For an orthonormal wavelet and its scaling function, set ${\varphi }_{j,k}(x)=2^{j/2}\varphi (2^{j}x-k)$ and similarly for $\psi$. The coarse scaling functions together with nonnegative-resolution wavelets form an orthonormal basis of $L^2(\mathbb{R})$. The coefficients are the corresponding L2 inner products. The series converges in the L2 norm, and the Parseval identity for a Hilbertian basis equates squared norm with the sum of squared coefficients. Pointwise convergence requires additional hypotheses.

A discrete wavelet transform computes multiscale approximation and detail coefficients from sampled data. For the Haar wavelet, replace each adjacent pair $(u,v)$ byThe approximation coefficient $a$ is $\sqrt{2}$ times the pair's arithmetic mean; it is not the ordinary mean. The detail coefficient $d$ records their signed contrast. Repeat on the approximation coefficients until only one remains, keeping every detail coefficient.

For a worked example, start with the vector $(2,4,6,8)$. The first pass gives approximation coefficients $(3\sqrt{2},7\sqrt{2})$ and detail coefficients $(-\sqrt{2},-\sqrt{2})$. Transforming the remaining pair gives the coarse coefficient $10$ and the half-to-half detail $-4$. In the order coarse value, coarse detail, then fine details, the transform isThis is an orthogonal matrix transformation. Each pair preserves $u^2+v^2=a^2+d^2$, so the full transform preserves the squared Euclidean norm:The inverse function for each pair is $u=(a+d)/\sqrt{2}$, $v=(a-d)/\sqrt{2}$. Reversing the passes therefore recovers every original value. This squared-norm bookkeeping is often called signal energy; it need not represent physical energy. ^[ref]

Continuous and discrete normalizations differ. If $n=2^J$ data values are interpreted as the heights of a step function on $n$ equal cells of $[0,1]$, its normalized L2 inner product coefficients are the discrete coefficients divided by $\sqrt{n}$. For our four values they are $(5,-2,-1/\sqrt{2},-1/\sqrt{2})$, as the Haar projection example shows. The same normalization is used by the wavelet regression estimator on $[0,1]$.

Under independent Gaussian noise with a common variance, orthogonality preserves that coefficient noise variance and independence. It does not make every arbitrary correlated signal independent: this statistical conclusion uses the Gaussian model and its equal-variance independent inputs.

If a wave packet experiences a fractional nonlinear change $\chi ={\tau }_w/{\tau }_{\mathrm{nl}}\ll 1$ per interaction, and successive interaction phases are decorrelated, $N$ changes add as a random walk of size $\sqrt{N}\chi$. Order-one transfer needs $N\sim {\chi }^{-2}$ interactions, giving the displayed cascade time. Phase coherence or resonant constraints can alter this estimate; weakness alone does not justify random addition.

For a complex random wave amplitude, this two-point expectation retains the coherent field and uses a complex conjugate on the second amplitude. It is not a normalized wave-field correlation coefficient. Subtracting $\mathbb{E}[E(\mathbf{r})]\mathbb{E}[E({\mathbf{r}}^{\prime }){]}^{\ast }$ gives the centered covariance function. At coincident points the uncentered wave-field correlation equals the mean wave intensity. Under second-order stationarity it depends only on the separation and has a spectral measure of a stationary random field. A deterministic field is included by taking a degenerate ensemble.

For constant wave speed $c>0$, the whole-line Cauchy problemhas the unique classical solution, for $f\in C^2(\mathbb{R})$ and $g\in C^1(\mathbb{R})$,The first term evolves displacement released from rest; the second evolves initial velocity. The factor $1/(2c)$ converts an integral of velocity over distance into displacement. On a finite string the formula applies before the backward characteristic curves reach an endpoint, or globally after an extension that enforces the boundary conditions. ^[ref]

For a completely smooth example, choose a bump function with compact support:where $a>0$ sets the half-width and $A$ sets the initial peak. ThenAt $t=0$ the two copies coincide, recovering peak $A$. At $t=a/(2c)$, $u(0,t)=A{e}^{-1/3}\simeq 0.716531A$. At $t=3a/(2c)$ the pulses are separated, centred at $x=\pm 3a/2$, each of width $2a$ and peak $A/2$. Zero initial velocity forces equal left and right contributions; a purely right-moving initial profile instead requires $g=-c{f}^{\prime }$. Thus “released from rest” and “launched to the right” describe different Cauchy data.

The domain of dependence of $(x_{\ast },t_{\ast })$, for $t_{\ast }\ge 0$, is contained in $[x_{\ast }-c{t}_{\ast },x_{\ast }+c{t}_{\ast }]$ on the initial line. More precisely, displacement contributes only through the two endpoints, while initial velocity contributes throughout the interval. With $c=1$, $x_{\ast }=0.5$ and $t_{\ast }=1.5$, that interval is $[-1,2]$. Changing either initial datum outside it cannot change $u(0.5,1.5)$. If both initial data vanish outside $[-a,a]$, the solution vanishes outside $[-a-ct,a+ct]$: this is finite propagation speed. The expanding interval bounds possible influence; it need not be filled by nonzero displacement, as the separated bumps demonstrate.

Historically, d'Alembert studied the vibrating-string problem in “Recherches sur la courbe que forme une corde tendue mise en vibration”, in the Berlin Academy's volume for the year 1747, pages 214–219. The memoir introduces the differential equation and integrates it through two arbitrary one-variable functions. The original geometric problem makes the formula's enduring appeal clear: arbitrary initial shapes can be handled without first decomposing them into normal modes. ^[ref]

For a brief derivation, use characteristic coordinates $\xi =x-ct$ and $\eta =x+ct$. The chain rule gives $u_{tt}-c^2{u}_{xx}=-4c^2{u}_{\xi \eta }$, so $u=F(\xi )+G(\eta )$. The initial conditions yield $F+G=f$ and $-c{F}^{\prime }+c{G}^{\prime }=g$. Integrating these relations gives the displayed formula; the arbitrary integration constants cancel in their sum.

Damping dissipates the energy of an oscillation, reducing its amplitude. A linearly damped string applies this mechanism to each normal mode.

Why can stronger damping make a return to equilibrium slower? A weak brake lets a spring exchange potential energy with kinetic energy many times; a very strong brake makes the final motion creep. Critical damping marks the boundary between these behaviours. The distinction depends on the force law: the classification below concerns a single damped harmonic oscillator with viscous damping, rather than every mechanism that removes energy.

Let $x(t)$ be displacement from equilibrium, $m>0$ the moving mass, $k>0$ the spring constant, and $b\ge 0$ the constant coefficient of viscous damping. With no driving force, Newton's second law givesThe two initial conditions specify displacement and velocity. Define the undamped eigenfrequency as an angular frequency ${\omega }_0=\sqrt{k/m}$ and the damping ratio as $\zeta =b/(2\sqrt{mk})$. The ordinary differential equation becomes $\ddot{x}+2\zeta {\omega }_0\dot{x}+{\omega }_0^{2}x=0$. A constant weight in a vertical spring merely shifts equilibrium; it is already absorbed in the definition of $x$.^[ref]

For example, the ideal parameters $m=1\mathrm{kg}$ and $k=4\mathrm{N}{\mathrm{m}}^{-1}$ give ${\omega }_0=2{\mathrm{s}}^{-1}$ and critical damping at $b=4\mathrm{kg}{\mathrm{s}}^{-1}$. Values $b=1,4,8\mathrm{kg}{\mathrm{s}}^{-1}$ give damping ratios $0.25,1,2$, respectively. These are model choices, not measured values. For release from rest with $x_0>0$, set $X=x/x_0$ and $\tau ={\omega }_{0}t$; then $X^{\prime \prime }+2\zeta X^{\prime }+X=0$, where primes now denote derivatives with respect to $\tau$.

The left panel uses identical initial conditions $X(0)=1$, $X^{\prime }(0)=0$. Underdamping produces repeated crossings of equilibrium; critical damping and overdamping give positive decreasing responses. The right panel keeps critical damping but changes initial velocity: $X^{\prime }(0)=-2$ gives one crossing, although there is no oscillation. Both panels are exact model predictions.

For a spherically symmetric function, $v=ru$ reduces the radial wave equation to $v_{tt}=c^2{v}_{rr}$. A time-harmonic separated solution has radial numerator $C\sin (kr)+D\cos (kr)$, with $k=\omega /c$. Finiteness at the origin excludes $D$, and the prescribed finite origin value fixes $C$. This is a standing combination of incoming and outgoing waves. A purely outgoing point-source wave instead has a $1/r$ singularity and must be specified by its source strength, not by a finite value at the origin.

For small quasi-geostrophic approximation waves on a locally uniform basic zonal flow with nonzero meridional potential-vorticity gradient $Q_y$, this signed quadratic wave activity obeys ${\mathcal{A}}_t+\nabla \cdot \mathbf{F}=\overline{q^{\prime }\mathcal{D}}/Q_y$, where $\mathbf{F}$ is the Eliassen–Palm flux and $\mathcal{D}$ is a PV source. Multiply the linear PV equation by $q^{\prime }/Q_y$ and use the Taylor identity for quasi-geostrophic flux to obtain it. In a locally uniform conservative wave packet, $\mathbf{F}=\mathcal{A}{\mathbf{c}}_g$, so the flux follows the group velocity, with its sign determined by $Q_y$.

For a small-amplitude conservative wave packet in a slowly varying fluid flow, its intrinsic energy divided by intrinsic frequency is transported along its group velocity, with the basic advection added in the laboratory frame. A stationary background conserves laboratory angular frequency, but a changing mean velocity changes the intrinsic one through the Doppler effect; intrinsic energy alone need not be conserved. Geometrical spreading and dissipation enter the action-flux balance.

The moments of the mutation count on a neutral coalescent tree give this variance for ${\hat{\theta }}_W=S/H_{n-1}$. Since $H_{n-1}\sim \log n$ and the sum of inverse squares is bounded, it tends to zero. Together with unbiasedness, this gives mean-square convergence to the mutation parameter, so it is a consistent estimator.

For fixed $\theta >0$, a geometric mutation count in a coalescent epoch with mean ${\mu }_i=\theta /i$ has third moment ${\mu }_i+6{\mu }_i^2+6{\mu }_i^3$. The centered absolute third moment is at most $4\mathbb{E}[Y_i^3]+4{\mu }_i^3$, so its sum is $O(\log n)$. The total variance is asymptotic to $\theta \log n$, making the Lyapunov condition with $\delta =1$ hold. Rescaling by the harmonic sum gives the displayed limit. At $\theta =0$ the estimator is identically zero and the limit is degenerate. The variance of the Watterson estimator decreases only logarithmically with sample size at a fixed nonrecombining locus.

The local skin friction coefficient normalizes wall shear stress by the outer dynamic pressure:For the ideal Blasius boundary layer on one face of length $L$, averaging the local skin friction coefficient gives ${\overline{c}}_f\simeq 1.328230/\sqrt{{\mathrm{Re}}_L}$. The factor of two follows from integrating $x^{-1/2}$, not from counting both plate faces. Drag on two equally exposed faces doubles when the reference area remains that of one face.

A trail is a walk whose edge sequence $e_1,\dots ,e_m$ has no repetitions. Its vertices may repeat, including its first and last vertices. The zero-length walk is a trail because it repeats no edge. Every graph path is a trail, but the converse fails: the trail in the shared example isIts six edges $BC,CA,AD,DE,EA,AF$ are pairwise distinct, although $A$ occurs twice. Visiting a junction again and traversing a road segment again are different events.^[ref]

In an undirected graph, traversing $AB$ and then $BA$ repeats the same edge; direction of travel does not give it a new identity. In a multigraph with distinct parallel edges $p,q$ joining $X$ and $Y$, the closed walk $(X,p,Y,q,X)$ is a trail, whereas $(X,p,Y,p,X)$ is not. Both have the same vertex sequence $X,Y,X$. In a directed graph, separate directed edges $X\to Y$ and $Y\to X$ may instead be used once each; record their identities and obey their directions. Thus testing a trail means checking the actual edge records, not merely checking consecutive endpoint labels.

For a finite graph, a trail has length at most $|E|$; unrestricted walks can have arbitrarily large length whenever a positive-length closed walk exists, since it can be repeated. An Eulerian trail is the special trail that uses every edge, so its length is exactly $|E|$. Our length-six trail omits $AB$ and is therefore not an Eulerian trail. In a road-inspection model, a trail describes servicing its selected segments once while allowing returns to junctions; requiring service of every segment leads to the Eulerian trail question. If extra travel between serviced segments is necessary, the complete journey can be a walk even when each segment is serviced only once. This distinction keeps the service requirement separate from the route actually travelled.

For a simple symmetric random walk starting at zero, let $T_1$ be its first hitting time of $1$. The infinite mean first passage of a simple symmetric random walk shows that $T_1<\infty$ almost surely but $\mathbb{E}[T_1]=\infty$. Since $S_{T_1}=1$, assuming a finite expected value for $T_1$ would let Wald's equation assert $1=0\mathbb{E}[T_1]=0$, a contradiction. The formal expression $0\cdot \infty$ is not an allowed use of the theorem.^[ref]

Compare $T_{\pm 2}=\inf \{n\ge 0:|S_n|=2\}$. From every surviving position in $\{-1,0,1\}$, three consecutive positive increments force absorption within three steps. Each fresh block has probability $1/8$ of this event, so $\mathbb{P}(T_{\pm 2}>3k)\le (7/8{)}^k$ and $\mathbb{E}[T_{\pm 2}]\le 24$. This establishes finiteness before invoking the theorem. If $e_j$ is the average remaining time from position $j$, the law of total expectation givesso $\mathbb{E}[T_{\pm 2}]=4$. Symmetry gives equal exit probabilities at $-2$ and $2$, and $\mathbb{E}[S_{T_{\pm 2}}]=0$, as Wald's equation predicts. This is a small gambler's ruin problem with two absorbing barriers.

Even for $T_1$, each bounded stopping time $T_1\wedge n$ satisfies $\mathbb{E}[S_{T_1\wedge n}]=0$. The stopped process tends to $1$ almost surely, but its expected value does not tend to $1$. Rare long negative excursions retain the balancing contribution. The existing stopped-walk martingale with almost sure but not L1 convergence explains the related failure of uniform integrability; almost sure convergence alone does not permit exchanging a limit and an expected value.

Let $X_1,X_2$ be independent fair signs, each taking $-1$ and $1$ with equal probability. Set $N=1$ if $X_2=1$, and $N=2$ if $X_2=-1$. Deciding whether $N\le 1$ requires the second sign, so $N$ is not a stopping time for the natural filtration. The four equally likely possibilities giveBoth the increments and the count are bounded, but the identity fails because the second contribution is retained only when it is negative. This is selection using an unobserved outcome, rather than a legitimate sequential decision. Boundedness alone cannot replace the stopping time hypothesis.^[ref]

Consider a stipulated repeated-attempt model, with independent Bernoulli trials of success probability $p\in (0,1]$. An unsuccessful attempt costs three units of work, and a successful attempt costs one. Let $B_i$ be its success indicator function, $X_i=3-2B_i$ its cost, and $N=\inf \{i\ge 1:B_i=1\}$ the first success. Observing $X_i$ reveals $B_i$, so $N$ is a stopping time for the costs' natural filtration. The geometric distribution gives $\mathbb{P}(N\ge i)=(1-p{)}^{i-1}$ and $\mathbb{E}[N]=1/p$, including the deterministic case $p=1$. ThusAt $p=1/4$, the average is four attempts and ten units of work. An independent check uses the pathwise identity $S_N=3(N-1)+1$: there are $N-1$ failures and one final success. No success probabilities or costs in this example are measured data.

Here $N$ is not independent of the observed costs. In particular, conditioning on $N=n$ forces every preceding cost to be three and the final cost to be one, so $\mathbb{E}[S_N\mid N=n]=3n-2$, generally different from $n\mathbb{E}[X_1]=(5/2)n$ at $p=1/4$. Nevertheless the unconditional identity holds. This distinction explains why conditioning on the total count as though the original cost law remained unchanged is a faulty proof.

An ideal straight line vortex in a two-dimensional incompressible flow has circulation $\Gamma$ and induces velocityIts stream function, with $\mathbf{u}=({\psi }_y,-{\psi }_x)$, is $\psi =-\Gamma \log |\mathbf{r}-{\mathbf{r}}_0|/(2\pi )$. Positive $\Gamma$ means counterclockwise circulation. In an ideal point-vortex description, the vortex position is advected by the regular velocity produced by the other vortices and the boundaries; its own singular velocity is excluded.

The evolution of a vorticity distribution together with the velocity field it induces. A transport equation advances vorticity; diagnostic inversion recovers velocity from it and the specified boundary conditions.

For density operators supported on mutually orthogonal subspaces, the eigenvalues of their mixture are $p_i{\lambda }_{ij}$, where ${\lambda }_{ij}$ are the eigenvalues of ${\rho }_i$. Substitution into the definition of Von Neumann entropy gives the displayed decomposition into Shannon entropy of the label and the mean entropy within each block. Zero eigenvalues use $0\log 0=0$.