Author: Tinku Tara

Question Number 219374 by SdC355 last updated on 23/Apr/25 Commented by SdC355 last updated on 23/Apr/25 $$\overset{\rightarrow} {\boldsymbol{\mathcal{S}}}\left({u},{v}\right)=\begin{cases}{\left(\mathrm{2}+{v}\centerdot\mathrm{sin}\left({u}\right)\right)\mathrm{sin}\left(\mathrm{2}\pi{v}\right)}\\{{v}\centerdot\mathrm{cos}\left({u}\right)}\\{\left(\mathrm{2}+{v}\centerdot\mathrm{sin}\left({u}\right)\right)\mathrm{cos}\left(\mathrm{2}\pi{v}\right)+\left(\mathrm{2}{v}−\mathrm{2}\right)}\end{cases} \\ $$$${u}\in\left[−\pi,\pi\right]\:,\:{v}\in\left[\mathrm{0},\frac{\pi}{\mathrm{2}}\right] \\ $$$$\mathrm{and}\:\mathrm{vector}\:\mathrm{field}\:\overset{\rightarrow} {\boldsymbol{\mathrm{F}}}\left({x},{y},{z}\right)=−{x}\overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\mathrm{1}} −{y}\overset{\rightarrow}…

Question Number 219360 by SdC355 last updated on 23/Apr/25 $$\mathrm{prove}\:\underset{{k}=−\infty} {\overset{\:\infty} {\sum}}\:{J}_{{k}} \left({z}\right)=\mathrm{1} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com

Question Number 219356 by SdC355 last updated on 23/Apr/25 $$\int_{\mathrm{0}} ^{\:\infty} \:\:\frac{\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \left({z}\right)}{{z}^{\mathrm{2}} }{e}^{−{zt}} \:\mathrm{d}{z} \\ $$ Answered by breniam last updated on 23/Apr/25 $$=−\underset{\mathrm{0}}…

Question Number 219358 by SdC355 last updated on 23/Apr/25 $$\underset{{h}=−\infty} {\overset{\infty} {\sum}}{J}_{\nu} \left({h}\right)=??\:,\:\nu\in\mathbb{Z}\backslash\left\{\mathrm{2}\mathbb{Z}\right\} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com