A complex series k=1 Qk is said to converge absolutely if k=1 laklconverges.Theorem (4.1.3)If-,Qkconvergesabsolutely,then it converes.(The converse isgenerally nottrue)Proof.By Cauchy criterion,given >O, there is an N>0, such that n≥Nimplies that Zn+p+ axl <e, = ,....By the triangle inequality, wehaven+pn+pZ [axl <e,p=1,2, ..7Qk=n+1n+口thus by Cauchy criterion kiQkconverges.FCV&ITAugust19,20198/50MineLiIGUni.ofSci&Tech)
Seqences of complex numbers A complex series P∞ k=1 αk is said to converge absolutely if P∞ k=1 |αk| converges. Theorem (4.1.3) If P∞ k=1 αk converges absolutely, then it converes. (The converse is generally not true) Proof. By Cauchy criterion, given ε > 0, there is an N > 0, such that n ≥ N implies that Pn+p k=n+1 |αk| < ε, p = 1, 2, · · · . By the triangle inequality, we have nX +p k=n+1 αk ≤ nX +p k=n+1 |αk| < ε, p = 1, 2, · · · thus by Cauchy criterion P∞ k=1 αk converges. Ming Li (Changsha Uni. of Sci & Tech) FCV & IT August 19, 2019 8 / 50
xnumbersSeoencSequenceandseriesof complexfunctionsDefinition(convergepointwise)Suppose that fn :A→C is a sequence of functions all defined on the setA. The sequence is said to converge pointwise, iff for each z E A thesequence(fn(z))converges.The limitdefinesanewfuntionf(z)on ADefinition(uniformlyconverge)A sequence fn:A→C of functions defined on a set A is said to convergeuniformlyto a function f,if foranye>O,thereisanN,suchthatn≥N implies that |fn(z)-f(z)/< e, for all zE A. This is writtenfn→funiformlyonA.=1 9k(2) converge pointwise= Sn(z) =h=1 9k(z) converge pointwisek=1 gk(z) converge uniformly=Sn(z) = h=1 9k(z) converge uniformly.Remark:uniformly converge=→ converge pointwise, but convergepointwiseuniformlyconvergeFCV&ITAugust19,20199/50aUni:of Sci&Tech)MineLilChs
Seqences of complex numbers Sequence and series of complex functions Definition (converge pointwise) Suppose that fn : A → C is a sequence of functions all defined on the set A. The sequence is said to converge pointwise, iff for each z ∈ A the sequence {fn(z)} converges. The limit defines a new funtion f(z) on A. Definition (uniformly converge) A sequence fn : A → C of functions defined on a set A is said to converge uniformly to a function f, if for any ε > 0, there is an N, such that n ≥ N implies that |fn(z) − f(z)| < ε, for all z ∈ A. This is written fn → f uniformly on A. P∞ k=1 gk(z) converge pointwise= Sn(z) = Pn k=1 P gk(z) converge pointwise. ∞ k=1 gk(z) converge uniformly=Sn(z) = Pn k=1 gk(z) converge uniformly. Remark: uniformly converge ⇒ converge pointwise, but converge pointwise 6⇒ uniformly converge. Ming Li (Changsha Uni. of Sci & Tech) FCV & IT August 19, 2019 9 / 50
TheCauchycriterion ispresentedasfollowingTheorem (4.1.4)()Asequencefn(z)convergesuniformlyonA,iffforanye>O,thereis an N, such that n ≥ N implies that Ifn(z) - fn+p(z)l < e for allzEAandallp=1.2.3,..(i) A series k=1 converges uniformly on A, iff for all e > 0, there is anN,such thatn≥N implies thatn+2g(a=n+1forall zEAand all p=1.2,3.*.FCV&IT10/50Uni.ofSci&Tech)August19,2019SLIC
Seqences of complex numbers The Cauchy criterion is presented as following Theorem (4.1.4) (i) A sequence fn(z) converges uniformly on A, iff for any ε > 0, there is an N, such that n ≥ N implies that |fn(z) − fn+p(z)| < ε for all z ∈ A and all p = 1, 2, 3, · · · . (ii) A series P∞ k=1 converges uniformly on A, iff for all ε > 0, there is an N, such that n ≥ N implies that nX +p k=n+1 gk(z) < ε for all z ∈ A and all p = 1, 2, 3, · · · . Ming Li (Changsha Uni. of Sci & Tech) FCV & IT August 19, 2019 10 / 50
xnumbersTheorem(WeierstrassMtest)Let fn be a sequence of functions on a set A C C.Suppose that there is asequence of real constantsMn≥O,suchthat(0)Ifn(z)I≤MnforallzEA,(i)En=, Mn converges.Then n=1 fn converges absolutely and uniformly on A.Proof.Sinceo-, Mn converges for any e>O, there is an N, such that n ≥N2pimplies that h=n+1 Mk< e for all p= 1,2,3, .... Thus n ≥ N implesthatn+pn+pn+p (2)≤ k(2)≤fs(z)Mk<e.k=n+1+k=n+1口ByTheorem4.1.4wehavethedesiredresult.FCV&ITAugust 19,201911/50MineLIoUni.ofSci&Tech)
Seqences of complex numbers Theorem (Weierstrass M test) Let fn be a sequence of functions on a set A ⊂ C. Suppose that there is a sequence of real constants Mn ≥ 0, such that (i) |fn(z)| ≤ Mn for all z ∈ A, (ii) P∞ n=1 Mn converges. Then P∞ n=1 fn converges absolutely and uniformly on A. Proof. Since P∞ n=1 Mn converges for any ε > 0, there is an N, such that n ≥ N implies that Pn+p k=n+1 Mk < ε for all p = 1, 2, 3, · · · . Thus n ≥ N imples that nX +p k=n+1 fk(z) ≤ nX +p k=n+1 |fk(z)| ≤ nX +p k=n+1 fk(z)Mk < ε. By Theorem 4.1.4 we have the desired result. Ming Li (Changsha Uni. of Sci & Tech) FCV & IT August 19, 2019 11 / 50
niimharExample(4.1.2)Consider the series f(z) = -, . It will be shown that this seriesconverges uniformly on the setsA,=[z:z≤r] for each 0≤ r<1.Solution.Here fn(2)=and |fn(2)/≤ysincez≤r.Therefore,letMn=.By≤r" and Em=1 rn converges for 0 ≤r < 1.AThus n=, Mn converges, so by theWeierstrassMtest,thegiven series convergesruniformlyonArIt converges pointwise on A = [z : |z|< 1],sinceeachzEAliesinsomeAr,forrcloseenoughto1.Thisseriesdoesnotconveresuniformly on A.Indeed if itdid,then"wouldconvergeuniformlyon[O,1).Supposethis were true,thenV>o,N,such that n≥N,would imply2n+1an+panfor all E[0,1),p=1,2,..n+1nn+pFCV&IT12/50Uni.ofSci&Tech)August 19.2019Ming Li(Char
Seqences of complex numbers Example (4.1.2) Consider the series f(z) = P∞ n=1 z n n . It will be shown that this series converges uniformly on the sets Ar = {z : |z| ≤ r} for each 0 ≤ r < 1. Ar A 1 r O x y Solution.Here fn(z) = z n n and |fn(z)| ≤ r n n since |z| ≤ r. Therefore, let Mn = r n n . By r n n ≤ r n and P∞ n=1 r n converges for 0 ≤ r < 1. Thus P∞ n=1 Mn converges, so by the Weierstrass M test, the given series converges uniformly on Ar. It converges pointwise on A = {z : |z| < 1}, since each z ∈ A lies in some Ar, for r close enough to 1.This series does not converes uniformly on A. Indeed if it did, then P x n n would converge uniformly on [0, 1). Suppose this were true, then ∀ ε > 0, ∃ N, such that n ≥ N, would imply x n n + x n+1 n + 1 + · · · + x n+p n + p < ε for all x ∈ [0, 1), p = 1, 2, · · · Ming Li (Changsha Uni. of Sci & Tech) FCV & IT August 19, 2019 12 / 50