1 files changed, 4063 insertions, 12 deletions

diff --git a/šola/ana1/teor.lyx b/šola/ana1/teor.lyx
index 6ff52cd..28f0397 100644
--- a/šola/ana1/teor.lyx
+++ b/šola/ana1/teor.lyx
@@ -66,11 +66,20 @@ theorems-ams-extended
 \output_sync 0
 \bibtex_command default
 \index_command default
-\float_placement class
+\float_placement H
 \float_alignment class
 \paperfontsize default
 \spacing single
-\use_hyperref false
+\use_hyperref true
+\pdf_bookmarks true
+\pdf_bookmarksnumbered false
+\pdf_bookmarksopen false
+\pdf_bookmarksopenlevel 1
+\pdf_breaklinks false
+\pdf_pdfborder false
+\pdf_colorlinks false
+\pdf_backref false
+\pdf_pdfusetitle true
 \papersize default
 \use_geometry true
 \use_package amsmath 1
@@ -2399,8 +2408,20 @@ Naj bo
  zato je tudi sama omejena.
 \end_layout
 
-\begin_layout Claim*
-Naj bosta 
+\begin_layout Theorem*
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hypertarget{pmkdlim}{Naj bosta}
+\end_layout
+
+\end_inset
+
+ 
 \begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
 \end_inset
 
@@ -3902,11 +3923,6 @@ Podzaporedje zaporedja
 \end_inset
 
  strogo naraščajoča funkcija.
- Definicijsko območje 
-\begin_inset Formula $\varphi$
-\end_inset
-
- mora vsebovati števno neskončno elementov.
 \end_layout
 
 \begin_layout Theorem*
@@ -4206,7 +4222,7 @@ status open
 
 \begin_layout Plain Layout
 Racionalnih števil je števno mnogo,
- zato jih lahko linearno uredimo in oštevilčimo
+ zato jih lahko linearno uredimo in oštevilčimo.
 \end_layout
 
 \end_inset
@@ -4260,12 +4276,4047 @@ Limita je stakališče,
 .
 \end_layout
 
+\begin_layout Theorem*
+\begin_inset Formula $S$
+\end_inset
+
+ je stekališče 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}\Leftrightarrow S$
+\end_inset
+
+ je limita nekega podzaporedja 
+\begin_inset Formula $a_{n}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Dokazujemo ekvivalenco.
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Leftarrow\right)$
+\end_inset
+
+ Očitno.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Rightarrow\right)$
+\end_inset
+
+ Definirajmo 
+\begin_inset Formula $\forall m\in\mathbb{N}:U_{m}\coloneqq\left(S-\frac{1}{m},S+\frac{1}{m}\right)$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $S$
+\end_inset
+
+ stekališče,
+ 
+\begin_inset Formula $\forall m\in\mathbb{N}\exists a_{k_{m}}\in U_{m}$
+\end_inset
+
+.
+ Podzaporedje 
+\begin_inset Formula $\left(a_{k_{m}}\right)_{m\in\mathbb{N}}$
+\end_inset
+
+ konvergira k 
+\begin_inset Formula $S$
+\end_inset
+
+,
+ kajti 
+\begin_inset Formula $\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow\left|a_{k_{n}}-S\right|<\frac{1}{n}<\varepsilon$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
 \begin_layout Corollary*
-sssssssssss
+Če je 
+\begin_inset Formula $L$
+\end_inset
+
+ limita nekega zaporedja,
+ je 
+\begin_inset Formula $L$
+\end_inset
+
+ edino njegovo stekališče.
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $a_{n}\to L$
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $S$
+\end_inset
+
+ stekališče za 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+.
+ Po izreku zgoraj je 
+\begin_inset Formula $S$
+\end_inset
+
+ limita nekega podzaporedja 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+.
+ Toda limita vsakega podzaporedja je enaka limiti zaporedja,
+ iz katerega to podzaporedje izhaja,
+ če ta limita obstaja.
+ Potemtakem je 
+\begin_inset Formula $S=L$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hypertarget{bw}{Bolzano-Weierstraß}
+\end_layout
+
+\end_inset
+
+.
+ Eksistenčni izrek.
+ Vsako omejeno zaporedje v realnih številih ima kakšno stekališče v 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Označimo 
+\begin_inset Formula $m_{0}\coloneqq\inf_{n\in\mathbb{N}}a_{n},M_{0}\coloneqq\sup_{n\in\mathbb{N}}a_{n},I_{0}\coloneqq\left[m_{0},M_{0}\right]$
+\end_inset
+
+.
+ Očitno je 
+\begin_inset Formula $\forall n\in\mathbb{N}:a_{n}\in I_{0}$
+\end_inset
+
+.
+ Interval 
+\begin_inset Formula $I_{0}$
+\end_inset
+
+ razdelimo na dve polovici:
+ 
+\begin_inset Formula $I_{0}=\left[m_{0},\frac{m_{0}+M_{0}}{2}\right]\cup\left[\frac{m_{0}+M_{0}}{2},M_{0}\right]$
+\end_inset
+
+.
+ Izberemo polovico (vsaj ena obstaja),
+ v kateri leži neskončno mnogo členov,
+ in jo označimo z 
+\begin_inset Formula $I_{1}$
+\end_inset
+
+.
+ Spet jo razdelimo na pol in z 
+\begin_inset Formula $I_{2}$
+\end_inset
+
+ označimo tisto polovico,
+ v kateri leži neskončno mnogo členov.
+ Postopek ponavljamo in dobimo zaporedje zaprtih intervalov 
+\begin_inset Formula $\left(I_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ in velja 
+\begin_inset Formula $I_{0}\supset I_{1}\supset I_{2}\supset\cdots$
+\end_inset
+
+ ter 
+\begin_inset Formula $\left|I_{n}\right|=2^{-n}\left|I_{0}\right|$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Označimo sedaj 
+\begin_inset Formula $I_{n}\eqqcolon\left[l_{n},d_{n}\right]$
+\end_inset
+
+.
+ Iz konstrukcije je očitno,
+ da 
+\begin_inset Formula $\left(l_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ narašča in 
+\begin_inset Formula $\left(d_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ pada ter da sta obe zaporedji omejeni.
+ Posledično 
+\begin_inset Formula $\exists l\coloneqq\lim_{n\to\infty}l_{n},d\coloneqq\lim_{n\to\infty}d_{n}$
+\end_inset
+
+.
+ Iz 
+\begin_inset Formula $l_{n}\leq l\leq d\leq d_{n}$
+\end_inset
+
+ sledi ocena 
+\begin_inset Formula $d-l\leq l_{n}-d_{n}=\left|I_{n}\right|=2^{-n}\left|I_{0}\right|$
+\end_inset
+
+,
+ kar konvergira k 0.
+ Posledično 
+\begin_inset Formula $d=l$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Treba je pokazati še,
+ da je 
+\begin_inset Formula $d=l$
+\end_inset
+
+ stekališče za 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+.
+ Vzemimo poljuben 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $l=\lim_{n\to\infty}l_{n}\Rightarrow\exists n_{1}\in\mathbb{N}\ni:l_{n_{1}}>l-\varepsilon$
+\end_inset
+
+ in ker je 
+\begin_inset Formula $d=\lim_{n\to\infty}d_{n}\Rightarrow\exists n_{2}\in\mathbb{N}\ni:d_{n_{2}}<d-\varepsilon$
+\end_inset
+
+.
+ Torej 
+\begin_inset Formula $\left[l_{n_{1}},d_{n_{2}}\right]\subset\left(l-\varepsilon,d+\varepsilon\right)$
+\end_inset
+
+.
+ Torej za 
+\begin_inset Formula $n_{0}\coloneqq\max\left\{ n_{1},n_{2}\right\} $
+\end_inset
+
+ velja 
+\begin_inset Formula $I_{n_{0}}=\left[l_{n_{0}},d_{n_{n}}\right]\subset\left(l-\varepsilon,d+\varepsilon\right)$
+\end_inset
+
+.
+ Ker 
+\begin_inset Formula $I_{n_{0}}$
+\end_inset
+
+ po konstrukciji vsebuje neskončno mnogo elementov,
+ jih torej tudi 
+\begin_inset Formula $\left(l-\varepsilon,d+\varepsilon\right)$
+\end_inset
+
+ oziroma poljubno majhna okolica 
+\begin_inset Formula $d=l$
+\end_inset
+
+,
+ torej je 
+\begin_inset Formula $d=l$
+\end_inset
+
+ stekališče za 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+.
 \end_layout
 
 \begin_layout Corollary*
-sssssssssss
+Če je 
+\begin_inset Formula $s\in\mathbb{R}$
+\end_inset
+
+ edino stekališče omejenega zaporedja 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+,
+ tedaj je 
+\begin_inset Formula $s=\lim_{n\to\infty}a_{n}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $s$
+\end_inset
+
+ stekališče 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+.
+ PDDRAA 
+\begin_inset Formula $a_{n}\not\to s$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $\exists\varepsilon>0\ni:$
+\end_inset
+
+ izven 
+\begin_inset Formula $\left(s-\varepsilon,s+\varepsilon\right)$
+\end_inset
+
+ se nahaja neskončno mnogo členov zaporedja.
+ Ti členi sami zase tvorijo omejeno zaporedje,
+ ki ima po 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{bw}{B.-W.}
+\end_layout
+
+\end_inset
+
+ izreku stekališče.
+ Slednje gotovo ne more biti enako 
+\begin_inset Formula $s$
+\end_inset
+
+,
+ torej imamo vsaj dve stekališči,
+ kar je v je v 
+\begin_inset Formula $\rightarrow\!\leftarrow$
+\end_inset
+
+ s predpostavko.
+\end_layout
+
+\begin_layout Definition*
+Pravimo,
+ da ima realno zaporedje:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+stekališče v 
+\begin_inset Formula $\infty$
+\end_inset
+
+,
+ če 
+\begin_inset Formula $\forall M>0:\left(M,\infty\right)$
+\end_inset
+
+ vsebuje neskončno mnogo členov zapopredja
+\end_layout
+
+\begin_layout Itemize
+limito v 
+\begin_inset Formula $\infty$
+\end_inset
+
+,
+ če 
+\begin_inset Formula $\forall M>0:\left(M,\infty\right)$
+\end_inset
+
+ vsebuje vse člene zaporedja od nekega indeksa dalje
+\end_layout
+
+\begin_layout Standard
+in podobno za 
+\begin_inset Formula $-\infty$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Remark*
+Povezava s pojmom realnega stekališča/limite:
+ okolice 
+\begin_inset Quotes gld
+\end_inset
+
+točke
+\begin_inset Quotes grd
+\end_inset
+
+ 
+\begin_inset Formula $\infty$
+\end_inset
+
+ so intervali oblike 
+\begin_inset Formula $\left(M,\infty\right)$
+\end_inset
+
+.
+ To je smiselno,
+ saj biti 
+\begin_inset Quotes gld
+\end_inset
+
+blizu 
+\begin_inset Formula $\infty$
+\end_inset
+
+
+\begin_inset Quotes grd
+\end_inset
+
+ pomeni bizi zelo velik,
+ kar je ravno biti v 
+\begin_inset Formula $\left(M,\infty\right)$
+\end_inset
+
+za poljubno velik 
+\begin_inset Formula $M$
+\end_inset
+
+.
+ 
+\begin_inset Quotes gld
+\end_inset
+
+Okolica točke 
+\begin_inset Formula $\infty$
+\end_inset
+
+
+\begin_inset Quotes grd
+\end_inset
+
+ so torej vsi intervali oblike 
+\begin_inset Formula $\left(M,\infty\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsection
+Limes superior in limes inferior
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+realno zaporedje.
+ Tvorimo novo zaporedje 
+\begin_inset Formula $s_{n}\coloneqq\sup\left\{ a_{k};k\geq n\right\} $
+\end_inset
+
+.
+ Očitno je padajoče (
+\begin_inset Formula $s_{1}\geq s_{2}\geq s_{3}\geq\cdots$
+\end_inset
+
+),
+ ker je supremum množice vsaj supremum njene stroge podmnožice.
+ Zaporedje 
+\begin_inset Formula $\left(s_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ ima limito,
+ ki ji rečemo limes superior oziroma zgornja limita zaporedja 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ in označimo 
+\begin_inset Formula $\limsup_{n\to\infty}a_{n}=\overline{\lim_{n\to\infty}}a_{n}\coloneqq\lim_{n\to\infty}s_{n}$
+\end_inset
+
+ in velja,
+ da leži v 
+\begin_inset Formula $\mathbb{R}\cup\left\{ -\infty,\infty\right\} $
+\end_inset
+
+.
+ Podobno definiramo tudi limes inferior oz.
+ spodnjo limito zaporedja:
+ 
+\begin_inset Formula $\liminf_{n\to\infty}a_{n}=\underline{\lim_{n\to\infty}}a_{n}\coloneqq\lim_{n\to\infty}\left(\inf_{k\geq n}a_{k}\right)=\sup_{n\in\mathbb{N}}\left(\inf_{k\geq n}a_{k}\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Remark*
+Za razliko od običajne limite,
+ ki lahko ne obstaja,
+ 
+\begin_inset Formula $\limsup$
+\end_inset
+
+ in 
+\begin_inset Formula $\liminf$
+\end_inset
+
+ vedno obstajata.
+\end_layout
+
+\begin_layout Claim*
+\begin_inset Formula $\limsup_{n\to\infty}a_{n}$
+\end_inset
+
+ je največje stekališče zaporedja 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ in 
+\begin_inset Formula $\liminf_{n\to\infty}$
+\end_inset
+
+ najmanjše.
+\end_layout
+
+\begin_layout Proof
+Označimo 
+\begin_inset Formula $s\coloneqq\limsup_{n\to\infty}a_{n}$
+\end_inset
+
+.
+ Za 
+\begin_inset Formula $\liminf$
+\end_inset
+
+ je dokaz analogen in ga ne bomo pisali.
+ Dokazujemo,
+ da je 
+\begin_inset Formula $s$
+\end_inset
+
+ stekališče in 
+\begin_inset Formula $\forall t>s:t$
+\end_inset
+
+ ni stekališče.
+ Ločimo primere:
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $s\in\mathbb{R}$
+\end_inset
+
+ Naj bo 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+ poljuben.
+ Ker
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+Infimum padajočega konvergentnega zaporedja je očitno njegova limita.
+\end_layout
+
+\end_inset
+
+ je 
+\begin_inset Formula $s=\inf s_{n}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\exists n_{0}\in\mathbb{N}\ni:s_{n_{0}}\in[s,s+\varepsilon)$
+\end_inset
+
+.
+ Ker 
+\begin_inset Formula $\left(s_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ pada proti 
+\begin_inset Formula $s$
+\end_inset
+
+,
+ sledi 
+\begin_inset Formula $\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow s_{n}\in[s,s+\varepsilon)$
+\end_inset
+
+.
+ Po definiciji 
+\begin_inset Formula $s_{n}$
+\end_inset
+
+ velja 
+\begin_inset Formula $\forall n\in\mathbb{N}\exists N\left(n\right)\geq n\ni:s_{n}-\varepsilon<a_{N\left(n\right)}$
+\end_inset
+
+.
+ Torej imamo 
+\begin_inset Formula $s-\varepsilon\leq s_{n}-\varepsilon<a_{N\left(n\right)}\leq s_{n}<s+\varepsilon$
+\end_inset
+
+ (zadnji neenačaj za 
+\begin_inset Formula $n\geq n_{0}$
+\end_inset
+
+),
+ skratka 
+\begin_inset Formula $a_{N\left(n\right)}-s<\varepsilon$
+\end_inset
+
+ oziroma 
+\begin_inset Formula $\forall n\geq n_{0}:\left|a_{N\left(n\right)}-s\right|<\varepsilon$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $N\left(n\right)\geq n$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $\left\{ N\left(n\right);n\in\mathbb{N}\right\} $
+\end_inset
+
+ neskončna množica,
+ torej je neskončno mnogo členov v poljubni okolici 
+\begin_inset Formula $s$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Treba je dokazati še,
+ da 
+\begin_inset Formula $\forall t>s:t$
+\end_inset
+
+ ni stekališče.
+ Naj bo 
+\begin_inset Formula $t>s$
+\end_inset
+
+.
+ Označimo 
+\begin_inset Formula $\delta\coloneqq t-s>0$
+\end_inset
+
+.
+ Po definiciji
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $s$
+\end_inset
+
+ je limita zaporedja 
+\begin_inset Formula $\left(s_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+,
+ zato v poljubno majhni okolici obstaja tak 
+\begin_inset Formula $s_{n_{1}}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $s_{n_{1}}$
+\end_inset
+
+ torej tu najdemo v 
+\begin_inset Formula $[s,s+\frac{\delta}{2})$
+\end_inset
+
+.
+\end_layout
+
+\end_inset
+
+ 
+\begin_inset Formula $s$
+\end_inset
+
+ 
+\begin_inset Formula $\exists n_{1}\in\mathbb{N}\ni:s\leq s_{n_{1}}<s+\frac{\delta}{2}<s+t$
+\end_inset
+
+.
+ Ker 
+\begin_inset Formula $\left(s_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ pada proti 
+\begin_inset Formula $s$
+\end_inset
+
+,
+ sledi 
+\begin_inset Formula $\forall n\geq n_{1}:s\leq s_{n}<s+\frac{\delta}{2}$
+\end_inset
+
+.
+ Po definiciji
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $s_{n}$
+\end_inset
+
+ je supremum členov od vključno 
+\begin_inset Formula $n$
+\end_inset
+
+ dalje
+\end_layout
+
+\end_inset
+
+ 
+\begin_inset Formula $s_{n}$
+\end_inset
+
+ sledi 
+\begin_inset Formula $\forall n\geq n_{1}:a_{n}\leq s+\frac{\delta}{2}$
+\end_inset
+
+.
+ Za takšne 
+\begin_inset Formula $n$
+\end_inset
+
+ je 
+\begin_inset Formula $\left|t-a_{n}\right|=t-a_{n}\geq t-\left(s+\frac{\delta}{2}\right)=\frac{\delta}{2}$
+\end_inset
+
+.
+ Torej v 
+\begin_inset Formula $\frac{\delta}{2}-$
+\end_inset
+
+okolici točke 
+\begin_inset Formula $t$
+\end_inset
+
+ leži kvečjemu končno mnogo členov zaporedja oziroma členi 
+\begin_inset Formula $\left(a_{1},a_{2},\dots,a_{n_{1}-1}\right)$
+\end_inset
+
+.
+ Torej 
+\begin_inset Formula $t$
+\end_inset
+
+ ni stekališče za 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $s=\infty$
+\end_inset
+
+ Naj bo 
+\begin_inset Formula $M>0$
+\end_inset
+
+ poljuben.
+ Ker je 
+\begin_inset Formula $s=\inf s_{n}$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $\forall n\in\mathbb{N}:s_{n}=\infty$
+\end_inset
+
+.
+ Po definiciji 
+\begin_inset Formula $s_{n}=\infty$
+\end_inset
+
+ velja 
+\begin_inset Formula $\forall n\in\mathbb{N}\exists N\left(n\right):a_{N\left(n\right)}>M$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $N\left(n\right)\geq n$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $\left\{ N\left(n\right);n\in\mathbb{N}\right\} $
+\end_inset
+
+ neskončna množica,
+ torej je neskončno mnogo členov v 
+\begin_inset Formula $\left(M,\infty\right)$
+\end_inset
+
+ za poljuben 
+\begin_inset Formula $M$
+\end_inset
+
+,
+ torej je 
+\begin_inset Formula $s=\infty$
+\end_inset
+
+ res stekališče.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Večjih stekališč od 
+\begin_inset Formula $\infty$
+\end_inset
+
+ očitno ni.
+\end_layout
+
+\end_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $s=-\infty$
+\end_inset
+
+ Naj bo 
+\begin_inset Formula $m<0$
+\end_inset
+
+ poljuben.
+ Ker je 
+\begin_inset Formula $s=\inf s_{n}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\exists n_{0}\in\mathbb{N}\ni:s_{n_{0}}\in\left(-\infty,m\right)$
+\end_inset
+
+ Ker 
+\begin_inset Formula $\left(s_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ pada proti 
+\begin_inset Formula $s=-\infty$
+\end_inset
+
+,
+ sledi 
+\begin_inset Formula $\forall n\in\mathbb{N}:n\geq n_{0}:s_{n}\in\left(-\infty,m\right)$
+\end_inset
+
+.
+ Po definiciji 
+\begin_inset Formula $s_{n}$
+\end_inset
+
+ velja 
+\begin_inset Formula $\forall n\in\mathbb{N}:a_{n}\in\left(-\infty,m\right)$
+\end_inset
+
+.
+ Ker je za poljuben 
+\begin_inset Formula $m$
+\end_inset
+
+ neskončno mnogo členov v 
+\begin_inset Formula $\left(-\infty,m\right)$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $s=-\infty$
+\end_inset
+
+ res stekališče.
+\end_layout
+
+\end_deeper
+\begin_layout Subsection
+Cauchyjev pogoj
+\end_layout
+
+\begin_layout Definition*
+Zaporedje 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ ustreza Cauchyjevemu pogoju (oz.
+ je Cauchyjevo),
+ če 
+\begin_inset Formula $\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\ni:\forall m,n\in\mathbb{N}:m,n\geq n_{0}\Rightarrow\left|a_{m}-a_{n}\right|<\varepsilon$
+\end_inset
+
+.
+ ZDB Dovolj pozni členi so si poljubno blizu.
+\end_layout
+
+\begin_layout Claim*
+Zaporedje v 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+ je konvergentno 
+\begin_inset Formula $\Leftrightarrow$
+\end_inset
+
+ je Cauchyjevo.
+\end_layout
+
+\begin_layout Proof
+Dokazujemo ekvivalenco.
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Rightarrow\right)$
+\end_inset
+
+ Če 
+\begin_inset Formula $a_{n}\to L$
+\end_inset
+
+,
+ tedaj 
+\begin_inset Formula $\left|a_{m}-a_{n}\right|=\left|\left(a_{m}-L\right)+\left(L-a_{n}\right)\right|\leq\left|a_{m}-\varepsilon\right|+\left|a_{n}-\varepsilon\right|$
+\end_inset
+
+.
+ Cauchyjev pogoj sledi iz definicije limite za 
+\begin_inset Formula $\frac{\varepsilon}{2}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Leftarrow\right)$
+\end_inset
+
+ Če je zaporedje Cauchyjevo,
+ je omejeno:
+ 
+\begin_inset Formula $\exists n_{0}\in\mathbb{N}\ni:\forall m,n\in\mathbb{N}:m,n\geq n_{0}\Rightarrow\left|a_{m}-a_{n}\right|\leq1$
+\end_inset
+
+.
+ V posebnem,
+ 
+\begin_inset Formula $m=n_{0}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\left|a_{n_{0}}-a_{n}\right|\leq1$
+\end_inset
+
+ oziroma 
+\begin_inset Formula $\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow a_{n}\in\left[a_{n_{0}}-1,a_{n_{0}}+1\right]$
+\end_inset
+
+.
+ Preostali členi tvorijo končno veliko množico,
+ ki ima 
+\begin_inset Formula $\min$
+\end_inset
+
+ in 
+\begin_inset Formula $\max$
+\end_inset
+
+,
+ torej je 
+\begin_inset Formula $\left\{ a_{k};k\in\mathbb{N}\right\} =\left\{ a_{1},a_{2},\dots,a_{n_{0}-1}\right\} \cup\left\{ a_{k};k\in\mathbb{N},k\geq n_{0}\right\} $
+\end_inset
+
+ tudi omejena.
+ Po 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{bw}{izreku od prej}
+\end_layout
+
+\end_inset
+
+ sledi,
+ da ima zaporedje stekališče 
+\begin_inset Formula $s$
+\end_inset
+
+.
+ Dokažimo,
+ da je 
+\begin_inset Formula $s=\lim_{n\to\infty}a_{n}$
+\end_inset
+
+.
+ Vzemimo poljuben 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ Cauchyjevo,
+ 
+\begin_inset Formula $\exists n_{0}\in\mathbb{N}\ni:\forall m,n\in\mathbb{N}:m,n\geq n_{0}\Rightarrow\left|a_{m}-a_{n}\right|<\frac{\varepsilon}{2}$
+\end_inset
+
+.
+ Po definiciji 
+\begin_inset Formula $s$
+\end_inset
+
+ 
+\begin_inset Formula $\exists n_{1}\geq n_{0}\ni:\left|a_{n_{1}}-s\right|<\frac{\varepsilon}{2}$
+\end_inset
+
+.
+ Sledi 
+\begin_inset Formula $\forall n\geq n_{0}:\left|a_{n}-s\right|=\left|a_{n}-s+s-a_{n_{1}}\right|\leq\left|a_{n}-s\right|+\left|s-a_{n_{1}}\right|<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Remark*
+Moč izreka je v tem,
+ da lahko konvergenco preverjamo tudi tedaj,
+ ko nimamo kandidatov za limito.
+\end_layout
+
+\begin_layout Section
+Številske vrste
+\end_layout
+
+\begin_layout Standard
+Kako sešteti neskončno mnogo števil?
+ Nadgradimo pristop končnih vsot na neskončne vsote!
+\end_layout
+
+\begin_layout Definition*
+Imejmo zaporedje 
+\begin_inset Formula $\left(a_{k}\right)_{k\in\mathbb{N}},a_{k}\in\mathbb{R}$
+\end_inset
+
+.
+ Izraz 
+\begin_inset Formula $\sum_{j=1}^{\infty}a_{j}$
+\end_inset
+
+ se imenuje vrsta s členi 
+\begin_inset Formula $a_{j}$
+\end_inset
+
+.
+ Pomen izraza opredelimo na naslednjo način:
+\end_layout
+
+\begin_layout Definition*
+Tvorimo novo zaporedje,
+ pravimo mu zaporedje delnih vsot vrste:
+ 
+\begin_inset Formula $s_{1}=a_{1}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $s_{2}=a_{1}+a_{2}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $s_{3}=a_{1}+a_{2}+a_{3}$
+\end_inset
+
+,
+ ...,
+ 
+\begin_inset Formula $s_{n}=a_{1}+a_{2}+\cdots+a_{n}=\sum_{j=1}^{n}a_{j}$
+\end_inset
+
+ —
+ številu 
+\begin_inset Formula $s_{n}$
+\end_inset
+
+ pravimo 
+\begin_inset Formula $n-$
+\end_inset
+
+ta delna vsota.
+\end_layout
+
+\begin_layout Definition*
+Vrsta je konvergentna,
+ če je v 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+ konvergentno zaporedje 
+\begin_inset Formula $\left(s_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+.
+ Številu 
+\begin_inset Formula $s=\lim_{n\to\infty}s_{n}$
+\end_inset
+
+ tedaj pravimo vsota vrste in pišemo 
+\begin_inset Formula $s\eqqcolon\sum_{j=1}^{\infty}a_{j}$
+\end_inset
+
+.
+ Pojem neskončne vsote torej prevedemo na pojem limite pridruženega zaporedja delnih vsot.
+ Včasih vrsto (kot operacijo) enačimo z njeno vsoto (izidom operacije).
+\end_layout
+
+\begin_layout Definition*
+Če vrsta ni konvergentna,
+ rečemo,
+ da je divergentna.
+ Enako,
+ če je 
+\begin_inset Formula $s\in\left\{ \pm\infty\right\} $
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Primeri vrst.
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $a_{n}=\frac{1}{2^{n}}$
+\end_inset
+
+,
+ torej zaporedje 
+\begin_inset Formula $\frac{1}{2},\frac{1}{4},\frac{1}{8},\dots$
+\end_inset
+
+.
+ Ali se sešteje v 1?
+ Velja 
+\begin_inset Formula $s=\lim_{n\to\infty}\sum_{j=1}^{n}a_{j}$
+\end_inset
+
+.
+ Pišimo 
+\begin_inset Formula $q=\frac{1}{2}$
+\end_inset
+
+,
+ tedaj 
+\begin_inset Formula $a_{n}=q^{n}$
+\end_inset
+
+ in 
+\begin_inset Formula 
+\[
+s_{n}=q+q^{2}+q^{3}+\cdots+q^{n}=q\left(1+q+q^{2}+\cdots+q^{n-1}\right)=q\frac{\left(1+q+q^{2}+\cdots+q^{n-1}\right)\left(1-q\right)}{1-q}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=q\frac{\left(1+q+q^{2}+\cdots+q^{n-1}\right)-\left(q+q^{2}+q^{3}+\cdots+q^{n}\right)}{1-q}=q\frac{1-q^{n}}{1-q}=\frac{q}{1-q}\left(1-q^{n}\right)
+\]
+
+\end_inset
+
+Izračunajmo 
+\begin_inset Formula $\lim_{n\to\infty}s_{n}=\lim_{n\to\infty}\frac{q}{1-q}\left(1-\cancelto{0}{q^{n}}\right)=\frac{q}{1-q}$
+\end_inset
+
+ (velja,
+ ker 
+\begin_inset Formula $q\in\left(-1,1\right)$
+\end_inset
+
+),
+ torej je 
+\begin_inset Formula $s=\sum_{n=1}^{\infty}q^{n}=\frac{q}{1-q}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Geometrijska vrsta (splošno).
+ Naj bo 
+\begin_inset Formula $q\in\mathbb{R}$
+\end_inset
+
+.
+ Vrsta 
+\begin_inset Formula $\sum_{j=0}^{\infty}q^{j}$
+\end_inset
+
+ se imenuje geometrijska vrsta.
+ Velja 
+\begin_inset Formula $s=\lim_{n\to\infty}\sum_{j=0}^{n}q^{j}$
+\end_inset
+
+ in 
+\begin_inset Formula $s_{n}=1+q+q^{2}+q^{3}+\cdots+q^{n}$
+\end_inset
+
+.
+ Če je 
+\begin_inset Formula $q=1$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $s_{n}=n+1$
+\end_inset
+
+,
+ sicer množimo izraz z 
+\begin_inset Formula $\left(1-q\right)$
+\end_inset
+
+:
+\begin_inset Formula 
+\[
+\left(1+q+q^{2}+\cdots+q^{n}\right)\left(1-q\right)=\left(1+q+q^{2}+\cdots+q^{n}\right)-\left(q+q^{2}+q^{3}+\cdots+q^{n+1}\right)=1-q^{n+1}
+\]
+
+\end_inset
+
+torej 
+\begin_inset Formula $s_{n}=\frac{1-q^{n+1}}{1-q}$
+\end_inset
+
+ in vrsta konvergira 
+\begin_inset Formula $\Leftrightarrow q\not=1$
+\end_inset
+
+ in 
+\begin_inset Formula $\lim_{n\to\infty}\frac{1-q^{n+1}}{1-q}\exists$
+\end_inset
+
+ v 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+.
+ To pa se zgodi natanko za 
+\begin_inset Formula $q\in\left(-1,1\right)$
+\end_inset
+
+,
+ takrat je 
+\begin_inset Formula $\lim_{n\to\infty}\frac{1-\cancelto{0}{q^{n+1}}}{1-q}=\frac{1}{1-q}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Harmonična vrsta.
+ Je vrsta 
+\begin_inset Formula $\sum_{j=1}^{\infty}\frac{1}{j}$
+\end_inset
+
+.
+ Velja 
+\begin_inset Formula $\frac{1}{j}\underset{j\to\infty}{\longrightarrow}0$
+\end_inset
+
+,
+ toda vrsta divergira.
+ Dokaz sledi kmalu malce spodaj.
+\end_layout
+
+\end_deeper
+\begin_layout Question*
+Kako lahko enostavno določimo,
+ ali dana vrsta konvergira?
+\end_layout
+
+\begin_layout Subsection
+Konvergenčni kriteriji
+\end_layout
+
+\begin_layout Theorem*
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hypertarget{cauchyvrste}{Cauchyjev pogoj}
+\end_layout
+
+\end_inset
+
+.
+ Vrsta 
+\begin_inset Formula $\sum_{j=1}^{\infty}a_{j}$
+\end_inset
+
+ je konvergentna 
+\begin_inset Formula $\Leftrightarrow$
+\end_inset
+
+ delne vrste ustrezajo Cauchyjevemu pogoju;
+ 
+\begin_inset Formula $\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\forall n,m\in\mathbb{N}:n,m\geq n_{0}\Rightarrow\left|s_{m}-s_{n}\right|=\left|\sum_{j=n+1}^{m}a_{j}\right|<\varepsilon$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Corollary*
+\begin_inset Formula $\sum_{j=1}^{\infty}a_{j}$
+\end_inset
+
+ konvergira 
+\begin_inset Formula $\Rightarrow\lim_{j\to\infty}a_{j}=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Uporabimo izrek zgoraj za 
+\begin_inset Formula $n=m-1$
+\end_inset
+
+:
+ 
+\begin_inset Formula $\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow\left|s_{n}-s_{n+1}\right|=\left|a_{n}\right|<\varepsilon$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Vrsti 
+\begin_inset Formula $\sum_{j=1}^{\infty}\cos n$
+\end_inset
+
+ in 
+\begin_inset Formula $\sum_{j=1}^{\infty}\sin n$
+\end_inset
+
+ divergirata,
+ saj smo videli,
+ da členi ne ene ne druge ne konvergirajo nikamor,
+ torej tudi ne proti 0,
+ kar je potreben pogoj za konvergenco vrste.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+Harmonična vrsta divergira.
+ Protiprimer Cauchyjevega pogoja:
+ Naj bo 
+\begin_inset Formula $\varepsilon=\frac{1}{4}$
+\end_inset
+
+.
+ Tedaj ne glede na izbiro 
+\begin_inset Formula $n_{0}$
+\end_inset
+
+ najdemo:
+\begin_inset Formula 
+\[
+s_{2n}-s_{n}=\sum_{j=n+1}^{2n}\frac{1}{j}=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}>\frac{1}{2n}+\frac{1}{2n}+\cdots+\frac{1}{2n}=\frac{1}{2}
+\]
+
+\end_inset
+
+Dokaz divergence brez Cauchyjevega pogoja:
+ 
+\begin_inset Formula $s_{2^{n}}=a_{1}+\sum_{j=1}^{n}\left(s_{2^{j}}-s_{s^{j-1}}\right)>1+\frac{n}{2}$
+\end_inset
+
+ in 
+\begin_inset Formula $\lim_{n\to\infty}1+\frac{n}{2}=\infty$
+\end_inset
+
+.
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+Geometrični argument za divergenco:
+ TODO XXX FIXME DODAJ
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Theorem*
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hypertarget{pk}{Primerjalni kriterij}
+\end_layout
+
+\end_inset
+
+.
+ Naj bosta 
+\begin_inset Formula $\sum_{n=1}^{\infty}a_{n}$
+\end_inset
+
+ in 
+\begin_inset Formula $\sum_{n=1}^{\infty}b_{n}$
+\end_inset
+
+ vrsti z nenegativnimi členi.
+ Naj bo 
+\begin_inset Formula $\forall k\geq k_{0}:a_{k}\leq b_{k}$
+\end_inset
+
+ (od nekod naprej) —
+ pravimo,
+ da je 
+\begin_inset Formula $\sum_{n=1}^{\infty}b_{n}$
+\end_inset
+
+ majoranta za 
+\begin_inset Formula $\sum_{n=1}^{\infty}a_{n}$
+\end_inset
+
+ od nekod naprej.
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Če 
+\begin_inset Formula $\sum_{n=1}^{\infty}b_{n}$
+\end_inset
+
+ konvergira,
+ tedaj tudi 
+\begin_inset Formula $\sum_{n=1}^{\infty}a_{n}$
+\end_inset
+
+ konvergira.
+\end_layout
+
+\begin_layout Itemize
+Če 
+\begin_inset Formula $\text{\ensuremath{\sum_{n=1}^{\infty}a_{n}=\infty}}$
+\end_inset
+
+,
+ tedaj tudi 
+\begin_inset Formula $\sum_{n=1}^{\infty}b_{n}=\infty$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Example*
+Videli smo,
+ da 
+\begin_inset Formula $\sum_{k=1}^{\infty}\frac{1}{k}$
+\end_inset
+
+ divergira.
+ Kaj pa 
+\begin_inset Formula $\sum_{k=1}^{\infty}\frac{1}{k^{2}}$
+\end_inset
+
+?
+ Preverimo naslednje in uporabimo primerjalni kriterij:
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $\forall k\in\mathbb{N}:\frac{1}{k^{2}}\leq\frac{2}{k\left(k+1\right)}$
+\end_inset
+
+?
+ Računajmo 
+\begin_inset Formula $k^{2}\geq\frac{k\left(k+1\right)}{2}\sim k\geq\frac{k+1}{2}\sim\frac{k}{2}\geq\frac{1}{2}$
+\end_inset
+
+.
+ Velja,
+ ker 
+\begin_inset Formula $k\in\mathbb{N}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Vrsta 
+\begin_inset Formula $\sum_{k=1}^{\infty}\frac{2}{k\left(k+1\right)}$
+\end_inset
+
+ konvergira?
+ Opazimo 
+\begin_inset Formula $\frac{1}{k}-\frac{1}{k+1}=\frac{k+1}{k\left(k+1\right)}-\frac{k}{k\left(k+1\right)}=\frac{1}{k\left(k+1\right)}$
+\end_inset
+
+.
+ Za delne vsote vrste 
+\begin_inset Formula $\sum_{k=1}^{\infty}\frac{1}{k\left(k+1\right)}$
+\end_inset
+
+ velja:
+\begin_inset Formula 
+\[
+\sum_{k=1}^{n}\frac{1}{k\left(k+1\right)}=\sum_{k=1}^{n}\left(\frac{1}{k}-\frac{1}{k+1}\right)=1-\frac{1}{n+1}\underset{n\to\infty}{\longrightarrow}1,
+\]
+
+\end_inset
+
+torej 
+\begin_inset Formula $\sum_{k=1}^{\infty}\frac{2}{k\left(k+1\right)}=2$
+\end_inset
+
+.
+ Posledično po primerjalnem kriteriju tudi 
+\begin_inset Formula $\sum_{k=1}^{\infty}\frac{1}{k^{2}}$
+\end_inset
+
+ konvergira.
+ Izkaže se 
+\begin_inset Formula $\sum_{k=1}^{\infty}\frac{1}{k^{2}}=\frac{\pi^{2}}{6}\approx1,645$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Theorem*
+Kvocientni oz.
+ d'Alembertov kriterij.
+ Za vrsto s pozitivnimi členi 
+\begin_inset Formula $\sum_{n=1}^{\infty}a_{n}$
+\end_inset
+
+ definirajmo 
+\begin_inset Formula $D_{n}\coloneqq\frac{a_{n+1}}{a_{n}}$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $\exists n_{0}\in\mathbb{N},q\in\left(0,1\right)\forall n\geq n_{0}:D_{n}\leq q\Longrightarrow\sum_{n=1}^{\infty}a_{n}<\infty$
+\end_inset
+
+ (vrsta konvergira)
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\exists n_{0}\in\mathbb{N}\forall n\geq n_{0}:D_{n}\geq1\Longrightarrow\sum_{n=1}^{\infty}a_{n}=\infty$
+\end_inset
+
+ (vrsta divergira)
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\exists D=\lim_{n\to\infty}D_{n}\in\mathbb{R}\Longrightarrow$
+\end_inset
+
+
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset CommandInset label
+LatexCommand label
+name "enu:kvocientni3a"
+
+\end_inset
+
+
+\begin_inset Formula $D<1\Longrightarrow\sum_{n=1}^{\infty}a_{n}<\infty$
+\end_inset
+
+ (vrsta konvergira)
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $D>1\Longrightarrow\sum_{n=1}^{\infty}a_{n}=\infty$
+\end_inset
+
+ (vrsta divergira)
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $D=1\Longrightarrow$
+\end_inset
+
+ s tem kriterijem ne moremo določiti konvergence.
+\end_layout
+
+\end_deeper
+\end_deeper
+\begin_layout Proof
+Razlaga.
+ 
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $\forall n>n_{0}:D_{n}\leq q$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\frac{a_{n+1}}{a_{n}}\leq q\sim a_{n+1}\leq qa_{n}$
+\end_inset
+
+ in hkrati 
+\begin_inset Formula $\text{\ensuremath{\frac{a_{n+2}}{a_{n+1}}\leq q\sim a_{n+2}\leq qa_{n+1}}}$
+\end_inset
+
+,
+ torej skupaj 
+\begin_inset Formula $a_{n+2}\leq qa_{n+1}\leq qqa_{n}=q^{2}a_{n}$
+\end_inset
+
+,
+ sledi 
+\begin_inset Formula $q_{n+2}\leq q^{2}a_{n}$
+\end_inset
+
+ in 
+\begin_inset Formula $\forall k\in\mathbb{N}:q_{n+k}\leq q^{k}a_{n}$
+\end_inset
+
+.
+ Vrsto smo majorizirali z geometrijsko vrsto,
+ ki ob 
+\begin_inset Formula $q\in\left(0,1\right)$
+\end_inset
+
+ konvergira po primerjalnem kriteriju,
+ zato tudi naša vrsta konvergira.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\forall n>n_{0}:\frac{a_{n+1}}{a_{n}}\geq D\geq1$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $a_{n+1}\geq a_{n}$
+\end_inset
+
+ in hkrati 
+\begin_inset Formula $a_{n+2}\geq a_{n+1}$
+\end_inset
+
+,
+ torej skupaj 
+\begin_inset Formula $a_{n+2}\geq a_{n}$
+\end_inset
+
+,
+ sledi 
+\begin_inset Formula $\forall k\in\mathbb{N}:a_{n+k}\geq a_{n}$
+\end_inset
+
+.
+ Naša vrsta torej majorizira konstantno vrsto,
+ ki očitno divergira;
+ 
+\begin_inset Formula $\sum_{k=n_{0}}^{\infty}a_{k}\geq\sum_{k=n_{0}}^{\infty}a_{n}=0$
+\end_inset
+
+.
+ Potemtakem tudi naša vrsta divergira.
+ Poleg tega niti ne velja 
+\begin_inset Formula $a_{k}\underset{k\to\infty}{\longrightarrow}0$
+\end_inset
+
+,
+ torej vrsta gotovo divergira.
+\end_layout
+
+\begin_layout Enumerate
+Enako kot 1 in 2.
+\end_layout
+
+\end_deeper
+\begin_layout Example*
+Za 
+\begin_inset Formula $x>0$
+\end_inset
+
+ definiramo 
+\begin_inset Formula $e^{x}=\sum_{k=0}^{\infty}\frac{x^{k}}{k!}$
+\end_inset
+
+.
+ Vrsta res konvergira po točki 
+\begin_inset CommandInset ref
+LatexCommand ref
+reference "enu:kvocientni3a"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+D_{n}=\frac{\frac{x^{n+1}}{\left(n+1\right)!}}{\frac{x^{n}}{n!}}=\frac{x^{n+1}n!}{x^{n}\left(n+1\right)!}=\frac{x}{n+1}\underset{n\to\infty}{\longrightarrow}0
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Theorem*
+Korenski oz.
+ Cauchyjev kriterij.
+ Naj bo 
+\begin_inset Formula $\sum_{k=1}^{\infty}a_{k}$
+\end_inset
+
+ vrsta z nenegativnimi členi.
+ Naj bo 
+\begin_inset Formula $c_{n}\coloneqq\sqrt[n]{a_{n}}$
+\end_inset
+
+.ž
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $\exists n_{0}\in\mathbb{N},q\in\left(0,1\right)\forall n>n_{0}:c_{n}\leq q\Longrightarrow\sum_{k=1}^{\infty}a_{k}<\infty$
+\end_inset
+
+ (vrsta konvergira)
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\exists n_{0}\in\mathbb{N}\forall n>n_{0}:c_{n}\geq1\Longrightarrow\sum_{k=1}^{\infty}a_{k}=\infty$
+\end_inset
+
+ (vrsta divergira)
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\exists c=\lim_{n\to\infty}c_{n}\in\mathbb{R}\Longrightarrow$
+\end_inset
+
+
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $c<1\Longrightarrow\sum_{k=1}^{\infty}a_{k}<\infty$
+\end_inset
+
+ (vrsta konvergira)
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $c>1\Longrightarrow\sum_{k=1}^{\infty}a_{k}=\infty$
+\end_inset
+
+ (vrsta divergira)
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $c=1\Longrightarrow$
+\end_inset
+
+ s tem kriterijem ne moremo določiti konvergence.
+\end_layout
+
+\end_deeper
+\end_deeper
+\begin_layout Proof
+Skica dokazov.
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+Velja 
+\begin_inset Formula $\forall n>n_{0}:c_{n}\leq q$
+\end_inset
+
+.
+ To pomeni 
+\begin_inset Formula $\sqrt[n]{a_{n}}\leq q$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $a_{n}\leq q^{n}$
+\end_inset
+
+ in 
+\begin_inset Formula $a_{n+1}\leq q^{n+1}$
+\end_inset
+
+,
+ torej je vrsta majorizirana z geometrijsko vrsto 
+\begin_inset Formula $\sum_{n=1}^{\infty}q^{n}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Velja 
+\begin_inset Formula $\forall n>n_{0}:c_{n}\geq1$
+\end_inset
+
+.
+ To pomeni 
+\begin_inset Formula $\sqrt[n]{a_{n}}\geq1$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $a_{n}\geq1$
+\end_inset
+
+,
+ torej je vrsta majorizirana s konstantno in zato divergentno vrsto 
+\begin_inset Formula $\sum_{n=1}^{\infty}1$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Enako kot 1 in 2.
+\end_layout
+
+\end_deeper
+\begin_layout Subsection
+Alternirajoče vrste
+\end_layout
+
+\begin_layout Definition*
+Vrsta je alternirajoča,
+ če je predznak naslednjega člena nasproten predznaku tega člena.
+ ZDB 
+\begin_inset Formula $\forall n\in\mathbb{N}:\sgn a_{n+1}=-\sgn a_{n}$
+\end_inset
+
+,
+ kjer je 
+\begin_inset Formula $\sgn:\mathbb{R}\to\left\{ -1,0,1\right\} $
+\end_inset
+
+ s predpisom 
+\begin_inset Formula $\sgn a=\begin{cases}
+-1 & ;a<0\\
+1 & ;a>0\\
+0 & ;a=0
+\end{cases}$
+\end_inset
+
+.
+ ZDB 
+\begin_inset Formula $\forall n\in\mathbb{N}:a_{n+1}a_{n}\leq0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+Leibnizov konvergenčni kriterij.
+ Naj bo 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ padajoče zaporedje in 
+\begin_inset Formula $\lim_{n\to\infty}a_{n}=0$
+\end_inset
+
+.
+ Tedaj vrsta 
+\begin_inset Formula $\sum_{k=1}^{\infty}\left(-1\right)^{k}a_{k}$
+\end_inset
+
+ konvergira.
+ Če je 
+\begin_inset Formula $s\coloneqq\sum_{k=1}^{\infty}\left(-1\right)^{k}a_{k}$
+\end_inset
+
+ in 
+\begin_inset Formula $s_{n}\coloneqq\sum_{k=1}^{\infty}\left(-1\right)^{k}a_{k}$
+\end_inset
+
+,
+ tedaj 
+\begin_inset Formula $\left|s-s_{k}\right|\leq a_{n+1}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Skica dokaza.
+ Vidimo,
+ da delne vsote 
+\begin_inset Formula $s_{2n}$
+\end_inset
+
+ padajo k 
+\begin_inset Formula $s''$
+\end_inset
+
+ in delne vsote 
+\begin_inset Formula $s_{2n-1}$
+\end_inset
+
+ naraščajo k 
+\begin_inset Formula $s'$
+\end_inset
+
+.
+ Toda ker 
+\begin_inset Formula $s_{2n}-s_{2n-1}=a_{2n}$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $s'=s''$
+\end_inset
+
+.
+ Limita razlike dveh zaporedij je razlika limit teh dveh zaporedij,
+ torej 
+\begin_inset Formula $s'=s''=s$
+\end_inset
+
+.
+ 
+\begin_inset Formula $s$
+\end_inset
+
+ je supremum lihih in infimum sodih vsot.
+ 
+\begin_inset Formula $\left|s-s_{n}\right|\leq\left|s_{n+1}-s_{n}\right|=a_{n+1}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Harmonična vrsta 
+\begin_inset Formula $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots\to\infty$
+\end_inset
+
+,
+ toda alternirajoča harmonična vrsta 
+\begin_inset Formula $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots\to\log2$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsection
+Absolutno konvergentne vrste
+\end_layout
+
+\begin_layout Definition*
+Vrsta 
+\begin_inset Formula $\sum_{n=1}^{\infty}a_{n}$
+\end_inset
+
+ je absolutno konvergentna,
+ če je 
+\begin_inset Formula $\sum_{n=1}^{\infty}\left|a_{n}\right|$
+\end_inset
+
+ konvergentna.
+\end_layout
+
+\begin_layout Theorem*
+Absolutna konvergenca 
+\begin_inset Formula $\Rightarrow$
+\end_inset
+
+ konvergenca.
+\end_layout
+
+\begin_layout Proof
+Uporabimo 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{cauchyvrste}{Cauchyjev pogoj za konvergenco vrst}
+\end_layout
+
+\end_inset
+
+ in trikotniško neenakost.
+\begin_inset Formula 
+\[
+\left|s_{m}-s_{n}\right|=\left|\sum_{j=n+1}^{m}a_{j}\right|\leq\sum_{j=n+1}^{m}\left|a_{j}\right|<\varepsilon
+\]
+
+\end_inset
+
+za 
+\begin_inset Formula $m,n\geq n_{0}$
+\end_inset
+
+ za nek 
+\begin_inset Formula $n_{0}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Remark*
+Obrat ne velja,
+ protiprimer je alternirajoča harmonična vrsta.
+\end_layout
+
+\begin_layout Subsection
+Pogojno konvergentne vrste
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $\sum_{k=0}^{\infty}2-\sum_{k=0}^{\infty}1\not=\sum_{k=0}^{\infty}\left(2-1\right)$
+\end_inset
+
+,
+ temveč 
+\begin_inset Formula $\infty-\infty=$
+\end_inset
+
+ nedefinirano.
+\end_layout
+
+\begin_layout Question*
+Ross-Littlewoodov paradoks.
+ Ali smemo zamenjati vrstni red seštevanja,
+ če imamo neskončno mnogo sumandov?
+\end_layout
+
+\begin_layout Standard
+Najprej vprašanje natančneje opredelimo in vpeljimo orodja za njegovo obravnavo.
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $\mathcal{M}\subset\mathbb{N}$
+\end_inset
+
+.
+ Permutacija 
+\begin_inset Formula $\mathcal{M}$
+\end_inset
+
+ je vsaka bijektivna preslikava 
+\begin_inset Formula $\pi:\mathcal{M}\to\mathcal{M}$
+\end_inset
+
+.
+ Če je 
+\begin_inset Formula $\mathcal{M}=\left\{ a_{1},\dots,a_{n}\right\} $
+\end_inset
+
+ končna množica,
+ tedaj 
+\begin_inset Formula $\pi$
+\end_inset
+
+ označimo s tabelo:
+\begin_inset Formula 
+\[
+\left(\begin{array}{ccc}
+a_{1} & \cdots & a_{n}\\
+\pi\left(a_{1}\right) & \cdots & \pi\left(a_{n}\right)
+\end{array}\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula 
+\[
+\pi=\left(\begin{array}{ccccc}
+1 & 2 & 3 & 4 & 5\\
+5 & 3 & 1 & 4 & 2
+\end{array}\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Vrsta 
+\begin_inset Formula $\sum_{n=1}^{\infty}a_{n}$
+\end_inset
+
+ je brezpogojno konvergentna,
+ če za vsako permutacijo 
+\begin_inset Formula $\pi:\mathbb{N}\to\mathbb{N}$
+\end_inset
+
+ vrsta 
+\begin_inset Formula $\sum_{n=1}^{\infty}\pi\left(a_{n}\right)$
+\end_inset
+
+ konvergira in vsota ni odvisna od 
+\begin_inset Formula $\pi$
+\end_inset
+
+.
+ Vrsta je pogojno konvergentna,
+ če je konvergentna,
+ toda ne brezpogojno.
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\cdots$
+\end_inset
+
+ je pogojno konvergentna,
+ ker pri seštevanju z vrstnim redom,
+ pri katerem tisočim pozitivnim členom sledi en negativen in njemu zopet tisoč pozitivnih itd.,
+ vrsta ne konvergira.
+\end_layout
+
+\begin_layout Theorem*
+Absolutna konvergenca 
+\begin_inset Formula $\Leftrightarrow$
+\end_inset
+
+ Brezpogojna konvergenca
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Theorem*
+Riemannov sumacijski izrek.
+ Če je vrsta pogojno konvergentna,
+ tedaj 
+\begin_inset Formula $\forall x\in\mathbb{R}\cup\left\{ \pm\infty\right\} \exists$
+\end_inset
+
+ permutacija 
+\begin_inset Formula $\pi:\mathbb{N}\to\mathbb{N}\ni:\sum_{n=1}^{\infty}a_{\pi\left(n\right)}=x$
+\end_inset
+
+.
+ ZDB Končna vsota je lahko karkoli,
+ če lahko poljubno spremenimo vrstni red seštevanja.
+ Prav tako obstaja taka permutacija 
+\begin_inset Formula $\pi$
+\end_inset
+
+,
+ pri kateri 
+\begin_inset Formula $\sum_{n=1}^{\infty}a_{\pi\left(n\right)}$
+\end_inset
+
+ nima vsote ZDB delne vsotee ne konvergirajo.
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}}{n}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Funkcijske vrste
+\end_layout
+
+\begin_layout Standard
+Tokrat poskušamo seštevati funkcije.
+ V prejšnjem razdelku seštevamo le realna števila.
+ Funkcijska vrsta,
+ če je 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ zaporedije funkcij 
+\begin_inset Formula $X\to\mathbb{R}$
+\end_inset
+
+ in 
+\begin_inset Formula $x$
+\end_inset
+
+ zunanja konstanta,
+ izgleda takole:
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula 
+\[
+\sum_{n=1}^{\infty}a_{n}\left(x\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $X$
+\end_inset
+
+ neka množica in 
+\begin_inset Formula $\Phi=\left\{ \varphi_{n}:X\to\mathbb{R},n\in\mathbb{N}\right\} $
+\end_inset
+
+ družina funkcij.
+\end_layout
+
+\begin_layout Definition*
+Pravimo,
+ da funkcije 
+\begin_inset Formula $\varphi_{n}$
+\end_inset
+
+ konvergirajo po točkah na 
+\begin_inset Formula $X$
+\end_inset
+
+,
+ če je 
+\begin_inset Formula $\forall x\in X$
+\end_inset
+
+ zaporedje 
+\begin_inset Formula $\left(\varphi_{n}\left(x\right)\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ konvergentno.
+\end_layout
+
+\begin_layout Definition*
+Označimo limito s 
+\begin_inset Formula $\varphi\left(x\right)$
+\end_inset
+
+.
+ ZDB to pomeni,
+ da 
+\begin_inset Formula 
+\[
+\forall\varepsilon>0,x\in X:\exists n_{0}=n_{0}\left(\varepsilon,x\right)\in\mathbb{N}\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow\left|\varphi_{n}\left(x\right)-\varphi\left(x\right)\right|<\varepsilon.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Pravimo,
+ da funkcije 
+\begin_inset Formula $\varphi_{n}$
+\end_inset
+
+ konvergirajo enakomerno na 
+\begin_inset Formula $X$
+\end_inset
+
+,
+ če 
+\begin_inset Formula 
+\[
+\forall\varepsilon>0\exists n_{0}=n_{0}\left(\varepsilon\right)\in\mathbb{N}\forall x\in X,n\in\mathbb{N}:n\geq n_{0}\Rightarrow\left|\varphi_{n}\left(x\right)-\varphi\left(x\right)\right|\leq\varepsilon
+\]
+
+\end_inset
+
+ oziroma ZDB 
+\begin_inset Formula 
+\[
+\forall\varepsilon>0\exists n_{0}=n_{0}\left(\varepsilon\right)\in\mathbb{N}\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow\sup_{x\in X}\left|\varphi_{n}\left(x\right)-\varphi\left(x\right)\right|\leq\varepsilon.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Poudariti je treba,
+ da je pri konvergenci po točkah 
+\begin_inset Formula $n_{0}$
+\end_inset
+
+ lahko odvisen od 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+ in 
+\begin_inset Formula $x$
+\end_inset
+
+,
+ pri enakomerni konvergenci pa le od 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Note*
+Očitno enakomerna konvergenca implicira konvergenco po točkah,
+ obratno pa ne velja.
+\end_layout
+
+\begin_layout Example*
+Za 
+\begin_inset Formula $n\in\mathbb{N}$
+\end_inset
+
+ definiramo 
+\begin_inset Formula $\varphi_{n}:\left[0,1\right]\to\left[0,1\right]$
+\end_inset
+
+ s predpisom 
+\begin_inset Formula $\varphi_{n}\left(x\right)=x^{n}$
+\end_inset
+
+.
+ Tedaj obstaja 
+\begin_inset Formula $\varphi\left(x\right)\coloneqq\lim_{n\to\infty}\varphi_{n}\left(x\right)=\begin{cases}
+0 & ;x\in[0,1)\\
+1 & ;x=1
+\end{cases}$
+\end_inset
+
+.
+ Torej po definiciji velja 
+\begin_inset Formula $\varphi_{n}\to\varphi$
+\end_inset
+
+ po točkah,
+ toda ne velja 
+\begin_inset Formula $\varphi_{n}\to\varphi$
+\end_inset
+
+ enakomerno.
+ Za poljubno velik pas okoli 
+\begin_inset Formula $\varphi\left(x\right)$
+\end_inset
+
+ bodo še tako pozne funkcijske vrednosti 
+\begin_inset Formula $\varphi_{n}\left(x\right)$
+\end_inset
+
+ od nekega 
+\begin_inset Formula $x$
+\end_inset
+
+ dalje izven tega pasu.
+ Če bi 
+\begin_inset Formula $\varphi_{n}\to\varphi$
+\end_inset
+
+ enakomerno,
+ tedaj bi za poljuben 
+\begin_inset Formula $\varepsilon\in\left(0,1\right)$
+\end_inset
+
+ in dovolj pozne 
+\begin_inset Formula $n$
+\end_inset
+
+ (večje od nekega 
+\begin_inset Formula $n_{0}\in\mathbb{N}$
+\end_inset
+
+) veljalo 
+\begin_inset Formula $\forall x\in\left[0,1\right]:\left|\varphi_{n}\left(x\right)-\varphi\left(x\right)\right|<\varepsilon$
+\end_inset
+
+.
+ To je ekvivalentno 
+\begin_inset Formula $\forall x\in\left(0,1\right):\left|x^{n}\right|<\varepsilon\Leftrightarrow n\log x<\log\varepsilon\Leftrightarrow n>\frac{\log\varepsilon}{\log x}$
+\end_inset
+
+.
+ Toda 
+\begin_inset Formula $\lim_{x\nearrow1}\frac{\log\varepsilon}{\log x}=\infty$
+\end_inset
+
+,
+ zato tak 
+\begin_inset Formula $n$
+\end_inset
+
+ ne obstaja.
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $X$
+\end_inset
+
+ neka množica in 
+\begin_inset Formula $\left(f_{j}:X\to\mathbb{R}\right)_{j\in\mathbb{N}}$
+\end_inset
+
+ dano zaporedje funkcij.
+ Pravimo,
+ da funkcijska vrsta 
+\begin_inset Formula $\sum_{j=1}^{\infty}f_{j}$
+\end_inset
+
+ konvergira po točkah na 
+\begin_inset Formula $X$
+\end_inset
+
+,
+ če 
+\begin_inset Formula $\forall x\in X:\sum_{j=1}^{\infty}f_{j}\left(x\right)<0$
+\end_inset
+
+ (številska vrsta je konvergentna).
+ ZDB to pomeni,
+ da funkcijsko zaporedje delnih vsot 
+\begin_inset Formula $s_{n}\coloneqq\sum_{j=1}^{n}f_{j}$
+\end_inset
+
+ konvergira po točkah na 
+\begin_inset Formula $X$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Funkcijska vrsta 
+\begin_inset Formula $s=\sum_{j=1}^{\infty}$
+\end_inset
+
+ konvergira enakomerno na 
+\begin_inset Formula $X$
+\end_inset
+
+,
+ če funkcijsko zaporedje delnih vsot 
+\begin_inset Formula $s_{n}\coloneqq\sum_{j=1}^{n}f_{j}$
+\end_inset
+
+ konvergira enakomerno na 
+\begin_inset Formula $X$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Funkcija oblike 
+\begin_inset Formula $x\mapsto\sum_{j=1}^{\infty}f_{j}\left(x\right)$
+\end_inset
+
+ se imenuje funkcijska vrsta.
+\end_layout
+
+\begin_layout Exercise*
+Dokaži,
+ da 
+\begin_inset Formula $\sum_{n=1}^{\infty}x^{n}$
+\end_inset
+
+ ne konvergira enakomerno!
+ Vrsta konvergira po točkah le na intervalu 
+\begin_inset Formula $x\in\left(0,1\right)$
+\end_inset
+
+,
+ za druge 
+\begin_inset Formula $x$
+\end_inset
+
+ divergira.
+ Ko fiksiramo zunanjo konstanto,
+ gre za geometrijsko vrsto.
+ Delna vsota 
+\begin_inset Formula $\sum_{j=1}^{n}x^{j}=\frac{x\left(1-x^{n}\right)}{1-x}$
+\end_inset
+
+.
+ Velja 
+\begin_inset Formula $\lim_{n\to\infty}\frac{x\left(1-x^{n}\right)}{1-x}=x\lim_{n\to\infty}\frac{1-\cancelto{0}{x^{n}}}{1-x}=\frac{x}{1-x}$
+\end_inset
+
+.
+ Sedaj prevedimo,
+ ali 
+\begin_inset Formula $\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\forall x\in\left(-1,1\right),n\geq n_{0}:\left|\frac{x\left(1-x^{n}\right)}{1-x}-\frac{x}{1-x}\right|<\varepsilon$
+\end_inset
+
+.
+ Za začetekk si oglejmo le 
+\begin_inset Formula $x>0$
+\end_inset
+
+.
+ Ker je tedaj 
+\begin_inset Formula $\frac{x\left(1-x^{n}\right)}{1-x}<\frac{x}{1-x}$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $\left|\frac{x\left(1-x^{n}\right)}{1-x}-\frac{x}{1-x}\right|=\frac{x}{1-x}-\frac{x\left(1-x^{n}\right)}{1-x}=\frac{\cancel{x-x+}x^{n+1}}{1-x}$
+\end_inset
+
+.
+ Računajmo sedaj 
+\begin_inset Formula $\frac{x^{n+1}}{1-x}<\varepsilon\sim x^{n+1}<\varepsilon\left(1-x\right)\sim\left(n+1\right)\log x<\log\left(\varepsilon\left(1-x\right)\right)\sim n+1>\frac{\log\left(\varepsilon\left(1-x\right)\right)}{\log x}\sim n>\frac{\log\left(\varepsilon\left(1-x\right)\right)}{\log x}-1$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $n$
+\end_inset
+
+ odvisen od 
+\begin_inset Formula $x$
+\end_inset
+
+,
+ vsota ni enakomerno konvergentna.
+\end_layout
+
+\begin_layout Standard
+Poseben primer funkcijskih vrst so funkcijske vrste funkcij oblike 
+\begin_inset Formula $f_{j}=b_{j}\cdot x^{j}$
+\end_inset
+
+,
+ torej potence (monomi).
+\end_layout
+
+\begin_layout Definition*
+Potenčna vrsta je funkcijska vrsta oblike 
+\begin_inset Formula $\sum_{j=1}^{\infty}b_{j}\cdot x^{j}$
+\end_inset
+
+,
+ kjer so a 
+\begin_inset Formula $\left(b_{j}\right)_{j\in\mathbb{N}}$
+\end_inset
+
+ dana realna števila.
+\end_layout
+
+\begin_layout Theorem*
+Cauchy-Hadamard.
+ Za vsako potenčno vrsto obstaja konvergenčni radij 
+\begin_inset Formula $R\in\left[0,\infty\right]\ni:$
+\end_inset
+
+
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+vrsta absolutno konvergira za 
+\begin_inset Formula $\left|x\right|<R$
+\end_inset
+
+,
+\end_layout
+
+\begin_layout Itemize
+vrsta divergira za 
+\begin_inset Formula $\left|x\right|>R$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Theorem*
+Velja 
+\begin_inset Formula $\text{\ensuremath{\frac{1}{R}=\limsup_{k\to\infty}\sqrt[k]{\left|b_{k}\right|}}}$
+\end_inset
+
+,
+ kjer vzamemo 
+\begin_inset Formula $\frac{1}{0}\coloneqq\infty$
+\end_inset
+
+ in 
+\begin_inset Formula $\frac{1}{\infty}\coloneqq0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Rezultat že poznamo za zelo poseben primer 
+\begin_inset Formula $\forall j\in\mathbb{N}:b_{j}=1$
+\end_inset
+
+ (geometrijska vrsta).
+ Ideja dokaza je,
+ da konvergenco vsake potenčne vrste opišemo s pomočjo geometrijske vrste.
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Konvergenca:
+ Za 
+\begin_inset Formula $x=0$
+\end_inset
+
+ vrsta očitno konvergira,
+ zato privzamemo 
+\begin_inset Formula $x\not=0$
+\end_inset
+
+.
+ Definirajmo 
+\begin_inset Formula $R$
+\end_inset
+
+ s formulo iz definicije (
+\begin_inset Formula $R=\frac{1}{\limsup_{k\to\infty}\sqrt[k]{\left|b_{k}\right|}}$
+\end_inset
+
+).
+ Naj bo 
+\begin_inset Formula $x$
+\end_inset
+
+ tak,
+ da 
+\begin_inset Formula $\left|x\right|<R\leq\infty$
+\end_inset
+
+ (sledi 
+\begin_inset Formula $R>0$
+\end_inset
+
+).
+ Naj bo 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+.
+ Tedaj po definiciji 
+\begin_inset Formula $R$
+\end_inset
+
+ velja 
+\begin_inset Formula $\sqrt[k]{\left|b_{k}\right|}\leq\frac{1}{R}+\varepsilon$
+\end_inset
+
+ za vse dovolj velike 
+\begin_inset Formula $k$
+\end_inset
+
+.
+ Za take 
+\begin_inset Formula $k$
+\end_inset
+
+ sledi
+\begin_inset Formula 
+\[
+\left|b_{k}\right|\left|x\right|^{k}\leq\left(\left(\frac{1}{R}+\varepsilon\right)\left|x\right|\right)^{k}.
+\]
+
+\end_inset
+
+Opazimo,
+ da je desna stran neenačbe člen geometrijske vrste,
+ s katero majoriziramo vrsto iz absolutnih vrednosti členov naše vrste.
+ Preverimo,
+ da desna stran konvergira.
+ Konvergira,
+ kadar 
+\begin_inset Formula $\left(\frac{1}{R}+\varepsilon\right)\left|x\right|<1$
+\end_inset
+
+ oziroma 
+\begin_inset Formula $\varepsilon<\frac{1}{\left|x\right|}-\frac{1}{R}$
+\end_inset
+
+.
+ Po 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{pk}{primerjalnem kriteriju}
+\end_layout
+
+\end_inset
+
+ torej naša vrsta absolutno konvergira.
+\end_layout
+
+\begin_layout Itemize
+Divergenca:
+ Vzemimo poljuben 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+.
+ Po definciji 
+\begin_inset Formula $R$
+\end_inset
+
+ sledi,
+ da je 
+\begin_inset Formula $\sqrt[k]{\left|b_{k}\right|}\geq\frac{1}{R}-\varepsilon$
+\end_inset
+
+ za vse dovolj velike 
+\begin_inset Formula $k$
+\end_inset
+
+.
+ Za take 
+\begin_inset Formula $k$
+\end_inset
+
+ sledi
+\begin_inset Formula 
+\[
+\left|b_{k}\right|\left|x\right|^{k}\geq\left(\left(\frac{1}{R}-\varepsilon\right)\left|x\right|\right)^{k}.
+\]
+
+\end_inset
+
+Opazimo,
+ da je desna stran neenačbe člen geometrijske vrste,
+ ki je majorizirana z vrsto iz absolutnih vrednosti členov naše vrste.
+ Desna stran divergira,
+ ko 
+\begin_inset Formula $\left(\frac{1}{R}-\varepsilon\right)\left|x\right|=1$
+\end_inset
+
+ oziroma 
+\begin_inset Formula $\varepsilon=\frac{1}{R}-\frac{1}{\left|x\right|}$
+\end_inset
+
+,
+ zato tudi naša vrsta divergira.
+\end_layout
+
+\end_deeper
+\begin_layout Example*
+Primer konvergenčnega radija potenčne vrste od prej:
+ 
+\begin_inset Formula $\sum_{j=1}^{\infty}x^{j}$
+\end_inset
+
+.
+ Velja 
+\begin_inset Formula $\forall j\in\mathbb{N}:b_{j}=1$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $R=\frac{1}{\limsup_{j\to\infty}\sqrt[k]{\left|b_{k}\right|}}=1$
+\end_inset
+
+,
+ torej po zgornjem izreku vrsta konvergira za 
+\begin_inset Formula $x\in\left(-1,1\right)$
+\end_inset
+
+ in divergira za 
+\begin_inset Formula $x\not\in\left[-1,1\right]$
+\end_inset
+
+.
+ Ročno lahko še preverimo,
+ da divergira tudi v 
+\begin_inset Formula $\left\{ -1,1\right\} $
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Zveznost
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+TODO XXX FIXME PREVERI ŠE V profesrojevih PDFJIH,
+ recimo dodaj dokaz zveznosti x^2
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Ideja:
+ Izdelati želimo formulacijo,
+ s katero preverimo,
+ le lahko z dovolj majhno spremembo 
+\begin_inset Formula $x$
+\end_inset
+
+ povzročimo majhno spremembo funkcijske vrednosti.
+\end_layout
+
+\begin_layout Example*
+Primer nezvezne funkcije je 
+\begin_inset Formula $f\left(x\right)=\begin{cases}
+0 & ;0\leq x<1\\
+1 & ;x=1
+\end{cases}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $D\subseteq\mathbb{R},a\in D$
+\end_inset
+
+ in 
+\begin_inset Formula $f:D\to\mathbb{R}$
+\end_inset
+
+.
+ Pravimo,
+ da je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna v 
+\begin_inset Formula $a$
+\end_inset
+
+,
+ če 
+\begin_inset Formula $\forall\varepsilon>0\exists\delta>0\forall x\in D:\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(a\right)\right|<\varepsilon$
+\end_inset
+
+.
+ 
+\begin_inset Formula $f$
+\end_inset
+
+ je zvezna na množici 
+\begin_inset Formula $x\subseteq D$
+\end_inset
+
+,
+ če je zvezna na vsaki točki v 
+\begin_inset Formula $D$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hypertarget{kzzz}{Karakterizacija zveznosti z zaporedji}
+\end_layout
+
+\end_inset
+
+.
+ Naj bodo 
+\begin_inset Formula $D,a,f$
+\end_inset
+
+ kot prej.
+ Velja:
+ 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna v 
+\begin_inset Formula $a\Leftrightarrow\forall\left(a_{n}\right)_{n\in\mathbb{N}},a_{n}\in D:\lim_{n\to\infty}a_{n}=a\Rightarrow\lim_{n\to\infty}f\left(a_{n}\right)=f\left(a\right)$
+\end_inset
+
+ ZDB 
+\begin_inset Formula $f$
+\end_inset
+
+ je zvezna v 
+\begin_inset Formula $a$
+\end_inset
+
+,
+ če za vsako k 
+\begin_inset Formula $a$
+\end_inset
+
+ konvergentno zaporedje na domeni velja,
+ da funkcijske vrednosti členov zaporedja konvergirajo k funkcijski vrednosti 
+\begin_inset Formula $a$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Dokazujemo ekvivalenco.
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Rightarrow\right)$
+\end_inset
+
+ Predpostavimo,
+ da je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna v 
+\begin_inset Formula $a$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\forall\varepsilon>0\exists\delta>0\forall x\in D:\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(a\right)\right|<\varepsilon$
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ poljubno zaporedje na 
+\begin_inset Formula $D$
+\end_inset
+
+,
+ ki konvergira k 
+\begin_inset Formula $a$
+\end_inset
+
+,
+ se pravi 
+\begin_inset Formula $\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow\left|a-a_{n}\right|<\varepsilon$
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+ poljuben.
+ Vsled zveznosti 
+\begin_inset Formula $f$
+\end_inset
+
+ velja,
+ da je 
+\begin_inset Formula $\left|f\left(a_{n}\right)-f\left(a\right)\right|<\varepsilon$
+\end_inset
+
+ za vse take 
+\begin_inset Formula $a_{n}$
+\end_inset
+
+,
+ da velja 
+\begin_inset Formula $\left|a_{n}-a\right|<\delta$
+\end_inset
+
+ za neko 
+\begin_inset Formula $\delta\in\mathbb{R}$
+\end_inset
+
+.
+ Ker je zaporedje 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ konvergentno k 
+\begin_inset Formula $a$
+\end_inset
+
+,
+ so vsi členi po nekem 
+\begin_inset Formula $n_{0}$
+\end_inset
+
+ v 
+\begin_inset Formula $\delta-$
+\end_inset
+
+okolici 
+\begin_inset Formula $a$
+\end_inset
+
+,
+ torej velja pogoj 
+\begin_inset Formula $\left|a_{n}-a\right|<\delta$
+\end_inset
+
+,
+ torej velja 
+\begin_inset Formula $\left|f\left(a_{n}\right)-f\left(a\right)\right|<\varepsilon$
+\end_inset
+
+ za vse 
+\begin_inset Formula $n\geq n_{0}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Leftarrow\right)$
+\end_inset
+
+ PDDRAA 
+\begin_inset Formula $f$
+\end_inset
+
+ ni zvezna v 
+\begin_inset Formula $a$
+\end_inset
+
+.
+ Da pridemo do protislovja,
+ moramo dokazati,
+ da 
+\begin_inset Formula $\exists\left(a_{n}\right)_{n\in\mathbb{N}},a_{n}\in D\ni:\lim_{n\to\infty}a_{n}=a$
+\end_inset
+
+,
+ a vendar 
+\begin_inset Formula $\lim_{n\to\infty}f\left(a_{n}\right)\not=f\left(a\right)$
+\end_inset
+
+.
+ Ker 
+\begin_inset Formula $f$
+\end_inset
+
+ ni zvezna,
+ velja,
+ da 
+\begin_inset Formula $\exists\varepsilon>0\forall\delta>0\exists x\in D\ni:\left|x-a\right|<\delta\wedge\left|f\left(x\right)-f\left(a\right)\right|\geq\varepsilon$
+\end_inset
+
+.
+ Izberimo 
+\begin_inset Formula $\forall n\in\mathbb{N}:\delta_{n}\coloneqq\frac{1}{n}$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $\forall n\in\mathbb{N}\exists\varepsilon>0,x\in D\eqqcolon x_{n}\ni:\left|x_{n}-a\right|<\frac{1}{n}\wedge\left|f\left(x_{n}\right)-f\left(a\right)\right|\geq\varepsilon$
+\end_inset
+
+.
+ S prvim argumentom konjunkcije smo poskrbeli za to,
+ da je naše konstruiramo zaporedje 
+\begin_inset Formula $\left(x_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ konvergentno k 
+\begin_inset Formula $a$
+\end_inset
+
+.
+ Konstruirali smo zaporedje,
+ pri katerem so funkcijske vrednosti za vsak 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+ izven 
+\begin_inset Formula $\varepsilon-$
+\end_inset
+
+okolice 
+\begin_inset Formula $f\left(a\right)$
+\end_inset
+
+,
+ torej zaporedje ne konvergira k 
+\begin_inset Formula $f\left(a\right)$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Theorem*
+Naj bo 
+\begin_inset Formula $f:D\to\mathbb{R}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $f$
+\end_inset
+
+ je zvezna na 
+\begin_inset Formula $D\Leftrightarrow$
+\end_inset
+
+ za vsako odprto množico 
+\begin_inset Formula $V\subset\mathbb{R}$
+\end_inset
+
+ je 
+\begin_inset Formula $f^{-1}\left(V\right)$
+\end_inset
+
+ spet odprta množica
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+Za funkcijo 
+\begin_inset Formula $f:D\to V$
+\end_inset
+
+ za 
+\begin_inset Formula $X\subseteq V$
+\end_inset
+
+ definiramo 
+\begin_inset Formula $f^{-1}\left(X\right)\coloneqq\left\{ x\in D;f\left(x\right)\in V\right\} \subseteq D$
+\end_inset
+
+.
+\end_layout
+
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Dokazujemo ekvivalenco.
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Leftarrow\right)$
+\end_inset
+
+ Predpostavimo,
+ da za vsako odprto množico 
+\begin_inset Formula $V\subset\mathbb{R}$
+\end_inset
+
+ je 
+\begin_inset Formula $f^{-1}\left(V\right)$
+\end_inset
+
+ spet odprta množica.
+ Dokazujemo,
+ da je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna na 
+\begin_inset Formula $D$
+\end_inset
+
+.
+ Naj bosta 
+\begin_inset Formula $a\in D,\varepsilon>0$
+\end_inset
+
+ poljubna.
+ Naj bo 
+\begin_inset Formula $V\coloneqq\left(f\left(a\right)-\varepsilon,f\left(a\right)+\varepsilon\right)$
+\end_inset
+
+ odprta množica.
+ Po predpostavki sledi,
+ da je 
+\begin_inset Formula $f^{-1}\left(V\right)$
+\end_inset
+
+ spet odprta.
+ Ker je 
+\begin_inset Formula $a\in f^{-1}\left(V\right)$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $a\in V$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $V$
+\end_inset
+
+ odprta,
+ 
+\begin_inset Formula $\exists\delta>0\ni:\left(a-\delta,a+\delta\right)\in V$
+\end_inset
+
+.
+ Torej 
+\begin_inset Formula $\forall x\in D:\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(a\right)\right|<\varepsilon$
+\end_inset
+
+,
+ torej je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna na 
+\begin_inset Formula $D$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Rightarrow\right)$
+\end_inset
+
+ Predpostavimo,
+ da je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna na 
+\begin_inset Formula $D$
+\end_inset
+
+,
+ to pomeni 
+\begin_inset Formula $\forall a\in D\forall\varepsilon>0\exists\delta>0\forall x\in D:\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(a\right)\right|<\varepsilon$
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $V$
+\end_inset
+
+ poljubna odprta podmnožica 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+ in naj bo 
+\begin_inset Formula $a\in f^{-1}\left(V\right)$
+\end_inset
+
+ poljuben (torej 
+\begin_inset Formula $f\left(a\right)\in V$
+\end_inset
+
+).
+ Ker je 
+\begin_inset Formula $f\left(a\right)\in V$
+\end_inset
+
+,
+ ki je odprta,
+ 
+\begin_inset Formula $\exists\varepsilon>0\ni:\left(f\left(a\right)-\varepsilon,f\left(a\right)+\varepsilon\right)\subseteq V$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna v 
+\begin_inset Formula $a$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\exists\delta>0\forall x\in D:\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(a\right)\right|<\varepsilon$
+\end_inset
+
+,
+ torej je tudi neka odprta okolica 
+\begin_inset Formula $f\left(a\right)$
+\end_inset
+
+ v 
+\begin_inset Formula $f^{-1}\left(V\right)$
+\end_inset
+
+.
+ Ker je bil 
+\begin_inset Formula $a$
+\end_inset
+
+ poljuben,
+ je 
+\begin_inset Formula $f^{-1}\left(V\right)$
+\end_inset
+
+ odprta,
+ ker je bila 
+\begin_inset Formula $V$
+\end_inset
+
+ poljubna,
+ je izrek dokazan.
+\end_layout
+
+\end_deeper
+\begin_layout Theorem*
+Naj bosta 
+\begin_inset Formula $f,g:D\to\mathbb{R}$
+\end_inset
+
+ zvezni v 
+\begin_inset Formula $a\in D$
+\end_inset
+
+.
+ Tedaj so v 
+\begin_inset Formula $a$
+\end_inset
+
+ zvezne tudi funkcije 
+\begin_inset Formula $f+g,f-g,f\cdot g$
+\end_inset
+
+ in 
+\begin_inset Formula $f/g$
+\end_inset
+
+,
+ slednja le,
+ če je 
+\begin_inset Formula $g\left(a\right)\not=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Ker je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna v 
+\begin_inset Formula $a$
+\end_inset
+
+ po 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{kzzz}{izreku o karakterizaciji zveznosti z zaporedji}
+\end_layout
+
+\end_inset
+
+ velja za vsako 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}},\forall n\in\mathbb{N}:a_{n}\subset D,\lim_{n\to\infty}a_{n}=a$
+\end_inset
+
+ tudi 
+\begin_inset Formula $\lim_{n\to\infty}f\left(a_{n}\right)=f\left(a\right)$
+\end_inset
+
+.
+ Po 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{pmkdlim}{izreku iz poglavja o zaporedjih}
+\end_layout
+
+\end_inset
+
+ velja,
+ da 
+\begin_inset Formula $f\left(a_{n}\right)*g\left(a_{n}\right)\to\left(f*g\right)\left(a_{n}\right)$
+\end_inset
+
+ za 
+\begin_inset Formula $*\in\left\{ +,-,\cdot,/\right\} $
+\end_inset
+
+.
+ Zopet uporabimo izrek o karakterizaciji zveznosti z zaporedji,
+ ki pove,
+ da so tudi 
+\begin_inset Formula $f*g$
+\end_inset
+
+ za 
+\begin_inset Formula $*\in\left\{ +,-,\cdot,/\right\} $
+\end_inset
+
+ zvezne v 
+\begin_inset Formula $a$
+\end_inset
+
+.
+ Pri deljenju velja omejitev 
+\begin_inset Formula $f\left(a\right)\not=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+Če sta 
+\begin_inset Formula $D,E\subseteq\mathbb{R}$
+\end_inset
+
+ in 
+\begin_inset Formula $f:D\to E$
+\end_inset
+
+ in 
+\begin_inset Formula $g:E\to\mathbb{R}$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $g\circ f:D\to\mathbb{R}$
+\end_inset
+
+.
+ Hkrati pa,
+ če je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna v 
+\begin_inset Formula $a$
+\end_inset
+
+ in 
+\begin_inset Formula $g$
+\end_inset
+
+ zvezna v 
+\begin_inset Formula $f\left(a\right)$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $g\circ f$
+\end_inset
+
+ zvezna v 
+\begin_inset Formula $a$
+\end_inset
+
+.
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+Velja 
+\begin_inset Formula $\left(g\circ f\right)\left(x\right)=g\left(f\left(x\right)\right)$
+\end_inset
+
+.
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Proof
+Vzemimo poljubno 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}\subseteq D$
+\end_inset
+
+,
+ da 
+\begin_inset Formula $a_{n}\to a\in D$
+\end_inset
+
+.
+ Zopet uporabimo 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{kzzz}{izrek o karakterizaciji zveznosti z zaporedji}
+\end_layout
+
+\end_inset
+
+:
+ ker je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna,
+ velja 
+\begin_inset Formula $f\left(a_{n}\right)\to f\left(a\right)$
+\end_inset
+
+ in ker je 
+\begin_inset Formula $g$
+\end_inset
+
+ zvezna,
+ velja 
+\begin_inset Formula $g\left(f\left(a_{n}\right)\right)\to g\left(f\left(a\right)\right)$
+\end_inset
+
+.
+ Potemtakem 
+\begin_inset Formula $\left(g\circ f\right)\left(a_{n}\right)\to\left(g\circ f\right)\left(a\right)$
+\end_inset
+
+ in po istem izreku je 
+\begin_inset Formula $g\circ f$
+\end_inset
+
+ zvezna na 
+\begin_inset Formula $D$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+Vsi polinomi so zvezni na 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Vzemimo 
+\begin_inset Formula $p\left(x\right)=\sum_{k=0}^{n}a_{k}k^{k}$
+\end_inset
+
+.
+ Uporabimo prejšnji izrek.
+ Polinom je sestavljen iz vsote konstantne funkcije,
+ zmnožene z identiteto,
+ ki je s seboj 
+\begin_inset Formula $n-$
+\end_inset
+
+krat množena.
+ Ker vsota in množenje ohranjata zveznost,
+ je treba dokazati le,
+ da je 
+\begin_inset Formula $f\left(x\right)=x$
+\end_inset
+
+ zvezna in da so 
+\begin_inset Formula $\forall c\in\mathbb{R}:f\left(x\right)=c$
+\end_inset
+
+ zvezne.
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $f\left(x\right)=x$
+\end_inset
+
+ Ali velja 
+\begin_inset Formula $\forall\varepsilon>0\exists\delta>0\ni:\forall x\in\mathbb{R}:\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(a\right)\right|<\varepsilon$
+\end_inset
+
+?
+ Da,
+ velja.
+ Vzamemo lahko katerokoli 
+\begin_inset Formula $\delta\in(0,\varepsilon]$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $f\left(x\right)=c$
+\end_inset
+
+ Naj bo 
+\begin_inset Formula $c\in\mathbb{R}$
+\end_inset
+
+ poljuben.
+ Tu je 
+\begin_inset Formula $\left|f\left(x\right)-f\left(a\right)\right|=\left|c-c\right|=0$
+\end_inset
+
+,
+ torej je desna stran implikacije vedno resnična,
+ torej je implikacija vedno resnična.
+\end_layout
+
+\end_deeper
+\begin_layout Theorem*
+Vse elementarne funkcije so na njihovih definicijskih območjih povsod zvezne.
+ To so:
+ polinomi,
+ potence,
+ racionalne funkcije,
+ koreni,
+ eksponentne funkcije,
+ logaritmi,
+ trigonometrične,
+ ciklometrične in kombinacije neskončno mnogo naštetih,
+ spojenih s 
+\begin_inset Formula $+,-,\cdot,/,\circ$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Tega izreka ne bomo dokazali.
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $f\left(x\right)\coloneqq\log\left(\sin^{3}x+\frac{1}{8}\right)+\frac{1}{\sqrt[4]{x-7}}$
+\end_inset
+
+ je zvezna povsod,
+ kjer je definirana.
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $\varepsilon>0,a\in\mathbb{R}$
+\end_inset
+
+ in 
+\begin_inset Formula $f:\left(a-\varepsilon,a+\varepsilon\right)\setminus\left\{ a\right\} \to\mathbb{R}$
+\end_inset
+
+.
+ Pravimo,
+ da je 
+\begin_inset Formula $L\in\mathbb{R}$
+\end_inset
+
+ limita 
+\begin_inset Formula $f$
+\end_inset
+
+ v točki 
+\begin_inset Formula $a$
+\end_inset
+
+ (zapišemo 
+\begin_inset Formula $L=\lim_{x\to a}f\left(x\right)$
+\end_inset
+
+),
+ če za vsako zaporedje 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}},a_{n}\in\left(a-\varepsilon,a+\varepsilon\right)\setminus\left\{ a\right\} $
+\end_inset
+
+,
+ za katero velja 
+\begin_inset Formula $a_{n}\to a$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $f\left(a_{n}\right)\to L$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+ZDB če 
+\begin_inset Formula $\forall\varepsilon>0\exists\delta>0\ni:\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-L\right|<\varepsilon$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+ZDB če za 
+\begin_inset Formula $\overline{f}:\left(a-\varepsilon,a+\varepsilon\right)\to\mathbb{R}$
+\end_inset
+
+ s predpisom 
+\begin_inset Formula $\overline{f}\left(x\right)\coloneqq\begin{cases}
+f\left(x\right) & ;x\in\left(a-\varepsilon,a+\varepsilon\right)\setminus\left\{ a\right\} \\
+L & ;x\in a
+\end{cases}$
+\end_inset
+
+ velja,
+ da je zvezna v 
+\begin_inset Formula $a$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Note*
+Vrednost 
+\begin_inset Formula $f\left(a\right)$
+\end_inset
+
+,
+ če sploh obstaja,
+ nima vloge pri vrednosti limite.
+\end_layout
+
+\begin_layout Corollary*
+Naj bo 
+\begin_inset Formula $a\in D\subseteq\mathbb{R}$
+\end_inset
+
+ in 
+\begin_inset Formula $f:D\to\mathbb{R}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $f$
+\end_inset
+
+ je zvezna v 
+\begin_inset Formula $a\Leftrightarrow\lim_{x\to a}f\left(x\right)=f\left(a\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Naj bo 
+\begin_inset Formula $f:\mathbb{R}\setminus\left\{ 0\right\} \to\mathbb{R}$
+\end_inset
+
+ s predpisom 
+\begin_inset Formula $x\mapsto\frac{\sin x}{x}$
+\end_inset
+
+.
+ Zanima nas,
+ ali obstaja 
+\begin_inset Formula $\lim_{x\to0}f\left(x\right)$
+\end_inset
+
+.
+ Grafični dokaz.
+\end_layout
+
+\begin_layout Example*
+\begin_inset Float figure
+placement document
+alignment document
+wide false
+sideways false
+status open
+
+\begin_layout Plain Layout
+TODO XXX FIXME SKICA S TKZ EUCLID,
+ glej ZVZ III/ANA1P1120/str.8
+\end_layout
+
+\begin_layout Plain Layout
+\begin_inset Caption Standard
+
+\begin_layout Plain Layout
+Skica.
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Očitno velja 
+\begin_inset Formula $\triangle ABD\subset$
+\end_inset
+
+ krožni izsek 
+\begin_inset Formula $DAB\subset\triangle ABC$
+\end_inset
+
+,
+ torej za njihove ploščine velja 
+\begin_inset Formula 
+\[
+\frac{\sin x}{2}\leq\frac{x}{2\pi}\cdot x=\frac{x}{2}\leq\frac{\tan x}{2}\quad\quad\quad\quad/\cdot\frac{2}{\sin x}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+1\leq\frac{x}{\sin x}\leq\frac{1}{\cos x}\quad\quad\quad\quad/\lim_{x\to0}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\lim_{x\to0}1\leq\lim_{x\to0}\frac{x}{\sin x}\leq\lim_{x\to0}\frac{1}{\cos x}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+1\leq\lim_{x\to0}\frac{x}{\sin x}\leq1
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\lim_{x\to0}\frac{x}{\sin x}=1
+\]
+
+\end_inset
+
+Da naš sklep res potrdimo,
+ je potreben spodnji izrek.
+\end_layout
+
+\begin_layout Theorem*
+Če za 
+\begin_inset Formula $f,g,h:D\to\mathbb{R}$
+\end_inset
+
+ velja za 
+\begin_inset Formula $a\in D$
+\end_inset
+
+:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $\exists\varepsilon>0\forall x\in\left(a-\varepsilon,a+\varepsilon\right)\setminus\left\{ a\right\} :f\left(x\right)\leq g\left(x\right)\leq h\left(x\right)$
+\end_inset
+
+ in hkrati
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $\exists\lim_{x\to a}f\left(x\right),\lim_{x\to a}h\left(x\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $\lim_{x\to a}f\left(x\right)=\lim_{x\to a}h\left(x\right)=1$
+\end_inset
+
+,
+ tedaj tudi 
+\begin_inset Formula $\exists\lim_{x\to a}g\left(x\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $\lim_{x\to a}g\left(x\right)=1$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Proof
+TODO XXX FIXME DOKAZ V SKRIPTi
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $\lim_{x\to0}\frac{1}{1+e^{1/x}}$
+\end_inset
+
+ ne obstaja.
+ Zakaj?
+ Izračunajmo levo in desno limito:
+\begin_inset Formula 
+\[
+\lim_{x\searrow0}\frac{1}{1+e^{1/x}}=0,\lim_{x\nearrow0}\frac{1}{1+e^{1/x}}=1
+\]
+
+\end_inset
+
+Toda 
+\begin_inset Formula $\exists\lim_{x\to a}f\left(x\right)\Leftrightarrow\exists\lim_{x\nearrow a}f\left(x\right)\wedge\exists\lim_{x\searrow a}f\left(x\right)\wedge\lim_{x\nearrow a}f\left(x\right)=\lim_{x\searrow a}f\left(x\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Označimo 
+\begin_inset Formula $\lim_{x\searrow a}f\left(x\right)\eqqcolon f\left(a+0\right),\lim_{x\nearrow a}f\left(x\right)\eqqcolon f\left(a-0\right)$
+\end_inset
+
+.
+ Če 
+\begin_inset Formula $\exists f\left(a+0\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $\exists f\left(a-0\right)$
+\end_inset
+
+,
+ vendar 
+\begin_inset Formula $f\left(a+0\right)\not=f\left(a-0\right)$
+\end_inset
+
+,
+ pravimo,
+ da ima 
+\begin_inset Formula $f$
+\end_inset
+
+ v točki 
+\begin_inset Formula $a$
+\end_inset
+
+ 
+\begin_inset Quotes gld
+\end_inset
+
+skok
+\begin_inset Quotes grd
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Corollary*
+sssssssssss 
 \end_layout
 
 \begin_layout Corollary*