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Author: Tinku Tara

Does-anyone-know-how-to-prove-this-V-dxdydz-1-x-4-y-4-z-4-4-1-4-4-4-where-V-is-the-unit-cube-0-1-3-Thankyou-

Question Number 197376 by megrex last updated on 15/Sep/23 $${Does}\:{anyone}\:{know}\:{how}\:{to}\:{prove}\:{this}? \\ $$$$\:\:\:\:\:\:\:\:\:\:\int\int\int_{{V}} \:\frac{{dxdydz}}{\mathrm{1}+{x}^{\mathrm{4}} +{y}^{\mathrm{4}} +{z}^{\mathrm{4}} }\:=\frac{\Gamma^{\mathrm{4}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)}{\mathrm{4}^{\mathrm{4}} } \\ $$$${where}\:{V}\:{is}\:{the}\:{unit}\:{cube}\:\left[\mathrm{0},\mathrm{1}\right]^{\mathrm{3}} \\ $$$${Thankyou}. \\ $$$$ \\…

lim-x-0-1-cosxcos2x-cos-nx-x-2-n-n-1-2n-1-12-

Question Number 197359 by sniper237 last updated on 14/Sep/23 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{\mathrm{1}−{cosxcos}\mathrm{2}{x}…{cos}\left({nx}\right)}{{x}^{\mathrm{2}} }\:=\:\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{12}}\: \\ $$ Commented by universe last updated on 16/Sep/23 Answered by witcher3 last…

calcul-n-1-oo-1-n-2n-1-n-n-1-

Question Number 197349 by SANOGO last updated on 14/Sep/23 $${calcul}\: \\ $$$$\underset{{n}=\mathrm{1}} {\overset{+{oo}} {\sum}}\left(−\mathrm{1}\right)^{{n}\:} \frac{\mathrm{2}{n}+\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)} \\ $$ Answered by MM42 last updated on 14/Sep/23 $$\frac{\mathrm{2}{n}+\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)}=\frac{\mathrm{1}}{{n}}+\frac{\mathrm{1}}{{n}+\mathrm{1}}…

d-dt-dx-dt-1-2-g-g-g-g-dx-dt-dx-dt-0-

Question Number 197344 by MathedUp last updated on 14/Sep/23 $$\frac{\mathrm{d}\:\:}{\mathrm{d}{t}}\centerdot\frac{\mathrm{d}{x}^{\boldsymbol{\lambda}} }{\mathrm{d}{t}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{g}^{\boldsymbol{\lambda\alpha}} \left(\partial_{\boldsymbol{\mu}} ^{\:} \mathrm{g}_{\boldsymbol{\alpha\nu}} +\partial_{\boldsymbol{\nu}} ^{\:} \mathrm{g}_{\boldsymbol{\alpha\mu}} −\partial_{\boldsymbol{\alpha}} ^{\:} \mathrm{g}_{\boldsymbol{\mu\nu}} \right)\frac{\mathrm{d}{x}^{\boldsymbol{\mu}} }{\mathrm{d}{t}}\centerdot\frac{\mathrm{d}{x}^{\boldsymbol{\nu}} }{\mathrm{d}{t}}=\mathrm{0} \\ $$…

Question-197362

Question Number 197362 by sonukgindia last updated on 14/Sep/23 Answered by MM42 last updated on 15/Sep/23 $$\frac{{e}^{{x}} −\mathrm{1}}{{e}^{{x}} +\mathrm{1}}={u}\Rightarrow{e}^{{x}} {u}+{u}={e}^{{x}} −\mathrm{1}\Rightarrow{e}^{{x}} \left({u}−\mathrm{1}\right)=−\mathrm{1}−{u} \\ $$$${e}^{{x}} =\frac{\mathrm{1}+{u}}{\mathrm{1}−{u}}\Rightarrow{f}\left({x}\right)=\left(\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\right)^{\mathrm{2}}…

Question-197346

Question Number 197346 by Amidip last updated on 14/Sep/23 Answered by som(math1967) last updated on 14/Sep/23 $$\:\frac{{a}+{b}}{{a}−{b}}=\frac{{tan}\left(\theta+\phi\right)}{{tan}\left(\theta−\phi\right)} \\ $$$$\:\frac{{a}+{b}}{{a}−{b}}=\frac{{sin}\left(\theta+\phi\right){cos}\left(\theta−\phi\right)}{{sin}\left(\theta−\phi\right){cos}\left(\theta+\phi\right)} \\ $$$$\frac{\mathrm{2}{a}}{\mathrm{2}{b}}=\frac{{sin}\mathrm{2}\theta}{{sin}\mathrm{2}\phi}\:\:\left[{using}\:{componendo\&dividendo}\right] \\ $$$${asin}\mathrm{2}\phi={bsin}\mathrm{2}\theta \\ $$$${a}^{\mathrm{2}}…

Question-197310

Question Number 197310 by sonukgindia last updated on 13/Sep/23 Answered by Frix last updated on 13/Sep/23 $$\left(\mathrm{1}\right)\:{z}^{\mathrm{2}} −\mathrm{7}{z}+\mathrm{11}=\mathrm{1} \\ $$$$\Rightarrow\:{z}=\mathrm{2}\vee{z}=\mathrm{5} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{ln}\:\left({z}^{\mathrm{2}} −\mathrm{3}{z}+\mathrm{e}^{\mathrm{2}} \right)\:=\mathrm{0} \\…