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Suppose the moment-generating function $M_X$ is finite on an open interval around zero. Its cumulant-generating function $K_X=\log M_X$ satisfies $K_X(0)=0$, $K_X^{\prime }(0)=\mathbb{E}X=\mu$ and $K_X^{\prime \prime }(0)=\mathrm{Var}(X)={\sigma }^2$. Indeed,and $M_X(0)=1$. The Taylor expansion gives the displayed formula, with little o notation describing the remainder. Finiteness in a neighborhood justifies differentiation: polynomial factors times $e^{tX}$ are dominated on a smaller interval by constants times two slightly larger exponential moments.

Let $Z_i$ be independent Rademacher random variables and observe $Y_i=g(x_0+h{Z}_i)+{ϵ}_i$, where the noises are independent of the $Z_i$ and of one another, with zero mean and variance ${\sigma }^2$. Assume the baseline $g(x_0)$ is known exactly and $|g^{\prime \prime }|\le M$ on the sampled interval. The estimatoruses function values on both sides of $x_0$. A first-order Taylor expansion givesThe average remainder therefore has squared magnitude at most $M^2{h}^2/4$, while the independent zero-mean noise contribution has variance ${\sigma }^2/(N{h}^2)$. Their cross term vanishes after conditioning on the sample directions, provingThe noise assumptions and exact baseline belong to this bound. Correlated or biased errors, or an independently estimated baseline, require their own error calculation.

An ideal electric dipole at the origin with smooth electric dipole moment $\mathbf{p}(t)$ has Lorenz gauge electric potential and magnetic vector potential, for $r>0$,For a short proof, place electric charges $sq$ at $s\mathbf{p}/(2q)$, $s=\pm 1$, and let $q\to \infty$ with $\mathbf{p}$ fixed. The corresponding retarded times are $t_0+s\mathbf{n}\cdot \mathbf{p}(t_0)/(2qc)+O(q^{-2})$. Each Liénard–Wiechert potential has denominator $r-s(\mathbf{n}\cdot \mathbf{p}+r\mathbf{n}\cdot {\mathbf{p}}^{\prime }/c)/(2q)+O(q^{-2})$. The opposite $q/r$ terms cancel, leaving the displayed electric potential. The products of signed electric charge and velocity add, giving the displayed magnetic vector potential. Bounded time derivatives on the relevant interval justify the Taylor expansions.

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.

For the minor circular arc cut off by a chord of length $0\le L\le 2R$ on a circle of radius $R>0$, the sagitta is the maximum perpendicular distance from the chord to that arc:The perpendicular bisector of the chord passes through its midpoint and the center. The Pythagorean theorem gives the center-to-chord distance $\sqrt{R^2-(L/2{)}^2}$; subtracting it from $R$ proves the formula. The major circular arc instead gives $R+\sqrt{R^2-L^2/4}$, so choosing the arc matters.

For $L/R\ll 1$, a Taylor expansion gives $s=L^2/(8R)+O(L^4/R^3)$. In a uniform magnetic field perpendicular to a charged track's motion, $R=p/(|q|B)$, so the sagitta measures the momentum $p$.

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 positive $a,b$, fixed real $p,q$, and $f(a,b)=a^p{b}^q$, change the inputs to $a(1+h),b(1+j)$ with $h,j$ small. The ratio of the new value to the old one is exactly $(1+h{)}^p(1+j{)}^q$. Applying the binomial series to each factor and multiplying proves the displayed Taylor expansion. Positive inputs and $1+h,1+j>0$ make arbitrary real exponents well defined. A percentage change of $k$ in $a$ means $h=k/100$, not $h=k$. The mixed term $pqhj$ is required even though there is no such term in either one-variable expansion separately.

For prescribed one-dimensional velocity $u=u_0\sin (kx-\omega t)$, let $\alpha =k{X}_0$, $\theta =\alpha -\omega t$ and expand the Lagrangian trajectory as $x=X_0+u_{0}a+u_0^{2}b$. The Taylor expansion of $\dot{x}=u(x,t)$ gives $\dot{a}=\sin \theta$ and $\dot{b}=ka\cos \theta$. Initial position $X_0$ requires $a=(\cos \theta -\cos \alpha )/\omega$ andThe secular velocity $k{u}_0^2/(2\omega )$ is Stokes drift, even though the Eulerian time average vanishes. This fixed-time expansion requires $|k{u}_0/\omega |\ll 1$ and is not uniform once the secular displacement changes the phase by order unity.

For a twice differentiable functional, the second variation at $u$ in an admissible direction $\eta$ isIts contribution to the Taylor expansion is ${ϵ}^2{\delta }^{2}E[u;\eta ]/2$. At a stationary local minimum it must be nonnegative in every direction. A negative direction rules out a minimum, but mere directional strict positivity in an infinite-dimensional space is weaker than a uniform coercive bound and does not alone supply a sufficient-minimum theorem.

For zero-endpoint functions on $(0,L)$, considerAt $u=0$, the second variation is ${\int }_0^L(\eta \prime {}^2-\lambda {\eta }^2)dx$. The Wirtinger inequality makes this coercively positive when $0\le \lambda <{\pi }^2/L^2$. At equality, $\eta =\sin (\pi x/L)$ is a zero direction; above it, this same direction has negative second variation. Thus lower-order terms can destroy minimality despite a positive highest-derivative coefficient. The Legendre necessary condition, Jacobi equation and conjugate point organize these tests for general variational integrals.

A seismic travel-time fold caustic occurs where two branches of a seismic ray family merge in the mapping from seismic ray parameter $p$ to surface range $X$. At an ordinary fold,A local Taylor expansion gives $X-X_c\simeq C(p-p_c{)}^2$. On one side of $X_c$ there are two nearby rays; at the fold they merge, and beyond it that pair no longer exists. This is the local geometry underlying a boundary of seismic travel-time triplication.

In a horizontally layered model the travel times also merge, since $dT/dp=pdX/dp$ for the ray family. Geometrical spreading then makes a simple ray-amplitude estimate singular because the same ray bundle is compressed into a small range interval. Actual seismic waves have finite wavelength; their interference requires a wave approximation near the fold instead of interpreting the geometrical singularity as infinite measured motion.

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.

Consider a smooth delayed compartmental model with susceptible and infected populations $S,I$, incidence $rI(t-\tau )S(t-\tau )$, and removal rate $aI(t)$, where $r,a>0$. The zero-delay equations are $S^{\prime }=-rIS$ and $I^{\prime }=rIS-aI$. A Taylor expansion of the incidence givesSubstitute the zero-delay equations only in the first-order correction to obtainThe remainder is controlled on a compact time interval when the required second derivatives are bounded. The same approximated transfer enters the susceptible and infected equations with opposite signs, preserving their compartmental model transfer balance. Defining the zero-delay dynamics explicitly is essential: different population equations would give a different correction.

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 first-order Taylor expansion of an electrical circuit's response about an operating point. It describes small changes in electric potential difference and electric current; the neglected higher-order terms limit the useful signal range.

Space focusing in time-of-flight mass spectrometry reduces differences in arrival time caused by the initial positions of otherwise equivalent ions. An ion starting farther from the exit of an accelerating region spends longer there but can acquire greater kinetic energy, allowing it to catch up during the drift. Electrode geometry and electric fields are chosen so that these effects compensate at the detector.^[ref]

A simple idealized derivation makes the compensation explicit. Consider positive ions of mass $m$ and electric charge $q$, initially at rest, with distance $a$ remaining in a uniform accelerating electric field $E$. Let $L$ be the field-free distance beyond its exit. Their total flight time isThe first term is acceleration time; the second is drift time. Setting the derivative $dt/da$ to zero at a reference position $a_0$ givesFor example, $a_0=1\mathrm{cm}$ gives a first-order space focus at $L=2\mathrm{cm}$ in this simplified single-field model. Nearby position errors then affect arrival time only at higher order in the Taylor expansion. Practical two-field sources allow the focus to be placed at more useful drift distances. Initial velocity spread is a separate contribution to peak width, motivating time-lag focusing.

For a three-dimensional zero-temperature parabolic Fermi gas, let $\zeta =(N_{\uparrow }-N_{\downarrow })/N$ and $E_F={\hslash }^2(3{\pi }^{2}N/V{)}^{2/3}/(2m)$. Suppose the interaction energy per electron is $U{N}_{\uparrow }{N}_{\downarrow }/N^2=U(1-{\zeta }^2)/4$. The total energy per electron isIts Taylor expansion has quadratic coefficient $E_F/3-U/4$ and positive quartic coefficient $E_F/81$. Hence the unpolarized state loses stability at $U_c=4E_F/3\propto (N/V{)}^{2/3}$. This $U$ has units of energy. It differs from a contact coupling $g$ defined by interaction-energy density $g{n}_{\uparrow }{n}_{\downarrow }$, for which $g=U/n$ and $g_c=4E_F/(3n)$. Normalization must be fixed before comparing critical couplings. For $U\le 4E_F/3$, convexity of $x^{-1/3}$ implies $f^{\prime \prime }(\zeta )=(E_F/3)[(1+\zeta {)}^{-1/3}+(1-\zeta {)}^{-1/3}]-U/2\ge 0$, so the unpolarized minimum is a global minimum; a first-order polarized minimum does not pre-empt this threshold in the stated model.