral + rightIntegral;

}

April 18, 2023 OSU/CSE 541 116

• Idea is that if we evaluate the function at certain points, and sum with certain weights, we will get a more accurate integral

• Evaluation points and weights are pre-computed and tabulated

• Basic form:

1

11

)()(n

iii xfcdxxfI

ci : weighting factors xi : sampling points selected optimally

Guassian Quadrature

New!!

April 18, 2023 OSU/CSE 541 117

• Note that the interval is between –1 and 1• For other intervals, a change of variables is used

to transfer the problem so that it utilizes the interval [-1, 1]

• This is a linear transform, such that for t[a,b]:

• We have for x[-1,1]:

b

adttf )(

2

)( abxabt

ab

abtx

2

Guassian Quadrature

April 18, 2023 OSU/CSE 541 118

• As t = a x = -1• As t = b x = 1

dxab

dt2

)(

2

)()(

abxabftf

dxabxab

fab

dttfb

a

1

1 2

)(

2

)()(

Guassian Quadrature

April 18, 2023 OSU/CSE 541 119

• Basic form of Gaussian quadrature:

• For n=2, we have:

• This leads to 4 unknowns: c1, c2, x1, and x2

– two unknown weights (c1, c2)

– two unknown sampling points (x1, x2)

1

11

)()(n

iii xfcdxxfI

2211 xfcxfcI

Guassian Quadrature

April 18, 2023 OSU/CSE 541 120

Guassian Quadrature

• What we need now, are four known values for the equation.

• If we had these, we could then attempt to solve for the four unknowns.

• Let’s make it work for polynomials!!!

April 18, 2023 OSU/CSE 541 121

• In particular, let’s look at these simple polynomials: – Constant

• f(x)=1

– Linear• f(x)=x

– Quadratic• f(x)=x2

– Cubic• f(x)=x3

Guassian Quadrature

April 18, 2023 OSU/CSE 541 122

Guassian Quadrature

• Recalling the formula: – Constant

• f(x)=1

– Linear• f(x)=x

– Quadratic• f(x)=x2

– Cubic• f(x)=x3

322

3112211

1

1

3

222

2112211

1

1

2

22112211

1

1

212211

1

1

0

3

2

0

21

xcxcxfcxfcdxx

xcxcxfcxfcdxx

xcxcxfcxfcxdx

ccxfcxfcdx

2211 xfcxfcI

April 18, 2023 OSU/CSE 541 123

• We can now solve for our unknowns:– Note, this is not an easy problem and will not be

covered in this class.

577.03

1

577.03

1

1

2

1

21

x

x

cc

Guassian Quadrature

April 18, 2023 OSU/CSE 541 124

Guassian Quadrature

• This yields the two point Gauss-Legendre formula

3

1

3

1ffI

April 18, 2023 OSU/CSE 541 125

Guassian Quadrature

• This is exact for all polynomials up to and including degree 3 (cubics).

1 1 1 1 13 2 3 2

1 1 1 1 1

3 3 2 2

1

3

1

3

1 1 1 1 1 11 1

3 3 3 3 3 3

3 2

ax bx cx d dx a x dx b x dx c xdx d dx

a b c d

ax bx cx d

April 18, 2023 OSU/CSE 541 126

x

f(x)

-1 1

f(0.577) f(-0.577)

-0.577 0.577

1

1)577.0()577.0()( ffdxxf

Guassian Quadrature

April 18, 2023 OSU/CSE 541 127

Example

5432 400900675200252.0)( xxxxxxf

dxabxab

fab

dttfb

a

1

1 2

)(

2

)()(

dttf

dtt

fdxxfI

1

1

1

1

8.0

0

4.04.04.0

2

08.0)08.0(

2

)08.0()(

• Integrate f(x) from a = 0 to b = 0.8

• Transform from [0, 0.8] to [-1, 1]

April 18, 2023 OSU/CSE 541 128

• Solving

• And substituting for the 2-point formula:


dtttt

tt

dttfI

1

1 543

2

1

1

)4.04.0(400)4.04.0(900)4.04.0(675

)4.04.0(200)4.04.0(252.04.0

4.04.04.0

3/1t

82257778.130583723.151674055.0 I

dttfI 1

14.0

Example

April 18, 2023 OSU/CSE 541 129

• Recall the basic form:

• Let’s look at n=3.

• We now have 6 unknowns: c1, c2, c3,x1, x2, and x3

– three unknown weights (c1, c2 , c3)– three unknown sampling points (x1, x2 , x3)

1

11

)()(n

iii xfcdxxfI

332211 xfcxfcxfcI

Higher-order Gaussian Quadrature

April 18, 2023 OSU/CSE 541 130

Use 6 equations - constant, linear, quadratic, cubic, 4th order and 5th order to find those unknowns

533

522

511332211

1

1

5

433

422

411332211

1

1

4

333

322

311332211

1

1

3

233

222

211332211

1

1

2

332211332211

1

1

321332211

1

1

0

5

2

0

3

2

0

21

xcxcxcxfcxfcxfcdxx

xcxcxcxfcxfcxfcdxx

xcxcxcxfcxfcxfcdxx

xcxcxcxfcxfcxfcdxx

xcxcxcxfcxfcxfcxdx

cccxfcxfcxfcdx

April 18, 2023 OSU/CSE 541 131

• Can solve these equations (or have some one smarter than us, like Guass solve them).

• Produces the three point Gauss-Legendre formula

– Exact for polynomials up to and including degree 5(because using 5th degree polynomial)

77459669.05/3 0 77459669.05/3

9/5 9/8 9/5

221

321

xxx

ccc

332211 xfcxfcxfcI


April 18, 2023 OSU/CSE 541 132

1

1)775.0(

9

5)0.0(

9

8)775.0(

9

5)( fffdxxf

-1 1 x

f(0.775)

f(-0.775)

-0.775 0.775

f(0)


April 18, 2023 OSU/CSE 541 133

5432 400900675200252.0)( xxxxxxf

Integrate from a = 0 to b = 0.8

Transform from [0, 0.8] to [-1, 1]

dtttt

tt

dxxfI

1

1 543

2

8.0

0

)4.04.0(400)4.04.0(900)4.04.0(675

)4.04.0(200)4.04.0(252.0

Example

replace -0.4 with +0.4

April 18, 2023 OSU/CSE 541 134

5

3

9

50

9

8

5

3

9

5fffI

64053334.1

485987599.0873244444.0281301290.0

I

Substitute into the transform equation and get


Example

• Using the 3-point Gauss-Legendre formula:

April 18, 2023 OSU/CSE 541 135

Can develop higher order Gauss-Legendre forms using

nn xfcxfcxfcI ...2211

Values for c’s and x’s are tabulated

Use the same transformation to map interval onto [-1, 1]

Gaussian Quadrature

April 18, 2023 OSU/CSE 541 136

1713245.0

3607616.0

4679139.0

4679139.0

3607616.0

1713245.0

500.23692688

050.47862867

890.56888888

050.47862867

500.23692688

510.34785484

490.65214515

490.65214515

510.34785484

560.55555555

890.88888888

560.55555555

0.1

0.1

9324695140

6612093860

2386191860

2386191860

6612093860

9324695140

90617984590

53846931010

00000000000

53846931010

90617984590

86113631160

33998104360

33998104360

86113631160

77459666920

00000000000

77459666920

57735026920

57735026920

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

2 3 4 5 6nci

xi

nn xfcxfcxfcdxxfI ...)( 2211

1

1

April 18, 2023 OSU/CSE 541 137

Gaussian Quadrature

• Requires function evaluations at non-uniformly spaced points within the integration interval– not appropriate for cases where the function is

unknown– not suited for dealing with tabulated data that

appear in many engineering problems

• If the function is known, its efficiency can be a decided advantage

April 18, 2023 OSU/CSE 541 138

Gaussian Quadrature

• Problems:– If we add more data points, like doubling the

number of sample points.

numerical integration roger crawfis. june 9, 2015osu/cse 5412 quadrature we talk in terms of...

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