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For a concrete comparison, take $f(x)=x^2$ on $[0,6]$. The integral is $I={\int }_0^6{x}^{2}dx=72$, whereas the unit-width left and right Riemann sums are $L=55$ and $R=91$. The exact integer power sum formula givesThe endpoint choice explains the opposite signs of the $18$ correction. The remaining correction is one in both cases.

The same method is useful far beyond squares of numbers. Substituting $B_0$, $B_1$, $B_2$ and $B_4$ into the integer power sum formula gives the exact fourth-power resultAt $N=10$, this isThus a finite polynomial calculation replaces ten additions, or a million additions if $N$ is a million. Although the individual corrections can be rational numbers, the result is the exact integer sum.

There is also a useful numerical integration interpretation. The average of the left and right Riemann sums is the composite trapezoidal rule. For $x^2$, the endpoint corrections cancel but its value is $73$, one above the true integral. Subtracting the $B_2$ correction makes this particular polynomial example exact. For $f(x)=x^4$, the two interior corrections above can be written $B_2(f^{\prime }(N)-f^{\prime }(0))/2!$ and $B_4(f^{\prime \prime \prime }(N)-f^{\prime \prime \prime }(0))/4!$. These are exact identities for these polynomials; extending such corrections to general functions requires a separate error analysis.

Electrolysis uses an external source of electrical work to drive redox chemical reactions at the electrodes of an electrolytic cell. Its characteristic application is forcing a chemical reaction in a direction that is nonspontaneous under the specified conditions: a discharging galvanic cell supplies electrical work, whereas electrolysis consumes electrical work. This makes electrolysis useful for producing otherwise difficult-to-obtain chemical substances, charging a rechargeable battery, and depositing a metal coating on a conducting object.^[ref]

Always identify the anode by oxidation and the cathode by reduction. In ordinary driven two-terminal operation, the source withdraws electrons from the positive anode and delivers electrons to the negative cathode. In a discharging galvanic cell, the anode is negative and the cathode positive. Thus terminal signs change with operating mode, while the reaction definitions remain valid. Through the electrolyte, ions transport electric charge; they need not all be consumed at an electrode. The electrolyte and electrolytic cell articles explain this transport and its conventional current direction.

For a net chemical equation transferring $z$ electrons per reaction event, the molar reaction Gibbs free energy obeyswhere $F$ is the Faraday constant and $E_{\mathrm{chem}}$ is the reversible cell electromotive force for the reaction in the written direction. A nonspontaneous direction has ${\Delta }_{r}G>0$ and $E_{\mathrm{chem}}<0$ at fixed temperature and pressure. The ideal opposing voltage magnitude is ${\Delta }_{r}G/(zF)$; sustaining a finite electric current generally needs additional drive because of electrical resistance and finite electrode reaction rates. This thermodynamic voltage concerns the actual composition, rather than necessarily standard conditions.^[ref] Not every use of electrolysis increases bulk chemical Gibbs free energy: transferring copper between equivalent surfaces can have zero bulk ${\Delta }_{r}G$ while the source still supplies dissipated electrical work. See the existing electrolytic cell example.

A particularly revealing experiment measures copper transfer instead of merely looking for a coating. Two separated copper electrodes stand in aqueous $\mathrm{CuS}{\mathrm{O}}_4$ electrolyte in a vessel, with insulated leads to a regulated direct-current power supply, a series ammeter and a switch. A balance reads each electrode's mass before and after a timed run. The Institute of Physics describes copper cells with ammeters, adjustable current and removable cathodes; its comparison uses three similar cells at $1\mathrm{A}$ for ten minutes, $0.5\mathrm{A}$ for ten minutes, and $0.5\mathrm{A}$ for twenty minutes. The specified observation to test is that deposited copper mass is proportional to electric current times time. These are documented experimental settings, not measurements generated here.^[ref]

The ideal half-reactions areEach deposited copper chemical atom requires two electrons, so $z=2$. The anode loses copper and the cathode gains it; ideally their mass changes have equal magnitudes, and the bulk copper cation inventory stays constant. Local concentrations near the electrodes can nevertheless differ. The sulfate anions participate in ionic transport without appearing in these net half-reactions; electronic electric current in the wires becomes ionic electric current in the electrolyte through the two interfaces.^[ref]

Clean, rinse, dry and identify the electrodes before their initial weighing. After interrupting the electric current, rinse away adhering electrolyte, dry again and weigh without rubbing off the deposit. KIT's university practical explicitly calls for distilled-water rinsing, careful drying and weighing both electrodes; it warns that loose copper grains can be lost during handling.^[ref] Useful additional controls are repeated drying and weighing until the reading stabilizes, and a matched unpowered immersion followed by the same rinse/dry cycle. These controls test retained liquid, residual dissolved salts, handling loss and chemical changes unrelated to passed electric charge; they are proposed checks, not additional observations reported by KIT. Keep immersed electrode area, spacing, electrolyte composition and temperature comparable across runs, and keep the electrodes from touching.

Record the electric current as a function of time, together with the exact switch-on and switch-off times. Taking $I(t)\ge 0$ in the chosen deposition direction, the transported electric charge isFor constant electric current, $Q=I(t_1-t_0)$; with sampled readings, use numerical integration rather than assuming that one ammeter reading remained constant. The two equal-charge Institute of Physics settings each give $600\mathrm{C}$, while its shorter $0.5\mathrm{A}$ setting gives $300\mathrm{C}$. With $M_{\mathrm{Cu}}=63.546\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$, the mass per amount of substance tabulated by KIT, and $F\approx 96485.33212\mathrm{C}\mathrm{mo}{\mathrm{l}}^{-1}$, ideal Faraday's laws of electrolysis predictThese are calculated predictions for complete conversion of the measured electric charge into retained copper, not experimental results. For a stable process, doubling the electric charge doubles the deposited mass; equal electric charge gives equal deposited mass even when the electric current and duration differ.

The balance reading is a test of the whole model. Competing reduction of water or other chemical species can divert electric current from copper deposition; transient electrode capacitance can also consume part of a short pulse without producing copper. A fragile deposit can detach even when its formation used the expected electric charge. Conversely, retained electrolyte or salts can increase an apparent mass gain. Comparing anode loss with cathode gain helps detect such problems but does not identify every cause. A slope below $M_{\mathrm{Cu}}/(2F)$ alone cannot distinguish competing chemical reactions from lost deposit. Current efficiency measures the fraction of transported electric charge assigned to a specified electrode reaction, rather than the fraction of electrical energy stored chemically.^[ref]

The beautiful connection is that an electrical measurement counts chemical events: integrating electric current and reading a balance provide two ways of following the same electron stoichiometry. This makes electrolysis useful for controlling coating quantity and for testing a charge-to-amount of substance relation. It does not determine coating uniformity, adhesion or the required voltage, which depend on transport and electrode chemical kinetics as well.

Function evaluation determines the function output $f(x)$ for a specified function input $x$ in the domain of a function. Numerical integration uses these values at selected quadrature nodes.

Let $V_h\subset V$ be a conforming finite element space. Suppose the exact solution satisfies $a(u,v)=\ell (v)$ for every $v\in V$ and its Galerkin method approximation satisfies $a(u_h,v_h)=\ell (v_h)$ for every $v_h\in V_h$, using the same bilinear form and linear functional. Subtracting the two identities gives Galerkin orthogonality:This identity itself does not require symmetry. Its interpretation as an orthogonal projection in an energy norm does require a bounded symmetric coercive bilinear form: then $a$ is an inner product, $e=u-u_h$ is perpendicular to every available discrete correction, andConsequently $u_h$ is the unique closest point in $V_h$ in the energy norm. This Pythagorean theorem in a function space explains the beauty of the Galerkin method: integrated local equations compute a global best approximation. The Céa lemma supplies the corresponding estimate in a chosen reference norm.

For the finite element solution of the one-dimensional Poisson equation, $a(w,v)={\int }_0^1{w}^{\prime }{v}^{\prime }dx$ and $\Vert w{\Vert }_a=\Vert w^{\prime }{\Vert }_{L^2(0,1)}$. Testing the error against the $i$th piecewise-linear hat function on a uniform mesh givesIn this model $e$ vanishes at every node, so each of these integrals is individually zero, although $e$ is positive inside every element. Galerkin orthogonality controls the derivative-weighted balance; it does not say that the pointwise error vanishes.

The exact identity assumes conformity and exact evaluation of the continuous forms, including any numerical integration that is used. If assembly instead uses $a_h$ and ${\ell }_h$, subtraction gives the consistency defectThus inexact numerical integration need not preserve the projection. In the Petrov-Galerkin method the error is tested against a different test space, and for a nonsymmetric bilinear form the displayed identity need not describe perpendicularity in any energy norm.

If the observed samples are $y_i=f(x_i)+{\epsilon }_i$, the resulting quadrature rule changes by $\Delta Q={\sum }_i{w}_i{\epsilon }_i$. Bounded sample perturbations $|{\epsilon }_i|\le \delta$ therefore satisfyFor ordinary numerical integration with nonnegative quadrature weights exact on constant functions, the bound is $(b-a)\delta$, independent of panel refinement. A common bias ${\epsilon }_i=\delta$ attains it: the quadrature error due to the function approximation may vanish while this sample error persists. Interpolating a fixed data set onto a finer grid likewise does not create new observations or remove its uncertainty.

With new samples whose errors are independent random variables of zero expectation and common variance ${\sigma }^2$, a different model givesFor equal panels this is ${\sigma }^2(b-a{)}^2/n$ for the composite midpoint quadrature rule and ${\sigma }^2{h}^2(n-1/2)$ for the composite trapezoidal rule. These variance reductions require the stated independence and unbiasedness assumptions; a common bias or correlated sample errors need not shrink. The two displayed perturbation formulas follow directly from the triangle inequality and the variance additivity for independent random variables, respectively.

Numerical integration approximates a definite integral using finitely many sampled function values. For an integrand $f$ on a finite interval $[a,b]$, $a<b$, a quadrature rule has the formThe quadrature nodes $x_j$ tell us where to evaluate $f$; the quadrature weights $w_j$ turn those values into an approximation to the accumulated signed area. The exact definite integral depends on the integrand throughout the interval, whereas the quadrature rule sees only these samples. Integrating a chosen interpolant exactly still generally approximates the original integrand's integral.^[ref]

The fundamental theorem of calculus evaluates $I(f)$ as $F(b)-F(a)$ when an antiderivative $F$ is available. Numerical integration is useful when no convenient antiderivative is known, or each evaluation of the integrand requires an expensive computation. The task then becomes choosing informative quadrature nodes and controlling the quadrature error with as few evaluations as possible. Fixed data also permit a quadrature rule, but refinement cannot supply new information between observations.^[ref]

For an exact comparison, take the polynomial $f(x)=x^2$ on $[0,1]$. Its antiderivative $x^3/3$ gives $I(f)=1/3$. On a partition of an interval into $n\ge 1$ equal panels, the composite midpoint quadrature rule uses $n$ midpoint samples, each with quadrature weight $1/n$; the composite trapezoidal rule uses $n+1$ endpoint samples, with outer quadrature weights $1/(2n)$ and interior quadrature weights $1/n$. The existing derivations under these two quadrature rules giveIn particular, $M_1=1/4$ and $T_1=1/2$ already bracket $1/3$; two panels give $M_2=5/16$ and $T_2=3/8$. The convex function $x^2$ explains the opposite signs. A midpoint rectangle can cross the curve within its panel, so its total signed area, rather than pointwise ordering, determines the underestimate.

Define the signed approximation error as $E_n=Q_n(f)-I(f)$. In this example the absolute errors are $1/(12n^2)$ and $1/(6n^2)$: doubling $n$ reduces each by exactly four. A plot with both axes on a logarithmic scale therefore has slope $-2$, expressing order two with respect to panel refinement. This is an exact calculation for this polynomial, including $n=1$, rather than an observed rate extrapolated from a few computed answers.

For a general integrand, the composite midpoint quadrature rule and composite trapezoidal rule give second-order quadrature error bounds when $f$ has two continuous derivatives on $[a,b]$ and a known bound on $|f^{\prime \prime }|$. These hypotheses describe behavior between samples. A formal error formula containing an unknown derivative is not a numerical accuracy certificate or a justified sample count. Finite samples do not certify quadrature accuracy makes the missing information explicit. Errors in the evaluations and roundoff error must also be considered; noise sensitivity of a quadrature rule treats the former.

More elaborate choices exploit other information: Gaussian quadrature obtains high polynomial exactness from carefully chosen quadrature nodes, and Simpson's rule uses a quadratic interpolant. Adaptive numerical integration compares approximations and concentrates samples where they disagree, often saving evaluations. Such a difference is an error estimate, not a certified bound without further assumptions: both approximations can miss the same feature.^[ref]

The dual space $V^{\ast }$ consists of all real linear functionals on $V$. If $v_1,\dots ,v_d$ is a basis, its dual basis ${\epsilon }^1,\dots ,{\epsilon }^d$ is characterized by ${\epsilon }^i(v_j)={\delta }_{ij}$; explicitly ${\epsilon }^i$ extracts the $i$th coordinate. The natural map $J:V\to V^{\ast \ast }$ is $J(v)(\varphi )=\varphi (v)$. It is linear and injective, because the coordinate functionals separate nonzero vectors, and both spaces have dimension $d$. Hence it is an isomorphism. Its definition uses no choice of basis, which is the meaning of “naturally” here.

Evaluation $e_x(p)=p(x)$ is linear, so belongs to $V_n^{\ast }$. For distinct nodes $x_i$, define the Lagrange interpolation polynomialsThey lie in $V_n$ and obey $e_{x_j}({\ell }_i)={\delta }_{ij}$. Applying a putative relation $\sum a_i{e}_{x_i}=0$ to ${\ell }_j$ gives $a_j=0$, so the $n+1$ evaluations are independent and form a basis of $V_n^{\ast }$. The polynomials ${\ell }_i$ give their dual basis in $V_n$, using the canonical identification with $V_n^{\ast \ast }$.

Every $p\in V_n$ has the interpolation expansion $p(t)={\sum }_{i}p(x_i){\ell }_i(t)$: the difference has $n+1$ distinct roots but degree at most $n$. Integrate to get the numerical integration weights ${\lambda }_i={\int }_{-1}^1{\ell }_i(t)dt$. For the specified symmetric five nodes, symmetry gives weights $(a,b,c,b,a)$. Exactness on $1,t^2,t^4$ givesOdd moments vanish automatically. Solving yields Boole's rule

Use the prescribed-speed time parametrization of a curve. On a regular part traversed in increasing $u$, let $q(u)=\Vert F^{\prime }(u)\Vert >0$ and assume the prescribed speed $v(F(u))$ is positive. The chain rule for velocity givesIntegrate this scalar ordinary differential equation with an adaptive Runge-Kutta method, arranging output at the exact times $t_k=t_0+k\Delta t$. Evaluate $P_k=F(u(t_k))$. Internal integration steps may be shorter than $\Delta t$; output interpolation must meet the same error tolerance, so variable numerical steps do not create variable frame times.

An equivalent implementation precomputesby adaptive numerical integration. This function is strictly increasing. Bracket each $t_k$ in the table and invert using safeguarded Newton method or interval bisection; monotonicity makes each solution unique. Stop at $u=1$, with a shorter final time interval if the total travel time is not a multiple of $\Delta t$. This accounts for changes in both the curve's parameter speed and the physical speed, unlike equal increments in $u$.

If the specified speed vanishes, examine the reciprocal-speed integral at that event. It may diverge, giving asymptotic approach, or be integrable, giving arrival in finite time. A scalar speed does not determine whether motion reverses or how it resumes after stopping; such behavior needs the physical dynamics or a chosen continuation. The positive-speed formula applies between these events, with the direction changed if the traversal reverses.

Use the Bromwich inversion formula in the real variable $\omega$, with $p=\gamma +i\omega$:When the relevant Laplace transforms and spatial integral transforms are explicit, every sample of this integrand can be evaluated directly, without a time-stepping approximation to the partial differential equation. Split the initial condition integral at $y=x$: the two pieces of $R_q$ are linear combinations of $e^{\pm qy}$, so truncated spatial exponential transforms suffice. For real data the negative-frequency half is the complex conjugate of the positive-frequency half.

A practical numerical integration is to truncate to $|\omega |\le \Omega$, apply an adaptive quadrature rule, and independently increase $\Omega$ and refine the quadrature. Choose $\gamma >0$ large enough to remain to the right of all singularities, but avoid an unnecessarily large $e^{\gamma t}$. Evaluate hyperbolic function ratios in scaled form; for example,This prevents overflow when $\mathrm{Re}qL$ is large. The boundary contributions at an interior point are damped by $e^{-x\mathrm{Re}q}$ or $e^{-(L-x)\mathrm{Re}q}$, with $\mathrm{Re}q\sim \sqrt{|\omega |/2}$. Near an endpoint more frequencies are needed; at the endpoint itself use the prescribed Dirichlet boundary condition. The initial condition resolvent has a generally only $O(1/\omega )$ tail. Treat its oscillatory tail accurately, subtract a known transform with the same leading term, or evaluate that contribution separately with the heat kernel.

An equally useful check is the eigenfunction expansion. DefineThe causal modal formula isIt follows either from the heat kernel formula or directly from integration by parts against a sine eigenfunction. Exponential-transform formulas evaluate the time integral explicitly when available. The initial condition contribution has a Gaussian function cutoff in $n$ for each positive time. Nonzero endpoint forcing produces a more slowly convergent sine series tail. A boundary lifting $\ell (x,t)=(1-x/L)g_0(t)+(x/L)e^{aL}h(t)$ gives $w=\ell +z$, where $z$ has zero Dirichlet boundary conditions and forcing $-{\ell }_t-a^2\ell$; expanding $z$ gives better convergence and imposes the endpoint values exactly. At very small times the method of images is often more efficient than retaining many modes.

Check convergence by increasing both the frequency cutoff and quadrature resolution, and compare with a separately truncated causal modal or image-kernel evaluation. Causality in finite-time Laplace contour inversion means a leftward contour deformation must respect both resolvent operator poles and growth of the transformed data. In particular, the transform of data extended by zero after $T$ can grow like $e^{-pT}$ in the left half-plane: for $t<T$ one cannot close that contour indiscriminately and discard its large arc. The modal forcing integral with upper limit $t$ avoids this causality error.

Conditional independence givesFor each satellite, Bayes' theorem gives $P(d_s\mid m)\propto P(m\mid d_s)/P(m)$. ThereforeIf ${\hat{p}}_0(m)$ is a kernel density estimator for the simulated marginal masses and ${\hat{p}}_s(m)$ estimates the posterior based on satellite $s$, thenThe one-dimensional integrals can be evaluated by numerical integration on a common mass grid.

Use two plane waves, $|\mathbf{k}⟩$ and $|\mathbf{k}-\mathbf{G}⟩$, as a basis near their Bragg degeneracy. The potential has Fourier coefficients $V_{\pm \mathbf{G}}$; its zero Fourier coefficient vanishes. All other plane waves are off resonance and are neglected in the leading nearly-free electron model. With ${\epsilon }_{\mathbf{k}}={\hslash }^2{k}^2/(2m_e)$, the Schrödinger equation reduces to the Hermitian matrix problemSetting the characteristic polynomial equal to zero givesSolving this quadratic equation givesFor the real cosine potential its coefficient $V_G$ is real; the absolute-value form also allows an arbitrary choice of origin and corresponding Fourier phase.

At the Brillouin-zone boundary, $\mathbf{k}\cdot \mathbf{G}=G^2/2$, the free energies are equal. The two standing waves have different probability densities relative to the lattice potential and are split byRepeating this avoided crossing at the reciprocal lattice boundaries joins smooth branches into energy bands. In one dimension a nonzero Fourier coefficient opens a gap at the corresponding crossing. In higher dimensions a direct gap at one wavevector need not give a common forbidden energy interval at every wavevector: different bands may still overlap in energy. The present two-dimensional example remains metallic even though the crossings are locally gapped.

For the rectangular lattice, choose real-space primitive vectors $a\hat{\mathbf{x}}$ and $b\hat{\mathbf{y}}$. The reciprocal lattice vectors areDefine the half-widths of the first Brillouin zone by $K=\pi /a$ and $L=\pi /b$. Their values and the zone areA primitive reciprocal cell can be drawn with a corner at the origin; its centred translate is the first Brillouin zone here. Its area is $4{\pi }^2/(ab)=37.9600\mathrm{n}{\mathrm{m}}^{-2}$.

Including both spin states, counting the occupied free-electron states per unit area givesConsequentlyThe occupied disk has precisely the area of the first Brillouin zone, but their shapes differ: $K<k_F<L$. It protrudes through both vertical zone boundaries and does not reach the horizontal ones. Thus two electrons per cell do not imply a filled first band in the absence of a separating band gap.

Fold the protruding caps by $\pm {\mathbf{G}}_x$. The original circle and the translated circle intersect at $k_x=K$, $k_y=\pm \sqrt{k_F^2-K^2}$. The potential mixes these states and replaces each crossing by an avoided crossing. The lower branch reconnects into two open sheets running across successive zones in the $k_x$ direction; the upper branch forms a closed lens-shaped pocket about a vertical zone boundary. In a reduced-zone sketch its two halves sit at opposite edges, but periodic identification makes them one closed sheet. This is rectangular-lattice Fermi-surface reconstruction.

The largest pocket in this weak-potential construction is the lens-shaped intersection of two disks enclosed by the intersecting free circles. Each has radius $k_F$ and their centres are separated by $2K$. If $\theta =\arccos (K/k_F)$, two sectors minus their two triangles give the lens bound on a reconstructed Fermi pocket:Opening a gap raises the upper branch at fixed chemical potential and shrinks its pocket. The reported quantum oscillation area satisfiesIt is therefore consistent with this closed sheet. Onsager quantization gives $F=\hslash A_k/(2\pi e)\simeq 601\mathrm{T}$. An orbit on this closed contour can return to itself under the Lorentz force and occupy quantized Landau levels. The open sheets do not enclose a finite orbit area for a perpendicular field in this two-dimensional model. The observation is consistent with the reconstructed pocket, without uniquely determining a completely arbitrary band structure.

For a quantitative rough coupling estimate, use the supplied fixed-chemical-potential approximation and retain the two waves coupled by ${\mathbf{G}}_x$. WriteTheir upper eigenvalue isSetting it equal to $C{k}_F^2$ gives the pocket boundaryIts intersection with $k_y=0$ occurs at $q=\pm q_m$, whereThe two-wave estimate of a Bragg-gap coupling from a pocket area therefore usesAt $v=0$ this reproduces the lens bound on a reconstructed Fermi pocket; the pocket disappears at $v=C(k_F^2-K^2)=0.23786\mathrm{eV}$. Numerical integration gives $A(0.075\mathrm{eV})=6.22\mathrm{n}{\mathrm{m}}^{-2}$ and $A(0.100\mathrm{eV})=5.54\mathrm{n}{\mathrm{m}}^{-2}$. Matching the measured area gives $v\simeq 0.0931\mathrm{eV}$, so the justified rough result isThis estimate neglects remote plane-wave admixtures and other Fourier coefficients. It also holds the chemical potential fixed as requested; an exact fixed-density calculation would determine it self-consistently after reconstruction. The numerical root is a way to implement the approximate model, not a measurement of $V_G$ to four significant figures.

The pressure coefficient compares a local static pressure with a specified reference static pressure, using a fixed reference dynamic pressure:The reference mass density, speed, pressure and location must accompany a reported pressure coefficient. Keep the same $q_{\infty }$ throughout a surface distribution; replacing it by local $\rho U^2/2$ gives a different quantity and fails at a stagnation point. The definition itself requires neither Bernoulli's principle nor inviscid flow, so measured pressure coefficients remain useful when those idealizations fail.

Positive $C_p$ means $p>p_{\infty }$; zero means equality; negative $C_p$ means $p<p_{\infty }$. This sign does not specify the sign of absolute pressure. For stipulated $p_{\infty }=101325\mathrm{Pa}$ absolute and $q_{\infty }=240\mathrm{Pa}$, $C_p=-3$ gives $p=100605\mathrm{Pa}$ absolute: a $720\mathrm{Pa}$ deficit, not a negative absolute pressure. If $p_{\infty }$ equals atmospheric pressure, the numerator is gauge pressure; a wind tunnel's reference static pressure need not equal room atmospheric pressure. These are reference choices, not different physical pressures.

Under the same-height, constant-mass density $\rho ={\rho }_{\infty }$, steady flow and inviscid flow hypotheses of the Bernoulli equation, and with the same Bernoulli invariant as the reference flow,The first equation immediately gives the second after subtracting $p_{\infty }$ and dividing by $q_{\infty }$. Hence a lossless stagnation point has $C_p=1$, $U=U_{\infty }$ gives $C_p=0$, and $U=2U_{\infty }$ gives $C_p=-3$. For a real stationary wall satisfying the no-slip boundary condition, the velocity is zero right at the wall; inserting that wall velocity in an inviscid flow formula would wrongly predict $C_p=1$ everywhere. A thin boundary layer approximation instead relates the wall pressure to the outer-flow pressure and uses the outer velocity where the Bernoulli equation applies.

For potential flow around a circular cylinder of radius $a$ with zero circulation, the surface tangential velocity is $u_{\theta }=-2U_{\infty }\sin \theta$. Thus the ideal pressure coefficient isWith $\theta$ measured from downstream, the front is at $\pi$ and the rear at zero. Both ideal stagnation points have $C_p=1$, while the sides at $\pi /2$ and $3\pi /2$ have $C_p=-3$. This exact model is a useful baseline for an experiment; its complete rearward pressure recovery is not a prediction for a separated real fluid wake.

To obtain a force, integrate the local normal contributions with their directions. Let $\mathbf{n}$ be the unit normal vector from the body into the fluid and $\Sigma$ its closed wetted surface:The second equality removes a spatially constant reference pressure over the closed surface; it does not remove a nonuniform hydrostatic pressure or justify dropping the reference term on an isolated open patch. A negative $C_p$ contributes outwards in this reference-subtracted calculation, even though a positive absolute pressure still pushes inwards locally. Project ${\mathbf{F}}_p$ onto the chosen lift and drag directions. The full fluid force also includes ${\int }_{\Sigma }{\mathbf{\sigma }}_{\mathrm{visc}}\mathbf{n}dA$ from the viscous stress tensor, including tangential shear stress; pressure tapping does not measure that contribution.

For the two-dimensional cylinder, force per unit span follows from $dA=ad\theta$ per unit span and $\mathbf{n}=(\cos \theta ,\sin \theta )$:Substituting $C_p=1-4{\sin }^2\theta$ gives $(0,0)$: opposite points have identical pressure coefficients and opposite normal vectors. Zero drag despite a strongly nonuniform pressure is the instructive limitation associated with D'Alembert's paradox. Real viscosity, boundary layers and flow separation can prevent rearward pressure recovery and produce pressure drag. Thus the measured curve can test the model and estimate a pressure contribution without proving that total drag vanishes.

A motivating measurement is Virginia Tech's Experiment 3 - Flow Past a Circular Cylinder. Its open-jet wind tunnel holds a rotatable $140\mathrm{mm}$ cylinder with end plates and 36 one-millimetre pressure tappings at ten-degree intervals. A reference Pitot-static tube supplies $p_{0\infty }$ and $p_{\infty }$ to one digital manometer; a second connects a surface tap to its positive port and reference static pressure to its negative port. Record the simultaneous readings:The last equality uses the low-speed stopping relation, not an additional speed measurement. The manual requires sealed tubing and zero checks with the tunnel off. It also provides a water-column manometer: read each column's initial level, and subtract the independently measured reference-to-atmosphere pressure from the tap-to-atmosphere readings. Record tap positions after rotation. Compare the two instruments and check spanwise uniformity; end plates do not guarantee two-dimensional flow. ^[ref]

The inferential distinction matters: the manometers directly measure pressure differences, the ratio estimates pressure coefficients, and a surface integral estimates the pressure part of the force. Measure room absolute pressure and temperature if converting the reference difference to speed through the ideal gas law $\rho =p_{\mathrm{abs}}/(RT)$; use thermodynamic temperature in that equation. For force reconstruction, cover the complete contour and use numerical integration with the tap geometry. Finite tap spacing, unknown end effects, probe disturbance and fluctuations limit that reconstruction; tubes and instruments also average or delay changing pressures. Retain the signed raw readings, repeated readings and reference readings so that these limits can be assessed. A small reference difference magnifies any zero error in the ratio, rather than producing a reliably large $C_p$.

A quadrature node is a point at which the integrand is sampled by a quadrature rule. Under the affine map $x=(a+b)/2+(b-a)t/2$ from $[-1,1]$ to $[a,b]$, a reference quadrature node $t_i$ becomes $x_i=(a+b)/2+(b-a)t_i/2$; for ordinary numerical integration, the corresponding quadrature weight is multiplied by $(b-a)/2$.

A quadrature rule approximates an integral by a finite weighted sum of sampled function values:The quadrature nodes $x_i$ specify where to sample, and the quadrature weights $w_i$ specify how much each sample contributes. The density $\rho (x)$ specifies the target integral, whereas the discrete quadrature weights belong to the approximation. Ordinary numerical integration has $\rho =1$. Gaussian quadrature and the midpoint quadrature rule choose interior quadrature nodes; the trapezoidal rule includes endpoints.^[ref]

For ordinary numerical integration, exactness on the constant function $1$ requires ${\sum }_i{w}_i=b-a$; exactness on $x$ additionally requires ${\sum }_i{w}_i{x}_i=(b^2-a^2)/2$. These two checks make the quadrature rule exact for every affine function. For example, both the midpoint quadrature rule and trapezoidal rule give $6$ for ${\int }_0^2(2x+1)dx$. Testing progressively higher polynomial degrees defines the degree of precision of a quadrature rule. Exactness describes a class of integrands, rather than promising exact results for every function.

Composite midpoint quadrature rules and composite trapezoidal rules apply a simple quadrature rule on every panel of a partition of an interval. This buys accuracy by shrinking the panels while keeping the elementary formula fixed. For continuous functions sampled exactly, their quadrature errors tend to zero as the largest panel width tends to zero. The familiar second-order bounds require stronger assumptions about derivatives. The signed approximation error used here is $Q(f)-I_{\rho }(f)$; some sources define quadrature error with the opposite sign, so the convention must accompany any formula.

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.

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.