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$(Microsoft PowerPoint - 03_Gruppi_Alcani [modalit\340 compatibilit ...$

Φ y ′ = A(t) y ′ = A(t)y + b(t) u(0) = y0 u(t) = R(t, t0)y0 + t t0 R(t, τ)b(τ) dτ Φ R(t, τ) = Φ(t)Φ −1 (τ) n y (n) + ... + any = Q(x) y(x) = c1y1(x) + ... + cnyn(x) ck ∈ R ck ck = ck(x) n n c ′ k (x) ⎧ ⎪⎨ ⎪⎩ c ′ 1 (x)y1(x) + ... + c ′ n(x)yn(x) = 0, c ′ 1 (x)y′ 1 (x) + ... + c′ n(x)y ′ n(x) = 0, c ′ 1 (x)y(n−1) 1 (x) + ... + c ′ n(x)y (n−1) n (x) = 0, c ′ 1 (x)y(n) 1 (x) + ... + c′ n(x)y (n) n (x) = Q(x). ⎛ y1(x) ⎜ y ⎜ ⎝ ... yn(x) ′ 1 (x) ... y′ y n(x) (n−1) ⎞ ⎛ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ (x) ⎠ ⎝ 1 (x) ... y (n−1) n y (n) 1 (x) ... y(n) n (x) c ′ 1 (x) c ′ 2 (x) c ′ n−1 c ′ n ⎞ ⎛ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = ⎜ ⎟ ⎜ ⎠ ⎝ 0 0 0 Q(x) y1(x), ..., yn(x) c ′ k (x) ck(x) ¯y(x) = c1(x)y1(x) + ... + cn(x)yn(x), y(x) + ¯y(x) = (c1(x) + c1)y1(x) + ... + (cn(x) + cn)yn(x), c1, ..., cn ∈ R. y ′′ + a(t)y ′ + b(t)y = f(t), ˜y = c1(t)y1(t) + c2(t)y2(t) y1(t) y2(t) y ′′ + a(t)y ′ + b(t)y = 0. ˜y ′ = c ′ 1y1 + c ′ 2y2 + c1y ′ 1 + c2y ′ 2. c ′ 1y1 + c ′ 2y2 = 0 ˜y ′ = c1y ′ 1 + c2y ′ 2 ˜y ′′ = c ′ 1y ′ 1 + c ′ 2y ′ 2 + c1y ′′ 1 + c2y ′′ 2. (c ′ 1y ′ 1 + c ′ 2y ′ 2 + c1y ′′ 1 + c2y ′′ 2) + a(c1y ′ 1 + c2y ′ 2) + b(c1y1 + c2y2) = f c1(y ′′ 1 + ay ′ 1 + by1) + c2(y ′′ 2 + ay ′ 2 + by2) + (c ′ 1y ′ 1 + c ′ 2y ′ 2) = f. ⎞ ⎟ ⎠
y1 y2 c ′ 1y ′ 1 + c ′ 2y ′ 2 = f c ′ 1 c ′ 2 c ′ 1y1 + c ′ 2y2 = 0 c ′ 1y′ 1 + c′ 2y′ 2 = f. y1 y2 y ′ 1 y′ 2 y1 y2 −y2f c ′ 1 = y ′ 2y1 − y ′ 1y2 y1f c ′ 2 = y ′ 2y1 − y ′ 1y2 c ′ 1 c ′ 2 ˙z = Az + b(t) R 2 A = 5 3 2 3 , b(t) = t e −t 2 ˙z = Az A A u ∈ R 2 Au = λu λ ∈ C u det(λid R 2 − A) = 0 0 = det λ − 5 −3 −2 λ − 3 . = (λ − 5)(λ − 3) − 6 = λ 2 − 8λ + 9 = 0, λ1 = 4 + √ 7 λ2 = 4 − √ 7 u1, u2 (λiid R 2 − A)ui = λi − 5 −3 −2 λi − 3 ui = 0, i = 1, 2. −2u x 1 + (1 + √ 7)u y 1 = 0, −2ux2 + (1 − √ 7)u y 2 = 0 ui = (ux i , uy i ) u1 = (1 + √ 7, 2) u2 = (1 − √ 7, 2) P u1 u2 √ 1 + 7 P = 2 √ 1 − 7 , 2 √ 1/(2 7) P = −1/(2 √ √ (−1 + 7)/(4 7) √ 7) (1 + √ 7)/(4 √ , 7)
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
Page 30 and 31:
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
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X = {(ρ, θ) ∈ R 2 : ρ > 0, 0 <
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arcsin sin θ = θ θ ∈] − π
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I R r : I → Rd γ µ : r
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γ lunghezza(γ) = |r ′ (θ)|
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p = 2 |(x, y)| −p dx dy = R 2 \B
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ρ 3 cos θ = aρ 2 (cos 2 θ − s
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Γ := {(x, y) ∈ R 2 : (x 2 + y 2
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x Γ Q2 Q3 x Γ \
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(x 2 + y 2 )(y 2 + x(x + 1)) = 4xy
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4 ±1, ±2, ±4 ±1, ±2 4
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−2 + 8t − 4t 2 t = 1/2(2
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a ∈ R R 2 (x 2 + y 2 ) 3 = 4
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(x 2 + y 2 ) 3 = 4x 2 y 2
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a > 0 x = at at2 , y = . 1
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cos(π − α) = − cos(α) π/4
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I := F · ˆn dσ ˆn Σ =
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S ψ(x, y) = (ψ1, ψ2, ψ3) = (
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C = {(x, y, z) : (x, y) ∈ D, λ <
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ˆn = ∇G = ((∇G)1, (∇G)2, (
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2π 0 cos θ sin θ dθ = 1 2 2π
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L − F · ˆn dσ = − L− √
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K ⊆ R n χK : R n → {0, 1}
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f [1, +∞[ [0, 1] [0
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f α k = 1 x α + k α , sα n = n
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D R X n 1 1 X C ℓ
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γ2(1/2) = (1/8, 3/4) γ2(1) = (0,
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C = −B ∂yu(x, y) = C/x 1 + y2 y
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dy N(x, y) = − dx M(x, y) N M
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2xy dx + (x 2 + 1) dy = 0 (x 2 + y
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x − c = arcsin(2y − 1) 0 < y
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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
Page 206 and 207: I := arctan y dx − xy dy γ γ
Page 208 and 209: f(ρ) f ′ (ρ) = 4ρ3 + 36ρ 8
Page 210 and 211: Jac ϕ(θ, y) = ⎛ ⎜ ⎝ − y
Page 212 and 213: z = (x, y) ˙z = Az + B(t)
Page 214 and 215: Q0 Q2 Q1 Q1
Page 216 and 217: ˙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
Page 223 and 224: α > 0 lim (x,y)→(0,0) (x,y)=(0,0
Page 225 and 226: 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,
Page 233 and 234: y ′ = 2xy − x y(0) = 3/2
Page 235 and 236: Γ f(x, y) = 0 f : R 2 → R
Page 237 and 238: ∇f(x0, y0) = (0, 0) (x0, y0)
Page 239 and 240: ϕ : I ×J → R 3 I, J R
Page 241 and 242: ψ(a, t) = ψ(b, t) t ∈]c, d[
Page 243 and 244: [
Page 245 and 246: Y K Ω R × Ω I R f :
Page 247 and 248: 1 ω(x, y) = M(x, y) dx + N(x, y)
Page 249 and 250: M(x, y) dx + N(x, y) dy = 0 1 M, N
Page 251: x, y z dy dz
Page 254 and 255: f(x) = N(x) Ak1 = + D(x) x − xk
Page 258 and 259: det P = 4 √ 7 = 0 P −1 AP
Page 260 and 261: 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
Page 266 and 267: η2 g g(η) > 0 y0 < η <
Page 269 and 270: 2 × 2 F ∈ R[x] n
Page 271: 2 × 2 t ↦→ x(
Page 274 and 275: ∆ < 0 α = −d 2c ω = √ |
Page 276 and 277: a = 0 b = 0 c = 0 ⎧ ⎪⎨ b
Page 279 and 280: SO(3) U(3) := {O ∈
Page 281: SO(3) Aξ = ω × ξ ξ
Page 284 and 285: x 2 + y 2 = 1 x 2 − y 2 = 1
Page 286 and 287: 1 arcsin(x) + x − x2 dx = √ 1
Page 288 and 289: z 1 + cos z cos = ± 2 2 z 1
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