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Let $h=\Delta x$. A one-dimensional Taylor expansion givesAdding the three coordinate directions shows that the exact solution has stencil defectThis is the local defect of the seven-point Dirichlet Laplacian. Its inverse has max-norm size $O(h^{-2})$: this follows from the discrete maximum principle, for example by comparison with the grid restriction of $x_1(1-x_1)+x_2(1-x_2)+x_3(1-x_3)$, whose stencil is $-6h^2$. Therefore the grid error is $O(h^2)$.

The question writes this error as $O(h^{p+1})$, soIn the more usual terminology, the finite difference method is second-order accurate.

For $|z|\ll r$, the gravitational potential has the Taylor expansionHence the vertical gravity is $-{\partial }_z\Phi \simeq -{\Omega }_K^{2}z$, where ${\Omega }_K^2=G{M}_{\star }/r^3$. Vertical hydrostatic equilibrium balances a pressure gradient of order $P/H$ against $\rho {\Omega }_K^{2}H$. Since $c_s^2\sim P/\rho$,The thin disk condition $H/r\ll 1$ is therefore equivalent to a highly supersonic orbital speed.

At latitude ${\theta }_0$, resolve the planetary angular velocity asFor velocity $(u,v,w)$, the Coriolis acceleration isThe traditional approximation drops the terms involving the horizontal rotation component $\Omega \cos {\theta }_0$. The retained horizontal force is thereforeIts direct velocity-scale requirement isFor an incompressible flow, $W/U=O(H/L)$, so a sufficient condition istogether with the small aspect ratio that makes hydrostatic pressure the leading vertical momentum balance. The approximation consequently becomes delicate near the equator.

The beta plane is the local Taylor expansionwhereand $a$ is the planetary radius. It requires a local Cartesian region,and $H/a\ll 1$. Away from the equator, treating the variation as a perturbation of an f-plane also requires $\beta L/|f_0|\ll 1$; near the equator one instead retains the linear term as the leading Coriolis parameter.

DefineIf $A$ and $B$ commute, the Helmholtz equation operator factorizes asWhen $n$ varies with $x$, the exact product also contains the commutator $[A,B]$; neglecting it assumes longitudinal changes are slow. The forward-propagating factor isbecause it admits $e^{i{k}_{0}x}$ in a uniform medium.

Put $\psi =E{e}^{i{k}_{0}x}$. The one-way equation becomesForthe Taylor expansion $(1+Q{)}^{1/2}=1+Q/2+O(Q^2)$ givesThus the parabolic wave equation is

The expansion is accurate under the paraxial approximation: transverse wavenumbers satisfy $|k_z|/k_0\ll 1$, the envelope varies slowly on the carrier scale, the refractive-index contrast is weak enough for $Q^2$ to be negligible, and $n$ varies slowly in $x$ so $[A,B]$ is small. The one-way factor discards backward propagation and reflection; the square-root expansion additionally discards large-angle and higher-order diffraction, and it does not accurately represent strongly evanescent components or abrupt longitudinal interfaces.

The maximum boundary slope is $s_{\max }=h_0{k}_T$, so subcritical internal-wave reflection requiresor equivalently

Take $k,m>0$ without loss of generality, putand write the incident field as the imaginary part of ${\tilde{w}}_i{e}^{i(kx+mz-\omega t)}$. The kinematic boundary condition on $z=h(x)$ isFor a flat boundary the reflected wave is $-{\tilde{w}}_i\sin (kx-mz-\omega t)$. Expanding the boundary condition in a Taylor expansion about $z=0$ creates the topographic sidebands of an internal gravity wave. With upward-radiating vertical wavenumber $-\mu |p_j|$, their complex amplitudes through second order areThus the general compact result isFor the convenient nondegenerate case $k>2k_T$, all displayed sideband wavenumbers are positive. Defining ${\Theta }_j=p_{j}x-\mu p_{j}z-\omega t$, the same answer is the explicitly real formulaValidity requires a linear incident wave, an inviscid uniformly stratified bulk, an outgoing-radiation condition, strict separation from critical slopes, and small boundary excursions for every retained mode, in particular $h_0{k}_T\ll 1$ and $\mu h_0{\max }_{|j|\le 2}|p_j|\ll 1$. A vanishing $p_j$ is a degenerate zero-horizontal-wavenumber case and must be treated by taking the corresponding zero-amplitude limit rather than dividing by $p_j$.

For a Poisson random variable $S_L$ of mean $L\lambda >0$ and $0<\beta <1$, the scaled cumulant-generating function is $\lambda L^{1-\beta }(e^{\theta L^{(\beta -1)/2}}-1-\theta L^{(\beta -1)/2})$. Its Taylor expansion tends to $\lambda {\theta }^2/2$ for every real $\theta$. The Gärtner–Ellis theorem applies because the limit is finite and differentiable everywhere. Its Legendre-Fenchel transform is the displayed quadratic good rate function.

With ${\mu }_i=e^{\beta x_i}$, define $U={\sum }_i{x}_i(Y_i-{\mu }_i)$ and $I={\sum }_i{x}_i^2{\mu }_i$. The estimating equation and a Taylor expansion give $\hat{\beta }-{\beta }_0\approx I({\beta }_0{)}^{-1}U({\beta }_0)$. If responses are independent with correct means and $\mathrm{Var}(Y_i)=\varphi {\mu }_i$, then $EU=0$ and $\mathrm{Var}U=\varphi I$. Consequently the first-order estimator variance is $\varphi /I$, even if the working Poisson distribution is wrong. Regular asymptotics require growing information and control of leverage and the Taylor remainder.

A smooth real rotation-invariant potential is constant on every positive-radius circle, so its logarithmic-radius function $u$ is smooth on $\mathbb{R}$. Off the origin, $f_{z\bar{z}}=e^{-t}{u}^{\prime \prime }(t)$. Since $f$ is smooth on the entire plane, this coefficient extends smoothly through the origin. A finite Taylor expansion, rather than a convergent series assumption, gives $f(z)=f(0)+b|z{|}^2+O(|z{|}^4)$.

A Runge-Kutta method advances an initial value problem $y^{\prime }=f(t,y)$, $y(t_0)=y_0$, by sampling its vector field at several trial states within one time step. These samples are Runge-Kutta stages; a weighted combination of their slopes produces the next state. The construction improves on the single tangent line used by the Euler method while requiring evaluations of $f$, rather than explicit higher derivatives of the solution. It applies to scalar equations and systems of ordinary differential equations alike. The coefficient choices distinguish methods such as Heun's method, the classical fourth-order Runge-Kutta method and implicit Runge-Kutta methods.

For $s$ stages, write $A=(a_{ij})$ for the stage matrix, $c_i$ for relative time nodes and $b_i$ for output weights. A time step of length $h$ computesHere $k_i$ is a slope, with units of state per time; $h{k}_i$ is a state increment. Another common convention writes $K_i=h{k}_i$, so its stage states are $Y_i=y_n+{\sum }_j{a}_{ij}{K}_j$ and its output is $y_n+{\sum }_i{b}_i{K}_i$. These are equivalent conventions: including $h$ both inside $K_i$ and again in the output would count the time step twice. The textbook formulation uses the increment convention.^[ref]

Usually the Butcher tableau obeys $c_i={\sum }_j{a}_{ij}$, called internal consistency. Stage $i$ then combines slopes over an effective time $c_{i}h$. The second-order conditions with independent Runge-Kutta abscissae explain what changes when time nodes are specified independently. A Runge-Kutta stage is generally an auxiliary approximation, not a separate accepted solution value at its node. Several stages can even have the same time node but different state arguments, as in the classical fourth-order Runge-Kutta method.

The useful idea is to approximate the integral $y(t_n+h)-y(t_n)={\int }_{t_n}^{t_n+h}f(t,y(t))dt$ without knowing the intermediate exact trajectory. Trial states provide places to sample $f$, and Butcher order conditions make the resulting combination agree with the exact Taylor expansion. More stages alone do not guarantee a higher order of a numerical method: their coefficients must satisfy the appropriate conditions. Embedded Runge-Kutta pairs reuse these same samples for a second output and a practical local error estimator.

Let $f$ have continuous second partial derivatives near an interior critical point $x_{\ast }$ of an open subset of ${\mathbb{R}}^n$, with $\nabla f(x_{\ast })=0$. The Hessian matrix gives the following sufficient conditions:

A positive semidefinite or negative semidefinite Hessian matrix with a zero eigenvalue does not settle the classification. The higher-order test for separated extrema gives two quartic examples with exactly the same zero Hessian. These conditions apply to an interior critical point; a boundary optimum or a constrained optimization problem requires attention to the allowed directions. ^[ref]

The proof outline is the Taylor expansionFor a positive-definite matrix, the smallest eigenvalue supplies a positive bound proportional to $\Vert h{\Vert }^2$, which dominates the remainder for small enough $h$. Negate the function for the local maximum case. For opposite-sign eigenvalues, choose the two corresponding eigenvectors: the leading terms have opposite signs along their lines.

There are also necessary conditions. At an interior local minimum, the Hessian matrix is positive semidefinite; at an interior local maximum, it is negative semidefinite. Restriction to each straight line reduces this to the one-variable second derivative condition. Since the Laplacian is the trace of the Hessian, it is nonnegative at such a local minimum and nonpositive at such a local maximum. The sign of the Laplacian alone is insufficient: $x^2-2y^2$ has negative Laplacian but a saddle point of a scalar function at the origin.

For fixed $\alpha \in [0,1)$, design $x_i=i/n$, $m\in C^2[0,1]$, a nonzero nonnegative symmetric piecewise $C^1$ regression kernel supported on $[-1,1]$, and $h\to 0$, $n{h}^2\to \infty$, write ${\mu }_{r,\alpha }={\int }_{-\alpha }^1{u}^{r}K(u)du$ and ${\beta }_r={\mu }_{r,\alpha }/{\mu }_{0,\alpha }$. On the rescaled grid of spacing $1/(nh)$, the Riemann sum error for $u^{r}K(u)$ is $O((nh{)}^{-1})$, by bounded variation. The normalized first and second local moments are therefore $h{\beta }_1+O(n^{-1})$ and $h^2{\beta }_2+O(h/n)$. A uniform second-order Taylor expansion gives the displayed bias of an estimator, since $n^{-1}=o(h^2)$. Equivalently it is $h{\beta }_1{m}^{\prime }(0)+h^2(\alpha {\beta }_1+{\beta }_2/2)m^{\prime \prime }(0)+o(h^2)$. The first-order term explains the generic boundary bias of a local constant estimator.

For a general nonautonomous Runge-Kutta method, Taylor expansion of stage $i$ gives $k_i=f+h[c_i{f}_t+(Ae{)}_i{f}_{y}f]+O(h^2)$. Matching the exact increment $hf+h^2(f_t+f_{y}f)/2$ gives the three displayed conditions. When internal consistency $c=Ae$ is imposed, the last two coincide. Internal consistency is a usual tableau convention, but is not logically necessary for order two of a generalized tableau.

For $L(f)=f^{\prime }(0)+\frac{3}{2}f(0)-2f(1)+\frac{1}{2}f(2)$ on $C^3[0,2]$, the Peano kernel isThe Taylor expansion with integral remainder gives $L(f)={\int }_0^{2}K(t)f^{(3)}(t)dt$, because $L$ annihilates polynomials of degree at most two. Since $K\ge 0$ and ${\int }_0^{2}K=1/3$, the sharp bound is $|L(f)|\le \frac{1}{3}\Vert f^{(3)}{\Vert }_{\infty }$. Equality holds for $f(x)=x^3/6$.

Simple harmonic motion is motion in which the acceleration is proportional to the displacement from equilibrium and points back towards it:Here $x$ is measured from equilibrium, and $\omega$ is the constant angular frequency. For a point mass attached to an ideal spring, Hooke's law gives the restoring force $F=-kx$, with stiffness $k>0$, and Newton's second law yields $\omega =\sqrt{k/m}$. The minus sign matters: for $x>0$ the force is negative, and for $x<0$ it is positive. A restoring force need not oppose velocity; on the return towards equilibrium it increases speed.

The general solution isThe amplitude $A\ge 0$ and oscillation phase $\phi$ encode the initial conditions. We use $P$ for the oscillation period to distinguish it from kinetic energy $T$. For $A>0$, the oscillation period is independent of amplitude, the property of isochronous harmonic motion. At $A=0$ the body remains at equilibrium. ^[ref]

Choose $m=0.5\mathrm{kg}$, $k=2N/m$, and release the mass from rest at displacement $A=0.3\mathrm{m}$. Then $\omega =2{\mathrm{s}}^{-1}$, $P=\pi \mathrm{s}$, and $\phi =0$. Writing time $t$ in seconds,At $t=0$ the velocity vanishes, but the acceleration is $-1.2m/{\mathrm{s}}^2$: a turning point is not a state of rest that persists. At $t=\pi /4\mathrm{s}$, the body crosses equilibrium with velocity $-0.6m/s$ and zero acceleration. At $t=\pi /2\mathrm{s}$ it reaches the opposite turning point, $x=-0.3\mathrm{m}$.

Taking potential energy zero at equilibrium, the same solution givesAt each turning point all mechanical energy is potential energy; at each equilibrium crossing it is kinetic energy. Both repeat twice during one complete position cycle because squaring removes the sign. At $t=\pi /8\mathrm{s}$, for example, each contributes $0.045\mathrm{J}$.

The physical position trace covers a line segment, $-0.3\le x\le 0.3$ in metres. In the phase plane, the paired position and velocity instead trace the ellipseThis is precisely $m{v}^2/2+k{x}^2/2=E$, rewritten: the single conserved mechanical energy fixes the amplitude and the whole ellipse. It does not fix the starting oscillation phase. At $x=0$ both $v=-0.6m/s$ and $v=+0.6m/s$ lie on it; position alone cannot distinguish them. Replacing $v$ by momentum $p=mv$ gives the equivalent phase-space trajectory. The arrows show increasing time, clockwise in $(x,v)$ coordinates.

This closed ellipse assumes a time-independent spring and no damping or applied drive. Adding a resisting force $-bv$, $b>0$, gives $m\ddot{x}+b\dot{x}+kx=0$ and $dE/dt=-b{v}^2\le 0$. An underdamped damped harmonic oscillator spirals inward in the phase plane, crossing successively smaller energy contours. The ideal harmonic oscillator is also useful beyond springs: near a smooth potential energy minimum with $V^{\prime \prime }(0)>0$, its Taylor expansion begins with $V(0)+V^{\prime \prime }(0)x^2/2$. For sufficiently small displacements, this gives the same restoring force to leading order, explaining why small oscillations share one solvable pattern.

Compare expected utility of $W(1+\varphi )$, where $\mathbb{E}\varphi =0$, with the certainty equivalent $W(1-\psi )$. A second-order Taylor expansion gives $U(W)+W^2{U}^{\prime \prime }(W)\mathbb{E}{\varphi }^2/2$ for the risky side and $U(W)-W{U}^{\prime }(W)\psi$ for the sure side. Equality gives the displayed proportional risk premium. This requires a genuinely small-shock family with controlled remainders; small variance alone does not control rare large wealth shocks.

A strong stability preserving Runge-Kutta method writes its stages as convex combinations of previously computed states and suitable Forward Euler method steps. It transfers any convex-functional nonincrease property of the forward step, such as a norm bound, under a proportionally scaled time-step restriction.

For example, if $E_k(U)=U+kF(U)$ is nonexpansive in a norm, then $Y=E_k(U)$ and $U_{\mathrm{new}}=(U+E_k(Y))/2$ is also nonexpansive: the triangle inequality bounds the new difference by one half of the initial difference plus one half of the twice-advanced difference. Both are at most the initial difference. A Taylor expansion yields $U_{\mathrm{new}}=U+kF(U)+k^2{F}^{\prime }(U)F(U)/2+O(k^3)$, so this is a second-order Runge-Kutta method. The forward-step hypothesis must hold on the stage states; preservation of one chosen convex bound is not the same as unconditional B-stability.

For a displaced wall $y=ϵh_1(x,t)$, write fluid velocity as $\mathbf{u}=ϵ{\mathbf{u}}_1+{ϵ}^2{\mathbf{u}}_2+\cdots$ and prescribed wall velocity as $ϵ{\mathbf{v}}_1+{ϵ}^2{\mathbf{v}}_2+\cdots$. A Taylor expansion about $y=0$ gives ${\mathbf{u}}_1(0)={\mathbf{v}}_1$ and ${\mathbf{u}}_2(0)={\mathbf{v}}_2-h_1{\partial }_y{\mathbf{u}}_1(0)$. The second-order correction converts first-order velocity gradients into a mean boundary flow.

Differential gravitational acceleration, multiplied by the affected mass, across a spatially extended body. In the linear approximation the acceleration difference is $T\Delta \mathbf{r}$, where $T$ is the tidal tensor. Stellar tidal forces limit the domain of a circumplanetary orbit; sufficiently strong ones can produce tidal disruption.

To isolate a tidal force, subtract the gravitational acceleration shared by the body as a whole. Let a reference test particle follow $\ddot{\mathbf{R}}=\mathbf{g}(\mathbf{R})$, and let another have position $\mathbf{R}+\mathbf{x}$. In a translating accelerated reference frame with parallel, nonrotating axes,for a source point mass, when $|\mathbf{x}|/R\ll 1$. Here $\mu =GM$ is its standard gravitational parameter and the last term uses Big O notation. This Taylor expansion removes the common inward pull and keeps the spatial variation. The near side has an excess pull toward the source; the far side has a deficit of inward pull, which is an outward relative acceleration. Both sides can stretch apart while both still accelerate toward the source.^[ref]

For a material element of mass $m$, the residual external force in this translating accelerated reference frame is $m[\mathbf{g}(\mathbf{R}+\mathbf{x})-\mathbf{g}(\mathbf{R})]$. Actual deformation also depends on elasticity, pressure and self-gravity. A rigid body approximation suppresses deformation but can still have internal forces that resist the tidal force. The point-mass tidal stretching and squeezing pattern describes freely falling, noninteracting test particles, rather than promising the same shape for every solid, star or ocean. A rotating reference frame introduces centrifugal acceleration and Coriolis acceleration, which must be included separately.

The dimensionless product of surface-wave amplitude and wavenumber, $ka$, measures orbital displacement relative to the wavelength. Small wave steepness permits a Taylor expansion of the velocity sampled by moving parcels. For a deep-water gravity wave, orbital amplitudes and the local steepness decrease exponentially with depth.

A real voltmeter has finite input electrical resistance $R_d$; a galvanometer also has a finite winding electrical resistance. Away from balance, its electric current changes both intermediate electric potentials. This is voltmeter loading when the detector is a voltmeter. Model the detector as a passive resistor, with no offset or injected electric current. Thévenin's theorem supplies the full response without pretending the connected bridge remains two independent voltage dividers.

With the detector removed, the equivalent source is $V_{\mathrm{oc}}$. Replace the ideal supply by a wire to calculate the source electrical resistance seen between $B,C$: nodes $A,D$ merge, making $R_1,R_2$ parallel and $R_3,R_x$ parallel. ConsequentlySince the attenuation factor is strictly positive for $R_d>0$, the null remains $R_2{R}_3=R_1{R}_x$. A smaller detector electrical resistance reduces the measured electric potential difference off balance but does not shift the ideal null. Its electric current sensitivity instead depends on $1/(R_{\mathrm{th}}+R_d)$; detector gain and resolution also matter in deciding how closely a null can be located.

In the plotted example, $R_x=1.100\mathrm{k\Omega }$ gives $V_{\mathrm{oc}}=-119.048\mathrm{mV}$ and $R_{\mathrm{th}}=1.02381\mathrm{k\Omega }$. A detector with $R_d=1.000\mathrm{k\Omega }$ instead reads $V_d=-58.824\mathrm{mV}$. Using that loaded reading in the unloaded voltage divider formula would underestimate the deviation of $R_x$ from $1\mathrm{k\Omega }$. Balancing the bridge avoids this particular systematic measurement error.

The local derivative makes the sensitivity explicit. For $R_1=R_2=R_3=R$ and $R_x=R+\delta R$, the first-order Taylor expansion isAt the plotted null, the open-circuit slope is $-1.25mV/\Omega$, compared with $-0.625mV/\Omega$ for $R_d=1\mathrm{k\Omega }$. Increasing the supply improves this ideal sensitivity but also increases Joule heating. A finite source electrical resistance requires recomputing $R_{\mathrm{th}}$; the displayed parallel-pair expression assumes an ideal supply.