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Question Number 192791 by Mastermind last updated on 27/May/23
If n is a positive integer, prove that 2^n Γ(n+(1/2)) = 1.3.5...(2n−1)(√π). Help!
$$\mathrm{If}\:\mathrm{n}\:\mathrm{is}\:\mathrm{a}\:\mathrm{positive}\:\mathrm{integer},\:\mathrm{prove}\:\mathrm{that}\: \\ $$$$\mathrm{2}^{\mathrm{n}} \Gamma\left(\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}}\right)\:=\:\mathrm{1}.\mathrm{3}.\mathrm{5}…\left(\mathrm{2n}−\mathrm{1}\right)\sqrt{\pi}. \\ $$$$ \\ $$$$\mathrm{Help}! \\ $$
Answered by Mathspace last updated on 27/May/23
montrons que Γ(n+(1/2))=((1.3.5...(2n−1))/2^n )(√π) par recurence sur n n=1 Γ((3/2))=((√π)/2) vraie car Γ((3/2))=Γ(1+(1/2))=(1/2)Γ((1/2))=((√π)/2) supposons la relation vraie pourn on a Γ(n+1+(1/2)) =Γ(1+n+(1/2))=(n+(1/2))Γ(n+(1/2)) =(n+(1/2))×((1.3.5...(2n−1))/2^n )(√π) =((1.3.5....(2n−1)(2n+1))/2^(n+1) )(√π) la relation est vraie a lordre n+1.
$${montrons}\:{que}\:\Gamma\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\mathrm{1}.\mathrm{3}.\mathrm{5}…\left(\mathrm{2}{n}−\mathrm{1}\right)}{\mathrm{2}^{{n}} }\sqrt{\pi} \\ $$$${par}\:{recurence}\:{sur}\:{n} \\ $$$${n}=\mathrm{1}\:\:\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)=\frac{\sqrt{\pi}}{\mathrm{2}} \\ $$$${vraie}\:{car} \\ $$$$\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)=\Gamma\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\sqrt{\pi}}{\mathrm{2}} \\ $$$${supposons}\:{la}\:{relation}\:{vraie}\:{pourn} \\ $$$${on}\:{a}\:\Gamma\left({n}+\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\Gamma\left(\mathrm{1}+{n}+\frac{\mathrm{1}}{\mathrm{2}}\right)=\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)×\frac{\mathrm{1}.\mathrm{3}.\mathrm{5}…\left(\mathrm{2}{n}−\mathrm{1}\right)}{\mathrm{2}^{{n}} }\sqrt{\pi} \\ $$$$=\frac{\mathrm{1}.\mathrm{3}.\mathrm{5}….\left(\mathrm{2}{n}−\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{2}^{{n}+\mathrm{1}} }\sqrt{\pi} \\ $$$${la}\:{relation}\:{est}\:{vraie}\:{a}\:{lordre} \\ $$$${n}+\mathrm{1}. \\ $$
Commented by Mastermind last updated on 28/May/23
you mean it should be (2n+1) not (2n−1)? besides pls use english
$$\mathrm{you}\:\mathrm{mean}\:\mathrm{it}\:\mathrm{should}\:\mathrm{be}\:\left(\mathrm{2n}+\mathrm{1}\right)\:\mathrm{not}\:\left(\mathrm{2n}−\mathrm{1}\right)? \\ $$$$\mathrm{besides}\:\mathrm{pls}\:\mathrm{use}\:\mathrm{english} \\ $$$$ \\ $$

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