Menu Close

Question-123823

Tinku Tara 0 Comments
Pin




Question Number 123823 by Dwaipayan Shikari last updated on 28/Nov/20
Answered by mindispower last updated on 29/Nov/20
(1+x)^a =1+Σ_(n≥1) (1/(n!))Π_(k=0) ^(n−1) (a−k).x^n (1−ksin^2 (x))^(−(1/2)) =1+Σ(1/(n!))Π_(k=0) ^(n−1) (−(1/2)−k).sin^(2n) (x) =1+Σ_(n≥1) (((−1)^n )/(n!2^n ))Π_(k≤n−1) (1+2k).k^n sin^(2n) (x) =1+Σ_(n≥1) (((−1)^n (2n−1)!!)/(n!.2^n ))k^n sin^(2n) (x) ∫_0 ^(π/2) (dx/( (√(1−ksin^2 (x)))))=∫_0 ^(π/2) (1+Σ_(n≥1) (((−1)^n )/(n!.2^n ))(2n−1)!!.k^n sin^(2n) (x))dx S_n =(π/2)+Σ(((−1)^n (2n−1)!!.k^n )/(2^n n!))∫_0 ^(π/2) sin^(2n) (x)dx ∫_0 ^(π/2) sin^a (x)=(1/2).2∫_0 ^(π/2) sin^(2((a/2)+(1/2))−1) cos^(2(1/2)−1) (x)dx (1/2)β((a/2)+(1/2),(1/2))=((Γ((a/2)+(1/2))Γ((1/2)))/(2Γ(1+(a/2)))) , valid for Re(a)>−1 s_n =(π/2)+Σ_(n≥1) (((−1)^n (2n−1)!!)/(2^n .n!)).k^n .(1/2)((Γ(n+(1/2))Γ((1/2)))/(Γ(1+n))) Γ(n+(1/2))=Π_(k=0) ^(n−1) (n−(1/2)−k)Γ((1/2)),Γ((1/2))=(√π) Γ(1+n)=n! ⇔ S_n =(π/2)+Σ_(n≥1) (((−1)^n (2n−1)!!)/(2^n n!))k^n .(1/2)Π_(k≤n−1) (((2n−1−2k)/2))Γ((1/2))Γ((1/2)).(1/(n!)) =(π/2)+Σ_(n≥1) (((−1)^n (2n−1)!!.(2n−1)!!)/(2^n n!.2.2^n n!))k^n Γ^2 ((1/2)) =(π/2)+Σ_(n≥1) ((((2n−1)!!)/(n!)))^2 (((−k)^n π)/2^(2n+1) ) somthing missing not,Σ_(n≥0) ((((2n−1)!!)/(n!)))^2 .... by comparison⇒(−k)^n =256^n ⇒(−k)^n =256^n e^(2iπm) ⇒−k=256e^((2iπm)/n) ,∀n tru n→∞,−k=256⇒k=−256 (√k)=+_− 16i
$$\left(\mathrm{1}+{x}\right)^{{a}} =\mathrm{1}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{{n}!}\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({a}−{k}\right).{x}^{{n}} \\ $$$$\left(\mathrm{1}−{ksin}^{\mathrm{2}} \left({x}\right)\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} =\mathrm{1}+\Sigma\frac{\mathrm{1}}{{n}!}\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left(−\frac{\mathrm{1}}{\mathrm{2}}−{k}\right).{sin}^{\mathrm{2}{n}} \left({x}\right) \\ $$$$=\mathrm{1}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!\mathrm{2}^{{n}} }\underset{{k}\leqslant{n}−\mathrm{1}} {\prod}\left(\mathrm{1}+\mathrm{2}{k}\right).{k}^{{n}} {sin}^{\mathrm{2}{n}} \left({x}\right) \\ $$$$=\mathrm{1}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} \left(\mathrm{2}{n}−\mathrm{1}\right)!!}{{n}!.\mathrm{2}^{{n}} }{k}^{{n}} {sin}^{\mathrm{2}{n}} \left({x}\right) \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dx}}{\:\sqrt{\mathrm{1}−{ksin}^{\mathrm{2}} \left({x}\right)}}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{1}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!.\mathrm{2}^{{n}} }\left(\mathrm{2}{n}−\mathrm{1}\right)!!.{k}^{{n}} {sin}^{\mathrm{2}{n}} \left({x}\right)\right){dx} \\ $$$${S}_{{n}} =\frac{\pi}{\mathrm{2}}+\Sigma\frac{\left(−\mathrm{1}\right)^{{n}} \left(\mathrm{2}{n}−\mathrm{1}\right)!!.{k}^{{n}} }{\mathrm{2}^{{n}} {n}!}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{2}{n}} \left({x}\right){dx} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{{a}} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{2}\left(\frac{{a}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)−\mathrm{1}} {cos}^{\mathrm{2}\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} \left({x}\right){dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\beta\left(\frac{{a}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\Gamma\left(\frac{{a}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}\Gamma\left(\mathrm{1}+\frac{{a}}{\mathrm{2}}\right)}\:,\:{valid}\:{for}\:{Re}\left({a}\right)>−\mathrm{1} \\ $$$${s}_{{n}} =\frac{\pi}{\mathrm{2}}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} \left(\mathrm{2}{n}−\mathrm{1}\right)!!}{\mathrm{2}^{{n}} .{n}!}.{k}^{{n}} .\frac{\mathrm{1}}{\mathrm{2}}\frac{\Gamma\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\mathrm{1}+{n}\right)} \\ $$$$\Gamma\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)=\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({n}−\frac{\mathrm{1}}{\mathrm{2}}−{k}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right),\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\sqrt{\pi} \\ $$$$\Gamma\left(\mathrm{1}+{n}\right)={n}! \\ $$$$\Leftrightarrow \\ $$$${S}_{{n}} =\frac{\pi}{\mathrm{2}}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} \left(\mathrm{2}{n}−\mathrm{1}\right)!!}{\mathrm{2}^{{n}} {n}!}{k}^{{n}} .\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}\leqslant{n}−\mathrm{1}} {\prod}\left(\frac{\mathrm{2}{n}−\mathrm{1}−\mathrm{2}{k}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right).\frac{\mathrm{1}}{{n}!} \\ $$$$=\frac{\pi}{\mathrm{2}}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} \left(\mathrm{2}{n}−\mathrm{1}\right)!!.\left(\mathrm{2}{n}−\mathrm{1}\right)!!}{\mathrm{2}^{{n}} {n}!.\mathrm{2}.\mathrm{2}^{{n}} {n}!}{k}^{{n}} \Gamma^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\frac{\pi}{\mathrm{2}}+\underset{{n}\geqslant\mathrm{1}} {\sum}\left(\frac{\left(\mathrm{2}{n}−\mathrm{1}\right)!!}{{n}!}\right)^{\mathrm{2}} \frac{\left(−{k}\right)^{{n}} \pi}{\mathrm{2}^{\mathrm{2}{n}+\mathrm{1}} } \\ $$$${somthing}\:{missing} \\ $$$${not},\underset{{n}\geqslant\mathrm{0}} {\sum}\left(\frac{\left(\mathrm{2}{n}−\mathrm{1}\right)!!}{{n}!}\right)^{\mathrm{2}} …. \\ $$$${by}\:{comparison}\Rightarrow\left(−{k}\right)^{{n}} =\mathrm{256}^{{n}} \\ $$$$\Rightarrow\left(−{k}\right)^{{n}} =\mathrm{256}^{{n}} {e}^{\mathrm{2}{i}\pi{m}} \\ $$$$\Rightarrow−{k}=\mathrm{256}{e}^{\frac{\mathrm{2}{i}\pi{m}}{{n}}} ,\forall{n}\:{tru}\: \\ $$$${n}\rightarrow\infty,−{k}=\mathrm{256}\Rightarrow{k}=−\mathrm{256} \\ $$$$\sqrt{{k}}=\underset{−} {+}\mathrm{16}{i} \\ $$$$ \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 29/Nov/20
excellent sir ...
$${excellent}\:{sir}\:… \\ $$
Commented by Dwaipayan Shikari last updated on 29/Nov/20
This is a question from′′ Brilliant ′′
$${This}\:{is}\:\:{a}\:{question}\:{from}''\:{Brilliant}\:'' \\ $$$$ \\ $$
Commented by Dwaipayan Shikari last updated on 29/Nov/20
https://brilliant.org/problems/elliptic-integral
Commented by mindispower last updated on 29/Nov/20
we can use (x+1)!!=(x+1)(x−1)!! ⇒1!!=1.(−1)!! ⇒(−1)!!=1 withe that n=0 ((((−1)!!)/(0!)))^2 .((256^0 .π)/2^(2.0+1) )=(π/2) we are donne
$${we}\:{can}\:{use}\:\left({x}+\mathrm{1}\right)!!=\left({x}+\mathrm{1}\right)\left({x}−\mathrm{1}\right)!! \\ $$$$\Rightarrow\mathrm{1}!!=\mathrm{1}.\left(−\mathrm{1}\right)!! \\ $$$$\Rightarrow\left(−\mathrm{1}\right)!!=\mathrm{1} \\ $$$${withe}\:{that} \\ $$$${n}=\mathrm{0} \\ $$$$\left(\frac{\left(−\mathrm{1}\right)!!}{\mathrm{0}!}\right)^{\mathrm{2}} .\frac{\mathrm{256}^{\mathrm{0}} .\pi}{\mathrm{2}^{\mathrm{2}.\mathrm{0}+\mathrm{1}} }=\frac{\pi}{\mathrm{2}} \\ $$$${we}\:{are}\:{donne} \\ $$$$ \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *