15 files changed, 22185 insertions, 59 deletions

diff --git a/šola/ana1/kolokvij2.lyx b/šola/ana1/kolokvij2.lyx
index a057288..4d94e99 100644
--- a/šola/ana1/kolokvij2.lyx
+++ b/šola/ana1/kolokvij2.lyx
@@ -1,5 +1,5 @@
-#LyX 2.3 created this file. For more info see http://www.lyx.org/
-\lyxformat 544
+#LyX 2.4 created this file. For more info see https://www.lyx.org/
+\lyxformat 620
 \begin_document
 \begin_header
 \save_transient_properties true
@@ -17,11 +17,11 @@
 enumitem
 theorems-ams
 \end_modules
-\maintain_unincluded_children false
+\maintain_unincluded_children no
 \language slovene
 \language_package default
-\inputencoding auto
-\fontencoding global
+\inputencoding auto-legacy
+\fontencoding auto
 \font_roman "default" "default"
 \font_sans "default" "default"
 \font_typewriter "default" "default"
@@ -29,7 +29,9 @@ theorems-ams
 \font_default_family default
 \use_non_tex_fonts false
 \font_sc false
-\font_osf 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
@@ -63,7 +65,9 @@ theorems-ams
 \suppress_date false
 \justification false
 \use_refstyle 1
+\use_formatted_ref 0
 \use_minted 0
+\use_lineno 0
 \index Index
 \shortcut idx
 \color #008000
@@ -86,42 +90,20 @@ theorems-ams
 \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
-List s formulami za 2.
- kolokvij Analize 1
-\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 Standard
 \begin_inset ERT
 status open
@@ -161,15 +143,18 @@ begin{multicols}{2}
 \begin_inset Formula $\log_{a}1=0$
 \end_inset
 
-, 
+,
+ 
 \begin_inset Formula $\log_{a}a=1$
 \end_inset
 
-, 
+,
+ 
 \begin_inset Formula $\log_{a}a^{x}=x$
 \end_inset
 
-, 
+,
+ 
 \begin_inset Formula $a^{\log_{a}x}=x$
 \end_inset
 
@@ -187,7 +172,8 @@ begin{multicols}{2}
 \begin_inset Formula $D=b^{2}-4ac$
 \end_inset
 
-, 
+,
+ 
 \begin_inset Formula $x_{1,2}=\frac{-b\pm\sqrt{D}}{2a}$
 \end_inset
 
@@ -205,7 +191,8 @@ begin{multicols}{2}
 \begin_inset Formula $zw=\left(ac-bd\right)+\left(ad+bc\right)i$
 \end_inset
 
-, 
+,
+ 
 \begin_inset Formula $\vert zw\vert=\vert z\vert\vert w\vert$
 \end_inset
 
@@ -241,7 +228,8 @@ begin{multicols}{2}
 \begin_inset Formula $z^{2}=a^{2}+2abi-b^{2}$
 \end_inset
 
-, 
+,
+ 
 \begin_inset Formula $z^{3}=a^{3}-3ab^{2}+\left(3a^{2}b-b^{3}\right)i$
 \end_inset
 
@@ -259,7 +247,8 @@ begin{multicols}{2}
 \begin_inset Formula $z^{n}=r^{3}\left(\cos\left(3\phi\right)+i\sin\left(3\phi\right)\right)$
 \end_inset
 
-, 
+,
+ 
 \begin_inset Formula $\phi=\arctan\frac{\Im z}{\Re z}$
 \end_inset
 
@@ -316,7 +305,8 @@ je konv.
 \end_layout
 
 \begin_layout Standard
-Vrsta je konv., če je konv.
+Vrsta je konv.,
+ če je konv.
  njeno zap.
  delnih vsot.
 \end_layout
@@ -343,7 +333,8 @@ n+1; & q=1
 \series bold
 Primerjalni krit.
 \series default
-: 
+:
+ 
 \begin_inset Formula $\sum_{1}^{\infty}a_{k}$
 \end_inset
 
@@ -389,11 +380,13 @@ majoranta
 \series bold
 Kvocientni
 \series default
-: 
+:
+ 
 \begin_inset Formula $a_{k}>0$
 \end_inset
 
-, 
+,
+ 
 \begin_inset Formula $D_{n}\coloneqq\frac{a_{n}+1}{a_{n}}$
 \end_inset
 
@@ -419,11 +412,13 @@ Kvocientni
 \begin_inset Formula $\exists D\coloneqq\lim_{n\to\infty}D_{n}$
 \end_inset
 
-: 
+:
+ 
 \begin_inset Formula $\vert D\vert<1\Longrightarrow$
 \end_inset
 
-konv., 
+konv.,
+ 
 \begin_inset Formula $\vert D\vert>1\Longrightarrow div.$
 \end_inset
 
@@ -435,7 +430,9 @@ konv.,
 \series bold
 Korenski
 \series default
-: Kot Kvocientni, le da 
+:
+ Kot Kvocientni,
+ le da 
 \begin_inset Formula $D_{n}\coloneqq\sqrt[n]{a_{n}}$
 \end_inset
 
@@ -447,7 +444,8 @@ Korenski
 \series bold
 Leibnizov
 \series default
-: 
+:
+ 
 \begin_inset Formula $a_{n}\to0\Longrightarrow\sum_{1}^{\infty}\left(\left(-1\right)^{k}a_{k}\right)<\infty$
 \end_inset
 
@@ -476,11 +474,13 @@ Pri konv.
 \begin_inset Formula $x$
 \end_inset
 
-, pri enakomerni ni.
+,
+ pri enakomerni ni.
 \end_layout
 
 \begin_layout Standard
-Potenčna vrsta: 
+Potenčna vrsta:
+ 
 \begin_inset Formula $\sum_{j=1}^{\infty}b_{j}x^{j}$
 \end_inset
 
@@ -495,7 +495,8 @@ Potenčna vrsta:
 \end_inset
 
 abs.
- konv., 
+ konv.,
+ 
 \begin_inset Formula $\vert x\vert>R\Longrightarrow$
 \end_inset
 
@@ -719,7 +720,8 @@ divergira
 \end_layout
 
 \begin_layout Standard
-Krožnica: 
+Krožnica:
+ 
 \begin_inset Formula $\left(x-p\right)^{2}+\left(y-q\right)^{2}=r^{2}$
 \end_inset
 
@@ -727,7 +729,8 @@ Krožnica:
 \end_layout
 
 \begin_layout Standard
-Elipsa: 
+Elipsa:
+ 
 \begin_inset Formula $\frac{\left(x-p\right)^{2}}{a^{2}}+\frac{\left(y-q\right)^{2}}{b^{2}}=1$
 \end_inset
 
@@ -800,7 +803,8 @@ Odvod
 \begin_inset Formula $\frac{f'g-fg'}{g^{2}}$
 \end_inset
 
-, 
+,
+ 
 \begin_inset Formula $g\not=0$
 \end_inset
 
@@ -1260,7 +1264,8 @@ Zvezna
 \begin_inset Formula $\sup$
 \end_inset
 
-, je omejena in doseže vse funkcijske vrednosti na 
+,
+ je omejena in doseže vse funkcijske vrednosti na 
 \begin_inset Formula $\left[f\left(a\right),f\left(b\right)\right]$
 \end_inset
 
@@ -1275,7 +1280,8 @@ Zvezna
 \begin_inset Formula $I$
 \end_inset
 
-, če 
+,
+ če 
 \begin_inset Formula $\forall\varepsilon>0\exists\delta_{\left(\varepsilon\right)}>0\ni:\forall x,y\in I:\left|x-y\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(y\right)\right|<\varepsilon$
 \end_inset
 
@@ -1290,7 +1296,8 @@ Zvezna
 \begin_inset Formula $I$
 \end_inset
 
-, če 
+,
+ če 
 \begin_inset Formula $\forall\varepsilon>0\forall x\in I\exists\delta_{\left(x,\varepsilon\right)}>0\ni:\forall x,y\in I:\left|x-y\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(y\right)\right|<\varepsilon$
 \end_inset
 
diff --git a/šola/ana1/prak.lyx b/šola/ana1/prak.lyx
new file mode 100644
index 0000000..b6b21e7
--- /dev/null
+++ b/šola/ana1/prak.lyx
@@ -0,0 +1,1609 @@
+#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{siunitx}
+\usepackage{pgfplots}
+\usepackage{listings}
+\usepackage{multicol}
+\sisetup{output-decimal-marker = {,}, quotient-mode=fraction, output-exponent-marker=\ensuremath{\mathrm{3}}}
+\DeclareMathOperator{\ctg}{ctg}
+\end_preamble
+\use_default_options true
+\begin_modules
+enumitem
+theorems-ams
+\end_modules
+\maintain_unincluded_children no
+\language slovene
+\language_package default
+\inputencoding auto-legacy
+\fontencoding auto
+\font_roman "default" "default"
+\font_sans "default" "default"
+\font_typewriter "default" "default"
+\font_math "auto" "auto"
+\font_default_family default
+\use_non_tex_fonts false
+\font_sc false
+\font_roman_osf false
+\font_sans_osf false
+\font_typewriter_osf false
+\font_sf_scale 100 100
+\font_tt_scale 100 100
+\use_microtype false
+\use_dash_ligatures true
+\graphics default
+\default_output_format default
+\output_sync 0
+\bibtex_command default
+\index_command default
+\float_placement class
+\float_alignment class
+\paperfontsize default
+\spacing single
+\use_hyperref false
+\papersize default
+\use_geometry true
+\use_package amsmath 1
+\use_package amssymb 1
+\use_package cancel 1
+\use_package esint 1
+\use_package mathdots 1
+\use_package mathtools 1
+\use_package mhchem 1
+\use_package stackrel 1
+\use_package stmaryrd 1
+\use_package undertilde 1
+\cite_engine basic
+\cite_engine_type default
+\biblio_style plain
+\use_bibtopic false
+\use_indices false
+\paperorientation portrait
+\suppress_date false
+\justification false
+\use_refstyle 1
+\use_formatted_ref 0
+\use_minted 0
+\use_lineno 0
+\index Index
+\shortcut idx
+\color #008000
+\end_index
+\leftmargin 1cm
+\topmargin 1cm
+\rightmargin 1cm
+\bottommargin 2cm
+\headheight 1cm
+\headsep 1cm
+\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 Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+newcommand
+\backslash
+euler{e}
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+setlength{
+\backslash
+columnseprule}{0.2pt}
+\backslash
+begin{multicols}{2}
+\end_layout
+
+\end_inset
+
+
+\begin_inset Formula $\log_{a}1=0$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\log_{a}a=1$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\log_{a}a^{x}=x$
+\end_inset
+
+,
+ 
+\begin_inset Formula $a^{\log_{a}x}=x$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $\binom{n}{k}\coloneqq\frac{n!}{k!\left(n-k\right)!}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\log_{a}x^{n}=n\log_{a}x$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $D=b^{2}-4ac$
+\end_inset
+
+,
+ 
+\begin_inset Formula $x_{1,2}=\frac{-b\pm\sqrt{D}}{2a}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right)$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $zw=\left(ac-bd\right)+\left(ad+bc\right)i$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\vert zw\vert=\vert z\vert\vert w\vert$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $\arg\left(zw\right)=\arg z+\arg w$
+\end_inset
+
+ (kot)
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $z\overline{z}=a^{2}-\left(bi\right)^{2}=a^{2}+b^{2}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $\left(\cos\phi+i\sin\phi\right)$
+\end_inset
+
+
+\begin_inset Formula $\left(\cos\psi+i\sin\psi\right)=\cos\left(\phi+\psi\right)+i\sin\left(\phi+\psi\right)$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $z^{2}=a^{2}+2abi-b^{2}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $z^{3}=a^{3}-3ab^{2}+\left(3a^{2}b-b^{3}\right)i$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $(a+b)^{n}=\sum_{k=0}^{n}{n \choose k}ab^{n-k}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $z^{n}=r^{n}\left(\cos\left(n\phi\right)+i\sin\left(n\phi\right)\right)$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\phi=\arctan\frac{\Im z}{\Re z}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Odprta množica ne vsebuje robnih točk.
+ Zaprta vsebuje vse.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $\sin\left(x\pm y\right)=\sin x\cdot\cos y\pm\sin y\cdot\cos x$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $\cos\left(x\pm y\right)=\cos x\cdot\cos y\mp\sin y\cdot\sin x$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $\tan\left(x\pm y\right)=\frac{\tan x\pm\tan y}{1\text{\ensuremath{\mp\tan}x\ensuremath{\cdot\tan y}}}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $a_{n}$
+\end_inset
+
+je konv.
+ 
+\begin_inset Formula $\Longleftrightarrow$
+\end_inset
+
+ 
+\begin_inset Formula $\forall\varepsilon>0:\exists n_{0}\ni:\forall n,m:n_{0}<n<m\wedge\vert a_{n}-a_{m}\vert<\varepsilon$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $\euler^{1/k}\coloneqq\lim_{n\to\infty}\left(1+\frac{1}{nk}\right)^{n}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Vrsta je konv.,
+ če je konv.
+ njeno zap.
+ delnih vsot.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $s_{n}=\begin{cases}
+\frac{1-q^{n+1}}{1-q}; & q\not=1\\
+n+1; & q=1
+\end{cases}$
+\end_inset
+
+.
+ Geom.
+ vrsta konv.
+ 
+\begin_inset Formula $\Longleftrightarrow q\in\left(-1,1\right)$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Primerjalni krit.
+\series default
+:
+ 
+\begin_inset Formula $\sum_{1}^{\infty}a_{k}$
+\end_inset
+
+ konv.
+ 
+\begin_inset Formula $\wedge$
+\end_inset
+
+ 
+\begin_inset Formula $b_{k}\leq a_{k}$
+\end_inset
+
+za 
+\begin_inset Formula $k>n_{0}$
+\end_inset
+
+ 
+\begin_inset Formula $\wedge$
+\end_inset
+
+ vrsti sta navzdol omejeni 
+\begin_inset Formula $\Longrightarrow$
+\end_inset
+
+ 
+\begin_inset Formula $\sum_{1}^{\infty}b_{k}$
+\end_inset
+
+ konv.
+ 
+\begin_inset Formula $\sum_{1}^{\infty}a_{k}$
+\end_inset
+
+ rečemo 
+\shape italic
+majoranta
+\shape default
+.
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Kvocientni
+\series default
+:
+ 
+\begin_inset Formula $a_{k}>0$
+\end_inset
+
+,
+ 
+\begin_inset Formula $D_{n}\coloneqq\frac{a_{n}+1}{a_{n}}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $\forall n<n_{0}:D_{n}\in\left(0,1\right)\Longrightarrow\sum_{1}^{\infty}a_{k}<\infty$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $\forall n<n_{0}:D_{n}\geq1\Longrightarrow\sum_{1}^{\infty}a_{k}=\infty$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+Če 
+\begin_inset Formula $\exists D\coloneqq\lim_{n\to\infty}D_{n}$
+\end_inset
+
+:
+ 
+\begin_inset Formula $\vert D\vert<1\Longrightarrow$
+\end_inset
+
+konv.,
+ 
+\begin_inset Formula $\vert D\vert>1\Longrightarrow div.$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Korenski
+\series default
+:
+ Kot Kvocientni,
+ le da 
+\begin_inset Formula $D_{n}\coloneqq\sqrt[n]{a_{n}}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Leibnizov
+\series default
+:
+ 
+\begin_inset Formula $a_{n}\to0\Longrightarrow\sum_{1}^{\infty}\left(\left(-1\right)^{k}a_{k}\right)<\infty$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Absolutna konvergenca 
+\begin_inset Formula $\left(\sum_{1}^{\infty}\vert a_{n}\vert<\infty\right)$
+\end_inset
+
+ 
+\begin_inset Formula $\Longrightarrow$
+\end_inset
+
+ konvergenca
+\end_layout
+
+\begin_layout Standard
+Pri konv.
+ po točkah je 
+\begin_inset Formula $n_{0}$
+\end_inset
+
+ odvisen od 
+\begin_inset Formula $x$
+\end_inset
+
+,
+ pri enakomerni ni.
+\end_layout
+
+\begin_layout Standard
+Potenčna vrsta:
+ 
+\begin_inset Formula $\sum_{j=1}^{\infty}b_{j}x^{j}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $R^{-1}=\limsup_{k\to\infty}\sqrt[k]{\vert b_{k}\vert}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $\vert x\vert<R\Longrightarrow$
+\end_inset
+
+abs.
+ konv.,
+ 
+\begin_inset Formula $\vert x\vert>R\Longrightarrow$
+\end_inset
+
+divergira
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $\lim_{x\to a}\left(\alpha f\left(x\right)\right)=\alpha\lim_{x\to a}f\left(x\right)$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Tabular
+<lyxtabular version="3" rows="4" columns="4">
+<features tabularvalignment="middle">
+<column alignment="center" valignment="top">
+<column alignment="center" valignment="top">
+<column alignment="center" valignment="top">
+<column alignment="center" valignment="top">
+<row>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+
+\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 $\sin$
+\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 $\cos$
+\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 $\tan$
+\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 $30^{\circ}=\frac{\pi}{6}$
+\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 $\frac{1}{2}$
+\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 $\frac{\sqrt{3}}{2}$
+\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 $\frac{\sqrt{3}}{3}$
+\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 $45^{\circ}=\frac{\pi}{4}$
+\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 $\frac{\sqrt{2}}{2}$
+\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 $\frac{\sqrt{2}}{2}$
+\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
+1
+\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 $60^{\circ}=\frac{\pi}{3}$
+\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 $\frac{\sqrt{3}}{2}$
+\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 $\frac{1}{2}$
+\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 $\sqrt{3}$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+</row>
+</lyxtabular>
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Krožnica:
+ 
+\begin_inset Formula $\left(x-p\right)^{2}+\left(y-q\right)^{2}=r^{2}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Elipsa:
+ 
+\begin_inset Formula $\frac{\left(x-p\right)^{2}}{a^{2}}+\frac{\left(y-q\right)^{2}}{b^{2}}=1$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Tabular
+<lyxtabular version="3" rows="8" columns="4">
+<features tabularvalignment="middle">
+<column alignment="center" valignment="top">
+<column alignment="center" valignment="top">
+<column alignment="center" valignment="top">
+<column alignment="center" valignment="top">
+<row>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+Izraz
+\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
+Odvod
+\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
+Izraz
+\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
+Odvod
+\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 $\frac{f}{g}$
+\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 $\frac{f'g-fg'}{g^{2}}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $g\not=0$
+\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 $f\left(g\right)$
+\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 $f'\left(g\right)g'$
+\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 $\tan x$
+\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 $\cos^{-2}x$
+\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 $\cot x$
+\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 $-sin^{-2}x$
+\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 $a^{x}$
+\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 $a^{x}\text{\ensuremath{\ln a}}$
+\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 $x^{x}$
+\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 $x^{x}\left(1+\ln x\right)$
+\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 $\log_{a}x$
+\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 $\frac{1}{x\ln a}$
+\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 $f^{-1}\left(a\right)$
+\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 $\frac{1}{f'\left(f^{-1}\left(a\right)\right)}$
+\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 $\arcsin x$
+\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 $\left(1-x^{2}\right)^{-\frac{1}{2}}$
+\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 $\arccos x$
+\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 $-\left(1-x^{2}\right)^{-\frac{1}{2}}$
+\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 $\arctan x$
+\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 $\frac{1}{1+x^{2}}$
+\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 $\text{arccot\,}x$
+\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 $-\frac{1}{1+x^{2}}$
+\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 $x^{n}$
+\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 $nx^{n-1}$
+\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
+
+\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
+
+\end_layout
+
+\end_inset
+</cell>
+</row>
+</lyxtabular>
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $f''\left(I\right)>0\Leftrightarrow f$
+\end_inset
+
+ konveksna na 
+\begin_inset Formula $I$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $f''\left(I\right)<0\Leftrightarrow f$
+\end_inset
+
+ konkavna na 
+\begin_inset Formula $I$
+\end_inset
+
+
+\begin_inset Formula 
+\[
+ab>0\wedge a<b\Leftrightarrow a^{-1}>b^{-1},\quad ab<0\wedge a<b\Leftrightarrow a^{-1}<b^{-1}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula 
+\[
+\lim_{x\to0}\frac{\sin x}{x}=1\quad\quad\tan\phi=\left|\frac{k_{1}-k_{2}}{1+k_{1}k_{2}}\right|
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula 
+\[
+\lim_{x\to0}x\ln x=0
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula 
+\[
+f\text{ zv.+odv.@ }\left[a,b\right]\Rightarrow\exists\xi\in\left[a,b\right]\ni:f\left(b\right)-f\left(a\right)=f'\left(\xi\right)\left(b-a\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula 
+\[
+T_{f,a,n}\left(x\right)=\sum_{k=0}^{n}\frac{f^{\left(k\right)}\left(a\right)}{k!}\left(x-a\right)^{k}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $f\text{\ensuremath{\in C^{n+1}}}$
+\end_inset
+
+ na odprtem 
+\begin_inset Formula $I\subset\mathbb{R}\Rightarrow\forall a,x\in I\exists c\in\left(\min\left\{ a,x\right\} ,\max\left\{ a,x\right\} \right)\ni:f\left(x\right)-T_{f,a,n}\left(x\right)=R_{f,a,n}\left(x\right)=\frac{f^{\left(n+1\right)}\left(c\right)}{\left(n+1\right)!}$
+\end_inset
+
+
+\begin_inset Formula $\left(x-a\right)^{n+1}.\text{ Posledično velja tudi takale ocena:}$
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\exists M>0\forall x\in I:\left|f^{\left(n+1\right)}\right|\leq M\Rightarrow R_{f,a,n}\left(x\right)=\frac{M}{\left(n+1\right)!}\left|x-a\right|^{n+1}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula 
+\[
+R=\lim_{n\to\infty}\left|\frac{c_{n}}{c_{n+1}}\right|,\quad R=\lim_{n\to\infty}\frac{1}{\sqrt[n]{\left|c_{n}\right|}}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Zvezna 
+\begin_inset Formula $\text{f}$
+\end_inset
+
+ na zaprtem intervalu 
+\begin_inset Formula $\left[a,b\right]$
+\end_inset
+
+ doseže 
+\begin_inset Formula $\inf$
+\end_inset
+
+ in 
+\begin_inset Formula $\sup$
+\end_inset
+
+,
+ je omejena in doseže vse funkcijske vrednosti na 
+\begin_inset Formula $\left[f\left(a\right),f\left(b\right)\right]$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $f:I\to\mathbb{R}$
+\end_inset
+
+ je enakomerno zvezna na 
+\begin_inset Formula $I$
+\end_inset
+
+,
+ če 
+\begin_inset Formula $\forall\varepsilon>0\exists\delta_{\left(\varepsilon\right)}>0\ni:\forall x,y\in I:\left|x-y\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(y\right)\right|<\varepsilon$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $f:I\to\mathbb{R}$
+\end_inset
+
+ je zvezna na 
+\begin_inset Formula $I$
+\end_inset
+
+,
+ če 
+\begin_inset Formula $\forall\varepsilon>0\forall x\in I\exists\delta_{\left(x,\varepsilon\right)}>0\ni:\forall x,y\in I:\left|x-y\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(y\right)\right|<\varepsilon$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Zvezna 
+\begin_inset Formula $f$
+\end_inset
+
+ na kompaktni množici je enakomerno zvezna.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula 
+\[
+f'\left(x\right)=\lim_{x\to0}\frac{f\left(x+h\right)-f\left(x\right)}{h}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula 
+\[
+\sinh x=\frac{e^{x}-e^{-x}}{2},\quad\cosh x=\frac{e^{x}+e^{-x}}{2}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Paragraph
+Uporabne vrste
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $\sin x=\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}}{\left(2n+1\right)!}x^{2n+1}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\cos x=\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}}{\left(2n\right)!}x^{2n}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\sinh x=\sum_{n=0}^{\infty}\frac{x^{2n+1}}{\left(2n+1\right)!}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $e^{x}=\sum_{x=0}^{\infty}\frac{x^{n}}{n!}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\left(1+x\right)^{\alpha}=\sum_{n=0}^{\infty}\frac{x^{n}}{n!}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\frac{1}{1-x}=\sum_{n=0}^{\infty}x^{n}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\ln\left(1+x\right)=\sum_{n=1}^{\infty}\left(-1\right)^{n+1}\frac{x^{n}}{n}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\ln\left(1-x\right)=-\sum_{n=1}^{\infty}\frac{x^{n}}{n}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Paragraph
+Razcep racionalnih
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula 
+\[
+\frac{p\left(x\right)}{\left(x-a\right)^{3}}=\frac{A}{x-a}+\frac{B}{\left(x-1\right)^{2}}+\frac{C}{\left(x-1\right)^{3}}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\frac{p\left(x\right)}{\left(x-a\right)\left(x-b\right)^{2}}=\frac{A}{x-a}+\frac{B}{x-b}+\frac{C}{\left(x-b\right)^{2}}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\frac{p\left(x\right)}{\left(x-a\right)\left(x^{2}-b\right)}=\frac{A}{x-a}+\frac{Bx-C}{x^{2}-b}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Paragraph
+Integrali
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula 
+\[
+\int\frac{1}{x^{2}+a^{2}}dx=\frac{1}{a}\arctan\frac{x}{a}+C
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\int\frac{1}{x^{2}-a^{2}}dx=\frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right|+C
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\int\frac{1}{a^{2}-x^{2}}dx=\frac{1}{2a}\ln\left|\frac{a+x}{a-x}\right|+C
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\int\frac{1}{ax+b}dx=\frac{1}{a}\ln\left|ax+b\right|+C
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\int\left(ax+b\right)^{n}dx=\frac{\left(ax+b\right)^{n+1}}{a\left(n+1\right)}+C
+\]
+
+\end_inset
+
+
+\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
+
+
+\begin_inset Formula 
+\[
+\int\frac{1}{\sin^{2}\left(x\right)}dx=-\ctg\left(x\right)
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\int\frac{1}{\cos^{2}\left(x\right)}=\tan\left(x\right)
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\int\frac{1}{\sqrt{a^{2}+x^{2}}}dx=\ln\left|x+\sqrt{x^{2}+a^{2}}\right|
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\int\frac{1}{\sqrt{x^{2}-a^{2}}}dx=\ln\left|x+\sqrt{x^{2}-a^{2}}\right|
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\int\sqrt{a^{2}+x^{2}}dx=\frac{1}{2}\left(x\sqrt{a^{2}+x^{2}}+a^{2}\ln\left(\sqrt{a^{2}+x^{2}}+x\right)\right)
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\int\sqrt{a^{2}-x^{2}}dx=\frac{1}{2}\left(x\sqrt{a^{2}-x^{2}}+a^{2}\arctan\left(\frac{x}{\sqrt{a^{2}-x^{2}}}\right)\right)
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\int\frac{A}{x-a}dx=A\ln\left|x-a\right|
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\int\frac{A}{\left(x-a\right)^{n}}dx=\frac{-A}{n-1}\cdot\frac{1}{\left(x-a\right)^{n-1}}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\int\frac{Bx+C}{x^{2}+bx+c}=\frac{B}{2}\ln\left|x^{2}+bx+c\right|+\frac{2C-Bb}{\sqrt{-D}}\arctan\left(\frac{2x+b}{\sqrt{-D}}\right)
+\]
+
+\end_inset
+
+ In velja 
+\begin_inset Formula $D=b^{2}-4c$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Pri 
+\begin_inset Formula $\int\sin\left(x\right)^{p}\cos\left(x\right)^{q}dx$
+\end_inset
+
+ lih 
+\begin_inset Formula $q$
+\end_inset
+
+ substituiramo 
+\begin_inset Formula $t=\cos\left(x\right)$
+\end_inset
+
+,
+ lih 
+\begin_inset Formula $p$
+\end_inset
+
+ pa 
+\begin_inset Formula $t=\sin\left(x\right)$
+\end_inset
+
+.
+ Pri sodih nižamo stopnje s formulo dvonega kota.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+https://en.wikipedia.org/wiki/List_of_integrals_of_rational_functions
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+end{multicols}
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\end_body
+\end_document
diff --git a/šola/ana1/teor.lyx b/šola/ana1/teor.lyx
new file mode 100644
index 0000000..104ba6c
--- /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
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+\use_default_options true
+\begin_modules
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+\maintain_unincluded_children no
+\language slovene
+\language_package default
+\inputencoding auto-legacy
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+\index Index
+\shortcut idx
+\color #008000
+\end_index
+\leftmargin 2cm
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+\end_header
+
+\begin_body
+
+\begin_layout Title
+Teorija Analize 1 —
+ IŠRM 2023/24
+\end_layout
+
+\begin_layout Author
+
+\noun on
+Anton Luka Šijanec
+\end_layout
+
+\begin_layout Date
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+today
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Abstract
+Povzeto po zapiskih s predavanj profesorja Oliverja Dragičevića.
+\end_layout
+
+\begin_layout Standard
+\begin_inset CommandInset toc
+LatexCommand tableofcontents
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Section
+Števila
+\end_layout
+
+\begin_layout Definition*
+Množica je matematični objekt,
+ ki predstavlja skupino elementov.
+ Če element 
+\begin_inset Formula $a$
+\end_inset
+
+ pripada množici 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ pišemo 
+\begin_inset Formula $a\in A$
+\end_inset
+
+,
+ sicer pa 
+\begin_inset Formula $a\not\in A$
+\end_inset
+
+.
+ Množica 
+\begin_inset Formula $B$
+\end_inset
+
+ je podmnožica množice 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ pišemo 
+\begin_inset Formula $B\subset A$
+\end_inset
+
+,
+ če 
+\begin_inset Formula $\forall b\in B:b\in A$
+\end_inset
+
+.
+ Presek 
+\begin_inset Formula $B$
+\end_inset
+
+ in 
+\begin_inset Formula $C$
+\end_inset
+
+ označimo 
+\begin_inset Formula $B\cap C\coloneqq\left\{ x;x\in B\wedge x\in C\right\} $
+\end_inset
+
+.
+ Unijo 
+\begin_inset Formula $B$
+\end_inset
+
+ in 
+\begin_inset Formula $C$
+\end_inset
+
+ označimo 
+\begin_inset Formula $B\cup C\coloneqq\left\{ x;x\in B\vee x\in C\right\} $
+\end_inset
+
+.
+ Razliko/komplement 
+\begin_inset Quotes gld
+\end_inset
+
+
+\begin_inset Formula $B$
+\end_inset
+
+ manj/brez 
+\begin_inset Formula $C$
+\end_inset
+
+
+\begin_inset Quotes grd
+\end_inset
+
+ označimo 
+\begin_inset Formula $B\setminus C\coloneqq\left\{ x;x\in B\wedge x\not\in C\right\} $
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsection
+Realna števila
+\end_layout
+
+\begin_layout Standard
+Množico realnih števil označimo 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+.
+ V njej obstajata binarni operaciji seštevanje 
+\begin_inset Formula $a+b$
+\end_inset
+
+ in množenje 
+\begin_inset Formula $a\cdot b$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsubsection
+Lastnosti seštevanja
+\end_layout
+
+\begin_layout Axiom
+Komutativnost:
+ 
+\begin_inset Formula $\forall a,b\in\mathbb{R}:a+b=b+a$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Axiom
+Asociativnost:
+ 
+\begin_inset Formula $\forall a,b,c\in\mathbb{R}:a+\left(b+c\right)=\left(a+b\right)+c$
+\end_inset
+
+,
+ torej je 
+\begin_inset Formula $a+\cdots+z$
+\end_inset
+
+ dobro definiran izraz.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Axiom
+Obstoj enote:
+ 
+\begin_inset Formula $\exists0\in\mathbb{R}\forall a\in\mathbb{R}:a+0=a$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Axiom
+Obstoj inverzov:
+ 
+\begin_inset Formula $\forall a\in\mathbb{R}\exists b\in\mathbb{R}\ni a+b=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Claim*
+Inverz je enoličen.
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $a,b,c\in\mathbb{R}$
+\end_inset
+
+ in 
+\begin_inset Formula $a+b=0$
+\end_inset
+
+ in 
+\begin_inset Formula $a+c=0$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $b=b+0=b+a+c=0+c=c$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Corollary*
+Inverz je funkcija in aditivni inverz 
+\begin_inset Formula $a$
+\end_inset
+
+ označimo z 
+\begin_inset Formula $-a$
+\end_inset
+
+.
+ Pri zapisu 
+\begin_inset Formula $a+\left(-b\right)$
+\end_inset
+
+ običajno 
+\begin_inset Formula $+$
+\end_inset
+
+ izpustimo in pišemo 
+\begin_inset Formula $a-b$
+\end_inset
+
+,
+ čemur pravimo odštevanje 
+\begin_inset Formula $b$
+\end_inset
+
+ od 
+\begin_inset Formula $a$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+\begin_inset Formula $\forall a\in\mathbb{R}:a=-\left(-a\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $b=-a$
+\end_inset
+
+ in 
+\begin_inset Formula $c=-b=-\left(-a\right)$
+\end_inset
+
+.
+ Tedaj velja 
+\begin_inset Formula $c-a=c+b=-\left(-a\right)-a=0$
+\end_inset
+
+ in 
+\begin_inset Formula $a=0+a=c-a+a=c+\left(-a\right)+a=c+0=c=-\left(-a\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+\begin_inset Formula $-\left(b+c\right)=-b-c$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Proof
+Velja 
+\begin_inset Formula $b+c+\left(-b-c\right)=b+c+\left(\left(-b\right)+\left(-c\right)\right)=b+\left(-b\right)+c+\left(-c\right)=0$
+\end_inset
+
+,
+ torej je 
+\begin_inset Formula $b+c$
+\end_inset
+
+ inverz od 
+\begin_inset Formula $\left(-b-c\right)$
+\end_inset
+
+,
+ torej je 
+\begin_inset Formula $-\left(b+c\right)=-b-c$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Subsubsection
+Lastnosti množenja
+\end_layout
+
+\begin_layout Axiom
+Komutativnost:
+ 
+\begin_inset Formula $\forall a,b\in\mathbb{R}:ab=ba$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Axiom
+Asociativnost:
+ 
+\begin_inset Formula $\forall a,b,c\in\mathbb{R}:a\left(bc\right)=\left(ab\right)c$
+\end_inset
+
+,
+ torej je 
+\begin_inset Formula $a\cdots z$
+\end_inset
+
+ dobro definiran izraz.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Axiom
+Obstoj enote:
+ 
+\begin_inset Formula $\exists1\in\mathbb{R}\forall a\in\mathbb{R}:a1=a$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Axiom
+Obstoj inverzov:
+ 
+\begin_inset Formula $\forall a\in\mathbb{R}\setminus\left\{ 0\right\} \exists b\in\mathbb{R}\setminus\left\{ 0\right\} \ni:ab=1$
+\end_inset
+
+
+\end_layout
+
+\begin_deeper
+\begin_layout Claim*
+Inverz je enoličen.
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $a,b,c\in\mathbb{R}\setminus\left\{ 0\right\} $
+\end_inset
+
+ in 
+\begin_inset Formula $ab=1$
+\end_inset
+
+ in 
+\begin_inset Formula $ac=1$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $b=b1=bac=1c=c$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Corollary*
+Inverz je funkcija in multiplikativni inverz 
+\begin_inset Formula $a$
+\end_inset
+
+ označimo z 
+\begin_inset Formula $a^{-1}$
+\end_inset
+
+.
+ Pri zapisu 
+\begin_inset Formula $a\cdot b^{-1}$
+\end_inset
+
+ lahko 
+\begin_inset Formula $\cdot$
+\end_inset
+
+ izpustimo in pišemo 
+\begin_inset Formula $a/b$
+\end_inset
+
+,
+ čemur pravimo deljenje 
+\begin_inset Formula $a$
+\end_inset
+
+ z 
+\begin_inset Formula $b$
+\end_inset
+
+ za neničeln 
+\begin_inset Formula $b$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Subsubsection
+Skupne lastnosti v 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Axiom
+\begin_inset Formula $1\not=0$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Axiom
+Distributivnost:
+ 
+\begin_inset Formula $\forall a,b,c\in\mathbb{R}:\left(a+b\right)c=ac+bc$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Paragraph
+Urejenost 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Realna števila delimo na pozitivna 
+\begin_inset Formula $\mathbb{R}_{+}\coloneqq\left\{ x\in\mathbb{R};x>0\right\} $
+\end_inset
+
+,
+ negativna 
+\begin_inset Formula $\mathbb{R}_{-}\coloneqq\left\{ x\in\mathbb{R};x<0\right\} $
+\end_inset
+
+ in ničlo 
+\begin_inset Formula $0$
+\end_inset
+
+.
+ Če je 
+\begin_inset Formula $x\in\mathbb{\mathbb{R}}_{+}\cup\left\{ 0\right\} $
+\end_inset
+
+,
+ pišemo 
+\begin_inset Formula $x\geq0$
+\end_inset
+
+,
+ če je 
+\begin_inset Formula $x\in\mathbb{R}_{-}\cup\left\{ 0\right\} $
+\end_inset
+
+,
+ pišemo 
+\begin_inset Formula $x\leq0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Axiom
+Če je 
+\begin_inset Formula $a\not=0$
+\end_inset
+
+,
+ je natanko eno izmed 
+\begin_inset Formula $\left\{ a,-a\right\} $
+\end_inset
+
+ pozitivno,
+ imenujemo ga absolutna vrednost 
+\begin_inset Formula $a$
+\end_inset
+
+ (pišemo 
+\begin_inset Formula $\left|a\right|$
+\end_inset
+
+),
+ in natanko eno negativno,
+ pišemo 
+\begin_inset Formula $-\left|a\right|$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+\begin_inset Formula $\left|0\right|=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Za 
+\begin_inset Formula $a,b\in\mathbb{R}$
+\end_inset
+
+ se 
+\begin_inset Formula $\left|a-b\right|$
+\end_inset
+
+ imenuje razdalja.
+\end_layout
+
+\begin_layout Axiom
+\begin_inset Formula $\forall a,b\in\mathbb{R}:a,b>0\Rightarrow\left(a+b>0\right)\wedge\left(ab>0\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Za 
+\begin_inset Formula $a,b\in\mathbb{R}$
+\end_inset
+
+:
+ 
+\begin_inset Formula $a$
+\end_inset
+
+ je večje od 
+\begin_inset Formula $b$
+\end_inset
+
+,
+ oznaka 
+\begin_inset Formula $a>b\Leftrightarrow a-b>0$
+\end_inset
+
+.
+ 
+\begin_inset Formula $a$
+\end_inset
+
+ je manjše od 
+\begin_inset Formula $b$
+\end_inset
+
+,
+ oznaka 
+\begin_inset Formula $a<b\Leftrightarrow a-b<0$
+\end_inset
+
+.
+ Podobno 
+\begin_inset Formula $\leq$
+\end_inset
+
+ in 
+\begin_inset Formula $\geq$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+Trikotniška neenakost.
+ 
+\begin_inset Formula $\forall a,b\in\mathbb{R}$
+\end_inset
+
+:
+ 
+\begin_inset Formula $\left|\left|a\right|-\left|b\right|\right|\leq\left|a+b\right|\leq\left|a\right|+\left|b\right|$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Dokažimo desni neenačaj.
+ Vemo 
+\begin_inset Formula $ab\leq\left|ab\right|$
+\end_inset
+
+ in 
+\begin_inset Formula $\left|a\right|=\sqrt{a^{2}}$
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $a^{2}+2ab+b^{2}\leq\left|a\right|^{2}+2\left|a\right|\left|b\right|+\left|b\right|^{2}$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\left(a+b\right)^{2}\leq\left(\left|a\right|+\left|b\right|\right)^{2}$
+\end_inset
+
+,
+ korenimo:
+ 
+\begin_inset Formula $\left|a+b\right|\leq\left|a\right|+\left|b\right|$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsubsection
+Intervali
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $a<b$
+\end_inset
+
+.
+ Označimo odprti interval 
+\begin_inset Formula $\left(a,b\right)\coloneqq\left\{ x\in\mathbb{R};a<x<b\right\} $
+\end_inset
+
+,
+ zaprti 
+\begin_inset Formula $\left[a,b\right]\coloneqq\left\{ x\in\mathbb{R};a\leq x\leq b\right\} $
+\end_inset
+
+,
+ polodprti 
+\begin_inset Formula $(a,b]\coloneqq\left\{ x\in\mathbb{R};a<x\leq b\right\} $
+\end_inset
+
+ in podobno 
+\begin_inset Formula $[a,b)$
+\end_inset
+
+.
+ 
+\begin_inset Formula $\left(a,\infty\right)\coloneqq\left\{ x\in\mathbb{R};x>a\right\} $
+\end_inset
+
+ in podobno 
+\begin_inset Formula $[a,\infty)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsection
+Temeljne številske podmnožice
+\end_layout
+
+\begin_layout Subsubsection
+Naravna števila 
+\begin_inset Formula $\mathbb{N}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+\begin_inset Formula $\mathbb{N}\coloneqq\left\{ 1,1+1,1+1+1,1+1+1+1,\dots\right\} $
+\end_inset
+
+
+\end_layout
+
+\begin_layout Paragraph
+Matematična indukcija
+\end_layout
+
+\begin_layout Standard
+Če je 
+\begin_inset Formula $A\subseteq\mathbb{N}$
+\end_inset
+
+ in velja 
+\begin_inset Formula $1\in A$
+\end_inset
+
+ (baza) in 
+\begin_inset Formula $a\in A\Rightarrow a+1\in A$
+\end_inset
+
+ (korak),
+ tedaj 
+\begin_inset Formula $A=\mathbb{N}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+\begin_inset Formula $1+2+3+\cdots+n=\frac{n\left(n+1\right)}{2}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+\begin_inset Formula $A\coloneqq\left\{ n\in\mathbb{N};\text{velja trditev za }n\right\} $
+\end_inset
+
+.
+ Dokažimo 
+\begin_inset Formula $A=\mathbb{N}$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Baza:
+ 
+\begin_inset Formula $1=\frac{1\cdot2}{2}=1$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Korak:
+ Predpostavimo 
+\begin_inset Formula $1+2+3+\cdots+n=\frac{n\left(n+1\right)}{2}$
+\end_inset
+
+.
+ Prištejmo 
+\begin_inset Formula $n+1$
+\end_inset
+
+:
+ 
+\begin_inset Formula 
+\[
+1+2+3+\cdots+n+\left(n+1\right)=\frac{n\left(n+1\right)}{2}+\left(n+1\right)=\frac{n\left(n+1\right)}{2}+\frac{2\left(n+1\right)}{2}=\frac{\left(n+2\right)\left(n+1\right)}{2}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Subsubsection
+Cela števila 
+\begin_inset Formula $\mathbb{Z}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Množica 
+\begin_inset Formula $\mathbb{N}$
+\end_inset
+
+ je zaprta za seštevanje in množenje,
+ torej 
+\begin_inset Formula $\forall a,b\in\mathbb{N}:a+b\in\mathbb{N}\wedge ab\in\mathbb{N}$
+\end_inset
+
+,
+ ni pa zaprta za odštevanje,
+ ker recimo 
+\begin_inset Formula $5-3\not\in\mathbb{N}$
+\end_inset
+
+.
+ Zapremo jo za odštevanje in dobimo množico 
+\begin_inset Formula $\mathbb{Z}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+\begin_inset Formula $\mathbb{Z}\coloneqq\left\{ a-b;b,a\in\mathbb{N}\right\} $
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsubsection
+Racionalna števila 
+\begin_inset Formula $\mathbb{Q}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Najmanjša podmnožica 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+,
+ ki vsebuje 
+\begin_inset Formula $\mathbb{Z}$
+\end_inset
+
+ in je zaprta za deljenje,
+ je 
+\begin_inset Formula $\mathbb{Q}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+\begin_inset Formula $\mathbb{Q}\coloneqq\left\{ a/b;a\in\mathbb{Z},b\in\mathbb{Z}\setminus\left\{ 0\right\} \right\} $
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Velja 
+\begin_inset Formula $\mathbb{N}\subset\mathbb{Z}\subset\mathbb{Q}\subset\mathbb{R}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+Za 
+\begin_inset Formula $a\in\mathbb{Q}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $b\not\in\mathbb{Q}$
+\end_inset
+
+ velja 
+\begin_inset Formula $a+b\not\in\mathbb{Q}$
+\end_inset
+
+ in 
+\begin_inset Formula $a\not=0\Rightarrow ab\not\in\mathbb{Q}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+PDDRAA 
+\begin_inset Formula $a+b\in\mathbb{Q}$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $a+b-a\in\mathbb{Q}$
+\end_inset
+
+,
+ tedaj 
+\begin_inset Formula $b\in\mathbb{Q}$
+\end_inset
+
+,
+ kar je 
+\begin_inset Formula $\rightarrow\!\leftarrow$
+\end_inset
+
+.
+ PDDRAA 
+\begin_inset Formula $ab\in\mathbb{Q}$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $\frac{ab}{a}\in\mathbb{Q}$
+\end_inset
+
+,
+ tedaj 
+\begin_inset Formula $b\in\mathbb{Q}$
+\end_inset
+
+,
+ kar je 
+\begin_inset Formula $\rightarrow\!\leftarrow$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsection
+\begin_inset CommandInset label
+LatexCommand label
+name "subsec:Omejenost-množic"
+
+\end_inset
+
+Omejenost množic
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $A\subset\mathbb{R}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $A$
+\end_inset
+
+ je navzgor omejena 
+\begin_inset Formula $\Leftrightarrow\exists m\in\mathbb{R}\forall a\in A:a\leq m$
+\end_inset
+
+.
+ Takemu 
+\begin_inset Formula $m$
+\end_inset
+
+ pravimo zgornja meja.
+ Najmanjši zgornji meji 
+\begin_inset Formula $A$
+\end_inset
+
+ pravimo supremum ali natančna zgornja meja množice 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ označimo 
+\begin_inset Formula $\sup A$
+\end_inset
+
+.
+ Če je zgornja meja 
+\begin_inset Formula $A$
+\end_inset
+
+ (
+\begin_inset Formula $m$
+\end_inset
+
+) element 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ je maksimum množice 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ označimo 
+\begin_inset Formula $m=\max A$
+\end_inset
+
+.
+ Če množica ni navzgor omejena,
+ pišemo 
+\begin_inset Formula $\sup A=\infty$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Če 
+\begin_inset Formula $s=\sup A\in\mathbb{R}$
+\end_inset
+
+,
+ mora veljati 
+\begin_inset Formula $\forall a\in A:a\leq s$
+\end_inset
+
+ in 
+\begin_inset Formula $\forall\varepsilon>0\exists b\in A\ni:b>s-\varepsilon$
+\end_inset
+
+,
+ torej za vsak neničeln 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+ 
+\begin_inset Formula $s-\varepsilon$
+\end_inset
+
+ ni več natančna zgornja meja za 
+\begin_inset Formula $A$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $A\subset\mathbb{R}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $A$
+\end_inset
+
+ je navzdol omejena 
+\begin_inset Formula $\Leftrightarrow\exists m\in\mathbb{R}\forall a\in A:a\geq m$
+\end_inset
+
+.
+ Takemu 
+\begin_inset Formula $m$
+\end_inset
+
+ pravimo spodnja meja.
+ Največji spodnji meji 
+\begin_inset Formula $A$
+\end_inset
+
+ pravimo infimum ali natančna spodnja meja množice 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ označimo 
+\begin_inset Formula $\inf A$
+\end_inset
+
+.
+ Če je spodnja meja 
+\begin_inset Formula $A$
+\end_inset
+
+ (
+\begin_inset Formula $m$
+\end_inset
+
+) element 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ je minimum množice 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ označimo 
+\begin_inset Formula $m=\min A$
+\end_inset
+
+.
+ Če množica ni navzdol omejena,
+ pišemo 
+\begin_inset Formula $\inf A=-\infty$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Množica 
+\begin_inset Formula $A\subset\mathbb{R}$
+\end_inset
+
+ je omejena,
+ če je hkrati navzgor in navzdol omejena.
+\end_layout
+
+\begin_layout Axiom
+\begin_inset CommandInset label
+LatexCommand label
+name "axm:Dedekind.-Vsaka-navzgor"
+
+\end_inset
+
+Dedekind.
+ Vsaka navzgor omejena množica v 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+ ima natančno zgornjo mejo v 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Remark*
+Za 
+\begin_inset Formula $\mathbb{Q}$
+\end_inset
+
+ aksiom 
+\begin_inset CommandInset ref
+LatexCommand ref
+reference "axm:Dedekind.-Vsaka-navzgor"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+ ne velja.
+ Če 
+\begin_inset Formula $B\subset\mathbb{Q}$
+\end_inset
+
+,
+ se lahko zgodi,
+ da 
+\begin_inset Formula $\sup B\not\in\mathbb{Q}$
+\end_inset
+
+.
+ Primer:
+ 
+\begin_inset Formula $B\coloneqq\left\{ q\in\mathbb{Q};q^{2}\leq2\right\} $
+\end_inset
+
+.
+ 
+\begin_inset Formula $\sup B=\sqrt{2}\not\in\mathbb{Q}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+
+\end_layout
+
+\begin_layout Subsection
+Decimalni zapis
+\end_layout
+
+\begin_layout Definition*
+\begin_inset Formula $\forall x\in\mathbb{R}^{+}\exists!m\in\mathbb{N}\cup\left\{ 0\right\} ,d_{1},d_{2},\dots\in\left\{ 0..9\right\} $
+\end_inset
+
+,
+ ki število natančno določajo.
+ Pišemo 
+\begin_inset Formula $x=m,d_{1}d_{2}\dots$
+\end_inset
+
+.
+ Natančno določitev mislimo v smislu:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $m\leq x<m+1$
+\end_inset
+
+ —
+ s tem se izognemo dvojnemu zapisu 
+\begin_inset Formula $1=0,\overline{9}$
+\end_inset
+
+ in 
+\begin_inset Formula $1=1,\overline{0}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $[m,m+1)$
+\end_inset
+
+ razdelimo na 10 enako dolgih polodprtih intervalov 
+\begin_inset Formula $I_{0},\dots,I_{9}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $x$
+\end_inset
+
+ leži na natanko enem izmed njih,
+ indeks njega je 
+\begin_inset Formula $d_{1}$
+\end_inset
+
+.
+ Nadaljujemo tako,
+ da 
+\begin_inset Formula $I_{d_{1}}$
+\end_inset
+
+ razdelimo zopet na 10 delov itd.
+\end_layout
+
+\end_deeper
+\begin_layout Definition*
+Števila 
+\begin_inset Formula $x\in\mathbb{R^{-}}$
+\end_inset
+
+ pišemo tako,
+ da zapišemo decimalni zapis števila 
+\begin_inset Formula $-x$
+\end_inset
+
+ in predenj zapišemo 
+\begin_inset Formula $-$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Če se decimalke v zaporedju 
+\begin_inset Formula $\left(d_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ ponavljajo,
+ uporabimo periodični zapis,
+ denimo 
+\begin_inset Formula $5,01\overline{763}\in\mathbb{Q}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsection
+Kompleksna števila
+\end_layout
+
+\begin_layout Definition*
+Vpeljimo število 
+\begin_inset Formula $i$
+\end_inset
+
+ z lastnostjo 
+\begin_inset Formula $i^{2}=-1$
+\end_inset
+
+,
+ da je 
+\begin_inset Formula $i$
+\end_inset
+
+ rešitev enačbe 
+\begin_inset Formula $x^{2}+1=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+\begin_inset Formula $i\not\in\mathbb{R}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Proof
+Sicer bi veljajo 
+\begin_inset Formula $i^{2}\geq0$
+\end_inset
+
+,
+ kar po definiciji ne velja.
+\end_layout
+
+\begin_layout Definition*
+Kompleksna števila so 
+\begin_inset Formula $\mathbb{C}\coloneqq\left\{ a+bi;a,b\in\mathbb{R}\right\} $
+\end_inset
+
+.
+ 
+\begin_inset Formula $bi$
+\end_inset
+
+ je še nedefinirano,
+ zato za kompleksna števila definirano seštevanje in množenje za 
+\begin_inset Formula $z=a+bi$
+\end_inset
+
+ in 
+\begin_inset Formula $w=c+di$
+\end_inset
+
+:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $z+w\coloneqq\left(a+c\right)+\left(b+d\right)i$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $zw\coloneqq\left(a+bi\right)\left(c+di\right)=ac+adi+bic+bidi=ac+adi+bic-bd=\left(ac-bd\right)+\left(ad+bc\right)i$
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Definition*
+Definiramo še konjugirano vrednost 
+\begin_inset Formula $z\in\mathbb{C}$
+\end_inset
+
+:
+ 
+\begin_inset Formula $\overline{z}\coloneqq a-bi$
+\end_inset
+
+ in označimo 
+\begin_inset Formula $\left|z\right|^{2}=z\overline{z}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+\begin_inset Formula $z\overline{z}=a^{2}+b^{2}\geq0$
+\end_inset
+
+ za 
+\begin_inset Formula $z=a+bi$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+\begin_inset Formula $\left(a+bi\right)\left(a-bi\right)=a^{2}+abi-bia-bibi=a^{2}+b^{2}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Velja,
+ da je 
+\begin_inset Formula $\mathbb{R}\subset\mathbb{C}$
+\end_inset
+
+ v smislu identifikacije 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+ z množico 
+\begin_inset Formula $\mathbb{C}$
+\end_inset
+
+:
+ 
+\begin_inset Formula $\mathbb{R}=\left\{ a+0i;a\in\mathbb{R}\right\} $
+\end_inset
+
+,
+ torej smo 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+ razširili v 
+\begin_inset Formula $\mathbb{C}$
+\end_inset
+
+,
+ kjer ima vsak polinom vedno rešitev.
+\end_layout
+
+\begin_layout Subsubsection
+Deljenje v 
+\begin_inset Formula $\mathbb{C}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Za 
+\begin_inset Formula $w,z\in\mathbb{C},w\not=0$
+\end_inset
+
+ iščemo 
+\begin_inset Formula $x\in\mathbb{C}\ni:wx=z$
+\end_inset
+
+.
+ Ločimo dva primera:
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $w\in\mathbb{R}\setminus\left\{ 0\right\} $
+\end_inset
+
+:
+ definiramo 
+\begin_inset Formula $x=\frac{z}{w}\coloneqq\frac{a}{w}+\frac{b}{w}i$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $w\in\mathbb{C}\setminus\left\{ 0\right\} $
+\end_inset
+
+ (splošno):
+ 
+\begin_inset Formula $wx=z\overset{/\cdot\overline{w}}{\Longrightarrow}w\overline{w}x=z\overline{w}\Rightarrow\left|w\right|^{2}x=z\overline{w}\Rightarrow x=\frac{z\overline{w}}{\left|w\right|^{2}}$
+\end_inset
+
+,
+ z 
+\begin_inset Formula $\left|w\right|^{2}$
+\end_inset
+
+ pa znamo deliti,
+ ker je realen.
+\end_layout
+
+\begin_layout Subsubsection
+Lastnosti v 
+\begin_inset Formula $\mathbb{C}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $+$
+\end_inset
+
+ in 
+\begin_inset Formula $\cdot$
+\end_inset
+
+ sta komutativni,
+ asociativni,
+ distributivni,
+ 
+\begin_inset Formula $0$
+\end_inset
+
+ je aditivna enota,
+ 
+\begin_inset Formula $1$
+\end_inset
+
+ je multiplikativna.
+\end_layout
+
+\begin_layout Definition*
+Za 
+\begin_inset Formula $z=a+bi$
+\end_inset
+
+ vpeljemo 
+\begin_inset Formula $\Re z=a$
+\end_inset
+
+ in 
+\begin_inset Formula $\Im z=b$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Remark*
+Opazimo 
+\begin_inset Formula $\Re z=\frac{z+\overline{z}}{2}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\Im z=\frac{z-\overline{z}}{2}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Remark*
+\begin_inset Formula $\mathbb{C}$
+\end_inset
+
+ si lahko predstavljamo kot urejene pare;
+ 
+\begin_inset Formula $a+bi$
+\end_inset
+
+ ustreza paru 
+\begin_inset Formula $\left(a,b\right)$
+\end_inset
+
+.
+ Tako 
+\begin_inset Formula $\mathbb{C}$
+\end_inset
+
+ enačimo/identificiramo z 
+\begin_inset Formula $\mathbb{R}^{2}\coloneqq\left\{ \left(a,b\right);a,b\in\mathbb{R}\right\} $
+\end_inset
+
+,
+ s čimer dobimo geometrično predstavitev 
+\begin_inset Formula $\mathbb{C}$
+\end_inset
+
+ kot vektorje v 
+\begin_inset Formula $\mathbb{R}^{2}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Za 
+\begin_inset Formula $z=a+bi$
+\end_inset
+
+,
+ predstavljen z vektorjem s komponentami 
+\begin_inset Formula $\left(a,b\right)$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $a=\left|z\right|\cos\varphi$
+\end_inset
+
+ in 
+\begin_inset Formula $v=\left|z\right|\sin\varphi$
+\end_inset
+
+.
+ Kotu 
+\begin_inset Formula $\varphi$
+\end_inset
+
+ pravimo argument kompleksnega števila 
+\begin_inset Formula $z$
+\end_inset
+
+,
+ oznaka 
+\begin_inset Formula $\arg z$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Corollary*
+\begin_inset Formula $z=$
+\end_inset
+
+
+\begin_inset Formula $\left|z\right|\left(\cos\varphi+i\sin\varphi\right)$
+\end_inset
+
+.
+ Velja
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+TODO DOPISATI ZAKAJ (v bistvu še jaz ne vem).
+ ne razumem.
+\end_layout
+
+\end_inset
+
+ 
+\begin_inset Formula $\left(\cos\varphi+i\sin\varphi\right)\left(\cos\psi+i\sin\psi\right)=\cos\left(\varphi+\psi\right)+i\sin\left(\varphi+\psi\right)$
+\end_inset
+
+,
+ zato lahko pišemo 
+\begin_inset Formula $e^{i\varphi}=\cos\varphi+i\sin\varphi$
+\end_inset
+
+.
+ Množenje kompleksnh števil 
+\begin_inset Formula $z=\left|z\right|e^{i\varphi}$
+\end_inset
+
+ in 
+\begin_inset Formula $w=\left|w\right|e^{i\psi}$
+\end_inset
+
+ vrne število 
+\begin_inset Formula $zw$
+\end_inset
+
+,
+ za katero velja:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $\left|zw\right|=\left|z\right|\left|w\right|$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $\arg zw=\arg z+\arg w$
+\end_inset
+
+ (do periode 
+\begin_inset Formula $2\pi$
+\end_inset
+
+ natančno)
+\end_layout
+
+\end_deeper
+\begin_layout Section
+Zaporedja
+\end_layout
+
+\begin_layout Definition*
+Funkcija 
+\begin_inset Formula $a:\mathbb{N}\to\mathbb{R}$
+\end_inset
+
+ se imenuje realno zaporedje,
+ oznaka 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $a_{n}$
+\end_inset
+
+ je funkcijska vrednost pri 
+\begin_inset Formula $n$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $a_{n}=n$
+\end_inset
+
+:
+ 
+\begin_inset Formula $1,2,3,\dots$
+\end_inset
+
+;
+ 
+\begin_inset Formula $a_{n}=\left(-1\right)^{n}n^{2}$
+\end_inset
+
+:
+ 
+\begin_inset Formula $-1,4,-9,16,-25,\dots$
+\end_inset
+
+;
+ 
+\begin_inset Formula $a_{n}=\cos\left(\frac{\pi}{2}n\right)=0,-1,0,1,0,-1,\dots$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Zaporedje lahko podamo rekurzivno.
+ Podamo prvi člen ali nekaj prvih členov in pravilo,
+ kako iz prejšnjih členov dobiti naslednje.
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $a_{1}=0,a_{n+1}=a_{n}+n$
+\end_inset
+
+ da zaporedje 
+\begin_inset Formula $a_{n}=\frac{n\left(n+1\right)}{2}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $a_{0}=0,a_{n+1}=\sqrt{b+a_{n}}$
+\end_inset
+
+ da zaporedje 
+\begin_inset Formula $0,\sqrt{b},\sqrt{b+\sqrt{b}},\sqrt{b+\sqrt{b+\sqrt{b}}},\dots$
+\end_inset
+
+.
+ Fibbonacijevo zaporedje:
+ 
+\begin_inset Formula $a_{1}=a_{2}=1,a_{n+1}=a_{n}+a_{n-1}$
+\end_inset
+
+ da zaporedje 
+\begin_inset Formula $1,1,2,3,5,8,\dots$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsection
+Posebni tipi zaporedij
+\end_layout
+
+\begin_layout Definition*
+Zaporedje 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ je aritmetično,
+ če 
+\begin_inset Formula $\exists k\in\mathbb{R}\forall n\in\mathbb{N}:a_{n+1}-a_{n}=k$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $a_{n+1}=a_{n}+k=a_{1}+nd$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Zaporedje 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ je geometrično,
+ če 
+\begin_inset Formula $\exists\lambda\in\mathbb{R}\forall n\in\mathbb{N}:a_{n+1}=a_{n}\lambda$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $a_{n}=\lambda^{n-1}a_{1}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Zaporedje 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ je navzdol oz.
+ navzgor omejeno,
+ če je množica vseh členov tega taporedja navzgor oz.
+ navzdol omejena (glej 
+\begin_inset CommandInset ref
+LatexCommand ref
+reference "subsec:Omejenost-množic"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+).
+ Podobno z množico členov definiramo supremum,
+ infimum,
+ maksimum in infimum zaporedja.
+\end_layout
+
+\begin_layout Definition*
+Zaporedje je naraščajoče,
+ če 
+\begin_inset Formula $\forall n\in\mathbb{N}:a_{n+1}\geq a_{n}$
+\end_inset
+
+,
+ padajoče,
+ če 
+\begin_inset Formula $\forall n\in\mathbb{N}:a_{n+1}\leq a_{n}$
+\end_inset
+
+,
+ strogo naraščajoče,
+ če 
+\begin_inset Formula $\forall n\in\mathbb{N}:a_{n+1}>a_{n}$
+\end_inset
+
+,
+ strogo padajoče podobno,
+ monotono,
+ če je naraščajoče ali padajoče in strogo monotono,
+ če je strogo naraščajoče ali strogo padajoče.
+\end_layout
+
+\begin_layout Subsection
+Limita zaporedja
+\end_layout
+
+\begin_layout Definition*
+Množica 
+\begin_inset Formula $U\subseteq\mathbb{R}$
+\end_inset
+
+ je odprta,
+ če 
+\begin_inset Formula $\forall u\in U\exists r>0\ni:\left(u-r,u+r\right)\subseteq U$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Množica 
+\begin_inset Formula $U\subseteq\mathbb{R}$
+\end_inset
+
+ je zaprta,
+ če je 
+\begin_inset Formula $U^{\mathcal{C}}\coloneqq\mathbb{R}\setminus U$
+\end_inset
+
+ odprta.
+\end_layout
+
+\begin_layout Claim*
+Odprt interval je odprta množica.
+\end_layout
+
+\begin_layout Proof
+Za poljubna 
+\begin_inset Formula $a,b\in\mathbb{R}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $b>a$
+\end_inset
+
+,
+ naj bo 
+\begin_inset Formula $u\in\left(a,b\right)$
+\end_inset
+
+ poljuben.
+ Ustrezen 
+\begin_inset Formula $r$
+\end_inset
+
+ je 
+\begin_inset Formula $\min\left\{ \left|r-a\right|,\left|r-b\right|\right\} $
+\end_inset
+
+,
+ da je 
+\begin_inset Formula $\left(u-r,u+r\right)\subseteq U$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+Zaprt interval je zaprt.
+\end_layout
+
+\begin_layout Proof
+Naj bosta 
+\begin_inset Formula $a,b\in\mathbb{R}$
+\end_inset
+
+ poljubna in 
+\begin_inset Formula $b>a$
+\end_inset
+
+.
+ Dokazujemo,
+ da je 
+\begin_inset Formula $\left[a,b\right]$
+\end_inset
+
+ zaprt,
+ torej da je 
+\begin_inset Formula $\left[a,b\right]^{\mathcal{C}}=\left(-\infty,a\right)\cup\left(b,\infty\right)$
+\end_inset
+
+ odprta množica.
+ Za poljuben 
+\begin_inset Formula $u\in\left[a,b\right]^{\mathcal{C}}$
+\end_inset
+
+ velja,
+ da je bodisi 
+\begin_inset Formula $\in\left(-\infty,a\right)$
+\end_inset
+
+ bodisi 
+\begin_inset Formula $\left(b,\infty\right)$
+\end_inset
+
+,
+ kajti 
+\begin_inset Formula $\left(-\infty,a\right)\cap\left(b,\infty\right)=\emptyset$
+\end_inset
+
+.
+ Po prejšnji trditvi v obeh primerih velja 
+\begin_inset Formula $\exists r>0\ni:\left(u-r,u+r\right)\subseteq U$
+\end_inset
+
+,
+ torej je 
+\begin_inset Formula $\left[a,b\right]^{\mathcal{C}}$
+\end_inset
+
+ res odprta,
+ torej je 
+\begin_inset Formula $\left[a,b\right]$
+\end_inset
+
+ res zaprta.
+\end_layout
+
+\begin_layout Definition*
+Množica 
+\begin_inset Formula $B$
+\end_inset
+
+ je okolica točke 
+\begin_inset Formula $t\in\mathbb{R}$
+\end_inset
+
+,
+ če vsebuje kakšno odprto množico 
+\begin_inset Formula $U$
+\end_inset
+
+,
+ ki vsebuje 
+\begin_inset Formula $t$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $t\in U^{\text{odp.}}\subseteq B\subseteq\mathbb{R}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+\begin_inset Formula $L\in\mathbb{R}$
+\end_inset
+
+ je limita zaporedja 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{R}}$
+\end_inset
+
+ 
+\begin_inset Formula $\Leftrightarrow\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow\left|a_{n}-L\right|<\varepsilon$
+\end_inset
+
+.
+ ZDB 
+\begin_inset Formula $\forall V$
+\end_inset
+
+ okolica 
+\begin_inset Formula $L\in\mathbb{R}\exists n_{0}\in\mathbb{N}\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow a_{n}\in V$
+\end_inset
+
+,
+ pravimo,
+ da 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ konvergira k 
+\begin_inset Formula $L$
+\end_inset
+
+ in pišemo 
+\begin_inset Formula $L\coloneqq\lim_{n\to\infty}a_{n}$
+\end_inset
+
+ ali drugače 
+\begin_inset Formula $a_{n}\underset{n\to\infty}{\longrightarrow}L$
+\end_inset
+
+.
+ Če zaporedje ima limito,
+ pravimo,
+ da je konvergentno,
+ sicer je divergentno.
+\end_layout
+
+\begin_layout Claim*
+Konvergentno zaporedje v 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+ ima natanko eno limito.
+\end_layout
+
+\begin_layout Proof
+Naj bosta 
+\begin_inset Formula $J$
+\end_inset
+
+ in 
+\begin_inset Formula $L$
+\end_inset
+
+ limiti zaporedja 
+\begin_inset Formula $\left(a_{n}\right)_{n\to\infty}$
+\end_inset
+
+.
+ Torej
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+Ko trdimo,
+ da obstaja 
+\begin_inset Formula $n_{0}$
+\end_inset
+
+,
+ še ne vemo,
+ ali sta za 
+\begin_inset Formula $L$
+\end_inset
+
+ in 
+\begin_inset Formula $J$
+\end_inset
+
+ ta 
+\begin_inset Formula $n_{0}$
+\end_inset
+
+ ista.
+ Ampak trditev še vedno velja,
+ ker lahko vzamemo večjega izmed njiju,
+ ako bi bila drugačna.
+\end_layout
+
+\end_inset
+
+ po definiciji 
+\begin_inset Formula $\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow\left|a_{n}-L\right|<\varepsilon\wedge\left|a_{n}-J\right|<\varepsilon$
+\end_inset
+
+.
+ Velja torej 
+\begin_inset Formula $\forall\varepsilon>0:\left|J-L\right|<\varepsilon$
+\end_inset
+
+.
+ PDDRAA 
+\begin_inset Formula $J\not=L$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $\left|J-L\right|\not=0$
+\end_inset
+
+,
+ naj bo 
+\begin_inset Formula $\left|J-L\right|=k$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $\exists\varepsilon>0:\left|J-L\right|\not<\varepsilon$
+\end_inset
+
+,
+ ustrezen 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+ je na primer 
+\begin_inset Formula $\frac{\left|J-L\right|}{2}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+\begin_inset CommandInset label
+LatexCommand label
+name "Konvergentno-zaporedje-v-R-je-omejeno"
+
+\end_inset
+
+Konvergentno zaporedje v 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+ je omejeno.
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $L=\lim_{n\to\infty}a_{n}$
+\end_inset
+
+.
+ Znotraj intervala 
+\begin_inset Formula $\left(L-1,L+1\right)$
+\end_inset
+
+ so vsi členi zaporedja razen končno mnogo (
+\begin_inset Formula $\left\{ a_{1},\dots,a_{n_{0}}\right\} $
+\end_inset
+
+).
+ 
+\begin_inset Formula $\left\{ a_{n}\right\} _{n\in\mathbb{N}}$
+\end_inset
+
+ je unija dveh omejenih množic;
+ 
+\begin_inset Formula $\left(L-1,L+1\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $\left\{ a_{1},\dots,a_{n_{0}}\right\} $
+\end_inset
+
+,
+ zato je tudi sama omejena.
+\end_layout
+
+\begin_layout Theorem*
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hypertarget{pmkdlim}{Naj bosta}
+\end_layout
+
+\end_inset
+
+ 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ in 
+\begin_inset Formula $\left(b_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ konvergentni zaporedji v 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+.
+ Tedaj so tudi 
+\begin_inset Formula $\left(a_{n}*b_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ konvergentna in velja 
+\begin_inset Formula $\lim_{n\to\infty}a_{n}*b_{n}=\lim_{n\to\infty}a_{n}*\lim_{n\to\infty}b_{n}$
+\end_inset
+
+ za 
+\begin_inset Formula $*\in\left\{ +,-,\cdot\right\} $
+\end_inset
+
+.
+ Če je 
+\begin_inset Formula $\lim_{n\to\infty}b_{n}\not=0$
+\end_inset
+
+,
+ isto velja tudi za deljenje.
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $a_{n}\to A$
+\end_inset
+
+ in 
+\begin_inset Formula $b_{n}\to B$
+\end_inset
+
+ oziroma 
+\begin_inset Formula $\forall\varepsilon>0\exists n_{1},n_{2}\ni:\left(n>n_{1}\Rightarrow\left|a_{n}-A\right|<\varepsilon\right)\wedge\left(n>n_{2}\Rightarrow\left|b_{n}-B\right|<\varepsilon\right)$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\forall\varepsilon>0\exists n_{0}=\max\left\{ n_{1},n_{2}\right\} \ni:n>n_{0}\Rightarrow\left|a_{n}-A\right|<\varepsilon\wedge\left|b_{n}-B\right|<\varepsilon$
+\end_inset
+
+.
+ Dokažimo za vse operacije:
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $+$
+\end_inset
+
+ Po predpostavki velja 
+\begin_inset Formula $\forall\varepsilon>0\exists n_{0}\ni:n>n_{0}\Rightarrow\left|a_{n}-A\right|+\left|a_{n}-B\right|<2\varepsilon$
+\end_inset
+
+.
+ Oglejmo si sedaj
+\begin_inset Formula 
+\[
+\left|\left(a_{n}+b_{n}\right)-\left(A+B\right)\right|=\left|\left(a_{n}-A\right)+\left(b_{n}-B\right)\right|\leq\left|a_{n}-A\right|+\left|b_{n}-B\right|
+\]
+
+\end_inset
+
+in uporabimo še prejšnjo trditev,
+ torej 
+\begin_inset Formula $\forall2\varepsilon\exists n_{0}\ni:\left|\left(a_{n}+b_{n}\right)-\left(A+B\right)\right|\leq2\varepsilon$
+\end_inset
+
+,
+ s čimer dokažemo 
+\begin_inset Formula $\left(a_{n}+b_{n}\right)\to A+B$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $-$
+\end_inset
+
+ Oglejmo si
+\begin_inset Formula 
+\[
+\left|\left(a_{n}-b_{n}\right)-\left(A-B\right)\right|=\left|a_{n}-b_{n}-A+B\right|=\left|\left(a_{n}-A\right)+\left(-\left(b_{n}-B\right)\right)\right|\leq\left|a_{n}-A\right|+\left|b_{n}-B\right|
+\]
+
+\end_inset
+
+in nato kot zgoraj.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\cdot$
+\end_inset
+
+ Oglejmo si 
+\begin_inset Formula 
+\[
+\left|a_{n}b_{n}-AB\right|=\left|a_{n}b_{n}-Ab_{n}+Ab_{n}-AB\right|=\left|\left(a_{n}-A\right)b_{n}+A\left(b_{n}-B\right)\right|\leq\left|a_{n}-A\right|\left|b_{n}\right|+\left|A\right|\left|b_{n}-B\right|.
+\]
+
+\end_inset
+
+Od prej vemo,
+ da sta zaporedji omejeni,
+ ker sta konvergentni,
+ zato 
+\begin_inset Formula $\exists M>0\forall n\in\mathbb{N}:\left|b_{n}\right|\leq M$
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+ poljuben 
+\begin_inset Formula $n_{1},n_{2}\in\mathbb{N}$
+\end_inset
+
+ taka,
+ da 
+\begin_inset Formula $n\geq n_{1}\Rightarrow\left|a_{n}-A\right|<\frac{\varepsilon}{2M}$
+\end_inset
+
+ in 
+\begin_inset Formula $n\geq n_{2}\Rightarrow\left|b_{n}-B\right|<\frac{\varepsilon}{2\left|A\right|}$
+\end_inset
+
+.
+ Tedaj za 
+\begin_inset Formula $n_{0}\coloneqq\max\left\{ n_{1},n_{2}\right\} $
+\end_inset
+
+ velja
+\begin_inset Formula $\left|a_{n}b_{n}-AB\right|\leq\left|a_{n}-A\right|\left|b_{n}\right|+\left|A\right|\left|b_{n}-B\right|<\frac{\varepsilon}{2M}M+\left|A\right|\frac{\varepsilon}{2\left|A\right|}=\varepsilon$
+\end_inset
+
+,
+ skratka 
+\begin_inset Formula $\left|a_{n}b_{n}-AB\right|<\varepsilon$
+\end_inset
+
+,
+ s čimer dokažemo 
+\begin_inset Formula $\left(a_{n}+b_{n}\right)\to A+B$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $/$
+\end_inset
+
+ Ker je 
+\begin_inset Formula $B\not=0$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\exists n_{0}\in\mathbb{N}\forall n\geq n_{0}:\left|b_{n}\right|\geq\frac{\left|B\right|}{2}>0$
+\end_inset
+
+.
+ ZDB vsi členi zaporedja razen končno mnogo so v poljubno majhni okolici 
+\begin_inset Formula $\left|B\right|$
+\end_inset
+
+.
+ Če torej vzamemo točko na polovici med 0 in 
+\begin_inset Formula $\left|B\right|$
+\end_inset
+
+,
+ to je 
+\begin_inset Formula $\frac{\left|B\right|}{2}$
+\end_inset
+
+,
+ bo neskončno mnogo absolutnih vrednosti členov večjih od 
+\begin_inset Formula $\frac{\left|B\right|}{2}$
+\end_inset
+
+.
+ Pri razumevanju pomaga številska premica.
+ Nadalje uporabimo predpostavko z 
+\begin_inset Formula $\varepsilon=\frac{\left|B\right|}{2}$
+\end_inset
+
+,
+ torej je za 
+\begin_inset Formula $n>n_{0}:$
+\end_inset
+
+ 
+\begin_inset Formula $\left|B-b_{n}\right|<\frac{\left|B\right|}{2}$
+\end_inset
+
+ in velja 
+\begin_inset Formula 
+\[
+\left|b_{n}\right|=\left|B-\left(B-b_{n}\right)\right|=\left|B+\left(-\left(B-b_{n}\right)\right)\right|\overset{\text{trik. neen.}}{\geq}\left|\left|B\right|-\left|B-b_{n}\right|\right|=\left|B\right|-\left|B-b_{n}\right|\overset{\text{predp.}}{>}\left|B\right|-\frac{\left|B\right|}{2}=\frac{\left|B\right|}{2},
+\]
+
+\end_inset
+
+skratka 
+\begin_inset Formula $\left|b_{n}\right|>\frac{\left|B\right|}{2}$
+\end_inset
+
+.
+ Če spet izpustimo končno začetnih členov,
+ velja
+\begin_inset Formula 
+\[
+\frac{a_{n}}{b_{n}}-\frac{A}{B}=\frac{a_{n}B-Ab_{n}}{b_{n}B}\overset{\text{prištejemo in odštejemo člen}}{=}\frac{\left(a_{n}-A\right)B+A\left(B-b_{n}\right)}{Bb_{n}}=\frac{1}{b_{n}}\left(a_{n}-A\right)+\frac{A/B}{b_{n}}\left(B-b_{n}\right)
+\]
+
+\end_inset
+
+sedaj uporabimo na obeh straneh absolutno vrednost:
+\begin_inset Formula 
+\[
+\left|\frac{a_{n}}{b_{n}}-\frac{A}{B}\right|=\left|\frac{1}{b_{n}}\left(a_{n}-A\right)+\frac{A}{Bb_{n}}\left(B-b_{n}\right)\right|\leq\frac{1}{\left|b_{n}\right|}\left|a_{n}-A\right|+\frac{\left|A\right|}{\left|B\right|\left|b_{n}\right|}\left|B-b_{n}\right|<\frac{2}{\left|B\right|}\left|a_{n}-A\right|+\frac{2\left|A\right|}{\left|B\right|^{2}}\left|B-B_{n}\right|
+\]
+
+\end_inset
+
+skratka 
+\begin_inset Formula $\left|\frac{a_{n}}{b_{n}}-\frac{A}{B}\right|<\frac{2}{\left|B\right|}\left|a_{n}-A\right|+\frac{2\left|A\right|}{\left|B\right|^{2}}\left|B-B_{n}\right|$
+\end_inset
+
+.
+ Opazimo,
+ da 
+\begin_inset Formula $\frac{2}{\left|B\right|}$
+\end_inset
+
+ in 
+\begin_inset Formula $\frac{2\left|A\right|}{\left|B\right|^{2}}$
+\end_inset
+
+ nista odvisna od 
+\begin_inset Formula $n$
+\end_inset
+
+.
+ Sedaj vzemimo poljuben 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+ in 
+\begin_inset Formula $n_{1},n_{2}\in\mathbb{N}$
+\end_inset
+
+ takšna,
+ da velja:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $n\geq n_{1}\Rightarrow\left|a_{n}-A\right|<\frac{\varepsilon\left|B\right|}{4}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $n\geq n_{2}\Rightarrow\left|b_{n}-B\right|<\frac{\varepsilon\left|B\right|^{2}}{4\left|A\right|}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Tedaj iz zgornje ocene sledi za 
+\begin_inset Formula $n\geq\max\left\{ n_{0},n_{1},n_{2}\right\} $
+\end_inset
+
+:
+\begin_inset Formula 
+\[
+\left|\frac{a_{n}}{b_{n}}-\frac{A}{B}\right|<\frac{\cancel{2}}{\cancel{\left|B\right|}}\cdot\frac{\varepsilon\cancel{\left|B\right|}}{\cancelto{2}{4}}+\frac{\cancel{2}\cancel{\left|A\right|}}{\cancel{\left|B\right|^{2}}}\cdot\frac{\varepsilon\cancel{\left|B\right|^{2}}}{\cancelto{2}{4}\cancel{\left|A\right|}}=\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon,
+\]
+
+\end_inset
+
+s čimer dokažemo 
+\begin_inset Formula $a_{n}/b_{n}\to A/B$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\end_deeper
+\begin_layout Example*
+Naj bo 
+\begin_inset Formula $a>0$
+\end_inset
+
+.
+ Izračunajmo
+\begin_inset Formula 
+\[
+\sqrt{a+\sqrt{a+\sqrt{a+\sqrt{\cdots}}}}\eqqcolon\alpha.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $\alpha$
+\end_inset
+
+ je torej 
+\begin_inset Formula $\lim_{n\to\infty}x_{n}$
+\end_inset
+
+,
+ kjer je 
+\begin_inset Formula $x_{0}=0,x_{1}=\sqrt{a},x_{2}=\sqrt{a+\sqrt{a}},x_{3}=\sqrt{a+\sqrt{a+\sqrt{a}}},\dots,x_{n+1}=\sqrt{a+x_{n}}$
+\end_inset
+
+.
+ Iz zadnjega sledi 
+\begin_inset Formula $x_{n+1}^{2}=a+x_{n}$
+\end_inset
+
+.
+ Če torej limita 
+\begin_inset Formula $\alpha\coloneqq\lim x_{n}$
+\end_inset
+
+ obstaja,
+ mora veljati 
+\begin_inset Formula $\alpha^{2}=a+\alpha$
+\end_inset
+
+ oziroma 
+\begin_inset Formula $\alpha_{1,2}=\frac{1\pm\sqrt{1+4a}}{2}$
+\end_inset
+
+.
+ Opcija z minusom ni mogoča,
+ ker je zaporedje 
+\begin_inset Formula $\left(x_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ očitno pozitivno.
+ Če torej limita obstaja (
+\series bold
+česar še nismo dokazali
+\series default
+),
+ je enaka 
+\begin_inset Formula $\frac{1+\sqrt{1+4a}}{2}$
+\end_inset
+
+,
+ za primer 
+\begin_inset Formula $a=2$
+\end_inset
+
+ je torej 
+\begin_inset Formula $\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{\cdots}}}}=2$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Remark*
+Lahko se zgodi,
+ da limita rekurzivno podanega zaporedja ne obstaja,
+ čeprav jo znamo izračunati,
+ če bi obstajala.
+ Na primer 
+\begin_inset Formula $y_{1}\coloneqq1$
+\end_inset
+
+,
+ 
+\begin_inset Formula $y_{n+1}=1-2y_{n}$
+\end_inset
+
+ nam da zaporedje 
+\begin_inset Formula $1,-1,3,-5,11,\dots$
+\end_inset
+
+,
+ kar očitno nima limite.
+ Če bi limita obstajala,
+ bi zanjo veljalo 
+\begin_inset Formula $\beta=1-2\beta$
+\end_inset
+
+ oz.
+ 
+\begin_inset Formula $3\beta=1$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\beta=\frac{1}{3}$
+\end_inset
+
+.
+ Navedimo torej nekaj zadostnih in potrebnih pogojev za konvergenco zaporedij.
+\end_layout
+
+\begin_layout Theorem*
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hypertarget{kmoz}{Konvergenca monotonega in omejenega zaporedja}
+\end_layout
+
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ monotono realno zaporedje.
+ Če narašča,
+ ima limito 
+\begin_inset Formula $\lim_{n\to\infty}a_{n}=\sup\left\{ a_{n},n\in\mathbb{N}\right\} $
+\end_inset
+
+.
+ Če pada,
+ ima limito 
+\begin_inset Formula $\lim_{n\to\infty}a_{n}=\inf\left\{ a_{n},n\in\mathbb{N}\right\} $
+\end_inset
+
+.
+ (
+\begin_inset Formula $\sup$
+\end_inset
+
+ in 
+\begin_inset Formula $\inf$
+\end_inset
+
+ imata lahko tudi vrednost 
+\begin_inset Formula $\infty$
+\end_inset
+
+ in 
+\begin_inset Formula $-\infty$
+\end_inset
+
+ —
+ zaporedje s tako limito ni konvergentno v 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+).
+\end_layout
+
+\begin_layout Proof
+Denimo,
+ da 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ narašča.
+ Pišimo 
+\begin_inset Formula $s\coloneqq\sup_{n\in\mathbb{N}}a_{n}$
+\end_inset
+
+.
+ Vzemimo poljuben 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $s-\varepsilon$
+\end_inset
+
+ ni zgornja meja za 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+,
+ zato 
+\begin_inset Formula $\exists n_{0}\in\mathbb{N}\ni:s-\varepsilon<a_{n_{0}}$
+\end_inset
+
+.
+ Ker pa je 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+naraščajoče,
+ sledi 
+\begin_inset Formula $\forall n\geq n_{0}:a_{n}\geq a_{n_{0}}>s-\varepsilon$
+\end_inset
+
+.
+ Hkrati je 
+\begin_inset Formula $a_{n}\leq s$
+\end_inset
+
+,
+ saj je 
+\begin_inset Formula $s$
+\end_inset
+
+ zgornja meja.
+ Torej 
+\begin_inset Formula $\forall n\geq n_{0}:a_{n}\in(s-\varepsilon,s]\subset\left(s-\varepsilon,s+\varepsilon\right)$
+\end_inset
+
+,
+ s čimer dokažemo konvergenco.
+\end_layout
+
+\begin_layout Proof
+Denimo sedaj,
+ da 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ pada.
+ Dokaz je povsem analogen.
+ Pišimo 
+\begin_inset Formula $m\coloneqq\inf_{n\in\mathbb{N}}a_{n}$
+\end_inset
+
+.
+ Vzemimo poljuben 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $m+\varepsilon$
+\end_inset
+
+ ni spodnja meja za 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+,
+ zato 
+\begin_inset Formula $\exists n_{0}\in\mathbb{N}\ni:m+\varepsilon>a_{n_{0}}$
+\end_inset
+
+.
+ Ker pa je 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ padajoče,
+ sledi 
+\begin_inset Formula $\forall n\geq n_{0}:a_{n}\leq a_{n_{0}}<m+\varepsilon$
+\end_inset
+
+.
+ Hkrati je 
+\begin_inset Formula $a_{n}\geq m$
+\end_inset
+
+,
+ saj je 
+\begin_inset Formula $m$
+\end_inset
+
+ spodnja meja.
+ Torej 
+\begin_inset Formula $\forall n\geq n_{0}:a_{n}\in[m,m+\varepsilon)\subset\left(m-\varepsilon,m+\varepsilon\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Corollary*
+Za monotono zaporedje velja,
+ da je v 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+ konvergentno natanko tedaj,
+ ko je omejeno.
+\end_layout
+
+\begin_layout Example*
+Naj bo,
+ kot prej,
+ 
+\begin_inset Formula $a>0$
+\end_inset
+
+ in 
+\begin_inset Formula $x_{0}=0,x_{n+1}=\sqrt{a+x_{n}}$
+\end_inset
+
+.
+ Dokažimo,
+ da je 
+\begin_inset Formula $\left(x_{n}\right)_{n}$
+\end_inset
+
+ konvergentno.
+ Dovolj je pokazati,
+ da je naraščajoče in navzgor omejeno.
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Naraščanje z indukcijo:
+ Baza:
+ 
+\begin_inset Formula $0=x_{0}>x_{1}=\sqrt{a}$
+\end_inset
+
+.
+ Dokažimo 
+\begin_inset Formula $x_{n+1}-x_{n}>0$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+\left(x_{n+1}-x_{n}\right)\left(x_{n+1}+x_{n}\right)=x_{n+1}^{2}-x_{n}^{2}=\left(a+x_{n}\right)-\left(a+x_{n-1}\right)=x_{n}-x_{n-1}
+\]
+
+\end_inset
+
+Ker je zaporedje pozitivno,
+ je 
+\begin_inset Formula $x_{n+1}+x_{n}>0$
+\end_inset
+
+.
+ Desna stran je po I.
+ P.
+ pozitivna,
+ torej tudi 
+\begin_inset Formula $x_{n+1}-x_{n}>0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Omejenost:
+ Če je zaporedje res omejeno,
+ je po zgornjem tudi konvergentno in je 
+\begin_inset Formula $\sup_{n\in\mathbb{N}}x_{n}=\lim_{n\to\infty}x_{n}=\frac{1+\sqrt{1+4a}}{2}\leq\frac{1+\sqrt{1+4a+4a^{2}}}{2}=\frac{1+\sqrt{\left(2a+1\right)^{2}}}{}=1+a$
+\end_inset
+
+.
+ Uganili smo neko zgornjo mejo.
+ Domnevamo,
+ da 
+\begin_inset Formula $\forall n\in\mathbb{N}:x_{n}\leq1+a$
+\end_inset
+
+.
+ Dokažimo to z indukcijo:
+ Baza:
+ 
+\begin_inset Formula $0=x_{0}<1+a$
+\end_inset
+
+.
+ Po I.
+ P.
+ 
+\begin_inset Formula $x_{n}>1+a$
+\end_inset
+
+.
+ Korak:
+\begin_inset Formula 
+\[
+x_{n+1}=\sqrt{x_{n}+a}\leq\sqrt{1+a+a}=\sqrt{1+2a}<\sqrt{1+2a+2a^{2}}=1+a
+\]
+
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Example*
+S tem smo dokazali,
+ da 
+\begin_inset Formula $\lim_{n\to\infty}x_{n}=\frac{1+\sqrt{1+4a}}{2}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+To lahko dokažemo tudi na alternativen način.
+ Vidimo,
+ da je edini kandidat za limito,
+ če obstaja 
+\begin_inset Formula $L=\frac{1+\sqrt{1+4a}}{2}$
+\end_inset
+
+ in da torej velja 
+\begin_inset Formula $L^{2}=a+L$
+\end_inset
+
+.
+ Preverimo,
+ da je 
+\begin_inset Formula $L$
+\end_inset
+
+ res limita:
+ 
+\begin_inset Formula 
+\[
+x_{n+1}-L=\sqrt{a+x_{n}}-L=\frac{\left(\sqrt{a+x_{n}}-L\right)\left(\sqrt{a+x_{n}}+L\right)}{\sqrt{a+x_{n}}+L}=\frac{\left(a+x_{n}\right)-L^{2}}{\sqrt{a+x_{n}}+L}=\frac{\left(a+x_{n}\right)-\left(a+L\right)}{\sqrt{a+x_{n}}+L}=\frac{x_{n}-L}{\sqrt{a+x_{n}}+L}.
+\]
+
+\end_inset
+
+Vpeljimo sedaj 
+\begin_inset Formula $y_{n}\coloneqq x_{n}-L$
+\end_inset
+
+.
+ Sledi 
+\begin_inset Formula $\left|y_{n+1}\right|\leq\frac{\left|y_{n}\right|}{\sqrt{a+x_{n}}+L}\leq\frac{\left|y_{n}\right|}{L}$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $\left|y_{0}\right|=L$
+\end_inset
+
+,
+ dobimo
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+Za razumevanje si oglej nekaj členov rekurzivnega zaporedje 
+\begin_inset Formula $y_{0}=L,y_{n}=\frac{\left|y_{n+1}\right|}{L}$
+\end_inset
+
+.
+ Začnemo z 1 in nato vsakič delimo z 
+\begin_inset Formula $L$
+\end_inset
+
+.
+\end_layout
+
+\end_inset
+
+ oceno 
+\begin_inset Formula $\left|y_{n}\right|\leq\frac{1}{L^{n-1}}$
+\end_inset
+
+ oziroma 
+\begin_inset Formula $\left|x_{n}-L\right|\leq\frac{1}{L^{n-1}}$
+\end_inset
+
+.
+ Ker iz definicije 
+\begin_inset Formula $L$
+\end_inset
+
+ sledi 
+\begin_inset Formula $L>1$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $L^{n}\to\infty$
+\end_inset
+
+ za 
+\begin_inset Formula $n\to\infty$
+\end_inset
+
+,
+ torej smo dokazali,
+ da 
+\begin_inset Formula $\left|x_{n}-L\right|$
+\end_inset
+
+ eksponentno pada proti 0 za 
+\begin_inset Formula $n\to\infty$
+\end_inset
+
+.
+ Eksponentno padanje 
+\begin_inset Formula $\left|x_{n}-L\right|$
+\end_inset
+
+ proti 0 je dovolj,
+ da rečemo,
+ da zaporedje konvergira k 
+\begin_inset Formula $L$
+\end_inset
+
+
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+a res,
+ vprašaj koga.
+ ne razumem.
+ zakaj.
+ TODO.
+\end_layout
+
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+\begin_inset Formula $\lim_{n\to\infty}\sin n$
+\end_inset
+
+ in 
+\begin_inset Formula $\lim_{n\to\infty}\cos n$
+\end_inset
+
+ ne obstajata.
+\end_layout
+
+\begin_layout Proof
+Pišimo 
+\begin_inset Formula $a_{n}=\sin n$
+\end_inset
+
+ in 
+\begin_inset Formula $b_{n}=\cos n$
+\end_inset
+
+.
+ Iz adicijskih izrekov dobimo 
+\begin_inset Formula $a_{n+1}=\sin\left(n+1\right)=\sin n\cos1+\cos n\sin1=a_{n}\cos1+b_{n}\sin1$
+\end_inset
+
+.
+ Torej 
+\begin_inset Formula $b_{n}=\frac{a_{n+1}-a_{n}\cos1}{\sin1}$
+\end_inset
+
+.
+ Torej če 
+\begin_inset Formula $\exists a\coloneqq\lim_{n\to\infty}a_{n},a\in\mathbb{R}$
+\end_inset
+
+,
+ potem tudi 
+\begin_inset Formula $\exists b\coloneqq\lim_{n\to\infty}b_{n},b\in\mathbb{R}$
+\end_inset
+
+.
+ Podobno iz adicijske formule za 
+\begin_inset Formula $\cos\left(n+1\right)$
+\end_inset
+
+ sledi 
+\begin_inset Formula $a_{n}=\frac{b_{n}\cos1-b_{n+1}}{\sin1}$
+\end_inset
+
+,
+ torej če 
+\begin_inset Formula $\exists b$
+\end_inset
+
+,
+ potem tudi 
+\begin_inset Formula $\exists a$
+\end_inset
+
+.
+ Iz obojega sledi,
+ da 
+\begin_inset Formula $\exists a\Leftrightarrow\exists b$
+\end_inset
+
+.
+ Posledično,
+ če 
+\begin_inset Formula $a$
+\end_inset
+
+ in 
+\begin_inset Formula $b$
+\end_inset
+
+ obstajata,
+ iz zgornjih obrazcev za 
+\begin_inset Formula $a_{n}$
+\end_inset
+
+ in 
+\begin_inset Formula $b_{n}$
+\end_inset
+
+ sledi,
+ da za
+\begin_inset Formula 
+\[
+\lambda=\frac{1-\cos1}{\sin1}\in\left(0,1\right)
+\]
+
+\end_inset
+
+velja 
+\begin_inset Formula $b=\lambda a$
+\end_inset
+
+ in 
+\begin_inset Formula $a=-\lambda b$
+\end_inset
+
+ in zato 
+\begin_inset Formula $b=\lambda\left(-\lambda b\right)$
+\end_inset
+
+ oziroma 
+\begin_inset Formula $1=-\lambda^{2}$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $-1=\lambda^{2}$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\lambda=i$
+\end_inset
+
+,
+ kar je v protislovju z 
+\begin_inset Formula $\lambda\in\left(0,1\right)$
+\end_inset
+
+.
+ Podobno za 
+\begin_inset Formula $a=-\lambda\left(\lambda a\right)$
+\end_inset
+
+ oziroma 
+\begin_inset Formula $1=-\lambda^{2}$
+\end_inset
+
+,
+ kar je zopet 
+\begin_inset Formula $\rightarrow\!\leftarrow$
+\end_inset
+
+.
+ Edina druga opcija je,
+ da je 
+\begin_inset Formula $a=b=0$
+\end_inset
+
+.
+ Hkrati pa vemo,
+ da 
+\begin_inset Formula $a_{n}^{2}+b_{n}^{2}=1$
+\end_inset
+
+,
+ zato 
+\begin_inset Formula $a+b=1$
+\end_inset
+
+,
+ kar ni mogoče za ničelna 
+\begin_inset Formula $a$
+\end_inset
+
+ in 
+\begin_inset Formula $b$
+\end_inset
+
+.
+ Torej 
+\begin_inset Formula $a$
+\end_inset
+
+ in 
+\begin_inset Formula $b$
+\end_inset
+
+ ne obstajata.
+\end_layout
+
+\begin_layout Subsection
+Eulerjevo število
+\end_layout
+
+\begin_layout Theorem*
+Bernoullijeva neenakost.
+ 
+\begin_inset Formula $\forall\alpha\leq1,n\in\mathbb{N}$
+\end_inset
+
+ velja 
+\begin_inset Formula $\left(1-\alpha\right)^{n}\geq1-n\alpha$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Z indukcijo na 
+\begin_inset Formula $n$
+\end_inset
+
+ ob fiksnem 
+\begin_inset Formula $\alpha$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Baza:
+ 
+\begin_inset Formula $n=1$
+\end_inset
+
+:
+ 
+\begin_inset Formula $\left(1-\alpha\right)^{1}=1-1\alpha$
+\end_inset
+
+.
+ Velja celo enakost.
+\end_layout
+
+\begin_layout Itemize
+I.
+ P.:
+ Velja 
+\begin_inset Formula $\left(1-\alpha\right)^{n}\geq1-n\alpha$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+Korak:
+ 
+\begin_inset Formula $\left(1-\alpha\right)^{n+1}=\left(1-\alpha\right)\left(1-\alpha\right)^{n}\geq\left(1-\alpha\right)$
+\end_inset
+
+
+\begin_inset Formula $\left(1-n\alpha\right)=1-n\alpha-\alpha-n\alpha^{2}=1-\left(n+1\right)\alpha-n\alpha^{2}\geq1-\left(n+1\right)\alpha$
+\end_inset
+
+,
+ torej ocena velja tudi za 
+\begin_inset Formula $n+1$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Definition*
+Vpeljimo oznaki:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Za 
+\begin_inset Formula $n\in\mathbb{N}$
+\end_inset
+
+ označimo 
+\begin_inset Formula $n!=1\cdot2\cdot3\cdot\cdots\cdot n$
+\end_inset
+
+ (pravimo 
+\begin_inset Formula $n-$
+\end_inset
+
+faktoriala oziroma 
+\begin_inset Formula $n-$
+\end_inset
+
+fakulteta).
+ Ker velja 
+\begin_inset Formula $n!=n\cdot\left(n-1\right)!$
+\end_inset
+
+ za 
+\begin_inset Formula $n\geq2$
+\end_inset
+
+,
+ je smiselno definirati še 
+\begin_inset Formula $0!=1$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Za 
+\begin_inset Formula $n,k\in\mathbb{N}$
+\end_inset
+
+ označimo še binomski simbol:
+ 
+\begin_inset Formula $\binom{n}{k}\coloneqq\frac{n!}{k!\left(n-k\right)!}$
+\end_inset
+
+ (pravimo 
+\begin_inset Formula $n$
+\end_inset
+
+ nad 
+\begin_inset Formula $k$
+\end_inset
+
+).
+\end_layout
+
+\begin_layout Itemize
+Če je 
+\begin_inset Formula $\left(a_{k}\right)_{k}$
+\end_inset
+
+ neko zaporedje (lahko tudi končno),
+ lahko pišemo 
+\begin_inset Formula $\sum_{k=1}^{n}a_{k}\coloneqq a_{1}+a_{2}+\cdots+a_{n}$
+\end_inset
+
+ (pravimo summa) in 
+\begin_inset Formula $\prod_{k=1}^{n}a_{k}\coloneqq a_{1}\cdot\cdots\cdot a_{n}$
+\end_inset
+
+ (pravimo produkt).
+\end_layout
+
+\end_deeper
+\begin_layout Example*
+\begin_inset Formula $\sum_{k=1}^{n}\frac{1}{k}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}$
+\end_inset
+
+ in 
+\begin_inset Formula $\prod_{k=1}^{n}k=1\cdot2\cdot3\cdot\cdots\cdot n=n!$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+Binomska formula.
+ 
+\begin_inset Formula $\forall a,b\in\mathbb{R},n\in\mathbb{N}:\left(a+b\right)^{n}=\sum_{k=0}^{n}\binom{n}{k}a^{k}b^{n-k}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Indukcija po 
+\begin_inset Formula $n$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Baza 
+\begin_inset Formula $n=1$
+\end_inset
+
+:
+ 
+\begin_inset Formula $\sum_{k=0}^{1}\binom{n}{k}a^{k}b^{n-k}=\binom{1}{0}a^{0}b^{1-0}+\binom{1}{1}a^{1}b^{1-1}=a+b=\left(a+b\right)^{1}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+I.
+ P.
+ 
+\begin_inset Formula $\sum_{k=0}^{n}\binom{n}{k}a^{k}b^{n-k}=\left(a+b\right)^{n}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+Korak:
+ 
+\begin_inset Formula 
+\[
+\left(a+b\right)^{n+1}=\left(a+b\right)\left(a+b\right)^{n}=\left(a+b\right)\sum_{k=0}^{n}\binom{n}{k}a^{k}b^{n-k}=\sum_{k=0}^{n}\binom{n}{k}a^{k+1}b^{n-k}+\sum_{k=0}^{n}\binom{n}{k}a^{k}b^{n-k+1}=
+\]
+
+\end_inset
+
+sedaj naj bo 
+\begin_inset Formula $m=k+1$
+\end_inset
+
+ v levem členu:
+\begin_inset Formula 
+\[
+=\sum_{m=1}^{n+1}\binom{n}{m-1}a^{m}b^{n-\left(m-1\right)}+\sum_{k=0}^{n}\binom{n}{k}a^{k}b^{n-k+1}=a^{n+1}+\sum_{k=1}^{n}\left[\binom{n}{k-1}+\binom{n}{k}\right]a^{k}b^{n-k+1}+b^{n+1}=
+\]
+
+\end_inset
+
+Sedaj obravnavajmo le izraz v oglatih oklepajih:
+\begin_inset Formula 
+\[
+\binom{n}{k-1}+\binom{n}{k}=\frac{n!}{\left(k-1\right)!\left(n-k+1\right)!}+\frac{n!}{k!\left(n-k\right)!}=\frac{kn!}{k!\left(n-k+1\right)!}+\frac{n!\left(n-k+1\right)}{k!\left(n-k+1\right)!}=\frac{kn!+n!\left(n-k+1\right)}{k!\left(n-k+1\right)!}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\frac{n!\left(\cancel{k+}n\cancel{-k}+1\right)}{k!\left(n+1-k\right)!}=\frac{n!\left(n+1\right)}{k!\left(n+1-k\right)!}=\frac{\left(n+1\right)!}{k!\left(n+1-k\right)}=\binom{n+1}{k}
+\]
+
+\end_inset
+
+in skratka dobimo 
+\begin_inset Formula $\binom{n}{k-1}+\binom{n}{k}=\binom{n+1}{k}$
+\end_inset
+
+.
+ Vstavimo to zopet v naš zgornji račun:
+\begin_inset Formula 
+\[
+\cdots=a^{n+1}+\sum_{k=1}^{n}\left[\binom{n}{k-1}+\binom{n}{k}\right]a^{k}b^{n-k+1}+b^{n+1}=a^{n+1}+\sum_{k=1}^{n}\binom{n+1}{k}a^{k}b^{n-k+1}+b^{n+1}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=a^{n+1}+\sum_{k=1}^{n}\binom{n+1}{k}a^{k}b^{n-k+1}+b^{n+1}=a^{n+1}+\sum_{k=0}^{n}\binom{n+1}{k}a^{k}b^{n-k+1}=\sum_{k=0}^{n+1}\binom{n+1}{k}a^{k}b^{n-k+1}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Theorem*
+Bernoulli.
+ Zaporedje 
+\begin_inset Formula $a_{n}\coloneqq\left(1+\frac{1}{n}\right)^{n}$
+\end_inset
+
+ je konvergentno.
+\end_layout
+
+\begin_layout Proof
+Dokazali bomo,
+ da je naraščajoče in omejeno.
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Naraščanje:
+ Dokazujemo,
+ da za 
+\begin_inset Formula $n\geq2$
+\end_inset
+
+ velja 
+\begin_inset Formula $a_{n}\geq a_{n-1}$
+\end_inset
+
+ oziroma 
+\begin_inset Formula 
+\[
+\left(1+\frac{1}{n}\right)^{n}\overset{?}{\geq}\left(1+\frac{1}{n-1}\right)^{n-1}=\left(\frac{n-1}{n-1}+\frac{1}{n-1}\right)^{n-1}=\left(\frac{n}{n-1}\right)^{n-1}=\left(\frac{n-1}{n}\right)^{1-n}=\left(1-\frac{1}{n}\right)^{1-n}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\left(1+\frac{1}{n}\right)^{n}\overset{?}{\geq}\left(1-\frac{1}{n}\right)^{1-n}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\left(1+\frac{1}{n^{2}}\right)^{n}=\left(\left(1+\frac{1}{n}\right)\left(1-\frac{1}{n}\right)\right)^{n}=\left(1+\frac{1}{n}\right)^{n}\left(1-\frac{1}{n}\right)^{n}\overset{?}{\geq}1-\frac{1}{n}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\left(1+\frac{1}{n^{2}}\right)^{n}\overset{?}{\geq}1-\frac{1}{n},
+\]
+
+\end_inset
+
+kar je poseben primer Bernoullijeve neenakosti za 
+\begin_inset Formula $\alpha=\frac{1}{n^{2}}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Omejenost:
+ Po binomski formuli je 
+\begin_inset Formula 
+\[
+a_{n}=\left(1+\frac{1}{n}\right)^{n}=\sum_{k=0}^{n}\binom{n}{k}\left(\frac{1}{n}\right)^{k}=\sum_{k=0}^{n}\frac{n!}{k!\left(n-k\right)!n^{k}}=1+1+\sum_{k=2}^{n}\frac{n!}{k!\left(n-k\right)!n^{k}}=2+\sum_{k=2}^{n}\frac{1}{k!}\cdot\frac{n\left(n-1\right)\left(n-2\right)\cdots\left(n-k+1\right)}{n^{k}}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=2+\sum_{k=2}^{n}\frac{1}{k!}\cdot\frac{n}{n}\cdot\frac{n-1}{n}\cdot\frac{n-2}{n}\cdot\cdots\cdot\frac{n-k+1}{n}=2+\sum_{k=2}^{n}\frac{1}{k!}\cdot\cancel{\left(1-0\right)}\cdot\left(1-\frac{1}{n}\right)\cdot\left(1-\frac{2}{n}\right)\cdot\cdots\cdot\left(1-\frac{k-1}{n}\right)=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=2+\sum_{k=2}^{n}\frac{1}{k!}\prod_{j=1}^{k-1}\left(1-\frac{j}{n}\right)<2+\sum_{k=2}^{n}\frac{1}{k!}<2+\sum_{k=2}^{n}\frac{1}{2^{k-1}}=\cdots
+\]
+
+\end_inset
+
+Opomnimo,
+ da je 
+\begin_inset Formula $1-\frac{j}{n}<1$
+\end_inset
+
+,
+ zato 
+\begin_inset Formula $\prod_{j=1}^{k-1}\left(1-\frac{j}{n}\right)<1$
+\end_inset
+
+ (prvi neenačaj) ter 
+\begin_inset Formula $k!=1\cdot2\cdot3\cdot\cdots\cdot k\geq1\cdot2\cdot2\cdot\cdots\cdot2=2^{k-1}$
+\end_inset
+
+ (drugi).
+ Sedaj si z indukcijo dokažimo 
+\begin_inset Formula $\sum_{k=2}^{n}\frac{1}{2^{k-1}}=1-\frac{1}{2^{n-1}}$
+\end_inset
+
+:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Baza:
+ 
+\begin_inset Formula $n=2$
+\end_inset
+
+:
+ 
+\begin_inset Formula $\frac{1}{2^{2-1}}=1-\frac{1}{2^{2-1}}=1-\frac{1}{2}=\frac{1}{2}$
+\end_inset
+
+.
+ Velja!
+\end_layout
+
+\begin_layout Itemize
+I.
+ P.:
+ 
+\begin_inset Formula $\sum_{k=2}^{n}\frac{1}{2^{k-1}}=1-\frac{1}{2^{n-1}}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+Korak:
+ 
+\begin_inset Formula $\sum_{k=2}^{n+1}\frac{1}{2^{k-1}}=1-\frac{1}{2^{n-1}}+\frac{1}{2^{n+1-1}}=1-2\cdot2^{-n}+2^{-n}=1+2^{-n}\left(1-2\right)=1+2^{-n}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+In nadaljujmo z računanjem:
+\begin_inset Formula 
+\[
+\cdots=2+1-\frac{1}{2^{n-1}}=3-\frac{1}{2^{n-1}},
+\]
+
+\end_inset
+
+s čimer dobimo zgornjo mejo 
+\begin_inset Formula $\forall n\in\mathbb{N}:a_{n}<3$
+\end_inset
+
+.
+ Ker je očitno 
+\begin_inset Formula $\forall n\in\mathbb{N}:a_{n}>0$
+\end_inset
+
+,
+ je torej zaporedje omejeno in ker je tudi monotono po 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{kmoz}{prejšnjem izreku}
+\end_layout
+
+\end_inset
+
+ konvergira.
+\end_layout
+
+\end_deeper
+\end_deeper
+\begin_layout Definition*
+Označimo število 
+\begin_inset Formula $e\coloneqq\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{n}$
+\end_inset
+
+ in ga imenujemo Eulerjevo število.
+ Velja 
+\begin_inset Formula $e\approx2,71828$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Remark*
+V dokazu vidimo moč izreka 
+\begin_inset Quotes gld
+\end_inset
+
+omejenost in monotonost 
+\begin_inset Formula $\Rightarrow$
+\end_inset
+
+ konvergenca
+\begin_inset Quotes grd
+\end_inset
+
+,
+ saj nam omogoča dokazati konvergentnost zaporedja brez kandidata za limito.
+ Jasno je,
+ da ne bi mogli vnaprej uganiti,
+ da je limita ravno 
+\shape italic
+transcendentno število
+\shape default
+ 
+\begin_inset Formula $e$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Podzaporedje zaporedja 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ je poljubno zaporedje oblike 
+\begin_inset Formula $\left(a_{\varphi\left(n\right)}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+,
+ kjer je 
+\begin_inset Formula $\varphi:\mathbb{N}\to\mathbb{N}$
+\end_inset
+
+ strogo naraščajoča funkcija.
+\end_layout
+
+\begin_layout Theorem*
+Če je 
+\begin_inset Formula $L=\lim_{n\to\infty}a_{n}$
+\end_inset
+
+,
+ tedaj je 
+\begin_inset Formula $L$
+\end_inset
+
+ tudi limita vsakega podzaporedja.
+\end_layout
+
+\begin_layout Proof
+Po predpostavki velja 
+\begin_inset Formula $\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\forall n\in\mathbb{N}:n\geq n_{0}:\left|a_{n}-L\right|<\varepsilon$
+\end_inset
+
+.
+ Vzemimo poljuben 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+.
+ Po predpostavki obstaja 
+\begin_inset Formula $n_{0}\in\mathbb{N}$
+\end_inset
+
+,
+ da bodo vsi členi zaporedja po 
+\begin_inset Formula $n_{0}-$
+\end_inset
+
+tem v 
+\begin_inset Formula $\left(L-\varepsilon,L+\varepsilon\right)$
+\end_inset
+
+.
+ Iz definicijskega območja 
+\begin_inset Formula $\varphi$
+\end_inset
+
+ vzemimo poljuben element 
+\begin_inset Formula $n_{1}$
+\end_inset
+
+,
+ da velja 
+\begin_inset Formula $n_{1}\geq n_{0}$
+\end_inset
+
+.
+ Gotovo obstaja,
+ ker je definicijsko območje števno neskončne moči in s pogojem 
+\begin_inset Formula $n_{1}\geq n_{0}$
+\end_inset
+
+ onemogočimo izbiro le končno mnogo elementov.
+ 
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+Če slednji ne obstaja,
+ je v 
+\begin_inset Formula $D_{\varphi}$
+\end_inset
+
+ končno mnogo elementov,
+ tedaj vzamemo 
+\begin_inset Formula $n_{1}\coloneqq\max D_{\varphi}+1$
+\end_inset
+
+ in je pogoj za limito izpolnjen na prazno.
+ Sicer pa v
+\end_layout
+
+\end_inset
+
+Velja 
+\begin_inset Formula $\forall n\in\mathbb{N}:n>n_{1}\Rightarrow\left|a_{\varphi n}-L\right|<\varepsilon$
+\end_inset
+
+,
+ ker je 
+\begin_inset Formula $\varphi$
+\end_inset
+
+ strogo naraščajoča in izbiramo le elemente podzaporedja,
+ ki so v izvornem zaporedju za 
+\begin_inset Formula $n_{0}-$
+\end_inset
+
+tim členom in zato v 
+\begin_inset Formula $\left(L-\varepsilon,L+\varepsilon\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $\lim_{n\to\infty}\frac{1}{2n+3}=\lim_{n\to\infty}\frac{1}{n}=0$
+\end_inset
+
+ za zaporedje 
+\begin_inset Formula $a_{n}=\frac{1}{n}$
+\end_inset
+
+ in podzaporedje 
+\begin_inset Formula $a_{\varphi n}$
+\end_inset
+
+,
+ kjer je 
+\begin_inset Formula $\varphi\left(n\right)=2n+3$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+Karakterizacija limite s podzaporedji.
+ Naj bo 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ realno zaporedje in 
+\begin_inset Formula $L\in\mathbb{R}$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $L=\lim_{n\to\infty}a_{n}\Leftrightarrow$
+\end_inset
+
+ za vsako podzaporedje 
+\begin_inset Formula $\left(a_{n_{k}}\right)_{k\in\mathbb{N}}$
+\end_inset
+
+ zaporedja 
+\begin_inset Formula $\left(a_{n}\right)_{n}$
+\end_inset
+
+ obstaja njegovo podzaporedje 
+\begin_inset Formula $\left(a_{n_{k_{l}}}\right)_{l\in\mathbb{N}}$
+\end_inset
+
+,
+ ki konvergira k 
+\begin_inset Formula $L$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Dokazujemo ekvivalenco:
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Rightarrow\right)$
+\end_inset
+
+ Dokazano poprej.
+ Limita se pri prehodu na podzaporedje ohranja.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Leftarrow\right)$
+\end_inset
+
+ PDDRAA 
+\begin_inset Formula $a_{n}\not\to L$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $\exists\varepsilon>0$
+\end_inset
+
+ in podzaporedje 
+\begin_inset Formula $\left(a_{n_{k}}\right)_{k\in\mathbb{N}}\ni:\forall k\in\mathbb{N}:\left|a_{n_{k}}-K\right|>\varepsilon$
+\end_inset
+
+ (*)
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+tu je na 
+\begin_inset Quotes gld
+\end_inset
+
+Zaporedja 2
+\begin_inset Quotes grd
+\end_inset
+
+ napaka,
+ neenačaj obrne v drugo smer
+\end_layout
+
+\end_inset
+
+.
+ Po predpostavki sedaj 
+\begin_inset Formula $\exists\left(a_{n_{k_{l}}}\right)_{l\in\mathbb{N}}\ni:\lim_{l\to\infty}a_{n_{k_{l}}}=L$
+\end_inset
+
+.
+ To pa je v protislovju z (*),
+ torej je začetna predpostavka 
+\begin_inset Formula $a_{n}\not\to L$
+\end_inset
+
+ napačna,
+ torej 
+\begin_inset Formula $a_{n}\to L$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Subsection
+Stekališča
+\end_layout
+
+\begin_layout Definition*
+Točka 
+\begin_inset Formula $s\in\mathbb{R}$
+\end_inset
+
+ je stekališče zaporedje 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}\subset\mathbb{R}$
+\end_inset
+
+,
+ če v vsaki okolici te točke leži neskončno členov zaporedja.
+\end_layout
+
+\begin_layout Remark*
+Pri limiti zahtevamo več;
+ da izven vsake okolice limite leži le končno mnogo členov.
+\end_layout
+
+\begin_layout Example*
+Primeri stekališč.
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $L=\lim_{n\to\infty}a_{n}\Rightarrow L$
+\end_inset
+
+ je stekališče za 
+\begin_inset Formula $a_{n}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $0,1,0,1,\dots$
+\end_inset
+
+ stekališči sta 
+\begin_inset Formula $\left\{ 0,1\right\} $
+\end_inset
+
+ in zaporedje nima limite (ni konvergentno)
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $1,1,2,1,2,3,1,2,3,4,\dots$
+\end_inset
+
+ ima neskončno stekališč,
+ 
+\begin_inset Formula $\mathbb{N}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $b_{n}=n-$
+\end_inset
+
+to racionalno število
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+Racionalnih števil je števno mnogo,
+ zato jih lahko linearno uredimo in oštevilčimo.
+\end_layout
+
+\end_inset
+
+ ima neskončno stekališč,
+ 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Remark*
+Limita je stakališče,
+ stekališče pa ni nujno limita.
+ Poleg tega,
+ če se spomnimo,
+ velja,
+ da vsota konvergentnih zaporedij konvergira k vsoti njunih limit,
+ ni pa nujno res,
+ da so stekališča vsote dveh zaporedij paroma vsote stekališč teh dveh zaporedij.
+ Primer:
+ 
+\begin_inset Formula $a_{n}=\left(-1\right)^{n}$
+\end_inset
+
+ in 
+\begin_inset Formula $b_{n}=-\left(-1\right)^{n}$
+\end_inset
+
+.
+ Njuni stekališči sta 
+\begin_inset Formula $\left\{ -1,1\right\} $
+\end_inset
+
+,
+ toda 
+\begin_inset Formula $a_{n}+b_{n}=0$
+\end_inset
+
+ ima le stekališče 
+\begin_inset Formula $\left\{ 0\right\} $
+\end_inset
+
+,
+ ne pa tudi 
+\begin_inset Formula $\left\{ 1,-1,2,-2\right\} $
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+\begin_inset Formula $S$
+\end_inset
+
+ je stekališče 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}\Leftrightarrow S$
+\end_inset
+
+ je limita nekega podzaporedja 
+\begin_inset Formula $a_{n}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Dokazujemo ekvivalenco.
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Leftarrow\right)$
+\end_inset
+
+ Očitno.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Rightarrow\right)$
+\end_inset
+
+ Definirajmo 
+\begin_inset Formula $\forall m\in\mathbb{N}:U_{m}\coloneqq\left(S-\frac{1}{m},S+\frac{1}{m}\right)$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $S$
+\end_inset
+
+ stekališče,
+ 
+\begin_inset Formula $\forall m\in\mathbb{N}\exists a_{k_{m}}\in U_{m}$
+\end_inset
+
+.
+ Podzaporedje 
+\begin_inset Formula $\left(a_{k_{m}}\right)_{m\in\mathbb{N}}$
+\end_inset
+
+ konvergira k 
+\begin_inset Formula $S$
+\end_inset
+
+,
+ kajti 
+\begin_inset Formula $\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow\left|a_{k_{n}}-S\right|<\frac{1}{n}<\varepsilon$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Corollary*
+Če je 
+\begin_inset Formula $L$
+\end_inset
+
+ limita nekega zaporedja,
+ je 
+\begin_inset Formula $L$
+\end_inset
+
+ edino njegovo stekališče.
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $a_{n}\to L$
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $S$
+\end_inset
+
+ stekališče za 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+.
+ Po izreku zgoraj je 
+\begin_inset Formula $S$
+\end_inset
+
+ limita nekega podzaporedja 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+.
+ Toda limita vsakega podzaporedja je enaka limiti zaporedja,
+ iz katerega to podzaporedje izhaja,
+ če ta limita obstaja.
+ Potemtakem je 
+\begin_inset Formula $S=L$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hypertarget{bw}{Bolzano-Weierstraß}
+\end_layout
+
+\end_inset
+
+.
+ Eksistenčni izrek.
+ Vsako omejeno zaporedje v realnih številih ima kakšno stekališče v 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Označimo 
+\begin_inset Formula $m_{0}\coloneqq\inf_{n\in\mathbb{N}}a_{n},M_{0}\coloneqq\sup_{n\in\mathbb{N}}a_{n},I_{0}\coloneqq\left[m_{0},M_{0}\right]$
+\end_inset
+
+.
+ Očitno je 
+\begin_inset Formula $\forall n\in\mathbb{N}:a_{n}\in I_{0}$
+\end_inset
+
+.
+ Interval 
+\begin_inset Formula $I_{0}$
+\end_inset
+
+ razdelimo na dve polovici:
+ 
+\begin_inset Formula $I_{0}=\left[m_{0},\frac{m_{0}+M_{0}}{2}\right]\cup\left[\frac{m_{0}+M_{0}}{2},M_{0}\right]$
+\end_inset
+
+.
+ Izberemo polovico (vsaj ena obstaja),
+ v kateri leži neskončno mnogo členov,
+ in jo označimo z 
+\begin_inset Formula $I_{1}$
+\end_inset
+
+.
+ Spet jo razdelimo na pol in z 
+\begin_inset Formula $I_{2}$
+\end_inset
+
+ označimo tisto polovico,
+ v kateri leži neskončno mnogo členov.
+ Postopek ponavljamo in dobimo zaporedje zaprtih intervalov 
+\begin_inset Formula $\left(I_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ in velja 
+\begin_inset Formula $I_{0}\supset I_{1}\supset I_{2}\supset\cdots$
+\end_inset
+
+ ter 
+\begin_inset Formula $\left|I_{n}\right|=2^{-n}\left|I_{0}\right|$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Označimo sedaj 
+\begin_inset Formula $I_{n}\eqqcolon\left[l_{n},d_{n}\right]$
+\end_inset
+
+.
+ Iz konstrukcije je očitno,
+ da 
+\begin_inset Formula $\left(l_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ narašča in 
+\begin_inset Formula $\left(d_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ pada ter da sta obe zaporedji omejeni.
+ Posledično 
+\begin_inset Formula $\exists l\coloneqq\lim_{n\to\infty}l_{n},d\coloneqq\lim_{n\to\infty}d_{n}$
+\end_inset
+
+.
+ Iz 
+\begin_inset Formula $l_{n}\leq l\leq d\leq d_{n}$
+\end_inset
+
+ sledi ocena 
+\begin_inset Formula $d-l\leq l_{n}-d_{n}=\left|I_{n}\right|=2^{-n}\left|I_{0}\right|$
+\end_inset
+
+,
+ kar konvergira k 0.
+ Posledično 
+\begin_inset Formula $d=l$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Treba je pokazati še,
+ da je 
+\begin_inset Formula $d=l$
+\end_inset
+
+ stekališče za 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+.
+ Vzemimo poljuben 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $l=\lim_{n\to\infty}l_{n}\Rightarrow\exists n_{1}\in\mathbb{N}\ni:l_{n_{1}}>l-\varepsilon$
+\end_inset
+
+ in ker je 
+\begin_inset Formula $d=\lim_{n\to\infty}d_{n}\Rightarrow\exists n_{2}\in\mathbb{N}\ni:d_{n_{2}}<d-\varepsilon$
+\end_inset
+
+.
+ Torej 
+\begin_inset Formula $\left[l_{n_{1}},d_{n_{2}}\right]\subset\left(l-\varepsilon,d+\varepsilon\right)$
+\end_inset
+
+.
+ Torej za 
+\begin_inset Formula $n_{0}\coloneqq\max\left\{ n_{1},n_{2}\right\} $
+\end_inset
+
+ velja 
+\begin_inset Formula $I_{n_{0}}=\left[l_{n_{0}},d_{n_{n}}\right]\subset\left(l-\varepsilon,d+\varepsilon\right)$
+\end_inset
+
+.
+ Ker 
+\begin_inset Formula $I_{n_{0}}$
+\end_inset
+
+ po konstrukciji vsebuje neskončno mnogo elementov,
+ jih torej tudi 
+\begin_inset Formula $\left(l-\varepsilon,d+\varepsilon\right)$
+\end_inset
+
+ oziroma poljubno majhna okolica 
+\begin_inset Formula $d=l$
+\end_inset
+
+,
+ torej je 
+\begin_inset Formula $d=l$
+\end_inset
+
+ stekališče za 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Corollary*
+Če je 
+\begin_inset Formula $s\in\mathbb{R}$
+\end_inset
+
+ edino stekališče omejenega zaporedja 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+,
+ tedaj je 
+\begin_inset Formula $s=\lim_{n\to\infty}a_{n}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $s$
+\end_inset
+
+ stekališče 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+.
+ PDDRAA 
+\begin_inset Formula $a_{n}\not\to s$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $\exists\varepsilon>0\ni:$
+\end_inset
+
+ izven 
+\begin_inset Formula $\left(s-\varepsilon,s+\varepsilon\right)$
+\end_inset
+
+ se nahaja neskončno mnogo členov zaporedja.
+ Ti členi sami zase tvorijo omejeno zaporedje,
+ ki ima po 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{bw}{B.-W.}
+\end_layout
+
+\end_inset
+
+ izreku stekališče.
+ Slednje gotovo ne more biti enako 
+\begin_inset Formula $s$
+\end_inset
+
+,
+ torej imamo vsaj dve stekališči,
+ kar je v je v 
+\begin_inset Formula $\rightarrow\!\leftarrow$
+\end_inset
+
+ s predpostavko.
+\end_layout
+
+\begin_layout Definition*
+Pravimo,
+ da ima realno zaporedje:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+stekališče v 
+\begin_inset Formula $\infty$
+\end_inset
+
+,
+ če 
+\begin_inset Formula $\forall M>0:\left(M,\infty\right)$
+\end_inset
+
+ vsebuje neskončno mnogo členov zapopredja
+\end_layout
+
+\begin_layout Itemize
+limito v 
+\begin_inset Formula $\infty$
+\end_inset
+
+,
+ če 
+\begin_inset Formula $\forall M>0:\left(M,\infty\right)$
+\end_inset
+
+ vsebuje vse člene zaporedja od nekega indeksa dalje
+\end_layout
+
+\begin_layout Standard
+in podobno za 
+\begin_inset Formula $-\infty$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Remark*
+Povezava s pojmom realnega stekališča/limite:
+ okolice 
+\begin_inset Quotes gld
+\end_inset
+
+točke
+\begin_inset Quotes grd
+\end_inset
+
+ 
+\begin_inset Formula $\infty$
+\end_inset
+
+ so intervali oblike 
+\begin_inset Formula $\left(M,\infty\right)$
+\end_inset
+
+.
+ To je smiselno,
+ saj biti 
+\begin_inset Quotes gld
+\end_inset
+
+blizu 
+\begin_inset Formula $\infty$
+\end_inset
+
+
+\begin_inset Quotes grd
+\end_inset
+
+ pomeni bizi zelo velik,
+ kar je ravno biti v 
+\begin_inset Formula $\left(M,\infty\right)$
+\end_inset
+
+za poljubno velik 
+\begin_inset Formula $M$
+\end_inset
+
+.
+ 
+\begin_inset Quotes gld
+\end_inset
+
+Okolica točke 
+\begin_inset Formula $\infty$
+\end_inset
+
+
+\begin_inset Quotes grd
+\end_inset
+
+ so torej vsi intervali oblike 
+\begin_inset Formula $\left(M,\infty\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsection
+Limes superior in limes inferior
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+realno zaporedje.
+ Tvorimo novo zaporedje 
+\begin_inset Formula $s_{n}\coloneqq\sup\left\{ a_{k};k\geq n\right\} $
+\end_inset
+
+.
+ Očitno je padajoče (
+\begin_inset Formula $s_{1}\geq s_{2}\geq s_{3}\geq\cdots$
+\end_inset
+
+),
+ ker je supremum množice vsaj supremum njene stroge podmnožice.
+ Zaporedje 
+\begin_inset Formula $\left(s_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ ima limito,
+ ki ji rečemo limes superior oziroma zgornja limita zaporedja 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ in označimo 
+\begin_inset Formula $\limsup_{n\to\infty}a_{n}=\overline{\lim_{n\to\infty}}a_{n}\coloneqq\lim_{n\to\infty}s_{n}$
+\end_inset
+
+ in velja,
+ da leži v 
+\begin_inset Formula $\mathbb{R}\cup\left\{ -\infty,\infty\right\} $
+\end_inset
+
+.
+ Podobno definiramo tudi limes inferior oz.
+ spodnjo limito zaporedja:
+ 
+\begin_inset Formula $\liminf_{n\to\infty}a_{n}=\underline{\lim_{n\to\infty}}a_{n}\coloneqq\lim_{n\to\infty}\left(\inf_{k\geq n}a_{k}\right)=\sup_{n\in\mathbb{N}}\left(\inf_{k\geq n}a_{k}\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Remark*
+Za razliko od običajne limite,
+ ki lahko ne obstaja,
+ 
+\begin_inset Formula $\limsup$
+\end_inset
+
+ in 
+\begin_inset Formula $\liminf$
+\end_inset
+
+ vedno obstajata.
+\end_layout
+
+\begin_layout Claim*
+\begin_inset Formula $\limsup_{n\to\infty}a_{n}$
+\end_inset
+
+ je največje stekališče zaporedja 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ in 
+\begin_inset Formula $\liminf_{n\to\infty}$
+\end_inset
+
+ najmanjše.
+\end_layout
+
+\begin_layout Proof
+Označimo 
+\begin_inset Formula $s\coloneqq\limsup_{n\to\infty}a_{n}$
+\end_inset
+
+.
+ Za 
+\begin_inset Formula $\liminf$
+\end_inset
+
+ je dokaz analogen in ga ne bomo pisali.
+ Dokazujemo,
+ da je 
+\begin_inset Formula $s$
+\end_inset
+
+ stekališče in 
+\begin_inset Formula $\forall t>s:t$
+\end_inset
+
+ ni stekališče.
+ Ločimo primere:
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $s\in\mathbb{R}$
+\end_inset
+
+ Naj bo 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+ poljuben.
+ Ker
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+Infimum padajočega konvergentnega zaporedja je očitno njegova limita.
+\end_layout
+
+\end_inset
+
+ je 
+\begin_inset Formula $s=\inf s_{n}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\exists n_{0}\in\mathbb{N}\ni:s_{n_{0}}\in[s,s+\varepsilon)$
+\end_inset
+
+.
+ Ker 
+\begin_inset Formula $\left(s_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ pada proti 
+\begin_inset Formula $s$
+\end_inset
+
+,
+ sledi 
+\begin_inset Formula $\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow s_{n}\in[s,s+\varepsilon)$
+\end_inset
+
+.
+ Po definiciji 
+\begin_inset Formula $s_{n}$
+\end_inset
+
+ velja 
+\begin_inset Formula $\forall n\in\mathbb{N}\exists N\left(n\right)\geq n\ni:s_{n}-\varepsilon<a_{N\left(n\right)}$
+\end_inset
+
+.
+ Torej imamo 
+\begin_inset Formula $s-\varepsilon\leq s_{n}-\varepsilon<a_{N\left(n\right)}\leq s_{n}<s+\varepsilon$
+\end_inset
+
+ (zadnji neenačaj za 
+\begin_inset Formula $n\geq n_{0}$
+\end_inset
+
+),
+ skratka 
+\begin_inset Formula $a_{N\left(n\right)}-s<\varepsilon$
+\end_inset
+
+ oziroma 
+\begin_inset Formula $\forall n\geq n_{0}:\left|a_{N\left(n\right)}-s\right|<\varepsilon$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $N\left(n\right)\geq n$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $\left\{ N\left(n\right);n\in\mathbb{N}\right\} $
+\end_inset
+
+ neskončna množica,
+ torej je neskončno mnogo členov v poljubni okolici 
+\begin_inset Formula $s$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Treba je dokazati še,
+ da 
+\begin_inset Formula $\forall t>s:t$
+\end_inset
+
+ ni stekališče.
+ Naj bo 
+\begin_inset Formula $t>s$
+\end_inset
+
+.
+ Označimo 
+\begin_inset Formula $\delta\coloneqq t-s>0$
+\end_inset
+
+.
+ Po definiciji
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $s$
+\end_inset
+
+ je limita zaporedja 
+\begin_inset Formula $\left(s_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+,
+ zato v poljubno majhni okolici obstaja tak 
+\begin_inset Formula $s_{n_{1}}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $s_{n_{1}}$
+\end_inset
+
+ torej tu najdemo v 
+\begin_inset Formula $[s,s+\frac{\delta}{2})$
+\end_inset
+
+.
+\end_layout
+
+\end_inset
+
+ 
+\begin_inset Formula $s$
+\end_inset
+
+ 
+\begin_inset Formula $\exists n_{1}\in\mathbb{N}\ni:s\leq s_{n_{1}}<s+\frac{\delta}{2}<s+t$
+\end_inset
+
+.
+ Ker 
+\begin_inset Formula $\left(s_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ pada proti 
+\begin_inset Formula $s$
+\end_inset
+
+,
+ sledi 
+\begin_inset Formula $\forall n\geq n_{1}:s\leq s_{n}<s+\frac{\delta}{2}$
+\end_inset
+
+.
+ Po definiciji
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $s_{n}$
+\end_inset
+
+ je supremum členov od vključno 
+\begin_inset Formula $n$
+\end_inset
+
+ dalje
+\end_layout
+
+\end_inset
+
+ 
+\begin_inset Formula $s_{n}$
+\end_inset
+
+ sledi 
+\begin_inset Formula $\forall n\geq n_{1}:a_{n}\leq s+\frac{\delta}{2}$
+\end_inset
+
+.
+ Za takšne 
+\begin_inset Formula $n$
+\end_inset
+
+ je 
+\begin_inset Formula $\left|t-a_{n}\right|=t-a_{n}\geq t-\left(s+\frac{\delta}{2}\right)=\frac{\delta}{2}$
+\end_inset
+
+.
+ Torej v 
+\begin_inset Formula $\frac{\delta}{2}-$
+\end_inset
+
+okolici točke 
+\begin_inset Formula $t$
+\end_inset
+
+ leži kvečjemu končno mnogo členov zaporedja oziroma členi 
+\begin_inset Formula $\left(a_{1},a_{2},\dots,a_{n_{1}-1}\right)$
+\end_inset
+
+.
+ Torej 
+\begin_inset Formula $t$
+\end_inset
+
+ ni stekališče za 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $s=\infty$
+\end_inset
+
+ Naj bo 
+\begin_inset Formula $M>0$
+\end_inset
+
+ poljuben.
+ Ker je 
+\begin_inset Formula $s=\inf s_{n}$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $\forall n\in\mathbb{N}:s_{n}=\infty$
+\end_inset
+
+.
+ Po definiciji 
+\begin_inset Formula $s_{n}=\infty$
+\end_inset
+
+ velja 
+\begin_inset Formula $\forall n\in\mathbb{N}\exists N\left(n\right):a_{N\left(n\right)}>M$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $N\left(n\right)\geq n$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $\left\{ N\left(n\right);n\in\mathbb{N}\right\} $
+\end_inset
+
+ neskončna množica,
+ torej je neskončno mnogo členov v 
+\begin_inset Formula $\left(M,\infty\right)$
+\end_inset
+
+ za poljuben 
+\begin_inset Formula $M$
+\end_inset
+
+,
+ torej je 
+\begin_inset Formula $s=\infty$
+\end_inset
+
+ res stekališče.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Večjih stekališč od 
+\begin_inset Formula $\infty$
+\end_inset
+
+ očitno ni.
+\end_layout
+
+\end_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $s=-\infty$
+\end_inset
+
+ Naj bo 
+\begin_inset Formula $m<0$
+\end_inset
+
+ poljuben.
+ Ker je 
+\begin_inset Formula $s=\inf s_{n}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\exists n_{0}\in\mathbb{N}\ni:s_{n_{0}}\in\left(-\infty,m\right)$
+\end_inset
+
+ Ker 
+\begin_inset Formula $\left(s_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ pada proti 
+\begin_inset Formula $s=-\infty$
+\end_inset
+
+,
+ sledi 
+\begin_inset Formula $\forall n\in\mathbb{N}:n\geq n_{0}:s_{n}\in\left(-\infty,m\right)$
+\end_inset
+
+.
+ Po definiciji 
+\begin_inset Formula $s_{n}$
+\end_inset
+
+ velja 
+\begin_inset Formula $\forall n\in\mathbb{N}:a_{n}\in\left(-\infty,m\right)$
+\end_inset
+
+.
+ Ker je za poljuben 
+\begin_inset Formula $m$
+\end_inset
+
+ neskončno mnogo členov v 
+\begin_inset Formula $\left(-\infty,m\right)$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $s=-\infty$
+\end_inset
+
+ res stekališče.
+\end_layout
+
+\end_deeper
+\begin_layout Subsection
+Cauchyjev pogoj
+\end_layout
+
+\begin_layout Definition*
+Zaporedje 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ ustreza Cauchyjevemu pogoju (oz.
+ je Cauchyjevo),
+ če 
+\begin_inset Formula $\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\ni:\forall m,n\in\mathbb{N}:m,n\geq n_{0}\Rightarrow\left|a_{m}-a_{n}\right|<\varepsilon$
+\end_inset
+
+.
+ ZDB Dovolj pozni členi so si poljubno blizu.
+\end_layout
+
+\begin_layout Claim*
+Zaporedje v 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+ je konvergentno 
+\begin_inset Formula $\Leftrightarrow$
+\end_inset
+
+ je Cauchyjevo.
+\end_layout
+
+\begin_layout Proof
+Dokazujemo ekvivalenco.
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Rightarrow\right)$
+\end_inset
+
+ Če 
+\begin_inset Formula $a_{n}\to L$
+\end_inset
+
+,
+ tedaj 
+\begin_inset Formula $\left|a_{m}-a_{n}\right|=\left|\left(a_{m}-L\right)+\left(L-a_{n}\right)\right|\leq\left|a_{m}-\varepsilon\right|+\left|a_{n}-\varepsilon\right|$
+\end_inset
+
+.
+ Cauchyjev pogoj sledi iz definicije limite za 
+\begin_inset Formula $\frac{\varepsilon}{2}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Leftarrow\right)$
+\end_inset
+
+ Če je zaporedje Cauchyjevo,
+ je omejeno:
+ 
+\begin_inset Formula $\exists n_{0}\in\mathbb{N}\ni:\forall m,n\in\mathbb{N}:m,n\geq n_{0}\Rightarrow\left|a_{m}-a_{n}\right|\leq1$
+\end_inset
+
+.
+ V posebnem,
+ 
+\begin_inset Formula $m=n_{0}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\left|a_{n_{0}}-a_{n}\right|\leq1$
+\end_inset
+
+ oziroma 
+\begin_inset Formula $\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow a_{n}\in\left[a_{n_{0}}-1,a_{n_{0}}+1\right]$
+\end_inset
+
+.
+ Preostali členi tvorijo končno veliko množico,
+ ki ima 
+\begin_inset Formula $\min$
+\end_inset
+
+ in 
+\begin_inset Formula $\max$
+\end_inset
+
+,
+ torej je 
+\begin_inset Formula $\left\{ a_{k};k\in\mathbb{N}\right\} =\left\{ a_{1},a_{2},\dots,a_{n_{0}-1}\right\} \cup\left\{ a_{k};k\in\mathbb{N},k\geq n_{0}\right\} $
+\end_inset
+
+ tudi omejena.
+ Po 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{bw}{izreku od prej}
+\end_layout
+
+\end_inset
+
+ sledi,
+ da ima zaporedje stekališče 
+\begin_inset Formula $s$
+\end_inset
+
+.
+ Dokažimo,
+ da je 
+\begin_inset Formula $s=\lim_{n\to\infty}a_{n}$
+\end_inset
+
+.
+ Vzemimo poljuben 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ Cauchyjevo,
+ 
+\begin_inset Formula $\exists n_{0}\in\mathbb{N}\ni:\forall m,n\in\mathbb{N}:m,n\geq n_{0}\Rightarrow\left|a_{m}-a_{n}\right|<\frac{\varepsilon}{2}$
+\end_inset
+
+.
+ Po definiciji 
+\begin_inset Formula $s$
+\end_inset
+
+ 
+\begin_inset Formula $\exists n_{1}\geq n_{0}\ni:\left|a_{n_{1}}-s\right|<\frac{\varepsilon}{2}$
+\end_inset
+
+.
+ Sledi 
+\begin_inset Formula $\forall n\geq n_{0}:\left|a_{n}-s\right|=\left|a_{n}-s+s-a_{n_{1}}\right|\leq\left|a_{n}-s\right|+\left|s-a_{n_{1}}\right|<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Remark*
+Moč izreka je v tem,
+ da lahko konvergenco preverjamo tudi tedaj,
+ ko nimamo kandidatov za limito.
+\end_layout
+
+\begin_layout Section
+Številske vrste
+\end_layout
+
+\begin_layout Standard
+Kako sešteti neskončno mnogo števil?
+ Nadgradimo pristop končnih vsot na neskončne vsote!
+\end_layout
+
+\begin_layout Definition*
+Imejmo zaporedje 
+\begin_inset Formula $\left(a_{k}\right)_{k\in\mathbb{N}},a_{k}\in\mathbb{R}$
+\end_inset
+
+.
+ Izraz 
+\begin_inset Formula $\sum_{j=1}^{\infty}a_{j}$
+\end_inset
+
+ se imenuje vrsta s členi 
+\begin_inset Formula $a_{j}$
+\end_inset
+
+.
+ Pomen izraza opredelimo na naslednjo način:
+\end_layout
+
+\begin_layout Definition*
+Tvorimo novo zaporedje,
+ pravimo mu zaporedje delnih vsot vrste:
+ 
+\begin_inset Formula $s_{1}=a_{1}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $s_{2}=a_{1}+a_{2}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $s_{3}=a_{1}+a_{2}+a_{3}$
+\end_inset
+
+,
+ ...,
+ 
+\begin_inset Formula $s_{n}=a_{1}+a_{2}+\cdots+a_{n}=\sum_{j=1}^{n}a_{j}$
+\end_inset
+
+ —
+ številu 
+\begin_inset Formula $s_{n}$
+\end_inset
+
+ pravimo 
+\begin_inset Formula $n-$
+\end_inset
+
+ta delna vsota.
+\end_layout
+
+\begin_layout Definition*
+Vrsta je konvergentna,
+ če je v 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+ konvergentno zaporedje 
+\begin_inset Formula $\left(s_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+.
+ Številu 
+\begin_inset Formula $s=\lim_{n\to\infty}s_{n}$
+\end_inset
+
+ tedaj pravimo vsota vrste in pišemo 
+\begin_inset Formula $s\eqqcolon\sum_{j=1}^{\infty}a_{j}$
+\end_inset
+
+.
+ Pojem neskončne vsote torej prevedemo na pojem limite pridruženega zaporedja delnih vsot.
+ Včasih vrsto (kot operacijo) enačimo z njeno vsoto (izidom operacije).
+\end_layout
+
+\begin_layout Definition*
+Če vrsta ni konvergentna,
+ rečemo,
+ da je divergentna.
+ Enako,
+ če je 
+\begin_inset Formula $s\in\left\{ \pm\infty\right\} $
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Primeri vrst.
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $a_{n}=\frac{1}{2^{n}}$
+\end_inset
+
+,
+ torej zaporedje 
+\begin_inset Formula $\frac{1}{2},\frac{1}{4},\frac{1}{8},\dots$
+\end_inset
+
+.
+ Ali se sešteje v 1?
+ Velja 
+\begin_inset Formula $s=\lim_{n\to\infty}\sum_{j=1}^{n}a_{j}$
+\end_inset
+
+.
+ Pišimo 
+\begin_inset Formula $q=\frac{1}{2}$
+\end_inset
+
+,
+ tedaj 
+\begin_inset Formula $a_{n}=q^{n}$
+\end_inset
+
+ in 
+\begin_inset Formula 
+\[
+s_{n}=q+q^{2}+q^{3}+\cdots+q^{n}=q\left(1+q+q^{2}+\cdots+q^{n-1}\right)=q\frac{\left(1+q+q^{2}+\cdots+q^{n-1}\right)\left(1-q\right)}{1-q}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=q\frac{\left(1+q+q^{2}+\cdots+q^{n-1}\right)-\left(q+q^{2}+q^{3}+\cdots+q^{n}\right)}{1-q}=q\frac{1-q^{n}}{1-q}=\frac{q}{1-q}\left(1-q^{n}\right)
+\]
+
+\end_inset
+
+Izračunajmo 
+\begin_inset Formula $\lim_{n\to\infty}s_{n}=\lim_{n\to\infty}\frac{q}{1-q}\left(1-\cancelto{0}{q^{n}}\right)=\frac{q}{1-q}$
+\end_inset
+
+ (velja,
+ ker 
+\begin_inset Formula $q\in\left(-1,1\right)$
+\end_inset
+
+),
+ torej je 
+\begin_inset Formula $s=\sum_{n=1}^{\infty}q^{n}=\frac{q}{1-q}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Geometrijska vrsta (splošno).
+ Naj bo 
+\begin_inset Formula $q\in\mathbb{R}$
+\end_inset
+
+.
+ Vrsta 
+\begin_inset Formula $\sum_{j=0}^{\infty}q^{j}$
+\end_inset
+
+ se imenuje geometrijska vrsta.
+ Velja 
+\begin_inset Formula $s=\lim_{n\to\infty}\sum_{j=0}^{n}q^{j}$
+\end_inset
+
+ in 
+\begin_inset Formula $s_{n}=1+q+q^{2}+q^{3}+\cdots+q^{n}$
+\end_inset
+
+.
+ Če je 
+\begin_inset Formula $q=1$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $s_{n}=n+1$
+\end_inset
+
+,
+ sicer množimo izraz z 
+\begin_inset Formula $\left(1-q\right)$
+\end_inset
+
+:
+\begin_inset Formula 
+\[
+\left(1+q+q^{2}+\cdots+q^{n}\right)\left(1-q\right)=\left(1+q+q^{2}+\cdots+q^{n}\right)-\left(q+q^{2}+q^{3}+\cdots+q^{n+1}\right)=1-q^{n+1}
+\]
+
+\end_inset
+
+torej 
+\begin_inset Formula $s_{n}=\frac{1-q^{n+1}}{1-q}$
+\end_inset
+
+ in vrsta konvergira 
+\begin_inset Formula $\Leftrightarrow q\not=1$
+\end_inset
+
+ in 
+\begin_inset Formula $\lim_{n\to\infty}\frac{1-q^{n+1}}{1-q}\exists$
+\end_inset
+
+ v 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+.
+ To pa se zgodi natanko za 
+\begin_inset Formula $q\in\left(-1,1\right)$
+\end_inset
+
+,
+ takrat je 
+\begin_inset Formula $\lim_{n\to\infty}\frac{1-\cancelto{0}{q^{n+1}}}{1-q}=\frac{1}{1-q}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Harmonična vrsta.
+ Je vrsta 
+\begin_inset Formula $\sum_{j=1}^{\infty}\frac{1}{j}$
+\end_inset
+
+.
+ Velja 
+\begin_inset Formula $\frac{1}{j}\underset{j\to\infty}{\longrightarrow}0$
+\end_inset
+
+,
+ toda vrsta divergira.
+ Dokaz sledi kmalu malce spodaj.
+\end_layout
+
+\end_deeper
+\begin_layout Question*
+Kako lahko enostavno določimo,
+ ali dana vrsta konvergira?
+\end_layout
+
+\begin_layout Subsection
+Konvergenčni kriteriji
+\end_layout
+
+\begin_layout Theorem*
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hypertarget{cauchyvrste}{Cauchyjev pogoj}
+\end_layout
+
+\end_inset
+
+.
+ Vrsta 
+\begin_inset Formula $\sum_{j=1}^{\infty}a_{j}$
+\end_inset
+
+ je konvergentna 
+\begin_inset Formula $\Leftrightarrow$
+\end_inset
+
+ delne vrste ustrezajo Cauchyjevemu pogoju;
+ 
+\begin_inset Formula $\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\forall n,m\in\mathbb{N}:n,m\geq n_{0}\Rightarrow\left|s_{m}-s_{n}\right|=\left|\sum_{j=n+1}^{m}a_{j}\right|<\varepsilon$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Corollary*
+\begin_inset Formula $\sum_{j=1}^{\infty}a_{j}$
+\end_inset
+
+ konvergira 
+\begin_inset Formula $\Rightarrow\lim_{j\to\infty}a_{j}=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Uporabimo izrek zgoraj za 
+\begin_inset Formula $n=m-1$
+\end_inset
+
+:
+ 
+\begin_inset Formula $\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow\left|s_{n}-s_{n+1}\right|=\left|a_{n}\right|<\varepsilon$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Vrsti 
+\begin_inset Formula $\sum_{j=1}^{\infty}\cos n$
+\end_inset
+
+ in 
+\begin_inset Formula $\sum_{j=1}^{\infty}\sin n$
+\end_inset
+
+ divergirata,
+ saj smo videli,
+ da členi ne ene ne druge ne konvergirajo nikamor,
+ torej tudi ne proti 0,
+ kar je potreben pogoj za konvergenco vrste.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+Harmonična vrsta divergira.
+ Protiprimer Cauchyjevega pogoja:
+ Naj bo 
+\begin_inset Formula $\varepsilon=\frac{1}{4}$
+\end_inset
+
+.
+ Tedaj ne glede na izbiro 
+\begin_inset Formula $n_{0}$
+\end_inset
+
+ najdemo:
+\begin_inset Formula 
+\[
+s_{2n}-s_{n}=\sum_{j=n+1}^{2n}\frac{1}{j}=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}>\frac{1}{2n}+\frac{1}{2n}+\cdots+\frac{1}{2n}=\frac{1}{2}
+\]
+
+\end_inset
+
+Dokaz divergence brez Cauchyjevega pogoja:
+ 
+\begin_inset Formula $s_{2^{n}}=a_{1}+\sum_{j=1}^{n}\left(s_{2^{j}}-s_{s^{j-1}}\right)>1+\frac{n}{2}$
+\end_inset
+
+ in 
+\begin_inset Formula $\lim_{n\to\infty}1+\frac{n}{2}=\infty$
+\end_inset
+
+.
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+Geometrični argument za divergenco:
+ TODO XXX FIXME DODAJ
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Theorem*
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hypertarget{pk}{Primerjalni kriterij}
+\end_layout
+
+\end_inset
+
+.
+ Naj bosta 
+\begin_inset Formula $\sum_{n=1}^{\infty}a_{n}$
+\end_inset
+
+ in 
+\begin_inset Formula $\sum_{n=1}^{\infty}b_{n}$
+\end_inset
+
+ vrsti z nenegativnimi členi.
+ Naj bo 
+\begin_inset Formula $\forall k\geq k_{0}:a_{k}\leq b_{k}$
+\end_inset
+
+ (od nekod naprej) —
+ pravimo,
+ da je 
+\begin_inset Formula $\sum_{n=1}^{\infty}b_{n}$
+\end_inset
+
+ majoranta za 
+\begin_inset Formula $\sum_{n=1}^{\infty}a_{n}$
+\end_inset
+
+ od nekod naprej.
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Če 
+\begin_inset Formula $\sum_{n=1}^{\infty}b_{n}$
+\end_inset
+
+ konvergira,
+ tedaj tudi 
+\begin_inset Formula $\sum_{n=1}^{\infty}a_{n}$
+\end_inset
+
+ konvergira.
+\end_layout
+
+\begin_layout Itemize
+Če 
+\begin_inset Formula $\text{\ensuremath{\sum_{n=1}^{\infty}a_{n}=\infty}}$
+\end_inset
+
+,
+ tedaj tudi 
+\begin_inset Formula $\sum_{n=1}^{\infty}b_{n}=\infty$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Example*
+Videli smo,
+ da 
+\begin_inset Formula $\sum_{k=1}^{\infty}\frac{1}{k}$
+\end_inset
+
+ divergira.
+ Kaj pa 
+\begin_inset Formula $\sum_{k=1}^{\infty}\frac{1}{k^{2}}$
+\end_inset
+
+?
+ Preverimo naslednje in uporabimo primerjalni kriterij:
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $\forall k\in\mathbb{N}:\frac{1}{k^{2}}\leq\frac{2}{k\left(k+1\right)}$
+\end_inset
+
+?
+ Računajmo 
+\begin_inset Formula $k^{2}\geq\frac{k\left(k+1\right)}{2}\sim k\geq\frac{k+1}{2}\sim\frac{k}{2}\geq\frac{1}{2}$
+\end_inset
+
+.
+ Velja,
+ ker 
+\begin_inset Formula $k\in\mathbb{N}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Vrsta 
+\begin_inset Formula $\sum_{k=1}^{\infty}\frac{2}{k\left(k+1\right)}$
+\end_inset
+
+ konvergira?
+ Opazimo 
+\begin_inset Formula $\frac{1}{k}-\frac{1}{k+1}=\frac{k+1}{k\left(k+1\right)}-\frac{k}{k\left(k+1\right)}=\frac{1}{k\left(k+1\right)}$
+\end_inset
+
+.
+ Za delne vsote vrste 
+\begin_inset Formula $\sum_{k=1}^{\infty}\frac{1}{k\left(k+1\right)}$
+\end_inset
+
+ velja:
+\begin_inset Formula 
+\[
+\sum_{k=1}^{n}\frac{1}{k\left(k+1\right)}=\sum_{k=1}^{n}\left(\frac{1}{k}-\frac{1}{k+1}\right)=1-\frac{1}{n+1}\underset{n\to\infty}{\longrightarrow}1,
+\]
+
+\end_inset
+
+torej 
+\begin_inset Formula $\sum_{k=1}^{\infty}\frac{2}{k\left(k+1\right)}=2$
+\end_inset
+
+.
+ Posledično po primerjalnem kriteriju tudi 
+\begin_inset Formula $\sum_{k=1}^{\infty}\frac{1}{k^{2}}$
+\end_inset
+
+ konvergira.
+ Izkaže se 
+\begin_inset Formula $\sum_{k=1}^{\infty}\frac{1}{k^{2}}=\frac{\pi^{2}}{6}\approx1,645$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Theorem*
+Kvocientni oz.
+ d'Alembertov kriterij.
+ Za vrsto s pozitivnimi členi 
+\begin_inset Formula $\sum_{n=1}^{\infty}a_{n}$
+\end_inset
+
+ definirajmo 
+\begin_inset Formula $D_{n}\coloneqq\frac{a_{n+1}}{a_{n}}$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $\exists n_{0}\in\mathbb{N},q\in\left(0,1\right)\forall n\geq n_{0}:D_{n}\leq q\Longrightarrow\sum_{n=1}^{\infty}a_{n}<\infty$
+\end_inset
+
+ (vrsta konvergira)
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\exists n_{0}\in\mathbb{N}\forall n\geq n_{0}:D_{n}\geq1\Longrightarrow\sum_{n=1}^{\infty}a_{n}=\infty$
+\end_inset
+
+ (vrsta divergira)
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\exists D=\lim_{n\to\infty}D_{n}\in\mathbb{R}\Longrightarrow$
+\end_inset
+
+
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset CommandInset label
+LatexCommand label
+name "enu:kvocientni3a"
+
+\end_inset
+
+
+\begin_inset Formula $D<1\Longrightarrow\sum_{n=1}^{\infty}a_{n}<\infty$
+\end_inset
+
+ (vrsta konvergira)
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $D>1\Longrightarrow\sum_{n=1}^{\infty}a_{n}=\infty$
+\end_inset
+
+ (vrsta divergira)
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $D=1\Longrightarrow$
+\end_inset
+
+ s tem kriterijem ne moremo določiti konvergence.
+\end_layout
+
+\end_deeper
+\end_deeper
+\begin_layout Proof
+Razlaga.
+ 
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $\forall n>n_{0}:D_{n}\leq q$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\frac{a_{n+1}}{a_{n}}\leq q\sim a_{n+1}\leq qa_{n}$
+\end_inset
+
+ in hkrati 
+\begin_inset Formula $\text{\ensuremath{\frac{a_{n+2}}{a_{n+1}}\leq q\sim a_{n+2}\leq qa_{n+1}}}$
+\end_inset
+
+,
+ torej skupaj 
+\begin_inset Formula $a_{n+2}\leq qa_{n+1}\leq qqa_{n}=q^{2}a_{n}$
+\end_inset
+
+,
+ sledi 
+\begin_inset Formula $q_{n+2}\leq q^{2}a_{n}$
+\end_inset
+
+ in 
+\begin_inset Formula $\forall k\in\mathbb{N}:q_{n+k}\leq q^{k}a_{n}$
+\end_inset
+
+.
+ Vrsto smo majorizirali z geometrijsko vrsto,
+ ki ob 
+\begin_inset Formula $q\in\left(0,1\right)$
+\end_inset
+
+ konvergira po primerjalnem kriteriju,
+ zato tudi naša vrsta konvergira.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\forall n>n_{0}:\frac{a_{n+1}}{a_{n}}\geq D\geq1$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $a_{n+1}\geq a_{n}$
+\end_inset
+
+ in hkrati 
+\begin_inset Formula $a_{n+2}\geq a_{n+1}$
+\end_inset
+
+,
+ torej skupaj 
+\begin_inset Formula $a_{n+2}\geq a_{n}$
+\end_inset
+
+,
+ sledi 
+\begin_inset Formula $\forall k\in\mathbb{N}:a_{n+k}\geq a_{n}$
+\end_inset
+
+.
+ Naša vrsta torej majorizira konstantno vrsto,
+ ki očitno divergira;
+ 
+\begin_inset Formula $\sum_{k=n_{0}}^{\infty}a_{k}\geq\sum_{k=n_{0}}^{\infty}a_{n}=0$
+\end_inset
+
+.
+ Potemtakem tudi naša vrsta divergira.
+ Poleg tega niti ne velja 
+\begin_inset Formula $a_{k}\underset{k\to\infty}{\longrightarrow}0$
+\end_inset
+
+,
+ torej vrsta gotovo divergira.
+\end_layout
+
+\begin_layout Enumerate
+Enako kot 1 in 2.
+\end_layout
+
+\end_deeper
+\begin_layout Example*
+Za 
+\begin_inset Formula $x>0$
+\end_inset
+
+ definiramo 
+\begin_inset Formula $e^{x}=\sum_{k=0}^{\infty}\frac{x^{k}}{k!}$
+\end_inset
+
+.
+ Vrsta res konvergira po točki 
+\begin_inset CommandInset ref
+LatexCommand ref
+reference "enu:kvocientni3a"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+D_{n}=\frac{\frac{x^{n+1}}{\left(n+1\right)!}}{\frac{x^{n}}{n!}}=\frac{x^{n+1}n!}{x^{n}\left(n+1\right)!}=\frac{x}{n+1}\underset{n\to\infty}{\longrightarrow}0
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Theorem*
+Korenski oz.
+ Cauchyjev kriterij.
+ Naj bo 
+\begin_inset Formula $\sum_{k=1}^{\infty}a_{k}$
+\end_inset
+
+ vrsta z nenegativnimi členi.
+ Naj bo 
+\begin_inset Formula $c_{n}\coloneqq\sqrt[n]{a_{n}}$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $\exists n_{0}\in\mathbb{N},q\in\left(0,1\right)\forall n>n_{0}:c_{n}\leq q\Longrightarrow\sum_{k=1}^{\infty}a_{k}<\infty$
+\end_inset
+
+ (vrsta konvergira)
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\exists n_{0}\in\mathbb{N}\forall n>n_{0}:c_{n}\geq1\Longrightarrow\sum_{k=1}^{\infty}a_{k}=\infty$
+\end_inset
+
+ (vrsta divergira)
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\exists c=\lim_{n\to\infty}c_{n}\in\mathbb{R}\Longrightarrow$
+\end_inset
+
+
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $c<1\Longrightarrow\sum_{k=1}^{\infty}a_{k}<\infty$
+\end_inset
+
+ (vrsta konvergira)
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $c>1\Longrightarrow\sum_{k=1}^{\infty}a_{k}=\infty$
+\end_inset
+
+ (vrsta divergira)
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $c=1\Longrightarrow$
+\end_inset
+
+ s tem kriterijem ne moremo določiti konvergence.
+\end_layout
+
+\end_deeper
+\end_deeper
+\begin_layout Proof
+Skica dokazov.
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+Velja 
+\begin_inset Formula $\forall n>n_{0}:c_{n}\leq q$
+\end_inset
+
+.
+ To pomeni 
+\begin_inset Formula $\sqrt[n]{a_{n}}\leq q$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $a_{n}\leq q^{n}$
+\end_inset
+
+ in 
+\begin_inset Formula $a_{n+1}\leq q^{n+1}$
+\end_inset
+
+,
+ torej je vrsta majorizirana z geometrijsko vrsto 
+\begin_inset Formula $\sum_{n=1}^{\infty}q^{n}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Velja 
+\begin_inset Formula $\forall n>n_{0}:c_{n}\geq1$
+\end_inset
+
+.
+ To pomeni 
+\begin_inset Formula $\sqrt[n]{a_{n}}\geq1$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $a_{n}\geq1$
+\end_inset
+
+,
+ torej je vrsta majorizirana s konstantno in zato divergentno vrsto 
+\begin_inset Formula $\sum_{n=1}^{\infty}1$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Enako kot 1 in 2.
+\end_layout
+
+\end_deeper
+\begin_layout Subsection
+Alternirajoče vrste
+\end_layout
+
+\begin_layout Definition*
+Vrsta je alternirajoča,
+ če je predznak naslednjega člena nasproten predznaku tega člena.
+ ZDB 
+\begin_inset Formula $\forall n\in\mathbb{N}:\sgn a_{n+1}=-\sgn a_{n}$
+\end_inset
+
+,
+ kjer je 
+\begin_inset Formula $\sgn:\mathbb{R}\to\left\{ -1,0,1\right\} $
+\end_inset
+
+ s predpisom 
+\begin_inset Formula $\sgn a=\begin{cases}
+-1 & ;a<0\\
+1 & ;a>0\\
+0 & ;a=0
+\end{cases}$
+\end_inset
+
+.
+ ZDB 
+\begin_inset Formula $\forall n\in\mathbb{N}:a_{n+1}a_{n}\leq0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+Leibnizov konvergenčni kriterij.
+ Naj bo 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ padajoče zaporedje in 
+\begin_inset Formula $\lim_{n\to\infty}a_{n}=0$
+\end_inset
+
+.
+ Tedaj vrsta 
+\begin_inset Formula $\sum_{k=1}^{\infty}\left(-1\right)^{k}a_{k}$
+\end_inset
+
+ konvergira.
+ Če je 
+\begin_inset Formula $s\coloneqq\sum_{k=1}^{\infty}\left(-1\right)^{k}a_{k}$
+\end_inset
+
+ in 
+\begin_inset Formula $s_{n}\coloneqq\sum_{k=1}^{\infty}\left(-1\right)^{k}a_{k}$
+\end_inset
+
+,
+ tedaj 
+\begin_inset Formula $\left|s-s_{k}\right|\leq a_{n+1}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Skica dokaza.
+ Vidimo,
+ da delne vsote 
+\begin_inset Formula $s_{2n}$
+\end_inset
+
+ padajo k 
+\begin_inset Formula $s''$
+\end_inset
+
+ in delne vsote 
+\begin_inset Formula $s_{2n-1}$
+\end_inset
+
+ naraščajo k 
+\begin_inset Formula $s'$
+\end_inset
+
+.
+ Toda ker 
+\begin_inset Formula $s_{2n}-s_{2n-1}=a_{2n}$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $s'=s''$
+\end_inset
+
+.
+ Limita razlike dveh zaporedij je razlika limit teh dveh zaporedij,
+ torej 
+\begin_inset Formula $s'=s''=s$
+\end_inset
+
+.
+ 
+\begin_inset Formula $s$
+\end_inset
+
+ je supremum lihih in infimum sodih vsot.
+ 
+\begin_inset Formula $\left|s-s_{n}\right|\leq\left|s_{n+1}-s_{n}\right|=a_{n+1}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Harmonična vrsta 
+\begin_inset Formula $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots\to\infty$
+\end_inset
+
+,
+ toda alternirajoča harmonična vrsta 
+\begin_inset Formula $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots\to\log2$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsection
+Absolutno konvergentne vrste
+\end_layout
+
+\begin_layout Definition*
+Vrsta 
+\begin_inset Formula $\sum_{n=1}^{\infty}a_{n}$
+\end_inset
+
+ je absolutno konvergentna,
+ če je 
+\begin_inset Formula $\sum_{n=1}^{\infty}\left|a_{n}\right|$
+\end_inset
+
+ konvergentna.
+\end_layout
+
+\begin_layout Theorem*
+Absolutna konvergenca 
+\begin_inset Formula $\Rightarrow$
+\end_inset
+
+ konvergenca.
+\end_layout
+
+\begin_layout Proof
+Uporabimo 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{cauchyvrste}{Cauchyjev pogoj za konvergenco vrst}
+\end_layout
+
+\end_inset
+
+ in trikotniško neenakost.
+\begin_inset Formula 
+\[
+\left|s_{m}-s_{n}\right|=\left|\sum_{j=n+1}^{m}a_{j}\right|\leq\sum_{j=n+1}^{m}\left|a_{j}\right|<\varepsilon
+\]
+
+\end_inset
+
+za 
+\begin_inset Formula $m,n\geq n_{0}$
+\end_inset
+
+ za nek 
+\begin_inset Formula $n_{0}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Remark*
+Obrat ne velja,
+ protiprimer je alternirajoča harmonična vrsta.
+\end_layout
+
+\begin_layout Subsection
+Pogojno konvergentne vrste
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $\sum_{k=0}^{\infty}2-\sum_{k=0}^{\infty}1\not=\sum_{k=0}^{\infty}\left(2-1\right)$
+\end_inset
+
+,
+ temveč 
+\begin_inset Formula $\infty-\infty=$
+\end_inset
+
+ nedefinirano.
+\end_layout
+
+\begin_layout Question*
+Ross-Littlewoodov paradoks.
+ Ali smemo zamenjati vrstni red seštevanja,
+ če imamo neskončno mnogo sumandov?
+\end_layout
+
+\begin_layout Standard
+Najprej vprašanje natančneje opredelimo in vpeljimo orodja za njegovo obravnavo.
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $\mathcal{M}\subset\mathbb{N}$
+\end_inset
+
+.
+ Permutacija 
+\begin_inset Formula $\mathcal{M}$
+\end_inset
+
+ je vsaka bijektivna preslikava 
+\begin_inset Formula $\pi:\mathcal{M}\to\mathcal{M}$
+\end_inset
+
+.
+ Če je 
+\begin_inset Formula $\mathcal{M}=\left\{ a_{1},\dots,a_{n}\right\} $
+\end_inset
+
+ končna množica,
+ tedaj 
+\begin_inset Formula $\pi$
+\end_inset
+
+ označimo s tabelo:
+\begin_inset Formula 
+\[
+\left(\begin{array}{ccc}
+a_{1} & \cdots & a_{n}\\
+\pi\left(a_{1}\right) & \cdots & \pi\left(a_{n}\right)
+\end{array}\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula 
+\[
+\pi=\left(\begin{array}{ccccc}
+1 & 2 & 3 & 4 & 5\\
+5 & 3 & 1 & 4 & 2
+\end{array}\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Vrsta 
+\begin_inset Formula $\sum_{n=1}^{\infty}a_{n}$
+\end_inset
+
+ je brezpogojno konvergentna,
+ če za vsako permutacijo 
+\begin_inset Formula $\pi:\mathbb{N}\to\mathbb{N}$
+\end_inset
+
+ vrsta 
+\begin_inset Formula $\sum_{n=1}^{\infty}\pi\left(a_{n}\right)$
+\end_inset
+
+ konvergira in vsota ni odvisna od 
+\begin_inset Formula $\pi$
+\end_inset
+
+.
+ Vrsta je pogojno konvergentna,
+ če je konvergentna,
+ toda ne brezpogojno.
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\cdots$
+\end_inset
+
+ je pogojno konvergentna,
+ ker pri seštevanju z vrstnim redom,
+ pri katerem tisočim pozitivnim členom sledi en negativen in njemu zopet tisoč pozitivnih itd.,
+ vrsta ne konvergira.
+\end_layout
+
+\begin_layout Theorem*
+Absolutna konvergenca 
+\begin_inset Formula $\Leftrightarrow$
+\end_inset
+
+ Brezpogojna konvergenca
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Theorem*
+Riemannov sumacijski izrek.
+ Če je vrsta pogojno konvergentna,
+ tedaj 
+\begin_inset Formula $\forall x\in\mathbb{R}\cup\left\{ \pm\infty\right\} \exists$
+\end_inset
+
+ permutacija 
+\begin_inset Formula $\pi:\mathbb{N}\to\mathbb{N}\ni:\sum_{n=1}^{\infty}a_{\pi\left(n\right)}=x$
+\end_inset
+
+.
+ ZDB Končna vsota je lahko karkoli,
+ če lahko poljubno spremenimo vrstni red seštevanja.
+ Prav tako obstaja taka permutacija 
+\begin_inset Formula $\pi$
+\end_inset
+
+,
+ pri kateri 
+\begin_inset Formula $\sum_{n=1}^{\infty}a_{\pi\left(n\right)}$
+\end_inset
+
+ nima vsote ZDB delne vsotee ne konvergirajo.
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}}{n}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Funkcijske vrste
+\end_layout
+
+\begin_layout Standard
+Tokrat poskušamo seštevati funkcije.
+ V prejšnjem razdelku seštevamo le realna števila.
+ Funkcijska vrsta,
+ če je 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ zaporedije funkcij 
+\begin_inset Formula $X\to\mathbb{R}$
+\end_inset
+
+ in 
+\begin_inset Formula $x$
+\end_inset
+
+ zunanja konstanta,
+ izgleda takole:
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula 
+\[
+\sum_{n=1}^{\infty}a_{n}\left(x\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $X$
+\end_inset
+
+ neka množica in 
+\begin_inset Formula $\Phi=\left\{ \varphi_{n}:X\to\mathbb{R},n\in\mathbb{N}\right\} $
+\end_inset
+
+ družina funkcij.
+\end_layout
+
+\begin_layout Definition*
+Pravimo,
+ da funkcije 
+\begin_inset Formula $\varphi_{n}$
+\end_inset
+
+ konvergirajo po točkah na 
+\begin_inset Formula $X$
+\end_inset
+
+,
+ če je 
+\begin_inset Formula $\forall x\in X$
+\end_inset
+
+ zaporedje 
+\begin_inset Formula $\left(\varphi_{n}\left(x\right)\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ konvergentno.
+\end_layout
+
+\begin_layout Definition*
+Označimo limito s 
+\begin_inset Formula $\varphi\left(x\right)$
+\end_inset
+
+.
+ ZDB to pomeni,
+ da 
+\begin_inset Formula 
+\[
+\forall\varepsilon>0,x\in X:\exists n_{0}=n_{0}\left(\varepsilon,x\right)\in\mathbb{N}\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow\left|\varphi_{n}\left(x\right)-\varphi\left(x\right)\right|<\varepsilon.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Pravimo,
+ da funkcije 
+\begin_inset Formula $\varphi_{n}$
+\end_inset
+
+ konvergirajo enakomerno na 
+\begin_inset Formula $X$
+\end_inset
+
+,
+ če 
+\begin_inset Formula 
+\[
+\forall\varepsilon>0\exists n_{0}=n_{0}\left(\varepsilon\right)\in\mathbb{N}\forall x\in X,n\in\mathbb{N}:n\geq n_{0}\Rightarrow\left|\varphi_{n}\left(x\right)-\varphi\left(x\right)\right|\leq\varepsilon
+\]
+
+\end_inset
+
+ oziroma ZDB 
+\begin_inset Formula 
+\[
+\forall\varepsilon>0\exists n_{0}=n_{0}\left(\varepsilon\right)\in\mathbb{N}\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow\sup_{x\in X}\left|\varphi_{n}\left(x\right)-\varphi\left(x\right)\right|\leq\varepsilon.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Poudariti je treba,
+ da je pri konvergenci po točkah 
+\begin_inset Formula $n_{0}$
+\end_inset
+
+ lahko odvisen od 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+ in 
+\begin_inset Formula $x$
+\end_inset
+
+,
+ pri enakomerni konvergenci pa le od 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Note*
+Očitno enakomerna konvergenca implicira konvergenco po točkah,
+ obratno pa ne velja.
+\end_layout
+
+\begin_layout Example*
+Za 
+\begin_inset Formula $n\in\mathbb{N}$
+\end_inset
+
+ definiramo 
+\begin_inset Formula $\varphi_{n}:\left[0,1\right]\to\left[0,1\right]$
+\end_inset
+
+ s predpisom 
+\begin_inset Formula $\varphi_{n}\left(x\right)=x^{n}$
+\end_inset
+
+.
+ Tedaj obstaja 
+\begin_inset Formula $\varphi\left(x\right)\coloneqq\lim_{n\to\infty}\varphi_{n}\left(x\right)=\begin{cases}
+0 & ;x\in[0,1)\\
+1 & ;x=1
+\end{cases}$
+\end_inset
+
+.
+ Torej po definiciji velja 
+\begin_inset Formula $\varphi_{n}\to\varphi$
+\end_inset
+
+ po točkah,
+ toda ne velja 
+\begin_inset Formula $\varphi_{n}\to\varphi$
+\end_inset
+
+ enakomerno.
+ Za poljubno velik pas okoli 
+\begin_inset Formula $\varphi\left(x\right)$
+\end_inset
+
+ bodo še tako pozne funkcijske vrednosti 
+\begin_inset Formula $\varphi_{n}\left(x\right)$
+\end_inset
+
+ od nekega 
+\begin_inset Formula $x$
+\end_inset
+
+ dalje izven tega pasu.
+ Če bi 
+\begin_inset Formula $\varphi_{n}\to\varphi$
+\end_inset
+
+ enakomerno,
+ tedaj bi za poljuben 
+\begin_inset Formula $\varepsilon\in\left(0,1\right)$
+\end_inset
+
+ in dovolj pozne 
+\begin_inset Formula $n$
+\end_inset
+
+ (večje od nekega 
+\begin_inset Formula $n_{0}\in\mathbb{N}$
+\end_inset
+
+) veljalo 
+\begin_inset Formula $\forall x\in\left[0,1\right]:\left|\varphi_{n}\left(x\right)-\varphi\left(x\right)\right|<\varepsilon$
+\end_inset
+
+.
+ To je ekvivalentno 
+\begin_inset Formula $\forall x\in\left(0,1\right):\left|x^{n}\right|<\varepsilon\Leftrightarrow n\log x<\log\varepsilon\Leftrightarrow n>\frac{\log\varepsilon}{\log x}$
+\end_inset
+
+.
+ Toda 
+\begin_inset Formula $\lim_{x\nearrow1}\frac{\log\varepsilon}{\log x}=\infty$
+\end_inset
+
+,
+ zato tak 
+\begin_inset Formula $n$
+\end_inset
+
+ ne obstaja.
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $X$
+\end_inset
+
+ neka množica in 
+\begin_inset Formula $\left(f_{j}:X\to\mathbb{R}\right)_{j\in\mathbb{N}}$
+\end_inset
+
+ dano zaporedje funkcij.
+ Pravimo,
+ da funkcijska vrsta 
+\begin_inset Formula $\sum_{j=1}^{\infty}f_{j}$
+\end_inset
+
+ konvergira po točkah na 
+\begin_inset Formula $X$
+\end_inset
+
+,
+ če 
+\begin_inset Formula $\forall x\in X:\sum_{j=1}^{\infty}f_{j}\left(x\right)<0$
+\end_inset
+
+ (številska vrsta je konvergentna).
+ ZDB to pomeni,
+ da funkcijsko zaporedje delnih vsot 
+\begin_inset Formula $s_{n}\coloneqq\sum_{j=1}^{n}f_{j}$
+\end_inset
+
+ konvergira po točkah na 
+\begin_inset Formula $X$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Funkcijska vrsta 
+\begin_inset Formula $s=\sum_{j=1}^{\infty}$
+\end_inset
+
+ konvergira enakomerno na 
+\begin_inset Formula $X$
+\end_inset
+
+,
+ če funkcijsko zaporedje delnih vsot 
+\begin_inset Formula $s_{n}\coloneqq\sum_{j=1}^{n}f_{j}$
+\end_inset
+
+ konvergira enakomerno na 
+\begin_inset Formula $X$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Funkcija oblike 
+\begin_inset Formula $x\mapsto\sum_{j=1}^{\infty}f_{j}\left(x\right)$
+\end_inset
+
+ se imenuje funkcijska vrsta.
+\end_layout
+
+\begin_layout Exercise*
+Dokaži,
+ da 
+\begin_inset Formula $\sum_{n=1}^{\infty}x^{n}$
+\end_inset
+
+ ne konvergira enakomerno!
+ Vrsta konvergira po točkah le na intervalu 
+\begin_inset Formula $x\in\left(0,1\right)$
+\end_inset
+
+,
+ za druge 
+\begin_inset Formula $x$
+\end_inset
+
+ divergira.
+ Ko fiksiramo zunanjo konstanto,
+ gre za geometrijsko vrsto.
+ Delna vsota 
+\begin_inset Formula $\sum_{j=1}^{n}x^{j}=\frac{x\left(1-x^{n}\right)}{1-x}$
+\end_inset
+
+.
+ Velja 
+\begin_inset Formula $\lim_{n\to\infty}\frac{x\left(1-x^{n}\right)}{1-x}=x\lim_{n\to\infty}\frac{1-\cancelto{0}{x^{n}}}{1-x}=\frac{x}{1-x}$
+\end_inset
+
+.
+ Sedaj prevedimo,
+ ali 
+\begin_inset Formula $\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\forall x\in\left(-1,1\right),n\geq n_{0}:\left|\frac{x\left(1-x^{n}\right)}{1-x}-\frac{x}{1-x}\right|<\varepsilon$
+\end_inset
+
+.
+ Za začetekk si oglejmo le 
+\begin_inset Formula $x>0$
+\end_inset
+
+.
+ Ker je tedaj 
+\begin_inset Formula $\frac{x\left(1-x^{n}\right)}{1-x}<\frac{x}{1-x}$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $\left|\frac{x\left(1-x^{n}\right)}{1-x}-\frac{x}{1-x}\right|=\frac{x}{1-x}-\frac{x\left(1-x^{n}\right)}{1-x}=\frac{\cancel{x-x+}x^{n+1}}{1-x}$
+\end_inset
+
+.
+ Računajmo sedaj 
+\begin_inset Formula $\frac{x^{n+1}}{1-x}<\varepsilon\sim x^{n+1}<\varepsilon\left(1-x\right)\sim\left(n+1\right)\log x<\log\left(\varepsilon\left(1-x\right)\right)\sim n+1>\frac{\log\left(\varepsilon\left(1-x\right)\right)}{\log x}\sim n>\frac{\log\left(\varepsilon\left(1-x\right)\right)}{\log x}-1$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $n$
+\end_inset
+
+ odvisen od 
+\begin_inset Formula $x$
+\end_inset
+
+,
+ vsota ni enakomerno konvergentna.
+\end_layout
+
+\begin_layout Standard
+Poseben primer funkcijskih vrst so funkcijske vrste funkcij oblike 
+\begin_inset Formula $f_{j}=b_{j}\cdot x^{j}$
+\end_inset
+
+,
+ torej potence (monomi).
+\end_layout
+
+\begin_layout Definition*
+Potenčna vrsta je funkcijska vrsta oblike 
+\begin_inset Formula $\sum_{j=1}^{\infty}b_{j}\cdot x^{j}$
+\end_inset
+
+,
+ kjer so a 
+\begin_inset Formula $\left(b_{j}\right)_{j\in\mathbb{N}}$
+\end_inset
+
+ dana realna števila.
+\end_layout
+
+\begin_layout Theorem*
+Cauchy-Hadamard.
+ Za vsako potenčno vrsto obstaja konvergenčni radij 
+\begin_inset Formula $R\in\left[0,\infty\right]\ni:$
+\end_inset
+
+
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+vrsta absolutno konvergira za 
+\begin_inset Formula $\left|x\right|<R$
+\end_inset
+
+,
+\end_layout
+
+\begin_layout Itemize
+vrsta divergira za 
+\begin_inset Formula $\left|x\right|>R$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Theorem*
+Velja 
+\begin_inset Formula $\text{\ensuremath{\frac{1}{R}=\limsup_{k\to\infty}\sqrt[k]{\left|b_{k}\right|}}}$
+\end_inset
+
+,
+ kjer vzamemo 
+\begin_inset Formula $\frac{1}{0}\coloneqq\infty$
+\end_inset
+
+ in 
+\begin_inset Formula $\frac{1}{\infty}\coloneqq0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Rezultat že poznamo za zelo poseben primer 
+\begin_inset Formula $\forall j\in\mathbb{N}:b_{j}=1$
+\end_inset
+
+ (geometrijska vrsta).
+ Ideja dokaza je,
+ da konvergenco vsake potenčne vrste opišemo s pomočjo geometrijske vrste.
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Konvergenca:
+ Za 
+\begin_inset Formula $x=0$
+\end_inset
+
+ vrsta očitno konvergira,
+ zato privzamemo 
+\begin_inset Formula $x\not=0$
+\end_inset
+
+.
+ Definirajmo 
+\begin_inset Formula $R$
+\end_inset
+
+ s formulo iz definicije (
+\begin_inset Formula $R=\frac{1}{\limsup_{k\to\infty}\sqrt[k]{\left|b_{k}\right|}}$
+\end_inset
+
+).
+ Naj bo 
+\begin_inset Formula $x$
+\end_inset
+
+ tak,
+ da 
+\begin_inset Formula $\left|x\right|<R\leq\infty$
+\end_inset
+
+ (sledi 
+\begin_inset Formula $R>0$
+\end_inset
+
+).
+ Naj bo 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+.
+ Tedaj po definiciji 
+\begin_inset Formula $R$
+\end_inset
+
+ velja 
+\begin_inset Formula $\sqrt[k]{\left|b_{k}\right|}\leq\frac{1}{R}+\varepsilon$
+\end_inset
+
+ za vse dovolj velike 
+\begin_inset Formula $k$
+\end_inset
+
+.
+ Za take 
+\begin_inset Formula $k$
+\end_inset
+
+ sledi
+\begin_inset Formula 
+\[
+\left|b_{k}\right|\left|x\right|^{k}\leq\left(\left(\frac{1}{R}+\varepsilon\right)\left|x\right|\right)^{k}.
+\]
+
+\end_inset
+
+Opazimo,
+ da je desna stran neenačbe člen geometrijske vrste,
+ s katero majoriziramo vrsto iz absolutnih vrednosti členov naše vrste.
+ Preverimo,
+ da desna stran konvergira.
+ Konvergira,
+ kadar 
+\begin_inset Formula $\left(\frac{1}{R}+\varepsilon\right)\left|x\right|<1$
+\end_inset
+
+ oziroma 
+\begin_inset Formula $\varepsilon<\frac{1}{\left|x\right|}-\frac{1}{R}$
+\end_inset
+
+.
+ Po 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{pk}{primerjalnem kriteriju}
+\end_layout
+
+\end_inset
+
+ torej naša vrsta absolutno konvergira.
+\end_layout
+
+\begin_layout Itemize
+Divergenca:
+ Vzemimo poljuben 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+.
+ Po definciji 
+\begin_inset Formula $R$
+\end_inset
+
+ sledi,
+ da je 
+\begin_inset Formula $\sqrt[k]{\left|b_{k}\right|}\geq\frac{1}{R}-\varepsilon$
+\end_inset
+
+ za vse dovolj velike 
+\begin_inset Formula $k$
+\end_inset
+
+.
+ Za take 
+\begin_inset Formula $k$
+\end_inset
+
+ sledi
+\begin_inset Formula 
+\[
+\left|b_{k}\right|\left|x\right|^{k}\geq\left(\left(\frac{1}{R}-\varepsilon\right)\left|x\right|\right)^{k}.
+\]
+
+\end_inset
+
+Opazimo,
+ da je desna stran neenačbe člen geometrijske vrste,
+ ki je majorizirana z vrsto iz absolutnih vrednosti členov naše vrste.
+ Desna stran divergira,
+ ko 
+\begin_inset Formula $\left(\frac{1}{R}-\varepsilon\right)\left|x\right|=1$
+\end_inset
+
+ oziroma 
+\begin_inset Formula $\varepsilon=\frac{1}{R}-\frac{1}{\left|x\right|}$
+\end_inset
+
+,
+ zato tudi naša vrsta divergira.
+\end_layout
+
+\end_deeper
+\begin_layout Example*
+Primer konvergenčnega radija potenčne vrste od prej:
+ 
+\begin_inset Formula $\sum_{j=1}^{\infty}x^{j}$
+\end_inset
+
+.
+ Velja 
+\begin_inset Formula $\forall j\in\mathbb{N}:b_{j}=1$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $R=\frac{1}{\limsup_{j\to\infty}\sqrt[k]{\left|b_{k}\right|}}=1$
+\end_inset
+
+,
+ torej po zgornjem izreku vrsta konvergira za 
+\begin_inset Formula $x\in\left(-1,1\right)$
+\end_inset
+
+ in divergira za 
+\begin_inset Formula $x\not\in\left[-1,1\right]$
+\end_inset
+
+.
+ Ročno lahko še preverimo,
+ da divergira tudi v 
+\begin_inset Formula $\left\{ -1,1\right\} $
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Zveznost
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+TODO XXX FIXME PREVERI ŠE V profesrojevih PDFJIH,
+ recimo dodaj dokaz zveznosti x^2
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Ideja:
+ Izdelati želimo formulacijo,
+ s katero preverimo,
+ če lahko z dovolj majhno spremembo 
+\begin_inset Formula $x$
+\end_inset
+
+ povzročimo majhno spremembo funkcijske vrednosti.
+\end_layout
+
+\begin_layout Example*
+Primer nezvezne funkcije je 
+\begin_inset Formula $f\left(x\right)=\begin{cases}
+0 & ;0\leq x<1\\
+1 & ;x=1
+\end{cases}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $D\subseteq\mathbb{R},a\in D$
+\end_inset
+
+ in 
+\begin_inset Formula $f:D\to\mathbb{R}$
+\end_inset
+
+.
+ Pravimo,
+ da je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna v 
+\begin_inset Formula $a$
+\end_inset
+
+,
+ če 
+\begin_inset Formula $\forall\varepsilon>0\exists\delta>0\forall x\in D:\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(a\right)\right|<\varepsilon$
+\end_inset
+
+.
+ 
+\begin_inset Formula $f$
+\end_inset
+
+ je zvezna na množici 
+\begin_inset Formula $x\subseteq D$
+\end_inset
+
+,
+ če je zvezna na vsaki točki v 
+\begin_inset Formula $D$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hypertarget{kzzz}{Karakterizacija zveznosti z zaporedji}
+\end_layout
+
+\end_inset
+
+.
+ Naj bodo 
+\begin_inset Formula $D,a,f$
+\end_inset
+
+ kot prej.
+ Velja:
+ 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna v 
+\begin_inset Formula $a\Leftrightarrow\forall\left(a_{n}\right)_{n\in\mathbb{N}},a_{n}\in D:\lim_{n\to\infty}a_{n}=a\Rightarrow\lim_{n\to\infty}f\left(a_{n}\right)=f\left(a\right)$
+\end_inset
+
+ ZDB 
+\begin_inset Formula $f$
+\end_inset
+
+ je zvezna v 
+\begin_inset Formula $a$
+\end_inset
+
+,
+ če za vsako k 
+\begin_inset Formula $a$
+\end_inset
+
+ konvergentno zaporedje na domeni velja,
+ da funkcijske vrednosti členov zaporedja konvergirajo k funkcijski vrednosti 
+\begin_inset Formula $a$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Dokazujemo ekvivalenco.
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Rightarrow\right)$
+\end_inset
+
+ Predpostavimo,
+ da je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna v 
+\begin_inset Formula $a$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\forall\varepsilon>0\exists\delta>0\forall x\in D:\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(a\right)\right|<\varepsilon$
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ poljubno zaporedje na 
+\begin_inset Formula $D$
+\end_inset
+
+,
+ ki konvergira k 
+\begin_inset Formula $a$
+\end_inset
+
+,
+ se pravi 
+\begin_inset Formula $\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow\left|a-a_{n}\right|<\varepsilon$
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+ poljuben.
+ Vsled zveznosti 
+\begin_inset Formula $f$
+\end_inset
+
+ velja,
+ da je 
+\begin_inset Formula $\left|f\left(a_{n}\right)-f\left(a\right)\right|<\varepsilon$
+\end_inset
+
+ za vse take 
+\begin_inset Formula $a_{n}$
+\end_inset
+
+,
+ da velja 
+\begin_inset Formula $\left|a_{n}-a\right|<\delta$
+\end_inset
+
+ za neko 
+\begin_inset Formula $\delta\in\mathbb{R}$
+\end_inset
+
+.
+ Ker je zaporedje 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ konvergentno k 
+\begin_inset Formula $a$
+\end_inset
+
+,
+ so vsi členi po nekem 
+\begin_inset Formula $n_{0}$
+\end_inset
+
+ v 
+\begin_inset Formula $\delta-$
+\end_inset
+
+okolici 
+\begin_inset Formula $a$
+\end_inset
+
+,
+ torej velja pogoj 
+\begin_inset Formula $\left|a_{n}-a\right|<\delta$
+\end_inset
+
+,
+ torej velja 
+\begin_inset Formula $\left|f\left(a_{n}\right)-f\left(a\right)\right|<\varepsilon$
+\end_inset
+
+ za vse 
+\begin_inset Formula $n\geq n_{0}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Leftarrow\right)$
+\end_inset
+
+ PDDRAA 
+\begin_inset Formula $f$
+\end_inset
+
+ ni zvezna v 
+\begin_inset Formula $a$
+\end_inset
+
+.
+ Da pridemo do protislovja,
+ moramo dokazati,
+ da 
+\begin_inset Formula $\exists\left(a_{n}\right)_{n\in\mathbb{N}},a_{n}\in D\ni:\lim_{n\to\infty}a_{n}=a$
+\end_inset
+
+,
+ a vendar 
+\begin_inset Formula $\lim_{n\to\infty}f\left(a_{n}\right)\not=f\left(a\right)$
+\end_inset
+
+.
+ Ker 
+\begin_inset Formula $f$
+\end_inset
+
+ ni zvezna,
+ velja,
+ da 
+\begin_inset Formula $\exists\varepsilon>0\forall\delta>0\exists x\in D\ni:\left|x-a\right|<\delta\wedge\left|f\left(x\right)-f\left(a\right)\right|\geq\varepsilon$
+\end_inset
+
+.
+ Izberimo 
+\begin_inset Formula $\forall n\in\mathbb{N}:\delta_{n}\coloneqq\frac{1}{n}$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $\forall n\in\mathbb{N}\exists\varepsilon>0,x\in D\eqqcolon x_{n}\ni:\left|x_{n}-a\right|<\frac{1}{n}\wedge\left|f\left(x_{n}\right)-f\left(a\right)\right|\geq\varepsilon$
+\end_inset
+
+.
+ S prvim argumentom konjunkcije smo poskrbeli za to,
+ da je naše konstruiramo zaporedje 
+\begin_inset Formula $\left(x_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ konvergentno k 
+\begin_inset Formula $a$
+\end_inset
+
+.
+ Konstruirali smo zaporedje,
+ pri katerem so funkcijske vrednosti za vsak 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+ izven 
+\begin_inset Formula $\varepsilon-$
+\end_inset
+
+okolice 
+\begin_inset Formula $f\left(a\right)$
+\end_inset
+
+,
+ torej zaporedje ne konvergira k 
+\begin_inset Formula $f\left(a\right)$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Theorem*
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hypertarget{kzsppom}{Karakterizacija zveznosti s pomočjo praslik odprtih množic}
+\end_layout
+
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $f:D\to\mathbb{R}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $f$
+\end_inset
+
+ je zvezna na 
+\begin_inset Formula $D\Leftrightarrow$
+\end_inset
+
+ za vsako odprto množico 
+\begin_inset Formula $V\subset\mathbb{R}$
+\end_inset
+
+ je 
+\begin_inset Formula $f^{-1}\left(V\right)$
+\end_inset
+
+ spet odprta množica
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+Za funkcijo 
+\begin_inset Formula $f:D\to V$
+\end_inset
+
+ za 
+\begin_inset Formula $X\subseteq V$
+\end_inset
+
+ definiramo 
+\begin_inset Formula $f^{-1}\left(X\right)\coloneqq\left\{ x\in D;f\left(x\right)\in V\right\} \subseteq D$
+\end_inset
+
+.
+\end_layout
+
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Dokazujemo ekvivalenco.
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Leftarrow\right)$
+\end_inset
+
+ Predpostavimo,
+ da za vsako odprto množico 
+\begin_inset Formula $V\subset\mathbb{R}$
+\end_inset
+
+ je 
+\begin_inset Formula $f^{-1}\left(V\right)$
+\end_inset
+
+ spet odprta množica.
+ Dokazujemo,
+ da je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna na 
+\begin_inset Formula $D$
+\end_inset
+
+.
+ Naj bosta 
+\begin_inset Formula $a\in D,\varepsilon>0$
+\end_inset
+
+ poljubna.
+ Naj bo 
+\begin_inset Formula $V\coloneqq\left(f\left(a\right)-\varepsilon,f\left(a\right)+\varepsilon\right)$
+\end_inset
+
+ odprta množica.
+ Po predpostavki sledi,
+ da je 
+\begin_inset Formula $f^{-1}\left(V\right)$
+\end_inset
+
+ spet odprta.
+ Ker je 
+\begin_inset Formula $a\in f^{-1}\left(V\right)$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $a\in V$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $V$
+\end_inset
+
+ odprta,
+ 
+\begin_inset Formula $\exists\delta>0\ni:\left(a-\delta,a+\delta\right)\in V$
+\end_inset
+
+.
+ Torej 
+\begin_inset Formula $\forall x\in D:\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(a\right)\right|<\varepsilon$
+\end_inset
+
+,
+ torej je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna na 
+\begin_inset Formula $D$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Rightarrow\right)$
+\end_inset
+
+ Predpostavimo,
+ da je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna na 
+\begin_inset Formula $D$
+\end_inset
+
+,
+ to pomeni 
+\begin_inset Formula $\forall a\in D\forall\varepsilon>0\exists\delta>0\forall x\in D:\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(a\right)\right|<\varepsilon$
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $V$
+\end_inset
+
+ poljubna odprta podmnožica 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+ in naj bo 
+\begin_inset Formula $a\in f^{-1}\left(V\right)$
+\end_inset
+
+ poljuben (torej 
+\begin_inset Formula $f\left(a\right)\in V$
+\end_inset
+
+).
+ Ker je 
+\begin_inset Formula $f\left(a\right)\in V$
+\end_inset
+
+,
+ ki je odprta,
+ 
+\begin_inset Formula $\exists\varepsilon>0\ni:\left(f\left(a\right)-\varepsilon,f\left(a\right)+\varepsilon\right)\subseteq V$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna v 
+\begin_inset Formula $a$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\exists\delta>0\forall x\in D:\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(a\right)\right|<\varepsilon$
+\end_inset
+
+,
+ torej je tudi neka odprta okolica 
+\begin_inset Formula $f\left(a\right)$
+\end_inset
+
+ v 
+\begin_inset Formula $f^{-1}\left(V\right)$
+\end_inset
+
+.
+ Ker je bil 
+\begin_inset Formula $a$
+\end_inset
+
+ poljuben,
+ je 
+\begin_inset Formula $f^{-1}\left(V\right)$
+\end_inset
+
+ odprta,
+ ker je bila 
+\begin_inset Formula $V$
+\end_inset
+
+ poljubna,
+ je izrek dokazan.
+\end_layout
+
+\end_deeper
+\begin_layout Theorem*
+Naj bosta 
+\begin_inset Formula $f,g:D\to\mathbb{R}$
+\end_inset
+
+ zvezni v 
+\begin_inset Formula $a\in D$
+\end_inset
+
+.
+ Tedaj so v 
+\begin_inset Formula $a$
+\end_inset
+
+ zvezne tudi funkcije 
+\begin_inset Formula $f+g,f-g,f\cdot g$
+\end_inset
+
+ in 
+\begin_inset Formula $f/g$
+\end_inset
+
+,
+ slednja le,
+ če je 
+\begin_inset Formula $g\left(a\right)\not=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Ker je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna v 
+\begin_inset Formula $a$
+\end_inset
+
+ po 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{kzzz}{izreku o karakterizaciji zveznosti z zaporedji}
+\end_layout
+
+\end_inset
+
+ velja za vsako 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}},\forall n\in\mathbb{N}:a_{n}\subset D,\lim_{n\to\infty}a_{n}=a$
+\end_inset
+
+ tudi 
+\begin_inset Formula $\lim_{n\to\infty}f\left(a_{n}\right)=f\left(a\right)$
+\end_inset
+
+.
+ Po 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{pmkdlim}{izreku iz poglavja o zaporedjih}
+\end_layout
+
+\end_inset
+
+ velja,
+ da 
+\begin_inset Formula $f\left(a_{n}\right)*g\left(a_{n}\right)\to\left(f*g\right)\left(a_{n}\right)$
+\end_inset
+
+ za 
+\begin_inset Formula $*\in\left\{ +,-,\cdot,/\right\} $
+\end_inset
+
+.
+ Zopet uporabimo izrek o karakterizaciji zveznosti z zaporedji,
+ ki pove,
+ da so tudi 
+\begin_inset Formula $f*g$
+\end_inset
+
+ za 
+\begin_inset Formula $*\in\left\{ +,-,\cdot,/\right\} $
+\end_inset
+
+ zvezne v 
+\begin_inset Formula $a$
+\end_inset
+
+.
+ Pri deljenju velja omejitev 
+\begin_inset Formula $f\left(a\right)\not=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+Če sta 
+\begin_inset Formula $D,E\subseteq\mathbb{R}$
+\end_inset
+
+ in 
+\begin_inset Formula $f:D\to E$
+\end_inset
+
+ in 
+\begin_inset Formula $g:E\to\mathbb{R}$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $g\circ f:D\to\mathbb{R}$
+\end_inset
+
+.
+ Hkrati pa,
+ če je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna v 
+\begin_inset Formula $a$
+\end_inset
+
+ in 
+\begin_inset Formula $g$
+\end_inset
+
+ zvezna v 
+\begin_inset Formula $f\left(a\right)$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $g\circ f$
+\end_inset
+
+ zvezna v 
+\begin_inset Formula $a$
+\end_inset
+
+.
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+Velja 
+\begin_inset Formula $\left(g\circ f\right)\left(x\right)=g\left(f\left(x\right)\right)$
+\end_inset
+
+.
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Proof
+Vzemimo poljubno 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}\subseteq D$
+\end_inset
+
+,
+ da 
+\begin_inset Formula $a_{n}\to a\in D$
+\end_inset
+
+.
+ Zopet uporabimo 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{kzzz}{izrek o karakterizaciji zveznosti z zaporedji}
+\end_layout
+
+\end_inset
+
+:
+ ker je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna,
+ velja 
+\begin_inset Formula $f\left(a_{n}\right)\to f\left(a\right)$
+\end_inset
+
+ in ker je 
+\begin_inset Formula $g$
+\end_inset
+
+ zvezna,
+ velja 
+\begin_inset Formula $g\left(f\left(a_{n}\right)\right)\to g\left(f\left(a\right)\right)$
+\end_inset
+
+.
+ Potemtakem 
+\begin_inset Formula $\left(g\circ f\right)\left(a_{n}\right)\to\left(g\circ f\right)\left(a\right)$
+\end_inset
+
+ in po istem izreku je 
+\begin_inset Formula $g\circ f$
+\end_inset
+
+ zvezna na 
+\begin_inset Formula $D$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+Vsi polinomi so zvezni na 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Vzemimo 
+\begin_inset Formula $p\left(x\right)=\sum_{k=0}^{n}a_{k}k^{k}$
+\end_inset
+
+.
+ Uporabimo prejšnji izrek.
+ Polinom je sestavljen iz vsote konstantne funkcije,
+ zmnožene z identiteto,
+ ki je s seboj 
+\begin_inset Formula $n-$
+\end_inset
+
+krat množena.
+ Ker vsota in množenje ohranjata zveznost,
+ je treba dokazati le,
+ da je 
+\begin_inset Formula $f\left(x\right)=x$
+\end_inset
+
+ zvezna in da so 
+\begin_inset Formula $\forall c\in\mathbb{R}:f\left(x\right)=c$
+\end_inset
+
+ zvezne.
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $f\left(x\right)=x$
+\end_inset
+
+ Ali velja 
+\begin_inset Formula $\forall\varepsilon>0\exists\delta>0\ni:\forall x\in\mathbb{R}:\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(a\right)\right|<\varepsilon$
+\end_inset
+
+?
+ Da,
+ velja.
+ Vzamemo lahko katerokoli 
+\begin_inset Formula $\delta\in(0,\varepsilon]$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $f\left(x\right)=c$
+\end_inset
+
+ Naj bo 
+\begin_inset Formula $c\in\mathbb{R}$
+\end_inset
+
+ poljuben.
+ Tu je 
+\begin_inset Formula $\left|f\left(x\right)-f\left(a\right)\right|=\left|c-c\right|=0$
+\end_inset
+
+,
+ torej je desna stran implikacije vedno resnična,
+ torej je implikacija vedno resnična.
+\end_layout
+
+\end_deeper
+\begin_layout Theorem*
+Vse elementarne funkcije so na njihovih definicijskih območjih povsod zvezne.
+ To so:
+ polinomi,
+ potence,
+ racionalne funkcije,
+ koreni,
+ eksponentne funkcije,
+ logaritmi,
+ trigonometrične,
+ ciklometrične in kombinacije neskončno mnogo naštetih,
+ spojenih s 
+\begin_inset Formula $+,-,\cdot,/,\circ$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Tega izreka ne bomo dokazali.
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $f\left(x\right)\coloneqq\log\left(\sin^{3}x+\frac{1}{8}\right)+\frac{1}{\sqrt[4]{x-7}}$
+\end_inset
+
+ je zvezna povsod,
+ kjer je definirana.
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $\varepsilon>0,a\in\mathbb{R}$
+\end_inset
+
+ in 
+\begin_inset Formula $f:\left(a-\varepsilon,a+\varepsilon\right)\setminus\left\{ a\right\} \to\mathbb{R}$
+\end_inset
+
+.
+ Pravimo,
+ da je 
+\begin_inset Formula $L\in\mathbb{R}$
+\end_inset
+
+ limita 
+\begin_inset Formula $f$
+\end_inset
+
+ v točki 
+\begin_inset Formula $a$
+\end_inset
+
+ (zapišemo 
+\begin_inset Formula $L=\lim_{x\to a}f\left(x\right)$
+\end_inset
+
+),
+ če za vsako zaporedje 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}},a_{n}\in\left(a-\varepsilon,a+\varepsilon\right)\setminus\left\{ a\right\} $
+\end_inset
+
+,
+ za katero velja 
+\begin_inset Formula $a_{n}\to a$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $f\left(a_{n}\right)\to L$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+ZDB če 
+\begin_inset Formula $\forall\varepsilon>0\exists\delta>0\ni:\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-L\right|<\varepsilon$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+ZDB če za 
+\begin_inset Formula $\overline{f}:\left(a-\varepsilon,a+\varepsilon\right)\to\mathbb{R}$
+\end_inset
+
+ s predpisom 
+\begin_inset Formula $\overline{f}\left(x\right)\coloneqq\begin{cases}
+f\left(x\right) & ;x\in\left(a-\varepsilon,a+\varepsilon\right)\setminus\left\{ a\right\} \\
+L & ;x\in a
+\end{cases}$
+\end_inset
+
+ velja,
+ da je zvezna v 
+\begin_inset Formula $a$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Note*
+Vrednost 
+\begin_inset Formula $f\left(a\right)$
+\end_inset
+
+,
+ če sploh obstaja,
+ nima vloge pri vrednosti limite.
+\end_layout
+
+\begin_layout Corollary*
+Naj bo 
+\begin_inset Formula $a\in D\subseteq\mathbb{R}$
+\end_inset
+
+ in 
+\begin_inset Formula $f:D\to\mathbb{R}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $f$
+\end_inset
+
+ je zvezna v 
+\begin_inset Formula $a\Leftrightarrow\lim_{x\to a}f\left(x\right)=f\left(a\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Kvadratna funkcija 
+\begin_inset Formula $f\left(x\right)=x^{2}$
+\end_inset
+
+ je zvezna.
+ Vzemimo poljuben 
+\begin_inset Formula $a\in\mathbb{R},\varepsilon>0$
+\end_inset
+
+.
+ Obstajati mora taka 
+\begin_inset Formula $\delta>0\ni:\forall x\in\mathbb{R}:\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(a\right)\right|<\varepsilon$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Podan imamo torej 
+\begin_inset Formula $a$
+\end_inset
+
+ in 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+,
+ želimo najti 
+\begin_inset Formula $\delta$
+\end_inset
+
+.
+ Želimo priti do neenakosti,
+ ki ima na manjši strani 
+\begin_inset Formula $\left|f\left(x\right)-f\left(a\right)\right|=\left|x^{2}-a^{2}\right|$
+\end_inset
+
+ in na večji strani nek izraz z 
+\begin_inset Formula $\left|x-a\right|$
+\end_inset
+
+,
+ da ta 
+\begin_inset Formula $\left|x-a\right|$
+\end_inset
+
+ nadomestimo z 
+\begin_inset Formula $\delta$
+\end_inset
+
+ in nato večjo stran enačimo z 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+,
+ da izrazimo 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+ v odvisnosti od 
+\begin_inset Formula $\delta$
+\end_inset
+
+ in 
+\begin_inset Formula $a$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Računajmo:
+ 
+\begin_inset Formula $\left|x^{2}-a^{2}\right|=\left|x-a\right|\left|x+a\right|$
+\end_inset
+
+.
+ Predelajmo izraz 
+\begin_inset Formula $\left|x+a\right|=\left|\left(x-a\right)+2a\right|\leq\left|x-a\right|+\left|2a\right|$
+\end_inset
+
+,
+ torej skupaj 
+\begin_inset Formula $\left|x^{2}-a^{2}\right|\leq\left|x-a\right|\left(\left|x-a\right|+\left|2a\right|\right)$
+\end_inset
+
+.
+ Sedaj nadomestimo 
+\begin_inset Formula $\left|x-a\right|$
+\end_inset
+
+ z 
+\begin_inset Formula $\delta$
+\end_inset
+
+:
+ 
+\begin_inset Formula $\left|x^{2}-a^{2}\right|\leq\delta\left(\delta+\left|2a\right|\right)$
+\end_inset
+
+.
+ Iščemo tak 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+,
+ da velja 
+\begin_inset Formula $\left|x^{2}-a^{2}\right|<\varepsilon$
+\end_inset
+
+,
+ zato enačimo 
+\begin_inset Formula $\delta\left(\delta+\left|2a\right|\right)=\varepsilon$
+\end_inset
+
+ in dobimo kvadratno enačbo 
+\begin_inset Formula $\delta^{2}+\left|2a\right|\delta-\varepsilon=0$
+\end_inset
+
+,
+ ki jo rešimo z obrazcem za ničle:
+\begin_inset Formula 
+\[
+\delta_{1,2}=\frac{-2\left|a\right|\pm\sqrt{4\left|a\right|^{2}-4\varepsilon}}{2}=-\left|a\right|\pm\sqrt{\left|a\right|^{2}-\varepsilon}
+\]
+
+\end_inset
+
+Toda ker iščemo le pozitivne 
+\begin_inset Formula $\delta$
+\end_inset
+
+,
+ je edina rešitev
+\begin_inset Formula 
+\[
+\delta=-\left|a\right|+\sqrt{\left|a\right|^{2}-\varepsilon}=\sqrt{\left|a\right|^{2}-\varepsilon}-\left|a\right|=\frac{\sqrt{\left|a\right|^{2}-\varepsilon}-\left|a\right|}{1}=\frac{\left(\sqrt{\left|a\right|^{2}-\varepsilon}-\left|a\right|\right)\left(\sqrt{\left|a\right|^{2}-\varepsilon}+\left|a\right|\right)}{\sqrt{\left|a\right|^{2}-\varepsilon}+\left|a\right|}=\frac{\varepsilon}{\sqrt{\left|a\right|^{2}-\varepsilon}+\left|a\right|}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $D\subset\mathbb{R},a\in\mathbb{R}\ni:\forall\varepsilon>0:D\cap\left(a,a+\varepsilon\right)\not=\emptyset$
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $f:D\to\mathbb{R}$
+\end_inset
+
+.
+ Število 
+\begin_inset Formula $L_{+}\in\mathbb{R}$
+\end_inset
+
+ je desna limita funkcije 
+\begin_inset Formula $f$
+\end_inset
+
+ v točki 
+\begin_inset Formula $a$
+\end_inset
+
+,
+ če 
+\begin_inset Formula $\forall\left(a_{n}\right)_{n\in\mathbb{N}}\subset D\cap\left(a,\infty\right):a_{n}\to a\Rightarrow f\left(a_{n}\right)\to L_{+}$
+\end_inset
+
+ ZDB če za vsako k 
+\begin_inset Formula $a$
+\end_inset
+
+ konvergentno zaporedje s členi desno od 
+\begin_inset Formula $a$
+\end_inset
+
+ velja,
+ da funkcijske vrednosti členov konvergirajo k 
+\begin_inset Formula $L_{+}$
+\end_inset
+
+.
+ Oznaka 
+\begin_inset Formula $L_{+}=\lim_{x\to a^{+}}f\left(x\right)=\lim_{x\searrow a}f\left(x\right)=f\left(a+0\right)$
+\end_inset
+
+.
+ Podobno definiramo tudi levo limito 
+\begin_inset Formula $L_{-}=\lim_{x\to a^{-}}f\left(x\right)=\lim_{x\nearrow a}f\left(x\right)=f\left(a-0\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+Naj bo 
+\begin_inset Formula $D\subset\mathbb{R}$
+\end_inset
+
+ in 
+\begin_inset Formula $a\in\mathbb{R}$
+\end_inset
+
+ da velja 
+\begin_inset Formula $\forall\varepsilon>0:D\cap\left(a,a-\varepsilon\right)\not=\emptyset\wedge D\cap\left(a,a+\varepsilon\right)\not=\emptyset$
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $f:D\to\mathbb{R}$
+\end_inset
+
+.
+ Velja 
+\begin_inset Formula $\exists\lim_{x\to a}f\left(x\right)\Leftrightarrow\exists\lim_{x\nearrow a}f\left(x\right)\wedge\exists\lim_{x\searrow a}f\left(x\right)\wedge\lim_{x\nearrow a}f\left(x\right)=\lim_{x\searrow a}f\left(x\right)$
+\end_inset
+
+ V tem primeru velja 
+\begin_inset Formula $\lim_{x\to a}f\left(x\right)=\lim_{x\nearrow a}f\left(x\right)=\lim_{x\searrow a}f\left(x\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Označimo 
+\begin_inset Formula $\lim_{x\searrow a}f\left(x\right)\eqqcolon f\left(a+0\right),\lim_{x\nearrow a}f\left(x\right)\eqqcolon f\left(a-0\right)$
+\end_inset
+
+.
+ Če 
+\begin_inset Formula $\exists f\left(a+0\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $\exists f\left(a-0\right)$
+\end_inset
+
+,
+ vendar 
+\begin_inset Formula $f\left(a+0\right)\not=f\left(a-0\right)$
+\end_inset
+
+,
+ pravimo,
+ da ima 
+\begin_inset Formula $f$
+\end_inset
+
+ v točki 
+\begin_inset Formula $a$
+\end_inset
+
+ 
+\begin_inset Quotes gld
+\end_inset
+
+skok
+\begin_inset Quotes grd
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $\lim_{x\to0}\frac{1}{1+e^{1/x}}$
+\end_inset
+
+ ne obstaja.
+ Zakaj?
+ Izračunajmo levo in desno limito:
+\begin_inset Formula 
+\[
+\lim_{x\searrow0}\frac{1}{1+e^{1/x}}=0,\lim_{x\nearrow0}\frac{1}{1+e^{1/x}}=1
+\]
+
+\end_inset
+
+Toda 
+\begin_inset Formula $\exists\lim_{x\to a}f\left(x\right)\Leftrightarrow\exists\lim_{x\nearrow a}f\left(x\right)\wedge\exists\lim_{x\searrow a}f\left(x\right)\wedge\lim_{x\nearrow a}f\left(x\right)=\lim_{x\searrow a}f\left(x\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Funkcija 
+\begin_inset Formula $f$
+\end_inset
+
+ je na intervalu 
+\begin_inset Formula $D$
+\end_inset
+
+ odsekoma zvezna,
+ če je zvezna povsod na 
+\begin_inset Formula $D$
+\end_inset
+
+,
+ razen morda v končno mnogo točkah,
+ v katerih ima skok.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+Naj bo 
+\begin_inset Formula $f:\mathbb{R}\setminus\left\{ 0\right\} \to\mathbb{R}$
+\end_inset
+
+ s predpisom 
+\begin_inset Formula $x\mapsto\frac{\sin x}{x}$
+\end_inset
+
+.
+ Zanima nas,
+ ali obstaja 
+\begin_inset Formula $\lim_{x\to0}f\left(x\right)$
+\end_inset
+
+.
+ Grafični dokaz.
+\end_layout
+
+\begin_layout Example*
+\begin_inset Float figure
+placement document
+alignment document
+wide false
+sideways false
+status open
+
+\begin_layout Plain Layout
+TODO XXX FIXME SKICA S TKZ EUCLID,
+ glej ZVZ III/ANA1P1120/str.8
+\end_layout
+
+\begin_layout Plain Layout
+\begin_inset Caption Standard
+
+\begin_layout Plain Layout
+Skica.
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Očitno velja 
+\begin_inset Formula $\triangle ABD\subset$
+\end_inset
+
+ krožni izsek 
+\begin_inset Formula $DAB\subset\triangle ABC$
+\end_inset
+
+,
+ torej za njihove ploščine velja 
+\begin_inset Formula 
+\[
+\frac{\sin x}{2}\leq\frac{x}{2\pi}\cdot x=\frac{x}{2}\leq\frac{\tan x}{2}\quad\quad\quad\quad/\cdot\frac{2}{\sin x}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+1\leq\frac{x}{\sin x}\leq\frac{1}{\cos x}\quad\quad\quad\quad/\lim_{x\to0}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\lim_{x\to0}1\leq\lim_{x\to0}\frac{x}{\sin x}\leq\lim_{x\to0}\frac{1}{\cos x}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+1\leq\lim_{x\to0}\frac{x}{\sin x}\leq1
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\lim_{x\to0}\frac{x}{\sin x}=1
+\]
+
+\end_inset
+
+Da naš sklep res potrdimo,
+ je potreben spodnji izrek.
+\end_layout
+
+\begin_layout Theorem*
+Če za 
+\begin_inset Formula $f,g,h:D\to\mathbb{R}$
+\end_inset
+
+ velja za 
+\begin_inset Formula $a\in D$
+\end_inset
+
+:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $\exists\varepsilon>0\forall x\in\left(a-\varepsilon,a+\varepsilon\right)\setminus\left\{ a\right\} :f\left(x\right)\leq g\left(x\right)\leq h\left(x\right)$
+\end_inset
+
+ in hkrati
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $\exists\lim_{x\to a}f\left(x\right),\lim_{x\to a}h\left(x\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $\lim_{x\to a}f\left(x\right)=\lim_{x\to a}h\left(x\right)\eqqcolon L$
+\end_inset
+
+,
+ tedaj tudi 
+\begin_inset Formula $\exists\lim_{x\to a}g\left(x\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $\lim_{x\to a}g\left(x\right)=L$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $A=A\left(x\right)\coloneqq\max\left\{ \left|f\left(x\right)-L\right|,\left|h\left(x\right)-L\right|\right\} $
+\end_inset
+
+.
+ Velja
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $g\left(x\right)-L\leq h\left(x\right)-L\leq\left|h\left(x\right)-L\right|\leq A\left(x\right)$
+\end_inset
+
+ in
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $L-g\left(x\right)\leq L-f\left(x\right)\leq\left|f\left(x\right)-L\right|\leq A\left(x\right)$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Proof
+Posledično 
+\begin_inset Formula $\left|g\left(x\right)-L\right|\leq A\left(x\right)$
+\end_inset
+
+.
+ Naj bo sedaj 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+ poljuben.
+ Tedaj velja 
+\begin_inset Formula $\exists\delta_{1}>0\ni:\left|x-a\right|<\delta_{1}\Rightarrow\left|f\left(x\right)-L\right|<\varepsilon$
+\end_inset
+
+ in 
+\begin_inset Formula $\exists\delta_{2}>0\ni:\left|x-a\right|<\delta_{2}\Rightarrow\left|h\left(x\right)-L\right|<\varepsilon$
+\end_inset
+
+.
+ Za 
+\begin_inset Formula $\delta\coloneqq\min\left\{ \delta_{1},\delta_{2}\right\} $
+\end_inset
+
+ torej velja 
+\begin_inset Formula $\left|x-a\right|<\delta\Rightarrow\left|g\left(x\right)-L\right|<\varepsilon$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsection
+Zvezne funkcije na kompaktnih množicah
+\end_layout
+
+\begin_layout Definition*
+Množica 
+\begin_inset Formula $K\subseteq\mathbb{R}$
+\end_inset
+
+ je kompaktna 
+\begin_inset Formula $\Leftrightarrow$
+\end_inset
+
+ je zaprta in omejena ZDB je unija zaprtih intervalov.
+\end_layout
+
+\begin_layout Theorem*
+Naj bo 
+\begin_inset Formula $K\subset\mathbb{R}$
+\end_inset
+
+ kompaktna in 
+\begin_inset Formula $f:K\to\mathbb{R}$
+\end_inset
+
+ zvezna.
+ Tedaj je 
+\begin_inset Formula $f$
+\end_inset
+
+ 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hypertarget{zfnkm}{omejena in doseže minimum in maksimum}
+\end_layout
+
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Primeri funkcij.
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $f_{1}\left(x\right)=\frac{1}{x}$
+\end_inset
+
+ na 
+\begin_inset Formula $I_{1}=(0,1]$
+\end_inset
+
+.
+ 
+\begin_inset Formula $f_{1}$
+\end_inset
+
+ je zvezna in 
+\begin_inset Formula $\lim_{x\to0}f_{1}\left(x\right)=\infty$
+\end_inset
+
+,
+ torej ni omejena,
+ a 
+\begin_inset Formula $I_{1}$
+\end_inset
+
+ ni zaprt.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $f_{2}\left(x\right)=\begin{cases}
+0 & ;x=0\\
+\frac{1}{x} & ;x\in(0,1]
+\end{cases}$
+\end_inset
+
+ ni omejena in je definirana na kompaktni množici,
+ a ni zvezna.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $f_{3}\left(x\right)=x$
+\end_inset
+
+ na 
+\begin_inset Formula $x\in\left(0,1\right)$
+\end_inset
+
+.
+ Je omejena,
+ ne doseže maksimuma,
+ a 
+\begin_inset Formula $D_{f_{3}}$
+\end_inset
+
+ ni kompaktna (ni zaprta).
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $f_{4}\left(x\right)=\begin{cases}
+x & ;x\in\left(0,1\right)\\
+\frac{1}{2} & ;x\in\left\{ 0,1\right\} 
+\end{cases}$
+\end_inset
+
+.
+ Velja 
+\begin_inset Formula $\sup f_{4}=1$
+\end_inset
+
+,
+ ampak ga ne doseže,
+ a ni zvezna
+\end_layout
+
+\end_deeper
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $K\subseteq\mathbb{R}$
+\end_inset
+
+ kompaktna in 
+\begin_inset Formula $f:K\to\mathbb{R}$
+\end_inset
+
+ zvezna.
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Omejenost navzgor:
+ PDDRAA 
+\begin_inset Formula $f$
+\end_inset
+
+ ni navzgor omejena.
+ Tedaj 
+\begin_inset Formula $\forall n\in\mathbb{N}\exists x_{n}\in K\ni:f\left(x_{n}\right)\geq n$
+\end_inset
+
+ (*).
+ Ker je 
+\begin_inset Formula $\left(x_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ omejeno zaporedje (vsi členi so na kompaktni 
+\begin_inset Formula $K$
+\end_inset
+
+),
+ ima stekališče,
+ recimo mu 
+\begin_inset Formula $s\in\mathbb{R}$
+\end_inset
+
+.
+ Vemo,
+ da tedaj obstaja podzaporedje 
+\begin_inset Formula $\left(x_{n_{k}}\right)_{k\in\mathbb{N}}\ni:s=\lim_{k\to\infty}x_{n_{k}}$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $K$
+\end_inset
+
+ tudi zaprta,
+ sledi 
+\begin_inset Formula $s\in K$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna na 
+\begin_inset Formula $K$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $f\left(s\right)=\lim_{k\to\infty}f\left(x_{n_{k}}\right)$
+\end_inset
+
+.
+ Toda po (*) sledi 
+\begin_inset Formula $\lim_{k\to\infty}f\left(x_{n_{k}}\right)=\infty$
+\end_inset
+
+,
+ zato 
+\begin_inset Formula $f\left(s\right)=\infty$
+\end_inset
+
+,
+ kar ni mogoče,
+ saj je 
+\begin_inset Formula $f\left(s\right)\in\mathbb{R}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $\rightarrow\!\leftarrow$
+\end_inset
+
+.
+ Torej je 
+\begin_inset Formula $f$
+\end_inset
+
+ navzgor omejena.
+\end_layout
+
+\begin_layout Itemize
+Omejenost navzdol:
+ PDDRAA 
+\begin_inset Formula $f$
+\end_inset
+
+ ni navzdol omejena.
+ Tedaj 
+\begin_inset Formula $\forall n\in\mathbb{N}\exists x_{n}\in K\ni:f\left(x_{n}\right)\leq-n$
+\end_inset
+
+ (*).
+ Ker je 
+\begin_inset Formula $\left(x_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ omejeno zaporedje (vsi členi so na kompaktni 
+\begin_inset Formula $K$
+\end_inset
+
+),
+ ima stekališče,
+ recimo mu 
+\begin_inset Formula $s\in\mathbb{R}$
+\end_inset
+
+.
+ Vemo,
+ da tedaj obstaja podzaporedje 
+\begin_inset Formula $\left(x_{n_{k}}\right)_{k\in\mathbb{N}}\ni:s=\lim_{k\to\infty}x_{n_{k}}$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $K$
+\end_inset
+
+ tudi zaprta,
+ sledi 
+\begin_inset Formula $s\in K$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna na 
+\begin_inset Formula $K$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $f\left(s\right)=\lim_{k\to\infty}f\left(x_{n_{k}}\right)$
+\end_inset
+
+.
+ Toda po (*) sledi 
+\begin_inset Formula $\lim_{k\to\infty}f\left(s_{n_{k}}\right)=-\infty$
+\end_inset
+
+,
+ zato 
+\begin_inset Formula $f\left(s\right)=-\infty$
+\end_inset
+
+,
+ kar ni mogoče,
+ saj je 
+\begin_inset Formula $f\left(s\right)\in\mathbb{R}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $\rightarrow\!\leftarrow$
+\end_inset
+
+.
+ Torej je 
+\begin_inset Formula $f$
+\end_inset
+
+ navzgor omejena.
+ 
+\end_layout
+
+\begin_layout Itemize
+Doseže maksimum:
+ Označimo 
+\begin_inset Formula $M\coloneqq\sup_{x\in K}f\left(x\right)$
+\end_inset
+
+.
+ Ravnokar smo dokazali,
+ da 
+\begin_inset Formula $M<\infty$
+\end_inset
+
+.
+ Po definiciji supremuma 
+\begin_inset Formula $\forall n\in\mathbb{N}\exists t_{n}\in K\ni:f\left(t_{n}\right)>M-\frac{1}{n}$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $K$
+\end_inset
+
+ omejena,
+ ima 
+\begin_inset Formula $\left(t_{n}\right)_{n}$
+\end_inset
+
+ stekališče in ker je zaprta,
+ velja 
+\begin_inset Formula $t\in K$
+\end_inset
+
+,
+ zato 
+\begin_inset Formula $\exists$
+\end_inset
+
+ podzaporedje 
+\begin_inset Formula $\left(t_{n_{j}}\right)_{j\in\mathbb{N}}\ni:t=\lim_{j\to\infty}t_{n_{j}}$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna,
+ velja 
+\begin_inset Formula $f\left(t\right)=\lim_{j\to\infty}f\left(t_{n_{j}}\right)$
+\end_inset
+
+.
+ Toda ker 
+\begin_inset Formula $f\left(t_{n_{j}}\right)>M-\frac{1}{n_{j}}\geq M-\frac{1}{j}$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $f\left(t\right)\geq M$
+\end_inset
+
+.
+ Hkrati po definiciji 
+\begin_inset Formula $M$
+\end_inset
+
+ velja 
+\begin_inset Formula $f\left(t\right)\leq M$
+\end_inset
+
+.
+ Sledi 
+\begin_inset Formula $M=f\left(t\right)$
+\end_inset
+
+ in zato 
+\begin_inset Formula $M=\max_{x\in K}f\left(x\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Doseže minimum:
+ Označimo 
+\begin_inset Formula $M\coloneqq\inf_{x\in K}f\left(x\right)$
+\end_inset
+
+.
+ Ko smo dokazali omejenost,
+ smo dokazali,
+ da 
+\begin_inset Formula $M>-\infty$
+\end_inset
+
+.
+ Po definiciji infimuma 
+\begin_inset Formula $\forall n\in\mathbb{N}\exists t_{n}\in K\ni:f\left(t_{n}\right)<M+\frac{1}{n}$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $K$
+\end_inset
+
+ omejena,
+ ima 
+\begin_inset Formula $\left(t_{n}\right)_{n}$
+\end_inset
+
+ stekališče in ker je zaprta,
+ velja 
+\begin_inset Formula $t\in K$
+\end_inset
+
+,
+ zato 
+\begin_inset Formula $\exists$
+\end_inset
+
+ podzaporedje 
+\begin_inset Formula $\left(t_{n_{j}}\right)_{j\in\mathbb{N}}\ni:t=\lim_{j\to\infty}t_{n_{j}}$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna,
+ velja 
+\begin_inset Formula $f\left(t\right)=\lim_{j\to\infty}f\left(t_{n_{j}}\right)$
+\end_inset
+
+.
+ Toda ker 
+\begin_inset Formula $f\left(t_{n_{j}}\right)<M-\frac{1}{n_{j}}\leq M-\frac{1}{j}$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $f\left(t\right)\leq M$
+\end_inset
+
+.
+ Hkrati po definiciji 
+\begin_inset Formula $M$
+\end_inset
+
+ velja 
+\begin_inset Formula $f\left(t\right)\geq M$
+\end_inset
+
+.
+ Sledi 
+\begin_inset Formula $M=f\left(t\right)$
+\end_inset
+
+ in zato 
+\begin_inset Formula $M=\min_{x\in K}f\left(x\right)$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Theorem*
+Naj bo 
+\begin_inset Formula $f:\left[a,b\right]\to\mathbb{R}$
+\end_inset
+
+ zvezna in 
+\begin_inset Formula $f\left(a\right)f\left(b\right)<0$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $\exists\xi\in\left(a,b\right)\ni:f\left(\xi\right)=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Interval 
+\begin_inset Formula $I_{0}=\left[a,b\right]$
+\end_inset
+
+ razpolovimo.
+ To pomeni,
+ da pogledamo levo in desno polovico intervala 
+\begin_inset Formula $I_{0}$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\left[a,\frac{a+b}{2}\right]$
+\end_inset
+
+ in 
+\begin_inset Formula $\left[\frac{a+b}{2},b\right]$
+\end_inset
+
+.
+ Če je 
+\begin_inset Formula $f\left(\frac{a+b}{2}\right)=0$
+\end_inset
+
+,
+ smo našli iskano točko 
+\begin_inset Formula $\xi$
+\end_inset
+
+,
+ sicer z 
+\begin_inset Formula $I_{1}$
+\end_inset
+
+ označimo katerokoli izmed polovic,
+ ki ima 
+\begin_inset Formula $f$
+\end_inset
+
+ v krajiščih različno predznačene funkcijske vrednosti.
+ Torej 
+\begin_inset Formula $I_{1}=\begin{cases}
+\left[a,\frac{a+b}{2}\right] & ;f\left(a\right)f\left(\frac{a+b}{2}\right)<0\\
+\left[\frac{a+b}{2},b\right] & ;f\left(\frac{a+b}{2}\right)f\left(b\right)<0
+\end{cases}$
+\end_inset
+
+.
+ S postopkom nadaljujemo.
+ Če v končno mnogo korakih najdemo 
+\begin_inset Formula $\xi$
+\end_inset
+
+,
+ da je 
+\begin_inset Formula $f\left(\xi\right)=0$
+\end_inset
+
+,
+ fino,
+ sicer pa dobimo zaporedje intervalov 
+\begin_inset Formula $I_{n}=\left[a_{n},b_{n}\right]\subset\left[a,b\right]=I_{0}\ni:$
+\end_inset
+
+
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $\forall n\in\mathbb{N}:\left|I_{n}\right|=2^{-n}\left|I_{0}\right|$
+\end_inset
+
+ in
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\forall n\in\mathbb{N}:I_{n+1}\subset I_{n}$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $a_{n+1}\geq a_{n}\wedge b_{n+1}\leq b_{n}$
+\end_inset
+
+,
+ in
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset CommandInset label
+LatexCommand label
+name "enu:različni-predznaki-istoležnih-clenov"
+
+\end_inset
+
+
+\begin_inset Formula $\forall n\in\mathbb{N}:f\left(a_{n}\right)f\left(b_{n}\right)<0$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Proof
+Ker sta zaporedji 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ in 
+\begin_inset Formula $\left(b_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ omejeni in monotoni,
+ imata po 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{kmoz}{izreku o konvergenci monotonih in omejenih zaporedij}
+\end_layout
+
+\end_inset
+
+ limiti 
+\begin_inset Formula $\alpha\coloneqq\lim_{n\to\infty}a_{n}=\sup_{n\in\mathbb{N}}a_{n}$
+\end_inset
+
+ in 
+\begin_inset Formula $\beta\coloneqq\lim_{n\to\infty}=\sup_{n\in\mathbb{N}}b_{n}$
+\end_inset
+
+ in 
+\begin_inset Formula $\alpha,\beta\in I_{0}$
+\end_inset
+
+,
+ ker je 
+\begin_inset Formula $I_{0}$
+\end_inset
+
+ zaprt.
+\end_layout
+
+\begin_layout Proof
+Sledi 
+\begin_inset Formula $\forall n\in\mathbb{N}:\left|\alpha-\beta\right|=\beta-\alpha\leq b_{n}-a_{n}=\left|I_{n}\right|=2^{-n}\left|I_{0}\right|$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\lim_{n\to\infty}\left|\alpha-\beta\right|=0\Rightarrow\alpha-\beta=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Ker je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna in 
+\begin_inset Formula $a_{n},b_{n},\xi\in I_{0}$
+\end_inset
+
+,
+ sledi 
+\begin_inset Formula $\lim_{n\to\infty}f\left(a_{n}\right)=f\left(\alpha\right)=f\left(\xi\right)=f\left(\beta\right)=\lim_{n\to\infty}f\left(b_{n}\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Po točki 
+\begin_inset CommandInset ref
+LatexCommand ref
+reference "enu:različni-predznaki-istoležnih-clenov"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+ velja 
+\begin_inset Formula $f\left(\alpha\right)f\left(\beta\right)\leq0$
+\end_inset
+
+.
+ Ker pa 
+\begin_inset Formula $f\left(\alpha\right)=f\left(\beta\right)$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $f\left(\alpha\right)=f\left(\beta\right)=f\left(\xi\right)=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Corollary*
+Naj bo 
+\begin_inset Formula $I=\left[a,b\right]$
+\end_inset
+
+ omejen zaprt interval 
+\begin_inset Formula $\in\mathbb{R}$
+\end_inset
+
+ in 
+\begin_inset Formula $f:I\to\mathbb{R}$
+\end_inset
+
+ zvezna.
+ Tedaj 
+\begin_inset Formula $\exists x_{-},x_{+}\in I\ni:\forall x\in I:f\left(x\right)\in\left[f\left(x_{-}\right),f\left(x_{+}\right)\right]$
+\end_inset
+
+ in 
+\begin_inset Formula $\forall y\in\left[f\left(x_{-}\right),f\left(x_{+}\right)\right]\exists x\in I\ni:y=f\left(x\right)$
+\end_inset
+
+ ZDB 
+\begin_inset Formula $f\left(I\right)=\left[f\left(x_{-}\right),f\left(x_{+}\right)\right]$
+\end_inset
+
+ ZDB zvezna funkcija na zaprtem intervalu 
+\begin_inset Formula $\left[a,b\right]$
+\end_inset
+
+ doseže vse funkcijske vrednosti na intervalu 
+\begin_inset Formula $\left[f\left(a\right),f\left(b\right)\right]$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Dokaz posledice.
+ Naj bo 
+\begin_inset Formula $y$
+\end_inset
+
+ poljuben.
+ Če je 
+\begin_inset Formula $y=f\left(x_{-}\right)$
+\end_inset
+
+,
+ smo našli 
+\begin_inset Formula $x=x_{-}$
+\end_inset
+
+.
+ Če je 
+\begin_inset Formula $y=f\left(x_{+}\right)$
+\end_inset
+
+,
+ smo našli 
+\begin_inset Formula $x=x_{+}$
+\end_inset
+
+.
+ Sicer pa je 
+\begin_inset Formula $f\left(x_{-}\right)<y<f\left(x_{+}\right)$
+\end_inset
+
+.
+ Oglejmo si funkcijo 
+\begin_inset Formula $g\left(x\right)\coloneqq f\left(x\right)-y$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $g\left(x_{-}\right)=f\left(x_{-}\right)-y<0$
+\end_inset
+
+ in 
+\begin_inset Formula $g\left(x_{+}\right)=f\left(x_{+}\right)-y>0$
+\end_inset
+
+ in 
+\begin_inset Formula $g$
+\end_inset
+
+ zvezna na 
+\begin_inset Formula $\left[x_{-}-y,x_{+}-y\right]$
+\end_inset
+
+,
+ torej po prejšnjem izreku 
+\begin_inset Formula $\exists x\in\left[x_{-}-y,x_{+}-y\right]\ni:g\left(x\right)=0$
+\end_inset
+
+,
+ kar pomeni ravno 
+\begin_inset Formula $f\left(x\right)=y$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+Naj bo 
+\begin_inset Formula $I$
+\end_inset
+
+ poljuben interval med 
+\begin_inset Formula $a,b\in\mathbb{R}\cup\left\{ -\infty,\infty\right\} $
+\end_inset
+
+ in 
+\begin_inset Formula $f:I\to\mathbb{R}$
+\end_inset
+
+ 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hypertarget{zism}{zvezna in strogo monotona}
+\end_layout
+
+\end_inset
+
+.
+ Tedaj je 
+\begin_inset Formula $f\left(I\right)$
+\end_inset
+
+ interval med 
+\begin_inset Formula $f\left(a+0\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $f\left(a-0\right)$
+\end_inset
+
+.
+ Inverzna funkcija 
+\begin_inset Formula $f^{-1}$
+\end_inset
+
+ je definirana na 
+\begin_inset Formula $f\left(I\right)$
+\end_inset
+
+ in zvezna.
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $f\coloneqq\arctan$
+\end_inset
+
+,
+ 
+\begin_inset Formula $I\coloneqq\left(-\infty,\infty\right)$
+\end_inset
+
+,
+ zvezna.
+ Naj bo 
+\begin_inset Formula $y\in f\left(I\right)$
+\end_inset
+
+ poljuben.
+ Tedaj 
+\begin_inset Formula $\exists!x\in I\ni:y=f\left(x\right)$
+\end_inset
+
+ in definiramo 
+\begin_inset Formula $x\coloneqq f^{-1}\left(x\right)$
+\end_inset
+
+.
+ 
+\begin_inset Formula $f^{-1}$
+\end_inset
+
+ obstaja in je spet zvezna.
+\end_layout
+
+\begin_layout Proof
+Ne bomo dokazali.
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+Označimo 
+\begin_inset Formula $g=f^{-1}:f\left(I\right)\to\mathbb{R}$
+\end_inset
+
+.
+ Uporabimo 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{kzsppom}{karakterizacijo zveznosti s pomočjo praslik odprtih množic}
+\end_layout
+
+\end_inset
+
+.
+ Dokazujemo torej,
+ da 
+\begin_inset Formula $\forall V^{\text{odp.}}\subset\mathbb{R}:g^{-1}\left(V\right)$
+\end_inset
+
+ je zopet odprta množica 
+\begin_inset Formula $\subseteq f\left(I\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Velja 
+\begin_inset Formula $g^{-1}\left(V\right)=\left\{ x\in f\left(I\right);g\left(x\right)\in V\right\} =\left\{ x\in f\left(I\right):\exists v\in V\cap I\ni:x=f\left(v\right)\right\} =f\left(V\cap I\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Torej dokazujemo 
+\begin_inset Formula $\forall V^{\text{odp.}}\subset\mathbb{R}:f\left(I\cap V\right)$
+\end_inset
+
+ je spet zopet odprta 
+\begin_inset Formula $\subseteq f\left(I\right)$
+\end_inset
+
+,
+ kar je ekvivalentno
+\begin_inset Formula 
+\[
+\forall y\in f\left(I\cap V\right)\exists\delta>0\ni:\left(y-\delta,y+\delta\right)\cap f\left(I\right)\subset f\left(I\cap V\right).
+\]
+
+\end_inset
+
+Pišimo 
+\begin_inset Formula $y=f\left(x\right),x\in I\cap V$
+\end_inset
+
+.
+ Privzemimo,
+ da 
+\begin_inset Formula $f$
+\end_inset
+
+ narašča (če pada,
+ ravnamo podobno).
+ Ker jer 
+\begin_inset Formula $ $
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsection
+Enakomerna zveznost
+\end_layout
+
+\begin_layout Definition*
+\begin_inset Formula $f:I\to\mathbb{R}$
+\end_inset
+
+ je 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hypertarget{ez}{enakomerno zvezna}
+\end_layout
+
+\end_inset
+
+ na 
+\begin_inset Formula $I$
+\end_inset
+
+,
+ če 
+\begin_inset Formula 
+\[
+\forall\varepsilon>0\exists\delta>0\forall x,y\in I:\left|x-y\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(y\right)\right|<\varepsilon.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Note*
+Primerjajmo to z definicijo 
+\begin_inset Formula $f:I\to\mathbb{R}$
+\end_inset
+
+ je (nenujno enakomerno) zvezna na 
+\begin_inset Formula $I$
+\end_inset
+
+,
+ če
+\begin_inset Formula 
+\[
+\forall\varepsilon>0,a\in I\exists\delta>0\forall x\in I:\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(y\right)\right|<\varepsilon.
+\]
+
+\end_inset
+
+Pri slednji definiciji je 
+\begin_inset Formula $\delta$
+\end_inset
+
+ odvisna od 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+ in 
+\begin_inset Formula $a$
+\end_inset
+
+,
+ pri enakomerni zveznosti pa le od 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $f\left(x\right)=\frac{1}{x}$
+\end_inset
+
+ ni enakomerno zvezna,
+ ker je 
+\begin_inset Formula $\delta$
+\end_inset
+
+ odvisen od 
+\begin_inset Formula $a$
+\end_inset
+
+.
+ Če pri fiksnem 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+ pomaknemo tisto pozitivno točko,
+ v kateri preizkušamo zveznost,
+ bolj v levo,
+ bo na neki točki potreben ožji,
+ manjši 
+\begin_inset Formula $\delta$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Theorem*
+Zvezna funkcija na kompaktni množici je enakomerno zvezna.
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $f:K\to\mathbb{R}$
+\end_inset
+
+ zvezna,
+ kjer je 
+\begin_inset Formula $K$
+\end_inset
+
+ kompaktna podmnožica 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+.
+ PDDRAA 
+\begin_inset Formula $f$
+\end_inset
+
+ ni enakomerno zvezna.
+ Zanikajmo definicijo 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{ez}{enakomerne zveznosti}
+\end_layout
+
+\end_inset
+
+:
+ 
+\begin_inset Formula $\exists\varepsilon>0\forall\delta>0\exists x_{\delta},y_{\delta}\in I:\left|x_{\delta}-y_{\delta}\right|<\delta\wedge\left|f\left(x_{\delta}\right)-f\left(y_{\delta}\right)\right|\geq\varepsilon$
+\end_inset
+
+.
+ 
+\begin_inset Formula $x,y$
+\end_inset
+
+ sta seveda lahko odvisna od 
+\begin_inset Formula $\delta$
+\end_inset
+
+ in 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+,
+ zato v subskriptu pišemo 
+\begin_inset Formula $\delta$
+\end_inset
+
+,
+ ki ji pripadata.
+ Ker smo dejali,
+ da to velja,
+ si oglejmo 
+\begin_inset Formula $\forall n\in\mathbb{N}:\delta_{n}\coloneqq\frac{1}{n}$
+\end_inset
+
+ in pripadajoči zaporedji 
+\begin_inset Formula $\left(x_{1/n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ in 
+\begin_inset Formula $\left(y_{1/n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $K$
+\end_inset
+
+ kompaktna,
+ ima zaporedje 
+\begin_inset Formula $\left(x_{1/n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ stekališče v 
+\begin_inset Formula $x\in K$
+\end_inset
+
+,
+ torej obstaja podzaporede 
+\begin_inset Formula $\left(x_{1/n_{k}}\right)_{k\in\mathbb{N}}$
+\end_inset
+
+,
+ ki konvergira k 
+\begin_inset Formula $x$
+\end_inset
+
+.
+ Podobno obstaja podzaporedje 
+\begin_inset Formula $\left(y_{1/n_{k_{l}}}\right)_{l\in\mathbb{N}}$
+\end_inset
+
+,
+ ki konvergira k 
+\begin_inset Formula $y\in K$
+\end_inset
+
+.
+ Pišimo sedaj 
+\begin_inset Formula $x_{l}\coloneqq x_{1/n_{k_{l}}}$
+\end_inset
+
+in 
+\begin_inset Formula $y_{l}\coloneqq y_{1/n_{k_{l}}}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Velja torej 
+\begin_inset Formula $x_{l}\to x$
+\end_inset
+
+ in 
+\begin_inset Formula $y_{l}\to y$
+\end_inset
+
+.
+ Sledi 
+\begin_inset Formula $\left|x-y\right|\leq\lim_{l\to\infty}\left(\left|x-x_{l}\right|+\left|x_{l}-y_{l}\right|+\left|y_{l}-y\right|\right)$
+\end_inset
+
+.
+ Levi in desni člen sta v limiti enaka 0 zaradi konvergence zaporedja,
+ srednji pa je manjši od 
+\begin_inset Formula $\frac{1}{j}$
+\end_inset
+
+ zaradi naše predpostavke (PDDRAA),
+ potemtakem je 
+\begin_inset Formula $x=y$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Zato 
+\begin_inset Formula $\lim_{l\to\infty}\left(f\left(x_{l}\right)-f\left(y_{l}\right)\right)=\lim_{l\to\infty}\left[\left(f\left(x_{l}\right)-f\left(x\right)\right)+\left(f\left(x\right)-f\left(y\right)\right)+\left(f\left(y\right)-f\left(y_{l}\right)\right)\right]$
+\end_inset
+
+.
+ Levi in desni člen sta v limiti enaka 0 zaradi konvergence zaporedja in zveznosti 
+\begin_inset Formula $f$
+\end_inset
+
+,
+ srednji pa je tudi 0,
+ ker 
+\begin_inset Formula $x=y$
+\end_inset
+
+,
+ potemtakem 
+\begin_inset Formula $f\left(x_{l}\right)-f\left(y_{l}\right)\to0$
+\end_inset
+
+,
+ kar je v protislovju z 
+\begin_inset Formula $\left|f\left(x_{l}\right)-f\left(y_{l}\right)\right|\geq\varepsilon$
+\end_inset
+
+ za fiksen 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+ in 
+\begin_inset Formula $\forall l\in\mathbb{N}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $\rightarrow\!\leftarrow$
+\end_inset
+
+,
+ 
+\begin_inset Formula $f$
+\end_inset
+
+ je enakomerno zvezna.
+\end_layout
+
+\begin_layout Corollary*
+En zaprt interval 
+\begin_inset Formula $\frac{1}{x}$
+\end_inset
+
+ bo enakomerno zvezen,
+ 
+\begin_inset Formula $\frac{1}{x}$
+\end_inset
+
+ sama po sebi kot 
+\begin_inset Formula $\left(0,\infty\right)\to\mathbb{R}$
+\end_inset
+
+ pa ni definirana na kompaktni množici.
+ Prav tako 
+\begin_inset Formula $\arcsin$
+\end_inset
+
+ in 
+\begin_inset Formula $x\mapsto\sqrt{x}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Odvod
+\end_layout
+
+\begin_layout Standard
+Najprej razmislek/ideja.
+ Odvod je hitrost/stopnja,
+ s katero se v danem trenutku neka količina spreminja.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Float figure
+placement document
+alignment document
+wide false
+sideways false
+status open
+
+\begin_layout Plain Layout
+TODO XXX FIXME SKICA S TKZ EUCLID (ali pa —
+ bolje —
+ s čim drugim),
+ glej PS zapiski/ANA1P FMF 2023-12-04.pdf
+\end_layout
+
+\begin_layout Plain Layout
+\begin_inset Caption Standard
+
+\begin_layout Plain Layout
+Skica.
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Radi bi določili naklon sekante,
+ torej naklon premice,
+ določene z 
+\begin_inset Formula $x$
+\end_inset
+
+ in neko bližnjo točko 
+\begin_inset Formula $x+h$
+\end_inset
+
+ na grafu funkcije,
+ ki je odvisen le od 
+\begin_inset Formula $x$
+\end_inset
+
+,
+ ne pa tudi od izbire 
+\begin_inset Formula $h$
+\end_inset
+
+.
+ Bližnjo točko pošljemo proti začetni —
+ 
+\begin_inset Formula $h$
+\end_inset
+
+ pošljemo proti 0.
+ Naklon izračunamo s izrazom 
+\begin_inset Formula $\frac{f\left(x+h\right)-f\left(x\right)}{h}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Odvod funkcije 
+\begin_inset Formula $f$
+\end_inset
+
+ v točki 
+\begin_inset Formula $x$
+\end_inset
+
+ označimo 
+\begin_inset Formula $f'\left(x\right)\coloneqq\lim_{h\to0}\frac{f\left(x+h\right)-f\left(x\right)}{h}$
+\end_inset
+
+.
+ Če limita obstaja v točki 
+\begin_inset Formula $x$
+\end_inset
+
+,
+ pravimo,
+ da je funkcija odvedljiva v 
+\begin_inset Formula $x$
+\end_inset
+
+.
+ Pravimo,
+ da je 
+\begin_inset Formula $f$
+\end_inset
+
+ odvedljiva na množici 
+\begin_inset Formula $I\subseteq\mathbb{R}$
+\end_inset
+
+,
+ če je odvedljiva na vsaki 
+\begin_inset Formula $t\in I$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Primeri odvodov preprostih funkcij.
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $f\left(x\right)=c,c\in\mathbb{R}$
+\end_inset
+
+ 
+\begin_inset Formula $f'\left(x\right)=\lim_{h\to0}\frac{\cancelto{c}{f\left(x+h\right)}-\cancelto{c}{f\left(x\right)}}{h}=0$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $f\left(x\right)=x$
+\end_inset
+
+ 
+\begin_inset Formula $f'\left(x\right)=\lim_{h\to0}\frac{f\left(x+h\right)-f\left(x\right)}{h}=\lim_{h\to0}\frac{x+h-x}{h}=1$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $f\left(x\right)=x^{2}$
+\end_inset
+
+ 
+\begin_inset Formula $f'\left(x\right)=\lim_{h\to0}\frac{f\left(x+h\right)-f\left(x\right)}{h}=\lim_{h\to0}\frac{x^{2}+2xh+h^{2}-x^{2}}{h}=\lim_{h\to0}\frac{2xh+h^{2}}{h}=\lim_{h\to0}2x+h=2x$
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Claim*
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hypertarget{op}{Odvod potence}
+\end_layout
+
+\end_inset
+
+.
+ Za poljuben 
+\begin_inset Formula $n\in\mathbb{N}$
+\end_inset
+
+ so funkcije 
+\begin_inset Formula $f\left(x\right)=x^{n}$
+\end_inset
+
+ odvedljive na 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+ in velja 
+\begin_inset Formula $f'\left(x\right)=nx^{n-1}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+\begin_inset Formula 
+\[
+\lim_{h\to0}\frac{f\left(x+h\right)-f\left(x\right)=\left(x+h\right)^{n}-x^{n}=\sum_{k=0}^{n}\binom{n}{k}h^{k}x^{n-k}-x^{n}=\cancel{x^{n}}+nhx^{n-1}+\sum_{k=2}^{n}\binom{n}{k}h^{k}x^{n-k}\cancel{-x^{n}}}{h}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\lim_{h\to0}\frac{nhx^{n-1}+\sum_{k=2}^{n}\binom{n}{k}h^{k}x^{n-k}}{h}=\lim_{h\to0}\frac{\cancel{h}\left(nx^{n-1}+\sum_{k=2}^{n}\binom{n}{k}h^{k-1}x^{n-k}\right)}{\cancel{h}}=\lim_{h\to0}\left(nx^{n-1}+\sum_{k=2}^{n}\binom{n}{k}h^{k-1}x^{n-k}\right)=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=nx^{n-1}+\cancel{\lim_{h\to0}\sum_{k=2}^{n}\binom{n}{k}\cancelto{0}{h^{k-1}}x^{n-k}}=nx^{n-1}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Claim*
+\begin_inset Formula $\sin'=\cos$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\cos'=-\sin$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Proof
+Najprej dokažimo 
+\begin_inset Formula $\sin'=\cos$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+\lim_{h\to\infty}\frac{\sin\left(x+h\right)-\sin\left(x\right)=\sin x\cos h+\sin h\cos x-\sin x=\sin x\left(\cos h-1\right)+\sin h\cos x}{h}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\lim_{h\to0}\left(\sin x\frac{\cos h-1}{h}+\cos x\frac{\sin h}{h}\right)=\lim_{h\to0}\left(\sin x\frac{\left(\cos h-1\right)\left(\cos h+1\right)=\cos^{2}h-1=-\sin^{2}h}{h\left(\cos h+1\right)}+\cos x\frac{\sin h}{h}\right)=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\lim_{h\to0}\left(\sin x\frac{-\sin h}{h}\cdot\frac{\sin h}{\cos h+1}+\cos x\frac{\sin h}{h}\right)=\lim_{h\to0}\cancelto{1}{\frac{\sin h}{h}}\left(\cos x-\cancel{\sin x\frac{\cancelto{0}{\sin h}}{\cos h+1}}\right)=\cos x
+\]
+
+\end_inset
+
+Sedaj dokažimo še 
+\begin_inset Formula $\cos'=-\sin$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+\lim_{h\to0}\frac{\cos\left(x+h\right)-\cos\left(x\right)=\cos x\cos h-\sin x\sin h-\cos x=\cos x\left(\cos h-1\right)-\sin x\sin h}{h}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\lim_{h\to0}\left(\cos x\frac{\cos h-1}{h}-\sin x\frac{\sin h}{h}\right)=\lim_{h\to0}\left(\cos x\frac{\left(\cos h-1\right)\left(\cos h+1\right)=\cos^{2}h-1=-\sin^{2}h}{h\left(\cos h+1\right)}-\sin x\frac{\sin h}{h}\right)=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\lim_{h\to0}\left(\cos x\frac{-\sin h}{h}\cdot\frac{\sin^{}h}{\cos h+1}-\sin x\frac{\sin h}{h}\right)=\lim_{h\to0}\left(\cancelto{1}{\frac{\sin h}{h}}\left(\cancel{-\cos x\frac{\cancelto{0}{\sin h}}{\cos h+1}}-\sin x\right)\right)=-\sin x
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Fact*
+Od prej vemo 
+\begin_inset Formula $\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{n}=e$
+\end_inset
+
+ (limita zaporedja).
+ Velja tudi 
+\begin_inset Formula $\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^{x}=e$
+\end_inset
+
+ (funkcijska limita).
+ Ne bomo dokazali.
+\end_layout
+
+\begin_layout Claim*
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hypertarget{oef}{Odvod eksponentne funkcije}
+\end_layout
+
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $a>0$
+\end_inset
+
+ in 
+\begin_inset Formula $f\left(x\right)=a^{x}$
+\end_inset
+
+.
+ Tedaj je 
+\begin_inset Formula $f'\left(x\right)=a^{x}\ln a$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+\begin_inset Formula 
+\[
+\lim_{h\to0}\frac{f\left(x+h\right)-f\left(x\right)=a^{x}a^{h}-a^{x}}{h}=\lim_{h\to0}a^{x}\frac{a^{h}-1}{h}=\cdots
+\]
+
+\end_inset
+
+Sedaj pišimo 
+\begin_inset Formula $\frac{1}{z}\coloneqq a^{h}-1$
+\end_inset
+
+.
+ Ulomek 
+\begin_inset Formula $\frac{a^{h}-1}{h}$
+\end_inset
+
+ namreč ni odvisen od 
+\begin_inset Formula $x$
+\end_inset
+
+.
+ Sedaj
+\begin_inset Formula 
+\[
+a^{h}-1=\frac{1}{z}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+a^{h}=\frac{1}{z}+1
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+h=\log_{a}\left(\frac{1}{z}+1\right)
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+h=\frac{\ln\left(\frac{1}{z}+1\right)}{\ln a}
+\]
+
+\end_inset
+
+Nadaljujmo s prvotnim računom,
+ ločimo primere:
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $a>1,h\searrow0$
+\end_inset
+
+ Potemtakem 
+\begin_inset Formula $a^{h}-1\searrow0$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\frac{1}{z}\searrow0$
+\end_inset
+
+,
+ sledi 
+\begin_inset Formula $z\nearrow\infty$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+\cdots=a^{x}\lim_{z\to\infty}\frac{\frac{1}{z}}{\frac{\ln\left(\frac{1}{z}+1\right)}{\ln a}}=a^{x}\lim_{z\to\infty}\frac{\frac{1}{z}\ln a}{\ln\left(\frac{1}{z}+1\right)}=a^{x}\lim_{z\to\infty}\frac{\ln a}{\ln\left(\frac{1}{z}+1\right)^{z}}=a^{x}\lim_{z\to\infty}\frac{\ln a}{\ln e}=a^{x}\ln a
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $a>1,h\nearrow0$
+\end_inset
+
+ Potemtakem 
+\begin_inset Formula $a^{h}-1\nearrow0$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\frac{1}{z}\nearrow0$
+\end_inset
+
+,
+ sledi 
+\begin_inset Formula $z\searrow-\infty$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+\cdots=a^{x}\lim_{z\to-\infty}\frac{\ln a}{\ln\left(\frac{1}{z}+1\right)^{z}}=a^{x}\lim_{z\to-\infty}\frac{\ln a}{\ln\cancelto{e}{\left(\frac{1}{z}+1\right)^{z}}}=a^{x}\ln a
+\]
+
+\end_inset
+
+Kajti 
+\begin_inset Formula $\lim_{x\to\infty}\left(1+\frac{k}{x}\right)^{x}=e^{k}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $a\in(0,1]$
+\end_inset
+
+ Podobno kot zgodaj,
+ bodisi 
+\begin_inset Formula $z\nearrow\infty$
+\end_inset
+
+ bodisi 
+\begin_inset Formula $z\searrow-\infty$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Claim*
+Če je 
+\begin_inset Formula $f$
+\end_inset
+
+ odvedljiva v točki 
+\begin_inset Formula $x$
+\end_inset
+
+,
+ je tam tudi zvezna.
+\end_layout
+
+\begin_layout Proof
+Predpostavimo,
+ da obstaja limita 
+\begin_inset Formula $\lim_{h\to0}\frac{f\left(x+h\right)-f\left(x\right)}{h}$
+\end_inset
+
+.
+ Želimo dokazati 
+\begin_inset Formula $f\left(x\right)=\lim_{t\to x}f\left(t\right)$
+\end_inset
+
+.
+ Računajmo:
+\begin_inset Formula 
+\[
+f\left(x\right)=\lim_{t\to x}f\left(t\right)
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+0=\lim_{t\to x}f\left(t\right)-f\left(x\right)
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+0=\lim_{h\to0}f\left(x+h\right)-f\left(x\right)
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+0=\lim_{h\to0}\left(f\left(x+h\right)-f\left(x\right)\right)=\lim_{h\to0}\left(\frac{f\left(x+h\right)-f\left(x\right)}{h}\cdot h\right)
+\]
+
+\end_inset
+
+Limita obstaja,
+ čim obstajata 
+\begin_inset Formula $\lim_{h\to0}\frac{f\left(x+h\right)-f\left(x\right)}{h}$
+\end_inset
+
+,
+ ki obstaja po predpostavki,
+ in 
+\begin_inset Formula $\lim_{h\to0}h$
+\end_inset
+
+,
+ ki obstaja in ima vrednost 
+\begin_inset Formula $0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $f\left(x\right)=\left|x\right|=\sqrt{x^{2}}$
+\end_inset
+
+.
+ Je zvezna,
+ ker je kompozitum zveznih funkcij,
+ toda v 
+\begin_inset Formula $0$
+\end_inset
+
+ ni odvedljiva,
+ kajti 
+\begin_inset Formula $\lim_{h\to0}\frac{f\left(0+h\right)-f\left(0\right)}{h}=\lim_{h\to0}\frac{\left|h\right|-0}{h}=\lim_{h\to0}\sgn h$
+\end_inset
+
+.
+ Limita ne obstaja,
+ ker 
+\begin_inset Formula $-1=\lim_{h\nearrow0}\sgn h\not=\lim_{h\searrow0}\sgn h=1$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+Naj bosta 
+\begin_inset Formula $f,g$
+\end_inset
+
+ odvedljivi v 
+\begin_inset Formula $x\in\mathbb{R}$
+\end_inset
+
+.
+ Tedaj so 
+\begin_inset Formula $f+g,f-g,f\cdot g,f/g$
+\end_inset
+
+ (slednja le,
+ če 
+\begin_inset Formula $g\left(x\right)\not=0$
+\end_inset
+
+) in velja 
+\begin_inset Formula $\left(f\pm g\right)'=f'\pm g'$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\left(fg\right)'=f'g+fg'$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\left(f/g\right)'=\frac{f'g-fg'}{g^{2}}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Dokažimo vse štiri trditve.
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $f+g$
+\end_inset
+
+ Velja 
+\begin_inset Formula $\left(f+g\right)\left(x\right)=f\left(x\right)+g\left(x\right)$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+\left(f+g\right)'\left(x\right)=\lim_{h\to0}\frac{\left(f+g\right)\left(x+h\right)-\left(f+g\right)\left(x\right)=f\left(x+h\right)+g\left(x+h\right)-f\left(x\right)-g\left(x\right)}{h}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\lim_{h\to0}\left(\frac{f\left(x+h\right)-f\left(x\right)}{h}+\frac{g\left(x+h\right)-g\left(x\right)}{h}\right)=f\left(x\right)'+g\left(x\right)'
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $-f$
+\end_inset
+
+ Naj bo 
+\begin_inset Formula $g=-f$
+\end_inset
+
+.
+ 
+\begin_inset Formula $g'\left(x\right)=\lim_{h\to0}\frac{g\left(x+h\right)-g\left(x\right)}{h}=\lim_{h\to0}\frac{-f\left(x+h\right)+f\left(x\right)}{h}=-\lim_{h\to0}\frac{f\left(x+h\right)-f\left(x\right)}{h}=-f\left(x\right)'$
+\end_inset
+
+,
+ zato
+\begin_inset Formula 
+\[
+\left(f-g\right)'\left(x\right)=\left(f+\left(-g\right)\right)'\left(x\right)=f'\left(x\right)+\left(-g\right)'\left(x\right)=f'\left(x\right)-g'\left(x\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $f\cdot g$
+\end_inset
+
+ Velja 
+\begin_inset Formula $\left(fg\right)\left(x\right)=f\left(x\right)g\left(x\right)$
+\end_inset
+
+.
+ Prištejemo in odštejemo isti izraz (v oglatih oklepajih).
+\begin_inset Formula 
+\[
+\left(fg\right)'\left(x\right)=\lim_{h\to0}\frac{\left(fg\right)\left(x+h\right)-\left(fg\right)\left(x\right)=f\left(x+h\right)g\left(x+h\right)-f\left(x\right)g\left(x\right)+\left[f\left(x\right)g\left(x+h\right)-f\left(x\right)g\left(x+h\right)\right]}{h}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\lim_{h\to0}\frac{g\left(x+h\right)\left(f\left(x+h\right)-f\left(x\right)\right)+f\left(x\right)\left(g\left(x+h\right)-g\left(x\right)\right)}{h}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\lim_{h\to0}\frac{f\left(x+h\right)-f\left(x\right)}{h}\cancelto{g\left(x\right)}{g\left(x+h\right)}+\lim_{h\to0}\frac{g\left(x+h\right)-g\left(x\right)}{h}f\left(x\right)=f'\left(x\right)g\left(x\right)+g'\left(x\right)f\left(x\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $f/g$
+\end_inset
+
+ Velja 
+\begin_inset Formula $\left(f/g\right)\left(x\right)=f\left(x\right)/g\left(x\right)$
+\end_inset
+
+.
+ Prištejemo in odštejemo isti izraz (v oglatih oklepajih).
+\begin_inset Formula 
+\[
+\left(f/g\right)'\left(x\right)=\lim_{h\to0}\frac{\left(f/g\right)\left(x+h\right)-\left(f/g\right)\left(x\right)=\frac{f\left(x+h\right)}{g\left(x+h\right)}-\frac{f\left(x\right)}{g\left(x\right)}=\frac{f\left(x+h\right)g\left(x\right)}{g\left(x+h\right)g\left(x\right)}-\frac{f\left(x\right)g\left(x+h\right)}{g\left(x\right)g\left(x+h\right)}=\frac{f\left(x+h\right)g\left(x\right)-f\left(x\right)g\left(x+h\right)}{g\left(x\right)g\left(x+h\right)}}{h}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\lim_{h\to0}\frac{f\left(x+h\right)g\left(x\right)-f\left(x\right)g\left(x+h\right)=f\left(x+h\right)g\left(x\right)-f\left(x\right)g\left(x+h\right)+\left[f\left(x\right)g\left(x\right)-f\left(x\right)g\left(x\right)\right]}{hg\left(x\right)g\left(x+h\right)}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\lim_{h\to0}\left(\frac{f\left(x+h\right)-f\left(x\right)}{h}\cdot\frac{g\left(x\right)}{g\left(x\right)g\left(x+h\right)}-\frac{g\left(x+h\right)-g\left(x\right)}{h}\cdot\frac{f\left(x\right)}{g\left(x\right)g\left(x+h\right)}\right)=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\lim_{h\to0}\left(\left(\frac{1}{g\left(x\right)g\left(x+h\right)}\right)\left(\frac{f\left(x+h\right)-f\left(x\right)}{h}g\left(x\right)-\frac{g\left(x+h\right)-g\left(x\right)}{h}f\left(x\right)\right)\right)=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\frac{1}{g^{2}\left(x\right)}\left(f'\left(x\right)g\left(x\right)-g'\left(x\right)f\left(x\right)\right)=\frac{f'\left(x\right)g\left(x\right)-f\left(x\right)g'\left(x\right)}{g^{2}\left(x\right)}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Example*
+\begin_inset Formula $\tan'\left(x\right)=\left(\frac{\sin\left(x\right)}{\cos\left(x\right)}\right)'=\frac{\sin'\left(x\right)\cos\left(x\right)-\sin\left(x\right)\cos'\left(x\right)}{\cos^{2}\left(x\right)}=\frac{\cos^{2}\left(x\right)+\sin^{2}\left(x\right)}{\cos^{2}\left(x\right)}=\cos^{-2}\left(x\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hypertarget{ok}{Odvod kompozituma}
+\end_layout
+
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $f$
+\end_inset
+
+ odvedljiva v 
+\begin_inset Formula $x$
+\end_inset
+
+ in 
+\begin_inset Formula $g$
+\end_inset
+
+ odvedljiva v 
+\begin_inset Formula $f\left(x\right)$
+\end_inset
+
+.
+ Tedaj je 
+\begin_inset Formula $g\circ f$
+\end_inset
+
+ odvedljiva v 
+\begin_inset Formula $x$
+\end_inset
+
+ in velja 
+\begin_inset Formula $\left(g\circ f\right)\left(x\right)=g'\left(f\left(x\right)\right)\cdot f'\left(x\right)$
+\end_inset
+
+ (opomba:
+ 
+\begin_inset Formula $\left(g\circ f\right)\left(x\right)=g\left(f\left(x\right)\right)$
+\end_inset
+
+).
+\end_layout
+
+\begin_layout Proof
+Označimo 
+\begin_inset Formula $a\coloneqq f\left(x\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $\delta_{h}\coloneqq f\left(x+h\right)-f\left(x\right)$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $f\left(x+h\right)\coloneqq a+\delta\left(h\right)$
+\end_inset
+
+.
+ 
+\begin_inset Formula 
+\[
+\lim_{h\to0}\frac{\left(g\circ f\right)\left(x+h\right)-\left(g\circ f\right)\left(x\right)=g\left(f\left(x+h\right)\right)-g\left(f\left(x\right)\right)=g\left(a+\delta_{h}\right)-g\left(a\right)}{h}=\lim_{h\to0}\frac{g\left(a+\delta_{h}\right)-g\left(a\right)}{\delta_{h}}\cdot\frac{\delta_{h}}{h}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\lim_{h\to0}\frac{g\left(a+\delta_{h}\right)-g\left(a\right)}{\delta_{h}}\cdot\frac{f\left(x+h\right)-f\left(x\right)}{h}=\cdots
+\]
+
+\end_inset
+
+Ker je 
+\begin_inset Formula $f$
+\end_inset
+
+ odvedljiva v 
+\begin_inset Formula $x$
+\end_inset
+
+,
+ je v 
+\begin_inset Formula $x$
+\end_inset
+
+ zvezna,
+ zato sledi 
+\begin_inset Formula $h\to0\Rightarrow\delta_{h}\to0$
+\end_inset
+
+,
+ torej
+\begin_inset Formula 
+\[
+\cdots=g'\left(a\right)\cdot f'\left(x\right)=g'\left(f\left(x\right)\right)\cdot f'\left(x\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $\varphi\left(x\right)=\sin\left(x^{2}\right)=\left(g\circ f\right)\left(x\right),f\left(x\right)=x^{2},g\left(x\right)=\sin x$
+\end_inset
+
+ in velja 
+\begin_inset Formula $\varphi'\left(x\right)=g'\left(f\left(x\right)\right)f'\left(x\right)=\sin'\left(x^{2}\right)\left(x^{2}\right)'=2x\cos\left(x^{2}\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $\psi\left(x\right)=\sin^{2}\left(x\right)=\left(g\circ f\right)\left(x\right),f\left(x\right)=\sin,g\left(x\right)=x^{2}$
+\end_inset
+
+ in velja 
+\begin_inset Formula $\psi'\left(x\right)=g'\left(f\left(x\right)\right)f'\left(x\right)=2\sin x\cos x=\sin2x$
+\end_inset
+
+ (sinus dvojnega kota)
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $\delta'\left(x\right)=\sin\left(e^{x^{2}}\right)=\sin\left(e^{\left(x^{2}\right)}\right)=\left(g\circ h\circ f\right)\left(x\right),g\left(x\right)=\sin x,h\left(x\right)=e^{x},f\left(x\right)=x^{2}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $\delta'\left(x\right)=\cos\left(e^{x^{2}}\right)e^{x^{2}}2x$
+\end_inset
+
+,
+ kajti 
+\begin_inset Formula $\left(e^{x}\right)'=e^{x}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Funkcija 
+\begin_inset Formula $f:I\subseteq\mathbb{R}\to\mathbb{R}$
+\end_inset
+
+ je zvezno odvedljiva na 
+\begin_inset Formula $I$
+\end_inset
+
+,
+ če je na 
+\begin_inset Formula $I$
+\end_inset
+
+ odvedljiva in je 
+\begin_inset Formula $f'$
+\end_inset
+
+ na 
+\begin_inset Formula $I$
+\end_inset
+
+ zvezna.
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $f\left(x\right)=\begin{cases}
+x^{2}\sin\frac{1}{x} & ;x\not=0\\
+0 & ;x=0
+\end{cases}$
+\end_inset
+
+ je na 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+ odvedljiva,
+ a ne zvezno.
+ Odvedljivost na 
+\begin_inset Formula $\mathbb{R}\setminus\left\{ 0\right\} $
+\end_inset
+
+ je očitna,
+ preverimo še odvedljivost v 
+\begin_inset Formula $0$
+\end_inset
+
+:
+\begin_inset Formula 
+\[
+f'\left(0\right)=\lim_{h\to0}\frac{f\left(h\right)-0}{h}=\lim_{h\to0}\frac{h^{\cancel{2}}\sin\frac{1}{h}}{\cancel{h}}=\lim_{h\to0}h\sin\frac{1}{h}=0,
+\]
+
+\end_inset
+
+ker 
+\begin_inset Formula $h$
+\end_inset
+
+ pada k 0,
+ 
+\begin_inset Formula $\sin\frac{1}{h}$
+\end_inset
+
+ pa je omejen z 1.
+ Velja torej
+\begin_inset Formula 
+\[
+f'\left(x\right)=\begin{cases}
+2x\sin\frac{1}{x}-\cos\frac{1}{x} & ;x\not=0\\
+0 & ;x=0
+\end{cases}
+\]
+
+\end_inset
+
+Preverimo nezveznost v 
+\begin_inset Formula $0$
+\end_inset
+
+.
+ Spodnja limita ne obstaja.
+\begin_inset Formula 
+\[
+\lim_{x\to0}\left(\cancel{2x\sin\frac{1}{x}}-\cos\frac{1}{x}\right)=-\lim_{x\to0}\cos\frac{1}{x}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Theorem*
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hypertarget{oi}{Odvod inverza}
+\end_layout
+
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $f$
+\end_inset
+
+ strogo monotona v okolici 
+\begin_inset Formula $a$
+\end_inset
+
+,
+ v 
+\begin_inset Formula $a$
+\end_inset
+
+ odvedljiva in naj bo 
+\begin_inset Formula $f'\left(a\right)\not=0$
+\end_inset
+
+.
+ Tedaj bo inverzna funkcija,
+ definirana v okolici 
+\begin_inset Formula $b=f\left(a\right)$
+\end_inset
+
+ v 
+\begin_inset Formula $b$
+\end_inset
+
+ odvedljiva in veljalo bo 
+\begin_inset Formula $\left(f^{-1}\right)'\left(b\right)=\frac{1}{f'\left(a\right)}=\frac{1}{f'\left(f^{-1}\left(b\right)\right)}.$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Proof
+Ker je 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{zism}{zvezna in strogo monotona}
+\end_layout
+
+\end_inset
+
+ na okolici 
+\begin_inset Formula $a$
+\end_inset
+
+,
+ inverz na okolici 
+\begin_inset Formula $f\left(a\right)$
+\end_inset
+
+ obstaja in velja 
+\begin_inset Formula $f\left(x\right)=s\Leftrightarrow x=f^{-1}\left(x\right)$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $f^{-1}\left(f\left(x\right)\right)=x$
+\end_inset
+
+ za 
+\begin_inset Formula $x$
+\end_inset
+
+ v okolici 
+\begin_inset Formula $a$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Uporabimo formulo za 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{ok}{odvod kompozituma}
+\end_layout
+
+\end_inset
+
+ in velja 
+\begin_inset Formula 
+\[
+\left(f^{-1}\left(f\left(x\right)\right)\right)'=\left(f^{-1}\right)'\left(f\left(x\right)\right)\cdot f'\left(x\right)=\left(x\right)'=1
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\left(f^{-1}\right)'\left(f\left(x\right)\right)=\frac{1}{f'\left(x\right)}
+\]
+
+\end_inset
+
+Vstavimo 
+\begin_inset Formula $x=f^{-1}\left(y\right)$
+\end_inset
+
+ in dobimo za vsak 
+\begin_inset Formula $y$
+\end_inset
+
+ blizu 
+\begin_inset Formula $f\left(a\right)$
+\end_inset
+
+:
+\begin_inset Formula 
+\[
+\left(f^{-1}\right)'\left(y\right)=\frac{1}{f'\left(f^{-1}\left(y\right)\right)}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+Nekaj primerov odvodov inverza.
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $g\left(x\right)=\sqrt[n]{x}=x^{\frac{1}{n}}$
+\end_inset
+
+ za 
+\begin_inset Formula $n\in\mathbb{N},x>0$
+\end_inset
+
+.
+ Velja 
+\begin_inset Formula $g=f^{-1}$
+\end_inset
+
+ za 
+\begin_inset Formula $f\left(x\right)=x^{n}$
+\end_inset
+
+.
+ Uporabimo formulo za 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{op}{odvod potence}
+\end_layout
+
+\end_inset
+
+ in zgornji izrek.
+ Velja 
+\begin_inset Formula $f'\left(x\right)=nx^{n-1}$
+\end_inset
+
+ in 
+\begin_inset Formula $f^{-1}=\sqrt[n]{x}$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+g'\left(x\right)=\left(f^{-1}\right)'\left(x\right)=\frac{1}{f'\left(f^{-1}\left(x\right)\right)}=\frac{1}{f'\left(\sqrt[n]{x}\right)}=\frac{1}{n\sqrt[n]{x}^{n-1}}=\frac{1}{nx^{\frac{n-1}{n}=1-\frac{1}{n}}}=\frac{1}{n}x^{\frac{1}{n}-1}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $h\left(x\right)=\sqrt[n]{x^{m}}=x^{\frac{m}{n}}=g\left(x\right)^{m}$
+\end_inset
+
+ za 
+\begin_inset Formula $n,m\in\mathbb{N},x>0$
+\end_inset
+
+.
+ Uporabimo formulo za 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{op}{odvod potence}
+\end_layout
+
+\end_inset
+
+ in 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{ok}{kompozituma}
+\end_layout
+
+\end_inset
+
+ in zgornji primer.
+ Velja 
+\begin_inset Formula $g'\left(x\right)=\frac{1}{n}x^{\frac{1}{n}-1}$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula 
+\[
+h'\left(x\right)=mg\left(x\right)^{m-1}\cdot g'\left(x\right)=m\left(x^{\frac{1}{n}}\right)^{m-1}\cdot\frac{1}{n}x^{\frac{1}{n}-1}=\frac{m}{n}x^{\frac{m-1}{n}+\frac{1}{n}-1}=\frac{m}{n}x^{\frac{m}{n}-1}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset CommandInset label
+LatexCommand label
+name "enu:Izkaže-se,-da"
+
+\end_inset
+
+Izkaže se,
+ da velja celo 
+\begin_inset Formula $\forall x>0,\alpha\in\mathbb{R}:\left(x^{\alpha}\right)'=\alpha x^{\alpha-1}$
+\end_inset
+
+.
+ Mi smo dokazali le za 
+\begin_inset Formula $\alpha\in\mathbb{Q}$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Logaritmi,
+ inverz 
+\begin_inset Formula $e^{x}$
+\end_inset
+
+.
+ Gre za 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{oef}{odvod eksponentne funkcije}
+\end_layout
+
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\left(a^{x}\right)=a^{x}\ln a$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $\left(e^{x}\right)=e^{x}\ln e=e^{x}$
+\end_inset
+
+.
+ Uporavimo 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{oi}{odvod inverza}
+\end_layout
+
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\left(f^{-1}\right)'\left(x\right)=\frac{1}{f'\left(f^{-1}\left(x\right)\right)}$
+\end_inset
+
+ in za 
+\begin_inset Formula $g\left(x\right)=\log x$
+\end_inset
+
+ uporabimo 
+\begin_inset Formula $g\left(x\right)=f^{-1}\left(x\right)$
+\end_inset
+
+,
+ kjer je 
+\begin_inset Formula $f\left(x\right)=e^{x}$
+\end_inset
+
+:
+\begin_inset Formula 
+\[
+\log'\left(x\right)=\left(\left(e^{x}\right)^{-1}\right)'\left(x\right)=\frac{1}{e^{\log x}}=\frac{1}{x}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $g\left(x\right)=\arcsin x$
+\end_inset
+
+ za 
+\begin_inset Formula $x\in\left[-1,1\right]$
+\end_inset
+
+,
+ torej je 
+\begin_inset Formula $g=f^{-1}$
+\end_inset
+
+,
+ kjer je 
+\begin_inset Formula $f=\sin$
+\end_inset
+
+ za 
+\begin_inset Formula $x\in\left[\frac{-\pi}{2},\frac{\pi}{2}\right]$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+g'\left(f\left(x\right)\right)=\frac{1}{f'\left(x\right)}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+g'\left(\sin x\right)=\frac{1}{\cos x}
+\]
+
+\end_inset
+
+Ker velja 
+\begin_inset Formula $\sin^{2}x+\cos^{2}x=1$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $\cos^{2}x=1-\sin^{2}x$
+\end_inset
+
+,
+ sledi 
+\begin_inset Formula $\cos x=\sqrt{1-\sin^{2}x}$
+\end_inset
+
+,
+ torej nadaljujemo:
+\begin_inset Formula 
+\[
+g'\left(\sin x\right)=\frac{1}{\sqrt{1-\sin^{2}x}}
+\]
+
+\end_inset
+
+Sedaj zamenjamo 
+\begin_inset Formula $\sin x$
+\end_inset
+
+ s 
+\begin_inset Formula $t$
+\end_inset
+
+ in dobimo:
+\begin_inset Formula 
+\[
+g'\left(t\right)=\frac{1}{\sqrt{1-t^{2}}}=\arcsin^{2}t
+\]
+
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Subsection
+Diferencial
+\end_layout
+
+\begin_layout Standard
+Fiksirajmo funkcijo 
+\begin_inset Formula $f$
+\end_inset
+
+ in točko 
+\begin_inset Formula $a\in\mathbb{R}$
+\end_inset
+
+,
+ v okolici katere je 
+\begin_inset Formula $f$
+\end_inset
+
+ definirana.
+ Želimo oceniti vrednost funkcije 
+\begin_inset Formula $f$
+\end_inset
+
+ v bližini točke 
+\begin_inset Formula $a$
+\end_inset
+
+ z linearno funkcijo – to je 
+\begin_inset Formula $y\left(x\right)=\lambda x$
+\end_inset
+
+ za neki 
+\begin_inset Formula $\lambda\in\mathbb{R}$
+\end_inset
+
+.
+ ZDB Iščemo najboljši linearni približek,
+ odvisen od 
+\begin_inset Formula $h$
+\end_inset
+
+,
+ za 
+\begin_inset Formula $f\left(a+h\right)-f\left(a\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $f$
+\end_inset
+
+ definirana v okolici točke 
+\begin_inset Formula $a\in\mathbb{R}$
+\end_inset
+
+.
+ Diferencial funkcije 
+\begin_inset Formula $f$
+\end_inset
+
+ v točki 
+\begin_inset Formula $a$
+\end_inset
+
+ je linearna preslikava 
+\begin_inset Formula $df\left(a\right):\mathbb{R}\to\mathbb{R}$
+\end_inset
+
+ z zahtevo
+\begin_inset Formula 
+\[
+\lim_{h\to0}\frac{\left|f\left(a+h\right)-f\left(a\right)-df\left(a\right)\left(h\right)\right|}{\left|h\right|}=0.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Note*
+\begin_inset Formula 
+\[
+\lim_{h\to0}\frac{f\left(a+h\right)-f\left(a\right)-df\left(a\right)\left(h\right)}{h}=0=\lim_{h\to0}\left(\frac{f\left(a+h\right)-f\left(a\right)}{h}-\frac{\left(df\left(a\right)\right)\left(h\right)}{h}\right)=
+\]
+
+\end_inset
+
+Upoštevamo linearnost preslikave
+\begin_inset Formula 
+\[
+=\lim_{h\to0}\frac{f\left(a+h\right)-f\left(a\right)}{h}-df\left(a\right)=f'\left(a\right)-df\left(a\right)=0
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+f'\left(a\right)=df\left(a\right)
+\]
+
+\end_inset
+
+Torej 
+\begin_inset Formula $f\left(a+h\right)-f\left(a\right)\approx df\left(a\right)\left(h\right)$
+\end_inset
+
+ – najboljši linearni približek za 
+\begin_inset Formula $f\left(a+h\right)-f\left(h\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Uporaba diferenciala.
+ 
+\begin_inset Formula $a$
+\end_inset
+
+ je točka,
+ v kateri znamo izračunati funkcijsko vrednost,
+ 
+\begin_inset Formula $a+h$
+\end_inset
+
+ pa je točka,
+ v kateri želimo približek funkcijske vrednosti.
+ Izračunajmo približek 
+\begin_inset Formula $\sqrt{2}$
+\end_inset
+
+:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $f\left(x\right)=\sqrt{x}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $a+h=2$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $a=2,25$
+\end_inset
+
+,
+ 
+\begin_inset Formula $h=-0,25$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $f\left(a\right)=\sqrt{a}=1,5$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $f'\left(s\right)=\frac{1}{2\sqrt{x}}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $f\left(a=2,25\right)=\frac{1}{3}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $f\left(2\right)\approx f\left(a\right)+f'\left(2,25\right)\cdot h=1,5-0,25\cdot\frac{1}{3}=\frac{3}{2}-\frac{1}{4}\cdot\frac{1}{3}=\frac{17}{12}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Preizkus:
+ 
+\begin_inset Formula $\left(\frac{17}{12}\right)^{2}=\frac{289}{144}=2+\frac{1}{144}$
+\end_inset
+
+ ...
+ Absolutna napaka 
+\begin_inset Formula $\frac{1}{144}$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $I\subseteq\mathbb{R}$
+\end_inset
+
+ interval in 
+\begin_inset Formula $f:I\to\mathbb{R}$
+\end_inset
+
+ odvedljiva povsod na 
+\begin_inset Formula $I$
+\end_inset
+
+.
+ Vzemimo 
+\begin_inset Formula $a\in I$
+\end_inset
+
+.
+ Če je v 
+\begin_inset Formula $a$
+\end_inset
+
+ odvedljiva tudi 
+\begin_inset Formula $f'$
+\end_inset
+
+,
+ pišemo 
+\begin_inset Formula $f''\left(a\right)=\left(f'\left(a\right)\right)'$
+\end_inset
+
+.
+ Podobno pišemo tudi višje odvode:
+ 
+\begin_inset Formula $f^{\left(1\right)}\left(a\right)=f'\left(a\right)$
+\end_inset
+
+,
+ 
+\begin_inset Formula $f^{\left(n+1\right)}=\left(f^{\left(n\right)}\right)'$
+\end_inset
+
+,
+ 
+\begin_inset Formula $f^{\left(0\right)}\left(a\right)=f\left(a\right)$
+\end_inset
+
+,
+ 
+\begin_inset Formula $f^{\left(2\right)}\left(a\right)=f''\left(a\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Note*
+Pomen besede 
+\begin_inset Quotes gld
+\end_inset
+
+odvod
+\begin_inset Quotes grd
+\end_inset
+
+:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Odvod v dani točki:
+ 
+\begin_inset Formula $f'\left(a\right)$
+\end_inset
+
+ za fiksen 
+\begin_inset Formula $a\in\mathbb{R}$
+\end_inset
+
+ ali
+\end_layout
+
+\begin_layout Itemize
+Funkcija,
+ ki vsaki točki 
+\begin_inset Formula $x\in\mathbb{R}$
+\end_inset
+
+ priredi 
+\begin_inset Formula $f'\left(x\right)$
+\end_inset
+
+ po zgornji definiciji.
+\end_layout
+
+\end_deeper
+\begin_layout Definition*
+\begin_inset Formula $C^{n}\left(I\right)$
+\end_inset
+
+ je množica funkcije 
+\begin_inset Formula $f:I\to\mathbb{R}$
+\end_inset
+
+,
+ da 
+\begin_inset Formula $\forall x\in I\exists f'\left(x\right),f''\left(x\right),f^{\left(3\right)},\dots,f^{\left(n\right)}\left(x\right)$
+\end_inset
+
+ in da so 
+\begin_inset Formula $f,f',f'',f^{\left(3\right)},\dots,f^{\left(n\right)}$
+\end_inset
+
+ zvezna funkcije na 
+\begin_inset Formula $I$
+\end_inset
+
+.
+ (seveda če obstaja 
+\begin_inset Formula $j-$
+\end_inset
+
+ti odvod,
+ obstaja tudi zvezen 
+\begin_inset Formula $j-1-$
+\end_inset
+
+ti odvod).
+ ZDB je to množica funkcij,
+ ki imajo vse odvode do 
+\begin_inset Formula $n$
+\end_inset
+
+ in so le-ti zvezni.
+ ZDB to so vse 
+\begin_inset Formula $n-$
+\end_inset
+
+krat zvezno odvedljive funkcije na intervalu 
+\begin_inset Formula $I$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Označimo 
+\begin_inset Formula $C^{\infty}\left(I\right)\coloneqq\bigcap_{n=1}^{\infty}C^{n}\left(I\right)$
+\end_inset
+
+ – to so neskončnokrat odvedljive funkcije na intervalu 
+\begin_inset Formula $I$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Note*
+Intuitivno
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+Baje.
+ Jaz sem itak do vsega skeptičen.
+\end_layout
+
+\end_inset
+
+ velja 
+\begin_inset Formula $C^{1}\left(I\right)\supset C^{2}\left(I\right)\supset C^{3}\left(I\right)\supset C^{4}\left(I\right)\supset\cdots$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Nekaj primerov.
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Polimomi 
+\begin_inset Formula $\subset C^{\infty}\left(\mathbb{R}\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $f\left(x\right)=\left|x\right|^{3}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $f'\left(x\right)=\begin{cases}
+3x^{2} & ;x\geq0\\
+-3x^{2} & ;x<0
+\end{cases}=3x^{2}\sgn x$
+\end_inset
+
+,
+ 
+\begin_inset Formula $f''\left(x\right)=\begin{cases}
+6x & ;x\geq0\\
+-6x & ;x<0
+\end{cases}=6x\sgn x$
+\end_inset
+
+,
+ 
+\begin_inset Formula $f'''\left(x\right)=\begin{cases}
+6 & ;x>0\\
+-6 & ;x<0
+\end{cases}=6\sgn x$
+\end_inset
+
+ in v 
+\begin_inset Formula $0$
+\end_inset
+
+ ni odvedljiva,
+ zato 
+\begin_inset Formula $f\in C^{2}\left(\mathbb{R}\right)$
+\end_inset
+
+ a 
+\begin_inset Formula $f\not\in C^{3}\left(\mathbb{R}\right)$
+\end_inset
+
+,
+ ker 
+\begin_inset Formula $\exists f''$
+\end_inset
+
+ in je zvezna na 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+,
+ a 
+\begin_inset Formula $f'''$
+\end_inset
+
+ sicer obstaja,
+ a ni zvezna na 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+.
+ Velja pa 
+\begin_inset Formula $f\in C^{\infty}\left(\mathbb{R}\setminus\left\{ 0\right\} \right)$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Theorem*
+Rolle.
+ Naj bo 
+\begin_inset Formula $f:\left[a,b\right]\to\mathbb{R}$
+\end_inset
+
+ za 
+\begin_inset Formula $a,b\in\mathbb{R}$
+\end_inset
+
+ zvezna na 
+\begin_inset Formula $\left[a,b\right]$
+\end_inset
+
+ in odvedljiva na 
+\begin_inset Formula $\left(a,b\right)$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+f\left(a\right)=f\left(b\right)\Longrightarrow\exists\alpha\in\left(a,b\right)\ni:f'\left(\alpha\right)=0
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Proof
+Sumimo,
+ da je ustrezna 
+\begin_inset Formula $\alpha$
+\end_inset
+
+ tista,
+ ki je 
+\begin_inset Formula $\max$
+\end_inset
+
+ ali 
+\begin_inset Formula $\min$
+\end_inset
+
+ od 
+\begin_inset Formula $f$
+\end_inset
+
+ na 
+\begin_inset Formula $I$
+\end_inset
+
+.
+ 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{zfnkm}{Ker}
+\end_layout
+
+\end_inset
+
+ je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna na 
+\begin_inset Formula $\left[a,b\right]$
+\end_inset
+
+ (kompaktni množici),
+ 
+\begin_inset Formula $\exists\alpha_{1}\in\left[a,b\right],\alpha_{2}\in\left[a,b\right]\ni:f\left(\alpha_{1}\right)=\max f\left(\left[a,b\right]\right)\wedge f\left(\alpha_{2}\right)=\min f\left(\left[a,b\right]\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Če je 
+\begin_inset Formula $\left\{ \alpha_{1},\alpha_{2}\right\} \subseteq\left\{ a,b\right\} $
+\end_inset
+
+,
+ je 
+\begin_inset Formula $f\left(\alpha_{1}\right)=f\left(\alpha_{2}\right)$
+\end_inset
+
+ in je v tem primeru 
+\begin_inset Formula $f$
+\end_inset
+
+ konstanta (
+\begin_inset Formula $\exists!c\in\mathbb{R}\ni:f\left(x\right)=c$
+\end_inset
+
+),
+ ki je odvedljiva in ima povsod odvod nič.
+\end_layout
+
+\begin_layout Proof
+Sicer pa 
+\begin_inset Formula $\left\{ \alpha_{1},\alpha_{2}\right\} \not\subseteq\left\{ a,b\right\} $
+\end_inset
+
+.
+ Tedaj ločimo dva primera:
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\alpha_{1}\in\left(a,b\right)$
+\end_inset
+
+ To pomeni,
+ da je globalni maksimum na odprtem intervalu.
+ Trdimo,
+ da je v lokalnem maksimumu odvod 0.
+ Dokaz:
+\begin_inset Formula 
+\[
+f'\left(\alpha_{1}\right)=\lim_{h\to0}\frac{f\left(\alpha_{1}+h\right)-f\left(\alpha_{1}\right)}{h}
+\]
+
+\end_inset
+
+Za 
+\begin_inset Formula $a_{1}$
+\end_inset
+
+ (maksimum) velja 
+\begin_inset Formula $f\left(\alpha_{1}+h\right)-f\left(\alpha_{1}\right)\leq0$
+\end_inset
+
+ (čim se pomaknemo izven točke,
+ v kateri je maksimum,
+ je funkcijska vrednost nižja).
+ Potemtakem velja
+\begin_inset Formula 
+\[
+\frac{f\left(\alpha_{1}+h\right)-f\left(\alpha_{1}\right)}{h}\quad\begin{cases}
+\leq0 & ;h>0\\
+\geq0 & ;h<0
+\end{cases}
+\]
+
+\end_inset
+
+Ker je funkcija odvedljiva na odprtem intervalu,
+ sta leva in desna limita enaki.
+\begin_inset Formula 
+\[
+0\geq\lim_{h\searrow0}\frac{f\left(\alpha_{1}+h\right)-f\left(\alpha_{1}\right)}{h}=\lim_{h\nearrow0}\frac{f\left(\alpha_{1}+h\right)-f\left(\alpha_{1}\right)}{h}\geq0
+\]
+
+\end_inset
+
+Sledi
+\begin_inset Formula 
+\[
+\lim_{h\to0}\frac{f\left(\alpha_{1}+h\right)-f\left(\alpha_{1}\right)}{h}=f'\left(x\right)=0
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\alpha_{2}\in\left(a,b\right)$
+\end_inset
+
+ To pomeni,
+ da je globalni minimum na odprtem intervalu.
+ Trdimo,
+ da je v lokalnem minimumu odvod 0.
+ Dokaz je podoben tistemu za lokalni maksimum.
+\end_layout
+
+\begin_layout Theorem*
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hypertarget{lagrange}{Lagrange}
+\end_layout
+
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $f:\left[a,b\right]\to\mathbb{R}$
+\end_inset
+
+ za 
+\begin_inset Formula $a,b\in\mathbb{R}$
+\end_inset
+
+ zvezna na 
+\begin_inset Formula $\left[a,b\right]$
+\end_inset
+
+ in odvedljiva na 
+\begin_inset Formula $\left(a,b\right)$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+\exists\alpha\in\left(a,b\right)\ni:f\left(b\right)-f\left(a\right)=f'\left(\alpha\right)\left(b-a\right)\sim\frac{f\left(b\right)-f\left(a\right)}{b-a}=f'\left(\alpha\right)
+\]
+
+\end_inset
+
+ZDB na neki točki na grafu funkcije je tangenta na graf funkcije vzporedna premici,
+ ki jo določata točki 
+\begin_inset Formula $\left(a,f\left(a\right)\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $\left(b,f\left(b\right)\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Za dokaz Lagrangevega uporabimo Rolleov izrek.
+ Splošen primer prevedemo na primer 
+\begin_inset Formula $h\left(a\right)=h\left(b\right)$
+\end_inset
+
+ tako,
+ da od naše splošne funkcije 
+\begin_inset Formula $f$
+\end_inset
+
+ odštejemo linearno funkcijo 
+\begin_inset Formula $g$
+\end_inset
+
+,
+ da bo veljalo 
+\begin_inset Formula $\left(f-g\right)\left(a\right)=\left(f-g\right)\left(b\right)$
+\end_inset
+
+.
+ Za funkcijo 
+\begin_inset Formula $g\left(x\right)$
+\end_inset
+
+ mora veljati naslednje:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $\exists k,n\in\mathbb{R}\ni:f\left(x\right)=kx+n$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $g\left(a\right)=0$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $g\left(b\right)=f\left(b\right)-f\left(a\right)$
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Proof
+Opazimo,
+ da mora biti koeficient funkcije 
+\begin_inset Formula $g$
+\end_inset
+
+ enak 
+\begin_inset Formula $\frac{f\left(b\right)-f\left(a\right)}{b-a}$
+\end_inset
+
+,
+ vertikalni odklon pa tolikšen,
+ da ima funkcija 
+\begin_inset Formula $g$
+\end_inset
+
+ v 
+\begin_inset Formula $a$
+\end_inset
+
+ ničlo:
+\begin_inset Formula 
+\[
+\frac{f\left(b\right)-f\left(a\right)}{b-a}a+n=0
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+n=-\frac{f\left(b\right)-f\left(a\right)}{b-a}a
+\]
+
+\end_inset
+
+Našli smo funkcijo 
+\begin_inset Formula $g\left(x\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}\left(x-a\right)$
+\end_inset
+
+.
+ Funkcija 
+\begin_inset Formula $\left(f-g\right)$
+\end_inset
+
+ sedaj ustreza pogojem za Rolleov izrek,
+ torej 
+\begin_inset Formula $\exists\alpha\in\left[a,b\right]\ni:\left(f-g\right)'\left(\alpha\right)=0\Leftrightarrow g'\left(\alpha\right)=f'\left(\alpha\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$
+\end_inset
+
+,
+ kar smo želeli dokazati.
+\end_layout
+
+\begin_layout Corollary*
+Naj bo 
+\begin_inset Formula $I\subseteq\mathbb{R}$
+\end_inset
+
+ nenujno zaprt niti omejen in 
+\begin_inset Formula $f:I\to\mathbb{R}$
+\end_inset
+
+ odvedljiva na 
+\begin_inset Formula $I$
+\end_inset
+
+.
+ Tedaj je 
+\begin_inset Formula $f$
+\end_inset
+
+ Lipschitzova.
+ Lipschitzove funkcije so enakomerno zvezne.
+\end_layout
+
+\begin_layout Proof
+Po Lagrangeu velja 
+\begin_inset Formula $\forall x,y\in I\exists\alpha\in\left(x,y\right)\ni:f\left(x\right)-f\left(y\right)=f'\left(\alpha\right)\left(x-y\right)$
+\end_inset
+
+.
+ Potemtakem 
+\begin_inset Formula $\left|f\left(x\right)-f\left(y\right)\right|=\left|f'\left(\alpha\right)\right|\left|x-y\right|\leq\sup_{\beta\in\left(x,y\right)}\left|f'\left(\beta\right)\right|\left|x-y\right|$
+\end_inset
+
+.
+ Torej 
+\begin_inset Formula $\exists M>0\forall x,y:\left|f\left(x\right)-f\left(y\right)\right|\leq M\left|x-y\right|$
+\end_inset
+
+,
+ enakomerno zveznost pa dobimo tako,
+ da 
+\begin_inset Formula $\delta\left(\varepsilon\right)=\frac{\varepsilon}{\sup_{\beta\in I}\left|f'\left(\beta\right)\right|}$
+\end_inset
+
+.
+ Računajmo.
+ Naj bo 
+\begin_inset Formula $M=\sup_{\beta\in I}\left|f'\left(\beta\right)\right|$
+\end_inset
+
+,
+ ki obstaja.
+\begin_inset Formula 
+\[
+\forall x,y:\left|f\left(x\right)-f\left(y\right)\right|\leq M\left|x-y\right|
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\forall x,y:\left|x-y\right|<\frac{\varepsilon}{\sup_{\beta\in I}\left|f'\left(\beta\right)\right|}\Rightarrow\left|f\left(x\right)-f\left(y\right)\right|\leq M\left|x-y\right|<\cancel{\sup_{\beta\in I}\left|f'\left(\beta\right)\right|}\frac{\varepsilon}{\cancel{\sup_{\beta\in I}\left|f'\left(\beta\right)\right|}}<\varepsilon
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\forall\varepsilon\exists\delta\left(\varepsilon\right)\forall x,y:\left|x-y\right|<\delta\left(\varepsilon\right)\Rightarrow\left|f\left(x\right)-f\left(y\right)\right|<\varepsilon
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Note*
+Lipschnitzovim funkcijam pravimo tudi Hölderjeve funkcije reda 1.
+ 
+\begin_inset Formula $f$
+\end_inset
+
+ je Hölderjeva funkcija reda 
+\begin_inset Formula $r$
+\end_inset
+
+,
+ če velja 
+\begin_inset Formula $\exists M>0\forall x,y\in I:\left|f\left(x\right)-f\left(y\right)\right|\leq M\left|x-y\right|^{r}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+Naj bo 
+\begin_inset Formula $I$
+\end_inset
+
+ odprti interval,
+ 
+\begin_inset Formula $f:I\to\mathbb{R}$
+\end_inset
+
+ odvedljiva.
+ Tedaj:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $f$
+\end_inset
+
+ narašča na 
+\begin_inset Formula $I\Leftrightarrow f'\geq0$
+\end_inset
+
+ na 
+\begin_inset Formula $I$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $f$
+\end_inset
+
+ pada na 
+\begin_inset Formula $I\Leftrightarrow f'\leq0$
+\end_inset
+
+ na 
+\begin_inset Formula $I$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $f$
+\end_inset
+
+ strogo narašča na 
+\begin_inset Formula $I\Leftarrow f'>0$
+\end_inset
+
+ na 
+\begin_inset Formula $I$
+\end_inset
+
+.
+ Protiprimer,
+ da ni 
+\begin_inset Formula $\Leftrightarrow:f\left(x\right)=x^{3}$
+\end_inset
+
+,
+ ki strogo narašča,
+ toda 
+\begin_inset Formula $f'\left(0\right)=0$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $f$
+\end_inset
+
+ strogo pada na 
+\begin_inset Formula $I\Leftarrow f'<0$
+\end_inset
+
+ na 
+\begin_inset Formula $I$
+\end_inset
+
+.
+ Protiprimer,
+ da ni 
+\begin_inset Formula $\Leftrightarrow:f\left(x\right)=-x^{3}$
+\end_inset
+
+,
+ ki strogo pada,
+ toda 
+\begin_inset Formula $f'\left(0\right)=0$
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Proof
+Dokažimo le 
+\begin_inset Formula $f$
+\end_inset
+
+ narašča na 
+\begin_inset Formula $I\Leftrightarrow f'\geq0$
+\end_inset
+
+ na 
+\begin_inset Formula $I$
+\end_inset
+
+.
+ Drugo točko dokažemo podobno.
+ Dokazujemo ekvivalenco:
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Leftarrow\right)$
+\end_inset
+
+ 
+\begin_inset Formula $f'\geq0\Rightarrow f$
+\end_inset
+
+ narašča.
+ Vzemimo poljubna 
+\begin_inset Formula $t_{1}<t_{2}\in I$
+\end_inset
+
+.
+ Po Lagrangeu 
+\begin_inset Formula $\exists\alpha\in\left(t_{1},t_{2}\right)\ni:f\left(t_{2}\right)-f\left(t_{1}\right)=f'\left(\alpha\right)\left(t_{2}-t_{1}\right)$
+\end_inset
+
+.
+ Ker je po predpostavki 
+\begin_inset Formula $f'\left(\alpha\right)\geq0$
+\end_inset
+
+ in 
+\begin_inset Formula $t_{2}-t_{1}>0$
+\end_inset
+
+,
+ je tudi 
+\begin_inset Formula $f\left(t_{2}\right)-f\left(t_{1}\right)\geq0$
+\end_inset
+
+ in zato 
+\begin_inset Formula $f\left(t_{2}\right)\geq f\left(t_{1}\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Rightarrow\right)$
+\end_inset
+
+ 
+\begin_inset Formula $f$
+\end_inset
+
+ narašča 
+\begin_inset Formula $\Rightarrow f'\geq0$
+\end_inset
+
+.
+ Velja 
+\begin_inset Formula $f'\left(x\right)=\lim_{h\to0}\frac{f\left(x+h\right)-f\left(x\right)}{h}$
+\end_inset
+
+.
+ Po predpostavki je 
+\begin_inset Formula $f\left(x+h\right)-f\left(x\right)\geq0$
+\end_inset
+
+,
+ čim je 
+\begin_inset Formula $h>0$
+\end_inset
+
+,
+ in 
+\begin_inset Formula $f\left(x+h\right)-f\left(x\right)\leq0$
+\end_inset
+
+,
+ čim je 
+\begin_inset Formula $h<0$
+\end_inset
+
+.
+ Torej je ulomek vedno nenegativen.
+\end_layout
+
+\end_deeper
+\begin_layout Subsection
+Konveksnost in konkavnost
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $I\subseteq\mathbb{R}$
+\end_inset
+
+ interval in 
+\begin_inset Formula $f:I\to\mathbb{R}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $f$
+\end_inset
+
+ je konveksna na 
+\begin_inset Formula $I$
+\end_inset
+
+,
+ če 
+\begin_inset Formula $\forall a,b\in I$
+\end_inset
+
+ daljica 
+\begin_inset Formula $\left(a,f\left(a\right)\right),\left(b,f\left(b\right)\right)$
+\end_inset
+
+ leži nad grafom 
+\begin_inset Formula $f$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Enačba premice,
+ ki vsebuje to daljico,
+ se glasi (razmislek je podoben kot pri 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{lagrange}{Lagrangevem izreku}
+\end_layout
+
+\end_inset
+
+)
+\begin_inset Formula 
+\[
+g\left(x\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}\left(x-a\right)+f\left(a\right)
+\]
+
+\end_inset
+
+Za konveksno funkcijo torej velja 
+\begin_inset Formula $\forall a,b\in I:\forall x\in\left(a,b\right):f\left(x\right)\leq\frac{f\left(b\right)-f\left(a\right)}{b-a}\left(x-a\right)+f\left(a\right)$
+\end_inset
+
+ oziroma
+\begin_inset Formula 
+\[
+\frac{f\left(x\right)-f\left(a\right)}{x-a}\leq\frac{f\left(b\right)-f\left(a\right)}{b-a}
+\]
+
+\end_inset
+
+Vsak 
+\begin_inset Formula $x$
+\end_inset
+
+ na intervalu lahko zapišemo kot 
+\begin_inset Formula $x=a+t\left(b-a\right)$
+\end_inset
+
+ za nek 
+\begin_inset Formula $t\in\left(0,1\right)$
+\end_inset
+
+.
+ Tedaj je 
+\begin_inset Formula $x-a=t\left(b-a\right)$
+\end_inset
+
+ in konveksnost se glasi
+\begin_inset Formula 
+\[
+\forall a,b\in I:\forall t\in\left(0,1\right):f\left(a+t\left(b-a\right)\right)\leq\frac{f\left(b\right)-f\left(a\right)}{\cancel{b-a}}t\cancel{\left(b-a\right)}+f\left(a\right)
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+f\left(a+t\left(b-a\right)\right)=f\left(a+tb-ta\right)=f\left(\left(1-t\right)a+tb\right)\leq tf\left(b\right)-tf\left(a\right)+f\left(a\right)=\left(1-t\right)f\left(a\right)+tf\left(b\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Konveksna kombinacija izrazov 
+\begin_inset Formula $a,b$
+\end_inset
+
+ je izraz oblike 
+\begin_inset Formula $\left(1-t\right)a+tb$
+\end_inset
+
+ za 
+\begin_inset Formula $t\in\left(0,1\right)$
+\end_inset
+
+.
+ Potemtakem je ZDB definicija konveksnosti 
+\begin_inset Formula $\forall a,b\in I:$
+\end_inset
+
+ funkcijska vrednost konveksne kombinacije 
+\begin_inset Formula $a,b$
+\end_inset
+
+ je kvečjemu konveksna kombinacija funkcijskih vrednosti 
+\begin_inset Formula $a,b$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Konkavnost pa je definirana tako,
+ da povsod obrnemo predznake,
+ torej daljica leži pod grafom 
+\begin_inset Formula $f$
+\end_inset
+
+ ZDB 
+\begin_inset Formula $\forall a,b\in I:f\left(\left(1-t\right)a+tb\right)\geq\left(1-t\right)f\left(a\right)+tf\left(b\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $f\left(x\right)=\sin x$
+\end_inset
+
+,
+ 
+\begin_inset Formula $I=\left[-\pi,0\right]$
+\end_inset
+
+.
+ Je konveksna.
+ Se vidi iz grafa.
+ Preveriti analitično bi bilo težko.
+\end_layout
+
+\begin_layout Example*
+Formulirajmo drugačen pogoj za konveksnost.
+ Naj bo spet 
+\begin_inset Formula $f:I\to\mathbb{R}$
+\end_inset
+
+,
+ kjer je 
+\begin_inset Formula $I$
+\end_inset
+
+ interval.
+ 
+\begin_inset Formula $f$
+\end_inset
+
+ je konveksna 
+\begin_inset Formula 
+\[
+\Leftrightarrow\forall a,b\in I\forall x\in\left(a,b\right):\frac{f\left(x\right)-f\left(a\right)}{x-a}\leq\frac{f\left(b\right)-f\left(a\right)}{b-a}.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+Sedaj glejmo le poljuben 
+\begin_inset Formula $a$
+\end_inset
+
+.
+ Po prejšnjem pogoju moramo gledati še vse poljubne 
+\begin_inset Formula $b$
+\end_inset
+
+,
+ večje od 
+\begin_inset Formula $a$
+\end_inset
+
+ (ker le tako lahko konstruiramo interval).
+ Za 
+\begin_inset Formula $b$
+\end_inset
+
+ in 
+\begin_inset Formula $a$
+\end_inset
+
+ mora biti diferenčni kvocient večji od diferenčnega kvocienta 
+\begin_inset Formula $x$
+\end_inset
+
+ in 
+\begin_inset Formula $a$
+\end_inset
+
+ za poljuben 
+\begin_inset Formula $x$
+\end_inset
+
+.
+ Ta pogoj pa je ekvivalenten temu,
+ da diferenčni kvocient 
+\begin_inset Formula $x$
+\end_inset
+
+ in 
+\begin_inset Formula $a$
+\end_inset
+
+ s fiksnim 
+\begin_inset Formula $a$
+\end_inset
+
+ in čedalje večjim 
+\begin_inset Formula $x$
+\end_inset
+
+ narašča,
+ torej je pogoj za konveksnost tudi:
+\begin_inset Formula 
+\[
+\forall a\in I\forall x>a:g_{a}\left(x\right)=\frac{f\left(x\right)-f\left(a\right)}{x-a}\text{ je naraščajoča funkcija}.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Corollary*
+Naj bo 
+\begin_inset Formula $f$
+\end_inset
+
+ konveksna na odprtem intervalu 
+\begin_inset Formula $I$
+\end_inset
+
+.
+ 
+\begin_inset Formula $\forall a\in I$
+\end_inset
+
+ obstajata funkciji
+\begin_inset Formula 
+\[
+\left(D_{+}f\right)\left(a\right)=\lim_{x\searrow a}g_{a}\left(x\right)=\inf_{x\in I,x>a}g_{a}\left(x\right)\text{ (desni odvod \ensuremath{f} v \ensuremath{a})}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\left(D_{-}f\right)\left(a\right)=\lim_{x\nearrow a}g_{a}\left(x\right)=\sup_{x\in I,x<a}g_{a}\left(x\right)\text{ (levi odvod \ensuremath{f} v \ensuremath{a})}
+\]
+
+\end_inset
+
+in obe sta naraščajoči na 
+\begin_inset Formula $I$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Obstoj sledi iz monotonosti 
+\begin_inset Formula $g_{a}\left(a\right)$
+\end_inset
+
+,
+ kajti 
+\begin_inset Formula $\lim_{x\searrow a}g_{a}\left(x\right)=\lim_{x\searrow a}\frac{f\left(x\right)-f\left(a\right)}{x-a}$
+\end_inset
+
+ in enako za levo limito.
+ Diferenčni kvocient mora namreč biti naraščajoč.
+ S tem smo dokazali,
+ da je vsaka konveksna funkcija zvezna
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+Ni pa vsaka konveksna funkcija odvedljiva,
+ protiprimer je 
+\begin_inset Formula $f\left(x\right)=\left|x\right|$
+\end_inset
+
+.
+\end_layout
+
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Naj bodo 
+\begin_inset Formula $x_{1},x_{2},x\in I\ni:x_{1}<x_{2}<x$
+\end_inset
+
+.
+ Pomagaj si s skico
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+TODO DORIŠI SKICO ZVZ VII/ANA1UČ/str.
+ 13
+\end_layout
+
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $f$
+\end_inset
+
+ konveksna,
+ sledi 
+\begin_inset Formula $g_{x}\left(x_{1}\right)\leq g_{x}\left(x_{2}\right)$
+\end_inset
+
+.
+ Ker 
+\begin_inset Formula $\forall s,t\in\mathbb{R}:g_{s}\left(t\right)=g_{t}\left(s\right)$
+\end_inset
+
+,
+ lahko našo neenakost zapišemo kot 
+\begin_inset Formula $g_{x_{1}}\left(x\right)\leq g_{x_{2}}\left(x\right)$
+\end_inset
+
+.
+ Sledi (desni neenačaj iz 
+\begin_inset Formula $g_{x_{1}}\left(x\right)\leq g_{x_{2}}\left(x\right)$
+\end_inset
+
+,
+ levi neenačaj pa ker 
+\begin_inset Formula $g$
+\end_inset
+
+ narašča):
+\begin_inset Formula 
+\[
+\left(D_{+}\left(f\right)\right)\left(x_{1}\right)=\inf_{x\in I,x>x_{1}}g_{x_{1}}\left(x\right)\leq\inf_{x\in I,x>x_{2}}g_{x_{1}}\left(x\right)\leq\inf_{x\in I,x>x_{2}}g_{x_{2}}\left(x\right)=\left(D_{+}\left(f\right)\right)\left(x_{2}\right)
+\]
+
+\end_inset
+
+Podobno dokažemo
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+DOPIŠI KAKO!
+ TODO XXX FIXME
+\end_layout
+
+\end_inset
+
+,
+ da 
+\begin_inset Formula $D_{-}$
+\end_inset
+
+ narašča.
+\end_layout
+
+\begin_layout Theorem*
+Naj bo 
+\begin_inset Formula $f:I^{\text{odp.}}\to\mathbb{R}$
+\end_inset
+
+ dvakrat odvedljiva.
+ Tedaj je 
+\begin_inset Formula $f$
+\end_inset
+
+ konveksna 
+\begin_inset Formula $\Leftrightarrow\forall x\in I:f''\left(x\right)\geq0$
+\end_inset
+
+ in 
+\begin_inset Formula $f$
+\end_inset
+
+ konkavna 
+\begin_inset Formula $\Leftrightarrow\forall x\in I:f''\left(x\right)\leq0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Dokazujemo ekvivalenco za konveksnost (konkavnost podobno).
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Rightarrow\right)$
+\end_inset
+
+ Po predpostavki je 
+\begin_inset Formula $f$
+\end_inset
+
+ konveksna in dvakrat odvedljiva,
+ torej je odvedljiva in sta levi in desni odvod enaka,
+ po prejšnji posledici pa levi in desni odvod naraščata,
+ torej 
+\begin_inset Formula $f'$
+\end_inset
+
+ narašča.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Leftarrow\right)$
+\end_inset
+
+ Naj bo 
+\begin_inset Formula $f''\geq0$
+\end_inset
+
+.
+ Vzemimo 
+\begin_inset Formula $x,a\in I$
+\end_inset
+
+.
+ Po Lagrangeu 
+\begin_inset Formula $\exists\xi\text{ med \ensuremath{x} in \ensuremath{a}}\ni:f\left(x\right)-f\left(x\right)=f'\left(\xi\right)\left(x-a\right)$
+\end_inset
+
+.
+ Iz predpostavke 
+\begin_inset Formula $f''>0$
+\end_inset
+
+ sledi,
+ da 
+\begin_inset Formula $f'$
+\end_inset
+
+ narašča.
+ Če je 
+\begin_inset Formula $x>\xi>a$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $f'\left(\xi\right)\geq f'\left(a\right)$
+\end_inset
+
+,
+ zato 
+\begin_inset Formula $f'\left(\xi\right)\left(x-a\right)\geq f'\left(a\right)\left(x-a\right)$
+\end_inset
+
+.
+ Če je 
+\begin_inset Formula $x<\xi<a$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $f'\left(\xi\right)\left(x-a\right)\leq f'\left(a\right)\left(x-a\right)$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Subsection
+Ekstremi funkcij ene spremenljivke
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $I\subseteq\mathbb{R}$
+\end_inset
+
+ odprt interal,
+ 
+\begin_inset Formula $a\in I$
+\end_inset
+
+ in 
+\begin_inset Formula $f:I\to\mathbb{R}$
+\end_inset
+
+.
+ Pravimo,
+ da ima 
+\begin_inset Formula $f$
+\end_inset
+
+ v točki 
+\begin_inset Formula $a$
+\end_inset
+
+ lokalni minimum,
+ če 
+\begin_inset Formula $\exists\delta>0\ni:\min\left\{ f\left(x\right);\forall x\in\left(a-\delta,a+\delta\right)\right\} =f\left(a\right)$
+\end_inset
+
+.
+ Pravimo,
+ da ima 
+\begin_inset Formula $f$
+\end_inset
+
+ v točki 
+\begin_inset Formula $a$
+\end_inset
+
+ lokalni maksimum,
+ če 
+\begin_inset Formula $\exists\delta>0\ni:\max\left\{ f\left(x\right);\forall x\in\left(a-\delta,a+\delta\right)\right\} =f\left(a\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+Če je 
+\begin_inset Formula $f:I\to\mathbb{R}$
+\end_inset
+
+ odvedljiva in ima v 
+\begin_inset Formula $a$
+\end_inset
+
+ lokalni minimum/maksimum,
+ tedaj je 
+\begin_inset Formula $f'\left(a\right)=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Glej dokaz Rolleovega izreka.
+\end_layout
+
+\begin_layout Definition*
+\begin_inset Formula $f$
+\end_inset
+
+ ima v 
+\begin_inset Formula $a$
+\end_inset
+
+ ekstrem,
+ če ima v 
+\begin_inset Formula $a$
+\end_inset
+
+ lokalni minimum ali lokalni maksimum.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Če je 
+\begin_inset Formula $f'\left(a\right)=0$
+\end_inset
+
+,
+ pravimo,
+ da ima 
+\begin_inset Formula $f$
+\end_inset
+
+ v 
+\begin_inset Formula $a$
+\end_inset
+
+ stacionarno točko.
+\end_layout
+
+\begin_layout Theorem*
+Naj bo 
+\begin_inset Formula $I\subseteq\mathbb{R}$
+\end_inset
+
+ odprt interval,
+ 
+\begin_inset Formula $a\in I$
+\end_inset
+
+ in 
+\begin_inset Formula $f:I\to\mathbb{R}$
+\end_inset
+
+ dvakrat odvedljiva ter naj bo 
+\begin_inset Formula $f'\left(a\right)=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $f''\left(a\right)>0\Rightarrow$
+\end_inset
+
+ v 
+\begin_inset Formula $a$
+\end_inset
+
+ ima 
+\begin_inset Formula $f$
+\end_inset
+
+ lokalni minimum
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $f''\left(a\right)<0\Rightarrow$
+\end_inset
+
+ v 
+\begin_inset Formula $a$
+\end_inset
+
+ ima 
+\begin_inset Formula $f$
+\end_inset
+
+ lokalni maksimum
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $f''\left(a\right)=0\Rightarrow$
+\end_inset
+
+ nedoločeno
+\end_layout
+
+\end_deeper
+\begin_layout Proof
+Sledi iz 
+\begin_inset Formula $f''>0\Rightarrow$
+\end_inset
+
+ stroga konveksnost in 
+\begin_inset Formula $f''<0\Rightarrow$
+\end_inset
+
+ stroga konkavnost.
+\end_layout
+
+\begin_layout Subsection
+L'Hopitalovo pravilo
+\end_layout
+
+\begin_layout Standard
+Kako izračunati 
+\begin_inset Formula $\lim_{x\to a}\frac{f\left(x\right)}{g\left(x\right)}$
+\end_inset
+
+?
+\end_layout
+
+\begin_layout Standard
+Če so funkcije zvezne v 
+\begin_inset Formula $a$
+\end_inset
+
+ in 
+\begin_inset Formula $g\left(a\right)\not=0$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $\lim_{x\to a}\frac{f\left(x\right)}{g\left(x\right)}=\frac{f\left(a\right)}{g\left(a\right)}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Če imata funkciji v 
+\begin_inset Formula $a$
+\end_inset
+
+ limito in 
+\begin_inset Formula $\lim_{x\to a}g\left(x\right)\not=0$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $\lim_{x\to a}\frac{f\left(x\right)}{g\left(x\right)}=\frac{\lim_{x\to a}f\left(x\right)}{\lim_{x\to a}g\left(x\right)}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Če 
+\begin_inset Formula $\lim_{x\to a}g\left(x\right)=\infty$
+\end_inset
+
+ in je na neki okolici 
+\begin_inset Formula $a$
+\end_inset
+
+ 
+\begin_inset Formula $f\left(x\right)$
+\end_inset
+
+ omejena,
+ velja 
+\begin_inset Formula $\lim_{x\to a}\frac{f\left(x\right)}{g\left(x\right)}=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Če 
+\begin_inset Formula $\lim_{x\to a}f\left(x\right)=0$
+\end_inset
+
+ in je na neki okolici 
+\begin_inset Formula $a$
+\end_inset
+
+ 
+\begin_inset Formula $g\left(x\right)$
+\end_inset
+
+ navzdol omejena več od nič ali navzgor omejena manj od nič,
+ velja 
+\begin_inset Formula $\lim_{x\to a}\frac{f\left(x\right)}{g\left(x\right)}=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Zanimivi primeri pa so,
+ ko 
+\begin_inset Formula $\lim_{x\to a}f\left(x\right)=\lim_{x\to a}g\left(x\right)=0$
+\end_inset
+
+ ali pa ko 
+\begin_inset Formula $\lim_{x\to a}f\left(x\right)=\infty$
+\end_inset
+
+ in hkrati 
+\begin_inset Formula $\lim_{x\to a}g\left(x\right)=\infty$
+\end_inset
+
+,
+ na primer 
+\begin_inset Formula $\lim_{x\to0}\frac{x}{x}$
+\end_inset
+
+ ali pa 
+\begin_inset Formula $\lim_{x\to0}\frac{x^{2}}{x}$
+\end_inset
+
+ ali pa 
+\begin_inset Formula $\lim_{x\to0}\frac{x}{x^{2}}$
+\end_inset
+
+.
+ Tedaj uporabimo L'Hopitalovo pravilo.
+\end_layout
+
+\begin_layout Theorem*
+Če velja hkrati:
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset CommandInset label
+LatexCommand label
+name "enu:Eno-izmed-slednjega:"
+
+\end_inset
+
+Eno izmed slednjega:
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $\lim_{x\to a}f\left(x\right)=\lim_{x\to a}g\left(x\right)=0$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\lim_{x\to a}f\left(x\right)=\infty$
+\end_inset
+
+ in hkrati 
+\begin_inset Formula $\lim_{x\to a}g\left(x\right)=\infty$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\lim_{x\to a}f\left(x\right)=-\infty$
+\end_inset
+
+ in hkrati 
+\begin_inset Formula $\lim_{x\to a}g\left(x\right)=-\infty$
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+\begin_inset Formula $f,g$
+\end_inset
+
+ v okolici 
+\begin_inset Formula $a$
+\end_inset
+
+ odvedljivi
+\end_layout
+
+\end_deeper
+\begin_layout Theorem*
+Potem 
+\begin_inset Formula $\exists L\coloneqq\lim_{x\to a}\frac{f'\left(x\right)}{g'\left(x\right)}\Rightarrow\exists\lim_{x\to a}\frac{f\left(x\right)}{g\left(x\right)}$
+\end_inset
+
+ in ta limita je enaka 
+\begin_inset Formula $L$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Ne bomo dokazali.
+\end_layout
+
+\begin_layout Example*
+Nekaj primerov uporabe L'Hopitalovega pravila.
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula 
+\[
+\lim_{x\to0}x^{x}=\lim_{x\to0}e^{lnx^{x}}=\lim_{x\to0}e^{x\ln x}=e^{\lim_{x\to0}x\ln x}
+\]
+
+\end_inset
+
+Računajmo 
+\begin_inset Formula $\lim_{x\to0}x\ln x$
+\end_inset
+
+ z L'Hopitalom.
+ Potrebujemo ulomek.
+ Ideja:
+ množimo števec in imenovalec z 
+\begin_inset Formula $x$
+\end_inset
+
+,
+ tedaj bi dobili 
+\begin_inset Formula $\lim_{x\to0}\frac{x^{2}\ln x}{x}$
+\end_inset
+
+.
+ Toda v tem primeru števec in imenovalec ne ustrezata pogoju 
+\begin_inset CommandInset ref
+LatexCommand ref
+reference "enu:Eno-izmed-slednjega:"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+ za L'Hopitalovo pravilo.
+ Druga ideja:
+ množimo števec in imenovalec z 
+\begin_inset Formula $\left(\ln x\right)^{-1}$
+\end_inset
+
+,
+ tedaj dobimo 
+\begin_inset Formula $\lim_{x\to0}\frac{x}{\left(\ln x\right)^{-1}}\overset{\text{L'H}}{=}\lim_{x\to0}\frac{1}{\frac{-1}{\log^{2}x}\cdot\frac{1}{x}}=\lim_{x\to0}-x\log^{2}x$
+\end_inset
+
+,
+ kar je precej komplicirano.
+ Tretja ideja:
+ množimo števec in imenovalec z 
+\begin_inset Formula $x^{-1}$
+\end_inset
+
+,
+ tedaj števec in imenovalec divergirata k 
+\begin_inset Formula $-\infty$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+\lim_{x\to0}\frac{\ln x}{x^{-1}}\overset{\text{L'H}}{=}\lim_{x\to0}\frac{\left(\ln x\right)'}{\left(x^{-1}\right)'}=\lim_{x\to0}\frac{x^{-1}}{-x^{-2}}=\lim_{x\to0}-x=0
+\]
+
+\end_inset
+
+Potemtakem 
+\begin_inset Formula $\lim_{x\to0}x^{x}=e^{0}=1$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\lim_{x\to0}\frac{1-\cos x}{x^{2}}$
+\end_inset
+
+.
+ Obe strani ulomkove črte konvergirata k 
+\begin_inset Formula $0$
+\end_inset
+
+.
+ Prav tako ko enkrat že uporabimo L'H.
+\begin_inset Formula 
+\[
+\lim_{x\to0}\frac{1-\cos x}{x^{2}}\overset{\text{L'H}}{=}\lim_{x\to0}\frac{\sin x}{2x}\overset{\text{L'H}}{=}\lim_{x\to0}\frac{\cos x}{2}=\frac{1}{2}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Section
+Taylorjev izrek in Taylorjeva formula
+\end_layout
+
+\begin_layout Standard
+Naj bo 
+\begin_inset Formula $f$
+\end_inset
+
+ v okolici 
+\begin_inset Formula $a$
+\end_inset
+
+ dovoljkrat odvedljiva.
+ Želimo aproksimirati 
+\begin_inset Formula $f\left(a+h\right)$
+\end_inset
+
+ s polinomi danega reda 
+\begin_inset Formula $n$
+\end_inset
+
+.
+ Iščemo polinome reda 
+\begin_inset Formula $n$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $n=0$
+\end_inset
+
+ konstante.
+ 
+\begin_inset Formula $f\left(a+h\right)\approx f\left(a\right)$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $n=1$
+\end_inset
+
+ linearne funkcije.
+ 
+\begin_inset Formula $f\left(a+h\right)\sim f\left(a\right)+f'\left(a\right)h$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $n=2$
+\end_inset
+
+ ...
+ Želimo najti 
+\begin_inset Formula $a_{0},a_{1},a_{2}\in\mathbb{R}$
+\end_inset
+
+,
+ odvisne le od 
+\begin_inset Formula $f$
+\end_inset
+
+ in 
+\begin_inset Formula $a$
+\end_inset
+
+,
+ za katere 
+\begin_inset Formula $f\left(a+b\right)\approx a_{0}+a_{1}h+a_{2}h^{2}$
+\end_inset
+
+.
+ Ko govorimo o aproksimaciji,
+ mislimo take koeficiente,
+ da se približek najbolje prilega dejanski funkcijski vrednosti,
+ v smislu,
+ da
+\begin_inset Formula 
+\[
+\lim_{h\to0}\frac{f\left(a+h\right)-\left(a_{0}+a_{1}h+a_{2}h^{2}\right)}{h^{2}}=0
+\]
+
+\end_inset
+
+
+\begin_inset Formula $a_{0}$
+\end_inset
+
+ izvemo takoj,
+ kajti 
+\begin_inset Formula $\lim_{h\to0}f\left(a+h\right)-\left(a_{0}+a_{1}h+a_{2}h^{2}\right)=0=f\left(a\right)-\left(a_{0}+0h+0h^{2}\right)=f\left(a\right)-a_{0}=0$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $a_{0}=f\left(a\right)$
+\end_inset
+
+.
+ Za preostale koeficiente uporabimo L'Hopitalovo pravilo,
+ ki pove,
+ da zadošča,
+ da je
+\begin_inset Formula 
+\[
+\lim_{h\to0}\frac{f'\left(a+h\right)-\left(0+a_{1}+a_{2}h\right)}{2h}=0
+\]
+
+\end_inset
+
+Zopet glejmo števec in vstavimo 
+\begin_inset Formula $h=0$
+\end_inset
+
+:
+ 
+\begin_inset Formula $f'\left(a\right)-a_{1}=0\Rightarrow f'\left(a\right)=a_{1}$
+\end_inset
+
+.
+ Spet uporabimo L'H:
+\begin_inset Formula 
+\[
+\lim_{h\to0}\frac{f''\left(a+h\right)-\left(0+0+2a_{2}\right)}{2}
+\]
+
+\end_inset
+
+Vstavimo 
+\begin_inset Formula $h=0$
+\end_inset
+
+ v 
+\begin_inset Formula $f''\left(a+h\right)-2a_{2}$
+\end_inset
+
+ in dobimo 
+\begin_inset Formula $2a_{2}=f''\left(a\right)$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $a_{2}=\frac{f''\left(a\right)}{2}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $n=3$
+\end_inset
+
+ Ugibamo,
+ da je najboljši kubični približek
+\begin_inset Formula 
+\[
+f\left(a+h\right)\approx h\mapsto f\left(a\right)+f'\left(a\right)h+\frac{f''\left(a\right)}{2}h^{2}+\frac{f'''\left(a\right)}{6}h^{3}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Theorem*
+Taylor.
+ Naj bo 
+\begin_inset Formula $n\in\mathbb{N}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $I$
+\end_inset
+
+ interval 
+\begin_inset Formula $\subseteq\mathbb{R}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $a\in I$
+\end_inset
+
+,
+ 
+\begin_inset Formula $f:I\to\mathbb{R}$
+\end_inset
+
+ 
+\begin_inset Formula $n-$
+\end_inset
+
+krat odvedljiva v točki 
+\begin_inset Formula $a$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $\exists g_{n}:I-a\to\mathbb{R}\ni:$
+\end_inset
+
+
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $I-a$
+\end_inset
+
+ pomeni interval 
+\begin_inset Formula $I$
+\end_inset
+
+ pomaknjen v levo za 
+\begin_inset Formula $a$
+\end_inset
+
+.
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $f\left(a+h\right)=\sum_{j=0}^{n}\frac{f^{\left(j\right)}\left(a\right)}{n!}h^{j}+g_{n}\left(h\right)h^{n}$
+\end_inset
+
+ in
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $\lim_{h\to0}g_{n}\left(h\right)=0$
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Theorem*
+Sedaj pišimo 
+\begin_inset Formula $x=a+h$
+\end_inset
+
+.
+ Tedaj se izrek glasi:
+ 
+\begin_inset Formula $\exists\tilde{g_{n}}:I\to\mathbb{R}\ni:$
+\end_inset
+
+
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $f\left(x\right)=\sum_{j=0}^{n}\frac{f^{\left(j\right)}\left(a\right)}{n!}\left(x-a\right)^{j}+\tilde{g_{n}}\left(x\right)\left(x-a\right)^{n}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $\lim_{x\to a}\tilde{g_{n}}\left(x\right)=0$
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Theorem*
+Tedaj označimo 
+\begin_inset Formula $T_{n,f,a}\left(x\right)=\sum_{j=0}^{n}\frac{f^{\left(j\right)}\left(a\right)}{n!}\left(x-a\right)^{j}$
+\end_inset
+
+ (pravimo 
+\begin_inset Formula $n-$
+\end_inset
+
+ti taylorjev polinom za 
+\begin_inset Formula $f$
+\end_inset
+
+ okrog točke 
+\begin_inset Formula $a$
+\end_inset
+
+) in 
+\begin_inset Formula $R_{n,f,a}\left(x\right)=\tilde{g_{n}}\left(x\right)\left(x-a\right)^{n}$
+\end_inset
+
+ (pravimo ostanek/napaka).
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Theorem*
+Če je 
+\begin_inset Formula $f$
+\end_inset
+
+ 
+\begin_inset Formula $\left(n+1\right)-$
+\end_inset
+
+krat odvedljiva na odprtem intervalu 
+\begin_inset Formula $I\subseteq\mathbb{R}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $a\in I$
+\end_inset
+
+,
+ tedaj 
+\begin_inset Formula $\forall b\in I\exists\alpha\in I\text{ med \ensuremath{a} in \ensuremath{b}}\ni:R_{n}\left(b\right)=\frac{f^{\left(n+1\right)}\left(\alpha\right)}{\left(n+1\right)!}\left(b-a\right)^{n+1}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Označimo 
+\begin_inset Formula $T_{n}\left(x\right)=f\left(a\right)+f'\left(a\right)\left(x-a\right)+\frac{f''\left(a\right)}{2!}\left(x-a\right)^{2}+\cdots+\frac{f^{\left(n\right)}\left(a\right)}{n!}\left(x-a\right)^{n}$
+\end_inset
+
+ torej 
+\begin_inset Formula $n-$
+\end_inset
+
+ti taylorjev polinom in naj bo 
+\begin_inset Formula $K$
+\end_inset
+
+ tako število,
+ da velja 
+\begin_inset Formula $f\left(b\right)-T_{n}\left(b\right)=K\left(b-a\right)^{n+1}$
+\end_inset
+
+.
+ Definirajmo 
+\begin_inset Formula $F\left(x\right)=f\left(x\right)-T_{n}\left(x\right)-K\left(x-a\right)^{n+1}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hypertarget{velja}{Velja}
+\end_layout
+
+\end_inset
+
+ 
+\begin_inset Formula $T_{n}^{\left(k\right)}\left(a\right)=f\left(a\right)$
+\end_inset
+
+ za 
+\begin_inset Formula $k\leq n$
+\end_inset
+
+,
+ kajti 
+\begin_inset Formula $\frac{d\sum_{j=1}^{n}\frac{f^{\left(j\right)}\left(a\right)}{n!}\left(h\right)^{n}}{dh}=\frac{f^{\left(j\right)}\left(a\right)n!}{n!}\cdot1=f^{\left(j\right)}\left(a\right)$
+\end_inset
+
+.
+ Vsi členi z eksponentom,
+ manjšim od 
+\begin_inset Formula $k$
+\end_inset
+
+,
+ se odvajajo v 0,
+ točno pri eksponentu 
+\begin_inset Formula $k$
+\end_inset
+
+ se člen odvaja v konstanto,
+ pri višjih členih pa ostane potencirana spremenljivka,
+ ki je 
+\begin_inset Formula $0$
+\end_inset
+
+ (tu mislimo odstopanje od 
+\begin_inset Formula $a$
+\end_inset
+
+,
+ označeno s 
+\begin_inset Formula $h$
+\end_inset
+
+),
+ torej se ti členi tudi izničijo.
+\end_layout
+
+\begin_layout Proof
+Zato 
+\begin_inset Formula $\forall k\leq n:F^{\left(k\right)}\left(a\right)=0$
+\end_inset
+
+.
+ Nadalje velja 
+\begin_inset Formula $F\left(a\right)=F\left(b\right)=0$
+\end_inset
+
+,
+ ker smo pač tako definirali funkcijo 
+\begin_inset Formula $F$
+\end_inset
+
+,
+ zato obstaja po Rolleovem izreku tak 
+\begin_inset Formula $\alpha_{1}$
+\end_inset
+
+,
+ da velja 
+\begin_inset Formula $F'\left(\alpha_{1}\right)=0$
+\end_inset
+
+.
+ Po Rolleovem izreku nadalje obstaja tak 
+\begin_inset Formula $\alpha_{2}$
+\end_inset
+
+ med 
+\begin_inset Formula $a$
+\end_inset
+
+ in 
+\begin_inset Formula $\alpha_{1}$
+\end_inset
+
+,
+ da velja 
+\begin_inset Formula $F''\left(\alpha_{2}\right)=0$
+\end_inset
+
+.
+ Spet po Rolleovem izreku obstaja tak 
+\begin_inset Formula $\alpha_{3}$
+\end_inset
+
+ med 
+\begin_inset Formula $a$
+\end_inset
+
+ in 
+\begin_inset Formula $\alpha_{2}$
+\end_inset
+
+,
+ da velja 
+\begin_inset Formula $F'''\left(\alpha_{3}\right)=0$
+\end_inset
+
+.
+ Postopek lahko ponavljamo in dobimo tak 
+\begin_inset Formula $\alpha=\alpha_{n+1}$
+\end_inset
+
+,
+ da velja 
+\begin_inset Formula $F^{\left(n+1\right)}\left(\alpha\right)=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Ker je 
+\begin_inset Formula $\forall x\in I:T_{n}^{\left(n+1\right)}\left(x\right)=0$
+\end_inset
+
+ (očitno,
+ isti argument kot v 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{velja}{drugem odstavku dokaza}
+\end_layout
+
+\end_inset
+
+),
+ to pomeni 
+\begin_inset Formula $f^{\left(n+1\right)}\left(\alpha\right)=\left(K\left(x-a\right)^{n+1}\right)^{\left(n+1\right)}=K\left(n+1\right)!$
+\end_inset
+
+.
+ Torej je 
+\begin_inset Formula $K=\frac{f^{\left(n+1\right)}\left(\alpha\right)}{\left(n+1\right)!}$
+\end_inset
+
+ in zato 
+\begin_inset Formula $f\left(b\right)=T_{n}\left(b\right)+\frac{f^{\left(n+1\right)}\left(\alpha\right)}{\left(n+1\right)!}\left(b-a\right)^{\left(n+1\right)}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Corollary*
+Če je 
+\begin_inset Formula $\left(n+1\right)-$
+\end_inset
+
+ti odvod omejen na 
+\begin_inset Formula $I$
+\end_inset
+
+,
+ t.
+ j.
+ 
+\begin_inset Formula $\exists M>0\forall x\in I:\left|f^{\left(n+1\right)}\left(x\right)\right|\leq M$
+\end_inset
+
+,
+ lahko ostanek eksplicitno ocenimo,
+ in sicer 
+\begin_inset Formula $\left|R_{n}\left(x\right)\right|\leq\frac{M}{\left(n+1\right)!}\left|x-a\right|^{n+1}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Kaj pa se zgodi,
+ ko 
+\begin_inset Formula $n$
+\end_inset
+
+ pošljemo v neskončnost?
+ Iskali bi aproksimacije s 
+\begin_inset Quotes gld
+\end_inset
+
+polinomi neskončnega reda
+\begin_inset Quotes grd
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Če je 
+\begin_inset Formula $f\in C^{\infty}$
+\end_inset
+
+ v okolici točke 
+\begin_inset Formula $a\in\mathbb{R}$
+\end_inset
+
+.
+ Tedaj definiramo Taylorjevo vrsto 
+\begin_inset Formula $f$
+\end_inset
+
+ v okolici točke 
+\begin_inset Formula $a$
+\end_inset
+
+:
+ 
+\begin_inset Formula $T_{f,a}\left(x\right)\coloneqq\sum_{j=0}^{\infty}\frac{f^{\left(j\right)}\left(a\right)}{j!}\left(x-a\right)^{j}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Question*
+Ali Taylorjeva vrsta konvergira oziroma kje konvergira?
+ Kakšna je zveza s 
+\begin_inset Formula $f\left(x\right)$
+\end_inset
+
+?
+ Kakšen je 
+\begin_inset Formula $R_{f,a}$
+\end_inset
+
+?
+\end_layout
+
+\begin_layout Standard
+Oglejmo si potenčne vrste (
+\begin_inset Formula $\sum_{j=0}^{\infty}b_{k}x^{k}$
+\end_inset
+
+) kot poseben primer funkcijskih vrst (
+\begin_inset Formula $\sum_{j=0}^{\infty}a_{k}\left(x\right)$
+\end_inset
+
+).
+ Vemo,
+ da ima potenčna vrsta konvergenčni radij 
+\begin_inset Formula $R$
+\end_inset
+
+.
+ Za 
+\begin_inset Formula $x\in\left(-R,R\right)$
+\end_inset
+
+ konvergira,
+ za 
+\begin_inset Formula $x\in\left[-R,R\right]^{C}$
+\end_inset
+
+ divergira.
+\end_layout
+
+\begin_layout Theorem*
+Naj ima potenčna vrsta 
+\begin_inset Formula $f\left(x\right)=\sum_{k=0}^{\infty}b_{k}x^{k}$
+\end_inset
+
+ konvergenčni radij 
+\begin_inset Formula $R$
+\end_inset
+
+.
+ Tedaj ima tudi 
+\begin_inset Formula $g\left(x\right)=\sum_{k=1}kb_{k}x^{k-1}$
+\end_inset
+
+ konvergenčni radij 
+\begin_inset Formula $R$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+\begin_inset Formula 
+\[
+\frac{1}{R_{g}}=\limsup_{k\to\infty}\sqrt[k]{\left|ka_{k}\right|}=\limsup_{k\to\infty}\sqrt[k]{\left|k\right|\left|a_{k}\right|}=\limsup_{k\to\infty}\sqrt[k]{\left|k\right|}\sqrt[k]{\left|a_{k}\right|}=\cdots
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Proof
+\begin_inset Formula 
+\[
+\lim_{k\to\infty}\sqrt[k]{\left|k\right|}=\lim_{k\to\infty}k^{1/k}=e^{\lim_{k\to\infty}\frac{1}{k}\ln k}\overset{\text{L'H}}{=}e^{\lim_{k\to\infty}\frac{\frac{1}{k}}{k}}=e^{\lim_{k\to\infty}\cancelto{0}{\frac{1}{k^{2}}}}=e^{0}=1
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\cdots=\limsup_{k\to\infty}1\cdot\sqrt[k]{\left|a_{k}\right|}=\frac{1}{R_{f}}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Corollary*
+Če ima potenčna vrsta 
+\begin_inset Formula $f$
+\end_inset
+
+ konvergenčni radij 
+\begin_inset Formula $R>0$
+\end_inset
+
+,
+ tedaj je 
+\begin_inset Formula $f\in C^{\infty}\left(\left(-R,R\right)\right)$
+\end_inset
+
+ in velja 
+\begin_inset Formula $a_{k}=\frac{f^{\left(k\right)}\left(0\right)}{k!}$
+\end_inset
+
+,
+ potem velja 
+\begin_inset Formula $g=f'$
+\end_inset
+
+ (iz izreka zgoraj).
+ Razlaga:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $f\left(x\right)=\sum_{k=0}^{\infty}a_{k}x^{k}=\sum_{k=0}^{\infty}\frac{f^{\left(k\right)}\left(0\right)}{k!}x^{k}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $f'\left(x\right)=\sum_{k=1}^{\infty}ka_{k}x^{k-1}=\sum_{k=1}^{\infty}\frac{kf^{\left(k\right)}\left(0\right)}{k!}x^{k-1}=\sum_{k=1}^{\infty}\frac{f^{\left(k\right)}\left(0\right)}{\left(k-1\right)!}x^{k-1}$
+\end_inset
+
+ (
+\begin_inset Formula $k$
+\end_inset
+
+ začne z 
+\begin_inset Formula $1$
+\end_inset
+
+,
+ ker se 
+\begin_inset Formula $k=0$
+\end_inset
+
+ člen odvaja v konstanto 
+\begin_inset Formula $0$
+\end_inset
+
+)
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $g\left(x\right)=\sum_{k=1}^{\infty}ka_{k}x^{k-1}=\sum_{k=1}^{\infty}\frac{kf^{\left(k\right)}\left(0\right)}{k!}x^{k-1}=\sum_{k=1}^{\infty}\frac{f^{\left(k\right)}\left(0\right)}{\left(k-1\right)!}x^{k-1}=f'\left(x\right)$
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Definition*
+Funkcija 
+\begin_inset Formula $f:J\to\mathbb{R}$
+\end_inset
+
+ (
+\begin_inset Formula $J$
+\end_inset
+
+ je interval 
+\begin_inset Formula $\subseteq\mathbb{R}$
+\end_inset
+
+) je realno analitična,
+ če se jo da okoli vsake točke 
+\begin_inset Formula $c\in J$
+\end_inset
+
+ razviti v potenčno vrsto,
+ torej če 
+\begin_inset Formula $f\left(x\right)=\sum_{k=0}^{\infty}\frac{f^{\left(k\right)}\left(c\right)}{k!}\left(x-c\right)^{k}$
+\end_inset
+
+ za 
+\begin_inset Formula $x$
+\end_inset
+
+ blizu 
+\begin_inset Formula $c$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+\begin_inset Formula $f\in C^{\infty}\Rightarrow f$
+\end_inset
+
+ je realno analitična.
+ Protiprimer je 
+\begin_inset Formula $f\left(x\right)=e^{\frac{-1}{\left|x\right|}}$
+\end_inset
+
+.
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+TODO XXX FIXME ZAKAJ?,
+ ne razumem
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+Primeri Taylorjevih vrst.
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $f\left(x\right)=e^{x}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $n-$
+\end_inset
+
+ti tayorjev polinom za 
+\begin_inset Formula $f\left(x\right)$
+\end_inset
+
+ okoli 
+\begin_inset Formula $0$
+\end_inset
+
+:
+ 
+\begin_inset Formula $T_{n,e^{x},0}\left(x\right)=1+x+\frac{x^{2}}{2}+\frac{x^{3}}{6}+\cdots+\frac{x^{n}}{n!}$
+\end_inset
+
+ in velja 
+\begin_inset Formula $e^{x}=T_{n,e^{x},0}\left(x\right)+R_{n,e^{x},0}\left(x\right)$
+\end_inset
+
+,
+ kjer 
+\begin_inset Formula $\lim_{n\to\infty}R_{n,e^{x},0}\left(x\right)=0$
+\end_inset
+
+.
+ Ne bomo dokazali.
+ Sledi 
+\begin_inset Formula $e^{x}=\sum_{k=0}^{\infty}\frac{x^{k}}{k!}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\sin x=x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\cdots+\left(-1\right)^{k}\frac{x^{2k+1}}{\left(2k+1\right)!}$
+\end_inset
+
+.
+ Opazimo sode eksponente in opazimo učinek odvajanja:
+ 
+\begin_inset Formula $\cos,-\sin,-\cos,\sin,\cos,-\sin,\dots$
+\end_inset
+
+.
+ Členi vrste 
+\begin_inset Formula $\sin x$
+\end_inset
+
+ v 
+\begin_inset Formula $x=0$
+\end_inset
+
+ so:
+ 
+\begin_inset Formula $1,0,-1,0,1,0,-1,\dots$
+\end_inset
+
+.
+ Opazimo izpadanje vsakega drugega člena.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\cos x=1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-\cdots+\left(-1\right)^{k}\frac{x^{2k}}{\left(2k\right)!}$
+\end_inset
+
+.
+ Opazimo sode eksponente.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $f\left(x\right)=\log\left(1-x\right)$
+\end_inset
+
+.
+ A lahko to funkcijo razvijemo v taylorjevo vrsto okoli točke 0?
+\begin_inset Float table
+placement document
+alignment document
+wide false
+sideways false
+status open
+
+\begin_layout Plain Layout
+\align center
+\begin_inset Tabular
+<lyxtabular version="3" rows="7" columns="3">
+<features tabularvalignment="middle">
+<column alignment="center" valignment="top" width="0pt">
+<column alignment="center" valignment="top" width="0pt">
+<column alignment="center" valignment="top" width="0pt">
+<row>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $k$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $f^{\left(k\right)}\left(x\right)$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $f^{\left(k\right)}\left(0\right)$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $0$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $\log\left(1-x\right)$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $0$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $1$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $\frac{-1}{1-x}$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $-1$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $2$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $\frac{-1}{\left(1-x\right)^{2}}$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $-1$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $3$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $\frac{-2}{\left(1-x\right)^{3}}$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $-2$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $\cdots$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $\cdots$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $\cdots$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $n$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $\frac{-\left(n-1\right)!}{\left(1-x\right)^{n}}$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $-\left(n-1\right)!$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+</row>
+</lyxtabular>
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Plain Layout
+\begin_inset Caption Standard
+
+\begin_layout Plain Layout
+Razvijanje 
+\begin_inset Formula $\log\left(1-x\right)$
+\end_inset
+
+ okoli točke 
+\begin_inset Formula $0$
+\end_inset
+
+.
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Velja 
+\begin_inset Formula $f\left(x\right)=\log\left(1-x\right)=\sum_{k=1}^{\infty}\frac{-\left(k-1\right)!}{k!}x^{k}=-\sum_{k=1}^{\infty}\frac{x^{k}}{k}$
+\end_inset
+
+ za 
+\begin_inset Formula $\left|x\right|<1$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Section
+Integrali
+\end_layout
+
+\begin_layout Standard
+Radi bi definirali ploščino 
+\begin_inset Formula $P=\left\{ \left(x,t\right)\in\mathbb{R}^{2};x\in\left[a,b\right],t\in\left[0,f\left(x\right)\right]\right\} $
+\end_inset
+
+ za funkcijo 
+\begin_inset Formula $f:\left[a,b\right]\to[0,\infty)$
+\end_inset
+
+.
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+TODO XXX FIXME skica ANA1P FMF 2024-01-09/str.3
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $P$
+\end_inset
+
+ aproksimiramo s pravokotniki,
+ katerih ploščino smo predhodno definirali takole:
+\end_layout
+
+\begin_layout Definition*
+Ploščina pravokotnika s stranicama 
+\begin_inset Formula $c$
+\end_inset
+
+ in 
+\begin_inset Formula $d$
+\end_inset
+
+ je 
+\begin_inset Formula $c\cdot d$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Najprej diskusija.
+ Naj bo 
+\begin_inset Formula $t_{j}$
+\end_inset
+
+ delitev 
+\begin_inset Formula $\left[a,b\right]$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $a=t_{0}<t_{1}<\cdots<t_{n}=b$
+\end_inset
+
+.
+ Ne zahtevamo ekvidistančne delitve,
+ torej take,
+ pri kateri bi bile razdalje enake.
+ Kako naj definiramo višine pravokotnikov,
+ katerih stranice so delilne točke 
+\begin_inset Formula $t_{n}$
+\end_inset
+
+?
+\end_layout
+
+\begin_layout Standard
+Lahko tako,
+ da na vsakem intervalu 
+\begin_inset Formula $\left[t_{i},t_{i+1}\right]$
+\end_inset
+
+ izberemo nek 
+\begin_inset Formula $\xi_{i}$
+\end_inset
+
+,
+ pravokotnicova osnovnica bode 
+\begin_inset Formula $t_{i+1}-t_{i}$
+\end_inset
+
+,
+ njegova višina pa 
+\begin_inset Formula $f\left(\xi_{i}\right)$
+\end_inset
+
+.
+ Ploščina 
+\begin_inset Formula $P$
+\end_inset
+
+ pod grafom funkcije je približno enaka vsoti ploščin teh pravokotnikov,
+ torej 
+\begin_inset Formula $\sum_{k=1}^{n}f\left(\xi_{k}\right)\left(t_{k}-t_{k-1}\right)=R\left(f,\vec{t},\vec{\xi}\right)$
+\end_inset
+
+,
+ kjer je 
+\begin_inset Formula $\vec{t}$
+\end_inset
+
+ delitev in 
+\begin_inset Formula $\vec{\xi}$
+\end_inset
+
+ izbira točk na intervalih delitve.
+ Temu pravimo Riemannova vsota za 
+\begin_inset Formula $f$
+\end_inset
+
+,
+ ki pripada delitvi 
+\begin_inset Formula $\vec{t}$
+\end_inset
+
+ in izboru 
+\begin_inset Formula $\vec{\xi}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Če je 
+\begin_inset Formula $D\coloneqq\left\{ \left[t_{j+1},t_{j}\right];j=\left\{ 1..n\right\} \right\} $
+\end_inset
+
+ delitev za 
+\begin_inset Formula $\left[a,b\right]$
+\end_inset
+
+,
+ definiramo tako oznako 
+\begin_inset Formula $\left|D\right|_{\infty}\coloneqq\max_{j=\left\{ 1..n\right\} }\left(t_{j}-t_{j-1}\right)=\max_{I\in D}\left(\left|I\right|\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Če 
+\begin_inset Formula $\exists A\in\mathbb{R}\ni:$
+\end_inset
+
+ za poljubno fine delitve (
+\begin_inset Formula $\left|D\right|_{\infty}=\infty^{-1}$
+\end_inset
+
+) 
+\begin_inset Formula $D$
+\end_inset
+
+ se pripadajoče Riemannove vsote malo razlikujejo od 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ pravimo številu 
+\begin_inset Formula $A$
+\end_inset
+
+ ploščina lika 
+\begin_inset Formula $P$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Sedaj pa še formalna definicija.
+\end_layout
+
+\begin_layout Definition*
+Naj bodo 
+\begin_inset Formula $f,D,\xi$
+\end_inset
+
+ kot prej in 
+\begin_inset Formula $I\in\mathbb{R}$
+\end_inset
+
+ realno število.
+ Če 
+\begin_inset Formula $\forall\varepsilon>0\exists\delta>0\ni:$
+\end_inset
+
+
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $\forall$
+\end_inset
+
+ delitev 
+\begin_inset Formula $D\ni:\left|D\right|_{\infty}<\delta$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $\forall$
+\end_inset
+
+ nabor 
+\begin_inset Formula $\xi=\xi_{1},\dots,\xi_{n}$
+\end_inset
+
+,
+ pripadajoč delitvi 
+\begin_inset Formula $D$
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Definition*
+velja 
+\begin_inset Formula $\left|R\left(f,D,\xi\right)-I\right|<\varepsilon\Longrightarrow I$
+\end_inset
+
+ je določen integral 
+\begin_inset Formula $f$
+\end_inset
+
+ na intervalu 
+\begin_inset Formula $\left[a,b\right]$
+\end_inset
+
+ in je po definiciji ploščina lika 
+\begin_inset Formula $P$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Če tak 
+\begin_inset Formula $I$
+\end_inset
+
+ obstaja,
+ kar ni 
+\emph on
+a priori
+\emph default
+,
+ pravimo,
+ da je 
+\begin_inset Formula $f$
+\end_inset
+
+ integrabilna na 
+\begin_inset Formula $\left[a,b\right]$
+\end_inset
+
+ in pišemo 
+\begin_inset Formula $I=\int_{a}^{b}f\left(x\right)dx$
+\end_inset
+
+.
+ Temu pravimo Riemannov integral funkcije 
+\begin_inset Formula $f$
+\end_inset
+
+ na 
+\begin_inset Formula $\left[a,b\right]$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Darbouxove vsote.
+ Imamo torej delitev 
+\begin_inset Formula $D=\left\{ \left[t_{j-1},t_{j}\right];j\in\left\{ 1..n\right\} ;t_{0}=1,t_{n}=b\right\} $
+\end_inset
+
+ delitev za 
+\begin_inset Formula $J=\left[a,b\right]$
+\end_inset
+
+ in 
+\begin_inset Formula $f:J\to\mathbb{R}$
+\end_inset
+
+.
+ Imamo tudi množico izbranih točk 
+\begin_inset Formula $\xi=\left\{ \xi_{j}\in\left[t_{j-1},t_{j}\right];j\in\left\{ 1..n\right\} \right\} $
+\end_inset
+
+ in 
+\begin_inset Formula $R\left(f,D,\xi\right)=\sum_{j=1}^{n}f\left(\xi_{j}\right)\left(t_{j}-t_{j-1}\right)$
+\end_inset
+
+.
+ Ocenimo 
+\begin_inset Formula $f\left(\xi_{j}\right)$
+\end_inset
+
+:
+ 
+\begin_inset Formula $\inf_{x\in\left[t_{j-1},t_{j}\right]}f\left(x\right)\leq f\left(\xi_{j}\right)\leq\sup_{x\in\left[t_{j-1},t_{j}\right]}f\left(x\right)$
+\end_inset
+
+.
+ Definirali smo 
+\begin_inset Formula $\int_{a}^{b}f\left(x\right)dx$
+\end_inset
+
+ kot limito Riemannovih vsot s kakršnokoli delitvijo in izbiro 
+\begin_inset Formula $\xi$
+\end_inset
+
+,
+ zato lahko pišemo 
+\begin_inset Formula $\forall j\in\left\{ 1..n\right\} :\inf_{x\in\left[t_{j-1},t_{j}\right]}f\left(x\right)=f\left(\xi_{j}\right)=\sup_{x\in\left[t_{j-1},t_{j}\right]}f\left(x\right)$
+\end_inset
+
+.
+ Zato lahko limito Riemannovih vsot obravnavamo neodvisno od 
+\begin_inset Formula $\xi$
+\end_inset
+
+:
+\begin_inset Formula 
+\[
+s\left(f,D\right)\coloneqq\sum_{j=1}^{n}\left(\inf_{x\in D_{j}}f\left(x\right)\right)\left(t_{j}-t_{j-1}\right)\leq R\left(f,D,\xi\right)\leq\sum_{j=1}^{n}\left(\sup_{x\in D_{j}}f\left(x\right)\right)\left(t_{j}-f_{j-1}\right)\eqqcolon S\left(f,D\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Definirali smo dva nova pojma,
+ spodnjo Darbouxovo vsoto 
+\begin_inset Formula $s\left(f,D\right)$
+\end_inset
+
+ in zgornjo Darbouxovo vsoto 
+\begin_inset Formula $S\left(f,D\right)$
+\end_inset
+
+ in velja 
+\begin_inset Formula $s\left(f,D\right)\leq R\left(f,D,\xi\right)\leq S\left(f,D\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Naj bosta 
+\begin_inset Formula $D$
+\end_inset
+
+ in 
+\begin_inset Formula $D'$
+\end_inset
+
+ delitvi za interval 
+\begin_inset Formula $J$
+\end_inset
+
+.
+ Pravimo,
+ da je 
+\begin_inset Formula $D'$
+\end_inset
+
+ finejša od 
+\begin_inset Formula $D$
+\end_inset
+
+,
+ če je ima 
+\begin_inset Formula $D'$
+\end_inset
+
+ vse delilne točke,
+ ki jih ima 
+\begin_inset Formula $D$
+\end_inset
+
+ in poleg njih še vsaj kakšno.
+ Označimo 
+\begin_inset Formula $D\subset D'$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+Naj bo 
+\begin_inset Formula $D\subset D'$
+\end_inset
+
+ (
+\begin_inset Formula $D'$
+\end_inset
+
+ finejša od 
+\begin_inset Formula $D$
+\end_inset
+
+).
+ Oglejmo si 
+\begin_inset Formula $s\left(f,D\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $s\left(f,D'\right)$
+\end_inset
+
+.
+ Tedaj velja 
+\begin_inset Formula $s\left(f,D\right)\leq s\left(f,D'\right)$
+\end_inset
+
+,
+ ker je infimum po manjši množici lahko le večji —
+ s finejšo delitvijo smo vsaj neko množico (delitveni interval) razdelili na dva dela.
+ Za zgornjo Darbouxovo vsoto velja obratno,
+ torej 
+\begin_inset Formula $S\left(f,D\right)\geq S\left(f,D'\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Theorem*
+Za poljubni različni delitvi 
+\begin_inset Formula $D_{1},D_{2}$
+\end_inset
+
+ intervala 
+\begin_inset Formula $J$
+\end_inset
+
+ velja 
+\begin_inset Formula $s\left(f,D_{1}\right)\leq S\left(f,D_{2}\right)$
+\end_inset
+
+ ZDB Katerakoli spodnja Darbouxova vsota je kvečjemu tolikšna kot katerakoli zgornja.
+\end_layout
+
+\begin_layout Proof
+Označimo z 
+\begin_inset Formula $D_{1}\cup D_{2}$
+\end_inset
+
+ delitev,
+ ki vsebuje vse delilne točke tako 
+\begin_inset Formula $D_{1}$
+\end_inset
+
+ kot tudi 
+\begin_inset Formula $D_{2}$
+\end_inset
+
+.
+ Očitno velja,
+ da sta 
+\begin_inset Formula $D_{1}\subset D_{1}\cup D_{2}$
+\end_inset
+
+ in 
+\begin_inset Formula $D_{2}\subset D_{1}\cup D_{2}$
+\end_inset
+
+.
+ Po prejšnjem izreku veljata leva in desna neenakost,
+ srednja pa iz definicije (očitno).
+\begin_inset Formula 
+\[
+s\left(f,D_{1}\right)\leq s\left(f,D_{1}\cup D_{2}\right)\leq S\left(f,D_{1}\cup D_{2}\right)\leq S\left(f,D_{2}\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $f:J\to\mathbb{R}$
+\end_inset
+
+ omejena.
+ Označimo 
+\begin_inset Formula $s\left(f\right)\coloneqq\sup_{\text{vse možne delitve }D}s\left(f,D\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $S\left(f\right)\coloneqq\inf_{\text{vse možne delitve }D}S\left(f,D\right)$
+\end_inset
+
+.
+ Funkcija 
+\begin_inset Formula $f:J\to\mathbb{R}$
+\end_inset
+
+ je Riemannovo,
+ če 
+\begin_inset Formula $s\left(f\right)=S\left(f\right)$
+\end_inset
+
+ oziroma če 
+\begin_inset Formula $\forall\varepsilon>0\exists$
+\end_inset
+
+ delitev 
+\begin_inset Formula $D$
+\end_inset
+
+ na 
+\begin_inset Formula $J\ni:S\left(f,D\right)-s\left(f,D\right)<\varepsilon$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Note*
+Integrabilnost 
+\begin_inset Formula $f$
+\end_inset
+
+ ne pomeni,
+ da 
+\begin_inset Formula $\exists D\ni:s\left(f,D\right)=S\left(f,D\right)$
+\end_inset
+
+.
+ Ni namreč nujno,
+ da množica vsebuje svoj supremum.
+ Primer:
+ za 
+\begin_inset Formula $f\left(x\right)=x$
+\end_inset
+
+ velja 
+\begin_inset Formula $\forall D:S\left(f,D\right)>s\left(f,D\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+Vsaka zvezna funkcija je integrabilna na 
+\begin_inset Formula $J$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+ poljuben.
+ Po definiciji 
+\begin_inset Formula $S\left(f,D\right)-s\left(f,D\right)=\sum_{j=1}^{n}\left(\sup_{x\in D_{j}}f\left(x\right)-\inf_{x\in D_{j}}f\left(x\right)\right)\left(t_{j}-t_{j-1}\right)$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna,
+ je na zaprtem 
+\begin_inset Formula $J=\left[a,b\right]$
+\end_inset
+
+ enakomerno zvezna,
+ torej 
+\begin_inset Formula $\exists\delta>0\forall x_{1},x_{2}\in J:\left|x_{1}-x_{2}\right|<\delta\Rightarrow\left|f\left(x_{1}\right)-f\left(x_{2}\right)\right|<\frac{\varepsilon}{b-a}$
+\end_inset
+
+.
+ Izberimo tako delitev 
+\begin_inset Formula $D$
+\end_inset
+
+,
+ da je 
+\begin_inset Formula $\forall j\in\left\{ 1..\left|D\right|\right\} :t_{j}-t_{j-1}<\delta$
+\end_inset
+
+.
+ Tedaj bo veljalo 
+\begin_inset Formula $\sum_{j=1}^{n}\left(\sup_{x\in D_{j}}f\left(x\right)-\inf_{x\in D_{j}}f\left(x\right)\right)\left(t_{j}-t_{j-1}\right)<\sum_{j=1}^{n}\frac{\varepsilon}{b-a}\left(t_{j}-t_{j-1}\right)=\frac{\varepsilon\left(b-a\right)}{b-a}=\varepsilon$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Skratka dokazali smo 
+\begin_inset Formula $S\left(f,D\right)-s\left(f,D\right)<\varepsilon$
+\end_inset
+
+ za poljuben 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+,
+ torej je funkcija Riemannovo integrabilna po zgornji definiciji.
+\end_layout
+
+\begin_layout Definition*
+\begin_inset Formula $A\subset\mathbb{R}$
+\end_inset
+
+ ima mero 
+\begin_inset Formula $0$
+\end_inset
+
+,
+ če 
+\begin_inset Formula $\forall\varepsilon>0\exists$
+\end_inset
+
+ družina intervalov 
+\begin_inset Formula $I_{j}\ni:A\subset\bigcup I_{j}\wedge\sum\left|I_{j}\right|<\varepsilon$
+\end_inset
+
+.
+ Primer:
+ vse števne in končne množice.
+\end_layout
+
+\begin_layout Theorem*
+Funkcija 
+\begin_inset Formula $f$
+\end_inset
+
+ je integrabilna na intervalu 
+\begin_inset Formula $J\Leftrightarrow\left\{ x\in J;f\text{ ni zvezna v }x\right\} $
+\end_inset
+
+ ima mero 
+\begin_inset Formula $0$
+\end_inset
+
+.
+ ZDB če ima množica točk z definicijskega območja 
+\begin_inset Formula $f$
+\end_inset
+
+,
+ v katerih 
+\begin_inset Formula $f$
+\end_inset
+
+ ni zvezna,
+ mero 
+\begin_inset Formula $0$
+\end_inset
+
+ (recimo če je teh točk končno mnogo),
+ je 
+\begin_inset Formula $f$
+\end_inset
+
+ integrabilna.
+\end_layout
+
+\begin_layout Fact*
+Označimo z 
+\begin_inset Formula $I\left(J\right)$
+\end_inset
+
+ množico vseh integrabilnih funkcij na intervalu 
+\begin_inset Formula $J$
+\end_inset
+
+.
+ 
+\begin_inset Formula $I\left(J\right)$
+\end_inset
+
+ je vektorski prostor za množenje s skalarji iz 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+.
+ Naj bodo 
+\begin_inset Formula $f,g\in I\left(J\right),\lambda\in\mathbb{R}$
+\end_inset
+
+.
+ Velja aditivnost 
+\begin_inset Formula $f\left(x\right)+g\left(x\right)\in J\left(I\right)$
+\end_inset
+
+,
+ kajti 
+\begin_inset Formula $\int_{a}^{b}\left(f\left(x\right)+g\left(x\right)\right)dx=\int_{a}^{b}\left(f\left(x\right)\right)dx+\int_{a}^{b}\left(g\left(x\right)\right)dx$
+\end_inset
+
+ in homogenost 
+\begin_inset Formula $\int_{a}^{b}\lambda f\left(x\right)dx=\lambda\int_{a}^{b}f\left(x\right)dx$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+Če je 
+\begin_inset Formula $f$
+\end_inset
+
+ integrabilna na 
+\begin_inset Formula $J=\left[a,b\right]$
+\end_inset
+
+ in je 
+\begin_inset Formula $c\in J$
+\end_inset
+
+,
+ tedaj je 
+\begin_inset Formula $f$
+\end_inset
+
+ integrabilna na 
+\begin_inset Formula $\left[a,c\right]$
+\end_inset
+
+ in 
+\begin_inset Formula $\left[c,b\right]$
+\end_inset
+
+ in velja 
+\begin_inset Formula $\int_{a}^{b}f\left(x\right)dx=\int_{a}^{c}f\left(x\right)dx+\int_{c}^{b}f\left(x\right)dx$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Theorem*
+Če sta 
+\begin_inset Formula $f,g$
+\end_inset
+
+ na 
+\begin_inset Formula $J$
+\end_inset
+
+ integrabilni funkciji in če je 
+\begin_inset Formula $\forall x\in J:f\left(x\right)\leq g\left(x\right)$
+\end_inset
+
+,
+ tedaj 
+\begin_inset Formula $\int_{a}^{b}f\left(x\right)dx\leq\int_{a}^{b}f\left(x\right)dx$
+\end_inset
+
+.
+ Posledično velja ob isti predpostavki 
+\begin_inset Formula $\left|\int_{a}^{b}f\left(x\right)dx\right|\leq\int_{a}^{b}\left|f\left(x\right)\right|dx$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Če je 
+\begin_inset Formula $f$
+\end_inset
+
+ integrabilna na 
+\begin_inset Formula $J=\left[a,b\right]$
+\end_inset
+
+,
+ definiramo povprečje 
+\begin_inset Formula $f$
+\end_inset
+
+ na 
+\begin_inset Formula $J$
+\end_inset
+
+ s predpisom 
+\begin_inset Formula 
+\[
+\left\langle f\right\rangle _{J}\coloneqq\frac{\int_{a}^{b}f\left(x\right)dx}{b-a}\in\mathbb{R}.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Theorem*
+Velja 
+\begin_inset Formula $\inf_{x\in J}f\left(x\right)\leq\left\langle f\right\rangle _{J}\leq\sup_{x\in J}f\left(x\right)$
+\end_inset
+
+.
+ Če je 
+\begin_inset Formula $f:J\to\mathbb{R}$
+\end_inset
+
+ zvezna,
+ 
+\begin_inset Formula $\exists\xi\in J\ni:f\left(\xi\right)=\left\langle f\right\rangle _{J}$
+\end_inset
+
+ (izrek o vmesni vrednosti).
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $f:J\to\mathbb{R}$
+\end_inset
+
+ dana funkcija.
+ Nedoločeni integral 
+\begin_inset Formula $f$
+\end_inset
+
+ je takšna funkcija 
+\begin_inset Formula $F$
+\end_inset
+
+,
+ če obstaja,
+ 
+\begin_inset Formula $\ni:F'=f\sim\forall x\in J:F'\left(x\right)=f\left(x\right)$
+\end_inset
+
+.
+ Pišemo tudi 
+\begin_inset Formula $Pf$
+\end_inset
+
+ ali 
+\begin_inset Formula $\mathbb{P}f$
+\end_inset
+
+ in pravimo,
+ da je 
+\begin_inset Formula $F=Pf$
+\end_inset
+
+ primitivna funkcija za 
+\begin_inset Formula $f$
+\end_inset
+
+.
+ Velja 
+\begin_inset Formula $P\left(f+g\right)=Pf+Pg$
+\end_inset
+
+ (aditivnost odvoda) in 
+\begin_inset Formula $P\left(\lambda f\right)=\lambda Pf$
+\end_inset
+
+ (homogenost odvoda).
+\end_layout
+
+\begin_layout Definition*
+Nedoločeni integral je na intervalu določen do aditivne konstante natančno.
+ Če je 
+\begin_inset Formula $F'_{1}=f=F_{2}'$
+\end_inset
+
+ na intervalu 
+\begin_inset Formula $J$
+\end_inset
+
+ oziroma če na 
+\begin_inset Formula $J$
+\end_inset
+
+ velja 
+\begin_inset Formula $\left(F_{1}-F_{2}\right)'=0$
+\end_inset
+
+,
+ potem 
+\begin_inset Formula $F_{1}-F_{2}=c$
+\end_inset
+
+ oziroma 
+\begin_inset Formula $F_{1}=F_{2}+c$
+\end_inset
+
+ za neko konstanto 
+\begin_inset Formula $c\in\mathbb{R}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Označimo 
+\begin_inset Formula $F\left(x\right)=Pf\left(x\right)=\int f\left(x\right)dx$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+Integracija po delih 
+\begin_inset Formula $\sim$
+\end_inset
+
+ per partes.
+ Velja 
+\begin_inset Formula $\int f\left(x\right)g'\left(x\right)dx=f\left(x\right)g\left(x\right)-\int f'\left(x\right)g\left(x\right)dx$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Izhaja iz odvoda produkza 
+\begin_inset Formula $\left(fg\right)'=f'g+fg'$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+Naj bo 
+\begin_inset Formula $f$
+\end_inset
+
+ integrabilna na 
+\begin_inset Formula $J$
+\end_inset
+
+.
+ Definirajmo 
+\begin_inset Formula $F\left(x\right)=\int_{a}^{x}f\left(t\right)dt$
+\end_inset
+
+.
+ Velja 
+\begin_inset Formula $\left|F\left(x_{1}\right)-F\left(x_{2}\right)\right|=$
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\left|\int_{a}^{x_{1}}f\left(t\right)dt-\int_{a}^{x_{2}}f\left(t\right)dt\right|=\left|\int_{a}^{x_{1}}f\left(t\right)dt+\int_{x_{2}}^{a}f\left(t\right)dt\right|=\left|\int_{x_{2}}^{x_{1}}f\left(t\right)dt\right|=\left|\int_{x_{1}}^{x_{2}}f\left(t\right)dt\right|\leq\int_{x_{1}}^{x_{2}}f\left(t\right)dt
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Theorem*
+Osnovni izrek analize/fundamental theorem of calcusus.
+ Naj bo 
+\begin_inset Formula $f:\left[a,b\right]\to\mathbb{R}$
+\end_inset
+
+ zvezna in 
+\begin_inset Formula $F\left(x\right)=\int_{a}^{x}f\left(t\right)dt$
+\end_inset
+
+.
+ Tedaj je 
+\begin_inset Formula $F$
+\end_inset
+
+ odvedljiva na 
+\begin_inset Formula $J$
+\end_inset
+
+ in velja 
+\begin_inset Formula $F'\left(x\right)=f\left(x\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+\begin_inset Formula 
+\[
+F\left(x+h\right)-F\left(x\right)=\int_{x}^{x+h}f\left(t\right)dt\quad\quad\quad\quad/:h
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\frac{F\left(x+h\right)-F\left(x\right)}{h}=\frac{\int_{x}^{x+h}f\left(t\right)dt}{h}=\left\langle f\right\rangle _{\left[x,x+h\right]}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+F'\left(x\right)=\lim_{h\to0}\left\langle f\right\rangle _{x,x+h}=f\left(x\right).
+\]
+
+\end_inset
+
+
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+glej ANA1P FMF 2024-01-15.pdf/str.
+ 5 za dokaz,
+ ki ga ne razumem,
+ zakaj je 
+\begin_inset Formula $\lim_{h\to0}\left\langle f\right\rangle _{\left[x,x+h\right]}-f\left(x\right)=0$
+\end_inset
+
+...
+ ampak sej to je nekak očitno
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Corollary*
+Naj bo 
+\begin_inset Formula $f:\left[a,b\right]\to\mathbb{R}$
+\end_inset
+
+ zvezna in 
+\begin_inset Formula $G=Pf$
+\end_inset
+
+ (
+\begin_inset Formula $G'=f$
+\end_inset
+
+).
+ Tedaj je 
+\begin_inset Formula $\int_{a}^{b}f\left(x\right)dx=G\left(b\right)-G\left(a\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $F\left(x\right)=\int_{a}^{x}f\left(t\right)dt$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $F'=f=G'$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $\left(F-G\right)'=0\Rightarrow F-G=c\in\mathbb{R}$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $G\left(x\right)=F\left(x\right)+c$
+\end_inset
+
+,
+ sledi 
+\begin_inset Formula $G\left(a\right)=F\left(a\right)=0$
+\end_inset
+
+ po definiciji 
+\begin_inset Formula $F$
+\end_inset
+
+,
+ torej je 
+\begin_inset Formula $G\left(a\right)=c$
+\end_inset
+
+.
+ Sledi 
+\begin_inset Formula $F\left(x\right)=G\left(x\right)-G\left(a\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $F\left(b\right)=G\left(b\right)-G\left(a\right)$
+\end_inset
+
+ in zato 
+\begin_inset Formula $F\left(b\right)=\int_{a}^{b}f\left(t\right)dt$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsection
+Iskanje primitivne funkcije
+\end_layout
+
+\begin_layout Itemize
+Uganemo jo
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $P\left(x^{n}\right)=\frac{x^{n+1}}{n+1}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $P\left(e^{x}\right)=e^{x}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $P\left(\sin x\right)=-\cos x$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $P\left(\ln x\right)=x\left(\ln x-1\right)$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Theorem*
+Substitucija/uvedba nove spremenljivke
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+ne razumem.
+ mogoče bom v naslednjem življenju.
+\end_layout
+
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $F\left(x\right)$
+\end_inset
+
+ nedoločeni integral funkcije 
+\begin_inset Formula $f\left(x\right)$
+\end_inset
+
+ ter 
+\begin_inset Formula $\phi\left(x\right)$
+\end_inset
+
+ odvedljiva funkcija.
+ Potem velja
+\begin_inset Formula 
+\[
+F\left(\phi\left(t\right)\right)=\int f\left(\phi\left(t\right)\right)\phi'\left(t\right)dx
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Proof
+Formula je posledica odvoda kompozituma:
+\begin_inset Formula 
+\[
+\left(F\left(\phi\left(t\right)\right)\right)'=F'\left(\phi\left(t\right)\right)\phi'\left(t\right)=f\left(\phi\left(t\right)\right)\phi'\left(t\right)
+\]
+
+\end_inset
+
+integrirajmo levo in desno stran:
+\begin_inset Formula 
+\[
+\int\left(F\left(\phi\left(t\right)\right)\right)'dt=\int f\left(\phi\left(t\right)\right)\phi'\left(t\right)dt.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsection
+Izlimitirani integrali
+\end_layout
+
+\begin_layout Standard
+Doslej smo računali določene integrale omejene funkcije na omejenem intervalu,
+ torej 
+\begin_inset Formula $\int_{a}^{b}f\left(x\right)dx$
+\end_inset
+
+.
+ Kaj pa neomejen interval,
+ torej 
+\begin_inset Formula $\lim_{b\to\infty}\int_{a}^{b}f\left(x\right)dx$
+\end_inset
+
+?
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $f:[a,\infty)\to\mathbb{R}$
+\end_inset
+
+ in naj bo 
+\begin_inset Formula $\forall m>a:f$
+\end_inset
+
+ integrabilna na 
+\begin_inset Formula $\left[a,-m\right]$
+\end_inset
+
+.
+ Če 
+\begin_inset Formula $\exists\lim_{m\to\infty}\int_{a}^{m}f\left(x\right)dx$
+\end_inset
+
+,
+ pracimo,
+ da integral 
+\begin_inset Formula $\int_{a}^{\infty}f\left(x\right)dx$
+\end_inset
+
+ konvergira,
+ sicer pa divergira.
+ Označimo 
+\begin_inset Formula $\int_{a}^{\infty}f\left(x\right)dx\coloneqq\lim_{m\to\infty}\int_{a}^{m}f\left(x\right)dx$
+\end_inset
+
+.
+ Podobno definiramo 
+\begin_inset Formula $\int_{-\infty}^{a}f\left(x\right)dx$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Pomemben primer.
+ 
+\begin_inset Formula $\int_{1}^{\infty}x^{\alpha}dx=?$
+\end_inset
+
+.
+ 
+\begin_inset Formula $\int_{1}^{M}x^{\alpha}dx=\frac{M^{\alpha+1}}{\alpha+1}-\frac{1}{\alpha+1}=\frac{M^{\alpha+1}-1}{\alpha+1}$
+\end_inset
+
+.
+ Torej 
+\begin_inset Formula $\exists\lim_{M\to\infty}\int_{1}^{M}x^{\alpha}dx\Leftrightarrow\alpha\not=-1$
+\end_inset
+
+.
+ Poglejmo,
+ kaj se zgodi v 
+\begin_inset Formula $\alpha=-1$
+\end_inset
+
+:
+ 
+\begin_inset Formula $\int_{1}^{\infty}x^{-1}dx=\ln M-\ln1=\ln M$
+\end_inset
+
+.
+ Toda 
+\begin_inset Formula $\lim_{n\to\infty}\ln M=\infty$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\int_{1}^{\infty}x^{-1}dx$
+\end_inset
+
+ divergira.
+\end_layout
+
+\begin_layout Definition*
+\begin_inset Formula $\int_{a}^{\infty}f\left(x\right)dx$
+\end_inset
+
+ je absolutno konvergenten,
+ če je 
+\begin_inset Formula $\int_{a}^{\infty}\left|f\left(x\right)\right|dx<\infty$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Fact*
+Velja 
+\begin_inset Formula $\int_{a}^{\infty}\left|f\left(x\right)\right|dx<0\Rightarrow\int_{a}^{\infty}f\left(x\right)dx<\infty$
+\end_inset
+
+.
+ Velja 
+\begin_inset Formula $\left|\int_{a}^{\infty}f\left(x\right)dx\right|\leq\int_{a}^{\infty}\left|f\left(x\right)\right|dx$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Ali je predpostavka,
+ da je 
+\begin_inset Formula $f$
+\end_inset
+
+ omejena,
+ sploh potrebna?
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $f:[a,b)\to\mathbb{R}\ni:\forall c<b:f$
+\end_inset
+
+ integrabilna na 
+\begin_inset Formula $\left[a,c\right]$
+\end_inset
+
+.
+ V točki 
+\begin_inset Formula $b$
+\end_inset
+
+ je 
+\begin_inset Formula $f$
+\end_inset
+
+ lahko neomejena.
+ Če 
+\begin_inset Formula $\exists$
+\end_inset
+
+ končna limita 
+\begin_inset Formula $\lim_{c\to b}\int_{a}^{c}f\left(x\right)dx$
+\end_inset
+
+,
+ je integral 
+\begin_inset Formula $\int_{a}^{b}f\left(x\right)dx$
+\end_inset
+
+ konvergenten,
+ sicer je divergenten.
+ Podobno definiramo,
+ če je funkcija definirana na intervalu 
+\begin_inset Formula $(a,b]$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $\int_{0}^{1}x^{\alpha}dx$
+\end_inset
+
+.
+ Za 
+\begin_inset Formula $\alpha<0$
+\end_inset
+
+ ima graf 
+\begin_inset Formula $x^{\alpha}$
+\end_inset
+
+ v 
+\begin_inset Formula $x=0$
+\end_inset
+
+ pol.
+ Računajmo
+\begin_inset Formula 
+\[
+\lim_{\varepsilon\to0}\int_{\varepsilon}^{1}x^{\alpha}dx=\lim_{\varepsilon\to0}\frac{x^{\alpha+1}}{\alpha+1}\vert_{\varepsilon}^{1}=\lim_{\varepsilon\to0}\left(\frac{1}{\alpha+1}-\frac{\varepsilon^{\alpha+1}}{\alpha+1}\right)=\lim_{\varepsilon\to0}\frac{1-\varepsilon^{\alpha+1}}{\alpha+1}=\lim_{\varepsilon\to0}\frac{1-\cancelto{0}{e^{\left(\alpha+1\right)\ln\varepsilon}}}{\alpha+1}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+Pridobimo pogoj 
+\begin_inset Formula $\alpha\not=-1$
+\end_inset
+
+ (imenovalec) in 
+\begin_inset Formula $\alpha+1>0$
+\end_inset
+
+ (da bo 
+\begin_inset Formula $\left(\alpha+1\right)\ln\varepsilon\to-\infty$
+\end_inset
+
+),
+ torej skupaj s predpostavko 
+\begin_inset Formula $\alpha\in\left(-1,0\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Torej 
+\begin_inset Formula $\int_{0}^{1}x^{\alpha}dx=\frac{1}{\alpha+1}$
+\end_inset
+
+ za 
+\begin_inset Formula $\alpha\in\left(-1,0\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsection
+Uporaba integrala
+\end_layout
+
+\begin_layout Itemize
+Ploščine:
+ 
+\begin_inset Formula $f\geq0$
+\end_inset
+
+ na 
+\begin_inset Formula $J=\left[a,b\right]$
+\end_inset
+
+ in je 
+\begin_inset Formula $f\in I\left(J\right)$
+\end_inset
+
+,
+ je ploščina lika med 
+\begin_inset Formula $x$
+\end_inset
+
+ osjo in grafom 
+\begin_inset Formula $f$
+\end_inset
+
+ definirana kot 
+\begin_inset Formula $\int_{a}^{b}f\left(x\right)dx$
+\end_inset
+
+.
+ Če 
+\begin_inset Formula $f$
+\end_inset
+
+ ni pozitivna,
+ pa je 
+\begin_inset Formula $\int_{a}^{b}f\left(x\right)dx=pl\left(L_{1}\right)-pl\left(L_{2}\right)$
+\end_inset
+
+,
+ kjer je 
+\begin_inset Formula $L_{1}$
+\end_inset
+
+ lik nad 
+\begin_inset Formula $x$
+\end_inset
+
+ osjo in 
+\begin_inset Formula $L_{2}$
+\end_inset
+
+ lik pod 
+\begin_inset Formula $x$
+\end_inset
+
+ osjo.
+\end_layout
+
+\begin_layout Example*
+Ploščina kroga:
+ Enačba krožnice je 
+\begin_inset Formula $x^{2}+y^{2}=r^{2}$
+\end_inset
+
+ za 
+\begin_inset Formula $r>0$
+\end_inset
+
+.
+ 
+\begin_inset Formula $y=\sqrt{r^{2}-x^{2}}$
+\end_inset
+
+.
+ Ploščina kroga z radijem 
+\begin_inset Formula $r$
+\end_inset
+
+ je torej 
+\begin_inset Formula $2\int_{-r}^{r}\sqrt{r^{2}-x^{2}}dx=\cdots=\pi r^{2}$
+\end_inset
+
+.
+\end_layout
+
+\end_body
+\end_document
diff --git a/šola/ana1/teor3.lyx b/šola/ana1/teor3.lyx
new file mode 100644
index 0000000..97befd1
--- /dev/null
+++ b/šola/ana1/teor3.lyx
@@ -0,0 +1,1238 @@
+#LyX 2.3 created this file. For more info see http://www.lyx.org/
+\lyxformat 544
+\begin_document
+\begin_header
+\save_transient_properties true
+\origin unavailable
+\textclass article
+\use_default_options true
+\maintain_unincluded_children false
+\language slovene
+\language_package default
+\inputencoding utf8
+\fontencoding global
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+\use_hyperref false
+\papersize default
+\use_geometry false
+\use_package amsmath 1
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+\cite_engine basic
+\cite_engine_type default
+\use_bibtopic false
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+\paperorientation portrait
+\suppress_date false
+\justification true
+\use_refstyle 1
+\use_minted 0
+\index Index
+\shortcut idx
+\color #008000
+\end_index
+\secnumdepth 3
+\tocdepth 3
+\paragraph_separation indent
+\paragraph_indentation default
+\is_math_indent 0
+\math_numbering_side default
+\quotes_style english
+\dynamic_quotes 0
+\papercolumns 1
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+\paperpagestyle default
+\tracking_changes false
+\output_changes false
+\html_math_output 0
+\html_css_as_file 0
+\html_be_strict false
+\end_header
+
+\begin_body
+
+\begin_layout Title
+Rešen tretji izpit teorije Analize 1 — IŠRM 2023/24
+\end_layout
+
+\begin_layout Abstract
+Izpit je potekal v petek, 30.
+ avgusta 2024 od desete
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+Avtor tega besedila je na izpit zamudil poldrugo uro.
+\end_layout
+
+\end_inset
+
+ do dvanajste ure.
+ Nosilec predmeta je 
+\noun on
+Oliver Dragičević
+\noun default
+.
+ Naloge in rešitve sem po spominu spisal 
+\noun on
+Anton Luka Šijanec
+\noun default
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\left[15\right]$
+\end_inset
+
+
+\begin_inset Newline newline
+\end_inset
+
+Podaj natančne definicije naslednjih pojmov:
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+limita zaporedja, stekališče zaporedja
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+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
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $L$
+\end_inset
+
+ je limita 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}\sim L=\lim_{n\to\infty}a_{n}\Leftrightarrow\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\forall n>n_{0}:\left|a_{n}-L\right|<\varepsilon$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $L$
+\end_inset
+
+ je stekališče 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}\Leftrightarrow\forall\varepsilon>0\exists\mathcal{A}\subseteq\mathbb{N},\left|\mathcal{A}\right|=\left|\mathcal{\mathbb{N}}\right|\ni:\left\{ a_{n};n\in\mathcal{A}\right\} \subseteq\left(L-\varepsilon,L+\varepsilon\right)$
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+vsota (neskončne) konvergentne vrste
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Naj bo 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ poljubno zaporedje.
+ 
+\begin_inset Formula $\sum_{n=1}^{\infty}a_{n}\coloneqq\lim_{n\to\infty}\sum_{k=1}^{n}a_{n}$
+\end_inset
+
+.
+ Če limita obstaja, je vrsta 
+\begin_inset Formula $\sum_{n=1}^{\infty}a_{n}$
+\end_inset
+
+ konvergentna in njena vsota je enaka tej limiti.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Cauchyjev pogoj za zaporedja
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Naj bo 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ realno zaporedje.
+ Konvergentno je natanko tedaj, ko ustreza Cauchyjevemu pogoju: 
+\begin_inset Formula $\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\forall m,n\geq n_{0}:\left|a_{n}-a_{m}\right|<\varepsilon$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+odprte, zaprte, omejene, kompaktne množice v 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+Množica 
+\begin_inset Formula $\mathcal{A}$
+\end_inset
+
+ je odprta, ko 
+\begin_inset Formula $\forall a\in\mathcal{A}\exists\varepsilon>0\ni:\left(a-\varepsilon,a+\varepsilon\right)\subseteq\mathcal{A}$
+\end_inset
+
+, ko za vsako točko množice obstaja neka njena okolica, ki je podmnožica
+ množice 
+\begin_inset Formula $\mathcal{A}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Množica 
+\begin_inset Formula $\mathcal{A}$
+\end_inset
+
+ je zaprta, ko je 
+\begin_inset Formula $\mathcal{A}^{\mathcal{C}}\coloneqq\mathbb{R}\setminus\mathcal{A}$
+\end_inset
+
+ odprta.
+\end_layout
+
+\begin_layout Enumerate
+Množica 
+\begin_inset Formula $\mathcal{A}$
+\end_inset
+
+ je omejena, ko 
+\begin_inset Formula $\exists m,M\in\mathbb{R}\forall a\in\mathcal{A}:a\leq M\wedge a\geq m$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Množica 
+\begin_inset Formula $\mathcal{A}$
+\end_inset
+
+ je kompaktna 
+\begin_inset Formula $\Leftrightarrow\mathcal{A}$
+\end_inset
+
+ zaprta 
+\begin_inset Formula $\wedge$
+\end_inset
+
+ 
+\begin_inset Formula $\mathcal{A}$
+\end_inset
+
+ omejena.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+limita funkcije v dani točki
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Naj bodo 
+\begin_inset Formula $a\in\mathbb{R}$
+\end_inset
+
+, 
+\begin_inset Formula $\mathcal{D}$
+\end_inset
+
+ okolica 
+\begin_inset Formula $a$
+\end_inset
+
+ in 
+\begin_inset Formula $f:\mathcal{D}\setminus\left\{ a\right\} \to\mathbb{R}$
+\end_inset
+
+ poljubne.
+ 
+\begin_inset Formula $L\in\mathbb{R}$
+\end_inset
+
+ je limita 
+\begin_inset Formula $f$
+\end_inset
+
+ v točki 
+\begin_inset Formula $a\sim L=\lim_{x\to a}f\left(x\right)\Leftrightarrow\forall\varepsilon>0\exists\delta>0\forall x\in\mathcal{D}\setminus\left\{ a\right\} :\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-L\right|<\varepsilon$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+zveznost funkcije
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Naj bodo 
+\begin_inset Formula $\mathcal{D}\subseteq\mathbb{R}$
+\end_inset
+
+, 
+\begin_inset Formula $a\in\mathcal{D}$
+\end_inset
+
+ in 
+\begin_inset Formula $f:\mathcal{D}\to\mathbb{R}$
+\end_inset
+
+ poljubne.
+ 
+\begin_inset Formula $f$
+\end_inset
+
+ je zvezna v 
+\begin_inset Formula $a\Leftrightarrow\forall\varepsilon>0\exists\delta>0\forall x\in\mathcal{D}:\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(a\right)\right|<\varepsilon$
+\end_inset
+
+ .
+ 
+\begin_inset Formula $f$
+\end_inset
+
+ je zvezna na množici 
+\begin_inset Formula $\mathcal{A}$
+\end_inset
+
+, če je zvezna na vsaki točki množice 
+\begin_inset Formula $\mathcal{A}$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+odvedljivost funkcije
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Naj bodo 
+\begin_inset Formula $a\in\mathbb{R}$
+\end_inset
+
+, 
+\begin_inset Formula $\mathcal{D}\subseteq\mathbb{R}$
+\end_inset
+
+, 
+\begin_inset Formula $f:\mathcal{D}\to\mathbb{R}$
+\end_inset
+
+ poljubne.
+ 
+\begin_inset Formula $f$
+\end_inset
+
+ je odvedljiva v 
+\begin_inset Formula $a\text{\ensuremath{\Leftrightarrow\lim_{h\to0}\frac{f\left(a+h\right)-f\left(a\right)}{h}}}\in\mathbb{R}$
+\end_inset
+
+, ZDB ko obstaja slednja limita.
+ Tedaj definiramo 
+\begin_inset Quotes eld
+\end_inset
+
+odvod funkcije 
+\begin_inset Formula $f$
+\end_inset
+
+ v točki 
+\begin_inset Formula $a$
+\end_inset
+
+
+\begin_inset Quotes erd
+\end_inset
+
+: 
+\begin_inset Formula $f'\left(a\right)=\lim_{h\to0}\frac{f\left(a+h\right)-f\left(a\right)}{h}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $f$
+\end_inset
+
+ je odvedljiva na množici 
+\begin_inset Formula $\mathcal{A}$
+\end_inset
+
+, če je odvedljiva na vsaki točki množice 
+\begin_inset Formula $\mathcal{A}$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+določen integral realne funkcije na zaprtem omejenem intervalu.
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+Naj bodo 
+\begin_inset Formula $a,b\in\mathbb{R}$
+\end_inset
+
+ in 
+\begin_inset Formula $f:\left[a,b\right]\to\mathbb{R}$
+\end_inset
+
+ poljubne.
+\end_layout
+
+\begin_layout Enumerate
+Definirajmo pojem delitve 
+\begin_inset Formula $\left[a,b\right]$
+\end_inset
+
+.
+ Delitev so točke 
+\begin_inset Formula $t_{0},\dots,t_{n}$
+\end_inset
+
+, da velja 
+\begin_inset Formula $a=t_{0}<t_{1}<\cdots<t_{n}=b$
+\end_inset
+
+ za nek 
+\begin_inset Formula $n\in\mathbb{N}$
+\end_inset
+
+.
+ Točke identificiramo z delilnimi intervali takole: 
+\begin_inset Formula $D_{n}=\left[t_{n-1},t_{n}\right]$
+\end_inset
+
+.
+ Delitev torej identificiramo z množico teh dedlilnih intervalov: 
+\begin_inset Formula $D=\left\{ D_{k};\forall k\in\left\{ 1..n\right\} \right\} $
+\end_inset
+
+.
+ Definiramo tudi velikost delitve: 
+\begin_inset Formula $\left|D_{\infty}\right|=\max_{k\in\left\{ 1..n\right\} }\left|D_{k}\right|$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Definirajmo pojem izbire za dano delitev.
+ Naj bo 
+\begin_inset Formula $D$
+\end_inset
+
+ delitev.
+ Pripadajoča izbira so take izbirne točke 
+\begin_inset Formula $\xi_{1},\dots,\xi_{n}$
+\end_inset
+
+, da velja 
+\begin_inset Formula $\forall k\in\left\{ 1..n\right\} :\xi_{k}\in D_{k}$
+\end_inset
+
+.
+ Množico teh izbirnih točk označimo z 
+\begin_inset Formula $\xi\coloneqq\left\{ \xi_{k};\forall k\in\left\{ 1..n\right\} \right\} $
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $f$
+\end_inset
+
+ je integrabilna na 
+\begin_inset Formula $\left[a,b\right]$
+\end_inset
+
+, če 
+\begin_inset Formula $\exists I\in\mathbb{R}\forall\varepsilon>0\exists\delta>0\forall$
+\end_inset
+
+ delitev 
+\begin_inset Formula $D\forall$
+\end_inset
+
+ izbiro 
+\begin_inset Formula $\xi$
+\end_inset
+
+, pripadajočo delitvi 
+\begin_inset Formula $D:\left|D_{\infty}\right|<\delta\Rightarrow\left|\sum_{k=1}^{n}\left|D_{k}\right|f\left(\xi\right)-I\right|<\varepsilon$
+\end_inset
+
+.
+ Tedaj pravimo, da je 
+\begin_inset Formula $I$
+\end_inset
+
+ določen integral 
+\begin_inset Formula $f$
+\end_inset
+
+ na 
+\begin_inset Formula $\left[a,b\right]$
+\end_inset
+
+ in pišemo 
+\begin_inset Formula $I\eqqcolon\int_{a}^{b}f\left(x\right)dx$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\end_deeper
+\begin_layout Enumerate
+\begin_inset Formula $\left[15\right]$
+\end_inset
+
+
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+Pojasni princip matematične indukcije.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Naj bo 
+\begin_inset Formula $\left(P_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ zaporedje logičnih vrednosti/izjav/izrazov.
+ Če velja
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $P_{1}$
+\end_inset
+
+ drži in hkrati
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\forall n\in\mathbb{N}:P_{n}$
+\end_inset
+
+ drži 
+\begin_inset Formula $\Rightarrow P_{n+1}$
+\end_inset
+
+ drži,
+\end_layout
+
+\begin_layout Standard
+potem velja 
+\begin_inset Formula $\forall n\in\mathbb{N}:P_{n}$
+\end_inset
+
+ drži.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Z matematično indukcijo dokaži
+\begin_inset Formula 
+\[
+\forall n\in\mathbb{N}:1+2+\cdots+n=\frac{n\left(n+1\right)}{2}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+Baza 
+\begin_inset Formula $n=1$
+\end_inset
+
+: 
+\begin_inset Formula $1=\frac{1\left(1+1\right)}{2}$
+\end_inset
+
+ Velja.
+\end_layout
+
+\begin_layout Enumerate
+Indukcijska predpostavka: 
+\begin_inset Formula $1+2+\cdots+n=\frac{n\left(n+1\right)}{2}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Korak 
+\begin_inset Formula $n\to n+1$
+\end_inset
+
+:
+\begin_inset Formula 
+\[
+1+2+\cdots+n+\cancel{n+1}\overset{?}{=}\frac{\left(n+1\right)\left(n+1+1\right)}{2}=\frac{n^{2}+2n+n+2}{2}=\frac{n\left(n+1\right)}{2}+\cancel{n+1}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+1+2+\cdots+n\overset{\text{I.P.}}{=}\frac{n\left(n+1\right)}{2}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+Sklep: 
+\begin_inset Formula $\forall n\in\mathbb{N}:1+2+\cdots+n=\frac{n\left(n+1\right)}{2}$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\end_deeper
+\begin_layout Enumerate
+\begin_inset Formula $\left[25\right]$
+\end_inset
+
+
+\begin_inset Newline newline
+\end_inset
+
+Naj bosta 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ in 
+\begin_inset Formula $\left(b_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ realni konvergentni zaporedji.
+ Dokaži, da je 
+\begin_inset Formula $c_{n}\coloneqq a_{n}b_{n}$
+\end_inset
+
+ prav tako konvergentno zaporedje.
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Označimo 
+\begin_inset Formula $\lim_{n\to\infty}a_{n}\eqqcolon A$
+\end_inset
+
+ in 
+\begin_inset Formula $\lim_{n\to\infty}b_{n}\eqqcolon B$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Uganemo, da je 
+\begin_inset Formula $\lim_{n\to\infty}a_{n}b_{n}=AB$
+\end_inset
+
+.
+ To moramo sedaj dokazati.
+\end_layout
+
+\begin_layout Itemize
+Dokazujemo, da 
+\begin_inset Formula $\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\forall n\geq n_{0}:\left|a_{n}b_{n}-AB\right|<\varepsilon\sim\left|a_{n}b_{n}+a_{n}B-a_{n}B-AB\right|=\left|a_{n}\left(b_{n}-B\right)+B\left(a_{n}-A\right)\right|<\varepsilon$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+Ker po trikotniški neenakosti velja 
+\begin_inset Formula $\left|a_{n}\left(b_{n}-B\right)+B\left(a_{n}-A\right)\right|\leq\left|a_{n}\right|\left|b_{n}-B\right|+\left|B\right|\left|a_{n}-A\right|$
+\end_inset
+
+, je dovolj za poljuben 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+ dokazati
+\begin_inset Formula 
+\[
+\exists n_{0}\in\mathbb{N}\forall n\geq n_{0}:\left|a_{n}\right|\left|b_{n}-B\right|+\left|B\right|\left|a_{n}-A\right|<\varepsilon
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+Ker je 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ konvergentno, 
+\begin_inset Formula $\exists n_{1}\in\mathbb{N}\forall n\geq n_{1}:\left|a_{n}-A\right|<\frac{\varepsilon}{2\left|a\right|}$
+\end_inset
+
+, kjer je 
+\begin_inset Formula $a$
+\end_inset
+
+ zgornja meja zaporedja 
+\begin_inset Formula $a_{n}$
+\end_inset
+
+.
+ Slednje je omejeno, ker je konvergentno.
+\end_layout
+
+\begin_layout Itemize
+Ker je 
+\begin_inset Formula $\left(b_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ konvergentno, 
+\begin_inset Formula $\exists n_{2}\in\mathbb{N}\forall n\geq n_{1}:\left|b_{n}-B\right|<\frac{\varepsilon}{2\left|B\right|}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Tedaj za 
+\begin_inset Formula $n_{0}\coloneqq\max\left\{ n_{1},n_{2}\right\} $
+\end_inset
+
+ velja
+\begin_inset Formula 
+\[
+\left|a_{n}\right|\left|b_{n}-B\right|+\left|B\right|\left|a_{n}-A\right|<\frac{\varepsilon\left|a\right|}{2\left|a_{n}\right|}+\frac{\varepsilon}{2}\leq\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon
+\]
+
+\end_inset
+
+in izrek je dokazan.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+\begin_inset Formula $\left[?\right]$
+\end_inset
+
+
+\begin_inset Newline newline
+\end_inset
+
+Dokaži, da je zvezna realna funkcija na zaprtem intervalu omejena.
+ Natančno navedi vse izreke, ki jih pri tem dokazu uporabiš.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Naj bodo 
+\begin_inset Formula $a,b\in\mathbb{R}$
+\end_inset
+
+ in zvezna 
+\begin_inset Formula $f:\left[a,b\right]\to\mathbb{R}$
+\end_inset
+
+ poljubne.
+\end_layout
+
+\begin_layout Itemize
+Dokaz, da je 
+\begin_inset Formula $f$
+\end_inset
+
+ omejena navzgor.
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+PDDRAA 
+\begin_inset Formula $f$
+\end_inset
+
+ ni navzgor omejena.
+ Tedaj 
+\begin_inset Formula $\forall n\in\mathbb{N}\exists x_{n}\in\left[a,b\right]\ni:f\left(x_{n}\right)\geq n$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Ker je 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ na zaprti množici, je omejeno zaporedje, torej ima stekališče.
+ Recimo mu 
+\begin_inset Formula $s\in\mathbb{R}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Ker je 
+\begin_inset Formula $\left[a,b\right]$
+\end_inset
+
+ zaprta, je 
+\begin_inset Formula $s\in\left[a,b\right]$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Ker je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna na 
+\begin_inset Formula $\left[a,b\right]$
+\end_inset
+
+ in s tem v 
+\begin_inset Formula $s$
+\end_inset
+
+, velja 
+\begin_inset Formula $\lim_{n\to\infty}f\left(x_{n}\right)=f\left(s\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Po konstrukciji 
+\begin_inset Formula $\left(x_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ velja 
+\begin_inset Formula $\lim_{n\to\infty}f\left(x_{n}\right)=\infty$
+\end_inset
+
+, torej 
+\begin_inset Formula $f\left(s\right)=\infty$
+\end_inset
+
+, kar ni mogoče, saj 
+\begin_inset Formula $f\left(s\right)\in\mathbb{R}$
+\end_inset
+
+ po predpostavki.
+ 
+\begin_inset Formula $\rightarrow\!\leftarrow$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Predpostavka 
+\begin_inset Quotes eld
+\end_inset
+
+
+\begin_inset Formula $f$
+\end_inset
+
+ ni navzgor omejena
+\begin_inset Quotes erd
+\end_inset
+
+ ne velja, torej smo dokazali, da je 
+\begin_inset Formula $f$
+\end_inset
+
+ navzgor omejena.
+\end_layout
+
+\end_deeper
+\begin_layout Itemize
+Dokaz, da je 
+\begin_inset Formula $f$
+\end_inset
+
+ omejena navzdol.
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+PDDRAA 
+\begin_inset Formula $f$
+\end_inset
+
+ ni navzdol omejena.
+ Tedaj 
+\begin_inset Formula $\forall n\in\mathbb{N}\exists x_{n}\in\left[a,b\right]\ni:f\left(x_{n}\right)\leq-n$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Ker je 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{\mathbb{N}}}$
+\end_inset
+
+ na zaprti množici, je omejeno zaporedje, torej ima stekališče.
+ Recimo mu 
+\begin_inset Formula $s\in\mathbb{R}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Ker je 
+\begin_inset Formula $\left[a,b\right]$
+\end_inset
+
+ zaprta, je 
+\begin_inset Formula $s\in\left[a,b\right]$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Ker je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna na 
+\begin_inset Formula $\left[a,b\right]$
+\end_inset
+
+ in s tem v 
+\begin_inset Formula $s$
+\end_inset
+
+, velja 
+\begin_inset Formula $\lim_{n\to\infty}f\left(x_{n}\right)=f\left(s\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Po konstrukciji 
+\begin_inset Formula $\left(x_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ velja 
+\begin_inset Formula $\lim_{n\to\infty}f\left(x_{n}\right)=-\infty$
+\end_inset
+
+, torej 
+\begin_inset Formula $f\left(s\right)=-\infty$
+\end_inset
+
+, kar ni mogoče, saj 
+\begin_inset Formula $f\left(s\right)\in\mathbb{R}$
+\end_inset
+
+ po predpostavki.
+ 
+\begin_inset Formula $\rightarrow\!\leftarrow$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Predpostavka 
+\begin_inset Quotes eld
+\end_inset
+
+
+\begin_inset Formula $f$
+\end_inset
+
+ ni navzdol omejena
+\begin_inset Quotes erd
+\end_inset
+
+ ne velja, torej smo dokazali, da je 
+\begin_inset Formula $f$
+\end_inset
+
+ navzdol omejena.
+\end_layout
+
+\end_deeper
+\begin_layout Itemize
+Ker je 
+\begin_inset Formula $f$
+\end_inset
+
+ omejena navzgor in navzdol, je omejena.
+\end_layout
+
+\begin_layout Itemize
+Uporabljeni izreki.
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Zaporedje 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ s členi na kompaktni množici je omejeno.
+\end_layout
+
+\begin_layout Itemize
+Omejeno zaporedje ima stekališče.
+\end_layout
+
+\begin_layout Itemize
+Če je 
+\begin_inset Formula $s\in\mathbb{R}$
+\end_inset
+
+ stekališče zaporedja 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+, obstaja konvergentno podzaporedje 
+\begin_inset Formula $\left(a_{n_{k}}\right)_{k\in\mathbb{N}}$
+\end_inset
+
+, da je 
+\begin_inset Formula $s$
+\end_inset
+
+ njegova limita.
+\end_layout
+
+\begin_layout Itemize
+Množica je kompaktna natanko tedaj, ko vsebuje limite vseh konvergentnih
+ zaporedij s členi v njej.
+\end_layout
+
+\begin_layout Itemize
+Funkcija 
+\begin_inset Formula $f$
+\end_inset
+
+ je zvezna v 
+\begin_inset Formula $s$
+\end_inset
+
+, če za vsako k 
+\begin_inset Formula $s$
+\end_inset
+
+ konvergentno zaporedje velja, da njegovi s 
+\begin_inset Formula $f$
+\end_inset
+
+ preslikani členi konvergirajo v 
+\begin_inset Formula $f\left(s\right)$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\end_deeper
+\begin_layout Enumerate
+\begin_inset Formula $\left[?\right]$
+\end_inset
+
+
+\begin_inset Newline newline
+\end_inset
+
+Za realno funkcijo ene spremenljivke dokaži verižno pravilo.
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Naj bodo 
+\begin_inset Formula $\mathcal{D},\mathcal{E},\mathcal{F}\subseteq\mathbb{R}$
+\end_inset
+
+, 
+\begin_inset Formula $x\in\mathcal{D}$
+\end_inset
+
+ in 
+\begin_inset Formula $f:\mathcal{D}\to\mathcal{E}$
+\end_inset
+
+, 
+\begin_inset Formula $g:\mathcal{E}\to\mathcal{F}$
+\end_inset
+
+ poljubne.
+ 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
+
+.
+\end_layout
+
+\begin_layout Itemize
+Dokažimo, da je 
+\begin_inset Formula $g\circ f$
+\end_inset
+
+ odvedljiva v 
+\begin_inset Formula $x$
+\end_inset
+
+ in da velja 
+\begin_inset Formula 
+\[
+\left(g\circ f\right)'\left(x\right)=g'\left(f\left(x\right)\right)f'\left(x\right).
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+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
+
+.
+ Potemtakem 
+\begin_inset Formula $f\left(x+h\right)=\delta_{h}+a$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+\left(g\circ f\right)'\left(x\right)=\lim_{h\to0}\frac{g\left(f\left(x+h\right)\right)-g\left(f\left(x\right)\right)=g\left(\delta_{h}+a\right)-g\left(a\right)}{h}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\lim_{h\to0}\frac{g\left(\delta_{h}+a\right)-g\left(a\right)}{\delta_{h}}\cdot\frac{\delta_{h}}{h}=\lim_{h\to0}\frac{g\left(\delta_{h}+a\right)-g\left(a\right)}{\delta_{h}}\cdot\frac{f\left(x+h\right)-f\left(x\right)}{h}=\cdots
+\]
+
+\end_inset
+
+Ker je 
+\begin_inset Formula $f$
+\end_inset
+
+ v 
+\begin_inset Formula $x$
+\end_inset
+
+ odvedljiva, je v 
+\begin_inset Formula $x$
+\end_inset
+
+ zvezna, zato sledi 
+\begin_inset Formula $h\to0\Rightarrow\delta_{h}\to0$
+\end_inset
+
+.
+\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
+
+\end_deeper
+\end_body
+\end_document
diff --git a/šola/aps1/dn/osvetlitev/Makefile b/šola/aps1/dn/osvetlitev/Makefile
new file mode 100644
index 0000000..2d78386
--- /dev/null
+++ b/šola/aps1/dn/osvetlitev/Makefile
@@ -0,0 +1,4 @@
+program: resitev.cpp
+	g++ -Wall -Wextra -pedantic -Wformat-security -std=c++20 -o$@ $<
+clean:
+	rm program
diff --git a/šola/aps1/dn/osvetlitev/in.txt b/šola/aps1/dn/osvetlitev/in.txt
new file mode 100644
index 0000000..76b90fe
--- /dev/null
+++ b/šola/aps1/dn/osvetlitev/in.txt
@@ -0,0 +1,9 @@
+30
+7
+10 2
+23 2
+14 1
+4 1
+14 4
+11 5
+1 2
diff --git a/šola/aps1/dn/osvetlitev/resitev.cpp b/šola/aps1/dn/osvetlitev/resitev.cpp
new file mode 100644
index 0000000..0a17f31
--- /dev/null
+++ b/šola/aps1/dn/osvetlitev/resitev.cpp
@@ -0,0 +1,46 @@
+#include <stdio.h>
+#include <stdlib.h>
+#include <stdbool.h>
+struct event {
+	int pos;
+	bool tip; // true za začetek, false za konec
+};
+int compar_events (const void * a, const void * b) {
+	if (((struct event *) a)->pos == ((struct event *) b)->pos)
+		return 0;
+	if (((struct event *) a)->pos < ((struct event *) b)->pos)
+		return -1;
+	return 1;
+}
+int main (void) {
+	struct event events[20000];
+	int M, N, x, d;
+	scanf("%d %d", &M, &N);
+	for (int i = 0; i < N; i++) {
+		scanf("%d %d", &x, &d);
+		events[2*i].pos = x-d >= 0 ? x-d : 0;
+		events[2*i].tip = true;
+		events[2*i+1].pos = x+d <= M ? x+d : M;
+		events[2*i+1].tip = false;
+	}
+	qsort(events, 2*N, sizeof events[0], compar_events);
+	int osv = 0;
+	int depth = 0;
+	int start;
+	for (int i = 0; i < 2*N; i++) {
+		// fprintf(stderr, "pos=%d\ttip=%d\n", events[i].pos, events[i].tip);
+		if (events[i].tip == true) {
+			if (depth == 0)
+				start = events[i].pos;
+			depth++;
+		}
+		if (events[i].tip == false) {
+			depth--;
+			if (depth == 0)
+				osv += events[i].pos - start;
+		}
+	}
+	if (depth != 0)
+		fprintf(stderr, "depth == %d\n", depth);
+	printf("%d\n", M-osv);
+}
diff --git a/šola/aps1/dn/zlivanje/in.txt b/šola/aps1/dn/zlivanje/in.txt
new file mode 100644
index 0000000..eaca3bf
--- /dev/null
+++ b/šola/aps1/dn/zlivanje/in.txt
@@ -0,0 +1,26 @@
+25 3 2
+13
+18
+7
+8
+17
+3
+16
+9
+10
+11
+11
+0
+2
+19
+14
+5
+6
+15
+4
+5
+12
+3
+18
+1
+3
diff --git a/šola/aps1/dn/zlivanje/out.txt b/šola/aps1/dn/zlivanje/out.txt
new file mode 100644
index 0000000..6def2f9
--- /dev/null
+++ b/šola/aps1/dn/zlivanje/out.txt
@@ -0,0 +1 @@
+0 2 3 3 4 5 5 6 7 8 9 10 11 11 12 13 14 15 16 17 18 18 19 1 3
diff --git a/šola/aps1/dn/zlivanje/resitev.cpp b/šola/aps1/dn/zlivanje/resitev.cpp
new file mode 100644
index 0000000..8926019
--- /dev/null
+++ b/šola/aps1/dn/zlivanje/resitev.cpp
@@ -0,0 +1,39 @@
+#include <sys/param.h>
+#include <stdio.h>
+#include <stdlib.h>
+#include <stdbool.h>
+int compar_long (const void * a, const void * b) {
+	if (*(long *)a < *(long *)b)
+		return -1;
+	return *(long *)a > *(long *) b;
+}
+int main (void) {
+	long N, K, A;
+	scanf("%ld %ld %ld", &N, &K, &A);
+	long * d = (long *) malloc(N*sizeof *d);
+	long čet = 0;
+	long lastidx = 0;
+	long long četkončno = 1;
+	for (long i = 0; i < A && četkončno <= 2000000; i++)	// pravzaprav četkončno := K**A,
+		četkončno *= K;		// toda C nima int potence
+	// fprintf(stderr, "aaaaaaaa %ld\n", četkončno);
+	for (long i = 0; i < N; i++) {
+		scanf("%ld", d+i);
+		if (i && d[i-1] > d[i])
+			if (++čet >= četkončno) {
+				qsort(d+lastidx, i-lastidx, sizeof d[0], compar_long);
+				čet = 0;
+				lastidx = i;
+			}
+	}
+	qsort(d+lastidx, N-lastidx, sizeof d[0], compar_long);
+	bool devica = true;
+	for (long i = 0; i < N; i++) {
+		if (devica)
+			devica = false;
+		else
+			printf(" ");
+		printf("%ld", d[i]);
+	}
+	printf("\n");
+}
diff --git a/šola/citati.bib b/šola/citati.bib
new file mode 100644
index 0000000..f553435
--- /dev/null
+++ b/šola/citati.bib
@@ -0,0 +1,242 @@
+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}
+}
+@misc{rfc4086,
+	series = {Request for Comments},
+	number = 4086,
+	howpublished = {RFC 4086},
+	publisher = {RFC Editor},
+	doi = {10.17487/RFC4086},
+	url = {https://www.rfc-editor.org/info/rfc4086},
+	author = {Donald E. Eastlake 3rd and Steve Crocker and Jeffrey I. Schiller},
+	title = {{Randomness Requirements for Security}},
+	pagetotal = 48,
+	year = 2005,
+	month = jun,
+	abstract = {Security systems are built on strong cryptographic algorithms that foil pattern analysis attempts. However, the security of these systems is dependent on generating secret quantities for passwords, cryptographic keys, and similar quantities. The use of pseudo-random processes to generate secret quantities can result in pseudo-security. A sophisticated attacker may find it easier to reproduce the environment that produced the secret quantities and to search the resulting small set of possibilities than to locate the quantities in the whole of the potential number space. Choosing random quantities to foil a resourceful and motivated adversary is surprisingly difficult. This document points out many pitfalls in using poor entropy sources or traditional pseudo-random number generation techniques for generating such quantities. It recommends the use of truly random hardware techniques and shows that the existing hardware on many systems can be used for this purpose. It provides suggestions to ameliorate the problem when a hardware solution is not available, and it gives examples of how large such quantities need to be for some applications. This document specifies an Internet Best Current Practices for the Internet Community, and requests discussion and suggestions for improvements.},
+}
diff --git a/šola/p2/dn/DN06a_63230317.c b/šola/p2/dn/DN06a_63230317.c
new file mode 100644
index 0000000..a57b548
--- /dev/null
+++ b/šola/p2/dn/DN06a_63230317.c
@@ -0,0 +1,18 @@
+#include <stdio.h>
+int globina (int * a, int * b) {
+	int globina[2];
+	if (*a)
+		globina[0] = globina(a+*a, b);
+	if (*b)
+		globina[1] = globina(a, b+*b);
+	
+}
+int main (void) {
+	int n;
+	scanf("%d\n", &n);
+	int a[n];
+	int b[n];
+	for (int i = 0; i < n; i++)
+		scanf("%d %d\n", &a[i], &b[i]);
+	printf("%d\n", globina(a, b));
+}
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..13a6f2b
--- /dev/null
+++ b/šola/članki/dht/dokument.lyx
@@ -0,0 +1,2563 @@
+#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 paper
+\begin_preamble
+% for subfigures/subtables
+\usepackage[caption=false,font=footnotesize]{subfig}
+\usepackage{textcomp}
+\usepackage[
+    type={CC},
+    modifier={by-sa},
+    version={3.0},
+]{doclicense}
+\usepackage{bera}% optional: just to have a nice mono-spaced font
+\usepackage{listings}
+\usepackage{xcolor}
+\lstset{
+	extendedchars=true,
+	literate={č}{{\v{c}}}1 {ž}{{\v{z}}}1 {š}{{\v{s}}}1,
+}
+
+\colorlet{punct}{red!60!black}
+\definecolor{background}{HTML}{EEEEEE}
+\definecolor{delim}{RGB}{20,105,176}
+\colorlet{numb}{magenta!60!black}
+
+\lstdefinelanguage{json}{
+    basicstyle=\normalfont\ttfamily,
+    numbers=left,
+    numberstyle=\scriptsize,
+    stepnumber=1,
+    numbersep=8pt,
+    showstringspaces=false,
+    breaklines=true,
+    frame=lines,
+    backgroundcolor=\color{background},
+    literate=
+     *{0}{{{\color{numb}0}}}{1}
+      {1}{{{\color{numb}1}}}{1}
+      {2}{{{\color{numb}2}}}{1}
+      {3}{{{\color{numb}3}}}{1}
+      {4}{{{\color{numb}4}}}{1}
+      {5}{{{\color{numb}5}}}{1}
+      {6}{{{\color{numb}6}}}{1}
+      {7}{{{\color{numb}7}}}{1}
+      {8}{{{\color{numb}8}}}{1}
+      {9}{{{\color{numb}9}}}{1}
+      {:}{{{\color{punct}{:}}}}{1}
+      {,}{{{\color{punct}{,}}}}{1}
+      {\{}{{{\color{delim}{\{}}}}{1}
+      {\}}{{{\color{delim}{\}}}}}{1}
+      {[}{{{\color{delim}{[}}}}{1}
+      {]}{{{\color{delim}{]}}}}{1},
+}
+\end_preamble
+\options journal
+\use_default_options false
+\maintain_unincluded_children no
+\language slovene
+\language_package babel
+\inputencoding utf8
+\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 false
+\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 "Kaj prenašamo s protokolom BitTorrent?"
+\pdf_author "Anton Lula Šijanec"
+\pdf_subject "Računalniška omrežja"
+\pdf_keywords "podazdeljena razpršilna tabela, porazdeljeni sistemi, omrežje P2P, podatkovno rudarjenje, BitTorrent"
+\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
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+\columnsep 0.5cm
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+\tocdepth 3
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+\html_be_strict false
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+\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
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+<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 Standard
+\begin_inset Float table
+placement document
+alignment document
+wide false
+sideways false
+status open
+
+\begin_layout Plain Layout
+\begin_inset Tabular
+<lyxtabular version="3" rows="4" columns="5">
+<features tabularvalignment="middle">
+<column alignment="center" valignment="top">
+<column alignment="center" valignment="top">
+<column alignment="center" valignment="top">
+<column alignment="center" valignment="top">
+<column alignment="center" valignment="top">
+<row>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+mesto
+\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
+datum
+\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
+dni
+\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
+torrentov
+\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
+sek./torrent
+\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
+T-2,
+ FTTH,
+ SI
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+1.-2.
+ '23
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+16
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+47863
+\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
+29
+\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
+GRNET,
+ VPS,
+ GR
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+1.-2.
+ '23
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+31
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+412846
+\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
+6
+\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
+T-2,
+ FTTH,
+ SI
+\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
+6.
+ '24
+\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
+5
+\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
+62110
+\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
+7
+\end_layout
+
+\end_inset
+</cell>
+</row>
+</lyxtabular>
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Plain Layout
+\begin_inset Caption Standard
+
+\begin_layout Plain Layout
+Omrežne lokacije in časovna obdobja zajemov.
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Plain Layout
+
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Metapodatki prvega zajema opisujejo 3084321 datotek v skupni velikosti 259 TiB,
+ metapodatki drugega zajema 17101702 datotek v velikosti 1881 TiB in metapodatki tretjega zajema 3725125 datotek v velikosti 345 TiB.
+ 
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+januarja in februarja 2023:
+\end_layout
+
+\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_deeper
+\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
+
+\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
+after "5.2"
+key "rfc4086"
+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
index ede5e1b..5be42f0 100644
--- a/šola/članki/dht/makefile
+++ b/šola/članki/dht/makefile
@@ -1,7 +1,13 @@
-all: Dht_example_SVG.svg
+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
+	rm *.svg *.png *.tsv