Question Number 9275 by geovane10math last updated on 27/Nov/16 $$\mathrm{About}\:\mathrm{the}\:\mathrm{Euler}-\mathrm{Mascheroni}\:\mathrm{Constant}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\gamma\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}\:−\:{x}}\:+\:\frac{\mathrm{1}}{\mathrm{ln}\:{x}}\:{dx} \\ $$$$\mathrm{We}\:\mathrm{can}\:\mathrm{see}\:\mathrm{that}\: \\ $$$$\:\mathrm{1}−\:{x}\:\neq\:\mathrm{0}\:\Leftrightarrow\:\mathrm{1}\:\neq\:{x}\:; \\ $$$${x}\:=\:\mathrm{0}\:\rightarrow\:\mathrm{ln}\:{x}\:\nexists\:. \\ $$$$\mathrm{If}\:{x}\:\neq\:\mathrm{0}\:\mathrm{and}\:{x}\:\neq\:\mathrm{1},\:\mathrm{in}\:\mathrm{the}\:\mathrm{Cartesian}\:\mathrm{Plane}, \\ $$$$\mathrm{this}\:\mathrm{function}\:\mathrm{has}\:\mathrm{singularity}\:{x}=\mathrm{0}\:{and}\:{x}=\mathrm{1}. \\…
Question Number 9274 by geovane10math last updated on 27/Nov/16 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{e}^{{t}} \:=\:{t} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 140344 by mathdave last updated on 06/May/21 $${it}\:{is}\:{assumed}\:{that}\:{when}\:{children}\:{are}\:{born}\:{they}\: \\ $$$${are}\:{equally}\:{likely}\:{to}\:{be}\:{boys}\:{or}\:{girls}. \\ $$$${what}\:{is}\:{the}\:{probability}\:{that}\:{a}\:{family} \\ $$$${of}\:{four}\:{children}\:{constains} \\ $$$$\left({a}\right){three}\:{boys}\:{and}\:{girl} \\ $$$$\left({b}\right){two}\:{boys}\:{and}\:{two}\:{girls} \\ $$$$\:{Mr}\:{W}\:{pls}\:{help}\:{out} \\ $$ Answered…
Question Number 140339 by mohammad17 last updated on 06/May/21 $$\int\frac{{dx}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{4}} }} \\ $$ Answered by Dwaipayan Shikari last updated on 06/May/21 $$\int\frac{{dx}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{4}} }}=\underset{{n}\geqslant\mathrm{0}} {\sum}\int\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} }{{n}!}\left(−{x}^{\mathrm{4}}…
Question Number 74802 by liki last updated on 30/Nov/19 Commented by liki last updated on 30/Nov/19 $$…{plz}\:{i}\:{need}\:{help}\:;{mr}\:{w}\:{and}\:{mr}\:{mind}\:{is}\:{power} \\ $$$$,\:{or}\:{anyeone}\:{to}\:{assist}\:{me}! \\ $$ Commented by abdomathmax last…
Question Number 9255 by geovane10math last updated on 26/Nov/16 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{e}^{{i}\pi} \:+\:\mathrm{1}\:=\:\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{e}^{{i}\pi} \:=\:−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({e}^{{i}\pi} \right)^{\mathrm{2}} \:=\:\left(−\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{e}^{\mathrm{2}{i}\pi} \:=\:\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({e}^{\mathrm{2}{i}\pi} \right)^{{i}} \:=\:\mathrm{1}^{{i}}…
Question Number 74782 by naka3546 last updated on 30/Nov/19 Commented by abdomathmax last updated on 01/Dec/19 $${let}\:{f}\left({x}\right)=\left\{\left(\frac{\mathrm{1}}{\mathrm{8}}\right)^{{x}} \:+\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{x}} \right\}^{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} \:\Rightarrow \\ $$$${ln}\left({f}\left({x}\right)\right)=\frac{\mathrm{1}}{{x}^{\mathrm{2}} }{ln}\left\{\:\left(\frac{\mathrm{1}}{\mathrm{8}}\right)^{{x}} \:+\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{x}}…
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Question Number 9237 by geovane10math last updated on 25/Nov/16 $${About}\:{the}\:{Euler}-{Mascheroni}\:{constant} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\gamma\:=\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}}\:−\:\mathrm{ln}\:{n}\right) \\ $$$${Why}\:{the}\:{limit}\:{converges}\:{if} \\ $$$$\:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}}\:−\:\mathrm{ln}\:{n}\right)\:=\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}}\:−\:\underset{{n}\rightarrow\infty}…
Question Number 9232 by geovane10math last updated on 24/Nov/16 $$\mathrm{1}\:+\:\mathrm{2}\:+\:\mathrm{3}\:+\:\mathrm{4}\:+…+\:{n}\:=\:\frac{{n}\left({n}\:+\:\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} \:+\mathrm{3}^{\mathrm{2}} \:+…+\:{n}^{\mathrm{2}} \:=\:\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}} \\ $$$$\mathrm{1}^{{x}} \:+\:\mathrm{2}^{{x}} \:+\:\mathrm{3}^{{x}} \:+…+\:=\:{n}^{{x}} \:=\:\mathrm{Formula}? \\ $$$${x}\:\in\:\mathbb{R} \\…