Partial Fraction Expansion via Differentiation
2.2
2002/03/04
2004/08/10 10:02:59.199 GMT-5
Charlet
Reedstrom
charlet@rice.edu
Charlet
Reedstrom
charlet@rice.edu
(Blank Abstract)
Introduction and Terminology
With the tools of
complex functions
we may produce complex polynomials:
f
z
z
m
c
m
-
1
z
m
1
…
c
1
z
c
0
We say that such an f is of
order m.
We shall often find it convenient to represent polynomials as
the product of their factors, namely:
f
z
z
λ
1
d
1
z
λ
2
d
2
…
z
λ
h
d
h
Each
λ
j
is a root of degree
d
j
. Here
h is the number of
distinct roots of
f. We call
λ
j
a simple root when
d
j
1
. We can observe the appearance of ratios of
polynomials or so called rational functions.
Suppose
q
z
f
z
g
z
is rational, that
f
is of order at most
m
1
while
g
is of order
m
with the simple roots
λ
1
…
λ
m
.
Simple Roots
It should come as no surprise that such a
q
should admit a Partial Fraction Expansion
q
z
j
1
m
q
k
z
λ
j
One uncovers the
q
j
by first multiplying each side by
z
λ
j
and then letting
z
tend to
λ
j
. For example, if
1
z
2
1
q
1
z
i
q
2
z
i
then multiplying each side by
z
i
produces
1
z
i
q
1
q
2
z
i
z
i
Now, in order to isolate
q
1
it is clear that we should set
z
i
.
So doing we find
q
1
i
2
.
In order to find
q
1
we multiply the above
equality by
z
i
and then set
z
i
. So doing we find
q
2
i
2
, and so
1
z
2
1
i
2
z
i
i
2
z
i
Returning to the general case, we encode the above in the
simple formula
q
j
z
λ
j
z
λ
j
q
z
You should be able to use this to confirm that
z
z
1
2
12
z
1
12
z
1
Multiple Roots
q
1
,
1
and
q
1
,
2
in the expansion
z
2
z
1
2
q
1
,
1
z
1
q
1
,
2
z
1
2
Arguing as above it seems wise to multiply through by
z
1
2
and so arrive at
z
2
q
1
,
1
z
1
q
1
,
2
On setting
z
-1
this gives
q
1
,
2
1
.
With
q
1
,
2
computed the previous equation takes the simple form
z
1
q
1
,
1
z
1
and so
q
1
,
1
1
as well. Hence:
z
2
z
1
2
1
z
1
1
z
1
2
This latter step grows more cumbersome for roots of higher
degree. Let us consider
z
2
2
z
1
3
q
1
,
1
z
1
q
1
,
2
z
1
2
q
1
,
3
z
1
3
The first step is still correct: multiply through by the factor
at its highest degree, here
3
. This leaves us with
z
2
2
q
1
,
1
z
1
2
q
1
,
2
z
1
q
1
,
3
Setting
z
-1
again produces the last coefficient, here
q
1
,
3
1
. We are left however with one equation in two
unknowns. Well, not really one equation, for the previous
equation is to hold for all
z
.
We exploit this by taking two derivatives, with respect to
z
, of this equation. This produces
2
z
2
2
q
1
,
1
z
1
q
1
,
2
and
2
q
1
,
1
The latter of course needs no comment. We derive
q
1
,
2
from the former by setting
z
-1
. This example will permit us to derive a simple
expression for a partial fraction expansion.
Partial Fraction Expansion The partial
fraction expansion of a general proper rational function
q
f
g
where
g
has
h
distinct roots
λ
1
…
λ
h
of respective degrees
d
1
…
d
h
can be written as
q
z
j
1
h
k
1
d
j
q
j
,
k
z
λ
j
k
and it can be noted, as above, that
q
j
,
k
is the coefficient of
z
λ
j
d
j
k
in the rational function
r
j
z
q
z
z
λ
j
d
j
. Hence,
q
j
,
k
may be computed by setting
z
λ
j
in the ratio of the
d
j
k
th derivative of
r
j
to
d
j
k
. That is,
q
j
,
k
z
λ
j
1
d
j
k
z
d
j
k
z
λ
j
d
j
q
z
.
Connection to Residues
With respect to the previous definition observe that if we choose
r
j
so small that
λ
j
is the only zero of
g
encircled by
R
j
C
λ
j
r
j
then by Cauchy's Theorem