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k → +∞ ⌊x⌋ x lim k→0 + +∞ 1 (−1) ⌊x⌋ | sin πx|1/k 1/x2 x = 1/2 + k k ∈ N N := {x = 1/2 + k : k ∈ N} lim k→0 + +∞ 1 (−1) x 2 ⌊x⌋ | sin πx|1/k x 2 dx, dx, = 0. gk(x) = k/π 1 + k2 k ∈ N \ {0} x2 gk > 0 gk(x) dx = 1 R ε > 0 gk {x : |x| > ε} f gk(x)f(x) dx = f(0). R lim k→+∞ gk > 0 x ∈ R k ≥ 1 gk(x) dx = 1 k dx π 1 + x2 1 dy = = 1. k2 π 1 + y2 R R R ε > 0 0 < gk(x) = gk(−x) < gk(ε) x > ε gk(ε) → 0 + k → +∞ {x : |x| > ε} gk(x)f(x) dx = 1 dy f(y/k) dy π 1 + y2 k f(y/k)/(1+y 2 ) f∞/(1+y 2 ) lim k→∞ gk(x)f(x) dx = 1 π dy f(0) lim f(y/k) dy = k→∞ 1 + y2 π R y = kx 1 k +∞ 1/k R sin x 1 dx = x2 k +∞ 1 lim k→∞ k 1/k +∞ 1 R sin x dx. x2 R sin(y/k) y2 /k2 dy k = +∞ sin(y/k) 1 y2 dy R dy = f(0). 1 + y2 1/y 2 α > 0 [0, +∞[ ∞ 1 Fα(x) = xα . + kα k=1
f α k = 1 x α + k α , sα n = n f α k . f α k > 0 sαn Fα +∞ f 0 α k +∞ 0 Fα(x) dx = lim n→∞ 1 (x) dx = kα +∞ 0 +∞ 0 sn(x) dx = lim dx (x/k) α 1 = + 1 kα +∞ 0 k=1 n +∞ f n→∞ k=1 0 α k (x) dx k dy yα 1 = + 1 kα−1 +∞ 0 α > 1 n +∞ lim f n→∞ k=1 0 α k (x) dx = lim n 1 n→∞ k k=1 α−1 +∞ dy 0 yα + 1 = +∞ 0 dy y α + 1 lim dy y α + 1 n n→∞ k=1 1 k α−1 α − 1 > 1 α > 2 Fα [0, +∞[ α > 2 k ∈ N \ {0} fk(x) = k 3 (x − k) 2 χ [k−1/k,k+1/k](x) fk R R lim k→∞ fk(x) dx = lim k→∞ R fk(x) dx. K R R > 0 |x| < R x ∈ K k > R + 1 K ∩ [k − 1/k, k + 1/k] = ∅ fk(x) = 0 x ∈ K K fk R fk(x) dx = k+1/k k k−1/k 3 (x − k) 2 1/k dx = −1/k fk 2/3 k 3 y 2 1/k dy = 2 k 0 3 y 2 dy = 2 3 , ϕ ∈ C∞ (R) S = {x ∈ R : ϕ(x) = 0} − log |x|ϕ R ′ (x) dx = lim ε→0 + ϕ(x) R\[−ε,ε] x dx. log |x|ϕ ′ (x) ≤ ϕ ′ ∞ log |x|χS(x) log |x|χS(x) L 1 0 /∈ S | log |x|| ¯ S L 1 1 0 | log |x|| dx = 1,
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R R n
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2 × 2 SO(
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R a
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R F G R F G ¯ F ⊇ G
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R n x, y ∈ Rn x = (x1, ...
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(X, τ) D ⊆ X X
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f : D → R D ⊆ Rn γ : [
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A =]0, +∞[×]0, +∞[ f : A →
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lim (x,y)→(0,0) (x,y)=(0,0) |xy|
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E E E f −1 ([1, +∞[)
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s ↦→ 1 − e −s s ≥ 0 1
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Fn : [0, +∞[→ R Fn(ρ) = 2n ρ
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I =]a, b[ R {fn}n∈N fn : I
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K [0, +∞[ R > 0 B(0, R) ⊇
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an = 0 n = 2k an = 4/(πn2 ) n
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5 1 18 = + 2 4 π2 ∞ n=0 ∞ n=0
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g(x) := x(π−x) x ∈ [0, π]
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X Y T : X → Y T ℓ >
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X K E X a, b ∈ R f : E × [a
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|f(x, y)| ≤ log(1 + 3y
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(x, y) ∈ R 2 xy =
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X, Y D ⊆ X f : D → Y u ∈
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H A 2 × 2 λ 2
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(2kπ, π/2+2hπ) (π+2kπ, 3π/2+2
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∂xf(x, y, z) = −2 cos x sin x =
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(0, 0) f : R 2 → R f(x, y) =
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α ∈ R (0, 0) f(x, y) = 2
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Cα > 0 α /∈ [−1/2, −1/4]
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x = cos θ y = sin θ θ ∈ [−π
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γ(t) = (t, 1/t) t → ±∞ L(x
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⎧ ⎪⎨ 1 + λ(2x − 2) = 0, 1
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f(x, y) = x 2 + y 2 V = {(x, y)
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⎧⎪ 2x + λ(2x + y) + 2µx = 0
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R 2 D ⊆ R 2 f : D → R (x0,
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x = 1 y = −1 z = 2 y z
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⎛ 1 ⎜ 3 F (m1, m2, p1, p2) =
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k = 3√ 4 ξ ζ
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x = ρ cos θ y = ρ sin θ ρ 2 (
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f(x, y, z) = ez+x2 f + αx + y
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ϕ(1) = 0 ϕ ′ (1) = − ∂xf(1
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y 3 − xy 2 + x 2 y = x − x 3
Page 98 and 99: X = {(ρ, θ) ∈ R 2 : ρ > 0, 0 <
Page 100 and 101: arcsin sin θ = θ θ ∈] − π
Page 102 and 103: I R r : I → Rd γ µ : r
Page 104 and 105: γ lunghezza(γ) = |r ′ (θ)|
Page 106 and 107: p = 2 |(x, y)| −p dx dy = R 2 \B
Page 108 and 109: ρ 3 cos θ = aρ 2 (cos 2 θ − s
Page 111 and 112: Γ := {(x, y) ∈ R 2 : (x 2 + y 2
Page 113 and 114: x Γ Q2 Q3 x Γ \
Page 115 and 116: (x 2 + y 2 )(y 2 + x(x + 1)) = 4xy
Page 117 and 118: 4 ±1, ±2, ±4 ±1, ±2 4
Page 119 and 120: −2 + 8t − 4t 2 t = 1/2(2
Page 121 and 122: a ∈ R R 2 (x 2 + y 2 ) 3 = 4
Page 123: (x 2 + y 2 ) 3 = 4x 2 y 2
Page 127 and 128: a > 0 x = at at2 , y = . 1
Page 129: cos(π − α) = − cos(α) π/4
Page 132 and 133: I := F · ˆn dσ ˆn Σ =
Page 134 and 135: S ψ(x, y) = (ψ1, ψ2, ψ3) = (
Page 136 and 137: C = {(x, y, z) : (x, y) ∈ D, λ <
Page 138 and 139: ˆn = ∇G = ((∇G)1, (∇G)2, (
Page 140 and 141: 2π 0 cos θ sin θ dθ = 1 2 2π
Page 142 and 143: L − F · ˆn dσ = − L− √
Page 145 and 146: K ⊆ R n χK : R n → {0, 1}
Page 147: f [1, +∞[ [0, 1] [0
Page 151 and 152: D R X n 1 1 X C ℓ
Page 153 and 154: γ2(1/2) = (1/8, 3/4) γ2(1) = (0,
Page 155 and 156: C = −B ∂yu(x, y) = C/x 1 + y2 y
Page 157 and 158: dy N(x, y) = − dx M(x, y) N M
Page 159 and 160: 2xy dx + (x 2 + 1) dy = 0 (x 2 + y
Page 161 and 162: x − c = arcsin(2y − 1) 0 < y
Page 163: h(x, y) = 1/(|x| √ x2 − 1)
Page 166 and 167: ˙x = e t−x /x x(α) = 1 α ∈ R
Page 168 and 169: ˙x = x 2 /(1 − tx) x(0) = α α
Page 170 and 171: y ′′′ − 6y ′′ + 11y ′
Page 172 and 173: c ′ 1 (x) cos x + c′ 2 (x) sin
Page 175 and 176: y ′ + y 1 x = tan x sin x y ′
Page 177 and 178: λ 2 − 4λ − 5 = 0, A
Page 179 and 180: F (x, y, ˙y) = 0, F : A → R A
Page 181 and 182: F (x, y, p) = 0 c = 0 dy = p dx
Page 183: y = xy ′ + f(y ′ ) F (x, y, p
Page 186 and 187: y → 0 −∞
Page 188 and 189: x > 0 x y = ± 1 + x x < 0
Page 191 and 192: [0, π] utt + 2ut − uxx = 0,
Page 193 and 194: e |uk(x, t)| = C −t sin( 4k(1
Page 195: ∞ n=1 bne −x sin nx = x(
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T ˙ Tn(t) = −5n 2 Tn(t) Tn(
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⎧ −utt + 3uxx = 0 ]0, π[×]0,
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∆ = 0 λ = −1 µ1 = µ2 = −
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x = cos θ y = sin θ F F (co
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I := arctan y dx − xy dy γ γ
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f(ρ) f ′ (ρ) = 4ρ3 + 36ρ 8
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Jac ϕ(θ, y) = ⎛ ⎜ ⎝ − y
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z = (x, y) ˙z = Az + B(t)
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Q0 Q2 Q1 Q1
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˙x = t − x 2 x(0) = a a ∈ R
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2π xt sin t F (x) = e dt x ∈ R
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xz 3 dx dy dz, T T y
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α > 0 lim (x,y)→(0,0) (x,y)=(0,0
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F (0, 1) = 0 ∂yF (x, y) = 3y 2
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(u + v) D 2 + (u − v) 2 1 + 2(u2
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α = = 2 3 π. 2π 0
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F D z = 0 (0,
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y ′ = 2xy − x y(0) = 3/2
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Γ f(x, y) = 0 f : R 2 → R
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∇f(x0, y0) = (0, 0) (x0, y0)
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ϕ : I ×J → R 3 I, J R
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ψ(a, t) = ψ(b, t) t ∈]c, d[
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[
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Y K Ω R × Ω I R f :
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1 ω(x, y) = M(x, y) dx + N(x, y)
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M(x, y) dx + N(x, y) dy = 0 1 M, N
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x, y z dy dz
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f(x) = N(x) Ak1 = + D(x) x − xk
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Φ y ′ = A(t) y ′ = A(t)
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det P = 4 √ 7 = 0 P −1 AP
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y ′′ (t) + py ′ (t) + qy(t) =
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y(0) = c1 + d1 = 1 ˙y(t) = (−c1
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a ′ x + b ′ y + c ′ = 0
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η2 g g(η) > 0 y0 < η <
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2 × 2 F ∈ R[x] n
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2 × 2 t ↦→ x(
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∆ < 0 α = −d 2c ω = √ |
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a = 0 b = 0 c = 0 ⎧ ⎪⎨ b
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SO(3) U(3) := {O ∈
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SO(3) Aξ = ω × ξ ξ
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x 2 + y 2 = 1 x 2 − y 2 = 1
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1 arcsin(x) + x − x2 dx = √ 1
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z 1 + cos z cos = ± 2 2 z 1
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