Symbolic evaluation of determinants and rhombus tilings of holey - PowerPoint PPT Presentation

Generalization Definition: For n, s, t ∈ ❩ , n � 1 , and µ an indeterminate, we define D s,t ( n ) to be the following ( n × n ) -determinant: � � µ + i + j − 2 �� D s,t ( n ) := det δ i,j + j s � i<s + n t � j<t + n � � µ + i + j + s + t − 4 �� = det δ i + s − 1 ,j + t − 1 + j + t − 1 1 � i,j � n 10 / 37

Generalization Definition: For n, s, t ∈ ❩ , n � 1 , and µ an indeterminate, we define D s,t ( n ) to be the following ( n × n ) -determinant: � � µ + i + j − 2 �� D s,t ( n ) := det δ i,j + j s � i<s + n t � j<t + n � � µ + i + j + s + t − 4 �� = det δ i + s − 1 ,j + t − 1 + j + t − 1 1 � i,j � n Known special cases: ◮ closed form for D 0 , 0 ( n ) (Andrews 1979) ◮ closed form for D 1 , 1 (2 n ) /D 1 , 1 (2 n − 1) (Andrews 1980) ◮ monstrous conjecture for D 1 , 1 ( n ) (K-T 2013) 10 / 37

Desnanot-Jacobi-Carroll Identity (DJC) � � Theorem. Let m i,j i,j ∈ ❩ be an infinite sequence and denote by M s,t ( n ) the determinant of the ( n × n ) -matrix whose upper left � � entry is m s,t , more precisely the matrix m i,j s � i<s + n,t � j<t + n . Then: M s,t ( n ) M s +1 ,t +1 ( n − 2) = M s,t ( n − 1) M s +1 ,t +1 ( n − 1) − M s +1 ,t ( n − 1) M s,t +1 ( n − 1) . 11 / 37

Desnanot-Jacobi-Carroll Identity (DJC) � � Theorem. Let m i,j i,j ∈ ❩ be an infinite sequence and denote by M s,t ( n ) the determinant of the ( n × n ) -matrix whose upper left � � entry is m s,t , more precisely the matrix m i,j s � i<s + n,t � j<t + n . Then: M s,t ( n ) M s +1 ,t +1 ( n − 2) = M s,t ( n − 1) M s +1 ,t +1 ( n − 1) − M s +1 ,t ( n − 1) M s,t +1 ( n − 1) . Schematically: × = × − × 11 / 37

DJC for D 1 , 1 ( n ) × = × − × By (DJC) we obtain a recurrence equation for D 1 , 1 ( n ) : D 0 , 0 ( n + 1) D 1 , 1 ( n − 1) = D 0 , 0 ( n ) D 1 , 1 ( n ) − D 1 , 0 ( n ) D 0 , 1 ( n ) . 12 / 37

DJC for D 1 , 1 ( n ) × = × − × By (DJC) we obtain a recurrence equation for D 1 , 1 ( n ) : D 0 , 0 ( n + 1) D 1 , 1 ( n − 1) = D 0 , 0 ( n ) D 1 , 1 ( n ) − D 1 , 0 ( n ) D 0 , 1 ( n ) . We rewrite it slightly: D 1 , 1 ( n ) = D 0 , 0 ( n + 1) D 1 , 1 ( n − 1) + D 1 , 0 ( n ) D 0 , 1 ( n ) . D 0 , 0 ( n ) D 0 , 0 ( n ) � �� � = R 0 , 0 ( n ) − → Hence we need to know D 1 , 0 ( n ) and D 0 , 1 ( n ) . 12 / 37

DJC for D 1 , 1 ( n ) × = × − × By (DJC) we obtain a recurrence equation for D 1 , 1 ( n ) : D 0 , 0 ( n + 1) D 1 , 1 ( n − 1) = D 0 , 0 ( n ) D 1 , 1 ( n ) − D 1 , 0 ( n ) D 0 , 1 ( n ) . We rewrite it slightly: D 1 , 1 ( n ) = D 0 , 0 ( n + 1) D 1 , 1 ( n − 1) + D 1 , 0 ( n ) D 0 , 1 ( n ) . D 0 , 0 ( n ) D 0 , 0 ( n ) � �� � = R 0 , 0 ( n ) − → Hence we need to know D 1 , 0 ( n ) and D 0 , 1 ( n ) . Question: What is the combinatorial interpretation of the general determinant D s,t ( n ) ? 12 / 37

Lindstr¨ om-Gessel-Viennot Lemma Let G be a directed acyclic graph and consider base vertices A = { a 1 , . . . , a n } and destination vertices B = { b 1 , . . . , b n } . 13 / 37

Lindstr¨ om-Gessel-Viennot Lemma Let G be a directed acyclic graph and consider base vertices A = { a 1 , . . . , a n } and destination vertices B = { b 1 , . . . , b n } . For each path P , let ω ( P ) be the product of its edge weights. Let � e ( a, b ) = ω ( P ) and P : a → b   e ( a 1 , b 1 ) e ( a 1 , b 2 ) · · · e ( a 1 , b n ) e ( a 2 , b 1 ) e ( a 2 , b 2 ) · · · e ( a 2 , b n )     M =  . . . . ...   . . . . . .  e ( a n , b 1 ) e ( a n , b 2 ) · · · e ( a n , b n ) 13 / 37

Lindstr¨ om-Gessel-Viennot Lemma Let G be a directed acyclic graph and consider base vertices A = { a 1 , . . . , a n } and destination vertices B = { b 1 , . . . , b n } . For each path P , let ω ( P ) be the product of its edge weights. Let � e ( a, b ) = ω ( P ) and P : a → b   e ( a 1 , b 1 ) e ( a 1 , b 2 ) · · · e ( a 1 , b n ) e ( a 2 , b 1 ) e ( a 2 , b 2 ) · · · e ( a 2 , b n )     M =  . . . . ...   . . . . . .  e ( a n , b 1 ) e ( a n , b 2 ) · · · e ( a n , b n ) Then the determinant of M is the signed sum over all n -tuples P = ( P 1 , . . . , P n ) of non-intersecting paths from A to B : n � � det( M ) = sign( σ ( P )) ω ( P i ) . i =1 ( P 1 ,...,P n ): A → B where σ denotes a permutation that is applied to B . 13 / 37

Lindstr¨ om-Gessel-Viennot Lemma Application: In our context, the lemma implies the following. Look at the determinant without the Kronecker-Delta: � µ + i + j + s + t − 4 � det . j + t − 1 1 � i,j � n It counts n -tuples of non-intersecting paths in the lattice ◆ 2 : ◮ The starting points are (0 , t ) , (0 , t + 1) , . . . , (0 , t + n − 1) . ◮ The end points are ( µ + s − 2 , 0) , . . . , ( µ + s + n − 3 , 0) . ◮ The allowed steps are (1 , 0) and (0 , − 1) . 14 / 37

Non-intersecting Lattice Paths n + t - 1 t + 1 t μ + s - 2 μ + n + s - 3 For 1 � i, j � n the number of paths from (0 , t + j − 1) to � µ + i + j + s + t − 4 � ( µ + s + i − 3 , 0) is given by , which is precisely the j + t − 1 ( i, j ) -entry of our matrix. 15 / 37

Lattice Paths — Rhombus Tilings n + t - 1 t + 1 t μ + s - 2 μ + n + s - 3 16 / 37

Lattice Paths — Rhombus Tilings n + t - 1 t + 1 t μ + s - 2 μ + n + s - 3 16 / 37

Lattice Paths — Rhombus Tilings n + t - 1 t + 1 t μ + s - 2 μ + n + s - 3 16 / 37

Lattice Paths — Rhombus Tilings n + t - 1 t + 1 t μ + s - 2 μ + n + s - 3 16 / 37

Determinant with Kronecker-Delta From the Laplace expansion one immediately sees that � � · · · b 1 ,j + 1 b 1 ,j +1 · · · � � � � · · · b 2 ,j b 2 ,j +1 + 1 · · · � � = � � . . � � . . . . � � � � · · · b 1 ,j b 1 ,j +1 · · · � � � � · · · b 2 ,j − 1 b 2 ,j +1 + 1 · · · � � � � · · · b 2 ,j b 2 ,j +1 + 1 · · · � � � � ± . . � � � � . . . . . . � � � � . . . . � � By applying this procedure recursively, one obtains � ( − 1) ( s − t ) ·| I | det � � M I D s,t ( n ) = ( s � t ) , I + s − t I ⊆{ 1 ,...,n − s + t } where M I J denotes the matrix that is obtained by deleting all rows with indices in I and all columns with indices in J from the matrix �� µ + i + j + s + t − 4 �� . j + t − 1 1 � i,j � n 17 / 37

Kronecker-Deltas on the Main Diagonal General formula: � ( − 1) ( s − t ) ·| I | det � � M I D s,t ( n ) = ( s � t ) I + s − t I ⊆{ 1 ,...,n − s + t } Special case: If s = t we obtain � � � M I D s,s ( n ) = det , I I ⊆{ 1 ,...,n } i.e., D s,s ( n ) is the sum of principal minors of the binomial matrix. 18 / 37

Kronecker-Deltas on the Main Diagonal General formula: � ( − 1) ( s − t ) ·| I | det � � M I D s,t ( n ) = ( s � t ) I + s − t I ⊆{ 1 ,...,n − s + t } Special case: If s = t we obtain � � � M I D s,s ( n ) = det , I I ⊆{ 1 ,...,n } i.e., D s,s ( n ) is the sum of principal minors of the binomial matrix. Hence: D s,s ( n ) counts all k -tuples of non-intersecting lattice paths, where k runs from 0 to n , and where the start and end points are given by the same k -subset. 18 / 37

Kronecker-Deltas on the Main Diagonal s = 2 t = 2 n = 6 µ = 4 19 / 37

Kronecker-Deltas on the Main Diagonal s = 2 t = 2 n = 6 µ = 4 19 / 37

Kronecker-Deltas on the Main Diagonal s = 2 t = 2 n = 6 µ = 4 19 / 37

Kronecker-Deltas on the Main Diagonal s = 2 t = 2 n = 6 µ = 4 19 / 37

Kronecker-Deltas on the Main Diagonal s = 2 t = 2 n = 6 µ = 4 19 / 37

Rhombus Tilings Finding: The determinant D s,s ( n ) counts ◮ rhombus tilings ◮ of a hexagon with a funny-shaped hole (“holey hexagon”) ◮ that are cyclically symmetric. ◮ The hole has the shape of a triangle (of size µ − 2 ) with “boundary lines” (of length s ) sticking out of its corners. 20 / 37

Rhombus Tilings Finding: The determinant D s,s ( n ) counts ◮ rhombus tilings ◮ of a hexagon with a funny-shaped hole (“holey hexagon”) ◮ that are cyclically symmetric. ◮ The hole has the shape of a triangle (of size µ − 2 ) with “boundary lines” (of length s ) sticking out of its corners. Remark: This combinatorial interpretation is due to Krattenthaler and Ciucu (at least for s = 0 ). 20 / 37

Rhombus Tilings Finding: The determinant D s,s ( n ) counts ◮ rhombus tilings ◮ of a hexagon with a funny-shaped hole (“holey hexagon”) ◮ that are cyclically symmetric. ◮ The hole has the shape of a triangle (of size µ − 2 ) with “boundary lines” (of length s ) sticking out of its corners. Remark: This combinatorial interpretation is due to Krattenthaler and Ciucu (at least for s = 0 ). Example: For s = t = 1 , n = 2 , and µ = 3 we obtain � � 4 6 � � � D 1 , 1 (2) µ → 3 = � = 20 . � � � 4 11 � 20 / 37

Cyclically Symmetric Rhombus Tilings of a Holey Hexagon 21 / 37

Off-Diagonal Kronecker-Deltas Now let’s look at the situation s � = t . General formula: � ( − 1) ( s − t ) ·| I | det � � M I + t − s D s,t ( n ) = ( t � s ) I I ⊆{ 1 ,...,n + s − t } 22 / 37

Off-Diagonal Kronecker-Deltas Now let’s look at the situation s � = t . General formula: � ( − 1) ( s − t ) ·| I | det � � M I + t − s D s,t ( n ) = ( t � s ) I I ⊆{ 1 ,...,n + s − t } Remark: If s − t is odd, we perform a weighted count with weights +1 and − 1 , according to the length of the tuples of paths. 22 / 37

Off-Diagonal Kronecker-Deltas s = 1 t = 3 n = 6 µ = 5 23 / 37

Off-Diagonal Kronecker-Deltas s = 1 t = 3 n = 6 µ = 5 23 / 37

Off-Diagonal Kronecker-Deltas s = 1 t = 3 n = 6 µ = 5 23 / 37

Off-Diagonal Kronecker-Deltas s = 1 t = 3 n = 6 µ = 5 23 / 37

Off-Diagonal Kronecker-Deltas s = 1 t = 3 n = 6 µ = 5 23 / 37

Off-Diagonal Kronecker-Deltas s = 1 t = 3 n = 6 µ = 5 23 / 37

Off-Diagonal Kronecker-Deltas Example: Shapes for different choices of the parameters s = 5 , t = 1 , n = 5 , µ = 4 s = − 1 , t = 2 , n = 6 , µ = 6 24 / 37

◗ Back to D 1 , 1 ( n ) Recall: We wanted to evaluate D 1 , 1 ( n ) using (DJC): D 1 , 1 ( n ) = D 0 , 0 ( n + 1) D 1 , 1 ( n − 1) + D 1 , 0 ( n ) D 0 , 1 ( n ) . D 0 , 0 ( n ) D 0 , 0 ( n ) � �� � = R 0 , 0 ( n ) 25 / 37

◗ Back to D 1 , 1 ( n ) Recall: We wanted to evaluate D 1 , 1 ( n ) using (DJC): D 1 , 1 ( n ) = D 0 , 0 ( n + 1) D 1 , 1 ( n − 1) + D 1 , 0 ( n ) D 0 , 1 ( n ) . D 0 , 0 ( n ) D 0 , 0 ( n ) � �� � = R 0 , 0 ( n ) Problem: We need to evaluate some other (simpler) determinants. For example, we can show that D 1 , 0 (2 n ) = 0 = D 0 , 1 (2 n ) for all n . 25 / 37

◗ Back to D 1 , 1 ( n ) Recall: We wanted to evaluate D 1 , 1 ( n ) using (DJC): D 1 , 1 ( n ) = D 0 , 0 ( n + 1) D 1 , 1 ( n − 1) + D 1 , 0 ( n ) D 0 , 1 ( n ) . D 0 , 0 ( n ) D 0 , 0 ( n ) � �� � = R 0 , 0 ( n ) Problem: We need to evaluate some other (simpler) determinants. For example, we can show that D 1 , 0 (2 n ) = 0 = D 0 , 1 (2 n ) for all n . ◮ Compute the (nontrivial) nullspace of D 0 , 1 (2 n ) for n � 15 . 25 / 37

Back to D 1 , 1 ( n ) Recall: We wanted to evaluate D 1 , 1 ( n ) using (DJC): D 1 , 1 ( n ) = D 0 , 0 ( n + 1) D 1 , 1 ( n − 1) + D 1 , 0 ( n ) D 0 , 1 ( n ) . D 0 , 0 ( n ) D 0 , 0 ( n ) � �� � = R 0 , 0 ( n ) Problem: We need to evaluate some other (simpler) determinants. For example, we can show that D 1 , 0 (2 n ) = 0 = D 0 , 1 (2 n ) for all n . ◮ Compute the (nontrivial) nullspace of D 0 , 1 (2 n ) for n � 15 . ◮ It has always dim. 1: ker( D 0 , 1 (2 n )) = � c n � for c n ∈ ◗ ( µ ) 2 n . 25 / 37

Back to D 1 , 1 ( n ) Recall: We wanted to evaluate D 1 , 1 ( n ) using (DJC): D 1 , 1 ( n ) = D 0 , 0 ( n + 1) D 1 , 1 ( n − 1) + D 1 , 0 ( n ) D 0 , 1 ( n ) . D 0 , 0 ( n ) D 0 , 0 ( n ) � �� � = R 0 , 0 ( n ) Problem: We need to evaluate some other (simpler) determinants. For example, we can show that D 1 , 0 (2 n ) = 0 = D 0 , 1 (2 n ) for all n . ◮ Compute the (nontrivial) nullspace of D 0 , 1 (2 n ) for n � 15 . ◮ It has always dim. 1: ker( D 0 , 1 (2 n )) = � c n � for c n ∈ ◗ ( µ ) 2 n . ◮ Normalize each generator c n (last component = 1 ). 25 / 37

Back to D 1 , 1 ( n ) Recall: We wanted to evaluate D 1 , 1 ( n ) using (DJC): D 1 , 1 ( n ) = D 0 , 0 ( n + 1) D 1 , 1 ( n − 1) + D 1 , 0 ( n ) D 0 , 1 ( n ) . D 0 , 0 ( n ) D 0 , 0 ( n ) � �� � = R 0 , 0 ( n ) Problem: We need to evaluate some other (simpler) determinants. For example, we can show that D 1 , 0 (2 n ) = 0 = D 0 , 1 (2 n ) for all n . ◮ Compute the (nontrivial) nullspace of D 0 , 1 (2 n ) for n � 15 . ◮ It has always dim. 1: ker( D 0 , 1 (2 n )) = � c n � for c n ∈ ◗ ( µ ) 2 n . ◮ Normalize each generator c n (last component = 1 ). ◮ “Guess” recurrence equations for the bivariate sequence c n,j . 25 / 37

Back to D 1 , 1 ( n ) Recall: We wanted to evaluate D 1 , 1 ( n ) using (DJC): D 1 , 1 ( n ) = D 0 , 0 ( n + 1) D 1 , 1 ( n − 1) + D 1 , 0 ( n ) D 0 , 1 ( n ) . D 0 , 0 ( n ) D 0 , 0 ( n ) � �� � = R 0 , 0 ( n ) Problem: We need to evaluate some other (simpler) determinants. For example, we can show that D 1 , 0 (2 n ) = 0 = D 0 , 1 (2 n ) for all n . ◮ Compute the (nontrivial) nullspace of D 0 , 1 (2 n ) for n � 15 . ◮ It has always dim. 1: ker( D 0 , 1 (2 n )) = � c n � for c n ∈ ◗ ( µ ) 2 n . ◮ Normalize each generator c n (last component = 1 ). ◮ “Guess” recurrence equations for the bivariate sequence c n,j . ◮ Use the holonomic systems approach (Zeilberger) to prove that D 0 , 1 (2 n ) · c n = 0 for all n . 25 / 37

The HOLONOMIC ANSATZ II. Automatic DISCOVERY(!) and PROOF(!!) of Holonomic Determinant Evaluations 26 / 37

The HOLONOMIC ANSATZ II. Automatic DISCOVERY(!) and PROOF(!!) of Holonomic Determinant Evaluations (D. Zeilberger, Annals of Combinatorics 11 :241–247, 2007) 26 / 37

The HOLONOMIC ANSATZ II. Automatic DISCOVERY(!) and PROOF(!!) of Holonomic Determinant Evaluations (D. Zeilberger, Annals of Combinatorics 11 :241–247, 2007) Algorithmic method to prove determinant evaluations of the form det A n = b n ( n � 1) where 26 / 37

The HOLONOMIC ANSATZ II. Automatic DISCOVERY(!) and PROOF(!!) of Holonomic Determinant Evaluations (D. Zeilberger, Annals of Combinatorics 11 :241–247, 2007) Algorithmic method to prove determinant evaluations of the form det A n = b n ( n � 1) where ◮ A n = ( a i,j ) 1 � i,j � n is an n × n matrix, 26 / 37

The HOLONOMIC ANSATZ II. Automatic DISCOVERY(!) and PROOF(!!) of Holonomic Determinant Evaluations (D. Zeilberger, Annals of Combinatorics 11 :241–247, 2007) Algorithmic method to prove determinant evaluations of the form det A n = b n ( n � 1) where ◮ A n = ( a i,j ) 1 � i,j � n is an n × n matrix, ◮ a i,j is a bivariate holonomic sequence, not depending on n , 26 / 37

The HOLONOMIC ANSATZ II. Automatic DISCOVERY(!) and PROOF(!!) of Holonomic Determinant Evaluations (D. Zeilberger, Annals of Combinatorics 11 :241–247, 2007) Algorithmic method to prove determinant evaluations of the form det A n = b n ( n � 1) where ◮ A n = ( a i,j ) 1 � i,j � n is an n × n matrix, ◮ a i,j is a bivariate holonomic sequence, not depending on n , linear recurrences polynomial coefficients finitely many initial values 26 / 37

The HOLONOMIC ANSATZ II. Automatic DISCOVERY(!) and PROOF(!!) of Holonomic Determinant Evaluations (D. Zeilberger, Annals of Combinatorics 11 :241–247, 2007) Algorithmic method to prove determinant evaluations of the form det A n = b n ( n � 1) where ◮ A n = ( a i,j ) 1 � i,j � n is an n × n matrix, ◮ a i,j is a bivariate holonomic sequence, not depending on n , ◮ b n � = 0 for all n � 1 . 26 / 37

Expansion Formula   a 1 , 1 a 1 , 2 a 1 , 3 · · · a 1 ,n           a 2 , 1 a 2 , 2 a 2 , 3 · · · a 2 ,n          a 3 , 1 a 3 , 2 a 3 , 3 · · · a 3 ,n        . . . .   . . . .  . . . .          a n, 1 a n, 2 a n, 3 · · · a n,n 27 / 37

Expansion Formula       ( A − 1 a 1 , 1 a 1 , 2 a 1 , 3 · · · a 1 ,n n ) 1 ,n 0                         ( A − 1       a 2 , 1 a 2 , 2 a 2 , 3 · · · a 2 ,n n ) 2 ,n 0                         ( A − 1 =  a 3 , 1 a 3 , 2 a 3 , 3 · · · a 3 ,n   n ) 3 ,n   0                    . . . . . .       . . . . . .  . . . .   .   .                          ( A − 1 a n, 1 a n, 2 a n, 3 · · · a n,n n ) n,n 1 27 / 37

Expansion Formula       ( − 1) n +1 det A ( n, 1) n a 1 , 1 a 1 , 2 a 1 , 3 · · · a 1 ,n 0       det A n             ( − 1) n +2 det A ( n, 2)       n       a 2 , 1 a 2 , 2 a 2 , 3 · · · a 2 ,n 0       det A n             ( − 1) n +3 det A ( n, 3)       n =  a 3 , 1 a 3 , 2 a 3 , 3 · · · a 3 ,n     0        det A n             . . . . . .       . . . . . .  . . . .   .   .                    ( − 1) 2 n det A ( n,n )       n a n, 1 a n, 2 a n, 3 · · · a n,n 1 det A n ◮ A ( i,j ) : matrix A n with row i and column j deleted n 27 / 37

Expansion Formula       ( − 1) n +1 M n, 1 a 1 , 1 a 1 , 2 a 1 , 3 · · · a 1 ,n 0       det A n                   ( − 1) n +2 M n, 2       a 2 , 1 a 2 , 2 a 2 , 3 · · · a 2 ,n 0       det A n                ( − 1) n +3 M n, 3    =  a 3 , 1 a 3 , 2 a 3 , 3 · · · a 3 ,n     0        det A n             . . . . . .       . . . . . .  . . . .   .   .                    M n,n       a n, 1 a n, 2 a n, 3 · · · a n,n 1 det A n ◮ A ( i,j ) : matrix A n with row i and column j deleted n ◮ M i,j = det A ( i,j ) is the ( i, j ) -minor of A n n 27 / 37

Expansion Formula       ( − 1) n +1 M n, 1 a 1 , 1 a 1 , 2 a 1 , 3 · · · a 1 ,n 0       M n,n                   ( − 1) n +2 M n, 2       a 2 , 1 a 2 , 2 a 2 , 3 · · · a 2 ,n 0       M n,n                ( − 1) n +3 M n, 3    =  a 3 , 1 a 3 , 2 a 3 , 3 · · · a 3 ,n     0        M n,n             . . . . . .       . . . . . .  . . . .   .   .                    det A n       a n, 1 a n, 2 a n, 3 · · · a n,n 1 M n,n ◮ A ( i,j ) : matrix A n with row i and column j deleted n ◮ M i,j = det A ( i,j ) is the ( i, j ) -minor of A n n 27 / 37

Expansion Formula       ( − 1) n +1 M n, 1 a 1 , 1 a 1 , 2 a 1 , 3 · · · a 1 ,n 0       M n,n                   ( − 1) n +2 M n, 2       a 2 , 1 a 2 , 2 a 2 , 3 · · · a 2 ,n 0       M n,n                ( − 1) n +3 M n, 3    =  a 3 , 1 a 3 , 2 a 3 , 3 · · · a 3 ,n     0        M n,n             . . . . . .       . . . . . .  . . . .   .   .                    det A n       a n, 1 a n, 2 a n, 3 · · · a n,n 1 det A n − 1 ◮ A ( i,j ) : matrix A n with row i and column j deleted n ◮ M i,j = det A ( i,j ) is the ( i, j ) -minor of A n n 27 / 37

Expansion Formula       a 1 , 1 a 1 , 2 a 1 , 3 · · · a 1 ,n c n, 1 0                               a 2 , 1 a 2 , 2 a 2 , 3 · · · a 2 ,n c n, 2 0                         =  a 3 , 1 a 3 , 2 a 3 , 3 · · · a 3 ,n   c n, 3   0                    . . . . . .       . . . . . .  . . . .   .   .                    det A n       a n, 1 a n, 2 a n, 3 · · · a n,n c n,n = 1 det A n − 1 ◮ A ( i,j ) : matrix A n with row i and column j deleted n ◮ M i,j = det A ( i,j ) is the ( i, j ) -minor of A n n ◮ Define c n,j := ( − 1) n + j M n,j /M n,n 27 / 37

Expansion Formula       a 1 , 1 a 1 , 2 a 1 , 3 · · · a 1 ,n c n, 1 0                               a 2 , 1 a 2 , 2 a 2 , 3 · · · a 2 ,n c n, 2 0                         =  a 3 , 1 a 3 , 2 a 3 , 3 · · · a 3 ,n   c n, 3   0                    . . . . . .       . . . . . .  . . . .   .   .                    det A n       a n, 1 a n, 2 a n, 3 · · · a n,n c n,n = 1 det A n − 1 ◮ A ( i,j ) : matrix A n with row i and column j deleted n ◮ M i,j = det A ( i,j ) is the ( i, j ) -minor of A n n ◮ Define c n,j := ( − 1) n + j M n,j /M n,n ◮ We obtain � n j =1 a i,j c n,j = δ i,n (det A n ) / (det A n − 1 ) 27 / 37

Determinant Evaluation: Proof by Induction . Problem: Prove that det A n = 1 � i,j � n a i,j = b n for all n ∈ ◆ . det 28 / 37

Determinant Evaluation: Proof by Induction . Problem: Prove that det A n = 1 � i,j � n a i,j = b n for all n ∈ ◆ . det Base case: verify that a 1 , 1 = b 1 . 28 / 37

Determinant Evaluation: Proof by Induction . Problem: Prove that det A n = 1 � i,j � n a i,j = b n for all n ∈ ◆ . det Base case: verify that a 1 , 1 = b 1 . Induction hypothesis: assume that det A n − 1 = b n − 1 � = 0 . 28 / 37

Determinant Evaluation: Proof by Induction . Problem: Prove that det A n = 1 � i,j � n a i,j = b n for all n ∈ ◆ . det Base case: verify that a 1 , 1 = b 1 . Induction hypothesis: assume that det A n − 1 = b n − 1 � = 0 . Induction step: the assumption implies that the linear system       a 1 , 1 · · · a 1 ,n − 1 a 1 ,n c n, 1 0 . . . . . ...  . . .   .   .  . . . . .        =       a n − 1 , 1 · · · a n − 1 ,n − 1 a n − 1 ,n c n,n − 1 0      0 · · · 0 1 c n,n 1 28 / 37

Determinant Evaluation: Proof by Induction . Problem: Prove that det A n = 1 � i,j � n a i,j = b n for all n ∈ ◆ . det Base case: verify that a 1 , 1 = b 1 . Induction hypothesis: assume that det A n − 1 = b n − 1 � = 0 . Induction step: the assumption implies that the linear system       a 1 , 1 · · · a 1 ,n − 1 a 1 ,n c n, 1 0 . . . . . ...  . . .   .   .  . . . . .        =       a n − 1 , 1 · · · a n − 1 ,n − 1 a n − 1 ,n c n,n − 1 0      0 · · · 0 1 c n,n 1 has a unique solution, namely c n,j = ( − 1) n + j M n,j /M n,n . 28 / 37

Determinant Evaluation: Proof by Induction . Problem: Prove that det A n = 1 � i,j � n a i,j = b n for all n ∈ ◆ . det Base case: verify that a 1 , 1 = b 1 . Induction hypothesis: assume that det A n − 1 = b n − 1 � = 0 . Induction step: the assumption implies that the linear system       a 1 , 1 · · · a 1 ,n − 1 a 1 ,n c n, 1 0 . . . . . ...  . . .   .   .  . . . . .        =       a n − 1 , 1 · · · a n − 1 ,n − 1 a n − 1 ,n c n,n − 1 0      0 · · · 0 1 c n,n 1 has a unique solution, namely c n,j = ( − 1) n + j M n,j /M n,n . Now use c n,j to do Laplace expansion of A n w.r.t. the last row: n n � � ( − 1) n + j M n,j a n,j = det A n = M n,n c n,j a n,j . � �� � j =1 j =1 = det A n − 1 28 / 37

Determinant Evaluation: Proof by Induction . Problem: Prove that det A n = 1 � i,j � n a i,j = b n for all n ∈ ◆ . det Base case: verify that a 1 , 1 = b 1 . Induction hypothesis: assume that det A n − 1 = b n − 1 � = 0 . Induction step: the assumption implies that the linear system       a 1 , 1 · · · a 1 ,n − 1 a 1 ,n c n, 1 0 . . . . . ...  . . .   .   .  . . . . .        =       a n − 1 , 1 · · · a n − 1 ,n − 1 a n − 1 ,n c n,n − 1 0      0 · · · 0 1 c n,n 1 has a unique solution, namely c n,j = ( − 1) n + j M n,j /M n,n . Now use c n,j to do Laplace expansion of A n w.r.t. the last row: n n � � ( − 1) n + j M n,j a n,j = det A n = M n,n c n,j a n,j . � �� � j =1 j =1 = det A n − 1 The induction step is completed by proving that this is equal to b n . 28 / 37

Recipe for the Holonomic Ansatz Problem: Given a i,j and b n � = 0 . Show that det ( a i,j ) 1 � i,j � n = b n . Method: “Pull out of the hat” a function c n,j and prove c n,n = 1 ( n � 1) , n � c n,j a i,j = 0 (1 � i < n ) , j =1 n b n � c n,j a n,j = ( n � 1) . b n − 1 j =1 Then det ( a i,j ) 1 � i,j � n = b n holds. 29 / 37

◆ The Magician’s Trick Problem: One cannot expect to be able to compute c n,j explicitly (at least not for symbolic n ) 30 / 37

◆ The Magician’s Trick Problem: One cannot expect to be able to compute c n,j explicitly (at least not for symbolic n ) Question: How can we define a candidate for the function c n,j ? 30 / 37

◆ The Magician’s Trick Problem: One cannot expect to be able to compute c n,j explicitly (at least not for symbolic n ) Question: How can we define a candidate for the function c n,j ? Solution: ◮ Hope that c n,j is holonomic (may be the case or not). 30 / 37

◆ The Magician’s Trick Problem: One cannot expect to be able to compute c n,j explicitly (at least not for symbolic n ) Question: How can we define a candidate for the function c n,j ? Solution: ◮ Hope that c n,j is holonomic (may be the case or not). ◮ Work with an implicit (recursive) definition of c n,j . 30 / 37

The Magician’s Trick Problem: One cannot expect to be able to compute c n,j explicitly (at least not for symbolic n ) Question: How can we define a candidate for the function c n,j ? Solution: ◮ Hope that c n,j is holonomic (may be the case or not). ◮ Work with an implicit (recursive) definition of c n,j . ◮ The values of c n,j can be computed for concrete n, j ∈ ◆ . 30 / 37

The Magician’s Trick Problem: One cannot expect to be able to compute c n,j explicitly (at least not for symbolic n ) Question: How can we define a candidate for the function c n,j ? Solution: ◮ Hope that c n,j is holonomic (may be the case or not). ◮ Work with an implicit (recursive) definition of c n,j . ◮ The values of c n,j can be computed for concrete n, j ∈ ◆ . ◮ If recurrences exist they can be guessed automatically 30 / 37

The Magician’s Trick Problem: One cannot expect to be able to compute c n,j explicitly (at least not for symbolic n ) Question: How can we define a candidate for the function c n,j ? Solution: ◮ Hope that c n,j is holonomic (may be the case or not). ◮ Work with an implicit (recursive) definition of c n,j . ◮ The values of c n,j can be computed for concrete n, j ∈ ◆ . ◮ If recurrences exist they can be guessed automatically Example: For D 0 , 0 (2 n ) we obtain the following holonomic system of recurrence relations for c n,j . 30 / 37

{ ( j + µ +2 n − 3)(2 µj 6 +8 nj 6 − 2 j 6 +3 µ 2 j 5 − 48 n 2 j 5 − 12 µj 5 − 24 nj 5 +9 j 5 + µ 3 j 4 + 48 n 3 j 4 − 11 µ 2 j 4 − 84 µn 2 j 4 + 204 n 2 j 4 + 21 µj 4 − 20 µ 2 nj 4 + 38 µnj 4 − 10 nj 4 − 11 j 4 + 216 n 4 j 3 − 2 µ 3 j 3 + 312 µn 3 j 3 − 408 n 3 j 3 + 7 µ 2 j 3 + 28 µ 2 n 2 j 3 + 122 µn 2 j 3 − 198 n 2 j 3 − 2 µj 3 − 9 µ 3 nj 3 + 68 µ 2 nj 3 − 113 µnj 3 + 78 nj 3 − 3 j 3 − 864 n 5 j 2 − 756 µn 4 j 2 + 432 n 4 j 2 − µ 3 j 2 − 112 µ 2 n 3 j 2 − 308 µn 3 j 2 + 600 n 3 j 2 + 11 µ 2 j 2 − 3 µ 3 n 2 j 2 − 66 µ 2 n 2 j 2 + 189 µn 2 j 2 − 168 n 2 j 2 − 23 µj 2 − 2 µ 4 nj 2 + 15 µ 3 nj 2 − 28 µ 2 nj 2 + 33 µnj 2 − 34 nj 2 + 13 j 2 + 864 n 6 j + 432 µn 5 j + 432 n 5 j − 144 µ 2 n 4 j + 1116 µn 4 j − 1104 n 4 j + 2 µ 3 j − 88 µ 3 n 3 j + 384 µ 2 n 3 j − 392 µn 3 j − 36 n 3 j − 10 µ 2 j − 14 µ 4 n 2 j + 45 µ 3 n 2 j + 40 µ 2 n 2 j − 317 µn 2 j + 270 n 2 j + 14 µj − µ 5 nj + 3 µ 4 nj + 17 µ 3 nj − 89 µ 2 nj + 112 µnj − 42 nj − 6 j + 432 µn 6 − 864 n 6 + 432 µ 2 n 5 − 1080 µn 5 +432 n 5 +144 µ 3 n 4 − 324 µ 2 n 4 − 156 µn 4 +456 n 4 +20 µ 4 n 3 − 18 µ 3 n 3 − 220 µ 2 n 3 + 470 µn 3 − 204 n 3 + µ 5 n 2 + 3 µ 4 n 2 − 37 µ 3 n 2 + 57 µ 2 n 2 + 36 µn 2 − 60 n 2 +2 µ 4 n − 18 µ 3 n +54 µ 2 n − 62 µn +24 n ) c n,j − ( j + µ − 3)(2 j + µ − 3)( j − 2 n + 1)( µ + 4 n − 1)( j 4 + 2 µj 3 − 6 j 3 + µ 2 j 2 − 12 n 2 j 2 − 9 µj 2 − 6 µnj 2 + 6 nj 2 + 13 j 2 − 3 µ 2 j − 12 µn 2 j + 36 n 2 j + 13 µj − 6 µ 2 nj + 24 µnj − 18 nj − 12 j + 2 µ 2 − 2 µ 2 n 2 + 20 µn 2 − 24 n 2 − 6 µ − µ 3 n + 11 µ 2 n − 22 µn + 12 n + 4) c n,j +1 + 2(2 j + µ − 2) n (2 n +1)( − j +2 n +1)( − j +2 n +2)( j + µ +2 n − 1)( µ +4 n − 3)( µ + 4 n − 1) c n +1 ,j , − ( j +1)(2 j + µ )( j − 2 n )( j + µ +2 n − 3) c n,j +(4 j 4 +8 µj 3 − 8 j 3 + 5 µ 2 j 2 − 8 n 2 j 2 − 5 µj 2 − 4 µnj 2 +12 nj 2 − 8 j 2 + µ 3 j +2 µ 2 j − 8 µn 2 j +8 n 2 j − 15 µj − 4 µ 2 nj + 16 µnj − 12 nj + 12 j + µ 3 − 3 µ 2 − 2 µ 2 n 2 + 16 n 2 − 2 µ − µ 3 n + 3 µ 2 n + 8 µn − 24 n +8) c n,j +1 − ( j + µ − 2)(2 j + µ − 2)( j − 2 n +2)( j + µ +2 n − 1) c n,j +2 } 31 / 37

{ ( j + µ +2 n − 3)(2 µj 6 +8 nj 6 − 2 j 6 +3 µ 2 j 5 − 48 n 2 j 5 − 12 µj 5 − 24 nj 5 +9 j 5 + µ 3 j 4 + 48 n 3 j 4 − 11 µ 2 j 4 − 84 µn 2 j 4 + 204 n 2 j 4 + 21 µj 4 − 20 µ 2 nj 4 + 38 µnj 4 − 10 nj 4 − 11 j 4 + 216 n 4 j 3 − 2 µ 3 j 3 + 312 µn 3 j 3 − 408 n 3 j 3 + 7 µ 2 j 3 + 28 µ 2 n 2 j 3 + 122 µn 2 j 3 − 198 n 2 j 3 − 2 µj 3 − 9 µ 3 nj 3 + 68 µ 2 nj 3 − 113 µnj 3 + 78 nj 3 − 3 j 3 − 864 n 5 j 2 − 756 µn 4 j 2 + 432 n 4 j 2 − µ 3 j 2 − 112 µ 2 n 3 j 2 − 308 µn 3 j 2 + 600 n 3 j 2 + 11 µ 2 j 2 − 3 µ 3 n 2 j 2 − 66 µ 2 n 2 j 2 + 189 µn 2 j 2 − 168 n 2 j 2 − 23 µj 2 − 2 µ 4 nj 2 + 15 µ 3 nj 2 − 28 µ 2 nj 2 + 33 µnj 2 − 34 nj 2 + 13 j 2 + 864 n 6 j + 432 µn 5 j + 432 n 5 j − 144 µ 2 n 4 j + 1116 µn 4 j − 1104 n 4 j + 2 µ 3 j − 88 µ 3 n 3 j + 384 µ 2 n 3 j − 392 µn 3 j − 36 n 3 j − 10 µ 2 j − 14 µ 4 n 2 j + 45 µ 3 n 2 j + 40 µ 2 n 2 j − 317 µn 2 j + 270 n 2 j + 14 µj − µ 5 nj + 3 µ 4 nj + 17 µ 3 nj − 89 µ 2 nj + 112 µnj − 42 nj − 6 j + 432 µn 6 − 864 n 6 + 432 µ 2 n 5 − 1080 µn 5 +432 n 5 +144 µ 3 n 4 − 324 µ 2 n 4 − 156 µn 4 +456 n 4 +20 µ 4 n 3 − 18 µ 3 n 3 − 220 µ 2 n 3 + 470 µn 3 − 204 n 3 + µ 5 n 2 + 3 µ 4 n 2 − 37 µ 3 n 2 + 57 µ 2 n 2 + 36 µn 2 − 60 n 2 +2 µ 4 n − 18 µ 3 n +54 µ 2 n − 62 µn +24 n ) c n,j − ( j + µ − 3)(2 j + µ − 3)( j − 2 n + 1)( µ + 4 n − 1)( j 4 + 2 µj 3 − 6 j 3 + µ 2 j 2 − 12 n 2 j 2 − 9 µj 2 − 6 µnj 2 + 6 nj 2 + 13 j 2 − 3 µ 2 j − 12 µn 2 j + 36 n 2 j + 13 µj − 6 µ 2 nj + 24 µnj − 18 nj − 12 j + 2 µ 2 − 2 µ 2 n 2 + 20 µn 2 − 24 n 2 − 6 µ − µ 3 n + 11 µ 2 n − 22 µn + 12 n + 4) c n,j +1 + 2(2 j + µ − 2) n (2 n +1)( − j +2 n +1)( − j +2 n +2)( j + µ +2 n − 1)( µ +4 n − 3)( µ + 4 n − 1) c n +1 ,j , − ( j +1)(2 j + µ )( j − 2 n )( j + µ +2 n − 3) c n,j +(4 j 4 +8 µj 3 − 8 j 3 + 5 µ 2 j 2 − 8 n 2 j 2 − 5 µj 2 − 4 µnj 2 +12 nj 2 − 8 j 2 + µ 3 j +2 µ 2 j − 8 µn 2 j +8 n 2 j − 15 µj − 4 µ 2 nj + 16 µnj − 12 nj + 12 j + µ 3 − 3 µ 2 − 2 µ 2 n 2 + 16 n 2 − 2 µ − µ 3 n + 3 µ 2 n + 8 µn − 24 n +8) c n,j +1 − ( j + µ − 2)(2 j + µ − 2)( j − 2 n +2)( j + µ +2 n − 1) c n,j +2 } 31 / 37

Back to D 1 , 1 ( n ) Using Zeilberger’s method, we obtain product formulas for the missing determinants. D 1 , 1 ( n ) = R 0 , 0 ( n ) D 1 , 1 ( n − 1) + D 1 , 0 ( n ) D 0 , 1 ( n ) . D 0 , 0 ( n ) Since D 0 , 1 ( n ) = D 1 , 0 ( n ) = 0 for even n , the recurrence simplifies: D 1 , 1 ( n ) = R 0 , 0 ( n ) D 1 , 1 ( n − 1) ( n even ) . For odd n we obtain D 1 , 1 ( n ) = �� n − 1 � �� n − 1 � j =1 R 1 , 0 ( j ) 2 j =1 R 0 , 1 ( j ) 2 = R 0 , 0 ( n ) D 1 , 1 ( n − 1) + ( µ − 1) 2 � n − 1 j =1 R 0 , 0 ( j ) ( n − 1) / 2 = R 0 , 0 ( n ) D 1 , 1 ( n − 1) + ( µ − 1) R 1 , 0 ( j ) R 0 , 1 ( j ) � R 0 , 0 (2 j − 1) R 0 , 0 (2 j ) . 2 j =1 32 / 37

Main Result Theorem. Let µ be an indeterminate and let ρ k be defined as ρ 0 ( a, b ) = a and ρ k ( a, b ) = b for k > 0 . If n is an odd positive integer then � � ( n +1) / 2 � � µ − 1 1 � 3 k − 2 D 1 , 1 ( n ) = ρ k 4( µ − 2) , � µ � 2 + k − 1 (2 k − 1)! 2 2 k =0 k − 1   � µ 2 � � � 2 + 2 j + 1 k − 1 µ + 2 j + 1 � 2 j − 1 j − 1 × � µ   � � � 2 + j + 1 j j − 1 2 j − 1 j =1   � µ � µ � � � � 2 ( n − 1) / 2 2 + 2 j − 1 2 + 2 j + 3 µ + 2 j � j 2 j 2 j +1 ×   � µ � � � � � 2 2 + j + 1 j j + 1 j = k 2 j j +1 j If n is an even positive integer then. . . [similar formula] 33 / 37

More Results We can give closed-form evaluations of some infinite one-dimensional families of D s,t ( n ) . t \ s · · · − 3 − 2 − 1 0 6 · · · 1 2 3 4 5 . . . . . . . . . . . . 6 D A C 5 F B E 4 D A C 3 F B E 2 D A C 1 F B E C E C E C · · · 0 D A B A B A B A · · · A ′ 0 − 1 0 0 0 0 0 0 · · · A ′ 0 − 2 0 0 0 0 0 0 0 · · · A ′ 0 − 3 0 0 0 0 0 0 0 0 · · · . ... ... . . . . . . . . . . ... . . . . . . . . . . . . . . . . . . . 34 / 37

Example of an Infinite Family (A) Family A: can be reduced to the base case D 0 , 0 ( n ) : � D 2 r, 0 ( n ) = D 0 , 0 ( n − 2 r ) � µ → µ +6 r 35 / 37

Example of an Infinite Family (A) Family A: can be reduced to the base case D 0 , 0 ( n ) : � D 2 r, 0 ( n ) = D 0 , 0 ( n − 2 r ) � µ → µ +6 r 35 / 37

Example of an Infinite Family (A) Family A: can be reduced to the base case D 0 , 0 ( n ) : � D 2 r, 0 ( n ) = D 0 , 0 ( n − 2 r ) � µ → µ +6 r 35 / 37