复旦数学分析教案03函数项级数的一致收敛性

1212331∑DS∞=1)(nnxun(x)Sn(x)=xE∑=nkkxu1)({Sn(x)}DS(x)∑∞=1)(nnxuS(x)=S∞→nlimn(x)xD{Sn(x)}u1(x)=S1(x)un+1(x)=Sn+1(x)Sn(x)(n=12...)∑∞=1)(nnxu{Sn(x)}{Sn(x)}∑{S∞=1)(nnxun(x)}2un(x)(Sn(x))(Riemann)()??(a)un(x)(Sn(x))D=S(x)(S∑∞=1)(nnxu∞→nlimn(x)=S(x))()S(x)Dx0DS(x)=S(x0limxx→0)=0limxx→∑∞=1)(nnxu∑∞=→1)(lim0nnxxxu0limxx→∞→nlimSn(x)=S∞→nlim0limxx→n(x).Sn(x)=xn{Sn(x)}(11]S(x)=S∞→nlimn(x)=.⎩⎨⎧=−.1,1,11,0xxnSn(x)(11]()S(x)x=1(x=1)bun(x)(Sn(x)D=S(x)(S∑∞=1)(nnxu∞→nlimn(x)=S(x))()S(x)DxddS(x)un(x)(Sn(x))()xdd∑∞=1)(nnxu=∑∞=1)(ddnnxuxxdd∞→nlimSn(x)=∞→nlimxddSn(x).1()()Sn(x)=nnxsin{Sn(x)}()S(x)=0S(x)S'(x)=0{S'(x)}S'(x)=ncosnxS'(x)(x=0S'n(0)=n0)cun(x)(Sn(x))[ab]⊂DRiemann∑=S(x)(S∞=1)(nnxu∞→nlimn(x)=S(x))()S(x)[ab]Riemannu∫baxxSd)(n(x)(Sn(x))()∫=∑∞=bannxxud)(1∑∫∞=1d)(nbanxxu∫∞→banlimSn(x)dx=dx∞→nlim∫banxS)(()RiemannRiemannSn(x)=⎩⎨⎧⋅.,0,!,1xnxxnSn(x)=0S(x)=S∞→nlimn(x)=0xpqpNqZnpSn(x)=1S(x)=S∞→nlimn(x)=1{Sn(x)}S(x)DirichletSn(x)RiemannS(x)RiemannSn(x)=nx(1x2)n{Sn(x)}[01]S(x)=0nSn(x)S(x)[01]Riemann∫10d)(xxSn==∫−102d)1(xxnxn2n∫−−1022)1d()1(xxn=)1(2+nn/(n)∫10d)(xxS3{Sn(x)}D()S(x)x0D{Sn(x)}S(x0)N0N=N(x0,)nNSn(x0)S(x0)N(x0,)x0DN(x0,)x0Sn(x)DS(x)x0DN=N()nNSn(x)S(x)xD?N(){Sn(x)}DS(x)D{Sn(x)}xD0DN()nN()Sn(x)S(x)xD{Sn(x)}DS(x)Sn(x)S(x)D⇒Sn(x)S(x)⇔∀0∃N∀nN∀xDD⇒Sn(x)S(x).xD{S∑∞=1)(nnxun(x)}Sn(x)=DS(x)∑DS(x)∑=nkkxu1)(∞=1)(nnxu∑DS(x)⇔∀0∃N∀nN∀xD∞=1)(nnxu∑S(x)=S=nkkxu1)(n(x)S(x)0NN()nN()y=Sn(x)xD{(xy)xDS(x)yS(x)+Sn=221xnx+{Sn(x)}()S(x)=0Sn(x)S(x)=221||xnx+n210N=ε21nNSn(x)S(x)n21x(){Sn(x)}()S(x)=00Nε21nNy=Sn(x)x(){(xy)y}Sn(x)=xn(10.1.2){Sn(x)}01)01Sn(x)S(x)=xnnxlnlnεN=N(x,)xlnlnεx1xlnlnεx[01]N=N(){Sn(x)}01ny=xn()01){(xy)x[01]0y01)001Sn(x)S(x)=xnnN=N()=ρεlnlnnNSn(x)S(x)nx[0]{Sn(x)}0(1ny=xn0x{(xy)0x0y}5{Sn(x)}DS(x)d(SnS)=SD∈xsupn(x)S(x){Sn(x)}DS(x)∞→nlimd(SnS)=0{Sn(x)}DS(x)0NN()nNSn(x)S(x)2εxDnNd(SnS)2ε∞→nlimd(SnS)=0d(S∞→nlimnS)=00NN()nNd(SnS)Sn(x)S(x)xD{Sn(x)}DS(x)5Sn(x)=221xnx+x()Sn(x)S(x)=221||xnx+n21x=n1d(SnS)=n210(n)6Sn(x)=xnx[01)d(SnS)=x10sup≤≤xn=1/0(n){Sn(x)}01Sn(x)=221xnnx+{Sn(x)}(0)S(x)=0Sn(x)S(x)=221xnnx+21x=n1d(SnS)=21/0(n){Sn(x)}(0)(4)ny=221xnnx+x=n1{(xy)0xy21}{Sn(x)}x=0x=0{Sn(x)}A(0A)Sn(x)S(x)=221xnnx+(221xnnx+)'=22222)1()1(xnxnn+−nρ1d(SnS)=221ρ+ρnn0(n){Sn(x)}AS(x)=0{Sn(x)}(0){Sn(x)}(0)6Sn(x)=xn{Sn(x)}01)0⊂01{Sn(x)}(Sn(x)=xn)01)DDSn(x)=(1x)xn{Sn(x)}01S(x)=0Sn(x)S(x)=(1x)xn[(1x)xn]'=[n(n1)x1−nxSn(x)S(x)x=1+nnd(SnS)=(11+nn)(1+nn)n=(1+nn)(1n1)n0(n){Sn(x)}01S(x)=0.{Sn(x)}DS(x){Sn(x)}DS(x){xn}xnD(S∞→nlimn(xn)S(xn))=0.{Sn(x)}DS(x)d(SnS)=SD∈xsupn(x)S(x)0(n).{xn}xnDSn(xn)S(xn)d(SnS)0(n).{Sn(x)}DS(x){xn}xnDSn(xn)S(xn)/0(n){xn}DS(x)∀0∃N∀nN∀xDSn(x)S(x).{Sn(x)}DS(x)∃0∀N0∃nN∃xDS0εn(x)S(x)0εN11∃n11∃D()S()1nx1nS1nx1nx0εN2n1∃n2n1∃D()S()2nx2nS2nx2nx0ε……NkNk1∃NkNk1∃D()S()knxknSknxknx0ε……mN+{n1n2...nk...}xmD{xn}xnDknx()S()knSknxknx0ε∞→nlim(Sn(xn)S(xn))=026Sn(x)=xnx[01)xn=1n1[01)Sn(x)S(x)=(1n1)ne1(n){Sn(x)}01)S(x)=010.1.8S(x)=221xnnx+x(0)xn=n1Sn(xn)S(xn)=21{Sn(x)}(0)S(x)=0.Sn(x)=nx(1x2)nx[01](4){Sn(x)}01S(x)=0xn=n1Sn(xn)S(xn)=(121n)n1(n){Sn(x)}01S(x)=0412324