From daab103c6258be3773c05077091ec6e010f4924b Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?Anton=20Luka=20=C5=A0ijanec?= <anton@sijanec.eu>
Date: Thu, 15 Aug 2024 00:42:02 +0200
Subject: =?UTF-8?q?za=C4=8Denjam=2015.=20predvanje,=20git=20push,=20grem?=
 =?UTF-8?q?=20spat?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 "\305\241ola/ana1/teor.lyx" | 4143 ++++++++++++++++++++++++++++++++++++++++++-
 1 file changed, 4119 insertions(+), 24 deletions(-)

diff --git "a/\305\241ola/ana1/teor.lyx" "b/\305\241ola/ana1/teor.lyx"
index e44036f..af865e2 100644
--- "a/\305\241ola/ana1/teor.lyx"
+++ "b/\305\241ola/ana1/teor.lyx"
@@ -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
@@ -10101,7 +10127,20 @@ Primeri odvodov preprostih funkcij.
 
 \end_deeper
 \begin_layout Claim*
-Za poljuben 
+\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
 
@@ -10234,7 +10273,20 @@ Od prej vemo
 \end_layout
 
 \begin_layout Claim*
-Naj bo 
+\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
 
@@ -10686,7 +10738,20 @@ Dokažimo vse štiri trditve.
 \end_layout
 
 \begin_layout Theorem*
-Naj bo 
+\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
 
@@ -10887,7 +10952,41 @@ x^{2}\sin\frac{1}{x} & ;x\not=0\\
 :
 \begin_inset Formula 
 \[
-f'\left(0\right)=\lim_{h\to0}\frac{f\left(h\right)-0}{h}=\lim_{h\to0}\cdots\text{NADALJUJEM JUTRI}
+f'\left(0\right)=\lim_{h\to0}\frac{f\left(h\right)-0}{h}=\lim_{h\to0}\frac{h^{\cancel{2}}\sin\frac{1}{h}}{\cancel{h}}=\lim_{h\to0}h\sin\frac{1}{h}=0,
+\]
+
+\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
@@ -10895,43 +10994,4039 @@ f'\left(0\right)=\lim_{h\to0}\frac{f\left(h\right)-0}{h}=\lim_{h\to0}\cdots\text
 
 \end_layout
 
-\begin_layout Standard
-\begin_inset Separator plain
+\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 Corollary*
-sssssssssss
+\begin_layout Proof
+Ker je 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{zism}{zvezna in strogo monotona}
 \end_layout
 
-\begin_layout Corollary*
-sssssssssss
+\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 Corollary*
-sssssssssss
+\begin_layout Proof
+Uporabimo formulo za 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{ok}{odvod kompozituma}
 \end_layout
 
-\begin_layout Corollary*
-sssssssssss
+\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 Corollary*
-sssssssssss
+\begin_layout Example*
+Nekaj primerov odvodov inverza.
 \end_layout
 
-\begin_layout Corollary*
-sssssssssss
+\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
 
-\begin_layout Corollary*
-sssssssssss
+\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_layout Corollary*
-sssssssssss
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $h\left(x\right)=\sqrt[n]{x^{m}}=x^{\frac{m}{n}}=g\left(x\right)^{m}$
+\end_inset
+
+ za 
+\begin_inset Formula $n,m\in\mathbb{N},x>0$
+\end_inset
+
+.
+ Uporabimo formulo za 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{op}{odvod potence}
+\end_layout
+
+\end_inset
+
+ in 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{ok}{kompozituma}
+\end_layout
+
+\end_inset
+
+ in zgornji primer.
+ Velja 
+\begin_inset Formula $g'\left(x\right)=\frac{1}{n}x^{\frac{1}{n}-1}$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula 
+\[
+h'\left(x\right)=mg\left(x\right)^{m-1}\cdot g'\left(x\right)=m\left(x^{\frac{1}{n}}\right)^{m-1}\cdot\frac{1}{n}x^{\frac{1}{n}-1}=\frac{m}{n}x^{\frac{m-1}{n}+\frac{1}{n}-1}=\frac{m}{n}x^{\frac{m}{n}-1}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset CommandInset label
+LatexCommand label
+name "enu:Izkaže-se,-da"
+
+\end_inset
+
+Izkaže se,
+ da velja celo 
+\begin_inset Formula $\forall x>0,\alpha\in\mathbb{R}:\left(x^{\alpha}\right)'=\alpha x^{\alpha-1}$
+\end_inset
+
+.
+ Mi smo dokazali le za 
+\begin_inset Formula $\alpha\in\mathbb{Q}$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Logaritmi,
+ inverz 
+\begin_inset Formula $e^{x}$
+\end_inset
+
+.
+ Gre za 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{oef}{odvod eksponentne funkcije}
+\end_layout
+
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\left(a^{x}\right)=a^{x}\ln a$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $\left(e^{x}\right)=e^{x}\ln e=e^{x}$
+\end_inset
+
+.
+ Uporavimo 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{oi}{odvod inverza}
+\end_layout
+
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\left(f^{-1}\right)'\left(x\right)=\frac{1}{f'\left(f^{-1}\left(x\right)\right)}$
+\end_inset
+
+ in za 
+\begin_inset Formula $g\left(x\right)=\log x$
+\end_inset
+
+ uporabimo 
+\begin_inset Formula $g\left(x\right)=f^{-1}\left(x\right)$
+\end_inset
+
+,
+ kjer je 
+\begin_inset Formula $f\left(x\right)=e^{x}$
+\end_inset
+
+:
+\begin_inset Formula 
+\[
+\log'\left(x\right)=\left(\left(e^{x}\right)^{-1}\right)'\left(x\right)=\frac{1}{e^{\log x}}=\frac{1}{x}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $g\left(x\right)=\arcsin x$
+\end_inset
+
+ za 
+\begin_inset Formula $x\in\left[-1,1\right]$
+\end_inset
+
+,
+ torej je 
+\begin_inset Formula $g=f^{-1}$
+\end_inset
+
+,
+ kjer je 
+\begin_inset Formula $f=\sin$
+\end_inset
+
+ za 
+\begin_inset Formula $x\in\left[\frac{-\pi}{2},\frac{\pi}{2}\right]$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+g'\left(f\left(x\right)\right)=\frac{1}{f'\left(x\right)}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+g'\left(\sin x\right)=\frac{1}{\cos x}
+\]
+
+\end_inset
+
+Ker velja 
+\begin_inset Formula $\sin^{2}x+\cos^{2}x=1$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $\cos^{2}x=1-\sin^{2}x$
+\end_inset
+
+,
+ sledi 
+\begin_inset Formula $\cos x=\sqrt{1-\sin^{2}x}$
+\end_inset
+
+,
+ torej nadaljujemo:
+\begin_inset Formula 
+\[
+g'\left(\sin x\right)=\frac{1}{\sqrt{1-\sin^{2}x}}
+\]
+
+\end_inset
+
+Sedaj zamenjamo 
+\begin_inset Formula $\sin x$
+\end_inset
+
+ s 
+\begin_inset Formula $t$
+\end_inset
+
+ in dobimo:
+\begin_inset Formula 
+\[
+g'\left(t\right)=\frac{1}{\sqrt{1-t^{2}}}=\arcsin^{2}t
+\]
+
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Subsection
+Diferencial
+\end_layout
+
+\begin_layout Standard
+Fiksirajmo funkcijo 
+\begin_inset Formula $f$
+\end_inset
+
+ in točko 
+\begin_inset Formula $a\in\mathbb{R}$
+\end_inset
+
+,
+ v okolici katere je 
+\begin_inset Formula $f$
+\end_inset
+
+ definirana.
+ Želimo oceniti vrednost funkcije 
+\begin_inset Formula $f$
+\end_inset
+
+ v bližini točke 
+\begin_inset Formula $a$
+\end_inset
+
+ z linearno funkcijo – to je 
+\begin_inset Formula $y\left(x\right)=\lambda x$
+\end_inset
+
+ za neki 
+\begin_inset Formula $\lambda\in\mathbb{R}$
+\end_inset
+
+.
+ ZDB Iščemo najboljši linearni približek,
+ odvisen od 
+\begin_inset Formula $h$
+\end_inset
+
+,
+ za 
+\begin_inset Formula $f\left(a+h\right)-f\left(a\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $f$
+\end_inset
+
+ definirana v okolici točke 
+\begin_inset Formula $a\in\mathbb{R}$
+\end_inset
+
+.
+ Diferencial funkcije 
+\begin_inset Formula $f$
+\end_inset
+
+ v točki 
+\begin_inset Formula $a$
+\end_inset
+
+ je linearna preslikava 
+\begin_inset Formula $df\left(a\right):\mathbb{R}\to\mathbb{R}$
+\end_inset
+
+ z zahtevo
+\begin_inset Formula 
+\[
+\lim_{h\to0}\frac{\left|f\left(a+h\right)-f\left(a\right)-df\left(a\right)\left(h\right)\right|}{\left|h\right|}=0.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Note*
+\begin_inset Formula 
+\[
+\lim_{h\to0}\frac{f\left(a+h\right)-f\left(a\right)-df\left(a\right)\left(h\right)}{h}=0=\lim_{h\to0}\left(\frac{f\left(a+h\right)-f\left(a\right)}{h}-\frac{\left(df\left(a\right)\right)\left(h\right)}{h}\right)=
+\]
+
+\end_inset
+
+Upoštevamo linearnost preslikave
+\begin_inset Formula 
+\[
+=\lim_{h\to0}\frac{f\left(a+h\right)-f\left(a\right)}{h}-df\left(a\right)=f'\left(a\right)-df\left(a\right)=0
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+f'\left(a\right)=df\left(a\right)
+\]
+
+\end_inset
+
+Torej 
+\begin_inset Formula $f\left(a+h\right)-f\left(a\right)\approx df\left(a\right)\left(h\right)$
+\end_inset
+
+ – najboljši linearni približek za 
+\begin_inset Formula $f\left(a+h\right)-f\left(h\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Uporaba diferenciala.
+ 
+\begin_inset Formula $a$
+\end_inset
+
+ je točka,
+ v kateri znamo izračunati funkcijsko vrednost,
+ 
+\begin_inset Formula $a+h$
+\end_inset
+
+ pa je točka,
+ v kateri želimo približek funkcijske vrednosti.
+ Izračunajmo približek 
+\begin_inset Formula $\sqrt{2}$
+\end_inset
+
+:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $f\left(x\right)=\sqrt{x}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $a+h=2$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $a=2,25$
+\end_inset
+
+,
+ 
+\begin_inset Formula $h=-0,25$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $f\left(a\right)=\sqrt{a}=1,5$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $f'\left(s\right)=\frac{1}{2\sqrt{x}}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $f\left(a=2,25\right)=\frac{1}{3}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $f\left(2\right)\approx f\left(a\right)+f'\left(2,25\right)\cdot h=1,5-0,25\cdot\frac{1}{3}=\frac{3}{2}-\frac{1}{4}\cdot\frac{1}{3}=\frac{17}{12}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Preizkus:
+ 
+\begin_inset Formula $\left(\frac{17}{12}\right)^{2}=\frac{289}{144}=2+\frac{1}{144}$
+\end_inset
+
+ ...
+ Absolutna napaka 
+\begin_inset Formula $\frac{1}{144}$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $I\subseteq\mathbb{R}$
+\end_inset
+
+ interval in 
+\begin_inset Formula $f:I\to\mathbb{R}$
+\end_inset
+
+ odvedljiva povsod na 
+\begin_inset Formula $I$
+\end_inset
+
+.
+ Vzemimo 
+\begin_inset Formula $a\in I$
+\end_inset
+
+.
+ Če je v 
+\begin_inset Formula $a$
+\end_inset
+
+ odvedljiva tudi 
+\begin_inset Formula $f'$
+\end_inset
+
+,
+ pišemo 
+\begin_inset Formula $f''\left(a\right)=\left(f'\left(a\right)\right)'$
+\end_inset
+
+.
+ Podobno pišemo tudi višje odvode:
+ 
+\begin_inset Formula $f^{\left(1\right)}\left(a\right)=f'\left(a\right)$
+\end_inset
+
+,
+ 
+\begin_inset Formula $f^{\left(n+1\right)}=\left(f^{\left(n\right)}\right)'$
+\end_inset
+
+,
+ 
+\begin_inset Formula $f^{\left(0\right)}\left(a\right)=f\left(a\right)$
+\end_inset
+
+,
+ 
+\begin_inset Formula $f^{\left(2\right)}\left(a\right)=f''\left(a\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Note*
+Pomen besede 
+\begin_inset Quotes gld
+\end_inset
+
+odvod
+\begin_inset Quotes grd
+\end_inset
+
+:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Odvod v dani točki:
+ 
+\begin_inset Formula $f'\left(a\right)$
+\end_inset
+
+ za fiksen 
+\begin_inset Formula $a\in\mathbb{R}$
+\end_inset
+
+ ali
+\end_layout
+
+\begin_layout Itemize
+Funkcija,
+ ki vsaki točki 
+\begin_inset Formula $x\in\mathbb{R}$
+\end_inset
+
+ priredi 
+\begin_inset Formula $f'\left(x\right)$
+\end_inset
+
+ po zgornji definiciji.
+\end_layout
+
+\end_deeper
+\begin_layout Definition*
+\begin_inset Formula $C^{n}\left(I\right)$
+\end_inset
+
+ je množica funkcije 
+\begin_inset Formula $f:I\to\mathbb{R}$
+\end_inset
+
+,
+ da 
+\begin_inset Formula $\forall x\in I\exists f'\left(x\right),f''\left(x\right),f^{\left(3\right)},\dots,f^{\left(n\right)}\left(x\right)$
+\end_inset
+
+ in da so 
+\begin_inset Formula $f,f',f'',f^{\left(3\right)},\dots,f^{\left(n\right)}$
+\end_inset
+
+ zvezna funkcije na 
+\begin_inset Formula $I$
+\end_inset
+
+.
+ (seveda če obstaja 
+\begin_inset Formula $j-$
+\end_inset
+
+ti odvod,
+ obstaja tudi zvezen 
+\begin_inset Formula $j-1-$
+\end_inset
+
+ti odvod).
+ ZDB je to množica funkcij,
+ ki imajo vse odvode do 
+\begin_inset Formula $n$
+\end_inset
+
+ in so le-ti zvezni.
+ ZDB to so vse 
+\begin_inset Formula $n-$
+\end_inset
+
+krat zvezno odvedljive funkcije na intervalu 
+\begin_inset Formula $I$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Označimo 
+\begin_inset Formula $C^{\infty}\left(I\right)\coloneqq\bigcap_{n=1}^{\infty}C^{n}\left(I\right)$
+\end_inset
+
+ – to so neskončnokrat odvedljive funkcije na intervalu 
+\begin_inset Formula $I$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Note*
+Intuitivno
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+Baje.
+ Jaz sem itak do vsega skeptičen.
+\end_layout
+
+\end_inset
+
+ velja 
+\begin_inset Formula $C^{1}\left(I\right)\supset C^{2}\left(I\right)\supset C^{3}\left(I\right)\supset C^{4}\left(I\right)\supset\cdots$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Nekaj primerov.
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Polimomi 
+\begin_inset Formula $\subset C^{\infty}\left(\mathbb{R}\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $f\left(x\right)=\left|x\right|^{3}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $f'\left(x\right)=\begin{cases}
+3x^{2} & ;x\geq0\\
+-3x^{2} & ;x<0
+\end{cases}=3x^{2}\sgn x$
+\end_inset
+
+,
+ 
+\begin_inset Formula $f''\left(x\right)=\begin{cases}
+6x & ;x\geq0\\
+-6x & ;x<0
+\end{cases}=6x\sgn x$
+\end_inset
+
+,
+ 
+\begin_inset Formula $f'''\left(x\right)=\begin{cases}
+6 & ;x>0\\
+-6 & ;x<0
+\end{cases}=6\sgn x$
+\end_inset
+
+ in v 
+\begin_inset Formula $0$
+\end_inset
+
+ ni odvedljiva,
+ zato 
+\begin_inset Formula $f\in C^{2}\left(\mathbb{R}\right)$
+\end_inset
+
+ a 
+\begin_inset Formula $f\not\in C^{3}\left(\mathbb{R}\right)$
+\end_inset
+
+,
+ ker 
+\begin_inset Formula $\exists f''$
+\end_inset
+
+ in je zvezna na 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+,
+ a 
+\begin_inset Formula $f'''$
+\end_inset
+
+ sicer obstaja,
+ a ni zvezna na 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+.
+ Velja pa 
+\begin_inset Formula $f\in C^{\infty}\left(\mathbb{R}\setminus\left\{ 0\right\} \right)$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Theorem*
+Rolle.
+ Naj bo 
+\begin_inset Formula $f:\left[a,b\right]\to\mathbb{R}$
+\end_inset
+
+ za 
+\begin_inset Formula $a,b\in\mathbb{R}$
+\end_inset
+
+ zvezna na 
+\begin_inset Formula $\left[a,b\right]$
+\end_inset
+
+ in odvedljiva na 
+\begin_inset Formula $\left(a,b\right)$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+f\left(a\right)=f\left(b\right)\Longrightarrow\exists\alpha\in\left(a,b\right)\ni:f'\left(\alpha\right)=0
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Proof
+Sumimo,
+ da je ustrezna 
+\begin_inset Formula $\alpha$
+\end_inset
+
+ tista,
+ ki je 
+\begin_inset Formula $\max$
+\end_inset
+
+ ali 
+\begin_inset Formula $\min$
+\end_inset
+
+ od 
+\begin_inset Formula $f$
+\end_inset
+
+ na 
+\begin_inset Formula $I$
+\end_inset
+
+.
+ 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{zfnkm}{Ker}
+\end_layout
+
+\end_inset
+
+ je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna na 
+\begin_inset Formula $\left[a,b\right]$
+\end_inset
+
+ (kompaktni množici),
+ 
+\begin_inset Formula $\exists\alpha_{1}\in\left[a,b\right],\alpha_{2}\in\left[a,b\right]\ni:f\left(\alpha_{1}\right)=\max f\left(\left[a,b\right]\right)\wedge f\left(\alpha_{2}\right)=\min f\left(\left[a,b\right]\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Če je 
+\begin_inset Formula $\left\{ \alpha_{1},\alpha_{2}\right\} \subseteq\left\{ a,b\right\} $
+\end_inset
+
+,
+ je 
+\begin_inset Formula $f\left(\alpha_{1}\right)=f\left(\alpha_{2}\right)$
+\end_inset
+
+ in je v tem primeru 
+\begin_inset Formula $f$
+\end_inset
+
+ konstanta (
+\begin_inset Formula $\exists!c\in\mathbb{R}\ni:f\left(x\right)=c$
+\end_inset
+
+),
+ ki je odvedljiva in ima povsod odvod nič.
+\end_layout
+
+\begin_layout Proof
+Sicer pa 
+\begin_inset Formula $\left\{ \alpha_{1},\alpha_{2}\right\} \not\subseteq\left\{ a,b\right\} $
+\end_inset
+
+.
+ Tedaj ločimo dva primera:
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\alpha_{1}\in\left(a,b\right)$
+\end_inset
+
+ To pomeni,
+ da je globalni maksimum na odprtem intervalu.
+ Trdimo,
+ da je v lokalnem maksimumu odvod 0.
+ Dokaz:
+\begin_inset Formula 
+\[
+f'\left(\alpha_{1}\right)=\lim_{h\to0}\frac{f\left(\alpha_{1}+h\right)-f\left(\alpha_{1}\right)}{h}
+\]
+
+\end_inset
+
+Za 
+\begin_inset Formula $a_{1}$
+\end_inset
+
+ (maksimum) velja 
+\begin_inset Formula $f\left(\alpha_{1}+h\right)-f\left(\alpha_{1}\right)\leq0$
+\end_inset
+
+ (čim se pomaknemo izven točke,
+ v kateri je maksimum,
+ je funkcijska vrednost nižja).
+ Potemtakem velja
+\begin_inset Formula 
+\[
+\frac{f\left(\alpha_{1}+h\right)-f\left(\alpha_{1}\right)}{h}\quad\begin{cases}
+\leq0 & ;h>0\\
+\geq0 & ;h<0
+\end{cases}
+\]
+
+\end_inset
+
+Ker je funkcija odvedljiva na odprtem intervalu,
+ sta leva in desna limita enaki.
+\begin_inset Formula 
+\[
+0\geq\lim_{h\searrow0}\frac{f\left(\alpha_{1}+h\right)-f\left(\alpha_{1}\right)}{h}=\lim_{h\nearrow0}\frac{f\left(\alpha_{1}+h\right)-f\left(\alpha_{1}\right)}{h}\geq0
+\]
+
+\end_inset
+
+Sledi
+\begin_inset Formula 
+\[
+\lim_{h\to0}\frac{f\left(\alpha_{1}+h\right)-f\left(\alpha_{1}\right)}{h}=f'\left(x\right)=0
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\alpha_{2}\in\left(a,b\right)$
+\end_inset
+
+ To pomeni,
+ da je globalni minimum na odprtem intervalu.
+ Trdimo,
+ da je v lokalnem minimumu odvod 0.
+ Dokaz je podoben tistemu za lokalni maksimum.
+\end_layout
+
+\begin_layout Theorem*
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hypertarget{lagrange}{Lagrange}
+\end_layout
+
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $f:\left[a,b\right]\to\mathbb{R}$
+\end_inset
+
+ za 
+\begin_inset Formula $a,b\in\mathbb{R}$
+\end_inset
+
+ zvezna na 
+\begin_inset Formula $\left[a,b\right]$
+\end_inset
+
+ in odvedljiva na 
+\begin_inset Formula $\left(a,b\right)$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+\exists\alpha\in\left(a,b\right)\ni:f\left(b\right)-f\left(a\right)=f'\left(\alpha\right)\left(b-a\right)\sim\frac{f\left(b\right)-f\left(a\right)}{b-a}=f'\left(\alpha\right)
+\]
+
+\end_inset
+
+ZDB na neki točki na grafu funkcije je tangenta na graf funkcije vzporedna premici,
+ ki jo določata točki 
+\begin_inset Formula $\left(a,f\left(a\right)\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $\left(b,f\left(b\right)\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Za dokaz Lagrangevega uporabimo Rolleov izrek.
+ Splošen primer prevedemo na primer 
+\begin_inset Formula $h\left(a\right)=h\left(b\right)$
+\end_inset
+
+ tako,
+ da od naše splošne funkcije 
+\begin_inset Formula $f$
+\end_inset
+
+ odštejemo linearno funkcijo 
+\begin_inset Formula $g$
+\end_inset
+
+,
+ da bo veljalo 
+\begin_inset Formula $\left(f-g\right)\left(a\right)=\left(f-g\right)\left(b\right)$
+\end_inset
+
+.
+ Za funkcijo 
+\begin_inset Formula $g\left(x\right)$
+\end_inset
+
+ mora veljati naslednje:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $\exists k,n\in\mathbb{R}\ni:f\left(x\right)=kx+n$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $g\left(a\right)=0$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $g\left(b\right)=f\left(b\right)-f\left(a\right)$
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Proof
+Opazimo,
+ da mora biti koeficient funkcije 
+\begin_inset Formula $g$
+\end_inset
+
+ enak 
+\begin_inset Formula $\frac{f\left(b\right)-f\left(a\right)}{b-a}$
+\end_inset
+
+,
+ vertikalni odklon pa tolikšen,
+ da ima funkcija 
+\begin_inset Formula $g$
+\end_inset
+
+ v 
+\begin_inset Formula $a$
+\end_inset
+
+ ničlo:
+\begin_inset Formula 
+\[
+\frac{f\left(b\right)-f\left(a\right)}{b-a}a+n=0
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+n=-\frac{f\left(b\right)-f\left(a\right)}{b-a}a
+\]
+
+\end_inset
+
+Našli smo funkcijo 
+\begin_inset Formula $g\left(x\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}\left(x-a\right)$
+\end_inset
+
+.
+ Funkcija 
+\begin_inset Formula $\left(f-g\right)$
+\end_inset
+
+ sedaj ustreza pogojem za Rolleov izrek,
+ torej 
+\begin_inset Formula $\exists\alpha\in\left[a,b\right]\ni:\left(f-g\right)'\left(\alpha\right)=0\Leftrightarrow g'\left(\alpha\right)=f'\left(\alpha\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$
+\end_inset
+
+,
+ kar smo želeli dokazati.
+\end_layout
+
+\begin_layout Corollary*
+Naj bo 
+\begin_inset Formula $I\subseteq\mathbb{R}$
+\end_inset
+
+ nenujno zaprt niti omejen in 
+\begin_inset Formula $f:I\to\mathbb{R}$
+\end_inset
+
+ odvedljiva na 
+\begin_inset Formula $I$
+\end_inset
+
+.
+ Tedaj je 
+\begin_inset Formula $f$
+\end_inset
+
+ Lipschitzova.
+ Lipschitzove funkcije so enakomerno zvezne.
+\end_layout
+
+\begin_layout Proof
+Po Lagrangeu velja 
+\begin_inset Formula $\forall x,y\in I\exists\alpha\in\left(x,y\right)\ni:f\left(x\right)-f\left(y\right)=f'\left(\alpha\right)\left(x-y\right)$
+\end_inset
+
+.
+ Potemtakem 
+\begin_inset Formula $\left|f\left(x\right)-f\left(y\right)\right|=\left|f'\left(\alpha\right)\right|\left|x-y\right|\leq\sup_{\beta\in\left(x,y\right)}\left|f'\left(\beta\right)\right|\left|x-y\right|$
+\end_inset
+
+.
+ Torej 
+\begin_inset Formula $\exists M>0\forall x,y:\left|f\left(x\right)-f\left(y\right)\right|\leq M\left|x-y\right|$
+\end_inset
+
+,
+ enakomerno zveznost pa dobimo tako,
+ da 
+\begin_inset Formula $\delta\left(\varepsilon\right)=\frac{\varepsilon}{\sup_{\beta\in I}\left|f'\left(\beta\right)\right|}$
+\end_inset
+
+.
+ Računajmo.
+ Naj bo 
+\begin_inset Formula $M=\sup_{\beta\in I}\left|f'\left(\beta\right)\right|$
+\end_inset
+
+,
+ ki obstaja.
+\begin_inset Formula 
+\[
+\forall x,y:\left|f\left(x\right)-f\left(y\right)\right|\leq M\left|x-y\right|
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\forall x,y:\left|x-y\right|<\frac{\varepsilon}{\sup_{\beta\in I}\left|f'\left(\beta\right)\right|}\Rightarrow\left|f\left(x\right)-f\left(y\right)\right|\leq M\left|x-y\right|<\cancel{\sup_{\beta\in I}\left|f'\left(\beta\right)\right|}\frac{\varepsilon}{\cancel{\sup_{\beta\in I}\left|f'\left(\beta\right)\right|}}<\varepsilon
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\forall\varepsilon\exists\delta\left(\varepsilon\right)\forall x,y:\left|x-y\right|<\delta\left(\varepsilon\right)\Rightarrow\left|f\left(x\right)-f\left(y\right)\right|<\varepsilon
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Note*
+Lipschnitzovim funkcijam pravimo tudi Hölderjeve funkcije reda 1.
+ 
+\begin_inset Formula $f$
+\end_inset
+
+ je Hölderjeva funkcija reda 
+\begin_inset Formula $r$
+\end_inset
+
+,
+ če velja 
+\begin_inset Formula $\exists M>0\forall x,y\in I:\left|f\left(x\right)-f\left(y\right)\right|\leq M\left|x-y\right|^{r}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+Naj bo 
+\begin_inset Formula $I$
+\end_inset
+
+ odprti interval,
+ 
+\begin_inset Formula $f:I\to\mathbb{R}$
+\end_inset
+
+ odvedljiva.
+ Tedaj:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $f$
+\end_inset
+
+ narašča na 
+\begin_inset Formula $I\Leftrightarrow f'\geq0$
+\end_inset
+
+ na 
+\begin_inset Formula $I$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $f$
+\end_inset
+
+ pada na 
+\begin_inset Formula $I\Leftrightarrow f'\leq0$
+\end_inset
+
+ na 
+\begin_inset Formula $I$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $f$
+\end_inset
+
+ strogo narašča na 
+\begin_inset Formula $I\Leftarrow f'>0$
+\end_inset
+
+ na 
+\begin_inset Formula $I$
+\end_inset
+
+.
+ Protiprimer,
+ da ni 
+\begin_inset Formula $\Leftrightarrow:f\left(x\right)=x^{3}$
+\end_inset
+
+,
+ ki strogo narašča,
+ toda 
+\begin_inset Formula $f'\left(0\right)=0$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $f$
+\end_inset
+
+ strogo pada na 
+\begin_inset Formula $I\Leftarrow f'<0$
+\end_inset
+
+ na 
+\begin_inset Formula $I$
+\end_inset
+
+.
+ Protiprimer,
+ da ni 
+\begin_inset Formula $\Leftrightarrow:f\left(x\right)=-x^{3}$
+\end_inset
+
+,
+ ki strogo pada,
+ toda 
+\begin_inset Formula $f'\left(0\right)=0$
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Proof
+Dokažimo le 
+\begin_inset Formula $f$
+\end_inset
+
+ narašča na 
+\begin_inset Formula $I\Leftrightarrow f'\geq0$
+\end_inset
+
+ na 
+\begin_inset Formula $I$
+\end_inset
+
+.
+ Drugo točko dokažemo podobno.
+ Dokazujemo ekvivalenco:
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Leftarrow\right)$
+\end_inset
+
+ 
+\begin_inset Formula $f'\geq0\Rightarrow f$
+\end_inset
+
+ narašča.
+ Vzemimo poljubna 
+\begin_inset Formula $t_{1}<t_{2}\in I$
+\end_inset
+
+.
+ Po Lagrangeu 
+\begin_inset Formula $\exists\alpha\in\left(t_{1},t_{2}\right)\ni:f\left(t_{2}\right)-f\left(t_{1}\right)=f'\left(\alpha\right)\left(t_{2}-t_{1}\right)$
+\end_inset
+
+.
+ Ker je po predpostavki 
+\begin_inset Formula $f'\left(\alpha\right)\geq0$
+\end_inset
+
+ in 
+\begin_inset Formula $t_{2}-t_{1}>0$
+\end_inset
+
+,
+ je tudi 
+\begin_inset Formula $f\left(t_{2}\right)-f\left(t_{1}\right)\geq0$
+\end_inset
+
+ in zato 
+\begin_inset Formula $f\left(t_{2}\right)\geq f\left(t_{1}\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Rightarrow\right)$
+\end_inset
+
+ 
+\begin_inset Formula $f$
+\end_inset
+
+ narašča 
+\begin_inset Formula $\Rightarrow f'\geq0$
+\end_inset
+
+.
+ Velja 
+\begin_inset Formula $f'\left(x\right)=\lim_{h\to0}\frac{f\left(x+h\right)-f\left(x\right)}{h}$
+\end_inset
+
+.
+ Po predpostavki je 
+\begin_inset Formula $f\left(x+h\right)-f\left(x\right)\geq0$
+\end_inset
+
+,
+ čim je 
+\begin_inset Formula $h>0$
+\end_inset
+
+,
+ in 
+\begin_inset Formula $f\left(x+h\right)-f\left(x\right)\leq0$
+\end_inset
+
+,
+ čim je 
+\begin_inset Formula $h<0$
+\end_inset
+
+.
+ Torej je ulomek vedno nenegativen.
+\end_layout
+
+\end_deeper
+\begin_layout Subsection
+Konveksnost in konkavnost
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $I\subseteq\mathbb{R}$
+\end_inset
+
+ interval in 
+\begin_inset Formula $f:I\to\mathbb{R}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $f$
+\end_inset
+
+ je konveksna na 
+\begin_inset Formula $I$
+\end_inset
+
+,
+ če 
+\begin_inset Formula $\forall a,b\in I$
+\end_inset
+
+ daljica 
+\begin_inset Formula $\left(a,f\left(a\right)\right),\left(b,f\left(b\right)\right)$
+\end_inset
+
+ leži nad grafom 
+\begin_inset Formula $f$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Enačba premice,
+ ki vsebuje to daljico,
+ se glasi (razmislek je podoben kot pri 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{lagrange}{Lagrangevem izreku}
+\end_layout
+
+\end_inset
+
+)
+\begin_inset Formula 
+\[
+g\left(x\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}\left(x-a\right)+f\left(a\right)
+\]
+
+\end_inset
+
+Za konveksno funkcijo torej velja 
+\begin_inset Formula $\forall a,b\in I:\forall x\in\left(a,b\right):f\left(x\right)\leq\frac{f\left(b\right)-f\left(a\right)}{b-a}\left(x-a\right)+f\left(a\right)$
+\end_inset
+
+ oziroma
+\begin_inset Formula 
+\[
+\frac{f\left(x\right)-f\left(a\right)}{x-a}\leq\frac{f\left(b\right)-f\left(a\right)}{b-a}
+\]
+
+\end_inset
+
+Vsak 
+\begin_inset Formula $x$
+\end_inset
+
+ na intervalu lahko zapišemo kot 
+\begin_inset Formula $x=a+t\left(b-a\right)$
+\end_inset
+
+ za nek 
+\begin_inset Formula $t\in\left(0,1\right)$
+\end_inset
+
+.
+ Tedaj je 
+\begin_inset Formula $x-a=t\left(b-a\right)$
+\end_inset
+
+ in konveksnost se glasi
+\begin_inset Formula 
+\[
+\forall a,b\in I:\forall t\in\left(0,1\right):f\left(a+t\left(b-a\right)\right)\leq\frac{f\left(b\right)-f\left(a\right)}{\cancel{b-a}}t\cancel{\left(b-a\right)}+f\left(a\right)
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+f\left(a+t\left(b-a\right)\right)=f\left(a+tb-ta\right)=f\left(\left(1-t\right)a+tb\right)\leq tf\left(b\right)-tf\left(a\right)+f\left(a\right)=\left(1-t\right)f\left(a\right)+tf\left(b\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Konveksna kombinacija izrazov 
+\begin_inset Formula $a,b$
+\end_inset
+
+ je izraz oblike 
+\begin_inset Formula $\left(1-t\right)a+tb$
+\end_inset
+
+ za 
+\begin_inset Formula $t\in\left(0,1\right)$
+\end_inset
+
+.
+ Potemtakem je ZDB definicija konveksnosti 
+\begin_inset Formula $\forall a,b\in I:$
+\end_inset
+
+ funkcijska vrednost konveksne kombinacije 
+\begin_inset Formula $a,b$
+\end_inset
+
+ je kvečjemu konveksna kombinacija funkcijskih vrednosti 
+\begin_inset Formula $a,b$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Konkavnost pa je definirana tako,
+ da povsod obrnemo predznake,
+ torej daljica leži pod grafom 
+\begin_inset Formula $f$
+\end_inset
+
+ ZDB 
+\begin_inset Formula $\forall a,b\in I:f\left(\left(1-t\right)a+tb\right)\geq\left(1-t\right)f\left(a\right)+tf\left(b\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $f\left(x\right)=\sin x$
+\end_inset
+
+,
+ 
+\begin_inset Formula $I=\left[-\pi,0\right]$
+\end_inset
+
+.
+ Je konveksna.
+ Se vidi iz grafa.
+ Preveriti analitično bi bilo težko.
+\end_layout
+
+\begin_layout Example*
+Formulirajmo drugačen pogoj za konveksnost.
+ Naj bo spet 
+\begin_inset Formula $f:I\to\mathbb{R}$
+\end_inset
+
+,
+ kjer je 
+\begin_inset Formula $I$
+\end_inset
+
+ interval.
+ 
+\begin_inset Formula $f$
+\end_inset
+
+ je konveksna 
+\begin_inset Formula 
+\[
+\Leftrightarrow\forall a,b\in I\forall x\in\left(a,b\right):\frac{f\left(x\right)-f\left(a\right)}{x-a}\leq\frac{f\left(b\right)-f\left(a\right)}{b-a}.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+Sedaj glejmo le poljuben 
+\begin_inset Formula $a$
+\end_inset
+
+.
+ Po prejšnjem pogoju moramo gledati še vse poljubne 
+\begin_inset Formula $b$
+\end_inset
+
+,
+ večje od 
+\begin_inset Formula $a$
+\end_inset
+
+ (ker le tako lahko konstruiramo interval).
+ Za 
+\begin_inset Formula $b$
+\end_inset
+
+ in 
+\begin_inset Formula $a$
+\end_inset
+
+ mora biti diferenčni kvocient večji od diferenčnega kvocienta 
+\begin_inset Formula $x$
+\end_inset
+
+ in 
+\begin_inset Formula $a$
+\end_inset
+
+ za poljuben 
+\begin_inset Formula $x$
+\end_inset
+
+.
+ Ta pogoj pa je ekvivalenten temu,
+ da diferenčni kvocient 
+\begin_inset Formula $x$
+\end_inset
+
+ in 
+\begin_inset Formula $a$
+\end_inset
+
+ s fiksnim 
+\begin_inset Formula $a$
+\end_inset
+
+ in čedalje večjim 
+\begin_inset Formula $x$
+\end_inset
+
+ narašča,
+ torej je pogoj za konveksnost tudi:
+\begin_inset Formula 
+\[
+\forall a\in I\forall x>a:g_{a}\left(x\right)=\frac{f\left(x\right)-f\left(a\right)}{x-a}\text{ je naraščajoča funkcija}.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Corollary*
+Naj bo 
+\begin_inset Formula $f$
+\end_inset
+
+ konveksna na odprtem intervalu 
+\begin_inset Formula $I$
+\end_inset
+
+.
+ 
+\begin_inset Formula $\forall a\in I$
+\end_inset
+
+ obstajata funkciji
+\begin_inset Formula 
+\[
+\left(D_{+}f\right)\left(a\right)=\lim_{x\searrow a}g_{a}\left(x\right)=\inf_{x\in I,x>a}g_{a}\left(x\right)\text{ (desni odvod \ensuremath{f} v \ensuremath{a})}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\left(D_{-}f\right)\left(a\right)=\lim_{x\nearrow a}g_{a}\left(x\right)=\sup_{x\in I,x<a}g_{a}\left(x\right)\text{ (levi odvod \ensuremath{f} v \ensuremath{a})}
+\]
+
+\end_inset
+
+in obe sta naraščajoči na 
+\begin_inset Formula $I$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Obstoj sledi iz monotonosti 
+\begin_inset Formula $g_{a}\left(a\right)$
+\end_inset
+
+,
+ kajti 
+\begin_inset Formula $\lim_{x\searrow a}g_{a}\left(x\right)=\lim_{x\searrow a}\frac{f\left(x\right)-f\left(a\right)}{x-a}$
+\end_inset
+
+ in enako za levo limito.
+ Diferenčni kvocient mora namreč biti naraščajoč.
+ S tem smo dokazali,
+ da je vsaka konveksna funkcija zvezna
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+Ni pa vsaka konveksna funkcija odvedljiva,
+ protiprimer je 
+\begin_inset Formula $f\left(x\right)=\left|x\right|$
+\end_inset
+
+.
+\end_layout
+
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Naj bodo 
+\begin_inset Formula $x_{1},x_{2},x\in I\ni:x_{1}<x_{2}<x$
+\end_inset
+
+.
+ Pomagaj si s skico
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+TODO DORIŠI SKICO ZVZ VII/ANA1UČ/str.
+ 13
+\end_layout
+
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $f$
+\end_inset
+
+ konveksna,
+ sledi 
+\begin_inset Formula $g_{x}\left(x_{1}\right)\leq g_{x}\left(x_{2}\right)$
+\end_inset
+
+.
+ Ker 
+\begin_inset Formula $\forall s,t\in\mathbb{R}:g_{s}\left(t\right)=g_{t}\left(s\right)$
+\end_inset
+
+,
+ lahko našo neenakost zapišemo kot 
+\begin_inset Formula $g_{x_{1}}\left(x\right)\leq g_{x_{2}}\left(x\right)$
+\end_inset
+
+.
+ Sledi (desni neenačaj iz 
+\begin_inset Formula $g_{x_{1}}\left(x\right)\leq g_{x_{2}}\left(x\right)$
+\end_inset
+
+,
+ levi neenačaj pa ker 
+\begin_inset Formula $g$
+\end_inset
+
+ narašča):
+\begin_inset Formula 
+\[
+\left(D_{+}\left(f\right)\right)\left(x_{1}\right)=\inf_{x\in I,x>x_{1}}g_{x_{1}}\left(x\right)\leq\inf_{x\in I,x>x_{2}}g_{x_{1}}\left(x\right)\leq\inf_{x\in I,x>x_{2}}g_{x_{2}}\left(x\right)=\left(D_{+}\left(f\right)\right)\left(x_{2}\right)
+\]
+
+\end_inset
+
+Podobno dokažemo
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+DOPIŠI KAKO!
+ TODO XXX FIXME
+\end_layout
+
+\end_inset
+
+,
+ da 
+\begin_inset Formula $D_{-}$
+\end_inset
+
+ narašča.
+\end_layout
+
+\begin_layout Theorem*
+Naj bo 
+\begin_inset Formula $f:I^{\text{odp.}}\to\mathbb{R}$
+\end_inset
+
+ dvakrat odvedljiva.
+ Tedaj je 
+\begin_inset Formula $f$
+\end_inset
+
+ konveksna 
+\begin_inset Formula $\Leftrightarrow\forall x\in I:f''\left(x\right)\geq0$
+\end_inset
+
+ in 
+\begin_inset Formula $f$
+\end_inset
+
+ konkavna 
+\begin_inset Formula $\Leftrightarrow\forall x\in I:f''\left(x\right)\leq0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Dokazujemo ekvivalenco za konveksnost (konkavnost podobno).
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Rightarrow\right)$
+\end_inset
+
+ Po predpostavki je 
+\begin_inset Formula $f$
+\end_inset
+
+ konveksna in dvakrat odvedljiva,
+ torej je odvedljiva in sta levi in desni odvod enaka,
+ po prejšnji posledici pa levi in desni odvod naraščata,
+ torej 
+\begin_inset Formula $f'$
+\end_inset
+
+ narašča.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Leftarrow\right)$
+\end_inset
+
+ Naj bo 
+\begin_inset Formula $f''\geq0$
+\end_inset
+
+.
+ Vzemimo 
+\begin_inset Formula $x,a\in I$
+\end_inset
+
+.
+ Po Lagrangeu 
+\begin_inset Formula $\exists\xi\text{ med \ensuremath{x} in \ensuremath{a}}\ni:f\left(x\right)-f\left(x\right)=f'\left(\xi\right)\left(x-a\right)$
+\end_inset
+
+.
+ Iz predpostavke 
+\begin_inset Formula $f''>0$
+\end_inset
+
+ sledi,
+ da 
+\begin_inset Formula $f'$
+\end_inset
+
+ narašča.
+ Če je 
+\begin_inset Formula $x>\xi>a$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $f'\left(\xi\right)\geq f'\left(a\right)$
+\end_inset
+
+,
+ zato 
+\begin_inset Formula $f'\left(\xi\right)\left(x-a\right)\geq f'\left(a\right)\left(x-a\right)$
+\end_inset
+
+.
+ Če je 
+\begin_inset Formula $x<\xi<a$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $f'\left(\xi\right)\left(x-a\right)\leq f'\left(a\right)\left(x-a\right)$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Subsection
+Ekstremi funkcij ene spremenljivke
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $I\subseteq\mathbb{R}$
+\end_inset
+
+ odprt interal,
+ 
+\begin_inset Formula $a\in I$
+\end_inset
+
+ in 
+\begin_inset Formula $f:I\to\mathbb{R}$
+\end_inset
+
+.
+ Pravimo,
+ da ima 
+\begin_inset Formula $f$
+\end_inset
+
+ v točki 
+\begin_inset Formula $a$
+\end_inset
+
+ lokalni minimum,
+ če 
+\begin_inset Formula $\exists\delta>0\ni:\min\left\{ f\left(x\right);\forall x\in\left(a-\delta,a+\delta\right)\right\} =f\left(a\right)$
+\end_inset
+
+.
+ Pravimo,
+ da ima 
+\begin_inset Formula $f$
+\end_inset
+
+ v točki 
+\begin_inset Formula $a$
+\end_inset
+
+ lokalni maksimum,
+ če 
+\begin_inset Formula $\exists\delta>0\ni:\max\left\{ f\left(x\right);\forall x\in\left(a-\delta,a+\delta\right)\right\} =f\left(a\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+Če je 
+\begin_inset Formula $f:I\to\mathbb{R}$
+\end_inset
+
+ odvedljiva in ima v 
+\begin_inset Formula $a$
+\end_inset
+
+ lokalni minimum/maksimum,
+ tedaj je 
+\begin_inset Formula $f'\left(a\right)=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Glej dokaz Rolleovega izreka.
+\end_layout
+
+\begin_layout Definition*
+\begin_inset Formula $f$
+\end_inset
+
+ ima v 
+\begin_inset Formula $a$
+\end_inset
+
+ ekstrem,
+ če ima v 
+\begin_inset Formula $a$
+\end_inset
+
+ lokalni minimum ali lokalni maksimum.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Če je 
+\begin_inset Formula $f'\left(a\right)=0$
+\end_inset
+
+,
+ pravimo,
+ da ima 
+\begin_inset Formula $f$
+\end_inset
+
+ v 
+\begin_inset Formula $a$
+\end_inset
+
+ stacionarno točko.
+\end_layout
+
+\begin_layout Theorem*
+Naj bo 
+\begin_inset Formula $I\subseteq\mathbb{R}$
+\end_inset
+
+ odprt interval,
+ 
+\begin_inset Formula $a\in I$
+\end_inset
+
+ in 
+\begin_inset Formula $f:I\to\mathbb{R}$
+\end_inset
+
+ dvakrat odvedljiva ter naj bo 
+\begin_inset Formula $f'\left(a\right)=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $f''\left(a\right)>0\Rightarrow$
+\end_inset
+
+ v 
+\begin_inset Formula $a$
+\end_inset
+
+ ima 
+\begin_inset Formula $f$
+\end_inset
+
+ lokalni minimum
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $f''\left(a\right)<0\Rightarrow$
+\end_inset
+
+ v 
+\begin_inset Formula $a$
+\end_inset
+
+ ima 
+\begin_inset Formula $f$
+\end_inset
+
+ lokalni maksimum
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $f''\left(a\right)=0\Rightarrow$
+\end_inset
+
+ nedoločeno
+\end_layout
+
+\end_deeper
+\begin_layout Proof
+Sledi iz 
+\begin_inset Formula $f''>0\Rightarrow$
+\end_inset
+
+ stroga konveksnost in 
+\begin_inset Formula $f''<0\Rightarrow$
+\end_inset
+
+ stroga konkavnost.
+\end_layout
+
+\begin_layout Subsection
+L'Hopitalovo pravilo
+\end_layout
+
+\begin_layout Standard
+Kako izračunati 
+\begin_inset Formula $\lim_{x\to a}\frac{f\left(x\right)}{g\left(x\right)}$
+\end_inset
+
+?
+\end_layout
+
+\begin_layout Standard
+Če so funkcije zvezne v 
+\begin_inset Formula $a$
+\end_inset
+
+ in 
+\begin_inset Formula $g\left(a\right)\not=0$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $\lim_{x\to a}\frac{f\left(x\right)}{g\left(x\right)}=\frac{f\left(a\right)}{g\left(a\right)}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Če imata funkciji v 
+\begin_inset Formula $a$
+\end_inset
+
+ limito in 
+\begin_inset Formula $\lim_{x\to a}g\left(x\right)\not=0$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $\lim_{x\to a}\frac{f\left(x\right)}{g\left(x\right)}=\frac{\lim_{x\to a}f\left(x\right)}{\lim_{x\to a}g\left(x\right)}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Če 
+\begin_inset Formula $\lim_{x\to a}g\left(x\right)=\infty$
+\end_inset
+
+ in je na neki okolici 
+\begin_inset Formula $a$
+\end_inset
+
+ 
+\begin_inset Formula $f\left(x\right)$
+\end_inset
+
+ omejena,
+ velja 
+\begin_inset Formula $\lim_{x\to a}\frac{f\left(x\right)}{g\left(x\right)}=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Če 
+\begin_inset Formula $\lim_{x\to a}f\left(x\right)=0$
+\end_inset
+
+ in je na neki okolici 
+\begin_inset Formula $a$
+\end_inset
+
+ 
+\begin_inset Formula $g\left(x\right)$
+\end_inset
+
+ navzdol omejena več od nič ali navzgor omejena manj od nič,
+ velja 
+\begin_inset Formula $\lim_{x\to a}\frac{f\left(x\right)}{g\left(x\right)}=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Zanimivi primeri pa so,
+ ko 
+\begin_inset Formula $\lim_{x\to a}f\left(x\right)=\lim_{x\to a}g\left(x\right)=0$
+\end_inset
+
+ ali pa ko 
+\begin_inset Formula $\lim_{x\to a}f\left(x\right)=\infty$
+\end_inset
+
+ in hkrati 
+\begin_inset Formula $\lim_{x\to a}g\left(x\right)=\infty$
+\end_inset
+
+,
+ na primer 
+\begin_inset Formula $\lim_{x\to0}\frac{x}{x}$
+\end_inset
+
+ ali pa 
+\begin_inset Formula $\lim_{x\to0}\frac{x^{2}}{x}$
+\end_inset
+
+ ali pa 
+\begin_inset Formula $\lim_{x\to0}\frac{x}{x^{2}}$
+\end_inset
+
+.
+ Tedaj uporabimo L'Hopitalovo pravilo.
+\end_layout
+
+\begin_layout Theorem*
+Če velja hkrati:
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset CommandInset label
+LatexCommand label
+name "enu:Eno-izmed-slednjega:"
+
+\end_inset
+
+Eno izmed slednjega:
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $\lim_{x\to a}f\left(x\right)=\lim_{x\to a}g\left(x\right)=0$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\lim_{x\to a}f\left(x\right)=\infty$
+\end_inset
+
+ in hkrati 
+\begin_inset Formula $\lim_{x\to a}g\left(x\right)=\infty$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\lim_{x\to a}f\left(x\right)=-\infty$
+\end_inset
+
+ in hkrati 
+\begin_inset Formula $\lim_{x\to a}g\left(x\right)=-\infty$
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+\begin_inset Formula $f,g$
+\end_inset
+
+ v okolici 
+\begin_inset Formula $a$
+\end_inset
+
+ odvedljivi
+\end_layout
+
+\end_deeper
+\begin_layout Theorem*
+Potem 
+\begin_inset Formula $\exists L\coloneqq\lim_{x\to a}\frac{f'\left(x\right)}{g'\left(x\right)}\Rightarrow\exists\lim_{x\to a}\frac{f\left(x\right)}{g\left(x\right)}$
+\end_inset
+
+ in ta limita je enaka 
+\begin_inset Formula $L$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Ne bomo dokazali.
+\end_layout
+
+\begin_layout Example*
+Nekaj primerov uporabe L'Hopitalovega pravila.
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula 
+\[
+\lim_{x\to0}x^{x}=\lim_{x\to0}e^{lnx^{x}}=\lim_{x\to0}e^{x\ln x}=e^{\lim_{x\to0}x\ln x}
+\]
+
+\end_inset
+
+Računajmo 
+\begin_inset Formula $\lim_{x\to0}x\ln x$
+\end_inset
+
+ z L'Hopitalom.
+ Potrebujemo ulomek.
+ Ideja:
+ množimo števec in imenovalec z 
+\begin_inset Formula $x$
+\end_inset
+
+,
+ tedaj bi dobili 
+\begin_inset Formula $\lim_{x\to0}\frac{x^{2}\ln x}{x}$
+\end_inset
+
+.
+ Toda v tem primeru števec in imenovalec ne ustrezata pogoju 
+\begin_inset CommandInset ref
+LatexCommand ref
+reference "enu:Eno-izmed-slednjega:"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+ za L'Hopitalovo pravilo.
+ Druga ideja:
+ množimo števec in imenovalec z 
+\begin_inset Formula $\left(\ln x\right)^{-1}$
+\end_inset
+
+,
+ tedaj dobimo 
+\begin_inset Formula $\lim_{x\to0}\frac{x}{\left(\ln x\right)^{-1}}\overset{\text{L'H}}{=}\lim_{x\to0}\frac{1}{\frac{-1}{\log^{2}x}\cdot\frac{1}{x}}=\lim_{x\to0}-x\log^{2}x$
+\end_inset
+
+,
+ kar je precej komplicirano.
+ Tretja ideja:
+ množimo števec in imenovalec z 
+\begin_inset Formula $x^{-1}$
+\end_inset
+
+,
+ tedaj števec in imenovalec divergirata k 
+\begin_inset Formula $-\infty$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+\lim_{x\to0}\frac{\ln x}{x^{-1}}\overset{\text{L'H}}{=}\lim_{x\to0}\frac{\left(\ln x\right)'}{\left(x^{-1}\right)'}=\lim_{x\to0}\frac{x^{-1}}{-x^{-2}}=\lim_{x\to0}-x=0
+\]
+
+\end_inset
+
+Potemtakem 
+\begin_inset Formula $\lim_{x\to0}x^{x}=e^{0}=1$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\lim_{x\to0}\frac{1-\cos x}{x^{2}}$
+\end_inset
+
+.
+ Obe strani ulomkove črte konvergirata k 
+\begin_inset Formula $0$
+\end_inset
+
+.
+ Prav tako ko enkrat že uporabimo L'H.
+\begin_inset Formula 
+\[
+\lim_{x\to0}\frac{1-\cos x}{x^{2}}\overset{\text{L'H}}{=}\lim_{x\to0}\frac{\sin x}{2x}\overset{\text{L'H}}{=}\lim_{x\to0}\frac{\cos x}{2}=\frac{1}{2}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Section
+Taylorjev izrek in Taylorjeva formula
+\end_layout
+
+\begin_layout Standard
+Naj bo 
+\begin_inset Formula $f$
+\end_inset
+
+ v okolici 
+\begin_inset Formula $a$
+\end_inset
+
+ dovoljkrat odvedljiva.
+ Želimo aproksimirati 
+\begin_inset Formula $f\left(a+h\right)$
+\end_inset
+
+ s polinomi danega reda 
+\begin_inset Formula $n$
+\end_inset
+
+.
+ Iščemo polinome reda 
+\begin_inset Formula $n$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $n=0$
+\end_inset
+
+ konstante.
+ 
+\begin_inset Formula $f\left(a+h\right)\approx f\left(a\right)$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $n=1$
+\end_inset
+
+ linearne funkcije.
+ 
+\begin_inset Formula $f\left(a+h\right)\sim f\left(a\right)+f'\left(a\right)h$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $n=2$
+\end_inset
+
+ ...
+ Želimo najti 
+\begin_inset Formula $a_{0},a_{1},a_{2}\in\mathbb{R}$
+\end_inset
+
+,
+ odvisne le od 
+\begin_inset Formula $f$
+\end_inset
+
+ in 
+\begin_inset Formula $a$
+\end_inset
+
+,
+ za katere 
+\begin_inset Formula $f\left(a+b\right)\approx a_{0}+a_{1}h+a_{2}h^{2}$
+\end_inset
+
+.
+ Ko govorimo o aproksimaciji,
+ mislimo take koeficiente,
+ da se približek najbolje prilega dejanski funkcijski vrednosti,
+ v smislu,
+ da
+\begin_inset Formula 
+\[
+\lim_{h\to0}\frac{f\left(a+h\right)-\left(a_{0}+a_{1}h+a_{2}h^{2}\right)}{h^{2}}=0
+\]
+
+\end_inset
+
+
+\begin_inset Formula $a_{0}$
+\end_inset
+
+ izvemo takoj,
+ kajti 
+\begin_inset Formula $\lim_{h\to0}f\left(a+h\right)-\left(a_{0}+a_{1}h+a_{2}h^{2}\right)=0=f\left(a\right)-\left(a_{0}+0h+0h^{2}\right)=f\left(a\right)-a_{0}=0$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $a_{0}=f\left(a\right)$
+\end_inset
+
+.
+ Za preostale koeficiente uporabimo L'Hopitalovo pravilo,
+ ki pove,
+ da zadošča,
+ da je
+\begin_inset Formula 
+\[
+\lim_{h\to0}\frac{f'\left(a+h\right)-\left(0+a_{1}+a_{2}h\right)}{2h}=0
+\]
+
+\end_inset
+
+Zopet glejmo števec in vstavimo 
+\begin_inset Formula $h=0$
+\end_inset
+
+:
+ 
+\begin_inset Formula $f'\left(a\right)-a_{1}=0\Rightarrow f'\left(a\right)=a_{1}$
+\end_inset
+
+.
+ Spet uporabimo L'H:
+\begin_inset Formula 
+\[
+\lim_{h\to0}\frac{f''\left(a+h\right)-\left(0+0+2a_{2}\right)}{2}
+\]
+
+\end_inset
+
+Vstavimo 
+\begin_inset Formula $h=0$
+\end_inset
+
+ v 
+\begin_inset Formula $f''\left(a+h\right)-2a_{2}$
+\end_inset
+
+ in dobimo 
+\begin_inset Formula $2a_{2}=f''\left(a\right)$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $a_{2}=\frac{f''\left(a\right)}{2}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $n=3$
+\end_inset
+
+ Ugibamo,
+ da je najboljši kubični približek
+\begin_inset Formula 
+\[
+f\left(a+h\right)\approx h\mapsto f\left(a\right)+f'\left(a\right)h+\frac{f''\left(a\right)}{2}h^{2}+\frac{f'''\left(a\right)}{6}h^{3}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Theorem*
+Taylor.
+ Naj bo 
+\begin_inset Formula $n\in\mathbb{N}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $I$
+\end_inset
+
+ interval 
+\begin_inset Formula $\subseteq\mathbb{R}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $a\in I$
+\end_inset
+
+,
+ 
+\begin_inset Formula $f:I\to\mathbb{R}$
+\end_inset
+
+ 
+\begin_inset Formula $n-$
+\end_inset
+
+krat odvedljiva v točki 
+\begin_inset Formula $a$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $\exists g_{n}:I-a\to\mathbb{R}\ni:$
+\end_inset
+
+
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $I-a$
+\end_inset
+
+ pomeni interval 
+\begin_inset Formula $I$
+\end_inset
+
+ pomaknjen v levo za 
+\begin_inset Formula $a$
+\end_inset
+
+.
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $f\left(a+h\right)=\sum_{j=0}^{n}\frac{f^{\left(j\right)}\left(a\right)}{n!}h^{j}+g_{n}\left(h\right)h^{n}$
+\end_inset
+
+ in
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $\lim_{h\to0}g_{n}\left(h\right)=0$
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Theorem*
+Sedaj pišimo 
+\begin_inset Formula $x=a+h$
+\end_inset
+
+.
+ Tedaj se izrek glasi:
+ 
+\begin_inset Formula $\exists\tilde{g_{n}}:I\to\mathbb{R}\ni:$
+\end_inset
+
+
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $f\left(x\right)=\sum_{j=0}^{n}\frac{f^{\left(j\right)}\left(a\right)}{n!}\left(x-a\right)^{j}+\tilde{g_{n}}\left(x\right)\left(x-a\right)^{n}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $\lim_{x\to a}\tilde{g_{n}}\left(x\right)=0$
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Theorem*
+Tedaj označimo 
+\begin_inset Formula $T_{n,f,a}\left(x\right)=\sum_{j=0}^{n}\frac{f^{\left(j\right)}\left(a\right)}{n!}\left(x-a\right)^{j}$
+\end_inset
+
+ (pravimo 
+\begin_inset Formula $n-$
+\end_inset
+
+ti taylorjev polinom za 
+\begin_inset Formula $f$
+\end_inset
+
+ okrog točke 
+\begin_inset Formula $a$
+\end_inset
+
+) in 
+\begin_inset Formula $R_{n,f,a}\left(x\right)=\tilde{g_{n}}\left(x\right)\left(x-a\right)^{n}$
+\end_inset
+
+ (pravimo ostanek/napaka).
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Theorem*
+Če je 
+\begin_inset Formula $f$
+\end_inset
+
+ 
+\begin_inset Formula $\left(n+1\right)-$
+\end_inset
+
+krat odvedljiva na odprtem intervalu 
+\begin_inset Formula $I\subseteq\mathbb{R}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $a\in I$
+\end_inset
+
+,
+ tedaj 
+\begin_inset Formula $\forall b\in I\exists\alpha\in I\text{ med \ensuremath{a} in \ensuremath{x}}\ni:R_{n}\left(b\right)=\frac{f^{\left(n+1\right)}\left(\alpha\right)}{\left(n+1\right)!}\left(b-a\right)^{n+1}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Označimo 
+\begin_inset Formula $T_{n}\left(x\right)=f\left(a\right)+f'\left(a\right)\left(x-a\right)+\frac{f''\left(a\right)}{2!}\left(x-a\right)^{2}+\cdots+\frac{f^{\left(n\right)}\left(a\right)}{n!}\left(x-a\right)^{n}$
+\end_inset
+
+ torej 
+\begin_inset Formula $n-$
+\end_inset
+
+ti taylorjev polinom in naj bo 
+\begin_inset Formula $K$
+\end_inset
+
+ tako število,
+ da velja 
+\begin_inset Formula $f\left(b\right)-T_{n}\left(b\right)=K\left(b-a\right)^{n+1}$
+\end_inset
+
+.
+ Definirajmo 
+\begin_inset Formula $F\left(x\right)=f\left(x\right)-T_{n}\left(x\right)-K\left(x-a\right)^{n+1}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hypertarget{velja}{Velja}
+\end_layout
+
+\end_inset
+
+ 
+\begin_inset Formula $T_{n}^{\left(k\right)}\left(a\right)=f\left(a\right)$
+\end_inset
+
+ za 
+\begin_inset Formula $k\leq n$
+\end_inset
+
+,
+ kajti 
+\begin_inset Formula $\frac{d\sum_{j=1}^{n}\frac{f^{\left(j\right)}\left(a\right)}{n!}\left(h\right)^{n}}{dh}=\frac{f^{\left(j\right)}\left(a\right)n!}{n!}\cdot1=f^{\left(j\right)}\left(a\right)$
+\end_inset
+
+.
+ Vsi členi z eksponentom,
+ manjšim od 
+\begin_inset Formula $k$
+\end_inset
+
+,
+ se odvajajo v 0,
+ točno pri eksponentu 
+\begin_inset Formula $k$
+\end_inset
+
+ se člen odvaja v konstanto,
+ pri višjih členih pa ostane potencirana spremenljivka,
+ ki je 
+\begin_inset Formula $0$
+\end_inset
+
+ (tu mislimo odstopanje od 
+\begin_inset Formula $a$
+\end_inset
+
+,
+ označeno s 
+\begin_inset Formula $h$
+\end_inset
+
+),
+ torej se ti členi tudi izničijo.
+\end_layout
+
+\begin_layout Proof
+Zato 
+\begin_inset Formula $\forall k\leq n:F^{\left(k\right)}\left(a\right)=0$
+\end_inset
+
+.
+ Nadalje velja 
+\begin_inset Formula $F\left(a\right)=F\left(b\right)=0$
+\end_inset
+
+,
+ ker smo pač tako definirali funkcijo 
+\begin_inset Formula $F$
+\end_inset
+
+,
+ zato obstaja po Rolleovem izreku tak 
+\begin_inset Formula $\alpha_{1}$
+\end_inset
+
+,
+ da velja 
+\begin_inset Formula $F'\left(\alpha_{1}\right)=0$
+\end_inset
+
+.
+ Po Rolleovem izreku nadalje obstaja tak 
+\begin_inset Formula $\alpha_{2}$
+\end_inset
+
+ med 
+\begin_inset Formula $a$
+\end_inset
+
+ in 
+\begin_inset Formula $\alpha_{1}$
+\end_inset
+
+,
+ da velja 
+\begin_inset Formula $F''\left(\alpha_{2}\right)=0$
+\end_inset
+
+.
+ Spet po Rolleovem izreku obstaja tak 
+\begin_inset Formula $\alpha_{3}$
+\end_inset
+
+ med 
+\begin_inset Formula $a$
+\end_inset
+
+ in 
+\begin_inset Formula $\alpha_{2}$
+\end_inset
+
+,
+ da velja 
+\begin_inset Formula $F'''\left(\alpha_{3}\right)=0$
+\end_inset
+
+.
+ Postopek lahko ponavljamo in dobimo tak 
+\begin_inset Formula $\alpha=\alpha_{n+1}$
+\end_inset
+
+,
+ da velja 
+\begin_inset Formula $F^{\left(n+1\right)}\left(\alpha\right)=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Ker je 
+\begin_inset Formula $\forall x\in I:T_{n}^{\left(n+1\right)}\left(x\right)=0$
+\end_inset
+
+ (očitno,
+ isti argument kot v 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{velja}{drugem odstavku dokaza}
+\end_layout
+
+\end_inset
+
+),
+ to pomeni 
+\begin_inset Formula $f^{\left(n+1\right)}\left(\alpha\right)=\left(K\left(x-a\right)^{n+1}\right)^{\left(n+1\right)}=K\left(n+1\right)!$
+\end_inset
+
+.
+ Torej je 
+\begin_inset Formula $K=\frac{f^{\left(n+1\right)}\left(\alpha\right)}{\left(n+1\right)!}$
+\end_inset
+
+ in zato 
+\begin_inset Formula $f\left(b\right)=T_{n}\left(b\right)+\frac{f^{\left(n+1\right)}\left(\alpha\right)}{\left(n+1\right)!}\left(b-a\right)^{\left(n+1\right)}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Corollary*
+Če je 
+\begin_inset Formula $\left(n+1\right)-$
+\end_inset
+
+ti odvod omejen na 
+\begin_inset Formula $I$
+\end_inset
+
+,
+ t.
+ j.
+ 
+\begin_inset Formula $\exists M>0\forall x\in I:\left|f^{\left(n+1\right)}\left(x\right)\right|\leq M$
+\end_inset
+
+,
+ lahko ostanek eksplicitno ocenimo,
+ in sicer 
+\begin_inset Formula $\left|R_{n}\left(x\right)\right|\leq\frac{M}{\left(n+1\right)!}\left|x-a\right|^{n+1}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Kaj pa se zgodi,
+ ko 
+\begin_inset Formula $n$
+\end_inset
+
+ pošljemo v neskončnost?
+ Iskali bi aproksimacije s 
+\begin_inset Quotes gld
+\end_inset
+
+polinomi neskončnega reda
+\begin_inset Quotes grd
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Če je 
+\begin_inset Formula $f\in C^{\infty}$
+\end_inset
+
+ v okolici točke 
+\begin_inset Formula $a\in\mathbb{R}$
+\end_inset
+
+.
+ Tedaj definiramo Taylorjevo vrsto 
+\begin_inset Formula $f$
+\end_inset
+
+ v okolici točke 
+\begin_inset Formula $a$
+\end_inset
+
+:
+ 
+\begin_inset Formula $T_{f,a}\left(x\right)\coloneqq\sum_{j=0}^{\infty}\frac{f^{\left(j\right)}\left(a\right)}{j!}\left(x-a\right)^{j}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Question*
+Ali Taylorjeva vrsta konvergira oziroma kje konvergira?
+ Kakšna je zveza s 
+\begin_inset Formula $f\left(x\right)$
+\end_inset
+
+?
+ Kakšen je 
+\begin_inset Formula $R_{f,a}$
+\end_inset
+
+?
+\end_layout
+
+\begin_layout Standard
+Oglejmo si potenčne vrste (
+\begin_inset Formula $\sum_{j=0}^{\infty}b_{k}x^{k}$
+\end_inset
+
+) kot poseben primer funkcijskih vrst (
+\begin_inset Formula $\sum_{j=0}^{\infty}a_{k}\left(x\right)$
+\end_inset
+
+).
+ Vemo,
+ da ima potenčna vrsta konvergenčni radij 
+\begin_inset Formula $R$
+\end_inset
+
+.
+ Za 
+\begin_inset Formula $x\in\left(-R,R\right)$
+\end_inset
+
+ konvergira,
+ za 
+\begin_inset Formula $x\in\left[-R,R\right]^{C}$
+\end_inset
+
+ divergira.
+\end_layout
+
+\begin_layout Theorem*
+Naj ima potenčna vrsta 
+\begin_inset Formula $f\left(x\right)=\sum_{k=0}^{\infty}b_{k}x^{k}$
+\end_inset
+
+ konvergenčni radij 
+\begin_inset Formula $R$
+\end_inset
+
+.
+ Tedaj ima tudi 
+\begin_inset Formula $g\left(x\right)=\sum_{k=1}kb_{k}x^{k-1}$
+\end_inset
+
+ konvergenčni radij 
+\begin_inset Formula $R$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+\begin_inset Formula 
+\[
+\frac{1}{R_{g}}=\limsup_{k\to\infty}\sqrt[k]{\left|ka_{k}\right|}=\limsup_{k\to\infty}\sqrt[k]{\left|k\right|\left|a_{k}\right|}=\limsup_{k\to\infty}\sqrt[k]{\left|k\right|}\sqrt[k]{\left|a_{k}\right|}=\cdots
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Proof
+\begin_inset Formula 
+\[
+\lim_{k\to\infty}\sqrt[k]{\left|k\right|}=\lim_{k\to\infty}k^{1/k}=e^{\lim_{k\to\infty}\frac{1}{k}\ln k}\overset{\text{L'H}}{=}e^{\lim_{k\to\infty}\frac{\frac{1}{k}}{k}}=e^{\lim_{k\to\infty}\cancelto{0}{\frac{1}{k^{2}}}}=e^{0}=1
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\cdots=\limsup_{k\to\infty}1\cdot\sqrt[k]{\left|a_{k}\right|}=\frac{1}{R_{f}}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Corollary*
+Če ima potenčna vrsta 
+\begin_inset Formula $f$
+\end_inset
+
+ konvergenčni radij 
+\begin_inset Formula $R>0$
+\end_inset
+
+,
+ tedaj je 
+\begin_inset Formula $f\in C^{\infty}\left(\left(-R,R\right)\right)$
+\end_inset
+
+ in velja 
+\begin_inset Formula $a_{k}=\frac{f^{\left(k\right)}\left(0\right)}{k!}$
+\end_inset
+
+,
+ potem velja 
+\begin_inset Formula $g=f'$
+\end_inset
+
+ (iz izreka zgoraj).
+ Razlaga:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $f\left(x\right)=\sum_{k=0}^{\infty}a_{k}x^{k}=\sum_{k=0}^{\infty}\frac{f^{\left(k\right)}\left(0\right)}{k!}x^{k}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $f'\left(x\right)=\sum_{k=1}^{\infty}ka_{k}x^{k-1}=\sum_{k=1}^{\infty}\frac{kf^{\left(k\right)}\left(0\right)}{k!}x^{k-1}=\sum_{k=1}^{\infty}\frac{f^{\left(k\right)}\left(0\right)}{\left(k-1\right)!}x^{k-1}$
+\end_inset
+
+ (
+\begin_inset Formula $k$
+\end_inset
+
+ začne z 
+\begin_inset Formula $1$
+\end_inset
+
+,
+ ker se 
+\begin_inset Formula $k=0$
+\end_inset
+
+ člen odvaja v konstanto 
+\begin_inset Formula $0$
+\end_inset
+
+)
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $g\left(x\right)=\sum_{k=1}^{\infty}ka_{k}x^{k-1}=\sum_{k=1}^{\infty}\frac{kf^{\left(k\right)}\left(0\right)}{k!}x^{k-1}=\sum_{k=1}^{\infty}\frac{f^{\left(k\right)}\left(0\right)}{\left(k-1\right)!}x^{k-1}=f'\left(x\right)$
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Definition*
+Funkcija 
+\begin_inset Formula $f:J\to\mathbb{R}$
+\end_inset
+
+ (
+\begin_inset Formula $J$
+\end_inset
+
+ je interval 
+\begin_inset Formula $\subseteq\mathbb{R}$
+\end_inset
+
+) je realno analitična,
+ če se jo da okoli vsake točke 
+\begin_inset Formula $c\in J$
+\end_inset
+
+ razviti v potenčno vrsto,
+ torej če 
+\begin_inset Formula $f\left(x\right)=\sum_{k=0}^{\infty}\frac{f^{\left(k\right)}\left(c\right)}{k!}\left(x-c\right)^{k}$
+\end_inset
+
+ za 
+\begin_inset Formula $x$
+\end_inset
+
+ blizu 
+\begin_inset Formula $c$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+\begin_inset Formula $f\in C^{\infty}\Rightarrow f$
+\end_inset
+
+ je realno analitična.
+ Protiprimer je 
+\begin_inset Formula $f\left(x\right)=e^{\frac{-1}{\left|x\right|}}$
+\end_inset
+
+.
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+TODO XXX FIXME ZAKAJ?,
+ ne razumem
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+Primeri Taylorjevih vrst.
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $f\left(x\right)=e^{x}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $n-$
+\end_inset
+
+ti tayorjev polinom za 
+\begin_inset Formula $f\left(x\right)$
+\end_inset
+
+ okoli 
+\begin_inset Formula $0$
+\end_inset
+
+:
+ 
+\begin_inset Formula $T_{n,e^{x},0}\left(x\right)=1+x+\frac{x^{2}}{2}+\frac{x^{3}}{6}+\cdots+\frac{x^{n}}{n!}$
+\end_inset
+
+ in velja 
+\begin_inset Formula $e^{x}=T_{n,e^{x},0}\left(x\right)+R_{n,e^{x},0}\left(x\right)$
+\end_inset
+
+,
+ kjer 
+\begin_inset Formula $\lim_{n\to\infty}R_{n,e^{x},0}\left(x\right)=0$
+\end_inset
+
+.
+ Ne bomo dokazali.
+ Sledi 
+\begin_inset Formula $e^{x}=\sum_{k=0}^{\infty}\frac{x^{k}}{k!}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\sin x=x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\cdots+\left(-1\right)^{k}\frac{x^{2k+1}}{\left(2k+1\right)!}$
+\end_inset
+
+.
+ Opazimo sode eksponente in opazimo učinek odvajanja:
+ 
+\begin_inset Formula $\cos,-\sin,-\cos,\sin,\cos,-\sin,\dots$
+\end_inset
+
+.
+ Členi vrste 
+\begin_inset Formula $\sin x$
+\end_inset
+
+ v 
+\begin_inset Formula $x=0$
+\end_inset
+
+ so:
+ 
+\begin_inset Formula $1,0,-1,0,1,0,-1,\dots$
+\end_inset
+
+.
+ Opazimo izpadanje vsakega drugega člena.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\cos x=1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-\cdots+\left(-1\right)^{k}\frac{x^{2k}}{\left(2k\right)!}$
+\end_inset
+
+.
+ Opazimo sode eksponente.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $f\left(x\right)=\log\left(1-x\right)$
+\end_inset
+
+.
+ A lahko to funkcijo razvijemo v taylorjevo vrsto okoli točke 0?
+\begin_inset Float table
+placement document
+alignment document
+wide false
+sideways false
+status open
+
+\begin_layout Plain Layout
+\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
 
 \end_body
-- 
cgit v1.2.3