习题十二
1.写出下列级数的一般项:
1+
(1)
13
+
15
+
17
+
;
(2)2
3
+
x 2⋅4
5
+
9
2⋅4⋅6
+
x
2
2⋅4⋅6⋅8
+
;
a
(3)3
-
a
5
+
a
7
71
-
a
9
+
;
解:(1)
U n =
2n -1;
n
U n =
(2)
x 2
(2n )!! ;
n +1
2n +1; (3)
2.求下列级数的和:
∞
U n =(-1)
a
2n +1
(1)
∑(
n =1
1
x +n -1)(x +n )(x +n +1);
∞
(2)
∑n =1
+15
3
;
1
(3)5
+
15
2
+
;
u n ==
解:(1)
1
(x +n -1)(x +n )(x +n +1)
111⎛⎫
- ⎪
2⎝(x +n -1)(x +n )(x +n )(x +n +1)⎭
S n =
11111⎛
-+-
2⎝x (x +1)(x +1)(x +2)(x +1)(x +2)(x +2)(x +3)
1
+ +
(x +n -1)(x +n )
-
⎫
(x +n )(x +n +1)⎪⎭
1
从而
111⎛⎫
-= ⎪
2⎝x (x +1)(x +n )(x +n +1)⎭
lim S n =
因此
n →∞
11
2x (x +1),故级数的和为2x (x +1)
(2)
因为
U n =
-
S n =
-
-
-1-
+1-
+
-
-
+
==
从而所以n →∞
+ +
-
lim S n =1-
S n =
15+15
2
1-
+ +
15
n
n
1⎡⎛1⎫⎤⎢1- ⎪⎥5⎣⎝5⎭⎦=
11-
5n
1⎡1⎫⎤⎛=⎢1- ⎪⎥4⎣⎝5⎭⎦
(3)因为
4,即级数的和为4. 从而n →∞
3.判定下列级数的敛散性:
∞
lim S n =
11
(1)
∑n =1
;
1
(2) 1⋅6
+2323
16⋅11++23
33
+
111⋅16
+ +
1
(5n -4)(5n +1)
23
n n
+
;
2
(3) 3
-
- +(-1) +
n -1
+
;
1
(4)5
+
;
S n =
解:
(1) 从而n →∞
+
-1
,故级数发散.
+ +
-
=
lim S n =+∞
S n ==
(2) 从而n →∞
1⎛1111111⎫1-+-+-+ +- ⎪5⎝661111165n -45n +1⎭1⎛1⎫ 1-⎪5⎝5n +1⎭
12
3的等比级数,且|q |
lim S n =
1
5,故原级数收敛,其和为5.
q =-
(3)此级数为
U n =
(4)
∵
lim U =1≠0n →∞n ,故级数发散.
4.利用柯西审敛原理判别下列级数的敛散性:
∞
(1)
∑
n =1∞
(-1)
n ⎛
1
n +1
∞
; (2)
∑
n =1
cos nx 2
n
;
(3) n =1
解:(1)当P 为偶数时,
+-⎪∑
⎝3n +13n +23n +3⎭
11⎫
.
n +1+U n +2+ +U n +p =
(-1)
1n +11
n +2
n +1
-
+1
(-1)
n +3
n +2+
1
+
(-1)
n +4
n +3- -
+ +1n +p
(-1)
n +p +1
n +p
=
n +2n +3
=
111⎛⎫1⎫⎛1
-- - ---⎪ ⎪
n +1⎝n +2n +3⎭⎝n +p -2n +p -1⎭n +p 1n +1
当P 为奇数时,
n +1+U n +2+ +U n +p =
(-1)
1n +11
n +2
n +1
-
+1
(-1)
n +3
n +2+
1
+
(-1)
n +4
n +3- +
+ +1n +p
(-1)
n +p +1
n +p
=
n +2n +3
=
11⎫⎛1⎫⎛1
-- - --⎪ ⎪
n +1⎝n +2n +3⎭⎝n +p -1n +p ⎭1n +1
1n +1
因而,对于任何自然数P ,都有
U n +1+U n +2+ +U n +p
N =
∀ε>0,取
⎡1⎤
+1⎢⎣ε⎥⎦
,则当n >N 时,对任何自然数P 恒有
n +1+U n +2+ +U n +p
成立,由
∞
柯西审敛原理知,级数n =1
(2)对于任意自然数P ,都有
∑
(-1)
n
n +1
收敛.
n +1+U n +2+ +U n +p =≤
cos (n +1)x
212
n +1
n +1
+
cos (n +2)x
22
n +2
+ +
cos (n +p )x
2
n +p
+
12
n +2
+ +
1
n +p
1⎛1⎫
1-n +1 p ⎪2⎝2⎭
=
11-
2=
1⎛1⎫
1- n p ⎪2⎝2⎭12
n
1⎤⎡
log 2⎢+U n +2+ +U n +p 0(0N 时,对任意的自然数P 都有n +1
成立,由柯西审敛原理知,该级数收敛.
(3)取P =n ,则
n +1+U n +2+ +U n +p
111111⎛⎫
= +-+ ++-⎪
3⋅2n +13⋅2n +23⋅2n +3⎝3(n +1)+13(n +1)+23(n +1)+3⎭≥≥>
13(n +1)+1
n 6(n +1)112
+ +
13⋅2n +1
12,则对任意的n ∈N ,都存在P =n 所得从而取
理知,原级数发散.
5.用比较审敛法判别下列级数的敛散性.
ε0=
1
n +1+U n +2+ +U n +p >ε0
,由柯西审敛原
1
(1)4⋅6
+
15⋅7
2
+ +1+31+3
2
1
(n +3)(n +5)
+ +
1+n 1+n
2
+
;
1+
(2)
∞
1+21+2
++
∞
(3)
∑
n =
1∞
sin 1
π3;
n n
∑
(4)
n =1∞
1n
;
(5)
∑1+a
n =1
(a >0)
;
(6)
∑(2
n =1
-1
)
.
U n =
解:(1)∵
1
(n +3)(n +5)
1n
2
∞
而
∑
n =1
1
n 收敛,由比较审敛法知
1+n 1+n
2
∞
2
∑U
n =1
n
收敛.
(2)∵
∞
U n =
≥
1+n n +n
2
=
1n
而
∑
n =1
1
n 发散,由比较审敛法知,原级数发散.
sin lim
(3)∵
∞
πn
∞
n
sin
=lim π⋅
n →∞
πn n
n →∞
13
π3π
=π
而
∑3
n =1
π
n
收敛,故
∑
n =1
sin
3也收敛.
n
U n =
(4)
∵
∞
=
1
3
n 2
∑
而
n =1
1
3
∞
n 2收敛,故
∑
n =1
1
收敛.
∞
(5)当a >1时,
U n =
1a ,而=121≠0
n
1+a
n
∑
n =1
1a
n
∞
收敛,故
∑1+a
n =1
1
n
也收敛.
当a =1时,
lim U n =lim
n →∞
n →∞
,级数发散.
n
当0
综上所述,当a >1时,原级数收敛,当0
1
lim U n =lim
n →∞
n →∞
1+a
=1≠0
x (6)由知
6.用比值判别法判别下列级数的敛散性:
x →0∞
lim
2-1
x
lim
2n -11n
=ln 2
x →∞
=ln 2
而
∞
∞
∑
n =1
1
n 发散,由比较审敛法知
∑(2
n =1
∞
1n
-1
)
发散.
(1)
∑
n =1
n
2n
3;
(2)
∑3
n =1
n !
n
+1;
3
(3)1⋅2
∞
+
n
3
22
2⋅2n
n
+
3
33
3⋅2
+ +
3
n n
n ⋅2
+
;
(1)
∑
n =1
2⋅n !
2
U n
3,解:(1)
由比值审敛法知,级数收敛.
n
n →∞
U n =
n
2
lim
U n +1
=lim
(n +1)
3
n +1
n →∞
⋅
3n
n 2
=
13
,
lim
U n +1U n
n →∞
(n +1)! 3+1=lim n +1⋅n →∞3+1n !
=lim (n +1)⋅
n →∞
n
3+13
n +1
n
+1
=+∞(2)
所以原级数发散.
lim
U n +1U n
n →∞
=lim
33n
n +1
n +1
n →∞
(n +1)⋅2
2(n +1)
⋅
n ⋅23
n
n
=lim =3
n →∞
2(3)
所以原级数发散.
>1
lim
U n +1U n
n →∞
=lim
2
n +1
⋅(n +1)!
n +1n
n →∞
(n +1)
⋅
n
n
n
2⋅n !
⎛n ⎫
=lim 2 ⎪n →∞
⎝n +1⎭=2lim
11⎫⎛
1+⎪
n ⎭⎝
n →∞
n
=
2e
(4)
故原级数收敛.
7.用根值判别法判别下列级数的敛散性:
⎛⎫
⎪∑ 3n +1⎝⎭; n =1(1)
⎛n ⎫
⎪∑
(3) n =1⎝3n -1⎭
∞
n
∞
5n
n ∞
∑
(2)
n =1
1
[ln (n +1)]
n
;
∞
2n -1
;
⎛b ⎫∑ a ⎪
(4) n =1⎝n ⎭,其中a n →a (n →∞),a n ,b ,a 均为正数.
lim
解:
(1)
故原级数发散.
n →∞
=lim
5n 3n +1
n →∞
=
53
>1
,
lim
(2)
故原级数收敛.
n →∞
=lim
1ln (n +1)
2-
n →∞
=0
,
1n
lim
(3)
故原级数收敛.
n →∞
⎛n ⎫
=lim ⎪n →∞
⎝3n -1⎭
=
19
,
lim
(4)
n →∞
b b =lim =n →∞a a , n
b b b
当b a 时,a >1,原级数发散;当b =a 时,a =1,无法判定其敛散性. 8.判定下列级数是否收敛?若收敛,是绝对收敛还是条件收敛?
1-
(1)
1+
1-
1∞
;
(2)
∑(-1)
n =1
n -1
1ln (n +1);
11111111
⋅-⋅2+⋅3-⋅4+
5353(3) 5353;
∞
(4)
∑(-1)
n =1∞
n -1
2
n
2
∞
n ! ;
n
(5)
∑(-1)
n =1
n -1
1n
α
(α∈R )
;
111⎫(-1)⎛
1+++ +⎪∑
23n ⎭n . (6) n =1⎝
U n =(
-1)
解:(1)
n -1
∞
∞
n =1
∑
U n
∞
1
>
∞
n
lim
n →∞
=0
,由莱布
∑
尼茨判别法级数收敛,又
n =1
n =
∞
∑
n =1
1n
2
是P
∑
n =1
发散,故原级数条件收敛.
U n =(-1)
(2)
n -1
1ln (n +1),
∑
n -1
(-1)
n -1
111ln (n +1)
ln (n +1)为交错级数,且ln (n +1)
>
1ln (n +2),1n +1
lim
1ln (n +1)
∞
n →∞
=0
,由莱布尼茨判别法知原级数收敛,但由于
U n =
≥
所以,
∑
n =1
U n
发散,所以原级数条件收敛.
n -1
(3)
∞
U n =(-1)
1
5⋅3民,显然
n
∞∞
∑
n =1
n =
∑5⋅3
n =1
1
n
=
1
∞
∑3
5
n =1
1
n
∞
,而
∑3
n =1
1
n
是收敛的等比级数,故
∑
n =1
n
收敛,所以原级数绝对收敛.
lim
(4)因为故可得∴n →∞
U n +1U n
n →∞
=lim
2
2n +1
n →∞
n +1
=+∞
. ,
n +1>n
,得n →∞
lim n ≠0
lim U n ≠0
,原级数发散.
∞
(5)当α>1时,由级数
∑
n =1
1n
∞
α
收敛得原级数绝对收敛.
n -1
当0
∞
∑(-1)
n =1
n -1
1n
α
1
满足条件:n
∞
α
>
1
α
(n +1);n →∞n α
lim
1
=0
,由莱布尼茨判别法
知级数收敛,但这时当α≤0时,n →∞
∑
n =1
(-1)
1n
α
=
∑
n =1
1n
α
发散,所以原级数条件收敛.
lim U n ≠0
,所以原级数发散.
111⎛1+++ +
23n (6)由于⎝
∞
1⎫1
⋅>⎪
n ⎭n
而
∞
∑
n =1
1
n 发散,由此较审敛法知级数
∑
n =1
111⎛
1+++ +
23n ⎝⎫(-1)
⎪⋅
n ⎭
n
发散.
111⎛
U n = 1+++ +
23n ⎝记⎫1
⎪⋅
⎭n ,则
1111⎫⎛11⎫⎛
U n -U n +1= 1+++ +⎪ --⎪2
23n ⎭⎝n n +1⎭(n +1)⎝
11111⎫⎛
= 1+++ +⎪-2
23n ⎭n (n +1)(n +1)⎝
111⎛⎫1⎫⎛11
-= ++ +⎪+ 2⎪
n ⎭n (n +1)⎝n (n +1)(n +1)⎭⎝23>0
即U n >U n +1
lim U n =lim
n →∞
n →∞
1⎛111⎫
1+++ + ⎪n ⎝23n ⎭
n
=
又
1
⎰n
1x
d x
1
由t →+∞t
lim
1
⎰
t 0
1
d x =lim t =0t →+∞1x
∞
知n →∞
9.判别下列函数项级数在所示区间上的一致收敛性.
∞
lim U n
111⎛1+++ + ∑=0
23n ,由莱布尼茨判别法,原级数n =1⎝
n
∞
⎫(-1)
⎪⋅
n 收敛,而且是条件收敛.⎭
n
(1)
∑(
n =1∞
x
n -1)! ,x ∈[-3,3];
n
(2)
∑
n =1
x
n 2
n ,x ∈[0,1];
(3)
∑
n =1
sin nx 3
,x ∈(-∞,+∞) ;
(4)
∞
∑
n =1
e
-nx
n ! ,|x |
∞
∑
(5)
n =1
x ∈(-∞,+∞)
x
解:(1)∵
n
(n -1)!
≤
3
n
(n -1)!
,x ∈[-3,3],
∞
而由比值审敛法可知
∑(
n =1
3
n
n -1)! 收敛,所以原级数在 [-3,3]上一致收敛.
x
(2)∵n
n 2
≤
1
n ,x ∈[0,1],
2
∞
而
∑
n =1
1
n 收敛,所以原级数在[0,1]上一致收敛. sin nx 31
n
2
(3)∵
∞
n
≤
1
3,x ∈(-∞,+∞) ,
n
而
∑3
n =1
是收敛的等比级数,所以原级数在(-∞,+∞) 上一致收敛.
-nx
e
≤
e
5n
(4)因为n !
n ! ,x ∈(-5,5) ,
∞
由比值审敛法可知
∑
n =1
e
5n
n ! 收敛,故原级数在(-5,5) 上一致收敛.
5
(5)
∞
1
5
≤
1
n 3,x ∈(-∞,+∞) ,
∑
而
n =1
n 3是收敛的P -级数,所以原级数在(-∞,+∞) 上一致收敛.
∞
10.若在区间Ⅰ上,对任何自然数n .都有|U n (x )|≤V n (x ) ,则当
∞
∑V
n =1
n
(x )
在Ⅰ上一致收敛时,级数
∑U
n =1
n
(x )
在这区间Ⅰ上也一致收敛.
∞
证:由n =1在Ⅰ上一致收敛知, ∀ε>0,∃N (ε)>0,使得当n >N 时,∀x ∈Ⅰ有 |V n +1(x )+V n +2(x )+…+V n +p (x )|
于是,∀ε>0,∃N (ε)>0,使得当n >N 时,∀x ∈Ⅰ有
|U n +1(x )+U n +2(x )+…+U n +p (x )|≤V n +1(x )+V n +2(x )+…+V n +p (x ) ≤|V n +1(x )+V n +2(x )+…+V n +p (x )|
∞
∞
n
∑V
n
(x )
因此,级数n =1在区间Ⅰ上处处收敛,由x 的任意性和与x 的无关性,可知收敛.
11.求下列幂级数的收敛半径及收敛域:
∑U
(x )
∑U
n =1
n
(x )
在Ⅰ上一致
(1)x +2x 2+3x 3+…+nx n +…;
∞
⎛x ⎫
⎪∑n ! n ⎝⎭; (2)n =1
∞
∞
n
(3)
∑
n =1
x
2n -1
2n -1;
(4)
∑
n =1
(x -1)
2
n
n ⋅2n ;
ρ=lim
解:(1)因为
∞
a n +1a n
n →∞
=lim
n +1n
n
n →∞
=1
,所以收敛半径
∞
R =
1
ρ
=1
收敛区间为(-1,1) ,而当x =±1时,
级数变为
∑(-1)
n =1
n
n
,由
lim (-1) n ≠0
x →n
知级数
n
∑(-1)
n =1
n
n
发散,所以级数的收敛域为(-1,1) .
n
-1
ρ=lim
(2)因为
a n +1a n
n →∞
=lim
(n +1)! (n +1)
n +1
n →∞
n
⎡⎛⎛⎫1⎫⎤
⋅=lim ⎪=lim ⎢ 1+⎪⎥n →∞n ! n →∞⎝n +1⎭n ⎭⎦⎣⎝
n n
=e
-1
R =
所以收敛半径
1
ρ
=e
,收敛区间为(-e,e) .
∞
当x =e时,级数变为
∑
n =1
e n n
n
n
1
lim
;应用洛必达法则求得
(1+x )x -e
x
x →0
=-
e 2,故有
⎛a n +1⎫1
-1⎪=-
n →∞2⎝a n ⎭由拉阿伯判别法知,级数发散;易知x =-e 时,级数也发散,故收敛域为(-e,e) .
(3)级数缺少偶次幂项.根据比值审敛法求收敛半径.
lim
U n +1U n
n →∞
=lim
x
2n +1
n →∞
2n +12n -12n +1
⋅
2n -1x
22n -1
=lim =x
2
n →∞
⋅x
所以当x 21即|x |>1时,级数发散,故收敛半径R =1.
1
∞
当x =1时,级数变为
∞
∑
n =1
1
2n -1,当x =-1时,级数变为
∞
∞
∑
n =1
-12n -1,由
1
lim =>0n →∞12
n
知,
∑
n =1
1
2n -1发散,从而
∑
n =1
-1
2n -1也发散,故原级数的收敛域为(-1,1) .
∞
(4)令t =x -1,则级数变为n =1n ⋅2n ,因为
所以收敛半径为R =1.收敛区间为 -1
n →∞
∞
∑
t
2
n
ρ=lim
a n +1a n
=lim
n ⋅2n
2
n →∞
(n +1)⋅2(n +1)
2
=1
当t =1时,级数n =12n 收敛,当t =-1时,级数敛.
所以,原级数收敛域为 0≤x ≤2,即[0,2]
12.利用幂级数的性质,求下列级数的和函数:
∞
∑
1
3
∞
∑(-1)
n =1
n
1
2⋅n 为交错级数,由莱布尼茨判别法知其收
3
∞
n +2
(1)
∑nx
n =1
; (2)
n +3
∑
n =0
x
2n +2
2n +1;
∞
lim
(n +1)x
n +2
nx 解:(1)由知,当|x |=
从而发散,故级数的收敛域为(-1,1) .
∞
∞
n +2
∞
n -1
n →∞
=x
∑nx
n +2
的通项不趋于0,
∞
S (x )=
记
∑nx
n =1
∞
=x
3
∑nx
n =1
易知
∑nx
n =1
n -1
S 1(x )=
的收敛域为(-1,1) ,记
∑nx
n =1
n -1
则
⎰
x 0
S 1(x )=
∑
n =1
x =
n
x 1-x
⎛x
S 1(x )=
⎝1-x 于是
lim
(2)由
3'x 1⎫
S (x )=⎪=22
(1-x ),所以(1-x )⎭
(x
x
2n +4
n →∞
2n +3
⋅
2n +1x
2n +2
=x
2
知,原级数当|x |
∞
S (x )=
域为(-1,1) ,记
∞
∑
n =0
x
2n +2∞
2n +1
=x ∑
n =0∞
x
2n +1∞
2n +1,易知级数n =02n +1收敛域为(-1,1) ,记=
11-x ,
12ln 1+x
1-x ,S 1(0)=0,所以
2
∑
x
2n +1
S 1(x )=
∑
n =0
x
2n +1
2n +1,则
12ln x
S 1'(x )=
∑
n =0
x
2n
故
⎰
x 0
S 1'(x )d x =
1+x 1-x 即1+x
S 1(x )-S 1(0)=
21-x
13.将下列函数展开成x 的幂级数,并求展开式成立的区间: (1)f (x )=ln(2+x ) ; (2)f (x )=cos2x ;
S (x )=xS 1(x )=
ln
(x
f (
x )=
(3)f (x )=(1+x )ln(1+x ) ;
(4)
x
2
12
;
(5)
f (x )=
x 3+x ;
2
(6)
f (x )=(e x -e -x )
1
2
;
f (x )=
(7)f (x )=ex cos x ;
(8)
(2-x ).
x ⎫x ⎫⎛⎛
f (x )=ln (2+x )=ln 2 1+⎪=ln 2+ln 1+⎪
⎝2⎭⎝2⎭ 解:(1)
∞
ln (1+x )=
由于
∑(-1)
n =0∞
n
n
x
n
n +1,(-1
n +1
n +1
x ⎫⎛
ln 1+⎪=
2⎭故⎝
∑(-1)
n =0
(n +1)2
∞
,(-2≤x ≤2)
ln (2+x )=ln 2+
因此
∑(-1)
n =0
n
x
n +1
n +1
(n +1)2
,(-2≤x ≤2)
(2)
f (x )=cos x =
∞
2
1+cos 2x
2
cos x =
由
∑(-1)
n =0
∞
n
x
2n
(2n )! ,(-∞
n
cos 2x =
得
所以
∑(-1)
n =02
(2x )
2n
∞
(2n )!
12
n
=
∑(-1)
n =0
n
4⋅x
n 2n
(2n )!
f (x ) =cos x =
=12+1
∞
12
+cos 2x 4⋅x
n
2n
(-1)∑2
n =0
(2n )! ,(-∞
x
n +1
(3)f (x )=(1+x )ln(1+x )
∞
ln (1+x )=
由
所以
∑(-1)
n =0
n
(n +1),(-1≤x ≤1)
∞
f (x )=(1+x )∑(-1)
n =0
∞
n
x
n +1
n +1
∞
=
∑(-1)
n =0
∞n =1∞
n
x
n +1
n +1
n
+
∑(-1)
n =0
∞
n
x
n +2
n +1
n +1
=x +
∑(-1)∑
n =1∞
x
n +1
n +1
n
+
∑(-1)
n =1n +1
x
n +1
n ⋅x
n +1
=x +
(-1)n +(-1)(-1)
2
n -1
(n +1)
n (n +1)
n (n +1)
2
=x +
∑
n =1
x
n +1
(-1≤x ≤1)
f (
x )=
(4)
=x ∞
=1+
∑(-1)
n =1
n
(2n -1)!! (2n )!!
x
2n
(-1≤x ≤1)
⎛
f (x )=x 1+
⎝故
2
∞
∞
∑
n =1
(-1)
n
n
(2n -1)!! (2n )!! (2n )!!
x
x
2n
⎫
⎪⎭
=x +
2
∑
n =1
(-1)
(2n -1)!!
2(n +1)
(-1≤x ≤1)
f (x )=
x 3x
⋅
11+
∞
x
2
3
n
2
n ⎛x ⎫
=⋅∑(-1) ⎪3n =0⎝3⎭
∞
=
(5)
∞
x
∑(
-1)
n =0
n
x
2n +1n +1
3
(x
e =
(6)由
∑
n =0
x
n
n ! ,x ∈(-∞,+∞)
n
n
∞
e
得
-x
=
∑
n =0
(-1)⋅x
n !
,x ∈(-∞,+∞)
f (x )=
12
(e x -e -x )
∞
1⎛∞x n = ∑-2⎝n =0n ! =12
∞
∑
n =0
n
n n
(-1)x ⎫
⎪n ! ⎭
∑⎡⎣1-(-1)
n =0∞
⎤⎦⋅
x
n
n !
=
所以
∑(
n =0
x
2n +1
2n +1)!
x
x ∈(-∞, +∞)
(1+i )x
x
(7)因为e cos x 为e (cos x +i sin x )=e
的实部,
∞
e
(1+i )x
=
∑
n =0∞
1n ! x
n
[(1+i )x ]
(
1+i )
n
n
=
∑
n =0∞
n !
n
=
∑
n =0∞
x ππ⎫⎤cos +i sin ⎪⎥n ! 44⎭⎦x
n
n
n
=
而
∑
n =0
n πn π⎫⎛⋅22 cos +i sin ⎪n ! ⎝44⎭
n
∞
22cos
n !
n π⋅x n
(-∞
e cos x =
取上式的实部.得
x
∑
n =0
1
(8)由于(1-x )
2
∞
=
∑nx
n =0
n -1
|x |
2
f (x )=
而
1
4⎛x ⎫ 1-⎪⎝2⎭,所以
∞
n -1
∞
⋅
1
⎛x ⎫
f (x )=⋅∑n ⎪
4n =0⎝2⎭
14.将
2
1
=
∑
n =0
n ⋅x 2
n -1
n +1
(|x |
f (x )=1
1
x +3x +2展开成(x +4)的幂级数. =
1x +1
-
1x +2
2
解:x +3x +2而
1x +1
=
1-3+(x +4)131⋅1-
∞
=-
1x +43
n
⎛x +4⎫
=-∑ ⎪
3n =0⎝3⎭
∞
⎛x +4⎫
⎝3⎭
=-∑
n =0
(x +4)
3
n +1
n
(-7
又
1x +2
=
1-2+(x +4)1211-
∞
=-
1x +42
n
⎛x +4⎫
=-∑ ⎪
2n =0⎝2⎭
∞
⎛x +4⎫
⎝2⎭
=-∑
n =0
(x +4)
2
2
n +1
n
(-6
f (x )=
1x +3x +2
∞
=-∑
n =0∞
(x +4)
3
n +1
n
∞
+
∑
n =0
(x +4)
2
n +1
n
=
所以
1⎛1-∑ n +1n +1
3n =0⎝2
⎫n
()x +4⎪⎭
(-6
15.将函数f (
x )=解:因为
(x -1) 的幂级数. m (m -1)
2!
2
(1+x )=1+
所以
m
m 1!
x +x + +
m (m -1) (m -n +1)
n !
x +
n
(-1
f (
x )=
3
=[1+(x -1)]2
3⎛33⎛3⎫⎫⎛3⎫⎛3⎫
-1-1-2 ⎪ ⎪ ⎪ -n +1⎪2⎝22⎝22n ⎭⎭⎝2⎭⎝2⎭
=1+(x -1)+(x -1)+ +(x -1)+
1! 2! n !
3
(-1
即
f (x )=1+
∞
32
(x -1)+
3⋅12⋅2!
2
(x -1)+
n
2
3⋅1⋅(-1)2⋅3!
3
(x -1)+
3
3⋅1⋅(-1)⋅(-3) (-2n +5)
2⋅n !
n
(x -1)+
n
=1+
∑
n =1
3⋅1⋅(-1)⋅(5-2n )
2⋅n !
n
(x -1)(0
16.利用函数的幂级数展开式,求下列各数的近似值: (1)ln3(误差不超过0.0001); (2)cos20(误差不超过0.0001)
352n -1
⎛⎫x x x
ln =2 x +++ ++ ⎪
352n -1⎝⎭,x ∈(-1,1) 解:(1)1-x
1+x
1+x
令1-x
=3
,可得
x =
12
∈(-1,1)
,
111⎡1⎤2+++ ++ ln 3=ln =2⎢352n -1⎥123⋅25⋅2(2n -1)⋅2⎣⎦1-
2故
又
1+
1
11⎡⎤
++ r n =2⎢2n +12n +3⎥(2n +3)⋅2⎣(2n +1)⋅2⎦=
2
(2n +1)2
2
2n +1
2n +12n +1
⎡⎤(2n +1)⋅2(2n +1)⋅21+++ ⎢⎥2n +32n +5
()⋅2()⋅22n +32n +5⎣⎦
(2n +1)2
2
2n +1
11⎛⎫1+++ ⎪24⎝22⎭⋅11-
14
(2n +1)2
1
2n +1
3(2n +1)2
1
r 5
3⨯11⨯2故 3⨯13⨯2
因而取n =6则
r 6
1
10
=
2n -2
≈0.00003
.
111⎛1⎫
ln 3=2 +≈1.0986++ +3511⎪
5⋅211⋅2⎭⎝23⋅2
⎛π⎫⎛π⎫ ⎪ ⎪
πn ⎝90⎭⎝90⎭0
cos 2=cos =1-+- +(-1)
902! 4! (2)⎛π⎫
⎪⎝90⎭
∵
2
24
⎛π⎫ ⎪⎝90⎭
+
(2n )!
2n
2!
≈6⨯10
-4
⎛π⎫
⎪⎝90⎭
;
4
4!
≈10
-8
cos 2≈1-
⎛π⎫ ⎪⎝90⎭2!
2
≈1-0.0006≈0.9994
故
17.利用被积函数的幂级数展开式,求定积分
⎰
0.50
arctan x
x
d x
(误差不超过0.001)的近似值.
arctan x =x -
解:由于
x
3
3
+
x
5
5
- +(-1)
n
x
2n +1
2n +1
+
,(-1≤x ≤1)
⎰
0.50
arctan x
x
d x =
⎰
0.50
242n
⎡⎤x x x
()1-+- ++ -1⎢⎥d x
352n +1⎣⎦
0.5
357
⎛⎫x x x = x -+-+ ⎪
92549⎝⎭
=
故
12
-1
1
92⋅1
5
⋅
1
3
+
1
252
⋅
1
5
-
11
492⋅1
7
⋅
1
7
+
1
而92
⋅
1
3
≈0.0139
,252
≈0.0013
,492
≈0.0002
.
⎰因此
∞
0.50
arctan x
x
n +
1n
n
d x ≈
12
-
1
92
⋅
1
3
+
1
252
⋅
1
5
≈0.487
18.判别下列级数的敛散性:
∑
n =1
n
(1)
∞
1⎫⎛n + ⎪
n ⎭; ⎝
∞
(2)n =1
∑
nx ⎫⎛
n cos ⎪
3⎭⎝
2
n
2
;
∑
n =1
ln (n +2)1⎫⎛
3+ ⎪
n ⎭⎝
n
n +
n
(3) .
1n
⎛n 2⎫
>= n n 2⎪1⎫1⎫⎝1+n ⎭⎛⎛ n +⎪ n +⎪
n ⎭n ⎭⎝解:(1)∵⎝
n
n
n
⎡⎛⎛n ⎫-1⎫
lim =lim 1+⎢⎪ 22⎪n →∞n →∞
1+n ⎝⎭1+n ⎝⎭⎣而
2∞
n
n
-n
-(1+n 2
)
⎤1+n 2
=1≠0⎥⎦
⎛n 2⎫
∑ 2⎪
故级数n =1⎝1+n ⎭发散,由比较审敛法知原级数发散.
0
(2)∵
nx ⎫⎛
n cos ⎪
3⎭⎝
2
n
2
≤
n 2
∞
n
∞
由比值审敛法知级数
∑
n =1
n
n
2收敛,由比较审敛法知,原级数n =1
∑
nx ⎫⎛
n cos ⎪
3⎭⎝
2
n
2
收敛.
0
(3)∵
ln (n +2)1⎫⎛
3+ ⎪
n ⎭⎝
n
ln (n +2)
3
n
lim
U n +1U n
n →∞
=lim ==1313
ln (n +3)3
n +1
n →∞
⋅
3
n
ln (n +2)
lim
ln (n +3)ln (n +2)
∞
n →∞
由
∞
知级数
∑
n =1
ln (n +2)
3
2
n
∑
n =1
ln (n +2)1⎫⎛
3+⎪
n ⎭收敛. ⎝
n
收敛,由比较审敛法知,原级数
∞
19.若
n →∞
lim n U n lim n U n
2
存在,证明:级数
∑U
n =1
n
收敛.
证:∵n →∞存在,∴∃M >0,使|n 2U n |≤M ,
M
即n 2|U n |≤M ,|U n |≤n
∞
2
n
而
∑
n =1
M n
2
∞
收敛,故
∞
∑U
n =12
绝对收敛.
∞
20.证明,若n =1
∑U n
1n
收敛,则
2
∑
n =1
U n
n 绝对收敛.
2
U n
证:∵n
∞
U n +
⋅U n ≤
∞
1=12U n +
2
=
2
1
2
2
2n
⋅
1
2
而由
∞
∑U n
n =1
收敛,
∑
n =1
1
n 收敛,知
∞
11⎫U n ⎛12
U +⋅∑ 2n 2n 2⎪∑
⎭收敛,故n =1n 收敛, n =1⎝
∞
因而
∑
n =1
U n
n 绝对收敛.
∞
∞
n
∞
n
21.若级数
∑a
n =1
与
∑b
n =1
都绝对收敛,则函数项级数
∑(a
n =1
n
cos nx +b n sin nx )
在R 上一致收敛.
证:U n (x )=a n cos nx +b n sin nx ,∀x ∈R 有
U n (x )=a n cos nx +b n sin nx ≤a n cos nx +b n sin nx ≤a n +b n
∞
∞
n
由于
∑a
n =1
与
∑b
n =1
n
都绝对收敛,故级数
∞
∑(a
n =1
n
∞
n
+b n
)
收敛.
由魏尔斯特拉斯判别法知,函数项级数n =1
22.计算下列级数的收敛半径及收敛域:
∑(a
cos nx +b n sin nx )
在R 上一致收敛.
∞
⎛∑ (1) n =1⎝
∞
∞
+1⎫n
⎪x
n +1⎭;
n
n
(2)
∑
n =1
sin
π2
n
(x +1)
n
;
(3)
∑
n =1
(x -1)
n ⋅2
2
n
ρ=lim
a n +1a n
n +2⎭+
1
n +1
n
n →∞
⎛
=lim n →∞
⎝=lim
解:
(1)
⎛⋅ ⎝n
n →∞
-1
n +2⋅e =
⎛⎛n +1⎫
⋅lim ⋅lim ⎪n →∞
⎝n +2⎭n →∞⎝
⎫
n
=e
R =
∴
1
ρ
=
3,
⎛∑ x =±
3又当时,级数变为n =1⎝
∞
+1⎫⎛⎫
±⎪ ⎪=
n +1⎭⎝3⎭
n n
⎫n ⎛3n +
(
)±1∑ ⎪n =1⎝3n +3⎭,
∞
n
⎛3n +⎫
lim ⎪=e n →∞
⎝3n +3⎭因为
x =±
所以当
n
≠0
3,级数发散,故原级数的收敛半径
R =
3,收敛域(
-3
, 3) .
ρ=lim
(2)
a n +1a n
sin =lim
n →∞
πn +1
π
=lim n →∞
n +1
n →∞
sin
π2
n
π
n
=
12
2
R =
故
1
ρ
=2
,
lim sin
n →∞
π2
n
sin
⋅2=lim π
n →∞n
π2
n n
π2π
n
=π≠0
.
又∵
∞
2所以当(x +1)=±2时,级数n =1发散,
从而原级数的收敛域为-2
∑
sin
(x +1)
n
ρ=lim
(3)
a n +1a n
n →∞
=lim
n ⋅2
2
2n n +1
n →∞
(n +1)⋅2
1
2
=
12
∞
∴R =2,收敛区间-2
∞
当x =-1时,级数变为
因此原级数的收敛域为[-1,3].
n =1
∑
x 0
(-1)
n
n ,其绝对收敛,当x =3时,级数变为n =1n ,收敛.
∑
1
2
23.将函数
F (x )=
⎰
arctan t
t
n
d t
展开成x 的幂级数.
∞
arctan t =
解:由于
x 0
x 0
∑(-1)
n =0
t
2n +1
2n +1
F (x )=
=
⎰
arctan t
t
d t =
n
⎰∑(-1)
0n =0
∞
x
∞
n
t
2n
2n +1
n
d t x
2n +1
2
∞
2n +1n =0n =0所以
24.判别下列级数在指定区间上的一致收敛性:
∞
∑⎰
(-1)
t
2n
d t =
∑(-1)
(2n +1)
(|x |≤1)
(1)n =1x +3
∞
∑
(-1)
n n
∞
,x ∈[-3,+∞) ;
(2)n =1x
∑
n
n
,x ∈(2,+∞) ;
(3)
∑(
n =1
n
2222
x +n )⎡⎣x +(n +1)⎤⎦,x ∈(-∞,+∞) ;
(-1)
∞
n n
=
1x +3
∞
解:(1)考虑n ≥2时,当x ≥-3时,有x +3
n
n
13-3
n
13
n -1
而
∑3
n =1
1
n -1
收敛,由魏尔斯特拉斯判别法知,级数
∑
n =1
(-1)
x +3在[-3,+∞) 上一致收敛.
n
n
(2)当x >2时,有x
n
=
n 2
n
n +1
n →∞
lim n +1
n
n
=
12
知级数
∞
由
2
∑
n =1
n
n
∞
2收敛,由魏尔斯特拉斯判别法知,级数n =1x 在(2,+∞) 上一致收敛.
∑
n
n
n
(3)∀x ∈R 有
∞
22
(x 2+n 2)⎡⎣x +(n +1)⎤⎦
≤
n n ⋅(n +1)
∞
22
n n
4
=
1n
3
而n =1
25.求下列级数的和函数:
∞
∑
1
22223⎤x +()n +1n 收敛,由魏尔斯特拉斯判别法知,级数n =1(x +n )⎡⎣⎦在(-∞,+∞) 上一致收敛.
∑
n
(1)
∑(-1)
n =1
n -1
x
2n ∞
2n -1; (2)n =02n +1;
∞
∑
x
2n +1
∞
(3)n =1
∑(
n n -1)!
x
n -1
; (4)n =1
∑n (
x
n
n +1).
∞
解:(1)可求得原级数的收敛半径R =1,且当|x |=1时,级数
域为[-1,1]
∞
∑(-1)
n =1
n -1
1
2n -1是收敛的交错级数,故收敛
S (x )=
记
∑(-1)
n =1
n -1
x
2n ∞
2n -1
∞
=x ∑(-1)
n =12n -2
n -1
x
2n -1
2n -11
2
=xS 1(x )
S 1'(x )=
则S 1(0)=0,
∑
n =1
(-1)
n -1
x =
1+x
所以 即S 1(x )=arctanx ,所以S (x )=x arctan x ,x ∈[-1,1].
∞
S 1(x )-S 1(0)=
⎰
x 0
11+x
2
d x =arctan x
S (x )=
(2)可求得原级数的收敛半径R =1,且当|x |=1时,原级数发散.记
∞
∑
n =0
x
2n +1
2n +1则
S '(x )=
1-x
x x 111+x 11+x 'S ()d x =d x =ln S ()-S ()=ln x x 0⎰0⎰01-x 2
21-x ,即21-x ,S (0)=0
n =0
∑x
2n
=
1
2
所以
S (x )=
12
ln
1+x 1-x ,(|x |
n +1
n →∞
lim
a
n +1n
=lim
a
n →∞
n ! n
=0
S (x )=
知收敛域为(-∞,+∞) .记
∞
∞
(3)由
(n -1)!
∞
∑(
n =1
n n -1)!
x
n -1
则
⎰
x 0
S (x )d x =
∑(
n =1
x
n
n -1)!
=x ∑
n =1
x
n -1
(n -1)!
=x e
x
,所以
S (x )=(x e x )'=(1+x )e ,(-∞
x
1
lim
(4)由
n →∞
(n +1)(n +2)
=1
1
n (n +1)
知收敛半径R =1,当x =1时,级数变为
∞
n
∞
∑
n =1
11
n (n +1),由n (n +1)
1n
2
知级数收敛,当x =-1时,级数变为
∞
∑n (
n =1
(-1)
n +1)是收敛的交错级数,故收敛域为[-1,1].
∞
S (x )=
记
∑n (
n =1
∞
x
n
n +1)则S (0)=0,
n -1
xS (x )=
∑n (
n =1
x
n +1
n +1),
[xS (x )]''=
x
∑
n =1
x =
1
1-x (x ≠1)
(x )]''d x =-ln (1-x )[xS
所以⎰0
()()即[xS x ]=-ln 1-x
'
⎰0[xS (x )]d x =⎰0-ln (1-x )d x =(1-x )ln (1-x )+x
即xS (x )=(1-x )ln (1-x )+x
x
x
'
⎛1⎫
S (x )=1+ -1⎪ln (1-x )
⎝x ⎭当x ≠0时,,又当x =1时,可求得S (1)=1 1⎫⎛
lim S (x )=lim 1-⎪=1n →∞n →∞⎝n +1⎭(∵)
综上所述
⎧0,
⎪
S (x )=⎪1,
⎨
⎛1⎫⎪
1+ -1⎪ln (1-x ), ⎪⎝x ⎭⎩
x =0x =1
x ∈[-1, 0) (0,1)
⎧2
f (x )=⎨3
⎩x 26.设f (x ) 是周期为2π的周期函数,它在(-π,π]上的表达式为
-π
试问f (x ) 的傅里叶级数在x =-π处收敛于何值?
解:所给函数满足狄利克雷定理的条件,x =-π是它的间断点,在x =-π处,f (x ) 的傅里叶级数收敛于
-+
f (-π)+f (-π)
2
=
12
[π3+2]=
12
(2+π3)
⎧-1
f (x )=⎨2
⎩x 27.写出函数
-π≤x ≤00
的傅里叶级数的和函数.
解:f (x ) 满足狄利克雷定理的条件,根据狄利克雷定理,在连续点处级数收敛于f (x ) ,在间断点x =0,x =±π
-+
f (0)+f (0)
处,分别收敛于
-+
f (-π)+f (-π)
2
=-
12,
-+
f (π)+f (π)
2
=
π-12
,
2
2
=
π-12
,综上所述和函数.
2
28.写出下列以2π为周期的周期函数的傅里叶级数,其中f (x ) 在[-π,π)上的表达式为:
⎧-1⎪2x ⎪⎪1
S (x )=⎨-
⎪2⎪π2-1⎪⎩2
-π
⎧π⎪⎪4
f (x )=⎨
⎪-π⎪⎩4(1)
(2)
0≤x
f (x )=x
2
(-π≤x ≤π);
-π≤x
π2π2, ,
⎧π⎪-2, ⎪⎪
f (x )=⎨x ,
⎪⎪π⎪, ⎩2(3)
x
f (x )=cos
2(4)
≤x
(-π≤x ≤π)
.
解:(1)函数f (x ) 满足狄利克雷定理的条件,x =n π,n ∈z 是其间断点,在间断占处f (x ) 的傅里叶级数收敛于
f (0
+
)+f (0)
-
π=
⎛π⎫+ -⎪4⎝4⎭
2=1
2
a n =
1π
=0
,在x ≠n π,有
⎰
π-π
f
(x )cos nx d x
1⎛π⎫
-cos nx d x +⎰ ⎪π-π⎝4⎭π
⎰
π0
π4
cos nx d x =0
b n =
1π
⎰
π-π
1⎛π⎫
f (x )sin nx d x =⎰ -⎪sin nx d x +
π-π⎝4⎭πn =2, 4, 6, , n =1, 3, 5, .
1
⎰
π
π4
sin nx d x
⎧0,
⎪=⎨1⎪, ⎩n
∞
于是f (x ) 的傅里叶级数展开式为
f (x )=
(x ≠n π)
(2)函数f (x ) 在(-∞,+∞) 上连续,故其傅里叶级数在(-∞,+∞) 上收敛于f (x ) ,注意到f (x ) 为偶函数,从而f (x )cos nx 为偶函数,f (x )sin nx 为奇函数,于是
n =1
∑2n -1sin (2n -1)x
π
1
b n =a n =
1π1π
⎰
-ππ-π
f (x )sin nx d x =0f
,
a 0=
π0
2
1π
⎰
π-π
x d x =
2
2π3
2
,
n
⎰
(x )cos nx d x
∞
n
=
2π
⎰
x cos nx d x =(-1)⋅
4
n (n =1,2,…)
2
所以,f (x ) 的傅里叶级数展开式为:
f (x )=
π
2
(-∞
(3)函数在x =(2n +1)π (n ∈z ) 处间断,在间断点处,级数收敛于0,当x ≠(2n +1)π时,由f (x ) 为奇函数,有a n =0,(n =0,1,2,…)
n =1
3
+
∑(-1)
⋅
4n
2
cos nx
b n =
2π
⎰
π0
π
2⎡2
f (x )sin nx d x =⎢⎰x sin nx d x +
π⎣0
⎰
ππ2
π
⎤
sin nx d x ⎥2⎦
=-
所以
1n
(-1)
∞
n
+
2n π
n +1
2
sin
n π22n π
2
(n =1, 2, )
sin
f
(x )=∑⎢(-1)
n =1
⎡⎣
⋅
1n
+
n π⎤
sin nx 2⎥⎦ (x ≠(2n +1)π,n ∈z )
2作为以2π为周期的函数时,处处连续,故其傅里叶级数收敛于f (x ) ,注意到f (x ) 为(4)因为
偶函数,有b n =0(n =1,2,…) ,
f
(x )=cos
x
a n ==
11
⎰π
π
π-ππ0
cos
x 2
cos nx d x =
⎰π
2
π0
cos
x 2
cos nx d x ⎫⎤
⎪x ⎥d x ⎭⎦
π
⎰
⎡1⎛cos n + ⎢
2⎝⎣1⎫⎛x +cos n -⎪
2⎭⎝
⎡1⎛
sin n + 1⎢2⎝=⎢
1π⎢
n +⎢2⎣=(-1)
n +1
1⎫⎤⎫⎛
x sin n -⎪ ⎪x ⎥
2⎭⎭⎝
⎥+
1⎥n -
⎥2⎦0
4⎛1⎫
⎪2
π⎝4n -1⎭4π
∞
(n =0,1, 2, )
所以f (x ) 的傅里叶级数展开式为:
f (x )=
2π
+
∑(-1)
n =1
n +1
cos nx
4n -1 x ∈[-π,π] x
2
29. 将下列函数f (x ) 展开为傅里叶级数:
f
(1)(2)
(x )=
π4
-
x 2
(-π
f (x )=sin x
a 0=
1π
(0≤
π
x ≤2π)
解:(1)
⎰
-π
f (x )cos nx d x =
1
x ⎫π⎛π-d x =⎪⎰
π-π⎝42⎭2
π
a n =
=b n =
1
⎛πx
-⎰-π
π⎝42
π1⎫
cos nx d x =⎪
4⎭
⎰
π-π
cos nx d x -
⎰2π
1
π-π
x cos nx d x
14n 1
[sin nx ]-π-0=0
π
π
(n =1, 2, )
x ⎛π
-⎰
π-π⎝42
n
1⎫
sin nx d x =⎪
4⎭
⎰
π-π
sin nx d x -
⎰2π
1
π-π
x sin nx d x
=(-1)⋅f
1n
∞
4n =1n (-π
(2)所给函数拓广为周期函数时处处连续, 因此其傅里叶级数在[0,2π]上收敛于f (x ) ,注意到f (x ) 为偶函数,
(x )=
π
+
∑(-1)
π-ππ0
n
sin nx
a 0==
有b n =0,
1
⎰π
2π
f (x )cos 0x d x =
4π
1
⎰π
π-π
sin x d x
⎰
sin x d x =
a n ===
⎰π
1
2
π0π0
f
(x )cos nx d x =
⎰π
2
π-π
sin x cos nx d x
⎰π
⎡⎣sin (n +1)x -sin (n -1)x ⎤⎦d x
⎡1+(-1)n ⎤2⎦π(n -1)⎣
n =1, 3, 5, n =2, 4, 6,
-2
⎧0, ⎪
-4=⎨
,
⎪π(n 2-1)⎩
所以
f (x )=
2π
∞
+
∑
n =1
-4cos 2nx π(4n -1)
2
(0≤x ≤2π)
30. 设f (x )=x +1(0≤x ≤π),试分别将f (x ) 展开为正弦级数和余弦级数. 解:将f (x ) 作奇延拓,则有a n =0 (n =0,1,2,…)
b n =
2π
⎰
π0
f (x )sin nx d x =
n
2π
⎰(x +1)sin nx d x
π
21-(-1)(1+π)=⋅πn f (x )=
n
n n =1从而
若将f (x ) 作偶延拓,则有b n =0 (n =1,2,…)
∑π
2
∞
1-(-1)(1+π)
sin nx
(0
a n =
2π
⎰
π0
f
(x )cos nx d x
=
2π
⎰(x +1)cos nx d x
π
从而
n =2, 4, 6 ⎧0,
⎪=⎨-4
⎪2, n =1, 3, 5, ⎩n π1π2π
a 0=⎰f (x )d x =⎰(x +1)d x =π+2
π-ππ0
π+24∞cos (2n -1)x
f (x )=-∑2
2πn =1(2n -1)
(0≤x ≤π)
∞
31. 将f (x )=2+|x | (-1≤x ≤1) 展开成以2为周期的傅里叶级数,并由此求级数n =1n 的和.
解:f (x ) 在(-∞,+∞) 内连续,其傅里叶级数处处收敛,由f (x ) 是偶函数,故b n =0,(n =1,2,…)
∑
1
2
a 0=
a n =
⎰⎰
1-11-1
f (x )d x =2⎰
f
10
(2+x )d x =5
10
(x )cos nx d x =2⎰
n =2, 4, 6 n =1, 3, 5,
(2+x )cos nx d x
⎧0, ⎪
=⎨-4
,
⎪n π2⎩()
所以
f
(x )=
52
-
∞
4π
2
∞
∑
n =1
cos (2n -1)πx
(2n -1)
2
2
,x ∈[-1,1]
∑
取x =0得,
∞
n =1∞
1
(2n -1)
1
2
=
∞
π
2
8
,故
∑
n =1
1n
2
=
∑
n =12
(2n -1)
π
+
∑
n =1
1
(2n )
2
=
1
∞
∑4
1n
2
+
π
2
n =1
8
∞
6 所以n =1n
32. 将函数f (x )=x -1(0≤x ≤2) 展开成周期为4的余弦级数.
解:将f (x ) 作偶延拓,作周期延拓后函数在(-∞,+∞) 上连续, 则有b n =0 (n =1,2,3,…) a 0=
1
∑
1
=
⎰2
12
2
2-2
f
(x )d x =
⎰(x -1)d x =0
2
a n ==
4
⎰
2-2
f
(x )cos
n
n πx 2
d x =
⎰0(x -1)cos
2
n πx 2
d x
n π
2
[(-1)-1]
n =2, 4, 6, n =1, 3, 5,
⎧0, ⎪=⎨8
-, ⎪22⎩n π
f
故
(x )=-
8π
2
∞
∑
n =1
1
(2n -1)
2
⋅cos
12
(2n -1)πx
2,
s (x )=
(0≤x ≤2)
⎧x , ⎪⎪
f (x )=⎨
⎪2-2x , ⎪⎩33. 设
0≤x ≤12
a 02
∞
+
∑a
n =1
n
cos n πx
,-∞
a n =2⎰
10
⎛5⎫
s -⎪
f (x )cos n πx d x
,求⎝2⎭. 5
2处间断,所以
解:先对f (x ) 作偶延拓到[-1,1],再以2为周期延拓到(-∞,+∞) 将f (x ) 展开成余弦级数而得到 s (x ) ,延拓后
x =-
f (x ) 在
+
⎛5-⎫⎤1⎡⎛1+⎫⎛1-⎫⎤⎛5⎫1⎡⎛5⎫
s -⎪=⎢f -⎪+f -⎪⎥=⎢f -⎪+f -⎪⎥
22222⎝2⎭2⎢⎢⎭⎝⎭⎥⎭⎝⎭⎥⎣⎝⎦⎣⎝⎦
=
1⎛1⎫3+1 ⎪=2⎝2⎭4
∞
34. 设函数f (x )=x (0≤x
2
s (x )=
∑b
n =1
n
sin n πx
,-∞
b n =2⎰
1
⎛1⎫
s -⎪
f (x )sin n πx d x
(n =1,2,3,…),求⎝2⎭.
解:先对f (x ) 作奇延拓到,[-1,1],再以2为周期延拓到(-∞,+∞) ,并将f (x ) 展开成正弦级数得到s (x ) ,延拓
x =-
后f (x ) 在
1
2处连续,故.
2
1⎛1⎫⎛1⎫⎛1⎫
s -⎪=-f -⎪=- -⎪=-
4. ⎝2⎭⎝2⎭⎝2⎭
35. 将下列各周期函数展开成为傅里叶级数,它们在一个周期内的表达式分别为:
1⎫⎛1
-≤x
0≤x
解:(1) f (x ) 在(-∞,+∞) 上连续,故其傅里叶级数在每一点都收敛于f (x ) ,由于f (x ) 为偶函数,有b n =0 (n =1,2,3,…)
1
1
f
(x )=⎨
⎧2x +1, -3≤x
a 0=2⎰21f
-2
(x )d x =4⎰02(1-x 2)d x =
1
116,
1
a n =2⎰21f
-2
(x )cos2n πx d x =4⎰02(1-x 2)cos2n πx d x
2
=
所以
(-1)
2
n +1
n π11121
(n =1, 2, )
1π
2
∞
f
(x )=
a 0=
+
3-3
∑
n =1
(-1)
n
2
n +1
cos 2n πx
(-∞
30
(2)
⎰3
f
1⎡0
f (x )d x =(2x +1)d x +
⎣⎰-33⎢
⎰
d x ⎤=-1
⎥⎦,
a n ===
1313
⎰
3-30
(x )cos
n πx 3
d x d x +
13
⎰-3(2x +1)cos
6
n πx 3
⎰
30
cos
n πx 3
d x
⎡1-(-1)n ⎤, (n =1, 2, 3, )22⎦n π⎣
13n πx b n =⎰f (x )sin d x
3-33==13
⎰-3(2x +1)sin
(-1)
-12
n +1
n πx 3
d x +
13
⎰
30
sin
n πx 3
d x
6n π
, (n =1, 2, )
而函数f (x ) 在x =3(2k +1),k =0,±1,±2,…处间断,故
∞
f
(x )=
+
n πx n πx ⎫n n +16⎧6⎡⎤1--1cos +-1sin ()()⎬∑⎨n 2π2⎣⎦3n π3⎭(x ≠3(2k +1),k =0,±1,±2,…) n =1⎩
36. 把宽为τ,高为h ,周期为T 的矩形波(如图所示)展开成傅里叶级数的复数形式
.
解:根据图形写出函数关系式
⎧⎪0, ⎪⎪
u (t )=⎨h ,
⎪⎪⎪0, ⎩
--
T 2
≤t
T 2
τ
2
τ
2
τ
2
τ
2
≤t ≤
T
c 0=c n ==1T
12l 12l
τ
⎰⎰
l -l l -l
u (t )d t =u (t )e
-i 2n πT
t
-i n πl
1T
t
⎰
2T -2
u (t )d t =1T
T
1T
τ
⎰τ
-2-i 2n πT
2
h d t =
t
h τT
d t =
⎰
2T -2
u (t )e
τ
d t
⎫t ⎪⎭
⎰τ
-2
2
h e
h ⎛-T ⎫2-i
d t =⋅ ⎪⎰τe
T ⎝2n πi ⎭-2
τ
2n πT
t
2n π⎛d -i
T ⎝
h ⎛
= -
⎝2n πi
h τT
h n πτ⎫⎡-i 2n πt ⎤2
=sin ⎪⎣e T ⎦-τ
n πT ⎭2
h π
∞
故该矩形波的傅里叶级数的复数形式为
u (t )=
+
∑
n =-∞
n ≠0
1n
sin
n πτT
e
-i
2n πT
t
22,…) (-∞
-37. 设f (x ) 是周期为2的周期函数,它在[-1,1]上的表达式为f (x )=ex ,试将f (x ) 展成傅里叶级数的复数形式. 解:函数f (x ) 在x ≠2k +1,k =0,±1,±2处连续.
t ≠±
τ
, ±
3τ
c n ==-
⎰2l
1
l -l
f
(x )e
-i
n πl
x
d x =
1
1
⎰2
1-1
e e
-x -in πx
d x
12(1+n πi )
-1
[e -(1+n πi )x ]-1
n
=
e -e 2
⋅(-1)⋅
n
11+n πi 1-n πi
2
=sinh 1⋅(-1)⋅
1+(n π)
n
故f (x ) 的傅里叶级数的复数形式为
∞
in πx
f (x )=sinh 1∑
n =-∞
(-1)⋅(1-in π)
1+(n π)
2
e
(x ≠2k +1,k =0,±1,±2,…)
⎧A ,
f (t )=⎨
⎩0, 38. 求矩形脉冲函数
F (ω)=
0≤t ≤T 其他
的傅氏变换
-i ωx
解:
39. 求下列函数的傅里叶积分:
⎰
+∞-∞
f (t )e
-i ωx
d t =
⎰
T 0
A e d t =
A (1-e -i ωx )
i ω
⎧e -t ,
f (t )=⎨
⎩0, (1)
t ≥0t
⎧-1, ⎪
f (t )=⎨1,
⎪0, ⎩(2)
F (ω)=
=
解:(1)
-1
-i ωt
⎰
+∞-∞
f (t )e =
d t =
⎰
+∞0
e
-t
⋅e
-i ωt
d t
11+ωi
1-i ω1+ω
2
f (t )=
==
⎰2π⎰2π
11
1
+∞-∞+∞-∞
F (ω)e
i ωt
d ω=
⎰2π
d t
1
+∞-∞
1-i ω1+ω
2
e
i ωt
d ω
cos ωt +ωsin ωt
1+ω1+ω
22
⎰π
+∞0
cos ωt +ωsin ωt
+∞-∞0-1
-i ωt
d t
F (ω)=
==
(2)
⎰⎰
f (t )e
d t
(-e -i ωt )d t +
i ω
⎰
10
e
-i ωt
d t
2(1-cos ω)
i ωt
+∞-∞
f (t )=
===
⎰2π
1π1π2π
1
F (ω)e
d ω=
⎰2π
1
+∞-∞
2(1-cos ω)
i ω
e
i ωt
d ω
⎰⎰
+∞-∞+∞-∞+∞0
(1-cos ω)
i ω
(cos ωt +i sin ωt )d ω
d ωd ω
(1-cos ω)sin ωt
ω
(1-cos ω)sin ωt
⎰
40. 求如图所示的三角形脉冲函数的频谱函数
.
ω
(t ≠0,1)
2E ⎧E +t ⎪⎪T
f (t )=⎨
⎪E -2E t ⎪⎩T 解:F (ω)=
-
T 2
≤t ≤0T 2
0
⎰
+∞-∞0
f (t )e
-i ωt
2E ⎛=⎰T E +-
T 2⎝=
⎫-i ωt
d t +t ⎪e ⎭
T
⎰
20
2E ⎛ E -⎝T ⎫-i ωt
d t t ⎪e ⎭
4E ⎛ωT ⎫
1-cos ⎪2
T ω⎝2⎭
习题十二
1.写出下列级数的一般项:
1+
(1)
13
+
15
+
17
+
;
(2)2
3
+
x 2⋅4
5
+
9
2⋅4⋅6
+
x
2
2⋅4⋅6⋅8
+
;
a
(3)3
-
a
5
+
a
7
71
-
a
9
+
;
解:(1)
U n =
2n -1;
n
U n =
(2)
x 2
(2n )!! ;
n +1
2n +1; (3)
2.求下列级数的和:
∞
U n =(-1)
a
2n +1
(1)
∑(
n =1
1
x +n -1)(x +n )(x +n +1);
∞
(2)
∑n =1
+15
3
;
1
(3)5
+
15
2
+
;
u n ==
解:(1)
1
(x +n -1)(x +n )(x +n +1)
111⎛⎫
- ⎪
2⎝(x +n -1)(x +n )(x +n )(x +n +1)⎭
S n =
11111⎛
-+-
2⎝x (x +1)(x +1)(x +2)(x +1)(x +2)(x +2)(x +3)
1
+ +
(x +n -1)(x +n )
-
⎫
(x +n )(x +n +1)⎪⎭
1
从而
111⎛⎫
-= ⎪
2⎝x (x +1)(x +n )(x +n +1)⎭
lim S n =
因此
n →∞
11
2x (x +1),故级数的和为2x (x +1)
(2)
因为
U n =
-
S n =
-
-
-1-
+1-
+
-
-
+
==
从而所以n →∞
+ +
-
lim S n =1-
S n =
15+15
2
1-
+ +
15
n
n
1⎡⎛1⎫⎤⎢1- ⎪⎥5⎣⎝5⎭⎦=
11-
5n
1⎡1⎫⎤⎛=⎢1- ⎪⎥4⎣⎝5⎭⎦
(3)因为
4,即级数的和为4. 从而n →∞
3.判定下列级数的敛散性:
∞
lim S n =
11
(1)
∑n =1
;
1
(2) 1⋅6
+2323
16⋅11++23
33
+
111⋅16
+ +
1
(5n -4)(5n +1)
23
n n
+
;
2
(3) 3
-
- +(-1) +
n -1
+
;
1
(4)5
+
;
S n =
解:
(1) 从而n →∞
+
-1
,故级数发散.
+ +
-
=
lim S n =+∞
S n ==
(2) 从而n →∞
1⎛1111111⎫1-+-+-+ +- ⎪5⎝661111165n -45n +1⎭1⎛1⎫ 1-⎪5⎝5n +1⎭
12
3的等比级数,且|q |
lim S n =
1
5,故原级数收敛,其和为5.
q =-
(3)此级数为
U n =
(4)
∵
lim U =1≠0n →∞n ,故级数发散.
4.利用柯西审敛原理判别下列级数的敛散性:
∞
(1)
∑
n =1∞
(-1)
n ⎛
1
n +1
∞
; (2)
∑
n =1
cos nx 2
n
;
(3) n =1
解:(1)当P 为偶数时,
+-⎪∑
⎝3n +13n +23n +3⎭
11⎫
.
n +1+U n +2+ +U n +p =
(-1)
1n +11
n +2
n +1
-
+1
(-1)
n +3
n +2+
1
+
(-1)
n +4
n +3- -
+ +1n +p
(-1)
n +p +1
n +p
=
n +2n +3
=
111⎛⎫1⎫⎛1
-- - ---⎪ ⎪
n +1⎝n +2n +3⎭⎝n +p -2n +p -1⎭n +p 1n +1
当P 为奇数时,
n +1+U n +2+ +U n +p =
(-1)
1n +11
n +2
n +1
-
+1
(-1)
n +3
n +2+
1
+
(-1)
n +4
n +3- +
+ +1n +p
(-1)
n +p +1
n +p
=
n +2n +3
=
11⎫⎛1⎫⎛1
-- - --⎪ ⎪
n +1⎝n +2n +3⎭⎝n +p -1n +p ⎭1n +1
1n +1
因而,对于任何自然数P ,都有
U n +1+U n +2+ +U n +p
N =
∀ε>0,取
⎡1⎤
+1⎢⎣ε⎥⎦
,则当n >N 时,对任何自然数P 恒有
n +1+U n +2+ +U n +p
成立,由
∞
柯西审敛原理知,级数n =1
(2)对于任意自然数P ,都有
∑
(-1)
n
n +1
收敛.
n +1+U n +2+ +U n +p =≤
cos (n +1)x
212
n +1
n +1
+
cos (n +2)x
22
n +2
+ +
cos (n +p )x
2
n +p
+
12
n +2
+ +
1
n +p
1⎛1⎫
1-n +1 p ⎪2⎝2⎭
=
11-
2=
1⎛1⎫
1- n p ⎪2⎝2⎭12
n
1⎤⎡
log 2⎢+U n +2+ +U n +p 0(0N 时,对任意的自然数P 都有n +1
成立,由柯西审敛原理知,该级数收敛.
(3)取P =n ,则
n +1+U n +2+ +U n +p
111111⎛⎫
= +-+ ++-⎪
3⋅2n +13⋅2n +23⋅2n +3⎝3(n +1)+13(n +1)+23(n +1)+3⎭≥≥>
13(n +1)+1
n 6(n +1)112
+ +
13⋅2n +1
12,则对任意的n ∈N ,都存在P =n 所得从而取
理知,原级数发散.
5.用比较审敛法判别下列级数的敛散性.
ε0=
1
n +1+U n +2+ +U n +p >ε0
,由柯西审敛原
1
(1)4⋅6
+
15⋅7
2
+ +1+31+3
2
1
(n +3)(n +5)
+ +
1+n 1+n
2
+
;
1+
(2)
∞
1+21+2
++
∞
(3)
∑
n =
1∞
sin 1
π3;
n n
∑
(4)
n =1∞
1n
;
(5)
∑1+a
n =1
(a >0)
;
(6)
∑(2
n =1
-1
)
.
U n =
解:(1)∵
1
(n +3)(n +5)
1n
2
∞
而
∑
n =1
1
n 收敛,由比较审敛法知
1+n 1+n
2
∞
2
∑U
n =1
n
收敛.
(2)∵
∞
U n =
≥
1+n n +n
2
=
1n
而
∑
n =1
1
n 发散,由比较审敛法知,原级数发散.
sin lim
(3)∵
∞
πn
∞
n
sin
=lim π⋅
n →∞
πn n
n →∞
13
π3π
=π
而
∑3
n =1
π
n
收敛,故
∑
n =1
sin
3也收敛.
n
U n =
(4)
∵
∞
=
1
3
n 2
∑
而
n =1
1
3
∞
n 2收敛,故
∑
n =1
1
收敛.
∞
(5)当a >1时,
U n =
1a ,而=121≠0
n
1+a
n
∑
n =1
1a
n
∞
收敛,故
∑1+a
n =1
1
n
也收敛.
当a =1时,
lim U n =lim
n →∞
n →∞
,级数发散.
n
当0
综上所述,当a >1时,原级数收敛,当0
1
lim U n =lim
n →∞
n →∞
1+a
=1≠0
x (6)由知
6.用比值判别法判别下列级数的敛散性:
x →0∞
lim
2-1
x
lim
2n -11n
=ln 2
x →∞
=ln 2
而
∞
∞
∑
n =1
1
n 发散,由比较审敛法知
∑(2
n =1
∞
1n
-1
)
发散.
(1)
∑
n =1
n
2n
3;
(2)
∑3
n =1
n !
n
+1;
3
(3)1⋅2
∞
+
n
3
22
2⋅2n
n
+
3
33
3⋅2
+ +
3
n n
n ⋅2
+
;
(1)
∑
n =1
2⋅n !
2
U n
3,解:(1)
由比值审敛法知,级数收敛.
n
n →∞
U n =
n
2
lim
U n +1
=lim
(n +1)
3
n +1
n →∞
⋅
3n
n 2
=
13
,
lim
U n +1U n
n →∞
(n +1)! 3+1=lim n +1⋅n →∞3+1n !
=lim (n +1)⋅
n →∞
n
3+13
n +1
n
+1
=+∞(2)
所以原级数发散.
lim
U n +1U n
n →∞
=lim
33n
n +1
n +1
n →∞
(n +1)⋅2
2(n +1)
⋅
n ⋅23
n
n
=lim =3
n →∞
2(3)
所以原级数发散.
>1
lim
U n +1U n
n →∞
=lim
2
n +1
⋅(n +1)!
n +1n
n →∞
(n +1)
⋅
n
n
n
2⋅n !
⎛n ⎫
=lim 2 ⎪n →∞
⎝n +1⎭=2lim
11⎫⎛
1+⎪
n ⎭⎝
n →∞
n
=
2e
(4)
故原级数收敛.
7.用根值判别法判别下列级数的敛散性:
⎛⎫
⎪∑ 3n +1⎝⎭; n =1(1)
⎛n ⎫
⎪∑
(3) n =1⎝3n -1⎭
∞
n
∞
5n
n ∞
∑
(2)
n =1
1
[ln (n +1)]
n
;
∞
2n -1
;
⎛b ⎫∑ a ⎪
(4) n =1⎝n ⎭,其中a n →a (n →∞),a n ,b ,a 均为正数.
lim
解:
(1)
故原级数发散.
n →∞
=lim
5n 3n +1
n →∞
=
53
>1
,
lim
(2)
故原级数收敛.
n →∞
=lim
1ln (n +1)
2-
n →∞
=0
,
1n
lim
(3)
故原级数收敛.
n →∞
⎛n ⎫
=lim ⎪n →∞
⎝3n -1⎭
=
19
,
lim
(4)
n →∞
b b =lim =n →∞a a , n
b b b
当b a 时,a >1,原级数发散;当b =a 时,a =1,无法判定其敛散性. 8.判定下列级数是否收敛?若收敛,是绝对收敛还是条件收敛?
1-
(1)
1+
1-
1∞
;
(2)
∑(-1)
n =1
n -1
1ln (n +1);
11111111
⋅-⋅2+⋅3-⋅4+
5353(3) 5353;
∞
(4)
∑(-1)
n =1∞
n -1
2
n
2
∞
n ! ;
n
(5)
∑(-1)
n =1
n -1
1n
α
(α∈R )
;
111⎫(-1)⎛
1+++ +⎪∑
23n ⎭n . (6) n =1⎝
U n =(
-1)
解:(1)
n -1
∞
∞
n =1
∑
U n
∞
1
>
∞
n
lim
n →∞
=0
,由莱布
∑
尼茨判别法级数收敛,又
n =1
n =
∞
∑
n =1
1n
2
是P
∑
n =1
发散,故原级数条件收敛.
U n =(-1)
(2)
n -1
1ln (n +1),
∑
n -1
(-1)
n -1
111ln (n +1)
ln (n +1)为交错级数,且ln (n +1)
>
1ln (n +2),1n +1
lim
1ln (n +1)
∞
n →∞
=0
,由莱布尼茨判别法知原级数收敛,但由于
U n =
≥
所以,
∑
n =1
U n
发散,所以原级数条件收敛.
n -1
(3)
∞
U n =(-1)
1
5⋅3民,显然
n
∞∞
∑
n =1
n =
∑5⋅3
n =1
1
n
=
1
∞
∑3
5
n =1
1
n
∞
,而
∑3
n =1
1
n
是收敛的等比级数,故
∑
n =1
n
收敛,所以原级数绝对收敛.
lim
(4)因为故可得∴n →∞
U n +1U n
n →∞
=lim
2
2n +1
n →∞
n +1
=+∞
. ,
n +1>n
,得n →∞
lim n ≠0
lim U n ≠0
,原级数发散.
∞
(5)当α>1时,由级数
∑
n =1
1n
∞
α
收敛得原级数绝对收敛.
n -1
当0
∞
∑(-1)
n =1
n -1
1n
α
1
满足条件:n
∞
α
>
1
α
(n +1);n →∞n α
lim
1
=0
,由莱布尼茨判别法
知级数收敛,但这时当α≤0时,n →∞
∑
n =1
(-1)
1n
α
=
∑
n =1
1n
α
发散,所以原级数条件收敛.
lim U n ≠0
,所以原级数发散.
111⎛1+++ +
23n (6)由于⎝
∞
1⎫1
⋅>⎪
n ⎭n
而
∞
∑
n =1
1
n 发散,由此较审敛法知级数
∑
n =1
111⎛
1+++ +
23n ⎝⎫(-1)
⎪⋅
n ⎭
n
发散.
111⎛
U n = 1+++ +
23n ⎝记⎫1
⎪⋅
⎭n ,则
1111⎫⎛11⎫⎛
U n -U n +1= 1+++ +⎪ --⎪2
23n ⎭⎝n n +1⎭(n +1)⎝
11111⎫⎛
= 1+++ +⎪-2
23n ⎭n (n +1)(n +1)⎝
111⎛⎫1⎫⎛11
-= ++ +⎪+ 2⎪
n ⎭n (n +1)⎝n (n +1)(n +1)⎭⎝23>0
即U n >U n +1
lim U n =lim
n →∞
n →∞
1⎛111⎫
1+++ + ⎪n ⎝23n ⎭
n
=
又
1
⎰n
1x
d x
1
由t →+∞t
lim
1
⎰
t 0
1
d x =lim t =0t →+∞1x
∞
知n →∞
9.判别下列函数项级数在所示区间上的一致收敛性.
∞
lim U n
111⎛1+++ + ∑=0
23n ,由莱布尼茨判别法,原级数n =1⎝
n
∞
⎫(-1)
⎪⋅
n 收敛,而且是条件收敛.⎭
n
(1)
∑(
n =1∞
x
n -1)! ,x ∈[-3,3];
n
(2)
∑
n =1
x
n 2
n ,x ∈[0,1];
(3)
∑
n =1
sin nx 3
,x ∈(-∞,+∞) ;
(4)
∞
∑
n =1
e
-nx
n ! ,|x |
∞
∑
(5)
n =1
x ∈(-∞,+∞)
x
解:(1)∵
n
(n -1)!
≤
3
n
(n -1)!
,x ∈[-3,3],
∞
而由比值审敛法可知
∑(
n =1
3
n
n -1)! 收敛,所以原级数在 [-3,3]上一致收敛.
x
(2)∵n
n 2
≤
1
n ,x ∈[0,1],
2
∞
而
∑
n =1
1
n 收敛,所以原级数在[0,1]上一致收敛. sin nx 31
n
2
(3)∵
∞
n
≤
1
3,x ∈(-∞,+∞) ,
n
而
∑3
n =1
是收敛的等比级数,所以原级数在(-∞,+∞) 上一致收敛.
-nx
e
≤
e
5n
(4)因为n !
n ! ,x ∈(-5,5) ,
∞
由比值审敛法可知
∑
n =1
e
5n
n ! 收敛,故原级数在(-5,5) 上一致收敛.
5
(5)
∞
1
5
≤
1
n 3,x ∈(-∞,+∞) ,
∑
而
n =1
n 3是收敛的P -级数,所以原级数在(-∞,+∞) 上一致收敛.
∞
10.若在区间Ⅰ上,对任何自然数n .都有|U n (x )|≤V n (x ) ,则当
∞
∑V
n =1
n
(x )
在Ⅰ上一致收敛时,级数
∑U
n =1
n
(x )
在这区间Ⅰ上也一致收敛.
∞
证:由n =1在Ⅰ上一致收敛知, ∀ε>0,∃N (ε)>0,使得当n >N 时,∀x ∈Ⅰ有 |V n +1(x )+V n +2(x )+…+V n +p (x )|
于是,∀ε>0,∃N (ε)>0,使得当n >N 时,∀x ∈Ⅰ有
|U n +1(x )+U n +2(x )+…+U n +p (x )|≤V n +1(x )+V n +2(x )+…+V n +p (x ) ≤|V n +1(x )+V n +2(x )+…+V n +p (x )|
∞
∞
n
∑V
n
(x )
因此,级数n =1在区间Ⅰ上处处收敛,由x 的任意性和与x 的无关性,可知收敛.
11.求下列幂级数的收敛半径及收敛域:
∑U
(x )
∑U
n =1
n
(x )
在Ⅰ上一致
(1)x +2x 2+3x 3+…+nx n +…;
∞
⎛x ⎫
⎪∑n ! n ⎝⎭; (2)n =1
∞
∞
n
(3)
∑
n =1
x
2n -1
2n -1;
(4)
∑
n =1
(x -1)
2
n
n ⋅2n ;
ρ=lim
解:(1)因为
∞
a n +1a n
n →∞
=lim
n +1n
n
n →∞
=1
,所以收敛半径
∞
R =
1
ρ
=1
收敛区间为(-1,1) ,而当x =±1时,
级数变为
∑(-1)
n =1
n
n
,由
lim (-1) n ≠0
x →n
知级数
n
∑(-1)
n =1
n
n
发散,所以级数的收敛域为(-1,1) .
n
-1
ρ=lim
(2)因为
a n +1a n
n →∞
=lim
(n +1)! (n +1)
n +1
n →∞
n
⎡⎛⎛⎫1⎫⎤
⋅=lim ⎪=lim ⎢ 1+⎪⎥n →∞n ! n →∞⎝n +1⎭n ⎭⎦⎣⎝
n n
=e
-1
R =
所以收敛半径
1
ρ
=e
,收敛区间为(-e,e) .
∞
当x =e时,级数变为
∑
n =1
e n n
n
n
1
lim
;应用洛必达法则求得
(1+x )x -e
x
x →0
=-
e 2,故有
⎛a n +1⎫1
-1⎪=-
n →∞2⎝a n ⎭由拉阿伯判别法知,级数发散;易知x =-e 时,级数也发散,故收敛域为(-e,e) .
(3)级数缺少偶次幂项.根据比值审敛法求收敛半径.
lim
U n +1U n
n →∞
=lim
x
2n +1
n →∞
2n +12n -12n +1
⋅
2n -1x
22n -1
=lim =x
2
n →∞
⋅x
所以当x 21即|x |>1时,级数发散,故收敛半径R =1.
1
∞
当x =1时,级数变为
∞
∑
n =1
1
2n -1,当x =-1时,级数变为
∞
∞
∑
n =1
-12n -1,由
1
lim =>0n →∞12
n
知,
∑
n =1
1
2n -1发散,从而
∑
n =1
-1
2n -1也发散,故原级数的收敛域为(-1,1) .
∞
(4)令t =x -1,则级数变为n =1n ⋅2n ,因为
所以收敛半径为R =1.收敛区间为 -1
n →∞
∞
∑
t
2
n
ρ=lim
a n +1a n
=lim
n ⋅2n
2
n →∞
(n +1)⋅2(n +1)
2
=1
当t =1时,级数n =12n 收敛,当t =-1时,级数敛.
所以,原级数收敛域为 0≤x ≤2,即[0,2]
12.利用幂级数的性质,求下列级数的和函数:
∞
∑
1
3
∞
∑(-1)
n =1
n
1
2⋅n 为交错级数,由莱布尼茨判别法知其收
3
∞
n +2
(1)
∑nx
n =1
; (2)
n +3
∑
n =0
x
2n +2
2n +1;
∞
lim
(n +1)x
n +2
nx 解:(1)由知,当|x |=
从而发散,故级数的收敛域为(-1,1) .
∞
∞
n +2
∞
n -1
n →∞
=x
∑nx
n +2
的通项不趋于0,
∞
S (x )=
记
∑nx
n =1
∞
=x
3
∑nx
n =1
易知
∑nx
n =1
n -1
S 1(x )=
的收敛域为(-1,1) ,记
∑nx
n =1
n -1
则
⎰
x 0
S 1(x )=
∑
n =1
x =
n
x 1-x
⎛x
S 1(x )=
⎝1-x 于是
lim
(2)由
3'x 1⎫
S (x )=⎪=22
(1-x ),所以(1-x )⎭
(x
x
2n +4
n →∞
2n +3
⋅
2n +1x
2n +2
=x
2
知,原级数当|x |
∞
S (x )=
域为(-1,1) ,记
∞
∑
n =0
x
2n +2∞
2n +1
=x ∑
n =0∞
x
2n +1∞
2n +1,易知级数n =02n +1收敛域为(-1,1) ,记=
11-x ,
12ln 1+x
1-x ,S 1(0)=0,所以
2
∑
x
2n +1
S 1(x )=
∑
n =0
x
2n +1
2n +1,则
12ln x
S 1'(x )=
∑
n =0
x
2n
故
⎰
x 0
S 1'(x )d x =
1+x 1-x 即1+x
S 1(x )-S 1(0)=
21-x
13.将下列函数展开成x 的幂级数,并求展开式成立的区间: (1)f (x )=ln(2+x ) ; (2)f (x )=cos2x ;
S (x )=xS 1(x )=
ln
(x
f (
x )=
(3)f (x )=(1+x )ln(1+x ) ;
(4)
x
2
12
;
(5)
f (x )=
x 3+x ;
2
(6)
f (x )=(e x -e -x )
1
2
;
f (x )=
(7)f (x )=ex cos x ;
(8)
(2-x ).
x ⎫x ⎫⎛⎛
f (x )=ln (2+x )=ln 2 1+⎪=ln 2+ln 1+⎪
⎝2⎭⎝2⎭ 解:(1)
∞
ln (1+x )=
由于
∑(-1)
n =0∞
n
n
x
n
n +1,(-1
n +1
n +1
x ⎫⎛
ln 1+⎪=
2⎭故⎝
∑(-1)
n =0
(n +1)2
∞
,(-2≤x ≤2)
ln (2+x )=ln 2+
因此
∑(-1)
n =0
n
x
n +1
n +1
(n +1)2
,(-2≤x ≤2)
(2)
f (x )=cos x =
∞
2
1+cos 2x
2
cos x =
由
∑(-1)
n =0
∞
n
x
2n
(2n )! ,(-∞
n
cos 2x =
得
所以
∑(-1)
n =02
(2x )
2n
∞
(2n )!
12
n
=
∑(-1)
n =0
n
4⋅x
n 2n
(2n )!
f (x ) =cos x =
=12+1
∞
12
+cos 2x 4⋅x
n
2n
(-1)∑2
n =0
(2n )! ,(-∞
x
n +1
(3)f (x )=(1+x )ln(1+x )
∞
ln (1+x )=
由
所以
∑(-1)
n =0
n
(n +1),(-1≤x ≤1)
∞
f (x )=(1+x )∑(-1)
n =0
∞
n
x
n +1
n +1
∞
=
∑(-1)
n =0
∞n =1∞
n
x
n +1
n +1
n
+
∑(-1)
n =0
∞
n
x
n +2
n +1
n +1
=x +
∑(-1)∑
n =1∞
x
n +1
n +1
n
+
∑(-1)
n =1n +1
x
n +1
n ⋅x
n +1
=x +
(-1)n +(-1)(-1)
2
n -1
(n +1)
n (n +1)
n (n +1)
2
=x +
∑
n =1
x
n +1
(-1≤x ≤1)
f (
x )=
(4)
=x ∞
=1+
∑(-1)
n =1
n
(2n -1)!! (2n )!!
x
2n
(-1≤x ≤1)
⎛
f (x )=x 1+
⎝故
2
∞
∞
∑
n =1
(-1)
n
n
(2n -1)!! (2n )!! (2n )!!
x
x
2n
⎫
⎪⎭
=x +
2
∑
n =1
(-1)
(2n -1)!!
2(n +1)
(-1≤x ≤1)
f (x )=
x 3x
⋅
11+
∞
x
2
3
n
2
n ⎛x ⎫
=⋅∑(-1) ⎪3n =0⎝3⎭
∞
=
(5)
∞
x
∑(
-1)
n =0
n
x
2n +1n +1
3
(x
e =
(6)由
∑
n =0
x
n
n ! ,x ∈(-∞,+∞)
n
n
∞
e
得
-x
=
∑
n =0
(-1)⋅x
n !
,x ∈(-∞,+∞)
f (x )=
12
(e x -e -x )
∞
1⎛∞x n = ∑-2⎝n =0n ! =12
∞
∑
n =0
n
n n
(-1)x ⎫
⎪n ! ⎭
∑⎡⎣1-(-1)
n =0∞
⎤⎦⋅
x
n
n !
=
所以
∑(
n =0
x
2n +1
2n +1)!
x
x ∈(-∞, +∞)
(1+i )x
x
(7)因为e cos x 为e (cos x +i sin x )=e
的实部,
∞
e
(1+i )x
=
∑
n =0∞
1n ! x
n
[(1+i )x ]
(
1+i )
n
n
=
∑
n =0∞
n !
n
=
∑
n =0∞
x ππ⎫⎤cos +i sin ⎪⎥n ! 44⎭⎦x
n
n
n
=
而
∑
n =0
n πn π⎫⎛⋅22 cos +i sin ⎪n ! ⎝44⎭
n
∞
22cos
n !
n π⋅x n
(-∞
e cos x =
取上式的实部.得
x
∑
n =0
1
(8)由于(1-x )
2
∞
=
∑nx
n =0
n -1
|x |
2
f (x )=
而
1
4⎛x ⎫ 1-⎪⎝2⎭,所以
∞
n -1
∞
⋅
1
⎛x ⎫
f (x )=⋅∑n ⎪
4n =0⎝2⎭
14.将
2
1
=
∑
n =0
n ⋅x 2
n -1
n +1
(|x |
f (x )=1
1
x +3x +2展开成(x +4)的幂级数. =
1x +1
-
1x +2
2
解:x +3x +2而
1x +1
=
1-3+(x +4)131⋅1-
∞
=-
1x +43
n
⎛x +4⎫
=-∑ ⎪
3n =0⎝3⎭
∞
⎛x +4⎫
⎝3⎭
=-∑
n =0
(x +4)
3
n +1
n
(-7
又
1x +2
=
1-2+(x +4)1211-
∞
=-
1x +42
n
⎛x +4⎫
=-∑ ⎪
2n =0⎝2⎭
∞
⎛x +4⎫
⎝2⎭
=-∑
n =0
(x +4)
2
2
n +1
n
(-6
f (x )=
1x +3x +2
∞
=-∑
n =0∞
(x +4)
3
n +1
n
∞
+
∑
n =0
(x +4)
2
n +1
n
=
所以
1⎛1-∑ n +1n +1
3n =0⎝2
⎫n
()x +4⎪⎭
(-6
15.将函数f (
x )=解:因为
(x -1) 的幂级数. m (m -1)
2!
2
(1+x )=1+
所以
m
m 1!
x +x + +
m (m -1) (m -n +1)
n !
x +
n
(-1
f (
x )=
3
=[1+(x -1)]2
3⎛33⎛3⎫⎫⎛3⎫⎛3⎫
-1-1-2 ⎪ ⎪ ⎪ -n +1⎪2⎝22⎝22n ⎭⎭⎝2⎭⎝2⎭
=1+(x -1)+(x -1)+ +(x -1)+
1! 2! n !
3
(-1
即
f (x )=1+
∞
32
(x -1)+
3⋅12⋅2!
2
(x -1)+
n
2
3⋅1⋅(-1)2⋅3!
3
(x -1)+
3
3⋅1⋅(-1)⋅(-3) (-2n +5)
2⋅n !
n
(x -1)+
n
=1+
∑
n =1
3⋅1⋅(-1)⋅(5-2n )
2⋅n !
n
(x -1)(0
16.利用函数的幂级数展开式,求下列各数的近似值: (1)ln3(误差不超过0.0001); (2)cos20(误差不超过0.0001)
352n -1
⎛⎫x x x
ln =2 x +++ ++ ⎪
352n -1⎝⎭,x ∈(-1,1) 解:(1)1-x
1+x
1+x
令1-x
=3
,可得
x =
12
∈(-1,1)
,
111⎡1⎤2+++ ++ ln 3=ln =2⎢352n -1⎥123⋅25⋅2(2n -1)⋅2⎣⎦1-
2故
又
1+
1
11⎡⎤
++ r n =2⎢2n +12n +3⎥(2n +3)⋅2⎣(2n +1)⋅2⎦=
2
(2n +1)2
2
2n +1
2n +12n +1
⎡⎤(2n +1)⋅2(2n +1)⋅21+++ ⎢⎥2n +32n +5
()⋅2()⋅22n +32n +5⎣⎦
(2n +1)2
2
2n +1
11⎛⎫1+++ ⎪24⎝22⎭⋅11-
14
(2n +1)2
1
2n +1
3(2n +1)2
1
r 5
3⨯11⨯2故 3⨯13⨯2
因而取n =6则
r 6
1
10
=
2n -2
≈0.00003
.
111⎛1⎫
ln 3=2 +≈1.0986++ +3511⎪
5⋅211⋅2⎭⎝23⋅2
⎛π⎫⎛π⎫ ⎪ ⎪
πn ⎝90⎭⎝90⎭0
cos 2=cos =1-+- +(-1)
902! 4! (2)⎛π⎫
⎪⎝90⎭
∵
2
24
⎛π⎫ ⎪⎝90⎭
+
(2n )!
2n
2!
≈6⨯10
-4
⎛π⎫
⎪⎝90⎭
;
4
4!
≈10
-8
cos 2≈1-
⎛π⎫ ⎪⎝90⎭2!
2
≈1-0.0006≈0.9994
故
17.利用被积函数的幂级数展开式,求定积分
⎰
0.50
arctan x
x
d x
(误差不超过0.001)的近似值.
arctan x =x -
解:由于
x
3
3
+
x
5
5
- +(-1)
n
x
2n +1
2n +1
+
,(-1≤x ≤1)
⎰
0.50
arctan x
x
d x =
⎰
0.50
242n
⎡⎤x x x
()1-+- ++ -1⎢⎥d x
352n +1⎣⎦
0.5
357
⎛⎫x x x = x -+-+ ⎪
92549⎝⎭
=
故
12
-1
1
92⋅1
5
⋅
1
3
+
1
252
⋅
1
5
-
11
492⋅1
7
⋅
1
7
+
1
而92
⋅
1
3
≈0.0139
,252
≈0.0013
,492
≈0.0002
.
⎰因此
∞
0.50
arctan x
x
n +
1n
n
d x ≈
12
-
1
92
⋅
1
3
+
1
252
⋅
1
5
≈0.487
18.判别下列级数的敛散性:
∑
n =1
n
(1)
∞
1⎫⎛n + ⎪
n ⎭; ⎝
∞
(2)n =1
∑
nx ⎫⎛
n cos ⎪
3⎭⎝
2
n
2
;
∑
n =1
ln (n +2)1⎫⎛
3+ ⎪
n ⎭⎝
n
n +
n
(3) .
1n
⎛n 2⎫
>= n n 2⎪1⎫1⎫⎝1+n ⎭⎛⎛ n +⎪ n +⎪
n ⎭n ⎭⎝解:(1)∵⎝
n
n
n
⎡⎛⎛n ⎫-1⎫
lim =lim 1+⎢⎪ 22⎪n →∞n →∞
1+n ⎝⎭1+n ⎝⎭⎣而
2∞
n
n
-n
-(1+n 2
)
⎤1+n 2
=1≠0⎥⎦
⎛n 2⎫
∑ 2⎪
故级数n =1⎝1+n ⎭发散,由比较审敛法知原级数发散.
0
(2)∵
nx ⎫⎛
n cos ⎪
3⎭⎝
2
n
2
≤
n 2
∞
n
∞
由比值审敛法知级数
∑
n =1
n
n
2收敛,由比较审敛法知,原级数n =1
∑
nx ⎫⎛
n cos ⎪
3⎭⎝
2
n
2
收敛.
0
(3)∵
ln (n +2)1⎫⎛
3+ ⎪
n ⎭⎝
n
ln (n +2)
3
n
lim
U n +1U n
n →∞
=lim ==1313
ln (n +3)3
n +1
n →∞
⋅
3
n
ln (n +2)
lim
ln (n +3)ln (n +2)
∞
n →∞
由
∞
知级数
∑
n =1
ln (n +2)
3
2
n
∑
n =1
ln (n +2)1⎫⎛
3+⎪
n ⎭收敛. ⎝
n
收敛,由比较审敛法知,原级数
∞
19.若
n →∞
lim n U n lim n U n
2
存在,证明:级数
∑U
n =1
n
收敛.
证:∵n →∞存在,∴∃M >0,使|n 2U n |≤M ,
M
即n 2|U n |≤M ,|U n |≤n
∞
2
n
而
∑
n =1
M n
2
∞
收敛,故
∞
∑U
n =12
绝对收敛.
∞
20.证明,若n =1
∑U n
1n
收敛,则
2
∑
n =1
U n
n 绝对收敛.
2
U n
证:∵n
∞
U n +
⋅U n ≤
∞
1=12U n +
2
=
2
1
2
2
2n
⋅
1
2
而由
∞
∑U n
n =1
收敛,
∑
n =1
1
n 收敛,知
∞
11⎫U n ⎛12
U +⋅∑ 2n 2n 2⎪∑
⎭收敛,故n =1n 收敛, n =1⎝
∞
因而
∑
n =1
U n
n 绝对收敛.
∞
∞
n
∞
n
21.若级数
∑a
n =1
与
∑b
n =1
都绝对收敛,则函数项级数
∑(a
n =1
n
cos nx +b n sin nx )
在R 上一致收敛.
证:U n (x )=a n cos nx +b n sin nx ,∀x ∈R 有
U n (x )=a n cos nx +b n sin nx ≤a n cos nx +b n sin nx ≤a n +b n
∞
∞
n
由于
∑a
n =1
与
∑b
n =1
n
都绝对收敛,故级数
∞
∑(a
n =1
n
∞
n
+b n
)
收敛.
由魏尔斯特拉斯判别法知,函数项级数n =1
22.计算下列级数的收敛半径及收敛域:
∑(a
cos nx +b n sin nx )
在R 上一致收敛.
∞
⎛∑ (1) n =1⎝
∞
∞
+1⎫n
⎪x
n +1⎭;
n
n
(2)
∑
n =1
sin
π2
n
(x +1)
n
;
(3)
∑
n =1
(x -1)
n ⋅2
2
n
ρ=lim
a n +1a n
n +2⎭+
1
n +1
n
n →∞
⎛
=lim n →∞
⎝=lim
解:
(1)
⎛⋅ ⎝n
n →∞
-1
n +2⋅e =
⎛⎛n +1⎫
⋅lim ⋅lim ⎪n →∞
⎝n +2⎭n →∞⎝
⎫
n
=e
R =
∴
1
ρ
=
3,
⎛∑ x =±
3又当时,级数变为n =1⎝
∞
+1⎫⎛⎫
±⎪ ⎪=
n +1⎭⎝3⎭
n n
⎫n ⎛3n +
(
)±1∑ ⎪n =1⎝3n +3⎭,
∞
n
⎛3n +⎫
lim ⎪=e n →∞
⎝3n +3⎭因为
x =±
所以当
n
≠0
3,级数发散,故原级数的收敛半径
R =
3,收敛域(
-3
, 3) .
ρ=lim
(2)
a n +1a n
sin =lim
n →∞
πn +1
π
=lim n →∞
n +1
n →∞
sin
π2
n
π
n
=
12
2
R =
故
1
ρ
=2
,
lim sin
n →∞
π2
n
sin
⋅2=lim π
n →∞n
π2
n n
π2π
n
=π≠0
.
又∵
∞
2所以当(x +1)=±2时,级数n =1发散,
从而原级数的收敛域为-2
∑
sin
(x +1)
n
ρ=lim
(3)
a n +1a n
n →∞
=lim
n ⋅2
2
2n n +1
n →∞
(n +1)⋅2
1
2
=
12
∞
∴R =2,收敛区间-2
∞
当x =-1时,级数变为
因此原级数的收敛域为[-1,3].
n =1
∑
x 0
(-1)
n
n ,其绝对收敛,当x =3时,级数变为n =1n ,收敛.
∑
1
2
23.将函数
F (x )=
⎰
arctan t
t
n
d t
展开成x 的幂级数.
∞
arctan t =
解:由于
x 0
x 0
∑(-1)
n =0
t
2n +1
2n +1
F (x )=
=
⎰
arctan t
t
d t =
n
⎰∑(-1)
0n =0
∞
x
∞
n
t
2n
2n +1
n
d t x
2n +1
2
∞
2n +1n =0n =0所以
24.判别下列级数在指定区间上的一致收敛性:
∞
∑⎰
(-1)
t
2n
d t =
∑(-1)
(2n +1)
(|x |≤1)
(1)n =1x +3
∞
∑
(-1)
n n
∞
,x ∈[-3,+∞) ;
(2)n =1x
∑
n
n
,x ∈(2,+∞) ;
(3)
∑(
n =1
n
2222
x +n )⎡⎣x +(n +1)⎤⎦,x ∈(-∞,+∞) ;
(-1)
∞
n n
=
1x +3
∞
解:(1)考虑n ≥2时,当x ≥-3时,有x +3
n
n
13-3
n
13
n -1
而
∑3
n =1
1
n -1
收敛,由魏尔斯特拉斯判别法知,级数
∑
n =1
(-1)
x +3在[-3,+∞) 上一致收敛.
n
n
(2)当x >2时,有x
n
=
n 2
n
n +1
n →∞
lim n +1
n
n
=
12
知级数
∞
由
2
∑
n =1
n
n
∞
2收敛,由魏尔斯特拉斯判别法知,级数n =1x 在(2,+∞) 上一致收敛.
∑
n
n
n
(3)∀x ∈R 有
∞
22
(x 2+n 2)⎡⎣x +(n +1)⎤⎦
≤
n n ⋅(n +1)
∞
22
n n
4
=
1n
3
而n =1
25.求下列级数的和函数:
∞
∑
1
22223⎤x +()n +1n 收敛,由魏尔斯特拉斯判别法知,级数n =1(x +n )⎡⎣⎦在(-∞,+∞) 上一致收敛.
∑
n
(1)
∑(-1)
n =1
n -1
x
2n ∞
2n -1; (2)n =02n +1;
∞
∑
x
2n +1
∞
(3)n =1
∑(
n n -1)!
x
n -1
; (4)n =1
∑n (
x
n
n +1).
∞
解:(1)可求得原级数的收敛半径R =1,且当|x |=1时,级数
域为[-1,1]
∞
∑(-1)
n =1
n -1
1
2n -1是收敛的交错级数,故收敛
S (x )=
记
∑(-1)
n =1
n -1
x
2n ∞
2n -1
∞
=x ∑(-1)
n =12n -2
n -1
x
2n -1
2n -11
2
=xS 1(x )
S 1'(x )=
则S 1(0)=0,
∑
n =1
(-1)
n -1
x =
1+x
所以 即S 1(x )=arctanx ,所以S (x )=x arctan x ,x ∈[-1,1].
∞
S 1(x )-S 1(0)=
⎰
x 0
11+x
2
d x =arctan x
S (x )=
(2)可求得原级数的收敛半径R =1,且当|x |=1时,原级数发散.记
∞
∑
n =0
x
2n +1
2n +1则
S '(x )=
1-x
x x 111+x 11+x 'S ()d x =d x =ln S ()-S ()=ln x x 0⎰0⎰01-x 2
21-x ,即21-x ,S (0)=0
n =0
∑x
2n
=
1
2
所以
S (x )=
12
ln
1+x 1-x ,(|x |
n +1
n →∞
lim
a
n +1n
=lim
a
n →∞
n ! n
=0
S (x )=
知收敛域为(-∞,+∞) .记
∞
∞
(3)由
(n -1)!
∞
∑(
n =1
n n -1)!
x
n -1
则
⎰
x 0
S (x )d x =
∑(
n =1
x
n
n -1)!
=x ∑
n =1
x
n -1
(n -1)!
=x e
x
,所以
S (x )=(x e x )'=(1+x )e ,(-∞
x
1
lim
(4)由
n →∞
(n +1)(n +2)
=1
1
n (n +1)
知收敛半径R =1,当x =1时,级数变为
∞
n
∞
∑
n =1
11
n (n +1),由n (n +1)
1n
2
知级数收敛,当x =-1时,级数变为
∞
∑n (
n =1
(-1)
n +1)是收敛的交错级数,故收敛域为[-1,1].
∞
S (x )=
记
∑n (
n =1
∞
x
n
n +1)则S (0)=0,
n -1
xS (x )=
∑n (
n =1
x
n +1
n +1),
[xS (x )]''=
x
∑
n =1
x =
1
1-x (x ≠1)
(x )]''d x =-ln (1-x )[xS
所以⎰0
()()即[xS x ]=-ln 1-x
'
⎰0[xS (x )]d x =⎰0-ln (1-x )d x =(1-x )ln (1-x )+x
即xS (x )=(1-x )ln (1-x )+x
x
x
'
⎛1⎫
S (x )=1+ -1⎪ln (1-x )
⎝x ⎭当x ≠0时,,又当x =1时,可求得S (1)=1 1⎫⎛
lim S (x )=lim 1-⎪=1n →∞n →∞⎝n +1⎭(∵)
综上所述
⎧0,
⎪
S (x )=⎪1,
⎨
⎛1⎫⎪
1+ -1⎪ln (1-x ), ⎪⎝x ⎭⎩
x =0x =1
x ∈[-1, 0) (0,1)
⎧2
f (x )=⎨3
⎩x 26.设f (x ) 是周期为2π的周期函数,它在(-π,π]上的表达式为
-π
试问f (x ) 的傅里叶级数在x =-π处收敛于何值?
解:所给函数满足狄利克雷定理的条件,x =-π是它的间断点,在x =-π处,f (x ) 的傅里叶级数收敛于
-+
f (-π)+f (-π)
2
=
12
[π3+2]=
12
(2+π3)
⎧-1
f (x )=⎨2
⎩x 27.写出函数
-π≤x ≤00
的傅里叶级数的和函数.
解:f (x ) 满足狄利克雷定理的条件,根据狄利克雷定理,在连续点处级数收敛于f (x ) ,在间断点x =0,x =±π
-+
f (0)+f (0)
处,分别收敛于
-+
f (-π)+f (-π)
2
=-
12,
-+
f (π)+f (π)
2
=
π-12
,
2
2
=
π-12
,综上所述和函数.
2
28.写出下列以2π为周期的周期函数的傅里叶级数,其中f (x ) 在[-π,π)上的表达式为:
⎧-1⎪2x ⎪⎪1
S (x )=⎨-
⎪2⎪π2-1⎪⎩2
-π
⎧π⎪⎪4
f (x )=⎨
⎪-π⎪⎩4(1)
(2)
0≤x
f (x )=x
2
(-π≤x ≤π);
-π≤x
π2π2, ,
⎧π⎪-2, ⎪⎪
f (x )=⎨x ,
⎪⎪π⎪, ⎩2(3)
x
f (x )=cos
2(4)
≤x
(-π≤x ≤π)
.
解:(1)函数f (x ) 满足狄利克雷定理的条件,x =n π,n ∈z 是其间断点,在间断占处f (x ) 的傅里叶级数收敛于
f (0
+
)+f (0)
-
π=
⎛π⎫+ -⎪4⎝4⎭
2=1
2
a n =
1π
=0
,在x ≠n π,有
⎰
π-π
f
(x )cos nx d x
1⎛π⎫
-cos nx d x +⎰ ⎪π-π⎝4⎭π
⎰
π0
π4
cos nx d x =0
b n =
1π
⎰
π-π
1⎛π⎫
f (x )sin nx d x =⎰ -⎪sin nx d x +
π-π⎝4⎭πn =2, 4, 6, , n =1, 3, 5, .
1
⎰
π
π4
sin nx d x
⎧0,
⎪=⎨1⎪, ⎩n
∞
于是f (x ) 的傅里叶级数展开式为
f (x )=
(x ≠n π)
(2)函数f (x ) 在(-∞,+∞) 上连续,故其傅里叶级数在(-∞,+∞) 上收敛于f (x ) ,注意到f (x ) 为偶函数,从而f (x )cos nx 为偶函数,f (x )sin nx 为奇函数,于是
n =1
∑2n -1sin (2n -1)x
π
1
b n =a n =
1π1π
⎰
-ππ-π
f (x )sin nx d x =0f
,
a 0=
π0
2
1π
⎰
π-π
x d x =
2
2π3
2
,
n
⎰
(x )cos nx d x
∞
n
=
2π
⎰
x cos nx d x =(-1)⋅
4
n (n =1,2,…)
2
所以,f (x ) 的傅里叶级数展开式为:
f (x )=
π
2
(-∞
(3)函数在x =(2n +1)π (n ∈z ) 处间断,在间断点处,级数收敛于0,当x ≠(2n +1)π时,由f (x ) 为奇函数,有a n =0,(n =0,1,2,…)
n =1
3
+
∑(-1)
⋅
4n
2
cos nx
b n =
2π
⎰
π0
π
2⎡2
f (x )sin nx d x =⎢⎰x sin nx d x +
π⎣0
⎰
ππ2
π
⎤
sin nx d x ⎥2⎦
=-
所以
1n
(-1)
∞
n
+
2n π
n +1
2
sin
n π22n π
2
(n =1, 2, )
sin
f
(x )=∑⎢(-1)
n =1
⎡⎣
⋅
1n
+
n π⎤
sin nx 2⎥⎦ (x ≠(2n +1)π,n ∈z )
2作为以2π为周期的函数时,处处连续,故其傅里叶级数收敛于f (x ) ,注意到f (x ) 为(4)因为
偶函数,有b n =0(n =1,2,…) ,
f
(x )=cos
x
a n ==
11
⎰π
π
π-ππ0
cos
x 2
cos nx d x =
⎰π
2
π0
cos
x 2
cos nx d x ⎫⎤
⎪x ⎥d x ⎭⎦
π
⎰
⎡1⎛cos n + ⎢
2⎝⎣1⎫⎛x +cos n -⎪
2⎭⎝
⎡1⎛
sin n + 1⎢2⎝=⎢
1π⎢
n +⎢2⎣=(-1)
n +1
1⎫⎤⎫⎛
x sin n -⎪ ⎪x ⎥
2⎭⎭⎝
⎥+
1⎥n -
⎥2⎦0
4⎛1⎫
⎪2
π⎝4n -1⎭4π
∞
(n =0,1, 2, )
所以f (x ) 的傅里叶级数展开式为:
f (x )=
2π
+
∑(-1)
n =1
n +1
cos nx
4n -1 x ∈[-π,π] x
2
29. 将下列函数f (x ) 展开为傅里叶级数:
f
(1)(2)
(x )=
π4
-
x 2
(-π
f (x )=sin x
a 0=
1π
(0≤
π
x ≤2π)
解:(1)
⎰
-π
f (x )cos nx d x =
1
x ⎫π⎛π-d x =⎪⎰
π-π⎝42⎭2
π
a n =
=b n =
1
⎛πx
-⎰-π
π⎝42
π1⎫
cos nx d x =⎪
4⎭
⎰
π-π
cos nx d x -
⎰2π
1
π-π
x cos nx d x
14n 1
[sin nx ]-π-0=0
π
π
(n =1, 2, )
x ⎛π
-⎰
π-π⎝42
n
1⎫
sin nx d x =⎪
4⎭
⎰
π-π
sin nx d x -
⎰2π
1
π-π
x sin nx d x
=(-1)⋅f
1n
∞
4n =1n (-π
(2)所给函数拓广为周期函数时处处连续, 因此其傅里叶级数在[0,2π]上收敛于f (x ) ,注意到f (x ) 为偶函数,
(x )=
π
+
∑(-1)
π-ππ0
n
sin nx
a 0==
有b n =0,
1
⎰π
2π
f (x )cos 0x d x =
4π
1
⎰π
π-π
sin x d x
⎰
sin x d x =
a n ===
⎰π
1
2
π0π0
f
(x )cos nx d x =
⎰π
2
π-π
sin x cos nx d x
⎰π
⎡⎣sin (n +1)x -sin (n -1)x ⎤⎦d x
⎡1+(-1)n ⎤2⎦π(n -1)⎣
n =1, 3, 5, n =2, 4, 6,
-2
⎧0, ⎪
-4=⎨
,
⎪π(n 2-1)⎩
所以
f (x )=
2π
∞
+
∑
n =1
-4cos 2nx π(4n -1)
2
(0≤x ≤2π)
30. 设f (x )=x +1(0≤x ≤π),试分别将f (x ) 展开为正弦级数和余弦级数. 解:将f (x ) 作奇延拓,则有a n =0 (n =0,1,2,…)
b n =
2π
⎰
π0
f (x )sin nx d x =
n
2π
⎰(x +1)sin nx d x
π
21-(-1)(1+π)=⋅πn f (x )=
n
n n =1从而
若将f (x ) 作偶延拓,则有b n =0 (n =1,2,…)
∑π
2
∞
1-(-1)(1+π)
sin nx
(0
a n =
2π
⎰
π0
f
(x )cos nx d x
=
2π
⎰(x +1)cos nx d x
π
从而
n =2, 4, 6 ⎧0,
⎪=⎨-4
⎪2, n =1, 3, 5, ⎩n π1π2π
a 0=⎰f (x )d x =⎰(x +1)d x =π+2
π-ππ0
π+24∞cos (2n -1)x
f (x )=-∑2
2πn =1(2n -1)
(0≤x ≤π)
∞
31. 将f (x )=2+|x | (-1≤x ≤1) 展开成以2为周期的傅里叶级数,并由此求级数n =1n 的和.
解:f (x ) 在(-∞,+∞) 内连续,其傅里叶级数处处收敛,由f (x ) 是偶函数,故b n =0,(n =1,2,…)
∑
1
2
a 0=
a n =
⎰⎰
1-11-1
f (x )d x =2⎰
f
10
(2+x )d x =5
10
(x )cos nx d x =2⎰
n =2, 4, 6 n =1, 3, 5,
(2+x )cos nx d x
⎧0, ⎪
=⎨-4
,
⎪n π2⎩()
所以
f
(x )=
52
-
∞
4π
2
∞
∑
n =1
cos (2n -1)πx
(2n -1)
2
2
,x ∈[-1,1]
∑
取x =0得,
∞
n =1∞
1
(2n -1)
1
2
=
∞
π
2
8
,故
∑
n =1
1n
2
=
∑
n =12
(2n -1)
π
+
∑
n =1
1
(2n )
2
=
1
∞
∑4
1n
2
+
π
2
n =1
8
∞
6 所以n =1n
32. 将函数f (x )=x -1(0≤x ≤2) 展开成周期为4的余弦级数.
解:将f (x ) 作偶延拓,作周期延拓后函数在(-∞,+∞) 上连续, 则有b n =0 (n =1,2,3,…) a 0=
1
∑
1
=
⎰2
12
2
2-2
f
(x )d x =
⎰(x -1)d x =0
2
a n ==
4
⎰
2-2
f
(x )cos
n
n πx 2
d x =
⎰0(x -1)cos
2
n πx 2
d x
n π
2
[(-1)-1]
n =2, 4, 6, n =1, 3, 5,
⎧0, ⎪=⎨8
-, ⎪22⎩n π
f
故
(x )=-
8π
2
∞
∑
n =1
1
(2n -1)
2
⋅cos
12
(2n -1)πx
2,
s (x )=
(0≤x ≤2)
⎧x , ⎪⎪
f (x )=⎨
⎪2-2x , ⎪⎩33. 设
0≤x ≤12
a 02
∞
+
∑a
n =1
n
cos n πx
,-∞
a n =2⎰
10
⎛5⎫
s -⎪
f (x )cos n πx d x
,求⎝2⎭. 5
2处间断,所以
解:先对f (x ) 作偶延拓到[-1,1],再以2为周期延拓到(-∞,+∞) 将f (x ) 展开成余弦级数而得到 s (x ) ,延拓后
x =-
f (x ) 在
+
⎛5-⎫⎤1⎡⎛1+⎫⎛1-⎫⎤⎛5⎫1⎡⎛5⎫
s -⎪=⎢f -⎪+f -⎪⎥=⎢f -⎪+f -⎪⎥
22222⎝2⎭2⎢⎢⎭⎝⎭⎥⎭⎝⎭⎥⎣⎝⎦⎣⎝⎦
=
1⎛1⎫3+1 ⎪=2⎝2⎭4
∞
34. 设函数f (x )=x (0≤x
2
s (x )=
∑b
n =1
n
sin n πx
,-∞
b n =2⎰
1
⎛1⎫
s -⎪
f (x )sin n πx d x
(n =1,2,3,…),求⎝2⎭.
解:先对f (x ) 作奇延拓到,[-1,1],再以2为周期延拓到(-∞,+∞) ,并将f (x ) 展开成正弦级数得到s (x ) ,延拓
x =-
后f (x ) 在
1
2处连续,故.
2
1⎛1⎫⎛1⎫⎛1⎫
s -⎪=-f -⎪=- -⎪=-
4. ⎝2⎭⎝2⎭⎝2⎭
35. 将下列各周期函数展开成为傅里叶级数,它们在一个周期内的表达式分别为:
1⎫⎛1
-≤x
0≤x
解:(1) f (x ) 在(-∞,+∞) 上连续,故其傅里叶级数在每一点都收敛于f (x ) ,由于f (x ) 为偶函数,有b n =0 (n =1,2,3,…)
1
1
f
(x )=⎨
⎧2x +1, -3≤x
a 0=2⎰21f
-2
(x )d x =4⎰02(1-x 2)d x =
1
116,
1
a n =2⎰21f
-2
(x )cos2n πx d x =4⎰02(1-x 2)cos2n πx d x
2
=
所以
(-1)
2
n +1
n π11121
(n =1, 2, )
1π
2
∞
f
(x )=
a 0=
+
3-3
∑
n =1
(-1)
n
2
n +1
cos 2n πx
(-∞
30
(2)
⎰3
f
1⎡0
f (x )d x =(2x +1)d x +
⎣⎰-33⎢
⎰
d x ⎤=-1
⎥⎦,
a n ===
1313
⎰
3-30
(x )cos
n πx 3
d x d x +
13
⎰-3(2x +1)cos
6
n πx 3
⎰
30
cos
n πx 3
d x
⎡1-(-1)n ⎤, (n =1, 2, 3, )22⎦n π⎣
13n πx b n =⎰f (x )sin d x
3-33==13
⎰-3(2x +1)sin
(-1)
-12
n +1
n πx 3
d x +
13
⎰
30
sin
n πx 3
d x
6n π
, (n =1, 2, )
而函数f (x ) 在x =3(2k +1),k =0,±1,±2,…处间断,故
∞
f
(x )=
+
n πx n πx ⎫n n +16⎧6⎡⎤1--1cos +-1sin ()()⎬∑⎨n 2π2⎣⎦3n π3⎭(x ≠3(2k +1),k =0,±1,±2,…) n =1⎩
36. 把宽为τ,高为h ,周期为T 的矩形波(如图所示)展开成傅里叶级数的复数形式
.
解:根据图形写出函数关系式
⎧⎪0, ⎪⎪
u (t )=⎨h ,
⎪⎪⎪0, ⎩
--
T 2
≤t
T 2
τ
2
τ
2
τ
2
τ
2
≤t ≤
T
c 0=c n ==1T
12l 12l
τ
⎰⎰
l -l l -l
u (t )d t =u (t )e
-i 2n πT
t
-i n πl
1T
t
⎰
2T -2
u (t )d t =1T
T
1T
τ
⎰τ
-2-i 2n πT
2
h d t =
t
h τT
d t =
⎰
2T -2
u (t )e
τ
d t
⎫t ⎪⎭
⎰τ
-2
2
h e
h ⎛-T ⎫2-i
d t =⋅ ⎪⎰τe
T ⎝2n πi ⎭-2
τ
2n πT
t
2n π⎛d -i
T ⎝
h ⎛
= -
⎝2n πi
h τT
h n πτ⎫⎡-i 2n πt ⎤2
=sin ⎪⎣e T ⎦-τ
n πT ⎭2
h π
∞
故该矩形波的傅里叶级数的复数形式为
u (t )=
+
∑
n =-∞
n ≠0
1n
sin
n πτT
e
-i
2n πT
t
22,…) (-∞
-37. 设f (x ) 是周期为2的周期函数,它在[-1,1]上的表达式为f (x )=ex ,试将f (x ) 展成傅里叶级数的复数形式. 解:函数f (x ) 在x ≠2k +1,k =0,±1,±2处连续.
t ≠±
τ
, ±
3τ
c n ==-
⎰2l
1
l -l
f
(x )e
-i
n πl
x
d x =
1
1
⎰2
1-1
e e
-x -in πx
d x
12(1+n πi )
-1
[e -(1+n πi )x ]-1
n
=
e -e 2
⋅(-1)⋅
n
11+n πi 1-n πi
2
=sinh 1⋅(-1)⋅
1+(n π)
n
故f (x ) 的傅里叶级数的复数形式为
∞
in πx
f (x )=sinh 1∑
n =-∞
(-1)⋅(1-in π)
1+(n π)
2
e
(x ≠2k +1,k =0,±1,±2,…)
⎧A ,
f (t )=⎨
⎩0, 38. 求矩形脉冲函数
F (ω)=
0≤t ≤T 其他
的傅氏变换
-i ωx
解:
39. 求下列函数的傅里叶积分:
⎰
+∞-∞
f (t )e
-i ωx
d t =
⎰
T 0
A e d t =
A (1-e -i ωx )
i ω
⎧e -t ,
f (t )=⎨
⎩0, (1)
t ≥0t
⎧-1, ⎪
f (t )=⎨1,
⎪0, ⎩(2)
F (ω)=
=
解:(1)
-1
-i ωt
⎰
+∞-∞
f (t )e =
d t =
⎰
+∞0
e
-t
⋅e
-i ωt
d t
11+ωi
1-i ω1+ω
2
f (t )=
==
⎰2π⎰2π
11
1
+∞-∞+∞-∞
F (ω)e
i ωt
d ω=
⎰2π
d t
1
+∞-∞
1-i ω1+ω
2
e
i ωt
d ω
cos ωt +ωsin ωt
1+ω1+ω
22
⎰π
+∞0
cos ωt +ωsin ωt
+∞-∞0-1
-i ωt
d t
F (ω)=
==
(2)
⎰⎰
f (t )e
d t
(-e -i ωt )d t +
i ω
⎰
10
e
-i ωt
d t
2(1-cos ω)
i ωt
+∞-∞
f (t )=
===
⎰2π
1π1π2π
1
F (ω)e
d ω=
⎰2π
1
+∞-∞
2(1-cos ω)
i ω
e
i ωt
d ω
⎰⎰
+∞-∞+∞-∞+∞0
(1-cos ω)
i ω
(cos ωt +i sin ωt )d ω
d ωd ω
(1-cos ω)sin ωt
ω
(1-cos ω)sin ωt
⎰
40. 求如图所示的三角形脉冲函数的频谱函数
.
ω
(t ≠0,1)
2E ⎧E +t ⎪⎪T
f (t )=⎨
⎪E -2E t ⎪⎩T 解:F (ω)=
-
T 2
≤t ≤0T 2
0
⎰
+∞-∞0
f (t )e
-i ωt
2E ⎛=⎰T E +-
T 2⎝=
⎫-i ωt
d t +t ⎪e ⎭
T
⎰
20
2E ⎛ E -⎝T ⎫-i ωt
d t t ⎪e ⎭
4E ⎛ωT ⎫
1-cos ⎪2
T ω⎝2⎭