In This question, the following appears:
$$[0,2] \times \left[0,4-\sqrt{3}\right] \times \left[0,4-\sqrt{3}\right] \times \mathop{\Large\times}_{i=1}^{k-3}[0,2]$$
Thinking that that last $\times$ should be somewhat comparable in size to $\sum$ or $\prod$ or $\bigoplus$ or $\bigcup,$ I tried changing it to this: $$ [0,2] \times \left[0,4-\sqrt{3}\right] \times \left[0,4-\sqrt{3}\right] \times \mathop{\Huge\times}_{i=1}^{k-3}[0,2] $$ The subscript $i=1$ definitely seems too high. Contrast these: $$ \begin{array}{c} \displaystyle \sum_{i=1}^{k-3} [0,2] \qquad\qquad \mathop{\Huge\times}_{i=1}^{k-3} [0,2] \qquad \qquad \sum_{i=1}^{k-3} [0,2] \\ \hline \end{array} $$ Can this be corrected?
\varprod
is available in MathJax. (At least in the version currently available on MO and SE.) Neither is\operatornamewithlimits
. $\endgroup$\raisebox
work here? … After experimentation, now edited out, no. Well, if it did, then\raisebox{-experimentally determined amount}
might be the easiest solution. But it doesn't.\bigtimes
also doesn't work. $\endgroup$\Huge
cannot be used inside a math formula in the first place (so this is a convenience feature of MathJax), but if you replace it with the correct\hbox{\Huge$\times$}
, it will behave exactly as seen here: even if wrapped in\mathop
, it will be set too high, unless you use\vcenter
to correct it. $\endgroup$\varprod
pointed out by @MartinSleziak (which is also not a bug report, but may be something that MathJax developers are willing to act on—by "at least for bug reports" I didn't mean "this is a bug report", but rather that, while they weren't necessarily here for this, they might act on it anyway). $\endgroup$