Let $S=U_z>0$, $N^2=-(g/{\rho }_0){\overline{\rho }}_z>0$, $q^2=\overline{u_i^{\prime }{u}_i^{\prime }}$, and constant correlations $C_u=-\overline{u^{\prime }{w}^{\prime }}/q^2$ and $C_{\rho }^2={\overline{{\rho }^{\prime }{w}^{\prime }}}^2/(q^2\overline{{\rho }^{\prime 2}})$. Using $ϵ=q^3/L_u$ and $\chi =\overline{{\rho }^{\prime 2}}q/L_{\rho }$, the scalar-variance balance implies $\overline{{\rho }^{\prime }{w}^{\prime }}=-C_{\rho }^2{L}_{\rho }q{\overline{\rho }}_z$. Substituting in the energy budget gives the displayed quadratic for $q>0$.

Write $a=C_{\rho }^2{L}_{\rho }{N}^2/(C_{u}S)$ and $b=1/(L_u{C}_{u}S)$. The ratio of total sinks to production is $R=a/q+bq$. Positive equilibria exist only for $4ab\le 1$, equivalent to $\mathrm{Ri}\le L_u{C}_u^2/(4L_{\rho }{C}_{\rho }^2)$. The larger-$q$ root has flux Richardson number $(1-\sqrt{1-4ab})/2$; the smaller root has the plus sign. Under the local kinetic-energy evolution the larger root is stable and the smaller root unstable. The critical value is closure-dependent and is not the universal inviscid linear-stability threshold.

Two local isometries from a connected Riemannian manifold that have the same value and differential at one point agree everywhere. They preserve geodesics, so their equality follows on a normal neighborhood from $f({\exp }_{p}v)={\exp }_{f(p)}(d{f}_{p}v)$. The set of points where both value and differential agree is consequently open, and continuity makes it closed. Connectedness completes the argument.

For spacelike-separated measurement regions, a sufficient specification $\lambda$ of their relevant past should screen each outcome from additional information in the other wing. In a two-setting experiment a Bell-local model has $P(A,B\mid a,b,\lambda )=P(A\mid a,\lambda )P(B\mid b,\lambda )$. Together with measurement independence and bounded outcomes, this implies the CHSH inequality. Bell local causality is stronger than quantum no-signalling and is not an assumption that every outcome was predetermined.

Outcome independence requires $P(A,B\mid a,b,\lambda )=P(A\mid a,b,\lambda )P(B\mid a,b,\lambda )$. It says that, once settings and the proposed complete state are fixed, learning the distant outcome does not change the local outcome probability. Combining it with parameter independence gives Bell factorization. Entangled quantum states generally fail outcome independence even though their observable marginal distributions obey quantum no-signalling.

If a successful preparation flag is fixed before freely selected Bell inequality settings, conditioning a local hidden-variable theory on that flag only replaces its hidden-variable distribution by the conditional distribution. The later response functions remain local and the distribution is independent of the later settings, so the CHSH inequality still holds. Setting-dependent rejection during the Bell measurements does not satisfy this argument.

For an order-$p$ one-step method with leading numerical error $C{h}^{p+1}$, two half steps have leading error $C{h}^{p+1}/2^p$. The displayed difference estimates exact minus the finer approximation. Adding it implements Richardson extrapolation. It requires a common starting value, the smooth local expansion, and accuracy of all algebraic solves; it can fail across a nonsmooth event.

Two Runge-Kutta methods share a stage matrix and nodes but use different output weights. An order-$p$ and order-$q$ pair with $q<p$ gives an output difference generally of order $h^{q+1}$, estimating the lower-order defect. Shared stage evaluations reduce the cost. The adaptive step-size controller exponent must reflect this estimator order, rather than blindly the higher reported solution order.

Using the slope convention for Runge-Kutta stages, an embedded Runge-Kutta pair computesIts Butcher tableau has two output rows, $b^T$ and ${\hat{b}}^T$. Evaluating $f$ is often the expensive part of solving an ordinary differential equation; after the shared slopes have been computed, the extra row needs only another linear combination.^[ref]

Write $\Phi$ for the exact endpoint from the current numerical state, and define signed local defects ${\ell }_q=\Phi -Y^{[q]}$ and ${\ell }_p=\Phi -Y^{[p]}$. Then $d={\ell }_q-{\ell }_p$. If ${\ell }_q=C_q{h}^{q+1}+O(h^{q+2})$ and ${\ell }_p=O(h^{p+1})$, with $C_q\ne 0$, the difference reproduces the leading lower-order defect. It does not reproduce the leading higher-order defect. Here the powers use big O notation and a smooth Taylor expansion; a vanishing leading coefficient or accidental cancellation can make the estimate misleading.

A common choice accepts $Y^{[p]}$ after testing $d$ with absolute and relative error tolerances. With the scale approximately unchanged on a retry, the adaptive step-size controller uses exponent $1/(q+1)$. Accepting a better endpoint does not change the power in the quantity being tested. The Euler-Heun embedded pair makes this visible without a large coefficient table.^[ref]

A predictor and corrector give distinct approximations from the same local history. Their difference, scaled by their leading error constants and orders, can estimate a local defect. Raw differences without that calibration need not estimate the corrected solution's error. Variable-step multistep methods also require updated coefficients and controlled history handling.

Let $S$ exclude $-I$, let $C(S)$ be its Pauli centralizer, and let $\mathcal{Z}=\{\pm I,\pm iI\}$. A set of Pauli errors is jointly correctable exactly when every product $E_a^{†}{E}_b$ belongs to $\mathcal{Z}S$ or lies outside $C(S)$. Outside the centralizer its code compression is zero by anticommutation; in $\mathcal{Z}S$ it acts as a scalar. A centralizing operator outside $\mathcal{Z}S$ has zero trace on the code but is unitary there, so cannot be scalar. This gives precisely the Knill--Laflamme condition. Scalar phases must be retained when formulating the group criterion.

For acquisition of a mutex, the guarantee that every persistent requester eventually succeeds, under specified scheduling assumptions and with owners eventually releasing. Some software threads repeatedly succeeding while another waits forever violates starvation freedom without violating mutual exclusion. A finite bound on the wait, and service in arrival order, are additional requirements.

Let customer $k$ enter a queue at $a_k$ and leave at $d_k$, and count it in $N(t)$ for $a_k\le t<d_k$. On a finite observation interval $[0,T]$, the integral of the queue population is exactly the sum of the lengths of the overlaps of customer visits with that interval, as displayed above; $x_{+}=\max (x,0)$ is the positive part. This deterministic identity includes visits already in progress at time zero and visits still unfinished at time $T$. It needs no probability distribution for arrivals or service times. The area argument is also developed in Richard C. Larson and Amedeo R. Odoni, Urban Operations Research, section 4.4.

For a fully computed example, start empty and observe through $T=6$ minutes. Three customers arrive at $a=(0,1,4)$ minutes, begin service at $b=(0,2,4)$ minutes, and depart at $d=(2,3,5)$ minutes. These times describe a single first come first served server with service times $(2,1,1)$ minutes. Their customer sojourn times are $(2,2,1)$ minutes, and their queueing delays are $(0,1,0)$ minutes. The queue population is respectively $1,2,1,0,1,0$ on the consecutive intervals $[0,1),[1,2),[2,3),[3,4),[4,5),[5,6]$, soThe same area is $2+2+1=5$ when counted by customer. Hence the finite-window averages satisfyThe waiting area is one customer-minute, giving $L_q=1/6$ and $W_q=1/3$ minute; $(1/2)(1/3)=1/6$. The server works for four minutes, so its server utilization is $4/6=2/3$ and $L-L_q=5/6-1/6=2/3$. Repeating this six-minute pattern indefinitely makes all these values long-run averages, without a Poisson process of arrivals.

An observation window that cuts visits short requires the overlap terms, rather than full customer sojourn times. Starting empty, let $A(T)$ count admissions before $T$ and write $w_k=d_k-a_k$. ThenFor the example cut at $T=2.5$ minutes, the area is $1+2+0.5=3.5$ customer-minutes, whereas the full durations of the two admitted customers sum to $4$; the remaining half-minute of the second visit is $R(2.5)=0.5$. A precise sufficient sample path condition for the limiting Little's law is $A(T)/T\to {\lambda }_{\mathrm{eff}}\in (0,\infty )$, $A(T{)}^{-1}{\sum }_{a_k<T}{w}_k\to W<\infty$, and $R(T)/T\to 0$. Dividing the identity by $T$ then gives $L={\lambda }_{\mathrm{eff}}W$. Equating admission and departure rates additionally uses $N(T)/T\to 0$, by the balance $N(T)=A(T)-D(T)$ for an initially empty queue, where $D(T)$ counts all departures.

If a graph is $2k$-connected and has a graph minor $H$ satisfying $2\delta (H)\ge |H|+4k-1$, it is $k$-linked. Write $d=|H|-1-\delta (H)$, so $|H|\ge 2d+4k+1$. The rooted branch-set reduction replaces the minor model by $|H|-k$ disjoint connected sets, the first $2k$ containing any prescribed terminals separately, each missing at most $d$ of the remaining sets. Every prescribed pair then has at least $k+1$ common adjacent unused sets, allowing the $k$ connections to be selected successively. The reduction uses rooted separations of a graph, contraction of edges outside distinct branch sets, and a final Hall marriage theorem matching; the separation condition prevents a contraction from destroying access to the model.

On vertices $0,\dots ,7$, join cyclic indices differing by one or two modulo eight. The resulting planar graph has a square face bounded in order by $0,2,4,6$. Deleting at most three vertices cannot disconnect it: each break in the surviving cyclic chain requires two consecutive deleted vertices, and two separate breaks require four deletions. It is therefore four-connected. Disjoint paths joining $0$ to $4$ and $2$ to $6$ would cross in the complementary disk of the square face, contrary to the Jordan curve theorem. Thus four-connectivity does not guarantee two-linkage.

Choose a checkerboard coloring of the complementary regions of a link diagram. The Tait graph has one vertex for each black region and one edge through each crossing joining its adjacent black regions. The other coloring gives its planar dual. For a connected alternating link diagram, all-A and all-B smoothing circles correspond to the regions of its two checkerboard colors; one-circle smoothing states correspond to spanning trees.

Replace each crossing of a diagram by a saddle joining its A smoothing above the projection sphere to its B smoothing below; cap the upper and lower state circles. For a connected classical link diagram this constructs a connected closed orientable surface. Its Euler characteristic is $s_A+s_B-c(D)$, so $s_A+s_B\le c(D)+2$. For an alternating diagram the surface has genus zero.

A diagram is A-adequate if changing a single smoothing of its all-A state joins two different state circles, and B-adequate analogously. It is adequate if both conditions hold. In an adequate diagram the top and bottom degree terms of the Kauffman bracket arise uniquely from these two states. Reduced alternating diagrams are adequate because their Tait graphs and duals have no loop edges.

A link diagram is alternating if overpassing and underpassing crossings alternate along every component. Its checkerboard smoothing choices are consistent across all crossings of a connected projection. An alternating knot diagram is the one-component case.

For $d\ge 2$, nonzero roots define projective-point coordinates $s=X^{q-1}$, satisfying $s^N+s+T=0$, where $N=(q^d-1)/(q-1)$. This polynomial is irreducible over ${\mathbb{F}}_q(T)$. After fixing one root $s$, the remaining coordinates satisfy $R(Y,s)=(Y^N-s^N)/(Y-s)+1$. Substitute $z=Y/s$ and $h=1/s$: the equation becomes $h^{N-1}+(z^N-1)/(z-1)=0$. At a nontrivial $N$th root of unity the second term has a simple zero, so the polynomial is Eisenstein over the algebraic closure's rational function field. This proves irreducibility of $R$ and hence two-transitivity on projective points.

For even $d=2n$ with $n\ge 2$, the roots of $X^{q^d}+X^{q^2}+TX$ form an $n$-dimensional vector space over ${\mathbb{F}}_{q^2}$. Ratios $(c\alpha )/\alpha$ for a nonzero root $\alpha$ show that the splitting field contains every $c\in {\mathbb{F}}_{q^2}$. The subgroup fixing that quadratic constant extension is normal of index two and lies in $GL(n,q^2)$; applying projective transitivity over ${\mathbb{F}}_{q^2}(T)$ makes it contain $SL(n,q^2)$. It therefore normalizes that special linear group, while the full group acts semilinearly.

On a smooth projective curve of genus two, $D=K+P+Q$ has three sections and no base points by the Riemann-Roch theorem. Its map to ${\mathbb{P}}^2$ is nondegenerate and has mapping degree times image degree four. The only nonbirational possibility is a double cover of a conic. Every degree-two divisor with two sections is canonical, again by Riemann-Roch theorem, so that possibility forces $D\sim 2K$, or $P+Q\sim K$. If this equivalence fails, the image is a quartic and the map is birational. The quartic is singular because a smooth plane quartic has genus three.