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Vary the curve by $\delta x^c$, fixing endpoints. Integration by parts in the kinetic functional gives its Euler-Lagrange equationsMultiplying by $g^{cd}/2$ and symmetrizing the velocity product givesThere is no term proportional to $\dot{x}$ from parameter reparametrization, so this is the affine geodesic equation.

In units $c=1$, the static spherical vacuum metric is Schwarzschild with $f(r)=1-2GM/r$. For example, starting from $-e^{2\Phi }d{t}^2+e^{2\Lambda }d{r}^2+r^{2}d{\Omega }^2$, the vacuum radial equations give $[r(1-e^{-2\Lambda }){]}^{\prime }=0$ and ${\Phi }^{\prime }+{\Lambda }^{\prime }=0$. Asymptotic flatness and the Newtonian mass fix $e^{2\Phi }=e^{-2\Lambda }=1-2GM/r$, giving the displayed metric. Time and azimuthal coordinates are cyclic in its geodesic Lagrangian, soNormalize the velocity norm to $-k$, with $k=1$ for proper-time parametrized massive motion and $k=0$ for null motion. In the equatorial plane this gives ${\dot{r}}^2+f(k+h^2/r^2)=E^2$. For a nonradial orbit, $h\ne 0$ and $u=1/r$ imply $\dot{r}=-hdu/d\varphi$. ThereforeDifferentiating with $f=1-2GMu$ yieldsIt extends across isolated turning points by continuity. The original PDF's later $f=1-GMu$ is inconsistent with this claimed equation: that choice gives $u^{\prime \prime }+u=kGM/(2h^2)+(3GM/2)u^2$. The intended Schwarzschild factor is $1-2GMu$, which we retain.

For massive motion put $\ell =h^2/(GM)$. The Newtonian solution is $u_0=(1+e\cos \varphi )/\ell$. The resonant cosine forcing in $3GM{u}_0^2$ is $6GMe\cos \varphi /{\ell }^2$, producing the secular correction $3GMe\varphi \sin \varphi /{\ell }^2$. Resumming it into a shifted phase gives the precessing-ellipse approximationFor finite eccentricity there are also bounded corrections of this same small order: one particular correction is $3GM(1+e^2/2)/{\ell }^2-GM{e}^2\cos (2\varphi )/(2{\ell }^2)$, besides adjustable homogeneous terms. Thus the simple displayed ellipse captures the secular precession rather than the complete first-order shape correction. Consecutive periapses are separated by $2\pi /\alpha$, so the Schwarzschild perihelion precession per radial orbit is approximately $6\pi GM/\ell$, forward in the direction of motion.

Replace $\phi$ by $\phi +ϵ\eta$, with $\eta$ vanishing on the boundary, and differentiate the action at zero. Integrate the derivative of $\eta$ by parts:The coefficient of $\eta$ is the functional derivative $\delta S/\delta \phi$.

For the electromagnetic action, $\delta F_{ab}={\partial }_a\delta A_b-{\partial }_b\delta A_a$. Antisymmetry combines these terms and integration by parts givesThus the inhomogeneous Maxwell equations are ${\partial }_a{F}^{ab}={\mu }_0{J}^b$. The definition of $F$ also gives the homogeneous equations ${\partial }_{[a}{F}_{bc]}=0$, equivalently ${\partial }_a{\tilde{F}}^{ab}=0$. These are the two four-vector Maxwell systems.

Unrestricted variation of the alternative component-gradient action instead gives $\square A^b={\mu }_0{J}^b$. This is equivalent to the first Maxwell system only when supplemented by the Lorenz gauge ${\partial }_a{A}^a=0$, becauseThe alternative action is not by itself gauge-invariant; the additional condition must be supplied consistently, not silently omitted.

Expand the tensor square: $F^{ab}{F}_{ab}=2[{\partial }^b{A}^a{\partial }_b{A}_a-{\partial }^b{A}^a{\partial }_a{A}_b]$. HenceIntegrating this cross term by parts twice expresses its integral as a boundary term plus $\int ({\partial }_a{A}^a{)}^2{d}^{4}x$. Under Lorenz gauge it is therefore purely a boundary term. With fixed or vanishing boundary contributions, the two actions have the same variational equations once the additional gauge condition is included. This is the Lorenz-gauge reduction of the electromagnetic action.

A fixed-axis diatomic molecule has one bond-stretch vibrational coordinate $q$, reduced mass $m_r$, and angular frequency $\omega$. In the classical harmonic approximation its internal Hamiltonian is $p^2/(2m_r)+m_r{\omega }^2{q}^2/2$. The phase-space contribution per molecule isThe Gaussian integrals cancel the reduced mass. Extending the small bond displacement to the whole real line is the usual harmonic approximation. There is no rotational factor because the molecular orientation is fixed.

For $N$ identical dilute molecules of total mass $M$ in volume $V$, translation supplies $V(2\pi MkT/h^2{)}^{3/2}$. Including indistinguishability givesUsing Stirling's approximation and defining $Q=(V/N)(2\pi MkT/h^2{)}^{3/2}kT/(\hslash \omega )$,The $7/2$ is $1+5/2$: the logarithm contains $T^{5/2}$, combining three translational quadratic degrees with two vibrational ones. This is a classical result, valid when vibration is thermally classical rather than quantum frozen.

Label the two displacements in cell $n$ by $u_n,v_n$, with cell length $2a$. Choose the $K$ spring within a cell and the $G$ spring between cells. The equations areA Bloch wave $(u_n,v_n)=(u,v)e^{i(2aqn-\omega t)}$ gives a two-by-two eigenvalue problem, whose determinant isThus the alternating-spring chain dispersion relation isPeriodic boundaries require $e^{2iaqN}=1$, so $q=\pi j/(Na)$ modulo $\pi /a$. There are $N$ distinct wavenumbers in the Brillouin zone $-\pi /(2a)\le q<\pi /(2a)$ and two branches, accounting for $2N$ modes. The minus branch is acoustic, with zero frequency and in-phase motion at the centre; the plus branch is optical, with opposite cell displacements.

Expanding the square root near zero givesAt the zone boundary the square root is $K-G$, so the acoustic maximum is $\sqrt{2G/m}$ and the optical minimum is $\sqrt{2K/m}$. The frequency gap is thereforeQuantizing each harmonic mode gives phonons, bosonic excitations of energy $\hslash {\omega }_{\pm }(q)$ and crystal momentum $\hslash q$ defined modulo a reciprocal lattice vector. Their group velocity is $d\omega /dq$; acoustic phonons carry long-wavelength sound, while optical phonons have a nonzero centre frequency. There is one longitudinal polarization per branch here, and phonon number is not conserved in thermal equilibrium.

The Pauli matrices obey $(\mathbf{n}\cdot \mathbf{\sigma }{)}^2=I$. Expanding the exponential into even and odd powers gives the time-evolution operatorFor orthogonal states the identity contribution vanishes, so the transition probability is $|⟨{\chi }^{\prime }|\mathbf{n}\cdot \mathbf{\sigma }|\chi ⟩{|}^2{\sin }^2(\gamma Bt/2)$. For spin up to spin down in a static field $(B_x,0,B_z)$, only ${\sigma }_x$ has the required matrix element. Hence

For the weak rotating transverse field, take $H_0=-\hslash \gamma B_z{\sigma }_z/2$. The unperturbed energies give $E_{\downarrow }-E_{\uparrow }=\hslash \gamma B_z$. The perturbation matrix element is $⟨\downarrow |V(t)|\uparrow ⟩=-\hslash \gamma A{e}^{i\alpha t}/2$. Thus first-order perturbation theory givesSquaring its magnitude yieldsThe hypothesis is $|\gamma A|\ll |\Omega |$, excluding resonance. At $\alpha =0$ this is the leading weak-$A$ expansion of the exact static-field formula with $B_x=A$. For extremely long times, a small higher-order frequency correction can accumulate phase, so first-order perturbation is not a uniform approximation on arbitrarily long time scales.

Away from the turning points, the Liouville–Green approximation gives in the allowed regionThe bound-state condition discards the growing exponential in each forbidden region, leavingThese approximations require variation slow compared with the local wavelength, and must be replaced near $a,b$ by Airy approximations.

Near $a$, set $z=-{\lambda }^{2/3}{q}^{\prime }(a{)}^{1/3}(x-a)$. The local differential equation becomes ${\psi }_{zz}-z\psi =0$. Decay on the left selects Airy function $\mathrm{Ai}(z)$ rather than its growing companion. Comparing the supplied asymptotics across the turning point yieldsAt $b$, use $z={\lambda }^{2/3}[-q^{\prime }(b){]}^{1/3}(x-b)$; decay on the right similarly yields $2C_R{q}^{-1/4}\cos (S_b(x)-\pi /4)$, where $S_b=\lambda {\int }_x^b\sqrt{q}ds$. Let $J=S_a+S_b=\lambda {\int }_a^b\sqrt{q}ds$. The left expression has equal cosine and sine coefficients in the variable $S_a$; the right expression has those coefficients proportional to $\cos (J-\pi /4)$ and $\sin (J-\pi /4)$. They can describe the same nonzero wave only when $J-\pi /4=\pi /4+n\pi$. Therefore the WKB quantization condition isThe two simple turning points contribute the two quarter-phase shifts, giving the half-integer correction.

For a $C^2$ function, let $M(r)=(4\pi {)}^{-1}{\int }_{S^2}u(y+r\omega )d\omega$ be its spherical mean. Differentiation and the divergence theorem giveIf $u$ is harmonic, this derivative is zero; since $M(0)=u(y)$ by continuity, every spherical mean equals the value at the centre. Integrating $4\pi r^{2}M(r)$ over $0<r<R$ gives the same statement for the volume mean on a ball. This proves the mean value property for harmonic functions.

If $u$ is subharmonic, $\Delta u\ge 0$, so $M^{\prime }(r)\ge 0$. ThusThese are the corresponding mean value inequalities. A subharmonic function has the maximum principle on a ball: apply the second-derivative test to $u+ϵ|x-y{|}^2$, whose Laplacian is strictly positive and hence cannot have an interior maximum, then let $ϵ\downarrow 0$.

For $w=|\varphi {|}^2$, the given equation says $\Delta \varphi =iV\varphi$. Since $V$ is real,The subharmonic maximum principle gives $w(x)\le {\max }_{\partial B}w$ throughout the ball. Taking square roots yieldsNo sign condition on the real potential $V$ is needed.

The two-dimensional Brouwer fixed-point theorem states that every continuous map from a closed filled triangle to itself has a fixed point. Use the triangle $\Delta =\{x\in {\mathbb{R}}^3:x_i\ge 0,{\sum }_i{x}_i=1\}$, which lies in a two-dimensional affine plane. Strict positivity of the matrix entries ensures $(Ax{)}_i>0$ for every $x\in \Delta$. Thereforeis a continuous self-map of $\Delta$, in fact into its interior. Its fixed point $x$ satisfies $Ax=\lambda x$, where $\lambda ={\sum }_i(Ax{)}_i>0$. Since $x=T(x)$ lies in the interior, $A$ has an eigenvector all of whose entries are strictly positive. This is positive matrix eigenvector from simplex normalization; it uses strict positivity, not merely nonnegative entries.

Let $P=\mathbb{E}{\Delta }_n^2$ in equilibrium, with ${\Delta }_n=X_n-{\hat{X}}_n$. The innovation is $Z_{n+1}=Y_{n+1}-{\hat{X}}_n={\Delta }_n+{\eta }_{n+1}$. The given filter givesOrthogonality to the new innovation requires $(1-H)P-H=0$, hence $H=P/(P+1)$. Since the variables are jointly Gaussian, this orthogonality is independence. Stationary variance gives $P=(1-H{)}^{2}P+1+H^2=P/(P+1)+1$, soThe added process noise reflects the fact that the new observation concerns $X_n$, not $X_{n+1}$.

For the control calculation choose a quadratic value coefficient $R>0$. Conditional minimization of $x^2+u^2+R\mathbb{E}[(x+u+\epsilon {)}^2]$ gives $K=R/(1+R)$ and the Riccati equation $R=1+R-R^2/(1+R)$, hence $R^2=R+1$. Separation replaces $x$ by its conditional estimate, givingTo determine the average cost without solving an additional state covariance, complete the square:With $u=-K{\hat{X}}_n$, the final square is $K^2{\Delta }_n^2$. In equilibrium the expected telescoping value term is zero, so the minimal cost isThe identity also proves optimality: conditional mean minimization of the square uniquely selects the stated control. Here $1-K\in (0,1)$ ensures stable closed-loop estimates, making the equilibrium averaging valid. This is the Scalar delayed-observation LQG regulator.

In the Black-Scholes model, the stock follows geometric Brownian motion $dS=\mu Sdt+\sigma SdW$, with constant volatility, and a risk-free account earns constant rate $r$. Frictionless continuous self-financing trading and no arbitrage lead to risk-neutral drift $r$. With $\tau =T-t$, risk-neutral lognormality givesThis is the Black-Scholes digital option formula, obtained by discounting $P(S_T>K)$, not the stock-weighted probability of an ordinary call. Differentiation gives the Delta hedgewhere $\Phi ,\phi$ are the standard normal distribution and density. At fixed $s\ne K$, the density term decays faster than ${\tau }^{-1/2}$ grows, so delta tends to zero. At $s=K$, $d_2\to 0$ and $\Delta \sim [K\sigma \sqrt{2\pi \tau }{]}^{-1}$ diverges. Thus a narrowing, increasingly large delta spike forms around the strike. The terminal payoff is discontinuous: frequent large hedge adjustments near the strike, transaction costs, jumps, discrete trading, and small model or price errors make practical replication difficult. A fixed path avoiding the strike and the at-the-strike limit need not behave alike.

Assume a common support, a differentiable likelihood, and dominated differentiation under the integrals. The score function $S_{\theta }={\partial }_{\theta }\log f(X;\theta )$ has mean zero. Its second moment is the Fisher information $I(\theta )$. If $T$ is unbiased, differentiation of ${\mathbb{E}}_{\theta }T=\theta$ gives ${\mathbb{E}}_{\theta }(T{S}_{\theta })=1$, hence ${\mathrm{Cov}}_{\theta }(T,S_{\theta })=1$. The Cauchy-Schwarz inequality yieldsthe Cramér-Rao lower bound, assuming positive finite information.

Equality holds precisely when the centred estimator and score are linearly dependent almost surely. The covariance fixes the factor, giving $S_{\theta }=I(\theta )(T-\theta )$. If the same unbiased estimator is efficient for every parameter on a regular interval, integrate this identity in $\theta$:Exponentiation gives $f(x;\theta )=h(x)\exp (A(\theta )T(x)-B(\theta ))$, an exponential family. This proves efficient unbiased estimator implies exponential family under the stated common-support regularity, rather than asserting it for nonregular support-dependent models.