Question Number 73340 by Raxreedoroid last updated on 10/Nov/19
![Is there any pi/product notation rules I discovered some such as: Π_(k=a) ^b [k]=((b!)/((a−1)!)) Π_(k=a) ^b [c]=c^(b−a+1) Π_(k=a) ^b [c∙k]=Π_(k=a) ^b [c]∙Π_(k=a) ^b [k] Π_(k=a) ^b [k+c]=Π_(k=a+c) ^(b+c) [k] But what about this Π_(k=a) ^b [c∙k+d]](Q73340.png)
$$\mathrm{Is}\:\mathrm{there}\:\mathrm{any}\:\mathrm{pi}/\mathrm{product}\:\mathrm{notation}\:\mathrm{rules} \\ $$$$\mathrm{I}\:\mathrm{discovered}\:\mathrm{some}\:\mathrm{such}\:\mathrm{as}: \\ $$$$\underset{{k}={a}} {\overset{{b}} {\prod}}\left[{k}\right]=\frac{{b}!}{\left({a}−\mathrm{1}\right)!} \\ $$$$\underset{{k}={a}} {\overset{{b}} {\prod}}\left[{c}\right]={c}^{{b}−{a}+\mathrm{1}} \\ $$$$\underset{{k}={a}} {\overset{{b}} {\prod}}\left[{c}\centerdot{k}\right]=\underset{{k}={a}} {\overset{{b}} {\prod}}\left[{c}\right]\centerdot\underset{{k}={a}} {\overset{{b}} {\prod}}\left[{k}\right] \\ $$$$\underset{{k}={a}} {\overset{{b}} {\prod}}\left[{k}+{c}\right]=\underset{{k}={a}+{c}} {\overset{{b}+{c}} {\prod}}\left[{k}\right] \\ $$$$\mathrm{But}\:\mathrm{what}\:\mathrm{about}\:\mathrm{this} \\ $$$$\underset{{k}={a}} {\overset{{b}} {\prod}}\left[{c}\centerdot{k}+{d}\right] \\ $$
Commented by mathmax by abdo last updated on 10/Nov/19
![what means [...]?](Q73368.png)
$${what}\:{means}\:\left[...\right]? \\ $$
Commented by MJS last updated on 10/Nov/19
![Π_(k=a) ^b [c×k+d] let d=c×e Π_(k=a) ^b [c×(k+e)]=(((b+e)!)/((a+e−1)!))c^(1−a+b) = =Π_(k=a) ^b [c]×Π_(k=a) ^b [k+e] if (d/c)=e∈Q\Z Π_(k=a) ^b [k+(d/c)]=(a+(d/c))((Γ(b+(d/c)+1))/(Γ(a+(d/c)+1)))](Q73370.png)
$$\underset{{k}={a}} {\overset{{b}} {\prod}}\left[{c}×{k}+{d}\right] \\ $$$$\mathrm{let}\:{d}={c}×{e} \\ $$$$\underset{{k}={a}} {\overset{{b}} {\prod}}\left[{c}×\left({k}+{e}\right)\right]=\frac{\left({b}+{e}\right)!}{\left({a}+{e}−\mathrm{1}\right)!}{c}^{\mathrm{1}−{a}+{b}} = \\ $$$$=\underset{{k}={a}} {\overset{{b}} {\prod}}\left[{c}\right]×\underset{{k}={a}} {\overset{{b}} {\prod}}\left[{k}+{e}\right] \\ $$$$\mathrm{if}\:\frac{{d}}{{c}}={e}\in\mathbb{Q}\backslash\mathbb{Z} \\ $$$$\underset{{k}={a}} {\overset{{b}} {\prod}}\left[{k}+\frac{{d}}{{c}}\right]=\left({a}+\frac{{d}}{{c}}\right)\frac{\Gamma\left({b}+\frac{{d}}{{c}}+\mathrm{1}\right)}{\Gamma\left({a}+\frac{{d}}{{c}}+\mathrm{1}\right)} \\ $$