<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[11pt,notitlepage]{report}

\frenchspacing % aktivuje pouĹľitĂ­ nÄ›kterĂ˝ch ÄŤeskĂ˝ch typografickĂ˝ch pravidel

\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc} % nastavuje pouĹľitĂ© kĂłdovĂˇnĂ­, uĹľivatelĂ© Windows zamÄ›nĂ­ utf8 za cp1250
\usepackage{amsthm,amsmath,amssymb,amscd,mathrsfs,enumerate,multicol,color,url}%,showkeys,layout}
\usepackage{verbatim} %pro vynechavani comment
\usepackage[top=1cm, left=1cm, right=1cm]{geometry} % nastavenĂ­ danĂ© velikosti okrajĹŻ
\pagestyle{empty}
\newtheorem*{theorem}{V\v{e}ta}
\theoremstyle{definition}
\newtheorem*{defin}{Definice}
\newtheorem*{pozn}{Pozn\'amka}
\newcommand\emphdef[1]{\textbf{#1}}
\def\st{\operatorname{st}}

\def\ce{\mathbb C} 
\def\en{\mathbb N}
\def\zet{\mathbb Z}
\def\qe{\mathbb Q} 
\def\er{\mathbb R}
\def\R{\er}
\def\N{\en}
\def\C{\mathcal C}
\def \sumaO {\sum_{n=0}^\infty}
\def\setsep{:\;}
\def\nec{Nech\v t\ }
\def\eqdef{\stackrel{\text{\tiny def}}{=}}
\def\Mod{\operatorname{mod}}
\def\lp{\displaystyle\lim_{n\to\infty}}
\newcommand\lx[1][0]{\displaystyle\lim_{x\to #1}}
\def \cotg {\operatorname{cotg}}
\def \tg {\operatorname{tg}}
\def \sgn {\operatorname{sgn}}
\def \arctg {\operatorname{arctg}}
\def \arccotg {\operatorname{arccotg}}
\def \Re {\operatorname{Re}}
\def \Im {\operatorname{Im}}
\def \suma {\sum_{n=1}^\infty}
\def \sumad {\sum_{n=2}^\infty}
\def \dx {\, \mathrm{d} x}
\def \dy {\, \mathrm{d} y}
\def \P {\mathcal{P}}
\newcommand\pdx[1][f]{\frac{\partial #1}{\partial x}}
\newcommand\pdy[1][f]{\frac{\partial #1}{\partial y}}
\newcommand\pdz[1][f]{\frac{\partial #1}{\partial z}}
\newcommand\dl[1]{\, \mathrm{d} \lambda^{#1}}
\def\x{\mathbf{x}}
\def\m{\mathbf{m}}

\newtheorem*{thm}{V\v{e}ta}

\newcounter{Priklad}
\addtocounter{Priklad}{1}
\newcommand{\pr}{\quad\alph{Priklad})\;\addtocounter{Priklad}{1}}
\newcommand{\resetpr}{\setcounter{Priklad}{1}}

\title{Kalkulus 3, MFF - p\v{r}'iklady}  % tyto dvÄ› poloĹľky jsou zde v podstatÄ› formĂˇlnÄ›, ve skuteÄŤnosti nejsou nikde 
\author{Marek C\'uth}

\begin{document}
\centerline{\MakeUppercase{1. cvi\v{c}en\'i}}\ \\[5pt]
\textbf{1. Najd\v{e}te re\'alnou a imagin\'arn\'i \v{c}\'ast komplexn\'ich \v{c}\'isel}\\[5pt] \resetpr
\pr $\frac{1-i}{1+i}$ \pr $\frac{2}{1-5i}$ \pr $(1+i\sqrt{2})^3$ \pr $\frac{2+i}{i} + \frac{i}{i+1} - \frac{2i+1}{i-1}$ \pr $\overline{\frac{3+4i}{1-2i}}$\\[5pt]
\textbf{2. Ur\v{c}ete absolutn\'i hodnotu}
\[\left|\frac{|4-3i|+i}{3-2i}\right|\]
\textbf{3. Zapi\v{s}te v goniometrick\'em tvaru}\\[5pt] \resetpr
\pr $-5$ \pr $-3-3i$ \pr $10-10i$ \pr $-2+5i$ \pr $-\cos\tfrac{\pi}{7} + i\sin\tfrac{\pi}{7}$\\[5pt]
\textbf{4. Najd\v{e}te re\'alnou a imagin\'arn\'i \v{c}\'ast n\'asleduj\'ic\'ich hodnot funkc\'i}\\[5pt] \resetpr
\pr $\sin(2+i)$ \pr $\cos(2i)$ \pr $\tg(2-i)$ \pr $\cotg(\tfrac{\pi}{4} - i\log 3)$\\[5pt]

	\begin{center}--------------------------------------------------------------------------\end{center}
	V\'ysledky: \textbf{1.}V\'ysledky ve tvaru $\Re z$; $\Im z$: \resetpr \pr $0$; $-1$ \pr $\tfrac{1}{13}$; $\tfrac{5}{13}$ \pr $-5$; $\sqrt{2}$ \pr $1$; $0$ \pr $-1$; $-2$\\
	\textbf{2.} $\sqrt{2}$\\
	\textbf{3.}\resetpr \pr $5(\cos \pi + i\sin \pi)$ \pr $3\sqrt{2}(\cos(-\tfrac{3}{4}\pi) + i\sin(-\tfrac{3}{4}\pi))$ \pr $10\sqrt{2}(\cos(-\tfrac{\pi}{4}) + i\sin(-\tfrac{\pi}{4}))$ \pr $\sqrt{29}(\cos \arccos \tfrac{-2}{\sqrt{29}} + i \sin \arccos \tfrac{-2}{\sqrt{29}})$ \pr $1(\cos(\tfrac{6}{7}\pi) + i\sin(\tfrac{6}{7}\pi))$\\
	\textbf{4.}V\'ysledky ve tvaru $\Re z$; $\Im z$ (pou\v{z}\'iv\'ame funkce $\sinh(x) = \frac{e^x - e^{-x}}{2}$ a $\cosh(x) = \frac{e^x + e^{-x}}{2}$):\resetpr \pr $\sin 2\cdot \cosh 1$; $\cos 2\cdot \sinh 1$; \pr $\cosh 2$; $0$ \pr $\frac{\sin2\cdot\cos 2}{\cos^2 2 + \sinh^2 1}$; $\frac{-\sinh 1\cdot\cosh 1}{\cos^2 2 + \sinh^2 1}$ \pr $\tfrac{9}{41}$; $\tfrac{40}{41}$
	\newpage
	\centerline{\MakeUppercase{2. cvi\v{c}en\'i}}\ \\[5pt]	
	\textbf{1. Najd\v{e}te ``v\v{s}echny hodnoty komplexn\'ich odmocnin'' (tj. v\v{s}echna \v{r}e\v{s}en\'i rovnice $z^n=a$, pokud je v zad\'an\'i uvedeno $\sqrt[n]{a}$):}\resetpr \pr $\sqrt[3]{1}$ \pr $\sqrt{i-1}$\\[5pt]
	\textbf{2. Najd\v{e}te v\v{s}echna \v{r}e\v{s}en\'i n\'asleduj\'ic\'ich rovnic v $\ce$:}\resetpr \pr $iz^2 - 3z + 4i = 0$ \pr $z^2 - 6iz - 9 = 0$\\ \pr $\sin z + \cos z = 10$ \pr $\sin z - \cos z = i$ \pr $\cos z = \tfrac{3}{4}i$\\[5pt]
	\textbf{3. Najd\v{e}te re\'alnou a imagin\'arn\'i \v{c}\'ast n\'asleduj\'ic\'ich hodnot funkc\'i}\\[5pt] \resetpr \pr $i^i$ \pr $(-1)^{\sqrt{2}}$ \pr $2^i$ \pr $(-2i)^{-2i}$
	
	\begin{center}--------------------------------------------------------------------------\end{center}
	V\'ysledky: 
	\textbf{1.}\resetpr \pr $1,-\tfrac{1}{2}+i\tfrac{\sqrt{3}}{2},-\tfrac{1}{2}-i\tfrac{\sqrt{3}}{2}$ \pr $\sqrt[4]{2}(\cos \tfrac{3}{8}\pi + i\sin \tfrac{3}{8}\pi),\sqrt[4]{2}(\cos \tfrac{11}{8}\pi + i\sin \tfrac{11}{8}\pi)$\\
	\textbf{2.}\resetpr \pr $i,-4i$ \pr $3i$ \pr $\tfrac{\pi}{4} + 2k\pi + i\log(5\sqrt{2}+7),\tfrac{\pi}{4} + 2k\pi + i\log(5\sqrt{2}-7)$\\ \pr $\tfrac{\pi}{4} + 2k\pi - i\log\tfrac{\sqrt{3}-1}{\sqrt{2}},-\tfrac{3\pi}{4} + 2k\pi - i\log\tfrac{\sqrt{3}+1}{\sqrt{2}}$ \pr $(4k+1)\pi + i\log 2,(4k-1)\pi - i\log 2$\\
	\textbf{3.}V\'ysledky ve tvaru $\Re z$; $\Im z$: \resetpr \pr $e^{-\pi/2};0$ \pr $\cos(\sqrt{2}\pi);\sin(\sqrt{2}\pi)$ \pr $\cos(\log2);\sin(\log2)$\\ \pr $e^{-\pi}\cos(-2\log 2);e^\pi\sin(-2\log 2)$
	\newpage
	\centerline{\MakeUppercase{3. cvi\v{c}en\'i}}\ \\[5pt]	
	\textbf{1. Ve kter\'ych bodech maj\'i n\'asleduj\'ic\'i funkce derivaci podle komplexn\'i prom\v{e}nn\'e?}\resetpr \pr $|z| $\pr $|(\Re z)^2 - (\Im z)^2| + 2i|\Re z \cdot \Im z|$ \pr $|z|^2 + i \Im(z^2)$\\[5pt]
	\textbf{2. Pomoc\'i definice k\v{r}ivkov\'eho integr\'alu spo\v{c}t\v{e}te $\int_\varphi f$, kde:}\resetpr \pr $f(z)=\Re z$, $\varphi$ je orientovan\'a \'use\v{c}ka $[0,1+i]$ \pr $f(z)=\Im z$, $\varphi$ je kladn\v{e} orientovan\'a polokru\v{z}nice $|z|=1$, $0\leq \arg z\leq \pi$ \pr $f(z)=|z|$, $\varphi$ je orientovan\'a \'use\v{c}ka $[0,2-i]$ \pr $f(z)=z^2$, $\varphi$ je oblouk paraboly $y=1-x^2$ s po\v{c}\'ate\v{c}n\'im bodem $-1$ a koncov\'ym $1$ \pr $f(z)=|z|z$, $\varphi$ je kladn\v{e} orientovan\'y obvod ``\v{c}tvrtkru\v{z}nicov\'e v\'yse\v{c}e'' $|z|=1$, $0\leq \arg z\leq \tfrac{\pi}{2}$ (viz. obr\'azek na tabuli) \pr $f(z)=|z|\bar{z}$, $\varphi$ je kladn\v{e} orientovan\'y obvod ``osminokru\v{z}nicov\'e v\'yse\v{c}e'' $|z|=1$, $0\leq \arg z\leq \tfrac{\pi}{4}$ (viz. obr\'azek na tabuli)
	
	\begin{center}--------------------------------------------------------------------------\end{center}
	V\'ysledky: 
	\textbf{1.}\resetpr \pr v \v{z}\'adn\'em bod\v{e} \pr v bodech $z$, pro kter\'e plat\'i $0&lt;\Im z &lt; \Re z$, $\Re z &lt; \Im z &lt; 0$, $0&lt;-\Re z &lt; \Im z$ nebo $\Im z &lt; -\Re z &lt; 0$ \pr v bodech re\'aln\'e osy\\
	\textbf{2.}\resetpr \pr $\tfrac{1+i}{2}$ \pr $-\tfrac{\pi}{2}$ \pr $\tfrac{\sqrt{5}}{2}(2-i)$ \pr $\tfrac{2}{3}$ \pr $\tfrac{1}{3}$ \pr $\tfrac{\pi}{4}i$
	\newpage
	\centerline{\MakeUppercase{4. cvi\v{c}en\'i}}\ \\[5pt]
	\textbf{1. S vyu\v{z}it\'im Cauchyovy v\v{e}ty, znalosti primitivn\'ich funkc\'i a definice spo\v{c}t\v{e}te $\int_\varphi f$, pokud:}\resetpr \\
	\pr $f(z) = -\tfrac{1}{z} + \tfrac{1}{2}\tfrac{1}{z-1} + \tfrac{1}{2}\tfrac{1}{z+1}$ a $\varphi$ je kladn\v{e} orientovan\'a kru\v{z}nice o polom\v{e}ru $\tfrac{1}{2}$ a st\v{r}edu (a2) $0$ (a3) $1$ (a4) $2$;\\
	\pr $f(z) = \tfrac{1}{z^2(z^2-1)}$ a $\varphi$ je kladn\v{e} orientovan\'a kru\v{z}nice o polom\v{e}ru $\tfrac{1}{2}$ a st\v{r}edu (b1) $-1$ (b2) $0$ (b3) $1$ (b4) $2$\\[5pt]
	\textbf{2. Pomoc\'i Cauchyova vzorce spo\v{c}t\v{e}te integr\'aly $\int_\varphi f$, kde $\varphi$ je kladn\v{e} orientovan\'a kru\v{z}nice o st\v{r}edu $0$ a polom\v{e}ru $2$;}\resetpr
	\pr $f(z) = \tfrac{e^z}{z^2-1}$ \pr $f(z) = \tfrac{e^z - z}{z^2 - 1}$ \pr $\tfrac{ze^z}{(z+1)^3}$
	
	\begin{center}--------------------------------------------------------------------------\end{center}
	V\'ysledky: 
	\textbf{1.}\resetpr \pr (a2) $-2\pi i$, (a3) $\pi i$, (a4) $0$; \pr (b1) $-\pi i$, (b2) $0$, (b3) $\pi i$, (b4) $0$\\
	\textbf{2.}\resetpr \pr $\pi i (e-\tfrac{1}{e})$ \pr $\pi i (e-\tfrac{1}{e})$ \pr $\tfrac{\pi i}{e}$
	\newpage
	\centerline{\MakeUppercase{5. cvi\v{c}en\'i}}\ \\[5pt]	
	\textbf{1. Najd\v{e}te Laurent\r{u}v rozvoj funkce $\tfrac{1}{(z+1)(z+2)}$ v maxim\'aln\'ich mezikru\v{z}\'ich o st\v{r}edech $-1$ a $1$.}\\[3pt]	
	\textbf{2. Zjist\v{e}te obor konvergence Laurentovy \v{r}ady $\sum_{n=-\infty}^\infty \tfrac{1}{2^{n^2}}(z-1)^n$.}\\[3pt]
	\textbf{3. Najd\v{e}te Laurentovu \v{r}adu funkce $\tfrac{1}{(z-i)^2}$ v mezikru\v{z}\'i $P(1,\sqrt{2},\infty)$.}\\[3pt]
	\textbf{4. Najd\v{e}te Laurent\r{u}v rozvoj funkc\'i v maxim\'aln\'ich mezikru\v{z}\'ich o uveden\'ych st\v{r}edech:}\ \resetpr \\	
	\pr $z^3e^{1/z}$, $0$
	\pr $\tfrac{1}{(z-2)(z-1)^2}$, $1$ a $2$
	\pr $\tfrac{1}{z^2 - 1}$, $1$, $-1$ a $0$
	\pr $\tfrac{1}{z-z^3}$, $0$ a $1$
	
	\begin{center}--------------------------------------------------------------------------\end{center}
	V\'ysledky: 
	\textbf{1.} v $P(-1,0,1)$: $\sum_{n=-1}^\infty (-1)^{n+1}(z+1)^n$, v $P(-1,1,\infty)$: $\sum_{n=2}^\infty \tfrac{(-1)^n}{(z+1)^n}$;\\ v $P(1,0,2)$: $\sum_{n=0}^\infty (-1)^{n}\left(\tfrac{1}{2^{n+1}} - \tfrac{1}{3^{n+1}}\right)(z-1)^n$, v $P(1,2,3)$: $\sum_{n=1}^\infty \tfrac{(-2)^{n-1}}{(z-1)^n} + \sum_{n=0}^\infty \tfrac{(-1)^{n+1}}{3^{n+1}}(z-1)^n$,\\ v $P(1,3,\infty)$: $\sum_{n=1}^\infty \tfrac{(-1)^{n-1}(2^{n-1} - 3^{n-1})}{(z-1)^n}$\\
	\textbf{2.} $P(1,0,\infty)$\\
	\textbf{3.} $\sum_{n=0}^\infty (-1)^{n}(1-i)^{n}\tfrac{n+1}{(z-1)^{n+2}}$\\
	\textbf{4.}\resetpr \pr v $P(0,0,\infty)$: $\sum_{n=-\infty}^3 \tfrac{1}{(3-n)!}z^n$ \pr v $P(1,0,1)$: $\sum_{n=0}^\infty (z-1)^n - \tfrac{1}{z-1} - \tfrac{1}{(z-1)^2}$,  v $P(1,1,\infty)$: $\sum_{n=3}^\infty \tfrac{1}{(z-1)^n}$; v $P(2,0,1)$: $\tfrac{1}{z-2} + \sum_{n=0}^\infty (-1)^{n+1}(n+2)(z-2)^n$, v $P(2,1,\infty)$: $\sum_{n=-\infty}^{-1} (-1)^{n+1}(z-2)^n$
	\newpage
	\centerline{\MakeUppercase{6. cvi\v{c}en\'i}}\ \\[5pt]	
	\textbf{1. Najd\v{e}te a klasifikujte izolovan\'e singularity n\'asleduj\'ic\'ich funkc\'i (v\v{c}etn\v{e} bodu $\infty$):}\ \resetpr \\
	\pr $\tfrac{z}{e^z+1}$ \pr $\sin \tfrac{\pi}{z^2}$ \pr $\tfrac{1}{e^z-1} - \tfrac{1}{\sin z}$ \pr $z^2e^{1/z}$ \pr $\sin z$ \pr $\tfrac{\sin z}{z^2}$\\[5pt]
	\textbf{2. Najd\v{e}te izolovan\'e singularity n\'asleduj\'ic\'ich funkc\'i a spo\v{c}t\v{e}te p\v{r}\'islu\v{s}n\'a rezidua:}\ \resetpr \\
	\pr $\sin \tfrac{1}{1-z}$ \pr $\tfrac{1}{z^3-z^5}$ \pr $\tfrac{z^2 + z - 1}{z^2(z-1)}$ \pr $\tfrac{\sin 2z}{(z+1)^3}$ \pr $\tfrac{e^z}{z^2(z^2 + 9)}$ \pr $(z+1)e^{\tfrac{z-1}{z}}$ \pr $\sin \tfrac{z}{z+1}$
	\begin{center}--------------------------------------------------------------------------\end{center}
	V\'ysledky: 
	\textbf{1.}\resetpr \pr p\'oly n\'asobnosti $1$ v bodech $\{(2k+1)\pi i\colon k\in\zet\}$, v $\infty$ nen\'i izolovan\'a singularita \pr v bod\v{e} $0$ podstatn\'a singularita, v $\infty$ odstraniteln\'a singularita \pr v bod\v{e} $0$ odstraniteln\'a singularita, v bodech $\{k\pi\colon k\in\zet\}\cup \{2k\pi i\colon k\in\zet\}$ p\'ol n\'asobnosti $1$, v $\infty$ nen\'i izolovan\'a singularita \pr v bod\v{e} $0$ podstatn\'a singularita, v bod\v{e} $\infty$ p\'ol n\'asobnosti $2$ \pr v $\infty$ podstatn\'a singularita \pr v bod\v{e} $0$ p\'ol n\'asobnosti $1$, v bod\v{e} $\infty$ podstatn\'a singularita\\
	\textbf{2.}\resetpr \pr $1$, reziduum $-1$ \pr $0$, reziduum $1$; $1$, reziduum $-\tfrac{1}{2}$; $-1$, reziduum $-\tfrac{1}{2}$ \pr $0$, reziduum $0$; $1$, reziduum $1$ \pr $-1$, reziduum $2\sin 2$ \pr $0$, reziduum $\tfrac{1}{9}$; $3i$, reziduum $\tfrac{ie^{3i}}{54}$; $-3i$, reziduum $-\tfrac{ie^{-3i}}{54}$ \pr $0$, reziduum $-\tfrac{e}{2}$ \pr $-1$, reziduum $-\cos 1$
	\newpage
	\centerline{\MakeUppercase{9. cvi\v{c}en\'i}}\ \\[5pt]	
	\textbf{1. SpoÄŤtÄ›te $\mathcal{L}(f)$ pro zadanĂ© funkce $f$:}
	\resetpr \pr $f(t) = e^{-3t}\cos(2t+\tfrac{\pi}{2}) + \sin(3(t-\pi))$ \pr $5\cdot 2^{-t} - 4t\cdot 3^t$ \pr $f(t) = t\sin(2t)$ \pr $4\sin^2 3t + te^{3t}\cos 2t$\\[3pt]
	\textbf{2. NaleznÄ›te alespoĹ jednu funkci z $\mathcal{L}^{-1}(g(s))$ pro zadanĂ© funkce $g$:}\resetpr
	\pr $g(s) = \tfrac{2s+3}{s^2+4s+3}$, $s&gt;0$ \pr $\tfrac{4s+7}{s^2+16}$, $s&gt;0$ \pr $g(s) = \tfrac{4s+5}{s^2 + 4s + 13}$, $s&gt;0$ \pr $g(s) = \tfrac{s^2}{(s-2)^3}$, $s&gt;2$ \pr $\tfrac{s+3}{s(s^2+1)} + \tfrac{1}{(s-2)^3}$, $s&gt;2$ \pr $g(s)=\tfrac{1}{s^2(s^2+1)}$, $s&gt;0$ \pr $\tfrac{2s+3}{s^2+4}$, $s&gt;0$ \pr $g(s) = \tfrac{3s-5}{s^2+2s+5}$, $s&gt;-1$ \pr $g(s) = \tfrac{2s+3}{(s+1)^3}$, $s&gt;-1$ \pr $g(s) = \tfrac{1}{s^2+4s+3}$, $s&gt;-1$ \pr $g(s) = \tfrac{s+3}{s^2(s^2+1)}$, $s&gt;0$\\[3pt]
	\textbf{3. SpoÄŤtÄ›te $\mathcal{L}(f)$ pro zadanĂ© funkce $f$:}\\ \resetpr \pr $f(t) = \begin{cases} 2 &amp; t\in[3,5]\\ 0 &amp; t\notin [3,5]\end{cases}$ \pr $f(t) = \begin{cases} 4-t &amp; t\in[1,4]\\ 0 &amp; t\notin [1,4]\end{cases}$ \pr $f(t) = \begin{cases} \sin t &amp; t\in[0,\pi]\\ 0 &amp; t\geq \pi\end{cases}$\\
	\pr $f(t) = \begin{cases} 1 &amp; t\in(1,3]\\ 0 &amp; t\notin (1,3]\end{cases}$ \pr $f(t) = \begin{cases} 2t &amp; t\in[0,1]\\ 2 &amp; t &gt; 1\end{cases}$ \pr $f(t) = \begin{cases} e^{-t} &amp; t\in[0,2]\\ 0 &amp; t&gt;2\end{cases}$\\[3pt]
	\textbf{4. NaleznÄ›te alespoĹ jednu funkci z $\mathcal{L}^{-1}(g(s))$ pro zadanĂ© funkce $g$:}\resetpr \pr $g(s) = \tfrac{2}{s(s-1)}e^{-s}$, $s&gt;1$ \pr $g(s) = \tfrac{3s+2-6e^{-\pi s}}{s^2+4}$, $s&gt;0$ \pr $g(s) = \tfrac{2}{s^2}(1-e^{-s} - se^{-3s})$, $s&gt;0$ \pr $g(s) = \tfrac{1}{s^2+9}(3-3e^{-\tfrac{\pi}{2}s}-(3+s)e^{-\tfrac{\pi}{6}s})$, $s&gt;0$
	\begin{center}--------------------------------------------------------------------------\end{center}
	V\'ysledky: 
	\textbf{1.}\resetpr \pr Hint: pou\v{z}ijte sou\v{c}tov\'y vzorec pro funkce cosinus a sinus, v\'ysledek: $\tfrac{-2}{(s+3)^2+4} - \tfrac{3}{s^2+9}$, $s&gt;0$ \pr $\tfrac{5}{s+\log 2} - \tfrac{4}{(s-\log 3)^2}$, $s&gt;\log 3$ \pr $\tfrac{4s}{(s^2+4)^2}$, $s&gt;0$ \pr $\tfrac{2}{s} - \tfrac{2s}{s^2+36} + \tfrac{s^2-6s+5}{((s-3)^2 + 4)^2}$, $s&gt;3$\\
	\textbf{2.}\resetpr \pr $\tfrac{3}{2}e^{-3t} + \tfrac{1}{2}e^{-t}$, $t\geq 0$ \pr $4\cos 4t + \tfrac{7}{4}\sin 4t$, $t\geq 0$ \pr $e^{-2t}(4\cos 3t - \sin 3t)$, $t\geq 0$ \pr $e^{2t}(1+4t+2t^2)$, $t\geq 0$ \pr $\sin t + 3 - 3\cos t + \tfrac{1}{2}t^2e^{2t}$, $t\geq 0$ \pr $t-\sin t$, $t\geq 0$ \pr $2\cos 2t + \tfrac{3}{2}\sin 2t$, $t\geq 0$ \pr $e^{-t}(3\cos 2t - 4\sin 2t)$, $t\geq 0$ \pr $e^{-t}(\tfrac{1}{2}t^2 + 2t)$, $t\geq 0$ \pr $\tfrac{1}{2}(e^{-t} - e^{-3t})$, $t\geq 0$ \pr $1+3t-\cos t - 3\sin t$, $t\geq 0$\\
	\textbf{3.}\resetpr \pr $\tfrac{2}{s}(e^{-3s} - e^{-5s})$, $s&gt;0$ \pr $\tfrac{1}{s^2}(3se^{-s} + e^{-4s} - e^{-s})$, $s&gt;0$ \pr $\tfrac{1+e^{-\pi s}}{s^2+1}$, $s&gt;0$ \pr $\tfrac{1}{s}(e^{-s} - e^{-3s})$, $s&gt;0$ \pr $\tfrac{2}{s^2}(1-e^{-s})$, $s&gt;0$ \pr $\tfrac{1}{s+1}(1-e^{-2(s+1)})$, $s&gt;-1$\\
	\textbf{4.}\resetpr \pr $f(t) = \begin{cases} 0 &amp; t\in[0,1]\\ 2(e^{t-1}-1) &amp; t&gt;1\end{cases}$ \pr $f(t) = \begin{cases} 3\cos 2t + \sin 2t &amp; t\in[0,\pi]\\ 3\cos 2t -2 \sin 2t &amp; t&gt;\pi\end{cases}$ \pr $f(t) = \begin{cases} 2t &amp; t\in[0,1]\\ 2 &amp; t\in (1,3]\\ 0 &amp; t&gt;3\end{cases}$\\ \pr $f(t) = \begin{cases} \sin 3t &amp; t\in[0,\tfrac{\pi}{6}]\\ \cos 3t &amp; t\in (\tfrac{\pi}{6},\tfrac{\pi}{2}]\\ 0 &amp; t&gt;\tfrac{\pi}{2}\end{cases}$
	\newpage
\centerline{\MakeUppercase{10. cvi\v{c}en\'i}}\ \\[5pt]	
\textbf{1. Nalezn\v{e}te re\'alnou funkci $y$, kter\'a spl\v{n}uje n\'asleduj\'ic\'i rovnice pro v\v{s}echna $t\in I$:}\\
	\resetpr \pr $y''(t) + 2y'(t) + 2y(t) = \chi_{(1,\infty)}(t)\cdot e^{-t},\qquad y'(0)=y(0)=0,\qquad I=(-\infty,\infty)\setminus\{1\}$\\
	\pr $y''(t) + 4y(t) = 2\cos 2t,\qquad y(0)=0,\,y'(0)=4,\qquad I=(-\infty,\infty)\setminus\{1\}$\\
	\pr $y''(t) + y(t) = e^{-t},\qquad y(0)-y_0,\, y'(0)=y_1,\qquad I=(-\infty,\infty)$\\
	\pr $y'(t) + 2y(t) + 10\int_0^t y(x) \mathrm{d}x = 0,\qquad y(0)=1,\qquad I=[0,\infty)$\\
	\pr $y'(t) + 6y(t) + 9\int_0^t y(x) \mathrm{d}x = 0,\qquad y(0)=1,\qquad I=[0,\infty)$\\
	\pr $ty''(t) - y'(t) = -1,\qquad y(0)=1,\,y(1)=2,\qquad I=(-\infty,\infty)$
	\begin{center}--------------------------------------------------------------------------\end{center}
	V\'ysledky: 
	\textbf{1.}\resetpr \pr $\chi_{(1,\infty)}(t)\cdot e^{-t}(1-\cos(t-1))$ \pr $\tfrac{1}{2}(4+t)\sin 2t$ \pr $\tfrac{1}{2}e^{-t} + (y_0-\tfrac{1}{2})\cos t + (y_1 + \tfrac{1}{2})\sin t$\\ \pr $e^{-t}(\cos 3t - \tfrac{1}{3}\sin 3t)$ \pr $e^{-3t}(1-3t)$ \pr $t+t^2$
	\newpage
	\centerline{\MakeUppercase{11. cvi\v{c}en\'i}}\ \\[5pt]	
	\textbf{1. SpoÄŤtÄ›te $\mathcal{F}(f)$ pro zadanĂ© funkce $f$:}
	\resetpr \pr $f(x) = \begin{cases} 1 &amp; x\in (0,1)\\ 0&amp;  \text{jinak}\end{cases}$ \pr $f(x) = \begin{cases} x &amp; x\in (-1,1)\\ 0&amp;  \text{jinak}\end{cases}$\\
	\pr $f(x) = \begin{cases} 1-|x| &amp; x\in (-1,1)\\ 0&amp;  \text{jinak}\end{cases}$ \pr $f(x) = \frac{1}{(x^2+a^2)^2}$\quad $(a&gt;0)$ \pr $f(x) = \frac{\sqrt{2}x}{\sqrt{\pi}(1+x^2)^2}$\\[3pt]
	\textbf{2. Nech\v{t} $a&gt;0$, $b\in\er$, $\mathcal{F}(e^{-ax}\chi_{[0,\infty)}) = \frac{1}{\sqrt{2\pi}}\frac{1}{a+it}$. Ur\v{c}ete $\mathcal{F}(e^{-ax}e^{ibx}\chi_{[0,\infty)})$.}\\[3pt]
	\textbf{3. Nech\v{t} pro $a&gt;0$ platĂ­ $\mathcal{F}(\chi_{[-a,a]})(t) = \sqrt{\frac{2}{\pi}}\frac{\sin at}{t}$ pro $t\neq 0$. Ur\v{c}ete $\mathcal{F}(\cos(x)\chi_{[-\tfrac{\pi}{2},\tfrac{\pi}{2}]})(t)$ pro $|t|\neq 1$.}\\[3pt]
	\textbf{4. Nech\v{t} platĂ­ $\mathcal{F}(e^{-|x|})(t) = \sqrt{\frac{2}{\pi}}\frac{1}{1+t^2}$. Ur\v{c}ete $\mathcal{F}(\cos(x)e^{-|x|})$.}\\[3pt]
	
	\begin{center}--------------------------------------------------------------------------\end{center}
	V\'ysledky: 
	\textbf{1.}\resetpr \pr $\frac{1}{\sqrt{2\pi}}\frac{1-e^{-it}}{it}$ \pr $-i\sqrt{\frac{2}{\pi}}\frac{\sin t - t\cos t}{t^2}$ pro $t\neq 0$, $\mathcal{F}(f)(0) = 0$ \pr $\sqrt{\frac{2}{\pi}}\frac{1-\cos t}{t^2}$ pro $t\neq 0$, $\mathcal{F}(f)(0) = \frac{1}{\sqrt{2\pi}}$ \pr $\sqrt{\frac{\pi}{2}}e^{-a|t|}\frac{1+a|t|}{2a^3}$ \pr $-it\frac{e^{-|t|}}{2}$\\
	\textbf{2.} $\frac{1}{\sqrt{2\pi}}\frac{1}{a+i(t-b)}$\quad
	\textbf{3.} $\sqrt{\frac{2}{\pi}}\frac{\cos\tfrac{t\pi}{2}}{1-t^2}$ (Hint: $\cos x = \frac{e^{ix} + e^{-ix}}{2}$)\quad
	\textbf{4.} $\sqrt{\frac{2}{\pi}} \frac{t^2+2}{t^4+4}$
	\newpage
	\centerline{\MakeUppercase{12. cvi\v{c}en\'i}}\ \\[5pt]
	\textbf{1. Pro zadan\'e oper\'atory $F$ sestavte p\v{r}\'islu\v{s}nou Euler-Lagrangeovu rovnici a pomoc\'i n\'i ur\v{c}ete, kter\'e funkce $y$ mohou b\'yti lok\'aln\'imi extr\'emy oper\'atoru $F$:}\\ \resetpr 
	\pr \[F(y) = \int_2^3 12xy(x) + y'(x)^2 \dx,\quad y(2) = 1,\, y(3) = 27.\]
	\pr \[F(y) = \int_0^1 x + y(x)^2 + 3y'(x) \dx,\quad y(0) = 0,\, y(1) = 1.\]
	\pr \[F(y) = \int_{-1}^1 y(x)^2 + x^2y'(x) \dx,\quad y(-1) = -1,\, y(1) = 1.\]
	\pr \[F(y) = \int_1^2 y'(x) \dx,\quad y(1) = 0,\, y(2) = 0.\]
	\pr \[F(y) = \int_0^{\pi/2} y(x)^2 - y'(x)^2 \dx,\quad y(0) = 2,\, y(\tfrac{\pi}{2}) = 1.\]
	\pr \[F(y) = \int_0^1 xy'(x) + y'(x)^2 \dx,\quad y(0) = 13,\, y(1) = 13+\tfrac{3}{4}.\]
	\pr \[F(y) = \int_0^1 2y(x)e^x + y(x)^2 + y'(x)^2 \dx,\quad y(0) = 0,\, y(1) = \tfrac{1}{e}.\]
	\pr \[F(y) = \int_2^3 2y'(x)^3 \dx,\quad y(2) = y(3) = 4.\]
	\begin{center}--------------------------------------------------------------------------\end{center}
	V\'ysledky: 
	\textbf{1.}\resetpr \pr $y(x) = x^3 + 7x - 21$ \pr lok\'aln\'i extr\'emy neexistuj\'i \pr $y(x) = x$ \pr Euler-Lagrangeov\v{e} rovnici vyhovuj\'i v\v{s}echny funkce $y$ \pr $y(x) = \sin x + 2\cos x$ \pr $y(x) = -\tfrac{x^2}{4} + x + 13$ \pr $y(x) = e^{-x} - e^x + xe^x$ \pr $y(x) = -x+6$
	\iffalse
	\newpage
	\begin{defin}[Line\'arn\'i rovnice 2. Ĺ™Ăˇdu s konstantn\'imi koeficienty]\emphdef{LineĂˇrnĂ­ diferenciĂˇlnĂ­ rovnicĂ­ druhĂ©ho Ĺ™Ăˇdu s konstantn\'imi koeficienty} rozumĂ­me rovnici tvaru
\begin{equation}\label{eq:linRadDvaKonst}
	y'' + py' + qy = r(x),
\end{equation}
kde $p, q\in\er$ a $r$ je funkce spojit\'a na intervalu $(a,b)$.

\emphdef{Homogenn\'i rovnic\'i} pĹ™Ă­sluĹˇnou k rovnici \eqref{eq:linRadDvaKonst} rozumĂ­me rovnici
\begin{equation}\label{eq:linRadDvaKonstHomogen}
	y'' + py' + qy = 0.
\end{equation}
\end{defin}

\begin{pozn} MaximĂˇlnĂ­ Ĺ™eĹˇenĂ­ rovnice \eqref{eq:linRadDvaKonstHomogen} jsou definovĂˇna na celĂ©m $\er$.
\end{pozn}

\begin{defin}\emph{CharakteristickĂ˝m polynomem} rovnice \eqref{eq:linRadDvaKonstHomogen} rozumĂ­me polynom
	\[\chi(\lambda) := \lambda^2 + p\lambda + q.\]
\end{defin}

\begin{theorem}[tvar fundamentĂˇlnĂ­ho systĂ©mu]Nechtâ€™ $\lambda_1. \lambda_2$ jsou ko\v{r}eny charakteristick\'eho polynomu rovnice \eqref{eq:linRadDvaKonstHomogen}.
	\begin{itemize}
		\item Pokud jsou $\lambda_1$, $\lambda$ r\r{u}zn\'e re\'aln\'e ko\v{r}eny, pak fundamentĂˇlnĂ­ systĂ©m Ĺ™eĹˇenĂ­ rovnice \eqref{eq:linRadDvaKonstHomogen} tvoĹ™Ă­ funkce $y_1(x) = e^{\lambda_1x}$ a $y_2(x) = e^{\lambda_2x}$.
		\item Pokud $\lambda_1 = \lambda_2$, pak fundamentĂˇlnĂ­ systĂ©m Ĺ™eĹˇenĂ­ rovnice \eqref{eq:linRadDvaKonstHomogen} tvoĹ™Ă­ funkce $y_1(x) = e^{\lambda_1x}$ a $y_2(x) = xe^{\lambda_1x}$.
		\item Pokud jsou $\lambda_1 = \alpha + \beta i$, $\lambda_2 = \alpha - \beta i$ r\r{u}zn\'e komplexn\'i ko\v{r}eny, pak fundamentĂˇlnĂ­ systĂ©m Ĺ™eĹˇenĂ­ rovnice \eqref{eq:linRadDvaKonstHomogen} tvoĹ™Ă­ funkce $y_1(x) = e^{\alpha x}\sin(\beta x)$ a $y_2(x) = e^{\alpha x}\cos(\beta x)$.
	\end{itemize}
\end{theorem}
	
\begin{theorem}[speciĂˇlnĂ­ pravĂˇ strana]
	Nechtâ€™
	\[r(x) = e^{ax}\cdot (P(x)\cos bx + Q(x)\sin bx),\quad x\in\er,\]
	kde $a, b\in\er$ a $P, Q$ jsou polynomy. Pak existuje \v{r}e\v{s}en\'i rovnice \eqref{eq:linRadDvaKonst} ve tvaru
	\[y_p(x) = x^me^{ax}(R(x)\sin bx + S(x)\cos bx),\quad x\in\er,\]
	kde $R, S$ jsou polynomy stupnÄ› ne vÄ›tĹˇĂ­ho neĹľ $\max\{\st P, \st Q\}$ a $m\in\en\cup\{0\}$ udĂˇvĂˇ, jakou nĂˇsobnost mĂˇ ÄŤĂ­slo $a + bi$ jakoĹľto koĹ™en charakteristick\'eho polynomu pĹ™Ă­sluĹˇnĂ© homogennĂ­ rovnice.
\end{theorem}\ \\[10pt]
-----------------------------------------\ \\[10pt]
OznaÄŤme
\[
	M:=\{y\in\C^1([a,b])\setsep y(a)=A, y(b)=B\}
\]
a uva\v{z}ujme zobrazen\'i $F:M\to \er$ definovan\'e pĹ™edpisem
\[
	F(y):=\int_a^b f(x,y(x),y'(x)) \dx,\quad y\in M.
\]
Argument funkce f znaÄŤĂ­me $(x, y, z)$. V
zĂˇjmu pĹ™ehlednosti zĂˇpisu budeme dĂˇle pouĹľĂ­vat znaÄŤenĂ­ $f_z = \frac{\partial f}{\partial z}$, atd.
\begin{theorem}[Euler-Lagrangeova rovnice]
	NechĹĄ $y_0\in M$ je bodem lokĂˇlnĂ­ho extrĂ©mu zobrazenĂ­ $F$. Pak funkce
	\[
		[a,b]\ni x\mapsto f_z(x,y_0(x),y_0'(x))
	\]
je spojitÄ› diferencovatelnĂˇ na $[a,b]$ a splĹuje \emphdef{Euler-Langrangeovu rovnici}
	\begin{equation}\label{eq:ELpodminky}
		f_y(x,y_0(x),y_0'(x)) - \frac{\mathrm{d}}{\mathrm{d} x}f_z(x,y_0(x),y_0'(x)) = 0,\quad x\in[a,b].
	\end{equation}
\end{theorem}

\begin{theorem}[o regularitÄ› minimizĂ©ru]
	NechĹĄ $y_0\in M$ je bodem lokĂˇlnĂ­ho extrĂ©mu zobrazenĂ­ $F$ a $x_0\in(a,b)$ je bod splĹujĂ­cĂ­
	\[
		f_{zz}(x_0,y_0(x_0),y_0'(x_0)) \neq 0.
	\]
	Pak existuje $\delta &gt; 0$ takovĂ©, Ĺľe $y_0\in\C^2((x_0-\delta,x_0+\delta))$.
\end{theorem}
\fi
\end{document}</pre></body></html>