Exercices corrigés sur les séries de Fourier

2π f :R→R
f(x) = π− |x|]−π, π]f
2π
f:R→Rf(x) = x2[0,2π[f
f:R→R2π
f(x) = 1x∈]0, π[
0x=π.
f
f
∞
k=0
(−1)k
2k+ 1,
∞
k=0
1
(2k+ 1)2,
∞
n=1
1
n2,
∞
n=1
(−1)n−1
n2.
f:R→R2π f(x) = exx∈]−π, π]
f
f
∞
n=1
(−1)n
n2+ 1,
∞
n=1
1
n2+ 1.
f:R→R2π
f(x) = (x−π)2, x ∈[0,2π[.
f
f
∞
n=1
(−1)n
n2,
∞
n=1
1
n2.

f:R→R2π α
x′(t) + αx(t) = f(t).
2π x(t)f(t)
α= 1
f(t) = 
t−π
22t∈[0, π[,
−t−3π
22
+π2
2t∈[π, 2π[.
C(R/2πZ)R
C2π f1, f2∈C(R/2πZ)
(f1~f2)(x) = 1
2ππ
−π
f1(x−y)f2(y) dy.
k∈N∗φk:R→C
φk(x) = 1
k
k−1
l=0
l
m=−l
eimx.
φk(x) = 
1
k
1−cos kx
1−cos xx̸∈ 2πZ,
k x ∈2πZ.
φk
k(2π)−1π
−πφk(x) dx= 1
ε∈]0, π[ (2π)−1|x|∈[ε,π]φk(x) dx→0k→ ∞
f∈C(R/2πZ)f~φkfR
f~φk

f bn
an(f) = 2
ππ
0
f(t) cos(nt) dt= 2 π
0
cos(nt) dt−2
ππ
0
tcos(nt) dt=2
πn21−(−1)nn̸= 0,
π n = 0.
SF (f)(t) = π
2+
k≥1
4
π(2k+ 1)2cos (2k+ 1)t.
fRf
R
f
a0(f) = 1
π2π
0
f(t) dt=1
π2π
0
t2dt=1
πt3 32π
0
=8π2
3,
n≥1
an(f) = 1
π2π
0
t2cos(nt) dt
=1
πt2sin(nt) n2π
0
−2π
0
2tsin(nt)
ndt
=−2
πt−cos(nt) n22π
0
+2π
0
cos(nt)
n2dt
=−2
π−2π n2+sin(nt) n32π
0
=4
n2
bn(f) = 1
π2π
0
t2sin(nt) dt
=1
π−t2cos(nt) n2π
0
+2π
0
2tcos(nt)
ndt
=1
π−4π2 n+2tsin(nt) n22π
0
−22π
0
sin(nt)
n2dt
=−4π
n+2 πcos(nt) n32π
0
=−4π
n.
SF (f)(t) = 4π2
3+ 4 
n≥1cos(nt)
n2−πsin(nt) n.

f SF (f)t
f(t+) + f(t−)
2=f(t)t̸∈ 2πZ,
2π2t∈2πZ.
f an= 0 n∈Nn≥1
bn(f) = 2
ππ
0
sin(nt) dt=−cos(nt)
nπ
0
=2
π
1−(−1)n
n=
4
nπ n ,
0n .
f
SF (f)(t) =
∞
k=0
4
(2k+ 1)πsin (2k+ 1)t.
f SF (f)
tf(t+) + f(t−)
2=f(t).
f
t=π/2
sin (2k+ 1)t= sin π
2+kπ= (−1)k,
∞
k=0
(−1)k
(2k+ 1) =π 4fπ 2=π 4.
f
1
2ππ
−π
|f(t)|2dt=1
2
∞
n=1
|bn(f)|2=8
π2
∞
k=0
1
(2k+ 1)2,
∞
k=0
1
(2k+ 1)2=π2 8.
∞
n=1
1
n2=
∞
k=0
1
(2k+ 1)2+
∞
k=1
1
(2k)2=π2 8+1 4
∞
n=1
1
n2,
∞
n=1
1
n2=4 3
π2
8=π2 6.
∞
n=1
(−1)n−1
n2+
∞
k=1
1
(2k)2=
∞
k=1
1
(2k+ 1)2=π2 8,
∞
n=1
(−1)n−1
n2=π2 8−1 4
π2
6=π2 12 .

cn(f) = 1
2ππ
−π
ete−int dt
=1
2ππ
−π
e(1−in)tdt
=1
2πe(1−in)t 1−in π
−π
=1
2πe(1−in)π−e−(1−in)π 1−in 
=1
2π(−1)neπ−e−π 1−in 
=sh π
π
(−1)n
1−in.
SF (f)(t) = c0(f) + 
n≥1cn(f)eint +c−n(f)e−int
f(t)t∈]−π, π[f(π+) + f(π−)/2 = ch π t =π
sh π
π
n∈Z
(−1)n
1−ineint =ett∈]−π, π[
ch π t =π.
t= 0
1 = sh π
π
n∈Z
(−1)n
1−in,
π
sh π= 1 +
∞
n=1
(−1)n1
1−in +1 1 + in=−1 +
∞
n=0
2(−1)n
1 + n2,
∞
n=0
(−1)n
1 + n2=1 2π sh π+ 1.
t=π
ch π=sh π
π
n∈Z
1
1−in,
π
th π=
n∈Z
1
1−in =−1 +
∞
n=0
2
1 + n2,
∞
n=0
1
1 + n2=1 2π th π+ 1.