%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%                                                                          %%%
%%% file:      s5.tex                                                        %%%
%%% author:    Tatiana Jajcayová, Ján Komara                                 %%%
%%% email:     Jan.Komara@fmph.uniba.sk                                      %%%
%%% copyright: public domain                                                 %%%
%%% version:   2018/12/08                                                    %%%
%%%                                                                          %%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\documentclass[12pt,a4paper]{article}

\usepackage[slovak]{babel}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}

\usepackage{amsmath}
\usepackage{amssymb}

\usepackage{lmodern}
\usepackage{indentfirst}

\renewcommand{\labelenumi}{{\normalfont (\alph{enumi})}}
\renewcommand{\labelenumii}{{\normalfont \roman{enumii}.}}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%                                                                          %%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{document}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%                                                                          %%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\title{5. domáca úloha z predmetu \\ 1-AIN-160 Matematika (3) ZS 2018/19}

\author{Tatiana Jajcayová \and Ján Komara}

%\date{}
\date{\today}

\maketitle

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%                                                                          %%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section*{1. príklad}

Vyriešte nasledujúce rekurentné vzťahy:
\begin{enumerate}
\item
$a_{n} - 2a_{n-1} - 2a_{n-2} = 0\ \text{pre}\ n \geq 2, a_{0} = 1, a_{1} = 3$.
\item 
$a_{n} - 6a_{n-1} + 9a_{n-2} = 0\ \text{pre}\ n \geq 2, a_{0} = 5, a_{1} = 12$.
\item 
$a_{n+3} - 5a_{n+2} + 8a_{n+1} - 4a_n = 0\ \text{pre}\ n \geq 0, 
 a_{0} = 0, a_{1} = 3, a_{2} = 11$.
\end{enumerate}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%                                                                          %%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\subsection*{Riešenie 1. príkladu}

Táto časť obsahuje riešenie 1. príkladu.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%                                                                          %%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section*{2. príklad}

Vieme, že $a_{0} = 0, a_{1} = 1, a_{2} = 4$ a $a_{3} = 37$. 
Vyriešte rekurentný vzťah $a_{n+2} + ba_{n+1} + ca_{n} = 0$, 
kde $b, c$ sú konštanty. Určite $b, c$ a riešte pre $a_{n}$.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%                                                                          %%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\subsection*{Riešenie 2. príkladu}

Táto časť obsahuje riešenie 2. príkladu.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%                                                                          %%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section*{3. príklad}

Koľko slov dĺžky $n \geq 0$ nad abecedou
\(
\{ 0,1,2,3,4,5,6,7,8,9 \}
\)
má párny počet výskytov znaku $0$.
Nájdite rekurentný vzťah vyjadrujúci tento počet.
Uveďte vhodné počiatočné podmienky.
Rekurentnú rovnicu vyriešte.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%                                                                          %%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\subsection*{Riešenie 3. príkladu}

Táto časť obsahuje riešenie 3. príkladu.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%                                                                          %%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section*{4. príklad}

Uvažujme slová vytvorené zo znakov abecedy $\{a, b, c, d, e, f \}$. 
Nech $b_n$ je počet slov dĺžky $n \geq 0$, 
ktoré neobsahujú dve po sebe idúce spoluhlásky.
Nájdite rekurentný vzťah pre postupnosť $b_n$. 
Uveďte vhodné počiatočné podmienky. 
Rekurentnú rovnicu vyriešte.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%                                                                          %%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\subsection*{Riešenie 4. príkladu}

Táto časť obsahuje riešenie 4. príkladu.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%                                                                          %%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section*{5. príklad}

Vyriešte nasledujúce rekurentné vzťahy:
\begin{enumerate}
\item 
$a_{n+2} - 5 a_{n+1} + 4 a_n  = 3 - 2^n, a_{0} = 1/2, a_{1} = 3$.
\item 
$c_{n+1} - 2 c_{n} = 2^{n+1} + 3n, c_{0} = 1$.
\end{enumerate}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%                                                                          %%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\subsection*{Riešenie 5. príkladu}

Táto časť obsahuje riešenie 5. príkladu.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%                                                                          %%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\end{document}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%                                                                          %%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
