pdf (it, 2.898,97 KB, 25/01/13)

π 4 /96
3π 2 = 9
4 π2 + 72
π2 ∞
n=0
1
.
(2n + 1) 4
f : R → R 2π
2 t ∈ [−π, 0[
f(t) =
−1 t ∈ [0, π[
∞
n=0
1
.
(2n + 1) 2
f ∈ L 2 (−π, π)
f 2 L2 π
=
−π
|f(t)| 2 dt =
0
−π
|f(t)| 2 dt +
π
0
|f(t)| 2 dt = 4π + π = 5π < +∞.
f ∈ L2 (−π, π)
a0 = 1
π
f(t) dt =
π −π
1
0
f(t) dt +
π −π
1
π
f(t) dt = 2 − 1 = 1
π 0
ak = 1
π
f(t) cos kt dt =
π −π
1
0
f(t) cos kt dt +
π −π
1
π
f(t) cos kt dt
π 0
= 2
0
t=0 sin kt
−
π
k
1
t=π sin kt
= 0
π k
bk = 1
π
= 2
π
cos kt dt −
−π
1
π
cos kt dt =
π 0
2
π
π
0
f(t) sin kt dt = 1
π
−π
−π
0
sin kt dt −
−π
1
π
sin kt dt =
π 0
2
π
f(t) sin kt dt + 1
π
− cos kt
k
t=−π
π
0
t=0 t=−π
= − 2
1
(1/k − cos(−kπ)) + (cos(kπ)/k − 1/k)
π π
= 1
3
(−2 + 2 cos(kπ) + cos(kπ) − 1) = (cos kπ − 1).
πk πk
bk = 0 k bk = −6/(πk) k
f(t) sin kt dt
− 1
π
t=0
− cos kt
g(t) = f(t) − 1/2 3/2 t ∈ [−π, 0[ −3/2
t ∈ [0, π[ ak = 0 k > 0
π
π
g(t) cos kt dt = f(t) − 1
π
cos kt dt = f(t) cos kt dt,
2
−π
f(t) = a0
2 +
∞
ak cos kt + bk sin kt = 1 6
−
2 π
k=0
1
2π
π
−π
−π
|f(t)| 2 dt = a2 0
4
∞
k=0
−π
1
2
+ ak + b
2
2 k ,
k
sin ((2k + 1)t)
.
2k + 1
t=π
t=0