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With the endpoint values zero, move the new-level differences to the left and keep the old-level differences on the right. Then $B$ is tridiagonal with diagonal $1+\mu$ and off-diagonal entries $-\mu /2$, while $C$ has diagonal $1-\mu$ and off-diagonal entries $\mu /2$. Thus $B{u}^{n+1}=C{u}^n$.

Part (a) supplies their common orthonormal eigenbasis. Their eigenvalues are $1+\mu (1-\cos {\theta }_k)$ and $1-\mu (1-\cos {\theta }_k)$. For $\mu \ge 0$, $B$ is invertible, and the step matrix $B^{-1}C$ has eigenvaluesSince the eigenbasis is orthonormal, $\Vert (B^{-1}C{)}^n{\Vert }_2={\max }_k|{\eta }_k{|}^n\le 1$, uniformly in the number of steps and mesh. Hence the method is unconditionally stable for every $\mu \ge 0$, which includes all physical positive time steps. For negative $\mu$, every nonzero modal parameter has modulus larger than one whenever its denominator is nonzero; if a denominator vanishes the step is undefined. Thus the algebraic stable range is exactly $[0,\infty )$. This proof uses simultaneous matrix diagonalization, not Fourier stability analysis.

For ${\theta }_k=k\pi /(M+1)$, extend the proposed vector by $v_0=v_{M+1}=0$. The sine addition identity gives $v_{i-1}+v_{i+1}=2\cos {\theta }_k{v}_i$. Thus row-by-row multiplication yieldsThese vectors do not depend on $a,b$. To prove they form a basis even when eigenvalues of $A$ coincide, apply the calculation first to the symmetric matrix with $a=0,b=1$, whose eigenvalues $2\cos {\theta }_k$ are distinct. Its $M$ nonzero eigenvectors are orthogonal and form a basis. Their squared lengths are $(M+1)/2$, so multiplying each by $\sqrt{2/(M+1)}$ gives a common orthonormal eigenbasis for every matrix in the family.

Independence of $x$ and incompressibility give $v_y=0$ for the secondary velocity $(u,v,0)$. No penetration at the upper and lower walls then forces $v=0$. The radial momentum equation, retaining the given centrifugal acceleration, is $-w^2/R=-p_x/\rho +\nu u^{\prime \prime }$. The transverse pressure gradient is constant in $y$ at this order. Integrating twice and absorbing the resulting quadratic coefficient into $C$ givesThe odd integration term vanishes because both no-slip conditions are equal. The remaining no-slip and zero-flux conditions areThe second equation is half of ${\int }_{-a}^{a}u(y)dy=0$. Optionally solving gives $C=-66K{a}^4/7$ and $D=5K{a}^6/7$. With $s=y/a$ the velocity is $K{a}^6(1-s^2)(s^4-4s^2+5/7)$, so the ratio to the primary velocity is bounded, including its limiting value at the walls, by a constant times $G{a}^4/(\rho {\nu }^{2}R)$. A sufficient parameter ordering isEquivalently the last condition is $\mathrm{Re}a/R\ll 1$ for $\mathrm{Re}=G{a}^3/(\rho {\nu }^2)$. It keeps the secondary circulation weak; the neglected radial return flow and side-wall structure supply closure outside this central wide-channel approximation.

The steady axial Navier-Stokes equation for the straight-pipe approximation is $0=G/\rho +\nu w^{\prime \prime }(y)$. No-slip boundary conditions are $w(\pm a)=0$. Integrating twice givesIgnoring the radial variation is justified away from the side walls by the given ordering $a\ll b$.

European put-call parity givesFor the American call, there are no dividends and the riskless rate is positive, so early exercise sacrifices the benefit of deferring the strike. Directly, at stock 30 its continuation value is $(4/5)[(1/6)30+(5/6)21]=18$, exceeding its exercise value 15; at stock 12 it is $(4/5)(5/6)=2/3$, exceeding exercise value zero. At time zero continuation is $86/15>0$. Thus $C_A(0)=C_E(0)=86/15$, with no optimal early exercise. European parity must not be applied using the American put price $3/2$.

At the upper time-one node, both exercise and continuation values are zero. At the lower node, immediate exercise gives three while continuation gives $2/3$, so exercise is strictly better. At time zero immediate exercise gives zero, whereas continuing givesThus continue at time zero and exercise at time one after a down move; at the upper node, exercise or continued holding is immaterial. One initial replicating portfolio for these optimal stopping values is $\Delta =-1/6$, $B=4$, costing $3/2$ and producing zero at stock 30 and three at stock 12.

The European put has terminal values $0,0,0,5$. Discounted backward expectation gives value zero at stock 30 and $(4/5)(5/6\cdot 0+1/6\cdot 5)=2/3$ at stock 12. ThereforeFor replication, denote by $B$ the money currently in the bank, which grows to $RB$ next period. At time zero hold $\Delta =-1/27$ shares and $B=8/9$, costing $15\Delta +B=1/3$. At stock 30 rebalance to zero shares and zero bank balance. At stock 12 rebalance to $\Delta =-5/6$, $B=32/3$. This costs $12(-5/6)+32/3=2/3$ and gives terminal wealth zero at 16 and five at 10. The initial portfolio produces exactly the two node values, so these rebalancings are self-financing.

At each nondegenerate binary node, no arbitrage requires the riskless gross return $R=1+r$ to lie strictly between the two stock returns. The three intervals are $(4/5,2)$ at time zero, $(6/5,3/2)$ at the upper node, and $(5/6,4/3)$ at the lower node. Their intersection givesAt $R=5/4$, the upper-branch risk-neutral probabilities are $(RS-S_d)/(S_u-S_d)$. They are $3/8$ initially, $1/6$ after the up move, and $5/6$ after the down move. All are strictly between zero and one, so the martingale measure is equivalent and unique. The four terminal probabilities in the PDF's order $45,36,16,10$ are $1/16,5/16,25/48,5/48$.

Write $X=\theta +Z$, with $Z\sim N(0,1)$ independent of the prior $\theta \sim N(0,{\tau }^{-1})$. Then $\lambda X-\theta =(\lambda -1)\theta +\lambda Z$, soCompleting the square in the joint density gives $\theta \mid X\sim N(X/(1+\tau ),1/(1+\tau ))$. Thus the Bayes rule is $d_{1/(1+\tau )}$, with Bayes risk $1/(1+\tau )$. The rule $d_1=X$ has integrated risk one for every $\tau >0$, so its excess Bayes risk is $\tau /(1+\tau )\to 0$ as $\tau \downarrow 0$. The priors remain proper at every positive $\tau$. Therefore $d_1$ is extended Bayes, without treating the limiting improper flat prior as a proper Bayes prior.

For a prior $\Pi$ and loss $L$, the Bayes risk of a rule $d$ is $r(\Pi ,d)=\int {\mathbb{E}}_{\theta }L(\theta ,d(X))\Pi (d\theta )$. A Bayes decision rule minimizes this integrated risk. An Extended Bayes rule is a rule whose excess over the optimal Bayes risk can be made arbitrarily small by a suitable proper prior: for every $\epsilon >0$ some $\Pi$ has $r(\Pi ,d)\le {\inf }_{d^{\prime }}r(\Pi ,d^{\prime })+\epsilon$, with finite risks.

Condition on the observation and use iterated expectation. A rule is Bayes precisely when it chooses a minimizer of the posterior expected loss almost surely, up to the usual measurability requirement. Under squared Euclidean loss, with a finite posterior second moment, write $m=\mathbb{E}[\theta \mid X]$. The identityshows that the Bayes rule is the posterior mean.

Reading the $j$th component givesThe chain has the interpretation of a rate-$R$ Poisson clock which independently redraws a colour from $\pi$, allowing a redraw of the same colour. No redraw by time $t$ has probability $e^{-Rt}$ and retains the initial state; at least one redraw gives distribution $\pi$.

The vector ${\delta }_i-\pi$ has component sum zero, so part (c) applies to it. Part (b) gives ${\pi }^{T}P(t)={\pi }^T$. Decompose ${\delta }_i^T={\pi }^T+({\delta }_i^T-{\pi }^T)$ to obtainThis is the full transition-law row, showing how the initial colour survives with exponentially decreasing weight.

If $x^{T}1=0$, then $x^{T}Q=-R{x}^T$. Induction gives $x^T{Q}^n=(-R{)}^n{x}^T$. Summing the matrix-exponential series yieldsThus deviations of total mass zero decay with the single rate $R$.

Write $Q=1r^T-RI$, with $r=R\pi$. Then ${\pi }^{T}1=1$ andThus $\pi$ is a stationary probability vector, a left eigenvector for eigenvalue zero. If any probability vector $\nu$ is stationary, the same calculation gives ${\nu }^{T}Q=r^T-R{\nu }^T=0$, forcing $\nu =\pi$. Hence the equilibrium is unique, even if some $r_j$ vanish and the chain is not irreducible; those zero-rate states are transient or inaccessible rather than extra stationary classes.

For a finite-state continuous-time Markov chain with generator matrix $Q$, $P(t)=e^{tQ}$. The exponential solves both Kolmogorov equations, $P^{\prime }=QP=PQ$, with initial condition $P(0)=I$.

Put $M=\Vert f{\Vert }_{\infty }>0$ and $M_0=\mu (E)$. There is a positive-measure set on which $|f|>M/2$, so $M_0>0$ and each $M_n$ is positive; boundedness and finite measure make each finite. For $n\ge 1$, apply Cauchy-Schwarz to $|f{|}^{(n-1)/2}$ and $|f{|}^{(n+1)/2}$ to getTherefore the positive ratios $r_n=M_{n+1}/M_n$ are nondecreasing. Telescoping gives $M_n/M_0=r_0\cdots r_{n-1}\le r_n^n$, while $|f{|}^{n+1}\le M|f{|}^n$ almost everywhere gives $r_n\le M$. HenceFor $0<\epsilon <M$, the set $A_{\epsilon }=\{|f|>M-\epsilon \}$ has positive measure by the definition of essential supremum. Thus $\Vert f{\Vert }_n\ge (M-\epsilon )\mu (A_{\epsilon }{)}^{1/n}$, whereas $\Vert f{\Vert }_n\le M\mu (E{)}^{1/n}$. Letting $n\to \infty$ and then $\epsilon \downarrow 0$ shows $\mu (E{)}^{-1/n}\Vert f{\Vert }_n\to M$. The preceding squeeze yields $M_{n+1}/M_n\to \Vert f{\Vert }_{\infty }$. This proves the ratio limit rather than assuming it from the moment-root limit.

The Minkowski inequality states the triangle inequality for every $L^p$ norm with $1\le p\le \infty$. Apply it to $f_n=(f_n-f)+f$ and then to $f=(f-f_n)+f_n$. The two bounds combine toThus norm convergence implies convergence of the norms themselves.

For $1\le p<\infty$, the Lp norm is $\Vert f{\Vert }_p=({\int }_E|f{|}^{p}d\mu {)}^{1/p}$; for $p=\infty$ it is the essential supremum of $|f|$. The Lp space consists of measurable functions with finite norm, identifying functions equal almost everywhere. This identification makes the quantity a norm rather than only a seminorm. Convergence in $L^p$ means $\Vert f_n-f{\Vert }_p\to 0$, including the essential-supremum norm when $p=\infty$.

Parametrize the torus by $F(u,v)=((2+\cos u)\cos v,(2+\cos u)\sin v,\sin u)$. Its outward Gauss map isThe derivative $N_u$ has unit length, while $N_v$ has length $|\cos u|$ and is perpendicular to $N_u$. Thus rank drops precisely at $u=\pi /2$ or $3\pi /2$. These are the two circles of height $z=1$ and $z=-1$ and radius two, and their images are the north and south poles. Therefore the critical values are $(0,0,\pm 1)$.

For any nonpolar target point, its third coordinate determines two values of $u$, one with positive and one with negative $\cos u$. For each, the longitude fixes a unique $v$ modulo $2\pi$, with a shift of $\pi$ when $\cos u$ changes sign. Thus every regular value has exactly two preimages, one on the outer and one on the inner side of the torus. ConsequentlyThis explicit two-sheet description justifies the count without relying on a sketch.

Assume $X$ is a compact smooth $n$-manifold without boundary and $Y$ a connected smooth $n$-manifold without boundary. For a regular value $y$, the inverse function theorem makes $f^{-1}(y)$ discrete, and compactness makes it finite. Define ${\mathrm{deg}}_{2}f=|f^{-1}(y)|(\mathrm{mod}2)$. No orientations are required.

The parity is independent of the regular value: join two regular values by a path and perturb the path in its interior to be transverse to $f$. Its inverse image is a compact one-dimensional manifold with boundary consisting of the two endpoint fibres. Every compact one-manifold has an even number of boundary points, so the two counts have the same parity. The same argument proves homotopy invariance. For noncompact $X$ one instead needs a proper map to retain compactness of the relevant inverse images.

If $f$ omitted a point of $Y$, that point would be a regular value with empty fibre and hence degree zero. Independence on the connected target gives the contradiction. Therefore a nonzero degree modulo two forces surjectivity.