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, ⋅ ⋅ ⋅),