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For a smooth planar incompressible flow, use $u={\psi }_y$, $v=-{\psi }_x$, so its scalar vorticity is $\omega =-\Delta \psi$. If $\omega =\lambda \psi$ for a constant $\lambda$, the flow solves the steady constant-density Euler equations with a suitable pressure.

Indeed the vector identity for the convective acceleration givesTherefore the pressure can be chosen so that the Bernoulli function satisfies $p/\rho +|\mathbf{u}{|}^2/2=C-\lambda {\psi }^2/2$. Its gradient is $-\omega \nabla \psi$, exactly the steady momentum equation. It is constant along each streamline, since the stream function is constant there.

For example $\psi =\sin x\cos y$ satisfies $\omega =2\psi$. Its pressure is $p_0-\rho {\sin }^{2}x/2+\rho {\sin }^{2}y/2$. Thus a rotational flow can have an explicit pressure and a streamline-dependent Bernoulli constant.

With $u={\psi }_y$ and $v=-{\psi }_x$, this steady incompressible solution of the Euler equations has velocity $(-\sin x\sin y,-\cos x\cos y)$ and vorticity $2\psi$. Its pressure is $C+\rho (\cos 2x-\cos 2y)/4$. The Bernoulli function is $C_0-{\psi }^2$, so it is constant along streamlines but generally varies between them.

Every continuous map from a space to a compact Hausdorff space extends continuously to its Stone-Čech compactification. Uniqueness follows from density of the original space and the Hausdorff property of the target. Two Hausdorff compactifications with this property have comparison maps both ways; their composites fix the dense original space and therefore are identity maps, proving equivalence of the compactifications.

Production of the acidic component of gastric fluid. Vagus nerve activity is one of several regulatory influences.

A probability plot compares order statistics with plotting positions or model quantiles to diagnose whether a proposed probability distribution describes data. If the model can be transformed into a linear identity $T(F(x))=a+bS(x)$, plot $T(i/(n+1))$ against $S(X_{(i)})$. Approximate linearity supports that model's shape; systematic curvature suggests a mismatch.

The plotting position $i/(n+1)$ comes from the exact mean of the $i$th uniform order statistic, by the probability integral transform. It does not assert equality between the expectation of a nonlinear transformation and that transformation of the expectation. Fitted slope and intercept can estimate model parameters, but the visual plot alone is not a calibrated hypothesis test.

The parameter space of a statistical model is the set of allowed statistical parameter values indexing its probability distributions. For an exponential distribution parametrized by its mean, $\Theta =(0,\infty )$; for a Bernoulli distribution, $\Theta =[0,1]$. Intersecting a confidence interval with the known parameter space cannot remove the true statistical parameter and therefore preserves its coverage probability.

A finite VC dimension gives simultaneous control of empirical misclassification risk for every member of a hypothesis class. Sufficient independent sample size is $A{\eta }^{-2}(v\log (2/\eta )+\log (2/\delta ))$. The displayed event implies an empirical risk minimization rule has misclassification risk at most ${\inf }_{h}R(h)+2\eta$, including for unrealizable labels.^[ref]

A concept class $C$ is weakly PAC learnable using a hypothesis class $H$ when a learning procedure has a fixed edge $\gamma >0$ such that, for every target in $C$, every input probability distribution, and every confidence parameter $\delta$, sufficiently many independent target-labelled examples let it return an $h\in H$ satisfying the displayed bound with probability at least $1-\delta$. The example bound must be polynomial in $1/\delta$ under this definition. Computational efficiency is an additional requirement when explicitly imposed. This distribution-wide guarantee should be distinguished from merely obtaining an advantage on one finite training sample.

Randomized response masks which of two complementary binary questions a person answered. Suppose the sensitive property has probability $p$, and independently the direct question is selected with known probability $x$, while the complementary question is selected otherwise. Truthful answers then have yes probability $q=xp+(1-x)(1-p)$. The investigator observes only the answer, preserving uncertainty about an individual's status while permitting population statistical estimation when $x\ne 1/2$.

For $n$ independent responses, $Y\sim \mathrm{Bin}(n,q)$, and the affine estimator $\tilde{p}=(Y/n-1+x)/(2x-1)$ is an unbiased estimator. Its variance is $q(1-q)/[n(2x-1{)}^2]$, by the mean and variance of a binomial distribution. Equivalently the variance is $p(1-p)/n+x(1-x)/[n(2x-1{)}^2]$, exposing the extra sampling uncertainty caused by the masking. At $x=1/2$, all $p$ produce $q=1/2$ and identifiability fails. At $x=0$ or $1$, there is no masking and the extra variance vanishes.

Because $0\le p\le 1$, maximum likelihood estimation gives $\hat{p}=\min (1,\max (0,\tilde{p}))$. This follows by maximizing the binomial likelihood over the interval $q\in [\min (x,1-x),\max (x,1-x)]$ and then inverting its monotone affine parametrization. Clipping can introduce bias: for $n=1$, $x=3/4$ and $p=0$, the clipped estimate has mean $1/4$.

For ${\eta }_{tt}+{\lambda }^2{\eta }_{xxxx}=0$ with Lorentzian displacement $ϵa/[\pi (x^2+{ϵ}^2)]$ and zero initial velocity, the Fourier transform is $a{e}^{-ϵ|k|}\cos (\lambda k^{2}t)$. Along a fixed ray $x=Vt$, the two phases $Vk\pm \lambda k^2$ have saddles $k=\mp V/(2\lambda )$ and curvatures $\pm 2\lambda$. Applying stationary phase and adding the conjugate contributions gives the displayed leading term. The finite pulse width supplies the attenuation factor; it may be dropped only if $ϵ|V|/\lambda \ll 1$. The expansion is an additive $o(t^{-1/2})$ statement, which remains meaningful at zeros of the leading cosine.

The stationary endpoint contributes $c{e}^{3i\pi /4}\sqrt{\pi }/(4x^{3/2})$ by method of steepest descent along $t=e^{i\pi /4}r$. Its ordinary $x^{-1/2}$ term vanishes because $a(0)=0$. A nonstationary upper endpoint with $a(b)\ne 0$ contributes $a(b)e^{ix{b}^2}/(2ibx)$ and therefore dominates the stationary contribution. The stationary phase method must retain both this boundary term and the first nonzero local-amplitude term.

A configuration of a system at one moment. In a transition system, a state is an element of its state space $S$, and a transition relation specifies permitted changes of state.

An open star-shaped set: there is a point $p$ such that every line segment from $p$ to another point of the set lies entirely in the set. In the complex plane, this geometric condition supplies the star-domain form of the Cauchy integral theorem. The positive-axis slit plane is star-shaped about $-1$, even though it is not convex.

Suppose each subtree of a binary expression tree is evaluated completely before its sibling, leaves one result on an operand stack, and is not spilled or interleaved. If the child depths are $p,q$, evaluating the first child needs $p$ slots and keeping its result while evaluating the second needs $1+q$. Reversing the order gives $\max (q,1+p)$. If either order preserves the expression's meaning, the minimum is $\max (p,q)$ when $p\ne q$ and $p+1$ when $p=q$. For example, if $p>q$, integrality gives $p\ge q+1$, so evaluating the deeper child first attains the lower bound $p$. If $p=q$, the second child needs its own $p$ slots plus one saved result regardless of order. Induction over the binary expression tree proves optimality in this evaluation model.

For a noncommutative operator, an instruction swapping the top two stack entries or a reverse-operand opcode permits the same ordering while preserving operand positions. Reordering still requires absence of order-sensitive side effects. Reassociation and interleaving are separate transformations and are not part of this optimality claim.

Code whose operand-producing instructions precede the instruction that consumes those operands. A stack machine can translate a binary program expression by producing its left operand, producing its right operand, and then applying the operation to the two top values.