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diff --git a/šola/ana1/teor.lyx b/šola/ana1/teor.lyx new file mode 100644 index 0000000..f716983 --- /dev/null +++ b/šola/ana1/teor.lyx @@ -0,0 +1,16315 @@ +#LyX 2.4 created this file. For more info see https://www.lyx.org/ +\lyxformat 620 +\begin_document +\begin_header +\save_transient_properties true +\origin unavailable +\textclass article +\begin_preamble +\usepackage{hyperref} +\usepackage{siunitx} +\usepackage{pgfplots} +\usepackage{listings} +\usepackage{multicol} +\sisetup{output-decimal-marker = {,}, quotient-mode=fraction, output-exponent-marker=\ensuremath{\mathrm{3}}} +\usepackage{amsmath} +\usepackage{tikz} +\newcommand{\udensdash}[1]{% + \tikz[baseline=(todotted.base)]{ + \node[inner sep=1pt,outer sep=0pt] (todotted) {#1}; + \draw[densely dashed] (todotted.south west) -- (todotted.south east); + }% +}% +\DeclareMathOperator{\Lin}{\mathcal Lin} +\DeclareMathOperator{\rang}{rang} +\DeclareMathOperator{\sled}{sled} +\DeclareMathOperator{\Aut}{Aut} +\DeclareMathOperator{\red}{red} +\DeclareMathOperator{\karakteristika}{char} +\DeclareMathOperator{\Ker}{Ker} +\DeclareMathOperator{\Slika}{Ker} +\DeclareMathOperator{\sgn}{sgn} +\DeclareMathOperator{\End}{End} +\DeclareMathOperator{\n}{n} +\DeclareMathOperator{\Col}{Col} +\usepackage{algorithm,algpseudocode} +\providecommand{\corollaryname}{Posledica} +\usepackage[slovenian=quotes]{csquotes} +\end_preamble +\use_default_options true +\begin_modules +enumitem +theorems-ams +theorems-ams-extended +\end_modules +\maintain_unincluded_children no +\language slovene +\language_package default +\inputencoding auto-legacy +\fontencoding auto +\font_roman "default" "default" +\font_sans "default" "default" +\font_typewriter "default" "default" +\font_math "auto" "auto" +\font_default_family default +\use_non_tex_fonts false +\font_sc false +\font_roman_osf false +\font_sans_osf false +\font_typewriter_osf false +\font_sf_scale 100 100 +\font_tt_scale 100 100 +\use_microtype false +\use_dash_ligatures true +\graphics default +\default_output_format default +\output_sync 0 +\bibtex_command default +\index_command default +\float_placement H +\float_alignment class +\paperfontsize default +\spacing single +\use_hyperref true +\pdf_bookmarks true +\pdf_bookmarksnumbered false +\pdf_bookmarksopen false +\pdf_bookmarksopenlevel 1 +\pdf_breaklinks false +\pdf_pdfborder false +\pdf_colorlinks false +\pdf_backref false +\pdf_pdfusetitle true +\papersize default +\use_geometry true +\use_package amsmath 1 +\use_package amssymb 1 +\use_package cancel 1 +\use_package esint 1 +\use_package mathdots 1 +\use_package mathtools 1 +\use_package mhchem 1 +\use_package stackrel 1 +\use_package stmaryrd 1 +\use_package undertilde 1 +\cite_engine basic +\cite_engine_type default +\biblio_style plain +\use_bibtopic false +\use_indices false +\paperorientation portrait +\suppress_date false +\justification false +\use_refstyle 1 +\use_formatted_ref 0 +\use_minted 0 +\use_lineno 0 +\index Index +\shortcut idx +\color #008000 +\end_index +\leftmargin 2cm +\topmargin 2cm +\rightmargin 2cm +\bottommargin 2cm +\headheight 2cm +\headsep 2cm +\footskip 1cm +\secnumdepth 3 +\tocdepth 3 +\paragraph_separation indent +\paragraph_indentation default +\is_math_indent 0 +\math_numbering_side default +\quotes_style german +\dynamic_quotes 0 +\papercolumns 1 +\papersides 1 +\paperpagestyle default +\tablestyle default +\tracking_changes false +\output_changes false +\change_bars false +\postpone_fragile_content false +\html_math_output 0 +\html_css_as_file 0 +\html_be_strict false +\docbook_table_output 0 +\docbook_mathml_prefix 1 +\end_header + +\begin_body + +\begin_layout Title +Teorija Analize 1 — + IŠRM 2023/24 +\end_layout + +\begin_layout Author + +\noun on +Anton Luka Šijanec +\end_layout + +\begin_layout Date +\begin_inset ERT +status open + +\begin_layout Plain Layout + + +\backslash +today +\end_layout + +\end_inset + + +\end_layout + +\begin_layout Abstract +Povzeto po zapiskih s predavanj profesorja Oliverja Dragičevića. +\end_layout + +\begin_layout Standard +\begin_inset CommandInset toc +LatexCommand tableofcontents + +\end_inset + + +\end_layout + +\begin_layout Section +Števila +\end_layout + +\begin_layout Definition* +Množica je matematični objekt, + ki predstavlja skupino elementov. + Če element +\begin_inset Formula $a$ +\end_inset + + pripada množici +\begin_inset Formula $A$ +\end_inset + +, + pišemo +\begin_inset Formula $a\in A$ +\end_inset + +, + sicer pa +\begin_inset Formula $a\not\in A$ +\end_inset + +. + Množica +\begin_inset Formula $B$ +\end_inset + + je podmnožica množice +\begin_inset Formula $A$ +\end_inset + +, + pišemo +\begin_inset Formula $B\subset A$ +\end_inset + +, + če +\begin_inset Formula $\forall b\in B:b\in A$ +\end_inset + +. + Presek +\begin_inset Formula $B$ +\end_inset + + in +\begin_inset Formula $C$ +\end_inset + + označimo +\begin_inset Formula $B\cap C\coloneqq\left\{ x;x\in B\wedge x\in C\right\} $ +\end_inset + +. + Unijo +\begin_inset Formula $B$ +\end_inset + + in +\begin_inset Formula $C$ +\end_inset + + označimo +\begin_inset Formula $B\cup C\coloneqq\left\{ x;x\in B\vee x\in C\right\} $ +\end_inset + +. + Razliko/komplement +\begin_inset Quotes gld +\end_inset + + +\begin_inset Formula $B$ +\end_inset + + manj/brez +\begin_inset Formula $C$ +\end_inset + + +\begin_inset Quotes grd +\end_inset + + označimo +\begin_inset Formula $B\setminus C\coloneqq\left\{ x;x\in B\wedge x\not\in C\right\} $ +\end_inset + +. +\end_layout + +\begin_layout Subsection +Realna števila +\end_layout + +\begin_layout Standard +Množico realnih števil označimo +\begin_inset Formula $\mathbb{R}$ +\end_inset + +. + V njej obstajata binarni operaciji seštevanje +\begin_inset Formula $a+b$ +\end_inset + + in množenje +\begin_inset Formula $a\cdot b$ +\end_inset + +. +\end_layout + +\begin_layout Subsubsection +Lastnosti seštevanja +\end_layout + +\begin_layout Axiom +Komutativnost: + +\begin_inset Formula $\forall a,b\in\mathbb{R}:a+b=b+a$ +\end_inset + +. +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Axiom +Asociativnost: + +\begin_inset Formula $\forall a,b,c\in\mathbb{R}:a+\left(b+c\right)=\left(a+b\right)+c$ +\end_inset + +, + torej je +\begin_inset Formula $a+\cdots+z$ +\end_inset + + dobro definiran izraz. +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Axiom +Obstoj enote: + +\begin_inset Formula $\exists0\in\mathbb{R}\forall a\in\mathbb{R}:a+0=a$ +\end_inset + +. +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Axiom +Obstoj inverzov: + +\begin_inset Formula $\forall a\in\mathbb{R}\exists b\in\mathbb{R}\ni a+b=0$ +\end_inset + +. +\end_layout + +\begin_deeper +\begin_layout Claim* +Inverz je enoličen. +\end_layout + +\begin_layout Proof +Naj bo +\begin_inset Formula $a,b,c\in\mathbb{R}$ +\end_inset + + in +\begin_inset Formula $a+b=0$ +\end_inset + + in +\begin_inset Formula $a+c=0$ +\end_inset + +. + Tedaj +\begin_inset Formula $b=b+0=b+a+c=0+c=c$ +\end_inset + +. +\end_layout + +\begin_layout Corollary* +Inverz je funkcija in aditivni inverz +\begin_inset Formula $a$ +\end_inset + + označimo z +\begin_inset Formula $-a$ +\end_inset + +. + Pri zapisu +\begin_inset Formula $a+\left(-b\right)$ +\end_inset + + običajno +\begin_inset Formula $+$ +\end_inset + + izpustimo in pišemo +\begin_inset Formula $a-b$ +\end_inset + +, + čemur pravimo odštevanje +\begin_inset Formula $b$ +\end_inset + + od +\begin_inset Formula $a$ +\end_inset + +. +\end_layout + +\begin_layout Claim* +\begin_inset Formula $\forall a\in\mathbb{R}:a=-\left(-a\right)$ +\end_inset + +. +\end_layout + +\begin_layout Proof +Naj bo +\begin_inset Formula $b=-a$ +\end_inset + + in +\begin_inset Formula $c=-b=-\left(-a\right)$ +\end_inset + +. + Tedaj velja +\begin_inset Formula $c-a=c+b=-\left(-a\right)-a=0$ +\end_inset + + in +\begin_inset Formula $a=0+a=c-a+a=c+\left(-a\right)+a=c+0=c=-\left(-a\right)$ +\end_inset + +. +\end_layout + +\begin_layout Claim* +\begin_inset Formula $-\left(b+c\right)=-b-c$ +\end_inset + + +\end_layout + +\begin_layout Proof +Velja +\begin_inset Formula $b+c+\left(-b-c\right)=b+c+\left(\left(-b\right)+\left(-c\right)\right)=b+\left(-b\right)+c+\left(-c\right)=0$ +\end_inset + +, + torej je +\begin_inset Formula $b+c$ +\end_inset + + inverz od +\begin_inset Formula $\left(-b-c\right)$ +\end_inset + +, + torej je +\begin_inset Formula $-\left(b+c\right)=-b-c$ +\end_inset + +. +\end_layout + +\end_deeper +\begin_layout Subsubsection +Lastnosti množenja +\end_layout + +\begin_layout Axiom +Komutativnost: + +\begin_inset Formula $\forall a,b\in\mathbb{R}:ab=ba$ +\end_inset + +. +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Axiom +Asociativnost: + +\begin_inset Formula $\forall a,b,c\in\mathbb{R}:a\left(bc\right)=\left(ab\right)c$ +\end_inset + +, + torej je +\begin_inset Formula $a\cdots z$ +\end_inset + + dobro definiran izraz. +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Axiom +Obstoj enote: + +\begin_inset Formula $\exists1\in\mathbb{R}\forall a\in\mathbb{R}:a1=a$ +\end_inset + +. +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Axiom +Obstoj inverzov: + +\begin_inset Formula $\forall a\in\mathbb{R}\setminus\left\{ 0\right\} \exists b\in\mathbb{R}\setminus\left\{ 0\right\} \ni:ab=1$ +\end_inset + + +\end_layout + +\begin_deeper +\begin_layout Claim* +Inverz je enoličen. +\end_layout + +\begin_layout Proof +Naj bo +\begin_inset Formula $a,b,c\in\mathbb{R}\setminus\left\{ 0\right\} $ +\end_inset + + in +\begin_inset Formula $ab=1$ +\end_inset + + in +\begin_inset Formula $ac=1$ +\end_inset + +. + Tedaj +\begin_inset Formula $b=b1=bac=1c=c$ +\end_inset + +. +\end_layout + +\begin_layout Corollary* +Inverz je funkcija in multiplikativni inverz +\begin_inset Formula $a$ +\end_inset + + označimo z +\begin_inset Formula $a^{-1}$ +\end_inset + +. + Pri zapisu +\begin_inset Formula $a\cdot b^{-1}$ +\end_inset + + lahko +\begin_inset Formula $\cdot$ +\end_inset + + izpustimo in pišemo +\begin_inset Formula $a/b$ +\end_inset + +, + čemur pravimo deljenje +\begin_inset Formula $a$ +\end_inset + + z +\begin_inset Formula $b$ +\end_inset + + za neničeln +\begin_inset Formula $b$ +\end_inset + +. +\end_layout + +\end_deeper +\begin_layout Subsubsection +Skupne lastnosti v +\begin_inset Formula $\mathbb{R}$ +\end_inset + + +\end_layout + +\begin_layout Axiom +\begin_inset Formula $1\not=0$ +\end_inset + + +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Axiom +Distributivnost: + +\begin_inset Formula $\forall a,b,c\in\mathbb{R}:\left(a+b\right)c=ac+bc$ +\end_inset + + +\end_layout + +\begin_layout Paragraph +Urejenost +\begin_inset Formula $\mathbb{R}$ +\end_inset + + +\end_layout + +\begin_layout Standard +Realna števila delimo na pozitivna +\begin_inset Formula $\mathbb{R}_{+}\coloneqq\left\{ x\in\mathbb{R};x>0\right\} $ +\end_inset + +, + negativna +\begin_inset Formula $\mathbb{R}_{-}\coloneqq\left\{ x\in\mathbb{R};x<0\right\} $ +\end_inset + + in ničlo +\begin_inset Formula $0$ +\end_inset + +. + Če je +\begin_inset Formula $x\in\mathbb{\mathbb{R}}_{+}\cup\left\{ 0\right\} $ +\end_inset + +, + pišemo +\begin_inset Formula $x\geq0$ +\end_inset + +, + če je +\begin_inset Formula $x\in\mathbb{R}_{-}\cup\left\{ 0\right\} $ +\end_inset + +, + pišemo +\begin_inset Formula $x\leq0$ +\end_inset + +. +\end_layout + +\begin_layout Axiom +Če je +\begin_inset Formula $a\not=0$ +\end_inset + +, + je natanko eno izmed +\begin_inset Formula $\left\{ a,-a\right\} $ +\end_inset + + pozitivno, + imenujemo ga absolutna vrednost +\begin_inset Formula $a$ +\end_inset + + (pišemo +\begin_inset Formula $\left|a\right|$ +\end_inset + +), + in natanko eno negativno, + pišemo +\begin_inset Formula $-\left|a\right|$ +\end_inset + +. +\end_layout + +\begin_layout Definition* +\begin_inset Formula $\left|0\right|=0$ +\end_inset + +. +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Definition* +Za +\begin_inset Formula $a,b\in\mathbb{R}$ +\end_inset + + se +\begin_inset Formula $\left|a-b\right|$ +\end_inset + + imenuje razdalja. +\end_layout + +\begin_layout Axiom +\begin_inset Formula $\forall a,b\in\mathbb{R}:a,b>0\Rightarrow\left(a+b>0\right)\wedge\left(ab>0\right)$ +\end_inset + +. +\end_layout + +\begin_layout Definition* +Za +\begin_inset Formula $a,b\in\mathbb{R}$ +\end_inset + +: + +\begin_inset Formula $a$ +\end_inset + + je večje od +\begin_inset Formula $b$ +\end_inset + +, + oznaka +\begin_inset Formula $a>b\Leftrightarrow a-b>0$ +\end_inset + +. + +\begin_inset Formula $a$ +\end_inset + + je manjše od +\begin_inset Formula $b$ +\end_inset + +, + oznaka +\begin_inset Formula $a<b\Leftrightarrow a-b<0$ +\end_inset + +. + Podobno +\begin_inset Formula $\leq$ +\end_inset + + in +\begin_inset Formula $\geq$ +\end_inset + +. +\end_layout + +\begin_layout Claim* +Trikotniška neenakost. + +\begin_inset Formula $\forall a,b\in\mathbb{R}$ +\end_inset + +: + +\begin_inset Formula $\left|\left|a\right|-\left|b\right|\right|\leq\left|a+b\right|\leq\left|a\right|+\left|b\right|$ +\end_inset + +. +\end_layout + +\begin_layout Proof +Dokažimo desni neenačaj. + Vemo +\begin_inset Formula $ab\leq\left|ab\right|$ +\end_inset + + in +\begin_inset Formula $\left|a\right|=\sqrt{a^{2}}$ +\end_inset + +. + Naj bo +\begin_inset Formula $a^{2}+2ab+b^{2}\leq\left|a\right|^{2}+2\left|a\right|\left|b\right|+\left|b\right|^{2}$ +\end_inset + +, + torej +\begin_inset Formula $\left(a+b\right)^{2}\leq\left(\left|a\right|+\left|b\right|\right)^{2}$ +\end_inset + +, + korenimo: + +\begin_inset Formula $\left|a+b\right|\leq\left|a\right|+\left|b\right|$ +\end_inset + +. +\end_layout + +\begin_layout Subsubsection +Intervali +\end_layout + +\begin_layout Definition* +Naj bo +\begin_inset Formula $a<b$ +\end_inset + +. + Označimo odprti interval +\begin_inset Formula $\left(a,b\right)\coloneqq\left\{ x\in\mathbb{R};a<x<b\right\} $ +\end_inset + +, + zaprti +\begin_inset Formula $\left[a,b\right]\coloneqq\left\{ x\in\mathbb{R};a\leq x\leq b\right\} $ +\end_inset + +, + polodprti +\begin_inset Formula $(a,b]\coloneqq\left\{ x\in\mathbb{R};a<x\leq b\right\} $ +\end_inset + + in podobno +\begin_inset Formula $[a,b)$ +\end_inset + +. + +\begin_inset Formula $\left(a,\infty\right)\coloneqq\left\{ x\in\mathbb{R};x>a\right\} $ +\end_inset + + in podobno +\begin_inset Formula $[a,\infty)$ +\end_inset + +. +\end_layout + +\begin_layout Subsection +Temeljne številske podmnožice +\end_layout + +\begin_layout Subsubsection +Naravna števila +\begin_inset Formula $\mathbb{N}$ +\end_inset + + +\end_layout + +\begin_layout Definition* +\begin_inset Formula $\mathbb{N}\coloneqq\left\{ 1,1+1,1+1+1,1+1+1+1,\dots\right\} $ +\end_inset + + +\end_layout + +\begin_layout Paragraph +Matematična indukcija +\end_layout + +\begin_layout Standard +Če je +\begin_inset Formula $A\subseteq\mathbb{N}$ +\end_inset + + in velja +\begin_inset Formula $1\in A$ +\end_inset + + (baza) in +\begin_inset Formula $a\in A\Rightarrow a+1\in A$ +\end_inset + + (korak), + tedaj +\begin_inset Formula $A=\mathbb{N}$ +\end_inset + +. +\end_layout + +\begin_layout Claim* +\begin_inset Formula $1+2+3+\cdots+n=\frac{n\left(n+1\right)}{2}$ +\end_inset + +. +\end_layout + +\begin_layout Proof +\begin_inset Formula $A\coloneqq\left\{ n\in\mathbb{N};\text{velja trditev za }n\right\} $ +\end_inset + +. + Dokažimo +\begin_inset Formula $A=\mathbb{N}$ +\end_inset + +. +\end_layout + +\begin_deeper +\begin_layout Itemize +Baza: + +\begin_inset Formula $1=\frac{1\cdot2}{2}=1$ +\end_inset + +. +\end_layout + +\begin_layout Itemize +Korak: + Predpostavimo +\begin_inset Formula $1+2+3+\cdots+n=\frac{n\left(n+1\right)}{2}$ +\end_inset + +. + Prištejmo +\begin_inset Formula $n+1$ +\end_inset + +: + +\begin_inset Formula +\[ +1+2+3+\cdots+n+\left(n+1\right)=\frac{n\left(n+1\right)}{2}+\left(n+1\right)=\frac{n\left(n+1\right)}{2}+\frac{2\left(n+1\right)}{2}=\frac{\left(n+2\right)\left(n+1\right)}{2} +\] + +\end_inset + + +\end_layout + +\end_deeper +\begin_layout Subsubsection +Cela števila +\begin_inset Formula $\mathbb{Z}$ +\end_inset + + +\end_layout + +\begin_layout Standard +Množica +\begin_inset Formula $\mathbb{N}$ +\end_inset + + je zaprta za seštevanje in množenje, + torej +\begin_inset Formula $\forall a,b\in\mathbb{N}:a+b\in\mathbb{N}\wedge ab\in\mathbb{N}$ +\end_inset + +, + ni pa zaprta za odštevanje, + ker recimo +\begin_inset Formula $5-3\not\in\mathbb{N}$ +\end_inset + +. + Zapremo jo za odštevanje in dobimo množico +\begin_inset Formula $\mathbb{Z}$ +\end_inset + +. +\end_layout + +\begin_layout Definition* +\begin_inset Formula $\mathbb{Z}\coloneqq\left\{ a-b;b,a\in\mathbb{N}\right\} $ +\end_inset + + +\end_layout + +\begin_layout Subsubsection +Racionalna števila +\begin_inset Formula $\mathbb{Q}$ +\end_inset + + +\end_layout + +\begin_layout Standard +Najmanjša podmnožica +\begin_inset Formula $\mathbb{R}$ +\end_inset + +, + ki vsebuje +\begin_inset Formula $\mathbb{Z}$ +\end_inset + + in je zaprta za deljenje, + je +\begin_inset Formula $\mathbb{Q}$ +\end_inset + +. +\end_layout + +\begin_layout Definition* +\begin_inset Formula $\mathbb{Q}\coloneqq\left\{ a/b;a\in\mathbb{Z},b\in\mathbb{Z}\setminus\left\{ 0\right\} \right\} $ +\end_inset + +. +\end_layout + +\begin_layout Standard +Velja +\begin_inset Formula $\mathbb{N}\subset\mathbb{Z}\subset\mathbb{Q}\subset\mathbb{R}$ +\end_inset + +. +\end_layout + +\begin_layout Claim* +Za +\begin_inset Formula $a\in\mathbb{Q}$ +\end_inset + +, + +\begin_inset Formula $b\not\in\mathbb{Q}$ +\end_inset + + velja +\begin_inset Formula $a+b\not\in\mathbb{Q}$ +\end_inset + + in +\begin_inset Formula $a\not=0\Rightarrow ab\not\in\mathbb{Q}$ +\end_inset + +. +\end_layout + +\begin_layout Proof +PDDRAA +\begin_inset Formula $a+b\in\mathbb{Q}$ +\end_inset + +. + Tedaj +\begin_inset Formula $a+b-a\in\mathbb{Q}$ +\end_inset + +, + tedaj +\begin_inset Formula $b\in\mathbb{Q}$ +\end_inset + +, + kar je +\begin_inset Formula $\rightarrow\!\leftarrow$ +\end_inset + +. + PDDRAA +\begin_inset Formula $ab\in\mathbb{Q}$ +\end_inset + +. + Tedaj +\begin_inset Formula $\frac{ab}{a}\in\mathbb{Q}$ +\end_inset + +, + tedaj +\begin_inset Formula $b\in\mathbb{Q}$ +\end_inset + +, + kar je +\begin_inset Formula $\rightarrow\!\leftarrow$ +\end_inset + +. +\end_layout + +\begin_layout Subsection +\begin_inset CommandInset label +LatexCommand label +name "subsec:Omejenost-množic" + +\end_inset + +Omejenost množic +\end_layout + +\begin_layout Definition* +Naj bo +\begin_inset Formula $A\subset\mathbb{R}$ +\end_inset + +. + +\begin_inset Formula $A$ +\end_inset + + je navzgor omejena +\begin_inset Formula $\Leftrightarrow\exists m\in\mathbb{R}\forall a\in A:a\leq m$ +\end_inset + +. + Takemu +\begin_inset Formula $m$ +\end_inset + + pravimo zgornja meja. + Najmanjši zgornji meji +\begin_inset Formula $A$ +\end_inset + + pravimo supremum ali natančna zgornja meja množice +\begin_inset Formula $A$ +\end_inset + +, + označimo +\begin_inset Formula $\sup A$ +\end_inset + +. + Če je zgornja meja +\begin_inset Formula $A$ +\end_inset + + ( +\begin_inset Formula $m$ +\end_inset + +) element +\begin_inset Formula $A$ +\end_inset + +, + je maksimum množice +\begin_inset Formula $A$ +\end_inset + +, + označimo +\begin_inset Formula $m=\max A$ +\end_inset + +. + Če množica ni navzgor omejena, + pišemo +\begin_inset Formula $\sup A=\infty$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Če +\begin_inset Formula $s=\sup A\in\mathbb{R}$ +\end_inset + +, + mora veljati +\begin_inset Formula $\forall a\in A:a\leq s$ +\end_inset + + in +\begin_inset Formula $\forall\varepsilon>0\exists b\in A\ni:b>s-\varepsilon$ +\end_inset + +, + torej za vsak neničeln +\begin_inset Formula $\varepsilon$ +\end_inset + + +\begin_inset Formula $s-\varepsilon$ +\end_inset + + ni več natančna zgornja meja za +\begin_inset Formula $A$ +\end_inset + +. +\end_layout + +\begin_layout Definition* +Naj bo +\begin_inset Formula $A\subset\mathbb{R}$ +\end_inset + +. + +\begin_inset Formula $A$ +\end_inset + + je navzdol omejena +\begin_inset Formula $\Leftrightarrow\exists m\in\mathbb{R}\forall a\in A:a\geq m$ +\end_inset + +. + Takemu +\begin_inset Formula $m$ +\end_inset + + pravimo spodnja meja. + Največji spodnji meji +\begin_inset Formula $A$ +\end_inset + + pravimo infimum ali natančna spodnja meja množice +\begin_inset Formula $A$ +\end_inset + +, + označimo +\begin_inset Formula $\inf A$ +\end_inset + +. + Če je spodnja meja +\begin_inset Formula $A$ +\end_inset + + ( +\begin_inset Formula $m$ +\end_inset + +) element +\begin_inset Formula $A$ +\end_inset + +, + je minimum množice +\begin_inset Formula $A$ +\end_inset + +, + označimo +\begin_inset Formula $m=\min A$ +\end_inset + +. + Če množica ni navzdol omejena, + pišemo +\begin_inset Formula $\inf A=-\infty$ +\end_inset + +. +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Definition* +Množica +\begin_inset Formula $A\subset\mathbb{R}$ +\end_inset + + je omejena, + če je hkrati navzgor in navzdol omejena. +\end_layout + +\begin_layout Axiom +\begin_inset CommandInset label +LatexCommand label +name "axm:Dedekind.-Vsaka-navzgor" + +\end_inset + +Dedekind. + Vsaka navzgor omejena množica v +\begin_inset Formula $\mathbb{R}$ +\end_inset + + ima natančno zgornjo mejo v +\begin_inset Formula $\mathbb{R}$ +\end_inset + +. +\end_layout + +\begin_layout Remark* +Za +\begin_inset Formula $\mathbb{Q}$ +\end_inset + + aksiom +\begin_inset CommandInset ref +LatexCommand ref +reference "axm:Dedekind.-Vsaka-navzgor" +plural "false" +caps "false" +noprefix "false" +nolink "false" + +\end_inset + + ne velja. + Če +\begin_inset Formula $B\subset\mathbb{Q}$ +\end_inset + +, + se lahko zgodi, + da +\begin_inset Formula $\sup B\not\in\mathbb{Q}$ +\end_inset + +. + Primer: + +\begin_inset Formula $B\coloneqq\left\{ q\in\mathbb{Q};q^{2}\leq2\right\} $ +\end_inset + +. + +\begin_inset Formula $\sup B=\sqrt{2}\not\in\mathbb{Q}$ +\end_inset + +. +\end_layout + +\begin_layout Example* + +\end_layout + +\begin_layout Subsection +Decimalni zapis +\end_layout + +\begin_layout Definition* +\begin_inset Formula $\forall x\in\mathbb{R}^{+}\exists!m\in\mathbb{N}\cup\left\{ 0\right\} ,d_{1},d_{2},\dots\in\left\{ 0..9\right\} $ +\end_inset + +, + ki število natančno določajo. + Pišemo +\begin_inset Formula $x=m,d_{1}d_{2}\dots$ +\end_inset + +. + Natančno določitev mislimo v smislu: +\end_layout + +\begin_deeper +\begin_layout Itemize +\begin_inset Formula $m\leq x<m+1$ +\end_inset + + — + s tem se izognemo dvojnemu zapisu +\begin_inset Formula $1=0,\overline{9}$ +\end_inset + + in +\begin_inset Formula $1=1,\overline{0}$ +\end_inset + +. +\end_layout + +\begin_layout Itemize +\begin_inset Formula $[m,m+1)$ +\end_inset + + razdelimo na 10 enako dolgih polodprtih intervalov +\begin_inset Formula $I_{0},\dots,I_{9}$ +\end_inset + +. + +\begin_inset Formula $x$ +\end_inset + + leži na natanko enem izmed njih, + indeks njega je +\begin_inset Formula $d_{1}$ +\end_inset + +. + Nadaljujemo tako, + da +\begin_inset Formula $I_{d_{1}}$ +\end_inset + + razdelimo zopet na 10 delov itd. +\end_layout + +\end_deeper +\begin_layout Definition* +Števila +\begin_inset Formula $x\in\mathbb{R^{-}}$ +\end_inset + + pišemo tako, + da zapišemo decimalni zapis števila +\begin_inset Formula $-x$ +\end_inset + + in predenj zapišemo +\begin_inset Formula $-$ +\end_inset + +. +\end_layout + +\begin_layout Definition* +Če se decimalke v zaporedju +\begin_inset Formula $\left(d_{n}\right)_{n\in\mathbb{N}}$ +\end_inset + + ponavljajo, + uporabimo periodični zapis, + denimo +\begin_inset Formula $5,01\overline{763}\in\mathbb{Q}$ +\end_inset + +. +\end_layout + +\begin_layout Subsection +Kompleksna števila +\end_layout + +\begin_layout Definition* +Vpeljimo število +\begin_inset Formula $i$ +\end_inset + + z lastnostjo +\begin_inset Formula $i^{2}=-1$ +\end_inset + +, + da je +\begin_inset Formula $i$ +\end_inset + + rešitev enačbe +\begin_inset Formula $x^{2}+1=0$ +\end_inset + +. +\end_layout + +\begin_layout Claim* +\begin_inset Formula $i\not\in\mathbb{R}$ +\end_inset + + +\end_layout + +\begin_layout Proof +Sicer bi veljajo +\begin_inset Formula $i^{2}\geq0$ +\end_inset + +, + kar po definiciji ne velja. +\end_layout + +\begin_layout Definition* +Kompleksna števila so +\begin_inset Formula $\mathbb{C}\coloneqq\left\{ a+bi;a,b\in\mathbb{R}\right\} $ +\end_inset + +. + +\begin_inset Formula $bi$ +\end_inset + + je še nedefinirano, + zato za kompleksna števila definirano seštevanje in množenje za +\begin_inset Formula $z=a+bi$ +\end_inset + + in +\begin_inset Formula $w=c+di$ +\end_inset + +: +\end_layout + +\begin_deeper +\begin_layout Itemize +\begin_inset Formula $z+w\coloneqq\left(a+c\right)+\left(b+d\right)i$ +\end_inset + + +\end_layout + +\begin_layout Itemize +\begin_inset Formula $zw\coloneqq\left(a+bi\right)\left(c+di\right)=ac+adi+bic+bidi=ac+adi+bic-bd=\left(ac-bd\right)+\left(ad+bc\right)i$ +\end_inset + + +\end_layout + +\end_deeper +\begin_layout Definition* +Definiramo še konjugirano vrednost +\begin_inset Formula $z\in\mathbb{C}$ +\end_inset + +: + +\begin_inset Formula $\overline{z}\coloneqq a-bi$ +\end_inset + + in označimo +\begin_inset Formula $\left|z\right|^{2}=z\overline{z}$ +\end_inset + +. +\end_layout + +\begin_layout Claim* +\begin_inset Formula $z\overline{z}=a^{2}+b^{2}\geq0$ +\end_inset + + za +\begin_inset Formula $z=a+bi$ +\end_inset + +. +\end_layout + +\begin_layout Proof +\begin_inset Formula $\left(a+bi\right)\left(a-bi\right)=a^{2}+abi-bia-bibi=a^{2}+b^{2}$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Velja, + da je +\begin_inset Formula $\mathbb{R}\subset\mathbb{C}$ +\end_inset + + v smislu identifikacije +\begin_inset Formula $\mathbb{R}$ +\end_inset + + z množico +\begin_inset Formula $\mathbb{C}$ +\end_inset + +: + +\begin_inset Formula $\mathbb{R}=\left\{ a+0i;a\in\mathbb{R}\right\} $ +\end_inset + +, + torej smo +\begin_inset Formula $\mathbb{R}$ +\end_inset + + razširili v +\begin_inset Formula $\mathbb{C}$ +\end_inset + +, + kjer ima vsak polinom vedno rešitev. +\end_layout + +\begin_layout Subsubsection +Deljenje v +\begin_inset Formula $\mathbb{C}$ +\end_inset + + +\end_layout + +\begin_layout Standard +Za +\begin_inset Formula $w,z\in\mathbb{C},w\not=0$ +\end_inset + + iščemo +\begin_inset Formula $x\in\mathbb{C}\ni:wx=z$ +\end_inset + +. + Ločimo dva primera: +\end_layout + +\begin_layout Itemize +\begin_inset Formula $w\in\mathbb{R}\setminus\left\{ 0\right\} $ +\end_inset + +: + definiramo +\begin_inset Formula $x=\frac{z}{w}\coloneqq\frac{a}{w}+\frac{b}{w}i$ +\end_inset + + +\end_layout + +\begin_layout Itemize +\begin_inset Formula $w\in\mathbb{C}\setminus\left\{ 0\right\} $ +\end_inset + + (splošno): + +\begin_inset Formula $wx=z\overset{/\cdot\overline{w}}{\Longrightarrow}w\overline{w}x=z\overline{w}\Rightarrow\left|w\right|^{2}x=z\overline{w}\Rightarrow x=\frac{z\overline{w}}{\left|w\right|^{2}}$ +\end_inset + +, + z +\begin_inset Formula $\left|w\right|^{2}$ +\end_inset + + pa znamo deliti, + ker je realen. +\end_layout + +\begin_layout Subsubsection +Lastnosti v +\begin_inset Formula $\mathbb{C}$ +\end_inset + + +\end_layout + +\begin_layout Standard +\begin_inset Formula $+$ +\end_inset + + in +\begin_inset Formula $\cdot$ +\end_inset + + sta komutativni, + asociativni, + distributivni, + +\begin_inset Formula $0$ +\end_inset + + je aditivna enota, + +\begin_inset Formula $1$ +\end_inset + + je multiplikativna. +\end_layout + +\begin_layout Definition* +Za +\begin_inset Formula $z=a+bi$ +\end_inset + + vpeljemo +\begin_inset Formula $\Re z=a$ +\end_inset + + in +\begin_inset Formula $\Im z=b$ +\end_inset + +. +\end_layout + +\begin_layout Remark* +Opazimo +\begin_inset Formula $\Re z=\frac{z+\overline{z}}{2}$ +\end_inset + +, + +\begin_inset Formula $\Im z=\frac{z-\overline{z}}{2}$ +\end_inset + +. +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Remark* +\begin_inset Formula $\mathbb{C}$ +\end_inset + + si lahko predstavljamo kot urejene pare; + +\begin_inset Formula $a+bi$ +\end_inset + + ustreza paru +\begin_inset Formula $\left(a,b\right)$ +\end_inset + +. + Tako +\begin_inset Formula $\mathbb{C}$ +\end_inset + + enačimo/identificiramo z +\begin_inset Formula $\mathbb{R}^{2}\coloneqq\left\{ \left(a,b\right);a,b\in\mathbb{R}\right\} $ +\end_inset + +, + s čimer dobimo geometrično predstavitev +\begin_inset Formula $\mathbb{C}$ +\end_inset + + kot vektorje v +\begin_inset Formula $\mathbb{R}^{2}$ +\end_inset + +. +\end_layout + +\begin_layout Definition* +Za +\begin_inset Formula $z=a+bi$ +\end_inset + +, + predstavljen z vektorjem s komponentami +\begin_inset Formula $\left(a,b\right)$ +\end_inset + +, + velja +\begin_inset Formula $a=\left|z\right|\cos\varphi$ +\end_inset + + in +\begin_inset Formula $v=\left|z\right|\sin\varphi$ +\end_inset + +. + Kotu +\begin_inset Formula $\varphi$ +\end_inset + + pravimo argument kompleksnega števila +\begin_inset Formula $z$ +\end_inset + +, + oznaka +\begin_inset Formula $\arg z$ +\end_inset + +. +\end_layout + +\begin_layout Corollary* +\begin_inset Formula $z=$ +\end_inset + + +\begin_inset Formula $\left|z\right|\left(\cos\varphi+i\sin\varphi\right)$ +\end_inset + +. + Velja +\begin_inset Foot +status open + +\begin_layout Plain Layout +TODO DOPISATI ZAKAJ (v bistvu še jaz ne vem). + ne razumem. +\end_layout + +\end_inset + + +\begin_inset Formula $\left(\cos\varphi+i\sin\varphi\right)\left(\cos\psi+i\sin\psi\right)=\cos\left(\varphi+\psi\right)+i\sin\left(\varphi+\psi\right)$ +\end_inset + +, + zato lahko pišemo +\begin_inset Formula $e^{i\varphi}=\cos\varphi+i\sin\varphi$ +\end_inset + +. + Množenje kompleksnh števil +\begin_inset Formula $z=\left|z\right|e^{i\varphi}$ +\end_inset + + in +\begin_inset Formula $w=\left|w\right|e^{i\psi}$ +\end_inset + + vrne število +\begin_inset Formula $zw$ +\end_inset + +, + za katero velja: +\end_layout + +\begin_deeper +\begin_layout Itemize +\begin_inset Formula $\left|zw\right|=\left|z\right|\left|w\right|$ +\end_inset + + +\end_layout + +\begin_layout Itemize +\begin_inset Formula $\arg zw=\arg z+\arg w$ +\end_inset + + (do periode +\begin_inset Formula $2\pi$ +\end_inset + + natančno) +\end_layout + +\end_deeper +\begin_layout Section +Zaporedja +\end_layout + +\begin_layout Definition* +Funkcija +\begin_inset Formula $a:\mathbb{N}\to\mathbb{R}$ +\end_inset + + se imenuje realno zaporedje, + oznaka +\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$ +\end_inset + +. + +\begin_inset Formula $a_{n}$ +\end_inset + + je funkcijska vrednost pri +\begin_inset Formula $n$ +\end_inset + +. +\end_layout + +\begin_layout Example* +\begin_inset Formula $a_{n}=n$ +\end_inset + +: + +\begin_inset Formula $1,2,3,\dots$ +\end_inset + +; + +\begin_inset Formula $a_{n}=\left(-1\right)^{n}n^{2}$ +\end_inset + +: + +\begin_inset Formula $-1,4,-9,16,-25,\dots$ +\end_inset + +; + +\begin_inset Formula $a_{n}=\cos\left(\frac{\pi}{2}n\right)=0,-1,0,1,0,-1,\dots$ +\end_inset + + +\end_layout + +\begin_layout Standard +Zaporedje lahko podamo rekurzivno. + Podamo prvi člen ali nekaj prvih členov in pravilo, + kako iz prejšnjih členov dobiti naslednje. +\end_layout + +\begin_layout Example* +\begin_inset Formula $a_{1}=0,a_{n+1}=a_{n}+n$ +\end_inset + + da zaporedje +\begin_inset Formula $a_{n}=\frac{n\left(n+1\right)}{2}$ +\end_inset + +. + +\begin_inset Formula $a_{0}=0,a_{n+1}=\sqrt{b+a_{n}}$ +\end_inset + + da zaporedje +\begin_inset Formula $0,\sqrt{b},\sqrt{b+\sqrt{b}},\sqrt{b+\sqrt{b+\sqrt{b}}},\dots$ +\end_inset + +. + Fibbonacijevo zaporedje: + +\begin_inset Formula $a_{1}=a_{2}=1,a_{n+1}=a_{n}+a_{n-1}$ +\end_inset + + da zaporedje +\begin_inset Formula $1,1,2,3,5,8,\dots$ +\end_inset + + +\end_layout + +\begin_layout Subsection +Posebni tipi zaporedij +\end_layout + +\begin_layout Definition* +Zaporedje +\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$ +\end_inset + + je aritmetično, + če +\begin_inset Formula $\exists k\in\mathbb{R}\forall n\in\mathbb{N}:a_{n+1}-a_{n}=k$ +\end_inset + +. + Tedaj +\begin_inset Formula $a_{n+1}=a_{n}+k=a_{1}+nd$ +\end_inset + +. +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Definition* +Zaporedje +\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$ +\end_inset + + je geometrično, + če +\begin_inset Formula $\exists\lambda\in\mathbb{R}\forall n\in\mathbb{N}:a_{n+1}=a_{n}\lambda$ +\end_inset + +. + Tedaj +\begin_inset Formula $a_{n}=\lambda^{n-1}a_{1}$ +\end_inset + +. +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Definition* +Zaporedje +\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$ +\end_inset + + je navzdol oz. + navzgor omejeno, + če je množica vseh členov tega taporedja navzgor oz. + navzdol omejena (glej +\begin_inset CommandInset ref +LatexCommand ref +reference "subsec:Omejenost-množic" +plural "false" +caps "false" +noprefix "false" +nolink "false" + +\end_inset + +). + Podobno z množico členov definiramo supremum, + infimum, + maksimum in infimum zaporedja. +\end_layout + +\begin_layout Definition* +Zaporedje je naraščajoče, + če +\begin_inset Formula $\forall n\in\mathbb{N}:a_{n+1}\geq a_{n}$ +\end_inset + +, + padajoče, + če +\begin_inset Formula $\forall n\in\mathbb{N}:a_{n+1}\leq a_{n}$ +\end_inset + +, + strogo naraščajoče, + če +\begin_inset Formula $\forall n\in\mathbb{N}:a_{n+1}>a_{n}$ +\end_inset + +, + strogo padajoče podobno, + monotono, + če je naraščajoče ali padajoče in strogo monotono, + če je strogo naraščajoče ali strogo padajoče. +\end_layout + +\begin_layout Subsection +Limita zaporedja +\end_layout + +\begin_layout Definition* +Množica +\begin_inset Formula $U\subseteq\mathbb{R}$ +\end_inset + + je odprta, + če +\begin_inset Formula $\forall u\in U\exists r>0\ni:\left(u-r,u+r\right)\subseteq U$ +\end_inset + +. +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Definition* +Množica +\begin_inset Formula $U\subseteq\mathbb{R}$ +\end_inset + + je zaprta, + če je +\begin_inset Formula $U^{\mathcal{C}}\coloneqq\mathbb{R}\setminus U$ +\end_inset + + odprta. +\end_layout + +\begin_layout Claim* +Odprt interval je odprta množica. +\end_layout + +\begin_layout Proof +Za poljubna +\begin_inset Formula $a,b\in\mathbb{R}$ +\end_inset + +, + +\begin_inset Formula $b>a$ +\end_inset + +, + naj bo +\begin_inset Formula $u\in\left(a,b\right)$ +\end_inset + + poljuben. + Ustrezen +\begin_inset Formula $r$ +\end_inset + + je +\begin_inset Formula $\min\left\{ \left|r-a\right|,\left|r-b\right|\right\} $ +\end_inset + +, + da je +\begin_inset Formula $\left(u-r,u+r\right)\subseteq U$ +\end_inset + +. +\end_layout + +\begin_layout Claim* +Zaprt interval je zaprt. +\end_layout + +\begin_layout Proof +Naj bosta +\begin_inset Formula $a,b\in\mathbb{R}$ +\end_inset + + poljubna in +\begin_inset Formula $b>a$ +\end_inset + +. + Dokazujemo, + da je +\begin_inset Formula $\left[a,b\right]$ +\end_inset + + zaprt, + torej da je +\begin_inset Formula $\left[a,b\right]^{\mathcal{C}}=\left(-\infty,a\right)\cup\left(b,\infty\right)$ +\end_inset + + odprta množica. + Za poljuben +\begin_inset Formula $u\in\left[a,b\right]^{\mathcal{C}}$ +\end_inset + + velja, + da je bodisi +\begin_inset Formula $\in\left(-\infty,a\right)$ +\end_inset + + bodisi +\begin_inset Formula $\left(b,\infty\right)$ +\end_inset + +, + kajti +\begin_inset Formula $\left(-\infty,a\right)\cap\left(b,\infty\right)=\emptyset$ +\end_inset + +. + Po prejšnji trditvi v obeh primerih velja +\begin_inset Formula $\exists r>0\ni:\left(u-r,u+r\right)\subseteq U$ +\end_inset + +, + torej je +\begin_inset Formula $\left[a,b\right]^{\mathcal{C}}$ +\end_inset + + res odprta, + torej je +\begin_inset Formula $\left[a,b\right]$ +\end_inset + + res zaprta. +\end_layout + +\begin_layout Definition* +Množica +\begin_inset Formula $B$ +\end_inset + + je okolica točke +\begin_inset Formula $t\in\mathbb{R}$ +\end_inset + +, + če vsebuje kakšno odprto množico +\begin_inset Formula $U$ +\end_inset + +, + ki vsebuje +\begin_inset Formula $t$ +\end_inset + +, + torej +\begin_inset Formula $t\in U^{\text{odp.}}\subseteq B\subseteq\mathbb{R}$ +\end_inset + +. +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Definition* +\begin_inset Formula $L\in\mathbb{R}$ +\end_inset + + je limita zaporedja +\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{R}}$ +\end_inset + + +\begin_inset Formula $\Leftrightarrow\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow\left|a_{n}-L\right|<\varepsilon$ +\end_inset + +. + ZDB +\begin_inset Formula $\forall V$ +\end_inset + + okolica +\begin_inset Formula $L\in\mathbb{R}\exists n_{0}\in\mathbb{N}\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow a_{n}\in V$ +\end_inset + +, + pravimo, + da +\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$ +\end_inset + + konvergira k +\begin_inset Formula $L$ +\end_inset + + in pišemo +\begin_inset Formula $L\coloneqq\lim_{n\to\infty}a_{n}$ +\end_inset + + ali drugače +\begin_inset Formula $a_{n}\underset{n\to\infty}{\longrightarrow}L$ +\end_inset + +. + Če zaporedje ima limito, + pravimo, + da je konvergentno, + sicer je divergentno. +\end_layout + +\begin_layout Claim* +Konvergentno zaporedje v +\begin_inset Formula $\mathbb{R}$ +\end_inset + + ima natanko eno limito. +\end_layout + +\begin_layout Proof +Naj bosta +\begin_inset Formula $J$ +\end_inset + + in +\begin_inset Formula $L$ +\end_inset + + limiti zaporedja +\begin_inset Formula $\left(a_{n}\right)_{n\to\infty}$ +\end_inset + +. + Torej +\begin_inset Foot +status open + +\begin_layout Plain Layout +Ko trdimo, + da obstaja +\begin_inset Formula $n_{0}$ +\end_inset + +, + še ne vemo, + ali sta za +\begin_inset Formula $L$ +\end_inset + + in +\begin_inset Formula $J$ +\end_inset + + ta +\begin_inset Formula $n_{0}$ +\end_inset + + ista. + Ampak trditev še vedno velja, + ker lahko vzamemo večjega izmed njiju, + ako bi bila drugačna. +\end_layout + +\end_inset + + po definiciji +\begin_inset Formula $\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow\left|a_{n}-L\right|<\varepsilon\wedge\left|a_{n}-J\right|<\varepsilon$ +\end_inset + +. + Velja torej +\begin_inset Formula $\forall\varepsilon>0:\left|J-L\right|<\varepsilon$ +\end_inset + +. + PDDRAA +\begin_inset Formula $J\not=L$ +\end_inset + +. + Tedaj +\begin_inset Formula $\left|J-L\right|\not=0$ +\end_inset + +, + naj bo +\begin_inset Formula $\left|J-L\right|=k$ +\end_inset + +. + Tedaj +\begin_inset Formula $\exists\varepsilon>0:\left|J-L\right|\not<\varepsilon$ +\end_inset + +, + ustrezen +\begin_inset Formula $\varepsilon$ +\end_inset + + je na primer +\begin_inset Formula $\frac{\left|J-L\right|}{2}$ +\end_inset + +. +\end_layout + +\begin_layout Claim* +\begin_inset CommandInset label +LatexCommand label +name "Konvergentno-zaporedje-v-R-je-omejeno" + +\end_inset + +Konvergentno zaporedje v +\begin_inset Formula $\mathbb{R}$ +\end_inset + + je omejeno. +\end_layout + +\begin_layout Proof +Naj bo +\begin_inset Formula $L=\lim_{n\to\infty}a_{n}$ +\end_inset + +. + Znotraj intervala +\begin_inset Formula $\left(L-1,L+1\right)$ +\end_inset + + so vsi členi zaporedja razen končno mnogo ( +\begin_inset Formula $\left\{ a_{1},\dots,a_{n_{0}}\right\} $ +\end_inset + +). + +\begin_inset Formula $\left\{ a_{n}\right\} _{n\in\mathbb{N}}$ +\end_inset + + je unija dveh omejenih množic; + +\begin_inset Formula $\left(L-1,L+1\right)$ +\end_inset + + in +\begin_inset Formula $\left\{ a_{1},\dots,a_{n_{0}}\right\} $ +\end_inset + +, + zato je tudi sama omejena. +\end_layout + +\begin_layout 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 + +.ž +\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 $\forall b\in I\exists\alpha\in I\text{ med \ensuremath{a} in \ensuremath{x}}\ni:R_{n}\left(b\right)=\frac{f^{\left(n+1\right)}\left(\alpha\right)}{\left(n+1\right)!}\left(b-a\right)^{n+1}$ +\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 integrabilna, + 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 diff --git a/šola/citati.bib b/šola/citati.bib new file mode 100644 index 0000000..3c34475 --- /dev/null +++ b/šola/citati.bib @@ -0,0 +1,227 @@ +Osebna bibtex knjižnica citatov. +Canonical: ~/projects/r/šola/citati.bib od 2024-07-28 +Stara različica je v ~/projects/sola-gimb-4/citati.bib + +@online{hpstrings, + author = {Nave, Carl Rod}, + title = {{HyperPhysics Concepts: Standing Waves on a String}}, + year = {2016}, + url = {http://hyperphysics.phy-astr.gsu.edu./hbase/Waves/string.html}, + urldate = {2016-11-09} +} +@online{cohen01, + author = {Cohen, Bram}, + title = {BitTorrent - a new P2P app}, + year = {2001}, + url = {http://finance.groups.yahoo.com/group/decentralization/message/3160}, + urldate = {2007-04-15}, + note = "Internet Archive" +} +@online{harrison07, + author = {Harrison, David}, + url = {http://www.bittorrent.org/beps/bep_0000.html}, + urldate = {2023-02-28}, + year = {2008}, + title = {Index of BitTorrent Enhancement Proposals} +} +@online{norberg08, + author = {Loewenstern, Andrew and Norberg, Arvid}, + url = {https://www.bittorrent.org/beps/bep_0005.html}, + urldate = {2023-02-28}, + year = {2020}, + title = {DHT Protocol} +} +@online{jones15, + author = {Jones, Ben}, + url = {https://torrentfreak.com/bittorrents-dht-turns-10-years-old-150607/}, + urldate = {2023-02-28}, + year = {2015}, + title = {BitTorrent’s DHT Turns 10 Years Old} +} +@online{muo11, + author = {Mukherjee, Abhijeet}, + url = {https://www.makeuseof.com/tag/btdigg-trackerless-torrent/}, + urldate = {2023-02-28}, + year = {2011}, + title = {BTDigg: A Trackerless Torrent Search Engine} +} +@online{evseenko11, + author = {Evseenko, Nina}, + url = {http://btdig.com}, + urldate = {2023-02-28}, + year = {2011}, + title = {Btdigg BitTorrent DHT search engine} + +} +@online{griffin17, + author = {Griffin, Andrew}, + url = {https://www.independent.co.uk/tech/torrent-website-download-safe-legal-privacy-i-know-what-you-friends-spying-a7504266.html}, + urldate = {2023-02-28}, + year = {2017}, + title = {'I Know What You Download': Website claims to let people see everything their friends have torrented} +} +@online{ikwyd, + url = {http://iknowwhatyoudownload.com}, + urldate = {2023-02-28}, + title = {I know what you download} +} +@online{cohen08, + author = {Choen, Bram}, + url = {https://www.bittorrent.org/beps/bep_0003.html}, + urldate = {2023-02-28}, + year = {2017}, + title = {The BitTorrent Protocol Specification} +} +@online{norberg09, + author = {Norberg, Arvid}, + url = {https://www.bittorrent.org/beps/bep_0029.html}, + urldate = {2023-02-28}, + year = {2017}, + title = {uTorrent transport protocol} +} +@online{hazel08, + author = {Hazel, Greg and Norberg, Arvid}, + url = {https://www.bittorrent.org/beps/bep_0009.html}, + urldate = {2023-02-28}, + year = {2017}, + title = {Extension for Peers to Send Metadata Files} +} +@online{strigeus08, + author = {Norberg, Arvid and Strigeus, Ludvig and Hazel, Greg}, + url = {https://www.bittorrent.org/beps/bep_0010.html}, + urldate = {2023-02-28}, + year = {2017}, + title = {Extension Protocol} +} +@online{v2, + author = {Cohen, Bram}, + url = {https://www.bittorrent.org/beps/bep_0052.html}, + urldate = {2023-02-28}, + year = {2017}, + title = {The BitTorrent Protocol Specification v2} +} +@incollection{maymounkov2002kademlia, + title={Kademlia: A peer-to-peer information system based on the xor metric}, + author={Maymounkov, Petar and Mazieres, David}, + booktitle={Peer-to-Peer Systems: First InternationalWorkshop, IPTPS 2002 Cambridge, MA, USA, March 7--8, 2002 Revised Papers}, + pages={53--65}, + year={2002}, + publisher={Springer} +} +@inproceedings{pezoa2016foundations, + title={Foundations of JSON schema}, + author={Pezoa, Felipe and Reutter, Juan L and Suarez, Fernando and Ugarte, Mart{\'\i}n and Vrgo{\v{c}}, Domagoj}, + booktitle={Proceedings of the 25th International Conference on World Wide Web}, + pages={263--273}, + year={2016}, + organization={International World Wide Web Conferences Steering Committee} +} +@online{harrison08, + author = {Harrison, David}, + title = {Private Torrents}, + year = {2008}, + url = {https://www.bittorrent.org/beps/bep_0027.html}, + urldate = {2023-02-28}, +} +@techreport{Eastlake2001, + added-at = {2011-05-12T13:55:38.000+0200}, + author = {Eastlake, D. and Jones, P.}, + biburl = {https://www.bibsonomy.org/bibtex/2d2dcad25e57d68db18495f062e955cae/voj}, + interhash = {276256be69453291def2a6fdbed1389d}, + intrahash = {d2dcad25e57d68db18495f062e955cae}, + keywords = {hash identifier sha1}, + month = {9}, + number = 3174, + organization = {IETF}, + publisher = {IETF}, + series = {RFC}, + timestamp = {2013-01-06T13:13:36.000+0100}, + title = {{US Secure Hash Algorithm 1 (SHA1)}}, + type = {RFC}, + year = 2001 +} +@online{dis, + author = {Gams, Matjaž and others}, + year = {2008}, + title = {DIS Slovarček}, + url = {https://dis-slovarcek.ijs.si/}, + urldate = {2023-02-28} +} +@online{rhilip20, + author = {Rhilip}, + year = {2020}, + urldate = {2023-02-28}, + url = {https://github.com/Rhilip/Bencode}, + title = {A pure and simple PHP library for encoding and decoding Bencode data} +} +@conference{Kluyver2016jupyter, + Title = {Jupyter Notebooks -- a publishing format for reproducible computational workflows}, + Author = {Thomas Kluyver and Benjamin Ragan-Kelley and Fernando P{\'e}rez and Brian Granger and Matthias Bussonnier and Jonathan Frederic and Kyle Kelley and Jessica Hamrick and Jason Grout and Sylvain Corlay and Paul Ivanov and Dami{\'a}n Avila and Safia Abdalla and Carol Willing}, + Booktitle = {Positioning and Power in Academic Publishing: Players, Agents and Agendas}, + Editor = {F. Loizides and B. Schmidt}, + Organization = {IOS Press}, + Pages = {87 - 90}, + Year = {2016} +} +@online{sampleih, + author = {The 8472}, + url = {https://www.bittorrent.org/beps/bep_0051.html}, + title = {DHT Infohash Indexing}, + urldate = {2023-02-28}, + year = {2017} +} +@Article{Hunter:2007, + Author = {Hunter, J. D.}, + Title = {Matplotlib: A 2D graphics environment}, + Journal = {Computing in Science \& Engineering}, + Volume = {9}, + Number = {3}, + Pages = {90--95}, + abstract = {Matplotlib is a 2D graphics package used for Python for + application development, interactive scripting, and publication-quality + image generation across user interfaces and operating systems.}, + publisher = {IEEE COMPUTER SOC}, + doi = {10.1109/MCSE.2007.55}, + year = 2007 +} +@online{norberg14, + author = {Norberg, Arvid}, + title = {DHT Security extension}, + url = {https://www.bittorrent.org/beps/bep_0042.html}, + urldate = {2023-02-28}, + year = {2016} +} +@InProceedings{10.1007/3-540-45748-8_24, + author="Douceur, John R.", + editor="Druschel, Peter + and Kaashoek, Frans + and Rowstron, Antony", + title="The Sybil Attack", + booktitle="Peer-to-Peer Systems", + year="2002", + publisher="Springer Berlin Heidelberg", + address="Berlin, Heidelberg", + pages="251--260", + abstract="Large-scale peer-to-peer systems face security threats from faulty or hostile remote computing elements. To resist these threats, many such systems employ redundancy. However, if a single faulty entity can present multiple identities, it can control a substantial fraction of the system, thereby undermining this redundancy. One approach to preventing these ``Sybil attacks'' is to have a trusted agency certify identities. This paper shows that, without a logically centralized authority, Sybil attacks are always possible except under extreme and unrealistic assumptions of resource parity and coordination among entities.", + isbn="978-3-540-45748-0" +} +@book{cedilnik12, + place={Radovljica}, + title={Matematični Priročnik}, + publisher={Didakta}, + author={Cedilnik, Anton and Razpet, Nada and Razpet, Marko}, + year={2012} +} +@online{wikihashuniformity, + author="{Wikipedia contributors}", + title={Hash function}, + url={https://w.wiki/An3L}, + urldate={2024-07-29}, + year={2024} +} +@online{maxmindgeoip2, + author={MaxMind}, + title={GeoLite2 Free Geolocation Data}, + url={https://dev.maxmind.com/geoip/geolite2-free-geolocation-data}, + urldate={2024-07-31}, + year={2024} diff --git a/šola/članki/dht/.gitignore b/šola/članki/dht/.gitignore new file mode 100644 index 0000000..abd26b3 --- /dev/null +++ b/šola/članki/dht/.gitignore @@ -0,0 +1,3 @@ +*.svg +*.png +*.tsv diff --git a/šola/članki/dht/dokument.lyx b/šola/članki/dht/dokument.lyx new file mode 100644 index 0000000..af668ed --- /dev/null +++ b/šola/članki/dht/dokument.lyx @@ -0,0 +1,2306 @@ +#LyX 2.4 created this file. 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+\graphics default +\default_output_format default +\output_sync 0 +\bibtex_command bibtex +\index_command default +\float_placement H +\float_alignment class +\paperfontsize default +\spacing single +\use_hyperref true +\pdf_title "Your Title" +\pdf_author "Your Name" +\pdf_bookmarks true +\pdf_bookmarksnumbered true +\pdf_bookmarksopen true +\pdf_bookmarksopenlevel 1 +\pdf_breaklinks false +\pdf_pdfborder true +\pdf_colorlinks false +\pdf_backref false +\pdf_pdfusetitle false +\pdf_quoted_options "pdfpagelayout=OneColumn, pdfnewwindow=true, pdfstartview=XYZ, plainpages=false" +\papersize default +\use_geometry true +\use_package amsmath 1 +\use_package amssymb 1 +\use_package cancel 1 +\use_package esint 1 +\use_package mathdots 1 +\use_package mathtools 1 +\use_package mhchem 1 +\use_package stackrel 1 +\use_package stmaryrd 1 +\use_package undertilde 1 +\cite_engine basic +\cite_engine_type default +\biblio_style plain +\use_bibtopic false +\use_indices false +\paperorientation portrait +\suppress_date false +\justification true +\use_refstyle 0 +\use_formatted_ref 0 +\use_minted 0 +\use_lineno 0 +\index Index +\shortcut idx +\color #008000 +\end_index +\leftmargin 0.5cm +\topmargin 0.5cm +\rightmargin 0.5cm +\bottommargin 1.25cm +\headheight 0.5cm +\headsep 0.5cm +\footskip 0.5cm +\columnsep 0.5cm +\secnumdepth 3 +\tocdepth 3 +\paragraph_separation indent +\paragraph_indentation default +\is_math_indent 0 +\math_numbering_side default +\quotes_style german +\dynamic_quotes 0 +\papercolumns 2 +\papersides 1 +\paperpagestyle default +\tablestyle default +\tracking_changes false +\output_changes false +\change_bars false +\postpone_fragile_content false +\html_math_output 0 +\html_css_as_file 0 +\html_be_strict false +\docbook_table_output 0 +\docbook_mathml_prefix 1 +\end_header + +\begin_body + +\begin_layout Title +Kaj prenašamo s protokolom BitTorrent? +\end_layout + +\begin_layout Author +Anton Luka Šijanec, + +\begin_inset CommandInset href +LatexCommand href +name "anton@sijanec.eu" +target "anton@sijanec.eu" +type "mailto:" +literal "true" + +\end_inset + + +\end_layout + +\begin_layout Institution +Fakulteta za računalništvo in informatiko Univerze v Ljubljani +\end_layout + +\begin_layout Abstract +V članku predstavimo metodo za učinkovito in za omrežje neinvazivno metodo prenašanja metapodatkov iz pomožnega omrežja Kademlia mainline DHT protokola BitTorrent za izmenjavo datotek. + Sledi pregled/analiza z opisano metodo pridobljenih metapodatkov o datotekah na voljo v omrežju BitTorrent. +\end_layout + +\begin_layout Abstract +Porazdeljene razpršilne tabele (angl. + distributed hash table) so razpršilne tabele, + ki podatke, + ponavadi so to dokumenti, + strukturirani kot vrednost in njej pripadajoči ključ, + hranijo distribuirano na več vozliščih, + na katerih se podatki shranjujejo. + V računalniških sistemih se DHT uporablja za hrambo podatkov v omrežjih P2P (angl. + peer to peer), + kjer se podatki vseh uporabnikov enakomerno porazdelijo med vozlišča in so tako decentralizirani in preprosto dostopni članom omrežja. + Ker se podatki izmenjujejo znotraj omrežja na vozliščih, + ki z izvorom in destinacijo podatkov niso povezani, + jih lahko vozlišča v velikih količinah shranjujejo za potrebe statistične analize omrežja. +\end_layout + +\begin_layout Abstract +V raziskavi preverimo praktično zmožnost pridobivanja velike količine podatkov v omrežju BitTorrent za P2P izmenjavo datotek, + nato še analiziramo pridobljene podatke. + Vsaka poizvedba po seznamu imetnikov datotek vsebuje ključ podatka v DHT in se prenese preko okoli +\begin_inset Formula $\log_{2}n$ +\end_inset + + vozlišč, + kjer je +\begin_inset Formula $n$ +\end_inset + + število vseh uporabnikov v omrežju. + Ker vsaka poizvedba obišče tako veliko število vozlišč, + lahko med poizvedbo eno drugače nepovezano vozlišče prejme veliko obstoječih ključev v omrežju, + ki jih lahko uporabi za prenos metapodatkov v omrežju BitTorrent. +\end_layout + +\begin_layout Abstract +Osredotočili smo se le na na pridobivanje metapodatkov v omrežju BitTorrent, + samih datotek, + na katere se le-ti metapodatki sklicujejo in so v omrežju na voljo, + ker jih ponujajo drugi računalniki, + pa tako vsled tehničnih (njihove ogromne skupne velikosti) kot tudi pravnih razlogov (avtorsko zaščitena in protizakonita vsebina) nismo prenašali. + Metapodatki konceptualno sicer niso shranjeni v DHT (namesto metapodatkov o datotekah so v omrežju shranjeni seznami računalnikov, + od katerih si metapodatke lahko prenesemo), + vendar odkrivanje njihovega obstoja omogoči DHT. +\end_layout + +\begin_layout Abstract +S pridobljenimi metapodatki ugotovimo, + kateri odjemalci so najpopularnejši ter kakšna je razporeditev vsebine glede na tip datotek, + ki je na voljo preko protokola BitTorrent. +\end_layout + +\begin_layout Keywords +porazdeljena razpršilna tabela, + porazdeljeni sistemi, + omrežje P2P, +\end_layout + +\begin_layout Keywords +podatkovno rudarjenje, + BitTorrent +\end_layout + +\begin_layout Section +Introduction +\end_layout + +\begin_layout Subsection +Distribucija datotek po principu P2P +\end_layout + +\begin_layout Standard +Koncept P2P (angl. + +\shape italic +peer-to-peer +\shape default +) predstavlja alternativen način distribucije identičnih datotek večim odjemalcem. + Namesto enega strežnika, + ki iste podatke odjemalcem pošlje vsakič znova, + v omrežjih P2P za distribucijo datotek vsak odjemalec podatke tako prejema kot tudi pošilja. + Odjemalec prejeto vsebino tudi sam deli drugim te vsebine željanim odjemalcem, + s čimer razbremeni ostale odjemalce. +\end_layout + +\begin_layout Standard +Odjemalec za druge izve s pomočjo centralnega strežnika ali pa drugačnega signalizacijskega protokola. + Ker se povezujejo neposredno, + medsebojno poznajo svoje internetne naslove. +\end_layout + +\begin_layout Subsection +Protokol BitTorrent +\end_layout + +\begin_layout Standard +Od 2008 +\begin_inset CommandInset citation +LatexCommand cite +key "harrison07" +literal "false" + +\end_inset + + je eden izmed popularnejših protokolov za P2P distribucijo BitTorrent, + razvit že 2001 +\begin_inset CommandInset citation +LatexCommand cite +key "cohen01" +literal "false" + +\end_inset + +. + Zaradi razširljive zasnove ga je moč dopolnjevati — + dodajati nove funkcije. + Sprva je BitTorrent temeljil na centralnih strežnikih za koordinacijo rojev, + od leta 2005 pa z uvedbo protokola DHT lahko deluje povsem neodvisno. +\begin_inset CommandInset citation +LatexCommand cite +key "jones15" +literal "false" + +\end_inset + + +\end_layout + +\begin_layout Standard +\begin_inset Float table +placement document +alignment document +wide false +sideways false +status open + +\begin_layout Plain Layout +\noindent +\align center +\begin_inset Tabular +<lyxtabular version="3" rows="8" columns="3"> +<features tabularvalignment="middle" tabularwidth="100col%"> +<column alignment="center" valignment="top" width="0pt"> +<column alignment="center" valignment="top" width="0pt"> +<column alignment="center" valignment="top" width="0pt" varwidth="true"> +<row> +<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none"> +\begin_inset Text + +\begin_layout Plain Layout +Pojem +\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 +Angleško +\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 +Razlaga +\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 +soležnik +\end_layout + +\end_inset +</cell> +<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none"> +\begin_inset Text + +\begin_layout Plain Layout +peer +\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 +odjemni program na računalniku, + za povezavo nanj potrebujemo IP naslov in vrata +\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 +roj +\begin_inset CommandInset citation +LatexCommand cite +key "dis" +literal "false" + +\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 +swarm +\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 +več soležnikov, + ki prenašajo datoteke nekega torrenta +\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 +torrent/ +\begin_inset Newline newline +\end_inset + +metainfo +\end_layout + +\end_inset +</cell> +<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none"> +\begin_inset Text + +\begin_layout Plain Layout +torrent/ +\begin_inset Newline newline +\end_inset + +metainfo +\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 +datoteka z metapodatki datotek; + imena, + velikosti, + zgoščene vrednosti in drugo +\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 +sledilnik +\end_layout + +\end_inset +</cell> +<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none"> +\begin_inset Text + +\begin_layout Plain Layout +tracker +\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 +koordinacijski strežnik z naslovi soležnikov v rojih +\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 +košček +\end_layout + +\end_inset +</cell> +<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none"> +\begin_inset Text + +\begin_layout Plain Layout +piece +\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 +delček datoteke konstantne dolžine +\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 +infohash +\end_layout + +\end_inset +</cell> +<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none"> +\begin_inset Text + +\begin_layout Plain Layout +infohash +\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 +zgoščena vrednost serializiranih podatkov pod ključem +\family typewriter +info +\family default + v torrentu, + ki unikatno opišejo ključne metapodatke o torrentu +\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 +objavi +\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 +announce +\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 +pošiljanje obvestila v DHT ali na sledilnik, + da se odjemalec želi priključiti nekemu roju +\end_layout + +\end_inset +</cell> +</row> +</lyxtabular> + +\end_inset + + +\begin_inset Caption Standard + +\begin_layout Plain Layout +Slovar pojmov BitTorrenta +\end_layout + +\end_inset + + +\end_layout + +\end_inset + + +\end_layout + +\begin_layout Standard +Za prenos je treba poznati metapodatke o obstoječih datotekah, + ki so shranjeni v t. + i. + obliki .torrent, + strojno berljivi z bencoding serializirani datoteki. + Vsebujejo vsaj imena in poti datotek ter njihove zgoščene vrednosti, + ime torrenta, + lastnosti prenosa in velikost koščka. +\end_layout + +\begin_layout Standard +V raziskavi ne iščemo soležnikov s sledilniki in ne prenašamo datotek, + temveč samo prenašamo in analiziramo metapodatke. +\end_layout + +\begin_layout Subsection +Protokol Kademlia mainline DHT +\end_layout + +\begin_layout Standard +V BitTorrent je za iskanje soležnikov v roju uporabljen DHT (angl. + +\shape italic +distributed hash table +\shape default +), + ki odpravi odvisnost od sledilnika. +\end_layout + +\begin_layout Standard +\begin_inset Float table +placement document +alignment document +wide false +sideways false +status open + +\begin_layout Plain Layout +\noindent +\align center +\begin_inset Tabular +<lyxtabular version="3" rows="6" columns="3"> +<features tabularvalignment="middle" tabularwidth="100col%"> +<column alignment="center" valignment="top" width="0pt"> +<column alignment="center" valignment="top" width="0pt"> +<column alignment="center" valignment="top" width="0pt" varwidth="true"> +<row> +<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none"> +\begin_inset Text + +\begin_layout Plain Layout +Pojem +\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 +Angleško +\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 +Razlaga +\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 +vozlišče +\begin_inset CommandInset citation +LatexCommand cite +key "dis" +literal "false" + +\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 +node +\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 +odjemni program na računalniku +\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 +usmerjevalna +\begin_inset Newline newline +\end_inset + +tabela +\begin_inset CommandInset citation +LatexCommand cite +key "dis" +literal "false" + +\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 +routing +\begin_inset Newline newline +\end_inset + +table +\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 +seznam vozlišč (IP, + vrata, + ID), + ki ga hrani posamezno vozlišče +\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 +ID vozlišča +\end_layout + +\end_inset +</cell> +<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none"> +\begin_inset Text + +\begin_layout Plain Layout +node ID +\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 +160-bitna ob zagonu generirana naključna vozlišču pripadajoča številka +\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 +merilo +\begin_inset Newline newline +\end_inset + +razdalje +\end_layout + +\end_inset +</cell> +<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none"> +\begin_inset Text + +\begin_layout Plain Layout +distance +\begin_inset Newline newline +\end_inset + +metric +\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 +funkcija (XOR), + ki izrazi razdaljo med vozliščema kot 160-bitno številko +\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 +koš +\begin_inset CommandInset citation +LatexCommand cite +key "dis" +literal "false" + +\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 +bucket +\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 +element usmerjevalne tabele, + ki glede na merilo razdalje vsebuje bližnja vozlišča +\end_layout + +\end_inset +</cell> +</row> +</lyxtabular> + +\end_inset + + +\begin_inset Caption Standard + +\begin_layout Plain Layout +Slovar pojmov Kademlie. + Slovenski prevodi niso ustaljeni. +\end_layout + +\end_inset + + +\end_layout + +\end_inset + + +\end_layout + +\begin_layout Standard +Na visokem nivoju gre za abstraktno razpršilno tabelo, + shranjeno porazdeljeno na velikem omrežju vozlišč. + Podpira naslednji operaciji +\begin_inset CommandInset citation +LatexCommand cite +key "norberg08" +literal "false" + +\end_inset + +: +\end_layout + +\begin_layout Paragraph + +\family typewriter +get_peers +\end_layout + +\begin_layout Standard +Vrne seznam soležnikov (IP naslov in vrata) za torrent, + opisan z njegovim infohashom. +\end_layout + +\begin_layout Paragraph + +\family typewriter +announce +\begin_inset Note Note +status open + +\begin_layout Plain Layout +qbittorrent pravi sporoči, + v deluge in transmission nisem našel prevoda +\end_layout + +\end_inset + + +\end_layout + +\begin_layout Standard +V seznam soležnikov za torrent, + opisan z njegovim infohashom, + vstavi IP naslov in vrata pošiljatelja zahteve. +\end_layout + +\begin_layout Standard +V raziskavi s sodelovanjem v DHT prestrezamo obstoječe ključe v razpršilni tabeli, + z njimi pridobimo soležnike, + od katerih prenesemo metapodatke o torrentih za kasnejšo analizo. +\end_layout + +\begin_layout Standard +\begin_inset Float figure +placement document +alignment document +wide false +sideways false +status open + +\begin_layout Plain Layout +\begin_inset Graphics + filename /root/projects/sola-gimb-4/inf/rn/predst/dht.svg + width 100col% + +\end_inset + + +\end_layout + +\begin_layout Plain Layout +\begin_inset Caption Standard + +\begin_layout Plain Layout +Shematski prikaz DHT +\end_layout + +\end_inset + + +\end_layout + +\begin_layout Plain Layout + +\end_layout + +\end_inset + + +\end_layout + +\begin_layout Section +Opis standardov +\end_layout + +\begin_layout Paragraph +Serializacija bkodiranje +\family typewriter +(bencoding) +\end_layout + +\begin_layout Standard +V BEP-0003 +\begin_inset CommandInset citation +LatexCommand cite +key "cohen08" +literal "false" + +\end_inset + + je opisan bencoding. + Z njim je serializirana večina struktur BitTorrenta in Kademlie. + Bkodiranje je podobno bolj znanemu JSONu +\begin_inset CommandInset citation +LatexCommand cite +key "pezoa2016foundations" +literal "false" + +\end_inset + + — + vsebuje štiri podatkovne tipe: + niz, + število, + seznam in slovar. +\end_layout + +\begin_layout Paragraph +Datoteka metainfo/.torrent +\end_layout + +\begin_layout Standard +Za distribucijo vsebine s protokolom BitTorrent ustvarimo .torrent datoteko, + bkodiran slovar z metapodatki, + nujnimi za prenos datotek. + Za raziskavo so pomembni metapodatki pod ključem +\family typewriter +info +\family default +: +\begin_inset CommandInset citation +LatexCommand cite +key "cohen08" +literal "false" + +\end_inset + + +\end_layout + +\begin_layout Itemize + +\family typewriter +private +\family default +: + prepoved objavljanja preko DHT, + le preko sledilnikov (ti torrenti niso zajeti v raziskavi) +\begin_inset CommandInset citation +LatexCommand cite +key "harrison08" +literal "false" + +\end_inset + + +\end_layout + +\begin_layout Itemize + +\family typewriter +name +\family default +: + ime torrenta oz. + datoteke za enodatotečne torrente +\end_layout + +\begin_layout Itemize + +\family typewriter +piece length +\family default +: + velikost koščka — + datoteke so spojene skupaj in razdeljene na enako velike koščke +\end_layout + +\begin_layout Itemize + +\family typewriter +pieces +\family default +: + niz dolžine +\begin_inset Formula $20n$ +\end_inset + + ( +\begin_inset Formula $n$ +\end_inset + + je število koščkov) s SHA-1 vrednostmi koščkov +\begin_inset CommandInset citation +LatexCommand cite +key "Eastlake2001" +literal "false" + +\end_inset + + +\end_layout + +\begin_layout Itemize + +\family typewriter +length +\family default +: + dolžina datoteke za enodatotečne torrente +\end_layout + +\begin_layout Itemize + +\family typewriter +files +\family default +: + seznam datotek v večdatotečnem torrentu — + vsaka datoteka je predstavljena kot slovar z +\family typewriter +length +\family default + in +\family typewriter +path +\family default +. +\end_layout + +\begin_layout Standard +Kdor pozna infohash, + lahko od soležnika prenese metainfo in posledično tudi pripadajoče datoteke. + Roj najde in se vanj vključi s poizvedbo v DHT, + saj je infohash ključ v tej razpršilni tabeli +\begin_inset CommandInset citation +LatexCommand cite +key "hazel08" +literal "false" + +\end_inset + +. + Infohash običajno oblikujemo v t. + i. + magnetno povezavo (magnet URI): + +\family typewriter + magnet:?dn= +\series bold +ime torrenta +\series default +&xt=urn:btih: +\series bold +infohash +\end_layout + +\begin_layout Standard +Druga različica BitTorrenta ima drugačno metainfo strukturo s podobnimi podatki. + Uporablja SHA-256 in namesto +\family typewriter +pieces +\family default + uporablja +\family typewriter +merkle hash tree +\family default + +\begin_inset CommandInset citation +LatexCommand cite +key "v2" +literal "false" + +\end_inset + + za zgoščene vrednosti datotek. +\end_layout + +\begin_layout Paragraph +Graf DHT +\end_layout + +\begin_layout Standard +DHT vzdržuje sezname soležnikov v roju vseh obstoječih torrentov. + Vozlišča komunicirajo preko UDP in so del velikega usmerjenega grafa, + vsako s +\begin_inset Formula $K\log_{2}n$ +\end_inset + + (konstanta +\begin_inset Formula $K=8$ +\end_inset + +, + +\begin_inset Formula $n$ +\end_inset + + je število vseh vozlišč na svetu) povezavami — + vpisi v usmerjevalno tabelo. +\end_layout + +\begin_layout Standard +Vozlišče skrbi za urejeno usmerjevalno tabelo dosegljivih +\begin_inset Foot +status open + +\begin_layout Plain Layout +dvosmerna komunikacija zaradi NAT ni samoumevna +\end_layout + +\end_inset + + vozlišč v koših; + +\begin_inset Formula $i$ +\end_inset + +ti koš hrani do +\begin_inset Formula $K$ +\end_inset + + med +\begin_inset Formula $2^{i}$ +\end_inset + + in +\begin_inset Formula $2^{i-1}$ +\end_inset + + po merilu XOR oddaljenih vozlišč, + torej je shranjenih veliko bližnjih in malo zelo oddaljenih vozlišč. +\begin_inset CommandInset citation +LatexCommand cite +key "maymounkov2002kademlia" +literal "false" + +\end_inset + + +\end_layout + +\begin_layout Paragraph +Poizvedbe po grafu +\end_layout + +\begin_layout Standard +Sprehod po grafu med poljubnima vozliščema je torej dolg v povprečju +\begin_inset Formula $\log n$ +\end_inset + + ( +\begin_inset Formula $n$ +\end_inset + + kot prej). + Roj torrenta z infohashom +\begin_inset Formula $x$ +\end_inset + + je shranjen na vozliščih z ID blizu +\begin_inset Formula $x$ +\end_inset + +, + tedaj ima poizvedba po soležnikih/objavljanje soležnika časovno kompleksnost +\begin_inset Formula $O\left(\log n\right)$ +\end_inset + +. + Za pridobitev seznama soležnikov torrenta pošljemo bkodiran UDP paket tipa +\family typewriter +get_peers +\family default + +\begin_inset Formula $t$ +\end_inset + + +\begin_inset Foot +status open + +\begin_layout Plain Layout +odvisno od implementacije +\end_layout + +\end_inset + + vozliščem iz usmerjevalne tabele, + ki so blizu infohasha. + Pozvana vozlišča odgovorijo s seznamom do +\begin_inset Formula $K$ +\end_inset + + temu infohashu najbližjih vozlišč in seznamom soležnikov za ta torrent, + če ga imajo. + Novodobljenim vozliščem spet pošljemo poizvedbo +\family typewriter +get_peers +\family default + in postopek nadaljujemo, + dokler ne najdemo nekaj infohashu najbližjih vozlišč. + V tista pošiljamo objave in od njih še naprej prejemamo informacije o roju. +\end_layout + +\begin_layout Standard +\begin_inset Float figure +placement document +alignment document +wide false +sideways false +status open + +\begin_layout Plain Layout +\align center +\begin_inset Graphics + filename Dht_example_SVG.svg + width 50col% + +\end_inset + + +\end_layout + +\begin_layout Plain Layout +\begin_inset Caption Standard + +\begin_layout Plain Layout +\begin_inset CommandInset label +LatexCommand label +name "fig:Usmerjevalna-tabela-za" + +\end_inset + +Usmerjevalna tabela za vozlišče 110 (koši so osenčeni) +\end_layout + +\end_inset + + +\end_layout + +\begin_layout Plain Layout + +\end_layout + +\end_inset + + +\end_layout + +\begin_layout Section +Metode +\end_layout + +\begin_layout Standard +Vsaka poizvedba obišče +\begin_inset Formula $\log n$ +\end_inset + + vozlišč, + torej vsako vozlišče v DHT prejema ogromno ključev — + infohashov. + V raziskavi v C spišemo program travnik, + nepopolno implementacijo odjemalca BitTorrent s poudarkom na DHT. + Osredotočimo se na zajem metapodatkov iz ključev, + ki jih prejmemo s sodelovanjem v omrežju. +\end_layout + +\begin_layout Standard +Ko vozlišče, + na katerem teče travnik, + prejme paket z infohashom, + ga doda v seznam željenih torrentov. + Neprestano posodablja roje torrentov znanih infohashov in se povezuje na soležnike iz njih, + dokler mu ne uspe prenesti metapodatkov torrenta oziroma dokler ne obupa/preteče 256 sekundni TTL torrenta. + Izdeluje .torrent datoteke z najdenimi metapodatki in internetnim naslovom ter ime programske opreme soležnika, + od katerega je metapodatke prejel. + Ne objavlja se v roj, + ker ne redistribuira niti metapodatkov niti datotek. +\end_layout + +\begin_layout Standard +Da program prvič začne sodelovati z omrežjem — + da ga sosednja vozlišča vpišejo v svoje usmerjevalne tabele — + prenese metapodatke vgrajenega torrenta +\family typewriter +Big Buck Bunny +\family default +. +\end_layout + +\begin_layout Standard +Za implementacijo spišemo knjižnico za bkodiranje in bdekodiranje, + knjižnico za DHT in nekaj funkcij za prenos metapodatkov od soležnikov preko TCP. +\end_layout + +\begin_layout Standard +Za obdelavo dobljenih torrent datotek uporabimo Jupyter Notebook +\begin_inset CommandInset citation +LatexCommand cite +key "Kluyver2016jupyter" +literal "false" + +\end_inset + + in spišemo preprosto knjižnico za razčlenjevanje .torrent metainfo datotek, + ki jih generira travnik. +\end_layout + +\begin_layout Section +Rezultati +\end_layout + +\begin_layout Subsection +Zajem +\end_layout + +\begin_layout Standard +Podatke smo zajemali iz različnih lokacij in v različnih časovnih obdobjih: +\end_layout + +\begin_layout Itemize +januarja in februarja 2023: +\end_layout + +\begin_deeper +\begin_layout Itemize +16 dni: + domači optični priključek v Sloveniji (T-2): + 47863 torrentov (3084321 datotek, + 259 TiB, + 29 sekund na torrent) +\begin_inset Note Note +status open + +\begin_layout Plain Layout +travnik +\end_layout + +\end_inset + + +\end_layout + +\begin_layout Itemize +31 dni VPS v Grčiji (grNet) +\begin_inset Note Note +status open + +\begin_layout Plain Layout +1 +\end_layout + +\end_inset + +: + +\family roman +\series medium +\shape up +\size normal +\emph off +\nospellcheck off +\bar no +\strikeout off +\xout off +\uuline off +\uwave off +\noun off +\color none +412846 +\family default +\series default +\shape default +\size default +\emph default +\nospellcheck default +\bar default +\strikeout default +\xout default +\uuline default +\uwave default +\noun default +\color inherit + torrentov (17101702 datotek, + 1881 TiB, + 6 sekund na torrent) +\begin_inset Note Note +status open + +\begin_layout Plain Layout +okeanos +\end_layout + +\end_inset + + +\begin_inset Note Note +status open + +\begin_layout Plain Layout +XX dni VPS v Grčiji (grNet) 2: + 342220 torrentov () +\begin_inset Note Note +status open + +\begin_layout Plain Layout +oliwerix +\end_layout + +\end_inset + + +\end_layout + +\end_inset + + +\end_layout + +\end_deeper +\begin_layout Itemize +5 dni junija 2024 na domačem optičnem priključku v Sloveniji (T-2): + 62110 torrentov (3725125 datotek, + 345 TiB, + 7 sekund na torrent) +\begin_inset Note Note +status open + +\begin_layout Plain Layout +2024b +\end_layout + +\end_inset + + +\end_layout + +\begin_layout Subsubsection +Primer strukture torrent datoteke z metapodatki +\end_layout + +\begin_layout Standard +Spodaj je iz bencoding v JSON pretvorjena metainfo datoteka prevzetega torrenta z infohashom +\family typewriter +696802a16728636cd72617e4cd7b64e3ca314e71 +\family default +. +\end_layout + +\begin_layout Standard +\begin_inset ERT +status open + +\begin_layout Plain Layout + + +\backslash +lstinputlisting[language=json,firstnumber=1, + numbers=none, + breaklines=true, + basicstyle= +\backslash +tiny]{../../../../sola-gimb-4/inf/rn/predst/torrent.json} +\end_layout + +\end_inset + + +\end_layout + +\begin_layout Subsection +Analiza +\end_layout + +\begin_layout Subsubsection +Odjemalci, + od katerih so bili prejeti torrenti +\end_layout + +\begin_layout Standard +Imenom programom odstranimo različico in jim ročno normaliziramo ime +\begin_inset Foot +status open + +\begin_layout Plain Layout +\begin_inset ERT +status open + +\begin_layout Plain Layout + + +\backslash +textmu Torrent +\end_layout + +\end_inset + + se sicer pojavi dvakrat, + enkrat ima znak mikro, + enkrat pa grško črko mu. + Unicode namreč ta dva znaka, + ki sicer izgledata identično, + hrani pod dvema različnima kodama. +\end_layout + +\end_inset + + ter prikažemo njihovo gostoto v populaciji. +\begin_inset CommandInset citation +LatexCommand cite +key "Hunter:2007" +literal "false" + +\end_inset + + +\end_layout + +\begin_layout Standard +\begin_inset Float figure +placement document +alignment document +wide false +sideways false +status open + +\begin_layout Plain Layout +\begin_inset Graphics + filename /root/projects/sola-gimb-4/inf/rn/dok/odjemalci_1_ods.png + width 100col% + +\end_inset + + +\end_layout + +\begin_layout Plain Layout +\begin_inset Caption Standard + +\begin_layout Plain Layout +Reprezentacija odjemalcev, + ki predstavljajo vsaj en odstotek populacije, + na logaritemski skali +\end_layout + +\end_inset + + +\end_layout + +\begin_layout Plain Layout + +\end_layout + +\end_inset + + +\end_layout + +\begin_layout Subsubsection +Različice odjemalca +\family typewriter +qBittorrent +\family default + skozi čas +\end_layout + +\begin_layout Standard +Primerjava porazdelitve različic po zgornji analizi najpopularnejšega odjemalca na +\begin_inset Formula $\log_{10}$ +\end_inset + + skali pokaže višanje različic skozi čas. + +\begin_inset Note Note +status open + +\begin_layout Plain Layout +Različice smo razvrstili s pythonsko funkcijo +\family typewriter +packaging.version.Version +\family default +. +\end_layout + +\end_inset + + V obeh letih smo prejeli torrente od skupno 88 različnih inačic qBittorrenta. + V 2023 smo največ torrentov prejeli od odjemalcev različice 4.5.0, + v 2024 pa od odjemalcev različice 4.6.3. + +\begin_inset Float figure +placement document +alignment document +wide false +sideways false +status open + +\begin_layout Plain Layout +\begin_inset Note Note +status open + +\begin_layout Plain Layout +\begin_inset Graphics + filename /root/projects/r/šola/članki/dht/verzije2324.png + width 100col% + +\end_inset + + +\end_layout + +\end_inset + + +\begin_inset Graphics + filename /root/projects/r/šola/članki/dht/verzije2324promil.png + width 100col% + +\end_inset + + +\end_layout + +\begin_layout Plain Layout +\begin_inset Caption Standard + +\begin_layout Plain Layout +Primerjava distribucij različic odjemalca +\family typewriter +qBittorrent +\family default + med 2023 (plavo) in 2024 (roza), + ki predstavljajo vsaj promil populacije (delež). +\end_layout + +\end_inset + + +\end_layout + +\end_inset + + +\end_layout + +\begin_layout Subsubsection +Geolokacija IP naslovov odjemalcev +\end_layout + +\begin_layout Standard +Z uporabo podatkovne zbirke MaxMind GeoLite2 +\begin_inset CommandInset citation +LatexCommand cite +key "maxmindgeoip2" +literal "false" + +\end_inset + + IP naslovom, + od katerih smo prejeli torrente, + določimo izvirno državo. +\end_layout + +\begin_layout Standard +\begin_inset Float figure +placement document +alignment document +wide false +sideways false +status open + +\begin_layout Plain Layout +\begin_inset Graphics + filename countries_procent.png + width 100col% + +\end_inset + + +\end_layout + +\begin_layout Plain Layout +\begin_inset Caption Standard + +\begin_layout Plain Layout +Reprezentativnost držav, + iz katerih smo prenesli metainfo, + na linearni skali. + Prikazane so le države, + iz katerih izvira vsaj odstotek populacije. +\end_layout + +\end_inset + + +\end_layout + +\begin_layout Plain Layout + +\end_layout + +\end_inset + + +\end_layout + +\begin_layout Subsubsection +Predstavnost ključev v prejetih slovarjih +\family typewriter +info +\end_layout + +\begin_layout Standard +Poleg standardnih obveznih nekateri torrenti vsebujejo tudi dodatne metapodatke v slovarju info. + Pogostost slednjih prikazuje spodnji grafikon. +\end_layout + +\begin_layout Standard +\begin_inset Float figure +placement document +alignment document +wide false +sideways false +status open + +\begin_layout Plain Layout +\begin_inset Graphics + filename /root/projects/sola-gimb-4/inf/rn/dok/vsi_ključi.png + width 100col% + +\end_inset + + +\end_layout + +\begin_layout Plain Layout +\begin_inset Caption Standard + +\begin_layout Plain Layout +Reprezentacija ključev v slovarju +\family typewriter +info +\family default + na logaritemski skali +\end_layout + +\end_inset + + +\end_layout + +\begin_layout Plain Layout + +\end_layout + +\end_inset + + +\end_layout + +\begin_layout Subsubsection +Tipi datotek, + ki se prenašajo v torrentih +\end_layout + +\begin_layout Standard +Iz končnice datoteke izvemo tip datoteke. + Vsakemu torrentu priredimo reprezentativen tip, + tisti, + ki po velikosti prevladuje v torrentu. + Glede na število torrentov z nekim reprezentativnim tipom kvantificiramo pogostost tega datotečnega tipa za tipe, + ki zavzemajo vsaj promil populacije. +\end_layout + +\begin_layout Standard +\begin_inset Float figure +placement document +alignment document +wide false +sideways false +status open + +\begin_layout Plain Layout +\begin_inset Graphics + filename /root/projects/sola-gimb-4/inf/rn/dok/reprezentativni_.1_ods.png + width 100col% + +\end_inset + + +\end_layout + +\begin_layout Plain Layout +\begin_inset Caption Standard + +\begin_layout Plain Layout +Reprezentativni tipi torrentov, + ki predstavljajo vsaj en promil populacije, + na logaritemski skali +\end_layout + +\end_inset + + +\end_layout + +\begin_layout Plain Layout + +\end_layout + +\end_inset + + +\end_layout + +\begin_layout Standard +Razvidno je, + da je večina torrentov namenjena prenosu videovsebin, + zvočnih datotek in stisnjenih arhivov. +\end_layout + +\begin_layout Standard +Če bi za določilo pojavnosti tipa uporabili število datotek, + bi prevladovali tipi vsebin, + ki so ponavadi preneseni kot kopica datotek, + denimo slike (diagram v prilogi na sliki +\begin_inset CommandInset ref +LatexCommand ref +reference "fig:Pojavnost-tipa-kot" +plural "false" +caps "false" +noprefix "false" +nolink "false" + +\end_inset + +), + če pa bi za določilo pojavnosti tipa uporabili velikost datotek tega tipa, + pa bi prevladovali tisti tipi, + ki zasedajo več prostora. + V tem primeru bi npr. + videovsebine zaradi svoje velikosti občutno presegale digitalne knjige (diagram v prilogi na sliki +\begin_inset CommandInset ref +LatexCommand ref +reference "fig:Pojavnost-tipa-kot-velikost" +plural "false" +caps "false" +noprefix "false" +nolink "false" + +\end_inset + +). +\end_layout + +\begin_layout Standard +\begin_inset Note Note +status open + +\begin_layout Subsubsection +\begin_inset CommandInset label +LatexCommand label +name "subsec:Porazdeljenost-infohashov" + +\end_inset + +Porazdeljenost infohashov +\end_layout + +\begin_layout Plain Layout +Zaradi delovanja poizvedb v DHT pričakujemo, + da je porazdelitev infohashov po celotnem sprektru števil z intervala +\begin_inset Formula $\left[0,2^{160}-1\right]$ +\end_inset + + gostejša okoli IDja vozlišča, + ki ga je med prenašanjem imelo naše iskalno vozlišče. + IDje smo izbrali naključno na vsaki merilni lokaciji in jih med meritvijo tudi nekajkrat zamenjali, + zato spodnjem grafikonu opazimo vrhove tam, + kjer so bili naši IDji med zajemom podatkov. +\end_layout + +\begin_layout Plain Layout +\begin_inset Float figure +placement document +alignment document +wide false +sideways false +status open + +\begin_layout Plain Layout +\begin_inset Caption Standard + +\begin_layout Plain Layout +Razporeditev pridobljenih infohashov na spektru vseh infohashov +\end_layout + +\end_inset + + +\end_layout + +\end_inset + + +\end_layout + +\end_inset + + +\end_layout + +\begin_layout Subsection +Diskusija +\end_layout + +\begin_layout Paragraph +Statistična kvaliteta vzorca +\end_layout + +\begin_layout Standard +\begin_inset Note Note +status open + +\begin_layout Plain Layout +V razdelku +\begin_inset CommandInset ref +LatexCommand ref +reference "subsec:Porazdeljenost-infohashov" +plural "false" +caps "false" +noprefix "false" +nolink "false" + +\end_inset + + opazimo neenakomerno porazdeljenost infohashov, + kar je posledica načina vzorčenja torrentov. + Vsa populacija ima namreč zaradi delovanja zgoščevalne funkcije SHA-1 homogeno porazdeljene infohashe. + Kljub temu trdimo, + da pridobljen vzorec torrentov dobro predstavlja celotno populacijo. + +\end_layout + +\end_inset + +Zaradi lastnosti uniformne porazdelitve zgoščevalne funkcije +\begin_inset CommandInset citation +LatexCommand cite +key "wikihashuniformity" +literal "false" + +\end_inset + + mesto infohasha na intervalu vseh možnih infohashov ni odvisno od metapodatkov. + Kot posledico načina vzorčenja z DHT pričakujemo, + da je porazdelitev infohashov prejetih torrentov po celotnem sprektru števil z intervala +\begin_inset Formula $\left[0,2^{160}-1\right]$ +\end_inset + + gostejša okoli IDja vozlišča, + ki ga je med prenašanjem imelo naše iskalno vozlišče. + IDje smo tekom raziskave izbrali naključno na vsaki merilni lokaciji in jih med meritvijo tudi nekajkrat zamenjali. + Kljub temu je vsled nepovezanosti vsebine in infohasha vzorec še vedno statistično reprezentativen. + Zajem ne more biti pristranski glede na metapodatke, + ker nikjer v procesu zajema ne obravnavamo torrentov glede na metainfo, + temveč le glede na infohash. +\end_layout + +\begin_layout Paragraph +Težava z zajemom podatkov +\end_layout + +\begin_layout Standard +Vsled majhne velikosti UDP paketov DHT glavno ozko grlo pri zajemu predstavlja število paketov, + ki jih mrežna oprema lahko posreduje v sekundi. + Domača optična povezava dopušča do okoli 2000 paketov na sekundo na naključno porazdeljene IP naslove, + odjemno mesto na VPS pa je imelo to omejitev veliko višjo, + zato smo tam v istem časovnem intervalu shranili veliko več torrent datotek. +\end_layout + +\begin_layout Paragraph +Etičnost in legitimnost rudarjenja podatkov +\end_layout + +\begin_layout Standard +Čeprav gre za izrazito osebne podatke, + se morajo uporabniki BitTorrent omrežja zavedati, + da so njihovi prenosi +\shape italic +a priori +\shape default + javni, + tudi če jih nihče aktivno ne zajema. + Nekateri BitTorrent odjemalci uporabnike ob prvem zagonu o tem obvestijo. +\end_layout + +\begin_layout Section +Priloge +\end_layout + +\begin_layout Standard +Izvorna koda programa travnik in ipynb datotek za analizo podatkov je na voljo na +\begin_inset CommandInset href +LatexCommand href +name "http://ni.šijanec.eu./sijanec/travnik" +target "http://ni.šijanec.eu./sijanec/travnik" +literal "false" + +\end_inset + +. +\end_layout + +\begin_layout Standard +Korpus zajetih metapodatkov je na voljo na +\begin_inset CommandInset href +LatexCommand href +name "rsync://b.sjanec.eu./travnik" +target "rsync://b.sijanec.eu./travnik" +type "other" +literal "false" + +\end_inset + +. +\end_layout + +\begin_layout Standard +\begin_inset Float figure +placement document +alignment document +wide false +sideways false +status open + +\begin_layout Plain Layout +\begin_inset Graphics + filename po_številu_datotek.png + width 100col% + +\end_inset + + +\end_layout + +\begin_layout Plain Layout +\begin_inset Caption Standard + +\begin_layout Plain Layout +\begin_inset CommandInset label +LatexCommand label +name "fig:Pojavnost-tipa-kot" + +\end_inset + +Pojavnost tipa kot število datotek +\end_layout + +\end_inset + + +\end_layout + +\begin_layout Plain Layout + +\end_layout + +\end_inset + + +\end_layout + +\begin_layout Standard +\begin_inset Float figure +placement document +alignment document +wide false +sideways false +status open + +\begin_layout Plain Layout +\begin_inset Graphics + filename /root/projects/sola-gimb-4/inf/rn/dok/po_velikosti_datotek.png + width 100col% + +\end_inset + + +\end_layout + +\begin_layout Plain Layout +\begin_inset Caption Standard + +\begin_layout Plain Layout +\begin_inset CommandInset label +LatexCommand label +name "fig:Pojavnost-tipa-kot-velikost" + +\end_inset + +Pojavnost tipa kot velikost datotek (tipi, + ki zavzamejo vsaj odstotek populacije) +\end_layout + +\end_inset + + +\end_layout + +\begin_layout Plain Layout + +\end_layout + +\end_inset + + +\end_layout + +\begin_layout Standard +\begin_inset CommandInset bibtex +LatexCommand bibtex +btprint "btPrintCited" +bibfiles "/root/projects/r/šola/citati" +options "IEEEtran" +encoding "default" + +\end_inset + + +\end_layout + +\begin_layout Section* +Viri slik +\end_layout + +\begin_layout Itemize +Slika +\begin_inset CommandInset ref +LatexCommand ref +reference "fig:Usmerjevalna-tabela-za" +plural "false" +caps "false" +noprefix "false" +nolink "false" + +\end_inset + +: + Limaner: + nespremenjena, + izvorna pod CC BY-SA +\end_layout + +\begin_layout Section* +Dovoljenje +\end_layout + +\begin_layout Standard +\begin_inset ERT +status open + +\begin_layout Plain Layout + + +\backslash +doclicenseImage[imagewidth=2cm] +\end_layout + +\end_inset + + +\begin_inset Note Note +status open + +\begin_layout Plain Layout +TODO: + preštej datoteke v oliwerix, + še enkrat nariši vse grafe upoštevajoč vse torrente, + primerjaj verzije med travnik in 2024b +\end_layout + +\end_inset + + +\end_layout + +\end_body +\end_document diff --git a/šola/članki/dht/makefile b/šola/članki/dht/makefile new file mode 100644 index 0000000..5be42f0 --- /dev/null +++ b/šola/članki/dht/makefile @@ -0,0 +1,13 @@ +all: Dht_example_SVG.svg bt.svg + +Dht_example_SVG.svg: + wget https://upload.wikimedia.org/wikipedia/commons/6/63/Dht_example_SVG.svg + +infohash.png: # tole je treba pognati na strežniku b + ./infohash.sh + +bt.svg: + wget -O- https://upload.wikimedia.org/wikipedia/commons/0/09/BitTorrent_network.svg | sed -e s/Downloader/Soležnik/ -e s/Uploader/Soležnik/ > bt.svg + +clean: + rm *.svg *.png *.tsv |