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ID:34428915
大小:152.50 KB
页数:13页
时间:2019-03-06
《数学物理方法答案2new》由会员上传分享,免费在线阅读,更多相关内容在教育资源-天天文库。
1、16、试求下列级数的收敛半径。nn!n!nza.z;b.nz;c.nnab0,0。n0n0nn0aibn1zn1!zn!nn!解:a.当limlimlimz1时,级数收敛。nnzznn!!nnn!当limz1时,级数发散。n亦即当z1时,级数收敛。而当z1时,级数发散。于是收敛半径R1。nn1nnnn!1nn!1n1b.Rlimlimlimlim1e。n1nnnnnn1!1nn
2、1!nnnn11nn22nn2nc.Rlim,Ralimnibalimb。nnnnan111又因为maxab,a22nnb2n22nmaxab,,且lim22n1,n122nn2n故limabmaxa,b。n于是所求级数的收敛半径Rmaxab,。22nn22aab或:n,RlimRlim。22nnnanabn12n22bab22nn22aba当ab时,Rlimlima,
3、22nn2nnnabb1a2na22ab22nn22abb当ab时,Rlimlimb,22nn2nnnaba1bRmaxab,17、将下列函数按z的幂展开,并指明收敛范围1b.2。cosznn2n22nn12zz1221解:b.coszz1cos2,cos2zz,2nn002!nn2!n212nn112z2。coszz22n0n!18
4、、将下列函数按z1的幂展开,并指出收敛范围。zza.cosz;b.;c.。2z2zz25解:a.coszzcos11cos1cosz1sin1sinz1。nn2nn2111z11zcosz1,sinz1,n02!nn021!nnn22nn111zz11coszcos1sin1z1。nn002!nn21!n2nn21n1cos1
5、cos1,1sin1cos1。2222nn1cos1cos12222nn1coszz1z1nn002!nn21!ncos12nz1z1。n0n!nnnn或:令fzzcos,则fzcosz,f1cos1,22nncos1f1nn2所以coszz1z1z1。n
6、n00nn!!z221b.11zz223z113nn21zz1nn11121n1z1333nn0032zz11z11c.2222zz25zzz141414z11112244zz1111222z11nn令t,11tt21tn02nnn211z11znz121n
7、,11z2z1nn0024212nn22nnzz1111zz11从而2nnzz254nn00444nn21nn211zz11nn11nn0044n122nn1n1zz11z124n0nn21n2n11zz11进一步,nn11nn00441111nnnn1122
8、122nnnnn12zz1131nz1nnn奇数22偶数0n22111nnz122n所以231nz1z12。zz25n0n2219、将下列函数在指定的环域内展成罗朗级数。z1a.,01z,1z2zz(1)zz11212解:a.。2222zz(1)(
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