Orient the edges of a comparison tree and set ${\eta }_{ij}=\log {\theta }_i-\log {\theta }_j$. After anchoring one vertex strength, these edge quantities are unconstrained coordinates. The Bradley-Terry model probability is the logistic function of ${\eta }_{ij}$, and its binomial log-likelihood contribution is $m[p{\eta }_{ij}-\log (1+e^{{\eta }_{ij}})]$. Differentiating gives the fitted odds $p/(1-p)$.

For observed neighbor win fractions $p_0,\dots ,p_{n-1}\in (0,1)$ on the path graph $0,1,\dots ,n$, the normalized Bradley-Terry model estimate with ${\hat{\theta }}_n=1$ is ${\hat{\theta }}_i={\prod }_{k=i}^{n-1}{p}_k/(1-p_k)$. This is backward propagation of empirical odds along the path. Every edge proportion is fitted exactly, and strict concavity in edge log-ratios proves that this is the unique global maximum.

A Boolean variable takes one of two truth values, false or true, encoded as zero or one. It is an input to a Boolean formula; a Boolean literal uses the Boolean variable either positively or with logical negation. A Boolean valuation assigns one truth value to each distinct Boolean variable, so three distinct Boolean variables give eight possible complete assignments regardless of how often each appears.

A Boolean literal is a Boolean variable $u$ or its logical negation $\neg u$. The two Boolean literals have opposite truth values: a Boolean valuation $v$ gives $v(\neg u)=1-v(u)$ when false and true are encoded as zero and one. Thus $u=0$ makes the positive Boolean literal $u$ false and the negative Boolean literal $\neg u$ true. A Boolean clause combines Boolean literals, and a Boolean-pair colouring gadget represents a Boolean variable and its logical negation by two vertices with opposite Boolean colours.

A Boolean clause is an expression ${\ell }_1\vee \cdots \vee {\ell }_k$ formed from Boolean literals, true when at least one Boolean literal is true. The empty Boolean clause is false. Repeating a Boolean literal does not change the truth value. A CNF Boolean formula is a logical conjunction of these Boolean clauses.

For $(\neg x\vee z)$, the only failing pair is $(x,z)=(1,0)$: its two Boolean literals then both have truth value zero. The one-Boolean literal Boolean clause $(x)$ requires $x=1$. A Boolean clause containing both $x$ and $\neg x$ is a tautology, since one of those Boolean literals is true under every Boolean valuation. These observations separate Boolean clause satisfaction from the simultaneous satisfaction required by SAT.

A Boolean formula is in conjunctive normal form when it is a logical conjunction of Boolean clauses. It is true precisely when every Boolean clause is true under the same Boolean valuation; an empty logical conjunction is true. For example, $(x\vee \neg y)\wedge (z)$ has two Boolean clauses with widths two and one. Width counts Boolean literal occurrences within a Boolean clause, whereas the Boolean clause count counts the conjuncts. 2-SAT bounds width by two, and 3-SAT bounds width by three. A Boolean formula with an empty Boolean clause is an unsatisfiable Boolean formula.

Subtracting the quadratic Taylor polynomial at the rescaling centre makes the function, gradient and Hessian matrix vanish there. Dividing by $A{r}^{2+\mu }$ preserves the normalized Hölder oscillation of the Hessian matrix while turning a small forcing-to-oscillation ratio into a harmonic limiting equation. The three vanished Taylor coefficients provide local bounds needed by the Arzelà-Ascoli theorem.

For a doubly stochastic matrix with a fractional entry, form the row-column bipartite graph of entries strictly between zero and one. Every incident vertex has degree at least two, so there is an even cycle. A matrix supported on that cycle with alternating entries $+1,-1$ has zero row and column sums. A sufficiently small positive multiple can be both added and subtracted while preserving entry bounds. The original matrix is then the midpoint of two distinct doubly stochastic matrices, so it is not an extreme point.

For a doubly stochastic matrix, the Hall marriage theorem supplies a perfect matching in its positive-entry support. Subtract the smallest matched entry times the corresponding permutation matrix. The residual has equal row and column sums and smaller support unless it vanishes. Repeating gives a convex combination of permutation matrices. With $r$ initially positive entries, there are at most $r-n+1$ terms, since a positive-mass residual has at least $n$ positive entries. Elementary augmenting-path matching gives $O(n^5)$ arithmetic operations.

The biharmonic equation is ${\nabla }^{4}f=0$, where the biharmonic operator is the square of the Laplacian. In planar Stokes flow, taking the curl of the momentum equation and using a Cartesian streamfunction produces this equation. Its decaying spatial Fourier series modes have form $(A+By)e^{-|k|y}{e}^{ikx}$.

The clamped problem uses the weak solution identity $\int \Delta u\Delta v=\int fv$ for every $v$ in the clamped second-order Sobolev space. For $f\in L^2(U)$, the clamped Hessian identity and the Poincaré inequality make this a bounded coercive bilinear form. The Lax-Milgram theorem gives a unique weak solution without a mean-zero condition on $f$.

An uncorrected nonnegative symmetric kernel density estimator can leak an order-one amount across a jump in a probability density function. If its transition region has width proportional to the smoothing bandwidth $h$, the integrated mean squared error includes squared bias of an estimator of order $h$, rather than the order $h^4$ familiar for twice-smooth interior densities. For the unit-width uniform window and the exponential distribution of rate one, the integrated squared bias of an estimator is $h/12+o(h)$.

For $M_j=F_j{A}_n{M}_0$ and $M_0\ne 0$, the action map in the display is injective. If $a\in F_j$ kills $M_j$, each commutator $[a,z]$ with a generator lies in $F_{j-1}$ and kills $M_{j-1}$. Induction makes all these commutators zero, so $a$ is scalar; since it kills $M_0$, it is zero. This finite-step faithfulness does not require the whole module to be finite-dimensional.

For independent fair signs on a two-sided Bernoulli shift, the coordinate observable $f(y)=y_0$ has unit norm and pairwise orthogonal orbit values. Their pairwise squared distance is two, excluding a finite small net in global $L^2$. Its pullback along a factor map can nevertheless be a relatively almost periodic observable, since it is fiber-constant and takes only two values.

For $W=n^2-⟨n^2⟩$, the covariance function is $⟨n^2{n}^{\prime 2}⟩-⟨n^2⟩⟨n^{\prime 2}⟩$. It involves fourth moments and is not determined by the two-point correlation of $n$ in general. For a centered Gaussian pair the fourth-moment identity reduces it to $2⟨n{n}^{\prime }{⟩}^2$. This extra Gaussian hypothesis must be stated before using that reduction.

Heavy elements provide many ionic transitions at temperatures where hydrogen and helium have lost their most effective line emitters. Metal-line cooling can therefore increase the plasma cooling function substantially and broaden its temperature features. Low-energy fine-structure transitions also permit cooling below the hydrogen atomic threshold when the relevant ions and electrons are present.

A diffusion remaining in $(-\pi /2,\pi /2)$ with the displayed equation is a bounded martingale and has a terminal limit. Its expected total quadratic variation is finite by its bounded second moments. An interior limit would leave its squared diffusion coefficient bounded away from zero and force infinite quadratic variation. Thus it converges to an endpoint, and the preserved mean determines the two endpoint probabilities.

The unscaled residual deviance of an exponential dispersion family compares its fitted mean with the saturated statistical model at a common dispersion parameter. Its normalization removes the factor $1/\varphi$ in the log-likelihood. In Poisson regression and binomial regression, $\varphi =1$, so it is twice the saturated-minus-fitted log-likelihood. Under suitable regularity, $D/\varphi$ has approximately a chi-squared distribution with the residual degrees of freedom; this is not an automatic accurate approximation for sparse responses. Nested deviance reductions with estimated dispersion lead to approximate F-tests after scaling by the dispersion estimate.

In the two-component alpha-Omega dynamo with $|f(t)|\le 1$, apply the weighted energy estimate for two coupled modes with $a=|{\alpha }_0|$, $c=|k\Omega |$, and $d=\eta k^2$. The field-amplitude growth rate is at most $\sqrt{|{\alpha }_0\Omega k|}-\eta k^2$. Optimizing over $k$ gives the upper constant $3/2^{8/3}$ multiplying $({\alpha }_0^2{\Omega }^2/\eta {)}^{1/3}$. The admissible choice $f=1$ gives a lower constant $3/2^{10/3}$, so the maximum possible growth has the displayed scaling. An arbitrary chosen modulation need not grow. The statement concerns the model with freely adjustable wavenumber; physical scale-separation restrictions can constrain that optimization.

The alpha tensor is the linear map from a uniform test magnetic field to the corresponding mean-field electromotive force. Its components satisfy ${\mathcal{E}}_i={\alpha }_{ij}{B}_{0j}$. Anisotropic helical flows can produce a real symmetric rank-two response, such as ${\alpha }_0\mathrm{diag}(1,1,0)$.