Esercizi svolti di analisi reale e complessa - Dipartimento di ...

log(1 + x) = x
0
1
1+tdt |t| < 1
1 1+t = ∞ k=0 (−1)ktk
−t
|x| < 1
log(1+x) =
x
0
1
dt =
1 + t
+∞
x
k=0
0
(−1) k t k ∞
k xk+1
dt = (−1)
k + 1 =
∞
(−1)
log(1−x) = − ∞ k=1 xk
k
x −x
log 1+x
1−x =
log (1 + x) − log (1 − x)
log
1 + x
1 − x =
∞
(−1)
k=1
k+1 xk
k +
∞
k=1
k=1
k=0
j=0
k=1
xk k =
∞
[(−1) k+1 +1] xk
∞ x
= 2
k 2j+1
2j + 1 .
+∞ α
k=0
( α ∈ R \ Z )
lim
k→∞
k
x k
k+1 xk
α
k + 1
α
= lim
k!α(α − 1) . . . (α − k)
k→∞ (k
+ 1)!α(α − 1) . . . (α − k + 1) = lim
α − k
k→∞ k + 1 = 1
k
lim sup α
k
1
k
= 1
R = 1
Dk (1+x) α |x=0 = α(α−1) . . . (α−k+1)(1+x)α−k |x=0 = α(α−1) . . . (α−k+1)
PN (x, 0) =
N α
k=0 x
k
k
∀ |x| < 1 (1 + x) α =
+∞ α
k=0 x
k
k
|(1 + x) α − PN−1(x, 0)| N→∞
−→ 0
x ∈ (−1, 1) |x| < θ < 1 ∃ ɛ > 0 θ(1 + ɛ) < 1
1
θ < 1 lim sup α k
= 1
N0 ∀ N ≥ N0 α
N
lim sup
1
N
≤ 1 + ɛ
k
k .