Successive jumps occur at times separated by $T=6$ s. Between $-3$ and $3$ s the plotted function is $f(t)=2-t$, periodically extended, with maximum $5$ and minimum $-1$. Its mean is therefore $2$ m, and $f(t)-2=-t$ is an odd function over that centered period.

For the Fourier series convention$a_0=4$ m and $a_r=0$ for every $r\ge 1$; the nonconstant content consists of sine terms. If $a_0$ denotes the mean directly in the alternative convention $f=a_0+\sum \cdots$, its value is $2$ m. Stating the convention resolves the otherwise ambiguous notation.

For completeness, integrating $-t$ against the sine basis gives $b_r=6(-1{)}^r/(\pi r)$ m. At a jump the Fourier series converges to the midpoint $2$ m. The nonzero mean does not destroy the oddness argument for the mean-subtracted function; declaring the original function itself odd would incorrectly force its mean to vanish.

Regular timing supplies a common interval and a well-defined frequency grid for a standard discrete Fourier transform and Fast Fourier transform. It also makes sample-index lags correspond to fixed physical delays in cross-correlation, autocorrelation and digital filtering. These methods estimate frequency content and characteristic periods, separate instrument or environmental noise from signals, identify arrivals and time delays, and compare ground-motion spectra.

A uniform interval establishes a Nyquist frequency; frequencies above it alias unless filtered before acquisition. Interpolating irregular data onto a regular grid does not undo prior aliasing or fill large gaps with new measurements, and it can alter spectra and create correlated errors. Methods that directly use irregular times are possible, but the ordinary FFT and fixed-lag filter formulas cannot simply treat the original irregular observations as evenly spaced.

Sampling at $0.2$ Hz means $\Delta t=5$ s. The target times are therefore $5$, $10$ and $15$ s, all within the supplied observation range. For bracketing observations, linear interpolation gives $d(t)=d_i+(d_{i+1}-d_i)(t-t_i)/(t_{i+1}-t_i)$. ThusThe estimated displacements are $7.00$, $12.99$ and $8.98\mu \mathrm{m}$. Interpolation assumes approximately linear behavior within each interval; it does not recover any unobserved high-frequency motion.

For an integrable initial profile $f(y)$, use its Fourier transform with the conventionThe transformed diffusion equation is ${\partial }_t\hat{T}=-\kappa k^2\hat{T}$, soEquivalently the heat-kernel convolution isThis also accommodates bounded piecewise-continuous initial profiles; it recovers $f$ at continuity points as $t\downarrow 0$. For periodic initial data use a Fourier series and multiply each mode by $e^{-\kappa k_r^{2}t}$; the constant mean mode remains unchanged. “Any function” needs the usual existence/integrability or growth conditions: an arbitrarily fast-growing initial function need not give a finite heat-kernel integral.

The diffusion equation is linear, so superposition givesEach term separately satisfies the equation, and their sum recovers the stated initial data at $t=0$. Its e-folding times are ${\tau }_i={\lambda }_i^2/(4{\pi }^2\kappa )$, so ${\lambda }_1<{\lambda }_2$ implies ${\tau }_1<{\tau }_2$. The shorter-wavelength oscillation loses a given fraction of its amplitude first. It approaches zero exponentially rather than disappearing at a finite instant. This preferential damping explains smoothing of fine-scale temperature structure.

Choose $t_1=1/(\kappa k^2)$ to make the sketch explicit. The three profiles areThey have the same wavelength, zero crossings and phase, with amplitudes $A$, $0.3679A$ and $0.1353A$. Diffusion reduces the temperature gradient and amplitude; it does not advect or change this individual mode’s wavelength.

Write $k=2\pi /\lambda$. For $T=A{e}^{\alpha t}\cos (ky)$, direct partial differentiation givesHence $T_t=\kappa T_{yy}$ exactly when $\alpha =-\kappa k^2$, proving the claimed diffusion equation solution. At $t=0$, $e^0=1$, so $T(0,y)=A\cos (2\pi y/\lambda )$. The infinite spatial domain needs no end-boundary conditions. For positive thermal diffusivity the amplitude decreases, as described by the thermal diffusion rate of a Fourier mode.

For a constant-mass, well-mixed box the mean residence time is stored water mass divided by its total outgoing water-mass transport. Using the source figure’s endpoints,At water steady state, each denominator can equivalently be replaced by that box’s summed inflow. If transports are specified as volume rates, use volume rather than mass in the numerator, or convert with density consistently. The drawn maximum basin depths alone do not supply layer volumes, plan areas or transport magnitudes, so no numerical residence times follow. Internal recycling counts as exit from a particular box; these times are not the time to escape the combined sea system.

Let $S_{Ms},S_{Md},S_{Bs},S_{Bd}$ be salt mass fractions, with reservoir water masses denoted by the corresponding $M$ symbols. Let $S_A$ be Atlantic inflow salinity and $S_D$ the Danube inflow salinity, approximately zero for fresh water. With the diagram endpoint convention from part i, instantaneous mixing givesEvery outgoing salt transport uses the salinity of its donor box. The water masses are constant because the water balances are at steady state, even while the salinities evolve. If reservoir water masses varied, the left side would instead be $d(M_j{S}_j)/dt$.

Summing yields $d({\sum }_j{M}_j{S}_j)/dt=F_1{S}_A+F_9{S}_D-F_2{S}_{Ms}$, a direct whole-system salt check. Fresh water carries essentially no salt, but its dilution effect is still present through the water balance and the resulting transport terms.

There are four well-mixed reservoirs: Mediterranean surface/deep $(M_s,M_d)$ and Black Sea surface/deep $(B_s,B_d)$. Use the actual drawing’s arrow endpoints: $F_2$ leaves $M_s$, $F_5$ enters $M_d$ from $B_s$, and $F_6$ starts in $M_s$ and enters $B_d$. With $F_i$ as positive water-mass transport rates, the four steady conservation of mass equations areAdding them cancels every internal transport and gives the whole-system requirement $F_1+F_9=F_2$. There is no stated evaporation, precipitation or other external transport to add.

The schematic endpoints are unusual for the conventional two-layer exchange picture: its Gibraltar outflow is drawn in the surface box, and its Black Sea surface inflow descends into the Mediterranean deep box. Moving $F_2$ to $M_d$, $F_5$ to $M_s$, and the source of $F_6$ to $M_d$ would define a different, more conventional idealization. The displayed equations consistently follow the supplied drawing; the same endpoint convention is used in the following two parts.

Permitted; returns "C". Both receiver and argument have static type C, so the most-specific applicable overload is show(C). That method is defined in C and returns its own one-letter result without calling super. The six results, in source order, are therefore AB, AB, AB, BC, BC, C.

Permitted; returns "BC". The receiver's static type is C, but the argument's static type is only B. Thus show(C) is inapplicable; among show(A) and show(B), the latter is more specific. Its implementation is C.show(B), whose superclass call returns "B" before "C" is appended. The argument happens to reference a C object, but method overloading uses its static type.

Permitted; returns "BC". Through static type B, both show(A) and show(B) accept an argument of type C; method overloading selects the more specific show(B). The receiver's runtime type is C, so dynamic dispatch invokes C.show(B). Its super.show(o) calls B.show(B), returning "B", and the override appends "C". The show(C) overload is not visible through receiver type B.

Permitted; returns "AB". A receiver of static type B exposes show(A) and show(B), but the argument's static type A is not a subtype of B. Only show(A) is applicable. Its effective implementation on the C receiver is B.show(A), which concatenates the superclass result "A" and "B".

Permitted; returns "AB". Through a receiver of static type A, only show(A) is visible. The argument's static type C is a subtype of A, so it can be passed to that method. Dynamic dispatch selects B.show(A) on the receiver object, exactly as in part (i). The argument's more specific type cannot expose overloads absent from the receiver's static type.

Permitted; returns "AB". The receiver's static type is A, which offers show(A). The argument also has static type A, so that signature is selected. All three variables refer to the same object of runtime type C; the most-derived implementation of show(A) is inherited from B. Its super.show(o) calls A.show(A), producing "A", and it appends "B". C overrides show(B), not this selected signature.

The strongly connected components areThe directed cycle $1\to 3\to 5\to 1$ makes its three vertices mutually reachable. Vertex $6$ has no outgoing edge. Vertex $2$ reaches that cycle but has no return path from it, and vertex $4$ reaches the other components but has no incoming path from them. These facts exclude every further merger.

The only possibilities are $f[2]\in \{10,11\}$. Vertex $2$ reaches all of $1,3,5,6$. In depth-first search, all vertices reachable from an active vertex must have been discovered and finished before its own finishing event, except for active ancestors. None of these four can be an ancestor of $2$, because none can reach $2$. Thus all four are finished before $2$ finishes.

Vertex $4$ cannot already be finished at that point: its outgoing edge to $2$ would have caused $2$ to be visited before it could finish. It also cannot be a descendant of $2$, since $2$ cannot reach $4$. Therefore $4$ is either still undiscovered or an active ancestor. In the first case the only remaining events are discovery and finishing of $4$, giving $f[2]=10$. In the second case the only remaining event is finishing of $4$, giving $f[2]=11$.

Both occur. Starting the outer traversal with $2$ leaves $4$ undiscovered and gives time $10$. Starting with $4$ and taking its edge to $2$ first makes $4$ an ancestor and gives time $11$. The argument does not assume any particular order on the other edges.

Increment a single clock on each discovery and finish, starting with clock zero. Lexicographic depth-first search givesThe first search tree has discovery edges $1\to 3$, $3\to 5$, $3\to 6$. The edge $5\to 1$ returns to an active ancestor. Once that tree finishes, vertex $2$ is a new root whose neighbour $1$ is already finished, and vertex $4$ is the final root. This accounts for all twelve events in the DFS discovery and finishing times.

Use source $1$, as in the preceding part. The Dijkstra algorithm extracts vertices in the order $1,3,4,2$. Distances after relaxing the outgoing edges of each extracted vertex areThe shortest path to $2$ is $1\to 3\to 4\to 2$, with weight $2+1+1=4$; the direct route to $2$ and the route via $3\to 2$ are both heavier.