Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

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)

$$\Gamma\left(\mathrm{x}+\mathrm{4}\right)=\left(\mathrm{x}+\mathrm{3}\right)\left(\mathrm{x}+\mathrm{2}\right)\left(\mathrm{x}+\mathrm{1}\right)\mathrm{x}\Gamma\left(\mathrm{x}\right) \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com