Question Number 206885 by efronzo1 last updated on 29/Apr/24 Answered by BaliramKumar last updated on 29/Apr/24 $${x}\:=\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{5}}{\mathrm{8}}\right)\:\approx\:\mathrm{32}° \\ $$ Answered by Rasheed.Sindhi last updated…
Question Number 206881 by efronzo1 last updated on 29/Apr/24 $$\:\:\cancel{\underline{\underbrace{\boldsymbol{{x}}}}} \\ $$ Answered by Skabetix last updated on 29/Apr/24 $$\sum_{{n}=\mathrm{0}} ^{\infty} \frac{{x}^{{n}} }{{n}!}={e}^{{x}} \rightarrow{here}\:{x}\:=\:\mathrm{1}\:{so}\:\mathrm{2}\:+\:\frac{\mathrm{1}}{\mathrm{2}!}+…={e} \\…
Question Number 206882 by efronzo1 last updated on 29/Apr/24 $$\:\:{f}\left({x}\right)=\mathrm{tan}\:^{\mathrm{2}} {x}\:\sqrt{\mathrm{tan}\:{x}\sqrt[{\mathrm{3}}]{\mathrm{tan}\:{x}\sqrt[{\mathrm{4}}]{\mathrm{tan}\:{x}\sqrt[{\mathrm{5}}]{\mathrm{tan}\:{x}\sqrt{…}}}}} \\ $$$$\:{f}\:'\left(\frac{\pi}{\mathrm{4}}\right)=? \\ $$ Answered by MM42 last updated on 29/Apr/24 $${f}\left({x}\right)={tan}^{\mathrm{2}} {x}×{tan}^{\frac{\mathrm{1}}{\mathrm{2}}} {x}×{tan}^{\frac{\mathrm{1}}{\mathrm{6}}}…
Question Number 206861 by MrGHK last updated on 28/Apr/24 $$\underset{\boldsymbol{\mathrm{k}}=\mathrm{1}} {\overset{\infty} {\boldsymbol{\sum}}}\frac{\mathrm{1}}{\boldsymbol{\mathrm{k}}^{\mathrm{2}} }\underset{\boldsymbol{\mathrm{n}}=\mathrm{0}} {\overset{\infty} {\boldsymbol{\sum}}}\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{1}+\boldsymbol{\mathrm{n}}} }\left(\frac{\boldsymbol{\Gamma}\left(\boldsymbol{\mathrm{n}}+\mathrm{1}\right)\boldsymbol{\Gamma}\left(\boldsymbol{\mathrm{k}}+\mathrm{1}\right)\boldsymbol{\mathrm{H}}_{\boldsymbol{\mathrm{n}}+\boldsymbol{\mathrm{k}}+\mathrm{1}} }{\boldsymbol{\Gamma}\left(\boldsymbol{\mathrm{n}}+\boldsymbol{\mathrm{k}}+\mathrm{2}\right)}\right)=??? \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 206862 by hardmath last updated on 28/Apr/24 Commented by SWPlaysMC last updated on 29/Apr/24 $$\mathrm{Easiest}\:\mathrm{way}\:\mathrm{I}\:\mathrm{thought}\:\mathrm{of}\:\mathrm{proving}\:\mathrm{this}\:\left(\mathrm{I}\:\mathrm{partially}\:\mathrm{solved}\:\mathrm{this}\right) \\ $$$$\mathrm{Let}\:{a}=\mathrm{1},\:{b}=\mathrm{1},\:\mathrm{and}\:{c}=\mathrm{1} \\ $$$$\left(\mathrm{Hint}:\:\mathrm{if}\:\mathrm{any}\:\mathrm{one}\:\mathrm{of}\:\mathrm{these}\:\mathrm{is}\:\mathrm{0}\:\mathrm{then}\:\mathrm{expression}\:\mathrm{is}\:\mathrm{theoretically}\:\mathrm{undefined}\:\mathrm{due}\:\mathrm{to}\:\boldsymbol{\div}\:\mathrm{by}\:\mathrm{0}\right) \\ $$$$\frac{\sqrt{\frac{\mathrm{1}}{\mathrm{1}}}+\frac{\mathrm{1}}{\mathrm{1}}}{\:\sqrt{\frac{\mathrm{1}}{\mathrm{1}}}+\sqrt{\frac{\mathrm{1}}{\mathrm{1}}}}+\frac{\sqrt{\frac{\mathrm{1}}{\mathrm{1}}}+\frac{\mathrm{1}}{\mathrm{1}}}{\:\sqrt{\frac{\mathrm{1}}{\mathrm{1}}}+\sqrt{\frac{\mathrm{1}}{\mathrm{1}}}}+\frac{\sqrt{\frac{\mathrm{1}}{\mathrm{1}}}+\frac{\mathrm{1}}{\mathrm{1}}}{\:\sqrt{\frac{\mathrm{1}}{\mathrm{1}}}+\sqrt{\frac{\mathrm{1}}{\mathrm{1}}}}\geqslant\mathrm{3} \\ $$$$\mathrm{so}\:\mathrm{all}\:\sqrt{\frac{\mathrm{1}}{\mathrm{1}}}=\mathrm{1}\:\mathrm{and}\:\frac{\mathrm{1}}{\mathrm{1}}=\mathrm{1},\:\mathrm{thus}\:\mathrm{substitute}\:\mathrm{all}\:\mathrm{nums}\:\mathrm{and}\:\mathrm{denoms}\:\mathrm{with}\:\mathrm{1}+\mathrm{1}\:\mathrm{like}\:\mathrm{so}…
Question Number 206873 by lmcp1203 last updated on 29/Apr/24 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 206874 by lmcp1203 last updated on 29/Apr/24 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 206868 by depressiveshrek last updated on 28/Apr/24 $$\mathrm{If}\:{A},\:{B}\:\mathrm{and}\:{A}+{B}\:\mathrm{are}\:\mathrm{non}−\mathrm{singular} \\ $$$$\mathrm{square}\:\mathrm{matrices},\:\mathrm{prove}\:\mathrm{that}\:{A}^{−\mathrm{1}} +{B}^{−\mathrm{1}} \\ $$$$\mathrm{is}\:\mathrm{also}\:\mathrm{non}−\mathrm{singular}. \\ $$ Answered by aleks041103 last updated on 28/Apr/24 $${A}^{−\mathrm{1}}…
Question Number 206869 by universe last updated on 28/Apr/24 Commented by Frix last updated on 28/Apr/24 $$\mathrm{Answer}\:\mathrm{is}\:\mathrm{e}^{{a}_{\mathrm{1}} } −\mathrm{1}\:\mathrm{for}\:{a}_{\mathrm{1}} \in\mathbb{R}\backslash\left\{\mathrm{0}\right\} \\ $$ Commented by universe…
Question Number 206845 by hardmath last updated on 27/Apr/24 $$\mathrm{Find}:\:\:\:\frac{\infty!}{\infty^{\infty} }\:=\:? \\ $$ Commented by A5T last updated on 27/Apr/24 $${Do}\:{you}\:{wish}\:{to}\:{find}\:{this}:\:\underset{{n}\rightarrow\infty} {{lim}}\frac{{n}!}{{n}^{{n}} }\:? \\ $$…