For temporal normal-mode stability of a viscous parallel shear flow, any unstable three-dimensional mode maps by the Squire transformation to an unstable two-dimensional mode at a no larger Reynolds number. Hence the first temporal modal instability occurs among two-dimensional disturbances. The theorem does not exclude three-dimensional transient growth or forced responses.

For a uniform spherical source of radius $a$, the origin-regular radial Poisson equation gives $\nabla f={\rho }_{0}x/3$ inside and $\nabla f={\rho }_0{a}^{3}x/(3|x{|}^3)$ outside. Integrating radially, imposing $f\to 0$ at infinity and continuity at $|x|=a$, givesThe potential and its normal derivative are continuous at the interface, so its distributional Laplacian has no surface delta contribution. The classical equation holds off the interface and the source equation holds almost everywhere and distributionally.

An arborescence is a rooted tree with each edge oriented away from its root. There is a unique directed path from the root to every vertex; equivalently the root has indegree zero and every other vertex has indegree one. Reversing all edges gives an in-arborescence, with unique directed paths toward the root.

For $n\ge 2$, every $n$-vertex arborescence embeds in every tournament on $2n-2$ vertices. The half-sparse median-order embedding proves this by repeatedly extending an outward leaf. Reversing the tournament and the arborescence gives the corresponding in-arborescence statement. The restriction $n\ge 2$ matters: a one-vertex arborescence needs a nonempty host, while $2n-2=0$ would give an empty host.

For a source-rooted oriented tree, let $b$ count edges pointing towards the root and let $c$ count nonempty connected components formed by those edges. A half-sparse median-order embedding exists in $2n+2(b-c)$ vertices. Remove forward leaves, or remove a terminal backward component of order $m$ and replace it temporarily by $2m-2$ outward leaves. Their images contain an in-arborescence copy of the component. Choose the direction and root to minimize $(b-c,b)$ lexicographically; this makes the root a source and gives $b-c\le (n-3)/2$ unless the tree is already an arborescence. The stated bound follows.

A multiplicative set $S$ of regular elements of a ring admits a right ring of fractions exactly when it satisfies the right Ore condition. The resulting localization embeds the original ring, inverts $S$, and expresses every element as $a{s}^{-1}$. For denominators that can be zero divisors, an additional denominator reversibility condition is required.

A periodic solution is a solution of a differential equation which is a periodic function: for some $T>0$, $x(t+T)=x(t)$ at every time. This demands repetition of the state, rather than merely repetition of the coefficients in the differential equation. A nonconstant periodic solution of an autonomous differential equation traces a periodic orbit; constant equilibrium points are also periodic functions, so they must be excluded when asserting absence of periodic orbits.

For an explicit example, consider the initial value problemIts coefficient has period $T=2\pi$. Taking a logarithmic derivative and then applying integration givesDirect differentiation verifies both the differential equation and the initial condition. After one coefficient period,Thus the repeating coefficient generally produces a repeating shape with a changed amplitude. With $b=1$, the three cases $a=-0.2,0,+0.2$ have one-period factors approximately $0.2846,1,3.5136$, respectively. Only $a=0$ gives a periodic solution with this initial condition. For $a<0$ the positive solution of a differential equation tends to zero, and for $a>0$ it tends to infinity; neither can have any positive period. The zero solution of a differential equation, obtained with a different initial condition, is periodic for every $a$.

The exponential function separates two roles:For $b=1$, even the decaying case $a=-0.2$ initially rises, because $y^{\prime }(0)=0.8>0$. The growing case $a=0.2$ has intervals of decline, because $a+\cos t$ becomes negative. This is the distinction between the instantaneous logarithmic derivative $y^{\prime }/y$ and the Floquet growth rate measured over many periods.

More generally, for a real continuous periodic function $q$ of period $T$, the scalar linear differential equation $y^{\prime }=q(t)y$ hasA nonzero solution of a differential equation is $T$-periodic exactly when $m=1$, or equivalently when the coefficient has zero integral over a period. This scalar calculation is the simplest example of a Floquet multiplier.^[ref]

A first-order ordinary differential equation with complex independent and dependent variables is solved by a holomorphic function on an ordinary domain, and its analytic continuation can have branch points. A holomorphic right-hand side gives local existence and uniqueness by the integral contraction mapping used in the Picard-Lindelöf theorem.

The six Painlevé equations are distinguished nonlinear second-order ordinary differential equations whose solutions have no movable branch points: singularities whose positions depend on the initial data can be poles, but not branching singularities. Their solutions are called Painlevé transcendents. The first equation is $y^{\prime \prime }=6y^2+x$, while the Painlevé II equation is $y^{\prime \prime }=2y^3+xy+\alpha$. The six families include parameters and admit special solutions at particular parameter values; they are not merely six individual functions.^[ref]

A second-order ordinary differential equation involves a second derivative as its highest derivative. In explicit form $y^{\prime \prime }=f(x,y,y^{\prime })$, introducing $v=y^{\prime }$ converts it into the first-order system $y^{\prime }=v$, $v^{\prime }=f(x,y,v)$. Under local regularity and Lipschitz continuity in the dependent variables, an initial value problem is specified by both $y(x_0)$ and $y^{\prime }(x_0)$. The linear case is a second-order linear differential equation.

A first integral of $y^{\prime }=f(t,y)$ is a differentiable function $I(t,y)$ whose value is constant along solutions through every time and state in its domain. The chain rule makes this equivalent to the displayed pointwise identity. For $I(y)=y^{T}Sy$ with symmetric $S$, it becomes $y^{T}Sf(t,y)=0$. This identity at the implicit stage is what proves exact Runge-Kutta conservation of quadratic invariants for the implicit midpoint rule. Invariance asserted only on the image of one initial-time flow needs that image to contain the stage under consideration.

Ordinal arithmetic defines addition, multiplication and exponentiation of ordinals through order types or transfinite recursion. The usual operations need not be commutative. Natural sum and natural product instead combine Cantor normal forms and are commutative; the Hessenberg natural sum is the natural addition operation.

The additively indecomposable ordinals are exactly the powers ${\omega }^{\delta }$.

A Von Neumann ordinal $x$ is finite exactly when $x$ and every member of $x$ are either zero or successor ordinals. Every finite ordinal has that property. Conversely, an infinite ordinal is either $\omega$, which is a nonzero limit, or is greater than $\omega$ and contains that nonzero limit ordinal. The test can be expressed by a bounded first-order formula: ordinality uses transitivity and membership comparability, and a nonempty ordinal is a successor exactly when it has a greatest member. This defines the class of natural numbers without quantifying over arbitrary infinite inductive sets.

A finite ordinal is the order type of a finite well-order; in the von Neumann representation it is one of $0,1,2,\dots$. They are exactly the ordinals $x$ such that $x$ and each member of $x$ are either zero or successor ordinals. If an ordinal is infinite, it either is the nonzero limit $\omega$ or contains $\omega$ as a member, so it fails this test. Thus the finite ordinals give the usual representation of the natural numbers.

For independent identically distributed observations with continuous cumulative distribution function $F$, fix the maximum $Y_n$ of the first $n\ge 1$ observations and let $\tau$ be the first subsequent observation exceeding it. The probability integral transform gives $U=F(Y_n)$ density $n{u}^{n-1}$ on $(0,1)$. Conditional on $U=u$, $\tau$ has a geometric distribution of parameter $1-u$. ThereforeThe masses sum to one, so the waiting time is finite almost surely, but the tail-sum formula for expectation gives $E\tau =\infty$. These conclusions require only continuity of $F$, not a density or a strictly increasing distribution function.

For $n$ independent and identically distributed rate-$\lambda$ exponential random variables, let $M_0=0$ and $M_1<\cdots <M_n$ be their order statistics. The gaps $D_i=M_i-M_{i-1}$ are independent and have rates $(n-i+1)\lambda$. After the first clock rings, the memoryless property makes the surviving clocks fresh independent exponentials, independently of the elapsed minimum and the winning label. Induction gives the result. Equivalently, the ordered joint probability density is $n!{\lambda }^n\exp [-\lambda {\sum }_i{m}_i]$ on $0<m_1<\cdots <m_n$; changing to gaps gives a product of exponential densities because ${\sum }_i{m}_i={\sum }_i(n-i+1)D_i$.

For two independent random variables with the uniform distribution on $[0,1]$, put $U=\min (X,Y)$ and $V=\max (X,Y)$. Then $P(U>u)=(1-u{)}^2$, so $\mathbb{E}U=1/3$. Since $U+V=X+Y$ and $UV=XY$, $\mathbb{E}V=2/3$ and $\mathbb{E}(UV)=1/4$. Thus their covariance is $1/36$, although the original variables are independent. Rescaling the interval to $[a,b]$ multiplies the covariance by $(b-a{)}^2$.

For independent and identically distributed random variables, suppose normalized sample maxima have limiting distribution function $G$. At a threshold $a_{n}x+b_n$, where $x$ is a continuity point of $G$ and $G(x)>0$, the exceedance count has limiting Poisson distribution with parameter $-\log G(x)$. The $r$th largest observation lies below the threshold exactly when at most $r-1$ observations strictly exceed it, including in samples with ties. Summing the first $r$ limiting Poisson probabilities gives the displayed limit for every fixed positive integer $r$. The equivalence follows from $F(a_{n}x+b_n{)}^n\to G(x)$ and $-\log (1-p)\sim p$ as $p\downarrow 0$.

The rank of an observation records where that observation lies among the order statistics of a statistical sample. When all values are distinct, $r_i=1+\#\{j:x_j<x_i\}$ and $x_i=x_{(r_i)}$. For $(0.7,0.1,0.9,0.4,0.2)$, the ranks of observations in original order are $(4,1,5,3,2)$: the first observation is fourth smallest, while the fourth observation is third smallest.

Ties require a convention. A tied block occupying sorted positions $a,\dots ,b$ can assign each member the average rank of an observation $(a+b)/2$, the minimum position $a$, or the maximum position $b$; an explicit rule can also break ties into distinct positions. R's official documentation describes these choices and uses average ranks by default. ^[ref].

For example, $(0.1,0.4,0.4,0.9)$ has order statistics $(0.1,0.4,0.4,0.9)$. The two $0.4$ observations occupy positions $2$ and $3$; average ranks of observations are $(1,2.5,2.5,4)$, whereas minimum-position ranks are $(1,2,2,4)$. A fractional average rank is a summary of tied positions, rather than an index of one order statistic. Without omissions, distinct ranks and average ranks sum to $n(n+1)/2$, since averaging each tied block preserves its sum; minimum-position ranks need not.

Wilcoxon signed-rank tests rank the absolute nonzero paired differences, while Wilcoxon rank sum tests rank the pooled observations from two samples; these are different rankings.

For an iid sample of size $n\ge 2$ from a continuous cumulative distribution function, the probability integral transform makes the transformed sample uniform. Integrating the joint density of its minimum and maximum gives range density $n(n-1)r^{n-2}(1-r)$ on $0<r<1$. Its distribution function is $n{r}^{n-1}-(n-1)r^n$. The result describes population content between the extreme order statistics, independently of the underlying continuous distribution.