数下-工数复习第六章多元函数微分学

第六章元函数微分f(Pf(x，y)

P0(x0y0D的聚点.Ao正数，总存在正数P(x，y)∈DU(P0,(x(xx)2(yy00

|f(P)–A|=|f(x，y)–A|<成立，那么就称常数Af(x，y)当(x，y)(x0,y0

(x,y)(x0,y0

f(xyAf(x，y)→A((x，y)(x0,y0

f(P)

或f(P)→A(PP0(x,y)(x0,y0

f(x,y)

(x0,y0)

zf(xy)Df(xy)Df(xy)D上的连续函数f(xy)P0(x0,y0P0(x0,y0f(xy)的间断点

p

f(P)f(P0)概念z

f(x,y)，z

f(xx,y)f(x, 说 1对x求导视y为常数，几何意义也说明了这个问z=f(xy)M0x0,y0)的偏导数有下述几何意义fx(x0y0zf(xyyy0的交线在点M0处的切线M0x轴的斜率.fy(x0y0z

f(x,y)与平面x=x0的交线M0M0Tyy轴的斜率2(x0,y0

时y0可先代入（因此可能简化函数）x求偏导数存可微 ，偏导数连续可微

fff=f连

z=f(xy)D内具有偏导数z

f(x,y),z

fy(xyz=f(xy)的二阶偏导数。按照对变量求导次序的不同有下列四个二阶偏z2z

z 2z

fxx(x,y),yxxy

fxy(x, z

z 2zxyyx

fyx(x,y),yyy2

fyy(x, zf(x,y)，uu(x,y)，vv(x,zfuf

zfuf u

v

u

vd(uv)du

d(uv)udv

duvduv F(x,y,z)

zz(xyzFxz

x y

(x,y,）z

xx0

yy0z

法平面方xt(xx0ytyy0zt(zz0xy y

z

zxx0yy0z 法平面方xx0yyy0z(zz0F(x,z,y)0FxFyyxFzzx0 3）

(1,yx,zxG(x,y,z) GxGyyxGzzx00 切平面Fx|p(xx0Fy|pyy0Fz|p(zz0 xx0

yy0z000000

z

f(x,y)F

f(x,y)zn(fx,f切平面fx(xx0fyyy0zz0xx0yy0z f 1xx(u,*yy(uv)（参数方程形式zz(u,,,ijk切线v1

(y,z)(z,x)(x,y)nv1v2

(u,v),(u,v),(u,v)

uu(x,y, l

xcos

ycos

z

gradul（l方向投影

uuux,y,z

graduux,y,z

ijkdivijk

rotAAzz

f(x,求驻点

驻点2

2z

2z PAx2BxyCy2A0A0

AC0minu3

f(x,y,

Lf(xyz(xyz

(x,y,z)（一）1zf(xyfx(x0y0)0fy(x0y0)0f(xy(x0y0dzx0y0)0f(xy0xx0f(x0yyy0 (C)②④ 2f(xy在点(0,0)

[f(x,y)f(0,0)](x,

f(x,0f(0,0)0，且

f(0,y)f(0,0) y limfx(x,0fx(0,0)0，且limfy(0yfy(0,0

(x,

f(x,y)f(0,0)

yx2x23证明极限y

x3x6x

2证当(xyykx3趋于(0,0)x3 x3 xxy

6

2x0x0

6

26 x01x01xk去不同值而取不同值。故极限x2x2

x3x6x

24z

(x2y2f(x,y)

x2y20, x2y20,（1）讨论f(x,y)的连续 （2）求fx(x,y),fy(x,（3）讨论fx(x,y),fy(x,y)的连续 （4）求df(x,解:(1)当(xy)0时，显然连续,x2limf(x,y)lim(x2x2

0

fy

yf(xy在(0,0)(2)当(xy)(0,0x2x21fx(xx2x21

xf(0,0xx

x2f(xx2x

xx2x21同理fy(x,x2x21

，fy(0,0)x2当(xy)(0,0fx(x,yfy(xy显然连续，当(xyx2x2x2xlimf(x,y)x2x2x

f(xyy

y x2x(0,0)fy(xx2x当(xy)(0,0df(x,y)fx(x,y)dxfy(x,x2x2x2x2=x2x2当(xy)(0,0

x2x2

dxx2x2x2x2x2x2由于limf(xyf(0,0)fx(0,0)xx2x2

sin 0x0y

x2所以df(0,0fx(0,0)dxfyx2

x0yf(xy在点(0,0)fx(0,0)2fy(0,0)1(A)df(0,0)2dx(B)f(xy在(0,0)zf(x, 在(0,0)点处切线的方向向量为jx(D)x0y

f(x,y)必存 (选考虑二元函数

f(xy)的下面4个性质①f(xy)在(x0y0)连续②f(xy)(x0y0)两个偏导数连续;f(xy在(x0y0)f(xy在(x0y0)(A)(B) (选f(xy)|xy|(xy，其中(x,y在点(0,0)(x,y)在什么条件下，fx(x,y)，fy(x,y)存在 ((x,y)0(x,y)在什么条件下，f(x,y)在(0,0)可微 ((x,y)0 1zf(xy,xx2y(xy2z

t)dtf和

x2

y

tu，则x2y

y

t)dtxy2(u)(du)x2yz

f1yf2

y)(xy2)y2(x2y)2xy2z

f1y(f11xf12x)x2f2fx2(f21xf22x)2y(xy2xy3(xy2)2x(x2y)2x3y(x2

y19x2y20有形如ux 解设ry，则u(rxu

r

y

2uy2

2y (r)

u r1

2u

1 (r)

x

y22y1

y2

2yx2x既 x2x 亦 (1r2)2r

(r)c1arctanr 故 u(x)c1arctanx

（c1,c2为任意常数3zf(x2y22z 2z 1 x2y2xxz

yzx2解:设zf(r)，x2 2

rx dz dz

z

z

dz dr2z

dr

d

r r 同理可求y2dr2z

，解得zc1cosrc2sinrrx2zx2

x2y2x2x2

x2z

f(t,et

，其中 具有一阶连续偏导数,

2z。2 （2xf2x3y(fexyf 已知函数uu(xyx2y2xy0，试确定参数a,b变换u(x,y)V(x,y)eaxby下不出现一阶偏导数项 （a1,b1设函数z

f(xy在点(1,1

3(x)

f(x,f(xxd

x

1yy(x)zz(xzxf(xyF(xyz)0所确定的函数,fFdzdy 解:zxf(xyF(xyz)0x

dz(

xf)Fyxfdx

fx(1 )

Fyxf

(Fxf

F(fxf FF F z

Fyxf2设函数u定,求du

f(xyzzz(xyxexyeyzez解:duudxudyx求导, u

fz，exxexzezzez ex

z

ex

u

ey所以xezzezx

fxfzezzez

fyfzez exxex eyyey所以dufxfzezzezdxfyfzezzez 3xyzlnyexz1，根据隐函敌存在定理，存在点(0,l,1)的一个邻zz(xyy(xzzz(xxxyzzz(xxxyzyy(x4zz(xy

0:c,f(xyf22

f

2

f22

2xyxy(x

y2d2证明:f(x, c为一直线,当且仅当

0f(xyx求导ffdy0dyfxx y

f dy( xy

dx

f

d2 dy d2 fxx2fxydxfyy(dx)fydx2

d2必要性：若(x,

0，由（1）

2

dyxy

(dy)2yy

2

2

2

(f

20

f

(f

f22

f

2

f22

2xyxy(x

y2

d2

2

dyxyd2

(dy)2yy比较（1）式,fy

0fy0

0，即(xy)c5z(xyF(xzyz0xzyzz xF(11zF(z1z0x2y y xyzFx2

xzFy2xzx xF1xzyzz

，由对称性得yzy xF1

x x二练习1.设函数z

f(u)u(uyp(t)dtuxyf(u),(up(u),(u连续，且(u)1pyzp(xz 例1求函数uxyezxtyt22ztt3M(1,1,0

t1，曲线在点MM M 其方向余弦为cos

，cos

cos3

coscos 11121 33

12xtyt2zt3x2y3z0垂直的切线方程,ux2y2z2z 解:设切点参数为t0，则s1,2t0,3t0n1,2,3，即00，所以t0切点(1,1,1s1,2,3)x1y1z

123s0 123

cos cos 0 0

3x2y2z23上与(1,1,1点处的切平面平行的切平面方程,ux22y23z2在点(1,1,1x2y2z2300n2(Fx,Fy,Fz)(x,y,z)(2x0,2y0,2z0)//(x0,y0,z000n1n2x0y0z0x0y0z0切平面方程

(x1)(y1)(z1)n0

1,1,

333 3333uucosucosu 3

0 0

xyb Lxayz30在平面上，而平面z(1,2,5)，求a,b之值

yx22.函数zx2 （D）可 （选2x22y2z212上求一点,f(xyz)x2y2z2

12

1,021f(xy在点(0,0)y

f(x,y)f(x2y2 1(A)点(0,0)f(xy极值点(B)点(0,0)f(xy(C)点(0,0)是f(x,y)极小值点(D)由条件不能确定(0,0)是否为极值 (选21f(0,0xy时,选例3f(xy预(x,y均为可微函数,且(xy)0，已知(x0y0)f(xy在约束条件(x,y)0下的一个极值点,下列选项正确的是（A）fx(x0y0)0fy(x0y0)（C）fx(x0y0)0fy(x0y0)

（B）fx(x0y0)0fy(x0y0（D）fx(x0y0)0fy(x0y0)例4zf(0,00

f(xy的全增量z2x3)x2y4)y

（1）z的极值；(2)zx2y225（3）zx2y225z2x3z2y4zx23xyzy)2y 所以yy24yc

zx23xy24y

，由f(0,0

，得c0zx23xy24z2x3 x

2z

2z

2z （1）由 A

2，B

0，C zy2y4

y1

ACB240A20z3,26.252（2）L(x,y,)x2y23x4y(x2y2Lx2x32x x3 x3y2y2Ly2y42y0得y2y2x2y225

4，z(3,4)

，及

，z(3,4)例5F(x,yz处处有连续的偏导数，并且三个偏导数在任何一点不同时等于SF(xyz0不含原点，M(x0y0z0是曲面上距原点最近的点。求证该M(x0,y0,z0)的法线经过原点。证明只须证明曲面在点M(x0,y0,z0)处的法向量平行于OM(x0,y0,z0) 极值问题minf(x,y,z)x2y2F(x,y,z)f(xyzS上任一点(xyz构造辅助函数L(xyz)f(xyzF(xy

按照题意f(xyz)在点M(x0y0z0x0y0z0(x0,y0(x0,y0,z(x0,y0,z0

2

(x0,y0,z0(x0,y0,z0(x0,y0,z0FFF于是向量(x0y0z0与x,y,z

(x,y,z00另一方面，曲面SF(xyz)

在点M(x0y0z0

处的法向量为FFFx,y,z

SM(x0y0z0M(x0y0z0 (x,y,z00与向量OMx0y0z01.设zz(x,y)是由x26xy10y22yzz2180确定的函数,求z(x,y)的极值 (极小值z(9,3)3，极大值z(9,3)3)2.zf(xy的全微分dz2xdx2ydyf(1,12f(xy D(xy|

(f(0,2f(0,22f(1,0f(1,03xoyDxy|x2y2xy75}，小山的高度函数为h(xy75x2y2M(x0y0D上一点，问h(xyg(x0,y0g(x0,y0Dx2y2xy75上找出使（1）g(xy达到(y2x)2(x2y 00 (y2x)i(y2x)2(x2y 00 (x0,y0 x2y22z24.已知曲线Cxy3z

，求Cxoy第七章元数量值函数积分多元数量值函数积分的概念与性nnf(M)d＝I＝d

f(Mi)ii1f(M)在几何形体f(M)在f(M)在有界闭几何形体f(M)在上必可积。11dd.2[f(M)g(M)]d

f(M)d

g(M)d 3f(M)df(M)df(M)d

f(M)g(MMmf(M)在闭几何形体f(M)在闭几何形体上连续，则在上至少存在一点M0 f(M)df 二重积 f(x,y)d=limf(,)0iD三重积分f(xyz)dv

f(i,i,i)

0i对弧长的曲线积分

f(x,y)ds＝ f( 0 f(x,y,z)ds f(,,0nf(x,y,z)dS＝limf(i,i,i)Sin

累次积分积之。说明：1（D）12选择积分次序要合适，若先yxI1

xsinydx322

ex2,cosx2,sinx2sinxx

21计算2

y2 2y

2y y

12

y

dy0

dy0xdx30y 12y2dey21y2ey22

y 26 6

14e4ey221(15e46 0 2计算二重积分dxsinxdy4dx2sinx 2 2 y 2 2 I1

sin2ydx

2)dy

y

dy (2 1xb例3计算

（a,b0

xyyxyy0

b=dxxydydyxydx

dy

ay

a4

xxxD

dxdyDx2y21xy1Dx r(cossin

2dxdy2d

Dx

42(cossin1)d 例5 (|x|D:|x||解：原式4|x|dxdy41dx1xxdyD D16求(xy)dxdyD是圆心(a,bRD(xy)dxdy((xaybab))dxdya 7计算xydxdyD是双纽线(x2y222xyD解：双纽线(x2y222xy的极坐标方程为r2sin原式22d

sin

rcosrsinrdr 例8 D:x2y2

1 1 解 xdxdyD:x2y2

ydxdyD:x2y2

(x2D:x2y22

)dxdy2

d0rdr9f(x在0x1f(x)01xf21 1

1f21010f1 证明：只须证明0f(x)dx0xf(x)dx0xf(x)dx0f=1f2(x)dx1yf(y)dy1xf2(x)dx1f(y)dxf2(x)f(y)(y D=f2(y)f(x)(xy)dxdy1f(x)f(y)(yx)(f(x)f(y))d 210求|x2y21|dxdyDx2y29D解Dx2y21D:1x2y2 原式(1x2y2dxdy(x2y21)dxdy 11求emax{x2y200y0x1y解：emax{x2y2}dex2dey2d0x1y

ex

ey

dxe100y

12计算二重积分[xy]dxdyDxy|0x2,0yD解:xyjj1,2,3,4DDk(k1,2,3,4，则[xy]k1k1,2,3,4，因而[xy]dxdy0dxdydxdy2dxdy3dxdy32331

f(x连续,f(0)1F(t)

f(x2y2x2y2t

t0)F000解F(t2dtf(r2rdr2tf(r2rdrF(t2f(t2000F(0)limF(t)F(0)lim2f(t2)2f(0)

t

t2y计算二重积分ydxdyDx2，y0，y22yD （4 2计箅二重积分|yx2|dxdy，其中D{(x,y)||x|1,0y （11 设D={(x,y)|x2y2 2,x0,y0},[1x2y2]表示不超过1x2y2的最大数,计箅二重积分xy[1x2y2]dxdy (3 f(x在[a,bf(x)0，证明bf(x)dxb1dxb aff(x在(0,f(t)e4t2

x2y24t

f

dxdyx2x2f(t

f(t)e4t2(4t2 y2( z2(x,y 投影

f(x,y,z)dvadxy(x)dyz(x,y)

f(x,y,z)dvc2dzf(x,y,z z柱面坐 f(x,y,z)dxdydz＝f(cos,sin,z)dd 球面坐标f(xyz)dxdydzf(rsincosrsinsinrcos)r2sin 一般方法f(xyz)dxdydzF(uvw|J|

V其 F(u,v,w)

Vf(x(uvwy(uvwz(uvw

y21求(x2y2z)dV:由曲线

x

z解:x2y22z4 2 43I (x2y2z)dxdydzd (r2z)rdr 3 Dz:xy2448 dx2y2(x2y2z)dzd rdrr2(r448

z)dz

2563 Dxy:x2y2 R2x2例2计算三重积分(3x25y27z2)dV，R2x21解设x2y2z2R2，则由3x25y27z2z1(3x25y27z2)dV1(3x25y27z21 21由轮换对称性x2dVy2dV 故原式1(3x25y27z2215(x2y2z22152dsindRr2r2dr2 0P0距离平方成正比(比例常数k>0)，求球体质心位置。0解:设球体为Ω,球心为原点，P(0,0Rx2y2z2R2(xyz

xy zk(x2y2(zR)2 2Rz k(x2y2(zR)2

(x2y2z2RzR22R(x22z2

8

R2dsindRr2r2drR2

4求f(xyz)dV，其中x2y2z2(xyz)2f(x,y,z)

当zx23x2x23x23解：原式(x2y2z22xy2yz2xz)dV

4cos

4cos

3sin

r4dr

d2sin3

r3cosdr 例5设函数f(x)连续且于零f(x2y2z2F(t):x2y2z2t

f(x2y2t，G(t)D:x2y2ttf(x2y2D:x2y2tF(t在区间(0,2证明当t0F(t)G(t2

f(x2 2dsindtf(r2 2f(r 000（1） 0002dtf(r2 tf(r2tf(tt2F(t)t2

)0f

)r(t0F(t在(0, 0f

)rdrt 20ft

2

tf t2 t2

t2（2）t0F(t)G(tt2

0f

0f

只需证

(t)

tf(rt

)r

tf(rt

f(r

)rdr

0 又(0)0，所以(t)0

(t)

f(t2)0

f(r2)(tr)2dr

:x2y2z2R2

z

及2

1:x2y2z2R211x0y0z01(A)xdV4

ydV4 （C）zdV4 （D）xyzdV4

2.计算f(xyz)dVx2y2za)2a2x2f(x,y,x2

,当zx2,0x2

a41x22x22f几何解释：1.

12.第一型曲线积分Lf(xy)dsf(xy0xOyL为zf(x,y)的柱面面积。xy

ds

x2y2

yy(x)dsds1y2(xxr()(dx)2(dy)2rr()(dx)2(dy)2

r2r2x yzy

ds

x2y2z2 z(x

ds

1y2z2dx(此类空间曲线常以隐式方程形式出现xdsdxy

ds x2y2z2 1I(zy)dsc为 xyz解曲线cx2y2z2R2xyz0xyzzdsydsxds1(xyz)ds10ds 3 3y2dsx2dsz2ds1(x2y2z2)ds1R2ds1R22R2 3

3 I(zy2)ds2R3 x2y2z2R2x0y0z0的边界曲线的质心,

(3

4R

4R

1zz(xy dS

1z2z2 f(x,y,z)dS f(x,y,z(x,y)1z2z2 yy(zx)，则 dS

1y2y2 f(x,y,z)dS f(x,y(z,x),z1y2y2 Dyzxxyz dS（一）

1x2x2 f(x,y,z)dS f(x(y,z),y,z1x2x2 1求(x2y2z2xy2x2yz)dS，其中x2y2

(0z1zx222解：由对称性，得xy2dS1zx222 原式 (2(x2y2) x2y2)2dxdy Dxy:xy3例2设曲面:|x||y||z|1，则(x|y|)ds (4 333S为球面(xa)2yb)2zc)21(xyS解(xyz)dSxaybzc)]dS(ab 由于球面(xa)2yb)2zc)21xaybzc(xa)dS(yb)dS(zc)dS (xyz)dS＝(abc)dS4(ab 例4计算曲面积分((2x3x2cos2y3y2cos3(z21cos)dsz1x2y2（z0），coscosco为曲面向余弦（cos0z解：设1x2y21原式

(6x22x6y22y6z)dV2

2

1r =6(x

z)dV3

(rz)dz323（二）1S:x2y2z2a2z0SS1

xdS4 （B）xdS4 （C）zdS4 （D）xyzdS4

x2计算曲面的质量其中为锥面z 在柱体x2y22x2任意一点(x,y,z)处的面密度函数为该点到xoy面的距离 （ 29 （4(abc)R2

I(xyS

S:(xa)2(ya)2(za)24.设半径为R的球面的球心在定球面x2y2z2a2(a0)上，问R为何值时，球面在定球面内部的面积最大？ （R4a）38曲面积向量值函数在有向曲线上的积分第二型曲线积

w|F||l|cosF变力沿曲线运动

dw|F|dsPdxQdy，则WLPdxQdy平面曲线LPdxQdy，空间曲线LPdxQdyRdz，性质LP(x,y)dx＋Q(x,y)dy＝t1{P[x(t),y(t)]x(t)Q[x(t), P

ydxdyLPdxQdyL的取正向的边界曲线，D为单连通区域，P，QDLP(x,y),Q(x,y及DLPdxQdy＝0DCLPdxQdy（3）存在u(x,y（3）存在u(x,ydu＝P(x,y)dxQ(x,y)dyPdxQdyduu（4）P

D内恒成立 （2）1I[ycosxy]dxy)sinx1]dy，这里yOm OmA是位于连接O(0,0A(,的线段OA下方的任一光滑曲线。且OmA与OA所围图2.ABBOB(0,I

(QP Om0(()cosx)dx0(1)dy22 xdy

例2L

x2

L为上半椭圆a2

1(abA(a,0)B(0,b)C(a,0)

y2 (x2y2

x，积分与路径无关，取l

（上半圆xacos即yasin原式

xdyydx

xdyydx

0(a2cos2ta2sin2t)dtallx2 2 a2all3设(xy),v(xyD:x2y22x2yD D曲线C上u(xy)x,v(xy)yxy)vxyD xy)vxy)u]dxdyuvdxuvdyxydx (yx)dxdy[(y1)(x1)2]dxdy2dxdy 4D为曲线Cr1cosA，C C函数uu(xyDx2y21，证明ndsACn是uDuds(ucosucos)ds(ucosucos)dsudyuC

C

C

C (x2y2)dxdydxdyA20d

rdr 2 2x4例5设f(u)存在连续的导数，且0f(u)duA0，L为半圆周2x4B(2,0。计算

f(x2y2)(xdxX(xy

f(x2y2)x,Y(x,y)

f(x2y2y，f(x2y2)(xdxydy)

XdxX(xy),Y(xyY2xyf(x2y2) f(x2y2)(xdxydy)＝

f(x2y2)(xdx2f(x2)xdx14f(u)du0

2

(xay)dx(x

为某函数的全微分,求常数a。a计算L

xdyydxL:ABCA(1,0x2y21B(1,04x2到Dxy|0x,0y}LDxesinydyyesinxdxxesinydyyesin xesinydyyesinxdx2L计算曲线积分I

xdybxbxaL

(ab0ab)L是以点(1,1)为中心，2222R(R

2为半径的圆周，取逆时针方向。R

时I0R

时，I

xdyydxA（常数，其中(xL是绕原点(0,0) y

xdyCC为任一不过原点也不包围原点的正向闭曲线,证明(xy2C当(1)4时,求(x)及A （(x)4x2,A向量值函数在有向曲面上的积 流量Q|v|Scos(nv)vsdQvdsPdydzQdzdxv(P(x,y,z),Q(x,y,z),R(x,y,vdSPdydzQdzdxS SS S R

x

zdvPdydzQdzdxRdxdy R或x

zdv(PcosQcosRcos 这里是的整个边界曲面的外测coscoscos是在点(x,yz)向余弦

xdydzydzdx

1I

3(x2y2z2

a2

1PQR0

x2y2z22 I

xdydzydzdxzdxdy

3dv 3

3

xdydz

例2

x2y2

，其中

是由曲面x

R及两平zRzR(R0解：设123依次为xy

xyxy

xyxy

(R)2dxdyx2y21

x2y22 R2

x2y2R

x2y2R2

x2y2R

x2y2 x2y2

R2

dydz

R2

R2RR2R

2dydz

R2y2

RDyzRR

RR 3If(xyzx)dydz2f(xyzy)dzdxf(xyzz)dxdyfxyz1被坐标面所截部分，上侧。F(x,yz)xyz1I((f(x,y,z)x)Fx(2f(x,y,z)y)Fy(f(x,y,z) ((f(x,y,z)x)1(2f(x,y,z)y)1(f(x,y,z) (xyz)dxdydxdy 级数的知识框级数的概念与性1u1＋u2＋u3＋＋un＋＝un sn＝ui称为部分和,若limsns称无穷级数un

uns，则kun收敛到ks unvn收敛到s,,则级数(unvn收敛到s

如果级数un(u1un)(un1un)(un1

)

k un收敛,则limun

数项级 小收，小发大比较法 lim 根植法：limnanl,l1收，l1

f(x)dxf(n) 交错级数：莱布尼兹判别法，un1un,limun任意项级数

函数项收敛半径R

（一） n

n 1结论1若an

an（如an

（2

n

1n

n

（3）

a收敛，则

a2收敛（如

(1)n

（4）

a2收敛nn

n

n

nn则n则a3nn （B）24对例2证明若偶函数f(x)x0的某邻域内有二阶导数，且f1

f(0)1，则n n1 f(xf(0

f(x)f(0)f(0)x

f(0)x2o(x2)f11f(0)

o1，

f11

f(0)

o1

f(0)n

n2 n

n2

|u

f11~f(0)

f1即 n n2， n 绝对收敛 n1 3设{un}，{cn}，满足cu

0

1发散，则u n n1n n1u

n n，满足cnncn1a（常数a0，且 收敛，则un也收un

c

c

n1u

n（1）

n

n n

11

n

n

ccunnn（2）由条件 unn

a0，得

c

cu，所以uc1u1un

n

n

n n

1 nc1收敛，则uc n1

n

n 4设正项数列{an}单调减少且

an发散，证明

n

n

an{a}单调减少且(1)nalim

a0n有

an a

n1

n 所以un1n1 n1

(anan1，而(anan1

nSna1an1a1ann例5xnnx10nxnaa1时。级数xnfn(1)n

nn(x)xnnx1，x(0,f(x)nxn1n0n

fn(0)10xnnx10

(0,1

0x

1f n1

1xn n

n6设an和bnn1n1bn收敛时，n

n

n

n 也收敛；当anbnn

nan1bn1 a2b2,a3b3,,

b1

a1

an

a

ban

1 有比较法知，当bnan也收敛；当anbnn

n

n

n

7设正项数列{a}单调减少，且(1)na发散，试问

n

n1an解由于an0{an}limanan

a0n若a0，则由莱布尼兹判别法有(1)na收敛，与假 ，故a0。于nn 1 1 1，从而 a a a

a

，而a

1

（二） a0p0limnpen1a1，若级数a （p2

n

n设annn

n

ununn

n

(u2n

)n

un1

0设an4tann0（1）求n

nan

1（2）证明：对任意的常数0，级数n

1 1

设a12an12an

15.y5.

nyx

且y(0)且

n1

n

y111散性

n1

n nn讨论级数p的敛散性nn|a|1时，级数发散；当a1时，若0p1p1级数收敛；当a1时，若0p1p1时级数绝对收敛) 两个正项级数anbnnn（n1,2,，讨论两个级数敛散n

n

an

（当an发散时，bn必发散；当bn收敛时，an必收敛n

n

n

ny2exsin

（x0）x

e1关于幂级n

R22(RR内ax（且内闭一致收敛nnf(n)(0) f(n)(x f(x)~ x或 0(xx0n n 1）1

1xx2x3(1)nxn

1x12）1

1xx2x3xn1xe

1x22

xnn

(x3sinxx3

x(2n

(x2cosx12

x4

x2n

(x2ln(1x)x22

3

4

n

(1x(1x)m1mxm(m1)x2m(m1)(mn1)xn (1x变换之后使用公式，求导，积分公式可和型和式求导（或积分）于原式结合1求n

(xn

1，即当|x3|12x4x2n

发散，故收敛域为[2,4n nn2设axnx2处条件收敛，则nn

(x1)2nx0n n（A）（B）（C）（D）敛散性由具体{an}13f(xx23x4x111

1

1x23x

5x

x11 1 1

(x1)n，xx

(x1)

31x13

3n 1 1 1

(x1)n，xx

(x1)

21x12

2n 1 (1)n n0x23x453n1n04（1）求n

2n

(x

，xR1|x|1S(x)nxn1nS(x)dx

x

nxn1dx

xn

，故S(x)

x

，|x|xx

0

1x

n（2）求nn R1x[1,1S(x)x，则S(x

xn1n1

n

1xS(x)S(xS(0)x

，1xS0 n（3）求nn

1 解：在（2）中，令x 2

n

n

S()ln25求n

(1)nn1x2n1(2n解：收敛域(,S(x)n

(1)nn1x2n1(2n 1 20S(x)dx 2n

1从而S(x) (sinxxcos12

x x2nn例8求135(2n1n x2nn解：收敛域(,S(x)135(2n1n

2x41S(x)1xS即

xS(x)e2

xe2xS(0) x（二）n若幂级数ann

R0，则liman11n

n limn1不存在，则幂级数an

n

n 若幂级数axn收敛域为[1,1，则幂级数naxn的收敛域为[1,1nn

n若幂级数n

aax收敛域为[1,1，则幂级数nn

anxn的收敛域为[1,1（D）n1 nf(x)arctan12xx的幂级数，并求级数2n1n (1)n

2n