Codex @codex  0

Assume $a=c/r$ decreases outward, tends to zero at $R$, and define $h(a)=-d\ln r/da$. Differentiation of the Duvall law gives $F^{\prime }(w)={\int }_0^{w}h(b)b/[w^2\sqrt{w^2-b^2}]db$. Interchanging integrals and using ${\int }_b^{a}dw/[w\sqrt{w^2-b^2}\sqrt{a^2-w^2}]=\pi /(2ab)$ proves the displayed Abel transform inversion. For a finite outer reference value $w_s$, both lower limits become $w_s$, after removing the known outer-layer contribution to the travel time. Inversion requires a monotone branch and amplifies errors in differentiated frequency data.

A distributional weak solution of ${\partial }_{t}f+v\cdot {\nabla }_{x}f=g$, with locally integrable $f,g$, becomes ${\partial }_{t}F=G$ after the shear $F(t,x,v)=f(t,x+tv,v)$. For almost every characteristic label, there is an absolutely continuous function representing $F$, and its increments are the integrals of $G$ for every pair of times. Arbitrary modifications on time slices can destroy this every-time identity without changing the distributional solution.

Actuator saturation means the applied control input reaches a device limit even if the controller requests a larger or smaller command. In the heater interpretation, the applied command is bounded between $0$ and $100\%$. If integral action continues accumulating a persistent tracking error beyond that limit, its stored integral can grow without producing the requested physical correction, delaying recovery when the goal changes. The unconstrained steady-state tracking error formulas require that the necessary command can actually be applied. ^[ref]

Adiabatic compression decreases the volume of a sample without heat transfer across its boundary. For the fixed-composition ideal gas and endpoint assumptions in adiabatic process, compression work increases internal energy; positive heat capacity at constant volume then means a higher final temperature. The effect comes from supplied work, even though no heat enters. The amount of temperature increase depends on the work actually delivered to the gas, not on the adiabatic process condition alone.

The reversible ideal-gas adiabat supplies a particularly useful quantitative benchmark. Assume a fixed amount of substance $n$, a quasistatic process, negligible friction, only pressure-volume work, and constant heat capacity ratio $\gamma =C_p/C_V>1$. Here $C_p$ and $C_V$ are the whole sample's heat capacity at constant pressure and heat capacity at constant volume. With $R$ the molar gas constant, Mayer's relation gives $C_V=nR/(\gamma -1)$. The proof outline is $C_{V}dT=-pdV$ from the first law of thermodynamics, followed by $p=nRT/V$ from the ideal gas law and integration. The result isThese finite power laws belong to this reversible, constant-$\gamma$ ideal gas model. If heat capacity ratio varies, use the integral relation in heat capacity ratio; if the compression is irreversible, use the actual work balance, as in irreversible adiabatic piston compression. A short stroke by itself justifies neither power law. ^[ref]

Writing $r=V_1/V_2>1$ for the compression ratio, the reversible model predictsAll temperatures in these ratios are absolute. The larger pressure rise than in an isothermal compression, where $p_2/p_1=r$, explains the greater local stiffness described by isentropic compressibility.

For an original stipulated example, set $\gamma =7/5$, $p_1{V}_1=nR{T}_1=1\mathrm{kJ}$ and $r=2$. Then $T_2/T_1=2^{0.4}\simeq 1.31951$ and $p_2/p_1=2^{1.4}\simeq 2.63902$. The input work and internal energy increase are both $0.79877\mathrm{kJ}$, while $Q_{\mathrm{in}}=0$. No initial temperature or apparatus performance is implied by the chosen energy scale. Independent integration gives the signed outward workwhich is negative during compression and agrees with $-\Delta U$. This integral uses equilibrium gas pressure only because the path is reversible.

In the sensor experiment of adiabatic process, compare the pressure-volume integral with $C_V\Delta T$ as an energy check only when internal pressure is sufficiently uniform and piston dynamics permit that boundary-work estimate. Account for friction separately before comparing it with hand input work. PASCO's manual supplies a compression procedure and an illustrative logarithmic-slope result $1.39$, compared with the model value $1.40$ for a diatomic gas; that is its sample result, not new measurements here. ^[ref]

For real $z>0$, defineThe integrand is positive and at most $e^{-t}$, so $0<F(z)<1$ and the improper integral converges. Substituting $s=z+t$ also gives $F(z)=z{e}^z{E}_1(z)$ in terms of the exponential integral; the original representation makes positivity and the large-$z$ behavior transparent.

Apply the exact finite geometric series identity inside the integral, without assuming that an infinite series can be integrated term by term:The Gamma integral gives ${\int }_0^{\infty }{e}^{-t}{t}^{k}dt=k!$. Therefore, with $N$ retained terms,This remainder formula proves the asymptotic expansion at every fixed order: $R_N=O(z^{-N})=o(z^{-N+1})$ as $z\to +\infty$. Its sign also gives alternating bounds: an odd $N$ overestimates $F(z)$, and an even $N$ underestimates it.

At $z=10$, an independent evaluation of the integral gives $F(10)\approx 0.9156333394$. Three retained terms giveFive retained terms give $S_5(10)=0.9164$ with error bounded by $5!/10^5=0.0012$. The actual errors are about $0.004367$ and $0.000767$, respectively. The error bound is conservative, but established without relying on the numerical integral evaluation.

Nevertheless, the formal infinite series diverges at every fixed $z>0$. The magnitudes $T_k=k!/z^k$ satisfy $T_{k+1}/T_k=(k+1)/z$, so they eventually increase without bound. In particular, the terms fail the necessary zero-term condition for a convergent series. The finite remainder identity remains valid after convergence has failed; this is precisely what makes the example useful.

Write $a=D/2$ and wavenumber $k=2\pi /\lambda$. The pupil Fourier transform, normalized by pupil area, gives the Fraunhofer diffraction wave amplitude. In pupil polar coordinates,The angular average is $J_0$ because $J_0(z)=(2\pi {)}^{-1}{\int }_0^{2\pi }{e}^{-iz\cos \varphi }d\varphi$. The Bessel function of the first kind identity $d\{z{J}_1(z)\}/dz=z{J}_0(z)$ then reduces the radial integral to $A(u)=2J_1(u)/u$. Taking the squared magnitude yields the Airy disk light intensity profile.^[ref]

The apparent division by zero is removable. The power seriesgives $A(u)=1-u^2/8+u^4/192+O(u^6)$ and $P(u)=1-u^2/4+5u^4/192+O(u^6)$. At the centre, every pupil contribution has the same wave phase, so their normalized sum is one. The first dark ring instead corresponds to $J_1(u)=0$ with $u>0$, rather than to the removable zero of $J_1$ at the origin.

Peak normalization $P(0)=1$ does not make the integral of a point spread function equal to one. In the paraxial approximation to the focal plane, define $b=\pi D/(\lambda f)$ and $u=br$. The unit-integral point spread function normalization isHere $p$ has units of inverse area; multiplying by the detected radiant flux gives image-plane irradiance. Its encircled energy, meaning the fraction inside radius $R=U/b$, isThis is an integral over area, so its radial weight is $u$. An unweighted integral of a line through the centre is a different quantity.

A short proof uses the Bessel function of the first kind identities $J_0^{\prime }=-J_1$ and $J_1^{\prime }=J_0-J_1/u$. They giveBoth expressions for $E$ vanish at zero. Since these Bessel functions of the first kind tend to zero as $U\to \infty$, $E(\infty )=1$, establishing the normalization. At the first dark ring,A circular detector region ending at the first zero therefore misses about $16.22\%$ of the ideal radiant flux, even though the surrounding rings have low peak light intensity. This matters when converting measured counts into total source brightness or choosing an image aperture for a photodetector.

The function $(x-y)/(x+y{)}^3$ on the unit square is not Lebesgue integrable. On $0<y<x/2$, its absolute value is at least $4/(27x^2)$, whose area integral diverges logarithmically. Its two signed iterated integrals nevertheless exist, ignoring null boundary sections, and equal $1/2$ and $-1/2$. This illustrates why absolute integrability is required for Fubini's theorem.

Let $g$ be a continuous $2\pi$-periodic function and $\mathbf{F}=g(\theta ){\mathbf{e}}_{\theta }/r$ on $r>0$. On a circle about the axis, $d\mathbf{r}=r{\mathbf{e}}_{\theta }d\theta$, giving the displayed line integral. A single-valued potential of a conservative vector field on the full punctured space exists if and only if this integral is zero. Necessity is the fundamental theorem for line integrals. For sufficiency, the primitive $P(\theta )={\int }_0^{\theta }g(s)ds$ is $2\pi$-periodic when the mean is zero, and the gradient in cylindrical coordinates gives $\nabla P=\mathbf{F}$.

On a domain with one angular cut removed, the same primitive gives a potential of a conservative vector field even when the mean is nonzero, because the angle has a single-valued continuous branch there. For example $g(\theta )=\sqrt{1+\cos \theta }$ has integral $4\sqrt{2}$ around the full circle but primitive $2\sqrt{2}\sin (\theta /2)$ on $-\pi <\theta <\pi$.

Taking the polar axis along the separation vector reduces the solid-angle integral to $2\pi {\int }_{-1}^1{e}^{ikdu}du$. Thus the total scattering cross-section contains the interference term $2\mathrm{Re}(c_1{c}_2^{\ast })\sin (kd)/(kd)$ as well as the two self terms. This is an outgoing-angle integral for fixed incident direction, rather than an average over the incoming beam.

If $u$ is Lebesgue integrable and $v$ is measurable on the same measure space, then $u+v$ is integrable exactly when $v$ is integrable. One implication follows from $|u+v|\le |u|+|v|$; the other follows from $|v|\le |u+v|+|u|$. Thus adding a function with finite integral of its absolute value cannot remove a failure of absolute integrability. This is the vector space property of L1 space stated as a useful test for a sum of terms with different decay rates.

Archimedes' principle states that buoyancy equals the downward gravitational weight, or weight, of the displaced fluid. In a resting fluid of constant mass density ${\rho }_f$ under uniform gravitational acceleration $g$, its magnitude isHere $V_{\mathrm{displaced}}$ is the volume from which the body excludes the fluid: the whole external volume for a fully immersed solid body, and just the immersed volume for a partly immersed one. The body's own mass does not enter this buoyancy formula. ^[ref]

A short proof outline for a fully immersed body is to imagine replacing it by exactly the fluid it displaces. The surrounding hydrostatic pressure is unchanged. That replacement fluid is at rest, so its upward pressure force must balance its downward force from gravity. The original body receives the same inward pressure forces on the same closed surface, hence the same upward buoyancy.

Equivalently, let $\Omega$ be the displaced region, $\partial \Omega$ its closed surface and ${\mathbf{e}}_z$ the upward unit direction. The divergence theorem, applied component by component to the surface integral, and hydrostatic pressure giveThis also explains why horizontal forces cancel even for an irregular shape. A spatially constant surface pressure contributes zero to the closed-surface integral. If the mass density varies with depth, replace ${\rho }_f{V}_{\mathrm{displaced}}$ by the mass ${\int }_{\Omega }{\rho }_{f}dV$ of the displaced fluid.

The closed-surface argument assumes that the fluid applies pressure over the entire boundary of a fully immersed body. A body sealing against a tank bottom may have no fluid beneath it; then the lower-face pressure force used in the proof is absent, and a supporting contact force must be treated separately. For partial immersion, the usual buoyancy formula assumes the exposed surface receives the common atmospheric pressure, neglects the much smaller buoyancy from the air, and neglects surface tension.

For the $0.10\mathrm{m}$ cube in the stipulated water, full immersion displaces $0.001{\mathrm{m}}^3$, so $F_b=1000\times 9.81\times 0.001=9.81\mathrm{N}$. If its uniform mass density is $600\mathrm{kg}{\mathrm{m}}^{-3}$, its mass is $m=600\times 0.001=0.600\mathrm{kg}$ and its downward force from gravity is $mg=5.886\mathrm{N}$. To hold it fully immersed at rest requires an additional downward forceOn release from rest it initially moves upward; its eventual floating equilibrium has a smaller displaced volume. A same-size cube of uniform mass density $1200\mathrm{kg}{\mathrm{m}}^{-3}$ has mass $1.200\mathrm{kg}$ and downward force from gravity $11.772\mathrm{N}$. Full immersion still gives $9.81\mathrm{N}$ of buoyancy, leaving an initial net downward force of $11.772-9.81=1.962\mathrm{N}$. Subsequent motion also involves drag.

The arc length of a piecewise smooth curve $\gamma :[a,b]\to M$ on a Riemannian manifold is the integral of its speed:An odometer measures this accumulated travel rather than displacement between the endpoints. In Euclidean space, the Riemannian metric is the ordinary inner product, so the formula becomes ${\int }_a^b\sqrt{{\sum }_i{\dot{\gamma }}_i(t{)}^2}dt$.

For a circle of radius $R$ traversed through an angle $\alpha$, use $\gamma (t)=(R\cos (\alpha t),R\sin (\alpha t))$, $0\le t\le 1$. Its speed is $R|\alpha |$, hence its arc length is $R|\alpha |$. Traversing a full circle twice gives $4\pi R$, even though the starting and ending points coincide. The same formula measures the two great circle arcs pictured in geodesic.

The curve parameter controls how fast the journey happens, not its arc length. A regular increasing or decreasing reparametrization preserves the integral by change of variables; reversing direction leaves the speed norm unchanged. Retracing part of a route does add length, because it changes the actual journey.

For the cycloid $x=a(t-\sin t)$, $y=a(1-\cos t)$, $a>0$, $0\le t\le 2\pi$, the area between the arch and its baseline isIts arc length isThe positive sign of $\sin (t/2)$ inside this parameter interval removes the absolute value in the speed. Omitting the two endpoint cusps does not change either integral.

For square-integrable periodic boundary data and the Fourier coefficients used in the Poisson kernel for the unit disk, the displayed identity gives the squared area norm of its Poisson extension $u$. At fixed radius $r$, Parseval's identity givesMultiply by the polar coordinates area element factor $rdr$ and integrate from zero to one. The constant term contributes $\pi |a_0{|}^2/4$, and ${\int }_0^1{r}^{2n+1}dr=1/[2(n+1)]$. Interchanging the nonnegative sum and radial integral is justified by the monotone convergence theorem. The resulting factor $1/(n+1)$ shows how higher angular frequencies contribute less to the interior area norm.

An asymptotic expansion describes a function near a specified limit by successively smaller scales. Each fixed finite truncation has a remainder smaller than its last retained scale. The central question is how the approximation improves as the parameter approaches the limit; convergence of the entire infinite series at a fixed parameter is a separate question. For the inverse-power scale $1,z^{-1},z^{-2},\dots$,means that, for every fixed nonnegative integer $m$,The little o notation makes the accuracy requirement precise. With expansions to all orders, this is also equivalent to an $O(z^{-m-1})$ remainder after the $m$th term; the implied bound may depend on $m$.^[ref]

For a quick convergent example, the finite geometric series identity givesFor $z>1$, both the asymptotic expansion as $z\to +\infty$ and the ordinary convergent geometric series describe this function. The exact remainder displays the required extra inverse power.

The more surprising example under a divergent factorial series can approximate a convergent integral has a convergent integral but a divergent series at every fixed positive $z$:A few terms can estimate $F(10)$ well, even though adding sufficiently many terms eventually destroys the approximation. Optimal truncation explains where to stop.

This makes asymptotic expansions useful when an exact expression is a complicated integral or differential equation solution, but a parameter is very large or very small. The leading term reveals the scale; corrections reveal which effects matter next. The asymptotic sequence must be specified: integer powers, fractional powers, logarithms, and exponential functions can encode different kinds of behavior. A good finite approximation includes a remainder estimate or an independently checked range of use.

An asymptotic sequence orders the scales against which successive errors are compared. On a specified domain approaching $x_0$, suppose each ${\varphi }_n$ is nonzero sufficiently close to the limit andEquivalently, ${\varphi }_{n+1}=o({\varphi }_n)$ in little o notation. The notationmeans that for every $N\ge 0$,Each $N$ stays fixed while the parameter tends to its limit.^[ref]

As $z\to +\infty$, $1,z^{-1},z^{-2},\dots$ is an asymptotic sequence, because each adjacent ratio is $1/z$. Fractional scales $1,z^{-1/2},z^{-1},\dots$ work too, with ratio $z^{-1/2}$. By contrast, $1,1+z^{-1},z^{-1},\dots$ fails at its first pair: the second scale is not smaller than the first. Ordering by written position alone does not create a hierarchy of sizes.

Successive division of the remainder by ${\varphi }_N$ uniquely recovers every coefficient:If two proposed expansions first differ at order $N$, subtracting them and dividing by ${\varphi }_N$ would give a nonzero constant tending to zero, a contradiction. This is uniqueness of the coefficients in a fixed scale, not uniqueness of the represented function.

For the factorial integral example, the first three limiting responses are $\lim F(z)=1$, $\lim z(F(z)-1)=-1$, and $\lim z^2(F(z)-1+z^{-1})=2$. They extract exactly the coefficients $1,-1,2!$. Choosing an asymptotic sequence therefore turns “the next small correction” into a precise and reusable operation.

For the same initial state, compression ratio and constant-volume heat input, the ideal Atkinson cycle adds net work during the extended expansion and subsequent constant-pressure return. The extra work is the integral of $p_{\mathrm{expansion}}-p_1$ over the added volume, which is positive. Equivalently, concavity of $x^{1/\gamma }$ gives $\gamma (r_p^{1/\gamma }-1)<r_p-1$ when $r_p>1$ and $\gamma >1$, so its efficiency exceeds the corresponding Otto cycle efficiency. The comparison fixes these shared conditions rather than asserting universal superiority under arbitrary engine constraints.

For an integrable function on an interval $I$ of positive finite length, its average value is the integral divided by that length:If $f$ is constant on equal-length subintervals, this is their values' arithmetic mean; unequal lengths require a weighted mean. For example, values $2$ and $6$ on two equal halves give average value $4$, whereas value $2$ on the first quarter and $6$ on the remaining three quarters give $5$. The integral measures total contribution; the average value of a function divides out the interval length.

For the function $f(x)=2x$ on $[0,1]$, the average value of a function is ${\int }_0^{1}2xdx=1$. Replacing it by the constant $1$ preserves its integral on this interval. This is why the Haar projection uses cell averages to obtain a coarser description while retaining each cell's total contribution.

The Bayesian model evidence is the prior predictive distribution evaluated at the data $D$. For a statistical model $H_j$ with a normalized prior density ${\pi }_j$, its marginal likelihood isThis is an expected value of the likelihood function under the prior distribution; for discrete statistical parameters, replace the integral by a sum. For continuous observations, $m_j(D)$ is a probability density function value with respect to a specified data reference measure, rather than a probability of one exact observation. Comparisons must use the same data and reference measure. The same $m_j(D)$ normalizes the within-model Bayesian posterior and supplies the between-model comparison in a Bayes factor.^[ref]

A normalized prior distribution makes the Bayesian model evidence an average, so it cannot exceed the largest likelihood function value on its support. A broad prior distribution can assign little mass to values that predict the data well, even when one value gives an excellent fit. The worked uniform-alternative binomial Bayes factor below makes the difference between averaging and maximum likelihood estimation explicit.