Electronic Journal: Southwest Journal of Pure and Applied Mathematics Internet: http://rattler.cameron.edu/swjpam.html ISSN 1083-0464 Vol. No. 02, 1997, pp. 12–15. Submitted: June 1, 1997. Published: December 30, 1997.

A FACTORED FORM OF COMPLEX ANALYTIC FUNCTIONS AND

ITS APPLICATION TO FINDING PARTIAL FRACTION DECOMPOSITIONS

L.F. Mahrt^{1}, and F. Szidarovszky^{2}

Abstract. A fast algorithm is introduced to find the coefficients of the partial fraction decomposition of proper rational functions. Instead of solving a large-scale system of linear algebraic equations with full coefficient matrix or differentiating repeatedly rational functions, a special system of linear algebraic equations is solved, where the coefficient matrix is lower triangular. Therefore, there is no need for elimination, the solutions are obtained by forward substitution making the procedure very effective. The method is based on a special representation of analytic functions which has further applications in constructing iteration methods for solving nonlinear equations.

A.M.S. (MOS) Subject classification Codes. 26C15

Key Words and Phrases. Rational functions, partial fractions.

1. INTRODUCTION.

Special function representations have been shown to be very useful in all kinds of applications. This paper introduces a new factored form of analytic complex functions, which can be applied to derive a fast algorithm to find partial fraction decompositions for rational functions with high multiplicity poles as well as in constructing iteration schemes for solving nonlinear equations. Integrating rational functions and solving time-invariant linear systems by the transformation method require the determination of the partial fraction decomposition of proper rational functions ([2],[3],[4]). In addition, several computer methods for solving polynomial equations are based on manipulations of rational functions ([1]). Partial fraction decomposition make the application of such methods easier to apply.

In the literature two methods are prevalent for performing partial fraction decompositions. In the case of first method, known as the algebraic method, a large system of linear equations is solved, which can be very difficult, unless the coefficient matrix has a special structure. In the second method, called the residue method, repeated derivatives of rational functions have to be determined, which is sometimes very complicated. The algorithm presented in this paper also requires the solution of a large system of linear algebraic equations; however, the coefficient matrix is of lower-triangular type and therefore there is no need for elimination, the solution can be easily obtained by forward substitution.

This paper develops as follows. In section 2 a unique representation theorem will be proved for analytic functions. This will be the basis for determining the partial fraction coefficients. The detailed algorithm will be presented in section 3, and section 4 concludes the paper with an illustrative example.

2. A REPRESENTATION THEOREM.

Let n ≥ 1 be an integer, and assume that functions P and Q are analytic on an open complex domain D. Let η ∈ D be a given point and assume that for k = 0, 1, …, n − 1, Q^{(k)}(η) ≠ 0.

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copyright \begin{matrix}c \\ ◯ \end{matrix}1997 by Cameron University

1Hughes Aircraft Company, Tucson, Arizona, E-mail address: ifmahrt@ccgate.hac.com

2Systems and Industrial Engineering Department, University of Arizona, Tucson, Arizona 85721, U.S.A.

E-mail address: szidar@sie.arizona.edu

Theorem 1. There exists a unique sequence a_{0}, A_{1}, …, A_{n−1} of complex numbers and a unique analytic complex function P_{1} such that

P (s) = Q(s) · [A_{0} + A_{1}(s − η) + ⋯ + A_{n−1}(s − η)^{n−1}] + P_{1}(s) · (s − η)^{n}.  (1)

Proof. Define R(s) = A_{0} + A_{1}(s − η) + ⋯ + A_{n−1}(s − η)^{n−1}, and assume that decomposition (1) exists. For k = 0, 1, …, n − 1, differentiate both sides of equation (1)k− times and substitute s = η to have

P^{(k)}(η) = ∑_{i=0}^{k} \binom{k}{i}Q^{(k−i)}(η) · \frac{d^{i}}{ds^{i}} [A_{0} + A_{1}(s − η) + ⋯ + A_{n−1}(s − η)^{n−1}]_{s=η}

+ ∑_{i=0}^{k} \binom{k}{i}P_{1}^{k−i}(η) · \frac{d^{i}}{ds^{i}} [(s − η)^{n}]_{s=η}

= ∑_{i=0}^{k} \binom{k}{i}Q^{k−i}(η) · A_{i} · i!.

Introducing the quantities t_{ki} = \frac{k!}{(k−i)!}Q^{(k−i)}(η) the above equations can be rewritten as a system of linear algebraic equations

\underline{p} = \underline{T} · \underline{a}  (2)

where

\underline{T} = [ \begin{matrix}t_{00} & 0 & … & 0 \\ t_{10} & t_{11} & … & 0 \\ ⋮ & ⋮ & ... & ⋮ \\ t_{n−10} & t_{n−11} & … & t_{n−1n−1} \end{matrix} ] , \underline{p} = [ P (η) ; P^{(1)}(η) ; ⋮ ; P^{(n−1)}(η) ]

and

\underline{a} = [ A_{0} ; A_{1} ; ⋮ ; A_{n−1} ]

The assumption Q^{(k)}(η) ≠ 0fork = 0.1…. , n − 1 quarantees that the diagonal elements of \underline{T} are nonzero, therefore system (2) has a unique solution which can be obtained by simple forward substitution.

For a given pair P, Q of analytic functions consider the coefficients A_{0}, A_{1}, …, A_{n−1} obtained by solving system (2). The construction of these constants implies that for k = 0, 1, …, n − 1

P^{(k)}(η) = \frac{d^{k}}{ds^{k}} [Q(s) · (A_{0} + A_{1}(s − η) + ⋯ + A_{n−1}(s − η)^{n−1})]_{s=η}

Therefore function P (s) − Q(s) · R(s) is also analytic and η is its root with multiplicity at least n. Hence, with some analytic function P_{1},

P (s) − Q(s) · R(s) = P_{1}(s) · (s − η)^{n},  (3)

which completes the proof.♠

Remark 1. From equation (2) we have

A_{0} = \frac{P (η)}{t_{00}},  (4)

and for k = 0, 1, …, n − 1

A_{k} = \frac{1}{t_{kk}} (P^{(k)}(η) − ∑_{i=0}^{k−1} t_{ki} · A_{i}) .  (5)

In addition, from equation (3) we conclude that

p_{1}(s) = \frac{P (s) − Q(s) · R(s)}{(s − η)^{n}}.  (6)

Remark 2.. Assume now that P and Q are polynomials of degree p and q, respectively. Then the degree of P − Q · R is at most m = max{p, q + n − 1}, and η is its root with multiplicity of at least n. Therefore P_{1} is also a polynomial with degree at most m − n. Furthermore, if we introduce the notation f_{n} = P − Q · R, then

P_{1}(s) = ∑_{k=0}^{p_{1}} \frac{f_{n}^{(n+k)}(η)}{(n + k)}! · (s − η)^{k},  (7)

where p_{1} is the degree of P_{1}.

Notice that in Theorem 1, functions P, Q, and P_{1} do not need to be polynomials. Representation (1) has many applications, and one of them will be elaborated in the next section. Other applications in constructing Newton- like methods and other iteration schemes of fast convergence rates will be discussed in a future paper.

3. PARTIAL FRACTION DECOMPOSITION.

Any proper rational function can be written as

\frac{P (s)}{(s − η)^{n}Q(s)} = \frac{B_{0}}{(s − η)} + \frac{B_{1}}{(s − η)^{2}} + ⋯ + \frac{B_{n−1}}{(s − η)^{n}} + \frac{P_{1}(s)}{Q(s)}  (8)

where Q(η) ≠ 0 and P_{1} is a polynomial of degree less than that of Q. Multiply both sides of equation (8) by the common denominator to have

P (s) = Q(s) · (A_{0} + A_{1}(s − η) + ⋯ + A_{n−1}(s − η)^{n−1}) + P_{1}(s) · (s − η)^{n},

which is exactly the same as (1) with A_{i} = B_{n−l−i}(i = 0, 1, …, n − 1). If function (8) is proper rational, then p < n + q. Hence, the algorithm of the previous section is applicable to find A_{0}, A_{1}, …, A_{n−1} and P_{1}. The algorithm has to be then repeated for a pole of the decompostion remainder rational function \frac{P_{1}(s)}{Q(s)}, and so on, until all poles of the original rational function have been exhausted.

4. AN ILLUSTRATIVE EXAMPLE.

Consider the rational function

\frac{4s^{8} − 33s^{7} + 101s^{6} − 146s^{5} + 94s4 + 94s^{3} − 178s^{2} − 166s + 66}{(s − 3)^{4} · (s^{5} + 2s^{3} + 4s^{2} + 2s + 3)}

where we have

\begin{aligned}P (s) & = 4s^{8} − 33s^{7} + 101s^{6} − 146s^{5} + 94s^{3} − 178s^{2} − 166s + 66, \\ Q(s) & = s^{5} + 2s^{3} + 4s^{2} + 2s + 3, \end{aligned}

η = 3, n = 4, q = 5 and p = 8. Simple differentiation shows that Q^{(0)}(3) = 342, Q^{(1)}(3) = 485, Q^{(2)}(3) = 584, Q^{(3)}(3) = 552, and P^{(0)}(3) = 342, P^{(1)}(3) = 1169, P^{(2)}(3) = 4576, P^{(3)}(3) = 20994. Therefore

\underline{T} = ( \begin{matrix}342 & o & o & o \\ 485 & 342 & 0 & 0 \\ 584 & 970 & 684 & 0 \\ 552 & 1752 & 2910 & 2052 \end{matrix} ) , and\underline{p} = [ \begin{matrix}342 \\ 1169 \\ 4576 \\ 20994 \end{matrix} ]

which implies that the solution of equation (2) is given by

\underline{a} = [ \begin{matrix}1 \\ 2 \\ 3 \\ 4 \end{matrix} ]

so A_{0} = 1, A_{1} = 2, A_{2} = 3, and A_{3} = 4. Hence,

R(s) = 1 + 2(s − 3) + 3(s − 3)^{2} + 4(s − 3)^{3} = 4s^{3} − 33s^{2} + 92s − 86,

and therefore

f_{4}(s) = P (s) − Q(s) · R(s) = s^{6} − 10s^{5} − 48s^{3} + 81s^{2} − 270s + 324.

The degree of P_{1} is 6 − 4 = 2, and simple differentiation shows that

f_{4}^{(4)}(3) = 456, f_{4}^{(5)}(3) = 960, f_{4}^{(6)}(3) = 720.

From equation (7) we conclude that

\begin{aligned}P_{1}(s) & = \frac{456}{4!}(s − 3)^{0} + \frac{960}{5!}(s − 3)^{1} + \frac{720}{6!}(s − 3)^{2} \\ & = 19 + 8(s − 3) + (s − 3)^{2} = s^{2} + 2s + 4, \end{aligned}

that is, the partial fraction decomposition (8) is as follows:

\frac{4}{(s − 3)} + \frac{3}{(s − 3)^{2}} + \frac{2}{(s − 3)^{3}} + \frac{1}{(s − 3)^{4}} + \frac{s^{2} + 2s + 4}{s^{5} + 2s^{3} + 4s^{2} + 2s + 3},

since B_{0} = A_{3} = 4, B_{1} = A_{2} = 3, B_{2} = A_{1} = 2, and B_{3} = A_{0} = 1. We finally mention that a subsequent paper we will provide more numerical examples and applications to

validate our results and compare them to those of earlier methods.

References

1. Argyros, I.K., An Iterative Solution of the Polynomial Equation in Banach Space, Bulletin of the Institute of Mathematics Academia Sinica 15 NO.4 (1987), 403-410.

2. Davis, H.T., Introduction to Nonlinear Differential and Integral Equations (1962), Dover Publications, New York.

3. La Page, W.R., Complex Variables and the Laplace Transform for Engineers (1961), McGraw-Hill, New York.

4. Szidarovszky, F. and A.T. Bahill, Linear Systems Theory (1992), CRC Press, Boca Raton.