From 2d52f3013988535890981ad0dbd95a41810137e0 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?Anton=20Luka=20=C5=A0ijanec?= <anton@sijanec.eu>
Date: Tue, 6 Aug 2024 01:35:27 +0200
Subject: =?UTF-8?q?ana1=20u=C4=8Denje=20vrsar=20lahko=20no=C4=8D?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 "\305\241ola/ana1/teor.lyx" | 4300 +++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 4300 insertions(+)
 create mode 100644 "\305\241ola/ana1/teor.lyx"

(limited to 'šola/ana1/teor.lyx')

diff --git "a/\305\241ola/ana1/teor.lyx" "b/\305\241ola/ana1/teor.lyx"
new file mode 100644
index 0000000..6ff52cd
--- /dev/null
+++ "b/\305\241ola/ana1/teor.lyx"
@@ -0,0 +1,4300 @@
+#LyX 2.4 created this file. For more info see https://www.lyx.org/
+\lyxformat 620
+\begin_document
+\begin_header
+\save_transient_properties true
+\origin unavailable
+\textclass article
+\begin_preamble
+\usepackage{hyperref}
+\usepackage{siunitx}
+\usepackage{pgfplots}
+\usepackage{listings}
+\usepackage{multicol}
+\sisetup{output-decimal-marker = {,}, quotient-mode=fraction, output-exponent-marker=\ensuremath{\mathrm{3}}}
+\usepackage{amsmath}
+\usepackage{tikz}
+\newcommand{\udensdash}[1]{%
+    \tikz[baseline=(todotted.base)]{
+        \node[inner sep=1pt,outer sep=0pt] (todotted) {#1};
+        \draw[densely dashed] (todotted.south west) -- (todotted.south east);
+    }%
+}%
+\DeclareMathOperator{\Lin}{\mathcal Lin}
+\DeclareMathOperator{\rang}{rang}
+\DeclareMathOperator{\sled}{sled}
+\DeclareMathOperator{\Aut}{Aut}
+\DeclareMathOperator{\red}{red}
+\DeclareMathOperator{\karakteristika}{char}
+\DeclareMathOperator{\Ker}{Ker}
+\DeclareMathOperator{\Slika}{Ker}
+\DeclareMathOperator{\sgn}{sgn}
+\DeclareMathOperator{\End}{End}
+\DeclareMathOperator{\n}{n}
+\DeclareMathOperator{\Col}{Col}
+\usepackage{algorithm,algpseudocode}
+\providecommand{\corollaryname}{Posledica}
+\usepackage[slovenian=quotes]{csquotes}
+\end_preamble
+\use_default_options true
+\begin_modules
+enumitem
+theorems-ams
+theorems-ams-extended
+\end_modules
+\maintain_unincluded_children no
+\language slovene
+\language_package default
+\inputencoding auto-legacy
+\fontencoding auto
+\font_roman "default" "default"
+\font_sans "default" "default"
+\font_typewriter "default" "default"
+\font_math "auto" "auto"
+\font_default_family default
+\use_non_tex_fonts false
+\font_sc false
+\font_roman_osf false
+\font_sans_osf false
+\font_typewriter_osf false
+\font_sf_scale 100 100
+\font_tt_scale 100 100
+\use_microtype false
+\use_dash_ligatures true
+\graphics default
+\default_output_format default
+\output_sync 0
+\bibtex_command default
+\index_command default
+\float_placement class
+\float_alignment class
+\paperfontsize default
+\spacing single
+\use_hyperref false
+\papersize default
+\use_geometry true
+\use_package amsmath 1
+\use_package amssymb 1
+\use_package cancel 1
+\use_package esint 1
+\use_package mathdots 1
+\use_package mathtools 1
+\use_package mhchem 1
+\use_package stackrel 1
+\use_package stmaryrd 1
+\use_package undertilde 1
+\cite_engine basic
+\cite_engine_type default
+\biblio_style plain
+\use_bibtopic false
+\use_indices false
+\paperorientation portrait
+\suppress_date false
+\justification false
+\use_refstyle 1
+\use_formatted_ref 0
+\use_minted 0
+\use_lineno 0
+\index Index
+\shortcut idx
+\color #008000
+\end_index
+\leftmargin 2cm
+\topmargin 2cm
+\rightmargin 2cm
+\bottommargin 2cm
+\headheight 2cm
+\headsep 2cm
+\footskip 1cm
+\secnumdepth 3
+\tocdepth 3
+\paragraph_separation indent
+\paragraph_indentation default
+\is_math_indent 0
+\math_numbering_side default
+\quotes_style german
+\dynamic_quotes 0
+\papercolumns 1
+\papersides 1
+\paperpagestyle default
+\tablestyle default
+\tracking_changes false
+\output_changes false
+\change_bars false
+\postpone_fragile_content false
+\html_math_output 0
+\html_css_as_file 0
+\html_be_strict false
+\docbook_table_output 0
+\docbook_mathml_prefix 1
+\end_header
+
+\begin_body
+
+\begin_layout Title
+Teorija Analize 1 —
+ IŠRM 2023/24
+\end_layout
+
+\begin_layout Author
+
+\noun on
+Anton Luka Šijanec
+\end_layout
+
+\begin_layout Date
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+today
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Abstract
+Povzeto po zapiskih s predavanj profesorja Oliverja Dragičevića.
+\end_layout
+
+\begin_layout Standard
+\begin_inset CommandInset toc
+LatexCommand tableofcontents
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Section
+Števila
+\end_layout
+
+\begin_layout Definition*
+Množica je matematični objekt,
+ ki predstavlja skupino elementov.
+ Če element 
+\begin_inset Formula $a$
+\end_inset
+
+ pripada množici 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ pišemo 
+\begin_inset Formula $a\in A$
+\end_inset
+
+,
+ sicer pa 
+\begin_inset Formula $a\not\in A$
+\end_inset
+
+.
+ Množica 
+\begin_inset Formula $B$
+\end_inset
+
+ je podmnožica množice 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ pišemo 
+\begin_inset Formula $B\subset A$
+\end_inset
+
+,
+ če 
+\begin_inset Formula $\forall b\in B:b\in A$
+\end_inset
+
+.
+ Presek 
+\begin_inset Formula $B$
+\end_inset
+
+ in 
+\begin_inset Formula $C$
+\end_inset
+
+ označimo 
+\begin_inset Formula $B\cap C\coloneqq\left\{ x;x\in B\wedge x\in C\right\} $
+\end_inset
+
+.
+ Unijo 
+\begin_inset Formula $B$
+\end_inset
+
+ in 
+\begin_inset Formula $C$
+\end_inset
+
+ označimo 
+\begin_inset Formula $B\cup C\coloneqq\left\{ x;x\in B\vee x\in C\right\} $
+\end_inset
+
+.
+ Razliko/komplement 
+\begin_inset Quotes gld
+\end_inset
+
+
+\begin_inset Formula $B$
+\end_inset
+
+ manj/brez 
+\begin_inset Formula $C$
+\end_inset
+
+
+\begin_inset Quotes grd
+\end_inset
+
+ označimo 
+\begin_inset Formula $B\setminus C\coloneqq\left\{ x;x\in B\wedge x\not\in C\right\} $
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsection
+Realna števila
+\end_layout
+
+\begin_layout Standard
+Množico realnih števil označimo 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+.
+ V njej obstajata binarni operaciji seštevanje 
+\begin_inset Formula $a+b$
+\end_inset
+
+ in množenje 
+\begin_inset Formula $a\cdot b$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsubsection
+Lastnosti seštevanja
+\end_layout
+
+\begin_layout Axiom
+Komutativnost:
+ 
+\begin_inset Formula $\forall a,b\in\mathbb{R}:a+b=b+a$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Axiom
+Asociativnost:
+ 
+\begin_inset Formula $\forall a,b,c\in\mathbb{R}:a+\left(b+c\right)=\left(a+b\right)+c$
+\end_inset
+
+,
+ torej je 
+\begin_inset Formula $a+\cdots+z$
+\end_inset
+
+ dobro definiran izraz.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Axiom
+Obstoj enote:
+ 
+\begin_inset Formula $\exists0\in\mathbb{R}\forall a\in\mathbb{R}:a+0=a$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Axiom
+Obstoj inverzov:
+ 
+\begin_inset Formula $\forall a\in\mathbb{R}\exists b\in\mathbb{R}\ni a+b=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Claim*
+Inverz je enoličen.
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $a,b,c\in\mathbb{R}$
+\end_inset
+
+ in 
+\begin_inset Formula $a+b=0$
+\end_inset
+
+ in 
+\begin_inset Formula $a+c=0$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $b=b+0=b+a+c=0+c=c$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Corollary*
+Inverz je funkcija in aditivni inverz 
+\begin_inset Formula $a$
+\end_inset
+
+ označimo z 
+\begin_inset Formula $-a$
+\end_inset
+
+.
+ Pri zapisu 
+\begin_inset Formula $a+\left(-b\right)$
+\end_inset
+
+ običajno 
+\begin_inset Formula $+$
+\end_inset
+
+ izpustimo in pišemo 
+\begin_inset Formula $a-b$
+\end_inset
+
+,
+ čemur pravimo odštevanje 
+\begin_inset Formula $b$
+\end_inset
+
+ od 
+\begin_inset Formula $a$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+\begin_inset Formula $\forall a\in\mathbb{R}:a=-\left(-a\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $b=-a$
+\end_inset
+
+ in 
+\begin_inset Formula $c=-b=-\left(-a\right)$
+\end_inset
+
+.
+ Tedaj velja 
+\begin_inset Formula $c-a=c+b=-\left(-a\right)-a=0$
+\end_inset
+
+ in 
+\begin_inset Formula $a=0+a=c-a+a=c+\left(-a\right)+a=c+0=c=-\left(-a\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+\begin_inset Formula $-\left(b+c\right)=-b-c$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Proof
+Velja 
+\begin_inset Formula $b+c+\left(-b-c\right)=b+c+\left(\left(-b\right)+\left(-c\right)\right)=b+\left(-b\right)+c+\left(-c\right)=0$
+\end_inset
+
+,
+ torej je 
+\begin_inset Formula $b+c$
+\end_inset
+
+ inverz od 
+\begin_inset Formula $\left(-b-c\right)$
+\end_inset
+
+,
+ torej je 
+\begin_inset Formula $-\left(b+c\right)=-b-c$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Subsubsection
+Lastnosti množenja
+\end_layout
+
+\begin_layout Axiom
+Komutativnost:
+ 
+\begin_inset Formula $\forall a,b\in\mathbb{R}:ab=ba$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Axiom
+Asociativnost:
+ 
+\begin_inset Formula $\forall a,b,c\in\mathbb{R}:a\left(bc\right)=\left(ab\right)c$
+\end_inset
+
+,
+ torej je 
+\begin_inset Formula $a\cdots z$
+\end_inset
+
+ dobro definiran izraz.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Axiom
+Obstoj enote:
+ 
+\begin_inset Formula $\exists1\in\mathbb{R}\forall a\in\mathbb{R}:a1=a$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Axiom
+Obstoj inverzov:
+ 
+\begin_inset Formula $\forall a\in\mathbb{R}\setminus\left\{ 0\right\} \exists b\in\mathbb{R}\setminus\left\{ 0\right\} \ni:ab=1$
+\end_inset
+
+
+\end_layout
+
+\begin_deeper
+\begin_layout Claim*
+Inverz je enoličen.
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $a,b,c\in\mathbb{R}\setminus\left\{ 0\right\} $
+\end_inset
+
+ in 
+\begin_inset Formula $ab=1$
+\end_inset
+
+ in 
+\begin_inset Formula $ac=1$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $b=b1=bac=1c=c$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Corollary*
+Inverz je funkcija in multiplikativni inverz 
+\begin_inset Formula $a$
+\end_inset
+
+ označimo z 
+\begin_inset Formula $a^{-1}$
+\end_inset
+
+.
+ Pri zapisu 
+\begin_inset Formula $a\cdot b^{-1}$
+\end_inset
+
+ lahko 
+\begin_inset Formula $\cdot$
+\end_inset
+
+ izpustimo in pišemo 
+\begin_inset Formula $a/b$
+\end_inset
+
+,
+ čemur pravimo deljenje 
+\begin_inset Formula $a$
+\end_inset
+
+ z 
+\begin_inset Formula $b$
+\end_inset
+
+ za neničeln 
+\begin_inset Formula $b$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Subsubsection
+Skupne lastnosti v 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Axiom
+\begin_inset Formula $1\not=0$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Axiom
+Distributivnost:
+ 
+\begin_inset Formula $\forall a,b,c\in\mathbb{R}:\left(a+b\right)c=ac+bc$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Paragraph
+Urejenost 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Realna števila delimo na pozitivna 
+\begin_inset Formula $\mathbb{R}_{+}\coloneqq\left\{ x\in\mathbb{R};x>0\right\} $
+\end_inset
+
+,
+ negativna 
+\begin_inset Formula $\mathbb{R}_{-}\coloneqq\left\{ x\in\mathbb{R};x<0\right\} $
+\end_inset
+
+ in ničlo 
+\begin_inset Formula $0$
+\end_inset
+
+.
+ Če je 
+\begin_inset Formula $x\in\mathbb{\mathbb{R}}_{+}\cup\left\{ 0\right\} $
+\end_inset
+
+,
+ pišemo 
+\begin_inset Formula $x\geq0$
+\end_inset
+
+,
+ če je 
+\begin_inset Formula $x\in\mathbb{R}_{-}\cup\left\{ 0\right\} $
+\end_inset
+
+,
+ pišemo 
+\begin_inset Formula $x\leq0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Axiom
+Če je 
+\begin_inset Formula $a\not=0$
+\end_inset
+
+,
+ je natanko eno izmed 
+\begin_inset Formula $\left\{ a,-a\right\} $
+\end_inset
+
+ pozitivno,
+ imenujemo ga absolutna vrednost 
+\begin_inset Formula $a$
+\end_inset
+
+ (pišemo 
+\begin_inset Formula $\left|a\right|$
+\end_inset
+
+),
+ in natanko eno negativno,
+ pišemo 
+\begin_inset Formula $-\left|a\right|$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+\begin_inset Formula $\left|0\right|=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Za 
+\begin_inset Formula $a,b\in\mathbb{R}$
+\end_inset
+
+ se 
+\begin_inset Formula $\left|a-b\right|$
+\end_inset
+
+ imenuje razdalja.
+\end_layout
+
+\begin_layout Axiom
+\begin_inset Formula $\forall a,b\in\mathbb{R}:a,b>0\Rightarrow\left(a+b>0\right)\wedge\left(ab>0\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Za 
+\begin_inset Formula $a,b\in\mathbb{R}$
+\end_inset
+
+:
+ 
+\begin_inset Formula $a$
+\end_inset
+
+ je večje od 
+\begin_inset Formula $b$
+\end_inset
+
+,
+ oznaka 
+\begin_inset Formula $a>b\Leftrightarrow a-b>0$
+\end_inset
+
+.
+ 
+\begin_inset Formula $a$
+\end_inset
+
+ je manjše od 
+\begin_inset Formula $b$
+\end_inset
+
+,
+ oznaka 
+\begin_inset Formula $a<b\Leftrightarrow a-b<0$
+\end_inset
+
+.
+ Podobno 
+\begin_inset Formula $\leq$
+\end_inset
+
+ in 
+\begin_inset Formula $\geq$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+Trikotniška neenakost.
+ 
+\begin_inset Formula $\forall a,b\in\mathbb{R}$
+\end_inset
+
+:
+ 
+\begin_inset Formula $\left|\left|a\right|-\left|b\right|\right|\leq\left|a+b\right|\leq\left|a\right|+\left|b\right|$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Dokažimo desni neenačaj.
+ Vemo 
+\begin_inset Formula $ab\leq\left|ab\right|$
+\end_inset
+
+ in 
+\begin_inset Formula $\left|a\right|=\sqrt{a^{2}}$
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $a^{2}+2ab+b^{2}\leq\left|a\right|^{2}+2\left|a\right|\left|b\right|+\left|b\right|^{2}$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\left(a+b\right)^{2}\leq\left(\left|a\right|+\left|b\right|\right)^{2}$
+\end_inset
+
+,
+ korenimo:
+ 
+\begin_inset Formula $\left|a+b\right|\leq\left|a\right|+\left|b\right|$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsubsection
+Intervali
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $a<b$
+\end_inset
+
+.
+ Označimo odprti interval 
+\begin_inset Formula $\left(a,b\right)\coloneqq\left\{ x\in\mathbb{R};a<x<b\right\} $
+\end_inset
+
+,
+ zaprti 
+\begin_inset Formula $\left[a,b\right]\coloneqq\left\{ x\in\mathbb{R};a\leq x\leq b\right\} $
+\end_inset
+
+,
+ polodprti 
+\begin_inset Formula $(a,b]\coloneqq\left\{ x\in\mathbb{R};a<x\leq b\right\} $
+\end_inset
+
+ in podobno 
+\begin_inset Formula $[a,b)$
+\end_inset
+
+.
+ 
+\begin_inset Formula $\left(a,\infty\right)\coloneqq\left\{ x\in\mathbb{R};x>a\right\} $
+\end_inset
+
+ in podobno 
+\begin_inset Formula $[a,\infty)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsection
+Temeljne številske podmnožice
+\end_layout
+
+\begin_layout Subsubsection
+Naravna števila 
+\begin_inset Formula $\mathbb{N}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+\begin_inset Formula $\mathbb{N}\coloneqq\left\{ 1,1+1,1+1+1,1+1+1+1,\dots\right\} $
+\end_inset
+
+
+\end_layout
+
+\begin_layout Paragraph
+Matematična indukcija
+\end_layout
+
+\begin_layout Standard
+Če je 
+\begin_inset Formula $A\subseteq\mathbb{N}$
+\end_inset
+
+ in velja 
+\begin_inset Formula $1\in A$
+\end_inset
+
+ (baza) in 
+\begin_inset Formula $a\in A\Rightarrow a+1\in A$
+\end_inset
+
+ (korak),
+ tedaj 
+\begin_inset Formula $A=\mathbb{N}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+\begin_inset Formula $1+2+3+\cdots+n=\frac{n\left(n+1\right)}{2}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+\begin_inset Formula $A\coloneqq\left\{ n\in\mathbb{N};\text{velja trditev za }n\right\} $
+\end_inset
+
+.
+ Dokažimo 
+\begin_inset Formula $A=\mathbb{N}$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Baza:
+ 
+\begin_inset Formula $1=\frac{1\cdot2}{2}=1$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Korak:
+ Predpostavimo 
+\begin_inset Formula $1+2+3+\cdots+n=\frac{n\left(n+1\right)}{2}$
+\end_inset
+
+.
+ Prištejmo 
+\begin_inset Formula $n+1$
+\end_inset
+
+:
+ 
+\begin_inset Formula 
+\[
+1+2+3+\cdots+n+\left(n+1\right)=\frac{n\left(n+1\right)}{2}+\left(n+1\right)=\frac{n\left(n+1\right)}{2}+\frac{2\left(n+1\right)}{2}=\frac{\left(n+2\right)\left(n+1\right)}{2}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Subsubsection
+Cela števila 
+\begin_inset Formula $\mathbb{Z}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Množica 
+\begin_inset Formula $\mathbb{N}$
+\end_inset
+
+ je zaprta za seštevanje in množenje,
+ torej 
+\begin_inset Formula $\forall a,b\in\mathbb{N}:a+b\in\mathbb{N}\wedge ab\in\mathbb{N}$
+\end_inset
+
+,
+ ni pa zaprta za odštevanje,
+ ker recimo 
+\begin_inset Formula $5-3\not\in\mathbb{N}$
+\end_inset
+
+.
+ Zapremo jo za odštevanje in dobimo množico 
+\begin_inset Formula $\mathbb{Z}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+\begin_inset Formula $\mathbb{Z}\coloneqq\left\{ a-b;b,a\in\mathbb{N}\right\} $
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsubsection
+Racionalna števila 
+\begin_inset Formula $\mathbb{Q}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Najmanjša podmnožica 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+,
+ ki vsebuje 
+\begin_inset Formula $\mathbb{Z}$
+\end_inset
+
+ in je zaprta za deljenje,
+ je 
+\begin_inset Formula $\mathbb{Q}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+\begin_inset Formula $\mathbb{Q}\coloneqq\left\{ a/b;a\in\mathbb{Z},b\in\mathbb{Z}\setminus\left\{ 0\right\} \right\} $
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Velja 
+\begin_inset Formula $\mathbb{N}\subset\mathbb{Z}\subset\mathbb{Q}\subset\mathbb{R}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+Za 
+\begin_inset Formula $a\in\mathbb{Q}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $b\not\in\mathbb{Q}$
+\end_inset
+
+ velja 
+\begin_inset Formula $a+b\not\in\mathbb{Q}$
+\end_inset
+
+ in 
+\begin_inset Formula $a\not=0\Rightarrow ab\not\in\mathbb{Q}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+PDDRAA 
+\begin_inset Formula $a+b\in\mathbb{Q}$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $a+b-a\in\mathbb{Q}$
+\end_inset
+
+,
+ tedaj 
+\begin_inset Formula $b\in\mathbb{Q}$
+\end_inset
+
+,
+ kar je 
+\begin_inset Formula $\rightarrow\!\leftarrow$
+\end_inset
+
+.
+ PDDRAA 
+\begin_inset Formula $ab\in\mathbb{Q}$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $\frac{ab}{a}\in\mathbb{Q}$
+\end_inset
+
+,
+ tedaj 
+\begin_inset Formula $b\in\mathbb{Q}$
+\end_inset
+
+,
+ kar je 
+\begin_inset Formula $\rightarrow\!\leftarrow$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsection
+\begin_inset CommandInset label
+LatexCommand label
+name "subsec:Omejenost-množic"
+
+\end_inset
+
+Omejenost množic
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $A\subset\mathbb{R}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $A$
+\end_inset
+
+ je navzgor omejena 
+\begin_inset Formula $\Leftrightarrow\exists m\in\mathbb{R}\forall a\in A:a\leq m$
+\end_inset
+
+.
+ Takemu 
+\begin_inset Formula $m$
+\end_inset
+
+ pravimo zgornja meja.
+ Najmanjši zgornji meji 
+\begin_inset Formula $A$
+\end_inset
+
+ pravimo supremum ali natančna zgornja meja množice 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ označimo 
+\begin_inset Formula $\sup A$
+\end_inset
+
+.
+ Če je zgornja meja 
+\begin_inset Formula $A$
+\end_inset
+
+ (
+\begin_inset Formula $m$
+\end_inset
+
+) element 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ je maksimum množice 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ označimo 
+\begin_inset Formula $m=\max A$
+\end_inset
+
+.
+ Če množica ni navzgor omejena,
+ pišemo 
+\begin_inset Formula $\sup A=\infty$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Če 
+\begin_inset Formula $s=\sup A\in\mathbb{R}$
+\end_inset
+
+,
+ mora veljati 
+\begin_inset Formula $\forall a\in A:a\leq s$
+\end_inset
+
+ in 
+\begin_inset Formula $\forall\varepsilon>0\exists b\in A\ni:b>s-\varepsilon$
+\end_inset
+
+,
+ torej za vsak neničeln 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+ 
+\begin_inset Formula $s-\varepsilon$
+\end_inset
+
+ ni več natančna zgornja meja za 
+\begin_inset Formula $A$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $A\subset\mathbb{R}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $A$
+\end_inset
+
+ je navzdol omejena 
+\begin_inset Formula $\Leftrightarrow\exists m\in\mathbb{R}\forall a\in A:a\geq m$
+\end_inset
+
+.
+ Takemu 
+\begin_inset Formula $m$
+\end_inset
+
+ pravimo spodnja meja.
+ Največji spodnji meji 
+\begin_inset Formula $A$
+\end_inset
+
+ pravimo infimum ali natančna spodnja meja množice 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ označimo 
+\begin_inset Formula $\inf A$
+\end_inset
+
+.
+ Če je spodnja meja 
+\begin_inset Formula $A$
+\end_inset
+
+ (
+\begin_inset Formula $m$
+\end_inset
+
+) element 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ je minimum množice 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ označimo 
+\begin_inset Formula $m=\min A$
+\end_inset
+
+.
+ Če množica ni navzdol omejena,
+ pišemo 
+\begin_inset Formula $\inf A=-\infty$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Množica 
+\begin_inset Formula $A\subset\mathbb{R}$
+\end_inset
+
+ je omejena,
+ če je hkrati navzgor in navzdol omejena.
+\end_layout
+
+\begin_layout Axiom
+\begin_inset CommandInset label
+LatexCommand label
+name "axm:Dedekind.-Vsaka-navzgor"
+
+\end_inset
+
+Dedekind.
+ Vsaka navzgor omejena množica v 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+ ima natančno zgornjo mejo v 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Remark*
+Za 
+\begin_inset Formula $\mathbb{Q}$
+\end_inset
+
+ aksiom 
+\begin_inset CommandInset ref
+LatexCommand ref
+reference "axm:Dedekind.-Vsaka-navzgor"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+ ne velja.
+ Če 
+\begin_inset Formula $B\subset\mathbb{Q}$
+\end_inset
+
+,
+ se lahko zgodi,
+ da 
+\begin_inset Formula $\sup B\not\in\mathbb{Q}$
+\end_inset
+
+.
+ Primer:
+ 
+\begin_inset Formula $B\coloneqq\left\{ q\in\mathbb{Q};q^{2}\leq2\right\} $
+\end_inset
+
+.
+ 
+\begin_inset Formula $\sup B=\sqrt{2}\not\in\mathbb{Q}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+
+\end_layout
+
+\begin_layout Subsection
+Decimalni zapis
+\end_layout
+
+\begin_layout Definition*
+\begin_inset Formula $\forall x\in\mathbb{R}^{+}\exists!m\in\mathbb{N}\cup\left\{ 0\right\} ,d_{1},d_{2},\dots\in\left\{ 0..9\right\} $
+\end_inset
+
+,
+ ki število natančno določajo.
+ Pišemo 
+\begin_inset Formula $x=m,d_{1}d_{2}\dots$
+\end_inset
+
+.
+ Natančno določitev mislimo v smislu:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $m\leq x<m+1$
+\end_inset
+
+ —
+ s tem se izognemo dvojnemu zapisu 
+\begin_inset Formula $1=0,\overline{9}$
+\end_inset
+
+ in 
+\begin_inset Formula $1=1,\overline{0}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $[m,m+1)$
+\end_inset
+
+ razdelimo na 10 enako dolgih polodprtih intervalov 
+\begin_inset Formula $I_{0},\dots,I_{9}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $x$
+\end_inset
+
+ leži na natanko enem izmed njih,
+ indeks njega je 
+\begin_inset Formula $d_{1}$
+\end_inset
+
+.
+ Nadaljujemo tako,
+ da 
+\begin_inset Formula $I_{d_{1}}$
+\end_inset
+
+ razdelimo zopet na 10 delov itd.
+\end_layout
+
+\end_deeper
+\begin_layout Definition*
+Števila 
+\begin_inset Formula $x\in\mathbb{R^{-}}$
+\end_inset
+
+ pišemo tako,
+ da zapišemo decimalni zapis števila 
+\begin_inset Formula $-x$
+\end_inset
+
+ in predenj zapišemo 
+\begin_inset Formula $-$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Če se decimalke v zaporedju 
+\begin_inset Formula $\left(d_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ ponavljajo,
+ uporabimo periodični zapis,
+ denimo 
+\begin_inset Formula $5,01\overline{763}\in\mathbb{Q}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsection
+Kompleksna števila
+\end_layout
+
+\begin_layout Definition*
+Vpeljimo število 
+\begin_inset Formula $i$
+\end_inset
+
+ z lastnostjo 
+\begin_inset Formula $i^{2}=-1$
+\end_inset
+
+,
+ da je 
+\begin_inset Formula $i$
+\end_inset
+
+ rešitev enačbe 
+\begin_inset Formula $x^{2}+1=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+\begin_inset Formula $i\not\in\mathbb{R}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Proof
+Sicer bi veljajo 
+\begin_inset Formula $i^{2}\geq0$
+\end_inset
+
+,
+ kar po definiciji ne velja.
+\end_layout
+
+\begin_layout Definition*
+Kompleksna števila so 
+\begin_inset Formula $\mathbb{C}\coloneqq\left\{ a+bi;a,b\in\mathbb{R}\right\} $
+\end_inset
+
+.
+ 
+\begin_inset Formula $bi$
+\end_inset
+
+ je še nedefinirano,
+ zato za kompleksna števila definirano seštevanje in množenje za 
+\begin_inset Formula $z=a+bi$
+\end_inset
+
+ in 
+\begin_inset Formula $w=c+di$
+\end_inset
+
+:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $z+w\coloneqq\left(a+c\right)+\left(b+d\right)i$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $zw\coloneqq\left(a+bi\right)\left(c+di\right)=ac+adi+bic+bidi=ac+adi+bic-bd=\left(ac-bd\right)+\left(ad+bc\right)i$
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Definition*
+Definiramo še konjugirano vrednost 
+\begin_inset Formula $z\in\mathbb{C}$
+\end_inset
+
+:
+ 
+\begin_inset Formula $\overline{z}\coloneqq a-bi$
+\end_inset
+
+ in označimo 
+\begin_inset Formula $\left|z\right|^{2}=z\overline{z}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+\begin_inset Formula $z\overline{z}=a^{2}+b^{2}\geq0$
+\end_inset
+
+ za 
+\begin_inset Formula $z=a+bi$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+\begin_inset Formula $\left(a+bi\right)\left(a-bi\right)=a^{2}+abi-bia-bibi=a^{2}+b^{2}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Velja,
+ da je 
+\begin_inset Formula $\mathbb{R}\subset\mathbb{C}$
+\end_inset
+
+ v smislu identifikacije 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+ z množico 
+\begin_inset Formula $\mathbb{C}$
+\end_inset
+
+:
+ 
+\begin_inset Formula $\mathbb{R}=\left\{ a+0i;a\in\mathbb{R}\right\} $
+\end_inset
+
+,
+ torej smo 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+ razširili v 
+\begin_inset Formula $\mathbb{C}$
+\end_inset
+
+,
+ kjer ima vsak polinom vedno rešitev.
+\end_layout
+
+\begin_layout Subsubsection
+Deljenje v 
+\begin_inset Formula $\mathbb{C}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Za 
+\begin_inset Formula $w,z\in\mathbb{C},w\not=0$
+\end_inset
+
+ iščemo 
+\begin_inset Formula $x\in\mathbb{C}\ni:wx=z$
+\end_inset
+
+.
+ Ločimo dva primera:
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $w\in\mathbb{R}\setminus\left\{ 0\right\} $
+\end_inset
+
+:
+ definiramo 
+\begin_inset Formula $x=\frac{z}{w}\coloneqq\frac{a}{w}+\frac{b}{w}i$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $w\in\mathbb{C}\setminus\left\{ 0\right\} $
+\end_inset
+
+ (splošno):
+ 
+\begin_inset Formula $wx=z\overset{/\cdot\overline{w}}{\Longrightarrow}w\overline{w}x=z\overline{w}\Rightarrow\left|w\right|^{2}x=z\overline{w}\Rightarrow x=\frac{z\overline{w}}{\left|w\right|^{2}}$
+\end_inset
+
+,
+ z 
+\begin_inset Formula $\left|w\right|^{2}$
+\end_inset
+
+ pa znamo deliti,
+ ker je realen.
+\end_layout
+
+\begin_layout Subsubsection
+Lastnosti v 
+\begin_inset Formula $\mathbb{C}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $+$
+\end_inset
+
+ in 
+\begin_inset Formula $\cdot$
+\end_inset
+
+ sta komutativni,
+ asociativni,
+ distributivni,
+ 
+\begin_inset Formula $0$
+\end_inset
+
+ je aditivna enota,
+ 
+\begin_inset Formula $1$
+\end_inset
+
+ je multiplikativna.
+\end_layout
+
+\begin_layout Definition*
+Za 
+\begin_inset Formula $z=a+bi$
+\end_inset
+
+ vpeljemo 
+\begin_inset Formula $\Re z=a$
+\end_inset
+
+ in 
+\begin_inset Formula $\Im z=b$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Remark*
+Opazimo 
+\begin_inset Formula $\Re z=\frac{z+\overline{z}}{2}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\Im z=\frac{z-\overline{z}}{2}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Remark*
+\begin_inset Formula $\mathbb{C}$
+\end_inset
+
+ si lahko predstavljamo kot urejene pare;
+ 
+\begin_inset Formula $a+bi$
+\end_inset
+
+ ustreza paru 
+\begin_inset Formula $\left(a,b\right)$
+\end_inset
+
+.
+ Tako 
+\begin_inset Formula $\mathbb{C}$
+\end_inset
+
+ enačimo/identificiramo z 
+\begin_inset Formula $\mathbb{R}^{2}\coloneqq\left\{ \left(a,b\right);a,b\in\mathbb{R}\right\} $
+\end_inset
+
+,
+ s čimer dobimo geometrično predstavitev 
+\begin_inset Formula $\mathbb{C}$
+\end_inset
+
+ kot vektorje v 
+\begin_inset Formula $\mathbb{R}^{2}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Za 
+\begin_inset Formula $z=a+bi$
+\end_inset
+
+,
+ predstavljen z vektorjem s komponentami 
+\begin_inset Formula $\left(a,b\right)$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $a=\left|z\right|\cos\varphi$
+\end_inset
+
+ in 
+\begin_inset Formula $v=\left|z\right|\sin\varphi$
+\end_inset
+
+.
+ Kotu 
+\begin_inset Formula $\varphi$
+\end_inset
+
+ pravimo argument kompleksnega števila 
+\begin_inset Formula $z$
+\end_inset
+
+,
+ oznaka 
+\begin_inset Formula $\arg z$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Corollary*
+\begin_inset Formula $z=$
+\end_inset
+
+
+\begin_inset Formula $\left|z\right|\left(\cos\varphi+i\sin\varphi\right)$
+\end_inset
+
+.
+ Velja
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+TODO DOPISATI ZAKAJ (v bistvu še jaz ne vem).
+ ne razumem.
+\end_layout
+
+\end_inset
+
+ 
+\begin_inset Formula $\left(\cos\varphi+i\sin\varphi\right)\left(\cos\psi+i\sin\psi\right)=\cos\left(\varphi+\psi\right)+i\sin\left(\varphi+\psi\right)$
+\end_inset
+
+,
+ zato lahko pišemo 
+\begin_inset Formula $e^{i\varphi}=\cos\varphi+i\sin\varphi$
+\end_inset
+
+.
+ Množenje kompleksnh števil 
+\begin_inset Formula $z=\left|z\right|e^{i\varphi}$
+\end_inset
+
+ in 
+\begin_inset Formula $w=\left|w\right|e^{i\psi}$
+\end_inset
+
+ vrne število 
+\begin_inset Formula $zw$
+\end_inset
+
+,
+ za katero velja:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $\left|zw\right|=\left|z\right|\left|w\right|$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $\arg zw=\arg z+\arg w$
+\end_inset
+
+ (do periode 
+\begin_inset Formula $2\pi$
+\end_inset
+
+ natančno)
+\end_layout
+
+\end_deeper
+\begin_layout Section
+Zaporedja
+\end_layout
+
+\begin_layout Definition*
+Funkcija 
+\begin_inset Formula $a:\mathbb{N}\to\mathbb{R}$
+\end_inset
+
+ se imenuje realno zaporedje,
+ oznaka 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $a_{n}$
+\end_inset
+
+ je funkcijska vrednost pri 
+\begin_inset Formula $n$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $a_{n}=n$
+\end_inset
+
+:
+ 
+\begin_inset Formula $1,2,3,\dots$
+\end_inset
+
+;
+ 
+\begin_inset Formula $a_{n}=\left(-1\right)^{n}n^{2}$
+\end_inset
+
+:
+ 
+\begin_inset Formula $-1,4,-9,16,-25,\dots$
+\end_inset
+
+;
+ 
+\begin_inset Formula $a_{n}=\cos\left(\frac{\pi}{2}n\right)=0,-1,0,1,0,-1,\dots$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Zaporedje lahko podamo rekurzivno.
+ Podamo prvi člen ali nekaj prvih členov in pravilo,
+ kako iz prejšnjih členov dobiti naslednje.
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $a_{1}=0,a_{n+1}=a_{n}+n$
+\end_inset
+
+ da zaporedje 
+\begin_inset Formula $a_{n}=\frac{n\left(n+1\right)}{2}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $a_{0}=0,a_{n+1}=\sqrt{b+a_{n}}$
+\end_inset
+
+ da zaporedje 
+\begin_inset Formula $0,\sqrt{b},\sqrt{b+\sqrt{b}},\sqrt{b+\sqrt{b+\sqrt{b}}},\dots$
+\end_inset
+
+.
+ Fibbonacijevo zaporedje:
+ 
+\begin_inset Formula $a_{1}=a_{2}=1,a_{n+1}=a_{n}+a_{n-1}$
+\end_inset
+
+ da zaporedje 
+\begin_inset Formula $1,1,2,3,5,8,\dots$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsection
+Posebni tipi zaporedij
+\end_layout
+
+\begin_layout Definition*
+Zaporedje 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ je aritmetično,
+ če 
+\begin_inset Formula $\exists k\in\mathbb{R}\forall n\in\mathbb{N}:a_{n+1}-a_{n}=k$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $a_{n+1}=a_{n}+k=a_{1}+nd$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Zaporedje 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ je geometrično,
+ če 
+\begin_inset Formula $\exists\lambda\in\mathbb{R}\forall n\in\mathbb{N}:a_{n+1}=a_{n}\lambda$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $a_{n}=\lambda^{n-1}a_{1}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Zaporedje 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ je navzdol oz.
+ navzgor omejeno,
+ če je množica vseh členov tega taporedja navzgor oz.
+ navzdol omejena (glej 
+\begin_inset CommandInset ref
+LatexCommand ref
+reference "subsec:Omejenost-množic"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+).
+ Podobno z množico členov definiramo supremum,
+ infimum,
+ maksimum in infimum zaporedja.
+\end_layout
+
+\begin_layout Definition*
+Zaporedje je naraščajoče,
+ če 
+\begin_inset Formula $\forall n\in\mathbb{N}:a_{n+1}\geq a_{n}$
+\end_inset
+
+,
+ padajoče,
+ če 
+\begin_inset Formula $\forall n\in\mathbb{N}:a_{n+1}\leq a_{n}$
+\end_inset
+
+,
+ strogo naraščajoče,
+ če 
+\begin_inset Formula $\forall n\in\mathbb{N}:a_{n+1}>a_{n}$
+\end_inset
+
+,
+ strogo padajoče podobno,
+ monotono,
+ če je naraščajoče ali padajoče in strogo monotono,
+ če je strogo naraščajoče ali strogo padajoče.
+\end_layout
+
+\begin_layout Subsection
+Limita zaporedja
+\end_layout
+
+\begin_layout Definition*
+Množica 
+\begin_inset Formula $U\subseteq\mathbb{R}$
+\end_inset
+
+ je odprta,
+ če 
+\begin_inset Formula $\forall u\in U\exists r>0\ni:\left(u-r,u+r\right)\subseteq U$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Množica 
+\begin_inset Formula $U\subseteq\mathbb{R}$
+\end_inset
+
+ je zaprta,
+ če je 
+\begin_inset Formula $U^{\mathcal{C}}\coloneqq\mathbb{R}\setminus U$
+\end_inset
+
+ odprta.
+\end_layout
+
+\begin_layout Claim*
+Odprt interval je odprta množica.
+\end_layout
+
+\begin_layout Proof
+Za poljubna 
+\begin_inset Formula $a,b\in\mathbb{R}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $b>a$
+\end_inset
+
+,
+ naj bo 
+\begin_inset Formula $u\in\left(a,b\right)$
+\end_inset
+
+ poljuben.
+ Ustrezen 
+\begin_inset Formula $r$
+\end_inset
+
+ je 
+\begin_inset Formula $\min\left\{ \left|r-a\right|,\left|r-b\right|\right\} $
+\end_inset
+
+,
+ da je 
+\begin_inset Formula $\left(u-r,u+r\right)\subseteq U$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+Zaprt interval je zaprt.
+\end_layout
+
+\begin_layout Proof
+Naj bosta 
+\begin_inset Formula $a,b\in\mathbb{R}$
+\end_inset
+
+ poljubna in 
+\begin_inset Formula $b>a$
+\end_inset
+
+.
+ Dokazujemo,
+ da je 
+\begin_inset Formula $\left[a,b\right]$
+\end_inset
+
+ zaprt,
+ torej da je 
+\begin_inset Formula $\left[a,b\right]^{\mathcal{C}}=\left(-\infty,a\right)\cup\left(b,\infty\right)$
+\end_inset
+
+ odprta množica.
+ Za poljuben 
+\begin_inset Formula $u\in\left[a,b\right]^{\mathcal{C}}$
+\end_inset
+
+ velja,
+ da je bodisi 
+\begin_inset Formula $\in\left(-\infty,a\right)$
+\end_inset
+
+ bodisi 
+\begin_inset Formula $\left(b,\infty\right)$
+\end_inset
+
+,
+ kajti 
+\begin_inset Formula $\left(-\infty,a\right)\cap\left(b,\infty\right)=\emptyset$
+\end_inset
+
+.
+ Po prejšnji trditvi v obeh primerih velja 
+\begin_inset Formula $\exists r>0\ni:\left(u-r,u+r\right)\subseteq U$
+\end_inset
+
+,
+ torej je 
+\begin_inset Formula $\left[a,b\right]^{\mathcal{C}}$
+\end_inset
+
+ res odprta,
+ torej je 
+\begin_inset Formula $\left[a,b\right]$
+\end_inset
+
+ res zaprta.
+\end_layout
+
+\begin_layout Definition*
+Množica 
+\begin_inset Formula $B$
+\end_inset
+
+ je okolica točke 
+\begin_inset Formula $t\in\mathbb{R}$
+\end_inset
+
+,
+ če vsebuje kakšno odprto množico 
+\begin_inset Formula $U$
+\end_inset
+
+,
+ ki vsebuje 
+\begin_inset Formula $t$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $t\in U^{\text{odp.}}\subseteq B\subseteq\mathbb{R}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+\begin_inset Formula $L\in\mathbb{R}$
+\end_inset
+
+ je limita zaporedja 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{R}}$
+\end_inset
+
+ 
+\begin_inset Formula $\Leftrightarrow\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow\left|a_{n}-L\right|<\varepsilon$
+\end_inset
+
+.
+ ZDB 
+\begin_inset Formula $\forall V$
+\end_inset
+
+ okolica 
+\begin_inset Formula $L\in\mathbb{R}\exists n_{0}\in\mathbb{N}\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow a_{n}\in V$
+\end_inset
+
+,
+ pravimo,
+ da 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ konvergira k 
+\begin_inset Formula $L$
+\end_inset
+
+ in pišemo 
+\begin_inset Formula $L\coloneqq\lim_{n\to\infty}a_{n}$
+\end_inset
+
+ ali drugače 
+\begin_inset Formula $a_{n}\underset{n\to\infty}{\longrightarrow}L$
+\end_inset
+
+.
+ Če zaporedje ima limito,
+ pravimo,
+ da je konvergentno,
+ sicer je divergentno.
+\end_layout
+
+\begin_layout Claim*
+Konvergentno zaporedje v 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+ ima natanko eno limito.
+\end_layout
+
+\begin_layout Proof
+Naj bosta 
+\begin_inset Formula $J$
+\end_inset
+
+ in 
+\begin_inset Formula $L$
+\end_inset
+
+ limiti zaporedja 
+\begin_inset Formula $\left(a_{n}\right)_{n\to\infty}$
+\end_inset
+
+.
+ Torej
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+Ko trdimo,
+ da obstaja 
+\begin_inset Formula $n_{0}$
+\end_inset
+
+,
+ še ne vemo,
+ ali sta za 
+\begin_inset Formula $L$
+\end_inset
+
+ in 
+\begin_inset Formula $J$
+\end_inset
+
+ ta 
+\begin_inset Formula $n_{0}$
+\end_inset
+
+ ista.
+ Ampak trditev še vedno velja,
+ ker lahko vzamemo večjega izmed njiju,
+ ako bi bila drugačna.
+\end_layout
+
+\end_inset
+
+ po definiciji 
+\begin_inset Formula $\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow\left|a_{n}-L\right|<\varepsilon\wedge\left|a_{n}-J\right|<\varepsilon$
+\end_inset
+
+.
+ Velja torej 
+\begin_inset Formula $\forall\varepsilon>0:\left|J-L\right|<\varepsilon$
+\end_inset
+
+.
+ PDDRAA 
+\begin_inset Formula $J\not=L$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $\left|J-L\right|\not=0$
+\end_inset
+
+,
+ naj bo 
+\begin_inset Formula $\left|J-L\right|=k$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $\exists\varepsilon>0:\left|J-L\right|\not<\varepsilon$
+\end_inset
+
+,
+ ustrezen 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+ je na primer 
+\begin_inset Formula $\frac{\left|J-L\right|}{2}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+\begin_inset CommandInset label
+LatexCommand label
+name "Konvergentno-zaporedje-v-R-je-omejeno"
+
+\end_inset
+
+Konvergentno zaporedje v 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+ je omejeno.
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $L=\lim_{n\to\infty}a_{n}$
+\end_inset
+
+.
+ Znotraj intervala 
+\begin_inset Formula $\left(L-1,L+1\right)$
+\end_inset
+
+ so vsi členi zaporedja razen končno mnogo (
+\begin_inset Formula $\left\{ a_{1},\dots,a_{n_{0}}\right\} $
+\end_inset
+
+).
+ 
+\begin_inset Formula $\left\{ a_{n}\right\} _{n\in\mathbb{N}}$
+\end_inset
+
+ je unija dveh omejenih množic;
+ 
+\begin_inset Formula $\left(L-1,L+1\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $\left\{ a_{1},\dots,a_{n_{0}}\right\} $
+\end_inset
+
+,
+ zato je tudi sama omejena.
+\end_layout
+
+\begin_layout Claim*
+Naj bosta 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ in 
+\begin_inset Formula $\left(b_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ konvergentni zaporedji v 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+.
+ Tedaj so tudi 
+\begin_inset Formula $\left(a_{n}*b_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ konvergentna in velja 
+\begin_inset Formula $\lim_{n\to\infty}a_{n}*b_{n}=\lim_{n\to\infty}a_{n}*\lim_{n\to\infty}b_{n}$
+\end_inset
+
+ za 
+\begin_inset Formula $*\in\left\{ +,-,\cdot\right\} $
+\end_inset
+
+.
+ Če je 
+\begin_inset Formula $\lim_{n\to\infty}b_{n}\not=0$
+\end_inset
+
+,
+ isto velja tudi za deljenje.
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $a_{n}\to A$
+\end_inset
+
+ in 
+\begin_inset Formula $b_{n}\to B$
+\end_inset
+
+ oziroma 
+\begin_inset Formula $\forall\varepsilon>0\exists n_{1},n_{2}\ni:\left(n>n_{1}\Rightarrow\left|a_{n}-A\right|<\varepsilon\right)\wedge\left(n>n_{2}\Rightarrow\left|b_{n}-B\right|<\varepsilon\right)$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\forall\varepsilon>0\exists n_{0}=\max\left\{ n_{1},n_{2}\right\} \ni:n>n_{0}\Rightarrow\left|a_{n}-A\right|<\varepsilon\wedge\left|b_{n}-B\right|<\varepsilon$
+\end_inset
+
+.
+ Dokažimo za vse operacije:
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $+$
+\end_inset
+
+ Po predpostavki velja 
+\begin_inset Formula $\forall\varepsilon>0\exists n_{0}\ni:n>n_{0}\Rightarrow\left|a_{n}-A\right|+\left|a_{n}-B\right|<2\varepsilon$
+\end_inset
+
+.
+ Oglejmo si sedaj
+\begin_inset Formula 
+\[
+\left|\left(a_{n}+b_{n}\right)-\left(A+B\right)\right|=\left|\left(a_{n}-A\right)+\left(b_{n}-B\right)\right|\leq\left|a_{n}-A\right|+\left|b_{n}-B\right|
+\]
+
+\end_inset
+
+in uporabimo še prejšnjo trditev,
+ torej 
+\begin_inset Formula $\forall2\varepsilon\exists n_{0}\ni:\left|\left(a_{n}+b_{n}\right)-\left(A+B\right)\right|\leq2\varepsilon$
+\end_inset
+
+,
+ s čimer dokažemo 
+\begin_inset Formula $\left(a_{n}+b_{n}\right)\to A+B$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $-$
+\end_inset
+
+ Oglejmo si
+\begin_inset Formula 
+\[
+\left|\left(a_{n}-b_{n}\right)-\left(A-B\right)\right|=\left|a_{n}-b_{n}-A+B\right|=\left|\left(a_{n}-A\right)+\left(-\left(b_{n}-B\right)\right)\right|\leq\left|a_{n}-A\right|+\left|b_{n}-B\right|
+\]
+
+\end_inset
+
+in nato kot zgoraj.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\cdot$
+\end_inset
+
+ Oglejmo si 
+\begin_inset Formula 
+\[
+\left|a_{n}b_{n}-AB\right|=\left|a_{n}b_{n}-Ab_{n}+Ab_{n}-AB\right|=\left|\left(a_{n}-A\right)b_{n}+A\left(b_{n}-B\right)\right|\leq\left|a_{n}-A\right|\left|b_{n}\right|+\left|A\right|\left|b_{n}-B\right|.
+\]
+
+\end_inset
+
+Od prej vemo,
+ da sta zaporedji omejeni,
+ ker sta konvergentni,
+ zato 
+\begin_inset Formula $\exists M>0\forall n\in\mathbb{N}:\left|b_{n}\right|\leq M$
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+ poljuben 
+\begin_inset Formula $n_{1},n_{2}\in\mathbb{N}$
+\end_inset
+
+ taka,
+ da 
+\begin_inset Formula $n\geq n_{1}\Rightarrow\left|a_{n}-A\right|<\frac{\varepsilon}{2M}$
+\end_inset
+
+ in 
+\begin_inset Formula $n\geq n_{2}\Rightarrow\left|b_{n}-B\right|<\frac{\varepsilon}{2\left|A\right|}$
+\end_inset
+
+.
+ Tedaj za 
+\begin_inset Formula $n_{0}\coloneqq\max\left\{ n_{1},n_{2}\right\} $
+\end_inset
+
+ velja
+\begin_inset Formula $\left|a_{n}b_{n}-AB\right|\leq\left|a_{n}-A\right|\left|b_{n}\right|+\left|A\right|\left|b_{n}-B\right|<\frac{\varepsilon}{2M}M+\left|A\right|\frac{\varepsilon}{2\left|A\right|}=\varepsilon$
+\end_inset
+
+,
+ skratka 
+\begin_inset Formula $\left|a_{n}b_{n}-AB\right|<\varepsilon$
+\end_inset
+
+,
+ s čimer dokažemo 
+\begin_inset Formula $\left(a_{n}+b_{n}\right)\to A+B$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $/$
+\end_inset
+
+ Ker je 
+\begin_inset Formula $B\not=0$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\exists n_{0}\in\mathbb{N}\forall n\geq n_{0}:\left|b_{n}\right|\geq\frac{\left|B\right|}{2}>0$
+\end_inset
+
+.
+ ZDB vsi členi zaporedja razen končno mnogo so v poljubno majhni okolici 
+\begin_inset Formula $\left|B\right|$
+\end_inset
+
+.
+ Če torej vzamemo točko na polovici med 0 in 
+\begin_inset Formula $\left|B\right|$
+\end_inset
+
+,
+ to je 
+\begin_inset Formula $\frac{\left|B\right|}{2}$
+\end_inset
+
+,
+ bo neskončno mnogo absolutnih vrednosti členov večjih od 
+\begin_inset Formula $\frac{\left|B\right|}{2}$
+\end_inset
+
+.
+ Pri razumevanju pomaga številska premica.
+ Nadalje uporabimo predpostavko z 
+\begin_inset Formula $\varepsilon=\frac{\left|B\right|}{2}$
+\end_inset
+
+,
+ torej je za 
+\begin_inset Formula $n>n_{0}:$
+\end_inset
+
+ 
+\begin_inset Formula $\left|B-b_{n}\right|<\frac{\left|B\right|}{2}$
+\end_inset
+
+ in velja 
+\begin_inset Formula 
+\[
+\left|b_{n}\right|=\left|B-\left(B-b_{n}\right)\right|=\left|B+\left(-\left(B-b_{n}\right)\right)\right|\overset{\text{trik. neen.}}{\geq}\left|\left|B\right|-\left|B-b_{n}\right|\right|=\left|B\right|-\left|B-b_{n}\right|\overset{\text{predp.}}{>}\left|B\right|-\frac{\left|B\right|}{2}=\frac{\left|B\right|}{2},
+\]
+
+\end_inset
+
+skratka 
+\begin_inset Formula $\left|b_{n}\right|>\frac{\left|B\right|}{2}$
+\end_inset
+
+.
+ Če spet izpustimo končno začetnih členov,
+ velja
+\begin_inset Formula 
+\[
+\frac{a_{n}}{b_{n}}-\frac{A}{B}=\frac{a_{n}B-Ab_{n}}{b_{n}B}\overset{\text{prištejemo in odštejemo člen}}{=}\frac{\left(a_{n}-A\right)B+A\left(B-b_{n}\right)}{Bb_{n}}=\frac{1}{b_{n}}\left(a_{n}-A\right)+\frac{A/B}{b_{n}}\left(B-b_{n}\right)
+\]
+
+\end_inset
+
+sedaj uporabimo na obeh straneh absolutno vrednost:
+\begin_inset Formula 
+\[
+\left|\frac{a_{n}}{b_{n}}-\frac{A}{B}\right|=\left|\frac{1}{b_{n}}\left(a_{n}-A\right)+\frac{A}{Bb_{n}}\left(B-b_{n}\right)\right|\leq\frac{1}{\left|b_{n}\right|}\left|a_{n}-A\right|+\frac{\left|A\right|}{\left|B\right|\left|b_{n}\right|}\left|B-b_{n}\right|<\frac{2}{\left|B\right|}\left|a_{n}-A\right|+\frac{2\left|A\right|}{\left|B\right|^{2}}\left|B-B_{n}\right|
+\]
+
+\end_inset
+
+skratka 
+\begin_inset Formula $\left|\frac{a_{n}}{b_{n}}-\frac{A}{B}\right|<\frac{2}{\left|B\right|}\left|a_{n}-A\right|+\frac{2\left|A\right|}{\left|B\right|^{2}}\left|B-B_{n}\right|$
+\end_inset
+
+.
+ Opazimo,
+ da 
+\begin_inset Formula $\frac{2}{\left|B\right|}$
+\end_inset
+
+ in 
+\begin_inset Formula $\frac{2\left|A\right|}{\left|B\right|^{2}}$
+\end_inset
+
+ nista odvisna od 
+\begin_inset Formula $n$
+\end_inset
+
+.
+ Sedaj vzemimo poljuben 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+ in 
+\begin_inset Formula $n_{1},n_{2}\in\mathbb{N}$
+\end_inset
+
+ takšna,
+ da velja:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $n\geq n_{1}\Rightarrow\left|a_{n}-A\right|<\frac{\varepsilon\left|B\right|}{4}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $n\geq n_{2}\Rightarrow\left|b_{n}-B\right|<\frac{\varepsilon\left|B\right|^{2}}{4\left|A\right|}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Tedaj iz zgornje ocene sledi za 
+\begin_inset Formula $n\geq\max\left\{ n_{0},n_{1},n_{2}\right\} $
+\end_inset
+
+:
+\begin_inset Formula 
+\[
+\left|\frac{a_{n}}{b_{n}}-\frac{A}{B}\right|<\frac{\cancel{2}}{\cancel{\left|B\right|}}\cdot\frac{\varepsilon\cancel{\left|B\right|}}{\cancelto{2}{4}}+\frac{\cancel{2}\cancel{\left|A\right|}}{\cancel{\left|B\right|^{2}}}\cdot\frac{\varepsilon\cancel{\left|B\right|^{2}}}{\cancelto{2}{4}\cancel{\left|A\right|}}=\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon,
+\]
+
+\end_inset
+
+s čimer dokažemo 
+\begin_inset Formula $a_{n}/b_{n}\to A/B$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\end_deeper
+\begin_layout Example*
+Naj bo 
+\begin_inset Formula $a>0$
+\end_inset
+
+.
+ Izračunajmo
+\begin_inset Formula 
+\[
+\sqrt{a+\sqrt{a+\sqrt{a+\sqrt{\cdots}}}}\eqqcolon\alpha.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $\alpha$
+\end_inset
+
+ je torej 
+\begin_inset Formula $\lim_{n\to\infty}x_{n}$
+\end_inset
+
+,
+ kjer je 
+\begin_inset Formula $x_{0}=0,x_{1}=\sqrt{a},x_{2}=\sqrt{a+\sqrt{a}},x_{3}=\sqrt{a+\sqrt{a+\sqrt{a}}},\dots,x_{n+1}=\sqrt{a+x_{n}}$
+\end_inset
+
+.
+ Iz zadnjega sledi 
+\begin_inset Formula $x_{n+1}^{2}=a+x_{n}$
+\end_inset
+
+.
+ Če torej limita 
+\begin_inset Formula $\alpha\coloneqq\lim x_{n}$
+\end_inset
+
+ obstaja,
+ mora veljati 
+\begin_inset Formula $\alpha^{2}=a+\alpha$
+\end_inset
+
+ oziroma 
+\begin_inset Formula $\alpha_{1,2}=\frac{1\pm\sqrt{1+4a}}{2}$
+\end_inset
+
+.
+ Opcija z minusom ni mogoča,
+ ker je zaporedje 
+\begin_inset Formula $\left(x_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ očitno pozitivno.
+ Če torej limita obstaja (
+\series bold
+česar še nismo dokazali
+\series default
+),
+ je enaka 
+\begin_inset Formula $\frac{1+\sqrt{1+4a}}{2}$
+\end_inset
+
+,
+ za primer 
+\begin_inset Formula $a=2$
+\end_inset
+
+ je torej 
+\begin_inset Formula $\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{\cdots}}}}=2$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Remark*
+Lahko se zgodi,
+ da limita rekurzivno podanega zaporedja ne obstaja,
+ čeprav jo znamo izračunati,
+ če bi obstajala.
+ Na primer 
+\begin_inset Formula $y_{1}\coloneqq1$
+\end_inset
+
+,
+ 
+\begin_inset Formula $y_{n+1}=1-2y_{n}$
+\end_inset
+
+ nam da zaporedje 
+\begin_inset Formula $1,-1,3,-5,11,\dots$
+\end_inset
+
+,
+ kar očitno nima limite.
+ Če bi limita obstajala,
+ bi zanjo veljalo 
+\begin_inset Formula $\beta=1-2\beta$
+\end_inset
+
+ oz.
+ 
+\begin_inset Formula $3\beta=1$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\beta=\frac{1}{3}$
+\end_inset
+
+.
+ Navedimo torej nekaj zadostnih in potrebnih pogojev za konvergenco zaporedij.
+\end_layout
+
+\begin_layout Theorem*
+Naj bo 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ monotono realno zaporedje.
+ Če narašča,
+ ima limito 
+\begin_inset Formula $\lim_{n\to\infty}a_{n}=\sup\left\{ a_{n},n\in\mathbb{N}\right\} $
+\end_inset
+
+.
+ Če pada,
+ ima limito 
+\begin_inset Formula $\lim_{n\to\infty}a_{n}=\inf\left\{ a_{n},n\in\mathbb{N}\right\} $
+\end_inset
+
+.
+ (
+\begin_inset Formula $\sup$
+\end_inset
+
+ in 
+\begin_inset Formula $\inf$
+\end_inset
+
+ imata lahko tudi vrednost 
+\begin_inset Formula $\infty$
+\end_inset
+
+ in 
+\begin_inset Formula $-\infty$
+\end_inset
+
+ —
+ zaporedje s tako limito ni konvergentno v 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+).
+\end_layout
+
+\begin_layout Proof
+Denimo,
+ da 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ narašča.
+ Pišimo 
+\begin_inset Formula $s\coloneqq\sup_{n\in\mathbb{N}}a_{n}$
+\end_inset
+
+.
+ Vzemimo poljuben 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $s-\varepsilon$
+\end_inset
+
+ ni zgornja meja za 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+,
+ zato 
+\begin_inset Formula $\exists n_{0}\in\mathbb{N}\ni:s-\varepsilon<a_{n_{0}}$
+\end_inset
+
+.
+ Ker pa je 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+naraščajoče,
+ sledi 
+\begin_inset Formula $\forall n\geq n_{0}:a_{n}\geq a_{n_{0}}>s-\varepsilon$
+\end_inset
+
+.
+ Hkrati je 
+\begin_inset Formula $a_{n}\leq s$
+\end_inset
+
+,
+ saj je 
+\begin_inset Formula $s$
+\end_inset
+
+ zgornja meja.
+ Torej 
+\begin_inset Formula $\forall n\geq n_{0}:a_{n}\in(s-\varepsilon,s]\subset\left(s-\varepsilon,s+\varepsilon\right)$
+\end_inset
+
+,
+ s čimer dokažemo konvergenco.
+\end_layout
+
+\begin_layout Proof
+Denimo sedaj,
+ da 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ pada.
+ Dokaz je povsem analogen.
+ Pišimo 
+\begin_inset Formula $m\coloneqq\inf_{n\in\mathbb{N}}a_{n}$
+\end_inset
+
+.
+ Vzemimo poljuben 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $m+\varepsilon$
+\end_inset
+
+ ni spodnja meja za 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+,
+ zato 
+\begin_inset Formula $\exists n_{0}\in\mathbb{N}\ni:m+\varepsilon>a_{n_{0}}$
+\end_inset
+
+.
+ Ker pa je 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ padajoče,
+ sledi 
+\begin_inset Formula $\forall n\geq n_{0}:a_{n}\leq a_{n_{0}}<m+\varepsilon$
+\end_inset
+
+.
+ Hkrati je 
+\begin_inset Formula $a_{n}\geq m$
+\end_inset
+
+,
+ saj je 
+\begin_inset Formula $m$
+\end_inset
+
+ spodnja meja.
+ Torej 
+\begin_inset Formula $\forall n\geq n_{0}:a_{n}\in[m,m+\varepsilon)\subset\left(m-\varepsilon,m+\varepsilon\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Corollary*
+Za monotono zaporedje velja,
+ da je v 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+ konvergentno natanko tedaj,
+ ko je omejeno.
+\end_layout
+
+\begin_layout Example*
+Naj bo,
+ kot prej,
+ 
+\begin_inset Formula $a>0$
+\end_inset
+
+ in 
+\begin_inset Formula $x_{0}=0,x_{n+1}=\sqrt{a+x_{n}}$
+\end_inset
+
+.
+ Dokažimo,
+ da je 
+\begin_inset Formula $\left(x_{n}\right)_{n}$
+\end_inset
+
+ konvergentno.
+ Dovolj je pokazati,
+ da je naraščajoče in navzgor omejeno.
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Naraščanje z indukcijo:
+ Baza:
+ 
+\begin_inset Formula $0=x_{0}>x_{1}=\sqrt{a}$
+\end_inset
+
+.
+ Dokažimo 
+\begin_inset Formula $x_{n+1}-x_{n}>0$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+\left(x_{n+1}-x_{n}\right)\left(x_{n+1}+x_{n}\right)=x_{n+1}^{2}-x_{n}^{2}=\left(a+x_{n}\right)-\left(a+x_{n-1}\right)=x_{n}-x_{n-1}
+\]
+
+\end_inset
+
+Ker je zaporedje pozitivno,
+ je 
+\begin_inset Formula $x_{n+1}+x_{n}>0$
+\end_inset
+
+.
+ Desna stran je po I.
+ P.
+ pozitivna,
+ torej tudi 
+\begin_inset Formula $x_{n+1}-x_{n}>0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Omejenost:
+ Če je zaporedje res omejeno,
+ je po zgornjem tudi konvergentno in je 
+\begin_inset Formula $\sup_{n\in\mathbb{N}}x_{n}=\lim_{n\to\infty}x_{n}=\frac{1+\sqrt{1+4a}}{2}\leq\frac{1+\sqrt{1+4a+4a^{2}}}{2}=\frac{1+\sqrt{\left(2a+1\right)^{2}}}{}=1+a$
+\end_inset
+
+.
+ Uganili smo neko zgornjo mejo.
+ Domnevamo,
+ da 
+\begin_inset Formula $\forall n\in\mathbb{N}:x_{n}\leq1+a$
+\end_inset
+
+.
+ Dokažimo to z indukcijo:
+ Baza:
+ 
+\begin_inset Formula $0=x_{0}<1+a$
+\end_inset
+
+.
+ Po I.
+ P.
+ 
+\begin_inset Formula $x_{n}>1+a$
+\end_inset
+
+.
+ Korak:
+\begin_inset Formula 
+\[
+x_{n+1}=\sqrt{x_{n}+a}\leq\sqrt{1+a+a}=\sqrt{1+2a}<\sqrt{1+2a+2a^{2}}=1+a
+\]
+
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Example*
+S tem smo dokazali,
+ da 
+\begin_inset Formula $\lim_{n\to\infty}x_{n}=\frac{1+\sqrt{1+4a}}{2}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+To lahko dokažemo tudi na alternativen način.
+ Vidimo,
+ da je edini kandidat za limito,
+ če obstaja 
+\begin_inset Formula $L=\frac{1+\sqrt{1+4a}}{2}$
+\end_inset
+
+ in da torej velja 
+\begin_inset Formula $L^{2}=a+L$
+\end_inset
+
+.
+ Preverimo,
+ da je 
+\begin_inset Formula $L$
+\end_inset
+
+ res limita:
+ 
+\begin_inset Formula 
+\[
+x_{n+1}-L=\sqrt{a+x_{n}}-L=\frac{\left(\sqrt{a+x_{n}}-L\right)\left(\sqrt{a+x_{n}}+L\right)}{\sqrt{a+x_{n}}+L}=\frac{\left(a+x_{n}\right)-L^{2}}{\sqrt{a+x_{n}}+L}=\frac{\left(a+x_{n}\right)-\left(a+L\right)}{\sqrt{a+x_{n}}+L}=\frac{x_{n}-L}{\sqrt{a+x_{n}}+L}.
+\]
+
+\end_inset
+
+Vpeljimo sedaj 
+\begin_inset Formula $y_{n}\coloneqq x_{n}-L$
+\end_inset
+
+.
+ Sledi 
+\begin_inset Formula $\left|y_{n+1}\right|\leq\frac{\left|y_{n}\right|}{\sqrt{a+x_{n}}+L}\leq\frac{\left|y_{n}\right|}{L}$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $\left|y_{0}\right|=L$
+\end_inset
+
+,
+ dobimo
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+Za razumevanje si oglej nekaj členov rekurzivnega zaporedje 
+\begin_inset Formula $y_{0}=L,y_{n}=\frac{\left|y_{n+1}\right|}{L}$
+\end_inset
+
+.
+ Začnemo z 1 in nato vsakič delimo z 
+\begin_inset Formula $L$
+\end_inset
+
+.
+\end_layout
+
+\end_inset
+
+ oceno 
+\begin_inset Formula $\left|y_{n}\right|\leq\frac{1}{L^{n-1}}$
+\end_inset
+
+ oziroma 
+\begin_inset Formula $\left|x_{n}-L\right|\leq\frac{1}{L^{n-1}}$
+\end_inset
+
+.
+ Ker iz definicije 
+\begin_inset Formula $L$
+\end_inset
+
+ sledi 
+\begin_inset Formula $L>1$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $L^{n}\to\infty$
+\end_inset
+
+ za 
+\begin_inset Formula $n\to\infty$
+\end_inset
+
+,
+ torej smo dokazali,
+ da 
+\begin_inset Formula $\left|x_{n}-L\right|$
+\end_inset
+
+ eksponentno pada proti 0 za 
+\begin_inset Formula $n\to\infty$
+\end_inset
+
+.
+ Eksponentno padanje 
+\begin_inset Formula $\left|x_{n}-L\right|$
+\end_inset
+
+ proti 0 je dovolj,
+ da rečemo,
+ da zaporedje konvergira k 
+\begin_inset Formula $L$
+\end_inset
+
+
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+a res,
+ vprašaj koga.
+ ne razumem.
+ zakaj.
+ TODO.
+\end_layout
+
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+\begin_inset Formula $\lim_{n\to\infty}\sin n$
+\end_inset
+
+ in 
+\begin_inset Formula $\lim_{n\to\infty}\cos n$
+\end_inset
+
+ ne obstajata.
+\end_layout
+
+\begin_layout Proof
+Pišimo 
+\begin_inset Formula $a_{n}=\sin n$
+\end_inset
+
+ in 
+\begin_inset Formula $b_{n}=\cos n$
+\end_inset
+
+.
+ Iz adicijskih izrekov dobimo 
+\begin_inset Formula $a_{n+1}=\sin\left(n+1\right)=\sin n\cos1+\cos n\sin1=a_{n}\cos1+b_{n}\sin1$
+\end_inset
+
+.
+ Torej 
+\begin_inset Formula $b_{n}=\frac{a_{n+1}-a_{n}\cos1}{\sin1}$
+\end_inset
+
+.
+ Torej če 
+\begin_inset Formula $\exists a\coloneqq\lim_{n\to\infty}a_{n},a\in\mathbb{R}$
+\end_inset
+
+,
+ potem tudi 
+\begin_inset Formula $\exists b\coloneqq\lim_{n\to\infty}b_{n},b\in\mathbb{R}$
+\end_inset
+
+.
+ Podobno iz adicijske formule za 
+\begin_inset Formula $\cos\left(n+1\right)$
+\end_inset
+
+ sledi 
+\begin_inset Formula $a_{n}=\frac{b_{n}\cos1-b_{n+1}}{\sin1}$
+\end_inset
+
+,
+ torej če 
+\begin_inset Formula $\exists b$
+\end_inset
+
+,
+ potem tudi 
+\begin_inset Formula $\exists a$
+\end_inset
+
+.
+ Iz obojega sledi,
+ da 
+\begin_inset Formula $\exists a\Leftrightarrow\exists b$
+\end_inset
+
+.
+ Posledično,
+ če 
+\begin_inset Formula $a$
+\end_inset
+
+ in 
+\begin_inset Formula $b$
+\end_inset
+
+ obstajata,
+ iz zgornjih obrazcev za 
+\begin_inset Formula $a_{n}$
+\end_inset
+
+ in 
+\begin_inset Formula $b_{n}$
+\end_inset
+
+ sledi,
+ da za
+\begin_inset Formula 
+\[
+\lambda=\frac{1-\cos1}{\sin1}\in\left(0,1\right)
+\]
+
+\end_inset
+
+velja 
+\begin_inset Formula $b=\lambda a$
+\end_inset
+
+ in 
+\begin_inset Formula $a=-\lambda b$
+\end_inset
+
+ in zato 
+\begin_inset Formula $b=\lambda\left(-\lambda b\right)$
+\end_inset
+
+ oziroma 
+\begin_inset Formula $1=-\lambda^{2}$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $-1=\lambda^{2}$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\lambda=i$
+\end_inset
+
+,
+ kar je v protislovju z 
+\begin_inset Formula $\lambda\in\left(0,1\right)$
+\end_inset
+
+.
+ Podobno za 
+\begin_inset Formula $a=-\lambda\left(\lambda a\right)$
+\end_inset
+
+ oziroma 
+\begin_inset Formula $1=-\lambda^{2}$
+\end_inset
+
+,
+ kar je zopet 
+\begin_inset Formula $\rightarrow\!\leftarrow$
+\end_inset
+
+.
+ Edina druga opcija je,
+ da je 
+\begin_inset Formula $a=b=0$
+\end_inset
+
+.
+ Hkrati pa vemo,
+ da 
+\begin_inset Formula $a_{n}^{2}+b_{n}^{2}=1$
+\end_inset
+
+,
+ zato 
+\begin_inset Formula $a+b=1$
+\end_inset
+
+,
+ kar ni mogoče za ničelna 
+\begin_inset Formula $a$
+\end_inset
+
+ in 
+\begin_inset Formula $b$
+\end_inset
+
+.
+ Torej 
+\begin_inset Formula $a$
+\end_inset
+
+ in 
+\begin_inset Formula $b$
+\end_inset
+
+ ne obstajata.
+\end_layout
+
+\begin_layout Subsection
+Eulerjevo število
+\end_layout
+
+\begin_layout Theorem*
+Bernoullijeva neenakost.
+ 
+\begin_inset Formula $\forall\alpha\leq1,n\in\mathbb{N}$
+\end_inset
+
+ velja 
+\begin_inset Formula $\left(1-\alpha\right)^{n}\geq1-n\alpha$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Z indukcijo na 
+\begin_inset Formula $n$
+\end_inset
+
+ ob fiksnem 
+\begin_inset Formula $\alpha$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Baza:
+ 
+\begin_inset Formula $n=1$
+\end_inset
+
+:
+ 
+\begin_inset Formula $\left(1-\alpha\right)^{1}=1-1\alpha$
+\end_inset
+
+.
+ Velja celo enakost.
+\end_layout
+
+\begin_layout Itemize
+I.
+ P.:
+ Velja 
+\begin_inset Formula $\left(1-\alpha\right)^{n}\geq1-n\alpha$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+Korak:
+ 
+\begin_inset Formula $\left(1-\alpha\right)^{n+1}=\left(1-\alpha\right)\left(1-\alpha\right)^{n}\geq\left(1-\alpha\right)$
+\end_inset
+
+
+\begin_inset Formula $\left(1-n\alpha\right)=1-n\alpha-\alpha-n\alpha^{2}=1-\left(n+1\right)\alpha-n\alpha^{2}\geq1-\left(n+1\right)\alpha$
+\end_inset
+
+,
+ torej ocena velja tudi za 
+\begin_inset Formula $n+1$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Definition*
+Vpeljimo oznaki:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Za 
+\begin_inset Formula $n\in\mathbb{N}$
+\end_inset
+
+ označimo 
+\begin_inset Formula $n!=1\cdot2\cdot3\cdot\cdots\cdot n$
+\end_inset
+
+ (pravimo 
+\begin_inset Formula $n-$
+\end_inset
+
+faktoriala oziroma 
+\begin_inset Formula $n-$
+\end_inset
+
+fakulteta).
+ Ker velja 
+\begin_inset Formula $n!=n\cdot\left(n-1\right)!$
+\end_inset
+
+ za 
+\begin_inset Formula $n\geq2$
+\end_inset
+
+,
+ je smiselno definirati še 
+\begin_inset Formula $0!=1$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Za 
+\begin_inset Formula $n,k\in\mathbb{N}$
+\end_inset
+
+ označimo še binomski simbol:
+ 
+\begin_inset Formula $\binom{n}{k}\coloneqq\frac{n!}{k!\left(n-k\right)!}$
+\end_inset
+
+ (pravimo 
+\begin_inset Formula $n$
+\end_inset
+
+ nad 
+\begin_inset Formula $k$
+\end_inset
+
+).
+\end_layout
+
+\begin_layout Itemize
+Če je 
+\begin_inset Formula $\left(a_{k}\right)_{k}$
+\end_inset
+
+ neko zaporedje (lahko tudi končno),
+ lahko pišemo 
+\begin_inset Formula $\sum_{k=1}^{n}a_{k}\coloneqq a_{1}+a_{2}+\cdots+a_{n}$
+\end_inset
+
+ (pravimo summa) in 
+\begin_inset Formula $\prod_{k=1}^{n}a_{k}\coloneqq a_{1}\cdot\cdots\cdot a_{n}$
+\end_inset
+
+ (pravimo produkt).
+\end_layout
+
+\end_deeper
+\begin_layout Example*
+\begin_inset Formula $\sum_{k=1}^{n}\frac{1}{k}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}$
+\end_inset
+
+ in 
+\begin_inset Formula $\prod_{k=1}^{n}k=1\cdot2\cdot3\cdot\cdots\cdot n=n!$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+Binomska formula.
+ 
+\begin_inset Formula $\forall a,b\in\mathbb{R},n\in\mathbb{N}:\left(a+b\right)^{n}=\sum_{k=0}^{n}\binom{n}{k}a^{k}b^{n-k}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Indukcija po 
+\begin_inset Formula $n$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Baza 
+\begin_inset Formula $n=1$
+\end_inset
+
+:
+ 
+\begin_inset Formula $\sum_{k=0}^{1}\binom{n}{k}a^{k}b^{n-k}=\binom{1}{0}a^{0}b^{1-0}+\binom{1}{1}a^{1}b^{1-1}=a+b=\left(a+b\right)^{1}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+I.
+ P.
+ 
+\begin_inset Formula $\sum_{k=0}^{n}\binom{n}{k}a^{k}b^{n-k}=\left(a+b\right)^{n}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+Korak:
+ 
+\begin_inset Formula 
+\[
+\left(a+b\right)^{n+1}=\left(a+b\right)\left(a+b\right)^{n}=\left(a+b\right)\sum_{k=0}^{n}\binom{n}{k}a^{k}b^{n-k}=\sum_{k=0}^{n}\binom{n}{k}a^{k+1}b^{n-k}+\sum_{k=0}^{n}\binom{n}{k}a^{k}b^{n-k+1}=
+\]
+
+\end_inset
+
+sedaj naj bo 
+\begin_inset Formula $m=k+1$
+\end_inset
+
+ v levem členu:
+\begin_inset Formula 
+\[
+=\sum_{m=1}^{n+1}\binom{n}{m-1}a^{m}b^{n-\left(m-1\right)}+\sum_{k=0}^{n}\binom{n}{k}a^{k}b^{n-k+1}=a^{n+1}+\sum_{k=1}^{n}\left[\binom{n}{k-1}+\binom{n}{k}\right]a^{k}b^{n-k+1}+b^{n+1}=
+\]
+
+\end_inset
+
+Sedaj obravnavajmo le izraz v oglatih oklepajih:
+\begin_inset Formula 
+\[
+\binom{n}{k-1}+\binom{n}{k}=\frac{n!}{\left(k-1\right)!\left(n-k+1\right)!}+\frac{n!}{k!\left(n-k\right)!}=\frac{kn!}{k!\left(n-k+1\right)!}+\frac{n!\left(n-k+1\right)}{k!\left(n-k+1\right)!}=\frac{kn!+n!\left(n-k+1\right)}{k!\left(n-k+1\right)!}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\frac{n!\left(\cancel{k+}n\cancel{-k}+1\right)}{k!\left(n+1-k\right)!}=\frac{n!\left(n+1\right)}{k!\left(n+1-k\right)!}=\frac{\left(n+1\right)!}{k!\left(n+1-k\right)}=\binom{n+1}{k}
+\]
+
+\end_inset
+
+in skratka dobimo 
+\begin_inset Formula $\binom{n}{k-1}+\binom{n}{k}=\binom{n+1}{k}$
+\end_inset
+
+.
+ Vstavimo to zopet v naš zgornji račun:
+\begin_inset Formula 
+\[
+\cdots=a^{n+1}+\sum_{k=1}^{n}\left[\binom{n}{k-1}+\binom{n}{k}\right]a^{k}b^{n-k+1}+b^{n+1}=a^{n+1}+\sum_{k=1}^{n}\binom{n+1}{k}a^{k}b^{n-k+1}+b^{n+1}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=a^{n+1}+\sum_{k=1}^{n}\binom{n+1}{k}a^{k}b^{n-k+1}+b^{n+1}=a^{n+1}+\sum_{k=0}^{n}\binom{n+1}{k}a^{k}b^{n-k+1}=\sum_{k=0}^{n+1}\binom{n+1}{k}a^{k}b^{n-k+1}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Theorem*
+Bernoulli.
+ Zaporedje 
+\begin_inset Formula $a_{n}\coloneqq\left(1+\frac{1}{n}\right)^{n}$
+\end_inset
+
+ je konvergentno.
+\end_layout
+
+\begin_layout Proof
+Dokazali bomo,
+ da je naraščajoče in omejeno.
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Naraščanje:
+ Dokazujemo,
+ da za 
+\begin_inset Formula $n\geq2$
+\end_inset
+
+ velja 
+\begin_inset Formula $a_{n}\geq a_{n-1}$
+\end_inset
+
+ oziroma 
+\begin_inset Formula 
+\[
+\left(1+\frac{1}{n}\right)^{n}\overset{?}{\geq}\left(1+\frac{1}{n-1}\right)^{n-1}=\left(\frac{n-1}{n-1}+\frac{1}{n-1}\right)^{n-1}=\left(\frac{n}{n-1}\right)^{n-1}=\left(\frac{n-1}{n}\right)^{1-n}=\left(1-\frac{1}{n}\right)^{1-n}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\left(1+\frac{1}{n}\right)^{n}\overset{?}{\geq}\left(1-\frac{1}{n}\right)^{1-n}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\left(1+\frac{1}{n^{2}}\right)^{n}=\left(\left(1+\frac{1}{n}\right)\left(1-\frac{1}{n}\right)\right)^{n}=\left(1+\frac{1}{n}\right)^{n}\left(1-\frac{1}{n}\right)^{n}\overset{?}{\geq}1-\frac{1}{n}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\left(1+\frac{1}{n^{2}}\right)^{n}\overset{?}{\geq}1-\frac{1}{n},
+\]
+
+\end_inset
+
+kar je poseben primer Bernoullijeve neenakosti za 
+\begin_inset Formula $\alpha=\frac{1}{n^{2}}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Omejenost:
+ Po binomski formuli je 
+\begin_inset Formula 
+\[
+a_{n}=\left(1+\frac{1}{n}\right)^{n}=\sum_{k=0}^{n}\binom{n}{k}\left(\frac{1}{n}\right)^{k}=\sum_{k=0}^{n}\frac{n!}{k!\left(n-k\right)!n^{k}}=1+1+\sum_{k=2}^{n}\frac{n!}{k!\left(n-k\right)!n^{k}}=2+\sum_{k=2}^{n}\frac{1}{k!}\cdot\frac{n\left(n-1\right)\left(n-2\right)\cdots\left(n-k+1\right)}{n^{k}}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=2+\sum_{k=2}^{n}\frac{1}{k!}\cdot\frac{n}{n}\cdot\frac{n-1}{n}\cdot\frac{n-2}{n}\cdot\cdots\cdot\frac{n-k+1}{n}=2+\sum_{k=2}^{n}\frac{1}{k!}\cdot\cancel{\left(1-0\right)}\cdot\left(1-\frac{1}{n}\right)\cdot\left(1-\frac{2}{n}\right)\cdot\cdots\cdot\left(1-\frac{k-1}{n}\right)=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=2+\sum_{k=2}^{n}\frac{1}{k!}\prod_{j=1}^{k-1}\left(1-\frac{j}{n}\right)<2+\sum_{k=2}^{n}\frac{1}{k!}<2+\sum_{k=2}^{n}\frac{1}{2^{k-1}}=\cdots
+\]
+
+\end_inset
+
+Opomnimo,
+ da je 
+\begin_inset Formula $1-\frac{j}{n}<1$
+\end_inset
+
+,
+ zato 
+\begin_inset Formula $\prod_{j=1}^{k-1}\left(1-\frac{j}{n}\right)<1$
+\end_inset
+
+ (prvi neenačaj) ter 
+\begin_inset Formula $k!=1\cdot2\cdot3\cdot\cdots\cdot k\geq1\cdot2\cdot2\cdot\cdots\cdot2=2^{k-1}$
+\end_inset
+
+ (drugi).
+ Sedaj si z indukcijo dokažimo 
+\begin_inset Formula $\sum_{k=2}^{n}\frac{1}{2^{k-1}}=1-\frac{1}{2^{n-1}}$
+\end_inset
+
+:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Baza:
+ 
+\begin_inset Formula $n=2$
+\end_inset
+
+:
+ 
+\begin_inset Formula $\frac{1}{2^{2-1}}=1-\frac{1}{2^{2-1}}=1-\frac{1}{2}=\frac{1}{2}$
+\end_inset
+
+.
+ Velja!
+\end_layout
+
+\begin_layout Itemize
+I.
+ P.:
+ 
+\begin_inset Formula $\sum_{k=2}^{n}\frac{1}{2^{k-1}}=1-\frac{1}{2^{n-1}}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+Korak:
+ 
+\begin_inset Formula $\sum_{k=2}^{n+1}\frac{1}{2^{k-1}}=1-\frac{1}{2^{n-1}}+\frac{1}{2^{n+1-1}}=1-2\cdot2^{-n}+2^{-n}=1+2^{-n}\left(1-2\right)=1+2^{-n}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+In nadaljujmo z računanjem:
+\begin_inset Formula 
+\[
+\cdots=2+1-\frac{1}{2^{n-1}}=3-\frac{1}{2^{n-1}},
+\]
+
+\end_inset
+
+s čimer dobimo zgornjo mejo 
+\begin_inset Formula $\forall n\in\mathbb{N}:a_{n}<3$
+\end_inset
+
+.
+ Ker je očitno 
+\begin_inset Formula $\forall n\in\mathbb{N}:a_{n}>0$
+\end_inset
+
+,
+ je torej zaporedje omejeno in ker je tudi monotono po prejšnjem izreku konvergira.
+\end_layout
+
+\end_deeper
+\end_deeper
+\begin_layout Definition*
+Označimo število 
+\begin_inset Formula $e\coloneqq\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{n}$
+\end_inset
+
+ in ga imenujemo Eulerjevo število.
+ Velja 
+\begin_inset Formula $e\approx2,71828$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Remark*
+V dokazu vidimo moč izreka 
+\begin_inset Quotes gld
+\end_inset
+
+omejenost in monotonost 
+\begin_inset Formula $\Rightarrow$
+\end_inset
+
+ konvergenca
+\begin_inset Quotes grd
+\end_inset
+
+,
+ saj nam omogoča dokazati konvergentnost zaporedja brez kandidata za limito.
+ Jasno je,
+ da ne bi mogli vnaprej uganiti,
+ da je limita ravno 
+\shape italic
+transcendentno število
+\shape default
+ 
+\begin_inset Formula $e$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Podzaporedje zaporedja 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ je poljubno zaporedje oblike 
+\begin_inset Formula $\left(a_{\varphi\left(n\right)}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+,
+ kjer je 
+\begin_inset Formula $\varphi:\mathbb{N}\to\mathbb{N}$
+\end_inset
+
+ strogo naraščajoča funkcija.
+ Definicijsko območje 
+\begin_inset Formula $\varphi$
+\end_inset
+
+ mora vsebovati števno neskončno elementov.
+\end_layout
+
+\begin_layout Theorem*
+Če je 
+\begin_inset Formula $L=\lim_{n\to\infty}a_{n}$
+\end_inset
+
+,
+ tedaj je 
+\begin_inset Formula $L$
+\end_inset
+
+ tudi limita vsakega podzaporedja.
+\end_layout
+
+\begin_layout Proof
+Po predpostavki velja 
+\begin_inset Formula $\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\forall n\in\mathbb{N}:n\geq n_{0}:\left|a_{n}-L\right|<\varepsilon$
+\end_inset
+
+.
+ Vzemimo poljuben 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+.
+ Po predpostavki obstaja 
+\begin_inset Formula $n_{0}\in\mathbb{N}$
+\end_inset
+
+,
+ da bodo vsi členi zaporedja po 
+\begin_inset Formula $n_{0}-$
+\end_inset
+
+tem v 
+\begin_inset Formula $\left(L-\varepsilon,L+\varepsilon\right)$
+\end_inset
+
+.
+ Iz definicijskega območja 
+\begin_inset Formula $\varphi$
+\end_inset
+
+ vzemimo poljuben element 
+\begin_inset Formula $n_{1}$
+\end_inset
+
+,
+ da velja 
+\begin_inset Formula $n_{1}\geq n_{0}$
+\end_inset
+
+.
+ Gotovo obstaja,
+ ker je definicijsko območje števno neskončne moči in s pogojem 
+\begin_inset Formula $n_{1}\geq n_{0}$
+\end_inset
+
+ onemogočimo izbiro le končno mnogo elementov.
+ 
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+Če slednji ne obstaja,
+ je v 
+\begin_inset Formula $D_{\varphi}$
+\end_inset
+
+ končno mnogo elementov,
+ tedaj vzamemo 
+\begin_inset Formula $n_{1}\coloneqq\max D_{\varphi}+1$
+\end_inset
+
+ in je pogoj za limito izpolnjen na prazno.
+ Sicer pa v
+\end_layout
+
+\end_inset
+
+Velja 
+\begin_inset Formula $\forall n\in\mathbb{N}:n>n_{1}\Rightarrow\left|a_{\varphi n}-L\right|<\varepsilon$
+\end_inset
+
+,
+ ker je 
+\begin_inset Formula $\varphi$
+\end_inset
+
+ strogo naraščajoča in izbiramo le elemente podzaporedja,
+ ki so v izvornem zaporedju za 
+\begin_inset Formula $n_{0}-$
+\end_inset
+
+tim členom in zato v 
+\begin_inset Formula $\left(L-\varepsilon,L+\varepsilon\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $\lim_{n\to\infty}\frac{1}{2n+3}=\lim_{n\to\infty}\frac{1}{n}=0$
+\end_inset
+
+ za zaporedje 
+\begin_inset Formula $a_{n}=\frac{1}{n}$
+\end_inset
+
+ in podzaporedje 
+\begin_inset Formula $a_{\varphi n}$
+\end_inset
+
+,
+ kjer je 
+\begin_inset Formula $\varphi\left(n\right)=2n+3$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+Karakterizacija limite s podzaporedji.
+ Naj bo 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ realno zaporedje in 
+\begin_inset Formula $L\in\mathbb{R}$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $L=\lim_{n\to\infty}a_{n}\Leftrightarrow$
+\end_inset
+
+ za vsako podzaporedje 
+\begin_inset Formula $\left(a_{n_{k}}\right)_{k\in\mathbb{N}}$
+\end_inset
+
+ zaporedja 
+\begin_inset Formula $\left(a_{n}\right)_{n}$
+\end_inset
+
+ obstaja njegovo podzaporedje 
+\begin_inset Formula $\left(a_{n_{k_{l}}}\right)_{l\in\mathbb{N}}$
+\end_inset
+
+,
+ ki konvergira k 
+\begin_inset Formula $L$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Dokazujemo ekvivalenco:
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Rightarrow\right)$
+\end_inset
+
+ Dokazano poprej.
+ Limita se pri prehodu na podzaporedje ohranja.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Leftarrow\right)$
+\end_inset
+
+ PDDRAA 
+\begin_inset Formula $a_{n}\not\to L$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $\exists\varepsilon>0$
+\end_inset
+
+ in podzaporedje 
+\begin_inset Formula $\left(a_{n_{k}}\right)_{k\in\mathbb{N}}\ni:\forall k\in\mathbb{N}:\left|a_{n_{k}}-K\right|>\varepsilon$
+\end_inset
+
+ (*)
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+tu je na 
+\begin_inset Quotes gld
+\end_inset
+
+Zaporedja 2
+\begin_inset Quotes grd
+\end_inset
+
+ napaka,
+ neenačaj obrne v drugo smer
+\end_layout
+
+\end_inset
+
+.
+ Po predpostavki sedaj 
+\begin_inset Formula $\exists\left(a_{n_{k_{l}}}\right)_{l\in\mathbb{N}}\ni:\lim_{l\to\infty}a_{n_{k_{l}}}=L$
+\end_inset
+
+.
+ To pa je v protislovju z (*),
+ torej je začetna predpostavka 
+\begin_inset Formula $a_{n}\not\to L$
+\end_inset
+
+ napačna,
+ torej 
+\begin_inset Formula $a_{n}\to L$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Subsection
+Stekališča
+\end_layout
+
+\begin_layout Definition*
+Točka 
+\begin_inset Formula $s\in\mathbb{R}$
+\end_inset
+
+ je stekališče zaporedje 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}\subset\mathbb{R}$
+\end_inset
+
+,
+ če v vsaki okolici te točke leži neskončno členov zaporedja.
+\end_layout
+
+\begin_layout Remark*
+Pri limiti zahtevamo več;
+ da izven vsake okolice limite leži le končno mnogo členov.
+\end_layout
+
+\begin_layout Example*
+Primeri stekališč.
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $L=\lim_{n\to\infty}a_{n}\Rightarrow L$
+\end_inset
+
+ je stekališče za 
+\begin_inset Formula $a_{n}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $0,1,0,1,\dots$
+\end_inset
+
+ stekališči sta 
+\begin_inset Formula $\left\{ 0,1\right\} $
+\end_inset
+
+ in zaporedje nima limite (ni konvergentno)
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $1,1,2,1,2,3,1,2,3,4,\dots$
+\end_inset
+
+ ima neskončno stekališč,
+ 
+\begin_inset Formula $\mathbb{N}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $b_{n}=n-$
+\end_inset
+
+to racionalno število
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+Racionalnih števil je števno mnogo,
+ zato jih lahko linearno uredimo in oštevilčimo
+\end_layout
+
+\end_inset
+
+ ima neskončno stekališč,
+ 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Remark*
+Limita je stakališče,
+ stekališče pa ni nujno limita.
+ Poleg tega,
+ če se spomnimo,
+ velja,
+ da vsota konvergentnih zaporedij konvergira k vsoti njunih limit,
+ ni pa nujno res,
+ da so stekališča vsote dveh zaporedij paroma vsote stekališč teh dveh zaporedij.
+ Primer:
+ 
+\begin_inset Formula $a_{n}=\left(-1\right)^{n}$
+\end_inset
+
+ in 
+\begin_inset Formula $b_{n}=-\left(-1\right)^{n}$
+\end_inset
+
+.
+ Njuni stekališči sta 
+\begin_inset Formula $\left\{ -1,1\right\} $
+\end_inset
+
+,
+ toda 
+\begin_inset Formula $a_{n}+b_{n}=0$
+\end_inset
+
+ ima le stekališče 
+\begin_inset Formula $\left\{ 0\right\} $
+\end_inset
+
+,
+ ne pa tudi 
+\begin_inset Formula $\left\{ 1,-1,2,-2\right\} $
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Corollary*
+sssssssssss
+\end_layout
+
+\begin_layout Corollary*
+sssssssssss
+\end_layout
+
+\begin_layout Corollary*
+sssssssssss
+\end_layout
+
+\begin_layout Corollary*
+sssssssssss
+\end_layout
+
+\begin_layout Corollary*
+sssssssssss
+\end_layout
+
+\begin_layout Corollary*
+sssssssssss
+\end_layout
+
+\begin_layout Corollary*
+sssssssssss
+\end_layout
+
+\begin_layout Corollary*
+sssssssssss
+\end_layout
+
+\begin_layout Corollary*
+sssssssssss
+\end_layout
+
+\end_body
+\end_document
-- 
cgit v1.2.3