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 $$\mathrm{prove}\:\mathrm{the}\:\mathrm{Following}\:\mathrm{Equation}. \\ $$$$\:{J}_{\nu} \left({z}\right)\:\mathrm{and}\:{Y}_{\nu} \left({z}\right)\:\mathrm{are}\:\:\mathrm{Bessel}\:\mathrm{function} \\ $$$${J}_{−\nu−\frac{\mathrm{1}}{\mathrm{2}}} \left({z}\right)=\left(−\mathrm{1}\right)^{\nu+\mathrm{1}} {Y}_{\nu+\frac{\mathrm{1}}{\mathrm{2}}} \left({z}\right) \\ $$$${Y}_{−\nu−\frac{\mathrm{1}}{\mathrm{2}}} \left({z}\right)=\left(−\mathrm{1}\right)^{\nu} {J}_{\nu+\frac{\mathrm{1}}{\mathrm{2}}} \left({z}\right) \\…
Question Number 212646 by MrGaster last updated on 20/Oct/24 $$ \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\underset{{j}={i}} {\overset{{i}} {\sum}}\frac{{i}\left({i}+{j}\right)}{\left({n}^{\mathrm{2}} +{i}^{\mathrm{2}} \right)\left({n}^{\mathrm{2}} +{j}^{\mathrm{2}} \right)} \\ $$ Commented by…
Question Number 212627 by MrGaster last updated on 19/Oct/24 $$ \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\sqrt{\mathrm{1}\centerdot\mathrm{2}}}{{n}^{\mathrm{2}} +\mathrm{1}}+\frac{\sqrt{\mathrm{2}\centerdot\mathrm{3}}}{{n}^{\mathrm{2}} +\mathrm{2}}+\ldots+\frac{\sqrt{{n}\left({n}+\mathrm{1}\right)}}{{n}^{\mathrm{2}} +{n}}\right) \\ $$ Answered by mehdee7396 last updated on 19/Oct/24…
Question Number 212618 by MrGaster last updated on 19/Oct/24 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{{n}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\frac{{k}^{\mathrm{2}} +\mathrm{3}{k}+\mathrm{2}}{{k}^{\mathrm{2}} +\mathrm{2}{k}+\mathrm{1}}\right)^{\frac{\mathrm{1}}{{n}}} \\ $$ Commented by mehdee7396 last updated on…
Question Number 212615 by MrGaster last updated on 19/Oct/24 $$\mathrm{Verify}\:\mathrm{the}\:\mathrm{equation}:\mathrm{tan}\:{nx}=\underset{{k}=\mathrm{1}} {\overset{\left[\frac{{n}+\mathrm{1}}{\mathrm{2}}\right]} {\sum}}\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \begin{pmatrix}{{n}}\\{\mathrm{2}{k}−\mathrm{1}}\end{pmatrix}\:\mathrm{tan}^{\mathrm{2}{k}−\mathrm{1}} {x}/\underset{{k}=\mathrm{0}} {\overset{\left[\frac{{n}}{\mathrm{2}}\right]} {\sum}}\left(−\mathrm{1}\right)^{{k}} \begin{pmatrix}{{n}}\\{\mathrm{2}{k}}\end{pmatrix}\mathrm{tan}^{\mathrm{2}{k}} {x} \\ $$ Terms of Service Privacy Policy…