Third Week Lectures 7-9

Transcription

1 Third Week Lectures 7-9 Lecture 7 Series Give two umbers, we ca add them to get aother umber. Repeatedly carryig out this operatio allows us to talk about sums of ay fiitely may umbers. We would like to talk about sum of ifiitely may umbers as well. A atural way to do this is to label the give umbers, take sums of first of them ad look at the limit of the sequece of umbers so obtaied. Thus give a (coutable) collectio of umbers, the first step is to label them to get a sequece {s }. I the secod step, we form aother sequece: the sequece of partial sums t = k= s k. Observe that the first sequece {s } ca be recovered completely from the secod oe {t }. The third step is to assig a limit to the secod sequece provided the limit exists. This etire process is coied uder a sigle term series. However, below, we shall stick to the popular defiitio of a series. Defiitio 4 By a series of real or complex umbers we mea a formal ifiite sum: s := s 0 + z + + s + 20

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 t := z k. We say the series s is coverget to k= the sum s if the associated sequece {t } of partial sums is coverget to s. 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. It should be oted that that eve if s is fiite, it is ot obtaied via a arithmetic operatio of takig sums of members of {s } but by takig the limit of the associated sequece {t } of partial sums. Sice displayig all elemets of {t } allows us to recover the origial sequece {s } by the formula s = t + t results that we formulate for sequeces have their couterpart for series ad vice versa ad hece i priciple we eed to do this oly for oe of them. For example, we ca talk a series which is the sum of two series a, b viz. (a + b ) ad if both a, b are coverget to fiite sums the the sum series (a + b ) is coverget to the sum of the their sums. Nevertheless, it is good to go through these otios. For example the Cauchy s criterio for the covergece of the sequece {t } ca be coverted ito Theorem 4 A series s is coverget to a fiite sum iff for every ǫ > 0 there exists 0 such that m k= s < ǫ, for all m, 0. As a corollary we obtai Corollary If s is coverget to a fiite sum the s 0. Of course the coverse does ot hold as see by the harmoic series. 2

3 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 λz = λ z ; (z + w ) = z + w. (5) Theorem 5 A series of positive terms a is coverget iff the sequece of parial sums is bouded. Theorem 6 Compariso Test (a) If a c for all 0 for some 0, ad c c is cgt the a is coverget. (b) If a b 0 for all 0 for some 0 ad b diverges implies a diverges. The geometric series is the mother of all series: Theorem 7 Geometric Series If 0 x < the sum x = x. If x >, the the series diverges. Here the partial sum sequece is give by t = x+ x x. Theorem 8 The series is cgt ad its sum is deoted by e. We! have, 2 < e < 3. Proof: For 2, we have, 2 < t = ++ 2! + +! < < + /2 = 3. Theorem 9 lim ( + ) = e. Proof: Put t = k=0, r k! = ( + ). The r = + + 2! ( )!! < t. 22

4 Therefore lim sup r e. O the other had, for a fixed m if m, we have r + + 2! ( ) + + ( ) ( m ). m! Therefore lim if r t m. Sice this true for all m, we get lim if r 3. Remark 5 The rapidity with which this sequece coverges is estimated by cosiderig: e t = ( + )! + ( + 2)! + < ( + )! [+ + + ( + ) 2+ ] =!. Thus 0 < e t <!. Corollary 2 e is irratioal. Proof: Assume o the cotrary that e = p q. The q!e ad q!t q are both itegers. O the other had 0 < q!e q!t q < q which is absurd. Defiitio 5 A series a is said to be absolutely coverget if the series a is coverget to a fiite limit. Theorem 20 Suppose {a } is a decreasig sequece of positive terms, the =0 a is cgt iff k 2k a 2 k is cgt. Proof: Put t = k= a k, T = k= 2k a 2 k. Check that t t 2 T + a 0 ad 2a 0 + a + T 2t 2. Theorem 2 p < iff p >. Corollary 3 The harmoic series is diverget. 23

5 Theorem 22 The series =2 is coverget iff p >. ( l ) p Theorem 23 Ratio Test: If {a } is a sequece of positive terms such that lim sup a + a = r <, the a is coverget. If a + a for all 0 for some 0, the a is diverget. Proof: To see the first part, choose s so that r < s <. The there exists N such that a + a < s for all N. This implies a N+k < a N s k, k. Sice the geometric series k sk is coverget, the covergece of a follows. The secod part is obvious sice a caot coverge to 0. Theorem 24 Root Test For sequece {a } of positive terms, put l = lim sup a. The (a) l < = a <. (b) l > = a =. (c)l = the series a ca be fiite or ifiite. Proof: Choose l < r < ad the a iteger N such that a < r for all N. Therefore a < r ad we ca ow compare with the geometric series. The proof of (b) is also similar. (c) is demostrated by the series ad. 2 Remark 6 As compared to ratio test, root test is more powerful, i the sese, whereever ratio test is coclusive so is root test. Also there are cases whe ratio test fails but root test holds. However, ratio test is easier to apply. Example 3 Put a 2+ = 2 +, a 2 = 3. The lim if a + a = lim 2 3 = 0; lim if a = lim 2 24 = 3 3.

6 a lim sup + ( a = lim 3 ) = ; lim sup 2+ 2 a = lim = 2 2. The ratio test caot be applied. The root test gives the covergece. The followig theorem proves the claim that we have made i the above remark. Theorem 25 For ay sequece {a } of positive terms, lim if a + a lim if a lim sup a lim sup a + a. Lecture 8 Example 4. Let z = x + ıy,. Show that z z = x + ıy iff x x ad y y. 2. Telescopig: Give a sequece {x } defie the differece sequece a := x x +. The show that the series a is coverget iff the sequece {x } is coverget ad i that case, a = x 0 lim x. Defiitio 6 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 ( ). The otio of absolute covergece plays a very importat role throughout the study of covergece of series. Theorem 26 Let z be a absolutely coverget series. The every rearragemet z σ of the series is also absolutely coverget, ad hece coverget. Moreover, each such rearragemet coverges to the same sum. 25

7 Proof: Recall that a rearragemet z σ of z is obtaied by takig a bijectio σ : N N.) Let z = z. The oly thig that eeds a proof at this stage is that z σ() = z. Let us deote the partial sums s = k=0 a k t = k=0 a σ(k). Sice a is absolutely coverget give ǫ > 0 there is a N such that m k= a k < ǫ for all m N. Pick up N large eough so that {, 2,..., N} {σ(), σ(2),...σ(n )}. The for N, we have s t k=n+ a < ǫ. Therefore, lim s = lim t. Riema s rearragmet Theorem: Let a be a coverget series of real umbers which is ot absolutely coverget. Give α β, there exists a rearragemets a τ() of a with partial sums t such that lim if t = α; lim sup t = β. We are ot goig to prove this. See [R] for a proof. Example 5. Let {z } be a bouded sequece ad w is a absolutely coverget series. Show that z w is absolutely coverget. 2. Abel s Test: For ay sequece of complex umbers {a }, defie S 0 = 0 ad S = k= a k,. Let {b } be ay sequece of complex umbers. (i) Prove Abels Idetity: a k b k = k=m k=m S k (b k b k+ ) S m b m + S b, m. (LHS = (S k S k )b k = m S kb k m S kb k+ = RHS.) (ii) Show that a b is coverget if the series k S k(b k b k+ ) 26

8 is coverget ad lim S b exits.(put m =.) (iii) Abel s Test: Let a be a coverget series ad {b } be a bouded mootoic sequece of real umbers. The show that a b is coverget. (Solutio: (b b + ) is coverget by Telescopig ad absolutely, sice {b } is mootoic. The series a is coverget ad hece {S } is bouded. By the previous exercise, the product series is coverget. Sice both S ad b are coverget S b is coverget. Therefore, (ii) applies. 3. Dirichlet s Test: Let a be such that the partial sums are bouded ad let {b } be a mootoic sequece tedig to zero. The show that a b is coverget. (Arguemets are already there i above eaxmple) 4. Derive the followig Leibiz s test from Dirichlet s Test: If {c } is a mootoic sequece covergig to 0 the the alteratig series ( ) c is coverget. (Take a = ( ) ad b = c i Dirichlet s test.) 5. Geeralize the Leibiz s test as follows: If {c } is a mootoic sequece covergig to 0 ad ζ is complex umber such that ζ =, ζ, the ζ a is coverget. Exercise 8 Show that if a is coverget the the followig sequeces are all coverget. (a) a p,p > 0; (b) a log p ; (c) a ; (d) Show that for ay p > 0, ad for every real umber x, coverget. ( + ) a ; si x p is 27

9 Lecture 9 Defiitio 7 Give two series a, b, the Cauchy product of these two series is defied to be c, where c = k=0 a kb k. Theorem 27 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. (6) Proof: Cosider the remaider after terms of the correspodig absolute series: Clearly, 0 k R = k c k ( k a k ; R = k a k )( l b k. b l ) R 0 T 0. Thus the partial sums of the series c k k forms a mootoically icreasig sequece which is bouded above. Therefore the series c is absolutely coverget. Further, ( ( ) c k a k) b k R 0 T + + T 0 R +, k 2 k k sice the terms that remai o the LHS after cacellatio are of the form a k b l where either k + or l +. Upo takig the limit as, we obtai (6). Remark 7 This theorem is true eve if oe of the two series is absolutely coverget ad the other is coverget. For a proof of this, see [R]. 28

10 Defiitio 8 By a formal power series i oe variable t over K, we mea a sum of the form a t, a K. =0 Note that for this defiitio to make sese, the sequece {a } ca be iside ay set. However, we shall restrict this ad assume that the sequeces are take iside field K. Let K[[t]] deote the set of all formal power series a t i t with coefficiets 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, i.e., K[t] K[[t]]. Just like polyomials, we ca add two power series term-by-term ad we ca also multiply them by scalars, viz., a t + b t := (a + b )t ; α( a t ) := αa t. Verified that the above two operatios make K[[t]] ito a vector space over K. Further, we ca eve multiply two formal power series: ( )( ) a t b t := c t, where, c = k=0 a kb k. This product is called the Cauchy product. Oe ca directly check that K[[t]] is the a commutative rig with the multiplicative idetity beig the power series := a t where, a 0 = ad a = 0,. Together with the vector space structure, K[[t]] is actually a K-algebra.) Observe that the rig of polyomials i t forms a subrig of K[[t]]. What we are ow iterested i is to get ice fuctios out of power series. 29

11 Observe that, if p(t) is a polyomial over K the by the method of substitutio, it defies a fuctio a p(a), from K to K. It is customary to deote this map by p(t) itself. However, due to the ifiite ature of the sum ivolved, give a power series P ad a poit a K, the substitutio P(a) may ot make sese i geeral. This is the reaso why we have to treat power series with a little more care, via the otio of covergece. Defiitio 9 A formal power series P(t) = a z is said to be coverget at z 0 C if the sequece {s }, where, s = a k z0 k is coverget. I that case we write P(z 0 ) = lim s for this limit. Puttig t = a z 0, this just meas that the series of complex umbers t is coverget. Remark 8 Observe that every power series is coverget at 0. Defiitio 20 A power series is said to be a coverget power series, if it is coverget at some poit z 0 0. The followig few theorems, which are attributed to Cauchy-Hadamard ad Abel 2, are most fudametal i the theory of coverget power series. Theorem 28 Cauchy-Hadamard Formula: Let P = 0 a t be a power series over C. Put L = lim sup a ad R = L k=0 with the Jacques Hadamard( ) was a Frech Mathematicia who was the most ifluetial mathematicia of his days, worked i several areas of mathematics such as complex aalysis, aalytic umber theory, partial differetial equatios, hydrodyamics ad logic. 2 Niels Herik Abel ( ) was a Norwegia, who died youg uder deprivatio. At the age of 2, 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. 30

12 covetio = ; = 0. The 0 (a) for all 0 < r < R, the series P(t) is absolutely ad uiformly coverget i z r ad (b) for all z > R the series is diverget. Proof: (a) Let 0 < r < R. Choose r < s < R. The /s > /R = L ad hece by property (Limsup-I), we must have 0 such that for all 0, a < /s. Therefore, for all z r, a z < (r/s), 0. Sice r/s <, by Weierstrass majorat criterio, (Theorem??), it follows that P(z) is absolutely ad uiformly coverget. (b) Suppose z > R. We fix s such that z > s > R. The /s < /R = L, ad hece by property (Limsup-II), there exist ifiitely may j, for which j aj > /s. This meas that a j z j > ( z /s) j >. It follows that the th term of the series a z does ot coverge to 0 ad hece the series is diverget. Defiitio 2 Give a power series a t, R = sup { z : a z < } is called the radius of covergece of the series. The above theorem gives you the formula for R. Remark 9 Observe that if P(t) is coverget for some z, the the radius of covergece of P is at least z. The secod part of the theorem gives you the formula for it. This is called the Cauchy-Hadamard formula. It is implicit i this theorem that the the 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. This disc is called the disc of covergece of the power series. Observe that the theorem does ot say aythig about the covergece of the series at poits o the boudary z = R. The examples below will tell you that ay thig ca happe. 3

13 Example 6 The series t, t, all have radius of covergece. The first oe is ot coverget at ay poit of the 2 boudary of the disc of covergece z =. The secod is coverget at all the poits of the boudary except at z = (Dirichlet s test) ad the last oe is coverget at all the poits of the boudary (compare with ). These examples clearly illustrate that the boudary behavior 2 of a power series eeds to be studied more carefully. Assigemet 3 Solutios to be submitted o st Sept. Wedesday morig.. Let z be a coverget series of complex umbers such that the real part R(z ) 0 for all. If z2 is also coverget, show that z 2 is coverget. 2. For 0 θ < 2π ad for ay α R, defie the closed sector S(α, θ) with spa θ by S(α, θ) = {re(β) : r 0 & α β α + θ}. t Let z be a coverget series. If z S(α, θ),, where θ < π, the show that z is coverget. 3. Give a direct proof of the fact that 0 z C. Use this to prove that z! is coverget for all lim sup! =. 4. If P(t), Q(t) are two coverget power series with radius of covergece r ad s respectively, show that the radius of covergece of (P(t) + Q(t) is at least mi{r, s}. 32

14 5. Let p(t) = a 0 +a t+ be a formal power series with coefficiets i C. Show that there exists aother power series q(t) = b 0 +b t+ such that p(t)q() = iff a 0 0. I this case, show that q is uique. (q is called the multiplicative iverse of p.) Write dow formula for b i terms of a 0, a, 6. Hemachadra Numbers For ay positive iteger, let H deote the umber of patters you may be able to produce o a drum i a fixed duratio of beats. For istace, Dha dhi dhi the first Dha takes two syllables whereas the followig two Dhi s take oe syllable each. Clearly H = ad H 2 = 2. Hemachadra 3 oted that sice the last syllable is either of oe beat or two beats it follows that H = H +H 2 for all 3. These umbers were kow to Idia poets, musicias ad percussioists as Hemachadra umbers. Defie F 0 = 0, F = ad F = F + F 2, 2. Note that F = H, 2. These F are called Fiboacci umbers. 4 (Thus the first few Fiboacci umbers are 0,,, 2, 3, 5, 8, 3, 2, 34,....) Form the formal power series F(z) = F z (7) =0 (a) Show that ( t t 2 )F(t) = t. (b) Put S w (t) := + wt + w 2 t 2 +. Fid α, β R such that F(t) = S α (t)s β (t)t. 3 Hemachadra Suri (089-75) was bor i Dhadhuka, Gujarat. He was a Jai mok ad was a adviser to kig Kumarapala. His work i early cetury is already based o eve earlier works of Gopala. 4 Leoardo Pisao (Fiboacci) was bor i Pisa, Italy (75-250) whose book Liber abbaci itroduced the Hidu-Arabic decimal system to the wester world. He discovered these umbers at least 50 years later tha Hemachadra s record. 33

15 (c) Show that F + = j=0 α j β j = α+ β + α β = 5 (α + β + ) 7. Summability Let F = {P α (t) = a α, t, : α Λ} C[[t]] be a family of formal power series i oe variable with complex coefficiets. We say F is summable if for every 0 the set Λ = {α Lambda : A α, 0} is fiite. I this case, we defie the sum of this family to be the elemet p(t) = ( a α, )t. α Λ Let ow A(t) = a t, B(t) = b t be ay two power series. Prove or disprove the followig statemets. (a) The Cauchy product AB is the sum of the family {a b m t m+ : m, 0}. (b) If {A j (t)} is a summable family the for ay B the family {A j B} is summable. (c) If b 0 0 the the family {a B } is summable. 34