CIS541_07_Integration.ppt

Numerical Integration
Roger Crawfis

September 19, 2022 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

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

What does an integral represent?
Basic definition of an integral:
Motivation
 
b
a
dx
x
f area
)
(   
d
c
b
a
dxdy
x
f volume
)
(

 




n
k
k
b
a n
x
x
f
dx
x
f
1
)
(
lim
)
(
sum of height  width
f(x)
x
n
a
b
x



where

• 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
a
dx
x
f
I )
(
dx
e
x
x x
5
.
0
2
0
5
.
0
1
)
1
cos(
2
 



• 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
i
i E
x
f
w
dx
x
f
I 

  

)
(
)
(
1
Reimann Integral Theorem

• The most common numerical integration formula
is based on equally spaced data points.
• Divide [x0 , xn] into n intervals (n1)

n
x
x
dx
x
f
0
)
(
Partitioning the Integral
1 2
0
0 1 1
( ) ( ) ( ) ( )
n
n
n
x
x x
x
x
x x x
f x dx f x f x f x

   
   

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
1
0
( )
n
x n
i i i
i
x
f x M x x



 


 
1
sup ( ):
i i i
M f x x x x
   Supremum - least
upper bound

Upper Sums
• Graphically:
x0 x1 x2 x4
x3

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
1
0
( )
n
x n
i i i
i
x
f x m x x



 


 
1
inf ( ):
i i i
m f x x x x
  
Infimum - greatest
lower bound

Lower Sums
• Graphically
x0 x1 x2 x4
x3

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 1
0 0
( )
n
x
n n
i i i i i i
i i
x
m x x f x M x x
 
 
 
   
 


Bounding the Integral
• Graphically
x0 x1 x2 x4
x3

• Halving each interval (much like Lab1):
x0 x3 x5 x9
x7

• One more time:
x0 x5 x7 x11
x9

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 x9
x7

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?

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.
m
m
m x
a
x
a
x
a
a
x
p 



 ...
)
( 2
2
1
0

• 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 2
0 0 0
( ) ( ) ( ) ... ( )
n m m m n
n
m n mn
x x x x
m m m
x x x x
f x dx p x dx p x dx p x dx
  
 
   
   
)
(
3/8
s
Simpson'
cubic
3
)
(
1/3
s
Simpson'
quadratic
2
)
(
Trapezoid
linear
1
Error
Formula
Polynomial
4
4
2
h
O
h
O
h
O
m
Newton-Cotes Formulas

• 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
a
dx
x
p
dx
x
f
I 1
        
a
x
a
b
a
f
b
f
a
f
x
p 




1
Trapezoid Rule

Trapezoid Rule
      
 










b
a
dx
a
x
a
b
a
f
b
f
a
f
I
     
 
b
f
a
f
a
b
I 


2
a b
f(a)
f(b)
height
average
width

I
Trapezoid Rule

Trapezoid Rule
• Improvement?
x0 x1 x2 x4
x3

• 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)
    2
3
12
)
(
12
1
h
f
a
b
h
f
Et 
 







 )
( 2
h
O
Trapezoid Rule Error

2
)
( x
e
x
f 

Integrate from a = 0 to b = 2
     
   
0183
.
1
)
(
1
)
0
(
)
2
(
2
)
0
2
(
2
0
4
2
0
2














e
e
f
f
b
f
a
f
a
b
dx
e
I x
Use trapezoidal rule:
Example

Estimate error:   3
12
1
h
f
Et 




Where h = b - a and a <  < b
Don’t know  - use average value
2
)
4
2
(
)
( 2 x
e
x
x
f 





2564
.
0
)
2
(
2
)
0
(







f
f
   
  58
.
0
2
2
0
12
23









f
f
E
E a
t
2
0
2 


h
Example

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

• 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
a
b
h


     


 







b
h
b
h
a
h
a
h
a
a
dx
x
f
dx
x
f
dx
x
f
I ...
2
Composite Trapezoid Rule

• Substituting the trapezoid rule for each integral.
• Results in the Composite Trapezoid Formula:
     
 
   
 
 
   
 
 
   
 
2
2
2
2 2
....
2
...
a h a h b
a a h b h
a h a a h a h
f a f a h f a h f a h
b b h
f b h f b
I f x dx f x dx f x dx
 
 
    
      
 
   
   
  
     








 


b
f
ih
a
f
a
f
h
I
n
i
1
1
2
2

     
 
     
n
b
f
ih
a
f
a
f
a
b
b
f
ih
a
f
a
f
h
I
n
i
n
i
2
2
2
2
1
1
1
1





















width
Average height
• Think of this as the width times the average
height.

• 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
a
b
f
h
a
b
Ea







 2
3
2
12
12
f 

)
( 2
h
O
Error

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
  5
4
3
2
200
810
730
140
20
3
.
0 x
x
x
x
x
x
f 






     
 
26
.
11
2
81
.
3
22
.
34
2
.
0
8
.
0
2







b
f
a
f
a
b
I
Example
• A single application of the Trapezoid rule.
• Error:
  3
12
1
a
b
f
Et 



 

Example
• We don’t know  so approximate with
average f
  3
2
4000
9720
4380
280 x
x
x
x
f 






  4
3
2
1000
3240
2190
280
20 x
x
x
x
x
f 





 
6
.
131
2
.
0
8
.
0
)
2
.
0
(
)
8
.
0
(
2
.
0
8
.
0
8
.
0
2
.
0














f
f
dx
f
x
f

   
  
3
2
2
3
12 12
1
131.6 0.8 0.2 2.37
12
t
b a h b a
E f f
n
 
 
 
    
Example
• The error can thus be estimated as:

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%

• Use intervals (0.2,0.4),(0.4,0.6),(0.6,0.8):
– (n = 3, h = 0.2)
 
     
       
   
 
57
.
12
6
22
.
34
16
.
30
93
.
13
2
31
.
3
6
.
0
)
3
)(
2
(
8
.
0
6
.
0
4
.
0
2
2
.
0
2
.
0
8
.
0
2
2
1
1


















f
f
f
f
n
b
f
ih
a
f
a
f
a
b
I
n
i
True value of integral is 12.82
Using Three Intervals

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%

             
   
 
76
.
12
12
22
.
34
22
.
35
16
.
30
18
.
22
93
.
13
34
.
7
2
31
.
3
6
.
0
)
6
)(
2
(
8
.
0
7
.
0
6
.
0
5
.
0
4
.
0
3
.
0
2
2
.
0
2
.
0
8
.
0
















f
f
f
f
f
f
f
I
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

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%

Multi-dimensional Integration
• Consider a two-dimensional case.
 
1 1 1
0
0 0 0
1
0 0
0 0
0 0
, ,
,
,
,
n
i
i
n
i
i
n n
i j
i j
n n
i j
i j
i
f x y dxdy A f y dy
n
i
A f y dy
n
i j
A A f
n n
i j
A A f
n n


 
 
 
  
 
 
  
 
 
  
 
 
  
 

 
 
 


• 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, 1
1
2 {1, 1} 1, , 1
4
4 1, , 2 1, , 2
ij
i n j n
i n j n
A
i n j n
n
i n j n
 


   

 
   

    

1
2
2
2
2
2
1
1 2 2 2 2 2 1

• 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.

• 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 hh/2.
– Remember this for Monte-Carlo Integration.
     
2
2
2 2 k k
k
O h O n O n O N


  
 
    
 
   

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 h0.
• Called Romberg Integration

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
2
2
n
h
i
h
I f a f a ih f b


 
   
 
 


Adding More Intervals
• This is called the Recursive Trapezoid
Rule in the book.
• We have n 2n and hh/2.
 
2 1
1
2
1 1
1 0
1
0
( ) 2 ( )
4 2
( ) 2 2 ( )
4 2
2
2 4 2
n
h
i
n n
i i
n
h
i
h h
I f a f a i f b
h h
f a f a ih f a ih f b
I h h
f a ih


 
 


 
 
   
 
 
 
 
 
 
      
 
 
 
 
 
 
   
 
 
 
 

 


• 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
)
(
)
( 1
1
n
h
O
h
A
A 

)
(
)
( 2
2
n
h
O
h
A
A 

Recall Richardson Extrapolation

• Multiplying the 2nd eqn by (h1/h2)n
and
subtracting from the 1st eqn:
• The new error term is generally O(h1
n+1
) or
O(h1
n+2
).
1
)
(
)
(
2
1
1
2
2
1


















 n
n
h
h
h
A
h
A
h
h
A
Richardson Extrapolation

• The true integral value can be written
• This is true for any iteration
   
h
E
h
I
I 

       
2
2
1
1 h
E
h
I
h
E
h
I
I 




• For example: Using (n = 2)
• where c is a constant
• Therefore:
 
  2
2
2
1
2
1
h
h
h
E
h
E

f
ch
E 

 2
[error = O(h2)]
order of error in
trapezoidal rule

• This leads to:
• For integration, we have:
    2
2
2
1
2
1
h
h
h
E
h
E 
       
2
2
2
2
2
1
2
1 h
E
h
I
h
h
h
E
h
I 



• Solving for E(h2):
• And plugging back into the estimated
integral.
     
2
2
2
1
2
1
2
1
h
h
h
I
h
I
h
E



   
2
2 h
E
h
I
I 


• Leads to:
• Letting h2 = h1 /2
 
 
   
 
1
2
2
2
1
2
1
/
1
h
I
h
I
h
h
h
I
I 



     
 
   
1
2
1
2
2
2
3
1
3
4
1
2
1
h
I
h
I
I
h
I
h
I
h
I
I







• 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.
   
l
m h
I
h
I
I
15
1
15
16


Romberg Integration

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.
   
l
m h
I
h
I
I
63
1
63
64



• 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
k
j k j k
j k j k j k j k
k k
I I
I I I I

 

   
 
Romberg Integration

Romberg Integration
• For example, j = 1, k = 1 leads to
 
1,0 0,0 1
1,1 1
4 4 1
3 3 2 3
I I h
I I I h
  
  
 
 

• Consider the function:
• Integrate from a = 0 to b = 0.8
• Using the trapezoidal rule yields the
following results:
5
4
3
2
400
900
675
200
25
2
.
0
)
( x
x
x
x
x
x
f 





Example

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
4



I
(j=1, k=1)
k
j
k = 0 k = 1
j = 0
j = 1
j = 2

2 4
segments ( ) ( )
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
4



I (j=2, k=1)
k = 1
k = 0
Example

Example
2 4
segments ( ) ( )
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 = 2
k = 1
64053334
.
1
)
3674667
.
1
(
15
1
)
62346667
.
1
(
15
16



I

2 4 6
segments ( ) ( ) ( )
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
j
k = 3
k = 2
k = 1
Example

2 4 6 8
segments ( ) ( ) ( ) ( )
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
64



I (j=3, k=3)
k
j
k = 3
Example
• Better and better results can be obtained by
continuing this

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!!!

Higher-Order Polynomials
• Recall:
)
(
3/8
s
Simpson'
cubic
3
)
(
1/3
s
Simpson'
quadratic
2
)
(
Trapezoid
linear
1
Error
Formula
Polynomial
4
4
2
h
O
h
O
h
O
m
1 1 2
1
0 0
1 2
0 0 0
( ) ( ) ( ) ... ( )
n m m m n
n
m n mn
x x x x
m m m
x x x x
f x dx p x dx p x dx p x dx
  
 
   
   

• If we use a 2nd order polynomial (need 3
points or 2 intervals):
– Lagrange form.
  
  
    
  
 
  
  
  dx
x
f
x
x
x
x
x
x
x
x
x
f
x
x
x
x
x
x
x
x
x
f
x
x
x
x
x
x
x
x
I
x
x




















 
2
1
2
0
2
1
0
1
2
1
0
1
2
0
0
2
0
1
0
2
1
2
0





 

2
2
0
1
x
x
x
Simpson’s 1/3 Rule

• Requiring equally-spaced intervals:
  
 
 
  
  
 
  
  
 
2
0
0 0 0 0
0 1
0 0
2
2 2
2
2
x
x
x x h x x h x x x x h
I f x f x
h h h h
x x x x h
f x dx
h h
       
 

  


  
 



• Integrate and simplify:
0
2
4
6
8
10
12
3 5 7 9 11 13 15
     
 
2
1
0 4
3
x
f
x
f
x
f
h
I 


2
a
b
h


Quadratic
Polynomial

• If we use a = x0 and b = x2, and
x1 = (b+a)/2
       
6
4 2
1
0 x
f
x
f
x
f
a
b
I




width
average height

• Error for Simpson’s 1/3 rule
Integrates a cubic exactly:
 
  0
4


f
 
     
 

 4
5
4
5
2880
90
f
a
b
f
h
Et





2
a
b
h


)
( 4
h
O

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.

• 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

• 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
x
dx
x
f
dx
x
f
dx
x
f
I
2
4
2
2
0
...
 
       
1 2
0
1,3,5 2,4,6
4 2
3
n n
i j n
i j
f x f x f x f x
I b a
n
 
 
  
 
 
n+1 data points, an odd number

• Odd coefficients receive a weight of 4, even
receive a weight of 2.
• Doesn’t seem very fair, does it?
coefficients on
numerator
1
4 1
1 4
1
1
4 1
i = 0
i = n

• The error can be estimated by:
• If n is doubled, h  h/2 and Ea  Ea/16
  )
4
(
4
)
4
(
5
180
180
f
h
a
b
f
nh
Ea



)
4
(
f is the average 4th derivative
)
( 4
h
O
Error Estimate

• Integrate from a = 0 to b = 2.
• Use Simpson’s 1/3 rule:
   
   
2
2
0 1 2
0
0 1 4
1
4 ( )
3
1
0 4 (1) 2
3
1
( 4 ) 0.82994
3
x
I e dx h f x f x f x
f f f
e e e

 
   
 
 
  
 
 
   

2
)
( x
e
x
f 

2
1
2
0
1
2
2
1
0 








 b
x
b
a
x
a
x
a
b
h
Example

• Error estimate:
• Where h = b - a and a <  < b
• Don’t know 
– use average value
 
 

4
5
90
f
h
Et 

 
 
   
   
 
4 4 4
5 5
0 1 2
4
1 1
90 90 3
t a
f x f x f x
E E f
 
 
 
    
Example

• Let’s look at the polynomial again:
– From a = 0 to b = 0.8
5
4
3
2
400
900
675
200
25
2
.
0
)
( x
x
x
x
x
x
f 





8
.
0
4
.
0
2
0
4
.
0
2
2
1
0 








 b
x
b
a
x
a
x
a
b
h
   
 
   
 
36746667
.
1
8
.
0
)
4
.
0
(
4
0
3
)
4
.
0
(
)
(
4
3
1
)
( 2
1
0
2
0







 
f
f
f
x
f
x
f
x
f
h
dx
x
f
I
Another Example

• Actual Error: (using the known exact value)
• Estimate error: (if the exact value is not
available)
• Where a <  < b.
 
 

4
5
90
f
h
Et 

27306666
.
0
36746667
.
1
-
1.64053334 

E 16%
Error

• Compute the fourth-derivative
• Matches actual error pretty well.
Error
x
x
f 48000
21600
)
(
)
4
(



 
   
  27306667
.
0
4
.
0
90
4
.
0
90
4
.
0 4
5
1
4
5





 f
x
f
E
E a
t
middle point

• If we use 4 segments instead of 1:
– x = [0.0 0.2 0.4 0.6 0.8]
0.2
b a
h
n

 
232
.
0
)
8
.
0
(
464
.
3
)
6
.
0
(
456
.
2
)
4
.
0
(
288
.
1
)
2
.
0
(
2
.
0
)
0
(





f
f
f
f
f
 
       
 
6234667
.
1
12
232
.
0
)
456
.
2
(
2
)
464
.
3
288
.
1
(
4
2
.
0
8
.
0
)
4
)(
3
(
)
8
.
0
(
)
6
.
0
(
4
)
4
.
0
(
2
)
2
.
0
(
4
)
0
(
0
8
.
0
3
2
4
1
5
,
3
,
1
2
6
,
4
,
2
0

















 




f
f
f
f
f
n
x
f
x
f
x
f
x
f
a
b
I
n
n
i
n
j
j
i
Example Continued

• Actual Error: (using the known exact value)
• Estimate error: (if the exact value is not
available)
01706667
.
0
6234667
.
1
-
1.64053334 

E 1%
 
   
  0085
.
0
4
.
0
90
2
.
0
90
2
.
0 4
5
2
4
5






 f
x
f
E
E a
t
middle point
Error

Error
• Actual is twice the estimated, why?
• Recall:
x
x
f 48000
21600
)
(
)
4
(



 
 
(4) (4)
0,0.8
(4)
max ( ) (0) 21600
(0.4) 2400
x
f x f
f

  


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 4
0.2 0.2
0 0.0768
90 90
a
E f f

   

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.

• Simpson’s 3/8 rule uses a third order polynomial
– need 3 intervals (4 data points)
3
3
2
2
1
0
3 )
(
)
( x
a
x
a
x
a
a
x
p
x
f 




   

 

3
0
3
0
3
x
x
x
x
dx
x
p
dx
x
f
I

• Determine a’s with Lagrange polynomial
• For evenly spaced points
3
a
b
h


       
 
3
2
1
0 3
3
8
3
x
f
x
f
x
x
f
h
I 




• Same order as 1/3 Rule.
– More function evaluations.
– Interval width, h, is smaller.
• Integrates a cubic exactly:

 
  0
4


f
 
 
4
5
3
80
t
E h f 
  )
( 4
h
O
Error

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

• 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

• 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

Adaptive Simpson’s Scheme
• Recall Simpson’s 1/3 Rule:
• Where initially, we have a=x0 and b=x2.
• Subdividing the integral into two:
     
 
2
1
0 4
3
x
f
x
f
x
f
h
I 


         
1 2 3
4 2 4
6
h
I f a f x f x f x f b
    
 
 

• We want to keep subdividing, until we
reach a desired error tolerance, .
• Mathematically:
       
           
1
1 2 3
4
3
4 2 4
6
b
a
b
a
h
f x dx f a f x f b
h
f x dx f a f x f x f x f b


 
   
 
 
 
 
 
     
 
 
 
 



• 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
h
f x dx f a f x f x and
h
f x dx f x f x f b where
a b
c x


 
   
 
 
 
 
 
   
 
 
 
 

 



• 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?

• 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.

• 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 b
S a b f a f f b and
b a
E a b f
  
   
 
  
 
 
 
 

 
   
 


• 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
 
 

• Now, since
• We can solve for E(2) in terms of E(1).
     
   
5 5
2 4 4
5
4 1
4
1 / 2 1 / 2
90 2 90 2
1 1 1
2 90 2 16
h h
E f f
h
f E
   
  
   
   
   
  
   
   
 
   
2
, ,
E E a c E c b
 

• Finally, using the identity:
• We have:
• Plugging into our definition:
       
1 1 2 2
I S E S E
   
         
2 1 1 2 2
15
S S E E E
   
         
 
2 2 2 2 1
1
15
I S E S S S
    

• Our error criteria is thus:
• Simplifying leads to the termination
formula:
     
 
2 2 1
1
15
I S S S 
   
   
 
2 1
15
S S 
 

• What happens graphically:

1
2 15
S subd v
S i ide

  

1
2 15
2
subdivide
S
S

  

 
2 1
2
1
15
S
I S S
  
1
2 15
4
S o
S d ne

  

1
2 15
4
subdivide
S
S

  

 
2 1
2
1
15
S
I S S
  

I=Ileft + Iright
Iright =Ileft + Iright

• We gradually capture the difficult spots.

Adaptive Simpson’s Code
• Simple Recursive Program
static 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 ) / 2.0
Real 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;

Adaptive Simpson’s Code
if( absf( twoLevel – oneLevel) < 15.0*error_tolerance) // Desired solution reached
return 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;
}

• 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
1
1
)
(
)
(
n
i
i
i x
f
c
dx
x
f
I
ci : weighting factors
xi : sampling points selected optimally
Guassian Quadrature
New!!

• 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
a
dt
t
f )
(
2
)
( a
b
x
a
b
t




a
b
a
b
t
x




2
Guassian Quadrature

• As t = a  x = -1
• As t = b  x = 1
dx
a
b
dt
2
)
( 






 



2
)
(
)
(
a
b
x
a
b
f
t
f
dx
a
b
x
a
b
f
a
b
dt
t
f
b
a 
 





 




1
1 2
)
(
2
)
(
)
(
Guassian Quadrature

• 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
1
1
)
(
)
(
n
i
i
i x
f
c
dx
x
f
I
   
2
2
1
1 x
f
c
x
f
c
I 

Guassian Quadrature

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!!!

• 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

Guassian Quadrature
• Recalling the formula:
– Constant
• f(x)=1
– Linear
• f(x)=x
– Quadratic
• f(x)=x2
– Cubic
• f(x)=x3
   
   
   
    3
2
2
3
1
1
2
2
1
1
1
1
3
2
2
2
2
1
1
2
2
1
1
1
1
2
2
2
1
1
2
2
1
1
1
1
2
1
2
2
1
1
1
1
0
3
2
0
2
1
x
c
x
c
x
f
c
x
f
c
dx
x
x
c
x
c
x
f
c
x
f
c
dx
x
x
c
x
c
x
f
c
x
f
c
xdx
c
c
x
f
c
x
f
c
dx




























   
2
2
1
1 x
f
c
x
f
c
I 


• We can now solve for our unknowns:
– Note, this is not an easy problem and will not be
covered in this class.
577
.
0
3
1
577
.
0
3
1
1
2
1
2
1








x
x
c
c
Guassian Quadrature

Guassian Quadrature
• This yields the two point Gauss-Legendre
formula















3
1
3
1
f
f
I

Guassian Quadrature
• This is exact for all polynomials up to and
including degree 3 (cubics).
 
 
 
1 1 1 1 1
3 2 3 2
1 1 1 1 1
3 3 2 2
1
3
1
3
1 1 1 1 1 1
1 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
    

      
     
  
           
       
     
           
   
           
 
   

    
  

x
f(x)
-1 1
f(0.577)
f(-0.577)
-0.577 0.577




1
1
)
577
.
0
(
)
577
.
0
(
)
( f
f
dx
x
f
Guassian Quadrature

Example
5
4
3
2
400
900
675
200
25
2
.
0
)
( x
x
x
x
x
x
f 





dx
a
b
x
a
b
f
a
b
dt
t
f
b
a 
 





 




1
1 2
)
(
2
)
(
)
(
 dt
t
f
dt
t
f
dx
x
f
I












 





1
1
1
1
8
.
0
0
4
.
0
4
.
0
4
.
0
2
0
8
.
0
)
0
8
.
0
(
2
)
0
8
.
0
(
)
(
• Integrate f(x) from a = 0 to b = 0.8
• Transform from [0, 0.8] to [-1, 1]

• Solving
• And substituting for the 2-point formula:
 
dt
t
t
t
t
t
dt
t
f
I























1
1 5
4
3
2
1
1
)
4
.
0
4
.
0
(
400
)
4
.
0
4
.
0
(
900
)
4
.
0
4
.
0
(
675
)
4
.
0
4
.
0
(
200
)
4
.
0
4
.
0
(
25
2
.
0
4
.
0
4
.
0
4
.
0
4
.
0
3
/
1


t
82257778
.
1
30583723
.
1
51674055
.
0 


I
 dt
t
f
I 

1
1
4
.
0
Example

• 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
1
1
)
(
)
(
n
i
i
i x
f
c
dx
x
f
I
     
3
3
2
2
1
1 x
f
c
x
f
c
x
f
c
I 


Higher-order Gaussian Quadrature

Use 6 equations - constant, linear, quadratic, cubic,
4th order and 5th order to find those unknowns
     
     
     
     
     
      5
3
3
5
2
2
5
1
1
3
3
2
2
1
1
1
1
5
4
3
3
4
2
2
4
1
1
3
3
2
2
1
1
1
1
4
3
3
3
3
2
2
3
1
1
3
3
2
2
1
1
1
1
3
2
3
3
2
2
2
2
1
1
3
3
2
2
1
1
1
1
2
3
3
2
2
1
1
3
3
2
2
1
1
1
1
3
2
1
3
3
2
2
1
1
1
1
0
5
2
0
3
2
0
2
1
x
c
x
c
x
c
x
f
c
x
f
c
x
f
c
dx
x
x
c
x
c
x
c
x
f
c
x
f
c
x
f
c
dx
x
x
c
x
c
x
c
x
f
c
x
f
c
x
f
c
dx
x
x
c
x
c
x
c
x
f
c
x
f
c
x
f
c
dx
x
x
c
x
c
x
c
x
f
c
x
f
c
x
f
c
xdx
c
c
c
x
f
c
x
f
c
x
f
c
dx























































• 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
.
0
5
/
3
0
77459669
.
0
5
/
3
9
/
5
9
/
8
9
/
5
2
2
1
3
2
1










x
x
x
c
c
c
     
3
3
2
2
1
1 x
f
c
x
f
c
x
f
c
I 








1
1
)
775
.
0
(
9
5
)
0
.
0
(
9
8
)
775
.
0
(
9
5
)
( f
f
f
dx
x
f
-1 1
x
f(0.775)
f(-0.775)
-0.775 0.775
f(0)

5
4
3
2
400
900
675
200
25
2
.
0
)
( x
x
x
x
x
x
f 





Integrate from a = 0 to b = 0.8
Transform from [0, 0.8] to [-1, 1]
 
dt
t
t
t
t
t
dx
x
f
I





















1
1 5
4
3
2
8
.
0
0
)
4
.
0
4
.
0
(
400
)
4
.
0
4
.
0
(
900
)
4
.
0
4
.
0
(
675
)
4
.
0
4
.
0
(
200
)
4
.
0
4
.
0
(
25
2
.
0
Example
replace -0.4 with +0.4

  



















5
3
9
5
0
9
8
5
3
9
5
f
f
f
I
64053334
.
1
485987599
.
0
873244444
.
0
281301290
.
0




I
Substitute into the transform equation and get
Example
• Using the 3-point Gauss-Legendre formula:

Can develop higher order Gauss-Legendre forms
using
     
n
n x
f
c
x
f
c
x
f
c
I 


 ...
2
2
1
1
Values for c’s and x’s are tabulated
Use the same transformation to map interval onto
[-1, 1]
Gaussian Quadrature

1713245
.
0
3607616
.
0
4679139
.
0
4679139
.
0
3607616
.
0
1713245
.
0
50
0.23692688
05
0.47862867
89
0.56888888
05
0.47862867
50
0.23692688
51
0.34785484
49
0.65214515
49
0.65214515
51
0.34785484
56
0.55555555
89
0.88888888
56
0.55555555
0
.
1
0
.
1
932469514
0
661209386
0
238619186
0
238619186
0
661209386
0
932469514
0
9061798459
0
5384693101
0
0000000000
0
5384693101
0
9061798459
0
8611363116
0
3399810436
0
3399810436
0
8611363116
0
7745966692
0
0000000000
0
7745966692
0
5773502692
0
5773502692
0
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.









2 3 4 5 6
n
ci
xi
     
n
n x
f
c
x
f
c
x
f
c
dx
x
f
I 



 
...
)
( 2
2
1
1
1
1

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

Gaussian Quadrature
• Problems:
– If we add more data points, like doubling the
number of sample points.

CIS541_07_Integration.ppt

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CIS541_07_Integration.ppt