\documentclass{article} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \usepackage{amssymb} \usepackage{amsmath} \setcounter{MaxMatrixCols}{10} %TCIDATA{OutputFilter=LATEX.DLL} %TCIDATA{Version=5.00.0.2570} %TCIDATA{} %TCIDATA{Created=Monday, June 25, 2001 17:45:57} %TCIDATA{LastRevised=Sunday, April 25, 2004 20:27:50} %TCIDATA{} %TCIDATA{} %TCIDATA{CSTFile=On line bluem.cst} %TCIDATA{PageSetup=72,72,72,72,0} %TCIDATA{AllPages= %H=36 %F=36,\PARA{038
\QTR{small}{Matematick\U{e1} anal\U{fd}za I online - D\U{f4}kazy\dotfill \thepage }} %} \newtheorem{theorem}{Theorem} \newtheorem{acknowledgement}[theorem]{Acknowledgement} \newtheorem{algorithm}[theorem]{Algorithm} \newtheorem{axiom}[theorem]{Axiom} \newtheorem{case}[theorem]{Case} \newtheorem{claim}[theorem]{Claim} \newtheorem{conclusion}[theorem]{Conclusion} \newtheorem{condition}[theorem]{Condition} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{criterion}[theorem]{Criterion} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{exercise}[theorem]{Exercise} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{notation}[theorem]{Notation} \newtheorem{problem}[theorem]{Problem} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \newtheorem{solution}[theorem]{Solution} \newtheorem{summary}[theorem]{Summary} \newenvironment{proof}[1][Proof]{\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \input{tcilatex} \begin{document} \author{A. U. Thor} \title{Lab Report} \date{The Date } \maketitle \begin{abstract} A Laboratory report created with Scientific Notebook \end{abstract} \section{Vlastnosti Laplaceovej transform\'{a}cie} \subsection{D\^{o}kaz vety o Laplaceovom obraze deriv\'{a}cie funkcie.} \textbf{D\^{o}kaz: }Pou\v{z}ijeme k-kr\'{a}t integr\'{a}ciu per partes a dostaneme: \begin{equation*} \mathbf{L}\left[ f^{\left( k\right) }\left( t\right) \right] =\int_{0}^{\infty }f^{\left( k\right) }\left( t\right) e^{-pt}dt=\lim_{t\longrightarrow \infty }f^{\left( k-1\right) }\left( t\right) e^{-pt}-f^{\left( k-1\right) }\left( 0+\right) +p\int_{0}^{\infty }f^{\left( k-1\right) }\left( t\right) e^{-pt}dt= \end{equation*}% \begin{equation*} =p\int_{0}^{\infty }f^{\left( k-1\right) }\left( t\right) e^{-pt}dt-f^{\left( k-1\right) }\left( 0+\right) =p^{2}\int_{0}^{\infty }f^{\left( k-2\right) }\left( t\right) e^{-pt}dt-pf^{\left( k-2\right) }\left( 0+\right) -f\left( k-1\right) \left( 0+\right) =\dots = \end{equation*}% \begin{equation*} =p^{k}F\left( p\right) -p^{k-1}f\left( 0+\right) -p^{k-2}f\,^{\prime }\left( 0+\right) -\dots -f^{\left( k-1\right) }\left( 0+\right) , \end{equation*}% pritom sme pou\v{z}ili fakt, \v{z}e ak $\func{Re}p$ $>$ $\alpha _{0},$ potom \begin{equation*} \lim_{t\longrightarrow \infty }f^{\left( j\right) }\left( t\right) e^{-pt}=0,\,\forall \,j=0,1,2,\dots ,k-1.\,\blacksquare \end{equation*} \begin{center} \begin{tabular}{|c|} \hline {\small %TCIMACRO{\hyperref{Sp\"{a}\v{t}}{}{}{K44.tex#7}}% %BeginExpansion \msihyperref{Sp\"{a}\v{t}}{}{}{K44.tex#7}% %EndExpansion } \\ \hline \end{tabular} \end{center} \rule{6.5in}{0.04in} \textsl{Matematick\'{a} anal\'{y}za III} \end{document}