Ma/CS 6a Class 22: Power Series

Transcription

1 Ma/CS 6a Class 22: Power Series By Ada Sheffer Power Series Mooial: ax i. Polyoial: a 0 + a 1 x + a 2 x a x. Power series: A x = a 0 + a 1 x + a 2 x 2 + Also called foral power series, because we do ot thik about the eaig of x. 1

2 Sus ad Products We defie sus ad products of power series as i the case of polyoials. A x = a 0 + a 1 x + a 2 x 2 + B x = b 0 + b 1 x + b 2 x 2 + A x + B x = a 0 + b 0 + a 1 + b 1 x + a 2 + b 2 x 2 + A x B x = a 0 b 0 + a 1 b 0 + a 0 b 1 x + a 2 b 0 + a 1 b 1 + a 0 b 2 x 2 + More Sus ad Products We defie sus ad products of power series as i the case of polyoials. A x = a 0 + a 1 x + a 2 x 2 + B x = b 0 + b 1 x + b 2 x 2 + C x = A x + B x. D x = A x B(x). c i = a i + b i. i d i = σ j=0 a j b i j. 2

3 Coutativity A group G is said to be coutative or Abelia if every a, b G satisfy ab = ba. Coutative or ot? Itegers uder additio atrices uder additio. S uder copositio of perutatios. Syetries of the square. Rigs A rig is a set R together with two biary operatios + ad that satisfy The set R is a coutative group uder +. The operatio satisfies the closure, associativity, ad idetity properties. Distributive laws. For ay a, b, c R, we have a b + c = ab + ac, a + b c = ac + bc. 3

4 Is this a Rig? The set Z 4 = 0,1,2,3 uder additio ad ultiplicatio od 4. We already kow that Z 4 uder additio od 4 is a coutative group. For ultiplicatio, we have closure, associativity, ad idetity. Stadard additio ad ultiplicatio are distributive, so the sae holds uder od 4. Is this a Rig? #2 2 2 atrices with real etries, uder atrix additio ad ultiplicatio. 2 2 atrices with real etries uder additio for a coutative group. For ultiplicatio, we have closure, associativity, ad idetity. Matrix additio ad ultiplicatio are distributive. 4

5 Polyoial Rig The polyoial rig R x is the set of polyoials i x with coefficiets i R ad stadard additio ad ultiplicatio. The set of polyoials uder additio is a group. Properties of ultiplicatio: Closure. The product of two polyoials i R x is a polyoial i R x. Associativity. By the associativity of the stadard ultiplicatio. Idetity. We have 1 R x. Stadard additio ad ultiplicatio are distributive. Rig of Power Series The power series rig R[ x ] is the set of power series i x with coefficiets i R uder additio ad ultiplicatio. The set of power series uder additio is a group. Properties of Closure. By our defiitio, the product of two power series is a power series. Associativity. Our defiitio of ultiplicatio is associative. Idetity. We have 1 R[ x ]. Distributivity is ot hard to verify. 5

6 What is Missig? I R x ad R[ x ], why do t we have groups with respect to the operatio? A ultiplicative iverse does ot always exist! I R x oly costat polyoials have a iverse. What about R [x]? Is there a iverse of 1 x? 1 x A x = 1. A x = 1 + x + x 2 + x 3 + Iverses i a Power Series Theore. A power series A x = a 0 + a 1 x + a 2 x 2 + R[ x ] has a iverse if ad oly if a 0 0. Proof. First assue that A x has a iverse B x. a 0 + a 1 x + a 2 x 2 + (b 0 + b 1 x + b 2 x 2 6

7 Proof Cot. Proof (cot.). Assue that a 0 0. We eed to solve a 0 b 0 = 1, a 1 b 0 + a 0 b 1 = 0, a 2 b 0 + a 1 b 1 + a 0 b 2 = 0, Sice a 0 0, we obtai the solutio b 0 = a 1 0, b 1 = a 1 b 0 a 1 0, 1 b 2 = a 1 b 1 + a 2 b 0 a 0 More o Iverses Notatio. The iverse of A x is soeties writte as A x 1 or as 1 A x. For exaple, we have 1 + x 1 = 1 + x 1 x 1 x = 1 + x 1 + x + x 2 + = 1 + 2x + 2x 2 + 2x 3 + 7

8 Recall: The Bioial Theore By the bioial theore, for 1 we have x + 1 = x i i 1 j 0 i,j i+j= = 0 x + 1 x x What about the case where is egative? Negative Powers What is x + 1 1? x + 1 a 0 + a 1 x + a 2 x 2 + = 1, x = 1 x + x 2 x 3 + What is x + 1 2? x = x = 1 x + x 2 x 3 + (1 x + x 2 x 3 8

9 Negative Powers Forula Theore. For ay positive iteger, x + 1 = x. Exaples. x = σ 1 x = 1 x + x 2 x 3 + x = σ x = 1 2x + 3x 2 4x 3 + Proof We look for the coefficiet of x k i x + 1 = x = 1 x To siplify, we replace y = x ad have 1 y = 1 y 1 = y The proble turs to: I how ay ways ca we write k as a su of o-egative itegers... 9

10 Proof (cot.) I how ay ways ca we write k as a su of o-egative itegers? This is equivalet to choosig 1 cells i a array of + k 1 cells. There are + k 1 1 = + k 1 k x x x x x ways. 1 or y 3 or y 3 0 or y 0 Cocludig the Proof + k 1 There are ways to write k as k a su of at ost positive itegers. The coefficiet of y k i is 1 y = 1 y 1 = + k 1 k x + 1 =. This iplies k 0 1 k + k 1 k y x k. 10

11 Cobiig Both Cases? For itegers 1 we have x + 1 = x. 0 For itegers 1 we have x + 1 = Ca we cobie the two forulas ito oe? x. Geeralizig the Bioial Coefficiets Give a iteger ad a positive iteger, we defie 0 = 1 ad = 1 + 1! This defiitio subsues the stadard bioial coefficiets (with, 0). If, this is idetical to If >, we have = 0.!!!.. 11

12 Negative Bioial Nubers Give positive itegers ad, we have =! + 1 ( + 1) = 1! = Cobiig Both Cases For itegers 1 we have x + 1 = 0 For itegers 1 we have x + 1 = x x. Either way, we have x + 1 = x. 12

13 A Variat Proble. Fid the value of 1 + ax for ay iteger ad a R. Solutio. Substitute y = ax. We have 1 + y = y. By brigig x back, we have 1 + ax = a x. A Exaple Proble. Write the power series of 2 + x 1 3x + 3x 2 x 3. Solutio. First, 1 3x + 3x 2 x 3 = 1 x 3. By the previous theore, we have 1 x 3 = 2 + x 1 x 3 = x = x. x. 13

14 The Ed 14