INTERMATHS, VOL. 4, NO. 1 (2023), 48–66
https://doi.org/10.22481/intermaths.v4i1.12294
Article
cb licença creative commons
On compact explicit formulas of the partial
fraction decomposition and applications
Leandro Bezerra de Lima
a,∗
and Rachidi Mustapha
a
a
Federal University of Mato Grosso do Sul (UFMS), Campo Grande - MS, Brazil
* Correspondence: leandro.lima@ufms.br
Abstract: This study concerns another approach for computing the scalars
A
(k)
i
of the partial
fraction decomposition
F
(
x
) =
R(x)
Q(x)
=
R(x)
Q
s
i=1
(x − γ
i
)
m
i
=
s
P
i=1
m
i
P
k=1
A
(k)
i
(x−γ
i
)
k
,
where
R
(
x
) and
Q
(
x
) are
polynomials of real or complex coefficients, with
deg
(
R
)
< deg
(
Q
). More precisely, we provide a method
to exhibit compact explicit formulas of the scalars
A
(k)
i
(1
≤ i ≤ s,
1
≤ k ≤ m
i
). Some illustrative
special cases and several examples are furnished, to show the efficiency of this new approach. Finally,
concluding remarks and perspectives are presented.
Keywords: Partial fraction decomposition, Derivation of higher order, Explicit formula.
Classification MSC: Primary 26A33; 30C15 Secondary 26C99
1 Introduction
Since a long time, the partial fraction decomposition (or expansion) represents a
fundamental topic, with several applications in various fields of pure and applied
mathematics such that differential equations, control theory, and other fields of applied
sciences and engineering. In addition, the partial fraction decomposition is also an
important tool in the teaching areas such that Calculus, Laplace transform, etc. It is
worth noting that the literature is very vast on this very rich topic.
A well known theoretical theorem in algebra asserts that every rational function has
a unique partial fraction decomposition (see, for example, [
3
,
4
,
10
,
11
]). Moreover,
several methods have been improved in the literature for computing the partial fraction
decomposition. Generally, there are two usual approaches for establishing the numerators
of the partial fractions, namely, for computing the scalars
A
(k)
i
. The technique of the first
approach, known as the "Method of undetermined coefficients", consists first in reducing
Submitted 26 March 2023; Accepted 10 June 2023; Available online 30 June 2023.
ISSN 2675-8318 Copyright ©2023 INTERMATHS. Published by Edições UESB. This is an Open Access article under the CC BY 4.0 license.
to the same denominator the partial fraction decomposition, of the given rational
function. Second, since the denominators on both sides are the same, the numerators
must also be the same. Then, we equalize the similar coefficients (corresponding to
the same power of
x
) of the two polynomials of the numerators on either side of the
equality
i
. Therefore, the scalars
A
(k)
i
can be found by solving a system of linear equations.
The second approach is based on the application of the Heaviside’s "cover-up method",
which necessitate substitutions to establish the scalars
A
(k)
i
, of the partial fraction
decomposition, in the case with single poles
γ
j
(1
≤ j ≤ deg
(
Q
)). For multiple poles
case
γ
j
(1
≤ j ≤ s
), with
m
j
≥
2 for some
j
, successive differentiation are applied, for
calculating the scalars
A
(k)
i
. Despite that, this topic continue to attract much attention,
and there has been recent developments in the computation aspect of the scalars
A
(k)
i
,
for general rational functions (see for example, [3, 4, 6, 15]) as well as for some special
cases (see, for example, [
5
,
6
,
12
–
14
]). Meanwhile, the approaches and methods for
decomposing a rational function into partial fractions are computationally intensive,
especially when the multiplicities of roots of the denominator are higher.
In this paper we establish another approach for providing the explicit formulas for the
scalars
A
(k)
i
of the partial fraction decomposition of the rational functions
F
(
x
) =
R(x)
Q(x)
,
where
R
(
x
),
Q
(
x
) are polynomials in
R
[
X
] or
C
[
X
], such that (without loss of generality)
the degree of
R
is less than the degree of
Q
and are mutually prime. The essence of our
approach requires a computational process, based on two known results of the literature.
More precisely, suppose that
R
and
Q
are mutually prime and
Q
(
x
) =
Q
s
j=1
(
x − γ
j
)
m
j
,
where each root
γ
j
is of multiplicity
m
j
≥
1. We develop a computational process,
which allows us to present a new method, for exhibiting compact explicit formulas of
the partial fraction decomposition,
F (x) =
R(x)
Q(x)
=
R(x)
Q
s
i=1
(x − γ
i
)
m
i
=
s
X
i=1
m
i
X
k=1
A
(k)
i
(x − γ
i
)
k
.
Our main goal is to give a new compact explicit formulas for the scalars
A
(k)
i
(1
≤ i ≤ s
,
1
≤ k ≤ m
j
). As a consequence, some applications and several illustrative examples are
presented, in order to show the efficiency of our approach.
This study is organized as follows. For reason of clarity and conciseness, Section 2
is devoted to the two fundamental results, representing the basic tools of our method.
Section 3 is concerned with the generic case, where a compact explicit formula of the
i
It is known that two polynomials are equal, if and only if, the coefficients at the corresponding powers of x are equal.
L. B. Lima; R. Mustapha INTERMATHS, 4(1), 48–66, June 2023 | 49
partial fraction decomposition is given for the special case
F
(
x
) =
1
Q(x)
. In Section 4
we study the general case of partial fraction decomposition, where the explicit formulas
of the partial fraction decomposition of the generic case plays a central key. In addition,
two special cases are provided. Results of Sections 3 and 4 are illustrated by significant
examples, in order to show the efficiency of our approach. Finally, in Section 5 we give
some concluding remarks and perspective.
For reason of clarity, in the sequel we suppose that for the rational fraction
F
(
x
) =
R(x)
Q(x)
, the two polynomials are relatively prime, namely, the great common divisor of
R(x) and q(x) is equal to 1.
2 Two fundamental results
In this section we consider the rational fraction
F
(
x
) =
R(x)
Q(x)
, where
Q
(
x
) =
x
r
+
a
0
x
r−1
+
· · ·
+
a
r−1
, and without loss of generality we suppose
deg
(
R
)
< deg
(
Q
). Let
γ
i
(1
≤ i ≤ s
) be the distinct roots (real or complex) of the polynomial
Q
(
x
), with
multiplicities m
i
(1 ≤ i ≤ s), respectively. Then, we have Q(x) =
Q
s
i=1
(x − γ
i
)
m
i
.
In sequel of our study, the next known result, will play a central role for computing
the scalars A
(k)
i
.
Theorem 2.1. Let
F
(
x
) =
R(x)
Q(x)
be a rational function such that
Q
(
x
) =
Q
s
i=1
(
x − γ
i
)
m
i
and
deg
(
R
)
< deg
(
Q
) =
m
1
+
...
+
m
s
, where the
γ
i
(1
≤ i ≤ s
), with
γ
i
̸
=
γ
j
for
i ̸
=
j
,
are real or complex numbers. Then, the partial fractional decomposition of
F
(
x
), is given
by,
F (x) =
R(x)
Q
s
i=1
(x − γ
i
)
m
i
=
s
X
i=1
m
i
X
k=1
A
(k)
i
(x − γ
i
)
k
,
where
A
(k)
i
=
1
(m
i
− k)!
"
R(x)
Q
s
j=1,j̸=i
(x − γ
j
)
m
j
#
(m
i
−k)
x=γ
i
.
Here [
f
(
x
)]
(n)
x=γ
=
d
n
f(x)
dx
n
|
x=γ
=
f
(n)
(
γ
), which means the value of the derivation of order
n of the function f at x = γ.
Result of Theorem 2.1 is well known in the literature and it has been demonstrated
by various algebraic and analytical methods (see, for example, [3, 4, 15]).
The second classical result, which will allow us to reach our goal, concerns a gen-
eralization, relating to the derivation of order
d ≥
2 of a product of differentiable
50 | https://doi.org/10.22481/intermaths.v4i1.12294 L. B. Lima; R. Mustapha
functions.
Theorem 2.2. Let
f
1
,
f
2
,...,
f
s
be derivable functions, until the order
d
. Then, we
have,
s
Y
j=1
f
j
(d)
=
X
h
1
+...+h
s
=d
d
h
1
...h
s
!
s
Y
j=1
f
(h
j
)
j
, (1)
where f
(k)
means the derivation de order k of the function f and
d
h
1
...h
s
!
=
d!
h
1
! ... h
s
!
.
Theorem 2.2 is not common in books of Calculus or real analysis. In fact, it represents
a generalization of the well- known formula of the Calculus (
fg
)
′
=
f
′
g
+
fg
′
. A known
generalization, of this former expression, has been established for
s
= 2 and
d ≥
2, and
it is given by,
(fg)
(d)
=
X
h
1
+h
2
=d
d
h
1
h
2
!
f
(h
1
)
g
(h
2
)
=
d
X
h=0
d
h
!
f
(h)
g
(d−h)
. (2)
The former Formula Eq. (2) can be established by induction, analogously to that which
makes it possible to establish the Newton’s binomial formula, namely, (
a
+
b
)
n
=
P
n
k=0
n
k
a
k
b
n−k
. Similarly, Formula Eq. (1) can be also established by induction, by
adapting the proof of the classical formula
[a
1
+ a
2
+ . . . a
s
]
(d)
=
X
h
1
+...+h
s
=d
d
h
1
...h
s
!
s
Y
j=1
a
(h
j
)
j
.
Theorem 2.2 and Formula Eq. (2), have been studied in [
1
,
8
]. They are considered
as generalization of Leibniz’s rule for differentiation.
3 Partial fraction decomposition: Generic case and special cases
A Study of the generic case In this section we consider a monic (or unitary) polynomial
Q
(
x
) =
x
r
+
a
0
x
r−1
+
· · ·
+
a
r−1
. Let
γ
i
(1
≤ i ≤ s
) be the distinct roots of
Q
(
x
), with
multiplicities
m
i
(1
≤ i ≤ s
), respectively. Then, we have
Q
(
x
) =
s
Y
i=1
(
x − γ
i
)
m
i
.
For the
partial fractional decomposition of
F
(
x
) =
1
Q(x)
, a direct application of Theorem 2.1
(with R(x) = 1) shows easily that we have,
F (x) =
1
Q
s
i=1
(x − γ
i
)
m
i
=
s
X
i=1
m
i
X
k=1
A
(k)
i
(x − γ
i
)
k
. (3)
L. B. Lima; R. Mustapha INTERMATHS, 4(1), 48–66, June 2023 | 51
with
A
(k)
i
=
1
(m
i
− k)!
"
1
Q
s
j=1,j̸=i
(x − γ
j
)
m
j
#
(m
i
−k)
x=γ
i
. (4)
In the aim to obtain the compact explicit expression of the scalars
A
(k)
i
, let apply
Theorem 2.2 by taking
f
j
(
x
) =
1
(x − γ
j
)
m
j
= (
x − γ
j
)
−m
j
. More precisely, we apply the
preceding Formula Eq. (4) to the family of functions H
i
(x) (1 ≤ i ≤ s) defined by,
H
i
(x) =
s
Y
j=1,j̸=i
f
j
(x) =
s
Y
j=1,j̸=i
(x − γ
j
)
−m
j
, for 1 ≤ i ≤ s.
For reason of clarity and simplicity, Γ
i,d
i
(k)
=
{
[
h
j
]
i
= (
h
1
, ..., h
i−1
, h
i+1
, ..., h
s
);
P
1≤j̸=i≤s
h
j
=
d
i
(
k
)
}
and we set
d
i
(
k
) =
m
i
− k
. Then, the main formula of Theorem 2.2, allows us to
derive that,
[H
i
(x)]
(d
i
(k))
x=γ
i
=
X
[h
j
]
i
∈Γ
i,k
d
i
(k)
h
1
..., h
i−1
, h
i+1
, ...h
s
!
s
Y
j=1,j̸=i
f
(h
j
)
j
(γ
i
), (5)
where
d
i
(k)
h
1
...
ˆ
h
i
...h
s
!
=
d
i
(k)!
h
1
!...h
i−1
!h
i+1
!...h
s
!
. It is well known that the derivative
f
(h
j
)
j
(
x
) =
[(
x − γ
j
)
−m
j
]
(h
j
)
is given explicitly by
f
(h
j
)
j
(
x
) = (
−
1)
h
j
(h
j
+ m
j
− 1)!
(m
j
− 1)!
(
x − γ
j
)
−m
j
−h
j
.
Thus, we derive that
f
(h
j
)
j
(
γ
i
) = (
−
1)
h
j
(h
j
+ m
j
− 1)!
(m
j
− 1)!
(
γ
i
− γ
j
)
−m
j
−h
j
.
By substitution of
this former formula of
f
(h
j
)
j
(
x
) in Expression Eq. (5) of [
H
i
(
x
)]
(d
i
(k))
x=γ
i
, a straightforward
computation allows us to deduce that
[H
i
(x)]
(d
i
(k))
x=γ
i
=
X
[h
j
]
i
∈Γ
i,k
d
i
(k)
h
1
..., h
i−1
, h
i+1
, ...h
s
!
s
Y
j=1,j̸=i
f
(h
j
)
j
(γ
i
).
Therefore, we get
[H
i
(x)]
(d
i
(k))
x=γ
i
=
X
[h
j
]
i
∈Γ
i,k
d
i
(k)
h
1
..., h
i−1
, h
i+1
, ...h
s
!
s
Y
j=1,j̸=i
(−1)
h
j
(h
j
+ m
j
− 1)!
(m
j
− 1)!
(γ
i
−γ
j
)
−m
j
−h
j
.
And a direct computation shows that, we have,
[H
i
(x)]
(d
i
(k))
x=γ
i
=
X
Γ
i,d
i
(k)
(−1)
m
i
−k
(m
i
− k)!
s
Y
j=1,j̸=i
h
j
+ m
j
− 1
h
j
!
(γ
i
− γ
j
)
−m
j
−h
j
,
52 | https://doi.org/10.22481/intermaths.v4i1.12294 L. B. Lima; R. Mustapha
or equivalently,
[H
i
(x)]
(d
i
(k))
x=γ
i
= (−1)
m
i
−k
(m
i
− k)!
X
[h
j
]
i
∈Γ
i,d
i
(k)
Ω
i
([h
j
]
i
; [γ
j
]), (6)
where [h
j
]
i
= (h
1
, ..., h
i−1
, h
i+1
, ..., h
s
), [γ
j
] = (γ
1
, ..., γ
s
) and
Ω
i
([h
j
]
i
; [γ
j
]) =
s
Y
j=1,j̸=i
"
h
j
+ m
j
− 1
h
j
!
(γ
i
− γ
j
)
−m
j
−h
j
#
. (7)
Consequently, first, we observe that Expressions Eq. (3)-Eq. (4), Theorem 2.1 and
Theorem 2.2, imply that the partial fraction decomposition of the function
F
(
x
) =
Q
(
x
)
−1
is given by,
1
Q(x)
=
s
X
i=1
m
i
X
k=1
A
(k)
i
(x − γ
i
)
k
=
s
X
i=1
m
i
X
k=1
1
(m
i
− k)!
[H
i
(x)]
(d
i
(k))
x=γ
i
(x − γ
i
)
k
.
Second, taking into account Expressions Eq. (6)-Eq. (7), we get
A
(k)
i
=
1
(m
i
−k)!
[
H
i
(
x
)]
(d
i
(k))
x=γ
i
=
1
(m
i
−k)!
× (−1)
m
i
−k
(m
i
− k)!
P
[h
j
]
i
∈Γ
i,d
i
(k)
Ω
i
([h
j
]
i
; [γ
j
]), which implies that we have,
A
(k)
i
=
1
(m
i
− k)!
[H
i
(x)]
(d
i
(k))
x=γ
i
= (−1)
m
i
−k
X
[h
j
]
i
∈Γ
i,d
i
(k)
Ω
i
([h
j
]
i
; [γ
j
]),
where the scalars
Ω
i
([
h
j
]
i
; [
γ
j
]) are as in Expression Eq. (7). In summary, the previous
discussion shows that the main result of this section, can be formulated as follows.
Theorem 3.1. Under the preceding data the partial fraction decomposition of the rational
function F (x) =
1
Q(x)
, where Q(x) =
s
Y
i=1
(x − γ
i
)
m
i
, is given by,
F (x) =
1
Q(x)
=
s
X
i=1
m
i
X
k=1
(−1)
m
i
−k
X
[h
j
]
i
∈Γ
i,d
i
(k)
Ω
i
([h
j
]
i
; [γ
j
])
1
(x − γ
i
)
k
, (8)
where Γ
i,d
i
(k)
=
{
[
h
j
]
i
= (
h
1
, ..., h
i−1
, h
i+1
, ..., h
s
);
P
1≤j̸=i≤s
h
j
=
m
i
− k}
and the
Ω
i
([
h
j
]
i
; [
γ
j
]) are yielded by Expression Eq. (7), namely, we have
F
(
x
) =
1
Q(x)
=
s
X
i=1
m
i
X
k=1
A
(k)
i
(x − γ
i
)
k
, where
A
(k)
i
= (−1)
m
i
−k
X
[h
j
]
i
∈Γ
i,d
i
(k)
Ω
i
([h
j
]
i
; [γ
j
]). (9)
L. B. Lima; R. Mustapha INTERMATHS, 4(1), 48–66, June 2023 | 53
In Section 4, Expressions Eq. (8)-Eq. (9) of Theorem 3.1, will be utilized for studying
the general case of the partial fraction decomposition of a rational function
F
(
x
) =
R(x)
Q(x)
,
where
Q
(
x
) =
Q
s
i=1
(
x − γ
i
)
m
i
and
R
(
x
) is a polynomial, such that (without loss of
generality)
deg
(
R
)
< deg
(
Q
) =
m
1
+
m
2
+
. . .
+
m
s
. That is, the results of this
subsection combined with Theorem 2.2, will be extensively used for providing explicit
compact formulas, for the scalars
A
(k)
i
, of the partial fraction decomposition in the
general setting.
Finally, it is important to note that a formula analogous to Eq. (8), has been highlighted
in [2], using another approach.
B Special cases and illustrative examples This subsection is devoted to illustrate
the efficiency of the result of the main Theorem 3.1. We study with more details the
special case
s
= 2, along with various numerical examples. To this aim, we start by the
following proposition.
Proposition 3.2. Consider the polynomial
Q
(
x
) = (
x − γ
1
)
m
1
(
x − γ
2
)
m
2
, where
γ
1
,
γ
2
are real or complex numbers and
m
1
,
m
2
are positive integer. Then, the partial fraction
decomposition of the rational function F (x) =
1
Q(x)
, is given by,
F (x) =
1
Q(x)
=
m
1
X
k=1
A
(k)
1
(x − γ
1
)
k
+
m
2
X
k=1
A
(k)
2
(x − γ
2
)
k
, (10)
where
A
(k)
1
= (−1)
m
1
−k
Ω
2,k
(γ
1
, γ
2
) = (−1)
m
1
−k
m
2
+m
1
−k−1
m
2
−k
(γ
2
− γ
1
)
−(m
1
+m
2
)+k
,
A
(k)
2
= (−1)
m
2
−k
Ω
2,k
(γ
1
, γ
2
) = (−1)
m
2
−k
m
1
+m
2
−k−1
m
2
−k
(γ
2
− γ
1
)
−(m
1
+m
2
)+k
.
(11)
Proof. We apply result of Theorem 3.2. We have
A
(k)
1
= (
−
1)
m
1
−k
P
Γ
1,k
Ω
1,k
(
γ
1
, γ
2
),
where Γ
1,k
=
{
(
h
2
);
h
2
=
m
1
−k}
=
{m
1
−k}
, which implies that, we have
Ω
1,k
(
γ
1
, γ
2
) =
h
2
+ m
2
− 1
h
2
!
(γ
1
− γ
2
)
−h
2
−m
2
. Therefore, we obtain,
Ω
1,k
(γ
1
, γ
2
) =
m
1
+ m
2
− (k + 1)
m
1
− k
!
(γ
1
− γ
2
)
−(m
1
+m
2
)+k
.
Similarly, since
A
(k)
2
= (
−
1)
m
2
−k
P
Γ
2,k
Ω
2,k
(
γ
1
, γ
2
) and Γ
2,k
=
{
(
h
1
);
h
1
=
m
2
− k}
=
{m
2
− k}
, we derive
Ω
2,k
(
γ
1
, γ
2
) =
h
1
+ m
1
− 1
h
1
!
(
γ
2
− γ
1
)
−h
1
−m
1
, which implies that
54 | https://doi.org/10.22481/intermaths.v4i1.12294 L. B. Lima; R. Mustapha
we have,
Ω
2,k
(γ
1
, γ
2
) =
m
1
+ m
2
− (k + 1)
m
2
− k
!
(γ
2
− γ
1
)
−(m
1
+m
2
)+k
.
Therefore, we obtain the partial fraction decomposition Eq. (10), with
A
(k)
1
and
A
(k)
2
given by Eq. (11). □
Suppose that
m
1
=
m
2
= 1 and consider the function
F
(
x
) =
1
(x − γ
1
)(x − γ
2
)
,
where
γ
1
and
γ
2
are in
K
(
K
=
R
or
C
), with
γ
1
̸
=
γ
2
. Then, its partial fraction
decomposition is given by
F
(
x
) =
A
(1)
1
x − γ
1
+
A
(1)
2
x−γ
2
. Let apply our process for computing
the scalars
A
(1)
1
and
A
(1)
2
. Since
m
1
= 1 and
m
2
= 1, we can show that Γ
1,1
=
{
1
−
1
}
=
{
0
}
and Γ
2,1
=
{
1
−
1
}
=
{
0
}
. Hence, we get
Ω
1,1
(
γ
1
, γ
2
) =
1+1−1−1
1−1
(
γ
1
−
γ
2
)
−(1+1)+1
=
1
γ
1
−γ
2
and
Ω
1,2
(
γ
1
, γ
2
) =
1+1−1−1
1−1
(
γ
2
− γ
1
)
−(1+1)+1
=
1
γ
2
−γ
1
.
Therefore, we
have
A
(1)
1
= (
−
1)
1−1
P
Γ
1,1
Ω
1,1
(
γ
1
, γ
2
) =
1
γ
1
−γ
2
and
A
(2)
1
= (
−
1)
1−1
P
Γ
1,2
Ω
1,1
(
γ
1
, γ
2
) =
1
γ
2
−γ
1
. Accordingly, we state the following corollary.
Corollary 3.3. Let
γ
1
and
γ
2
be in
K
(
K
=
R
or
C
), with
γ
1
̸
=
γ
2
. Then, the partial
fraction decomposition of the rational functions F (x) =
1
(x − γ
1
)(x − γ
2
)
, is given by,
F (x) =
1
(x − γ
1
)(x − γ
2
)
=
1
γ
1
− γ
2
1
x − γ
1
+
1
γ
2
− γ
1
1
x − γ
2
. (12)
Let consider the following illustrative numerical example of Expression Eq. (12).
Example 3.4. Let
Q
(
x
) = (
x −
8)(
x −
6), and consider the rational fration
F
(
x
) =
1
(x − 8)(x − 6)
. Here we have
γ
1
= 8,
γ
2
= 6. And by applying Expression Eq. (12), we
show that the partial fraction decomposition of the rational function
F
(
x
) =
1
(x − 8)(x − 6)
,
is given by, F (x) =
1
2
.
1
x − 8
−
1
2
.
1
x − 6
.
This decomposition is consistent with the direct calculation, which shows the efficiency
of Theorem 3.1 and its Corollary 3.3.
Suppose
m
1
= 2 and
m
2
= 1 and consider the rational function
F
(
x
) =
1
(x − γ
1
)
2
(x − γ
2
)
,
where
γ
1
,
γ
2
in
K
(
K
=
R
or
C
), with
γ
1
̸
=
γ
2
. Then, its partial fraction decomposition
can be written under the form
F
(
x
) =
A
(1)
1
x − γ
1
+
A
(2)
1
(x − γ
1
)
2
+
A
(1)
2
x − γ
2
. Let apply our process
for computing the scalars
A
(1)
1
,
A
(2)
1
and
A
(1)
2
. Since
m
1
= 2 and
m
2
= 1, we can show that
Γ
1,1
=
{
1
−
1
}
=
{
0
}
, Γ
1,2
=
{
1
−
1
}
=
{
0
}
and Γ
2,1
=
{
1
−
1
}
=
{
0
}
. A direct application
of Expression Eq. (11), allows us to show that
Ω
1,1
(
γ
1
, γ
2
) =
2 + 1 − (1 + 1)
2 − 1
!
(
γ
1
−
γ
2
)
−(2+1)+1
=
1
(γ
1
− γ
2
)
2
,
Ω
1,2
(
γ
1
, γ
2
) =
2 + 1 − (2 + 1)
2 − 2
!
(
γ
1
− γ
2
)
−(2+1)+2
=
1
γ
1
− γ
2
L. B. Lima; R. Mustapha INTERMATHS, 4(1), 48–66, June 2023 | 55
and
Ω
2,1
(
γ
2
, γ
1
) =
2 + 1 − (1 + 1)
1 − 1
!
(
γ
2
− γ
1
)
−(1+2)+1
=
1
(γ
2
− γ
1
)
2
.
Therefore, have
A
(1)
1
= (
−
1)
2−1
X
Γ
1,1
Ω
1,1
(
γ
1
, γ
2
) =
−
1
(γ
1
− γ
2
)
2
,
A
(2)
1
= (
−
1)
2−2
X
Γ
1,2
Ω
1,2
(
γ
1
, γ
2
) =
1
γ
1
− γ
2
and
A
(1)
2
= (
−
1)
1−1
X
Γ
2,1
Ω
2,1
(
γ
1
, γ
2
) =
1
(γ
2
− γ
1
)
2
. Finally, we can state the following
corollary.
Corollary 3.5. Let
γ
1
and
γ
2
be in
K
(
K
=
R
or
C
), with
γ
1
̸
=
γ
2
. Then, the partial
fraction decomposition of the rational functions F (x) =
1
(x − γ
1
)
2
(x − γ
2
)
, is given by,
F (x) = −
1
(γ
1
− γ
2
)
2
.
1
x − γ
1
+
1
γ
1
− γ
2
.
1
(x − γ
1
)
2
+
1
(γ
2
− γ
1
)
2
.
1
x − γ
2
. (13)
Let consider the following illustrative numerical example of Expression Eq. (13).
Example 3.6. Consider
F
(
x
) =
1
(x − 8)
2
(x − 6)
. Since
γ
1
= 8,
γ
2
= 6,
m
1
= 2 and
m
2
= 1, a direct computation implies that we have
A
(1)
1
=
−
X
Γ
1,1
Ω
1,1
(8
,
6) =
−
1
4
,
A
(2)
1
=
X
Γ
1,1
Ω
1,2
(8
,
6) =
1
2
and
A
(1)
2
=
X
Γ
2,1
Ω
2,1
(6
,
8) =
−
1
2
.
Therefore, Expression Eq. (13) implies
that the partial fraction decomposition of the rational function
F
(
x
) =
1
(x − 8)
2
(x − 6)
, is
given by, F (x) = −
1
4
.
1
x − 8
+
1
2
.
1
(x − 8)
2
+
1
4
.
1
x−6
.
This partial fraction decomposition is consistent with the direct calculation, using
other known methods, which shows the efficiency of Theorem 3.1 and its Corollary 3.5.
Suppose that
m
1
= 2 and
m
2
= 3 and consider the function
F
(
x
) =
1
(x − γ
1
)
2
(x − γ
2
)
3
,
where
γ
1
and
γ
2
are in
K
(
K
=
R
or
C
), with
γ
1
̸
=
γ
2
, then, its partial fraction
decomposition is given by
F
(
x
) =
A
(1)
1
x−8
+
A
(2)
1
(x−8)
2
+
A
(1)
2
x−6
+
A
(2)
2
(x−6)
2
+
A
(3)
2
(x−6)
3
. Let apply our
process for computing the scalars
A
(1)
1
,
A
(2)
1
,
A
(1)
2
,
A
(2)
2
and
A
(2)
2
. Since
m
1
= 2 and
m
2
= 3,
we can show that Γ
1,1
=
{
2
−
1
}
=
{
1
}
, Γ
1,2
=
{
2
−
2
}
=
{
0
}
, Γ
2,1
=
{
3
−
1
}
=
{
2
}
,
Γ
2,2
=
{
3
−
2
}
=
{
1
}
and Γ
2,3
=
{
3
−
3
}
=
{
0
}
. An analogous straightforward
computation allows us to establish that,
Ω
1,1
(γ
1
, γ
2
) =
2+3−(1+1)
2−1
(γ
1
− γ
2
)
−(2+3)+1
=
3
(γ
1
−γ
2
)
4
,
Ω
1,2
(γ
1
, γ
2
) =
2+3−(2+1)
2−2
(γ
1
− γ
2
)
−(2+3)+2
=
1
(γ
1
−γ
2
)
3
,
Ω
2,1
(γ
2
, γ
1
) =
2+3−(1+1)
3−1
(γ
2
− γ
1
)
−(2+3)+1
= 3.(γ
2
− γ
1
)
−4
=
3
(γ
2
−γ
1
)
4
,
Ω
2,2
(γ
2
, γ
1
) =
2+3−(2+1)
3−2
(γ
2
− γ
1
)
−(2+3)+2
= 2.(γ
2
− γ
1
)
−3
=
2
(γ
2
−γ
1
)
3
,
Ω
2,3
(γ
2
, γ
1
) =
2+3−(3+1)
3−3
(γ
2
− γ
1
)
−(2+3)+3
= 1.(γ
2
− γ
1
)
−2
=
1
(γ
2
−γ
1
)
2
.
56 | https://doi.org/10.22481/intermaths.v4i1.12294 L. B. Lima; R. Mustapha
Since
A
(1)
1
= (
−
1)
2−1
X
Γ
1,1
Ω
1,1
(
γ
2
, γ
1
),
A
(2)
1
= (
−
1)
2−2
X
Γ
1,2
Ω
1,2
(
γ
2
, γ
1
),
A
(1)
2
= (
−
1)
3−1
X
Γ
2,1
Ω
2,1
(
γ
2
, γ
1
),
A
(2)
2
= (−1)
3−2
X
Γ
2,2
Ω
2,2
(γ
2
, γ
1
) and A
(3)
2
= (−1)
3−3
X
Γ
2,3
Ω
2,3
(γ
2
, γ
1
), we get the corollary.
Corollary 3.7. Let
γ
1
and
γ
2
be in
K
(
K
=
R
or
C
), with
γ
1
̸
=
γ
2
. Then, the partial
fraction decomposition of the rational functions
F
(
x
) =
1
(x − γ
1
)
2
(x − γ
2
)
3
is written
under the form,
F (x) =
A
(1)
1
x − γ
1
+
A
(2)
1
(x − γ
1
)
2
+
A
(1)
2
x − γ
2
+
A
(2)
2
(x − γ
2
)
2
+
A
(3)
2
(x − γ
2
)
3
, (14)
where
A
(1)
1
= −
3
(γ
1
−γ
2
)
4
, A
(2)
1
=
1
(γ
1
−γ
2
)
3
,
A
(1)
2
=
3
(γ
2
−γ
1
)
4
, A
(2)
2
= −
2
(γ
2
−γ
1
)
3
, A
(3)
2
=
1
(γ
2
−γ
1
)
2
.
(15)
The following numerical example is an illustrative application of Expressions Eq. (14)-
Eq. (15).
Example 3.8. Consider the rational fraction
F
(
x
) =
1
(x − 8)
2
(x − 6)
3
. We show easily
that
γ
1
= 8,
γ
2
= 6,
m
1
= 2 and
m
2
= 3. Then, application of Expression Eq. (15),
allows us to deduce that
A
(1)
1
=
−
3
16
,
A
(2)
1
=
1
8
,
A
(1)
2
=
3
16
,
A
(2)
2
=
1
4
and
A
(3)
2
=
1
4
.
Therefore, Expression Eq. (14) implies that the partial fraction decomposition of the
rational function F (x) =
1
(x − 8)
2
(x − 6)
3
is given by,
F (x) = −
3
16
.
1
x − 8
+
1
8
.
1
(x − 8)
2
+
3
16
.
x − 6
+
1
4
.
1
(x − 6)
2
+
1
4
.
1
(x − 6)
3
.
This decomposition is consistent with the direct calculation, utilizing the usual methods,
which shows the efficiency of Theorem 3.1 and its Corollary 3.7.
Note that, some special cases of polynomials
Q
(
x
) have been studied in the literature,
but without establishing a compact explicit formulas for the quantities A
(k)
i
.
The generic special case
s
= 3, will be treated in the next section, as a particular case
on the general sitting (see Subsection C).
4 General setting and special cases
A General cases Following Theorem 2.1, for every polynomial
R
(
x
) of degree
< m
1
+
. . .
+
m
s
, and
γ
i
∈ K
(1
≤ i ≤ s
), with
γ
i
̸
=
γ
j
, the partial fractional decomposition
L. B. Lima; R. Mustapha INTERMATHS, 4(1), 48–66, June 2023 | 57
of the rational fraction
F
(
x
) =
R(x)
Q
s
i=1
(x − γ
i
)
m
i
is given by
F
(
x
) =
R(x)
Q
s
i=1
(x − γ
i
)
m
i
=
s
X
i=1
m
i
X
k=1
A
(k)
i
(x − γ
i
)
k
, where
A
(k)
i
=
1
(m
i
− k)!
"
R(x)
Q
s
j=1,j̸=i
(x − γ
j
)
m
j
#
(m
i
−k)
x=γ
i
.
Application of Theorem 2.2, or more precisely applying the formula Eq. (2), to the
function F
i
(x) = R(x)H
i
(x), permits us to get,
F
(d)
i
(x) = [R × H
i
]
(d)
(x) =
X
k
1
+k
2
=d
d
k
1
k
2
!
R
(k
1
)
(x)H
(k
2
)
i
(x),
for every
d ≥
0. Since
R
(k)
= 0 for
deg
(
R
) =
p < k ≤ d
, the preceding expression takes
the form
F
(d)
i
(x) = [R × H
i
]
(d)
(x) =
min(d,p)
X
k=0
d
k
!
R
(k)
(x)H
(d−k)
i
(x).
Therefore, the first amelioration of Theorem 2.1 is given in the following preliminary
proposition.
Proposition 4.1. Let
R
(
x
) and
Q
(
x
) =
Q
s
i=1
(
x−γ
i
)
m
i
be a polynomials with coefficients
in
K
(
K
=
R
or
C
); where 0
≤ p
=
deg
(
R
)
< m
1
+
. . .
+
m
s
and
γ
i
∈ K
(1
≤ i ≤ s
), with
γ
i
̸
=
γ
j
for
i ̸
=
j
. Then, the partial fraction decomposition of of the rational function
F (x) =
R(x)
Q
s
i=1
(x − γ
i
)
m
i
, is given by
R(x)
Q
s
i=1
(x − γ
i
)
m
i
=
s
X
i=1
m
i
X
k=1
A
(k)
i
(x − γ
i
)
k
, where
A
(k)
i
=
1
(m
i
− k)!
min(m
i
−k,p)
X
h=0
m
i
− k
h
!
R
(h)
(γ
i
)H
(m
i
−k−h)
i
(γ
i
). (16)
Now by combining results of Theorem 3.1 and Proposition 4.1, we can establish the
compact explicit expression the partial fraction decomposition. That is, the substitution
of the expression of [
H
i
(
x
)]
(d
i
(k))
x=γ
i
given by Eq. (6)-Eq. (7) in the formula Eq. (16) of
A
(k)
i
,
permits to have,
A
(k)
i
=
1
(m
i
− k)!
min(m
i
−k,p)
X
h=0
m
i
− k
h
!
R
(h)
(γ
i
)H
(m
i
−k−h)
i
(γ
i
),
58 | https://doi.org/10.22481/intermaths.v4i1.12294 L. B. Lima; R. Mustapha
or equivalently,
A
(k)
i
=
min(m
i
−k,p)
X
h=0
X
[h
j
]
i
∈Γ
i,m
i
−k−h
(−1)
m
i
−k−h
(m
i
− k − h)!
m
i
− k
h
!
R
(h)
(γ
i
)Ω
i
([h
j
]
i
; [γ
j
]),
where [
h
j
]
i
= (
h
1
, ..., h
i−1
, h
i+1
, ..., h
s
, [
γ
j
] = (
γ
1
, ..., γ
s
) and
Ω
i
([
h
j
]
i
; [
γ
j
]) is given as in
Expression Eq. (7). Therefore, the explicit formula of the partial fraction decomposition
is given in the next result.
Theorem 4.2. Let
Q
(
x
) and
R
(
x
) be two polynomials such that
Q
(
x
) =
Q
s
i=1
(
x − γ
i
)
m
i
,
where
γ
i
̸
=
γ
j
for
i ̸
=
j
(1
≤ i ≤ s
) and
deg
(
R
)
< m
1
+
. . .
+
m
s
. Then, the partial
fraction decomposition of F (x) =
R(x)
Q(x)
, is given by,
F (x) =
R(x)
Q
s
i=1
(x − γ
i
)
m
i
=
s
X
i=1
m
i
X
k=1
min(m
i
−k,p)
X
d=0
X
[h
j
]
i
∈Γ
i,m
i
−k−d
Ψ
d,i,k
(γ
1
, ....γ
s
)
1
(x − γ
i
)
k
,
where Γ
i,r
= {[h
j
]
i
= (h
1
, ..., h
i−1
, h
i+1
, ..., h
s
);
P
1≤j̸=i≤s
h
j
= r} and
Ψ
d,i,k
(γ
1
, ....γ
s
) = (−1)
m
i
−k−h
(m
i
− k − h)!
m
i
− k
h
!
R
(h)
(γ
i
)Ω
i
([h
j
]
i
; [γ
j
]), (17)
such that [
h
j
]
i
= (
h
1
, ..., h
i−1
, h
i+1
, ..., h
s
, [
γ
j
] = (
γ
1
, ..., γ
s
) and
Ω
i
([
h
j
]
i
; [
γ
j
]) is given by
Expression Eq. (7). In other words, the scalars
A
(k)
i
of partial fraction decomposition
of the rational fraction
F
(
x
) =
R(x)
Q(x)
=
s
X
i=1
m
i
X
k=1
A
(k)
i
(x − γ
i
)
k
,
are expressed under the explicit
form,
A
(k)
i
=
min(m
i
−k,p)
X
d=0
X
[h
j
]
i
∈Γ
i,m
i
−k−d
Ψ
d,i,k
(γ
1
, ....γ
s
), (18)
where the Ψ
d,i,k
(γ
1
, ....γ
s
) are given by Eq. (17).
When the degree of the polynomial
R
(
x
) is
p
= 0, we derive easily the result of
Theorem 3.1. Indeed, for
p
=
deg
(
R
) = 0, we have
R
(
x
) =
c
(constant), thus Expression
Eq. (18) takes the form Eq. (9), up to a multiplicative constant by c.
B Special case
s
= 2 and illustrative examples Suppose that
F
(
x
) =
R(x)
Q(x)
=
R(x)
(x − γ
1
)
m
1
(x − γ
2
)
m
2
,
where 0
≤ p
=
deg R < m
1
+
m
2
. Then, the partial fraction
decomposition of F (x) is given by F (x) =
m
1
X
k=1
A
(k)
1
(x − γ
1
)
k
+
m
2
X
k=1
A
(k)
2
(x − γ
2
)
k
.
L. B. Lima; R. Mustapha INTERMATHS, 4(1), 48–66, June 2023 | 59
Let
d
1
(
k
) =
max
(
m
1
− k, p
),
d
2
(
k
) =
max
(
m
2
− k, p
) and set Γ
1,d
=
{h
2
;
h
2
=
d} for
0
≤ d ≤ d
1
(
k
) and Γ
2,d
=
{h
1
;
h
1
=
d}, for
0
≤ d ≤ d
2
(
k
)
.
Then, the scalars
A
(k)
i
, i = 1, 2, are given by,
A
(k)
1
=
min(m
1
−k,p)
X
d=0
X
h
2
∈Γ
1,m
1
−k−d
(−1)
m
1
−k−d
(m
1
− k − d)!
m
1
− k
d
!
R
(d)
(γ
1
)Ω
1
(h
2
; [γ
j
]),
A
(k)
2
=
min(m
2
−k,p)
X
d=0
X
h
1
∈Γ
2,m
2
−k−d
(−1)
m
2
−k−d
(m
2
− k − d)!
m
2
− k
d
!
R
(d)
(γ
2
)Ω
2
(h
1
; [γ
j
]),
where
Ω
1
(
h
2
; [
γ
j
]) =
h
2
+m
2
−1
h
2
(
γ
1
− γ
2
)
−m
2
−h
2
and
Ω
2
(
h
1
; [
γ
j
]) =
h
1
+m
1
−1
h
1
(
γ
2
−
γ
1
)
−m
1
−h
1
. In summary, we have the following proposition.
Proposition 4.3. Consider the rational fraction
F
(
x
) =
R(x)
Q(x)
, where
Q
(
x
)(
x−γ
1
)
m
1
(
x−
γ
2
)
m
2
= with 0
≤ p
=
deg R < m
1
+
m
2
. Then, the partial fraction decomposition of
F (x) is given by F (x) =
m
1
X
k=1
A
(k)
1
(x − γ
1
)
k
+
m
2
X
k=1
A
(k)
2
(x − γ
2
)
k
, where
A
(k)
i
=
min(m
1
−k,p)
X
d=0
X
h
2
∈Γ
1,m
i
−k−d
(−1)
m
i
−k−d
(m
i
−k−d)!
m
i
− k
d
!
R
(d)
(γ
i
)Ω
i
([(h
1
, h
2
)]
i
; [γ
j
]),
for
i
= 1
,
2, with
Ω
1
(
h
2
; [
γ
j
]) =
h
2
+ m
2
− 1
h
2
!
(
γ
1
− γ
2
)
−m
2
−h
2
and
Ω
2
(
h
1
; [
γ
j
]) =
h
1
+ m
1
− 1
h
1
!
(γ
2
− γ
1
)
−m
1
−h
1
.
Proposition 4.3 is deduced as a particular case of Theorem 3.1. We will illustrate the
steps of its application on the following particular case. Let
F
(
x
) =
ax
2
+ bx + c
(x − γ
1
)
2
(x − γ
2
)
,
where (
a, b
)
̸
= (0
,
0),
γ
1
̸
=
γ
2
. Comparing with the general case, we show that
m
1
= 2,
m
1
= 1 and R(x) = ax
2
+ bx + c, of degree 1 ≤ p ≤ 2. Then, we have,
F (x) =
ax
2
+ bx + c
(x − γ
1
)
2
(x − γ
2
)
=
A
(1)
1
(x − γ
1
)
+
A
(2)
1
(x − γ
2
)
2
+
A
(1)
2
(x − γ
1
)
. (19)
Let exhibit the expressions of A
(1)
1
, A
(2)
1
and A
(1)
2
using the compact formula Eq. (18).
Computation of
A
(1)
1
and
A
(2)
1
. Following the general case
m
1
= 2 and
i
= 1, we have
A
(k)
1
=
1
(m
1
− k)!
min(m
2
−k,1)
X
d=0
m
1
− k
d
!
R
(d)
(γ
1
)H
(m
1
−k−d)
1
(γ
1
),
60 | https://doi.org/10.22481/intermaths.v4i1.12294 L. B. Lima; R. Mustapha
where
H
(m
1
−k−d)
1
(
γ
1
) = (
−
1)
m
1
−k−d
(
m
1
− k − d
)!
Ω
1
([
h
j
]
1
; [
γ
j
]). For
k
= 1 we have
m
1
− 1 = 1 and min(m
1
− 1, 1) = 1. Thus, we derive,
A
(1)
1
=
1
1!
1
X
d=0
1
d
!
R
(d)
(γ
1
)H
(1−d)
1
(γ
1
) = R
(1)
(γ
1
)H
(0)
1
(γ
1
) + R
(0)
(γ
1
)H
(1)
1
(γ
1
).
Since m
2
= 1, Γ
1,0
= {h
2
; h
2
= 0} and Γ
1,1
= {h
2
; h
2
= 1}, we get,
H
(0)
1
(γ
1
) = (−1)
0
0!Ω
1
(0, γ
1
) =
0 + m
2
− 1
0
!
(γ
1
− γ
2
)
−m
2
=
1
γ
1
− γ
2
,
H
(1)
1
(γ
1
) = (−1)
1
1!Ω
1
(1, γ
1
) =
1 + m
2
− 1
1
!
(γ
1
− γ
2
)
−m
2
−1
= −
1
(γ
1
− γ
2
)
2
.
Therefore, it follows that,
A
(1)
1
= R
(1)
(γ
1
)H
(0)
1
(γ
1
) + R
(0)
(γ
1
)H
(1)
1
(γ
1
) =
2aγ
1
+ b
γ
1
− γ
2
−
aγ
2
1
+ bγ
1
+ c
(γ
1
− γ
2
)
2
.
Since R(x) = ax
2
+ bx + c we derive R(γ
1
) = aγ
2
1
+ bγ
1
+ c and R
(1)
(γ
1
) = 2aγ
1
+ b.
For
k
= 2, we show that
m
1
− k
= 2
−
2 = 0,
min
(
m
1
−
2
,
1) =
min
(0
,
1) =
0 and Γ
1,0
=
{h
2
;
h
2
= 0
}
. Hence, we have
A
(2)
1
(
γ
1
) =
1
(2−2)!
R
(0)
(
γ
1
)
H
(0)
1
(
γ
1
) =
R
(0)
(
γ
1
)
H
(0)
1
(
γ
1
). Therefore, we derive that
A
(2)
1
(
γ
1
) = 0!(
−
1)
0
R
(0)
(
γ
1
)
Ω
1
(0
, γ
1
) =
R(γ
1
)
0 + m
1
− 1
0
!
(γ
1
− γ
2
)
−1
=
R(γ
1
)
γ
1
− γ
2
, and it ensues that,
A
(1)
1
=
aγ
1
+ b
γ
1
− γ
2
−
aγ
2
1
+ bγ
1
+ c
(γ
1
− γ
2
)
2
and A
(2)
1
(γ
1
) =
aγ
2
1
+ bγ
1
+ c
γ
1
− γ
2
. (20)
Computation of
A
(1)
2
. Following the general case, we show that
m
2
= 1 and
i
= 2,
witch implies that
m
1
−
1 = 0 and
min
(
m
1
−
1
,
1) = 0 and Γ
2,0
=
{h
2
;
h
2
= 0
}
.
Hence, we have
A
(1)
2
=
1
(1−1)!
R
(0)
(
γ
2
)
H
(0)
2
(
γ
2
) =
R
(
γ
2
)
H
(0)
2
(
γ
2
), which permits to have
A
(1)
2
=
R
(
γ
2
)0!(
−
1)
0
Ω
2
(0;
γ
2
) =
R
(
γ
2
)
0 + m
2
− 1
0
!
(
γ
2
− γ
1
)
−m
2
−0
=
R(γ
2
)
(γ
2
− γ
1
)
2
,
namely,
we have,
A
(1)
2
=
aγ
2
1
+ bγ
1
+ c
(γ
2
− γ
1
)
2
. (21)
In summary, we can formulate the following proposition.
Proposition 4.4. The partial fraction decomposition of
F
(
x
) =
ax + b
(x − γ
1
)
2
(x − γ
2
)
is
given by
F
(
x
) =
A
(1)
1
(x − γ
1
)
+
A
(2)
1
(x − γ
2
)
2
+
A
(1)
2
(x − γ
1
)
, where
A
(1)
1
,
A
(2)
1
and
A
(1)
2
are given as in
L. B. Lima; R. Mustapha INTERMATHS, 4(1), 48–66, June 2023 | 61
expressions Eq. (20)-Eq. (21).
To better illustrate the efficiency of the preceding special case, namely, Proposition
4.4, we apply the formulas Eq. (20)-Eq. (21) to the following numerical cases.
Example 4.5. Suppose that
F
(
x
) =
2x + 1
(x − 8)
2
(x − 6)
.
We show easily that
R
(
x
) = 2
x
+ 1,
γ
1
= 8,
γ
2
= 6. Hence, we get
R
′
(8) = 2,
γ
1
− γ
2
= 2,
R
(8) = 17 and
R
(6) = 13. Now,
application of Expressions Eq. (20)-Eq. (21), implies that we have,
A
(1)
1
=
2
2
−
17
4
=
−
13
4
, A
(2)
1
=
17
2
and A
(1)
2
=
13
4
.
Therefore, we get the partial fraction decomposition of
the rational function F (x) as follows F (x) = −
13
4(x − 8)
+
17
2(x − 8)
2
+
13
4(x − 6)
.
A similar process can be applied to the following numerical example.
Example 4.6. Suppose that
F
(
x
) =
x
2
+ x + 1
(x − 8)
2
(x − 6)
.
We show easily that
R
(
x
) =
x
2
+
x
+ 1 and
R
′
(
x
) = 2
x
+ 1. Since
γ
1
= 8,
γ
2
= 6, we have
γ
1
− γ
2
= 2,
R
(8) = 73,
R
′
(8) = 17 and
R
(6) = 43. A direct calculation, applying Expressions Eq. (20)-Eq. (21),
shows that
A
(1)
1
=
17
2
−
73
4
=
−
39
4
, A
(2)
1
=
73
2
and A
(1)
2
=
43
4
.
Therefore, the partial
fraction decomposition, is given as follows F (x) = −
39
4(x − 8)
+
73
2(x − 8)
2
+
43
4(x − 6)
.
Since theses numerical examples are simple, their partial fraction decomposition can
be obtained by means of other usual methods. However, our formula Eq. (18) of Theorem
4.2, allows us to compute compact explicit expressions for the coefficients
A
(1)
1
,
A
(2)
1
and
A
(1)
2
, of the partial fraction decomposition of Expression Eq. (19). And these compact
explicit formulas Eq. (20)-Eq. (21), applied to the preceding two numerical examples,
allows us to get their associated partial fraction decomposition.
C Special case
s
= 3 and illustrative examples Suppose that
F
(
x
) =
R(x)
Q(x)
=
R(x)
(x−γ
1
)
m
1
(x−γ
2
)
m
2
(x−γ
3
)
m
3
,
where 0
≤ p
=
deg R < m
1
+
m
2
+
m
3
. Then, the partial
fraction decomposition of F (x) is given by,
F (x) =
m
1
X
k=1
A
(k)
1
(x − γ
1
)
k
+
m
2
X
k=1
A
(k)
2
(x − γ
2
)
k
+
m
3
X
k=1
A
(k)
3
(x − γ
3
)
k
.
Let
d
i
(
k
) =
max
(
m
i
− k, p
), for
i
= 1
,
2
,
3. For every 0
≤ d ≤ d
i
(
k
), we set Γ
1,d
=
{
(
h
2
, h
3
);
h
2
+
h
3
=
d}
, Γ
2,d
=
{
(
h
1
, h
3
);
h
1
+
h
3
=
d}
, Γ
3,d
=
{
(
h
1
, h
2
);
h
1
+
h
2
=
d}.
Then, the result of Theorem 3.1, namely, Expression Eq. (18), a direct computation
62 | https://doi.org/10.22481/intermaths.v4i1.12294 L. B. Lima; R. Mustapha
shows that the scalars A
(k)
i
, i = 1, 2, 3, are given by,
A
(k)
1
=
min(m
1
−k,p)
X
d=0
X
(h
2
,h
3
)∈Γ
1,m
1
−k−d
(−1)
m
1
−k−d
(m
1
− k − d)!
m
1
− k
d
!
R
(d)
(γ
1
)Ω
1
((h
2
, h
3
); [γ
j
]),
A
(k)
2
=
min(m
2
−k,p)
X
d=0
X
(h
1
,h
3
)∈Γ
2,m
2
−k−d
(−1)
m
2
−k−d
(m
1
− k − d)!
m
2
− k
d
!
R
(d)
(γ
2
)Ω
2
(h
1
, h
3
); [γ
j
]),
A
(k)
3
=
min(m
3
−k,p)
X
d=0
X
(h
1
,h
2
)∈Γ
3,m
3
−k−d
(−1)
m
1
−k−d
(m
3
− k − d)!
m
3
− k
d
!
R
(d)
(γ
3
)Ω
3
(h
1
, h
2
); [γ
j
]),
where the
Ω
i
(
h
1
, h
2
); [
γ
j
]), for
i
= 1
,
2
,
3, are given as in Eq. (7). Accordingly, we can
state the following proposition.
Proposition 4.7. The partial fraction decomposition of
F
(
x
) =
R(x)
(x − γ
1
)
m
1
(x − γ
2
)
m
2
(x − γ
3
)
m
3
,
where 0
≤ p
=
deg R < m
1
+
m
2
+
m
3
is given by
F
(
x
) =
m
1
X
k=1
A
(k)
1
(x − γ
1
)
k
+
m
2
X
k=1
A
(k)
2
(x − γ
2
)
k
+
m
3
X
k=1
A
(k)
3
(x − γ
3
)
k
, such that the A
(k)
i
, for i = 1, 2, 3, are given by,
A
(k)
i
=
min(m
i
−k,p)
X
d=0
X
(h
2
,h
3
)∈Γ
1,m
1
−k−d
Λ
i,d
R
(d)
(γ
1
)Ω
i
((h
2
, h
3
); [γ
j
]), (22)
where Λ
i,d
= (
−
1)
m
i
−k−d
(
m
i
− k − d
)!
m
i
−k
d
and the
Ω
i
(
h
1
, h
2
); [
γ
j
]), for
i
= 1
,
2
,
3, are
given as in Eq. (7), namely,
Ω
i
([h
j
]
i
; [γ
j
]) =
3
Y
j=1,j̸=i
"
h
j
+ m
j
− 1
h
j
!
(γ
i
− γ
j
)
−m
j
−h
j
#
, for i = 1, 2, 3. (23)
Suppose that R(x) = C ∈ K (constant), then, Expression Eq. (22) takes the form,
A
(k)
i
= (−1)
m
i
−k
(m
i
− k)!C × Ω
i
((h
2
, h
3
); [γ
j
]).
Example 4.8. Consider the rational fraction
F
(
x
) =
2x + 1
(x − 8)(x − 6)(x − 4)
.
We show
easily that
R
(
x
) = 2
x
+ 1,
γ
1
= 8,
γ
2
= 6,
γ
3
= 4, which implies that
R
(8) = 17,
R
(6) = 13 and
R
(4) = 9. Now, application of Expressions Eq. (22)-Eq. (23), implies
that
A
(1)
1
=
17
8
, A
(1)
2
=
−
13
4
and A
(1)
3
=
9
8
.
Therefore, the partial fraction decomposition
of F (x), is given as follows F (x) =
17
8(x − 8)
−
13
4(x − 6)
+
9
8(x − 4)
.
Example 4.9. Suppose that
F
(
x
) =
2x + 1
(x − 8)(x − 6)
2
(x − 4)
. We show easily that
R
(
x
) =
L. B. Lima; R. Mustapha INTERMATHS, 4(1), 48–66, June 2023 | 63
2
x
+ 1 and
R
′
(
x
) = 2,
γ
1
= 8,
γ
2
= 6,
γ
3
= 4, which permits to get
R
(8) = 17,
R
(6) = 13,
R
(4) = 9 and
R
′
(6) = 2. By applying Expressions Eq. (22)-Eq. (23), we
obtain,
A
(1)
1
=
17
16
, A
(1)
2
=
−
1
2
A
(2)
2
=
−
13
4
and A
(1)
3
=
−
9
16
.
Therefore, the partial
fraction decomposition of F (x) is F (x) =
17
16(x − 8)
−
1
2(x − 6)
−
13
4(x − 6)
2
−
9
16(x − 4)
.
Example 4.10. Suppose that
F
(
x
) =
2x + 1
(x − 8)(x − 6)
2
(x − 4)
3
. Since
R
(
x
) = 2
x
+ 1,
R
′
(
x
) = 2 and
R
′′
(
x
) = 0 ,
γ
1
= 8,
γ
2
= 6,
γ
3
= 4, we get
R
(8) = 17,
R
(6) = 13,
R
(4) = 9,
R
′
(8) = 2,
R
′
(6) = 2,
R
′
(4) = 2 and
R
′′
(4) = 0. A straightforward compu-
tation, utilizing Expressions Eq. (22)-Eq. (23), allows us to obtain
A
(1)
1
=
17
256
, A
(1)
2
=
11
16
, A
(2)
2
=
−
13
16
, A
(1)
3
=
−
193
256
, A
(2)
3
=
−
53
64
and A
(3)
3
=
−
9
16
.
Therefore, the partial
fraction decomposition of
F
(
x
) is
F
(
x
) =
17
256(x − 8)
+
11
16(x − 6)
−
13
16(x − 6)
2
−
193
256(x − 4)
−
53
64(x − 4)
2
−
9
16(x − 4)
3
.
5 Discussion, Concluding remarks and perspective
The partial fraction decomposition of the general case
R(x)
Q(x)
, where the polynomial
Q
(
x
) =
Q
s
j=1
(
x − λ
j
)
m
j
, have been largely studied in the literature, by using various
methods and techniques. In this study we have proposed an approach to determine the
partial fraction decomposition, where there is no need to resort to other techniques. It
seems to us that the compact explicit formulas, considered for the calculation of the
coefficients
A
(k)
i
(1
≤ i ≤ s
, 1
≤ k ≤ m
j
), of the partial fractions decomposition, require
direct computation. This approach represents another method to determine the partial
fractions decomposition, apart from the usual techniques. In addition, comparing with
the literature, in the best of our knowledge, we show that there is no explicit formula
for the scalars A
k
i
.
As a perspective, it seems important to us to highlight these results by a study of
an educational nature. Indeed, calculating the coefficients
A
(k)
i
, is not an easy task for
undergraduate. In the diversity of the usual methods, the choice of the adequate one is
not easy for the student. It seems interesting that the compact formulas of this approach,
can be proposed to students to allow them to acquire another method to determine the
partial fractions decomposition, using other techniques. Moreover, the elaboration of
algorithms will allow (as in [
9
]) to facilitate the use of the software such as MATLAB,
which will make it possible to better enhance the content of future educational research,
of this new approach.
64 | https://doi.org/10.22481/intermaths.v4i1.12294 L. B. Lima; R. Mustapha
Acknowledgments. The authors express their sincere thanks to the referees for their
valuable suggestions. The authors express their sincere thanks to the INMA and the UFMS
for their valuable support and encouragements. Special thanks to Prof. Dr. M. Turine, Profa.
Dra. C. Ítavo and Prof. Dr. B. Dias.
Disclosure statement. The authors declare no conflict of interest in the writing of the
manuscript, or in the decision to publish the results.
ORCID
Leandro Bezerra de Lima https://orcid.org/0000-0002-7128-0789
Mustapha Rachidi https://orcid.org/0000-0002-8210-7383
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