Proof of the Partial Fractions Theorem for Quadratic FactorsDate: 10/14/2001 at 00:54:30 From: William Subject: Proof of the Partial Fractions Theorem for Quadratic Factors Hello, I was reviewing a chapter in my calculus book about Integration using Partial Fractions. The concept seems to be clear for linear factors and repeated linear factors, but why is it that when you have a non-reducible quadratic factor, you have to let the numerator of the partial fraction be Ax+B? Could you please show me a proof or explain me why it is Ax+B instead of just A? Thanks. Date: 10/14/2001 at 12:30:54 From: Doctor Fenton Subject: Re: Proof of the Partial Fractions Theorem for Quadratic Factors Hi William, Thanks for writing to Dr. Math. Essentially, the reason you have to allow such terms is that the theorem isn't true unless you do. Partial fractions "undoes" the operation of adding rational functions by finding a common denominator. To decompose P(x)/Q(x), you first make sure the rational function is "proper": P(x)/Q(x) is proper if the degree of P(x) is strictly less than the degree of Q(x). If the degree of P(x) is greater than or equal to the degree of Q(x), you carry out long division A(x) ------- Q(x) | P(x) : ---- R(x) getting a quotient of A(x) (a polynomial) and a remainder R(x), whose degree is (strictly) less than the degree of Q(x), so that P(x) R(x) ---- = A(x) + ---- Q(x) Q(x) , and you apply partial fractions decomposition to R(x)/Q(x). When decomposing a proper fraction into partial fractions, you must allow enough proper fractions to occur in the decomposition. For example, Ax+B --------- x^2+Cx+D is a proper fraction, and you must allow for its occurrence. There is no way to write x A --------- = ------- x^2 + 1 x^2 + 1 . If this equation were true for all x, you could multiply by x^2+1 (which is never 0) and get x = A, which means that x can have only one value, contradicting the claim that the equation is true for ALL x. So, if you don't allow for enough types of proper fractions in the decomposition into partial fractions, you won't be able to decompose the original rational function. However, you can prove that if you do allow partial fraction terms of the form Ax+B Ex+F -------- and ---------- (for repeated irreducible factors) x^2+Cx+D (x^2+Cx+D)^k then the decomposition is always possible. Does this answer your question adequately? If you still have questions, please write back and I will ry to explain further. - Doctor Fenton, The Math Forum http://mathforum.org/dr.math/ Date: 10/14/2001 at 20:02:07 From: (anonymous) Subject: Re: Proof of the Partial Fractions Theorem for Quadratic Factors Hi, Thanks for your explanation. I asked my Calculus teacher the same question and he told me that in order to understand a formal Proof of the 'Partial Fractions Theorem for Irreducible Quadratic Factors' I must have knowledge of Abstract Algebra. He told me this proof went beyond the scope of our Calculus class. I would like to see that theorem. I have searched the Internet and I haven't been able to find it. Do you think you could show it to me? Thanks, William Date: 10/15/2001 at 21:01:46 From: Doctor Fenton Subject: Re: Proof of the Partial Fractions Theorem for Quadratic Factors Hi again William, I looked up the proof in B. L. van der Waerden's _Modern Algebra_, and I also found one in Birkhoff and MacLane's _A Survey of Modern Algebra_. I don't think I had ever actually read a proof of the result. Every calculus book I have is content to show the method and state that there is a proof. Actually, it's not too hard to follow, but it depends upon some division properties of polynomials that may not be familiar. The first is the Division Algorithm, which says that given two polynomials f(x) and g(x), you can divide f(x) by g(x) to get a quotient q(x) and a remainder r(x) whose degree is strictly less than the degree of the divisor g(x): that is, f(x) = g(x)q(x) + r(x), and 0 <= deg(r(x)) < deg(g(x) . Next, by repeated applications of this algorithm, you obtain the Euclidean Algorithm for finding the greatest common divisor d(x) of two polynomials f(x) and g(x). The key idea we need is that d(x) can be written d(x) = f(x)r(x) + g(x)s(x) . If you're not familiar with these ideas, you'll need to read up on them. See Euclidean and Division Algorithms http://mathforum.org/dr.math/problems/terry11.26.97.html You might also want to review these ideas for integers first: see The Official Euclidean Algorithm http://mathforum.org/dr.math/problems/julie.11.16.00.html and Explaining the Euclidean Algorithm http://mathforum.org/dr.math/problems/megan10.27.98.html Now, suppose we have a rational function f(x) ---- , where the degree of f is less than the degree of a(x), a(x) where the denominator a(x) factors into a(x) = g(x)h(x), where g(x) and h(x) have no common factors. If they have no common factors, then their greatest common divisor must be [the constant polynomial] 1, and by the Euclidean Algorithm, we can find polynomials c(x) and d(x) such that 1 = c(x)g(x) + d(x)h(x) . Multiplying by f(x) gives f(x) = [f(x)c(x)]g(x) + [f(x)d(x)]h(x) . Let G =degree of g(x), H =degree(h), so degree(f) < G + H. If the degree of f(x)c(x) is H or larger, then use the Division Algorithm to write f(x)c(x) = q(x)h(x) + r(x) where degree(r(x)) < H . Then f(x) = r(x)g(x) + [q(x)g(x) + f(x)d(x)]h(x) . The degree of f(x) < H + G, and the degree of r(x)g(x) < H + G, so the remaining term [q(x)g(x) + f(x)d(x)]h(x) must also have degree < H + G . Since h(x) has degree H, [q(x)g(x) + f(x)d(x)] must have degree less than G. All this shows us that we can write f(x) = r(x)g(x) + s(x)h(x) , with degree(r(x)) < H and degree(s(x)) < G . Then f(x) r(x)g(x) + s(x)h(x) ---- = ------------------- a(x) g(x)h(x) r(x) s(x) = ---- + ---- h(x) g(x) . That's the basic decomposition. If h(x) or g(x) factors into a product of polynomials with no common factors, you can repeat this process. Notice that the two partial fractions are proper: the numerator has degree less than the denominator. You can continue the decomposition until all your fractions are of the form t(x) ------ , p(x)^k because as long as you have different factors in the denominator, you can keep applying the above decomposition. For repeated factors like this, suppose degree(p(x))=P, so since this rational function is proper, degree(t(x)) < P*k . Using the Division Algorithm, divide t(x) by p(x)^(k-1) to get t(x) = s1(x)p(x)^(k-1) + r1(x), and degree(r1(x)) < P*(k-1); then divide r1(x) by p(x)^(k-2): r1(x) = s2(x)p(x)^(k-2) + r2(x), with degree(r2(x)) < P*(k-2); and so on, until you reach r(k-2)(x) = s(k-1)(x)p(x) + r(k-1)(x) , degree(r(k-1)(x))< P . Let sk(x) = r(k-1)(x) . Note that all the quotients (the s1(x), s2(x), ,..., etc.) must have degree less than P (compare the degrees on the two sides of each equation for the s's). Then t(x) = s1(x)*p(x)^(k-1) + s2(x)*p(x)^(k-2) + ... + s(k-1)(x)p(x)+ sk(x)(x) so t(x) s1(x) s2(x) s3(x) s(k-1)(x) s(k)(x) ------ = ----- + ------ + ------ + ... + --------- + ------- p(x)^k p(x) p(x)^2 p(x)^3 p(x)^(k-1) p(x)^k . For real polynomials, the polynomials p(x) must be linear or irreducible quadratics. So there's the proof. If you want to study these issues in more detail, the Birkhoff and MacLane book is very readable. If you have further questions, please write again. - Doctor Fenton, The Math Forum http://mathforum.org/dr.math/ |
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