Equality in the triangle inequality holds exactly when the two displacement vectors $a-z$ and $z-b$ point in the same direction. Thus $z$ lies on the closed line segment joining $a$ and $b$. Conversely, every point of that segment satisfiesThe solution set is therefore precisely that segment in the complex plane.

The triangle inequality givesHence the left-hand side of the proposed equation is always at least $|a-b|$. There are no solutions when $c<|a-b|$.

The hypergeometric exponent differences give $1-c=m$, $c-a-b=-m$, and exponents $a,b$ at infinity. Hence one may takeso a solution is ${}_2{F}_1(n,1-n;1-m;(1-z)/2)$, with the prefactor from part (i) when returning to the original dependent variable.

Set $z\prime =(1-z)/2$, sending $1,-1,\infty$ to $0,1,\infty$. Then rescale the dependent variable byThis shifts the exponent pairs to $(0,m)$ at zero and $(0,-m)$ at one, while leaving the exponents at infinity unchanged.

For a null vector, the perfect fluid in general relativity tensor givesbecause $g_{\alpha \beta }{V}^{\alpha }{V}^{\beta }=0$. For phantom energy, $\rho +P=(1+w)\rho <0$. A nonzero $u_{\alpha }{V}^{\alpha }$ therefore makes the contraction negative, so phantom energy violates the null energy condition.

For example, choose $w=-5/3$. A schematic solution can join the pressureless matter law $a\propto t^{2/3}$ to the phantom-dominated law $a\propto (t^{\ast }-t{)}^{-1}$. It begins at the Big Bang, bends down during matter domination, changes to accelerated expansion when phantom energy takes over, and has a vertical asymptote at the finite Big Rip time $t^{\ast }$.

Separate stress-energy conservation and the constant equation-of-state parameter $w$ give the constant-equation-of-state density scalingBecause $w<-1$, the phantom energy density grows with the scale factor, whereas the pressureless matter density decays as $a^{-3}$. Phantom energy therefore dominates at late times. With $\Lambda =0$, the Friedmann equation becomesIntegrating and absorbing the integration constant into the finite time $t^{\ast }$ givesThe exponent is negative, so $a(t)\to \infty$ as $t\to t^{\ast -}$. This is a Big Rip.

Contracting the Ricci tensor gives the Ricci scalarHence the time-time component of the Einstein tensor is $G_{00}=R_{00}-\frac{1}{2}{g}_{00}R=3H^2$. A comoving perfect fluid in general relativity has $T_{00}=\rho$, so the time-time Einstein field equations give the flat Friedmann equation

For the Friedmann-Lemaître-Robertson-Walker metric in Cartesian spatial coordinates, the only time-dependent components are $g_{ij}=a^2{\delta }_{ij}$. The Christoffel symbol formula therefore givesSubstituting these into the definition of the Ricci tensor givesThus, with no summation over $i$, the requested components are ${\Gamma }^0{}_{ii}=a\dot{a}$, ${\Gamma }^i{}_{0i}={\Gamma }^i{}_{i0}=\dot{a}/a$, and $R_{ii}=a\ddot{a}+2{\dot{a}}^2$.

For the Dirac comb, Poisson summation gives $U_l=1$ for every integer $l$. Hence every crossing at $k=l\pi /a$ opens a gap of width $2\lambda$ to first order, producing an infinite sequence of allowed bands between consecutive gaps.

For $V=2\lambda \cos (2\pi x/a)$, only $U_{\pm 1}=1$ are nonzero. To first order there is one gap of width $2\lambda$ at the first Brillouin-zone boundary $k=\pm \pi /a$; higher gaps vanish at this order. The allowed bands are the folded free-electron parabolas with that interval removed.

The two period-five orbits areDirect use of the two affine branches verifies every arrow.

For the stated map, the graph cycles give the period-two orbitand the period-four orbit

If $V$ is the defining representation, ${\chi }_V=2t$. The character of ${\mathrm{Sym}}^{3}V$ is $(2t{)}^3-2(2t)=8t^3-4t$. Therefore $8t^3$ is the character of ${\mathrm{Sym}}^{3}V\oplus V\oplus V$.

For $-1<x<1$, the centralizer of the diagonal representative is the maximal torus $T$. Hence the orbit is $SU(2)/T$. Identifying $SU(2)$ with $S^3$ makes this quotient the Hopf quotient $S^3/S^1\cong S^2$.

$SU(2)$ consists of unitary $2\times 2$ matrices of determinant one. Every element is unitarily diagonalizable with eigenvalues $e^{\pm i\theta }$, and two are conjugate exactly when they have the same trace. Thus $t(g)=\mathrm{tr}(g)/2=\cos \theta$ bijects conjugacy classes with $[-1,1]$.

Because the shell lies well inside its Schwarzschild radius once it reaches $r=2E$, an event horizon forms and the final exterior is a black hole of mass $E$. In the ideal instantaneous-shell model the collapse ends at a spacetime singularity.

Inside the spherical shell the enclosed mass vanishes by the spherical vacuum solution, so spacetime is flat. In the same ingoing coordinates,valid for $r\ge 0$ on the interior side before the shell reaches the centre.

Outside the ingoing null shell the enclosed mass is $E$, so use the ingoing ingoing Eddington-Finkelstein coordinatesThis is the Schwarzschild metric of mass $E$, regular across $r=2E$ and valid for $r>0$ on the exterior side of the shell.

Let $\zeta =e^{2\pi i/7}$ and let ${\lambda }_k(a)={\zeta }^k$. Conjugation by $b$ partitions the six nontrivial characters of $H=⟨a⟩$ into the two orbitsInducing one representative from each orbit gives two irreducible degree-three characters $\psi ,\overline{\psi }$. On the five classes represented by $1,a,a^3,b,b^2$ their values arewith the two cyclotomic sums interchanged for $\overline{\psi }$. Their squared degrees, together with the three linear characters, sum toso these are all irreducible characters.