Question Number 151699 by iloveisrael last updated on 22/Aug/21 $$\:\:\:\:\underset{{x}^{\mathrm{3}} +\mathrm{2022}{x}−\mathrm{2021}=\mathrm{0}} {\sum}\left(\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\right)\:=? \\ $$ Answered by mr W last updated on 22/Aug/21 $${x}^{\mathrm{3}} +\mathrm{2022}{x}−\mathrm{2021}=\mathrm{0} \\…
Question Number 20619 by Tinkutara last updated on 29/Aug/17 $${Solve}\:{the}\:{equation}\:{z}^{{n}−\mathrm{1}} \:=\:\bar {{z}}\:\left({n}\:\in\:{N}\right) \\ $$ Answered by ajfour last updated on 29/Aug/17 $$\mid{z}\mid=\mathrm{1}\: \\ $$$$\:{z}^{{n}} =\mid{z}\mid^{\mathrm{2}}…
Question Number 151673 by mathdanisur last updated on 22/Aug/21 Answered by Kamel last updated on 22/Aug/21 $${L}=\underset{{n}\rightarrow+\infty} {{lim}}\underset{{k}={n}} {\overset{\mathrm{2}{n}} {\prod}}\frac{\pi}{\pi−{Arctan}\left(\frac{\mathrm{1}}{{k}}\right)}=\underset{{n}\rightarrow+\infty} {{lim}}\underset{{k}={n}} {\overset{\mathrm{2}{n}} {\prod}}\frac{\pi{k}}{\pi{k}−\mathrm{1}} \\ $$$$\:\:\:=\underset{{n}\rightarrow+\infty}…
Question Number 86141 by TawaTawa1 last updated on 27/Mar/20 $$\mathrm{A}\:\mathrm{number}\:\mathrm{n}\:\mathrm{leaves}\:\mathrm{a}\:\mathrm{remainder}\:\mathrm{of}\:\:\mathrm{22}\:\:\mathrm{when}\:\mathrm{divided}\:\mathrm{by}\:\mathrm{24}\:\mathrm{and} \\ $$$$\mathrm{remainder}\:\:\mathrm{30}\:\:\mathrm{when}\:\mathrm{divided}\:\mathrm{by}\:\:\mathrm{33}.\:\:\mathrm{Find}\:\mathrm{the}\:\mathrm{least}\:\mathrm{possible} \\ $$$$\mathrm{value}\:\mathrm{of}\:\:\mathrm{n} \\ $$ Commented by mr W last updated on 27/Mar/20 $${there}\:{is}\:{no}\:{such}\:{number}!…
Question Number 151660 by mathdanisur last updated on 22/Aug/21 Answered by Olaf_Thorendsen last updated on 22/Aug/21 $$\mathrm{ln}\left({e}+\mathrm{sin}{kx}\right)\:=\:\mathrm{1}+\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{sin}{kx}}{{e}}\right)\:\underset{\mathrm{0}} {\sim}\:\mathrm{1}+\frac{{kx}}{{e}} \\ $$$$\frac{\mathrm{1}−\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\mathrm{ln}\left({e}+\mathrm{sin}{kx}\right)}{{x}}\:\underset{\mathrm{0}} {\sim}\:\frac{\mathrm{1}−\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\left(\mathrm{1}+\frac{{kx}}{{e}}\right)}{{x}}\:\:\left(\mathrm{1}\right)…
Question Number 151636 by Tawa11 last updated on 22/Aug/21 Commented by mr W last updated on 22/Aug/21 $${do}\:{you}\:{know}\:{the}\:{psi}−{function}\:{and} \\ $$$$\psi\left(\mathrm{1}+{z}\right)=−\gamma+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}+{z}}\right)\:? \\ $$$${if}\:{not},\:{then}\:{you}\:{should}\:{learn}\:{this}\:{at} \\…
Question Number 151638 by mathdanisur last updated on 22/Aug/21 $$\underset{\:\mathrm{0}} {\overset{\:\mathrm{2}\boldsymbol{\pi}} {\int}}\left(\mathrm{1}\:-\:\mathrm{cos}\boldsymbol{\mathrm{x}}\right)^{\mathrm{10}} \:\mathrm{cos}\left(\mathrm{10x}\right)\:\mathrm{dx}\:=\:? \\ $$ Answered by Olaf_Thorendsen last updated on 22/Aug/21 $$\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \left(\mathrm{1}−\mathrm{cos}{x}\right)^{\mathrm{10}}…
Question Number 20551 by Tinkutara last updated on 28/Aug/17 $${Find}\:{the}\:{minimum}\:{value}\:{of} \\ $$$$\mid{a}\:+\:{b}\omega\:+\:{c}\omega^{\mathrm{2}} \mid,\:{where}\:{a},\:{b}\:{and}\:{c}\:{are}\:{all} \\ $$$${not}\:{equal}\:{integers}\:{and}\:\omega\left(\neq\mathrm{1}\right)\:{is}\:{a}\:{cube} \\ $$$${root}\:{of}\:{unity}. \\ $$ Commented by ajfour last updated on…
Question Number 20552 by ajfour last updated on 28/Aug/17 $${The}\:{roots}\:{of}\:{the}\:{equation}\: \\ $$$$\:\left(\mathrm{3}−{x}\right)^{\mathrm{4}} +\left(\mathrm{2}−{x}\right)^{\mathrm{4}} =\left(\mathrm{5}−\mathrm{2}{x}\right)^{\mathrm{4}} \:{are} \\ $$$$\left({a}\right)\:{all}\:{real}\:\:\:\:\left({b}\right)\:{all}\:{imaginary} \\ $$$$\left({c}\right)\:{two}\:{real}\:{and}\:{two}\:{imaginary} \\ $$$$\left({d}\right){none}\:{of}\:{the}\:{above}\:. \\ $$ Answered by…
Show-that-if-z-1-z-2-z-3-z-4-0-and-z-1-z-2-0-then-the-complex-numbers-z-1-z-2-z-3-z-4-are-concyclic-
Question Number 20549 by Tinkutara last updated on 28/Aug/17 $${Show}\:{that}\:{if}\:{z}_{\mathrm{1}} {z}_{\mathrm{2}} \:+\:{z}_{\mathrm{3}} {z}_{\mathrm{4}} \:=\:\mathrm{0}\:{and}\:{z}_{\mathrm{1}} \:+ \\ $$$${z}_{\mathrm{2}} \:=\:\mathrm{0},\:{then}\:{the}\:{complex}\:{numbers}\:{z}_{\mathrm{1}} , \\ $$$${z}_{\mathrm{2}} ,\:{z}_{\mathrm{3}} ,\:{z}_{\mathrm{4}} \:{are}\:{concyclic}. \\…