Codex @codex  0

For the uniform noninteracting electron model with two spin states and kinetic energy of nonrelativistic particles $E=p^2/(2m_e)$, count all available one-particle states up to an energy $E\ge 0$:This extensive density of states has units of inverse energy; $g(E)/V$ is the density of states per unit volume. Availability and occupation are different: $g(E)$ is positive above $E_F$, but at zero temperature those states are empty. The occupied energy distribution is $g(E)f(E)$, with the Fermi-Dirac distribution reduced to $f(E)=1$ below $E_F$ and $f(E)=0$ above it. The convention at the exact boundary does not affect a continuum integral.

With the dimensionless quantity $x=E/E_F$, the occupied count per $dx$, divided by the total particle number, is $3\sqrt{x}/2$ for $0<x<1$. Its area is one. Weighting this by $E$ gives the internal energy and its mean per particle:Thus the highest occupied energy and the mean energy differ. For the Fermi energy of a model electron gas, $U/N\simeq 1.0155\mathrm{eV}$. At fixed $N$, $E_F\propto V^{-2/3}$, so the derivative $P=-(\partial U/\partial V{)}_N$ gives the electron degeneracy pressure $P=2n{E}_F/5$. The same state-counting geometry therefore predicts a nonzero pressure at zero temperature.

For an integrable periodic function of period $P>0$, its integral over an interval of length $P$ is independent of the starting point. Write $a=mP+r$ with $m$ an integer and $0\le r<P$. By a change of variables and the defining identity of a periodic function,No differentiability is required. This also allows the Fourier coefficients of a periodic function to be computed on any interval of one full period.

As a quadrature rule, the trapezoidal rule replaces an integrand by the linear interpolation joining its two endpoint values:For nonnegative endpoint values, this is the area of a trapezoid; in general it is a signed area. The quadrature nodes are $a,b$, each with quadrature weight $(b-a)/2$. Unlike the midpoint quadrature rule, it uses endpoint samples, which can be reused between adjacent panels.

Every affine function $f(x)=\alpha x+\beta$ agrees with its linear interpolation, so the trapezoidal rule integrates it exactly. On $[0,2]$, $f(x)=2x+1$ gives $T(f)=(2/2)(1+5)=6=I(f)$. On $[0,1]$, the polynomial $f(x)=x^2$ instead gives $T(f)=1/2$, exceeding $I(f)=1/3$ by $1/6$. Its degree of precision of a quadrature rule is exactly one.

If $f\in C^2[a,b]$, its signed approximation error satisfiesA convex function lies below its endpoint chord, so the trapezoidal rule overestimates its integral; a concave function gives the reverse inequality. The error formula assumes that the second derivative exists and is a continuous function throughout the closed interval.^[ref] A short derivation is the identityobtained by two integrations by parts. The nonnegative Peano kernel has integral $(b-a{)}^3/12$, and the weighted mean value theorem for integrals gives the formula.

The trapezoidal rule for an ordinary differential equation averages vector fields at the old and new states and is A-stable.

This time-stepping use applies the same trapezoidal rule to the integral of the derivative in an ordinary differential equation. That connection explains its place under implicit Runge-Kutta methods; its quadrature behavior and the nonlinear issue in trapezoidal rule fails B-stability concern different uses of the same endpoint averaging.

A trigonometric substitution is integration by substitution using a trigonometric function to simplify an integrand, especially a square root. For example, $x=a\sin u$, with $a>0$ and $-\pi /2\le u\le \pi /2$, changes $\sqrt{a^2-x^2}$ to $a\cos u$ and $dx$ to $a\cos udu$. The interval restriction fixes the sign of the square root. Similarly $x=a\tan u$ simplifies $\sqrt{a^2+x^2}$. The transformed integral must include the differential and, for a definite integral, the new limits.

An instantaneous reversal is an idealization. A real traveller changes velocity over a finite time, follows a smooth world line, and has finite proper acceleration. Special relativity still calculates the elapsed proper time using the home inertial frame and the same integral; general relativity is not needed merely because the traveller accelerates in otherwise flat spacetime. The clock hypothesis identifies an ideal clock's reading with this geometric proper time.

One explicit smooth trip between the same $O$ and $R$ starts and ends at rest in the home inertial frame. ChooseIts velocity reverses continuously at $T/2$, and its proper acceleration is bounded because $v_{\max }<c$ and $dv/dt$ is finite. With $v_{\max }=0.6c$ and $T=10\mathrm{yr}$, numerical integration givesThis comparison fixes the reunion spacetime events and maximum speed, not the turnaround distance: the smooth trip reaches $v_{\max }T/\pi \simeq 1.910$ light-years, whereas the ideal trip reaches $3$ light-years. The smooth traveller spends less time at high speed and therefore accumulates more proper time than the ideal traveller, while still accumulating less than the home clock.

For a sufficiently short smooth reversal replacing the corner, the proper time approaches the piecewise inertial result. A reversal interval of home duration $\delta t$ contributes between zero and $\delta t$ to the clock reading; shrinking that interval cannot create a finite extra loss concentrated at the corner. The finite difference survives in the long coasting legs. Keeping the same reversal manoeuvre while lengthening those legs changes the age difference. Conversely, two travellers following mirrored paths with the same speed history accumulate equal proper times, even though both accelerate. Thus acceleration distinguishes paths and enables reunion here, but an amount of acceleration alone does not determine the age difference.

The changing simultaneity assignment also becomes continuous. At a traveller spacetime event $(t,x(t))$ with instantaneous velocity $v(t)$, the momentarily comoving inertial frame assigns home time $t_H=t-v(t)x(t)/c^2$. Differentiating givesDuring the reversal at positive $x$, the last term advances the assigned home time rapidly. It describes the traveller's changing relativity of simultaneity, not a locally measured alteration of the home clock's ticking.

Work in Minkowski spacetime, neglect gravity, and choose the stay-at-home clock's inertial frame. Let departure be $O=(t,x)=(0,0)$ and reunion be $R=(T,0)$. The home clock follows $x_H(t)=0$. A traveller moves outward with velocity $+v$ and returns with velocity $-v$, where $0<v<c$ and $c$ is the speed of light. The ideal world line has a sharp turnaround at $P=(T/2,vT/2)$:Each straight leg lies in an inertial frame, but no single traveller inertial frame covers both legs.

For any differentiable timelike curve in this home inertial frame, its clock accumulates proper time through the integralApply it separately to the traveller's two legs. With the Lorentz factor $\gamma =(1-v^2/c^2{)}^{-1/2}$,For $v=0.6c$ and $T=10\mathrm{yr}$, the Lorentz factor is $\gamma =1.25$. The turnaround is $3$ light-years away at home time $5\mathrm{yr}$. The traveller records $4\mathrm{yr}$ outward and $4\mathrm{yr}$ returning: the reunited clocks read $10\mathrm{yr}$ and $8\mathrm{yr}$. These are illustrative model values, not experimental measurements.

The same integral gives a short geometric explanation: $\sqrt{1-v(t{)}^2/c^2}\le 1$, so every such competing timelike curve between these fixed spacetime events has $\Delta \tau \le T$. Equality requires zero velocity throughout, apart from sets of zero duration. Thus the straight home world line maximizes proper time in this Minkowski spacetime problem. This statement concerns fixed endpoints and an inertial home clock; in a gravitational field, compare paths using the appropriate metric tensor.

Project a bosonic field on a ring onto angular-momentum modes zero and one. The kinetic level spacing is $\mathcal{E}={\hslash }^2/(2m{R}^2)$. The contact-interaction integral enforces angular momentum conservation, leaving two same-mode terms and four mixed terms. The displayed projected Quantum Hamiltonian follows by normal ordering. Couplings to discarded modes belong to the full model and can limit the two-mode approximation.

Take a closed system consisting of one mole of a classical monatomic gas, idealized as an ideal gas with constant heat capacity at constant volume $C_V=3nR/2$. Stipulate amount of substance $n=1\mathrm{mol}$ and molar gas constant $R=8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ for this calculation. Set initial temperature $T_0=300\mathrm{K}$ and volume $V_0=0.010{\mathrm{m}}^3$, and define $E_0=nR{T}_0=2.4942\mathrm{kJ}$ and $p_0=E_0/V_0=249.42\mathrm{kPa}$. The specified equilibrium endpoints areThus $B$ has volume $0.030{\mathrm{m}}^3$, temperature $600\mathrm{K}$ and pressure $166.28\mathrm{kPa}$. The fixed material inventory and ideal gas law make these complete endpoint specifications for this model. All numbers below are calculated values, rather than measurements of a particular gas.

Use a frictionless piston, ignore bulk mechanical energy changes, and allow only pressure-volume work. Each step is the limit of a reversible thermodynamic process: adjust loads by infinitesimal pressure differences, and supply heat through infinitesimal temperature differences. Heating with changing temperature requires a succession of matching thermal reservoirs; contact with a single hotter heat bath would give an irreversible finite difference. On a fixed-volume leg the piston is locked. These assumptions justify using the actual equilibrium path's pressure inAn endpoint chord is not the path; substituting a guessed straight line would compute a different thermodynamic process.^[ref]

The upper route heats at fixed volume from $A$ to $C=(V_0,2T_0,2p_0)$, then undergoes isothermal expansion at $2T_0$ to $B$. The lower route undergoes isothermal expansion at $T_0$ from $A$ to $D=(3V_0,T_0,p_0/3)$, then heats at fixed volume to $B$. The tuples use the same $(V,T,p)$ ordering as the endpoints. Vertical legs do zero expansion work, and the two curved legs have different pressures at every shared intermediate volume.

The internal energy of an ideal gas gives the same endpoint difference for both routes:On either fixed-volume heating leg, $W_{\mathrm{by}}=0$ and $Q=3E_0/2$. On each isothermal expansion, $\Delta U=0$, so the first law of thermodynamics gives $Q=W_{\mathrm{by}}$. The ideal gas law and the integral of $1/V$ then giveHere $\ln$ is the natural logarithm. Rounded to $0.001\mathrm{kJ}$, the upper route has $(Q,W_{\mathrm{by}},\Delta U)=(9.222,5.480,3.741)\mathrm{kJ}$ and the lower route has $(6.481,2.740,3.741)\mathrm{kJ}$. Subtracting separately rounded entries can differ by $0.001\mathrm{kJ}$; the unrounded balances are exact in the stipulated model.

The shaded area between the curves explains the entire difference:Following the upper route forward and the lower route backward makes a thermodynamic cycle with this positive net work and matching net heat. Its internal energy change is zero. The two open routes themselves are reversible, so their different work outputs should not be mistaken for a loss caused by entropy production: they receive different heat transfers at different temperatures.

Two other state functions provide independent endpoint checks. With enthalpy $H=5nRT/2$ in this model,To obtain the entropy change along either route, integrate $C_{V}dT/T$ on its heating leg, giving $C_V\ln 2$, and $nRdV/V$ on its isothermal expansion, giving $nR\ln 3$. Reordering the legs changes their heat totals but leaves the sum of their entropy changes unchanged. In particular, neither route's total $Q$ divided by its final temperature equals $\Delta S$, and neither total $Q$ equals $\Delta H$: both routes include an isothermal expansion, rather than a single fixed-pressure heating step.

For a binomial distribution count, compare a point null $\theta =1/2$ with the uniform alternative prior distribution on $(0,1)$. The alternative Bayesian model evidence isso the exact Bayes factor favouring the point null is $(n+1)\left(\genfrac{}{}{0ex}{}{n}{y}\right)2^{-n}$. With standardized deviation $z=2(y-n/2)/\sqrt{n}$, the local normal distribution approximation is $B_{01}\approx \sqrt{2n/\pi }{e}^{-z^2/2}$. At fixed $z$ the frequentist p-value remains approximately constant, while the factor grows like $\sqrt{n}$. This demonstrates the prior-width mechanism behind the Jeffreys-Lindley paradox for a Gaussian point null in a discrete sampling model.

For a separate original example, change the point null to $H_0:p=1/2$ and take $H_U$ to have a uniform prior $p\sim \mathrm{Beta}(1,1)$, fixed before the ten Bernoulli trials. Here $p$ is drawn once under $H_U$ and shared by all ten trials: they are independent conditional on $p$. For one particular ordered sequence with eight heads and two tails, the Euler beta function givesThus the Bayes factor for the uniform alternative against the point null isThe binomial distribution count formulation gives $m_0(K=8)=45/1024$ and $m_U(K=8)=45/495=1/11$. Their ratio is again $1024/495$; these count masses must not be combined with the ordered-sequence masses above. The general $B_{01}$ formula at the start of this article puts the point null in the numerator, so it equals $B_{0U}$ here.

In contrast, maximum likelihood estimation gives $\hat{p}=0.8$: differentiating the log-likelihood yields $8/p-2/(1-p)=0$, and the endpoints have zero likelihood. Its peak-to-null likelihood ratio isThis peak is chosen after seeing the data, so it is not the Bayes factor of the prespecified $H_U$. The Bayesian model evidence for $H_U$ averages $L(p)$ over its entire uniform prior; it is smaller than the peak. It is also smaller than $L(0.7)=0.7^{8}0.3^2\approx 0.00518832$. Against the same $p=0.5$ null, a point alternative fixed at $p=0.7$ would give a Bayes factor about $5.31284$, greater than $2.06869$. An integrated alternative need not outperform a point alternative.

For a general proper beta distribution prior distribution $\mathrm{Beta}(a,b)$ with $a,b>0$, the same ordered-sequence marginal likelihood is $B(8+a,2+b)/B(a,b)$. This follows by substituting the normalized prior density into the integral, and shows exactly where the parameter prior distribution changes the Bayes factor. Holding the parameter prior distributions fixed while changing model prior odds leaves that Bayes factor unchanged.

If Riemann integrable functions $f_n:[a,b]\to \mathbb{R}$ converge by uniform convergence to $f$, then $f$ is Riemann integrable andThis concerns one fixed compact interval, with finite length.^[ref]

Here is a proof outline that also establishes Riemann integrability of the uniform limit. Assume $a<b$, and put $\ell =b-a$ and $E_N=\Vert f_N-f{\Vert }_{\infty }$. One bounded $f_N$ and finite $E_N$ show that $f$ is a bounded function. For any partition of an interval $P$, its upper Darboux sum and lower Darboux sum satisfyGiven $\eta >0$, choose $N$ with $E_N<\eta /(4\ell )$, then a partition of an interval with $U(f_N,P)-L(f_N,P)<\eta /2$. The displayed gap is below $\eta$, proving Riemann integrability. Finally, $|f_n-f|\le E_n$ and monotonicity of the Riemann integral give the stated error estimate. When $a=b$, every integral is zero.

For the polynomial functions $f_n(x)=x^n$ on $[0,0.8]$, direct integration and the uniform bound giveAt $n=21$, the exact integral is approximately $0.0003353953$, while the general error bound is approximately $0.0073786976$. The bound certifies the result without an exact antiderivative. On $[0,1]$, these same integrals still tend to zero, although there is no uniform convergence: sufficient conditions need not be necessary. On an unbounded interval, even uniform convergence can fail to control total area, as uniformly vanishing functions can retain a nonzero integral shows.

Uniform convergence alone also fails to preserve derivatives: $f_n(x)=\sin (n^{2}x)/n$ converges uniformly to zero on $[-1,1]$, while $f_n^{\prime }(0)=n$. The small amplitude does not control the slopes of the sine oscillations.

Prescribe constant magnetization $\mathbf{M}=M\hat{\mathbf{z}}$, with $M>0$, inside a circular cylinder of radius $R$ and length $L$, surrounded by zero magnetization. In cylindrical coordinates, the outward unit normal on the curved surface is $\hat{\mathbf{r}}$, soThe endcaps have outward unit normals $+\hat{\mathbf{z}}$ and $-\hat{\mathbf{z}}$. Their cross products with magnetization vanish, so ${\mathbf{K}}_{\mathrm{b}}=0$ on each endcap. The side bound current circulates counterclockwise when viewed from $+z$, by the right-hand rule. Zero endcap surface current density does not imply zero endcap magnetic field.

For an original example, let $M=800A/m$, $R=0.020\mathrm{m}$, and $L=0.10\mathrm{m}$. The volume and total magnetic dipole moment areCheck the result independently using the equivalent bound current: a side strip of axial length $dz$ forms a current loop with $dI=Mdz$ and magnetic dipole moment $d\mathbf{m}=\pi R^{2}Mdz\hat{\mathbf{z}}$. Its integral over $L$ gives the same $M\pi R^{2}L\hat{\mathbf{z}}$. In particular, $ML=80\mathrm{A}$ is the sum of the equivalent loop electric currents across a full axial cut; it is not a surface current density.

The side surface current density has exactly the geometry of a continuum solenoid with $nI=M$, where $n$ is solenoid turn density. In the infinite-cylinder idealization with no applied magnetic field, the long-solenoid approximation therefore becomesThe limit is useful well away from the ends of a sufficiently long circular cylinder. A finite circular cylinder instead has fringing magnetic field and a demagnetizing field. For this prescribed uniform magnetization, integrate the on-axis magnetic field of a circular current loop, or substitute $nI=M$ in the on-axis magnetic field of a finite solenoid, to obtain the central magnetic field:Our example has $L/(2R)=2.5$ and $B_z(0)/({\mu }_{0}M)=5/\sqrt{29}\approx 0.92848$, rather than unity. With the rounded vacuum permeability ${\mu }_0\approx 4\pi \times 10^{-7}\mathrm{T}m/A$, this is approximately $0.9334\mathrm{mT}$, compared with the infinite-cylinder value $1.0053\mathrm{mT}$. These are model calculations for a prescribed magnetization, not measured values or a solution for the self-consistent magnetization of an arbitrary material.

Suppose $M_X$ is finite on a neighborhood of zero, $0<c<1$, and $X$ is independent of each of $R,R^{\prime }$, with $cX+R$ and $cX+R^{\prime }$ both having the law of a random variable $X$. Independence and the Tonelli theorem first give $M_X(t)=M_X(ct)M_R(t)$ as an extended nonnegative integral. Since $0<M_X(ct)<\infty$ and $M_X(t)<\infty$, both remainder moment-generating functions are finite and equal to the displayed ratio on the same neighborhood. The uniqueness theorem for moment-generating functions gives equality of their probability distributions. No independence between $R$ and $R^{\prime }$ is required.

A vanishing moment is a cancellation condition: positive and negative parts of an analyzing function balance when weighted by a monomial. Having $q$ vanishing moments means that the first $q$ such balances hold, starting with the unweighted integral:Consequently a wavelet ignores local polynomial trends of degree of a polynomial less than $q$. A smooth feature is close to its Taylor polynomial, so only the small Taylor remainder contributes to its detail coefficient. A jump discontinuity cannot be approximated this way across its edge; its jump response of a localized analyzing function can remain large relative to details from a smooth region.^[ref]

These are ordinary absolutely convergent integrals, not a prescription for assigning values to divergent expressions. A sufficient assumption isIt makes $\psi$ and all $x^r\psi$ for $r<q$ integrable functions. Being square-integrable alone does not ensure these weighted integrals exist. For a square-integrable function with compact support, the Cauchy-Schwarz inequality guarantees every such weighted integral is finite. Saying exactly $q$ vanishing moments additionally requires an existing, nonzero next integral $\int x^q\psi (x)dx$; the usual phrase “has $q$” allows more cancellation.

Use the Fourier transform convention $\hat{\psi }(\omega )=\int \psi (x)e^{-i\omega x}dx$. If $\psi$ and $x^r\psi$ for $r<q$ are integrable functions, moment differentiation of the Fourier transform yields a $C^{q-1}$ Fourier transform andThe dominated convergence theorem justifies these derivatives. Under these assumptions, $q$ vanishing moments are equivalent to vanishing derivatives through order $q-1$ at zero. For $q>1$ this gives a Peano zero $\hat{\psi }(\omega )=o(|\omega {|}^{q-1})$; for $q=1$ it gives $\hat{\psi }(\omega )=o(1)$. It does not by itself justify a $q$th derivative or an $O(|\omega {|}^q)$ estimate.

If also $\int |x{|}^q|\psi (x)|dx<\infty$, subtract the degree-$(q-1)$ Taylor polynomial of $e^{-i\omega x}$ inside the integral. Its Taylor remainder gives the stronger boundFor an integrable function with compact support, every weighted integral exists and its Fourier transform extends to an entire function. The language “zero of order at least $q$” is then literal analytic function terminology. If the next integral is nonzero, the order is exactly $q$. The smooth-mask vanishing-moment criterion relates these cancellations to a refinement mask under its additional hypotheses.^[ref]

Velocity is the vector giving the rate of change of position with time, measured in a specified reference frame. For a differentiable path $\mathbf{r}(t)$, its instantaneous velocity and its average velocity between two times areThe displacement in the numerator retains direction. In Cartesian coordinates, each vector component is a derivative, for example $v_x=dx/dt$; its sign tells whether the corresponding coordinate is increasing or decreasing. The SI unit of velocity is $m/s$. ^[ref]

For a concrete reversal, let $x(t)=(1\mathrm{m})+(1m/{\mathrm{s}}^2)t^2-(2m/s)t$ for $0\le t\le 2\mathrm{s}$. Its signed velocity is $v_x(t)=(2m/{\mathrm{s}}^2)t-2m/s$: at $t=0$ it is $-2m/s$, at $t=1\mathrm{s}$ it is zero, and at $t=2\mathrm{s}$ it is $+2m/s$. The opposite endpoint velocities describe opposite directions, although the endpoint speeds agree. The position moves from $1\mathrm{m}$ to $0$ and back to $1\mathrm{m}$.

Recovering displacement from a measured velocity is an integration problem: $\mathbf{r}(t_2)-\mathbf{r}(t_1)={\int }_{t_1}^{t_2}\mathbf{v}(t)dt$, with the integral taken component by component. In the reversal example, ${\int }_0^{2\mathrm{s}}{v}_x(t)dt=0$: the negative and positive contributions cancel. This distinction is useful in navigation, where the final position depends on direction as well as speed.

A regular phase-mixed distribution $F(H,p_{\varphi },I)$ in an axisymmetric Stäckel potential is even separately in $v_{\lambda }$ and $v_{\mu }$, because these integrals depend on those components only through squares. Odd integration then gives all three mixed second moments zero, even if the $p_{\varphi }$ dependence produces azimuthal streaming. Bare stationarity alone is insufficient in degenerate resonant potentials with additional integrals. For example the isotropic harmonic Hamiltonian has the extra conserved $K=v_x{v}_y+{\omega }^{2}xy$; a positive Gaussian $\exp (-aH+ϵK)$ with $a>|ϵ|>0$ is a steady tracer and has nonzero $⟨v_x{v}_y⟩$. Its harmonic gravitational potential is a Stäckel potential in these spheroidal coordinates. This tracer need not itself source that harmonic gravitational potential. The alignment claim thus needs the phase-mixed three-integral hypothesis, not merely the weak form of Jeans theorem.

The Verdet constant is the coefficient relating Faraday rotation to the magnetic flux density path integral. Its SI units are $\mathrm{rad}{\mathrm{T}}^{-1}{\mathrm{m}}^{-1}$. It depends on material, wavelength and temperature; a sign requires the stated viewing convention.^[ref]

The name commemorates Émile Verdet. For nonuniform material or temperature, use $\theta =\int V(\ell )B_{\parallel }(\ell )d\ell$ when the local scalar description remains valid. A single central magnetic field reading cannot generally replace this weighted integral. Strong additional linear birefringence or appreciable circular dichroism requires following the full polarization of an electromagnetic wave, rather than treating the output as merely a rotated line.

Write the quadratic polynomial asIts function graph is a parabola with vertex of a parabola $(h,k)$. Equal offsets from $h$ give equal values, since $p(h+t)=p(h-t)=a{t}^2+k$. If $a>0$, the minimum is $k$; if $a<0$, the maximum is $k$. In both cases it is attained only at $x=h$.

For the nonmonic quadratic polynomialthe negative leading coefficient reverses the inequality: $-2(x-2{)}^2\le 0$. Its maximum is $3$ at $x=2$. Its coefficients are different from those of the positive example, but the same form exposes the shape of both function graphs.

Completing the square also simplifies an integral containing the exponential function:The substitution $u=x+3$ removes the linear term and leaves the standard Gaussian integral. The same shift that locates the minimum makes this integral recognizable.

Thermal power generated per unit volume. Its integral over a region is the region's total heat generation rate.

A vortex line is an integral curve of the vorticity vector field: its tangent is parallel to $\mathbf{\omega }=\nabla \times \mathbf{u}$. In a two-dimensional incompressible flow, these lines are perpendicular to the flow plane and the vorticity equation gives $D\omega /Dt=0$. Since material cross-sectional area is preserved, the flux of vorticity through a material vortex tube, equal to its circulation, is constant.