Question Number 61625 by Sharath Kumar last updated on 05/Jun/19 $$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+…}}}}}= \\ $$ Answered by MJS last updated on 05/Jun/19 $${x}=\mathrm{1}+\frac{\mathrm{1}}{{x}}\:\wedge\:{x}>\mathrm{1}\:\Rightarrow\:{x}=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$ Commented by…
Question Number 192687 by Mastermind last updated on 24/May/23 $$\frac{\mathrm{1}}{\pi}\int_{\mathrm{0}} ^{\mathrm{2}\pi} \mathrm{xcos}\left(\mathrm{nx}\right)\mathrm{dx} \\ $$$$ \\ $$$$\mathrm{Help}! \\ $$ Answered by Subhi last updated on 24/May/23…
Question Number 192648 by pascal889 last updated on 24/May/23 $$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{g}}\left(\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)\right) \\ $$$$\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)=\boldsymbol{\mathrm{x}}^{\mathrm{3}} −\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{3}\boldsymbol{\mathrm{x}} \\ $$$$\boldsymbol{\mathrm{g}}\left(\boldsymbol{\mathrm{x}}\right)=\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\mathrm{2}\boldsymbol{\mathrm{x}}+\mathrm{1} \\ $$ Answered by Skabetix last updated on…
Question Number 192636 by pascal889 last updated on 23/May/23 $${lim}_{{x}\rightarrow\mathrm{1}\:} \left(\mathrm{3}{x}^{\mathrm{2}} \:−\mathrm{7}{x}+\mathrm{3}\right)^{\mathrm{10}} \: \\ $$$${find}\:{the}\:{limit} \\ $$ Answered by Subhi last updated on 24/May/23 $$\mathrm{1}…
Question Number 127080 by Dwaipayan Shikari last updated on 26/Dec/20 $$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{e}^{−\phi{n}} +\frac{{e}^{\mathrm{2}\pi{n}} −{e}^{−\mathrm{2}\phi{n}} \:}{\mathrm{2}{e}^{−\phi{n}} +\frac{{e}^{\mathrm{2}\pi{n}} −{e}^{−\mathrm{2}\phi{n}} }{\mathrm{2}{e}^{−\phi{n}} +\frac{{e}^{\mathrm{2}\pi{n}} −{e}^{−\mathrm{2}\phi{n}} }{\mathrm{2}{e}^{−\mathrm{2}\phi{n}} …}}}} \\ $$…
Question Number 61510 by Tony Lin last updated on 03/Jun/19 $${S}_{\mathrm{1}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\sqrt{\left(\mathrm{16}{n}−\mathrm{16}{k}\right)\left(\mathrm{16}{n}+\mathrm{16}{k}\right)} \\ $$$${S}_{\mathrm{2}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\sqrt{\left(\mathrm{16}{k}−\mathrm{16}\right)\left(\mathrm{16}{k}+\mathrm{16}\right)} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{S}_{\mathrm{1}} +{S}_{\mathrm{2}} }{{n}^{\mathrm{2}} }=?…
Question Number 192570 by mechanics last updated on 21/May/23 Answered by cortano12 last updated on 21/May/23 $$\:\mid\mathrm{x}^{\mathrm{2}} −\mathrm{4}\mid<\mathrm{5} \\ $$$$\:\left(\mathrm{x}^{\mathrm{2}} −\mathrm{9}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)<\mathrm{0} \\ $$$$\:\left(\mathrm{x}+\mathrm{3}\right)\left(\mathrm{x}−\mathrm{3}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)<\mathrm{0}…
Question Number 192544 by Mastermind last updated on 20/May/23 $$\mathrm{If}\:\mathrm{f}\left(\mathrm{t}\right)\:=\:\mathrm{2}\left(\mathrm{e}^{\mathrm{t}} \:−\:\mathrm{1}\right) \\ $$$$\left.\mathrm{a}\right)\:\mathrm{Does}\:\mathrm{f}\left(\mathrm{t}\right)\:\mathrm{exists}\:\mathrm{for}\:\mathrm{all}\:\mathrm{n}\:? \\ $$$$\left.\mathrm{b}\right)\:\mathrm{If}\:\mathrm{it}\:\mathrm{exist},\:\mathrm{does}\:\mathrm{it}\:\mathrm{converge}\:? \\ $$$$\left.\mathrm{c}\right)\:\mathrm{If}\:\mathrm{the}\:\mathrm{sequence}\:\mathrm{converge},\:\mathrm{does}\:\mathrm{the} \\ $$$$\mathrm{limit}\:\mathrm{converge}\:? \\ $$$$\left.\mathrm{d}\right)\:\mathrm{Is}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{uniques}\:? \\ $$ Commented by…
Question Number 127004 by Dastronomer last updated on 26/Dec/20 Commented by Tinku Tara last updated on 26/Dec/20 http://www.tinkutara.com/equationeditorhelp.html Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 126977 by Dwaipayan Shikari last updated on 25/Dec/20 $$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}^{\mathrm{7}} }{\mathrm{7}^{{n}} } \\ $$ Commented by MJS_new last updated on 25/Dec/20 $$=\frac{\mathrm{285929}}{\mathrm{11664}}…