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1 Theoreticl foundtions of Gussin qudrture 1 Inner product vector spce Definition 1. A vector spce (or liner spce) is set V = {u, v, w,...} in which the following two opertions re defined: (A) Addition of vectors: u + v V, which stisfies the properties (A 1 ) ssocitivity: u + (v + w) = (u + v) + w u, v, w in V ; (A 2 ) existence of zero vector: 0 V such tht u + 0 = u u V ; (A ) existence of n opposite element: u V ( u) V such tht u + ( u) = 0; (A 4 ) commuttivity: u + v = v + u u, v in V ; (B) Multipliction of number nd vector: αu V for α R, which stisfies the properties (B 1 ) α(u + v) = αu + αv α R, u, v V ; (B 2 ) (α + β)u = αu + βu α, β R, u V ; (B ) (αβ)u = α(βu) α, β R, u V ; (B 4 ) 1u = u u V. Definition 2. An inner product liner spce is liner spce V with n opertion (, ) stisfying the properties () (u, v) = (v, u) u, v in V ; (b) (u + v, w) = (u, w) + (v, w) u, v, w in V ; (c) (αu, v) = α(u, v) α R, u, v, w in V ; (d) (u, u) 0 u V ; moreover, (u, u) = 0 if nd only if u = 0. Exmple. The stndrd inner product of the vectors u = (u 1, u 2,..., u d ) R d nd v = (v 1, v 2,..., v d ) R d is given by (u, v) = d u i v i. 1

2 Exmple. Let G be symmetric positive-definite mtrix, for exmple G = (g ij ) = Then one cn define sclr product corresponding to G by (u, v) := d d u i g ij v j. j=1 Remrk. In n inner product liner spce, one cn define the norm of vector by The fmous Cuchy-Schwrz inequlity reds u := (u, u). (u, v) u v. Think bout the mening of this inequlity in R. Exercise. Find the norm of the vector u = (, 0, 4) using the stndrd inner product in R nd then by using the inner product in R defined through the mtrix G. A very importnt exmple. Consider the set of ll polynomils of degree no greter thn 4, where the opertions ddition of vectors nd multipliction of number nd vector re defined in the stndrd wy, nmely: if P nd Q re such polynomils, P (x) = p 4 x 4 + p x + p 2 x 2 + p 1 x + p 0, Q(x) = q 4 x 4 + q x + q 2 x 2 + q 1 x + q 0, then their sum, P + Q is given by (P + Q)(x) = (p 4 + q 4 )x 4 + (p + q )x + (p 2 + q 2 )x 2 + (p 1 + q 1 )x + (p 0 + q 0 ), nd, for α R, the product αp is defined by (αp )(x) = (αp 4 )x 4 + (αp )x + (αp 2 )x 2 + (αp 1 )x + (αp 0 ). Then this set of polynomils is vector spce of dimension 5. One cn tke for bsis in this spce the set of polynomils E 0 (x) := 1, E 1 (x) := x, E 2 (x) := x 2, E (x) := x, E 4 (x) := x 4. This, however, is only one of the infinitely mny bses in this spce. For exmple, the set of vectors G 0 (x) := x 1, G 1 (x) := x + 1, G 2 (x) := x 2 + x +, 2

3 G (x) := x + x 2 4, G 4 (x) := x 4 x 2x is perfectly good bsis. (Note tht I clled G 0, G 1,..., G n vectors to emphsize tht wht is importnt for us is the structure of vector spce nd not so much the fct tht these vectors re polynomils.) Any vector (i.e., polynomil of degree 4) cn be represented in unique wy in ny bsis, for exmple, the polynomil P (x) = x 4 5x 2 + x + 7 cn be written s P = E 4 5 E 2 + E E 0, or, lterntively, s P = G 4 G + 4 G 2 11 G G 0. 2 Inner product in the spce of polynomils One cn define n inner product structure in the spce of polynomils in mny different wys. Let V n (, b) stnd for the spce of polynomils of degree n defined for x [, b]. Most of the theory we will develop works lso if = nd/or b =. Let w : [, b] R be weight function, i.e., function stisfying the following properties: () the integrl w(x) dx exists; (b) w(x) 0 for ll x [, b], nd w(x) cn be zero only t isolted points in [, b] (in prticulr, w(x) cnnot be zero in n intervl of nonzero length). We define sclr product in V n (, b) by (P, Q) := P (x) Q(x) w(x) dx ; (1) if the intervl (, b) is of infinite length, then one hs to tke w such tht this integrl exists for ll P nd Q in V n (, b). Let V n (, b; w) stnds for the inner product liner spce of polynomils of degree n defined on [, b], nd sclr product defined by (1). Exmple. The Legendre polynomils re fmily of polynomils P 0, P 1, P 2,... such tht P n is polynomil of degree n defined for x [ 1, 1], with leding coefficients equl to 1 ( leding re the coefficients of the highest powers of x) nd such tht P n nd P m re orthogonl for n m in the sense of the following inner product: (P n, P m ) = 1 1 P n (x) P m (x) dx. In other words, the polynomils P 0, P 1, P 2,..., P n constitute n orthogonl bsis of the spce V n ( 1, 1; w(x) 1). Here re the first severl Legendre polynomils: P 0 (x) = 1, P 1 (x) = x, P 2 (x) = x 2 1, P (x) = x 5 x, P 4(x) = x x2 + 5,....

4 Sometime Legendre polynomils re normlized in different wy: P 0 (x) = 1, P1 (x) = x, P2 (x) = 1 2 (x2 1), P (x) = 1 2 (5x x), P4 (x) = 1 8 (5x4 0x 2 + ),... ; check tht P n is proportionl to P n for ll the polynomils given here. Exercise. Check tht ech of the first five Legendre polynomils is orthogonl to ll other Legendre polynomils in the exmple bove. Exmple. The Hermite polynomils re fmily of polynomils H 0, H 1, H 2,... such tht H n is polynomil of degree n defined for x R, normlized in such wy tht (H n, H n ) = 2 n n! π nd (H n, H m ) = 0 for n m, where the inner product is defined s follows: (H n, H m ) = H n (x) H m (x) e x2 dx. In other words, the polynomils H 0, H 1, H 2,..., H n constitute n orthogonl bsis of the spce V n (, ; e x2 ). Here re the first five Hermite polynomils: H 0 (x) = 1, H 1 (x) = 2x, H 2 (x) = 4x 2 2, H (x) = 8x 12x, H 4 (x) = 16x 4 48x Gussin qudrture Theorem 1. Let w be weight function on [, b], let n be positive integer, nd let G 0, G 1,..., G n be n orthogonl fmily of polynomils with degree of G k equl to k for ech k = 0, 1,..., n. In other words, G 0, G 1,..., G n form n orthogonl bsis of the inner product liner spce V n (, b; w). Let x 1, x 2,..., x n be the roots of G n, nd define L i (x) := n j=1, j i x x j x i x j for i = 1, 2,..., n. Then the corresponding Gussin qudrture formul is given by where I(f) := f(x) w(x) dx I n (f) := w i := L i (x) w(x) dx. w i f(x i ), The formul I n (f) hs degree of precision exctly 2n 1, which mens tht I n (x k ) = I(x k ) for k = 0, 1,..., 2n 1, but there is polynomil Q of degree 2n for which I n (Q) I(Q). 4

5 Proof: We will prove tht the qudrture formul I n (f) = w i f(x i ) (where the weights w i re given by the formul bove) hs degree of precision exctly 2n 1 in three steps: Step 1: The degree of precision of the qudrture formul is n 1. Let V n 1 stnd for the liner spce of ll polynomils of degree not exceeding n 1 defined for x [, b]. We need to prove tht our qudrture formul is exct for ny R V n 1. The polynomils L i (x), i = 1, 2,..., n re Lgrnge polynomils of degree n 1 such tht L i (x k ) = δ ik for ll k = 1, 2,..., n. This implies tht R(x i ) L i (x) is the Lgrnge polynomil of degree n 1 tht interpoltes the polynomil R t the n points x 1, x 2,..., x n. Reclling tht R is polynomil of degree of degree n 1, we conclude tht R(x) = R(x i ) L i (x) x [, b]. Then we hve I(R) = = R(x) w(x) dx = [ ] R(x i ) L i (x) w(x) dx [ ] R(x i ) L i (x) w(x) dx = R(x i ) w i = I n (R), therefore the qudrture formul is exct for ny polynomil R of degree up to nd including n 1, or, in other words, the degree of precision of the qudrture formul is n 1. Step 2: The degree of precision of the qudrture formul is 2n 1. Let P be polynomil of degree 2n 1, i.e., P V 2n 1. Divide P by G n (which is polynomil of degree n to obtin P (x) = Q(x) G n (x) + R(x), where Q V n 1 nd R V n 1. Since the polynomils G 0, G 1, G 2,..., G n 1 form bsis of the liner spce V n 1, we cn write Q in the form n 1 Q(x) = q i G i (x). 5

6 Then I(P ) = I(Q G n + R) = = [ b n 1 = 0 + [Q(x) G n (x) + R(x)] w(x) dx ] q i G i (x) G n (x) w(x) dx + R(x) w(x) dx = I(R), R(x) w(x) dx becuse ll polynomils G 0, G 1,..., G n 1 re orthogonl to G n (with respect to the sclr product defined with the weight function w): (G i, G n ) = G i (x) G n (x) w(x) dx = 0 i = 0, 1, 2,..., n 1. Now recll tht R is polynomil of degree n 1 to conclude tht I(R) = I n (R) ccording to wht we proved in Step 1. Since x i re roots of the polynomil G n, it follows tht P (x i ) = Q(x i ) G n (x i ) + R(x i ) = 0 + R(x i ) = R(x i ), which implies tht I n (R) = I n (P ). Combining these results, we obtin I(P ) = I n (P ) P V 2n 1. Step : The degree of precision of the qudrture formul is 2n 1. We lredy know tht the degree of precision is t lest 2n 1, so to prove tht it is exctly 2n 1, it is enough to find one polynomil of degree 2n for which the qudrture formul is not exct. Note tht the polynomil G 2 n(x) hs degree exctly 2n, nd tht I(G 2 n) = G 2 n(x) w(x) dx > 0 (2) becuse the integrnd, G 2 n(x) w(x), is nonnegtive nd cn be zero only t isolted points ( polynomil like G 2 n hs t most 2n roots, nd w cn vnish only t isolted points by the definition of weight function). But, since x i re roots of G n for i = 1, 2,..., n, the qudrture formul gives I n (G 2 n) = w i G 2 n(x i ) = 0. () Compring (2) nd (), we see tht the qudrture formul is not exct for G 2 n. 6

7 In the proof of the bove theorem, we implicitly used tht G n hs exctly n simple roots, ll of which re inside the intervl (, b). This property is estblished rigorously in the following lemm. Lemm 1. Let w be weight function on [, b], let n be positive integer, nd let G 0, G 1,..., G n be fmily of polynomils such tht the degree of G k is equl to k for ech k = 0, 1,..., n, nd (G i, G j ) = 0 if i j, where the sclr product is defined by (1). Then for ech k = 0, 1, 2,..., n, the polynomil G k hs exctly k rel roots which re simple nd lie in the intervl (, b). Proof: First we will prove the existence of rel roots for G k, nd then we will count those roots. Step 1: Existence of rel roots of G k. Since the degree of G 0 is zero, it follows tht G 0 (x) = α x [, b] for some constnt α 0. (If you don t understnd why α 0, look t pge 87 of the book.) Then, for k 1, 0 = G 0 (x) G k (x) w(x) dx = α G k (x) w(x) dx, where the first equlity holds due to orthogonlity. However, since w is positive on [, b] except possibly being equl to 0 t isolted points, nd the integrl G k(x) w(x) dx is equl to zero, we conclude tht the function G k must chnge sign in (, b). Since G k is continuous function (ll polynomils re continuous functions), the Intermedite Vlue Theorem gurntees tht G k hs rel root somewhere in (, b). Step 2: Counting the roots of G k. Suppose tht G k chnges sign t exctly j points r 1, r 2,..., r j in (, b) such tht < r 1 < r 2 < < r j < b. Without loss of generlity, ssume tht G k (x) > 0 for x (, r 1 ). Then G k lterntes sign on (r 1, r 2 ), (r 2, r ),..., (r j, b). Define the uxiliry function Z(x) := ( 1) j j (x r j ) V j. By construction, Z(x) nd G k (x) hve the sme sign for ll x [, b]. From this it follows tht Z(x) G k (x) w(x) dx > 0. (4) 7

8 Suppose tht j < k. Since the polynomils G 0, G 1, G 1,..., G j form bsis for V j, there exist constnts z 0, z 1, z 2,..., z j such tht Z(x) = j z i G j (x). Substituting this expression into (4), we obtin [ b j ] Z(x) G k (x) w(x) dx = z i G j (x) G k (x) w(x) dx = j z i G j (x) G k (x) w(x) dx = 0, where we hve used tht j < k, which implies tht the polynomils G j nd G k re orthogonl. The lst equlity clerly contrdicts (4)! Hence the ssumption tht j < k ws wrong, i.e., we must hve tht j k. However, since the degree of the polynomil G k is k, G k cnnot hve more thn k roots, which implies tht j = k. Exmple: As n exmple of ppliction of the bove Theorem, let us rederive the qudrture formul we derived in clss, nmely, 1 ( f(x) dx f 1 ) ( ) 1 + f. 1 This formul cn be obtined by using the Legendre polynomils P k defined on pge. Tke the polynomils P 0 (x) = 1, P 1 (x) = x, nd P 2 (x) = x 2 1, which re n orthogonl bsis in the liner spce V 2 ( 1, 1; w(x) 1). Since n = 2, we expect tht the formul tht we will obtin will hve degree of precision 2n 1 = 2(2) 1 =. As in the Theorem bove, define x 1 = 1 nd x 2 = 1 to be the zeros of the polynomil P 2, define the polynomils (of degree n 1 = 2 1 = 1) nd the weights w 1 = 1 1 L 1 (x) w(x) dx = L 1 (x) = x x 2 x 1 x 2, L 2 (x) = x x 1 x 2 x 1, 1 1 x x 2 x 1 x 2 1 dx = 1 x 1 x 2 nd, similrly, w 2 = 1 1 L 2(x) w(x) dx = = 1, to obtin I 2 (f) = 2 w i f(x i ) = f ( 1 ) ( ) 1 + f ( ) x x 2x. x= 1 = 1, As n exercise, check tht the degree of ccurcy of this qudrture formul is indeed. 8

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