數值方法 2008, applied mathematics ndhu 1 simpson rule composite simpson rule
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數值方法2008, Applied Mathematics NDHU 1
Simpson ruleComposite Simpson rule
數值方法2008, Applied Mathematics NDHU 2
Simpson rule for numerical integration
b
aab
fbf
bafaf
abdxxf
baCf
5)4(
4
)(2880
)()()
2(4)(
6)(
withexists b)(a,in number a then ],,[ If
數值方法2008, Applied Mathematics NDHU 3
a=1.2;b=2;x=linspace(a,b);plot(x,sin(x));hold on
y=sin(x) for x within [1.2 2]
數值方法2008, Applied Mathematics NDHU 4
Quadratic polynomial
c=0.5*(a+b);x=[a b c]; y=sin(x);p=polyfit(x,y,2);z=linspace(a,b);plot(z,polyval(p,z) ,'r')
• Approximate sin using a quadratic polynomial
數值方法2008, Applied Mathematics NDHU 5
a=1;b=3;x=linspace(a,b);plot(x,sin(x))hold on;c=0.5*(a+b);x=[a b c]; y=sin(x);p=polyfit(x,y,2);z=linspace(a,b);plot(z,polyval(p,z) ,'r')
Approximate sin(x) within [1 3] by a quadratic polynomial
數值方法2008, Applied Mathematics NDHU 6
a=1;b=3;x=linspace(a,b);plot(x,sin(x))hold on;c=0.5*(a+b);x=[a b ]; y=sin(x);p=polyfit(x,y,1);z=linspace(a,b);plot(z,polyval(p,z) ,'r')
Approximate sin(x) within [1 3] by a line
1 1.5 2 2.5 30.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
數值方法2008, Applied Mathematics NDHU 7
1 1.5 2 2.5 30.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Strategy I : Apply Trapezoid rule to calculate area under a lineStrategy II : Apply Simpson rule to calculate area under a second order polynomial
Strategy II is more accurate and general than strategy I, since a line is a special case of quadratic polynomial
數值方法2008, Applied Mathematics NDHU 8
Simpson rule
Original task: integration of f(x) within [a,b]
c=(a+b)/2 Numerical task
Approximate f(x) within [a,b] by a quadratic polynomial, p(x)
Integration of p(x) within [a,b]
數值方法2008, Applied Mathematics NDHU 9
Blue: f(x)=sin(x) within [1,3] Red: a quadratic polynomial that pass (1,sin(1)), (2,sin(2)),(3,sin(3))
數值方法2008, Applied Mathematics NDHU 10
))2
()((23
a-b
Area(I)
abfaf
3
ab
))2
()2
((23
a-b
Area(II)
abf
abf
))()2
((23
a-b
Area(III)
bfab
f
2
ab
• The area under a quadratic polynomial is a sum of area I, II and III
數值方法2008, Applied Mathematics NDHU 11
• The area under a quadratic polynomial is a sum of area I, II and III
• Partition [a,b] to three equal-size intervals, and use the high of the middle point c to produce three Trapezoids
• Use the composite Trapezoid rule to determine area I, II and III h/2*(f(a)+f(c) +f(c)+f(c)+ f(c)+f(b)) • substitute h=(b-a)/3 and c=(a+b)/2• area I + II + III = (b-a)/6 *(f(a)+4*f((a+b)/2) +f(b))
數值方法2008, Applied Mathematics NDHU 12
))2
()((23
a-b
Area(I)
abfaf
3
ab
))2
()2
((23
a-b
Area(II)
abf
abf
))()2
((23
a-b
Area(III)
bfab
f
))()2
(4)((23
a-b
Area(III)Area(II)Area(I)
bfab
faf
2
ab
數值方法2008, Applied Mathematics NDHU 13
passes )(
)( 2
xp
Pxp ))(,( bfb ))
2(,
2(
baf
ba
b
adxxp )(
that Show
)()
2(4)(
6bf
bafaf
ab
))(,( afa
數值方法2008, Applied Mathematics NDHU 14
)()()( :
))((
))(()(
))((
))(()(
))((
))(()( )(
cf,p(c)bf,p(b)afp(a)proof
bcac
bxaxcf
cbab
cxaxbf
caba
cxbxafxp
))(,( afa
))(,( bfb
2)),(,(
baccfc
數值方法2008, Applied Mathematics NDHU 15
3
1
)02)(12(
)1)(0(
))((
))((
3
4
)01)(21(
)2)(0(
))((
))((
3
1
)10)(20(
)1)(2(
))((
))(2(
))((
))((
2
))((
))(()(
))((
))(()(
))((
))(()(
))((
))(()(
))((
))(()(
))((
))(()( )(
2
0
2
0
2
0
2
hdttt
hdxcbab
cxax
hdttt
hdxbcac
bxax
hdttt
hdxcaba
haxhaxdx
caba
cxbx
hac
hab
dxbcac
bxaxcfdx
cbab
cxaxbfdx
caba
cxbxaf
dxbcac
bxaxcf
cbab
cxaxbf
caba
cxbxafdxxp
b
a
b
a
ha
a
b
a
b
a
b
a
b
a
b
a
b
a
))()(4)((
3bfcfaf
h
))()(4)((6
)(
))()(4)((3
bfcfafab
bfcfafh
數值方法2008, Applied Mathematics NDHU 16
dttt
hdthh
htht
dxcaba
haxhaxdx
caba
cxbx
h
ha
a
b
a
2
0
2
0
2
)10)(20(
)0)(2(
)0)(20(
))(2(
))((
))(2(
))((
))((
-1 -0.5 0 0.5 1 1.5 2 2.5 3-0.5
0
0.5
1
1.5
2
2.5
3
Shift Lagrange polynomialRescale Lagrange polynomial
數值方法2008, Applied Mathematics NDHU 17
Shortcut to Simpson 1/3 rule
))()2
(4)((6
)()()()(
)(
2
2
bfba
fafab
IIIQIIQIQdxxp
Pp
CBxAxxp
b
a
1. Partition [a,b] to three equal intervals2. Draw three trapezoids, Q(I), Q(II) and Q(III)3. Calculate the sum of areas of the three trapezoids
數值方法2008, Applied Mathematics NDHU 18
Proof
))()2
(4)((6
23
)()(
))()((4))()((2))()((2)(
1)(
23
2
2
2
2
bfba
fafab
CxxB
xA
dxCBxAxdxxp
Pp
CBxAx
bxaxcfcxaxbfcxbxafba
xp
b
a
b
a
b
a
數值方法2008, Applied Mathematics NDHU
19
Composite Simpson rule
Partition [a,b] into n interval Integrate each interval by Simpson
rule h=(b-a)/2n
1
0
)22(
2)(
)(
n
i
hia
iha
b
a
dxxf
dxxf
數值方法2008, Applied Mathematics NDHU 20
Apply Simpson sule to each interval
)))22(())12((4)2((3
)()22(
2
hiafhiafihafh
dxxfhia
iha
))()2
(4)((6
)(
bfba
fafab
dxxpb
a
數值方法2008, Applied Mathematics NDHU 21
Composite Simpson rule
1
0
1
0
)22(
2
)))22(())12((4)2((3
)()(
n
i
n
i
hia
iha
b
a
hiafhiafihafh
dxxfdxxf
數值方法2008, Applied Mathematics NDHU 22
1
0
1
0
)22(
2
)))22(())12((4)2((3
)()(
n
i
n
i
hia
iha
b
a
hiafhiafihafh
dxxfdxxf
Set nSet fx
h=(b-a)/(2*n)ans=0;
for i=0:n-1
aa=a+2*i*h;cc=aa+h;bb=cc+h;
ans=ans+h/3*(fx(aa)+4*fx(cc)+fx(bb))
exit
數值方法2008, Applied Mathematics NDHU 23
Composite Simpson rule
1
112
1
12
23210
1
0
)(3
4)(
3
2))()((
3
)(...)(4)(2)(4)(3
)))22(())12((4)2((3
)(
2,...,1,0,
n
kk
n
kk
n
n
i
b
a
k
xfh
xfh
bfafh
xfxfxfxfxfh
hiafhiafihafh
dxxf
njjhax h=(b-a)/2n
數值方法2008, Applied Mathematics NDHU 24
Composite Simpson rule
1
112
1
12 )(
3
4)(
3
2))()((
3
)(
2,...,1,0,
n
kk
n
kk
b
a
k
xfh
xfh
bfafh
dxxf
njjhax h=(b-a)/2n
Simpson's Rule for Numerical Integration