How to split equation into a table and under square root?

Question

I need to split the following equation which is into a table, in the second column. As the equation is under square root, I find it difficul to split. I tried to use \begin{split} environment but it does not work properly within a table. I also tried to change \sqrt[] into a power ^{1/2} but I couldn't find a solution. This is the equation.

P_{ij}^{[n]}=\sqrt[]{\frac{1}{C}\cdot \sum_{(xy) s.t. u_{xy} \in I_{ij}}\biggl[\biggl(\frac{1}{C}\cdot \sum_{(xy) s.t. u_{xy} \in I_{ij}}w_{(ij),(xy)}^{[n]}\biggr)-w_{(ij),(xy)}^{[n]}\biggr]^2}

Thanks for your help!

This is the table

\begin{table}[htbp]
\centering
\caption{Some of the rules available to create new images.}
\label{tabella1}
\renewcommand{\arraystretch}{3}
\begin{tabular}{cc}
    \toprule
    \textbf{Name} & \textbf{Equation} \\
    \midrule
    Mean & $P_{ij}^{[n]}=\frac{1}{C}\cdot \sum_{(xy) s.t. u_{xy} \in I_{ij}}w_{(ij),(xy)}^{[n]}$ \\
    \midrule
    Variance & $P_{ij}^{[n]}=\sqrt[]{\frac{1}{C}\cdot \sum_{(xy) s.t. u_{xy} \in I_{ij}}\biggl[\biggl(\frac{1}{C}\cdot \sum_{(xy) s.t. u_{xy} \in I_{ij}}w_{(ij),(xy)}^{[n]}\biggr)-w_{(ij),(xy)}^{[n]}\biggr]^2}$ \\
    \midrule
    \text{Maximum} &$ P_{ij}^{[n]}=\max_{(xy) s.t. u_{xy} \in I_{ij}}\biggl(w_{(ij),(xy)}^{[n]}\biggr)$ \\
    \midrule
    \text{Minimum} &$ P_{ij}^{[n]}=\min_{(xy) s.t. u_{xy} \in I_{ij}}\biggl(w_{(ij),(xy)}^{[n]}\biggr)$ \\
    \bottomrule
\end{tabular}
\renewcommand{\arraystretch}{1}

\end{table}

Answer 1

You haven't said how wide your page is and I would remove the superfluous (xy) s.t. clauses to look more like

However assuming default article class width and keeping the s.t then this appears to fit:

\documentclass{article}

\usepackage{amsmath,booktabs}

\begin{document}

\begin{table}[htbp]
\centering
\caption{Some of the rules available to create new images.}
\label{tabella1}
\renewcommand{\arraystretch}{3}
\begin{tabular}{cc}
    \toprule
    \textbf{Name} & \textbf{Equation} \\
    \midrule
    Mean & $\displaystyle
P_{ij}^{[n]}=\frac{1}{C}\cdot \sum_{\substack{(xy)\\\text{s.t. }u_{xy} \in I_{ij}}}w_{(ij),(xy)}^{[n]}$ \\
    \midrule
    Variance & $\displaystyle
P_{ij}^{[n]}=\sqrt[]{\frac{1}{C}\cdot
\!\sum_{\substack{(xy)\\\text{s.t. }u_{xy} \in I_{ij}}}\biggl[\biggl(\frac{1}{C}\cdot
\!\sum_{\substack{(xy)\\\text{s.t. }u_{xy} \in I_{ij}}}w_{(ij),(xy)}^{[n]}\biggr)-w_{(ij),(xy)}^{[n]}\biggr]^2}$ \\
    \midrule
    \text{Maximum} &$\displaystyle
 P_{ij}^{[n]}=\max_{\substack{(xy)\\\text{s.t. }u_{xy} \in I_{ij}}}\biggl(w_{(ij),(xy)}^{[n]}\biggr)$ \\
    \midrule
    \text{Minimum} &$\displaystyle
 P_{ij}^{[n]}=\min_{\substack{(xy)\\\text{s.t. }u_{xy} \in I_{ij}}}\biggl(w_{(ij),(xy)}^{[n]}\biggr)$ \\
    \bottomrule
\end{tabular}

\end{table}
\end{document}

Answer 2

(Remark: This answer is based on the original form of the OP's query, which contained just one formula. The OP subsequently posted an table with four [4] equations.)

One of the following forms? Observe that the "s.t." qualifiers don't seem to be needed, and neither are the \cdot terms. In the first row, the \smash instructions in the square root expression along with the subsequent \vphantom directive serve to keep the surd from having to become excessively tall. Observe that neither the \smash directives nor the \vphantom instruction are needed in the second form. And, do observe that I've rearranged some of the terms to make it easy on readers to notice that P_{ij}^{[n]} is a standard deviation.

\documentclass{article}
\usepackage{mathtools} % for \smashoperator macro
%% macros for two terms that occur several times:
\newcommand\wterm{w_{(ij),(xy)}^{[n]}}
\newcommand\sumterm{\smashoperator{\sum_{\substack{(xy)\\ u_{xy} \in I_{ij}}}}}

\begin{document}
\begin{align*}
P_{ij}^{[n]}
% first form
&=\sqrt{   \frac{1}{C} \smash{\sumterm}\biggl[ \wterm 
  -\frac{1}{C} \Bigl(\,\smash{\sumterm} \wterm\Bigr) \biggr]^2} \vphantom{\sumterm} \\
% second form
&=\Biggl\{ \frac{1}{C}\sumterm \biggl[ \wterm 
  - \frac{1}{C} \Bigl(\,\sumterm\wterm\Bigr) \biggr]^2 \Biggr\}^{1/2}
\end{align*}
\end{document}