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The Taylor expansions about $t_n$ areAdding them cancels the odd powers of $h$ and givesThus the centred finite difference has truncation error $O(h^2)$, so $\boxed{\alpha =2}$.

Using the Taylor expansionwe obtainWhen cooling is negligible, the milk lowers the temperature by one unit at $t=1$. The hot-water term begins at $t=2$ and raises it at one unit per unit time, so one further unit of time is needed to erase the deficit.

Let $\rho =a/\sqrt{3}$ be the distance from the origin to each fixed charge, and denote their position vectors by ${\mathbf{R}}_j$. For a small planar displacement $\mathbf{r}$, the Taylor expansioncan be summed using the threefold symmetry relationsThe electric potential due to the fixed charges is consequentlyThe moving charge has electrostatic potential energy of point charges $U=q\Phi$. Because the quadratic coefficient is positive, the origin is a strict local minimum of $U$, and the equilibrium is stable.

Write $E=|\mathbf{E}|$ and $\hat{\mathbf{E}}=\mathbf{E}/E$. The constant electric force gives momentumCombining $\mathbf{p}=\gamma m\dot{\mathbf{x}}$ with the relativistic energy-momentum relation givesThe dimensional analysis from part (a) confirms that $qEt$ has the dimensions of momentum, so $qEt/(mc)$ is dimensionless and the expression has the dimensions of velocity.

The speed increases monotonically from zero and approaches the speed of light asymptotically:Integrating in the direction of the field givesAt small times, the Taylor expansion of the square root gives the nonrelativistic limitwhich is motion from rest with the constant acceleration $q\mathbf{E}/m$.

For unit mass, the Lagrangian in cylindrical coordinates isIntroduce shearing sheet coordinates by $r=r_0+x$ and $\phi ={\Omega }_{0}t+y/r_0$. Evaluate all derivatives of $\Phi$ at $(r_0,0)$. The reference circular orbit obeys ${\Phi }_r=r_0{\Omega }_0^2$, while midplane symmetry gives ${\Phi }_z={\Phi }_{rz}=0$. Its Taylor expansion isExpanding the kinetic energy to the same order givesThe terms linear in $x$ cancel by circular-orbit balance. Discard the constant and the term $r_0{\Omega }_0\dot{y}$ by total-time-derivative invariance of a Lagrangian; these do not change the Euler-Lagrange equations. The particle Lagrangian in a shearing sheet is thereforewith shearing-sheet tidal potentialThe second form follows from ${\Phi }_r=r{\Omega }^2$, the orbital shear parameter $q_0=-d\log \Omega /d\log r$ and the vertical epicyclic frequency ${\Omega }_z^2={\Phi }_{zz}$.

Expand $\psi =ϵ{\psi }_1+{ϵ}^2{\psi }_2+\cdots$ and $U=ϵU_1+{ϵ}^2{U}_2+\cdots$. A Taylor expansion of the sheet boundary conditions about $y=0$ givesThe first-order Fourier mode is ${\psi }_1=f(y)\sin \theta$, with $({\partial }_y^2-1{)}^{2}f=0$, $f(0)=1$, and $f^{\prime }(0)=0$. Decay excludes the growing exponentials, leaving $f=(a+by)e^{-y}$. The two boundary values give $a=b=1$:There is no first-order mean boundary motion. Boundedness excludes mean shear, so $\boxed{U_1=0}$.

The electric potential of a point charge $Q$ at the origin isFor the two charges forming the dipole,The first-order Taylor expansion at large $|\mathbf{x}|$ isand therefore, with the electric dipole moment $\mathbf{p}=Q\mathbf{d}$,to leading order. Taking the interaction of a second dipole with the corresponding electric field gives the stated electric dipole-dipole interaction

Let the lattice spacing be $d$ and write the central dipole asIts two horizontal neighbours have moment $p(\cos \theta ,-\sin \theta )$. For either one, the expression in parentheses, after extracting $d^{-3}$, isIts two vertical neighbours have moment $p(-\cos \theta ,\sin \theta )$, and each contributes insteadAdding all four nearest-neighbour interactions and using the Pythagorean trigonometric identity givesThe angle has cancelled, so the energy is independent of $\theta$.

Write $\ell (x,\theta )=\log f(x,\theta )$ and let ${\dot{\ell }}_{{\theta }_0}(x)$ be its score function. A second-order Taylor expansion around ${\theta }_0$ gives, for fixed $h$,where every ${\theta }_{i,n}$ lies between ${\theta }_0$ and ${\theta }_0+h/\sqrt{n}$. The mean-zero score identity, the Fisher information matrix, and the multivariate central limit theorem giveThe assumed regularity and a uniform law of large numbers giveConsequently, by the Slutsky theorem,

For the stated Gaussian shift model, direct expansion of the two normal distribution densities givesSince $I({\theta }_0)X\sim N(0,I({\theta }_0))$, this has exactly the limiting distribution above. Equivalently, both limits are normal with mean $-\frac{1}{2}{h}^{T}I({\theta }_0)h$ and variance $h^{T}I({\theta }_0)h$. This is the local asymptotic normality of the model at ${\theta }_0$.

Let the pair potential be $u(r)$, so that in the dilute limit only one interacting pair need be retained at a time. The leading Mayer f-function expansion isFor each pair, translation of its center coordinate supplies $V$ and the other $N-2$ particles supply $V^{N-2}$. Neglecting boundary corrections,The Taylor expansion $\log (1+x)=x+O(x^2)$ and $N(N-1)\simeq N^2$ in the thermodynamic limit giveUsing $P=-(\partial F/\partial V{)}_{T,N}$ and the ideal gas free energy,With number density $n=N/V$, this is $P/(nT)=1+B(T)n+O(n^2)$, so $B(T)$ is precisely the second virial coefficient.

This is the potential of Starobinsky inflation. Put $b=\sqrt{2/3}/M_{\mathrm{Pl}}$ and $y=e^{-b\varphi }$, so $V=V_0(1-y{)}^2$. Its potential slow-roll parameter and second potential slow-roll parameter areAt large positive $\varphi$, $y\ll 1$, so both ${ϵ}_V$ and $|{\eta }_V|$ are small and slow-roll inflation is possible. Near the minimum, a Taylor expansion gives $1-y\simeq b\varphi$ and henceThey are large for $|\varphi |\ll M_{\mathrm{Pl}}$, so that region cannot sustain slow roll.

Near the minimum, the Taylor expansion of the potential isNeglecting Hubble friction during one short oscillation reduces the Klein-Gordon equation to the harmonic oscillator equationThe virial theorem gives $⟨{\dot{\varphi }}^2/2⟩=⟨V⟩$, so $⟨P_{\varphi }⟩=0$. Restoring the slow cosmological damping in the averaged cosmological perfect-fluid continuity equation givesThe coherently oscillating inflaton therefore behaves as pressureless matter.

At $T_{\nu }/a\ll m_{\nu }$, the Taylor expansion of the relativistic energy is $\sqrt{p^2+m_{\nu }^2}=m_{\nu }+p^2/(2m_{\nu })+O(p^4/m_{\nu }^3)$. The number density isThe ratio of the next momentum moment to this one isThereforeso $\boxed{F=15\zeta (5)/(2\zeta (3))\simeq 6.47}$.

Put $x=kH$, with $H=c_s/\Omega$ and $Q=c_s\Omega /(\pi G\Sigma )$. The dispersion relation becomesThe two neutral roots areReal roots enclosing a range with ${\omega }^2<0$ exist precisely whenwhich is the Toomre stability criterion for axisymmetric instability. For $Q\ll 1$, the Taylor expansion of the square root gives

Along an exact autonomous-ODE solution, $f(y)=y^{\prime }$ andMove every term to the left and substitute the Taylor expansions about $t_n$. The coefficients of $h^j{y}^{(j)}(t_n)$ vanish for $0\le j\le 4$, while the first nonzero coefficient isThus the local defect is $O(h^5)$. At $h=0$ the first characteristic polynomial iswhich satisfies the root condition for a multistep method. The method is therefore zero-stable and has

Write the perturbation as $\eta e^{ikx+st}$ and use the same factor for all velocity and pressure amplitudes. At the upper interface, the linearized kinematic boundary condition, zero tangential traction, and normal-stress balance areThe first term in the normal stress is the linearization of the attractive disjoining pressure $V/h^3$, while the second is the stabilizing capillary pressure. Symmetry about $y=0$ makes $u$ even and $v$ odd, and supplies the lower-interface conditions.

For a two-dimensional Fourier mode, the Papkovich–Neuber representation is equivalently expressed by the odd biharmonic stream function for planar Stokes flowThe tangential-stress condition at $y=h_0$, with $K=k{h}_0$, giveswhereas the kinematic condition gives $iB\sinh K=s\eta$. The associated normal traction isEquating this with the linearized interfacial traction yields the dispersion relationThis is the Van der Waals rupture instability of a viscous sheet.

For $\Gamma =0$, the growth rate is positive, starts from $3V/(4\mu h_0^3)$ as $K\to 0$, and decreases to zero like $3V/(2\mu h_0^{3}K)$ as $K\to \infty$. For $\Gamma =1$, it has the same long-wave limit, vanishes at $K=1$, and is negative for $K>1$: surface tension damps wavelengths shorter than the cutoff. Long waves feel the attractive interaction but require coherent flow over a large distance; at short wavelengths viscous resistance suppresses the clean-film instability, while capillarity adds direct decay. A finite film size, fluid inertia, surrounding-fluid stresses, gravity, surface viscosity, and failure of the continuum disjoining pressure law can shift the observable most unstable wavelength.

During a growing thin spot, interfacial flow stretches the surface and dilutes its surfactant, thereby increasing the local surface tension above ${\gamma }_0$. Adjacent less-stretched regions retain more surfactant and lower tension. The resulting surface-tension gradient pulls toward the thin spot and opposes the outward flow that drives thinning. This is surfactant stabilization of film rupture; in the strong limit the surfaces behave almost as immobile boundaries.

For strong surfactant and $\Gamma \gg 1$, instability requires $K=O({\Gamma }^{-1/2})$. The Taylor expansionreduces the supplied relation toIt is maximal atThus the characteristic rupture time is $s_{\max }^{-1}=4\mu {\gamma }_0{h}_0^5/(3V^2)$. Surfactant greatly extends the life of a soap bubble while its film is moderately thick, but the $h_0^{-5}$ growth-rate dependence predicts rapid final rupture after drainage has made the film sufficiently thin.

Let ${\rho }_s$ be the grain density and $g^{\prime }=g/g_0$. Vertical force balance for the poroelastic aquifer giveswith compression taken negative in ${\sigma }_e=E({\varphi }_0-\varphi )/{\varphi }_0$. HenceThe zero-effective-stress condition $\varphi (h)={\varphi }_0$ then givesThis identifies the poroelastic compaction length.

The water volume per unit horizontal distance isFor $h\ll {\ell }_c$, a Taylor expansion yieldsStorage conservation is $V_t+q_x=R$. Integrating from river to divide and using $q(L,t)=0$ gives the river influxFor $g^{\prime }=1+ϵe^{i\omega t}$ and a groundwater profile that changes little during one tide, the direct storage oscillation has magnitudeWith typical thickness $H$ and mean recharge discharge ${\overline{Q}}_r=RL$,Using the steady balance $R\sim K{H}^2/L^2$ gives the equivalent estimate $|\delta Q_r|/{\overline{Q}}_r\sim ϵ\omega {\varphi }_0{L}^2/(2{\ell }_{c}K)$.

The ordinary least squares equations for model 2 areFor the Gamma generalized linear model, $V(\mu )={\mu }^2$ and $d\mu /d\eta =\mu$, so its score equation under the logarithmic link isWhen $T_i$ is close to ${\mu }_i$,by the first-order Taylor expansion of the exponential function. The Gamma score equations then become the model-2 normal equations, so their coefficient estimates are close.

Write $Y=B_{\varphi }^2$, $x=R^2/L^2$, and retain the frozen axial field $B_z=(B_0/{\rho }_0)\rho$. The radial magnetostatic equilibrium equation now includes both the gradient of the toroidal magnetic pressure and the inward magnetic tension:For $\rho ={\rho }_1{e}^{-x}$, cylindrical gravity givesMultiplication by the integrating factor $R^2$ therefore producesThe integration constant must vanish for regularity on the axis. Direct integration givesIts Taylor expansion at the axis isso the regular field behaves as $B_{\varphi }=O(R)$. At large radius,Nonnegativity at infinity requiresThis condition is also sufficient: at the limiting value, the braces divided by $K_{\mathrm{eff}}{L}^2$ reduce to $2e^{-x}[1-(1+x)e^{-x}]\ge 0$, and decreasing $L^2$ only increases them. Thus the toroidal magnetic field is real and regular at every radius exactly in the stated range.

The Taylor expansion of the square root isSubstitution into the exact reflection coefficient givesConsequentlyThis is exactly the reflected field furnished by both the Born approximation and the term linear in $\alpha$ in the Rytov approximation. Thus the exact, Born, and Rytov fields agree through first order. If the Rytov exponential is retained without re-expansion, its higher powers generate spatial harmonics $e^{i(px+(2m-1)q_{0}z)}$; those terms are part of the approximation and should not be confused with the exact interface's single reflected wave.

For a slowly varying envelope $E=\psi e^{-ikx}$, the paraxial approximation to the Helmholtz equation isBecause $n=1+\mu W$ and ${\mu }^2\ll 1$, passage through a sufficiently thin phase screen producesIts modulus is one at the screen exit. Beyond the screen, $n=1$, so the parabolic wave equation is $E_x=i{E}_{zz}/(2k)$. A Taylor expansion in propagation distance givesSincewe findIt follows thator, equivalently, $|E{|}^2=1-x{\varphi }^{\prime \prime }/k+O(x^2)$. Thus random phase curvature produces local focusing and defocusing: free-space diffraction converts phase fluctuations into amplitude fluctuations immediately after the screen.