Tutorato Analisi II esercizi

Criterio del rapporto

lim_k→∞ |a_k+1 / a_k| = L R = 1/L

Criterio della radice

lim_k→∞ k√|a_k| = L R = 1/L

Criterio di Leibniz

1. lim_k→∞ q_k = 0

2. decrescente → converge

Esercizi

1. ∑_k=3 (5k / 2^k) (x - 3)^k R = 2 I = (-2, 2) x₀ = 3
2. ∑ k! (x + 2)^k R = 0 I = [-2, ?]
3. ∑ (1 / √(1 + k²)) x^k x = 0 R = 1 I = [1, 1]

Σ ⎦_k=1^∞ ∅ (^3k + 1)^k

R=∞

Σ ⎦_k=0^∞ ∅ (^1k2x)^k

x₀=1 R=2 I=[2,6]

Σ ⎦_k=0^∞ ∅ (^1k2x)^k

(^(-1)kk2) Converge

Σ ⎦_k=0^∞ (∅²)^k = Σ_{1k² Converge}

Σ ⎦_k=1^∞ (^1k2)(-^{2k) = Σ_(-1)kk2 Converge}

Σ ⎦_k=1^{∞ 1k2} Converge

Σ ⎦_k=1^∞ (^1k)(x+5)^k

x₀=5

I=[6,-4]

Σ ⎦_k=3^∞ (x^3k2)^k Converge

Σ ⎦_k=1^{∞ 1k} (∅^1k2)^k Converge

Σ ⎦_k=1^∞ (^1√(k+2k))x^k

x₀=0 R=2 I=(-2,2)

lim_k→∞

⎦_k=∞ ^(√k+2k)1 Approx. ½

R(x,y) = √x² + y²

4 - x² ≥ 0

4 - y²

R(x,y) = log (1 - |x|) - log (2 - |y|)

{

1 - |x| > 0

2 - |y| > 0

{

|x| < 1

|y| < 2

3/9/14

5. \( f(x, y) = (1 - 2x) e^{x-y^2} \)

\(\frac{\partial f}{\partial x} = -2 e^{x-y^2} + (1 - 2x) e^{x-y^2} \)

\(\frac{\partial f}{\partial y} = ( \frac{-2y}{e^{x-y^2}}) \)

\(\frac{\partial f}{\partial x} = e^{x-y^2} (-2x-1) \)

(1-2x) - 2xe^{x-y^2} = (1-2x - 2x)e^{x-y^2} \)

\(\Delta = 0 \Rightarrow \left\{ \begin{array}{c} -2x-1=0 \\ (1-2x) - 2y = 0 \end{array} \right. \quad \Rightarrow \left\{ \begin{array}{c} -1 - x = 1 \\ (1+1-2y) = 0 \end{array} \right.\)

\(\left( -\frac{1}{2}, 1 \right) \quad massimo \)

\(H \left( \frac{1}{2}, 0 \right) = \begin{pmatrix} -2 e^{-\frac{1}{2}} & 0 \\ 0 & -4e^{-\frac{1}{2}} \end{pmatrix} \quad = \quad \begin{pmatrix} -2e^{-\frac{1}{2}} & 0 \\ 0 & -4 e^{\frac{-1}{2}} \end{pmatrix}\)

\(\frac{\partial^2 f}{\partial x^2} = -2 e^{x-y^2} - 2 e^{x-y^2} + (1 - 2x) e^{x-y^2} \quad \Rightarrow -ae^{-\frac{1}{2}} + 2 = 2 \rightarrow (1-e^{-\frac{1}{2}}) \)

\(\frac{\partial^2 f}{\partial y^2} = (1-2x) \cdot \left[ 2 \cdot e^{x-y^2} - 2y(-2y) e^{x-y^2} \right] \quad \Rightarrow 2 \cdot (1-2e^{-\frac{1}{2}}) \)

5-

f(x) = x² + x · e^x

u = x₂

z = x

w = e^x

g(u, z, w) = u + sen(z) · w

h = g ° f = x² + e^x sen(x)

(x³ + e^x sen x)

h(x) ∈ R - R

h(x) = 3x² + e^x sen x + e^x cos x

∂f₁ / ∂x = ?x

∂f₂ / ∂x = 1

∂f₃ / ∂x = e^x

Dg(u, z, w) = (∂g / ∂x, ∂g / ∂y, ∂g / ∂w)

( 1 ) ( z ) sen(z) 1

z

( 1 ) ( z 4w cos(z) sen(z) )

e^x

= 2x + 4u cos z + e^xsen z = 2x · (x) + x²e^x cos x ,

+ e^xsen x

= 3x² + e^xcos x + e^xsen x

RISOLUZIONE

∬∬ (3 ρ cosθ - ¹/₉ ρ³ sen³θ) a b ρ dθ dρ =

= ∫₀^2π ∫₀¹ 3ρ cosθ - (¹/₃ ρ senθ)^{2 1}/₃ ρ dρ dθ = const.

= ∫₀^2π ∫₀¹ 3ρ cosθ ∫₀^1/27 ρ³ sen²θ dθ dρ = *

[cosθ perde qualche parte su simmetrico]

* = - ∫₀^2π sen2θdθ ∫₀^1/27 ρ³dρ = -^π/₄ ⋅ ¹/₂₇

= - ∫₀^2π senθdθ ∫₀^{1 1}/₂₇ ρ³dρ = -^π/₂₇ x

3

D = { (x, y) ∈ℝ²; 0 ≤ x ≤ 1, x ≤ y ≤ 2^x }

φ(x,y) = x y

∬_ω x y dx dy = ∫₀¹ dx x ∫_x^2x y dy =

= ∫₀¹ x dx [ ^y2/₂ ]_x^2x = ∫₀¹ x dx [ ^2xx/₂ - ^x2/₂ ] =

= ∫₀¹ 2^{x x} dx - ∫₀^{1 x3}/₂ dx = 2 [ ³/_ln2 [3/2] ]¹/₀ - ¹/₂ [ ^x4/₄ ]¹/₀ =

= 2 ⋅ ¹/₃ - ¹/₂ = ¹/₉ = 2 - ³/₈ -^{8 ⋅ 1 ⋅ 3}/₂₄ = ^{16 - 3 ⋅ 3}/₂₄

∋ Θ

z = 0

(x, y, z) ∈ ℝ³

x² + y² + z² ≤ 3, x² + y² ≤ 1, z ≥ 0

(0, 0, 4/6)

x_b = 1^∯ _|J|∬x dx dy dz = 0 = y_b

∬ _{√3 - x² - y²} ∫ 1 dz dx dy = ∬ √3 - (x² + y²) dx dy

x = ρ cos Θ

0 ≤ ρ ≤ √1

dx dy = ρ dρ dΘ

y = ρ sin Θ

0 ≤ Θ ≤ 2π

∬ ₀ ^2π √3 - (ρ²) ρ dρ dΘ = ⌠₀ ⌠^√3 ρ³ ρ dρ ⌡ dΘ =

-1/2 ⌠₀ (3 - ρ²)^1/2 dρ ⌠^2π dΘ = -1/2 [2(√3 - ρ²)^3/2/3] ^2π = 3,157

2/3 (27 - 16√2)

z_b = ∬ _A ⌠_z dx dy dz/∬¹_|J|

(√3 - x² - y²) = 1/∬₄ (√z) dz dx dy

1/8_,157⟦1/2 ⌠_A ⌠^1/2 √3 - x² - y² dx dy

⌠₀ ⌠^2π (3 - ρ²) ρ dρ dΘ

-1/8_,157⌠₀ ⌠^2π -ρ(9 - ρ²) ρ dρ dΘ = π/2 ⟪1/⌡₈

∫∫ 6 z + ∫∫ –2 x = 27 -

x = a ρ cosθ

y = b ρ x cosθ

z = c ρ cosθ

&frac;{3}{2} ρ sinθ

e^kz dè = e b ×

-∫ π⊃2; ρ⊃2; sinθ χ y φ

∫∫ &frac;{3}{2} 1 (-ω)

s1 = &frag;{1}{3} z = 27

-∫∫ -2 φ

∫∫ φ⊃2; sin y φ

+ 8 ∫∫ (

-8/2 ∫∫

π ∫∫

-8/2 &frac{1}{4}

+ 8/2 (

φ⊃2; 1

-8 1 (