A coeval binary has components formed at the same time. A population of such binary stars can still have a distribution of system ages. If component masses have independence and the state at age $t$ is determined separately for each component, the states have conditional independence given $t$. If $g(t)$ and $w(t)$ are the conditional probabilities of two states, the unconditional unordered pair fractions are $\mathbb{E}[g(t{)}^2]$ and $2\mathbb{E}[g(t)w(t)]$, not generally products of marginal state fractions.

A detached binary has both components lying inside their respective Roche lobes, so neither currently transfers matter by Roche-lobe overflow. A detached system may still exchange matter through a stellar wind.

A binary binary mass ratio is the quotient of its component masses. For mass-transfer stability, specify the donor-to-accretor convention $q=M_d/M_a$ explicitly, since the inverse convention also occurs. Mass-ratio reversal changes which component is more massive.

For a circular Kepler orbit, neglecting stellar spin, the orbital angular momentum of a binary star is $J=\mu a^2\Omega =\mu \sqrt{GMa}$, where $M$ is total mass, $\mu$ is the reduced mass, and $a$ is separation. Kepler's third law also gives $J=G^{2/3}{P}^{1/3}{M}_1{M}_2/[(2\pi {)}^{1/3}{M}^{1/3}]$. Distances from the center of mass are $a{M}_2/M$ and $a{M}_1/M$; adding their individual orbital angular momenta proves the formula.

For a real rectangular matrix $S$, maximize $p^{T}Sq$ over separate sign choices for $p$ and $q$. With $x=(p,q)$, the symmetric block matrix $A=\frac{1}{2}\left(\begin{array}{cc}0& S\\ S^T& 0\end{array}\right)$ satisfies $x^{T}Ax=p^{T}Sq$. Its zero diagonal blocks allow the Krivine rounding scheme to linearize the cross-block expected objective.

A bimatrix game is nondegenerate if every mixed strategy with $k$ positive coordinates has at most $k$ pure best responses for the other player. This condition prevents ties in the leaving-variable choice along the usual Lemke-Howson algorithm path and allows unique continuation once the dropped label has been selected.

A square bimatrix game is symmetric when exchanging players exchanges their payoffs: its payoff matrices are $P,Q$ with $Q=P^T$. If $(x,y)$ is a Nash equilibrium, then $(y,x)$ is also a Nash equilibrium. A symmetric bimatrix game can still have asymmetric equilibria, so the two strategy vectors must be kept separate in the Lemke-Howson algorithm.

A Type I sum in analytic number theory has the form ${\sum }_{d\le U}{a}_d{\sum }_{m\in I_d}f(dm)$, with a short weighted variable and a relatively long inner sum whose coefficients are simple. The Vaughan identity produces these sums when the Möbius function is truncated.

A Type II sum has the form ${\sum }_{d\sim D}{\sum }_{m\sim M}{a}_d{b}_{m}f(dm)$ with two nontrivially weighted variables, both long enough for a Cauchy-Schwarz inequality and a large sieve estimate. The Vaughan identity supplies this bilinear structure for sums weighted by the Von Mangoldt function.

A real symmetric bilinear form is positive-definite when $B(v,v)>0$ for every nonzero vector. It defines an inner product and the associated norm $\sqrt{B(v,v)}$. Symmetry is part of this definition: positive diagonal values by themselves do not imply symmetry.^[ref]

The identity matrix $A=\mathrm{diag}(1,1)$ gives $B(v,w)=v_1{w}_1+v_2{w}_2$ and $B(v,v)=v_1^2+v_2^2$. For $v=(3,4)$ this is $25$, giving norm $5$. A weighted example, $A=\mathrm{diag}(4,1)$, gives norm $\sqrt{4x^2+y^2}$: the first coordinate costs twice as much in length. Such inner products let geometry assign different weights to different directions.

Every positive-definite bilinear form is nondegenerate: a vector in its radical of a bilinear form has $B(v,v)=0$, so positivity forces $v=0$. The converse fails, as $\mathrm{diag}(1,-1)$ shows. Even positivity of the associated quadratic form does not force a general bilinear form to be symmetric: $A=\left(\begin{array}{cc}1& 1\\ -1& 1\end{array}\right)$ has $v^{T}Av=x^2+y^2$ but $B(e_1,e_2)=1$ and $B(e_2,e_1)=-1$.

The complex Killing form is not positive-definite on the entire complex Cartan subalgebra; its restriction to the Euclidean subspace of a Cartan subalgebra is a real positive-definite form.

For real $f$, a candidate $u\in U\subset C(K,\mathbb{R})$, and active set $E=\{x:|f(x)-u(x)|=\Vert f-u{\Vert }_{\infty }\}$, the candidate is best precisely when every $v\in U$ has some $x\in E$ with $(f(x)-u(x))v(x)\le 0$. A direction with strictly matching error signs on the whole active set gives a uniformly improving small perturbation by compactness. Conversely one oppositely signed active value already prevents improvement. This criterion yields the Chebyshev alternation theorem.

For a fixed degree bound, two best approximating polynomials have a best approximating average. At each active point of that average their errors agree, hence the two polynomials agree. There must be at least two more active points than the degree: otherwise polynomial interpolation constructs a strictly improving sign-matching perturbation. The difference has too many zeros and must vanish. Finite-dimensional coefficient compactness also gives existence.

The spherical second-kind function is $n_{\ell }(z)=\sqrt{\pi /(2z)}{Y}_{\ell +1/2}(z)$. Together with the first-kind Spherical Bessel function, it spans the radial free-wave solutions; $n_{\ell }(z)\sim -\cos (z-\ell \pi /2)/z$ at large positive $z$.

For fixed $0<\alpha <\pi /2$, the Bessel function of the first kind has an interior nondegenerate stationary point in its cosine integral. The stationary phase method gives the displayed oscillatory leading term and an additive $O(n^{-3/2})$ remainder. The approximation loses uniformity near $\alpha =0$, where the stationary point reaches the endpoint and the curvature vanishes. For negative integer-order arguments, parity supplies a factor $(-1{)}^n$.

For steady flow of constant-density fluid in a uniformly rotating frame with conservative body forces, the displayed quantity is constant along each relative streamline. The centrifugal potential contributes the negative quadratic term, and the Coriolis force contributes no work because it is perpendicular to the relative velocity. The result follows by dotting the rotating Euler equations for an inviscid fluid with that velocity.

With the streamfunction convention $(u,v)=({\psi }_y,-{\psi }_x)$, constant scalar vorticity gives $\mathbf{u}\times \mathbf{\omega }=-\omega \nabla \psi$. For steady inviscid constant-density flow without body force, this integrates tothroughout a connected fluid region. Reversing the streamfunction convention changes the sign of its term; the declared convention matters.

For nonnegative rewards, a nonnegative value satisfying the maximizing Bellman equation dominates every competing policy, even for $\beta \ge 1$. Iterate the one-step inequality, discard the nonnegative terminal value, then apply the monotone convergence theorem to the accumulated reward.

The reduced BCS pairing Hamiltonian retains scattering of opposite-momentum, opposite-spin Cooper pairs:A mean-field approximation introduces the BCS anomalous average and pairing field ${\Delta }_k=-{\sum }_{k^{\prime }}{V}_{k,k^{\prime }}⟨b_{k^{\prime }}⟩$. The subtraction $-\sum V_{k,k^{\prime }}⟨b_k^{†}⟩⟨b_{k^{\prime }}⟩$ corrects the double counting of the mean-field interaction energy.

The BCS anomalous average is the pair expectation $⟨b_k⟩=⟨c_{-k\downarrow }{c}_{k\uparrow }⟩$. In the BCS ground state it is $u_k{v}_k={\Delta }_k/(2E_k)$ for real gap at zero temperature. At temperature $T$ it becomes ${\Delta }_k\tanh (E_k/(2k_{B}T))/(2E_k)$. Substitution into the pairing-field definition gives the BCS gap equation.

For real gap and the particle-hole block $\left(\begin{array}{cc}{\xi }_k& -{\Delta }_k\\ -{\Delta }_k& -{\xi }_k\end{array}\right)$, the BCS coherence factors obey $u_k^2=(1+{\xi }_k/E_k)/2$, $v_k^2=(1-{\xi }_k/E_k)/2$ and $u_k{v}_k={\Delta }_k/(2E_k)$, where $E_k=\sqrt{{\xi }_k^2+{\Delta }_k^2}$. They diagonalize the Bogoliubov--de Gennes Hamiltonian and give the empty/pair amplitudes of the BCS ground state. Their relative sign depends on the pairing convention.

For an original check, write $x={\xi }_k/\Delta$ with constant real $\Delta >0$. At $x=0$, the BCS coherence factors give $u_k^2=v_k^2=1/2$. At $x=1$, they give $(u_k^2,v_k^2)=(0.853553,0.146447)$; at $x=-1$ these values exchange places. Thus the zero-temperature occupation $v_k^2$ rounds the normal Fermi sea's step over an energy interval of order $\Delta$. This mixing is strongest at the Fermi surface, where the BCS quasiparticle spectrum has its minimum.