19.Appendix C.Polynomials and rational functions

This version: 22/10/2004

Appendix C

Polynomials and rational functions

We shall be frequently be encountering and manipulating polynomials, so it will be helpful to have on hand some basic facts concerning such objects. That polynomials might be useful to us can be seen by noting that we have already defined the characteristic polynomial which will be very important to us in these notes.

		Contents
C.1	Polynomials . . . . . . . . . . . . .	. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 609
C.2	Rational functions . . . . . . . . . .	. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 611

C.1 Polynomials

We denote by F[ξ] the set of polynomials in indeterminant ξ and with coe cients in F, where F is either R or C. Thus a typical element of F[ξ] looks like

P (ξ) = akξk + ak−1ξk−1 + · · · + a1ξ + a0

(C.1)

where a0, . . . , ak F with ak 6= 0. We call k the degree of P which we denote by deg(P ). You should not think of ξ as being an element of F, but rather as just being a placeholder. If we wish to plug in values from F into P we shall generally say when we do this. Of course, you can add and multiply polynomials in just the ways with which you are familiar.

A polynomial P of the form (C.1) is monic if ak = 1; that is, a monic polynomial is one where the coe cient of the highest power of the indeterminant is +1. For example, the characteristic polynomial is always a monic polynomial.1 Given two polynomials P1, P2 F[ξ] their least common multiple (LCM ) is the unique monic polynomial Q F[ξ] of least degree with the property that Q = P1R1 and Q = P2R2 for some R1, R2 F[ξ].

C.1 Examples In each of these examples, the polynomials may be thought of as in either R[ξ] or C[ξ].

1. If

P1(ξ) = ξ + 2, P2(ξ) = 2ξ − 3,

then the LCM of P1 and P2 is Q(ξ) = ξ2 − 12 ξ − 32 .

1This is why we defined PA(λ) by det(λIn − A) instead of by det(A − λIn).

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2.	If
	P1(ξ) = ξ2 + 2ξ + 1, P2(ξ) = ξ + 1,
	then the LCM of P1 and P2 is Q(ξ) = ξ2 + 2ξ + 1.
	If P F[ξ] is given by
	P (ξ) = akξk + ak−1ξk−1 + · · · + a1ξ + a0,

a root for P is a number α F with the property that

akαk + ak−1αk−1 + · · · + a1α + a0 = 0.

We denote the set of roots of P by spec(P ), mirroring our notation for eigenvalues for matrices. If α is a root of P then there exists Q F[ξ] so that P (ξ) = (ξ − α)Q(ξ). If α is a root of P and if m is the unique integer with the property that P (ξ) = (ξ − α)mQ(ξ) where α is not a root of Q, then m is the multiplicty of the root. If F = C, a root α C is (1) in the positive half-plane if Re(α) > 0, (2) in the negative half-plane if Re(α) < 0, and

(3) on the imaginary axis if Re(α) = 0. We denote the positive half-plane by C+ and the negative half-plane by C−. We shall often also denote the imaginary axis by iR (meaning {iω | ω R}). By C+ we mean the positive half-plane along with the imaginary axis, and similarly be C− we mean the negative half-plane along with the imaginary axis.

The greatest common divisor (GCD) of two polynomials P1, P2 F[ξ] is the unique monic polynomial T of greatest degree so that P1 = T Q1 and P2 = T Q2 for some Q1, Q2 F[ξ].

C.2 Examples In each of these examples, the polynomials may be thought of as in either R[ξ] or C[ξ].

1. If

P1(ξ) = ξ2 + 1, P2(ξ) = 3ξ

then the GCD of P1	and P2	is T (ξ) = 1.
2. If	P1(ξ) = 3ξ2 + 6ξ + 3, P2(ξ) = 2ξ2 − 6ξ − 8,
then the GCD of P1	and P2	is T (ξ) = ξ + 1.

If the GCD of polynomials P1 and P2 is T (ξ) = 1, then P1 and P2 are said to be coprime. A polynomial P F[ξ] with deg(P ) > 0 is irreducible if there are no polynomials P1, P2 F[ξ], both with degree less than P , with the property that P = P1P2. Thus an irreducible polynomial cannot be factored.

C.3 Examples 1. By the Fundamental Theorem of Algebra, the only irreducible monic polynomials in C[ξ] are of the form P (ξ) = ξ + a for some a C.

2.As we know, there are polynomials of degree two in R[ξ] which are irreducible. For example P (ξ) = ξ2 + 1 is irreducible in R[ξ], but not in C[ξ]. The irreducible monic polynomials in R[ξ] are of the form

(a)P (ξ) = ξ + a for some a R or

(b)P (ξ) = ξ2 + bξ + c where b, c R satisfy b2 − 4c < 0.

One can easily show that an irreducible monic polynomial of degree 2 in R[ξ] must have the form (ξ − σ)2 + ω2 for σ R and ω > 0 (see Exercise EC.1).

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The following result is related to the Euclidean algorithm for F[ξ] which, you will recall, states that for any polynomials P, Q F[ξ] there exists polynomials F, R F[ξ] with the properties

1.P = F Q + R and

2.deg(D) < deg(Q).

That is to say, a polynomial can be written as a product of a given polynomial and a remainder. With this in mind, we can prove the following result.

C.4 Lemma Let P1, P2, F F[ξ] be polynomials with P1 and P2 coprime. Then there exists Q1, Q2 F[s] so that

(i)	deg(Q1) < deg(P2) and
(ii)	Q1P1 + Q2P2 = F .
		˜	˜
Proof Since P1 and P2 are coprime it follows that there exists Q1			, Q2 R[s] so that
	˜	˜
	Q1P1	+ Q2P2 = 1.
Therefore we have
	˜	˜
	(Q1F )P1	+ (Q2F )P2 = F,
from which it follows that for any G R[s] we have
	˜	˜
	(Q1F − P2G)P1	+ (Q2F + P1G)P2 = F.
The Euclidean algorithm asserts that there exists a G, R F[ξ] so that
	˜
	Q1F = GP2 + R
and deg(R) < deg(P2). That is to say, there exists G F[ξ] so that
	˜
	deg(Q1F − P2G) < deg(P2).
	˜	˜
Choosing this G and then defining Q1 = Q1F −P2G and Q2 = Q2F +P1G gives the result.

C.2 Rational functions

We also wish to talk about objects which are quotients of polynomials. A rational function over F with indeterminate ξ is a quotient of two elements of F[ξ]. Thus we write a rational function over F as

R(ξ) = ND((ξξ)), N, D F[ξ].

Thus

R(ξ) = akξk + ak−1ξk−1 + · · · + a1ξ + a0 b`ξ` + b`−1ξ`−1 + · · · + b1ξ + b0

where a0, . . . , ak, b0, . . . , b` F, and not all the bi’s are zero. We denote the set of rational functions over F with indeterminate ξ by F(ξ). One should take care not to unduly concern oneself about things like whether the rational function blows up for certain values of ξ where

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D(ξ) = 0. As a polynomial, the only polynomial which is zero is the zero polynomial. If P F[ξ] is a non-zero polynomial, then the two rational functions

R1 =	P N	,	R2 =	N
	P D			D

will in fact represent the same rational function. This is exactly the same thing we do when we say that 13 and 26 are the same rational number. Note that for R F(ξ) there are unique coprime monic polynomials N, D F[ξ] so that

	R = a		N				(C.2)
			D
for some a F.
C.5 Example Consider the rational function
R(ξ) =	2ξ2 − 2ξ − 12					.
	3ξ2 − 15ξ + 18
We can write this as		ξ + 2
R(ξ) = 2
	3 ξ − 2
which is the unique representation of the form (C.2).

Given a rational function R F(ξ) with a F and N, D F[ξ] defined by (C.2), we call (aN, D) the canonical fractional representative of R. We will be frequently in need of this simple concept, so shall abbreviate if c.f.r.. A rational function R with c.f.r. (N, D) is proper if deg(N) ≤ deg(D), and strictly proper if deg(N) < deg(D). A rational function which is not proper is improper.

Let R F(ξ) which we write in its unique representation (C.2) for some coprime monic polynomials N, D F[ξ]. A zero of R is defined to be a root of N and a pole of R is defined to be a root of D. Note that in this way we get around the problem of the “function” R not being defined at poles. Two rational functions R1, R2 F(ξ) are coprime if they have no common zeros.

The final thing we do is provide a discussion of the so-called “partial fraction expansion.” Recall that the idea here is to take a rational function and expand it as a sum of rational functions whose denominators are powers of an irreducible polynomial. Thus, for example

ξ3 − 3ξ + 2	=	1		+ 5	1	+	19 1				+	1		1	.
ξ3 − 5ξ2 + 3ξ + 9 1					(ξ − 3)2			4 ξ − 3 4 ξ + 1

It is hard to come across an accurate description of how this is done, so let us provide one here.

C.6 Theorem Let R F(ξ) and suppose that

R = ND

where N, D F[ξ] are coprime, and take D to be monic. There exists

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613

(i)m irreducible monic polynomials D1, . . . , Dm F[ξ],

(ii)positive integers j1, . . . , jm,

(iii)j1 + · · · + jm polynomials N1,1(x), . . . , N1,j1 , . . . , Nm,1, . . . , Nm,jm F[ξ], and

(iv)a polynomial Q F[ξ] of degree deg(N)−deg(D) (take Q = 0 if deg(N)−deg(D) < 0),

with the properties

(v)D1, . . . , Dm are coprime (i.e., distinct),

(vi)deg(Ni,k) < deg(Di) for i = 1, . . . , m and k = 1, . . . , ji,

(vii) Ni,k and Di are coprime for i = 1, . . . , m and k = 1, . . . , ji, and

m ji

(viii) R = X X Ni,k + Q.

i=1 k=1 (Di)k

Furthermore the objects described in (i)–(iv) are the unique such objects with the properties (v)–(viii).

The expression in part (viii) is called the partial fraction expansion of R. The proof of this is straightforward, but requires some buildup, and we refer to [Lang 1984, Theorems 5.2 and 5.3].

It will turn out that we are primarily interested in the case when deg(N) ≤ deg(D), and in this case Q will be a constant, possibly zero, when deg(N) = deg(D), and zero when deg(N) < deg(D). For a rational function R C(ξ) which satisfies our degree condition, Theorem C.6 tells us that we may write

m	jm
X X −
R(ξ) =		βi,k	+ β	(C.3)
		(ξ αi)k
i=1 k=1

for some uniquely defined complex numbers β, αi, i = 1, . . . , m, and βi,k, i = 1, . . . , m, k = 1, . . . , ji. For R R(ξ), things are a bit more complicated. We may write

r ji

βi,k

m `i

ai,kξ + bi,k

X X

X X

+ b

(C.4)

−

αi)k

−

σi)2

i

R(ξ) = i=1 k=1 (ξ

+ i=1 k=1 (ξ

+ ω2

k

for real numbers αi, i = 1, . . . , r, βi,k, i

= 1, . . . , r, k

= 1, . . . , ji, ai,k, bi,k, i = 1, . . . , m,

k = 1, . . . , `i, σi, ωi, i = 1, . . . , `, and b.

This sort of leaves open how we compute the constants in the partial fraction expansion

for R = N/D. We shall say here how to do it when R		C	(ξ). In this case α1, . . . , αm are
				ji	Q(ξ) where
the poles of R, and ji is the multiplicity of the pole αi. That is, D(ξ) = (ξ − αi)
Q and (ξ − αi) are coprime. It turns out that
1	dk−ji				(C.5)
βi,ji = (k − ji)! dξk−ji (ξ − α)kR(ξ) ξ=αi

As usual, this is self-explanatory in examples.

C.7 Examples 1. First let us show that one cannot dispense with the constant term if the numerator and denominator polynomials have the same degree. If we take

R(ξ) = ξ + 1 ξ + 2

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then its partial fraction expansion is

R(ξ) = 1 −

1

.

ξ + 2

2.

We take

5ξ + 4

R(ξ) =

.

ξ2 + ξ − 2

The first thing to do is factor the denominator: ξ2 + ξ − 2 = (ξ − 1)(ξ + 2). Thus in the

parlance of (C.3) we have α1 = 1 and α2 = −2. These roots have multiplicity 1 and so

j1 = j2 = 1. By (C.5) we then have

β1,1 = (ξ − 1)

ξ + 4

9

= 3

(ξ −51)(ξ + 2) ξ=1 =

3

and

5ξ + 4

−6

β

= (ξ + 2)

=

= 2.

2,1

(ξ

− 1)(ξ + 2) ξ=−2

−3

Thus the partial fraction expansion is

R(ξ) =

3

+

2

.

ξ − 1

ξ + 2

3.

We take

−3ξ2 + 5ξ + 2

R(ξ) =

.

ξ

3 − 3ξ2 + ξ − 3

The roots of the denominator polynomial are α1 = 3, α2 = i, and α3 = −i. Since we have

complex roots, there will be di erent partial fraction expansions, depending on whether

we are thinking of R C(ξ) or R R(ξ). Let us take the complex case first. Using (C.5)

we have

β

= (ξ

3)

−3ξ2 + 5ξ + 2

=

1

−

(ξ + 3)(ξ − i)(ξ + i) ξ=3

−

1,1

3ξ2 + 5ξ + 2

β

= (ξ

−

i)

−

= 1 + i

2,1

(ξ + 3)(ξ − i)(ξ + i)

ξ=i

−

2

3ξ2 + 5ξ + 2

β

3,1

= (ξ + i)

−

=

−

1

−

i .

(ξ + 3)(ξ − i)(ξ + i) ξ=−i

2

Thus the partial fraction expansion over C is

1

1 − 2i

1 + 2i

R(ξ) = −

ξ − 3

−

ξ − i

−

ξ + i

.

The partial fraction expansion over R turns out to be

1

2ξ + 1

R(ξ) = −

−

.

ξ − 3

ξ2 + 1

Thus, employing the symbols in (C.4) we have α1 = 3, σ1 = 0, and ω1 = 1. The easiest way to determine this is to compute the complex partial fraction expansion, and then recombine the complex conjugate pairs over a common denominator.

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4. We take		2ξ2 + 1
	R(ξ) =		.
		ξ3 + 3ξ2 + 3ξ + 1

The root of the denominator polynomial is −1 which has multiplicity 3. We use (C.5) to get

1 d2

2ξ2 + 1

ξ

+

2

−

β1,1 =

d

(ξ + 1)3

3

2ξ + 1

ξ= 1

= 2

2 dξ2

3ξ2 + 3ξ + 1

β1,2

=

(ξ + 1)3

=

4

dξ

ξ3 + 3ξ22 + 3ξ + 1

ξ=−1

−

2ξ

+ 1

β1,3 = (ξ + 1)3

= 3.

ξ3 + 3ξ2 + 3ξ + 1

ξ=−1

Thus the partial fraction expansion is

2

−

4

3

R(ξ) =

+

.

ξ + 1

(ξ + 1)2

(ξ + 1)3

Heavyside coverup?

616 C Polynomials and rational functions 22/10/2004

Exercises

EC.1	Let P (ξ) R[ξ] be monic,	irreducible, and degree two. Show that there exists σ					R
			2	2
EC.2	and ω > 0 so that P (ξ) = (ξ − σ)			+ ω	.	¯
	Note that R[ξ] is naturally a subset of C[ξ]. If P (ξ) R[ξ], denote by P (ξ) the same
	polynomial, except thought of as being in C[ξ]. Show that polynomials P1(ξ), P2(ξ)
		¯		¯	(ξ) are coprime.
	R[ξ] are coprime if and only if P1(ξ), P2
EC.3	For
	P (ξ) = ξn + pn−1ξn−1 + · · · + p1ξ + p0 F[ξ], F {R, C},

show that sum of the roots of P , counting multiplicities, is equal to −pn−1.

EC.4 Determine the c.f.r. of the following rational functions:

ξ+ 1

(a)3ξ2 + 6;

	(b)	−3ξ2 + 6ξ + 9		;
		ξ3 + 5ξ2 + 7ξ + 3
	(c)	2ξ + 2
			;
		(ξ + 1)3
	(d)	2ξ2 + 4			.
		3ξ3 + 9ξ2 + 3ξ + 9

EC.5 Determine the partial fraction expansion of the following rational functions (for functions with complex poles, determine both the real and complex partial fraction expansions):

(a)ξ2 − 1;

ξ+ 2

	(b)	3ξ + 4		;
		ξ2 + 3ξ + 2
	(c)	2ξ2 − ξ + 1			;
		2ξ3 + 18ξ2 + 48ξ + 32
	(d)	ξ2 + 2
			.
		(ξ2 + 1)2

EC.6 Let R C(ξ). Show that R R(ξ) if and only if R(s) = R(¯s) for every s C which is not a pole for R.

Exercises for Chapter C

617