Analisi 1 Esercizi

1) NUMERI COMPLESSI

1. i² = -1
2. z = a + ib
3. |z| = √(a² + b²)
4. z̅ = a - ib
5. Re(z) = a
6. Im(z) = b
7. z₁ = z₂ ⇔ a₁ = a₂, b₁ = b₂
8. z · z̅ = |z|²
9. |z₁| = |z̅₁|
10. z · z̅ ≠ 0 ⇔ z ≠ 0
11. z = z̅ ⇔ Re(z) ≠ 0
12. z̅ = z
• z · z₂ = |z|² ⇔ Re(z) ≠ 0

Problema 1)

z² = 4 + 4i

|z|² = 8

a² - b² = 4

2ab = 4

b = ±2

Verifica:

-2b + 8 + i(2ab - 4) = 0

Problema 2)

z = a + ib

|z| = √(a² + b²)

Problema 3)

z + 3 = |z| = i

z = a + ib

Problema 4)

Nota: L'immaginario coniugato varia con l'alfabeto

5)

z cos(θ) + ρ sin(θ) [ 1 ] = 0

z = ρ [ cos(θ) + i sin(θ) ]

ρ [ cos(θ) + i sin(θ) ] + ρ [ cos (ϕ) + i sin (ϕ) ] = 0

ρ [ cos(θ) + i sin(θ) + cos(ϕ) + i sin(ϕ) ] = 0

ρ [ cos(θ) + cos(ϕ) ] = -ρ [ sin(θ) + sin(ϕ) ] = 0

cos(θ) + cos(ϕ) = 0

sin(θ) + sin(ϕ) = 0

cos( θ - ϕ ) = 0

sin( θ ) - sin (ϕ) = 0

z₀ = 1

Verification: 1⋅1 = 1 = 1 √

ℓ₃ = 1 + (cos(0) + i sin(0))

Verification : 1⋅1 + (1) = 0=0 √

Verification: -1+(-1) = 0=0 √

6)

z₂z₃ + z₂ = 1

3)

|z| ²[ a+1] + 4 = 0

z = ρ [ cos (θ) + i sin (θ) ]

ρ² [cos(3θ) + i sin (3θ)] cos (30)

3θ = a KΠp

z₃ = cos(θ) + i sin(θ)

5)

lim_x→0 ^{ex + log_e(11})⁄_(x)

lim_x→0 ^{ex + log_e(1+x) - 1}⁄_{2(cosh x - 1) - sech x}

D:

• cosh x = 1 + x²⁄₂ + o(x²)
• sech x = 1 - x²⁄₂ + o(x²)

N:

• e^x = 1 + x + x²⁄₂ + o(x²)
• log(1 + x) = x - x²⁄₂ + x³⁄₃ + o(x³)

In sintesi:

lim_x→0 ^{(x - x3⁄₂ + x2⁄₂ - x3⁄₂ + x3⁄₃)}⁄_{(x - x²⁄2)(x²⁄2)+ o(x³)} = ^x3⁄₃⁄_x³⁄6 = -¹⁄₆

b)

lim_x→0 ^{ln(1 + arcsin x) - sen(ln(1 + x))}⁄_x

D:

• arctg x = x - o(x³)

N:

• ln(1 + arcsin x) = arcsin x + arcsin^x⁄₂ - o(x²)

con arcsin x = x + o(x)

sen(ln(1 + x)) = ln(1 + x) + o(x)

con ln(1 + x) = x - x²⁄₂ + o(x⁴)

Rottitiamo:

lim_x→0 ^{x - 1⁄₂₂ - [- x2⁄₂]}⁄^{x3 + [-x⁄₂(x - x2⁄₂)x2⁄₂] + o(x4) = 0}

2)

lim_x→0 ^arct√x

N:

arct√x = x - √x

In sintesi:

lim_x→0 (1 + 1⁄₃(x) ¹⁄₁₈ = - ͠=^1.3)

1.2) g(x) =

x

x - 1

x² - 5x + 3

D = (-∞,

lim x→0

lim x→+∞

lim x→1_-

lim x→1₊

x → x ²

f(x) =

x = x - 1

A. ∅ da sopra

Studio del segno g(x) > 0

x² - 5x + 3 > 0

x² - 5x + 3

= -5 ± √13 / 2

x² - 10x + 6 + 2x + 3 - 2x² - 5x

2x³ + x² - 15x - 6

a_n = (arc (m² + 3)/(m² + 3) ) exp ( arc(1) ) ) = 0, o(1/m²)

lim_n→∞ exp [ (m³ - 3)/(m - 1) ] ln [ 1/m + 1/m² + 1/m² + o(1/m²) ]

lim_n→∞ exp [ 2/m + 1/m + 1/m² + o(1/m²) ]

lim_n→∞ exp [ 7/m³ ]

lim_n→∞ exp [ (m³ - 7)/(m - 1) ] exp [ e² - 3n - 7 - lim_n→∞ exp[m³ - 3/m - 1)] ]

lim_n→∞ exp [ (1 / m - 1/m³)

anm = [m + 1 / m³] [exp [m / 2 (m+3)] arc [m² - m] ]

anm = [m + m³] [1 1/m o(1/m)] + (o(1/m)

= 1/m + o(1/m) } [+1/m + o(1/m)] (+1/m ]

o(1/m) u(1/4 ov

m pair [* (1,1) ] + o(1)] +

m² = [ 0.1 .] + o(1)] ]

3)

lim     arccos    M + 1 + m² m_{→ +∞}                   3m + 1 + 2

arccos           arccos                                       arccos m²            m(1+ ¹) 3         (3 - ^λ) arccos = ½ + ½ + ¹ √₁ = n  (^+∞-1)     2    2/m²                  m  (m^-1)

4)

f(x) = arctg¹(x+1)  - x 2 E     x ≠ 1    [0, -1] = (0, 1]  2

x →+∞     arctg(^{x - 1}) = x = arctg(^x(i!)1)&sup>∞  arctg(^1-xt(i!)) - x = (sup>arctg₁) pe ½/^x = 1/2 = ½ = -x + 1 = (i  x)  ↔  0(

½ &to; (a)     ∞

f'(k)  =  ⟨⟨∠∠∠&<(e)(1 + ^x)_{x - 1}

f'(k) ≥ 0

x² ≤  x

f(x) = exp(...) ₂ / (x+1)² - exp(...) _2x+1 / 2√(...)

= exp (x⁄x-1) √x+2 √(2x² - 3x + 1) / 2x (x+1)^3/2

f'(x) = ± ∞

g(x) = exp (x⁄x-1) √x+2 √3x² - 6x -1) 4x (1+x)^3/2 (1+x)³

5) ∫^∞ ln(x) / (x-1)² dx = ∫^∞ ln (1+t) / t² dt = [-2t^-1/2 ln (1+t)]^∞ -∫^∞ 2t^-1/2 / 1+t dt

t = y²

= 0 + ∫₁^∞ 1 / 1+y²

b)∑_m=1^∞ ln (m^α a^β+1 / m^α + m^β+2) α, β > 0

a, β = max {a, β} ∑ diverge

∞ > β α > 1 β ≥ 1 conv.

-∞ < α β ≥ 1 α = 1 conv.

an = ln (m^α/nm^β+2) - ln 1n