6 files changed, 9727 insertions, 21 deletions

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
index 2b6057d..104ba6c 100644
--- a/šola/ana1/teor.lyx
+++ b/šola/ana1/teor.lyx
@@ -6089,7 +6089,7 @@ Korenski oz.
 \begin_inset Formula $c_{n}\coloneqq\sqrt[n]{a_{n}}$
 \end_inset
 
-.ž
+.
 \end_layout
 
 \begin_deeper
@@ -8698,7 +8698,20 @@ Naj bo
 \begin_inset Formula $f$
 \end_inset
 
- omejena in doseže minimum in maksimum.
+ 
+\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*
@@ -9468,7 +9481,20 @@ Naj bo
 \begin_inset Formula $f:I\to\mathbb{R}$
 \end_inset
 
- zvezna in strogo monotona.
+ 
+\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
@@ -9620,7 +9646,20 @@ Enakomerna zveznost
 \begin_inset Formula $f:I\to\mathbb{R}$
 \end_inset
 
- je enakomerno zvezna na 
+ 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
 
@@ -9700,6 +9739,1124 @@ Pri slednji definiciji je
 \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
 
@@ -9710,36 +10867,5448 @@ Pri slednji definiciji je
 
 \end_layout
 
-\begin_layout Corollary*
-sssssssssss
+\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 Corollary*
-sssssssssss
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
 \end_layout
 
-\begin_layout Corollary*
-sssssssssss
+\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*
-sssssssssss
+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*
-sssssssssss
+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*
-sssssssssss
+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*
-sssssssssss
+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*
-sssssssssss
+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
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
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+\use_geometry false
+\use_package amsmath 1
+\use_package amssymb 1
+\use_package cancel 1
+\use_package esint 1
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+\use_package mathtools 1
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+\cite_engine_type default
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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
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+\tracking_changes false
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+\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/citati.bib b/šola/citati.bib
index 3c34475..f553435 100644
--- a/šola/citati.bib
+++ b/šola/citati.bib
@@ -225,3 +225,18 @@ Stara različica je v ~/projects/sola-gimb-4/citati.bib
 	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/dokument.lyx b/šola/članki/dht/dokument.lyx
index af668ed..13a6f2b 100644
--- a/šola/članki/dht/dokument.lyx
+++ b/šola/članki/dht/dokument.lyx
@@ -87,8 +87,10 @@
 \paperfontsize default
 \spacing single
 \use_hyperref true
-\pdf_title "Your Title"
-\pdf_author "Your Name"
+\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
@@ -1431,14 +1433,262 @@ Zajem
 \end_layout
 
 \begin_layout Standard
-Podatke smo zajemali iz različnih lokacij in v različnih časovnih obdobjih:
+Podatke smo zajemali iz različnih lokacij in v različnih časovnih obdobjih.
 \end_layout
 
-\begin_layout Itemize
+\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_deeper
 \begin_layout Itemize
 16 dni:
  domači optični priključek v Sloveniji (T-2):
@@ -1457,6 +1707,7 @@ travnik
 
 \end_layout
 
+\begin_deeper
 \begin_layout Itemize
 31 dni VPS v Grčiji (grNet)
 \begin_inset Note Note
@@ -1551,6 +1802,11 @@ status open
 
 \end_layout
 
+\end_inset
+
+
+\end_layout
+
 \begin_layout Subsubsection
 Primer strukture torrent datoteke z metapodatki
 \end_layout
@@ -2064,7 +2320,8 @@ nolink "false"
 Zaradi lastnosti uniformne porazdelitve zgoščevalne funkcije
 \begin_inset CommandInset citation
 LatexCommand cite
-key "wikihashuniformity"
+after "5.2"
+key "rfc4086"
 literal "false"
 
 \end_inset