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The Rao-Blackwell theorem says that an unbiased finite-variance estimator $\delta (X)$ of a parameter function $\tau (\theta )$ can be improved using a sufficient statistic $T$. DefineBecause the conditional distribution given $T$ is parameter-independent, this is a statistic computable without knowing $\theta$. The tower property of conditional expectation gives ${\mathbb{E}}_{\theta }{\delta }^{\ast }={\mathbb{E}}_{\theta }\delta =\tau (\theta )$, so it remains unbiased. The law of total variance givesHence the improved estimator has no larger variance or mean squared error:Equality holds precisely when $\delta ={\delta }^{\ast }(T)$ almost surely under the relevant parameter. More generally, even without unbiasedness, conditional Jensen inequality shows that conditioning on a sufficient statistic cannot increase risk for any convex loss in the estimate, provided the expectations exist. The squared-loss result above is the usual variance form of the theorem.

A statistic $T$ is a sufficient statistic for $\theta$ if the conditional distribution of the data given $T$ does not depend on $\theta$. For a dominated family with density or mass function $f_{\theta }(x)$, the Fisher-Neyman factorization theorem states that this is equivalent towhere $h$ is nonnegative and independent of $\theta$, and $g_{\theta }$ is nonnegative and depends on the data only through $T$.

For a discrete sample space, suppose first that the factorization holds. If ${\mathbb{P}}_{\theta }(T=t)>0$, summing over the fibre gives ${\mathbb{P}}_{\theta }(T=t)=g_{\theta }(t){\sum }_{T(x)=t}h(x)$, soThis conditional distribution is independent of $\theta$, proving that $T$ is a sufficient statistic. Conversely, let $q_t(x)$ be the common conditional mass function guaranteed by the sufficient statistic property, for every fibre attainable under at least one parameter. Set $h(x)=q_{T(x)}(x)$ and $g_{\theta }(t)={\mathbb{P}}_{\theta }(T=t)$. Then ${\mathbb{P}}_{\theta }(X=x)=g_{\theta }(T(x))h(x)$. Fibres never attained under any parameter can be assigned any conditional distribution, since their multiplying probability is always zero. This proves both directions of the discrete criterion, including possible parameter-dependent supports.

With ${\rho }_0=a$, ${\rho }_1=\theta$, ${\rho }_2=1$, the constant order condition gives $a=-1-\theta$. The next three order conditions for a linear multistep method giveSolving this linear system yields the unique coefficients of order at least three:The characteristic polynomial controlling zero-stability factors asThe other root is $-1-\theta$. Requiring its modulus to be at most one gives $-2\le \theta \le 0$, but $\theta =-2$ produces a repeated root at one and violates the root condition for a multistep method. At $\theta =0$, the roots $1,-1$ are both simple, so that endpoint is allowed. By the Dahlquist equivalence theorem, the method converges precisely forThe fourth defect coefficient is $C_4=\theta$, so the method has order exactly three for nonzero $\theta$. At $\theta =0$ it is the two-step Simpson method, of order four; thus the extra-order endpoint has correctly remained in the answer.

The Dahlquist equivalence theorem states that a linear multistep method has convergence of a numerical method if and only if it is consistent and zero-stable, under the standard assumptions for the initial value problem and convergent starting values. Consistency of a numerical method means order at least one, orThe root condition for a multistep method expresses zero-stability: every root of $\rho$ has modulus at most one, and every root with modulus one is simple. Usual hypotheses include a well-posed ordinary differential equation with $f$ continuous in time and Lipschitz continuous in the solution variable, a valid step equation, and starting approximations converging to the exact values.

Insert an exact solution into the linear multistep method and use $f(t,y(t))=y^{\prime }(t)$. Its local residual isExpanding both terms by Taylor theorem givesConsequently $C_0=\cdots =C_p=0$ is sufficient for order at least $p$. It is also necessary: test the residual on the polynomials $y(t)=1,t,\dots ,t^p$, evaluating at $t=0$. For $y(t)=t^k$, it equals $h^k{C}_k$ (and equals $C_0$ for the constant), which can be $O(h^{p+1})$ only if $C_k=0$. These functions can be realized as exact solutions of suitable smooth ordinary differential equations, so this is a genuine necessity test. Thus the order conditions areThis is a statement about local order; zero-stability is the separate condition that turns it into convergent approximation.

Convergence of a numerical method on a fixed interval $[0,T]$ means that, as the step size tends to zero and the starting approximations tend to the exact initial values,For a vector-valued ordinary differential equation, replace the absolute value by a norm. The interval remains fixed as $h\to 0$; a small error at only one step is insufficient.

The order of a numerical method is measured by its defect on a smooth exact solution. With the unscaled residual convention for a linear multistep method, order at least $p$ means local residual $O(h^{p+1})$; equivalently, the residual divided by $h$ is $O(h^p)$. Order exactly $p$ means the next coefficient does not vanish in general. For a zero-stable method, under the usual smoothness and Lipschitz continuity assumptions and with starting errors $O(h^p)$, this yields global error $O(h^p)$ on the fixed interval. Local order by itself does not imply convergence without stability.

Parametrize a streamline by $\mathbf{r}(s)$ with $d\mathbf{r}/ds=\mathbf{u}(\mathbf{r}(s))$. The chain rule givesTherefore $H$ is constant along every streamline. A different regular parametrization merely multiplies this derivative by a scalar and gives the same conclusion. For time-dependent flow, the claim concerns streamlines at a fixed time, rather than Lagrangian trajectories.

Let $\mathbf{\omega }=\nabla \times \mathbf{u}$ be the vorticity. In component notation, contracting two Levi-Civita symbols givesThe first term is ${\partial }_i(|\mathbf{u}{|}^2/2)$ and the second is the $i$th component of the convective acceleration. Rearranging proves the velocity identity

Differentiating the image charge potential givesAt the conducting plane, $E_x(d^{-})=\mu d/[\pi {\epsilon }_0(d^2+y^2)]$. With outward conductor normal $-{\mathbf{e}}_x$, the surface charge density is thereforeIt is independent of $z$ by translational symmetry. As a check, integrating across the plane gives ${\int }_{-\infty }^{\infty }\sigma (y,z)dy=-\mu$, the total induced charge per unit length, opposite to the real line charge.

The force on the real line charge is due to the induced conductor field, represented by the image charge; its own singular field must be omitted. The opposite image at distance $2d$ produces at the origin a field toward the conductor,Multiplying by the charge per unit length gives the attractive force per unit length:The attraction holds for either sign of $\mu$. The geometry assumes $d>0$, so the real line lies outside the conductor.

The method of images uses the real line charge at the origin and the opposite image charge at $(2d,0)$. LetSuperposing their logarithmic electrostatic potentials gives, in the physical region $x<d$,At $x=d$, $r_1=r_2$, so $\varphi =0$. Its tangential derivatives also vanish on that plane, giving $E_y=E_z=0$, exactly the electrostatic boundary conditions at a conductor. The image lies outside the physical domain, so it creates no extra physical source there. Together with the appropriate behaviour at infinity, uniqueness of Poisson equation identifies this field with the grounded-conductor solution.

For the conductor in electrostatic equilibrium, the interior electric field is zero and the electrostatic potential is constant. Earthing fixes that constant to zero. Thus the conductor boundary conditions at $x=d$ areThe normal electric field need not vanish; it is supported by a surface charge density. The outward normal from this conductor into the exterior is $\mathbf{n}=-{\mathbf{e}}_x$, so the normal jump condition isThis sign convention will determine the sign of the induced surface charge density below.

By cylindrical symmetry the electric field of an infinite line charge is radial, independent of $z$ and angle. Use a coaxial cylindrical Gaussian surface of radius $r$ and length $L$. The end caps have zero flux, so Gauss's law gives $2\pi rL{E}_r=\mu L/{\epsilon }_0$. Hence the field and electrostatic potential areHere $r_0>0$ is an arbitrary reference length, and $\mathbf{E}=-\nabla \varphi$. One cannot choose zero electrostatic potential at infinity for a single infinite line charge, because its logarithmic potential diverges there; the arbitrary reference handles the additive constant.

In electrostatics, the relevant Maxwell equations reduce to $\nabla \cdot \mathbf{E}=\rho /{\epsilon }_0$ and $\nabla \times \mathbf{E}=0$. For a volume $V$ with outward unit normal $\mathbf{n}$, integrate the first equation and apply the divergence theorem:This is the integral form of Gauss's law, with $Q_V$ the enclosed electric charge. The zero curl is the accompanying condition allowing an electrostatic potential locally, and globally when the domain permits one.

For a bound state, $0<E<U_0$: the Hamiltonian operator is nonnegative, and an energy at or above the exterior potential does not give a nonzero normalizable wavefunction. The time-independent Schrödinger equation and the hard wall at zero therefore giveContinuity of the wavefunction and its derivative at the finite step gives $k\cot (ka)=-\kappa$. Put $z=ka$ and $z_0=a\sqrt{2m{U}_0}/\hslash$. The matching equation becomesIt requires $\cot z<0$, so solutions lie in the intervals $((j+1/2)\pi ,(j+1)\pi )$ for $j=0,1,\dots$. Squaring, with that sign restriction retained, gives $z_0=z/|\sin z|$. On each such interval this function increases strictly from $(j+1/2)\pi$ to infinity: its derivative is $(|\sin z|-z\mathrm{sgn}(\sin z)\cos z)/{\sin }^{2}z>0$, because the second term in the numerator is positive there. Each interval therefore contributes exactly one bound state if and only if $z_0>(j+1/2)\pi$. These are the bound-state thresholds for a square well with one hard wall.

Exactly one normalizable bound state exists if and only ifThe printed strict upper inequality misses the threshold case. At the upper endpoint, the prospective second state has $E=U_0$ and $\kappa =0$. Its exterior solution is constant or linear, hence is not square-integrable unless zero; the zero exterior solution and matching would force the entire solution to vanish. Thus there is still exactly one bound state at that endpoint. The strict upper inequality is valid if threshold values are additionally excluded, but that exclusion is not part of the stated assumptions.