Power Series, Taylor Series

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1 CHAPTER 5 Power Series, Tylor Series In Chpter 4, we evluted complex integrls directly by using Cuchy s integrl formul, which ws derived from the fmous Cuchy integrl theorem. We now shift from the pproch of Cuchy nd Gourst to nother pproch of evluting complex integrls, tht is, evluting them by residue integrtion. This pproch, discussed in Chpter 6, first requires thorough understnding of power series nd, in prticulr, Tylor series. (To develop the theory of residue integrtion, we still use Cuchy s integrl theorem!) In this chpter, we focus on complex power series nd in prticulr Tylor series. They re nlogs of rel power series nd Tylor series in clculus. Section 5. discusses convergence tests for complex series, which re quite similr to those for rel series. Thus, if you re fmilir with convergence tests from clculus, you my use Sec. 5. s reference section. The min results of this chpter re tht complex power series represent nlytic functions, s shown in Sec. 5.3, nd tht, conversely, every nlytic function cn be represented by power series, clled Tylor series, s shown in Sec The lst section (5.5) on uniform convergence is optionl. Prerequisite: Chps. 3, 4. Sections tht my be omitted in shorter course: 5., 5.5. References nd Answers to Problems: App. Prt D, App Sequences, Series, Convergence Tests The bsic concepts for complex sequences nd series nd tests for convergence nd divergence re very similr to those concepts in (rel) clculus. Thus if you feel t home with rel sequences nd series nd wnt to tke for grnted tht the rtio test lso holds in complex, skip this section nd go to Section 5.2. Sequences The bsic definitions re s in clculus. An infinite sequence or, briefly, sequence, is obtined by ssigning to ech positive integer n number z n, clled term of the sequence, nd is written z, z 2, Á or {z, z 2, Á } or briefly {z n }. We my lso write z or z 2, z 3, Á, z, Á or strt with some other integer if convenient. A rel sequence is one whose terms re rel. 67

2 672 CHAP. 5 Power Series, Tylor Series Convergence. A convergent sequence z, z 2, Á is one tht hs limit c, written lim z n c or simply z n : c. n: By definition of limit this mens tht for every P we cn find n N such tht () ƒ z n c ƒ P for ll n N; geometriclly, ll terms z n with n N lie in the open disk of rdius P nd center c (Fig. 36) nd only finitely mny terms do not lie in tht disk. [For rel sequence, () gives n open intervl of length 2P nd rel midpoint c on the rel line s shown in Fig. 362.] A divergent sequence is one tht does not converge. y c x c c c + Fig. 36. Convergent complex sequence Fig Convergent rel sequence x EXAMPLE Convergent nd Divergent Sequences The sequence {i n >n} {i, is convergent with limit. The sequence {i n 2, i>3, 4 } {i,, i,, Á, Á } } is divergent, nd so is {z with z n ( i) n n }. EXAMPLE 2 Sequences of the Rel nd the Imginry Prts The sequence {z with z is 6i, 3 4 4i, 8 9 i>3, 5 6 3i, Á n x n iy n >n 2 n } i(2 4>n). (Sketch it.) It converges with the limit c 2i. Observe tht {x n } hs the limit Re c nd {y n } hs the limit 2 Im c. This is typicl. It illustrtes the following theorem by which the convergence of complex sequence cn be referred bck to tht of the two rel sequences of the rel prts nd the imginry prts. THEOREM Sequences of the Rel nd the Imginry Prts A sequence of complex numbers (where 2, Á z, z 2, Á, z n, Á z n x n iy n n ) converges to c ib if nd only if the sequence of the rel prts converges to nd the sequence of the imginry prts y, y 2, Á x, x 2, Á, converges to b. PROOF Convergence z n : c ib implies convergence x n : nd y n : b becuse if ƒ z n c ƒ P, then lies within the circle of rdius P bout c ib, so tht (Fig. 363) z n ƒ x n ƒ P, ƒ y n b ƒ P. Conversely, if x n : nd y n : b s n :, then for given P we cn choose N so lrge tht, for every n N, ƒ x n ƒ P 2, ƒ y n b ƒ P 2.

3 SEC. 5. Sequences, Series, Convergence Tests 673 y y b + b + 2 b c b c b 2 b + x 2 () (b) Fig Proof of Theorem + 2 x These two inequlities imply tht z n x n iy n lies in squre with center c nd side P. Hence, must lie within circle of rdius P with center c (Fig. 363b). Series z n Given sequence z, z 2, Á, z m, Á, we my form the sequence of the sums s z, s 2 z z 2, s 3 z z 2 z 3, Á nd in generl (2) s n z z 2 Á z n (n, 2, Á ). Here (3) s n is clled the nth prtil sum of the infinite series or series m z m z z 2 Á. The z, z 2, Á re clled the terms of the series. (Our usul summtion letter is n, unless we need n for nother purpose, s here, nd we then use m s the summtion letter.) A convergent series is one whose sequence of prtil sums converges, sy, lim s n s. n: Then we write s m z m z z 2 Á nd cll s the sum or vlue of the series. A series tht is not convergent is clled divergent series. If we omit the terms of from (3), there remins s n (4) R n z n z n2 z n3 Á. This is clled the reminder of the series (3) fter the term z n. Clerly, if (3) converges nd hs the sum s, then s s n R n, thus R n s s n. Now s n : s by the definition of convergence; hence R n :. In pplictions, when s is unknown nd we compute n pproximtion s n of s, then ƒ R n ƒ is the error, nd R n : mens tht we cn mke ƒ R n ƒ s smll s we plese, by choosing n lrge enough.

4 674 CHAP. 5 Power Series, Tylor Series An ppliction of Theorem to the prtil sums immeditely reltes the convergence of complex series to tht of the two series of its rel prts nd of its imginry prts: THEOREM 2 Rel nd Imginry Prts A series (3) with z converges nd hs the sum if nd only if x converges nd hs the sum u nd y y 2 Á x 2 Á m x m iy m s u iv converges nd hs the sum v. Tests for Convergence nd Divergence of Series Convergence tests in complex re prcticlly the sme s in clculus. We pply them before we use series, to mke sure tht the series converges. Divergence cn often be shown very simply s follows. THEOREM 3 Divergence If series z converges, then lim z z 2 Á m. Hence if this does not hold, m: the series diverges. PROOF If z z 2 Á converges, with the sum s, then, since z m s m s m, lim z m lim (s m s m ) lim s m lim s m s s. m: m: m: m: CAUTION! z m : is necessry for convergence but not sufficient, s we see from the hrmonic series Á, which stisfies this condition but diverges, s is shown in clculus (see, for exmple, Ref. [GenRef] in App. ). The prcticl difficulty in proving convergence is tht, in most cses, the sum of series is unknown. Cuchy overcme this by showing tht series converges if nd only if its prtil sums eventully get close to ech other: THEOREM 4 Cuchy s Convergence Principle for Series A series z z 2 Á is convergent if nd only if for every given P (no mtter how smll) we cn find n N (which depends on P, in generl) such tht (5) ƒ z for every n N nd p, 2, Á n z n2 Á z np ƒ P The somewht involved proof is left optionl (see App. 4). Absolute Convergence. A series z z 2 Á is clled bsolutely convergent if the series of the bsolute vlues of the terms is convergent. m ƒ z m ƒ ƒ z ƒ ƒ z 2 ƒ Á

5 SEC. 5. Sequences, Series, Convergence Tests 675 If z converges but ƒ z diverges, then the series z z 2 Á ƒ ƒ z 2 ƒ Á z 2 Á is clled, more precisely, conditionlly convergent. EXAMPLE 3 A Conditionlly Convergent Series The series Á converges, but only conditionlly since the hrmonic series diverges, s mentioned bove (fter Theorem 3). If series is bsolutely convergent, it is convergent. This follows redily from Cuchy s principle (see Prob. 29). This principle lso yields the following generl convergence test. THEOREM 5 Comprison Test If series z is given nd we cn find convergent series with nonnegtive rel terms such tht ƒ z ƒ b, ƒ z 2 ƒ b 2, Á b b 2 Á z 2 Á, then the given series converges, even bsolutely. PROOF By Cuchy s principle, since b b 2 Á converges, for ny given P we cn find n N such tht b n Á b np P for every n N nd p, 2, Á. From this nd ƒ z ƒ b, ƒ z 2 ƒ b 2, Á we conclude tht for those n nd p, ƒ z n ƒ Á ƒ z np ƒ b n Á b np P. Hence, gin by Cuchy s principle, ƒ z converges, so tht z z 2 Á ƒ ƒ z 2 ƒ Á is bsolutely convergent. A good comprison series is the geometric series, which behves s follows. THEOREM 6 Geometric Series The geometric series (6*) m q m q q 2 Á converges with the sum >( q) if ƒ q ƒ nd diverges if ƒ q ƒ. PROOF If ƒ q ƒ, then ƒ q m ƒ nd Theorem 3 implies divergence. Now let ƒ q ƒ. The nth prtil sum is s n q Á q n. From this, qs n q Á q n q n.

6 676 CHAP. 5 Power Series, Tylor Series On subtrction, most terms on the right cncel in pirs, nd we re left with s n qs n ( q)s n q n. Now q since q, nd we my solve for s n, finding (6) s n q n q n q q q. Since ƒ q ƒ, the lst term pproches zero s n :. Hence if ƒ q ƒ, the series is convergent nd hs the sum >( q). This completes the proof. Rtio Test This is the most importnt test in our further work. We get it by tking the geometric series s comprison series b b 2 Á in Theorem 5: THEOREM 7 Rtio Test If series z with z n (n, 2, Á z 2 Á ) hs the property tht for every n greter thn some N, (7) z n ` z ` q n (n N) (where q is fixed), this series converges bsolutely. If for every n N, (8) the series diverges. ` z n z n ` (n N), PROOF If (8) holds, then ƒ z n ƒ ƒ z n ƒ for n N, so tht divergence of the series follows from Theorem 3. If (7) holds, then ƒ z n ƒ ƒ z n ƒ q for n N, in prticulr, ƒ z N2 ƒ ƒ z N ƒ q, ƒ z N3 ƒ ƒ z N2 ƒ q ƒ z N ƒ q 2, etc., nd in generl, ƒ z Np ƒ ƒ z N ƒ q p. Since q, we obtin from this nd Theorem 6 ƒ z N ƒ ƒ z N2 ƒ ƒ z N3 ƒ Á ƒ z N ƒ ( q q 2 Á ) ƒ z N ƒ q. Absolute convergence of z z 2 Á now follows from Theorem 5.

7 SEC. 5. Sequences, Series, Convergence Tests 677 CAUTION! The inequlity (7) implies ƒ z n >z n ƒ, but this does not imply convergence, s we see from the hrmonic series, which stisfies z n >z n n>(n ) for ll n but diverges. If the sequence of the rtios in (7) nd (8) converges, we get the more convenient THEOREM 8 Rtio Test z n If series z with z n (n, 2, Á z 2 Á ) is such tht lim ` n: z ` L, n then: () If L, the series converges bsolutely. (b) If L, the series diverges. (c) If L, the series my converge or diverge, so tht the test fils nd permits no conclusion. PROOF () We write k n ƒ z n >z n ƒ nd let L b. Then by the definition of limit, the k n must eventully get close to b, sy, for ll n greter thn some N. Convergence of z z 2 Á k n q 2 b now follows from Theorem 7. (b) Similrly, for L c we hve k n for ll (sufficiently lrge), which implies divergence of z z 2 Á 2 c n N* by Theorem 7. (c) The hrmonic series 2 3 Á hs z n >z n n>(n ), hence L, nd diverges. The series Á hs z n z n n2 (n ) 2, hence lso L, but it converges. Convergence follows from (Fig. 364) s n 4 Á n 2 n so tht s, s 2, Á is bounded sequence nd is monotone incresing (since the terms of the series re ll positive); both properties together re sufficient for the convergence of the rel sequence s, s 2, Á. (In clculus this is proved by the so-clled integrl test, whose ide we hve used.) dx x 2 2 n, y y = x 2 Are Are 9 Are 6 Are Fig Convergence of the series Á x

8 678 CHAP. 5 Power Series, Tylor Series EXAMPLE 4 Rtio Test Is the following series convergent or divergent? (First guess, then clculte.) n ( 75i) n ( 75i) n! 2! ( 75i)2 Á Solution. By Theorem 8, the series is convergent, since z n ` ` ƒ 75i ƒ n >(n )! ƒ 75i ƒ 25 z n ƒ 75i ƒ n >n! n n : L. EXAMPLE 5 Theorem 7 More Generl Thn Theorem 8 Let nd b n >2 3n n i>2 3n. Is the following series convergent or divergent? b b Á i 2 i 8 6 i Á Solution. The rtios of the bsolute vlues of successive terms re 2, 4, 2, 4, Á. Hence convergence follows from Theorem 7. Since the sequence of these rtios hs no limit, Theorem 8 is not pplicble. Root Test The rtio test nd the root test re the two prcticlly most importnt tests. The rtio test is usully simpler, but the root test is somewht more generl. THEOREM 9 Root Test If series z z 2 Á is such tht for every n greter thn some N, (9) 2 n ƒ z n ƒ q (n N) (where q is fixed), this series converges bsolutely. If for infinitely mny n, () 2 n ƒ z n ƒ, the series diverges. PROOF If (9) holds, then ƒ z for ll Hence the series ƒ z ƒ ƒ z 2 ƒ Á n ƒ q n n N. converges by comprison with the geometric series, so tht the series z z 2 Á converges bsolutely. If () holds, then for infinitely mny n. Divergence of z z 2 Á ƒ z n ƒ now follows from Theorem 3. CAUTION! Eqution (9) implies 2 n ƒ z n ƒ, but this does not imply convergence, s we see from the hrmonic series, which stisfies 2 n >n (for n ) but diverges.

9 SEC. 5. Sequences, Series, Convergence Tests 679 If the sequence of the roots in (9) nd () converges, we more conveniently hve THEOREM Root Test If series z z 2 Á is such tht lim then: n: 2n ƒ z n ƒ L, () The series converges bsolutely if L. (b) The series diverges if L. (c) If L, the test fils; tht is, no conclusion is possible. PROBLEM SET 5. SEQUENCES Is the given sequence z, z 2, Á, z n, Á 2. bounded? Convergent? Find its limit points. Show your work in detil.. z 2. z n (3 4i) n n ( i) 2n >2 n 2. >n! 3. z 4. z n ( 2i) n n np>(4 2ni) 5. z n () n i 6. z n (cos npi)>n 7. z 8. z n [( 3i)> ] n n n 2 i>n 2 9. z n (3 3i) n.. CAS EXPERIMENT. Sequences. Write progrm for grphing complex sequences. Use the progrm to discover sequences tht hve interesting geometric properties, e.g., lying on n ellipse, spirling to its limit, hving infinitely mny limit points, etc. 2. Addition of sequences. If converges with the limit l nd z*, z* 2, Á z, z 2, Á converges with the limit show tht z z*, z 2 z* 2, Á l*, is convergent with the limit l l*. 3. Bounded sequence. Show tht complex sequence is bounded if nd only if the two corresponding sequences of the rel prts nd of the imginry prts re bounded. 4. On Theorem. Illustrte Theorem by n exmple of your own. 5. On Theorem 2. Give nother exmple illustrting Theorem SERIES Is the given series convergent or divergent? Give reson. Show detils. (2 3i) n (i) n n! ln n n 8. n 2 i n 9. 4 b n z n sin ( n 4 np) i n2 n i n n 2 i n n n () n ( i) 2n (2n)! n n n n 3n 2 2i (p pi) 2n (2n )! (3i) n n! n n i n n n i 26. Significnce of (7). Wht is the difference between (7) nd just stting ƒ z n >z n ƒ? 27. On Theorems 7 nd 8. Give nother exmple showing tht Theorem 7 is more generl thn Theorem CAS EXPERIMENT. Series. Write progrm for computing nd grphing numeric vlues of the first n prtil sums of series of complex numbers. Use the progrm to experiment with the rpidity of convergence of series of your choice. 29. Absolute convergence. Show tht if series converges bsolutely, it is convergent. 3. Estimte of reminder. Let so tht the series z z 2 Á ƒ z n>z n ƒ q, converges by the rtio test. Show tht the reminder R n z n z n2 Á stisfies the inequlity ƒ R n ƒ ƒ z n ƒ >( q). Using this, find how mny terms suffice for computing the sum s of the series n n i 2 n n with n error not exceeding.5 nd compute s to this ccurcy.

10 68 CHAP. 5 Power Series, Tylor Series 5.2 Power Series The student should py close ttention to the mteril becuse we shll show how power series ply n importnt role in complex nlysis. Indeed, they re the most importnt series in complex nlysis becuse their sums re nlytic functions (Theorem 5, Sec. 5.3), nd every nlytic function cn be represented by power series (Theorem, Sec. 5.4). A power series in powers of z z is series of the form () n n (z z ) n (z z ) 2 (z z ) 2 Á where z is complex vrible,,, Á re complex (or rel) constnts, clled the coefficients of the series, nd z is complex (or rel) constnt, clled the center of the series. This generlizes rel power series of clculus. If z, we obtin s prticulr cse power series in powers of z: (2) n n z n z 2 z 2 Á. Convergence Behvior of Power Series Power series hve vrible terms (functions of z), but if we fix z, then ll the concepts for series with constnt terms in the lst section pply. Usully series with vrible terms will converge for some z nd diverge for others. For power series the sitution is simple. The series () my converge in disk with center z or in the whole z-plne or only t z. We illustrte this with typicl exmples nd then prove it. EXAMPLE Convergence in Disk. Geometric Series The geometric series z n z z 2 Á n converges bsolutely if ƒ z ƒ nd diverges if ƒ z ƒ (see Theorem 6 in Sec. 5.). EXAMPLE 2 Convergence for Every z The power series (which will be the Mclurin series of in Sec. 5.4) e z n z n n! 2 z z 2! z 3 3! Á is bsolutely convergent for every z. In fct, by the rtio test, for ny fixed z, z n >(n )! ` ` z n >n! ƒ z ƒ n : s n :.

11 ` SEC. 5.2 Power Series 68 EXAMPLE 3 Convergence Only t the Center. (Useless Series) The following power series converges only t z, but diverges for every z, s we shll show. n!z n z 2z 2 6z 3 Á n In fct, from the rtio test we hve (n )!z n ` (n ) ƒ z ƒ : n!z n s n : (z fixed nd ). THEOREM Convergence of Power Series () Every power series () converges t the center z. (b) If () converges t point z z z, it converges bsolutely for every z closer to z thn z, tht is, ƒ z z ƒ ƒ z z ƒ. See Fig (c) If () diverges t z z 2, it diverges for every z frther wy from z thn z 2. See Fig y Divergent Conv. z z z 2 x Fig Theroem PROOF () For z z the series reduces to the single term. (b) Convergence t z z gives by Theorem 3 in Sec. 5. n (z z ) n : s n :. This implies boundedness in bsolute vlue, ƒ n (z z ) n ƒ M for every n,, Á. Multiplying nd dividing by (z z ) n n (z z ) n we obtin from this ƒ n (z z ) n ƒ ` n (z z ) n z z n n z z z z b ` M ` z z `. Summtion over n gives (3) n ƒ n (z z ) n ƒ M n z z ` z z ` n. Now our ssumption ƒ z z ƒ ƒ z z ƒ implies tht ƒ (z z )>(z z ) ƒ. Hence the series on the right side of (3) is converging geometric series (see Theorem 6 in

12 682 CHAP. 5 Power Series, Tylor Series Sec. 5.). Absolute convergence of () s stted in (b) now follows by the comprison test in Sec. 5.. (c) If this were flse, we would hve convergence t z 3 frther wy from z thn z 2. This would imply convergence t z 2, by (b), contrdiction to our ssumption of divergence t z 2. Rdius of Convergence of Power Series Convergence for every z (the nicest cse, Exmple 2) or for no z z (the useless cse, Exmple 3) needs no further discussion, nd we put these cses side for moment. We consider the smllest circle with center z tht includes ll the points t which given power series () converges. Let R denote its rdius. The circle ƒ z z ƒ R (Fig. 366) is clled the circle of convergence nd its rdius R the rdius of convergence of (). Theorem then implies convergence everywhere within tht circle, tht is, for ll z for which (4) ƒ z z ƒ R z (the open disk with center nd rdius R). Also, since R is s smll s possible, the series () diverges for ll z for which (5) ƒ z z ƒ R. No generl sttements cn be mde bout the convergence of power series () on the circle of convergence itself. The series () my converge t some or ll or none of the points. Detils will not be importnt to us. Hence simple exmple my just give us the ide. y Divergent Conv. R z x Fig Circle of convergence EXAMPLE 4 Behvior on the Circle of Convergence On the circle of convergence (rdius R in ll three series), S z n >n 2 converges everywhere since S >n 2 converges, S z n >n converges t (by Leibniz s test) but diverges t, S z n diverges everywhere.

13 SEC. 5.2 Power Series 683 Nottions R nd R. To incorporte these two excluded cses in the present nottion, we write R if the series () converges for ll z (s in Exmple 2), R if () converges only t the center z z (s in Exmple 3). These re convenient nottions, but nothing else. Rel Power Series. In this cse in which powers, coefficients, nd center re rel, formul (4) gives the convergence intervl ƒ x x ƒ R of length 2R on the rel line. Determintion of the Rdius of Convergence from the Coefficients. For this importnt prcticl tsk we cn use THEOREM 2 Rdius of Convergence R Suppose tht the sequence ƒ n > n ƒ, n, 2, Á, converges with limit L*. If L*, then R ; tht is, the power series () converges for ll z. If L* (hence L* ), then (6) R lim (Cuchy Hdmrd formul ). L* ` ` n: n If ƒ n > n ƒ :, then R (convergence only t the center z ). n PROOF For () the rtio of the terms in the rtio test (Sec. 5.) is n (z z ) n n ` n (z z ) n ` ` ` ƒ z z ƒ. n The limit is L L* ƒ z z ƒ. Let L*, thus L*. We hve convergence if L L* ƒ z z ƒ, thus ƒ z z ƒ >L*, nd divergence if ƒ z z ƒ >L*. By (4) nd (5) this shows tht >L* is the convergence rdius nd proves (6). If L*, then L for every z, which gives convergence for ll z by the rtio test. If ƒ n > n ƒ :, then ƒ n > n ƒƒz z ƒ for ny z z nd ll sufficiently lrge n. This implies divergence for ll z z by the rtio test (Theorem 7, Sec. 5.). Formul (6) will not help if L* does not exist, but extensions of Theorem 2 re still possible, s we discuss in Exmple 6 below. EXAMPLE 5 Rdius of Convergence (2n)! By (6) the rdius of convergence of the power series (z 3i)n is 2 (n!) n R lim n: c (2n!) (2n 2)! ^ (n!) 2 ((n )!) 2 d lim n: c (2n!) (2n 2)! ((n )!)2 (n!) 2 d lim n: (n ) 2 (2n 2)(2n ) 4. The series converges in the open disk ƒ z 3i ƒ 4 of rdius 4 nd center 3i. Nmed fter the French mthemticins A. L. CAUCHY (see Sec. 2.5) nd JACQUES HADAMARD ( ). Hdmrd mde bsic contributions to the theory of power series nd devoted his lifework to prtil differentil equtions.

14 684 CHAP. 5 Power Series, Tylor Series EXAMPLE 6 Extension of Theorem 2 Find the rdius of convergence R of the power series n c () n 2 n d z n 3 2 z 2 4 b z2 8 z3 2 6 b z4 Á. ƒ ƒ ƒ ƒ ƒ ƒ Solution. The sequence of the rtios 6, 2(2 4 ), >(8(2 4 )), Á does not converge, so tht Theorem 2 is of no help. It cn be shown tht (6*) R >L, L lim 2 n n. n: This still does not help here, since does not converge becuse 2 n n 2 n >2 n n ) 2 for odd n, wheres for even n we hve 2 n ƒ n ƒ 2 n 2 >2 n : s n :, so tht 2n ƒ n ƒ hs the two limit points nd. It cn further be shown tht 2, the gretest limit point of the sequence { 2 n (6**) R >l l ƒ n ƒ }. Here l I, so tht R. Answer. The series converges for ƒ z ƒ. Summry. Power series converge in n open circulr disk or some even for every z (or some only t the center, but they re useless); for the rdius of convergence, see (6) or Exmple 6. Except for the useless ones, power series hve sums tht re nlytic functions (s we show in the next section); this ccounts for their importnce in complex nlysis. PROBLEM SET 5.2. Power series. Are nd z 2 z 3 Á >z z z 2 Á z z 3>2 n(n ) 8. power series? Explin. (z pi)n 9. (z i) 2n n! 3 n n n 2. Rdius of convergence. Wht is it? Its role? Wht (z 2i) n motivtes its nme? How cn you find it?.. 2 i n 5i b n zn 3. Convergence. Wht re the only bsiclly different n n possibilities for the convergence of power series? () n n On Exmples 3. Extend them to power series in z n 3. 6 n (z i) 4n 8 n n n powers of z 4 3pi. Extend Exmple to the cse of rdius of convergence 6. () n (2n)! Powers z 2n. Show tht if S n z n hs rdius of 4 n (n!) 2 (z 2i) n 2 2n (n!) 2 z 2n n n convergence R (ssumed finite), then S n z 2n hs (3n)! 2 n rdius of convergence R n (n!) 3 z n n(n ) z2n 6 8 RADIUS OF CONVERGENCE Find the center nd the rdius of convergence. () n (2n)! z 4 n (z ) n 2 pb n n 2n n n n n 2() n 8. p(2n )n! z2n n 9. CAS PROJECT. Rdius of Convergence. Write progrm for computing R from (6), (6*), or (6**), in

15 SEC. 5.3 Functions Given by Power Series 685 this order, depending on the existence of the limits needed. Test the progrm on some series of your choice such tht ll three formuls (6), (6*), nd (6**) will come up. 2. TEAM PROJECT. Rdius of Convergence. () Understnding (6). Formul (6) for R contins ƒ n > n ƒ, not ƒ n > n ƒ. How could you memorize this by using qulittive rgument? (b) Chnge of coefficients. Wht hppens to R ( R ) if you (i) multiply ll by k, n (ii) multiply ll by k n n, (iii) replce n by > n? Cn you think of n ppliction of this? (c) Understnding Exmple 6, which extends Theorem 2 to nonconvergent cses of n > n. Do you understnd the principle of mixing by which Exmple 6 ws obtined? Mke up further exmples. (d) Understnding (b) nd (c) in Theorem. Does there exist power series in powers of z tht converges t z 3 i nd diverges t z 3 6i? Give reson. 5.3 Functions Given by Power Series Here, our min gol is to show tht power series represent nlytic functions. This fct (Theorem 5) nd the fct tht power series behve nicely under ddition, multipliction, differentition, nd integrtion ccounts for their usefulness. To simplify the formuls in this section, we tke z nd write () n n z n. There is no loss of generlity becuse series in powers of ẑ z with ny z cn lwys be reduced to the form () if we set ẑ z z. Terminology nd Nottion. If ny given power series () hs nonzero rdius of convergence R (thus R ), its sum is function of z, sy f (z). Then we write (2) f (z) n n z n z 2 z 2 Á ( ƒ z ƒ R). We sy tht f (z) is represented by the power series or tht it is developed in the power series. For instnce, the geometric series represents the function f (z) >( z) in the interior of the unit circle ƒ z ƒ. (See Theorem 6 in Sec. 5..) Uniqueness of Power Series Representtion. This is our next gol. It mens tht function f (z) cnnot be represented by two different power series with the sme center. We clim tht if f (z) cn t ll be developed in power series with center z, the development is unique. This importnt fct is frequently used in complex nlysis (s well s in clculus). We shll prove it in Theorem 2. The proof will follow from THEOREM Continuity of the Sum of Power Series If function f (z) cn be represented by power series (2) with rdius of convergence R, then f (z) is continuous t z.

16 686 CHAP. 5 Power Series, Tylor Series PROOF From (2) with z we hve f (). Hence by the definition of continuity we must show tht lim z: f (z) f (). Tht is, we must show tht for given P there is d such tht ƒ z ƒ d implies ƒ f (z) ƒ P. Now (2) converges bsolutely for ƒ z ƒ r with ny r such tht r R, by Theorem in Sec Hence the series n ƒ n ƒ r n r ƒ n ƒ r n n converges. Let S be its sum. ( S is trivil.) Then for ƒ z ƒ r, ƒ f (z) ƒ 2 n n z n 2 ƒ z ƒ n ƒ n ƒƒzƒ n ƒ z ƒ ƒ n ƒ r n ƒ z ƒ S n nd ƒ z ƒ S P when ƒ z ƒ d, where d is less thn r nd less thn P>S. Hence ƒ z ƒ S ds (P>S)S P. This proves the theorem. From this theorem we cn now redily obtin the desired uniqueness theorem (gin ssuming z without loss of generlity): THEOREM 2 Identity Theorem for Power Series. Uniqueness Let the power series nd b b z b 2 z 2 Á z 2 z 2 Á both be convergent for ƒ z ƒ R, where R is positive, nd let them both hve the sme sum for ll these z. Then the series re identicl, tht is, b, b, 2 b 2, Á. Hence if function f (z) cn be represented by power series with ny center z, this representtion is unique. PROOF We proceed by induction. By ssumption, z 2 z 2 Á b b z b 2 z 2 Á ( ƒ z ƒ R). The sums of these two power series re continuous t z, by Theorem. Hence if we consider ƒ z ƒ nd let z : on both sides, we see tht the ssertion is true for Now ssume tht for n,, Á b : n. n b n, m. Then on both sides we my omit the terms tht re equl nd divide the result by z m ( ); this gives m m2 z m3 z 2 Á b m b m2 z b m3 z 2 Á. Similrly s before by letting z : we conclude from this tht m b m. This completes the proof. Opertions on Power Series Interesting in itself, this discussion will serve s preprtion for our min gol, nmely, to show tht functions represented by power series re nlytic.

17 R SEC. 5.3 Functions Given by Power Series 687 Termwise ddition or subtrction of two power series with rdii of convergence nd R 2 yields power series with rdius of convergence t lest equl to the smller of R nd R 2. Proof. Add (or subtrct) the prtil sums s n nd s n * term by term nd use lim (s n s* n ) lim s n lim s* n. Termwise multipliction of two power series nd f (z) k k z k z Á g(z) m b m z m b b z Á mens the multipliction of ech term of the first series by ech term of the second series nd the collection of like powers of z. This gives power series, which is clled the Cuchy product of the two series nd is given by b ( b b )z ( b 2 b 2 b )z 2 Á We mention without proof tht this power series converges bsolutely for ech z within the smller circle of convergence of the two given series nd hs the sum s(z) f (z)g(z). For proof, see [D5] listed in App.. n ( b n b n Á n b )z n. Termwise differentition nd integrtion of power series is permissible, s we show next. We cll derived series of the power series () the power series obtined from () by termwise differentition, tht is, (3) n n n z n 2 2 z 3 3 z 2 Á. THEOREM 3 Termwise Differentition of Power Series The derived series of power series hs the sme rdius of convergence s the originl series. PROOF This follows from (6) in Sec. 5.2 becuse lim n: n ƒ n ƒ (n ) ƒ n ƒ lim n: n n lim n: ` n n ` lim n: ` n n ` or, if the limit does not exist, from (6**) in Sec. 5.2 by noting tht 2 n n : s n :.

18 688 CHAP. 5 Power Series, Tylor Series EXAMPLE Appliction of Theorem 3 Find the rdius of convergence R of the following series by pplying Theorem 3. n 2 b zn z 2 3z 3 6z 4 z 5 Á. n2 Solution. Differentite the geometric series twice term by term nd multiply the result by z 2 >2. This yields the given series. Hence R by Theorem 3. THEOREM 4 Termwise Integrtion of Power Series The power series n n n zn z 2 z2 2 3 z3 Á obtined by integrting the series z 2 z 2 Á term by term hs the sme rdius of convergence s the originl series. The proof is similr to tht of Theorem 3. With the help of Theorem 3, we estblish the min result in this section. Power Series Represent Anlytic Functions THEOREM 5 Anlytic Functions. Their Derivtives A power series with nonzero rdius of convergence R represents n nlytic function t every point interior to its circle of convergence. The derivtives of this function re obtined by differentiting the originl series term by term. All the series thus obtined hve the sme rdius of convergence s the originl series. Hence, by the first sttement, ech of them represents n nlytic function. PROOF () We consider ny power series () with positive rdius of convergence R. Let f (z) be its sum nd f (z) the sum of its derived series; thus (4) f (z) nd f (z) n n z n n z n. n n We show tht f (z) is nlytic nd hs the derivtive f (z) in the interior of the circle of convergence. We do this by proving tht for ny fixed z with ƒ z ƒ R nd z : the difference quotient [ f (z z) f (z)]> z pproches f (z). By termwise ddition we first hve from (4) f (z z) f (z) (5) f (z) z n2 n c (z z)n z n nz n d. z Note tht the summtion strts with 2, since the constnt term drops out in tking the difference f (z z) f (z), nd so does the liner term when we subtrct f (z) from the difference quotient.

19 SEC. 5.3 Functions Given by Power Series 689 (b) We clim tht the series in (5) cn be written (6) n2 n z[(z z) n2 2z(z z) n3 Á (n 2)z n3 (z z) (n )z n2 ]. The somewht technicl proof of this is given in App. 4. (c) We consider (6). The brckets contin n terms, nd the lrgest coefficient is n. Since (n ) 2 n(n ), we see tht for ƒ z ƒ R nd ƒ z z ƒ R, R R, the bsolute vlue of this series (6) cnnot exceed (7) ƒ zƒ ƒ n ƒ n(n )R n2. n2 This series with n insted of ƒ n ƒ is the second derived series of (2) t z R nd converges bsolutely by Theorem 3 of this section nd Theorem of Sec Hence our present series (7) converges. Let the sum of (7) (without the fctor ƒ zƒ ) be K (R ). Since (6) is the right side of (5), our present result is f (z z) f (z) ` z f (z) ` ƒ z ƒ K(R ). Letting z : nd noting tht R ( R) is rbitrry, we conclude tht f (z) is nlytic t ny point interior to the circle of convergence nd its derivtive is represented by the derived series. From this the sttements bout the higher derivtives follow by induction. Summry. The results in this section show tht power series re bout s nice s we could hope for: we cn differentite nd integrte them term by term (Theorems 3 nd 4). Theorem 5 ccounts for the gret importnce of power series in complex nlysis: the sum of such series (with positive rdius of convergence) is n nlytic function nd hs derivtives of ll orders, which thus in turn re nlytic functions. But this is only prt of the story. In the next section we show tht, conversely, every given nlytic function f (z) cn be represented by power series, clled Tylor series nd being the complex nlog of the rel Tylor series of clculus. PROBLEM SET 5.3. Reltion to Clculus. Mteril in this section generlizes clculus. Give detils. 2. Termwise ddition. Write out the detils of the proof on termwise ddition nd subtrction of power series. 3. On Theorem 3. Prove tht n n : s n :, s climed. 4. Cuchy product. Show tht ( z) 2 (n )z n n () by using the Cuchy product, (b) by differentiting suitble series. 5 5 RADIUS OF CONVERGENCE BY DIFFERENTIATION OR INTEGRATION Find the rdius of convergence in two wys: () directly by the Cuchy Hdmrd formul in Sec. 5.2, nd (b) from series of simpler terms by using Theorem 3 or Theorem 4. n(n ) 5. (z 2i) n 6. 2 n n2 n n (z 2i) 2n n(n ) zn n n n () n 2n z 2n 2p b 5 n

20 69 CHAP. 5 Power Series, Tylor Series n nk n n n n n2 (2) n n(n )(n 2) z2n n k b z 2 b n 3 n n(n ) 7 n (z 2) 2n 2n(2n ) n n z 2n2 c n k b d z nk k n m m b zn 4 n n(n ) 3 n (z i) n 6 2 APPLICATIONS OF THE IDENTITY THEOREM Stte clerly nd explicitly where nd how you re using Theorem Even functions. If f (z) in (2) is even (i.e., f (z) f (z)), show tht n for odd n. Give exmples. 7. Odd function. If f (z) in (2) is odd (i.e., f (z) f (z)), show tht n for even n. Give exmples. 8. Binomil coefficients. Using ( z) p ( z) q ( z) pq, obtin the bsic reltion r p n b q q b p b. r n r n 9. Find pplictions of Theorem 2 in differentil equtions nd elsewhere. 2. TEAM PROJECT. Fiboncci numbers. 2 () The Fiboncci numbers re recursively defined by if n, 2, Á, n n n. Find the limit of the sequence ( n > n ). (b) Fiboncci s rbbit problem. Compute list of, Á, 2. Show tht is the number of pirs of rbbits fter 2 months if initilly there is pir nd ech pir genertes pir per month, beginning in the second month of existence (no deths occurring). (c) Generting function. Show tht the generting function of the Fiboncci numbers is f (z) >( z z 2 ); tht is, if power series () represents this f (z), its coefficients must be the Fiboncci numbers nd conversely. Hint. Strt from f (z)( z z 2 ) nd use Theorem Tylor nd Mclurin Series The Tylor series 3 of function f (z), the complex nlog of the rel Tylor series is () f (z) n (z z ) n where n n n! f (n) (z ) or, by (), Sec. 4.4, (2) n 2pi C f (z*) n dz*. (z* z ) In (2) we integrte counterclockwise round simple closed pth C tht contins z in its interior nd is such tht f (z) is nlytic in domin contining C nd every point inside C. A Mclurin series 3 is Tylor series with center z. 2 LEONARDO OF PISA, clled FIBONACCI ( son of Bonccio), bout 8 25, Itlin mthemticin, credited with the first renissnce of mthemtics on Christin soil. 3 BROOK TAYLOR (685 73), English mthemticin who introduced rel Tylor series. COLIN MACLAURIN ( ), Scots mthemticin, professor t Edinburgh.

21 SEC. 5.4 Tylor nd Mclurin Series 69 (3) The reminder of the Tylor series () fter the term n (z z ) n is R n (z) (z z ) n 2pi (proof below). Writing out the corresponding prtil sum of (), we thus hve C f (z*) (z* z ) n (z* z) dz* (4) f (z) f (z ) z z! (z z ) n n! f r(z ) (z z ) 2 f s(z ) 2! Á f (n) (z ) R n (z). This is clled Tylor s formul with reminder. We see tht Tylor series re power series. From the lst section we know tht power series represent nlytic functions. And we now show tht every nlytic function cn be represented by power series, nmely, by Tylor series (with vrious centers). This mkes Tylor series very importnt in complex nlysis. Indeed, they re more fundmentl in complex nlysis thn their rel counterprts re in clculus. THEOREM Tylor s Theorem Let f (z) be nlytic in domin D, nd let z z be ny point in D. Then there exists precisely one Tylor series () with center z tht represents f (z). This representtion is vlid in the lrgest open disk with center z in which f (z) is nlytic. The reminders R n (z) of () cn be represented in the form (3). The coefficients stisfy the inequlity (5) ƒ n ƒ M r n where M is the mximum of ƒ f (z) ƒ on circle ƒ z z ƒ r in D whose interior is lso in D. PROOF The key tool is Cuchy s integrl formul in Sec. 4.3; writing z nd z* insted of z nd z (so tht z* is the vrible of integrtion), we hve (6) f (z) 2pi C f (z*) z* z dz*. z lies inside C, for which we tke circle of rdius r with center z nd interior in D (Fig. 367). We develop >(z* z) in (6) in powers of z z. By stndrd lgebric mnipultion (worth remembering!) we first hve (7) z* z z* z (z z ) (z* z ) z z. b z* z

22 692 CHAP. 5 Power Series, Tylor Series y z r z z* C Fig Cuchy formul (6) x For lter use we note tht since z* is on C while z is inside C, we hve (7*) z z ` `. z* z (Fig. 367). To (7) we now pply the sum formul for finite geometric sum (8*) q Á q n q n q n q q q (q ), which we use in the form (tke the lst term to the other side nd interchnge sides) (8) q q Á q n q n q. Applying this with q (z z )>(z* z ) to the right side of (7), we get z* z c z z z z 2 b z* z z* z z* z Á z z n b d z* z We insert this into (6). Powers of z z do not depend on the vrible of integrtion z*, so tht we my tke them out from under the integrl sign. This yields f (z) 2pi C z* z z z n b. z* z f (z*) dz* z z z* z 2pi (z z Á ) n 2pi with R n (z) given by (3). The integrls re those in (2) relted to the derivtives, so tht we hve proved the Tylor formul (4). Since nlytic functions hve derivtives of ll orders, we cn tke n in (4) s lrge s we plese. If we let n pproch infinity, we obtin (). Clerly, () will converge nd represent f (z) if nd only if (9) lim n: R n (z). C f (z*) 2 dz* Á (z* z ) C f (z*) (z* z ) n dz* R n(z)

23 SEC. 5.4 Tylor nd Mclurin Series 693 We prove (9) s follows. Since z* lies on C, wheres z lies inside C (Fig. 367), we hve ƒ z* z ƒ. Since f (z) is nlytic inside nd on C, it is bounded, nd so is the function f (z*)>(z* z), sy, f (z*) ` ` M z* z for ll z* on C. Also, C hs the rdius r ƒ z* z ƒ nd the length 2pr. Hence by the ML-inequlity (Sec. 4.) we obtin from (3) () ƒ R n ƒ ƒ z z n ƒ 2p f (z*) ` (z* z ) n (z* z) dz* ` C ƒ z z n ƒ 2p M n r 2pr M z z ` `. n r Now ƒ z z ƒ r becuse z lies inside C. Thus ƒ z z ƒ >r, so tht the right side pproches s n :. This proves tht the Tylor series converges nd hs the sum f (z). Uniqueness follows from Theorem 2 in the lst section. Finlly, (5) follows from n in () nd the Cuchy inequlity in Sec This proves Tylor s theorem. Accurcy of Approximtion. We cn chieve ny pressinged ccurcy in pproximting f (z) by prtil sum of () by choosing n lrge enough. This is the prcticl use of formul (9). Singulrity, Rdius of Convergence. On the circle of convergence of () there is t lest one singulr point of f (z), tht is, point z c t which f (z) is not nlytic (but such tht every disk with center c contins points t which f (z) is nlytic). We lso sy tht f (z) is singulr t c or hs singulrity t c. Hence the rdius of convergence R of () is usully equl to the distnce from z to the nerest singulr point of f (z). (Sometimes R cn be greter thn tht distnce: Ln z is singulr on the negtive rel xis, whose distnce from z i is, but the Tylor series of Ln z with center z i hs rdius of convergence 2.) Power Series s Tylor Series Tylor series re power series of course! Conversely, we hve THEOREM 2 Reltion to the Previous Section A power series with nonzero rdius of convergence is the Tylor series of its sum. PROOF Given the power series f (z) (z z ) 2 (z z ) 2 3 (z z ) 3 Á.

24 694 CHAP. 5 Power Series, Tylor Series Then f (z ). By Theorem 5 in Sec. 5.3 we obtin f r(z) 2 2 (z z ) 3 3 (z z ) 2 Á, f s(z) # 2(z z ) Á, thus thus f r(z ) f s(z ) 2! 2 nd in generl f (n) (z ) n! n. With these coefficients the given series becomes the Tylor series of f (z) with center z. Comprison with Rel Functions. One surprising property of complex nlytic functions is tht they hve derivtives of ll orders, nd now we hve discovered the other surprising property tht they cn lwys be represented by power series of the form (). This is not true in generl for rel functions; there re rel functions tht hve derivtives of ll orders but cnnot be represented by power series. (Exmple: f (x) exp (>x 2 ) if x nd f () ; this function cnnot be represented by Mclurin series in n open disk with center becuse ll its derivtives t re zero.) Importnt Specil Tylor Series These re s in clculus, with x replced by complex z. Cn you see why? (Answer. The coefficient formuls re the sme.) EXAMPLE Geometric Series Let f (z) >( z). Then we hve f (n) (z) n!>( z) n, f (n) () n!. Hence the Mclurin expnsion of >( z) is the geometric series () z z n z z 2 Á n ( ƒ z ƒ ). f (z) is singulr t z ; this point lies on the circle of convergence. EXAMPLE 2 Exponentil Function We know tht the exponentil function (Sec. 3.5) is nlytic for ll z, nd (e z )r e z. Hence from () with z we obtin the Mclurin series e z (2) e z n z n n! 2 z z 2! Á. This series is lso obtined if we replce x in the fmilir Mclurin series of by z. Furthermore, by setting z iy in (2) nd seprting the series into the rel nd imginry prts (see Theorem 2, Sec. 5.) we obtin e iy (iy) n n! () k y2k (2k)! i () k y 2k (2k )!. n k Since the series on the right re the fmilir Mclurin series of the rel functions cos y nd sin y, this shows tht we hve rediscovered the Euler formul k e x (3) e iy cos y i sin y. e z Indeed, one my use (2) for defining nd derive from (2) the bsic properties of e z. For instnce, the differentition formul (e z )r e z follows redily from (2) by termwise differentition.

25 SEC. 5.4 Tylor nd Mclurin Series 695 EXAMPLE 3 Trigonometric nd Hyperbolic Functions By substituting (2) into () of Sec. 3.6 we obtin (4) cos z n sin z n z2n () n (2n)! z 2n () n (2n )! 2 z 2! z 4 4! Á 3 z z 3! z 5 5! Á. When z x these re the fmilir Mclurin series of the rel functions cos x nd sin x. Similrly, by substituting (2) into (), Sec. 3.6, we obtin (5) cosh z n sinh z n z 2n (2n)! z 2n (2n )! 2 z 2! z 4 4! Á 3 z z 3! z 5 5! Á. EXAMPLE 4 Logrithm From () it follows tht (6) Ln ( z) z z2 2 z 3 3 Á ( ƒ z ƒ ). Replcing z by z nd multiplying both sides by, we get (7) Ln ( z) Ln z z2 z3 z 2 3 Á ( ƒ z ƒ ). By dding both series we obtin (8) Ln z z 3 z 2 z 3 z 5 5 Á b ( ƒ z ƒ ). Prcticl Methods The following exmples show wys of obtining Tylor series more quickly thn by the use of the coefficient formuls. Regrdless of the method used, the result will be the sme. This follows from the uniqueness (see Theorem ). EXAMPLE 5 Substitution Find the Mclurin series of f (z) >( z 2 ). Solution. By substituting z 2 for z in () we obtin (9) ( ƒ z ƒ ). z 2 (z 2 (z 2 ) n () n z 2n z 2 z 4 z 6 Á ) n n

26 ` 696 CHAP. 5 Power Series, Tylor Series EXAMPLE 6 Integrtion Find the Mclurin series of f (z) rctn z. Solution. We hve f r(z) >( z 2 ). Integrting (9) term by term nd using f () we get () n rctn z 2n z2n z z 3 3 z n 5 5 Á ( ƒ z ƒ ); this series represents the principl vlue of ƒ u ƒ p>2. w u iv rctn z defined s tht vlue for which EXAMPLE 7 Development by Using the Geometric Series Develop >(c z) in powers of z z, where c z. Solution. This ws done in the proof of Theorem, where c z*. The beginning ws simple lgebr nd then the use of () with z replced by (z z )>(c z ): This series converges for c z c z (z z ) (c z ) z z c z b c z n z z z z 2 b c z c z c z Á b. z z n b c z z z c z `, tht is, ƒ z z ƒ ƒ c z ƒ. EXAMPLE 8 Binomil Series, Reduction by Prtil Frctions Find the Tylor series of the following function with center z. f (z) 2z2 9z 5 z 3 z 2 8z 2 Solution. We develop f (z) in prtil frctions nd the first frction in binomil series (2) ( z) m ( z)m n m n b zn m(m ) m(m )(m 2) mz z 2 z 3 Á 2! 3! with m 2 nd the second frction in geometric series, nd then dd the two series term by term. This gives f (z) (z 2) 2 2 z 3 [3 (z )] (z ) 9 [ 3 (z b )]2 2 (z ) 9 n n n b z b 3 n z n b 2 c ()n (n ) n 3 n2 2 n d (z ) n 3 23 (z ) 54 8 (z ) (z )3 Á. We see tht the first series converges for ƒ z ƒ 3 nd the second for ƒ z ƒ 2. This hd to be expected becuse >(z 2) 2 is singulr t 2 nd 2>(z 3) t 3, nd these points hve distnce 3 nd 2, respectively, from the center z. Hence the whole series converges for ƒ z ƒ 2.

27 SEC. 5.4 Tylor nd Mclurin Series 697 PROBLEM SET 5.4. Clculus. Which of the series in this section hve you discussed in clculus? Wht is new? 2. On Exmples 5 nd 6. Give ll the detils in the derivtion of the series in those exmples. 3 MACLAURIN SERIES Find the Mclurin series nd its rdius of convergence. 3. sin 2z 2 4. z 2 z z 4 3iz 7. cos 2 2 z 8. sin 2 z 9. exp t 2 2 b dt. z exp (z 2 ) exp (t 2 ) dt z 4 HIGHER TRANSCENDENTAL FUNCTIONS Find the Mclurin series by termwise integrting the integrnd. (The integrls cnnot be evluted by the usul methods of clculus. They define the error function erf z, sine integrl Si(z), nd Fresnel integrls 4 S(z) nd C(z), which occur in sttistics, het conduction, optics, nd other pplictions. These re specil so-clled higher trnscendentl functions.) z z. S(z) sin t 2 dt 2. C(z) cos t 2 dt z 3. erf z 2 z e t2 dt 4. Si(z) p 5. CAS Project. sec, tn. () Euler numbers. The Mclurin series (2) sec z E E 2 2! z2 E 4 4! z4 Á defines the Euler numbers E 2n. Show tht E, E 2, E 4 5, E 6 6. Write progrm tht computes the E 2n from the coefficient formul in () or extrcts them s list from the series. (For tbles see Ref. [GenRef], p. 8, listed in App..) (b) Bernoulli numbers. The Mclurin series (22) z e z B z B 2 2! z2 B 3 3! z3 Á sin t t dt defines the Bernoulli numbers B n. Using undetermined coefficients, show tht B (23) 2, B 2 6, B 3, B 4 3, B 5, B 6 42, Á. Write progrm for computing B n. (c) Tngent. Using (), (2), Sec. 3.6, nd (22), show tht tn z hs the following Mclurin series nd clculte from it tble of B, Á, B 2 : (24) tn z 2i e 2iz 4i e 4iz i n () n 22n (2 2n ) (2n)! 6. Inverse sine. Developing > 2 z 2 nd integrting, show tht rcsin z z 2 b z3 # # 4 b z5 5 B 2n z 2n. # 3 # 5 2 # 4 # 6 b z7 7 Á ( ƒ z ƒ ). Show tht this series represents the principl vlue of rcsin z (defined in Tem Project 3, Sec. 3.7). 7. TEAM PROJECT. Properties from Mclurin Series. Clerly, from series we cn compute function vlues. In this project we show tht properties of functions cn often be discovered from their Tylor or Mclurin series. Using suitble series, prove the following. () The formuls for the derivtives of e z, cos z, sin z, cosh z, sinh z. nd Ln ( z) (b) 2 (eiz e iz ) cos z (c) sin z for ll pure imginry z iy 8 25 TAYLOR SERIES Find the Tylor series with center z nd its rdius of convergence. 8. >z, z i 9. >( z), z i 2. cos 2 z, z p>2 2. sin z, z p>2 22. cosh (z pi), z pi 23. >(z i) 2, z i 24. e z(z2), z 25. sinh (2z i), z i>2 4 AUGUSTIN FRESNEL ( ), French physicist nd engineer, known for his work in optics.

28 698 CHAP. 5 Power Series, Tylor Series 5.5 Uniform Convergence. Optionl We know tht power series re bsolutely convergent (Sec. 5.2, Theorem ) nd, s nother bsic property, we now show tht they re uniformly convergent. Since uniform convergence is of generl importnce, for instnce, in connection with termwise integrtion of series, we shll discuss it quite thoroughly. To define uniform convergence, we consider series whose terms re ny complex functions f (z), f (z), Á () m f m (z) f (z) f (z) f 2 (z) Á. (This includes power series s specil cse in which f m (z) m (z z ) m.) We ssume tht the series () converges for ll z in some region G. We cll its sum s (z) nd its nth prtil sum s n (z); thus s n (z) f (z) f (z) Á f n (z). Convergence in G mens the following. If we pick z z in G, then, by the definition of convergence t z, for given P we cn find n N (P) such tht ƒ s (z ) s n (z ) ƒ P for ll n N (P). If we pick in G, keeping P s before, we cn find n N 2 (P) such tht z 2 ƒ s (z 2 ) s n (z 2 ) ƒ P for ll n N 2 (P), nd so on. Hence, given n P, to ech z in G there corresponds number N z (P). This number tells us how mny terms we need (wht s n we need) t z to mke ƒ s (z) s n (z) ƒ smller thn P. Thus this number N z (P) mesures the speed of convergence. Smll N z (P) mens rpid convergence, lrge N z (P) mens slow convergence t the point z considered. Now, if we cn find n N (P) lrger thn ll these N z (P) for ll z in G, we sy tht the convergence of the series () in G is uniform. Hence this bsic concept is defined s follows. D E F I N I T I O N Uniform Convergence A series () with sum s (z) is clled uniformly convergent in region G if for every P we cn find n N N (P), not depending on z, such tht ƒ s (z) s n (z) ƒ P for ll n N (P) nd ll z in G. Uniformity of convergence is thus property tht lwys refers to n infinite set in the z-plne, tht is, set consisting of infinitely mny points. EXAMPLE Geometric Series Show tht the geometric series z z 2 Á is () uniformly convergent in ny closed disk ƒ z ƒ r, (b) not uniformly convergent in its whole disk of convergence ƒ z ƒ.

29 ` SEC. 5.5 Uniform Convergence. Optionl 699 Solution. () For z in tht closed disk we hve ƒ z ƒ r (sketch it). This implies tht > ƒ z ƒ >( r). Hence (remember (8) in Sec. 5.4 with q z) ƒ s(z) s n (z) ƒ 2 mn Since r, we cn mke the right side s smll s we wnt by choosing n lrge enough, nd since the right side does not depend on z (in the closed disk considered), this mens tht the convergence is uniform. (b) For given rel K (no mtter how lrge) nd n we cn lwys find z in the disk ƒ z ƒ such tht z n z m 2 2 z 2 r n r. z n n z ` ƒ z ƒ ƒ z ƒ K, simply by tking z close enough to. Hence no single N (P) will suffice to mke ƒ s (z) s n (z) ƒ smller thn given P throughout the whole disk. By definition, this shows tht the convergence of the geometric series in ƒ z ƒ is not uniform. This exmple suggests tht for power series, the uniformity of convergence my t most be disturbed ner the circle of convergence. This is true: THEOREM Uniform Convergence of Power Series A power series (2) m m (z z ) m with nonzero rdius of convergence R is uniformly convergent in every circulr disk ƒ z z ƒ r of rdius r R. PROOF For ƒ z z ƒ r nd ny positive integers n nd p we hve (3) ƒ n (z z ) n Á np (z z ) np ƒ ƒ n ƒ r n Á ƒ np ƒ r np. Now (2) converges bsolutely if ƒ z z ƒ r R (by Theorem in Sec. 5.2). Hence it follows from the Cuchy convergence principle (Sec. 5.) tht, n P being given, we cn find n N (P) such tht ƒ n ƒ r n Á ƒ np ƒ r np P for n N (P) nd p, 2, Á. From this nd (3) we obtin ƒ n (z z ) n Á np (z z ) np ƒ P for ll z in the disk ƒ z z every nd every p, 2, Á ƒ r, n N (P),. Since N (P) is independent of z, this shows uniform convergence, nd the theorem is proved. Thus we hve estblished uniform convergence of power series, the bsic concern of this section. We now shift from power series to rbitry series of vrible terms nd exmine uniform convergence in this more generl setting. This will give deeper understnding of uniform convergence.