General Expression for Partial FractionsDate: 02/09/99 at 04:01:38 From: TOMMY Subject: Partial fractions Could you please tell me the general algorithm to convert an expression into partial fraction form? Thank you! Date: 02/09/99 at 10:56:56 From: Doctor Rob Subject: Re: Partial fractions Thanks for writing to Ask Dr. Math! This applies to the quotient of two polynomials, g(x)/f(x). If g(x) has degree at least as large as that of f(x), then divide: g(x) = q(x)*f(x) + r(x), deg(r) < deg(f) or r = 0 g(x)/f(x) = q(x) + r(x)/f(x) Then you want to express r(x)/f(x) as a sum of fractions, and the degree of r(x) is smaller than that of f(x). This means that you can assume without loss of generality that g(x) has degree less than that of f(x). We let n = degree(f). Now if f(x) has real coefficients, we know that we can factor it as follows: t f(x) = K * PRODUCT (x - a[i])^e[i] * i=1 u PRODUCT (x^2 - b[j]*x + c[j])^d[j] j=1 where the a's, b's, c's, and K are all real numbers, and the d's and e's are positive integers. The a's are all the real roots of f(x), and the quadratic factors have as roots all the complex conjugate pairs of roots of f(x). Then the partial fraction theorem says that there are constants A[i,r], B[j,s], and C[j,s] such that t e[i] g(x)/f(x) = SUM SUM A[i,r]/(x-a[i])^r + i=1 r=1 u d[j] SUM SUM (B[j,s]*x+C[j,s])/(x^2-b[j]*x+c[j])^s j=1 s=1 This may be the "general expression" to which you were referring in your question. Next, multiply through by f(x), g(x) = SUM SUM A[i,r]*[f(x)/(x-a[i])^r] + SUM (B[j,s]*x+C[j,s])*[f(x)/(x^2-b[j]*x+c[j])^s] and do the indicated divisions, which will leave no remainders. Now there are two ways to find the A's, B's, and C's. The first is to expand everything in sight into a polynomial, and gather together all terms involving the same powers of x. Then you will have an equation stating the equality of g(x) on the left and a polynomial in x on the right side. For them to be equal, the coefficients of each power of x on each side must be equal, and this gives you n real linear equations in the n real unknown A's, B's, and C's. These you can solve by various methods, for example, Gaussian elimination. The second way is to substitute for x in the resulting equation n different values. Each will give you a real linear equation relating the values of the real unknown A's, B's, and C's. Again you can solve these by various methods, for example, Gaussian elimination. Example: Write (x+1)/x^4 using partial fractions. Then f(x) = x^4 so t = 1, u = 0, a[1] = 0, e[1] = 4, and the theorem says that there exist constants A[1,1], A[1,2], A[1,3], and A[1,4] such that (x+1)/x^4 = A[1,1]/x + A[1,2]/x^2 + A[1,3]/x^3 + A[1,4]/x^4 Multiplying through by x^4 and dividing out, you get x + 1 = A[1,1]*x^3 + A[1,2]*x^2 + A[1,3]*x + A[1,4] Now equating coefficients on both sides of the equation, we get A[1,1] = 0 A[1,2] = 0 A[1,3] = 1 A[1,4] = 1 This system of linear equations is already solved for us, so we are done, and (x+1)/x^4 = 1/x^3 + 1/x^4 If instead we substitute x = -1, 0, 1, and 2, we would get the system of equations 0 = -A[1,1] + A[1,2] - A[1,3] + A[1,4] 1 = A[1,4] 2 = A[1,1] + A[1,2] + A[1,3] + A[1,4] 3 = 8*A[1,1] + 4*A[1,2] + 2*A[1,2] + A[1,4] These have, of course, the same solutions as before. If the coefficients of f(x) are complex numbers, then factor f(x) completely over the complex numbers, obtaining t f(x) = K * PRODUCT (x-a[i])^e[i] i=1 where the a's are complex roots of f(x), the e's are positive integers, and K is a complex number. Then the partial fraction theorem says that you can find constants A[i,r] such that t e[i] g(x)/f(x) = SUM SUM A[i,r]/(x-a[i])^r i=1 r=1 This may be the "general expression" to which you were referring in your question. Now proceed as before, with u = 0. Now the n A's will be complex numbers, in general, that are found by solving n linear equations with complex coefficients. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
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