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Question Number 48718 by Abdo msup. last updated on 27/Nov/18
let I_n =∫_0 ^1 (1−t^2 )^n dt 1) calculate I_n by recurrence 2)find the value of Σ_(k=0) ^n (((−1)^k )/(2k+1))C_n ^k
$${let}\:{I}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{{n}} {dt} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{I}_{{n}} \:{by}\:{recurrence} \\ $$$$\left.\mathrm{2}\right){find}\:{the}\:{value}\:{of}\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{2}{k}+\mathrm{1}}{C}_{{n}} ^{{k}} \\ $$
Commented by Abdo msup. last updated on 27/Nov/18
we have I_n =∫_0 ^1 (1−t^2 )^(n−1) (1−t^2 )dt =∫_0 ^1 (1−t^2 )^(n−1) −∫_0 ^1 t^2 (1−t^2 )^(n−1) dt but ∫_0 ^1 (1−t^2 )^(n−1) dt = I_(n−1) by parts ∫_0 ^1 t^2 (1−t^2 )^(n−1) dt =−(1/2) ∫_0 ^1 t(−2t)(1−t^2 )^(n−1) dt u =t and v^′ =(−2t)(1−t^2 )^(n−1) =−(1/2){ [ (1/n)(1−t^2 )^n t]_0 ^1 −∫_0 ^1 (1/n)(1−t^2 )^n dt} =(1/(2n)) I_n ⇒I_n = I_(n−1) −(1/(2n)) I_n ⇒ (1+(1/(2n)))I_n =I_(n−1) ⇒((2n+1)/(2n)) I_n =I_(n−1) ⇒I_n =((2n)/(2n+1)) I_(n−1) ⇒ Π_(k=1) ^n I_k =Π_(k=1) ^n ((2k)/(2k+1)) Π_(k=1) ^n I_(k−1) ⇒ I_n =Π_(k=1) ^n ((2k)/(2k+1)) I_o but I_0 =1 ⇒I_n =((Π_(k=1) ^n (2k))/(Π_(k=1) ^n (2k+1))) =((2^n n!)/(3.5.7.....(2n+1))) =((2^n n!)/(2.3.4.5...(2n)(2n+1))) ×2^n n! =((2^(2n) (n!)^2 )/((2n+1)!))
$${we}\:{have}\:{I}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{{n}−\mathrm{1}} \left(\mathrm{1}−{t}^{\mathrm{2}} \right){dt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{{n}−\mathrm{1}} \:−\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{\mathrm{2}} \left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{{n}−\mathrm{1}} {dt} \\ $$$${but}\:\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{{n}−\mathrm{1}} {dt}\:=\:{I}_{{n}−\mathrm{1}} \\ $$$${by}\:{parts}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{t}^{\mathrm{2}} \left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{{n}−\mathrm{1}} {dt} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{t}\left(−\mathrm{2}{t}\right)\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{{n}−\mathrm{1}} {dt}\:\:\:\:{u}\:={t}\:{and}\:{v}^{'} =\left(−\mathrm{2}{t}\right)\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{{n}−\mathrm{1}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\left\{\:\left[\:\frac{\mathrm{1}}{{n}}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{{n}} {t}\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}}{{n}}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{{n}} {dt}\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{n}}\:{I}_{{n}} \:\Rightarrow{I}_{{n}} =\:{I}_{{n}−\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{2}{n}}\:{I}_{{n}} \:\Rightarrow \\ $$$$\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{n}}\right){I}_{{n}} ={I}_{{n}−\mathrm{1}} \:\Rightarrow\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{2}{n}}\:{I}_{{n}} ={I}_{{n}−\mathrm{1}} \:\Rightarrow{I}_{{n}} =\frac{\mathrm{2}{n}}{\mathrm{2}{n}+\mathrm{1}}\:{I}_{{n}−\mathrm{1}} \:\Rightarrow \\ $$$$\prod_{{k}=\mathrm{1}} ^{{n}} \:{I}_{{k}} =\prod_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{2}{k}}{\mathrm{2}{k}+\mathrm{1}}\:\prod_{{k}=\mathrm{1}} ^{{n}} \:{I}_{{k}−\mathrm{1}} \:\Rightarrow \\ $$$${I}_{{n}} =\prod_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{2}{k}}{\mathrm{2}{k}+\mathrm{1}}\:{I}_{{o}} \:\:\:\:\:{but}\:{I}_{\mathrm{0}} =\mathrm{1}\:\Rightarrow{I}_{{n}} =\frac{\prod_{{k}=\mathrm{1}} ^{{n}} \left(\mathrm{2}{k}\right)}{\prod_{{k}=\mathrm{1}} ^{{n}} \:\left(\mathrm{2}{k}+\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{2}^{{n}} {n}!}{\mathrm{3}.\mathrm{5}.\mathrm{7}…..\left(\mathrm{2}{n}+\mathrm{1}\right)}\:=\frac{\mathrm{2}^{{n}} {n}!}{\mathrm{2}.\mathrm{3}.\mathrm{4}.\mathrm{5}…\left(\mathrm{2}{n}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}\:×\mathrm{2}^{{n}} {n}! \\ $$$$=\frac{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!} \\ $$$$ \\ $$
Commented by Abdo msup. last updated on 27/Nov/18
2)we have I_n =∫_0 ^1 (1−t^2 )^n dt =∫_0 ^1 Σ_(k=0) ^n C_n ^k (−t^2 )^k dt =Σ_(k=0) ^n (−1)^k C_n ^k ∫_0 ^1 t^(2k) dt =Σ_(k=0) ^n (((−1)^k )/(2k+1)) C_n ^k ⇒ Σ_(k=0) ^n (((−1)^k )/(2k+1)) C_n ^k =((2^(2n) (n!)^2 )/((2n+1)!)) .
$$\left.\mathrm{2}\right){we}\:{have}\:\:{I}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{{n}} {dt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \:\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\left(−{t}^{\mathrm{2}} \right)^{{k}} \:{dt}\:=\sum_{{k}=\mathrm{0}} ^{{n}} \:\left(−\mathrm{1}\right)^{{k}} \:{C}_{{n}} ^{{k}} \:\int_{\mathrm{0}} ^{\mathrm{1}} \:{t}^{\mathrm{2}{k}} \:{dt} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}} \frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{2}{k}+\mathrm{1}}\:{C}_{{n}} ^{{k}} \:\:\Rightarrow\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\:\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{2}{k}+\mathrm{1}}\:{C}_{{n}} ^{{k}} \:\:\:=\frac{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 27/Nov/18
∫_0 ^1 (1−t^2 )^n dt t=sinθ dt=cosθdθ ∫_0 ^(π/2) cos^(2n) θ×cosθdθ ∫_0 ^(π/2) (sinθ)^(2×(1/2)−1) ×(cosθ)^(2n+2−1) dθ using gamma beta function =((⌈((1/2))×⌈(2n+2))/(2⌈(2n+2+(1/2)))) =(((√π) ×⌈(2n+2))/(2⌈(2n+(5/2)))) formula 2∫_0 ^(π/2) sin^(2p−1) θcos^(2q−1) θdθ =((⌈(p)⌈(q))/(⌈(p+q)))
$$\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{{n}} {dt} \\ $$$${t}={sin}\theta\:\:{dt}={cos}\theta{d}\theta \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {cos}^{\mathrm{2}{n}} \theta×{cos}\theta{d}\theta \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left({sin}\theta\right)^{\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} ×\left({cos}\theta\right)^{\mathrm{2}{n}+\mathrm{2}−\mathrm{1}} {d}\theta \\ $$$${using}\:{gamma}\:{beta}\:{function} \\ $$$$=\frac{\lceil\left(\frac{\mathrm{1}}{\mathrm{2}}\right)×\lceil\left(\mathrm{2}{n}+\mathrm{2}\right)}{\mathrm{2}\lceil\left(\mathrm{2}{n}+\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}}\right)} \\ $$$$=\frac{\sqrt{\pi}\:×\lceil\left(\mathrm{2}{n}+\mathrm{2}\right)}{\mathrm{2}\lceil\left(\mathrm{2}{n}+\frac{\mathrm{5}}{\mathrm{2}}\right)} \\ $$$${formula}\:\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{2}{p}−\mathrm{1}} \theta{cos}^{\mathrm{2}{q}−\mathrm{1}} \theta{d}\theta \\ $$$$=\frac{\lceil\left({p}\right)\lceil\left({q}\right)}{\lceil\left({p}+{q}\right)} \\ $$

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