Partial FractionsDate: 01/29/98 at 07:51:00 From: yanzhen Subject: Partial fractions Dear Dr. Maths, How do I express the following in partial fractions? 3 (i) --------- 1-(x^3) 2x (ii) --------- (x^3)+1 Thank you for your time! Date: 01/29/98 at 14:00:48 From: Doctor Rob Subject: Re: Partial fractions First you have to factor the denominators into linear or quadratic factors. In this case 1 - x^3 = (1 - x)*(1 + x + x^2) 1 + x^3 = (1 + x)*(1 - x + x^2) Those factors will be the denominators of the partial fractions. The numerators will be of lower degree, with unknown constant coefficients, so the numerator of a fraction with a degree-1 denominator will just be an unknown constant, and the numerator of a fraction with a degree-2 denominator will be a degree-1 polynomial with unknown constant coefficients. Thus 3/(1-x^3) = A/(1-x) + (C+B*x)/(1+x+x^2) where A, B, and C are constants, to be determined. This must be an identity, that is, true for all values of x. Now clear fractions, and you get 3 = A*(1+x+x^2) + (C+B*x)*(1-x) Expand and collect terms, bringing everything to one side: 0 = -3 + A + A*x + A*x^2 + C - C*x + B*x - B*x^2 0 = (A - B)*x^2 + (A + B - C)*x + (A + C - 3) This must still be an identity. Now there are two ways to proceed. First, if this is true for all x values, we can pick a few and substitute them in, and get a system of linear equations in A, B, and C to solve: x = 0: A + C - 3 = 0. x = 1: 3*A - 3 = 0. x = -1: A - 2*B + 2*C - 3 = 0. From the second equation A = 1, so from the first equation, C = 2, and then from the last equation, B = 1. The second method is observe that if this is an identity, then the coefficients of each power of x must all be zero. This gives the system of equations A - B = 0, A + B - C = 0, A + C - 3 = 0. The solution is again A = B = 1, C = 2. Either way, you have the identity 3/(1-x^3) = 1/(1-x) + (x+2)/(x^2+x+1) The second problem is similar. Using the above as a model, you should be able to solve it yourself. -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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