An explicit linear multistep method that integrates a polynomial extrapolation of the past derivative values. The displayed two-step member is second order in time: the linear interpolant through $f_{n-1},f_n$, integrated over the next step, gives coefficients $3/2,-1/2$. For a spatially discretized partial differential equation, the spatial error must also be included in the total local defect.

The exponential-symbol order criterion for a multistep method gives leading defect $-(1+5a)z^3/12$, so the family has order two except at $a=-1/5$, where it has order three. The root condition for a multistep method holds exactly for $-1\le a<1$, and the convergent A-stable members are exactly $0\le a<1$. At $a=0$ it reduces to the trapezoidal rule after removal of a zero root. At $a=1$ a repeated root one destroys convergence of a numerical method of the original recurrence despite a common polynomial factor and a formally vanishing second-order residual.

For a consistent zero-stable real linear $k$-step method with constant coefficients and only first-derivative evaluations, the attainable order is at most $k+1$ for odd $k$ and $k+2$ for even $k$. For an explicit method the bound is $k$. These are order barriers under the root condition for a multistep method, not bounds on multiderivative methods or on arbitrary variable-step formulations. The Second Dahlquist barrier adds the stronger bound of order two for irreducible A-stable methods.

Integrating a quadratic interpolant of the derivative over two steps gives this linear multistep method, which has order four. Its local truncation error is $-h^5{y}^{(5)}(t_{n+1})/90+O(h^7)$ and its zero-stability polynomial is $\rho (\zeta )={\zeta }^2-1$, with simple roots $1,-1$. The Dahlquist equivalence theorem therefore gives fourth-order convergence with sufficiently accurate starts. This attains the even two-step first Dahlquist barrier. Its undamped parasitic zero-step mode need not have favorable long-time absolute stability; finite-time convergence and stiff damping are separate requirements.

Take $\sigma (\zeta )=\zeta [(5+\alpha ){\zeta }^2-(4+8\alpha )\zeta +11-5\alpha ]/6$. The root condition for a multistep method holds exactly for $-1<\alpha <1$. Since ${\rho }^{\prime }(1)=\sigma (1)=2(1-\alpha )$, these and only these parameters give a convergent method. Its exponential-symbol order criterion for a multistep method has defect $-(\alpha +5)z^4/12+O(z^5)$, giving formal order three except at $\alpha =-5$, where the first defect is $z^5/10$. None of the convergent family is A-stable: for a parasitic root $r=\alpha +i\sqrt{1-{\alpha }^2}$, perturbation of $\rho (r(z))-z\sigma (r(z))=0$ gives $\mathrm{Re}[r^{\prime }(0)/r]=(\alpha -1)/12<0$. A small negative real $z$ therefore moves it outside the unit disk. At $\alpha =1$, cancellation would remove a double unit root, but the original recurrence retains that unstable parasitic behavior; it is not legitimately replaced by the reduced Backward Euler recurrence for arbitrary starting values.

A system of linear inequalities requires several linear inequalities to hold simultaneously. The notation $Ax\le b$ compares coordinates componentwise, giving one inequality per row of the matrix $A$. Its solution set is the intersection of the corresponding half-spaces. For non-strict inequalities it is a linear polyhedron, which may be empty. For example, $x\ge 1$, $y\ge 1$, $x+y\le 1$ are inconsistent: adding their equivalent forms $-x\le -1$, $-y\le -1$, $x+y\le 1$ gives $0\le -1$. A Farkas certificate for linear inequalities generalizes this cancellation.

If chosen logarithms of algebraic numbers are linearly independent over $\mathbb{Q}$, then they, together with one, are linearly independent over $\overline{\mathbb{Q}}$. More generally a nonzero homogeneous linear combination with algebraic coefficients is transcendental. Quantitative lower bounds for integer coefficients are a separate, effective aspect of logarithmic-form theory.

Under the convention that a register emits each shifted-out symbol, its first $d$ output symbols are exactly its initial fill. Thus a binary register of length $d$ has $2^d-1$ distinct outputs from nonzero fills. For lengths $r<s$, at least one longer-register output is absent from the shorter register, guaranteeing a pair of unequal nonzero outputs. The two sets need not be disjoint: recurrences $s_{n+1}=s_n$ and $s_{n+2}=s_n$ both generate constant ones. Nor must they intersect: the length-one recurrence and $s_{n+2}=s_{n+1}+s_n$ have no common nonzero output, since constant ones fail the second recurrence.

A displacement field $\mathbf{u}(\mathbf{x})$ gives the change in position of each material point from a reference configuration.

A bending moment is an internal torque transmitted across a cut through a beam or elastic plate. For a slender linear elastic beam, $M=EI\kappa$ up to sign convention, where $E$ is Young's modulus, $I$ the second moment of area, and $\kappa$ the centreline curvature. The corresponding bending response of an elastic plate is described by its bending stiffness and curvature tensor.

An elastic stiffness tensor maps a small displacement gradient tensor to the stress tensor. Minor symmetries in each index pair ensure symmetric stress and dependence only on symmetric strain; major symmetry $c_{ijpq}=c_{pqij}$ follows from an elastic energy and underlies reciprocity. Stability requires positive elastic energy on nonzero strains. In an isotropic material, $c_{ijpq}=\lambda {\delta }_{ij}{\delta }_{pq}+\mu ({\delta }_{ip}{\delta }_{jq}+{\delta }_{iq}{\delta }_{jp})$, with Lamé parameters $\lambda ,\mu$.

Antiplane shear has displacement normal to a plane and independent of that normal coordinate. In an isotropic linear elastic body, ${\sigma }_{13}=\mu w_{,1}$ and ${\sigma }_{23}=\mu w_{,2}$, together with their symmetric partners, are the only nonzero stress tensor entries. Static balance without body force reduces to the Laplace equation for $w$.

The bulk modulus relates hydrostatic compressive pressure to fractional volume compression. In isotropic three-dimensional linear elasticity, the spherical part of the stress tensor is $\kappa \mathrm{tr}(\epsilon )I$, so uniform dilation has strain energy density $\kappa (\mathrm{tr}\epsilon {)}^2/2$. Positive bulk modulus and positive shear modulus make the isotropic elastic stiffness tensor positive on nonzero symmetric strains.

For a homogeneous comparison elastic stiffness tensor $C^0$, the elastic strain Green operator maps an elastic polarization to minus the induced infinitesimal strain tensor. In a bounded domain the induced displacement field has zero boundary data; in an infinite medium it is the decaying perturbation. The defining static elastic equilibrium equation is $\nabla \cdot (C^0\mathrm{sym}\nabla u+\tau )=0$. The bounded-domain operator is self-adjoint, annihilates constant elastic polarizations, has zero mean output, and satisfies $\Gamma C^0\Gamma =\Gamma$.

Apply the divergence theorem to ${\partial }_k(x_i{\sigma }_{jk})$ and use momentum balance ${\sigma }_{jk,k}=\rho (a_j-b_j)$ and stress symmetry to obtain the displayed mean stress identity. Applying the same theorem to the infinitesimal strain tensor gives $V{\overline{e}}_{ij}=\frac{1}{2}{\int }_S(u_i{n}_j+u_j{n}_i)dS$. The acceleration term cannot be dropped during a transient.

For a nonrelativistic polytropic equation of state $P=K{\rho }^{4/3}$ during matter domination, $c_s^2\propto a^{-1}$ and the comoving Jeans wavenumber is constant. The leading density equation becomes $\ddot{\delta }+4\dot{\delta }/(3t)+[2(q-1)/(3t^2)]\delta =0$, giving the displayed powers. A growing mode exists for $q<1$; at $q=1$ one mode is constant. At $q=25/24$ the repeated root has the second solution $t^{-1/6}\ln t$. For $q>25/24$ both real solutions oscillate in $\ln t$ with envelope $t^{-1/6}$. Thus the instability threshold and the onset of oscillatory solutions are distinct in an expanding background.

For a flat Newtonian expanding background $R\propto t^{2/3}$ with polytropic equation of state $p=K{\rho }^{\gamma }$, a comoving density contrast satisfies $\ddot{\delta }+4\dot{\delta }/(3t)+[{\Lambda }^2{t}^{-2\nu -2}-2/(3t^2)]\delta =0$. Here $\Lambda =kc(t_0)t_0^{\gamma -1/3}$ and $\nu =\gamma -4/3$. For $\nu >0$, substitution $z=\Lambda /(\nu t^{\nu })$ and $\delta =t^{-1/6}y(z)$ gives the Bessel equation of order $\lambda$. A universally valid basis uses the Bessel function of the first kind and Bessel function of the second kind, $t^{-1/6}{J}_{\lambda }(z)$ and $t^{-1/6}{Y}_{\lambda }(z)$. When $\lambda$ is not an integer, $J_{-\lambda }$ can replace $Y_{\lambda }$. For $4/3<\gamma <5/3$, the early oscillatory envelope is $t^{(3\gamma -5)/6}$; late modes have leading powers $t^{2/3}$ and $t^{-1}$. Late growth requires a nonzero coefficient of the growing branch.

A linear code over the two-element finite field. Its dimension $r$ gives $2^r$ codewords. Its weight enumerator in the convention $W_C(s,t)={\sum }_{c\in C}{s}^{w(c)}{t}^{n-w(c)}$ is symmetric in $s,t$ exactly when the all-ones word belongs to $C$: adding that word complements every bit and replaces weight $w$ by $n-w$. Conversely symmetry makes the coefficient of $s^n$ equal the coefficient of $t^n$, which is one.

For a linear code of length $n$ and dimension $k$ over ${\mathbb{F}}_q$, the code rate is $R=k/n$. There are $q^k$ possible messages represented by $q^k$ codewords of length $n$. More generally a block code $C\subseteq {\mathbb{F}}_q^n$ has rate ${\log }_q|C|/n$. This is a dimensionless redundancy measure, distinct from information transmitted per unit physical time.

Let $1\le k\le n\le q$ and let $a_1,\dots ,a_n$ be distinct elements of a finite field ${\mathbb{F}}_q$. The Reed-Solomon code consists of evaluation vectors $(f(a_1),\dots ,f(a_n))$ for polynomials $f$ of polynomial degree less than $k$. It is a linear code with parameters $[n,k,n-k+1]$ and therefore a maximum distance separable code.

Indeed, evaluation is an injective linear map: a nonzero polynomial of polynomial degree at most $k-1$ cannot vanish at all $n\ge k$ evaluation points. The same polynomial root bound shows that every nonzero codeword has at least $n-k+1$ nonzero coordinates. Equality is attained by a polynomial vanishing at any chosen $k-1$ evaluation points. Thus the minimum Hamming distance is exactly $n-k+1$, allowing correction of up to $\lfloor (n-k)/2\rfloor$ unknown symbol errors.

Primitive cyclic Reed-Solomon codes use all nonzero field elements, ordered as powers of a primitive root of unity. An arbitrary evaluation set need not give a cyclic code.