Question Number 133228 by mnjuly1970 last updated on 20/Feb/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:….{advanced}\:\:\:\:{calculus}…. \\ $$$$\:\:\:{prove}\:\:{that}\::: \\ $$$$\:\:\:\:\:\:\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\Gamma\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)\psi\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}^{{n}} .{n}!}=−\sqrt{\mathrm{2}\pi}\:\left(\gamma+{ln}\left(\mathrm{2}\right)\right)…. \\ $$$$ \\ $$ Answered by Dwaipayan Shikari…
Question Number 2158 by Yozzis last updated on 05/Nov/15 $${Find}\:{the}\:{ratio},\:{over}\:{one}\:{revolution},\:{of}\:{the}\:{distance}\:{moved}\:{by} \\ $$$${a}\:{wheel}\:{rolling}\:{on}\:{a}\:{flat}\:{surface}\:{to}\:{the}\:{distance}\:{traced}\:{out}\:{by} \\ $$$${a}\:{point}\:{on}\:{its}\:{circumference}.\: \\ $$ Commented by ssahoo last updated on 06/Nov/15 $$\mathrm{Wheel}\:\mathrm{distance}=\mathrm{2}\pi{r} \\…
Question Number 133231 by SOMEDAVONG last updated on 20/Feb/21 $$\mathrm{Give}\:\mathrm{S}_{\mathrm{n}} =\frac{\mathrm{n}+\mathrm{1}}{\mathrm{2}^{\mathrm{n}+\mathrm{1}} }\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{2}^{\mathrm{k}} }{\mathrm{k}}\:\:.\mathrm{Find}\:\underset{\mathrm{n}\rightarrow+\propto} {\mathrm{lim}S}_{\mathrm{n}} \:. \\ $$ Answered by Ar Brandon last updated…
Question Number 67692 by mr W last updated on 30/Aug/19 Answered by mr W last updated on 30/Aug/19 Commented by mr W last updated on…
Question Number 2157 by Yozzi last updated on 05/Nov/15 $${Evaluate}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}+\frac{\mathrm{2}^{\mathrm{3}} }{\mathrm{1}!}+\frac{\mathrm{3}^{\mathrm{3}} }{\mathrm{2}!}+\frac{\mathrm{4}^{\mathrm{3}} }{\mathrm{3}!}+… \\ $$$${by}\:{considering}\:{the}\:{series}\:{expansion} \\ $$$${of}\:{an}\:{expression}\:{of}\:{the}\:{form}\:{P}\left({x}\right){e}^{{x}} \\ $$$${where}\:{P}\left({x}\right)\:{is}\:{a}\:{suitably}\:{chosen} \\ $$$${polynomial}\:{in}\:{x}.\: \\ $$$$…
Question Number 67688 by Rio Michael last updated on 30/Aug/19 $${A}\:{relation}\:\mathbb{R}\:{defined}\:{by}\:\:\:_{\left({x},{y}\right)} {R}_{\left({u},{v}\right)} \:\Leftrightarrow\:\:{v}^{\mathrm{2}} −{y}^{\mathrm{2}} \:=\:{u}^{\mathrm{2}} −{x}^{\mathrm{2}} \\ $$$${show}\:{that}\:{R}\:{is}\:{an}\:{equivalent}\:{Relation}. \\ $$ Commented by Prithwish sen last…
Question Number 67686 by Rio Michael last updated on 30/Aug/19 $${given}\:{that}\:{the}\:{roots}\:{of}\:{the}\:{equation}\:\:\mathrm{4}{x}^{\mathrm{2}} \:+\:\mathrm{6}{x}\:+\:\mathrm{9}\:=\mathrm{0}\:{are}\:\:\lambda\:{and}\:\delta\:\:{where}\: \\ $$$$\:\lambda\:=\:\left(\mathrm{1}\:+\:\alpha^{\mathrm{2}} \:+\beta^{\mathrm{2}} \right)\:\:{and}\:\:\delta\:=\:\alpha^{\mathrm{3}} \:+\:\beta^{\mathrm{3}} \\ $$$${find}\:{an}\:{equation}\:{whose}\:{roots}\:{are}\: \\ $$$$\:\:\frac{\mathrm{1}}{\alpha\lambda}\:{and}\:\:\frac{\mathrm{1}}{\beta\delta} \\ $$ Commented by…
Question Number 67687 by Rio Michael last updated on 30/Aug/19 $${find}\:{the}\:{range}\:{of}\:{values}\:{of}\: \\ $$$$\:\:\mid\frac{{x}^{\mathrm{2}} −\mathrm{9}}{\mathrm{3}_{\:} }\mid=\:\frac{\mathrm{9}−{x}^{\mathrm{2}} }{\mathrm{3}} \\ $$$$ \\ $$ Commented by mr W last…
Question Number 67684 by Rio Michael last updated on 30/Aug/19 $${given}\:{the}\:{function}\: \\ $$$${f}\left({x}\right)\:=\begin{cases}{{x}^{\mathrm{2}} \:\:,\:{for}\:\:\:\mathrm{0}\leqslant\:{x}<\:\mathrm{2}}\\{{ax}\:+\:\mathrm{3},\:{for}\:\:\mathrm{2}\leqslant\:{x}\:<\:\mathrm{4}}\end{cases} \\ $$$${is}\:{periodic}\:{of}\:{period}\:\:\mathrm{4},\:{and}\:{is}\:{continuous}. \\ $$$$\left.{a}\right)\:{Find}\:\:{the}\:{value}\:{of}\:\:{a}. \\ $$$$\left.{b}\right)\:{Find}\:{the}\:{valu}\:{of}\:\:{f}\left(\mathrm{6}\right) \\ $$$$\left.{c}\right)\:{sketch}\:{the}\:{graph}\:{for}\:{y}\:={f}\left({x}\right). \\ $$$${help}\:{me}\:{please},\:{for}\:{the}\:{graph}\:{i}\:{don}'{t}\:{know}\:{wbere}\:{to}\:{put}\:\:{y}={x}^{\mathrm{2}} \:{and}\:{y}\:=\:{ax}\:+\:\mathrm{3}\:{and}…
Question Number 133222 by john_santu last updated on 20/Feb/21 $$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{\mathrm{x}^{\mathrm{3}} \:\mathrm{dx}}{\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{3}} +\mathrm{3x}−\mathrm{5}} \\ $$ Answered by liberty last updated on 20/Feb/21 $$\:\mathrm{I}\:=\:\underset{\mathrm{0}} {\overset{\mathrm{1}}…