A relation showing that the Grünwald-Letnikov and generalized Cauchy
derivatives are equal is deduced confirming the validity of a well known
conjecture. Integral representations for both direct and reverse fractional
differences are presented. From these the fractional derivative is readily
obtained generalizing the Cauchy integral formula

Coito, Fernando

Duarte Ortigueira, Manuel

Fractional Calculus and Applied Analysis

Bulgarian Digital Mathematics Library at IMI-BAS

FROM DIFFERENCES TO DERIVATIVESManuel Duarte Ortigueira, Fernando CoitoAbstractA relation showing that the Gru¨nwald-Letnikov and generalized Cauchyderivatives are equal is deduced confirming the validity of a well knownconjecture. Integral representations for both direct and reverse fractionaldifferences are presented. From these the fractional derivative is readilyobtained generalizing the Cauchy integral formula.Mathematics Subject Classification: 26A33Key Words and Phrases: fractional difference; Gru¨nwald-Letnikov deriva-tive; generalized Cauchy derivative1. IntroductionIn [1] Diaz and Osler presented a brief insight into the fractional differ-ences. They proposed an integral formulation for the differences and con-jectured about the possibility of using them for defining fractional deriva-tives. This problem was also discussed in a round table held at the Interna-tional Conference on ”Transform Methods & Special Functions, Varna’96”as stated by Kiryakova [3].In this paper, the validity of such conjecture is proved. In fact, we willenounce and later prove the following statement [5]:The Gru¨nwald-Letnikov fractional derivative is equal to the generalizedCauchy derivative.460 M. D. Ortigueira, F. CoitoWe must refer that we will not address the existence problem. We aremainly interested in obtaining generalization for a well known formalism.For the proof, we begin by introducing the fractional differences, consid-ering two cases: direct and reverse1. From these we obtain the direct andreverse Gru¨nwald-Letnikov fractional derivatives. For those differences, in-tegral representations will be proposed following a Diaz and Osler’s idea.The integration path is a U shaped contour that encircles all the poles and”closes” at infinite. From these representations we obtain the fractionaldifferintegrals by using the asymptotic properties of the Gamma function.As we will show there are infinite ways of computing a fractional deriva-tive: we only have to choose a branch cut line and fix a suitable integrationpath. If α is a positive integer, we have an integrand with a pole: this isthe well known Cauchy formula. It is important to remark that accordinglyto the equality of the Cauchy and the Gru¨nwald-Letnikov derivatives, theintegration path has necessarily a form like we referred above, unless α is apositive integer. This means that the derivative definitions based in finitestraight line segments are not valid.A theorem stating the main result of the paper is presented in Section 2.In Section 3, we will introduce the general formulation for the differences,considering two cases: direct and reverse. For these, integral representationswill be proposed. From these representations we obtain in Section 4 thederivative integrals by using the properties of the Gamma function. At lastwe will present some conclusions.2. Main resultThe main result of this paper can be stated in a theorem that we willpresent below. Consider the U shaped contour represented in Figure 1.Let f(z) be a complex variable function analytic in the region inside andcontinuous on that contour. The generalized Cauchy derivative is given by[4-6]:Dαdf(z) =Γ(α+ 1)2pii∫Cdf(w)1(w − z)α+1dw, (1)where we assume that the branch cut line is inside the above referred ana-lyticity region, and Cd is a U shaped contour encircling the branch cut line.By now, we consider a branch cut line (with the corresponding contour) in1We avoid the common used backward and forward since can be misleading.FROM DIFFERENCES TO DERIVATIVES 461Figure 1: U shaped contourthe left hand complex plane or in the lower imaginary half axis. Now, weintroduce the direct Gru¨nwald-Letnikov derivative given by:Dαdf(z) = limh→0+∑∞k=0 (−1)k(αk)f(z − kh)hα, (2)where h is any complex in the right hand complex plane. Making a substi-tution h→ −h, we obtain the reverse Gru¨nwald-Letnikov derivative.Dαr f(z) = (−1)α limh→0+∑∞k=0 (−1)k(αk)f(z + kh)hα. (3)Both expressions agree with the usual derivative definition when α is apositive integer. Moreover, expression (2) corresponds to the so-called left-hand sided Gru¨nwald-Letnikov fractional derivative while (3) has the extrafactor (−1)α, when compared with the right-hand sided Gru¨nwald-Letnikovfractional derivative [5]. We maintain that this factor must be retainedand call the pairs defined by (2) and (3) as Gru¨nwald-Letnikov fractional462 M. D. Ortigueira, F. Coitoderivatives2. We assume also that the points z−nh, n = 0, 1, . . ., are on thebranch cut line.Theorem . Under the above stated conditions we have:limh→0+∑∞k=0 (−1)k(αk)f(z − kh)hα=Γ(α+ 1)2pii∫Cdf(w)1(w − z)α+1 . (4)This theorem remains valid for the reverse Gru¨nwald-Letnikov fractionalderivative provided that we choose a branch cut line in the right handcomplex plane or in the upper imaginary half axis (see Figure 3). Thisresult is very important since it establishes a bridge between two differentformulations both generalizing classical results. Other definitions that arenot logically deduced from (4), although allowing similar results, shouldbe considered as pseudo-derivatives. Although all the above relations areformally valid for any α ∈ R (or even for α ∈ C), in the following we willspeak always of derivative. Before we prove the theorem, we will presentthe integral representations for the fractional differences.3. Differences3.1. DefinitionsLet f(z) be a function of complex variable and introduce 4d and 4r asfinite ”direct” and ”reverse” differences defined by:4df(z) = f(z)− f(z − h) (5)and4rf(z) = f(z + h)− f(z) (6)with h ∈ C and, as before, we assume that Re(h) > 0. The repeated use ofthe above definitions leads to4Nd f(z) =N∑k=0(−1)k(Nk)f(z − kh) (7)2These derivatives were proposed first by Liouville [2]. So, they should be calledLiouville-Gru¨nwald-Letnikov derivatives.FROM DIFFERENCES TO DERIVATIVES 463and4Nr f(z) = (−1)NN∑k=0(−1)k(Nk)f(z + kh) (8)where(Nk)are the binomial coefficients. These definitions are readilyextended to the fractional order case [1]:4αdf(z) =∞∑k=0(−1)k(αk)f(z − kh) (9)and4αr f(z) = (−1)α∞∑k=0(−1)k(αk)f(z + kh). (10)This formulation remains valid in the negative integer order case. Letα = −N (N a positive integer). As it is well known from the Z Transformtheory, the following relation holds if k ≥ 0∞∑n=0(n+ 1)k qn =k!(1− q)k+1 for |q| < 1. (11)Introducing the Pochhammer symbol for the shifted factorial, (a)k =a(a+ 1)(a+ 2) · · · (a+ k − 1) and putting k = N − 1, we obtain easily:(1− z)−N =∞∑n=0(n+ 1)N−1(N − 1)! zn for |z| < 1 (12)leading to4−Nd f(z) =∞∑n=0(n+ 1)N−1(N − 1)! · f(z − nh). (13)For the reverse case, we have:4−Nr f(z) = (z − 1)−N= (−1)N∞∑n=0(n+ 1)N−1(N − 1)! · f(z + nh). (14)As(n+ 1)N−1 =(n+N − 1)!n!=(N − 1)!(N)nn!(15)464 M. D. Ortigueira, F. Coitoand(−a)nn!= (−1)n(an), (16)we have(n+ 1)N−1 = (N − 1)! (−1)n( −Nn). (17)So, we can write:4−Nd f(z) =∞∑n=0(−1)n( −Nn)f(z − nh). (18)For the anti-causal case, we have:4−Nr f(z) = (−1)N∞∑n=0(−1)n( −Nn)f(z + nh). (19)As it can be seen, these expressions are the ones we obtain by puttingα = −N into (9) and (10). So, the relations (9) and (10) are representationsfor the differences of any order.3.2. Integral representations3.2.1. Positive integer orderConsider first the positive integer order case. Assume that f(z) isanalytic inside and on a closed integration path that includes the pointst = z − kh in the direct case and t = z + kh in the corresponding reversecase, with k = 0, 1, . . . , N . The results stated in (9) and (10) can be inter-preted in terms of the residue theorem3. In fact, they can be considered as12pii∑residues in the computation of the integral of a function with polesat t = z − kh and t = z + kh, k = 0, 1, 2, . . .. As it can be seen by directverification, we have [1]:N∑k=0(−1)k(Nk)f(z − kh) = N !2piih·∫Cdf(w)∏Nk=0(w−zh + k)dw (20)3It is important to remark that the poles are simple and that this case can be deducedwithout the use of the derivative notion.FROM DIFFERENCES TO DERIVATIVES 465Figure 2: integration paths and poles for the integral representation ofinteger order differencesandN∑k=0(−1)k(Nk)f(z + kh) = − N !2piih·∫Crf(w)∏Nk=0(z−wh + k)dw. (21)We must remark that the binomial coefficients appear naturally whencomputing the residues.Introducing the Pochhammer symbol, we can rewrite the above formulaeas:4Nd f(z) =N !2piih∫Cdf(w)(w−zh)N+1dw (22)and4Nr f(z) =(−1)N+1N !2piih∫Crf(w)(z−wh)N+1dw. (23)Attending to the relation between the Pochhammer symbol and theGamma function:Γ(z + n) = (z)nΓ(z), (24)466 M. D. Ortigueira, F. Coitowe can write:4Nd f(z) =N !2piih∫Cdf(w)Γ(w−zh)Γ(w−zh +N + 1)dw (25)and4Nr f(z) =(−1)N+1N !2piih∫Crf(w)Γ(z−wh)Γ(z−wh +N + 1)dw. (26)This is correct and is coherent with (20) and (21), because the Gammafunction Γ(z) has poles at the negative integers (z = −n). The correspond-ing residues are equal to (−1)n /n!, [5]. Although both the Gamma functionshave infinite poles, outside the contour they cancel out and the integrand isanalytic. We should also remark that the direct and reverse differences arenot formally equal.3.2.2. Fractional orderConsider the fractional order differences defined in (9) and (10). It isnot hard to see that we are in presence of a situation similar to the positiveinteger case, excepting the fact of having infinite poles. So we have to usean integration path that encircles all the poles. This can be done with a Ushaped contour like those shown in Figure 3.With the suitable adaptations, we obtain from (25) and (26):4αdf(z) =Γ (α+ 1)2piih∫Cf(w)Γ(w−zh)Γ(w−zh + α+ 1)dw (27)and4αr f(z) =(−1)α+1 Γ (α+ 1)2piih∫Cf(w)Γ(z−wh)Γ(z−wh + α+ 1)dw. (28)Remark that one turns into the other with the substitution h → −h,justifying the assumption we made in Section 2.3.2.3. Two propertiesIn the following, we shall be concerned with the fractional order caseand will consider the direct case only. The other one is similar.FROM DIFFERENCES TO DERIVATIVES 467Figure 3: integration paths and poles for the integral representation offractional order differences3.2.3.1. Repeated differencingWe are going to study the effect of a sequential application of the dif-ference operator 4. We have4βd [4αdf(z)] = Γ(β+1)Γ(α+1)(2piih)2∫C∫C f(w)Γ(w−sh )Γ(w−sh +α+1)dwΓ( s−zh )Γ( s−zh +β+1)ds.(29)Permuting the integrations, we have:4βd [4αdf(z)] = Γ(β+1)Γ(α+1)(2piih)2∫C f(w)∫CΓ(w−sh )Γ(w−sh +α+1)Γ( s−zh )Γ( s−zh +β+1)dsdw.(30)468 M. D. Ortigueira, F. CoitoAs, by the residue theoremΓ(β+1)2piih∫CΓ(w−sh )Γ(w−sh +α+1)Γ( s−zh )Γ( s−zh +β+1)ds= 1h∑∞n=0(−1)nn!Γ(β+1)·Γ(w−zh +n)Γ(w−zh +α+1+n)Γ(β−n+1)= 1hΓ(w−zh )Γ(w−zh +α+1)∑∞n=0(w−zh )n(−β)n(w−zh +α+1)n= 1hΓ(w−zh )Γ(w−zh +α+1)2F1(w−zh ,−β, w−zh + α+ 1, 1), (31)where 2F1 is the Gauss hypergeometric function. If α+ β+1 > 0, we have:Γ(β+1)2piih∫CΓ(w−sh )Γ(w−sh +α+1)Γ( s−zh )Γ( s−zh +β+1)ds= 1hΓ(w−zh )Γ(α+β+1)Γ(w−zh +α+β+1)Γ(α+1), (32)leading to the conclusion that:4βd [4αdf(z)] =Γ (α+ β + 1)2piih∫Cf(w)Γ(w−zh)Γ(w−zh + α+ β + 1)dw (33)and4βd [4αdf(z)] = 4α+βd f(z), (34)provided that α+β+1 > 0. It is not difficult to see that the above operationis commutative. The condition α+ β +1 > 0 is restrictive, since we cannothave β ≤ −α − 1. However, we must remark that (27) and (28) are validfor every α ∈ R. The same happens with (α+ β) in (33). This means thatwe can use (33) with every α, β ∈ R, at least formally.3.2.3.2. InversionPutting α = −β into (33), we obtain:4−αd [4αdf(z)] = 4αd[4−αd f(z)] = 12pii∫Cf(w)1w − z dw = f(z) (35)as we would expect. So the operation of differencing is invertible. Thismeans that we can write:f(z) =Γ (α+ 1)2piih∫C4αdf(w)Γ(w−zh)Γ(w−zh − α+ 1)dw (36)FROM DIFFERENCES TO DERIVATIVES 469in the direct case. In the reverse case, we will have:f(z) =Γ (α+ 1)2piih∫C4αr f(w)Γ(z−wh)Γ(z−wh − α+ 1)dw, (37)according to (27).4. Derivatives4.1. On Gru¨nwald-Letnikov derivativesThe Gru¨nwald-Letnikov derivatives are obtained from (9) and (10) bydividing them by hα and performing the limit h → 0+ 4. Although we arenot concerned here with existence problems, we must refer that in generalwe can have the direct derivative without existing the reverse one and vice-versa. For example, let us apply both definitions to the function f(z) =eaz. If Re(a) > 0, expression (2) converges to Dαdf(z) = aαeaz, while (3)diverges. On the other hand, if f(z) = e−az equation (2) diverges while (3)converges to Dαr f(z) = (−a)α e−az. This suggests that (2) and (3) shouldbe adopted for right and left functions5, respectively in agreement withLiouville reasoning [2]. In particular, for the functions such that f(z) = 0for Re(z) < 0 or f(z) = 0 for Re(z) > 0.4.2. Generalizing the Cauchy formulaThe ratio of two gamma functions Γ(s+a)Γ(s+b) has an interesting expansion[6]:Γ(s+ a)Γ(s+ b)= sa−b[1 +N∑1cks−k +O (s−N−1)] (38)as |s| → ∞, uniformly in every sector that excludes the negative real half-axis. The coefficients in the series can be expressed in terms of Bernoullipolynomials. Their knowledge is not important here.Consider (27) and (28) again. Let |h| < ² ∈ R, where ² is a smallnumber. This allows us to write:4αdf(z) =Γ (α+ 1)2piih∫Cdf(w)1(w−zh)α+1dw + hα+2g1(h) (39)40+ means that Re(h) > 05We say that f(t) is a right [left] function if f(−∞) = 0 [f(+∞) = 0].470 M. D. Ortigueira, F. Coitoand4αr f(z) =(−1)α+1 Γ (α+ 1)2piih∫Crf(w)1(z−wh)α+1dw + hα+2g2(h), (40)where Cd and Cr are the contours represented in Figure 3. The g1(h) andg2(h) are functions that assume a finite value near zero. So, the fractionalincremental ratia are given by:4αdf(z)hα=Γ (α+ 1)2pii∫Cdf(w)1(w − z)α+1dw (41)and4αr f(z)hα=Γ (α+ 1)2pii∫Crf(w)1(w − z)α+1dw. (42)Allowing h → 0+, we obtain the direct and reverse generalized Cauchyderivatives:Dαdf(z) =Γ (α+ 1)2pii∫Cdf(w)1(w − z)α+1dw (43)andDαr f(z) =Γ (α+ 1)2pii∫Crf(w)1(w − z)α+1dw. (44)If α = N , both the derivatives are equal and coincide with the usualCauchy definition. In the fractional case we have different solutions, sincewe are using a different integration path. Remark that (43) and (44) areformally the same. They differ only in the integration path. Thus we candefine a generalized Cauchy derivative by:Dαf(z) =Γ (α+ 1)2pii∫Cf(w)1(w − z)α+1dw, (45)where C is any half straight line starting at z.5. ConclusionLet γ be a half straight line beginning at any z ∈ C and let h ∈ γ. TheGru¨nwald-Letnikov fractional derivative is equal to the generalized Cauchyderivative provided that it is computed along an integration path surround-ing the above straight line.FROM DIFFERENCES TO DERIVATIVES 471Figure 4: Gru¨nwald-Letnikov derivative vs Cauchy derivativeReferences[1] J. B. D i a z, T. J. O s l e r, Differences of fractional order. Math-ematics of Computation 28, No 125 (1974), 185-202.[2] S. D u g o w s o n, Les diffe´rentielles me´taphysiques. Ph.D. Thesis,Universite´ Paris Nord (1994).[3] V. K i r y a k o v a, A long standing conjecture failed?. Proceedings ofthe 2nd International Workshop “Transform Methods & Special Func-tions, Varna’96”, IMI- BAS, Sofia (1996), 579-588.[4] K. N i s h i m o t o, Fractional Calculus. Descartes Press Co., Ko-riyama (1989).[5] M. D. Ortigueira, Fractional differences integral representation andits use to define fractional derivatives. Proceedings of the Fifth EU-ROMECH Nonlinear Dynamics Conference (ENOC-2005), Eindhoven(2005), 1333-1337.[6] S. G. S a m k o, A. A. K i l b a s, O. I. M a r i c h e v, Frac-tional Integrals and Derivatives - Theory and Applications. Gordon& Breach Science Publishers, New York (1987).UNINOVA and Faculty of Sciences and TechnologyUniversidade Nova de LisboaQuinta da Torre Received: December 27, 20042825-114 Monte da Caparica, PORTUGALe-mails: mdortigueira@uninova.pt and fjvc@fct.unl.pt

From Differences to Derivatives

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From Differences to Derivatives

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