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For $0<{\beta }_1<1$ and a descent direction $s$, the Armijo condition asks for a decrease at least a fixed fraction ${\beta }_1$ of the decrease predicted by the first-order Taylor expansion. Put $d=\nabla f(x{)}^{T}s<0$. Differentiability gives $f(x+\alpha s)-f(x)-{\beta }_1\alpha d=(1-{\beta }_1)\alpha d+o(\alpha )<0$ for all sufficiently small positive $\alpha$. Thus the condition is satisfiable without a second derivative.

Asymptotic analysis describes what remains dominant as an argument approaches a specified limit. It compares functions or sequences by scale: how much of a complicated expression can be replaced by a simpler one, and how large is the omitted part? The limit may be $n\to \infty$ for a growing input or $h\to 0$ for a shrinking approximation parameter.

For example, $n^2+10^{5}n$ is eventually dominated by its quadratic term, since division by $n^2$ gives $1+10^5/n\to 1$. But at $n=100$, the linear term contributes $10^7$, compared with only $10^4$ from the quadratic term. A statement about the limiting scale and a good approximation at a particular input are different claims.

Three useful comparisons keep their strengths distinct. Big O notation supplies an eventual upper bound up to a constant multiplier; big theta notation supplies both an upper bound and a lower bound up to constants; asymptotic equivalence makes the ratio tend to one. Little o notation says that a term becomes negligible relative to a specified scale. These distinctions make approximations checkable instead of merely calling one expression small.

The same language connects time complexity with Taylor remainders. In an operation-count model, one asks how a cost scales with input size. In a Taylor expansion, one asks how the approximation error scales with the increment. Neither use requires interpreting the unspecified constant as one. The asymptotic comparison of exponential functions gives another example: dividing by the largest contributing exponential reveals the terms that disappear.

A peak has zero slope at its center, so a single peak-height reading does not tell feedback control which way to tune. Instead sample the two sides at ordinary frequencies ${\nu }_c\pm d$, where $d>0$ is a chosen modulation offset, and subtract:For an even signal around the true center ${\nu }_0$, write $ϵ={\nu }_c-{\nu }_0$. A Taylor expansion gives $E({\nu }_0+ϵ)=2S^{\prime }({\nu }_0+d)ϵ+O({ϵ}^3)$. On the descending side of a positive central peak $S^{\prime }({\nu }_0+d)<0$, so the update ${\nu }_c\mapsto {\nu }_c+gE$ has restoring sign for $g>0$ and is locally stable in this ideal discrete update when $0<-2g{S}^{\prime }({\nu }_0+d)<2$. A detector dip requires the opposite sign. These conditions follow from linearizing the update, not from assuming any particular electronics.

For the ideal signal $S=A{\cos }^2[\pi (\nu -{\nu }_0)T]+S_{\mathrm{bg}}$, choose $d=1/(4T)$. Then $E\simeq -2\pi ATϵ$. With $T=5\mathrm{ms}$ and $A=1$, a center error of $+1\mathrm{Hz}$ gives exact $E=-\sin (0.01\pi )\simeq -0.03141$. The sign directs the update toward lower frequency. Constant background cancels, but background drift between samples or asymmetric neighboring spectral lines can bias the zero. An initial scan matters because other fringes can also supply zeros.

The Bennett rate function has Taylor expansion $\varphi (u)=u^2/2-u^3/6+O(u^4)$ near zero. It is the optimized exponent in the Chernoff bound based on the moment-generating function estimate for the Bennett inequality.

A transformed-response regression estimates the conditional mean $m$ on the transformed scale. Inverting that mean does not generally estimate the original-scale mean, because nonlinear transformations do not commute with expected values. For a smooth inverse $g$ and centered transformed error of variance ${\sigma }^2$, a second-order Taylor expansion gives the displayed local correction. For $g(z)=z^q$ at positive $m$, it is $m^q+\frac{1}{2}q(q-1)m^{q-2}{\sigma }^2$. The variance is the response's residual variance, not the sampling variance of the estimated mean. This approximation requires errors small enough that the inverse is meaningful over the relevant range; it is not an exact moment formula for arbitrary transformed noise.

Bootstrap ratio bias with unequal cluster sizes arises because the pooled mean in a cluster bootstrap is a ratio with a random denominator. Let the empirical distribution over the $K$ observed groups assign mass $1/K$ to each pair $(n_j,T_j)$ of group size and response sum. One bootstrap replicate draws $K$ such pairs and computes ${\overline{T}}^{\ast }/{\overline{n}}^{\ast }$. Its conditional expectation need not equal $r=\overline{T}/\overline{n}$, even though the numerator and denominator averages are individually unbiased for their empirical means.

A second-order Taylor expansion of $f(t,n)=t/n$ explains the leading effect. Write $v_n$ and $c_{Tn}$ for the empirical variance of one group size and empirical covariance of one group size with its response sum, using denominator $K$. The conditional bootstrap bias is approximatelyThe second derivatives are $f_{nn}=2\overline{T}/{\overline{n}}^3$ and $f_{tn}=-1/{\overline{n}}^2$; the variances/covariance of the averages bring in $1/K$. This delta method approximation requires a denominator sufficiently concentrated away from zero and control of the remainder. It is not an exact identity for a few highly unequal groups.

For an original three-group example, take $(n,T)=(1,0),(2,2),(4,16)$. Then $r=18/7\approx 2.57143$, $\overline{n}=7/3$, $v_n=14/9$, and $c_{Tn}=26/3$. The second-order bias estimate is $-2/7\approx -0.28571$. Enumerating all $3^3=27$ ordered bootstrap draws instead gives mean about $2.24797$, hence exact conditional bias about $-0.32346$. The approximation identifies the direction and mechanism but misses a visible remainder at this small group count.

Equal group sizes remove the random denominator and this particular ratio bias. If every group has the same response mean, $T_j=r{n}_j$, every replicate ratio is exactly $r$ regardless of size. Neither case justifies replacing a row-weighted estimand with an equally weighted group mean in general; see unequal-cluster weighting in a cluster bootstrap. The primary Stata specification samples entire clusters and allows replicate row counts to vary.^[ref]

For bounded measurable $g$ with $Pg=0$, the formula $p_t=p(1+tg)$ defines a normalized positive probability density function for $|t|\Vert g{\Vert }_{\infty }<1$. A uniform Taylor expansion of the square root proves differentiability in quadratic mean with score function $g$: the squared remainder is $O(t^{4}P{g}^4)$. Such paths realize all bounded centered directions of an unrestricted density model.

Differentiate Bragg's law at fixed correctly assigned diffraction order:Divide the second equality by the first. For small signed changes,The Bragg angle error $\delta \theta$ must be in radians. An error quoted for detector deflection $2\theta$ must first be halved. The expression is a first-order Taylor expansion; an unknown diffraction order is a discrete assignment error, not an infinitesimal perturbation.

For error bounds, add magnitudes. For independent zero-mean random errors with standard deviations ${\sigma }_d,{\sigma }_{\theta }$, variance addition instead givesCorrelated errors require a covariance term. A shared calibration bias does not become independent merely because many line positions are measured.

With a chosen angular error $10$ arcseconds, $\delta \theta =4.84814\times 10^{-5}\mathrm{rad}$, the angular contribution is $8.3972\times 10^{-5}$ at $30^{\circ }$ and $1.7646\times 10^{-5}$ at $70^{\circ }$. Larger Bragg angles reduce this contribution because $\cot \theta$ decreases. For a nearby spectral doublet, the inverse relation is $\Delta \theta \simeq \tan \theta \Delta \lambda /\lambda$: a fixed fractional separation spreads farther apart angularly at large Bragg angle.

This does not permit unlimited improvement. The geometry requires $n\lambda \le 2d$; a desired wavelength might not have an accessible large-angle reflection with the chosen crystal. Higher diffraction orders can reach larger Bragg angles but their strengths depend on the crystal lattice structure factor. Finite crystal response, slit widths and source size remain relevant. An error in $d$, including uncorrected thermal expansion, remains even when the angular term is small. Very near $90^{\circ }$, the leading angular derivative vanishes and the second-order Taylor expansion matters.

A central finite difference uses sampled values symmetrically around an evaluation point. Common scaled derivative formulas areFor sufficiently smooth functions, they approximate $f^{\prime }(x)$ and $f^{\prime \prime }(x)$ with error $O(h^2)$. Taylor expansion explains the accuracy: symmetry cancels the unwanted even or odd powers. The leading errors are $h^2{f}^{(3)}(x)/6$ for $D_0$ and $h^2{f}^{(4)}(x)/12$ for $D_{xx}$ when the next derivatives permit these expansions.

At unit spacing and $x=0$, the second formula becomes $f(-1)-2f(0)+f(1)$, with finite difference coefficients $1,-2,1$. Symmetry is a local approximation property; it does not by itself establish stability of a time update or provide values beyond the boundary. Those require von Neumann stability analysis, an energy estimate, or an appropriate boundary closure of a difference scheme.

Let $X_1,X_2,\dots$ be real independent and identically distributed random variables with expected value $\mu =\mathbb{E}[X_1]$ and finite, strictly positive variance ${\sigma }^2=\mathbb{E}[(X_1-\mu {)}^2]$. Write $S_n={\sum }_{j=1}^n{X}_j$ and ${\overline{X}}_n=S_n/n$ for their sum and sample mean. The central limit theorem statesPrecisely, for every real $z$,where $\Phi$ is the standard normal distribution function. The conclusion is convergence in distribution to the standard normal distribution. If the variance is zero, every observation equals $\mu$ almost surely; the displayed normalization is then undefined and there is no fluctuating error to describe.

The striking feature is that the observations need not have a normal distribution. They may be binary, bounded, or strongly skewed. Their detailed probability distribution disappears from this limit except through the expected value and variance. The subtraction removes predictable growth: $\mathbb{E}[S_n]=n\mu$. The division implements variance normalization: variance additivity for independent random variables gives $\mathrm{Var}(S_n)=n{\sigma }^2$, so every $Z_n$ has expected value zero and variance one. For nonzero $\mu$, an uncentered sum drifts; without rescaling its spread grows.

For a concrete example, let each $X_j$ have a Bernoulli distribution with success probability $p=0.2$. Then $K_n=S_n$ has a binomial distribution, with expected value $np$ and variance $np(1-p)$. Its exact probability mass function and its normalized count areHere $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ is a binomial coefficient. Individual trials are conspicuously skewed, yet their centered, rescaled counts approach the same standard normal distribution as sums of fair trials.

At $n=100$, $K_n$ has expected value $20$ and standard deviation $4$. The exact binomial distribution gives $\mathbb{P}(16\le K_n\le 24)=0.740141\dots$. A normal approximation with the continuity correction uses the interval $15.5$ to $24.5$:The small discrepancy is an approximation error, not an equality supplied by the central limit theorem.

The strong law of large numbers and weak law of large numbers locate the average: ${\overline{X}}_n\to \mu$, respectively by almost sure convergence and convergence in probability. The central limit theorem magnifies that shrinking error and describes its shape. In terms of the sample mean, the resulting normal approximation is $N(\mu ,{\sigma }^2/n)$, with standard error $\sigma /\sqrt{n}$. This explains why four times as many independent observations halve the standard error. When $\sigma$ is known, ${\overline{X}}_n\pm 1.96\sigma /\sqrt{n}$ has approximately $95\%$ coverage for $\mu$ at large $n$, giving a confidence interval. This coverage refers to repeated samples, not a guarantee for an individual interval.

A useful application is the Monte Carlo method: estimate a success probability by the proportion of successful simulated trials. Nicholas Metropolis and S. Ulam's original paper, "The Monte Carlo Method" (1949), illustrates this with games of solitaire and estimation of a region's volume by randomly selected points, and discusses why estimates require probabilistic error control. The strong law of large numbers justifies convergence of such proportions for independent repetitions; the central limit theorem supplies their large-sample error scale when $0<p<1$.^[ref]

There are three useful limits to this picture:

A historical original-paper reference is J. W. Lindeberg's "Eine neue Herleitung des Exponentialgesetzes in der Wahrscheinlichkeitsrechnung" (1922), published in Mathematische Zeitschrift 15, pp. 211–225. The Lindeberg condition and Lindeberg-Feller central limit theorem below describe a version for sums whose summands need not share a probability distribution.^[ref]

A brief proof outline uses the characteristic function of $Y=(X_1-\mu )/\sigma$. Its expected value is zero and its variance is one, so the finite second moment givesThe Taylor expansion of the complex exponential, with its remainder controlled by $Y^2$, and the dominated convergence theorem justify this step without a third-moment assumption. Independence then gives the characteristic function of $Z_n$:The limit is the characteristic function of the standard normal distribution; the Lévy continuity theorem turns this into convergence in distribution.

Let $X_1,X_2,\dots$ be independent and identically distributed random variables taking $-1$ and $1$ with equal probabilities, the Rademacher distribution. The two opposite complex exponentials give ${\phi }_X(t)=(e^{it}+e^{-it})/2=\cos t$. Their expected value is zero and their variance is one. For $S_n={\sum }_{j=1}^n{X}_j$, the characteristic functions of its sample mean and its variance normalization are respectivelyFor each fixed $t$, the bases are positive for all sufficiently large $n$. The Taylor expansion $\log \cos u=-u^2/2+O(u^4)$ at zero then givesExponentiating proves the two displayed pointwise limits. The first limit is the characteristic function of the constant zero; the second is that of the standard normal distribution. The Lévy continuity theorem gives the corresponding convergence in distribution statements. Averaging has variance $1/n$, whereas dividing the sum by $\sqrt{n}$ preserves variance one. This sign calculation illustrates why the law of large numbers and the central limit theorem use different scales; it is not a proof for arbitrary probability distributions.

The charge-transfer efficiency is the fraction of a packet delivered into its intended next site or time slot in one transfer, under stated conditions. It is distinct from quantum efficiency, which concerns conversion of incident photons into collected charge carriers. Define the complementary inefficiency as $ϵ=1-{\eta }_{\mathrm{tr}}$.

In an original model with constant $0\le {\eta }_{\mathrm{tr}}\le 1$, let $N_j$ denote the mean number of original packet electrons still delivered on time after $j$ transfers. Assuming no reinjection into that packet,Repeated substitution proves the result; for different efficiencies the fraction is ${\prod }_{j=1}^n{\eta }_j$. For $ϵ\ll 1$, a Taylor expansion of $\ln (1-ϵ)$ gives ${\eta }_{\mathrm{tr}}^n\simeq e^{-nϵ}$, while the simpler $1-nϵ$ requires $nϵ\ll 1$.

Choose $N_{\mathrm{in}}=10000$ and $n=2000$. For ${\eta }_{\mathrm{tr}}=0.9999$, the expected on-time output is $8187.23$ electrons; for ${\eta }_{\mathrm{tr}}=0.99999$, it is $9801.99$. These are model means, so fractional electron counts do not describe an individual packet. Even a small per-transfer imperfection can imprint distance-dependent image errors.

If every original electron independently stays on time with probability $p={\eta }_{\mathrm{tr}}^n$, a fixed initial count gives a binomial distribution with variance $N_{\mathrm{in}}p(1-p)$. An initial Poisson distribution with mean $\mu$ instead gives an on-time Poisson distribution with mean and variance $\mu p$, by Poisson thinning. A deterministic attenuation formula alone does not specify noise. Real trapping can correlate packets and depends on timing, illumination and temperature; the independent constant-efficiency model then fails.

A common-origin pressure-tap position error shifts several inferred tap positions together relative to a fixed physical reference. It is a shared contribution to correlated position errors of pressure taps, not an independent placement error for every hole. Whether it affects a pressure-derived load depends on which references are held fixed.

For dimensionless chord coordinate $\xi$, suppose the uncertain displacement is the same scalar $e$ for every tap. A first-order Taylor expansion of the pressure coefficient gives $\delta C_{p,i}\simeq C_p^{\prime }({\xi }_i)e$. For a reconstruction $Q={\sum }_i{w}_i{C}_{p,i}$ with fixed quadrature weights,The position covariance matrix is $u_e^{2}11^T$, a rank-one matrix. Squaring each tap's contribution before adding would discard the shared covariances. The covariance terms belong in the general propagation of uncertainty.^[ref]

An original locally linear example takes eight equal weights $w_i=1/8$, equal local slopes $C_p^{\prime }=20$ and $u_e=0.002$. The reconstructed coefficient has standard uncertainty $u_Q=0.040$. Incorrect independence would give $0.040/\sqrt{8}\approx 0.0141$. If a two-dimensional load per span is $L^{\prime }=qcQ$ with stipulated dynamic pressure $q=500$ Pa and airfoil chord $c=0.20$ m, these become $4.0$ and $1.41$ N/m. The chosen slopes describe a sensitivity model, not measured airfoil data.

Opposite-signed sensitivities can cancel the shared contribution. Conversely, repeating coordinate readings without independently checking the misplaced datum cannot average away its fixed error. A proposed test measures an independent reference mark and deliberately translates the assumed origin in the reconstruction, comparing the observed load sensitivity with the derivative above. If moving coordinates also changes the integration weights, those additional derivatives must be included.

Pressure-tap datum cancellation is a different situation. When the tap, leading edge and trailing edge are all expressed through the same affine map $u=a+bx$, with $b>0$,The common origin and scale cancel exactly. Merely translating a shared coordinate system causes no physical tap misregistration. Affine pressure-tap coordinate calibration must therefore state whether the datum is shared or independently fixed before assigning a common-origin uncertainty.

Let ${\Gamma }_9$ have center weight $-10/3$, axial weights $2/3$ and diagonal weights $1/6$, and let $M_h$ have center weight $2/3$ and axial weights $1/12$. Taylor expansion gives $h^{-2}{\Gamma }_{9}u=\Delta u+h^2{\Delta }^{2}u/12+O(h^4)$ and $M_{h}u=u+h^2\Delta u/12+O(h^4)$. On a smooth solution of the constant-coefficient Helmholtz equation, the second-order term becomes $h^2\Delta (\Delta u+\lambda u)/12=0$. Thus the normalized PDE residual is fourth order, while the original unscaled stencil residual is sixth order. This local cancellation alone does not prove uniform global accuracy at or near a resonance.

For $f\in C^2[a,b]$, divide $[a,b]$ into cells of widths $h_i$ and midpoints $m_i$. The composite midpoint quadrature rule is $Q(f)={\sum }_i{h}_{i}f(m_i)$. Taylor's theorem around each midpoint givesThe odd linear term integrates to zero; integrating $\frac{1}{2}\Vert f^{\prime \prime }{\Vert }_{\infty }(x-m_i{)}^2$ gives $h_i^3/24$. Consequently, with $h={\max }_i{h}_i$,Thus each cell has a cubic quadrature error, while the composite error is second order in the largest cell width. For $f(x)=x^2$ the constant is attained, since the second-order Taylor expansion is exact. Endpoint singularities that make $f^{\prime \prime }$ unbounded require a separate estimate or a smoothing change of variables.

Let $f$ be a finite convex function on a neighbourhood of $x\in {\mathbb{R}}^d$, and suppose its first partial derivatives exist at $x$. For any fixed vector $h$, convexity givesThe equality uses the one-dimensional first-order Taylor expansion along each coordinate line; no continuity of second partial derivatives is assumed. Applying this upper bound to $-h$ and using $f(x)\le [f(x+th)+f(x-th)]/2$ gives the matching lower bound. Hence the displayed directional expansion holds. This supplies the expansion used in the supporting inequality for a differentiable convex function even when continuous second partial derivatives have not been postulated.

The cumulant expansion is the Taylor expansion of the logarithm of a moment-generating function:If the moment-generating function is finite on a neighbourhood of zero, the logarithm is analytic on a sufficiently small neighbourhood and this is a convergent local expansion. The first terms areA formal series of moments is not automatically a convergent representation of a moment-generating function. Finitely many moments can instead justify finite-order expansions of the characteristic function and its logarithm near zero; exponential integrability is a separate hypothesis.

For a centered normal distribution of variance ${\sigma }^2$, $K_X(t)={\sigma }^2{t}^2/2$, so all cumulants beyond order two vanish. For a Poisson distribution of mean $\lambda$, $K_X(t)=\lambda (e^t-1)$, so every cumulant equals $\lambda$. independent sums add their cumulant-generating functions, making cumulants additive.

For a fluctuating interaction $V$, expanding a Boltzmann weight gives $\log \mathbb{E}{e}^{-\beta gV}=-\beta g\mathbb{E}V+{\beta }^2{g}^2\mathrm{Var}(V)/2+\cdots$ whenever the exponential moment exists near $g=0$. The logarithm thus organizes the expansion into cumulants rather than raw moments.

Detector volume averaging is the spatial averaging of a nonuniform radiation field over a detector's finite sensitive region. A reading associated with the detector centre need not equal the local absorbed dose at that point. This limits spatial resolution even if the detector position and electronic readout are exact.

For an idealized linear response, let $D(\mathbf{x})$ be the absorbed dose field in the undisturbed medium and let $w(\mathbf{u})\ge 0$ be a normalized spatial weighting function, with $\int w(\mathbf{u})d\mathbf{u}=1$. Model the spatially averaged quantity byThe model separates averaging from the change in the radiation field caused by inserting the detector. A real ionization chamber can require both effects to be considered. The IAEA definition compares a point dose with the average over the detector's sensitive volume, both evaluated in water with the detector absent.^[ref]

For an original one-dimensional example, suppose the detector averages uniformly over a full width $a>0$. ThenAn ideal sharp edge $D(x)=D_{0}H(x)$, where $H$ is the Heaviside step function, is recorded asThus a $4\mathrm{mm}$ detector gives a $2.4\mathrm{mm}$ distance between the 20-percent and 80-percent response points, although the stipulated field edge has zero transition width. A $1\mathrm{mm}$ detector gives $0.6\mathrm{mm}$ in the same model.

For a smooth field, a Taylor expansion under the symmetric uniform weighting gives $\overline{D}(x)=D(x)+a^2{D}^{\prime \prime }(x)/24+O(a^4)$. A constant slope averages to its centre value; curvature produces the leading local discrepancy. A detector-volume-averaging correction factor therefore depends on the field shape and detector geometry, rather than only on the detector's nominal volume.

A proposed apparatus test scans a stable radiation field with detectors of different effective widths or orientations, keeping depth and irradiation geometry fixed. Changes in the recorded edge width test spatial response, while a uniform region checks relative sensitivity. The IAEA's geometric weighting examples show why a projected detector area may need nonuniform weighting; finer scan steps alone cannot remove averaging within each reading.^[ref]

Two Runge-Kutta methods share a stage matrix and nodes but use different output weights. An order-$p$ and order-$q$ pair with $q<p$ gives an output difference generally of order $h^{q+1}$, estimating the lower-order defect. Shared stage evaluations reduce the cost. The adaptive step-size controller exponent must reflect this estimator order, rather than blindly the higher reported solution order.

Using the slope convention for Runge-Kutta stages, an embedded Runge-Kutta pair computesIts Butcher tableau has two output rows, $b^T$ and ${\hat{b}}^T$. Evaluating $f$ is often the expensive part of solving an ordinary differential equation; after the shared slopes have been computed, the extra row needs only another linear combination.^[ref]

Write $\Phi$ for the exact endpoint from the current numerical state, and define signed local defects ${\ell }_q=\Phi -Y^{[q]}$ and ${\ell }_p=\Phi -Y^{[p]}$. Then $d={\ell }_q-{\ell }_p$. If ${\ell }_q=C_q{h}^{q+1}+O(h^{q+2})$ and ${\ell }_p=O(h^{p+1})$, with $C_q\ne 0$, the difference reproduces the leading lower-order defect. It does not reproduce the leading higher-order defect. Here the powers use big O notation and a smooth Taylor expansion; a vanishing leading coefficient or accidental cancellation can make the estimate misleading.

A common choice accepts $Y^{[p]}$ after testing $d$ with absolute and relative error tolerances. With the scale approximately unchanged on a retry, the adaptive step-size controller uses exponent $1/(q+1)$. Accepting a better endpoint does not change the power in the quantity being tested. The Euler-Heun embedded pair makes this visible without a large coefficient table.^[ref]

The Euler-Heun 1(2) pair combines the explicit Euler method and Heun's method. Here 1(2) names their respective order of a numerical method; the accepted endpoint in this article is the second-order one. Its shared Runge-Kutta stages and endpoints areThe first output row in this Butcher tableau is Heun's method; the second is the explicit Euler method. Both outputs cost two evaluations of $f$ together, rather than one for Euler method plus two for a separately computed Heun's method.

For the initial value problem $y^{\prime }=y$, $y(0)=1$, take $h=0.2$. The first slope is $k_1=1$; the stage state is $1+0.2k_1=1.2$, giving $k_2=1.2$. ThereforeThe exact exponential function gives signed local defects approximately $0.02140275816$ for the explicit Euler method and $0.00140275816$ for Heun's method. Their difference is exactly $0.02$. Thus the pair difference estimates the larger Euler defect; calling it the actual Heun defect would miss by a factor about fourteen here.

More generally, a step from $y_n$ on $y^{\prime }=y$ hasThese original Taylor expansions separate three quantities: the accepted Heun's method has local truncation error $O(h^3)$ and global error $O(h^2)$ under suitable numerical stability, while the tested local error estimator is $O(h^2)$. For $E=|d|/s$ with $s$ held approximately fixed, use $h_{\mathrm{new}}=\eta h{E}^{-1/2}$. A cube-root update would model a different quantity, the actual higher-order local defect.