同济第六版《高等数学》教(学)案-第11章-无穷级数

word第十一

无穷级教目：P

e,sin

)

教重：

e

x

,sinxln(1))

教难1/

§11常项数的念性质nn23nu

n

nn

i23ni

nn

nn

{}n

limsnn

nns

123n

2/

nnnn{}n

nnnnnn

nn

aq

n

2

n

n

nnsn

2

a11

limnn

nn

limsnn

n

nna3/

n1n111n1n111s

n

n

n

n

n

n

n

(

limsnn

1nn

1n

23()2nlimlim(1n

4/

1n111111n11111

n

1(n

23()3nn

limlim(1n

n

1n

1nn

n

1

n

n

nn

nn

nlimn

lim(un1212nn

nn5/

n

u)nn

n

n2

n

)nn

nnnn

u)nn

nlimn

n

nnlim[(u))112nnlim[(u1n2nlim(nn

性3

12n

10000

1123(

14n(

nn

推级收的要条如nn

unn

如

nn

unn

6/

111111111n

nn

s

limsnn

ulim(s)limnnnnn0nn

证n

n3

n

1n

n

1n

limsnn

lims2n

()2nnn

snnnn2n

lim()nnn

n

1n

§112常项数审敛一正级与其敛

7/

n

n

n

nnnn

n

nnnnn

n

nnnnnn

n

n

nn

n

unnnnn8/

1111[1111[]]

n

n

n

nunnn讨

n

n

n2ppnp

(

1nn

n

1n

n

1n

1nnnnnx(npn

]

[n

1(npnp

]

]p

113np

]

(

s

(

]

[n

1(npnp

可

n

1n

时

1np

9/

dx[]]dx[]]n

1npn

n

1n

n

1n

npnx(p

]

[n

1(npnp

]

n

1n

[n

1(npnp

]

1p2p3np

]

(

limslim[1nnn

1(n

p

]

[n

1(npnp

]

n

1n

n

(

111(n(n2

10/47

llllvnnn

n

11n3n

limn

nn

n

n

n

limn

nn

nn

limn

nn

n

nn

nnnnnnnn

l

u32n

1判

sin

1n

11/47

lim

sin

n

1n

sin

1n

1n2

)

lim

ln(1)22

n

1n

ln(1n

1n

)

nlimnn

limn

nn

limn

nn

limn

nn

12/47

limlimnnnlimnlimlimnnnlimn1limn

nn

limn

nn

111

limn

1nn

10210

!

limn

(10nn!n10

n

1n

limn

nn

limn

(2(2

1(2n

n2n

1(2n

n

1n2

13/47

limn

nn

limn

1(2n

n

1n2

的nlimnu

limn

nun

limnu

limnu

limn

112n

s近s

u

limn

14/47

n12111()2s所n12111()2r

11((nn

1((n

(

n

22n

n

limnlim

2

12

limnulimnun

limn

p

u(0n

判

ln(1n

1n2

)

~(nn

lim2ulimnln(1limn2nn

判

n)nn

3limn2unn

limnn

32

)limnn

2

1

15/47

n

n

n

n

n

n

1n

n

n

1nn

n

n

n

limunn

rn

n

n

n

nn

limnnrn项ssnnnn

snn

n

n证

n

n

1n

16/47

11n1uu11n1uu

un

limu

n

1n

(n

n

n

n

1n

u

n判

n

nan

17/47

(1)111(1)(1)111(1)()()

sinna1n2n2

n

1n

n

sinn

n

nan

判

n

n

12nn

n

(1)2

2

limnu|lim(1e22

limunn

n

n

12nn

n

§113幂级一函项数的念uI上

()

x

()0

n

()

()

18/47

()(x()(xu(x)

()

x

nn

()

x(n

()

u(nulims(x))nn

()

n

()

的项是的n

limr)

n19/47

xxnxnn

11!

na

(11

x

xx

xaxnx是

x

设

limx

ax

1,

n

|an

nxn

n0

|n|x00

20/47

x|annnxnxx|annnxnxx

M|nx0

n

xxnnax

|n

xnn

xx0

|an00

|nx0

|0

M

xx

nx

n

R存

x都

lim|

21/47

0

xn

lim|n

nn

0

lim|n

nn

x

|n

axnan

nn

|n

anan

x|

1

(

x323

求

(

lim1

22/47

11lim11lim

R

)n

!

x

2x2!!求

1!

1

a(nn!nnn!

求

!xn

lim|

(|limn!

求

(2)!(!)

x

2

不

()n()nlimx()nn

)!(!)

2

23/47

xxxxx()n

x

12

12

)()

[2(x2([(2(2n)!x2n(!)2

(22)(2(2

x

求

(nn

tnn

n12

1(tnnnn

x

x

(

)x

x

nn

xxnnxxnn

)

bb

bn24/47

(x)dxx)(xax(x)dxx)(xaxxs()))x)dx)dxan幂

xn

幂nn

0

xn

n

幂

n

ann

nn

幂

幂x

n

x

n

幂

a)x)naxnn

求

x

(x

x

xs()

x

[(x

到25/47

111111111111xs()

x

11

)

(x)ln(1

ln(1)0(xxx

xs()

x

[x

1

)

(x)ln(1)

ln(1)0(x)x1

求

x

(x

x

xs()

x

[x

1

)(

0

()

ln(1)0(x)x1

S(limS(x)2

)x(x)x126/47

F((0)FF((0)F(

F

11

(

1n

(

s(

12n

§11函展成幂数一泰级xf)fx)0

)(x0

f

02!

(x)0

2

f((x)!

(x)

(x)

R()n

f

(n((

()0

n

x之的

)f(x)

)0

ff(n)x()202!!

()

27/47

0000f(x)f0

x00

f

f0()2(x2!3!

3

f(n(x)0(x)n!

xR0limR(x((

f()fx)

f

f()(x)()0(x2!!

snnnn)f

f

(n)2xnn!

xxn

28/47

11

2

f2!

fnn!

xx处

f

f

f(n(0)22!!

Rlim(x)nnn

f

(n(

n

f()f(0)

f

(2!!

1n!

(xn

|nn|x((

29/47

11|x2x11|x2x5x2n11

limn

x(

R(x

x

x2n!

f

()

sin(

f

3xx2n3!5!n

x|(

(n]x(nn

0x3!(2

x2n!

将

x的

m(m2!

30/47

3xn2x2x33xn2x2x3xnxxdx[(1)

(mm(22!!

n

将xsin3!5!(2

cosx2!(2)!

将

(

2

x的

11

11

将x11

x

11

2

到x34

)]

1

xn

31/47

2x)()sin())2x)()sin())x)(x)()4(1))n1)3xx将x(x)4

x)][cos())]444

142!4

1443!4

x

42!3!4

将

fx)

x

1x

f()

111xx(x2(1))4

n

n

(1(x84n

n

(

(

x22

(

1x

x2

11

1x21x4

nn

nn

(x2(x4

nn

(2(4

xx24

32/47

11xx4n2x3xn5x11xx4n2x3xn5x111111

2

x

x2n!

sin3!5!(2

cos2!(2)!

34

(1

m(m(m22!n!

n

§115函数幂数展式应一近计

5

240

240

)1/5

1153

24053

112333

5r

12383125316

[1)2588133/47

111112x3xn2x3xln)xx111112x3xn2x3xln)xx1111113

24054

)

5

2.9926

ln

ln

ln223n

r

1n

10

34x)341115

11

xx33ln23553

34/47

r2112)0.3333311119sin1sinr2112)0.3333311119sin1sin53[1)9

211131147000009

1113335373

111333

53

77

ln

3

180201203!20

x

1120203!7!sin

20

r2

1

115120300000

20

35/47

11n1111111n111112

0

dx0

0.56419

求

2

0

dx

e

2

()()2()1!!n

2

dx

2

2nn!

2

(n!

2n

dx(1

226

|r4

28

2

dx

(1)0.52952242

10

0

10x

36/47

x24xdxx24xdxlimx

sinxx

1sin10x

11730000

1x0x

nunn

u)

u22n

(u)n

n37/47

11z(iy)11z(iy)yyyy11ixixixcos()xixix12!n!

z

112!!

iy

12!!

113!5!

x可z的esin+

12i

)

38/47

0n0n§11.7傅叶数一三级角函系正性12

a0

n

(acossin)n

cosnxdx

cos

sin

cosnxdx

1

dx

cos2nxdx

nxdx

39/47

aaf()0coskx)2

f(nxdx

a02

nxdx

[

nxdx

kxcosnxdx]

fnxdx

a

1

f(x)dx

an

1

f()cos

bn

1

fxaa2

(acosnxnx?如一的x

12

[(x(0)]

设

40/47

f()11111n1cosnxf()11111n1cosnx[]4411f()[((x(010[f(xf(x22

af(x)cos(0

nbn

1

f(nxdx

1

(nxdx

1

[]0n

1cos1

2n

n3,5,n0n6,f()[sinxsin3xsin(2k32设0

122

41/47

1x[]bf()sinnxdxxsin[n1x[]bf()sinnxdxxsin[n(sincos3sin3sin4x1f)24a0

1

f(x

xdx

2

a

1

f(x

11sin1n

(12nn6,111cossinn

]0

cosnn

(

13145

cos5)5

将0

a0

1

f(x)

1

(dx

1

xdx

an

1

f(xnxdx

1

()cosnxdx

1

(cosn

n5,204,42/47

42n042n0bn

1

fx

1

()sin

10

f()

x22

12

cos5

(bf()sin

n

bsinnxn

a

2

f(

(a02

nn

[f([

bn

2

fx

2

sinnxdx43/47

[211ntEE]4[]2[211ntEE]4[]

]n

fx)xsin2sin3x2n

将()t2

an

2

(ntdt

2

Esinntdt2[sin(n))2[

cos()tcos(t2

4n

ut)(nt)24n

将

bn

2

fx)sin

2

(nxdx

2x