Let $R,S$ commute. On an open set where $\{R^n\}$ is equicontinuous, so is $\{S\circ R^n\}$, by the rational map is Lipschitz in the chordal metric bound. Since this equals $\{R^n\circ S\}$, equicontinuity descends through the open map $S$: near $S(x)$ every nearby value has a preimage near $x$, and the corresponding estimates hold for all $n$. Thus $S(F(R))\subseteq F(R)$. A degree-at-least-two Julia set contains at least three points, all omitted by the iterates of $S$ on $F(R)$. The three-value Montel theorem gives $F(R)\subseteq F(S)$. Exchanging the maps proves equality of their Fatou and Julia sets. The degree hypothesis excludes Möbius examples with only one Julia point.

For $1<k<n-1$, the maximal cliques of $J(n,k)$ are the families containing a fixed $(k-1)$-subset, of size $n-k+1$, and the families contained in a fixed $(k+1)$-subset, of size $k+1$. To see why these exhaust the possibilities, write two adjacent vertices as $D\cup \{a\}$ and $D\cup \{b\}$. A common adjacent vertex either contains $D$ or lies in $D\cup \{a,b\}$. A vertex of the first kind outside that union is not adjacent to a vertex of the second kind omitting an element of $D$, so a clique cannot mix these alternatives. For $J(n,3)$ and $n\ge 5$, $n\ne 6$, the clique sizes $n-2$ and four distinguish the families containing pairs. Their incidence recovers $J(n,2)$, whose stars of size $n-1$ distinguish the individual ground-set points from its triangular cliques.

One edge of a Johnson graph replaces one element of a $k$-subset, so it can increase intersection with a fixed target by at most one. Replacing each element outside the target by one missing target element gives a path reaching this lower bound. Equal-distance ordered pairs have equal sizes of intersection, the two differences and the remaining ground set; a permutation matching these four regions proves distance transitivity.

The Jacobson radical is the intersection of the ideals ${\mathrm{Ann}}_R(S)$ over all simple modules, where $\mathrm{Ann}$ is the annihilator of a module, so it is a two-sided ideal. If $y\in J(R)$, the left ideal $R(1-y)$ cannot be proper, since a containing maximal left ideal would also contain $y$ and hence $1$. Thus $c(1-y)=1$ for some $c$. Now $c=1+cy$ also differs from $1$ by a radical element and has a left inverse $d$. Multiplying $c(1-y)=1$ by $d$ gives $d=1-y$, and hence $(1-y)c=1$. This proves the unit assertion. Conversely, if $x$ is outside a maximal left ideal $L$, then $ax+l=1$ for some $a\in R$, $l\in L$. Consequently $1-ax\in L$ cannot be a unit, proving the reverse direction.

The intersection of maximal left ideals is the intersection of annihilators of simple modules on the left, hence is a two-sided ideal. If $j$ belongs to it, $1-rj$ has a left inverse for every $r$. Its left inverse also differs from $1$ by a radical element and has a left inverse; this makes the original inverse two-sided. The identity $(1-ba{)}^{-1}=1+b(1-ab{)}^{-1}a$ then shows that $1-jr$ is invertible too. If $j$ were outside a maximal right ideal $I$, one would have $1-jr\in I$ for some $r$, a contradiction. Applying the same argument to the opposite ring proves equality.

For a complex unital Banach algebra $A$, the unit criterion for the Jacobson radical implies that every radical element is a quasinilpotent element, because $1-x/\lambda$ is invertible for every nonzero $\lambda$. Conversely, if a two-sided ideal consists of quasinilpotent elements, then $yx$ is quasinilpotent for every member $x$ and every $y\in A$. Hence $1+yx$ is invertible and $x$ lies in the Jacobson radical. The maximality statement concerns ideals contained in the quasinilpotent set, not the whole set itself.

For an odd positive nonsquare integer $a$, there are infinitely many primes $p$ with Legendre symbol $(a/p)=-1$. Choose a residue $n$ modulo $a$ with Jacobi symbol $(n/a)=-1$ by making it a quadratic nonresidue at one prime occurring to odd multiplicity and a square at every other prime. The Chinese remainder theorem also imposes $n\equiv 1(\mathrm{mod}4)$ and $n\equiv 1$ at any finite list of previously found primes. Then Jacobi reciprocity gives $(a/n)=-1$, so a prime factor of $n$ has $(a/p)=-1$ and is outside that list.

For $p\ge 1$ and real-valued $f\in C^p({\mathbb{T}}^n)$, there is a real trigonometric polynomial with coordinate frequencies $|k_j|\le N$ whose uniform error is at most $C_{n,p}{N}^{-p}\Vert f{\Vert }_{C^p}$. One obtains the multivariable statement by applying bounded one-dimensional Jackson approximation operators successively in each coordinate and telescoping their errors. Each operator commutes with differentiation in the other coordinates, and its uniform operator norm is bounded independently of $N$; the total error is bounded by a constant times $N^{-p}{\sum }_j\Vert {\partial }_j^{p}f{\Vert }_{\infty }$. For complex-valued $f$, approximate its real and imaginary parts separately; the resulting trigonometric polynomial may have complex coefficients.

The jackknife variance estimator estimates the sampling variance of the full-data estimator, using the deletion estimates and their mean from jackknife resampling:The square root is the estimated standard error. The factor $n-1$ multiplies the divisor-$n$ variance of the deletion estimates: those estimates vary less than estimates from new independent experiments because almost all their observations are shared. Ryan Tibshirani's ^[ref] give the same scaling in terms of the divisor-$(n-1)$ sample variance of the deletion estimates.

For the sample mean, this formula has an exact algebraic simplification. With $s_n^2$ denoting the usual sample variance,and thereforeUnder the independent and identically distributed assumption and a finite second moment, $s_n^2$ is an unbiased estimator of the population variance ${\sigma }^2$. Consequently ${\hat{V}}_J$ is an unbiased estimator of the sample mean's sampling variance ${\sigma }^2/n$. The equality ${\hat{V}}_J=s_n^2/n$ itself holds for any observed data with $n>1$; its sampling interpretation uses the stated assumptions.

For $(1,2,3,4,10)$, the sum of squared deviations from the sample mean is $50$, giving sample variance $s_5^2=50/4=12.5$. The deletion sample means have squared deviations totaling $3.125$, soHere $12.5$ estimates the variance of individual observations, while $2.5$ estimates the sampling variance of their sample mean. These answer different questions.

For more general statistics, validity requires regularity beyond the deletion formula, such as a sufficiently smooth function of sample means with suitable moments. Ordinary delete-one jackknife resampling can fail for nonsmooth quantiles, including the median; see Larry Wasserman's ^[ref]. With statistical dependence, deleting individual observations need not reproduce the relevant sampling variance; the deletion scheme must reflect that statistical dependence. Shalizi's ^[ref] distinguish the basic independent and identically distributed case from extensions.

The classical Iwasawa main conjecture, now a theorem over $\mathbb{Q}$, identifies characteristic ideals of appropriate cyclotomic Iwasawa modules with ideals generated by p-adic L-function power series. For a nontrivial even character in the p-ramified convention, ${\mathrm{char}}_{\Lambda }{X}_{\chi }=(g_{\chi })$ with $g_{\chi }((1+p{)}^{1-s}-1)=L_p(\chi ,s)$. The Kummer reflection in Iwasawa theory accounts for the reflected odd class-group convention.

Spherical symmetry makes moments over a ball isotropic tensors. Odd moments vanish under $x\mapsto -x$, while the second and fourth moments areFor the second moment, a tensor contraction gives $3\alpha =\int r^{2}dV=4\pi a^5/5$. For the fourth, complete index symmetry makes the three coefficients equal and a double tensor contraction gives $15\beta =\int r^{4}dV=4\pi a^7/7$. These identities turn polynomial volume integrals into scalar radial integrals and are useful for inertia and rotational averages.

Assume the radial moment integral exists. Rotation through $\varphi$ about the third axis multiplies $x_1+i{x}_2$ by $e^{i\varphi }$ and preserves the region, volume element and radial weight. The integral equals $e^{in\varphi }$ times itself for every angle, so it vanishes. For degree two and four this also follows by contracting the isotropic sphere moments with the complex null vector $(1,i,0)$, whose bilinear square is zero. Odd moments vanish under inversion.

Away from the origin the Hessian matrix of $r^{-1}$ is $(3x_k{x}_l-r^2{\delta }_{kl})/r^5$. Multiplication by $x_i{x}_j$ makes the singularity locally integrable. Rotational invariance and the symmetries in each index pair reduce the integral to $A{\delta }_{ij}{\delta }_{kl}+B({\delta }_{ik}{\delta }_{jl}+{\delta }_{il}{\delta }_{jk})$. The Laplacian trace gives $3A+2B=0$, while contracting the two pairs as $k=i,l=j$ gives $3A+12B={\int }_{r<a}2/rdV=4\pi a^2$. Solving these two equations yields the displayed expression.

Writing an isophote's radial residual about its fitted ellipse as $\delta r=a_4\cos 4\theta +\cdots$, positive $a_4$ extends the major and minor axes and gives a pointed, disky shape; negative $a_4$ extends diagonal directions and gives a boxy shape. The sign stated here is for radial residuals, not an unconverted brightness residual. Massive slowly rotating elliptical galaxies commonly have boxy isophotes, while embedded rotating disks commonly make lower-mass elliptical galaxies disky.

The area of a surface of revolution about the horizontal axis is $2\pi \int |y|\sqrt{1+y^{\prime 2}}dx$. For a curve below the axis with endpoints $(\pm a,0)$ and fixed arc length, its stationarity is equivalent to stationarity of $\int (y+\lambda )\sqrt{1+y^{\prime 2}}dx$. The Beltrami identity gives $(y+\lambda )/\sqrt{1+y^{\prime 2}}=C$. Choosing $C=1/k$, imposing the endpoints and integrating the length gives the displayed curve. The function $\sinh s/s$ increases strictly from $1$ to infinity on $s>0$, so every $l>2a$ determines exactly one positive $k$. Stationarity alone does not establish a global minimum.

Zeros and poles describe the positive and negative local orders of a meromorphic function. Locally $f(z)=(z-a{)}^{m}g(z)$ with $g$ holomorphic and nonzero at $a$: $m>0$ is a zero, $m<0$ is a pole, and $m=0$ is a regular nonzero value. A simple pole has $m=-1$, and the order of vanishing records $m$.

Let $H(\sigma )={\sum }_{\{x,y\}}{J}_{xy}{\sigma }_x{\sigma }_y+{\sum }_x{h}_x{\sigma }_x$ on a finite graph, with each undirected edge counted once, $J_{xy}\ge 0$, and Ising spins ${\sigma }_x\in \{-1,1\}$. Here $H$ is a dimensionless log-weight: the probability mass function is $\pi (\sigma )=e^{H(\sigma )}/Z$. If $E$ denotes physical energy, the convention is $H=-\beta E$, not $H=E$.

Order configurations coordinatewise, so $(\sigma \vee \tau {)}_x=\max ({\sigma }_x,{\tau }_x)$ and $(\sigma \wedge \tau {)}_x=\min ({\sigma }_x,{\tau }_x)$. Define $U=\{x:{\sigma }_x=1,{\tau }_x=-1\}$ and $W=\{x:{\sigma }_x=-1,{\tau }_x=1\}$. An edge contributes zero to the displayed difference unless it joins $U$ to $W$. On such an edge the joined and met configurations each have spin product $+1$, whereas both original configurations have product $-1$. Its contribution is therefore $4J_{xy}$. Each field term cancels because the coordinatewise maximum and minimum preserve the sum of the two spins. Consequently,Exponentiating gives $\pi (\sigma \vee \tau )\pi (\sigma \wedge \tau )\ge \pi (\sigma )\pi (\tau )$, the FKG lattice condition. The corresponding FKG inequality implies positive association of random variables: increasing observables have nonnegative covariance. The sign convention matters: using a positive ferromagnetic log-weight as though it were physical energy in $e^{-\beta E}$ would reverse which couplings favor alignment.

With the convention $H=J{\sum }_{⟨ij⟩}{\sigma }_i{\sigma }_j-h{\sum }_i{\sigma }_i$, positive $J$ favors opposite neighboring Ising spins. On a bipartite graph, the Néel order parameter measures opposite sublattice magnetizations; it is distinct from the uniform magnetization. Changing the signs of spins on one sublattice maps the zero-field model to the ferromagnetic Ising model.

A fixed Ising boundary condition pins the Ising spins on a specified boundary, leaving the interior distributed with the original interaction weights. The boundary can be the inner graph vertex boundary of a finite box, or an exterior neighboring set if that convention is explicitly chosen. Conditioning on pinned Ising spins preserves the FKG lattice condition of a ferromagnetic Ising model. The FKG inequality then makes interior expectations of increasing observables nondecreasing in the pinned boundary Ising spins.

For the infinite one-dimensional coupling matrix $J_{ij}=J{e}^{-\kappa |i-j|}$ with $J,\kappa >0$, its Fourier symbol is $J\sinh \kappa /(\cosh \kappa -\cos q)$. Its inverse is nearest-neighbor even though the original interaction is not. The inverse quadratic form equals $[2J\sinh \kappa {]}^{-1}{\sum }_j(m_j-m_{j+1}{)}^2+[\tanh (\kappa /2)/J]{\sum }_j{m}_j^2$. Finite open chains add endpoint corrections; a periodized coupling gives the same translation-invariant formula.