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第十二章级12.1x2(1)fn(x) ,x(x2xfn(x)sinnⅰ)x(l,l) ⅱ)x(,)1fn(x)1nx,x(0,1)1fn(x)1nxⅰ)x[a,) a0,ⅱ)x(0,)n2xfn(x)1n3x3ⅰ)x[a,) a0,ⅱ)x(0,)

(x)

,xnxn

x[0,fn(x)1xnⅰ)x[0,b]

b0,ⅱ)x[0,1],ⅲ)x[a,)

a0nf(x)xnx2n,x[0,nnf(x)xnxn1,x[0,n

(x)xlnx,x(0,1) f

(x)1ln1enx,x(,)nnf(x)e(xn)2nⅰ)x[ll,ⅱ)x.解(1)x,x20x2

fn(x)x2x2

x

f(x)x2n2fn(x)fx2n2

x

1n1n2 x21 x21x2n21

N11nNx

fn(x)f(x)x2因此fn(x) 在x(,)一致收敛于f(x)x2(2)x(,),

f(x)limsinx0

f(x)nⅰ)0n

fn(x)f(x)

xnxl xnxlln

,故Nl1,当nN

fn(xf(x),因此

(x)sinn在(llf(x)0 ⅱ)020NnN11,xn2 fn(xn)f(xn)sin2120

(x)sinx在(n

f(x)0(3)x(0,1)1

fn(x)

xx

11

f(x)(n)020NnN1Nxnn(01fn

)f(xn)

11

111111

(x)

1

在(01)不一致收敛到fx

x(0,),fn(x)1nx0

f(x)(n)1ⅰ)01

fn(x)f(x)1nx1nana只须n1

N11nN1

fn(xf(x)

x[a,fn(x)1nx在[a,)(a0f(x)0 1ⅱ)020,NnN1Nxnn0,1fn(xn)f(xn)

1nn

20

(x)

1

在(0,f(x)0x n2x

x(0,),fn(x)1n3

1

0

f(x)(n)ⅰ)0n2x2 fn(x)f(x)n3

只须n1

N11nN

fn(xf(x)

x[a,n2xfn(x)1n31

在[a,)(a0f(x)010ⅱ)020,NnN1Nxnn0,0f

)f(x)

n2

1n2x

1n3 fn(x)1n3

在(0,f(x)0 (6)fn(x)xn11n

x

f(x)(n

x[0,1]01n1nfn(x)f(x)N21nN

x 1n fn(x)f(x)x[01fn(x)xn1在[01]f(x)x

iix

iiixa,,axnxfn(x)1

0,0x121,x1 2

f(x)(n)ⅰ)0

x1xn fn(x)f(x)1xn

b,x[0,b](b1)Nlogb1n

fn(xf(x)x[0bb1)xnfn(x)1

在[0bb1f(x)0ⅱ)

10,N,nN1N,x1n1n1

[01]f(x)f

)210 0

1 2

(x)

xn1

0,0x11,x在[011,xⅲ)0fn(x)f(x)

1 xn1xn1xn1xn

1,x[a,](a1)anNloga1nN时,

fn(x)f(x)

,所以fn(x)

xn在1xn[a,a1f(x)0

f(x)lim(xnx2n)01n1n

f(x),x[0,1]

10,N,nN1N,x

[01]fn

)f(xn)

11 112nf(x)xnx2n在[01]f(x)012n

f(x)lim(xnxn1)0

f(x),x[0,1]nf(x)f(x)xnxn1n

g(x)xn1nn

g(x)0x

,且0x

g(x)0

x1g(x)0x

n

达到[01]上的最大值,于是x[01nn

n n g(x) 1

n1

n n

n

n 0N11,则当nN

fn(xf(x)对[01xnf(x)xnxn1在[01]f(x)0n(10)x(0,1),

(x)xlnx0

f(x)(n) 0,当0x1时,由于limx 0,故0,使当0

xnn xnxnx N11,则当nxn

时,有0x1x

x01

xlnx在(01f(x)0 x0

(x)1ln1enx0(n)nx0

(0) ln20(n)1n1x0时,由于x1lnenx1ln1enx1lnenxenx1ln2x,而 1nln2xx(n)fn(x)x(n

(x1ln1enx在(fx0x0 x,x00

f(xf(x)1ln21,取N11nN时,

fn(xf(x)xf

(x)1ln1enx在(nf(x

f(x)lim

0

f(x)2nⅰ)0x[ll时,f(xf(x)e(xn)nnl后,由于e

exl)2,故当nl

f(xf(x)e(xl)2nnf(x)f(x)nn

,只须n

lN

lnl1,则当n

f(xf(x)x[llf(x)e(xn)2在[llf(x)0

0,N121

nN1N

n

fn

)f

)11 nf(x)e(xn)2在(f(x)0nfn(x)(n1,2在[ab上有界,并且fn(x)在[abfn(x在[a,b证 由于fn(x)(n1,2,)在[a,b]上有界,故nN,Mn0,使x[a,b],有

fn(x)Mnfn(x在[abf(x10NnN

fn(xf(x)1

fN1(xf(x)1f(x)

fN1(x)1MN11,从而fn(x)f(x)1MN12对一切nN成立MmaxM1,M2,MN,MN120,则xa,bnfn(x)Mfn(x在[abf(x定义于(abfn(x)

n

(n1,2,)fn(x在(abf(x证明由于nf(x1nf(x)nf(xf(x)1n

fn(x)

nf(x)n

f因此limf(x)f(xxab0,由于1

(xf(x)0n 1N11,nN

fn(x)f(x)nfn(xf(x)对(abxfn

(x在(a,f(xf(x在(abf(xf(x)n[f(x1)f(x)] 求证:在闭区间[,abfn(xf(xf(x1)f证

(x)n[f(x1)f(x)] n

f(x)(n)0f(x在(abf(x在[,ab致连续,故0x,x,

x

f(xf(x).f(x)f(x)n[f(x1)f(x)]fx

f(x1)f(x),0n

,即n ,就有

(xf(x)N11,则当nN时

fnxfxx[,abfn(xf(xf1(x在[abRiemannxfn1(x)xfn(x在[ab

fn

(n1,2,)证明f1x在[abRiemann可积,故必有界,即M0,x[abxf1x)Mxxf2(x)x

f1

f1(t)dtM(xa)f(x) f(t)dt

xf(t)dt M(ta)dt M(xa)2xx x

(x)1M(xa)n,n1,2,故x[a,b] (x)1M(xa)n1M(ba)n0(n) fn1(x)0(n.即n

fn(x)0

f(x),x[a,nN

(x01M(ba)n0(n,故0NnN时

fn1(x0fn(x在[ab问参数

f(x)nxenx

n1,2,nn在闭区间[0,1]收敛?在闭区间[0,1]一致收敛?使 f(x)dx可在积分号下取极限解,当x0时fn(x0,n12,当0x1时fn(x)

xenx

enx

0(n),故不论参数fn(x)nxenx在闭区间[01f(x)0nf(x)n[enxenx(n)nenx(1nx)n

(xx1

(1)n

ne

1e1nx[0,1],f(x)n1e1nn故当1时,因为n1e10(nf(x在[01f(x)0n当1

e10,N,nN1n,

[011nn1nfn

)f

)nn

ne

1e1e110fn(x在[01不一致收敛.100fn

(x)dx

1nxenxdxn1

1xd(enx)n1[en0

0n1[en1en1]n2en(n1)n2 2lim0fn(x)dx02lim0fn(x)dx12时, lim0fn(x)dx

f(x)dx0,因此当0

lim0fn(x)dxn7.f(x)nxenx2(n1,2在闭区间[01n

1 f f n证明当0x1f(x)nxenx20n)x0f(x)0n1n1

f 0

fn(x)dx00dx01 1f(x)dx nxenxdx

1enx2d(nx2)1(1en)

0

lim1fn(x)dxlim1(1en)101

fn(x)dxn

n

08.

fn(x)(n1,2在(fn(x在(f(xf(x在(证明fn(x在(f(x,故0N,当nNfn(xf(x)3xfN1(x在(0,0,x1,x2(x1

fN1(x1)fN1(x2)3从而x1x2(,只要x1x2,就f(x1)f(x2)f(x1)fN1(x1)fN1(x1)fN1(x2)fN1(x2)f(x2 f(x1)fN1(x1)

fN1(x1)fN1(x2)

fN1(x2)f(x2)333f(x在(9fn(x)[ab上的连续函数序列,且fn(x)一致收敛于f(x);又xn[a,bn1,2),满足limxnx0,求证

fn(xn)

f(x0) 证明f(x是[ab上的连续函数,因而00x[abx

f(x)f(x0)2又limxnx0xn[ab],故对上述0N1nN1xn而当nN1

f(xn)f(x0)2而fn(x)在[abf(x,故对上述0N2,当nN2x[a,b]

f(x)f(x) NmaxN1N2,则当nN fn(xn)f(xn)2 f(xn)f(x0)2所以,当nNfn(xn)f(x0)

fn(xn)f(xn)f(xn)f(x0 因此

fn(xn)

f(x0)

fn(xn)f(xn)

f(xn)f(x0)22设fn(x)是在(abf(xx0ab

fn(x)an(n1,2,)liman

f(x

n

fn(x)limxx0

fn(x)证明由于fn(x)是在(abf(x,故由函数列Cauchy0N,当n,mNfn(x)fm

(x)2

fn(x)an(n1,2xx0

2

Cauchy收敛准则,知limnn

a,故0N1,当nN1ana3又fn(x)是在(abf(x,故对上述0N2,当nN2xab

f(x)f(x) NmaxN1N2fN

(x)f(x)3

aN

a3

fN1(x)aN1,故0xabx

xabx

fN1(x)aN13f(x)a

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

(x)

Nf(x)a,即limn

Nfn(x)limxx0

Nfn(x)

N

fn(x)(n1,2)在[abRiemann可积,且fn(x)在[ab一致收敛于f(xf(x在[abRiemann证明由于fn(x)在[abf(x,故0N,当nNf(x)f(x) x[a,b]f

N

(x)f(x) 4(bx[a,b]fN

4(b

f(x)

fN

(x) 4(bx[a,b]ifN1(x在[abRiemann0,0,对一切分划,当分划的小区间的最大长度时,就有inn

(N1)

i2i其中N1)MN1)mN1)

(x)

(x)(i1,2,,n)

xi1

N

NMi

xi1x

f(xmi

f(xiMimiMM(N1) ,m

m(N1) , 4(b 4(b

(N1)

i1,2,,n 故当

2(bnnxnn

((N1)

x

2(b

nn

2(b

22(b

(ba)f(x在[abRiemann§12.2 xn1(1) 21 n (2)n

12x(1)n1x(3)2n (4)

11xn1 .n1a2nxn解(1)x1xn1xxxn1xx

xn xn而

当x1时收敛,故 2n在x1时绝对收敛11x1

xn1xx11xxn1xx1

1x1

xn n1x

x1时收敛,故11

x1

xx

时,级数的一般项分别为

11的绝对收敛区域为(

1,得1x1或1x1,因而当1x1 n

x2x x2xx

n n12x x1x3时,由于2x

绝对收敛 单调上升且有界,由n

n1 n

判别法知n12x1

收敛,即

n1n

绝对收敛.所以绝对收敛域为(,1)1111

,)31,得x11x0,因而x11x0时,1x (1)n1x , (n),故级数发散1x 2n11xx0时,级数为2n111 111x0

1,因而级数(1)n 绝对收敛,

1x

2n(1)n1x减有界,由Abel判别法,这时级数2n 绝对收敛(1)n1x

11x所以级数2n

绝对收敛域(0),条件收敛域x0,收敛域[0,)

11x

1

1a1

a2,由于级数 a2n1a2nxnxn1a2nxnxnxx0收敛,故

1 x0x0时,级数为1nn发nn发

1a2nxn

当a1时由于

xn1a2nx2 n(1x2n

因而

1 x发散.n1a2nxnna1R0a1时,收敛域为(1)(1x)xn,x[0,(1)n1x(2)(1x2

,x(,)

(x) (1x)xkk0

1xns(x)0

0x1x1

(n)3n41020,N,nN1N,xn3n41

[01] 3 sn(xn)s(xn)1 4

1420所以(1x)xn在[01]

n 21

2 n(1)k1x

2n

1x 1

(2)sn(x)(1x2

1x2

1 k n1 x (1)n1 x

1

1x22x21(1x2)ns(x)2x

(n) x 由于sn(xs(x)2x2x(x2n1 n21)(x2n1 n2

f(xf(x)

(2x2)2(1x2n21n求得f(x)的稳定点x0,x ,可判定x0n21nn21nf(x0fn21n

f nf n21n1)n21nn21n(n21n2)n21nn21nn21n

2n2n21n3

(当n1时0(n)

(4n

sn(x)s(x)

f(x)

f

2n1n1)2

2

(n1 故0N1nNsn(xs(x)x

(1)n1xsn(x)在(一致收敛于s(x)

(1x2

在(x敛于和函数s(x) 2x(1)

sinnx,x(,)33n4x(2)1n4x2,x(,)

(1)n(1enxn2x

,x[0,);sinnx,x(2,)n1x2n 1n5x

,x(,)

(xxn2n2

),1x22x2enx,x[0,);

xnlnnx

,x[0,x2x2nx21(n n2

,x(,)x(10)n,xr1x(11)

,x[a,),an4n4x

4x成立,而

4知级数

n3n4x3n4xx2xn4x2xn4x

n1n (2)由 1n4x

2n2x成立,而2n2 故级数1n4x2在((1)n(1enxn2(1)n(1enxn2x(3)[0

2x[0,成立,故nn

n2x(4)

xx

2n

(n2)x2,而级数

1

sinnx在(2,n1

n1x2nnn

对x(,)一致地成立所以 1n5x

2n2 2n

n11n5x(nn2

(xn

xn)

n(x

xn)

n2

1x22且由于

(n(n1)2 n2(n

2(n

n2nn2n

n2

因而

(xxn2n2

在x2121(7)当x0时,enx1nx1n2x21n2x2,所以enx ,故x0时 n2x2x x0显然也成立,故由

2 2 2 2

x2enx在[0

n1n2

(8)f(x)xlnxf(x)lnx1x1x1e1limxlnx0f(1)0f(xx[01

(f(0)0.因而x[01f(x1e|xnlnnx (xln 所

ennenn由D′AlembertM

xnlnnx

x2x2nx21(n(9)因为 n2

( 1)(

1x2n2x21(x2n2x21(nx21 x2n21(n

n1 1n

n1 1n n(n

(n

,x(,)而 收敛,故原级数在(,)一致收敛.n2(nnnn n

n11

xn

x

n

n

n r n

n1r

x因而nxr1x(11)

1nxlimln1

0

1nx x x[a,

(a1N

nN时n

1,从而当nNln(1nx) xn

1an

a而a

收敛,因而

在[a,

(a1cosn2n2x2

,x(,);(2)

sinxsinnn

,x[0,2];x(3) ,x(1,x (4)nsinx,x(,)2nsin1

x(0,) 3n3n2e3n2ex

(1)xn

,xann

,x[1,0]

,x[1,1]2n解(1)由于级数cos2n ncos2k k2

3

2sin

3n2x有 ,因而在(,)一致有界,对每一固定的x(,3n2x n2xn2x

一致趋向于0,因而有判别法,知

cosn2x2 在(n2x2由于级数sinxsinnxnsinxsinkx

x 2n1 x 2n1k

k

2222 2222nx有界2,因而在[0,2]一致有界,对每个固定的x[0,2nx

n在[02]

n nx

sinxsinnn 级数(1n的部分和序列(1k有界1,因而在(1,

k

列xnx1,单调递减且一致趋向于0,故xn在(1,

bn(x)

nn,n

(x) ,则显然n1bxnsin x

(x) k2

1,而an(x)

nsin

对每个x

x),有1n1nsinn

0,因此

(x)

nsin

在(0 判别法,知nsinx在(13n 12 12 13n2n

2n ,而级数 收敛,故2n 3n

x3

n1x3

3nx0,)绝对收敛,从而收敛.但由于010NnN1N1 22xn3n01 2213nn13nn

1 因而级数的一般项2 在x(0,)不一致趋于0因而2nsin1在 1

3n

3n3n2ex取bn(x)(1) ,an3n2ex bk(x)(1)

(n)k k3n2ex3n2ex

x:xa1an(x)1

0(n)

3n2e3n2ex3a(x) x

(1)3n23n2ex

一致趋于0,因此,函数项级数 3n3n2exxax(1xn)1xx(1xn)1xnkk取an(x) ,bn(x)x,则n,有bk(x)nkk

2 xnnn而an(x) 单调下降且趋于0,因此在[1,0]一致趋于零,所以nn

在[10取

(x)(1)nx2n1

(x)

2n

,则

(x)单调下降且在[1,1]一致nnn

nnk

nn(x)(1)kx2k1k

x3(1(x3(1(x2)n

2Dirichlet

在[112n证明级数

(1)n x在(x并非绝对nx x收敛;而级敛(1x2n虽在(证明

n1

,则

b(x)

1bn

(

an

n

nk n

kan(x)

n

对每个x单调递减,且由于

a(x)10n)

(1)n x在(一nx致收敛.但xN,当nN时,有nx2,所以,当nN

1nxn1nx

11

(1)n 发散,即

(1)n x并非绝对收敛.n1

nx

nx x而对级数(1x2nx0x0

x x(1x x(1x2 (1x2)n

1 x所以级数(1x2nx x

x

1s(x)n1(1x2

1x21

1x

sn(x)s(x)1(1x2)n1(1x2)n lim1

1nn

e3,故N1n

1

1nn

3.0

10,N11nnmaxN1,N1N,xn1n

sn(xn)s(xn)11

1

30n1 n x因此(1x2n在(设每一项n(x)都是[a,b]上的单调函数,如果n(x)在[a,b]的端点为绝敛,那么这级数在[ab证明由于n

都是[ab上的单调函数,不妨设为单调增函数,则x[ab], (a)(x)(b),因此x[ab(x)max(a),( n

在[ab

,

收敛,由判nn法,知级数n(x)在[a,b]一致收敛 若un(xun(x)cn(xxX,并且cn(x)X 证明un(xX证明由于cn(xXCauchy原理,0xNn n只要nN,p,x[ab

ck(x)ck(x),因此当nN时,pk kx[ab

nuk(x)k

nuk(x)k

nck(x,同样由Cauchykun(xX上一致收敛;又由于xXun(x)cn

,而cn(x)un(x§12.3(1)xn

1x1;n xnn1

1x1;xnn(3)2n

x1(4)(xn)(xn1)

0x (5)1n2x2

x0(6)

n,nn

x (7)1n4x2

x0 x(8)(1x2)n

x解(1)x011

1

r1xrs(xxnxr连续,由和函数的连续性知s(xxrx0x011的任意性知级数所表示的函数在(1,1(2)

[1,

r0

r

,则在[11

(x)xn

b(x)

xk

a(x)1an

,则n, n

n1 n

nnkn

k 因而在[1,

nn1n

xns(x)n

在[1,rs(x在[1,rs(xx0[1,rx0[11)xnnxnxnnxn

xn

n n

nn

判别法知nnx在区间[11n2x1

在[1,1]连续,因 nn1n

(4)

x0,),有

1

收敛,因此(xn)(xn

n(n

n1n2 (xn)(xn1在(0,(xn)(xn1在(0, 续,由连续性定理,级数(xn)(xn1所表示的函数在(0,(5)x0:

00

xx ,因为22收敛,故 21n2x2 1n2 n2 n 1n

x x

xxx

0

1n2x x0

0的任意性,知级数1n2x2x0nn

在(

1n,级数1

nnnn

()一致收敛,因而级数

n33nn

n1n3所表示的函数在3(x0x000,使0

x0x1n4x

nn3nnn3

1 而n3收敛,故1n4x2x一致收敛,又级数的每一项1n4x2x 续,故级数1n4x2xx0x0

0 任意性知级数1n4x2x0s(x)

x 1x21

1x0x0s(0)01x xs(xx0x0间断,且为可去间断点,即级数(1x2n在(00,x0f(x)

sinnx在(由于级数的每一项sinnx在(内连续,且x1 而1收敛,使用判别法,知级数sin 而在 n1 f(x)

sinnx在( 2又2

ncosnx n2

cosnx对每一个n在(1cosnx

f(x)

1cosn2

, n2

f(x

n1n2

enx1设f(x) 21f(xx0f(xx0证明(1)x0时,nx1,故nenx1

1

1由1收敛及由

enx

在[0)一致收敛,级数的每一项

都在n1n2

n11n2

1 enx1[0,)连续,因此f(x) 2在[0,)1enx

(2)x00,)r0rx0,而1n2

1

,由于n,当x[r,

ne1n

ne1n

1

e2e

(n(n nenr 1n

enx 故 enr收敛,由判别法知

在[r,n11n

n11n2

1n2

[r,

x(0, f

1

)连续特别在0连续由

x0(k

(1)knkenx若设

(x)

1

x0存在,即

f(x)x

k次可微,则(1)knkenx

(1)k1nkx00,同样00rx0,而

1n2

1[r,连续,且由nx[r,((1)k1nk1e1n

nk1n2

nr及(n(n1)k1 nke1(n 1n

1, nk

(1)k1nk1知级数 1

M

1n2

在[r,因此有f(kx)

f(k1)(x)

(1)k1nk1

存在且连续于[r,),特别地在点x00f(xk1x00f(xk1f(xx0证明nenx在(0,0证 x0,0,使x0,在x[,)时级数的项nenx连续,且0 nenxnennenMnenx在[ 0所以nenx在[连续,特别地在0

,)连续,由

0的任意性,知nenx在(0,设un(x在(a,b

un(x)(n1,2)在[abun(x在[a,bs(x)un(x在[ab证明(1)0un(x在(ab内一致收敛,Nx无关,nN,pxab

un1(x)un2(x)unp(x)2由于un(x(n1,2)在[abxaxbun1(a)

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