Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2013, Article ID 310679, 11 pages
http://dx.doi.org/10.1155/2013/310679
Research Article
𝑁-Dimensional Fractional Lagrange’s Inversion Theorem
F. A. Abd El-Salam1,2
1
2
Department of Mathematics, Faculty of Science, Taibah University, Al-Madinah, Saudi Arabia
Department of Astronomy, Faculty of Science, Cairo University, Cairo 12613, Egypt
Correspondence should be addressed to F. A. Abd El-Salam; f.a.abdelsalam@gmail.com
Received 10 November 2012; Revised 14 January 2013; Accepted 20 January 2013
Academic Editor: Ciprian A. Tudor
Copyright © 2013 F. A. Abd El-Salam. This is an open access article distributed under the Creative Commons Attribution License,
which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
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 of 𝑁-dimensional polyadics
is derived. A fractional 𝑁-dimensional Lagrange inversion theorem is proved.
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-differentiating 𝑥. 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 spacetime 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
2
Abstract and Applied Analysis
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 LagrangeBürmann formula, giving the Taylor series expansion of the
inverse function. Suppose that 𝑧 = 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
to obtain a fractional form of the Lagrange expansion and
some generalizations.
2. Basic Definitions and Theorems
Definition 1. By 𝐷, 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
𝑥
𝑎; that is, 𝐽𝑎 𝑓(𝑥) = ∫𝑎 𝑓(𝑡)𝑑𝑡 for all 𝑎 ≤ 𝑥 ≤ 𝑏.
Definition 2. By 𝐷𝑛 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. Let 𝑓 be Riemann integrable on [𝑎, 𝑏]. Then, for
𝑎 ≤ 𝑥 ≤ 𝑏 and 𝑛 ∈ N, one has
𝑥
1
𝐽𝑎𝑛 𝑓(𝑥) =
∫ (𝑥 − 𝑡)𝑛−1 𝑓(𝑡)𝑑𝑡, 𝑛 ∈ N. (2)
(𝑛 − 1)! 𝑎
Definition 4. The operator J𝑎𝛼 , defined on Lebesgue space
𝐿 1 [𝑎, 𝑏], denotes the Riemann-Liouville fractional operator
of order 𝛼. That is,
𝑥
1
J𝑎𝛼 𝑓(𝑥) =
∫ (𝑥 − 𝑡)𝛼−1 𝑓(𝑡)𝑑𝑡, 𝑎 ≤ 𝑥 ≤ 𝑏, 𝛼 ∈ R.
Γ (𝛼) 𝑎
(3)
Remark 5. It is evident that J𝑎𝛼 ≡ 𝐽𝑎𝑛 , for all 𝛼 ∈ 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 J𝑎𝛼 𝑓(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,
J𝑎𝛼 (J𝑎𝛽 𝑓(𝑥)) = J𝑎𝛽 (J𝑎𝛼 𝑓(𝑥))
= J𝑎𝛼+𝛽 𝑓(𝑥) ,
𝛼, 𝛽 > 0, 𝑓(𝑥) ∈ 𝐿 1 [𝑎, 𝑏] .
(4)
Definition 6. Let 𝛼 ∈ R+ and let 𝑚 = ⌈𝛼⌉, The RiemannLiouville fractional differential operator of order 𝛼 is defined
as such that. Then, D𝛼𝑎 = 𝐷𝑎𝑚 J𝑎𝑚−𝛼 . That is,
D𝛼𝑎 𝑓(𝑥) =
1
𝑑 𝑚 𝑥
( ) ∫ (𝑥 − 𝑡)𝑚−𝛼−1
Γ (𝑚 − 𝛼) 𝑑𝑥
𝑎
× 𝑓(𝑡) 𝑑𝑡,
(5)
𝑎 ≤ 𝑥 ≤ 𝑏, 𝛼 ∈ R.
+
Lemma 7. Let 𝛼 ∈ R and let m ∈ N such that > 𝛼. Then,
D𝛼𝑎 = D𝑎𝑚 J𝑎𝑚−𝛼 .
𝑤 = 𝑔 (𝑧)
𝑛
∞
= 𝑎 + ∑ (lim (
𝑛=1
𝑤→𝑎
𝑛
(𝑧 − 𝑓 (𝑎)) 𝑑𝑛−1
𝑤−𝑎
(
)) .
)
𝑛!
𝑑𝑤𝑛−1 𝑓(𝑤) − 𝑓(𝑎)
(1)
In this work, we will apply the concepts of fractional calculus
Proof. Since 𝑚 > 𝛼 yields 𝑚 ≥ ⌈𝛼⌉.
Thus,
𝐷𝑚 J𝑎𝑚−𝛼 = 𝐷⌈𝛼⌉ 𝐷𝑚−⌈𝛼⌉ 𝐽𝑚−⌈𝛼⌉ 𝐽⌈𝛼⌉−𝛼
= 𝐷⌈𝛼⌉+𝑚−⌈𝛼⌉−𝑚+⌈𝛼⌉−⌈𝛼⌉+𝛼 = 𝐷𝛼 = D𝛼𝑎 .
(6)
Abstract and Applied Analysis
3
Theorem 8. Let (𝛼 ≥ 0) ∈ R+ . Then, for every 𝑓(𝑥) ∈ 𝐿 1 [𝑎, 𝑏],
D𝑎𝛼 J𝑎𝛼 𝑓(𝑥) = 𝑓(𝑥).
Proof. For 𝛼 = 0, both operator, are the identity. For 𝛼 > 0,
let 𝑚 ≥ ⌈𝛼⌉; then,
J𝛼𝑎 𝑓 (𝑥) = 𝐷𝑎𝑚 J𝑚
D𝛼𝑎 J𝛼𝑎 𝑓 (𝑥) = 𝐷𝑎𝑚 J𝑚−𝛼
𝑎
𝑎 𝑓 (𝑥)
= 𝐷𝑎𝑚 𝐽𝑎𝑚 𝑓(𝑥) = 𝑓(𝑥) .
(7)
To prove (𝐷-1) we use the relation
D𝛼𝑎 = 𝐷⌈𝛼⌉ J𝑎⌈𝛼⌉−𝛼 ,
−𝛼
𝑘!Γ(𝛼) (𝛼 + 𝑘) ( ) = (−1)𝑘 Γ (𝑘 + 1 + 𝛼) .
𝑘
(11)
This allows us to rewrite the statement (𝐼-1) as
∞
(𝑥 − 𝑎)𝑘+⌈𝛼⌉−𝛼
⌈𝛼⌉ − 𝛼
𝐷𝑘 𝑓 (𝑥) .
J𝑎⌈𝛼⌉−𝛼 𝑓(𝑥) = ∑ (
)
𝑘
Γ (𝑘 + 1 + ⌈𝛼⌉ − 𝛼) 𝑎
𝑘=0
Corollary 9. Let 𝑓 be analytic in (𝑎−ℎ, 𝑎+ℎ) for some ℎ > 0,
and let (𝛼 ≥ 0) ∈ R+ .
Then,
(𝐼-1)
J𝑎𝛼 𝑓(𝑥)
(12)
Differentiating ⌈𝛼⌉ times with respect to 𝑥, we find
∞
⌈𝛼⌉ − 𝛼
𝐷𝑎⌈𝛼⌉ J𝑎⌈𝛼⌉−𝛼 𝑓(𝑥) = D𝛼𝑎 𝑓(𝑥) = ∑ (
)
𝑘
∞
(−1)𝑚 (𝑥 − 𝑎)𝑘+𝛼 𝑘
𝐷 𝑓(𝑥) ,
=∑
𝑘! (𝛼 + 𝑘) Γ(𝛼) 𝑎
𝑘=0
𝑘=0
ℎ
∀𝑎 ≤ 𝑥 < 𝑎 + ,
2
×
(𝑥 − 𝑎)𝑘+𝛼 𝑘
𝐷 𝑓(𝑎) ,
Γ (𝑘 + 1 + 𝛼) 𝑎
𝑘=0
∞
𝑘−𝛼
𝛼 (𝑥 − 𝑎)
𝐷𝑘 𝑓(𝑥) ,
(𝐷-1) D𝑎𝛼 𝑓(𝑥) = ∑ ( )
𝑘 (𝑘 + 1 − 𝛼) 𝑎
The classical version of Leibniz’ formula yields
(8)
⌈𝛼⌉
∞
1
⌈𝛼⌉ − 𝛼
⌈𝛼⌉
)
D𝛼𝑎 𝑓(𝑥) = ∑ (
∑( )
𝑘
𝑗
Γ (𝑘 + 1 + ⌈𝛼⌉ − 𝛼)
which yields
𝑘−𝛼
(𝑥 − 𝑎)
𝐷𝑎𝑘 𝑓(𝑎) ,
+
1
−
𝛼)
(𝑘
𝑘=0
𝑘+𝑗−𝛼
⌈𝛼⌉
∞
]
𝛼 − ⌈𝛼⌉
⌈𝛼⌉ [(𝑥 − 𝑎)
D𝛼𝑎 𝑓(𝑥) = ∑ (
) ∑( )
𝑗 Γ(𝑘 + 1 + 𝑗 − 𝛼)
𝑘
The binomial coefficients for 𝛼 ∈ R and 𝑘 ∈ N are defined as
𝛼!
=
.
𝑘! (𝛼 − 𝑘)!
(9)
Γ(𝑘 + 1)
(𝑥 − 𝑎)𝑘+𝛼 .
(𝛼 + 𝑘 + 1)
(15)
× (𝑥) 𝐷𝑎𝑘+𝑗 𝑓.
𝜇
By definition, ( 𝑗 ) = 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
∞ ∞
D𝛼𝑎 𝑓(𝑥) = ∑ ∑ (
Proof. For the first two statements (𝐼-1), (𝐼-2) and we use the
definition of the Riemann-Liouville integral operator J𝑎𝛼 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:
J𝑎𝛼 (𝑥 − 𝑎)𝑘 =
𝑗=0
𝑘=0
∀𝑎 ≤ 𝑥 < 𝑎 + ℎ.
𝛼 (𝛼 − 1) (𝛼 − 2) ⋅ ⋅ ⋅ (𝛼 − 𝑘 + 1)
𝛼
( )=
𝑘
𝑘!
(14)
× 𝐷𝑎⌈𝛼⌉−𝑗 [(∙ − 𝑎)𝑘+⌈𝛼⌉−𝛼 ] (𝑥) 𝐷𝑎𝑘+𝑗 𝑓,
ℎ
∀𝑎 ≤ 𝑥 < 𝑎 + ,
2
(𝐷-2) D𝑎𝛼 𝑓(𝑥) = ∑
𝑗=0
𝑘=0
𝑘=0
∞
(13)
× 𝐷𝑎⌈𝛼⌉ [(∙ − 𝑎)𝑘+⌈𝛼⌉−𝛼 𝐷𝑎𝑘 𝑓](𝑥) .
∞
(𝐼-2) J𝑎𝛼 𝑓(𝑥) = ∑
∀𝑎 ≤ 𝑥 < 𝑎 + ℎ,
1
Γ(𝑘 + 1 + ⌈𝛼⌉ − 𝛼)
𝑙=0 𝑘=0
𝑙−𝛼
𝛼 − ⌈𝛼⌉
⌈𝛼⌉ [(𝑥 − 𝑎) ]
)(
)
(𝑥) 𝐷𝑎𝑙 𝑓.
𝑘
𝑙 − 𝑘 Γ(𝑙 + 1 − 𝛼)
(16)
∞
∞
𝑙
Using the fact that ∑∞
𝑙=0 ∑𝑘=0 = ∑𝑙=0 ∑𝑘=0 ,
D𝛼𝑎 𝑓(𝑥)
𝑙−𝛼
𝛼 − ⌈𝛼⌉
⌈𝛼⌉ [(𝑥 − 𝑎) ]
=∑∑(
)(
)
(𝑥) 𝐷𝑎𝑙 𝑓.
𝑘
𝑙 − 𝑘 Γ(𝑙 + 1 − 𝛼)
∞
𝑙
𝑙=0 𝑘=0
(17)
(10)
(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 J𝑎𝛼
follows immediately from the second statement.
And the explicit calculation yields
𝑙
𝛼 − ⌈𝛼⌉
⌈𝛼⌉
⌈𝛼⌉
)(
) = ( ),
∑(
𝑙
𝑘
𝑙−𝑘
𝑘=0
thus, (𝐷-1) follows directly.
(18)
4
Abstract and Applied Analysis
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 Definition 6
to obtain the fractional form of Lagrange’s expansion as
follows.
Let 𝑧 be a function of (𝜁, 𝜀) and in terms of another
function 𝜇 such that
𝑧 = 𝑧 (𝜁, 𝜀) ≡ 𝜁 + 𝜀𝜇 (𝑧) .
The integral over 𝜉 then gives 𝛿(𝜁 − 𝜏) = 0 for all 𝜁 − 𝜏 = 0
and we have
𝑛
𝑛 (𝜀𝜇 (𝜁))
1 ∞
𝑓(𝜁) (1 − 𝜀 𝜁 D𝛼𝑎 𝜇 (𝜁))
𝑓 (𝑧) =
∑ ( D𝛼 )
Γ (𝛼) 𝑛=0 𝜁 𝑎
𝑛!
=
𝑛
× 𝑓(𝜁) (𝜇 (𝜁)) (𝜁 D𝛼𝑎 𝜇 (𝜁))]
(19)
=
Then, for any function 𝑓
𝑛
1 ∞
𝑛
∑ ( D𝛼 ) [𝜀𝑛 (𝜇 (𝜁)) 𝑓(𝜁) − 𝜀𝑛+1
𝑛!Γ (𝛼) 𝑛=0 𝜁 𝑎
𝑛
1 ∞
𝑛
∑ ( D𝛼 ) [𝜀𝑛 (𝜇 (𝜁)) 𝑓(𝜁) − 𝜀𝑛+1 𝑓(𝜁) Γ
𝑛!Γ(𝛼) 𝑛=0 𝜁 𝑎
𝑛
× (1 + 𝛼) (𝜇 (𝜁)) (𝜇(𝜁)1−𝛼 )]
∞
𝑑𝑓 (𝜁)
𝜀𝑛 𝑑𝑛−1
[𝜇𝑛 (𝜁)
]
𝑛−1
𝑑𝜁
𝑛=1 𝑛! 𝑑𝜁
𝑓(𝑧) = 𝑓(𝜁) + ∑
(20)
=
for small 𝜀. If 𝑓 is the identity
𝑛
1 ∞
𝑛
∑ ( D𝛼 ) [𝜀𝑛 (𝜇 (𝜁)) 𝑓(𝜁) − 𝜀𝑛+1
𝑛!Γ (𝛼) 𝑛=0 𝜁 𝑎
× 𝑓(𝜁) Γ(1 + 𝛼) (𝜇(𝜁)𝑛+1−𝛼 )]
∞ 𝑛
𝜀 𝑑𝑛−1 𝑛
[𝜇 (𝜁)] ,
𝑧=𝜁+∑
𝑛−1
𝑛=1𝑛! 𝑑𝜁
(21)
then
=
𝑛
𝑛 𝜀
𝜀𝑛+1 Γ(1 + 𝛼)
1 ∞
𝑛
∑ (𝜁 D𝛼𝑎 ) [ (𝜇 (𝜁)) 𝑓(𝜁) −
Γ (𝛼) 𝑛=0
𝑛!
Γ(𝑛 + 1 + 𝛼)
𝑛+1
× {𝜁 D𝛼𝑎 [𝑓(𝜁) (𝜇 (𝜁))
∞
𝑑𝑓 (𝜁)
𝜀𝑛 𝑑𝑛−1
𝑓(𝑧) = 𝑓(𝜁) + ∑
[𝜇𝑛 (𝜁)
].
𝑛−1
𝑛!
𝑑𝜁
𝑑𝜁
𝑛=1
]
𝑛+1
(22)
−𝜁 D𝛼𝑎 𝑓(𝜁) (𝜇 (𝜁))
(26)
This classical result can be obtained using the following
integral:
𝑓(𝑧) = ∫ 𝛿 (𝜀𝜇 (𝜏) − 𝜏 + 𝜁) 𝑓(𝜏) (1 − 𝜀𝜇󸀠 (𝜏)) 𝑑𝜏
(23)
Now we are going to introduce a fractional form of the
Lagrange inversion formula.
Rewrite the integral (23) in the fractional form as
On extracting the first term out of summation, set 𝑛+1 = 𝑘 ⇒
𝑛 = 𝑘 − 1, and rearranging the terms then gives the result:
𝑓(𝑧) =
𝑧
1
=
∫ (𝑧 − 𝜏)𝛼−1 𝛿 (𝜀𝜇 (𝜏) − 𝜏 + 𝜁)
Γ (𝛼) 𝑎
(24)
4. Generalized Fractional Lagrange’s
Expansion in One Variable
We can the Lagrange’s expansion in more than one unknown
function 𝜇𝑖 (𝑧) as
× 𝑓(𝜏) (1 − 𝜀𝜏 D𝛼𝑎 𝜇 (𝜏)) 𝑑𝜏.
Writing the delta function as an integral, we have
=
𝑛
𝑧 = 𝑧 (𝜁, 𝜀𝑖 ) ≡ 𝜁 + ∑𝜀𝑖 𝜇𝑖 (𝑧) .
𝜋
1
∫ ∫ (𝑧 − 𝜏)𝛼−1 𝑒(𝑖𝜉[𝜀𝜇(𝜏)−𝜏+𝜁])
2𝜋Γ (𝛼) 𝑎 −𝜋
× 𝑓 (𝜏) (1 − 𝜀𝜏 D𝛼𝑎 𝜇 (𝜏)) 𝑑𝜉𝑑𝜏
∞
𝑧
𝜋
𝑛 𝑖𝑘(𝜁−𝜏)
(𝑖𝜉𝜀𝜇 (𝜏)) 𝑒
1
∑∫ ∫
2𝜋Γ (𝛼) 𝑛=0 𝑎 −𝜋
𝑛!
This classical result can be obtained using the following
integral:
(𝑧 − 𝜏)𝛼−1
𝑛 𝑖𝜉(𝜁−𝜏)
∞
𝑧 𝜋 (𝜀𝜇 (𝜏)) 𝑒
1
𝑛
∑(𝑧 D𝛼𝑎 ) ∫ ∫
2𝜋Γ (𝛼) 𝑛=0
𝑛!
𝑎 −𝜋
(28)
𝑖=1
× 𝑓(𝜏) (1 − 𝜀𝜏 D𝛼𝑎 𝜇 (𝜏)) 𝑑𝜉 𝑑𝜏
=
(27)
𝑘
× 𝑓(𝑧) (1 − 𝜀𝑧 D𝛼𝑎 𝜇 (𝑧))
𝑓(𝑧) =
𝑘
∞
𝑘−1 𝛼 (𝜀 )
1
𝑓(𝜁) + ∑(𝜁 D𝛼𝑎 )
Γ(𝛼)
Γ (𝑘 + 𝛼)
𝑘=0
× [(𝜇 (𝜁)) 𝜁 D𝛼𝑎 𝑓(𝜁)] .
𝑓(𝑧) = 𝑧 J𝑎𝛼 𝛿 (𝜀𝜇 (𝑧) − 𝑧 + 𝜁)
𝑧
}].
(𝑧 − 𝜏)𝛼−1
× 𝑓 (𝜏) (1 − 𝜀𝜏 D𝛼𝑎 𝜇 (𝜏)) 𝑑𝜉 𝑑𝜏.
𝑛
𝑛
𝑖=1
𝑖=1
𝑓(𝑧) = ∫ 𝛿 (∑𝜀𝑖 𝜇𝑖 (𝜏) − 𝜏 + 𝜁) 𝑓(𝜏) (1 − ∑𝜀𝑖 𝜇𝑖󸀠 (𝜏)) 𝑑𝜏.
(25)
(29)
Now, we are going to introduce a fractional form of the
Lagrange inversion formula.
Abstract and Applied Analysis
5
𝑛
𝑛
Rewrite the integral (29) in the fractional form as
∞
×(∑𝜀𝑖 𝜇𝑖 (𝜁)) × (𝜁 𝐷𝑎𝛼 )∑𝜀𝑖 𝜇𝑖 (𝜁)]
𝑖=1
𝑖=1
𝑛
𝑛
𝑓(𝑧) = 𝑧 J𝑎𝛼 𝛿 (∑𝜀𝑖 𝜇𝑖 (𝑧) − 𝑧 + 𝜁)
=
𝑖=1
𝑛
𝑛
1 ∞
∑ (𝜁 D𝛼𝑎 ) [(∑𝜀𝑖 𝜇𝑖 (𝜁)) 𝑓(𝜁)
𝑛!Γ (𝛼) 𝑛=0
𝑖=1
𝑛
𝑛
𝑛
× 𝑓(𝑧) (1 − 𝑧 D𝛼𝑎 ∑𝜀𝑖 𝜇𝑖 (𝑧))
𝑛
−𝑓 (𝜁) Γ(1 + 𝛼) (∑𝜀𝑖 𝜇𝑖 (𝜁)) (∑𝜀𝑖 [𝜇𝑖 (𝜁)]
𝑖=1
𝑛
𝑧
1
=
∫ (𝑧 − 𝜏)𝛼−1 𝛿 (∑𝜀𝑖 𝜇𝑖 (𝜏) − 𝜏 + 𝜁)
Γ (𝛼) 𝑎
𝑖=1
𝑖=1
(30)
𝑛
=
𝑛
𝑛
1 ∞
∑ (𝜁 D𝛼𝑎 ) [(∑𝜀𝑖 𝜇𝑖 (𝜁)) 𝑓(𝜁) − 𝑓 (𝜁)
𝑛!Γ(𝛼) 𝑛=0
𝑖=1
𝑛
× Γ (1 + 𝛼)
𝑖=1
𝑛
𝑛+1−𝛼
× (∑𝜀𝑖 [𝜇𝑖 (𝜁)]
𝑛
𝑧 𝜋
𝑛
1
∫ ∫ (𝑧 − 𝜏)𝛼−1 𝑒(𝑖𝜉[∑𝑖=1 𝜀𝑖 𝜇𝑖 (𝜏)−𝜏+𝜁]) 𝑓(𝜏)
2𝜋Γ (𝛼) 𝑎 −𝜋
𝑛
1 𝑛
1 ∞
=
∑ (𝜁 D𝛼𝑎 ) [ (∑𝜀𝑖 𝜇𝑖 (𝜁)) 𝑓(𝜁)
Γ (𝛼) 𝑛=0
𝑛! 𝑖=1
[
−
𝑛
× (1 − 𝜏 D𝛼𝑎 ∑𝜀𝑖 𝜇𝑖 (𝜏)) 𝑑𝜉 𝑑𝜏
𝑛+1
𝑛
𝑛
1 ∞ 𝑧 𝜋 1
=
(𝑖𝜉∑𝜀𝑖 𝜇𝑖 (𝜏))
∑∫ ∫
2𝜋Γ(𝛼) 𝑛=0 𝑎 −𝜋 𝑛!
𝑖=1
×𝑒
(𝑧 − 𝜏)
Γ (1 + 𝛼)
Γ (𝑛 + 1 + 𝛼)
𝑛
{
× {𝜁 D𝛼𝑎 [𝑓(𝜁)(∑𝜀𝑖 𝜇𝑖 (𝜁))
𝑖=1
[
{
𝑖=1
𝛼−1
𝑛
× (1 − 𝜏 𝐷𝑎𝛼 ∑𝜀𝑖 𝜇𝑖 (𝜏)) 𝑑𝜉 𝑑𝜏
=
𝑖=1
𝑛
𝑧 𝜋
1 𝑛
1 ∞
𝑛
(∑𝜀𝑖 𝜇𝑖 (𝜏))
=
∑ (𝑧 D𝛼𝑎 ) ∫ ∫
2𝜋Γ(𝛼) 𝑛=0
𝑎 −𝜋 𝑛! 𝑖=1
∞
𝑛
1
∑ (𝜁 D𝛼𝑎 )
Γ (𝑛 + 1) Γ (𝛼) Γ (𝑛 + 1 + 𝛼) 𝑛=0
𝑛
𝑛
× [Γ (𝑛 + 1 + 𝛼) (∑𝜀𝑖 𝜇𝑖 (𝜁)) 𝑓(𝜁) − Γ (𝑛 + 1)
𝑖=1
[
× 𝑒𝑖𝜉(𝜁−𝜏) (𝑧 − 𝜏)𝛼−1 𝑓(𝜏)
𝑛+1
𝑛
{
× Γ (1 + 𝛼) {𝜁 D𝛼𝑎 [𝑓(𝜁) (∑𝜀𝑖 𝜇𝑖 (𝜁))
𝑖=1
{
[
𝑛
× (1 −𝜏 D𝛼𝑎 ∑𝜀𝑖 𝜇𝑖 (𝜏)) 𝑑𝜉 𝑑𝜏.
𝑖=1
(31)
𝑛
−𝜁 D𝛼𝑎 𝑓(𝜁) (∑𝜀𝑖 𝜇𝑖 (𝜁))
𝑖=1
The integral over 𝜉 then gives 𝛿(𝜁 − 𝜏) = 0 for all 𝜁 − 𝜏 = 0
and we have
𝑛
𝑛
𝑛 1
1 ∞
𝑓(𝑧) =
∑ (𝜁 D𝛼𝑎 ) (∑𝜀𝑖 𝜇𝑖 (𝜁))
Γ (𝛼) 𝑛=0
𝑛! 𝑖=1
× 𝑓(𝜁) (1 − 𝜁 D𝛼𝑎 ∑𝜀𝑖 𝜇𝑖 (𝜁))
𝑖=1
=
𝑛
1 ∞
∑ (𝜁 𝐷𝑎𝛼 ) [(∑𝜀𝑖 𝜇𝑖 (𝜁)) 𝑓 (𝜁) − 𝑓 (𝜁)
𝑛!Γ(𝛼) 𝑛=0
𝑖=1
]
]
𝑛+1
}
]
} .
}]
(32)
On extracting the first term out of summation, set 𝑛+1 = 𝑘 ⇒
𝑛 = 𝑘 − 1, and rearranging the terms then gives the result
𝑓(𝑧) =
𝑛
]
}
]
}
}]
𝑖=1
𝑛
]
𝑛+1
−𝜁 D𝛼𝑎 𝑓(𝜁)(∑𝜀𝑖 𝜇𝑖 (𝜁))
𝑓(𝜏)
𝑛
)]
𝑖=1
Writing the delta function as an integral, we have
𝑖𝑘(𝜁−𝜏)
)]
𝑖=1
× 𝑓(𝜏) (1 − 𝜏 D𝛼𝑎 ∑𝜀𝑖 𝜇𝑖 (𝜏)) 𝑑𝜏.
𝑓(𝑧) =
1−𝛼
∞
𝑘−1
1
𝛼
𝑓(𝜁) + ∑ (𝜁 D𝛼𝑎 )
Γ (𝛼)
Γ (𝑘 + 𝛼)
𝑘=0
𝑛
𝑘
× [(∑𝜀𝑖 𝜇𝑖 (𝜁)) 𝜁 D𝛼𝑎 𝑓 (𝜁)] .
]
[ 𝑖=1
(33)
6
Abstract and Applied Analysis
5. Vector and Tensor Definitions and Notation
For the treatment in higher dimensions, consider the 𝑁dimensional space with orthogonal unit base vectors ̂𝑒𝑘 , (𝑘 =
1, 2, . . . , 𝑛):
= 1 for
̂𝑒𝑖 ⋅ ̂𝑒𝑗 = 𝛿𝑖𝑗 {
= 0 for
𝑖=𝑗
𝑖 ≠ 𝑗
(𝑖, 𝑗 = 1, 2, . . . , 𝑛) .
(34)
triadics, and tetradics [21]. The following defined scalar
products then follow quite naturally from (34):
𝑛
𝐴 ⋅ 𝐵 ≡ ∑𝐴 𝑖 𝐵𝑖
𝑖=1
𝑛 𝑛
𝐴𝐴 : 𝐵𝐵 ≡ 𝐴 ⋅ (𝐴 ⋅ 𝐵𝐵) = ∑ ∑𝐴 𝑖 𝐴 𝑗 𝐵𝑗 𝐵𝑖 ,
𝑖=1 𝑗=1
Let 𝜁, 𝑧 , and the function 𝜇(𝑧) be 𝑁-dimensional vectors in
this space such that
𝑛 𝑛
𝑛
𝐴𝐴𝐴 : 𝐵𝐵𝐵 ≡ 𝐴 ⋅ [𝐴 ⋅ (𝐴 ⋅ 𝐵𝐵𝐵)] = ∑∑ ∑ 𝐴 𝑖 𝐴 𝑗 𝐴 𝐾 𝐵𝑘 𝐵𝑗 𝐵𝑖 ,
𝑖=1 𝑗=1𝑘=1
𝑧 = 𝑧 (𝜁, 𝜀) = 𝜁 + 𝜀𝜇 (𝑧) ,
(35)
(41)
and, in general, define the 𝑛th scalar product:
where
𝑛
𝜁 = ∑̂𝑒𝑘 𝜁𝑘 ,
𝑘=1
𝑛
𝑧 = ∑̂𝑒𝑘 𝑧𝑘 ,
𝑘=1
𝑛
𝜇 (𝑧) = ∑̂𝑒𝑘 𝜇𝑘 .
(36)
𝑘=1
𝑛
Definition 10. Let Ω be a domain of R . Let 𝐹(𝜁, 𝜀) ∈ 𝐴𝐶 (Ω)
is a scalar function that has absolutely continuous derivatives
up to order (𝑛 − 1) on; then fractional gradient is defined as
∇𝜁𝛼 𝐹 (𝜁, 𝜀)
=
𝛼
𝜁 D𝑎 𝐹 (𝜁, 𝜀)
=
𝑖𝑛 =1
(42)
Particular examples of 𝑛th order tensors to be used are
(𝑛)
[𝜇 (𝜁)]
≡ ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
𝜇 (𝜁) 𝜇 (𝜁) ⋅ ⋅ ⋅ 𝜇 (𝜁)
𝑛 times
𝑛
𝛼
𝑒𝑠
𝜁𝑠 D𝑎 𝐹 (𝜁𝑠 , 𝜀) ̂
𝑛
(43)
𝑛
= ∑ ∑ ⋅ ⋅ ⋅ ∑ ̂𝑒𝑖1 ⋅ ⋅ ⋅ ̂𝑒𝑖𝑛 𝜇𝑖1 (𝜁) ⋅ ⋅ ⋅ 𝜇𝑖𝑛 (𝜁) .
1
𝜕 𝑚 𝜁
= ̂𝑒𝑠
( ) ∫ (𝜁 − 𝑡)𝑚−𝛼−1 𝑓(𝜏) 𝑑𝜏,
Γ (𝑚 − 𝛼) 𝜕𝜁𝑠
𝑎
𝑎 ≤ 𝜁 ≤ 𝑏,
𝑛
𝑛
𝑖1 =1 𝑖2 =1
For any arbitrary differentiable function 𝐹(𝜁, 𝜀), we can
introduce the following fractional gradient operator as ∇𝜁𝛼 .
𝑛
𝑛
𝑛
𝐴(𝑛) ( ) 𝐵(𝑛) ≡ ∑ ∑ ⋅ ⋅ ⋅ ∑ 𝐴 𝑖1 𝐴 𝑖2 ⋅ ⋅ ⋅ 𝐴 𝑖𝑛 𝐵𝑖𝑛 𝐵𝑖𝑛−1 ⋅ ⋅ ⋅ 𝐵𝑖1 .
⋅
𝛼 ∈ R,
(37)
𝑖1 =1 𝑖2 =1
𝑖𝑛 =1
Theorem 11. Assume that 𝛼1 , 𝛼2 , . . . , 𝛼𝑛 ≥ 0, and let Ψ ∈
𝐿 1 [𝑎, 𝑏]. Then,
J𝑎𝛼1 J𝑎𝛼2 ⋅ ⋅ ⋅ J𝑎𝛼𝑛 𝐹(𝑥) = J𝛼1 +𝛼2 +⋅⋅⋅+𝛼𝑛 𝐹(𝑥)
(44)
where the partial derivatives are taken holding all other
components of the argument fixed.
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].
6. The 𝑁-Dimensional Polyadics
(𝑛th-Order Tensors)
Proof. We have
J𝑎𝛼 𝐹(𝑥) =
For arbitrary 𝑛-dimensional vectors
𝑛
𝐴 ≡ ∑̂𝑒𝑘 𝐴 𝑘 ,
𝑘=1
(45)
(38)
𝑘=1
(39)
𝑖=1 𝑗=1
𝑛 times
J𝑎𝛼1 J𝑎𝛼2 ⋅ ⋅ ⋅ J𝑎𝛼𝑛 𝐹(𝑥) =
𝑥
1
1
1
𝛼 −1
⋅⋅⋅
∫ (𝑥 − 𝑡1 ) 1
Γ (𝛼1 ) Γ (𝛼2 )
Γ (𝛼𝑛 ) 𝑎
𝑎
𝑡𝑛−2
𝑎
(𝑛)
𝐵
≡ ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
𝐵𝐵𝐵 ⋅ ⋅ ⋅ 𝐵.
𝑛 times
(40)
We might call 𝐴(𝑛) and 𝐵(𝑛) “polyadics,” since the special
cases for 𝑛 = 2, 3, and 4 are known, respectively, as dyadics,
𝛼2 −1
× ∫ (𝑡1 − 𝑡2 )
×∫
to define the 𝑛th-order tensors:
≡ ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
𝐴𝐴𝐴 ⋅ ⋅ ⋅ 𝐴,
Thus, we can write
𝑡1
𝑛 𝑛
𝐴𝐵 ≡ ∑ ∑̂𝑒𝑖̂𝑒𝑗 𝐴 𝑖 𝐵𝑗
𝐴
𝑎 ≤ 𝑥 ≤ 𝑏, 𝛼 ∈ R
𝑛
𝐵 ≡ ∑̂𝑒𝑘 𝐵𝑘 ,
we use an extension of the notion of an 𝑛-dimensional dyadic
(second-order tensor):
(𝑛)
𝑥
1
∫ (𝑥 − 𝑡)𝛼−1 𝐹(𝑡) 𝑑𝑡,
Γ(𝛼) 𝑎
×∫
𝑡𝑛−1
𝑎
⋅⋅⋅
𝛼𝑛−1 −1
(𝑡𝑛−2 − 𝑡𝑛−1 )
𝛼𝑛 −1
(𝑡𝑛−1 − 𝑡𝑛 )
× 𝐹(𝑡𝑛 ) 𝑑𝑡𝑛 𝑑𝑡𝑛−1 ⋅ ⋅ ⋅ 𝑑𝑡2 𝑑𝑡1 .
(46)
Abstract and Applied Analysis
7
Using Fubini’s theorem to interchange the order of integration yields
J𝑎𝛼1 J𝑎𝛼2 ⋅ ⋅ ⋅ J𝑎𝛼𝑛 𝐹 (𝑥) =
1
Γ(𝛼1 ) Γ(𝛼2 ) ⋅ ⋅ ⋅ Γ(𝛼𝑛 )
𝑥
𝑥
𝑥
𝑎
𝑡1
× (𝑡1 − 𝑡2 )
J𝑎𝛼1 J𝑎𝛼2 ⋅ ⋅ ⋅ J𝑎𝛼𝑛 𝐹 (𝑥) =
𝑥
× ∫ ∫ ⋅⋅⋅∫
∫ (𝑥 − 𝑡1 )
𝑡𝑛−1
𝛼2 −1
1
Iterating the Euler Beta integral ∫0 (1 − 𝑥)𝛼1 −1 𝑥𝛼2 − 1 𝑑𝑥 =
Γ(𝛼1 )Γ(𝛼2 )/Γ(𝛼1 + 𝛼2 ) yields
1
Γ(𝛼1 + 𝛼2 + ⋅ ⋅ ⋅ + 𝛼𝑛 )
𝑥
𝛼1 −1
𝛼1 +𝛼2 +⋅⋅⋅+𝛼𝑛
× ∫ 𝐹(𝑡𝑛 ) (𝑥 − 𝑡𝑛 )
𝑡𝑛
𝑎
= J𝛼1 +𝛼2 +⋅⋅⋅+𝛼𝑛 𝐹(𝑥) .
⋅⋅⋅
(49)
𝛼𝑛−1 − 1
× (𝑡𝑛−2 − 𝑡𝑛−1 )
𝛼𝑛 −1
hold almost everywhere on [𝑎, 𝑏].
Moreover, by the classical theorems on parameter integrals, if Ψ ∈ 𝐶[𝑎, 𝑏], then also J𝑎𝛼 Ψ ∈ 𝐶[𝑎, 𝑏], and therefore
J𝑎𝛼1 J𝑎𝛼2 ⋅ ⋅ ⋅ J𝑎𝛼𝑛 Ψ ∈ 𝐶[𝑎, 𝑏], and J𝑎𝛼1 +𝛼2 +⋅⋅⋅+𝛼𝑛 Ψ ∈ 𝐶[𝑎, 𝑏] too.
Thus, since these two continuous functions coincide almost
everywhere, they must coincide everywhere. Finally, if Ψ ∈
𝐿 1 [𝑎, 𝑏] and 𝛼1 + 𝛼2 + ⋅ ⋅ ⋅ + 𝛼𝑛 ≥ 1, we have, by the result
above
× (𝑡𝑛−1 − 𝑡𝑛 )
× 𝐹 (𝑡𝑛 ) 𝑑𝑡1 𝑑𝑡2 ⋅ ⋅ ⋅ 𝑑𝑡𝑛−1 𝑑𝑡𝑛
=
1
Γ (𝛼1 ) Γ (𝛼2 ) ⋅ ⋅ ⋅ Γ (𝛼𝑛 )
𝑛
𝑥
𝑠=2
𝑡𝑠−1
𝑥
𝛼1 −1
× ∫ 𝐹(𝑡𝑛 ) ∏ ∫ (𝑥 − 𝑡1 )
𝑎
𝛼𝑠 −1
× (𝑡𝑠−1 − 𝑡𝑠 )
𝑑𝑡𝑛
J𝑎𝛼1 J𝑎𝛼2 ⋅ ⋅ ⋅ J𝑎𝛼𝑛 𝐹 (𝑥) = J𝛼1 +𝛼2 +⋅⋅⋅+𝛼𝑛 −1 𝐽1 𝐹(𝑥)
𝑑𝑡𝑠 𝑑𝑡𝑛 .
(50)
(47)
The substitutions 𝑡𝑠 = 𝑡𝑛 +𝑦𝑠−1 (𝑥−𝑡𝑛 ), 𝑠 = 2, 3, . . . , 𝑛, and 𝑛 =
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:
J𝑎𝛼1 J𝑎𝛼2 ⋅ ⋅ ⋅ J𝑎𝛼𝑛 𝐹 (𝑥) =
𝑥
1
∫ 𝐹 (𝑡𝑛 )
Γ(𝛼1 ) Γ(𝛼2 ) ⋅ ⋅ ⋅ Γ(𝛼𝑛 ) 𝑎
𝑛
1
𝑠=2
0
𝛼𝑠 −1
(𝑥 − 𝑡𝑛 )𝑑𝑦𝑠−1]𝑑𝑡𝑛
𝑥
1
∫ 𝐹(𝑡𝑛 )
Γ(𝛼1 ) Γ(𝛼2 ) ⋅ ⋅ ⋅ Γ(𝛼𝑛 ) 𝑎
1
𝑠=2
0
= D𝛼𝑎 1 D𝛼𝑎 2 ⋅ ⋅ ⋅ D𝛼𝑎 𝑛 J𝛼𝑎 1 +𝛼2 +⋅⋅⋅+ 𝛼𝑛 Ψ
𝛼1 −1
𝛼𝑠 −1
×(𝑦𝑠−1 (𝑥 − 𝑡𝑛 ))
= 𝐷⌈𝛼1 ⌉ J𝑎⌈𝛼1 ⌉−𝛼1 𝐷⌈𝛼2 ⌉ J𝑎⌈𝛼2 ⌉−𝛼2 ⋅ ⋅ ⋅ 𝐷⌈𝛼𝑛 ⌉ J𝑎⌈𝛼𝑛 ⌉−𝛼𝑛
(𝑥 − 𝑡𝑛 )𝑑𝑦𝑠−1]𝑑𝑡𝑛
1
=
Γ(𝛼1 ) Γ(𝛼2 ) ⋅ ⋅ ⋅ Γ(𝛼𝑛 )
𝑥
𝛼1 −1
𝑎
1
𝛼1 −1
× [∫ (1 − 𝑦1 )
0
× J𝛼𝑎 1 +𝛼2 +⋅⋅⋅+ 𝛼𝑛 Ψ
= 𝐷⌈𝛼1 ⌉ J𝑎⌈𝛼1 ⌉−𝛼1 𝐷⌈𝛼2 ⌉ J𝑎⌈𝛼2 ⌉−𝛼2 ⋅ ⋅ ⋅ 𝐷⌈𝛼𝑛 ⌉ J𝑎⌈𝛼𝑛 ⌉
× J𝑎𝛼1 +𝛼2 +⋅⋅⋅+ 𝛼𝑛−1 Ψ
× ∏ ∫ 𝐹(𝑡𝑛 ) (𝑥 − 𝑡2 )
𝑠=2
Proof. The key for proof is using the semigroup property
of the integral operators, the assumption on 𝐹, and the
definition of the Riemann-Liouville differential operator:
D𝛼𝑎 1 D𝛼𝑎 2 ⋅ ⋅ ⋅ D𝛼𝑎 𝑛 𝐹
× ∏ [∫ ((𝑥 − 𝑡2 ) (1 − 𝑦1 ))
𝑛
(51)
𝛼1 −1
×(𝑦𝑠−1 (𝑥 − 𝑡𝑛 ))
𝑛
Theorem 12. Assume that 𝛼1 , 𝛼2 , . . . , 𝛼𝑛 ≥ 0. Moreover let Ψ ∈
𝐿 1 [𝑎, 𝑏]. Then,
D𝛼a 1 D𝛼a 2 ⋅ ⋅ ⋅ D𝛼a n 𝐹 = D𝛼a 1 +𝛼n +⋅⋅⋅𝛼n 𝐹.
× ∏ [∫ ((𝑥 − 𝑡2 ) (1 − 𝑦1 ))
=
almost everywhere. Since 𝐽1 𝐹(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.
𝛼𝑠
(𝑥 − 𝑡𝑛 )
𝑦𝑠−1 𝛼𝑠 −1 𝑑𝑦𝑠−1 ] 𝑑𝑡𝑛 .
(48)
= 𝐷⌈𝛼1 ⌉ J𝑎⌈𝛼1 ⌉−𝛼1 𝐷⌈𝛼2 ⌉ J𝑎⌈𝛼2 ⌉−𝛼2 ⋅ ⋅ ⋅ 𝐷⌈𝛼𝑛−1 ⌉ J𝑎⌈𝛼𝑛−1 ⌉−𝛼𝑛−1
× J𝛼𝑎 1 +𝛼2 +⋅⋅⋅+ 𝛼𝑛−1 Ψ
= 𝐷⌈𝛼1 ⌉ J𝑎⌈𝛼1 ⌉−𝛼1 𝐷⌈𝛼2 ⌉ J𝑎⌈𝛼2 ⌉−𝛼2 ⋅ ⋅ ⋅ 𝐷⌈𝛼𝑛−1 ⌉ J𝑎⌈𝛼𝑛−1 ⌉
× J𝛼𝑎 1 +𝛼2 +⋅⋅⋅+ 𝛼𝑛−2 Ψ
8
Abstract and Applied Analysis
= 𝐷⌈𝛼1 ⌉ J𝑎⌈𝛼1 ⌉−𝛼1 𝐷⌈𝛼2 ⌉ J𝑎⌈𝛼2 ⌉−𝛼2 ⋅ ⋅ ⋅ 𝐷⌈𝛼𝑛−2 ⌉ J𝑎⌈𝛼𝑛−2 ⌉−𝛼𝑛−2
× J𝑎𝛼1 +𝛼2 +⋅⋅⋅+ 𝛼𝑛−2 Ψ
𝑚
×(
𝜁
𝜕
) ∫ (𝜁 − 𝜏)𝑚−𝛼𝑛 −1 𝐹 (𝜏) 𝑑𝜏
𝜕𝜁𝑖𝑛
𝑎
𝑛
𝑛
𝑛
= ∑ ∑ ⋅ ⋅ ⋅ ∑ ̂𝑒𝑖1 ̂𝑒𝑖2 ⋅ ⋅ ⋅ ̂𝑒𝑖𝑛
..
.
𝑖1 =1 𝑖2 =1
=𝐷
⌈𝛼1 ⌉
J𝑎⌈𝛼1 ⌉ Ψ
= Ψ.
(52)
The proof that D𝛼𝑎 1 +𝛼𝑛 +⋅⋅⋅+𝛼𝑛 𝐹 = Ψ is quite straightforward
D𝑎𝛼1 +𝛼𝑛 +⋅⋅⋅𝛼𝑛 𝐹 = 𝐷⌈𝛼1 +𝛼𝑛 +⋅⋅⋅𝛼𝑛 ⌉
𝑚
𝑚
𝑚
𝜕
𝜕
𝜕
) (
) ⋅⋅⋅(
)
×(
𝜕𝜁𝑖1
𝜕𝜁𝑖2
𝜕𝜁𝑖𝑛
𝜁
J𝑎𝛼1 +𝛼2 +⋅⋅⋅+𝛼𝑛 Ψ
𝑎
(53)
𝜁
× ∫ (𝜁 − 𝜏)𝑚−𝛼2 −1 𝐹(𝜏) 𝑑𝜏 ⋅ ⋅ ⋅
𝑎
𝜁
D𝑎𝛼1 +𝛼𝑛 +....𝛼𝑛 𝐹 = 𝐷⌈𝛼1 +𝛼𝑛 +....𝛼𝑛 ⌉
×
1
Γ (𝑚 − 𝛼1 ) Γ (𝑚 − 𝛼2 ) ⋅ ⋅ ⋅ Γ (𝑚 − 𝛼𝑛 )
× ∫ (𝜁 − 𝜏)𝑚−𝛼1 −1 𝐹(𝜏) 𝑑𝜏
× J𝑎⌈𝛼1 +𝛼𝑛 +⋅⋅⋅𝛼𝑛 ⌉−𝛼1 −𝛼𝑛 −⋅⋅⋅−𝛼𝑛
×
×
𝑖𝑛 =1
J𝑎⌈𝛼1 +𝛼𝑛 +....𝛼𝑛 ⌉ Ψ
× ∫ (𝜁 − 𝜏)𝑚−𝛼𝑛 −1 𝐹(𝜏) 𝑑𝜏
= Ψ.
𝑛
𝑎
𝑛
𝑛
= ∑ ∑ ⋅ ⋅ ⋅ ∑ ̂𝑒𝑖1 ̂𝑒𝑖2 ⋅ ⋅ ⋅ ̂𝑒𝑖𝑛
Thus,
D𝛼𝑎 1 +𝛼𝑛 +⋅⋅⋅+𝛼𝑛 𝐹 = D𝛼𝑎 1 D𝛼𝑎 2 ⋅ ⋅ ⋅ D𝛼𝑎 𝑛 𝐹.
𝑖1 =1 𝑖2 =1
(54)
Theorem 13. Assume that 𝛼1 , 𝛼2 , . . . 𝛼𝑛 ≥ 0. Moreover, let
Ψ ∈ 𝐿 1 [𝑎, 𝑏] and 𝐹 = J𝛼𝑎 1 +𝛼2 +⋅⋅⋅+ 𝛼𝑛 Ψ. And let ∇𝜁𝛼 be the
fractional gradient operator. Then,
×
𝑖𝑛 =1
1
𝜕𝑛
)
(
Γ (𝑛𝑚 − 𝛼1 − 𝛼2 ⋅ ⋅ ⋅ − 𝛼𝑛 ) 𝜕𝜁𝑖1 ⋅ ⋅ ⋅ 𝜕𝜁𝑖2
𝜁
𝜁
× ∫ (𝜁 − 𝜏)𝑚−𝛼1 −1 𝐹(𝜏) 𝑑𝜏 ∫ (𝜁 − 𝜏)𝑚−𝛼2 −1
𝑎
𝑎
𝜁
× 𝐹(𝜏) 𝑑𝜏 ⋅ ⋅ ⋅ ∫ (𝜁 − 𝜏)
𝐹(𝜏) 𝑑𝜏.
(56)
𝑛
𝑛
= ∑ ∑ ⋅ ⋅ ⋅ ∑ ̂𝑒𝑖1 ̂𝑒𝑖2 ⋅ ⋅ ⋅ ̂𝑒𝑖𝑛
𝑖1 =1 𝑖2 =1
𝑖𝑛 =1
1
Γ (𝑛𝑚 − 𝛼1 − 𝛼2 ⋅ ⋅ ⋅ − 𝛼𝑛 )
𝑚
×(
𝑚−𝛼𝑛 −1
𝑎
∇𝜁𝛼(𝑛) 𝐹 (𝜁)
𝑛
𝑚
𝜁
𝜕𝑛
𝑛𝑚−𝛼1 −𝛼2 −⋅⋅⋅−𝛼𝑛
) ∫ 𝐹(𝜏𝑛 ) (𝜁 − 𝜏𝑛 )
𝑑𝜏𝑛 .
𝜕𝜁𝑖1 ⋅ ⋅ ⋅ 𝜕𝜁𝑖2
𝑎
(55)
Proof. By the assumption on 𝐹 and the successive application of fractional gradient operator, we have ∇𝜁𝛼(𝑛) 𝐹(𝜁) ≡
𝛼
𝛼 𝛼
∇𝜁 1 ∇𝜁 2 ⋅ ⋅ ⋅ ∇𝜁 𝑛 𝐹(𝜁):
𝑛
∇𝜁𝛼(𝑛) 𝐹 (𝜁) = ∑ ̂𝑒𝑖1
𝑖1 =1
𝜁
𝑎
𝑖2 =1
𝑚
1
𝜕
)
(
Γ (𝑚 − 𝛼2 ) 𝜕𝜁𝑖2
𝜁
× ∫ (𝜁 − 𝜏)𝑚−𝛼2 −1 𝐹 (𝜏) 𝑑𝜏 ⋅ ⋅ ⋅
𝑎
𝑛
× ∑ ̂𝑒𝑖𝑛
𝑖𝑛 =1
𝑛
𝑛
𝑖1 =1 𝑖2 =1
×
×
𝑖𝑛 =1
1
𝜕𝑛
)
(
Γ (𝑛𝑚 − 𝛼1 − 𝛼2 ⋅ ⋅ ⋅ − 𝛼𝑛 ) 𝜕𝜁𝑖1 ⋅ ⋅ ⋅ 𝜕𝜁𝑖2
𝜁
𝑛𝑚−𝛼1 −𝛼2 −⋅⋅⋅−𝛼𝑛
∫ 𝐹 (𝜏𝑛 ) (𝜁 − 𝜏𝑛 )
𝑎
𝑚
𝑑𝜏𝑛 .
(57)
7. Fractional Taylor Expansion of a Function
of 𝑁-Dimensional Polyadics
× ∫ (𝜁 − 𝜏)𝑚−𝛼1 −1 𝐹 (𝜏) 𝑑𝜏
𝑛
𝑛
∇𝜁𝛼(𝑛) 𝐹 (𝜁) = ∑ ∑ ⋅ ⋅ ⋅ ∑ ̂𝑒𝑖1 ̂𝑒𝑖2 ⋅ ⋅ ⋅ ̂𝑒𝑖𝑛
𝑚
1
𝜕
)
(
Γ (𝑚 − 𝛼1 ) 𝜕𝜁𝑖1
× ∑ ̂𝑒𝑖2
Thus using the theorem, the results follow directly:
1
Γ (𝑚 − 𝛼𝑛 )
We have the classical Taylor expansion for the 𝑚-independent
variables as
𝑓 (𝑥1 , 𝑥2 , . . . , 𝑥𝑚 )
∞
𝑛
1 𝑚
𝜕
= ∑ (∑ (𝑥𝑖 − 𝑥𝑖0 )
)
𝜕𝑥𝑖
𝑛=1 𝑛! 𝑖=1
󵄨󵄨
󵄨󵄨
× 𝑓(𝑥1 , 𝑥2 , . . . , 𝑥𝑚 ) 󵄨󵄨󵄨
󵄨󵄨
󵄨𝑥1 =𝑥10 , 𝑥2 =𝑥20 , ⋅⋅⋅𝑥𝑚 =𝑥𝑚0
(58)
Abstract and Applied Analysis
9
In the light of the above definitions and theorems, we can state
the following theorem.
Definition 14. Let 𝑓(𝑥𝑖 ) = 𝑓(𝑥1 , . . . , 𝑥𝑛 ) ∈ 𝐴𝑚 (Ω), where
Ω = ∏𝑛𝑖=1 (𝑎𝑖 , 𝑏𝑖 ) ⊂ R𝑛 ; then the fractional Riemann-Liouville
multiple integrals and partial derivatives with respect to 𝑥𝑖
are;
Thus, by definition of D𝑛𝑎𝑘 ,𝑥𝑘 ,
𝑖
𝑖
J𝑛𝑎𝑘 ,𝑥𝑘 D𝑛𝑎𝑘 ,𝑥𝑘 = 𝑓(𝑥𝑖 ) = J𝑛𝑎𝑘 ,𝑥𝑘 𝐷𝑥𝑚𝑘 J𝑚−𝑛
𝑎𝑘 ,𝑥𝑘 𝑓(𝑥𝑖 )
𝑖
𝑖
𝑖
𝑖
𝑖
𝑖
𝑖
×∫
𝑎𝑘𝑠
𝑚−1
𝑖
𝑖
(𝑥𝑖 − 𝑎𝑖 )
lim 𝐷𝑥𝑘 𝑘
𝑖
𝑧𝑖 → 𝑎𝑖+
𝑘!
𝑘=0
𝑖
× J𝑚−𝑛
𝑎𝑘 ,𝑥𝑘 𝑓 (𝑧𝑖 )
𝑖
𝑠
𝑥
𝑖=1
𝑡𝑠−1
𝑓 (𝑡𝑖 ) ∏ ∫
𝑖
𝑖
𝑖
(𝑥𝑘𝑖 − 𝑡𝑘𝑖 )
(59)
𝑥𝑖
𝑛−𝛼−1
𝑎𝑖
𝑚
𝑚−1 J𝑛
𝑎𝑘𝑖 ,𝑥𝑘𝑖 𝐷𝑥𝑘𝑖 (𝑥𝑖
= 𝑓 (𝑥𝑖 ) = ∑
𝑘
− 𝑎𝑖 )
𝑖
𝑖
𝑖
+ J𝑛𝑎𝑘 ,𝑥𝑘 𝐷𝑥𝑚𝑘 𝐽𝑎𝑚𝑘 ,𝑥𝑘 Ψ (𝑥𝑖 ) .
𝑑𝑡𝑖 .
𝑖
𝑖
𝑖
𝑖
𝑖
𝐷𝑥𝑚𝑘 annihilates every summand in the sum. And due to
𝑖
Theorem 8, we obtain
J𝑛𝑎𝑘 ,𝑥𝑘 D𝑛𝑎𝑘 ,𝑥𝑘 𝑓 (𝑥𝑖 ) = J𝑛𝑎𝑘 ,𝑥𝑘 = Ψ (𝑥𝑖 ) .
𝑖
𝑖
𝑖
𝑖
𝑖
(64)
𝑖
Next, we apply the operator D𝑚−𝑛
𝑎𝑘 ,𝑥𝑘 to (63), finding
𝑖
(𝑥𝑖 − 𝑎𝑖 )
lim J𝑚−𝑛 = 𝑓(𝑧𝑖 )
Γ (𝑛 − 𝑚 − 1) 𝑧𝑖 → 𝑎𝑖+ 𝑎𝑘𝑖 ,𝑥𝑘𝑖
𝑘+𝑛−𝑚
(𝑥𝑖 − 𝑎𝑖 )
lim 𝐷𝑥𝑘 𝑘 J𝑚−𝑛
+∑
𝑎𝑘𝑖 ,𝑥𝑘𝑖
𝑖
𝑧𝑖 → 𝑎𝑖+
Γ
+
𝑛
−
𝑚
−
1)
(𝑘
𝑘=0
× 𝑓(𝑧𝑖 ) +
(63)
𝑘!
𝑘=0
𝑧𝑖 → 𝑎𝑖
𝑛−𝑚
𝑚−1
𝑖
× lim + 𝐷𝑥𝑘 𝑘 J𝑚−𝑛
𝑎𝑘 ,𝑥𝑘 𝑓(𝑧𝑖 )
Theorem 15. Let 𝑛 > 0 and 𝑚 = ⌊𝑛⌋ + 1. Assume that 𝑓(𝑥𝑖 ) is
𝑛
𝑚
𝑛
such that J𝑛−𝑚
𝑎 𝑓(𝑥𝑖 ) ∈ 𝐴 (Ω), where Ω = ∏𝑖=1 (𝑎𝑖 , 𝑏𝑖 ) ⊂ R
is the domain of 𝑓. Then,
𝑓(𝑥𝑖 ) =
J𝑛𝑎𝑘 ,𝑥𝑘 D𝑛𝑎𝑘 ,𝑥𝑘
𝑖
𝑖
𝑖
𝑖
𝜕 𝑛
1
)(
)
Γ (𝑛 − 𝛼)
𝜕𝑥𝑖
× ∫ 𝑓(𝑡𝑖 ) (𝑥𝑖 − 𝑡𝑖 )
𝑖
+𝐽𝑎𝑚𝑘 ,𝑥𝑘 Ψ (𝑥𝑖 ) ] ,
𝛼−1
× 𝑑𝑡𝑘1 ⋅ ⋅ ⋅ 𝑑𝑡𝑘𝑠
D𝛼𝑎𝑘 ,𝑥𝑘 𝑓(𝑥𝑖 ) = (
𝑖
𝑘
= J𝑛𝑎𝑘 ,𝑥𝑘 𝐷𝑥𝑚𝑘 [ ∑
1 𝑠 𝑥𝑘1
) ∫ ⋅⋅⋅
J𝛼𝑎𝑘 ,𝑥𝑘 𝑓 (𝑥𝑖 ) = (
𝑖
𝑖
Γ (𝛼)
𝑎𝑘1
𝑥𝑘𝑠
𝑖
𝑚−1 D𝑚−𝑛 (𝑥𝑖
𝑎𝑘𝑖 ,𝑥𝑘𝑖
(60)
𝑓 (𝑥𝑖 ) = ∑
𝑖
𝑘
− 𝑎𝑖 )
lim 𝐷𝑥𝑘 𝑘 J𝑚−𝑛
𝑎𝑘 ,𝑥𝑘
𝑘!
𝑘=0
𝑧𝑖 → 𝑎𝑖+
𝑖
𝑖
𝑖
𝑚
= 𝑓 (𝑧𝑖 ) + D𝑚−𝑛
𝑎𝑘 ,𝑥𝑘 𝐽𝑎𝑘 ,𝑥𝑘 Ψ (𝑥𝑖 )
𝑖
J𝑛𝑎𝑘 ,𝑥𝑘 D𝑛𝑎𝑘 ,𝑥𝑘 𝑓(𝑥𝑖 ) .
𝑖
𝑖
𝑖
𝑖
𝑖
𝑚−1 D𝑚−𝑛 (𝑥𝑖
𝑎𝑘𝑖 ,𝑥𝑘𝑖
= ∑
𝑖
𝑘
− 𝑎𝑖 )
𝑘!
𝑘=0
Proof. Because of our assumption about 𝑓 that implies the
J𝑎𝑚−𝑛
𝑓, there exists some Ψ ∈ 𝐿 1 (Ω) such
continuity of 𝐷𝑥𝑚−1
𝑘𝑖
𝑘𝑖 ,𝑥𝑘𝑖
that
𝑖
lim 𝐷𝑥𝑘 𝑘 J𝑚−𝑛
𝑎𝑘 ,𝑥𝑘
𝑧𝑖 → 𝑎𝑖+
𝑖
𝑖
𝑖
(65)
𝑚
= 𝑓 (𝑧𝑖 ) + 𝐷𝑥1 𝑘 J1−𝑚+𝑛
𝑎𝑘 ,𝑥𝑘 𝐽𝑎𝑘 ,𝑥𝑘 Ψ (𝑥𝑖 )
𝑖
𝑖
𝑖
𝑖
𝑖
𝑘+𝑛−𝑚
𝑚−1
(𝑥𝑖 − 𝑎𝑖 )
lim 𝐷𝑥𝑘 𝑘 J𝑚−𝑛
𝑎𝑘𝑖 ,𝑥𝑘𝑖
𝑖
𝑧𝑖 → 𝑎𝑖+
Γ
+
𝑛
−
𝑚
−
1)
(𝑘
𝑘=0
= ∑
𝑚−1 𝑚−𝑛
1
𝐷𝑥𝑚−1
J𝑚−𝑛
𝑎𝑘 ,𝑥𝑘 𝑓 (𝑥𝑖 ) = 𝐷𝑥𝑘 J𝑎𝑘 ,𝑥𝑘 𝑓 (𝑎𝑖 ) + 𝐽𝑎𝑘 ,𝑥𝑘 Ψ (𝑥𝑖 ) .
𝑘
𝑖
𝑖
𝑖
𝑖
𝑖
𝑖
𝑖
𝑖
= 𝑓 (𝑧𝑖 ) + J𝑚−𝑛
𝑎𝑘 ,𝑥𝑘 Ψ (𝑥𝑖 ) .
(61)
This is a classical partial differential equation of order 𝑚 − 1
for J𝑎𝑚−𝑛 𝑓(𝑥𝑖 ); its solution is easily seen to be of the form:
𝑖
Using (64), we obtain
J𝑛𝑎𝑘 ,𝑥𝑘 Ψ (𝑥𝑖 ) = J𝑎𝑛𝑘 ,𝑥𝑘 D𝑛𝑎𝑘 ,𝑥𝑘 𝑓 (𝑥𝑖 )
𝑖
𝑚−1
𝑖
𝑖
= 𝑓 (𝑧𝑖 ) +
𝐽𝑎𝑚𝑘 ,𝑥𝑘 Ψ (𝑥𝑖 ) .
𝑖
𝑖
𝑖
𝑖
𝑖
𝑖
𝑖
𝑚−1
𝑘
(𝑥𝑖 − 𝑎𝑖 )
lim 𝐷𝑥𝑘 𝑘 J𝑚−𝑛
𝑎𝑘𝑖 ,𝑥𝑘𝑖
𝑖
𝑧𝑖 → 𝑎𝑖+
𝑘!
𝑘=0
J𝑎𝑚−𝑛
𝑓(𝑥𝑖 ) = ∑
𝑘 ,𝑥𝑘
𝑖
(62)
𝑘+𝑛−𝑚
(𝑥𝑖 − 𝑎𝑖 )
= 𝑓 (𝑥𝑖 ) − ∑
lim 𝐷𝑥𝑘 𝑘
𝑖
𝑧𝑖 → 𝑎𝑖+
Γ
+
𝑛
−
𝑚
−
1)
(𝑘
𝑘=0
× J𝑚−𝑛
𝑎𝑘 ,𝑥𝑘 𝑓 (𝑧𝑖 ) .
𝑖
𝑖
(66)
10
Abstract and Applied Analysis
Upon extracting the first term out of summation, we can write
Proof. It is possible to invert or solve the equation for 𝐹(𝑧) in a
neighborhood of 𝐹(𝑎). Using the fractional Taylor expansion.
𝑛−𝑚
(𝑥𝑖 − 𝑎𝑖 )
lim J𝑚−𝑛 𝑓 (𝑧𝑖 )
𝑓 (𝑥𝑖 ) =
Γ (𝑛 − 𝑚 − 1) 𝑧𝑖 → 𝑎𝑖+ 𝑎𝑘𝑖 ,𝑥𝑘𝑖
∞ ∞
𝑘=0 𝑛=0
𝑘+𝑛−𝑚
𝑚−1
(𝑥𝑖 − 𝑎𝑖 )
lim 𝐷𝑥𝑘 𝑘 J𝑎𝑚−𝑛
𝑘𝑖 ,𝑥𝑘𝑖
𝑖
𝑧𝑖 → 𝑎𝑖+
Γ(𝑘
+
𝑛
−
𝑚
−
1)
𝑘=1
+∑
𝑖
𝑖
× lim + ([𝜇(𝑤 − 𝑎)𝑛−𝑚 ]
(67)
𝑖
𝑤→𝑎
Extracting the first term out of the summations, we obtain the
result directly
× 𝑓 (𝑥𝑖 ) .
𝐹 (𝑧) =
This result can be written conveniently in the Polyadics
notation 𝐴(𝑛) , extending the upper limit of summation to ∞
to absorb the remainder, as
1
Γ
+
𝑛
− 𝑚 − 1)
(𝑘
𝑘=0
×
𝑧(𝑛) → 𝑎(𝑛)
J𝑚−𝑛
𝑎𝑘𝑖 ,𝑥𝑘𝑖
∞
∞
𝜀𝑛
𝑛=0 Γ (𝑘 + 𝛼 − 𝑚 − 1)
+∑ ∑
𝑘=1
𝑤→𝑎
𝐹 (𝑟(𝑛) ) = ∑
lim
1
J𝑚−𝛼 = 𝐹(𝑎)
Γ (𝛼 − 𝑚 − 1) 𝑎𝑘𝑖 ,𝑥𝑘𝑖
× lim + ([𝜇(𝑤 − 𝑎)𝑛−𝑚 ]
∞
×
𝑘
(71)
𝑛
( ) ∇𝜁𝛼(𝑛) ) J𝑎𝑚−𝛼
𝑘𝑖 ,𝑥𝑘𝑖
⋅
(𝑛)
× 𝐹 (𝑤) .
× 𝑓 (𝑧𝑖 ) + J𝑛𝑎𝑘 ,𝑥𝑘 D𝑛𝑎𝑘 ,𝑥𝑘
𝑖
𝜀𝑛
Γ (𝑘 + 𝛼 − 𝑚 − 1)
𝐹(𝑧) = ∑ ∑
(𝑛)
(72)
𝑘
𝑛
( ) ∇𝜁𝛼(𝑛) ) J𝑚−𝛼
𝑎𝑘𝑖 ,𝑥𝑘𝑖
⋅
× 𝐹 (𝑤) .
((𝑟(𝑛) − 𝑎(𝑛) )
+
𝑛−𝑚
𝑘
𝑛
( ) ∇𝜁𝛼(𝑛) )
⋅
(68)
Acknowledgments
(𝑛)
= 𝐹 (𝑧 ) .
8. Fractional 𝑁-Dimensional Lagrange
Expansion
Theorem 16. Let 𝐹(𝑧) = 𝐹(𝑤) + 𝜇(𝑧, 𝜀), 𝐹, 𝜇 ∈ 𝐴𝑚 (Ω) such
that
The author is deeply indebted and thankful to the deanship of the scientific research and its helpful and distinct
team of employees at Taibah University, Al-Madinah AlMunawwarah, Saudi Arabia This research work was supported by a Grant no. 3017/1434. Also he wishes to express
his deep gratitude and indebtedness to the referees of his
manuscript for all very fruitful discussions, constructive
comments, remarks, and suggestions.
∞
𝜀𝑛
(𝑛) 𝑛
[𝜇 (𝑧)] ( ) ∇𝑤(𝑛) 𝐹 (𝑤) ,
⋅
𝑛=1 𝑛!
𝐹(𝑧) = 𝐹(𝑤) + ∑
(69)
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:
𝐹 (𝑧) =
1
J𝑚−𝛼 = 𝐹(𝑎)
Γ (𝛼 − 𝑚 − 1) 𝑎𝑘𝑖 ,𝑥𝑘𝑖
∞
∞
𝜀𝑛
𝑛=0 Γ (𝑘 + 𝛼 − 𝑚 − 1)
+∑ ∑
𝑘=1
(𝑛)
× lim + ([𝜇(𝑤 − 𝑎)𝑛−𝑚 ]
𝑤→𝑎
× J𝑚−𝛼
𝑎𝑘 ,𝑥𝑘 = 𝐹 (𝑤) ,
𝑖
𝑖
where 𝛼 > 0 and 𝑚 = ⌊𝛼⌋ + 1.
(70)
𝑘
𝑛
( ) ∇𝜁𝛼(𝑛) )
⋅
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