Journal of Computational and Applied Mathematics 279 (2015) 57–66

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Journal of Computational and Applied

Mathematics
journal homepage: www.elsevier.com/locate/cam

On conformable fractional calculus

Thabet Abdeljawad ∗
Department of Mathematics and General Sciences, Prince Sultan University-Riyadh-KSA, Saudi Arabia
Department of Mathematics, Çankaya University, 06530, Ankara, Turkey

article info abstract

Article history: Recently, the authors Khalil et al. (2014) introduced a new simple well-behaved defini-
Received 6 January 2014 tion of the fractional derivative called conformable fractional derivative. In this article we
Received in revised form 27 July 2014 proceed on to develop the definitions there and set the basic concepts in this new simple
interesting fractional calculus. The fractional versions of chain rule, exponential functions,
Keywords: Gronwall’s inequality, integration by parts, Taylor power series expansions, Laplace trans-
Left and right conformable fractional
forms and linear differential systems are proposed and discussed.
derivatives
Left and right conformable fractional
© 2014 Elsevier B.V. All rights reserved.
integrals
Fractional Taylor power series expansion
Fractional exponential function
Fractional Gronwall’s inequality
Chain rule

1. Introduction

The fractional calculus [1–3] attracted many researches in the last and present centuries. The impact of this fractional
calculus in both pure and applied branches of science and engineering started to increase substantially during the last two
decades apparently. Many researches started to deal with the discrete versions of this fractional calculus benefitting from
the theory of time scales (see [4–8] and the references therein. The main idea behind setting this fractional calculus is
summarized into two approaches. The first approach is the Riemann–Liouville which based on iterating the integral operator
n times and then replaced it by one integral via the famous Cauchy formula where then n! is changed to the Gamma function
and hence the fractional integral of noninteger order is defined. Then integrals were used to define Riemann and Caputo
fractional derivatives. The second approach is the Grünwald–Letnikov approach which based on iterating the derivative n
times and then fractionalizing by using the Gamma function in the binomial coefficients. The obtained fractional derivatives
in this calculus seemed complicated and lost some of the basic properties that usual derivatives have such as the product
rule and the chain rule. However, the semigroup properties of these fractional operators behave well in some cases. Recently,
the author in [9] define a new well-behaved simple fractional derivative called ‘‘the conformable fractional derivative’’
depending just on the basic limit definition of the derivative. Namely, for a function f : (0, ∞) → R the (conformable)
fractional derivative of order 0 < α ≤ 1 of f at t > 0 was defined by

f (t + ϵ t 1−α ) − f (t )
Tα f (t ) = lim , (1)
ϵ→0 ϵ
and the fractional derivative at 0 is defined as (Tα f )(0) = limt →0+ (Tα f )(t ).

∗ Correspondence to: Department of Mathematics and General Sciences, Prince Sultan University-Riyadh-KSA, Saudi Arabia.
E-mail address: tabdeljawad@psu.edu.sa.

http://dx.doi.org/10.1016/j.cam.2014.10.016
0377-0427/© 2014 Elsevier B.V. All rights reserved.
58 T. Abdeljawad / Journal of Computational and Applied Mathematics 279 (2015) 57–66

They then define the fractional derivative of higher order (i.e. of order α > 1) as we will see below in next sections. They
also define the fractional integral of order 0 < α ≤ 1 only. They then proved the product rule, the fractional mean value
tα
theorem, and solved some (conformable) fractional differential equations where the fractional exponential function e α
played an important rule. While in case of well-known fractional calculus Mittag-Leffler functions generalized exponential
functions. In this article we continue to settle the basic definitions and concepts of this new theory motivated by the fact
that there are certain functions which do not have Taylor power series representation or their Laplace transform cannot be
calculated and so forth, but will be possible to do so by the help of the theory of this conformable fractional calculus. The
article is organized as follows: In Section 2 the left and right (conformable) fractional derivatives and fractional integrals
of higher orders are defined, the fractional chain rule and Gronwall inequality are obtained and the action of fractional
derivatives and integrals to each other are discussed. The conformable and sequential conformable fractional derivatives
of higher orders are discussed at the end points as well. In Section 3, two kinds of fractional integration by parts formulas
when 0 < α ≤ 1 are obtained where the usual integration by parts formulas in usual cases are reobtained when α → 1. In
Section 4, the fractional power series expansions for certain functions that do not have Taylor power series representation in
usual calculus are obtained and the fractional Taylor inequality is proved. Finally, in Section 5 the fractional Laplace transform
is defined and used to solve a conformable fractional linear differential equation, where also the fractional Laplace of certain
basic functions are calculated.

2. Basic definitions and tools

Definition 2.1. The (left) fractional derivative starting from a of a function f : [a, ∞) → R of order 0 < α ≤ 1 is defined
by

f (t + ϵ(t − a)1−α ) − f (t )
(Tαa f )(t ) = lim . (2)
ϵ→0 ϵ
When a = 0 we write Tα . If (Tα f )(t ) exists on (a, b) then (Tαa f )(a) = limt →a+ (Tαa f )(t ).
The (right) fractional derivative of order 0 < α ≤ 1 terminating at b of f is defined by

f (t + ϵ(b − t )1−α ) − f (t )
(bα Tf )(t ) = − lim . (3)
ϵ→0 ϵ
If (b Tα f )(t ) exists on (a, b) then (b Tα f )(b) = limt →b− (b Tα f )(t ).
Note that if f is differentiable then (Tαa f )(t ) = (t − a)1−α f ′ (t ) and (bα Tf )(t ) = −(b − t )1−α f ′ (t ). It is clear that the
conformable fractional derivative of the constant function is zero. Conversely, if Tα f (t ) = 0 on an interval (a, b) then by the
help of conformable fractional mean value theorem proved in [9] we can easily show that f (x) = 0 for all x ∈ (a, b). Also, by
the help of the fractional mean value theorem there we can show that if the conformable fractional derivative of a function
f on an interval (a, b) is positive (negative) then the graph of f is increasing (decreasing) there.

Notation. (Iαa f )(t ) = a f (x)dα(x, a) = a (x − a)α−1 f (x)dx. When a = 0 we write dα(x). Similarly, in the right case we have
t t

(b Iα f )(t ) = t f (x)dα(b, x) = t (b − x)α−1 f (x)dx. The operators Iαa and b Iα are called conformable left and right fractional
b b
integrals of order 0 < α ≤ 1.
In the higher order case we can generalize to the following:

Definition 2.2. Let α ∈ (n, n + 1], and set β = α − n. Then, the (left) fractional derivative starting from a of a function
f : [a, ∞) → R of order α , where f (n) (t ) exists, is defined by

(Tαa f )(t ) = (Tβa f (n) )(t ). (4)

When a = 0 we write Tα .
The (right) fractional derivative of order α terminating at b of f is defined by

(bα T f )(t ) = (−1)n+1 (bβ Tf (n) )(t ). (5)

Note that if α = n + 1 then β = 1 and the fractional derivative of f becomes f (n+1) (t ). Also when n = 0 (or α ∈ (0, 1))
then β = α and the definition coincides with those in Definition 2.1.

Lemma 2.1 ([9]). Assume that f : [a, ∞) → R is continuous and 0 < α ≤ 1. Then, for all t > a we have
Tαa Iαa f (t ) = f (t ).

In the right case we can similarly prove:

Lemma 2.2. Assume that f : (−∞, b] → R is continuous and 0 < α ≤ 1. Then, for all t < b we have
b
Tα b Iα f (t ) = f (t ).
 T. Abdeljawad / Journal of Computational and Applied Mathematics 279 (2015) 57–66 59

Next we give the definition of left and right fractional integrals of any order α > 0.

Definition 2.3. Let α ∈ (n, n + 1] then the left fraction integral of order α starting at a is defined by
 t
1
(Iαa f )(t ) = Ina+1 ((t − a)β−1 f ) = (t − x)n (x − a)β−1 f (x)dx (6)
n! a

t
Notice that if α = n + 1 then β = α − n = n + 1 − n = 1 and hence (Iαa f )(t ) = (Ian+1 f )(t ) = n1! a (t − x)n f (x)dx, which
is by means of Cauchy formula the iterative integral of f , n + 1 times over (a, t ].
Recalling that the left Riemann–Liouville fractional integral of order α > 0 starting from a is defined by
 t
1
(Iaα f )(t ) = (t − s)α−1 f (s)ds, (7)
Γ (α) a
we see that (Iαa f )(t ) = (Iaα f )(t ) for α = n + 1, n = 0, 1, 2, . . ..
Γ (µ)
Example 2.4. Recalling that [1] (Iαa (t − a)µ−1 )(x) = Γ (µ+α) (x − a)α+µ−1 , α, µ > 0, we can calculate the (conformable)
fractional integral of (t − a)µ of order α ∈ (n, n + 1]. Indeed, if µ ∈ R such that α + µ − n > 0 then

Γ (α + µ − n)
(Iαa (t − a)µ )(x) = (Ina+1 (t − a)µ+α−n−1 )(x) = (x − a)α+µ . (8)
Γ (α + µ + 1)
Analogously, we can find the (conformable) right fractional integral of such functions. Namely,

Γ (α + µ − n)
(b Iα (b − t )µ )(x) = (b In+1 (t − a)µ+α−n−1 )(x) = (b − x)α+µ , (9)
Γ (α + µ + 1)
where µ ∈ R such that α + µ − n > 0.
From the above discussion, we notice that the Riemann fractional integrals and conformable fractional integrals of
polynomial functions are different up to a constant multiple and coincide for natural orders.
The following semigroup property relates the composition operator Iµ Iα and the operator Iα+µ .

Proposition 2.3. Let f : [a, ∞) → R be a function and 0 < α, µ ≤ 1 be such that 1 < α + µ ≤ 2. Then

tµ t

1 t
(Iα Iµ f )(t ) = (Iα f )(t ) + (Iα+µ f )(t ) − sα+µ−2 f (s)ds. (10)
µ µ µ 0

Proof. Interchanging the order of integrals and noting that

 t
(Iα+µ f )(t ) = (I2 sα+µ−2 f (s))(t ) = (t − s)sα+µ−2 ds, (11)
0

we see that
 t  t1

α−1 µ−1
(Iα Iµ f )(t ) = f (s)s ds t1 dt1
0 0
t t
  
µ−1
= f (s)sα−1 t1 dt1 ds
0 s
t
tµ µ
 
s
= f (s)sα−1 − ds
0 µ µ
µ   t 
t 1 α+µ−2
= (Iα f )(t ) + (Iα+µ f )(t ) − t s f (s)ds .  (12)
µ µ 0

Notice that if in (10) we let α, µ → 1 we verify that (I1 I1 f )(t ) = (I2 f )(t ).
Recalling the action of the Q -operator on fractional integrals (Qf (t ) = f (a + b − t ), f : [a, b] → R) on Riemann left
and right fractional integrals, we see that:

Q Iaα f (t ) = b Iα Qf (t ). (13)
Indeed, for α ∈ (n, n + 1] we have

QIαa f (t ) = Q Ian+1 ((t − a)α−n−1 f (t )) = b In+1 ((b − t )α−n−1 f (a + b − t )) = b Iα Qf (t ), (14)

Now we give the generalized version of Lemma 2.1.
60 T. Abdeljawad / Journal of Computational and Applied Mathematics 279 (2015) 57–66

Lemma 2.4. Assume that f : [a, ∞) → R such that f (n) (t ) is continuous and α ∈ (n, n + 1]. Then, for all t > a we have
Tαa Iαa f (t ) = f (t ).

Proof. From the definition we have

dn
   n 
d a β−1
Tα Iα f (t ) = Tβ
a a a
I f (t ) = Tβ
a
n α
a
I ((t − a)
n n+1
f (t )) = Tβa (I1a ((t − a)β−1 f (t ))) (15)
dt dt
That it is Taα Iαa f (t ) = Tβa Iβa f (t ) and hence the result follows by Lemma 2.1. 
Similarly we can generalize Lemma 2.2. Indeed,

Lemma 2.5. Assume that f : (−∞, b] → R such that f (n) (t ) is continuous and α ∈ (n, n + 1]. Then, for all t < b we have
b
Tα b Iα f (t ) = f (t ).

Lemma 2.6. Let f , h : [a, ∞) → R be functions such that Tαa exists for t > a, f is differentiable on (a, ∞) and Tαa f (t ) =
(t − a)1−α h(t ). Then h(t ) = f ′ (t ) for all t > a.
The proof follows by definition and setting h = ϵ(t − a)1−α so that h → 0 as ϵ → 0. As a result of Lemma 2.6 we can
state

Corollary 2.7. Let f : [a, b) → R be such that (Iαa Tαa )f (t ) exists for b > t > a. Then, f (t ) is differentiable on (a, b).

Lemma 2.8. Let f : (a, b) → R be differentiable and 0 < α ≤ 1. Then, for all t > a we have

Iαa Tαa (f )(t ) = f (t ) − f (a). (16)

Proof. Since f is differentiable then by the help of Theorem 2.1 (6) in [9] we have
 t  t
Iαa Tαa (f )(t ) = (x − a)α−1 Tα (f )(x)dx = (x − a)α−1 (x − a)1−α f ′ (x)dx = f (t ) − f (a).  (17)
a a

Lemma 2.8 can be generalized for the higher order as follows.

Proposition 2.9. Let α ∈ (n, n + 1] and f : [a, ∞) → R be (n + 1) times differentiable for t > a. Then, for all t > a we have

f (k) (a)(t − a)k

n

Iαa Tαa (f )(t ) = f (t ) − . (18)
k=0
k!

Proof. From definition and Theorem 2.1 (6) in [9] we have

Iαa Taα (f )(t ) = Ina+1 ((t − a)β−1 Tβa f (n) (t )) = Ina+1 ((t − a)β−1 (t − a)1−β f (n+1) (t )) = Ina+1 f (n+1) (t ). (19)
Then integration by parts gives (20). 
Analogously, in the right case we have

Proposition 2.10. Let α ∈ (n, n + 1] and f : (−∞, b] → R be (n + 1) times differentiable for t < b. Then, for all t < b we
have
n
 (−1)k f (k) (b)(b − t )k
b
Iα b Tα (f )(t ) = f (t ) − . (20)
k=0
k!

In particular, if n = 0 or 0 < α ≤ 1, then b Iα b Tα (f )(t ) = f (t ) − f (b).

Theorem 2.11 (Chain Rule). Assume f , g : (a, ∞) → R be (left) α -differentiable functions, where 0 < α ≤ 1. Let h(t ) =
f (g (t )). Then h(t ) is (left) α -differentiable and for all t with t ̸= a and g (t ) ̸= 0 we have

(Tαa h)(t ) = (Tαa f )(g (t )) · (Tαa g )(t ) · g (t )α−1 . (21)

If t = a we have

(Tαa h)(a) = lim (Tαa f )(g (t )) · (Tαa g )(t ) · g (t )α−1 . (22)

t →a+
 T. Abdeljawad / Journal of Computational and Applied Mathematics 279 (2015) 57–66 61

Proof. By setting u = t + ϵ(t − a)1−α in the definition and using continuity of g we see that
f (g (u)) − f (g (t )) 1−α
Tαa h(t ) = = lim t
u→t (u − t )
f (g (u)) − f (g (t )) g (u) − g (t ) 1−α
= lim · lim t
u→t (g (u) − g (t )) u→t u−t
f (g (u)) − f (g (t ))
= lim · g (t )1−α · Tαa g (t ) · g (t )α−1
g (u)→g (t ) (g (u) − g (t ))
= (Tαa f )(g (t )) · (Tαa g )(t ) · g (t )α−1 .  (23)

Proposition 2.12. Let f : [a, ∞) → ∞ be twice differentiable on (a, ∞) and 0 < α, β ≤ 1 such that 1 < α + β ≤ 2. Then

(Tαa Tβa f )(t ) = Tα+β

a
f (t ) + (1 − β)(t − a)−β Tαa f (t ). (24)

Proof. By the fractional product rule and that f is twice differentiable we have
d
(Tαa Tβa f )(t ) = t 1−α [t 1−β (t − a)−β f ′ (t )]
dt
= t 1−α [t 1−β f ′′ (t ) + (1 − β)(t − a)−β f ′ (t )]
a
= Tα+β f (t ) + (1 − β)(t − a)−β Tαa f (t ).  (25)

Note that in (24) if we let α, β → 1 then we have Tαa Tβ f (t ) = T2 f (t ) = f ′′ (t ).

Next we prove a fractional version of Gronwall inequality which will be useful is studying stability of (conformable)
fractional systems.

Theorem 2.13. Let r be a continuous, nonnegative function on an interval J = [a, b] and δ and k be nonnegative constants such
that
 t
r (t ) ≤ δ + kr (s)(s − a)α−1 ds (t ∈ J ).
a

Then for all t ∈ J

(t −a)α
r (t ) ≤ δ ek α .

kr (s)(s − a)α−1 ds = δ + Iαa (kr (s))(t ). Then R(a) = δ and R(t ) ≥ r (t ), and
t
Proof. Define R(t ) = δ + a

Tαa R(t ) − kR(t ) = kr (t ) − kR(t ) ≤ kr (t ) − kr (t ) = 0. (26)

(t −a)α
Multiply (26) by K (t ) = By the help of chain rule in Theorem 2.11 we see that Tαa K (t ) = −kK (t ) and hence by the
e −k α .
product rule we conclude that Tα (K (t )R(t )) ≤ 0. Since K (t )R(t ) is differentiable on (a, b) then Lemma 2.8 implies that
a

Iαa Tαa (K (t )R(t )) = K (t )R(t ) − K (a)R(a) = K (t )R(t ) − δ ≤ 0. (27)

Hence
δ (t −a)α
r (t ) ≤ R(t ) ≤ = δ ek α (28)
K (t )
which completes the proof. 

Finally, in this section we discuss the conformable fractional derivative at a in the left case and at b in the right case for
some smooth functions. Let n − 1 < α < n and assume f : [a, ∞) → R be such that f (n) (t ) exists and continuous. Then,
(n−1)
(Taα f )(t ) = (Tα+
a
1 −n f )(t ) = (t − a)n−α f (n) (t ) and thus (Taα f )(a) = limt →a+ (t − a)n−α f (n) (t ) = 0. Similarly, in the right
case we have ( Tα f )(b) = limt →b− (b − t )n−α f (n) (t ) = 0, for (−∞, b] → R with f (n) (t ) exists and continuous. Now, let
b

0 < α < 1 and n ∈ {1, 2, 3, . . .} then the left (right) sequential conformable fractional derivative of order n is defined by
(n) a
Tα f (t ) = Tαa Tαa . . . Tαa f (t ) (29)
  
n−times
62 T. Abdeljawad / Journal of Computational and Applied Mathematics 279 (2015) 57–66

and
b
Tα(n) f (t ) = b Tα b Tα . . . b Tα f (t ), (30)
  
n−times

respectively. If f : [a, ∞) → R is second continuously differentiable and 0 < α ≤ 1

2
then direct calculations show that

(1 − α)(t − a)1−2α f ′ (t ) + (t − a)2−2α f ′′ (t ) if t > a,


(2) a
Tα (t ) = Tαa Tαa f (t ) =
0 if t = a.

Similarly, in the right case, for f : (−∞, b] → R is second continuously differentiable and 0 < α ≤ 1
2
then direct
calculations show that
(1 − α)(b − t )1−2α f ′ (t ) + (b − t )2−2α f ′′ (t ) if t < b,

b
Tα(2) (t ) = Tα Tα f (t ) =
b b
0 if t = b.
This shows that the second order sequential conformable fractional derivative may not be continuous even f is second
continuously differentiable for 21 < α < 1. If we proceed inductively, then we can see that if f is n-continuously continuously
differentiable and 0 < α ≤ 1n then the n-th order sequential conformable fractional derivative is continuous and vanishes
at the end points (a in the left case and b in the right case).

3. Integration by parts

Theorem 3.1. Let f , g : [a, b] → R be two functions such that fg is differentiable. Then
 b  b
f (x)Tα (g )(x)dα(x, a) = fg | −
a b
a g (x)Tαa (f )(x)dα(x, a) (31)
a a

The proof followed by Lemma 2.8 applied to fg and Theorem 2.1 (3) in [9].
The following integration by parts formula is given by means of left and right fractional integrals.

Proposition 3.2. Let f , g : [a, b] → R be functions and 0 < α ≤ 1. Then

 b  b
(Iαa f )(t )g (t )dα (b, t ) = f (t )(b Iα g )(t )dα (t , a). (32)
a a

Proof. From definition we get

 b  b  t
(Iαa f )(t )g (t )dα (t , a) = ( (x − a)α−1 f (x)dx)g (t )(b − t )α−1 dt . (33)
a a a

Interchanging the order of integrals we reach at

 b  b
(Iαa f )(t )g (t )dα (b, t ) = f (x)(b Iα g )(x)dα (x, a)
a a

which completes the proof. 

Next we employ Proposition 3.2 to prove an integration by parts formula by means of left and right fractional derivatives.

Theorem 3.3. Let f , g : [a, b] → R be differentiable functions and 0 < α ≤ 1. Then

 b  b
(Tαa f )(t )g (t )dα (t , a) = f (t )(b Tα g )(t )dα (b, t ) + f (t )g (t )|ba . (34)
a a

Proof. By Proposition 2.10 and that g is differentiable, we have

 b  b  b
(Tα f )(t )g (t )dα (t , a) =
a
(Tα f )(t ) Iα Tα g (t )dα (t , a) + g (b)
a b b
(Tαa f )(t )dα (t , a). (35)
a a a

Applying Proposition 3.2 leads to

 b  b
(Tαa f )(t )g (t )dα (t , a) = (Iαa Tαa f )(t )b Tα g (t )dα (b, t ) + g (b)(Iαa Tαa f )(a). (36)
a a

Then the proof is completed by the help of Lemma 2.8 by substituting (Iαa Tαa f )(t ) = f (t ) − f (a) using that f is differentiable
and by the help of Proposition 2.10 and that g is differentiable by substituting (b Iα b Tα g )(t ) = g (t ) − g (b). 
 T. Abdeljawad / Journal of Computational and Applied Mathematics 279 (2015) 57–66 63

Remark 3.1. Notice that if in Theorem 3.1 or Theorem 3.3 we let α → 1 then we obtain the integration by parts formula in
usual calculus, where we have to note that dα (t , a) → dt , dα (b, t ) → dt , Tαa f (t ) → f ′ (t ) and b Tα f (t ) → −f ′ (t ) as α → 1.

In Theorems 3.1 and 3.3 we needed some differentiability conditions. We next define some function spaces on which the
obtained integration by parts formulas are still valid.

Definition 3.1. For 0 < α ≤ 1 and an interval [a, b] define

Iα ([a, b]) = {f : [a, b] → R : f (x) = (Iαa ψ)(x) + f (a), for some ψ ∈ Lα (a)},
and
α
I ([a, b]) = {g : [a, b] → R : g (x) = (b Iα ϕ)(x) + g (b), for some ϕ ∈ Lα (b)},
where

Lα (a) = {ψ : [a, b] → R} : (Iαa ψ)(x) exists for all x ∈ [a, b]},

and

Lα (b) = {ϕ : [a, b] → R} : (b Iα ϕ)(x) exists for all x ∈ [a, b]}.

Lemma 3.4. Let f , g : [a, b] → R be functions and 0 < α ≤ 1. Then

(a) If f is left (g is right) α -differentiable then f ∈ Iα ([a, b]) (g ∈ α I ([a, b])).
(b) If f ∈ Iα ([a, b]) with f (x) = (Iαa ψ)(x) + f (a) where ψ is continuous then ψ(x) = Tαa f (x) and (Iαa Tαa f )(x) = f (x) − f (a).
(c) If g ∈ α I ([a, b]) with g (x) = (b Iα ϕ)(x)+ g (b) where ϕ is continuous then ϕ(x) = b Tα g (x) and (b Iα b Tα g )(x) = g (x)− g (b).

Proof. The proof of (a) follows by Lemma 2.8 and Proposition 2.10 by choosing ψ(t ) = Tαa f and ϕ(t ) = b Tα g. The proof
of (b) follows by Lemma 2.1 and the fact that the left α -derivative of constant function is zero. The proof of (c) follows by
Lemma 2.2 and the fact that the right α -derivative of constant function is zero. 

Theorem 3.5. Let f , g : [a, b] → R be functions such that f ∈ Iα ([a, b]) with ψ(t ) is continuous and g ∈ α I ([a, b]) with ϕ(t )
is continuous and 0 < α ≤ 1. Then
 b  b
(Tα f )(t )g (t )dα (t , a) =
a
f (t )(b Tα g )(t )dα (b, t ) + f (t )g (t )|ba . (37)
a a

Proof. The proof is similar to that in Theorem 3.3 where we make use of (b) and (c) in Lemma 3.4. 

4. Fractional power series expansions

Certain functions, being not infinitely differentiable at some point, do not have Taylor power series expansion there.
In this section we set the fractional power series expansions so that those functions will have fractional power series
expansions.

Theorem 4.1. Assume f is an infinitely α -differentiable function, for some 0 < α ≤ 1 at a neighborhood of a point t0 . Then f
has the fractional power series expansion:

(Tα0 f )(k) (t0 )(t − t0 )kα

∞ t

t0 < t < t0 + R1/α , R > 0.


f (t ) = , (38)
k=0
α k k!

Here (Tα0 f )(k) (t0 ) means the application of the fractional derivative k times.
t

Proof. Assume f (t ) = c0 + c1 (t − t0 )α + c2 (t − t0 )2α + c3 (t − t0 )3α + · · · , t0 < t < t0 + R1/α , R > 0. 

t0
t t (Tα f )(t0 )
Then, f (t0 ) = c0 . Apply Tα0 to f and evaluate at t0 we see that (Tα0 f )(t0 ) = c1 α and hence c1 = α
. Proceeding
inductively and applying Tα0 to f n-times and evaluating at t0 we see that (Tα0 f )(n) (t0 ) = cn .α(2α) · · · (nα) = α n .n! and
t t

hence

(Tα0 f )(n) (t0 )

t

cn = . (39)
α n .n!
Hence (38) is obtained and the proof is completed.
64 T. Abdeljawad / Journal of Computational and Applied Mathematics 279 (2015) 57–66

Proposition 4.2 (Fractional Taylor Inequality). Assume f is an infinitely α -differentiable function, for some 0 < α ≤ 1 at a
neighborhood of a point t0 has the Taylor power series representation (38) such that |(Tαa f )n+1 | ≤ M , M > 0 for some n ∈ N.
Then, for all (t0 , t0 + R)

M
|Rαn (t )| ≤ (t − t0 )α(n+1) , (40)
α n+1 (n + 1)!
t t
(Tα0 f )(k) (t0 )(t −t0 )kα (Tα0 f )(k) (t0 )(t −t0 )kα
where Rαn (t ) = = f (x) − nk=0
∞ 
k=n+1 α k k! α k k!
.
t
The proof is similar to that in usual calculus by applying Iα0 instead of integration.
(t −t0 )α
Example 4.1. Consider the fractional exponential function f (t ) = e α , where 0 < α < 1. The function f (t ) is clearly
not differentiable at t0 and thus it does not have Taylor power series representation about t0 . However, (Tα0 f )(n) (t0 ) = 1 for
t

all n and hence

∞
 (t − t0 )kα
f (t ) = . (41)
k =0
α k k!
The ratio test shows that this series is convergent to f on the interval [t0 , ∞).
(t −t0 )α (t −t0 )α
Example 4.2. The functions g (t ) = sin α
and h(t ) = sin α
do not have Taylor power series expansions about t =
(t −t0 )α α
= cos (t −αt0 )
t
t0 for 0 < α < 1 since they are not differentiable there. However, by the help of Eq. (38) and that Tα0 sin α
(t −t0 )α α
= − sin (t −αt0 ) we can see that
t
and Tα0 cos α

(t − t0 )α ∞
(t − t0 )(2k+1)α
sin = (−1)k (2k+1) , t ∈ [t0 , ∞) (42)
α k=0
α (2k + 1)!
and

(t − t0 )α ∞
(t − t0 )(2k)α
cos = (−1)k (2k) , t ∈ [t0 , ∞). (43)
α k=0
α (2k)!

Example 4.3. The function f (x) = 1

α does not have Taylor power series representation about t = 0 for 0 < α < 1, since
1− tα
it is not differentiable there. However, by the help of Eq. (38) we can see that
∞
1
t αk ,

tα
= t ∈ [0, 1) (44)
1− α k=0

or more generally,
∞
1
(t − t0 )αk ,

(t −t0 )α
= t ∈ [t0 , t0 + 1). (45)
1− α k=0

Remark 4.1. In case the function f is defined on (−∞, a) and not differentiable at a then we search for its (conformal) right
fractional order derivatives a Tα at a for some 0 < α ≤ 1 and use it for our fractional Taylor series on some (a − R, a), R > 0.
(a−t )α (a−t )α
For example the functions α , sin α and so on.

5. The fractional Laplace transform

In this section we will define the fractional Laplace transform and use it to solve some linear fractional equations to
produce the fractional exponential function. Then, we use the method of successive approximation to verify the solution by
making use of the fractional power series representation discussed in the above section. Also we shall calculate the Laplace
transform for certain (fractional) type functions.

Definition 5.1. Let t0 ∈ R , 0 < α ≤ 1 and f : [t0 , ∞) → be real valued function. Then the fractional Laplace transform of
order α starting from a of f is defined by
∞ (t −t0 )α ∞ (t −t0 )α
 
Ltα0 {f (t )}(s) = Fαt0 (s) = e −s α f (t )dα(t , t0 ) = e −s α f (t )(t − t0 )α−1 dt . (46)
t0 t0
 T. Abdeljawad / Journal of Computational and Applied Mathematics 279 (2015) 57–66 65

Theorem 5.1. Let a ∈ R , 0 < α ≤ 1 and f : (a, ∞) → be differentiable real valued function. Then

Laα {Tα (f )(t )}(s) = sFα (s) − f (a). (47)

Proof. The proof followed by definition, Theorem 2.1(6) in [9] and the usual integration by parts. 

Example 5.2. Consider the conformable fractional initial value problem:

(Tαa y)(t ) = λy(t ), y(a) = y0 , t > a, (48)

where the solution is assumed to be differentiable on (a, ∞).
Apply the operator Iαa to the above equation to obtain

y(t ) = y0 + λ(Iαa y)(t ). (49)

Then
yn+1 = y0 + λ(Iαa yn )(t ), n = 0, 1, 2, . . . . (50)
For n = 0 we see that
(t − a)α (t − a)α
 
y1 = y0 + λy0 = y0 1 + λ . (51)
α α
For n = 1 we see that

(t − a)α (t − a)2α
 
y2 = y0 1 + λ + λ2 . (52)
α α(2α)
If we proceed inductively we conclude that
n
 λk (t − a)kα
yn = y0 . (53)
k=0
α k k!
Letting n → ∞ we see that
∞
 λk (t − a)kα
y(t ) = y0 . (54)
k=0
α k k!
(t −a)α
Which is clearly the fractional Taylor power series representation of the (fractional) exponential function y0 eλ α .
The following lemma relates the fractional Laplace transform to the usual Laplace transform.
t t
Lemma 5.2. Let f : [t0 , ∞) → R be a function such that Lα0 {f (t )}(s) = Fα0 (s) exists. Then

Fαt0 (s) = L{f (t0 + (α t )1/α )}(s), (55)

∞
where L{g (t )}(s) = 0
e−st g (t )dt .
(t −t0 )α
The proof follows easily by setting u = α
.

Example 5.3. In this example we calculate the fractional Laplace for certain functions.
t
• Lα0 {1}(s) = 1s , s > 0
Γ (1+ α1 )
• Lα0 {t }(s) = L{t0 + (α t )1/α }(s) = + α 1/α
t t0
s 1 , s > 0.
s1+ α
α p/α
• L0α {t p }(s) = s1+p/α
Γ (1 + α1 ), s > 0.
tα
• L0α {e α }(s) = , s > 1.
1
s −1
α
• L0α {sin ω tα }(s) = L{sin ωt }(s) = 1
ω 2 +s 2
.
tα
• Lα {cos ω α }(s) = L{cos ωt }(s) =
0 s
ω 2 +s 2
.
t0 (t −t0 )α 1 tα α
• Lα e−k α
{ f (t )}(s) = L{e−kt f (t0 + (α t ) α )}. For example L0α {e−k α sin tα }(s) = L{e−kt sin t }(s) = (s+k1)2 +1 and
(t −t0 )α
Lα0 eλ α } = L{eλt } =
t 1
{ s−λ
.
66 T. Abdeljawad / Journal of Computational and Applied Mathematics 279 (2015) 57–66

Notice that in the above example there are some functions, with 0 < α < 1, whose usual Laplace is not easy to be
calculated. However, their fractional Laplace can be easily calculated.

Example 5.4. We use the fractional Laplace transform to verify the solution of the conformable fractional initial value
problem:

(Tαa y)(t ) = λy(t ), y(a) = y0 , t > a, (56)

where the solution is assumed to be differentiable on (a, ∞).
Apply Laα and use (47) to conclude that
y0
Laα {y(t )}(s) = , (57)
s−λ
(t −a)α
and hence, y(t ) = y0 eλ α .
Finally, we use the fractional fundamental exponential matrix to express the solution of (conformable) fractional linear
systems.
Consider the system

Tαa y(t ) = Ay(t ) + f(t ), 0 < α ≤ 1, (58)

where y, f : [a, b) → R are vector functions and A is an n × n matrix. The general solution of the fractional non-
n

homogeneous system (58) is expressed by

t
(t −a)α (t −a)α (s−a)α

y(t ) = eA α c+ eA α e −A α f(s)(s − a)1−α ds, (59)
a
(t −a)α ∞ Ak (t −a)kα
where eA α = k=0 α k k!
and c is a constant vector.

6. Some conclusions and comparisons

1. The conformable fractional derivative behaves well in the product rule and chain rule while complicated formulas appear
in case of usual fractional calculus.
2. The conformable fractional derivative of a constant function is zero while it is not the case for Riemann fractional
derivatives.
3. Mittag-Leffler functions play important rule in fractional calculus as a generalization to exponential functions while the
tα
fractional exponential function f (t ) = e α appears in case of conformable fractional calculus.
4. Conformable fractional derivatives, conformable chain rule, conformable integration by parts, conformable Gronwall’s
inequality, conformable exponential function, conformable Laplace transform and so forth, all tend to the corresponding
ones in usual calculus.
5. In case of usual calculus there some functions that do not have Taylor power series representations about certain points
but in the theory of conformable fractional they do have.
6. Open problem: Is it hard to fractionalize the conformable fractional calculus, either by iterating the conformable fractional
derivative (Grünwald–Letnikov approach) or by iterating the conformable fractional integral of order 0 < α ≤ 1
(Riemann approach)? Notice that when α = 0 we obtain Hadamard type fractional integrals.

References

[1] G. Samko, A.A. Kilbas, Marichev, Fractional Integrals and Derivatives: Theory and Applications, Gordon and Breach, Yverdon, 1993.
[2] I. Podlubny, Fractional Differential Equations, Academic Press, San Diego CA, 1999.
[3] A. Kilbas, M.H. Srivastava, J.J. Trujillo, Theory and Application of Fractional Differential Equations, in: North Holland Mathematics Studies, vol. 204,
2006.
[4] H.L. Gray, N.F. Zhang, On a new definition of the fractional difference, Math. Comp. 50 (182) (1988) 513–529.
[5] K.S. Miller, B. Ross, Fractional difference calculus, in: Proceedings of the International Symposium on Univalent Functions, Fractional Calculus and Their
Applications, Nihon University, Koriyama, Japan, (1989), 139–152.
[6] F.M. Atıcı, P.W. Eloe, Initial value problems in discrete fractional calculus, Proc. Amer. Math. Soc. 137 (2009) 981–989.
[7] T. Abdeljawad, F. Atıcı, On the definitions of nabla fractional operators, Abstr. Appl. Anal. 2012 (2012) 13. http://dx.doi.org/10.1155/2012/406757.
Article ID 406757.
[8] T. Abdeljawad, Dual identities in fractional difference calculus within Riemann, Adv. Difference Equ. 2013 (2013) 36. arXiv:1112.5795
http://dx.doi.org/10.1186/1687-1847-2013-36.
[9] R. Khalil, M. Al Horani, A. Yousef, M. Sababheh, A new definition of fractional derivative, J. Comput. Appl. Math. 264 (2014) 65–70.