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5 1 18
= +
2 4 π2 ∞
n=0
∞
n=0
1
,
(2n + 1) 2
1 π2
=
(2n + 1) 2 8 .
u : R → R 2π
−t − π ≤ t < 0,
u(t) =
π 0 ≤ t < π.
u
∞
k=0
1
.
(2k + 1) 2
u |u(t)| [−π, π]
a0 = 1
π
π
−π
π
f(t) dt = 1
π
t=−π
0
−π
an = 1
f(t) cos(nt) dt =
π −π
1
π
= 1
t=0 −t sin(nt)
+
π n
1
nπ
−t dt + 1
π
0
−π
0
−π
π
0
π dt = π 3π
+ π =
2 2 .
−t cos(nt) dt + 1
π
π
sin(nt) dt + 1
[sin nt]t=π t=0
n
0
π cos(nt) dt
= − 1
n2 1 − (−1)n
[cos nt]t=0 t=−π = −
π n2 .
π
bn = 1
π
f(t) sin(nt) dt =
π −π
1
0
−t sin(nt) dt +
π −π
1
π
π sin(nt) dt
π 0
= 1
t=0 t cos(nt)
−
π n
1
0
cos(nt) dt −
nπ
1
[cos nt]t=π t=0
n
t=−π
= (−1)n
n − (−1)n − 1
=
n
1
n .
f(t) = a0
2 +
∞
n=1
−π
an cos(nx) + bn sin(nx) = 3π
4 +
∞
n=1
(−1) n − 1
n 2 π
cos(nx) +
sin nx
n .
u C ∞ u
u
xk = 2kπ k ∈ Z
π 0 (2k + 1)π k ∈ Z
0 u π/2
π
2
= 3π
4 −
∞ 1 − (−1) n
πn2 .
n=1