GivenU-n-0-1-x-n-1-x-dx-n-N-show-that-U-n-2-n-2-n-n-1-2n-3-

Question Number 98271 by Ar Brandon last updated on 12/Jun/20

$$\mathcal{G}\mathrm{ivenU}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{x}^{\mathrm{n}} \sqrt{\mathrm{1}−\mathrm{x}}\mathrm{dx}\:\:\mathrm{n}\in\mathbb{N},\:\mathrm{show}\:\mathrm{that} \\ $$$$\mathrm{U}_{\mathrm{n}} =\frac{\mathrm{2}^{\mathrm{n}+\mathrm{2}} \mathrm{n}!\left(\mathrm{n}+\mathrm{1}\right)}{\left(\mathrm{2n}+\mathrm{3}\right)!} \\ $$

Commented by Ar Brandon last updated on 12/Jun/20

Thanks �� But, it really isn't clear to me. Any explanations please ?��

Commented by PRITHWISH SEN 2 last updated on 12/Jun/20

then U_n = ((2n)/(2n+3)).((2(n−1))/(2n+1))......(4/7).(2/5).(2/3) =((2^n {n(n−1)(n−2)(n−3)......2.1}.2)/((2n+3)(2n+1).........7.5.3)) = ((2^(n+1) .n!)/((2n+3)!)) ×2^(n+1) {(n+1)n(n−1)........2.1} = ((2^(2n+2) n!(n+1)!)/((2n+3)!)) I got this please check

$$\mathrm{then}\:\mathrm{U}_{\mathrm{n}} =\:\frac{\mathrm{2n}}{\mathrm{2n}+\mathrm{3}}.\frac{\mathrm{2}\left(\mathrm{n}−\mathrm{1}\right)}{\mathrm{2n}+\mathrm{1}}……\frac{\mathrm{4}}{\mathrm{7}}.\frac{\mathrm{2}}{\mathrm{5}}.\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:=\frac{\mathrm{2}^{\mathrm{n}} \left\{\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)\left(\mathrm{n}−\mathrm{2}\right)\left(\mathrm{n}−\mathrm{3}\right)……\mathrm{2}.\mathrm{1}\right\}.\mathrm{2}}{\left(\mathrm{2n}+\mathrm{3}\right)\left(\mathrm{2n}+\mathrm{1}\right)………\mathrm{7}.\mathrm{5}.\mathrm{3}} \\ $$$$\:\:\:\:\:\:=\:\frac{\mathrm{2}^{\mathrm{n}+\mathrm{1}} .\mathrm{n}!}{\left(\mathrm{2n}+\mathrm{3}\right)!}\:×\mathrm{2}^{\mathrm{n}+\mathrm{1}} \left\{\left(\mathrm{n}+\mathrm{1}\right)\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)……..\mathrm{2}.\mathrm{1}\right\} \\ $$$$\:\:\:=\:\frac{\mathrm{2}^{\mathrm{2}\boldsymbol{\mathrm{n}}+\mathrm{2}} \boldsymbol{\mathrm{n}}!\left(\boldsymbol{\mathrm{n}}+\mathrm{1}\right)!}{\left(\mathrm{2}\boldsymbol{\mathrm{n}}+\mathrm{3}\right)!}\:\boldsymbol{\mathrm{I}}\:\boldsymbol{\mathrm{got}}\:\boldsymbol{\mathrm{this}}\:\boldsymbol{\mathrm{please}}\:\boldsymbol{\mathrm{check}} \\ $$

Commented by Ar Brandon last updated on 12/Jun/20

I′m stucked here. Can anyone help me out, please?

$$\mathrm{I}'\mathrm{m}\:\mathrm{stucked}\:\mathrm{here}.\:\mathrm{Can}\:\mathrm{anyone}\:\mathrm{help}\:\mathrm{me}\:\mathrm{out},\:\mathrm{please}? \\ $$

Commented by Ar Brandon last updated on 12/Jun/20

Commented by PRITHWISH SEN 2 last updated on 20/Jul/20

I did not do anything I just expanded your expression U_n =((2n)/(2n+3))U_(n−1) then U_(n−1) = ((2(n−1))/(2(n−1)+3)) U_(n−2) ⇒U_(n−1) = ((2(n−1))/(2n+1))U_(n−2) and likewise you can get by putting n= n−2,n−3 and so on now for U_1 =(2/5)U_0 andU_0 = ∫_0 ^1 x^0 (√(1−x))dx=(2/3) I think now it makes sense

$$\mathrm{I}\:\mathrm{did}\:\mathrm{not}\:\mathrm{do}\:\mathrm{anything}\:\mathrm{I}\:\mathrm{just}\:\mathrm{expanded}\:\mathrm{your}\:\mathrm{expression} \\ $$$$\mathrm{U}_{\mathrm{n}} =\frac{\mathrm{2n}}{\mathrm{2n}+\mathrm{3}}\mathrm{U}_{\mathrm{n}−\mathrm{1}} \:\:\:\:\:\mathrm{then}\:\mathrm{U}_{\mathrm{n}−\mathrm{1}} =\:\frac{\mathrm{2}\left(\mathrm{n}−\mathrm{1}\right)}{\mathrm{2}\left(\mathrm{n}−\mathrm{1}\right)+\mathrm{3}}\:\mathrm{U}_{\mathrm{n}−\mathrm{2}} \\ $$$$\Rightarrow\mathrm{U}_{\mathrm{n}−\mathrm{1}} =\:\frac{\mathrm{2}\left(\mathrm{n}−\mathrm{1}\right)}{\mathrm{2n}+\mathrm{1}}\mathrm{U}_{\mathrm{n}−\mathrm{2}} \:\:\mathrm{and}\:\mathrm{likewise}\:\mathrm{you}\:\mathrm{can}\:\mathrm{get} \\ $$$$\mathrm{by}\:\mathrm{putting}\:\mathrm{n}=\:\mathrm{n}−\mathrm{2},\mathrm{n}−\mathrm{3}\:\mathrm{and}\:\mathrm{so}\:\mathrm{on} \\ $$$$\mathrm{now}\:\mathrm{for}\:\mathrm{U}_{\mathrm{1}} =\frac{\mathrm{2}}{\mathrm{5}}\mathrm{U}_{\mathrm{0}} \:\:\mathrm{andU}_{\mathrm{0}} \:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{x}^{\mathrm{0}} \sqrt{\mathrm{1}−\mathrm{x}}\mathrm{dx}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\mathrm{I}\:\mathrm{think}\:\mathrm{now}\:\mathrm{it}\:\mathrm{makes}\:\mathrm{sense} \\ $$

Answered by smridha last updated on 12/Jun/20

let x=sin^2 𝛉 so 2∫_0 ^(𝛑/2) sin^((2n+1)) 𝛉cos^2 𝛉d𝛉 =((𝚪(n+1).𝚪((3/2)))/(𝚪(n+(5/2))))=((n!(√𝛑))/((2n+3).𝚪(n+(3/2)))) I think now you can easyly get your ans.

$$\boldsymbol{{let}}\:\boldsymbol{{x}}=\mathrm{sin}\:^{\mathrm{2}} \boldsymbol{\theta}\:\boldsymbol{{so}} \\ $$$$\mathrm{2}\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \boldsymbol{{sin}}^{\left(\mathrm{2}\boldsymbol{{n}}+\mathrm{1}\right)} \boldsymbol{\theta{cos}}^{\mathrm{2}} \boldsymbol{\theta{d}\theta} \\ $$$$=\frac{\boldsymbol{\Gamma}\left(\boldsymbol{{n}}+\mathrm{1}\right).\boldsymbol{\Gamma}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)}{\boldsymbol{\Gamma}\left(\boldsymbol{{n}}+\frac{\mathrm{5}}{\mathrm{2}}\right)}=\frac{\boldsymbol{{n}}!\sqrt{\boldsymbol{\pi}}}{\left(\mathrm{2}\boldsymbol{{n}}+\mathrm{3}\right).\boldsymbol{\Gamma}\left(\boldsymbol{{n}}+\frac{\mathrm{3}}{\mathrm{2}}\right)} \\ $$$$\boldsymbol{{I}}\:\boldsymbol{{think}}\:\boldsymbol{{now}}\:\boldsymbol{{you}}\:\boldsymbol{{can}}\:\boldsymbol{{easyly}} \\ $$$$\boldsymbol{{get}}\:\boldsymbol{{your}}\:\boldsymbol{{ans}}. \\ $$

Commented by PRITHWISH SEN 2 last updated on 13/Jun/20

$$\mathrm{wow}!\:\mathrm{excellent}\:\mathrm{sir}. \\ $$

Commented by smridha last updated on 13/Jun/20

$$\boldsymbol{{thanks}}.. \\ $$

Commented by PRITHWISH SEN 2 last updated on 13/Jun/20

$$\mathrm{welcome} \\ $$

Commented by Ar Brandon last updated on 13/Jun/20

Hum, thank you ��

Answered by Dwaipayan Shikari last updated on 01/Nov/20

∫_0 ^1 x^n (1−x)^(1/2) dx =∫_0 ^1 x^(n+1−1) x^((3/2)−1) dx =β(n+1,(3/2))=((Γ(n+1)Γ((3/2)))/(Γ(n+(5/2))))

$$\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}} \left(\mathrm{1}−{x}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} {dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}+\mathrm{1}−\mathrm{1}} {x}^{\frac{\mathrm{3}}{\mathrm{2}}−\mathrm{1}} {dx} \\ $$$$=\beta\left({n}+\mathrm{1},\frac{\mathrm{3}}{\mathrm{2}}\right)=\frac{\Gamma\left({n}+\mathrm{1}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)}{\Gamma\left({n}+\frac{\mathrm{5}}{\mathrm{2}}\right)} \\ $$

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