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Diffstat (limited to 'šola/ana1/teor.lyx')
| -rw-r--r-- | šola/ana1/teor.lyx | 4133 |
1 files changed, 4114 insertions, 19 deletions
diff --git a/šola/ana1/teor.lyx b/šola/ana1/teor.lyx index e44036f..af865e2 100644 --- a/šola/ana1/teor.lyx +++ b/šola/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 +10952,388 @@ 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 + + +\end_layout + +\begin_layout Theorem* +\begin_inset ERT +status open + +\begin_layout Plain Layout + + +\backslash +hypertarget{oi}{Odvod inverza} +\end_layout + +\end_inset + +. + Naj bo +\begin_inset Formula $f$ +\end_inset + + strogo monotona v okolici +\begin_inset Formula $a$ +\end_inset + +, + v +\begin_inset Formula $a$ +\end_inset + + odvedljiva in naj bo +\begin_inset Formula $f'\left(a\right)\not=0$ +\end_inset + +. + Tedaj bo inverzna funkcija, + definirana v okolici +\begin_inset Formula $b=f\left(a\right)$ +\end_inset + + v +\begin_inset Formula $b$ +\end_inset + + odvedljiva in veljalo bo +\begin_inset Formula $\left(f^{-1}\right)'\left(b\right)=\frac{1}{f'\left(a\right)}=\frac{1}{f'\left(f^{-1}\left(b\right)\right)}.$ +\end_inset + + +\end_layout + +\begin_layout Proof +Ker je +\begin_inset ERT +status open + +\begin_layout Plain Layout + + +\backslash +hyperlink{zism}{zvezna in strogo monotona} +\end_layout + +\end_inset + + na okolici +\begin_inset Formula $a$ +\end_inset + +, + inverz na okolici +\begin_inset Formula $f\left(a\right)$ +\end_inset + + obstaja in velja +\begin_inset Formula $f\left(x\right)=s\Leftrightarrow x=f^{-1}\left(x\right)$ +\end_inset + +, + torej +\begin_inset Formula $f^{-1}\left(f\left(x\right)\right)=x$ +\end_inset + + za +\begin_inset Formula $x$ +\end_inset + + v okolici +\begin_inset Formula $a$ +\end_inset + +. +\end_layout + +\begin_layout Proof +Uporabimo formulo za +\begin_inset ERT +status open + +\begin_layout Plain Layout + + +\backslash +hyperlink{ok}{odvod kompozituma} +\end_layout + +\end_inset + + in velja +\begin_inset Formula +\[ +\left(f^{-1}\left(f\left(x\right)\right)\right)'=\left(f^{-1}\right)'\left(f\left(x\right)\right)\cdot f'\left(x\right)=\left(x\right)'=1 +\] + +\end_inset + + +\begin_inset Formula +\[ +\left(f^{-1}\right)'\left(f\left(x\right)\right)=\frac{1}{f'\left(x\right)} +\] + +\end_inset + +Vstavimo +\begin_inset Formula $x=f^{-1}\left(y\right)$ +\end_inset + + in dobimo za vsak +\begin_inset Formula $y$ +\end_inset + + blizu +\begin_inset Formula $f\left(a\right)$ +\end_inset + +: +\begin_inset Formula +\[ +\left(f^{-1}\right)'\left(y\right)=\frac{1}{f'\left(f^{-1}\left(y\right)\right)} +\] + +\end_inset + + +\end_layout + +\begin_layout Example* +Nekaj primerov odvodov inverza. +\end_layout + +\begin_deeper +\begin_layout Enumerate +\begin_inset Formula $g\left(x\right)=\sqrt[n]{x}=x^{\frac{1}{n}}$ +\end_inset + + za +\begin_inset Formula $n\in\mathbb{N},x>0$ +\end_inset + +. + Velja +\begin_inset Formula $g=f^{-1}$ +\end_inset + + za +\begin_inset Formula $f\left(x\right)=x^{n}$ +\end_inset + +. + Uporabimo formulo za +\begin_inset ERT +status open + +\begin_layout Plain Layout + + +\backslash +hyperlink{op}{odvod potence} +\end_layout + +\end_inset + + in zgornji izrek. + Velja +\begin_inset Formula $f'\left(x\right)=nx^{n-1}$ +\end_inset + + in +\begin_inset Formula $f^{-1}=\sqrt[n]{x}$ +\end_inset + +. +\begin_inset Formula +\[ +g'\left(x\right)=\left(f^{-1}\right)'\left(x\right)=\frac{1}{f'\left(f^{-1}\left(x\right)\right)}=\frac{1}{f'\left(\sqrt[n]{x}\right)}=\frac{1}{n\sqrt[n]{x}^{n-1}}=\frac{1}{nx^{\frac{n-1}{n}=1-\frac{1}{n}}}=\frac{1}{n}x^{\frac{1}{n}-1} +\] + +\end_inset + + +\end_layout + +\begin_deeper +\begin_layout Enumerate +\begin_inset Formula $h\left(x\right)=\sqrt[n]{x^{m}}=x^{\frac{m}{n}}=g\left(x\right)^{m}$ +\end_inset + + za +\begin_inset Formula $n,m\in\mathbb{N},x>0$ +\end_inset + +. + Uporabimo formulo za +\begin_inset ERT +status open + +\begin_layout Plain Layout + + +\backslash +hyperlink{op}{odvod potence} +\end_layout + +\end_inset + + in +\begin_inset ERT +status open + +\begin_layout Plain Layout + + +\backslash +hyperlink{ok}{kompozituma} +\end_layout + +\end_inset + + in zgornji primer. + Velja +\begin_inset Formula $g'\left(x\right)=\frac{1}{n}x^{\frac{1}{n}-1}$ +\end_inset + +, + torej +\begin_inset Formula +\[ +h'\left(x\right)=mg\left(x\right)^{m-1}\cdot g'\left(x\right)=m\left(x^{\frac{1}{n}}\right)^{m-1}\cdot\frac{1}{n}x^{\frac{1}{n}-1}=\frac{m}{n}x^{\frac{m-1}{n}+\frac{1}{n}-1}=\frac{m}{n}x^{\frac{m}{n}-1} +\] + +\end_inset + + +\end_layout + +\begin_layout Enumerate +\begin_inset CommandInset label +LatexCommand label +name "enu:Izkaže-se,-da" + +\end_inset + +Izkaže se, + da velja celo +\begin_inset Formula $\forall x>0,\alpha\in\mathbb{R}:\left(x^{\alpha}\right)'=\alpha x^{\alpha-1}$ +\end_inset + +. + Mi smo dokazali le za +\begin_inset Formula $\alpha\in\mathbb{Q}$ +\end_inset + +. +\end_layout + +\end_deeper +\begin_layout Enumerate +Logaritmi, + inverz +\begin_inset Formula $e^{x}$ +\end_inset + +. + Gre za +\begin_inset ERT +status open + +\begin_layout Plain Layout + + +\backslash +hyperlink{oef}{odvod eksponentne funkcije} +\end_layout + +\end_inset + +, + torej +\begin_inset Formula $\left(a^{x}\right)=a^{x}\ln a$ +\end_inset + +. + Tedaj +\begin_inset Formula $\left(e^{x}\right)=e^{x}\ln e=e^{x}$ +\end_inset + +. + Uporavimo +\begin_inset ERT +status open + +\begin_layout Plain Layout + + +\backslash +hyperlink{oi}{odvod inverza} +\end_layout + +\end_inset + +, + torej +\begin_inset Formula $\left(f^{-1}\right)'\left(x\right)=\frac{1}{f'\left(f^{-1}\left(x\right)\right)}$ +\end_inset + + in za +\begin_inset Formula $g\left(x\right)=\log x$ +\end_inset + + uporabimo +\begin_inset Formula $g\left(x\right)=f^{-1}\left(x\right)$ +\end_inset + +, + kjer je +\begin_inset Formula $f\left(x\right)=e^{x}$ +\end_inset + +: +\begin_inset Formula +\[ +\log'\left(x\right)=\left(\left(e^{x}\right)^{-1}\right)'\left(x\right)=\frac{1}{e^{\log x}}=\frac{1}{x} \] \end_inset @@ -10895,43 +11341,3692 @@ f'\left(0\right)=\lim_{h\to0}\frac{f\left(h\right)-0}{h}=\lim_{h\to0}\cdots\text \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 -\begin_inset Separator plain +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 Corollary* -sssssssssss +\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 Corollary* -sssssssssss +\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 Corollary* -sssssssssss +\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_layout Corollary* -sssssssssss +\begin_deeper +\begin_layout Itemize +\begin_inset Formula $f\left(x\right)=\sqrt{x}$ +\end_inset + + +\end_layout + +\begin_layout Itemize +\begin_inset Formula $a+h=2$ +\end_inset + + +\end_layout + +\begin_layout Itemize +\begin_inset Formula $a=2,25$ +\end_inset + +, + +\begin_inset Formula $h=-0,25$ +\end_inset + + +\end_layout + +\begin_layout Itemize +\begin_inset Formula $f\left(a\right)=\sqrt{a}=1,5$ +\end_inset + + +\end_layout + +\begin_layout Itemize +\begin_inset Formula $f'\left(s\right)=\frac{1}{2\sqrt{x}}$ +\end_inset + +, + +\begin_inset Formula $f\left(a=2,25\right)=\frac{1}{3}$ +\end_inset + + +\end_layout + +\begin_layout Itemize +\begin_inset Formula $f\left(2\right)\approx f\left(a\right)+f'\left(2,25\right)\cdot h=1,5-0,25\cdot\frac{1}{3}=\frac{3}{2}-\frac{1}{4}\cdot\frac{1}{3}=\frac{17}{12}$ +\end_inset + +. +\end_layout + +\begin_layout Itemize +Preizkus: + +\begin_inset Formula $\left(\frac{17}{12}\right)^{2}=\frac{289}{144}=2+\frac{1}{144}$ +\end_inset + + ... + Absolutna napaka +\begin_inset Formula $\frac{1}{144}$ +\end_inset + +. +\end_layout + +\end_deeper +\begin_layout Definition* +Naj bo +\begin_inset Formula $I\subseteq\mathbb{R}$ +\end_inset + + interval in +\begin_inset Formula $f:I\to\mathbb{R}$ +\end_inset + + odvedljiva povsod na +\begin_inset Formula $I$ +\end_inset + +. + Vzemimo +\begin_inset Formula $a\in I$ +\end_inset + +. + Če je v +\begin_inset Formula $a$ +\end_inset + + odvedljiva tudi +\begin_inset Formula $f'$ +\end_inset + +, + pišemo +\begin_inset Formula $f''\left(a\right)=\left(f'\left(a\right)\right)'$ +\end_inset + +. + Podobno pišemo tudi višje odvode: + +\begin_inset Formula $f^{\left(1\right)}\left(a\right)=f'\left(a\right)$ +\end_inset + +, + +\begin_inset Formula $f^{\left(n+1\right)}=\left(f^{\left(n\right)}\right)'$ +\end_inset + +, + +\begin_inset Formula $f^{\left(0\right)}\left(a\right)=f\left(a\right)$ +\end_inset + +, + +\begin_inset Formula $f^{\left(2\right)}\left(a\right)=f''\left(a\right)$ +\end_inset + +. +\end_layout + +\begin_layout Note* +Pomen besede +\begin_inset Quotes gld +\end_inset + +odvod +\begin_inset Quotes grd +\end_inset + +: +\end_layout + +\begin_deeper +\begin_layout Itemize +Odvod v dani točki: + +\begin_inset Formula $f'\left(a\right)$ +\end_inset + + za fiksen +\begin_inset Formula $a\in\mathbb{R}$ +\end_inset + + ali +\end_layout + +\begin_layout Itemize +Funkcija, + ki vsaki točki +\begin_inset Formula $x\in\mathbb{R}$ +\end_inset + + priredi +\begin_inset Formula $f'\left(x\right)$ +\end_inset + + po zgornji definiciji. +\end_layout + +\end_deeper +\begin_layout Definition* +\begin_inset Formula $C^{n}\left(I\right)$ +\end_inset + + je množica funkcije +\begin_inset Formula $f:I\to\mathbb{R}$ +\end_inset + +, + da +\begin_inset Formula $\forall x\in I\exists f'\left(x\right),f''\left(x\right),f^{\left(3\right)},\dots,f^{\left(n\right)}\left(x\right)$ +\end_inset + + in da so +\begin_inset Formula $f,f',f'',f^{\left(3\right)},\dots,f^{\left(n\right)}$ +\end_inset + + zvezna funkcije na +\begin_inset Formula $I$ +\end_inset + +. + (seveda če obstaja +\begin_inset Formula $j-$ +\end_inset + +ti odvod, + obstaja tudi zvezen +\begin_inset Formula $j-1-$ +\end_inset + +ti odvod). + ZDB je to množica funkcij, + ki imajo vse odvode do +\begin_inset Formula $n$ +\end_inset + + in so le-ti zvezni. + ZDB to so vse +\begin_inset Formula $n-$ +\end_inset + +krat zvezno odvedljive funkcije na intervalu +\begin_inset Formula $I$ +\end_inset + +. +\end_layout + +\begin_layout Definition* +Označimo +\begin_inset Formula $C^{\infty}\left(I\right)\coloneqq\bigcap_{n=1}^{\infty}C^{n}\left(I\right)$ +\end_inset + + – to so neskončnokrat odvedljive funkcije na intervalu +\begin_inset Formula $I$ +\end_inset + +. +\end_layout + +\begin_layout Note* +Intuitivno +\begin_inset Foot +status open + +\begin_layout Plain Layout +Baje. + Jaz sem itak do vsega skeptičen. +\end_layout + +\end_inset + + velja +\begin_inset Formula $C^{1}\left(I\right)\supset C^{2}\left(I\right)\supset C^{3}\left(I\right)\supset C^{4}\left(I\right)\supset\cdots$ +\end_inset + +. +\end_layout + +\begin_layout Example* +Nekaj primerov. +\end_layout + +\begin_deeper +\begin_layout Itemize +Polimomi +\begin_inset Formula $\subset C^{\infty}\left(\mathbb{R}\right)$ +\end_inset + +. +\end_layout + +\begin_layout Itemize +\begin_inset Formula $f\left(x\right)=\left|x\right|^{3}$ +\end_inset + +, + +\begin_inset Formula $f'\left(x\right)=\begin{cases} +3x^{2} & ;x\geq0\\ +-3x^{2} & ;x<0 +\end{cases}=3x^{2}\sgn x$ +\end_inset + +, + +\begin_inset Formula $f''\left(x\right)=\begin{cases} +6x & ;x\geq0\\ +-6x & ;x<0 +\end{cases}=6x\sgn x$ +\end_inset + +, + +\begin_inset Formula $f'''\left(x\right)=\begin{cases} +6 & ;x>0\\ +-6 & ;x<0 +\end{cases}=6\sgn x$ +\end_inset + + in v +\begin_inset Formula $0$ +\end_inset + + ni odvedljiva, + zato +\begin_inset Formula $f\in C^{2}\left(\mathbb{R}\right)$ +\end_inset + + a +\begin_inset Formula $f\not\in C^{3}\left(\mathbb{R}\right)$ +\end_inset + +, + ker +\begin_inset Formula $\exists f''$ +\end_inset + + in je zvezna na +\begin_inset Formula $\mathbb{R}$ +\end_inset + +, + a +\begin_inset Formula $f'''$ +\end_inset + + sicer obstaja, + a ni zvezna na +\begin_inset Formula $\mathbb{R}$ +\end_inset + +. + Velja pa +\begin_inset Formula $f\in C^{\infty}\left(\mathbb{R}\setminus\left\{ 0\right\} \right)$ +\end_inset + +. +\end_layout + +\end_deeper +\begin_layout Theorem* +Rolle. + Naj bo +\begin_inset Formula $f:\left[a,b\right]\to\mathbb{R}$ +\end_inset + + za +\begin_inset Formula $a,b\in\mathbb{R}$ +\end_inset + + zvezna na +\begin_inset Formula $\left[a,b\right]$ +\end_inset + + in odvedljiva na +\begin_inset Formula $\left(a,b\right)$ +\end_inset + +. +\begin_inset Formula +\[ +f\left(a\right)=f\left(b\right)\Longrightarrow\exists\alpha\in\left(a,b\right)\ni:f'\left(\alpha\right)=0 +\] + +\end_inset + + +\end_layout + +\begin_layout Proof +Sumimo, + da je ustrezna +\begin_inset Formula $\alpha$ +\end_inset + + tista, + ki je +\begin_inset Formula $\max$ +\end_inset + + ali +\begin_inset Formula $\min$ +\end_inset + + od +\begin_inset Formula $f$ +\end_inset + + na +\begin_inset Formula $I$ +\end_inset + +. + +\begin_inset ERT +status open + +\begin_layout Plain Layout + + +\backslash +hyperlink{zfnkm}{Ker} +\end_layout + +\end_inset + + je +\begin_inset Formula $f$ +\end_inset + + zvezna na +\begin_inset Formula $\left[a,b\right]$ +\end_inset + + (kompaktni množici), + +\begin_inset Formula $\exists\alpha_{1}\in\left[a,b\right],\alpha_{2}\in\left[a,b\right]\ni:f\left(\alpha_{1}\right)=\max f\left(\left[a,b\right]\right)\wedge f\left(\alpha_{2}\right)=\min f\left(\left[a,b\right]\right)$ +\end_inset + +. +\end_layout + +\begin_layout Proof +Če je +\begin_inset Formula $\left\{ \alpha_{1},\alpha_{2}\right\} \subseteq\left\{ a,b\right\} $ +\end_inset + +, + je +\begin_inset Formula $f\left(\alpha_{1}\right)=f\left(\alpha_{2}\right)$ +\end_inset + + in je v tem primeru +\begin_inset Formula $f$ +\end_inset + + konstanta ( +\begin_inset Formula $\exists!c\in\mathbb{R}\ni:f\left(x\right)=c$ +\end_inset + +), + ki je odvedljiva in ima povsod odvod nič. +\end_layout + +\begin_layout Proof +Sicer pa +\begin_inset Formula $\left\{ \alpha_{1},\alpha_{2}\right\} \not\subseteq\left\{ a,b\right\} $ +\end_inset + +. + Tedaj ločimo dva primera: +\end_layout + +\begin_layout Labeling +\labelwidthstring 00.00.0000 +\begin_inset Formula $\alpha_{1}\in\left(a,b\right)$ +\end_inset + + To pomeni, + da je globalni maksimum na odprtem intervalu. + Trdimo, + da je v lokalnem maksimumu odvod 0. + Dokaz: +\begin_inset Formula +\[ +f'\left(\alpha_{1}\right)=\lim_{h\to0}\frac{f\left(\alpha_{1}+h\right)-f\left(\alpha_{1}\right)}{h} +\] + +\end_inset + +Za +\begin_inset Formula $a_{1}$ +\end_inset + + (maksimum) velja +\begin_inset Formula $f\left(\alpha_{1}+h\right)-f\left(\alpha_{1}\right)\leq0$ +\end_inset + + (čim se pomaknemo izven točke, + v kateri je maksimum, + je funkcijska vrednost nižja). + Potemtakem velja +\begin_inset Formula +\[ +\frac{f\left(\alpha_{1}+h\right)-f\left(\alpha_{1}\right)}{h}\quad\begin{cases} +\leq0 & ;h>0\\ +\geq0 & ;h<0 +\end{cases} +\] + +\end_inset + +Ker je funkcija odvedljiva na odprtem intervalu, + sta leva in desna limita enaki. +\begin_inset Formula +\[ +0\geq\lim_{h\searrow0}\frac{f\left(\alpha_{1}+h\right)-f\left(\alpha_{1}\right)}{h}=\lim_{h\nearrow0}\frac{f\left(\alpha_{1}+h\right)-f\left(\alpha_{1}\right)}{h}\geq0 +\] + +\end_inset + +Sledi +\begin_inset Formula +\[ +\lim_{h\to0}\frac{f\left(\alpha_{1}+h\right)-f\left(\alpha_{1}\right)}{h}=f'\left(x\right)=0 +\] + +\end_inset + + +\end_layout + +\begin_layout Labeling +\labelwidthstring 00.00.0000 +\begin_inset Formula $\alpha_{2}\in\left(a,b\right)$ +\end_inset + + To pomeni, + da je globalni minimum na odprtem intervalu. + Trdimo, + da je v lokalnem minimumu odvod 0. + Dokaz je podoben tistemu za lokalni maksimum. +\end_layout + +\begin_layout Theorem* +\begin_inset ERT +status open + +\begin_layout Plain Layout + + +\backslash +hypertarget{lagrange}{Lagrange} +\end_layout + +\end_inset + +. + Naj bo +\begin_inset Formula $f:\left[a,b\right]\to\mathbb{R}$ +\end_inset + + za +\begin_inset Formula $a,b\in\mathbb{R}$ +\end_inset + + zvezna na +\begin_inset Formula $\left[a,b\right]$ +\end_inset + + in odvedljiva na +\begin_inset Formula $\left(a,b\right)$ +\end_inset + +. +\begin_inset Formula +\[ +\exists\alpha\in\left(a,b\right)\ni:f\left(b\right)-f\left(a\right)=f'\left(\alpha\right)\left(b-a\right)\sim\frac{f\left(b\right)-f\left(a\right)}{b-a}=f'\left(\alpha\right) +\] + +\end_inset + +ZDB na neki točki na grafu funkcije je tangenta na graf funkcije vzporedna premici, + ki jo določata točki +\begin_inset Formula $\left(a,f\left(a\right)\right)$ +\end_inset + + in +\begin_inset Formula $\left(b,f\left(b\right)\right)$ +\end_inset + +. +\end_layout + +\begin_layout Proof +Za dokaz Lagrangevega uporabimo Rolleov izrek. + Splošen primer prevedemo na primer +\begin_inset Formula $h\left(a\right)=h\left(b\right)$ +\end_inset + + tako, + da od naše splošne funkcije +\begin_inset Formula $f$ +\end_inset + + odštejemo linearno funkcijo +\begin_inset Formula $g$ +\end_inset + +, + da bo veljalo +\begin_inset Formula $\left(f-g\right)\left(a\right)=\left(f-g\right)\left(b\right)$ +\end_inset + +. + Za funkcijo +\begin_inset Formula $g\left(x\right)$ +\end_inset + + mora veljati naslednje: +\end_layout + +\begin_deeper +\begin_layout Itemize +\begin_inset Formula $\exists k,n\in\mathbb{R}\ni:f\left(x\right)=kx+n$ +\end_inset + + +\end_layout + +\begin_layout Itemize +\begin_inset Formula $g\left(a\right)=0$ +\end_inset + + +\end_layout + +\begin_layout Itemize +\begin_inset Formula $g\left(b\right)=f\left(b\right)-f\left(a\right)$ +\end_inset + + +\end_layout + +\end_deeper +\begin_layout Proof +Opazimo, + da mora biti koeficient funkcije +\begin_inset Formula $g$ +\end_inset + + enak +\begin_inset Formula $\frac{f\left(b\right)-f\left(a\right)}{b-a}$ +\end_inset + +, + vertikalni odklon pa tolikšen, + da ima funkcija +\begin_inset Formula $g$ +\end_inset + + v +\begin_inset Formula $a$ +\end_inset + + ničlo: +\begin_inset Formula +\[ +\frac{f\left(b\right)-f\left(a\right)}{b-a}a+n=0 +\] + +\end_inset + + +\begin_inset Formula +\[ +n=-\frac{f\left(b\right)-f\left(a\right)}{b-a}a +\] + +\end_inset + +Našli smo funkcijo +\begin_inset Formula $g\left(x\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}\left(x-a\right)$ +\end_inset + +. + Funkcija +\begin_inset Formula $\left(f-g\right)$ +\end_inset + + sedaj ustreza pogojem za Rolleov izrek, + torej +\begin_inset Formula $\exists\alpha\in\left[a,b\right]\ni:\left(f-g\right)'\left(\alpha\right)=0\Leftrightarrow g'\left(\alpha\right)=f'\left(\alpha\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$ +\end_inset + +, + kar smo želeli dokazati. \end_layout \begin_layout Corollary* -sssssssssss +Naj bo +\begin_inset Formula $I\subseteq\mathbb{R}$ +\end_inset + + nenujno zaprt niti omejen in +\begin_inset Formula $f:I\to\mathbb{R}$ +\end_inset + + odvedljiva na +\begin_inset Formula $I$ +\end_inset + +. + Tedaj je +\begin_inset Formula $f$ +\end_inset + + Lipschitzova. + Lipschitzove funkcije so enakomerno zvezne. +\end_layout + +\begin_layout Proof +Po Lagrangeu velja +\begin_inset Formula $\forall x,y\in I\exists\alpha\in\left(x,y\right)\ni:f\left(x\right)-f\left(y\right)=f'\left(\alpha\right)\left(x-y\right)$ +\end_inset + +. + Potemtakem +\begin_inset Formula $\left|f\left(x\right)-f\left(y\right)\right|=\left|f'\left(\alpha\right)\right|\left|x-y\right|\leq\sup_{\beta\in\left(x,y\right)}\left|f'\left(\beta\right)\right|\left|x-y\right|$ +\end_inset + +. + Torej +\begin_inset Formula $\exists M>0\forall x,y:\left|f\left(x\right)-f\left(y\right)\right|\leq M\left|x-y\right|$ +\end_inset + +, + enakomerno zveznost pa dobimo tako, + da +\begin_inset Formula $\delta\left(\varepsilon\right)=\frac{\varepsilon}{\sup_{\beta\in I}\left|f'\left(\beta\right)\right|}$ +\end_inset + +. + Računajmo. + Naj bo +\begin_inset Formula $M=\sup_{\beta\in I}\left|f'\left(\beta\right)\right|$ +\end_inset + +, + ki obstaja. +\begin_inset Formula +\[ +\forall x,y:\left|f\left(x\right)-f\left(y\right)\right|\leq M\left|x-y\right| +\] + +\end_inset + + +\begin_inset Formula +\[ +\forall x,y:\left|x-y\right|<\frac{\varepsilon}{\sup_{\beta\in I}\left|f'\left(\beta\right)\right|}\Rightarrow\left|f\left(x\right)-f\left(y\right)\right|\leq M\left|x-y\right|<\cancel{\sup_{\beta\in I}\left|f'\left(\beta\right)\right|}\frac{\varepsilon}{\cancel{\sup_{\beta\in I}\left|f'\left(\beta\right)\right|}}<\varepsilon +\] + +\end_inset + + +\begin_inset Formula +\[ +\forall\varepsilon\exists\delta\left(\varepsilon\right)\forall x,y:\left|x-y\right|<\delta\left(\varepsilon\right)\Rightarrow\left|f\left(x\right)-f\left(y\right)\right|<\varepsilon +\] + +\end_inset + + +\end_layout + +\begin_layout Note* +Lipschnitzovim funkcijam pravimo tudi Hölderjeve funkcije reda 1. + +\begin_inset Formula $f$ +\end_inset + + je Hölderjeva funkcija reda +\begin_inset Formula $r$ +\end_inset + +, + če velja +\begin_inset Formula $\exists M>0\forall x,y\in I:\left|f\left(x\right)-f\left(y\right)\right|\leq M\left|x-y\right|^{r}$ +\end_inset + +. +\end_layout + +\begin_layout Claim* +Naj bo +\begin_inset Formula $I$ +\end_inset + + odprti interval, + +\begin_inset Formula $f:I\to\mathbb{R}$ +\end_inset + + odvedljiva. + Tedaj: +\end_layout + +\begin_deeper +\begin_layout Itemize +\begin_inset Formula $f$ +\end_inset + + narašča na +\begin_inset Formula $I\Leftrightarrow f'\geq0$ +\end_inset + + na +\begin_inset Formula $I$ +\end_inset + + +\end_layout + +\begin_layout Itemize +\begin_inset Formula $f$ +\end_inset + + pada na +\begin_inset Formula $I\Leftrightarrow f'\leq0$ +\end_inset + + na +\begin_inset Formula $I$ +\end_inset + + +\end_layout + +\begin_layout Itemize +\begin_inset Formula $f$ +\end_inset + + strogo narašča na +\begin_inset Formula $I\Leftarrow f'>0$ +\end_inset + + na +\begin_inset Formula $I$ +\end_inset + +. + Protiprimer, + da ni +\begin_inset Formula $\Leftrightarrow:f\left(x\right)=x^{3}$ +\end_inset + +, + ki strogo narašča, + toda +\begin_inset Formula $f'\left(0\right)=0$ +\end_inset + + +\end_layout + +\begin_layout Itemize +\begin_inset Formula $f$ +\end_inset + + strogo pada na +\begin_inset Formula $I\Leftarrow f'<0$ +\end_inset + + na +\begin_inset Formula $I$ +\end_inset + +. + Protiprimer, + da ni +\begin_inset Formula $\Leftrightarrow:f\left(x\right)=-x^{3}$ +\end_inset + +, + ki strogo pada, + toda +\begin_inset Formula $f'\left(0\right)=0$ +\end_inset + + +\end_layout + +\end_deeper +\begin_layout Proof +Dokažimo le +\begin_inset Formula $f$ +\end_inset + + narašča na +\begin_inset Formula $I\Leftrightarrow f'\geq0$ +\end_inset + + na +\begin_inset Formula $I$ +\end_inset + +. + Drugo točko dokažemo podobno. + Dokazujemo ekvivalenco: +\end_layout + +\begin_deeper +\begin_layout Labeling +\labelwidthstring 00.00.0000 +\begin_inset Formula $\left(\Leftarrow\right)$ +\end_inset + + +\begin_inset Formula $f'\geq0\Rightarrow f$ +\end_inset + + narašča. + Vzemimo poljubna +\begin_inset Formula $t_{1}<t_{2}\in I$ +\end_inset + +. + Po Lagrangeu +\begin_inset Formula $\exists\alpha\in\left(t_{1},t_{2}\right)\ni:f\left(t_{2}\right)-f\left(t_{1}\right)=f'\left(\alpha\right)\left(t_{2}-t_{1}\right)$ +\end_inset + +. + Ker je po predpostavki +\begin_inset Formula $f'\left(\alpha\right)\geq0$ +\end_inset + + in +\begin_inset Formula $t_{2}-t_{1}>0$ +\end_inset + +, + je tudi +\begin_inset Formula $f\left(t_{2}\right)-f\left(t_{1}\right)\geq0$ +\end_inset + + in zato +\begin_inset Formula $f\left(t_{2}\right)\geq f\left(t_{1}\right)$ +\end_inset + +. +\end_layout + +\begin_layout Labeling +\labelwidthstring 00.00.0000 +\begin_inset Formula $\left(\Rightarrow\right)$ +\end_inset + + +\begin_inset Formula $f$ +\end_inset + + narašča +\begin_inset Formula $\Rightarrow f'\geq0$ +\end_inset + +. + Velja +\begin_inset Formula $f'\left(x\right)=\lim_{h\to0}\frac{f\left(x+h\right)-f\left(x\right)}{h}$ +\end_inset + +. + Po predpostavki je +\begin_inset Formula $f\left(x+h\right)-f\left(x\right)\geq0$ +\end_inset + +, + čim je +\begin_inset Formula $h>0$ +\end_inset + +, + in +\begin_inset Formula $f\left(x+h\right)-f\left(x\right)\leq0$ +\end_inset + +, + čim je +\begin_inset Formula $h<0$ +\end_inset + +. + Torej je ulomek vedno nenegativen. +\end_layout + +\end_deeper +\begin_layout Subsection +Konveksnost in konkavnost +\end_layout + +\begin_layout Definition* +Naj bo +\begin_inset Formula $I\subseteq\mathbb{R}$ +\end_inset + + interval in +\begin_inset Formula $f:I\to\mathbb{R}$ +\end_inset + +. + +\begin_inset Formula $f$ +\end_inset + + je konveksna na +\begin_inset Formula $I$ +\end_inset + +, + če +\begin_inset Formula $\forall a,b\in I$ +\end_inset + + daljica +\begin_inset Formula $\left(a,f\left(a\right)\right),\left(b,f\left(b\right)\right)$ +\end_inset + + leži nad grafom +\begin_inset Formula $f$ +\end_inset + +. +\end_layout + +\begin_layout Definition* +Enačba premice, + ki vsebuje to daljico, + se glasi (razmislek je podoben kot pri +\begin_inset ERT +status open + +\begin_layout Plain Layout + + +\backslash +hyperlink{lagrange}{Lagrangevem izreku} +\end_layout + +\end_inset + +) +\begin_inset Formula +\[ +g\left(x\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}\left(x-a\right)+f\left(a\right) +\] + +\end_inset + +Za konveksno funkcijo torej velja +\begin_inset Formula $\forall a,b\in I:\forall x\in\left(a,b\right):f\left(x\right)\leq\frac{f\left(b\right)-f\left(a\right)}{b-a}\left(x-a\right)+f\left(a\right)$ +\end_inset + + oziroma +\begin_inset Formula +\[ +\frac{f\left(x\right)-f\left(a\right)}{x-a}\leq\frac{f\left(b\right)-f\left(a\right)}{b-a} +\] + +\end_inset + +Vsak +\begin_inset Formula $x$ +\end_inset + + na intervalu lahko zapišemo kot +\begin_inset Formula $x=a+t\left(b-a\right)$ +\end_inset + + za nek +\begin_inset Formula $t\in\left(0,1\right)$ +\end_inset + +. + Tedaj je +\begin_inset Formula $x-a=t\left(b-a\right)$ +\end_inset + + in konveksnost se glasi +\begin_inset Formula +\[ +\forall a,b\in I:\forall t\in\left(0,1\right):f\left(a+t\left(b-a\right)\right)\leq\frac{f\left(b\right)-f\left(a\right)}{\cancel{b-a}}t\cancel{\left(b-a\right)}+f\left(a\right) +\] + +\end_inset + + +\begin_inset Formula +\[ +f\left(a+t\left(b-a\right)\right)=f\left(a+tb-ta\right)=f\left(\left(1-t\right)a+tb\right)\leq tf\left(b\right)-tf\left(a\right)+f\left(a\right)=\left(1-t\right)f\left(a\right)+tf\left(b\right) +\] + +\end_inset + + +\end_layout + +\begin_layout Definition* +Konveksna kombinacija izrazov +\begin_inset Formula $a,b$ +\end_inset + + je izraz oblike +\begin_inset Formula $\left(1-t\right)a+tb$ +\end_inset + + za +\begin_inset Formula $t\in\left(0,1\right)$ +\end_inset + +. + Potemtakem je ZDB definicija konveksnosti +\begin_inset Formula $\forall a,b\in I:$ +\end_inset + + funkcijska vrednost konveksne kombinacije +\begin_inset Formula $a,b$ +\end_inset + + je kvečjemu konveksna kombinacija funkcijskih vrednosti +\begin_inset Formula $a,b$ +\end_inset + +. +\end_layout + +\begin_layout Definition* +Konkavnost pa je definirana tako, + da povsod obrnemo predznake, + torej daljica leži pod grafom +\begin_inset Formula $f$ +\end_inset + + ZDB +\begin_inset Formula $\forall a,b\in I:f\left(\left(1-t\right)a+tb\right)\geq\left(1-t\right)f\left(a\right)+tf\left(b\right)$ +\end_inset + +. +\end_layout + +\begin_layout Example* +\begin_inset Formula $f\left(x\right)=\sin x$ +\end_inset + +, + +\begin_inset Formula $I=\left[-\pi,0\right]$ +\end_inset + +. + Je konveksna. + Se vidi iz grafa. + Preveriti analitično bi bilo težko. +\end_layout + +\begin_layout Example* +Formulirajmo drugačen pogoj za konveksnost. + Naj bo spet +\begin_inset Formula $f:I\to\mathbb{R}$ +\end_inset + +, + kjer je +\begin_inset Formula $I$ +\end_inset + + interval. + +\begin_inset Formula $f$ +\end_inset + + je konveksna +\begin_inset Formula +\[ +\Leftrightarrow\forall a,b\in I\forall x\in\left(a,b\right):\frac{f\left(x\right)-f\left(a\right)}{x-a}\leq\frac{f\left(b\right)-f\left(a\right)}{b-a}. +\] + +\end_inset + + +\end_layout + +\begin_layout Example* +Sedaj glejmo le poljuben +\begin_inset Formula $a$ +\end_inset + +. + Po prejšnjem pogoju moramo gledati še vse poljubne +\begin_inset Formula $b$ +\end_inset + +, + večje od +\begin_inset Formula $a$ +\end_inset + + (ker le tako lahko konstruiramo interval). + Za +\begin_inset Formula $b$ +\end_inset + + in +\begin_inset Formula $a$ +\end_inset + + mora biti diferenčni kvocient večji od diferenčnega kvocienta +\begin_inset Formula $x$ +\end_inset + + in +\begin_inset Formula $a$ +\end_inset + + za poljuben +\begin_inset Formula $x$ +\end_inset + +. + Ta pogoj pa je ekvivalenten temu, + da diferenčni kvocient +\begin_inset Formula $x$ +\end_inset + + in +\begin_inset Formula $a$ +\end_inset + + s fiksnim +\begin_inset Formula $a$ +\end_inset + + in čedalje večjim +\begin_inset Formula $x$ +\end_inset + + narašča, + torej je pogoj za konveksnost tudi: +\begin_inset Formula +\[ +\forall a\in I\forall x>a:g_{a}\left(x\right)=\frac{f\left(x\right)-f\left(a\right)}{x-a}\text{ je naraščajoča funkcija}. +\] + +\end_inset + + \end_layout \begin_layout Corollary* -sssssssssss +Naj bo +\begin_inset Formula $f$ +\end_inset + + konveksna na odprtem intervalu +\begin_inset Formula $I$ +\end_inset + +. + +\begin_inset Formula $\forall a\in I$ +\end_inset + + obstajata funkciji +\begin_inset Formula +\[ +\left(D_{+}f\right)\left(a\right)=\lim_{x\searrow a}g_{a}\left(x\right)=\inf_{x\in I,x>a}g_{a}\left(x\right)\text{ (desni odvod \ensuremath{f} v \ensuremath{a})} +\] + +\end_inset + + +\begin_inset Formula +\[ +\left(D_{-}f\right)\left(a\right)=\lim_{x\nearrow a}g_{a}\left(x\right)=\sup_{x\in I,x<a}g_{a}\left(x\right)\text{ (levi odvod \ensuremath{f} v \ensuremath{a})} +\] + +\end_inset + +in obe sta naraščajoči na +\begin_inset Formula $I$ +\end_inset + +. +\end_layout + +\begin_layout Proof +Obstoj sledi iz monotonosti +\begin_inset Formula $g_{a}\left(a\right)$ +\end_inset + +, + kajti +\begin_inset Formula $\lim_{x\searrow a}g_{a}\left(x\right)=\lim_{x\searrow a}\frac{f\left(x\right)-f\left(a\right)}{x-a}$ +\end_inset + + in enako za levo limito. + Diferenčni kvocient mora namreč biti naraščajoč. + S tem smo dokazali, + da je vsaka konveksna funkcija zvezna +\begin_inset Foot +status open + +\begin_layout Plain Layout +Ni pa vsaka konveksna funkcija odvedljiva, + protiprimer je +\begin_inset Formula $f\left(x\right)=\left|x\right|$ +\end_inset + +. +\end_layout + +\end_inset + +. +\end_layout + +\begin_layout Proof +Naj bodo +\begin_inset Formula $x_{1},x_{2},x\in I\ni:x_{1}<x_{2}<x$ +\end_inset + +. + Pomagaj si s skico +\begin_inset Foot +status open + +\begin_layout Plain Layout +TODO DORIŠI SKICO ZVZ VII/ANA1UČ/str. + 13 +\end_layout + +\end_inset + +. + Ker je +\begin_inset Formula $f$ +\end_inset + + konveksna, + sledi +\begin_inset Formula $g_{x}\left(x_{1}\right)\leq g_{x}\left(x_{2}\right)$ +\end_inset + +. + Ker +\begin_inset Formula $\forall s,t\in\mathbb{R}:g_{s}\left(t\right)=g_{t}\left(s\right)$ +\end_inset + +, + lahko našo neenakost zapišemo kot +\begin_inset Formula $g_{x_{1}}\left(x\right)\leq g_{x_{2}}\left(x\right)$ +\end_inset + +. + Sledi (desni neenačaj iz +\begin_inset Formula $g_{x_{1}}\left(x\right)\leq g_{x_{2}}\left(x\right)$ +\end_inset + +, + levi neenačaj pa ker +\begin_inset Formula $g$ +\end_inset + + narašča): +\begin_inset Formula +\[ +\left(D_{+}\left(f\right)\right)\left(x_{1}\right)=\inf_{x\in I,x>x_{1}}g_{x_{1}}\left(x\right)\leq\inf_{x\in I,x>x_{2}}g_{x_{1}}\left(x\right)\leq\inf_{x\in I,x>x_{2}}g_{x_{2}}\left(x\right)=\left(D_{+}\left(f\right)\right)\left(x_{2}\right) +\] + +\end_inset + +Podobno dokažemo +\begin_inset Foot +status open + +\begin_layout Plain Layout +DOPIŠI KAKO! + TODO XXX FIXME +\end_layout + +\end_inset + +, + da +\begin_inset Formula $D_{-}$ +\end_inset + + narašča. +\end_layout + +\begin_layout Theorem* +Naj bo +\begin_inset Formula $f:I^{\text{odp.}}\to\mathbb{R}$ +\end_inset + + dvakrat odvedljiva. + Tedaj je +\begin_inset Formula $f$ +\end_inset + + konveksna +\begin_inset Formula $\Leftrightarrow\forall x\in I:f''\left(x\right)\geq0$ +\end_inset + + in +\begin_inset Formula $f$ +\end_inset + + konkavna +\begin_inset Formula $\Leftrightarrow\forall x\in I:f''\left(x\right)\leq0$ +\end_inset + +. +\end_layout + +\begin_layout Proof +Dokazujemo ekvivalenco za konveksnost (konkavnost podobno). +\end_layout + +\begin_deeper +\begin_layout Labeling +\labelwidthstring 00.00.0000 +\begin_inset Formula $\left(\Rightarrow\right)$ +\end_inset + + Po predpostavki je +\begin_inset Formula $f$ +\end_inset + + konveksna in dvakrat odvedljiva, + torej je odvedljiva in sta levi in desni odvod enaka, + po prejšnji posledici pa levi in desni odvod naraščata, + torej +\begin_inset Formula $f'$ +\end_inset + + narašča. +\end_layout + +\begin_layout Labeling +\labelwidthstring 00.00.0000 +\begin_inset Formula $\left(\Leftarrow\right)$ +\end_inset + + Naj bo +\begin_inset Formula $f''\geq0$ +\end_inset + +. + Vzemimo +\begin_inset Formula $x,a\in I$ +\end_inset + +. + Po Lagrangeu +\begin_inset Formula $\exists\xi\text{ med \ensuremath{x} in \ensuremath{a}}\ni:f\left(x\right)-f\left(x\right)=f'\left(\xi\right)\left(x-a\right)$ +\end_inset + +. + Iz predpostavke +\begin_inset Formula $f''>0$ +\end_inset + + sledi, + da +\begin_inset Formula $f'$ +\end_inset + + narašča. + Če je +\begin_inset Formula $x>\xi>a$ +\end_inset + +, + velja +\begin_inset Formula $f'\left(\xi\right)\geq f'\left(a\right)$ +\end_inset + +, + zato +\begin_inset Formula $f'\left(\xi\right)\left(x-a\right)\geq f'\left(a\right)\left(x-a\right)$ +\end_inset + +. + Če je +\begin_inset Formula $x<\xi<a$ +\end_inset + +, + velja +\begin_inset Formula $f'\left(\xi\right)\left(x-a\right)\leq f'\left(a\right)\left(x-a\right)$ +\end_inset + +. +\end_layout + +\end_deeper +\begin_layout Subsection +Ekstremi funkcij ene spremenljivke +\end_layout + +\begin_layout Definition* +Naj bo +\begin_inset Formula $I\subseteq\mathbb{R}$ +\end_inset + + odprt interal, + +\begin_inset Formula $a\in I$ +\end_inset + + in +\begin_inset Formula $f:I\to\mathbb{R}$ +\end_inset + +. + Pravimo, + da ima +\begin_inset Formula $f$ +\end_inset + + v točki +\begin_inset Formula $a$ +\end_inset + + lokalni minimum, + če +\begin_inset Formula $\exists\delta>0\ni:\min\left\{ f\left(x\right);\forall x\in\left(a-\delta,a+\delta\right)\right\} =f\left(a\right)$ +\end_inset + +. + Pravimo, + da ima +\begin_inset Formula $f$ +\end_inset + + v točki +\begin_inset Formula $a$ +\end_inset + + lokalni maksimum, + če +\begin_inset Formula $\exists\delta>0\ni:\max\left\{ f\left(x\right);\forall x\in\left(a-\delta,a+\delta\right)\right\} =f\left(a\right)$ +\end_inset + +. +\end_layout + +\begin_layout Theorem* +Če je +\begin_inset Formula $f:I\to\mathbb{R}$ +\end_inset + + odvedljiva in ima v +\begin_inset Formula $a$ +\end_inset + + lokalni minimum/maksimum, + tedaj je +\begin_inset Formula $f'\left(a\right)=0$ +\end_inset + +. +\end_layout + +\begin_layout Proof +Glej dokaz Rolleovega izreka. +\end_layout + +\begin_layout Definition* +\begin_inset Formula $f$ +\end_inset + + ima v +\begin_inset Formula $a$ +\end_inset + + ekstrem, + če ima v +\begin_inset Formula $a$ +\end_inset + + lokalni minimum ali lokalni maksimum. +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Definition* +Če je +\begin_inset Formula $f'\left(a\right)=0$ +\end_inset + +, + pravimo, + da ima +\begin_inset Formula $f$ +\end_inset + + v +\begin_inset Formula $a$ +\end_inset + + stacionarno točko. +\end_layout + +\begin_layout Theorem* +Naj bo +\begin_inset Formula $I\subseteq\mathbb{R}$ +\end_inset + + odprt interval, + +\begin_inset Formula $a\in I$ +\end_inset + + in +\begin_inset Formula $f:I\to\mathbb{R}$ +\end_inset + + dvakrat odvedljiva ter naj bo +\begin_inset Formula $f'\left(a\right)=0$ +\end_inset + +. +\end_layout + +\begin_deeper +\begin_layout Itemize +\begin_inset Formula $f''\left(a\right)>0\Rightarrow$ +\end_inset + + v +\begin_inset Formula $a$ +\end_inset + + ima +\begin_inset Formula $f$ +\end_inset + + lokalni minimum +\end_layout + +\begin_layout Itemize +\begin_inset Formula $f''\left(a\right)<0\Rightarrow$ +\end_inset + + v +\begin_inset Formula $a$ +\end_inset + + ima +\begin_inset Formula $f$ +\end_inset + + lokalni maksimum +\end_layout + +\begin_layout Itemize +\begin_inset Formula $f''\left(a\right)=0\Rightarrow$ +\end_inset + + nedoločeno +\end_layout + +\end_deeper +\begin_layout Proof +Sledi iz +\begin_inset Formula $f''>0\Rightarrow$ +\end_inset + + stroga konveksnost in +\begin_inset Formula $f''<0\Rightarrow$ +\end_inset + + stroga konkavnost. +\end_layout + +\begin_layout Subsection +L'Hopitalovo pravilo +\end_layout + +\begin_layout Standard +Kako izračunati +\begin_inset Formula $\lim_{x\to a}\frac{f\left(x\right)}{g\left(x\right)}$ +\end_inset + +? +\end_layout + +\begin_layout Standard +Če so funkcije zvezne v +\begin_inset Formula $a$ +\end_inset + + in +\begin_inset Formula $g\left(a\right)\not=0$ +\end_inset + +, + velja +\begin_inset Formula $\lim_{x\to a}\frac{f\left(x\right)}{g\left(x\right)}=\frac{f\left(a\right)}{g\left(a\right)}$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Če imata funkciji v +\begin_inset Formula $a$ +\end_inset + + limito in +\begin_inset Formula $\lim_{x\to a}g\left(x\right)\not=0$ +\end_inset + +, + velja +\begin_inset Formula $\lim_{x\to a}\frac{f\left(x\right)}{g\left(x\right)}=\frac{\lim_{x\to a}f\left(x\right)}{\lim_{x\to a}g\left(x\right)}$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Če +\begin_inset Formula $\lim_{x\to a}g\left(x\right)=\infty$ +\end_inset + + in je na neki okolici +\begin_inset Formula $a$ +\end_inset + + +\begin_inset Formula $f\left(x\right)$ +\end_inset + + omejena, + velja +\begin_inset Formula $\lim_{x\to a}\frac{f\left(x\right)}{g\left(x\right)}=0$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Če +\begin_inset Formula $\lim_{x\to a}f\left(x\right)=0$ +\end_inset + + in je na neki okolici +\begin_inset Formula $a$ +\end_inset + + +\begin_inset Formula $g\left(x\right)$ +\end_inset + + navzdol omejena več od nič ali navzgor omejena manj od nič, + velja +\begin_inset Formula $\lim_{x\to a}\frac{f\left(x\right)}{g\left(x\right)}=0$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Zanimivi primeri pa so, + ko +\begin_inset Formula $\lim_{x\to a}f\left(x\right)=\lim_{x\to a}g\left(x\right)=0$ +\end_inset + + ali pa ko +\begin_inset Formula $\lim_{x\to a}f\left(x\right)=\infty$ +\end_inset + + in hkrati +\begin_inset Formula $\lim_{x\to a}g\left(x\right)=\infty$ +\end_inset + +, + na primer +\begin_inset Formula $\lim_{x\to0}\frac{x}{x}$ +\end_inset + + ali pa +\begin_inset Formula $\lim_{x\to0}\frac{x^{2}}{x}$ +\end_inset + + ali pa +\begin_inset Formula $\lim_{x\to0}\frac{x}{x^{2}}$ +\end_inset + +. + Tedaj uporabimo L'Hopitalovo pravilo. +\end_layout + +\begin_layout Theorem* +Če velja hkrati: +\end_layout + +\begin_deeper +\begin_layout Enumerate +\begin_inset CommandInset label +LatexCommand label +name "enu:Eno-izmed-slednjega:" + +\end_inset + +Eno izmed slednjega: +\end_layout + +\begin_deeper +\begin_layout Enumerate +\begin_inset Formula $\lim_{x\to a}f\left(x\right)=\lim_{x\to a}g\left(x\right)=0$ +\end_inset + + +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $\lim_{x\to a}f\left(x\right)=\infty$ +\end_inset + + in hkrati +\begin_inset Formula $\lim_{x\to a}g\left(x\right)=\infty$ +\end_inset + + +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $\lim_{x\to a}f\left(x\right)=-\infty$ +\end_inset + + in hkrati +\begin_inset Formula $\lim_{x\to a}g\left(x\right)=-\infty$ +\end_inset + + +\end_layout + +\end_deeper +\begin_layout Enumerate +\begin_inset Formula $f,g$ +\end_inset + + v okolici +\begin_inset Formula $a$ +\end_inset + + odvedljivi +\end_layout + +\end_deeper +\begin_layout Theorem* +Potem +\begin_inset Formula $\exists L\coloneqq\lim_{x\to a}\frac{f'\left(x\right)}{g'\left(x\right)}\Rightarrow\exists\lim_{x\to a}\frac{f\left(x\right)}{g\left(x\right)}$ +\end_inset + + in ta limita je enaka +\begin_inset Formula $L$ +\end_inset + +. +\end_layout + +\begin_layout Proof +Ne bomo dokazali. +\end_layout + +\begin_layout Example* +Nekaj primerov uporabe L'Hopitalovega pravila. +\end_layout + +\begin_deeper +\begin_layout Enumerate +\begin_inset Formula +\[ +\lim_{x\to0}x^{x}=\lim_{x\to0}e^{lnx^{x}}=\lim_{x\to0}e^{x\ln x}=e^{\lim_{x\to0}x\ln x} +\] + +\end_inset + +Računajmo +\begin_inset Formula $\lim_{x\to0}x\ln x$ +\end_inset + + z L'Hopitalom. + Potrebujemo ulomek. + Ideja: + množimo števec in imenovalec z +\begin_inset Formula $x$ +\end_inset + +, + tedaj bi dobili +\begin_inset Formula $\lim_{x\to0}\frac{x^{2}\ln x}{x}$ +\end_inset + +. + Toda v tem primeru števec in imenovalec ne ustrezata pogoju +\begin_inset CommandInset ref +LatexCommand ref +reference "enu:Eno-izmed-slednjega:" +plural "false" +caps "false" +noprefix "false" +nolink "false" + +\end_inset + + za L'Hopitalovo pravilo. + Druga ideja: + množimo števec in imenovalec z +\begin_inset Formula $\left(\ln x\right)^{-1}$ +\end_inset + +, + tedaj dobimo +\begin_inset Formula $\lim_{x\to0}\frac{x}{\left(\ln x\right)^{-1}}\overset{\text{L'H}}{=}\lim_{x\to0}\frac{1}{\frac{-1}{\log^{2}x}\cdot\frac{1}{x}}=\lim_{x\to0}-x\log^{2}x$ +\end_inset + +, + kar je precej komplicirano. + Tretja ideja: + množimo števec in imenovalec z +\begin_inset Formula $x^{-1}$ +\end_inset + +, + tedaj števec in imenovalec divergirata k +\begin_inset Formula $-\infty$ +\end_inset + +. +\begin_inset Formula +\[ +\lim_{x\to0}\frac{\ln x}{x^{-1}}\overset{\text{L'H}}{=}\lim_{x\to0}\frac{\left(\ln x\right)'}{\left(x^{-1}\right)'}=\lim_{x\to0}\frac{x^{-1}}{-x^{-2}}=\lim_{x\to0}-x=0 +\] + +\end_inset + +Potemtakem +\begin_inset Formula $\lim_{x\to0}x^{x}=e^{0}=1$ +\end_inset + +. +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $\lim_{x\to0}\frac{1-\cos x}{x^{2}}$ +\end_inset + +. + Obe strani ulomkove črte konvergirata k +\begin_inset Formula $0$ +\end_inset + +. + Prav tako ko enkrat že uporabimo L'H. +\begin_inset Formula +\[ +\lim_{x\to0}\frac{1-\cos x}{x^{2}}\overset{\text{L'H}}{=}\lim_{x\to0}\frac{\sin x}{2x}\overset{\text{L'H}}{=}\lim_{x\to0}\frac{\cos x}{2}=\frac{1}{2} +\] + +\end_inset + + +\end_layout + +\end_deeper +\begin_layout Section +Taylorjev izrek in Taylorjeva formula +\end_layout + +\begin_layout Standard +Naj bo +\begin_inset Formula $f$ +\end_inset + + v okolici +\begin_inset Formula $a$ +\end_inset + + dovoljkrat odvedljiva. + Želimo aproksimirati +\begin_inset Formula $f\left(a+h\right)$ +\end_inset + + s polinomi danega reda +\begin_inset Formula $n$ +\end_inset + +. + Iščemo polinome reda +\begin_inset Formula $n$ +\end_inset + +. +\end_layout + +\begin_layout Labeling +\labelwidthstring 00.00.0000 +\begin_inset Formula $n=0$ +\end_inset + + konstante. + +\begin_inset Formula $f\left(a+h\right)\approx f\left(a\right)$ +\end_inset + + +\end_layout + +\begin_layout Labeling +\labelwidthstring 00.00.0000 +\begin_inset Formula $n=1$ +\end_inset + + linearne funkcije. + +\begin_inset Formula $f\left(a+h\right)\sim f\left(a\right)+f'\left(a\right)h$ +\end_inset + + +\end_layout + +\begin_layout Labeling +\labelwidthstring 00.00.0000 +\begin_inset Formula $n=2$ +\end_inset + + ... + Želimo najti +\begin_inset Formula $a_{0},a_{1},a_{2}\in\mathbb{R}$ +\end_inset + +, + odvisne le od +\begin_inset Formula $f$ +\end_inset + + in +\begin_inset Formula $a$ +\end_inset + +, + za katere +\begin_inset Formula $f\left(a+b\right)\approx a_{0}+a_{1}h+a_{2}h^{2}$ +\end_inset + +. + Ko govorimo o aproksimaciji, + mislimo take koeficiente, + da se približek najbolje prilega dejanski funkcijski vrednosti, + v smislu, + da +\begin_inset Formula +\[ +\lim_{h\to0}\frac{f\left(a+h\right)-\left(a_{0}+a_{1}h+a_{2}h^{2}\right)}{h^{2}}=0 +\] + +\end_inset + + +\begin_inset Formula $a_{0}$ +\end_inset + + izvemo takoj, + kajti +\begin_inset Formula $\lim_{h\to0}f\left(a+h\right)-\left(a_{0}+a_{1}h+a_{2}h^{2}\right)=0=f\left(a\right)-\left(a_{0}+0h+0h^{2}\right)=f\left(a\right)-a_{0}=0$ +\end_inset + +, + torej +\begin_inset Formula $a_{0}=f\left(a\right)$ +\end_inset + +. + Za preostale koeficiente uporabimo L'Hopitalovo pravilo, + ki pove, + da zadošča, + da je +\begin_inset Formula +\[ +\lim_{h\to0}\frac{f'\left(a+h\right)-\left(0+a_{1}+a_{2}h\right)}{2h}=0 +\] + +\end_inset + +Zopet glejmo števec in vstavimo +\begin_inset Formula $h=0$ +\end_inset + +: + +\begin_inset Formula $f'\left(a\right)-a_{1}=0\Rightarrow f'\left(a\right)=a_{1}$ +\end_inset + +. + Spet uporabimo L'H: +\begin_inset Formula +\[ +\lim_{h\to0}\frac{f''\left(a+h\right)-\left(0+0+2a_{2}\right)}{2} +\] + +\end_inset + +Vstavimo +\begin_inset Formula $h=0$ +\end_inset + + v +\begin_inset Formula $f''\left(a+h\right)-2a_{2}$ +\end_inset + + in dobimo +\begin_inset Formula $2a_{2}=f''\left(a\right)$ +\end_inset + +, + torej +\begin_inset Formula $a_{2}=\frac{f''\left(a\right)}{2}$ +\end_inset + +. +\end_layout + +\begin_layout Labeling +\labelwidthstring 00.00.0000 +\begin_inset Formula $n=3$ +\end_inset + + Ugibamo, + da je najboljši kubični približek +\begin_inset Formula +\[ +f\left(a+h\right)\approx h\mapsto f\left(a\right)+f'\left(a\right)h+\frac{f''\left(a\right)}{2}h^{2}+\frac{f'''\left(a\right)}{6}h^{3} +\] + +\end_inset + + +\end_layout + +\begin_layout Theorem* +Taylor. + Naj bo +\begin_inset Formula $n\in\mathbb{N}$ +\end_inset + +, + +\begin_inset Formula $I$ +\end_inset + + interval +\begin_inset Formula $\subseteq\mathbb{R}$ +\end_inset + +, + +\begin_inset Formula $a\in I$ +\end_inset + +, + +\begin_inset Formula $f:I\to\mathbb{R}$ +\end_inset + + +\begin_inset Formula $n-$ +\end_inset + +krat odvedljiva v točki +\begin_inset Formula $a$ +\end_inset + +. + Tedaj +\begin_inset Formula $\exists g_{n}:I-a\to\mathbb{R}\ni:$ +\end_inset + + +\begin_inset Foot +status open + +\begin_layout Plain Layout +\begin_inset Formula $I-a$ +\end_inset + + pomeni interval +\begin_inset Formula $I$ +\end_inset + + pomaknjen v levo za +\begin_inset Formula $a$ +\end_inset + +. +\end_layout + +\end_inset + + +\end_layout + +\begin_deeper +\begin_layout Itemize +\begin_inset Formula $f\left(a+h\right)=\sum_{j=0}^{n}\frac{f^{\left(j\right)}\left(a\right)}{n!}h^{j}+g_{n}\left(h\right)h^{n}$ +\end_inset + + in +\end_layout + +\begin_layout Itemize +\begin_inset Formula $\lim_{h\to0}g_{n}\left(h\right)=0$ +\end_inset + + +\end_layout + +\end_deeper +\begin_layout Theorem* +Sedaj pišimo +\begin_inset Formula $x=a+h$ +\end_inset + +. + Tedaj se izrek glasi: + +\begin_inset Formula $\exists\tilde{g_{n}}:I\to\mathbb{R}\ni:$ +\end_inset + + +\end_layout + +\begin_deeper +\begin_layout Itemize +\begin_inset Formula $f\left(x\right)=\sum_{j=0}^{n}\frac{f^{\left(j\right)}\left(a\right)}{n!}\left(x-a\right)^{j}+\tilde{g_{n}}\left(x\right)\left(x-a\right)^{n}$ +\end_inset + + +\end_layout + +\begin_layout Itemize +\begin_inset Formula $\lim_{x\to a}\tilde{g_{n}}\left(x\right)=0$ +\end_inset + + +\end_layout + +\end_deeper +\begin_layout Theorem* +Tedaj označimo +\begin_inset Formula $T_{n,f,a}\left(x\right)=\sum_{j=0}^{n}\frac{f^{\left(j\right)}\left(a\right)}{n!}\left(x-a\right)^{j}$ +\end_inset + + (pravimo +\begin_inset Formula $n-$ +\end_inset + +ti taylorjev polinom za +\begin_inset Formula $f$ +\end_inset + + okrog točke +\begin_inset Formula $a$ +\end_inset + +) in +\begin_inset Formula $R_{n,f,a}\left(x\right)=\tilde{g_{n}}\left(x\right)\left(x-a\right)^{n}$ +\end_inset + + (pravimo ostanek/napaka). +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Theorem* +Če je +\begin_inset Formula $f$ +\end_inset + + +\begin_inset Formula $\left(n+1\right)-$ +\end_inset + +krat odvedljiva na odprtem intervalu +\begin_inset Formula $I\subseteq\mathbb{R}$ +\end_inset + +, + +\begin_inset Formula $a\in I$ +\end_inset + +, + tedaj +\begin_inset Formula $\forall b\in I\exists\alpha\in I\text{ med \ensuremath{a} in \ensuremath{x}}\ni:R_{n}\left(b\right)=\frac{f^{\left(n+1\right)}\left(\alpha\right)}{\left(n+1\right)!}\left(b-a\right)^{n+1}$ +\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* -sssssssssss +Če je +\begin_inset Formula $\left(n+1\right)-$ +\end_inset + +ti odvod omejen na +\begin_inset Formula $I$ +\end_inset + +, + t. + j. + +\begin_inset Formula $\exists M>0\forall x\in I:\left|f^{\left(n+1\right)}\left(x\right)\right|\leq M$ +\end_inset + +, + lahko ostanek eksplicitno ocenimo, + in sicer +\begin_inset Formula $\left|R_{n}\left(x\right)\right|\leq\frac{M}{\left(n+1\right)!}\left|x-a\right|^{n+1}$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Kaj pa se zgodi, + ko +\begin_inset Formula $n$ +\end_inset + + pošljemo v neskončnost? + Iskali bi aproksimacije s +\begin_inset Quotes gld +\end_inset + +polinomi neskončnega reda +\begin_inset Quotes grd +\end_inset + +. +\end_layout + +\begin_layout Definition* +Če je +\begin_inset Formula $f\in C^{\infty}$ +\end_inset + + v okolici točke +\begin_inset Formula $a\in\mathbb{R}$ +\end_inset + +. + Tedaj definiramo Taylorjevo vrsto +\begin_inset Formula $f$ +\end_inset + + v okolici točke +\begin_inset Formula $a$ +\end_inset + +: + +\begin_inset Formula $T_{f,a}\left(x\right)\coloneqq\sum_{j=0}^{\infty}\frac{f^{\left(j\right)}\left(a\right)}{j!}\left(x-a\right)^{j}$ +\end_inset + +. +\end_layout + +\begin_layout Question* +Ali Taylorjeva vrsta konvergira oziroma kje konvergira? + Kakšna je zveza s +\begin_inset Formula $f\left(x\right)$ +\end_inset + +? + Kakšen je +\begin_inset Formula $R_{f,a}$ +\end_inset + +? +\end_layout + +\begin_layout Standard +Oglejmo si potenčne vrste ( +\begin_inset Formula $\sum_{j=0}^{\infty}b_{k}x^{k}$ +\end_inset + +) kot poseben primer funkcijskih vrst ( +\begin_inset Formula $\sum_{j=0}^{\infty}a_{k}\left(x\right)$ +\end_inset + +). + Vemo, + da ima potenčna vrsta konvergenčni radij +\begin_inset Formula $R$ +\end_inset + +. + Za +\begin_inset Formula $x\in\left(-R,R\right)$ +\end_inset + + konvergira, + za +\begin_inset Formula $x\in\left[-R,R\right]^{C}$ +\end_inset + + divergira. +\end_layout + +\begin_layout Theorem* +Naj ima potenčna vrsta +\begin_inset Formula $f\left(x\right)=\sum_{k=0}^{\infty}b_{k}x^{k}$ +\end_inset + + konvergenčni radij +\begin_inset Formula $R$ +\end_inset + +. + Tedaj ima tudi +\begin_inset Formula $g\left(x\right)=\sum_{k=1}kb_{k}x^{k-1}$ +\end_inset + + konvergenčni radij +\begin_inset Formula $R$ +\end_inset + +. +\end_layout + +\begin_layout Proof +\begin_inset Formula +\[ +\frac{1}{R_{g}}=\limsup_{k\to\infty}\sqrt[k]{\left|ka_{k}\right|}=\limsup_{k\to\infty}\sqrt[k]{\left|k\right|\left|a_{k}\right|}=\limsup_{k\to\infty}\sqrt[k]{\left|k\right|}\sqrt[k]{\left|a_{k}\right|}=\cdots +\] + +\end_inset + + +\end_layout + +\begin_layout Proof +\begin_inset Formula +\[ +\lim_{k\to\infty}\sqrt[k]{\left|k\right|}=\lim_{k\to\infty}k^{1/k}=e^{\lim_{k\to\infty}\frac{1}{k}\ln k}\overset{\text{L'H}}{=}e^{\lim_{k\to\infty}\frac{\frac{1}{k}}{k}}=e^{\lim_{k\to\infty}\cancelto{0}{\frac{1}{k^{2}}}}=e^{0}=1 +\] + +\end_inset + + +\begin_inset Formula +\[ +\cdots=\limsup_{k\to\infty}1\cdot\sqrt[k]{\left|a_{k}\right|}=\frac{1}{R_{f}} +\] + +\end_inset + + \end_layout \begin_layout Corollary* -sssssssssss +Če ima potenčna vrsta +\begin_inset Formula $f$ +\end_inset + + konvergenčni radij +\begin_inset Formula $R>0$ +\end_inset + +, + tedaj je +\begin_inset Formula $f\in C^{\infty}\left(\left(-R,R\right)\right)$ +\end_inset + + in velja +\begin_inset Formula $a_{k}=\frac{f^{\left(k\right)}\left(0\right)}{k!}$ +\end_inset + +, + potem velja +\begin_inset Formula $g=f'$ +\end_inset + + (iz izreka zgoraj). + Razlaga: +\end_layout + +\begin_deeper +\begin_layout Itemize +\begin_inset Formula $f\left(x\right)=\sum_{k=0}^{\infty}a_{k}x^{k}=\sum_{k=0}^{\infty}\frac{f^{\left(k\right)}\left(0\right)}{k!}x^{k}$ +\end_inset + + +\end_layout + +\begin_layout Itemize +\begin_inset Formula $f'\left(x\right)=\sum_{k=1}^{\infty}ka_{k}x^{k-1}=\sum_{k=1}^{\infty}\frac{kf^{\left(k\right)}\left(0\right)}{k!}x^{k-1}=\sum_{k=1}^{\infty}\frac{f^{\left(k\right)}\left(0\right)}{\left(k-1\right)!}x^{k-1}$ +\end_inset + + ( +\begin_inset Formula $k$ +\end_inset + + začne z +\begin_inset Formula $1$ +\end_inset + +, + ker se +\begin_inset Formula $k=0$ +\end_inset + + člen odvaja v konstanto +\begin_inset Formula $0$ +\end_inset + +) +\end_layout + +\begin_layout Itemize +\begin_inset Formula $g\left(x\right)=\sum_{k=1}^{\infty}ka_{k}x^{k-1}=\sum_{k=1}^{\infty}\frac{kf^{\left(k\right)}\left(0\right)}{k!}x^{k-1}=\sum_{k=1}^{\infty}\frac{f^{\left(k\right)}\left(0\right)}{\left(k-1\right)!}x^{k-1}=f'\left(x\right)$ +\end_inset + + +\end_layout + +\end_deeper +\begin_layout Definition* +Funkcija +\begin_inset Formula $f:J\to\mathbb{R}$ +\end_inset + + ( +\begin_inset Formula $J$ +\end_inset + + je interval +\begin_inset Formula $\subseteq\mathbb{R}$ +\end_inset + +) je realno analitična, + če se jo da okoli vsake točke +\begin_inset Formula $c\in J$ +\end_inset + + razviti v potenčno vrsto, + torej če +\begin_inset Formula $f\left(x\right)=\sum_{k=0}^{\infty}\frac{f^{\left(k\right)}\left(c\right)}{k!}\left(x-c\right)^{k}$ +\end_inset + + za +\begin_inset Formula $x$ +\end_inset + + blizu +\begin_inset Formula $c$ +\end_inset + +. +\end_layout + +\begin_layout Claim* +\begin_inset Formula $f\in C^{\infty}\Rightarrow f$ +\end_inset + + je realno analitična. + Protiprimer je +\begin_inset Formula $f\left(x\right)=e^{\frac{-1}{\left|x\right|}}$ +\end_inset + +. +\begin_inset Foot +status open + +\begin_layout Plain Layout +TODO XXX FIXME ZAKAJ?, + ne razumem +\end_layout + +\end_inset + + +\end_layout + +\begin_layout Example* +Primeri Taylorjevih vrst. +\end_layout + +\begin_deeper +\begin_layout Enumerate +\begin_inset Formula $f\left(x\right)=e^{x}$ +\end_inset + +. + +\begin_inset Formula $n-$ +\end_inset + +ti tayorjev polinom za +\begin_inset Formula $f\left(x\right)$ +\end_inset + + okoli +\begin_inset Formula $0$ +\end_inset + +: + +\begin_inset Formula $T_{n,e^{x},0}\left(x\right)=1+x+\frac{x^{2}}{2}+\frac{x^{3}}{6}+\cdots+\frac{x^{n}}{n!}$ +\end_inset + + in velja +\begin_inset Formula $e^{x}=T_{n,e^{x},0}\left(x\right)+R_{n,e^{x},0}\left(x\right)$ +\end_inset + +, + kjer +\begin_inset Formula $\lim_{n\to\infty}R_{n,e^{x},0}\left(x\right)=0$ +\end_inset + +. + Ne bomo dokazali. + Sledi +\begin_inset Formula $e^{x}=\sum_{k=0}^{\infty}\frac{x^{k}}{k!}$ +\end_inset + +. +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $\sin x=x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\cdots+\left(-1\right)^{k}\frac{x^{2k+1}}{\left(2k+1\right)!}$ +\end_inset + +. + Opazimo sode eksponente in opazimo učinek odvajanja: + +\begin_inset Formula $\cos,-\sin,-\cos,\sin,\cos,-\sin,\dots$ +\end_inset + +. + Členi vrste +\begin_inset Formula $\sin x$ +\end_inset + + v +\begin_inset Formula $x=0$ +\end_inset + + so: + +\begin_inset Formula $1,0,-1,0,1,0,-1,\dots$ +\end_inset + +. + Opazimo izpadanje vsakega drugega člena. +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $\cos x=1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-\cdots+\left(-1\right)^{k}\frac{x^{2k}}{\left(2k\right)!}$ +\end_inset + +. + Opazimo sode eksponente. +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $f\left(x\right)=\log\left(1-x\right)$ +\end_inset + +. + A lahko to funkcijo razvijemo v taylorjevo vrsto okoli točke 0? +\begin_inset Float table +placement document +alignment document +wide false +sideways false +status open + +\begin_layout Plain Layout +\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 |