International Mathematical Forum, Vol. 7, 2012, no. 31, 1517 - 1535
Two Proofs of the Existence and Uniqueness
of the Partial Fraction Decomposition
William T. Bradley and William J. Cook1
Appalachian State University
Mathematical Sciences – Walker Hall
121 Bodenheimer Dr. Boone, NC 28608, USA
Abstract
Two proofs of the existence and uniqueness of the partial fraction
decomposition of a (real) rational function are presented. The first proof
of existence and uniqueness uses only elementary facts from linear algebra. The second existence proof uses the Euclidean algorithm and
applies to all Euclidean domains. While existence can be proven for
any Euclidean domain, additional hypotheses are required to establish
uniqueness. The second uniqueness proof applies to polynomials with
arbitrary field coefficients as well as the ring of integers. A partial fraction decompostion contains a whole and a fractional part. We show
that the whole part will vanish for proper fractions of polynomials with
field coefficients. In the final section we present a method which uses
the partial fraction decomposition to solve linear differential equations
with constant coefficients. Then we show how the partial fraction decomposition gives a refinement of Chinese remaindering which in turn
proves the existence of Hermite interpolation polynomials.
Mathematics Subject Classification: 13F07
Keywords: partial fraction decomposition, euclidean domain
1
Introduction
The partial fraction decomposition is the main tool which allows one to integrate any rational function. This is why it is usually introduced when learning
elementary integral calculus. The decomposition’s next appearance is usually
in an introductory course on solving differential equations when it is again used
to integrate rational functions. In particular, one uses it to solve the logistic
1
cookwj@appstate.edu
1518
W. T. Bradley, W. J. Cook
equation after separating variables. When Laplace transforms are covered,
the decomposition will reappear during calculations of inverse transforms. After these instances, the partial fraction decomposition often fades into distant
memory.
If one looks at any number of standard calculus texts or even many nonstandard texts, techniques for computing the partial fraction decomposition of
real rational functions is often presented. However, such texts never discuss or
prove why such a decomposition exists. We suggest this happens because such
textbooks are typically focused on teaching how to use techniques. Proofs
of many facts are pushed off to a future course in analysis. This is where
the partial fraction decomposition slips into the cracks. Analysts view this as
pure algebra (which it is) so it should not be addressed in a course on analysis. Algebraists typically view partial fractions as a technique only good for
integrating and thus a problem for analysts. So, outside of the realm of symbolic computation, the partial fraction decomposition tends to never be fully
discussed.
In this paper we provide a simple accessible proof of the existence and
uniqueness of the partial fraction decomposition which requires only a few
facts from elementary linear algebra (see section 2). The second proof found
in sections 3 (existence) and 4 (uniqueness) relies on the Euclidean algorithm.
The existence proof works in any Euclidean domain while the uniqueness only
holds for certain Euclidean domains. These proofs would fit in any course
covering basic ring theory. In the final section we discuss some (less often
seen) applications of the partial fraction decomposition including solutions of
nonhomogeneous linear differential equations with constant coefficients, the
Chinese remainder theorem, and Hermite interpolation.
Both authors would like to thank the National Science Foundation (NSF)
for partial support (NSF grant STEP 0756928). We would also like to thank
the members of our research cluster: Hadi Marrow, Hannah Gilmore, and Kyle
Kelley for many helpful discussions.
2
A Linear Algebraic Proof
As noted in the introduction, partial fraction decompositions are primarily
useful when dealing with rational functions. In this section we present a proof
of the existence and uniqueness of the partial fraction decomposition for real
polynomials using only elementary facts from linear algebra.
Suppose we are given two real polynomials f (x), g(x) ∈ R[x] where g(x) =
0. If the degree of f (x) is bigger than the degree of g(x), we should start by
performing polynomial long division. From long division we obtain (unique)
real polynomials q(x), r(x) ∈ R[x] such that f (x) = q(x)g(x) + r(x) where
1519
Two proofs of existence and uniqueness
r(x)
f (x)
= q(x) +
. Therefore, we
g(x)
g(x)
may assume that our rational function is a proper fraction. Also, for convenience assume that g(x) is monic (its leading coefficient is 1).
We will begin our proof by noting that the existence of a partial fraction
decomposition for f (x)/g(x) is equivalent to f (x) belonging to the span of a
certain set of polynomials.
Any real non-constant real polynomial can be factored into linear and
quadratic factors. In particular, suppose that g(x) is factored as follows:
deg(r(x)) < deg(g(x)) or r(x) = 0 and so
g(x) = 1 (x)k1 2 (x)k2 · · · m (x)km · q1 (x)s1 q2 (x)s2 · · · qn (x)sn
(2.1)
where i (x) = x−ri are distinct linear factors, qj (x) = x2 +aj x+bj are distinct
irreducible quadratic factors (which have complex roots), and the exponents
ki and sj are positive integers.
A partial fraction decomposition of f (x)/g(x) is
i
i
Aij
Bij x + Cij
f (x) +
=
j
g(x)
(x − ri )
qi (x)j
i=1 j=1
i=1 j=1
m
k
n
s
(2.2)
where all Aij , Bij , and Cij are real numbers. Multiplying both sides by g(x)
clears the denominators.
f (x) =
ki
m i=1
i
g(x)
g(x)
g(x)
Aij
+
Bij x
+ Cij
j
j
(x − ri )
qi (x)
qi (x)j
j=1
i=1 j=1
n
s
(2.3)
Notice that since g(x) is divisible by these linear and quadratic factors, we
g(x)
g(x)
have
∈
R[x]
for
all
i
=
1,
.
.
.
,
m
and
j
=
1,
.
.
.
,
k
and
∈ R[x]
i
(x − ri )j
qi (x)j
for all i = 1, . . . , n and j = 1, . . . si . Therefore, finding a partial fraction decomposition for f (x)/g(x) is equivalent to writing f (x) as a linear combination
of the members of the set S where
g(x) g(x) g(x)
S=
,x
1 ≤ i ≤ m, 1 ≤ j ≤ ki ∪
1 ≤ i ≤ n, 1 ≤ j ≤ si
(x − ri )j
qi (x)j qi (x)j
(2.4)
First, note that the cardinality of S is N = k1 + · · · + kn + 2s1 + · · · + 2sn =
deg(g(x)). Consider PN = {h(x) ∈ R[x] | deg(h(x)) < N or h(x) = 0}, the
vector space of all real polynomials of degree less than N = deg(g(x)) and recall
that dim(PN ) = N = deg(g(x)). Also, S ⊆ PN since all of the polynoimials in
S are constructed by knocking out factors from g(x) or knocking out at least
a degree 2 polynomial from g(x) and then multiplying by x (which still results
in a polynomial of degree than less g(x)).
1520
W. T. Bradley, W. J. Cook
Therefore, showing that all polynomials f (x) (of degree less than g(x))
have a partial fraction decomposition is equivalent to showing PN = Span(S).
But since |S| = N = dim(PN ), this is equivalent to showing S is a basis for
PN . Now since S has N elements (it is the correct size), we know S is a basis
if and only if it is linearly independent. To prove this we need a few lemmas.
Lemma 2.1. Let h(x), g(x) ∈ R[x] (real polynomials), k a positive integer,
and r ∈ R such that g(r) = 0. If A1 , . . . , Ak ∈ R such that h(x)(x − r)k +
A1 g(x)(x − r)k−1 + A2 g(x)(x − r)k−2 + · · · + Ak g(x) = 0, then A1 = A2 = · · · =
Ak = 0.
Proof: If we evaluate h(x)(x − r)k + A1 g(x)(x − r)k−1 + A2 g(x)(x − r)k−2 +
· · · + Ak g(x) = 0 at x = r, we get h(r)(0)k + A1 g(r)(0)k−1 + · · · + Ak−1 g(r)(0) +
Ak g(r) = 0. So that Ak g(r) = 0. However, g(r) = 0 therefore, Ak = 0.
Now our equation reads h(x)(x−r)k +A1 g(x)(x−r)k−1 +· · ·+Ak−1g(x)(x−
r) = 0. Factoring out and canceling x−r, gives us h(x)(x−r)k−1 + A1 g(x)(x−
r)k−2 + · · · + Ak−1 g(x) = 0.
Therefore, applying the same argument again we find that Ak−1 = 0. Continuing in this fashion we find that A1 = A2 = · · · = Ak = 0. The next lemma (which is quite similar to Lemma 2.1) will deal with irreducible quadratic factors. Remember that irreducible quadratic polynomials
have √
a conjugate pair of (complex) roots. Here we let i denote the imaginary
root −1.
Lemma 2.2. Let h(x), g(x) ∈ R[x] (real polynomials), s be a positive integer,
and q(x) ∈ R[x] an irreducible quadratic factor with roots z = a+bi, z̄ = a−bi ∈
C. In addition suppose that g(z) = 0 and g(z̄) = 0. If B1 , . . . , Bk , C1 , . . . , Ck ∈
R[x] such that h(x)q(x)s + (B1 x + C1 )g(x)q(x)s−1 + (B2 x + C2 )g(x)q(x)s−2 +
· · ·+(Bk x+Ck )g(x) = 0, then B1 = B2 = · · · = Bk = C1 = C2 = · · · = Ck = 0.
Proof: If we evaluate h(x)q(x)s +(B1 x+C1 )g(x)q(x)s−1 +(B2 x+C2 )g(x)q(x)s−2 +
· · · + (Bs x + Cs )g(x) = 0 at both x = z and x = z̄, we get h(z)(0)s +
(B1 z + C1 )g(z)(0)s−1 + · · · + (Bs−1 z + Cs−1 )g(z)(0) + (Bs z + Cs )g(z) = 0 and
h(z̄)(0)s +(B1 z̄+C1 )g(z̄)(0)s−1 +· · ·+(Bs−1 z̄+Cs−1 )g(z̄)(0)+(Bs z̄+Cs )g(z̄) = 0
so that (Bs z + Cs )g(z) = 0 and (Bs z̄ + Cs )g(z̄) = 0. Thus Bs z + Cs = 0 and
Bs z̄ + Cs = 0 since g(z) = 0 and g(z̄) = 0. Next, expand recall z = a + bi and
z̄ = a − bi so we have aBs + bBs i + Cs = 0 and aBs − bBs i + Cs = 0. Subtracting equations yields, 2bBs i = 0 and so Bs = 0 since b = 0 (if b = 0 then
our quadratic factor q(x) has a real root z = a + bi = a which contradicts the
assumption that q(x) is irreducible). Since Bs = 0 we get Cs = Bs z + Cs = 0.
Therefore, we have that Bs = Cs = 0.
Now our equation reads h(x)q(x)s +(B1 x+C1 )g(x)q(x)s−1 +(B2 x+C2 )g(x)q(x)s−2 +
· · · + (Bs−1 x + Cs−1 )q(x)g(x) = 0. Factoring out and canceling the common
1521
Two proofs of existence and uniqueness
factor q(x), gives us h(x)q(x)s−1 + (B1 x + C1 )g(x)q(x)s−2 + · · · + (Bs−1 x +
Cs−1 )g(x) = 0
We can now apply the same argument again and find that Bs−1 = Cs−1 = 0.
Continuing in this fashion we have B1 = B2 = · · · = Bs = C1 = C2 = · · · =
Cs = 0. Proposition 2.3. The set S is linearly independent.
Proof: Suppose that
ki
m Aij
i=1 j=1
si
n g(x)
g(x)
g(x)
+
B
x
+C
= 0.
ij
ij
(x − ri )j i=1 j=1
qi (x)j
qi (x)j
First, we focus on the factor x − rt (for some 1 ≤ t ≤ m). Let gt (x) =
g(x)
. Notice that all of the elements in S have the factor (x − rt )kt
(x − rt )kt
g(x)
g(x)
g(x)
except
,
,...,
. Therefore,
2
x − rt (x − rt )
(x − rt )kt
ki
m i=1
i=t
i
g(x)
g(x)
g(x)
+
B
x
+
C
= h(x)(x − rt )kt
ij
ij
j
j
(x − ri )j
q
(x)
q
(x)
i
i
i=1 j=1
n
Aij
j=1
where h(x) =
ki
m i=1
i=t
Note that
s
i
gt (x)
gt (x)
gt (x)
+
B
x
+
C
.
ij
ij
j
j
(x − ri )j
q
(x)
q
(x)
i
i
i=1 j=1
n
Aij
j=1
g(x)
= gt (x)(x − rt )kt −j and gt (rt ) = 0. Therefore,
j
(x − rt )
ki
m kt
j=1
kt
j=1
s
g(x)
Atj (x−r
j
t)
+
i=1 j=1
ki
m i=1
i=t
j=1
g(x)
Aij (x−r
j +
i)
g(x)
Aij (x−r
j +
i)
Atj gt (x)(x − rt )kt −j +
si
n i=1 j=1
si
n i=1 j=1
g(x)
Bij x qg(x)
j + Cij q (x)j
i (x)
i
=
g(x)
Bij x qg(x)
j + Cij q (x)j
i (x)
i
=
h(x)(x − rt )kt
= 0
Applying Lemma 2.1, we have that At1 = At2 = · · · = Atkt = 0. Therefore,
Aij = 0 for all i and j.
Next, consider a quadratic factor, qt (x) (where 1 ≤ t ≤ n). As before, let
g(x)
g(x) g(x) g(x)
st
.
Everything
in
S
has
the
factor
q
(x)
except
,
gt (x) =
,x
,
t
s
qt (x) t
qt (x) qt (x) qt (x)2
si
n g(x)
g(x)
g(x)
g(x)
g(x)
Bij x
+Cij
= h(x)qt (x)st
x qt (x)2 , . . . , qt (x)st , x qt (x)st . Therefore,
j
j
q
(x)
q
(x)
i
i
i=1 j=1
i=t
1522
where h(x) =
W. T. Bradley, W. J. Cook
si
n i=1
i=t
Note that
j=1
Bij x
gt (x)
gt (x)
+
C
ij
qi (x)j
qi (x)j
g(x)
= gt (x)qt (x)st −j , gt (z) = 0, and gt (z̄) = 0. Therefore,
qt (x)j
si
n st
j=1
st
j=1
g(x)
Btj x qg(x)
j + Ctj q (x)j
t (x)
t
+
i=1 j=1
si
n i=1
i=t
Btj xgt (x)qt (x)st −j + Ctj gt (x)qt (x)st −j +
j=1
g(x)
Bij x qg(x)
j + Cij q (x)j
i (x)
i
=
g(x)
Bij x qg(x)
j + Cij q (x)j
i (x)
i
=
h(x)qt (x)st
= 0
Applying Lemma 2.2, we find that Bt = Ct = 0. Therefore, Bij = Cij = 0 for
all i and j. Therefore, S is linearly independent. The linear independence of S ⊂ PN and the fact that |S| = dim(PN )
implies S is a basis for (and thus spans) PN . Therefore, the partial fraction
decomposition exists for real polynomials.
Theorem 2.4. Let f (x), g(x) ∈ R[x] with g(x) = 0. Then f (x)/g(x) possesses a unique partial fraction decomposition (up to rearrangement of terms).
Proof: We just need to show uniqueness. Suppose g(x) factors as before and
ki
si
ki
m n m Ãij
Aij
Bij x + Cij
f (x)
= h̃(x)+
h(x)+
+
=
+
j
j
(x − ri ) i=1 j=1 qi (x)
g(x)
(x − ri )j
i=1 j=1
i=1 j=1
si
n B̃ij x + C̃ij
where h(x), h̃(x) ∈ R[x] and Aij , Ãij , Bij , B̃ij , Cij , C̃ij ∈ R
qi (x)j
i=1 j=1
f (x)
. Then clearing the denomig(x)
nators and moving everything to the left hand side we have that g(x)(h(x) −
ki
si
m n g(x)
g(x)
h̃(x))+
(Aij − Ãij )
+
((Bij − B̃ij )x+(Cij − C̃ij ))
=
i
j
(x
−
r
)
q
(x)
j
i
i=1 j=1
i=1 j=1
are two partial fraction decompositions for
0. In both of the summations, each term has a degree strictly less than g(x).
Therefore, if h(x) − h̃(x) = 0, the degree of the left hand side of the equation
would be at least N = deg(g(x)). But this is impossible since it is equal to 0.
Therefore, h(x) − h̃(x) = 0 and so h(x) = h̃(x). This means that
ki
m (Aij − Aij )
i=1 j=1
i
g(x)
g(x)
+
((Bij − Bij )x + (Cij − Cij ))
=0
i
(x − rj )
qi (x)j
i=1 j=1
n
s
Two proofs of existence and uniqueness
1523
and so using the linear independence of S, we have that Aij − Ãij = 0, Bij −
B̃ij = 0, and Cij = C̃ij = 0 for all i and j. Thus Aij = Ãij , Bij = B̃ij , and
Cij = C̃ij for all i and j. Therefore, the decomposition of f (x)/g(x) is unique.
Remark 2.5. The proofs in this section apply equally well to polynomials
whose coefficients lie in other fields. However, the appearance of irreducible
factors of degree greater than 2 requires more complicated formulae and more
tedious calculations. We will see how to avoid these unpleasantries in the next
section.
3
Existence in Euclidean Domains
While the proofs in the previous section used only results and concepts from
elementary linear algebra, there is something unsatisfying about having to deal
with linear and quadratic factors separately (as is often the case when working
with polynomials over the field of real numbers). Here we present a second
existence proof using the Euclidean algorithm. In fact, this proof works in the
context of an arbitrary Euclidean domain. However, while a decomposition
will always exist in a Euclidean domain, it is not necessarily unique. We will
take a look at the uniqueness of such decompositions in section 4. Please
note that a similar proof of the existence of partial fraction decompositions
in Euclidean domains can be found in [PW]. We begin by reviewing a few
definitions.
Definition 3.1. An integral domain R equipped with a norm δ : R − {0} →
Z≥0 (δ(r) is a non-negative integer for each r = 0 in R) is called a Euclidean
Domain if and only if
E1 For each a, b ∈ R such that b = 0 there exists q, r ∈ R such that a = bq + r
and either r = 0 or δ(r) < δ(b).
E2 δ(a) ≤ δ(ab) for all a, b ∈ R − {0}.
Remark 3.2. The definition of a Euclidean domain varies from text to text.
However, these differences are (for the most part) superficial. In particular,
the requirement E2 is often omitted. However, Roger in [Ro] proved that E2
is unnecessary. Specifically, if an integral domain R posseses a norm N and
satisfies E1, then when R’s norm is replaced by δ(a) = min{N(ax) | x ∈ R −
{0}}, it becomes a Euclidean domain (in the sense of definition 3.1).
For what follows we will assume R is a Euclidean domain with norm δ :
R − {0} → Z≥0 . Recall that every integral domain can be imbedded in its
1524
W. T. Bradley, W. J. Cook
field of fractions. Let F be R’s field of fractions. Suppose that f /g ∈ F (where
f, g ∈ R) we say that f /g is a proper fraction if either f = 0 or δ(f ) < δ(g).
Recall that all Euclidean domains are principal ideal domains – that is –
given any ideal I R, there exists some a ∈ R such that I = (a) = {ra | r ∈ R}.
Definition 3.3. Let f /g ∈ F where f, g ∈ R and g = 0. Also, suppose that
g = ak11 ak22 · · · ak where ki ∈ Z>0 (each ki is a positive integer) and every pair
ai and aj is relatively prime (when i = j).
ki
rij
f
Then
=q+
j where q, rij ∈ R is a partial fraction decompog
a
i=1 j=1 i
sition of f /g relative to g = ak11 · · · ak if for each 1 ≤ i ≤ and 1 ≤ j ≤ ki
either rij = 0 or δ(rij ) < δ(ai ).
If g = ak11 · · · ak is a prime factorization of g (a factorization of g into
irreducibles), then we simply call this a partial fraction decomposition of f /g.
Remark 3.4. Consider the special case R = R[x] (real polynomials) and so
F = R(x) is the field of real rational polynomials. In R[x] we have δ = deg
(the degree of the polynomial). Consider a partial fraction decomposition of
f /g (so g is factored into irreducibles a1 (x), a2 (x), . . . , a (x)).
If ai (x) is a linear factor, then for each 1 ≤ j ≤ ki , ri,j (x) = 0 or rij has
degree 0 so in either case, it is a constant. If ai (x) is an irreducible quadratic
factor, then for each 1 ≤ j ≤ ki either ri,j (x) = 0 or rij has degree 0 or 1, so
rij (x) is a constant or a linear polynomial. Thus defintion 3.3 includes our old
definition as a special case.
Proposition 3.5. Let f, g ∈ R where g = 0. In addition suppose that g = ab
f
A B
where a and b are relatively prime. Then = +
for some A, B ∈ R.
g
a
b
Proof: Since a and b are relatively prime elements of a Euclidean domain,
there exist c, d ∈ R such that bd + ac = 1. Therefore, f = b(f d) + a(f c) and
thus
f
b(f d) + a(f c)
b(f d) a(f c)
fd fc
=
=
+
=
+
g
ab
ab
ab
a
b
So A = f d and B = f c are the desired elements of R. Proposition 3.6. Let f, g ∈ R where g = 0 and g = ak11 ak22 · · · ak where
A1
f
= k1 +
a1 , a2 , . . . , a are pairwise relatively prime and ki ∈ Z>0 . Then
g
a1
A
A2
+ · · · + k for some A1 , A2 , . . . , A ∈ R.
ak22
a
Two proofs of existence and uniqueness
1525
Proof: Since a1 is relatively prime with each of ai (2 ≤ i ≤ ), we have that
ak11 and ak22 · · · ak are relatively prime (if not they would share a prime factor,
so a1 would share a prime factor with some ai contradicting our assumption
that a1 and ai are relatively prime.)
B1
f
A1
where A1 , B1 ∈
Applying Proposition 3.5, we obtain = k1 + k2 k3
g
a1 a2 a3 · · · ak R. Applying Proposition 3.5 again to the pair ak22 and ak33 · · · ak , we obtain
A1
B1
A2
B2
f
A2
= k1 + k2 +
= k2 + k3
and so
A2 , B2 ∈ R such that k2
k
k
g
a2
a1
a2
a2 · · · a
a3 · · · a
B2
. Continuing in this fashion, we obtain A1 , A2 , . . . , A ∈ R such that
ak33 · · · ak A1
A2
A
f
= k1 + k2 + · · · + k . g
a1
a2
a
Remark 3.7. At this point we should note that both of these propositions (and
their proofs) work equally well for principal ideal domains (PIDs). If R is a
PID and a, b ∈ R, then (a, b) = {as+bt | s, t ∈ R} is an ideal. Thus (a, b) = (d)
for some d ∈ R. It is easy to show that d is a greatest common divisor of a
and b. Thus if a and b are relatively prime, we would have s, t ∈ R such that
as + bt = 1. However, in a PID we are just guaranteed existence whereas in a
Euclidean domain, we have an algorithm for computing s and t.
Also, note that these propositions do not follow for unique factorization
domains (UFDs). In a UFD if a and b are relatively prime, one cannot always
find c and d such that ac + bd = 1.
Proposition 3.8. Let A, a ∈ R, a = 0, and k ∈ Z>0 . Then there exists
q1 , r1 , . . . , rk ∈ R such that for each i = 1, . . . , k, either ri = 0 or δ(ri ) < δ(a)
A
r1 r2
rk
and k = q1 + + 2 + · · · + k .
a
a
a
a
Proof: Since R is a Euclidean domain, we can apply the division algorithm
for R to A and a. Thus, there exists qk , rk ∈ R such that A = aqk + rk where
rk = 0 or δ(rk ) < δ(a). Now apply the division algorithm to qk and a and get
qk−1 , rk−1 ∈ R such that qk = aqk−1 +rk−1 and either rk−1 = 0 or δ(rk−1 ) < δ(a).
Continuing in this fashion, we obtain q1 , q2 , . . . , qk , r1 , r2 , . . . , rk ∈ R where for
each i = 1, . . . , k either ri = 0 or δ(ri ) < δ(a) and A = aqk + rk , qi+1 = aqi + ri
for i = 1, . . . , k − 1. Therefore,
A
aqk + rk
qk
rk
aqk−1 + rk−1 rk
qk−1
rk−1
rk
=
= k−1 + k =
+ k = k−2 + k−1 + k = . . .
k
k
k−1
p
a
a
a
a
a
a
a
a
r1 r2
rk
= q1 + + 2 + · · · + k
a
a
a
1526
W. T. Bradley, W. J. Cook
Theorem 3.9. Given f, g ∈ R where 0 = g = ak11 · · · ak and a1 , . . . , a ∈ R
are pairwise relatively prime, then f /g has a partial fraction decomposition
relative to a1 , . . . , a . In particular, f /g has a partial fraction decomposition.
Proof: First, we apply Proposition 3.6 to decompose f /g into fractions whose
denominators are ak11 , . . . , ak . Then we apply Proposition 3.8 to each term.
Finally, adding the q1 ’s together, we obtain the desired decomposition. If we put the proofs of Propositions 3.6 and 3.8 together, we get an algorithm for computing partial fraction decompositions. In particular a decomposition can be computed by running the Euclidean algorithm times and then
performing k1 + · · · + k divisions with remainder.
Remark 3.10. Although propositions 3.5 and 3.6 work over an PID, Proposition 3.8 requires a division algorithm (thus a Euclidean domain). Notice that
the very concept of a partial fraction decomposition becomes unclear over a
PID since we have no well defined way of deciding if numerators are “smaller”
than denominators.
4
Uniqueness in Euclidean Domains
We have shown that a partial fraction decomposition exists for any element
of the fraction field of a Euclidean domain. Naturally we now turn to the
question of uniqueness.
Recall that we already proved uniqueness in the context of real polynomials
(see section 2). It turns out that uniqueness of the decomposition does not
follow for all Euclidean domains. It even fails for integers. An easy example
1
−1
is 0 + = 1 +
. We only have unique decompositions in certain Euclidean
2
2
domains.
After dealing with uniqueness, we turn to the related question of when q = 0
ki
rij
f
in a partial fraction decomposition = q +
j . For real polynomials
g
a
i=1 j=1 i
we already know that q = 0 in the partial fraction decomposition of f (x)/g(x)
exactly when f (x)/g(x) is a proper fraction (i.e. deg(f (x)) < deg(g(x))).
Considering the example above, we see that an analogous statement does not
hold for integers.
Throughout the remainder of this section R is a Euclidean domain with
norm δ : R − {0} → Z≥0 . Also, f, g ∈ R and g = ak11 ak22 · · · ak where
a1 , . . . , a ∈ R are pairwise relatively prime. In this section all partial fraction
decompositions are relative to g = ak11 ak22 · · · ak .
If the division algorithm in R does not produce unique quotients and remainders, one should not expect partial fraction decompositions to be unique.
1527
Two proofs of existence and uniqueness
However, if one requires uniqueness for quotients and remainders, we will essentially be reduced to dealing with the case of polynomials.
Proposition 4.1. Let R be a Euclidean domain.
1. Quotients and remainders are unique in R if and only if δ(a + b) ≤
max{δ(a), δ(b)} for all a, b ∈ R − {0} such that a + b = 0.
2. If quotients and remainders are unique, then the units of R along with
zero form a field, say K. Moreover, either R = K or R is isomorphic to
K[x].
Proof: [Ra] and [J] As a first step towards proving the uniqueness of the partial fraction decomposition when R has unique quotients and remainders, we will show that
the sum of two partial fraction decomposition (with common denominators) is
again a partial fraction decomposition.
Lemma 4.2. Suppose that we have unique quotients and remainders in R and
f, h ∈ R.
ki
ki
rij
r̃ij
h
If fg = q +
and
=
q̃
+
are partial fraction decompositions,
g
aj
aj
i=1 j=1
then
f +h
g
i
= (q + q̃) +
ki
i=1 j=1
i=1 j=1
rij +r̃ij
aji
i
is a partial fraction decomposition.
Proof: For each i and j either rij + r̃ij = 0 or we have one the following cases:
rij = 0 and r̃ij = 0 Thus we have δ(rij + r̃ij ) = δ(rij ) < δ(ai ).
rij = 0 and r̃ij = 0 Thus we have δ(rij + r̃ij ) = δ(r̃ij ) < δ(ai ).
rij = 0 and r̃ij = 0] Thus we have δ(rij + r̃ij ) ≤ max{δ(rij ), δ(r̃ij )} < δ(ai ).
Theorem 4.3. When R has unique quotients and remainders, partial fraction
decompositions are unique.
Proof: Suppose q +
ki
i=1 j=1
rij
aji
=
f
g
compositions of f /g. Now, q − q̃ +
= q̃ +
ki
i=1 j=1
ki
i=1 j=1
rij −r̃ij
aji
=
r̃ij
aji
are partial fraction de-
f −f
g
=
0
g
= 0. By Lemma
4.2 the expression on the left hand side of the equation is a partial fraction
decomposition of 0/g.
1528
W. T. Bradley, W. J. Cook
Let us relabel , k1 , . . . , k , and a1 , . . . , a (reordering terms if necessary)
so that the sum over i no longer includes sums over j which are identically
zero and each sum over j stops at the last rij − r̃ij = 0. The equation q −
ki
ki
r̃ij −rij
is
equivalent
to
g
·
(q
−
q̃)
=
(r̃ij − rij ) · âij , where
q̃ =
aj
i
i=1 j=1
ki−1 ki −j ki+1
g
k1
âij = aj = a1 . . . ai−1
ai ai+1
i
ki
i=1 j=1
. . . ak .
(rij − r̃ij ) · âij +
g · (q − q̃) +
i=1
i=m
j=1
km
−1
j=1
Thus, we have
(rmj − r̃mj ) · âmj = (r̃mkm − rmkm ) · âmkm .
Notice that am divides the left hand side since am divides g and am divides
each âij except âmkm . Thus am divides the right hand side as well. Since am
is relatively prime to all ai for i = m, it cannot divide âmkm so it must divide
r̃mkm − rmkm . So there exists some u ∈ R such that am u = r̃mkm − rmkm . Now
r̃mkm − rmkm = 0 since our summation stopped at the last nonzero term and so
δ(r̃mkm − rmkm ) < δ(am ) by the definition of a partial fraction decomposition.
Therefore, δ(am ) ≤ δ(am u) = δ(r̃mkm − rmkm ) < δ(am ) which is impossible.
Therefore, all r̃ij − rij = 0 and so r̃ij = rij for all i, j. Thus q − q̃ = 0 and
so q = q̃ as well. Corollary 4.4. Partial fraction decompositions are unique for polynomials
with field coefficients.
Proposition 4.5. In Z, if we only allow non-negative remainders for division
and thus only allow nonnegative integers in the numerators of the fractional
part of partial fraction decompositions, then decompositions are unique.
Proof: If we force non-negative remainders, then 0 ≤ r < |a| and 0 ≤ r̃ < |a|
implies that |r − r̃| ≤ max{|r|, |r̃|} < |a| and the proof above applies. 1
77
2
1
+ 2 + is the unique decomposition of
2 2
3
12
(if we only allow nonnegative integers in numerators).
Example 4.6. 5 +
Beyond having unique quotients and remainders we need an additional
assumption to guarantee that proper fractions have q = 0 in their partial
fraction decompositions.
Theorem 4.7. Suppose R has unique quotients and remainders and for all
x, y, z ∈ R − {0} with δ(y) < δ(z) we have δ(xy) < δ(xz).
ki
rij
is a
If f /g is a proper fraction (i.e. δ(f ) < δ(g)) and fg = q +
aj
i=1 j=1
partial fraction decomposition, then q = 0.
i
1529
Two proofs of existence and uniqueness
Proof:
Clearing denominators, f = q · g +
ki
rij âij where âij =
g
aji
i=1 j=1
ki−1 ki −j ki+1
k
=
. . . ai−1 ai ai+1 . . . a . Suppose q = 0 so q · g = 0, then
ki
ki
f−
= δ(q · g).
j=1 rij âij = q · g and so δ f −
j=1 rij âij
i=1
i=1
Notice that g = âij aji and δ(rij ) < δ(ai ) ≤ δ(aji ). Thus by assumption δ(rij âij ) < δ(âij aji ) = δ(g). Next, recall that units do not effect the
ak11
norm’s value and that
and remain
R hasunique quotients
we have assumed
k
k
i
i
=
ders. Therefore, δ f −
rij âij ≤ max δ(f ), δ −
rij âij
i=1 j=1
i=1 j=1
ki
max δ(f ), δ
rij âij
≤ max{δ(f ), δ(rij âij )} < δ(g) since δ(f ) <
i=1 j=1
ij
δ(g) and δ(rij ĝij ) < δ(g) for all ij.
ki
= δ(q·g) ≥ δ(g) and so δ(g) < δ(g) which
However, δ f −
j=1 rij âij
i=1
is impossible. Therefore, q must be 0. Corollary 4.8. If R has unique quotients and remainders and either δ(xy) =
δ(x)δ(y) and δ(x) > 0 for all x, y ∈ R − {0} or δ(xy) ≤ δ(x) + δ(y) for
all x, y ∈ R − {0}, then q = 0 in partial fraction decompositions of proper
fractions.
Corollary 4.9. For polynomials with field coefficients, deg(f ) < deg(g) implies q = 0.
Remark 4.10. In light of proposition 4.1, we have q = 0 for proper fractions
whenever R has unique quotients and remainders.
Corollary 4.11. Let R = Z and suppose a1 , . . . , a ∈ Z>0 . Also, suppose
ki
rij
f
=q+
is a partial fraction decomposition with rij ≥ 0 for all ij.
g
aj
i=1 j=1
i
Then q = 0 if and only if f /g is nonnegative proper fraction.
Proof: We know by Proposition 4.5 that partial fraction decompositions are
unique under the assumption that rij ≥ 0.
Since |xy| = |x||y| for all x, y ∈ Z−{0}, notice that the proof of Theorem 4.7
applies for nonnegative integers. Thus if f /g is a nonnegative proper fraction,
then q = 0. [Note that if negative integers are allowed, |x − y| ≤ max{|x|, |y|}
may fail to hold. Thus the proof of Theorem 4.7 does not work for negative
integers.]
1530
W. T. Bradley, W. J. Cook
If f /g is nonnegative and f ≥ g (non-proper), then there exists (unique)
q·g+r
r
f
=q+ .
q, r ∈ Z such that f = q · g + r and 0 ≤ r < g. Thus =
g
g
g
Now as we have just shown, the partial fraction decomposition of r/g (which
is nonnegative and proper) has the property that “q = 0”. Thus the nonfractional part of the decomposition of f /g is q. Since f /g is not proper,
q = 0.
ki
rij
Finally if f /g is negative and rij ≥ 0, then q = fg −
< 0. Therefore,
aj
i=1 j=1
q = 0. i
1
1 1
Example 4.12. − = −1+ + is a partial fraction decomposition of −1/6.
6
2 3
Notice that q = −1 = 0 since −1/6 < 0. It is interesting to note that
− 16 = −1
+ 13 is also a partial fraction decomposition. This time q = 0.
2
However, we needed a negative numerator to get this.
5
A Few Applications
As mentioned in the introduction, the partial fraction decomposition of a real
rational function is usually discussed in one’s introduction to integral calculus and ordinary differential equations. This decomposition is primarily used
to facilitate the integration of rational functions or compute inverse Laplace
transforms. For example,
2x3 + 3x2 + 2x − 1
dx =
(x2 + 1)(x + 1)2
−1
2x
1
+
+C
dx = ln(x2 + 1) +
2
+ 1 (x + 1)
x+1
x2
In differential equations, one solves the logistic equation by first separating
variables and then uses the partial fraction decomposition to help integrate a
certain rational function.
Sometimes the partial fraction decomposition is used as an algebraic “trick”
for dealing with certain expressions. For example, it lets us see that the following series telescopes (and thus lets us find its sum):
∞
n=1
1
1
1
=
−
=
n(n + 1) n=1 n n + 1
∞
1 1
−
1 2
+
1 1
−
2 3
+
1 1
−
3 4
+··· = 1
We now turn to a few lesser known applications of the partial fraction
decomposition.
Two proofs of existence and uniqueness
5.1
1531
Linear Differential Equations with Constant Coefficients
Another interesting application to the field of differential equations is that
of inverting certain linear differential operators. The following technique will
allow us to solve any (homogeneous or nonhomogeneous) linear differential
equation with constant coefficients. In a typical differential equations course
one is taught to solve such equations using variation of parameters (or in
certain special cases using the method of undetermined coefficients). We offer
this alternative (see for example [I] section 6.2).
Suppose we have a linear differential equation with constant coefficients
y (n) + an−1 y (n−1) + · · · + a2 y + a1 y + a0 y = g(t). Then letting L = D n +
an−1 D n−1 + · · ·+ a2 D 2 + a1 D + a0 where D = d/dt, we can rewrite the equation
as L[y] = g(t). If L−1 existed we could write y = L−1 [g(t)] and “solve”
the equation. However, L has a nontrivial kernel (there are homogeneous
solutions), so technically L−1 does not exist. Let us try to attach some meaning
to L−1 anyway.
First, factor L. This can be done since we are dealing with constant coefficients and all operators commute. We will factor over the field of complex
number to simplify notation. L = (D − r1 )k1 (D − r2 )k2 · · · (D − r )k . Next,
treat D as a formal variable and perform the partial fraction decomposition on
ki
1/L. So that L−1 =
sij (D − ri )−j for some complex numbers sij . This
i=1 j=1
implies the following formal relation:
1=
ki
sij (D − r1 )k1 · · · (D − ri−1 )ki−1 (D − ri )ki −j (D − ri+1 )ki+1 · · · (D − r )k
i=1 j=1
(5.1)
Since the relation holds among formal variables, it also holds among operators. Note that in terms of operators, the left hand side is the identity
operator.
Consider the special case (D − r)k [y] = g(t). If k = 1, then we have
(D − r)[y] = y − ry = g(t). This is a first order linear differential equation which is easily solved using an integrating factor. In this case y =
ert e−rt g(t) dt. If k = 2, we have (D − r)2 [y] = (D − r)[(D − r)[y]] =
g(t). So (D − r)[y]
= ert e−rt g(t) dt
and−rtapplying this formula again yields
rt
−rt rt
−rt
rt
e e
e g(t) dt dt = e
e g(t) dt. In general, the solution of
y=e
k
(D − r) [y] = g(t) is
y = ert
· · · e−rt g(t) dt · · · dt dt.
k−fold
(5.2)
1532
W. T. Bradley, W. J. Cook
ki
sij yij . Then
· · · e−ri t g(t) dt · · · dt dt and y =
i=1 j=1
j−fold
k
i
sij yij
L[y] = (D − r1 )k1 · · · (D − r )k
ri t
Let yij = e
i=1 j=1
=
ki
sij (D − r1 )k1 · · · (D − ri−1 )ki−1 (D − ri )ki −j (D − ri+1 )ki+1 · · · (D −
i=1 j=1
ki
sij (D − r1 )k1 · · · (D − ri−1 )ki−1 (D − ri )ki −j (D −
r )k (D − ri )j [yij ] =
i=1 j=1
ri+1 )ki+1 · · · (D − r )k [g(t)] = g(t)
The second equality follows from linearity and the fact that these operators
commute, the third equality follows from 5.2, and the fourth equality follows
from 5.1.
We should note that each j-fold indefinite integral will add j arbitrary
constants to the solution. Also, integrals coming from the same repeated
factor of the linear operator can introduce redundant constants. With a little
more care this can be avoided.
Example 5.1. Consider y + y − 5y + 3y = g(t). Then L = D 3 + D 2 −
5D + 3 = (D − 1)2 (D + 3). The partial fraction decomposition of 1/L is
1
(−1/16)
(1/4)
1
(1/16)
=
=
+
+
2
2
L
(D − 1) (D + 3)
D−1
(D − 1)
D+3
1
(D − 1)−1 [g(t)] + 14 (D − 1)−2 [g(t)] +
So our solution looks like y = L−1 [y] = − 16
1
(D + 3)−1 [g(t)]. Thus the general solution is
16
1 t
1 t
1 −3t
−t
−t
y=− e
e g(t) dt + e
e g(t) dt dt + e
e3t g(t) dt
16
4
16
Remark 5.2. If we want to avoid complex functions appearing in our solution,
we will also need to know how to invert second order differential operators
related to irreducible quadratic factors.
√
Consider (D−z)(D−z̄) where z = a+bi and z̄ = a−bi and i = −1. It can
be shown by performing the partial fraction decomposition on 1/[(D−z)(D−z̄)],
applying the previous formula for a single factor, and some algebra that the
solution of (D − z)(D − z̄)[y] = g(t) is
1 at
1 at
−at
y = e sin(bt) e cos(bt)g(t) dt − e cos(bt) e−at sin(bt)g(t) dt
b
b
To deal with irreducible quadratic factors with multiplicity two or greater,
one just iterates.
1533
Two proofs of existence and uniqueness
5.2
Chinese Remaindering and Hermite Interpolation
Once the partial fraction decomposition is established, it is easy to prove the
Chinese remainder theorem. In fact this gives a method for computing the
solution which the theorem claims exists.
Note that in a principal ideal domain R the congruence x ≡ y (mod z)
means that there exists some k ∈ R such that x = y + zk (x and y are off by
a multiple of z).
Theorem 5.3 (Chinese Remainder Theorem). Let R be a principal ideal
domain with a1 , . . . , a ∈ R pairwise relatively prime elements. In addition
suppose that b1 , . . . , b ∈ R. Then the system of congruences x ≡ bi (mod ai )
for i = 1, . . . , has a simultaneous solution in R.
Proof: By Proposition 3.6 there exists A1 , . . . , A ∈ R such that
1
A1
A
=
+···+
a1 · · · a
a1
a
where this equality holds in the field of fractions of R. Note: By Remark 3.7
proposition 3.6 holds over all PIDs not just Euclidean domains.
Clearing denominators we have 1 = A1 a2 · · · a + A2 a1 a3 · · · a + · · · +
A a1 · · · a−1 . Let ci = Ai a1 · · · ai−1 ai+1 · · · a and so 1 = c1 + c2 + · · · + c .
Notice that cj ≡ 0 (mod ai ) when i = j since cj contains ai as a factor for
i = j. Likewise, bj cj ≡ 0 (mod ai ). Therefore, 1 = c1 + · · · + c ≡ ci (mod ai )
for each i and so x = b1 c1 + · · · + b c ≡ bi ci ≡ bi · 1 (mod ai ) . Therefore, x = b1 c1 + · · · + b c ∈ R is a simultaneous solution of the congruences
x ≡ bi (mod ai ) where i = 1, . . . , . This result can be somewhat refined over a Euclidean domain. Let a1 , . . . , a
be pairwise relatively prime elements of the Euclidean domain R and ki ∈ Z>0
(positive integers). Let bij ∈ R where i = 1, . . . , and j = 1, . . . , ki. Then
there exists rij ∈ R such that
i
1
rij
=
k
k1
aj
a1 · · · a
i=1 j=1 i
k
where each rij = 0 or δ(rij ) < δ(ai ). Clearing denominators this gives us
ki
ki−1 ki −j ki+1
rij âij where âij = ak11 · · · ai−1
ai ai+1 · · · ak .
1=
i=1 j=1
Consider the following element of R:
ki
ki
rim âim
bij aj−1
x=
i
i=1 j=1
m=j
(5.3)
1534
W. T. Bradley, W. J. Cook
Fix 1 ≤ s ≤ and 1 ≤ t ≤ ks . Then since aks s divides âij for all j = 1, . . . , ki
when i = s and also t ≤ ks
, we have âij ≡ 0 (mod ats ) for i = s. Therefore,
ks
ks
x≡
bsj aj−1
rsm âsm (mod ats ).
s
j=1
m=j
ks −m+j−1
and so aj−1
In addition notice that aj−1
s âsm is divisible by as
s âsm ≡
ks
0 (mod as ) when ks − m + j − 1 ≥ ks . This is equivalent to m ≤ j − 1.
Therefore, we can change the lower limit in the sum over m from m = j to
m = 1
without changing the
value of x modulo ats (for all t ≤ ks ). Thus
ks
ks
rsm âsm (mod ats ).
x≡
bsj aj−1
i
j=1
m=1
ki
ks
rsm âsm (mod ats ). Therefore,
ki
ki
ks · 1 (mod ats ). In summary, x =
rim âim
bsj aj−1
x≡
bij aj−1
i
i
Finally note that 1 =
i=1 m=1
rim âim ≡
m=1
j=1
i=1 j=1
simultaneously solves x ≡
t
m=j
t
bsm am
s (mod as ) for all s = 1, . . . , and t =
m=1
1, . . . , ks .
This refinement of Chinese remaindering gives us the following theorem.
Theorem 5.4 (Hermite Interpolation). Given real numbers xi , fij ∈ R
where i = 1, . . . , , j = 0, . . . , ki − 1, and xi = xj for i = j, there exists a
polynomial f (x) ∈ R[x] such that f (x) = 0 or deg(f (x)) ≤ k1 + k2 + · · ·+ k − 1
and f (j) (xi ) = fij for all i = 1, . . . , and j = 0, . . . , ki − 1.
Proof: Let g(x) = (x − x1 )k1 · · · (x − x )k and ĝij (x) = g(x)/(x − xi )j . We
can compute the partial fraction decomposition of 1/g(x), clear denominators,
ki
and get 1 =
rij ĝij (x) where rij ∈ R (since either rij = 0 or deg(ri j) <
i=1 j=1
deg(x − xi ) = 1).
By the above discussion, f (x) =
k
i −1
i=1 j=0
multaneously solves the congruences f (x) ≡
fij
(x
j!
t−1
− xi )j
m=j+1
fsm
(x − xs )m
m!
m=0
+ f2s2 (x
ki
rim ĝim (x)
si-
(mod (x − xs )t )
f
s −1)
− xs )2 + · · · + (ks(ks −1)!
(x −
In particular f (x) = fs0 + fs1 (x − xs )
ks −1
ks
is the Taylor expansion of f (x) about x = xs (thus
+ O (x − xs )
xs )
f (x) has the requisite derivatives).
In addition note that the degree of (x − xi )j ĝim (x) is k1 + · · · + ki−1 + (ki −
m + j) + ki+1 + · · · + k which never exceeds k1 + · · · + k − 1 since j < m. Two proofs of existence and uniqueness
1535
References
[DF]
D.S. Dummit and R.M. Foote, Abstract Algebra, 3rd ed., John Wiley
& Sons, Hoboken, 2004.
[I]
E.L. Ince, Ordinary Differential Equations, Dover, 1956.
[J]
M.A. Jodeit Jr., “Uniqueness in the Division Algorithm”, American
Mathematical Monthly, Vol. 74, 1967, No. 7, 835–836.
[PW]
R.W. Packard and E. Stephen, “Partial fractions in Euclidean domains”, Math. Mag. 62 (1989), no. 2, 115–118.
[Ra]
T.-S. Rhai, “A Characterization of Polynomial Domains Over a
Field”, American Mathematical Monthly, Vol. 69, 1962, No. 10, 981–
986.
[Ro]
K. Rogers, “The Axioms for Euclidean Domains”, American Mathematical Monthly, Vol. 78, 1971, No. 10, 1127–1128.
Received: October, 2011