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Move the first $n-k$ columns ofpast its last $k$ columns. This requires $k(n-k)$ column interchanges and produces the block diagonal matrixEach interchange reverses the determinant's sign, so part (a) gives

For $A=(a_{ij})\in M_n(\mathbb{C})$, the determinant isForany nonzero term in the Leibniz formula for determinants must match every row in the lower block to a column in the $N$ block. The remaining upper rows must then match the $M$ columns. The sum consequently factors into the determinant sums for the two diagonal blocks:

Letfor the walk on $G_n$. Decompose $G_n$ into its three outer copies of $G_{n-1}$ and trace the walk only when it passes between distinct corner vertices of these copies. The self-similarity and reflection symmetry of the Sierpinski graph make this trace the simple random walk on $G_1$. It requires an expected five transitions to hit the two target outer corners.

Each coarse transition is an excursion across a copy of $G_{n-1}$ and has mean duration $H_{n-1}$. Applying the strong Markov property at the coarse stopping times givesSince $H_1=5$, induction yields

View each of the three level-one triangles in $G_2$ as a copy of $G_1$. Observe the walk only when it moves from one corner of such a copy to a different corner. By symmetry, the next of the two other corners is equally likely, so this embedded chain is the simple random walk on the coarse graph $G_1$. Part (a) says it makes an expected five coarse transitions before reaching $B$ or $C$.

Within each level-one copy, the mean time for one coarse transition is again five by part (a). The Strong Markov property at successive coarse-corner hitting times therefore gives

Let $D,E,F$ be the midpoints of $AB,AC,BC$. WriteFirst-step analysis for the simple random walk givesbecause $A$ has neighbours $D,E$;because $D$ has neighbours $A,B,E,F$; andbecause $F$ has neighbours $B,C,D,E$. Solving,

The posterior expected loss at decision $\delta$ isAt continuity points its derivative isThus the risk is minimized at a posterior one-third quantile:More generally, any $\delta$ satisfyingis optimal. The quantile lies below the posterior median, reflecting the smaller penalty assigned to underestimation.

The probability of no recovery through month $M$ is the product of the conditional survival probabilities, soUnder quadratic loss, the Bayes estimator is the posterior mean. Posterior independence and the mean of a Beta distribution giveConsequently

LetThus $d_t$ patients recover during month $t$, while $r_t-d_t$ survive that month without recovery. The likelihood function is

The independent $\mathrm{Beta}(T,1)$ prior densities are proportional to $q_t^{T-1}$. Beta-binomial conjugacy therefore leaves the coordinates posteriorly independent, with

The constant is sharp because continuous functions bounded by one can approximate $\mathrm{sgn}K$ arbitrarily closely outside an arbitrarily small interval around zero. Choose $f^{\prime \prime \prime }$ to be such an approximation and integrate three times to obtain $f\in C^3[-1,1]$. ThenNo smaller constant can therefore satisfy the inequality for every $f$.

The Peano kernel theorem says that if a continuous linear functional $L$ annihilates polynomials of degree below $m$, thenHere $m=3$ andApplying $L$ to $(x-t{)}_{+}^2/2$ givesHenceThe smallest candidate is therefore

If the bound is to hold for all $f\in C^3[-1,1]$, the error functional must annihilate every polynomial of degree at most two, because such a polynomial has $f^{\prime \prime \prime }=0$. Applying it to $1,x,x^2$ givesThereforeso $\eta (f)=f(-1)-2f(0)+f(1)$ is the central finite difference approximation.

The mass flux is density times the volume flux:Thus

The net axial pressure force on a length-$L$ fluid cylinder isFrom part (a),Multiplying this wall shear stress by the lateral area $2\pi RL$ givesConsequentlywhich is the required steady force balance.

For the steady, fully developed ansatz $\mathbf{u}=w(r){\mathbf{e}}_z$, both ${\partial }_t\mathbf{u}$ and $(\mathbf{u}\cdot \nabla )\mathbf{u}$ vanish. The axial Navier-Stokes equation reduces toTwo integrations giveRegularity at the axis forces $C_1=0$, and the no-slip boundary condition $w(R)=0$ gives the Hagen-Poiseuille flowFor flow in the positive $z$ direction, $\partial p/\partial z<0$.

Take the wire at $(a,0)$ to carry $+\lambda$ and that at $(-a,0)$ to carry $-\lambda$. Ifthen, up to one additive constant,Thus an equipotential with $k=2\pi {\epsilon }_0\varphi /\lambda$ satisfies $r_{-}/r_{+}=e^k$. For $k\ne 0$, completing the square gives the Apollonius circleIts centre is $(a\coth k,0)$ and its radius is $a/|\sinh k|$. Positive and negative values give nested circles around the positive and negative wires. The electric field is orthogonal to these circles and points from the positive wire toward the negative wire. For $\varphi =0$, the limiting equipotential is the straight line $x=0$.

Direct superposition givesIn the limit $a\to 0$ with $\lambda a=p$, the potential and field become those of a two-dimensional electric dipole:and

For a coaxial Gaussian cylinder of radius $r$ and length $L$, symmetry makes the electric field radial and constant on the curved surface. Gauss's law givessoSince $\mathbf{E}=-\nabla \varphi$, a potential relative to an arbitrary reference radius $r_0$ isAn infinite line charge has no finite convention $\varphi (\infty )=0$.

At each fixed time, $|\psi (x,t){|}^2$ is a bell-shaped Gaussian centred at $x=h(t)$, with width determined by $\mathrm{Re}F(t)>0$. Completing the square as in part (d) makes the density symmetric about $h(t)$. Therefore its position expectation value isThe sketch is a Gaussian peak whose centre oscillates between the two turning positions found in part (e), while its width varies periodically through $\mathrm{Re}F(t)$.

Writing $\theta =\omega t$ andthe supplied identities giveUsing $A=m\omega /(2\hslash )$ in part (d),The peak executes simple harmonic motion about the origin with angular frequency $\omega$ and amplitude $\sqrt{\hslash /(m\omega )}/\sinh a$.

The probability density isCompleting the square shows that, when $\mathrm{Re}F>0$,where