24 Apr
2014
24 Apr
'14
5 a.m.
Hans Hagen
indeed. this is what (sort of) happens in case of a plugged in renderer:
\sqrt{#1} -> ...\mathstylehbox{#1}... -> ...\hbox{\stylecommand #1}...
when #a = a \over b the style gets applied to the a only as \over creates two (pseudo) groups i.e. the \over isolates a and b
\sqrt{{k\over m}}\quad \sqrt{\displaystyle{k\over m}}
would work ok
Ah, I understand. The extra-braces solution, \sqrt{{k\over m}}, is simple and I can retain my \over habits. -Sanjoy