feat(analisi): aggiunge gli appunti della lezione del 17/03/2023

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parent 0f21dbc536
commit 36f398add3

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\documentclass[11pt]{article}
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\usepackage[italian]{babel}
\title{\textbf{Note del corso di Analisi Matematica 1}}
\author{Gabriel Antonio Videtta}
\date{17 marzo 2023}
\begin{document}
\maketitle
\begin{center}
\Large \textbf{Successioni per ricorsione}
\end{center}
\begin{remark}
Sia $X$ l'insieme delle successioni a valori reali che soddisfano una data
eq.~ricorsiva lineare ed omogenea di ordine $k$ (ossia che coinvolge
$k$ precedenti elementi di una successione). \\
\li $X$ è uno spazio vettoriale su $\RR$. \\
\li $T : X \to \RR^n$, $(x_n) \mapsto (x_0, ..., x_{k-1})^\top$ è
un isomorfismo, e quindi $\dim X = k$. \\
\li Si può facilmente individuare una base naturale di $X$, costituita dagli
elementi della forma $\vec{x_i} = T\inv(\vec{e_ {i + 1}})$ con $i = 0, ..., k - 1$,
dove $\vec{x_i}$ rappresenta una successione di $X$ dove l'$i$-esimo elemento
è pari a $1$ e gli altri, tra $0$ e $k-1$, sono nulli.
\end{remark}
\begin{remark}
Le eq.~differenziali ordinarie si possono approssimare
ad eq.~su differenze finite (e questa considerazione è
alla base della grande somiglianza tra i concetti sviluppati
sia per queste che per quelle).
\end{remark}
\begin{example} (ricondursi a un caso discreto)
Si consideri un'eq.~differenziale omogenea lineare del primo
ordine su $x(t)$. Si può approssimare $t$ con $nh$, dato
$h$ piccolo, e così scrivere $x_n = x(nh) \approx x(t)$.
Così, allora, $x_{n+1} = x((n+1)h) = x(t + h)$. Conseguentemente $h
x'(t) \approx x(t + h) - x(t) \approx x_{n+1} - x_n$. \\
Si provi a risolvere, per esempio, l'eq.~ differenziale $x'(t) = x(t)$.
Sostituendo, si ottiene $x_{n+1} - x_n = h x_n$, da cui
si ricava l'eq.~ricorsiva $x_{n+1} = (1 + h) x_n$. Allora
$x(nh) = x_n = (1 + h)^n \underbrace{x(0)}_c = (1 + h)^n c$. \\
In effetti $x(t) = \displaystyle \lim_{h \to 0} (1 + h)^n c =
\lim_{h \to 0} \left[(1 + h)^{\frac{1}{h}}\right]^t c = c e^t$,
la famiglia di soluzioni dell'eq.~differenziale originale.
\end{example}
\begin{example} (metodo delle bisettrici)
Sia data la sequente successione:
\[ (x_n) = \begin{cases} x_n = x_{n-1}^4, \\ x_0 = \frac12. \end{cases} \]
Si consideri allora il sistema di funzioni:
\[ \begin{cases}
f(x) = x^4, \\ y = x,
\end{cases} \]
ossia i punti fissi di $f(x)$. Si può disegnare facilmente
la successione mediante il seguente algoritmo: si prenda
$x_0$ sull'asse delle ascisse, e si valuti $f(x_0) = x_1$ collegando
il punto $(x_0, 0)$ a $(x_0, x_1)$,
alla fine ricollegato sulla bisettrice al punto $(x_1, x_1)$;
si colleghi $(x_1, x_1)$ a $(x_1, x_2 = f(x_1))$ e quest'ultimo a
$(x_2, x_2)$, etc. Si sarà allora
disegnato in modo grafico la successione, e considerando
i blocchi che connettono $(x_{n-1}, x_{n-1})$, $(x_{n-1}, x_n)$
e $(x_n, x_n)$, si potrà facilmente intuire che $x_n \tendston \infty$ per $x_0 > 1$, che $x_n \tendston 1$ per $x_0 = 1$,
e che $x_n \tendston 0$ per $x_0 < 1$. Quindi nel caso
dell'esempio, $x_n \tendston 0$.
\begin{figure}[H]
\centering
\includegraphics[width=0.6\textwidth]{esempio.eps}
\caption{Applicazione dell'algoritmo con $x_0 = 1,0001$.}
\label{fig:my_label}
\end{figure}
\end{example}
\begin{example}
Riprendendo l'esempio precedente, si può ora provare
a dimostrare formalmente i risultati ottenuti.
Sempre graficamente, si intuisce che $(x_n)$ sarà
decrescente, e quindi che ammetterà limite (che,
in particolare, coinciderà con il suo estremo inferiore). \\
Si dimostra quindi, per prima cosa, che $(x_n)$ è
decrescente, e che vale $0 \leq x_n \leq \frac{1}{2}$.
Si procede per induzione: se $n=0$, la tesi è già
verificata; se la tesi è vera fino a $n-1$, allora
$x_n = \underbrace{x_{n-1}^4}_{\geq 0} \leq \left(\frac{1}{2}\right)^4 = \frac{1}{16} \leq \frac{1}{2}$. Quindi
$(x_n)$ è decrescente, e poiché $0$ ne è minorante,
varrà in particolare che $\ell = \lim_{n \to \infty} x_n \in [0, \frac{1}{2}]$. \\
Si mostra che $\ell$ deve essere un punto fisso di
$f$: poiché $x_n \tendston \ell$, anche $x_{n+1} \tendston
\ell$ (essendone una sottosuccessione); inoltre, poiché
$x_{n+1} = x_n^4$, $x_{n+1} \tendston \ell^4$. Poiché il limite è unico, deve allora valere $\ell = \ell^4 = f(\ell)$. Poiché gli unici punti di fissi di $f$ sono
$0$ e $1$, e $1$ non è
minorante di $(x_n)$,
deve valere che $\ell = 0$. \\
Se invece $x_0$ fosse stato
maggiore di $1$, si sarebbe
dimostrato che $(x_n)$ era
strettamente crescente, e
dunque avrebbe ammesso comunque
limite; tale limite non sarebbe
potuto essere né $0$$1$,
dacché non sarebbero stati maggioranti
di $(x_n)$, né tantomeno
$-\infty$. Allora tale limite
avrebbe dovuto essere,
forzatamente, $\infty$.
\end{example}
\begin{example}
Si consideri adesso la
successione:
\[ \begin{cases}
x_0 = 2, \\
x_{n+1} = \frac{x_n}{2} + \frac{1}{x_n}.
\end{cases} \]
Applicando lo stesso ragionamento
di prima, si considera $f(x) = \frac{x}{2} + \frac{2}{x}$. È sufficiente
dimostrare che $(x_n)$ è tale che
$\sqrt{2} \leq x_n \leq 2$
$\forall n \in \NN$ (dove $\sqrt{2}$
è l'unico punto fisso di $f(x)$) per
concludere immediatamente che
il limite di tale successione è
proprio $\sqrt{2}$.
\end{example}
\begin{example}
Si consideri l'eq.~ricorsiva $x_n = \frac{1}{x_{n-1}^2}$,
con $x_0 > 1$.
Qualsiasi disegno si faccia, si osserverà una "spirale"
nella configurazione della successione: si ipotizzerà
dunque che $x_n$ non ammetterà limite. Si distinguono
dal disegno due sottosuccessioni: $x_{2n}$ e $x_{2n+1}$,
che, rispettivamente, obbediranno a due eq.~ricorsive,
$x_{2(n+1)} = x_{2n}^4$ e $x_{2(n+1) + 1} = x_{2n + 1}^4$,
ossia la successione analizzata in uno scorso esempio. \\
\begin{figure}[H]
\centering
\includegraphics[width=0.6\textwidth]{esempio2.eps}
\caption{Applicazione del metodo della bisettrice con $x_0 = 1,0001$.}
\label{fig:my_label}
\end{figure}
Poiché $x_0 > 1$, $x_1 = \frac{1}{x_0^2} < 1$. Allora
$x_{2n} \tendston \infty$, mentre $x_{2n+1} \tendston
0$: poiché una sottosuccessione deve tendere allo
stesso limite della successione da cui deriva, ed il
limite è unico, si conclude che $(x_n)$ non ammette
limite.
\end{example}
\end{document}

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\newcommand{\FFpn}{\mathbb{F}_{p^n}}
\newcommand{\FFp}[1]{\mathbb{F}_{p^{#1}}}
\newcommand{\KKx}{\mathbb{K}[x]}
\newcommand{\QQx}{\mathbb{Q}[x]}
\newcommand{\RRx}{\mathbb{R}[x]}
\newcommand{\ZZi}{\mathbb{Z}[i]}
\newcommand{\ZZom}{\mathbb{Z}[\omega]}
\newcommand{\ZZpx}{\mathbb{Z}_p[x]}
\newcommand{\ZZsqrt}[1]{\mathbb{Z}[\sqrt{#1}]}
\newcommand{\ZZx}{\mathbb{Z}[x]}
\newcommand{\ii}{\mathbf{i}}
\newcommand{\jj}{\mathbf{j}}
\newcommand{\kk}{\mathbf{k}}
\newcommand{\valalpha}{\varphi_\alpha}
\newcommand{\Frob}{\mathcal{F}}
\newcommand{\Frobexp}{\mathcal{F}{\mkern 1.5mu}}
\newcommand{\dual}[1]{#1^{*}}
\newcommand{\LL}[2]{\mathcal{L} \left(#1, \, #2\right)}
\newcommand{\M}[1]{\mathcal{M}_{#1}\left(\KK\right)}
\newcommand{\nsg}{\mathrel{\unlhd}}
\renewcommand{\vec}[1]{\underline{#1}}
\newcommand{\hatpi}{\hat{\pi}}
\newcommand{\hatpip}{\hat{\pi}_p}
% evan.sty original commands
\newcommand{\cbrt}[1]{\sqrt[3]{#1}}
\newcommand{\floor}[1]{\left\lfloor #1 \right\rfloor}
\newcommand{\ceiling}[1]{\left\lceil #1 \right\rceil}
\newcommand{\mailto}[1]{\href{mailto:#1}{\texttt{#1}}}
\newcommand{\eps}{\varepsilon}
\newcommand{\vocab}[1]{\textbf{\color{blue}\sffamily #1}}
\providecommand{\alert}{\vocab}
\newcommand{\catname}{\mathsf}
\providecommand{\arc}[1]{\wideparen{#1}}
% From H113 "Introduction to Abstract Algebra" at UC Berkeley
\newcommand{\CC}{\mathbb C}
\newcommand{\FF}{\mathbb F}
\newcommand{\NN}{\mathbb N}
\newcommand{\QQ}{\mathbb Q}
\newcommand{\RR}{\mathbb R}
\newcommand{\ZZ}{\mathbb Z}
\DeclareMathOperator{\Aut}{Aut}
\DeclareMathOperator{\Inn}{Inn}
\DeclareMathOperator{\Syl}{Syl}
\DeclareMathOperator{\Gal}{Gal}
\DeclareMathOperator{\GL}{GL}
\DeclareMathOperator{\SL}{SL}
%From Kiran Kedlaya's "Geometry Unbound"
\newcommand{\abs}[1]{\left\lvert #1 \right\rvert}
\newcommand{\norm}[1]{\left\lVert #1 \right\rVert}
\newcommand{\dang}{\measuredangle} %% Directed angle
\newcommand{\ray}[1]{\overrightarrow{#1}}
\newcommand{\seg}[1]{\overline{#1}}
% From M275 "Topology" at SJSU
\newcommand{\id}{\mathrm{id}}
\newcommand{\taking}[1]{\xrightarrow{#1}}
\newcommand{\inv}{^{-1}}
\DeclareMathOperator{\ord}{ord}
\newcommand{\defeq}{\overset{\mathrm{def}}{=}}
\newcommand{\defiff}{\overset{\mathrm{def}}{\iff}}
% From the USAMO .tex files
\newcommand{\dg}{^\circ}
\newcommand{\liff}{\leftrightarrow}
\newcommand{\lthen}{\rightarrow}
\newcommand{\opname}{\operatorname}
\newcommand{\surjto}{\twoheadrightarrow}
\newcommand{\injto}{\hookrightarrow}
\DeclareMathOperator{\Char}{char}
\DeclareMathOperator{\Dom}{Dom}
\DeclareMathOperator{\Fix}{\textit{Fix}\,}
\DeclareMathOperator{\End}{End}
\DeclareMathOperator{\existsone}{\exists !}
\DeclareMathOperator{\Hom}{Hom}
\DeclareMathOperator{\Imm}{Imm}
\DeclareMathOperator{\Ker}{Ker}
\DeclareMathOperator{\rank}{rank}
\DeclareMathOperator{\MCD}{MCD}
\DeclareMathOperator{\Mor}{Mor}
\DeclareMathOperator{\mcm}{mcm}
\DeclareMathOperator{\Sym}{Sym}
\DeclareMathOperator{\tr}{tr}
\let\oldemptyset\emptyset
\let\emptyset\varnothing
\let\oldcirc\circ
\let\circ\undefined
\DeclareMathOperator{\circ}{\oldcirc}
\let\oldexists\exists
\let\exists\undefined
\DeclareMathOperator{\exists}{\oldexists}
\let\oldforall\forall
\let\forall\undefined
\DeclareMathOperator{\forall}{\oldforall}
\let\oldnexists\nexists
\let\nexists\undefined
\DeclareMathOperator{\nexists}{\oldnexists}
\let\oldland\land
\let\land\undefined
\DeclareMathOperator{\land}{\oldland}
\let\oldlnot\lnot
\let\lnot\undefined
\DeclareMathOperator{\lnot}{\oldlnot}
\let\oldlor\lor
\let\lor\undefined
\DeclareMathOperator{\lor}{\oldlor}
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