PARTIAL FRACTION DECOMPOSITION

Transcription

1 PARTIAL FRACTION DECOMPOSITION LARRY SUSANKA 1. Fcts bout Polynomils nd Nottion We must ssemble some tools nd nottion to prove the existence of the stndrd prtil frction decomposition, used s n integrtion technique in most Clculus texts. If f is generic polynomil, Deg(f) will denote the degree of f. If f is ny polynomil let c(f) be the leding coefficient of f. f is clled monic if c(f) = 1. If p nd q re two nonzero polynomils let GCF(p, q) denote the monic polynomil which is the gretest common fctor of p nd q. The theorem sttes tht when f nd re polynomils with rel coefficients nd no common fctors nd Deg(f) < Deg() then the frction f/ cn be written in prticulr form which is menble to methods of integrtion lerned previously. For convenience, we presume without loss tht is monic. We ssume tht is fctored into the product of powers of distinct monic terms which hve rel coefficients nd re either liner or irreducible qudrtic. The form gurnteed by the theorem is tht f/ cn be written s (possibly lengthy) sum of terms of the form µ (x β) n or µx + γ (x 2 + δx + ɛ) n where (x β)n nd (x 2 + δx + ɛ) n re fctors of nd ech x 2 + δx + ɛ is irreducible (tht is, it hs complex roots) nd ll the coefficients nd constnts involved re rel. This theorem doesn t tell you how to find the µ or µx + γ for ech summnd, nor does it mention the existence of (unique) fctoriztion of the denomintor into the product of irreducible powers. The existence of the fctoriztion follows from the Fundmentl Theorem of Algebr, though the proofs re not constructive in the sense tht they don t tell you how to find these fctors in the wy tht the qudrtic formul does for second degree polynomils. We will describe generl proof technique nd two theorems from lgebr. The proof technique is clled Proof by Induction. If you hve sequence of propositions P 1, P 2, P 3,... indexed by the nturl numbers you my conclude tht Dte: November 13,

2 2 LARRY SUSANKA every proposition on this list is true provided you cn show (I) P 1 is true nd (II) If P k is true for ll k < n then P n must be true. The reson we cn drw this conclusion is tht if some of the P j re flse there would hve to be some smllest integer n for which P n is flse. Tht n cnnot be 1 by (I), nd (II) then implies P n is true. This contrdiction implies tht the premise tht led to it, nmely tht there re flse P j, must itself be flse. In other words ll P n re true. With this in hnd, we give the two lgebric results. First, is the Eucliden lgorithm. It sttes tht if p nd h re polynomils nd h 0 there re polynomils q nd r with r = 0 or Deg(r) < Deg(h) nd p = qh+r. Second, if the gretest common fctor of two nonzero polynomils g nd h is w then there re polynomils c nd d for which w = cg + dh. In tht cse, it follows tht c nd d shre no common fctor. The specil cse we will cre bout is when w = 1: tht is, g nd h shre no common fctor. You cn either ccept these two results, or follow the rguments for them given now. The first result is obviously true whenever p = 0 or if Deg(p) < Deg(h) since we cn choose q = 0 nd r = p. If the first result were ever flse, there would be nonzero pir p nd h exhibiting this, nd for which Deg(p) Deg(h) is s smll s possible. Tht difference cnnot be negtive, by the discussion we just gve. So the difference n = Deg(p) Deg(h) must be positive or 0. Then w = x n c(p) c(h) h p is 0 or it is polynomil tht hs degree less thn the degree of p, so w = 0 or Deg(w) Deg(h) < n nd the first result pplies: tht is, there re polynomils nd b with b = 0 or Deg(b) < Deg(h) nd w = h + b. But then h + b = w = x n c(p) c(h) h p so p = ( x n c(p) c(h) ) h + ( b). But this represents p in the form required by the first result nd p nd h were specificlly picked becuse they were mong the pirs of polynomils for which the first result fils. Since tht is impossible, it must be tht the first result never fils. Tht is, the first result is true for ll p, h pirs. The rgument my be recognized s (very slightly disguised) Proof by Induction, in this cse induction on Deg(p) Deg(h). The proof of the second lgebric result cn lso be hndled by Induction. If p nd q re nonzero polynomils let S(p, q) = { p + bq, b re polynomils }. Every member of this set of polynomils is divisible by GCF (p, q) nd the clim of the second result is tht GCF (p, q) is ctully member of S(p, q). If q is the constnt polynomil, the second result is obviously true. If the second result were to ever fil it would hve to fil for some p, q pir where q hs smllest degree mong ll the filed pirs.

3 PARTIAL FRACTION DECOMPOSITION 3 By the minimlity ssumption on q we hve Deg(q) Deg(p) so there re polynomils c nd r with r = 0 or Deg(r) < Deg(q) nd p = cq + r. r cnnot be 0 becuse in tht cse p is multiple of q nd the second result is obviously true in tht cse. It must then be tht GCF (q, r) = GCF (p, q) since every shred fctor of p nd q must be shred by p nd r, nd conversely. Since the degree of r is less thn tht of q the second result is true for the q, r pir nd there re polynomils x nd y for which xq + yr = GCF (q, r) = GCF (p, q). But then xq + y(p cq) = yp + (x yc)q = GCF (p, q) contrry to choice of the p, q pir. In other words, there cn be no pir for which the second result fils. It must lwys be true. 2. The Proof Prt One We will suppose tht sttement P n is: The Prtil Frction Theorem is true when f hs the form ζ ζx + µ (i) (x µ) n or (ii) (x 2 + δx + ɛ) n. P 1 is true: tht is, the theorem is true for f/ of the form (i) ζ x µ or (ii) ζx + µ x 2 + δx + ɛ becuse (i) nd (ii) re lredy in the form specified in the theorem. We now presume tht sttement P k is true for ll k < n nd some n exceeding 1. This is prt (II) of the Proof by Induction process. Let s exmine P n. In cse (i) there is polynomil q nd rel constnt r with f = q(x µ) + r. Then f q(x µ) + r q = (x µ) n (x µ) n = (x µ) n 1 + r (x µ) n. The lst frction is in the form specified in the theorem, while the previous term cn be plced in the pproprite form by inductive hypothesis (II). In cse (ii) there re polynomils q nd (t most) first degree r with f = q(x 2 + δx + ɛ) + r. This time we hve: f (x 2 + δx + ɛ) n = q(x2 + δx + ɛ) + r q (x 2 + δx + ɛ) n = (x 2 + δx + ɛ) n 1 + r (x 2 + δx + ɛ) n. Agin, the lst frction is in the form specified in the theorem, while the previous term cn be plced in the pproprite form by inductive hypothesis (II). So we hve proved the theorem when is power of fctors of these two specified types.

4 4 LARRY SUSANKA 3. The Proof Prt Two Let s exmine now the sitution in the generl theorem where nth degree monic polynomil hs nontrivil fctoriztion = gh where g nd h re monic nd hve no common fctors nd Deg(f) < n. So g will hve ll of some of the irreducible fctors of nd h will hve ll of the remining irreducible fctors. Deg(f) < Deg(g) + Deg(h) = n nd both Deg(g) nd Deg(h) re t lest 1. Since the gretest common fctor of g nd h is 1 there re polynomils c nd d for which 1 = cg + dh. Note tht c cn shre no fctors with h, while d cn shre no fctors with g. Then f f(cg + dh) = = fcg + fdh = fc h + fd g. It is possible tht both Deg(fc) < Deg(h) nd Deg(fd) < Deg(g) in which cse both frctions on the end of the line bove re in the form to which the theorem pplies, but ech denomintor hs strictly fewer irreducible fctor types in the denomintor thn hs. On the other hnd, it my be tht one of Deg(fc) Deg(h) or Deg(fd) Deg(g) holds. We will ssume tht Deg(fc) Deg(h) nd the other cse is hndled identiclly. If Deg(fc) Deg(h) then fc = qh + r for certin polynomils q nd r where Deg(r) < Deg(h). Note tht this implies tht r cn shre no fctors with h becuse c shres no fctor with h so ny shred fctor of r nd h would hve to be shred lso with f, contrdicting the ssumption tht f/ is in lowest terms. We now hve f f(cg + dh) (qh + r)g + dhf rg + (q + df)h = = =. Note tht q +df cn shre no fctor with g, gin becuse f/ is in lowest terms. The left-hnd side of the lst numertor hs degree strictly less thn Deg(h) + Deg(g), nd since f lso hs degree less thn Deg(h) + Deg(g) we must hve Deg( (q + df)h ) less thn Deg(h) + Deg(g) too. So Deg(q + df) < Deg(g) nd (by its definition) Deg(r) < Deg(h). Now we hve f rg + (q + df)h = = r h + q + df g nd both of these frctions re in the form to which the theorem pplies, but with strictly fewer types of irreducible fctors in the denomintor thn hs. 4. The Proof Prt Three We will suppose this time tht sttement P n is the ssertion The Prtil Frction Theorem is true when hs n different types of irreducible fctors. P 1 is true: tht is the content of The Proof Prt One, bove. We hve in hnd, therefore, step (I) of the Proof by Induction process.

5 PARTIAL FRACTION DECOMPOSITION 5 We now presume tht sttement P k is true for ll k < n nd some n exceeding 1. This is the setup for step (II) of the Proof by Induction process. Let s look t P n. In The Proof Prt Two bove we showed tht ny frction in the form of the theorem whose denomintor hs n different types of irreducible fctors cn be split into the sum of two frctions of the type to which the theorem pplies, but with strictly fewer irreducible fctor types in their denomintors. But ech of these two terms cn be plced in the pproprite form gurnteed by the theorem by inductive hypothesis (II). So their sum cn be s well. We hve verified the two steps of the Proof by Induction process, nd conclude tht the theorem is true for ny number of irreducible fctors.