Introduction to Fourier Series
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Transcript Introduction to Fourier Series
Numerical Methods
Part: Simpson
Integration.
3
8
Rule For
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Lecture # 1
Chapter 07.08: Simpson 3 8 Rule
For Integration.
Major: All Engineering Majors
Authors: Duc Nguyen
http://numericalmethods.eng.usf.edu
Numerical Methods for STEM undergraduates
4/24/2020
http://numericalmethods.eng.usf.edu
5
Introduction
The main objective in this chapter is to develop
appropriated formulas for obtaining
the integral
expressed in the following form:
b
I f ( x)dx
a
(1)
where f (x) is a given function.
Most (if not all) of the developed formulas for integration
is based on a simple concept of replacing a given
(oftently complicated) function f (x) by a simpler
function (usually a polynomial function) fi (x ) where i
6 represents the order of the polynomial function.
In the previous chapter, it has been explained and
illustrated that Simpsons 1/3 rule for integration can
be derived by replacing the given function f (x )
with the 2nd –order (or quadratic) polynomial function
fi ( x) f 2 ( x), defined as:
f 2 ( x) a0 a1x a2 x
7
2
(2)
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8
In a similar fashion, Simpson 3 8 rule for integration
can be derived by replacing the given function f (x)
with the 3rd-order (or cubic) polynomial (passing
through 4 known data points) function fi ( x) f3 ( x)
defined as f ( x) a a x a x 2 a x 3
3
0
1
2
3
a0
a
(3)
1
2
3
1, x, x , x
a
2
a3
which can also be symbolically represented in Figure 1.
Method 1
The unknown coefficients a0 , a1 , a2 and a3 (in Eq. (3))
can be obtained by substituting 4 known coordinate
data points x0 , f x0 , x1 , f x1 , x2 , f x2 and x3 , f x3
into Eq. (3), as following
9
f ( x0 ) a0 a1x0 a2 x02 a3 x02
f ( x1 ) a0 a1x1 a2 x12 a3 x12
(4)
f ( x2 ) a0 a1x2 a2 x22 a3 x22
2
2
f ( x3 ) a0 a1x3 a2 x3 http://numericalmethods.eng.usf.edu
a3 x3
Eq. (4) can be expressed in matrix notation as
1
1
1
1
x0
x02
x1
x12
x2
x22
x32
x3
x03 a0 f x0
3 a f x
x1 1 1
3 a f x
x2 2 2
3 a f x
x3 3 3
(5)
The above Eq. (5) can be symbolically represented as
A44 a41 f 41
10
(6)
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Thus,
a0
a
1
1
a A f
a2
a3
(7)
Substituting Eq. (7) into Eq. (3), one gets
1
2 3
f3 x 1, x, x , x A f
11
(8)
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Remarks
As indicated in Figure 1, one has
x0 a
b a 2a b
x1 a h a
3
3
2b 2a a 2b
x2 a 2h a
3
3
3b 3a
x3 a 3h a
b
3
(9)
With
the help from MATLAB [2], the unknown vector
a (shown in Eq. 7) can be solved.
12
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Method 2
Using Lagrange interpolation, the cubic polynomial
function f
x that passes through 4 data points
i 3
(see Figure 1) can be explicitly given as
x x1 x x2 x x3 f x x x0 x x2 x x3 f x
0
1
x0 x1 x0 x2 x0 x3
x1 x0 x1 x2 x1 x3
x x0 x x1 x x3
x x0 x x1 x x2
f x3
f x3
x2 x0 x2 x1 x2 x3
x3 x0 x3 x1 x3 x2
f3 x
(10)
13
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Simpsons 8 Rule For
Integration
3
Thus, Eq. (1) can be calculated as (See Eqs. 8, 10 for
Method 1 and Method 2, respectively):
b
b
a
a
I f x dx f3 x dx
Integrating the right-hand-side
equations, one obtains
of
the
above
f x0 3 f x1 3 f x2 f x3
I b a
(11)
8
14
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ba
Since h
hence b a 3h , and the above
3
equation becomes:
3h
I f x0 3 f x1 3 f x2 f x3
8
(12)
The error introduced by the Simpson 3/8 rule can be
derived as [Ref. 1]:
(b a)
Et
f , where
6480
5
15
a b
(13)
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Example 1 (Single Simpson 3 8 rule)
Compute
b 30
140,000
I 2000 ln
9.8 x dx,
140,000 2100 x
a 8
16
by using a single segment Simpson 3 8 rule
Solution
In this example:
b a 30 8
h
7.3333
3
3
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140000
x0 8 f x0 2000 ln
9.8 8 177.2667
140000 2100 8
x1 x0 h 8 7.3333 15.3333
140000
f x1 2000 ln 140000 2100 15.3333 9.8 15.3333 372.4629
17
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x2 x0 2h 8 2(7.3333) 22.6666
140000
f
x
2000
ln
9.8 22.6666 608.8976
2
140000 2100 22.6666
x3 x0 3h 8 3(7.3333) 30
140000
f
x
2000
ln
9.8 30 901.6740
3
140000 2100 30
18
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Applying Eq. (12), one has:
3
I 7.3333 177.2667 3 372.4629 3 608.8976 901.6740
8
I 11063 .3104
The “exact” answer can be computed as
I exact 11061 .34
19
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3. Multiple Segments for
Simpson 3 8 Rule
Using "n" = number of equal (small) segments, the
width "h" can be defined as
ba
h
3
(14)
Notes:
n = multiple of 3 = number of small "h" segments
20
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The integral, shown in Eq. (1), can be expressed as
b
b
a
a
I f x dx f3 x dx
I
21
x3
x6
xn b
x0 a
x3
x n 3
f3 x dx f3 x dx ........ f3 x dx
(15)
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Substituting Simpson 3 8 rule (See Eq. 12) into
Eq. (15), one gets
3h f x0 3 f x1 3 f x2 f x3 f x3 3 f x4 3 f x5 f x6
I
8 ..... f xn 3 3 f xn 2 3 f xn 1 f xn
(16)
n2
n 1
n 3
3h
I f x0 3 f xi 3 f xi 2 f xi f xn
8
i 1, 4,7,..
i 2,5,8,..
i 3,6,9,..
(17)
22
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Example 2 (Multiple segments Simpson 3 8 rule)
b 30
140,000
Compute I 2000 ln
9.8 x dx,
140,000 2100 x
a 8
3
using Simple 8 multiple segments rule, with number
(of "h" ) segments = n = 6 (which corresponds to 2
“big” segments).
23
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Solution
In this example, one has (see Eq. 14):
30 8
h
3.6666
6
x0 , f x0 8,177.2667
x1, f x1 11.6666,270.4104; where x1 x0 h 8 3.6666
11.6666
x2 , f x2 15.3333,372.4629; where x2 x0 2h 15.3333
x3 , f x3 19,484.7455; where x3 x0 3h 19
24
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x4 , f x4 22.6666,608.8976; where x4 x0 4h 22.6666
x5 , f x5 26.3333,746.9870; where x5 x0 5h 26.3333
x6 , f x6 30,901.6740; where x6 x0 6h 30
25
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Applying Eq. (17), one obtains:
n 24
n 15
n 3 3
3
I 3.6666 177.2667 3 f xi 3 f xi 2 f xi 901.6740
8
i 1, 4,..
i 2,5,..
i 3,6,..
177.2667 3270.4104 608.8976 3372.4629 746.9870
I 1.3750
2484.7455 901.6740
I 11,601.4696
26
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3
1
Example 3 (Mixed, multiple segments Simpson 3 and 8
rules)
b 30
140,000
Compute I 2000 ln
9.8 x dx,
140,000 2100 x
a 8
using Simpson 1/3 rule (with n1 4 small segments), and
Simpson 3/8 rule (with n2 3 small segments).
Solution:
In this example, one has:
27
ba ba
30 8
h
3.1429
n
n1 n2 4 3
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x0 a 8
x1 x0 1h 8 3.1429 11.1429
x2 x0 2h 8 23.1429 14.2857 Simpson 1 rule
3
x3 x0 3h 8 33.1429 17.4286
x4 x0 4h 8 43.1429 20.5714
x5 x0 5h 8 53.1429 23.7143
x6 x0 6h 8 63.1429 26.8571
x7 x0 7h 8 73.1429 30
28
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140,000
f x0 8 2000 ln
9.8 8 177.2667
140,000 2100 8
Similarly:
f x1 11.1429 256 .5863
f x2 342 .3241
f x3 435 .2749
f x4 536 .3909
f x5 646 .8260
f x6 767 .9978
29
f x7 901 .6740
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For multiple segments n1 first 4 segments
using Simpson 1 rule, one obtains (See Eq. 19):
3
n1 1 3
n1 2 2
h
I1 f x0 4 f xi 2 f xi f xn1
3
i 1,3,...
i 2,...
3.1429
I1
177 .2667 4256 .5863 435 .2749 2342 .3241 536 .3909
3
I1 4364 .1197
30
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For multiple segments n2 last 3 segments
using Simpson 3/8 rule, one obtains (See Eq. 17):
n2 2 1
n2 1 2
n2 3 0
3h
I 2 f x0 3 f xi 3 f xi 2 f xi f xn1
8
i 1,3,...
i 2,...
i 3 , 6 ,...
3
I 2 3.1429 177 .2667 3256 .5863 3342 .3241 skip! 435 .2749
8
I 2 6697 .2748
The mixed (combined) Simpson 1/3 and 3/8 rules give:
I I1 I 2 4364 .1197 6697 .2748
I 11,061.3946
31
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Remarks:
(a) Comparing the truncated error of Simpson 1/3 rule
b a
5
Et
2880
f
(18)
With Simple 3/8 rule (See Eq. 13), the latter seems to
offer slightly more accurate answer than the former.
However, the cost associated with Simpson 3/8 rule
(using 3rd order polynomial function) is significant
higher than the one associated with Simpson 1/3 rule
(using 2nd order polynomial function).
32
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(b) The number of multiple segments that can be used
in the conjunction with Simpson 1/3 rule is 2,4,6,8,..
(any even numbers).
h
I1 f x0 4 f x1 f x2 f x2 4 f x3 f x4 ..... f xn 2 4 f xn 1 f xn
3
n 1
n2
h
I 2 f x0 4 f xi 2 f xi f xn
3
i 1,3,...
i 2, 4, 6...
33
(19)
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However, Simpson 3/8 rule can be used with the
number of segments equal to 3,6,9,12,.. (can be
either certain odd or even numbers).
(c) If the user wishes to use, say 7 segments, then the
mixed Simpson 1/3 rule (for the first 4 segments),
and Simpson 3/8 rule (for the last 3 segments).
34
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4. Computer Algorithm For Mixed Simpson
1/3 and 3/8
rule For Integration
Based on the earlier discussions on (Single and Multiple
segments) Simpson 1/3 and 3/8 rules, the following
“pseudo” step-by-step mixed Simpson rules can be given
as
Step 1 User’s input information, such as
Given function f (x), integral limits " a, b",
n1= number of small, “h” segments, in conjunction with
Simpson 1/3 rule.
35
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n2 = number of small, “h” segments, in conjunction with
Simpson 3/8 rule.
Notes:
n1 = a multiple of 2 (any even numbers)
n2 = a multiple of 3 (can be certain odd, or even
numbers)
36
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Step 2
Compute n n1 n2
ba
h
n
x0 a
x1 a 1h
x2 a 2h
.
.
xi a ih
.
.
37
xn a nh b
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Step 3
Compute “multiple segments” Simpson 1/3 rule (See
Eq. 19)
n1 1
n1 2
h
I1 f x0 4 f xi 2 f xi f xn
3
i 1,3,...
i 2, 4,6...
(19, repeated)
38
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Step 4
Compute “multiple segments” Simpson 3/8 rule (See
Eq. 17)
n2 2
n2 1
n2 3
3h
I 2 f x0 3 f xi 3 f xi 2 f xi f xn2
8
i 1, 4,7...
i 2,5,8...
i 3,6,9,...
(17, repeated)
Step 5
(20)
I I1 I 2
and print out the final approximated answer for I.
39
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THE END
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Acknowledgement
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undergraduate
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This material is based upon work supported by the National
Science Foundation under Grant # 0717624. Any opinions,
findings, and conclusions or recommendations expressed in
this material are those of the author(s) and do not necessarily
reflect the views of the National Science Foundation.
The End - Really