Question Number 123823 by Dwaipayan Shikari last updated on 28/Nov/20 | ||
Answered by mindispower last updated on 29/Nov/20 | ||
$$\left(\mathrm{1}+{x}\right)^{{a}} =\mathrm{1}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{{n}!}\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({a}−{k}\right).{x}^{{n}} \\ $$$$\left(\mathrm{1}−{ksin}^{\mathrm{2}} \left({x}\right)\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} =\mathrm{1}+\Sigma\frac{\mathrm{1}}{{n}!}\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left(−\frac{\mathrm{1}}{\mathrm{2}}−{k}\right).{sin}^{\mathrm{2}{n}} \left({x}\right) \\ $$$$=\mathrm{1}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!\mathrm{2}^{{n}} }\underset{{k}\leqslant{n}−\mathrm{1}} {\prod}\left(\mathrm{1}+\mathrm{2}{k}\right).{k}^{{n}} {sin}^{\mathrm{2}{n}} \left({x}\right) \\ $$$$=\mathrm{1}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} \left(\mathrm{2}{n}−\mathrm{1}\right)!!}{{n}!.\mathrm{2}^{{n}} }{k}^{{n}} {sin}^{\mathrm{2}{n}} \left({x}\right) \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dx}}{\:\sqrt{\mathrm{1}−{ksin}^{\mathrm{2}} \left({x}\right)}}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{1}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!.\mathrm{2}^{{n}} }\left(\mathrm{2}{n}−\mathrm{1}\right)!!.{k}^{{n}} {sin}^{\mathrm{2}{n}} \left({x}\right)\right){dx} \\ $$$${S}_{{n}} =\frac{\pi}{\mathrm{2}}+\Sigma\frac{\left(−\mathrm{1}\right)^{{n}} \left(\mathrm{2}{n}−\mathrm{1}\right)!!.{k}^{{n}} }{\mathrm{2}^{{n}} {n}!}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{2}{n}} \left({x}\right){dx} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{{a}} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{2}\left(\frac{{a}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)−\mathrm{1}} {cos}^{\mathrm{2}\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} \left({x}\right){dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\beta\left(\frac{{a}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\Gamma\left(\frac{{a}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}\Gamma\left(\mathrm{1}+\frac{{a}}{\mathrm{2}}\right)}\:,\:{valid}\:{for}\:{Re}\left({a}\right)>−\mathrm{1} \\ $$$${s}_{{n}} =\frac{\pi}{\mathrm{2}}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} \left(\mathrm{2}{n}−\mathrm{1}\right)!!}{\mathrm{2}^{{n}} .{n}!}.{k}^{{n}} .\frac{\mathrm{1}}{\mathrm{2}}\frac{\Gamma\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\mathrm{1}+{n}\right)} \\ $$$$\Gamma\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)=\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({n}−\frac{\mathrm{1}}{\mathrm{2}}−{k}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right),\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\sqrt{\pi} \\ $$$$\Gamma\left(\mathrm{1}+{n}\right)={n}! \\ $$$$\Leftrightarrow \\ $$$${S}_{{n}} =\frac{\pi}{\mathrm{2}}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} \left(\mathrm{2}{n}−\mathrm{1}\right)!!}{\mathrm{2}^{{n}} {n}!}{k}^{{n}} .\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}\leqslant{n}−\mathrm{1}} {\prod}\left(\frac{\mathrm{2}{n}−\mathrm{1}−\mathrm{2}{k}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right).\frac{\mathrm{1}}{{n}!} \\ $$$$=\frac{\pi}{\mathrm{2}}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} \left(\mathrm{2}{n}−\mathrm{1}\right)!!.\left(\mathrm{2}{n}−\mathrm{1}\right)!!}{\mathrm{2}^{{n}} {n}!.\mathrm{2}.\mathrm{2}^{{n}} {n}!}{k}^{{n}} \Gamma^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\frac{\pi}{\mathrm{2}}+\underset{{n}\geqslant\mathrm{1}} {\sum}\left(\frac{\left(\mathrm{2}{n}−\mathrm{1}\right)!!}{{n}!}\right)^{\mathrm{2}} \frac{\left(−{k}\right)^{{n}} \pi}{\mathrm{2}^{\mathrm{2}{n}+\mathrm{1}} } \\ $$$${somthing}\:{missing} \\ $$$${not},\underset{{n}\geqslant\mathrm{0}} {\sum}\left(\frac{\left(\mathrm{2}{n}−\mathrm{1}\right)!!}{{n}!}\right)^{\mathrm{2}} .... \\ $$$${by}\:{comparison}\Rightarrow\left(−{k}\right)^{{n}} =\mathrm{256}^{{n}} \\ $$$$\Rightarrow\left(−{k}\right)^{{n}} =\mathrm{256}^{{n}} {e}^{\mathrm{2}{i}\pi{m}} \\ $$$$\Rightarrow−{k}=\mathrm{256}{e}^{\frac{\mathrm{2}{i}\pi{m}}{{n}}} ,\forall{n}\:{tru}\: \\ $$$${n}\rightarrow\infty,−{k}=\mathrm{256}\Rightarrow{k}=−\mathrm{256} \\ $$$$\sqrt{{k}}=\underset{−} {+}\mathrm{16}{i} \\ $$$$ \\ $$$$ \\ $$ | ||
Commented by mnjuly1970 last updated on 29/Nov/20 | ||
$${excellent}\:{sir}\:... \\ $$ | ||
Commented by Dwaipayan Shikari last updated on 29/Nov/20 | ||
$${This}\:{is}\:\:{a}\:{question}\:{from}''\:{Brilliant}\:'' \\ $$$$ \\ $$ | ||
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 | ||
$${we}\:{can}\:{use}\:\left({x}+\mathrm{1}\right)!!=\left({x}+\mathrm{1}\right)\left({x}−\mathrm{1}\right)!! \\ $$$$\Rightarrow\mathrm{1}!!=\mathrm{1}.\left(−\mathrm{1}\right)!! \\ $$$$\Rightarrow\left(−\mathrm{1}\right)!!=\mathrm{1} \\ $$$${withe}\:{that} \\ $$$${n}=\mathrm{0} \\ $$$$\left(\frac{\left(−\mathrm{1}\right)!!}{\mathrm{0}!}\right)^{\mathrm{2}} .\frac{\mathrm{256}^{\mathrm{0}} .\pi}{\mathrm{2}^{\mathrm{2}.\mathrm{0}+\mathrm{1}} }=\frac{\pi}{\mathrm{2}} \\ $$$${we}\:{are}\:{donne} \\ $$$$ \\ $$ | ||