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For fixed $\theta >0$, the central limit theorem givesThe weak law of large numbers gives $\overline{X}\to \theta$, so Slutsky theorem permits replacing $\theta$ in the standard error by $\overline{X}$. With $z=z_{0.025}\approx 1.96$, an approximate 95% confidence interval isThe lower endpoint may be truncated at zero because the parameter is nonnegative. This is a large-sample normal approximation for fixed positive $\theta$; an all-zero or very sparse sample does not justify the plug-in normal approximation. At the degenerate parameter value zero, every observation is zero almost surely.

Write $S={\sum }_i{X}_i$. The likelihood function for the Poisson distribution is proportional to $e^{-n\theta }{\theta }^S$. For $S>0$, differentiation of the log-likelihood gives $-n+S/\theta =0$, and its second derivative is negative. For $S=0$, the likelihood decreases with $\theta$ and has its maximum at the boundary zero. HenceThe estimator is unbiased, since $\mathbb{E}\hat{\theta }=n^{-1}{\sum }_i\mathbb{E}{X}_i=\theta$. Its variance is $\theta /n$. If the parameter space excludes zero, an all-zero sample gives a supremum as $\theta \downarrow 0$ rather than an attained maximum.

The Peano kernel theorem says that if a continuous linear error functional $L$ annihilates all polynomials of degree less than $r$ and permits the required interchange with integration, then for $f\in C^r[a,b]$,It follows by applying $L$ to the integral-remainder Taylor formula with integral remainder. Here choose $r=2$ andThis functional annihilates constants and linear functions (indeed quadratics too), and its Peano kernel isThus ${\int }_0^3|K|d\theta =1/2$ and $\max |K|=1/3$. The integral representation givesFor the first inequality, continuous functions $f^{\prime \prime }$ with $\Vert f^{\prime \prime }{\Vert }_{\infty }\le 1$ can approximate $\mathrm{sgn}K$ away from its zeros, making $L(f)$ arbitrarily close to $\int |K|$. Such derivatives integrate twice to admissible $C^2$ functions, proving no smaller $\alpha$ works. For the second, use continuous $f^{\prime \prime }$ of $L^1$ norm one concentrated near $\theta =1$ or $2$, where $|K|=1/3$, with the sign of $K$ there. This makes $|L(f)|$ approach $1/3$, proving optimality of $\beta$.

For $f=x^3$, $L(f)=2$, $\max |f^{\prime \prime }|=18$ and $\Vert f^{\prime \prime }{\Vert }_1=27$. Both right-hand bounds equal nine, so both inequalities hold.

Evaluating the Lagrange interpolation polynomial coefficients at five, and their derivatives at one, givesBoth rules are exact for every polynomial of degree at most two. For $f(x)=x^3$, the first gives $95$, while the true value is $125$; the second gives $4$, while the true derivative is $3$. Thus the respective approximation errors, estimate minus true value, are $-30$ and $1$. There is no contradiction: a cubic lies outside the exactness class. The first formula extrapolates beyond the interpolation nodes, so its error can be substantial.

The Lagrange interpolation polynomial basis for nodes $0,2,3$ givesEach polynomial is one at its own node and zero at the other two. For a polynomial $f$ of degree at most two, the difference between $f$ and this interpolation polynomial has degree at most two and three distinct roots, so it is identically zero. This also proves uniqueness of the quadratic coefficients.

Put $\kappa =|k|>0$. Profiles satisfying the rigid-wall conditions areThe kinematic conditions at zero giveConsequently ${\phi }_{1t}(0)$ has amplitude ${\omega }^2{\eta }_0\coth (\kappa h_1)/\kappa$, whereas ${\phi }_{2t}(0)$ has amplitude $-{\omega }^2{\eta }_0\coth (\kappa h_2)/\kappa$. Substituting into the dynamic condition yields the dispersion relation for interfacial gravity waves between rigid boundaries:It is positive because the heavier fluid is below. As $k\to 0$, ${\omega }^2\sim g({\rho }_2-{\rho }_1)k^2/({\rho }_1/h_1+{\rho }_2/h_2)$, while in the deep-layer limit it becomes $g({\rho }_2-{\rho }_1)|k|/({\rho }_1+{\rho }_2)$. The spatially uniform mode has zero restoring frequency.

Linearize about the flat interface at rest, $\eta =0$, ${\phi }_j=0$. Both potentials still satisfy Laplace equation, now on the fixed strips $0<y<h_1$ and $-h_2<y<0$, with the same rigid-wall conditions. Discarding products of perturbations and evaluating the interface conditions at $y=0$ givesFor a normal mode proportional to $e^{i(kx-\omega t)}$, the vertical profiles satisfy ${\Phi }_j^{\prime \prime }-k^2{\Phi }_j=0$, the interface velocity is $-i\omega \eta$, and the wall-normal derivatives vanish. Smallness of the amplitude relative to wavelength and layer depths justifies this linearization.

Use the standard ideal-fluid assumptions of incompressible flow, inviscid flow, gravity $-g{\mathbf{e}}_y$, and no surface tension. Irrotational flow gives ${\mathbf{u}}_j=\nabla {\phi }_j$ in each layer, and incompressibility gives Laplace equationThe rigid-wall boundary conditions are ${\phi }_{1y}=0$ at $y=h_1$ and ${\phi }_{2y}=0$ at $y=-h_2$. At the moving interface, each fluid has the same normal velocity as the interface:The dynamic condition is continuity of pressure, $p_1=p_2$ at $y=\eta$. Bernoulli equation in each fluid reads ${\phi }_{jt}+\frac{1}{2}|\nabla {\phi }_j{|}^2+p_j/{\rho }_j+gy=C_j(t)$. Choosing the potential gauges to subtract a common reference pressure, the interface condition becomesTangential velocities need not agree across an inviscid interface.

For a moving closed circuit $C(t)$ bounding an oriented surface $S(t)$, Faraday's law isHere $\mathbf{E}$ is the electric field, $\mathbf{B}$ the magnetic field, $\mathbf{v}$ the local velocity of a circuit element, and ${\mathcal{E}}_{\mathrm{em}}$ the electromagnetic electromotive force per unit charge around the circuit. The line orientation and normal $\mathbf{n}$ obey the right-hand rule. The flux derivative is a total derivative and includes both changing field and changing surface. The term $\mathbf{v}\times \mathbf{B}$ is the motional electromotive force. A battery contributes an additional non-electromagnetic source electromotive force.

The agreement is in the energy levels, not in electron trajectories. The Bohr model assumes a particle on a circular orbit, with orbital angular momentum quantized as $N\hslash$. The Coulomb bound states here are spatially distributed wavefunctions with no definite classical orbit. Their probability density is spherically symmetric, and their orbital angular momentum is zero because $\ell =0$. Their radial distributions also depend on $N$ and can have nodes. A real radial wavefunction times a global stationary phase has zero probability current, unlike the circulating point charge in the Bohr picture.

Use the physical Coulomb bound state boundary condition: the spherically symmetric wavefunction is regular at the origin and square integrable with measure $4\pi r^{2}dr$. For $E<0$, setAt large $r$, the decaying branch behaves exponentially as $e^{-Kr}$. Substituting $\psi =f{e}^{-Kr}$ into the radial Schrödinger equation givesThe regular solution has a power series $f(r)={\sum }_{j\ge 0}{a}_j{r}^j$, with recurrenceA nonzero regular solution has $a_0\ne 0$. If the series does not terminate, its coefficients eventually have one sign after choosing an overall real phase. For any fixed $0<\delta <1/2$ and sufficiently large $j$, the ratio of consecutive magnitudes is at least $2K(1-\delta )/(j+2)$. The tail therefore grows at least as a positive constant times $e^{2K(1-\delta )r}$ divided by a power of $r$; a finite polynomial of earlier terms cannot cancel it. Multiplication by $e^{-Kr}$ leaves exponential growth, contradicting normalizability.

Thus the series must terminate at degree $N-1$, where $KN=\lambda$ and $N$ is a positive integer. Conversely, termination gives a regular polynomial times $e^{-\lambda r/N}$, which is normalizable. ThereforeThese are the $\ell =0$ members of the hydrogen spectrum. Regularity at the origin is essential: square integrability alone on the punctured radial interval would also admit a singular $1/r$ branch. Such a branch is not a physical eigenfunction of the ordinary three-dimensional Coulomb Hamiltonian; its singularity introduces an inadmissible point-source contribution. In the reduced radial variable $u=r\psi$, the physical condition is $u(0)=0$.

For the wave equation, introduce characteristic coordinates $\xi =x-ct$, $\eta =x+ct$ when $c\ne 0$. The equation becomes $u_{\xi \eta }=0$, so $u=F(x-ct)+G(x+ct)$. The initial conditions imply $F+G=\varphi$ and $-c{F}^{\prime }+c{G}^{\prime }=0$. Constants can be absorbed into $F,G$, giving the D'Alembert formulaIf $\varphi$ vanishes outside $[-\alpha ,\alpha ]$, then $u(x,t)=0$ whenever $|x|>\alpha +|c|t$. Thus Property P fails: the wave equation has finite propagation speed $|c|$. No arbitrarily short time can transmit the disturbance arbitrarily far. If $c=0$, the zero initial velocity instead gives $u(x,t)=\varphi (x)$, and the conclusion still holds.

Taking the spatial Fourier transform of the diffusion equation gives ${\tilde{\theta }}_t=-D{k}^2\tilde{\theta }$ and $\tilde{\theta }(k,0)=\tilde{g}(k)$. Hence $\tilde{\theta }=e^{-D{k}^{2}t}\tilde{g}$. Fourier inversion and the Gaussian transform from part (a), or direct evaluation of the inverse Gaussian integral, give the heat kernelSubstituting the integral for $\tilde{g}$ and interchanging absolutely integrable terms produces the convolution solutionFor the stated nonnegative compactly supported data, understood to be nonzero on a set of positive measure, the kernel is strictly positive everywhere. Therefore $\theta (x,t)>0$ for every real $x$ and every $t>0$. Property P holds: diffusion has infinite propagation speed. The distant value can be extremely small, but is not zero. For continuous initial data, any nontrivial nonnegative $g$ automatically has the required positive-measure set.

Use the Fourier transform $\tilde{f}(k)={\int }_{\mathbb{R}}f(x)e^{-ikx}dx$. For a rapidly decaying smooth function, integration by parts gives $\widetilde{f^{\prime }}=ik\tilde{f}$, and differentiation under the integral gives $\widetilde{xf}=id\tilde{f}/dk$. Both are justified here by Gaussian decay. Since $f^{\prime }=-2a^{2}xf$,The Gaussian integral supplied in the hint gives $\tilde{f}(0)=\sqrt{\pi }/a$. Solving this first-order equation yieldsThe calculation states the transform normalization and the two differentiation properties explicitly.

Write $z_j=c+R{e}^{i{\theta }_j}$ with ${\theta }_1<{\theta }_2<{\theta }_3<{\theta }_4<{\theta }_1+2\pi$. In the identitythe phase factors cancel in the cross-ratio. In our convention the two numerator sine factors are negative, while the denominator factors have opposite signs. Thus $r<0$, and the exchanged cross-ratio $1-r$ is greater than one. Consequently $|r|+1=1-r=|1-r|$.

Taking absolute values of the two formulas in part (a) givesMultiplying $|r|+1=|1-r|$ by the common positive denominator proves Ptolemy theorem:The product of the diagonals equals the sum of the products of opposite sides.