Trapezoidal Rule, n = 1, x 0 = a, x 1 = b, h = b a. f (x)dx = h 2 (f (x 0) + f (x 1 )) h3
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- Hilary Long
- 5 years ago
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1 Trpezoidl Rule, n = 1, x 0 =, x 1 = b, h = b f (x)dx = h 2 (f (x 0) + f (x 1 )) h3 12 f (ξ).
2 Simpson s Rule: n = 3, x 0 =, x 1 = +b 2, x 2 = b, h = b 2. Qudrture Rule, double node t x 1 P 3 (x)dx = f (x 0 ) +f (x 1 ) (x x 1 ) 2 (x x 2 ) (x 0 x 1 ) 2 (x 0 x 2 ) dx + f (x 2) (x x 0 )(x x 2 ) (x 1 x 0 )(x 1 x 2 ) +f (x x 0 )(x x 1 )(x x 2 ) (x 1 ) dx (x 1 x 0 )(x 1 x 2 ) (x x 1 ) 2 (x x 0 ) (x 2 x 1 ) 2 (x 2 x 0 ) dx ( 1 (x x ) 1)(2x 1 x 0 x 2 ) dx (x 1 x 0 )(x 1 x 2 )
3 Simpson s Rule: n = 3, x 0 =, x 1 = +b 2, x 2 = b, h = b 2. Qudrture Rule, double node t x 1 P 3 (x)dx = f (x 0 ) +f (x 1 ) (x x 1 ) 2 (x x 2 ) (x 0 x 1 ) 2 (x 0 x 2 ) dx + f (x 2) (x x 0 )(x x 2 ) (x 1 x 0 )(x 1 x 2 ) +f (x x 0 )(x x 1 )(x x 2 ) (x 1 ) dx (x 1 x 0 )(x 1 x 2 ) = h 3 (f (x 0) + 4f (x 1 ) + f (x 2 )). (x x 1 ) 2 (x x 0 ) (x 2 x 1 ) 2 (x 2 x 0 ) dx ( 1 (x x ) 1)(2x 1 x 0 x 2 ) dx (x 1 x 0 )(x 1 x 2 ) Stroke of luck: f (x 1 ) does not pper in qudrture
4 Simpson s Rule: f (x) = P 3 (x) + 1 4! f (4) (ξ(x))(x x 0 )(x x 1 ) 2 (x x 2 ). Qudrture Rule: f (x) dx = P 3 (x) dx + 1 4! = h 3 (f (x 0) + 4f (x 1 ) + f (x 2 )) + 1 4! f (4) (ξ(x))(x x 0 )(x x 1 ) 2 (x x 2 ) dx f (4) (ξ(x))(x x 0 )(x x 1 ) 2 (x x 2 ) dx h 3 (f (x 0) + 4f (x 1 ) + f (x 2 )). Qudrture Error: 1 4! = f (4) (ξ) 24 f (4) (ξ(x))(x x 0 )(x x 1 ) 2 (x x 2 ) dx (x x 0 )(x x 1 ) 2 (x x 2 ) dx = f (4) (ξ) 90 h 5.
5 Simpson s Rule: n = 3, x 0 =, x 1 = +b 2, x 2 = b, h = b 2. f (x)dx = h 3 (f (x 0) + 4f (x 1 ) + f (x 2 )) f (4) (ξ) 90 h 5.
6 Simpson s Rule: n = 3, x 0 =, x 1 = +b 2, x 2 = b, h = b 2. f (x)dx = h 3 (f (x 0) + 4f (x 1 ) + f (x 2 )) f (4) (ξ) 90 h 5.
7 Simpson s Rule: n = 3, x 0 =, x 1 = +b 2, x 2 = b, h = b 2. f (x)dx = h 3 (f (x 0) + 4f (x 1 ) + f (x 2 )) f (4) (ξ) 90 h 5. Wrong plot: slopes should mtch t midpoint.
8 Simpson s Rule: n = 3, x 0 =, x 1 = +b 2, x 2 = b, h = b 2. f (x)dx = h 3 (f (x 0) + 4f (x 1 ) + f (x 2 )) f (4) (ξ) 90 h 5. Right: slopes mtch t midpoint.
9 Exmple: pproximte 2 0 f (x)dx: Simpson wins
10 Degree of precision Degree of precision: the lrgest positive integer n such tht qudrture formul is exct for x k, for ech k = 0, 1,, n.
11 Degree of precision Degree of precision: the lrgest positive integer n such tht qudrture formul is exct for x k, for ech k = 0, 1,, n. Impliction: qudrture formul is exct for ll polynomils of degree t most n.
12 Degree of precision Degree of precision: the lrgest positive integer n such tht qudrture formul is exct for x k, for ech k = 0, 1,, n. Impliction: qudrture formul is exct for ll polynomils of degree t most n. Simplifiction: only need to verify exctness on intervl [0, 1].
13 Degree of precision Degree of precision: the lrgest positive integer n such tht qudrture formul is exct for x k, for ech k = 0, 1,, n. Impliction: qudrture formul is exct for ll polynomils of degree t most n. Simplifiction: only need to verify exctness on intervl [0, 1]. DoP = 1 for Trpezoidl Rule, DoP = 3 for Simpson.
14 Simpson s Rule: n = 3, x 0 =, x 1 = +b 2, x 2 = b, h = b 2. f (x)dx = h 3 (f (x 0) + 4f (x 1 ) + f (x 2 )) f (4) (ξ) 90 h 5. Degree of precision = 3
15 Composite Simpson s Rule (n = 2m, x j = + j h, h = b n, 0 j n) f (x)dx = = m x2j f (x)dx x 2(j 1) ( m h ( f (x2(j 1) ) + 4f (x 2j 1 ) + f (x 2j ) ) ) f (4) (ξ j ) h
16 Composite Simpson s Rule (n = 2m, x j = + j h, h = b n, 0 j n) (x)dx = f h f () = h f () = h f () m 1 m 1 m 1 m m f (x 2j ) + 4 f (x 2j 1 ) + f (b) h5 f (4) (ξ j ) 90 m f (x 2j ) + 4 f (x 2j 1 ) + f (b) h5 m 90 f (4) (µ) m f (x 2j ) + 4 f (x 2j 1 ) + f (b) (b )h4 180 f (4) (µ)
17 Trpezoidl Rule: n = 1, x 0 =, x 1 = b, h = b. f (x)dx = h 2 (f (x 0) + f (x 1 )) f (ξ) 12 h3. Degree of precision = 1
18 Composite Trpezoidl Rule (x j = + j h, h = b n, 0 j n) f (x)dx = = n xj f (x)dx x j 1 n ( h 2 (f (x j 1) + f (x j )) f ) (ξ j ) h 3. 12
19 Composite Trpezoidl Rule (x j = + j h, h = b n, 0 j n) (x)dx = f h n 1 n f () + 2 f (x j ) + f (b) h3 f (ξ j ) 2 12 = h n 1 f () + 2 f (x j ) + f (b) h3 n 2 12 f (µ) = h n 1 f () + 2 f (x j ) + f (b) (b )h2 f (µ) 2 12 For the sme work, Composite Simpson yields twice s mny correct digits.
20 Composite Simpson s Rule, exmple Determine vlues of h for n pproximtion error ɛ = 10 5 when pproximting π 0 sin(x) dx with Composite Simpson.
21 Composite Simpson s Rule, exmple Determine vlues of h for n pproximtion error ɛ = 10 5 when pproximting π 0 sin(x) dx with Composite Simpson. Solution: f (4) (µ) = sin(µ) 1, Error = π h f (4) (µ) π5 180n 4. Choosing or h = π 2m π 5 ( π 180n 4 ɛ, leding to n π 180ɛ with m 11. )
22 Composite Simpson s Rule, exmple 2 = Determine vlues of h for n pproximtion error ɛ = 10 5 when pproximting π 0 sin(x) dx with Composite Simpson. Solution: f (4) (µ) = sin(µ) 1, Error = π h f (4) (µ) π5 180n 4. Choosing or h = π 2m π 0 π π 0 ( π 180n 4 ɛ, leding to n π 180ɛ with m 11. For n = 2m = 22, sin(x) dx π 10 ( ) jπ 2 sin sin(x) dx π sin ) ( ) (2j 1)π sin 22 ( ) jπ (Trpezoidl) 22
23 Composite Simpson s Rule: Round-Off Error Stbility (n = 2m, x j = + j h, h = b n, 0 j n) (x)dx f h f () def = I(f ). m 1 f (x 2j ) + 4 m f (x 2j 1 ) + f (b)
24 Composite Simpson s Rule: Round-Off Error Stbility (n = 2m, x j = + j h, h = b n, 0 j n) (x)dx f h f () def = I(f ). m 1 Assume round-off error model: f (x 2j ) + 4 m f (x 2j 1 ) + f (b) f (x i ) = f (x i ) + e i, e i ɛ, i = 0, 1,, n.
25 Composite Simpson s Rule: Round-Off Error Stbility (n = 2m, x j = + j h, h = b n, 0 j n) (x)dx f h f () def = I(f ). m 1 Assume round-off error model: f (x 2j ) + 4 m f (x 2j 1 ) + f (b) f (x i ) = f (x i ) + e i, e i ɛ, i = 0, 1,, n. I(f ) = I( f ) + h m 1 m e e 2j + 4 e 2j 1 + e n. 3 I(f ) I( f ) h m 1 m e e 2j + 4 e 2j 1 + e n 3 hnɛ = (b )ɛ (numericlly stble!!!)
26 Composite Trpezoidl Rule: Round-Off Error Stbility (x j = + j h, h = b n, 0 j n) (x)dx f h n 1 f () + 2 f (x j ) + f (b) 2 def = I(f ).
27 Composite Trpezoidl Rule: Round-Off Error Stbility (x j = + j h, h = b n, 0 j n) (x)dx f h n 1 f () + 2 f (x j ) + f (b) 2 def = I(f ). Assume round-off error model: f (x i ) = f (x i ) + e i, e i ɛ, i = 0, 1,, n.
28 Composite Trpezoidl Rule: Round-Off Error Stbility (x j = + j h, h = b n, 0 j n) (x)dx f h n 1 f () + 2 f (x j ) + f (b) 2 def = I(f ). Assume round-off error model: f (x i ) = f (x i ) + e i, e i ɛ, i = 0, 1,, n. I(f ) = I( f ) + h n 1 e e j + e n. 2 I(f ) I( f ) h n 1 e e j + e n 2 hnɛ = (b )ɛ (numericlly stble!!!)
29 Recursive Composite Trpezoidl: with h k = (b )/2 k 1. f (x)dx h 2 n 1 f () + 2 f (x j ) + f (b) (b )h2 f (µ) 12
30 Recursive Composite Trpezoidl: with h k = (b )/2 k 1. f (x)dx h 2 book ==== h 2 def === R k,1 + n 1 f () + 2 f (x j ) + f (b) (b )h2 f (µ) 12 n 1 f () + 2 f (x j ) + f (b) + K j h 2j. K j h 2j, for n = 2 k.
31 Recursive Composite Trpezoidl: with h k = (b )/2 k 1. f (x)dx R 1,1 = h 1 2 h 2 book ==== h 2 def === R k,1 + n 1 f () + 2 f (x j ) + f (b) (b )h2 f (µ) 12 n 1 f () + 2 f (x j ) + f (b) + K j h 2j. K j h 2j, for n = 2 k. (f () + f (b)) = b 2 (f () + f (b)),
32 Recursive Composite Trpezoidl: with h k = (b )/2 k 1. f (x)dx R 1,1 = h 1 2 h 2 book ==== h 2 def === R k,1 + n 1 f () + 2 f (x j ) + f (b) (b )h2 f (µ) 12 n 1 f () + 2 f (x j ) + f (b) + K j h 2j. K j h 2j, for n = 2 k. (f () + f (b)) = b 2 (f () + f (b)), R 2,1 = h 2 2 (f () + f (b) + 2f ( + h 2)) = 1 2 (R 1,1 + h 1 f ( + h 2 )),
33 Recursive Composite Trpezoidl: with h k = (b )/2 k 1. f (x)dx R 1,1 = h 1 2 h 2 book ==== h 2 def === R k,1 + n 1 f () + 2 f (x j ) + f (b) (b )h2 f (µ) 12 n 1 f () + 2 f (x j ) + f (b) + K j h 2j. K j h 2j, for n = 2 k. (f () + f (b)) = b 2 (f () + f (b)), R 2,1 = h 2 2 (f () + f (b) + 2f ( + h 2)) = 1 2 (R 1,1 + h 1 f ( + h 2 )),.. R k,1 = 1 2 k 2 R k 1,1 + h k 1 f ( + (2j 1)h k ), k = 2,, log 2 2 n.
34
35 Romberg Extrpoltion Tble O(h 2 k ) O(h4 k ) O(h6 k ) O(h8 k ) R 1,1 R 2,1 R2,2 R 3,1 R3,2 R3,3 R 4,1 R 4,2 R 4,3 R 4,4
36 Romberg Extrpoltion for π 0 sin(x) dx, n = 1, 2, 22, 2 3, 2 4, 2 5. R 1,1 = π (sin(0) + sin(π)) = 0, 2 R 2,1 = 1 (R 1,1 + π sin( π ) 2 2 ) = , R 3,1 = 1 R 2,1 + π 2 (2j 1)π sin( = , R 4,1 = 1 R 3,1 + π 4 (2j 1)π sin( = , R 5,1 = 1 R 4,1 + π 8 (2j 1)π sin( = , R 6,1 = 1 R 5,1 + π 2 4 (2j 1)π sin( =
37 Romberg Extrpoltion, π 0 sin(x) dx = function evlutions used in the tble.
38 Recursive Composite Simpson: f (x)dx m 1 h f () + 2 f (x 2j ) (b )h4 f (4) (µ) 12 m f (x 2j 1 ) + f (b)
39 Recursive Composite Simpson: f (x)dx exists ==== h 3 m 1 h f () + 2 f (x 2j ) (b )h4 f (4) (µ) 12 + f () + 2 K j h 2j. j=2 def === R k,1 + m 1 f (x 2j ) + 4 m f (x 2j 1 ) + f (b) m f (x 2j 1 ) + f (b) K j h 2j, for n = 2 k. j=2
40 Recursive Composite Simpson: with h k = (b )/2 k 1. f (x)dx R k,1 + K j h 2j, for n = 2 k. j=2
41 Recursive Composite Simpson: with h k = (b )/2 k 1. R 1,1 = b 6 f (x)dx R k,1 + K j h 2j, for n = 2 k. j=2 (f () + 4S 1 + f (b)), S 1 = f (( + b)/2),
42 Recursive Composite Simpson: with h k = (b )/2 k 1. R 1,1 = b 6. T k = f (x)dx R k, k 1 K j h 2j, for n = 2 k. j=2 (f () + 4S 1 + f (b)), S 1 = f (( + b)/2), f ( + (2j 1)h k ), R k,1 = h k 3 (f () + 2S k 1 + 4T k + f (b)), S k = S k 1 + T k, k = 2,, log 2 n.
43 Romberg Extrpoltion Tble, Simpson Rule O(h 4 k ) O(h6 k ) O(h8 k ) O(h10 k ) R 1,1 R 2,1 R2,2 R 3,1 R3,2 R3,3 R 4,1 R 4,2 R 4,3 R 4,4
44 Tricks of the Trde, f (x)dx Composite Simpson/Trpezoidl rules: Adding more equi-spced points. Romberg extrpoltion: Obtin higher order rules from lower order rules. Adptive qudrtures: Adding more points only when necessry. qud function of mtlb: combintion of ll three.
45 Adptive Qudrture Methods: step-size mtters y(x) = e 3x sin 4x. Oscilltion for smll x; nerly 0 for lrger x. Mechnicl engineering (spring nd shock bsorber systems) Electricl engineering (circuit simultions) y(x) behves different for smll x nd for lrge x.
![(, b), Simpson on [, b] (f () + 4f ( +](/docs-images/90/102029249/images/46-1.jpg "h) + f (b)), h = b 2.")
46 Adptive Qudrture (I) where S(, b) = h 3 f (x)dx = S(, b) h5 90 f (4) (ξ), ξ (, b), Simpson on [, b] (f () + 4f ( + h) + f (b)), h = b 2. Composite Simpson
47 Adptive Qudrture (II) f (x)dx = S(, b) h5 90 f (4) (ξ), ξ (, b), where f (x)dx = +b 2 f (x)dx + f (x)dx +b 2 = S(, + b 2 ) + S( + b 2, b) (h/2)5 90 f (4) (ξ 1 ) (h/2)5 90 f (4) (ξ 2 ) = S(, + b 2 ) + S( + b 2, b) 1 16 ( h 5 90 ξ 1 (, + b 2 ), ξ 2 ( + b 2, b), ξ (, b). ) f (4) ( ξ),
48 Adptive Qudrture (III) f (x)dx = S(, + b 2 ) + S( + b 2, b) 1 16 = S(, b) h5 90 f (4) (ξ) ( h 5 90 ) f (4) ( ξ)
49 Adptive Qudrture (III) f (x)dx = S(, + b 2 ) + S( + b 2, b) 1 16 ( h 5 90 ) f (4) ( ξ) = S(, b) h5 90 f (4) (ξ) S(, b) h5 90 f (4) ( ξ).
50 Adptive Qudrture (III) ( h 5 f (x)dx = S(, + b 2 ) + S( + b 2, b) ( h 5 90 ) f (4) ( ξ) = S(, b) h5 90 f (4) (ξ) S(, b) h5 90 f (4) ( ξ). ) f (4) ( ξ) 16 ( S(, b) S(, + b 15 2 ) S( + b ) 2, b),
51 Adptive Qudrture (III) ( h 5 f (x)dx = S(, + b 2 ) + S( + b 2, b) ( h 5 90 ) f (4) ( ξ) = S(, b) h5 90 f (4) (ξ) S(, b) h5 90 f (4) ( ξ). ) f (4) ( ξ) 16 ( S(, b) S(, + b 15 2 ) S( + b ) 2, b), f (x)dx S(, + b 2 ) S( + b ( ) 2, b) = 1 h 5 f (4) ( ξ) b S(, b) S(, 2 ) S( + b 2, b).
52 Adptive Qudrture (IV) For given tolernce τ, if 1 15 then S(, +b 2 + b S(, b) S(, 2 ) S( + b 2, b) τ, +b ) + S( 2, b) is sufficiently ccurte pproximtion to f (x)dx; otherwise recursively develop qudrtures on (, +b 2 ) nd, b), respectively. ( +b 2
53 AdptQud(f, [, b], τ) for computing f (x) dx compute S(, b), S(, +b 2 if 1 15 ), S( +b 2, b), + b S(, b) S(, 2 ) S( + b 2, b) τ, return S(, +b +b 2 ) + S( 2, b). else return AdptQud(f, [, + b 2 + b ], τ/2)+adptqud(f, [, b], τ/2). 2
54 Adptive Simpson (I)
55 Adptive Simpson (II)
56 Adptive Simpson, exmple Integrl 3 1 f (x) dx, f (x) = 100 x 2 sin Tolernce τ = ( ) 10. x
57 function qud(f, [, b], τ) of mtlb For given tolernce τ, composite Simpson: S(, b), S(, +b +b 2 ) nd S( 2, b). Romberg extrpoltion: if Q 1 = S(, + b 2 )+S( + b 2, b), Q = Q (Q 1 S(, b)). return Q else return qud(f, [, + b 2 Q Q 1 τ, + b ], τ/2) + qud(f, [, b], τ/2). 2
58 Gussin Qudrture (I) Trpezoidl nodes x 1 =, x 2 = b unlikely best choices. Likely better node choices.
59 Gussin Qudrture (II) Given n > 0, choose both distinct nodes x 1,, x n [ 1, 1] nd weights c 1,, c n, so qudrture 1 1 f (x) dx n c j f (x j ), (1) gives the gretest degree of precision (DoP).
60 Gussin Qudrture (II) Given n > 0, choose both distinct nodes x 1,, x n [ 1, 1] nd weights c 1,, c n, so qudrture 1 1 f (x) dx n c j f (x j ), (1) gives the gretest degree of precision (DoP). 2n totl number of prmeters in qudrture, could choose 2n monomils f (x) = 1, x, x 2,, x 2n 1 in eqution (1).
61 Gussin Qudrture (II) Given n > 0, choose both distinct nodes x 1,, x n [ 1, 1] nd weights c 1,, c n, so qudrture 1 1 f (x) dx n c j f (x j ), (1) gives the gretest degree of precision (DoP). 2n totl number of prmeters in qudrture, could choose 2n monomils f (x) = 1, x, x 2,, x 2n 1 in eqution (1). directly solving eqution (1) cn be very hrd.
62 Gussin Qudrture, n = 2 (I) Consider Gussin qudrture 1 1 f (x) dx c 1 f (x 1 ) + c 2 f (x 2 ).
63 Gussin Qudrture, n = 2 (I) Consider Gussin qudrture 1 1 f (x) dx c 1 f (x 1 ) + c 2 f (x 2 ). Choose prmeters c 1, c 2 nd x 1 < x 2 so tht Gussin qudrture is exct for f (x) = 1, x, x 2, x 3 : 1 1 f (x) dx = c 1 f (x 1 ) + c 2 f (x 2 ),
64 Gussin Qudrture, n = 2 (I) Consider Gussin qudrture 1 1 f (x) dx c 1 f (x 1 ) + c 2 f (x 2 ). Choose prmeters c 1, c 2 nd x 1 < x 2 so tht Gussin qudrture is exct for f (x) = 1, x, x 2, x 3 : 1 1 f (x) dx = c 1 f (x 1 ) + c 2 f (x 2 ), or 2 = 2 3 = dx = c 1 + c 2, 0 = x 2 dx = c 1 x c 2 x 2 2, 0 = x dx = c 1 x 1 + c 2 x 2, x 3 dx = c 1 x c 2 x 3 2.
65 Gussin Qudrture, n = 2 (II) x 1 < x 2, c 1 x 1 = c 2 x 2, c 1 x 3 1 = c 2 x 3 2, implying x 2 1 = x 2 2. Thus x 1 = x 2 nd c 1 = c 2.
66 Gussin Qudrture, n = 2 (II) x 1 < x 2, c 1 x 1 = c 2 x 2, c 1 x 3 1 = c 2 x 3 2, implying x 2 1 = x 2 2. Thus x 1 = x 2 nd c 1 = c 2. c 1 + c 2 = 2, c 1 x c 2 x 2 2 = 2 3, which implies c 1 = c 2 = 1, x 2 = 1 3. Gussin qudrture for n = f (x) dx f ( 1 3 ) + f ( 1 3 ), exct for f (x) = 1, x, x 2, x 3, but not for f (x) = x 4.
67 Legendre
68 Legendre Legendre polynomils: P 0 (x) = 1, P 1 (x) = x. Bonnet s recursive formul for n 1: (n + 1)P n+1 (x) = (2n + 1)xP n (x) np n 1 (x).
69 Legendre Legendre polynomils: P 0 (x) = 1, P 1 (x) = x. Bonnet s recursive formul for n 1: (n + 1)P n+1 (x) = (2n + 1)xP n (x) np n 1 (x).
70 P n (x) hs degree exctly n. Legendre polynomils re orthogonl polynomils: 1 1 P m (x) P n (x) dx = 0, whenever m < n.
71 P n (x) hs degree exctly n. Legendre polynomils re orthogonl polynomils: 1 1 P m (x) P n (x) dx = 0, whenever m < n. Let Q(x) be ny polynomil of degree < n. Then Q(x) is liner combintion of P 0 (x), P 1 (x),, P n (x): Q(x) = α 0 P 0 (x) + α 1 P 1 (x) + + α n 1 P n 1 (x).
72 P n (x) hs degree exctly n. Legendre polynomils re orthogonl polynomils: 1 1 P m (x) P n (x) dx = 0, whenever m < n. Let Q(x) be ny polynomil of degree < n. Then Q(x) is liner combintion of P 0 (x), P 1 (x),, P n (x): Q(x) = α 0 P 0 (x) + α 1 P 1 (x) + + α n 1 P n 1 (x) Q(x)P n (x) dx = α 0 P 0 (x)p n (x) dx + α 1 P 1 (x)p n (x) dx = α n 1 P n 1 (x)p n (x) dx 1
73 Gussin Qudrture: Definition Theorem: P n (x) hs exctly n distinct roots 1 < x 1 < x 2 < < x n < 1. Define: Gussin qudrture 1 1 f (x) dx c 1 f (x 1 ) + c 2 f (x 2 ) + + c n f (x n ),
74 Gussin Qudrture: Definition Theorem: P n (x) hs exctly n distinct roots 1 < x 1 < x 2 < < x n < 1. Define: Gussin qudrture 1 1 f (x) dx c 1 f (x 1 ) + c 2 f (x 2 ) + + c n f (x n ), Choose: for i = 1,, n, 1 1 def c i == L i (x) dx = x x j dx. 1 1 x i x j j i
75 Gussin Qudrture: Definition Theorem: P n (x) hs exctly n distinct roots 1 < x 1 < x 2 < < x n < 1. Define: Gussin qudrture 1 1 f (x) dx c 1 f (x 1 ) + c 2 f (x 2 ) + + c n f (x n ), Choose: for i = 1,, n, 1 1 def c i == L i (x) dx = x x j dx. 1 1 x i x j j i Qudrture exct for polynomils of degree t most n 1.
76 Theorem: DoP of Gussin Qudrture = 2n 1 Gussin qudrture, with roots of P n (x) x 1, x 2,, x n : 1 1 f (x) dx c 1 f (x 1 ) + c 2 f (x 2 ) + + c n f (x n ),
77 Theorem: DoP of Gussin Qudrture = 2n 1 Gussin qudrture, with roots of P n (x) x 1, x 2,, x n : 1 1 f (x) dx c 1 f (x 1 ) + c 2 f (x 2 ) + + c n f (x n ), Let P(x) be ny polynomil of degree t most 2n 1. Then P(x) = Q(x)P n (x) + R(x), (Polynomil Division) where Q(x), R(x) polynomils of degree t most n 1.
78 Theorem: DoP of Gussin Qudrture = 2n 1 Gussin qudrture, with roots of P n (x) x 1, x 2,, x n : 1 1 f (x) dx c 1 f (x 1 ) + c 2 f (x 2 ) + + c n f (x n ), Let P(x) be ny polynomil of degree t most 2n 1. Then P(x) = Q(x)P n (x) + R(x), (Polynomil Division) 1 1 where Q(x), R(x) polynomils of degree t most n 1. P(x) dx = 1 = 0 + Q(x)P n (x) dx R(x) dx 1 1 R(x) dx = c 1 R(x 1 ) + c 2 R(x 2 ) + + c n R(x n ) (qud exct for R(x)) = c 1 P(x 1 ) + c 2 P(x 2 ) + + c n P(x n ). (qud exct for P(x))
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