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Question-212816

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}}…

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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}…

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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) \\…

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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…