Partial Fractions Expansion [expository notes]

2 Pages·2007·0.064 MB·English

by  David G. Wagner

Partial Fractions Expansion
Theorem: Let F(x) = P(x)/Q(x) be a quotient of polynomials in which degP <
degQ and the constant term of Q(x) is 1. Factor the denominator to obtain its
inverse roots:
Q(x) = (1−λ x)d1(1−λ x)d2···(1−λ x)ds
1 2 s
where λ ,...,λ are distinct nonzero complex numbers and d +···+d = d = degQ.
1 s 1 s
Then there are d complex numbers:
C(1),C(2),...,C(d1); C(1),C(2),...,C(d2); ... ; C(1),C(2),...,C(ds);
1 1 1 2 2 2 s s s
(which are uniquely determined) such that
F(x) = P(x) = Xs Xds Ci(j) .
Q(x) (1−λ x)j
i=1j=1 i
Proof: Consider the set V of all quotients of polynomials P(x)/Q(x) in which
Q
degP < d = degQ, where the denominator is some polynomial which remains fixed
throughout the proof and satisfies the hypothesis of the theorem. It is easily seen
that V is a vector space over the field C, since if P(x) and R(x) both have degree
Q
less than d and α ∈ C then P(x)+αR(x) has degree less than d. It is also clear that
the vectors
1 x x2 xd−1
, , , ...,
Q(x) Q(x) Q(x) Q(x)
span V as a vector space over C. Therefore, the dimension of V is at most d.
Q Q
Now we claim that for every 1 ≤ i ≤ s and 1 ≤ j ≤ d , the quotient 1/(1−λ x)j is
i i
in V . This is because
Q
1 (1−λ x)di−jQ (1−λ x)dh
= i h6=i h
(1−λ x)j Q(x)
i
and the numerator has degree d−j ≤ d−1 < d.
The essential point in the proof is that the set of vectors
( )
1
B = : 1 ≤ i ≤ s and 1 ≤ j ≤ d
(1−λ x)j i
i
in V is linearly independant. From this we can conclude that the dimension of V
Q Q
is at least d +···+d = d. It then follows that dimV = d and that B is a basis for
1 s Q
V . Therefore, every vector in V can be written uniquely as a linear combination of
Q Q
vectors in B. That is exactly what the Partial Fractions Theorem is claiming.
It remains only to show that B is a linearly independant set. Consider any linear
combination of vectors in B which gives the zero vector:
Xs Xds Ci(j) = 0. (1)
(1−λ x)j
i=1j=1 i
We must show that C(j) = 0 for all 1 ≤ i ≤ s and 1 ≤ j ≤ d . Suppose not. Then
i i
there is some coefficient C(t) 6= 0 with 1 ≤ p ≤ s and 1 ≤ t ≤ d which also satisfies
p p
C(t+1) = ··· = C(dp) = 0. Now multiply equation (1) by (1−λ x)t. Separating out
p p p
the terms with i = p and using the fact that C(t+1) = ··· = C(dp) = 0, we see that
p p
Xt C(j)(1−λ x)t−j +XXdi C(j)(1−λpx)t = 0.
p p i (1−λ x)j
j=1 i6=pj=1 i
The left-hand side of this equation is a rational function of the variable x which does
not have a pole at the point x = 1/λ , so we can substitute this value for x. But
p
every term has a factor of (1−λ x) except for the term with i = p and j = t. Thus,
p
this equation becomes
C(t) = 0
p
after substituting the value x = 1/λ . But this contradicts our choice of p and t. This
p
contradiction shows that all the coefficients C(j) in equation (1) must be zero, and it
i
follows that B is linearly independant.
Since B is a set of d linearly independant vectors in a vector space V of dimension
Q
at most d, it follows that B is a basis for V , and the proof is complete. (cid:4)
Q