Determinants and Their Applications in Mathematical Physics
146 4. Particular Determinants Proof. Apply the Jacobi identity (Section 3.6.1) toAr, wherer≥i+1: ∣ ∣ ∣ ∣ ∣ A (r) ij A (r) ir A ...
4.11 Hankelians 4 147 Proof. By interchanging first rows and then columns in a suitable manner it is easy to show that En= ∣ ∣ ∣ ...
148 4. Particular Determinants Theorem 4.52. (xG ′ n ) ′ =Kn(h)x h P 2 n (x, h), where Pn(x, h)= D h+n [x n (1 +x) h+n− 1 ] (h+n ...
4.11 Hankelians 4 149 where eij(x)= (1 +x) i+j+1 −x i+j+1 i+j+1 . (4.11.61) Theorem 4.53. The polynomial determinantE satisfies ...
150 4. Particular Determinants Note thatUij=Vijin general. Since ψ ′ m=−mψm−^1 , ψ 0 =x+c, (4.11.68) it follows that V ′ =V 11 ...
4.11 Hankelians 4 151 =Dc n ∑ r=0 frc r =f 1 + n ∑ r=2 rfrc r− 1 . (4.11.75) Hence, f 1 = [ Dc{c n U(x, c − 1 )} ] c=0 =Dc [ c n ...
152 4. Particular Determinants The identity xD n [x n− 1 (1 +x) n ]=nD n− 1 [x n (1 +x) n− 1 ] (4.11.81) can be proved by showin ...
4.12 Hankelians 5 153 4.12 Hankelians 5 Notes in orthogonal and other polynomials are given in Appendices A.5 and A.6. Hankelian ...
154 4. Particular Determinants The theorem follows from (4.12.2) and (4.12.3) after replacing xby x − 1 . In the next theorem, ...
4.12 Hankelians 5 155 = 1 2 n+1 πi ∫ C (ζ 2 −1) n (ζ−x) n+1 dζ (putζ=x+2t) = 1 2 n+1πi ∫ C′ [(x+1+2t)(x−1+2t)] n (2t)n+1 dt = 1 ...
156 4. Particular Determinants = n+r ∑ σ=r ( n n−σ )( n n−σ+r ) u σ v n−σ+r =(uv) r n+r ∑ σ=r ( n σ )( n σ−r ) u σ−r v n−σ . (4. ...
4.12 Hankelians 5 157 =2 n− 1 n ∏ i=2 (uv) i− 1 =2 n− 1 (uv) 1+2+3+···+n− 1 =2 −(n−1) 2 (x 2 −1) 1 2 n(n−1) . which completes th ...
158 4. Particular Determinants Hn=|Si+j− 2 |n=λnx n(n+1)/ 2 , Jn=|Ai+j− 2 |n=λnx n(n−1)/ 2 , where λn= [ 1! 2! 3!···(n−1)!] 2 . ...
4.12 Hankelians 5 159 D n τ {Qn(t, t)}= ∣ ∣C n(t)C 1 (t)C 2 (t)···Cn− 1 (t) ∣ ∣ n =(−1) n− 1 ∣ ∣C 1 (t)C 2 (t)···Cn− 1 (t)Cn(t) ...
160 4. Particular Determinants Reverting toxand referring to (4.12.17), xDx [ Gn− 1 Fn− 1 ] = FnGn− 2 F 2 n− 1 , (4.12.28) where ...
4.12 Hankelians 5 161 From (4.12.28) and (4.12.33), xu ′ n− 1 = ( Gn− 1 Fn− 1 ) 2 ( Fn Gn )( Gn− 2 Gn− 1 )( Gn Gn− 1 ) , unu ′ n ...
162 4. Particular Determinants EvaluatingEnfor small values ofn, it is found that E 1 = x 1 −x ,E 2 = 1! 2 x 3 (1−x) 4 ,E 3 = [1 ...
4.12 Hankelians 5 163 wherefis an arbitrary function oft. Then, it is proved that Section 6.5.2 on Toda equations that D 2 (logA ...
164 4. Particular Determinants = ∞ ∑ r=0 (p)re (r+p)t r! , where (p)r=p(p+ 1)(p+2)···(p+r−1) (4.12.52) and denote the correspond ...
4.13 Hankelians 6 165 Substituting this formula into (4.12.50) withAn→E (p) n andE (1) n =f, E (p) n = ( e t 1 −e t )p n− 1 ∏ r= ...
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