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+171.576 347.9 L +171.576 55.204 L +464.271 55.204 L +S +GR +%%Trailer +%%Pages: 1 +%%EOF diff --git a/Analisi I/LaTeX/2023-03-17, Successioni per ricorsione/main.tex b/Analisi I/LaTeX/2023-03-17, Successioni per ricorsione/main.tex new file mode 100644 index 0000000..4847611 --- /dev/null +++ b/Analisi I/LaTeX/2023-03-17, Successioni per ricorsione/main.tex @@ -0,0 +1,171 @@ +\documentclass[11pt]{article} +\usepackage{personal_commands} +\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 $0$ né $1$, + 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} diff --git a/Analisi I/LaTeX/2023-03-17, Successioni per ricorsione/personal_commands.sty b/Analisi I/LaTeX/2023-03-17, Successioni per ricorsione/personal_commands.sty new file mode 100644 index 0000000..74be1e0 --- /dev/null +++ b/Analisi I/LaTeX/2023-03-17, Successioni per ricorsione/personal_commands.sty @@ -0,0 +1,181 @@ +\ProvidesPackage{personal_commands} + +\usepackage{amsmath,amssymb} +\usepackage{amsfonts} +\usepackage{amsthm} +\usepackage{amssymb} +\usepackage{amsopn} +\usepackage{mathtools} +\usepackage{marvosym} +\usepackage{floatflt} +\usepackage{graphicx} +\usepackage{float} + +\hfuzz=\maxdimen +\tolerance=10000 +\hbadness=10000 + +\newcommand{\cororef}[1]{\textit{Corollario \ref{#1}}} +\newcommand{\exref}[1]{\textit{Esercizio \ref{#1}}} +\newcommand{\propref}[1]{\textit{Proposizione \ref{#1}}} +\newcommand{\lemmaref}[1]{\textit{Lemma \ref{#1}}} +\newcommand{\thref}[1]{\textit{Teorema \ref{#1}}} +\newcommand{\li}[0]{$\blacktriangleright\;\;$} + +\newcommand{\tendsto}[1]{\xrightarrow[\text{$x \to #1$}]{}} +\newcommand{\tendston}[0]{\xrightarrow[\text{$n \to \infty$}]{}} + +\setlength\parindent{0pt} + +% Personal commands +\newcommand{\card}[1]{\left|#1\right|} +\newcommand{\nsqrt}[2]{\!\sqrt[#1]{#2}\,} +\newcommand{\zeroset}{\{0\}} +\newcommand{\setminuszero}{\setminus \{0\}} + +\newenvironment{solution} + {\begin{proof}[Soluzione]} + {\end{proof}} + +\theoremstyle{definition} + +\newtheorem{corollary}{Corollario} +\newtheorem*{definition}{Definizione} +\newtheorem*{example}{Esempio} +\newtheorem{exercise}{Esercizio} +\newtheorem{lemma}{Lemma} +\newtheorem*{remark}{Osservazione} +\newtheorem*{proposition}{Proposizione} +\newtheorem{theorem}{Teorema} + +\newcommand{\BB}{\mathcal{B}} +\newcommand{\HH}{\mathbb{H}} +\newcommand{\KK}{\mathbb{K}} +\newcommand{\ZZp}{\mathbb{Z}_p} + +\newcommand{\CCx}{\mathbb{C}[x]} +\newcommand{\FFpp}{\mathbb{F}_p} +\newcommand{\FFpd}{\mathbb{F}_{p^d}} +\newcommand{\FFpm}{\mathbb{F}_{p^m}} +\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} \ No newline at end of file diff --git a/Analisi I/PDF/Parte teorica/Successioni/2023-03-17, Successioni per ricorsione.pdf b/Analisi I/PDF/Parte teorica/Successioni/2023-03-17, Successioni per ricorsione.pdf new file mode 100644 index 0000000..078ee9e Binary files /dev/null and b/Analisi I/PDF/Parte teorica/Successioni/2023-03-17, Successioni per ricorsione.pdf differ