Question Number 4650 by FilupSmith last updated on 18/Feb/16
$$\mathrm{In}\:\mathrm{mathematics}\:\mathrm{I}\:\mathrm{have}\:\mathrm{worked}\:\mathrm{with}: \\ $$$$\Sigma\:\mathrm{and}\:\Pi. \\ $$$${e}.{g}. \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{x}} {\sum}}{k}\:=\:\frac{\mathrm{1}}{\mathrm{2}}{x}\left({x}+\mathrm{1}\right) \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{x}} {\prod}}{k}\:=\:{k}! \\ $$$$ \\ $$$$\mathrm{I}\:\mathrm{have}\:\mathrm{never}\:\mathrm{worked}\:\mathrm{with}\:\coprod.\:\mathrm{I}\:\mathrm{am}\:\mathrm{curious} \\ $$$$\mathrm{as}\:\mathrm{to}\:\mathrm{what}\:\mathrm{it}\:\mathrm{is}.\:\mathrm{Is}\:\mathrm{it}\:\mathrm{another}\:\mathrm{notation}\:\mathrm{form}? \\ $$$$\mathrm{Does}\:\underset{{k}=\mathrm{1}} {\overset{{x}} {\coprod}}{k}\:\mathrm{exist}? \\ $$
Commented by Yozzii last updated on 19/Feb/16
Commented by Yozzii last updated on 19/Feb/16
$${Category}\:{theory};\:{source}:\:{wikipedia} \\ $$
Commented by Yozzii last updated on 19/Feb/16
Commented by Yozzii last updated on 19/Feb/16
Commented by FilupSmith last updated on 19/Feb/16
$$\mathrm{Very}\:\mathrm{interesting}! \\ $$
Commented by 2closedStringsMeet last updated on 19/Feb/16
$${Awesome}!\:{Indeed}\:{very}\:{interesting}. \\ $$
Answered by 2closedStringsMeet last updated on 19/Feb/16
$${it}\:{is}\:{likely}\:{the}\:{disjuntive}\:{sum}\:{operator}.\: \\ $$$${Logical}\:{operator}\:\left[{or}\right]…\: \\ $$
Commented by FilupSmith last updated on 19/Feb/16
$$\mathrm{I}\:\mathrm{see}.\:\mathrm{Thanks}! \\ $$