Graphics, Figures & Tables ⇒ Adjust Width of two Tables

[0texel]

How can I get effect like this. I know this is a two tables but I need to fix the width of columns to expand it on width of page.

Code: Select all

\begin{center}
     \begin{tabular}{  l  c  c  r }
     \textbf{Rząd} & \textbf{Tradycyjna nazwa} & \textbf{Wzór} & \textbf{Błąd} \\
     \end{tabular}
\end{center}
\newpage
\begin{center}
     \begin{tabular}{  c  c  c  p{5cm} }
1 & wzór trapezów  & $\frac{h}{2}(f_{0}+f_{1})$ & $-\frac{h^3}{12}f^{(2)}(\xi)$  \\  
2 & wzór Simpsona & $\frac{h}{3}(f_0+4f_1+f_2)$ & $-\frac{h^5}{90}f^{(4)}(\xi)$  \\ 
3 & reguła 3/8 & $\frac{3h}{8}(f_0+3f_1+3f_2+f_3)$ & $-\frac{3h^5}{80}f^{(4)}(\xi)$ \\  
4 & wzór Boole'a często błędnie\footnotemark  & $\frac{2h}{45}(7f_0+32f_1+12f_2+32f_3+7f_4)$ & $-\frac{8h^7}{945}f^{(6)}(\xi)$ \\ 
  & nazywany wzorem Bode'a & & \\
     \end{tabular}
\end{center}

[localghost]

I would not allow a page break between header and body of this table (it's only one). So, why not use just one table and let if float?

To make the table occupy the complete text width use tabularx as shown below.

Code: Select all

\documentclass[11pt,a4paper,polish]{article}
\usepackage[T1]{fontenc}
\usepackage{selinput}
\SelectInputMappings{
  oacute={ó},
  lslash={ł},
}
\usepackage{babel}
\usepackage[margin=2cm]{geometry}
\usepackage[font=small,labelfont=bf,tableposition=top]{caption}
\usepackage{tabularx}

\newcolumntype{C}{>{\centering\arraybackslash}X}

\begin{document}
  \begin{table}[!ht]
    \caption{A dummy table}
    \label{tab:dummy}
    \centering
    \begin{tabularx}{\linewidth}{cCCc}
     \textbf{Rząd} & \textbf{Tradycyjna nazwa} & \textbf{Wzór} & \textbf{Błąd} \\
      1 & wzór trapezów  & $\frac{h}{2}(f_{0}+f_{1})$ & $-\frac{h^3}{12}f^{(2)}(\xi)$  \\  
      2 & wzór Simpsona & $\frac{h}{3}(f_0+4f_1+f_2)$ & $-\frac{h^5}{90}f^{(4)}(\xi)$  \\
      3 & reguła 3/8 & $\frac{3h}{8}(f_0+3f_1+3f_2+f_3)$ & $-\frac{3h^5}{80}f^{(4)}(\xi)$ \\  
      4 & wzór Boole'a często błędnie\footnotemark  & $\frac{2h}{45}(7f_0+32f_1+12f_2+32f_3+7f_4)$ & $-\frac{8h^7}{945}f^{(6)}(\xi)$ \\
      & nazywany wzorem Bode'a & & \\
    \end{tabularx}
  \end{table}
\end{document}

Best regards and welcome to the board
Thorsten