# N-Dimensional Fractional  Lagrange's Inversion Theorem

Authors: F. A. Abd El-Salam
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

Using Riemann-Liouville fractional differential operator, a fractional extension of the Lagrange inversion theorem and related formulas are developed. The required basic definitions, lemmas,  and theorems in the fractional calculus are presented. A fractional form of Lagrange's expansion for one implicitly defined independent variable is obtained. Then, a fractional version of Lagrange's expansion in more than one unknown function is generalized. For extending the treatment in higher dimensions, some relevant vectors and tensors definitions and notations are presented. A fractional Taylor expansion of a function ofN-dimensional polyadics is derived.  A fractional N-dimensional  Lagrange inversion theorem is proved.


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## Body

## 1. Introduction

The fractional calculus (FC) may be considered as an old and yet novel topic. It dates back to the end of  the seventeenth century through the pioneering works of Leibniz, Euler, Lagrange, Abel, Liouville, and many others. In a letter to L'Hospital in 1695, Leibniz raised the possibility of generalizing the operation of differentiation to noninteger orders, and L'Hospital asked what would be the result of half-differentiatingx. Leibniz replied: It leads to a paradox, from which one day useful consequences will be drawn. The paradoxical aspects are due to the fact that there are several different ways of generalizing the differentiation operator to non-integer powers, leading to inequivalent results.The fractional calculus (FC) generalizes the ordinary differentiation and integration so as to include any arbitrary real or even complex order instead of being only the positive integers (see, e.g., Samko et al. [1], Kilbas, et al. [2], Magin [3], and Podlubny [4]).During the second half of the twentieth century till now, FC gained considerable popularity and importance.  Many authors have explored the world of FC giving new insight into many areas of scientific research in physics, mechanics, and mathematics. Miller and Ross [5] pointed out that there is hardly a field of science or engineering that has remained untouched by the new concepts of FC.Fractional derivatives provide an excellent as well as  very powerful tool for the description and modeling of many phenomena in nature. There are many applications where the fractional calculus can be widely used, for example, viscoelasticity, electrochemistry, diffusion processes, control theory, heat conduction, electricity, mechanics, chaos and fractals, turbulence, fluid dynamics, stochastic dynamical system, plasma physics and controlled thermonuclear fusion, nonlinear control theory, image processing, nonlinear biological systems, astrophysics, and so forth,  see for details [2–12] and the references therein.In a very good book by Baleanu et al. [13], readers were given  the possibility of finding very important mathematical tools for working with fractional models and solving fractional differential equations, such as a generalization of Stirling numbers in the framework of fractional calculus and a set of efficient numerical methods. Moreover, they introduced some applied topics, in particular, fractional variational methods which are used in physics, engineering, or economics. They also discussed the relationship between semi-Markov continuous-time random walks and the space-time fractional diffusion equation, which generalized the usual theory relating random walks to the diffusion equation.Debbouche and Baleanu [14] introduced a new concept called implicit evolution system to establish the existence results of mild and strong solutions of a class of fractional nonlocal nonlinear  integrodifferential system; then they proved the exact null controllability result of a class of fractional evolution nonlocal integrodifferential control systems in Banach space. As an application that illustrates their abstract results, they provided two examples.Babakhani and Baleanu [15] considered a class of nonlinear fractional order differential equations involving Caputo fractional derivative with lower terminal at 0 in order to study the existence solution satisfying the boundary conditions or satisfying the initial conditions. They derived  unique solution under Lipschitz condition. In order to illustrate their results they presented several examples.Finally and roughly speaking, the  fractional calculus may improve the smoothness properties of functions rather than the calculus with integer orders. The development of the FC theory is due to the contributions of many mathematicians such as Euler, Liouville, Riemann, and Letnikov. Several definitions of a fractional derivative have been proposed. These definitions include Riemann-Liouville, Grunwald-Letnikov,Weyl, Caputo, Marchaud, and Riesz fractional derivatives, see Miller and Ross [5] and Riewe [16]. Riemann-Liouville derivative is the most used generalization of the derivatives. It is based on the direct generalization of Cauchy's formula for calculating an n-fold or repeated integral, see Oldham and Spanier [17].In 1770,  Lagrange (1736–1813) published his power series solution of the implicit equation. However, his solution used cumbersome series expansions of logarithms. [18, 19]. This expansion was generalized by Bürmann [20–22]. There is a straightforward derivation using complex analysis and contour integration; the complex formal power series version is clearly a consequence of knowing the formula for polynomials; so the theory of analytic functions may be applied. Actually, the machinery from analytic function theory enters only in a formal way in this proof.  In 1780,  Laplace (1749–1827) published a simpler proof of the theorem, based on the relations between partial derivatives with respect to the variable and the parameter, see [23, 24],  Hermite (1822–1901) presented the most straightforward proof of the theorem by using contour integration [25–27].In mathematical analysis, this series expansions is known as Lagrange inversion theorem, also known as the Lagrange-Bürmann formula, giving the Taylor series expansion of the inverse function.  Suppose thatz=f(w), where  f  is analytic function at a point  a and f(a)≠0. Then, it is possible to invert or solve the equation for  w such that w=g(z) on a neighborhood of f(a), where g is analytic at the point f(a). This is also called reversion of series. The series expansion of  g is given by
(1)w=g(z)=a+∑n=1∞(limw→a((z-f(a))nn!dn-1dwn-1(w-af(w)-f(a))n)).
In this work, we will apply the concepts of fractional calculus to obtain a fractional form of the  Lagrange expansion and some generalizations.


## 2. Basic Definitions and Theorems

Definition 1.
ByD, we denote the operator that maps a differentiable function onto its integer derivative; that is, Df(x)=f′;  by Ja, we denote the integer integration operator that maps a function f, assumed to be (Riemann) integrable on the compact interval [a,b], onto its primitive centered at a; that is,  Jaf(x)=∫axf(t)dtforalla≤x≤b.Definition 2.
ByDn and Jan,n∈ℕ, we denote the n-fold iterates of D and Ja, respectively. Note that Dn  is the left inverse of  Jan  in a suitable space of functions.Lemma 3.
Letf be Riemann integrable on [a,b]. Then, for a≤x≤b and n∈ℕ, one has
(2)Janf(x)=1(n-1)!∫ax(x-t)n-1f(t)dt,n∈ℕ.Definition 4.
The operator𝒥aα, defined on Lebesgue space L1[a,b], denotes the Riemann-Liouville fractional operator of order α. That is,
(3)𝒥aαf(x)=1Γ(α)∫ax(x-t)α-1f(t)dt,a≤x≤b,α∈ℝ.Remark 5.
It is evident that𝒥aα≡Jan,forallα∈ℕ, except for the fact that we have extended the domain from Riemann integrable functions to Lebesgue integrable functions (which will not lead to any problems in our development). Moreover, in the case α≥1, it is obvious that the integral 𝒥aαf(x) exists for every x∈[a,b] because the integrand is the product of an integrable  function f and the continuous function (x-•)α-1. One important property of integer-order integral operators, is preserved by this generalization. That  is,
(4)𝒥aα(𝒥aβf(x))=𝒥aβ(𝒥aαf(x))=𝒥aα+βf(x),α,β>0,f(x)∈L1[a,b].Definition 6.
Letα∈ℝ+ and let m=⌈α⌉, The Riemann-Liouville fractional differential operator of order α is defined as  such that. Then, 𝒟aα=Dam𝒥am-α. That is,
(5)𝒟aαf(x)=1Γ(m-α)(ddx)m∫ax(x-t)m-α-1×f(t)dt,a≤x≤b,α∈ℝ.Lemma 7.
Letα∈ℝ+ and let m∈ℕ such that >α. Then, 𝒟aα=Dam𝒥am-α.Proof.
Sincem>α yields m≥⌈α⌉.
Thus,(6)Dm𝒥am-α=D⌈α⌉Dm-⌈α⌉Jm-⌈α⌉J⌈α⌉-α=D⌈α⌉+m-⌈α⌉-m+⌈α⌉-⌈α⌉+α=Dα=𝒟aα.Theorem 8.
Let(α≥0)∈ℝ+. Then, for every f(x)∈L1[a,b],  𝒟aα𝒥aαf(x)=f(x).Proof.
Forα=0, both operator, are the identity.  For α>0, let m≥⌈α⌉;  then,
(7)𝒟aα𝒥aαf(x)=Dam𝒥am-α𝒥aαf(x)=Dam𝒥amf(x)=DamJamf(x)=f(x).Corollary 9.
Letf be analytic in (a-h,a+h) for some h>0, and let (α≥0)∈ℝ+.
Then,(8)(I-1)𝒥aαf(x)=∑k=0∞(-1)m(x-a)k+αk!(α+k)Γ(α)Dakf(x),∀a≤x<a+h2,(I-2)𝒥aαf(x)=∑k=0∞(x-a)k+αΓ(k+1+α)Dakf(a),∀a≤x<a+h,(D-1)𝒟aαf(x)=∑k=0∞(αk)(x-a)k-α(k+1-α)Dakf(x),∀a≤x<a+h2,(D-2)𝒟aαf(x)=∑k=0∞(x-a)k-α(k+1-α)Dakf(a),∀a≤x<a+h.
The binomial coefficients  for α∈ℝ and k∈ℕ are  defined as
(9)(αk)=α(α-1)(α-2)⋯(α-k+1)k!=α!k!(α-k)!.Proof.
For the first two statements(I-1),(I-2) and we use the definition of the Riemann-Liouville  integral operator 𝒥aα  and expand f(t) into a power series about x.   Since  x∈[a,a+h/2), the power series converges in the entire interval of integration  and exchanges summation and integration. Then, we use the explicit representation for the fractional integral of the power function:
(10)𝒥aα(x-a)k=Γ(k+1)(α+k+1)(x-a)k+α.(I-1) follows immediately.  For the second statement, we proceed in a similar way; but we now expand the power series at a and not at x. This allows us again to conclude the convergence of the series in the required interval. The analyticity of 𝒥aα follows immediately from the second statement.
To prove(D-1) we use the relation(11)𝒟aα=D⌈α⌉𝒥a⌈α⌉-α,k!Γ(α)(α+k)(-αk)=(-1)kΓ(k+1+α).
This allows us to rewrite the statement  (I-1) as
(12)𝒥a⌈α⌉-αf(x)=∑k=0∞(⌈α⌉-αk)(x-a)k+⌈α⌉-αΓ(k+1+⌈α⌉-α)Dakf(x).
Differentiating ⌈α⌉ times with respect to x, we find
(13)Da⌈α⌉𝒥a⌈α⌉-αf(x)=𝒟aαf(x)=∑k=0∞(⌈α⌉-αk)×1Γ(k+1+⌈α⌉-α)×Da⌈α⌉[(•-a)k+⌈α⌉-αDakf](x).
The classical version of Leibniz’ formula yields
(14)𝒟aαf(x)=∑k=0∞(⌈α⌉-αk)1Γ(k+1+⌈α⌉-α)∑j=0⌈α⌉(⌈α⌉j)×Da⌈α⌉-j[(•-a)k+⌈α⌉-α](x)Dak+jf,
which yields
(15)𝒟aαf(x)=∑k=0∞(α-⌈α⌉k)∑j=0⌈α⌉(⌈α⌉j)[(x-a)k+j-α]Γ(k+1+j-α)×(x)Dak+jf.
By definition, (μj)=0   if μ∈ℕ and  μ<j. Thus, we may replace the upper limit in the inner sum by ∞ without changing the expression. The substitution j=l-k gives
(16)𝒟aαf(x)=∑l=0∞∑k=0∞(α-⌈α⌉k)(⌈α⌉l-k)[(x-a)l-α]Γ(l+1-α)(x)Dalf.
Using the fact that ∑l=0∞∑k=0∞=∑l=0∞∑k=0l,
(17)𝒟aαf(x)=∑l=0∞∑k=0l(α-⌈α⌉k)(⌈α⌉l-k)[(x-a)l-α]Γ(l+1-α)(x)Dalf.
And the explicit calculation yields
(18)∑k=0l(α-⌈α⌉k)(⌈α⌉l-k)=(⌈α⌉l),
thus, (D-1) follows directly.


## 3. Fractional Form of Lagrange's Expansion in One Variable

We can use the standard form of Lagrange's expansion for one implicitly defined independent variable and the Definition6 to obtain the fractional form of Lagrange's expansion as follows.Letz be a function of (ζ,ε) and in terms of another function  μ  such that
(19)z=z(ζ,ε)≡ζ+εμ(z).
Then, for any function f(20)f(z)=f(ζ)+∑n=1∞εnn!dn-1dζn-1[μn(ζ)df(ζ)dζ]
for small ε.  If  f is the identity
(21)z=ζ+∑n=1∞εnn!dn-1dζn-1[μn(ζ)],
then
(22)f(z)=f(ζ)+∑n=1∞εnn!dn-1dζn-1[μn(ζ)df(ζ)dζ].
This classical result can be obtained using the following integral:
(23)f(z)=∫δ(εμ(τ)-τ+ζ)f(τ)(1-εμ′(τ))dτ
Now we are going to introduce a fractional form of the Lagrange inversion formula.Rewrite the integral (23) in the fractional form as
(24)f(z)=z𝒥aαδ(εμ(z)-z+ζ)×f(z)(1-εz𝒟aαμ(z))=1Γ(α)∫az(z-τ)α-1δ(εμ(τ)-τ+ζ)×f(τ)(1-ετ𝒟aαμ(τ))dτ.
Writing the delta function as an integral, we have
(25)f(z)=12πΓ(α)∫az∫-ππ(z-τ)α-1e(iξ[εμ(τ)-τ+ζ])12πΓ(α)×f(τ)(1-ετ𝒟aαμ(τ))dξdτ=12πΓ(α)∑n=0∞‍∫az∫-ππ(iξεμ(τ))neik(ζ-τ)n!(z-τ)α-1=12πΓ(α)×f(τ)(1-ετ𝒟aαμ(τ))dξdτ=12πΓ(α)∑n=0∞(z𝒟aα)n∫az∫-ππ(εμ(τ))neiξ(ζ-τ)n!(z-τ)α-1=12πΓ(α)×f(τ)(1-ετ𝒟aαμ(τ))dξdτ.
The integral over  ξ then gives   δ(ζ-τ)=0forallζ-τ=0  and we have
(26)f(z)=1Γ(α)∑n=0∞(ζ𝒟aα)n(εμ(ζ))nn!f(ζ)(1-εζ𝒟aαμ(ζ))=1n!Γ(α)∑n=0∞(ζ𝒟aα)n[εn(μ(ζ))nf(ζ)-εn+1=1n!Γ(α)∑n=0∞(ζ𝒟aα)n×f(ζ)(μ(ζ))n(ζ𝒟aαμ(ζ))]=1n!Γ(α)∑n=0∞(ζ𝒟aα)n[εn(μ(ζ))nf(ζ)-εn+1f(ζ)Γ&1n!Γ(α)∑n=0∞(ζ𝒟aα)n×(1+α)(μ(ζ))n(μ(ζ)1-α)]=1n!Γ(α)∑n=0∞(ζ𝒟aα)n[εn(μ(ζ))nf(ζ)-εn+11n!Γ(α)∑n=0∞(ζ𝒟aα)n×f(ζ)Γ(1+α)(μ(ζ)n+1-α)]=1Γ(α)∑n=0∞(ζ𝒟aα)n[εnn!(μ(ζ))nf(ζ)-εn+1Γ(1+α)Γ(n+1+α)1Γ(α)∑n=0∞(ζ𝒟aα)n×{ζ𝒟aα[f(ζ)(μ(ζ))n+1]1Γ(α)∑n=0∞(ζ𝒟aα)n-ζ𝒟aαf(ζ)(μ(ζ))n+1}εn+1Γ(1+α)Γ(n+1+α)].
On extracting the first term out of summation, set   n+1=k⇒n=k-1,  and rearranging the terms then gives the result:
(27)f(z)=1Γ(α)f(ζ)+∑k=0∞(ζ𝒟aα)k-1α(εk)Γ(k+α)×[(μ(ζ))k𝒟aαζf(ζ)].


## 4. Generalized Fractional Lagrange's Expansion in One Variable

We can the Lagrange's expansion in more than one unknown functionμi(z) as
(28)z=z(ζ,εi)≡ζ+∑i=1nεiμi(z).
This classical result can be obtained using the following integral:
(29)f(z)=∫δ(∑i=1nεiμi(τ)-τ+ζ)f(τ)(1-∑i=1nεiμi′(τ))dτ.
Now, we are going to introduce a fractional form of the Lagrange inversion formula.Rewrite the integral (29) in the fractional form as
(30)f(z)=z𝒥aαδ(∑i=1nεiμi(z)-z+ζ)×f(z)(1-z𝒟aα∑i=1nεiμi(z))=1Γ(α)∫az(z-τ)α-1δ(∑i=1nεiμi(τ)-τ+ζ)×f(τ)(1-τ𝒟aα∑i=1nεiμi(τ))dτ.
Writing the delta function as an integral, we have
(31)f(z)=12πΓ(α)∫az∫-ππ(z-τ)α-1e(iξ[∑i=1nεiμi(τ)-τ+ζ])f(τ)12πΓ(α)×(1-𝒟τaα∑i=1nεiμi(τ))dξdτ=12πΓ(α)∑n=0∞∫az∫-ππ1n!(iξ∑i=1nεiμi(τ))n12πΓ(α)×eik(ζ-τ)(z-τ)α-1f(τ)×(1-Daατ∑i=1nεiμi(τ))dξdτ=12πΓ(α)∑n=0∞(z𝒟aα)n∫az∫-ππ1n!(∑i=1nεiμi(τ))n12πΓ(α)×eiξ(ζ-τ)(z-τ)α-1f(τ)12πΓ(α)×(1-τ𝒟aα∑i=1nεiμi(τ))dξdτ.
The integral over ξ then gives   δ(ζ-τ)=0forallζ-τ=0  and we have
(32)f(z)=1Γ(α)∑n=0∞(ζ𝒟aα)n1n!(∑i=1nεiμi(ζ))n×f(ζ)(1-𝒟aαζ∑i=1nεiμi(ζ))=1n!Γ(α)∑n=0∞(Daαζ)[(∑i=1nεiμi(ζ))nf(ζ)-f(ζ)×1n!Γ(α)∑n=0∞(Daαζ)×(∑i=1nεiμi(ζ))n×(Daαζ)∑i=1∞εiμi(ζ)]=1n!Γ(α)∑n=0∞(ζ𝒟aα)n[(∑i=1nεiμi(ζ))nf(ζ)-f(ζ)Γ(1+α)(∑i=1nεiμi(ζ))n(∑i=1nεi[μi(ζ)]1-α)]=1n!Γ(α)∑n=0∞(ζ𝒟aα)n[(∑i=1nεiμi(ζ))nf(ζ)-f(ζ)1n!Γ(α)∑n=0∞(ζ𝒟aα)n×Γ(1+α)=1n!Γ(α)∑n=0∞(ζ𝒟aα)n×(∑i=1nεi[μi(ζ)]n+1-α)]=1Γ(α)∑n=0∞(ζ𝒟aα)n[(∑i=1nεiμi(ζ))n+11n!(∑i=1nεiμi(ζ))nf(ζ)1Γ(α)∑n=0∞(ζ𝒟aα)n-Γ(1+α)Γ(n+1+α)=1Γ(α)∑n=0∞(ζ𝒟aα)n×{ζ𝒟aα[f(ζ)(∑i=1nεiμi(ζ))n+1]×=1Γ(α)∑n=0∞(ζ𝒟aα)n-ζ𝒟aαf(ζ)(∑i=1nεiμi(ζ))n+1}]=1Γ(n+1)Γ(α)Γ(n+1+α)∑n=0∞(ζ𝒟aα)n×[(∑i=1nεiμi(ζ))n+1Γ(n+1+α)(∑i=1nεiμi(ζ))nf(ζ)-Γ(n+1)×Γ(1+α){ζ𝒟aα[f(ζ)(∑i=1nεiμi(ζ))n+1]-ζ𝒟aαf(ζ)(∑i=1nεiμi(ζ))n+1}].
On extracting the first term out of summation, set   n+1=k⇒n=k-1,  and rearranging the terms then gives the result
(33)f(z)=1Γ(α)f(ζ)+∑k=0∞(ζ𝒟aα)k-1αΓ(k+α)×[(∑i=1nεiμi(ζ))k𝒟aαζf(ζ)].


## 5. Vector and Tensor Definitions and Notation

For the treatment in higher dimensions, consider theN-dimensional space with orthogonal unit base vectors  e^k,(k=1,2,…,n):
(34)e^i·e^j=δij{=1fori=j=0fori≠j(i,j=1,2,…,n).
Let  ζ,z, and the function  μ(z) be N-dimensional vectors in this space such that
(35)z=z(ζ,ε)=ζ+εμ(z),
where
(36)ζ=∑k=1ne^kζk,z=∑k=1ne^kzk,μ(z)=∑k=1ne^kμk.
For any arbitrary differentiable function F(ζ,ε), we can introduce the following fractional gradient operator as  ∇ζα.Definition 10.
LetΩ  be a domain of ℝn. Let F(ζ,ε)∈ACn(Ω) is a scalar function that has absolutely continuous derivatives up to order (n-1) on; then fractional gradient is  defined as(37)∇ζαF(ζ,ε)=ζ𝒟aαF(ζ,ε)=𝒟ζsaαF(ζs,ε)e^s=e^s1Γ(m-α)(∂∂ζs)m∫aζ(ζ-t)m-α-1f(τ)dτ,a≤ζ≤b,α∈ℝ,
where the partial derivatives are taken holding all other components of the argument fixed.


## 6. TheN-Dimensional Polyadics (nth-Order Tensors)

For arbitraryn-dimensional vectors
(38)A≡∑k=1ne^kAk,B≡∑k=1ne^kBk,
we use an extension of the notion of an n-dimensional dyadic (second-order tensor):
(39)AB≡∑i=1n∑j=1ne^ie^jAiBj
to define the nth-order tensors:
(40)A(n)≡AAA⋯A︸ntimes,B(n)≡BBB⋯B︸ntimes.
We might call A(n) and B(n) “polyadics,” since the special cases for  n=2,3,and4 are known, respectively, as dyadics, triadics, and tetradics [21]. The following defined scalar products then follow quite naturally from (34):
(41)A·B≡∑i=1nAiBiAA:BB≡A·(A·BB)=∑i=1n∑j=1nAiAjBjBi,AAA:BBB≡A·[A·(A·BBB)]=∑i=1n∑j=1n∑k=1nAiAjAKBkBjBi,
and, in general, define the  nth  scalar product:
(42)A(n)(n·)B(n)≡∑i1=1n∑i2=1n⋯∑in=1nAi1Ai2⋯AinBinBin-1⋯Bi1.
Particular examples of nth  order tensors to be used are
(43)[μ(ζ)](n)≡μ(ζ)μ(ζ)⋯μ(ζ)︸ntimes=∑i1=1n∑i2=1n⋯∑in=1ne^i1⋯e^inμi1(ζ)⋯μin(ζ).Theorem 11.
Assume thatα1,α2,…,αn≥0, and let Ψ∈L1[a,b].  Then,
(44)𝒥aα1𝒥aα2⋯𝒥aαnF(x)=𝒥α1+α2+⋯+αnF(x)
holds almost everywhere on [a,b]. If additionally Ψ∈C[a,b] or α1+α2+⋯+αn≥1, then the identity holds everywhere on [a,b].Proof.
We have(45)𝒥aαF(x)=1Γ(α)∫ax(x-t)α-1F(t)dt,a≤x≤b,α∈ℝ
Thus, we can write
(46)𝒥aα1𝒥aα2⋯𝒥aαnF(x)=1Γ(α1)1Γ(α2)⋯1Γ(αn)∫ax(x-t1)α1-1×∫at1(t1-t2)α2-1⋯×∫atn-2(tn-2-tn-1)αn-1-1×∫atn-1(tn-1-tn)αn-1×F(tn)dtndtn-1⋯dt2dt1.
Using  Fubini’s theorem to interchange the order of integration yields
(47)𝒥aα1𝒥aα2⋯𝒥aαnF(x)=1Γ(α1)Γ(α2)⋯Γ(αn)×∫ax∫t1x⋯∫tn-1x∫tnx(x-t1)α1-1×(t1-t2)α2-1⋯×(tn-2-tn-1)αn-1-1×(tn-1-tn)αn-1×F(tn)dt1dt2⋯dtn-1dtn=1Γ(α1)Γ(α2)⋯Γ(αn)×∫axF(tn)∏s=2n‍∫ts-1x(x-t1)α1-1×(ts-1-ts)αs-1dtsdtn.
The substitutions   ts=tn+ys-1(x-tn),s=2,3,…,n,andn=1,2,3,…  yields the new limits of integration as follows: when ts=x⇒x-tn=ys-1(x-tn)⇒ys-1=1, and when ts=tn⇒tn-tn=ys-1(x-tn)⇒ys-1=0:
(48)𝒥aα1𝒥aα2⋯𝒥aαnF(x)=1Γ(α1)Γ(α2)⋯Γ(αn)∫axF(tn)×∏s=2n[∫01((x-t2)(1-y1))α1-1×(ys-1(x-tn))αs-1(x-tn)dys-1∫axF(tn)]dtn=1Γ(α1)Γ(α2)⋯Γ(αn)∫axF(tn)×∏s=2n[∫01((x-t2)(1-y1))α1-1×(ys-1(x-tn))αs-1(x-tn)dys-1∫axF(tn)((x-t2)(1-y1))α1-1]dtn=1Γ(α1)Γ(α2)⋯Γ(αn)×∏s=2n‍∫axF(tn)(x-t2)α1-1(x-tn)αs×[∫01(1-y1)α1-1ys-1αs-1dys-1]dtn.
Iterating the Euler Beta integral ∫01(1-x)α1-1xα2-1dx=Γ(α1)Γ(α2)/Γ(α1+α2)  yields
(49)𝒥aα1𝒥aα2⋯𝒥aαnF(x)=1Γ(α1+α2+⋯+αn)×∫axF(tn)(x-tn)α1+α2+⋯+αndtn=𝒥α1+α2+⋯+αnF(x).
hold almost everywhere on [a,b].
Moreover, by the classical theorems on parameter integrals, ifΨ∈C[a,b], then also 𝒥aαΨ∈C[a,b], and therefore 𝒥aα1𝒥aα2⋯𝒥aαnΨ∈C[a,b],  and 𝒥aα1+α2+⋯+αnΨ∈C[a,b] too. Thus, since these two continuous functions coincide almost everywhere, they must coincide everywhere.  Finally, if Ψ∈L1[a,b] and α1+α2+⋯+αn≥1, we have, by the result above
(50)𝒥aα1𝒥aα2⋯𝒥aαnF(x)=𝒥α1+α2+⋯+αn-1J1F(x)
almost everywhere. Since J1F(x) is continuous, and once again we may conclude that the two functions on either side of the equality almost everywhere are continuous, thus they must be identical everywhere.Theorem 12.
Assume thatα1,α2,…,αn≥0. Moreover let Ψ∈L1[a,b].  Then,
(51)𝒟aα1𝒟aα2⋯𝒟aαnF=𝒟aα1+αn+⋯αnF.Proof.
The key for proof is using the semigroup property of the integral operators, the  assumption onF, and the definition of the Riemann-Liouville differential operator:
(52)𝒟aα1𝒟aα2⋯𝒟aαnF=𝒟aα1𝒟aα2⋯𝒟aαn𝒥aα1+α2+⋯+αnΨ=D⌈α1⌉𝒥a⌈α1⌉-α1D⌈α2⌉𝒥a⌈α2⌉-α2⋯D⌈αn⌉𝒥a⌈αn⌉-αn×𝒥aα1+α2+⋯+αnΨ=D⌈α1⌉𝒥a⌈α1⌉-α1D⌈α2⌉𝒥a⌈α2⌉-α2⋯D⌈αn⌉𝒥a⌈αn⌉×𝒥aα1+α2+⋯+αn-1Ψ=D⌈α1⌉𝒥a⌈α1⌉-α1D⌈α2⌉𝒥a⌈α2⌉-α2⋯D⌈αn-1⌉𝒥a⌈αn-1⌉-αn-1×𝒥aα1+α2+⋯+αn-1Ψ=D⌈α1⌉𝒥a⌈α1⌉-α1D⌈α2⌉𝒥a⌈α2⌉-α2⋯D⌈αn-1⌉𝒥a⌈αn-1⌉×𝒥aα1+α2+⋯+αn-2Ψ=D⌈α1⌉𝒥a⌈α1⌉-α1D⌈α2⌉𝒥a⌈α2⌉-α2⋯D⌈αn-2⌉𝒥a⌈αn-2⌉-αn-2×𝒥aα1+α2+⋯+αn-2Ψ⋮=D⌈α1⌉𝒥a⌈α1⌉Ψ=Ψ.
The proof that 𝒟aα1+αn+⋯+αnF=Ψ is quite straightforward
(53)𝒟aα1+αn+⋯αnF=D⌈α1+αn+⋯αn⌉×𝒥a⌈α1+αn+⋯αn⌉-α1-αn-⋯-αn×𝒥aα1+α2+⋯+αnΨ𝒟aα1+αn+....αnF=D⌈α1+αn+....αn⌉×𝒥a⌈α1+αn+....αn⌉Ψ=Ψ.
Thus,
(54)𝒟aα1+αn+⋯+αnF=𝒟aα1𝒟aα2⋯𝒟aαnF.Theorem 13.
Assume thatα1,α2,…αn≥0. Moreover, let Ψ∈L1[a,b] and F=𝒥aα1+α2+⋯+αnΨ.  And let ∇ζα  be the fractional gradient operator.  Then,
(55)∇ζα(n)F(ζ)=∑i1=1n‍∑i2=1n‍⋯∑in=1ne^i1e^i2⋯e^in1Γ(nm-α1-α2⋯-αn)×(∂n∂ζi1⋯∂ζi2)m∫aζF(τn)(ζ-τn)nm-α1-α2-⋯-αndτn.Proof.
By the  assumption onF and the successive application of fractional gradient operator, we have ∇ζα(n)F(ζ)≡∇ζα1∇ζα2⋯∇ζαnF(ζ):
(56)∇ζα(n)F(ζ)=∑i1=1ne^i11Γ(m-α1)(∂∂ζi1)m×∫aζ(ζ-τ)m-α1-1F(τ)dτ×∑i2=1n‍e^i21Γ(m-α2)(∂∂ζi2)m×∫aζ(ζ-τ)m-α2-1F(τ)dτ⋯×∑in=1ne^in1Γ(m-αn)×(∂∂ζin)m∫aζ(ζ-τ)m-αn-1F(τ)dτ=∑i1=1n‍∑i2=1n‍⋯∑in=1n‍e^i1e^i2⋯e^in×1Γ(m-α1)Γ(m-α2)⋯Γ(m-αn)×(∂∂ζi1)m(∂∂ζi2)m⋯(∂∂ζin)m×∫aζ(ζ-τ)m-α1-1F(τ)dτ×∫aζ(ζ-τ)m-α2-1F(τ)dτ⋯×∫aζ(ζ-τ)m-αn-1F(τ)dτ=∑i1=1n‍∑i2=1n‍⋯∑in=1ne^i1e^i2⋯e^in×1Γ(nm-α1-α2⋯-αn)(∂n∂ζi1⋯∂ζi2)m×∫aζ(ζ-τ)m-α1-1F(τ)dτ∫aζ(ζ-τ)m-α2-1×F(τ)dτ⋯∫aζ(ζ-τ)m-αn-1F(τ)dτ.
Thus using the theorem, the results follow directly:(57)∇ζα(n)F(ζ)=∑i1=1n‍∑i2=1n‍⋯∑in=1ne^i1e^i2⋯e^in×1Γ(nm-α1-α2⋯-αn)(∂n∂ζi1⋯∂ζi2)m×∫aζF(τn)(ζ-τn)nm-α1-α2-⋯-αndτn.


## 7. Fractional Taylor Expansion of a Function ofN-Dimensional Polyadics

We have the classical Taylor expansion for them-independent variables as
(58)f(x1,x2,…,xm)=∑n=1∞1n!(∑i=1m(xi-xi0)∂∂xi)n×f(x1,x2,…,xm)∑n=1∞1n!|x1=x10,x2=x20,⋯xm=xm0
In the light of the above definitions and theorems, we can state the following theorem.Definition 14.
Letf(xi)=f(x1,…,xn)∈Am(Ω),whereΩ=∏i=1n(ai,bi)⊂ℝn; then the fractional Riemann-Liouville  multiple integrals and partial derivatives with respect to xi are;
(59)𝒥aki,xkiαf(xi)=(1Γ(α))s∫ak1xk1⋯×∫aksxksf(ti)∏i=1s∫ts-1x(xki-tki)α-1×dtk1⋯dtks𝒟aki,xkiαf(xi)=(1Γ(n-α))(∂∂xi)n×∫aixif(ti)(xi-ti)n-α-1dti.Theorem 15.
Letn>0  and  m=⌊n⌋+1. Assume that f(xi) is such that 𝒥an-mf(xi)∈Am(Ω), where Ω=∏i=1n(ai,bi)⊂ℝn is the domain of  f. Then,
(60)f(xi)=(xi-ai)n-mΓ(n-m-1)limzi→ai+𝒥aki,xkim-n=f(zi)+∑k=0m-1(xi-ai)k+n-mΓ(k+n-m-1)limzi→ai+Dxkik𝒥aki,xkim-n×f(zi)+𝒥aki,xkin𝒟aki,xkinf(xi).Proof.
Because of our assumption aboutf that implies the continuity of  Dxkim-1𝒥aki,xkim-nf, there exists some Ψ∈L1(Ω) such that
(61)Dxkim-1𝒥aki,xkim-nf(xi)=Dxkim-1𝒥aki,xkim-nf(ai)+Jaki,xki1Ψ(xi).
This is a classical partial differential equation of order m-1 for 𝒥am-nf(xi); its solution is easily seen to be of the form:
(62)𝒥aki,xkim-nf(xi)=∑k=0m-1(xi-ai)kk!limzi→ai+Dxkik𝒥aki,xkim-n=f(zi)+Jaki,xkimΨ(xi).
Thus, by definition of  𝒟aki,xkin,
(63)𝒥aki,xkin𝒟aki,xkin=f(xi)=𝒥aki,xkinDxkim𝒥aki,xkim-nf(xi)=𝒥aki,xkinDxkim[∑k=0m-1(xi-ai)kk!limzi→ai+Dxkik×𝒥aki,xkim-nf(zi)+Jaki,xkimΨ(xi)∑k=0m-1(xi-ai)kk!],𝒥aki,xkin𝒟aki,xkin=f(xi)=∑k=0m-1𝒥aki,xkinDxkim(xi-ai)kk!×limzi→ai+Dxkik𝒥aki,xkim-nf(zi)+𝒥aki,xkinDxkimJaki,xkimΨ(xi).Dxkim annihilates every summand in the sum. And due to Theorem 8, we obtain
(64)𝒥aki,xkin𝒟aki,xkinf(xi)=𝒥aki,xkin=Ψ(xi).
Next, we apply the operator 𝒟aki,xkim-n to (63), finding
(65)f(xi)=∑k=0m-1𝒟aki,xkim-n(xi-ai)kk!limzi→ai+Dxkik𝒥aki,xkim-n=f(zi)+𝒟aki,xkim-nJaki,xkimΨ(xi)=∑k=0m-1𝒟aki,xkim-n(xi-ai)kk!limzi→ai+Dxkik𝒥aki,xkim-n=f(zi)+Dxki1𝒥aki,xki1-m+nJaki,xkimΨ(xi)=∑k=0m-1(xi-ai)k+n-mΓ(k+n-m-1)limzi→ai+Dxkik𝒥aki,xkim-n=f(zi)+𝒥aki,xkim-nΨ(xi).
Using (64), we obtain
(66)𝒥aki,xkinΨ(xi)=𝒥aki,xkin𝒟aki,xkinf(xi)=f(xi)-∑k=0m-1(xi-ai)k+n-mΓ(k+n-m-1)limzi→ai+Dxkik×𝒥aki,xkim-nf(zi).
Upon extracting the first term out of summation, we can write(67)f(xi)=(xi-ai)n-mΓ(n-m-1)limzi→ai+𝒥aki,xkim-nf(zi)+∑k=1m-1(xi-ai)k+n-mΓ(k+n-m-1)limzi→ai+Dxkik𝒥aki,xkim-n×f(zi)+𝒥aki,xkin𝒟aki,xkin×f(xi).This result can be written conveniently in the Polyadics notationA(n),  extending the upper limit of summation to ∞  to  absorb the remainder,  as
(68)F(r(n))=∑k=0∞1Γ(k+n-m-1)×limz(n)→a(n)+((r(n)-a(n))n-m(n·)∇ζα(n))k×𝒥aki,xkim-n=F(z(n)).


## 8. FractionalN-Dimensional Lagrange Expansion

Theorem 16.
LetF(z)=F(w)+μ(z,ε),F,μ∈Am(Ω) such that
(69)F(z)=F(w)+∑n=1∞εnn![μ(z)](n)(n·)∇w(n)F(w),
then it is still possible to invert or solve the equation for  F(z) such that w=F(z),F∈Am(Ω) on a neighborhood of f(a) using the fractional calculus as follows:
(70)F(z)=1Γ(α-m-1)𝒥aki,xkim-α=F(a)+∑k=1∞‍∑n=0∞‍εnΓ(k+α-m-1)×limw→a+([μ(w-a)n-m](n)(n·)∇ζα(n))k×𝒥aki,xkim-α=F(w),
where  α>0  and  m=⌊α⌋+1.Proof.
It is possible to invert or solve the equation forF(z) in a neighborhood of F(a). Using the fractional Taylor expansion.
(71)F(z)=∑k=0∞‍∑n=0∞‍εnΓ(k+α-m-1)×limw→a+([μ(w-a)n-m](n)(n·)∇ζα(n))k𝒥aki,xkim-α×F(w).
Extracting the first term out of the summations, we obtain the result directly
(72)F(z)=1Γ(α-m-1)𝒥aki,xkim-α=F(a)+∑k=1∞‍∑n=0∞‍εnΓ(k+α-m-1)×limw→a+([μ(w-a)n-m](n)(n·)∇ζα(n))k𝒥aki,xkim-α×F(w).



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*Source: 310679-2013-03-24.xml*