Determinants and Their Applications in Mathematical Physics
126 4. Particular Determinants a.K r 1 n(0) = (−1) r+1 n(r+n−1)! (r−1)!r!(n−r)! . b.K rs n(0) = (−1) r+s rs r+s− 1 ( n− 1 r− 1 ...
4.10 Henkelians 3 127 =V 1 −(r−1) ∑ j K j 1 h+r+j− 1 =V 1 −(r−1)δr 1 , 1 ≤r≤n. The second term is zero. The result follows. The ...
128 4. Particular Determinants = n− 1 ∑ i=0 (−1) i ( n− 1 i ) (h+n+i)! (h+i+ 1)! x h+i+1 = n ∑ j=1 (−1) j− 1 ( n− 1 j− 1 ) (h+n+ ...
4.10 Henkelians 3 129 Sn(x, h)=Kn(h)Vnn n ∑ j=1 Vnj(−x) j− 1 h+n+j− 1 Hence, ( h+n− 1 h ) Sn(x, h) Sn(0,h) =(−1) n+1 n ∑ j=1 Vnj ...
130 4. Particular Determinants Referring to the section on differences in Appendix A.8, φm=∆ m θ 0 so that B=A. The HankelianBar ...
4.10 Henkelians 3 131 = ∑ i,j,r,s AisAjr (2i−1)(2j−1) [p 2 x 2(i+j−1) +q 2 y 2(i+j−1) −1] = ∑ i,j,r,s (i+j−1)aijAisArj (2i−1)(2j ...
132 4. Particular Determinants The result is 1 A^2 [ p 2 V 2 (x)+q 2 V 2 (y)−W 2 ] = ∑ i ∑ j i+j− 1 (2i−1)(2j−1) A ij = 1 2 ∑ i ...
4.10 Henkelians 3 133 = cij 2 i− 1 . After removing the factor (2i−1) − 1 from rowi,1≤i≤n, the result is U= 2 n n! (2n)! ∣ ∣ ∣ ∣ ...
134 4. Particular Determinants Let A=|φm|n, 0 ≤m≤ 2 n− 2 , where φm= x 2 m+2 − 1 m+1 . Ais identical to|aij|n, whereaijis define ...
4.10 Henkelians 3 135 Now, perform the row and column operations R ′ i = i− 2 ∑ r=0 (−1) r ( i− 2 r ) Ri−r,i=n, n− 1 ,n− 2 ,..., ...
136 4. Particular Determinants Theorem. An=Kn(x 2 −1) n 2 ,Kn=Kn(0). Proof. φ 0 =x 2 − 1. Referring to Example A.3 (withc= 1) in ...
4.11 Hankelians 4 137 in which the column difference is 1 3 (v 3 −u 3 ). Let the determinant of the elements in the firstnrows a ...
138 4. Particular Determinants v-Numbers satisfy the identities n ∑ k=1 vnk i+k− 1 =1, 1 ≤i≤n, (4.11.3) vni n ∑ k=1 vnk (i+k−1)( ...
4.11 Hankelians 4 139 Qn=Qn(x)= [ x 2(i+j−1) i+j− 1 ] n . (4.11.12) BothKnandQnare Hankelians andQn(1) =Kn, the simple Hilbert m ...
140 4. Particular Determinants =Kn ( K − 1 n Qn−t 2 In ) =Kn ( S 2 n −t 2 In ) =Kn(Sn+tIn)(Sn−tIn) =KnHnHn. Corollary. B − 1 n ...
4.11 Hankelians 4 141 The only element which remains in columnnis a 1 in position (n+1,n). Hence, En+1=− ∣ ∣ ∣ ∣ ∣ ∣ ∣ h 11 h 12 ...
142 4. Particular Determinants Theorem 4.45 now follows from (4.11.24). Theorem 4.46. Gn=(−1) n− 1 vnnKnHnHn− 1 , whereGnis de ...
4.11 Hankelians 4 143 b. ∣ ∣ ∣ ∣ 1 (i+j−2)! ∣ ∣ ∣ ∣ = (−1) n(n−1)/ 2 1! 2! 3!···(n−2)! n!(n+ 1)!···(2n−2)! . The second determin ...
144 4. Particular Determinants = ∣ ∣ ∣ ∣ ( n+j− 2 n−i )∣ ∣ ∣ ∣ n =An, which proves (b). Exercises Apply similar methods to pro ...
4.11 Hankelians 4 145 where C ′ n=λn−^1 ,n−^1 Cn+ n− 1 ∑ j=1 λn− 1 ,j− 1 Cj = n− 1 ∑ j=1 λn− 1 ,j− 1 [ νj− 1 νj···νn+j− 3 νn+j− ...
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