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Consider the advection equation $u_t+a{u}_x=0$ with constant speed $a>0$, no sources, and periodic boundary conditions on $[0,1]$. Use $N$ equal cells of width $h=1/N$, time step $k>0$, and indices modulo $N$. The exact solution is the periodic translation $u(x,t)=u_0(x-at)$. At a face, the Riemann problem transports the left state to the right, so the two-state upwind numerical flux is $F(L,R)=aL$, with $F(U,U)=aU$.^[ref]

The finite volume method with this numerical flux and an explicit Euler method step givesHere $U_j^n$ represents a cell average; the recurrence also occurs in the upwind finite difference scheme. Impose the specific CFL condition $0<\nu \le 1$. Each updated cell average is then a convex combination of two old ones. This proves preservation of their initial minimum and maximum, including nonnegativity for nonnegative data, and contraction of the maximum norm of the difference of two solutions on this periodic grid. These statements concern this constant-speed scalar update with this time step.

For an original synthetic example, take $N=8$, $a=1$, $h=1/8$, $k=1/16$, hence Courant number $\nu =1/2$. Choose a function constant on each successive cell, with the following values, so these are exact initial cell averages. One step givesFor example, cell zero receives from cell seven across the periodic face: $U_0^1=(0+1)/2=1/2$. All eight face transfers cancel in the total by shared face flux cancellation. More generally this periodic recurrence conserves the total algebraically for any $\nu$, even though the CFL condition is needed for the convex combination argument. With the same initial data and $\nu =3/2$, one step gives $U_2^1=-1$ and $U_4^1=5/2$, despite preserving the total.

After four steps, $t=1/4$, the exact translated profile has moved two cells to the right. Repeated averaging instead gives the explicit binomial coefficient formulawith all indices modulo eight. The diagram compares these computed cell averages with the exact translated cell averages. Their integrals agree, but the numerical jumps spread across cells. The finite volume method is useful for jump discontinuities because it retains the integral balance even at a jump; conservation alone does not guarantee a sharp profile. At $\nu =1$ this particular update is an exact one-cell shift, whereas for $0<\nu <1$ it mixes neighboring states. No accuracy conclusion at a jump follows just from the conserved total.

Phase matching makes wave amplitudes emitted at different depths add coherently. Consider collinear plane waves in a uniform transparent material of length $L$, with fixed polarization of waves and nonzero selected nonlinear optical susceptibility. Assume negligible pump depletion, negligible absorption, an initially absent harmonic, and a wide beam for which diffraction is negligible. The pump's wave amplitude is then constant throughout the interaction.

The pump electric field has spatial wave phase $k_{\omega }z$; its quadratic source has spatial wave phase $2k_{\omega }z$. The freely propagating harmonic has wave number $k_{2\omega }$. Define the second-harmonic phase mismatch byFor the chosen complex-field convention, the outgoing harmonic wave amplitude, up to an overall constant and propagation phase, is proportional toChoosing the conjugate complex-field convention reverses the exponent's sign without changing the wave intensity. Taking the squared complex modulus givesHere $C$ is fixed for the chosen modes and material; its dimensions accommodate the stated wave intensity convention. The quadratic pump dependence follows because the source wave amplitude is proportional to $E_0^2$, whereas $I_{\omega }$ is proportional to $E_0^2$. The sinc function comes from summing contributions along the whole material, just as a finite aperture produces diffraction through addition of complex amplitudes.

At $\Delta k=0$, the integral is $L$ and the harmonic wave intensity grows as $L^2$. At $\Delta kL=2\pi$, all those contributions cancel. This cancellation needs no absorption: their wave phases wind through a full turn. At high conversion, the constant-pump assumption fails, so the $L^2$ law cannot continue indefinitely. Tight focusing, finite optical pulses, absorption, and transverse separation of modes also require a more complete model.

Optical dispersion generally makes $n_{2\omega }\ne n_{\omega }$. In a suitable crystal, birefringence allows modes with different polarization of waves to experience different refractive indices, making their selected values equal. The appropriate crystal direction and allowed nonlinear optical susceptibility component must both work.^[ref]

For a worked calculation, choose $L_0=1\mathrm{mm}$ and $\Delta k=\pi \mathrm{m}{\mathrm{m}}^{-1}$. At $L=L_0$, the harmonic wave intensity is $4/{\pi }^2\simeq 0.4053$ of its perfectly matched value. At $L=2L_0$ it vanishes in this model. More material can therefore reduce the output. If instead $\Delta k=0$, doubling $L$ multiplies output by four. The two curves retain that common normalization, so their different peak heights are meaningful.

Photometry weights visible electromagnetic radiation by standardized visual response functions; radiometry measures its energy and power without visual weighting.^[ref]

The question is how much light a source supplies under a stated visual weighting. A beam containing $1\mathrm{mW}$ at the defining frequency $540\mathrm{THz}$ has luminous flux $0.683\mathrm{lm}$. Moving the same radiant flux to a region of smaller spectral luminous efficacy reduces its luminous flux. Changing the optical spectrum changes this conversion even when its total power stays fixed. This makes photometry useful for comparing lighting, while radiometry remains necessary for an energy budget. The lumen counts luminous flux; the candela additionally specifies its distribution over solid angle.

Two measurement routes implement the weighting: measure an optical spectrum and evaluate a weighted integral, or use a photometer whose spectral responsivity approximates the prescribed weighting. Both require a specified collection geometry. Neither is a complete prediction of an individual's perception of a scene.^[ref]

A photon gas is a collection of photons described statistically as radiation. Here the equilibrium model is homogeneous, isotropic electromagnetic radiation in vacuum at temperature $T>0$: each photon has energy $E=c|\mathbf{p}|=h\nu$, with momentum $\mathbf{p}$ and frequency $\nu$, and there are two independent transverse states of electromagnetic polarization. An absorbing cavity with walls at a common temperature provides a way to establish this thermal equilibrium. The bulk formulas assume cavity dimensions much larger than the typical wavelength $hc/(k_{B}T)$, so the electromagnetic mode density in a box can replace individual cavity modes by a continuum. ^[ref]

The crucial constraint is whether photon emission and photon absorption, or other number-changing processes, can equilibrate the photon population. When they can, photon number is not an independently conserved quantity and the equilibrium photon chemical potential is ${\mu }_{\gamma }=0$. For each polarization and frequency, the mean occupation number then follows the Bose-Einstein distribution:If number-changing processes are ineffective but number-conserving interactions redistribute energy, a constrained Bose-Einstein distribution can instead have $f=[\exp ((h\nu -{\mu }_{\gamma })/(k_{B}T))-1{]}^{-1}$ with chemical potential ${\mu }_{\gamma }\le 0$ fixed by the number constraint. Thus thermal equilibrium with respect to energy exchange alone does not establish ${\mu }_{\gamma }=0$. The zero value can also persist in an already prepared Planck photon distribution during suitable subsequent evolution; continual photon emission is not required to preserve it. ^[ref]

Combining $f$ with the two-polarization density of states, $g_{\nu }/V=8\pi {\nu }^2/c^3$, gives the Planck photon distribution and the spectral photon energy density:Here $B_{\nu }$ is the spectral radiance in Planck's law; $u_{\nu }$ is stored energy density per unit frequency, not outgoing power per area. Their different units are connected by the full-sphere solid angle and the speed of light $c$. The Boltzmann constant is $k_B$, and $h$ is the Planck constant. ^[ref]

The equilibrium equation of state is particularly simple:The radiation constant $a_R$ is the same constant called $a_{\mathrm{rad}}$ in black-body radiation; $\sigma$ is the Stefan-Boltzmann constant. An outline of the energy density calculation substitutes $x=h\nu /(k_{B}T)$ and uses ${\int }_0^{\infty }{x}^3/(e^x-1)dx={\pi }^4/15$. The radiation pressure follows because the angular average of the squared directional component of photon momentum is $1/3$ for isotropy; the thermodynamic derivation of blackbody radiation pressure supplies an independent route. These formulas describe the radiation contribution, not the cavity walls. The isotropic massless relation $p=u/3$ survives a different isotropic spectrum, whereas the coefficient in $u=a_R{T}^4$ requires the zero-chemical potential Planck photon distribution.

For a computed example with stipulated $T=2.725\mathrm{K}$, the photon energy density is $u\simeq 4.17174\times 10^{-14}\mathrm{J}{\mathrm{m}}^{-3}$ and the radiation pressure is $p\simeq 1.39058\times 10^{-14}\mathrm{Pa}$. The photon number density in the International System of Units isHere $\zeta$ is the Riemann zeta function. The mean photon energy differs from the energy at the peak of a spectral curve: counting photons and weighting them by energy emphasize different frequencies. The calculations use the exact International System of Units values of the Planck constant, Boltzmann constant and speed of light. ^[ref]

A motivating observation is the cosmic microwave background spectrum measured by the Far Infrared Absolute Spectrophotometer (FIRAS) aboard COBE. Its apparatus used a polarizing Michelson interferometer, with one input viewing a roughly seven-degree sky patch and the other an internal blackbody. Scanning mirrors changed the optical path length difference; heat-sensitive detectors recorded the varying power, and Fourier transform spectroscopy recovered the spectral difference. Operation outside the terrestrial atmosphere removed atmospheric emission and absorption, while shielding and cooling the optics in a helium cryostat suppressed the instrument's own radiation. ^[ref]

Absolute calibration came from moving an external blackbody into the sky aperture, filling the same beam used for the sky. This was an iron-loaded epoxy absorber shaped as a re-entrant cone. Repeated reflections made it a close approximation to a blackbody, while thermometers supplied its temperature. Mather and colleagues explicitly evaluated reflections, temperature gradients, radiation leaking through the material and leakage around its edge. These are limits on how faithfully the calibrator realizes thermal equilibrium radiation, rather than details that can be eliminated merely by calling it black. ^[ref]

During sky observations the internal blackbody temperature was adjusted to nearly null the spectral difference, reducing gain and drift errors. Fixsen and colleagues' 1996 analysis used calibrated spectra over wavenumbers $2$ to $21\mathrm{c}{\mathrm{m}}^{-1}$, about $60$ to $630\mathrm{GHz}$. They separated the sky's spatially uniform component, its dipole and Galactic emission, and varied sky cuts and foreground templates as controls. The resulting cosmic microwave background spectrum agreed with a single Planck function, with reported RMS deviations below 50 parts per million of its peak brightness. This is a residual statistic over the analyzed spectrum, not a pointwise relative error bound at every frequency. ^[ref]

The observation motivates using a photon gas as a physical model rather than merely a formal counting exercise. Its interpretation has a clear boundary: agreement with Planck's law constrains departures from an equilibrium spectral shape within the calibrated band and foreground assumptions; it does not directly measure radiation pressure or entropy density of a photon gas. Nor does it establish that today's cosmic microwave background is continually re-equilibrating with matter. A previously thermal spectrum can retain its Planck function shape as expansion of the universe lowers every photon frequency in the same proportion. The usefulness of the model is that one measured spectral temperature, together with its equilibrium assumptions, predicts energy density, radiation pressure, photon number density and entropy density of a photon gas consistently.

The photopic luminous efficiency function $V(\lambda )$ is dimensionless and describes standardized relative sensitivity for photopic vision. It is normalized to $V(555\mathrm{nm})=1$ and tabulated over $360$–$830\mathrm{nm}$. The wavelength coordinate is in standard air, rather than vacuum; changing coordinates requires converting the spectral radiant flux density as well. The official prescription uses linear interpolation between the specified 1 nm entries.^[ref]

Write ${\lambda }_a$ for the standard-air wavelength at $540\mathrm{THz}$. With spectral radiant flux ${\Phi }_{e,\lambda }$ in $\mathrm{W}\mathrm{n}{\mathrm{m}}^{-1}$ and $d\lambda$ in nanometres, the luminous flux isThe integral has units of watts, and $K_m$ converts them to lumens. For radiation at a single wavelength ${\lambda }_0$ with total radiant flux $P$, this reduces to ${\Phi }_v=K_{m}V({\lambda }_0)P$. For a sufficiently narrow finite band, this is an approximation requiring nearly constant $V$ across the band.^[ref]

Picard iteration replaces a tentative trajectory by the accumulated integral of its proposed rate. For an initial value problem $y^{\prime }=f(t,y)$, $y(t_0)=y_0$, define the Picard map and its starting function byHere $y_0(t)$ is the constant starting trajectory; the same symbol without its argument denotes the initial state. A fixed point of $\mathcal{P}$ solves this integral equation, and the fundamental theorem of calculus gives the original ordinary differential equation.

For $y^{\prime }=y$, $y(0)=1$, successive integrals giveAt $t=1$, the first five values are $1,2,2.5,2.66666667,2.70833333$, while $e\approx 2.71828183$. The fourth polynomial misses by about $0.00994850$. On $0\le t\le 1$, every omitted term of the exponential series is nonnegative and increasing, so the largest error occurs at $t=1$. The tail tends to zero, giving uniform convergence to the exponential function on this interval.

The beauty of Picard iteration is that each proposed trajectory supplies a better rate history for the next one. For this example, one integration adds exactly the next term of the exponential series. In general, the integrals may be difficult to compute, but the construction still explains local existence and uniqueness.

The short contraction mapping argument needs two independent controls. Choose a time-state cylinder $[t_0-a,t_0+a]\times {\overline{B}}_r(y_0)\subset U$ on whichThe second bound is the uniform local Lipschitz condition. Set $J=[t_0-h,t_0+h]$ with $0<h\le a$, $Mh\le r$ and $Lh<1$. In the function space of continuous functions $J\to {\mathbb{R}}^d$, use the supremum norm $\Vert u{\Vert }_{\infty }={\sup }_{t\in J}\Vert u(t)\Vert$. On its closed ball $\Vert u-y_0{\Vert }_{\infty }\le r$, direct integral estimates giveThus the map preserves the closed ball and is a contraction mapping there. This closed ball is a complete metric space, so the contraction mapping theorem gives convergence to exactly one fixed point. Any other local solution of a differential equation stays in the same ball after shortening its interval, and hence agrees there.

For $y^{\prime }=y$, take $r=1$, so $|y|\le 2$ in the state ball and $M=2$, $L=1$. Choosing $h=1/2$ gives $Mh=1$ and $Lh=1/2$, with the actual solution satisfying $|e^{1/2}-1|\approx 0.64872127<1$. The plotted interval $[0,1]$ is longer than this simple guarantee: the explicitly computed exponential series converges there anyway. The condition $Lh<1$ is sufficient for this one-step contraction mapping argument, rather than a necessary lifespan bound.

An ideal PID controller combines proportional control, integral action, and a term proportional to the derivative of the tracking error. With $k_d=0$, it gives proportional-integral control. The integral stores past error, while the derivative responds to its current rate of change. Practical implementations must account for measurement noise, which an ideal differentiator can amplify. ^[ref].

A planar lamina is an ideal mass distribution confined to a plane, with surface mass density $\sigma$ and infinitesimal mass $dm=\sigma dA$. Its moment of inertia about an axis is the integral of squared perpendicular distance against this mass element. Three mutually perpendicular axes through one point, with two in the lamina's plane, obey the perpendicular axis theorem even when the density is nonuniform.

Now both plates at $z=0$ and $z=H$ are stationary and both obey the Navier slip boundary condition with the same slip length $b$. Drive steady flow of a Newtonian fluid in incompressible flow by the constant pressure gradient $G=-dp/dx>0$, with constant dynamic viscosity $\mu >0$ and no other streamwise body force. The planar Poiseuille flow profile and its volumetric flow rate per unit width areThe full volumetric flow rate for channel width $W$ is $Q=Wq$ when side-wall effects are negligible. Thus $q$ has units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$, while $Q$ has units ${\mathrm{m}}^3{\mathrm{s}}^{-1}$. Denote the value at $b=0$ by $q_0=G{H}^3/(12\mu )$.

The two wall checks expose the normal signs:Both Navier slip boundary conditions therefore hold exactly. The two walls exert backward tangential tractions of magnitude $GH/2$ on the fluid. Increasing $b$ at fixed $G$ adds a uniform velocity offset, leaving the velocity gradient and wall shear stress unchanged. This differs from Couette flow with a slipping lower wall, where upper-wall velocity is fixed and shear stress decreases.

For $H=10\mu \mathrm{m}$, $b=2\mu \mathrm{m}$, $G=10^3\mathrm{Pa}{\mathrm{m}}^{-1}$ and $\mu =10^{-3}\mathrm{Pa}\mathrm{s}$, the wall velocity is $10\mu \mathrm{m}{\mathrm{s}}^{-1}$, the centre velocity is $22.5\mu \mathrm{m}{\mathrm{s}}^{-1}$, and $q=1.83\times 10^{-10}{\mathrm{m}}^2{\mathrm{s}}^{-1}$. The enhancement is $q/q_0=2.2$. Equivalently, a prescribed volumetric flow rate needs only $1/2.2$ of the original pressure gradient. This makes slip length useful for estimating pumping requirements in small channels.

At fixed $H$, $G>0$ and $\mu$, $b\to 0$ recovers ordinary planar Poiseuille flow. As $b\to \infty$, the velocity and $q$ grow without bound in this model, although the varying part of the profile stays the same. At exactly zero tangential wall traction, no steady solution can balance this nonzero pressure gradient: $u^{\prime }(0)=u^{\prime }(H)=0$ conflicts with $\mu [u^{\prime }(H)-u^{\prime }(0)]=-GH$. Unlimited enhancement is consequently not a prediction for a real channel; finite wall friction, finite driving capacity and the validity of the assumed steady flow must be assessed. At fixed finite $q$ instead, $G\to 0$ and the profile approaches uniform velocity $q/H$.

A short derivation integrates the reduced Navier-Stokes equation $\mu u^{\prime \prime }=-G$ twice. Equal slip lengths and the opposite inward unit normals give $u=G[z(H-z)+bH]/(2\mu )$. Its integral over the gap supplies the enhancement factor $1+6b/H$.

On continuous functions on a compact interval, evaluation at an interior point is not continuous in the integral norm. Narrow triangular peaks have fixed point value and integral tending to zero. Subtracting such a peak from the constant function one gives functions vanishing at the point and converging in the integral norm to one. Thus the evaluation kernel is not closed, although it is closed in the uniform norm.

Here is a fully computed endpoint version, now using the L2 norm on the actual continuous functions $C([0,1])$. For each integer $n\ge 1$, set $f_n(x)=\max (1-nx,0)$. Its point evaluation at zero is $E(f_n)=f_n(0)=1$, whileConsequently $f_n\to 0$ in the L2 norm but $E(f_n)\nrightarrow 0$: point evaluation is discontinuous on this normed space. Quantitatively $|E(f_n)|/\Vert f_n{\Vert }_2=\sqrt{3n}$ is unbounded. At $n=100$, the L2 norm is approximately $0.057735$, the integral is $0.005$, and the point value is still $1$. In the supremum norm, $\Vert f_n{\Vert }_{\infty }=1$ and $|E(f)|\le \Vert f{\Vert }_{\infty }$, so $E$ has operator norm one instead.

Set $g_n=1-f_n$. Then $E(g_n)=0$ and $\Vert g_n-1{\Vert }_2=1/\sqrt{3n}\to 0$, although $E(1)=1$. Thus the kernel of $E$ fails to be a closed set. On the full L2 space, a further distinction is essential: its elements are equivalence classes under equality almost everywhere, and changing a representative at zero does not change its class. Arbitrary point evaluation is therefore not even well-defined there. Restricting first to continuous functions makes the discontinuity statement precise.^[ref]

A point spread function is an imaging system's response to an optical point source. Here it means the distribution of image light intensity: an ideal point object becomes a finite spot, often with faint rings, because of diffraction. An unresolved star and a sufficiently small bead emitting fluorescence are practical approximations to an optical point source. Their image tells us how nearby object details will overlap, and reveals blurring from optical aberrations, defocus, and, for an optical telescope, astronomical seeing.

A useful convention separates shape from throughput. At a fixed image plane, let $p(\mathbf{x})\ge 0$ be the point spread function with ${\int }_{{\mathbb{R}}^2}p(\mathbf{x})d^2\mathbf{x}=1$. Its units are inverse area. A point delivering detected radiant flux $F$ at image position ${\mathbf{x}}_0$ produces light intensity $Fp(\mathbf{x}-{\mathbf{x}}_0)$; this point spread function normalization makes the integral over the image equal to $F$. A profile divided by its own peak is convenient for comparing widths, but is a different point spread function normalization and does not by itself compare total radiant flux.

In incoherent imaging, different object points have negligible mutual spatial optical coherence over the measurement, so their image light intensities add. If the response also obeys translation invariance over the chosen field of view, the expected continuous image is the convolutionHere $O$ is the object brightness distribution expressed in image-scaled coordinates, with magnification already included, and $\tau$ is constant transmission. For $O(\mathbf{u})=F_0\delta (\mathbf{u}-{\mathbf{u}}_0)$, where $\delta$ is the two-dimensional Dirac delta function, this gives $I=\tau F_{0}p(\mathbf{x}-{\mathbf{u}}_0)$. A changing response across the field of view instead requires a position-dependent integral kernel $p(\mathbf{x};\mathbf{u})$; one measured point spread function is then only a local calibration.

In coherent imaging, the superposition principle applies to the complex wave amplitudes. With amplitude spread function $h$, the corresponding ideal model isThus $|h\ast E{|}^2$ generally differs from $|h{|}^2\ast |E{|}^2$: coherent contributions produce wave interference terms that depend on their relative wave phases. Their time average vanishes for mutually incoherent emitters, which explains the intensity convolution. Partial optical coherence requires the correlations between object points as well.^[ref]

The figure is an original diffraction model, not measured data. All panels have the same total radiant flux: each member of a pair contributes half the single-source radiant flux. Dashed curves show the separate light intensity contributions; the solid curve is their sum. The point spread functions have tails, so even sources satisfying the Rayleigh criterion overlap. As separation decreases, the image changes continuously; there is no abrupt disappearance of all information at one separation.

A motivating measured example is Measurement of point-spread function (PSF) for confocal fluorescence microscopy by InCheon Song and others (2005). A $543.8\mathrm{nm}$ laser, delivered through an optical fiber, excited fluorescent beads on a slide through a $100\times$, numerical aperture $1.3$ oil-immersion objective. A moving stage scanned the bead; emitted light returned through the objective, an emission filter and a pinhole to a photodiode. The readout was fluorescence light intensity against scanned position. Under the same setup, nominal bead diameters $0.17$, $0.5$, and $1\mu \mathrm{m}$ produced progressively broader measured spots. The size comparison controls for the finite object: the measured response is the convolution of bead emission with the effective point spread function, rather than a pure point response. Inverting this convolution model gave a response compared with the theoretical one. The smaller bead gave a weaker signal relative to noise; uncertainty in actual bead diameter also affected the restored width. These are limits of the calibration. In confocal microscopy, excitation and pinhole detection jointly determine the effective point spread function; this experiment does not measure the ideal widefield Airy disk plotted above.^[ref]

For a new calibration, useful additional controls are to measure detector background, avoid saturating the photodetector, repeat the scan to check stability, and compare isolated beads at several positions in the field of view. Record wavelength, numerical aperture, pinhole setting, sampling interval and optical focus. These are practical recommendations, not extra measurements claimed for the published experiment.

Point spread function normalization specifies how the response is scaled. Unit-integral scaling separates image shape from total radiant flux; unit-peak scaling compares shapes but does not make their total radiant fluxes equal.

For a sequence of functions $f_n:X\to \mathbb{R}$, pointwise convergence to a function $f:X\to \mathbb{R}$ meansIt is the ordinary limit of a sequence at each fixed input. The threshold $N=N(x,\epsilon )$ may depend on the input. Uniform convergence instead requires one threshold for the entire domain of a function.^[ref]

Consider the polynomial functions $f_n(x)=x^n$, for positive integers $n$, on the interval $[0,1]$. For $0\le x<1$ the geometric sequence $(x^n)$ tends to zero, while $f_n(1)=1$. ThusEach $f_n$ is a continuous function, but its pointwise limit is discontinuous at one. Holding an input still hides what happens near that endpoint. The analogous even-power example on $[-1,1]$ appears in the source.^[ref]

For $0<x<1$ and $0<\epsilon <1$, taking the logarithm of $x^n<\epsilon$ and dividing by the negative real number $\log x$ reverses the inequality:The floor function followed by adding one handles the strict inequality even when the ratio is an integer. For tolerance $\epsilon =0.01$, the first successful indices at $x=0.5$, $0.8$, and $0.99$ are respectively $7$, $21$, and $459$. At $x=0.5$, $2^{-6}=0.015625$ fails and $2^{-7}=0.0078125$ succeeds. At $x=0$ and $x=1$, the error from this pointwise limit is already zero, so $N=1$ works. Nevertheless, $N(x,\epsilon )$ grows without bound as $x\to 1$ from below. A finite list of accurate test inputs cannot establish uniform convergence.

An even sharper warning concerns integration. On $[0,1]$, define the continuous functionsTheir supports are $[0,1/n]$, with height $n$ at $1/(2n)$ and value zero at both ends. Every fixed $x>0$ eventually lies beyond the support, and $g_n(0)=0$ always, so pointwise convergence gives the zero function. Nevertheless, each integral is the area of a triangle: ${\int }_0^1{g}_n(x)dx=\frac{1}{2}(1/n)n=\frac{1}{2}$. Shrinking width compensates increasing height. Hence the limit of a sequence of these integrals differs from the integral of the pointwise limit; interchanging them requires additional conditions, such as uniform convergence preserves integrals on a compact interval.

Put $I=I(1)$. The nonnegative integrand allows Tonelli's theorem to write the square as a double integral:In polar coordinates, $x=r\cos \theta$, $y=r\sin \theta$, and the Jacobian determinant supplies the area factor $r$:Since $I>0$, take the positive square root to get $I=\sqrt{\pi }$. The computation proves finiteness as well as the value. For an explicit limiting argument, first integrate over disks of radius $R$; their integrals are $\pi (1-e^{-R^2})$, which increase to $\pi$ as $R\to \infty$. ^[ref]

The geometric surprise is that radial shells get longer as $r$ increases, even while the Gaussian function gets smaller. Its central height is largest at zero, but the radial contribution $2\pi r{e}^{-r^2}$ vanishes there and peaks at $r=1/\sqrt{2}$. Differentiating gives $2\pi e^{-r^2}(1-2r^2)$, which changes sign at that radius. Keeping the Jacobian determinant is what makes the one-dimensional integral and the plane geometry agree.

Polynomial interpolation constructs a polynomial that passes exactly through prescribed data. Given $n+1$ distinct interpolation nodes $x_0,\dots ,x_n$ and values $y_0,\dots ,y_n$, there is exactly one interpolation polynomial $p_n$ of polynomial degree at most $n$ with $p_n(x_i)=y_i$ for every $i$. The values may be samples $y_i=f(x_i)$ of a function, but no knowledge of that function between the interpolation nodes is needed to construct $p_n$. The Lagrange polynomial and Newton polynomial are two representations of this same answer.^[ref]

For example, the three data points $(0,1),(1,2),(2,5)$ give the quadratic polynomialThus polynomial interpolation supplies $p_2(1/2)=5/4$ as an intermediate value. Whether this approximates an unknown underlying function well is a separate question about polynomial interpolation error.

Both hypotheses in the uniqueness statement matter. The Lagrange cardinal polynomials prove existence constructively. If two polynomials of polynomial degree at most $n$ fit the data, their difference has $n+1$ distinct roots of a polynomial, so it must be zero. Repeated interpolation nodes with conflicting values have no solution, and repeated equal values provide no additional independent condition; prescribed derivatives instead lead to Hermite interpolation. Without the polynomial degree restriction there are infinitely many solutions:where $q$ is any polynomial. By the factor theorem, this describes every polynomial fitting the same data. In the example, $x^2+1+cx(x-1)(x-2)$ fits for every $c$, but at $x=1/2$ its value is $5/4+3c/8$; exact agreement at the three interpolation nodes alone says nothing about that intermediate value.

The usefulness of polynomial interpolation is that data become an explicit function whose values, derivatives, and integrals can be calculated. Differentiating or integrating the Lagrange cardinal polynomials produces differentiation weights from nodal evaluations or a quadrature rule. Exact passage through observations is the defining constraint. In polynomial regression, linear least squares instead chooses a polynomial degree and minimizes a sum of squared residuals, generally without passing through every observation. Interpolating noisy values reproduces their errors at the interpolation nodes and can amplify them between interpolation nodes; polynomial interpolation does not itself smooth measurement noise. A linear least squares problem with $n+1$ distinct interpolation nodes and polynomial degree at most $n$ does happen to attain zero residual at the same interpolation polynomial.

The variable $w=h^2$ converts the shallow V-channel lubrication flux equation into a porous medium equation:Indeed $h^4{h}_x=w^{3/2}{w}_x/2=(w^{5/2}{)}_x/5$. Its fixed-mass source-type solution gives $h\propto t^{-1/7}$ and support length proportional to $t^{2/7}$. Integrating the similarity equation once gives $C{h}^4{h}_x=-2x{h}^2/(7t)$, and hence $h=[3(X(t{)}^2-x^2)/(7Ct){]}_{+}^{1/3}$. The constant-volume similarity in a V-shaped channel fixes $X$ from the positive integral $J={\int }_{-1}^1(1-s^2{)}^{2/3}ds$: $X=(\alpha V/J{)}^{3/7}(7Ct/3{)}^{2/7}$.

For $p>0$ and a nonnegative integer $n$, integration by parts gives, when $n>0$,The endpoint term $[s(1-s^p{)}^n{]}_0^1$ vanishes. ThereforeRepeated application yields $J_n(p)={\prod }_{j=1}^{n}pj/(pj+1)$. The integration by substitution $w=s^p$ also identifies $J_n(p)=B(1/p,n+1)/p$ with an Euler beta function. A related weighted integral follows immediately from the same substitution:

The singular equation $2x{w}^{\prime \prime }-w^{\prime }+\lambda x^{2}w=0$ becomes $W^{\prime \prime }+(2\lambda /9)W=0$ under $z=x^{3/2}$, $w(x)=W(z)$. Its zero-endpoint finite-energy quotient transforms as $2{\int }_0^1{x}^{-1/2}(w^{\prime }{)}^{2}dx/{\int }_0^1{x}^{1/2}{w}^{2}dx=(9/2){\int }_0^1(W^{\prime }{)}^{2}dz/{\int }_0^1{W}^{2}dz$. The Poincaré inequality therefore proves the sharp minimum $9{\pi }^2/2$, attained by $w=\sin (\pi x^{3/2})$. This coordinate change turns an apparently singular weighted variational problem into the usual interval energy problem. The admissible eigenfunction is $O(x^{3/2})$, making the integration-by-parts boundary term $w{x}^{-1/2}{w}^{\prime }$ vanish and both energy integrals converge.

A power-law distribution uses a power law for a probability mass function or a probability density on a specified range. For a continuous model with $p_X(x)=C{x}^{-\beta }$ on $x\ge x_{\min }>0$ and zero below it, the total integral is finite exactly when $\beta >1$. Normalization then requires $C=(\beta -1)x_{\min }^{\beta -1}$, giving a Pareto distribution with shape parameter $\alpha =\beta -1$. Discrete models instead normalize a sum. The support and normalizing coefficient are part of the model, and a power law for its tail probability uses a different exponent from its probability density.

Two stationary no-slip surfaces with local gap $H(x)=h+x^2/(2a)$ transmit pressure-driven volume flux $q=-H^3{p}_x/(12\mu )$ per unit transverse length. Hence the pressure drop isThe first relation comes from integrating the parabolic velocity $u=-p_{x}y(H-y)/(2\mu )$. Substitution $x=\sqrt{2ah}\tan \theta$ evaluates the resistance integral using ${\int }_{-\pi /2}^{\pi /2}{\cos }^4\theta d\theta =3\pi /8$. The formula gives the leading resistance when $h\ll a$.