Question Number 3198 by Filup last updated on 07/Dec/15
$$\underset{{i}={a}} {\overset{{n}} {\prod}}{i}=?\:\:\:\:\:\mathrm{for}\:\mid{i}\mid<\mathrm{1},\:\:{a}\in\mathbb{R} \\ $$
Commented by Filup last updated on 07/Dec/15
$$=\frac{\mathrm{1}}{{a}}×\frac{\mathrm{1}}{{a}+\mathrm{1}}×…×\frac{\mathrm{1}}{{n}} \\ $$$$=\frac{\mathrm{1}}{\left(\frac{{n}!}{\left({a}−\mathrm{1}\right)!}\right)} \\ $$$$=\frac{\left({a}−\mathrm{1}\right)!}{{n}!} \\ $$$$ \\ $$$$\mathrm{correct}? \\ $$
Commented by prakash jain last updated on 07/Dec/15
$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{think}\:\mathrm{factorial}\:\mathrm{function}\:\mathrm{is} \\ $$$$\mathrm{defined}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}. \\ $$
Commented by Filup last updated on 07/Dec/15
$$\mathrm{W}{hat}\:\mathrm{about}\:\mathrm{in}\:\Gamma\:\mathrm{notation}? \\ $$
Commented by prakash jain last updated on 07/Dec/15
$$\mathrm{Let}\:\mathrm{us}\:\mathrm{say}\:{n}={i}+{k}\:{k}\in\mathbb{Z}^{+} \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{n}} {\prod}}{i}={a}\left({a}+\mathrm{1}\right)\left({a}+\mathrm{2}\right)…\left({a}+{k}\right) \\ $$
Commented by prakash jain last updated on 07/Dec/15
$$\mathrm{If}\:{n}={i}+{k},\:{k}\in\mathbb{Z}^{+} \\ $$$$\mathrm{then}\:\Gamma\left({n}\right)\:{and}\:\Gamma\left({i}\right)\:\mathrm{be}\:\mathrm{derived}\:\mathrm{with} \\ $$$$\Gamma\left({x}+\mathrm{1}\right)={x}\Gamma\left({x}\right) \\ $$