Question Number 220015 by SdC355 last updated on 04/May/25 $$\int_{\mathrm{0}} ^{\:\infty} \:\mid\mid{J}_{\nu} \left({r}\right)\mid\mid{e}^{−{rt}} \:\mathrm{d}{r}=\underset{{h}=\mathrm{1}} {\overset{\infty} {\sum}}\:\int_{{z}_{{h}} } ^{\:{z}_{{h}+\mathrm{1}} } \:{J}_{\nu} \left({r}\right){e}^{−{rt}} \mathrm{d}{r} \\ $$$${z}_{{j}} \:\mathrm{is}\:\mathrm{point}\:\mathrm{of}\:\:{J}_{\nu}…
Question Number 220066 by SdC355 last updated on 04/May/25 $$\mathrm{evaluate} \\ $$$$−\frac{\mathrm{csc}\left(\pi{s}\right)}{\boldsymbol{{i}}\pi}\int_{\:\boldsymbol{\mathcal{C}}} \:\left(−{t}\right)^{{s}−\mathrm{1}} {e}^{−{t}} \:\mathrm{d}{t}\:,\:\mathrm{path}\:\boldsymbol{\mathcal{C}};\left(−\infty,\infty\right) \\ $$$$−\frac{\boldsymbol{\Gamma}\left(\mathrm{1}−{s}\right)}{\mathrm{2}\pi\boldsymbol{{i}}}\:\int_{\:\boldsymbol{\mathcal{C}}} \:\frac{\left(−{t}\right)^{{s}−\mathrm{1}} }{{e}^{{t}} −\mathrm{1}}\:\mathrm{d}{t}\:,\:\mathrm{path}\:\boldsymbol{\mathcal{C}};\left(−\infty,\infty\right) \\ $$ Terms of Service…
Question Number 219996 by SdC355 last updated on 04/May/25 $$\mathrm{Solve}\:\mathrm{Equation} \\ $$$$\frac{\mathrm{d}{x}\left({t}\right)}{\mathrm{d}{t}}=\mathrm{2}{x}\left({t}\right)+{y}\left({t}\right) \\ $$$$\frac{\mathrm{d}{y}\left({t}\right)}{\mathrm{d}{t}}=−\mathrm{3}{y}\left({t}\right) \\ $$$$\begin{pmatrix}{{x}^{\left(\mathrm{1}\right)} \left({t}\right)}\\{{y}^{\left(\mathrm{1}\right)} \left({t}\right)}\end{pmatrix}=\begin{pmatrix}{\mathrm{2}}&{\:\:\:\:\mathrm{1}}\\{\mathrm{0}}&{−\mathrm{3}}\end{pmatrix}\begin{pmatrix}{{x}\left({t}\right)}\\{{y}\left({t}\right)}\end{pmatrix} \\ $$$$\mathrm{A}=\begin{pmatrix}{\mathrm{2}}&{\:\:\:\:\mathrm{1}}\\{\mathrm{0}}&{−\mathrm{3}}\end{pmatrix} \\ $$$$\mathrm{det}\left\{\mathrm{A}−\boldsymbol{\lambda}\mathrm{E}\right\}=\mathrm{0} \\ $$$$\mathrm{det}\left\{\begin{pmatrix}{\mathrm{2}}&{\:\:\:\:\mathrm{1}}\\{\mathrm{0}}&{−\mathrm{3}}\end{pmatrix}−\begin{pmatrix}{\boldsymbol{\lambda}}&{\mathrm{0}}\\{\mathrm{0}}&{\boldsymbol{\lambda}}\end{pmatrix}\right\}=\mathrm{0} \\…
Question Number 219890 by SdC355 last updated on 03/May/25 $$\mathrm{Find}\:\mathrm{Maxima}\: \\ $$$${x}+{y}\:\mathrm{where}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={r}^{\mathrm{2}} \:\left(\mathrm{use}\:\mathrm{Lagrange}\:\mathrm{Method}\right) \\ $$ Answered by MrGaster last updated on 03/May/25 $$\bigtriangledown\left({x}+{y}\right)=\lambda\bigtriangledown\left({x}^{\mathrm{2}}…
Question Number 219887 by SdC355 last updated on 03/May/25 $$\mathrm{what}\:\mathrm{is}\: \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)\uparrow\uparrow^{\infty} =?? \\ $$$${a}\uparrow\uparrow^{{m}} =\underset{{m}\:\mathrm{times}} {\underbrace{{a}^{{a}^{{a}^{{a}^{\iddots} } } } }}\:\:\left(\mathrm{aka}\:\mathrm{Knuth}'\mathrm{s}\:\mathrm{up}\:\mathrm{notation}\right) \\ $$ Answered…
Question Number 219872 by SdC355 last updated on 03/May/25 $$\mathrm{prove} \\ $$$$\int\:\:{Y}_{−\frac{\mathrm{3}}{\mathrm{2}}} \left({z}\right)\:\mathrm{d}{z}=\frac{\mathrm{4sin}\left({z}\right)+\frac{{z}\boldsymbol{\Gamma}\left(\frac{\mathrm{1}}{\mathrm{2}},−\boldsymbol{{i}}{z}\right)}{\:\sqrt{−\boldsymbol{{i}}{z}}}+\frac{{z}\boldsymbol{\Gamma}\left(\frac{\mathrm{1}}{\mathrm{2}},\boldsymbol{{i}}{z}\right)}{\:\sqrt{\boldsymbol{{i}}{z}}}}{\:\sqrt{\mathrm{2}\pi{z}}}+{C} \\ $$ Answered by MrGaster last updated on 03/May/25 $${Y}_{−\nu} =\left(−\mathrm{1}\right)^{\nu} {Y}_{\nu}…
Question Number 219870 by SdC355 last updated on 03/May/25 $$\int_{\mathrm{0}} ^{\:\infty} \:{K}_{\nu} \left({r}\right)\mathrm{d}{r} \\ $$$$\int_{\mathrm{0}} ^{\:\infty} \:{t}\centerdot{Y}_{\mathrm{0}} \left({t}\right)\mathrm{d}{t} \\ $$$$\int_{\mathrm{0}} ^{\:\infty} \:\:\frac{\mathrm{sin}\left({t}\right){e}^{−{kt}} }{{t}^{\mathrm{2}} +\rho^{\mathrm{2}} }\mathrm{d}{t}\:…
Question Number 219831 by SdC355 last updated on 02/May/25 $$\mathrm{prove} \\ $$$$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{g}\left({z}+{h}\right)}{\mathrm{g}\left({z}\right)}\right)^{\frac{\mathrm{1}}{{h}}} ={e}^{\frac{\mathrm{d}\:\:}{\mathrm{d}{z}}\:\mathrm{ln}\:\left(\mathrm{g}\left({z}\right)\right)} ={e}^{\frac{\mathrm{g}^{\left(\mathrm{1}\right)} \left({z}\right)}{\mathrm{g}\left({z}\right)}} \\ $$ Answered by MrGaster last updated on 04/May/25…
Question Number 219806 by SdC355 last updated on 02/May/25 $$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\left(\frac{\mathrm{cos}\left({x}+{h}\right)}{\mathrm{cos}\left({x}\right)}\right)^{\frac{\mathrm{1}}{{h}}} =?? \\ $$ Answered by fantastic last updated on 02/May/25 $${cos}\left({y}\right)^{\underset{{y}} {\mathrm{1}}} \\ $$…
Question Number 219800 by SdC355 last updated on 02/May/25 $${y}^{\left(\mathrm{2}\right)} \left({t}\right)=\left(\mathrm{1}−{e}^{{t}} \right){y}\left({t}\right)+{y}^{\left(\mathrm{1}\right)} \left({t}\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com