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}
(x,y) s,t
clauses add anything so is this output already narrow enough??