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Author: Tinku Tara

What-s-the-value-for-7-

Question Number 196205 by Mastermind last updated on 19/Aug/23 $$\mathrm{What}'\mathrm{s}\:\mathrm{the}\:\mathrm{value}\:\mathrm{for}\:!\mathrm{7}\:? \\ $$ Answered by AST last updated on 19/Aug/23 $$!\mathrm{7}=\mathrm{7}!\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}!}+\frac{\mathrm{1}}{\mathrm{2}!}−\frac{\mathrm{1}}{\mathrm{3}!}+\frac{\mathrm{1}}{\mathrm{4}!}−\frac{\mathrm{1}}{\mathrm{5}!}+\frac{\mathrm{1}}{\mathrm{6}!}−\frac{\mathrm{1}}{\mathrm{7}!}\right)=\mathrm{1854} \\ $$ Commented by AST…

Calcul-I-a-a-1-a-lnx-1-x-2-dx-a-gt-0-

Question Number 196173 by pticantor last updated on 19/Aug/23 $$\boldsymbol{{Calcul}}\:\boldsymbol{{I}}_{\boldsymbol{{a}}} =\int_{\boldsymbol{{a}}} ^{\frac{\mathrm{1}}{\boldsymbol{{a}}}} \frac{\boldsymbol{{lnx}}}{\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}} }\boldsymbol{{dx}}\:\left(\boldsymbol{{a}}>\mathrm{0}\right)\: \\ $$ Answered by Erico last updated on 19/Aug/23 $$\:\mathrm{J}_{{a}} =\:\underset{\:\mathrm{1}}…

Solve-y-3y-y-2-1-y-2-

Question Number 196199 by Frix last updated on 19/Aug/23 $$\mathrm{Solve}: \\ $$$${y}'''=\frac{\mathrm{3}{y}'\left({y}''\right)^{\mathrm{2}} }{\mathrm{1}+\left({y}'\right)^{\mathrm{2}} } \\ $$ Answered by aleks041103 last updated on 20/Aug/23 $$\left({ln}\left({y}''\right)\right)'=\frac{{y}'''}{{y}''}=\frac{\mathrm{3}{y}'{y}''}{\mathrm{1}+\left({y}'\right)^{\mathrm{2}} }=\frac{\mathrm{3}}{\mathrm{2}}\:\frac{\left(\left({y}'\right)^{\mathrm{2}}…

for-any-a-b-1-1-calculate-arccos-a-arcsin-b-arcsin-a-arcsin-b-arccos-a-arccos-b-

Question Number 196166 by pticantor last updated on 19/Aug/23 $$\boldsymbol{{for}}\:\boldsymbol{{any}}\:\boldsymbol{{a}},\boldsymbol{{b}}\in\left[−\mathrm{1},\mathrm{1}\right]\:\:\boldsymbol{{calculate}}:\: \\ $$$$\boldsymbol{{arccos}}\left(\boldsymbol{{a}}\right)+\boldsymbol{{arcsin}}\left(\boldsymbol{{b}}\right)=? \\ $$$$\boldsymbol{{arcsin}}\left(\boldsymbol{{a}}\right)+\boldsymbol{{arcsin}}\left(\boldsymbol{{b}}\right)=? \\ $$$$\boldsymbol{{arccos}}\left(\boldsymbol{{a}}\right)+\boldsymbol{{arc}}{cos}\left(\boldsymbol{{b}}\right)=? \\ $$ Commented by pticantor last updated on 19/Aug/23…

Question-196094

Question Number 196094 by leicianocosta last updated on 18/Aug/23 Answered by dimentri last updated on 18/Aug/23 $$\left(\mathrm{1}\right)\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{2}^{\mathrm{3}\left({n}−\mathrm{1}\right)} }{\mathrm{3}^{\mathrm{2}{n}} }\right)=\:\frac{\mathrm{1}}{\mathrm{9}}+\frac{\mathrm{8}}{\mathrm{81}}+\frac{\mathrm{64}}{\mathrm{729}}+\ldots \\ $$$$\:\:\:\:\:\:=\:\frac{\frac{\mathrm{1}}{\mathrm{9}}}{\mathrm{1}−\frac{\mathrm{8}}{\mathrm{9}}}\:=\:\mathrm{1}\: \\ $$…