Heun functions and combinatorial identities
aa r X i v : . [ m a t h . C A ] J a n Heun functions and combinatorial identities
Adina B˘arar ∗ , Gabriela Raluca Mocanu † , Ioan Ra¸sa ‡ Subject Class: 33E30, 94A17, 05A19.Keywords: Heun functions, entropies, combinatorial identities.
Abstract
We give closed forms for several families of Heun functions related to classicalentropies. By comparing two expressions of the same Heun function, we getseveral combinatorial identities generalizing some classical ones.
1. INTRODUCTION
Consider the general Heun equation (see, e.g., [12], [4], [2] and the referencestherein) u ′′ ( x ) + (cid:18) γx + δx − ǫx − a (cid:19) u ′ ( x ) + αβx − qx ( x − x − a ) u ( x ) = 0 , (1)where a / ∈ { , } , γ / ∈ { , − , − , . . . } and α + β + 1 = γ + δ + ǫ . Its solution u ( x ) normalized by the condition u (0) = 1 is called the (local) Heun function and is denoted by Hl ( a, q ; α, β ; γ, δ ; x ).The confluent Heun equation is u ′′ ( x ) + (cid:18) p + γx + δx − (cid:19) u ′ ( x ) + 4 pαx − σx ( x − u ( x ) = 0 , (2)where p = 0. The solution u ( x ) normalized by u (0) = 1 is called the confluentHeun function and is denoted by HC ( p, γ, δ, α, σ ; x ).It was proved in [5] that Hl (cid:18) , − n ; − n,
1; 1 , x (cid:19) = n X k =0 (cid:18)(cid:18) nk (cid:19) x k (1 − x ) n − k (cid:19) , (3) ∗ Technical University of Cluj-Napoca, Department of Mathematics, MemorandumuluiStreet 28, 400114, Cluj-Napoca, Romania † Romanian Academy, Cluj-Napoca Branch, Astronomical Institute, Cire¸silor Street 19,400487, Cluj-Napoca, Romania, [email protected] ‡ Technical University of Cluj-Napoca, Department of Mathematics, Memorandumului Street 28, 400114, Cluj-Napoca, Romania l (cid:18) , n ; 2 n,
1; 1 , − x (cid:19) = ∞ X k =0 (cid:18)(cid:18) n + k − k (cid:19) x k (1 + x ) − n − k (cid:19) , (4) HC (cid:18) n, , , , n ; x (cid:19) = ∞ X k =0 (cid:18) e − nx ( nx ) k k ! (cid:19) . (5)More general results, providing closed forms of the functions Hl (cid:0) , − nθ ; − n, θ ; γ, γ ; x (cid:1) and Hl (cid:0) , nθ ; 2 n, θ ; γ, γ ; x (cid:1) , and explicit expres-sions for some confluent Heun functions can be found in [6].In this paper we give closed forms for several families of Heun functions andconfluent Heun functions, extending (3), (4) and (5). Basic tools will be theresults of [3] and [10] concerning the derivatives of Heun functions, respectivelyconfluent Heun functions; see also [1] and [6].By comparing two expressions of the same Heun function, we get severalcombinatorial identities; very particular forms of them can be traced in theclassical book [9].Let us mention that the families of (confluent) Heun functions mentionedabove are naturally related to some classical entropies: see [1], [5], [6], [7], [8].Throughout the paper we shall use the notation( x ) := 1 , ( x ) k := x ( x + 1) . . . ( x + k − , k ≥ ,a nj := 4 − n (cid:18) jj (cid:19)(cid:18) n − jn − j (cid:19) , (6) r nj := (cid:18) nj (cid:19) − a nj . (7)
2. HEUN FUNCTIONS
Let αβ = 0 . As a consequence of the results of [3] we have (see [6, Prop. 1] and [6, (14)]): Hl (cid:18) ,
12 ( α + 2)( β + 2); α + 2 , β + 2; γ + 1 , γ + 1; x (cid:19) = (8) γαβ (1 − x ) − ddx Hl (cid:18) , αβ ; α, β ; γ, γ ; x (cid:19) . From [6, (6)], [6, (22)] and (6) we obtain Hl (cid:18) , − n ; − n,
1; 1 , x (cid:19) = n X j =0 a nj (1 − x ) j . (9)2 heorem 0.1. Let ≤ m ≤ n . Then Hl (cid:18) , (2 m + 1)( m − n ); 2( m − n ) , m + 1; m + 1 , m + 1; x (cid:19) (10)= 4 m (cid:18) nm (cid:19) − (cid:18) mm (cid:19) − n − m X j =0 (cid:18) m + jm (cid:19) a n,m + j (1 − x ) j = n − m X j =0 j (cid:18) n − mj (cid:19) ( m + 1 / j ( m + 1) j ( x − x ) j . Proof.
We shall prove the first equality by induction with respect to m .For m = 0, it follows from (9). Suppose that it is valid for a certain m < n .Then (8) implies Hl (cid:18) ,
12 (2 m + 3)( m + 1 − n ); 2( m + 1 − n ) , m + 3; m + 2 , m + 2; x (cid:19) = ( m + 1)(1 − x ) − m − n )(2 m + 1) ddx Hl (cid:18) , (2 m + 1)( m − n ); 2( m − n ) , m + 1; m + 1 , m + 1; x (cid:19) = ( m + 1)(1 − x ) − m − n )(2 m + 1) 4 m (cid:18) nm (cid:19) − (cid:18) mm (cid:19) − n − m X i =1 (cid:18) m + im (cid:19) a n,m + i ( − i )(1 − x ) i − = 4 m +1 (cid:18) nm + 1 (cid:19) − (cid:18) m + 2 m + 1 (cid:19) − n − m − X j =0 (cid:18) m + 1 + jm + 1 (cid:19) a n,m +1+ j (1 − x ) j , and so the desired equality is true for m + 1; this finishes the proof by induction.In order to prove that the first member and the last member of (10) areequal, it suffices to use [6, Th. 1] with γ = m + 1, θ = m + , and n replacedby n − m . Corollary 0.2.
Let ≤ i ≤ n − m , ≤ j ≤ n − m . Then n − m X j = i ( − j − i (cid:18) n − mj (cid:19) ( m + 1 / j ( m + 1) j (cid:18) ji (cid:19) == 4 m (cid:18) nm (cid:19) − (cid:18) mm (cid:19) − (cid:18) m + im (cid:19) a n,m + i , (11) n − m X i = j (cid:18) m + im (cid:19)(cid:18) ij (cid:19) a n,m + i = 4 − m (cid:18) nm (cid:19)(cid:18) mm (cid:19) ( m + 1 / j ( m + 1) j (cid:18) n − mj (cid:19) . (12) Proof.
It suffices to combine the last equality in (10) with( x − x ) j = 4 − j (cid:0) (1 − x ) − (cid:1) j , − x ) j = (cid:0) x − x ) (cid:1) j . Example 0.3.
For i = m = 0, (11) reduces to n X j =0 (cid:18) − (cid:19) j (cid:18) nj (cid:19)(cid:18) jj (cid:19) = 4 − n (cid:18) nn (cid:19) , (13)which is (3.85) in [9].For j = m = 0, (12) becomes n X i =0 (cid:18) ii (cid:19)(cid:18) n − in − i (cid:19) = 4 n , (14)which is (3.90) in [9].From [6, (7)], [6, (23)] and (6) we know that Hl (cid:18) , n + 1; 2 n + 2 ,
1; 1 , x (cid:19) = n X j =0 a nj (1 − x ) j − n − . (15) Theorem 0.4.
For m ≥ we have Hl (cid:18) , (2 m + 1)( m + n + 1); 2( m + n + 1) , m + 1; m + 1 , m + 1; x (cid:19) = (cid:18) n + mn (cid:19) − n X j =0 (cid:18) n + 2 m − j m (cid:19)(cid:18) n + m − jm (cid:19) − a nj (1 − x ) j − n − m − = (1 − x ) − n − m − n X j =0 j (cid:18) nj (cid:19) (1 / j ( m + 1) j ( x − x ) j . Proof.
As in the proof of Theorem 0.1, the first equality can be proved byinduction with respect to m , if we use (15) and (8). The equality of the firstmember and the last member follows from [6, Cor. 2] by choosing γ = m + 1, θ = m + 1 /
2, and replacing n by n + m + 1. Corollary 0.5.
Let ≤ i ≤ n , ≤ j ≤ n . Then n X j = i ( − j − i (cid:18) nj (cid:19) (1 / j ( m + 1) j (cid:18) ji (cid:19) = (cid:18) n + 2 m − i m (cid:19)(cid:18) n + mn (cid:19) − (cid:18) n + m − im (cid:19) − a ni , (16)4 X i = j (cid:18) n + 2 m − i m (cid:19)(cid:18) n + m − im (cid:19) − (cid:18) ij (cid:19) a ni = (cid:18) n + mn (cid:19)(cid:18) nj (cid:19) (1 / j ( m + 1) j . (17)The proof is similar to the proof of Corollary 0.2.For i = m = 0, (16) reduces to (13), i.e., (3.85) in [9].For j = m = 0, (17) reduces to (14), i.e., (3.90) in [9].Let again αβ = 0. According to the results of [3] (see [6, Prop. 1] and [6, (15)]),we have Hl (cid:18) ,
12 (2 γ − α )(2 γ − β ); 2 γ − α, γ − β ; γ + 1 , γ + 1; x (cid:19) = γαβ (1 − x ) α + β +1 − γ ddx Hl (cid:18) , αβ ; α, β ; γ, γ ; x (cid:19) . (18)Using (9), (18) and the above methods of proof, we obtain the followingidentities: Hl (cid:18) , (2 k + 1)( k − n ); 2( k − n ) , k + 1; 2 k + 1 , k + 1; x (cid:19) = 4 k (cid:18) n + kn (cid:19) − (cid:18) nk (cid:19) − n − k X i =0 (cid:18) n − i k (cid:19)(cid:18) ni (cid:19) r n,k + i (1 − x ) i = n − k X j =0 j (cid:18) n − kj (cid:19) ( k + 1 / j (2 k + 1) j ( x − x ) j , ≤ k ≤ n. (19)As a consequence of (19) one gets n − k X j = i ( − j − i (cid:18) n − kj (cid:19) ( k + 1 / j (2 k + 1) j (cid:18) ji (cid:19) (20)= 4 k (cid:18) n + kn (cid:19) − (cid:18) nk (cid:19) − (cid:18) n − i k (cid:19)(cid:18) ni (cid:19) r n,k + i , ≤ i ≤ n − k, n − k X i = j (cid:18) n − i k (cid:19)(cid:18) ni (cid:19)(cid:18) ij (cid:19) r n,k + i (21)= 4 − k (cid:18) n + kn (cid:19)(cid:18) nk (cid:19)(cid:18) n − kj (cid:19) ( k + 1 / j (2 k + 1) j , ≤ j ≤ n − k. For i = k = 0, (20) reduces to (13); for j = k = 0, (21) becomes (14).5 l (cid:18) , (2 k − k + n ); 2( k + n ) , k −
1; 2 k, k ; x (cid:19) = 2 k − (cid:18) n + k − k − (cid:19) − (cid:18) n − k − (cid:19) − n − k X i =0 (cid:18) n − i − k − (cid:19)(cid:18) n − i (cid:19) r n,k + i (1 − x ) − n +2 i = (1 − x ) − n n − k X j =0 j (cid:18) n − kj (cid:19) ( k + 1 / j (2 k ) j ( x − x ) j , ≤ k ≤ n. (22)From (22) we derive n − k X j = i ( − j − i (cid:18) n − kj (cid:19) ( k + 1 / j (2 k ) j (cid:18) ji (cid:19) (23)= 2 k − (cid:18) n + k − k − (cid:19) − (cid:18) n − k − (cid:19) − (cid:18) n − i − k − (cid:19)(cid:18) n − i (cid:19) r n,k + i , ≤ i ≤ n − k, n − k X i = j (cid:18) n − i − k − (cid:19)(cid:18) n − i (cid:19)(cid:18) ij (cid:19) r n,k + i (24)= 2 − k (cid:18) n + k − k − (cid:19)(cid:18) n − k − (cid:19)(cid:18) n − kj (cid:19) ( k + 1 / j (2 k ) j , ≤ j ≤ n − k. For i = 0, k = 1 and n replaced by n + 1, from (23) we obtain n X j =0 (cid:18) − (cid:19) j (cid:18) nj (cid:19)(cid:18) j + 1 j (cid:19) = 1( n + 1)4 n (cid:18) nn (cid:19) . (25)With j = 0, k = 1, and n replaced by n + 1, (24) yields n X i =0 ( i + 1) (cid:18) i + 2 i + 1 (cid:19)(cid:18) n − in − i (cid:19) = n + 12 4 n +1 . (26)It is a pleasant calculation to prove (25) and (26) directly.Using (15) and (18) we get Hl (cid:18) , (2 k + 1)( k + n + 1); 2( k + n + 1) , k + 1; 2 k + 1 , k + 1; x (cid:19) = 4 k (cid:18) n + kn (cid:19) − (cid:18) nk (cid:19) − n − k X j =0 (cid:18) k + 2 j j (cid:19)(cid:18) nk + j (cid:19) r nj (1 − x ) − k − − j = (1 − x ) − n − n − k X j =0 j (cid:18) n − kj (cid:19) ( k + 1 / j (2 k + 1) j ( x − x ) j , ≤ k ≤ n. (27)6aking into account that r n,n − j = r nj , from (27) we derive (20) and (21).Moreover, Hl (cid:18) , (2 k + 1)( k − n ); 2( k − n ) , k + 1; 2 k + 2 , k + 2; x (cid:19) = 4 k k + 1 n + 1 (cid:18) n + k + 1 n (cid:19) − (cid:18) nk (cid:19) − n − k X j =0 (cid:18) k + 2 j + 22 j (cid:19)(cid:18) n + 1 k + j + 1 (cid:19) r nj (1 − x ) n − k − j = n − k X j =0 j (cid:18) n − kj (cid:19) ( k + 1 / j (2 k + 2) j ( x − x ) j , ≤ k ≤ n. (28)From (28) we derive n − k X j = i ( − j − i (cid:18) n − kj (cid:19) ( k + 1 / j (2 k + 2) j (cid:18) ji (cid:19) , ≤ i ≤ n − k (29)= 4 k k + 1 n + 1 (cid:18) n + k + 1 n (cid:19) − (cid:18) nk (cid:19) − (cid:18) n − i + 22 k + 2 (cid:19)(cid:18) n + 1 i (cid:19) r n,k + i , n − k X i = j (cid:18) n − i + 22 k + 2 (cid:19)(cid:18) n + 1 i (cid:19)(cid:18) ij (cid:19) r n,k + i , ≤ j ≤ n − k (30)= 4 − k n + 12 k + 1 (cid:18) n + k + 1 n (cid:19)(cid:18) nk (cid:19)(cid:18) n − kj (cid:19) ( k + 1 / j (2 k + 2) j . For i = k = 0, (29) becomes n X j =0 (cid:18) − (cid:19) j (cid:18) n + 1 j + 1 (cid:19)(cid:18) jj (cid:19) = 2 n + 14 n (cid:18) nn (cid:19) . (31)Let us recall the formula (7.6) in [9]: n X j =0 (cid:18) − (cid:19) j (cid:18) nj (cid:19)(cid:18) jj (cid:19)(cid:18) j + hh (cid:19) − = 14 n (cid:18) n + 2 hn + h (cid:19)(cid:18) hh (cid:19) − . (32)For h = 1, (32) reduces to (31).For j = k = 0, (30) becomes n X i =0 (2 n − i + 1) (cid:18) ii (cid:19)(cid:18) n − in − i (cid:19) = ( n + 1)4 n , (33)which can be proved also directly. 7 . CONFLUENT HEUN FUNCTIONS The hypergeometric function v ( t ) = F ( α ; γ ; t ) satisfies (see [11, p. 336], [13,13.2.1]) v (0) = 1 and tv ′′ ( t ) + ( γ − t ) v ′ ( t ) − αv ( t ) = 0 . (34)Moreover (see [11, p. 338, 5.6], [13, 13.3.15]), F ( α + 1; γ + 1; t ) = γα ddt F ( α ; γ ; t ) . (35)With the above notation we have: Theorem 0.6.
For αp = 0 , the confluent Heun function HC ( p, γ, , α, pα ; x ) satisfies HC ( p, γ, , α, pα ; x ) = F ( α ; γ ; − px ) , (36) HC ( p, γ + 1 , , α + 1 , p ( α + 1); x ) = − γ pα ddx HC ( p, γ, , α, pα ; x ) , (37) HC ( p, γ + j, , α + j, p ( α + j ); x ) = ( − j ( γ ) j (4 p ) j ( α ) j d j dx j HC ( p, γ, , α, pα ; x ) , (38) for all integers j ≥ with ( α ) j = 0 .Proof. According to (2), the function u ( x ) = HC ( p, γ, , α, pα ; x ) satisfies u (0) = 1 and xu ′′ ( x ) + (4 px + γ ) u ′ ( x ) + 4 pαu ( x ) = 0 . (39)From (34) and (39) it is easy to deduce that u ( x ) = v ( − px ), and this entails(36).Now (37) is a consequence of (36) and (35); (38) can be proved by inductionwith respect to j . Let us remark that (37) coincides with (30) in [6]. Corollary 0.7.
Let K n ( x ) := HC (cid:0) n, , , , n ; x (cid:1) be the function given by (5) .Then K n ( x ) = F (cid:18)
12 ; 1; − nx (cid:19) , (40) K n ( x ) = 1 π Z − e − nx (1+ t ) dt √ − t . (41) Proof. (40) follows from (36) with α = 1 / γ = 1 and p = n . By using (40)and [11, p. 338, 5.9], [13, 13.4(i)] we get (41). Let us remark that (41) coincideswith (69) in [5].Using (38) with p = n , γ = 1, α = 1 / HC (cid:18) n, j + 1 , , j + 12 , n (2 j + 1); x (cid:19) = ( − j n j (cid:18) jj (cid:19) − K ( j ) n ( x ) , j ≥ . (42)8rom (42) and [6, (34)] we obtain K ( j ) n (0) = ( − n ) j (cid:18) jj (cid:19) , (43)which is (35) in [6].On the other hand, (41) implies (with t = sin ϕ ) K ( j ) n (0) = ( − n ) j π j X k =0 (cid:18) jk (cid:19) Z π/ − π/ sin k ϕdϕ = ( − n ) j [ j/ X i =0 (cid:18) j i (cid:19)(cid:18) ii (cid:19) − i . Combined with (43), this produces [ j/ X i =0 (cid:18) j i (cid:19)(cid:18) ii (cid:19) − i = 2 − j (cid:18) jj (cid:19) , which is (3.99) in [9].Finally, we give closed forms for some families of confluent Heun functions. Theorem 0.8. [(i)]1. For ≤ j ≤ n we have HC (cid:18) p, j + 12 , , j − n, p ( j − n ); x (cid:19) = (2 j )! j ! n − j X k =0 (cid:18) n − jk (cid:19) ( n − k )!(2 n − k )! (16 px ) n − j − k . (44)
2. More generally, for ≤ j ≤ n and λ > − , HC ( p, j + 1 + λ, , j − n, p ( j − n ); x ) (45)= ( λ + 1) j Γ( λ + 1)Γ( n + λ + 1) n − j X k =0 ( λ + n + 1 − k ) k (cid:18) n − jk (cid:19) (4 px ) n − j − k . Proof.
By using the relation between the function F and the Hermite polyno-mials (see [11, p. 340, 5.16], [11, p. 235, (4.51)], [13, 13.6.16]) we have F (cid:18) − n ; 12 ; x (cid:19) = n ! n X k =0 k !(2 n − k )! ( − x ) n − k . (46)From (36) and (46) it follows that HC (cid:18) p, , , − n, − pn ; x (cid:19) = n ! n X k =0 (16 px ) n − k k !(2 n − k )! . (47)9ow (44) is a consequence of (47) and (38).In order to prove (45), we need the relation between F and the Laguerrepolynomials (see [11, p. 340, 5.14], [13, 13.6.19]): F ( − n ; λ + 1; x ) = n !Γ( λ + 1)Γ( n + λ + 1) L λn ( x ) , λ > − , (48)where (see [11, p. 245, (4.61)], [13, 18.5.12]) L λn ( x ) = n X k =0 ( − k ( λ + k + 1) n − k k !( n − k )! x k . (49)From (36), (48) and (49) we get HC ( p, λ + 1 , , − n, − pn ; x ) = n !Γ( λ + 1)Γ( n + λ + 1) L λn ( − px ) . (50)Combined with (38), (50) produces(45), and this concludes the proof.
4. OTHER COMBINATORIAL IDENTITIES
Let us return to (11). Since( m + 1 / j ( m + 1) j = 4 − j (cid:18) m + 2 jm + j (cid:19)(cid:18) mm (cid:19) − , (51)it becomes n − m X j = i ( − j − i − j (cid:18) n − mj (cid:19)(cid:18) m + 2 jm + j (cid:19)(cid:18) ji (cid:19) = 4 m − n (cid:18) nm (cid:19) − (cid:18) m + im (cid:19)(cid:18) m + 2 im + i (cid:19)(cid:18) n − m − in − m − i (cid:19) . Set i + m = r , j = r − m + k , and replace n by n + r ; we get n X k =0 (cid:18) − (cid:19) k (cid:18) n + r − mn − k (cid:19)(cid:18) r + 2 kr + k (cid:19)(cid:18) r − m + kk (cid:19) (52)= 4 − n (cid:18) n + rm (cid:19) − (cid:18) rm (cid:19)(cid:18) rr (cid:19)(cid:18) nn (cid:19) , n ≥ , r ≥ m ≥ . Here are some particular cases of (52). r = m = n : n X k =0 (cid:18) − (cid:19) k (cid:18) nk (cid:19)(cid:18) n + 2 kn + k (cid:19) = 4 − n (cid:18) nn (cid:19) . = m : n X k =0 (cid:18) − (cid:19) k (cid:18) nk (cid:19)(cid:18) r + 2 kr + k (cid:19) = 4 − n (cid:18) n + rr (cid:19) − (cid:18) rr (cid:19)(cid:18) nn (cid:19) .m = 0 : n X k =0 (cid:18) − (cid:19) k (cid:18) n + rn − k (cid:19)(cid:18) r + 2 kr + k (cid:19)(cid:18) r + kk (cid:19) = 4 − n (cid:18) rr (cid:19)(cid:18) nn (cid:19) .r = n : n X k =0 (cid:18) − (cid:19) k (cid:18) n − mn − k (cid:19)(cid:18) n + 2 kn + k (cid:19)(cid:18) n − m + kk (cid:19) = 4 − n (cid:18) nm (cid:19) − (cid:18) nm (cid:19)(cid:18) nn (cid:19) .m = n : n X k =0 (cid:18) − (cid:19) k (cid:18) rn − k (cid:19)(cid:18) r + 2 kr + k (cid:19)(cid:18) r − n + kk (cid:19) = 4 − n (cid:18) n + rr (cid:19) − (cid:18) rn (cid:19)(cid:18) rr (cid:19)(cid:18) nn (cid:19) . Now let us return to (12); use (51), set j + m = r , i = r − m + k , and replace n by n + r . We get n X k =0 (cid:18) r + km (cid:19)(cid:18) r + k − mk (cid:19)(cid:18) r + 2 kr + k (cid:19)(cid:18) n − kn − k (cid:19) (53)= 4 n (cid:18) n + rm (cid:19)(cid:18) rr (cid:19)(cid:18) n + r − mn (cid:19) , n ≥ , r ≥ m ≥ . For r = m = n , (53) reduces to n X k =0 (cid:18) n + kn (cid:19)(cid:18) n + 2 kn + k (cid:19)(cid:18) n − kn − k (cid:19) = 4 n (cid:18) nn (cid:19) . Clearly, there are many other particular cases of (53).Several other particular combinatorial identities can be obtained startingwith other general formulas from the preceding sections, but we omit the details.
Acknowledgements
GM is partially supported by a grant of the Romanian Ministry of NationalEducation and Scientific Research, RDI Programme for Space Technology andAdvanced Research - STAR, project number 513, 118/14.11.2016.
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