Question Number 213846 by issac last updated on 18/Nov/24 $$\mathrm{path}\:\mathscr{C}\:\mathrm{is}\:\mathrm{closed} \\ $$$${f}\:\mathrm{is}\:\mathrm{regular}\:\mathrm{function}\:\mathrm{in}\:\mathrm{Path}\:\mathscr{C} \\ $$$${f}\:\mathrm{is}\:\mathrm{have}\:\mathrm{zero}\:\mathrm{point}\:\mathrm{and}\:\mathrm{poles}\:\mathrm{in}\:\mathscr{C} \\ $$$$\mathrm{show}\:\mathrm{that}\: \\ $$$$\oint_{\:\mathscr{C}} \:\frac{{f}\left({z}\right)}{{f}'\left({z}\right)}\:\mathrm{d}{z}=\mathrm{2}\pi\boldsymbol{{i}}\left({Z}−{P}\right) \\ $$$$\mathrm{and}\:\mathrm{if}\:\mathrm{poles}\:\mathrm{are}\:\mathrm{not}\:\mathrm{exist} \\ $$$$\mathrm{show}\:\mathrm{that} \\ $$$$\oint_{\:\mathscr{C}\:}…
Question Number 213790 by issac last updated on 16/Nov/24 $$\mathrm{So}\:\mathrm{Weird}…… \\ $$$$\int_{\mathrm{0}} ^{\:\infty} {J}_{\nu} \left({t}\right){e}^{−{st}} \mathrm{d}{t}=\frac{\left({s}+\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}\right)^{−\nu} }{\:\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}}\: \\ $$$${J}_{−\nu} \left({t}\right)=\left(−\mathrm{1}\right)^{\nu} {J}_{\nu} \left({t}\right)\:\: \\…
Question Number 213764 by Hanuda354 last updated on 16/Nov/24 Answered by mr W last updated on 16/Nov/24 $$\left(\mathrm{1}\right)\:\mathrm{4}−\mathrm{1}−\mathrm{3}−\mathrm{2} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{4}−\mathrm{3}−\mathrm{1}−\mathrm{2} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{1}−\mathrm{4}−\mathrm{3}−\mathrm{2} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{1}−\mathrm{4}−\mathrm{2}−\mathrm{3} \\…
Question Number 213725 by zakirullah last updated on 14/Nov/24 $${please}\:{prove}\:\frac{\mathrm{1}}{{x}}\:=\:{x}^{−\mathrm{1}} \\ $$ Answered by Rasheed.Sindhi last updated on 14/Nov/24 $$\:\:\:\:\:\frac{\mathrm{1}}{{x}}=\frac{{x}^{\mathrm{0}} }{{x}^{\mathrm{1}} }={x}^{\mathrm{0}−\mathrm{1}} ={x}^{−\mathrm{1}} \\ $$…
Question Number 213713 by issac last updated on 14/Nov/24 $$\mathrm{Does}\:\mathrm{Volume}\:\mathrm{integral}\: \\ $$$${V}=\pi\centerdot\int_{\mathrm{0}} ^{\:\infty} {J}_{\nu} ^{\:\mathrm{2}} \left({z}\right)\mathrm{d}{z}\:\mathrm{is}\:\mathrm{divergence}…?? \\ $$$${J}_{\nu} \left({z}\right)\:\mathrm{is}\:\nu\left(\mathrm{nu}\right)'\mathrm{th}\:\mathrm{Bessel}\:\mathrm{function} \\ $$ Terms of Service Privacy…
Question Number 213680 by issac last updated on 13/Nov/24 Commented by Frix last updated on 13/Nov/24 $$\mathrm{WolframAlpha} \\ $$$$\mathrm{It}'\mathrm{s}\:\mathrm{worthless}\:\mathrm{if}\:\mathrm{you}\:\mathrm{don}'\mathrm{t}\:\mathrm{understand}\:\mathrm{how} \\ $$$$\mathrm{to}\:\mathrm{get}\:\mathrm{there}. \\ $$ Commented by…
Question Number 213679 by issac last updated on 13/Nov/24 $${Q}\mathrm{213662} \\ $$$$…… \\ $$$$\mathrm{Not}\:\mathrm{easy}……. \\ $$$$\:\:_{{p}} \boldsymbol{\mathrm{F}}_{{q}} \left({z};\cancel{\underbrace{ }}\:\boldsymbol{\mathrm{a}},\boldsymbol{\mathrm{b}}\right)\:\mathrm{is}\:\mathrm{hypergeometric}\:\mathrm{function} \\ $$$$\mathrm{Li}_{\nu} \left({z}\right)\:\mathrm{is}\:\mathrm{Dilogarithm}\:\mathrm{function}. \\ $$ Terms…
Question Number 213648 by issac last updated on 12/Nov/24 $${a}_{{m}} =\underset{{h}={m}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{h}}\right)^{\mathrm{2}} =\frac{\pi^{\mathrm{2}} }{\mathrm{6}}−\underset{{h}=\mathrm{1}} {\overset{{m}} {\sum}}\:\left(\frac{\mathrm{1}}{{h}}\right)^{\mathrm{2}} \\ $$$${a}_{{m}} \approx\boldsymbol{\psi}^{\left(\mathrm{1}\right)} \left({m}+\mathrm{1}\right) \\ $$$$\boldsymbol{\psi}^{\left({n}\right)} \left({z}\right)=\frac{\mathrm{d}^{{n}+\mathrm{1}} \:\:}{\mathrm{d}{z}^{{n}+\mathrm{1}}…
Question Number 213604 by issac last updated on 10/Nov/24 $$\mathrm{show}\:\mathrm{that}\:\:\int_{\boldsymbol{\mathcal{C}}} \:{e}^{{z}^{\mathrm{3}} } \:\mathrm{d}{z}=\mathrm{0} \\ $$$$\mathrm{where}\:\boldsymbol{\mathcal{C}}\:\mathrm{is}\:\mathrm{any}\:\mathrm{simple}\:\mathrm{closed}\:\mathrm{contour}. \\ $$$$\:\:\:\: \\ $$$$\mathrm{Evaluate}\:\mathrm{the}\:\mathrm{integral} \\ $$$$\int_{\:{C}_{\mathrm{1}} } {f}\left({z}\right)\mathrm{d}{z}\:,\:\int_{\:{C}_{\mathrm{2}} } {f}\left({z}\right)\mathrm{d}{z}…
Question Number 213607 by tri26112004 last updated on 10/Nov/24 $${Give}\:{a},{b},{c}\:>\:\mathrm{0}\:{such}\:{that} \\ $$$${ab}+{bc}+{ca}={abc} \\ $$$${Prove}\:{that}\:: \\ $$$${a}+{b}+{c}\geqslant\mathrm{4}\left(\frac{{a}}{{bc}}+\frac{{b}}{{ca}}+\frac{{c}}{{ab}}\right)+\mathrm{5} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com