The generalized hypergeometric difference equation

Demonstr. Math. 2018; 51:62–75

Research Article

Open Access

Martin Bohner and Tom Cuchta*

The generalized hypergeometric difference equation https://doi.org/10.1515/dema-2018-0007 Received November 20, 2017; accepted March 27, 2018

Abstract: A difference equation analogue of the generalized hypergeometric differential equation is defined, its contiguous relations are developed, and its relation to numerous well-known classical special functions are demonstrated. Keywords: special functions, discrete hypergeometric series, delay difference equations, contiguous relation, generalized hypergeometric functions MSC: 33C20, 39A12

1 Introduction The Pochhammer symbol (a)k is defined for k ∈ N by (a)k = a(a + 1) . . . (a + k − 1) and for k = 0 by (a)0 = 1. We define the product notation (︃ k−1 )︃ (︃ m )︃ m ∏︁ ∏︁ ∏︁ As = As As . s=1,(k)

s=1

s=k+1

We use the forward difference operator ∆ defined by ∆f (t) = f (t + 1) − f (t). From the definition of ∆ it is easy to see that ∆[(−1)n (−t)n ] = n(−1)n−1 (−t)n−1 is the discrete analogue of the power rule of differentiation. The generalized hypergeometric series p Fq is defined by p Fq (a 1 , . . . , a p ; b 1 , . . . , b q ; t)

=

∞ ∑︁ (a1 )k . . . (a p )k t k . (b1 )k . . . (b q )k k!

(1)

k=0

The derivative of (1) is known [1, 5.2.2. (1)]: (a1 )n . . . (a p )n dn p Fq (a 1 , . . . , a p ; b q , . . . , b q ; t) = p Fq (a 1 , . . . , a p ; b 1 , . . . , b q ; t). dt n (b1 )n . . . (b q )n

(2)

d Let θ denote the operator θ = t . It is known [2, S46 (3)] that if y(t) = p Fq (a1 , . . . , a p ; b1 , . . . , b q ; t), then y dt satisfies the differential equation ⎡ ⎤ q q ∏︁ ∏︁ ⎣θ (θ + b j − 1) − z (θ + a i )⎦ y(t) = 0. (3) j=1

i=1

Two hypergeometric functions are called contiguous if one of their parameters a i or b i differs in one by ±1. We adopt the following notation: F = p Fq (a1 , . . . , a p ; b1 , . . . , b q ; t),

Martin Bohner: Department of Mathematics & Statistics, Missouri University of Science & Technology, Rolla, MO, USA 65409, E-mail: [email protected] *Corresponding Author: Tom Cuchta: Department of Computer Science and Math, Fairmont State University, Fairmont, WV, USA 26554, E-mail: [email protected] Open Access. © 2018 Martin Bohner and Tom Cuchta, published by De Gruyter. This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivs 4.0 License. Unauthenticated

Download Date | 6/23/18 5:19 AM

The generalized hypergeometric difference equation |

63

F(a i ±) = p Fq (a1 , . . . , a i ± 1, . . . , a p ; b q , . . . , b q ; t), and F(b i ±) = p Fq (a1 , . . . , a p ; b1 , . . . , b i ± 1, . . . , b q ; t). Following [3, p. 81] we adopt the following notations: Sn =

(a1 + n)(a2 + n) . . . (a p + n) , (b1 + n) . . . (b q + n) τ n,k =

(4)

Sn , ak + n

(5)

(a1 )n+1 . . . (a p )n+1 (a )n . . . (a p )n = Sn 1 , (b1 )n+1 . . . (b q )n+1 (b1 )n . . . (b q )n and

p ∏︁

Uj = bj

(6)

(a s − b j )

s=1 q ∏︁

. (b s − b j )

s=1,(j)

A total of 2p + q contiguous relations exist for p Fq , split into three classes [3, p. 85]: if p < q, then ⎧ k = 2, 3, . . . , p ⎪ ⎪ (a1 − a k )F = a1 F(a1 +) − a k F(a k +), ⎪ (a − b + 1)F = a F(a +) − (b − 1)F(b −), k = 1, 2, . . . , q ⎪ 1 1 1 k k k ⎪ ⎪ q ⎪ ∑︁ ⎪ ⎨ F = F(a −) + t W j,k F(b j +), k = 1, 2, . . . , p k j=1 ⎪ ⎪ q ⎪ ⎪ ∑︁ ⎪ ⎪ ⎪ a F = a F(a +) − t U j F(b j +). 1 1 ⎪ ⎩ 1

(7)

j=1

If p = q, then

⎧ (a1 − a k )F = a1 F(a1 +) − a k F(a k +), ⎪ ⎪ ⎪ (a1 − b k + 1)F = a1 F(a1 +) − (b k − 1)F(b k −), ⎪ ⎪ ⎪ q ⎪ ∑︁ ⎪ ⎨ F = F(a −) + t W j,k F(b j +), k j=1 ⎪ ⎪ q ⎪ ⎪ ∑︁ ⎪ ⎪ ⎪ (a + t)F = a F(a +) − t U j F(b j +). 1 1 ⎪ ⎩ 1

k = 2, 3, . . . , p k = 1, 2, . . . , q k = 1, 2, . . . , p

(8)

j=1

If p = q + 1, then ⎧ (a1 − a k )F = a1 F(a1 +) − a k F(a k +), k = 2, 3, . . . , p ⎪ ⎪ ⎪ (a1 − b k + 1)F = a1 F(a1 +) − (b k − 1)F(b k −), k = 1, 2, . . . , q ⎪ ⎪ ⎪ q ⎪ ∑︁ ⎪ ⎨ (1 − t)F = F(a −) + t W j,k F(b j +), k = 1, 2, . . . , p k j=1 ⎪ ⎪ q ⎪ ⎪ ∑︁ [︀ ]︀ ⎪ ⎪ ⎪ (1 − t)a + (A − B)t F = (1 − t)a F(a +) − t U j F(b j +), 1 1 1 ⎪ ⎩

(9)

j=1

where A =

p ∑︁ s=1

a s and B =

q ∑︁

bs .

s=1

Many classical ⎛ ⎞ special functions can be expressed in terms of p Fq . The classical exponential function ∫︁ t exp ⎝ p(τ)dτ⎠ satisfies the initial value problem 0

y′ (t) = p(t)y(t),

y(0) = 1.

(10)

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64 | Martin Bohner and Tom Cuchta Substituting αt for t in the relationship given by [4, (2.1.9)] yields e αt = 0 F0 (; ; αt).

(11)

The functions sin(at) and cos(at) satisfy the differential equation y′′ (t) = −a2 y(t). The relationship between sin(at) and 0 F1 is known [4, (2.1.7)] to be (︂ )︂ 3 a2 t2 sin(at) = at0 F1 ; ; − . 2 4

(12)

The relationship between cos(at) and 0 F1 is known [4, (2.1.8)] to be )︂ (︂ 1 a2 t2 . cos(at) = 0 F1 ; ; − 2 4

(13)

Classical Bessel functions Jn are defined by the series Jn (t) =

∞ ∑︁ k=0

(−1)k t2k+n . k!Γ(k + n + 1)22k+n

It is known [2, p. 108 (1)] that )︂ (︂ tn t2 , (14) F ; n + 1; − 0 1 4 2n Γ(n + 1) (︂ )︂ t2 2n n!J n (t) and consequently, y(t) = 0 F1 ; n + 1; − = satisfies the differential equation [2, p. 109 (4)] 4 tn Jn (t) =

t2 y′′ (t) + ty′ (t) + (t2 − n2 )y(t) = 0.

(15)

A “hypergeometric difference equation”, originally investigated in [5], is defined in [6] by (a2 x + b2 )y(x + 2) + (a1 x + b1 )y(x + 1) + (a0 x + b0 )y(x) = 0, and it is named this “because . . . its solutions can be expressed in terms of the hypergeometric series”, which is referring specifically to 2 F1 . The “basic hypergeometric series” (or “q-hypergeometric series”) is defined in [7] by the notation r ϕ s (a 1 , . . . , a r ; b 1 , . . . , b s ; q, z) =

∞ ∑︁ k=0

[︁ ]︁1+s−r k (a1 ; q)k . . . (a r ; q)k (−1)k q(2) zk . (q; q)k (b1 ; q)k . . . (b s ; q)k

The paper [8] defines a “discrete hypergeometric function” by r M s (a 1 , . . . , a r ; b 1 , . . . , b s ; q, z)

=

∞ ∑︁ (q a1 )k . . . (q a2 )k z(k) , (q)k (q b1 )k . . . (q b s )k k=0

where z(k) is a q-analogue of z k . The paper [9] speaks of “discrete analogues” of theorems related to classical hypergeometric functions, but the authors simply mean restricting the otherwise complex parameters and variables to nonnegative integers. Difference equations “of hypergeometric type” are discussed in [10] and are defined by the mixed forward and backwards difference equation σ(t)∇∆y(t) + τ(t)∆y(t) + λy(t) = 0, where σ and τ are first- or second-order polynomials. A Taylor theorem for the difference operator is known [11, Theorem 1.113]: f (t) =

∞ ∑︁ ∆ k f (0)(−1)k (−t)k . k!

(16)

k=0



65

Let p : Z → C be so that for all t ∈ Z, 1 + p(t) ≠ 0. The discrete exponential function e p : Z → C is defined to be the unique solution of the initial value problem ∆y(t) = p(t)y(t),

y(0) = 1.

(17)

In particular, if p ≡ α is constant and α ∈ C \ {−1}, then (16) shows e α (t) =

∞ ∑︁

αk

k=0

(−1)k (−t)k . k!

(18)

The discrete trigonometric functions sinα and cosα are often defined for α ∈ C \ {1, −1} [11, Definition 3.25] by e ip (t) + e−ip (t) e ip (t) − e−ip (t) and cosα (t) = . sinα (t) = 2i 2 Both functions are solutions of the difference equation ∆2 y(t) + α2 y(t) = 0.

(19)

It is easy to see that the function cosα obeys the discrete Taylor series cosα (t) =

∞ ∑︁ (−1)k (−t)

2k

(2k)!

k=0

,

(20)

and that the function sinα obeys the discrete Taylor series sinα (t) =

∞ ∑︁ (−1)k (−t)

2k+1

(2k + 1)!

k=0

.

(21)

Discrete Bessel functions, J n , were defined in [12] by (︂ )︂ (−1)n (−t)n n−t n+1−t J n (t) = F , ; n + 1; −1 . 2n n! 2 1 2 2

(22)

It was shown that this function obeys the difference equation t(t − 1)∆2 y(t − 2) + t∆y(t − 1) + t(t − 1)y(t − 2) − n2 y(t) = 0. For further results on J n , see also the recent paper [13] by Antonín Slavík.

2 Discrete hypergeometric series We define the discrete generalized hypergeometric series for p, q ∈ N0 by p F q (a 1 , . . . , a p ; b 1 , . . . , b q ; t, n, ξ )

=

∞ ∑︁ (a1 )k . . . (a p )k ξ k (−1)nk (−t)nk , k! (b1 )k . . . (b q )k

(23)

k=0

where a j ∈ C for j = 1, 2, . . . , q and b j ∈ C \ (−N0 ) for j ∈ N. If n = 0 in (23), then p F q becomes a constant value of p Fq independent of t: p F q (a 1 , . . . , a p ; b 1 , . . . , b q ; t, 0, ξ )

= p Fq (a1 , . . . , a p ; b1 , . . . , b q ; (−1)n ξ ).

If n = 1 in (23), then p F q (a 1 , . . . , a p ; b 1 , . . . , b q ; t, 1, ξ )

(︀ )︀ = p+1 Fq a1 , . . . , a p , −t; b1 , . . . , b q ; (−1)n ξ .

To get the relationship between p F q and p Fq for n ∈ N \ {1}, we begin with the following lemma.


66 | Martin Bohner and Tom Cuchta Lemma 1. We have (−t)nk = n

nk

)︂ n−1 (︂ ∏︁ −t + j n

j=0

k

for n ∈ N and k ∈ N0 . Proof. We compute (−t)nk = (−t)(−t + 1) . . . (−t + n)(−t + n + 1) . . . (−t + nk − 1) ⎡ ⎤⎡ ⎤ ⎡ ⎤ k−1 k−1 k−1 ∏︁ ∏︁ ∏︁ (︀ )︀ = ⎣ (−t + jn)⎦ ⎣ (−t + jn + 1)⎦ . . . ⎣ −t + jn + (n − 1) ⎦ j=0

⎡

j=0 k−1 ∏︁

⎣ = n nk

⎤⎡ (−t + jn)⎦ ⎣

j=0 k−1 ∏︁

⎤

⎡

(−t + jn + 1)⎦

⎣

j=0

j=0

k−1 ∏︁ (︀

⎤ )︀ −t + jn + (n − 1) ⎦

j=0

... nk nk nk ⎡ ⎤⎡ ⎤ ⎡ ⎤ )︂ ∏︁ )︂ )︂ k−1 (︂ k−1 (︂ k−1 (︂ ∏︁ ∏︁ −t −t + 1 −t + (n − 1) = n nk ⎣ + j ⎦⎣ +j ⎦...⎣ +j ⎦ n n n j=0 j=0 j=0 )︂ (︂ )︂ (︂ −t −t + (n − 1) = n nk ... n k m k )︂ (︂ n−1 ∏︁ −t + j = n nk , n k j=0

as was to be shown. Now we may express p F q in terms of p+n Fq for n ∈ N. Proposition 2. Let y(t) = p F q (a1 , . . . , a p ; b1 , . . . , b q ; t, n, ξ ). Then, (︂ )︂ −t −t + 1 −t + n − 1 y(t) = p+n Fq a1 , . . . , a p , , ,..., ; b1 , . . . , b q ; (−nξ )n . n n n

(24)

We now consider the convergence of the series defining p F q . Theorem 3. The series (23) converges for all t ∈ N. Now suppose t ∈ C \ N, then (23) converges when: i. p + n < q + 1, ii. p + n = q + 1 and |ξ |n n < 1, or iii. p + n = q + 1, |ξ |n n = 1, and (︃ q )︃ p n−1 ∑︁ ∑︁ 1 ∑︁ Re bk − ak − −t + k > 0. n k=1

k=1

k=0

Also, (23) diverges provided that iv. p + n > q + 1, or v. p + n = q + 1 and |ξ |n n > 1. Proof. If t ∈ N, then (−t)nk eventually terminates by definition. Assume t ∈ C \ N. For j = 1, . . . , p, we see (a j )k+1 (b j )k 1 = a j + k. Similarly, for j = 1, . . . , q, we see = . Now we compute (a j )k (b j )k+1 b j + k (−t)n(k+1) (−t)nk (−t + nk)n = = (−t + nk)(−t + nk + 1) . . . (−t + nk + n − 1). (−t)nk (−t)nk



67

Applying the ratio test, we take the limit as k → ∞ of ⃒ ⃒ ⃒ ⃒ ⃒ (a1 + k) . . . (a p + k)(nk − t) . . . (nk − 1 + n − t) ⃒ ⃒ n n k n+p + . . . ⃒ ⃒ ⃒ξ ⃒ = ⃒ξ ⃒ ⃒ ⃒ k q+1 + . . . ⃒ . (b1 + k) . . . (b q + k)k The limit is zero when p + n < q + 1, yielding convergence of p F q in that case. The limit is infinity when p + n > q + 1, yielding divergence in that case. If p + n = q + 1, then the limit is |ξ |n n and so when this quantity is less than 1 we get convergence and when it is greater than one we get divergence. The case |ξ |n n = 1 can be resolved by noting (24) and the well-known convergence properties of q+1 Fq [2, p. 74]. We now provide an analogue of (2). Theorem 4. We have p ∏︁

[︃ ∆

n

(︃ p Fq

a1 , . . . , a p

)︃]︃ ; t, 1, ξ

=

b1 , . . . , b q

j=1 q ∏︁

(a j )n

a1 + n, . . . , a p + n

(︃ ξ

n

p Fq

)︃ ; t, 1, ξ

.

b1 + n, . . . , b q + n

(bℓ )n

ℓ=1

Proof. We compute [︃ (︃ ∆

n

p Fq

a1 , . . . , a p

)︃]︃

[︃

; t, 1, ξ

=∆

b1 , . . . , b q

n

∞ ∑︁ (a1 )k . . . (a p )k ξ k (−1)k (−t)k k! (b1 )k . . . (b q )k

]︃

k=0

∞ ∑︁ (a1 )k . . . (a p )k ξ k (−1)k−n (−t)k−n = (b1 )k . . . (b q )k (k − n)! k=n

=

∞ ∑︁ (a1 ) k=0 p ∏︁

. . . (a p )k+n ξ k+n (−1)k (−t)k k! (b1 )k+n . . . (b q )k+n k+n

(a j )n

=

j=1 q ∏︁

(︃ ξ

n

p Fq

(bℓ )n

a1 + n, . . . , a p + n

)︃ ; t, 1, ξ

,

b1 + n, . . . , b q + n

ℓ=1

which is exactly the desired relation. Corollary 5. We have [︃ (︃ ∆

p Fq

a1 , . . . , a p

)︃]︃ ; t, 1, ξ

b1 , . . . , b q

a a . . . ap = 1 2 ξ p Fq b1 b2 . . . b p

(︃

a1 + 1, . . . , a p + 1

)︃ ; t, 1, ξ

.

b1 + 1, . . . , b q + 1

Define the function shift operator ρ by (ρf )(t) = f (t − 1), and define the operator Θ by Θ = tρ∆, which is a discrete analogue of θ. Lemma 6. We have Θ(−1)k (−t)k = k(−1)k (−t)k for k ∈ N. Proof. We calculate Θ(−1)k (−t)k

= tρ∆[(−1)k (−t)k ] = tρ[k(−1)k−1 (−t)k−1 ] = t[k(−1)k−1 (−t + 1)k−1 ] = k(−1)k (−t)k ,

which is exactly the desired relation.


68 | Martin Bohner and Tom Cuchta We now prove that p F q satisfies a difference equation that resembles a generalized form of (3). Theorem 7. Define y(t) = p F q (a1 , . . . , a p ; b1 , . . . , b q ; t, n, ξ ). Then y satisfies the difference equation ⎡ ⎤ )︂ )︂ p (︂ q (︂ ∏︁ ∏︁ 1 1 ⎣Θ Θ + b j − 1 − nξ (−1)n (−t)n ρ n Θ + a i ⎦ y(t) = 0. n n j=1

i=1

Proof. We first compute )︂ p (︂ ∏︁ 1 n n Θ + a i y(t) nξ (−1) (−t)n ρ n

)︂ p (︂ ∞ ∑︁ (a1 )k . . . (a p )k ξ k (−1)nk ∏︁ 1 = nξ (−1) (−t)n ρ Θ + a i (−t)nk k! n (b1 )k . . . (b q )k n

i=1

n

i=1

k=0

p ∞ ∑︁ (a1 )k . . . (a p )k ξ k (−1)nk (−t)nk ∏︁ = nξ (−1) (−t)n ρ (k + a i ) k! (b1 )k . . . (b q )k n

n

i=1

k=0

p ∞ ∑︁ (a1 )k . . . (a p )k ξ k (−1)nk (−t)nk+n ∏︁ n (k + a i ). = nξ (−1) k! (b1 )k . . . (b q )k i=1

k=0

Now we compute )︂ q (︂ ∏︁ 1 Θ Θ + bj − 1 y n

=Θ

j=1

)︂ q (︂ ∞ ∑︁ (a1 )k . . . (a p )k ξ k (−1)nk ∏︁ 1 Θ + b j − 1 (−t)nk k! n (b1 )k . . . (b q )k j=1

k=0

=Θ

∞ ∑︁

q ∏︁ (k + b j − 1)

(a1 )k . . . (a p )k

k=0

ξ k (−1)nk (−t)nk k! (b1 )k . . . (b q )k j=1

∞ ∑︁

(a1 )k . . . (a p )k ξ k (−1)nk (−t)nk k! (b1 )k−1 . . . (b q )k−1 k=0 ∞ k ∑︁ (a1 )k . . . (a p )k ξ (−1)nk−1 (−t)nk−1 = ntρ (b1 )k−1 . . . (b q )k−1 (k − 1)! k=1 ∞ k+1 ∑︁ (a1 ) . . . (a p ) (−1)nk+n−1 (−t)nk+n−1 k+1 k+1 ξ = ntρ k! (b1 )k . . . (b q )k

=Θ

k=0

= nξ (−t)ρ

p ∞ ∑︁ (a1 )k . . . (a p )k ξ k (−1)nk+n (−t)nk+n−1 ∏︁ (k + a i ) k! (b1 )k . . . (b q )k i=1

k=0

p ∞ ∑︁ (a1 )k . . . (a p )k ξ k (−1)nk+n (−t + 1)nk+n−1 ∏︁ (k + a i ) = nξ (−t) k! (b1 )k . . . (b q )k

= nξ (−1)

k=0 ∞ ∑︁

n

k=0

i=1

p (a1 )k . . . (a p )k ξ k (−1)nk (−t)nk+n ∏︁ (k + a i ), k! (b1 )k . . . (b q )k i=1

which is the same series as above, completing the proof. If n = 1 in Theorem 7, then we obtain a direct analogue of (3). Corollary 8. Let y(t) = p F q (a1 , . . . , a p ; b1 , . . . , b q ; t, 1, ξ ). Then y satisfies the difference equation ⎤ ⎡ q p ∏︁ ∏︁ (︀ )︀ ⎣Θ Θ + b j − 1 − ξtρ (Θ + a i )⎦ y(t) = 0. j=1

i=1

3 Contiguous relations We define F, F(a i ±), and F(b i ±) in the same way as F, F(a i ±), and F(b j ±), but we use p F q in place of p Fq . The following with n = 1 is an analogue of [2, p. 82 (12)].



69

Lemma 9. The following recurrence holds for j ∈ {1, 2, . . . , p}: (︂ )︂ 1 Θ + a j F = a j F(a j +). n

Proof. We calculate (︂ )︂ (︂ )︂ ∞ ∑︁ (a1 )k . . . (a p )k ξ k (−1)nk 1 1 Θ + aj F = Θ + a j (−t)nk = F(a j +), n k! n (b1 )k . . . (b q )k k=0

as was to be shown. The following with n = 1 is an analogue of [2, p. 82 (13)]. Lemma 10. The following recurrence holds for j ∈ {1, 2, . . . , q}: )︂ (︂ 1 Θ + b j − 1 F = (b j − 1)F(b j −). n

Proof. We calculate (︂ )︂ ∞ ∑︁ (a1 )k . . . (a p )k 1 ξ k (−1)nk (−t)nk Θ + bj − 1 F = = (b j − 1)F(b j −), n k! (b1 )k . . . (b j )k−1 . . . (b q )k k=0

as was to be shown. The following with n = 1 is an analogue of [2, p. 82 (14)]. Lemma 11. The following formula holds for j ∈ {2, 3, . . . , p}: (a1 − a j )F = a1 F(a1 +) − a j F(a j +). Proof. Applying Lemma 9 with j = 1 and j ∈ {2, . . . , p} yields (a1 − a j )F = (a1 F(a1 +) − ΘF) − (a j F(a j +) − ΘF) = a1 F(a1 +) − a j F(a j +), as was to be shown. The following with n = 1 is an analogue of [2, p. 82 (15)]. Lemma 12. The following formula holds for j ∈ {1, 2, . . . , q}: (a1 − b j + 1)F = a1 F(a1 +) − (b j − 1)F(b j −). Proof. Using Lemma 9 and Lemma 10, we get (a1 − b j + 1)F = a1 F(a1 +) − (b j − 1)F(b j −), as was to be shown. The following with n = 1 is an analogue of [2, p. 83 (16)]. Lemma 13. If p < q and the b i are pairwise different, then we have q

∑︁ 1 ΘF = ξ (−1)n (−t)n ρ n U j F(b j +). n j=1


70 | Martin Bohner and Tom Cuchta Proof. Using (6), we compute 1 ΘF n

[︂ ]︂ ∞ ∑︁ (a1 )k . . . (a p )k ξ k (−1)nk 1 Θ(−t)nk k! n (b1 )k . . . (b q )k k=0 ∞ k nk ∑︁ (a1 )k . . . (a p )k ξ (−1) = (−t)nk (b1 )k . . . (b q )k (k − 1)! k=1 ∞ ∑︁ (a1 )k+1 . . . (a p )k+1 ξ k+1 (−1)nk+n (−t)nk+n = k! (b1 )k+1 . . . (b q )k+1 k=0 ∞ ∑︁ (a ) . . . (a p )k ξ k+1 (−1)nk+n (−t)nk+n . = Sk 1 k k! (b1 )k . . . (b q )k =

(25)

k=0

If p < q, then it is known [2, p. 82] that S n =

q ∑︁ bj Uj . Therefore, bj + n j=1

⎛

⎞ q ∞ k+1 nk+n ∑︁ ∑︁ bj Uj 1 ⎝ ⎠ (a1 )k . . . (a p )k ξ (−1) ΘF = (−t)nk+n . n b j + k (b1 )k . . . (b q )k k! j=1

k=0

Note that (−t)nk+n = (−t)n (−t + n)nk and bj 1 1 = , (b j )k b j + k (b j + 1)k

(26)

so we obtain after interchanging sums, 1 ΘF n

= ξ (−1)n (−t)n ρ n

q ∑︁

U j F(b j +),

j=1

which is exactly the desired relation. The following lemma with n = 1 is an analogue of [2, p. 83 (17)]. Lemma 14. If p < q and the b j are pairwise different, then a1 F = a1 F(a1 +) − ξ (−1)n (−t)n ρ n

q ∑︁

U j F(b j +).

j=1

Proof. Lemma 9 with j = 1 says 1 ΘF + a1 F = a1 F(a1 +). n Combining this with Lemma 13 yields a1 F = a1 F(a1 +) − ξ (−1)n (−t)n ρ n

q ∑︁

U j F(b j +),

j=1

as was to be shown. The following with n = 1 is an analogue of [2, p. 83 (18)]. Lemma 15. When p = q, we have a1 F + ξ (−1)n (−t)n ρ n F = a1 F(a1 +) − ξ (−1)n (−t)n ρ n

q ∑︁

U j F(b j +).

j=1



Proof. By (25), we get

71

∞

∑︁ (a1 ) . . . (a p ) ξ k+1 (−1)nk+n 1 k k ΘF = Sk (−t)nk+n . n k! (b1 )k . . . (b q )k k=0

If p = q, then it is known [2, p. 83] that S n = 1 +

q ∑︁ bj Uj . Interchanging sums and taking into account (26), bj + n j=1

we find

q

∑︁ 1 ΘF = ξ (−1)n (−t)n ρ n F + ξ (−1)n (−t)n ρ n U j F(b j +). n j=1

Lemma 9 with j = 1 says 1 ΘF = a1 F(a1 +) − a1 F. n 1 Combining the previous two formulas to eliminate ΘF yields n a1 F + ξ (−1)n (−t)n ρ n F = a1 F(a1 +) − ξ (−1)n (−t)n ρ n

q ∑︁

U j F(b j +),

j=1

completing the proof. The following lemma with n = 1 is an analogue of [2, p. 84 (20)]. Lemma 16. The following formula holds for p ≤ q and i ∈ {1, 2, . . . , p} : F = F(a i −) + (−1)n (−t)n ρ n

q ∑︁

W j,i F(b j +).

j=1

Proof. We use Lemma 6 to compute 1 ΘF(a i −) n

= =

[︂ ]︂ ∞ ∑︁ (a1 )k . . . (a i − 1)k . . . (a p )k ξ k (−1)nk 1 Θ(−t)nk k! n (b1 )k . . . (b q )k k=0 ∞ k nk ∑︁ a − 1 (a1 ) . . . (a p ) ξ (−1) (−t) i

k=1

k

k

a i + n (b1 )k . . . (b q )k

nk

(k − 1)!

∞ ∑︁ a i − 1 (a1 )k+1 . . . (a p )k+1 ξ k+1 (−1)nk+n (−t)nk+n = a i + k (b1 )k+1 . . . (b q )k+1 k! k=0 ∞ k nk ∑︁ a − 1 (a1 ) . . . (a p ) i k+1 k+1 ξ (−1) (−t)nk = ξ (−1)n (−t)n ρ n . a i + k (b1 )k+1 . . . (b q )k+1 k! k=0

Formula (6) shows (a1 )k . . . (a p )k 1 (a ) . . . (a p )k = τ n,k 1 k . (b1 )k . . . (b q )k a k + n (b1 )k . . . (b q )k Since p ≤ q, it is known [2, p. 84] that τ n,k =

q ∑︁ b j W j,k . bj + n j=1

Therefore, we have

q

∑︁ 1 ΘF(a i −) = (a i − 1)(−1)n (−t)n ρ n W j,i F(b j +). n j=1

By Lemma 9, 1 ΘF(a i −) = (a i − 1)(F − F(a i −)). n 1 Therefore, using the preceding two formulas to eliminate ΘF(a i −), we complete the proof. n Unauthenticated Download Date | 6/23/18 5:19 AM

72 | Martin Bohner and Tom Cuchta The following theorem presents the contiguous relations for p F q analogous to (7), (8), and (9). Theorem 17. If p ≤ q, then there are a total of 2p + q contiguity relations for p F q . If p < q, then ⎧ (a1 − a k )F = a1 F(a1 +) − a k F(a k +), k = 2, 3, . . . , p ⎪ ⎪ ⎪ (a − b + 1)F = a F(a +) − (b − 1)F(b −), k = 1, 2, . . . , q ⎪ 1 1 1 k k k ⎪ ⎪ q ⎪ ∑︁ ⎪ ⎨ F = F(a −) + (−1)n (−t)n ρ n W j,k F(b j +), k = 1, 2, . . . , p k j=1 ⎪ ⎪ q ⎪ ⎪ ∑︁ ⎪ n n ⎪ ⎪ a F = a F(a +) − ξ (−1) (−t) ρ U j F(b j +). n 1 1 1 ⎪ ⎩

(27)

j=1

If p = q, then ⎧ (a1 − a k )F = a1 F(a1 +) − a k F(a k +), ⎪ ⎪ ⎪ (a1 − b k + 1)F = a1 F(a1 +) − (b k − 1)F(b k −), ⎪ ⎪ ⎪ q ⎪ ∑︁ ⎪ ⎨ F = F(a +) + t W j,k F(b j +), k j=1 ⎪ ⎪ q ⎪ ⎪ ∑︁ ⎪ ⎪ ⎪ (a + t)F = a F(a +) − t U j F(b j +). 1 1 ⎪ ⎩ 1

k = 2, 3, . . . , p k = 1, 2, . . . , q k = 1, 2, . . . , p

(28)

j=1

If p = q + 1, then F becomes a constant independent of t whose contiguous relations are exactly the first two of those in (9). Proof. Considering (24), there are a total of 2(p + n) + q + 1 = 2p + 2n + q + 1 contiguous relations of p+n Fq that −t + j for j = 0, . . . , n − 1 are not relevant under may be relevant. The n relations involving the parameters n the definition of contiguous relations for p F q . The first two relations in (27) and (28) are those of Lemma 11 and Lemma 12. Lemma 14 and Lemma 16 complete the proof for p < q. Lemma 15 and Lemma 16 complete the proof for p = q. If p = q + 1, then Theorem 3 guarantees that n = 0 and (2) shows that it is independent of t, completing the proof.

4 Discrete special functions in terms of discrete hypergeometric series The following proposition is immediate from (18) and is an analogue of (11). Proposition 18. For α ∈ C \ {−1}, we have e α (t) = 0 F0 (; ; t, 1, α). Now we show that 0 F0 satisfies numerous difference equations, generalizing (17). Theorem 19. The function y(t) = 0 F0 (; ; t, n, α) satisfies the difference equation t∆y(t − 1) − nα(−1)n (−t)n y(t − n) = 0. Proof. By Theorem 7, y satisfies Θy(t) − nα(−1)n (−t)n ρ n y(t) = 0. Substituting the definition of Θ yields tρ∆y(t) − nα(−1)n (−t)n ρ n y(t) = 0, and simplifying completes the proof.



73

The following formula is an analogue of (12) and follows immediately from (21). Proposition 20. For α ∈ C \ {−1, 1}, we have (︂ sinα (t) = αtρ0 F1

3 α2 ; ; t, 2, − 2 4

)︂ .

The following theorem is a modification and generalization of (19). (︂ )︂ α2 3 Theorem 21. The function y(t) = 0 F1 ; ; t, n, − satisfies the difference equation 2 4 (︂ )︂ nα2 (−1)n t(t − 1) 2 1 1 ∆ y(t − 2) + + t∆y(t − 1) + (−t)n y(t − n) = 0. n n 2 4 Proof. By Theorem 7, y satisfies (︂ Θ

1 1 Θ+ n 2

)︂ y(t) +

nα2 (−1)n (−t)n ρ n y(t) = 0. 4

Substituting in the definition of Θ yields [︂ ]︂ 1 1 nα2 (−1)n tρ∆ t∆y(t − 1) + + (−t)n ρ n y(t) = 0, n 2 4 hence t(t − 1) 2 ∆ y(t − 2) + n

(︂

1 1 + n 2

)︂ t∆y(t − 1) +

nα2 (−1)n (−t)n y(t − n) = 0. 4

This completes the proof. The following formula is an analogue of (13) and follows immediately from (20). Proposition 22. For all α ∈ C \ {−1, 1}, we have (︂ cosα (t) = 0 F1

1 α2 ; ; t, 2, − 2 4

)︂ .

The following theorem generalizes (19), which holds for n = 2. )︂ α2 1 Theorem 23. The function y(t) = 0 F1 ; ; t, n, − satisfies the difference equation 2 4 (︂ )︂ t(t − 1) 2 nα2 (−1)n 1 1 ∆ y(t − 2) + − t∆y(t − 1) + (−t)n y(t − n) = 0. n n 2 4 (︂

Proof. By Theorem 7, y satisfies (︂ Θ

1 1 Θ− n 2

)︂ y(t) +

nα2 (−1)n (−t)n ρ n y(t) = 0. 4

Substituting the definition of Θ yields [︂ ]︂ 1 1 nα2 (−1)n tρ∆ t∆y(t − 1) − y(t) + (−t)n y(t − n) = 0, n 2 4 or equivalently, t(t − 1) 2 ∆ y(t − 2) + n

(︂

1 1 − n 2

)︂ t∆y(t − 1) +

nα2 (−1)n (−t)n y(t − n) = 0, 4

as was to be shown.


74 | Martin Bohner and Tom Cuchta We now relate the discrete Bessel function to the discrete hypergeometric 0 F1 by finding an analogue of (14). Lemma 24. The discrete Bessel function satisfies (︂ )︂ (−1)n (−t)n 1 J n (t) = F ; n + 1; t − n, 2, − . 2n n! 0 1 4 Proof. By direct computation, we get (︂ )︂ (︂ )︂ n−t n+1−t (n − t)2k (t − n)2k = = . 2 2 22k 22k k k Therefore, (22) shows J n (t)

∞ (︀ n−t )︀ (︀ n+1−t )︀ (−1)k (−1)n (−t)n ∑︁ 2 k 2 k = 2n n! (n + 1)k k! k=0 ∞ n ∑︁ (−1) (−t)n (n − t)2k (−1)k = n 2 n! (n + 1)k 22k k! k=0 ∞ k (−1)n (−t)n ∑︁ (t − n)2k (− 14 ) = n 2 n! (n + 1)k k! k=0 (︂ )︂ n (−1) (−t)n 1 = F ; n + 1; t − n, 2, − , 2n n! 0 1 4

which is exactly the desired relation. Now we prove an analogue of (15). Theorem 25. The function y(t) =

(−1)n J n (t)2n n! satisfies the difference equation (−t)n t∆2 y(t − 1) + (2n + 1)∆y(t) + ty(t − 1) = 0.

Proof. By Lemma 24 and Theorem 7, y solves [︂ (︂ ]︂ )︂ (︂ )︂ 1 1 2 2 Θ (−1) (−t)2 ρ y(t) = 0, Θ + (n + 1) − 1 − 2 − 2 4 which becomes

tρ t(t − 1) 2 ∆(tρ∆y(t)) + ntρ∆y(t) + ρ y(t) = 0. 2 2

The product rule for ∆ shows ∆(tρ∆y(t)) = ∆(t∆y(t − 1)) = ∆y(t) + t∆2 y(t − 1). Therefore, y satisfies (︂ )︂ t(t − 1) 2 1 t(t − 1) ∆ y(t − 2) + n + t∆y(t − 1) + y(t − 2) = 0. 2 2 2 Dividing by t, replacing t with t + 1, and multiplying by 2 complete the proof.

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