For each i 0, and n i, H n,i is defined to be the i i connected minor with x n along the back diagonal.

Transcription

1 1 1 Hankel Determinants A sequence of indeterminants x 0, x 1,... is used to make a a Hankel array: x 0 x 1 x 2 x 3... x 1 x 2 x 3 x 4... x 2 x 3 x 4 x 5... x 3 x 4 x 5 x 6... For each i 0, and n i, H n,i is defined to be the i i connected minor with x n along the back diagonal. h n,i = det H n,i The c-table is the set of h n,i for n + 1 i. these determinants satisfy the Sylvester identity: h n,i 1 h n,i+1 = h n 1,i h n+1,i h 2 n,i 2 A pattern of locations in the c-table of the form is called a Sylvestor diamond. N W C E S n s = w e c 2 Suppose the following constellation occurs in the c-table, C is located as (s, i). NN NW N NE W W W C E EE SW S SE E SS Let Λ be the 3 3 matrix whose entries are at the positions: NN NE EE NW C SE W W SW SS 3 Gragg s identity is c det(λ) = c 2 (w e + n s) This identity is derived solely from Sylvester s identity (5 times). If x 0, x 1,... are indeterminates in the ring Z[x 0, x 1,...], then each Hankel det is an irreducible poly in the x 0, x 1,.... One c may be cancelled to give: 1

2 det(λ) = c(w e + n s) Consider a larger region in the c-table. Suppose A = (s, i) and Z = (t, i ) are locations as in the diagram: A... B B M..... W N Y Y 2... Z 4 Thus A is at the location of the i i matrix with x s on the back diagonal, and Z is at the location of the i i matrix with x t on the back diagonal. Let Λ be the matrix formed by putting the SW diagonal from A as the left hand column and the SE diagonal as the top row: Λ = a... b 2... n b 1... w... y m... y 1... z This matrix Λ is indexed so that Λ[1, 1] = a and Λ[j, k] = z. The notation Λ[g, h, i; l, m, n] will stand for the submatrix of Λ withe columns g, h, i and rows l, m, n For each pair of locations A and Z as above, there is a polynomial φ A,Z in Z[x 0, x 1,...]. These polynomials have the following property. Let j be a row index with 1 < j < j, and k be a columns index with 1 < k < k. For each pair of locations A and Z, there is a poly φ A,Z with the following property: Let j be a row index with 1 < j < j let k be a column index with 1 < k < k. Refer

3 to the diagram for the locations of B 1, B 2, Y 1, Y 2, W. Thus Λ[1, j, j; 1, k, k] = a b 2 n b 1 w y 2 m y 1 z 5 Theorem There are two expressions for the det of this matrix: (i) det Λ[1, j, j; 1, k, k] = y 1 φ A,Y2 + y 2 φ A,Y1 + zφ A,W wφ A,Z (ii) det Λ[1, j, j; 1, k, k] = b 1 φ B2,Z + b 2 φ B1,Z + aφ W,Z wφ A,Z We need to consider in more detail the region which has A and Z at opp corners of a rectangle. From Λ, 4 rows are seleted, containing A; B 1 (and C); Y 2 and V ; and Z. Four columns are selected containing A; B 2 (and C; Y 1 (and V ; Z. A... B 1... B C.... M N.... V Y 1... Y 2... Z 6 Let Λ bethe matrix by putting the SW diagonal emanating from A as the LH column and the SE diagonal as the top row. Λ = a. b n b 1. c v. y 2 m y 2. z

4 These four rows and columns define a matrix, which we call Γ. Thus a b 2.. Γ = b 1 c.... v y 2.. y 1 z 7 Polynomials Ω A,V,Z and Υ Z,C,A will be defined inductively depending on the locations A, C, V, and Z. These poly s and φ A,Z and φ Z,A are defined only when t s + j 1 is an even integer and j i 1. (i) If Z is on either SW or SE diagonal, then all four are 0 (ii) Suppose A = (s, i), C = V = (s, i + 2) and Z = (z, i + 4). Then let p be the poly p = (h s 1,i+2 h s+1,i+2 + (h s1,i+1 h s,i+3 ) These are the Hankel dets to W, E, N, and S of position C. Then Ω A,V,Z = v p = c p = Υ Z,C,A φ S,T = p = φ T,S (iii) Suppose inductively that φ S,T and φ T,S have been defined when the difference between S and T is strictly less than the difference between A and Z. Then Ω A,V,Z and Υ Z,C,A are defined by Ω A,V,Z = y 1 φ A,Y2 + y 2 φ A,Y1 zφ A,V [1, 3, 4; 1, 3, 4] Υ Z,C,A = b 1 φ Z,B2 + b 2 φ Z,B1 zφ Z,C [1, 2, 4; 1, 2, 4] 8 (iv) The expressions φ A,Z and φ Z,A are defined by Lemma 1 There is a polynomial identity φ A,Z = Ω A,V,Z v φ Z,A = Ω Z,C,A c Ω A,V,Z c = Υ Z,C,A v 8 Both c and v are Hankel dets and hence are irreducible polys in the x i. The identity of Lemma 1 implies that if V C, there is an equality φ A,Z = Ω A,V,Z v = Υ Z,C,A c = φ Z,A

5 The case V = C is handled by (ii) above. In this case Gragg s identity implies that Ω A,V,Z = [1, 2, 3; 1, 2, 3] = Υ Z,C,A 9 The proof of Lemma 1 makes use of the 8-term identity (below). Recall that Γ = a b 2.. b 1 c.... v y 2.. y 1 z The determinant of Γ[g, h, i; l, m, n] will be denoted [g, h, i; l, m, n]. The proof uses the 8-term identity among the 3 by 3 dets of Γ. Lemma 2 There is an 8 term identity: [2, 3, 4; 2, 3, 4] a [2, 3, 4; 1, 3, 4] b 2 [1, 3, 4; 2, 3, 4] b 1 + [1, 3, 4; 1, 3, 4] c [1, 2, 4; 1, 2, 4] w + [1, 2, 4; 1, 2, 3] y 2 + [1, 2, 3; 1, 2, 4] y 1 [1, 2, 3; 1, 2, 3] z = 0 The proof of 8-term id uses the cofactor expansion by the first and second rows, and by the third and fourth columns. After taking the sum, and cancelling common terms the id follows. Observe that the uncancelled terms are the cofactors of 8 entries of the matrix. Proof of Lemma 1 If Z is on either the SW or the SE diagonal from A, then φ A,Z, Ω A,V,Z, Υ Z,C,A, φ Z,A are all defined to be 0. For Z strictly within the SW and SE diagonals, teh proof is by induction on diff levels A and Z. Start at diff = 4. Thus A = (s, i) and Z = (s, i+4). Then C = V = (s, i + 2), c = v In this case φ A,Z, Ω A,V,Z, Υ Z,C,A, φ Z,A defined by (ii) and statement is Gragg s id.

6 Suppose diff A and Z is k, with k > 4. Then Ω A,V,Z c = y 1 φ A,y2 c + y 2 φ A,Y1 c zφ A,V c [1, 3, 4; 1, 3, 4] c = y 1 {b 1 φ Y2,B 2 + b 2 φ Y2,B 1 aφ Y2,C [1, 2, 3; 1, 2, 4]} +y 2 {b 1 φ Y2,C [1, 2, 3; 1, 2, 4] aφ Y1,C [1, 2, 4; 1, 2, 3]} z{b 1 φ V,B2 + b 2 φ V,B aφ V,C [1, 2, 3; 1, 2, 3]} [1, 3, 4; 1, 3, 4] c Υ Z,C,A v = b 1 φ Z,B2 v + b 2 φ Z,B1 v aφ Z,C v [1, 2, 4; 1, 2, 4] v = b 1 {y 1 φ B2,Y 2 + y 2 φ B2,Y 1 zφ B2,V [2, 3, 4; 1, 3, 4]} +b 2 {y 1 φ B1,Y 2 + y 2 φ B1,Y 1 zφ B1,V [1, 3, 4; 2, 3, 4]} a{y 1 φ C,Y2 + y 2 φ C,Y1 zφ C,V [2, 3, 4; 2, 3, 4]} [1, 2, 4; 1, 2, 4] v The equality Ω A,V,Z c = Υ Z,C,A v follows from the 8-term id, and repeated use (9 times) of φ S,T = φ T,S. It follows that Ω A,V,Z v = Υ Z,C,A c 11 The LHS depends only on the choice of rows and columns 1, 3, 4 of Γ, which are the rows 1, j, j and columns 1, k, k of Λ. Similarly, The RHS depends only on the choice of rows and columns 1, 2, 4 of Γ, which are the rows 1, j, j and columns 1, k, k of Λ. Consequently, the poly Ω A,V,Z v is the same for all choices of row j and column

7 k of Λ. Similarly, the fraction Υ Z,C,A is the same for all choices row j and column c k of Λ. This common quotient is the definition of φ A,Z = φ Z,A. A special case occurs when A = (s, i) C = (s, i + 2) = V, and Z = (s, i + 4). Then equation - becomes where φ A,Z = (we + ns) det[1, 2, 3; 1, 2, 3] = c Φ A,Z It follows from Lemma 1, that there is a polynomial identity 12 Example [1, 3, 4; 1, 3, 4] = vφ A,Z y 1 φ A,Y2 y 2 φ A,Y1 + zφ A,V Suppose a portion of the table is a n b n+1 c n+2 d n+3.. b n 1 c n d n+1 e n+2.. Λ = c n 2 d n 1 e n f n+1. d n 3 e n 2 f n 1 g n A special case of the 8-term identity is [2, 3, 4; 1, 2, 3] b n+1 [1, 3, 4; 1, 2, 3] c n + [1, 2, 4; 1, 2, 3] d n 1 [1, 2, 3; 1, 2, 3] e n 2 = 0 This can also be seen by obs that it is the exp of a 4 by 4 det with two equal columns. Gragg s id centered at C n and D n 1 are det [1, 2, 3; 1, 2, 3] = c n r det [2, 3, 4; 1, 2, 3] = d n 1 s where r and s are each of the form (we - ns) for appropriate Sylvester Diamonds centered at C n and D n 1 respectively. Using these identities, and rearranging terms, we have (r e n 2 [1, 3, 4; 1, 2, 3]) c n = (s b n+1 [1, 2, 4; 1, 2, 3]) d n 1

8 In the notation of Lemma 1, r = φ A,En and s = φ Fn 1,B n 1, and this equation would be Ω A,Dn 1,F n 1 c n = Υ Fn 1,C n,a d n 1 Let p 1 be the GCD of Ω A,Dn 1,F n 1 and Υ Fn 1,C n,a Then p 1 = φ A,Fn 1 = Ω A,D n 1,F n 1 d n 1 = Υ F n 1,C n,a c n = φ Fn 1,A and p 1 d n 1 = e n 2 s [1, 3, 4; 1, 2, 3] There is a similar expression e n+2 s [1, 2, 3; 1, 3, 4] = p 2 d n+1 For the location G n we have Ω An,En,G n = g r f n 1 p 2 f n+1 p 1 [1, 3, 4; 1, 3, 4] Reversing the roles of A n and G n there are polynomials r, q 1, and q 2 with Υ Gn,Cn,A n = a r b n 1 q 2 b n+1 q 1 [1, 2, 4; 1, 2, 4] A calculation similar to that in Lemma 1, shows that Ω An,En,G n c n = Υ Gn,Cn,A n e n In the notation of and, φ An,G n is the GCD of the LHS and the RHS. Thus φ An,G n = Ω A n,c n,g n e n = Υ G n,e n,a n c n 13 Suppose ɛ is the error at C n. Then the error at G n will be [ φan,g δ(g n ) = n + a n g n d n 3 d n+3 + a ] e 2 n e n 2 e n+2 b n 1 b n+1 c b n 1 b n+1 For any position X, let δ(x) be the change at X that results from a change of δ(c) at C. Assume the calculations are made in char 2, and that second order effects can be ignored.with these assumptions, [ φa,z az + mn δ(z) = + a ] φc,z + m 2 n 2 δ(c) b 1 b 2 c b 1 b 2 14 The FPA A fixed precision of p bits is chosen. A FPE is (m, e), where

9 (i) m is an odd integer of p bits; (ii) e is an integer exponent. Each entry in the c-table is a FPE, computed by the FPA. Specifically, s is computed as S = w e c2 n 16 Robbins Conjecture Suppose the c-table is computed by the FPA with mantissa length p, and suppose the largest exponent in the computed c-table is q. Then each computed expression is accurate in the lowest p q bits. The expression (m, e) corresponds to the polynomial (m p 1 t p m 2 t 2 + m 1 t + 1) t e In every case the mantissa has p bits, with the lowest bit always 1. N W C E S Neglecting second order effects, s = w e c2 n δ(s) = = e δ(w) + w δ(e) 2 δ(c) c n e δ(w) + w δ(e) s δ(n) n (we c2 ) δ(n) n 2 17 A portion of the c-table.

10 A B n 1. B n+1 C n 2. C n. C n+2 D n 3. D n 1. D n+1. D n+3 E n 4. E n 2. E n. E n+2. E n+4.. F n 3. F n 1. F n+1. F n G n..... Here the dots are locations at the centers of the Sylvester diamonds. 18 Proposition (p 21) Suppose that at location C n = (n, i), c n is changed to c n + ɛ. Assume that above C n the computations are exact, and that the lower levels are computed with high precision. Assume also that second order effects can be neglected. Let (1) r = max { ord(b n 1 ) + ord(b n+1 ) ord(b n 1 ) + ord(b n+1 ) + ord(c n ) ord(a n ) Then the change at any position in the c-table has order at least ord(ɛ) r Change at D n SI is used in the computation b n D n = c n 1 c n+1 Cn 2 = c n 1 c n+1 (c n + ɛ) 2 = c n 1 c n+1 c 2 n 2c n ɛ ɛ 2 = b n d n 2c n ɛ ɛ 2 D n = d n 2c n + ɛ 2 b n Change at D n 1 SI is used in the computation b n 1 d n 1 = c n 2 c n c 2 n 1

11 b n 1 D n 1 = c n 2 C n c 2 n 1 = c n 2 (c n + ɛ) c 2 n 1 = c n 2 c n + c n 2 ɛ c 2 n 1 = b n 1 d n 1 + c n 2 ɛ Changes along diagonals D n 1 = d n 1 + c n 2ɛ b n 1 The SW diagonal is {C n, D n 1, E n 2, F n 3,...} The SW diagonal is {C n, D n+1, E n+2, F n+3,...} The computation of e n 2 is Then c n 2 e n 2 = d n 3 d n 1 + d 2 n 2 (e n 2 ) = d ( n 3 dn 3 d n 1 = c n 2 (f n 3 ) = e n 4 d n 3 e n 2 = ) cn 2 = c n 2 b n 1 en 4 dn 3 d n 3 b n 1 ( dn 3 b n 1 ) = e n 4 b n 1 For each position on the diagonal emansating from C n, (h n k,i+k ) = h n k 1,i+k+1 b n+1 (h n+k,i+k ) = h n+k 1,i+k+1 b n+1 Change at E n. E n is defined by C n E n = D n 1 D n+1 + D 2 n. Let Y = D n 1 D n+1 D 2 n We must show Y is divisible by c n. b n 1 Y = (c n 2 C n c 2 n 1)D n+1 b n 1 D 2 n = c n 2 c n D n+1 c 2 n 1D n+1 b n 1 D 2 n = X 1 C n U where X 1 = c n 2 D n+1 and U = c 2 n 1D n+1 + b n 1 D 2 n

12 b n+1 U = c 2 n 1(C n c n+2 c 2 n+1) + b n+1 b n 1 D 2 n = c 2 n 1c n+2 C n c 2 n 1c 2 n+1 + b n 1 b n+1 D 2 n = X 2 C n + V where X 2 = c 2 n 1c n+2 and V c 2 n 1c 2 n+1 + b n 1 b n+1 D 2 n V = c 2 n 1c 2 n+1 + b n 1 b n+1 D 2 n = (b n D n + c 2 n) 2 + b n 1 b n+1 D 2 n = b 2 nd 2 n 2b n c n D 2 n c 4 n + b n 1 b n+1 D 2 n = C n X 3 + (a n C n + b n 1 b n+1 ɛ)d 2 n = C n X 4 + b n 1 b n+1 D 2 nɛ where X 4 = X 3 + a n D 2 n Change at E n 1 The computation of e n 1 is c n 1 e n 1 = D n 2 D n + D 2 n 1 Let Y = D n 2 D n D 2 n 1 We must show Y is divisible by c n 1. b n 2 Y = (c n 3 C n 1 c 2 n 2)D n b n 2 D 2 n 1 = c n 3 c n 1 D n c 2 n 2D n b n 2 D 2 n 1 = X 1 c n 1 U where X 1 = c n 3 D n and U = c 2 n 2D n + b n 2 D 2 n 1 b n U = c 2 n 2(C n 1 c n+1 C 2 n) + b n+2 b n D 2 n 1 = c 2 n 2c n+1 C n 1 c 2 n 2C 2 n + b n 2 b n D 2 n 1 = X 2 c n 1 + V where X 2 = c 2 n 2c n+1 and V c 2 n 2c 2 n + b n 2 b n D 2 n

13 V = c 2 n 2Cn 2 + b n 2 b n Dn 1 2 = (b n 1 D n 1 + c 2 n 1) 2 + b n 2 b n Dn 1 2 = b 2 n 1Dn 1 2 2b n 1 c n 1 Dn 1 2 c 4 n 1 + b n 2 b n Dn 1 2 = c n 1 X 3 + a n 1 c n 1 Dn 1 2 = c n 1 X 4 where X 4 = X 3 + a n D 2 n E n+1 is similar to E n 1. The first interesting case occurs at E n. This is a special case of the inductive step (later). It must be dealt with separately, because it is the starting point of an induction based on the diffeence between A and Z. The value of E n is c n e n = d n 1 d n 1 d 2 n (e n ) = d n 1 (d n+1 ) c n + d n+1 (d n 1 ) c n e n (c n ) = Let M be the matrix: dn 1 c n+2 b n 1 + d n+1 c n 2 b n+1 e n b n 1 b n+1 M = c n a n b n+1 c n+2 b n 1 c n d n+1 c n 2 d n 1 e n Gragg s identity centered at C n is det(m) = p c n, where p = (ns + ew), and n, s, w, e are the entries neighboring c n. Thus the change at E n can be rewritten as c n (e n ) = pc n a n c n e n + c n+2 c n c n 2 + a n d n 1 d n+1 c n b n 1 b n+1 p an e n + c n+2 c n 2 an d n 1 d n+1 = + b n 1 b n+1 c n b n 1 b n+1 The first fraction has order ord(ɛ) ord(b n 1 b n+1 )

14 The second fraction has order ord(ɛ + ord(a n )) ord(c n ) ord(b n 1 b n+1 ) Taken together, thses show that ord( (e n )) ord(ɛ) r 25 Change at Z Λ = a b n 1 n b 1 c n 3 n 2.. e m 1 m v y 2 m m y 1 z We want to show that [ φa,z az + mn (z) = + a ] φc,z + m 2 n 2 b 1 b 2 c b 1 b 2 The expression for (z) can be simplified as follows. For any two locationss and T let M and N be the locations at the opposite corners of the rectangle determined by S and T. let ψ S,T = φ S,T + mn With this substitution, the identity becomes y 1 ψ A,Y2 + y 2 ψ A,Y1 zψ A,V vψ A,Z = avz + ay 1 y 2 The expression for z becomes cψa,z acz + aψ C.Z (z) = cb 1 b 2

15 When Z is at location E ψ C,E = d n 1 d n+1 It was established that c 3 (e n ) = n + a n c n e n c n+2 c n c n 2 a n d n 1 d n+1 c n b n 1 b n+1 cn ψ An,E = n + a n c n e n + a n ψ Cn,En c n b n 1 b n+1 Start of induction based on diff between A and Z. The value of z is computed as z = y 1 y 2 y0 2 v 26 In Char 2, the change at C causes no change at Y 0 (to the first order effects). Inductively, assume that cψa,y1 acy 1 + aψ C,Y1 (y 1 ) = cb 1 b 2 cψa,y2 acy 2 + aψ C,Y2 (y 2 ) = cb 1 b 2 cψa,v acv + aψ C,V (v) = cb 1 b 2 (z) = = y1 δ(y 2 ) + y 2 δ(y 1 ) z (v) + v cy1 ψ A,Y2 y 1 acy 2 + ay 1 ψ C,Y2 + vcb 1 b 2 cy2 ψ A,Y1 y 2 acy 1 + ay 2 ψ C,Y1 + vcb 1 b 2 czψa,v zacv zaψ C,V vcb 1 b 2 cψa,z acz + aψ C,Z cb 1 b 2 SI is used in c = b 1b 2 b 2 0 a

16 where b 1 b 2 b 2 0 is computed first, and then the result is divided by a. An error can originate at C only if ord b 1 b 2 = ordb 2 0 Suppose this occurs. Each mantissa is an odd integer with p bits, where p = precision. Therefore ord (c) p + ord(b 1 ) + ord(b 2 ) ord(a) Let q = max{ord(a), ord(c)} the change δ(z) at Z satisfies ord (z) p q