Question Number 161822 by HongKing last updated on 22/Dec/21 $$\mathrm{Find}: \\ $$$$\Omega\:=\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\underset{\boldsymbol{\mathrm{k}}=\mathrm{1}} {\overset{\boldsymbol{\mathrm{n}}} {\sum}}\left[\underset{\boldsymbol{\mathrm{i}}=\mathrm{1}} {\overset{\boldsymbol{\mathrm{k}}} {\sum}}\left(\mathrm{i}\:+\:\frac{\mathrm{1}}{\mathrm{4}}\right)\right]^{\:-\mathrm{1}} =\:? \\ $$ Answered by qaz last updated…
Question Number 161817 by HongKing last updated on 22/Dec/21 Answered by aleks041103 last updated on 23/Dec/21 $${f}\left({x}\right)=\left({e}^{−\mathrm{1}} \circ{g}\circ{e}\right)\left({x}\right) \\ $$$${e}\left({x}\right)={x}^{\mathrm{2021}} \\ $$$${g}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{1}−{x}} \\ $$$${e}^{−\mathrm{1}} \left({x}\right)={x}^{\mathrm{1}/\mathrm{2021}}…
Question Number 161818 by HongKing last updated on 22/Dec/21 $$\mathrm{Show}\:\mathrm{that}: \\ $$$$\Phi\:=\underset{\:\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\sqrt{\frac{\mathrm{1}\:-\:\mathrm{x}^{\mathrm{2}} }{\mathrm{1}\:+\:\mathrm{x}^{\mathrm{2}} }}\:\mathrm{dx}\:=\:\frac{\sqrt{\pi}}{\mathrm{4}}\:\left(\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)}{\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)}\:-\:\mathrm{4}\:\frac{\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)}{\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)}\right) \\ $$$$\mathrm{where}:\:\Gamma-\mathrm{Gamma}\:\mathrm{function} \\ $$ Answered by Lordose last updated…
Question Number 30745 by abdo imad last updated on 25/Feb/18 $${let}\:\:{U}_{{n}} =\left\{{z}\in{C}/{z}^{{n}} =\mathrm{1}\right\}\:\:{simlify} \\ $$$${A}_{{n}} =\:\sum_{\alpha\in{U}_{{n}} } \:\left({x}+\alpha\right)^{{n}} \:{and}\:{B}_{{n}} =\sum_{\alpha\in\:{U}_{{n}} } \:\:\left({x}−\alpha\right)^{{n}} . \\ $$…
Question Number 30743 by abdo imad last updated on 25/Feb/18 $${decompose}\:{inside}\:{R}\left[{x}\right]\:\:{p}\left({x}\right)={x}^{\mathrm{2}{n}+\mathrm{1}} \:−\mathrm{1}\:{then}\:{find} \\ $$$$\prod_{{k}=\mathrm{1}} ^{{n}} \:{sin}\left(\:\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)\:. \\ $$ Commented by abdo imad last updated on…
Question Number 30744 by abdo imad last updated on 25/Feb/18 $${let}\:{p}\left({x}\right)=\:\left({x}−\mathrm{1}\right)^{{n}} \:−{x}^{{n}} \:+\mathrm{1}\:\:{with}\:{n}\:{integr}\:{find}\:{n} \\ $$$${in}\:{ordre}\:{that}\:{p}\left({x}\right)\:{have}\:{a}\:{double}\:{root}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 30742 by abdo imad last updated on 25/Feb/18 $${prove}\:{that}\:\forall{p}\:\in{N}\:\:{it}\:{exist}\:{one}\:{polynomial}\:{Q}_{\mathrm{2}{p}} \:/ \\ $$$${sin}\left(\mathrm{2}{p}+\mathrm{1}\right)\theta={sin}^{\mathrm{2}{p}+\mathrm{1}} \theta\:{Q}_{\mathrm{2}{p}} \:\left({cotan}\theta\right)\:{and}\:{degQ}_{\mathrm{2}{p}} =\mathrm{2}{p} \\ $$$$\left.\mathrm{2}\right)\:{prove}\:{that}\:\prod_{{k}=\mathrm{1}} ^{{p}} \:{tan}\left(\frac{{k}\pi}{\mathrm{2}{p}+\mathrm{1}}\right)=\sqrt{\mathrm{2}{p}+\mathrm{1}}\:. \\ $$$$ \\ $$…
Question Number 161815 by HongKing last updated on 22/Dec/21 Answered by Lordose last updated on 23/Dec/21 $$\Omega\left(\mathrm{x}\right)\:=\:\Gamma\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\Gamma\left(\mathrm{1}−\frac{\mathrm{x}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\mathrm{1}−\left(\frac{\mathrm{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)\right)\mathrm{sin}\left(\pi\mathrm{x}\right) \\ $$$$\Omega\left(\mathrm{x}\right)\:=\:\boldsymbol{\pi}\mathrm{csc}\left(\frac{\boldsymbol{\pi}\mathrm{x}}{\mathrm{2}}\right)\centerdot\boldsymbol{\pi}\mathrm{csc}\left(\frac{\boldsymbol{\pi}\left(\mathrm{1}+\mathrm{x}\right)}{\mathrm{2}}\right)\mathrm{sin}\left(\pi\mathrm{x}\right) \\ $$$$\Omega\left(\mathrm{x}\right)\:=\:\mathrm{2}\boldsymbol{\pi}^{\mathrm{2}} \\ $$$$\mathrm{x}^{\mathrm{2}} \:−\:\frac{\mathrm{4x}}{\Omega\left(\mathrm{x}\right)}\:+\:\frac{\mathrm{1}}{\boldsymbol{\pi}^{\mathrm{4}} }\:=\:\mathrm{0}…
Question Number 30739 by abdo imad last updated on 25/Feb/18 $${let}\:\left({u}_{{n}} \right)\:/\:{u}_{\mathrm{1}} =\mathrm{1}−{i}\:{and}\:\:\forall{p}\in\left\{\mathrm{2},\mathrm{3},…{n}\right\}\:{u}_{{p}} ={u}_{{p}−\mathrm{1}} {j}\:{with} \\ $$$${j}={e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \\ $$$$\left.\mathrm{1}\right){verify}\:{that}\:{u}_{\mathrm{1}} \:+{u}_{\mathrm{2}} \:+{u}_{\mathrm{3}} =\mathrm{0} \\ $$$$\left.\mathrm{2}\right){prove}\:{that}\:\forall{p}\in\:\left\{\mathrm{4},\mathrm{5},…,{n}\right\}\:\:{u}_{{p}} ={u}_{{p}−\mathrm{3}}…
Question Number 161800 by mathlove last updated on 22/Dec/21 $$\frac{\mathrm{1}^{\mathrm{2}} \centerdot\mathrm{2}!+\mathrm{2}^{\mathrm{2}} \centerdot\mathrm{3}!+\mathrm{3}^{\mathrm{2}} \centerdot\mathrm{4}!+\centerdot\centerdot\centerdot+{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)!−\mathrm{2}}{\left({n}+\mathrm{1}\right)!}=\mathrm{108} \\ $$$${n}=? \\ $$ Commented by Rasheed.Sindhi last updated on 22/Dec/21…