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an = 0 n = 2k an = 4/(πn2 ) n = 2k + 1
f = 4
∞ cos ((2k + 1)t)
π (2k + 1) 2 .
k=0
L2
S(t) = 2
∞ (1 − (−1)
π
n )
n2 cos nt,
n=1
t = kπ k ∈ Z f
S(t) = f(t)
t = kπ k ∈ Z f
S(t) = f(t)
S(0) = f(0) = π/2 π
2
∞
k=0
1 π2
=
(2k + 1) 2 8
= 4
π
∞
k=0
1
(2k + 1) 2
f : R → R 2π f(t) = 3(π + t)
t ∈ [−π, 0] f
∞ 1
(2n + 1) 4
L2 (0, 2π)
π
|f(t)|
−π
2 0
dt = 2 9(π + t)
−π
2
(π + t) 3 0
= 18
= 6π
3 −π
3 .
f
S(t) = a0
2 +
∞
ak cos(kt),
k=1
a0 = 1
π
f(t) dt =
π −π
2
0
f(t) dt =
π −π
2
0
3(π + t) dt =
π −π
6
(π + t) 2 0
= 3π
π 2
ak = 1
π
= 6
π
π
f(t) cos(kt) dt =
−π
2
π
(π + t) sin(kt)
0 −
k
6
π
−π
0
−π
0
−π
f(t) cos(kt) dt = 2
π
sin(kt)
k
dt = − 6
kπ
0
−π
−π
3(π + t) cos(kt) dt
− cos(kt)
k
0
−π
n=0
= 6
πk 2 (1 − (−1)k )
k ≥ 1 ak = 0 k = 2n ak = 12/(πk2 ) k = 2n + 1
S(t) = 3
∞ 12 cos((2n + 1)t)
π +
2 π (2n + 1) 2 .
1
2π
π
−π
n=0
|f(t)| 2 dt =
a0 2
4
+ 1
2
∞
k=1
a 2 k ,