复旦版数学分析答案

nTaaan={12}2nkknC∑==+=nknnknC02)11((1)(2)ABAB∪1TTa∈1TTTa∈212aa≠Ta∈313aa≠23aa≠{},,,,21naaaS=TS⊂2{},,,,21naaaA={},,,,21nbbbB=AB∪AB∪{},,,,,,,2211nnbababa=(1){}0=∅(2)a⊂{,,}abc(3){,}ab∈{,,}abc(4){,,{,}}abab={,}ab1}0{02a{,,}abc∈a{,,}abc13{,{ba,},}abc{}⊂ba,{,,}abc4}},{,,{bababa,{,}ab⊃}},{,,{baba{,}ab≠}},{,,{baba{,}ab(1)xx−+≤320(2)(3)01(4)0cotsin=xx1{}32|≤−xx2{}00|),(yxyx3{}Qxxx∈10|4⎭⎬⎫⎩⎨⎧∈+=Zkkxx,2|ππ(1)ABDABAD∩∪∩∪∩()()(=)C(2)()ABABCC∪∩=1)(DBAx∪∩∈Ax∈Bx∈Dx∈BAx∩∈DAx∩∈)()(DABAx∩∪∩∈)()()(DABADBA∩∪∩∪∩⊂)()(DABAx∩∪∩∈BAx∩∈DAx∩∈Ax∈Bx∈Dx∈)(DBAx∪∩∈)()()(DABADBA∩∪∩∪∩⊃22CBAx)(∪∈BAx∪∈Ax∈Bx∈CCBAx∩∈CCCBABA∩∪⊂)(CCBAx∩∈Ax∈Bx∈BAx∪∈CBAx)(∪∈CCCBABA∩∪⊃)((1)ABAC∪∪=BC=(2)ABAC∩∩=BC=1{}cbaA,,={}dcbB,,={}dcC,=ABAC∪∪=CB≠2{}cbaA,,={}edcB,,={}dcC,=ABAC∩∩=CB≠?(1)BAx∩∈⇔Ax∈Bx∈(2)BAx∪∈⇔Ax∈Bx∈1BAx∩∈⇔Ax∈Bx∈2BAx∪∈⇔Ax∈Bx∈31.},,{γβα=S,T??abc={,,}fST→2733=6!3=⎪⎩⎪⎨⎧cbafγβα:⎪⎩⎪⎨⎧bcafγβα:⎪⎩⎪⎨⎧acbfγβα:⎪⎩⎪⎨⎧cabfγβα:⎪⎩⎪⎨⎧bacfγβα:⎪⎩⎪⎨⎧abcfγβα:2.(1)[,[,]ab]01(2)(,)01(,−∞)+∞1]1,0[],[:→bafabaxyx−−=2),()1,0(:+∞−∞→f)cot()21tan(xxxππ−=−3.fg(1)yfu==()logau,ugx==()x23−;(2)yfu==()arcsinu,ugx==()ex(3)yfu==()u21−,ugx==()secx(4)yfu==()u,ugx==()xx−+111)3(log2−=xya()()+∞−∞−,33,∪),(+∞−∞2xy3arcsin=(]0,∞−⎥⎦⎤⎜⎝⎛2,0π3xytan=⎟⎠⎞⎜⎝⎛+−∈2,2ππππkkZk∪[)+∞,04411+−=xxy()[)+∞−∞−,11,∪[)()+∞,11,0∪4.(1)yx=+arcsin112;(2)321log(1)3ayx=−1uyarcsin=vu1=12+=xv2331uy=vualog=12−=xv5.(1)xyasinlog=1a(2)yx=cos(3)yx=−−432x(4)yxx=+2411()ππ)12(,2+∈kkZk∪(]0,∞−2⎥⎦⎤⎢⎣⎡+−∈22,22ππππkkZk∪[]1,03[1,4−]⎥⎦⎤⎢⎣⎡25,04()()+∞∞−,00,∪⎟⎟⎠⎞⎢⎣⎡+∞,22336.fg(1)fx()=2log()axgx()=2logax(2)fx()=22sectanxx−,gx()=1(3)fx()=sincos22xx+,gx()=11fg52fg3fg7.(1),fxxxx()+=−+−3235321fx()(2)13131+−=⎟⎠⎞⎜⎝⎛−xxxxf,fx()1tx=+33−=tx1)3(5)3(3)3(2)(23−−+−−−=ttttf977721223−+−=ttt9777212)(23−+−=xxxxf2txx=−11−=ttx()113113+−−−=tttttf1412−+=tt1412)(−+=xxxf8.fx()=+11x,,fffffffff121)(++=xxxff322)(++=xxxfff5332)(++=xxxffff(,−∞+∞)2)()(xfxf−+2)()(xfxf−−2)()()(xfxfxf−+=2)()(xfxf−−+ABCDyfx=()A=(,)03B=−(,)11C=(,)32D=(,)406[](](]⎪⎪⎩⎪⎪⎨⎧∈+−∈−∈+−=4,3823,125231,034xxxxxxyy(,)11Ox2x1.2.81.2.91.2.8fx()yfx=(),x∈[,]02[](]2210,121211,22xxyxxx⎧∈⎪⎪=⎨⎪−+−∈⎪⎩13.610.8(1.2.9)542Px[](](]⎪⎩⎪⎨⎧∈−∈−∈=11,92.112118.13329,598985,04.78)(xxxxxxxP13.[,[,[,]01]01]017[,]01⎩⎨⎧−=xxxxxf1)(81.(1)6(2)3+2?16nm=6226nm=mkm2=2223kn=nnm23+23+232+nm=222623nm=++252622−=nm612.Axx=≥{|}0⎭⎬⎫⎩⎨⎧=320|sinπxxB⎭⎬⎫⎩⎨⎧∈=+mnnmmnCN,0min=AAx∈∀Ax∈+1xx+1Amax12sinmax==πBBx∈∀⎥⎦⎤⎜⎝⎛∈∃2,0πααsin=xB∈2sinαx2sinαBmin9CmaxCminCmn∈∀Cmn∈+1Cmn∈++11111+++mnmnmnCmaxCmin3.AB,(1)AB∪(2)SxyxAyB=+∈∈{|,}1Ax∈∀1Mx≤Bx∈∀2Mx≤BAx∪∈∀{}21,maxMMx≤2Ax∈∀1Mx≤Bx∈∀2Mx≤Sx∈∀21MMx+≤4.STxxS=−∈{|}supS=Tinf−SSsup∈xTxxS=−∈{|}Sxsup≤−Sxsup−≥0εSy∈ε−SysupTy∈−ε+−−SysupSsup−TSTsupinf−=5.SsupABBA02−=ABεBSSx∈ABx−εASBA=6.SsupS≥infSsupS=infSS?Sx∈SxSsupinf≤≤SSinfsup≥supS=infSS107.2.1.18.S}3|{2∈=xxxQ(1)S(2)SQ1Spq∈∀0pq32⎟⎟⎠⎞⎜⎜⎝⎛pq2pq0r2234⎟⎟⎠⎞⎜⎜⎝⎛−+pqrr++⎟⎟⎠⎞⎜⎜⎝⎛=⎟⎟⎠⎞⎜⎜⎝⎛+rpqrpqrpq22223422++⎟⎟⎠⎞⎜⎜⎝⎛rrpqSrpq∈+SS2SQmnS=sup+∈Nnm,nm,20mn3i32⎟⎠⎞⎜⎝⎛mn10r32⎟⎠⎞⎜⎝⎛+rmnSrmn∈+mnS=supii32⎟⎠⎞⎜⎝⎛mn0r3422−⎟⎠⎞⎜⎝⎛−mnrr+−⎟⎠⎞⎜⎝⎛=⎟⎠⎞⎜⎝⎛−2222rrmnmnrmn3422+−⎟⎠⎞⎜⎝⎛rrmnrmn−SmnS=supSS111.⎭⎬⎫⎩⎨⎧++112nn{()}(.)−1099nn⎭⎬⎫⎩⎨⎧+−nn51⎭⎬⎫⎩⎨⎧++++3321nn⎭⎬⎫⎩⎨⎧nn32⎭⎬⎫⎩⎨⎧!3nn⎭⎬⎫⎩⎨⎧nnn!⎭⎬⎫⎩⎨⎧−+−+++−nnnnn21)1(211111)20(∀εε⎥⎦⎤⎢⎣⎡=ε2NNnε++nnn211022)10(∀εεlglg0.99Nε⎡⎤=⎢⎥⎣⎦Nnlglg0.99(1)(0.99)(0.99)nnεε−=3)20(∀εε⎥⎦⎤⎢⎣⎡=ε21N1Nn21εn252logNε⎡⎤=⎢⎥⎣⎦2Nn5n−2ε{}21,maxNNNn=15nnε−+4)10(∀εε⎥⎦⎤⎢⎣⎡=ε1NNnε+=+++nnnnn12121023511n222n333(12)2nnnnnC=+nnnn1)2)(1(86−−=0∀ε⎭⎬⎫⎩⎨⎧⎥⎦⎤⎢⎣⎡=ε1,11maxNNnεnnn13021265n55521321!53!3−−⎟⎠⎞⎜⎝⎛⋅⎟⎠⎞⎜⎝⎛⋅≤nnnn)30(∀εεlg351lg2Nε⎡⎤⎢⎥=+⎢⎢⎥⎣⎦⎥Nnε⎟⎠⎞⎜⎝⎛⋅−5213!30nnn72nmmnnn⎟⎠⎞⎜⎝⎛21!)10(∀εεlg241lg2Nε⎡⎤⎢⎥=+⎢⎥⎢⎥⎣⎦Nnlg112lg2Nmε−ε⎟⎠⎞⎜⎝⎛mnnn21!08nnnnnn121)1(211110−+−+++−)10(∀εε⎥⎦⎤⎢⎣⎡=ε1NNnε−+−+++−nnnnnn121)1(2111102.limn→∞21322322nn−+=limn→∞nnn21+=limn→∞()nnn212+−=limn→∞32nn+=1limn→∞xn=1,⎪⎩⎪⎨⎧−+=−nnnnnxnn,101,10∀ε⎥⎦⎤⎢⎣⎡=ε1NNnε+=−+−22221)23(37322312nnnn20∀ε⎥⎦⎤⎢⎣⎡=ε21NNn13ε++=−+nnnnnnn21112230∀ε⎥⎦⎤⎢⎣⎡=ε81NNnε++=−−+nnnnnnnn81)(221)(2224nnan+=+1230na221)1(23nnnnaCan++=+3nnnnnan3)1()13(2−+0∀ε⎥⎦⎤⎢⎣⎡=29εNNnε=−+nannn31235)10(∀εε⎭⎬⎫⎩⎨⎧⎥⎦⎤⎢⎣⎡⎥⎦⎤⎢⎣⎡=εε1lg,1max2NNnnε=−nxn11nε=−nnx10113.(1)ε0NnNxnnnε(2)ε,0xx1nxn−={}nx2⎪⎩⎪⎨⎧=nnnnxn1{}nx4.klimn→∞xn=alimn→∞xnk+=alimn→∞xn=a0∀εN∃Nn∀ε−axnε−+axknlimn→∞xnk+=alimn→∞xnk+=a0∀ε'N∃'Nn∀ε−+axkn14kNN+='Nn∀ε−axnlimn→∞xn=a5.=limn→∞xn2limn→∞xn21+=alimn→∞xn=a=limn→∞xn2limn→∞xn21+=a0∀ε1N∃1Nn∀ε−axn22N∃2Nn∀ε−+axn12{}12,2max21+=NNNNn∀ε−axn6.,xn≥0limn→∞xn=a≥0limn→∞xn=aaxax−≤−limn→∞xna=0∀εN∃Nn∀2ε−axnε−≤−nnnnaxax7.{}{}{}xnnnnyxynMyn≤{}xn0∀εN∃Nn∀MxnεNn∀εnnyx{}xnyn8.(1)limn→∞nn1131211⎟⎠⎞⎜⎝⎛++++(2)limn→∞⎜⎝⎛+11n+12n+++⎟⎟⎠⎞+nn1(3)limn→∞∑+=22)1(1nnkk(4)limn→∞135212462⋅⋅⋅⋅−⋅⋅⋅⋅()()nn1nnnn⎟⎠⎞⎜⎝⎛++++113121111lim=∞→nnn15limn→∞11312111=⎟⎠⎞⎜⎝⎛++++nn2⎜⎝⎛++11nnnn+12n+++11+⎟⎟⎠⎞+nnnn1lim=+∞→nnnn11lim=+∞→nnnlimn→∞⎜⎝⎛+11n+12n+++11=⎟⎟⎠⎞+nn3nnknnnnk221122222)1(+++=∑+=222lim=+∞→nnnlimn→∞∑+=22)1(1nnkk2=4)12)(12(2+−kkk121)2(642)12(5310+⋅⋅⋅⋅−⋅⋅⋅⋅nnn0121lim=+∞→nnlimn→∞0)2(642)12(531=⋅⋅⋅⋅−⋅⋅⋅⋅nn9.limn→∞34122nnn+−+1;limn→∞nnnnn3232323+−+−+1;limn→∞331313nnnn++++();limn→∞()