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+#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]{%
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+\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"
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+\cite_engine basic
+\cite_engine_type default
+\biblio_style plain
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+\paperorientation portrait
+\suppress_date false
+\justification false
+\use_refstyle 1
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+\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
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+\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 Theorem*
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hypertarget{pmkdlim}{Naj bosta}
+\end_layout
+
+\end_inset
+
+ 
+\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*
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hypertarget{kmoz}{Konvergenca monotonega in omejenega zaporedja}
+\end_layout
+
+\end_inset
+
+.
+ 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 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{kmoz}{prejšnjem izreku}
+\end_layout
+
+\end_inset
+
+ 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.
+\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 Theorem*
+\begin_inset Formula $S$
+\end_inset
+
+ je stekališče 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}\Leftrightarrow S$
+\end_inset
+
+ je limita nekega podzaporedja 
+\begin_inset Formula $a_{n}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Dokazujemo ekvivalenco.
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Leftarrow\right)$
+\end_inset
+
+ Očitno.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Rightarrow\right)$
+\end_inset
+
+ Definirajmo 
+\begin_inset Formula $\forall m\in\mathbb{N}:U_{m}\coloneqq\left(S-\frac{1}{m},S+\frac{1}{m}\right)$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $S$
+\end_inset
+
+ stekališče,
+ 
+\begin_inset Formula $\forall m\in\mathbb{N}\exists a_{k_{m}}\in U_{m}$
+\end_inset
+
+.
+ Podzaporedje 
+\begin_inset Formula $\left(a_{k_{m}}\right)_{m\in\mathbb{N}}$
+\end_inset
+
+ konvergira k 
+\begin_inset Formula $S$
+\end_inset
+
+,
+ kajti 
+\begin_inset Formula $\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow\left|a_{k_{n}}-S\right|<\frac{1}{n}<\varepsilon$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Corollary*
+Če je 
+\begin_inset Formula $L$
+\end_inset
+
+ limita nekega zaporedja,
+ je 
+\begin_inset Formula $L$
+\end_inset
+
+ edino njegovo stekališče.
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $a_{n}\to L$
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $S$
+\end_inset
+
+ stekališče za 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+.
+ Po izreku zgoraj je 
+\begin_inset Formula $S$
+\end_inset
+
+ limita nekega podzaporedja 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+.
+ Toda limita vsakega podzaporedja je enaka limiti zaporedja,
+ iz katerega to podzaporedje izhaja,
+ če ta limita obstaja.
+ Potemtakem je 
+\begin_inset Formula $S=L$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hypertarget{bw}{Bolzano-Weierstraß}
+\end_layout
+
+\end_inset
+
+.
+ Eksistenčni izrek.
+ Vsako omejeno zaporedje v realnih številih ima kakšno stekališče v 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Označimo 
+\begin_inset Formula $m_{0}\coloneqq\inf_{n\in\mathbb{N}}a_{n},M_{0}\coloneqq\sup_{n\in\mathbb{N}}a_{n},I_{0}\coloneqq\left[m_{0},M_{0}\right]$
+\end_inset
+
+.
+ Očitno je 
+\begin_inset Formula $\forall n\in\mathbb{N}:a_{n}\in I_{0}$
+\end_inset
+
+.
+ Interval 
+\begin_inset Formula $I_{0}$
+\end_inset
+
+ razdelimo na dve polovici:
+ 
+\begin_inset Formula $I_{0}=\left[m_{0},\frac{m_{0}+M_{0}}{2}\right]\cup\left[\frac{m_{0}+M_{0}}{2},M_{0}\right]$
+\end_inset
+
+.
+ Izberemo polovico (vsaj ena obstaja),
+ v kateri leži neskončno mnogo členov,
+ in jo označimo z 
+\begin_inset Formula $I_{1}$
+\end_inset
+
+.
+ Spet jo razdelimo na pol in z 
+\begin_inset Formula $I_{2}$
+\end_inset
+
+ označimo tisto polovico,
+ v kateri leži neskončno mnogo členov.
+ Postopek ponavljamo in dobimo zaporedje zaprtih intervalov 
+\begin_inset Formula $\left(I_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ in velja 
+\begin_inset Formula $I_{0}\supset I_{1}\supset I_{2}\supset\cdots$
+\end_inset
+
+ ter 
+\begin_inset Formula $\left|I_{n}\right|=2^{-n}\left|I_{0}\right|$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Označimo sedaj 
+\begin_inset Formula $I_{n}\eqqcolon\left[l_{n},d_{n}\right]$
+\end_inset
+
+.
+ Iz konstrukcije je očitno,
+ da 
+\begin_inset Formula $\left(l_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ narašča in 
+\begin_inset Formula $\left(d_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ pada ter da sta obe zaporedji omejeni.
+ Posledično 
+\begin_inset Formula $\exists l\coloneqq\lim_{n\to\infty}l_{n},d\coloneqq\lim_{n\to\infty}d_{n}$
+\end_inset
+
+.
+ Iz 
+\begin_inset Formula $l_{n}\leq l\leq d\leq d_{n}$
+\end_inset
+
+ sledi ocena 
+\begin_inset Formula $d-l\leq l_{n}-d_{n}=\left|I_{n}\right|=2^{-n}\left|I_{0}\right|$
+\end_inset
+
+,
+ kar konvergira k 0.
+ Posledično 
+\begin_inset Formula $d=l$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Treba je pokazati še,
+ da je 
+\begin_inset Formula $d=l$
+\end_inset
+
+ stekališče za 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+.
+ Vzemimo poljuben 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $l=\lim_{n\to\infty}l_{n}\Rightarrow\exists n_{1}\in\mathbb{N}\ni:l_{n_{1}}>l-\varepsilon$
+\end_inset
+
+ in ker je 
+\begin_inset Formula $d=\lim_{n\to\infty}d_{n}\Rightarrow\exists n_{2}\in\mathbb{N}\ni:d_{n_{2}}<d-\varepsilon$
+\end_inset
+
+.
+ Torej 
+\begin_inset Formula $\left[l_{n_{1}},d_{n_{2}}\right]\subset\left(l-\varepsilon,d+\varepsilon\right)$
+\end_inset
+
+.
+ Torej za 
+\begin_inset Formula $n_{0}\coloneqq\max\left\{ n_{1},n_{2}\right\} $
+\end_inset
+
+ velja 
+\begin_inset Formula $I_{n_{0}}=\left[l_{n_{0}},d_{n_{n}}\right]\subset\left(l-\varepsilon,d+\varepsilon\right)$
+\end_inset
+
+.
+ Ker 
+\begin_inset Formula $I_{n_{0}}$
+\end_inset
+
+ po konstrukciji vsebuje neskončno mnogo elementov,
+ jih torej tudi 
+\begin_inset Formula $\left(l-\varepsilon,d+\varepsilon\right)$
+\end_inset
+
+ oziroma poljubno majhna okolica 
+\begin_inset Formula $d=l$
+\end_inset
+
+,
+ torej je 
+\begin_inset Formula $d=l$
+\end_inset
+
+ stekališče za 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Corollary*
+Če je 
+\begin_inset Formula $s\in\mathbb{R}$
+\end_inset
+
+ edino stekališče omejenega zaporedja 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+,
+ tedaj je 
+\begin_inset Formula $s=\lim_{n\to\infty}a_{n}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $s$
+\end_inset
+
+ stekališče 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+.
+ PDDRAA 
+\begin_inset Formula $a_{n}\not\to s$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $\exists\varepsilon>0\ni:$
+\end_inset
+
+ izven 
+\begin_inset Formula $\left(s-\varepsilon,s+\varepsilon\right)$
+\end_inset
+
+ se nahaja neskončno mnogo členov zaporedja.
+ Ti členi sami zase tvorijo omejeno zaporedje,
+ ki ima po 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{bw}{B.-W.}
+\end_layout
+
+\end_inset
+
+ izreku stekališče.
+ Slednje gotovo ne more biti enako 
+\begin_inset Formula $s$
+\end_inset
+
+,
+ torej imamo vsaj dve stekališči,
+ kar je v je v 
+\begin_inset Formula $\rightarrow\!\leftarrow$
+\end_inset
+
+ s predpostavko.
+\end_layout
+
+\begin_layout Definition*
+Pravimo,
+ da ima realno zaporedje:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+stekališče v 
+\begin_inset Formula $\infty$
+\end_inset
+
+,
+ če 
+\begin_inset Formula $\forall M>0:\left(M,\infty\right)$
+\end_inset
+
+ vsebuje neskončno mnogo členov zapopredja
+\end_layout
+
+\begin_layout Itemize
+limito v 
+\begin_inset Formula $\infty$
+\end_inset
+
+,
+ če 
+\begin_inset Formula $\forall M>0:\left(M,\infty\right)$
+\end_inset
+
+ vsebuje vse člene zaporedja od nekega indeksa dalje
+\end_layout
+
+\begin_layout Standard
+in podobno za 
+\begin_inset Formula $-\infty$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Remark*
+Povezava s pojmom realnega stekališča/limite:
+ okolice 
+\begin_inset Quotes gld
+\end_inset
+
+točke
+\begin_inset Quotes grd
+\end_inset
+
+ 
+\begin_inset Formula $\infty$
+\end_inset
+
+ so intervali oblike 
+\begin_inset Formula $\left(M,\infty\right)$
+\end_inset
+
+.
+ To je smiselno,
+ saj biti 
+\begin_inset Quotes gld
+\end_inset
+
+blizu 
+\begin_inset Formula $\infty$
+\end_inset
+
+
+\begin_inset Quotes grd
+\end_inset
+
+ pomeni bizi zelo velik,
+ kar je ravno biti v 
+\begin_inset Formula $\left(M,\infty\right)$
+\end_inset
+
+za poljubno velik 
+\begin_inset Formula $M$
+\end_inset
+
+.
+ 
+\begin_inset Quotes gld
+\end_inset
+
+Okolica točke 
+\begin_inset Formula $\infty$
+\end_inset
+
+
+\begin_inset Quotes grd
+\end_inset
+
+ so torej vsi intervali oblike 
+\begin_inset Formula $\left(M,\infty\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsection
+Limes superior in limes inferior
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+realno zaporedje.
+ Tvorimo novo zaporedje 
+\begin_inset Formula $s_{n}\coloneqq\sup\left\{ a_{k};k\geq n\right\} $
+\end_inset
+
+.
+ Očitno je padajoče (
+\begin_inset Formula $s_{1}\geq s_{2}\geq s_{3}\geq\cdots$
+\end_inset
+
+),
+ ker je supremum množice vsaj supremum njene stroge podmnožice.
+ Zaporedje 
+\begin_inset Formula $\left(s_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ ima limito,
+ ki ji rečemo limes superior oziroma zgornja limita zaporedja 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ in označimo 
+\begin_inset Formula $\limsup_{n\to\infty}a_{n}=\overline{\lim_{n\to\infty}}a_{n}\coloneqq\lim_{n\to\infty}s_{n}$
+\end_inset
+
+ in velja,
+ da leži v 
+\begin_inset Formula $\mathbb{R}\cup\left\{ -\infty,\infty\right\} $
+\end_inset
+
+.
+ Podobno definiramo tudi limes inferior oz.
+ spodnjo limito zaporedja:
+ 
+\begin_inset Formula $\liminf_{n\to\infty}a_{n}=\underline{\lim_{n\to\infty}}a_{n}\coloneqq\lim_{n\to\infty}\left(\inf_{k\geq n}a_{k}\right)=\sup_{n\in\mathbb{N}}\left(\inf_{k\geq n}a_{k}\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Remark*
+Za razliko od običajne limite,
+ ki lahko ne obstaja,
+ 
+\begin_inset Formula $\limsup$
+\end_inset
+
+ in 
+\begin_inset Formula $\liminf$
+\end_inset
+
+ vedno obstajata.
+\end_layout
+
+\begin_layout Claim*
+\begin_inset Formula $\limsup_{n\to\infty}a_{n}$
+\end_inset
+
+ je največje stekališče zaporedja 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ in 
+\begin_inset Formula $\liminf_{n\to\infty}$
+\end_inset
+
+ najmanjše.
+\end_layout
+
+\begin_layout Proof
+Označimo 
+\begin_inset Formula $s\coloneqq\limsup_{n\to\infty}a_{n}$
+\end_inset
+
+.
+ Za 
+\begin_inset Formula $\liminf$
+\end_inset
+
+ je dokaz analogen in ga ne bomo pisali.
+ Dokazujemo,
+ da je 
+\begin_inset Formula $s$
+\end_inset
+
+ stekališče in 
+\begin_inset Formula $\forall t>s:t$
+\end_inset
+
+ ni stekališče.
+ Ločimo primere:
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $s\in\mathbb{R}$
+\end_inset
+
+ Naj bo 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+ poljuben.
+ Ker
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+Infimum padajočega konvergentnega zaporedja je očitno njegova limita.
+\end_layout
+
+\end_inset
+
+ je 
+\begin_inset Formula $s=\inf s_{n}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\exists n_{0}\in\mathbb{N}\ni:s_{n_{0}}\in[s,s+\varepsilon)$
+\end_inset
+
+.
+ Ker 
+\begin_inset Formula $\left(s_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ pada proti 
+\begin_inset Formula $s$
+\end_inset
+
+,
+ sledi 
+\begin_inset Formula $\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow s_{n}\in[s,s+\varepsilon)$
+\end_inset
+
+.
+ Po definiciji 
+\begin_inset Formula $s_{n}$
+\end_inset
+
+ velja 
+\begin_inset Formula $\forall n\in\mathbb{N}\exists N\left(n\right)\geq n\ni:s_{n}-\varepsilon<a_{N\left(n\right)}$
+\end_inset
+
+.
+ Torej imamo 
+\begin_inset Formula $s-\varepsilon\leq s_{n}-\varepsilon<a_{N\left(n\right)}\leq s_{n}<s+\varepsilon$
+\end_inset
+
+ (zadnji neenačaj za 
+\begin_inset Formula $n\geq n_{0}$
+\end_inset
+
+),
+ skratka 
+\begin_inset Formula $a_{N\left(n\right)}-s<\varepsilon$
+\end_inset
+
+ oziroma 
+\begin_inset Formula $\forall n\geq n_{0}:\left|a_{N\left(n\right)}-s\right|<\varepsilon$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $N\left(n\right)\geq n$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $\left\{ N\left(n\right);n\in\mathbb{N}\right\} $
+\end_inset
+
+ neskončna množica,
+ torej je neskončno mnogo členov v poljubni okolici 
+\begin_inset Formula $s$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Treba je dokazati še,
+ da 
+\begin_inset Formula $\forall t>s:t$
+\end_inset
+
+ ni stekališče.
+ Naj bo 
+\begin_inset Formula $t>s$
+\end_inset
+
+.
+ Označimo 
+\begin_inset Formula $\delta\coloneqq t-s>0$
+\end_inset
+
+.
+ Po definiciji
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $s$
+\end_inset
+
+ je limita zaporedja 
+\begin_inset Formula $\left(s_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+,
+ zato v poljubno majhni okolici obstaja tak 
+\begin_inset Formula $s_{n_{1}}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $s_{n_{1}}$
+\end_inset
+
+ torej tu najdemo v 
+\begin_inset Formula $[s,s+\frac{\delta}{2})$
+\end_inset
+
+.
+\end_layout
+
+\end_inset
+
+ 
+\begin_inset Formula $s$
+\end_inset
+
+ 
+\begin_inset Formula $\exists n_{1}\in\mathbb{N}\ni:s\leq s_{n_{1}}<s+\frac{\delta}{2}<s+t$
+\end_inset
+
+.
+ Ker 
+\begin_inset Formula $\left(s_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ pada proti 
+\begin_inset Formula $s$
+\end_inset
+
+,
+ sledi 
+\begin_inset Formula $\forall n\geq n_{1}:s\leq s_{n}<s+\frac{\delta}{2}$
+\end_inset
+
+.
+ Po definiciji
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $s_{n}$
+\end_inset
+
+ je supremum členov od vključno 
+\begin_inset Formula $n$
+\end_inset
+
+ dalje
+\end_layout
+
+\end_inset
+
+ 
+\begin_inset Formula $s_{n}$
+\end_inset
+
+ sledi 
+\begin_inset Formula $\forall n\geq n_{1}:a_{n}\leq s+\frac{\delta}{2}$
+\end_inset
+
+.
+ Za takšne 
+\begin_inset Formula $n$
+\end_inset
+
+ je 
+\begin_inset Formula $\left|t-a_{n}\right|=t-a_{n}\geq t-\left(s+\frac{\delta}{2}\right)=\frac{\delta}{2}$
+\end_inset
+
+.
+ Torej v 
+\begin_inset Formula $\frac{\delta}{2}-$
+\end_inset
+
+okolici točke 
+\begin_inset Formula $t$
+\end_inset
+
+ leži kvečjemu končno mnogo členov zaporedja oziroma členi 
+\begin_inset Formula $\left(a_{1},a_{2},\dots,a_{n_{1}-1}\right)$
+\end_inset
+
+.
+ Torej 
+\begin_inset Formula $t$
+\end_inset
+
+ ni stekališče za 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $s=\infty$
+\end_inset
+
+ Naj bo 
+\begin_inset Formula $M>0$
+\end_inset
+
+ poljuben.
+ Ker je 
+\begin_inset Formula $s=\inf s_{n}$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $\forall n\in\mathbb{N}:s_{n}=\infty$
+\end_inset
+
+.
+ Po definiciji 
+\begin_inset Formula $s_{n}=\infty$
+\end_inset
+
+ velja 
+\begin_inset Formula $\forall n\in\mathbb{N}\exists N\left(n\right):a_{N\left(n\right)}>M$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $N\left(n\right)\geq n$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $\left\{ N\left(n\right);n\in\mathbb{N}\right\} $
+\end_inset
+
+ neskončna množica,
+ torej je neskončno mnogo členov v 
+\begin_inset Formula $\left(M,\infty\right)$
+\end_inset
+
+ za poljuben 
+\begin_inset Formula $M$
+\end_inset
+
+,
+ torej je 
+\begin_inset Formula $s=\infty$
+\end_inset
+
+ res stekališče.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Večjih stekališč od 
+\begin_inset Formula $\infty$
+\end_inset
+
+ očitno ni.
+\end_layout
+
+\end_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $s=-\infty$
+\end_inset
+
+ Naj bo 
+\begin_inset Formula $m<0$
+\end_inset
+
+ poljuben.
+ Ker je 
+\begin_inset Formula $s=\inf s_{n}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\exists n_{0}\in\mathbb{N}\ni:s_{n_{0}}\in\left(-\infty,m\right)$
+\end_inset
+
+ Ker 
+\begin_inset Formula $\left(s_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ pada proti 
+\begin_inset Formula $s=-\infty$
+\end_inset
+
+,
+ sledi 
+\begin_inset Formula $\forall n\in\mathbb{N}:n\geq n_{0}:s_{n}\in\left(-\infty,m\right)$
+\end_inset
+
+.
+ Po definiciji 
+\begin_inset Formula $s_{n}$
+\end_inset
+
+ velja 
+\begin_inset Formula $\forall n\in\mathbb{N}:a_{n}\in\left(-\infty,m\right)$
+\end_inset
+
+.
+ Ker je za poljuben 
+\begin_inset Formula $m$
+\end_inset
+
+ neskončno mnogo členov v 
+\begin_inset Formula $\left(-\infty,m\right)$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $s=-\infty$
+\end_inset
+
+ res stekališče.
+\end_layout
+
+\end_deeper
+\begin_layout Subsection
+Cauchyjev pogoj
+\end_layout
+
+\begin_layout Definition*
+Zaporedje 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ ustreza Cauchyjevemu pogoju (oz.
+ je Cauchyjevo),
+ če 
+\begin_inset Formula $\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\ni:\forall m,n\in\mathbb{N}:m,n\geq n_{0}\Rightarrow\left|a_{m}-a_{n}\right|<\varepsilon$
+\end_inset
+
+.
+ ZDB Dovolj pozni členi so si poljubno blizu.
+\end_layout
+
+\begin_layout Claim*
+Zaporedje v 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+ je konvergentno 
+\begin_inset Formula $\Leftrightarrow$
+\end_inset
+
+ je Cauchyjevo.
+\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
+
+ Če 
+\begin_inset Formula $a_{n}\to L$
+\end_inset
+
+,
+ tedaj 
+\begin_inset Formula $\left|a_{m}-a_{n}\right|=\left|\left(a_{m}-L\right)+\left(L-a_{n}\right)\right|\leq\left|a_{m}-\varepsilon\right|+\left|a_{n}-\varepsilon\right|$
+\end_inset
+
+.
+ Cauchyjev pogoj sledi iz definicije limite za 
+\begin_inset Formula $\frac{\varepsilon}{2}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Leftarrow\right)$
+\end_inset
+
+ Če je zaporedje Cauchyjevo,
+ je omejeno:
+ 
+\begin_inset Formula $\exists n_{0}\in\mathbb{N}\ni:\forall m,n\in\mathbb{N}:m,n\geq n_{0}\Rightarrow\left|a_{m}-a_{n}\right|\leq1$
+\end_inset
+
+.
+ V posebnem,
+ 
+\begin_inset Formula $m=n_{0}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\left|a_{n_{0}}-a_{n}\right|\leq1$
+\end_inset
+
+ oziroma 
+\begin_inset Formula $\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow a_{n}\in\left[a_{n_{0}}-1,a_{n_{0}}+1\right]$
+\end_inset
+
+.
+ Preostali členi tvorijo končno veliko množico,
+ ki ima 
+\begin_inset Formula $\min$
+\end_inset
+
+ in 
+\begin_inset Formula $\max$
+\end_inset
+
+,
+ torej je 
+\begin_inset Formula $\left\{ a_{k};k\in\mathbb{N}\right\} =\left\{ a_{1},a_{2},\dots,a_{n_{0}-1}\right\} \cup\left\{ a_{k};k\in\mathbb{N},k\geq n_{0}\right\} $
+\end_inset
+
+ tudi omejena.
+ Po 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{bw}{izreku od prej}
+\end_layout
+
+\end_inset
+
+ sledi,
+ da ima zaporedje stekališče 
+\begin_inset Formula $s$
+\end_inset
+
+.
+ Dokažimo,
+ da je 
+\begin_inset Formula $s=\lim_{n\to\infty}a_{n}$
+\end_inset
+
+.
+ Vzemimo poljuben 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ Cauchyjevo,
+ 
+\begin_inset Formula $\exists n_{0}\in\mathbb{N}\ni:\forall m,n\in\mathbb{N}:m,n\geq n_{0}\Rightarrow\left|a_{m}-a_{n}\right|<\frac{\varepsilon}{2}$
+\end_inset
+
+.
+ Po definiciji 
+\begin_inset Formula $s$
+\end_inset
+
+ 
+\begin_inset Formula $\exists n_{1}\geq n_{0}\ni:\left|a_{n_{1}}-s\right|<\frac{\varepsilon}{2}$
+\end_inset
+
+.
+ Sledi 
+\begin_inset Formula $\forall n\geq n_{0}:\left|a_{n}-s\right|=\left|a_{n}-s+s-a_{n_{1}}\right|\leq\left|a_{n}-s\right|+\left|s-a_{n_{1}}\right|<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Remark*
+Moč izreka je v tem,
+ da lahko konvergenco preverjamo tudi tedaj,
+ ko nimamo kandidatov za limito.
+\end_layout
+
+\begin_layout Section
+Številske vrste
+\end_layout
+
+\begin_layout Standard
+Kako sešteti neskončno mnogo števil?
+ Nadgradimo pristop končnih vsot na neskončne vsote!
+\end_layout
+
+\begin_layout Definition*
+Imejmo zaporedje 
+\begin_inset Formula $\left(a_{k}\right)_{k\in\mathbb{N}},a_{k}\in\mathbb{R}$
+\end_inset
+
+.
+ Izraz 
+\begin_inset Formula $\sum_{j=1}^{\infty}a_{j}$
+\end_inset
+
+ se imenuje vrsta s členi 
+\begin_inset Formula $a_{j}$
+\end_inset
+
+.
+ Pomen izraza opredelimo na naslednjo način:
+\end_layout
+
+\begin_layout Definition*
+Tvorimo novo zaporedje,
+ pravimo mu zaporedje delnih vsot vrste:
+ 
+\begin_inset Formula $s_{1}=a_{1}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $s_{2}=a_{1}+a_{2}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $s_{3}=a_{1}+a_{2}+a_{3}$
+\end_inset
+
+,
+ ...,
+ 
+\begin_inset Formula $s_{n}=a_{1}+a_{2}+\cdots+a_{n}=\sum_{j=1}^{n}a_{j}$
+\end_inset
+
+ —
+ številu 
+\begin_inset Formula $s_{n}$
+\end_inset
+
+ pravimo 
+\begin_inset Formula $n-$
+\end_inset
+
+ta delna vsota.
+\end_layout
+
+\begin_layout Definition*
+Vrsta je konvergentna,
+ če je v 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+ konvergentno zaporedje 
+\begin_inset Formula $\left(s_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+.
+ Številu 
+\begin_inset Formula $s=\lim_{n\to\infty}s_{n}$
+\end_inset
+
+ tedaj pravimo vsota vrste in pišemo 
+\begin_inset Formula $s\eqqcolon\sum_{j=1}^{\infty}a_{j}$
+\end_inset
+
+.
+ Pojem neskončne vsote torej prevedemo na pojem limite pridruženega zaporedja delnih vsot.
+ Včasih vrsto (kot operacijo) enačimo z njeno vsoto (izidom operacije).
+\end_layout
+
+\begin_layout Definition*
+Če vrsta ni konvergentna,
+ rečemo,
+ da je divergentna.
+ Enako,
+ če je 
+\begin_inset Formula $s\in\left\{ \pm\infty\right\} $
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Primeri vrst.
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $a_{n}=\frac{1}{2^{n}}$
+\end_inset
+
+,
+ torej zaporedje 
+\begin_inset Formula $\frac{1}{2},\frac{1}{4},\frac{1}{8},\dots$
+\end_inset
+
+.
+ Ali se sešteje v 1?
+ Velja 
+\begin_inset Formula $s=\lim_{n\to\infty}\sum_{j=1}^{n}a_{j}$
+\end_inset
+
+.
+ Pišimo 
+\begin_inset Formula $q=\frac{1}{2}$
+\end_inset
+
+,
+ tedaj 
+\begin_inset Formula $a_{n}=q^{n}$
+\end_inset
+
+ in 
+\begin_inset Formula 
+\[
+s_{n}=q+q^{2}+q^{3}+\cdots+q^{n}=q\left(1+q+q^{2}+\cdots+q^{n-1}\right)=q\frac{\left(1+q+q^{2}+\cdots+q^{n-1}\right)\left(1-q\right)}{1-q}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=q\frac{\left(1+q+q^{2}+\cdots+q^{n-1}\right)-\left(q+q^{2}+q^{3}+\cdots+q^{n}\right)}{1-q}=q\frac{1-q^{n}}{1-q}=\frac{q}{1-q}\left(1-q^{n}\right)
+\]
+
+\end_inset
+
+Izračunajmo 
+\begin_inset Formula $\lim_{n\to\infty}s_{n}=\lim_{n\to\infty}\frac{q}{1-q}\left(1-\cancelto{0}{q^{n}}\right)=\frac{q}{1-q}$
+\end_inset
+
+ (velja,
+ ker 
+\begin_inset Formula $q\in\left(-1,1\right)$
+\end_inset
+
+),
+ torej je 
+\begin_inset Formula $s=\sum_{n=1}^{\infty}q^{n}=\frac{q}{1-q}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Geometrijska vrsta (splošno).
+ Naj bo 
+\begin_inset Formula $q\in\mathbb{R}$
+\end_inset
+
+.
+ Vrsta 
+\begin_inset Formula $\sum_{j=0}^{\infty}q^{j}$
+\end_inset
+
+ se imenuje geometrijska vrsta.
+ Velja 
+\begin_inset Formula $s=\lim_{n\to\infty}\sum_{j=0}^{n}q^{j}$
+\end_inset
+
+ in 
+\begin_inset Formula $s_{n}=1+q+q^{2}+q^{3}+\cdots+q^{n}$
+\end_inset
+
+.
+ Če je 
+\begin_inset Formula $q=1$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $s_{n}=n+1$
+\end_inset
+
+,
+ sicer množimo izraz z 
+\begin_inset Formula $\left(1-q\right)$
+\end_inset
+
+:
+\begin_inset Formula 
+\[
+\left(1+q+q^{2}+\cdots+q^{n}\right)\left(1-q\right)=\left(1+q+q^{2}+\cdots+q^{n}\right)-\left(q+q^{2}+q^{3}+\cdots+q^{n+1}\right)=1-q^{n+1}
+\]
+
+\end_inset
+
+torej 
+\begin_inset Formula $s_{n}=\frac{1-q^{n+1}}{1-q}$
+\end_inset
+
+ in vrsta konvergira 
+\begin_inset Formula $\Leftrightarrow q\not=1$
+\end_inset
+
+ in 
+\begin_inset Formula $\lim_{n\to\infty}\frac{1-q^{n+1}}{1-q}\exists$
+\end_inset
+
+ v 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+.
+ To pa se zgodi natanko za 
+\begin_inset Formula $q\in\left(-1,1\right)$
+\end_inset
+
+,
+ takrat je 
+\begin_inset Formula $\lim_{n\to\infty}\frac{1-\cancelto{0}{q^{n+1}}}{1-q}=\frac{1}{1-q}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Harmonična vrsta.
+ Je vrsta 
+\begin_inset Formula $\sum_{j=1}^{\infty}\frac{1}{j}$
+\end_inset
+
+.
+ Velja 
+\begin_inset Formula $\frac{1}{j}\underset{j\to\infty}{\longrightarrow}0$
+\end_inset
+
+,
+ toda vrsta divergira.
+ Dokaz sledi kmalu malce spodaj.
+\end_layout
+
+\end_deeper
+\begin_layout Question*
+Kako lahko enostavno določimo,
+ ali dana vrsta konvergira?
+\end_layout
+
+\begin_layout Subsection
+Konvergenčni kriteriji
+\end_layout
+
+\begin_layout Theorem*
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hypertarget{cauchyvrste}{Cauchyjev pogoj}
+\end_layout
+
+\end_inset
+
+.
+ Vrsta 
+\begin_inset Formula $\sum_{j=1}^{\infty}a_{j}$
+\end_inset
+
+ je konvergentna 
+\begin_inset Formula $\Leftrightarrow$
+\end_inset
+
+ delne vrste ustrezajo Cauchyjevemu pogoju;
+ 
+\begin_inset Formula $\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\forall n,m\in\mathbb{N}:n,m\geq n_{0}\Rightarrow\left|s_{m}-s_{n}\right|=\left|\sum_{j=n+1}^{m}a_{j}\right|<\varepsilon$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Corollary*
+\begin_inset Formula $\sum_{j=1}^{\infty}a_{j}$
+\end_inset
+
+ konvergira 
+\begin_inset Formula $\Rightarrow\lim_{j\to\infty}a_{j}=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Uporabimo izrek zgoraj za 
+\begin_inset Formula $n=m-1$
+\end_inset
+
+:
+ 
+\begin_inset Formula $\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow\left|s_{n}-s_{n+1}\right|=\left|a_{n}\right|<\varepsilon$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Vrsti 
+\begin_inset Formula $\sum_{j=1}^{\infty}\cos n$
+\end_inset
+
+ in 
+\begin_inset Formula $\sum_{j=1}^{\infty}\sin n$
+\end_inset
+
+ divergirata,
+ saj smo videli,
+ da členi ne ene ne druge ne konvergirajo nikamor,
+ torej tudi ne proti 0,
+ kar je potreben pogoj za konvergenco vrste.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+Harmonična vrsta divergira.
+ Protiprimer Cauchyjevega pogoja:
+ Naj bo 
+\begin_inset Formula $\varepsilon=\frac{1}{4}$
+\end_inset
+
+.
+ Tedaj ne glede na izbiro 
+\begin_inset Formula $n_{0}$
+\end_inset
+
+ najdemo:
+\begin_inset Formula 
+\[
+s_{2n}-s_{n}=\sum_{j=n+1}^{2n}\frac{1}{j}=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}>\frac{1}{2n}+\frac{1}{2n}+\cdots+\frac{1}{2n}=\frac{1}{2}
+\]
+
+\end_inset
+
+Dokaz divergence brez Cauchyjevega pogoja:
+ 
+\begin_inset Formula $s_{2^{n}}=a_{1}+\sum_{j=1}^{n}\left(s_{2^{j}}-s_{s^{j-1}}\right)>1+\frac{n}{2}$
+\end_inset
+
+ in 
+\begin_inset Formula $\lim_{n\to\infty}1+\frac{n}{2}=\infty$
+\end_inset
+
+.
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+Geometrični argument za divergenco:
+ TODO XXX FIXME DODAJ
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Theorem*
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hypertarget{pk}{Primerjalni kriterij}
+\end_layout
+
+\end_inset
+
+.
+ Naj bosta 
+\begin_inset Formula $\sum_{n=1}^{\infty}a_{n}$
+\end_inset
+
+ in 
+\begin_inset Formula $\sum_{n=1}^{\infty}b_{n}$
+\end_inset
+
+ vrsti z nenegativnimi členi.
+ Naj bo 
+\begin_inset Formula $\forall k\geq k_{0}:a_{k}\leq b_{k}$
+\end_inset
+
+ (od nekod naprej) —
+ pravimo,
+ da je 
+\begin_inset Formula $\sum_{n=1}^{\infty}b_{n}$
+\end_inset
+
+ majoranta za 
+\begin_inset Formula $\sum_{n=1}^{\infty}a_{n}$
+\end_inset
+
+ od nekod naprej.
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Če 
+\begin_inset Formula $\sum_{n=1}^{\infty}b_{n}$
+\end_inset
+
+ konvergira,
+ tedaj tudi 
+\begin_inset Formula $\sum_{n=1}^{\infty}a_{n}$
+\end_inset
+
+ konvergira.
+\end_layout
+
+\begin_layout Itemize
+Če 
+\begin_inset Formula $\text{\ensuremath{\sum_{n=1}^{\infty}a_{n}=\infty}}$
+\end_inset
+
+,
+ tedaj tudi 
+\begin_inset Formula $\sum_{n=1}^{\infty}b_{n}=\infty$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Example*
+Videli smo,
+ da 
+\begin_inset Formula $\sum_{k=1}^{\infty}\frac{1}{k}$
+\end_inset
+
+ divergira.
+ Kaj pa 
+\begin_inset Formula $\sum_{k=1}^{\infty}\frac{1}{k^{2}}$
+\end_inset
+
+?
+ Preverimo naslednje in uporabimo primerjalni kriterij:
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $\forall k\in\mathbb{N}:\frac{1}{k^{2}}\leq\frac{2}{k\left(k+1\right)}$
+\end_inset
+
+?
+ Računajmo 
+\begin_inset Formula $k^{2}\geq\frac{k\left(k+1\right)}{2}\sim k\geq\frac{k+1}{2}\sim\frac{k}{2}\geq\frac{1}{2}$
+\end_inset
+
+.
+ Velja,
+ ker 
+\begin_inset Formula $k\in\mathbb{N}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Vrsta 
+\begin_inset Formula $\sum_{k=1}^{\infty}\frac{2}{k\left(k+1\right)}$
+\end_inset
+
+ konvergira?
+ Opazimo 
+\begin_inset Formula $\frac{1}{k}-\frac{1}{k+1}=\frac{k+1}{k\left(k+1\right)}-\frac{k}{k\left(k+1\right)}=\frac{1}{k\left(k+1\right)}$
+\end_inset
+
+.
+ Za delne vsote vrste 
+\begin_inset Formula $\sum_{k=1}^{\infty}\frac{1}{k\left(k+1\right)}$
+\end_inset
+
+ velja:
+\begin_inset Formula 
+\[
+\sum_{k=1}^{n}\frac{1}{k\left(k+1\right)}=\sum_{k=1}^{n}\left(\frac{1}{k}-\frac{1}{k+1}\right)=1-\frac{1}{n+1}\underset{n\to\infty}{\longrightarrow}1,
+\]
+
+\end_inset
+
+torej 
+\begin_inset Formula $\sum_{k=1}^{\infty}\frac{2}{k\left(k+1\right)}=2$
+\end_inset
+
+.
+ Posledično po primerjalnem kriteriju tudi 
+\begin_inset Formula $\sum_{k=1}^{\infty}\frac{1}{k^{2}}$
+\end_inset
+
+ konvergira.
+ Izkaže se 
+\begin_inset Formula $\sum_{k=1}^{\infty}\frac{1}{k^{2}}=\frac{\pi^{2}}{6}\approx1,645$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Theorem*
+Kvocientni oz.
+ d'Alembertov kriterij.
+ Za vrsto s pozitivnimi členi 
+\begin_inset Formula $\sum_{n=1}^{\infty}a_{n}$
+\end_inset
+
+ definirajmo 
+\begin_inset Formula $D_{n}\coloneqq\frac{a_{n+1}}{a_{n}}$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $\exists n_{0}\in\mathbb{N},q\in\left(0,1\right)\forall n\geq n_{0}:D_{n}\leq q\Longrightarrow\sum_{n=1}^{\infty}a_{n}<\infty$
+\end_inset
+
+ (vrsta konvergira)
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\exists n_{0}\in\mathbb{N}\forall n\geq n_{0}:D_{n}\geq1\Longrightarrow\sum_{n=1}^{\infty}a_{n}=\infty$
+\end_inset
+
+ (vrsta divergira)
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\exists D=\lim_{n\to\infty}D_{n}\in\mathbb{R}\Longrightarrow$
+\end_inset
+
+
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset CommandInset label
+LatexCommand label
+name "enu:kvocientni3a"
+
+\end_inset
+
+
+\begin_inset Formula $D<1\Longrightarrow\sum_{n=1}^{\infty}a_{n}<\infty$
+\end_inset
+
+ (vrsta konvergira)
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $D>1\Longrightarrow\sum_{n=1}^{\infty}a_{n}=\infty$
+\end_inset
+
+ (vrsta divergira)
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $D=1\Longrightarrow$
+\end_inset
+
+ s tem kriterijem ne moremo določiti konvergence.
+\end_layout
+
+\end_deeper
+\end_deeper
+\begin_layout Proof
+Razlaga.
+ 
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $\forall n>n_{0}:D_{n}\leq q$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\frac{a_{n+1}}{a_{n}}\leq q\sim a_{n+1}\leq qa_{n}$
+\end_inset
+
+ in hkrati 
+\begin_inset Formula $\text{\ensuremath{\frac{a_{n+2}}{a_{n+1}}\leq q\sim a_{n+2}\leq qa_{n+1}}}$
+\end_inset
+
+,
+ torej skupaj 
+\begin_inset Formula $a_{n+2}\leq qa_{n+1}\leq qqa_{n}=q^{2}a_{n}$
+\end_inset
+
+,
+ sledi 
+\begin_inset Formula $q_{n+2}\leq q^{2}a_{n}$
+\end_inset
+
+ in 
+\begin_inset Formula $\forall k\in\mathbb{N}:q_{n+k}\leq q^{k}a_{n}$
+\end_inset
+
+.
+ Vrsto smo majorizirali z geometrijsko vrsto,
+ ki ob 
+\begin_inset Formula $q\in\left(0,1\right)$
+\end_inset
+
+ konvergira po primerjalnem kriteriju,
+ zato tudi naša vrsta konvergira.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\forall n>n_{0}:\frac{a_{n+1}}{a_{n}}\geq D\geq1$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $a_{n+1}\geq a_{n}$
+\end_inset
+
+ in hkrati 
+\begin_inset Formula $a_{n+2}\geq a_{n+1}$
+\end_inset
+
+,
+ torej skupaj 
+\begin_inset Formula $a_{n+2}\geq a_{n}$
+\end_inset
+
+,
+ sledi 
+\begin_inset Formula $\forall k\in\mathbb{N}:a_{n+k}\geq a_{n}$
+\end_inset
+
+.
+ Naša vrsta torej majorizira konstantno vrsto,
+ ki očitno divergira;
+ 
+\begin_inset Formula $\sum_{k=n_{0}}^{\infty}a_{k}\geq\sum_{k=n_{0}}^{\infty}a_{n}=0$
+\end_inset
+
+.
+ Potemtakem tudi naša vrsta divergira.
+ Poleg tega niti ne velja 
+\begin_inset Formula $a_{k}\underset{k\to\infty}{\longrightarrow}0$
+\end_inset
+
+,
+ torej vrsta gotovo divergira.
+\end_layout
+
+\begin_layout Enumerate
+Enako kot 1 in 2.
+\end_layout
+
+\end_deeper
+\begin_layout Example*
+Za 
+\begin_inset Formula $x>0$
+\end_inset
+
+ definiramo 
+\begin_inset Formula $e^{x}=\sum_{k=0}^{\infty}\frac{x^{k}}{k!}$
+\end_inset
+
+.
+ Vrsta res konvergira po točki 
+\begin_inset CommandInset ref
+LatexCommand ref
+reference "enu:kvocientni3a"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+D_{n}=\frac{\frac{x^{n+1}}{\left(n+1\right)!}}{\frac{x^{n}}{n!}}=\frac{x^{n+1}n!}{x^{n}\left(n+1\right)!}=\frac{x}{n+1}\underset{n\to\infty}{\longrightarrow}0
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Theorem*
+Korenski oz.
+ Cauchyjev kriterij.
+ Naj bo 
+\begin_inset Formula $\sum_{k=1}^{\infty}a_{k}$
+\end_inset
+
+ vrsta z nenegativnimi členi.
+ Naj bo 
+\begin_inset Formula $c_{n}\coloneqq\sqrt[n]{a_{n}}$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $\exists n_{0}\in\mathbb{N},q\in\left(0,1\right)\forall n>n_{0}:c_{n}\leq q\Longrightarrow\sum_{k=1}^{\infty}a_{k}<\infty$
+\end_inset
+
+ (vrsta konvergira)
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\exists n_{0}\in\mathbb{N}\forall n>n_{0}:c_{n}\geq1\Longrightarrow\sum_{k=1}^{\infty}a_{k}=\infty$
+\end_inset
+
+ (vrsta divergira)
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\exists c=\lim_{n\to\infty}c_{n}\in\mathbb{R}\Longrightarrow$
+\end_inset
+
+
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $c<1\Longrightarrow\sum_{k=1}^{\infty}a_{k}<\infty$
+\end_inset
+
+ (vrsta konvergira)
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $c>1\Longrightarrow\sum_{k=1}^{\infty}a_{k}=\infty$
+\end_inset
+
+ (vrsta divergira)
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $c=1\Longrightarrow$
+\end_inset
+
+ s tem kriterijem ne moremo določiti konvergence.
+\end_layout
+
+\end_deeper
+\end_deeper
+\begin_layout Proof
+Skica dokazov.
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+Velja 
+\begin_inset Formula $\forall n>n_{0}:c_{n}\leq q$
+\end_inset
+
+.
+ To pomeni 
+\begin_inset Formula $\sqrt[n]{a_{n}}\leq q$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $a_{n}\leq q^{n}$
+\end_inset
+
+ in 
+\begin_inset Formula $a_{n+1}\leq q^{n+1}$
+\end_inset
+
+,
+ torej je vrsta majorizirana z geometrijsko vrsto 
+\begin_inset Formula $\sum_{n=1}^{\infty}q^{n}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Velja 
+\begin_inset Formula $\forall n>n_{0}:c_{n}\geq1$
+\end_inset
+
+.
+ To pomeni 
+\begin_inset Formula $\sqrt[n]{a_{n}}\geq1$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $a_{n}\geq1$
+\end_inset
+
+,
+ torej je vrsta majorizirana s konstantno in zato divergentno vrsto 
+\begin_inset Formula $\sum_{n=1}^{\infty}1$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Enako kot 1 in 2.
+\end_layout
+
+\end_deeper
+\begin_layout Subsection
+Alternirajoče vrste
+\end_layout
+
+\begin_layout Definition*
+Vrsta je alternirajoča,
+ če je predznak naslednjega člena nasproten predznaku tega člena.
+ ZDB 
+\begin_inset Formula $\forall n\in\mathbb{N}:\sgn a_{n+1}=-\sgn a_{n}$
+\end_inset
+
+,
+ kjer je 
+\begin_inset Formula $\sgn:\mathbb{R}\to\left\{ -1,0,1\right\} $
+\end_inset
+
+ s predpisom 
+\begin_inset Formula $\sgn a=\begin{cases}
+-1 & ;a<0\\
+1 & ;a>0\\
+0 & ;a=0
+\end{cases}$
+\end_inset
+
+.
+ ZDB 
+\begin_inset Formula $\forall n\in\mathbb{N}:a_{n+1}a_{n}\leq0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+Leibnizov konvergenčni kriterij.
+ Naj bo 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ padajoče zaporedje in 
+\begin_inset Formula $\lim_{n\to\infty}a_{n}=0$
+\end_inset
+
+.
+ Tedaj vrsta 
+\begin_inset Formula $\sum_{k=1}^{\infty}\left(-1\right)^{k}a_{k}$
+\end_inset
+
+ konvergira.
+ Če je 
+\begin_inset Formula $s\coloneqq\sum_{k=1}^{\infty}\left(-1\right)^{k}a_{k}$
+\end_inset
+
+ in 
+\begin_inset Formula $s_{n}\coloneqq\sum_{k=1}^{\infty}\left(-1\right)^{k}a_{k}$
+\end_inset
+
+,
+ tedaj 
+\begin_inset Formula $\left|s-s_{k}\right|\leq a_{n+1}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Skica dokaza.
+ Vidimo,
+ da delne vsote 
+\begin_inset Formula $s_{2n}$
+\end_inset
+
+ padajo k 
+\begin_inset Formula $s''$
+\end_inset
+
+ in delne vsote 
+\begin_inset Formula $s_{2n-1}$
+\end_inset
+
+ naraščajo k 
+\begin_inset Formula $s'$
+\end_inset
+
+.
+ Toda ker 
+\begin_inset Formula $s_{2n}-s_{2n-1}=a_{2n}$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $s'=s''$
+\end_inset
+
+.
+ Limita razlike dveh zaporedij je razlika limit teh dveh zaporedij,
+ torej 
+\begin_inset Formula $s'=s''=s$
+\end_inset
+
+.
+ 
+\begin_inset Formula $s$
+\end_inset
+
+ je supremum lihih in infimum sodih vsot.
+ 
+\begin_inset Formula $\left|s-s_{n}\right|\leq\left|s_{n+1}-s_{n}\right|=a_{n+1}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Harmonična vrsta 
+\begin_inset Formula $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots\to\infty$
+\end_inset
+
+,
+ toda alternirajoča harmonična vrsta 
+\begin_inset Formula $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots\to\log2$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsection
+Absolutno konvergentne vrste
+\end_layout
+
+\begin_layout Definition*
+Vrsta 
+\begin_inset Formula $\sum_{n=1}^{\infty}a_{n}$
+\end_inset
+
+ je absolutno konvergentna,
+ če je 
+\begin_inset Formula $\sum_{n=1}^{\infty}\left|a_{n}\right|$
+\end_inset
+
+ konvergentna.
+\end_layout
+
+\begin_layout Theorem*
+Absolutna konvergenca 
+\begin_inset Formula $\Rightarrow$
+\end_inset
+
+ konvergenca.
+\end_layout
+
+\begin_layout Proof
+Uporabimo 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{cauchyvrste}{Cauchyjev pogoj za konvergenco vrst}
+\end_layout
+
+\end_inset
+
+ in trikotniško neenakost.
+\begin_inset Formula 
+\[
+\left|s_{m}-s_{n}\right|=\left|\sum_{j=n+1}^{m}a_{j}\right|\leq\sum_{j=n+1}^{m}\left|a_{j}\right|<\varepsilon
+\]
+
+\end_inset
+
+za 
+\begin_inset Formula $m,n\geq n_{0}$
+\end_inset
+
+ za nek 
+\begin_inset Formula $n_{0}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Remark*
+Obrat ne velja,
+ protiprimer je alternirajoča harmonična vrsta.
+\end_layout
+
+\begin_layout Subsection
+Pogojno konvergentne vrste
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $\sum_{k=0}^{\infty}2-\sum_{k=0}^{\infty}1\not=\sum_{k=0}^{\infty}\left(2-1\right)$
+\end_inset
+
+,
+ temveč 
+\begin_inset Formula $\infty-\infty=$
+\end_inset
+
+ nedefinirano.
+\end_layout
+
+\begin_layout Question*
+Ross-Littlewoodov paradoks.
+ Ali smemo zamenjati vrstni red seštevanja,
+ če imamo neskončno mnogo sumandov?
+\end_layout
+
+\begin_layout Standard
+Najprej vprašanje natančneje opredelimo in vpeljimo orodja za njegovo obravnavo.
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $\mathcal{M}\subset\mathbb{N}$
+\end_inset
+
+.
+ Permutacija 
+\begin_inset Formula $\mathcal{M}$
+\end_inset
+
+ je vsaka bijektivna preslikava 
+\begin_inset Formula $\pi:\mathcal{M}\to\mathcal{M}$
+\end_inset
+
+.
+ Če je 
+\begin_inset Formula $\mathcal{M}=\left\{ a_{1},\dots,a_{n}\right\} $
+\end_inset
+
+ končna množica,
+ tedaj 
+\begin_inset Formula $\pi$
+\end_inset
+
+ označimo s tabelo:
+\begin_inset Formula 
+\[
+\left(\begin{array}{ccc}
+a_{1} & \cdots & a_{n}\\
+\pi\left(a_{1}\right) & \cdots & \pi\left(a_{n}\right)
+\end{array}\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula 
+\[
+\pi=\left(\begin{array}{ccccc}
+1 & 2 & 3 & 4 & 5\\
+5 & 3 & 1 & 4 & 2
+\end{array}\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Vrsta 
+\begin_inset Formula $\sum_{n=1}^{\infty}a_{n}$
+\end_inset
+
+ je brezpogojno konvergentna,
+ če za vsako permutacijo 
+\begin_inset Formula $\pi:\mathbb{N}\to\mathbb{N}$
+\end_inset
+
+ vrsta 
+\begin_inset Formula $\sum_{n=1}^{\infty}\pi\left(a_{n}\right)$
+\end_inset
+
+ konvergira in vsota ni odvisna od 
+\begin_inset Formula $\pi$
+\end_inset
+
+.
+ Vrsta je pogojno konvergentna,
+ če je konvergentna,
+ toda ne brezpogojno.
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\cdots$
+\end_inset
+
+ je pogojno konvergentna,
+ ker pri seštevanju z vrstnim redom,
+ pri katerem tisočim pozitivnim členom sledi en negativen in njemu zopet tisoč pozitivnih itd.,
+ vrsta ne konvergira.
+\end_layout
+
+\begin_layout Theorem*
+Absolutna konvergenca 
+\begin_inset Formula $\Leftrightarrow$
+\end_inset
+
+ Brezpogojna konvergenca
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Theorem*
+Riemannov sumacijski izrek.
+ Če je vrsta pogojno konvergentna,
+ tedaj 
+\begin_inset Formula $\forall x\in\mathbb{R}\cup\left\{ \pm\infty\right\} \exists$
+\end_inset
+
+ permutacija 
+\begin_inset Formula $\pi:\mathbb{N}\to\mathbb{N}\ni:\sum_{n=1}^{\infty}a_{\pi\left(n\right)}=x$
+\end_inset
+
+.
+ ZDB Končna vsota je lahko karkoli,
+ če lahko poljubno spremenimo vrstni red seštevanja.
+ Prav tako obstaja taka permutacija 
+\begin_inset Formula $\pi$
+\end_inset
+
+,
+ pri kateri 
+\begin_inset Formula $\sum_{n=1}^{\infty}a_{\pi\left(n\right)}$
+\end_inset
+
+ nima vsote ZDB delne vsotee ne konvergirajo.
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}}{n}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Funkcijske vrste
+\end_layout
+
+\begin_layout Standard
+Tokrat poskušamo seštevati funkcije.
+ V prejšnjem razdelku seštevamo le realna števila.
+ Funkcijska vrsta,
+ če je 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ zaporedije funkcij 
+\begin_inset Formula $X\to\mathbb{R}$
+\end_inset
+
+ in 
+\begin_inset Formula $x$
+\end_inset
+
+ zunanja konstanta,
+ izgleda takole:
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula 
+\[
+\sum_{n=1}^{\infty}a_{n}\left(x\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $X$
+\end_inset
+
+ neka množica in 
+\begin_inset Formula $\Phi=\left\{ \varphi_{n}:X\to\mathbb{R},n\in\mathbb{N}\right\} $
+\end_inset
+
+ družina funkcij.
+\end_layout
+
+\begin_layout Definition*
+Pravimo,
+ da funkcije 
+\begin_inset Formula $\varphi_{n}$
+\end_inset
+
+ konvergirajo po točkah na 
+\begin_inset Formula $X$
+\end_inset
+
+,
+ če je 
+\begin_inset Formula $\forall x\in X$
+\end_inset
+
+ zaporedje 
+\begin_inset Formula $\left(\varphi_{n}\left(x\right)\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ konvergentno.
+\end_layout
+
+\begin_layout Definition*
+Označimo limito s 
+\begin_inset Formula $\varphi\left(x\right)$
+\end_inset
+
+.
+ ZDB to pomeni,
+ da 
+\begin_inset Formula 
+\[
+\forall\varepsilon>0,x\in X:\exists n_{0}=n_{0}\left(\varepsilon,x\right)\in\mathbb{N}\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow\left|\varphi_{n}\left(x\right)-\varphi\left(x\right)\right|<\varepsilon.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Pravimo,
+ da funkcije 
+\begin_inset Formula $\varphi_{n}$
+\end_inset
+
+ konvergirajo enakomerno na 
+\begin_inset Formula $X$
+\end_inset
+
+,
+ če 
+\begin_inset Formula 
+\[
+\forall\varepsilon>0\exists n_{0}=n_{0}\left(\varepsilon\right)\in\mathbb{N}\forall x\in X,n\in\mathbb{N}:n\geq n_{0}\Rightarrow\left|\varphi_{n}\left(x\right)-\varphi\left(x\right)\right|\leq\varepsilon
+\]
+
+\end_inset
+
+ oziroma ZDB 
+\begin_inset Formula 
+\[
+\forall\varepsilon>0\exists n_{0}=n_{0}\left(\varepsilon\right)\in\mathbb{N}\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow\sup_{x\in X}\left|\varphi_{n}\left(x\right)-\varphi\left(x\right)\right|\leq\varepsilon.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Poudariti je treba,
+ da je pri konvergenci po točkah 
+\begin_inset Formula $n_{0}$
+\end_inset
+
+ lahko odvisen od 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+ in 
+\begin_inset Formula $x$
+\end_inset
+
+,
+ pri enakomerni konvergenci pa le od 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Note*
+Očitno enakomerna konvergenca implicira konvergenco po točkah,
+ obratno pa ne velja.
+\end_layout
+
+\begin_layout Example*
+Za 
+\begin_inset Formula $n\in\mathbb{N}$
+\end_inset
+
+ definiramo 
+\begin_inset Formula $\varphi_{n}:\left[0,1\right]\to\left[0,1\right]$
+\end_inset
+
+ s predpisom 
+\begin_inset Formula $\varphi_{n}\left(x\right)=x^{n}$
+\end_inset
+
+.
+ Tedaj obstaja 
+\begin_inset Formula $\varphi\left(x\right)\coloneqq\lim_{n\to\infty}\varphi_{n}\left(x\right)=\begin{cases}
+0 & ;x\in[0,1)\\
+1 & ;x=1
+\end{cases}$
+\end_inset
+
+.
+ Torej po definiciji velja 
+\begin_inset Formula $\varphi_{n}\to\varphi$
+\end_inset
+
+ po točkah,
+ toda ne velja 
+\begin_inset Formula $\varphi_{n}\to\varphi$
+\end_inset
+
+ enakomerno.
+ Za poljubno velik pas okoli 
+\begin_inset Formula $\varphi\left(x\right)$
+\end_inset
+
+ bodo še tako pozne funkcijske vrednosti 
+\begin_inset Formula $\varphi_{n}\left(x\right)$
+\end_inset
+
+ od nekega 
+\begin_inset Formula $x$
+\end_inset
+
+ dalje izven tega pasu.
+ Če bi 
+\begin_inset Formula $\varphi_{n}\to\varphi$
+\end_inset
+
+ enakomerno,
+ tedaj bi za poljuben 
+\begin_inset Formula $\varepsilon\in\left(0,1\right)$
+\end_inset
+
+ in dovolj pozne 
+\begin_inset Formula $n$
+\end_inset
+
+ (večje od nekega 
+\begin_inset Formula $n_{0}\in\mathbb{N}$
+\end_inset
+
+) veljalo 
+\begin_inset Formula $\forall x\in\left[0,1\right]:\left|\varphi_{n}\left(x\right)-\varphi\left(x\right)\right|<\varepsilon$
+\end_inset
+
+.
+ To je ekvivalentno 
+\begin_inset Formula $\forall x\in\left(0,1\right):\left|x^{n}\right|<\varepsilon\Leftrightarrow n\log x<\log\varepsilon\Leftrightarrow n>\frac{\log\varepsilon}{\log x}$
+\end_inset
+
+.
+ Toda 
+\begin_inset Formula $\lim_{x\nearrow1}\frac{\log\varepsilon}{\log x}=\infty$
+\end_inset
+
+,
+ zato tak 
+\begin_inset Formula $n$
+\end_inset
+
+ ne obstaja.
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $X$
+\end_inset
+
+ neka množica in 
+\begin_inset Formula $\left(f_{j}:X\to\mathbb{R}\right)_{j\in\mathbb{N}}$
+\end_inset
+
+ dano zaporedje funkcij.
+ Pravimo,
+ da funkcijska vrsta 
+\begin_inset Formula $\sum_{j=1}^{\infty}f_{j}$
+\end_inset
+
+ konvergira po točkah na 
+\begin_inset Formula $X$
+\end_inset
+
+,
+ če 
+\begin_inset Formula $\forall x\in X:\sum_{j=1}^{\infty}f_{j}\left(x\right)<0$
+\end_inset
+
+ (številska vrsta je konvergentna).
+ ZDB to pomeni,
+ da funkcijsko zaporedje delnih vsot 
+\begin_inset Formula $s_{n}\coloneqq\sum_{j=1}^{n}f_{j}$
+\end_inset
+
+ konvergira po točkah na 
+\begin_inset Formula $X$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Funkcijska vrsta 
+\begin_inset Formula $s=\sum_{j=1}^{\infty}$
+\end_inset
+
+ konvergira enakomerno na 
+\begin_inset Formula $X$
+\end_inset
+
+,
+ če funkcijsko zaporedje delnih vsot 
+\begin_inset Formula $s_{n}\coloneqq\sum_{j=1}^{n}f_{j}$
+\end_inset
+
+ konvergira enakomerno na 
+\begin_inset Formula $X$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Funkcija oblike 
+\begin_inset Formula $x\mapsto\sum_{j=1}^{\infty}f_{j}\left(x\right)$
+\end_inset
+
+ se imenuje funkcijska vrsta.
+\end_layout
+
+\begin_layout Exercise*
+Dokaži,
+ da 
+\begin_inset Formula $\sum_{n=1}^{\infty}x^{n}$
+\end_inset
+
+ ne konvergira enakomerno!
+ Vrsta konvergira po točkah le na intervalu 
+\begin_inset Formula $x\in\left(0,1\right)$
+\end_inset
+
+,
+ za druge 
+\begin_inset Formula $x$
+\end_inset
+
+ divergira.
+ Ko fiksiramo zunanjo konstanto,
+ gre za geometrijsko vrsto.
+ Delna vsota 
+\begin_inset Formula $\sum_{j=1}^{n}x^{j}=\frac{x\left(1-x^{n}\right)}{1-x}$
+\end_inset
+
+.
+ Velja 
+\begin_inset Formula $\lim_{n\to\infty}\frac{x\left(1-x^{n}\right)}{1-x}=x\lim_{n\to\infty}\frac{1-\cancelto{0}{x^{n}}}{1-x}=\frac{x}{1-x}$
+\end_inset
+
+.
+ Sedaj prevedimo,
+ ali 
+\begin_inset Formula $\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\forall x\in\left(-1,1\right),n\geq n_{0}:\left|\frac{x\left(1-x^{n}\right)}{1-x}-\frac{x}{1-x}\right|<\varepsilon$
+\end_inset
+
+.
+ Za začetekk si oglejmo le 
+\begin_inset Formula $x>0$
+\end_inset
+
+.
+ Ker je tedaj 
+\begin_inset Formula $\frac{x\left(1-x^{n}\right)}{1-x}<\frac{x}{1-x}$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $\left|\frac{x\left(1-x^{n}\right)}{1-x}-\frac{x}{1-x}\right|=\frac{x}{1-x}-\frac{x\left(1-x^{n}\right)}{1-x}=\frac{\cancel{x-x+}x^{n+1}}{1-x}$
+\end_inset
+
+.
+ Računajmo sedaj 
+\begin_inset Formula $\frac{x^{n+1}}{1-x}<\varepsilon\sim x^{n+1}<\varepsilon\left(1-x\right)\sim\left(n+1\right)\log x<\log\left(\varepsilon\left(1-x\right)\right)\sim n+1>\frac{\log\left(\varepsilon\left(1-x\right)\right)}{\log x}\sim n>\frac{\log\left(\varepsilon\left(1-x\right)\right)}{\log x}-1$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $n$
+\end_inset
+
+ odvisen od 
+\begin_inset Formula $x$
+\end_inset
+
+,
+ vsota ni enakomerno konvergentna.
+\end_layout
+
+\begin_layout Standard
+Poseben primer funkcijskih vrst so funkcijske vrste funkcij oblike 
+\begin_inset Formula $f_{j}=b_{j}\cdot x^{j}$
+\end_inset
+
+,
+ torej potence (monomi).
+\end_layout
+
+\begin_layout Definition*
+Potenčna vrsta je funkcijska vrsta oblike 
+\begin_inset Formula $\sum_{j=1}^{\infty}b_{j}\cdot x^{j}$
+\end_inset
+
+,
+ kjer so a 
+\begin_inset Formula $\left(b_{j}\right)_{j\in\mathbb{N}}$
+\end_inset
+
+ dana realna števila.
+\end_layout
+
+\begin_layout Theorem*
+Cauchy-Hadamard.
+ Za vsako potenčno vrsto obstaja konvergenčni radij 
+\begin_inset Formula $R\in\left[0,\infty\right]\ni:$
+\end_inset
+
+
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+vrsta absolutno konvergira za 
+\begin_inset Formula $\left|x\right|<R$
+\end_inset
+
+,
+\end_layout
+
+\begin_layout Itemize
+vrsta divergira za 
+\begin_inset Formula $\left|x\right|>R$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Theorem*
+Velja 
+\begin_inset Formula $\text{\ensuremath{\frac{1}{R}=\limsup_{k\to\infty}\sqrt[k]{\left|b_{k}\right|}}}$
+\end_inset
+
+,
+ kjer vzamemo 
+\begin_inset Formula $\frac{1}{0}\coloneqq\infty$
+\end_inset
+
+ in 
+\begin_inset Formula $\frac{1}{\infty}\coloneqq0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Rezultat že poznamo za zelo poseben primer 
+\begin_inset Formula $\forall j\in\mathbb{N}:b_{j}=1$
+\end_inset
+
+ (geometrijska vrsta).
+ Ideja dokaza je,
+ da konvergenco vsake potenčne vrste opišemo s pomočjo geometrijske vrste.
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Konvergenca:
+ Za 
+\begin_inset Formula $x=0$
+\end_inset
+
+ vrsta očitno konvergira,
+ zato privzamemo 
+\begin_inset Formula $x\not=0$
+\end_inset
+
+.
+ Definirajmo 
+\begin_inset Formula $R$
+\end_inset
+
+ s formulo iz definicije (
+\begin_inset Formula $R=\frac{1}{\limsup_{k\to\infty}\sqrt[k]{\left|b_{k}\right|}}$
+\end_inset
+
+).
+ Naj bo 
+\begin_inset Formula $x$
+\end_inset
+
+ tak,
+ da 
+\begin_inset Formula $\left|x\right|<R\leq\infty$
+\end_inset
+
+ (sledi 
+\begin_inset Formula $R>0$
+\end_inset
+
+).
+ Naj bo 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+.
+ Tedaj po definiciji 
+\begin_inset Formula $R$
+\end_inset
+
+ velja 
+\begin_inset Formula $\sqrt[k]{\left|b_{k}\right|}\leq\frac{1}{R}+\varepsilon$
+\end_inset
+
+ za vse dovolj velike 
+\begin_inset Formula $k$
+\end_inset
+
+.
+ Za take 
+\begin_inset Formula $k$
+\end_inset
+
+ sledi
+\begin_inset Formula 
+\[
+\left|b_{k}\right|\left|x\right|^{k}\leq\left(\left(\frac{1}{R}+\varepsilon\right)\left|x\right|\right)^{k}.
+\]
+
+\end_inset
+
+Opazimo,
+ da je desna stran neenačbe člen geometrijske vrste,
+ s katero majoriziramo vrsto iz absolutnih vrednosti členov naše vrste.
+ Preverimo,
+ da desna stran konvergira.
+ Konvergira,
+ kadar 
+\begin_inset Formula $\left(\frac{1}{R}+\varepsilon\right)\left|x\right|<1$
+\end_inset
+
+ oziroma 
+\begin_inset Formula $\varepsilon<\frac{1}{\left|x\right|}-\frac{1}{R}$
+\end_inset
+
+.
+ Po 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{pk}{primerjalnem kriteriju}
+\end_layout
+
+\end_inset
+
+ torej naša vrsta absolutno konvergira.
+\end_layout
+
+\begin_layout Itemize
+Divergenca:
+ Vzemimo poljuben 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+.
+ Po definciji 
+\begin_inset Formula $R$
+\end_inset
+
+ sledi,
+ da je 
+\begin_inset Formula $\sqrt[k]{\left|b_{k}\right|}\geq\frac{1}{R}-\varepsilon$
+\end_inset
+
+ za vse dovolj velike 
+\begin_inset Formula $k$
+\end_inset
+
+.
+ Za take 
+\begin_inset Formula $k$
+\end_inset
+
+ sledi
+\begin_inset Formula 
+\[
+\left|b_{k}\right|\left|x\right|^{k}\geq\left(\left(\frac{1}{R}-\varepsilon\right)\left|x\right|\right)^{k}.
+\]
+
+\end_inset
+
+Opazimo,
+ da je desna stran neenačbe člen geometrijske vrste,
+ ki je majorizirana z vrsto iz absolutnih vrednosti členov naše vrste.
+ Desna stran divergira,
+ ko 
+\begin_inset Formula $\left(\frac{1}{R}-\varepsilon\right)\left|x\right|=1$
+\end_inset
+
+ oziroma 
+\begin_inset Formula $\varepsilon=\frac{1}{R}-\frac{1}{\left|x\right|}$
+\end_inset
+
+,
+ zato tudi naša vrsta divergira.
+\end_layout
+
+\end_deeper
+\begin_layout Example*
+Primer konvergenčnega radija potenčne vrste od prej:
+ 
+\begin_inset Formula $\sum_{j=1}^{\infty}x^{j}$
+\end_inset
+
+.
+ Velja 
+\begin_inset Formula $\forall j\in\mathbb{N}:b_{j}=1$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $R=\frac{1}{\limsup_{j\to\infty}\sqrt[k]{\left|b_{k}\right|}}=1$
+\end_inset
+
+,
+ torej po zgornjem izreku vrsta konvergira za 
+\begin_inset Formula $x\in\left(-1,1\right)$
+\end_inset
+
+ in divergira za 
+\begin_inset Formula $x\not\in\left[-1,1\right]$
+\end_inset
+
+.
+ Ročno lahko še preverimo,
+ da divergira tudi v 
+\begin_inset Formula $\left\{ -1,1\right\} $
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Zveznost
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+TODO XXX FIXME PREVERI ŠE V profesrojevih PDFJIH,
+ recimo dodaj dokaz zveznosti x^2
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Ideja:
+ Izdelati želimo formulacijo,
+ s katero preverimo,
+ če lahko z dovolj majhno spremembo 
+\begin_inset Formula $x$
+\end_inset
+
+ povzročimo majhno spremembo funkcijske vrednosti.
+\end_layout
+
+\begin_layout Example*
+Primer nezvezne funkcije je 
+\begin_inset Formula $f\left(x\right)=\begin{cases}
+0 & ;0\leq x<1\\
+1 & ;x=1
+\end{cases}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $D\subseteq\mathbb{R},a\in D$
+\end_inset
+
+ in 
+\begin_inset Formula $f:D\to\mathbb{R}$
+\end_inset
+
+.
+ Pravimo,
+ da je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna v 
+\begin_inset Formula $a$
+\end_inset
+
+,
+ če 
+\begin_inset Formula $\forall\varepsilon>0\exists\delta>0\forall x\in D:\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(a\right)\right|<\varepsilon$
+\end_inset
+
+.
+ 
+\begin_inset Formula $f$
+\end_inset
+
+ je zvezna na množici 
+\begin_inset Formula $x\subseteq D$
+\end_inset
+
+,
+ če je zvezna na vsaki točki v 
+\begin_inset Formula $D$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hypertarget{kzzz}{Karakterizacija zveznosti z zaporedji}
+\end_layout
+
+\end_inset
+
+.
+ Naj bodo 
+\begin_inset Formula $D,a,f$
+\end_inset
+
+ kot prej.
+ Velja:
+ 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna v 
+\begin_inset Formula $a\Leftrightarrow\forall\left(a_{n}\right)_{n\in\mathbb{N}},a_{n}\in D:\lim_{n\to\infty}a_{n}=a\Rightarrow\lim_{n\to\infty}f\left(a_{n}\right)=f\left(a\right)$
+\end_inset
+
+ ZDB 
+\begin_inset Formula $f$
+\end_inset
+
+ je zvezna v 
+\begin_inset Formula $a$
+\end_inset
+
+,
+ če za vsako k 
+\begin_inset Formula $a$
+\end_inset
+
+ konvergentno zaporedje na domeni velja,
+ da funkcijske vrednosti členov zaporedja konvergirajo k funkcijski vrednosti 
+\begin_inset Formula $a$
+\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
+
+ Predpostavimo,
+ da je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna v 
+\begin_inset Formula $a$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\forall\varepsilon>0\exists\delta>0\forall x\in D:\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(a\right)\right|<\varepsilon$
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ poljubno zaporedje na 
+\begin_inset Formula $D$
+\end_inset
+
+,
+ ki konvergira k 
+\begin_inset Formula $a$
+\end_inset
+
+,
+ se pravi 
+\begin_inset Formula $\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow\left|a-a_{n}\right|<\varepsilon$
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+ poljuben.
+ Vsled zveznosti 
+\begin_inset Formula $f$
+\end_inset
+
+ velja,
+ da je 
+\begin_inset Formula $\left|f\left(a_{n}\right)-f\left(a\right)\right|<\varepsilon$
+\end_inset
+
+ za vse take 
+\begin_inset Formula $a_{n}$
+\end_inset
+
+,
+ da velja 
+\begin_inset Formula $\left|a_{n}-a\right|<\delta$
+\end_inset
+
+ za neko 
+\begin_inset Formula $\delta\in\mathbb{R}$
+\end_inset
+
+.
+ Ker je zaporedje 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ konvergentno k 
+\begin_inset Formula $a$
+\end_inset
+
+,
+ so vsi členi po nekem 
+\begin_inset Formula $n_{0}$
+\end_inset
+
+ v 
+\begin_inset Formula $\delta-$
+\end_inset
+
+okolici 
+\begin_inset Formula $a$
+\end_inset
+
+,
+ torej velja pogoj 
+\begin_inset Formula $\left|a_{n}-a\right|<\delta$
+\end_inset
+
+,
+ torej velja 
+\begin_inset Formula $\left|f\left(a_{n}\right)-f\left(a\right)\right|<\varepsilon$
+\end_inset
+
+ za vse 
+\begin_inset Formula $n\geq n_{0}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Leftarrow\right)$
+\end_inset
+
+ PDDRAA 
+\begin_inset Formula $f$
+\end_inset
+
+ ni zvezna v 
+\begin_inset Formula $a$
+\end_inset
+
+.
+ Da pridemo do protislovja,
+ moramo dokazati,
+ da 
+\begin_inset Formula $\exists\left(a_{n}\right)_{n\in\mathbb{N}},a_{n}\in D\ni:\lim_{n\to\infty}a_{n}=a$
+\end_inset
+
+,
+ a vendar 
+\begin_inset Formula $\lim_{n\to\infty}f\left(a_{n}\right)\not=f\left(a\right)$
+\end_inset
+
+.
+ Ker 
+\begin_inset Formula $f$
+\end_inset
+
+ ni zvezna,
+ velja,
+ da 
+\begin_inset Formula $\exists\varepsilon>0\forall\delta>0\exists x\in D\ni:\left|x-a\right|<\delta\wedge\left|f\left(x\right)-f\left(a\right)\right|\geq\varepsilon$
+\end_inset
+
+.
+ Izberimo 
+\begin_inset Formula $\forall n\in\mathbb{N}:\delta_{n}\coloneqq\frac{1}{n}$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $\forall n\in\mathbb{N}\exists\varepsilon>0,x\in D\eqqcolon x_{n}\ni:\left|x_{n}-a\right|<\frac{1}{n}\wedge\left|f\left(x_{n}\right)-f\left(a\right)\right|\geq\varepsilon$
+\end_inset
+
+.
+ S prvim argumentom konjunkcije smo poskrbeli za to,
+ da je naše konstruiramo zaporedje 
+\begin_inset Formula $\left(x_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ konvergentno k 
+\begin_inset Formula $a$
+\end_inset
+
+.
+ Konstruirali smo zaporedje,
+ pri katerem so funkcijske vrednosti za vsak 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+ izven 
+\begin_inset Formula $\varepsilon-$
+\end_inset
+
+okolice 
+\begin_inset Formula $f\left(a\right)$
+\end_inset
+
+,
+ torej zaporedje ne konvergira k 
+\begin_inset Formula $f\left(a\right)$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Theorem*
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hypertarget{kzsppom}{Karakterizacija zveznosti s pomočjo praslik odprtih množic}
+\end_layout
+
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $f:D\to\mathbb{R}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $f$
+\end_inset
+
+ je zvezna na 
+\begin_inset Formula $D\Leftrightarrow$
+\end_inset
+
+ za vsako odprto množico 
+\begin_inset Formula $V\subset\mathbb{R}$
+\end_inset
+
+ je 
+\begin_inset Formula $f^{-1}\left(V\right)$
+\end_inset
+
+ spet odprta množica
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+Za funkcijo 
+\begin_inset Formula $f:D\to V$
+\end_inset
+
+ za 
+\begin_inset Formula $X\subseteq V$
+\end_inset
+
+ definiramo 
+\begin_inset Formula $f^{-1}\left(X\right)\coloneqq\left\{ x\in D;f\left(x\right)\in V\right\} \subseteq D$
+\end_inset
+
+.
+\end_layout
+
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Dokazujemo ekvivalenco.
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Leftarrow\right)$
+\end_inset
+
+ Predpostavimo,
+ da za vsako odprto množico 
+\begin_inset Formula $V\subset\mathbb{R}$
+\end_inset
+
+ je 
+\begin_inset Formula $f^{-1}\left(V\right)$
+\end_inset
+
+ spet odprta množica.
+ Dokazujemo,
+ da je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna na 
+\begin_inset Formula $D$
+\end_inset
+
+.
+ Naj bosta 
+\begin_inset Formula $a\in D,\varepsilon>0$
+\end_inset
+
+ poljubna.
+ Naj bo 
+\begin_inset Formula $V\coloneqq\left(f\left(a\right)-\varepsilon,f\left(a\right)+\varepsilon\right)$
+\end_inset
+
+ odprta množica.
+ Po predpostavki sledi,
+ da je 
+\begin_inset Formula $f^{-1}\left(V\right)$
+\end_inset
+
+ spet odprta.
+ Ker je 
+\begin_inset Formula $a\in f^{-1}\left(V\right)$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $a\in V$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $V$
+\end_inset
+
+ odprta,
+ 
+\begin_inset Formula $\exists\delta>0\ni:\left(a-\delta,a+\delta\right)\in V$
+\end_inset
+
+.
+ Torej 
+\begin_inset Formula $\forall x\in D:\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(a\right)\right|<\varepsilon$
+\end_inset
+
+,
+ torej je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna na 
+\begin_inset Formula $D$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Rightarrow\right)$
+\end_inset
+
+ Predpostavimo,
+ da je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna na 
+\begin_inset Formula $D$
+\end_inset
+
+,
+ to pomeni 
+\begin_inset Formula $\forall a\in D\forall\varepsilon>0\exists\delta>0\forall x\in D:\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(a\right)\right|<\varepsilon$
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $V$
+\end_inset
+
+ poljubna odprta podmnožica 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+ in naj bo 
+\begin_inset Formula $a\in f^{-1}\left(V\right)$
+\end_inset
+
+ poljuben (torej 
+\begin_inset Formula $f\left(a\right)\in V$
+\end_inset
+
+).
+ Ker je 
+\begin_inset Formula $f\left(a\right)\in V$
+\end_inset
+
+,
+ ki je odprta,
+ 
+\begin_inset Formula $\exists\varepsilon>0\ni:\left(f\left(a\right)-\varepsilon,f\left(a\right)+\varepsilon\right)\subseteq V$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna v 
+\begin_inset Formula $a$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\exists\delta>0\forall x\in D:\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(a\right)\right|<\varepsilon$
+\end_inset
+
+,
+ torej je tudi neka odprta okolica 
+\begin_inset Formula $f\left(a\right)$
+\end_inset
+
+ v 
+\begin_inset Formula $f^{-1}\left(V\right)$
+\end_inset
+
+.
+ Ker je bil 
+\begin_inset Formula $a$
+\end_inset
+
+ poljuben,
+ je 
+\begin_inset Formula $f^{-1}\left(V\right)$
+\end_inset
+
+ odprta,
+ ker je bila 
+\begin_inset Formula $V$
+\end_inset
+
+ poljubna,
+ je izrek dokazan.
+\end_layout
+
+\end_deeper
+\begin_layout Theorem*
+Naj bosta 
+\begin_inset Formula $f,g:D\to\mathbb{R}$
+\end_inset
+
+ zvezni v 
+\begin_inset Formula $a\in D$
+\end_inset
+
+.
+ Tedaj so v 
+\begin_inset Formula $a$
+\end_inset
+
+ zvezne tudi funkcije 
+\begin_inset Formula $f+g,f-g,f\cdot g$
+\end_inset
+
+ in 
+\begin_inset Formula $f/g$
+\end_inset
+
+,
+ slednja le,
+ če je 
+\begin_inset Formula $g\left(a\right)\not=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Ker je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna v 
+\begin_inset Formula $a$
+\end_inset
+
+ po 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{kzzz}{izreku o karakterizaciji zveznosti z zaporedji}
+\end_layout
+
+\end_inset
+
+ velja za vsako 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}},\forall n\in\mathbb{N}:a_{n}\subset D,\lim_{n\to\infty}a_{n}=a$
+\end_inset
+
+ tudi 
+\begin_inset Formula $\lim_{n\to\infty}f\left(a_{n}\right)=f\left(a\right)$
+\end_inset
+
+.
+ Po 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{pmkdlim}{izreku iz poglavja o zaporedjih}
+\end_layout
+
+\end_inset
+
+ velja,
+ da 
+\begin_inset Formula $f\left(a_{n}\right)*g\left(a_{n}\right)\to\left(f*g\right)\left(a_{n}\right)$
+\end_inset
+
+ za 
+\begin_inset Formula $*\in\left\{ +,-,\cdot,/\right\} $
+\end_inset
+
+.
+ Zopet uporabimo izrek o karakterizaciji zveznosti z zaporedji,
+ ki pove,
+ da so tudi 
+\begin_inset Formula $f*g$
+\end_inset
+
+ za 
+\begin_inset Formula $*\in\left\{ +,-,\cdot,/\right\} $
+\end_inset
+
+ zvezne v 
+\begin_inset Formula $a$
+\end_inset
+
+.
+ Pri deljenju velja omejitev 
+\begin_inset Formula $f\left(a\right)\not=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+Če sta 
+\begin_inset Formula $D,E\subseteq\mathbb{R}$
+\end_inset
+
+ in 
+\begin_inset Formula $f:D\to E$
+\end_inset
+
+ in 
+\begin_inset Formula $g:E\to\mathbb{R}$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $g\circ f:D\to\mathbb{R}$
+\end_inset
+
+.
+ Hkrati pa,
+ če je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna v 
+\begin_inset Formula $a$
+\end_inset
+
+ in 
+\begin_inset Formula $g$
+\end_inset
+
+ zvezna v 
+\begin_inset Formula $f\left(a\right)$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $g\circ f$
+\end_inset
+
+ zvezna v 
+\begin_inset Formula $a$
+\end_inset
+
+.
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+Velja 
+\begin_inset Formula $\left(g\circ f\right)\left(x\right)=g\left(f\left(x\right)\right)$
+\end_inset
+
+.
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Proof
+Vzemimo poljubno 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}\subseteq D$
+\end_inset
+
+,
+ da 
+\begin_inset Formula $a_{n}\to a\in D$
+\end_inset
+
+.
+ Zopet uporabimo 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{kzzz}{izrek o karakterizaciji zveznosti z zaporedji}
+\end_layout
+
+\end_inset
+
+:
+ ker je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna,
+ velja 
+\begin_inset Formula $f\left(a_{n}\right)\to f\left(a\right)$
+\end_inset
+
+ in ker je 
+\begin_inset Formula $g$
+\end_inset
+
+ zvezna,
+ velja 
+\begin_inset Formula $g\left(f\left(a_{n}\right)\right)\to g\left(f\left(a\right)\right)$
+\end_inset
+
+.
+ Potemtakem 
+\begin_inset Formula $\left(g\circ f\right)\left(a_{n}\right)\to\left(g\circ f\right)\left(a\right)$
+\end_inset
+
+ in po istem izreku je 
+\begin_inset Formula $g\circ f$
+\end_inset
+
+ zvezna na 
+\begin_inset Formula $D$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+Vsi polinomi so zvezni na 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Vzemimo 
+\begin_inset Formula $p\left(x\right)=\sum_{k=0}^{n}a_{k}k^{k}$
+\end_inset
+
+.
+ Uporabimo prejšnji izrek.
+ Polinom je sestavljen iz vsote konstantne funkcije,
+ zmnožene z identiteto,
+ ki je s seboj 
+\begin_inset Formula $n-$
+\end_inset
+
+krat množena.
+ Ker vsota in množenje ohranjata zveznost,
+ je treba dokazati le,
+ da je 
+\begin_inset Formula $f\left(x\right)=x$
+\end_inset
+
+ zvezna in da so 
+\begin_inset Formula $\forall c\in\mathbb{R}:f\left(x\right)=c$
+\end_inset
+
+ zvezne.
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $f\left(x\right)=x$
+\end_inset
+
+ Ali velja 
+\begin_inset Formula $\forall\varepsilon>0\exists\delta>0\ni:\forall x\in\mathbb{R}:\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(a\right)\right|<\varepsilon$
+\end_inset
+
+?
+ Da,
+ velja.
+ Vzamemo lahko katerokoli 
+\begin_inset Formula $\delta\in(0,\varepsilon]$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $f\left(x\right)=c$
+\end_inset
+
+ Naj bo 
+\begin_inset Formula $c\in\mathbb{R}$
+\end_inset
+
+ poljuben.
+ Tu je 
+\begin_inset Formula $\left|f\left(x\right)-f\left(a\right)\right|=\left|c-c\right|=0$
+\end_inset
+
+,
+ torej je desna stran implikacije vedno resnična,
+ torej je implikacija vedno resnična.
+\end_layout
+
+\end_deeper
+\begin_layout Theorem*
+Vse elementarne funkcije so na njihovih definicijskih območjih povsod zvezne.
+ To so:
+ polinomi,
+ potence,
+ racionalne funkcije,
+ koreni,
+ eksponentne funkcije,
+ logaritmi,
+ trigonometrične,
+ ciklometrične in kombinacije neskončno mnogo naštetih,
+ spojenih s 
+\begin_inset Formula $+,-,\cdot,/,\circ$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Tega izreka ne bomo dokazali.
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $f\left(x\right)\coloneqq\log\left(\sin^{3}x+\frac{1}{8}\right)+\frac{1}{\sqrt[4]{x-7}}$
+\end_inset
+
+ je zvezna povsod,
+ kjer je definirana.
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $\varepsilon>0,a\in\mathbb{R}$
+\end_inset
+
+ in 
+\begin_inset Formula $f:\left(a-\varepsilon,a+\varepsilon\right)\setminus\left\{ a\right\} \to\mathbb{R}$
+\end_inset
+
+.
+ Pravimo,
+ da je 
+\begin_inset Formula $L\in\mathbb{R}$
+\end_inset
+
+ limita 
+\begin_inset Formula $f$
+\end_inset
+
+ v točki 
+\begin_inset Formula $a$
+\end_inset
+
+ (zapišemo 
+\begin_inset Formula $L=\lim_{x\to a}f\left(x\right)$
+\end_inset
+
+),
+ če za vsako zaporedje 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}},a_{n}\in\left(a-\varepsilon,a+\varepsilon\right)\setminus\left\{ a\right\} $
+\end_inset
+
+,
+ za katero velja 
+\begin_inset Formula $a_{n}\to a$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $f\left(a_{n}\right)\to L$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+ZDB če 
+\begin_inset Formula $\forall\varepsilon>0\exists\delta>0\ni:\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-L\right|<\varepsilon$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+ZDB če za 
+\begin_inset Formula $\overline{f}:\left(a-\varepsilon,a+\varepsilon\right)\to\mathbb{R}$
+\end_inset
+
+ s predpisom 
+\begin_inset Formula $\overline{f}\left(x\right)\coloneqq\begin{cases}
+f\left(x\right) & ;x\in\left(a-\varepsilon,a+\varepsilon\right)\setminus\left\{ a\right\} \\
+L & ;x\in a
+\end{cases}$
+\end_inset
+
+ velja,
+ da je zvezna v 
+\begin_inset Formula $a$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Note*
+Vrednost 
+\begin_inset Formula $f\left(a\right)$
+\end_inset
+
+,
+ če sploh obstaja,
+ nima vloge pri vrednosti limite.
+\end_layout
+
+\begin_layout Corollary*
+Naj bo 
+\begin_inset Formula $a\in D\subseteq\mathbb{R}$
+\end_inset
+
+ in 
+\begin_inset Formula $f:D\to\mathbb{R}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $f$
+\end_inset
+
+ je zvezna v 
+\begin_inset Formula $a\Leftrightarrow\lim_{x\to a}f\left(x\right)=f\left(a\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Kvadratna funkcija 
+\begin_inset Formula $f\left(x\right)=x^{2}$
+\end_inset
+
+ je zvezna.
+ Vzemimo poljuben 
+\begin_inset Formula $a\in\mathbb{R},\varepsilon>0$
+\end_inset
+
+.
+ Obstajati mora taka 
+\begin_inset Formula $\delta>0\ni:\forall x\in\mathbb{R}:\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(a\right)\right|<\varepsilon$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Podan imamo torej 
+\begin_inset Formula $a$
+\end_inset
+
+ in 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+,
+ želimo najti 
+\begin_inset Formula $\delta$
+\end_inset
+
+.
+ Želimo priti do neenakosti,
+ ki ima na manjši strani 
+\begin_inset Formula $\left|f\left(x\right)-f\left(a\right)\right|=\left|x^{2}-a^{2}\right|$
+\end_inset
+
+ in na večji strani nek izraz z 
+\begin_inset Formula $\left|x-a\right|$
+\end_inset
+
+,
+ da ta 
+\begin_inset Formula $\left|x-a\right|$
+\end_inset
+
+ nadomestimo z 
+\begin_inset Formula $\delta$
+\end_inset
+
+ in nato večjo stran enačimo z 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+,
+ da izrazimo 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+ v odvisnosti od 
+\begin_inset Formula $\delta$
+\end_inset
+
+ in 
+\begin_inset Formula $a$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Računajmo:
+ 
+\begin_inset Formula $\left|x^{2}-a^{2}\right|=\left|x-a\right|\left|x+a\right|$
+\end_inset
+
+.
+ Predelajmo izraz 
+\begin_inset Formula $\left|x+a\right|=\left|\left(x-a\right)+2a\right|\leq\left|x-a\right|+\left|2a\right|$
+\end_inset
+
+,
+ torej skupaj 
+\begin_inset Formula $\left|x^{2}-a^{2}\right|\leq\left|x-a\right|\left(\left|x-a\right|+\left|2a\right|\right)$
+\end_inset
+
+.
+ Sedaj nadomestimo 
+\begin_inset Formula $\left|x-a\right|$
+\end_inset
+
+ z 
+\begin_inset Formula $\delta$
+\end_inset
+
+:
+ 
+\begin_inset Formula $\left|x^{2}-a^{2}\right|\leq\delta\left(\delta+\left|2a\right|\right)$
+\end_inset
+
+.
+ Iščemo tak 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+,
+ da velja 
+\begin_inset Formula $\left|x^{2}-a^{2}\right|<\varepsilon$
+\end_inset
+
+,
+ zato enačimo 
+\begin_inset Formula $\delta\left(\delta+\left|2a\right|\right)=\varepsilon$
+\end_inset
+
+ in dobimo kvadratno enačbo 
+\begin_inset Formula $\delta^{2}+\left|2a\right|\delta-\varepsilon=0$
+\end_inset
+
+,
+ ki jo rešimo z obrazcem za ničle:
+\begin_inset Formula 
+\[
+\delta_{1,2}=\frac{-2\left|a\right|\pm\sqrt{4\left|a\right|^{2}-4\varepsilon}}{2}=-\left|a\right|\pm\sqrt{\left|a\right|^{2}-\varepsilon}
+\]
+
+\end_inset
+
+Toda ker iščemo le pozitivne 
+\begin_inset Formula $\delta$
+\end_inset
+
+,
+ je edina rešitev
+\begin_inset Formula 
+\[
+\delta=-\left|a\right|+\sqrt{\left|a\right|^{2}-\varepsilon}=\sqrt{\left|a\right|^{2}-\varepsilon}-\left|a\right|=\frac{\sqrt{\left|a\right|^{2}-\varepsilon}-\left|a\right|}{1}=\frac{\left(\sqrt{\left|a\right|^{2}-\varepsilon}-\left|a\right|\right)\left(\sqrt{\left|a\right|^{2}-\varepsilon}+\left|a\right|\right)}{\sqrt{\left|a\right|^{2}-\varepsilon}+\left|a\right|}=\frac{\varepsilon}{\sqrt{\left|a\right|^{2}-\varepsilon}+\left|a\right|}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $D\subset\mathbb{R},a\in\mathbb{R}\ni:\forall\varepsilon>0:D\cap\left(a,a+\varepsilon\right)\not=\emptyset$
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $f:D\to\mathbb{R}$
+\end_inset
+
+.
+ Število 
+\begin_inset Formula $L_{+}\in\mathbb{R}$
+\end_inset
+
+ je desna limita funkcije 
+\begin_inset Formula $f$
+\end_inset
+
+ v točki 
+\begin_inset Formula $a$
+\end_inset
+
+,
+ če 
+\begin_inset Formula $\forall\left(a_{n}\right)_{n\in\mathbb{N}}\subset D\cap\left(a,\infty\right):a_{n}\to a\Rightarrow f\left(a_{n}\right)\to L_{+}$
+\end_inset
+
+ ZDB če za vsako k 
+\begin_inset Formula $a$
+\end_inset
+
+ konvergentno zaporedje s členi desno od 
+\begin_inset Formula $a$
+\end_inset
+
+ velja,
+ da funkcijske vrednosti členov konvergirajo k 
+\begin_inset Formula $L_{+}$
+\end_inset
+
+.
+ Oznaka 
+\begin_inset Formula $L_{+}=\lim_{x\to a^{+}}f\left(x\right)=\lim_{x\searrow a}f\left(x\right)=f\left(a+0\right)$
+\end_inset
+
+.
+ Podobno definiramo tudi levo limito 
+\begin_inset Formula $L_{-}=\lim_{x\to a^{-}}f\left(x\right)=\lim_{x\nearrow a}f\left(x\right)=f\left(a-0\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+Naj bo 
+\begin_inset Formula $D\subset\mathbb{R}$
+\end_inset
+
+ in 
+\begin_inset Formula $a\in\mathbb{R}$
+\end_inset
+
+ da velja 
+\begin_inset Formula $\forall\varepsilon>0:D\cap\left(a,a-\varepsilon\right)\not=\emptyset\wedge D\cap\left(a,a+\varepsilon\right)\not=\emptyset$
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $f:D\to\mathbb{R}$
+\end_inset
+
+.
+ Velja 
+\begin_inset Formula $\exists\lim_{x\to a}f\left(x\right)\Leftrightarrow\exists\lim_{x\nearrow a}f\left(x\right)\wedge\exists\lim_{x\searrow a}f\left(x\right)\wedge\lim_{x\nearrow a}f\left(x\right)=\lim_{x\searrow a}f\left(x\right)$
+\end_inset
+
+ V tem primeru velja 
+\begin_inset Formula $\lim_{x\to a}f\left(x\right)=\lim_{x\nearrow a}f\left(x\right)=\lim_{x\searrow a}f\left(x\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Označimo 
+\begin_inset Formula $\lim_{x\searrow a}f\left(x\right)\eqqcolon f\left(a+0\right),\lim_{x\nearrow a}f\left(x\right)\eqqcolon f\left(a-0\right)$
+\end_inset
+
+.
+ Če 
+\begin_inset Formula $\exists f\left(a+0\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $\exists f\left(a-0\right)$
+\end_inset
+
+,
+ vendar 
+\begin_inset Formula $f\left(a+0\right)\not=f\left(a-0\right)$
+\end_inset
+
+,
+ pravimo,
+ da ima 
+\begin_inset Formula $f$
+\end_inset
+
+ v točki 
+\begin_inset Formula $a$
+\end_inset
+
+ 
+\begin_inset Quotes gld
+\end_inset
+
+skok
+\begin_inset Quotes grd
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $\lim_{x\to0}\frac{1}{1+e^{1/x}}$
+\end_inset
+
+ ne obstaja.
+ Zakaj?
+ Izračunajmo levo in desno limito:
+\begin_inset Formula 
+\[
+\lim_{x\searrow0}\frac{1}{1+e^{1/x}}=0,\lim_{x\nearrow0}\frac{1}{1+e^{1/x}}=1
+\]
+
+\end_inset
+
+Toda 
+\begin_inset Formula $\exists\lim_{x\to a}f\left(x\right)\Leftrightarrow\exists\lim_{x\nearrow a}f\left(x\right)\wedge\exists\lim_{x\searrow a}f\left(x\right)\wedge\lim_{x\nearrow a}f\left(x\right)=\lim_{x\searrow a}f\left(x\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Funkcija 
+\begin_inset Formula $f$
+\end_inset
+
+ je na intervalu 
+\begin_inset Formula $D$
+\end_inset
+
+ odsekoma zvezna,
+ če je zvezna povsod na 
+\begin_inset Formula $D$
+\end_inset
+
+,
+ razen morda v končno mnogo točkah,
+ v katerih ima skok.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+Naj bo 
+\begin_inset Formula $f:\mathbb{R}\setminus\left\{ 0\right\} \to\mathbb{R}$
+\end_inset
+
+ s predpisom 
+\begin_inset Formula $x\mapsto\frac{\sin x}{x}$
+\end_inset
+
+.
+ Zanima nas,
+ ali obstaja 
+\begin_inset Formula $\lim_{x\to0}f\left(x\right)$
+\end_inset
+
+.
+ Grafični dokaz.
+\end_layout
+
+\begin_layout Example*
+\begin_inset Float figure
+placement document
+alignment document
+wide false
+sideways false
+status open
+
+\begin_layout Plain Layout
+TODO XXX FIXME SKICA S TKZ EUCLID,
+ glej ZVZ III/ANA1P1120/str.8
+\end_layout
+
+\begin_layout Plain Layout
+\begin_inset Caption Standard
+
+\begin_layout Plain Layout
+Skica.
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Očitno velja 
+\begin_inset Formula $\triangle ABD\subset$
+\end_inset
+
+ krožni izsek 
+\begin_inset Formula $DAB\subset\triangle ABC$
+\end_inset
+
+,
+ torej za njihove ploščine velja 
+\begin_inset Formula 
+\[
+\frac{\sin x}{2}\leq\frac{x}{2\pi}\cdot x=\frac{x}{2}\leq\frac{\tan x}{2}\quad\quad\quad\quad/\cdot\frac{2}{\sin x}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+1\leq\frac{x}{\sin x}\leq\frac{1}{\cos x}\quad\quad\quad\quad/\lim_{x\to0}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\lim_{x\to0}1\leq\lim_{x\to0}\frac{x}{\sin x}\leq\lim_{x\to0}\frac{1}{\cos x}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+1\leq\lim_{x\to0}\frac{x}{\sin x}\leq1
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\lim_{x\to0}\frac{x}{\sin x}=1
+\]
+
+\end_inset
+
+Da naš sklep res potrdimo,
+ je potreben spodnji izrek.
+\end_layout
+
+\begin_layout Theorem*
+Če za 
+\begin_inset Formula $f,g,h:D\to\mathbb{R}$
+\end_inset
+
+ velja za 
+\begin_inset Formula $a\in D$
+\end_inset
+
+:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $\exists\varepsilon>0\forall x\in\left(a-\varepsilon,a+\varepsilon\right)\setminus\left\{ a\right\} :f\left(x\right)\leq g\left(x\right)\leq h\left(x\right)$
+\end_inset
+
+ in hkrati
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $\exists\lim_{x\to a}f\left(x\right),\lim_{x\to a}h\left(x\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $\lim_{x\to a}f\left(x\right)=\lim_{x\to a}h\left(x\right)\eqqcolon L$
+\end_inset
+
+,
+ tedaj tudi 
+\begin_inset Formula $\exists\lim_{x\to a}g\left(x\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $\lim_{x\to a}g\left(x\right)=L$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $A=A\left(x\right)\coloneqq\max\left\{ \left|f\left(x\right)-L\right|,\left|h\left(x\right)-L\right|\right\} $
+\end_inset
+
+.
+ Velja
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $g\left(x\right)-L\leq h\left(x\right)-L\leq\left|h\left(x\right)-L\right|\leq A\left(x\right)$
+\end_inset
+
+ in
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $L-g\left(x\right)\leq L-f\left(x\right)\leq\left|f\left(x\right)-L\right|\leq A\left(x\right)$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Proof
+Posledično 
+\begin_inset Formula $\left|g\left(x\right)-L\right|\leq A\left(x\right)$
+\end_inset
+
+.
+ Naj bo sedaj 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+ poljuben.
+ Tedaj velja 
+\begin_inset Formula $\exists\delta_{1}>0\ni:\left|x-a\right|<\delta_{1}\Rightarrow\left|f\left(x\right)-L\right|<\varepsilon$
+\end_inset
+
+ in 
+\begin_inset Formula $\exists\delta_{2}>0\ni:\left|x-a\right|<\delta_{2}\Rightarrow\left|h\left(x\right)-L\right|<\varepsilon$
+\end_inset
+
+.
+ Za 
+\begin_inset Formula $\delta\coloneqq\min\left\{ \delta_{1},\delta_{2}\right\} $
+\end_inset
+
+ torej velja 
+\begin_inset Formula $\left|x-a\right|<\delta\Rightarrow\left|g\left(x\right)-L\right|<\varepsilon$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsection
+Zvezne funkcije na kompaktnih množicah
+\end_layout
+
+\begin_layout Definition*
+Množica 
+\begin_inset Formula $K\subseteq\mathbb{R}$
+\end_inset
+
+ je kompaktna 
+\begin_inset Formula $\Leftrightarrow$
+\end_inset
+
+ je zaprta in omejena ZDB je unija zaprtih intervalov.
+\end_layout
+
+\begin_layout Theorem*
+Naj bo 
+\begin_inset Formula $K\subset\mathbb{R}$
+\end_inset
+
+ kompaktna in 
+\begin_inset Formula $f:K\to\mathbb{R}$
+\end_inset
+
+ zvezna.
+ Tedaj je 
+\begin_inset Formula $f$
+\end_inset
+
+ 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hypertarget{zfnkm}{omejena in doseže minimum in maksimum}
+\end_layout
+
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Primeri funkcij.
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $f_{1}\left(x\right)=\frac{1}{x}$
+\end_inset
+
+ na 
+\begin_inset Formula $I_{1}=(0,1]$
+\end_inset
+
+.
+ 
+\begin_inset Formula $f_{1}$
+\end_inset
+
+ je zvezna in 
+\begin_inset Formula $\lim_{x\to0}f_{1}\left(x\right)=\infty$
+\end_inset
+
+,
+ torej ni omejena,
+ a 
+\begin_inset Formula $I_{1}$
+\end_inset
+
+ ni zaprt.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $f_{2}\left(x\right)=\begin{cases}
+0 & ;x=0\\
+\frac{1}{x} & ;x\in(0,1]
+\end{cases}$
+\end_inset
+
+ ni omejena in je definirana na kompaktni množici,
+ a ni zvezna.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $f_{3}\left(x\right)=x$
+\end_inset
+
+ na 
+\begin_inset Formula $x\in\left(0,1\right)$
+\end_inset
+
+.
+ Je omejena,
+ ne doseže maksimuma,
+ a 
+\begin_inset Formula $D_{f_{3}}$
+\end_inset
+
+ ni kompaktna (ni zaprta).
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $f_{4}\left(x\right)=\begin{cases}
+x & ;x\in\left(0,1\right)\\
+\frac{1}{2} & ;x\in\left\{ 0,1\right\} 
+\end{cases}$
+\end_inset
+
+.
+ Velja 
+\begin_inset Formula $\sup f_{4}=1$
+\end_inset
+
+,
+ ampak ga ne doseže,
+ a ni zvezna
+\end_layout
+
+\end_deeper
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $K\subseteq\mathbb{R}$
+\end_inset
+
+ kompaktna in 
+\begin_inset Formula $f:K\to\mathbb{R}$
+\end_inset
+
+ zvezna.
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Omejenost navzgor:
+ PDDRAA 
+\begin_inset Formula $f$
+\end_inset
+
+ ni navzgor omejena.
+ Tedaj 
+\begin_inset Formula $\forall n\in\mathbb{N}\exists x_{n}\in K\ni:f\left(x_{n}\right)\geq n$
+\end_inset
+
+ (*).
+ Ker je 
+\begin_inset Formula $\left(x_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ omejeno zaporedje (vsi členi so na kompaktni 
+\begin_inset Formula $K$
+\end_inset
+
+),
+ ima stekališče,
+ recimo mu 
+\begin_inset Formula $s\in\mathbb{R}$
+\end_inset
+
+.
+ Vemo,
+ da tedaj obstaja podzaporedje 
+\begin_inset Formula $\left(x_{n_{k}}\right)_{k\in\mathbb{N}}\ni:s=\lim_{k\to\infty}x_{n_{k}}$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $K$
+\end_inset
+
+ tudi zaprta,
+ sledi 
+\begin_inset Formula $s\in K$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna na 
+\begin_inset Formula $K$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $f\left(s\right)=\lim_{k\to\infty}f\left(x_{n_{k}}\right)$
+\end_inset
+
+.
+ Toda po (*) sledi 
+\begin_inset Formula $\lim_{k\to\infty}f\left(x_{n_{k}}\right)=\infty$
+\end_inset
+
+,
+ zato 
+\begin_inset Formula $f\left(s\right)=\infty$
+\end_inset
+
+,
+ kar ni mogoče,
+ saj je 
+\begin_inset Formula $f\left(s\right)\in\mathbb{R}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $\rightarrow\!\leftarrow$
+\end_inset
+
+.
+ Torej je 
+\begin_inset Formula $f$
+\end_inset
+
+ navzgor omejena.
+\end_layout
+
+\begin_layout Itemize
+Omejenost navzdol:
+ PDDRAA 
+\begin_inset Formula $f$
+\end_inset
+
+ ni navzdol omejena.
+ Tedaj 
+\begin_inset Formula $\forall n\in\mathbb{N}\exists x_{n}\in K\ni:f\left(x_{n}\right)\leq-n$
+\end_inset
+
+ (*).
+ Ker je 
+\begin_inset Formula $\left(x_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ omejeno zaporedje (vsi členi so na kompaktni 
+\begin_inset Formula $K$
+\end_inset
+
+),
+ ima stekališče,
+ recimo mu 
+\begin_inset Formula $s\in\mathbb{R}$
+\end_inset
+
+.
+ Vemo,
+ da tedaj obstaja podzaporedje 
+\begin_inset Formula $\left(x_{n_{k}}\right)_{k\in\mathbb{N}}\ni:s=\lim_{k\to\infty}x_{n_{k}}$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $K$
+\end_inset
+
+ tudi zaprta,
+ sledi 
+\begin_inset Formula $s\in K$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna na 
+\begin_inset Formula $K$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $f\left(s\right)=\lim_{k\to\infty}f\left(x_{n_{k}}\right)$
+\end_inset
+
+.
+ Toda po (*) sledi 
+\begin_inset Formula $\lim_{k\to\infty}f\left(s_{n_{k}}\right)=-\infty$
+\end_inset
+
+,
+ zato 
+\begin_inset Formula $f\left(s\right)=-\infty$
+\end_inset
+
+,
+ kar ni mogoče,
+ saj je 
+\begin_inset Formula $f\left(s\right)\in\mathbb{R}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $\rightarrow\!\leftarrow$
+\end_inset
+
+.
+ Torej je 
+\begin_inset Formula $f$
+\end_inset
+
+ navzgor omejena.
+ 
+\end_layout
+
+\begin_layout Itemize
+Doseže maksimum:
+ Označimo 
+\begin_inset Formula $M\coloneqq\sup_{x\in K}f\left(x\right)$
+\end_inset
+
+.
+ Ravnokar smo dokazali,
+ da 
+\begin_inset Formula $M<\infty$
+\end_inset
+
+.
+ Po definiciji supremuma 
+\begin_inset Formula $\forall n\in\mathbb{N}\exists t_{n}\in K\ni:f\left(t_{n}\right)>M-\frac{1}{n}$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $K$
+\end_inset
+
+ omejena,
+ ima 
+\begin_inset Formula $\left(t_{n}\right)_{n}$
+\end_inset
+
+ stekališče in ker je zaprta,
+ velja 
+\begin_inset Formula $t\in K$
+\end_inset
+
+,
+ zato 
+\begin_inset Formula $\exists$
+\end_inset
+
+ podzaporedje 
+\begin_inset Formula $\left(t_{n_{j}}\right)_{j\in\mathbb{N}}\ni:t=\lim_{j\to\infty}t_{n_{j}}$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna,
+ velja 
+\begin_inset Formula $f\left(t\right)=\lim_{j\to\infty}f\left(t_{n_{j}}\right)$
+\end_inset
+
+.
+ Toda ker 
+\begin_inset Formula $f\left(t_{n_{j}}\right)>M-\frac{1}{n_{j}}\geq M-\frac{1}{j}$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $f\left(t\right)\geq M$
+\end_inset
+
+.
+ Hkrati po definiciji 
+\begin_inset Formula $M$
+\end_inset
+
+ velja 
+\begin_inset Formula $f\left(t\right)\leq M$
+\end_inset
+
+.
+ Sledi 
+\begin_inset Formula $M=f\left(t\right)$
+\end_inset
+
+ in zato 
+\begin_inset Formula $M=\max_{x\in K}f\left(x\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Doseže minimum:
+ Označimo 
+\begin_inset Formula $M\coloneqq\inf_{x\in K}f\left(x\right)$
+\end_inset
+
+.
+ Ko smo dokazali omejenost,
+ smo dokazali,
+ da 
+\begin_inset Formula $M>-\infty$
+\end_inset
+
+.
+ Po definiciji infimuma 
+\begin_inset Formula $\forall n\in\mathbb{N}\exists t_{n}\in K\ni:f\left(t_{n}\right)<M+\frac{1}{n}$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $K$
+\end_inset
+
+ omejena,
+ ima 
+\begin_inset Formula $\left(t_{n}\right)_{n}$
+\end_inset
+
+ stekališče in ker je zaprta,
+ velja 
+\begin_inset Formula $t\in K$
+\end_inset
+
+,
+ zato 
+\begin_inset Formula $\exists$
+\end_inset
+
+ podzaporedje 
+\begin_inset Formula $\left(t_{n_{j}}\right)_{j\in\mathbb{N}}\ni:t=\lim_{j\to\infty}t_{n_{j}}$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna,
+ velja 
+\begin_inset Formula $f\left(t\right)=\lim_{j\to\infty}f\left(t_{n_{j}}\right)$
+\end_inset
+
+.
+ Toda ker 
+\begin_inset Formula $f\left(t_{n_{j}}\right)<M-\frac{1}{n_{j}}\leq M-\frac{1}{j}$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $f\left(t\right)\leq M$
+\end_inset
+
+.
+ Hkrati po definiciji 
+\begin_inset Formula $M$
+\end_inset
+
+ velja 
+\begin_inset Formula $f\left(t\right)\geq M$
+\end_inset
+
+.
+ Sledi 
+\begin_inset Formula $M=f\left(t\right)$
+\end_inset
+
+ in zato 
+\begin_inset Formula $M=\min_{x\in K}f\left(x\right)$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Theorem*
+Naj bo 
+\begin_inset Formula $f:\left[a,b\right]\to\mathbb{R}$
+\end_inset
+
+ zvezna in 
+\begin_inset Formula $f\left(a\right)f\left(b\right)<0$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $\exists\xi\in\left(a,b\right)\ni:f\left(\xi\right)=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Interval 
+\begin_inset Formula $I_{0}=\left[a,b\right]$
+\end_inset
+
+ razpolovimo.
+ To pomeni,
+ da pogledamo levo in desno polovico intervala 
+\begin_inset Formula $I_{0}$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\left[a,\frac{a+b}{2}\right]$
+\end_inset
+
+ in 
+\begin_inset Formula $\left[\frac{a+b}{2},b\right]$
+\end_inset
+
+.
+ Če je 
+\begin_inset Formula $f\left(\frac{a+b}{2}\right)=0$
+\end_inset
+
+,
+ smo našli iskano točko 
+\begin_inset Formula $\xi$
+\end_inset
+
+,
+ sicer z 
+\begin_inset Formula $I_{1}$
+\end_inset
+
+ označimo katerokoli izmed polovic,
+ ki ima 
+\begin_inset Formula $f$
+\end_inset
+
+ v krajiščih različno predznačene funkcijske vrednosti.
+ Torej 
+\begin_inset Formula $I_{1}=\begin{cases}
+\left[a,\frac{a+b}{2}\right] & ;f\left(a\right)f\left(\frac{a+b}{2}\right)<0\\
+\left[\frac{a+b}{2},b\right] & ;f\left(\frac{a+b}{2}\right)f\left(b\right)<0
+\end{cases}$
+\end_inset
+
+.
+ S postopkom nadaljujemo.
+ Če v končno mnogo korakih najdemo 
+\begin_inset Formula $\xi$
+\end_inset
+
+,
+ da je 
+\begin_inset Formula $f\left(\xi\right)=0$
+\end_inset
+
+,
+ fino,
+ sicer pa dobimo zaporedje intervalov 
+\begin_inset Formula $I_{n}=\left[a_{n},b_{n}\right]\subset\left[a,b\right]=I_{0}\ni:$
+\end_inset
+
+
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $\forall n\in\mathbb{N}:\left|I_{n}\right|=2^{-n}\left|I_{0}\right|$
+\end_inset
+
+ in
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\forall n\in\mathbb{N}:I_{n+1}\subset I_{n}$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $a_{n+1}\geq a_{n}\wedge b_{n+1}\leq b_{n}$
+\end_inset
+
+,
+ in
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset CommandInset label
+LatexCommand label
+name "enu:različni-predznaki-istoležnih-clenov"
+
+\end_inset
+
+
+\begin_inset Formula $\forall n\in\mathbb{N}:f\left(a_{n}\right)f\left(b_{n}\right)<0$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Proof
+Ker sta zaporedji 
+\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
+
+ omejeni in monotoni,
+ imata po 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{kmoz}{izreku o konvergenci monotonih in omejenih zaporedij}
+\end_layout
+
+\end_inset
+
+ limiti 
+\begin_inset Formula $\alpha\coloneqq\lim_{n\to\infty}a_{n}=\sup_{n\in\mathbb{N}}a_{n}$
+\end_inset
+
+ in 
+\begin_inset Formula $\beta\coloneqq\lim_{n\to\infty}=\sup_{n\in\mathbb{N}}b_{n}$
+\end_inset
+
+ in 
+\begin_inset Formula $\alpha,\beta\in I_{0}$
+\end_inset
+
+,
+ ker je 
+\begin_inset Formula $I_{0}$
+\end_inset
+
+ zaprt.
+\end_layout
+
+\begin_layout Proof
+Sledi 
+\begin_inset Formula $\forall n\in\mathbb{N}:\left|\alpha-\beta\right|=\beta-\alpha\leq b_{n}-a_{n}=\left|I_{n}\right|=2^{-n}\left|I_{0}\right|$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\lim_{n\to\infty}\left|\alpha-\beta\right|=0\Rightarrow\alpha-\beta=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Ker je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna in 
+\begin_inset Formula $a_{n},b_{n},\xi\in I_{0}$
+\end_inset
+
+,
+ sledi 
+\begin_inset Formula $\lim_{n\to\infty}f\left(a_{n}\right)=f\left(\alpha\right)=f\left(\xi\right)=f\left(\beta\right)=\lim_{n\to\infty}f\left(b_{n}\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Po točki 
+\begin_inset CommandInset ref
+LatexCommand ref
+reference "enu:različni-predznaki-istoležnih-clenov"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+ velja 
+\begin_inset Formula $f\left(\alpha\right)f\left(\beta\right)\leq0$
+\end_inset
+
+.
+ Ker pa 
+\begin_inset Formula $f\left(\alpha\right)=f\left(\beta\right)$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $f\left(\alpha\right)=f\left(\beta\right)=f\left(\xi\right)=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Corollary*
+Naj bo 
+\begin_inset Formula $I=\left[a,b\right]$
+\end_inset
+
+ omejen zaprt interval 
+\begin_inset Formula $\in\mathbb{R}$
+\end_inset
+
+ in 
+\begin_inset Formula $f:I\to\mathbb{R}$
+\end_inset
+
+ zvezna.
+ Tedaj 
+\begin_inset Formula $\exists x_{-},x_{+}\in I\ni:\forall x\in I:f\left(x\right)\in\left[f\left(x_{-}\right),f\left(x_{+}\right)\right]$
+\end_inset
+
+ in 
+\begin_inset Formula $\forall y\in\left[f\left(x_{-}\right),f\left(x_{+}\right)\right]\exists x\in I\ni:y=f\left(x\right)$
+\end_inset
+
+ ZDB 
+\begin_inset Formula $f\left(I\right)=\left[f\left(x_{-}\right),f\left(x_{+}\right)\right]$
+\end_inset
+
+ ZDB zvezna funkcija na zaprtem intervalu 
+\begin_inset Formula $\left[a,b\right]$
+\end_inset
+
+ doseže vse funkcijske vrednosti na intervalu 
+\begin_inset Formula $\left[f\left(a\right),f\left(b\right)\right]$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Dokaz posledice.
+ Naj bo 
+\begin_inset Formula $y$
+\end_inset
+
+ poljuben.
+ Če je 
+\begin_inset Formula $y=f\left(x_{-}\right)$
+\end_inset
+
+,
+ smo našli 
+\begin_inset Formula $x=x_{-}$
+\end_inset
+
+.
+ Če je 
+\begin_inset Formula $y=f\left(x_{+}\right)$
+\end_inset
+
+,
+ smo našli 
+\begin_inset Formula $x=x_{+}$
+\end_inset
+
+.
+ Sicer pa je 
+\begin_inset Formula $f\left(x_{-}\right)<y<f\left(x_{+}\right)$
+\end_inset
+
+.
+ Oglejmo si funkcijo 
+\begin_inset Formula $g\left(x\right)\coloneqq f\left(x\right)-y$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $g\left(x_{-}\right)=f\left(x_{-}\right)-y<0$
+\end_inset
+
+ in 
+\begin_inset Formula $g\left(x_{+}\right)=f\left(x_{+}\right)-y>0$
+\end_inset
+
+ in 
+\begin_inset Formula $g$
+\end_inset
+
+ zvezna na 
+\begin_inset Formula $\left[x_{-}-y,x_{+}-y\right]$
+\end_inset
+
+,
+ torej po prejšnjem izreku 
+\begin_inset Formula $\exists x\in\left[x_{-}-y,x_{+}-y\right]\ni:g\left(x\right)=0$
+\end_inset
+
+,
+ kar pomeni ravno 
+\begin_inset Formula $f\left(x\right)=y$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+Naj bo 
+\begin_inset Formula $I$
+\end_inset
+
+ poljuben interval med 
+\begin_inset Formula $a,b\in\mathbb{R}\cup\left\{ -\infty,\infty\right\} $
+\end_inset
+
+ in 
+\begin_inset Formula $f:I\to\mathbb{R}$
+\end_inset
+
+ 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hypertarget{zism}{zvezna in strogo monotona}
+\end_layout
+
+\end_inset
+
+.
+ Tedaj je 
+\begin_inset Formula $f\left(I\right)$
+\end_inset
+
+ interval med 
+\begin_inset Formula $f\left(a+0\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $f\left(a-0\right)$
+\end_inset
+
+.
+ Inverzna funkcija 
+\begin_inset Formula $f^{-1}$
+\end_inset
+
+ je definirana na 
+\begin_inset Formula $f\left(I\right)$
+\end_inset
+
+ in zvezna.
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $f\coloneqq\arctan$
+\end_inset
+
+,
+ 
+\begin_inset Formula $I\coloneqq\left(-\infty,\infty\right)$
+\end_inset
+
+,
+ zvezna.
+ Naj bo 
+\begin_inset Formula $y\in f\left(I\right)$
+\end_inset
+
+ poljuben.
+ Tedaj 
+\begin_inset Formula $\exists!x\in I\ni:y=f\left(x\right)$
+\end_inset
+
+ in definiramo 
+\begin_inset Formula $x\coloneqq f^{-1}\left(x\right)$
+\end_inset
+
+.
+ 
+\begin_inset Formula $f^{-1}$
+\end_inset
+
+ obstaja in je spet zvezna.
+\end_layout
+
+\begin_layout Proof
+Ne bomo dokazali.
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+Označimo 
+\begin_inset Formula $g=f^{-1}:f\left(I\right)\to\mathbb{R}$
+\end_inset
+
+.
+ Uporabimo 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{kzsppom}{karakterizacijo zveznosti s pomočjo praslik odprtih množic}
+\end_layout
+
+\end_inset
+
+.
+ Dokazujemo torej,
+ da 
+\begin_inset Formula $\forall V^{\text{odp.}}\subset\mathbb{R}:g^{-1}\left(V\right)$
+\end_inset
+
+ je zopet odprta množica 
+\begin_inset Formula $\subseteq f\left(I\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Velja 
+\begin_inset Formula $g^{-1}\left(V\right)=\left\{ x\in f\left(I\right);g\left(x\right)\in V\right\} =\left\{ x\in f\left(I\right):\exists v\in V\cap I\ni:x=f\left(v\right)\right\} =f\left(V\cap I\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Torej dokazujemo 
+\begin_inset Formula $\forall V^{\text{odp.}}\subset\mathbb{R}:f\left(I\cap V\right)$
+\end_inset
+
+ je spet zopet odprta 
+\begin_inset Formula $\subseteq f\left(I\right)$
+\end_inset
+
+,
+ kar je ekvivalentno
+\begin_inset Formula 
+\[
+\forall y\in f\left(I\cap V\right)\exists\delta>0\ni:\left(y-\delta,y+\delta\right)\cap f\left(I\right)\subset f\left(I\cap V\right).
+\]
+
+\end_inset
+
+Pišimo 
+\begin_inset Formula $y=f\left(x\right),x\in I\cap V$
+\end_inset
+
+.
+ Privzemimo,
+ da 
+\begin_inset Formula $f$
+\end_inset
+
+ narašča (če pada,
+ ravnamo podobno).
+ Ker jer 
+\begin_inset Formula $ $
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsection
+Enakomerna zveznost
+\end_layout
+
+\begin_layout Definition*
+\begin_inset Formula $f:I\to\mathbb{R}$
+\end_inset
+
+ je 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hypertarget{ez}{enakomerno zvezna}
+\end_layout
+
+\end_inset
+
+ na 
+\begin_inset Formula $I$
+\end_inset
+
+,
+ če 
+\begin_inset Formula 
+\[
+\forall\varepsilon>0\exists\delta>0\forall x,y\in I:\left|x-y\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(y\right)\right|<\varepsilon.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Note*
+Primerjajmo to z definicijo 
+\begin_inset Formula $f:I\to\mathbb{R}$
+\end_inset
+
+ je (nenujno enakomerno) zvezna na 
+\begin_inset Formula $I$
+\end_inset
+
+,
+ če
+\begin_inset Formula 
+\[
+\forall\varepsilon>0,a\in I\exists\delta>0\forall x\in I:\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(y\right)\right|<\varepsilon.
+\]
+
+\end_inset
+
+Pri slednji definiciji je 
+\begin_inset Formula $\delta$
+\end_inset
+
+ odvisna od 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+ in 
+\begin_inset Formula $a$
+\end_inset
+
+,
+ pri enakomerni zveznosti pa le od 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $f\left(x\right)=\frac{1}{x}$
+\end_inset
+
+ ni enakomerno zvezna,
+ ker je 
+\begin_inset Formula $\delta$
+\end_inset
+
+ odvisen od 
+\begin_inset Formula $a$
+\end_inset
+
+.
+ Če pri fiksnem 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+ pomaknemo tisto pozitivno točko,
+ v kateri preizkušamo zveznost,
+ bolj v levo,
+ bo na neki točki potreben ožji,
+ manjši 
+\begin_inset Formula $\delta$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Theorem*
+Zvezna funkcija na kompaktni množici je enakomerno zvezna.
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $f:K\to\mathbb{R}$
+\end_inset
+
+ zvezna,
+ kjer je 
+\begin_inset Formula $K$
+\end_inset
+
+ kompaktna podmnožica 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+.
+ PDDRAA 
+\begin_inset Formula $f$
+\end_inset
+
+ ni enakomerno zvezna.
+ Zanikajmo definicijo 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{ez}{enakomerne zveznosti}
+\end_layout
+
+\end_inset
+
+:
+ 
+\begin_inset Formula $\exists\varepsilon>0\forall\delta>0\exists x_{\delta},y_{\delta}\in I:\left|x_{\delta}-y_{\delta}\right|<\delta\wedge\left|f\left(x_{\delta}\right)-f\left(y_{\delta}\right)\right|\geq\varepsilon$
+\end_inset
+
+.
+ 
+\begin_inset Formula $x,y$
+\end_inset
+
+ sta seveda lahko odvisna od 
+\begin_inset Formula $\delta$
+\end_inset
+
+ in 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+,
+ zato v subskriptu pišemo 
+\begin_inset Formula $\delta$
+\end_inset
+
+,
+ ki ji pripadata.
+ Ker smo dejali,
+ da to velja,
+ si oglejmo 
+\begin_inset Formula $\forall n\in\mathbb{N}:\delta_{n}\coloneqq\frac{1}{n}$
+\end_inset
+
+ in pripadajoči zaporedji 
+\begin_inset Formula $\left(x_{1/n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ in 
+\begin_inset Formula $\left(y_{1/n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $K$
+\end_inset
+
+ kompaktna,
+ ima zaporedje 
+\begin_inset Formula $\left(x_{1/n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ stekališče v 
+\begin_inset Formula $x\in K$
+\end_inset
+
+,
+ torej obstaja podzaporede 
+\begin_inset Formula $\left(x_{1/n_{k}}\right)_{k\in\mathbb{N}}$
+\end_inset
+
+,
+ ki konvergira k 
+\begin_inset Formula $x$
+\end_inset
+
+.
+ Podobno obstaja podzaporedje 
+\begin_inset Formula $\left(y_{1/n_{k_{l}}}\right)_{l\in\mathbb{N}}$
+\end_inset
+
+,
+ ki konvergira k 
+\begin_inset Formula $y\in K$
+\end_inset
+
+.
+ Pišimo sedaj 
+\begin_inset Formula $x_{l}\coloneqq x_{1/n_{k_{l}}}$
+\end_inset
+
+in 
+\begin_inset Formula $y_{l}\coloneqq y_{1/n_{k_{l}}}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Velja torej 
+\begin_inset Formula $x_{l}\to x$
+\end_inset
+
+ in 
+\begin_inset Formula $y_{l}\to y$
+\end_inset
+
+.
+ Sledi 
+\begin_inset Formula $\left|x-y\right|\leq\lim_{l\to\infty}\left(\left|x-x_{l}\right|+\left|x_{l}-y_{l}\right|+\left|y_{l}-y\right|\right)$
+\end_inset
+
+.
+ Levi in desni člen sta v limiti enaka 0 zaradi konvergence zaporedja,
+ srednji pa je manjši od 
+\begin_inset Formula $\frac{1}{j}$
+\end_inset
+
+ zaradi naše predpostavke (PDDRAA),
+ potemtakem je 
+\begin_inset Formula $x=y$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Zato 
+\begin_inset Formula $\lim_{l\to\infty}\left(f\left(x_{l}\right)-f\left(y_{l}\right)\right)=\lim_{l\to\infty}\left[\left(f\left(x_{l}\right)-f\left(x\right)\right)+\left(f\left(x\right)-f\left(y\right)\right)+\left(f\left(y\right)-f\left(y_{l}\right)\right)\right]$
+\end_inset
+
+.
+ Levi in desni člen sta v limiti enaka 0 zaradi konvergence zaporedja in zveznosti 
+\begin_inset Formula $f$
+\end_inset
+
+,
+ srednji pa je tudi 0,
+ ker 
+\begin_inset Formula $x=y$
+\end_inset
+
+,
+ potemtakem 
+\begin_inset Formula $f\left(x_{l}\right)-f\left(y_{l}\right)\to0$
+\end_inset
+
+,
+ kar je v protislovju z 
+\begin_inset Formula $\left|f\left(x_{l}\right)-f\left(y_{l}\right)\right|\geq\varepsilon$
+\end_inset
+
+ za fiksen 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+ in 
+\begin_inset Formula $\forall l\in\mathbb{N}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $\rightarrow\!\leftarrow$
+\end_inset
+
+,
+ 
+\begin_inset Formula $f$
+\end_inset
+
+ je enakomerno zvezna.
+\end_layout
+
+\begin_layout Corollary*
+En zaprt interval 
+\begin_inset Formula $\frac{1}{x}$
+\end_inset
+
+ bo enakomerno zvezen,
+ 
+\begin_inset Formula $\frac{1}{x}$
+\end_inset
+
+ sama po sebi kot 
+\begin_inset Formula $\left(0,\infty\right)\to\mathbb{R}$
+\end_inset
+
+ pa ni definirana na kompaktni množici.
+ Prav tako 
+\begin_inset Formula $\arcsin$
+\end_inset
+
+ in 
+\begin_inset Formula $x\mapsto\sqrt{x}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Odvod
+\end_layout
+
+\begin_layout Standard
+Najprej razmislek/ideja.
+ Odvod je hitrost/stopnja,
+ s katero se v danem trenutku neka količina spreminja.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Float figure
+placement document
+alignment document
+wide false
+sideways false
+status open
+
+\begin_layout Plain Layout
+TODO XXX FIXME SKICA S TKZ EUCLID (ali pa —
+ bolje —
+ s čim drugim),
+ glej PS zapiski/ANA1P FMF 2023-12-04.pdf
+\end_layout
+
+\begin_layout Plain Layout
+\begin_inset Caption Standard
+
+\begin_layout Plain Layout
+Skica.
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Radi bi določili naklon sekante,
+ torej naklon premice,
+ določene z 
+\begin_inset Formula $x$
+\end_inset
+
+ in neko bližnjo točko 
+\begin_inset Formula $x+h$
+\end_inset
+
+ na grafu funkcije,
+ ki je odvisen le od 
+\begin_inset Formula $x$
+\end_inset
+
+,
+ ne pa tudi od izbire 
+\begin_inset Formula $h$
+\end_inset
+
+.
+ Bližnjo točko pošljemo proti začetni —
+ 
+\begin_inset Formula $h$
+\end_inset
+
+ pošljemo proti 0.
+ Naklon izračunamo s izrazom 
+\begin_inset Formula $\frac{f\left(x+h\right)-f\left(x\right)}{h}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Odvod funkcije 
+\begin_inset Formula $f$
+\end_inset
+
+ v točki 
+\begin_inset Formula $x$
+\end_inset
+
+ označimo 
+\begin_inset Formula $f'\left(x\right)\coloneqq\lim_{h\to0}\frac{f\left(x+h\right)-f\left(x\right)}{h}$
+\end_inset
+
+.
+ Če limita obstaja v točki 
+\begin_inset Formula $x$
+\end_inset
+
+,
+ pravimo,
+ da je funkcija odvedljiva v 
+\begin_inset Formula $x$
+\end_inset
+
+.
+ Pravimo,
+ da je 
+\begin_inset Formula $f$
+\end_inset
+
+ odvedljiva na množici 
+\begin_inset Formula $I\subseteq\mathbb{R}$
+\end_inset
+
+,
+ če je odvedljiva na vsaki 
+\begin_inset Formula $t\in I$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Primeri odvodov preprostih funkcij.
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $f\left(x\right)=c,c\in\mathbb{R}$
+\end_inset
+
+ 
+\begin_inset Formula $f'\left(x\right)=\lim_{h\to0}\frac{\cancelto{c}{f\left(x+h\right)}-\cancelto{c}{f\left(x\right)}}{h}=0$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $f\left(x\right)=x$
+\end_inset
+
+ 
+\begin_inset Formula $f'\left(x\right)=\lim_{h\to0}\frac{f\left(x+h\right)-f\left(x\right)}{h}=\lim_{h\to0}\frac{x+h-x}{h}=1$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $f\left(x\right)=x^{2}$
+\end_inset
+
+ 
+\begin_inset Formula $f'\left(x\right)=\lim_{h\to0}\frac{f\left(x+h\right)-f\left(x\right)}{h}=\lim_{h\to0}\frac{x^{2}+2xh+h^{2}-x^{2}}{h}=\lim_{h\to0}\frac{2xh+h^{2}}{h}=\lim_{h\to0}2x+h=2x$
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Claim*
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hypertarget{op}{Odvod potence}
+\end_layout
+
+\end_inset
+
+.
+ Za poljuben 
+\begin_inset Formula $n\in\mathbb{N}$
+\end_inset
+
+ so funkcije 
+\begin_inset Formula $f\left(x\right)=x^{n}$
+\end_inset
+
+ odvedljive na 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+ in velja 
+\begin_inset Formula $f'\left(x\right)=nx^{n-1}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+\begin_inset Formula 
+\[
+\lim_{h\to0}\frac{f\left(x+h\right)-f\left(x\right)=\left(x+h\right)^{n}-x^{n}=\sum_{k=0}^{n}\binom{n}{k}h^{k}x^{n-k}-x^{n}=\cancel{x^{n}}+nhx^{n-1}+\sum_{k=2}^{n}\binom{n}{k}h^{k}x^{n-k}\cancel{-x^{n}}}{h}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\lim_{h\to0}\frac{nhx^{n-1}+\sum_{k=2}^{n}\binom{n}{k}h^{k}x^{n-k}}{h}=\lim_{h\to0}\frac{\cancel{h}\left(nx^{n-1}+\sum_{k=2}^{n}\binom{n}{k}h^{k-1}x^{n-k}\right)}{\cancel{h}}=\lim_{h\to0}\left(nx^{n-1}+\sum_{k=2}^{n}\binom{n}{k}h^{k-1}x^{n-k}\right)=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=nx^{n-1}+\cancel{\lim_{h\to0}\sum_{k=2}^{n}\binom{n}{k}\cancelto{0}{h^{k-1}}x^{n-k}}=nx^{n-1}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Claim*
+\begin_inset Formula $\sin'=\cos$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\cos'=-\sin$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Proof
+Najprej dokažimo 
+\begin_inset Formula $\sin'=\cos$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+\lim_{h\to\infty}\frac{\sin\left(x+h\right)-\sin\left(x\right)=\sin x\cos h+\sin h\cos x-\sin x=\sin x\left(\cos h-1\right)+\sin h\cos x}{h}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\lim_{h\to0}\left(\sin x\frac{\cos h-1}{h}+\cos x\frac{\sin h}{h}\right)=\lim_{h\to0}\left(\sin x\frac{\left(\cos h-1\right)\left(\cos h+1\right)=\cos^{2}h-1=-\sin^{2}h}{h\left(\cos h+1\right)}+\cos x\frac{\sin h}{h}\right)=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\lim_{h\to0}\left(\sin x\frac{-\sin h}{h}\cdot\frac{\sin h}{\cos h+1}+\cos x\frac{\sin h}{h}\right)=\lim_{h\to0}\cancelto{1}{\frac{\sin h}{h}}\left(\cos x-\cancel{\sin x\frac{\cancelto{0}{\sin h}}{\cos h+1}}\right)=\cos x
+\]
+
+\end_inset
+
+Sedaj dokažimo še 
+\begin_inset Formula $\cos'=-\sin$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+\lim_{h\to0}\frac{\cos\left(x+h\right)-\cos\left(x\right)=\cos x\cos h-\sin x\sin h-\cos x=\cos x\left(\cos h-1\right)-\sin x\sin h}{h}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\lim_{h\to0}\left(\cos x\frac{\cos h-1}{h}-\sin x\frac{\sin h}{h}\right)=\lim_{h\to0}\left(\cos x\frac{\left(\cos h-1\right)\left(\cos h+1\right)=\cos^{2}h-1=-\sin^{2}h}{h\left(\cos h+1\right)}-\sin x\frac{\sin h}{h}\right)=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\lim_{h\to0}\left(\cos x\frac{-\sin h}{h}\cdot\frac{\sin^{}h}{\cos h+1}-\sin x\frac{\sin h}{h}\right)=\lim_{h\to0}\left(\cancelto{1}{\frac{\sin h}{h}}\left(\cancel{-\cos x\frac{\cancelto{0}{\sin h}}{\cos h+1}}-\sin x\right)\right)=-\sin x
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Fact*
+Od prej vemo 
+\begin_inset Formula $\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{n}=e$
+\end_inset
+
+ (limita zaporedja).
+ Velja tudi 
+\begin_inset Formula $\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^{x}=e$
+\end_inset
+
+ (funkcijska limita).
+ Ne bomo dokazali.
+\end_layout
+
+\begin_layout Claim*
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hypertarget{oef}{Odvod eksponentne funkcije}
+\end_layout
+
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $a>0$
+\end_inset
+
+ in 
+\begin_inset Formula $f\left(x\right)=a^{x}$
+\end_inset
+
+.
+ Tedaj je 
+\begin_inset Formula $f'\left(x\right)=a^{x}\ln a$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+\begin_inset Formula 
+\[
+\lim_{h\to0}\frac{f\left(x+h\right)-f\left(x\right)=a^{x}a^{h}-a^{x}}{h}=\lim_{h\to0}a^{x}\frac{a^{h}-1}{h}=\cdots
+\]
+
+\end_inset
+
+Sedaj pišimo 
+\begin_inset Formula $\frac{1}{z}\coloneqq a^{h}-1$
+\end_inset
+
+.
+ Ulomek 
+\begin_inset Formula $\frac{a^{h}-1}{h}$
+\end_inset
+
+ namreč ni odvisen od 
+\begin_inset Formula $x$
+\end_inset
+
+.
+ Sedaj
+\begin_inset Formula 
+\[
+a^{h}-1=\frac{1}{z}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+a^{h}=\frac{1}{z}+1
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+h=\log_{a}\left(\frac{1}{z}+1\right)
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+h=\frac{\ln\left(\frac{1}{z}+1\right)}{\ln a}
+\]
+
+\end_inset
+
+Nadaljujmo s prvotnim računom,
+ ločimo primere:
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $a>1,h\searrow0$
+\end_inset
+
+ Potemtakem 
+\begin_inset Formula $a^{h}-1\searrow0$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\frac{1}{z}\searrow0$
+\end_inset
+
+,
+ sledi 
+\begin_inset Formula $z\nearrow\infty$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+\cdots=a^{x}\lim_{z\to\infty}\frac{\frac{1}{z}}{\frac{\ln\left(\frac{1}{z}+1\right)}{\ln a}}=a^{x}\lim_{z\to\infty}\frac{\frac{1}{z}\ln a}{\ln\left(\frac{1}{z}+1\right)}=a^{x}\lim_{z\to\infty}\frac{\ln a}{\ln\left(\frac{1}{z}+1\right)^{z}}=a^{x}\lim_{z\to\infty}\frac{\ln a}{\ln e}=a^{x}\ln a
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $a>1,h\nearrow0$
+\end_inset
+
+ Potemtakem 
+\begin_inset Formula $a^{h}-1\nearrow0$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\frac{1}{z}\nearrow0$
+\end_inset
+
+,
+ sledi 
+\begin_inset Formula $z\searrow-\infty$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+\cdots=a^{x}\lim_{z\to-\infty}\frac{\ln a}{\ln\left(\frac{1}{z}+1\right)^{z}}=a^{x}\lim_{z\to-\infty}\frac{\ln a}{\ln\cancelto{e}{\left(\frac{1}{z}+1\right)^{z}}}=a^{x}\ln a
+\]
+
+\end_inset
+
+Kajti 
+\begin_inset Formula $\lim_{x\to\infty}\left(1+\frac{k}{x}\right)^{x}=e^{k}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $a\in(0,1]$
+\end_inset
+
+ Podobno kot zgodaj,
+ bodisi 
+\begin_inset Formula $z\nearrow\infty$
+\end_inset
+
+ bodisi 
+\begin_inset Formula $z\searrow-\infty$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Claim*
+Če je 
+\begin_inset Formula $f$
+\end_inset
+
+ odvedljiva v točki 
+\begin_inset Formula $x$
+\end_inset
+
+,
+ je tam tudi zvezna.
+\end_layout
+
+\begin_layout Proof
+Predpostavimo,
+ da obstaja limita 
+\begin_inset Formula $\lim_{h\to0}\frac{f\left(x+h\right)-f\left(x\right)}{h}$
+\end_inset
+
+.
+ Želimo dokazati 
+\begin_inset Formula $f\left(x\right)=\lim_{t\to x}f\left(t\right)$
+\end_inset
+
+.
+ Računajmo:
+\begin_inset Formula 
+\[
+f\left(x\right)=\lim_{t\to x}f\left(t\right)
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+0=\lim_{t\to x}f\left(t\right)-f\left(x\right)
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+0=\lim_{h\to0}f\left(x+h\right)-f\left(x\right)
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+0=\lim_{h\to0}\left(f\left(x+h\right)-f\left(x\right)\right)=\lim_{h\to0}\left(\frac{f\left(x+h\right)-f\left(x\right)}{h}\cdot h\right)
+\]
+
+\end_inset
+
+Limita obstaja,
+ čim obstajata 
+\begin_inset Formula $\lim_{h\to0}\frac{f\left(x+h\right)-f\left(x\right)}{h}$
+\end_inset
+
+,
+ ki obstaja po predpostavki,
+ in 
+\begin_inset Formula $\lim_{h\to0}h$
+\end_inset
+
+,
+ ki obstaja in ima vrednost 
+\begin_inset Formula $0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $f\left(x\right)=\left|x\right|=\sqrt{x^{2}}$
+\end_inset
+
+.
+ Je zvezna,
+ ker je kompozitum zveznih funkcij,
+ toda v 
+\begin_inset Formula $0$
+\end_inset
+
+ ni odvedljiva,
+ kajti 
+\begin_inset Formula $\lim_{h\to0}\frac{f\left(0+h\right)-f\left(0\right)}{h}=\lim_{h\to0}\frac{\left|h\right|-0}{h}=\lim_{h\to0}\sgn h$
+\end_inset
+
+.
+ Limita ne obstaja,
+ ker 
+\begin_inset Formula $-1=\lim_{h\nearrow0}\sgn h\not=\lim_{h\searrow0}\sgn h=1$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+Naj bosta 
+\begin_inset Formula $f,g$
+\end_inset
+
+ odvedljivi v 
+\begin_inset Formula $x\in\mathbb{R}$
+\end_inset
+
+.
+ Tedaj so 
+\begin_inset Formula $f+g,f-g,f\cdot g,f/g$
+\end_inset
+
+ (slednja le,
+ če 
+\begin_inset Formula $g\left(x\right)\not=0$
+\end_inset
+
+) in velja 
+\begin_inset Formula $\left(f\pm g\right)'=f'\pm g'$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\left(fg\right)'=f'g+fg'$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\left(f/g\right)'=\frac{f'g-fg'}{g^{2}}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Dokažimo vse štiri trditve.
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $f+g$
+\end_inset
+
+ Velja 
+\begin_inset Formula $\left(f+g\right)\left(x\right)=f\left(x\right)+g\left(x\right)$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+\left(f+g\right)'\left(x\right)=\lim_{h\to0}\frac{\left(f+g\right)\left(x+h\right)-\left(f+g\right)\left(x\right)=f\left(x+h\right)+g\left(x+h\right)-f\left(x\right)-g\left(x\right)}{h}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\lim_{h\to0}\left(\frac{f\left(x+h\right)-f\left(x\right)}{h}+\frac{g\left(x+h\right)-g\left(x\right)}{h}\right)=f\left(x\right)'+g\left(x\right)'
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $-f$
+\end_inset
+
+ Naj bo 
+\begin_inset Formula $g=-f$
+\end_inset
+
+.
+ 
+\begin_inset Formula $g'\left(x\right)=\lim_{h\to0}\frac{g\left(x+h\right)-g\left(x\right)}{h}=\lim_{h\to0}\frac{-f\left(x+h\right)+f\left(x\right)}{h}=-\lim_{h\to0}\frac{f\left(x+h\right)-f\left(x\right)}{h}=-f\left(x\right)'$
+\end_inset
+
+,
+ zato
+\begin_inset Formula 
+\[
+\left(f-g\right)'\left(x\right)=\left(f+\left(-g\right)\right)'\left(x\right)=f'\left(x\right)+\left(-g\right)'\left(x\right)=f'\left(x\right)-g'\left(x\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $f\cdot g$
+\end_inset
+
+ Velja 
+\begin_inset Formula $\left(fg\right)\left(x\right)=f\left(x\right)g\left(x\right)$
+\end_inset
+
+.
+ Prištejemo in odštejemo isti izraz (v oglatih oklepajih).
+\begin_inset Formula 
+\[
+\left(fg\right)'\left(x\right)=\lim_{h\to0}\frac{\left(fg\right)\left(x+h\right)-\left(fg\right)\left(x\right)=f\left(x+h\right)g\left(x+h\right)-f\left(x\right)g\left(x\right)+\left[f\left(x\right)g\left(x+h\right)-f\left(x\right)g\left(x+h\right)\right]}{h}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\lim_{h\to0}\frac{g\left(x+h\right)\left(f\left(x+h\right)-f\left(x\right)\right)+f\left(x\right)\left(g\left(x+h\right)-g\left(x\right)\right)}{h}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\lim_{h\to0}\frac{f\left(x+h\right)-f\left(x\right)}{h}\cancelto{g\left(x\right)}{g\left(x+h\right)}+\lim_{h\to0}\frac{g\left(x+h\right)-g\left(x\right)}{h}f\left(x\right)=f'\left(x\right)g\left(x\right)+g'\left(x\right)f\left(x\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $f/g$
+\end_inset
+
+ Velja 
+\begin_inset Formula $\left(f/g\right)\left(x\right)=f\left(x\right)/g\left(x\right)$
+\end_inset
+
+.
+ Prištejemo in odštejemo isti izraz (v oglatih oklepajih).
+\begin_inset Formula 
+\[
+\left(f/g\right)'\left(x\right)=\lim_{h\to0}\frac{\left(f/g\right)\left(x+h\right)-\left(f/g\right)\left(x\right)=\frac{f\left(x+h\right)}{g\left(x+h\right)}-\frac{f\left(x\right)}{g\left(x\right)}=\frac{f\left(x+h\right)g\left(x\right)}{g\left(x+h\right)g\left(x\right)}-\frac{f\left(x\right)g\left(x+h\right)}{g\left(x\right)g\left(x+h\right)}=\frac{f\left(x+h\right)g\left(x\right)-f\left(x\right)g\left(x+h\right)}{g\left(x\right)g\left(x+h\right)}}{h}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\lim_{h\to0}\frac{f\left(x+h\right)g\left(x\right)-f\left(x\right)g\left(x+h\right)=f\left(x+h\right)g\left(x\right)-f\left(x\right)g\left(x+h\right)+\left[f\left(x\right)g\left(x\right)-f\left(x\right)g\left(x\right)\right]}{hg\left(x\right)g\left(x+h\right)}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\lim_{h\to0}\left(\frac{f\left(x+h\right)-f\left(x\right)}{h}\cdot\frac{g\left(x\right)}{g\left(x\right)g\left(x+h\right)}-\frac{g\left(x+h\right)-g\left(x\right)}{h}\cdot\frac{f\left(x\right)}{g\left(x\right)g\left(x+h\right)}\right)=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\lim_{h\to0}\left(\left(\frac{1}{g\left(x\right)g\left(x+h\right)}\right)\left(\frac{f\left(x+h\right)-f\left(x\right)}{h}g\left(x\right)-\frac{g\left(x+h\right)-g\left(x\right)}{h}f\left(x\right)\right)\right)=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\frac{1}{g^{2}\left(x\right)}\left(f'\left(x\right)g\left(x\right)-g'\left(x\right)f\left(x\right)\right)=\frac{f'\left(x\right)g\left(x\right)-f\left(x\right)g'\left(x\right)}{g^{2}\left(x\right)}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Example*
+\begin_inset Formula $\tan'\left(x\right)=\left(\frac{\sin\left(x\right)}{\cos\left(x\right)}\right)'=\frac{\sin'\left(x\right)\cos\left(x\right)-\sin\left(x\right)\cos'\left(x\right)}{\cos^{2}\left(x\right)}=\frac{\cos^{2}\left(x\right)+\sin^{2}\left(x\right)}{\cos^{2}\left(x\right)}=\cos^{-2}\left(x\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hypertarget{ok}{Odvod kompozituma}
+\end_layout
+
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $f$
+\end_inset
+
+ odvedljiva v 
+\begin_inset Formula $x$
+\end_inset
+
+ in 
+\begin_inset Formula $g$
+\end_inset
+
+ odvedljiva v 
+\begin_inset Formula $f\left(x\right)$
+\end_inset
+
+.
+ Tedaj je 
+\begin_inset Formula $g\circ f$
+\end_inset
+
+ odvedljiva v 
+\begin_inset Formula $x$
+\end_inset
+
+ in velja 
+\begin_inset Formula $\left(g\circ f\right)\left(x\right)=g'\left(f\left(x\right)\right)\cdot f'\left(x\right)$
+\end_inset
+
+ (opomba:
+ 
+\begin_inset Formula $\left(g\circ f\right)\left(x\right)=g\left(f\left(x\right)\right)$
+\end_inset
+
+).
+\end_layout
+
+\begin_layout Proof
+Označimo 
+\begin_inset Formula $a\coloneqq f\left(x\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $\delta_{h}\coloneqq f\left(x+h\right)-f\left(x\right)$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $f\left(x+h\right)\coloneqq a+\delta\left(h\right)$
+\end_inset
+
+.
+ 
+\begin_inset Formula 
+\[
+\lim_{h\to0}\frac{\left(g\circ f\right)\left(x+h\right)-\left(g\circ f\right)\left(x\right)=g\left(f\left(x+h\right)\right)-g\left(f\left(x\right)\right)=g\left(a+\delta_{h}\right)-g\left(a\right)}{h}=\lim_{h\to0}\frac{g\left(a+\delta_{h}\right)-g\left(a\right)}{\delta_{h}}\cdot\frac{\delta_{h}}{h}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\lim_{h\to0}\frac{g\left(a+\delta_{h}\right)-g\left(a\right)}{\delta_{h}}\cdot\frac{f\left(x+h\right)-f\left(x\right)}{h}=\cdots
+\]
+
+\end_inset
+
+Ker je 
+\begin_inset Formula $f$
+\end_inset
+
+ odvedljiva v 
+\begin_inset Formula $x$
+\end_inset
+
+,
+ je v 
+\begin_inset Formula $x$
+\end_inset
+
+ zvezna,
+ zato sledi 
+\begin_inset Formula $h\to0\Rightarrow\delta_{h}\to0$
+\end_inset
+
+,
+ torej
+\begin_inset Formula 
+\[
+\cdots=g'\left(a\right)\cdot f'\left(x\right)=g'\left(f\left(x\right)\right)\cdot f'\left(x\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $\varphi\left(x\right)=\sin\left(x^{2}\right)=\left(g\circ f\right)\left(x\right),f\left(x\right)=x^{2},g\left(x\right)=\sin x$
+\end_inset
+
+ in velja 
+\begin_inset Formula $\varphi'\left(x\right)=g'\left(f\left(x\right)\right)f'\left(x\right)=\sin'\left(x^{2}\right)\left(x^{2}\right)'=2x\cos\left(x^{2}\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $\psi\left(x\right)=\sin^{2}\left(x\right)=\left(g\circ f\right)\left(x\right),f\left(x\right)=\sin,g\left(x\right)=x^{2}$
+\end_inset
+
+ in velja 
+\begin_inset Formula $\psi'\left(x\right)=g'\left(f\left(x\right)\right)f'\left(x\right)=2\sin x\cos x=\sin2x$
+\end_inset
+
+ (sinus dvojnega kota)
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $\delta'\left(x\right)=\sin\left(e^{x^{2}}\right)=\sin\left(e^{\left(x^{2}\right)}\right)=\left(g\circ h\circ f\right)\left(x\right),g\left(x\right)=\sin x,h\left(x\right)=e^{x},f\left(x\right)=x^{2}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $\delta'\left(x\right)=\cos\left(e^{x^{2}}\right)e^{x^{2}}2x$
+\end_inset
+
+,
+ kajti 
+\begin_inset Formula $\left(e^{x}\right)'=e^{x}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Funkcija 
+\begin_inset Formula $f:I\subseteq\mathbb{R}\to\mathbb{R}$
+\end_inset
+
+ je zvezno odvedljiva na 
+\begin_inset Formula $I$
+\end_inset
+
+,
+ če je na 
+\begin_inset Formula $I$
+\end_inset
+
+ odvedljiva in je 
+\begin_inset Formula $f'$
+\end_inset
+
+ na 
+\begin_inset Formula $I$
+\end_inset
+
+ zvezna.
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $f\left(x\right)=\begin{cases}
+x^{2}\sin\frac{1}{x} & ;x\not=0\\
+0 & ;x=0
+\end{cases}$
+\end_inset
+
+ je na 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+ odvedljiva,
+ a ne zvezno.
+ Odvedljivost na 
+\begin_inset Formula $\mathbb{R}\setminus\left\{ 0\right\} $
+\end_inset
+
+ je očitna,
+ preverimo še odvedljivost v 
+\begin_inset Formula $0$
+\end_inset
+
+:
+\begin_inset Formula 
+\[
+f'\left(0\right)=\lim_{h\to0}\frac{f\left(h\right)-0}{h}=\lim_{h\to0}\frac{h^{\cancel{2}}\sin\frac{1}{h}}{\cancel{h}}=\lim_{h\to0}h\sin\frac{1}{h}=0,
+\]
+
+\end_inset
+
+ker 
+\begin_inset Formula $h$
+\end_inset
+
+ pada k 0,
+ 
+\begin_inset Formula $\sin\frac{1}{h}$
+\end_inset
+
+ pa je omejen z 1.
+ Velja torej
+\begin_inset Formula 
+\[
+f'\left(x\right)=\begin{cases}
+2x\sin\frac{1}{x}-\cos\frac{1}{x} & ;x\not=0\\
+0 & ;x=0
+\end{cases}
+\]
+
+\end_inset
+
+Preverimo nezveznost v 
+\begin_inset Formula $0$
+\end_inset
+
+.
+ Spodnja limita ne obstaja.
+\begin_inset Formula 
+\[
+\lim_{x\to0}\left(\cancel{2x\sin\frac{1}{x}}-\cos\frac{1}{x}\right)=-\lim_{x\to0}\cos\frac{1}{x}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Theorem*
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hypertarget{oi}{Odvod inverza}
+\end_layout
+
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $f$
+\end_inset
+
+ strogo monotona v okolici 
+\begin_inset Formula $a$
+\end_inset
+
+,
+ v 
+\begin_inset Formula $a$
+\end_inset
+
+ odvedljiva in naj bo 
+\begin_inset Formula $f'\left(a\right)\not=0$
+\end_inset
+
+.
+ Tedaj bo inverzna funkcija,
+ definirana v okolici 
+\begin_inset Formula $b=f\left(a\right)$
+\end_inset
+
+ v 
+\begin_inset Formula $b$
+\end_inset
+
+ odvedljiva in veljalo bo 
+\begin_inset Formula $\left(f^{-1}\right)'\left(b\right)=\frac{1}{f'\left(a\right)}=\frac{1}{f'\left(f^{-1}\left(b\right)\right)}.$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Proof
+Ker je 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{zism}{zvezna in strogo monotona}
+\end_layout
+
+\end_inset
+
+ na okolici 
+\begin_inset Formula $a$
+\end_inset
+
+,
+ inverz na okolici 
+\begin_inset Formula $f\left(a\right)$
+\end_inset
+
+ obstaja in velja 
+\begin_inset Formula $f\left(x\right)=s\Leftrightarrow x=f^{-1}\left(x\right)$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $f^{-1}\left(f\left(x\right)\right)=x$
+\end_inset
+
+ za 
+\begin_inset Formula $x$
+\end_inset
+
+ v okolici 
+\begin_inset Formula $a$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Uporabimo formulo za 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{ok}{odvod kompozituma}
+\end_layout
+
+\end_inset
+
+ in velja 
+\begin_inset Formula 
+\[
+\left(f^{-1}\left(f\left(x\right)\right)\right)'=\left(f^{-1}\right)'\left(f\left(x\right)\right)\cdot f'\left(x\right)=\left(x\right)'=1
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\left(f^{-1}\right)'\left(f\left(x\right)\right)=\frac{1}{f'\left(x\right)}
+\]
+
+\end_inset
+
+Vstavimo 
+\begin_inset Formula $x=f^{-1}\left(y\right)$
+\end_inset
+
+ in dobimo za vsak 
+\begin_inset Formula $y$
+\end_inset
+
+ blizu 
+\begin_inset Formula $f\left(a\right)$
+\end_inset
+
+:
+\begin_inset Formula 
+\[
+\left(f^{-1}\right)'\left(y\right)=\frac{1}{f'\left(f^{-1}\left(y\right)\right)}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+Nekaj primerov odvodov inverza.
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $g\left(x\right)=\sqrt[n]{x}=x^{\frac{1}{n}}$
+\end_inset
+
+ za 
+\begin_inset Formula $n\in\mathbb{N},x>0$
+\end_inset
+
+.
+ Velja 
+\begin_inset Formula $g=f^{-1}$
+\end_inset
+
+ za 
+\begin_inset Formula $f\left(x\right)=x^{n}$
+\end_inset
+
+.
+ Uporabimo formulo za 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{op}{odvod potence}
+\end_layout
+
+\end_inset
+
+ in zgornji izrek.
+ Velja 
+\begin_inset Formula $f'\left(x\right)=nx^{n-1}$
+\end_inset
+
+ in 
+\begin_inset Formula $f^{-1}=\sqrt[n]{x}$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+g'\left(x\right)=\left(f^{-1}\right)'\left(x\right)=\frac{1}{f'\left(f^{-1}\left(x\right)\right)}=\frac{1}{f'\left(\sqrt[n]{x}\right)}=\frac{1}{n\sqrt[n]{x}^{n-1}}=\frac{1}{nx^{\frac{n-1}{n}=1-\frac{1}{n}}}=\frac{1}{n}x^{\frac{1}{n}-1}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $h\left(x\right)=\sqrt[n]{x^{m}}=x^{\frac{m}{n}}=g\left(x\right)^{m}$
+\end_inset
+
+ za 
+\begin_inset Formula $n,m\in\mathbb{N},x>0$
+\end_inset
+
+.
+ Uporabimo formulo za 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{op}{odvod potence}
+\end_layout
+
+\end_inset
+
+ in 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{ok}{kompozituma}
+\end_layout
+
+\end_inset
+
+ in zgornji primer.
+ Velja 
+\begin_inset Formula $g'\left(x\right)=\frac{1}{n}x^{\frac{1}{n}-1}$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula 
+\[
+h'\left(x\right)=mg\left(x\right)^{m-1}\cdot g'\left(x\right)=m\left(x^{\frac{1}{n}}\right)^{m-1}\cdot\frac{1}{n}x^{\frac{1}{n}-1}=\frac{m}{n}x^{\frac{m-1}{n}+\frac{1}{n}-1}=\frac{m}{n}x^{\frac{m}{n}-1}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset CommandInset label
+LatexCommand label
+name "enu:Izkaže-se,-da"
+
+\end_inset
+
+Izkaže se,
+ da velja celo 
+\begin_inset Formula $\forall x>0,\alpha\in\mathbb{R}:\left(x^{\alpha}\right)'=\alpha x^{\alpha-1}$
+\end_inset
+
+.
+ Mi smo dokazali le za 
+\begin_inset Formula $\alpha\in\mathbb{Q}$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Logaritmi,
+ inverz 
+\begin_inset Formula $e^{x}$
+\end_inset
+
+.
+ Gre za 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{oef}{odvod eksponentne funkcije}
+\end_layout
+
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\left(a^{x}\right)=a^{x}\ln a$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $\left(e^{x}\right)=e^{x}\ln e=e^{x}$
+\end_inset
+
+.
+ Uporavimo 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{oi}{odvod inverza}
+\end_layout
+
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\left(f^{-1}\right)'\left(x\right)=\frac{1}{f'\left(f^{-1}\left(x\right)\right)}$
+\end_inset
+
+ in za 
+\begin_inset Formula $g\left(x\right)=\log x$
+\end_inset
+
+ uporabimo 
+\begin_inset Formula $g\left(x\right)=f^{-1}\left(x\right)$
+\end_inset
+
+,
+ kjer je 
+\begin_inset Formula $f\left(x\right)=e^{x}$
+\end_inset
+
+:
+\begin_inset Formula 
+\[
+\log'\left(x\right)=\left(\left(e^{x}\right)^{-1}\right)'\left(x\right)=\frac{1}{e^{\log x}}=\frac{1}{x}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $g\left(x\right)=\arcsin x$
+\end_inset
+
+ za 
+\begin_inset Formula $x\in\left[-1,1\right]$
+\end_inset
+
+,
+ torej je 
+\begin_inset Formula $g=f^{-1}$
+\end_inset
+
+,
+ kjer je 
+\begin_inset Formula $f=\sin$
+\end_inset
+
+ za 
+\begin_inset Formula $x\in\left[\frac{-\pi}{2},\frac{\pi}{2}\right]$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+g'\left(f\left(x\right)\right)=\frac{1}{f'\left(x\right)}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+g'\left(\sin x\right)=\frac{1}{\cos x}
+\]
+
+\end_inset
+
+Ker velja 
+\begin_inset Formula $\sin^{2}x+\cos^{2}x=1$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $\cos^{2}x=1-\sin^{2}x$
+\end_inset
+
+,
+ sledi 
+\begin_inset Formula $\cos x=\sqrt{1-\sin^{2}x}$
+\end_inset
+
+,
+ torej nadaljujemo:
+\begin_inset Formula 
+\[
+g'\left(\sin x\right)=\frac{1}{\sqrt{1-\sin^{2}x}}
+\]
+
+\end_inset
+
+Sedaj zamenjamo 
+\begin_inset Formula $\sin x$
+\end_inset
+
+ s 
+\begin_inset Formula $t$
+\end_inset
+
+ in dobimo:
+\begin_inset Formula 
+\[
+g'\left(t\right)=\frac{1}{\sqrt{1-t^{2}}}=\arcsin^{2}t
+\]
+
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Subsection
+Diferencial
+\end_layout
+
+\begin_layout Standard
+Fiksirajmo funkcijo 
+\begin_inset Formula $f$
+\end_inset
+
+ in točko 
+\begin_inset Formula $a\in\mathbb{R}$
+\end_inset
+
+,
+ v okolici katere je 
+\begin_inset Formula $f$
+\end_inset
+
+ definirana.
+ Želimo oceniti vrednost funkcije 
+\begin_inset Formula $f$
+\end_inset
+
+ v bližini točke 
+\begin_inset Formula $a$
+\end_inset
+
+ z linearno funkcijo – to je 
+\begin_inset Formula $y\left(x\right)=\lambda x$
+\end_inset
+
+ za neki 
+\begin_inset Formula $\lambda\in\mathbb{R}$
+\end_inset
+
+.
+ ZDB Iščemo najboljši linearni približek,
+ odvisen od 
+\begin_inset Formula $h$
+\end_inset
+
+,
+ za 
+\begin_inset Formula $f\left(a+h\right)-f\left(a\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $f$
+\end_inset
+
+ definirana v okolici točke 
+\begin_inset Formula $a\in\mathbb{R}$
+\end_inset
+
+.
+ Diferencial funkcije 
+\begin_inset Formula $f$
+\end_inset
+
+ v točki 
+\begin_inset Formula $a$
+\end_inset
+
+ je linearna preslikava 
+\begin_inset Formula $df\left(a\right):\mathbb{R}\to\mathbb{R}$
+\end_inset
+
+ z zahtevo
+\begin_inset Formula 
+\[
+\lim_{h\to0}\frac{\left|f\left(a+h\right)-f\left(a\right)-df\left(a\right)\left(h\right)\right|}{\left|h\right|}=0.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Note*
+\begin_inset Formula 
+\[
+\lim_{h\to0}\frac{f\left(a+h\right)-f\left(a\right)-df\left(a\right)\left(h\right)}{h}=0=\lim_{h\to0}\left(\frac{f\left(a+h\right)-f\left(a\right)}{h}-\frac{\left(df\left(a\right)\right)\left(h\right)}{h}\right)=
+\]
+
+\end_inset
+
+Upoštevamo linearnost preslikave
+\begin_inset Formula 
+\[
+=\lim_{h\to0}\frac{f\left(a+h\right)-f\left(a\right)}{h}-df\left(a\right)=f'\left(a\right)-df\left(a\right)=0
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+f'\left(a\right)=df\left(a\right)
+\]
+
+\end_inset
+
+Torej 
+\begin_inset Formula $f\left(a+h\right)-f\left(a\right)\approx df\left(a\right)\left(h\right)$
+\end_inset
+
+ – najboljši linearni približek za 
+\begin_inset Formula $f\left(a+h\right)-f\left(h\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Uporaba diferenciala.
+ 
+\begin_inset Formula $a$
+\end_inset
+
+ je točka,
+ v kateri znamo izračunati funkcijsko vrednost,
+ 
+\begin_inset Formula $a+h$
+\end_inset
+
+ pa je točka,
+ v kateri želimo približek funkcijske vrednosti.
+ Izračunajmo približek 
+\begin_inset Formula $\sqrt{2}$
+\end_inset
+
+:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $f\left(x\right)=\sqrt{x}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $a+h=2$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $a=2,25$
+\end_inset
+
+,
+ 
+\begin_inset Formula $h=-0,25$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $f\left(a\right)=\sqrt{a}=1,5$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $f'\left(s\right)=\frac{1}{2\sqrt{x}}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $f\left(a=2,25\right)=\frac{1}{3}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $f\left(2\right)\approx f\left(a\right)+f'\left(2,25\right)\cdot h=1,5-0,25\cdot\frac{1}{3}=\frac{3}{2}-\frac{1}{4}\cdot\frac{1}{3}=\frac{17}{12}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Preizkus:
+ 
+\begin_inset Formula $\left(\frac{17}{12}\right)^{2}=\frac{289}{144}=2+\frac{1}{144}$
+\end_inset
+
+ ...
+ Absolutna napaka 
+\begin_inset Formula $\frac{1}{144}$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $I\subseteq\mathbb{R}$
+\end_inset
+
+ interval in 
+\begin_inset Formula $f:I\to\mathbb{R}$
+\end_inset
+
+ odvedljiva povsod na 
+\begin_inset Formula $I$
+\end_inset
+
+.
+ Vzemimo 
+\begin_inset Formula $a\in I$
+\end_inset
+
+.
+ Če je v 
+\begin_inset Formula $a$
+\end_inset
+
+ odvedljiva tudi 
+\begin_inset Formula $f'$
+\end_inset
+
+,
+ pišemo 
+\begin_inset Formula $f''\left(a\right)=\left(f'\left(a\right)\right)'$
+\end_inset
+
+.
+ Podobno pišemo tudi višje odvode:
+ 
+\begin_inset Formula $f^{\left(1\right)}\left(a\right)=f'\left(a\right)$
+\end_inset
+
+,
+ 
+\begin_inset Formula $f^{\left(n+1\right)}=\left(f^{\left(n\right)}\right)'$
+\end_inset
+
+,
+ 
+\begin_inset Formula $f^{\left(0\right)}\left(a\right)=f\left(a\right)$
+\end_inset
+
+,
+ 
+\begin_inset Formula $f^{\left(2\right)}\left(a\right)=f''\left(a\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Note*
+Pomen besede 
+\begin_inset Quotes gld
+\end_inset
+
+odvod
+\begin_inset Quotes grd
+\end_inset
+
+:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Odvod v dani točki:
+ 
+\begin_inset Formula $f'\left(a\right)$
+\end_inset
+
+ za fiksen 
+\begin_inset Formula $a\in\mathbb{R}$
+\end_inset
+
+ ali
+\end_layout
+
+\begin_layout Itemize
+Funkcija,
+ ki vsaki točki 
+\begin_inset Formula $x\in\mathbb{R}$
+\end_inset
+
+ priredi 
+\begin_inset Formula $f'\left(x\right)$
+\end_inset
+
+ po zgornji definiciji.
+\end_layout
+
+\end_deeper
+\begin_layout Definition*
+\begin_inset Formula $C^{n}\left(I\right)$
+\end_inset
+
+ je množica funkcije 
+\begin_inset Formula $f:I\to\mathbb{R}$
+\end_inset
+
+,
+ da 
+\begin_inset Formula $\forall x\in I\exists f'\left(x\right),f''\left(x\right),f^{\left(3\right)},\dots,f^{\left(n\right)}\left(x\right)$
+\end_inset
+
+ in da so 
+\begin_inset Formula $f,f',f'',f^{\left(3\right)},\dots,f^{\left(n\right)}$
+\end_inset
+
+ zvezna funkcije na 
+\begin_inset Formula $I$
+\end_inset
+
+.
+ (seveda če obstaja 
+\begin_inset Formula $j-$
+\end_inset
+
+ti odvod,
+ obstaja tudi zvezen 
+\begin_inset Formula $j-1-$
+\end_inset
+
+ti odvod).
+ ZDB je to množica funkcij,
+ ki imajo vse odvode do 
+\begin_inset Formula $n$
+\end_inset
+
+ in so le-ti zvezni.
+ ZDB to so vse 
+\begin_inset Formula $n-$
+\end_inset
+
+krat zvezno odvedljive funkcije na intervalu 
+\begin_inset Formula $I$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Označimo 
+\begin_inset Formula $C^{\infty}\left(I\right)\coloneqq\bigcap_{n=1}^{\infty}C^{n}\left(I\right)$
+\end_inset
+
+ – to so neskončnokrat odvedljive funkcije na intervalu 
+\begin_inset Formula $I$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Note*
+Intuitivno
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+Baje.
+ Jaz sem itak do vsega skeptičen.
+\end_layout
+
+\end_inset
+
+ velja 
+\begin_inset Formula $C^{1}\left(I\right)\supset C^{2}\left(I\right)\supset C^{3}\left(I\right)\supset C^{4}\left(I\right)\supset\cdots$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Nekaj primerov.
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Polimomi 
+\begin_inset Formula $\subset C^{\infty}\left(\mathbb{R}\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $f\left(x\right)=\left|x\right|^{3}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $f'\left(x\right)=\begin{cases}
+3x^{2} & ;x\geq0\\
+-3x^{2} & ;x<0
+\end{cases}=3x^{2}\sgn x$
+\end_inset
+
+,
+ 
+\begin_inset Formula $f''\left(x\right)=\begin{cases}
+6x & ;x\geq0\\
+-6x & ;x<0
+\end{cases}=6x\sgn x$
+\end_inset
+
+,
+ 
+\begin_inset Formula $f'''\left(x\right)=\begin{cases}
+6 & ;x>0\\
+-6 & ;x<0
+\end{cases}=6\sgn x$
+\end_inset
+
+ in v 
+\begin_inset Formula $0$
+\end_inset
+
+ ni odvedljiva,
+ zato 
+\begin_inset Formula $f\in C^{2}\left(\mathbb{R}\right)$
+\end_inset
+
+ a 
+\begin_inset Formula $f\not\in C^{3}\left(\mathbb{R}\right)$
+\end_inset
+
+,
+ ker 
+\begin_inset Formula $\exists f''$
+\end_inset
+
+ in je zvezna na 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+,
+ a 
+\begin_inset Formula $f'''$
+\end_inset
+
+ sicer obstaja,
+ a ni zvezna na 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+.
+ Velja pa 
+\begin_inset Formula $f\in C^{\infty}\left(\mathbb{R}\setminus\left\{ 0\right\} \right)$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Theorem*
+Rolle.
+ Naj bo 
+\begin_inset Formula $f:\left[a,b\right]\to\mathbb{R}$
+\end_inset
+
+ za 
+\begin_inset Formula $a,b\in\mathbb{R}$
+\end_inset
+
+ zvezna na 
+\begin_inset Formula $\left[a,b\right]$
+\end_inset
+
+ in odvedljiva na 
+\begin_inset Formula $\left(a,b\right)$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+f\left(a\right)=f\left(b\right)\Longrightarrow\exists\alpha\in\left(a,b\right)\ni:f'\left(\alpha\right)=0
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Proof
+Sumimo,
+ da je ustrezna 
+\begin_inset Formula $\alpha$
+\end_inset
+
+ tista,
+ ki je 
+\begin_inset Formula $\max$
+\end_inset
+
+ ali 
+\begin_inset Formula $\min$
+\end_inset
+
+ od 
+\begin_inset Formula $f$
+\end_inset
+
+ na 
+\begin_inset Formula $I$
+\end_inset
+
+.
+ 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{zfnkm}{Ker}
+\end_layout
+
+\end_inset
+
+ je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna na 
+\begin_inset Formula $\left[a,b\right]$
+\end_inset
+
+ (kompaktni množici),
+ 
+\begin_inset Formula $\exists\alpha_{1}\in\left[a,b\right],\alpha_{2}\in\left[a,b\right]\ni:f\left(\alpha_{1}\right)=\max f\left(\left[a,b\right]\right)\wedge f\left(\alpha_{2}\right)=\min f\left(\left[a,b\right]\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Če je 
+\begin_inset Formula $\left\{ \alpha_{1},\alpha_{2}\right\} \subseteq\left\{ a,b\right\} $
+\end_inset
+
+,
+ je 
+\begin_inset Formula $f\left(\alpha_{1}\right)=f\left(\alpha_{2}\right)$
+\end_inset
+
+ in je v tem primeru 
+\begin_inset Formula $f$
+\end_inset
+
+ konstanta (
+\begin_inset Formula $\exists!c\in\mathbb{R}\ni:f\left(x\right)=c$
+\end_inset
+
+),
+ ki je odvedljiva in ima povsod odvod nič.
+\end_layout
+
+\begin_layout Proof
+Sicer pa 
+\begin_inset Formula $\left\{ \alpha_{1},\alpha_{2}\right\} \not\subseteq\left\{ a,b\right\} $
+\end_inset
+
+.
+ Tedaj ločimo dva primera:
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\alpha_{1}\in\left(a,b\right)$
+\end_inset
+
+ To pomeni,
+ da je globalni maksimum na odprtem intervalu.
+ Trdimo,
+ da je v lokalnem maksimumu odvod 0.
+ Dokaz:
+\begin_inset Formula 
+\[
+f'\left(\alpha_{1}\right)=\lim_{h\to0}\frac{f\left(\alpha_{1}+h\right)-f\left(\alpha_{1}\right)}{h}
+\]
+
+\end_inset
+
+Za 
+\begin_inset Formula $a_{1}$
+\end_inset
+
+ (maksimum) velja 
+\begin_inset Formula $f\left(\alpha_{1}+h\right)-f\left(\alpha_{1}\right)\leq0$
+\end_inset
+
+ (čim se pomaknemo izven točke,
+ v kateri je maksimum,
+ je funkcijska vrednost nižja).
+ Potemtakem velja
+\begin_inset Formula 
+\[
+\frac{f\left(\alpha_{1}+h\right)-f\left(\alpha_{1}\right)}{h}\quad\begin{cases}
+\leq0 & ;h>0\\
+\geq0 & ;h<0
+\end{cases}
+\]
+
+\end_inset
+
+Ker je funkcija odvedljiva na odprtem intervalu,
+ sta leva in desna limita enaki.
+\begin_inset Formula 
+\[
+0\geq\lim_{h\searrow0}\frac{f\left(\alpha_{1}+h\right)-f\left(\alpha_{1}\right)}{h}=\lim_{h\nearrow0}\frac{f\left(\alpha_{1}+h\right)-f\left(\alpha_{1}\right)}{h}\geq0
+\]
+
+\end_inset
+
+Sledi
+\begin_inset Formula 
+\[
+\lim_{h\to0}\frac{f\left(\alpha_{1}+h\right)-f\left(\alpha_{1}\right)}{h}=f'\left(x\right)=0
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\alpha_{2}\in\left(a,b\right)$
+\end_inset
+
+ To pomeni,
+ da je globalni minimum na odprtem intervalu.
+ Trdimo,
+ da je v lokalnem minimumu odvod 0.
+ Dokaz je podoben tistemu za lokalni maksimum.
+\end_layout
+
+\begin_layout Theorem*
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hypertarget{lagrange}{Lagrange}
+\end_layout
+
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $f:\left[a,b\right]\to\mathbb{R}$
+\end_inset
+
+ za 
+\begin_inset Formula $a,b\in\mathbb{R}$
+\end_inset
+
+ zvezna na 
+\begin_inset Formula $\left[a,b\right]$
+\end_inset
+
+ in odvedljiva na 
+\begin_inset Formula $\left(a,b\right)$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+\exists\alpha\in\left(a,b\right)\ni:f\left(b\right)-f\left(a\right)=f'\left(\alpha\right)\left(b-a\right)\sim\frac{f\left(b\right)-f\left(a\right)}{b-a}=f'\left(\alpha\right)
+\]
+
+\end_inset
+
+ZDB na neki točki na grafu funkcije je tangenta na graf funkcije vzporedna premici,
+ ki jo določata točki 
+\begin_inset Formula $\left(a,f\left(a\right)\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $\left(b,f\left(b\right)\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Za dokaz Lagrangevega uporabimo Rolleov izrek.
+ Splošen primer prevedemo na primer 
+\begin_inset Formula $h\left(a\right)=h\left(b\right)$
+\end_inset
+
+ tako,
+ da od naše splošne funkcije 
+\begin_inset Formula $f$
+\end_inset
+
+ odštejemo linearno funkcijo 
+\begin_inset Formula $g$
+\end_inset
+
+,
+ da bo veljalo 
+\begin_inset Formula $\left(f-g\right)\left(a\right)=\left(f-g\right)\left(b\right)$
+\end_inset
+
+.
+ Za funkcijo 
+\begin_inset Formula $g\left(x\right)$
+\end_inset
+
+ mora veljati naslednje:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $\exists k,n\in\mathbb{R}\ni:f\left(x\right)=kx+n$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $g\left(a\right)=0$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $g\left(b\right)=f\left(b\right)-f\left(a\right)$
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Proof
+Opazimo,
+ da mora biti koeficient funkcije 
+\begin_inset Formula $g$
+\end_inset
+
+ enak 
+\begin_inset Formula $\frac{f\left(b\right)-f\left(a\right)}{b-a}$
+\end_inset
+
+,
+ vertikalni odklon pa tolikšen,
+ da ima funkcija 
+\begin_inset Formula $g$
+\end_inset
+
+ v 
+\begin_inset Formula $a$
+\end_inset
+
+ ničlo:
+\begin_inset Formula 
+\[
+\frac{f\left(b\right)-f\left(a\right)}{b-a}a+n=0
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+n=-\frac{f\left(b\right)-f\left(a\right)}{b-a}a
+\]
+
+\end_inset
+
+Našli smo funkcijo 
+\begin_inset Formula $g\left(x\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}\left(x-a\right)$
+\end_inset
+
+.
+ Funkcija 
+\begin_inset Formula $\left(f-g\right)$
+\end_inset
+
+ sedaj ustreza pogojem za Rolleov izrek,
+ torej 
+\begin_inset Formula $\exists\alpha\in\left[a,b\right]\ni:\left(f-g\right)'\left(\alpha\right)=0\Leftrightarrow g'\left(\alpha\right)=f'\left(\alpha\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$
+\end_inset
+
+,
+ kar smo želeli dokazati.
+\end_layout
+
+\begin_layout Corollary*
+Naj bo 
+\begin_inset Formula $I\subseteq\mathbb{R}$
+\end_inset
+
+ nenujno zaprt niti omejen in 
+\begin_inset Formula $f:I\to\mathbb{R}$
+\end_inset
+
+ odvedljiva na 
+\begin_inset Formula $I$
+\end_inset
+
+.
+ Tedaj je 
+\begin_inset Formula $f$
+\end_inset
+
+ Lipschitzova.
+ Lipschitzove funkcije so enakomerno zvezne.
+\end_layout
+
+\begin_layout Proof
+Po Lagrangeu velja 
+\begin_inset Formula $\forall x,y\in I\exists\alpha\in\left(x,y\right)\ni:f\left(x\right)-f\left(y\right)=f'\left(\alpha\right)\left(x-y\right)$
+\end_inset
+
+.
+ Potemtakem 
+\begin_inset Formula $\left|f\left(x\right)-f\left(y\right)\right|=\left|f'\left(\alpha\right)\right|\left|x-y\right|\leq\sup_{\beta\in\left(x,y\right)}\left|f'\left(\beta\right)\right|\left|x-y\right|$
+\end_inset
+
+.
+ Torej 
+\begin_inset Formula $\exists M>0\forall x,y:\left|f\left(x\right)-f\left(y\right)\right|\leq M\left|x-y\right|$
+\end_inset
+
+,
+ enakomerno zveznost pa dobimo tako,
+ da 
+\begin_inset Formula $\delta\left(\varepsilon\right)=\frac{\varepsilon}{\sup_{\beta\in I}\left|f'\left(\beta\right)\right|}$
+\end_inset
+
+.
+ Računajmo.
+ Naj bo 
+\begin_inset Formula $M=\sup_{\beta\in I}\left|f'\left(\beta\right)\right|$
+\end_inset
+
+,
+ ki obstaja.
+\begin_inset Formula 
+\[
+\forall x,y:\left|f\left(x\right)-f\left(y\right)\right|\leq M\left|x-y\right|
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\forall x,y:\left|x-y\right|<\frac{\varepsilon}{\sup_{\beta\in I}\left|f'\left(\beta\right)\right|}\Rightarrow\left|f\left(x\right)-f\left(y\right)\right|\leq M\left|x-y\right|<\cancel{\sup_{\beta\in I}\left|f'\left(\beta\right)\right|}\frac{\varepsilon}{\cancel{\sup_{\beta\in I}\left|f'\left(\beta\right)\right|}}<\varepsilon
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\forall\varepsilon\exists\delta\left(\varepsilon\right)\forall x,y:\left|x-y\right|<\delta\left(\varepsilon\right)\Rightarrow\left|f\left(x\right)-f\left(y\right)\right|<\varepsilon
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Note*
+Lipschnitzovim funkcijam pravimo tudi Hölderjeve funkcije reda 1.
+ 
+\begin_inset Formula $f$
+\end_inset
+
+ je Hölderjeva funkcija reda 
+\begin_inset Formula $r$
+\end_inset
+
+,
+ če velja 
+\begin_inset Formula $\exists M>0\forall x,y\in I:\left|f\left(x\right)-f\left(y\right)\right|\leq M\left|x-y\right|^{r}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+Naj bo 
+\begin_inset Formula $I$
+\end_inset
+
+ odprti interval,
+ 
+\begin_inset Formula $f:I\to\mathbb{R}$
+\end_inset
+
+ odvedljiva.
+ Tedaj:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $f$
+\end_inset
+
+ narašča na 
+\begin_inset Formula $I\Leftrightarrow f'\geq0$
+\end_inset
+
+ na 
+\begin_inset Formula $I$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $f$
+\end_inset
+
+ pada na 
+\begin_inset Formula $I\Leftrightarrow f'\leq0$
+\end_inset
+
+ na 
+\begin_inset Formula $I$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $f$
+\end_inset
+
+ strogo narašča na 
+\begin_inset Formula $I\Leftarrow f'>0$
+\end_inset
+
+ na 
+\begin_inset Formula $I$
+\end_inset
+
+.
+ Protiprimer,
+ da ni 
+\begin_inset Formula $\Leftrightarrow:f\left(x\right)=x^{3}$
+\end_inset
+
+,
+ ki strogo narašča,
+ toda 
+\begin_inset Formula $f'\left(0\right)=0$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $f$
+\end_inset
+
+ strogo pada na 
+\begin_inset Formula $I\Leftarrow f'<0$
+\end_inset
+
+ na 
+\begin_inset Formula $I$
+\end_inset
+
+.
+ Protiprimer,
+ da ni 
+\begin_inset Formula $\Leftrightarrow:f\left(x\right)=-x^{3}$
+\end_inset
+
+,
+ ki strogo pada,
+ toda 
+\begin_inset Formula $f'\left(0\right)=0$
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Proof
+Dokažimo le 
+\begin_inset Formula $f$
+\end_inset
+
+ narašča na 
+\begin_inset Formula $I\Leftrightarrow f'\geq0$
+\end_inset
+
+ na 
+\begin_inset Formula $I$
+\end_inset
+
+.
+ Drugo točko dokažemo podobno.
+ Dokazujemo ekvivalenco:
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Leftarrow\right)$
+\end_inset
+
+ 
+\begin_inset Formula $f'\geq0\Rightarrow f$
+\end_inset
+
+ narašča.
+ Vzemimo poljubna 
+\begin_inset Formula $t_{1}<t_{2}\in I$
+\end_inset
+
+.
+ Po Lagrangeu 
+\begin_inset Formula $\exists\alpha\in\left(t_{1},t_{2}\right)\ni:f\left(t_{2}\right)-f\left(t_{1}\right)=f'\left(\alpha\right)\left(t_{2}-t_{1}\right)$
+\end_inset
+
+.
+ Ker je po predpostavki 
+\begin_inset Formula $f'\left(\alpha\right)\geq0$
+\end_inset
+
+ in 
+\begin_inset Formula $t_{2}-t_{1}>0$
+\end_inset
+
+,
+ je tudi 
+\begin_inset Formula $f\left(t_{2}\right)-f\left(t_{1}\right)\geq0$
+\end_inset
+
+ in zato 
+\begin_inset Formula $f\left(t_{2}\right)\geq f\left(t_{1}\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Rightarrow\right)$
+\end_inset
+
+ 
+\begin_inset Formula $f$
+\end_inset
+
+ narašča 
+\begin_inset Formula $\Rightarrow f'\geq0$
+\end_inset
+
+.
+ Velja 
+\begin_inset Formula $f'\left(x\right)=\lim_{h\to0}\frac{f\left(x+h\right)-f\left(x\right)}{h}$
+\end_inset
+
+.
+ Po predpostavki je 
+\begin_inset Formula $f\left(x+h\right)-f\left(x\right)\geq0$
+\end_inset
+
+,
+ čim je 
+\begin_inset Formula $h>0$
+\end_inset
+
+,
+ in 
+\begin_inset Formula $f\left(x+h\right)-f\left(x\right)\leq0$
+\end_inset
+
+,
+ čim je 
+\begin_inset Formula $h<0$
+\end_inset
+
+.
+ Torej je ulomek vedno nenegativen.
+\end_layout
+
+\end_deeper
+\begin_layout Subsection
+Konveksnost in konkavnost
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $I\subseteq\mathbb{R}$
+\end_inset
+
+ interval in 
+\begin_inset Formula $f:I\to\mathbb{R}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $f$
+\end_inset
+
+ je konveksna na 
+\begin_inset Formula $I$
+\end_inset
+
+,
+ če 
+\begin_inset Formula $\forall a,b\in I$
+\end_inset
+
+ daljica 
+\begin_inset Formula $\left(a,f\left(a\right)\right),\left(b,f\left(b\right)\right)$
+\end_inset
+
+ leži nad grafom 
+\begin_inset Formula $f$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Enačba premice,
+ ki vsebuje to daljico,
+ se glasi (razmislek je podoben kot pri 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{lagrange}{Lagrangevem izreku}
+\end_layout
+
+\end_inset
+
+)
+\begin_inset Formula 
+\[
+g\left(x\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}\left(x-a\right)+f\left(a\right)
+\]
+
+\end_inset
+
+Za konveksno funkcijo torej velja 
+\begin_inset Formula $\forall a,b\in I:\forall x\in\left(a,b\right):f\left(x\right)\leq\frac{f\left(b\right)-f\left(a\right)}{b-a}\left(x-a\right)+f\left(a\right)$
+\end_inset
+
+ oziroma
+\begin_inset Formula 
+\[
+\frac{f\left(x\right)-f\left(a\right)}{x-a}\leq\frac{f\left(b\right)-f\left(a\right)}{b-a}
+\]
+
+\end_inset
+
+Vsak 
+\begin_inset Formula $x$
+\end_inset
+
+ na intervalu lahko zapišemo kot 
+\begin_inset Formula $x=a+t\left(b-a\right)$
+\end_inset
+
+ za nek 
+\begin_inset Formula $t\in\left(0,1\right)$
+\end_inset
+
+.
+ Tedaj je 
+\begin_inset Formula $x-a=t\left(b-a\right)$
+\end_inset
+
+ in konveksnost se glasi
+\begin_inset Formula 
+\[
+\forall a,b\in I:\forall t\in\left(0,1\right):f\left(a+t\left(b-a\right)\right)\leq\frac{f\left(b\right)-f\left(a\right)}{\cancel{b-a}}t\cancel{\left(b-a\right)}+f\left(a\right)
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+f\left(a+t\left(b-a\right)\right)=f\left(a+tb-ta\right)=f\left(\left(1-t\right)a+tb\right)\leq tf\left(b\right)-tf\left(a\right)+f\left(a\right)=\left(1-t\right)f\left(a\right)+tf\left(b\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Konveksna kombinacija izrazov 
+\begin_inset Formula $a,b$
+\end_inset
+
+ je izraz oblike 
+\begin_inset Formula $\left(1-t\right)a+tb$
+\end_inset
+
+ za 
+\begin_inset Formula $t\in\left(0,1\right)$
+\end_inset
+
+.
+ Potemtakem je ZDB definicija konveksnosti 
+\begin_inset Formula $\forall a,b\in I:$
+\end_inset
+
+ funkcijska vrednost konveksne kombinacije 
+\begin_inset Formula $a,b$
+\end_inset
+
+ je kvečjemu konveksna kombinacija funkcijskih vrednosti 
+\begin_inset Formula $a,b$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Konkavnost pa je definirana tako,
+ da povsod obrnemo predznake,
+ torej daljica leži pod grafom 
+\begin_inset Formula $f$
+\end_inset
+
+ ZDB 
+\begin_inset Formula $\forall a,b\in I:f\left(\left(1-t\right)a+tb\right)\geq\left(1-t\right)f\left(a\right)+tf\left(b\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $f\left(x\right)=\sin x$
+\end_inset
+
+,
+ 
+\begin_inset Formula $I=\left[-\pi,0\right]$
+\end_inset
+
+.
+ Je konveksna.
+ Se vidi iz grafa.
+ Preveriti analitično bi bilo težko.
+\end_layout
+
+\begin_layout Example*
+Formulirajmo drugačen pogoj za konveksnost.
+ Naj bo spet 
+\begin_inset Formula $f:I\to\mathbb{R}$
+\end_inset
+
+,
+ kjer je 
+\begin_inset Formula $I$
+\end_inset
+
+ interval.
+ 
+\begin_inset Formula $f$
+\end_inset
+
+ je konveksna 
+\begin_inset Formula 
+\[
+\Leftrightarrow\forall a,b\in I\forall x\in\left(a,b\right):\frac{f\left(x\right)-f\left(a\right)}{x-a}\leq\frac{f\left(b\right)-f\left(a\right)}{b-a}.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+Sedaj glejmo le poljuben 
+\begin_inset Formula $a$
+\end_inset
+
+.
+ Po prejšnjem pogoju moramo gledati še vse poljubne 
+\begin_inset Formula $b$
+\end_inset
+
+,
+ večje od 
+\begin_inset Formula $a$
+\end_inset
+
+ (ker le tako lahko konstruiramo interval).
+ Za 
+\begin_inset Formula $b$
+\end_inset
+
+ in 
+\begin_inset Formula $a$
+\end_inset
+
+ mora biti diferenčni kvocient večji od diferenčnega kvocienta 
+\begin_inset Formula $x$
+\end_inset
+
+ in 
+\begin_inset Formula $a$
+\end_inset
+
+ za poljuben 
+\begin_inset Formula $x$
+\end_inset
+
+.
+ Ta pogoj pa je ekvivalenten temu,
+ da diferenčni kvocient 
+\begin_inset Formula $x$
+\end_inset
+
+ in 
+\begin_inset Formula $a$
+\end_inset
+
+ s fiksnim 
+\begin_inset Formula $a$
+\end_inset
+
+ in čedalje večjim 
+\begin_inset Formula $x$
+\end_inset
+
+ narašča,
+ torej je pogoj za konveksnost tudi:
+\begin_inset Formula 
+\[
+\forall a\in I\forall x>a:g_{a}\left(x\right)=\frac{f\left(x\right)-f\left(a\right)}{x-a}\text{ je naraščajoča funkcija}.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Corollary*
+Naj bo 
+\begin_inset Formula $f$
+\end_inset
+
+ konveksna na odprtem intervalu 
+\begin_inset Formula $I$
+\end_inset
+
+.
+ 
+\begin_inset Formula $\forall a\in I$
+\end_inset
+
+ obstajata funkciji
+\begin_inset Formula 
+\[
+\left(D_{+}f\right)\left(a\right)=\lim_{x\searrow a}g_{a}\left(x\right)=\inf_{x\in I,x>a}g_{a}\left(x\right)\text{ (desni odvod \ensuremath{f} v \ensuremath{a})}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\left(D_{-}f\right)\left(a\right)=\lim_{x\nearrow a}g_{a}\left(x\right)=\sup_{x\in I,x<a}g_{a}\left(x\right)\text{ (levi odvod \ensuremath{f} v \ensuremath{a})}
+\]
+
+\end_inset
+
+in obe sta naraščajoči na 
+\begin_inset Formula $I$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Obstoj sledi iz monotonosti 
+\begin_inset Formula $g_{a}\left(a\right)$
+\end_inset
+
+,
+ kajti 
+\begin_inset Formula $\lim_{x\searrow a}g_{a}\left(x\right)=\lim_{x\searrow a}\frac{f\left(x\right)-f\left(a\right)}{x-a}$
+\end_inset
+
+ in enako za levo limito.
+ Diferenčni kvocient mora namreč biti naraščajoč.
+ S tem smo dokazali,
+ da je vsaka konveksna funkcija zvezna
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+Ni pa vsaka konveksna funkcija odvedljiva,
+ protiprimer je 
+\begin_inset Formula $f\left(x\right)=\left|x\right|$
+\end_inset
+
+.
+\end_layout
+
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Naj bodo 
+\begin_inset Formula $x_{1},x_{2},x\in I\ni:x_{1}<x_{2}<x$
+\end_inset
+
+.
+ Pomagaj si s skico
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+TODO DORIŠI SKICO ZVZ VII/ANA1UČ/str.
+ 13
+\end_layout
+
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $f$
+\end_inset
+
+ konveksna,
+ sledi 
+\begin_inset Formula $g_{x}\left(x_{1}\right)\leq g_{x}\left(x_{2}\right)$
+\end_inset
+
+.
+ Ker 
+\begin_inset Formula $\forall s,t\in\mathbb{R}:g_{s}\left(t\right)=g_{t}\left(s\right)$
+\end_inset
+
+,
+ lahko našo neenakost zapišemo kot 
+\begin_inset Formula $g_{x_{1}}\left(x\right)\leq g_{x_{2}}\left(x\right)$
+\end_inset
+
+.
+ Sledi (desni neenačaj iz 
+\begin_inset Formula $g_{x_{1}}\left(x\right)\leq g_{x_{2}}\left(x\right)$
+\end_inset
+
+,
+ levi neenačaj pa ker 
+\begin_inset Formula $g$
+\end_inset
+
+ narašča):
+\begin_inset Formula 
+\[
+\left(D_{+}\left(f\right)\right)\left(x_{1}\right)=\inf_{x\in I,x>x_{1}}g_{x_{1}}\left(x\right)\leq\inf_{x\in I,x>x_{2}}g_{x_{1}}\left(x\right)\leq\inf_{x\in I,x>x_{2}}g_{x_{2}}\left(x\right)=\left(D_{+}\left(f\right)\right)\left(x_{2}\right)
+\]
+
+\end_inset
+
+Podobno dokažemo
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+DOPIŠI KAKO!
+ TODO XXX FIXME
+\end_layout
+
+\end_inset
+
+,
+ da 
+\begin_inset Formula $D_{-}$
+\end_inset
+
+ narašča.
+\end_layout
+
+\begin_layout Theorem*
+Naj bo 
+\begin_inset Formula $f:I^{\text{odp.}}\to\mathbb{R}$
+\end_inset
+
+ dvakrat odvedljiva.
+ Tedaj je 
+\begin_inset Formula $f$
+\end_inset
+
+ konveksna 
+\begin_inset Formula $\Leftrightarrow\forall x\in I:f''\left(x\right)\geq0$
+\end_inset
+
+ in 
+\begin_inset Formula $f$
+\end_inset
+
+ konkavna 
+\begin_inset Formula $\Leftrightarrow\forall x\in I:f''\left(x\right)\leq0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Dokazujemo ekvivalenco za konveksnost (konkavnost podobno).
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Rightarrow\right)$
+\end_inset
+
+ Po predpostavki je 
+\begin_inset Formula $f$
+\end_inset
+
+ konveksna in dvakrat odvedljiva,
+ torej je odvedljiva in sta levi in desni odvod enaka,
+ po prejšnji posledici pa levi in desni odvod naraščata,
+ torej 
+\begin_inset Formula $f'$
+\end_inset
+
+ narašča.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Leftarrow\right)$
+\end_inset
+
+ Naj bo 
+\begin_inset Formula $f''\geq0$
+\end_inset
+
+.
+ Vzemimo 
+\begin_inset Formula $x,a\in I$
+\end_inset
+
+.
+ Po Lagrangeu 
+\begin_inset Formula $\exists\xi\text{ med \ensuremath{x} in \ensuremath{a}}\ni:f\left(x\right)-f\left(x\right)=f'\left(\xi\right)\left(x-a\right)$
+\end_inset
+
+.
+ Iz predpostavke 
+\begin_inset Formula $f''>0$
+\end_inset
+
+ sledi,
+ da 
+\begin_inset Formula $f'$
+\end_inset
+
+ narašča.
+ Če je 
+\begin_inset Formula $x>\xi>a$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $f'\left(\xi\right)\geq f'\left(a\right)$
+\end_inset
+
+,
+ zato 
+\begin_inset Formula $f'\left(\xi\right)\left(x-a\right)\geq f'\left(a\right)\left(x-a\right)$
+\end_inset
+
+.
+ Če je 
+\begin_inset Formula $x<\xi<a$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $f'\left(\xi\right)\left(x-a\right)\leq f'\left(a\right)\left(x-a\right)$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Subsection
+Ekstremi funkcij ene spremenljivke
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $I\subseteq\mathbb{R}$
+\end_inset
+
+ odprt interal,
+ 
+\begin_inset Formula $a\in I$
+\end_inset
+
+ in 
+\begin_inset Formula $f:I\to\mathbb{R}$
+\end_inset
+
+.
+ Pravimo,
+ da ima 
+\begin_inset Formula $f$
+\end_inset
+
+ v točki 
+\begin_inset Formula $a$
+\end_inset
+
+ lokalni minimum,
+ če 
+\begin_inset Formula $\exists\delta>0\ni:\min\left\{ f\left(x\right);\forall x\in\left(a-\delta,a+\delta\right)\right\} =f\left(a\right)$
+\end_inset
+
+.
+ Pravimo,
+ da ima 
+\begin_inset Formula $f$
+\end_inset
+
+ v točki 
+\begin_inset Formula $a$
+\end_inset
+
+ lokalni maksimum,
+ če 
+\begin_inset Formula $\exists\delta>0\ni:\max\left\{ f\left(x\right);\forall x\in\left(a-\delta,a+\delta\right)\right\} =f\left(a\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+Če je 
+\begin_inset Formula $f:I\to\mathbb{R}$
+\end_inset
+
+ odvedljiva in ima v 
+\begin_inset Formula $a$
+\end_inset
+
+ lokalni minimum/maksimum,
+ tedaj je 
+\begin_inset Formula $f'\left(a\right)=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Glej dokaz Rolleovega izreka.
+\end_layout
+
+\begin_layout Definition*
+\begin_inset Formula $f$
+\end_inset
+
+ ima v 
+\begin_inset Formula $a$
+\end_inset
+
+ ekstrem,
+ če ima v 
+\begin_inset Formula $a$
+\end_inset
+
+ lokalni minimum ali lokalni maksimum.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Če je 
+\begin_inset Formula $f'\left(a\right)=0$
+\end_inset
+
+,
+ pravimo,
+ da ima 
+\begin_inset Formula $f$
+\end_inset
+
+ v 
+\begin_inset Formula $a$
+\end_inset
+
+ stacionarno točko.
+\end_layout
+
+\begin_layout Theorem*
+Naj bo 
+\begin_inset Formula $I\subseteq\mathbb{R}$
+\end_inset
+
+ odprt interval,
+ 
+\begin_inset Formula $a\in I$
+\end_inset
+
+ in 
+\begin_inset Formula $f:I\to\mathbb{R}$
+\end_inset
+
+ dvakrat odvedljiva ter naj bo 
+\begin_inset Formula $f'\left(a\right)=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $f''\left(a\right)>0\Rightarrow$
+\end_inset
+
+ v 
+\begin_inset Formula $a$
+\end_inset
+
+ ima 
+\begin_inset Formula $f$
+\end_inset
+
+ lokalni minimum
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $f''\left(a\right)<0\Rightarrow$
+\end_inset
+
+ v 
+\begin_inset Formula $a$
+\end_inset
+
+ ima 
+\begin_inset Formula $f$
+\end_inset
+
+ lokalni maksimum
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $f''\left(a\right)=0\Rightarrow$
+\end_inset
+
+ nedoločeno
+\end_layout
+
+\end_deeper
+\begin_layout Proof
+Sledi iz 
+\begin_inset Formula $f''>0\Rightarrow$
+\end_inset
+
+ stroga konveksnost in 
+\begin_inset Formula $f''<0\Rightarrow$
+\end_inset
+
+ stroga konkavnost.
+\end_layout
+
+\begin_layout Subsection
+L'Hopitalovo pravilo
+\end_layout
+
+\begin_layout Standard
+Kako izračunati 
+\begin_inset Formula $\lim_{x\to a}\frac{f\left(x\right)}{g\left(x\right)}$
+\end_inset
+
+?
+\end_layout
+
+\begin_layout Standard
+Če so funkcije zvezne v 
+\begin_inset Formula $a$
+\end_inset
+
+ in 
+\begin_inset Formula $g\left(a\right)\not=0$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $\lim_{x\to a}\frac{f\left(x\right)}{g\left(x\right)}=\frac{f\left(a\right)}{g\left(a\right)}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Če imata funkciji v 
+\begin_inset Formula $a$
+\end_inset
+
+ limito in 
+\begin_inset Formula $\lim_{x\to a}g\left(x\right)\not=0$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $\lim_{x\to a}\frac{f\left(x\right)}{g\left(x\right)}=\frac{\lim_{x\to a}f\left(x\right)}{\lim_{x\to a}g\left(x\right)}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Če 
+\begin_inset Formula $\lim_{x\to a}g\left(x\right)=\infty$
+\end_inset
+
+ in je na neki okolici 
+\begin_inset Formula $a$
+\end_inset
+
+ 
+\begin_inset Formula $f\left(x\right)$
+\end_inset
+
+ omejena,
+ velja 
+\begin_inset Formula $\lim_{x\to a}\frac{f\left(x\right)}{g\left(x\right)}=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Če 
+\begin_inset Formula $\lim_{x\to a}f\left(x\right)=0$
+\end_inset
+
+ in je na neki okolici 
+\begin_inset Formula $a$
+\end_inset
+
+ 
+\begin_inset Formula $g\left(x\right)$
+\end_inset
+
+ navzdol omejena več od nič ali navzgor omejena manj od nič,
+ velja 
+\begin_inset Formula $\lim_{x\to a}\frac{f\left(x\right)}{g\left(x\right)}=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Zanimivi primeri pa so,
+ ko 
+\begin_inset Formula $\lim_{x\to a}f\left(x\right)=\lim_{x\to a}g\left(x\right)=0$
+\end_inset
+
+ ali pa ko 
+\begin_inset Formula $\lim_{x\to a}f\left(x\right)=\infty$
+\end_inset
+
+ in hkrati 
+\begin_inset Formula $\lim_{x\to a}g\left(x\right)=\infty$
+\end_inset
+
+,
+ na primer 
+\begin_inset Formula $\lim_{x\to0}\frac{x}{x}$
+\end_inset
+
+ ali pa 
+\begin_inset Formula $\lim_{x\to0}\frac{x^{2}}{x}$
+\end_inset
+
+ ali pa 
+\begin_inset Formula $\lim_{x\to0}\frac{x}{x^{2}}$
+\end_inset
+
+.
+ Tedaj uporabimo L'Hopitalovo pravilo.
+\end_layout
+
+\begin_layout Theorem*
+Če velja hkrati:
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset CommandInset label
+LatexCommand label
+name "enu:Eno-izmed-slednjega:"
+
+\end_inset
+
+Eno izmed slednjega:
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $\lim_{x\to a}f\left(x\right)=\lim_{x\to a}g\left(x\right)=0$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\lim_{x\to a}f\left(x\right)=\infty$
+\end_inset
+
+ in hkrati 
+\begin_inset Formula $\lim_{x\to a}g\left(x\right)=\infty$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\lim_{x\to a}f\left(x\right)=-\infty$
+\end_inset
+
+ in hkrati 
+\begin_inset Formula $\lim_{x\to a}g\left(x\right)=-\infty$
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+\begin_inset Formula $f,g$
+\end_inset
+
+ v okolici 
+\begin_inset Formula $a$
+\end_inset
+
+ odvedljivi
+\end_layout
+
+\end_deeper
+\begin_layout Theorem*
+Potem 
+\begin_inset Formula $\exists L\coloneqq\lim_{x\to a}\frac{f'\left(x\right)}{g'\left(x\right)}\Rightarrow\exists\lim_{x\to a}\frac{f\left(x\right)}{g\left(x\right)}$
+\end_inset
+
+ in ta limita je enaka 
+\begin_inset Formula $L$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Ne bomo dokazali.
+\end_layout
+
+\begin_layout Example*
+Nekaj primerov uporabe L'Hopitalovega pravila.
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula 
+\[
+\lim_{x\to0}x^{x}=\lim_{x\to0}e^{lnx^{x}}=\lim_{x\to0}e^{x\ln x}=e^{\lim_{x\to0}x\ln x}
+\]
+
+\end_inset
+
+Računajmo 
+\begin_inset Formula $\lim_{x\to0}x\ln x$
+\end_inset
+
+ z L'Hopitalom.
+ Potrebujemo ulomek.
+ Ideja:
+ množimo števec in imenovalec z 
+\begin_inset Formula $x$
+\end_inset
+
+,
+ tedaj bi dobili 
+\begin_inset Formula $\lim_{x\to0}\frac{x^{2}\ln x}{x}$
+\end_inset
+
+.
+ Toda v tem primeru števec in imenovalec ne ustrezata pogoju 
+\begin_inset CommandInset ref
+LatexCommand ref
+reference "enu:Eno-izmed-slednjega:"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+ za L'Hopitalovo pravilo.
+ Druga ideja:
+ množimo števec in imenovalec z 
+\begin_inset Formula $\left(\ln x\right)^{-1}$
+\end_inset
+
+,
+ tedaj dobimo 
+\begin_inset Formula $\lim_{x\to0}\frac{x}{\left(\ln x\right)^{-1}}\overset{\text{L'H}}{=}\lim_{x\to0}\frac{1}{\frac{-1}{\log^{2}x}\cdot\frac{1}{x}}=\lim_{x\to0}-x\log^{2}x$
+\end_inset
+
+,
+ kar je precej komplicirano.
+ Tretja ideja:
+ množimo števec in imenovalec z 
+\begin_inset Formula $x^{-1}$
+\end_inset
+
+,
+ tedaj števec in imenovalec divergirata k 
+\begin_inset Formula $-\infty$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+\lim_{x\to0}\frac{\ln x}{x^{-1}}\overset{\text{L'H}}{=}\lim_{x\to0}\frac{\left(\ln x\right)'}{\left(x^{-1}\right)'}=\lim_{x\to0}\frac{x^{-1}}{-x^{-2}}=\lim_{x\to0}-x=0
+\]
+
+\end_inset
+
+Potemtakem 
+\begin_inset Formula $\lim_{x\to0}x^{x}=e^{0}=1$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\lim_{x\to0}\frac{1-\cos x}{x^{2}}$
+\end_inset
+
+.
+ Obe strani ulomkove črte konvergirata k 
+\begin_inset Formula $0$
+\end_inset
+
+.
+ Prav tako ko enkrat že uporabimo L'H.
+\begin_inset Formula 
+\[
+\lim_{x\to0}\frac{1-\cos x}{x^{2}}\overset{\text{L'H}}{=}\lim_{x\to0}\frac{\sin x}{2x}\overset{\text{L'H}}{=}\lim_{x\to0}\frac{\cos x}{2}=\frac{1}{2}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Section
+Taylorjev izrek in Taylorjeva formula
+\end_layout
+
+\begin_layout Standard
+Naj bo 
+\begin_inset Formula $f$
+\end_inset
+
+ v okolici 
+\begin_inset Formula $a$
+\end_inset
+
+ dovoljkrat odvedljiva.
+ Želimo aproksimirati 
+\begin_inset Formula $f\left(a+h\right)$
+\end_inset
+
+ s polinomi danega reda 
+\begin_inset Formula $n$
+\end_inset
+
+.
+ Iščemo polinome reda 
+\begin_inset Formula $n$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $n=0$
+\end_inset
+
+ konstante.
+ 
+\begin_inset Formula $f\left(a+h\right)\approx f\left(a\right)$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $n=1$
+\end_inset
+
+ linearne funkcije.
+ 
+\begin_inset Formula $f\left(a+h\right)\sim f\left(a\right)+f'\left(a\right)h$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $n=2$
+\end_inset
+
+ ...
+ Želimo najti 
+\begin_inset Formula $a_{0},a_{1},a_{2}\in\mathbb{R}$
+\end_inset
+
+,
+ odvisne le od 
+\begin_inset Formula $f$
+\end_inset
+
+ in 
+\begin_inset Formula $a$
+\end_inset
+
+,
+ za katere 
+\begin_inset Formula $f\left(a+b\right)\approx a_{0}+a_{1}h+a_{2}h^{2}$
+\end_inset
+
+.
+ Ko govorimo o aproksimaciji,
+ mislimo take koeficiente,
+ da se približek najbolje prilega dejanski funkcijski vrednosti,
+ v smislu,
+ da
+\begin_inset Formula 
+\[
+\lim_{h\to0}\frac{f\left(a+h\right)-\left(a_{0}+a_{1}h+a_{2}h^{2}\right)}{h^{2}}=0
+\]
+
+\end_inset
+
+
+\begin_inset Formula $a_{0}$
+\end_inset
+
+ izvemo takoj,
+ kajti 
+\begin_inset Formula $\lim_{h\to0}f\left(a+h\right)-\left(a_{0}+a_{1}h+a_{2}h^{2}\right)=0=f\left(a\right)-\left(a_{0}+0h+0h^{2}\right)=f\left(a\right)-a_{0}=0$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $a_{0}=f\left(a\right)$
+\end_inset
+
+.
+ Za preostale koeficiente uporabimo L'Hopitalovo pravilo,
+ ki pove,
+ da zadošča,
+ da je
+\begin_inset Formula 
+\[
+\lim_{h\to0}\frac{f'\left(a+h\right)-\left(0+a_{1}+a_{2}h\right)}{2h}=0
+\]
+
+\end_inset
+
+Zopet glejmo števec in vstavimo 
+\begin_inset Formula $h=0$
+\end_inset
+
+:
+ 
+\begin_inset Formula $f'\left(a\right)-a_{1}=0\Rightarrow f'\left(a\right)=a_{1}$
+\end_inset
+
+.
+ Spet uporabimo L'H:
+\begin_inset Formula 
+\[
+\lim_{h\to0}\frac{f''\left(a+h\right)-\left(0+0+2a_{2}\right)}{2}
+\]
+
+\end_inset
+
+Vstavimo 
+\begin_inset Formula $h=0$
+\end_inset
+
+ v 
+\begin_inset Formula $f''\left(a+h\right)-2a_{2}$
+\end_inset
+
+ in dobimo 
+\begin_inset Formula $2a_{2}=f''\left(a\right)$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $a_{2}=\frac{f''\left(a\right)}{2}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $n=3$
+\end_inset
+
+ Ugibamo,
+ da je najboljši kubični približek
+\begin_inset Formula 
+\[
+f\left(a+h\right)\approx h\mapsto f\left(a\right)+f'\left(a\right)h+\frac{f''\left(a\right)}{2}h^{2}+\frac{f'''\left(a\right)}{6}h^{3}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Theorem*
+Taylor.
+ Naj bo 
+\begin_inset Formula $n\in\mathbb{N}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $I$
+\end_inset
+
+ interval 
+\begin_inset Formula $\subseteq\mathbb{R}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $a\in I$
+\end_inset
+
+,
+ 
+\begin_inset Formula $f:I\to\mathbb{R}$
+\end_inset
+
+ 
+\begin_inset Formula $n-$
+\end_inset
+
+krat odvedljiva v točki 
+\begin_inset Formula $a$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $\exists g_{n}:I-a\to\mathbb{R}\ni:$
+\end_inset
+
+
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $I-a$
+\end_inset
+
+ pomeni interval 
+\begin_inset Formula $I$
+\end_inset
+
+ pomaknjen v levo za 
+\begin_inset Formula $a$
+\end_inset
+
+.
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $f\left(a+h\right)=\sum_{j=0}^{n}\frac{f^{\left(j\right)}\left(a\right)}{n!}h^{j}+g_{n}\left(h\right)h^{n}$
+\end_inset
+
+ in
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $\lim_{h\to0}g_{n}\left(h\right)=0$
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Theorem*
+Sedaj pišimo 
+\begin_inset Formula $x=a+h$
+\end_inset
+
+.
+ Tedaj se izrek glasi:
+ 
+\begin_inset Formula $\exists\tilde{g_{n}}:I\to\mathbb{R}\ni:$
+\end_inset
+
+
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $f\left(x\right)=\sum_{j=0}^{n}\frac{f^{\left(j\right)}\left(a\right)}{n!}\left(x-a\right)^{j}+\tilde{g_{n}}\left(x\right)\left(x-a\right)^{n}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $\lim_{x\to a}\tilde{g_{n}}\left(x\right)=0$
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Theorem*
+Tedaj označimo 
+\begin_inset Formula $T_{n,f,a}\left(x\right)=\sum_{j=0}^{n}\frac{f^{\left(j\right)}\left(a\right)}{n!}\left(x-a\right)^{j}$
+\end_inset
+
+ (pravimo 
+\begin_inset Formula $n-$
+\end_inset
+
+ti taylorjev polinom za 
+\begin_inset Formula $f$
+\end_inset
+
+ okrog točke 
+\begin_inset Formula $a$
+\end_inset
+
+) in 
+\begin_inset Formula $R_{n,f,a}\left(x\right)=\tilde{g_{n}}\left(x\right)\left(x-a\right)^{n}$
+\end_inset
+
+ (pravimo ostanek/napaka).
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Theorem*
+Če je 
+\begin_inset Formula $f$
+\end_inset
+
+ 
+\begin_inset Formula $\left(n+1\right)-$
+\end_inset
+
+krat odvedljiva na odprtem intervalu 
+\begin_inset Formula $I\subseteq\mathbb{R}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $a\in I$
+\end_inset
+
+,
+ tedaj 
+\begin_inset Formula $\forall b\in I\exists\alpha\in I\text{ med \ensuremath{a} in \ensuremath{b}}\ni:R_{n}\left(b\right)=\frac{f^{\left(n+1\right)}\left(\alpha\right)}{\left(n+1\right)!}\left(b-a\right)^{n+1}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Označimo 
+\begin_inset Formula $T_{n}\left(x\right)=f\left(a\right)+f'\left(a\right)\left(x-a\right)+\frac{f''\left(a\right)}{2!}\left(x-a\right)^{2}+\cdots+\frac{f^{\left(n\right)}\left(a\right)}{n!}\left(x-a\right)^{n}$
+\end_inset
+
+ torej 
+\begin_inset Formula $n-$
+\end_inset
+
+ti taylorjev polinom in naj bo 
+\begin_inset Formula $K$
+\end_inset
+
+ tako število,
+ da velja 
+\begin_inset Formula $f\left(b\right)-T_{n}\left(b\right)=K\left(b-a\right)^{n+1}$
+\end_inset
+
+.
+ Definirajmo 
+\begin_inset Formula $F\left(x\right)=f\left(x\right)-T_{n}\left(x\right)-K\left(x-a\right)^{n+1}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hypertarget{velja}{Velja}
+\end_layout
+
+\end_inset
+
+ 
+\begin_inset Formula $T_{n}^{\left(k\right)}\left(a\right)=f\left(a\right)$
+\end_inset
+
+ za 
+\begin_inset Formula $k\leq n$
+\end_inset
+
+,
+ kajti 
+\begin_inset Formula $\frac{d\sum_{j=1}^{n}\frac{f^{\left(j\right)}\left(a\right)}{n!}\left(h\right)^{n}}{dh}=\frac{f^{\left(j\right)}\left(a\right)n!}{n!}\cdot1=f^{\left(j\right)}\left(a\right)$
+\end_inset
+
+.
+ Vsi členi z eksponentom,
+ manjšim od 
+\begin_inset Formula $k$
+\end_inset
+
+,
+ se odvajajo v 0,
+ točno pri eksponentu 
+\begin_inset Formula $k$
+\end_inset
+
+ se člen odvaja v konstanto,
+ pri višjih členih pa ostane potencirana spremenljivka,
+ ki je 
+\begin_inset Formula $0$
+\end_inset
+
+ (tu mislimo odstopanje od 
+\begin_inset Formula $a$
+\end_inset
+
+,
+ označeno s 
+\begin_inset Formula $h$
+\end_inset
+
+),
+ torej se ti členi tudi izničijo.
+\end_layout
+
+\begin_layout Proof
+Zato 
+\begin_inset Formula $\forall k\leq n:F^{\left(k\right)}\left(a\right)=0$
+\end_inset
+
+.
+ Nadalje velja 
+\begin_inset Formula $F\left(a\right)=F\left(b\right)=0$
+\end_inset
+
+,
+ ker smo pač tako definirali funkcijo 
+\begin_inset Formula $F$
+\end_inset
+
+,
+ zato obstaja po Rolleovem izreku tak 
+\begin_inset Formula $\alpha_{1}$
+\end_inset
+
+,
+ da velja 
+\begin_inset Formula $F'\left(\alpha_{1}\right)=0$
+\end_inset
+
+.
+ Po Rolleovem izreku nadalje obstaja tak 
+\begin_inset Formula $\alpha_{2}$
+\end_inset
+
+ med 
+\begin_inset Formula $a$
+\end_inset
+
+ in 
+\begin_inset Formula $\alpha_{1}$
+\end_inset
+
+,
+ da velja 
+\begin_inset Formula $F''\left(\alpha_{2}\right)=0$
+\end_inset
+
+.
+ Spet po Rolleovem izreku obstaja tak 
+\begin_inset Formula $\alpha_{3}$
+\end_inset
+
+ med 
+\begin_inset Formula $a$
+\end_inset
+
+ in 
+\begin_inset Formula $\alpha_{2}$
+\end_inset
+
+,
+ da velja 
+\begin_inset Formula $F'''\left(\alpha_{3}\right)=0$
+\end_inset
+
+.
+ Postopek lahko ponavljamo in dobimo tak 
+\begin_inset Formula $\alpha=\alpha_{n+1}$
+\end_inset
+
+,
+ da velja 
+\begin_inset Formula $F^{\left(n+1\right)}\left(\alpha\right)=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Ker je 
+\begin_inset Formula $\forall x\in I:T_{n}^{\left(n+1\right)}\left(x\right)=0$
+\end_inset
+
+ (očitno,
+ isti argument kot v 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{velja}{drugem odstavku dokaza}
+\end_layout
+
+\end_inset
+
+),
+ to pomeni 
+\begin_inset Formula $f^{\left(n+1\right)}\left(\alpha\right)=\left(K\left(x-a\right)^{n+1}\right)^{\left(n+1\right)}=K\left(n+1\right)!$
+\end_inset
+
+.
+ Torej je 
+\begin_inset Formula $K=\frac{f^{\left(n+1\right)}\left(\alpha\right)}{\left(n+1\right)!}$
+\end_inset
+
+ in zato 
+\begin_inset Formula $f\left(b\right)=T_{n}\left(b\right)+\frac{f^{\left(n+1\right)}\left(\alpha\right)}{\left(n+1\right)!}\left(b-a\right)^{\left(n+1\right)}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Corollary*
+Če je 
+\begin_inset Formula $\left(n+1\right)-$
+\end_inset
+
+ti odvod omejen na 
+\begin_inset Formula $I$
+\end_inset
+
+,
+ t.
+ j.
+ 
+\begin_inset Formula $\exists M>0\forall x\in I:\left|f^{\left(n+1\right)}\left(x\right)\right|\leq M$
+\end_inset
+
+,
+ lahko ostanek eksplicitno ocenimo,
+ in sicer 
+\begin_inset Formula $\left|R_{n}\left(x\right)\right|\leq\frac{M}{\left(n+1\right)!}\left|x-a\right|^{n+1}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Kaj pa se zgodi,
+ ko 
+\begin_inset Formula $n$
+\end_inset
+
+ pošljemo v neskončnost?
+ Iskali bi aproksimacije s 
+\begin_inset Quotes gld
+\end_inset
+
+polinomi neskončnega reda
+\begin_inset Quotes grd
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Če je 
+\begin_inset Formula $f\in C^{\infty}$
+\end_inset
+
+ v okolici točke 
+\begin_inset Formula $a\in\mathbb{R}$
+\end_inset
+
+.
+ Tedaj definiramo Taylorjevo vrsto 
+\begin_inset Formula $f$
+\end_inset
+
+ v okolici točke 
+\begin_inset Formula $a$
+\end_inset
+
+:
+ 
+\begin_inset Formula $T_{f,a}\left(x\right)\coloneqq\sum_{j=0}^{\infty}\frac{f^{\left(j\right)}\left(a\right)}{j!}\left(x-a\right)^{j}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Question*
+Ali Taylorjeva vrsta konvergira oziroma kje konvergira?
+ Kakšna je zveza s 
+\begin_inset Formula $f\left(x\right)$
+\end_inset
+
+?
+ Kakšen je 
+\begin_inset Formula $R_{f,a}$
+\end_inset
+
+?
+\end_layout
+
+\begin_layout Standard
+Oglejmo si potenčne vrste (
+\begin_inset Formula $\sum_{j=0}^{\infty}b_{k}x^{k}$
+\end_inset
+
+) kot poseben primer funkcijskih vrst (
+\begin_inset Formula $\sum_{j=0}^{\infty}a_{k}\left(x\right)$
+\end_inset
+
+).
+ Vemo,
+ da ima potenčna vrsta konvergenčni radij 
+\begin_inset Formula $R$
+\end_inset
+
+.
+ Za 
+\begin_inset Formula $x\in\left(-R,R\right)$
+\end_inset
+
+ konvergira,
+ za 
+\begin_inset Formula $x\in\left[-R,R\right]^{C}$
+\end_inset
+
+ divergira.
+\end_layout
+
+\begin_layout Theorem*
+Naj ima potenčna vrsta 
+\begin_inset Formula $f\left(x\right)=\sum_{k=0}^{\infty}b_{k}x^{k}$
+\end_inset
+
+ konvergenčni radij 
+\begin_inset Formula $R$
+\end_inset
+
+.
+ Tedaj ima tudi 
+\begin_inset Formula $g\left(x\right)=\sum_{k=1}kb_{k}x^{k-1}$
+\end_inset
+
+ konvergenčni radij 
+\begin_inset Formula $R$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+\begin_inset Formula 
+\[
+\frac{1}{R_{g}}=\limsup_{k\to\infty}\sqrt[k]{\left|ka_{k}\right|}=\limsup_{k\to\infty}\sqrt[k]{\left|k\right|\left|a_{k}\right|}=\limsup_{k\to\infty}\sqrt[k]{\left|k\right|}\sqrt[k]{\left|a_{k}\right|}=\cdots
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Proof
+\begin_inset Formula 
+\[
+\lim_{k\to\infty}\sqrt[k]{\left|k\right|}=\lim_{k\to\infty}k^{1/k}=e^{\lim_{k\to\infty}\frac{1}{k}\ln k}\overset{\text{L'H}}{=}e^{\lim_{k\to\infty}\frac{\frac{1}{k}}{k}}=e^{\lim_{k\to\infty}\cancelto{0}{\frac{1}{k^{2}}}}=e^{0}=1
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\cdots=\limsup_{k\to\infty}1\cdot\sqrt[k]{\left|a_{k}\right|}=\frac{1}{R_{f}}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Corollary*
+Če ima potenčna vrsta 
+\begin_inset Formula $f$
+\end_inset
+
+ konvergenčni radij 
+\begin_inset Formula $R>0$
+\end_inset
+
+,
+ tedaj je 
+\begin_inset Formula $f\in C^{\infty}\left(\left(-R,R\right)\right)$
+\end_inset
+
+ in velja 
+\begin_inset Formula $a_{k}=\frac{f^{\left(k\right)}\left(0\right)}{k!}$
+\end_inset
+
+,
+ potem velja 
+\begin_inset Formula $g=f'$
+\end_inset
+
+ (iz izreka zgoraj).
+ Razlaga:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $f\left(x\right)=\sum_{k=0}^{\infty}a_{k}x^{k}=\sum_{k=0}^{\infty}\frac{f^{\left(k\right)}\left(0\right)}{k!}x^{k}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $f'\left(x\right)=\sum_{k=1}^{\infty}ka_{k}x^{k-1}=\sum_{k=1}^{\infty}\frac{kf^{\left(k\right)}\left(0\right)}{k!}x^{k-1}=\sum_{k=1}^{\infty}\frac{f^{\left(k\right)}\left(0\right)}{\left(k-1\right)!}x^{k-1}$
+\end_inset
+
+ (
+\begin_inset Formula $k$
+\end_inset
+
+ začne z 
+\begin_inset Formula $1$
+\end_inset
+
+,
+ ker se 
+\begin_inset Formula $k=0$
+\end_inset
+
+ člen odvaja v konstanto 
+\begin_inset Formula $0$
+\end_inset
+
+)
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $g\left(x\right)=\sum_{k=1}^{\infty}ka_{k}x^{k-1}=\sum_{k=1}^{\infty}\frac{kf^{\left(k\right)}\left(0\right)}{k!}x^{k-1}=\sum_{k=1}^{\infty}\frac{f^{\left(k\right)}\left(0\right)}{\left(k-1\right)!}x^{k-1}=f'\left(x\right)$
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Definition*
+Funkcija 
+\begin_inset Formula $f:J\to\mathbb{R}$
+\end_inset
+
+ (
+\begin_inset Formula $J$
+\end_inset
+
+ je interval 
+\begin_inset Formula $\subseteq\mathbb{R}$
+\end_inset
+
+) je realno analitična,
+ če se jo da okoli vsake točke 
+\begin_inset Formula $c\in J$
+\end_inset
+
+ razviti v potenčno vrsto,
+ torej če 
+\begin_inset Formula $f\left(x\right)=\sum_{k=0}^{\infty}\frac{f^{\left(k\right)}\left(c\right)}{k!}\left(x-c\right)^{k}$
+\end_inset
+
+ za 
+\begin_inset Formula $x$
+\end_inset
+
+ blizu 
+\begin_inset Formula $c$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+\begin_inset Formula $f\in C^{\infty}\Rightarrow f$
+\end_inset
+
+ je realno analitična.
+ Protiprimer je 
+\begin_inset Formula $f\left(x\right)=e^{\frac{-1}{\left|x\right|}}$
+\end_inset
+
+.
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+TODO XXX FIXME ZAKAJ?,
+ ne razumem
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+Primeri Taylorjevih vrst.
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $f\left(x\right)=e^{x}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $n-$
+\end_inset
+
+ti tayorjev polinom za 
+\begin_inset Formula $f\left(x\right)$
+\end_inset
+
+ okoli 
+\begin_inset Formula $0$
+\end_inset
+
+:
+ 
+\begin_inset Formula $T_{n,e^{x},0}\left(x\right)=1+x+\frac{x^{2}}{2}+\frac{x^{3}}{6}+\cdots+\frac{x^{n}}{n!}$
+\end_inset
+
+ in velja 
+\begin_inset Formula $e^{x}=T_{n,e^{x},0}\left(x\right)+R_{n,e^{x},0}\left(x\right)$
+\end_inset
+
+,
+ kjer 
+\begin_inset Formula $\lim_{n\to\infty}R_{n,e^{x},0}\left(x\right)=0$
+\end_inset
+
+.
+ Ne bomo dokazali.
+ Sledi 
+\begin_inset Formula $e^{x}=\sum_{k=0}^{\infty}\frac{x^{k}}{k!}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\sin x=x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\cdots+\left(-1\right)^{k}\frac{x^{2k+1}}{\left(2k+1\right)!}$
+\end_inset
+
+.
+ Opazimo sode eksponente in opazimo učinek odvajanja:
+ 
+\begin_inset Formula $\cos,-\sin,-\cos,\sin,\cos,-\sin,\dots$
+\end_inset
+
+.
+ Členi vrste 
+\begin_inset Formula $\sin x$
+\end_inset
+
+ v 
+\begin_inset Formula $x=0$
+\end_inset
+
+ so:
+ 
+\begin_inset Formula $1,0,-1,0,1,0,-1,\dots$
+\end_inset
+
+.
+ Opazimo izpadanje vsakega drugega člena.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\cos x=1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-\cdots+\left(-1\right)^{k}\frac{x^{2k}}{\left(2k\right)!}$
+\end_inset
+
+.
+ Opazimo sode eksponente.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $f\left(x\right)=\log\left(1-x\right)$
+\end_inset
+
+.
+ A lahko to funkcijo razvijemo v taylorjevo vrsto okoli točke 0?
+\begin_inset Float table
+placement document
+alignment document
+wide false
+sideways false
+status open
+
+\begin_layout Plain Layout
+\align center
+\begin_inset Tabular
+<lyxtabular version="3" rows="7" columns="3">
+<features tabularvalignment="middle">
+<column alignment="center" valignment="top" width="0pt">
+<column alignment="center" valignment="top" width="0pt">
+<column alignment="center" valignment="top" width="0pt">
+<row>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $k$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $f^{\left(k\right)}\left(x\right)$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $f^{\left(k\right)}\left(0\right)$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $0$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $\log\left(1-x\right)$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $0$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $1$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $\frac{-1}{1-x}$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $-1$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $2$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $\frac{-1}{\left(1-x\right)^{2}}$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $-1$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $3$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $\frac{-2}{\left(1-x\right)^{3}}$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $-2$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $\cdots$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $\cdots$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $\cdots$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $n$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $\frac{-\left(n-1\right)!}{\left(1-x\right)^{n}}$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $-\left(n-1\right)!$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+</row>
+</lyxtabular>
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Plain Layout
+\begin_inset Caption Standard
+
+\begin_layout Plain Layout
+Razvijanje 
+\begin_inset Formula $\log\left(1-x\right)$
+\end_inset
+
+ okoli točke 
+\begin_inset Formula $0$
+\end_inset
+
+.
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Velja 
+\begin_inset Formula $f\left(x\right)=\log\left(1-x\right)=\sum_{k=1}^{\infty}\frac{-\left(k-1\right)!}{k!}x^{k}=-\sum_{k=1}^{\infty}\frac{x^{k}}{k}$
+\end_inset
+
+ za 
+\begin_inset Formula $\left|x\right|<1$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Section
+Integrali
+\end_layout
+
+\begin_layout Standard
+Radi bi definirali ploščino 
+\begin_inset Formula $P=\left\{ \left(x,t\right)\in\mathbb{R}^{2};x\in\left[a,b\right],t\in\left[0,f\left(x\right)\right]\right\} $
+\end_inset
+
+ za funkcijo 
+\begin_inset Formula $f:\left[a,b\right]\to[0,\infty)$
+\end_inset
+
+.
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+TODO XXX FIXME skica ANA1P FMF 2024-01-09/str.3
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $P$
+\end_inset
+
+ aproksimiramo s pravokotniki,
+ katerih ploščino smo predhodno definirali takole:
+\end_layout
+
+\begin_layout Definition*
+Ploščina pravokotnika s stranicama 
+\begin_inset Formula $c$
+\end_inset
+
+ in 
+\begin_inset Formula $d$
+\end_inset
+
+ je 
+\begin_inset Formula $c\cdot d$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Najprej diskusija.
+ Naj bo 
+\begin_inset Formula $t_{j}$
+\end_inset
+
+ delitev 
+\begin_inset Formula $\left[a,b\right]$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $a=t_{0}<t_{1}<\cdots<t_{n}=b$
+\end_inset
+
+.
+ Ne zahtevamo ekvidistančne delitve,
+ torej take,
+ pri kateri bi bile razdalje enake.
+ Kako naj definiramo višine pravokotnikov,
+ katerih stranice so delilne točke 
+\begin_inset Formula $t_{n}$
+\end_inset
+
+?
+\end_layout
+
+\begin_layout Standard
+Lahko tako,
+ da na vsakem intervalu 
+\begin_inset Formula $\left[t_{i},t_{i+1}\right]$
+\end_inset
+
+ izberemo nek 
+\begin_inset Formula $\xi_{i}$
+\end_inset
+
+,
+ pravokotnicova osnovnica bode 
+\begin_inset Formula $t_{i+1}-t_{i}$
+\end_inset
+
+,
+ njegova višina pa 
+\begin_inset Formula $f\left(\xi_{i}\right)$
+\end_inset
+
+.
+ Ploščina 
+\begin_inset Formula $P$
+\end_inset
+
+ pod grafom funkcije je približno enaka vsoti ploščin teh pravokotnikov,
+ torej 
+\begin_inset Formula $\sum_{k=1}^{n}f\left(\xi_{k}\right)\left(t_{k}-t_{k-1}\right)=R\left(f,\vec{t},\vec{\xi}\right)$
+\end_inset
+
+,
+ kjer je 
+\begin_inset Formula $\vec{t}$
+\end_inset
+
+ delitev in 
+\begin_inset Formula $\vec{\xi}$
+\end_inset
+
+ izbira točk na intervalih delitve.
+ Temu pravimo Riemannova vsota za 
+\begin_inset Formula $f$
+\end_inset
+
+,
+ ki pripada delitvi 
+\begin_inset Formula $\vec{t}$
+\end_inset
+
+ in izboru 
+\begin_inset Formula $\vec{\xi}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Če je 
+\begin_inset Formula $D\coloneqq\left\{ \left[t_{j+1},t_{j}\right];j=\left\{ 1..n\right\} \right\} $
+\end_inset
+
+ delitev za 
+\begin_inset Formula $\left[a,b\right]$
+\end_inset
+
+,
+ definiramo tako oznako 
+\begin_inset Formula $\left|D\right|_{\infty}\coloneqq\max_{j=\left\{ 1..n\right\} }\left(t_{j}-t_{j-1}\right)=\max_{I\in D}\left(\left|I\right|\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Če 
+\begin_inset Formula $\exists A\in\mathbb{R}\ni:$
+\end_inset
+
+ za poljubno fine delitve (
+\begin_inset Formula $\left|D\right|_{\infty}=\infty^{-1}$
+\end_inset
+
+) 
+\begin_inset Formula $D$
+\end_inset
+
+ se pripadajoče Riemannove vsote malo razlikujejo od 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ pravimo številu 
+\begin_inset Formula $A$
+\end_inset
+
+ ploščina lika 
+\begin_inset Formula $P$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Sedaj pa še formalna definicija.
+\end_layout
+
+\begin_layout Definition*
+Naj bodo 
+\begin_inset Formula $f,D,\xi$
+\end_inset
+
+ kot prej in 
+\begin_inset Formula $I\in\mathbb{R}$
+\end_inset
+
+ realno število.
+ Če 
+\begin_inset Formula $\forall\varepsilon>0\exists\delta>0\ni:$
+\end_inset
+
+
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $\forall$
+\end_inset
+
+ delitev 
+\begin_inset Formula $D\ni:\left|D\right|_{\infty}<\delta$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $\forall$
+\end_inset
+
+ nabor 
+\begin_inset Formula $\xi=\xi_{1},\dots,\xi_{n}$
+\end_inset
+
+,
+ pripadajoč delitvi 
+\begin_inset Formula $D$
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Definition*
+velja 
+\begin_inset Formula $\left|R\left(f,D,\xi\right)-I\right|<\varepsilon\Longrightarrow I$
+\end_inset
+
+ je določen integral 
+\begin_inset Formula $f$
+\end_inset
+
+ na intervalu 
+\begin_inset Formula $\left[a,b\right]$
+\end_inset
+
+ in je po definiciji ploščina lika 
+\begin_inset Formula $P$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Če tak 
+\begin_inset Formula $I$
+\end_inset
+
+ obstaja,
+ kar ni 
+\emph on
+a priori
+\emph default
+,
+ pravimo,
+ da je 
+\begin_inset Formula $f$
+\end_inset
+
+ integrabilna na 
+\begin_inset Formula $\left[a,b\right]$
+\end_inset
+
+ in pišemo 
+\begin_inset Formula $I=\int_{a}^{b}f\left(x\right)dx$
+\end_inset
+
+.
+ Temu pravimo Riemannov integral funkcije 
+\begin_inset Formula $f$
+\end_inset
+
+ na 
+\begin_inset Formula $\left[a,b\right]$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Darbouxove vsote.
+ Imamo torej delitev 
+\begin_inset Formula $D=\left\{ \left[t_{j-1},t_{j}\right];j\in\left\{ 1..n\right\} ;t_{0}=1,t_{n}=b\right\} $
+\end_inset
+
+ delitev za 
+\begin_inset Formula $J=\left[a,b\right]$
+\end_inset
+
+ in 
+\begin_inset Formula $f:J\to\mathbb{R}$
+\end_inset
+
+.
+ Imamo tudi množico izbranih točk 
+\begin_inset Formula $\xi=\left\{ \xi_{j}\in\left[t_{j-1},t_{j}\right];j\in\left\{ 1..n\right\} \right\} $
+\end_inset
+
+ in 
+\begin_inset Formula $R\left(f,D,\xi\right)=\sum_{j=1}^{n}f\left(\xi_{j}\right)\left(t_{j}-t_{j-1}\right)$
+\end_inset
+
+.
+ Ocenimo 
+\begin_inset Formula $f\left(\xi_{j}\right)$
+\end_inset
+
+:
+ 
+\begin_inset Formula $\inf_{x\in\left[t_{j-1},t_{j}\right]}f\left(x\right)\leq f\left(\xi_{j}\right)\leq\sup_{x\in\left[t_{j-1},t_{j}\right]}f\left(x\right)$
+\end_inset
+
+.
+ Definirali smo 
+\begin_inset Formula $\int_{a}^{b}f\left(x\right)dx$
+\end_inset
+
+ kot limito Riemannovih vsot s kakršnokoli delitvijo in izbiro 
+\begin_inset Formula $\xi$
+\end_inset
+
+,
+ zato lahko pišemo 
+\begin_inset Formula $\forall j\in\left\{ 1..n\right\} :\inf_{x\in\left[t_{j-1},t_{j}\right]}f\left(x\right)=f\left(\xi_{j}\right)=\sup_{x\in\left[t_{j-1},t_{j}\right]}f\left(x\right)$
+\end_inset
+
+.
+ Zato lahko limito Riemannovih vsot obravnavamo neodvisno od 
+\begin_inset Formula $\xi$
+\end_inset
+
+:
+\begin_inset Formula 
+\[
+s\left(f,D\right)\coloneqq\sum_{j=1}^{n}\left(\inf_{x\in D_{j}}f\left(x\right)\right)\left(t_{j}-t_{j-1}\right)\leq R\left(f,D,\xi\right)\leq\sum_{j=1}^{n}\left(\sup_{x\in D_{j}}f\left(x\right)\right)\left(t_{j}-f_{j-1}\right)\eqqcolon S\left(f,D\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Definirali smo dva nova pojma,
+ spodnjo Darbouxovo vsoto 
+\begin_inset Formula $s\left(f,D\right)$
+\end_inset
+
+ in zgornjo Darbouxovo vsoto 
+\begin_inset Formula $S\left(f,D\right)$
+\end_inset
+
+ in velja 
+\begin_inset Formula $s\left(f,D\right)\leq R\left(f,D,\xi\right)\leq S\left(f,D\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Naj bosta 
+\begin_inset Formula $D$
+\end_inset
+
+ in 
+\begin_inset Formula $D'$
+\end_inset
+
+ delitvi za interval 
+\begin_inset Formula $J$
+\end_inset
+
+.
+ Pravimo,
+ da je 
+\begin_inset Formula $D'$
+\end_inset
+
+ finejša od 
+\begin_inset Formula $D$
+\end_inset
+
+,
+ če je ima 
+\begin_inset Formula $D'$
+\end_inset
+
+ vse delilne točke,
+ ki jih ima 
+\begin_inset Formula $D$
+\end_inset
+
+ in poleg njih še vsaj kakšno.
+ Označimo 
+\begin_inset Formula $D\subset D'$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+Naj bo 
+\begin_inset Formula $D\subset D'$
+\end_inset
+
+ (
+\begin_inset Formula $D'$
+\end_inset
+
+ finejša od 
+\begin_inset Formula $D$
+\end_inset
+
+).
+ Oglejmo si 
+\begin_inset Formula $s\left(f,D\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $s\left(f,D'\right)$
+\end_inset
+
+.
+ Tedaj velja 
+\begin_inset Formula $s\left(f,D\right)\leq s\left(f,D'\right)$
+\end_inset
+
+,
+ ker je infimum po manjši množici lahko le večji —
+ s finejšo delitvijo smo vsaj neko množico (delitveni interval) razdelili na dva dela.
+ Za zgornjo Darbouxovo vsoto velja obratno,
+ torej 
+\begin_inset Formula $S\left(f,D\right)\geq S\left(f,D'\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Theorem*
+Za poljubni različni delitvi 
+\begin_inset Formula $D_{1},D_{2}$
+\end_inset
+
+ intervala 
+\begin_inset Formula $J$
+\end_inset
+
+ velja 
+\begin_inset Formula $s\left(f,D_{1}\right)\leq S\left(f,D_{2}\right)$
+\end_inset
+
+ ZDB Katerakoli spodnja Darbouxova vsota je kvečjemu tolikšna kot katerakoli zgornja.
+\end_layout
+
+\begin_layout Proof
+Označimo z 
+\begin_inset Formula $D_{1}\cup D_{2}$
+\end_inset
+
+ delitev,
+ ki vsebuje vse delilne točke tako 
+\begin_inset Formula $D_{1}$
+\end_inset
+
+ kot tudi 
+\begin_inset Formula $D_{2}$
+\end_inset
+
+.
+ Očitno velja,
+ da sta 
+\begin_inset Formula $D_{1}\subset D_{1}\cup D_{2}$
+\end_inset
+
+ in 
+\begin_inset Formula $D_{2}\subset D_{1}\cup D_{2}$
+\end_inset
+
+.
+ Po prejšnjem izreku veljata leva in desna neenakost,
+ srednja pa iz definicije (očitno).
+\begin_inset Formula 
+\[
+s\left(f,D_{1}\right)\leq s\left(f,D_{1}\cup D_{2}\right)\leq S\left(f,D_{1}\cup D_{2}\right)\leq S\left(f,D_{2}\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $f:J\to\mathbb{R}$
+\end_inset
+
+ omejena.
+ Označimo 
+\begin_inset Formula $s\left(f\right)\coloneqq\sup_{\text{vse možne delitve }D}s\left(f,D\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $S\left(f\right)\coloneqq\inf_{\text{vse možne delitve }D}S\left(f,D\right)$
+\end_inset
+
+.
+ Funkcija 
+\begin_inset Formula $f:J\to\mathbb{R}$
+\end_inset
+
+ je Riemannovo,
+ če 
+\begin_inset Formula $s\left(f\right)=S\left(f\right)$
+\end_inset
+
+ oziroma če 
+\begin_inset Formula $\forall\varepsilon>0\exists$
+\end_inset
+
+ delitev 
+\begin_inset Formula $D$
+\end_inset
+
+ na 
+\begin_inset Formula $J\ni:S\left(f,D\right)-s\left(f,D\right)<\varepsilon$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Note*
+Integrabilnost 
+\begin_inset Formula $f$
+\end_inset
+
+ ne pomeni,
+ da 
+\begin_inset Formula $\exists D\ni:s\left(f,D\right)=S\left(f,D\right)$
+\end_inset
+
+.
+ Ni namreč nujno,
+ da množica vsebuje svoj supremum.
+ Primer:
+ za 
+\begin_inset Formula $f\left(x\right)=x$
+\end_inset
+
+ velja 
+\begin_inset Formula $\forall D:S\left(f,D\right)>s\left(f,D\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+Vsaka zvezna funkcija je integrabilna na 
+\begin_inset Formula $J$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+ poljuben.
+ Po definiciji 
+\begin_inset Formula $S\left(f,D\right)-s\left(f,D\right)=\sum_{j=1}^{n}\left(\sup_{x\in D_{j}}f\left(x\right)-\inf_{x\in D_{j}}f\left(x\right)\right)\left(t_{j}-t_{j-1}\right)$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna,
+ je na zaprtem 
+\begin_inset Formula $J=\left[a,b\right]$
+\end_inset
+
+ enakomerno zvezna,
+ torej 
+\begin_inset Formula $\exists\delta>0\forall x_{1},x_{2}\in J:\left|x_{1}-x_{2}\right|<\delta\Rightarrow\left|f\left(x_{1}\right)-f\left(x_{2}\right)\right|<\frac{\varepsilon}{b-a}$
+\end_inset
+
+.
+ Izberimo tako delitev 
+\begin_inset Formula $D$
+\end_inset
+
+,
+ da je 
+\begin_inset Formula $\forall j\in\left\{ 1..\left|D\right|\right\} :t_{j}-t_{j-1}<\delta$
+\end_inset
+
+.
+ Tedaj bo veljalo 
+\begin_inset Formula $\sum_{j=1}^{n}\left(\sup_{x\in D_{j}}f\left(x\right)-\inf_{x\in D_{j}}f\left(x\right)\right)\left(t_{j}-t_{j-1}\right)<\sum_{j=1}^{n}\frac{\varepsilon}{b-a}\left(t_{j}-t_{j-1}\right)=\frac{\varepsilon\left(b-a\right)}{b-a}=\varepsilon$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Skratka dokazali smo 
+\begin_inset Formula $S\left(f,D\right)-s\left(f,D\right)<\varepsilon$
+\end_inset
+
+ za poljuben 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+,
+ torej je funkcija Riemannovo integrabilna po zgornji definiciji.
+\end_layout
+
+\begin_layout Definition*
+\begin_inset Formula $A\subset\mathbb{R}$
+\end_inset
+
+ ima mero 
+\begin_inset Formula $0$
+\end_inset
+
+,
+ če 
+\begin_inset Formula $\forall\varepsilon>0\exists$
+\end_inset
+
+ družina intervalov 
+\begin_inset Formula $I_{j}\ni:A\subset\bigcup I_{j}\wedge\sum\left|I_{j}\right|<\varepsilon$
+\end_inset
+
+.
+ Primer:
+ vse števne in končne množice.
+\end_layout
+
+\begin_layout Theorem*
+Funkcija 
+\begin_inset Formula $f$
+\end_inset
+
+ je integrabilna na intervalu 
+\begin_inset Formula $J\Leftrightarrow\left\{ x\in J;f\text{ ni zvezna v }x\right\} $
+\end_inset
+
+ ima mero 
+\begin_inset Formula $0$
+\end_inset
+
+.
+ ZDB če ima množica točk z definicijskega območja 
+\begin_inset Formula $f$
+\end_inset
+
+,
+ v katerih 
+\begin_inset Formula $f$
+\end_inset
+
+ ni zvezna,
+ mero 
+\begin_inset Formula $0$
+\end_inset
+
+ (recimo če je teh točk končno mnogo),
+ je 
+\begin_inset Formula $f$
+\end_inset
+
+ integrabilna.
+\end_layout
+
+\begin_layout Fact*
+Označimo z 
+\begin_inset Formula $I\left(J\right)$
+\end_inset
+
+ množico vseh integrabilnih funkcij na intervalu 
+\begin_inset Formula $J$
+\end_inset
+
+.
+ 
+\begin_inset Formula $I\left(J\right)$
+\end_inset
+
+ je vektorski prostor za množenje s skalarji iz 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+.
+ Naj bodo 
+\begin_inset Formula $f,g\in I\left(J\right),\lambda\in\mathbb{R}$
+\end_inset
+
+.
+ Velja aditivnost 
+\begin_inset Formula $f\left(x\right)+g\left(x\right)\in J\left(I\right)$
+\end_inset
+
+,
+ kajti 
+\begin_inset Formula $\int_{a}^{b}\left(f\left(x\right)+g\left(x\right)\right)dx=\int_{a}^{b}\left(f\left(x\right)\right)dx+\int_{a}^{b}\left(g\left(x\right)\right)dx$
+\end_inset
+
+ in homogenost 
+\begin_inset Formula $\int_{a}^{b}\lambda f\left(x\right)dx=\lambda\int_{a}^{b}f\left(x\right)dx$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+Če je 
+\begin_inset Formula $f$
+\end_inset
+
+ integrabilna na 
+\begin_inset Formula $J=\left[a,b\right]$
+\end_inset
+
+ in je 
+\begin_inset Formula $c\in J$
+\end_inset
+
+,
+ tedaj je 
+\begin_inset Formula $f$
+\end_inset
+
+ integrabilna na 
+\begin_inset Formula $\left[a,c\right]$
+\end_inset
+
+ in 
+\begin_inset Formula $\left[c,b\right]$
+\end_inset
+
+ in velja 
+\begin_inset Formula $\int_{a}^{b}f\left(x\right)dx=\int_{a}^{c}f\left(x\right)dx+\int_{c}^{b}f\left(x\right)dx$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Theorem*
+Če sta 
+\begin_inset Formula $f,g$
+\end_inset
+
+ na 
+\begin_inset Formula $J$
+\end_inset
+
+ integrabilni funkciji in če je 
+\begin_inset Formula $\forall x\in J:f\left(x\right)\leq g\left(x\right)$
+\end_inset
+
+,
+ tedaj 
+\begin_inset Formula $\int_{a}^{b}f\left(x\right)dx\leq\int_{a}^{b}f\left(x\right)dx$
+\end_inset
+
+.
+ Posledično velja ob isti predpostavki 
+\begin_inset Formula $\left|\int_{a}^{b}f\left(x\right)dx\right|\leq\int_{a}^{b}\left|f\left(x\right)\right|dx$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Če je 
+\begin_inset Formula $f$
+\end_inset
+
+ integrabilna na 
+\begin_inset Formula $J=\left[a,b\right]$
+\end_inset
+
+,
+ definiramo povprečje 
+\begin_inset Formula $f$
+\end_inset
+
+ na 
+\begin_inset Formula $J$
+\end_inset
+
+ s predpisom 
+\begin_inset Formula 
+\[
+\left\langle f\right\rangle _{J}\coloneqq\frac{\int_{a}^{b}f\left(x\right)dx}{b-a}\in\mathbb{R}.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Theorem*
+Velja 
+\begin_inset Formula $\inf_{x\in J}f\left(x\right)\leq\left\langle f\right\rangle _{J}\leq\sup_{x\in J}f\left(x\right)$
+\end_inset
+
+.
+ Če je 
+\begin_inset Formula $f:J\to\mathbb{R}$
+\end_inset
+
+ zvezna,
+ 
+\begin_inset Formula $\exists\xi\in J\ni:f\left(\xi\right)=\left\langle f\right\rangle _{J}$
+\end_inset
+
+ (izrek o vmesni vrednosti).
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $f:J\to\mathbb{R}$
+\end_inset
+
+ dana funkcija.
+ Nedoločeni integral 
+\begin_inset Formula $f$
+\end_inset
+
+ je takšna funkcija 
+\begin_inset Formula $F$
+\end_inset
+
+,
+ če obstaja,
+ 
+\begin_inset Formula $\ni:F'=f\sim\forall x\in J:F'\left(x\right)=f\left(x\right)$
+\end_inset
+
+.
+ Pišemo tudi 
+\begin_inset Formula $Pf$
+\end_inset
+
+ ali 
+\begin_inset Formula $\mathbb{P}f$
+\end_inset
+
+ in pravimo,
+ da je 
+\begin_inset Formula $F=Pf$
+\end_inset
+
+ primitivna funkcija za 
+\begin_inset Formula $f$
+\end_inset
+
+.
+ Velja 
+\begin_inset Formula $P\left(f+g\right)=Pf+Pg$
+\end_inset
+
+ (aditivnost odvoda) in 
+\begin_inset Formula $P\left(\lambda f\right)=\lambda Pf$
+\end_inset
+
+ (homogenost odvoda).
+\end_layout
+
+\begin_layout Definition*
+Nedoločeni integral je na intervalu določen do aditivne konstante natančno.
+ Če je 
+\begin_inset Formula $F'_{1}=f=F_{2}'$
+\end_inset
+
+ na intervalu 
+\begin_inset Formula $J$
+\end_inset
+
+ oziroma če na 
+\begin_inset Formula $J$
+\end_inset
+
+ velja 
+\begin_inset Formula $\left(F_{1}-F_{2}\right)'=0$
+\end_inset
+
+,
+ potem 
+\begin_inset Formula $F_{1}-F_{2}=c$
+\end_inset
+
+ oziroma 
+\begin_inset Formula $F_{1}=F_{2}+c$
+\end_inset
+
+ za neko konstanto 
+\begin_inset Formula $c\in\mathbb{R}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Označimo 
+\begin_inset Formula $F\left(x\right)=Pf\left(x\right)=\int f\left(x\right)dx$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+Integracija po delih 
+\begin_inset Formula $\sim$
+\end_inset
+
+ per partes.
+ Velja 
+\begin_inset Formula $\int f\left(x\right)g'\left(x\right)dx=f\left(x\right)g\left(x\right)-\int f'\left(x\right)g\left(x\right)dx$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Izhaja iz odvoda produkza 
+\begin_inset Formula $\left(fg\right)'=f'g+fg'$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+Naj bo 
+\begin_inset Formula $f$
+\end_inset
+
+ integrabilna na 
+\begin_inset Formula $J$
+\end_inset
+
+.
+ Definirajmo 
+\begin_inset Formula $F\left(x\right)=\int_{a}^{x}f\left(t\right)dt$
+\end_inset
+
+.
+ Velja 
+\begin_inset Formula $\left|F\left(x_{1}\right)-F\left(x_{2}\right)\right|=$
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\left|\int_{a}^{x_{1}}f\left(t\right)dt-\int_{a}^{x_{2}}f\left(t\right)dt\right|=\left|\int_{a}^{x_{1}}f\left(t\right)dt+\int_{x_{2}}^{a}f\left(t\right)dt\right|=\left|\int_{x_{2}}^{x_{1}}f\left(t\right)dt\right|=\left|\int_{x_{1}}^{x_{2}}f\left(t\right)dt\right|\leq\int_{x_{1}}^{x_{2}}f\left(t\right)dt
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Theorem*
+Osnovni izrek analize/fundamental theorem of calcusus.
+ Naj bo 
+\begin_inset Formula $f:\left[a,b\right]\to\mathbb{R}$
+\end_inset
+
+ zvezna in 
+\begin_inset Formula $F\left(x\right)=\int_{a}^{x}f\left(t\right)dt$
+\end_inset
+
+.
+ Tedaj je 
+\begin_inset Formula $F$
+\end_inset
+
+ odvedljiva na 
+\begin_inset Formula $J$
+\end_inset
+
+ in velja 
+\begin_inset Formula $F'\left(x\right)=f\left(x\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+\begin_inset Formula 
+\[
+F\left(x+h\right)-F\left(x\right)=\int_{x}^{x+h}f\left(t\right)dt\quad\quad\quad\quad/:h
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\frac{F\left(x+h\right)-F\left(x\right)}{h}=\frac{\int_{x}^{x+h}f\left(t\right)dt}{h}=\left\langle f\right\rangle _{\left[x,x+h\right]}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+F'\left(x\right)=\lim_{h\to0}\left\langle f\right\rangle _{x,x+h}=f\left(x\right).
+\]
+
+\end_inset
+
+
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+glej ANA1P FMF 2024-01-15.pdf/str.
+ 5 za dokaz,
+ ki ga ne razumem,
+ zakaj je 
+\begin_inset Formula $\lim_{h\to0}\left\langle f\right\rangle _{\left[x,x+h\right]}-f\left(x\right)=0$
+\end_inset
+
+...
+ ampak sej to je nekak očitno
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Corollary*
+Naj bo 
+\begin_inset Formula $f:\left[a,b\right]\to\mathbb{R}$
+\end_inset
+
+ zvezna in 
+\begin_inset Formula $G=Pf$
+\end_inset
+
+ (
+\begin_inset Formula $G'=f$
+\end_inset
+
+).
+ Tedaj je 
+\begin_inset Formula $\int_{a}^{b}f\left(x\right)dx=G\left(b\right)-G\left(a\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $F\left(x\right)=\int_{a}^{x}f\left(t\right)dt$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $F'=f=G'$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $\left(F-G\right)'=0\Rightarrow F-G=c\in\mathbb{R}$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $G\left(x\right)=F\left(x\right)+c$
+\end_inset
+
+,
+ sledi 
+\begin_inset Formula $G\left(a\right)=F\left(a\right)=0$
+\end_inset
+
+ po definiciji 
+\begin_inset Formula $F$
+\end_inset
+
+,
+ torej je 
+\begin_inset Formula $G\left(a\right)=c$
+\end_inset
+
+.
+ Sledi 
+\begin_inset Formula $F\left(x\right)=G\left(x\right)-G\left(a\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $F\left(b\right)=G\left(b\right)-G\left(a\right)$
+\end_inset
+
+ in zato 
+\begin_inset Formula $F\left(b\right)=\int_{a}^{b}f\left(t\right)dt$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsection
+Iskanje primitivne funkcije
+\end_layout
+
+\begin_layout Itemize
+Uganemo jo
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $P\left(x^{n}\right)=\frac{x^{n+1}}{n+1}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $P\left(e^{x}\right)=e^{x}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $P\left(\sin x\right)=-\cos x$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $P\left(\ln x\right)=x\left(\ln x-1\right)$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Theorem*
+Substitucija/uvedba nove spremenljivke
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+ne razumem.
+ mogoče bom v naslednjem življenju.
+\end_layout
+
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $F\left(x\right)$
+\end_inset
+
+ nedoločeni integral funkcije 
+\begin_inset Formula $f\left(x\right)$
+\end_inset
+
+ ter 
+\begin_inset Formula $\phi\left(x\right)$
+\end_inset
+
+ odvedljiva funkcija.
+ Potem velja
+\begin_inset Formula 
+\[
+F\left(\phi\left(t\right)\right)=\int f\left(\phi\left(t\right)\right)\phi'\left(t\right)dx
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Proof
+Formula je posledica odvoda kompozituma:
+\begin_inset Formula 
+\[
+\left(F\left(\phi\left(t\right)\right)\right)'=F'\left(\phi\left(t\right)\right)\phi'\left(t\right)=f\left(\phi\left(t\right)\right)\phi'\left(t\right)
+\]
+
+\end_inset
+
+integrirajmo levo in desno stran:
+\begin_inset Formula 
+\[
+\int\left(F\left(\phi\left(t\right)\right)\right)'dt=\int f\left(\phi\left(t\right)\right)\phi'\left(t\right)dt.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsection
+Izlimitirani integrali
+\end_layout
+
+\begin_layout Standard
+Doslej smo računali določene integrale omejene funkcije na omejenem intervalu,
+ torej 
+\begin_inset Formula $\int_{a}^{b}f\left(x\right)dx$
+\end_inset
+
+.
+ Kaj pa neomejen interval,
+ torej 
+\begin_inset Formula $\lim_{b\to\infty}\int_{a}^{b}f\left(x\right)dx$
+\end_inset
+
+?
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $f:[a,\infty)\to\mathbb{R}$
+\end_inset
+
+ in naj bo 
+\begin_inset Formula $\forall m>a:f$
+\end_inset
+
+ integrabilna na 
+\begin_inset Formula $\left[a,-m\right]$
+\end_inset
+
+.
+ Če 
+\begin_inset Formula $\exists\lim_{m\to\infty}\int_{a}^{m}f\left(x\right)dx$
+\end_inset
+
+,
+ pracimo,
+ da integral 
+\begin_inset Formula $\int_{a}^{\infty}f\left(x\right)dx$
+\end_inset
+
+ konvergira,
+ sicer pa divergira.
+ Označimo 
+\begin_inset Formula $\int_{a}^{\infty}f\left(x\right)dx\coloneqq\lim_{m\to\infty}\int_{a}^{m}f\left(x\right)dx$
+\end_inset
+
+.
+ Podobno definiramo 
+\begin_inset Formula $\int_{-\infty}^{a}f\left(x\right)dx$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Pomemben primer.
+ 
+\begin_inset Formula $\int_{1}^{\infty}x^{\alpha}dx=?$
+\end_inset
+
+.
+ 
+\begin_inset Formula $\int_{1}^{M}x^{\alpha}dx=\frac{M^{\alpha+1}}{\alpha+1}-\frac{1}{\alpha+1}=\frac{M^{\alpha+1}-1}{\alpha+1}$
+\end_inset
+
+.
+ Torej 
+\begin_inset Formula $\exists\lim_{M\to\infty}\int_{1}^{M}x^{\alpha}dx\Leftrightarrow\alpha\not=-1$
+\end_inset
+
+.
+ Poglejmo,
+ kaj se zgodi v 
+\begin_inset Formula $\alpha=-1$
+\end_inset
+
+:
+ 
+\begin_inset Formula $\int_{1}^{\infty}x^{-1}dx=\ln M-\ln1=\ln M$
+\end_inset
+
+.
+ Toda 
+\begin_inset Formula $\lim_{n\to\infty}\ln M=\infty$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\int_{1}^{\infty}x^{-1}dx$
+\end_inset
+
+ divergira.
+\end_layout
+
+\begin_layout Definition*
+\begin_inset Formula $\int_{a}^{\infty}f\left(x\right)dx$
+\end_inset
+
+ je absolutno konvergenten,
+ če je 
+\begin_inset Formula $\int_{a}^{\infty}\left|f\left(x\right)\right|dx<\infty$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Fact*
+Velja 
+\begin_inset Formula $\int_{a}^{\infty}\left|f\left(x\right)\right|dx<0\Rightarrow\int_{a}^{\infty}f\left(x\right)dx<\infty$
+\end_inset
+
+.
+ Velja 
+\begin_inset Formula $\left|\int_{a}^{\infty}f\left(x\right)dx\right|\leq\int_{a}^{\infty}\left|f\left(x\right)\right|dx$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Ali je predpostavka,
+ da je 
+\begin_inset Formula $f$
+\end_inset
+
+ omejena,
+ sploh potrebna?
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $f:[a,b)\to\mathbb{R}\ni:\forall c<b:f$
+\end_inset
+
+ integrabilna na 
+\begin_inset Formula $\left[a,c\right]$
+\end_inset
+
+.
+ V točki 
+\begin_inset Formula $b$
+\end_inset
+
+ je 
+\begin_inset Formula $f$
+\end_inset
+
+ lahko neomejena.
+ Če 
+\begin_inset Formula $\exists$
+\end_inset
+
+ končna limita 
+\begin_inset Formula $\lim_{c\to b}\int_{a}^{c}f\left(x\right)dx$
+\end_inset
+
+,
+ je integral 
+\begin_inset Formula $\int_{a}^{b}f\left(x\right)dx$
+\end_inset
+
+ konvergenten,
+ sicer je divergenten.
+ Podobno definiramo,
+ če je funkcija definirana na intervalu 
+\begin_inset Formula $(a,b]$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $\int_{0}^{1}x^{\alpha}dx$
+\end_inset
+
+.
+ Za 
+\begin_inset Formula $\alpha<0$
+\end_inset
+
+ ima graf 
+\begin_inset Formula $x^{\alpha}$
+\end_inset
+
+ v 
+\begin_inset Formula $x=0$
+\end_inset
+
+ pol.
+ Računajmo
+\begin_inset Formula 
+\[
+\lim_{\varepsilon\to0}\int_{\varepsilon}^{1}x^{\alpha}dx=\lim_{\varepsilon\to0}\frac{x^{\alpha+1}}{\alpha+1}\vert_{\varepsilon}^{1}=\lim_{\varepsilon\to0}\left(\frac{1}{\alpha+1}-\frac{\varepsilon^{\alpha+1}}{\alpha+1}\right)=\lim_{\varepsilon\to0}\frac{1-\varepsilon^{\alpha+1}}{\alpha+1}=\lim_{\varepsilon\to0}\frac{1-\cancelto{0}{e^{\left(\alpha+1\right)\ln\varepsilon}}}{\alpha+1}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+Pridobimo pogoj 
+\begin_inset Formula $\alpha\not=-1$
+\end_inset
+
+ (imenovalec) in 
+\begin_inset Formula $\alpha+1>0$
+\end_inset
+
+ (da bo 
+\begin_inset Formula $\left(\alpha+1\right)\ln\varepsilon\to-\infty$
+\end_inset
+
+),
+ torej skupaj s predpostavko 
+\begin_inset Formula $\alpha\in\left(-1,0\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Torej 
+\begin_inset Formula $\int_{0}^{1}x^{\alpha}dx=\frac{1}{\alpha+1}$
+\end_inset
+
+ za 
+\begin_inset Formula $\alpha\in\left(-1,0\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsection
+Uporaba integrala
+\end_layout
+
+\begin_layout Itemize
+Ploščine:
+ 
+\begin_inset Formula $f\geq0$
+\end_inset
+
+ na 
+\begin_inset Formula $J=\left[a,b\right]$
+\end_inset
+
+ in je 
+\begin_inset Formula $f\in I\left(J\right)$
+\end_inset
+
+,
+ je ploščina lika med 
+\begin_inset Formula $x$
+\end_inset
+
+ osjo in grafom 
+\begin_inset Formula $f$
+\end_inset
+
+ definirana kot 
+\begin_inset Formula $\int_{a}^{b}f\left(x\right)dx$
+\end_inset
+
+.
+ Če 
+\begin_inset Formula $f$
+\end_inset
+
+ ni pozitivna,
+ pa je 
+\begin_inset Formula $\int_{a}^{b}f\left(x\right)dx=pl\left(L_{1}\right)-pl\left(L_{2}\right)$
+\end_inset
+
+,
+ kjer je 
+\begin_inset Formula $L_{1}$
+\end_inset
+
+ lik nad 
+\begin_inset Formula $x$
+\end_inset
+
+ osjo in 
+\begin_inset Formula $L_{2}$
+\end_inset
+
+ lik pod 
+\begin_inset Formula $x$
+\end_inset
+
+ osjo.
+\end_layout
+
+\begin_layout Example*
+Ploščina kroga:
+ Enačba krožnice je 
+\begin_inset Formula $x^{2}+y^{2}=r^{2}$
+\end_inset
+
+ za 
+\begin_inset Formula $r>0$
+\end_inset
+
+.
+ 
+\begin_inset Formula $y=\sqrt{r^{2}-x^{2}}$
+\end_inset
+
+.
+ Ploščina kroga z radijem 
+\begin_inset Formula $r$
+\end_inset
+
+ je torej 
+\begin_inset Formula $2\int_{-r}^{r}\sqrt{r^{2}-x^{2}}dx=\cdots=\pi r^{2}$
+\end_inset
+
+.
+\end_layout
+
+\end_body
+\end_document