Let $\mathcal{A}$ be the uniform closure on $K$ of polynomials. It is a closed algebra: uniform convergence on a compact set is preserved under sums and products. For $\lambda \in \Lambda$, put $q(z)=(z-\lambda {)}^{-1}\in \mathcal{A}$ and $d=\mathrm{dist}(\lambda ,K)>0$. If $|\mu -\lambda |<d/2$, then $\mu \notin K$ andThe series converges uniformly, its ratio having modulus at most one half. Each partial sum is in $\mathcal{A}$, so its limit is too. Hence $\mu \in \Lambda$ in this neighbourhood. For empty $K$, approximation is vacuous and the assertion is immediate.

All poles lie outside some circle $|z|=R$ with $R>1$. The rational function is holomorphic in that disc, so its Taylor series at zero converges uniformly on $|z|\le 1$. Its polynomial partial sums therefore approximate uniformly on $K$. If there are no finite poles, the function is already a polynomial.

For an $n\times n$ matrix here, $D=I$. The sine vectors $v_j(i)=\sin (ij\pi /(n+1))$ give eigenvaluesThus every eigenvalue lies strictly between $1/2$ and $3/2$. For $0<\omega \le 4/3$, $0<\omega {\lambda }_j<2$, so $|1-\omega {\lambda }_j|<1$, including the upper endpoint $\omega =4/3$. Hence the iteration converges for the entire stated range.

The extreme eigenvalues are $1-a$ and $1+a$, with $a=\frac{1}{2}\cos (\pi /(n+1))$. The spectral radius is $\max (|1-\omega (1-a)|,|1-\omega (1+a)|)=|1-\omega |+a\omega$ for $\omega >0$. It decreases up to one and increases after one, since $0\le a<1$. ThereforeFor $n=1$ the same formula gives zero spectral radius and exact solution in one ordinary Jacobi step.

Positive definiteness of $A$ implies $D>0$. Let $B=D^{-1/2}A{D}^{-1/2}$, a symmetric positive definite matrix. The additional hypothesis says $2I-B=D^{-1/2}(2D-A)D^{-1/2}>0$, so all eigenvalues $\lambda$ of $B$ satisfy $0<\lambda <2$. The ordinary Jacobi error matrix $I-D^{-1}A$ is similar to $I-B$ and therefore has eigenvalues $1-\lambda \in (-1,1)$. Its spectral radius is less than one. The convergence criterion for stationary linear iterations consequently gives convergence from every starting vector.

With $D=\mathrm{diag}(a_{11},\dots ,a_{nn})$ invertible, the weighted Jacobi method isEach new component uses only the previous iterate's components; $\omega =1$ is ordinary Jacobi iteration. Errors evolve by $e^{(k+1)}=(I-\omega D^{-1}A)e^{(k)}$.

Balance the time derivative, cubic dispersion and quadratic transport using the cylindrical KdV equation similarity ansatz $q(x,t)=t^{-2/3}F(\xi )$, $\xi =x{t}^{-1/3}$ on $t>0$. For $t<0$ the same real cube-root convention can be used separately. Substitution and removal of the common factor $t^{-5/3}$ giveEvery term is a total derivative: $F{F}^{\prime }=(F^2/2{)}^{\prime }$ and $\xi F^{\prime }+F=(\xi F{)}^{\prime }$. Integration therefore gives the requested second-order equationwhere $C$ is a constant determined by any additional boundary data. No condition in the question forces $C=0$.

The Cole-Hopf transformation for the sign convention here is $Q=q_x/q$, with $q\ne 0$ and $q_t=q_{xx}$. Indeed $q_t/q=Q_x+Q^2$, soConversely a solution $Q$ determines a locally nonzero $q$ through $q_x=Qq$, $q_t=(Q_x+Q^2)q$; their compatibility is this Burgers equation. A spatially constant term can be removed by rescaling $q$ in time.

Set $\mu =q\psi$ in the heat-equation pair. Division by $q$ givesTheir compatibility can be checked without referring back to $q$: writing $A=ik-Q$, $B=-Q_x-Q^2-k^2$, we have ${\psi }_x=A\psi +1$, ${\psi }_t=B\psi +Q+ik$. The coefficient of $\psi$ in ${\psi }_{xt}-{\psi }_{tx}$ is $-Q_t+Q_{xx}+2Q{Q}_x$, and the constant term is $(ik-Q)(Q+ik)-Q_x-B=0$. Thus this linear auxiliary pair is compatible exactly for the required viscous Burgers equation. As in part (a), adjoining a constant component turns it into a homogeneous $2\times 2$ matrix Lax pair.

Let $E=e^{-ikx+k^{2}t}$. Expanding ${\partial }_t(Eq)+{\partial }_x(EX)$ shows that the required flux isIndeed $X_x-ikX=-q_{xx}-k^{2}q$, cancelling the $k^{2}q$ from the time derivative. Introduce a potential $\phi$ by ${\phi }_x=Eq$, ${\phi }_t=E(q_x+ikq)$. Equality of its mixed derivatives is precisely the conservation law. With $\phi =E\mu$, these equations become the linear auxiliary systemThis is a Lax pair in the affine-potential form used for linear evolution equations. If a homogeneous matrix pair is preferred, set $\Psi =(\mu ,1{)}^T$ and use ${\Psi }_x=\left(\begin{array}{cc}ik& q\\ 0& 0\end{array}\right)\Psi$, ${\Psi }_t=\left(\begin{array}{cc}-k^2& q_x+ikq\\ 0& 0\end{array}\right)\Psi$. Its zero-curvature condition has upper-right entry $q_t-q_{xx}$ and all other entries zero.

Here $dy/ds=1$, $dx/ds=u^2$, $du/ds=0$, so from initial parameter $\xi$ we get $y=s$, $x=\xi +{\xi }^{2}y$ and $u=\xi$. Solving for the branch agreeing with $u=x$ at $y=0$, and rationalizing the apparent division by $y$, givesThis formula is smooth on that open set, which contains every point of the initial line. It satisfies $x=u+y{u}^2$ and $1+2yu=\sqrt{1+4xy}>0$. Differentiation gives $u_x=1/(1+2yu)$, $u_y=-u^2/(1+2yu)$, verifying the equation and the data.

The transport field is $(u^2,1)$, with normal component one on $y=0$, so the initial line is everywhere non-characteristic. This explains why neighborhoods cover the entire initial line, although a single fixed-width strip need not work for unbounded $x$.

Characteristics satisfy $dx/ds=1$, $dy/ds=x$, $du/ds=0$. From $(\xi ,0)$ they are $x=\xi +s$, $y=\xi s+s^2/2$, $u=f(\xi )$, so $x^2-2y={\xi }^2$. The initial parameter is recovered on the two initial half-axes byThis is $C^1$ on the displayed open set, takes the prescribed data at $y=0,x\ne 0$, and substitution verifies $u_x+x{u}_y=0$. Each branch may extend separately beyond this chosen domain, but arbitrary data on both half-axes need not allow one common extension.

The initial line has normal $(0,1)$; the transport field is $(1,x)$, whose normal component is $x$. Thus it is non-characteristic exactly away from the origin. At the origin the equation itself would require $u_x(0,0)=f^{\prime }(0)=0$. For generic $f$ this fails, so a $C^1$ solution on a neighborhood of the entire initial line is not guaranteed. The parabola $y=x^2/2$ is the envelope where the recovery of $\xi$ degenerates. Characteristics can also meet both initial half-axes, imposing compatibility such as $f(\xi )=f(-\xi )$ on an attempted larger domain.

For the first-order quasilinear equation $a(x,u)\cdot \nabla u=b(x,u)$, suppose the coefficients and initial data $u=g$ on a smooth hypersurface $\Sigma$ are $C^1$. At an initial point $x_0$, require $a(x_0,g(x_0))\cdot n(x_0)\ne 0$ for a normal $n$ to $\Sigma$; this is precisely the condition that $\Sigma$ be a non-characteristic hypersurface. Then in a neighborhood of $x_0$ there is a unique $C^1$ solution taking the given data. Smooth coefficients and data give corresponding higher regularity.

The method of characteristics solves $\dot{x}=a(x,u)$, $\dot{u}=b(x,u)$ from each initial point. The non-characteristic condition makes the map from the initial coordinates and characteristic time to $x$ locally invertible, so these characteristic solutions define the asserted function.

Only paths whose first move is down have positive payoffs. On the down-up path the raised strike gives $992-960=32$, and on down-down it gives $992-384=608$. Their probabilities are $2/9$ and $1/9$. HenceThe payoff at the middle stock-price node is path dependent: the up-down path still has strike 864 and pays zero.

At the up node, both immediate exercise and continuation are zero. At the down node the exercise value is $864-576=288$, while the discounted continuation value is $R^{-1}(480/3)=120$. Exercise there is optimal. At time zero immediate exercise pays zero, so

The terminal European put pays zero at the upper and middle nodes and $864-384=480$ at the lower one. Its probability is $(1-p{)}^2=1/9$. Thus

Interpret the printed expression as $d^{\ast }(x)=x/({\sigma }^2\lambda )$. If $X\sim N(\theta ,{\sigma }^2)$, the Gaussian exponential-moment formula givesIts supremum is one, attained at zero. For every rule $d$, $L(0,d(X))=1$, so $R(0,d)=1$ and ${\sup }_{\theta }R(\theta ,d)\ge 1$. HenceIn the language of part (b), the point-mass prior ${\delta }_0$ makes every rule Bayes with Bayes risk one, and $d^{\ast }$ has exactly that worst-case risk.

A loss function $L(\theta ,a)$ quantifies the cost of taking action $a$ when the parameter is $\theta$. A decision rule $d$ is a measurable map from the observation space to the action space (or, for randomized rules, a conditional action distribution). Its risk function is $R(\theta ,d)={\mathbb{E}}_{\theta }L(\theta ,d(X))$. For a prior probability distribution $\pi$, its Bayes risk isA minimax rule minimizes ${\sup }_{\theta }R(\theta ,d)$.

All four permitted hops have rate one: $1\to 2$, $2\to 1$, $2\to 3$, and $3\to 1$. Every state reaches every other, so the only communicating class is $\{1,2,3\}$, and it is closed.

Solving $\pi Q=0$, ${\sum }_i{\pi }_i=1$, gives ${\pi }_1=2{\pi }_2$ and ${\pi }_3={\pi }_2$. ThusFor the transient probability, let $p_j(t)=P_{1j}(t)$. The forward equations and $p_1+p_2+p_3=1$ give $p_2^{\prime }=p_1-2p_2$, $p_1^{\prime }=1-2p_1$. From $p_1(0)=1$, $p_1(t)=1/2+e^{-2t}/2$. Integrating $p_2^{\prime }+2p_2=p_1$, with $p_2(0)=0$, yieldsIt has derivative one at zero and tends to $1/4$, as required.

A conservative Q-matrix has $q_{ij}\ge 0$ for $i\ne j$, finite exit rates $-q_{ii}={\sum }_{j\ne i}{q}_{ij}$, and hence row sums zero. It specifies the infinitesimal transition probabilities $P_{ij}(h)={\delta }_{ij}+h{q}_{ij}+o(h)$. For a finite state space,The first derivative at zero is the generator. On infinite state spaces one must additionally address nonexplosion and the domains needed for the forward and backward equations.

Split the overlapping products into two nonoverlapping families:Within each sum the summands depend on disjoint independent pairs and are independent identically distributed Bernoulli variables of mean $1/4$. The two families need not be independent of each other. The strong law of large numbers says that averages of independent identically distributed integrable random variables converge almost surely to their expectation. Applying it separately to each family and intersecting the two probability-one events gives

The first event is a tail event: for any $m$ it equals ${\bigcup }_{N\ge m}{\bigcap }_{n\ge N}\{X_n\le 0\}$, measurable in $\sigma (X_m,X_{m+1},\dots )$.

Convergence of the partial sums is also a tail event. For any $m$, subtracting the finite real sum $X_1+\cdots +X_{m-1}$ shows that convergence is equivalent to convergence of ${\sum }_{j=m}^n{X}_j$. The Cauchy criterion expresses the latter using countable unions and intersections of inequalities involving these tail sums, so it belongs to every tail sigma-algebra.

The event that the partial sums are nonpositive infinitely often is not generally a tail event. For a counterexample let $X_1$ take values $-1$ and one on a two-point probability space, each with positive probability, and let $X_n=0$ for $n\ge 2$. The event is then $\{X_1=-1\}$, while $\sigma (X_2,X_3,\dots )$ is trivial. The variables in this example are even independent. Its status can be different for particular sequences, but it is not tail-measurable in general.