Question Number 127186 by Dwaipayan Shikari last updated on 27/Dec/20 $$\int_{\mathrm{0}} ^{{a}} {e}^{−{x}^{\mathrm{2}} } {dx}=\frac{\sqrt{\pi}}{\mathrm{2}}−\frac{{e}^{−{a}^{\mathrm{2}} } }{\mathrm{2}{a}+\frac{\mathrm{1}}{{a}+\frac{\mathrm{2}}{\mathrm{2}{a}+\frac{\mathrm{3}}{{a}+\frac{\mathrm{4}}{\mathrm{2}{a}+…}}}}}\:\left({Prove}\right) \\ $$ Commented by Dwaipayan Shikari last updated…
Question Number 127187 by Dwaipayan Shikari last updated on 27/Dec/20 $$\mathrm{1}−\mathrm{5}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}} +\mathrm{9}\left(\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{3}} −\mathrm{13}\left(\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{3}}{\mathrm{4}}.\frac{\mathrm{5}}{\mathrm{6}}\right)^{\mathrm{3}} +..=\frac{\mathrm{2}}{\pi}\:\left({prove}\right) \\ $$ Commented by Dwaipayan Shikari last updated on 27/Dec/20 Terms…
Question Number 61646 by maxmathsup by imad last updated on 05/Jun/19 $${let}\:{f}\left({x}\right)\:={e}^{−{ax}} \:{arctan}\left(\mathrm{3}{x}\right)\:\:\:{with}\:{a}>\mathrm{0} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}^{\left({n}\right)} \left({x}\right)\:{and}\:{f}^{\left({n}\right)} \left(\mathrm{0}\right) \\ $$$$\left.\mathrm{2}\right)\:{developp}\:{f}\:\left({x}\right)\:{at}\:{integr}\:{serie}\:. \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\infty} \:{f}\left({x}\right){dx}\:. \\ $$…
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}…