Question Number 212798 by MrGaster last updated on 24/Oct/24 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{\frac{{x}^{\mathrm{2}} }{{e}^{{x}} }} =? \\ $$ Answered by mehdee7396 last updated on 24/Oct/24 $${lim}_{{x}\rightarrow\infty} \left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)=\mathrm{1}\:\:\:\:\&\:\:\:\:{lim}_{{x}\rightarrow\infty}…
Question Number 212788 by MrGaster last updated on 24/Oct/24 $$ \\ $$$${Let}\:{f}\left({x}\right)\:\mathrm{There}\:\mathrm{is}\:\mathrm{a}\:\mathrm{secondorder}\:\mathrm{continuoust} \\ $$$$\mathrm{derivaive},{t}=\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} },{g}\left({x},{y}\right)={f}\left(\frac{\mathrm{1}}{{r}}\right),\mathrm{ask}\:\frac{\partial^{\mathrm{2}} {g}}{\partial{x}^{\mathrm{2}} }+\frac{\partial^{\mathrm{2}} {g}}{\partial{y}^{\mathrm{2}} }. \\ $$ Terms of Service…
Question Number 212816 by Akayx last updated on 24/Oct/24 Answered by mahdipoor last updated on 24/Oct/24 $${f}\left({t}\right)=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{2}\left({t}−\mathrm{4}{k}\right)\left({u}_{\mathrm{4}{k}} −{u}_{\mathrm{4}{k}+\mathrm{2}} \right)= \\ $$$$\mathrm{2}\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left[\left({t}−\mathrm{4}{k}\right){u}_{\mathrm{4}{k}}…
Question Number 212809 by MrGaster last updated on 24/Oct/24 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{certificate}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\mathrm{csc}\frac{{x}}{\mathrm{2}^{{k}} }=\mathrm{cot}\frac{{x}}{\mathrm{2}^{{n}+\mathrm{1}} }−\mathrm{cot}\:{x} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 212765 by issac last updated on 23/Oct/24 $$\mathrm{i}\:\:\mathrm{generalized}\:\boldsymbol{\mathrm{Bessel}}\:\boldsymbol{\mathrm{function}}'\mathrm{s} \\ $$$$\mathrm{Laplace}\:\mathrm{Transform} \\ $$$$\mathrm{L}.\mathrm{T}{J}_{\nu} \left({z}\right)=\frac{\left({s}+\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}\right)^{−\nu} }{\:\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}}\:,\:{s}\in\left[\mathrm{0},\infty\right)\:,\:\nu\in\mathbb{R} \\ $$$$\mathrm{L}.\mathrm{T}\:{Y}_{\nu} \left({z}\right)=\frac{\mathrm{cot}\left(\pi\nu\right)\left({s}+\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}\right)^{−\nu} }{\:\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}}−\frac{\mathrm{csc}\left(\pi\nu\right)\left({s}+\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}\right)^{\nu}…
Question Number 212779 by issac last updated on 23/Oct/24 $$\mathrm{excuse}\:\mathrm{me}???? \\ $$$$\mathrm{am}\:\mathrm{i}\:\mathrm{invisible}??\: \\ $$$$\mathrm{why}\:\mathrm{isn}'\mathrm{t}\:\mathrm{anyone}\:\mathrm{answering}\:\mathrm{my}\:\mathrm{qusestion} \\ $$$$ \\ $$ Commented by mr W last updated on…
Question Number 212775 by Davidtim last updated on 23/Oct/24 $${vector}×{scalar}=? \\ $$$${vector}\:{or}\:{scalar}? \\ $$ Commented by A5T last updated on 23/Oct/24 $${Q}\mathrm{207812};\:{similar}\:{idea}. \\ $$ Terms…
Question Number 212716 by MrGaster last updated on 22/Oct/24 $$ \\ $$$$\:\:\:\:\:\:\:\mathrm{cos}\frac{\pi}{\mathrm{7}}+\mathrm{cos}\frac{\mathrm{3}\pi}{\mathrm{7}}+\mathrm{cos}\frac{\mathrm{5}\pi}{\mathrm{7}}=? \\ $$ Answered by golsendro last updated on 22/Oct/24 Terms of Service Privacy…
Question Number 212741 by mathocean1 last updated on 22/Oct/24 $${a},\:{b},{c}\:\in\:\mathbb{N}^{\ast} \\ $$$$ \\ $$$${Show}\:{that}\:{ab}<{c}\:\Rightarrow\:{a}+{b}\leqslant{c} \\ $$$$ \\ $$ Answered by A5T last updated on 22/Oct/24…
Question Number 212648 by issac last updated on 20/Oct/24 $$\mathrm{prove}\:\mathrm{the}\:\mathrm{Following}\:\mathrm{Equation}. \\ $$$$\:{J}_{\nu} \left({z}\right)\:\mathrm{and}\:{Y}_{\nu} \left({z}\right)\:\mathrm{are}\:\:\mathrm{Bessel}\:\mathrm{function} \\ $$$${J}_{−\nu−\frac{\mathrm{1}}{\mathrm{2}}} \left({z}\right)=\left(−\mathrm{1}\right)^{\nu+\mathrm{1}} {Y}_{\nu+\frac{\mathrm{1}}{\mathrm{2}}} \left({z}\right) \\ $$$${Y}_{−\nu−\frac{\mathrm{1}}{\mathrm{2}}} \left({z}\right)=\left(−\mathrm{1}\right)^{\nu} {J}_{\nu+\frac{\mathrm{1}}{\mathrm{2}}} \left({z}\right) \\…