an?12n?1n2?4??lim?lim??2.nn??an???n?1?2?42nR?1
所以,级数②的收敛半径是
??n1.2
??1??1?11t??t???22n?4也收敛.所以级数n?4n?022n?0又当时,级数②化为收敛;又当时,级数②化为
?11?t???,??22?. ②的收敛域是
?11??13??13?,?.2x?1???,?x??,???22?解得?44?,故原级数的收敛域为?44? (二)由
?x2n?1x2??x???1?n|x|?22n?12(1)如果,即时,则收敛; ?x2n?1x2??x???1?n|x?22n?12(2)(1)如果,即时,则发散,
所以,R?2.
??11?2n?1?(3)又在端点x??2处所以,收敛域为
发散.
??2,2
?
三、计算题(每小题5分,共45分)
f?x???ln?1??f?x?sin2x?lim?x?5lim23?141.已知x?0①,求x?0x.
解:由①式得
f?x???f?x?f?x?ln?1??sin2x?2x?lim2x5?lim?x?limsinx?0x?0exln3?1x?0xln33?1
?1f?x?lim2.2ln3x?0x②
由②式即可算得
f?x?lim2?10ln3.x?0x
?x?x?t?,?t2?y??ln(1?u)du???y?yx0?42.设函数由参数方程确定,其中x?x?t?是微分方程 d2ydx?x?2te?0x|?02dt在初始条件t?0下的特解,求dx.
dx?2te?x?0解:(一)微分方程dt为可分离变量型,可转化为 exdx?2tdt①
①两边积分得
xx2edx?2tdt?e?t?C??②
又将初始条件
x|t?0?0代入②,得C?1,因此
x?t??ln1?t2③
??dydy?dxdt(二)
dxln(1?t2).2t??1?t2ln1?t22tdt1?t2
????d2yd?dy?d?dy?1??????.2dx?dx?dt?dx?dxdxdt (三)
d1?t2ln1?t21?.?2tdt221?t2?1?t?1?ln(1?t).
????????22z?xf2x?y,ysinx,x?2,其中f具有二阶连续偏导数,求 43.设函数
???z?2z;2.?x?y
解: (一) ?z?2xf?x2?f1?.2?f2?ycosx?f3?2x??x (二)
?z?x2??f1??f2?sinx??y,所以
?2z2????1??f12??sinx??sinx?f21????1??f22??sinx?????f11?x2?y
44.计算反常积分解:
???01dx?x?2??x?3?
1?11x?2?1dx??dx?dx?dx?ln?c???x?2??x?3?????x?2x?3x?2x?3x?3??
所以
21???1x?2??x?22x?ln2dx?ln?limln?ln?limln|?0?x?2??x?3?x?30x???x?33x???1?33x
23?ln1?ln?ln.32 ?x2?y2?z2?6,?:?.45.求曲线?x?y?z?0.在点?1,?2,1?的切线. 解:方程组两边关于x求导,得:
1dydz?2x?2y?2z?0,??dxdx.??1?dy?dz?0.??dxdx①
将点?1,?2,1?代入(1),得:
dydz?2?4?2?0,||?x?1x?1?dxdx.?dydzdydz?1?|?|?0.?0,||x?1??1.?x?1?dxx?1dxx?1dxdx解之,有:
1,0,?1? 所以,切线向量为:s??x?1y?2z?1??.??1,?2,110?1故曲线在点的切线为:
46.设函数f?x?在正半轴?x?0?上有连续导数f??x?且f?1??2.若 在右半平面内沿任意闭合光滑曲线l,都有
?4xl3ydx?xf?x?dy?0
求函数f?x?.
3??Px,y?4xy,Q?x,y??xf?x?都是右半平面上的连续函数,由于在右半平 解:
面内沿任意闭合光滑曲线l,都有
34x?ydx?xf?x?dy?0l
故有
?P?Q??y?x
即
4x3?f?x??xf??x?
化简,得
f??x??1f?x??4x2x(1)
(1)为一阶线性微分方程,其通解为
11dx??xdx?2?xf?x??e4xe?c?????
1?e?lnx?4x2elnx?c??4x3dx?cx
11?x4?c?x3?c.xx(2)
???????代入条件f?1??2,得
c?1. 故
1f?x??x3?.x
47.求幂级数n?1??n?1?!x?nn?1的和函数.
解:(一)记
an?n?n?1?!,n?1,2,?,则
??liman?1n?1?lim2?0n??an??n?2nns?x????,故收敛半径为R???.收敛域为???,???.
(二)记
nxn?1,!n?1?n?1????x???.
??nn?1??1n?1?1n?1?1n?1s?x???x??x??x??xn?1!!!n?1?n?1?n?1?n?1?n?1n!n?1?n?1?则
?1?1n1??x?2xn?1n!x11?1n1n?1x??x?2?!xn?1n!xn?1?n?1???n!xn?2?1n
1??1n?1??1n????x?1??2??x?1?x?x?n?0n!?x?n?0n!?
1xxex?ex?11x?x?0??e?1?2e?1?x?2xxx.
????xex?ex?1ex1s?0??lims?x??lim?lim?2x?0x?0x?022. x又
所以
?xex??S(x)?????s?x?????x?1,x?0x21,x?02
解法二:记
xnxn?1,!n?1?n?1????x???.
11?11?xnnn?1x??x???0s?x?dx??????n?1!xn?1!xn?2n! n?1n?11?ex?1?xx 所以
?xx?ex?1?x?xex?1?ex?1?xxe?e?1?s?x?????2??xx??x2.
???????48.计算二重积分
I???exdxdy,DD23y?xy?x是第一象限中由直线和曲线所围成封闭区域.
2x??fx,y?e解:因为二重积分的被积函数,它适宜于“先对y,后对x”
?x3?y?x,D:??0?x?1.于是 ,故D可用不等式表示为
I???exdxdy??dx?3exdy??x?x3exdxD0x01x213x221x21??2
11x2112x22??xedx??xedx??0edx??0xde0022
????