As ${\Omega }_M\uparrow 1$, use $\mathrm{arsinh}z\sim z$ near zero. ConsequentlyThe flat dust-and-vacuum expansion tends to $a(t)=(3H_{0}t/2{)}^{2/3}$, the Einstein-de Sitter universe.

As ${\Omega }_M\downarrow 0$, use $\mathrm{arsinh}z=\log (z+\sqrt{1+z^2})\sim \log (2z)$ to obtainAt fixed Hubble constant, the age of a flat matter-Lambda universe diverges logarithmically. The limiting purely vacuum-dominated de Sitter spacetime has an exponential scale factor and no Big Bang at a finite value of cosmic time in this flat slicing.

If $A_0\notin K$, let $p$ be a nearest point of the closed convex cone $K$ to $A_0$. Existence follows by minimizing the continuous squared distance on a sufficiently large compact ball. The first-order minimization inequality is $(A_0-p{)}^{\top }(y-p)\le 0$ for every $y\in K$. Set $x=p-A_0$. Taking $y=0$ and $y=2p$, both in the cone, gives $x^{\top }p=0$. ConsequentlyThe cone's test-weight characterization again yields $x^{\top }{A}_1\ge 0$ almost surely. This is separation from a closed convex cone with the sign needed forThe strict benefit is already available at the initial date, even if the terminal payoff happens to be zero.

For the Heaviside step function forcing, the integral begins at zero when $t>0$:Before zero it vanishes. ThereforeThis causal response of a first-order relaxation equation is continuous at zero and approaches the new constant equilibrium exponentially.

Inserting the Dirac delta forcing into the integral gives the impulse responseIt is zero before the impulse and obeys the homogeneous equation afterward. Integrating the equation across zero gives $\alpha [y(0^{+})-y(0^{-})]=1$, exactly the jump $1/\alpha$ in this formula. Thus the formula also verifies the differential equation as a distribution; assigning a particular value at the jump does not affect that distribution.

Substitute the signed displacement $\lambda =p-\mathbf{n}\cdot \mathbf{c}$ into the Pythagorean theorem relation from part (a):A genuine circle requires $|p-\mathbf{n}\cdot \mathbf{c}|<a$. Equality gives a tangent plane and a single point, interpreted as radius zero; a larger distance gives no intersection. Under the stipulated circle intersection, the radicand is positive. Its centre is $\mathbf{c}+(p-\mathbf{n}\cdot \mathbf{c})\mathbf{n}$.

Set ${\mathbf{b}}_0=\mathbf{c}+h\mathbf{n}$, where $h=p-\mathbf{n}\cdot \mathbf{c}$. Then $\mathbf{n}\cdot {\mathbf{b}}_0=p$, so ${\mathbf{b}}_0$ lies in the plane. For any position $\mathbf{r}$ in that plane, $\mathbf{r}-{\mathbf{b}}_0$ is perpendicular to ${\mathbf{b}}_0-\mathbf{c}=h\mathbf{n}$. The Pythagorean theorem therefore givesOn the sphere the left side is $a^2$. Thus the intersection is the circle in the plane centred at ${\mathbf{b}}_0$, with squared radius $a^2-h^2$. Its centre must therefore be $\mathbf{b}={\mathbf{b}}_0$, andIn terms of the radii, $\lambda =\pm \sqrt{a^2-{\rho }^2}$, with the sign fixed by $\lambda =p-\mathbf{n}\cdot \mathbf{c}$. In the central-plane case $\rho =a$, both signs give zero. This is the center and radius of a sphere-plane intersection formula with its orientation made explicit.

The vector $\mathbf{r}-(\mathbf{r}\cdot \mathbf{u})\mathbf{u}$ is the perpendicular component of $\mathbf{r}$ relative to the axis $\mathbb{R}\mathbf{u}$, so its length is the distance from that axis. Hence the surface is an infinite circular cylinder of radius $\beta$ about the line through the origin parallel to $\mathbf{u}$. Here $\beta$ is the radius and $\mathbf{u}$ fixes the axis direction; reversing $\mathbf{u}$ leaves the same circular cylinder.

If ${\mathbf{v}}_1,{\mathbf{v}}_2$ are an orthonormal basis of the plane perpendicular to $\mathbf{u}$, a parameterization isThis verifies that all heights and all angular positions, but no interior points, belong to the surface.

For $\mathbf{r}\ne 0$, the dot product equation says that the angle $\theta$ between $\mathbf{r}$ and the oriented unit vector $\mathbf{u}$ satisfies $\cos \theta =\alpha$. Thus $\theta =\arccos \alpha$ is fixed. More explicitly, using the orthogonal projection from the introductory solution,Consequently the surface is one infinite nappe of a right circular cone, with apex at the origin, including the apex itself. The cone's axis points along $\mathrm{sgn}(\alpha )\mathbf{u}$, and its acute semi-angle is $\arccos |\alpha |$. Equivalently, $\alpha$ is the cosine of the angle from the oriented $\mathbf{u}$ axis. Squaring the original equation without retaining the sign of $s$ would incorrectly add the opposite nappe of a double circular cone. No base disk is included in this infinite lateral surface.

If $v=\pm u$, then $Au=u$. On the perpendicular plane, $u{u}^T$ vanishes and $[v{]}_{\times }$ is a signed right-angle rotation, whose square is minus the identity. Hence its complex eigenvalues are $\pm i$, and

If $u\cdot v=0$, use the positively oriented orthonormal basis $(u,v\times u,v)$. The cross product identities give $Au=u+v\times u$, $A(v\times u)=-u$ and $Av=0$. In this basis,Its characteristic polynomial is $\lambda ({\lambda }^2-\lambda +1)$, so

Gravitational lensing probes the cluster's projected mass through the deflection of background light. Strong gravitational lensing produces arcs and multiple images; weak gravitational lensing measures coherent distortions of many background galaxies. The lensing convergence is $\kappa =\Sigma /{\Sigma }_{\mathrm{crit}}$, withThus the image geometry and shear constrain the projected surface density $\Sigma$ without assuming the cluster is in dynamical or hydrostatic equilibrium. The reconstructed total mass again exceeds the visible stellar and gas contribution. Source-distance errors, line-of-sight structures and the mass-sheet degeneracy affect the reconstruction, but the independent agreement of lensing, galaxy motions and gas confinement is strong evidence for cluster dark matter.

The hot intracluster gas emits X-rays, which constrain its temperature and density profile. If its pressure is predominantly thermal and it is in hydrostatic equilibrium,Eliminating the pressure givesHere $\mu m_p$ is the mean mass per gas particle. A gas at the observed cluster temperatures would not remain confined without a deep gravitational potential; the inferred total mass exceeds the gas and stellar mass together. Nonthermal pressure, mergers, gas clumping and temperature-profile errors can bias this estimate, but do not turn X-ray luminosity itself into a direct measurement of total mass.

The three-gamma matrix $C{\gamma }^{mnp}$ is antisymmetric, and for three distinct indices it is nonzero. A pair of nonzero Grassmann components therefore produces a nonzero term exactly as in the scalar case. Consequently this three-form bilinear is not identically zero. Equivalently it is dual to the axial-vector Majorana spinor bilinear, which is allowed; it must not be confused with the vanishing vector in part (a).

Four-dimensional Hodge duality identifies ${\gamma }^{mn}{\gamma }_5$ with a nonzero convention-dependent multiple of ${ϵ}^{mnpq}{\gamma }_{pq}$. Every $C{\gamma }_{pq}$ is symmetric by the four-dimensional Majorana bilinear vanishing criterion; taking this linear combination preserves symmetry. HenceThe phase in the duality formula depends on the definition of ${\gamma }_5$ and orientation, but does not affect the vanishing result.

The matrix $C$ is antisymmetric, so the scalar Majorana spinor bilinear is not identically zero. For example, choose two independent odd Grassmann variables ${\lambda }_{\alpha },{\lambda }_{\beta }$ at a pair of indices with $C_{\alpha \beta }\ne 0$. Their contribution is $2C_{\alpha \beta }{\lambda }_{\alpha }{\lambda }_{\beta }$, which is nonzero as a Grassmann polynomial. This is why a Majorana mass term is possible.

A classical sequential probability ratio test addresses a fixed-start choice between two simple statistical hypotheses. It sums log-likelihood ratio increments until an upper boundary accepts the alternative or a lower boundary accepts the null; it then terminates that test. A CUSUM instead monitors indefinitely for a change at an unknown time. Its reset at zero discards sustained evidence from a previously satisfactory period, so a late change need not first undo all earlier negative scores. An upper boundary $h$ signals when the accumulated change evidence is sufficiently large.

A fast initial response CUSUM takes $0<X_0<h$, often $X_0=h/2$, in place of a zero start. Until its first reset or signal, it acts like a fixed-start sequential probability ratio test with a lower boundary at $-X_0$ and an upper boundary at $h-X_0$ for the subsequent cumulative increments. Upon reaching the lower boundary it resets and continues as an ordinary CUSUM. The head start gives faster detection if a change is already present when monitoring begins or restarts. It is useful after an intervention, a shutdown or a previous alarm when immediate renewed surveillance is desired. Its initial false-alarm behavior differs from the zero-start scheme, so the head start and signalling boundary must be calibrated together; the preceding numerical choice is a design convention, not a universal optimum.

Let $p_1$ be the alternative death probability. The odds ratio specifiesAssume $0<p_0<1$ and $k>0$. For observed batch count $y_t$, the common combinatorial factor in the two binomial likelihoods cancels, giving the CUSUM incrementThereforeFor $k>1$ large death counts add positive evidence of deterioration; for $k<1$ unusually small death counts add evidence for improvement. The score has nonpositive expectation under the null and nonnegative expectation under the specified alternative, as follows from the nonnegativity of Kullback-Leibler divergence. Independent, nonoverlapping batches avoid counting the same patient's outcome repeatedly.

The same pricing relation gives $x^{\top }{A}_0\ge 0$ whenever the terminal payment is nonnegative. Therefore negative initial cost also contradicts a strictly positive state-price density, even if the terminal payment is identically zero.

This second form is needed for directions invisible in the terminal payments. Let $L$ be the linear span of the essential values of $A_1$, equivalentlyIf $A_0\notin L$, let $v$ be its nonzero orthogonal projection on $L^{\perp }$ and set $x=-v$. Then $x^{\top }{A}_1=0$ almost surely while $x^{\top }{A}_0=-\Vert v{\Vert }^2<0$. Hence the second form exists in this case. Part (b) handles all remaining cases.

It is helpful to solve using the nonsingular total flux $J$ and volume flux $Q$ before recovering $M$. The relation $J=M/(1-Q^2/(R^{2}M))$ is quadratic:The plus sign selects the small-area pure-source branch. Its discriminant vanishes when $J=4Q^2/R^2$, givingThereforeThe specific upward plume and ambient momentum-flux contributions are equal, not opposite. The ambient's transported momentum has the opposite sign from its transported volume, so their product remains positive.

To show that the pure branch actually reaches this fold in finite height and that it is a genuine pole, introduceThen $s=q^2/m$ andThe pure-source condition is $j\sim (5/8{)}^{2/5}{q}^{4/5}$. Before the fold, $s<1/2$ and $m>2q^2$. Thus for every fixed positive $q_1$, the derivative $dj/dq$ is bounded above by a constant times $q^{-2}$ on $q\ge q_1$. If the branch continued for arbitrarily large $q$, $j$ would remain bounded, contradicting $j\ge 4q^2$. Hence it reaches $j=4q^2$ at a finite $q_o>0$. Its height is finite because $d\zeta /dq=(1-s)/(2\sqrt{m})$ is finite away from the source and behaves like a constant times $q^{-2/5}$ at the source, an integrable singularity.

For the numerator, write $m=j(1-s)$. ThenAt the fold $j=4q_o^2$, so $q_o^3\ge 5/256$. On the other hand the dimensionless numerator in the $m$ equation at $s=1/2$ isIt is strictly negative because $5/256>1/(64\sqrt{2})$. The denominator $1-2s$ approaches zero from above. Consequently $dM/dz\to -\infty$ at finite $z_o$, although the fluxes and velocities themselves remain finite. This proves the confined-plume momentum fold rather than merely noticing a possible zero denominator.

The dimensionless pure-source branch uniquely determines the remaining amplitudes. They can be specified without an elementary antiderivative by integrating the displayed scalar equation for $j(q)$ until $j=4q^2$, and settingA direct numerical integration gives $q_o\simeq 0.281610$, ${\zeta }_o\simeq 0.400049$, so the dimensional results for this specified entrainment convention areA convention using $\alpha w$ instead of $\alpha (w-v)$ would produce different numerical amplitudes; the relative-speed convention was stated in part (a).

Physically $z_o$ is a turnover or breakdown height for a slender steady plume surrounded by uniform counterflow. The opposing streams then occupy equal areas and have equal speeds; the single-valued small-area branch cannot continue smoothly. Above it one expects lateral spreading, stronger recirculation, mixing and unsteady buoyant structures rather than a continuation of the same slender top-hat column. A detailed flow or upper-layer evolution requires restoring pressure and stress dynamics and, eventually, the modified ambient stratification. The model does not predict an infinite physical velocity, nor does it require that all buoyant fluid forever remain below $z_o$.