— Partial Fractions Review —
Introduction.
The goal of
partial fraction decomposition
is to rewrite a rational function in a sim
pler form: split it into a sum of smaller rational functions where each denominator is an irreducible
polynomial. Here,
irreducible
means that it cannot be factored as a product of smaller polynomials.
Definition 1.
A
rational function
is a fraction where both the numerator and denominator are (real)
polynomials. If the numerator is a polynomial
P
(
s
) and the denominator is a polynomial
Q
(
s
), then
R
(
s
)
defined by
R
(
s
)
=
P
(
s
)
Q
(
s
)
=
p
0
+
p
1
s
+
p
2
s
2
+
· · ·
+
p
m
s
m
q
0
+
q
1
s
+
q
2
s
2
+
· · ·
+
q
‘
s
‘
,
p
i
,
q
i
∈
R
,
is a rational function.
Theorem 2
(Fundamental Theorem of Algebra)
.
Suppose P
(
s
)
=
p
0
+
p
1
s
+
p
2
s
2
+
· · ·
+
p
n
s
n
is a
polynomial of degree n with p
i
∈
R
. There exists a unique set of numbers
{
a
0
,
a
1
, . . . ,
a
k
} ⊆
C
such that
P
(
s
)
factors into a product of linear terms
P
(
s
)
=
a
0
k
Y
i
=
1
(
s

a
i
)
ε
i
=
a
0
(
s

a
1
)
ε
1
(
s

a
2
)
ε
2
. . .
(
s

a
k
)
ε
k
where the degrees match:
ε
1
+
ε
2
+
· · ·
+
ε
k
=
n,
ε
i
>
0
. In this case, each a
i
is a pole of P
(
s
)
.
However, if you are unwilling to deal with complex numbers, then you may be stuck with quadratic
factors that are not reducible in terms of reals (i.e., that are “irreducible over
R
”). For example, you can
factor
s
2
+
1
=
(
s

)(
s
+
) over
C
, but you cannot write
s
2
+
1 in terms of linear factors with real
coe
ffi
cients. More generally, you have the following:
Corollary 3.
Suppose P
(
s
)
=
p
0
+
p
1
s
+
p
2
s
2
+
· · ·
+
p
n
s
n
is a polynomial of degree n with p
i
∈
R
. There
exists a unique set of real numbers
{
a
0
,
a
1
, . . . ,
a
j
,
c
1
,
c
2
, . . . ,
c
k
}
such that P
(
s
)
factors into a product of
linear and irreducible quadratic terms
P
(
s
)
=
a
0
h
(
s

a
1
)
ε
1
(
s

a
2
)
ε
2
. . .
(
s

a
j
)
ε
j
i h
(
s
2
+
β
1
s
+
c
1
)
δ
1
(
s
2
+
β
2
s
+
c
2
)
δ
2
. . .
(
s
2
+
β
m
s
+
c
m
)
δ
m
i
where the degrees match:
ε
1
+
ε
2
+
· · ·
+
ε
j
+
2
δ
1
+
2
δ
2
+
· · ·
+
2
δ
m
=
n
,
ε
i
, δ
i
>
0
.
Here, each a
i
is a pole of P
(
s
)
, and each quadratic factor corresponds to a
pair
of complex conjugate
poles of P
(
s
)
.
Cor.
3
makes partial fractions a bit more obnoxious whenever there are irreducible quadratic factors.
Since we want to factor the denominator
Q
(
s
), let’s fix the notation and assume we have
Q
(
s
) factored
like so:
Q
(
s
)
=
a
0
j
Y
i
=
1
(
s

a
i
)
ε
i
·
m
Y
k
=
1
((
s

a
k
)
2
+
b
2
k
)
δ
k
(1)
Recall that one can always write the quadratic terms like this by completing the square:
s
2
+
β
i
s
+
c
i
=
s
2
+
β
i
s
+
β
i
2
2

β
i
2
2
!
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 Fall '08
 DICKEY
 Algebra, Fractions, Rational Functions, Fraction, Elementary algebra, Rational function, Partial fractions in integration

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