Question Number 219408 by CrispyXYZ last updated on 24/Apr/25 $${a}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}.\:{a}_{{n}} =\frac{\mathrm{2}}{{n}+\mathrm{3}}+\left(\mathrm{1}−\frac{\mathrm{1}}{{n}+\mathrm{3}}\right)^{\mathrm{2}} {a}_{{n}−\mathrm{1}} . \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{of}\:{a}_{{n}} . \\ $$ Answered by SdC355 last updated on…
Question Number 219388 by SdC355 last updated on 23/Apr/25 $$\int_{\mathrm{0}} ^{\:\infty} \:\:\frac{\mathrm{sin}^{\mathrm{2}} \left({u}\right)}{{u}^{\mathrm{2}} }\:\mathrm{d}{u}={I} \\ $$$${I}\left({t}\right)=\int_{\mathrm{0}} ^{\:\infty} \:\:\frac{\mathrm{sin}^{\mathrm{2}} \left({u}\right)}{{u}^{\mathrm{2}} }{e}^{−{ut}} \mathrm{d}{u}=\int_{\mathrm{0}} ^{\:\infty} \:\:\:\frac{\mathrm{1}}{{u}}\centerdot\frac{\mathrm{sin}^{\mathrm{2}} \left({u}\right)}{{u}}{e}^{−{ut}} \mathrm{d}{u}=…
Question Number 219323 by SdC355 last updated on 23/Apr/25 $${f}\left({s}\right)=\frac{\mathrm{1}}{\mathrm{2}\pi}\:\int\:\:{e}^{−\boldsymbol{{i}}{t}\left({s}−\alpha\right)} \:\mathrm{d}{t}\: \\ $$$$\int_{\mathrm{0}} ^{\:\infty} \int_{−\infty} ^{\:+\infty} \:\:{e}^{−\boldsymbol{{i}}{t}\left({s}−\alpha\right)} {e}^{−{sp}} \mathrm{d}{t}\mathrm{d}{s}=? \\ $$ Terms of Service Privacy…
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} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 219361 by SdC355 last updated on 23/Apr/25 $$\int_{−\infty} ^{+\infty} \:\:\frac{\mathrm{cos}\left({z}\right){e}^{−{z}} }{{z}^{\mathrm{2}} +\mathrm{1}}\:\mathrm{d}{z}=?? \\ $$ Answered by breniam last updated on 24/Apr/25 $$\underset{−\infty} {\overset{\mathrm{0}}…
Question Number 219357 by SdC355 last updated on 23/Apr/25 $$\int_{\mathrm{0}} ^{\:\infty} \:\:\:\frac{\mathrm{ln}\left({z}\right)}{{z}^{\mathrm{2}} +\mathrm{1}}\:\mathrm{d}{z} \\ $$ Answered by breniam last updated on 23/Apr/25 $$=\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{ln}\:\left({z}\right)}{{z}^{\mathrm{2}}…
Question Number 219356 by SdC355 last updated on 23/Apr/25 $$\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
Question Number 219359 by SdC355 last updated on 23/Apr/25 $$\mathrm{is}\:\underset{{k}=−\infty} {\overset{\infty} {\sum}}\:{J}_{\nu} \left({k}\right)= \\ $$$$\underset{{k}=−\infty} {\overset{\infty} {\sum}}\:\underset{{l}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{{l}} }{{l}!\left({l}+\nu\right)!}\left(\frac{{k}}{\mathrm{2}}\right)^{\mathrm{2}{l}+\nu} =\underset{{l}=\mathrm{0}} {\overset{\infty} {\sum}}\:\underset{{k}=−\infty} {\overset{\infty} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{{l}}…