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Numerical Methods
What computers can’t do
• Solve (by reasoning) general mathematical
problems  they can only repetitively
apply arithmetic primitives to input.
• Solve problems exactly.
• Represent all numbers. Only a finite subset
of the numbers between 0 and 1 can be
represented.
Numerical Integration
• In NA, take visual view of integration as
area under the curve
• Many integrals that occur in science or
engineering practice do not have a closed
form solution – must be solved using
numerical integration
Trapezoidal Rule
• The area under the curve from
[a, fa] to [b, fb] is initially approximated by
a trapezoid:
I1 = ( b – a ) * ( fa + fb ) / 2
Simple
trapezoid over
large interval is
prone to error.
Divide interval
into halves…
fc
I2 = ( b – a )/2 * ( fa + 2*fc + fb ) / 2
(Note that interior sides count twice, since they belong to 2 traps.)
Trapezoidal rule

b
a
f ( x)dx 
x
Tn 
[ f ( x0 )  2 f ( x1 )  2 f ( x2 )    2 f ( xn 1 )  f ( xn )]
2
ba
where x 
and
xi  a  i x
n
This gives us a better approximation
than either left or right rectangles.
Example Single Application of the Trapezoidal Rule
f(x) = 0.2 +25x – 200x2 + 675x3 – 900x4 + 400x5
Integrate f(x) from a=0 to b=0.8
b 0.8
True integral value : I 
 f ( x )dx  1.64053
a 0
Solution: f(a)=f(0) = 0.2 and f(b)=f(0.8) = 0.232
f (a )  f (b)
2
0.2  0.232
 0.8
 0.1728
2
Trapezoida l Rule : I  ( b  a )
which represents an error of :
E t  1.64053  0.1728  1.46773  t  89.5%
Simpson’s rule
•
•
•
•
.. Another approach
Rather than use straight line of best fit,
Use parabola of best fit (curves)
Converges more quickly
Simpson’s Rules
• More accurate estimate of an integral is obtained if
a high-order polynomial is used to connect the
points. These formulas are called Simpson’s rules.
Simpson’s 1/3 Rule: results when a 2nd order
Lagrange interpolating polynomial is used for f(x)
a=x0
b
b=x2
b
I   f ( x)dx   f 2 ( x)dx
a
Using
x1
where f 2 ( x ) is a second - order polynomial.
a
a  x0
b  x2
 ( x  x1 )( x  x2 )

( x  x0 )( x  x2 )
( x  x0 )( x  x1 )
I  
f ( x0 ) 
f ( x1 ) 
f ( x2 )dx
(
x

x
)(
x

x
)
(
x

x
)(
x

x
)
(
x

x
)(
x

x
)
0
1
0
2
1
0
1
2
2
0
2
1

x0 
x2
after integratio n and algebraic manipulation, the following formula results :
I
h
 f ( x0 )  4 f ( x1 )  f ( x2 )
3
h
ba
2

SIMPSON'S 1/3 RULE
The Multiple-Application Simpson’s 1/3 Rule
• Just as the trapezoidal rule, Simpson’s rule can be improved by dividing the
integration interval into a number of segments of equal width.
• However, it is limited to cases where values are equispaced, there are an even
number of segments and odd number of points.
h
I
ba
n
x2

x0
I
n  # of seg. a  x0
x4
f ( x)dx   f ( x)dx 
0
2
x2

b  xn
xn

f n  2 ( x)dx
xn2
h
 f ( x0 )  4 f ( x1 )  f ( x2 ) 
3
h
  f ( x2 )  4 f ( x3 )  f ( x4 )  
3
h
  f ( xn  2 )  4 f ( xn 1 )  f ( xn ) 
3
n 1
n2

h

I   f ( x0 )  4  f ( xi )  2  f ( x j )  f ( xn ) 
3
i 1, 3, 5...
j  2, 4, 6...

Simpson’s rule

b
a
f ( x)dx 
x
Sn 
[ f ( x0 )  4 f ( x1 )  2 f ( x2 )  4 f ( x3 )  
3
 2 f ( xn  2 )  4 f ( xn 1 )  f ( xn )]
ba
where n is even and x 
n
Simpson’s rule can also be interpreted as fitting parabolas to sections of
the curve.
Simpson’s rule will usually give a very good approximation with
relatively few subintervals.
Simpson’s 3/8 Rule
Fit a 3rd order Lagrange interpolating
polynomial to four points and integrate
b
b
a
a
Simpson’s 1/3 and 3/8 rules can be
applied in tandem to handle
multiple applications with odd
number of intervals
I   f ( x)dx   f 3 ( x)dx
3h
I   f ( x0 )  3 f ( x1 )  3 f ( x2 )  f ( x3 )
8
(b  a )
h 
3
I  (b  a)
f ( x0 )  3 f ( x1 )  3 f ( x2 )  f ( x3 )
8
13
Rn APPROXIMATIONEquation 2
• If we choose xi* to be the right endpoint,
xi* = xi and we have:

b
a
n
f ( x) dx  Rn   f ( xi ) x
– The approximation Rn
is called right endpoint
approximation.
i 1
• The M
figure
shows
APPROXIMATION
n
the midpoint
approximation Mn.
THE MIDPOINT RULE

b
a
f ( x) dx  M n
 x [ f ( x1 )  f ( x 2 )  ...  f ( x n )]
• wherex  b  a
n
and
xi  12 ( xi 1  xi )  midpoint of [ xi 1 , xi ]
APPROXIMATE
INTEGRATION
• Approximate the integral
with n = 5, using:
a. Trapezoidal Rule
Example 1

2
1
• b. Midpoint Rule
(1/ x) dx
APPROXIMATE
INTEGRATION
Example 1 a
• With n = 5, a = 1 and b = 2,
we have: ∆x = (2 – 1)/5 = 0.2
– So, the Trapezoidal Rule gives:

2
1
1
0.2
dx  T5 
[ f (1)  2 f (1.2)  2 f (1.4)
x
2
 2 f (1.6)  2 f (1.8)  f (2)]
2
2
2 1
1 2
 0.1 



 
 1 1.2 1.4 1.6 1.8 2 
 0.695635
APPROXIMATE
INTEGRATION Example 1 b
• The midpoints of the five subintervals
are: 1.1, 1.3, 1.5, 1.7, 1.9
APPROXIMATE
Example 1 b
INTEGRATION
• So, the Midpoint Rule gives:

2
1
1
dx  x [ f (1.1)  f (1.3)  f (1.5)
x
 f (1.7)  f (1.9)]
1 1
1
1
1
1 
 





5  1.1 1.3 1.5 1.7 1.9 
 0.691908
ERROR ESTIMATES
Example 3
a. Use the Midpoint Rule with n = 10 to
1 2
approximate the integral
x

0
e dx.
b. Give an upper bound for the error involved in
this approximation.
Example 3 a
ERROR ESTIMATES
• As a = 0, b = 1, and n = 10, the Midpoint Rule
gives:
1
e
0
x2
dx
 x [ f (0.05)  f (0.15)  ...  f (0.85)  f (0.95)]

e
e
e
e
e
 0.1  0.3025 0.4225 0.5625 0.7225 0.9025 
e
e
e
e

 e
 1.460393
0.0025
0.0225
0.0625
0.1225
0.2025