simpson rule composite simpson rule

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

aabfbfbafafabdxxf

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

()((23a-b

Area(I)abfaf

3ab

))2

()2

((23a-b

Area(II)abfabf

))()2

((23a-b

Area(III)

bfabf

2ab

• 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

()((23a-b

Area(I)abfaf

3ab

))2

()2

((23a-b

Area(II)abfabf

))()2

((23a-b

Area(III)

bfabf

))()2

(4)((23a-b

Area(III)Area(II)Area(I)

bfabfaf

2ab

數值方法2008, Applied Mathematics NDHU 13

passes )()( 2

xpPxp

))(,( bfb ))2

(,2

( bafba

b

adxxp )(

that Show

)()2

(4)(6

bfbafafab

))(,( afa

數值方法2008, Applied Mathematics NDHU 14

)()()( :))(())(()(

))(())(()(

))(())(()( )(

cf,p(c)bf,p(b)afp(a)proofbcacbxaxcf

cbabcxaxbf

cabacxbxafxp

))(,( afa

))(,( bfb

2)),(,( baccfc

數值方法2008, Applied Mathematics NDHU 15

31

)02)(12()1)(0(

))(())((

34

)01)(21()2)(0(

))(())((

31

)10)(20()1)(2(

))(())(2(

))(())((

2))(())(()(

))(())(()(

))(())(()(

))(())(()(

))(())(()(

))(())(()( )(

2

0

2

0

2

0

2

hdttthdxcbabcxax

hdttthdxbcacbxax

hdttthdxcaba

haxhaxdxcabacxbx

hachab

dxbcacbxaxcfdx

cbabcxaxbfdx

cabacxbxaf

dxbcacbxaxcf

cbabcxaxbf

cabacxbxafdxxp

b

a

b

a

ha

a

b

a

b

a

b

a

b

a

b

a

b

a

))()(4)((

3bfcfafh

))()(4)((6

)(

))()(4)((3

bfcfafab

bfcfafh

數值方法2008, Applied Mathematics NDHU 16

dttthdthhhtht

dxcaba

haxhaxdxcabacxbx

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

bfbafafab

IIIQIIQIQdxxp

PpCBxAxxp

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

bfbafafab

CxxBxA

dxCBxAxdxxp

PpCBxAx

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

)(

bfbafafab

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

xfhxfhbfafh

xfxfxfxfxfh

hiafhiafihafh

dxxf

njjhax h=(b-a)/2n

數值方法2008, Applied Mathematics NDHU 24

Composite Simpson rule

1

112

1

12 )(

34)(

32))()((

3

)(

2,...,1,0,

n

kk

n

kk

b

a

k

xfhxfhbfafh

dxxf

njjhax h=(b-a)/2n

Simpson's Rule for Numerical Integration