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Question Number 225656    Answers: 0   Comments: 0

Question Number 225652    Answers: 2   Comments: 0

∫_0 ^(π/2) e^(iπx) dx

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {e}^{{i}\pi{x}} \:{dx} \\ $$

Question Number 225651    Answers: 0   Comments: 0

Question Number 225629    Answers: 2   Comments: 0

Question Number 225625    Answers: 0   Comments: 4

quick short Q 2 things are mixed 1)both same volume 2)both same mass density⇒d_v and d_m d_v ? d_m [=,< or>]

$${quick}\:{short}\:{Q} \\ $$$$\mathrm{2}\:{things}\:{are}\:{mixed}\: \\ $$$$\left.\mathrm{1}\right){both}\:{same}\:{volume} \\ $$$$\left.\mathrm{2}\right){both}\:{same}\:{mass} \\ $$$${density}\Rightarrow{d}_{{v}} \:{and}\:{d}_{{m}} \\ $$$${d}_{{v}} \:?\:{d}_{{m}} \left[=,<\:{or}>\right] \\ $$

Question Number 225613    Answers: 7   Comments: 0

Question Number 225610    Answers: 0   Comments: 0

prove Gauss curvature K is intrinsic by showing K=(( determinant (((−(1/2)E_(vv) +F_(uv) −G_(uu) ),((1/2)E_u ),(F_u −(1/2)E_v )),(( F_v −(1/2)G_u ),( E),( F)),(( (1/2)G_v ),( F),( G)))− determinant ((( 0),((1/2)E_v ),((1/2)G_u )),(((1/2)E_v ),( E),( F)),(((1/2)G_v ),( F),( G))))/((EG−F^( 2) )^2 )) E,F,G is First Fundametal form of metric tensor.

$$\mathrm{prove} \\ $$$$\mathrm{Gauss}\:\mathrm{curvature}\:{K}\:\mathrm{is}\:\mathrm{intrinsic}\:\mathrm{by}\:\mathrm{showing} \\ $$$${K}=\frac{\begin{vmatrix}{−\frac{\mathrm{1}}{\mathrm{2}}{E}_{{vv}} +{F}_{{uv}} −{G}_{{uu}} }&{\frac{\mathrm{1}}{\mathrm{2}}{E}_{{u}} }&{{F}_{{u}} −\frac{\mathrm{1}}{\mathrm{2}}{E}_{{v}} }\\{\:\:\:\:\:\:\:\:\:\:\:\:{F}_{{v}} −\frac{\mathrm{1}}{\mathrm{2}}{G}_{{u}} }&{\:\:\:{E}}&{\:\:\:\:\:\:{F}}\\{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}{G}_{{v}} }&{\:\:\:{F}}&{\:\:\:\:\:{G}}\end{vmatrix}−\begin{vmatrix}{\:\:\:\:\mathrm{0}}&{\frac{\mathrm{1}}{\mathrm{2}}{E}_{{v}} }&{\frac{\mathrm{1}}{\mathrm{2}}{G}_{{u}} }\\{\frac{\mathrm{1}}{\mathrm{2}}{E}_{{v}} }&{\:\:\:\:{E}}&{\:\:\:\:{F}}\\{\frac{\mathrm{1}}{\mathrm{2}}{G}_{{v}} }&{\:\:\:\:\:{F}}&{\:\:\:{G}}\end{vmatrix}}{\left({EG}−{F}^{\:\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$${E},{F},{G}\:\mathrm{is}\:\mathrm{First}\:\mathrm{Fundametal}\:\mathrm{form}\:\mathrm{of}\:\mathrm{metric}\:\mathrm{tensor}. \\ $$

Question Number 225604    Answers: 3   Comments: 2

Question Number 225601    Answers: 0   Comments: 1

2^(2025) ×3^(2025)

$$\:\:\: \mathrm{2}^{\mathrm{2025}} ×\mathrm{3}^{\mathrm{2025}} \: \\ $$

Question Number 225581    Answers: 0   Comments: 0

prove Gauss curvature K intrinsic it′s the same thing as saying; Show that Gauss curvature K can only consist of First Fundamental Form and it′s Derivatives.

$$\mathrm{prove} \\ $$$$\mathrm{Gauss}\:\mathrm{curvature}\:{K}\:\mathrm{intrinsic} \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{the}\:\mathrm{same}\:\mathrm{thing}\:\mathrm{as}\:\mathrm{saying}; \\ $$$$\mathrm{Show}\:\mathrm{that}\:\mathrm{Gauss}\:\mathrm{curvature}\:{K}\:\mathrm{can}\:\mathrm{only}\:\mathrm{consist}\:\mathrm{of} \\ $$$$\mathrm{First}\:\mathrm{Fundamental}\:\mathrm{Form}\:\mathrm{and}\:\mathrm{it}'\mathrm{s}\:\mathrm{Derivatives}. \\ $$

Question Number 225599    Answers: 1   Comments: 2

Question Number 225568    Answers: 1   Comments: 1

A farmer produces seeds in packets for sale. The probability that a seed selected at random will grow is 0.8. If there are 20 seeds, what is the probability that less than 2 will not grow?

$${A}\:{farmer}\:{produces}\:{seeds}\:{in}\:{packets} \\ $$$${for}\:{sale}.\:{The}\:{probability}\:{that}\:{a}\:{seed} \\ $$$${selected}\:{at}\:{random}\:{will}\:{grow}\:{is}\:\mathrm{0}.\mathrm{8}. \\ $$$${If}\:{there}\:{are}\:\mathrm{20}\:{seeds},\:{what}\:{is}\:{the} \\ $$$${probability}\:{that}\:{less}\:{than}\:\mathrm{2}\:{will}\:{not} \\ $$$${grow}? \\ $$

Question Number 225563    Answers: 2   Comments: 4

Question Number 225506    Answers: 1   Comments: 2

Question Number 225503    Answers: 2   Comments: 0

find x∈C for r∈R\{0} (−r)^x =r

$${find}\:{x}\in\mathbb{C}\:{for}\:{r}\in\mathbb{R}\backslash\left\{\mathrm{0}\right\} \\ $$$$\left(−{r}\right)^{{x}} ={r} \\ $$

Question Number 225497    Answers: 9   Comments: 0

Question Number 225490    Answers: 0   Comments: 19

Question Number 225488    Answers: 0   Comments: 0

Diffenrantial Geometry.....=∧= Christoffel symbol Γ_(σμν) first kind and Γ_(μν) ^( σ) second kind... and Chritoffel symbol satisfy Γ_(μν) ^( σ) =(1/2)g^(σl) Γ_(lμν) Γ_(ijk) {θ,ρ}∈{i,j,k} Γ_(θθθ) , Γ_(ρρρ) , Γ_(θθρ) Γ_(θρθ) , Γ_(ρθρ) , Γ_(ρθθ) Γ_(θρρ) , Γ_(ρρθ) Γ_(jk) ^i =(1/2)g^(il) Γ_(ljk) g_(μν) = ((( r^2 ),( 0)),(( 0),(r^2 sin^2 (θ))) ) (g_(μν) )^(−1) =g^(μν) g^(μν) = ((( (1/r^2 )),( 0)),(( 0),(1/(r^2 sin^2 (θ)))) ) g_(θθ) =r^2 , g_(θρ) =0 , g_(ρθ) =0 , g_(ρρ) =r^2 sin^2 (θ) ∂_θ g_(θθ) =0 , ∂_ρ g_(θθ) =0 , ∂_θ g_(θρ) =0 , ∂_ρ g_(θρ) =0 ∂_θ g_(ρθ) =0 , ∂_ρ g_(ρθ) =0 , ∂_θ g_(ρρ) =2r^2 sin(θ)cos(θ) , ∂_ρ g_(ρρ) =0 Γ_(abc) =(1/2)(∂_b ^ g_(ac) +∂_c g_(ab) −∂_a g_(bc) ) Γ_(θθθ) =(1/2)(∂_θ g_(θθ) +∂_θ g_(θθ) −∂_θ g_(θθ) )=0 Γ_(ρρρ) =(1/2)(∂_ρ g_(ρρ) +∂_ρ g_(ρρ) −∂_ρ g_(ρρ) )=0 Γ_(θρρ) =(1/2)(∂_ρ g_(θρ) +∂_ρ g_(θρ) −∂_θ g_(ρρ) )=−r^2 sin(θ)cos(θ) Γ_(ρθρ) =(1/2)(∂_θ g_(ρρ) +∂_ρ g_(ρθ) −∂_ρ g_(θρ) )=r^2 sin(θ)cos(θ) Γ_(ρθθ) =(1/2)(∂_θ g_(ρθ) +∂_θ g_(ρθ) −∂_ρ g_(θθ) )=0 Γ_(θρθ) =(1/2)(∂_ρ g_(θθ) +∂_θ g_(θρ) −∂_θ g_(ρθ) )=0 Γ_(θθρ) =(1/2)(∂_θ g_(θρ) +∂_ρ g_(θθ) −∂_θ g_(θρ) )=0 Γ_(ρρθ) =(1/2)(∂_ρ g_(ρθ) +∂_θ g_(ρρ) −∂_ρ g_(ρθ) )=r^2 sin(θ)cos(θ) Γ_(jk) ^i =g^(il) Γ_(ljk) g^(θθ) =(1/r^2 ) , g^(θρ) =0 , g^(ρθ) =0 , g^(ρρ) =(1/(r^2 sin^2 (θ))) Γ_(θθ) ^θ =g^(θl) Γ_(lθθ) = { ((g^(θθ) Γ_(θθθ) =0)),((g^(θρ) Γ_(ρθθ) =0)) :}, Γ_(ρθ) ^θ =g^(θl) Γ_(lρθ) = { ((g^(θθ) Γ_(θρθ) =0)),((g^(θρ) Γ_(ρρθ) =0)) :} , Γ_(θρ) ^θ =g^(θl) Γ_(lθρ) = { ((g^(θθ) Γ_(θθρ) =0)),((g^(θρ) Γ_(ρθρ) =0)) :} , Γ_(ρρ) ^θ =g^(θl) Γ_(lρρ) = { ((g^(θθ) Γ_(θρρ) =−sin(θ)cos(θ))),((g^(θρ) Γ_(ρρρ) =0)) :} Γ_(θθ) ^ρ =g^(ρl) Γ_(lθθ) = { ((g^(ρθ) Γ_(θθθ) =0)),((g^(ρρ) Γ_(ρθθ) =0)) :} , Γ_(ρθ) ^ρ =g^(ρl) Γ_(lρθ) = { ((g^(ρθ) Γ_(θρθ) =0)),((g^(ρρ) Γ_(ρρθ) =cot(θ))) :}, Γ_(θρ) ^ρ =g^(ρl) Γ_(lθρ) = { ((g^(ρθ) Γ_(θθρ) =0)),((g^(ρρ) Γ_(ρθρ) =cot(θ))) :} , Γ_(ρρ) ^ρ =g^(ρl) Γ_(lρρ) = { ((g^(ρθ) Γ_(θρρ) =0)),((g^(ρρ) Γ_(ρρρ) =0)) :} ∴Γ_(ρρ) ^θ =−sin(θ)cos(θ) Γ_(μν) ^( ρ) =Γ_(νμ) ^ρ Γ_(ρθ) ^ρ =Γ_(θρ) ^ρ =cot(θ) R_(jkl) ^i =∂_k Γ_(jl) ^k −∂_l Γ_(jk) ^i +Γ_(km) ^i Γ_(jl) ^m −Γ_(lm) ^i Γ_(jk) ^m R_(θθθ) ^θ , R_(θρθ) ^θ , R_(ρθθ) ^θ , R_(ρρθ) ^θ , R_(θρρ) ^θ , R_(ρρρ) ^θ , R_(ρθρ) ^θ , R_(θρρ) ^θ R_(θθθ) ^ρ , R_(θρθ) ^ρ , R_(ρθθ) ^ρ , R_(ρρθ) ^ρ , R_(θρρ) ^ρ , R_(ρρρ) ^ρ , R_(ρθρ) ^ρ , R_(θρρ) ^ρ and Riemann metric tensor have symmetries R_(abcd) =−R_(bacd) R_(abcd) =−R_(abdc) R_(abcd) =R_(cdab) Non-Zero Γ_(μν) ^( ρ) Γ_(ρρ) ^( θ) =−sin(θ)cos(θ) , Γ_(ρθ) ^( ρ) , Γ_(θρ) ^( ρ) =cot(θ) {i,j,k,ℓ}∈{θ,ρ} R_(ρθρ) ^θ =∂_θ Γ_(ρρ) ^θ −∂_ρ Γ_(ρθ) ^θ +Γ_(θm) ^( θ) Γ_(ρρ) ^m −Γ_(ρm) ^θ Γ_(ρθ) ^m =sin^2 (θ) R_(αρθρ) =g_(αμ) R_(ρθρ) ^μ R_(θρθρ) =g_(θμ) R_(ρθρ) ^μ = { ((g_(θθ) R_(ρθρ) ^θ =r^2 sin^2 (θ))),((g_(θρ) R_(ρθρ) ^ρ =0)) :} R_(θρθρ) =r^2 sin^2 (θ) ∴ R_(ρθρ) ^θ =sin^2 (θ) , R_(θρθρ) =r^2 sin^2 (θ) ∼2-dimensional Riemann Manifold∼ R_(abcd) =K(g_(ac) g_(bd) −g_(ad) g_(bc) ) R_(μν) ^(Ricci) =Kg_(μν) K is gaussian curvature aka K=((detII_p )/(det I_p ))=((LM−N^2 )/(EG−F^2 )) Q225479

$$\mathrm{Diffenrantial}\:\mathrm{Geometry}.....=\wedge= \\ $$$$\mathrm{Christoffel}\:\mathrm{symbol}\: \\ $$$$\Gamma_{\sigma\mu\nu} \:\mathrm{first}\:\mathrm{kind}\:\mathrm{and}\:\Gamma_{\mu\nu} ^{\:\sigma} \:\mathrm{second}\:\mathrm{kind}... \\ $$$$\mathrm{and}\:\mathrm{Chritoffel}\:\mathrm{symbol}\:\mathrm{satisfy} \\ $$$$\Gamma_{\mu\nu} ^{\:\sigma} =\frac{\mathrm{1}}{\mathrm{2}}{g}^{\sigma{l}} \Gamma_{{l}\mu\nu} \\ $$$$\Gamma_{{ijk}} \:\:\left\{\theta,\rho\right\}\in\left\{{i},{j},{k}\right\} \\ $$$$\Gamma_{\theta\theta\theta} \:,\:\Gamma_{\rho\rho\rho} \:,\:\Gamma_{\theta\theta\rho} \\ $$$$\Gamma_{\theta\rho\theta} \:,\:\Gamma_{\rho\theta\rho} \:,\:\Gamma_{\rho\theta\theta} \\ $$$$\Gamma_{\theta\rho\rho} \:,\:\Gamma_{\rho\rho\theta} \\ $$$$\Gamma_{{jk}} ^{{i}} =\frac{\mathrm{1}}{\mathrm{2}}{g}^{{il}} \Gamma_{{ljk}} \\ $$$${g}_{\mu\nu} =\begin{pmatrix}{\:{r}^{\mathrm{2}} }&{\:\:\:\:\:\:\mathrm{0}}\\{\:\mathrm{0}}&{{r}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \left(\theta\right)}\end{pmatrix} \\ $$$$\left({g}_{\mu\nu} \right)^{−\mathrm{1}} ={g}^{\mu\nu} \\ $$$${g}^{\mu\nu} =\begin{pmatrix}{\:\frac{\mathrm{1}}{{r}^{\mathrm{2}} }}&{\:\:\:\:\:\:\:\:\mathrm{0}}\\{\:\mathrm{0}}&{\frac{\mathrm{1}}{{r}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \left(\theta\right)}}\end{pmatrix} \\ $$$${g}_{\theta\theta} ={r}^{\mathrm{2}} ,\:{g}_{\theta\rho} =\mathrm{0}\:,\:{g}_{\rho\theta} =\mathrm{0}\:,\:{g}_{\rho\rho} ={r}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \left(\theta\right) \\ $$$$\partial_{\theta} {g}_{\theta\theta} =\mathrm{0}\:,\:\partial_{\rho} {g}_{\theta\theta} =\mathrm{0}\:,\:\partial_{\theta} {g}_{\theta\rho} =\mathrm{0}\:,\:\partial_{\rho} {g}_{\theta\rho} =\mathrm{0} \\ $$$$\partial_{\theta} {g}_{\rho\theta} =\mathrm{0}\:,\:\partial_{\rho} {g}_{\rho\theta} =\mathrm{0}\:,\:\partial_{\theta} {g}_{\rho\rho} =\mathrm{2}{r}^{\mathrm{2}} \mathrm{sin}\left(\theta\right)\mathrm{cos}\left(\theta\right)\:,\:\partial_{\rho} {g}_{\rho\rho} =\mathrm{0} \\ $$$$\Gamma_{{abc}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\partial_{{b}} ^{\:} {g}_{{ac}} +\partial_{{c}} {g}_{{ab}} −\partial_{{a}} {g}_{{bc}} \right) \\ $$$$\Gamma_{\theta\theta\theta} =\frac{\mathrm{1}}{\mathrm{2}}\left(\partial_{\theta} {g}_{\theta\theta} +\partial_{\theta} {g}_{\theta\theta} −\partial_{\theta} {g}_{\theta\theta} \right)=\mathrm{0} \\ $$$$\Gamma_{\rho\rho\rho} =\frac{\mathrm{1}}{\mathrm{2}}\left(\partial_{\rho} {g}_{\rho\rho} +\partial_{\rho} {g}_{\rho\rho} −\partial_{\rho} {g}_{\rho\rho} \right)=\mathrm{0} \\ $$$$\Gamma_{\theta\rho\rho} =\frac{\mathrm{1}}{\mathrm{2}}\left(\partial_{\rho} {g}_{\theta\rho} +\partial_{\rho} {g}_{\theta\rho} −\partial_{\theta} {g}_{\rho\rho} \right)=−{r}^{\mathrm{2}} \mathrm{sin}\left(\theta\right)\mathrm{cos}\left(\theta\right) \\ $$$$\Gamma_{\rho\theta\rho} =\frac{\mathrm{1}}{\mathrm{2}}\left(\partial_{\theta} {g}_{\rho\rho} +\partial_{\rho} {g}_{\rho\theta} −\partial_{\rho} {g}_{\theta\rho} \right)={r}^{\mathrm{2}} \mathrm{sin}\left(\theta\right)\mathrm{cos}\left(\theta\right) \\ $$$$\Gamma_{\rho\theta\theta} =\frac{\mathrm{1}}{\mathrm{2}}\left(\partial_{\theta} {g}_{\rho\theta} +\partial_{\theta} {g}_{\rho\theta} −\partial_{\rho} {g}_{\theta\theta} \right)=\mathrm{0} \\ $$$$\Gamma_{\theta\rho\theta} =\frac{\mathrm{1}}{\mathrm{2}}\left(\partial_{\rho} {g}_{\theta\theta} +\partial_{\theta} {g}_{\theta\rho} −\partial_{\theta} {g}_{\rho\theta} \right)=\mathrm{0} \\ $$$$\Gamma_{\theta\theta\rho} =\frac{\mathrm{1}}{\mathrm{2}}\left(\partial_{\theta} {g}_{\theta\rho} +\partial_{\rho} {g}_{\theta\theta} −\partial_{\theta} {g}_{\theta\rho} \right)=\mathrm{0} \\ $$$$\Gamma_{\rho\rho\theta} =\frac{\mathrm{1}}{\mathrm{2}}\left(\partial_{\rho} {g}_{\rho\theta} +\partial_{\theta} {g}_{\rho\rho} −\partial_{\rho} {g}_{\rho\theta} \right)={r}^{\mathrm{2}} \mathrm{sin}\left(\theta\right)\mathrm{cos}\left(\theta\right) \\ $$$$\Gamma_{{jk}} ^{{i}} ={g}^{{il}} \Gamma_{{ljk}} \\ $$$${g}^{\theta\theta} =\frac{\mathrm{1}}{{r}^{\mathrm{2}} }\:,\:{g}^{\theta\rho} =\mathrm{0}\:,\:{g}^{\rho\theta} =\mathrm{0}\:,\:{g}^{\rho\rho} =\frac{\mathrm{1}}{{r}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \left(\theta\right)} \\ $$$$\Gamma_{\theta\theta} ^{\theta} ={g}^{\theta{l}} \Gamma_{{l}\theta\theta} =\begin{cases}{{g}^{\theta\theta} \Gamma_{\theta\theta\theta} =\mathrm{0}}\\{{g}^{\theta\rho} \Gamma_{\rho\theta\theta} =\mathrm{0}}\end{cases},\:\Gamma_{\rho\theta} ^{\theta} ={g}^{\theta{l}} \Gamma_{{l}\rho\theta} =\begin{cases}{{g}^{\theta\theta} \Gamma_{\theta\rho\theta} =\mathrm{0}}\\{{g}^{\theta\rho} \Gamma_{\rho\rho\theta} =\mathrm{0}}\end{cases}\:,\:\Gamma_{\theta\rho} ^{\theta} ={g}^{\theta{l}} \Gamma_{{l}\theta\rho} =\begin{cases}{{g}^{\theta\theta} \Gamma_{\theta\theta\rho} =\mathrm{0}}\\{{g}^{\theta\rho} \Gamma_{\rho\theta\rho} =\mathrm{0}}\end{cases}\:,\:\Gamma_{\rho\rho} ^{\theta} ={g}^{\theta{l}} \Gamma_{{l}\rho\rho} =\begin{cases}{{g}^{\theta\theta} \Gamma_{\theta\rho\rho} =−\mathrm{sin}\left(\theta\right)\mathrm{cos}\left(\theta\right)}\\{{g}^{\theta\rho} \Gamma_{\rho\rho\rho} =\mathrm{0}}\end{cases} \\ $$$$\Gamma_{\theta\theta} ^{\rho} ={g}^{\rho{l}} \Gamma_{{l}\theta\theta} =\begin{cases}{{g}^{\rho\theta} \Gamma_{\theta\theta\theta} =\mathrm{0}}\\{{g}^{\rho\rho} \Gamma_{\rho\theta\theta} =\mathrm{0}}\end{cases}\:,\:\Gamma_{\rho\theta} ^{\rho} ={g}^{\rho{l}} \Gamma_{{l}\rho\theta} \:=\begin{cases}{{g}^{\rho\theta} \Gamma_{\theta\rho\theta} =\mathrm{0}}\\{{g}^{\rho\rho} \Gamma_{\rho\rho\theta} =\mathrm{cot}\left(\theta\right)}\end{cases},\:\Gamma_{\theta\rho} ^{\rho} ={g}^{\rho{l}} \Gamma_{{l}\theta\rho} =\begin{cases}{{g}^{\rho\theta} \Gamma_{\theta\theta\rho} =\mathrm{0}}\\{{g}^{\rho\rho} \Gamma_{\rho\theta\rho} =\mathrm{cot}\left(\theta\right)}\end{cases}\:,\:\Gamma_{\rho\rho} ^{\rho} ={g}^{\rho{l}} \Gamma_{{l}\rho\rho} =\begin{cases}{{g}^{\rho\theta} \Gamma_{\theta\rho\rho} =\mathrm{0}}\\{{g}^{\rho\rho} \Gamma_{\rho\rho\rho} =\mathrm{0}}\end{cases} \\ $$$$\therefore\Gamma_{\rho\rho} ^{\theta} =−\mathrm{sin}\left(\theta\right)\mathrm{cos}\left(\theta\right) \\ $$$$\:\:\:\Gamma_{\mu\nu} ^{\:\rho} =\Gamma_{\nu\mu} ^{\rho} \\ $$$$\:\:\:\Gamma_{\rho\theta} ^{\rho} =\Gamma_{\theta\rho} ^{\rho} =\mathrm{cot}\left(\theta\right) \\ $$$$\:{R}_{{jkl}} ^{{i}} =\partial_{{k}} \Gamma_{{jl}} ^{{k}} −\partial_{{l}} \Gamma_{{jk}} ^{{i}} +\Gamma_{{km}} ^{{i}} \Gamma_{{jl}} ^{{m}} −\Gamma_{{lm}} ^{{i}} \Gamma_{{jk}} ^{{m}} \\ $$$${R}_{\theta\theta\theta} ^{\theta} \:,\:{R}_{\theta\rho\theta} ^{\theta} \:,\:{R}_{\rho\theta\theta} ^{\theta} \:,\:{R}_{\rho\rho\theta} ^{\theta} \:,\:{R}_{\theta\rho\rho} ^{\theta} \:,\:{R}_{\rho\rho\rho} ^{\theta} \:,\:{R}_{\rho\theta\rho} ^{\theta} \:,\:{R}_{\theta\rho\rho} ^{\theta} \\ $$$${R}_{\theta\theta\theta} ^{\rho} \:,\:{R}_{\theta\rho\theta} ^{\rho} \:,\:{R}_{\rho\theta\theta} ^{\rho} \:,\:{R}_{\rho\rho\theta} ^{\rho} \:,\:{R}_{\theta\rho\rho} ^{\rho} \:,\:{R}_{\rho\rho\rho} ^{\rho} \:,\:{R}_{\rho\theta\rho} ^{\rho} \:,\:{R}_{\theta\rho\rho} ^{\rho} \: \\ $$$$\mathrm{and}\:\mathrm{Riemann}\:\mathrm{metric}\:\mathrm{tensor}\:\mathrm{have}\:\mathrm{symmetries} \\ $$$${R}_{{abcd}} =−{R}_{{bacd}} \\ $$$${R}_{{abcd}} =−{R}_{{abdc}} \\ $$$${R}_{{abcd}} ={R}_{{cdab}} \\ $$$$\mathrm{Non}-\mathrm{Zero}\:\Gamma_{\mu\nu} ^{\:\rho} \: \\ $$$$\Gamma_{\rho\rho} ^{\:\theta} =−\mathrm{sin}\left(\theta\right)\mathrm{cos}\left(\theta\right)\:,\:\Gamma_{\rho\theta} ^{\:\rho} \:,\:\Gamma_{\theta\rho} ^{\:\rho} =\mathrm{cot}\left(\theta\right) \\ $$$$\left\{{i},{j},{k},\ell\right\}\in\left\{\theta,\rho\right\} \\ $$$${R}_{\rho\theta\rho} ^{\theta} =\partial_{\theta} \Gamma_{\rho\rho} ^{\theta} −\partial_{\rho} \Gamma_{\rho\theta} ^{\theta} +\Gamma_{\theta{m}} ^{\:\theta} \Gamma_{\rho\rho} ^{{m}} −\Gamma_{\rho{m}} ^{\theta} \Gamma_{\rho\theta} ^{{m}} \\ $$$$=\mathrm{sin}^{\mathrm{2}} \left(\theta\right) \\ $$$${R}_{\alpha\rho\theta\rho} =\mathrm{g}_{\alpha\mu} {R}_{\rho\theta\rho} ^{\mu} \\ $$$${R}_{\theta\rho\theta\rho} =\mathrm{g}_{\theta\mu} {R}_{\rho\theta\rho} ^{\mu} =\begin{cases}{{g}_{\theta\theta} {R}_{\rho\theta\rho} ^{\theta} ={r}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \left(\theta\right)}\\{{g}_{\theta\rho} {R}_{\rho\theta\rho} ^{\rho} =\mathrm{0}}\end{cases} \\ $$$${R}_{\theta\rho\theta\rho} ={r}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \left(\theta\right) \\ $$$$\therefore\:{R}_{\rho\theta\rho} ^{\theta} =\mathrm{sin}^{\mathrm{2}} \left(\theta\right)\:,\:{R}_{\theta\rho\theta\rho} ={r}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \left(\theta\right) \\ $$$$\sim\mathrm{2}-\mathrm{dimensional}\:\mathrm{Riemann}\:\mathrm{Manifold}\sim \\ $$$${R}_{{abcd}} ={K}\left({g}_{{ac}} {g}_{{bd}} −{g}_{{ad}} {g}_{{bc}} \right) \\ $$$${R}_{\mu\nu} ^{\mathrm{Ricci}} ={Kg}_{\mu\nu} \\ $$$${K}\:\mathrm{is}\:\mathrm{gaussian}\:\mathrm{curvature}\:\:\mathrm{aka}\:{K}=\frac{\mathrm{det}\mathbb{II}_{{p}} }{\mathrm{det}\:\mathbb{I}_{{p}} }=\frac{{LM}−{N}^{\mathrm{2}} }{{EG}−{F}^{\mathrm{2}} } \\ $$$${Q}\mathrm{225479} \\ $$

Question Number 225479    Answers: 1   Comments: 0

Let S;R^2 →R^3 Sphere Q1. Find metric tensor g_(μν) Q2. Find Riemann metric tensor R_(jkl) ^i Q.3 Find Ricci tensor R_(αβ) Q.4 Find Ricci Scalar R Christoffel symbol first kind Γ_(μνσ) =(1/2)(∂_ν ^ g_(μσ) +∂_σ g_(μν) −∂_μ g_(νσ) ) Christoffel symbol second kind Γ_(jk) ^i =g^(il) Γ_(ljk) =(1/2)g^(iσ) (∂_j ^ g_(lk) +∂_k g_(lj) −∂_l g_(jk) ) Riemann metric tensor R_(jkl) ^i =∂_k Γ_(jl) ^i −∂_l Γ_(jk) ^i +Γ_(km) ^i Γ_(jl) ^m −Γ_(lm) ^i Γ_(jk) ^m

$$\mathrm{Let}\:\mathcal{S};\mathbb{R}^{\mathrm{2}} \rightarrow\mathbb{R}^{\mathrm{3}} \:\mathrm{Sphere}\: \\ $$$${Q}\mathrm{1}.\:\mathrm{Find}\:\mathrm{metric}\:\mathrm{tensor}\:\mathrm{g}_{\mu\nu} \: \\ $$$${Q}\mathrm{2}.\:\mathrm{Find}\:\mathrm{Riemann}\:\mathrm{metric}\:\mathrm{tensor}\:{R}_{{jkl}} ^{{i}} \\ $$$${Q}.\mathrm{3}\:\:\mathrm{Find}\:\mathrm{Ricci}\:\mathrm{tensor}\:\mathrm{R}_{\alpha\beta} \\ $$$${Q}.\mathrm{4}\:\:\mathrm{Find}\:\mathrm{Ricci}\:\mathrm{Scalar}\:\mathcal{R} \\ $$$$\mathrm{Christoffel}\:\mathrm{symbol}\:\mathrm{first}\:\mathrm{kind}\: \\ $$$$\Gamma_{\mu\nu\sigma} =\frac{\mathrm{1}}{\mathrm{2}}\left(\partial_{\nu} ^{\:} \mathrm{g}_{\mu\sigma} +\partial_{\sigma} \mathrm{g}_{\mu\nu} −\partial_{\mu} \mathrm{g}_{\nu\sigma} \right) \\ $$$$\mathrm{Christoffel}\:\mathrm{symbol}\:\mathrm{second}\:\mathrm{kind} \\ $$$$\Gamma_{{jk}} ^{{i}} =\mathrm{g}^{{il}} \Gamma_{{ljk}} =\frac{\mathrm{1}}{\mathrm{2}}\mathrm{g}^{{i}\sigma} \left(\partial_{{j}} ^{\:} \mathrm{g}_{{lk}} +\partial_{{k}} \mathrm{g}_{{lj}} −\partial_{{l}} \mathrm{g}_{{jk}} \right) \\ $$$$\mathrm{Riemann}\:\mathrm{metric}\:\mathrm{tensor} \\ $$$${R}_{{jkl}} ^{{i}} =\partial_{{k}} \Gamma_{{jl}} ^{{i}} −\partial_{{l}} \Gamma_{{jk}} ^{{i}} +\Gamma_{{km}} ^{{i}} \Gamma_{{jl}} ^{{m}} −\Gamma_{{lm}} ^{{i}} \Gamma_{{jk}} ^{{m}} \\ $$

Question Number 225474    Answers: 1   Comments: 2

Question Number 225461    Answers: 1   Comments: 0

Which one is the oddest prime number?

$${Which}\:{one}\:{is}\:{the}\:{oddest}\:{prime}\:{number}? \\ $$

Question Number 225449    Answers: 2   Comments: 0

Question Number 225416    Answers: 1   Comments: 2

Question Number 225407    Answers: 4   Comments: 3

Question Number 225391    Answers: 0   Comments: 0

Σ_(n = 1) ^∞ (− 1)^(n − 1) ((H_n H_(2n) ^((2)) )/n) = ?

$$\:\:\:\:\:\:\:\underset{\mathrm{n}\:\:\:=\:\:\:\mathrm{1}} {\overset{\infty} {\sum}}\:\left(−\:\:\:\:\mathrm{1}\right)^{\mathrm{n}\:\:\:\:−\:\:\:\mathrm{1}} \:\frac{\mathrm{H}_{\mathrm{n}} \:\mathrm{H}_{\mathrm{2n}} ^{\left(\mathrm{2}\right)} }{\mathrm{n}}\:\:\:\:\:\:\:=\:\:\:\:\:? \\ $$

Question Number 225392    Answers: 2   Comments: 1

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