Abstract

The partial fraction decomposition technique is very useful in many areas including mathematics and engineering. In this paper we present a new and simple method on the partial fraction decomposition of proper rational functions which have completely factored denominators over or . The method is based on a recursive computation of the -adic polynomial in commutative algebra which is a generalization of the Taylor polynomial. Since its computation requires only simple algebraic operations, it does not require a computer algebra system to be programmed.

1. Introduction

Let be or and be a polynomial ring with the coefficients in . We also assume the rational function to be proper (i.e., the degree of denominator is greater than the degree of numerator) with the denominator factored completely over . Now, we will show how to apply our method to the following partial fraction decomposition over :We multiply through by the least common denominator to clear the fractions:Since , we can replace with a power series in . From Proposition 6(a) which we will prove later, we getSince we have , is equal to , which is the Taylor polynomial of order 2 for at . Similarly, is the Taylor polynomial of order 1 for at and we obtainHence

In [1], Ma et al. explained several recent approaches of pfd (partial fraction decomposition) and introduced a fast recursive method of pfd over . Our method is more algebraic and works for pfd of any proper rational functions over or by extending the concept of Taylor polynomials into -adic polynomials.

2. -Adic Polynomials and Main Theorem

In this section we introduce the completion of a commutative ring which is useful in commutative algebra and algebraic geometry. We also define the -adic polynomial of order .

Definition 1 (ch. 10 in [3]). For any commutative ring with unity and a proper ideal of , the completion of with respect to iswith the quotient maps . We also use if with .

If with a maximal ideal , then is the ring of formal power series. For with a prime ideal , is , the ring of -adic integers. Now we will define the -adic polynomial of order , which is crucial for our method.

Lemma 2 (-adic expansion and -adic polynomial of order ). Let be an irreducible polynomial in with and be the corresponding maximal ideal. Then, consider the following:(a)The natural map via is injective and factors through the localization of at the maximal ideal to .(b) induces(c) For , there exists the unique -adic expansion such thatwhere is a polynomial with .

Definition 3. For , , and , we define the -adic polynomial of order for , , and , respectively, as

Proof. (a) is true by the Krull intersection theorem for Noetherian domains (Corollary  10.18 in [3]) and (22.13) in [4]. (b) is true from Proposition  10.15 in [3].
Case (c). For , let be a representative of in . Now we will use induction on . For , let be the remainder of . Then is independent of choice of , , and . Assume that there exists with such that . The condition implies thatwith . Let and be the remainder and the quotient of . Then does not depend on choice of and we have . Since ,HenceThe uniqueness is clear by the construction.

If with , -adic polynomial of order for is the Taylor polynomial of order for at . Here we present the following main theorem.

Theorem 4 (main theorem). For and , let be irreducible in with and . Assume thatwith . Then

Proof. By taking the modulus of and applying Lemma 2(b), we have implies thatfrom which we complete the proof.

In the next section we will explain how to compute -adic polynomials with .

3. Formulas of -Adic Polynomials with and the Product of -Polynomials

For or , an irreducible polynomial has at most degree 2. Since we can convert and into and , respectively, by replacing with , we only present recursive formulas of -adic polynomials and -adic polynomials.

Lemma 5. For with , let with . Then

Proof. implies .

Proposition 6. We present two formulas for -adic polynomials with :(a)If with and , then and(b)If with and , then , and

Proof. Case  (a). From Lemma 5 with and , we haveBy comparing the coefficients of , one can findThus , where .
Case  (b). We can also prove part (b) by repeating the same argument with .

To compute the -adic polynomial of a polynomial, we need formulas of the Taylor shift.

Definition 7 (coefficient vector of a -adic polynomial of order ). For a given -adic polynomial of order , we define the corresponding coefficient vectors , , and , respectively, aswhere is the transpose of a row vector and (the floor function).

Definition 8 (the Taylor shift matrix). For with , matrix is

From the binomial expansion, , we obtain the following Taylor shift formulas for the coefficient vectors.

Proposition 9 (Taylor shift). For a polynomial with ,where and .

Remark 10. To compute the Taylor shift matrix which is a Toeplitz matrix, we may use recursive formulas given by

Proposition 11. We present three formulas for -adic polynomials with :
(a) (-Adic shift for ) for a polynomial with , the corresponding -adic polynomial of order , with , is determined byIf , then and (b) For with ,For , let and . Then with and (c) For , and ,For , and ,

Proof. Case (a). You can find a similar argument on p. 591 in [2]. For , it is clear. For with , we haveHence and can be computed by the Taylor shift (Proposition 9).
Cases (b) and (c). (b) and (c) are clear by direct computations.

To get -adic polynomials of arbitrary rational functions, we just multiply simple -adic polynomials which are computed using Propositions 6, 9, and 11. In general, the product of two -adic polynomials may carry a term likeBut if the coefficients of one of two -adic polynomials are in , then the product is carryless multiplication and the classic Cauchy product (convolution) formula still holds.

Proposition 12 (-adic Cauchy product formula). For , let , , and and assume . Thenwhere .

4. Example

Example 1. We compute the following partial fraction decomposition:(i): from Theorem 4, we getFrom the Taylor shift (Proposition 9), we have .From Proposition 6(b), we computeHence(ii) : using Proposition 11(b), we get . From Theorem 4, we haveFrom Proposition 6(a), we compute .Using Propositions 6(b) and 12, we getFrom Proposition 11(a), we computeUsing Proposition 12, we getHence

5. Main Algorithm (Algorithm 1)

For a given proper () fractionwhere and are relatively prime, respectively, we return (-adic polynomials of order ) and (-adic polynomial of order ) such that

  Output: for and for .
for to do
  Let and find using the Taylor shift (Proposition 9). Compute :
    using a binary splitting scheme (Lemma  4.1 on [2]), Proposition 6 and the -adic Cauchy product formula (Proposition 12).
end
for to do
  Let and . Using the Taylor shift (Proposition 9), find . Using the -adic shift
    (Proposition 11(a)), find
    where and in Proposition 11(b). Compute :
    using a binary splitting scheme (Lemma  4.1 on [2]), Proposition 6 and the -adic Cauchy product formula (Proposition 12).
(6) end

From the computation of Theorems  3.1 and   6.1 in [2], the complexity of the main algorithm (Algorithm 1) with simple algebraic operations is given by with , where , , and is any upper bound on the number of operations needed to multiply two -th degree polynomials. Since or , if an FFT (fast Fourier transform) is used, we can reduce the complexity to since with FFT (Ch. 2 in [5]). For , the best known complexity is in [2]. If is not big enough, our algorithm is still effective since it does not require computingwhich is an overhead of the algorithm in [2], where is or .

6. Discussion

The main theorem and Propositions 6, 9, 11, and 12 still hold for any field with characteristic zero. Since an irreducible polynomial over or has at most degree , our recursive formulas for cover all possible proper rational functions over or . If , we need to replace Propositions 6, 11(b), and 11(c) with the Extended Euclidean algorithm (Ch. 2 in [5]) which requires an algebra system to be implemented. To compute a -adic polynomial of a polynomial , we use the repeated division of by instead of Propositions 9 and 11(c). Then we obtain as the remainders.We can compute it quickly by applying the divide-and-conquer method (Problem 3.5 of [6]).

7. Conclusion

This paper shows that the partial fraction of a proper rational function with denominator factored completely over or can be found by computing suitable -adic polynomials with . We also present recursive formulas (Propositions 6, 9, 11, and 12) to compute -adic polynomials with . Since this algorithm only requires simple arithmetic operations, it can be implemented easily without a complex computer algebra system.

Conflicts of Interest

The authors declare that there are no conflicts of interest regarding the publication of this paper.

Acknowledgments

The work of Xin Zhang has been partially supported through The CUNY Research Scholars Program (CRSP) and US Department of Education Queensborough MSEIP (P120A140057). The authors thank Dr. Jeehoon Park and Dr. Kostas Stroumbakis for their valuable comments.