With unit lattice spacing and $\hslash =1$, a nearest-neighbour Heisenberg model has quadratic low-momentum ferromagnetic magnons and linear antiferromagnetic modes in the linear spin-wave approximation. The ferromagnetic polarized vacuum is an exact ground state, whereas the assumed antiferromagnetic ordered vacuum has quantum depletion of Néel order. In one dimension its logarithmic infrared divergence invalidates finite ordered magnetization and prevents the linear approximation from deciding the exact gap.

For a rotated nearest-neighbour chain, a full-zone sum of $JS(a_k^{†},a_{-k})M_k(a_k,a_{-k}^{†}{)}^T$ contains a normal-ordering constant $JSN$. Rewriting the classical constant as $-JNS(S+1)$ compensates it. One must not additionally halve the sum without changing mode counting.

For independent errors of common unknown variance in $Y_i=\beta x_i+{\epsilon }_i$, the ordinary least squares estimate is $\hat{\beta }={\sum }_i{x}_i{Y}_i/{\sum }_i{x}_i^2$. Fitting one parameter leaves $n-1$ residual degrees of freedom. Therefore $s^2={\sum }_i(Y_i-\hat{\beta }{x}_i{)}^2/(n-1)$ estimates the common error variance, and $s^2/{\sum }_i{x}_i^2$ estimates the slope variance. For errors with a Gaussian distribution, the slope's standardized deviation follows a Student's t-distribution with $n-1$ degrees of freedom. A common calibration bias does not appear as independent residual scatter.

Let $\mathbf{R}$ join the instantaneous position of charge q to the field point, with constant $\mathbf{\beta }=\mathbf{v}/c$. Transforming its rest-frame Coulomb field givesThe electric field is radial from the current position, reduced along the velocity and enhanced transversely. This present-position form is compatible with the causal retarded solution specifically because motion is uniform. There is no acceleration-dependent radiation field, although the moving charge has a magnetic field and transports electromagnetic energy.

For $m$ functions with sufficiently many derivatives, the $n$th derivative of their product distributes the $n$ differentiations among all factors. If factor $j$ is differentiated $k_j$ times, its term has the multinomial coefficient $n!/(k_1!\cdots k_m!)$, where the nonnegative integers $k_j$ sum to $n$. Induction on the number of factors proves this from the two-factor Leibniz rule: splitting off the final factor gives a binomial coefficient, which multiplies the previous multinomial coefficient to give the one displayed above. Equivalently, count the ways to choose which of the $n$ successive applications of the product rule differentiate each factor.

Substitute $y=x^{\sigma }{\sum }_{n\ge 0}{a}_n{x}^n$, $a_0\ne 0$, into the Legendre differential equation. The lowest powers give $\sigma (\sigma -1)=0$ and $\sigma (\sigma +1)a_1=0$; subsequent powers give the recurrence in the title. Thus exponent zero has independent even and odd coefficients, while exponent one requires $a_1=0$ and represents the odd solution. Zero is an ordinary point of the equation, so the exponent-one choice simply selects a solution that vanishes there.

The singular points at $\pm 1$ bound the generic radius of convergence about zero by one. Every nonterminating parity recurrence has successive-coefficient ratio tending to one in the variable $x^2$, giving radius exactly one. If $\lambda =N$ is a nonnegative integer, choosing the parity of $N$ makes the recurrence terminate at degree $N$, producing a Legendre polynomial with infinite radius. These are precisely the nonnegative parameters with a nonzero entire solution: even and odd parts of an entire solution would both have to be entire, so any nonzero nonterminating part is excluded.

For a simple polygon whose vertices are lattice points in ${\mathbb{Z}}^2$, its area is $A=I+B/2-1$, where $I$ counts interior lattice points and $B$ counts boundary lattice points.

A lattice cube is a cube whose vertices belong to a specified lattice. Its three edge vectors are mutually perpendicular lattice vectors of equal length. In the simple cubic lattice $a{\mathbb{Z}}^3$, an axis-aligned cube of side $a$ is a primitive unit cell.

Larger cubes can have lattice vertices while enclosing many primitive cells, so “lattice cube” is a geometric condition rather than a synonym for primitive cell. For a cube of side $Na$ in that example, its volume is $N^3$ times the primitive-cell volume. Consistent shared-boundary counting assigns one lattice point to the side-$a$ cell, even though eight corner points are visible in a closed drawing.

A lattice parallelepiped is a parallelepiped spanned by three linearly independent lattice vectors. If the vectors are a primitive basis of a Bravais lattice, translates of the corresponding half-open parallelepiped tile space without overlap, with one lattice point per cell under consistent boundary counting.

Its volume is $|{\mathbf{a}}_1\cdot ({\mathbf{a}}_2\times {\mathbf{a}}_3)|$. Choosing longer lattice vectors can instead produce a nonprimitive cell containing several lattice points; integer coordinates of its vertices alone do not prove primitiveness. The primitive version is useful for counting states and relating real-space volume to the volume of a reciprocal lattice cell.

For the ordinary Laguerre polynomial $L_n(t)={\sum }_{j=0}^n\left(\genfrac{}{}{0ex}{}{n}{j}\right)(-t{)}^j/j!$, apply the Laplace transform to each term, using $\mathcal{L}[t^j](p)=j!/p^{j+1}$ for $\mathrm{Re}p>0$. The binomial theorem givesThis proves the formula for every nonnegative integer $n$. In particular $L_2(t)=1-2t+t^2/2$ has transform $(p-1{)}^2/p^3$.

For a nonempty finite collection of Euclidean vectors, its sample mean $\bar{x}=n^{-1}{\sum }_i{x}_i$ uniquely minimizes the sum of squared Euclidean distances to a proposed centroid $z$. Expand each squared Euclidean norm around $\bar{x}$ and use ${\sum }_i(x_i-\bar{x})=0$ to cancel the cross term. The displayed identity leaves a nonnegative term $n\Vert z-\bar{x}{\Vert }^2$, equal to zero exactly at $z=\bar{x}$.

For two distinct centroids in a Euclidean space, equal squared Euclidean distances expand to the displayed affine hyperplane. It is their perpendicular bisector; each closed half-space consists of points at least as close to the centroid on that side. This proves weak linear separability after a two-centroid nearest-assignment step, allowing ties on the boundary. Coincident centroids are an essential exception: every point ties and the distance comparison supplies no nonzero normal vector.

An approximation to a Jacobian matrix obtained using finite differences instead of analytic partial derivatives. For a vector-valued function $F$, the forward-difference entry is $J_{ij}\approx (F_i(x+h{e}_j)-F_i(x))/h$, where $e_j$ is a coordinate basis vector. The step $h$ balances truncation and rounding errors; the resulting matrix can be used in a Newton root-finding iteration.

Compare an unknown allowed branch of parent $P$ with a reference allowed decay of mean lifetime ${\tau }_R$ and unit branching fraction. Its branching fraction is the parent mean lifetime ${\tau }_P$ times the inferred partial rate. The phase-space ratio $f_X/f_R$ and the total vector-plus-axial strengths $S_X/S_R$ enter separately. Approximating both strengths by their Fermi transition contributions and both phase-space integrals by $Q^5$ gives a useful scale, but does not establish a unique branch fraction if the axial strengths are unknown.

At Lick Observatory, the Hamilton spectrograph records starlight with an imprinted iodine reference. The reference tracks instrumental wavelength and profile changes during the science exposure. Barycentric correction and a time series of relative stellar shifts then support a dynamical companion model.

For a response proportional to $s/[(1+s{\tau }_1)(1+s{\tau }_2)]$ with positive time constants, its half-power width is $1/{\tau }_1+1/{\tau }_2$. Combining with the geometric-mean centre frequency gives the displayed quality factor; the arithmetic-geometric mean inequality proves the bound. Equality requires matched time constants and gives half-power frequencies $(\sqrt{2}\pm 1){\omega }_0$. Adjusting gains does not make this topology a high-Q resonant filter.

Suppose $f$ is a formal power series with invertible constant coefficient and $g$ satisfies $fg\equiv 1(\mathrm{mod}{x}^m)$. The updateis an inverse modulo $x^{2m}$, because $1-f{g}_{\mathrm{new}}=(1-fg{)}^2$ before truncation. Each update doubles the number of correct coefficients.

Start with the inverse of the constant coefficient and repeat, truncating all products to the required order. For $f=1-x$, the first approximations are $1$, $1+x$, and $1+x+x^2+x^3$. With a polynomial multiplication cost $M(n)$ satisfying the usual growth assumptions, computing the first $n$ coefficients takes $O(M(n))$ operations. The identity is algebraic and remains valid in characteristic two.

The difference in plane-wave propagation time between two antennas is the projected baseline along the source direction divided by the speed of light. Its sign depends on the ordering of the receivers. A monochromatic signal acquires relative phase $2\pi \nu {\tau }_g$; its change with Earth rotation is compensated by fringe stopping for the chosen reference direction.

The shortest useful projected baseline sets the lowest sampled spatial frequency, of order $b_{\min }/\lambda$, and hence an angular scale $\lambda /b_{\min }$. Operational maximum recoverable source sizes also depend on source shape, coverage, weighting and primary beam. The longest baseline sets a different scale, the synthesized angular resolution. The number of antennas alone does not fix either limiting baseline or supply zero-spacing total power.

A birefringent plate with retardation $\Gamma =t\Delta n$ gives $I/I_P={\sin }^2(2\alpha ){\sin }^2(\pi \Gamma /\lambda )$. At non-extinction azimuths, wavelengths satisfying $\Gamma =m\lambda$ are removed by a crossed analyser. At a principal-axis azimuth every wavelength is extinguished.