数学分析 第三版 下册 (欧阳光中 朱学研 著) 高等教育出版社 课后答案 144页

'课后答案网www.khdaw.com1n?êØ1Ü©ê?êÚ2ÂÈ©1ÊÙê?ê§1.ý£µþ4Úe41.y²µ(1)lim(xn+yn)6limxn+limynn→∞n→∞n→∞(2)lim(xn+yn)>limxn+limynn→∞n→∞n→∞y²µ(1)limxn,limynþkê§K{xn},{yn}þk..n→∞n→∞-αk=sup{xn},βk=sup{yn}"u´§n>k§kxn+yn6αk+βkn>kn>klsup{xn+yn}6αk+βkn>klim(xn+yn)=limsup{xn+yn}6limαk+limβk=limxn+limynn→∞k→∞n>kk→∞k→∞n→∞n→∞5µelimxn,limyn+∞.~Xµlimxn=+∞§¦"Xªm>{$k¿Â§KlimynØn→∞n→∞n→∞n→∞−∞.ù{xn}Ãþ.§{xn+yn}½Ãþ.§þã"Xªw,¤á¶elimxn=−∞§¦"Xªn→∞m>{$k¿Â§KlimynØ+∞.u´{yn}þk.§llim(xn+yn)=−∞§þã"Xªn→∞n→∞w,¤á.(2)Ïxn>inf{xn},yn>inf{yn}§xn+yn>inf{xn}+inf{yn}âe(.e.¥§Kinf{xn+yn}>inf{xn}+inf{yn}§linf{xn+yn}>inf{xn}+inf{yn}n>kn>kn>kKliminf{xn+yn}>liminf{xn}+inf{yn}=liminf{xn}+liminf{yn}k→∞n>kk→∞n>kn>kk→∞n>kk→∞n>k=lim(xn+yn)>limxn+limyn.n→∞n→∞n→∞2.xn>0,yn>0§y²µ(1)limxnyn6limxn·limynn→∞n→∞n→∞(2)limxnyn>limxn·limynn→∞n→∞n→∞y²µ(1)Ï06xn6sup{xn},06yn6sup{yn}§K06xnyn6sup{xn}·sup{yn}âþ(.´þ.¥§Kk06sup{xn·yn}6sup{xn}·sup{yn}l06sup{xn·yn}6sup{xn}·sup{yn}n>kn>kn>kKlimsup{xn·yn}6limsup{xn}·sup{yn}=limsup{xn}·limsup{yn}k→∞n>kk→∞n>kn>kk→∞n>kk→∞n>k=limxnyn6limxn·limyn.n→∞n→∞n→∞(2)Ïxn>inf{xn}>0,yn>inf{yn}>0§Kxnyn>inf{xn}·inf{yn}>0âe(.´e.¥§Kkinf{xn·yn}>inf{xn}·inf{yn}>0linf{xn·yn}>inf{xn}·inf{yn}>0n>kn>kn>kKliminf{xn·yn}>liminf{xn}·inf{yn}=liminf{xn}·liminf{yn}k→∞n>kk→∞n>kn>kk→∞n>kk→∞n>k=limxnyn>limxn·limynn→∞n→∞n→∞3.elimxn3§Ké?Ûê{yn}¤áµn→∞(1)lim(xn+yn)=limxn+limynn→∞n→∞n→∞ 课后答案网www.khdaw.com143(2)lim(xn·yn)=limxn·limyn§elimxn>0n→∞n→∞n→∞n→∞y²µlimxn=αn→∞elimyn=+∞(½−∞)§K(1)w,¤á.Ïα>0§K(2)w,¤á.n→∞Ølimyn=βkên→∞Ïlimyn=β§3{yn}f{yn}§¦limyn=ββ¤kÂñf4¥ö.kkn→∞k→∞qlimxn=α§Klimxn=α§lim(xn+yn)=α+β,lim(xn·yn)=αβkkkkkn→∞k→∞k→∞k→∞eyα+β{xn+yn}Âñf4¥ö(^y{)
b{xn+yn}Âñf{xn+yn}§¦limxn+yn=γ>α+βk0k0k0→∞k0k0
Klimyn=limxn+yn−limxn=γ−α>βk0→∞k0k0→∞k0k0k0→∞k0ùβ{yn}¤kÂñf4¥gñ.u´α+βÒ´{xn+yn}¤kÂñf4.Óny§α>0§α+β{xn+yn}Âñf4¥.llim(xn+yn)=α+β=limxn+limynn→∞n→∞n→∞lim(xn·yn)=αβ=limxn·limyn§elimxn>0n→∞n→∞n→∞n→∞4.¦eêþ4e4µ1(1)an=(n=1,2,···)2−n+(−1)n!1n(2)an=(−1)1+(n=1,2,···)nn(−1)(3)an=(n=1,2···)nnπ(4)an=sin(n=1,2,···)5)µ(1)§küä4fêµa2k→1,a2k+1→−1(k→∞)(k=1,2,3···)u´liman=1,liman=−1.n→∞n→∞(2)§küä4fêµa2k→1,a2k+1→−1(k→∞)(k=1,2,3···)u´liman=1,liman=−1.n→∞n→∞(3)Ïliman=0§liman=0,liman=0.n→∞n→∞n→∞2nπ2(4)−sinπ6sin6sinπ5552n=10k+2(k=1,2,···)§a10k+2→sinπ(k→∞)52n=10k−2(k=1,2,···)§a10k−2→−sinπ(k→∞)522u´liman=sinπ,liman=−sinπ.n→∞5n→∞5pp5.elimn|ann|=α§Klim|ak|=α0+nn→∞n→∞d?k0´?¿½ê.y²k0111n(1)Ï|ak|n=|ak|k0+n|ak|k0+n0+n0+n0+npk0k011nk0+n|ak0+nk0+nqlimk|=α§α>0§limln|ak|=0§lim|ak|=10+n0+n0+nn→∞pn→∞nn→∞d12K(1)§limn|ak|6α0+nn→∞p1nn(2)Ïlim|an|=α§3f{an}§¦lim|an|k=α§kkn→∞k→∞k0111nk−k0α>0§klim|an|nk−k0=lim|an|nk·lim|an|nk=αkkkk→∞k→∞k→∞ 课后答案网www.khdaw.compllimn|ak|>α0+nn→∞nÜ(1)(2)§α>0§(ؤá.ppk0+n1n(3)eα=0§Kw,klimn|ank0+nn|=0§llim|ak|=lim|ak|=00+n0+nn→∞n→∞n→∞u´d(Ø(.6.eliman=aN§kanN§kan1(n=1,2,···)n→∞ 课后答案网www.khdaw.com145§2.?êÂñ59ÙÄ51.?Øe?êñÑ5µ111(1)++···++···1·66·11(5n−4)(5n+1)23n(2)1+++···++···352n−1!!!111111(3)++++···+++···2322322n3n111(4)++···++···1·44·7(3n−2)(3n+1)πππ(5)cos+cos+cos+···345)µ"#!11111111111(1)ÏSn=++···+=1−+−+···+−=1−1·66·11(5n−4)(5n+1)566115n−45n+155n+1!111KlimSn=lim1−=n→∞n→∞55n+15111u´â½Â§?ê++···++···Âñ.1·66·11(5n−4)(5n+1)n1(2)Ïlim=6=0§?êuÑ.n→∞2n−12X∞1X∞1(3)duþÂñAÛ?ê§2n3nn=1n=1!X∞11X∞1X∞113dê?ê52§+=+=1+=2n3n2n2n22n=1n=1n=1"#!11111111111(4)ÏSn=++···+=1−+−+···+−=1−1·44·7(3n−2)(3n+1)34473n−23n+133n+1!111KlimSn=lim1−=n→∞n→∞33n+13111u´â½Â§?ê++···++···Âñ.1·44·7(3n−2)(3n+1)π(5)Ïlimcos=16=0§?êuÑ.n→∞n+22.|^ÜÂñnOe?ê´Âñ´uÑ.2n(1)a0+a1q+a2q+···+anq+···,|q|<1,|an|6A,(n=0,1,2,···)11111(2)1+−++−+···23456y²µ(1)Ïé?Ûg,êp§|Sn+p−Sn|=an+an+1qn+1+···+an+p−1qn+p−16a|qn|+aqn+1+···+aqn+p−16nqnn+1n+p−1p1−|q|nA|q|1−|q|n|q|pq|q|<1§K0<1−|q|<1§u´|Sn+p−Sn|0§N=ln/ln|q|§n>N§é?Ûp=1,2,3,···§Ao¤á|Sn+p−Sn|<ε2nUÂñn§?êa0+a1q+a2q+···+anq+···Âñ. 课后答案网www.khdaw.com!X∞111(2)d?ê+−3n+13n+23n+3n=010<ε0<§ØØnõ§e-p=n§Kk611111111|Sn+p−Sn|=|S2n−Sn|=+−+···++−>+−3n+13n+23n+36n−26n−16n3n+33n+3!1111111111111+···++−=++···+>++···+=>ε03n+36n6n6n3n+1n+22n32n2n2n6|{z}!nX∞111Ïd?ê+−uÑ.3n+13n+23n+3n=0X∞X∞3.k?êan(=zan>0)§Áy²eéÙ)Ò¤|¤?êÂñ§Kan½Âñ.n=1n=1X∞X∞y²µanÜ©Úê{Sn}§)Ò¤|¤?êAnn=1n=1X∞Ù¥An=ai+ai+···+ai§w,AnE?ê.n−1+1n−1+2nn=100ÙÜ©Úê{Sn}§Ù¥Sn=(a1+a2+···+ai)+(ai+···+ai)+···+(ai+···+ai)11+12n−1+1n0w,Sn>SnX∞000qAnÂñ§dĽn§{Sn}kþ.§=3M>0§¦Sn6M§lSn6Sn6M§`n=1²{Sn}kþ.X∞KdĽn§anÂñ.n=14.(½¦e?êÂñx.X∞1(1)(1+x)nn=0X∞n(2)(lnx)n=1)µ11(1)d?ê´ú""?ê§<1?êÂñ1+x1+xlÂñx<−2½x>0.(2)d?ê´ú"lnx"?ê§|lnx|<1?êÂñ1lÂñ1)1+ann=1X∞1(6)√n·nnn=1!nX∞1(7)2n+1n=1X∞1(8)[ln(n+1)]nn=1X∞2+(−1)n(9)2nn=1X∞πn(10)2sin3nn=1X∞nn(11)n!n=1X∞xn(12),(x>0)(1+x)(1+x2)···(1+xn)n=1!nX∞b(13)§Ù¥an→a,an,b,aê§a6=0ann=1)µ1√X∞n2+nn1(1)Ïlim=lim√=1§?ê´uÑn→∞1n→∞n2+nnn=1nX∞1Kd"O{§?ê√½uÑ.n2+nn=11un+1(2n+1)22n+12n−11(2)Ïlim=lim=lim=<1n→∞unn→∞1n→∞4(2n+1)4(2n−1)22n−1X∞1KdKO{§?êÂñ.(2n−1)·22n−1n=1√∞√n−n1Xn−n(3)Ïlim=90§?êuÑ.n→∞2n−122n−1n=1ππX∞πX∞π(4)Ïsin6§Âñ§?êsinÂñ.2n2n2n2nn=1n=1 课后答案网www.khdaw.com!n!n11X∞1X∞1(5)Ï6§Âñ§?êÂñ.1+anaa1+ann=1n=1!1nxlimxlnx11(6)Ïlimxx=limelnx=limexlnx=ex→+0=1§lim√=lim=1x→+0x→+0x→+0n→∞nnn→∞n1√∞n·nn1X1qlim=lim√=1§?ê´uÑn→∞1n→∞nnnn=1nX∞1Kd"O{§?ê√uÑ.n·nnn=1vu!n!nu11X∞1(7)Ïlimtn=lim=0<1§?êÂñ.n→∞2n+1n→∞2n+12n+1n=1s11X∞1n(8)Ïlim=lim=0<1§?êÂñ.n→∞[ln(n+1)]nn→∞ln(n+1)[ln(n+1)]nn=12+(−1)n3X∞3(9)Ï6?êÂñ2n2n2nn=1X∞2+(−1)nKâ"O{§?êÂñ.2nn=1!n!nπ2X∞2n(10)Ï0<2sin6π?êÂñ3n33n=1X∞πnKâ"O{§?ê2sinÂñ.3nn=1n+1(n+1)!nX∞nun+1(n+1)!1n(11)Ïlim=lim=lim1+=e>1§?êuÑ.nn→∞unn→∞nn→∞nn!n=1n!n+12nn+10<1,x>1½x=0un+1x/[(1+x)(1+x)···(1+x)(1+x)]x1(12)Ïlim=lim=lim=<1,x=1n→∞unn→∞xn/[(1+x)(1+x2)···(1+xn)]n→∞1+xn+12x<1,01=b>a§?êuѶaann=1!n!nbX∞1X∞1X∞1=1=b=a§I?Úä"~Xµ?ê√=uѶ?ê√=annnnn2n=1n=1n=1X∞1Âñ.n2n=1X∞X∞22.e?êunÂñ§y²unÂñ§Ù_XÛºn=1n=1X∞y²µÏunÂñ§Klimun=0n→∞n=12ε0=1§K3êN§n>N§k|un|<ε0=1=06un<1§u´06unN)§ 课后答案网www.khdaw.com149X∞2ld"O{§unÂñ.n=1X∞1X∞1X∞1X∞1Ù_Øý.~µÂñ§uѶÂñ§Âñ.n2nn4n2n=1n=1n=1n=1X∞X∞unX∞X∞X∞X∞3.unÚvnü?ê§lim=0§y²µvnÂñ§unÂñ.qevnuѧunXn→∞vnn=1n=1n=1n=1n=1n=1unX∞X∞Ûºelim=∞§@ounÚvnñÑ5mko"Xºn→∞vnn=1n=1y²µunX∞X∞(1)Ïlim=0§unÚvnü?ên→∞vnn=1n=1ununε0=1§K3êN§n>N§k<ε0=1=06<1§u´unN)vnvnX∞X∞qvnÂñ§Kd"O{§unÂñn=1n=1X∞X∞evnuѧKunUÂñ§UuÑn=1n=11X∞1X∞1n2~µuѧlim=0§Âñ¶nn→∞1n2n=1n=1n1X∞1X∞1√uѧlimn=0§uÑ.nn→∞1nn=1√n=1nunX∞X∞(2)Ïlim=∞§unÚvnü?ên→∞vnn=1n=1unG0=1§K3êN§n>N§k>G0=1§u´un>vn(n>N)vnX∞X∞X∞X∞X∞X∞eunÂñ§Kd"O{§vnÂñ¶evnuѧKunuѶeunuѧKvnñn=1n=1n=1n=1n=1n=1Ñ5ؽ.X∞X∞X∞X∞4.eü?êunÚvnuѧmax(un,vn)§min(un,vn)ü?êXÛºn=1n=1n=1n=1X∞X∞)µÏü?êunÚvnuѧun6max(un,vn)n=1n=1X∞Kd"O{§max(un,vn)uÑ.n=1X∞éumin(un,vn)ñÑ5ؽ.n=1!X∞1X∞1X∞11X∞1~µ§Ñuѧmin,=uѶn2nn2n2nn=1n=1n=1n=1!X∞1+(−1)nX∞1−(−1)nX∞1+(−1)n1−(−1)n§Ñuѧmin,=0+0+···+0+···%Âñ.2222n=1n=1n=15.|^?êÂñ7^y²µnn(1)lim=0n→∞(n!)2(2n)!(2)lim=0(a>1)n→∞an!y²µ 课后答案网www.khdaw.comX∞X∞nn(1)un=(n!)2n=1n=1n+1(n+1)!nun+1[(n+1)!]211Ïlim=lim=lim1+=0<1nn→∞unn→∞nn→∞n+1n(n!)2X∞nnnnKâKO{4/ª§Âñ§ld?êÂñ7^§lim=0(n!)2n→∞(n!)2n=1X∞X∞(2n)!(2)un=an!n=1n=1[2(n+1)]!2un+1a(n+1)!(2n+2)(2n+1)4(n+1)Ï0<==<(a>1)un2(n−1)!n+1n(2n)!aaan!k2n4(n+1)un+1lim=0(a>1,k∈N)§u´lim=0§llim=0n→∞ann→∞an+1n→∞unX∞(2n)!(2n)!KâKO{4/ª§Âñ§ld?êÂñ7^§lim=0an!n→∞an!n=16.?Øe?êÂñ5µX∞1(1)n·(lnn)pn=1X∞1(2)n·lnn·lnlnnn=1X∞1(3)n·(lnn)1+σlnlnnn=1X∞1(4)n·(lnn)p(lnlnn)qn=1)µ1(1)duØØpÛê§x¿©§¼êf(x)=Ñ´K4~§x(lnx)pZndx11−p(ln2),p>1limp=p−1n→∞2x(lnx)∞,p61p>1§?êÂñ¶p61§?êuÑ.1(2)f(x)=§f(x)x>3´4~¼ê.ZxlnxlnlnxndxX∞1lim=lim(lnlnlnn−lnlnln2)=∞§K?êuÑ.n→∞3xlnxlnlnxn→∞n·lnn·lnlnnn=1Z!ndx1111(3)Ïlim=lim−=(σ>0)n→∞2x(lnx)1+σn→∞σ(ln2)σ(lnn)σσ(ln2)σX∞1?êÂñ.n(lnn)1+σn=211X∞1q6§Kd"O{§?êÂñ.n·(lnn)1+σlnlnnn(lnn)1+σn·(lnn)1+σlnlnnn=11(4)-f(x)=§n63´4~¼ê.x(lnx)p(lnlnx)qZZ+∞dx+∞dtqÏ=3x(lnx)p(lnlnx)qlnln3e(p−1)ttqé?Ûq§p−1>0§È©Âñ§p−1<0§È©uѶp=1§eq>1§È©Âñ§eq61§È©uÑ. 课后答案网www.khdaw.com151dÜÈ©O{§?êñÑ5È©ñÑ5^K?êp>1Âñ¶p<1uѶp=1§q>1?êÂñ¶q61?êuÑ.X∞7.eun´Âñ?꧿ê{un}üNeü§y²limnun=0.n→∞n=1X∞X∞Xny²µÏunÂñ§S=un§Sn=ukn=1n=1k=1KlimSn=S=limS2n§u´lim(S2n−Sn)=0n→∞n→∞n→∞q{un}üNeü§KS2n−Sn=un+1+un+2+···+u2n>u2n+u2n+···+u2n=nu2nqun>0§K06nu2n6S2n−Sn§u´limnu2n=0§llim(2n)u2n=0n→∞n→∞2n+1qÏu2n+16u2n,un>0§K06(2n+1)u2n+16(2n+1)u2n=(2nu2n)→0(n→∞)2nu´lim(2n+1)u2n+1=0§llimnun=0n→∞n→∞8.y²KO{9Ù4/ª.y²µ(1)KO{µun+1uN+2uN+3uN+k+1Ïn>N§k6q<1§K6q,uN+26quN+1;6q,uN+36quN+2;···;6unuN+1uN+2uN+kKq,uN+k+16quN+k6···6quN+1X∞X∞kkÏq<1§KqÂñ§u´dÂñ?ê51§quN+1Âñ§ld"O{§k=1k=1X∞uN+kÂñk=2X∞2dÂñ?ê55§Vku1,u2,···,uN+1#?êunÂñ.n=1un+1uN+1en>N§>1§K>1,uN+1>uN§ù`²{uN}´üNOunuNX∞qun>0§Kun90(n→∞)§u´unuÑ.n=1(2)KO{4/ªµun(i)elim=¯r<1n→∞un−1d¢êÈ573ε0>0(§¦r<)r¯+ε0<1un+1dþ4½n1y²¥§kkur¯+ε0§u´½3êN(unun+1k¥eIN=)§¦n>N§k1n→∞un−1d¢êÈ573ε0>0(§¦r>r¯)−ε0>1un+1dþ4½n2y²¥§kkur+ε0§u´½3êN(unun+1k¥eIN=)§¦n>N§k>r+ε0>1§dKO{un?êuÑ.X∞X∞1un+1un+1X∞X∞1(iii)Þ~`²µun=§lim=lim=1§un=uѶnn→∞unn→∞unnn=1n=1n=1n=1X∞X∞1un+1un+1X∞X∞1un=§lim=lim=1§un=Âñ.n2n→∞unn→∞unn2n=1n=1n=1n=1 课后答案网www.khdaw.com§4.?¿?ê1.?Øe?êÂñ5£)^Âñ½ýéÂñ¤µ131313(1)−+−+−+···21022103231051111(2)1−+−+−···23!45!X∞lnnn−1(3)(−1)nn=2X∞n3n−1(4)(−1)2nn=1X∞nn+1(5)(−1)(n+1)2n=1X∞xn(6)(−1)sin(x6=0)nn=1111111(7)√−√+√−√+···+√−√+···2−12+13−13+1n−1n+1)µX∞1X∞1X∞3(1)ÏÂñ§Âñ§KÂñ2n102n−1102n−1n=1n=1n=1X∞1X∞1131313u´+Âñ§=++++++···Âñ§l?êýéÂñ.2n102n−12102210323105n=1n=1!X∞1X∞1(2)ÏuѧK−uÑn2nn=1n=1X∞1qé?ê(2n−1)!n=11X∞(2n+1)!11Ïlim=lim=0<1§KdKO{4/ª§?êÂñ1n→∞n→∞2n(2n+1)(2n1)!(2n−1)!n=1u´?êuÑ.X∞lnnX∞lnnn−1(3)Ï(−1)=nnn=2n=2lnnX∞1X∞lnnnqlim=limlnn=+∞uѧKd"O{§uÑ1n→∞n→∞nnnn=1!n=20lnxlnx1−lnxlnx0qf(x)=(x>3)§Kf(x)==<0(x>3)§u´f(x)=üNeü§lxxx2x()lnn3n>3üNeünlnx1lnnqlim=lim=0§Klim=0x→+∞xx→+∞xn→∞nX∞lnnn−1u´â4ÙZ[½n§(−1)^Âñ.nn=2X∞n3X∞n3n−1(4)(−1)=2n2nn=1n=13!3(n+1)1n+11X∞n32n+1Ïlim=lim=<1§KâKO{§Âñn→∞n3n→∞2n22n2nn=1X∞n3n−1l(−1)ýéÂñ.2nn=1 课后答案网www.khdaw.com153X∞nX∞nn+1(5)(−1)=(n+1)2(n+1)2n=1n=1n2X∞X∞(n+1)2n1nÏlim=lim=1uѧKuÑn→∞1n→∞(n+1)2n(n+1)2nn=1n=1x1−x0f(x)=(x>2)§Kf(x)=<0(x>2)§u´x>2§f(x)üNeü§l((x)+1)2(x+1)3nn>2üNeü(n+1)2nX∞nn+1qlim=0§Kâ4ÙZ[½n§(−1)Âñn→∞(n+1)2(n+1)2n=1X∞nn+1l(−1)^Âñ.(n+1)2n=1X∞xX∞xn(6)(−1)sin=sinnnn=1n=1sinxX∞1X∞xnÏ→|x|6=0(n→∞)uѧKsinuÑ1nnnn=1n=1xxπ+qé∀x∈R,x6=0§Ï→0(n→∞)§K3N∈Z§n>N§k0<<§u´n>nn2xxX∞xnN§sinxkÓÎÒsinnO~0§Kd4ÙZ[O{§(−1)sinÂnnnn=1ñX∞xnl(−1)sin^Âñ.nn=1!nX+111nX+12Xn1(7)Ü©Úê{Sn}§KS2n=√−√==2k−1k+1k−1kk=2k=2k=1u´limS2n=+∞§Kd?ê)Òuѧl?êuÑ.n→∞2.y²µe?ê)Ò¤¤?êÂñ§¿3Ó)ÒSÎÒÓ§@"K)Ò§d?√X∞(−1)[n]ê½Âñ¶¿d?êÂñ5.nn=1y²:X∞0(1)®#?êun=(u1+···+un1)+(un1+1+···+un2)+···+(unk−1+1+···+unk)+···Âñn=13Ó)ÒSÎÒÓXnXn00000uk=Sn,uk=Sn§KS1=Sn1,S2=Sn2,···,Sk=Snk,···k=1k=1X∞?êeInlnk−1nk§unÜ©ÚüNCz§=n=100un,···,unþ§kSk−1=SnSn>Sn=Skk−1+1kk−1kX∞X∞0Âñ§=limS0000®unk=limSk−1=S§KlimSn=S§u´unÂñ.k→∞k→∞n→∞n=1n=1√X∞(−1)[n](2)Änn=1111X∞222kn=k,k+1,···,k+2k(k=1,2,···)§ÃanÓÒ.PAk=++···+§(−1)Ak´k2k2+1k2+2kk=1?êZk2+2kZ(k+1)222dx111dxk+2k(k+1)Ï6++···+6=ln6Ak6ln§lk2−1xk2k2+1k2+2kk2xk2−1k2k→∞§Ak→0 课后答案网www.khdaw.comk2+2k(k+2)2k2+kX∞kqAk−Ak+1>ln−ln=ln>0§Kd4ÙZ[O{(−1)AkÂk2−1(k+1)2k2+k−2k=1ñ§l?êÂñ.3.?Øe?ê´ÄýéÂñ½^ÂñµX∞(−1)n(1)n+xn=1X∞sin(2nx)(2)n!n=1X∞sinnx(3),(00§üN~§lim=0§Kd4ÙZ[½n§Âñn+xn→∞n+xn+xn=1+x<0ØKê§Ïx½ê§Kn¿©§=3N∈Z§n>N§kn+x>X∞(−1)n11X∞(−1)n0§u´´?ê§düN~9lim=0§KÂñ§ln+xn+xn→∞n+xn+xn=N+1n=N+1X∞(−1)nÂñn+xn=1KxØKê§d?ê^Âñ.sin(2nx)111X∞1(n+1)!(2)Ï6§lim=lim=0<1§KdKO{§Âñ1n!n!n→∞n→∞n+1n!n!n=1X∞sin(2nx)X∞sin(2nx)2â"O{§Âñ§lýéÂñ.n!n!n=1n=1()Xncosx−cos2n+1x11(3)Ïsinkx=226êüNªu02sinxsinxnk=122X∞sinnxKd)á4O{§Âñ.nn=1()sinnxsin2nx1cos2nxXnsinx−sin(2n+1)x11q>=−cos2kx=69êünn2n2n2sinx|sinx|2nk=1Nªu0X∞cos2nxKd)á4O{§Âñ.nn=1!X∞1X∞1X∞1cos2nxX∞sinnxquѧKuѧu´−uѧluÑn2n2n2nnn=1n=1n=1n=1X∞sinnxK?ê(01§Ï6p>1Âñ§K?ê(0=+dfây²Âñ.npnp2np2np(2n)pn=1X∞cos2nxX∞cos2nxp−1K·2Âñ§=Âñ(2n)p2npn=1n=1!X∞1X∞1X∞1cos2nxq0x0Âñ.nx0nxn=1n=11!x−x0!x−x0(n+1)x−x0n111y²µÏx>x0§K==1−<1§K<1n+1n+1(n+1)x−x0nx−x0nx−x0161nx−x0()11u´êüNk.§61nx−x0nx−x0X∞anX∞anq?êÂñ§KdCO{§Âñ.nx0nxn=1n=1X∞X∞6.{nan}Âñ§n(an−an−1)Âñ§KanÂñ.n=1n=1y²µÏ{nan}Âñ§Ù4a()X∞Xnqn(an−an−1)Âñ§KÙÜ©Úêk(ak−ak−1)k4§Ù4Sn=1k=1XnnX−1qk(ak−ak−1)=(a1−a0)+2(a2−a1)+···+n(an−an−1)=nan−akk=1k=0nX−1XnnX−1Xn=ak=nan−k(ak−ak−1)§Klimak=limnan−limk(ak−ak−1)=a−Sn→∞n→∞n→∞k=0k=1k=0k=1X∞X∞X∞u´anÂñ§lan=an−a0Âñ.n=0n=1n=0 课后答案网www.khdaw.comX∞X∞X∞7.e(av−av−1)ýéÂñ§bvÂñ§@"avbvÂñ.v=1v=1v=1nX+mn+my²µ-Bn=bvv=n+1nX+pXp−1n+pn+idAbelC§avbv=an+pBn+Bn(an+i−an+i+1)v=n+1i=1nX+pXp−1avbv6|an+p|Bn+p+Bn+i|annn+i−an+i+1|v=n+1i=1"#nX+pXp−1-Hp=maxBn+1,Bn+2,···,Bn+p§Kkavbv6Hp|an+p|+|an+i−an+i+1|nnnnnv=n+1i=1X∞X∞X∞Ï|av−av−1|Âñ§(av−av−1)Âñ(av−av−1)=−a0+an§liman3n→∞v=1v=1v=1Ï3M>0§¦én§kXp−1|an+i−an+i+1|+|an+p|0,∃N∈Z§n>N§ép∈Z§kv=1εpHn<(5)MnX+pX∞d(??),(??)§n>N§kavbv<ε§ùL²?êavbvÂñv=n+1v=18.|^ÜÂñny²?ê4ÙZ[½n.y²µé?Ûg,êp§k|Sn+p−Sn|=(−1)n+2un+1+(−1)n+3un+2+···+(−1)n+p+1un+p=(−1)n+2(un+1−un+2+···+(−1)p−1un+p)=up−1un+1−un+2+···+(−1)n+ppóê§(un+1−un+2)+···+(un+p−1−un+p)>0pÛê§(un+1−un+2)+···+(un+p−2−un+p−1)+un+p>0oup−1un+p=up−1un+pn+1−un+2+···+(−1)n+1−un+2+···+(−1)qpóê§un+1−(un+2−un+3)−···−(un+p−2−un+p−1)−un+p6un+1pÛê§un+1−(un+2−un+3)−···−(un+p−1−un+p)6un+1p−1oun+1−un+2+···+(−1)un+p6un+1é?¿ε>0§Ïlimun=0§Klimun+1=0n→∞n→∞+u´73N∈Z§n>N§k|un+1−0|<ε§Kun+1<εpddn>N§é?Ûg,êpÑk|Sn+p−Sn|=un+1−un+2+···+(−1)un+p6un+1<εX∞n+1ldÜÂñn§(−1)unÂñ.n=1 课后答案网www.khdaw.com157§5.ýéÂñ?êÚ^Âñ?ê5X∞1v−1v−2v−11.|x|<1,|y|<1§y²(x+xy+···+y)=(1−x)(1−y)v=1y²µÏ|x|<1,|y|<1§KX∞1v−12vx=1+x+x+···+x+···=ýéÂñ(6)1−xv=1X∞1v−12vy=1+y+y+···+y+···=ýéÂñ(7)1−yv=1X∞X∞1v−1v−1(??)·(??)§xy=(1−x)(1−y)v=1v=1X∞X∞X∞v−1v−12v2vv−1v−2v−1qxy=(1+x+x+···+x+···)(1+y+y+···+y+···)=(x+xy+···+y)§v=1v=1v=1X∞1v−1v−2v−1K(x+xy+···+y)=.(1−x)(1−y)v=1X∞xnX∞ynX∞(x+y)n2.y²µ=n!n!n!n=0n=0n=0|x|n+1|x|X∞|x|n(n+1)!y²µÏlimn=lim=0<1§KâKO{4/ª§?êÂñn→∞|x|n→∞n+1n!n!n=0X∞|x|nu´?êýéÂñn!n=0X∞|y|nÓn§?êýéÂñn!n=0X∞xnX∞ynX∞¤=Cnn!n!n=0n=0n=0Xnxiyn−iynxyn−1xn1(x+y)n0n1n−1nnÙ¥Cn=·=+·+···+=(Cny+Cnxy+···+Cnx)=i!(n−i)!n!1!(n−1)!n!n!n!i=0X∞xnX∞ynX∞(x+y)nK=n!n!n!n=0n=0n=03.y²µ±Ñ^Âñ?êS?ꧦÙuÑ+∞.X∞y²µun^Âñn=1X∞X∞X∞X∞d½n1§vnÚwnÑuѧvnuÑ+∞§(−wn)uÑ−∞n=1n=1n=1n=1ÀuÑ+∞ê{βn}§=limβn=+∞n→∞X∞rvnU^Så5n=1m1§¦v1+v2+···+vm>β1+w11,m2§¦v1+v2+···+vm+vm+···+vm>β2+w1+w211+12/§¿©mk>mk−1§¦v1+v2+···+vm+···+vm+···+vm>βk+w1+w2+···+wk(k=12k3,4,···)ù/|ÚKµ(v1+···+vm−w1)+(vm+···+vm−w2)+···+(vm+···+vm−wk)+···(∗)11+12k−1+1kd?êw,?êS?êX∞Ï(∗))Ò?ê(vm+···+vm−wk)kgÜ©Úk−1+1kk=1(v1+···+vm−w1)+(vm+···+vm−w2)+···+(vm+···+vm−wk)>βk11+12k−1+1k 课后答案网www.khdaw.comlimβk=+∞k→∞X∞K(vm+···+vm−wk)uÑ+∞k−1+1kk=1duÑ?ê?¿)Ò§K±Ñ^Âñ?êS?ꧦÙuÑ+∞. 课后答案网www.khdaw.com159§6.á¦È1.?Øá¦ÈÂñ5µY∞n2−4(1)n2−1n=3Y∞(−1)n(2)an(a>0)n=1sY∞n+1(3)n+2n=0)µ2n−433(1)Ï=1−,n>3§−<0n2−1n2−1n2−1!3X∞1X∞3X∞3n2−1qlim=3Âñ§KÂñ§u´−Âñn→∞1n2n2−1n2−1n2n=3n=3n=3Y∞n2−4lâ½n2§Âñ.n2−1n=3X∞(−1)nX∞(−1)nX∞(−1)n(2)lnan=lna=lnannn=1n=1n=1X∞(−1)nX∞(−1)nY∞(−1)nÏ?ê4ÙZ[.?ê§KÙÂñ§u´?êlnanÂñ§lá¦ÈanÂnn=1n=1n=1ñ.ss12nn+11(3)duÜ©¦ÈPn=·····=→0(n→∞)23n+1n+2n+2sY∞n+1á¦ÈuÑu0.n+2n=0X∞Y∞22.y²µexnÂñ§KcosxnÂñ.n=1n=1!!2Y∞Y∞xnxnsinxnx222ny²µÏcosxn=1−2sin062sin62·=2222n=1n=1X∞X∞xn22qxnÂñ§K2sinÂñ2n=1n=1Y∞u´â½n2§cosxnÂñ.n=1!!X∞Y∞ππ3.y²µeαnýéÂñ§Ktan+αnÂñÙ¥|αn|<.44n=1!n=1!Y∞πY∞1+tanαnY∞2tanαny²µtan+αn==1+41−tanαn1−tanαnn=1n=1n=1∞2tanαnX1−tanαn2tanαnÏαnýéÂñ§Klimαn=0§u´lim=lim=2n→∞n→∞|αn|n→∞1−tanαnαnn=1X∞X∞X∞2tanαn2tanαndαnýéÂñ§Âñ§u´ýéÂñ1−tanαn1−tanαnn=1!n=1n=1Y∞πltan+αnýéÂñ.4n=1 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课后答案网www.khdaw.com179X∞X∞X∞5.eun(x)|un(x)|6cn(x)§¿cn(x)3XþÂñ§Kun(x)3Xþ½Âñýén=1n=1n=1Âñ.X∞y²µÏcn(x)3XþÂñn=1+KdÂñÜ¿^§é∀ε>0,∃N=N(ε)∈Z§¦n>N§éx∈XÚ?¿êp§k|cn+1(x)+cn+2(x)+···+cn+p(x)|<εX∞qun(x)|un(x)|6cn(x)n=1Kéþãε>0§êN=N(ε)§¦n>N§éx∈XÚþãêp§k|un+1(x)+un+2(x)+···+un+p(x)|6|un+1(x)|+|un+2(x)|+···+|un+p(x)|6|cn+1(x)+cn+2(x)+···+cn+p(x)|<εX∞dÂñÜ¿^§un(x)3XþÂñýéÂñ.n=1Zx6.f0(x)3[0,a]þëY§qfn(x)=fn−1(t)dt§y²{fn(x)}3[0,a]þÂñu".0y²µÏf0Z(x)3[0,a]þëY§KÙk.§=3M>0§k|f0(x)|6Mxqfn(x)=fn−1(t)dt§KZ0ZZxxx|f1(x)=f0(t)dt6|f0(t)|dt6Mdt=Mx6MaZ0Z0Z0xxxMx2Ma2|f2(x)=f1(t)dt6|f1(t)|dt6Mtdt=6.......................................................00022ZxZxZxtn−1xnan|fn(x)=fn−1(t)dt6|fn−1(t)|dt6Mdt=M6M000(n−1)!n!n!ananan+Ïlim=0=é∀ε>0,∃N∈Z§n>N§k−0=<εn→∞n!n!n!u´|fn(x)−0|0§N=§Kn>N§éx∈(−∞,+∞)§Ñk−0=6<εεn+x2n+x2n()1X∞1n−1K"ux∈(−∞,+∞)Âñu0§u´d)á4O{§(−1)3(−∞,+∞)Sn+x2n+x2n=1Âñ.X∞1X∞1n−1(−1)=n+x2n+x2n=1n=11X∞1X∞1n+x2Ïlim=1uѧKd"O{4/ª§uѧu´é?Ûx?êýén→∞1nn+x2nn=1n=1Âñ.X∞x2X∞x2=(1+x2)n(1+x2)nn=1n=1s21xn,x6=0é½x∈(−∞,+∞)§Ïlim2n=1+x2n→∞(1+x)0,x=0X∞x2dÜO{§3(−∞,+∞)Âñ§u´ýéÂñ.(1+x2)nn=1 课后答案网www.khdaw.comXnx21x6=0§Sn(x)==1−§S(x)=limSn(x)=1(1+x2)k(1+x2)nn→∞k=11,x6=0x=0§Sn(0)=0,S(0)=0§KS(x)=0,x=0X∞x2ÏSn(x)3(−∞,+∞)þëY§S(x)3(−∞,+∞)þØëY§K3(−∞,+∞)SØÂñ.(1+x2)nn=18.y²µXX∞(1)XJ|fn(x)|3[a,b]þÂñ§@"fn(x)3[a,b]þÂñ¶1XXX∞nnn+1(2)XJfn(x)3[a,b]þÂñ§|fn(x)|7Âñ§±(−1)(x−x),06x61~15`².y²µ(1)dÜOK9K§+=é∀ε>0,∃N=N(ε)∈Z§¦n>N§éx∈[a,b]Ú?¿p∈Z§k|fn+1(x)|+|fn+2(x)|+···+|fn+p(x)|<εl|fn+1(x)+fn+2(x)+···+fn+p(x)|6|fn+1(x)|+|fn+2(x)|+···+|fn+p(x)|<εX∞KâÂñÜOK§fn(x)3[a,b]þÂñ.1X∞nnn+1(2)~µ(−1)(x−x)3[0,1]þÂñ1nn+1nn+1nnn+1Ïx−x=0(x=0,1)¶00)(iii)1+ε6x<∞)µπ0,06x6πx6=(1)f(x)=limfn(x)=2n→∞π1,x=2nÏf(x)3[0,π]þØëY§fn(x)3[0,π]þëY§Kfn(x)=(sinx)3[0,π]þØÂñ. 课后答案网www.khdaw.com1810,x=0,π(2)(i)f(x)=limfn(x)=n→∞1,00§¦α0§Ke>1§u´lim=<1§KdKO{4/ª§?n→∞ne−nαeαX∞X∞−nα−nxêneÂñ§lâMO{§ne3[α,β]þÂñ.11X∞−nx−nxqne3[α,β]þëY§lne3[α,β]þëY1X∞−nxÏx0∈[α,β]§Kne3x0:ëY1X∞−nxdux0´(0,+∞)?¿:§ne3(0,+∞)SëY.1X∞sinnx12.y²¼êf(x)=3(−∞,+∞)SëY§¿këY¼ê.n31sinnx1X∞1X∞sinnxy²µÏ6Âñ§KâMO{§f(x)=3(−∞,+∞)Âñn3n3n3n311sinnxX∞sinnxq3(−∞,+∞)SëY§Kf(x)=3(−∞,+∞)SëYn3n3!1dsinnxcosnx=dxn3n2cosnx1X∞1X∞cosnxÏ6Âñ§KâMO{§3(−∞,+∞)Âñn2n2n2n21!1dX∞sinnxX∞cosnx0u´f(x)==dxn3n211cosnxX∞cosnxq3(−∞,+∞)SëY§K3(−∞,+∞)SëYn2n21 课后答案网www.khdaw.comX∞cosnx00=f(x)3(−∞,+∞)SëYf(x)=.n21X∞113.y²¼êζ(x)=3(1,+∞)ëY§¿këY¼ê.nx1X∞lnny²µ¦ê¤?ê−.ey§311?¿5§ζ(x)=−é11)ÑÂñ§ìþã§yµé?Ûnan=1(k)êk§ζ(x)31m§cos(2πy)=1§d?êتu0§Kcos(2πx)=cos(2πy)uѧu´πcos(2πx)u111Ñ−mmq3?Û«mSÑ3x=k+2h(h=0,1,2,···,2−1)ù:§kxêÜ©X∞sin(2nπx)K?ê3?Û«mSØUŦû.2n115.ky1−r2X∞n=1+2rcosnx1−2rcosx+r2n=1|r|<1¤á§ly²µZπ21−rdx=2π(|r|<1)−π1−2rcosx+r2.nny²µ|rcosnx|6|r|é∀x∈(−∞,+∞)ѤáX∞X∞nnÏ|r|<1§K|r|Âñ§u´dMO{§rcosnx3(−∞,+∞)SÂñn=1n=1 课后答案网www.khdaw.com183X∞nlf(x)=1+2rcosnxn=122Ï1−2rcosx+r6=0§þªüàÓ¦±1−2rcosx+r§K!X∞22n(1−2rcosx+r)f(x)=(1−2rcosx+r)1+2rcosnx"n=1#X∞X∞X∞2nn+1n+2=1−2rcosx+r+2rcosnx−2r(2cosnxcosx)+2rcosnx"n=1n=1n=1#X∞X∞X∞X∞2nn+1n+1n+2=1−2rcosx+r+2rcosnx−2rcos(n+1)x−2rcos(n−1)x+2rcosnx"n=1n=1!n=1n=1!#X∞X∞X∞X∞2nn+1n+12n+2=1−r+2rcosnx−2rcos(n+1)x+rcosx−2rcos(n−1)x−r+2rcosnxn=1n=1n=1n=12=1−r1−r21−r2X∞nu´f(x)===1+2rcosnx1−2rcosx+r21−2rcosx+r2n=1nduþªmà?ê3[−π,π]þÂñ§rcosnx3[−π,π]þëY§Kþª?ê±ÅÈ©§Zπ2ZπX∞!X∞Zπ1−rnndx=1+2rcosnxdx=2π+2rcosnxdx=2π.−π1−2rcosx+r2−π−πn=1n=116.^kCX½ny²)Z½n.+y²µÏ{Sn(x)}3[a,b]þÂñuS(x)§é∀ε>0,∀x∈[a,b],∃N(ε,x)∈Z§¦n>N(ε,x)§ÑAk|Sn(x)−S(x)|<ε§AOk|SN(ε,x)−S(x)|<εdSN(ε,x)(x)−S(x)3x:ëY§3x:mOx§¦|SN(ε,x)(y)−S(y)|,∀y∈Oxu´Oxx∈[a,b]¤[a,b]mCX(éà:a,bëYòÿ)âkCX½n§l¥ÀÑkmOx1,···,OxmÓCX[a,b]÷v|SN(ε,xi)(y)−S(y)|<ε,∀y∈Ox,i=1,2,···,miSmN=maxN(ε,xi)§Kn>N§é∀x∈[a,b]§d{Sn(x)}üN5ÚOx⊃[a,b]§73,ii6i6mi=1Ox§¦x∈Ox§k|Sn(x)−S(x)|6|SN(x)−S(x)|6|SN(ε,xi)(x)−S(x)|<εii={Sn(x)}3[a,b]þÂñuS(x).17.eSn(x)3c:ëY(n=1,2,3,···)§{Sn(c)}uѧK3?Ûm«m(c−δ,c)S(δ>0)§{Sn(x)}7ØÂñ.y²µ^y{.b3δ0>0§¦{Sn(x)}3(c−δ0,c)SÂñ++dÂñÜn§é∀ε>0,∃N(ε)∈Z§¦n>N(ε)§é∀x∈(c−δ0,c)Ú∀p∈Z§ÑAεk|Sn+p(x)−Sn(x)|<(∗)¤á2ÏzSn(x)3c:ëY§KSn+p(x)−Sn(x)3c:ëYu´lim[Sn+p(x)−Sn(x)]=Sn+p(c)−Sn(c)x→c−0ε3(∗)ªüà-x→c−0§|Sn+p(c)−Sn(c)|6<ε2dêÜÂñn§{Sn(c)}Âñ§®{Sn(c)}uÑgñbØ(§K3?Ûm«m(c−δ,c)S(δ>0)§{Sn(x)}7ØÂñ. 课后答案网www.khdaw.com§2.?ê1.¦e?êÂñ«mµX∞(2x)n(1)n!n=1X∞ln(n+1)n+1(2)xn+1n=1"!n#nX∞n+1(3)xnn=1∞n2Xx(4)2nn=1X∞[3+(−1)n]nn(5)xnn=1X∞3n+(−2)nn(6)(x+1)nn=1)µn2(1)an=n!anÏR=lim=+∞§KÙÂñ(−∞,+∞).n→∞an+1X∞ln(n+1)X∞lnnlnnn+1n(2)x=x§an=n+1nnn=1n=2(y+1)lnyy+1lnyandulim=limlim=1§KR=lim=1§u´ÙÂñ«my→+∞yln(y+1)y→+∞yy→+∞ln(y+1)n→∞an+1(−1,1)X∞lnnnnx=−1§?ê(−1)xn!n=2!()00lnx1−lnxlnxlnnÏ=x>3§<0§KüN~xx2xnlnnX∞lnnX∞lnnnnnnqlim=0§K?ê(−1)x4ÙZ[?ê§u´?ê(−1)xÂñn→∞nnnn=2n=2X∞lnnnx=1§?êxnn=2lnnX∞1X∞lnnnnÏlim=+∞§Kâ?ê"O{9?êuѧ?êxuÑn→∞nnnn=1n=2Kd?êÂñ[−1,1)."!n#n!n2!n2X∞n+1X∞11n(3)Ïx=1+x§Kan=1+nnnn=1n=1!p111qlimn|an|=e§KÙÂñ»R=§Âñ«m−,.n→∞eee!n2!n!n2!n1X∞1111nnx=±§?ê(±1)1+§Kun=(±1)1+enenen=1−1dâ7{K§lim|un|=e26=0n→∞!n2!n!X∞1111nK?ê(±1)1+uѧu´?êÂñ−,.neeen=1 课后答案网www.khdaw.com1851(4)an=2nrnp2n2n1dlim|an|x=|x|lim=|x|<1§ÙÂñ»R=1§Âñ«m(−1,1)n→∞n→∞2∞n2X(±1)|x|=1=x=±1§?êC2nn=1X∞n2X∞X∞n2(±1)1(±1)du?ê=Âñ§K?êýéÂñKÂñ2n2n2nn=1n=1n=1∞n2Xxl?êÂñ[−1,1].2nn=1nn[3+(−1)](5)an=rn!nnn[3+(−1)]111Ïlim=4§K?êÂñ»R=§Âñ«m−,n→∞n4441X∞[3+(−1)n]nX∞1X∞1x=§?êC=+4n·4n2k(2k+1)22k+1n=1k=1k=1X∞1é?ê(2k+1)22k+1k=11X∞(2k+3)22k+311Ïlim=<1§KâKO{§?êÂñk→∞14(2k+1)22k+1k=1(2k+1)22k+1X∞1X∞[3+(−1)n]nq?êuѧK?êuÑ2kn·4nk=1n=11X∞[3+(−1)n]nnÓ{§x=−§?ê(−1)uÑ4n·4nn=1!X∞[3+(−1)n]n11nK?êxÂñ−,.n44n=1nn3+(−2)(6)an=n!an1142Ïlim=§K?êÂñ»R=§Âñ«m−,−n→∞an+13333X∞nn!nX∞nX∞2
n43+(−2)1(−1)n3x=−§?êC(−1)=+3n3nn
n=1n=1n=1X∞2n3é?ênn=12
n+1X∞2
n/(n+1)23
3Ïlim2n=<1§KâKO{§Âñn→∞/n3n3n=1X∞(−1)n4q?êÂñ§Kx=−§?êÂñ¶n3n=12Ó{§x=−§?êuÑ3"!X∞3n+(−2)n42nK?ê(x+1)Âñ−,−.n33n=12.¦?êÂñ»µ!X∞11n(1)1++···+x2nn=1X(2n)!n(2)x(n!)2)µ 课后答案网www.khdaw.com11(1)an=1++···+r2rnn1n11√nÏ1=n·61++···+6n·1→1(n→∞)pn2nKlimn|an|=1§u´ÙÂñ»R=1.n→∞(2n)!(2)an=(n!)2a2n(n+1)11Ïlim=lim=§u´ÙÂñ»R=.n→∞an+1n→∞2(n+1)(2n+1)44X∞3n+(−2)nXXnnn3.(x+1)?êanxÂñ»R§bnxÂñ»Q§?Øe?êÂñnn=1»µX2n(1)anxXn(2)(an+bn)xXn(3)anbnx)µrpp1211√(1)lim2n|an√n|=lim|an|==§KÙÂñ»R1=R.n→∞n→∞RR(2)Anp=an+bnppp√p√ppKkn|Annnnnnnn√n|=|an+bn|6|an|+|bn|62max(|an|,|bn|)=2·max(|an,bn|)=2max(|an|,|bn|)nÏlim2=1n→∞1p√ppppppK=limn|Annnnnnnn|6lim{2max(|an|,|bn|)}=lim{max(|an|,|bn|)}=maxlim|an|,lim|an|=R2n→∞!n→∞n→∞n→∞n→∞11max,RQ1l§R2>!=min(R,Q).11max,RQ(3)Bnp=anbnpppKkn|Bnnnn|=|anbn|=|an|·|bn|1ppppp111u´=limn|Bnnnnn|=lim{|an|·|bn|6lim|an|·lim|bn|}=·=R3n→∞n→∞n→∞n→∞RQRQlR3>RQ.XXnn4.é¿©n§|an|6|bn|§@"?êanxÂñ»ØubnxÂñ».ppppy²µÏé¿©n§|annnnn|6|bn|§K|an|6|bn|§u´lim|an|6lim|bn|XXn→∞n→∞nn?êanxÂñ»R§?êbnxÂñ»Qpp11K0Q¶n→∞n→∞limn|an|limn|bn|ppn→∞n→∞0=limn|ann|6lim|bn|§KR=∞,Q6∞§u´R>Q¶n→∞pn→∞plimn|ann|6lim|bn|=∞§KR>0,Q=0§u´R>Qn→∞Xn→∞Xnnnþ§?êanxÂñ»ØubnxÂñ».5.y²?ê51Ú52.y²µ51.x(x0−R,x0+R)S?:§o±À00§½3õªQ(x)§¦||f(x)−Q(x)||=max|f(x)−Q(x)|0§kϕ(z)dz→(p→∞)πaz2y²µ(1)Ïϕ(x)3[a,b]þüNO¼ê§Kϕ(−t)3[−b,−a]þüN~¼êa=0,b<0§ϕ(−t)3[0,−b](−b>0)þüNO¼êZZZbsinpzbsinpz−bsinptéϕ(z)dzCþz=−t§Kϕ(z)=−ϕ(−t)dtazZaz0Zt−bsinptπbsinpzπKd)á4Ún§limϕ(−t)dt=ϕ(−0)=limϕ(z)dz=−ϕ(−0)Zp→∞0t2p→∞az21bsinpz1u´ϕ(z)dz→−ϕ(−0)(p→∞)πaz2ZZZbsinpz0sinpzbsinpz(2)Ïa<0,b>0§ϕ(x)3[a,b]þüNO¼ê§ϕ(z)dz=ϕ(z)dz+ϕ(z)dzZaZzaz0zasinpzπ0sinpzπâ(1)§limϕ(z)dz=−ϕ(−0)§Klimϕ(z)dz=ϕ(−0)p→∞0zZ2p→∞az2bsinpzπqd)á4Ún§limϕ(z)dt=ϕ(+0)p→∞0z2ZbsinpzπKlimϕ(z)dz=[ϕ(+0)+ϕ(−0)]p→∞az2Z1bsinpzϕ(−0)+ϕ(+0)u´ϕ(z)dz→(p→∞)πaz2 课后答案网www.khdaw.com§2.LpDC1.f(x)3(−∞,+∞)Sýéȧy²fb(ω)3(−∞,+∞)SëY.000000y²µé∀ω∈(−∞,+∞)§okA,A§¦ω∈[A,A]Z+∞Z+∞dufb(ω)=f(x)e−iωxdx6|f(x)|dx−∞−∞Z+∞−iωx000öÂñعëþω§ùL²È©fb(ω)=f(x)edx3[A,A]þÂñ−∞000âÂñÈ©ëY5§fb(ω)3[A,A]þëY§l3:ω?ëYdω?¿5§fb(ω)3(−∞,+∞)SëY.2.f(x)3(−∞,+∞)Sýéȧy²limfb(ω)=0.ω→∞Z+∞εy²µdf(x)3(−∞,+∞)Sýéȧéu?ε>0§3A>0§¦k|f(x)|dx0§¦ω>δ§ðk|mk|<ω3k=1k=1uZ´éþã¤Àδ§Zω>δZZ+∞A+∞+∞f(x)sinωxdx6f(x)sinωxdx+f(x)sinωxdx6ε=limf(x)sinωxdx=0ω→∞00A0Ùg§f(x)3«m[0,A]¥k×:§{Bå§Øk×:0Zηεu´é?ε>0§3η>0§¦k|f(x)|dx<03ZAεqf(x)3[η,A]þÃa:§A^þã(J3䧦ω>δ§ðkf(x)sinωxdx<η3ZZZZ+∞ηA+∞u´ω>δ§kf(x)sinωxdx6|f(x)|dx+f(x)sinωxdx+|f(x)|dx<εZ00ηA+∞=limf(x)sinωxdx=0ω→∞0Z+∞Ó{§f(x)3(−∞,+∞)Sýéȧþklimf(x)sinωxdx=0ω→∞Z−∞+∞Ó{yf(x)3(−∞,+∞)Sýéȧlimf(x)cosωxdx=0ω→∞−∞u´limfb(x)=0.ω→∞3.¦e¼êLpDCµπEsinω0x,|x|<(1)f(x)=ω0π0,|x|>ω0π0,−∞0,∃N∈Z§n>N§kr(Mn,M0)<εpn→∞=(xn−x0)2+(yn−y0)2<εu´½k|xn−x0|6r(Mn,M0)<ε,|yn−y0|6r(Mn,M0)<ε=xn→x0,yn→y0(n→∞)⇐pÏ(|x22222n−x0|+|yn−y0|)>|xn−x0|+|yn−y0|=06|xn−x0|+|yn−y0|6|xn−x0|+|yn−y0|pqxn→x0,yn→y0(n→∞)§K|xn−x0|2+|yn−y0|2→0(n→∞)=(xn,yn)→(x0,y0)(n→∞)2.y²µe²¡þ:{Mn}Âñ§K§k4.0y²µlimMn=M0§bqklimMn=M0n→∞n→∞εε+0d½Â§é∀ε>0,∃N∈Z§n>N§kr(Mn,M0)<,r(Mn,M0)<2200dnت§kr(M0,M0)6r(Mn,M0)+r(Mn,M0)<ε000qM0,M0½ü:§dε?¿5§r(M0,M0)=0=M0=M0.3.y²µeMn→M0(n→∞)§@o§?ÛfMn→M0.k+y²µÏMn→M0(n→∞)§Ké∀ε>0,∃N∈Z§n>N§kr(Mn,M0)<ε8K=N§Kék>K§knk>nK=nN>N§g,kr(Mn,M0)<ε=Mn→M0(k→∞).kk4.¦e:8ES:§:§>.:µ2(1)Ed÷vyx:(x,y)´E:¶÷vy=x:(x,y)´E>.:.222yyy222(2)÷v14:(x,y)´E:¶44422yy22÷vx+=1½x+=4:(x,y)´E>.:.442222(3)÷v01:(x,y)´E:¶:θ9÷22vx+y=1:(x,y)´E>.:.(4)dknê9ÃnêÈ5§²¡þ¤k:(x,y)Ñ´E>.:.5.y²µeM0´²¡:8Eà:§K3E¥3:Mn→M0(n→∞).1y²µ®M0´²¡:8Eà:§δn=§3O(M0,δ1)¥½3E:M16=M0¶3O(M0,δ2)¥½n3E:M2,M26=Mi(i6=0,1)1Xd?1e§:{Mn}(Mn6=Mi)(i=0,1,···,n−1)r(M0,Mn)0,∃N∈Z§n,m>N§kr(Mn,M0)<,r(Mm,M0)<22 课后答案网www.khdaw.com205dålnت§r(Mm,Mn)6r(Mn,M0)+r(Mm,M0)<ε⇐+:{Mn}÷vé∀ε>0,∃N∈Z§n,m>N§kr(Mn,Mm)<εò{Mn}©OÝKüI¶þ§ê{xn},{yn}Ï|xm−xn|0y61(2)½Â÷vتy6x+1:8(3)½Â²¡x+y<022(4)½Â÷vت2kπ6x+y6(2k+1)π(k=0,1,2,···):822222(5)½Â÷vتr6x+y+z6R:82.¦e4µ22x+y(1)limx→0|x|+|y|y→022x+y(2)limpx→0x2+y2+1−1y→0221+x+y(3)limx→0x2+y2y→033sin(x+y)(4)limx→0x2+y2y→022−(x+y)(5)lim(x+y)ex→+∞y→+∞yln(x+e)(6)limpx→1x2+y2y→0)µ22222x+y(|x|+|y|)x+y(1)Ï066=|x|+|y|lim(|x|+|y|)=0§Klim=0|x|+|y||x|+|y|x→0x→0|x|+|y|y→0y→022t√x+y(2)Ïlim√=lim(t+1+1)=2§Klimp=2t→+0t+1−1t→+0x→0x2+y2+1−1y→0221+t1+x+y(3)Ïlim=+∞§Klim=+∞t→+0tx→0x2+y2y→0sin(x3+y3)|x3+y3||x|3+|y|3|x|3|y|3(4)Ï0666=+6|x|+|y|lim(|x|+|y|)=0x2+y2x2+y2x2+y2x2+y2x2+y2x→0y→033sin(x+y)Klim=0x→0x2+y2y→02tt(5)Ïlim=0,lim=0t→+∞ett→+∞et"#2(x+y)xy22−(x+y)Klim(x+y)e=lim−2·=0x→+∞x→+∞e−(x+y)exeyy→+∞y→+∞yln(x+e)(6)limp=ln2x→1x2+y2y→0 课后答案网www.khdaw.com2073.Áyelimf(x,y)=A3§x?ÛaC§4limf(x,y)=ϕ(x)3§Kg43§y→ay→bx→buAµlimlimf(x,y)=limf(x,y)=Ax→ay→by→ax→b.22y²µÏ43§Ké∀ε>0,∃δ>0§|x−a|<δ,|y−b|<δ(x−a)+(y−b)6=0§ðk|f(x,y)−A|<εy30<|x−a|<δ¥½x§3þª¥-y→b§=|ϕ(x)−A|6ε§ùÒy²limϕ(x)=Ax→au´limlimf(x,y)=limϕ(x)=A=limf(x,y)x→ay→bx→ay→ax→b4.(1)ÁÞÑüg4Ø~f¶(2)ÁÞÑkg43~f¶(3)ÁÞÑ43§g4Ø3~f.)µx−y,x+y6=0(1)~µf(x,y)=x+y3:(0,0)g40,x+y=0x−ylimlimf(x,y)=lim=1,limlimf(x,y)=lim=−1x→0y→0x→0xy→0x→0y→0yKlimlimf(x,y)6=limlimf(x,y).x→0y→0y→0x→01xsin+yx(2)~µf(x,y)=3:(0,0)g4x+yylimlimf(x,y)=lim=1y→0x→0y→0y1xsinx1limlimf(x,y)=lim=limsinØ3.x→0y→0x→0xx→0x1xsin,y6=0(3)~µf(x,y)=y3:(0,0)g4Ú40,y=01Ï06|f(x,y)|=xsin6|x|§Klimf(x,y)=0=Ù43yx→0y→01limlimf(x,y)=0§y→0§xsin4Ø3§=limlimf(x,y)Ø3.y→0x→0yx→0y→05.?Øe¼ê3:(0,0)g4Ú4µ22xy(1)f(x,y)=x2y2+(x−y)211(2)f(x,y)=(x+y)·sin·sinxy)µ(1)limlimf(x,y)=0,limlimf(x,y)=0x→0y→0y→0x→022xkeUy=kx→04§Kklimf(x,y)=limy=kxx→0x2k2+(1−k)2x→0AO§©Ok6=19k=1§BØÓ4091§Ïdlimf(x,y)Ø3.x→0y→0(2)Ï06|f(x,y)|6|x+y|6|x|+|y|§Klimf(x,y)=0=Ù43x→0y→0!11111qlim(x+y)·sin·sinØ3x6=(k=±1,±2,···)§lim(x+y)·sin·sinØy→0xykπx→0xy!13y6=(k=±1,±2,···)kπ=limlimf(x,y)9limlimf(x,y)ÑØ3.x→0y→0y→0x→0 课后答案网www.khdaw.com6.?Øe¼êëYµ1(1)u=px2+y222(2)u=ln(1−x−y)1(3)u=sinxsiny1(4)u=ln(x−1)a+(y−b)2+(z−c)2)µ1(1)¼êu=p3:(0,0)ý§:(0,0)d¼êØëY:§Ød:þëY¶x2+y22222(2)üS:§=÷vx+y<1:¼êu=ln(1−x−y)ëY:¶(3)ëYx6=mπ,y6=nπ(m,n=0,±1,±2,···).(4)Ø:(a.b.c)þëY.7.y²¼ê2xy22,x+y6=0f(x,y)=x2+y2220,x+y=0©OéuzCþxÚy´ëY§"uCþëY¼ê.2axy²µk½y=a6=0§Kx¼êg(x)=f(x,a)=(−∞0§|x−x0|<δ1§εk|f(x,y0)−f(x0,y0)|<2qÏf(x,y)3GSéy÷voÊF[^§Ké?Û(x,y)∈G,(x,y0)∈G§k|f(x,y)−f(x,y0)|6L|y−y0| 课后答案网www.khdaw.com209εε-L|y−y0|<§K|y−y0|<2!2Lεδ=minδ1,§|x−x0|<δ,|y−y0|<䧽k2L|f(x,y)−f(x0,y0)|6|f(x,y)−f(x,y0)|+|f(x,y0)−f(x0,y0)|<ε=d¼ê3GSëY. 课后答案网www.khdaw.com1oÙêÚ©§1.êÚ©Vg1.¦e¼êêµ222(1)z=xln(x+y)xy(2)u=ex(3)z=xy+yy(4)u=arctanx222(5)u=x+y+z+2xy+2yz+2zxϕ−θ(6)u=ecos(θ+ϕ))µ"#22x2xy22(1)zx=2xln(x+y)+,zy=.x2+y2x2+y2xyxy(2)ux=ye,uy=xe.21x(y−1)(3)zx=y+,zy=.yy2yx(4)ux=−,uy=.x2+y2x2+y2(5)ux=2(x+y+z),uy=2(x+y+z),uz=2(x+y+z).ϕ−θϕ−θ(6)uϕ=e[cos(θ+ϕ)−sin(θ+ϕ)],uθ=−e[sin(θ+ϕ)+cos(θ+ϕ)].222.f(x,y)=xy−2y§¦fx(x,y),fy(x,y),fx(2,3),fy(0,0),fy(x,y)x=y.y=x222)µfx(x,y)=2xy,fy(x,y)=2xy−2,fx(2,3)=36,fy(0,0)=−2,fy(x,y)x=y=2xy−2y=x√√∂z∂z13.z=ln(x+y)§y²x+y=.∂x∂y2√√∂z1∂z1y²µÏz=ln(x+y)§K=√√√,=√√√∂x2x(x+y)∂y2y(x+y)∂z∂z1u´x+y=.∂x∂y24.¦e¼ê3½:(x0,y0)©µ4422(1)u=x+y−4xy,(0,0),(1,1)x(2)u=p,(1,0),(0,1)x2+y2!ππ(3)u=xsin(x+y),(0,0),,442(4)u=ln(x+y),(0,1),(1,1))µ2222(1)Ïdu=4x(x−2y)dx+4y(y−2x)dy§K3(0,0):du=0¶3(1,1):du=−4dx−4dy.2yxy(2)Ïdu=dx−dy§K33(x2+y2)2(x2+y2)23(1,0):du=0¶3(0,1):du=dx.(3)Ïdu=[sin(x+y)+xcos(x+y)]dx+xcos(x+y)dy§K!ππ3(0,0):du=0¶3,:du=dx.44 课后答案网www.khdaw.com211dx2y(4)Ïdu=+dy§Kx+y2x+y2dx3(0,1):du=dx+2dy¶3(1,1):du=+dy.25.¦e¼ê©µ22(1)u=sin(x+y)mn(2)u=x·yxy(3)u=ey(4)u=xp(5)u=x2+y2+z2222(6)u=ln(x+y+z))µ22(1)du=2cos(x+y)(xdx+ydy)m−1n−1(2)du=xy(mydx+nxdy)xy(3)du=e(ydx+xdy)y−1(4)du=x(ydx+xlnxdy)xdx+ydy+zdz(5)du=px2+y2+z22(xdx+ydy+zdz)(6)du=x2+y2+z2p6.y²µf(x,y)=|xy|3(0,0)ëY§fx(0,0),fy(0,0)3§3(0,0):Ø.py²µdlim|xy|=0§limf(x,y)=0=f(0,0)§Kf(x,y)3(0,0):ëYx→0x→0y→0y→0f(∆x,0)−f(0,0)f(0,∆y)−f(0,0)Ïfx(0,0)=lim=0,fy(0,0)=lim=0∆x→0∆x∆y→0∆yKfx(0,0),fyp(0,0)3f(x,y)=|xy|3(0,0):Ø.e§Kk∆f=fx(0,0)∆x+fy(0,0)∆y+o(ρ)=∆f=o(ρ)p∆f|∆x∆y|1Ä:P(x,y)÷y=xªu0§k=p=√6→0(ρ→0)gñ§u´bؤá§ρ∆x2+∆y22Kf(x,y)3(0,0):Ø.xy22p,x+y6=07.y²µf(x,y)=x2+y23(0,0):¥ëY§fx(x,y),fy(x,y)k.§3(0,0):220,x+y=0Ø.xy22y²µdup´Ð¼ê§3Ù½ÂS7ëY§Kf(x,y)3x+y6=0ëYx2+y2pxy|xy|x2+y2q00)∂z∂zpu´x+y=z=x2+y2.∂x∂y9.ϕψ´?¿¼ê§y²µ!!yyz=xϕ+ψxx 课后答案网www.khdaw.com222∂z∂z∂z22÷vx+2xy+y=0∂x2∂x∂y!∂y2!!!!∂zyyyyy∂zy1y0000y²µÏ=ϕ−ϕ−ψ,=ϕ+ψ∂xxxxx2x∂yxxx22222∂zy2yy∂zy1y∂z1100000000000000K=ϕ+ψ+ψ,=−ϕ−ψ−ψ,=ϕ+ψ∂x2x3x3x4∂x∂yx2x2x3∂y2xx2222∂z∂z∂z22u´x+2xy+y=0∂x2∂x∂y∂y210.u=ϕ(x+at)+ψ(x−at)§Ù¥ϕ,ψ´?¿g¼ê§¦y22∂u∂u2=a.∂t2∂x2y²µÏu=ϕ(x+at)+ψ(x−at)§ϕ,ψ´?¿g¼ê22∂u∂u∂u∂u0000200000000K=a(ϕ−ψ),=ϕ+ψ§u´=a(ϕ+ψ),=ϕ+ψ∂t∂x∂t2∂x222∂u∂u2l=a.∂t2∂x2 课后答案网www.khdaw.com217§3.d§(|)¤(½¼ê¦{1.¦de§¤(½¼êz=f(x,y)Úêµz(1)x+y+z=e(2)xyz=x+y+z)µz1ez(1)ü>"ux¦§1+zx=zxe§Kzx=§u´zx2=ez−1(1−ez)3zz1eeÓ{§zy=,zy2=,zxy=zyx=ez−1(1−ez)3(1−ez)3yz−1(2)ü>"ux¦§yz+xyzx=1+zx(∗)§Kzx=1−xy2yzx2y(yz−1)ò(∗)ªü>"ux¦§2yzx+xyzx2=zx2§Kzx2==1−xy(xy−1)2xz−12x(xz−1)2zÓ{§zy=,zy2=,zxy=zyx=1−xy(xy−1)2(xy−1)22.¦de§¤(½¼ê©½êµ∂z∂z(1)f(x+y,y+z,z+x)=0§¦,¶∂x∂y(2)z=f(xz,z−y)§¦dz¶∂z∂z(3)F(x−y,y−z,z−x)=0§¦,¶∂x∂y2∂z∂z∂z(4)F(x,x+y,x+y+z)=0§¦,,.∂x∂y∂x2)µf1+f3(1)ü>"ux¦§z=z(x,y)§f1+f2zx+f3(zx+1)=0§Kzx=−f2+f3f1+f2Ó{§zy=−f2+f3zf1dx−f2dy(2)ü੧dz=(xdz+zdx)f1+(dz−dy)f2§Kdz=1−xf1−f2F1−F3(3)ü>"ux¦§z=z(x,y)§F1−F2zx+F3(zx−1)=0§Kzx=F2−F3F2−F1Ó{§zy=F2−F3F1+F2+F3(4)ü>"ux¦§z=z(x,y)§F1+F2+F3(1+zx)=0(∗)§Kzx=−F33(∗)ªü>2"ux¦§F11+F12+F13(1+zx)+F21+F22+F23(1+zx)+zx2F3+(1+zx)[F13+F23+F33(1+zx)]=0122Kzx2=−3[F3(F11+2F12+F22)−2F3(F1+F2)(F13+F23)+F33(F1+F2)]F3F2+F3Ó{§zy=−F303.d§z=x+y·ϕ(z)(½¼êz=z(x,y)§1−yϕ(z)6=0§y²∂z∂z=ϕ(z)·∂y∂x0y²µ§ü੧z=z(x,y)§dz=dx+ϕ(z)dy+yϕ(z)dzdx+ϕ(z)dy0q1−yϕ(z)6=0§Kdz=1−yϕ0(z)∂zϕ(z)∂z1∂z∂zu´=,=§l=ϕ(z)·∂y1−yϕ0(z)∂x1−yϕ0(z)∂y∂x 课后答案网www.khdaw.com∂z∂z2224.y²d§ax+by+cz=Φ(x+y+z)¤½Â¼êz=z(x,y)÷v§(cy−bz)+(az−cx)=∂x∂ybx−ay§Ù¥Φ(u)´u¼ê§a,b,c~ê.y²µ§ü੧z=z(x,y)§Φ(u)´u¼ê0Kadx+bdy+cdz=2(xdx+ydy+zdz)Φ00∂z2xΦ−a∂z2yΦ−bu´=,=∂xc−2zΦ0∂yc−2zΦ0∂z∂zl(cy−bz)+(az−cx)=bx−ay∂x∂y∂z∂z5.ϕ?¿¼ê§y²d§ϕ(cx−az,cy−bz)=0¤½Â¼êz=z(x,y)÷va+b=c.∂x∂yy²µé§üà©O"ux,y¦§z=z(x,y)§cϕ1−aϕ1zx−bϕ2zx=0,−aϕ1zy+cϕ2−bϕ2zy=0∂zcϕ1∂zcϕ2u´=,=∂xaϕ1+bϕ2∂yaϕ1+bϕ2∂z∂zla+b=c.∂x∂y∂z∂z−1−16.y²d§F(x+zy,y+zx)=0¤(½¼êz=z(x,y)÷vx+y=z−xy.∂x∂yy²µé§üà©O"ux,y¦§z=z(x,y)§!!!!zxzxzzyzzyF11++F2−=0,F1−+F21+=0yxx2yy2x22∂zyzF2−xyF1∂zxzF1−xyF2u´=,=∂xx(xF1+yF2)∂yy(xF1+yF2)∂z∂zlx+y=z−xy.∂x∂y7.¦e§|¤(½¼êê½ê½©µ2x+y+z=0,dydzdy(1)¦,,¶x·y·z=1,dxdxdx2x+y=u+v,(2)xsinu¦du,dv¶=,ysinv2xu+yv=0,∂u∂u∂v∂v∂u(3)¦,,,,¶yu+xv=1,∂x∂y∂x∂y∂x∂yx=cosθcosϕ,∂z∂z(4)y=cosθsinϕ,¦,¶∂x∂yz=sinθ,u=f(u,x,v+y),∂u∂v(5)2¦,.v=g(u−x,u·y),∂x∂x)µdydz1++=0(1)éx¦§dxdx(∗)dydzyz+xz+xy=0dxdxdyy(z−x)dzz(x−y)éá¦)§=,=dxx(y−z)dxx(y−z)22dydz+=0(∗)ª2éx¦§dx2dx222dydzdydydzdydzdydzdzz+y+z+x·+xz+y+x·+xy=0dxdxdxdxdxdx2dxdxdxdx2d2y2zdy+2ydz+2xdy·dzdxdxdxdxéá§=dx2x(y−z)2222dydzdyyz[(x−y)+(x−z)+(y−z)]ò,§=dxdxdx2x2(z−y)3 课后答案网www.khdaw.com219u+v=x+ydu+dv=dx+dy(2)òªU©§ysinu=xsinvsinudy+ycosudu=sinvdx+xcosvdv1Kdu=[(sinv+xcosv)dx−(sinu−xcosv)dy]xcosv+ycosu1dv=[−(sinv−ycosu)dx+(sinu+ycosu)dy]xcosv+ycosuxdu+ydv=−udx−vdy(3)©§ydu+xdv=−vdx−udy11u´du=[(yv−xu)dx+(yu−xv)dy],dv=[(yu−xv)dx+(yv−xu)dy]x2−y2x2−y2∂uyv−xu∂uyu−xv∂vyu−xv∂uyv−xuK=,=,=,=∂xx2−y2∂yx2−y2∂xx2−y2∂xx2−y2222∂u(yux−v−xvx)(x−y)−2x(yu−xv)u´=∂x∂y(x2−y2)2222∂u∂v∂u2(xv+yv−2xyu)ò,§=∂x∂x∂x∂y(x2−y2)2∂θ∂ϕ1=−sinθ·cosϕ−cosθ·sinϕ(4)dx,yéx¦ê§∂x∂x∂θ∂ϕ0=−sinθ·sinϕ+cosθ·cosϕ∂x∂x∂θcosϕ∂ϕsinϕ∂z∂z∂θxK=−,=−§u´=·=−cotθcosϕ=−∂xsinθ∂xcosθ∂x∂θ∂xz∂zyÓn§=−∂yz∂u∂u∂v=f1+f2+f3∂x∂x!∂x(5)éx¦§∂v∂u∂v=g1−1+2vyg2∂x∂x∂x∂uf2(1−2vyg2)−g1f3∂vg1(f1+f2−1)K=,=.∂x(f1−1)(2vyg2−1)−g1f3∂x(f1−1)(2vyg2−1)−g1f3∂z∂z22338.§x=u+v,y=u+v,z=u+v½Âzx,y¼ê§¦,.∂x∂yx2222)µÏx−y=2uv§Kz=(u+v)(u−uv+v)=(3y−x)2∂z3∂z32u´=(y−x),=x.∂x2∂y29.x=rcosθ,y=rsinθ§C§dx22=y+kx(x+y)dtdy22=−x+ky(x+y)dt4I§.)µd§§x,y´t¼ê§l4ICr,θ´t¼ê§x=rcosθ,y=rsinθdxdrdθ=cosθ−rsinθüàét¦§dtdtdtdydrdθ=sinθ+rcosθdtdtdtdrdθ2dxdycosθ−rsinθ=rsinθ+krcosθ·ròx,y,,§|§dtdtdtdtdrdθ2sinθ+rcosθ=−rcosθ+krsinθ·rdtdtdrdθ3u´=kr,=−1.dtdt22∂z∂zuu10.x=ecosθ,y=esinθ§C§+=0.∂x2∂y2yuu22)µÏx=ecosθ,y=esinθ§Ku=ln(x+y),θ=arctanx 课后答案网www.khdaw.com∂ux∂uy∂θy∂θx∂u∂θ∂u∂θK=,=;=−,=§u´=,=−∂xx2+y2∂yx2+y2∂xx2+y2∂yx2+y2∂x∂y∂y∂x∂z∂z∂u∂z∂θ∂z∂z∂u∂z∂θq=·+·,=·+·∂x∂u∂x∂θ∂x∂y∂u∂y∂θ∂y!2!2222222∂z∂z∂u∂z∂u∂θ∂z∂θ∂z∂u∂z∂θK=+2··++·+·∂x2∂u2∂x∂θ∂u∂x∂x∂θ2∂x∂u∂x2∂θ∂x2!2!2222222∂z∂z∂u∂z∂u∂θ∂z∂θ∂z∂u∂z∂θ=+2··++·+·∂y2∂u2∂y∂θ∂u∂y∂y∂θ2∂y∂u∂y2∂θ∂y2!!!22∂u∂∂θ∂∂θ∂∂u∂uq===−=−∂x2∂x∂y∂y∂x∂y∂y∂y222∂θ∂θÓ{§=−∂x2∂y22222∂θ∂θ∂u∂uK+=+=0∂x2∂y2∂x2∂y2!2!2!2!2∂u∂u∂θ∂θ∂u∂θ∂u∂θq+=+,·=−·∂x∂y∂x∂y∂x∂x∂y∂y!2222∂z∂z∂z∂z−2uK+=e+=0∂x2∂y2∂u2∂θ222∂z∂z=+=0.∂u2∂θ222∂f∂f11.x=rcosθ,y=rsinθ§Kf(x,y)=Φ(r,θ)§^Φ"ur,θê5L«+.∂x2∂y2)µòf(x,y)=Φ(r,θ)"ur,覧∂f∂x∂f∂y∂Φ∂f∂f∂Φ·+·==cosθ+sinθ=∂x∂r∂y∂r∂r∂x∂y∂r∂f∂x∂f∂y∂Φ∂f∂f∂Φ·+·==−rsinθ+rcosθ=∂x∂θ∂y∂θ∂θ∂x∂y∂θ2222∂Φ∂f∂f∂f22K=cosθ+sin2θ+sinθ∂r2∂x2∂x∂y∂y2!2222∂Φ∂f∂f∂f∂f∂f222=rsinθ−sin2θ+cosθ−rcosθ−rsinθ∂θ2∂x2∂x∂y∂y2∂x∂y2222∂Φ1∂Φ∂f∂f1∂Φu´+=+−∂r2r2∂θ2∂x2∂y2r∂r2222∂f∂f∂Φ1∂Φ1∂Φl+=++∂x2∂y2∂r2r2∂θ2r∂r222∂z∂z∂zξη2212.x=e,y=e§C§ax+2bxy+cy=0(a,b,c~ê).∂x2∂x∂y∂y2dξ1dη1ξη)µÏx=e,y=e§Kξ=lnx,η=lny§u´=,=dxxdyy∂z1∂z∂z1∂zK=,=∂xx∂ξ∂yy∂η!!222222∂z1∂z∂z∂z1∂z∂z∂z1∂zu´=−,=−,=∂x2x2∂ξ2∂ξ∂y2y2∂η2∂η∂x∂yxy∂ξ∂η!!222∂z∂z∂z∂z∂z§§z{n§a−+2b+c−=0.∂ξ2∂ξ∂ξ∂η∂η2∂η∂z∂z2213.ξ=x,η=x+y§C§y−x=0.∂x∂y)µd§z´x,y¼ê§ξ,ηq´x,y¼ê§lzw¤´ÏL¥mCþξ,η"ux,yEܼê∂z∂z∂z∂z∂zu´=+2x,=2y∂x∂ξ∂η∂y∂η∂z∂z∂zÏy−x=y∂x∂y∂ξ 课后答案网www.khdaw.com221∂z∂z∂zÏy6≡0§Kdy−x=0§=0.∂x∂y∂ξ∂u∂u∂u14.ξ=x,η=y−x,ζ=z−x§C§++=0.∂x∂y∂z)µd§u´x,y,z¼ê§ξ,η,ζq´x,y,z¼ê§luw¤´ÏL¥mCþξ,η,ζ"ux,y.zEܼê∂u∂u∂u∂u∂u∂u∂u∂uu´=−−,=,=∂x∂ξ∂η∂ζ∂y∂η∂z∂ζ∂u∂u∂u∂zKd++=0§=0∂x∂y∂z∂ξ222∂u∂u∂u15.5Cξ=x+λ1y,η=x+λ2y§y3r§A+2B+C=0(A,B,C~ê§∂x2∂x∂y∂y22∂u22AC−B<0)C=0§y²λ1,λ2§Cλ+2Bλ+A=0üÉ¢.∂ξ∂ηy²µd§u´x,y¼ê§Ï±ruÀ±ξ,η¥mCþ"ux,yEܼê§u´∂u∂u∂u∂u∂u∂u=+,=λ1+λ2∂x∂ξ∂η∂y∂ξ∂η22222222∂u∂u∂u∂u∂u∂u∂u∂u22∂x2=∂ξ2+2∂ξ∂η+∂η2,∂y2=λ1∂ξ2+2λ1λ2∂ξ∂η+λ2∂η2,2222∂u∂u∂u∂u=λ1+(λ1+λ2)+λ2∂x∂y∂ξ2∂ξ∂η∂η2222222∂u∂u∂u∂u∂u∂u2ÏA+2B+C=(A+2Bλ1+Cλ1)+2[A+B(λ1+λ2)+Cλ1λ2]+(A+∂x2∂x∂y∂y2∂ξ2∂ξ∂η∂η222Bλ2+Cλ2)=02∂2u∂2u∂2u∂2uA+2Bλ1+Cλ1=02¦A+2B+C=0C=0§7LA+2Bλ2+Cλ2=0∂x2∂x∂y∂y2∂ξ∂ηA+B(λ1+λ2)+Cλ1λ26=02dcü§λ1,λ2´§Cλ+2Bλ+A=02d1n§λ16=λ2§Kλ1,λ2´Cλ+2Bλ+A=0üÉ¢2BA22qÏλ1+λ2=−,λ1λ1=§KA+B(λ1+λ2)+Cλ1λ2=(AC−B)6=0CCC2222∂u∂u∂u∂uu´§A+2B+C=035Cξ=x+λ1y,η=x+λ2ye(¢C=0§∂x2∂x∂y∂y2∂ξ∂η2λ1,λ2§Cλ+2Bλ+A=0üÉ¢.!22∂w∂w∂ϕ∂ψ∂ϕ∂ψ16.y².Ê.d§∆w≡+=03Czx=ϕ(u,v),y=ψ(u,v)§÷v=,=−e∂x2∂y2∂u∂v∂v∂u/G±ØC.y²µl§ω´x,y¼ê§x,y´u,v¼ê§Kw´±x,y¥mCþu,v¼ê∂w∂w∂ϕ∂w∂ψ∂w∂w∂ϕ∂w∂ψu´=·+·,=·+·∂u∂x∂u∂y∂u∂v∂x∂v∂y∂v!2!2222222∂w∂w∂ϕ∂w∂ϕ∂ψ∂w∂ψ∂w∂ϕ∂w∂ψ=+2··++·+·∂u2∂x2∂u∂x∂y∂u∂u∂y2∂u∂x∂u2∂y∂u2!2!2222222∂w∂w∂ϕ∂w∂ϕ∂ψ∂w∂ψ∂w∂ϕ∂w∂ψ=+2··++·+·∂v2∂x2∂v∂x∂y∂v∂v∂y2∂v∂x∂v2∂y∂v222222222∂ϕ∂ψ∂ϕ∂ψ∂ϕ∂ψ∂ϕ∂ψ∂ψ∂ϕ∂ψ∂ϕ5¿òz^=,=−§K=,=−,=−,=∂u∂v∂v∂u∂u2∂u∂v∂v2∂v∂u∂u2∂u∂v∂v2∂v∂u22∂w∂wò,§¿òþ㪧∂u2∂v2!"!!#222222∂w∂w∂w∂w∂ϕ∂ϕ+=++∂u2∂v2∂x2∂y2∂u∂v2222∂w∂w∂w∂wÏ+=0§K+=0∂x2∂y2∂u2∂v222∂w∂wùL².Ê.d§∆w≡+=03Czx=ϕ(u,v),y=ψ(u,v)e/G±ØC.∂x2∂y222∂u∂u217.ξ=x−at,η=x+at§C§=a.∂t2∂x2 课后答案网www.khdaw.com)µd§u´t,x¼ê§ξ,η´t,x¼ê§òuÀ±ξ,η¥mCþ"ut,x¼ê∂u∂u∂u∂u∂u∂uK=−a+a,=+∂t∂ξ∂η∂x∂ξ∂η22222222∂u∂u∂u∂u∂u∂u∂u∂u222=a−2a+a,=++∂t2∂ξ2∂ξ∂η∂η2∂x2∂ξ2∂ξ∂η∂η2222∂u∂u∂u22u´d=a§4a=0∂t2∂x2∂ξ∂η2∂uqa6≡0§K=0.∂ξ∂η18.gCêÚÏCêC§u,v#gCê§w=w(u,v)#ÏCêµ1122(1)u=x+y,v=+,w=lnz−(x+y)§C§xy∂z∂zy−x=(y−x)·z∂x∂yyz(2)u=x+y,v=,w=§C§xx222∂z∂z∂z−2+=0∂x2∂x∂y∂y2uu(3)x=u,y=,z=§C§1+uv1+u·w∂z∂z222x+y=z∂x∂yx(4)u=,v=x,w=xz−y§C§y2∂z∂z2y+2=∂y2∂yx)µ111(1)d®§du=2xdx+2ydy,dv=−dx−dy,dw=dz−dx−dyx2y2z∂w∂w,¡§dw=du+dv∂u∂v!1∂w∂w11Kdz−dx−dy=(2xdx+2ydy)+−dx−dyz∂u∂vx2y2!!∂wz∂w∂wz∂wn§dz=2xz−·+zdx+2yz−·+zdy∂ux2∂v∂uy2∂v!∂z∂zxy∂wòþª¤(½,§§z−=0∂x∂yy2x2∂v!xy∂wqz−6≡0§K=0.y2x2∂vy1z1(2)d®§du=dx+dy,dv=−dx+dy,dw=−dx+dzx2xx2x∂w∂w,¡§dw=du+dv∂u∂v!z1∂w∂wy1K−dx+dz=(dx+dy)+−dx+dyx2x∂u∂vx2x!!∂wy∂wz∂w∂wn§dz=x−·+dx+x+dy∂ux∂vx∂u∂v∂z∂wy∂wz∂z∂w∂wK=x−·+,=x+∂x∂ux∂vx∂y∂u∂v 课后答案网www.khdaw.com223∂z∂z∂w-R=−=w−(1+v)∂x∂y∂v!!!222∂z∂z∂z∂∂z∂z∂∂z∂z∂R∂R∂R∂u∂uK−2+=−−−=−=−+∂x2∂x∂y∂y2∂x∂x∂y∂y∂x∂y∂x∂y∂u∂x∂y!"#!22∂R∂v∂v∂∂wy1(1+v)∂w−=w−(1+v)−−==0∂v∂x∂y∂v∂vx2xx∂v222(1+v)∂wÏ6≡0§K§C=0.x∂v2uu1111(3)Ïx=u,y=,z=§Ku=x,v=−,w=−1+uv1+u·wyxzx1111u´du=dx,dv=dx−dy,dw=dx−dzx2y2x2z2∂w∂w,¡§dw=du+dv∂u∂v!11∂w∂w11Kdx−dz=dx+dx−dyx2z2∂u∂vx2y2!21∂w1∂wz∂w2n§dz=z−−·dx+·dyx2∂ux2∂vy2∂v∂z∂z∂w22òþª¤(½,§xz=0∂x∂y∂u∂wqxz6≡0§K=0.∂ux∂w∂z∂w∂w∂u∂w∂v∂ux∂v(4)Ïu=,v=x,w=xz−y§K=x−1,=·+·,=−,=0y∂y∂y∂y∂u∂y∂v∂y∂yy2∂y∂wx∂w∂z11∂wu´=−§K=−∂yy2∂u∂yxy2∂u22∂z∂z2x∂w2u´y+2=+=∂y2∂yxy3∂u2x2x∂wq6≡0§K§C=0.y3∂u2 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课后答案网www.khdaw.com§6.êÚFÝ221.¦u=x−xy+y3(1,1)?÷l=(cosα,sinα)ê.¿?Ú¦µ(1)3=þÙêk¶(2)3=þÙêk¶(3)3=þÙê0¶(4)¦uFÝ.)µÏux=2x−y,uy=−x+2y§Kux(1,1)=1,uy(1,1)=1!∂u∂u√πq=ux(1,1)cosα+uy(1,1)sinα§K=cosα+sinα=2sinα+∂l∂l4u´√√!π22√(1)α=§3l=,þÙêk2¶422√√!322√(2)α=−π§3l=−,−þÙêk−2¶422√√!√√!π32222(3)α=−,π§3l=,−½l=−,þÙê0¶442222(4)gradu=ux(1,1)i+uy(1,1)j=i+j.2.¦u=xyz3:M(1,1,1)§÷l=(2,−1,3)ê9FÝ.)µÏux=yz,uy=xz,uz=xy§K3(1,1,1):ux=uy=uz=1213qþl{ucosα=√,sinβ=−√,cosγ=√141414∂u2√K=ux(1,1,1)cosα+uy(1,1,1)cosβ+uz(1,1,1)cosγ=14¶gradu=i+j+k.∂l72223.¦êþ¼êu=x+2y+3z+xy+3x−2y−6z3O(0,0,0)9A(1,1,1)FÝ9Ù.)µÏux=2x+y+3,uy=4y+x−2,uz=6z−6K3O(0,0,0):µux=3,uy=−2,uz=−6§u´gradu=3i−2j+6k,|√gradu|=73A(1,1,1):µux=6,uy=3,uz=0§u´gradu=6i+3j,|gradu|=35.4.y²µ(1)grad(αu+βv)=αgradu+βgradv§Ù¥α,βÑ´~ê¶(2)grad(uv)=ugradv+vgradu¶0(3)gradF(u)=F(u)graduy²µ±¼ê~5y.-u=u(x,y),v=v(x,y)∂(αu+βv)∂u∂v∂(αu+βv)∂u∂v(1)Ï=α+β,=α+β∂x∂x∂x∂y!∂y∂y!!∂(αu+βv)∂(αu+βv)∂u∂v∂u∂vKgrad(αu+βv)=,=α,+β,=αgradu+βgradv.∂x∂y∂x∂y∂x∂y∂(uv)∂u∂v∂(uv)∂u∂v(2)Ï=v+u,=v+u∂x∂x∂x∂y!∂y∂y!!∂(uv)∂(uv)∂u∂u∂v∂vKgrad(uv)=,=v,+u,=ugradv+vgradu.∂x∂y∂x∂y∂x∂y!!!∂F∂F∂u∂u∂u∂u0000(3)gradF(u)=,=F(u),F(u)=F(u),=F(u)gradu∂x∂y∂x∂y∂x∂yõ¼ê¼êy.1rp5.y²grad=−§r=x2+y2+z2,r=xi+yj+zk.rr3∂rx∂ry∂rzy²µÏ=,=,=∂xr∂y!r∂zr!1d11∂r∂r∂r11rKgrad=gradr=−i+j+k=−·(xi+yj+zk)=−.rdrrr2∂x∂y∂zr2rr3 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课后答案网www.khdaw.com2291ÊÙ4Ú^4§1.4Ú¦{1.¦e¼ê4µ22(1)z=x−(y−1)2(2)z=(x−y+1)33(3)z=3axy−x−y(a>0)xπ(4)z=sinx+cosy+cos(x−y)066y2r22xy(5)z=xy·1−−(a,b>0)a2b2)µ(1)-zx=2x=0,zy=−2(y−1)=0Kx=0,y=1§u´:(0,1)U4:qzx2=2,zxy=0,zy2=−2§KA=2,B=0,C=−2§u´H=−4<0§ld¼êÃ4.(2)-zx=2(x−y+1)=0,zy=−2(x−y+1)=0K:©Ù3x−y+1=0þ§¼êUk4qA=2,B=−2,C=2§KH=0§I?ÚäÏéx−y+1=0þ:þkz=0§z>0ð¤áK¼êz3x−y+1=0þ:4z=0.22(3)-zx=3ay−3x=0,zy=3ax−3y=0x1=0x2=aK=3:(0,0),(a,a)?Uk4y1=0y2=aqzx2=−6x,zxy=3a,zy2=−6y2K3:(0,0)§A=0,B=3a,C=0§u´H=−9a<0§d¼êÃ4¶233:(a,a)§A=−6a<0,B=3a,C=−6a§u´H=27a>0§d¼êk4z=a.(4)-zx=cosx−sin(x−y)=0,zy=−siny+sin(x−y)=0!ππx3Kcosx=siny§u´y=−x§cosx−sin(x−y)=cos−sin2x−=2coscosx=02222!πx3ππππÏ06x6§Kcos6=0,cosx=0§u´x=,y=§=3:,?Uk42223636qzx2=−sinx−cos(x√−y),zxy=cos(x−y),zy2=−cosy−cos(x−!y)√3√9ππ3√KA=−3<0,B=,C=−3§u´H=>0§=3:,?¼êk4z=3.24362rr222222xyxyxyxy(5)-zx=y1−−−r=0,zy=x1−−−r=0a2b2x2y2a2b2x2y2a21−−b21−−a2b2a2b2aaaax2=√x3=−√x4=√x5=−√Kx1=03333y1=0bbbby2=√y3=−√y4=−√y5=√3!3!3!3!ababababu´3:P1(0,0),P2√,√,P3−√,−√,P4√,−√,P5−√,√?U4333333332223234424244422222244−3abxy+2bxy+3axyab−3abx+2bx−3aby+3abxy+2ayqzx2=3,zxy=3a4b21−x2−y22a4b41−x2−y22a2b2a2b22322233bxy−3abxy+2axyzy2=324x2y22ab1−a2−b2 课后答案网www.khdaw.com!!√√abab43b2√43a3:P2√,√,P3−√,−√?§A=−<0,B=−3,C=−33333a33b√3dH=4>0§u´¼êk4z=ab¶!!9√√abab43b2√43a3:P4√,−√,P5−√,√?§A=>0,B=−3,C=33333a33b√3dH=4>0§u´¼êk4z=−ab¶93:P1(0,0)?§A=0,B=1,C=0§dH=−1<0§u´¼êÃ4.yy2.y²¼êz=(1+e)cosx−yekáõ4§Ã4.yyyyy²µ-zx=−(1+e)sinx=0zy=ecosx−e−ye=0yÏ1+e6=0§Ksinx=0§u´x=kπ(k∈Z)yqe6=0§Kcosx−1−y=0=ky=cosx−1x1=2kπx2=(2k+1)πkóê§y=0¶kÛê§y=−2§KU4:(k=y1=0y2=−20,±1,±2,···)yyyyyqzx2=−(1+e)cosx,zxy=−esinx,zy2=ecosx−2e−yeK3:(2kπ,0)§A=−2<0,B=0,C=−1§dH=2>0§Kd¼êk4z=2!11113:((2k+1)π,−2)§A=1+,B=0,C=−§dH=−1+<0§Kd¼êÃ4e2e2e2e2yynþ§¼êz=(1+e)cosx−yekáõ4§Ã4.3.3®±2pn/¥¦Ñ¡Èn/.x+y+z)µn/>©Ox,y,z§KC=x+y+z=2p§D==p§u´z=2p−x−ypp2KS=D(D−x)(D−y)(D−z)=p(p−x)(p−y)(x+y−p)2ļêu=S=p(p−x)(p−y)(x+y−p),00,B=−4,C=18§u´H=2>0Kd¼êk4z=−5§l¡k$:(1,1,−5)22qx+y→+∞§z→+∞§¡Ãp:.25.®y=ax+bx+c§yÿ|êâ(xi,yi),i=1,2,···,n§|^¦{§¦Xêa,b,c¤÷vng§|.Xn22)µd®§ε=(yi−axi−bxi−c)§¦oε(a,b,c)§d47^§ki=1∂εXn∂εXn∂εXn2222=−2(yi−axi−bxi−c)xi=0,=−2(yi−axi−bxi−c)xi=0,=−2(yi−axi−bxi−c)=0∂a∂b∂ci=1!i=1!!i=1XnXnXnXnx4a+x3b+x2c=x2yiiiiii=1!i=1!i=1!i=1XnXnXnXn32=a,b,c÷veng§|µxia+xib+xic=xiyii=1!i=1!i=1i=1XnXnXn2xia+xib+nc=yii=1i=1i=126.y=x3[0,1]þ^oη=ax+b5O§âU¦§²ØÈ©Z12J(a,b)=(y−η)dx4¿ÂeZCqº0 课后答案网www.khdaw.com231Z1Z11a2a22222)µJ(a,b)=(y−η)dx=(x−ax−b)dx=++b−−b+ab005323ÀJa,b¦²ØÈ©J(a,b)4§d47^§k∂J12∂J2-=−+a+b=0,=−+a+2b=0∂a23∂b31Ka=1,b=−612u´y=x^η=x−5O§ZCq¦.6 课后答案网www.khdaw.com§2.^41.¦e¼ê3¤^e4µ22(1)f=x+y,ex+y=1¶222(2)f=x−2y+2z,ex+y+z=1¶1111(3)f=xyz,e++=(x>0,y>0,z>0,a>0)¶xyza11(4)f=+,ex+y=2¶xy222(5)f=xyz,ex+y+z=1,x+y+z=0.)µ22(1)¼êL=x+y+λ(x+y−1)√√22x1=x2=−Lx=1+2λx=0√2√222)§|Ly=1+2λy=0§y1=y2=−Lλ=x2+y2−1=02√√222λ1=−λ2=22qLx2=2√λ,Lxy√=0!,Ly2=2λ√√!22√22√222KdL,=−2(dx+dy)<0§u´¼ê3,?42¶2222√√!22√Ón§¼ê3−,−?4−2.22222(2)¼êL=x−2y+2z+λ(x+y+z−1)11x1=x2=−33Lx=1+2λx=022Ly1=−y2=y=−2+2λy=033)§|§Lz=2+2λz=022222Z1=z2=−Lλ=x+y+z−1=03333λ1=−λ2=22qLx2=Ly2=Lz2=2λ,Lxy==Lxz=Lyz=0!1222222KdL(x2,y2,z2)=3(dx+dy+dz)>0§u´¼ê3−,,−?4−3¶333!122Ón§¼ê3,−,?43.333!1111(3)¼êL=xyz+λ++−xyzaλLx=yz−=0x2λLy=xz−=0y24)§|§x=y=z=3a,λ=81aλLz=xy−=0z21111Lλ=++−=0xyzaqLx2(3a,3a,3a)=Ly2(3a,3a,3a)=Lz2(3a,3a,3a)=6a,Lxy(3a,3a,3a)=Lxz(3a,3a,3a)=Lyz(3a,3a,3a)=3a22222KdL(3a,3a,3a)=3a[(dx+dy+dz)+dx+dy+dz]>0§u´¼ê3(3a,3a,3a)?4327a.11(4)¼êL=++λ(x+y−2)xy 课后答案网www.khdaw.com2331Lx=−+λ=0x2)§|1§x=y=λ=1Ly=−+λ=0y2Lλ=x+y−2=0qLx2(1,1)=Ly2(1,1)=2,Lxy(1,1)=0222KdL(1,1)=2(dx+dy)>0§u´¼ê3(1,1)?42.222(5)¼êL=xyz+u(x+y+z−1)+v(x+y+z)Lx=yz+2ux+v=0Ly=xz+2uy+v=0)§|Lz=xy+2uz+v=0§222Lu=x+y+z−1=0Lv=x+y+z=0√√√√√√666666x1=x2=−x3=−x4=x5=x6=−√6√6√33√6√√6666666y1=y2=−y3=y4=−y5=−y6=6√√6√6√6√33√666666z1=−√3z2=3√z3=√6z4=−√6z5=√6z6=−√6666666u1=u2=−u3=u4=−u5=u6=−121212121212111111v1=v2=v3=v4=v5=v6=6666662222qdL=2u(dx+dy+dz)+2(zdxdy+ydxdz+xdydz)√62222K3:(x1,y1,z1)?§dL=(dx+dy+dz−4dxdy+2dxdz+2dydz)6222dx+y+z=1§2xdx+2ydy+2zdz=0§K3:(x1,y1,z1)?§kdx+dy=2d√z22qdx+y+z=0§dx+dy+dz=0§Kdx=−dy,dz=0§u´dL(x1,y1,z1)=6dx>0§√√√!√6666K¼ê3,,−?4−66318√6Ón§¼ê3(x3,y3,z3),(x5,y5,z5)?4−√186¼ê3(x2,y2,z2),(x4,y4,z4),(x6,y6,z6)?4.18mnp2.¦f=xyz3^x+y+z=a,a>0,m>0,n>0,p>0,x>0,y>0,z>0e.mnp)µÏx>0,y>0,z>0§Kf=xyz§lnf=mlnx+nlny+plnz§½,§I¦lnf4:§´f4:-L=mlnx+nlny+plnz+λ(x+y+z−a)mamx=Lx=x+λ=0m+n+pnnay=K)§Ly=y+λ=0§m+n+ppapz=Lz=+λ=0m+n+pzm+n+pLλ=x+y+z−a=0λ=−!amanapaK,,U4:m+n+pm+n+pm+n+p!mnpmnp2222qLx2=−x2,Lxy=Lyz=Lxz=0,Ly2=−y2,Lz2=−z2,dL=−x2dx−y2dy−z2dz<0!mnpmanapamnpm+n+p3,,?lnfk4§=fk4am+n+pm+n+pm+n+p(m+n+p)m+n+pmnpx+y=ax+z=ay+z=aqf=xnz(x,y,z)ªu>.§f→0§f4:z=0y=0x=0´§:.223.¦ýx+3y=12Sn/§¦Ù.>²1uý¶§¡È.22xy)µduK¥n/Suý√+=1´n/§.>²1u¶(23)24 课后答案网www.khdaw.comÙ.>¤éº:7´á¶þýº:(0,±2)n/,º:I(x,y)(x,y>0)§KÙSn/.>2x§py+2n/nº:IA(0,−2),B(x,y),C(−x,y)§dýé¡5§A(0,2),B(x,−y),C(−x,−y)´Ùº:22KS=x(y+2)§:(x,y)3ýx+3y=12þ22qÏd¯K´¦S=x(y+2)3^x+3y=12þ(x,y>0)22¼êL=x(y+2)+λ(x+3y−12)x=3Lx=y+2+2λx=0y=1K)§Ly=x+6λy=0§221Lλ=x+3y−12=0λ=−2u´Ùº:IA(0,2),B(3,−1),C(−3,−1)½A(0,−2),B(3,1),C(−3,1)Ïd¯K¢S¯K§73§K3(0,2),(3,−1),(−3,−1)½(0,−2),(3,1),(−3,1)?١ȧÙ9.24.Á¦Ôy=4xþ:§¦§x−y+4=0åC.12)µ¤¦:I(x,y)§K§åld=√|x−y+4|§Ù¥y=4x2x−y+4=0ò²¡©!müÜ©§¡x−y+4<0§m¡x−y+4>012Ôy=4x3m¡Ü©§Ï:(x,y)§åld=√(x−y+4)212-L=√(x−y+4)+λ(y−4x)21Lx=√−4λ=0x=12y=2K)§|1§Ly=−√+2λy=012λ=√242Lλ=y−4x=021dy222qLx2=0,Ly2=√,Lxy=0,dL(1,2)=Lx2dx+2Lxydxdy+Ly2dy=√>02222(1,2)4:§=:(1,2)ålC.225.Ô¡z=x+y²¡px+y+z=1¤ý§¦:ùýáål.)µâK¿§¦åld=x2+y2+z23^z=x2+y2,x+y+z=1222=¦u=x+y+z3^e22222L=x+y+z+λ(z−x−y)+γ(x+y+z−1)√√−1+3−1−3x1=x2=2√2√Lx=2x−2λx+γ=0−1+3−1−3Ly=2y−2λy+γ=0y1=y2=√2√2K)§|Lz=2z+λ+γ=0§z1=2−3,z2=2+322√√Lλ=z−x−y=0−53+953+9Lλ1=λ2=γ=x+y+z−1=03311√11√γ1=−7+3γ2=−7−3p√p√33u´d(x1,y1,z1)=9−53,d(x2,y2,z2)=9+53â¯K¢S¿Â§!á√ål3√!1+31+3√p√Kål::−,−,2+3ål§9+53¶22√√!−1+3−1+3√p√áål::,,2−3ål§9−53.226.¦m:(a,b,c)²¡Ax+By+Cz+D=0áål.p)µ(x,y,z)²¡Ax+By+Cz+D=0þ?:§K§(a,b,c):åld=(x−a)2+(y−b)2+(z−c)2§Ù¥(x,y,z)÷vAx+By+Cz+D=02Ïd>0§d⇐⇒d2222UK§A¦d=(x−a)+(y−b)+(z−c)3^Ax+By+Cz+D=0e4222-L=(x−a)+(y−b)+(z−c)+λ(Ax+By+Cz+D) 课后答案网www.khdaw.com2351x=a−λA2Lx=2(x−a)+λA=01Ly=b−λBK)§|y=2(y−b)+λB=0§2Lz=2(z−c)+λC=01z=c−λCLλ=Ax+By+Cz−D=022(Aa+Bb+Cc+D)λ=A2+B2+C2|Aa+Bb+Cc+D|u´d=√A2+B2+C2qx,y,z¥k?ªu∞§d→∞§Ïd3(x,y,z)(x−a)2+(y−b)2+(z−c)20,F(x0,y0−b)<0d^(1)§F(x,y)3«DþëY§Ï3η>0§¦|x−x0|<秽kF(x,y0+b)>0,F(x,y0−b)<0@"é∀x∈O(x0,η)§d¼êF(x,y)3[y0−b,y0+b]ëY59F(x,y0+b)>0,F(x,y0−b)<0â":3½n§73y∈(y0−b,y0+b)§¦F(x,y)=0duF(x,y)3[y0−b,y0+b]îüN§ly>y§F(x,y)>0¶y0(ε0,F(x1,y1−ε)<0KdFëY5§3O(x1,δ)⊂O(x0,a)§¦x∈O(x1,δ)§ðkF(x,y1+ε)>0,F(x,y1−ε)<0u´â":3½n§7ky∈O(y1,ε)§¦F(x,y)=0=y=f(x)=|x−x1|<δ§Òk|f(x)−f(x1)|=|y−y1|<ε=y=f(x)3x1:ëYdx1∈O(x0,a)?¿5§f(x)ëY¼ê.2222.¼êF(x,y)≡y−x(1−x)=03=:C/û½üëY§këYê¼êy=y(x).2223)µ¼êF(x,y)=y−x(1−x)3mëY§§êFx=4x−2x,Fy=2yëY22222dy−x(1−x)=0§ey=0§Kx(1−x)=0§)x=0,x=±1222qy>0,x>0§1−x>0=−16x61y6=0§Fy6=0dÛ¼ê3½n1§3?Û÷v{(x,y)|x|<1,x6=0,y2−x2(1−x2)=0}C/û½üëYkëYê¼êy=y(x).03.y²k¼êy=y(x)÷v§siny+sinhy=x§¿¦Ñêy(x).y²µ¼êF(x,y)=siny+sinhy−x3mëY§Fx=−1,Fy=cosy+coshyëYy−ye+eqcoshy=>1Ò3y=0¤á§dcosy=1§3¹e|cosy|612Ké:(x,y)§ðkFy=cosy+coshy>0§u´Fy6=0dÛ¼ê3½n1§3?Û÷vþã§:(x,y)§k¼ê÷v§siny+sinhy=xFx10Ù¼êy=−=.Fycosy+coshy4.D´:P0:(x0,y0,z0,u0,v0)§e(1)F(x0,y0,z0,u0,v0)=0,G(x0,y0,z0,u0,v0)=0¶(2)F,G"uCþê3D¥ëY¶FuFv(3)J=G3P0:Ø"¶uGvK3(x0,y0,z0)RS3é¼êu=f(x,y,z);v=g(x,y,z)÷vµ 课后答案网www.khdaw.com237(1)u0=f(x0,y0,z0),v0=g(x0,y0,z0)(2)F(x,y,z,f,g)≡0,G(x,y,z,f,g)≡0(3)u=f(x,y,z),v=g(x,y,z)3:(x0,y0,z0)RSkéCþê§∂f1D(F,G)∂f1D(F,G)∂f1D(F,G)=−,=−,=−∂xJD(x,v)∂yJD(y,v)∂zJD(z,v)∂g1D(F,G)∂g1D(F,G)∂g1D(F,G)=−,=−,=−∂xJD(u,x)∂yJD(u,y)∂zJD(u,z)y²µd^(3)§Fu,Fv¥k3P0:Øu0ØFv(P0)6=0§KdÛ¼ê3½n2§3P0:,S±rvlF(x,y,z,u,v)=0¥)Ñ5.v=ϕ(x,y,z,u)v0=ϕ(x0,y0,z0,u0)3(x0,y0,z0,u0),S´§äk"ux,y,z,uëYêrv=ϕ(x,y,z,u)G(x,y,z,u,v)¥G(x,y,z,u,ϕ(x,y,z,u))=ψ(x,y,z,u)!FuJψu=Gu+Gv·vu=Gu+Gv−=−FvFvdbFv(P0)6=03P0:J6=0§ψu(x0,y0,z0,u0)6=0Kd½n2§3(x0,y0,z0,u0),Sl§G=G(x,y,z,u,ϕ)≡ψ(x,y,z,u)=0¥)Ñu5.u=f(x,y,z)§3(x0,y0,z0),SkëYê§u0=f(x0,y0,z0)ru=f(x,y,z)ϕ(x,y,z,u)¥v=ϕ(x,y,z,f(x,y,z))=g(x,y,z)Kkg(x0,y0,z0)=ϕ(x0,y0,z0,u0)=v0u=f(x,y,z),v=g(x,y,z)=¤¦∂F∂F∂f∂F∂gF(x,y,z,u,v)=0∂x+∂u·∂x+∂v·∂x=0é§|üà"ux¦§G(x,y,z,u,v)=0∂G∂G∂f∂G∂g+·+·=0∂x∂u∂x∂v∂x∂f1D(F,G)∂g1D(F,G))§=−,=−∂xJD(x,v)∂xJD(x,u)∂f1D(F,G)∂f1D(F,G)∂g1D(F,G)∂g1D(F,G)Ón=−,=−,=−,=−∂yJD(y,v)∂zJD(z,v)∂yJD(u,y)∂zJD(u,z)5.ϕi(x)(i=1,2,···,n)´xëY¼ê§Gi(x1,···,xn)=Fi(ϕ1(x1),···,ϕn(xn))∂(G1,G2,···,Gn)Qn0y²=∆(ϕ(x1),···,ϕn(xn))ϕi(xi)∂(x1,x2,···,xn)i=1D(F1,F2,···,Fn)Ù¥∆(x1,x2,···,xn)=D(x1,x2,···,xn)Qn0000ϕi(xi)=ϕ1(x1)ϕ2(x2)···ϕn(xn).i=1y²µÏϕi(x)(i=1,2,···,n)´xëY¼ê§Gi(x1,···,xn)=Fi(ϕ1(x1),···,ϕn(xn))∂Gi∂Fi∂ϕj∂Fi0K=·=ϕi(xj)(i,j=1,2,···,n)∂xj∂ϕj∂xj∂ϕj∂G1∂G1∂G1∂F10∂F10∂F10···∂ϕϕ1(x1)ϕ2(x2)···ϕn(xn)∂x1∂x2∂xn1(x1)∂ϕ2(x2)∂ϕn(xn)∂G2∂G2∂G2∂F2∂F2∂F2∂(G1,G2,···,Gn)···ϕ10(x1)ϕ20(x2)···ϕn0(xn)u´=∂x1∂x2∂xn=∂ϕ1(x1)∂ϕ2(x2)∂ϕn(xn)=∂(x1,x2,···,xn).......................................................................................∂Gn∂Gn∂Gn∂Fn∂Fn∂Fn···000ϕ1(x1)ϕ2(x2)···ϕn(xn)∂x1∂x2∂xn∂ϕ1(x1)∂ϕ2(x2)∂ϕn(xn)∂F1∂F1∂F1···∂ϕ1(x1)∂ϕ2(x2)∂ϕn(xn)∂F2∂F2∂F2···QnD(F1,F2,···,Fn)ϕ0000=1(x1)ϕ2(x2)···ϕn(xn)∂ϕ1(x1)∂ϕ2(x2)∂ϕn(xn)=ϕi(xi)......................................i=1D(ϕ1(x1),ϕ2(x2),···,ϕn(xn))∂Fn∂Fn∂Fn···∂ϕ1(x1)∂ϕ2(x2)∂ϕn(xn)Qn0∆(ϕ(x1),···,ϕn(xn))ϕi(xi).i=16.F(x,y,z)këY꧿dF(x,y,z)=0(½z=f(x,y).?Øz=f(x,y)47Ú¿©^ 课后答案网www.khdaw.com.2¦d222x+y+z−2x+2y−4z−10=0¤(½z=f(x,y)4.zx=fx(x,y)=0y²µÏ¼êz=f(x,y)47^zy=fy(x,y)=0FxFyFx=0qzx=−,zy=−§KF(x,y,z)47^FzFzFy=0qÛ¼ê4¿©^w¼êaÓ§´¦ê^Û¼êpê¦{Fx=2x−2=0x=1-)§éAzz1=−2,z2=6Fy=2y+2=0y=−12222222∂zx−1∂z1+y∂z(x−1)+(2−z)∂z(1+y)+(2−z)∂z(x−1)(1+y)q=,=,=,=,=∂x2−z∂y2−z∂x2(2−z)3∂y2(2−z)3∂x∂y(2−z)3222∂z1∂z1∂z1111u´3:(1,−1,−2),=,=,=0§d·−0=>09>0§Kz1=−24¶∂x24∂y24∂x∂y4!4!164222∂z1∂z1∂z11113:(1,−1,6),=−,=−,=0§d−−−0=>09−<0§Kz2=64∂x24∂y24∂x∂y44164 课后答案网www.khdaw.com239§2.¼ê1ª5!¼ê"x=rcosθcosϕ1.y²y=rcosθsinϕ¼êÕáz=rsinθcosθcosϕ−rsinθcosϕ−rcosθsinϕD(x,y,z)y²µÏ=cosθsinϕ−rsinθsinϕrcosθcosϕ=−r2cosθD(r,θ,ϕ)sinθrcosθ0πD(x,y,z)K3r6=0θ6=kπ+«DS6=02D(r,θ,ϕ)u´â½n5§¼ê|3«DS¼êÕá.y1=x1+x2+···+xn2222.y²y2=x1+x2+···+xn¼ê"§¿ÑÙ¼ê"Xª.y3=x1x2+x1x3+···+xn−1xn1122y²µÏ3¼êϕ(t1,t2)=(t1−t2)§¦y3=ϕ(y1,y2)=(y1−y2)3nm(x1,x2,···,xn)S22¤ðª12K¼ê|3nm¥¼ê"§Ù¼ê"Xªy3=(y1−y2).23.e¼ê´Ä"ºx−yy−zz−x(1),,x−zy−xz−yxyz(2),,1−x−y−z1−x−y−z1−x−y−z)µ(1)Ïf1·f2·f3=−1§K¼ê".xyz(2)-f1(x,y,z)=,f2(x,y,z)=,f3(x,y,z)=1−x−y−z1−x−y−z1−x−y−z∂f1∂f1∂f11−y−zxx∂x∂y∂z(1−x−y−z)2(1−x−y−z)2(1−x−y−z)2∂f2∂f2∂f2y1−x−zyKJocobiÝ∂x∂y∂z=(1−x−y−z)2(1−x−y−z)2(1−x−y−z)2∂f3∂f3∂f3zz1−x−y∂x∂y∂z(1−x−y−z)2(1−x−y−z)2(1−x−y−z)2qdÝ3§Kâ½n6§¼ê|¼êÕá. 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课后答案网www.khdaw.com267p(3)Ïx2+y2+z2=2cz§Kx2+y2+(z−c)2=c2,z=c+c2−x2−y2xypcu´zx=−p,zy=−p§K1+zx2+zy2=pc2−x2−y2c2−x2−y2c2−x2−y2ZZZZZZc2πccr2u´dS=pdσ=dθ√dr=2πc.c2−x2−y200c2−r2Sσ(4)©üÜ©µppp√1Ü©µz=1,1+zx2+zy2=1¶1Ü©µz=x2+y2,1+zx2+zy2=2ZZZZZZ2π12π1√π√2233K(x+y)dS=dθrdr+dθ2rdr=(1+2).00002S(5)x=Rcosθ,y=Rsinθ,z=z(06θ2π,06z6H)2222222KE=xθ+yθ+zθ=R,FZZ=xθxz+Zyθyz+Zzθzz=0,G=xz+yz+zz=1√dS2πHRHu´EG−F2=R§l=dθdz=2πarctan.r200R2+z2RS1223.¦Ô¡z=(x+y),06z61þ.dÝρ=z.21pp)µÏz=(x2+y2)§Kz2222x=x,zy=y§u´1+zx+zy=1+x+y2ZZZZZZ√√1p12π2p2(1+63)KþM=ρdS=(x2+y2)1+x2+y2dxdy=dθr31+r2dr=π.220015Sx2+y262 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课后答案网www.khdaw.com2792222(1)(x+2xy−y)dx+(x−2xy−y)dy22(2)(2xcosy−ysinx)dx+(2ycosx−xsiny)dyab−by−ax(3)dx+dy+dzzzz2222(4)(x−2yz)dx+(y−2xz)dy+(z−2xy)dzxyxy(5)e[e(x−y+2)+y]dx+e[e(x−y)+1]dy)µ∂P∂Q∂P∂Q(1)Ï=2x−2y,=2x−2y§K=∂y∂x∂y∂xZxZyx3y322222u´u=(x+2xy−y)dx+(−y)dy+C=+xy−xy−+C0033222222(2)Ï(2xcosy−ysinx)dx+(2ycosx−xsiny)dy=cosydx+xdcosy+ydcosx+cosxdy=22d(xcosy+ycosx)22Ku=xcosy+ycosx+Cab−by−axzdx−xdzzdy−ydzax+by(3)Ïdx+dy+dz=a+b=dzzz2z2z2z1Ku=(ax+by)+Cz1222333(4)Ï(x−2yz)dx+(y−2xz)dy+(z−2xy)dz=(dx+dy+dz)−2(yzdx+xzdy+xydz)=!3333x+y+zd−2xyz31333Ku=(x+y+z)−2xyz+C3xyxyx+yx+yx(5)Ïe[e(x−y+2)+y]dx+e[e(x−y)+1]dy=(x−y)e(dx+dy)+2edx+d(ye)=x+yx+yxx+yxd((x−y)e)+ed(x+y)+d(ye)=d((x−y+1)e)+d(ye)x+yxKu=(x−y+1)e+ye+C4.yµ1xdy−ydxPdx+Qdy=2Ax2+2Bxy+Cy22(A,B,C~ê§AC−B>0)·Ü^µ∂P∂Q=∂y∂x¦µ"uÛ:(0,0)Ì~ê.yxy²µÏP=−,Q=2(Ax2+2Bxy+Cy2)2(Ax2+2Bxy+Cy2)22∂PCy−Ax∂QK==∂y2(Ax2+2Bxy+Cy2)2∂xIZπB
π22dt1Ctant+Cu´ω=Pdx+Qdy==√arctan√2+y2=1π2(Acos2t+2Bsintcost+Csin2t)AC−B2AC−B2x−π2−2π√,C>0=AC−B2π−√,C<0AC−B25.y²µZxdx+ydyx2+y2xdx+ydy"uÛ:(0,0)Ì~ê0§lÈ©´»Ã".x2+y2xy∂P2xy∂Qy²µÏP=,Q=§K=−=Ix2+y2x2+y2∂y(x2+y2)2∂xu´ω=Pdx+Qdy=0x2+y2=1 课后答案网www.khdaw.com(1)e4´lØ(0,0):§òÛ:(0,0)«D>.^^Cëå5§u´EëÏ«C¤üëÏ«I∂P∂Qxdx+ydyq=§Kdd^§=0∂y∂xlx2+y2Ixdx+ydy(2)e4´lÛ:(0,0)§Ï÷7Û:?4´È©uÌ~ê§K=0x2+y2lxdx+ydyoÈ©´»Ã".x2+y2 课后答案网www.khdaw.com281§3.|ØÐÚt−t1.H(t)=ea+eb§Ù¥a,b~þ§tëê§dH(1)¦dt2dH(2)y²=Hdt2t−t)µÏH(t)=ea+eb§a,b~þ§KdHt−t(1)=ea−ebdt!2dHddHt−t(2)==ea+eb=Hdt2dtdt!!ddAdBdC2.y²µ[A·(B×C)]=·(B×C)+A·×C+A·B×dtdtdtdty²µA=Axi+Ayj+Azk,B=Bxi+Byj+Bzk,C=Cxi+Cyj+CzkAxAyAzKA·(B×C)=BxByBz§éªüধmà^é1ª¦{K§CxCyCzAxAyAzAxAyAzAxAyAzdttt[A·(B×C)]=BxByBz+BxtBytBzt+BxByBzdtCxCyCzCxCyCzCxCyCz!!tttdAdBdC=·(B×C)+A·×C+A·B×dtdtdt3.a=3i+20j−15k§éeêþ|φ©O¦Ñgradφ9div(φa).222−1(1)φ=(x+y+z)2222(2)φ=x+y+z222(3)φ=ln(x+y+z))µ−x−y−z(1)gradφ=φxi+φyj+φzk=i+j+k333(x2+xy+y2)2(x2+xy+y2)2(x2+xy+y2)2−3x−20y+15zdiv(φa)=φdiva+gradφ·a=gradφ·a=3(x2+xy+y2)2(2)gradφ=2xi+2yj+2zk,div(φa)=6x+40y−30z2x2y2z6x+40y−30z(3)gradφ=i+j+k,div(φa)=x2+y2+z2x2+y2+z2x2+y2+z2x2+y2+z24.U(x,y,z)=xyz(1)¦U(x,y,z)3:P1(0,0,0),P2(1,1,1)9P3(2,1,1)?÷b=2i+3j−4kê¶(2)3þãn:?§U(M)êÛº(3)3þãn:?§¦divgradU(M)9rotgradU(M).)µ234(1)Ïb{ucosα=√,cosβ=√,cosγ=−√292929∂U1K=yzcosα+xzcosβ+xycosγ=√(2yz+3xz−4xy)∂b29√∂U∂U29∂Uu´3P1(0,0,0):=0¶3P2(1,1,1):=¶3P3(2,1,1):=0∂b∂b29∂b∂U(2)Ï=gradU·b0=|gradU|cos(gradU,b0)§Ù¥b0´büþ∂bpKU(M)ê|gradU|=y2z2+x2z2+x2y2√u´3P1(0,0,0):|gradU|=0¶3P2(1,1,1):|gradU|=3¶3P3(2,1,1):|gradU|=3 课后答案网www.khdaw.com(3)ÏgradU=yzi+xzj+xyk∂(yz)∂(xz)∂(xy)KdivgradU=++=0∂x∂y!∂z!!∂(xy)∂(xz)∂(yz)∂(xy)∂(xz)∂(yz)rotgradU=−i+−j+−k=0∂y∂z∂z∂x∂x∂yu´3þãn:?§divgradU(M)=0,rotgradU(M)=0.2222225.¦þa=xi+yj+zkBL¥x+y+z=1,x>0,y>0,z>06þ.ZZZZZπZ1222222π)µΦ=xdydz+ydxdz+zdxdy§zdxdy=dθ(1−r)rdr=008SSZZZZπ322aq/§©OXOZ,YOZ²¡ÝK§ydxdz=xdydz=§u´Φ=π.88SS6.¦a=yzi+xzj+xykÏLS6þ§222(1)SÎNx+y6a,06z6hý¡¶(2)S(1)¥ÎNþ.¡¶(3)S(1)¥ÎNL¡.)µZZZZZZZZ"#∂(yz)∂(xz)∂(xy)(3)andS=divadV=++dV=0∂x∂y∂zSVVu´þaBLÎNL¡6þ0ZZZZZZ2πa3(2)Ï3ÎNþ!e.¡an=xy§KandS=xydxdy=dθrsinθcosθdr=000Sþx2+y26a2ZZÓnandS=0§u´þaBLÎNþ.¡6þ0ZZSeZZZZZZZZ(1)ÏandS=andS+andS+andS§KandS=0.SSýSþSeSý!y7.¦a=gradarctan÷l6þµx22(1)l(x−2)+(y−2)=1,z=0¶22(2)lx+y=4,z=1.)µ!yyx(1)d®§ka=gradarctan=−i+jxx2+y2x2+y2xy∂x2+y2∂−x2+y2Krota=−k=0(Øx=y=0=Oz¶þ:)∂x∂y22Ïl:(x−2)+(y−2)=1,zI=0´ØZZ7z¶§ulþÜ¡S§¦SOz¶ØKâd÷iúª§k6þadl=n·rotadS=0lS22(2)Ïl:x+y=4,z=1§dlÐ7Oz¶^=±§~êc>0¿©(c<2)§¦lu²222¡z=cþ§3²¡z=cþ7Oz¶±lr:x+y=r,z=c§r¿©§¦ru2§±llr>.ÜþI¡S§¦SIOz¶ØZZdd÷iúª§a·dr+a·dr=n·rotadS=0§Ù¥−lrL«÷^l−lrIISu´6þa·dr=a·drllrIZ2πqlrëê§x=rcosθ,y=rsinθ,z=c§a·dr=dθ=2πIlr0la·dr=2π.l 课后答案网www.khdaw.com2838.¦þa=−yi+xj+ck(c~ê)6þµ22(1)÷±x+y=1,z=0¶22(2)÷±(x−2)+y=1,z=0.)µ22(1)Ïl:x+y=1,z=0§Kl=costi+sintj+0k(06t62π)IZ2πu´a·dl=dt§l¤¦6þa·dl=dt=2πl022(2)Ïl:(x−2)+y=1,z=0§Kl=(2+cost)i+sintj+0k(06t62π)IZ2πu´a·dl=(2cost+1)dt§l¤¦6þa·dl=(2cost+1)dt=2πl09.y²µ(1)rot(uA)=urotA+gradu×A¶(2)div(φa)=φdiva+gradφ·a(3)graddiva−rotrota=∆ay²µ!!∂Az∂Ay∂u∂u(1)Ïrotx(uA)=u−+Az−Ay=urotxA+(gradu×A)x∂y∂z∂y∂zÓ{§roty(uA)=urotyA+(gradu×A)y,rotz(uA)=urotzA+(gradu×A)zu´rot(uA)=urotA+gradu×A∂(φax)∂ax∂φ∂(φay)∂ay∂φ∂(φaz)∂az∂φ(2)Ï=φ+ax,=φ+ay,=φ+az∂x∂x∂x∂y∂y∂y∂z∂z∂zKdiv(φa)=φdiva+gradφ·a!∂ax∂ay∂az(3)Ïgraddiva=grad++∂x∂y∂z!!!222222222∂ax∂ay∂az∂ax∂ay∂az∂ax∂ay∂az=++i+++j+++k∂x2∂x∂y∂x∂z∂x∂y∂y2∂y∂z∂x∂z∂y∂z∂z2"!!!#∂az∂ay∂ax∂az∂ay∂axrotrota=rot−i+−j+−k∂y∂z∂z∂x∂x∂y!!!222222222222∂ay∂az∂ax∂ax∂ax∂az∂ay∂ay∂ax∂ay∂az∂az=+−−i++−−j++−−k∂x∂y∂x∂z∂y2∂z2∂x∂y∂y∂z∂x2∂z2∂x∂z∂y∂z∂x2∂y2Kgraddiva−rotrota=∆axi+∆ayj+∆azk=∆a10.¦rot(w×r)§w~¥þ§r¥»þ.)µw=(w1,w2,w3),r=(x,y,z)ijku´w×r=w1w2w3=(w2z−w3y)i+(w3x−w1z)j+(w1y−w2x)kxyzrot(w×r)=2w1i+2w2j+2w2k=2w11.y²a=yz(2x+y+z)i+xz(x+2y+z)j+xy(x+y+2z)kÅ|§¿¦Ù³¼ê.y²µém?:(x,y,z)§k()∂∂rota=[xy(x+y+2z)]−[xz(x+2y+z)]i∂y∂z()∂∂+[yz(2x+y+z)]−[xy(x+y+2z)]j∂z∂x()∂∂+[xz(x+2y+z)]−[yz(2x+y+z)]k∂x∂y=0KaÅ|du³φ÷vdφ=a·dl=axdx+aydy+azdz=d[(xyz(x+y+z)]KÙ³¼êu(x,y,z)=xyz(x+y+z)+C§Ù¥C?¿~ê. 课后答案网www.khdaw.com12.¦þa=r÷Úr=acosti+asintj+btk(06t62π)ã¤õ.)µÏa=xi+yj+Zzk,l:x=acost,y=asint,z=bt(062π)22K¤¦õW=xdx+ydy+zdz=2bπl13.φ¼ê§Oµgradφ(r),div(φ(r)r)9rot(φ(r)r).r00)µgradφ(r)=φ(r)gradr=φ(r)·r0div(φ(r)r)=φ(r)divr+r·gradφ(r)=3φ(r)+rφ(r)rot(φ(r)r)=φ(r)rotr+gradφ(r)×r=014.¦÷v^div(φ(r)r)=0¼êφ(r).0)µdþK§div(φ(r)r)=3φ(r)+rφ(r)0φ(r)30¦div(φ(r)r)=0§3φ(r)+rφ(r)=0==−φ(r)rcKφ(r)=(c~ê)r315.¦±eþÑÝ9^Ý(a,b~þ)µ(1)(a·r)b(2)a×r(3)φ(r)(a×r)(4)r×(a×r))µ(1)div[(a·r)b]=(a·r)divb+grad(a·r)·b=grad(a·r)·b=a·brot[(a·r)b]=(a·r)rotb+grad(a·r)×b=grad(a·r)×b=a×b(2)Ï(a×r)x=ayz−azy,(a×r)y=azx−axz,(a×r)z=axy−ayx∂∂∂Kdiv(a×r)=(ayz−azy)+(azx−axz)+(axy−ayx)=0∂x∂y∂zijk∂∂∂rot(a×r)==2a∂x∂y∂zayz−azyazx−axzaxy−ayx0(3)div[φ(r)(a×r)]=φ(r)div(a×r)+grad(φ(r))(a×r)=φ(r)(r·rota−a·rotr)+(rφ(r)·(a×r)=0rφ(r)0φ(r)·(a×r)=[r·(a×r)]=0rrrot[φ(r)(a×r)]=φ(r)rot(a×r)+grad(φ(r))×(a×r)=!0rφ(r)02φ(r)[−(axi+ayj+azk)+3a]+φ(r)×(a×r)=2φ(r)a+[ra−(a·r)r]rr2(4)Ïr×(a×r)=|r|a−(a·r)r22Kdiv[r×(a×r)]=|r|diva+grad|r|·a−[(r·a)divr+r·grad(a·r)]=(ax+ay+az)−4(xax+yay+zaz)122rot[r×(a×r)]=|r|rotr+grad|r|×a−(r·a)rotr−grad(r·a)×r=(r×a)−a×r.r16.y²±eªµ(1)grad(a·b)=a×rotb+b×rota+(b·∇)a+(a·∇)b(2)rot(a×b)=(b·∇)a−(a·∇)b+(divb)a−(diva)b(3)c·grad(a·b)=a·(c·∇)+b·(c·∇)a(4)(c·∇)(a×b)=a×(c·∇)b−b×(c·∇)ay²µ(1)grad(a·b)=grad(axbx+ayby+azbz)=!∂bx∂ax∂by∂ay∂bz∂azax+bx+ay+by+az+bzi+∂x∂x∂x∂x∂x∂x!∂bx∂ax∂by∂ay∂bz∂azax+bx+ay+by+az+bzj+∂y∂y∂y∂y∂y∂y 课后答案网www.khdaw.com285!∂bx∂ax∂by∂ay∂bz∂azax+bx+ay+by+az+bzk∂z∂z∂z∂z∂z∂z=a×rotb+b×rota+(b·∇)a+(a·∇)b!!∂∂∂∂∂∂(2)rot(a×b)=bx+by+bza−ax+ay+azb+(divb)a−(diva)b∂x∂y∂z∂x∂y∂z=(b·∇)a−(a·∇)b+(divb)a−(diva)b(3)c·grad(a·b)=!∂bx∂ax∂by∂ay∂bz∂azcxax+bx+ay+by+az+bz+∂x∂x∂x∂x∂x∂x!∂bx∂ax∂by∂ay∂bz∂azcyax+bx+ay+by+az+bz+∂y∂y∂y∂y∂y∂y!∂bx∂ax∂by∂ay∂bz∂azczax+bx+ay+by+az+bz∂z∂z∂z∂z∂z∂z=a·(c·∇)+b·(c·∇)a!∂∂∂(4)(c·∇)(a×b)=cx+cy+cz[(aybz−azby)i+(azbx−axbz)j+(axby−aybx)k]∂x∂y∂z∂ay∂bz∂az∂by∂ay∂bz∂az∂by=bzcx+aycx−bycx−azcx+bzcy+aycy−bycy−azcy+∂x∂x∂x∂x∂y∂y∂y∂y!∂ay∂bz∂az∂bybzcz+aycz−bycz−azczi+∂z∂z∂z∂z∂az∂bx∂ax∂bz∂az∂bx∂ax∂bzbxcx+azcx−bzcx−axcx+bxcy+azcy−bzcy−axcy+∂x∂x∂x∂x∂y∂y∂y∂y!∂az∂bx∂ax∂bzbxcz+azcz−bzcz−axczj+∂z∂z∂z∂z∂ax∂by∂ay∂bx∂ax∂by∂ay∂bxbycx+axcx−bxcx−aycx+bycy+axcy−bxcy−aycy+∂x∂x∂x∂x∂y∂y∂y∂y!∂ax∂by∂ay∂bxbycz+axcz−bxcz−ayczk∂z∂z∂z∂z=a×(c·∇)b−b×(c·∇)a217.ÁydivgradsinrL«¤F(r)/ª§¿ÑF(x).222222∂sinr∂sinr∂sinr22y²µdivgradsinr=∆sinr=++=!!∂x2∂y2!∂z2∂x∂y∂z12sin2r·+sin2r·+sin2r·=2cos2r+2sin2r·=(sin2r+rcos2r)∂xr∂yr∂zrrr2=F(r)=(sin2r+rcos2r)r22dsinx2KF(x)=divgradsinxr==2cos2xdx2218.y²µ|a|=~ê§k(a·∇)a=−a×rota.22y²µÏ|a|=~ê§Kgrad|a|=0d16K(1)§grad(a·a)=2[a×rota+(a·∇)a]=0K(a·∇)a=−a×rota.'