Solution

Solution

03/solutions #3/exercise-03-plik.tex

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+\documentclass{article}
+\usepackage[utf8]{inputenc}
+\usepackage{polski}
+\usepackage{amsmath}
+\begin{document}
+
+
+
+\begin{enumerate}
+\item Properly typeset the following command and properly refere to it in the text
+\[
+(\sum_{i_1,\dots,i_m} a_{i_1,\dots,i_m} ^{2m}{m+1} ^{\frac{m+1}{2m}} \leq C \sup\{ |\sum_{i_1,\dots, i_m} a_{i_1,\dots,i_m} x^1_{i_1}\dots x^m_{i_m}|: \|(x_i^k)_{i=1}^n \|_\infty\leq1,\ 1\leq k\leq m\},
+\]
+
+\item Properly typset the expression: $Re z$.
+
+\item Properly typeset indexes in the following sum:
+\[
+f(x)=\sum_{n=0, k=2}^\infty a_n^k
+\]
+
+\item Properly typeset the following theorem
+
+\noindent\textbf{Theorem 1} (Cauchy--Hadamard) \emph{The radius of convergence $R$ of the power series 
+\[
+\sum_{n=0}^\infty a_n(z-z_0)^n\ \ \ \ \ |z-z_0|<R
+\]
+can by calculated via the following formula
+\[
+\frac{1}{R}=limsup_{n\to\infty} \sqrt[n]{|a_n|}.
+\]}
+
+\noindent\textbf{Definition 2.} (Prime numbers) A number is called prime, if it is not compound. 
+
+\item Typeset the follwing matrix (display-style):
+\[
+\left\{\begin{array}{cc}
+         a_{11} & a_{12}\\
+         a_{21} & a_{22}
+       \end{array}
+     \right\}
+\]
+and in the text \[\left\{\begin{array}{cc}
+                            a_{11} & a_{12}\\
+                            a_{21} & a_{22}
+                          \end{array}
+     \right\}\]
+     Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do
+     eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut
+     enim ad minim veniam, quis nostrud exercitation ullamco laboris
+     nisi ut aliquip ex ea commodo consequat. Duis aute irure dolor in
+     reprehenderit in voluptate velit esse cillum dolore eu fugiat
+     nulla pariatur. Excepteur sint occaecat cupidatat non proident,
+     sunt in culpa qui officia deserunt mollit anim id est laborum.
+
+\end{enumerate}
+\end{document}