Numerical Integration - (4.3) k 1 bk 1 a k 1. k f x 1 k 1 bk 1 a k 1 9 hf a h hf b 0 1 b a 9 h h 3h b a.

Numerical Integration - (4.3) 1. The Degree of Accuracy of a Quadrature Formula:

The degree of accuracy of a quadrature formula Qf is the largest positive integer n such that



a b xk_{dx  Qx}k_{, k  0, 1, 2, . . . , n.} Example



a b fxdx  9 4hfa  h  34hfb, h   b− a

3 , 3h  b − a. Find the degree of

accuracy

of this quadrature formula Qf  9

4 hfa  h  34hfb. Consider fx  1, fx  x, fx  x2_{, . . . . It is known that}



a b xk_{dx } 1 k 1b k1_{− a}k1_. k fx 1 k 1b k1− ak1_ 9 4hfa  h  34hfb 0 1 b− a 9 4h 34h  3h  b − a 1 x 1 2 b 2 _{− a}2_ 9 4ha  h  34hb  3h4 3a  h  b  b − a 4 3a  b − a  b  b − a4 2a  2b  12b 2 _{− 1} 2 a 2 2 x2 1 3 b 3_{− 1} 3a 3 9 4ha  h 2 _{ 3} 4hb 2 _ 1 3b 3 ₋ 1 3a 3 3 x3 1 4 b 4_{− 1} 4a 4 9 4ha  h 3 _{ 3} 4hb 3 _{ − 1} 9 ba 3 _{ 1} 6 b 2_a2 ₋ 1 9 b 3_{a 5} 18 b 4 _{− 2} 9 a 4

So the degree of accuracy is 2

Example Suppose that the quadrature formula



0 2

fxdx  c0f0  c1f1  c2f2

is exact for all polynomials of degree less than and equal to 2. Find c0, c1and c2

Qf  c0f0  c1f1  c2f2 . fx  1, c0  c1  c2  2 1 fx  x, 0 c1 2c2  2 2 fx  x2_{, 0 c} 1  4c2  8 3 3 3 − 2: 2c2  2 3, c2  13 Substitute it into2 : c1  2 − 2 1 3  43 Substitute both c1and c2into1 :

c0  2 − 4 3 − 13  13



₀2 fxdx  1 3f0  43f1  13f2 2. Numerical Quadrature:



a b fxdx ≈

∑

i0 n aifxi.

What are a0, a1, . . . , an? What is the approximation error?

a. Using Lagrange Interpolating Polynomials: If fx 

∑

i0 n fxiLn,ix  f n1_c n  1!



_i0 n x − xi then



a b fxdx ≈

∑

i0 n fxi



a b Ln,ixdx  ai 



a b Ln,ixdx, i  0, 1, . . . , n

The corresponding approximation error is

Rint 



a b _fn1c n  1!



_i0 n x − xidx. Trapezoidal Rule:

Consider the 1st degree Lagrange interpolating polynomial P1x with data pairs a, fa and

b, fb . Let h  b − a. P1x  fa x − b a− b  fb x − ab− a  −fa x − bh  fb x − ah



a b P1xdx 



a b −fa x − b h  fb x − ah dx  − fa 2h x − b 2 _ fb 2h x − a 2 a b  1 2h fbh 2  fah2  h 2 fb  fa

Weighted Mean Value Theorem for Integrals: (Page 9, Theorem 1.11 Suppose that F is continuous on a, b ,



a b

gxdx exists, and gx does not change sign on a, b .

Then there exists a number c in a, b with



a b Fxgxdx  Fc



a b gxdx.

The corresponding approximation error:

Consider Fx  f ′′ c , and gx  x − ax − b. gx does not change sign on a, b .

Rtrap 



a b _f ′′ c  2! x − ax − bdx  12f ′′ c



a b x − ax − bdx  1 2f ′′ c



a b x − ax − a − hdx  1 2f ′′ c



a b x − a2 _{− x − ah dx}  1 2f ′′ c 1 3x − a 3_{− 1} 2x− a 2_h a b  1 2f ′′ c 1 3b − a 3 _{− 1} 2b− a 2_h  1 2f ′′ c 1 3h 3 − 1 2h 3  − 1 12f ′′ ch3

b. Using Taylor Polynomials and Difference Formulas:

Though we can derive an numerical quadrature using the formula:

∑

_in_0fxi



a b Ln,ixdx, the remainder formula Rint 



a b _fn1c n  1!



_i0 n x − xidx

may not be easy to use to derive the approximation error. To derive the approximation error for the Trapezoidal Rule, we use the Weighted Mean Value Theorem which has a condition: gx does not change sign on a, b . This condition is not always satisfied for f

n1_c

n  1!



i0 n

x − xi. For

example, to derive the approximation error for Simpson’s Rule (3 pairs of data, a 2nd degree Lagrange polynomial) Rint 



a b _f ′′′c 3! x − x0x − x1x − x2dx Here gx  1

3!x − x0x − x1x − x2 which does change sign over the interval x0, x2 . By Taylor Theorem, we know

fx  Pnx  Rnx, and



a b fxdx 



a b Pnxdx 



a b Rnxdx where Pnx  fxi  f ′ xi 1! x − xi  f ′′xi 2! x − xi 2 _ f ′′′ xi 3! x − xi 3 _{. . . }f n xi n! x − xi n Rnx  f n1 c n  1! x − xin1

The approximation error for



a b fxdx ≈



a b Pnxdx is Rint 



a b _f n1c n  1! x − xin1dx.

Observe that the sign of f

n1

c

n  1! x − xin1does not change on the interval a, b . So by the

Weighted Mean-Value Theorem we know there exists a number c in a, b such that



_ab _f n1c n  1! x − xin1dx  f n1_c̄ n  1!



a b x − xin1dx  fn1c̄ n  1! n 2 1 b− xi n2_{− a − x} in2

For the approximation error for Simpson’s Rule: choose xi  x1, and use x0, x1, x2 :

x0  a  x1 − h, x2  b  x1  h, h  b − a

2 Note that the interval a, b  x0, x2 is symmetric about x  x1. So,



x0 x2 x − x1mdx 



x0 x1 x − x1mdx



x1 x2 x − x1mdx  0 if m is odd 2



x1 x2 x − x1mdx  2 m 1 x2 − x1 m1 _ 2 m 1h m1 _{if m is even} Consider



a b fxdx ≈



a b P3xdx 



x0 x2 fx1  f ′ x1 1! x − x1  f ′′x1 2! x − x1 2 _ f ′′′ x1 3! x − x1 3 dx  2hfx1  f ′′ x1 2 2 3h 3

Approximate f ′′x1 by the difference formula:

f ′′x  fx  h − fx  fx − h

h2 − 1₁₂f4c1h

2

we have the numerical integral formula:

Qf  2hfx1  h 3 3 fx2 − fx1  fx0 h2  2hfx1  h₃fx2 − fx1  fx0  h 3fx0  4fx1  fx2 and the approximation error:

Rint 



x0 x2 f 4c₁ 4! x − x1 4 dx− h3 3 1 12f 4_c 2h2  f 4 c̄1 4! 2 5h 5 _{− h}5 36f 4_c 2  f4c 1 60 − 136 h 5 _{ − h}5 90f 4_c where c is in x0, x2 .

Numerical integration formulas and approximation errors:

Name Approximation formula Approximation error

Trapezoidal Rule h 2 fb  fa , h  b − a − 112f ′′ ch3 Simpson’s Rule h 3 fx0  4fx1  fx2 , h  b − a2 − h 5 90 f 4_c Example Approximate



0 /4

sinx2_{dx by the Trapezoidal Rule, and Simpson’s Rule . Estimate}

corresponding approximation errors.

0.6 0.4 0.2 0 0.5 0.375 0.25 0.125 0 y  sinx2_{, 0 ≤ x ≤}  4, dash yT, dashdots y Trapezoidal Rule: yT  sin  2 16 −0  4 −0 x − 0  0  4 sin  2 16 yS  −0. 151 162 9  0. 776 077 7x  0. 940 758 1x − 0. 392 699 12 h   4 − 0  4 , T  12 4 f0  f 4  8 0 sin  2 16  0. 227 164 h  1 2 4 − 0  8 , S  1 3  8 f0  4f 8  f 4  13  8 0 4 sin  2 64  sin  2 14  0. 165 250 46 True error: RT 



0 /4 sinx2_{dx − 0. 227 164  −7. 000 933 252 893 539 95  10}−2 RS 



0 /4 sinx2_{dx − 0. 165 250 46  8. 095 792 5  10}−3 Approximation errors: f ′′x  −4sin x2_x2 _{ 2 cos x}2 f ′′′x  −8cos x2_x3 _{− 12sin x}2_x

f4x  16sin x2_x4 − 48cosx2_x2 − 12 sin x2

0.8 0.6 0.4 0.2 0 2 1.5 1 0.5 0 x y x y y  f ′′x 0.8 0.6 0.4 0.2 0 25 20 15 10 5 0 x y x y y  |f4x|

f ′′x ≤ f ′′0  2, f 4x ≤ f 4  4 ≤ 16 sin 4 2 _ 4 4 − 48 cos  4 2 _ 4 2 − 12 sin  4 2 ≤ 28 Err_T  − h3 12f ′′  ≤ 2 12 4 3  8. 074 551  10−2 Err_S  − h5 90f4 ≤ 19028 8 5  2. 905 46  10−3 3. Newton-Cotes Formulas:

These formulas are derived using Lagrange interpolating polynomials.

Several numerical integration formulas are listed on Page 192-4: formulas (4.23)-(4.30). a. n  1 −point closed Newton - Cotes formulas:

x0  a, xn  b, h  b − a_n , xi  a  ih n # of pts # of sub-int. Formula (4.23)-(4.26) approx error c 1 2 1 h 1 2fx0  12fx1 − h 3 12 f ′′ c x0, x1 2 3 2 h 1 3 fx0  43fx1  13fx2 − h 5 90 f 4_{c x} 0, x2 3 4 3 h 3 8 fx0  98 fx1  98 fx2  38 fx3 − 3h 5 80 f 4_{c x} 0, x3 4 5 4 h 14 45 fx0  6445fx1  2445fx2  6445fx3  1445fx4 − 8h 7 945 f6c x0, x4 b. n  1 −point open Newton - Cotes formulas:

x−1  a, x0  a  h, xn  b − h, xn1  b, h  b − a_{n 2}, xi  xi−1  h.

The Midpoint formulas are open Newton - Cotes formulas.

n # of pts # of sub-int. Formula (4.27)-(4.30) approx error c 0 3 2 2hfx0 h 3 3 f ′′ c x−1, x1 1 4 3 h 3 2fx0  32fx1 3h3 4 f ′′ c x−1, x2 2 5 4 h 8 3fx0 − 43fx1  83fx2 14h5 45 f 4_{c x−1, x} 3 3 6 5 h 55 fx   5 fx   5 fx   55fx  95h5 f4c x−1, x

Example Use the Midpoint Rule (4.28) and4. 29 to approximate



0

/4

sinx2_{dx and estimate the}

approximation errors. a. (4.28): h  1 3 4  12, x0  12, x1  0  2 12  6 T1  3 2  12 fx0  fx1  8 sin  12 2  sin  6 2  0. 133 211 b. (4.29): h  1 4 4  16. xi  0  i 16 , i  0, 1, 2 T2  4h 3 2fx0 − fx1  2fx2  43 16 2 sin 16 2 − sin  8 2  2 sin 3 16 2  0. 158 022 True errors: Ra 



0 /4 sinx2_{dx − 0. 133 211  2. 394 366 7  10}−2 Rb 



0 /4 sinx2_{dx − 0. 158 022  8. 673 325 2  10}−4 Approximation errors: a. Ra  3 4 h 3_f ′′_{ ≤ 3} 42 12 3  2. 691 517  10−2 b. Rb  14h5 45 f 4  ≤ 14 4530 16 5  2. 723 869  10−3