Chapter 4: Seriessha Uni. of Sci & Tech)FCV&ITAugust19,20193/50MinsLlChar
Chapter 4: Series Ming Li (Changsha Uni. of Sci & Tech) FCV & IT August 19, 2019 3 / 50
s4.1 Series of complex numbers and series of complexfunctionsaUni.ofSci&Tech)FCV&ITAugust19,20194/50
§4.1 Series of complex numbers and series of complex functions Ming Li (Changsha Uni. of Sci & Tech) FCV & IT August 19, 2019 4 / 50
SeoennumberSeqencesofcomplexnumbersAsequenceQnofcomplexnumbersQ1=a1+ibi,Q2=a2+ib2,.,Qn=an+ibn，isdenoted by [an]Definition (Convergence)It is said to converge to a complex number , if for all e> O, there is anintegerN>O,suchthatn≥N impliesthatan-a0,thereisanN,suchthatn,m>N impliesthatan-Qm/<.limn-→oQn=α=a+ib iffliman=aandlimbn=b.n-→0n→xFCV&ITAugust 19,20195/50tha Uni.of Sci&Tech)MineLilChan
Seqences of complex numbers Seqences of complex numbers A sequence αn of complex numbers: α1 = a1 + ib1, α2 = a2 + ib2, · · · , αn = an + ibn, · · · is denoted by {αn} Definition (Convergence) It is said to converge to a complex number α, if for all ε > 0, there is an integer N > 0, such that n ≥ N implies that |αn − α| 0, there is an N, such that n, m ≥ N implies that |αn − αm| < ε. limn→∞ αn = α = a + ib iff limn→∞ an = a and limn→∞ bn = b. Ming Li (Changsha Uni. of Sci & Tech) FCV & IT August 19, 2019 5 / 50
numbersSeoerSeriesof complexnumbersAn infinite seriesisa1+a2++an+.*Sn =h=1 Qk is called the partial sums, it is a sequence.Definition (Convergence)the series k-, Qk is said to converges to S and we write k=, Qk = S,ifthesequenceofpartialsumsSnconvergestoSTheorem (Carchycriterion)k=i ak converges iff for every e>O, there is an N such that n ≥Nan+pimpliesthat<eforallp=1.2,3....一k=n+10kTheoremk=1 ak converges iff both k=1 ak and k=, bk convergeFCV&ITAugust 19,2019sha Uni. of Sci&Tech)6/50MingLi(Chan)
Seqences of complex numbers Series of complex numbers An infinite series is α1 + α2 + · · · + αn + · · · Sn = Pn k=1 αk is called the partial sums, it is a sequence. Definition (Convergence) the series P∞ k=1 αk is said to converges to S and we write P∞ k=1 αk = S, if the sequence of partial sums Sn converges to S. Theorem (Carchy criterion) P∞ k=1 αk converges iff for every ε > 0, there is an N such that n ≥ N implies that Pn+p k=n+1 αk < ε for all p = 1, 2, 3, · · · Theorem P∞ k=1 αk converges iff both P∞ k=1 ak and P∞ k=1 bk converge. Ming Li (Changsha Uni. of Sci & Tech) FCV & IT August 19, 2019 6 / 50
numbersExample(4.1.1)Determine whether the series k=1 (+)convergesSolution.No,because=divergesTheorem (4.1.2)If =1 ak converges, then ak → 0.Proof.Z=1 k converges, iff for every e >0, there is an N, such that n ≥ Nmple that +, fo all =1,3..Asa particularcaseof Cauchycriterionwithp=1口FCV&ITha Uni.of Sci &Tech)August19,20197/50MineLilCh
Seqences of complex numbers Example (4.1.1) Determine whether the series P∞ k=1 1 k + i 2 k converges. Solution. No, because P∞ k=1 1 k diverges. Theorem (4.1.2) If P∞ k=1 αk converges, then αk → 0. Proof. P∞ k=1 αk converges, iff for every ε > 0, there is an N, such that n ≥ N implies that Pn+p k=n+1 < ε, for all p = 1, 2, 3, · · · . As a particular case of Cauchy criterion with p = 1. Ming Li (Changsha Uni. of Sci & Tech) FCV & IT August 19, 2019 7 / 50
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 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)| 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≤ ro,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 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