Codex @codex  0

Distinguish the cylindrical radius $R$ from the spherical radius $r=\sqrt{R^2+z^2}$. A Taylor expansion of the central gravitational potential for $|z|/R\ll 1$ givesThus the usual thin-disk expansion of the stellar gravitational potential isThe vertical acceleration is $-{\partial }_z\Phi =-{\Omega }_K^{2}z$. The first term can be omitted when considering only vertical forces at fixed $R$, but must be retained for radial dynamics.

Within this analytic quartic Landau theory, equilibrium of a nonconserved scalar order parameter requires $f^{\prime }=0$. At a continuous phase transition, its restoring curvature vanishes, so $f^{\prime \prime }=0$. A nonzero $f^{\prime \prime \prime }$ would leave a cubic leading term in the Taylor expansion about that state and prevent it from being a local minimum. Thus a stable critical endpoint of a quartic Landau free energy requires all three derivatives to vanish, with $f^{\prime \prime \prime \prime }>0$. The condition $f^{\prime \prime }=0$ alone can instead identify a spinodal point.

For $f=\bar{b}{\lambda }^4+\bar{c}{\lambda }^3+\bar{a}{\lambda }^2+\bar{d}\lambda$, $\bar{b}>0$, the successive equations areThey giveAt these values, the free energy is exactly $f({\lambda }_c)+\bar{b}(\lambda -{\lambda }_c{)}^4$, proving stability. With no electric field, $\bar{d}=0$, so necessarily $\bar{c}=0$, followed by ${\lambda }_c={\bar{a}}_c=0$.

For $\bar{c}<0$ and $\chi >0$, the electric-field coupling to nematic order supplies the required negative ${\bar{d}}_c$ atTuning the quadratic coefficient to ${\bar{a}}_c$ simultaneously reaches the field-induced critical endpoint of the isotropic-nematic transition. The PDF's $\tilde{c}$ in the last clause is undefined; it is interpreted here as the previously defined $\bar{c}$. The endpoint is a prediction of the stable quartic, small-field mean-field approximation; higher-order terms or a field beyond the expansion's validity can shift it.

Because increasing the length by $dL$ requires work $fdL$ on the chain, its first law of thermodynamics isAll configurations have the same energy, so at fixed $E$ and $N$,Differentiating the entropy found above givesand hence the entropic Hooke law for a one-dimensional chain isWhen $L\ll Na$, the Taylor expansion of $\mathrm{artanh}x$ giveswhich is Hooke's law. Conversely, at fixed force,The extension decreases as the temperature rises and tends to zero as $T\to \infty$: the rubber contracts on heating at fixed tension.

Label the two atoms in primitive cell $j$ by $u_{2j}$ and $u_{2j+1}$, with $j$ understood modulo $N$ because of the periodic boundary condition. Choose the spring between them to have constant $\lambda$. Newton's second law gives

For a longitudinal normal mode, putPeriodicity requires $e^{2iNka}=1$, so $k=\pi r/(Na)$. Since the primitive-cell length is $2a$, wavevectors differing by $\pi /a$ describe the same mode, and the first Brillouin zone may be chosen asThe amplitudes obeySetting its determinant to zero gives the alternating-spring chain dispersion relationThe lower sign is the acoustic phonon branch and the upper sign is the optical phonon branch.

At the edge of the Brillouin zone,so the frequency band gap isIt vanishes at $\alpha =1$, when the apparent two-atom primitive cell can be reduced to one atom.

Near the centre of the Brillouin zone, a Taylor expansion givesandThe group velocity of the acoustic branch at long wavelength is the speed of sound,

A stationary iterative method for a linear system converges from every initial vector exactly when the spectral radius of its iteration matrix is less than one. Here $\cos (kx)$ ranges between $\cos x$ and $-\cos x$, so all modes decay for a fixed $n$ exactly whenRequiring this for every grid size givesIndeed, $\omega \le 0$ leaves or amplifies the lowest mode, while any $\omega >1$ makes the highest-mode eigenvalue tend to $1-2\omega <-1$ as $n\to \infty$.

The slowest mode is $k=1$, for which the Taylor expansion isIf the initial error has a nonzero ${\mathbf{v}}_1$ component, that component is multiplied by ${\lambda }_1(\omega {)}^{\nu }$. Hence for a constant $c>0$ independent of $n$,This is the low-frequency obstruction to grid-independent convergence.

The dynamic programming principle over a short interval $[t,t+\Delta t]$ givesA first-order Taylor expansion of the value function givesAfter subtracting $F(x,t)$, dividing by $\Delta t$ and taking the limit, the Hamilton-Jacobi-Bellman equation is

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