Closed Form Solutions of Linear Difference Equations Yongjae Cha - PowerPoint PPT Presentation

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions LbIK = z τ 2 + ( 2 + 2 v + 2 x ) τ − z Solutions: Modified Bessel functions of the first and second kind, I v + x ( z ) and K v + x ( − z ) LbJY = z τ 2 − ( 2 + 2 v + 2 x ) τ + z Solutions: Bessel functions of the first and second kind, J v + x ( z ) and Y v + x ( z ) LWW = τ 2 + ( z − 2 v − 2 x − 2 ) τ − v − x − 1 4 − v 2 − 2 vx − x 2 + n 2 Solution: Whittaker function W x , n ( z ) LWM = τ 2 ( 2 n + 2 v + 3 + 2 x ) + ( 2 z − 4 v − 4 x − 4 ) τ − 2 n + 1 + 2 v + 2 x Solution: Whittaker function M x , n ( z ) L 2 F 1 = ( z − 1 )( a + x + 1 ) τ 2 + ( − z + 2 − za − zx + 2 a + 2 x + zb − c ) τ − a + c − 1 − x Solution: Hypergeometric function 2 F 1 ( a + x , b ; c ; z ) ( 2 x + 3 + a + b )( a 2 − b 2 +( 2 x + a + b + 2 )( 2 x + 4 + a + b ) z ) Ljc = τ 2 − 1 τ + ( x + 1 + a )( x + 1 + b )( 2 x + 4 + a + b ) 2 ( x + 2 )( x + 2 + a + b )( 2 x + a + b + 2 ) ( x + 2 )( x + 2 + a + b )( 2 x + a + b + 2 ) Solution: Jacobian polynomial P a , b ( z ) x Lgd = τ 2 − ( 2 x + 3 ) z τ + x + 1 x + 2 x + 2 Solution: Legendre functions P x ( z ) and Q x ( z ) Lgr = τ 2 − 2 x + 3 + α − z τ + x + 1 + α x + 2 x + 2 Solution: Laguerre polynomial L ( α ) ( z ) x Lgb = τ 2 − 2 z ( m + x + 1 ) τ − 2 m + x x + 2 x + 2 Solution: Gegenbauer polynomial C m x ( z ) Lgr 1 = ( x + 2 ) τ 2 + ( x + z − b + 1 ) τ + z Solution: Laguerre polynomial L ( b − x ) ( z ) x Lkm = ( a + x + 1 ) τ 2 + ( − 2 a − 2 x − 2 + b − c ) τ + a + x + 1 − b Solution: Kummer’s function M ( a + x , b , c ) L 2 F 0 = τ 2 + ( − zb + zx + z + za − 1 ) τ + z ( b − x − 1 ) Solution: Hypergeometric function 2 F 0 ( a , b − x ; ; z ) Lge = ( x + 2 ) τ 2 + ( − ab − d + ( a + 1 )( 1 + x )) τ + ax − a ( b + d ) Solution: Sequences whose ordinary generating function is ( 1 + ax ) b ( 1 + bx ) d Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Questions Can we construct such table? 1 Yes How can we find the right base equation and the 2 parameter values? Local data Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Questions Can we construct such table? 1 Yes How can we find the right base equation and the 2 parameter values? Local data Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Main Algorithm Compute local data of L . 1 Compare the data with those in the table and find a base 2 equation that matches the data. If there is no such base equation then return ∅ . Compute candidate values for each parameters. 1 Construct a set cdd by plugging values found in step 1 to 2 corresponding parameters. For each L c ∈ cdd check if L ∼ gt L c and if so 3 Generate a basis of solutions or a solution of L c by plugging 1 in corresponding parameters. Apply the term transformation and the gauge 2 transformation to the result from 1. Return the result of step 2 as output and stop the algorithm. 3 Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Main Algorithm Compute local data of L . 1 Compare the data with those in the table and find a base 2 equation that matches the data. If there is no such base equation then return ∅ . Compute candidate values for each parameters. 1 Construct a set cdd by plugging values found in step 1 to 2 corresponding parameters. For each L c ∈ cdd check if L ∼ gt L c and if so 3 Generate a basis of solutions or a solution of L c by plugging 1 in corresponding parameters. Apply the term transformation and the gauge 2 transformation to the result from 1. Return the result of step 2 as output and stop the algorithm. 3 Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Main Algorithm Compute local data of L . 1 Compare the data with those in the table and find a base 2 equation that matches the data. If there is no such base equation then return ∅ . Compute candidate values for each parameters. 1 Construct a set cdd by plugging values found in step 1 to 2 corresponding parameters. For each L c ∈ cdd check if L ∼ gt L c and if so 3 Generate a basis of solutions or a solution of L c by plugging 1 in corresponding parameters. Apply the term transformation and the gauge 2 transformation to the result from 1. Return the result of step 2 as output and stop the algorithm. 3 Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Algorithm TryBessel : Input: L ∈ C ( x )[ τ ] Compute the local data of L v , z = z τ 2 + ( 2 + 2 v + 2 x ) τ − z 1 (Bessel recurrence). Compute local data of L that is invariant under ∼ gt . 2 Compare the local data of L v , z with that of L . 3 If compatible, compute v , z from this comparison. 4 Check if L ∼ gt L v , z , and if so, return solution(s). 5 Note: Step 1 is done only once, and then stored in a table. Remark: Checking L ∼ gt L v , z and computing the gt-transformation can only be done after we have found the values of the parameter v , z . Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Algorithm TryBessel : Input: L ∈ C ( x )[ τ ] Compute the local data of L v , z = z τ 2 + ( 2 + 2 v + 2 x ) τ − z 1 (Bessel recurrence). Compute local data of L that is invariant under ∼ gt . 2 Compare the local data of L v , z with that of L . 3 If compatible, compute v , z from this comparison. 4 Check if L ∼ gt L v , z , and if so, return solution(s). 5 Note: Step 1 is done only once, and then stored in a table. Remark: Checking L ∼ gt L v , z and computing the gt-transformation can only be done after we have found the values of the parameter v , z . Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Algorithm TryBessel : Input: L ∈ C ( x )[ τ ] Compute the local data of L v , z = z τ 2 + ( 2 + 2 v + 2 x ) τ − z 1 (Bessel recurrence). Compute local data of L that is invariant under ∼ gt . 2 Compare the local data of L v , z with that of L . 3 If compatible, compute v , z from this comparison. 4 Check if L ∼ gt L v , z , and if so, return solution(s). 5 Note: Step 1 is done only once, and then stored in a table. Remark: Checking L ∼ gt L v , z and computing the gt-transformation can only be done after we have found the values of the parameter v , z . Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Algorithm TryBessel : Input: L ∈ C ( x )[ τ ] Compute the local data of L v , z = z τ 2 + ( 2 + 2 v + 2 x ) τ − z 1 (Bessel recurrence). Compute local data of L that is invariant under ∼ gt . 2 Compare the local data of L v , z with that of L . 3 If compatible, compute v , z from this comparison. 4 Check if L ∼ gt L v , z , and if so, return solution(s). 5 Note: Step 1 is done only once, and then stored in a table. Remark: Checking L ∼ gt L v , z and computing the gt-transformation can only be done after we have found the values of the parameter v , z . Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Algorithm TryBessel : Input: L ∈ C ( x )[ τ ] Compute the local data of L v , z = z τ 2 + ( 2 + 2 v + 2 x ) τ − z 1 (Bessel recurrence). Compute local data of L that is invariant under ∼ gt . 2 Compare the local data of L v , z with that of L . 3 If compatible, compute v , z from this comparison. 4 Check if L ∼ gt L v , z , and if so, return solution(s). 5 Note: Step 1 is done only once, and then stored in a table. Remark: Checking L ∼ gt L v , z and computing the gt-transformation can only be done after we have found the values of the parameter v , z . Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Algorithm TryBessel : Input: L ∈ C ( x )[ τ ] Compute the local data of L v , z = z τ 2 + ( 2 + 2 v + 2 x ) τ − z 1 (Bessel recurrence). Compute local data of L that is invariant under ∼ gt . 2 Compare the local data of L v , z with that of L . 3 If compatible, compute v , z from this comparison. 4 Check if L ∼ gt L v , z , and if so, return solution(s). 5 Note: Step 1 is done only once, and then stored in a table. Remark: Checking L ∼ gt L v , z and computing the gt-transformation can only be done after we have found the values of the parameter v , z . Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Outline Difference Operator 1 Example 2 Transformations 3 Main Idea 4 Invariant Local Data 5 Finite Singularity Generalized Exponent 6 Liouvillian Special Functions 7 Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Invariant Local Data Question: If L ∼ gt L v , z , how to find v , z from L ? Need data that is invariant under ∼ gt Two sources Finite Singularities (valuation growths) 1 Singularity at ∞ (generalized exponents) 2 Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Invariant Local Data Question: If L ∼ gt L v , z , how to find v , z from L ? Need data that is invariant under ∼ gt Two sources Finite Singularities (valuation growths) 1 Singularity at ∞ (generalized exponents) 2 Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Invariant Local Data Question: If L ∼ gt L v , z , how to find v , z from L ? Need data that is invariant under ∼ gt Two sources Finite Singularities (valuation growths) 1 Singularity at ∞ (generalized exponents) 2 Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Finite Singularity: Valuation Growth Suppose L 1 ∼ g L 2 and G = r k ( x ) τ k + · · · + r 0 ( x ) , r i ( x ) ∈ C ( x ) Let u ( x ) = Γ( x ) ∈ V ( L 1 ) and v ( x ) = G ( u ( x )) is a non-zero element in V ( L 2 ) . Z -3 -2 -1 0 1 2 3 valuation of u ( x ) -1 -1 -1 -1 0 0 0 Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Finite Singularity: Valuation Growth Suppose L 1 ∼ g L 2 and G = r k ( x ) τ k + · · · + r 0 ( x ) , r i ( x ) ∈ C ( x ) Let u ( x ) = Γ( x ) ∈ V ( L 1 ) and v ( x ) = G ( u ( x )) is a non-zero element in V ( L 2 ) . Z -3 -2 -1 0 1 2 3 valuation of u ( x ) -1 -1 -1 -1 0 0 0 Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Finite Singularity: Valuation Growth Suppose L 1 ∼ g L 2 and G = r k ( x ) τ k + · · · + r 0 ( x ) , r i ( x ) ∈ C ( x ) Let u ( x ) = Γ( x ) ∈ V ( L 1 ) and v ( x ) = G ( u ( x )) is a non-zero element in V ( L 2 ) . Z -3 -2 -1 0 1 2 3 valuation of u ( x ) -1 -1 -1 -1 0 0 0 p l -3 p l -2 p l -1 p r +1 p r +2 p r +3 Z valuation of v ( x ) -1 -1 -1 0 0 0 Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Finite Singularity: Valuation Growth Suppose we have a difference equation a n ( x ) f ( x + n )+ a n − 1 ( x ) f ( x + n − 1 )+ · · · + a 0 ( x ) f ( x )+ a ( x ) = 0 a i ( x ) ∈ C [ x ] . To calculate f ( s + n ) with values of f ( s ) , . . . , f ( s + n − 1 ) , s ∈ C , f ( x + n ) = − a n − 1 ( x ) a n ( x ) f ( x + n − 1 ) − · · · − a 0 ( x ) a n ( x ) f ( x ) To calculate f ( s ) with values of f ( s + 1 ) , . . . , f ( s + n ) , s ∈ C , f ( x ) = − a n ( x ) a 0 ( x ) f ( x + n ) − · · · − a 1 ( x ) a 0 ( x ) f ( x + 1 ) Definition Let L = a n τ n + · · · + a 0 τ 0 with a i ∈ C [ x ] . q ∈ C is called a problem point of L if q is a root of the polynomial a 0 ( x ) a n ( x − n ) . p ∈ C / Z is called a finite singularity of L if it contains a problem point. Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Finite Singularity: Valuation Growth Suppose we have a difference equation a n ( x ) f ( x + n )+ a n − 1 ( x ) f ( x + n − 1 )+ · · · + a 0 ( x ) f ( x )+ a ( x ) = 0 a i ( x ) ∈ C [ x ] . To calculate f ( s + n ) with values of f ( s ) , . . . , f ( s + n − 1 ) , s ∈ C , f ( x + n ) = − a n − 1 ( x ) a n ( x ) f ( x + n − 1 ) − · · · − a 0 ( x ) a n ( x ) f ( x ) To calculate f ( s ) with values of f ( s + 1 ) , . . . , f ( s + n ) , s ∈ C , f ( x ) = − a n ( x ) a 0 ( x ) f ( x + n ) − · · · − a 1 ( x ) a 0 ( x ) f ( x + 1 ) Definition Let L = a n τ n + · · · + a 0 τ 0 with a i ∈ C [ x ] . q ∈ C is called a problem point of L if q is a root of the polynomial a 0 ( x ) a n ( x − n ) . p ∈ C / Z is called a finite singularity of L if it contains a problem point. Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Finite Singularity: Valuation Growth Suppose we have a difference equation a n ( x ) f ( x + n )+ a n − 1 ( x ) f ( x + n − 1 )+ · · · + a 0 ( x ) f ( x )+ a ( x ) = 0 a i ( x ) ∈ C [ x ] . To calculate f ( s + n ) with values of f ( s ) , . . . , f ( s + n − 1 ) , s ∈ C , f ( x + n ) = − a n − 1 ( x ) a n ( x ) f ( x + n − 1 ) − · · · − a 0 ( x ) a n ( x ) f ( x ) To calculate f ( s ) with values of f ( s + 1 ) , . . . , f ( s + n ) , s ∈ C , f ( x ) = − a n ( x ) a 0 ( x ) f ( x + n ) − · · · − a 1 ( x ) a 0 ( x ) f ( x + 1 ) Definition Let L = a n τ n + · · · + a 0 τ 0 with a i ∈ C [ x ] . q ∈ C is called a problem point of L if q is a root of the polynomial a 0 ( x ) a n ( x − n ) . p ∈ C / Z is called a finite singularity of L if it contains a problem point. Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Finite Singularity: Valuation Growth Suppose we have a difference equation a n ( x ) f ( x + n )+ a n − 1 ( x ) f ( x + n − 1 )+ · · · + a 0 ( x ) f ( x )+ a ( x ) = 0 a i ( x ) ∈ C [ x ] . To calculate f ( s + n ) with values of f ( s ) , . . . , f ( s + n − 1 ) , s ∈ C , f ( x + n ) = − a n − 1 ( x ) a n ( x ) f ( x + n − 1 ) − · · · − a 0 ( x ) a n ( x ) f ( x ) To calculate f ( s ) with values of f ( s + 1 ) , . . . , f ( s + n ) , s ∈ C , f ( x ) = − a n ( x ) a 0 ( x ) f ( x + n ) − · · · − a 1 ( x ) a 0 ( x ) f ( x + 1 ) Definition Let L = a n τ n + · · · + a 0 τ 0 with a i ∈ C [ x ] . q ∈ C is called a problem point of L if q is a root of the polynomial a 0 ( x ) a n ( x − n ) . p ∈ C / Z is called a finite singularity of L if it contains a problem point. Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Finite Singularity: Valuation Growth Definition Let u ( x ) ∈ C ( x ) be a non-zero meromorphic function. The valuation growth of u ( x ) at p = q + Z is n →∞ ( order of u ( x ) at x = n + q ) lim inf − lim inf n →∞ ( order of u ( x ) at x = − n + q ) Definition Let p ∈ C / Z and L be a difference operator. Then Min p ( L ) resp. Max p ( L ) is the minimum resp. maximum valuation growth at p , taken over all meromorphic solutions of L . Theorem If L 1 ∼ g L 2 then they have the same Min p , Max p for all p ∈ C / Z . Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Finite Singularity: Valuation Growth Definition Let u ( x ) ∈ C ( x ) be a non-zero meromorphic function. The valuation growth of u ( x ) at p = q + Z is n →∞ ( order of u ( x ) at x = n + q ) lim inf − lim inf n →∞ ( order of u ( x ) at x = − n + q ) Definition Let p ∈ C / Z and L be a difference operator. Then Min p ( L ) resp. Max p ( L ) is the minimum resp. maximum valuation growth at p , taken over all meromorphic solutions of L . Theorem If L 1 ∼ g L 2 then they have the same Min p , Max p for all p ∈ C / Z . Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Finite Singularity: Valuation Growth Theorem Max p − Min p is ∼ gt invariant for all p ∈ C / Z . Invariant data: Compute all p ∈ C / Z for which Max p � = Min p store [ p , Max p − Min p ] for all such p . Note: Since p ∈ C / Z and not in C , the parameters computed from such data are determined mod r Z for some r ∈ Q . Suppose we need parameter ν mod Z but find it mod 1 2 Z , then we need to check two cases. Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Finite Singularity: Valuation Growth Theorem Max p − Min p is ∼ gt invariant for all p ∈ C / Z . Invariant data: Compute all p ∈ C / Z for which Max p � = Min p store [ p , Max p − Min p ] for all such p . Note: Since p ∈ C / Z and not in C , the parameters computed from such data are determined mod r Z for some r ∈ Q . Suppose we need parameter ν mod Z but find it mod 1 2 Z , then we need to check two cases. Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Singularity at ∞ : Generalized Exponent Definition ∞ i � If τ − ct v ( 1 + r ) , with t = 1 / x , is right hand factor of L for a i t i = 1 some v ∈ 1 r Z , c ∈ C ∗ , a i ∈ C , r ∈ N , then the dominant term 1 ct v ( 1 + a 1 t r + · · · + a r t 1 ) is called a generalized exponent of L . We say two generalized exponents 1 g 1 = c 1 t v 1 ( 1 + a 1 t r + · · · + a r t 1 ) and 1 g 2 = c 2 t v 2 ( 1 + b 1 t r + · · · + b r t 1 ) are equivalent if c 1 = c 2 , v 1 = v 2 , a i = b i for i = 1 . . . r − 1 and a r ≡ b r mod 1 r Z and denote g 1 ∼ r g 2 Theorem Generalized exponents are invariant up to ∼ r under Gauge equivalence. Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Singularity at ∞ : Generalized Exponent Definition ∞ i � If τ − ct v ( 1 + r ) , with t = 1 / x , is right hand factor of L for a i t i = 1 some v ∈ 1 r Z , c ∈ C ∗ , a i ∈ C , r ∈ N , then the dominant term 1 ct v ( 1 + a 1 t r + · · · + a r t 1 ) is called a generalized exponent of L . We say two generalized exponents 1 g 1 = c 1 t v 1 ( 1 + a 1 t r + · · · + a r t 1 ) and 1 g 2 = c 2 t v 2 ( 1 + b 1 t r + · · · + b r t 1 ) are equivalent if c 1 = c 2 , v 1 = v 2 , a i = b i for i = 1 . . . r − 1 and a r ≡ b r mod 1 r Z and denote g 1 ∼ r g 2 Theorem Generalized exponents are invariant up to ∼ r under Gauge equivalence. Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Singularity at ∞ : Generalized Exponent Definition ∞ i � If τ − ct v ( 1 + r ) , with t = 1 / x , is right hand factor of L for a i t i = 1 some v ∈ 1 r Z , c ∈ C ∗ , a i ∈ C , r ∈ N , then the dominant term 1 ct v ( 1 + a 1 t r + · · · + a r t 1 ) is called a generalized exponent of L . We say two generalized exponents 1 g 1 = c 1 t v 1 ( 1 + a 1 t r + · · · + a r t 1 ) and 1 g 2 = c 2 t v 2 ( 1 + b 1 t r + · · · + b r t 1 ) are equivalent if c 1 = c 2 , v 1 = v 2 , a i = b i for i = 1 . . . r − 1 and a r ≡ b r mod 1 r Z and denote g 1 ∼ r g 2 Theorem Generalized exponents are invariant up to ∼ r under Gauge equivalence. Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Singularity at ∞ : Generalized Exponent Generalized exponents are not invariant under term-product. Definition Suppose ord ( L ) = 2 and let genexp ( L ) = { a 1 , a 2 } such that v ( a 1 ) ≥ v ( a 2 ) . Then we define the set of quotient of the two generalized exponents as if v ( a 1 ) > v ( a 2 ) � a 1 � Gquo ( L ) = and a 2 if v ( a 1 ) = v ( a 2 ) then we define � a 1 , a 2 � Gquo ( L ) = . a 2 a 1 Theorem If L 1 ∼ gt L 2 then Gquo ( L 1 ) = Gquo ( L 2 ) mod ∼ r Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Singularity at ∞ : Generalized Exponent Generalized exponents are not invariant under term-product. Definition Suppose ord ( L ) = 2 and let genexp ( L ) = { a 1 , a 2 } such that v ( a 1 ) ≥ v ( a 2 ) . Then we define the set of quotient of the two generalized exponents as if v ( a 1 ) > v ( a 2 ) � a 1 � Gquo ( L ) = and a 2 if v ( a 1 ) = v ( a 2 ) then we define � a 1 , a 2 � Gquo ( L ) = . a 2 a 1 Theorem If L 1 ∼ gt L 2 then Gquo ( L 1 ) = Gquo ( L 2 ) mod ∼ r Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Singularity at ∞ : Generalized Exponent Generalized exponents are not invariant under term-product. Definition Suppose ord ( L ) = 2 and let genexp ( L ) = { a 1 , a 2 } such that v ( a 1 ) ≥ v ( a 2 ) . Then we define the set of quotient of the two generalized exponents as if v ( a 1 ) > v ( a 2 ) � a 1 � Gquo ( L ) = and a 2 if v ( a 1 ) = v ( a 2 ) then we define � a 1 , a 2 � Gquo ( L ) = . a 2 a 1 Theorem If L 1 ∼ gt L 2 then Gquo ( L 1 ) = Gquo ( L 2 ) mod ∼ r Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Outline Difference Operator 1 Example 2 Transformations 3 Main Idea 4 Invariant Local Data 5 Finite Singularity Generalized Exponent 6 Liouvillian Special Functions 7 Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Liouvillian Solutions of Linear Difference Equations: Property Theorem (Hendriks Singer 1999) If L = a n τ n + · · · + a 0 τ 0 is irreducible then ∃ Liouvillian Solutions ⇐ ⇒ ∃ b 0 ∈ C ( x ) such that a n τ n + · · · + a 0 τ 0 τ n + b 0 τ 0 ∼ g Remark Operators of the form τ n + b 0 τ 0 are easy to solve, so if we know b 0 then we can solve L . Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Liouvillian Solutions of Linear Difference Equations: Property Theorem (Hendriks Singer 1999) If L = a n τ n + · · · + a 0 τ 0 is irreducible then ∃ Liouvillian Solutions ⇐ ⇒ ∃ b 0 ∈ C ( x ) such that a n τ n + · · · + a 0 τ 0 τ n + b 0 τ 0 ∼ g Remark Operators of the form τ n + b 0 τ 0 are easy to solve, so if we know b 0 then we can solve L . Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Liouvillian Solutions of Linear Difference Equations: The Problem Let L = a n τ n + · · · + a 0 τ 0 with a i ∈ C [ x ] and assume that L ∼ g τ n + b 0 τ 0 for some unknown b 0 ∈ C ( x ) . If we can find b 0 then we can solve τ n + b 0 τ 0 and hence solve L . Notation write b 0 = c φ where φ = monic poly monic poly and c ∈ C ∗ . Remark c is easy to compute, the main task is to compute φ . Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Liouvillian Solutions of Linear Difference Equations: The Problem Let L = a n τ n + · · · + a 0 τ 0 with a i ∈ C [ x ] and assume that L ∼ g τ n + b 0 τ 0 for some unknown b 0 ∈ C ( x ) . If we can find b 0 then we can solve τ n + b 0 τ 0 and hence solve L . Notation write b 0 = c φ where φ = monic poly monic poly and c ∈ C ∗ . Remark c is easy to compute, the main task is to compute φ . Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Liouvillian Solutions of Linear Difference Equations: The Problem Let L = a n τ n + · · · + a 0 τ 0 with a i ∈ C [ x ] and assume that L ∼ g τ n + b 0 τ 0 for some unknown b 0 ∈ C ( x ) . If we can find b 0 then we can solve τ n + b 0 τ 0 and hence solve L . Notation write b 0 = c φ where φ = monic poly monic poly and c ∈ C ∗ . Remark c is easy to compute, the main task is to compute φ . Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Liouvillian Solutions of Linear Difference Equations: Approach Remark Let L = a n τ n + · · · + a 0 τ 0 ∈ C [ x ][ τ ] then the finite singularities of L are Sing = { q + Z ∈ C / Z | q is root of a 0 a n } Theorem If q 1 + Z , . . . , q k + Z are the finite singularities then we may k n − 1 � � ( x − q i − j ) k i , j assume φ = with k i , j ∈ Z . i = 1 j = 0 At each finite singularity p i ∈ C / Z (where p i = q i + Z ) we 1 have to find n unknown exponents k i , 0 , . . . , k i , n − 1 . We can compute k i , 0 + · · · + k i , n − 1 from a 0 / a n . 2 Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Liouvillian Solutions of Linear Difference Equations: Approach Remark Let L = a n τ n + · · · + a 0 τ 0 ∈ C [ x ][ τ ] then the finite singularities of L are Sing = { q + Z ∈ C / Z | q is root of a 0 a n } Theorem If q 1 + Z , . . . , q k + Z are the finite singularities then we may k n − 1 � � ( x − q i − j ) k i , j assume φ = with k i , j ∈ Z . i = 1 j = 0 At each finite singularity p i ∈ C / Z (where p i = q i + Z ) we 1 have to find n unknown exponents k i , 0 , . . . , k i , n − 1 . We can compute k i , 0 + · · · + k i , n − 1 from a 0 / a n . 2 Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Liouvillian Solutions of Linear Difference Equations: Approach Remark Let L = a n τ n + · · · + a 0 τ 0 ∈ C [ x ][ τ ] then the finite singularities of L are Sing = { q + Z ∈ C / Z | q is root of a 0 a n } Theorem If q 1 + Z , . . . , q k + Z are the finite singularities then we may k n − 1 � � ( x − q i − j ) k i , j assume φ = with k i , j ∈ Z . i = 1 j = 0 At each finite singularity p i ∈ C / Z (where p i = q i + Z ) we 1 have to find n unknown exponents k i , 0 , . . . , k i , n − 1 . We can compute k i , 0 + · · · + k i , n − 1 from a 0 / a n . 2 Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Liouvillian Solutions of Linear Difference Equations: Approach Remark Let L = a n τ n + · · · + a 0 τ 0 ∈ C [ x ][ τ ] then the finite singularities of L are Sing = { q + Z ∈ C / Z | q is root of a 0 a n } Theorem If q 1 + Z , . . . , q k + Z are the finite singularities then we may k n − 1 � � ( x − q i − j ) k i , j assume φ = with k i , j ∈ Z . i = 1 j = 0 At each finite singularity p i ∈ C / Z (where p i = q i + Z ) we 1 have to find n unknown exponents k i , 0 , . . . , k i , n − 1 . We can compute k i , 0 + · · · + k i , n − 1 from a 0 / a n . 2 Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Example of Operator of order 2 with one finite singularity at p = Z Suppose L = a 2 τ 2 + a 1 τ + a 0 and that L ∼ g τ 2 + c · x k 0 ( x − 1 ) k 1 c can be computed from a 0 / a 2 1 k 0 + k 1 can be computed from a 0 / a 2 2 max { k 0 , k 1 } = Max Z ( L ) 3 min { k 0 , k 1 } = Min Z ( L ) 4 Items 2, 3, 4 determine k 0 , k 1 up to a permutation. Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Example of Operator of order 2 with one finite singularity at p = Z Suppose L = a 2 τ 2 + a 1 τ + a 0 and that L ∼ g τ 2 + c · x k 0 ( x − 1 ) k 1 c can be computed from a 0 / a 2 1 k 0 + k 1 can be computed from a 0 / a 2 2 max { k 0 , k 1 } = Max Z ( L ) 3 min { k 0 , k 1 } = Min Z ( L ) 4 Items 2, 3, 4 determine k 0 , k 1 up to a permutation. Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Example of Operator of order 2 with one finite singularity at p = Z Suppose L = a 2 τ 2 + a 1 τ + a 0 and that L ∼ g τ 2 + c · x k 0 ( x − 1 ) k 1 c can be computed from a 0 / a 2 1 k 0 + k 1 can be computed from a 0 / a 2 2 max { k 0 , k 1 } = Max Z ( L ) 3 min { k 0 , k 1 } = Min Z ( L ) 4 Items 2, 3, 4 determine k 0 , k 1 up to a permutation. Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Example from The On-Line Encyclopedia of Integer Sequences TM (OEIS TM ) A000246 =(1, 1, 1, 3, 9, 45, 225, 1575, 11025, 99225,...) Number of permutations in the symmetric group S n that have odd order. τ 2 − τ − x ( x + 1 ) Sing = { Z } and c = 1. At Z , min = 0 , max = 2 , sum = 2 So the exponents of x ··· ( x − 1 ) ··· must be a permutation of 0 , 2 Candidates of c φ are x 2 and ( x − 1 ) 2 . Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Example from The On-Line Encyclopedia of Integer Sequences TM (OEIS TM ) A000246 =(1, 1, 1, 3, 9, 45, 225, 1575, 11025, 99225,...) Number of permutations in the symmetric group S n that have odd order. τ 2 − τ − x ( x + 1 ) Sing = { Z } and c = 1. At Z , min = 0 , max = 2 , sum = 2 So the exponents of x ··· ( x − 1 ) ··· must be a permutation of 0 , 2 Candidates of c φ are x 2 and ( x − 1 ) 2 . Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Example from The On-Line Encyclopedia of Integer Sequences TM (OEIS TM ) τ 2 − x 2 is gauge equivalent to L Gauge transformation is τ + x . Basis of solutions of τ 2 − x 2 is { 2 x Γ( 1 2 x ) 2 , ( − 2 ) x Γ( 1 2 x ) 2 } Thus, Basis of solutions of L is { x 2 x Γ( 1 2 x ) 2 + 2 x + 1 Γ( 1 2 x + 1 2 ) 2 , x ( − 2 ) x Γ( 1 2 x ) 2 +( − 2 ) x + 1 Γ( 1 2 x + 1 2 ) 2 } Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Example from The On-Line Encyclopedia of Integer Sequences TM (OEIS TM ) τ 2 − x 2 is gauge equivalent to L Gauge transformation is τ + x . Basis of solutions of τ 2 − x 2 is { 2 x Γ( 1 2 x ) 2 , ( − 2 ) x Γ( 1 2 x ) 2 } Thus, Basis of solutions of L is { x 2 x Γ( 1 2 x ) 2 + 2 x + 1 Γ( 1 2 x + 1 2 ) 2 , x ( − 2 ) x Γ( 1 2 x ) 2 +( − 2 ) x + 1 Γ( 1 2 x + 1 2 ) 2 } Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Example from The On-Line Encyclopedia of Integer Sequences TM (OEIS TM ) τ 2 − x 2 is gauge equivalent to L Gauge transformation is τ + x . Basis of solutions of τ 2 − x 2 is { 2 x Γ( 1 2 x ) 2 , ( − 2 ) x Γ( 1 2 x ) 2 } Thus, Basis of solutions of L is { x 2 x Γ( 1 2 x ) 2 + 2 x + 1 Γ( 1 2 x + 1 2 ) 2 , x ( − 2 ) x Γ( 1 2 x ) 2 +( − 2 ) x + 1 Γ( 1 2 x + 1 2 ) 2 } Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Example of Operator of order 3 with one finite singularity at p = Z Suppose L = a 3 τ 3 + a 2 τ 2 + a 1 τ + a 0 and that L ∼ g τ 3 + c · x k 0 ( x − 1 ) k 1 ( x − 2 ) k 2 c can be computed from a 0 / a 3 1 k 0 + k 1 + k 2 can be computed from a 0 / a 3 2 max { k 0 , k 1 , k 2 } = Max Z ( L ) 3 min { k 0 , k 1 , k 2 } = Min Z ( L ) 4 Items 2, 3, 4 determine k 0 , k 1 , k 2 up to a permutation. Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Example of Operator of order 3 with one finite singularity at p = Z Suppose L = a 3 τ 3 + a 2 τ 2 + a 1 τ + a 0 and that L ∼ g τ 3 + c · x k 0 ( x − 1 ) k 1 ( x − 2 ) k 2 c can be computed from a 0 / a 3 1 k 0 + k 1 + k 2 can be computed from a 0 / a 3 2 max { k 0 , k 1 , k 2 } = Max Z ( L ) 3 min { k 0 , k 1 , k 2 } = Min Z ( L ) 4 Items 2, 3, 4 determine k 0 , k 1 , k 2 up to a permutation. Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Example of Operator of order 3 with one finite singularity at p = Z Suppose L = a 3 τ 3 + a 2 τ 2 + a 1 τ + a 0 and that L ∼ g τ 3 + c · x k 0 ( x − 1 ) k 1 ( x − 2 ) k 2 c can be computed from a 0 / a 3 1 k 0 + k 1 + k 2 can be computed from a 0 / a 3 2 max { k 0 , k 1 , k 2 } = Max Z ( L ) 3 min { k 0 , k 1 , k 2 } = Min Z ( L ) 4 Items 2, 3, 4 determine k 0 , k 1 , k 2 up to a permutation. Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Example with two finite singularities at Z and 1 2 + Z Suppose L = a 3 τ 3 + a 2 τ 2 + a 1 τ + a 0 is gauge equivalent to τ 3 + c · x k 0 ( x − 1 ) k 1 ( x − 2 ) k 2 · ( x − 1 2 ) l 0 ( x − 3 2 ) l 1 ( x − 5 2 ) l 2 c , k 0 + k 1 + k 2 , and l 0 + l 1 + l 2 can be computed from a 0 / a 3 1 min { k 0 , k 1 , k 2 } = Min Z ( L ) 2 max { k 0 , k 1 , k 2 } = Max Z ( L ) 3 min { l 0 , l 1 , l 2 } = Min 1 2 + Z ( L ) 4 max { l 0 , l 1 , l 2 } = Max 1 2 + Z ( L ) 5 This determines k 0 , k 1 , k 2 up to a permutation, and also l 0 , l 1 , l 2 up to a permutation. Worst case is 3 ! · 3 ! combinations (actually: 1/3 of that). Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Example with two finite singularities at Z and 1 2 + Z Suppose L = a 3 τ 3 + a 2 τ 2 + a 1 τ + a 0 is gauge equivalent to τ 3 + c · x k 0 ( x − 1 ) k 1 ( x − 2 ) k 2 · ( x − 1 2 ) l 0 ( x − 3 2 ) l 1 ( x − 5 2 ) l 2 c , k 0 + k 1 + k 2 , and l 0 + l 1 + l 2 can be computed from a 0 / a 3 1 min { k 0 , k 1 , k 2 } = Min Z ( L ) 2 max { k 0 , k 1 , k 2 } = Max Z ( L ) 3 min { l 0 , l 1 , l 2 } = Min 1 2 + Z ( L ) 4 max { l 0 , l 1 , l 2 } = Max 1 2 + Z ( L ) 5 This determines k 0 , k 1 , k 2 up to a permutation, and also l 0 , l 1 , l 2 up to a permutation. Worst case is 3 ! · 3 ! combinations (actually: 1/3 of that). Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Example with two finite singularities at Z and 1 2 + Z Suppose L = a 3 τ 3 + a 2 τ 2 + a 1 τ + a 0 is gauge equivalent to τ 3 + c · x k 0 ( x − 1 ) k 1 ( x − 2 ) k 2 · ( x − 1 2 ) l 0 ( x − 3 2 ) l 1 ( x − 5 2 ) l 2 c , k 0 + k 1 + k 2 , and l 0 + l 1 + l 2 can be computed from a 0 / a 3 1 min { k 0 , k 1 , k 2 } = Min Z ( L ) 2 max { k 0 , k 1 , k 2 } = Max Z ( L ) 3 min { l 0 , l 1 , l 2 } = Min 1 2 + Z ( L ) 4 max { l 0 , l 1 , l 2 } = Max 1 2 + Z ( L ) 5 This determines k 0 , k 1 , k 2 up to a permutation, and also l 0 , l 1 , l 2 up to a permutation. Worst case is 3 ! · 3 ! combinations (actually: 1/3 of that). Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Example with two finite singularities at Z and 1 2 + Z Suppose L = a 3 τ 3 + a 2 τ 2 + a 1 τ + a 0 is gauge equivalent to τ 3 + c · x k 0 ( x − 1 ) k 1 ( x − 2 ) k 2 · ( x − 1 2 ) l 0 ( x − 3 2 ) l 1 ( x − 5 2 ) l 2 c , k 0 + k 1 + k 2 , and l 0 + l 1 + l 2 can be computed from a 0 / a 3 1 min { k 0 , k 1 , k 2 } = Min Z ( L ) 2 max { k 0 , k 1 , k 2 } = Max Z ( L ) 3 min { l 0 , l 1 , l 2 } = Min 1 2 + Z ( L ) 4 max { l 0 , l 1 , l 2 } = Max 1 2 + Z ( L ) 5 This determines k 0 , k 1 , k 2 up to a permutation, and also l 0 , l 1 , l 2 up to a permutation. Worst case is 3 ! · 3 ! combinations (actually: 1/3 of that). Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Liouvillian Solutions of Linear Difference Equations: Example L = x τ 3 + τ 2 − ( x + 1 ) τ − x ( x + 1 ) 2 ( 2 x − 1 ) Sing = { Z , 1 2 + Z } and c = − 2. At Z , min = 0 , max = 1 , sum = 2 So the exponents of x ··· ( x − 1 ) ··· ( x − 2 ) ··· must be a permutation of 0 , 1 , 1 At 1 2 + Z , min = 0 , max = 1 , sum = 1 2 ) ··· must be a So the exponents of ( x − 1 2 ) ··· ( x − 3 2 ) ··· ( x − 5 permutation of 0 , 0 , 1 Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Liouvillian Solutions of Linear Difference Equations: Example L = x τ 3 + τ 2 − ( x + 1 ) τ − x ( x + 1 ) 2 ( 2 x − 1 ) Sing = { Z , 1 2 + Z } and c = − 2. At Z , min = 0 , max = 1 , sum = 2 So the exponents of x ··· ( x − 1 ) ··· ( x − 2 ) ··· must be a permutation of 0 , 1 , 1 At 1 2 + Z , min = 0 , max = 1 , sum = 1 2 ) ··· must be a So the exponents of ( x − 1 2 ) ··· ( x − 3 2 ) ··· ( x − 5 permutation of 0 , 0 , 1 Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Liouvillian Solutions of Linear Difference Equations: Example L = x τ 3 + τ 2 − ( x + 1 ) τ − x ( x + 1 ) 2 ( 2 x − 1 ) Sing = { Z , 1 2 + Z } and c = − 2. At Z , min = 0 , max = 1 , sum = 2 So the exponents of x ··· ( x − 1 ) ··· ( x − 2 ) ··· must be a permutation of 0 , 1 , 1 At 1 2 + Z , min = 0 , max = 1 , sum = 1 2 ) ··· must be a So the exponents of ( x − 1 2 ) ··· ( x − 3 2 ) ··· ( x − 5 permutation of 0 , 0 , 1 Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Liouvillian Solutions of Linear Difference Equations: Example L = x τ 3 + τ 2 − ( x + 1 ) τ − x ( x + 1 ) 2 ( 2 x − 1 ) Candidates of c φ are − 2 x 1 ( x − 1 ) 1 ( x − 2 ) 0 ( x − 1 / 2 ) 1 ( x − 3 / 2 ) 0 ( x − 5 / 2 ) 0 1 − 2 x 1 ( x − 1 ) 1 ( x − 2 ) 0 ( x − 1 / 2 ) 0 ( x − 3 / 2 ) 1 ( x − 5 / 2 ) 0 2 − 2 x 1 ( x − 1 ) 1 ( x − 2 ) 0 ( x − 1 / 2 ) 0 ( x − 3 / 2 ) 0 ( x − 5 / 2 ) 1 3 − 2 x 0 ( x − 1 ) 1 ( x − 2 ) 1 ( x − 1 / 2 ) 0 ( x − 3 / 2 ) 0 ( x − 5 / 2 ) 1 4 − 2 x 0 ( x − 1 ) 1 ( x − 2 ) 1 ( x − 1 / 2 ) 0 ( x − 3 / 2 ) 1 ( x − 5 / 2 ) 0 5 − 2 x 0 ( x − 1 ) 1 ( x − 2 ) 1 ( x − 1 / 2 ) 1 ( x − 3 / 2 ) 0 ( x − 5 / 2 ) 0 6 − 2 x 1 ( x − 1 ) 0 ( x − 2 ) 1 ( x − 1 / 2 ) 1 ( x − 3 / 2 ) 0 ( x − 5 / 2 ) 0 7 − 2 x 1 ( x − 1 ) 0 ( x − 2 ) 1 ( x − 1 / 2 ) 0 ( x − 3 / 2 ) 0 ( x − 5 / 2 ) 1 8 − 2 x 1 ( x − 1 ) 0 ( x − 2 ) 1 ( x − 1 / 2 ) 0 ( x − 3 / 2 ) 1 ( x − 5 / 2 ) 0 9 Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Liouvillian Solutions of Linear Difference Equations: Example L = x τ 3 + τ 2 − ( x + 1 ) τ − x ( x + 1 ) 2 ( 2 x − 1 ) Remark τ n − c φ ∼ g τ n − c τ k ( φ ) for k = 1 . . . n − 1 − 2 x 1 ( x − 1 ) 1 ( x − 2 ) 0 ( x − 1 / 2 ) 1 ( x − 3 / 2 ) 0 ( x − 5 / 2 ) 0 1 − 2 x 1 ( x − 1 ) 1 ( x − 2 ) 0 ( x − 1 / 2 ) 0 ( x − 3 / 2 ) 1 ( x − 5 / 2 ) 0 2 − 2 x 1 ( x − 1 ) 1 ( x − 2 ) 0 ( x − 1 / 2 ) 0 ( x − 3 / 2 ) 0 ( x − 5 / 2 ) 1 3 Only need to try 1, 2, 3, the others are redundant. Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Liouvillian Solutions of Linear Difference Equations: Example L = x τ 3 + τ 2 − ( x + 1 ) τ − x ( x + 1 ) 2 ( 2 x − 1 ) Remark τ n − c φ ∼ g τ n − c τ k ( φ ) for k = 1 . . . n − 1 − 2 x 1 ( x − 1 ) 1 ( x − 2 ) 0 ( x − 1 / 2 ) 1 ( x − 3 / 2 ) 0 ( x − 5 / 2 ) 0 1 − 2 x 1 ( x − 1 ) 1 ( x − 2 ) 0 ( x − 1 / 2 ) 0 ( x − 3 / 2 ) 1 ( x − 5 / 2 ) 0 2 − 2 x 1 ( x − 1 ) 1 ( x − 2 ) 0 ( x − 1 / 2 ) 0 ( x − 3 / 2 ) 0 ( x − 5 / 2 ) 1 3 Only need to try 1, 2, 3, the others are redundant. Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Liouvillian Solutions of Linear Difference Equations: Example L = x τ 3 + τ 2 − ( x + 1 ) τ − x ( x + 1 ) 2 ( 2 x − 1 ) τ 3 − 2 x ( x − 1 )( x − 1 / 2 ) is gauge equivalent to L Gauge transformation is τ + x − 1. Basis of solutions of τ 3 − 2 x ( x − 1 )( x − 1 / 2 ) is { ( ξ k ) x v ( x ) } for k = 0 . . . 2 x − 1 3 ) and ξ 3 = 1. where v ( x ) = 3 x 2 x / 3 Γ( x 3 )Γ( x − 1 3 )Γ( 2 Thus, Basis of solutions of L is { ( ξ k ) x + 1 v ( x + 1 ) + ( x − 1 )( ξ k ) x v ( x ) } for k = 0 . . . 2 Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Liouvillian Solutions of Linear Difference Equations: Example L = x τ 3 + τ 2 − ( x + 1 ) τ − x ( x + 1 ) 2 ( 2 x − 1 ) τ 3 − 2 x ( x − 1 )( x − 1 / 2 ) is gauge equivalent to L Gauge transformation is τ + x − 1. Basis of solutions of τ 3 − 2 x ( x − 1 )( x − 1 / 2 ) is { ( ξ k ) x v ( x ) } for k = 0 . . . 2 x − 1 3 ) and ξ 3 = 1. where v ( x ) = 3 x 2 x / 3 Γ( x 3 )Γ( x − 1 3 )Γ( 2 Thus, Basis of solutions of L is { ( ξ k ) x + 1 v ( x + 1 ) + ( x − 1 )( ξ k ) x v ( x ) } for k = 0 . . . 2 Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Liouvillian Solutions of Linear Difference Equations: Example L = x τ 3 + τ 2 − ( x + 1 ) τ − x ( x + 1 ) 2 ( 2 x − 1 ) τ 3 − 2 x ( x − 1 )( x − 1 / 2 ) is gauge equivalent to L Gauge transformation is τ + x − 1. Basis of solutions of τ 3 − 2 x ( x − 1 )( x − 1 / 2 ) is { ( ξ k ) x v ( x ) } for k = 0 . . . 2 x − 1 3 ) and ξ 3 = 1. where v ( x ) = 3 x 2 x / 3 Γ( x 3 )Γ( x − 1 3 )Γ( 2 Thus, Basis of solutions of L is { ( ξ k ) x + 1 v ( x + 1 ) + ( x − 1 )( ξ k ) x v ( x ) } for k = 0 . . . 2 Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Outline Difference Operator 1 Example 2 Transformations 3 Main Idea 4 Invariant Local Data 5 Finite Singularity Generalized Exponent 6 Liouvillian Special Functions 7 Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions LbIK = z τ 2 + ( 2 + 2 v + 2 x ) τ − z Solutions: Modified Bessel functions of the first and second kind, I v + x ( z ) and K v + x ( − z ) LbJY = z τ 2 − ( 2 + 2 v + 2 x ) τ + z Solutions: Bessel functions of the first and second kind, J v + x ( z ) and Y v + x ( z ) LWW = τ 2 + ( z − 2 v − 2 x − 2 ) τ − v − x − 1 4 − v 2 − 2 vx − x 2 + n 2 Solution: Whittaker function W x , n ( z ) LWM = τ 2 ( 2 n + 2 v + 3 + 2 x ) + ( 2 z − 4 v − 4 x − 4 ) τ − 2 n + 1 + 2 v + 2 x Solution: Whittaker function M x , n ( z ) L 2 F 1 = ( z − 1 )( a + x + 1 ) τ 2 + ( − z + 2 − za − zx + 2 a + 2 x + zb − c ) τ − a + c − 1 − x Solution: Hypergeometric function 2 F 1 ( a + x , b ; c ; z ) ( 2 x + 3 + a + b )( a 2 − b 2 +( 2 x + a + b + 2 )( 2 x + 4 + a + b ) z ) Ljc = τ 2 − 1 τ + ( x + 1 + a )( x + 1 + b )( 2 x + 4 + a + b ) 2 ( x + 2 )( x + 2 + a + b )( 2 x + a + b + 2 ) ( x + 2 )( x + 2 + a + b )( 2 x + a + b + 2 ) Solution: Jacobian polynomial P a , b ( z ) x Lgd = τ 2 − ( 2 x + 3 ) z τ + x + 1 x + 2 x + 2 Solution: Legendre functions P x ( z ) and Q x ( z ) Lgr = τ 2 − 2 x + 3 + α − z τ + x + 1 + α x + 2 x + 2 Solution: Laguerre polynomial L ( α ) ( z ) x Lgb = τ 2 − 2 z ( m + x + 1 ) τ − 2 m + x x + 2 x + 2 Solution: Gegenbauer polynomial C m x ( z ) Lgr 1 = ( x + 2 ) τ 2 + ( x + z − b + 1 ) τ + z Solution: Laguerre polynomial L ( b − x ) ( z ) x Lkm = ( a + x + 1 ) τ 2 + ( − 2 a − 2 x − 2 + b − c ) τ + a + x + 1 − b Solution: Kummer’s function M ( a + x , b , c ) L 2 F 0 = τ 2 + ( − zb + zx + z + za − 1 ) τ + z ( b − x − 1 ) Solution: Hypergeometric function 2 F 0 ( a , b − x ; ; z ) Lge = ( x + 2 ) τ 2 + ( − ab − d + ( a + 1 )( 1 + x )) τ + ax − a ( b + d ) Solution: Sequences whose ordinary generating function is ( 1 + ax ) b ( 1 + x ) d Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Special Functions: Functions and their Local Data Operator Val Gquo {− 1 4 T 2 z 2 ( 1 − ( 1 + 2 v ) T ) } LbIK {} { 1 4 T 2 z 2 ( 1 − ( 1 + 2 v ) T ) } LbJY {} √ √ √ √ { [ − n + 1 2 − v , 1 ] , [ n + 1 2 ( 1 − 1 2 ( 1 + 1 LWW 2 − v , 1 ] } {− 3 − 2 2 z ) T , − 3 + 2 2 z ) T } 2 2 { 1 − 2 √− zT − 2 zT 2 , 1 + 2 √− zT − 2 zT 2 } { [ − n + 1 2 − v , 1 ] , [ n + 1 LWM 2 − v , 1 ] } 1 L 2 F 1 { [ − a + c , 1 ] , [ − a , 1 ] } {− z − 1 ( 1 + ( 2 b − c ) T ) , ( − z + 1 )( 1 + ( − 2 b + c ) T ) } { 2 z 2 − 2 z z 2 − 1 − 1 , 2 z 2 + 2 z p p z 2 − 1 − 1 } Ljc { [ 0 , 1 ] , [ − a , 1 ] , [ − b , 1 ] , [ − a − b , 1 ] } { 2 z 2 − 2 z z 2 − 1 − 1 , 2 z 2 + 2 z p p z 2 − 1 − 1 } Lgd { [ 0 , 2 ] } { 1 + 2 √− zT − 2 zT 2 , 1 − 2 √− zT − 2 zT 2 } Lgr { [ 0 , 1 ] , [ − α, 1 ] } Lgr 1 { [0,1] } { zT ( 1 + 2 bT ) } z 2 + 1 − 2 z 2 − 1 , 2 z z 2 + 1 − 2 z 2 − 1 } p p Lgb { [ 0 , 1 ] , [ − 2 m , 1 ] } {− 2 z { 1 − 2 √ cT + 2 cT 2 , 1 + 2 √ cT + 2 cT 2 } Lkm { [ − a , 1 ] , [ − a + b , 1 ] } { T L 2 F 0 { [ b , 1 ] } z ( 1 + ( b − 2 a ) T ) } { a ( 1 + ( d − b ) T ) , 1 Lge { [ 0 , 1 ] , [ b + d , 1 ] } a ( 1 + ( − b − d ) T ) } Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Effectiveness of solver Found 10,659 sequences in OEIS TM that satisfy a second order recurrence but not a first order recurrence. 9,455 were reducible 161 irreducible Liouvillian 86 Bessel 330 Legendre 374 Hermite 21 Jacobi 8 Kummer 44 Laguerre 7 2 F 1 14 2 F 0 77 Generating function ( 1 + x ) a ( 1 + bx ) c 82 Not yet solved Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Example from The On-Line Encyclopedia of Integer Sequences TM (OEIS TM ) A096121 = (2, 8, 60, 816, 17520, 550080, 23839200,...) Number of full spectrum rook’s walks on a (2 x n) board. Difference Operator: τ 2 − ( 1 + x )( x + 2 ) τ − ( 1 + x )( x + 2 ) Val: {} Gquo: {− T 2 ( 1 − 3 T ) } Modified Bessel functions of the first and second kind, I v + x ( z ) and K v + x ( − z ) . Difference Operator: z τ 2 + ( 2 + 2 v + 2 x ) τ − z Val: {} Gquo: {− 1 4 T 2 z 2 ( 1 − ( 1 + 2 v ) T ) } Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Example from The On-Line Encyclopedia of Integer Sequences TM (OEIS TM ) A096121 = (2, 8, 60, 816, 17520, 550080, 23839200,...) Number of full spectrum rook’s walks on a (2 x n) board. Difference Operator: τ 2 − ( 1 + x )( x + 2 ) τ − ( 1 + x )( x + 2 ) Val: {} Gquo: {− T 2 ( 1 − 3 T ) } Modified Bessel functions of the first and second kind, I v + x ( z ) and K v + x ( − z ) . Difference Operator: z τ 2 + ( 2 + 2 v + 2 x ) τ − z Val: {} Gquo: {− 1 4 T 2 z 2 ( 1 − ( 1 + 2 v ) T ) } Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Example from The On-Line Encyclopedia of Integer Sequences TM (OEIS TM ) Comparing Gquo, {− T 2 ( 1 − 3 T ) } and {− 1 4 z 2 T 2 ( 1 − ( 1 + 2 v ) T ) } , we get candidates of z = { 2 , − 2 } and candidates of v = { 1 2 , 1 } We get four candidates to check ∼ gt , 2 τ 2 − ( 2 x + 4 ) τ − 2 , 2 τ 2 − ( 2 x + 3 ) τ − 2 2 τ 2 + ( 2 x + 4 ) τ − 2 , 2 τ 2 + ( 2 x + 3 ) τ − 2 . Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Example from The On-Line Encyclopedia of Integer Sequences TM (OEIS TM ) Comparing Gquo, {− T 2 ( 1 − 3 T ) } and {− 1 4 z 2 T 2 ( 1 − ( 1 + 2 v ) T ) } , we get candidates of z = { 2 , − 2 } and candidates of v = { 1 2 , 1 } We get four candidates to check ∼ gt , 2 τ 2 − ( 2 x + 4 ) τ − 2 , 2 τ 2 − ( 2 x + 3 ) τ − 2 2 τ 2 + ( 2 x + 4 ) τ − 2 , 2 τ 2 + ( 2 x + 3 ) τ − 2 . Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Example from The On-Line Encyclopedia of Integer Sequences TM (OEIS TM ) 2 τ 2 − ( 2 x + 4 ) τ − 2 ∼ gt L (When v = 1 , z = − 2) Term-transformation is x + 2 and gauge-transformation is 1. Applying gt-transformation to I 1 + x ( 2 ) and K 1 + x ( − 2 ) we get basis of a basis of solutions of L , { I 1 + x ( 2 )Γ( x + 2 ) , K 1 + x ( − 2 )Γ( x ) } Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Example from The On-Line Encyclopedia of Integer Sequences TM (OEIS TM ) 2 τ 2 − ( 2 x + 4 ) τ − 2 ∼ gt L (When v = 1 , z = − 2) Term-transformation is x + 2 and gauge-transformation is 1. Applying gt-transformation to I 1 + x ( 2 ) and K 1 + x ( − 2 ) we get basis of a basis of solutions of L , { I 1 + x ( 2 )Γ( x + 2 ) , K 1 + x ( − 2 )Γ( x ) } Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Example from The On-Line Encyclopedia of Integer Sequences TM (OEIS TM ) A000262 = (1, 1, 3, 13, 73, 501, 4051, 37633, 394353,...) Number of “sets of lists”: number of partitions of { 1 , .., n } into any number of lists. Difference Operator: τ 2 − ( 3 + 2 x ) τ + x ( x + 1 ) Val: { [ 0 , 2 ] } Gquo: { 1 − 2 T + 2 T 2 , 1 + 2 T + 2 T 2 } Laguerre polynomial L ( α ) x ( z ) . Difference Operator: Lgr = τ 2 − 2 x + 3 + α − z τ + x + 1 + α x + 2 x + 2 Val: { [ 0 , 1 ] , [ − α, 1 ] } Gquo: { 1 − 2 √− zT − 2 zT 2 , 1 + 2 √− zT − 2 zT 2 ) } Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Example from The On-Line Encyclopedia of Integer Sequences TM (OEIS TM ) A000262 = (1, 1, 3, 13, 73, 501, 4051, 37633, 394353,...) Number of “sets of lists”: number of partitions of { 1 , .., n } into any number of lists. Difference Operator: τ 2 − ( 3 + 2 x ) τ + x ( x + 1 ) Val: { [ 0 , 2 ] } Gquo: { 1 − 2 T + 2 T 2 , 1 + 2 T + 2 T 2 } Laguerre polynomial L ( α ) x ( z ) . Difference Operator: Lgr = τ 2 − 2 x + 3 + α − z τ + x + 1 + α x + 2 x + 2 Val: { [ 0 , 1 ] , [ − α, 1 ] } Gquo: { 1 − 2 √− zT − 2 zT 2 , 1 + 2 √− zT − 2 zT 2 ) } Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Example from The On-Line Encyclopedia of Integer Sequences TM (OEIS TM ) Comparing Gquo, { 1 − 2 T + 2 T 2 1 + 2 T + 2 T 2 } and { 1 − 2 √− zT − 2 zT 2 , 1 + 2 √− zT − 2 zT 2 ) } , we get z = − 1. Val = { [ 0 , 2 ] } is a special case of Lgr when α = 0. We get one candidate to check ∼ gt , τ 2 − 2 x + 4 x + 2 τ + x + 1 x + 2 Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Example from The On-Line Encyclopedia of Integer Sequences TM (OEIS TM ) Comparing Gquo, { 1 − 2 T + 2 T 2 1 + 2 T + 2 T 2 } and { 1 − 2 √− zT − 2 zT 2 , 1 + 2 √− zT − 2 zT 2 ) } , we get z = − 1. Val = { [ 0 , 2 ] } is a special case of Lgr when α = 0. We get one candidate to check ∼ gt , τ 2 − 2 x + 4 x + 2 τ + x + 1 x + 2 Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Example from The On-Line Encyclopedia of Integer Sequences TM (OEIS TM ) Comparing Gquo, { 1 − 2 T + 2 T 2 1 + 2 T + 2 T 2 } and { 1 − 2 √− zT − 2 zT 2 , 1 + 2 √− zT − 2 zT 2 ) } , we get z = − 1. Val = { [ 0 , 2 ] } is a special case of Lgr when α = 0. We get one candidate to check ∼ gt , τ 2 − 2 x + 4 x + 2 τ + x + 1 x + 2 Yongjae Cha Closed Form Solutions of Linear Difference Equations

Difference Operator Example Transformations Main Idea Invariant Local Data Liouvillian Special Functions Example from The On-Line Encyclopedia of Integer Sequences TM (OEIS TM ) Term-transformation is x and x τ − x 2 + 2 x gauge-transformation is x + 1 . x Applying gt-transformation to L ( 0 ) x ( − 1 ) , ( x + 1 ) L ( 0 ) x + 1 ( − 1 ) − ( x + 2 ) L ( 0 ) � � x ( − 1 ) Γ( x ) Yongjae Cha Closed Form Solutions of Linear Difference Equations