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

k
2k + 1
2n
= (−1)k
(2k + 1)!
n=0
2k + 1 2k + 1
=
2n 2k + 1 − 2n
= (−1)k 1
(2k + 1)! 2
2k+1
n=0
2k + 1
n
=
1
(2k + 1)!
= (−1) k
2 22k+1
e iπ = cos π + i sin π = cos 2 π
π π π π
+ i sin 2π = cos2 − sin2 + 2i cos sin
2 2 2 2 2 2 =
π cos π
2 = 0
2 π π
= − sin = cos2 − 1 = −1
2 2
φɛ R
f ∈ L 1 (R) C ∞ f ∈ C ∞ 0
+∞
φɛdx =
−∞
ɛ
γ
0
γ(x) ≡
φɛdx ≤ φɛ∞ɛ < ∞ .
x
−∞ φɛdx
+∞ .
φɛdx −∞
φɛ C ∞ φɛ ∀k ≥
1 γ (k) (x) = φ(k−1)
ɛ (x)
+∞
−∞ φɛdx
2k+1 2k + 1
n=0
= (1 + 1)
n
2k+1 = 22k+1
φɛ [0, ɛ]