lim za n n = z lim a n n.

Transcription

1 Lecture 6 Sequeces ad Series Defiitio 1 By a sequece i a set A, we mea a mappig f : N A. It is customary to deote a sequece f by {s } where, s := f(). A sequece {z } of (complex) umbers is said to be coverget to the limit w if for every ɛ > 0 there exists a iteger 0 such that for all 0, we have, z w < ɛ. It is easily see that the limit of a sequece i A C if it exists, is uique. But it eed ot belog to the set A. We use the followig two otatios lim z := w; OR z w, to represet the limit of the sequece {z }. If a sequece is ot coverget the it is said to be diverget. If {a }, {b } are coverget sequeces of real or complex umbers, the the sequeces {a +b } ad {za } are also so with limits give by lim (a + b ) = lim a + lim b ; lim za = z lim a. It is fairly obvious that if z = a + ıb where, a, b R, the z w iff a R(w) ad b I(w). The followig criterio for covergece of sequeces of real (complex) is oe of the immediate cosequeces of the costructio of real umbers. Theorem 1 (Cauchy ) A sequece {z } of umbers is coverget iff for every ɛ > 0 there exists 0 z z m < ɛ. such that for all, m 0, we have Theorem 2 Let f : X C be ay fuctio defied o a subset of C. For ay z X, f is cotiuous at z iff for every sequece z z, we have, f(z ) f(z). Defiitio 2 By a series of real (complex) umbers we mea a ifiite sum: z := z 0 + z + + z + 1

2 Of course, it is possible that there are oly fiitely may o zero terms here. The sequece of partial sums associated to the above series is defied to be s := z k. We say the series is coverget if the associated sequece k=1 {s } of partial sums is coverget. I that case, if s is the limit of this sequece, the we say s is the sum of the series ad write z := s. Oce agai it is immediate that if z ad w are coverget series the for ay complex umber λ, we have, λz ad (z + w ) are coverget ad yields: λz = λ z ; (z + w ) = z + Cauchy s covergece criterio ca be applied to series also. This Theorem 3 A series z of real or complex umbers is coverget iff for every ɛ > 0, there exists 0 such that for all 0 ad for all p 0, we have, z + z z +p < ɛ. Ideed, all otios ad results that we have for sequeces have correspodig otios ad results for series also, via the sequece of partial sums of the series. Thus, oce a result is established for a sequece, the correspodig result is available for series as well ad vice versa, without specifically metioig it. It follows that if a series is coverget, the its th term z teds to 0. However, this is ot a sufficiet coditio for covergece of the series, as illustrated by the series 1. Defiitio 3 A series z is said to be absolutely coverget if the series z is coverget. Agai, it is easily see that a absolutely coverget series is coverget, whereas the coverse is ot true as see with the stadard example ( 1) 1. The otio of absolute covergece plays a very importat role 2 w.

3 throughout the study of covergece of series. As a illustratio we shall obtai the followig useful result about the covergece of the product series. Defiitio 4 Give two series a, b the (Cauchy) product of these two series is defied to be c where c = k=0 a k b k. Theorem 4 If a, b are two absolutely coverget series the their Cauchy-product series is absolutely coverget ad its sum is equal to the product of the sums of the two series: c = ( ) ( a b ). (1) Proof: [ot doe i the class] We begi with the remark that if both the series are of o egative real umbers, the the assertio of the theorem is obvious. We shall use this i what follows. Cosider the remaider after ( 1) terms of the correspodig absolute series: Clearly, R = a k ; T = b k. k k c a k b l = R 0 T 0. 0 k 0 l 0 Therefore the series c is absolutely coverget. Further, k 2 c k a k b k k k R 0T +1 + T 0 R +1, sice the terms that remai o the LHS after cacellatio are of the form a k b l where either k +1 or l +1. Upo takig the limit as, we obtai (1). A importat property of a absolutely coverget series is: Theorem 5 Let z be a absolutely coverget series. rearragemet z σ of the series is also absolutely coverget. The every Recall that a rearragemet z σ permutatio σ : N N. Lecture 7 of z is obtaied by takig a 3

4 Uiform Covergece of Sequ. of fuctios We shall ow cosider the study of a family of sequeces, viz., let {f } be a sequece of fuctios takig complex values defied o some set A. The to each x A, we get a sequece {f (x)}. If each of these sequeces is coverget, the we get a fuctio f(x) = lim f (x). Defiitio 5 Let {f } be sequece of complex valued fuctios o a set A. We say that it is uiformly coverget o A to a fuctio f(x) if for every ɛ > 0 there exists 0, idepedet of x A such that for all 0, we have, f (x) f(x) < ɛ, for all x A. Remark 1 Observe that if {f } is uiformly coverget, the for each x A, we have, f (x) f(x). This is called poit-wise covergece of the sequece of fuctios. As see i the example below, poit-wise covergece does ot imply uiform covergece. However, it is fairly easy to see that this is so if A is a fiite set. Thus the iterestig case of uiform covergece occurs oly whe A itself is a ifiite set. The termiology is also adopted i a obvious way for series of fuctios via the associated sequeces. As i the case of ordiary covergece, we have Cauchy s criterio here also. Theorem 6 A sequece of complex valued fuctios {f } is uiformly coverget iff it is uiformly Cauchy i.e., give ɛ > 0 there exists 0, such that for all 0 ad for all x A, we have, f +p (x) f (x) < ɛ. Example 1 A simple example of a sequece which is poit-wise coverget but ot uiformly coverget is f : (0, ) R give by f (x) = 1/x. It is uiformly coverget i [α, ) for all α > 0 but ot so i (0, α). Example 2 The most useful series is the geometric series 1 + z + z 2 + 4

5 The sequece of partial sums is give by 1 + z + + z 1 = 1 z 1 z. As z 0 if z < 1, it follows that the geometric series poit-wise coverges to 1/(1 z) for all z < 1. I fact, if we take 0 < r < 1, the i the disc B r (0), the series is uiformly coverget. For, give ɛ > 0, choose 0 such that r 0 have, 1 z 1 z 1 < ɛ(1 r). The for all z < r ad 0, we 1 z = z 1 z z 0 1 z < ɛ Theorem 7 Weierstrass 1 M-test: Let a be a coverget series of positive terms. Suppose there exists M > 0 ad a iteger N such that f (x) < Ma for all N ad for all x A. The f is uiformly ad absolutely coverget i A. Proof: Give ɛ > 0 choose 0 > N such that a + a a +p < ɛ/m, for all 0. This is possible by Cauchy s criterio, sice a is coverget. The it follows that f (x) + + f +p (x) M(a + a +p ) < ɛ, for all 0 ad for all x A. Agai, by Cauchy s criterio, this meas that f is uiformly ad absolutely coverget. Remark 2 The series a i the above theorem is called a majorat for the series f. As a illustratio of the importace of uiform covergece, we shall prove: Theorem 8 Let {f } be a sequece of cotiuous fuctios defied ad uiformly coverget o a subset A of R or C. The the limit fuctio f(x) = lim f (x) is cotiuous o A. 1 Karl Weierstrass ( ) a Germa mathematicia is well kow for his perfect rigor. He clarified ay remaiig ambiguities i the otio of a fuctio, of derivatives, of miimum etc.. 5

6 Proof: [ot doe i the classlet x A be ay. I order to prove the cotiuity of f at x, give ɛ > 0 we should fid δ > 0 such that for all y A with y x < δ, we have, (f(y) f(x) < ɛ. So, by the uiform covergece, first we get 0 such that f 0 (y) f(y) < ɛ/3 for all y A. Sice f 0 is cotiuous at x, we also get δ > 0 such that for all y A with y x < δ, we have f 0 (y) f 0 (x) < ɛ/3. Now, usig triagle iequality, we get, f(y) f(x) f(y) f 0 (y) + f 0 (y) f 0 (x) + f 0 (x) f(x) < ɛ, wheever y A is such that y x < δ. ] Power series K = R, or C. Defiitio 6 By a formal power series i oe variable t over K, we mea a sum of the form =0 a t, a K. Observe that whe at most a fiite umber of a are o zero the above sum gives a polyomial. Thus, all polyomials i t are power series i t. We ca add two power series, by term-by-term additio ad we ca also multiply them by scalars, just like polyomials, viz., a t + b t := (a + b )t ; α( a t ) := αa t. Let K[[t]] deote the set of all formal power series i t over K. The set of polyomials is cotaied i K[[t]] ad is closed uder each of these operatios. Defiitio 7 A formal power series P (t) is said to be coverget if there exists a o zero umber z (real or complex) such that the series of complex umbers a z is coverget. We shall ow cosider a few theorems, which are attributed to Cauchy- Hadamard 2 ad Abel 3, are most fudametal i the theory of power series. 2 Jacques Hadamard( ) was a Frech Mathematicia who was the most ifluetial mathematicia of his days, worked several areas of mathematics such as complex aalysis, aalytic umber theory, partial differetial equatios, hydrodyamics ad logic. 3 Niels Herik Abel ( ) was a Norwegia, who died youg uder deprivatio. At the age of 6

7 It may be recalled that the limsup of a sequece of real umbers {b } is defied to be lim sup{b } := lim (sup{b, b +1,...}). This is also the same as the least upper boud of the set of limits of all coverget subsequeces of {b }. Theorem 9 (Cauchy-Hadamard) Let P = 0 a t be a power series over C. The (a) there exists a umber 0 R such that for all 0 < r < R, the series P (z) is absolutely ad uiformly coverget i z r ad for all z > R the series is diverget. (b) 1 R = lim sup a, with the covetio (0) 1 = ; ( ) 1 = 0. Proof: Ideed, all that we have to prove is to take R as give by (b) ad show that it satisfies (a). So, let 0 < r < R. Choose r < s < R. The 1/s > 1/R ad hece by the defiitio of limsup, we must have 0 such that for all 0, a < 1/s. Therefore, for all z r, a z < (r/s), 0. Sice r/s < 1, by Weierstrass majorat criterio, (theorem 7), it follows that P (z) is absolutely ad uiformly coverget. O the other had, suppose z > R. We fix s such that z > s > R. The 1/s < 1/R, hece, by the property of limsup, there exist ifiitely may j, for which j aj > 1/s. This meas that a j z j > (R/s) j. It follows that the sequece {a z } is ubouded ad hece the series a z is diverget. Remark 3 Observe that if P (z) is coverget for some z, the from the last part of (a), the radius of covergece of P is at least z. Defiitio 8 The umber R obtaied i the above theorem is called the radius of covergece of P (t). The secod part of the theorem gives you the formula for it. This is called the Cauchy-Hadamard formula. Thus the 21, he proved the impossibility of solvig a geeral quitic by radicals. He did ot get ay recogitio durig his life time for his ow famous works o covergece, o so called abelia itegrals, ad o elliptic fuctios. 7

8 collectio of all poits at which a give power series coverges cosists of a ope disc cetered at the origi ad perhaps some poits o the boudary of the disc. Observe that the theorem does ot say aythig about the covergece of the series at poits o the boudary z = R. Example 3 The series t, t, all have radius of covergece 2 1. The first oe is ot coverget at ay poit of the boudary of the disc of covergece z = 1. The secod is coverget at all the poits of the boudary except at z = 1 ad the last oe is coverget at all the poit of the boudary. t These examples clearly illustrate that the boudary behavior of a power series eeds to be studied more carefully. Defiitio 9 Give a power series P (t) = 0 a t, the derived series P (t) is defied by takig term-by-term differetiatio: P (t) = 1 a t 1. The series 0 a +1 t+1 is called the itegrated series. As a applicatio of Cauchy-Hadamard formula, we derive: Theorem 10 Ay give power series, its derived series ad its itegrated series all have the same radius of covergece. Proof: Let the radius of covergece of P (t) = a t, ad P (t) be r, r respectively. It is eough to prove that r = r. We will first show that r r. For this we may assume without loss of geerality that r > 0. Let 0 < r 1 < r. The a r1 r 1 a r 1 < r a r 1 <. 1 1 r r. 1 1 It follows that r r 1. Sice this is true for all 0 < r 1 < r this meas Now to show that r r, we ca assume that r > 0 ad let 0 < r 1 < r. Choose r 2 such that r 1 < r 2 < r. The for each 1 r1 1 ( ) r1 r2 M r2 r 1 r 2 r 1 where M = k 1 k r 1 k r 2 <, sice the radius of covergece of k kt k is 1. Therefore, a r1 1 M a r2 <. r

9 We coclude that r > r 1 ad sice this holds for all r 1 < r, it follows that r r. 9

10 Lecture 8 Remark 4 A power series with radius of covergece 0 is useless for us, for, it oly defies a fuctio at a poit. A power series P (t) with a positive radius of covergece R defies a cotiuous fuctio z p(z) i the disc of covergece B R (0), by theorem 8. Also, by shiftig the origi, we ca eve get cotiuous fuctios defied i B R (z 0 ), viz., by substitutig t = z z 0. Oe expects that fuctios which agree with a coverget power series i a small eighborhood of every poit will have properties aki to those of polyomials. Before takig up this study, let us make a defiitio. Defiitio 10 Let f : Ω C be a fuctio. We say f is aalytic at z 0 Ω if there exists r > 0 ad a power series P of radius of covergece r such that f(z) = P (z z 0 ) for all z B r (z 0 ). We say f is aalytic i Ω if it is so at each poit z 0 Ω. Remark 5 As see above it follows that a aalytic fuctio is cotiuous. O the other had, what is ot obvious is that if P (t) is a formal power series with positive radius of covergece r, the the fuctio f(z) = P (z z 0 ) is aalytic i z z 0 < r. (Observe that the defiitio oly says that it is aalytic at z 0.) This fact ca be directly proved by algebraic methods. However, we eed ot do this, as the result will follow easily from Taylor s theorem which we shall prove i due course. The ext theorem tells you that aalytic fuctios are holomorphic. Theorem 11 Abel: Let 0 a t be a power series of radius of covergece R > 0. The the fuctio defied by f(z) = a (z z 0 ) is complex differetiable i B r (z 0 ). Moreover the derivative of f is give by the derived series iside z z 0 < R. f (z) = a (z z 0 )

11 Proof: Without loss of geerality, we may assume that z 0 = 0. We already kow that the derived series is coverget i B R (0) ad hece defies a cotiuous fuctio g o it. We have to show that this fuctio g is the derivative of f at each poit of B R (0). So, fix a poit z B R (0). Let z < r < R ad let 0 h r z. Cosider the differece quotiet f(z + h) f(z) h g(z) = 1 u (h) (2) a z 1. Upo simplifi- where, we have put u (h) := a [(z + h) z ] h catio we get u (h) = a [(z + h) 1 + (z + h) 2 z + + (z + h)z 2 + z 1 z 1 ]. (3) We must show that give ɛ > 0, there exists δ > 0 such that for all 0 < h < δ, we have, f(z + h) f(z) h g(z) Sice z < r ad z + h < r, usig (3), it follows that < ɛ. (4) u (h) 2 a r 1. (5) Sice the derived series has radius of covergece R > r, it follows that we ca fid 0 such that 0 2 a r 1 < ɛ/2. (6) O the other had, agai usig (3), each u (h) is a polyomial i h which vaishes at h = 0. Therefore so does the fiite sum < 0 u (h). Hece by cotiuity, there exists δ such that h < δ implies that 0<< 0 2 a r 1 < ɛ/2. (7) Takig δ = mi{δ, r z } ad combiig (6) ad (7) yields (4). Remark 6 (i) It follows that a aalytic fuctio is repeatedly differetiable, with its th derivatives at z 0 give by!a. As a cosequece, we kow that, 11

12 at each poit the power series represetig the fuctio is uique. (ii) It is fairly obvious that the sum of ay two aalytic fuctios is agai a aalytic fuctio. The correspodig statemet about power series is that the sum of two formal power series is coverget with radius of covergece at least the miimum of the two radii of covergece. Similarly, the product of two aalytic fuctios is also aalytic. The idetity fuctio writte f(z) = z is clearly aalytic i the etire plae (take P (t) = z 0 + t to see that f is aalytic at z 0 ). Startig from this ad usig the above two observatios we ca deduce that ay polyomial fuctio is aalytic throughout the plae. (iii) Later o we shall prove that every holomorphic fuctio is aalytic, thus completig the picture. I particular, this will the prove that a fuctio which is complex differetiable oce i a ope set is complex differetiable ifiitely may times. Of course, such a result is far from beig true i the real differetiable case. Examples such as g(x) = x x illustrate the existece of fuctios which are -times real differetiable but ot (+1)-times. Also, there are fuctios which are C but ot real aalytic. (Take f(x) = 0 for x 0 ad = e 1/x2 for x > 0.) For the time beig, we are satisfied with gettig plety of examples of holomorphic fuctios via aalytic fuctios. That is the topic that we are goig to take up ow. 12