While solving an equation for a given algorithm [the equation is quite complex but it is similar to] $\prod_{n=1,n\neq l}^L\frac{1}{2+\lambda_n}$
I stumbled on a situation where $L=1$ and $l=1$. This means that the only value that $n$ can take is $1$ which is also forbidden given the conditioned $n\neq l$. i tried ignoring the product operator in that specific situation and just solving $\frac{1}{2+\lambda_n}$ for n=1. This produced a false result. this maybe a silly question but how should i act in this situation ? [side note: the value prduced by this expression is later inserted into a larger equation in order to solve the algorithm] thank you
Generally, the empty product has the value $1$.
Similarly, the empty sum has the value $0$.
There are times when these defaults do not make sense, but those times are context-dependent.