复旦数学分析教案04数学分析中一个反例的教学

WeierstrassWeierstrass1WeierstrassWeierstrassWeierstrassWeierstrassWeierstrassKarlWeierstrass1815189719Weierstrassδε−2Weierstrass1872()∑∞==0cos)(nnnxbaxfba101abVanDerWaerden1930(x)xx=1.26(x)=0.26x=3.67ϕϕϕ(x)=0.33ϕ(x)12/1)(≤ϕxyx,]21,[+∈kk]1,21[++kk|||)()(|yxyx−=−ϕϕVanDerWaerden)(xf=∑∞=ϕ010)10(nnnx.nnx10)10(ϕ≤n1021⋅∑∞=⋅01021nnWeierstrass),(+∞−∞∈x)(xf),(+∞−∞3xx)(xf)(xf10≤xx=0.a1a2anx0hm=⎩⎨⎧=−=−−,9,4,10,8,7,6,5,3,2,1,0,10mmmmaax=0.309546h1=h110−2=h210−3=h310−−4=h410−5=h510−−6=610−0→mh(∞→m)mmmhxfhxf)()(lim−+∞→mmhxfhxf)()(−+=∑∞=ϕ−+ϕ010)10())(10(nmnnmnhxhx∑−=ϕ−+ϕ=1010)10())(10(mnmnnmnhxhx∑∞=ϕ−+ϕ+mnmnnmnhxhx10)10())(10((10mn≥ϕn(xhm))=ϕ(10nx)=mn−10ϕ(10nx)mmhxfhxf)()(−+∑−=ϕ−+ϕ=1010)10())(10(mnmnnmnhxhx.1,,2,1,0−=mnxn10manm−,.10121mnnnaaaaax+=,)1(.)(10121±=++mnnmnaaaaahx10mhn(xhm)xn10]21,[+kk]1,21[++kkϕ((x))n10mhϕ(x)=n10±mnh10mmhxfhxf)()(−+=∑−=±101mnm∞→mlimmmhxfhxf)()(−+x)(xf45Weierstrass1Weierstrassmh1,,2,1,0−=mn)(10mnhx+xn10]21,[+kk]1,21[++kkmmhxfhxf)()(−+∑−=ϕ−+ϕ=1010)10())(10(mnmnnmnhxhx=∑−=±101mn2WeierstrassWeierstrass