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A quadrature weight multiplies a sampled integrand value in a quadrature rule. For ordinary numerical integration, its units match those of the integration variable. Positive quadrature weights that sum to $b-a$ make $Q(f)/(b-a)$ a weighted average of the samples; signed quadrature weights can magnify sample perturbations, as quantified by noise sensitivity of a quadrature rule.

Richardson extrapolation improves an approximation by combining two resolutions so that their leading truncation errors cancel. The useful surprise is that the size of that leading error need not be known: its scaling under refinement is enough. This lets an existing finite difference method or quadrature rule produce a more accurate answer from two runs, without first deriving a new method from scratch.

Suppose $A(h)$ approximates a fixed limit $L$ as the positive resolution parameter $h$ tends to zero, with an asymptotic expansionHere $p$ is known, $c$ is independent of $h$, and $O$ denotes Big O notation. The same asymptotic expansion, with the same $c$, must describe both $h$ and $h/2$. Nested grids make it natural to reuse samples, although nesting itself does not establish this error model. Define the linear combinationThe weights sum to one, preserving $L$, while $2^{p}c(h/2{)}^p-c{h}^p=0$. Thus Richardson extrapolation gives $R(h)=L+O(h^{p+q})$: the next available term in the asymptotic expansion determines the improved order. Justin Solomon's numerical analysis text develops this cancellation for finite differences and applies it to numerical integration.^[ref]

Geometrically, regard the approximation as a function of $t=h^p$. Its first two terms are a straight line $L+ct$. The two computed points lie at positive $t$; extending the line through them to $t=0$ gives $R(h)$. This is why the operation is extrapolation: it estimates a value outside the sampled parameter interval. The remaining curvature corresponds to the uncancelled terms of the asymptotic expansion.

Worked example: a centered derivative of sine. For the smooth function $f(x)=\sin x$, the exact derivative at $x=0$ is $L=\cos 0=1$. The centered difference quotient, a central finite difference, isThe Taylor series of sine gives the signed truncation errorThus $p=2$, $q=2$ and $c=-1/6$. Halving $h$ divides the leading truncation error by four. Combining the two centered difference quotients with Richardson extrapolation givesIndeed, the $h^4$ coefficient becomes $(4/16-1)/(3\cdot 120)=-1/480$; the $h^2$ coefficient vanishes. This is the even-power case of Richardson extrapolation discussed in Persson's authored lecture notes.^[ref]

For $h=0.4$ and $h/2=0.2$, the actual computations areThe absolute error drops from about $2.6454\times 10^{-2}$ for $A(0.4)$ to $5.3080\times 10^{-5}$ for $R(0.4)$, an improvement by roughly a factor of 498. Refining again gives $R(0.2)\approx 0.999996670632607$, whose absolute error is about $3.3294\times 10^{-6}$. The refined truncation errors decrease by nearly four for $A$ and nearly sixteen for $R$, reflecting their second and fourth orders.

The logarithmic scale on each axis turns power laws into straight lines. The dashed reference curves are $h^2/6$ and $h^4/480$, the magnitudes of the leading truncation errors. The plotted errors are computed with 70-digit decimal arithmetic, rather than clipped to a display floor; this isolates the asymptotic expansion from roundoff error.

Worked example: integrating an exponential. Approximate the definite integral of the exponential function on $[0,1]$:For numerical integration, the composite trapezoidal rule adds the areas of trapezoids between successive samples; this is the quadrature use of the endpoint average that also occurs in the ODE trapezoidal rule. On $n$ equal intervals, with $h=1/n$, defineSumming the geometric series gives $T_n=(e-1)\frac{h}{2}\frac{e^h+1}{e^h-1}$. Expanding the exponential function near zero yieldsThe leading truncation error again has $p=2$, $q=2$. On the uniform nested grids with $n$ and $2n$ intervals, Richardson extrapolation therefore givesThis linear combination is exactly composite Simpson's rule on the finer grid, whose spacing is $h/2$. On each coarse panel, an old endpoint receives weight $h/6$ and the new midpoint weight $2h/3$; shared old endpoints receive weight $h/3$. These are the usual Simpson's rule weights $1,4,2,\dots ,4,1$ multiplied by $h/6$. The equivalence is explained directly in LeMesurier's university notes.^[ref]

For two coarse panels and four fine panels,The absolute errors are about $0.0356493$, $0.00894008$ and $0.0000370135$, respectively. Reusing the fine-grid samples makes Richardson extrapolation especially attractive here: the resulting Simpson's rule needs no samples beyond those already used by $T_{2n}$.

When the improvement is justified. Richardson extrapolation requires the stated asymptotic expansion and the correct positive $p$. If the actual leading exponent is $r\ne p$, its coefficient after combination is $c(2^{p-r}-1)/(2^p-1)$, so the leading truncation error generally remains. Merely knowing $A(h)-L=O(h^p)$ does not supply a common coefficient $c$ or a higher-order remainder. A discontinuity of a function, a singular endpoint, or unrelated changes between grids can invalidate the assumed asymptotic expansion; refinement alone does not guarantee improvement for every numerical problem.

Roundoff error and errors in the sampled values survive the cancellation of truncation error. If the coarse and fine values have absolute perturbations bounded by $\epsilon$, the perturbation in $R(h)$ is bounded by $(2^p+1)\epsilon /(2^p-1)$; for $p=2$ this is $5\epsilon /3$. In floating-point arithmetic, a centered difference quotient can also amplify sample errors through division by $h$, so an indefinitely smaller $h$ need not be more accurate. Persson's notes describe this balance between truncation error and roundoff error.^[ref]

The same cancellation idea appears in step-doubling error estimation. For an order-$p$ one-step method for an ordinary differential equation, two half steps each contribute local truncation error of order $h^{p+1}$, so their accumulated leading local error has ratio $2^{-p}$ to one full step. This explains the denominator $2^p-1$ there; the appropriate exponent follows from that local error model, rather than automatically from a generic use of Richardson extrapolation.

For points $x_1,\dots ,x_N$ in $[0,1{)}^d$, star discrepancy compares their empirical frequencies in anchored boxes with Lebesgue measure:The half-open box convention makes boundary membership precise. Small star discrepancy means that every such box contains nearly its uniform share of points; it does not mean the points are independent random samples.

In one dimension, take midpoint points $x_i=(i-1/2)/N$. Between consecutive points, the empirical count is constant while the interval length increases linearly. Just before or after a point the largest difference is $1/(2N)$, so $D_N^{\ast }=1/(2N)$. Repeating a single point instead leaves large gaps and large discrepancy. The definition directly controls the integration error for indicators of anchored boxes and motivates deterministic sampling for numerical integration.

Surface pressure integration for airfoil lift reconstructs the pressure contribution to lift by summing the local surface pressure loads with their directions. With $\mathbf{n}$ pointing from a two-dimensional body into the fluid, the force per unit span from its closed contour $\Gamma$ isThe uniform reference pressure cancels on the closed contour. This is a vector line integral, not the sum of positive pressure magnitudes.^[ref]

For upper and lower surfaces that are graphs over $0\le x\le c$, their pressure contribution perpendicular to the chord isVertical end closures have no chord-normal pressure contribution. The pressure force parallel to the chord, $A_p^{\prime }$, still matters when resolving flow-relative lift. With flow direction $(\cos \alpha ,\sin \alpha )$ in aft/chord-normal axes, $L_p^{\prime }=N_p^{\prime }\cos \alpha -A_p^{\prime }\sin \alpha$. Equating $N_p^{\prime }$ with lift needs a suitable angle of attack approximation. Pressure integration also omits the viscous stress contribution.

Finite pressure tappings require numerical integration. A composite trapezoidal rule integrates the linearly interpolated samples exactly; it need not integrate the actual unsampled distribution exactly. For an original example, let $f(\xi )=C_{p,l}-C_{p,u}=0.8$ and place taps at $\xi =0,0.2,\dots ,1$. A second invented profile agrees at every tap but has an additional $1.2{\sin }^2[\pi (\xi -0.4)/0.2]$ on $0.4\le \xi \le 0.6$ and zero addition elsewhere. Both reconstructions give $0.8$, while the second true integral is $0.8+1.2(0.2/2)=0.92$.

At $q_{\infty }=500\mathrm{Pa}$ and $c=0.20\mathrm{m}$, the two stipulated normal pressure forces are $80$ and $92N/m$, although the six samples are identical. This quadrature ambiguity of pressure-tap data is an instance of how finite samples do not certify quadrature accuracy. Additional geometric or regularity information is needed to bound the missed contribution.

A motivating experimental comparison uses a force balance independently of the pressure reconstruction. Robert M. Pinkerton's 54-orifice distribution measurements provide a historical example of resolving the shape of a pressure field.^[ref] A modern proposed test repeats both readings at controlled Reynolds number and angle of attack, checks tap coordinates and pressure-channel response, and refines sampling near sharp features. Agreement in a single integrated load does not prove agreement in the entire pressure distribution or pitching moment.

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.

Stipulate $H_0=70\mathrm{km}{\mathrm{s}}^{-1}\mathrm{Mp}{\mathrm{c}}^{-1}$ and $c=299792.458\mathrm{km}{\mathrm{s}}^{-1}$, so $c/H_0=4282.7494\mathrm{Mpc}$. A megaparsec is $10^6$ parsecs. Neglect radiation in cosmology. Compare an accelerating flat model $({\Omega }_{m,0},{\Omega }_{\Lambda ,0})=(0.3,0.7)$, an open matter-only model $(0.3,0)$ and the Einstein-de Sitter universe $(1,0)$. Their present deceleration parameters are respectively $-0.55$, $0.15$ and $0.5$. For these models,Here $\chi$ is dimensionless; the radial null geodesic in FLRW spacetime supplies its integral and spatial curvature of an FLRW universe supplies $S_k$. The observer-to-source comoving radial distance is $(c/H_0)\chi$, whereas the transverse distance uses $S_k$. Omitting the latter distinction would bias the open-model comparison.

At $z=0.5$, numerical integration gives $\chi =0.440984335$ for the flat accelerating model and $\chi =0.392227378$ for the open matter-only model. ThusThe accelerating model is $0.21559$ mag fainter than the open matter-only prediction at the same redshift and standardized absolute magnitude, corresponding to $1.10438$ times the luminosity distance and $0.81991$ times the radiative flux. These are model calculations, not observations from either collaboration.

The Milne model has $q=0$, ${\Omega }_{m,0}={\Omega }_{\Lambda ,0}=0$ and ${\Omega }_{k,0}=1$, giving $\chi =\log (1+z)$ and $d_L=(c/H_0)(z+z^2/2)$. The Einstein-de Sitter luminosity-distance relation independently checks the flat matter-only integral. At $z=0.5$, the accelerating model exceeds the Milne model by only $0.12317$ mag, illustrating the dependence of a residual on the chosen reference. Over a longer redshift interval, luminosity distance integrates past deceleration as well as recent acceleration; a residual need not increase forever.

Multiplying $H_0$ by $f>0$ at fixed density fractions divides every $d_L$ by $f$ and shifts every distance modulus by $-5{\log }_{10}f$. The predicted apparent magnitude is unchanged if $M_{B,0}$ simultaneously increases by $5{\log }_{10}f$. This is the absolute magnitude–Hubble constant degeneracy. For example, $70\to 77\mathrm{km}{\mathrm{s}}^{-1}\mathrm{Mp}{\mathrm{c}}^{-1}$ shifts $\mu$ by $-0.20696$ mag and $M_{B,0}$ by $+0.20696$ mag. Relative curves at the same $H_0$ retain their separation; determining $H_0$ requires an external absolute magnitude calibration.