Based on the k-Mittag-Lefler function and the k-α-Exponential Function  we introduce families of functions that allows us define new fractional  trigonometric functions that contain the classical trigonometric  functions as particular case for some convenient election of parameters.  We study some elementary properties and obtain the Laplace transform  of some elements of the families.Fil: Cerutti, Ruben Alejandro. Universidad Nacional del Nordeste. Facultad de Ciencias Exactas Naturales y Agrimensura; ArgentinaFil: Luque, Luciano Leonardo. Universidad Nacional del Nordeste. Facultad de Ciencias Exactas Naturales y Agrimensura; Argentina. Consejo Nacional de Investigaciones Científicas y Técnicas; Argentin

Cerutti, Ruben Alejandro

Luque, Luciano Leonardo

International Journal of Contemporary Mathematical Sciences

English

Based on the k-Mittag-Lefler function and the k- -Exponential Function
we introduce families of functions that allows us define new fractional
trigonometric functions that contain the classical trigonometric
functions as particular case for some convenient election of parameters.
We study some elementary properties and obtain the Laplace transform
of some elements of the families.Fil: Cerutti, Rubén Alejandro. Universidad Nacional del Nordeste. Facultad de Ciencias Exactas y Naturales y Agrimensura; Argentina.Fil: Luque, Luciano Leonardo. Consejo Nacional de Investigaciones Científicas y Técnicas. Instituto de Modelado e Innovación Tecnológica; Argentina.Fil: Luque, Luciano Leonardo. Universidad Nacional del Nordeste. Facultad de Ciencias Exactas y Naturales y Agrimensura. Departamento de Matemática; Argentina

Cerutti, Rubén Alejandro

LAReferencia - Red Federada de Repositorios Institucionales de Publicaciones Científicas Latinoamericanas

k-Fractional trigonometric functions

Ruben A. Cerutti

Luciano L. Luque

Crossref

CONICET Digital

International Journal of Contemporary Mathematical SciencesVol. 9, 2014, no. 12, 569 - 578HIKARI Ltd, www.m-hikari.comhttp://dx.doi.org/10.12988/ijcms.2014.4885k-Fractional Trigonometric FunctionsRubén A. Cerutti and Luciano L. LuqueFaculty of Exact SciencesNational University of the NortheastAv. Libertad 5540 (3400); Corrientes, ArgentinaCopyright c© 2014 Rubén A. Cerutti and Luciano L. Luque. This is an open access articledistributed under the Creative Commons Attribution License, which permits unrestricteduse, distribution, and reproduction in any medium, provided the original work is properlycited.AbstractBased on the k-Mittag-Lefler function and the k-α-Exponential Func-tion we introduce families of functions that allows us define new frac-tional trigonometric functions that contain the classical trigonometricfunctions as particular case for some convenient election of parameters.We study some elementary properties and obtain the Laplace transformof some elements of the families.Keywords: Fractional Calculus, k-Calculus, Exponential FunctionI IntroductionAs is well known the classical exponential function, and from it, the classicaltrigonometric functions play an important role in the solution of ordinary dif-ferential equation with constant coefficients. Hence the importance of studyingit and its generalizations including the Mittag-Leﬄer function of one or twoparameters, the one introduced by Prajapati (c.f. [8]) or for us in [5] the calledk-Mittag-Leﬄer function given byEγk,α,β(z) =∞∑n=0(γ)n,k znΓk(nα + β)n!(I.1)570 Rubén A. Cerutti and Luciano L. Luquewhere (γ)n,k denote the k-Pochhammer symbol given by(γ)n,k = γ(γ + k)(γ + 2k)...(γ + (n− 1)k) (I.2)and Γk(z) is the k-Gamma Function due to Diaz and Pariguan (c.f.[4])Γk(z) =∫ ∞0tz−1e−tkk dt. (I.3)Elementary calculations allow us to establish the following relationship be-tween (γ)n,k and Γk(z)(γ)n,k =Γk(γ + nk)Γk(γ). (I.4)It can be seen that if in (I.1), is taken k = γ = α = β = 1, the classicalexponential function is obtainedE11,1,1(z) = ez. (I.5)By Euler equation can be established thatez = ex+iy = ex (cos y + i sin y) , (I.6)and considering the series expansion of the exponential function and the sineand cosine functions we havecos z =eiz + e−iz2=∞∑n=0(−1)nz2n(2n)!(I.7)sin z =eiz − e−iz2i=∞∑n=0(−1)nz2n+1(2n+ 1)!. (I.8)The extension to fractional calculus of the exponential function carries withit the extension of trigonometric functions. It may be seen the elementarypresentation (c.f.[1]) and the very interesting one given in [3] in which startingfrom a generalized exponential functioneλ(x−a)α = (x− a)α−1∞∑n=0λn(x− a)nαΓ [α(n+ 1)]= (x− a)α−1Eα,α [λ(x− a)α] , (I.9)where x > a, λ ∈ C, α ∈ R+, α ∈ R and Eα,β(x) is the two paramettersMittag-Leﬄer function.Eα,β(x) =∞∑n=0xnΓ(nα + β)(I.10)k-Fractional trigonometric functions 571may be extended to Fractional Calculus definitions of sine and cosine functionbycosα [λ(x− α)] =∞∑n=0(−1)nλ2n(x− a)(2n+1)α−1Γ [(2n+ 1)α](I.11)andsinα [λ(x− α)] =∞∑n=0(−1)nλ2n+1(x− a)(n+1)2α−1Γ [(n+ 1)2α](I.12)verifyingDαa+ [sinα(λx)] = λ cosα(λx) (I.13)andDαa+ [cosα(λx)] = −λ sinα(λx) (I.14)where Dαa+ denote the Riemann-Liouville fractional derivative of order α de-fined by(Dαa+f) (x) =[DnIn−αa+ f](x) (I.15)being D the usual derivative operator and Iαa+ the fractional Riemann-Liouvilleintegral operator of order α.Through this paper we will use frequentlyDγ [(x− a)α] = Γ(α + 1)Γ(α− γ + 1)(x− a)α−γ (I.16)(c.f. [2]), andΓ(x) =(1k)x−1Γk(kx) (I.17)(c.f.[4]).II Riemann-Liouville Fractional Derivative oforder ak of the k-α-Exponential FunctionIn this paragraph, based on the k-α-Exponential Function, new definitionsof trigonometric functions are presented and some elementary properties ofthem are studied. To do that we will begin computing the Riemann-Liouvillefractional derivative of order α/k of the k-α-Exponential Function.Let keazγ,α be the k-α-Exponential Function (c.f.[7]) given bykeazγ,α = zαk−1Eγk,α,α(azαk)= zαk−1∞∑n=0(γ)n,γ(azαk)nΓk [(n+ 1)α]n!. (II.1)572 Rubén A. Cerutti and Luciano L. LuqueTaking into account formulae (I.16) we haveDαk[keazγ,α]=∞∑n=1(γ)n,kanΓ[αk(n+ 1)]zαkn−1Γk [(n+ 1)α]n! Γ[αk(n+ 1)− αk] . (II.2)Reminding the relationship between the two Gamma Functions given by (I.17)we haveDαk[keazγ,α]=∞∑n=1(γ)n,kank1−αk(n+1)Γk [α(n+ 1)] zαkn−1Γk [α(n+ 1)]n! k1−[αk (n+1)−αk ]Γk [α(n+ 1)− α]=azαk−1kαk∞∑n=0(γ)n+1,k(azαk)(n+ 1)! Γk [α(n+ 1)]. (II.3)If in (II.3); k = γ = 1 is considered we obtain the formulae (23) of [2].When a = iλ, λ ∈ R, and x− a = t > 0, (II.1) can be writtenkeiλtγ,α = tαk−1k Eγα,α(iλtαk)=∞∑n=0(γ)n,kinλntαk(n+1)−1Γk [(n+ 1)α]n!(II.4)where i is the imaginary unit.By grouping summands according to the powers of the imaginary unit andthe parity of n, we obtainkeiλtγ,α =∞∑n=0(−1)n(γ)2n,kλ2ntαk (2n+1)−1Γk [(2n+ 1)α] (2n)!+ i∞∑n=0(−1)n(γ)2n+1,kλ2n+1t2αk (n+1)−1Γk [2α(n+ 1)] (2n+ 1)!.(II.5)Taking into account the above, we definek cosγ,α(λt) = Re(keiλtγ,α)(II.6)andk sinγ,α(λt) = Im(keiλtγ,α). (II.7)If γ = k = 1, coincides with formulaes (20) and (21) from [3]. It may beobserved that if we adopt k = γ = 1, we have1 cos1,α(λt) =∞∑n=0(−1)nλ2ntk(2n+1)−1Γk [(2n+ 1)α]and1 sin1,α(λt) =∞∑n=0(−1)nλ2n+1tk(2n+2)−1Γk [(2n+ 2)α]k-Fractional trigonometric functions 573which coincides when λ = 1 with the ones in [2]. Then, it may be writtenkeiλtγ,α = k cosγ,α(λt) + i k sinγ,α(λt). (II.8)From (II.6) and (II.7) when k = α = γ = 1 we have1 cos1,1(t) = cos(t) = Re(eit)and1 sin1,1(t) = sin(t) = Im(eit).Analogously, it may be obtainedke−iλtγ,α =k cosγ,α(λt)− i k sinγ,α(λt). (II.9)From (II.8) and (II.9) it resultk cosγ,α(λt) =keiλtγ,α + ke−iλtγ,α2(II.10)andk sinγ,α(λt) =keiλtγ,α − ke−iλtγ,α2i. (II.11)Can be clearly seen that when k = α = γ = 1 results the known expressionfor the classical trigonometric functions given by (I.7) and (I.8).From the defining formulae (II.10) and (II.11) may be derived the followingproperties:k cosγ,α(−λt) = k cosγ,α(λt) (II.12)k sinγ,α(−λt) = −k sinγ,α(λt) (II.13)k cos2γ,α(λt) + k sin2γ,α(λt) =k eiλtγ,α ke−iλtγ,α (II.14)By replacing k,α, γ and λ for 1 in (II.14) is obtained the well known relationcos2(t) + sin2(t) = 1. (II.15)III The Laplace Transform of the k-α-ExponentialFunctionWe begin by recalling the Laplace Transform of the function tαk(n+1)−1, givenbyL [tαk (n+1)−1] (s) = Γ [αk (n+ 1)]sαk((n+1))(III.1)574 Rubén A. Cerutti and Luciano L. Luque(c.f.[7]).The series expansion of the k-α-Exponential Function allows us evaluateL [keλtγ,α] = L[ ∞∑n=0(γ)n,k λntαk(n+1)−1Γk [(n+ 1)α]n!]=∞∑n=0(γ)n,k λnL [tαk (n+1)−1]Γk [(n+ 1)α]n!,and taking into account (III.1), it resultsL [keλtγ,α] = ∞∑n=0(γ)n,k λnΓ[αk(n+ 1)]n!Γk [(n+ 1)α] sαk(n+1)=∞∑n=0(γ)n,k k1−αkn−αk ann!sαknsαk=k1−αksαk∞∑n=0(γ)n,kn!(λ(ks)αk)n. (III.2)According the notation by [8] formula (2.22) we may writteL [keλtγ,α] = k1−αksαk(1− λ(ks)αk)−γ,k. (III.3)If in (III.3) we consider k = γ = 1 we haveL [eλtα ] = 1sα∞∑n=0(λsα)n=1sα − λ (III.4)(c.f. [6]).For the convergence of the series in (III.2) see [8] and also [9].To obtain the Laplace Transform of k cosγ,α(λt) and k sinγ,α(λt) takinginto account (II.10) and (II.11) we haveL [k cosγ,α(λt)] = 12{L [keiλtγ,α]+ L [ke−iλtγ,α ]} (III.5)and from (III.3) it resultsL [k cosγ,α(λt)] = k1−αk2sα{(1− λ(ks)αk)−γ,k+(1 +λ(ks)αk)−γ,k}. (III.6)When taken k = α = γ = 1; (III.6) reduces to the classical expressionL [cos(λt)] = ss2 + λ2(III.7)Analogous it may be obtainedL [sin(λt)] = ss2 + λ2(III.8)k-Fractional trigonometric functions 575IV The Ek,γ,αj FunctionMotivated by the result expressed by the formula (II.5) and looking for tobe verified the classical relationship between the derivative of the sine(cosine)with the cosine(sine), by considering the variation of family of the index of thek-Pochhammer symbol, we introduce the following family of Mittag-Leﬄertype.Definition 1 : Let α, γ ∈ C, Re(γ) > 0, Re(s) > 0, k > 0 and z ∈ C− {0},and j ∈ N0.Ek,γ,αj (λz) =∑n≥0(γ)n+j,kλnzαk(n+1)−1Γk(α(n+ 1))(n+ j)!. (IV.1)It can be seen that if in (IV.1) is considered j = 0, is obtained the functionk-α-Exponential (c.f.[7])keλzγ,α = zα/k−1Eγk,α,α(λzα/k), (IV.2)When j = 0 and k = 1 is obtained the function α-Exponencial functiongiven by (I.9).We will demostrate that the Ek,γ,αj function satisfies the following proper-ties:1. Ek,γ,α0 (λz) =k eλzγ,α.2. Dα/k[Ek,γ,α0 (λz)]= λk−α/kEk,γ,α1 (λz).3. Dα/k[Ek,γ,αj (λz)]= λk−α/kEk,γ,αj+1 (λz).4.(Dα/k)j (keλzγ,α)=(λk−α/k)j Ek,γ,αj (λz).Proof.The first one results from (IV.1) when j = 0. Moreover, it suffices to prove3) ,to see that also verified 2). Finaly we will show the property 3) by inductionon the index j.We will prove that 4) is verified. By using the definition and the relationship(I.16)Dα/k(Ek,γ,αj (λz))=∑n≥1(γ)n+jλnΓk [α(n+ 1)] (n+ j)!Dα/k(zα/k(n+1)−1)=∑n≥1(γ)n+jλnΓk [α(n+ 1)] (n+ j)!Γ(αk(n+ 1))Γ(αnk)zαkn−1576 Rubén A. Cerutti and Luciano L. LuqueNow, taking into account the relationship between the two gamma functions,the classical and k-Gamma, Γ(αnk) = k1−αnn Γ(αn), resultsDα/k(Ek,γ,αj (λz))=∑n≥1(γ)n+jλnΓk [α(n+ 1)] (n+ j)!k−αk Γ(α(n+ 1))Γ(αn)zαkn−1= k−αk∑n≥1(γ)n+j+1λn+1Γk [α(n+ 1)] (n+ j + 1)!zαkn−1= λk−α/kEk,γ,αj+1 (λz). (IV.3)We will prove the property 4) by inducction on j.In view the properties 1) y 2) we known that this relationship is verified toj = 1. Suppose it holds for j = n, we see that this implies that equality holdsfor j = n+ 1:(Dα/k)j+1(Ek,γ,α0 ) = Dα/k(Dα/k)j(Ek,γ,α0 ) = Dα/k[(λk−α/k)j Ek,γ,αj ]=(λk−α/k)jDα/k(Ek,γ,αj )=(λk−α/k)j+1 Ek,γ,αj+1 . (IV.4)that is what we wanted to prove. IV.1 Trigonometric FunctionsOn the basis of (I.1), looking for another generalization of the classical trigono-metric functions cosine and sine, we put byDefinition 2 : Let α, γ ∈ C, Re(γ) > 0, Re(s) > 0, k > 0 and z ∈ C − {0}y j ∈ N0j cosk,γ,α(λz) = Re{Ek,γ,αj (iλz)}j sink,γ,α(λz) = Im{Ek,γ,αj (iλz)} (IV.5)o equivalentlyj cosk,γ,α(λz) =∑n≥0(−1)n(γ)2n+j,kλ2nzα/k(2n+1)−1Γk [α(2n+ 1)] (2n+ j)!. (IV.6)j sink,γ,α(λz) =∑n≥0(−1)n(γ)2n+1+j,kλ2n+1zα/k(2n+2)−1Γk [α(2n+ 2)] (2n+ 1 + j)!. (IV.7)k-Fractional trigonometric functions 577The functions introduced in (IV.5) satisfies the following properties:Lemma: Given α, γ ∈ C, Re(γ) > 0, Re(s) > 0, k > 0 and z ∈ C− {0} yj ∈ N0, then1.Dα/k {j sink,γ,α(λz)} = λk−α/k j+1 cosk,γ,α(λz). (IV.8)2.Dα/k {j cosk,γ,α(λz)} = −λk−α/k j+1 sink,γ,α(λz). (IV.9)Proof. By using the definition (IV.1) and the properties demostrated inIV.1, we haveDα/k[Ek,γ,αj (iλz)]= λk−α/kEk,γ,αj+1 (iλz)= λk−α/k∑n≥0(γ)n+j+1,k(iλ)nzα/k(n+1)−1Γk [α(n+ 1)] (n+ j + 1)!. (IV.10)Grouping of course with the powers of imaginary unit, it resultsDα/k[Ek,γ,αj (iλz)]=− λk−α/k∑n≥0(−1)n (γ)2n+2+j,kλ2n+1zα/k(2n+2)−1Γk [α(2n+ 2)] (2n+ 2 + j)!++ i λk−α/k∑n≥0(−1)n (γ)2n+1+j,kλ2nzα/k(2n+1)−1Γk [α(2n+ 1)] (2n+ 1 + j)!. (IV.11)As Ek,γ,αj (iλz) =j cosk,γ,α(λz) + ij sink,γ,α(λz), results the assertion. References[1] Almusharrf, A. Development of Fractional Trigonometry and anapplications of Fractional Calculus to Pharmacokenetic Model. MatersTheses and Specialist Proyects. (2011). Paper 1048.http://digitalcommons.wku.edu/theses/1048.[2] Bonilla, B.; Ribero, M.; Trujillo, J. Linear Differential Equations offractional order. J. Sabatier et al. (eds.), Advances in FractionalCalculus: Theoretical Developments and Applications in Physics andEngineering. Springer. (2007).578 Rubén A. Cerutti and Luciano L. Luque[3] Bonilla, B.; Rivero, M. and Trujillo,J.; Fractional Differential Ecuationsas alternative models to nonlinear differential equations. AppliedMathematic and Computation. Applied Math. and Computation.187.(2007).[4] Diaz, R.; Pariguan, E. On Hypergeometric Functions andk-Pochhammer symbol. Divulgaciones Matematicas.Vol.15 2, (2007).[5] Dorrego, G.; Cerutti, R. The k-Mittag-Leﬄer Function. Int. J. Contemp.Math. Sciences, no 15, Vol. 7, (2012).[6] Kilvas, H.; Srivastava, H, and Trujillo, J. Theory and Application ofFractional Differential Equations. Elsevier.(2006).[7] Luque, L.; Cerutti, R.; The k-α-Exponential Function. Int. J. of Math.Analysis. Hikari. Vol.7, no 9-12, (2013).[8] Prajapati J.; Shukla, A. Decomposition of Generalized Mittag-Leﬄerfunction and its properties. Advances in Pure Math.. No. 2. (2012).DOI:10.4236/apm.2012.21003.[9] Shah, P.; Jana, R. Results on Generalized Mittag-Leﬄer Function viaLaplace Transform. Applied Math. Sciences. 7(12).(2013).Received: August 1, 2014

International Journal of Contemporary Mathematical SciencesVol. 9, 2014, no. 12, 569 - 578HIKARI Ltd, www.m-hikari.comhttp://dx.doi.org/10.12988/ijcms.2014.4885k-Fractional Trigonometric FunctionsRubén A. Cerutti and Luciano L. LuqueFaculty of Exact SciencesNational University of the NortheastAv. Libertad 5540 (3400); Corrientes, ArgentinaCopyright c© 2014 Rubén A. Cerutti and Luciano L. Luque. This is an open access articledistributed under the Creative Commons Attribution License, which permits unrestricteduse, distribution, and reproduction in any medium, provided the original work is properlycited.AbstractBased on the k-Mittag-Lefler function and the k-α-Exponential Func-tion we introduce families of functions that allows us define new frac-tional trigonometric functions that contain the classical trigonometricfunctions as particular case for some convenient election of parameters.We study some elementary properties and obtain the Laplace transformof some elements of the families.Keywords: Fractional Calculus, k-Calculus, Exponential FunctionI IntroductionAs is well known the classical exponential function, and from it, the classicaltrigonometric functions play an important role in the solution of ordinary dif-ferential equation with constant coefficients. Hence the importance of studyingit and its generalizations including the Mittag-Leffler function of one or twoparameters, the one introduced by Prajapati (c.f. [8]) or for us in [5] the calledk-Mittag-Leffler function given byEγk,α,β(z) =∞∑n=0(γ)n,k znΓk(nα + β)n!(I.1)570 Rubén A. Cerutti and Luciano L. Luquewhere (γ)n,k denote the k-Pochhammer symbol given by(γ)n,k = γ(γ + k)(γ + 2k)...(γ + (n− 1)k) (I.2)and Γk(z) is the k-Gamma Function due to Diaz and Pariguan (c.f.[4])Γk(z) =∫ ∞0tz−1e−tkk dt. (I.3)Elementary calculations allow us to establish the following relationship be-tween (γ)n,k and Γk(z)(γ)n,k =Γk(γ + nk)Γk(γ). (I.4)It can be seen that if in (I.1), is taken k = γ = α = β = 1, the classicalexponential function is obtainedE11,1,1(z) = ez. (I.5)By Euler equation can be established thatez = ex+iy = ex (cos y + i sin y) , (I.6)and considering the series expansion of the exponential function and the sineand cosine functions we havecos z =eiz + e−iz2=∞∑n=0(−1)nz2n(2n)!(I.7)sin z =eiz − e−iz2i=∞∑n=0(−1)nz2n+1(2n+ 1)!. (I.8)The extension to fractional calculus of the exponential function carries withit the extension of trigonometric functions. It may be seen the elementarypresentation (c.f.[1]) and the very interesting one given in [3] in which startingfrom a generalized exponential functioneλ(x−a)α = (x− a)α−1∞∑n=0λn(x− a)nαΓ [α(n+ 1)]= (x− a)α−1Eα,α [λ(x− a)α] , (I.9)where x > a, λ ∈ C, α ∈ R+, α ∈ R and Eα,β(x) is the two paramettersMittag-Leffler function.Eα,β(x) =∞∑n=0xnΓ(nα + β)(I.10)k-Fractional trigonometric functions 571may be extended to Fractional Calculus definitions of sine and cosine functionbycosα [λ(x− α)] =∞∑n=0(−1)nλ2n(x− a)(2n+1)α−1Γ [(2n+ 1)α](I.11)andsinα [λ(x− α)] =∞∑n=0(−1)nλ2n+1(x− a)(n+1)2α−1Γ [(n+ 1)2α](I.12)verifyingDαa+ [sinα(λx)] = λ cosα(λx) (I.13)andDαa+ [cosα(λx)] = −λ sinα(λx) (I.14)where Dαa+ denote the Riemann-Liouville fractional derivative of order α de-fined by(Dαa+f) (x) =[DnIn−αa+ f](x) (I.15)being D the usual derivative operator and Iαa+ the fractional Riemann-Liouvilleintegral operator of order α.Through this paper we will use frequentlyDγ [(x− a)α] = Γ(α + 1)Γ(α− γ + 1)(x− a)α−γ (I.16)(c.f. [2]), andΓ(x) =(1k)x−1Γk(kx) (I.17)(c.f.[4]).II Riemann-Liouville Fractional Derivative oforder ak of the k-α-Exponential FunctionIn this paragraph, based on the k-α-Exponential Function, new definitionsof trigonometric functions are presented and some elementary properties ofthem are studied. To do that we will begin computing the Riemann-Liouvillefractional derivative of order α/k of the k-α-Exponential Function.Let keazγ,α be the k-α-Exponential Function (c.f.[7]) given bykeazγ,α = zαk−1Eγk,α,α(azαk)= zαk−1∞∑n=0(γ)n,γ(azαk)nΓk [(n+ 1)α]n!. (II.1)572 Rubén A. Cerutti and Luciano L. LuqueTaking into account formulae (I.16) we haveDαk[keazγ,α]=∞∑n=1(γ)n,kanΓ[αk(n+ 1)]zαkn−1Γk [(n+ 1)α]n! Γ[αk(n+ 1)− αk] . (II.2)Reminding the relationship between the two Gamma Functions given by (I.17)we haveDαk[keazγ,α]=∞∑n=1(γ)n,kank1−αk(n+1)Γk [α(n+ 1)] zαkn−1Γk [α(n+ 1)]n! k1−[αk (n+1)−αk ]Γk [α(n+ 1)− α]=azαk−1kαk∞∑n=0(γ)n+1,k(azαk)(n+ 1)! Γk [α(n+ 1)]. (II.3)If in (II.3); k = γ = 1 is considered we obtain the formulae (23) of [2].When a = iλ, λ ∈ R, and x− a = t > 0, (II.1) can be writtenkeiλtγ,α = tαk−1k Eγα,α(iλtαk)=∞∑n=0(γ)n,kinλntαk(n+1)−1Γk [(n+ 1)α]n!(II.4)where i is the imaginary unit.By grouping summands according to the powers of the imaginary unit andthe parity of n, we obtainkeiλtγ,α =∞∑n=0(−1)n(γ)2n,kλ2ntαk(2n+1)−1Γk [(2n+ 1)α] (2n)!+ i∞∑n=0(−1)n(γ)2n+1,kλ2n+1t2αk(n+1)−1Γk [2α(n+ 1)] (2n+ 1)!.(II.5)Taking into account the above, we definek cosγ,α(λt) = Re(keiλtγ,α)(II.6)andk sinγ,α(λt) = Im(keiλtγ,α). (II.7)If γ = k = 1, coincides with formulaes (20) and (21) from [3]. It may beobserved that if we adopt k = γ = 1, we have1 cos1,α(λt) =∞∑n=0(−1)nλ2ntk(2n+1)−1Γk [(2n+ 1)α]and1 sin1,α(λt) =∞∑n=0(−1)nλ2n+1tk(2n+2)−1Γk [(2n+ 2)α]k-Fractional trigonometric functions 573which coincides when λ = 1 with the ones in [2]. Then, it may be writtenkeiλtγ,α = k cosγ,α(λt) + i k sinγ,α(λt). (II.8)From (II.6) and (II.7) when k = α = γ = 1 we have1 cos1,1(t) = cos(t) = Re(eit)and1 sin1,1(t) = sin(t) = Im(eit).Analogously, it may be obtainedke−iλtγ,α =k cosγ,α(λt)− i k sinγ,α(λt). (II.9)From (II.8) and (II.9) it resultk cosγ,α(λt) =keiλtγ,α + ke−iλtγ,α2(II.10)andk sinγ,α(λt) =keiλtγ,α − ke−iλtγ,α2i. (II.11)Can be clearly seen that when k = α = γ = 1 results the known expressionfor the classical trigonometric functions given by (I.7) and (I.8).From the defining formulae (II.10) and (II.11) may be derived the followingproperties:k cosγ,α(−λt) = k cosγ,α(λt) (II.12)k sinγ,α(−λt) = −k sinγ,α(λt) (II.13)k cos2γ,α(λt) + k sin2γ,α(λt) =k eiλtγ,α ke−iλtγ,α (II.14)By replacing k,α, γ and λ for 1 in (II.14) is obtained the well known relationcos2(t) + sin2(t) = 1. (II.15)III The Laplace Transform of the k-α-ExponentialFunctionWe begin by recalling the Laplace Transform of the function tαk(n+1)−1, givenbyL[tαk(n+1)−1] (s) = Γ [αk (n+ 1)]sαk((n+1))(III.1)574 Rubén A. Cerutti and Luciano L. Luque(c.f.[7]).The series expansion of the k-α-Exponential Function allows us evaluateL[keλtγ,α]= L[∞∑n=0(γ)n,k λntαk(n+1)−1Γk [(n+ 1)α]n!]=∞∑n=0(γ)n,k λnL[tαk(n+1)−1]Γk [(n+ 1)α]n!,and taking into account (III.1), it resultsL[keλtγ,α]=∞∑n=0(γ)n,k λnΓ[αk(n+ 1)]n!Γk [(n+ 1)α] sαk(n+1)=∞∑n=0(γ)n,k k1−αkn−αk ann!sαknsαk=k1−αksαk∞∑n=0(γ)n,kn!(λ(ks)αk)n. (III.2)According the notation by [8] formula (2.22) we may writteL[keλtγ,α]=k1−αksαk(1− λ(ks)αk)−γ,k. (III.3)If in (III.3) we consider k = γ = 1 we haveL[eλtα]=1sα∞∑n=0(λsα)n=1sα − λ(III.4)(c.f. [6]).For the convergence of the series in (III.2) see [8] and also [9].To obtain the Laplace Transform of k cosγ,α(λt) and k sinγ,α(λt) takinginto account (II.10) and (II.11) we haveL [k cosγ,α(λt)] =12{L[keiλtγ,α]+ L[ke−iλtγ,α]}(III.5)and from (III.3) it resultsL [k cosγ,α(λt)] =k1−αk2sα{(1− λ(ks)αk)−γ,k+(1 +λ(ks)αk)−γ,k}. (III.6)When taken k = α = γ = 1; (III.6) reduces to the classical expressionL [cos(λt)] = ss2 + λ2(III.7)Analogous it may be obtainedL [sin(λt)] = ss2 + λ2(III.8)k-Fractional trigonometric functions 575IV The Ek,γ,αj FunctionMotivated by the result expressed by the formula (II.5) and looking for tobe verified the classical relationship between the derivative of the sine(cosine)with the cosine(sine), by considering the variation of family of the index of thek-Pochhammer symbol, we introduce the following family of Mittag-Lefflertype.Definition 1 : Let α, γ ∈ C, Re(γ) > 0, Re(s) > 0, k > 0 and z ∈ C− {0},and j ∈ N0.Ek,γ,αj (λz) =∑n≥0(γ)n+j,kλnzαk(n+1)−1Γk(α(n+ 1))(n+ j)!. (IV.1)It can be seen that if in (IV.1) is considered j = 0, is obtained the functionk-α-Exponential (c.f.[7])keλzγ,α = zα/k−1Eγk,α,α(λzα/k), (IV.2)When j = 0 and k = 1 is obtained the function α-Exponencial functiongiven by (I.9).We will demostrate that the Ek,γ,αj function satisfies the following proper-ties:1. Ek,γ,α0 (λz) =k eλzγ,α.2. Dα/k[Ek,γ,α0 (λz)]= λk−α/kEk,γ,α1 (λz).3. Dα/k[Ek,γ,αj (λz)]= λk−α/kEk,γ,αj+1 (λz).4.(Dα/k)j (keλzγ,α)=(λk−α/k)j Ek,γ,αj (λz).Proof.The first one results from (IV.1) when j = 0. Moreover, it suffices to prove3) ,to see that also verified 2). Finaly we will show the property 3) by inductionon the index j.We will prove that 4) is verified. By using the definition and the relationship(I.16)Dα/k(Ek,γ,αj (λz))=∑n≥1(γ)n+jλnΓk [α(n+ 1)] (n+ j)!Dα/k(zα/k(n+1)−1)=∑n≥1(γ)n+jλnΓk [α(n+ 1)] (n+ j)!Γ(αk(n+ 1))Γ(αnk)zαkn−1576 Rubén A. Cerutti and Luciano L. LuqueNow, taking into account the relationship between the two gamma functions,the classical and k-Gamma, Γ(αnk) = k1−αnn Γ(αn), resultsDα/k(Ek,γ,αj (λz))=∑n≥1(γ)n+jλnΓk [α(n+ 1)] (n+ j)!k−αk Γ(α(n+ 1))Γ(αn)zαkn−1= k−αk∑n≥1(γ)n+j+1λn+1Γk [α(n+ 1)] (n+ j + 1)!zαkn−1= λk−α/kEk,γ,αj+1 (λz). (IV.3)We will prove the property 4) by inducction on j.In view the properties 1) y 2) we known that this relationship is verified toj = 1. Suppose it holds for j = n, we see that this implies that equality holdsfor j = n+ 1:(Dα/k)j+1(Ek,γ,α0 ) = Dα/k(Dα/k)j(Ek,γ,α0 ) = Dα/k[(λk−α/k)j Ek,γ,αj ]=(λk−α/k)jDα/k(Ek,γ,αj )=(λk−α/k)j+1 Ek,γ,αj+1 . (IV.4)that is what we wanted to prove. IV.1 Trigonometric FunctionsOn the basis of (I.1), looking for another generalization of the classical trigono-metric functions cosine and sine, we put byDefinition 2 : Let α, γ ∈ C, Re(γ) > 0, Re(s) > 0, k > 0 and z ∈ C − {0}y j ∈ N0j cosk,γ,α(λz) = Re{Ek,γ,αj (iλz)}j sink,γ,α(λz) = Im{Ek,γ,αj (iλz)} (IV.5)o equivalentlyj cosk,γ,α(λz) =∑n≥0(−1)n(γ)2n+j,kλ2nzα/k(2n+1)−1Γk [α(2n+ 1)] (2n+ j)!. (IV.6)j sink,γ,α(λz) =∑n≥0(−1)n(γ)2n+1+j,kλ2n+1zα/k(2n+2)−1Γk [α(2n+ 2)] (2n+ 1 + j)!. (IV.7)k-Fractional trigonometric functions 577The functions introduced in (IV.5) satisfies the following properties:Lemma: Given α, γ ∈ C, Re(γ) > 0, Re(s) > 0, k > 0 and z ∈ C− {0} yj ∈ N0, then1.Dα/k {j sink,γ,α(λz)} = λk−α/k j+1 cosk,γ,α(λz). (IV.8)2.Dα/k {j cosk,γ,α(λz)} = −λk−α/k j+1 sink,γ,α(λz). (IV.9)Proof. By using the definition (IV.1) and the properties demostrated inIV.1, we haveDα/k[Ek,γ,αj (iλz)]= λk−α/kEk,γ,αj+1 (iλz)= λk−α/k∑n≥0(γ)n+j+1,k(iλ)nzα/k(n+1)−1Γk [α(n+ 1)] (n+ j + 1)!. (IV.10)Grouping of course with the powers of imaginary unit, it resultsDα/k[Ek,γ,αj (iλz)]=− λk−α/k∑n≥0(−1)n (γ)2n+2+j,kλ2n+1zα/k(2n+2)−1Γk [α(2n+ 2)] (2n+ 2 + j)!++ i λk−α/k∑n≥0(−1)n (γ)2n+1+j,kλ2nzα/k(2n+1)−1Γk [α(2n+ 1)] (2n+ 1 + j)!. (IV.11)As Ek,γ,αj (iλz) =j cosk,γ,α(λz) + ij sink,γ,α(λz), results the assertion. References[1] Almusharrf, A. Development of Fractional Trigonometry and anapplications of Fractional Calculus to Pharmacokenetic Model. MatersTheses and Specialist Proyects. (2011). Paper 1048.http://digitalcommons.wku.edu/theses/1048.[2] Bonilla, B.; Ribero, M.; Trujillo, J. Linear Differential Equations offractional order. J. Sabatier et al. (eds.), Advances in FractionalCalculus: Theoretical Developments and Applications in Physics andEngineering. Springer. (2007).578 Rubén A. Cerutti and Luciano L. Luque[3] Bonilla, B.; Rivero, M. and Trujillo,J.; Fractional Differential Ecuationsas alternative models to nonlinear differential equations. AppliedMathematic and Computation. Applied Math. and Computation.187.(2007).[4] Diaz, R.; Pariguan, E. On Hypergeometric Functions andk-Pochhammer symbol. Divulgaciones Matematicas.Vol.15 2, (2007).[5] Dorrego, G.; Cerutti, R. The k-Mittag-Leffler Function. Int. J. Contemp.Math. Sciences, no 15, Vol. 7, (2012).[6] Kilvas, H.; Srivastava, H, and Trujillo, J. Theory and Application ofFractional Differential Equations. Elsevier.(2006).[7] Luque, L.; Cerutti, R.; The k-α-Exponential Function. Int. J. of Math.Analysis. Hikari. Vol.7, no 9-12, (2013).[8] Prajapati J.; Shukla, A. Decomposition of Generalized Mittag-Lefflerfunction and its properties. Advances in Pure Math.. No. 2. (2012).DOI:10.4236/apm.2012.21003.[9] Shah, P.; Jana, R. Results on Generalized Mittag-Leffler Function viaLaplace Transform. Applied Math. Sciences. 7(12).(2013).Received: August 1, 2014

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k-Fractional trigonometric functions

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