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The inclusive product of hadronic-current matrix elements, summed over final states with a Lorentz-invariant phase-space measure. Electromagnetic current conservation makes both indices transverse to the exchanged four-momentum.

Here $Q^2=-q^2>0$. In the leading massless collinear parton approximation, this kinematic variable equals the momentum fraction of the struck parton.

An infinite Dedekind-finite set is an infinite set with no subset that is a countably infinite set. The name Dedekind set is sometimes used for this combination of properties. Under the axiom of choice no such set exists; results about these sets must avoid obtaining arbitrary enumerations of finite subsets by unstated choices.

If $X$ is a Dedekind-finite set, its set of finite repetition-free sequences is also Dedekind-finite. An injective sequence of distinct finite lists would, by countable union of explicitly ordered finite lists without choice, either produce an injection $\mathbb{N}\to X$ or use only finitely many entries. The latter possibility is impossible because a fixed finite pool supports only finitely many repetition-free lists. If $X$ is infinite, the one-entry lists also show that the resulting set is infinite.

Every positive linear map between $M_2(\mathbb{C})$ and $M_2(\mathbb{C})$, or between $M_2(\mathbb{C})$ and $M_3(\mathbb{C})$ in either direction, is a decomposable positive map. The low-dimensional result is the ingredient that makes the positive partial transpose criterion sufficient in two-qubit and qubit-qutrit systems. Woronowicz's low-dimensional positive-map paper establishes the $M_2$ to $M_3$ case; taking adjoints gives the reverse direction.

In the normal linear model, assume $\epsilon \sim N_n(0,{\sigma }^{2}I)$ is independent of $X$. Independent rows of $X$ with multivariate normal distribution $N_p(0,I_p)$ give a simple random design for Debiased Lasso. Let $p\to \infty$, $\log p=o(n)$ and $s\log p=o(\sqrt{n})$, with fixed positive noise scale. Choose the Lasso penalty and all Nodewise Lasso penalties as sufficiently large constant multiples of $\sqrt{\log p/n}$, including the noise scale in the former. Uniform sample cross-correlations are bounded by this scale, so zero is the nodewise solution with probability tending to one. Column squared norms divided by $n$ converge uniformly to one. To check the Compatibility condition for the Lasso directly, put $\hat{\Sigma }=X^{T}X/n$ and $\eta ={\max }_{j,k}|{\hat{\Sigma }}_{jk}-1_{\{j=k\}}|$. On the Lasso cone condition, $\Vert \delta {\Vert }_1\le 4\Vert {\delta }_S{\Vert }_1$, and the Cauchy-Schwarz inequality givesFor $s\ge 1$, entrywise concentration of a Gaussian Gram matrix gives $\eta \le D_0\sqrt{\log p/n}$ with probability tending to one for sufficiently large fixed $D_0$. The stated sparsity condition implies $s\eta \to 0$ on this event, so one can take ${\varphi }^2\ge 1/2$ eventually. The conditional Gaussian tail bound for the noise score and the Basic inequality for the Lasso then give $\Vert \hat{\beta }-{\beta }^0{\Vert }_1\le Cs\sqrt{\log p/n}$ with probability tending to one. Consequently the debiased-Lasso remainder bound is at most $As\log p/\sqrt{n}$ with probability tending to one, for a fixed constant $A$. These assumptions allow both $p$ and $s$ to grow, for example $p=n^2$, $s=\lfloor n^{1/4}\rfloor$.

A perturbation proportional to $\sin (n\pi z)e^{st+ikx}$ has growth rateMinimizing the neutral curve $(k^2+n^2{\pi }^2{)}^2/k^2$ gives $n=1$, $k=\pi$ and $R_c=4{\pi }^2$. Thus the first convection rolls have horizontal wavelength twice the layer depth.

For a linear multistep method, convergence for every set of consistent starting values requires the root condition for a multistep method. Apply the recurrence to $y^{\prime }=0$ over $N$ steps. A root $\xi$ with $|\xi |>1$ gives the error mode $e_n=|\xi {|}^{-N}{\xi }^n$: its fixed-index starting errors vanish as $N\to \infty$, but $|e_N|=1$. A unit-modulus root of multiplicity $m\ge 2$ instead gives $e_n=N^{-(m-1)}{n}^{m-1}{\xi }^n$, with the same failure. Repeated roots inside the unit disk are harmless because their polynomial growth is dominated by exponential decay.

For an axisymmetric vector field independent of the axial coordinate and regular on the axis, Gauss's law for magnetism gives $B_R=0$. The magnetostatic Ampère-Maxwell equation then gives $j_R=0$, $j_{\varphi }=-c{B}_z^{\prime }/(4\pi )$, and $j_z=c(R{B}_{\varphi }{)}^{\prime }/(4\pi R)$ in Gaussian units. Radial magnetostatic equilibrium isFor enclosed axial electric current $I(R)$, the magnetostatic Ampère-Maxwell equation gives $B_{\varphi }=2I/(cR)$, henceThe first term contains pressure and axial magnetic pressure; the second combines toroidal magnetic pressure with the hoop force from magnetic tension.

In a Z-pinch, an axial electric current produces a toroidal magnetic field whose inward magnetic tension confines a plasma. A cylindrical magnetostatic equilibrium balances the resulting Lorentz force density against a pressure gradient.

For an odd prime $\ell$, the unique quadratic subfield of $\mathbb{Q}({\zeta }_{\ell })$ is the displayed field. A quadratic Gauss sum squares to $(-1{)}^{(\ell -1)/2}\ell$, and the cyclic Galois group has a unique subgroup of index two.

The prime $\ell$ is totally ramified in $\mathbb{Q}({\zeta }_{{\ell }^r})$, whose degree is $\phi ({\ell }^r)$. The displayed equality is an equality of ideals in its ring of integers of a number field, and the residue field is ${\mathbb{F}}_{\ell }$.

Every field $\mathbb{Q}({\zeta }_{59^r})$ contains $\mathbb{Q}(\sqrt{-59})$ and is totally ramified over it at the prime above $59$. The class number divisibility under total ramification therefore shows that its class number is divisible by three. Its ring of integers is not a principal ideal domain. The degrees $58\cdot 59^{r-1}$ strictly increase, giving infinitely many distinct examples.

In a unitary representation of a finite group, if a cyclic vector $v$ satisfies $\rho (g)v=av$ for some $a\ne 1$, then there are no invariant vectors. Indeed the averaging orthogonal projection $P=|G{|}^{-1}{\sum }_h\rho (h)$ satisfies $P\rho (g)=P$, so $Pv=aPv=0$. All translates of $v$ also project to zero, and they span the representation.

If $X$ has a continuous cumulative distribution function $F$, then $F(X)$ has the uniform distribution on $(0,1)$. Combined with inverse transform sampling, this constructs the monotone rearrangement $G^{-1}(F(X))$ to a target with quantile function $G^{-1}$.

Multiplying individual fields by constant phases changes the apparent phases of interaction coefficients. A physical CP-violating phase survives the available changes of convention, as in the three-generation Cabibbo-Kobayashi-Maskawa matrix.

Assume independent individual event times with positive time-constant hazard multipliers and a common baseline hazard whose cumulative hazard tends to infinity. Transforming each event time by that cumulative hazard then gives independent exponential event times with corresponding rates, so all event orders exist. Their complete-order probabilities select each next label proportionally to its remaining multiplier. Marginalizing over the position of a deleted label leaves the order law of the other exponential times, hence the same sequential formula without that label. This proves the deletion identity used when unobserved ranks are summed out. It does not supply a likelihood function for censoring times; ignoring censoring in observed survival analysis still needs independent censoring.

The compensator of a counting process $N(t)$ is a predictable increasing process $\Lambda (t)$ such that $N(t)-\Lambda (t)$ is a local martingale. For an observed event process with at-risk process $Y(t)$ and hazard function $h(t)$, it is $\Lambda (t)={\int }_0^{t}Y(u)h(u)du$ under independent censoring. It is cumulative intensity, rather than an event probability.

A family of actions is closed under counterterms when all required divergent local terms can be absorbed into that family. This does not imply that the full quantum effective action contains no additional finite effective interaction vertices or terms in its derivative expansion.

If every sequence of distinct elements of a positive-real set $A$ has arithmetic averages tending to zero, then $A$ is a countable set. For each positive integer $n$, the threshold set $\{a\in A:a\ge 1/n\}$ must be finite: an infinite sequence inside it would have averages at least $1/n$. The Archimedean property makes $A$ their union, so the countable union of countable sets theorem applies. The hypothesis concerns every distinct-element sequence, not merely one enumeration.