Question Number 153988 by mathdanisur last updated on 12/Sep/21 $$\mathrm{let}\:\:\mathrm{a};\mathrm{b}\:\:\mathrm{be}\:\mathrm{positive}\:\mathrm{real}\:\mathrm{numbers} \\ $$$$\mathrm{such}\:\mathrm{that}\:\:\mathrm{a}+\mathrm{b}=\mathrm{2}\:\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\frac{\mathrm{1}}{\mathrm{a}^{\boldsymbol{\mathrm{n}}} }\:+\:\frac{\mathrm{1}}{\mathrm{b}^{\boldsymbol{\mathrm{n}}} }\:\geqslant\:\mathrm{a}^{\boldsymbol{\mathrm{n}}+\mathrm{1}} \:+\:\mathrm{b}^{\boldsymbol{\mathrm{n}}+\mathrm{1}} \:\:;\:\:\forall\mathrm{n}\in\mathbb{N}^{\ast} \\ $$ Answered by metamorfose last updated…
Question Number 153989 by mathdanisur last updated on 12/Sep/21 $$\mathrm{find}\:\mathrm{all}\:\mathrm{functions}\:\:\mathrm{f}\::\:\mathbb{R}\rightarrow\mathbb{R}\:\:\mathrm{with}\:\mathrm{the} \\ $$$$\mathrm{property}\:\mathrm{that} \\ $$$$\mathrm{f}\left(\mathrm{f}\left(\mathrm{x}\right)\:+\:\mathrm{2y}\:=\:\mathrm{10x}\:+\:\mathrm{f}\left(\mathrm{f}\left(\mathrm{y}\right)-\mathrm{3x}\right)\right. \\ $$$$\mathrm{holds}\:\mathrm{for}\:\mathrm{all}\:\:\mathrm{a};\mathrm{b}\in\mathbb{R} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
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Question Number 22856 by FilupS last updated on 23/Oct/17 $$\mathrm{solve}:\:\mathrm{sin}\left({x}\right)=\mathrm{2},\:\:\:{x}\in\mathbb{C} \\ $$ Commented by Tinkutara last updated on 23/Oct/17 $${Write}\:{as}\:{e}^{{i}\theta} =\mathrm{cos}\:\theta+{i}\mathrm{sin}\:\theta\:{and} \\ $$$${e}^{−{i}\theta} =\mathrm{cos}\:\theta−{i}\mathrm{sin}\:\theta \\…
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Question Number 88372 by mr W last updated on 10/Apr/20 $${There}\:{are}\:{four}\:{boxes},\:{each}\:{of}\:{them} \\ $$$${contains}\:{exactly}\:{the}\:{same}\:{numbers}: \\ $$$$\mathrm{1},\mathrm{2},\mathrm{3},…,{n}. \\ $$$${Four}\:{different}\:{numbers}\:{are}\:{drawn} \\ $$$${from}\:{the}\:{boxes}\:{and}\:{multiplicated} \\ $$$${with}\:{each}\:{other}\:{to}\:{get}\:{a}\:{product}. \\ $$$${What}'{s}\:{the}\:{sum}\:{of}\:{all}\:{products}? \\ $$$$\underset{{a}\neq{b}\neq{c}\neq{d}}…
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Question Number 153897 by mathdanisur last updated on 11/Sep/21 $$\mathrm{Denote}\:\:\mathrm{x}_{\boldsymbol{\mathrm{n}}} \:\:\mathrm{is}\:\mathrm{the}\:\mathrm{unique}\:\mathrm{positive}\:\mathrm{root} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{following}\:\mathrm{equation}: \\ $$$$\mathrm{x}^{\boldsymbol{\mathrm{n}}} \:+\:\mathrm{x}^{\boldsymbol{\mathrm{n}}−\mathrm{1}} \:+\:…\:\mathrm{x}\:=\:\mathrm{n}\:+\:\mathrm{2} \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{sequence}\:\left(\mathrm{x}_{\boldsymbol{\mathrm{n}}} \right)\:\mathrm{converges} \\ $$$$\mathrm{to}\:\mathrm{a}\:\mathrm{positive}\:\mathrm{real}\:\mathrm{number}.\:\mathrm{Find}\:\mathrm{that} \\ $$$$\mathrm{limit}. \\…
Question Number 88360 by jagoll last updated on 10/Apr/20 $$\frac{\mathrm{e}}{\:\sqrt{\mathrm{e}}}\:×\:\frac{\sqrt[{\mathrm{3}\:\:}]{\mathrm{e}}}{\:\sqrt[{\mathrm{4}\:\:}]{\mathrm{e}}}\:×\:\frac{\sqrt[{\mathrm{5}\:\:}]{\mathrm{e}}}{\:\sqrt[{\mathrm{6}\:\:}]{\mathrm{e}}}\:×\:\frac{\sqrt[{\mathrm{7}\:\:}]{\mathrm{e}}}{\:\sqrt[{\mathrm{8}\:\:}]{\mathrm{e}}}×…=? \\ $$ Commented by john santu last updated on 10/Apr/20 $$=\:{e}^{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{6}}+…} \\ $$$$\left[\:\mathrm{ln}\:\left(\mathrm{1}+\mathrm{x}\right)\:=\:\mathrm{x}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{3}}−\frac{\mathrm{x}^{\mathrm{4}}…
Question Number 153899 by mathdanisur last updated on 11/Sep/21 $$\mathrm{Determine}\:\mathrm{whether}\:\mathrm{there}\:\mathrm{exists}\:\:\mathrm{2016} \\ $$$$\mathrm{distinct}\:\mathrm{prime}\:\mathrm{numbers}\:\:\mathrm{p}_{\mathrm{1}} ,\mathrm{p}_{\mathrm{2}} ,…,\mathrm{p}_{\mathrm{2016}} \\ $$$$\mathrm{and}\:\mathrm{positive}\:\mathrm{integer}\:\:\boldsymbol{\mathrm{n}}\:\:\mathrm{such}\:\mathrm{that}: \\ $$$$\underset{\boldsymbol{\mathrm{i}}=\mathrm{1}} {\overset{\mathrm{2016}} {\sum}}\:\frac{\mathrm{1}}{\mathrm{p}_{\boldsymbol{\mathrm{i}}} ^{\mathrm{2}} \:+\:\mathrm{1}}\:=\:\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} } \\ $$…