\documentclass[12pt,notitlepage]{report}

\frenchspacing % aktivuje použití některých českých typografických pravidel

\usepackage{amsthm,amsmath,amssymb,amscd,mathrsfs,hyperref}%,showkeys,layout}
\usepackage{verbatim} %pro vynechavani comment
\usepackage[top=1cm, left=1cm]{geometry} % nastavení dané velikosti okrajů
\pagestyle{empty}

\def\ce{\mathbb C}
\def\C{\mathcal{C}}
\def\en{\mathbb N} 
\def\zet{\mathbb Z}
\def\er{\mathbb R}
\def\qe{\mathbb Q}
\def\ov{\overline}
\newcommand\lx[1][0]{\displaystyle\lim_{x\to #1}}
\def \sgn {\operatorname{sgn}}
\def \dx {\, \mathrm{d} x}
\def \tg {\operatorname{tg}}
\def \arctg {\operatorname{arctg}}
\newcounter{Priklad}
\addtocounter{Priklad}{1}
\newcommand{\pr}{\quad\alph{Priklad})\;\addtocounter{Priklad}{1}}
\newcommand{\resetpr}{\setcounter{Priklad}{1}}
\def \suma {\sum_{n=1}^\infty}
\def \sumaO {\sum_{n=0}^\infty}
%\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}}

\title{Cvi\v{c}en\'i - Teorie m\'iry a integr\'alu, MFF}  % tyto dvě položky jsou zde v podstatě formálně, ve skutečnosti nejsou nikde 
\author{Marek C\'uth}

\begin{document}
 \centerline{\textbf{V\'YSLEDKY}}\ \\	
 \centerline{\MakeUppercase{i. konvergence lebesgueova integr\'alu}}\ \\
\textbf{1.}\resetpr \pr K (konverguje) \pr K \pr K \pr je roven $-\infty$ \pr K$\iff k\in(1,2)$ \pr K \pr K \pr K \pr K\\[10pt]
\textbf{2.}\resetpr \pr K$\iff \alpha < 2$ \pr K$\iff (\alpha > 1) \vee (\alpha = 1\;\wedge\;\beta < -1)$ \pr  K$\iff (\alpha > -1) \vee (\alpha = -1\;\wedge\;\beta < -1)$  \pr K$\iff \alpha^2\leq 1$ \pr K$\iff 0 < \beta < \alpha < 1$ \pr K$\iff \alpha\in(0,1]$ \pr K$\iff \alpha\in(-1,1)$ \pr K pro v\v{s}echna $\alpha\in\er$ \pr K$\iff \alpha > 1$ \pr K $\iff \alpha < 3/2$\\[20 pt]
\centerline{\MakeUppercase{ii. integrace posloupnost\'i a \v{r}ad funkc\'i}}\ \\
\textbf{1.}\resetpr \pr 0 \pr 0 \pr 1 \pr 0 \pr $-\infty$\\[10pt]
\textbf{2.}\resetpr \pr 0 \pr $\infty$ \pr $-\infty$ \pr 0 \pr 0\\[10pt]
\textbf{3.}\resetpr \pr $- \sumaO \tfrac{1}{(n+1)^2}$ \pr $\sumaO \tfrac{(-1)^n}{(n+1)^2}$ \pr n\'apov\v{e}da: $\tfrac{x}{e^x-1} = \tfrac{x}{e^x}\cdot\tfrac{1}{1-e^{-x}} = xe^{-x}\sumaO e^{-nx}$, v\'ysledek: $\sumaO \tfrac{1}{(n+1)^2}$ \pr  n\'apov\v{e}da: analogie p\v{r}edchoz\'iho p\v{r}\'ikladu,  v\'ysledek: $\sumaO \tfrac{(-1)^n}{(n+1)^2}$ \pr  n\'apov\v{e}da: $\tfrac{x^{p-1}}{1 + x^q} = x^{p-1}\sumaO (-x^{q})^n$,  v\'ysledek: $\sumaO \tfrac{(-1)^n}{p+nq}$ \pr n\'apov\v{e}da: pou\v{z}ijte Taylor\r{u}v rozvoj funkce cosinus, pak $e^{-x}\cos\sqrt{x} = \sumaO (-1)^ne^{-x}\tfrac{x^n}{(2n)!}$,  v\'ysledek: $\sumaO (-1)^n\tfrac{n!}{(2n)!}$ \pr  n\'apov\v{e}da: pou\v{z}ijeme rozvoj logaritmu, pak $\log x \log(1-x) = - \suma \tfrac{x^n}{n}\log x$,  v\'ysledek: $\suma \tfrac{1}{n(n+1)^2} = \suma\left(\tfrac{1}{n} - \tfrac{1}{n+1} - \tfrac{1}{(n+1)^2}\right) = 1 - \suma\tfrac{1}{(n+1)^2} = 2 - \tfrac{\pi^2}{6}$ \pr n\'apov\v{e}da: analogie p\v{r}edchoz\'iho p\v{r}\'ikladu,  v\'ysledek: $\suma \tfrac{(-1)^n}{n(n+1)^2} = \suma (-1)^n\left(\tfrac{1}{n} - \tfrac{1}{n+1} - \tfrac{1}{(n+1)^2}\right) \stackrel{(e)}{=} (- \log 2) - (\log 2 - 1) - \suma\tfrac{(-1)^n}{(n+1)^2} = 1 - 2\log 2 - (\suma \tfrac{1}{(n+1)^2} - 2\suma \tfrac{1}{(2n)^2}) = 1 - 2\log 2 - (\tfrac{\pi^2}{6} - 1 - 2\tfrac{1}{4}\suma \tfrac{1}{n^2}) = 2(1-\log 2) - \pi^2/6 + \pi^2/12 = 2(1 - \log 2) - \pi^2/12$ \pr n\'apov\v{e}da: $\tfrac{\sin ax}{e^x - 1} = (\sin ax)e^{-x}\sumaO e^{-nx}$, v\'ysledek: $\sumaO \tfrac{a}{a^2 + (n+1)^2}$ \pr $\sumaO \tfrac{1}{(2n+1)^2} = \tfrac{\pi^2}{8}$ \pr $\sum_{n=3}^\infty \tfrac{(-1)^{n+1}}{n^2} = -1+\tfrac 14 + \tfrac{\pi^2}{12}$ \pr $-\suma\tfrac{1}{n^2} = -\tfrac{\pi^2}{6}$ \pr $\suma \tfrac{1}{n\cdot n!}$ \pr $\sumaO (\tfrac{1}{(n+1)^2} + \tfrac{1}{(n+2)^2}) = \tfrac{\pi^2}{3} - 1$ \pr $\sumaO (-1)^n \tfrac{1}{8(n+1)} \stackrel{(e)}{=} \tfrac{\log 2}{8}$\\[20 pt]
\centerline{\MakeUppercase{iii. integr\'aly z\'avisl\'e na parametru}}\ \\
\textbf{1.}\resetpr defini\v{c}n\'i obory: \pr $[0,\infty)$ \pr $(0,\infty)$ \pr $\er$ \pr $\er\setminus\{0\}$ \pr $(2,\infty)$ \pr $(1,\infty)$ \pr $(0,1)$ \pr $(-1,\infty)$ \pr $(-1,\infty)$ \pr$\er$ \pr $(0,\infty)$\\[10pt]
\textbf{2.}\resetpr 
	\pr $\sgn a\cdot \tfrac \pi2\cdot \log(1 + |a|),\; a\in\er$
	\pr $\log(a + 1),\;a\in(-1,\infty)$
	\pr $\tfrac 12\log(a + 1),\;a\in(-1,\infty)$\\
	\pr $-\log(1+a),\;a\in(-1,\infty)$
	\pr $\arctg \tfrac ba,\; a>0$, $b\in\er$. Lze tak\'e zapsat: $-\arctg \tfrac ab + \tfrac \pi2$ pro $a,b>0$; $-\arctg \tfrac ab - \tfrac \pi2$ pro $a>0$, $b<0$; $0$ pro $a>0$, $b=0$
	\pr $\tfrac 12\log \tfrac ba,\; a,b>0$ nebo $a=b\neq 0$, $F(0,0)=0$
	\pr $\sgn a\cdot \tfrac{\pi}{2}\log\left(\tfrac ab\right),\; ab > 0$ nebo $a = b\neq 0$, $F(0,0)=0$
	\pr $\pi\arcsin a,\; a\in[-1,1]$
	\pr $\pi(\log(1+\sqrt{1-a^2}) - \log 2),\; a\in[-1,1]$\\
	\pr $\tfrac \pi2 \log(a + \sqrt{1+a^2}),\; a\in\er$
	\pr $\tfrac \pi2 \log(a + \sqrt{1+a^2}),\; a\in\er$
	\pr $\tfrac{\pi}{|b|}\log(|a| + |b|),\; b\neq 0$
	\pr $\ln\left(\frac{\sqrt{a^2+1}}{a}\right),\; a>0$\\[20pt]
\centerline{\MakeUppercase{iv. v\'icerozm\v{e}rn\'a integrace}}\ \\
\textbf{1.}\resetpr \pr $12 - \log 5$ \pr $1/2$ \pr $+\infty$ \pr $7$ \pr $2/3(a-b)\sqrt{ab}$\\[10pt]
\textbf{2.}\resetpr \pr $\pi/12$ \pr $1/3$ \pr $33/140$ \pr $R^532/45$ \pr $2$ \pr $1/3(\pi/3 + \sqrt{3}/2)$\\[10pt]
\textbf{3.}\resetpr \pr $2\pi$ \pr $2$ \pr $\displaystyle\frac{a^{10}b^6}{840c^{12}}$ \pr $4/27$ \pr $\tfrac{\sqrt{6}}{3}$\\[10pt]
\textbf{4.}\resetpr \pr $2a^2$ \pr $\pi a^23/4$ \pr $R^2(\pi/3 - \sqrt{3}/4)$ \pr $5/8$ \pr $a^2/60$ \pr $\tfrac{1}{2}(b-a)\log(q/p)$ \pr $\tfrac{\pi ab}{2}(a^2 + b^2)$\\[10pt]
\textbf{5.}\resetpr \pr $\sgn a\cdot\tfrac{\pi}{2}\log(a/b)$ \pr $\sqrt{\pi b} - \sqrt{\pi a}$ \pr $\tfrac{1}{2}\log(b/a)$ \pr $\log(b/a)$ \pr $\pi(|a|-|b|)$ \pr $\sqrt{\pi}(|a|-|b|)$ \pr $\log(\tfrac{b+1}{a+1})$\\[10pt]
\textbf{6.}\resetpr \pr $a^3\pi$ \pr $a^3/60$ \pr $\tfrac{2}{3}R^3(\pi - \tfrac{4}{3})$ \pr $2\pi^2Ra^2$ \pr $\displaystyle\frac{a^7b^4c\pi}{192h^9}$ \pr $\pi ca^2/2$ \pr $\pi^2\sqrt{3}$\\[10pt]
\textbf{7.}\resetpr \pr $\pi R^5 \tfrac{59}{480}$ \pr $\tfrac{4}{5}abc\pi$ \pr $\tfrac{13}{4}\pi$ \pr $\tfrac{\pi}{10}$
\end{document}