HOME David Terr Ph.D. Math, UC Berkeley 7.3. Partial Fractions One of the most important applications of linear algebra is in computing partial fraction expansions of rational functions. This is a very powerful technique used in computing integrals of rational functions in calculus. In this section we explain how to compute partial fraction expansions. Consider the rational function 1 / (1 - x2). We know the denominator factors as (1 - x)(1 + x). We also know that when we add two rational functions, one with denominator 1 - x and the other with denominator 1 + x, in general we get a rational function with denominator 1 - x2, because this is the least common multiple of the denominators of the original rational functions. It seems we should be then able to do this in reverse, that is, to break up a rational function with denominator 1 - x2 into two rational functions, one with deonominator 1 - x and the other with denominator 1 + x. In particular, we should be able to do so with 1 / (1 - x2). In other words, we would like to find constants A and B such that the following equation holds: To solve this equation, we multiply both sides by (1 - x2) = (1 - x)(1 + x), obtaining 1 = A(1 + x) + B(1 - x). Now this equation can only hold if it holds for both the linear and constant coefficients on each side. Equating constant coefficients yields the equation 1 = A + B. Similarly, equating linear coefficients yields the equation 0 = A - B. Thus we have the following system of linear equations in unknowns A and B: A + B = 1 A - B = 0 From Section (7.1) we know how to solve such a system. Rewriting it as a matrix yields Subtracting the first row from the second yields the matrix Dividing the second row by -2 yields Finally, subtracting the second row from the first yields Thus we see that A = B = 1/2, whence the partial fraction expansion becomes   Example 1: Express x / (1 - x2) as a sum of partial fractions. Solution: We solve this the same way as we did the previous example. We look for a solution of the form Multiplying the numerator and denominator each by 1 - x2 yields x = A(1 + x) + B(1 - x). Equating coefficients of 1 and x, we find A + B = 0 A - B = 1 which we write in matrix form as By applying elementary row operations to this matrix (we leave out the details), we arrive at the following matrix: This corresponds to the solution A = 1/2; B = -1/2, whence the partial fraction expansion is     Example 2: Compute the partial fraction expansion of (4x2 + 5x + 6) / (2x2 + 11x + 12). Solution: Since the degree of the numerator is not less than that of the denominator, we need to first divide the numerator by the denominator. (See Section P.6 to review how to do this.) Omitting the details, the result is 2 + (-17x - 18) / (2x2 + 11x + 12). Next we must factor the denominators. By trial and error or otherwise, we find that it is equal to (2x + 3)(x + 4). Thus we have Subtracting 2 from both sides and then multiplying both sides by 2x2 + 11x + 12, we find A(x + 4) + B(2x + 3) = -17x - 18. Equating coefficients of 1 and x yields the following system of linear equations in A and B: A + 2B = -17 4A + 3B = -18 Writing this as a matrix yields Once again omitting the details, the solution matrix is whence the solution is A = 3; B = -10, yielding the partial fraction expansion   Example 3: Compute the partial fraction expansion of (x2 - 3x + 5) / (x3 + x2 + x + 1). Solution: The denominator factors as (x + 1)(x2 + 1). Thus, we look for a partial fraction expansion of the form Multiplying both sides by x3 + x2 + x + 1 yields A(x2 + 1) + (Bx + C)(x + 1) = x2 - 3x + 5. Multiplying out the terms and gathering and equating coefficients yields the following system of linear equations in the three variables A, B and C: A + B = 1 B + C = -3 A + C = 5 This leads to the following matrix: Subtracting the first row from the third yields Adding the third row to the first now yields Adding the second row to the third yields Dividing the third row by 2, we get Subtracting the third row from the first gives us the matrix Finally, subtracting the third row from the second yields From this matrix, we read off the solution: A = 9/2; B = -7/2; C = 1/2. This gives us the following partial fraction expansion: Copyright © 2007-2009 - MathAmazement.com. All Rights Reserved.