将下列各周期函数展开成傅里叶级数(下面给出函数在一个...

习题11-8
1. 将下列各周期函数展开成傅里叶级数(下面给出函数在一个周期内的表达式):
(1))2
12
1(1)(2<≤--=x x x f ;
解 因为f (x )=1-x 2为偶函数, 所以b n =0(n =1, 2, ⋅ ⋅ ⋅), 而 611)1(4)1(2/1221
0221
020=-=-=⎰⎰dx x dx x a ,
⎰-=21022/1c o s )1(2/12dx x n x a n π
2
2
121
2
)1(2c o s )1(4π
πn x d x n x n +-=
-=⎰
(n =1, 2, ⋅ ⋅ ⋅),
由于f (x )在(-∞, +∞)内连续, 所以
∑
∞
=+-+=1
2
1
2
2c o s )1(1
1211)(n n x n n x f ππ
, x ∈(-∞, +∞).
(2)⎪⎪
⎩
⎪⎪⎨⎧
<≤-<≤<≤-=1
21
12
1
0 101 )(x x x x x f ;
解 2
1)(1
2
121
1
11
-=-+==⎰⎰⎰⎰--dx dx xdx dx x f a n ,
⎰⎰
⎰⎰-+==--1
2
121
1
11
c o s c o s c o s c o s )(x
d x n x d x n x d x n x x d x n x f a n ππππ
2
s i n 2])1(1[122πππ
n n n n +--= (n =1, 2, ⋅ ⋅ ⋅),
dx x n xdx n xdx n x xdx n x f b n ⎰⎰
⎰⎰-+==--1
2
1210
1
1
1
sin sin sin sin )(ππππ
π
ππ
n n n 12
c o s 2+-= (n =1, 2, ⋅ ⋅ ⋅).
而在(-∞, +∞)上f (x )的间断点为x =2k , 2
12+k , k =0, ±1, ±2, ⋅ ⋅ ⋅,
故 }s i n 2c o s
21c o s ]2s i n 2)1(1{[41)(122x n n n x n n n n x f n n
πππππππ
-++--+-=∑∞
=
(x ≠2k , 2
12+≠k x , k =0, ±1, ±2, ⋅ ⋅ ⋅).
(3)⎩⎨
⎧<≤<≤-+=3
0 1 03 12)(x x x x f .
解 1])12([3
1)(3
13
33
30-=++==⎰⎰⎰--dx dx x dx x f a ,
]3
c o s 3
c o s )12([3
13c o s )(3
13
3
3
3
⎰⎰⎰--++==dx x n dx x n x dx x n x f a n πππ
])1(1[62
2
n n --=
π
(n =1, 2, ⋅ ⋅ ⋅ ),
]3
s i n 3
s i n )12([3
13s i n )(3
13
3
33
⎰⎰⎰--++==dx x n dx x n x dx x n x f b n πππ
n n )1(6-=π
(n =1, 2, ⋅ ⋅ ⋅ ),
而在(-∞, +∞)上, f (x )的间断点为 x =3(2k +1), k =0, ±1, ±2, ⋅ ⋅ ⋅, 故
}3
s i n 6)1(3c o s ])1(1[6{21)(1122∑∞
=+-+--+-=n n n x
n n x n n x f ππππ,
(x ≠3(2k +1), k =0, ±1, ±2, ⋅ ⋅ ⋅).
2. 将下列函数分别展开成正弦级数和余弦级数:
(1)⎪⎩
⎪⎨
⎧≤≤-<≤=l
x x l l x x x f 2l
20 )(; 解 正弦级数:
对f (x )进行奇延拓, 则函数的傅氏系数为 a 0=0(n =0, 1, 2, ⋅ ⋅ ⋅),
2s i n 4]s i n )(s i n [22
22
121
0π
πππn n l dx l x n x l dx l x n x l b l n =-+=⎰⎰(n =1, 2, ⋅ ⋅ ⋅ )
故 ∑∞
==
1
2
2s i n
2s i n 14)(n l
x n n n
l
x f ππ
π, x ∈[0, l ].
余弦级数:
对f (x )进行偶延拓, 则函数的傅氏系数为
2])([22
121
00l dx x l xdx l a l
=-+=⎰⎰,
⎰⎰-+=l
n dx l x n x l dx l x n x l a 2
1210]cos )(cos [2ππ
])1(12
c o s 2[22
2n n n l ---=ππ
(n =1, 2, ⋅ ⋅ ⋅ ) b n =0(n =1, 2, ⋅ ⋅ ⋅ ), 故
l x n n n l l x f n n πππ
c o s ])1(12c o s 2[124)(1
2
2
∑∞
=---+=, x ∈[0, l ].
(2)f (x )=x 2(0≤x ≤2).
解 正弦级数:
对f (x )进行奇延拓, 则函数的傅氏系数为 a 0=0(n =0, 1, 2, ⋅ ⋅ ⋅),
]1)1[()
(168)1(2sin 2
2312
2--+-==+⎰n n n n n dx x n x b πππ,
故 2
s i n }]1)1[()(168)1{()(1
31x n n n x f n n n πππ∑∞
=+--+-=
2s i n }]1)1[(2)1({8
1231x
n n n n n n πππ∑∞
=+--+-=, x ∈[0, 2).
余弦级数:
对f (x )进行偶延拓, 则函数的傅氏系数为
3
82
22
20==⎰dx x a
2
2
2)
(16
)1(2cos 2
2ππn dx x n x a n n -==⎰(n =1, 2, ⋅ ⋅ ⋅),
b n =0(n =1, 2, ⋅ ⋅ ⋅),
故
2
c o s
)(16)1(34)(12x
n n x f n n ππ∑∞
=-+=
2
c o s
)1(16341
2
2x n n n n ππ
∑
∞
=-+=, x ∈[0, 2].。