Hypergeometric Series Solutions of Linear Operator Equations - PowerPoint PPT Presentation

Hypergeometric Series Solutions of Linear Operator Equations Qing-Hu Hou Center for Combinatorics Nankai University P .R. China Joint work with Yan-Ping Mu, Tianjin University of Technology HyperRep – p. 1/25

Contents Background HyperRep – p. 2/25

Contents Background Solving linear operator equations HyperRep – p. 2/25

Contents Background Solving linear operator equations Differential/Difference equations HyperRep – p. 2/25

Contents Background Solving linear operator equations Differential/Difference equations Recurrence relations HyperRep – p. 2/25

Background HyperRep – p. 3/25

Linear differential equations lve linear differential equations by means of power series. HyperRep – p. 4/25

Linear differential equations lve linear differential equations by means of power series. 2 x ( x − 1) y ′′ ( x ) + (7 x − 3) y ′ ( x ) + 2 y ( x ) = 0 HyperRep – p. 4/25

Linear differential equations lve linear differential equations by means of power series. 2 x ( x − 1) y ′′ ( x ) + (7 x − 3) y ′ ( x ) + 2 y ( x ) = 0 ∞ � c n x n ⇒ y ( x ) = k =0 HyperRep – p. 4/25

Linear differential equations lve linear differential equations by means of power series. 2 x ( x − 1) y ′′ ( x ) + (7 x − 3) y ′ ( x ) + 2 y ( x ) = 0 ∞ � c n x n ⇒ y ( x ) = k =0 ⇒ ( n + 1)(2 n + 3) c n +1 − ( n + 2)(2 n + 1) c n = 0 HyperRep – p. 4/25

Linear differential equations lve linear differential equations by means of power series. 2 x ( x − 1) y ′′ ( x ) + (7 x − 3) y ′ ( x ) + 2 y ( x ) = 0 ∞ � c n x n ⇒ y ( x ) = k =0 ⇒ ( n + 1)(2 n + 3) c n +1 − ( n + 2)(2 n + 1) c n = 0 ∞ 2( n + 1) � 2 n + 1 x n . ⇒ y ( x ) = n =0 HyperRep – p. 4/25

Linear differential equations Abramov and Petkovsˇ ek considered general polynomial sequences, especially ∞ � c k ( x − a ) k . y ( x ) = k =0 HyperRep – p. 5/25

Linear differential equations Abramov and Petkovsˇ ek considered general polynomial sequences, especially ∞ � c k ( x − a ) k . y ( x ) = k =0 Abramov, Paule and Petkovšek visited formal power series solutions and basic hypergeometric series solutions for q -difference equations. ∞ � c ( q k ) x k . y ( x ) = k =0 HyperRep – p. 5/25

Our aim Given a linear differential/difference equation L ( y ( x )) = 0 , find a hypergeometric series solution ∞ � y ( x ) = c k b k ( x ) . k =0 HyperRep – p. 6/25

Our aim Given a linear differential/difference equation L ( y ( x )) = 0 , find a hypergeometric series solution ∞ � y ( x ) = c k b k ( x ) . k =0 (1 − x 2 ) p ′′ ( x ) − xp ′ ( x ) + n 2 p ( x ) = 0 � � � n � k − n, n ( − n ) k ( n ) k � 1 − x 1 − x � � ⇒ p ( x ) = = 2 F 1 . � (1 / 2) k k ! 2 2 � 1 / 2 � k =0 HyperRep – p. 6/25

Solving linear operator equations HyperRep – p. 7/25

Suitable bases Let L be a linear operator acting on the ring K [ x ] . HyperRep – p. 8/25

Suitable bases Let L be a linear operator acting on the ring K [ x ] . We aim to find a basis { b k ( x ) } of K [ x ] such that HyperRep – p. 8/25

Suitable bases Let L be a linear operator acting on the ring K [ x ] . We aim to find a basis { b k ( x ) } of K [ x ] such that L ( b k ( x )) = A k b k ( x ) + B k b k − h ( x ) , ∀ k ∈ N , where A k , B k ∈ K and h is a fixed positive integer. HyperRep – p. 8/25

Suitable bases Let L be a linear operator acting on the ring K [ x ] . We aim to find a basis { b k ( x ) } of K [ x ] such that L ( b k ( x )) = A k b k ( x ) + B k b k − h ( x ) , ∀ k ∈ N , where A k , B k ∈ K and h is a fixed positive integer. We further require that b k ( x ) are monic b k − 1 ( x ) divides b k ( x ) HyperRep – p. 8/25

Suitable bases Let L be a linear operator acting on the ring K [ x ] . We aim to find a basis { b k ( x ) } of K [ x ] such that L ( b k ( x )) = A k b k ( x ) + B k b k − h ( x ) , ∀ k ∈ N , where A k , B k ∈ K and h is a fixed positive integer. We further require that b k ( x ) are monic b k − 1 ( x ) divides b k ( x ) Then b k ( x ) = ( x − x 1 )( x − x 2 ) · · · ( x − x k ) . HyperRep – p. 8/25

Suitable bases Let L be a linear operator acting on the ring K [ x ] . We aim to find a basis { b k ( x ) } of K [ x ] such that L ( b k ( x )) = A k b k ( x ) + B k b k − h ( x ) , ∀ k ∈ N , where A k , B k ∈ K and h is a fixed positive integer. We further require that b k ( x ) are monic b k − 1 ( x ) divides b k ( x ) Then b k ( x ) = ( x − x 1 )( x − x 2 ) · · · ( x − x k ) . Such bases are called suitable bases. HyperRep – p. 8/25

Searching for suitable bases We solve x 1 , . . . , x k for explicit integer k . Recall that L ( b k ( x )) = A k b k ( x ) + B k b k − h ( x ) . HyperRep – p. 9/25

Searching for suitable bases We solve x 1 , . . . , x k for explicit integer k . Recall that L ( b k ( x )) = A k b k ( x ) + B k b k − h ( x ) . Then A k = [ x k ] L ( b k ( x )) B k = [ x k − h ] � L ( b k ( x )) − A k b k ( x ) � and can be expressed in terms of x 1 , . . . , x k . HyperRep – p. 9/25

Searching for suitable bases We solve x 1 , . . . , x k for explicit integer k . Recall that L ( b k ( x )) = A k b k ( x ) + B k b k − h ( x ) . Then A k = [ x k ] L ( b k ( x )) B k = [ x k − h ] � L ( b k ( x )) − A k b k ( x ) � and can be expressed in terms of x 1 , . . . , x k . Comparing coefficients of x i , we obtain a system of polynomial equations on x 1 , . . . , x k . HyperRep – p. 9/25

Searching for suitable bases We solve x 1 , . . . , x k for explicit integer k . Recall that L ( b k ( x )) = A k b k ( x ) + B k b k − h ( x ) . Then A k = [ x k ] L ( b k ( x )) B k = [ x k − h ] � L ( b k ( x )) − A k b k ( x ) � and can be expressed in terms of x 1 , . . . , x k . Comparing coefficients of x i , we obtain a system of polynomial equations on x 1 , . . . , x k . Starting from k = 1 , we iteratively set up and solve the equations until reaching a certain degree k 0 . HyperRep – p. 9/25

Searching for suitable bases We solve x 1 , . . . , x k for explicit integer k . Recall that L ( b k ( x )) = A k b k ( x ) + B k b k − h ( x ) . Then A k = [ x k ] L ( b k ( x )) B k = [ x k − h ] � L ( b k ( x )) − A k b k ( x ) � and can be expressed in terms of x 1 , . . . , x k . Comparing coefficients of x i , we obtain a system of polynomial equations on x 1 , . . . , x k . Starting from k = 1 , we iteratively set up and solve the equations until reaching a certain degree k 0 . Finally, guess the general form of x k from the pattern. HyperRep – p. 9/25

Example L ( p ( x )) = (1 − x 2 ) p ′′ ( x ) − xp ′ ( x ) + n 2 p ( x ) . HyperRep – p. 10/25

Example L ( p ( x )) = (1 − x 2 ) p ′′ ( x ) − xp ′ ( x ) + n 2 p ( x ) . Take h = 1 and set b 0 ( x ) = 1 , b 1 ( x ) = x − x 1 . HyperRep – p. 10/25

Example L ( p ( x )) = (1 − x 2 ) p ′′ ( x ) − xp ′ ( x ) + n 2 p ( x ) . Take h = 1 and set b 0 ( x ) = 1 , b 1 ( x ) = x − x 1 . L ( b 1 ( x )) = ( n 2 − 1) x − n 2 x 1 = A 1 ( x − x 1 ) + B 1 . HyperRep – p. 10/25

Example L ( p ( x )) = (1 − x 2 ) p ′′ ( x ) − xp ′ ( x ) + n 2 p ( x ) . Take h = 1 and set b 0 ( x ) = 1 , b 1 ( x ) = x − x 1 . L ( b 1 ( x )) = ( n 2 − 1) x − n 2 x 1 = A 1 ( x − x 1 ) + B 1 . A 1 = ( n 2 − 1) and B 1 = − x 1 . HyperRep – p. 10/25

Example L ( p ( x )) = (1 − x 2 ) p ′′ ( x ) − xp ′ ( x ) + n 2 p ( x ) . Take h = 1 and set b 0 ( x ) = 1 , b 1 ( x ) = x − x 1 . L ( b 1 ( x )) = ( n 2 − 1) x − n 2 x 1 = A 1 ( x − x 1 ) + B 1 . A 1 = ( n 2 − 1) and B 1 = − x 1 . We do not obtain any equation on x 1 . HyperRep – p. 10/25

Example Set b 2 ( x ) = ( x − x 1 )( x − x 2 ) . HyperRep – p. 11/25

Example Set b 2 ( x ) = ( x − x 1 )( x − x 2 ) . ( n 2 − 4) x 2 − ( n 2 − 1)( x 1 + x 2 ) x + 2 + n 2 x 1 x 2 = A 2 ( x − x 1 )( x − x 2 ) + B 2 ( x − x 1 ) , HyperRep – p. 11/25

Example Set b 2 ( x ) = ( x − x 1 )( x − x 2 ) . ( n 2 − 4) x 2 − ( n 2 − 1)( x 1 + x 2 ) x + 2 + n 2 x 1 x 2 = A 2 ( x − x 1 )( x − x 2 ) + B 2 ( x − x 1 ) , A 2 = n 2 − 4 , x 1 x 2 = 3 x 2 and B 2 = − 3( x 1 + x 2 ) , 1 − 2 . HyperRep – p. 11/25

Example Set b 2 ( x ) = ( x − x 1 )( x − x 2 ) . ( n 2 − 4) x 2 − ( n 2 − 1)( x 1 + x 2 ) x + 2 + n 2 x 1 x 2 = A 2 ( x − x 1 )( x − x 2 ) + B 2 ( x − x 1 ) , A 2 = n 2 − 4 , x 1 x 2 = 3 x 2 and B 2 = − 3( x 1 + x 2 ) , 1 − 2 . For k ≥ 3 , we derive and x 1 = x 2 = · · · = x k = 1 x 1 = x 2 = · · · = x k = − 1 . HyperRep – p. 11/25

Example Set b 2 ( x ) = ( x − x 1 )( x − x 2 ) . ( n 2 − 4) x 2 − ( n 2 − 1)( x 1 + x 2 ) x + 2 + n 2 x 1 x 2 = A 2 ( x − x 1 )( x − x 2 ) + B 2 ( x − x 1 ) , A 2 = n 2 − 4 , x 1 x 2 = 3 x 2 and B 2 = − 3( x 1 + x 2 ) , 1 − 2 . For k ≥ 3 , we derive and x 1 = x 2 = · · · = x k = 1 x 1 = x 2 = · · · = x k = − 1 . Guess: b k ( x ) = ( x + 1) k b k ( x ) = ( x − 1) k . or HyperRep – p. 11/25

Verify suitable bases L ( b k ( x )) = A k b k ( x ) + B k b k − h ( x ) HyperRep – p. 12/25