Partial Fractions
ü
Copyright Brian G. Higgins (2003)
Introduction
The method of partial fractions allows one to simplify proper rational functions of the form
P
H
s
L ê
Q
H
s
L
. Partial
fraction decomposition of rational polynomials is very helpful for solving integrals and finding the inverse Laplace
Transform of polynomial functions.
The idea behind
partial fraction decomposition is founded in the addition of fractions:
1
ÅÅÅÅ
2
+
1
ÅÅÅÅ
5
=
5
ÅÅÅÅÅÅÅ
10
+
2
ÅÅÅÅÅÅÅ
10
=
7
ÅÅÅÅÅÅÅ
10
The partial fraction decomposition of
7
ÅÅÅÅÅ
10
is
7
ÅÅÅÅÅÅÅ
10
=
1
ÅÅÅÅ
2
+
1
ÅÅÅÅ
5
The same concept carries over for algebraic fractions
2
ÅÅÅÅ
s
+
s
ÅÅÅÅÅÅÅÅÅÅÅÅÅ
s
2
+
3
=
2
H
s
2
+
3
L
ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅ
s
H
s
+
3
L
+
s
2
ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅ
s
H
s
2
+
3
L
=
3
s
2
+
6
ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅ
s
H
s
2
+
3
L
Thus the partial fraction decomposition of
H
3
s
2
+
6
L ê
s
H
s
2
+
3
L
is
P
H
s
L
ÅÅÅÅÅÅÅÅÅÅÅÅÅ
Q
H
s
L
=
3
s
2
+
6
ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅ
s
H
s
2
+
3
L
=
2
ÅÅÅÅ
s
+
s
ÅÅÅÅÅÅÅÅÅÅÅÅÅ
s
2
+
3
In this notebook the theoretical basis for the method of partial fractions is reviewed using 4 key concepts from
calculus/algebra. We also show how
Mathematica
's
Apart
and
Factor
functions can be used to facilitate working with
partial fractions.
Mathematical Concepts
‡
Concept 1
The fundamental theorem of algebra(FTA) states that a nth degree polynomial
(1)
Q
H
s
L
=
a
n
s
n
+
a
n

1
s
n

1
…
+
a
1
s
+
a
0
has
n
roots, although some may be repeated. If
s
i
is a root of Q(s), then
H
s

s
i
L
is a factor of Q(s).
We say
s
i
is a root of multiplicity
m
i
of Q(s) if
m
i
is the largest positive integer such that
H
s

s
i
L
m
i
is a factor of Q(s).
These ideas can be expressed mathematically as
(2)
Q
H
s
L
=
A
‰
i
=
1
M
H
s

s
i
L
m
i
where
A
is a constant. Eqn(2) Is a mathematical statement of the FTA. In general the roots
s
i
are complex, and
m
i
is
the multiplicity of root
s
i
such that
(3)
n
=
‚
i
=
1
M
m
i
PartialFractions.nb
1
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m
i
=
1
for all i,
and
n
=
M
so that
(4)
Q
H
s
L
=
A
‰
i
=
1
n
H
s

s
i
L
‡
Example 1:
Roots of polynomials
Show that the following polynomials can be expressed as
(5)
Q
H
s
L
=
A
‰
i
=
1
M
H
s

s
i
L
m
i
(6)
(i)
s
3

s
,
(ii)
s
2
+
s

12
,
(iii)
8 s
2
+
12 s
3
+
6 s
4
+
s
5
(iv)
2
s
2
+
4
s
+
8
, (v)
16 s
+
8 s
3
+
s
5
For each case determine
A, n
, and
m
i
ü
Solution Problem (i)
The function for factoring a polynomial over the integers is Factor. We can use this function on the polynomial
s
3

s
as
follows:
Factor
@
s
3

s
D
H

1
+
s
L
s
H
1
+
s
L
Thus
A
=
1
,
n
=
3, m
i
=
1
Alternatively, we can use Solve to find the factors of the
polynomial equation:
Solve
@
s
3

s
ã
0
D
88
s
Ø 
1
<
,
8
s
Ø
0
<
,
8
s
Ø
1
<<
However, using Solve does not determine the factor A!
ü
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 Spring '10
 Franics
 Algebra, Fraction, Rational function, Partial fractions in complex analysis, Partial fractions in integration

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