Partial fraction decompositionFrom Wikipedia, the free encyclopedia

In algebra, the partial fraction decomposition or partial fraction expansion of a rational fraction (that is a fraction such that the numerator and the denominator are bothpolynomials) is the operation that consists in expressing the fraction as a sum of a polynomial (possibly zero) and one or several fractions with a simpler denominator.

The importance of the partial fraction decomposition lies in the fact that it provides an algorithm for computing the antiderivative of a rational function.

In symbols, one can use partial fraction expansion to change a rational fraction in the form

where ƒ and g are polynomials, into an expression of the form

where gj (x) are polynomials that are factors of g(x), and are in general of lower degree. Thus, the partial fraction decomposition may be seen as the inverse procedure of themore elementary operation of addition of rational fractions, which produces a single rational fraction with a numerator and denominator usually of high degree. The fulldecomposition pushes the reduction as far as it will go: in other words, the factorization of g is used as much as possible. Thus, the outcome of a full partial fraction expansionexpresses that fraction as a sum of fractions, where:

the denominator of each term is a power of an irreducible (not factorable) polynomial andthe numerator is a polynomial of smaller degree than that irreducible polynomial. To decrease the degree of the numerator directly, the Euclidean division can be used,but in fact if ƒ already has lower degree than g this isn't helpful.

Contents

1 Basic principles2 Application to symbolic integration3 Procedure

3.1 Illustration3.2 Residue method

4 Over the reals4.1 General result

5 Examples5.1 Example 15.2 Example 25.3 Example 35.4 Example 4 (residue method)5.5 Example 5 (limit method)

6 The role of the Taylor polynomial7 Fractions of integers8 Notes9 References10 External links

Basic principlesThe basic principles involved are quite simple; it is the algorithmic aspects that require attention in particular cases. On the other hand, the existence of a decomposition of acertain kind is an assumption in practical cases, and the principles should explain which assumptions are justified.

Assume a rational function in one indeterminate x has a denominator that factors as

over a field K (we can take this to be real numbers, or complex numbers). If P and Q have no common factor, then R may be written as

for some polynomials A(x) and B(x) over K. The existence of such a decomposition is a consequence of the fact that the polynomial ring over K is a principal ideal domain, sothat

for some polynomials C(x) and D(x) (see Bézout's identity).

Using this idea inductively we can write R(x) as a sum with denominators powers of irreducible polynomials. To take this further, if required, write:

as a sum with denominators powers of F and numerators of degree less than F, plus a possible extra polynomial. This can be done by the Euclidean algorithm, polynomialcase. The result is the following theorem:

Let ƒ and g be nonzero polynomials over a field K. Write g as a product of powers of distinct irreducible polynomials :

There are (unique) polynomials b and aij with deg aij < deg pi such that

If deg ƒ < deg g, then b = 0.

Therefore, when the field K is the complex numbers, we can assume that each pi has degree 1 (by the fundamental theorem of algebra) the numerators will be constant. WhenK is the real numbers, some of the pi might be quadratic, so in the partial fraction decomposition a quotient of a linear polynomial by a power of a quadratic might occur.

In the preceding theorem, one may replace "distinct irreducible polynomials" by "pairwise coprime polynomials that are coprime with their derivative". For example, the pimay be the factors of the square-free factorization of g. When K is the field of the rational numbers, as it is typically the case in computer algebra, this allows to replacefactorization by greatest common divisor to compute the partial fraction decomposition.

Application to symbolic integrationFor the purpose of symbolic integration, the preceding result may be refined into

Let ƒ and g be nonzero polynomials over a field K. Write g as a product of powers of pairwise coprime polynomials which have no multiple root in analgebraically closed field:

There are (unique) polynomials b and cij with deg cij < deg pi such that

where denotes the derivative of

This reduces the computation of the antiderivative of a rational function to the integration of the last sum, with is called the logarithmic part, because its antiderivative is alinear combination of logarithms. In fact, we have

There are various methods to compute above decomposition. The one that is the simplest to describe is probably the so-called Hermite's method. As the degree of cij isbounded by the degree of pi, and the degree of b is the difference of the degrees of f and g (if this difference is non negative; otherwise, b=0), one may write these unknownspolynomials as polynomials with unknown coefficients. Reducing the two members of above formula to the same denominator and writing that the coefficients of each powerof x are the same in the two numerators, one gets a system of linear equations which can be solved to obtain the desired values for the unknowns coefficients.

Procedure

Given two polynomials and , where the αi are distinct constants and deg P < n, partial fractions are generally obtainedby supposing that

and solving for the ci constants, by substitution, by equating the coefficients of terms involving the powers of x, or otherwise. (This is a variant of the method of undeterminedcoefficients.)

A more direct computation, which is strongly related with Lagrange interpolation consists in writing

where is the derivative of the polynomial .

This approach does not account for several other cases, but can be modified accordingly:

If deg P deg Q, then it is necessary to perform the Euclidean division of P by Q, using polynomial long division, giving P(x) = E(x) Q(x) + R(x) with deg R < n.Dividing by Q(x) this gives

and then seek partial fractions for the remainder fraction (which by definition satisfies deg R < deg Q).

If Q(x) contains factors which are irreducible over the given field, then the numerator N(x) of each partial fraction with such a factor F(x) in the denominator must besought as a polynomial with deg N < deg F, rather than as a constant. For example, take the following decomposition over R:

Suppose Q(x) = (x − α)rS(x) and S(α) ≠ 0. Then Q(x) has a zero α of multiplicity r, and in the partial fraction decomposition, r of the partial fractions will involve thepowers of (x − α). For illustration, take S(x) = 1 to get the following decomposition:

Illustration

In an example application of this procedure, (3x + 5)/(1 − 2x)2 can be decomposed in the form

Clearing denominators shows that 3x + 5 = A + B(1 − 2x). Expanding and equating the coefficients of powers of x gives

5 = A + B and 3x = −2Bx

Solving for A and B yields A = 13/2 and B = −3/2. Hence,

Residue method

Over the complex numbers, suppose ƒ(x) is a rational proper fraction, and can be decomposed into

Let

then according to the uniqueness of Laurent series, aij is the coefficient of the term (x − xi)−1 in the Laurent expansion of gij(x) about the point xi, i.e., its residue

This is given directly by the formula

or in the special case when xi is a simple root,

when

Note that P(x) and Q(x) may or may not be polynomials.

Over the realsPartial fractions are used in real-variable integral calculus to find real-valued antiderivatives of rational functions. Partial fraction decomposition of real rational functions isalso used to find their Inverse Laplace transforms. For applications of partial fraction decomposition over the reals, see

Application to symbolic integration, abovePartial fractions in Laplace transforms

General result

Let ƒ(x) be any rational function over the real numbers. In other words, suppose there exist real polynomials functions p(x) and q(x)≠ 0, such that

By dividing both the numerator and the denominator by the leading coefficient of q(x), we may assume without loss of generality that q(x) is monic. By the fundamentaltheorem of algebra, we can write

where a1,..., am, b1,..., bn, c1,..., cn are real numbers with bi2 − 4ci < 0, and j1,..., jm, k1,..., kn are positive integers. The terms (x − ai) are the linear factors of q(x) which

correspond to real roots of q(x), and the terms (xi2 + bix + ci) are the irreducible quadratic factors of q(x) which correspond to pairs of complex conjugate roots of q(x).

Then the partial fraction decomposition of ƒ(x) is the following:

Here, P(x) is a (possibly zero) polynomial, and the Air, Bir, and Cir are real constants. There are a number of ways the constants can be found.

The most straightforward method is to multiply through by the common denominator q(x). We then obtain an equation of polynomials whose left-hand side is simply p(x) andwhose right-hand side has coefficients which are linear expressions of the constants Air, Bir, and Cir. Since two polynomials are equal if and only if their correspondingcoefficients are equal, we can equate the coefficients of like terms. In this way, a system of linear equations is obtained which always has a unique solution. This solution canbe found using any of the standard methods of linear algebra. It can also be found with limits (see Example 5).

Examples

Example 1

Here, the denominator splits into two distinct linear factors:

so we have the partial fraction decomposition

Multiplying through by x2 + 2x − 3, we have the polynomial identity

Substituting x = −3 into this equation gives A = −1/4, and substituting x = 1 gives B = 1/4, so that

Example 2

After long-division, we have

Since (−4)2 − 4×8 = −16 < 0, the factor x2 − 4x + 8 is irreducible, and the partial fraction decomposition over the reals has the shape

Multiplying through by x3 − 4x2 + 8x, we have the polynomial identity

Taking x = 0, we see that 16 = 8A, so A = 2. Comparing the x2 coefficients, we see that 4 = A + B = 2 + B, so B = 2. Comparing linear coefficients, we see that −8 = −4A + C =−8 + C, so C = 0. Altogether,

The following example illustrates almost all the "tricks" one would need to use short of consulting a computer algebra system.

Example 3

After long-division and factoring the denominator, we have

The partial fraction decomposition takes the form

Multiplying through by (x − 1)3(x2 + 1)2 we have the polynomial identity

Taking x = 1 gives 4 = 4C, so C = 1. Similarly, taking x = i gives 2 + 2i = (Fi + G)(2 + 2i), so Fi + G = 1, so F = 0 and G = 1 by equating real and imaginary parts. With C = G= 1 and F = 0, taking x = 0 we get A − B + 1 − E − 1 = 0, thus E = A − B.

We now have the identity

Expanding and sorting by exponents of x we get

We can now compare the coefficients and see that

with A = 2 − D and −A −3 D =−4 we get A = D = 1 and so B = 0, furthermore is C = 1, E = A − B = 1, F = 0 and G = 1.

The partial fraction decomposition of ƒ(x) is thus

Alternatively, instead of expanding, one can obtain other linear dependences on the coefficients computing some derivatives at x=1 and at x=i in the above polynomialidentity. (To this end, recall that the derivative at x=a of (x−a)mp(x) vanishes if m > 1 and it is just p(a) if m=1.) Thus, for instance the first derivative at x=1 gives

that is 8 = 2B + 8 so B=0.

Example 4 (residue method)

Thus, f(z) can be decomposed into rational functions whose denominators are z+1, z−1, z+i, z−i. Since each term is of power one, −1, 1, −i and i are simple poles.

Hence, the residues associated with each pole, given by

,

are

,

respectively, and

.

Example 5 (limit method)

Limits can be used to find a partial fraction decomposition.[1]

First, factor the denominator:

The decomposition takes the form of

As , the A term dominates, so the right-hand side approaches . Thus, we have

As , the right-hand side is

Thus, .

At , . Therefore, .

The decomposition is thus .

The role of the Taylor polynomialThe partial fraction decomposition of a rational function can be related to Taylor's theorem as follows. Let

be real or complex polynomials; assume that

that

and that

Define also

Then we have

if, and only if, for each the polynomial is the Taylor polynomial of of order at the point :

Taylor's theorem (in the real or complex case) then provides a proof of the existence and uniqueness of the partial fraction decomposition, and a characterization of thecoefficients.

Sketch of the proof: The above partial fraction decomposition implies, for each 1 ≤ i ≤ r, a polynomial expansion

, as

so is the Taylor polynomial of , because of the unicity of the polynomial expansion of order , and by assumption .

Conversely, if the are the Taylor polynomials, the above expansions at each hold, therefore we also have

, as

which implies that the polynomial is divisible by

For also is divisible by , so we have in turn that is divisible by . Since we thenhave , and we find the partial fraction decomposition dividing by .

Fractions of integersThe idea of partial fractions can be generalized to other integral domains, say the ring of integers where prime numbers take the role of irreducible denominators. For example:

Notes1. ^ Bluman, George W. (1984). Problem Book for First Year Calculus. New York: Springer-Verlag. pp. 250–251.

References

Rao, K. R.; Ahmed, N. (1968). "Recursive techniques for obtaining the partial fraction expansion of a rational function". IEEE Trans. Educ. 11 (2). pp. 152–154.doi:10.1109/TE.1968.4320370 (http://dx.doi.org/10.1109%2FTE.1968.4320370).Henrici, Peter (1971). "An algorithm for the incomplete decomposition of a rational function into partial fractions". Z. f. Angew. Mathem. Physik 22 (4). pp. 751–755.doi:10.1007/BF01587772 (http://dx.doi.org/10.1007%2FBF01587772).Chang, Feng-Cheng (1973). "Recursive formulas for the partial fraction expansion of a rational function with multiple poles". Proc. IEEE 61 (8). pp. 1139–1140.doi:10.1109/PROC.1973.9216 (http://dx.doi.org/10.1109%2FPROC.1973.9216).Kung, H. T.; Tong, D. M. (1977). "Fast Algorithms for Partial Fraction Decomposition". SIAM Journal on Computing 6 (3): 582. doi:10.1137/0206042(http://dx.doi.org/10.1137%2F0206042).Eustice, Dan; Klamkin, M. S. (1979). "On the coefficients of a partial fraction decomposition" 86 (6). pp. 478–480. JSTOR 2320421(https://www.jstor.org/stable/2320421).Mahoney, J. J.; Sivazlian, B. D. (1983). "Partial fractions expansion: a review of computational methodology and efficiency". J. Comp. Appl. Math. 9. pp. 247–269.doi:10.1016/0377-0427(83)90018-3 (http://dx.doi.org/10.1016%2F0377-0427%2883%2990018-3).Miller, Charles D.; Lial, Margaret L.; Schneider, David I. (1990). Fundamentals of College Algebra (3rd ed. ed.). Addison-Wesley Educational Publishers, Inc. pp. 364–370. ISBN 0-673-38638-4.Westreich, David (1991). "partial fraction expansion without derivative evaluation". IEEE Trans. Circ. Syst. 38 (6). pp. 658–660. doi:10.1109/31.81863(http://dx.doi.org/10.1109%2F31.81863).Kudryavtsev, L. D. (2001), "Undetermined coefficients, method of" (http://www.encyclopediaofmath.org/index.php?title=u/u095160), in Hazewinkel, Michiel,Encyclopedia of Mathematics, Springer, ISBN 978-1-55608-010-4Velleman, Daniel J. (2002). "Partial fractions, binomial coefficients and the integral of an odd power of sec theta". Am. Math. Monthly 109 (8). pp. 746–749.JSTOR 3072399 (https://www.jstor.org/stable/3072399).Slota, Damian; Witula, Roman (2005). "Three brick method of the partial fraction decomposition of some type of rational expression". Lect. Not. Computer Sci. 33516.pp. 659–662. doi:10.1007/11428862_89 (http://dx.doi.org/10.1007%2F11428862_89).Kung, Sidney H. (2006). "Partial fraction decomposition by division". Coll. Math. J. 37 (2): 132–134. JSTOR 27646303 (https://www.jstor.org/stable/27646303).Witula, Roman; Slota, Damian (2008). "Partial fractions decompositions of some rational functions". Appl. Math. Comput. 197. pp. 328–336.doi:10.1016/j.amc.2007.07.048 (http://dx.doi.org/10.1016%2Fj.amc.2007.07.048). MR 2396331 (https://www.ams.org/mathscinet-getitem?mr=2396331).

External links

Weisstein, Eric W., "Partial Fraction Decomposition" (http://mathworld.wolfram.com/PartialFractionDecomposition.html), MathWorld.Blake, Sam. "Step-by-Step Partial Fractions" (http://calc101.com/webMathematica/partial-fractions.jsp).[1] (http://cajael.com/eng/control/LaplaceT/LaplaceT-1_Example_2_6_OGATA_4editio.php) Make partial fraction decompositions with Scilab.

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