Question Number 72117 by psyche last updated on 24/Oct/19 $${Please}\:{help}\:{with}\:{the}\:{proof}\:{of}\:{cauchy}\:{inequality}. \\ $$ Commented by Tanmay chaudhury last updated on 24/Oct/19 Terms of Service Privacy Policy…
Question Number 72112 by A8;15: last updated on 24/Oct/19 Answered by mind is power last updated on 24/Oct/19 $$\mathrm{x}=\frac{\mathrm{1}}{\mathrm{t}}\Rightarrow\mathrm{dx}=\frac{−\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} } \\ $$$$\frac{\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{899999999}} }}{\frac{\left(\mathrm{1}+\mathrm{t}^{\mathrm{9}} \right)^{\mathrm{1000000001}} }{\mathrm{t}^{\mathrm{900}\:\mathrm{000009}}…
Question Number 137648 by JulioCesar last updated on 05/Apr/21 Commented by mr W last updated on 05/Apr/21 $$\mathrm{2}+\sqrt{\mathrm{5}} \\ $$ Answered by mr W last…
Question Number 72111 by A8;15: last updated on 24/Oct/19 Commented by mathmax by abdo last updated on 24/Oct/19 $${let}\:{I}=\int_{\mathrm{0}} ^{\frac{{e}}{\pi}} \:\frac{{arctan}\left(\frac{\pi{x}}{{e}}\right)}{\pi{x}\:+{e}}{dx}\:{changement}\:\frac{\pi{x}}{{e}}\:={t}\:{give} \\ $$$${I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{arctant}}{\pi\left(\frac{{et}}{\pi}\right)+{e}}\frac{{e}}{\pi}{dt}\:=\frac{{e}}{\pi}\int_{\mathrm{0}}…
Question Number 137611 by JulioCesar last updated on 04/Apr/21 Answered by Dwaipayan Shikari last updated on 04/Apr/21 $$\gamma\:\:=\:\mathrm{0}.\mathrm{57721}\:\left({Euler}\:{Mascheroni}\:{constant}\right) \\ $$ Terms of Service Privacy Policy…
Question Number 6535 by Temp last updated on 01/Jul/16 $$\underset{{x}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{x}}\:=\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{5}}+… \\ $$$$\boldsymbol{\mathrm{Is}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{following}}\:\boldsymbol{\mathrm{true}}? \\ $$$$\therefore\Sigma\frac{\mathrm{1}}{{x}}=\Sigma\left(\frac{\mathrm{1}}{\mathrm{2}{x}}+\frac{\mathrm{1}}{\mathrm{2}{x}−\mathrm{1}}\right) \\ $$$$=\Sigma\left(\frac{\mathrm{2}{x}−\mathrm{1}+\mathrm{2}{x}}{\mathrm{2}{x}\left(\mathrm{2}{x}−\mathrm{1}\right)}\right) \\ $$$$=\Sigma\left(\frac{\mathrm{2}{x}−\mathrm{2}+\mathrm{1}+\mathrm{2}{x}}{\mathrm{2}{x}\left(\mathrm{2}{x}−\mathrm{1}\right)}\right) \\ $$$$=\Sigma\left(\frac{\mathrm{4}{x}−\mathrm{2}}{\mathrm{2}{x}\left(\mathrm{2}{x}−\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{2}{x}\left(\mathrm{2}{x}−\mathrm{1}\right)}\right) \\ $$$$=\Sigma\left(\frac{\mathrm{2}\left(\mathrm{2}{x}−\mathrm{1}\right)}{\mathrm{2}{x}\left(\mathrm{2}{x}−\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{2}{x}\left(\mathrm{2}{x}−\mathrm{1}\right)}\right) \\…
Question Number 6523 by Tawakalitu. last updated on 30/Jun/16 $$\sqrt{{x}^{{x}^{\mathrm{6}} } }\:\:\:=\:\:\:\mathrm{729} \\ $$$$ \\ $$$${find}\:\:{x} \\ $$ Commented by Rasheed Soomro last updated on…
Question Number 137588 by bramlexs22 last updated on 04/Apr/21 $${For}\:{a}\:{positive}\:{number}\:{n}\:,\:{let} \\ $$$${f}\left({n}\right)\:{be}\:{the}\:{value}\:{of}\: \\ $$$${f}\left({n}\right)=\frac{\mathrm{4}{n}+\sqrt{\mathrm{4}{n}^{\mathrm{2}} −\mathrm{1}}}{\:\sqrt{\mathrm{2}{n}+\mathrm{1}}\:+\sqrt{\mathrm{2}{n}−\mathrm{1}}} \\ $$$${calculate}\:{f}\left(\mathrm{1}\right)+{f}\left(\mathrm{2}\right)+{f}\left(\mathrm{3}\right)+…+{f}\left(\mathrm{40}\right). \\ $$ Answered by bemath last updated on…
Question Number 6512 by Rasheed Soomro last updated on 30/Jun/16 $${Find}\:{complex}\:{number}\:{whose}\:{additive}\: \\ $$$${inverse}\:{is}\:{equal}\:{to}\:{its}\:{multiplicative} \\ $$$${inverse}. \\ $$ Commented by Temp last updated on 30/Jun/16 $$\mathrm{What}\:\mathrm{is}\:\mathrm{additive}\:\mathrm{and}\:\mathrm{multiplicitive}…
Question Number 137584 by bramlexs22 last updated on 04/Apr/21 $${Given}\:\begin{cases}{{a}_{\mathrm{2}{n}} \:=\:{a}_{{n}} .{a}_{\mathrm{2}} \:+\mathrm{1}}\\{{a}_{\mathrm{2}{n}+\mathrm{1}} \:=\:{a}_{{n}} .{a}_{\mathrm{2}} \:−\mathrm{2}\:}\end{cases}\:{and} \\ $$$$\:\begin{cases}{{a}_{\mathrm{7}} \:=\:\mathrm{2}}\\{\mathrm{0}<{a}_{\mathrm{1}} <\mathrm{1}}\end{cases}.\:{Find}\:{a}_{\mathrm{25}} \:=? \\ $$$$ \\ $$…