Question Number 108239 by Dwaipayan Shikari last updated on 15/Aug/20
$$\left(\frac{\mathrm{7}}{\mathrm{2}}\right)!=? \\ $$$$ \\ $$
Answered by abdomsup last updated on 15/Aug/20
$$\Gamma\left({x}+{n}\right)\:=\Gamma\left({x}+{n}−\mathrm{1}\:+\mathrm{1}\right) \\ $$$$=\left({x}+{n}−\mathrm{1}\right)\Gamma\left({x}+{n}−\mathrm{2}\:+\mathrm{1}\right) \\ $$$$=\left({x}+{n}−\mathrm{1}\right)\left({x}+{n}−\mathrm{2}\right)\Gamma\left({x}+{n}−\mathrm{2}\right) \\ $$$$=\left({x}+{n}−\mathrm{1}\right)\left({x}+{n}−\mathrm{2}\right)\left({x}+{n}−\mathrm{3}\right)\Gamma\left({x}+{n}−\mathrm{3}\right) \\ $$$$\Rightarrow\left(\frac{\mathrm{7}}{\mathrm{2}}\right)!\:=\Gamma\left(\frac{\mathrm{7}}{\mathrm{2}}+\mathrm{1}\right) \\ $$$$=\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\:+\mathrm{4}\right)=\left(\frac{\mathrm{1}}{\mathrm{2}}\:+\mathrm{3}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{2}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{1}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{1}\right) \\ $$$$=\frac{\mathrm{7}}{\mathrm{2}}.\frac{\mathrm{5}}{\mathrm{2}}.\frac{\mathrm{3}}{\mathrm{2}}.\frac{\mathrm{1}}{\mathrm{2}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{35}.\mathrm{3}}{\mathrm{8}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:=\frac{\mathrm{105}}{\mathrm{8}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\Gamma\left({x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:{t}^{{x}−\mathrm{1}} \:{e}^{−{t}\:} \:{dt}\:\Rightarrow \\ $$$$\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{{e}^{−{t}} }{\:\sqrt{{t}}}{dt}\:=_{{t}={u}^{\mathrm{2}} } \:\int_{\mathrm{0}} ^{\infty} \:\frac{{e}^{−{u}^{\mathrm{2}} } }{{u}}\left(\mathrm{2}{u}\right){du} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \:{e}^{−{u}^{\mathrm{2}} } {du}\:=\mathrm{2}.\frac{\sqrt{\pi}}{\mathrm{2}}=\sqrt{\pi}\:\Rightarrow \\ $$$$\left(\frac{\mathrm{7}}{\mathrm{2}}\right)!\:=\frac{\mathrm{105}\sqrt{\pi}}{\mathrm{8}} \\ $$
Commented by abdomsup last updated on 15/Aug/20
$${sorry}\:\:\:\left(\frac{\mathrm{7}}{\mathrm{2}}\right)!\:=\frac{\mathrm{105}\sqrt{\pi}}{\mathrm{16}} \\ $$
Commented by abdomsup last updated on 15/Aug/20
$${generaly}\:\:{we}\:{have} \\ $$$$\Gamma\left({x}+{n}\right)=\left({x}+{n}−\mathrm{1}\right)\left({x}+{n}−\mathrm{2}\right) \\ $$$$….\left({x}+{n}−{k}\right)\Gamma\left({x}+{n}−{k}\right) \\ $$$${k}={n}\:\Rightarrow\Gamma\left({x}+{n}\right)=\left({x}+{n}−\mathrm{1}\right)\left({x}+{n}−\mathrm{2}\right) \\ $$$$……\left({x}+\mathrm{1}\right){x}\Gamma\left({x}\right) \\ $$$${ex}\:\:\:\:\:{n}=\mathrm{4}\:\Rightarrow \\ $$$$\Gamma\left({x}+\mathrm{4}\right)=\left({x}+\mathrm{3}\right)\left({x}+\mathrm{2}\right)\left({x}+\mathrm{1}\right)\Gamma\left({x}\right) \\ $$$${x}=\frac{\mathrm{1}}{{m}}\:\Rightarrow\Gamma\left(\frac{\mathrm{1}}{{m}}+\mathrm{4}\right) \\ $$$$=\left(\frac{\mathrm{1}}{{m}+\mathrm{3}}\right)\left(\frac{\mathrm{1}}{{m}}+\mathrm{2}\right)\left(\frac{\mathrm{1}}{{m}}+\mathrm{1}\right)\frac{\mathrm{1}}{{m}}\Gamma\left(\frac{\mathrm{1}}{{m}}\right) \\ $$
Commented by abdomsup last updated on 15/Aug/20
$${error}\:{of}\:{typo} \\ $$$$\Gamma\left(\frac{\mathrm{1}}{{m}}+\mathrm{4}\right)=\left(\frac{\mathrm{1}}{{m}}+\mathrm{3}\right)\left(\frac{\mathrm{1}}{{m}}+\mathrm{2}\right)\left(\frac{\mathrm{1}}{{m}}+\mathrm{1}\right)\frac{\mathrm{1}}{{m}}\Gamma\left(\frac{\mathrm{1}}{{m}}\right) \\ $$
Commented by mathmax by abdo last updated on 15/Aug/20
$$\Gamma\left(\mathrm{x}+\mathrm{4}\right)=\left(\mathrm{x}+\mathrm{3}\right)\left(\mathrm{x}+\mathrm{2}\right)\left(\mathrm{x}+\mathrm{1}\right)\mathrm{x}\Gamma\left(\mathrm{x}\right) \\ $$