title: Discrete fractional calculus and the Saalschutz theorem
authors: Rui A. C. Ferreira
text: 
  
  
  
  
  
  
  Grupo Física-Matemática, Faculdade de Ciências, Universidade de Lisboa,
  Av. Prof. Gama Pinto, 2, 1649-003, Lisboa, Portugal.
  
  
  
  raferreira@fc.ul.pt
  
  Rui A. C. Ferreira was supported by the “Fundação para a Ciência e a Tecnologia (FCT)" through the program “Stimulus of Scientific Employment, Individual Support-2017 Call" with reference CEECIND/00640/2017.
  
  
  
  
  [2000]Primary 39A12 , 33C20
  
  
  
  
  
  
  
  In this work we present a novel proof of the Saalschutz formula by using the theory of discrete fractional calculus. The proofs of some results within this theory, namely, the fractional power rule and the fractional Leibniz rule are revisited.
  
  
  Discrete fractional calculus and the Saalschutz theorem
      Rui A. C. Ferreira
      
  =======================================================
  
  
  
  
  
  
  § INTRODUCTION
  
  
  This work had its origin when the author found a (presumably) novel way to obtain the  Saalschutz formula (cf. <cit.>), namely,
  
  
      (c-a)_m(c-b)_m/(c)_m(c-a-b)_m=_3F_2(a,b,-m;c,1+a+b-c-m;1),
  
  using the discrete fractional calculus theory. In order to understand (<ref>), we introduce the following concepts:
  
  
  The Pochhammer symbol is defined, for x,y∈ℝ, by
  
      (x)_y={[ x(x+1)…(x+y-1)  y∈ℕ_1,;                1  y=0,;  Γ(x+y)/Γ(x) x,x+y∉ℕ^0;        0 x∈ℕ^0x+y∉ℕ^0. ].
  
  
  
  
  The function _3F_2 above, known as a hypergeometric function, is defined by
  
      _3F_2(a_1,a_2,a_3;b_1,b_2;z)=∑_k=0^∞(a_1)_k(a_2)_k(a_3)_k/(b_1)_k(b_2)_kz^n/n!,
  
  when the series converges.
  
  Meanwhile, while deriving (<ref>), we recast some results already known in the literature, namely the fractional power rule and the fractional Leibniz rule. In particular, we will present novel proofs for the fractional power rule and for the Leibniz rule respectively, pointing out some inconsistencies that we detected in the process (cf. Remark <ref> and Remark <ref> below).
  
  In this work we use the Miller–Ross fractional sum-difference operator introduced in 1988 in <cit.>. We give its definition in the next section, more precisely, in Definition <ref>. For this operator, we present and prove fractional power rules as well as the fractional Leibniz rule. These are the key ingredients to obtain the equality (<ref>), though with some restrictions in the parameters.
  
  The theory of discrete fractional calculus and fractional difference equations has been developed in recent years in several directions and is currently a hot topic of research (cf. for example <cit.>). This work aims at contributing to the development of the theory, namely, by showing how ancient results may be obtained from manipulating with (discrete) fractional calculus formulas.
  
  
  
  § DEFINITIONS AND PRELIMINARY RESULTS
  
  
  In this section we introduce some of the concepts used within the discrete fractional calculus (mainly for the “delta" Δ-case) theory as well as some basic facts about it.
  
  For a number a∈ℝ we put ℕ_a={a,a+1,…} and ℕ^a={…,a-1,a}. Sometimes we will also write ℕ_a^b={a,a+1,…,b}, where b=a+k with k∈ℕ_1. 
  
  
  Consider a function f:ℕ_a→ℝ.
  The forward difference operatorForward difference operator is defined by Δ [f](t)=f(t+1)-f(t), for t∈ℕ_a. Also, we define higher order differences recursively as Δ^n [f] (t)=Δ [Δ^n-1 f] (t), n∈ℕ_1, where Δ^0 is the identity operator, i.e. Δ^0 f(t)=f(t).
  
  
  
  The falling function is defined, for x,y∈ℝ, by
  
      (x)^y={[     x(x-1)…(x-y+1)  y∈ℕ_1,;                    1  y=0,; Γ(x+1)/Γ(x+1-y) x,x-y∉ℕ^-1;          0 x∉ℕ^-1x-y∈ℕ^-1. ].
  
  
  When they are defined, the following formulas
  
      t^α+β=(t-β)^αt^β,
  
  
      Δ[s^α](t)=α t^α-1,
  
  and
  
      (t+α-1)^α=(t)_α,
  
  hold, and will be widely used throughout this manuscript
  
  We now introduce the Miller and Ross fractional operator:
  
  
  Let a∈ℝ, ν∈ℝ\ℕ^0 and f:ℕ_a→ℝ. Then the fractional sum-difference operator of order ν is defined by
  
      Δ_a^-νf(t)=1/Γ(ν)∑_s=a^t-ν(t-(s+1))^ν-1f(s),   t∈ℕ_a+ν.
  
  
  Miller and Ross called (<ref>) the fractional sum of order ν if ν>0. We use here the nomenclature “sum-difference" to include the “diferences", that is, when one is considering ν<0.
  
  It is worth mentioning that Atici and Eloe (who built on the work of Miller and Ross roughly 20 years later <cit.>) defined the fractional sum of order ν>0 by (<ref>) but defined the fractional difference of order μ>0 by (in analogy with the Riemann–Liouville fractional derivative)
  
      Δ_a^μ f(t)=Δ^n[Δ_a^-(n-μ)f](t),   t∈ℕ_a+n-μ,
  
  where n∈ℕ_1 is such that n-1<μ≤ n. Let us briefly describe the difference between both concepts. To simplify, consider 0<μ<1. Then, the Miller–Ross fractional difference of order μ is
  
      ^M-RΔ_a^μf(t)=1/Γ(-μ)∑_s=a^t+μ(t-(s+1))^-μ-1f(s),   t∈ℕ_a-μ,
  
  while the Atici–Eloe one is given by
  
      ^A-EΔ_a^μ f(t)=Δ/Γ(1-μ)∑_s=a^t+μ-1(t-(s+1))^-μf(s),   t∈ℕ_a+1-μ.
  
  Let us expand (<ref>). We have (Δ_t stands for the derivative with respect to t),
  
      ^A-EΔ_a^μ f(t)   =1/Γ(1-μ)(∑_s=a^t+μ(t-s)^-μf(s)-∑_s=a^t+μ-1(t-s-1)^-μf(s))
         =1/Γ(1-μ)(∑_s=a^t+μ-1Δ_t(t-s-1)^-μf(s)+Γ(1-μ)f(t+μ))
         =1/Γ(-μ)∑_s=a^t+μ-1(t-s-1)^-μ-1f(s)+f(t+μ)
         =1/Γ(-μ)∑_s=a^t+μ(t-s-1)^-μ-1f(s),   t∈ℕ_a+1-μ.
  
  Therefore, we may say that the Miller–Ross fractional difference and the Atici–Eloe fractional difference coincide but have different domains of definition.
  
  
  
  
  
  § MAIN RESULTS
  
  
  We start with a binomial-type formula for which, for completeness, we present a proof.
  
  
  Let x,y∈ℝ and n∈ℕ_0. Then
  
      (x+y)^n=∑_k=0^nnkx^n-ky^k.
  
  
  
  
  We use induction on n. The case n=0 is obvious. Now, suppose (<ref>) holds for any real numbers x,y. Then,
  
      ∑_k=0^n+1n+1kx^n+1-ky^k=x^n+1+y^n+1
          +∑_k=1^n[nk+nk-1]x^n+1-ky^k
         =∑_k=0^nnkx^n+1-ky^k+∑_k=0^n-1nkx^n-ky^k+1+y^n+1
         =∑_k=0^nnkx^n-k(x-n+k)y^k+∑_k=0^nnkx^n-ky^k(y-k)
         =(x+y-n)(x+y)^n=(x+y)^n+1
          ,
  
  where we have used (<ref>) repeatedly. The proof is done.
  
  
  
  It can also be shown that,
  
      (x+y)_n=∑_k=0^nnk(x)_n-k(y)_k,
  
  for all x,y∈ℝ and n∈ℕ_0.
  
  We now proceed to present the fractional power rule. The proof is inspired by the work of Gray and Zhang <cit.>.
  
  
  Let a∈ℝ. Assume μ∈ℝ\ℕ^-1 and ν∈ℝ\ℕ^0. Then
  
      Δ_a+μ^-ν[(s-a)^μ](t)=Γ(μ+1)/(t-a-μ-ν)!(μ+ν+1)_t-a-μ-ν, t∈ℕ_a+μ+ν.
  
  
  
  
  Let t∈ℕ_a+μ+ν. Then
  
      Δ_a+μ^-ν[(s-a)^μ](t)   =1/Γ(ν)∑_s=a+μ^t-ν(t-(s+1))^ν-1(s-a)^μ
         =∑_s=a+μ^t-ν1/Γ(ν)Γ(t-s)/Γ(t-s+1-ν)Γ(s-a+1)/Γ(s-a+1-μ)
         =∑_s=0^t-a-μ-ν1/Γ(ν)Γ(t-s-a-μ)/Γ(t-s-a-μ+1-ν)Γ(s+μ+1)/Γ(s+1).
  
  Put n=t-a-μ-ν∈ℕ_0. Then we get from the previous equality,
  
      Δ_a+μ^-ν[(s-a)^μ](t)
              =∑_s=0^n1/Γ(ν)Γ(n-s+ν)/Γ(n-s+1)Γ(s+μ+1)/Γ(s+1)
         =Γ(μ+1)/n!∑_s=0^nn!/(n-s)!s!Γ(n-s+ν)/Γ(ν)Γ(s+μ+1)/Γ(μ+1)
         =Γ(μ+1)/n!∑_s=0^nns(ν)_n-s(μ+1)_s
         =Γ(μ+1)/n!(μ+ν+1)_n,
  
  where we have used (<ref>). Finally, substituting n by t-a-μ-ν, we obtain (<ref>).
  
  
  
  Let a∈ℝ. Assume μ∈ℝ\ℕ^-1 and ν∈ℝ\ℕ^0.
  
  If μ+ν∉ℕ^-1, then
  
      Δ_a+μ^-ν[(s-a)^μ](t)=Γ(μ+1)/Γ(μ+ν+1)(t-a)^μ+ν, t∈ℕ_a+μ+ν,
  
  while, if μ+ν∈ℕ^-1, then
  
      Δ_a+μ^-ν[(s-a)^μ](t)=0, t∈ℕ_a.
  
  
  
  
  If μ+ν∉ℕ^-1, then 
  
      (μ+ν+1)_t-a-μ-ν=Γ(t-a+1)/Γ(μ+ν+1),   t∈ℕ_a+μ+ν.
  
  Hence, (<ref>) becomes
  
      Δ_a+μ^-ν[(s-a)^μ](t)=Γ(μ+1)/Γ(μ+ν+1)Γ(t-a+1)/Γ(t-a+1-μ-ν),
  
  which is just (<ref>).
  
  If μ+ν∈ℕ^-1 then, using Definition <ref>, (μ+ν+1)_t-a-μ-ν=0 for t∈ℕ_a, and this concludes the proof.
  
  
  
  Note that, for the values of t such that the function f(t)=(t-a)^μ is well-defined, we have Δ^m f(t)=μ^m(t-a)^μ-m, for m∈ℕ_1. Formally, this formula is in accordance with (<ref>).
  
  
  
  The formula (<ref>) is in many contexts presented assuming only that μ∈ℝ\ℕ^-1 (cf., e.g.,  <cit.> and <cit.>). However, the authors usually present the falling function as
  
      t^ν=Γ(t+1)/Γ(t+1-ν),
  
  for any t, ν∈ℝ for which the right-hand side is well-defined, which is rather vague. In Corollary <ref> we clearly stated the set of values for the parameters a,μ,ν in which equality (<ref>) is valid.
  
  
  
  Some interesting consequences may be extracted from Corollary <ref>. Under the conditions of it, consider a=0, μ+ν∈ℕ^-1, and fix equality (<ref>). We then have,
  
      1/Γ(ν)∑_k=μ^t-νΓ(t-k)/Γ(t-k+1-ν)Γ(k+1)/Γ(k+1-μ)=0,   t∈ℕ_0,
  
  which is equivalent to
  
      ∑_k=0^t-(μ+ν)Γ(t-(k+μ))/Γ(t-(k+μ)+1-ν)Γ(k+μ+1)/Γ(k+μ+1-μ)=0.
  
  Let n=t-(μ+ν) and note that n∈ℕ_-(μ+ν). It follows from the previous equality, 
  
      ∑_k=0^nΓ(n+ν-k)/(n-k)!Γ(k+μ+1)/k!=0,
  
  or
  
      ∑_k=0^nnkΓ(n+ν-k)Γ(k+μ+1)=0, n∈ℕ_-(μ+ν).
  
  
  
  
  A word of caution: in the context of discrete fractional calculus it is essential to keep track of the domains of the fractional operators involved or one may be lead to some inconsistencies.  As an example, we now consider <cit.>. There, it is shown that (with a slight change in notation)
  
      ∇_a+1^α[(s-a)_p](t)=0.
  
  Nothing is written about the domain of the nabla operator ∇^α_a+1 (cf. (<ref>) below). When proving <cit.> the authors use the following definition for ∇^α_a+1 (which seems to derive from <cit.>):
  
      ∇_a+1^α[(s-a)_p](t)=1/Γ(-α)∑_j=a+1^t(t-j+1)_-α-1(j-a)_p.
  
  From <cit.> one can only assume that t∈ℕ_a+1. However, if we consider <cit.>, we get,
  
      ∇_1^3/2[(s)_1/2](1)=1/Γ(-3/2)∑_j=1^1(1-j+1)_-3/2-1(j)_1/2=Γ(3/2)
  
  and not zero as stated therein.
  
  We point out that we may derive a correct formula for the nabla case, i.e. for (<ref>), by using (<ref>). Indeed, for μ+ν=-m with m∈ℕ_1 and t∈ℕ_a, we have
  
      ∑_s=a+μ^t-ν(t-s-1)^ν-1(s-a)^μ=0
      ⇔   ∑_s=a^t-μ-ν(t-s-μ-1)^ν-1(s-a+μ)^μ=0
      ⇔   ∑_s=a^t+m(t-s+1+m)_ν-1(s-a+1)_μ=0
      ⇔   ∑_s=a+1^t+m+1(t+m+1-s+1)_ν-1(s-a)_μ=0.
  
  Therefore,
  
      1/Γ(-α)∑_j=a+1^t(t-j+1)_-α-1(j-a)_p=0,   t∈ℕ_a+1+m, α-p=m∈ℕ_1.
  
  To finalize this issue we test the previous formula with a=0, α=3/2, p=1/2 (hence, m=1), and t=2:
  
      ∑_j=1^2(2-j+1)_-3/2-1(j)_1/2=(2)_-5/2(1)_1/2+(1)_-5/2(2)_1/2=0.
  
  
  Now we need a preparation lemma for the Leibniz rule.
  
  
  Suppose that g:ℕ_a→ℝ and k∈ℕ_0, α∈ℝ. Then
  
      ∑_n=0^k(-1)^nknΔ^ng(t-α-n)=g(t-α-k),   t∈ℕ_a+α+k.
  
  
  
  
  We use induction on k. So, let k=0 and t∈ℕ_a+α. Then the equality trivially holds. Now, assume the equality holds for k∈ℕ_0 and let t∈ℕ_a+α+k+1. We have,
  
      ∑_n=0^k+1(-1)^nk+1nΔ^ng(t-α-n)
         =g(t-α)+(-1)^k+1Δ^k+1g(t-α-(k+1))
         +∑_n=1^k(-1)^n[kn+kn-1]Δ^ng(t-α-n)
         =∑_n=0^k(-1)^nknΔ^ng(t-α-n)-∑_n=0^k(-1)^nknΔ^nΔ g(t-1-α-n)
         =g(t-α-k)-Δ g(t-1-α-k)
         =g(t-α-(k+1)),
  
  and the proof is done.
  
  The following result was proved for the first time in the seminal work by Miller and Ross <cit.> and later in <cit.>. In both works it is assumed that the function g is defined in a union of discrete domains. Here, g is defined on ℕ_a.
  
  
  Suppose that f,g:ℕ_a→ℝ and α∈ℝ\ℕ^0. Then,
  
      Δ_a^-α[fg](t)=∑_n=0^t-α-a-αnΔ_a^-(α+n)f(t)·Δ^ng(t-α-n),   t∈ℕ_a+α.
  
  
  
  
  We start by fixing t∈ℕ_a+α. By Lemma <ref>, we have
  
      ∑_n=0^t-s-α(-1)^nt-s-αnΔ^ng(t-α-n)=g(s),   s∈ℕ_a^t-α.
  
  Therefore,
  
      Δ_a^-α[fg](t)
         =1/Γ(α)∑_s=a^t-α(t-(s+1))^α-1f(s)g(s)
         =1/Γ(α)∑_s=a^t-α∑_n=0^t-s-α(t-(s+1))^α-1(-1)^nt-s-αnf(s)Δ^ng(t-α-n)
         =1/Γ(α)∑_n=0^t-a-α∑_s=a^t-α-n(t-(s+1))^α-1(-1)^nt-s-αnf(s)Δ^ng(t-α-n).
  
  Now, using the equality (-1)^n=(-α)^n/(α)_n, we get from the previous deduction and after some cancellations that,
  
      1/Γ(α)∑_n=0^t-a-α∑_s=a^t-α-n(t-(s+1))^α-1(-1)^nt-s-αnf(s)Δ^ng(t-α-n)
         =∑_n=0^t-a-α(-α)^n/n!1/Γ(n+α)∑_s=a^t-(α+n)(t-(s+1))^α+n-1f(s)Δ^ng(t-α-n)
         =∑_n=0^t-a-α-αnΔ_a^-(α+n)f(t)·Δ^ng(t-α-n),
  
  which concludes the proof.
  
  
  As a first consequence of Theorem <ref> we prove the following result.
  
  
  Consider the parameters α,β,γ such that α∈ℝ\ℕ^0, β,β+γ∈ℝ\ℕ^-1. Then
  
      Γ(β+γ+1)/(t-(β+γ)-α)!(β+γ+α+1)_t-(β+γ)-α/Γ(β+1)
      
      =∑_n=0^t-α-β-γ-αn(α+β+n+1)_(t-α-β-γ-n)γ^n(t-α-n)^γ-n/(t-α-β-γ-n)!,   t∈ℕ_α+β+γ.
  
  
  
  
  Consider the function h(t)=t^β+γ on t∈ℕ_β+γ. By Theorem <ref>, we have
  
      Δ_β+γ^-αh(t)=Γ(β+γ+1)/(t-(β+γ)-α)!(β+γ+α+1)_t-(β+γ)-α, t∈ℕ_α+β+γ.
  
  Now consider the functions f,g:ℕ_β+γ→ℝ defined by
  
      f(t)=(t-γ)^β, g(t)=t^γ.
  
  Then for t∈ℕ_α+β+γ and n∈ℕ_0^t-(β+γ)-α,
  
      Δ_β+γ^-(α+n)f(t)   =Γ(β+1)/(t-α-β-γ-n)!(α+β+n+1)_(t-α-β-γ-n),
      Δ^n g(t)   =γ^nt^γ-n.
  
  Theorem <ref> now implies that,
  
      Γ(β+γ+1)/(t-(β+γ)-α)!(β+γ+α+1)_t-(β+γ)-α
      
      =∑_n=0^t-α-β-γ-αnΓ(β+1)/(t-α-β-γ-n)!(α+β+n+1)_(t-α-β-γ-n)γ^n(t-α-n)^γ-n,
  
  for t∈ℕ_α+β+γ, and (<ref>) follows immediately. 
  
  
  
  Suppose a,c∈ℝ\ℕ^0 with c-a-1∈ℝ\ℕ^-1, and b∈ℝ with c-a-b-1∈ℝ\ℕ^-1. Then for m∈ℕ_0
  
      (c-a)_m(c-b)_m/(c)_m(c-a-b)_m=_3F_2(a,b,-m;c,1+a+b-c-m;1).
  
  
  
  
  Consider the hypothesis of Proposition <ref>, together with α+β∈ℝ\ℕ^-1. Put a=α, b=-γ, c=1+α+β and m=t-α-β-γ.
  
  Then (<ref>) may be written as
  
      Γ(β+γ+1)/m!(β+γ+α+1)_m/Γ(β+1)
      
      =∑_n=0^t-α-β-γ-αn(α+β+n+1)_(t-α-β-γ-n)γ^n(t-α-n)^γ-n/(t-α-β-γ-n)!
      
      =Γ(1+α+β+m)/Γ(m+β+1)∑_n=0^m(α)_n/n!(-γ)_nΓ(m+β+1+γ-n)/Γ(α+β+n+1)(m-n)!,
  
  where we have used (-1)^n(α)_n=(-α)^n. Now, after some rearrangements, we get
  
      Γ(β+γ+1)/Γ(1+α+β+m)(β+γ+α+1)_m/Γ(β+1)Γ(m+β+1)
      
      =Γ(1+β+γ+m)/Γ(1+α+β)∑_n=0^m(α)_n(-γ)_n(-m)_n/(1+α+β)_n(-β-γ-m)_n1/n!,
  
  or
  
      Γ(β+γ+1)Γ(1+α+β)/Γ(1+α+β+m)(β+γ+α+1)_mΓ(m+β+1)/Γ(β+1)Γ(1+β+γ+m)
      
      =∑_n=0^m(α)_n(-γ)_n(-m)_n/(1+α+β)_n(-β-γ-m)_n1/n!.
  
  Now, note that
  
      (c-a)_m   =Γ(1+β+m)/Γ(1+β)
      
          (c-b)_m   =Γ(1+α+β+γ+m)/Γ(1+α+β+γ)
      
           (c)_m   =Γ(1+α+β+m)/Γ(1+α+β)
      
            (c-a-b)_m   =Γ(1+β+γ+m)/Γ(1+β+γ).
  
  The equality (<ref>) now follows from the arbitrariness of α,β,γ and t.
  
  
  
  
  § ACKNOWLEDGMENTS
  
  The author would like to thank the referees for their careful reading of the manuscript, and their corrections and suggestions which contributed to improve this article.
  
  
  
  amsplain
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