د. محمد مدحت موسى-Dr. Mohamed M. Mousa

أستاذ مشارك بقسم الرياضيات-Associate Professor of Mathematics

Partial fraction

Partial fraction decomposition

From Wikipedia, the free encyclopedia
  (Redirected from Partial fractions in integration)
Jump to navigationJump to search

In algebra, the partial fraction decomposition or partial fraction expansion of a rational function (that is, a fraction such that the numerator and the denominator are both polynomials) is an operation that consists of expressing the fraction as a sum of a polynomial (possibly zero) and one or several fractions with a simpler denominator.[1]

The importance of the partial fraction decomposition lies in the fact that it provides algorithms for various computations with rational functions, including the explicit computation of antiderivatives,[2] Taylor series expansionsinverse Z-transformsinverse Laplace transforms. The concept was discovered in 1702 by both Johann Bernoulli and Gottfried Leibnizindependently.[3]

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

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

where:

  • the denominatorgj (x), of each fraction is a power of an irreducible (not factorable into polynomials of positive degree) polynomial and
  • the numerator is a polynomial of smaller degree than this irreducible polynomial.

As factorization of polynomials may be difficult, a coarser decomposition is often preferred, which consists of replacing factorization by square-free factorization. This amounts to replace "irreducible" by "square-free" in the preceding description of the outcome.

Basic principles[edit]

If 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) and if in addition P and Q have no common factor, then by Bézout's identity for polynomials, there exist polynomials C(x) and D(x) such that , and

Thus

and hence R may be written as

where all numerators are polynomials.

Using this idea inductively the rational function R(x) can be written as a sum with denominators being powers of irreducible polynomials. To take this further, if required, write:

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

Theorem — Let f 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 f < deg g, then b = 0.

If K is field of complex numbers, the fundamental theorem of algebra implies that all pi have degree one, and all numerators  are constants. When K is the field of real numbers, some of the pi may be quadratic, so, in the partial fraction decomposition, quotients of linear polynomials by powers of quadratic polynomials may also occur.

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

Application to symbolic integration[edit]

For the purpose of symbolic integration, the preceding result may be refined into

Theorem — Let f 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 an algebraically 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, which is called the logarithmic part, because its antiderivative is a linear 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 is bounded 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 unknowns polynomials as polynomials with unknown coefficients. Reducing the two members of above formula to the same denominator and writing that the coefficients of each power of 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[edit]

Given two polynomials  and , where the αi are distinct constants and deg P < n, partial fractions are generally obtained by 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 undetermined coefficients.)

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

where  is the derivative of the polynomial .

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

  • If  then it is necessary to perform the Euclidean division of P by Q, using polynomial long division, giving P(x) = E(xQ(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 be sought 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 the powers of (x − α). For illustration, take S(x) = 1 to get the following decomposition:

Illustration[edit]

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[edit]

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

Let

then according to the uniqueness of Laurent seriesaij 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

Over the reals[edit]

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

General result[edit]

Let f(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 fundamental theorem of algebra, we can write

where a1,..., amb1,..., bnc1,..., cn are real numbers with bi2 − 4ci < 0, and j1,..., jmk1,..., 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 f(x) is the following:

Here, P(x) is a (possibly zero) polynomial, and the AirBir, 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) and whose right-hand side has coefficients which are linear expressions of the constants AirBir, and Cir. Since two polynomials are equal if and only if their corresponding coefficients 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 can be found using any of the standard methods of linear algebra. It can also be found with limits (see Example 5).

Examples[edit]

Example 1[edit]

Here, the denominator splits into two distinct linear factors:

so we have the partial fraction decomposition

Multiplying through by the denominator on the left-hand side gives us 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[edit]

After long-division, we have

The factor x2 − 4x + 8 is irreducible over the reals, as its discriminant (−4)2 − 4×8 = − 16 is negative. Thus 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 fraction can be completely decomposed using complex numbers. According to the fundamental theorem of algebra every complex polynomial of degree n has n (complex) roots (some of which can be repeated). The second fraction can be decomposed to:

Multiplying through by the denominator gives:

Equating the coefficients of x and the constant (with respect to x) coefficients of both sides of this equation, one gets a system of two linear equations in D and E, whose solution is

Thus we have a complete decomposition:

One may also compute directly AD and E with the residue method (see also example 4 below).

Example 3[edit]

This example illustrates almost all the "tricks" we might need to use, short of consulting a computer algebra system.

After long-division and factoring the denominator, we have

The partial fraction decomposition takes the form

Multiplying through by the denominator on the left-hand side we have the polynomial identity

Now we use different values of x to compute the coefficients:

Solving this we have:

Using these values we can write:

We compare the coefficients of x6 and x5 on both side and we have:

Therefore:

which gives us B = 0. Thus the partial fraction decomposition is given by:

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

that is 8 = 4B + 8 so B = 0.

Example 4 (residue method)[edit]

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)[edit]

Limits can be used to find a partial fraction decomposition.[4] Consider the following example:

First, factor the denominator which determines the decomposition:

Multiplying everything by , and taking the limit when , we get

On the other hand

and thus:

Multiplying by x and taking the limit when , we have

and

This implies A + B = 0 and so .

For x = 0, we get  and thus .

Putting everything together, we get the decomposition

Example 6 (integral)[edit]

Suppose we have the indefinite integral:

Before performing decomposition, it is obvious we must perform polynomial long division and factor the denominator. Doing this would result in:

Upon this, we may now perform partial fraction decomposition.

so:

.

Upon substituting our values, in this case, where x=1 to solve for B and x=-2 to solve for A, we will result in:

Plugging all of this back into our integral allows us to find the answer:

The role of the Taylor polynomial[edit]

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

be real or complex polynomials assume that

satisfies

Also define

Then we have

if, and only if, each 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 the coefficients.

Sketch of the proof[edit]

The above partial fraction decomposition implies, for each 1 ≤ i ≤ r, a polynomial expansion

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

which implies that the polynomial  is divisible by 

For  is also divisible by , so

is divisible by . Since

we then have

and we find the partial fraction decomposition dividing by .

Fractions of integers[edit]

The 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:

Notes[edit]

  1. ^ Larson, Ron (2016). Algebra & Trigonometry. Cengage Learning. ISBN 9781337271172.
  2. ^ Horowitz, Ellis. "Algorithms for partial fraction decomposition and rational function integration." Proceedings of the second ACM symposium on Symbolic and algebraic manipulation. ACM, 1971.
  3. ^ Grosholz, Emily (2000). The Growth of Mathematical Knowledge. Kluwer Academic Publilshers. p. 179. ISBN 978-90-481-5391-6.
  4. ^ Bluman, George W. (1984). Problem Book for First Year Calculus. New York: Springer-Verlag. pp. 250–251.

References[edit]

  • Rao, K. R.; Ahmed, N. (1968). "Recursive techniques for obtaining the partial fraction expansion of a rational function". IEEE Trans. Educ11 (2). pp. 152–154. doi:10.1109/TE.1968.4320370.
  • Henrici, Peter (1971). "An algorithm for the incomplete decomposition of a rational function into partial fractions". Z. Angew. Math. Phys22 (4). pp. 751–755. doi:10.1007/BF01587772.
  • Chang, Feng-Cheng (1973). "Recursive formulas for the partial fraction expansion of a rational function with multiple poles". Proc. IEEE61 (8). pp. 1139–1140. doi:10.1109/PROC.1973.9216.
  • Kung, H. T.; Tong, D. M. (1977). "Fast Algorithms for Partial Fraction Decomposition". SIAM Journal on Computing6 (3): 582. doi:10.1137/0206042.
  • Eustice, Dan; Klamkin, M. S. (1979). "On the coefficients of a partial fraction decomposition". American Mathematical Monthly86 (6). pp. 478–480. JSTOR 2320421.
  • Mahoney, J. J.; Sivazlian, B. D. (1983). "Partial fractions expansion: a review of computational methodology and efficiency". J. Comput. Appl. Math9. pp. 247–269. doi:10.1016/0377-0427(83)90018-3.
  • Miller, Charles D.; Lial, Margaret L.; Schneider, David I. (1990). Fundamentals of College Algebra (3rd 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. Syst38 (6). pp. 658–660. doi:10.1109/31.81863.
  • Kudryavtsev, L. D. (2001) [1994], "Undetermined coefficients, method of", in Hazewinkel, MichielEncyclopedia of Mathematics, Springer Science+Business Media B.V. / Kluwer Academic Publishers, ISBN 978-1-55608-010-4
  • Velleman, Daniel J. (2002). "Partial fractions, binomial coefficients and the integral of an odd power of sec theta". Amer. Math. Monthly109 (8). pp. 746–749. JSTOR 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.
  • Kung, Sidney H. (2006). "Partial fraction decomposition by division". Coll. Math. J37 (2): 132–134. JSTOR 27646303.
  • Witula, Roman; Slota, Damian (2008). "Partial fractions decompositions of some rational functions". Appl. Math. Comput197. pp. 328–336. doi:10.1016/j.amc.2007.07.048MR 2396331.

جامعة المجمعة

أهلاً ومرحباً بكم

كلية العلوم والدراسات الإنسانية

بحوطة سدير

قسم الرياضيات

التوقيت والتقويم





 








توقيت الصلاة بمدينة حوطة سدير


محرك بحث جوجل

للتواصل


  1. الهاتف : 0164044771

تحويلة: 4771


mm.mousa@mu.edu.sa

dr.eng.mmmm@gmail.com

(QR Code)

mailto:mm.mousa@mu.edu.sa


إعلانات

1- الاختبار الفصلى الثانى لمقرر التحليل العددى (يوم الاحد الموافق 3 / 7/ 1440 هـ)

2- الاختبار الفصلى الثانى لمقررحساب المتجهات (يوم الثلاثاء الموافق 5 / 7 / 1440 هـ)

الساعات المكتبية

الأثنين: 10 - 12

الثلاثاء: 8 - 10

الأربعاء: 8 - 10

أخبار الجامعة والكلية

أخبار الجامعة

أخبار الكلية


اللوائح الطلابية بجامعة المجمعة

روابط مفيدة على موقع الجامعة












مواقع التواصل الإجتماعى

آلة حاسبة

التقويم الجامعى

التقويم الجامعى 1440/1439




بعض الجوائز والتكريمات









إحصائية الموقع

عدد الصفحات: 258

البحوث والمحاضرات: 155

الزيارات: 68003