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A cover-up approach to partial fractions with linear or irreducible quadratic factors in the denominators
Applied Mathematics and Computation 219 (2012) 3855–3862
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Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc
A cover-up approach to partial fractions with linear or irreducible quadratic factors in the denominators Yiu-Kwong Man Department of Mathematics and Information Technology, HKIEd, 10 Lo Ping Road, Tai Po, New Territories, Hong Kong
a r t i c l e
i n f o
Keywords: Partial fraction decomposition Cover-up approach Irreducible quadratic factors
a b s t r a c t In this paper, a simple cover-up approach for computing partial fractions with linear or irreducible quadratic factors in the denominators is presented. It mainly involves polynomial divisions and substitutions only, without having to factorize the irreducible quadratic factors into linear factors with complex coefficients, to use differentiation or to solve a system of linear equations. Examples and its application to derive some useful formulas of partial fraction decompositions are included. Ó 2012 Elsevier Inc. All rights reserved.
1. Introduction Partial fraction decomposition is an important topic in the study of calculus, differential equations, control theory and some areas of pure or applied mathematics. One common approach to compute the partial fractions is to use the method of undetermined coefficients. Another approach is to apply the Heaviside’s cover-up technique, which uses substitutions to determine the unknown coefficients of the partial fractions with single poles, and successive differentiations to handle those with multiple poles (see [1]). However, the computations involved could be quite tedious for the latter case. Thus, there are a number of alternative approaches proposed to handle partial fractions of rational functions from time to time. For instance, one can refer to [3–12] for some recent developments on this topic. In [9], an improved Heaviside’s coverup approach was proposed to handle the case of multiple poles by substitutions and polynomial divisions only. However, the drawback is that this approach cannot be effectively applied to compute partial fractions with irreducible quadratic factors only. To fill-up this gap, we now present a further improved cover-up approach in this paper, for handling partial fractions with linear or irreducible quadratic factors by polynomial divisions and substitutions only, without having to factorize the irreducible quadratic factors into linear factors with complex coefficients, to use differentiation or to solve a system of linear equations. Examples and its applications to derive some useful formulas of partial fraction decompositions are included. 2. The cover-up approach The existence of partial fraction decompositions of proper rational functions is based on the following result, whose proof can be found in [2]. Theorem 1. Let F be a constant field and aðxÞ and bðxÞ be polynomials in F½x such that deg aðxÞ < deg bðxÞ and
bðxÞ ¼ ðx a1 Þn1 ðx as Þns ðx2 þ b1 x þ c1 Þm1 ðx2 þ br x þ cr Þmr
E-mail address: [email protected] 0096-3003/$ - see front matter Ó 2012 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.amc.2012.10.016
ð1Þ
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where n1 ; . . . ; ns ; m1 ; . . . ; mr are positive integers, a1 ; . . . ; as ; b1 ; . . . ; br , c1 ; . . . ; cr 2 F; ak are distinct, x2 þ bi x þ ci are distinct irreducible polynomials, 1 6 k 6 s and 1 6 i 6 r. Then aðxÞ=bðxÞ has a unique partial fraction expansion of the form
X s r ai;ni bi;mi x þ ci;mi aðxÞ X ai;1 bi;1 x þ ci;1 þ ¼ þ þ þ þ bðxÞ i¼1 ðx ai Þ ðx2 þ bi x þ ci Þ ðx ai Þni ðx2 þ bi x þ ci Þmi i¼1
ð2Þ
where ai;j ; bi;j ; ci;j 2 F. A cover-up approach for computing the unknown partial fractions is given by the theorem below, which was proposed and proved by the author in [11]. Theorem 2. The unknown partial fraction coefficients described in the above theorem can be computed by the procedure below. Case (I): Partial fractions with linear factors in the denominators.
Step 1 : aðxÞ ðx ai Þni ai;ni ¼ bðxÞ x¼ai Step 2 : ai;ni j ¼
"
j1 aðxÞ X ai;ni k bðxÞ k¼0 ðx ai Þni k
ð3Þ !
# ðx ai Þni j
ð4Þ
x¼ai
where 1 6 i 6 s and 1 6 j 6 ni 1. The cancellations of the common factors ðx ai Þni and ðx ai Þni j from the denominators of the rational functions concerned should be done before substituting x ¼ ai into the whole expressions inside the square brackets. Case (II): Partial fractions with irreducible quadratic factors in the denominators.
Step1 : bi;mi x þ ci;mi ¼
aðxÞ ðx2 þ bi x þ ci Þmi modðx2 þ bi x þ ci Þ bðxÞ
Step2 : bi;mi j x þ ci;mi j ¼
"
j1 aðxÞ X bi;mi k x þ ci;mi k bðxÞ k¼0 ðx2 þ bi x þ ci Þmi k
ð5Þ
!
# 2
ðx þ bi x þ ci Þ
mi j
modðx2 þ bi x þ ci Þ
ð6Þ
where 1 6 i 6 r and 1 6 j 6 mi 1. The cancellations of the common factors ðx2 þ bi x þ ci Þmi and ðx2 þ bi x þ ci Þmi j from the denominators of the rational functions concerned should be done before applying the operation ‘mod ðx2 þ bi x þ ci Þ’ to the whole expressions inside the square brackets. Some general remarks related to Theorem 2 are provided below: In Case (I), the first step is to eliminate ðx ai Þni in bðxÞ and include ðx ai Þ or its higher power to each partial fraction, except the one containing ai;ni . Hence, ai;ni can be found by substituting x ¼ ai into Eq. (3). This technique is the so-called cover-up technique as described in [1]. The second step is to exploit the uniqueness of partial fraction decompositions of rational functions. Suppose ai;1 ðxÞ=bi;1 ðxÞ is equal to aðxÞ=bðxÞ ai;ni =ðx ai Þni . We can use polynomial division to eliminate the common factor ðx ai Þ in ai;1 ðxÞ=bi;1 ðxÞ, so that the highest power of ðx ai Þ in bi;1 ðxÞ is equal to ðni 1Þ. Thus, the multiplicity of the pole ai is reduced by one. By repeating this process on the new rational function obtained at each time, we can determine the unknown coefficients ai;ni j successively by using Eq. (4). In Case (II), the first step is to eliminate ðx2 þ bi x þ ci Þmi in bðxÞ and include ðx2 þ bi x þ ci Þ or its higher power to each partial fraction, except the one containing bi;mi x þ ci;mi . Hence, bi;mi x þ ci;mi can be found by applying the operation ‘mod ðx2 þ bi x þ ci Þ’ to the whole expression, which is equivalent to finding the remainder when each polynomial involved is divided by ðx2 þ bi x þ ci Þ. If the result is not a linear polynomial, we can use polynomial division to express its numerator and denominator in the forms ðx2 þ bi x þ ci Þqi;1 ðxÞ þ ðui;1 x þ v i;1 Þ and ðx2 þ bi x þ ci Þqi;2 ðxÞ þ ðui;2 x þ v i;2 Þ, respectively. By applying ‘mod ðx2 þ bi x þ ci Þ’, only the linear parts ui;1 x þ v i;1 and ui;2 x þ v i;2 will be left. By multiplying pi x þ qi to each of them, where pi ¼ 1=ui;2 and qi ¼ ½bi ðv i;2 =ui;2 Þ=ui;2 , we can apply ‘mod ðx2 þ bi x þ ci Þ’ again to reduce ðui;2 x þ v i;2 Þðpi x þ qi Þ and ðui;1 x þ v i;1 Þðpi x þ qi Þ into a constant and a linear polynomial, respectively. Thus, bi;mi x þ ci;mi can be determined. The second step is to exploit the uniqueness of partial fraction decompositions of rational functions. Suppose ai;1 ðxÞ=bi;1 ðxÞ is equal to aðxÞ=bðxÞ ðbi;mi x þ ci;mi Þ=ðx2 þ bi x þ ci Þmi . We can use polynomial division to eliminate the common factor ðx2 þ bi x þ ci Þ in ai;1 ðxÞ=bi;1 ðxÞ, so that the highest power of ðx2 þ bi x þ ci Þ in bi;1 ðxÞ is equal toðmi 1Þ. Hence, the power of ðx2 þ bi x þ ci Þ is reduced by one. By repeating this process on the new rational function obtained at each time, we can determine the unknown coefficients bi;mi j x þ ci;mi j successively by using Eq. (6). It is worth to point out that the main ideas behind Theorem 2 can be regarded as a technique for rewriting the higher degree terms in terms of lower degree terms. For instance, if there are higher degree terms xk appeared in Case (II), where k P 2, then they can be reduced by applying the operation ‘mod ðx2 þ bi x þ ci Þ’ into a linear polynomial. Thus, this technique
Y.-K. Man / Applied Mathematics and Computation 219 (2012) 3855–3862
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extends the original Heaviside’s cover-up approach to handle partial fraction decompositions with linear or irreducible quadratic factors (with simple or multiple poles), without having to use differentiations. 3. Algorithms For the ease of hand calculations or implementation in computer algebra systems, we can rewrite Theorem 2 in an algorithmic forms below so that one can follow the calculation steps more easily. In brief, the first two algorithms are designed for handling Case (I) and Case (II) of Theorem 2, respectively. The third algorithm is a subprogram to be called by the second algorithm during the calculations. For convenience, we will use the abbreviations numer and denom to represent numerator and denominator, respectively. As the main ideas and results behind the algorithms have been described in Section 2 and the proof of Theorem 2 can be found in [11], so the formal proofs of the algorithms will not be provided here. Algorithm 1. Partial_Fraction_linear Input: A proper rational function a(x)/b(x), where aðxÞ; bðxÞ are polynomials, bðxÞ ¼ ðx a1 Þn1 ðx as Þns ; n1 ; . . . ; ns are positive integers and a1 ; . . . ; as are distinct constants. Output: The partial fraction coefficients ai;j as described in Theorem 2, where 1 6 i 6 s; 1 6 j 6 ni : Procedure: rðxÞ ¼ aðxÞ=bðxÞ For i = 1 to s do For j = 0 to ni 1 do dðxÞ ¼ denomðrðxÞÞ ðx ai Þni j r 1 ðxÞ ¼ numerðrðxÞÞ dðxÞ ai;ni j ¼ r1 ðai Þ sðxÞ ¼ rðxÞ
ai;n j i
ðxai Þni j
cðxÞ ¼ numerðsðxÞÞ ðx ai Þ dðxÞ ¼ denomðsðxÞÞ ðx ai Þ cðxÞ rðxÞ ¼ dðxÞ
end do end do return fai;j g where 1 6 i 6 s; 1 6 j 6 ni . end Algorithm 2. Partial_Fraction_quadratic Input: A proper rational function a(x)/b(x), where aðxÞ; bðxÞ are polynomials, bðxÞ ¼ ðx2 þ b1 x þ c1 Þm1 ðx2 þ br x þ cr Þmr ; b1 ; . . . ; br ; c1 ; . . . ; cr are constants;m1 ; . . . ; mr are positive integers and x2 þ bi x þ ci are distinct irreducible polynomials and 1 6 i 6 r. Output: The polynomials bi;j x þ ci;j as described in Theorem 2, where 1 6 i 6 r; 1 6 j 6 mi . Procedure: sðxÞ ¼ aðxÞ=bðxÞ For i = 1 to r do For j = 0 to mi 1 do dðxÞ ¼ denomðsðxÞÞ ðx2 þ bi x þ ci Þmi j s1 ðxÞ ¼ numerðsðxÞÞ dðxÞ sðxÞ ¼ s1 ðxÞmodðx2 þ bi x þ ci Þ if sðxÞ is a polynomial then bi;mi j x þ ci;mi j ¼ sðxÞ else bi;mi j x þ ci;mi j = Simp PolyðnumerðsðxÞÞ; denomðsðxÞÞ; x2 þ bi x þ ci Þ endif s1 ðxÞ ¼ sðxÞ
bi;m j xþci;m j i
i
ðx2 þbi xþci Þmi j
(continued on next page)
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cðxÞ ¼ numerðs1 ðxÞÞ ðx2 þ bi x þ ci Þ dðxÞ ¼ denomðs1 ðxÞÞ ðx2 þ bi x þ ci Þ cðxÞ sðxÞ ¼ dðxÞ
end do end do return fbi;j x þ ci;j g where 1 6 i 6 r; 1 6 j 6 mi : end Algorithm 3. Simp_Poly Input: Polynomials mðxÞ; nðxÞ and x2 þ bx þ c, where degðmðxÞÞ < degðnðxÞÞ. h i 2 Output: The result of mðxÞ nðxÞ modðx þ bx þ cÞ: Procedure: u1 x þ v 1 ¼ mðxÞmodðx2 þ bx þ cÞ sðxÞ ¼ nðxÞmodðx2 þ bx þ cÞ if sðxÞ is a constant then rðxÞ ¼ ðu1 x þ v 1 Þ=sðxÞ else u2 ¼ the coefficient of x in sðxÞ v 2 ¼ theconstanttermofsðxÞ p ¼ 1=u2 q ¼ ðb v 2 pÞ=u2 w1 x þ w2 ¼ ðu1 x þ v 1 Þðpx þ qÞmodðx2 þ bx þ cÞ s ¼ ðu2 x þ v 2 Þðpx þ qÞ mod ðx2 þ bx þ cÞ rðxÞ ¼ ðw1 x þ w2 Þ=s endif return rðxÞ end 4. Examples We now illustrate how to apply the cover-up technique to solve the problems below. Example 1. Find the partial fraction decomposition of the rational function
FðxÞ ¼ SOLUTION
3x2 6x 2 ðx 1Þ3
:
Suppose the partial fraction decomposition of FðxÞ is given by
FðxÞ ¼
a b c þ ; þ x 1 ðx 1Þ2 ðx 1Þ3
where a, b, c are unknown coefficients to be determined. Using the cover-up technique, we have
denomðFðxÞÞ ðx 1Þ3 ¼ ðx 1Þ3 ðx 1Þ3 ¼ 1 2 3x 6x 2 2 c¼ ¼ ð3x 6x 2Þ ¼ 3ð1Þ2 6ð1Þ 2 ¼ 5 1 x¼1 x¼1 5
3x2 6x þ 3
ð3x 3Þðx 1Þ ¼ ðx 1Þ ðx 1Þ3 ðx 1Þ3 3x 3 ¼ Note : Only one ðx 1Þ is removed at this stage: ðx 1Þ2
F 1 ðxÞ ¼ FðxÞ þ
3
¼
denomðF 1 ðxÞÞ ðx 1Þ2 ¼ ðx 1Þ2 ðx 1Þ2 ¼ 1 3x 3 ¼ ð3x 3Þ ¼ 3ð1Þ 3 ¼ 0 b¼ 1 x¼1 x¼1 3ðx 1Þ 3 F 2 ðxÞ ¼ F 1 ðxÞ 0 ¼ ¼ ðx 1Þ2 x 1 denomðF 2 ðxÞÞ ðx 1Þ ¼ ðx 1Þ ðx 1Þ ¼ 1 3 a¼ ¼3 j 1 x¼1
Y.-K. Man / Applied Mathematics and Computation 219 (2012) 3855–3862
Hence,
FðxÞ ¼
3 5 : x 1 ðx 1Þ3
Example 2. Find the partial fraction decomposition of the rational function
FðxÞ ¼ SOLUTION
3x þ 1 x2 ðx2 þ 1Þ2
:
Suppose the partial fraction decomposition of FðxÞ is given by
FðxÞ ¼
a b cx þ d ex þ f þ þ ; þ x x2 x2 þ 1 ðx2 þ 1Þ2
where a, b, c, d, e, f are unknown coefficients to be determined. Using the cover-up technique, we have
denomðFðxÞÞ ðx2 þ 1Þ2 ¼ x2 ðx2 þ 1Þ2 ðx2 þ 1Þ2 ¼ x2 3x þ 1 3x þ 1 modðx2 þ 1Þ ¼ ¼ 3x 1 ex þ f ¼ x2 1 3x þ 1 ð3x þ 1Þðx2 þ 1Þ 3x þ 1 F 1 ðxÞ ¼ FðxÞ þ ¼ ¼ 2 2 2 2 x ðx þ 1Þ ðx þ 1Þ x2 ðx2 þ 1Þ2 denomðF 1 ðxÞÞ ðx2 þ 1Þ ¼ x2 ðx2 þ 1Þ ðx2 þ 1Þ ¼ x2 3x þ 1 3x þ 1 modðx2 þ 1Þ ¼ ¼ 3x 1 cx þ d ¼ x2 1 3x þ 1 ð3x þ 1Þðx2 þ 1Þ 3x þ 1 ¼ ¼ F 2 ðxÞ ¼ F 1 ðxÞ þ 2 x þ1 x2 ðx2 þ 1Þ x2 denomðF 2 ðxÞÞ x2 ¼ x2 x2 ¼ 1 3x þ 1 ¼ ð3x þ 1Þ ¼ 3ð0Þ þ 1 ¼ 1 b¼ 1 x¼0 x¼0 1 3x 3 F 3 ðxÞ ¼ F 2 ðxÞ 2 ¼ 2 ¼ x x x denomðF 3 ðxÞÞ x ¼ x x ¼ 1 3 a¼ ¼3 1 x¼0
Hence,
FðxÞ ¼
3 1 3x þ 1 3x þ 1 þ : x x2 x2 þ 1 ðx2 þ 1Þ2
Example 3. Find the partial fraction decomposition of the rational function
FðxÞ ¼ SOLUTION
2x 1 : ðx2 þ 1Þðx2 þ x þ 1Þ
Suppose the partial fraction decomposition of FðxÞ is given by
FðxÞ ¼
ax þ b cx þ d þ ; x2 þ 1 x2 þ x þ 1
where a, b, c, d are unknown coefficients to be determined. Using the cover-up technique, we have
denomðFðxÞÞ ðx2 þ 1Þ ¼ ðx2 þ 1Þðx2 þ x þ 1Þ ðx2 þ 1Þ ¼ x2 þ x þ 1 2x 1 modðx2 þ 1Þ ax þ b ¼ 2 x þxþ1 2x 1 2x 1 x modðx2 þ 1Þ ¼ modðx2 þ 1Þ ¼ x x x 2 2x x 2 x modðx2 þ 1Þ ¼ ¼ xþ2 ¼ x2 1
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F 1 ðxÞ ¼ FðxÞ
xþ2 x3 3x2 x 3 ðx þ 3Þ ¼ ¼ x2 þ 1 ðx2 þ 1Þðx2 þ x þ 1Þ x2 þ x þ 1
denomðF 1 ðxÞÞ ðx2 þ x þ 1Þ ¼ ðx2 þ x þ 1Þ ðx2 þ x þ 1Þ ¼ 1 ðx þ 3Þ modðx2 þ x þ 1Þ ¼ ðx þ 3Þmodðx2 þ x þ 1Þ cx þ d ¼ 1 ¼ x 3 Hence,
FðxÞ ¼
xþ2 xþ3 : x 2 þ 1 x2 þ x þ 1
We now illustrate how to use the cover-up technique to derive some useful formulas for partial fraction decompositions below. Example 4. Prove the following identity
1 1 1 xþaþb ¼ ; 2 ðx aÞðx2 þ bx þ cÞ a þ ab þ c x a x2 þ bx þ c 2
where a, b, c are non-zero constants, b < 4c and a2 þ ab þ c – 0. 2 SOLUTION Let FðxÞ ¼ 1=ðx aÞðx þ bx þ cÞ. Suppose its partial fraction decomposition is given by
FðxÞ ¼
d ex þ f þ ; x a x2 þ bx þ c
where d, e, f are unknown coefficients to be determined. Using the cover-up technique, we have
denomðFðxÞÞ ðx2 þ bx þ cÞ ¼ ðx aÞðx2 þ bx þ cÞ ðx2 þ bx þ cÞ ¼ x a 1 modðx2 þ bx þ cÞ ex þ f ¼ xa 1 xþbþa modðx2 þ bx þ cÞ ¼ xa xþbþa xþaþb ðx þ a þ bÞ modðx2 þ bx þ cÞ ¼ 2 ¼ 2 x þ bx aða þ bÞ a þ ab þ c xþaþb x2 þ bx þ c ¼ 2 2 þ ab þ cÞðx þ bx þ cÞ ða þ ab þ cÞðx aÞðx2 þ bx þ cÞ 1 ¼ 2 ða þ ab þ cÞðx aÞ
F 1 ðxÞ ¼ FðxÞ þ
ða2
denomðF 1 ðxÞÞ ðx aÞ ¼ ða2 þ ab þ cÞðx aÞ ðx aÞ ¼ a2 þ ab þ c 1 1 ¼ d¼ 2 a þ ab þ c x¼a a2 þ ab þ c Alternatively, we can determine the value of d more directly as follows:
denomðFðxÞÞ ðx aÞ ¼ ðx aÞðx2 þ bx þ cÞ ðx aÞ ¼ x2 þ bx þ c 1 1 ¼ d¼ 2 x þ bx þ c x¼a a2 þ ab þ c Hence,
FðxÞ ¼
1 1 xþaþb : a2 þ ab þ c x a x2 þ bx þ c
Example 5. Derive a formula for the partial fraction decomposition of the following rational function
FðxÞ ¼
ax þ b ; ðx2 þ ax þ bÞðx2 þ cx þ dÞ
where a; b; a; b; c; d are constants, a – c; a2 4b < 0 and c2 4d < 0. SOLUTION Suppose the partial fraction decomposition of FðxÞ is given by
Y.-K. Man / Applied Mathematics and Computation 219 (2012) 3855–3862
FðxÞ ¼
cx þ d ex þ f þ ; x2 þ ax þ b x2 þ cx þ d
where c, d, e, f are unknown coefficients to be determined. Case (I): Assume b ¼ d. Using the cover-up technique, we have
denomðFðxÞÞ ðx2 þ ax þ bÞ ¼ ðx2 þ ax þ bÞðx2 þ cx þ dÞ ðx2 þ ax þ bÞ
¼ x2 þ cx þ d
ax þ b modðx2 þ ax þ bÞ x2 þ cx þ d ax þ b ax þ b modðx2 þ ax þ bÞ ¼ modðx2 þ ax þ bÞ ¼ ðc aÞx þ ðd bÞ ðc aÞx ax þ b x þ a modðx2 þ ax þ bÞ ¼ ðc aÞx x þ a 1 ax2 þ ðaa þ bÞx þ ba ¼ modðx2 þ ax þ bÞ ca x2 þ ax 1 1 ½aðax bÞ þ ðaa þ bÞx þ ba ¼ ðbx þ ba abÞ ¼ bða cÞ bða cÞ
cx þ d ¼
Similarly, applying the algorithm again or simply replacing a by c in the above steps, we can obtain
ex þ f ¼
1 ðbx þ bc abÞ: bða cÞ
Hence, the partial fraction decomposition of FðxÞ is given by
FðxÞ ¼
1 bx þ ba ab bx þ bc ab : 2 2 bða cÞ x þ ax þ b x þ cx þ d
Case (II): Assume b – d. Using the cover-up technique, we have
denomðFðxÞÞ ðx2 þ ax þ bÞ ¼ ðx2 þ ax þ bÞðx2 þ cx þ dÞ ðx2 þ ax þ bÞ
¼ x2 þ cx þ d
ax þ b modðx2 þ ax þ bÞ x2 þ cx þ d ax þ b modðx2 þ ax þ bÞ ¼ ðc aÞx þ ðd bÞ 2 3 ax þ b h i5modðx2 þ ax þ bÞ ¼4 ðc aÞ x þ cdb a 2 " #3 x þ a cdb ax þ b a 5 h i modðx2 þ ax þ bÞ ¼4 x þ a cdb ðc aÞ x þ cdb a a 2 3
cx þ d ¼
ðax þ bÞ x þ a cdb 6 7 a 2 ¼6 2 7 4 5modðx þ ax þ bÞ db db ðc aÞ x2 þ ax þ a ca ca 2 h i 3
db db 2 6aðx þ axÞ þ b a ca x þ b a ca 7 7modðx2 þ ax þ bÞ ¼6 4 5 2 db b ðc aÞ a cdb a ca h i 3 2 ab þ b a cdb x þ b a cdb a a 5modðx2 þ a þ xbÞ ¼4 2 aðd bÞ ðc aÞb ðdbÞ ca h i 3 2 x þ b ab þ b a cdb a cdb a a 5modðx2 þ ax þ bÞ ¼4 2 ad bc ðdbÞ ca h i b a cdb x þ ba ab b cdb a a ¼ 2 ad bc ðdbÞ ca
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Similarly, applying the algorithm again or simply replacing a by c and b by d in the above steps, we can obtain
ex þ f ¼
h i b a cdb x þ bc ad b cdb a a
ad bc ðdbÞ ca
2
:
Hence, the partial fraction decomposition of FðxÞ is given by
i h i 3 2h db db x þ bc ad b cdb b a cdb a a 1 4 b a ca x þ ba ab b ca 5; FðxÞ ¼ þ k x2 þ ax þ b x2 þ cx þ d where k ¼ ad bc ðd bÞ2 =ðc aÞ. 5. Conclusion In this paper, we have presented a simple and algorithmic cover-up approach for computing partial fraction decompositions with linear or irreducible quadratic factors in the denominators. This method has the advantage that there is no need to factorize the irreducible quadratic factors into linear factors with complex coefficients, to use differentiation or to solve a system of linear equations. It can be applied to derive some useful formulas for partial fraction decompositions, as illustrated in the last two examples. Due to its simplicity and algorithmic in nature, it is suitable for hand calculations or implementations in computer algebra systems for computing partial fraction decompositions. Acknowledgements The author is very grateful to the referees for their suggestions and comments. References [1] L.B. Norman, Discrete Mathematics, second ed., Oxford University Press, New York, 1990. [2] G. Chrystal, Algebra, Part I, Chelsea Publishing, New York, 1999. [3] A. de Paor, How to simplify differentiation when performing partial fraction expansion in the presence of a single multiple pole, Int. J. Elec. Eng. Educ. 40 (2003) 315–320. [4] S.H. Hou, S.H. Edwin, Hou, On partial fraction expansion with multiple poles, Int. J. Math. Educ. Sci. Technol. 35 (2004) 782–791. [5] D. Slota, R. Witula, Three bricks method of the partial fraction decomposition of some type of rational expression, Lect. Notes Comput. Sci. 3516 (2005) 659–662. [6] S.H. Kung, Partial fraction decomposition by division, College Math. J. 37 (2006) 132–134. [7] D.A. Rose, Partial fractions by substitutions, College Math. J. 38 (2007) 145–147. [8] R. Witula, D. Slota, Partial fractions decompositions of some rational functions, Appl. Math. Comput. 197 (2008) 328–336. [9] Y.K. Man, A simple algorithm for computing partial fraction expansions with multiple poles, Int. J. Math. Educ. Sci. Technol. 38 (2007) 247–251. [10] Y.K. Man, An improved Heaviside approach to partial fraction expansion and its applications, Int. J. Math. Educ. Sci. Technol. 40 (2009) 808–814. [11] Y.K. Man, On partial fraction decomposition of rational functions with irreducible quadratic factors in the denominators, in: Proceedings of the World Congress Engineering 2011, Vol. I, International Association of Engineers, London, 2011, pp. 237–239. [12] Y.K. Man, Allen Leung, Teaching a new method of partial fraction decomposition to senior secondary students: results and analysis from a pilot study, Int. J. Math. Teaching Learning (April) (2012) 1–18.
Contents lists available at SciVerse ScienceDirect
Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc
A cover-up approach to partial fractions with linear or irreducible quadratic factors in the denominators Yiu-Kwong Man Department of Mathematics and Information Technology, HKIEd, 10 Lo Ping Road, Tai Po, New Territories, Hong Kong
a r t i c l e
i n f o
Keywords: Partial fraction decomposition Cover-up approach Irreducible quadratic factors
a b s t r a c t In this paper, a simple cover-up approach for computing partial fractions with linear or irreducible quadratic factors in the denominators is presented. It mainly involves polynomial divisions and substitutions only, without having to factorize the irreducible quadratic factors into linear factors with complex coefficients, to use differentiation or to solve a system of linear equations. Examples and its application to derive some useful formulas of partial fraction decompositions are included. Ó 2012 Elsevier Inc. All rights reserved.
1. Introduction Partial fraction decomposition is an important topic in the study of calculus, differential equations, control theory and some areas of pure or applied mathematics. One common approach to compute the partial fractions is to use the method of undetermined coefficients. Another approach is to apply the Heaviside’s cover-up technique, which uses substitutions to determine the unknown coefficients of the partial fractions with single poles, and successive differentiations to handle those with multiple poles (see [1]). However, the computations involved could be quite tedious for the latter case. Thus, there are a number of alternative approaches proposed to handle partial fractions of rational functions from time to time. For instance, one can refer to [3–12] for some recent developments on this topic. In [9], an improved Heaviside’s coverup approach was proposed to handle the case of multiple poles by substitutions and polynomial divisions only. However, the drawback is that this approach cannot be effectively applied to compute partial fractions with irreducible quadratic factors only. To fill-up this gap, we now present a further improved cover-up approach in this paper, for handling partial fractions with linear or irreducible quadratic factors by polynomial divisions and substitutions only, without having to factorize the irreducible quadratic factors into linear factors with complex coefficients, to use differentiation or to solve a system of linear equations. Examples and its applications to derive some useful formulas of partial fraction decompositions are included. 2. The cover-up approach The existence of partial fraction decompositions of proper rational functions is based on the following result, whose proof can be found in [2]. Theorem 1. Let F be a constant field and aðxÞ and bðxÞ be polynomials in F½x such that deg aðxÞ < deg bðxÞ and
bðxÞ ¼ ðx a1 Þn1 ðx as Þns ðx2 þ b1 x þ c1 Þm1 ðx2 þ br x þ cr Þmr
E-mail address: [email protected] 0096-3003/$ - see front matter Ó 2012 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.amc.2012.10.016
ð1Þ
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where n1 ; . . . ; ns ; m1 ; . . . ; mr are positive integers, a1 ; . . . ; as ; b1 ; . . . ; br , c1 ; . . . ; cr 2 F; ak are distinct, x2 þ bi x þ ci are distinct irreducible polynomials, 1 6 k 6 s and 1 6 i 6 r. Then aðxÞ=bðxÞ has a unique partial fraction expansion of the form
X s r ai;ni bi;mi x þ ci;mi aðxÞ X ai;1 bi;1 x þ ci;1 þ ¼ þ þ þ þ bðxÞ i¼1 ðx ai Þ ðx2 þ bi x þ ci Þ ðx ai Þni ðx2 þ bi x þ ci Þmi i¼1
ð2Þ
where ai;j ; bi;j ; ci;j 2 F. A cover-up approach for computing the unknown partial fractions is given by the theorem below, which was proposed and proved by the author in [11]. Theorem 2. The unknown partial fraction coefficients described in the above theorem can be computed by the procedure below. Case (I): Partial fractions with linear factors in the denominators.
Step 1 : aðxÞ ðx ai Þni ai;ni ¼ bðxÞ x¼ai Step 2 : ai;ni j ¼
"
j1 aðxÞ X ai;ni k bðxÞ k¼0 ðx ai Þni k
ð3Þ !
# ðx ai Þni j
ð4Þ
x¼ai
where 1 6 i 6 s and 1 6 j 6 ni 1. The cancellations of the common factors ðx ai Þni and ðx ai Þni j from the denominators of the rational functions concerned should be done before substituting x ¼ ai into the whole expressions inside the square brackets. Case (II): Partial fractions with irreducible quadratic factors in the denominators.
Step1 : bi;mi x þ ci;mi ¼
aðxÞ ðx2 þ bi x þ ci Þmi modðx2 þ bi x þ ci Þ bðxÞ
Step2 : bi;mi j x þ ci;mi j ¼
"
j1 aðxÞ X bi;mi k x þ ci;mi k bðxÞ k¼0 ðx2 þ bi x þ ci Þmi k
ð5Þ
!
# 2
ðx þ bi x þ ci Þ
mi j
modðx2 þ bi x þ ci Þ
ð6Þ
where 1 6 i 6 r and 1 6 j 6 mi 1. The cancellations of the common factors ðx2 þ bi x þ ci Þmi and ðx2 þ bi x þ ci Þmi j from the denominators of the rational functions concerned should be done before applying the operation ‘mod ðx2 þ bi x þ ci Þ’ to the whole expressions inside the square brackets. Some general remarks related to Theorem 2 are provided below: In Case (I), the first step is to eliminate ðx ai Þni in bðxÞ and include ðx ai Þ or its higher power to each partial fraction, except the one containing ai;ni . Hence, ai;ni can be found by substituting x ¼ ai into Eq. (3). This technique is the so-called cover-up technique as described in [1]. The second step is to exploit the uniqueness of partial fraction decompositions of rational functions. Suppose ai;1 ðxÞ=bi;1 ðxÞ is equal to aðxÞ=bðxÞ ai;ni =ðx ai Þni . We can use polynomial division to eliminate the common factor ðx ai Þ in ai;1 ðxÞ=bi;1 ðxÞ, so that the highest power of ðx ai Þ in bi;1 ðxÞ is equal to ðni 1Þ. Thus, the multiplicity of the pole ai is reduced by one. By repeating this process on the new rational function obtained at each time, we can determine the unknown coefficients ai;ni j successively by using Eq. (4). In Case (II), the first step is to eliminate ðx2 þ bi x þ ci Þmi in bðxÞ and include ðx2 þ bi x þ ci Þ or its higher power to each partial fraction, except the one containing bi;mi x þ ci;mi . Hence, bi;mi x þ ci;mi can be found by applying the operation ‘mod ðx2 þ bi x þ ci Þ’ to the whole expression, which is equivalent to finding the remainder when each polynomial involved is divided by ðx2 þ bi x þ ci Þ. If the result is not a linear polynomial, we can use polynomial division to express its numerator and denominator in the forms ðx2 þ bi x þ ci Þqi;1 ðxÞ þ ðui;1 x þ v i;1 Þ and ðx2 þ bi x þ ci Þqi;2 ðxÞ þ ðui;2 x þ v i;2 Þ, respectively. By applying ‘mod ðx2 þ bi x þ ci Þ’, only the linear parts ui;1 x þ v i;1 and ui;2 x þ v i;2 will be left. By multiplying pi x þ qi to each of them, where pi ¼ 1=ui;2 and qi ¼ ½bi ðv i;2 =ui;2 Þ=ui;2 , we can apply ‘mod ðx2 þ bi x þ ci Þ’ again to reduce ðui;2 x þ v i;2 Þðpi x þ qi Þ and ðui;1 x þ v i;1 Þðpi x þ qi Þ into a constant and a linear polynomial, respectively. Thus, bi;mi x þ ci;mi can be determined. The second step is to exploit the uniqueness of partial fraction decompositions of rational functions. Suppose ai;1 ðxÞ=bi;1 ðxÞ is equal to aðxÞ=bðxÞ ðbi;mi x þ ci;mi Þ=ðx2 þ bi x þ ci Þmi . We can use polynomial division to eliminate the common factor ðx2 þ bi x þ ci Þ in ai;1 ðxÞ=bi;1 ðxÞ, so that the highest power of ðx2 þ bi x þ ci Þ in bi;1 ðxÞ is equal toðmi 1Þ. Hence, the power of ðx2 þ bi x þ ci Þ is reduced by one. By repeating this process on the new rational function obtained at each time, we can determine the unknown coefficients bi;mi j x þ ci;mi j successively by using Eq. (6). It is worth to point out that the main ideas behind Theorem 2 can be regarded as a technique for rewriting the higher degree terms in terms of lower degree terms. For instance, if there are higher degree terms xk appeared in Case (II), where k P 2, then they can be reduced by applying the operation ‘mod ðx2 þ bi x þ ci Þ’ into a linear polynomial. Thus, this technique
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extends the original Heaviside’s cover-up approach to handle partial fraction decompositions with linear or irreducible quadratic factors (with simple or multiple poles), without having to use differentiations. 3. Algorithms For the ease of hand calculations or implementation in computer algebra systems, we can rewrite Theorem 2 in an algorithmic forms below so that one can follow the calculation steps more easily. In brief, the first two algorithms are designed for handling Case (I) and Case (II) of Theorem 2, respectively. The third algorithm is a subprogram to be called by the second algorithm during the calculations. For convenience, we will use the abbreviations numer and denom to represent numerator and denominator, respectively. As the main ideas and results behind the algorithms have been described in Section 2 and the proof of Theorem 2 can be found in [11], so the formal proofs of the algorithms will not be provided here. Algorithm 1. Partial_Fraction_linear Input: A proper rational function a(x)/b(x), where aðxÞ; bðxÞ are polynomials, bðxÞ ¼ ðx a1 Þn1 ðx as Þns ; n1 ; . . . ; ns are positive integers and a1 ; . . . ; as are distinct constants. Output: The partial fraction coefficients ai;j as described in Theorem 2, where 1 6 i 6 s; 1 6 j 6 ni : Procedure: rðxÞ ¼ aðxÞ=bðxÞ For i = 1 to s do For j = 0 to ni 1 do dðxÞ ¼ denomðrðxÞÞ ðx ai Þni j r 1 ðxÞ ¼ numerðrðxÞÞ dðxÞ ai;ni j ¼ r1 ðai Þ sðxÞ ¼ rðxÞ
ai;n j i
ðxai Þni j
cðxÞ ¼ numerðsðxÞÞ ðx ai Þ dðxÞ ¼ denomðsðxÞÞ ðx ai Þ cðxÞ rðxÞ ¼ dðxÞ
end do end do return fai;j g where 1 6 i 6 s; 1 6 j 6 ni . end Algorithm 2. Partial_Fraction_quadratic Input: A proper rational function a(x)/b(x), where aðxÞ; bðxÞ are polynomials, bðxÞ ¼ ðx2 þ b1 x þ c1 Þm1 ðx2 þ br x þ cr Þmr ; b1 ; . . . ; br ; c1 ; . . . ; cr are constants;m1 ; . . . ; mr are positive integers and x2 þ bi x þ ci are distinct irreducible polynomials and 1 6 i 6 r. Output: The polynomials bi;j x þ ci;j as described in Theorem 2, where 1 6 i 6 r; 1 6 j 6 mi . Procedure: sðxÞ ¼ aðxÞ=bðxÞ For i = 1 to r do For j = 0 to mi 1 do dðxÞ ¼ denomðsðxÞÞ ðx2 þ bi x þ ci Þmi j s1 ðxÞ ¼ numerðsðxÞÞ dðxÞ sðxÞ ¼ s1 ðxÞmodðx2 þ bi x þ ci Þ if sðxÞ is a polynomial then bi;mi j x þ ci;mi j ¼ sðxÞ else bi;mi j x þ ci;mi j = Simp PolyðnumerðsðxÞÞ; denomðsðxÞÞ; x2 þ bi x þ ci Þ endif s1 ðxÞ ¼ sðxÞ
bi;m j xþci;m j i
i
ðx2 þbi xþci Þmi j
(continued on next page)
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cðxÞ ¼ numerðs1 ðxÞÞ ðx2 þ bi x þ ci Þ dðxÞ ¼ denomðs1 ðxÞÞ ðx2 þ bi x þ ci Þ cðxÞ sðxÞ ¼ dðxÞ
end do end do return fbi;j x þ ci;j g where 1 6 i 6 r; 1 6 j 6 mi : end Algorithm 3. Simp_Poly Input: Polynomials mðxÞ; nðxÞ and x2 þ bx þ c, where degðmðxÞÞ < degðnðxÞÞ. h i 2 Output: The result of mðxÞ nðxÞ modðx þ bx þ cÞ: Procedure: u1 x þ v 1 ¼ mðxÞmodðx2 þ bx þ cÞ sðxÞ ¼ nðxÞmodðx2 þ bx þ cÞ if sðxÞ is a constant then rðxÞ ¼ ðu1 x þ v 1 Þ=sðxÞ else u2 ¼ the coefficient of x in sðxÞ v 2 ¼ theconstanttermofsðxÞ p ¼ 1=u2 q ¼ ðb v 2 pÞ=u2 w1 x þ w2 ¼ ðu1 x þ v 1 Þðpx þ qÞmodðx2 þ bx þ cÞ s ¼ ðu2 x þ v 2 Þðpx þ qÞ mod ðx2 þ bx þ cÞ rðxÞ ¼ ðw1 x þ w2 Þ=s endif return rðxÞ end 4. Examples We now illustrate how to apply the cover-up technique to solve the problems below. Example 1. Find the partial fraction decomposition of the rational function
FðxÞ ¼ SOLUTION
3x2 6x 2 ðx 1Þ3
:
Suppose the partial fraction decomposition of FðxÞ is given by
FðxÞ ¼
a b c þ ; þ x 1 ðx 1Þ2 ðx 1Þ3
where a, b, c are unknown coefficients to be determined. Using the cover-up technique, we have
denomðFðxÞÞ ðx 1Þ3 ¼ ðx 1Þ3 ðx 1Þ3 ¼ 1 2 3x 6x 2 2 c¼ ¼ ð3x 6x 2Þ ¼ 3ð1Þ2 6ð1Þ 2 ¼ 5 1 x¼1 x¼1 5
3x2 6x þ 3
ð3x 3Þðx 1Þ ¼ ðx 1Þ ðx 1Þ3 ðx 1Þ3 3x 3 ¼ Note : Only one ðx 1Þ is removed at this stage: ðx 1Þ2
F 1 ðxÞ ¼ FðxÞ þ
3
¼
denomðF 1 ðxÞÞ ðx 1Þ2 ¼ ðx 1Þ2 ðx 1Þ2 ¼ 1 3x 3 ¼ ð3x 3Þ ¼ 3ð1Þ 3 ¼ 0 b¼ 1 x¼1 x¼1 3ðx 1Þ 3 F 2 ðxÞ ¼ F 1 ðxÞ 0 ¼ ¼ ðx 1Þ2 x 1 denomðF 2 ðxÞÞ ðx 1Þ ¼ ðx 1Þ ðx 1Þ ¼ 1 3 a¼ ¼3 j 1 x¼1
Y.-K. Man / Applied Mathematics and Computation 219 (2012) 3855–3862
Hence,
FðxÞ ¼
3 5 : x 1 ðx 1Þ3
Example 2. Find the partial fraction decomposition of the rational function
FðxÞ ¼ SOLUTION
3x þ 1 x2 ðx2 þ 1Þ2
:
Suppose the partial fraction decomposition of FðxÞ is given by
FðxÞ ¼
a b cx þ d ex þ f þ þ ; þ x x2 x2 þ 1 ðx2 þ 1Þ2
where a, b, c, d, e, f are unknown coefficients to be determined. Using the cover-up technique, we have
denomðFðxÞÞ ðx2 þ 1Þ2 ¼ x2 ðx2 þ 1Þ2 ðx2 þ 1Þ2 ¼ x2 3x þ 1 3x þ 1 modðx2 þ 1Þ ¼ ¼ 3x 1 ex þ f ¼ x2 1 3x þ 1 ð3x þ 1Þðx2 þ 1Þ 3x þ 1 F 1 ðxÞ ¼ FðxÞ þ ¼ ¼ 2 2 2 2 x ðx þ 1Þ ðx þ 1Þ x2 ðx2 þ 1Þ2 denomðF 1 ðxÞÞ ðx2 þ 1Þ ¼ x2 ðx2 þ 1Þ ðx2 þ 1Þ ¼ x2 3x þ 1 3x þ 1 modðx2 þ 1Þ ¼ ¼ 3x 1 cx þ d ¼ x2 1 3x þ 1 ð3x þ 1Þðx2 þ 1Þ 3x þ 1 ¼ ¼ F 2 ðxÞ ¼ F 1 ðxÞ þ 2 x þ1 x2 ðx2 þ 1Þ x2 denomðF 2 ðxÞÞ x2 ¼ x2 x2 ¼ 1 3x þ 1 ¼ ð3x þ 1Þ ¼ 3ð0Þ þ 1 ¼ 1 b¼ 1 x¼0 x¼0 1 3x 3 F 3 ðxÞ ¼ F 2 ðxÞ 2 ¼ 2 ¼ x x x denomðF 3 ðxÞÞ x ¼ x x ¼ 1 3 a¼ ¼3 1 x¼0
Hence,
FðxÞ ¼
3 1 3x þ 1 3x þ 1 þ : x x2 x2 þ 1 ðx2 þ 1Þ2
Example 3. Find the partial fraction decomposition of the rational function
FðxÞ ¼ SOLUTION
2x 1 : ðx2 þ 1Þðx2 þ x þ 1Þ
Suppose the partial fraction decomposition of FðxÞ is given by
FðxÞ ¼
ax þ b cx þ d þ ; x2 þ 1 x2 þ x þ 1
where a, b, c, d are unknown coefficients to be determined. Using the cover-up technique, we have
denomðFðxÞÞ ðx2 þ 1Þ ¼ ðx2 þ 1Þðx2 þ x þ 1Þ ðx2 þ 1Þ ¼ x2 þ x þ 1 2x 1 modðx2 þ 1Þ ax þ b ¼ 2 x þxþ1 2x 1 2x 1 x modðx2 þ 1Þ ¼ modðx2 þ 1Þ ¼ x x x 2 2x x 2 x modðx2 þ 1Þ ¼ ¼ xþ2 ¼ x2 1
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F 1 ðxÞ ¼ FðxÞ
xþ2 x3 3x2 x 3 ðx þ 3Þ ¼ ¼ x2 þ 1 ðx2 þ 1Þðx2 þ x þ 1Þ x2 þ x þ 1
denomðF 1 ðxÞÞ ðx2 þ x þ 1Þ ¼ ðx2 þ x þ 1Þ ðx2 þ x þ 1Þ ¼ 1 ðx þ 3Þ modðx2 þ x þ 1Þ ¼ ðx þ 3Þmodðx2 þ x þ 1Þ cx þ d ¼ 1 ¼ x 3 Hence,
FðxÞ ¼
xþ2 xþ3 : x 2 þ 1 x2 þ x þ 1
We now illustrate how to use the cover-up technique to derive some useful formulas for partial fraction decompositions below. Example 4. Prove the following identity
1 1 1 xþaþb ¼ ; 2 ðx aÞðx2 þ bx þ cÞ a þ ab þ c x a x2 þ bx þ c 2
where a, b, c are non-zero constants, b < 4c and a2 þ ab þ c – 0. 2 SOLUTION Let FðxÞ ¼ 1=ðx aÞðx þ bx þ cÞ. Suppose its partial fraction decomposition is given by
FðxÞ ¼
d ex þ f þ ; x a x2 þ bx þ c
where d, e, f are unknown coefficients to be determined. Using the cover-up technique, we have
denomðFðxÞÞ ðx2 þ bx þ cÞ ¼ ðx aÞðx2 þ bx þ cÞ ðx2 þ bx þ cÞ ¼ x a 1 modðx2 þ bx þ cÞ ex þ f ¼ xa 1 xþbþa modðx2 þ bx þ cÞ ¼ xa xþbþa xþaþb ðx þ a þ bÞ modðx2 þ bx þ cÞ ¼ 2 ¼ 2 x þ bx aða þ bÞ a þ ab þ c xþaþb x2 þ bx þ c ¼ 2 2 þ ab þ cÞðx þ bx þ cÞ ða þ ab þ cÞðx aÞðx2 þ bx þ cÞ 1 ¼ 2 ða þ ab þ cÞðx aÞ
F 1 ðxÞ ¼ FðxÞ þ
ða2
denomðF 1 ðxÞÞ ðx aÞ ¼ ða2 þ ab þ cÞðx aÞ ðx aÞ ¼ a2 þ ab þ c 1 1 ¼ d¼ 2 a þ ab þ c x¼a a2 þ ab þ c Alternatively, we can determine the value of d more directly as follows:
denomðFðxÞÞ ðx aÞ ¼ ðx aÞðx2 þ bx þ cÞ ðx aÞ ¼ x2 þ bx þ c 1 1 ¼ d¼ 2 x þ bx þ c x¼a a2 þ ab þ c Hence,
FðxÞ ¼
1 1 xþaþb : a2 þ ab þ c x a x2 þ bx þ c
Example 5. Derive a formula for the partial fraction decomposition of the following rational function
FðxÞ ¼
ax þ b ; ðx2 þ ax þ bÞðx2 þ cx þ dÞ
where a; b; a; b; c; d are constants, a – c; a2 4b < 0 and c2 4d < 0. SOLUTION Suppose the partial fraction decomposition of FðxÞ is given by
Y.-K. Man / Applied Mathematics and Computation 219 (2012) 3855–3862
FðxÞ ¼
cx þ d ex þ f þ ; x2 þ ax þ b x2 þ cx þ d
where c, d, e, f are unknown coefficients to be determined. Case (I): Assume b ¼ d. Using the cover-up technique, we have
denomðFðxÞÞ ðx2 þ ax þ bÞ ¼ ðx2 þ ax þ bÞðx2 þ cx þ dÞ ðx2 þ ax þ bÞ
¼ x2 þ cx þ d
ax þ b modðx2 þ ax þ bÞ x2 þ cx þ d ax þ b ax þ b modðx2 þ ax þ bÞ ¼ modðx2 þ ax þ bÞ ¼ ðc aÞx þ ðd bÞ ðc aÞx ax þ b x þ a modðx2 þ ax þ bÞ ¼ ðc aÞx x þ a 1 ax2 þ ðaa þ bÞx þ ba ¼ modðx2 þ ax þ bÞ ca x2 þ ax 1 1 ½aðax bÞ þ ðaa þ bÞx þ ba ¼ ðbx þ ba abÞ ¼ bða cÞ bða cÞ
cx þ d ¼
Similarly, applying the algorithm again or simply replacing a by c in the above steps, we can obtain
ex þ f ¼
1 ðbx þ bc abÞ: bða cÞ
Hence, the partial fraction decomposition of FðxÞ is given by
FðxÞ ¼
1 bx þ ba ab bx þ bc ab : 2 2 bða cÞ x þ ax þ b x þ cx þ d
Case (II): Assume b – d. Using the cover-up technique, we have
denomðFðxÞÞ ðx2 þ ax þ bÞ ¼ ðx2 þ ax þ bÞðx2 þ cx þ dÞ ðx2 þ ax þ bÞ
¼ x2 þ cx þ d
ax þ b modðx2 þ ax þ bÞ x2 þ cx þ d ax þ b modðx2 þ ax þ bÞ ¼ ðc aÞx þ ðd bÞ 2 3 ax þ b h i5modðx2 þ ax þ bÞ ¼4 ðc aÞ x þ cdb a 2 " #3 x þ a cdb ax þ b a 5 h i modðx2 þ ax þ bÞ ¼4 x þ a cdb ðc aÞ x þ cdb a a 2 3
cx þ d ¼
ðax þ bÞ x þ a cdb 6 7 a 2 ¼6 2 7 4 5modðx þ ax þ bÞ db db ðc aÞ x2 þ ax þ a ca ca 2 h i 3
db db 2 6aðx þ axÞ þ b a ca x þ b a ca 7 7modðx2 þ ax þ bÞ ¼6 4 5 2 db b ðc aÞ a cdb a ca h i 3 2 ab þ b a cdb x þ b a cdb a a 5modðx2 þ a þ xbÞ ¼4 2 aðd bÞ ðc aÞb ðdbÞ ca h i 3 2 x þ b ab þ b a cdb a cdb a a 5modðx2 þ ax þ bÞ ¼4 2 ad bc ðdbÞ ca h i b a cdb x þ ba ab b cdb a a ¼ 2 ad bc ðdbÞ ca
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Similarly, applying the algorithm again or simply replacing a by c and b by d in the above steps, we can obtain
ex þ f ¼
h i b a cdb x þ bc ad b cdb a a
ad bc ðdbÞ ca
2
:
Hence, the partial fraction decomposition of FðxÞ is given by
i h i 3 2h db db x þ bc ad b cdb b a cdb a a 1 4 b a ca x þ ba ab b ca 5; FðxÞ ¼ þ k x2 þ ax þ b x2 þ cx þ d where k ¼ ad bc ðd bÞ2 =ðc aÞ. 5. Conclusion In this paper, we have presented a simple and algorithmic cover-up approach for computing partial fraction decompositions with linear or irreducible quadratic factors in the denominators. This method has the advantage that there is no need to factorize the irreducible quadratic factors into linear factors with complex coefficients, to use differentiation or to solve a system of linear equations. It can be applied to derive some useful formulas for partial fraction decompositions, as illustrated in the last two examples. Due to its simplicity and algorithmic in nature, it is suitable for hand calculations or implementations in computer algebra systems for computing partial fraction decompositions. Acknowledgements The author is very grateful to the referees for their suggestions and comments. References [1] L.B. Norman, Discrete Mathematics, second ed., Oxford University Press, New York, 1990. [2] G. Chrystal, Algebra, Part I, Chelsea Publishing, New York, 1999. [3] A. de Paor, How to simplify differentiation when performing partial fraction expansion in the presence of a single multiple pole, Int. J. Elec. Eng. Educ. 40 (2003) 315–320. [4] S.H. Hou, S.H. Edwin, Hou, On partial fraction expansion with multiple poles, Int. J. Math. Educ. Sci. Technol. 35 (2004) 782–791. [5] D. Slota, R. Witula, Three bricks method of the partial fraction decomposition of some type of rational expression, Lect. Notes Comput. Sci. 3516 (2005) 659–662. [6] S.H. Kung, Partial fraction decomposition by division, College Math. J. 37 (2006) 132–134. [7] D.A. Rose, Partial fractions by substitutions, College Math. J. 38 (2007) 145–147. [8] R. Witula, D. Slota, Partial fractions decompositions of some rational functions, Appl. Math. Comput. 197 (2008) 328–336. [9] Y.K. Man, A simple algorithm for computing partial fraction expansions with multiple poles, Int. J. Math. Educ. Sci. Technol. 38 (2007) 247–251. [10] Y.K. Man, An improved Heaviside approach to partial fraction expansion and its applications, Int. J. Math. Educ. Sci. Technol. 40 (2009) 808–814. [11] Y.K. Man, On partial fraction decomposition of rational functions with irreducible quadratic factors in the denominators, in: Proceedings of the World Congress Engineering 2011, Vol. I, International Association of Engineers, London, 2011, pp. 237–239. [12] Y.K. Man, Allen Leung, Teaching a new method of partial fraction decomposition to senior secondary students: results and analysis from a pilot study, Int. J. Math. Teaching Learning (April) (2012) 1–18.