Question Number 3250 by Filup last updated on 08/Dec/15 $$\mathrm{It}\:\mathrm{is}\:\mathrm{known}\:\mathrm{that}: \\ $$$$\zeta\left({s}\right)=\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}{i}^{−{s}} \\ $$$$ \\ $$$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\zeta\left({s}\right)=\underset{{i}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\mathrm{1}−\frac{\mathrm{1}}{\pi\left({i}\right)^{{s}} }\right) \\ $$$$\mathrm{where}\:\pi\left({n}\right)={n}\mathrm{th}\:\mathrm{prime}…
Question Number 134097 by bobhans last updated on 27/Feb/21 $$\mathrm{In}\:\mathrm{a}\:\mathrm{square}\:\mathrm{ABCD}\:,\:\mathrm{a}\:\mathrm{triangle} \\ $$$$\mathrm{APQ}\:\mathrm{inscribed}\:\mathrm{in}\:\mathrm{it}.\:\mathrm{AP}=\mathrm{4}\:\mathrm{cm}, \\ $$$$\mathrm{PQ}=\mathrm{3}\:\mathrm{cm}\:\mathrm{and}\:\mathrm{AQ}=\mathrm{5}\:\mathrm{cm}.\:\mathrm{Point} \\ $$$$\mathrm{P}\:\mathrm{is}\:\mathrm{on}\:\mathrm{the}\:\mathrm{side}\:\mathrm{BC}\:\mathrm{and}\:\mathrm{point}\:\mathrm{Q} \\ $$$$\mathrm{is}\:\mathrm{on}\:\mathrm{side}\:\mathrm{CD}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{square}\:\mathrm{ABCD}. \\ $$ Answered by mr…
Question Number 132230 by pticantor last updated on 12/Feb/21 $$ \\ $$$$ \\ $$$$\boldsymbol{{S}}=\underset{\boldsymbol{{k}}=\mathrm{1}} {\overset{+\infty} {\sum}}\left(\frac{\mathrm{1}}{\:\sqrt{\boldsymbol{{k}}^{\mathrm{2}} −\mathrm{1}}}−\frac{\mathrm{1}}{\:\sqrt{\boldsymbol{{k}}^{\mathrm{2}} +\mathrm{1}}}\right) \\ $$$$\boldsymbol{{S}}\:={l}\:{or}\:\boldsymbol{{S}}=\infty\:\:??? \\ $$ Answered by JDamian…
Question Number 131847 by pticantor last updated on 09/Feb/21 $$\:\boldsymbol{{Let}}\:\left(\boldsymbol{\Omega},\digamma,\boldsymbol{{P}}\right)\:\boldsymbol{{be}}\:\boldsymbol{{a}}\:\boldsymbol{{probalistics}}\:\boldsymbol{{space}}\: \\ $$$$\boldsymbol{{show}}\:\boldsymbol{{that}},\: \\ $$$$\boldsymbol{{A}},\boldsymbol{{B}}\in\digamma \\ $$$$\boldsymbol{{if}}\:\left(\boldsymbol{{P}}\left(\boldsymbol{{A}}\mid\boldsymbol{{B}}\right)\leqslant{P}\left(\boldsymbol{{A}}\right)\:\boldsymbol{{and}}\:\boldsymbol{{P}}\left(\boldsymbol{{B}}\mid\boldsymbol{{A}}\right)\ll\boldsymbol{{P}}\left({B}\right)\right) \\ $$$${t}\boldsymbol{{h}}{en}\:\boldsymbol{{P}}\left({A}\mid\overset{\_} {\boldsymbol{{B}}}\right)\geqslant\boldsymbol{{P}}\left(\boldsymbol{{A}}\right) \\ $$$$ \\ $$ Answered by…
Question Number 131590 by pticantor last updated on 06/Feb/21 $$\boldsymbol{{show}}\:\boldsymbol{{that}} \\ $$$$\boldsymbol{{U}}_{\boldsymbol{{n}}} =\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\boldsymbol{{dx}}}{\mathrm{1}+\boldsymbol{{x}}+\boldsymbol{{x}}^{\mathrm{2}} +….+\boldsymbol{{x}}^{\boldsymbol{{n}}} }\: \\ $$$$\boldsymbol{{converges}}!! \\ $$ Answered by mathmax by…