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Because equilibrium reactions balance chemical potentials and photons have zero chemical potential,The rest-mass difference in the Maxwell--Boltzmann densities then supplies the Boltzmann factor $e^{-I_A/(k_{B}T)}$.

If $K\subset \mathbb{C}$ is compact with connected complement and $f$ is holomorphic near $K$, then for each $\epsilon >0$ there is a polynomial $p$ such that ${\sup }_K|p-f|<\epsilon$.

A Runge-Kutta method has classical order $p$ when its one-step map started from exact data agrees with the exact flow through powers $h^p$ for every sufficiently smooth ordinary differential equation. Its local truncation error is then $O(h^{p+1})$, with a nonzero coefficient at that order for some problem if $p$ is the exact finite order.

For example, forward Euler method on $y^{\prime }=y$, $y(0)=1$, returns $1+h$, while the exact solution is $e^h=1+h+h^2/2+\cdots$. The first omitted term is quadratic, so the local error is $O(h^2)$ and the method is first order. On a fixed time interval, Lipschitz continuity and a stable error-propagation estimate turn local $O(h^{p+1})$ defects into global error $O(h^p)$.^[ref]

Matching the Dahlquist test equation checks only the linear stability expansion. The nonlinear Butcher order conditions must also hold; agreement on one scalar equation alone does not prove classical order. Stage accuracy is a separate property, so the number or individual precision of Runge-Kutta stages should not be substituted for the order of the final update.

An implicit Runge-Kutta method defines at least one Runge-Kutta stage using an equation that involves that stage or a later unresolved stage. In the slope convention,An Explicit Runge-Kutta method has strictly lower-triangular $A$, so the slopes are computed sequentially. An implicit tableau instead requires one or more algebraic solves, usually nonlinear when the vector field is nonlinear. Existence, uniqueness and numerical accuracy of those solves are additional requirements; A-stability of the scalar test equation does not establish them.

For the implicit midpoint rule, $Y=y_n+(h/2)f(Y)$ and $y_{n+1}=y_n+hf(Y)$. On $y^{\prime }=-\lambda y$ with $\lambda >0$, the stage is $Y=y_n/(1+h\lambda /2)$ and the update factor is $(1-h\lambda /2)/(1+h\lambda /2)$. This linear example needs only one division, whereas a nonlinear $f$ generally requires an iterative solve. The stage solvability of an implicit Runge-Kutta method gives a sufficient small-step contraction condition.

The local truncation error is the defect left when exact solution values are inserted into a numerical update. For a one-step method started from the exact solution,A local error $O(h^{p+1})$ generally leads to global error $O(h^p)$ under a uniform stability bound.

A one-sheet hyperboloid is a connected, doubly ruled surface, diffeomorphic to a cylinder.

An object has rotational symmetry when a nontrivial rotation matrix leaves it unchanged.

Potential vorticity combines vorticity with vortex stretching and density stratification. It also applies to nonrotating stratified flow through Ertel potential vorticity. In the inviscid shallow water equations,where $\zeta$ is relative vorticity and $h$ is layer thickness.

The Coriolis parameter is $f=2\Omega \sin \phi$ for planetary rotation rate $\Omega$ and latitude $\phi$. It controls the vertical component of the Coriolis acceleration acting on horizontal flow.

A Baroclinic Rossby wave has nontrivial vertical structure. Constant stratification and rigid boundaries add a positive vertical-mode contribution to the denominator of its dispersion relation.

A Barotropic Rossby wave has no vertical shear. For horizontal wavenumbers $k,l$, its quasi-geostrophic frequency is $\omega =-\beta k/(k^2+l^2)$.

A common root of polynomials $f$ and $g$ is a scalar $a$ satisfying $f(a)=g(a)=0$.

A root is multiple when its multiplicity of a root is greater than one.

If a vehicle moving at speed $v$ ejects mass forward at relative speed $u>0$, thenAs $dM<0$, the vehicle slows down.

Let $X_1,\dots ,X_n$ be independent and identically distributed random variables with unique median $m$, and let their probability distribution have continuous probability density function $f$ with $f(m)>0$. For odd $n$, the sample median satisfiesThe influence function of the sample median has mean zero and variance $1/(4f(m{)}^2)$: each of its two nonzero values occurs with probability $1/2$. Its asymptotic linear representation, followed by the central limit theorem, gives the displayed limit.

For a standard normal distribution, $f(0)=1/\sqrt{2\pi }$, so the asymptotic variance of the unscaled sample median is $\pi /(2n)$. The sample mean instead has variance $1/n$ in that model. This quantifies an efficiency cost at the reference model without contradicting the sample median's bounded response to an outlier.

A median-of-means estimator divides a sample into $k$ disjoint groups, computes the sample mean in each group, and returns a sample median of those group means. A single extreme observation changes only one group mean, so it need not dominate the final answer.

Here is a finite-variance guarantee. Let the observations be independent and identically distributed random variables with mean $\mu$ and variance $v>0$. Take odd $k$, groups of equal size $m=\lfloor n/k\rfloor$, and discard any unused observations. By Chebyshev inequality, each group mean differs from $\mu$ by more than $2\sqrt{v/m}$ with probability at most $1/4$.

The groups are independent. If their sample median exceeds this error, at least half of the groups must be bad. Applying Hoeffding inequality to their bad-group indicators givesThus an odd $k\ge 8\log (1/\delta )$ gives confidence at least $1-\delta$, provided there are enough observations. If $n\ge 2k$, then $m\ge n/(2k)$, giving error at most $2\sqrt{2vk/n}$. This is a sub-Gaussian deviation bound under finite variance; it does not assert that the observations are sub-Gaussian random variables. The block argument above derives the bound from the two stated inequalities. Choosing too many groups makes the individual means noisy.

The Huber location estimator solves ${\sum }_i{\psi }_k(x_i-\theta )=0$, where ${\psi }_k$ clips its argument to $[-k,k]$.

More generally, fix a known scale $s>0$ and a dimensionless threshold $k>0$ before fitting. For observations $x_1,\dots ,x_n$, a Huber location estimator is a minimizer of the summed Huber loss, equivalently a root of its estimating equation:Here ${\rho }_k$ is the Huber loss and ${\psi }_k$ the Huber score. The physical clipping distance is $ks$, in the observations' units; the initial formula uses $s=1$. A large observation keeps contributing a bounded signed score rather than being discarded. The original clipped location criterion appears in Peter J. Huber's 1964 paper.^[ref]

For the original data $(0,1,2,3,4,40)$, take $s=1$ and $k=2$. At $\theta =2.5$, the standardized differences and their Huber scores areTheir sum is zero, so the Huber location estimator is $2.5$. Four scores remain in the central linear region, making the root unique. By comparison, the sample mean is $50/6\simeq 8.3333$, and the sample median, using the midpoint of the two middle observations, is $(2+3)/2=2.5$. Agreement with the sample median is a feature of these data, rather than an identity between the two estimators.

The geometry explains both existence and possible nonuniqueness. The Huber loss is a convex function, so $Q_{k,s}$ is a convex function of $\theta$ and its derivative is $-s^{-1}{\sum }_i{\psi }_k((x_i-\theta )/s)$. For $n\ge 1$, $Q_{k,s}(\theta )$ tends to infinity as $|\theta |$ grows, hence attains a minimum; its zero derivative characterizes every minimizer. Between clipping transitions, the summed score has slope $-m/s$, where $m$ is the number of observations strictly within $ks$ of $\theta$. A root with $m>0$ is unique. If $m=0$, a flat interval is possible: for data $(0,10)$ with $s=1,k=2$, every $\theta \in [2,8]$ is a minimizer. A complete single-valued estimator then needs a selection rule, such as the interval's midpoint.

A bounded Huber score limits one observation's contribution to the estimating equation. The influence function of a Huber location estimator additionally divides by the population's central mass, so the size of its bound depends on the reference probability distribution. Clipping does not by itself make the bias of an estimator zero or prove a universal breakdown point; the separate local influence versus replacement breakdown argument specifies what fixed scale permits. Estimating $s$ from the same data changes the procedure and requires separate analysis.

For a fixed sample $y$ of size $n$, let $b_n(m;y)$ be its finite-sample maximum bias after at most $m$ arbitrary replacements, with $m\in \{0,\dots ,n\}$. The largest-bounded convention used here isWhen breakdown is possible, the first-unbounded convention isThey differ by $1/n$. For five observations, the sample median remains bounded under any two replacements, while three replacements can make it unbounded. These conventions therefore give $2/5$ and $3/5$, respectively. Their common large-sample limit does not remove the need to specify the finite-sample convention. The finite-sample maximum bias article displays the first-unbounded form explicitly.

An influence function asks how much a population answer moves when an infinitesimal fraction of its probability distribution is relocated to a specified point. It distinguishes the unbounded response of the mean from the bounded response of the median and the Huber location estimator. The normal-location influence curves and finite contamination bias of normal location below make that difference quantitative.

Let $T$ be a real-valued statistical functional, $F$ a reference probability measure, and ${\delta }_x$ the Dirac measure concentrated at $x$. Along the epsilon-contamination neighborhood path $F_{\epsilon ,x}=(1-\epsilon )F+\epsilon {\delta }_x$, define the right derivativeThe name influence curve refers to the same object. The derivative is in the signed direction ${\delta }_x-F$, so the calculation removes baseline mass as well as adding a Dirac measure. The definition is a limit for each fixed $x$; it does not automatically give a remainder uniform over arbitrarily distant $x$.^[ref]

The statistical functional must be defined on these contaminated probability measures, and the right derivative must exist and be finite. For a mean, a finite first absolute moment suffices. For a median, require a unique baseline median and a probability density function continuous and strictly positive there. For a smooth M-estimator, require an isolated population root, justified differentiation of its expected value, and a nonzero derivative with respect to the fitted parameter. These are specific conditions for the formulas below. Existence of the pointwise influence function alone proves neither an asymptotic linear representation nor pathwise differentiability of a statistical functional; each needs control of its own remainder and admissible statistical paths.

When worsening is the adverse event, drug efficacy relative to a control group is