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[0, 1] fn = nxe−nx2 N N N N sN = fn(x) − fn+1(x) = fn(x) − fn+1(x) = f1(x) − fN+1(x), n=1 n=1 sN(x) = xe−x2 −(N +1)xe−(N+1)x2 x ∈ [0, 1] limn→∞(n + 1)xe−(n+1)x2 = 0 f(x) = xe−x2 n=1 sup |xe x∈[0,1] −x2 − sn| = sup |(N + 1)xe x∈[0,1] −(N+1)x2 | = sup |fn+1(x)| x∈[0,1] fN+1(0) = 0 fn+1(1) = (N + 1)e −(N+1) f ′ N+1(x) = (N + 1)e −(N+1)x2 − 2(N + 1) 2 x 2 e −(N+1)x2 = (N + 1)e −(N+1)x2 (1 − 2(N + 1)x 2 ), [0, 1] x = 1/ 2(N + 1) 1 fN+1 = 2(N + 1) (N + 1) e 2(N + 1) −1/2 → +∞ N → +∞ Γ := {(x, y) ∈ R 2 : 4(x 2 + y 2 − x) 3 = 27(x 2 + y 2 ) 2 }. Γ Pi = (xi, yi) i = 1, 2, 3, 4 P1 Γ P2 P3 P4 i = 1, 2, 3, 4 Γ y = ϕi(x) C 1 xi ϕi(xi) = yi h(x, y) = x 2 + y 2 Γ Γ Γ f(x, y) = 4(x 2 + y 2 − x) 3 − 27(x 2 + y 2 ) 2 f(x, −y) = f(x, y) Γ f(ρ cos θ, ρ sin θ) = 4ρ 3 (ρ − cos θ) 3 − 27ρ 4 = ρ 3 (4(ρ − cos θ) 3 − 27ρ). ρ > 0 4(ρ − cos θ) 3 = 27ρ 3√ 3 Γ = {(ρ cos θ, ρ sin θ) : 4ρ − 3 √ ρ = 3√ 4 cos θ, ρ ≥ 0, θ ∈ [0, 2π[} ∪ {(0, 0)}. f(0, 0) = 0 Γ f(0, y) = 4y 6 − 27y 4 = y 4 (4y 2 − 27), y = 0 y = ±3 √ 3/2 P3 = (0, 3 √ 3/2) P4 = (0, −3 √ 3/3) f(x, 0) = 4x 3 (x − 1) 3 = 27x 4 , x = 0 4(x−1) 3 −27x = 0 4x 3 −12x 2 −15x−4 = 0
4 ±1, ±2, ±4 ±1, ±2 4 x − 4 4x 3 −12x 2 −15x −4 x − 4 −4x 3 +16x 2 4x 2 + 4x + 1 4x 2 −15x −4x 2 +16x x −4 −x 4 0 4x 3 − 12x 2 − 15x − 4 = (x − 4)(4x 2 + 4x + 1) = (x − 4)(2x + 1) 2 , x = 4 x = −1/2 P1 = (−1/2, 0) P2 = (4, 0) f ∂xf(x, y) = 12(x 2 + y 2 − x) 2 (2x − 1) − 108x(x 2 + y 2 ) = 12((x 2 + y 2 − x) 2 (2x − 1) + 9x(x 2 + y 2 )) ∂yf(x, y) = 12y(2(x 2 + y 2 − x) 2 − 9(x 2 + y 2 )). P2 P3 P4 df(4, 0) = 12(144 · 7 − 36 · 4) dx = 144 · 72 dx (4, 0) x = 0 x = 4 df(0, 3 √ 3/2) = 729 16 (− dx + √ 3 dy) (0, 3 √ 3/2) −x + √ 3y = q q = 9/2 −x + √ 3y = 9/2. (0, −3 √ 3/2) x + √ 3y = −9/2 P3 P4 ∂yf(P3) = 0 ∂yf(P4) = 0 ∂yf(P1) = ∂yf(P2) = 0 ρ ≥ 0 − 3√ 4 ≤ 3√ 4ρ − 3 3√ ρ ≤ 3√ 4 z(ρ) = 3√ 4ρ − 3 3√ ρ z(0) = 0 ˙z(ρ) = 3√ 4 − ρ −2/3 > 0 z(ρ) ρ = 1/2 ¨z(ρ) > 0 z(1/2) = − 3√ 4 ρ > 1/2 3√ 4 ρ ρmax z(ρmax) = 3√ 4 θ = 0 (4, 0) ρ 1/2 ρ 4 ρ 2 16 f Γ ρ Γ H(ρ, θ) := z(ρ) − 3√ 4 cos θ = 0 ρ θ ˙z(ρ) = 0 ρ = 1/2 ρ = ρ(θ) C 1 dρ dθ (θ) = −∂θH(ρ(θ), θ) ∂ρH(ρ(θ), θ) = − 3√ sin θ 4 ˙z(ρ(θ) , θ = 0, π Γ ρ = 1/2 (4, 0) (0, 0) (−1/2, 0) (0, 0) ρ (4, 0) ρ
Page 3 and 4:
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) =
Page 65 and 66: α ∈ R (0, 0) f(x, y) = 2
Page 67: Cα > 0 α /∈ [−1/2, −1/4]
Page 70 and 71: x = cos θ y = sin θ θ ∈ [−π
Page 72 and 73: γ(t) = (t, 1/t) t → ±∞ L(x
Page 74 and 75: ⎧ ⎪⎨ 1 + λ(2x − 2) = 0, 1
Page 76 and 77: f(x, y) = x 2 + y 2 V = {(x, y)
Page 78 and 79: ⎧⎪ 2x + λ(2x + y) + 2µx = 0
Page 81 and 82: R 2 D ⊆ R 2 f : D → R (x0,
Page 83 and 84: x = 1 y = −1 z = 2 y z
Page 85: ⎛ 1 ⎜ 3 F (m1, m2, p1, p2) =
Page 88 and 89: k = 3√ 4 ξ ζ
Page 90 and 91: x = ρ cos θ y = ρ sin θ ρ 2 (
Page 92 and 93: f(x, y, z) = ez+x2 f + αx + y
Page 94 and 95: ϕ(1) = 0 ϕ ′ (1) = − ∂xf(1
Page 96 and 97: 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: (x 2 + y 2 )(y 2 + x(x + 1)) = 4xy
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 and 148: f [1, +∞[ [0, 1] [0
Page 149 and 150: f α k = 1 x α + k α , sα n = n
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)
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˙x = e t−x /x x(α) = 1 α ∈ R
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˙x = x 2 /(1 − tx) x(0) = α α
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y ′′′ − 6y ′′ + 11y ′
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c ′ 1 (x) cos x + c′ 2 (x) sin
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y ′ + y 1 x = tan x sin x y ′
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λ 2 − 4λ − 5 = 0, A
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F (x, y, ˙y) = 0, F : A → R A
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F (x, y, p) = 0 c = 0 dy = p dx
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y = xy ′ + f(y ′ ) F (x, y, p
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y → 0 −∞
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x > 0 x y = ± 1 + x x < 0
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[0, π] utt + 2ut − uxx = 0,
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e |uk(x, t)| = C −t sin( 4k(1
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∞ 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
Page 219 and 220:
2π xt sin t F (x) = e dt x ∈ R
Page 221 and 222:
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
Page 227 and 228:
(u + v) D 2 + (u − v) 2 1 + 2(u2
Page 229 and 230:
α = = 2 3 π. 2π 0
Page 231 and 232:
F D z = 0 (0,
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y ′ = 2xy − x y(0) = 3/2
Page 235 and 236:
Γ f(x, y) = 0 f : R 2 → R
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∇f(x0, y0) = (0, 0) (x0, y0)
Page 239 and 240:
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
Page 256 and 257:
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) =
Page 262 and 263:
y(0) = c1 + d1 = 1 ˙y(t) = (−c1
Page 264 and 265:
a ′ x + b ′ y + c ′ = 0
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η2 g g(η) > 0 y0 < η <
Page 269 and 270:
2 × 2 F ∈ R[x] n
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2 × 2 t ↦→ x(
Page 274 and 275:
∆ < 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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