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For a slowly moving particle in the weak-field approximation, $dt/d\tau =1+O(\Phi ,v^2)$ and terms containing two spatial velocities may be neglected. The spatial geodesic equation reduces toFor $g_{00}=-(1+2\Phi )$, the Levi-Civita connection isHence the Newtonian limit of general relativity is

The magnetization $M$ is magnetic dipole moment per unit volume. It produces bound volume and surface current densityStarting from the vector potential of magnetic dipoles and integrating by parts giveswhich has precisely the form generated by those bound currents.

Write the cylinder radius as $R$ and use cylindrical radius $\rho$. Ampère's law for the free axial current gives, inside the material,At $\rho =R$ this is tangential to the cylinder. Since $n=\hat{r}$,andFor $\mu >{\mu }_0$, the bound surface current therefore runs opposite to the free wire current.

The electric polarization $P$ is electric dipole moment per unit volume. A polarized body acts as the bound volume and surface charge densityIndeed, summing the scalar potential of its dipoles and integrating by parts givesThis is exactly the potential of ${\rho }_{\mathrm{b}}$ and ${\sigma }_{\mathrm{b}}$.

For the central free charge, Gauss's law for the electric displacement field givesThus on the sphere,

Expanding the trace in the spin basis givesso the transfer matrix for the one-dimensional Ising model reproduces the partition function.

In the ordered basis $s=+1,-1$,Its eigenvalues areSince ${\lambda }_{+}>|{\lambda }_{-}|$ at every positive temperature, the thermodynamic limit gives the Helmholtz free energy per spinThis is an analytic function of $T$ and $B$ for $T>0$, so the one-dimensional short-range Ising model has no finite-temperature phase transition.

Completing the square in the Hamilton-Jacobi-Bellman equation shows that the minimizing feedback isThis is a linear-quadratic optimal control problem, so set $F(x,t)=a(t)x^2$. The equation and terminal condition becomeThis Riccati equation is separable and givesConsequently the general optimal feedback law isWhen $c_t=c$ is constant,Along the corresponding optimal trajectory, $x_s=x(c+h-s)/(c+h-t)$, so the open-loop control is constant: $u_s^{\ast }=-x/(c+h-t)$ for $s\in [t,h]$.

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

Differentiating $f(\mu ,\sigma )=\mathbb{E}[U(\mu +\sigma Z)]$ twice gives the Hessian matrixFor every vector $(a,b)$,Thus the Hessian is a negative semidefinite matrix, andEquivalently, with $W=-U^{\prime \prime }(X)\ge 0$, its determinant isby the Cauchy-Schwarz inequality, while both diagonal entries are nonpositive.

Using the representation $X=\mu +\sigma Z$ and differentiation under the integral sign,because $U$ is strictly increasing. Similarly, part (a) givesbecause a twice differentiable concave function has $U^{\prime \prime }\le 0$. ThereforeIn mean-variance optimization, an expected-utility investor consequently prefers a larger mean at fixed standard deviation and a smaller variance at fixed mean. This is the usual risk-averse ordering of Gaussian payoffs.

Write $X=\mu +\sigma Z$, where $Z$ has the standard normal distribution. Apply Gaussian integration by parts to $g(z)=U^{\prime }(\mu +\sigma z)$:Since $X-\mu =\sigma Z$,The polynomial growth assumption and decay of the normal distribution density justify the integration by parts.

Substituting part (b) givesThe curvature bound gives $|A|\le hM/2$. Conditional on the $Z_i$ and the evaluation points, $B$ has mean zero, so the cross term has expectation zero. Independence of the normal random variables ${\epsilon }_i$ givesConsequently

The two terms are balanced byAt this value both terms equal $\sigma M/(2\sqrt{N})$, and hence

Ignoring the bounded second-order remainder, this is the linear regression modelThe ordinary least squares estimate through the known zero intercept isbecause $Z_i^2=1$. This is the randomized symmetric finite-difference derivative estimator.

The Taylor theorem with Lagrange remainder supplies a point ${\xi }_i$ between $x_0$ and $X_i=x_0+h{Z}_i$ such thatSince $Z_i^2=1$ and $Y_i=g(X_i)+{\epsilon }_i$,

DefineThe $Z_i$ are independent Rademacher random variables. Sendingto the machine therefore produces the required independent design points.

Because $X=\mathbb{N}$ is a countable set and $\mu (X)=1$, some $x\in X$ satisfies $\mu (\{x\})>0$. Invariance and the inclusion $\{T^{n}x\}\subseteq T^{-1}\{T^{n+1}x\}$ giveIf the forward orbit contained infinitely many distinct points, their disjoint singletons would each have measure at least $\mu (\{x\})>0$, contradicting that $\mu$ is a probability measure. The orbit therefore eventually enters a finite periodic orbit $Y$.

Now $T(Y)=Y$, so $Y\subseteq T^{-1}Y$. Since $\mu$ is invariant,Thus $Y$ belongs to the invariant sigma-algebra modulo null sets. It has positive measure, so the fact that $T$ is an ergodic transformation forcesThis is the ergodic invariant probability on a countable state space has finite cyclic support.

For every $x\in X$, invariance givesThus $T^{-1}\{x\}$ is nonempty, so $T$ is a surjective function. A surjection from a finite set to itself is a bijection. ThereforeThis is the finite full-support measure-preserving map is bijective principle.

Take the unit circle $X=\mathbb{R}/\mathbb{Z}$ with its Borel sigma-algebra and normalized Lebesgue measure, and letThis is the doubling map. Its two inverse branches halve length, so Lebesgue invariance of the doubling map gives $\mu (T^{-1}A)=\mu (A)$. Moreoverfor every point of the circle.

The Birkhoff ergodic theorem states that if $T$ is a measure-preserving transformation and $f\in L^1(\mu )$, thenwhere $\mathcal{I}$ is the invariant sigma-algebra. The limit has the same integral as $f$. If $T$ is an ergodic transformation, the limit is the constant ${\int }_{X}fd\mu$ almost everywhere.