An infinite connected tree is regular of degree $k$ if every vertex has exactly $k$ neighbours. Necessarily $k\ge 2$. For $k=2$ it is a two-sided infinite path. For $k\ge 3$, rooting it gives $k$ children at the root and $k-1$ children at each other vertex; this constructs the unique such tree up to graph isomorphism.

A tree considered up to graph isomorphism, with vertex names ignored. For example, all four labelled trees that are star graphs on $\{1,2,3,4\}$ represent one unlabelled tree. Dividing the total in Cayley's formula by $n!$ does not count unlabelled trees: different shapes can have different numbers of distinct labellings.

A tree whose vertices carry distinct labels from a fixed finite set, as a labelled combinatorial structure. Two labelled trees on the same vertex set are equal when their sets of edges are equal: any comparison must keep every label fixed. Cayley's formula counts these trees.

For $u_t=u_{xx}+u(k-u)$ with $k>0$, a decreasing travelling wave connecting $k$ to $0$ satisfies $w^{\prime }=p$, $p^{\prime }=-cp-w(k-w)$. If $c\ge 2\sqrt{k}$, choose $a>0$ with $a(c-a)\ge k$. The triangle $0\le w\le k$, $-aw\le p\le 0$ is forward invariant: on its sloping side, $(p+aw{)}^{\prime }=w[a(c-a)-k+w]\ge 0$. The descending unstable manifold of the saddle equilibrium $(k,0)$ stays in this triangle and tends to $(0,0)$, producing a positive heteroclinic orbit. Below the threshold the origin is a stable spiral for positive $c$, forcing sign changes; at $c=0$ conserved energy also excludes the connection.

Fisher's equation is a reaction-diffusion equation combining diffusion with logistic growth. For positive $D,r,K$, one-dimensional fronts joining $K$ to zero have minimum speed $2\sqrt{Dr}$. The phase-plane construction of a Fisher travelling front and the linearized leading edge explain this threshold.

A front is a spatial profile connecting different asymptotic states on its two sides. A travelling wave front moves with a constant speed; a stationary front has zero speed. In a one-dimensional stationary reduction it is a heteroclinic orbit between the corresponding homogeneous equilibria. Equal first-integral values are necessary for a stationary front in a scalar conservative spatial equation.

Find a maximum-weight Hamiltonian cycle in a complete weighted graph or complete directed graph. Nonnegative weights admit a cycle-cover patching half-approximation for Max-TSP without a triangle inequality. For integer weights, changing $c_{ij}$ to $M-c_{ij}$, with $M\ge \max c_{ij}$, changes an $n$-vertex tour's weight from $C$ to $nM-C$. Consequently the strict-threshold decision version is NP-complete: minimum-tour threshold $C\le L$ becomes maximum-tour threshold $C^{\prime }>nM-L-1$.

Use binary directed-edge variables $x_{ij}$, one incoming and one outgoing edge at each city, and integer order variables $1\le u_i\le n-1$ for cities other than a distinguished city. The constraints $u_i-u_j+(n-1)x_{ij}\le n-2$ for distinct nondistinguished cities force order increase along every selected edge away from the distinguished city. Summing around any cycle avoiding that city gives a contradiction, so the degree equations describe a single tour. There are $O(n^2)$ nonzero constraint entries apart from the degree equations, also totaling $O(n^2)$ entries. With costs at most $2^n$, the objective requires $O(n^3)$ bits and the remaining sparse constraint encoding $O(n^2\log n)$ bits. Omitting zero matrix entries is essential for this particular size bound.

Binary arc variables with one outgoing and one incoming arc at each vertex describe a cycle cover of a directed graph. For every nonempty proper vertex subset $S$, the displayed inequality forbids a directed cycle confined to $S$. Together these inequalities make the cycle cover one Hamiltonian cycle. They can be generated as violated cuts or used in branch and bound, rather than all being stored initially.

Add a dummy job $0$. Give the arc $0\to i$ cost equal to job $i$'s initial setup, the arc $i\to j$ its changeover cost, and every arc $i\to 0$ zero cost. A Hamiltonian cycle through the dummy represents a schedule, starting just after $0$ and ending just before it. The total processing time is constant on a single machine, so minimizing cycle cost minimizes completion time. Dropping subtour elimination constraints gives an assignment problem lower bound.

A 2-opt tour move removes two boundary arcs and reverses the intervening block to reconnect one tour. Such moves define a local-search neighbourhood; block reversals of length two include adjacent swaps and hence connect permutation schedules. For directed costs the asymmetric 2-opt reversal cost must include every reversed internal arc, not just the boundary changes.

Minimizing ${\sum }_{ij}{c}_{ij}{x}_{ij}$ over binary variables with one outgoing and one incoming edge at every graph vertex, and $x_{ii}=0$, is an assignment problem. Its feasible solutions are cycle covers of a directed graph, whereas a travelling salesman problem requires one Hamiltonian cycle. The relaxed minimum is therefore a lower bound on the tour cost. If its optimum contains a proper subtour $C$, every tour omits at least one edge of $C$. Branching into subproblems that respectively forbid each edge of $C$ preserves every possible tour. Solving each child assignment problem and pruning those whose bounds exceed the best tour yields a finite branch and bound algorithm. Equality also permits pruning when only one optimum is wanted.

Combine the untransmitted maternal and paternal alleles into a hypothetical control genotype, paired with the child's transmitted genotype. Even when two heterozygous parents have a heterozygous child and parent-specific phase is ambiguous, the combined untransmitted genotype is unambiguously heterozygous. Such pseudo-controls are matched family controls, not independent random individuals; Hardy-Weinberg principle is therefore not guaranteed merely by this construction.

If one parent is heterozygous and the other parent's genotype is missing, a homozygous child reveals the observed parent's transmitted allele, whereas a heterozygous child need not. Selecting only revealed transmissions can therefore bias the transmission disequilibrium test. For allele-1 frequency $p$ under random mating and fair transmission, the probabilities of an allele-1 or allele-2 homozygous child from an observed heterozygous parent are $p/2$ and $(1-p)/2$. Among selected homozygous children, the observed allele-1 transmission probability is $p$, not generally $1/2$. A complete-trio analysis avoids this reconstruction rule; methods retaining missing-parent families need an explicit likelihood function or a valid conditional adjustment. Family-based association tests using genotype data with uncertainty discusses this bias.

Bounded-input bounded-output stability means that every bounded plant input produces a bounded plant output from zero initial state. For a causal continuous-time linear time-invariant system with a proper rational transfer function, BIBO stability is equivalent to every pole of the reduced transfer function having strictly negative real part. Cancel common factors first: an apparent denominator root may be absent from the reduced transfer function. For $K/(\tau s+1)$ with $\tau >0$, the only pole is $-1/\tau$, so the first-order low-pass filter is BIBO stable. ^[ref]

Internal asymptotic stability also requires every unforced internal state to decay. A canceled factor can hide a growing internal state. For example, consider the linear system of ordinary differential equationsIts zero-state transfer function can be written $(s-1)/((s+1)(s-1))=1/(s+1)$ and has BIBO stability. Yet from zero initial conditions and a unit-step plant input, $x_2(t)=e^t-1$ grows without bound while the plant output remains bounded. The reduced transfer function conceals that second state. Even a transfer function with no visible cancellations can be implemented with additional hidden states; its poles alone do not establish internal asymptotic stability of an arbitrary implementation.

A proper rational transfer function is a rational function whose numerator polynomial has polynomial degree no greater than its denominator polynomial, after common factors are canceled. Every finite-dimensional continuous-time linear time-invariant system of the form $\dot{x}=Ax+Bu$, $y=Cx+Du$ has such a transfer function $G(s)=C(sI-A{)}^{-1}B+D$. ^[ref]

The continuous-time frequency response is the frequency response of a continuous-time linear time-invariant system. From zero initial state, a BIBO stable system driven by a sustained sinusoid $u(t)=A\cos (\omega t)$ approaches the steady responseAny decaying initial or switching transient is separate. The modulus $|G(i\omega )|$ is the amplitude gain; the complex argument is the shift in oscillation phase. Thus evaluating a transfer function at $s=i\omega$ measures both how much of an oscillation passes through and how far it lags.

For the first-order low-pass filter, with $K>0$, direct complex number arithmetic givesThe phase uses the inverse tangent branch between $-\pi /2$ and $\pi /2$. Low angular frequencies have gain close to $K$ and little lag. High angular frequencies have gain approximately $K/(\omega \tau )$ and lag approaching $\pi /2$. This is why the model is a first-order low-pass filter: it suppresses fast variation compared with slow variation. Rowell relates frequency response gain and phase to poles and then introduces Bode plots in MIT 2.004, lecture 32, sections 1 and 2.

At the corner angular frequency,For $K=2$ and $\tau =3\mathrm{s}$, ${\omega }_c=1/3\mathrm{rad}{\mathrm{s}}^{-1}$, the gain is $\sqrt{2}\approx 1.4142$, and the lag is $45$ degrees. The angular frequency $\omega$ is measured in radians per second; ordinary frequency $f$ is measured in hertz (cycles per second), with $\omega =2\pi f$. Consequently the same corner is $f_c=1/(2\pi \tau )\approx 0.0531\mathrm{Hz}$, not $1/3\mathrm{Hz}$.

A first-order low-pass filter with positive constant gain $K$ and time constant $\tau >0$ obeys the ordinary differential equationThe derivative term limits how quickly the plant output follows a changing plant input. A larger time constant means slower adjustment. This standard form, with $K=1$, is developed in Derek Rowell, MIT 2.004, lecture 20, section 1.1: first-order systems; the following calculation keeps the gain $K$ explicit.

Write $y_0=y(0)$. The Laplace transform of the derivative is $sY(s)-y_0$, soSetting the initial condition $y_0=0$ gives the transfer function $G(s)=K/(\tau s+1)$. Its single simple pole is $s=-1/\tau$, which has negative real part. The unforced response decays, and the input/output map has BIBO stability. The eventual gain for a constant plant input is $G(0)=K$.

For a unit step, represented by the Heaviside step function $u(t)=H(t)$, the Laplace transform is $U(s)=1/s$. From zero initial condition,Take $K=2$ and $\tau =3\mathrm{s}$, with dimensionless plant input and plant output. Then $y(t)=2(1-e^{-t/(3\mathrm{s})})$ rises toward $2$. After one time constant, at $t=3\mathrm{s}$, it reaches $2(1-e^{-1})\approx 1.2642$, or $63.2\%$ of its final value. At $t=9\mathrm{s}=3\tau$, it reaches $2(1-e^{-3})\approx 1.9004$, or $95.0\%$. These percentages describe the rise from zero, rather than an instantaneous jump to the target.

With a nonzero initial condition, the same step instead givesFor $K=2$ and $\tau =3\mathrm{s}$, compare $y_0=0$ with $y_0=1$. Their plant outputs differ by $e^{-t/(3\mathrm{s})}$ despite having the same transfer function and the same plant input. The extra exponential decay term must be supplied from the initial condition; $G$ and $u$ alone cannot determine it.

A transcritical bifurcation need not already have its persistent branch at coordinate zero. The smooth parameter-dependent shift $u=y-\mu$ turns the displayed crossing of $y=\pm \mu$ into the standard form with parameter $-2\mu$. The crossing branches exchange dynamical stability. A mere nonsmooth relabeling of stable and unstable branches as $\pm |\mu |$ would conceal this smooth exchange.

Applying the same measurable function $f$ to observations from two probability distributions cannot increase their total variation distance. Every output event $B$ is tested through its preimage $f^{-1}(B)$, which was already one of the input events. ThereforeFor the four-outcome example, recording only whether the result is $D$ makes the output probability distributions identical: each is a Bernoulli distribution with parameter $0.3$. The original total variation distance was $0.3$, but this coarse measurement leaves zero discrepancy. Discarding information can make two sources less distinguishable.

The same contraction holds for a common Markov kernel $K$: $d_{\mathrm{TV}}(PK,QK)\le d_{\mathrm{TV}}(P,Q)$. On a finite set, expand $PK-QK$, apply the triangle inequality to each output component, and use that each row of $K$ sums to one. A common randomized postprocessing step cannot increase distinguishability either.