数分答案-9到22补充包

第十四章级 三角级数与级(1)sinxsin2x,sinnx,是[0（2）sinxsin3x,sin2n1)x,是[0

上的正交系；1cosxcos2x,cosnx,是[01,sinxsin2x,sinnx,不是[0上的正交系．证明（1）m,nN,mn，有sinmxsinnxdx1[cos(mn)xcos(m 21

sin(mn)x

sin(mn)x

]0m m sinxsin2x,sinnx,是[0（2）knNkn2sin(2k1)xsin(2n1)xdx

2[cos2(kn)xcos2(kn012

k

1

00

sinxsin3x,sin2n1)x,

[0

2（3）由于m,nN,mncosmxcosnxdx1[cos(mn)xcos(m 21

sin(mn)x

sin(mn)x

]0nN

m

m 0cosnxdxnsinnx

0故1cosx,cos2x,cosnx,是[0 （4）sinxdxcos

201,sinxsin2x,sinnx,不是 [0上的正交求下列周期为2FouriernnPn(x(aicosixbisinixf(x)x3(x)f(x)cosx2f(x)eax(x)f(x)sin (x)f(x)xcosx(x)f(x)x0

x0,0xf(x)2x2(x)f(x)sgncosx（10）f(x)

2

(0x2)解（1）1 0Pn(x)dxa022a01

P(x)coskxdxak

kxdxak

0kn

kn1

P(x)coskxdxbk

kxdxbk

0kn

knPn(xFouriernkP(x)～0(coskxnk

sinkx)a(acoskx

k

nkknk

1x3coskxdx0，k0,1,2,， b1x3sinkxdx2(1)k1(26)，k1,2,， kf(xx3～2(1)n126sin

(x)n1 n2f(x是偶函数，故bn0n1,2．1 a0cos2dx1

(1)n1ancos2cosnxdx(4n2

，n1,2,，f(x)cosx～2

2

n14n1 1

2

ea)

a0a0f(x)dx

dx2

a0 (1)k

a

f(x)coskxdx

eaxcoskxdx(a2k2)

) a0

0

a0 (1)k

b

f(x)sinkxdx

eaxsinkxdx(a2k2)

) a0

0

a0k1,2,．所以，当x时 f(x)eax～ash(a)sh(a)a2k2(acoskxksinkx),a01

k

a0f(x是偶函数，故bn0n1,2，2 a00sinxdx1

n12 an

sinxcosnxdx2[(1)n1,

n1(n2f(x)sinx～2

1cosx

(1)n12

(x

n1f(x是奇函数，故an0n01,2，1

n12 1 bn0xcosxsinnxdx0x[sin(n1)xsin(n1)x]dx2(1)n

,n2n21n1,2,． (1)nf(x)xcosx～2sinx2n21sinnx

(x1 1 a0f(x)dx

xdx 21 1

(1)n1anf(x)cosnxdx

xcosnxdx

1 1

bnf(x)sinnxdx

xsinnxdx n因此f(x)x

x0,～

(1)n1 cosnx sinnx)0

0x

f(x为偶函数，故bn0n1,2，a2(2x2)dx32 2

an0

x)cosnxdx f(x)

~38

n2

(xf(x)sgncosx为偶函数，故bn0n1,2，2 a0

sgncosxdx (2dx(1)dx)0 2

an

sgncosxcosnxdx (2cosnxdx(1)cosnxdx)

(1)k

n2k1(2k

n2k

(k1,2,)4kf(x～2k1cos(2k1)xk1

12a0

f(x)dx

dx0212f(x)cosnxdx

2xcosnxdx0，n1,2,，

12f(x)sinnxdx

2xsinnxdx1，n1,2,，

所以，f(x)

(0x2) n1f(x以2为周期，在[（1）f(x)在[上满足f(x)f(x)a2m1b2m10m1,2,（2）如果函数

f(x)在[上满足f(x)f(x)a2mb2m0m1,2,1 证明（1）anf(xcosnxdx(f(xcosnxdx

f(x)cos00

010

f(x)cosnxdx(1)n

f(t)cos 1[1(1)n

f(x)cosnxdx 因此，当n2m1 1[1(1)2m1]

f(x)cos(2m1)xdx0，m1,2,

b1[1(1)n]

f(x)sinnxdx 所以，当n2m1

1[1(1)2m1]

f(x)sin(2m1)xdx0，m1,2, 1

0（2）anf(x)cosnxdx(f(x)cosnxdx0

f(x)cos00

1

f(x)cosnxdx(1)n

f(t)cos 1[1(1)n

f(x)cosnxdx 故当n2m时 1[1(1)2m]

f(x)cos2mxdx0，m1,2,

b1[1(1)n]

f(x)sinnxdx 因此，当n2m时，得 1[1(1)2m]

f(x)sin2mxdx0，m1,2,

级数的收敛（1）f(x)xsinx，x[,x2（2）f(x)

x[0,x[,解（1）f(x在(可微，而在(处处连续，故在[]f(x收Fourierf(xbn0n1,2，2 2a00f(x)dx

xsinxdx2a

f(x)cosnxdx

xsinxcos

xsin2xdx

n

n

1

22

n1 2所以，f(x)xsinx1 cosx2

cosnx，x[,n210（2）a1f(x)dx1 dxx2dx)1120 1

0anf(x)cosnxdx(cosnxdx00

cosnxdx) b

f

nxdx

2

)]

n1,2,，f(x)

26

cosnx

2

1)]sinnx}f(x在(f(00)f(02

102

f(0)

f(0)f(0)

1，2f(x)

26

cosnx

2

1)]sinnx}x(,0)(0,)

x

(1)n1sinn

(x)用逐项积分法求x2x3x4在(,中的Fourier展开式 n求级数 n

4nn1

(1)n1

解（1）2

0xdxnn

0sinnxdx nn

cosnx nn

n2

cosnx6

，x[,x2

3

，x[,3

0

x2dx x322

(1)n1sinnx

2(1)n1(

)sinnx，x(,)22

n

(

66xx4

0

3dx

(n

(

6)cosn

，x[,x423

8

，x[,（2）

(

1

，故只须求出

n

n1n

n1

n1n4x

，注意到

1 n1

1 ，n1

7

7n n

3（1）在(f(xexFourier 1求级数 21解（1）a1exdx1(ee 1

an

cosnxdx (e1

)，n1,2,，1

an

sinnxdx

1

(e

)，n1,2,，

所以f(x

(e

) 2(e11

f(x)ex在(

f(x)

(e

) 2(e11

)(cosnxnsinnx)，x(,)（2）x0，1e01sh2sh ， n11n2

1(11) 1．n11n 2 e f(x在[f()

f(，anbnf(xFouriera00,annbn,bn

(n1,2,)1 证明a0

f(x)dx

f

0a1f(x)cosnxdx1cosnxdf n 1 f(x)cos

f(x)sinnxdxn

f(x)sinnxdxnbnb1f(x)sinnxdx1sinnxdf n 1 f(x)sin

f(x)cosnxdxn

f(x)cosnxdxnan(n1,2,)

a0

cosnx

中的系数anbnmax{n3a

nn,nn

}MMnn证明因为max{nn

,

M

M

M，对一切n

nnnMn bMna0nnnMn

cosnx

sinnx

ancosnxbnsinnxxancosnxbnsinnx

2M由于级数2MMa0

nncosnxnn

sinnx在(n1 f(x，则由于逐次求导后的级数(nansinnxnbncosnx足nansinnxnbncosnxn(

bn)

，x(,)而级数

2M

sinnx

cosnx在(f(x在(f(x)(nansinnxnbncosnx)nnansinnxnbncosnx在(f(x在(设

(x)a0 k

coskx

T(x) T(x

sin(n12

dt 2

2证明象§14.12（1）类似地计算可得Tn(xFouriera0a0

ak

knkn

bk0

knkn

n1,2,，因而Tn(x)FouriernT(x)a0n k

coskx

而Tn(xFourier系数的前nT(x) T(x

sin(n12

dt 2

2f(x以2为周期，在(0,2bn0(n0证明f(xn0有

f(xsinnxdx存在．将[]n1 1

kbn f(x)sinnxdx

n2f(x)sin k

(kn1 (k1)

k 2nf(x)sinnxdx

f(x)sin

k 21 (k1) (k1) 2nf(x)sinnxdx 2nf(t

x)sinntdt]k

n

nnnxtn

1 (k1) bn 2n[f(x)f(x

k

nnnf(x2为周期，在(0,2)f(xf(x)0n[k12k1)2上sinnx0 非负，因而bn0(n0f(x以2

(02f(xan0(n0证明n0a12f(x)cos

f(x)dsin

n1(f(x)sinnx22f(x)sinnxdx)12[f(x)]sinnxdx nf(x2为周期，在(0,2上单调上升有界，故f(x2(0,2上单调减少有界，直接由上题结论，即知an0(n0f(xx0点满足Lipschitzf(xx0点连续．给出一个(0)证明f(xx0点满足阶(0Lipschitz00M0xx00 f(x)f(x)Mxx 因此，不妨设

x

00

f(x)f(x)Mxx

x

，取

,

0,x

M 0M f(x)f(x0)f(xx0

x0, lnf(x)lnln0

x

则由于

f(x)

0

f(0

f(xx连续．但0，由于limx

lnx0，所以

xln

，故

0，0当0

x

M

0Mx0

f(x)f(0)Mx01xln1ln对一切0xf(xx0Lipschitz1xln1lnf(x是以2[Sn(xf(x级数的前nnS(x)a0n k

coskx

sinkx)4f(x2t)f(x02 Sn(x)2 Dn(2t)dt，02Dn(tDirichlet证明

(x)

[f(xt)f(x

sin(n1

2sin2 [f(xt)f(x

t

2 [f(x2u)f(x2u)]D2 04f(x2t)f(x022 Dn(2t)dt．02f(x以2为周期，在(Fourierx0Sn(x0)f(x0)(n)证明f(x是以2Fourierx0Fourierx0S limS(x)S

n

)lim1nnn1k0

Sk

)S Fejerlim(x

f(x0，因此S

f(x0，即limSn(x0)n

f(x0)f(x以2Fourier系数全为0f(x)0证明x0f(xFourier系数全为0Fourierx00，由上题结论知其Fourierx0点又收敛于x0f(x)0

f(x0，因此f(x00，由f(x以2为周期，在[x0t

f(x0t)f(x0t)2limn(x0)Lf(xx0连续，则limn(x0)

f(x0n(x)1nn1

Sk(x)

证明Fejern(x0)

1

Sk(x0)

f(x0

n1tn1

2(n1)

sin

n1t

n1 t

f(x0t) dt

f(x0t) dt2(n1)

sin

sin

n1u

n1 t

f(x0 du

f(x0t) dt2(n1)

sin

sin

f

t)

f

n1t

dt2(n1)

sin 由Fejer核的性质

0

n1t2t

0

n1t2t

dtL

(n

(n

f

t)f

n1t

)L

L

dt (n1)0

0t

f(x0tf(x0t)L，知0（，使得只要20t

Lf(xf(x0t)f(x02

f

t)f

n1t

)L

L (n1)

f

t)f

n1t L (n1)

sin f

t)f

n1t L (n1)

sin

n1t2

n1tt Ⅰ(n1)0

2

dt2(n1)0

2

dt t2t

2

2t，因而当t[22时，22

2

，由此得到当t

2f

t)f

Ⅱ

L

(n

t

f

t)f

2 n n

L2

dtf(x在[绝对可积，因而M0f

t)f

L2

M1M (n1)32(n1)3Ⅱn(n1)32(n1)3 N2M，则当nN

，从而当nN

(x)LⅠ+Ⅱ

所以，limn(x0Lf(xx0Ln

f(x0limn(x0)n

f(x0) 任意区间上的级在区间(02lf(x)A,0xl0,lxf(x)xcosx，(,) f(x)x，(0,l) xf(x) 1

0x1x2,3x

2xa12l

1

解（1）

l

co(s)xdx

l0Acos

xdx

nb12lf(x)sinnxdx

lAsinnxdxA(1(1)n)，n1,2, l

l f(x)～A

A(1(1)n)sinnxA

2 sin(2n1)x． n1 n1(2n f(x在(02lf(x)A

2 sin(2n1)x，

x(0,l)(l,2l) n1(2n 由于f(x)xcosx是 2

的奇函数，因此

0，n0,1,2,．4 400bn2f(x)sin2nxdx2xcosxsin002 02x[sin(2n1)xsin(2n1)x]dx0

(4n

，n1,2,f(x在(2

216 (1)n1 f(x)(4n21)2sin2nx，x(2,2)2 2a0l0f(x)dxl0xdxl2lf(x)cos2nxdx2lxcos2nxdx0，n1,2, l

l 2lf(x)sin2nxdx

lxsin2nxdx

，n1,2, l

l f(x在(0lf(x)

l

1sin2nx，x(0,l)22

n1 a030f(x)dx3(0xdx11dx2(3x)dx)323f(x)cos2n 3 2(1xcos2nxdx2cos2nxdx3(3x)cos2n3

sin2n3

3n2

1)3

n1,2,23f(x)sin2n 3 (xsin12xdx 2cos(xsin12xdx 2cos133 n233f(x在[03

，n1,2,f(x)2{[2sin

(cos

1)]cos2n n2

2cos2n

sin2n)sin2n n2 2[2sin2n(x1) cos2n(x1)3 cos2nx]，

n2

n2 x[0,f(x)cosxf(x)x[x]．解（1）这是周期为f(x在(连续，逐段可微，又是偶函数，故bn0n1,2．4 4

4 000a02f(x)dx2cosxdx2cosxdx0004 400an2f(x)cos2nxdx2cosxcos002 02[cos(2n1)xcos(2n1)x]dx0

，n1,2,

f(x)cosx

2

x(,) n14n1 a020(x[x])dx20xdx an20(x[x])cos2nxdx20xcos2nxdx0，n1,2, bn20(x[x])sin2nxdx20xsin2nxdx

，n1,2,f(x)x[x]111sin ， )且 Z n1（1）f(x)sinx，0xx3（2）f(x)x3

0x22x4解Fourier2 a00sinxdx0

n1a2sinxcosnxdx

n1,2,0

[(1)n11],n1f(x)sinx

2

2

cosnx

24

cos2nx，x[0,]

n

n14n1222 a040f(x)dx2(0(1x)dx2(x3)dx)0a24f(x)cosnxdx1(2(1x)cosnxdx

4(x3)cosn 4

2

n2

[1(1)n]，

n1,2,f(x) [1(1)n]cosnx4

1cosnx，x(0,4)n1n2

n1n2 xf(x)2

，0xf(x)x2，0x2解（1）Fourierb

f(x)sinnxdx

[sin(n1)xsin(n

(4n2

n1,2,

f(x)cosx4 nx ， n14n22

2

bn2

f(x)

2xdx0

xdx

n3

1]n1,2,f(x)x2{8(1)n116[(1)n1]}sinnn1

n3 8

1[n22(1)n12(1)n2]sinnx，0x2n1 f(x)(x1)2在(0126(11 解Fourier

)a21f(x)dx21(x1)2dx21 11an20f(x)cosnxdx1

1)2cosnxdx

n2

，n1,2,

f(x)(x1)2141cos 2n1n2既使扩充f(x)的定义于()成为偶延拓后的周期为2F(xFourierF(xF(xF(0)f(0)1x0就有1141 ．，即 ．

n1n

n1n2f(xFourier

0x2f(x)