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7-2-




Question Number 108239 by Dwaipayan Shikari last updated on 15/Aug/20
((7/2))!=?
$$\left(\frac{\mathrm{7}}{\mathrm{2}}\right)!=? \\ $$$$ \\ $$
Answered by abdomsup last updated on 15/Aug/20
Γ(x+n) =Γ(x+n−1 +1)  =(x+n−1)Γ(x+n−2 +1)  =(x+n−1)(x+n−2)Γ(x+n−2)  =(x+n−1)(x+n−2)(x+n−3)Γ(x+n−3)  ⇒((7/2))! =Γ((7/2)+1)  =Γ((1/2) +4)=((1/2) +3)((1/2)+2)((1/2)+1)Γ((1/2)+1)  =(7/2).(5/2).(3/2).(1/2)Γ((1/2))  =((35.3)/8)Γ((1/2)) =((105)/8)Γ((1/2))  Γ(x) =∫_0 ^∞  t^(x−1)  e^(−t )  dt ⇒  Γ((1/2))=∫_0 ^∞  (e^(−t) /( (√t)))dt =_(t=u^2 )  ∫_0 ^∞  (e^(−u^2 ) /u)(2u)du  =2∫_0 ^∞  e^(−u^2 ) du =2.((√π)/2)=(√π) ⇒  ((7/2))! =((105(√π))/8)
$$\Gamma\left({x}+{n}\right)\:=\Gamma\left({x}+{n}−\mathrm{1}\:+\mathrm{1}\right) \\ $$$$=\left({x}+{n}−\mathrm{1}\right)\Gamma\left({x}+{n}−\mathrm{2}\:+\mathrm{1}\right) \\ $$$$=\left({x}+{n}−\mathrm{1}\right)\left({x}+{n}−\mathrm{2}\right)\Gamma\left({x}+{n}−\mathrm{2}\right) \\ $$$$=\left({x}+{n}−\mathrm{1}\right)\left({x}+{n}−\mathrm{2}\right)\left({x}+{n}−\mathrm{3}\right)\Gamma\left({x}+{n}−\mathrm{3}\right) \\ $$$$\Rightarrow\left(\frac{\mathrm{7}}{\mathrm{2}}\right)!\:=\Gamma\left(\frac{\mathrm{7}}{\mathrm{2}}+\mathrm{1}\right) \\ $$$$=\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\:+\mathrm{4}\right)=\left(\frac{\mathrm{1}}{\mathrm{2}}\:+\mathrm{3}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{2}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{1}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{1}\right) \\ $$$$=\frac{\mathrm{7}}{\mathrm{2}}.\frac{\mathrm{5}}{\mathrm{2}}.\frac{\mathrm{3}}{\mathrm{2}}.\frac{\mathrm{1}}{\mathrm{2}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{35}.\mathrm{3}}{\mathrm{8}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:=\frac{\mathrm{105}}{\mathrm{8}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\Gamma\left({x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:{t}^{{x}−\mathrm{1}} \:{e}^{−{t}\:} \:{dt}\:\Rightarrow \\ $$$$\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{{e}^{−{t}} }{\:\sqrt{{t}}}{dt}\:=_{{t}={u}^{\mathrm{2}} } \:\int_{\mathrm{0}} ^{\infty} \:\frac{{e}^{−{u}^{\mathrm{2}} } }{{u}}\left(\mathrm{2}{u}\right){du} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \:{e}^{−{u}^{\mathrm{2}} } {du}\:=\mathrm{2}.\frac{\sqrt{\pi}}{\mathrm{2}}=\sqrt{\pi}\:\Rightarrow \\ $$$$\left(\frac{\mathrm{7}}{\mathrm{2}}\right)!\:=\frac{\mathrm{105}\sqrt{\pi}}{\mathrm{8}} \\ $$
Commented by abdomsup last updated on 15/Aug/20
sorry   ((7/2))! =((105(√π))/(16))
$${sorry}\:\:\:\left(\frac{\mathrm{7}}{\mathrm{2}}\right)!\:=\frac{\mathrm{105}\sqrt{\pi}}{\mathrm{16}} \\ $$
Commented by abdomsup last updated on 15/Aug/20
generaly  we have  Γ(x+n)=(x+n−1)(x+n−2)  ....(x+n−k)Γ(x+n−k)  k=n ⇒Γ(x+n)=(x+n−1)(x+n−2)  ......(x+1)xΓ(x)  ex     n=4 ⇒  Γ(x+4)=(x+3)(x+2)(x+1)Γ(x)  x=(1/m) ⇒Γ((1/m)+4)  =((1/(m+3)))((1/m)+2)((1/m)+1)(1/m)Γ((1/m))
$${generaly}\:\:{we}\:{have} \\ $$$$\Gamma\left({x}+{n}\right)=\left({x}+{n}−\mathrm{1}\right)\left({x}+{n}−\mathrm{2}\right) \\ $$$$….\left({x}+{n}−{k}\right)\Gamma\left({x}+{n}−{k}\right) \\ $$$${k}={n}\:\Rightarrow\Gamma\left({x}+{n}\right)=\left({x}+{n}−\mathrm{1}\right)\left({x}+{n}−\mathrm{2}\right) \\ $$$$……\left({x}+\mathrm{1}\right){x}\Gamma\left({x}\right) \\ $$$${ex}\:\:\:\:\:{n}=\mathrm{4}\:\Rightarrow \\ $$$$\Gamma\left({x}+\mathrm{4}\right)=\left({x}+\mathrm{3}\right)\left({x}+\mathrm{2}\right)\left({x}+\mathrm{1}\right)\Gamma\left({x}\right) \\ $$$${x}=\frac{\mathrm{1}}{{m}}\:\Rightarrow\Gamma\left(\frac{\mathrm{1}}{{m}}+\mathrm{4}\right) \\ $$$$=\left(\frac{\mathrm{1}}{{m}+\mathrm{3}}\right)\left(\frac{\mathrm{1}}{{m}}+\mathrm{2}\right)\left(\frac{\mathrm{1}}{{m}}+\mathrm{1}\right)\frac{\mathrm{1}}{{m}}\Gamma\left(\frac{\mathrm{1}}{{m}}\right) \\ $$
Commented by abdomsup last updated on 15/Aug/20
error of typo  Γ((1/m)+4)=((1/m)+3)((1/m)+2)((1/m)+1)(1/m)Γ((1/m))
$${error}\:{of}\:{typo} \\ $$$$\Gamma\left(\frac{\mathrm{1}}{{m}}+\mathrm{4}\right)=\left(\frac{\mathrm{1}}{{m}}+\mathrm{3}\right)\left(\frac{\mathrm{1}}{{m}}+\mathrm{2}\right)\left(\frac{\mathrm{1}}{{m}}+\mathrm{1}\right)\frac{\mathrm{1}}{{m}}\Gamma\left(\frac{\mathrm{1}}{{m}}\right) \\ $$
Commented by mathmax by abdo last updated on 15/Aug/20
Γ(x+4)=(x+3)(x+2)(x+1)xΓ(x)
$$\Gamma\left(\mathrm{x}+\mathrm{4}\right)=\left(\mathrm{x}+\mathrm{3}\right)\left(\mathrm{x}+\mathrm{2}\right)\left(\mathrm{x}+\mathrm{1}\right)\mathrm{x}\Gamma\left(\mathrm{x}\right) \\ $$

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