Use $X_n\sim \mathrm{Unif}[0,1/n]$ from part (a). We already proved convergence in distribution to zero. For any prescribed $p\in [0,1]$, put $c_n=p/n$. Then $c_n\to 0$, and the cumulative distribution function satisfiesThis includes $p=0$ and $p=1$, since the uniform distribution assigns zero probability to each endpoint. The limit cumulative distribution function is discontinuous at zero, explaining why the conclusion of (ii) does not apply there.

The definition of convergence in distribution gives $F_n(t)\to F(t)$ at each continuity point $t$ of $F$. We must handle a moving argument rather than assume this convergence is uniform.

A cumulative distribution function has at most countably many discontinuities: each is a jump of positive size, and there are at most $r$ jumps of size at least $1/r$, since their sizes sum to at most one. Thus we can choose continuity points $u_j<c<v_j$ with $u_j\uparrow c$ and $v_j\downarrow c$. For fixed $j$, eventually $u_j<c_n<v_j$. The monotonicity of each cumulative distribution function impliesFirst let $n\to \infty$, using the definition of convergence in distribution at $u_j,v_j$. Then let $j\to \infty$, using continuity at $c$:No result relating different convergence of random variables has been assumed.

Fix $\delta >0$. Once $|c_n-c|<\delta$, the following event inclusions hold outside $\{|X_n-X|>\delta \}$:Taking probabilities gives the explicit boundsBy the assumed convergence in probability, the error tends to zero. ThereforeThe continuity of the cumulative distribution function $F$ at $c$ lets $\delta$ decrease to zero, proving $\boxed{F_n(c_n)\to F(c).}$ This proves the needed implication directly, without invoking any theorem comparing convergence of random variables.

To return the broad interior transport, the western boundary current has opposite sign and characteristic peakThe leading exponential profile retains the smaller interior offset, $v(0)\simeq (F_0/\beta )(1-L/{\delta }_W)\simeq -52.1\mathrm{m}{\mathrm{s}}^{-1}$. Because $\gamma L\simeq 0.593$ is not very small, the exact two-wall solution gives a more accurate linear-model peak:All three evaluations identify an intense southward western boundary current, with peak at $x=0,y=0$, of order tens of metres per second. They are predictions of the stated steady linear barotropic model. Nonlinear advection, stratification and the actual basin geometry could alter them substantially; the calculation does not establish that real Pluto winds attain these values.

In the open-basin Sverdrup balance approximation,The maximum is at $y=0$. Evaluating the exact finite-friction solution instead gives its largest northward velocity at the eastern side of the equator, about $4.96\mathrm{m}{\mathrm{s}}^{-1}$. Thus the requested scale is roughly $5\mathrm{m}{\mathrm{s}}^{-1}$ northward; the exact model also resolves variation across the open basin.

Along $y=0$, differentiating the approximate streamfunction givesin the leading small-friction limit. Hence the velocity is strongly southward at the west wall, crosses zero after a few Stommel boundary layer widths, and approaches a much weaker northward interior value $F_0/\beta$. The leading zero is $x\simeq {\delta }_W\ln (L/{\delta }_W)$ when $L/{\delta }_W\gg 1$.

For plotting or finite-friction evaluation, use the exact derivativeIt increases monotonically from its negative western value to its positive eastern value because $v^{\prime }(x)=-A[{\lambda }^2{C}_W{e}^{-\lambda x}+{\gamma }^2{C}_E{e}^{\gamma (x-L)}]>0$. Moreover ${\int }_0^{L}v(x)dx=\Psi (L)-\Psi (0)=0$, as required for a closed basin.

The streamlines are contours of $\psi$. Since $A<0$, the exact streamfunction is negative inside the rectangle and zero on its perimeter. It has one interior minimum, with closed anticlockwise streamlines: a broad northward interior flow returns southward along the west wall; flow is westward near the north edge and eastward near the south edge. The western parts of the streamlines are closely packed because the western boundary current is much faster.

The figure uses the exact ordinary differential equation solution, so it also remains consistent with both wall conditions when the asymptotic scale separation is only moderate. It depicts the stipulated dynamical model, not observed winds or a reconstruction of the source illustration.

Retaining the integrated energy flux from part (i), slow changes of depth and latitude obeyAt fixed depth and higher absolute latitude in the same hemisphere, $|f|=2\Omega |\sin \phi |$ increases, so $A_0\propto \sqrt{|f|}$ increases. The offshore trapping width $L_R=c/|f|$ narrows, concentrating the same transported energy closer to the coast.

This uses a local Kelvin-mode approximation on a slowly varying background, despite the original constant-$f$ derivation. The scale of variation must be long enough for the wave to adjust without substantial reflection, scattering or dissipation. It is not a rule to extrapolate through the equator, where the coastal trapping scale and hemisphere-dependent propagation rule change.

The Kelvin-wave coastal amplitude scaling follows from the flux. For a wave with peak coastal amplitude $A_0$, $\overline{v\eta }=(c/H)A_0^2{e}^{-2r/L_R}/2$. Integrating the supplied flux over the offshore coordinate gives its magnitudeThe direction is the alongshore propagation direction; the expression above is positive magnitude. For slowly varying depth, negligible dissipation/reflection and no other source or sink, the transported energy per time is conserved. At fixed latitude,The coastal amplitude increases on entering shallower water. This is stronger than the familiar fixed-width $H^{-1/4}$ shoaling law because the Kelvin mode's offshore width also contracts, $L_R\propto \sqrt{H}$. The comparison is a slowly varying trapped-mode result; abrupt topography can scatter or reflect energy and invalidate simple flux conservation for one outgoing mode. The amplitude panel in solution illustrates the scaling.

As $\omega \to \infty$, both capacitors approach short circuits. Thus $Z\to 0$, $W\to R_v>0$, and the unloaded right junction tends to the source's top-terminal voltage. HenceThe connected ideal ammeter carries limiting electric current $V_0/R$ from right to left because $Z_{\mathrm{th}}\to R/2$. Reversing the output polarity reverses both nonzero open-circuit answers, without changing their magnitudes.

At the balance angular frequency, $Z=W$ and the two voltage dividers have the same ratio. ThusThis zero detector electric current is the operational condition used to find $R_v$ and ${\omega }_0$.

At zero angular frequency, the lower right capacitor is an open circuit, while the upper right electrical impedance tends to $R_x$. With the detector removed, no electric current flows down that branch, so $V_R=V_0$ and $V_L=V_0/2$. ThereforeFor the connected ideal ammeter, the resulting electric current from right to left is $V_0/(R+2R_x)$, despite its zero terminal voltage. This follows from $I_A=V_{\mathrm{oc}}/Z_{\mathrm{th}}$ with $Z_{\mathrm{th}}=R/2+R_x$.

For electrical resistance $R_p>0$ in a parallel circuit with inductance $L>0$, add electrical admittances and invert:Conjugate impedance matching to the series RC impedance requiresIt is possible at a finite positive angular frequency exactly when $C{R}_p^2>L$, in which caseThe expression for $R$ follows by substituting this ${\omega }^2$ into the real part of $Z$. If $C{R}_p^2\le L$, no such finite matching angular frequency exists. Under literal equality, both passive inductive configurations fail by the sign of their imaginary parts.

Let the unknown series circuit contain electrical resistance $R_s>0$ and inductance $L>0$. Its electrical impedance is $Z=R_s+i\omega L$. Thus $A=R_s$, $B=L$, and conjugate impedance matching is possible withAt this angular frequency, the inductor and load capacitor have opposite electrical reactances of equal magnitude. Literal equality of their separate electrical impedances is impossible; for example, $R_s=L=C=1$ gives $1+i\omega$ versus $R-i/\omega$ for every $\omega >0$.

For the negative test electric charge, use the same change in the fixed point charge's electric potential:The external agent extracts energy, since the final Coulomb potential energy is lower. Again the path must avoid the fixed point charge, and the work done by the electric field is $-W_{-}$.

The fixed point charge produces electric potential $\varphi (r)=k_{e}q/r$, where $k_e=1/(4\pi {ϵ}_0)$. The external quasistatic work is the increase in Coulomb potential energy:The endpoints have radii $2d$ and $d$, irrespective of their opposite coordinate signs. One must use a path avoiding the singular point charge at the origin. The electric field is conservative on such paths, so the endpoint calculation suffices; a literal straight path through the origin is inadmissible. The work done by the electric field has the opposite sign.

When $\omega RC$ is of order one, the output consists of visibly curved exponential charging and discharging arcs, with neither long flat plateaus nor nearly straight sides. At $\omega RC=1$,The output is continuous and lags the abrupt input transitions; its precise shape is the periodic solution given in the parent part.

When $\omega RC\gg 1$, the capacitor voltage changes little during each half-cycle. Thus ${\dot{V}}_{\mathrm{out}}\simeq V_{\mathrm{in}}/(RC)$, giving an approximately triangular output. Its extrema areThe finite but small triangle is centered on zero for the chosen zero-mean input. A nonzero input mean shifts the output mean by the same amount.

When $\omega RC\ll 1$, each half-cycle lasts many time constants, so $M\simeq V_0$. The output nearly follows the two input levels, with exponentially rounded transitions lasting order $RC$. It remains continuous rather than making ideal vertical jumps.

A parallel connection has the same electric potential difference across each element, so the electrical admittances add. With $G=1/R$ and $B=\omega C-1/(\omega L)$,For a prescribed sinusoidal current, $\tilde{V}=Z_{\mathrm{p}}\tilde{I}$. Thus the voltage-minus-current phase isBelow ${\omega }_0=1/\sqrt{LC}$ the circuit is inductive, so electric potential difference leads current; above it the circuit is capacitive, so electric potential difference lags current. At ${\omega }_0$, the reactive branch currents cancel, $Y_{\mathrm{p}}=1/R$ and $\boxed{Z_{\mathrm{p}}=R}$. This is the impedance maximum for positive finite $R$, whereas the series circuit has its impedance minimum at the same frequency. It does not mean that either individual reactive branch current vanishes.