Esercizi Analisi 2

Esame 26/02/2021

2) \( f: \mathbb{R} \rightarrow \mathbb{R} \) con \( T = 4 \) definita da

\( \rho(x) = \begin{cases} -1 & x \in [-2, 0) \\-6 & x \in [0, 2) \end{cases} \)

a) \( \omega = \frac{2\pi}{T} = \frac{\pi}{2}, \, T = 4 \)

b) \( a_0 = \frac{1}{2} \left( \int_{-2}^{0} dx + \int_{0}^{2} (-6) dx \right) \)

\( = \frac{1}{2} \left( [-x]_{-2}^{0} + [-6x]_{0}^{2} \right) = \frac{1}{2} \left( 2 + 12 \right) = -7 \)

c) \( a_k = \frac{2}{T} \left( \int_{-2}^{0} \cos (k \omega x) dx + \int_{0}^{2} -6 \cos (k \omega x) dx \right) \)

\( = \frac{1}{2} \left( \left[ \frac{\sin(k \omega x)}{k \omega} \right]_{-2}^{0} - 6 \left[ \frac{\sin(k \omega x)}{k \omega} \right]_{0}^{2} \right) \)

\( = \frac{1}{2} \left( -\frac{\sin(\pi k)}{k \pi} + 6 \frac{\sin(\pi k)}{k \pi} \right) = 0 \)

d) \( b_k = \frac{2}{T} \left( -\int_{-2}^{0} \sin (k \omega x) + -6 \int_{0}^{2} \sin (k \omega x) \right) \)

\( = \frac{1}{2} \left( \left[ -\frac{\cos(k \omega x)}{k \omega} \right]_{-2}^{0} + 6 \left[ -\frac{\cos(k \omega x)}{k \omega} \right]_{0}^{2} \right) \)

\( = \frac{1}{2} \left( \frac{1}{k \omega} \left( \cos(0) - \cos(-\pi k) \right) + \frac{6}{k \omega} \left( \cos(\pi k) - \cos(0) \right) \right) \)

\( = \frac{1}{\pi k} \left( (1 - \cos(\pi k)) + \frac{6}{\pi k} (\cos(\pi k) - 1) \right) \)

\( = \frac{1}{\pi k} (1 - \cos(\pi k))\), \( \frac{6}{\pi k} (\cos(\pi k) - 1) \)

\(= \frac{1}{\pi k} (-\cos(\pi k) - 6 \frac{\cos(\pi k)}{\pi k}) = \ldots \)

\(= \frac{5}{\pi k} \, \cos(\pi k) \, \ldots \, \frac{10}{\pi k} \ldots \)

e) \( \mathcal{F}[\rho] (x) = -\frac{\xi}{2} + \sum_{k>1} \frac{5}{\pi k} (1 - \cos(\pi/4)) b_k(\cos(k \omega x) \)

\( = 1 - \frac{\xi}{2} + \sum_{k>1} \frac{10}{\pi k} \sin\left(\frac{\pi}{4} (k-1)x \right) \)

\( \ldots \) perodico, \, i \, k \, dispari

f) \( \mathcal{F}[\rho](x) \) converge puntualmente ad \( \rho(x) \) per \( x \neq 0, 2, -2, \, x \in [-2, 2] \)

\( \rho(-x) = \rho(x) \) pari

x ∈ J, quindi J = [-1; 1]

∃ f(ƒ⁻¹(x)) quanti questo punto medio =

= [f(0) + f(0)] / 2 - 2 / 2 ∀ x = 0, 2, -2

-10x - 3y - x _{(-27 + √655)}/28 = 0

-12y - 3x - _{(-17 + √655)}/2 = 0

y = -10x - 47x - √655x/7 / 3

-4 _{(23/7x - √655}) - 3x - _{(-17 + √655)}/2 - _{(-23x - √655)}/21

+ 92/7x + 4

Compito di 22/02

f_n(x) = n sin_{(6x - 4/n)} n ∈ N

Convergenza Puntuale

lim n sin_{(x + 4/n)} - 6x - 4/n = lim n 6x - 15/3 6x - 4

f: ℝ → ℝ

x → 6x - 4

Conv. Unif.

lim sup_x |f_n - f|: lim sup n sin_{(x/n) - 6x - 4} |

⊛ sup n sin_(x/n)≤ 6x - 6x

⊛ sup n - 6 = sup di zero = 1

gⁿ = 6 costante < 0 g(x) decrescente sup x ∈ ℝ

⊛

⊛ sup f - f_n 6x - 4 - 1 sin_{(6x - 4/n)} sup 6x - 4 - n

sempre crescente

sup x ∞

Esercizio 22/2

1. f_n(x) = ^{3x2 - 3n}/_{n(x² + 1)}, n ∈ ℕ

   Convergenza puntuale

   lim_n f_n(x) = ^{-3x2 - 3n}/_{n(x² + 1)} = ^-3/_{x² + 1}

   f: ℝ → ℝ

   ^x/_{x² + 1} - ³/_{x² + 1}

   Conv. Uniforme

   lim_sup = ^{-3x2 - 3n}/_{n(x² + 1)} + ³/_{n(x² + 1)}

   f_n - f → 0

   0 è punto di sup

   lim_n f_n(0) = lim_n ^-3x/₁

   sup_x {^-3/_{n(x² + 1)} + ^{3x2 + 3n}/_{n(x² + 1)}} = ^3x2/_{n(x² + 1)}

   gⁿ = 6x {x + x² + f_n} _{/((fn(x² + 1))²) + 3xlogx^{-42nx5 - 42nx + 2x5}((fn(x² + 1))²) > 0 se x < 0}

   lim lim ^3x2/_{n(x² + fn)} = ³/_{n → 0}

   => C'e conv uniforme su tutto ℝ

2. f: ℝ → ℝ T = 1

   f(x) = x², x ∈ [^-1/₂, 1/2]

   w = 2π 1/3