Devoir d`Analyse - Université de Caen

tan
sin cos tan cotan
sinh cosh tanh cotanh
Cf f(0) = 1 f(t) = et−1
tt6= 0
fC
0f
C
f(t) = 0 t∈2iπZ∗
B0= 1
Bk=−1
k+ 1
k−1
X
j=0 k+ 1
jBj,∀k>1
REE(t) =
+∞
X
k=0
Bk
k!tk
Bk06k66
k>1
k
X
j=0 k+ 1
jBj= 0
D(0, RE)f(t)E(t) = 1
E(t) = t
et−1
RE62π
Bk(x) =
k
X
j=0 k
jBjxk−j
E(t, x) =
+∞
X
k=0
Bk(x)
k!tkBk(0) = BkE(t, 0) = E(t)
Bk(x) 0 6k66

Bk(x)k
etxE(t) = E(t, x)
t∈D(0, RE)t6= 0
E(t, x) = tetx
et−1
k>1B0
k(x) = kBk−1(x)
t∈D(0, RE)E(t, x + 1) = E(t, x) + tetx
k>1Bk(x+ 1) = Bk(x) + kxk−1
k>0Bk(1) = Bk+δk,0
N
X
n=0
nk=1
k+ 1 (Bk+1(N+ 1) −Bk+1) = Bk+1(N) k+ 1 −Bk+1 k+ 1 +Nk.
N
X
n=0
nkN06k65
t∈D(0, RE)E(−t, −x) = E(t, x) + tetx
k>1 (−1)kBk(−x) = Bk(x) + kxk−1
B1
k>0bk(t) 2π bk(t) = Bkt
2π
[0,2π[
bk
k bk
Bk(x)0=kBk−1(x)
k>2x∈[0,1[
(−1)k/2(2π)kBk(x)
k!=−2
+∞
X
n=1
1
nkcos(2πnx)
k
(−1)(k−1)/2(2π)kBk(x)
k!=−2
+∞
X
n=1
1
nksin(2πnx)
k
k>1
+∞
X
n=1
1
n2k=(−1)k−1 2(2π)2kB2k (2k)!
+∞
X
n=1
1
n2kk= 1 2 3
k>1|B2k|
(2k)! 6π2 3(2π)−2k
E(t)|t|<2π RE= 2π

t∈D(0,2π)E(t) + t
2=t 2cotanh t 2
t∈D(0, π)tcotanh t=
+∞
X
k=0
22kB2k
(2k)!t2k
cotan(t) = icotanh(it)t∈C\πZ
t∈D(0, π)tcotan t=
+∞
X
k=0
(−1)k22kB2k
(2k)!t2k
tan(t) = cotan(t)−2 cotan(2t)t∈C\π
2Z
t∈D(0,π
2) tan t=
+∞
X
k=1
(−1)k−122k(22k−1) B2k
(2k)!t2k−1
tanh t=
+∞
X
k=1
22k(22k−1) B2k
(2k)!t2k−1