For fixed $A>\gamma >D$, let $x_0=1-\gamma /A$. The rare high-risk group's SIS model has $x=x_0+p{B}^2\gamma /[A^2(\gamma -D)]+O(p^2)$, whereas low-risk infection is $y=pB{x}_0/(\gamma -D)+O(p^2)$. Substitution into both equilibrium equations proves these coefficients. The leading basic reproduction number is $A/\gamma$ with correction $p{B}^2/[\gamma (A-D)]$, while global susceptibility tends to one. These expressions explain why an endemic core can coexist with a large susceptible majority; they are nonuniform when a group is near its own threshold.

If a thrice differentiable solution has both endpoint values and both endpoint first derivatives bounded in modulus by $N$, it obeys $|f|\le N$ and $|f^{\prime }|\le N$ throughout the interval. A positive interior maximum of $f$ would give $f^{\prime \prime }=f>0$, and a negative interior minimum would give $f^{\prime \prime }=f<0$, both impossible. For $w=f^{\prime }$, differentiation gives $w^{\prime \prime }=w+2w{w}^{\prime }$, so the same argument applies at an extremum of $w$. The resulting uniform bound and Lipschitz constant imply total boundedness of the solution family by the Arzelà-Ascoli theorem.

For a short exact sequence $0\to N\to M\to Q\to 0$, the middle module is Artinian exactly when both outer modules are Artinian. Submodules and quotients inherit the descending chain condition. Conversely, for a descending chain $M_i$, both $M_i\cap N$ and its image in $Q$ eventually stabilize. If $x\in M_i$, choose $y\in M_{i+1}$ with the same image in $Q$; then $x-y\in M_i\cap N=M_{i+1}\cap N$, giving $M_i=M_{i+1}$. Induction shows that finite direct sums of Artinian modules are Artinian.

For an injective module endomorphism $f:M\to M$ of an Artinian module, the chain $M\supseteq fM\supseteq f^{2}M\supseteq \cdots$ stabilizes. If $f^{n}M=f^{n+1}M$, then every $x$ satisfies $f^{n}x=f^{n+1}y$. Injectivity of $f^n$ gives $x=fy$, proving surjectivity. No finite-length or Noetherian hypothesis is needed.

Assume that every nontrivial one-dimensional Artin L-function over the intermediate fields is holomorphic and nonzero at one, while the trivial one is the base Dedekind zeta function with a simple pole. By Brauer induction, $\chi ={\sum }_j{n}_j{\mathrm{Ind}}_{H_j}^G{\psi }_j$ with integers $n_j$ and linear characters ${\psi }_j$. Compatibility of Artin L-functions with induced representations expresses $L_K(\chi ,s)$ as the corresponding product of $L_{L^{H_j}}({\psi }_j,s{)}^{n_j}$. Its order at one is $-{\sum }_j{n}_j⟨{\psi }_j,1_{H_j}⟩$. Frobenius reciprocity turns this sum into $-⟨\chi ,1_G⟩$. Hence every nontrivial irreducible character gives order zero and a holomorphic nonzero function near one. This local conclusion does not assert global Artin holomorphy.

Only the inertia group invariants enter an Artin L-function local factor. A Frobenius automorphism lift acts unambiguously on this subspace, since other lifts differ by inertia. For an unramified one-dimensional group representation, the factor is $(1-\chi ({\mathrm{Frob}}_{\mathfrak{p}})(N\mathfrak{p}{)}^{-s}{)}^{-1}$; for a ramified one-dimensional group representation, the invariant subspace is zero and the factor is one. This explains the zero values used at conductor primes by quadratic Dirichlet characters.

For $|\varphi |<1$ and $|\theta |<1$, the autoregressive moving-average model $(1-\varphi B)X_t=(1+\theta B){ϵ}_t$ has $c_0=d_0=1$ and the displayed coefficients for $j\ge 1$. These follow by multiplying the two geometric power series. If $\varphi +\theta =0$, all later coefficients vanish and the filtered process is white noise. Thus root restrictions on an uncancelled nonminimal representation are sufficient but need not be necessary for the process after common factor cancellation in an ARMA model.

A binary quadratic refinement of a fixed nonsingular alternating bilinear form $B$ has polar form $B$. The difference of two refinements is linear, so nondegeneracy gives the displayed unique parametrization by $v$. For $\dim V=2m$, refinements of plus and minus Arf invariant of a quadratic form have respectively $2^{m-1}(2^m+1)$ and $2^{m-1}(2^m-1)$ elements. Completing the square in the binary character sum gives $ϵ(Q_v)=ϵ(Q_0)(-1{)}^{Q_0(v)}$.

Assume a constant positive mass $m$, a twice differentiable trajectory in the fixed plane, a fixed origin in an inertial frame, and Newton's second law for the net force. The product rule gives the short proofThe term $\mathbf{v}\times \mathbf{v}$ vanishes. This is the planar angular momentum balance written in area language.^[ref]

On a connected time interval, areal velocity is constant if and only if ${\tau }_n=0$ throughout. Since both position and acceleration lie in the fixed plane, the net torque has only that normal component. For $r\ne 0$, zero net torque is therefore equivalent to net force parallel to $\mathbf{r}$: its instantaneous direction is central, possibly inward, outward or zero. This diagnoses the net force along this motion, not its dependence on position or time elsewhere. Opposing transverse forces may cancel.

A central force about a fixed origin gives conservation of angular momentum, irrespective of its radial dependence: an inverse-square law is sufficient but unnecessary. With nonzero conserved angular momentum, $\mathbf{r}\cdot \mathbf{L}=0$ fixes the orbital plane. Zero angular momentum gives radial motion and zero areal velocity; it does not select a unique orbital plane. The converse needs the fixed-plane assumption: a constant projected areal velocity in three dimensions only constrains one torque component. Also, $r=0$ makes the instantaneous torque vanish regardless of force and makes plane polar coordinates singular. A moving origin or changing plane requires additional terms.

Why infer an area law from astronomy? In the authored account The great Martian catastrophe and how Kepler fixed it by Owen Gingerich, a 13 August 1593 Mars observation has ecliptic longitude $346^{\circ }7.5^{\prime }$, against predicted $342^{\circ }{0}^{\prime }$ and $351^{\circ }26^{\prime }$. These are Earth-based directions with a date, not measured Sun-based areas. The observer's armillary sphere used graduated circles and sighting arms to read declination; reversing it allowed a second reading. Comparing stars at different altitudes exposed refraction, and a planet-based check revealed an erroneous solar correction table. The apparent Mars distance signal had been contaminated by that correction. Johannes Kepler subsequently corrected the observer's Earth orbit, then tested Mars trajectories. The account describes typical angular errors of $2^{\prime }$ or less ($120$ arcseconds): small residuals needed care.^[ref]

The Tufts class notes Kepler's Astronomia Nova and the Orbit of Mars explain the key limitation: an area requires a trajectory and distances, so Kepler's second law cannot be directly tested from angular observations alone. They describe checking the combined ellipse and area law against 28 dated observations around seven oppositions during 1583-1595, using an Earth model to predict Earth-based directions; agreement improved greatly but some discrepancies remained.^[ref]

The inferential chain is useful today too: recover relative positions and times, compare swept areas, and ask what forces are compatible with their evolution. Agreement with equal areas supports zero torque about the chosen origin under the assumptions below; it alone does not identify an inverse-square law. Recovering the radial force also needs the trajectory and its timing, as in inferring an inverse-square force from Kepler laws. The two diagrams here are calculated examples, not the historical measurements.

Take an illustrative elliptic Kepler orbit with semi-major axis $a=1$, orbital eccentricity $e=0.6$ and gravitational parameter $\mu =1$, in consistent dimensionless units. Its semi-minor axis is $b=a\sqrt{1-e^2}=0.8$, mean motion is $n=\sqrt{\mu /a^3}=1$, and specific angular momentum magnitude is $h=\sqrt{\mu a(1-e^2)}=0.8$. About the attracting focus,The two sectors below each span $\Delta t=0.5$, giving $\Delta A=0.2$. Near periapsis the short radius sweeps a wide angle; near apoapsis the long radius sweeps a narrow angle. Constant areal velocity therefore does not require constant angular velocity or constant speed.

To compute the sectors without mistaking equal angles for equal times, parametrize the ellipse by eccentric anomaly $E$:This follows by integrating $\frac{1}{2}(xdy-ydx)$. Kepler's equation gives $E-e\sin E=M=n(t-t_{\mathrm{p}})$, with mean anomaly $M$ measured from periapsis time $t_{\mathrm{p}}$. Consequently $\Delta A=abn\Delta t/2=h\Delta t/2$. The displayed sectors use mean anomaly intervals $[-0.25,0.25]$ and $[\pi -0.25,\pi +0.25]$, rather than equal eccentric anomaly increments.

Signed swept area counts the area traversed by a position vector, positively or negatively according to its direction of turning. For a continuously differentiable planar trajectory, its accumulated value relative to $t_0$ is the line integralA short segment contributes the oriented triangle area $\frac{1}{2}\mathbf{n}\cdot (\mathbf{r}\times d\mathbf{r})$; taking its time derivative gives areal velocity. Repeated turns count repeated area; a reversal subtracts it. Thus this quantity need not equal the area of the union of all previously swept regions. The origin matters: the relevant origin for Kepler's second law is the attracting focus of the ellipse, rather than its geometric centre.

If a domain of finite hyperbolic area $A$ contains a hyperbolic disc of radius $r$ in curvature $-1$, then $2\pi (\cosh r-1)\le A$, hence $r\le \mathrm{arcosh}(1+A/(2\pi ))$. For an interior point of a hyperbolic triangle, the open disc with radius equal to its distance from the boundary lies inside the triangle. Its area is at most $\pi$ by Gauss-Bonnet theorem, giving the elementary bound $r\le \mathrm{arcosh}(3/2)$; this need not be the sharp triangle inradius bound.

For returns $r_i=a_i+{\sum }_j{b}_{ji}{f}_j$, fund a risky position $v$ by a risk-free asset. Its zero-cost payoff is $(a-R1{)}^{T}v+{\sum }_j(Bv{)}_j{f}_j$. If $Bv=0$, a nonzero constant term would create an arbitrage after choosing the sign of $v$. Hence $a-R1$ annihilates $\ker B$ and lies in the row span of $B$. Taking expectations gives exact factor-premium pricing. Full row rank makes the coefficients unique; if exposures have smaller rank, only the effective spanned factors are identified.

Suppose asset returns have the exact representation $r=a+Bf$, allowing unrestricted long and short portfolios. A position $v$ with $1^{T}v=0$ and $B^{T}v=0$ costs zero and has constant financial payoff $a^{T}v$. Absence of arbitrage therefore forces $a^{T}v=0$, since either sign of $v$ is allowed. The orthogonal complement identity gives $a\in \mathrm{span}(1,\mathrm{col}B)$. Taking expectations proves the displayed pricing relation, with $\lambda$ equal to the coefficients of $B$ in $a$ plus $\mathbb{E}f$. When $[1B]$ has full column rank, ${\lambda }_0$ is the return of a unit-cost zero-exposure portfolio and ${\lambda }_j$ is the expected return of a zero-cost portfolio with exposure one to factor $j$ and zero to the others. Without that rank condition the coefficients need not be unique, and a traded zero-exposure portfolio need not exist.

For centered residual random variables ${ϵ}_i$ with covariance matrix $\Gamma$, the residual of a portfolio has variance $w^T\Gamma w$. Pairwise uncorrelated residuals with uniformly bounded variances and weights $|w_i|\le W/n$ give $\mathrm{Var}(w^Tϵ)\le s^2{W}^2/n$, hence convergence to zero in mean-square convergence. For nonnegative weights summing to one, the sharper bound is $s^{2}W/n$. With correlated residuals, $|w_i|\le W/n$ and ${\sum }_{i,j}|{\Gamma }_{ij}|=o(n^2)$ also suffice. Individual bounded variances alone do not: if every residual equals the same nondegenerate centered $Z$, every fully invested portfolio retains residual $Z$. A one-sided upper bound on signed weights is insufficient even for uncorrelated residuals because it allows a fixed large short selling.

On the torus ${\mathbb{T}}^2$ with normalized Haar measure, choose $c_N$ so that $\int K_N=1$. The kernel is a nonnegative trigonometric polynomial. Its base $2+\cos s+\cos t$ has unique maximum four at the origin. Outside any fixed neighborhood its maximum is $q<4$, while in a smaller neighborhood of positive measure it is at least some $q^{\prime }>q$. Normalization consequently bounds the exterior kernel by $C(q/q^{\prime }{)}^N$, which tends to zero. Convolution with these kernels gives uniform trigonometric polynomial approximation to every continuous function on the torus.

If $L>0$ and $f$ is bounded and continuous at zero, the positive kernel $n{e}^{-nx}$ has mass $1-e^{-nL}$ on $[0,L]$, tending to one, and mass at most $e^{-n\delta }$ outside $[0,\delta ]$. Split the integral of $f(x)-f(0)$ at $\delta$: continuity makes the near part arbitrarily small, and boundedness times the exponentially small tail controls the far part. This proves the displayed endpoint approximate identity. A shrinking cutoff $L_n=n^{-1/2}$ still captures mass $1-e^{-\sqrt{n}}$ and therefore gives the same limit.

For a bounded Fourier multiplier operator $T$ on $L^2$ and a bump function $\varphi$, the displayed convolution identity follows by multiplying Fourier transforms. The convergence holds in $L^2$ and almost everywhere by the approximate identity theorem, after accounting for the integral of the kernel. The limit is $T(f)$ when $\int \varphi =1$. Smoothness and compact support alone do not imply this normalization.

Weightlessness is the absence of supporting or constraining forces that would give a body an apparent weight. It does not require zero gravitational acceleration. A freely falling cabin and a small freely falling object inside share the same acceleration in a uniform field, so neither needs to support the other.

The circular orbit as continuous free fall gives a sustained example. Real spacecraft have small residual relative accelerations from tidal forces, atmospheric drag force and onboard disturbances; the ideal common-fall explanation does not assert that every point in an extended spacecraft experiences exactly the same Newtonian gravitational field. ^[ref]