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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]

For two unbiased estimators $A,B$ of the same parameter with variances $V_X,V_Y$ and covariance $C$, the fixed-weight estimator $wA+(1-w)B$ is unbiased and has variance $w^2{V}_X+(1-w{)}^2{V}_Y+2w(1-w)C$. Completing the square gives $w_{\ast }=(V_Y-C)/(V_X+V_Y-2C)$ and minimum variance $(V_X{V}_Y-C^2)/(V_X+V_Y-2C)$ when the denominator is positive. These are optimal among fixed linear unbiased combinations. Substituting weights estimated from the same observations need not preserve unbiasedness.

For standard Cauchy distribution probability density function $f$ and $k>0$, this estimates its upper-tail probability without sampling a heavy tail. The function $f(ku)$ decreases on $[0,1]$, while $f(k(1-u))$ increases, so their covariance is nonpositive. The paired estimator is unbiased and has no larger variance than an estimator using the same number $2r$ of independent probability density function evaluations.

For an antisymmetric payoff matrix, such a vector guarantees that an opponent playing rows cannot win positively against it. Antisymmetry also gives $p^{T}A\ge 0$, so the same mixed strategy guarantees nonnegative payoff as a row strategy and proves game value zero. Furthermore $p^{T}Ap=0$ forces $(Ap{)}_i=0$ whenever $p_i>0$. To prove uniqueness, use one certified strategy $q$: any other optimal $p$ satisfies $q^{T}Ap=-(Aq{)}^{T}p$, so a strict negative component of $Aq$ forces the corresponding probability in $p$ to vanish.

A harmonic conjugate $\varphi$ satisfying ${\varphi }_{,1}=w_{,2}$ and ${\varphi }_{,2}=-w_{,1}$ makes $F=\varphi +iw$ a holomorphic function. Its derivative packages the two antiplane shear stresses into the displayed complex field. On multiply connected domains a global potential additionally requires the appropriate periods to vanish; prescribed displacement jumps can instead be represented by branches of the potential.

For subsonic motion $|V|<c$, an antiplane shear displacement $w(x_1-Vt,x_2)$ satisfies ${\beta }^2{w}_{xx}+w_{x_2{x}_2}=0$. The displayed rescaling makes it a harmonic function of $(x,y)$, so locally $w=\mathrm{Re}H(x+iy)$ for a holomorphic function $H$. Set $F=\beta H$. The Cauchy-Riemann equations then give $\mu F^{\prime }=\beta {\sigma }_{13}-i{\sigma }_{23}$. Keeping this potential normalization explicit avoids losing a factor of $\beta$ in the dynamic stress field. At sonic or supersonic speeds the elliptic harmonic reduction does not apply.

A centred fan consists of straight characteristic curves emanating from one point. Stress and antiplane velocity are constant on each ray and depend on the polar angle $\varphi$. The radial tangent condition is $(\cos \varphi ,\sin \varphi )\parallel (-f_b,f_a)$, so $f_a\cos \varphi +f_b\sin \varphi =0$. A boundary ray carrying a traction-free crack-face stress $b=0$ has slope $\tan {\varphi }_b=-f_a/f_b$, interpreted as vertical when $f_b=0$. For the elliptic yield condition $a^2/A^2+b^2/B^2=1$, a forward fan has $a=-A^2\sin \varphi /H$, $b=B^2\cos \varphi /H$, $H=(A^2{\sin }^2\varphi +B^2{\cos }^2\varphi {)}^{1/2}$. The boundary rays are vertical; the adjacent rear regions have constant stresses $(\mp A,0)$. Velocity is a function of angle whose magnitude requires further loading data.

An endomorphism $J$ that is an antilinear map with $J^2=-I$ equips a complex vector space with a quaternionic structure: complex scalar multiplication supplies the quaternion generator $i$ and $J$ supplies $j$, satisfying $Ji=-iJ$. In finite dimensions the complex dimension must be even. Indeed, there is no $J$-invariant complex line: $Jv=cv$ for $v\ne 0$ would imply $-v=|c{|}^{2}v$. Consequently $v,Jv$ span a $J$-invariant complex two-plane; passing to its quotient vector space and repeating proves evenness.

In complex dimension four the invariant complex two-planes are quaternionic lines, forming the quaternionic projective space ${\mathbb{HP}}^1\cong S^4$. This is the Euclidean reality structure on twistor space used in twistor theory.

On a regular bipartite graph with coordination number $q$, replace each bond product by ${\sigma }_B{\sigma }_W\simeq M_W{\sigma }_B+M_B{\sigma }_W-M_B{M}_W$. For $N$ sites there are $Nq/2$ bonds, so the constant in the effective Hamiltonian is $-NqJ{M}_B{M}_W/2$. The mean-field self-consistency equations are $M_B=\tanh \beta (h+g-qJ{M}_W)$ and $M_W=\tanh \beta (h-g-qJ{M}_B)$. At zero fields the antisymmetric branch has a continuous mean-field Ising critical bifurcation at $k_B{T}_c=qJ$.

The stable bundle consists of the transverse directions exponentially contracted in positive time. The unstable bundle has the corresponding negative-time property. A positive time change of a smooth flow can tilt these bundles in the flow direction, even though it preserves the underlying orbits.

The unstable bundle consists of transverse directions exponentially contracted in negative time. Together with the stable bundle and the flow direction it gives the invariant splitting of an Anosov flow. Under a positive time change its graph coefficient is obtained from the convergent past integral $-\rho (x{)}^{-1}{\int }_{-\infty }^{0}d\rho (D{\varphi }_{r}v)dr$.

The Livšic theorem turns periodic-orbit data into a global derivative along a hyperbolic flow. Let ${\Phi }_t$ be a transitive Anosov flow on a compact smooth manifold, with generating vector field $F$, and let $h$ be a real Hölder continuous observable. There is a Hölder continuous function $u$, differentiable along flow lines, satisfyingif and only if ${\int }_0^{T}h({\Phi }_{t}p)dt=0$ for every periodic orbit $p$ of period $T$. The solution preserves the Hölder exponent.^[ref]

Necessity is immediate by putting ${\Phi }_{T}p=p$ in the identity. Two solutions differ by a continuous invariant function. Transitivity supplies a dense orbit; their difference is constant on that orbit, and continuity makes it constant everywhere. Thus the solution is unique up to an additive constant.

The harder sufficiency argument defines $u({\Phi }_t{x}_0)$ by integrating $h$ along a dense orbit. The Anosov closing lemma converts a sufficiently close return into a nearby periodic trajectory. Its vanishing integral and exponential shadowing bound the discrepancy between the endpoint values in terms of their distance. This gives Hölder control on the dense orbit and permits extension to the whole manifold.

When both the flow and $h$ are smooth, a separate regularity theorem gives smooth $u$. This does not follow merely from the preceding continuity argument.^[ref] The theorem is named for A. N. Livšic.