Partial fraction decomposition over the reals
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In mathematics, partial fractions are used in real-variable integral calculus to find real-valued antiderivatives of rational functions. The partial fraction decomposition of real rational functions is also used for Laplace transforms.
This article describes the general method for obtaining the partial fraction decomposition of any real rational function. The proof of the existence of the partial fraction decomposition over an arbitrary field is not given here. For a sketch of the general proof, see partial fraction.
For applications of partial fraction decomposition over the reals, see
First we state the general result, and then we offer several examples to illustrate the result and show how the method is used in practice.
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General result
Let f(x) be any rational function over the real numbers. In other words, suppose there exist real polynomials p(x) and q(x)≠ 0, such that
- <math>f(x) = \frac{p(x)}{q(x)}<math>
By removing 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
- <math>q(x) = (x-a_1)^{j_1}\cdots(x-a_m)^{j_m}(x^2+b_1x+c_1)^{k_1}\cdots(x^2+b_nx+c_n)^{k_n}<math>
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 f(x) is the following.
- <math>f(x) = \frac{p(x)}{q(x)} = P(x) + \sum_{i=1}^m\sum_{r=1}^{j_i} \frac{A_{ir}}{(x-a_i)^r} + \sum_{i=1}^n\sum_{r=1}^{k_i} \frac{B_{ir}x+C_{ir}}{(x^2+b_ix+c_i)^r}<math>
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) and whose 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 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. However, this is often not the best way to go when computing by pencil and paper, and there are other ways ("tricks") to obtain the constants.
The algorithm
The partial fraction decomposition of a real rational function can be obtained by the following procedure:
- Remove the leading coefficient of the denominator q(x) if necessary.
- Reduce the rational function as much as possible; i.e. cancel any obvious common factors of the numerator and denominator.
- Make sure the degree of the numerator p(x) is strictly smaller than the degree of the denominator q(x). If it is not, then use the division algorithm for polynomials and long-divide the denominator into the numerator. You will get a (nonzero) polynomial quotient and a remainder whose degree is now strictly smaller than the degree of the denominator. Take this remainder as your new numerator and continue with the following steps. (Don't forget the polynomial quotient when you're done!)
- Factor the denominator q(x) into linear and irreducible quadratic factors. This is not a straightforward task; however, you have some tools to use. If you find any real root a of q(x), then you know that (x - a) is a factor of q(x). Also, if you differentiate q(x) and use the Euclidean algorithm to find the greatest common divisor of q ' (x) and q(x), then any real root a of the greatest common divisor gives you a factor (x - a). You can look for ways to complete the square in different powers of x. Most importantly, you can use the rational root test to determine all possible rational roots of q(x).
- Express the partial fraction decomposition of f(x) as the sum of double sums above.
- Find the constants Air, Bir, and Cir by whatever methods possible. (These methods are described below.) Once you have found all the constants, you now have the partial fraction decomposition. (Don't forget the polynomial quotient!)
Some examples
Example 1
- <math>f(x)=\frac{1}{x^2+2x-3}<math>
Here, the denominator splits into two distinct linear factors:
- <math>q(x)=x^2+2x-3=(x+3)(x-1)<math>
so we have the partial fraction decomposition
- <math>f(x)=\frac{1}{x^2+2x-3}=\frac{A}{x+3}+\frac{B}{x-1}<math>
Multiplying through by x2 + 2x - 3, we have the polynomial identity
- <math>1=A(x-1)+B(x+3)<math>
Substituting x = -3 and x = 1 into this equation gives A = -1/4 and B = 1/4, so that
- <math>f(x)=\frac{1}{x^2+2x-3}=\frac{1}{4}\left(\frac{-1}{x+3}+\frac{1}{x-1}\right)<math>
Example 2
- <math>f(x)=\frac{x^3+16}{x^3-4x^2+8x}<math>
After long-division, we have
- <math>f(x)=1+\frac{4x^2-8x+16}{x^3-4x^2+8x}=1+\frac{4x^2-8x+16}{x(x^2-4x+8)}<math>
Since (-4)2-4(8) = -16 < 0, x2 - 4x + 8 is irreducible, and so
- <math>\frac{4x^2-8x+16}{x(x^2-4x+8)}=\frac{A}{x}+\frac{Bx+C}{x^2-4x+8}<math>
Multiplying through by x3 - 4x2 + 8x, we have the polynomial identity
- <math>4x^2-8x+16 = A(x^2-4x+8)+(Bx+C)x<math>
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,
- <math>f(x)=1+2\left(\frac{1}{x}+\frac{x}{x^2-4x+8}\right)<math>
The following example illustrates almost all the "tricks" one would need to use short of consulting a computer algebra system.
Example 3
- <math>f(x)=\frac{x^9-2x^6+2x^5-7x^4+13x^3-11x^2+12x-4}{x^7-3x^6+5x^5-7x^4+7x^3-5x^2+3x-1}<math>
After long-division and factoring, we have
- <math>f(x)=x^2+3x+4+\frac{x^5-2x^4+5x^3-5x^2+6x-1}{(x-1)^3(x^2+1)^2}<math>
The partial fraction decomposition takes the form
- <math>\frac{x^5-2x^4+5x^3-5x^2+6x-1}{(x-1)^3(x^2+1)^2}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{(x-1)^3}+\frac{Dx+E}{x^2+1}+\frac{Fx+G}{(x^2+1)^2}<math>
Multiplying through by (x - 1)3(x2 + 1)2 we have the polynomial identity
- <math>x^5-2x^4+5x^3-5x^2+6x-1<math>
- <math>=A(x-1)^2(x^2+1)^2+B(x-1)(x^2+1)^2+C(x^2+1)^2+(Dx+E)(x-1)^3(x^2+1)+(Fx+G)(x-1)^3<math>
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. We now have the identity
- <math>x^5-2x^4+5x^3-5x^2+6x-1<math>
- <math>=A(x-1)^2(x^2+1)^2+B(x-1)(x^2+1)^2+(x^2+1)^2+(Dx+E)(x-1)^3(x^2+1)+(x-1)^3<math>
Taking constant terms gives E = A - B + 1, taking leading coefficients gives A = -D, and taking x-coefficients gives B = 3 - D - 3E. Putting all of this together, E = A - B + 1 = -D - (3 - D - 3E) + 1 = 3E - 2, so E = 1 and A = B = -D. Now,
- <math>x^5-2x^4+5x^3-5x^2+6x-1<math>
- <math>=A(x-1)^2(x^2+1)^2+A(x-1)(x^2+1)^2+(x^2+1)^2+(-Ax+1)(x-1)^3(x^2+1)+(x-1)^3<math>
Taking x = -1 gives -20 = -8A - 20, so A = B = D = 0. The partial fraction decomposition of f(x) is thus
- <math>f(x)=x^2+3x+4+\frac{1}{(x-1)^3}+\frac{1}{x^2+1}+\frac{1}{(x^2+1)^2}<math>
General methods for finding coefficients
We present in the sequel a list of methods for finding the coefficients occurring in the r.h.s. sums of the above equation for f(x)=p(x)/q(x)=..., to which we shall refer to by (PFD).
For some of the methods it would be useful if p and/or q were already written in factored form.
Simple poles of multiplicity 1
If ai is a simple pole of f, i.e. ji=1, then one multiplies (PFD) by (x-ai) and takes the limit x→ai. On the r.h.s. only Ai1 "survives", which is therefore given by
- Ai1 = limx→ai (x-ai) f(x) = p(ai)/qi(ai)
where qi is q divided (simplified) by the factor (x - ai).
Leading term of a pole
Using the same idea, we determine the coefficients <math>A_{ij_i}<math> (multiplying in the highest negative power), by just replacing (x-ai) by its <math>j_i<math>-th power.
Leading term of a quadratic factor
Still using the same idea, but going to the complex root of x²+b x+c, one gets in the same way coefficients <math>B_{ik_i},C_{ik_i}<math> (from real and imaginary part of the previous equation for A which is now an equation for B x+C).
Other coefficients Aik
All coefficients relative to one pole can be obtained "in one step" by making a change of variables t = x - ai (i.e. replacing x by t + ai) and then (long) dividing the "new" p by the "new" q, with the "pure" power tji</sub> removed.
Method of derivatives
Instead of making the change of variables and using long division, one can obtain a recursive formula involving derivatives of p and qi at x = ai.
General methods
Besides the above methods to obtain a particular coefficient, there are several general methods to get one ore more equations involving one or several of the coefficients, which should then allow to determine the remaining unknown coefficients:
- method of particular values: this just consists in putting x equal to particular fixed values (0, ±1, ±2...), and thus obtaining numerical equations for the coefficients.
- method of limits: multiply both sides of (PFD), after subtracting known terms (especially the polynomial), by the highest possible power of x such that lim(x→∞) still is finite, and take this limit
- method of identification: the "final rescue", put all of the r.h.s. on the common denominator q(x) and identify coefficients of alike powers in the numerator on the r.h.s. and p(x) on the l.h.s.