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A Hamiltonian path preserving the history subspace can be analysed using its restricted spectral gap. If this gap is $\Omega (T^{-2})$ and the derivative has operator norm $O(1)$, a runtime criterion proportional to $\Vert \dot{H}{\Vert }^2/(ϵ{\Delta }^3)$ permits computational history state preparation in time $O(T^6/ϵ)$. Other invariant sectors do not cause leakage from a state initialized in the history subspace.

For a one-particle Hilbert space $H$, the bosonic or fermionic Fock space is ${\mathcal{F}}_{\pm }(H)={⨁}_{n=0}^{\infty }{H}^{{\otimes }_{\pm }n}$, using symmetric or antisymmetric tensor powers. Its zeroth summand is the one-dimensional vacuum sector. Creation operators and annihilation operators change particle number. The bosonic Fock space and fermionic Fock space implement the corresponding exchange statistics.

Optimization of a quadratic form over binary choices, here using signs $x_i\in \{-1,1\}$. Lifting $x{x}^T$ turns the objective into a matrix trace; dropping the rank-one condition gives a semidefinite relaxation of binary quadratic optimization.

Two primes differing by two form a twin-prime pair, for example $3,5$ or $11,13$. The twin-prime upper bound using Bombieri–Vinogradov limits their number without proving there are infinitely many.

The Fermat test looks for a coprime base $a$ with $a^{N-1}\not\equiv 1(\mathrm{mod}N)$. Such a witness proves compositeness by the Fermat little theorem, but Carmichael numbers pass for every coprime base.

A potential dipole in three dimensions is the potential flow $\mathbf{u}=\nabla \varphi$ with $\varphi =\mathbf{d}\cdot \mathbf{x}/r^3$. Away from the origin its potential is a harmonic function, so it is also a constant-pressure Stokes flow. Its velocity decays as $r^{-3}$ and its vorticity is zero. It is distinct from a stresslet, whose leading velocity decays as $r^{-2}$.

A two-dimensional incompressible line source with area flux $Q$ has radial velocity $Q/(2\pi r)$, obtained from flux conservation around a circle. It is irrotational away from the source, with potential $(Q/(2\pi ))\ln r$. For a circular expanding interface, $Q=2\pi a{a}_t$ and $u_r=a{a}_t/r$, whose radial strain is $-a{a}_t/r^2$. That strain supplies a normal viscous stress even though the velocity is a potential flow.

A real positive-semidefinite kernel satisfies $|k(x,y){|}^2\le k(x,x)k(y,y)$. This follows from positivity of its two-point kernel matrix. If a diagonal value is zero, the corresponding off-diagonal values must also vanish.

In the compact-domain spectral setting, a Mercer kernel is a continuous symmetric positive-semidefinite kernel on a compact space equipped with a finite full-support measure. Mercer's theorem yields nonnegative eigenvalues and an expansion $k(x,z)={\sum }_j{\lambda }_j{e}_j(x)e_j(z)$, giving features $(\sqrt{{\lambda }_j}{e}_j(x){)}_j$ in ${\ell }^2$. The finite-Gram positivity condition on arbitrary domains is more general than these spectral hypotheses; all such kernels still have a Reproducing kernel Hilbert space realization.

A family of $L\ge 1$ lines in ${\mathbb{R}}^3$ imposes at most $L(d+1)$ linear conditions on degree-at-most-$d$ polynomials, by polynomial restriction to a line. The coefficient space has dimension $\left(\genfrac{}{}{0ex}{}{d+3}{3}\right)$. Whenever $(d+2)(d+3)>6L$, a nonzero common vanishing polynomial exists. Taking $d=\lceil \sqrt{6L}\rceil$ gives degree less than $4\sqrt{L}$.

The even part of a polynomial $P$ is $(P(z)+P(-z))/2$. It retains exactly the terms whose exponents are even. Applied to the binomial theorem, it extracts sums of even-indexed binomial coefficients.

Let $V(p)$ be the number of sign changes in the list of nonzero coefficients of a real polynomial, arranged by increasing or decreasing degree. The positive roots of a polynomial, counted with multiplicity of a root, satisfyThe bound follows by induction on the nonzero terms after removing the lowest power of $x$. Differentiation removes the constant coefficient. If it and the next coefficient have opposite signs, the Rolle root count with multiplicities loses at most one root and the sign-change count loses one. If their signs agree, the derivative must have an extra root before the first positive root, so neither count needs that extra allowance. The parity statement follows by comparing the signs near zero and at positive infinity: roots of odd multiplicity reverse the sign, and roots of even multiplicity preserve it. Apply the same rule to $p(-x)$ to bound negative roots.

A symmetric polynomial is unchanged by permutations of its variables. The Fundamental theorem of symmetric polynomials expresses it as a polynomial in the elementary symmetric polynomials. The square of the Vandermonde determinant is an important example.

Two polynomials over a field are coprime if their only common factors are nonzero constants. Equivalently, their greatest common divisor is a unit in the polynomial ring. In one variable this is equivalent to a Bezout identity $Af+Bg=1$ with polynomial coefficients. Independent members of a pencil spanned by coprime polynomials are themselves coprime.

If a planar equilibrium with purely imaginary linear eigenvalues has radial return map $r\mapsto r+\kappa r^3+O(r^4)$, then $\kappa <0$ gives nonlinear attraction and $\kappa >0$ gives repulsion. Vanishing linear decay therefore does not imply a centre.

The astronomical unit is a length scale used for Solar System distances, approximately $1.5\times 10^{13}\mathrm{cm}$. Coarse estimates sometimes round it to $10^{13}\mathrm{cm}$. Numerical coefficients in planet formation estimates depend on which approximation is used.

For three non-collinear points with position vectors $\mathbf{a},\mathbf{b},\mathbf{c}$, a nonzero normal vector isThe plane is the set of $\mathbf{r}$ satisfying $(\mathbf{r}-\mathbf{a})\cdot \mathbf{N}=0$. Expanding the cross product and using the scalar triple product givesNon-collinearity is essential: otherwise $\mathbf{N}=0$ and the equation fails to select a unique plane.

A line $r=b+sm$ intersects a plane $(r-a)\cdot n=0$ exactly when the scalar equation $s(m\cdot n)=(a-b)\cdot n$ can be solved. If $m\cdot n\ne 0$, there is one point. If that product vanishes, the line is either contained in the plane or disjoint from it. Direction and normal vectors must be nonzero.

Planes with common unit normal $n$ have distance $|(a_1-a_2)\cdot n|$. Every joining displacement has this normal component, giving a lower bound by its absolute value; a displacement parallel to the normal attains the bound.

For a common nonzero normal vector $\mathbf{n}$, not necessarily of unit Euclidean norm, the planes $\mathbf{n}\cdot \mathbf{x}=d_1$ and $\mathbf{n}\cdot \mathbf{x}=d_2$ have distance $|d_2-d_1|/\Vert \mathbf{n}\Vert$. Thus $2x-y+2z=3$ and $2x-y+2z=9$ are distance $6/3=2$ apart. First put both plane equations into the same normal-vector scale: subtracting offsets from independently rescaled equations need not give a geometric distance.

The Felsenstein pruning algorithm computes a site likelihood function by sum-product belief propagation from leaves to root. For a finite-state Markov kernel $K$, the subtree likelihood function obeys $L_v(a)={\prod }_{w\text{ child of }v}{\sum }_{b}K(a,b)L_w(b)$, with observed-state indicators at leaves. Summing against the root probability distribution gives the site likelihood function. An outward pass gives edge posterior probabilities and expected transition counts for the expectation-maximization algorithm.