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Question Number 123823 by Dwaipayan Shikari last updated on 28/Nov/20

Answered by mindispower last updated on 29/Nov/20

$${(1+x)}^a=1+\sum _{n⩾1}\frac{1}{n!}\underset{k=0}{\prod ^{n-1}}(a-k).x^n$$$${(1-{ksin}^2\left(x\right))}^{-\frac{1}{2}}=1+\mathrm{\Sigma }\frac{1}{n!}\underset{k=0}{\prod ^{n-1}}(-\frac{1}{2}-k).{sin}^{2n}\left(x\right)$$$$=1+\sum _{n⩾1}\frac{{(-1)}^n}{n!2^n}\prod _{k⩽n-1}(1+2k).k^n{sin}^{2n}\left(x\right)$$$$=1+\sum _{n⩾1}\frac{{(-1)}^n(2n-1)!!}{n!.2^n}{k}^n{sin}^{2n}\left(x\right)$$$${\int }_0^{\frac{\pi }{2}}\frac{dx}{\sqrt{1-{ksin}^2\left(x\right)}}={\int }_0^{\frac{\pi }{2}}(1+\sum _{n⩾1}\frac{{(-1)}^n}{n!.2^n}(2n-1)!!.k^n{sin}^{2n}\left(x\right))dx$$$$S_n=\frac{\pi }{2}+\mathrm{\Sigma }\frac{{(-1)}^n(2n-1)!!.k^n}{2^{n}n!}{\int }_0^{\frac{\pi }{2}}{sin}^{2n}\left(x\right)dx$$$${\int }_0^{\frac{\pi }{2}}{sin}^a\left(x\right)=\frac{1}{2}.2{\int }_0^{\frac{\pi }{2}}{sin}^{2(\frac{a}{2}+\frac{1}{2})-1}{cos}^{2\frac{1}{2}-1}\left(x\right)dx$$$$\frac{1}{2}\beta (\frac{a}{2}+\frac{1}{2},\frac{1}{2})=\frac{\mathrm{\Gamma }(\frac{a}{2}+\frac{1}{2})\mathrm{\Gamma }\left(\frac{1}{2}\right)}{2\mathrm{\Gamma }(1+\frac{a}{2})},validforRe\left(a\right)>-1$$$$s_n=\frac{\pi }{2}+\sum _{n⩾1}\frac{{(-1)}^n(2n-1)!!}{2^n.n!}.k^n.\frac{1}{2}\frac{\mathrm{\Gamma }(n+\frac{1}{2})\mathrm{\Gamma }\left(\frac{1}{2}\right)}{\mathrm{\Gamma }(1+n)}$$$$\mathrm{\Gamma }(n+\frac{1}{2})=\underset{k=0}{\prod ^{n-1}}(n-\frac{1}{2}-k)\mathrm{\Gamma }\left(\frac{1}{2}\right),\mathrm{\Gamma }\left(\frac{1}{2}\right)=\sqrt{\pi }$$$$\mathrm{\Gamma }(1+n)=n!$$$$\iff$$$$S_n=\frac{\pi }{2}+\sum _{n⩾1}\frac{{(-1)}^n(2n-1)!!}{2^{n}n!}{k}^n.\frac{1}{2}\prod _{k⩽n-1}\left(\frac{2n-1-2k}{2}\right)\mathrm{\Gamma }\left(\frac{1}{2}\right)\mathrm{\Gamma }\left(\frac{1}{2}\right).\frac{1}{n!}$$$$=\frac{\pi }{2}+\sum _{n⩾1}\frac{{(-1)}^n(2n-1)!!.(2n-1)!!}{2^{n}n!.2.2^{n}n!}{k}^n{\mathrm{\Gamma }}^2\left(\frac{1}{2}\right)$$$$=\frac{\pi }{2}+\sum _{n⩾1}{\left(\frac{(2n-1)!!}{n!}\right)}^2\frac{{(-k)}^n\pi }{2^{2n+1}}$$$$somthingmissing$$$$not,\sum _{n⩾0}{\left(\frac{(2n-1)!!}{n!}\right)}^2....$$$$bycomparison\Rightarrow {(-k)}^n={256}^n$$$$\Rightarrow {(-k)}^n={256}^n{e}^{2i\pi m}$$$$\Rightarrow -k=256e^{\frac{2i\pi m}{n}},\mathrm{\forall }ntru$$$$n\to \mathrm{\infty },-k=256\Rightarrow k=-256$$$$\sqrt{k}=\underset{-}{+}16i$$$$$$$$$$

Commented by mnjuly1970 last updated on 29/Nov/20

$$excellentsir...$$

Commented by Dwaipayan Shikari last updated on 29/Nov/20

$$Thisisaquestion{from}^{″}Brilliant{}^{″}$$$$$$

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

$$wecanuse(x+1)!!=(x+1)(x-1)!!$$$$\Rightarrow 1!!=1.(-1)!!$$$$\Rightarrow (-1)!!=1$$$$withethat$$$$n=0$$$${\left(\frac{(-1)!!}{0!}\right)}^2.\frac{{256}^0.\pi }{2^{2.0+1}}=\frac{\pi }{2}$$$$wearedonne$$$$$$

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