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For $|z|>2$, expand directly in inverse powers:Therefore the outer Laurent series is

The partial fraction decomposition isFor $0<|z|<2$, expand the second term as a geometric series:Hence

The identity mapis a continuous bijection by $d_1\le d_{\infty }$. Its domain is compact by part (i), and its codomain is a Hausdorff space because every metric space is Hausdorff. The compact-to-Hausdorff continuous bijection theorem therefore makes the identity a homeomorphism. Consequently

HereThe L1 norm and uniform norm satisfyso the identity from the $d_{\infty }$ metric to the $d_1$ metric is one-Lipschitz and therefore continuous. The two metrics do not induce the same topology on all of $C[0,1]$. Forone has $d_1(f_n,0)=1/(2n)\to 0$ but $d_{\infty }(f_n,0)=1$. Thus $d_1$ convergence need not imply uniform convergence.

A map $F:(X,d_X)\to (Y,d_Y)$ is Lipschitz continuous if there is $L<\infty$ such thatfor all $x,x^{\prime }\in X$. Evaluation at $a\in [0,1]$ is one-Lipschitz because

Choose distinct $a_0,\dots ,a_n\in [0,1]$ and defineThe Vandermonde determinant is nonzero, so $T$ is a bijection. It is one-Lipschitz for the two uniform metrics. If ${\ell }_i$ are the associated Lagrange cardinal polynomials, thenand henceThus $T^{-1}$ is also Lipschitz.

Let ${\hat{\mathcal{P}}}_n$ be the polynomials whose values lie in $[-1,1]$. Its image under $T$ lies in the bounded cube $[-1,1{]}^{n+1}$. It is closed: if $T(p_j)\to v$, then the Lipschitz inverse gives uniform convergence $p_j\to p=T^{-1}(v)$, and passing to the limit pointwise preserves $|p(x)|\le 1$. By the Heine-Borel theorem, $T({\hat{\mathcal{P}}}_n)$ is compact, and the continuous inverse carries compactness back to ${\hat{\mathcal{P}}}_n$. Therefore

First solve $L\mathbf{y}=\mathbf{b}$ by forward substitution:Then solve $U\mathbf{x}=\mathbf{y}$ by backward substitution, obtainingDirect substitution verifies $A\mathbf{x}=\mathbf{b}$.

Gaussian elimination gives the LU decompositionThe prescribed diagonal entries of $L$ are visible, and direct multiplication gives $LU=A$.

Let $T_k$ be the expected capture time starting from $D_0=k$, so $T_0=0$. The Strong Markov property at successive first passages to lower levels and part (a) giveAt $D_n=1$, the next state is zero with probability $r/2$, one with probability $1/2$, and two with probability $(1-r)/2$. First-step analysis therefore givesorSolving,Substituting $\mu =2/(2q-1)$ now yieldswhich is the required expected capture time.

A random time $\tau$ is a stopping time for the natural filtration $({\mathcal{F}}_n)$ of a Markov chain when $\{\tau \le n\}\in {\mathcal{F}}_n$ for every $n$. The Strong Markov property says that, conditionally on $\tau <\infty$ and $X_{\tau }=i$, the process after $\tau$ is a fresh copy of the chain started from $i$, independent of the pre-$\tau$ history.

Before capture define the half-separationFor $D_n=k\ge 2$, independence of the two moves gives the transition probabilitiesIgnore the holding steps. The resulting embedded chain is a biased random walk that moves left with probability $q$ and right with probability $1-q$. Its expected number of moves needed to descend one level is $1/(2q-1)$; this follows either from first-step analysis or from its drift $1-2q$. A non-holding move occurs with probability $1/2$ at each time, so its mean waiting time is two. Hence, for every $m\ge 2$, the expected time to go from separation $2m$ to $2(m-1)$ is

Because $0\le U\le 1$, for fixed $t$ and sufficiently large $n$,ThereforeEquivalently, $n(1-\hat{\theta }/\theta )$ converges in distribution to an exponential distribution of rate one.

Put $c=\log (1/\alpha )$. The limiting probabilityand the certain inequality $\hat{\theta }\le \theta$ suggest the approximate confidence intervalSince $\mathbb{E}[\hat{\theta }]=n\theta /(n+1)$, its expected length isas required.

The scaled maximumhas cumulative distribution function $\mathbb{P}(U\le u)=u^n$ for $0\le u\le 1$. ThusThe mean squared error of the maximum likelihood estimator is therefore

Each $X_i$ from the uniform distribution on $[0,2\theta ]$ has mean $\theta$ and variance ${\theta }^2/3$. The sample mean is unbiased and has variance ${\theta }^2/(3n)$, hence

A statistic $T$ is sufficient for $\theta$ if the conditional distribution of the full sample given $T$ does not depend on $\theta$. It is minimal sufficient if it is a function of every sufficient statistic.

The likelihood factors asso the Fisher-Neyman factorization theorem shows that $M$, and hence its one-to-one transform $\hat{\theta }=M/2$, is sufficient. Moreover, for two samples $\mathbf{x}$ and $\mathbf{y}$, the ratio $L(\theta ;\mathbf{x})/L(\theta ;\mathbf{y})$ is independent of $\theta$ exactly when their maxima agree. The likelihood-ratio criterion for minimal sufficiency therefore shows that $M$ and $\hat{\theta }$ are minimal sufficient.

For $n\ge 2$, the sample mean $\tilde{\theta }$ is not sufficient: two samples can have the same mean but different maxima, and their likelihood ratio then depends on $\theta$ through the support indicators. It is consequently not minimal sufficient. For the degenerate special case $n=1$, the sample mean and maximum coincide and both conclusions reverse.

Let $M=X_{(n)}={\max }_i{X}_i$. The likelihood function isOn its feasible range $\theta \ge M/2$, this is decreasing in $\theta$. Therefore the maximum likelihood estimator is

The displayed Rodrigues' formula gives the Legendre polynomialA four-node Gaussian rule is exact through degree $2\cdot 4-1=7$. Solving $q_4(x)=0$ first for $x^2$ gives the four nodesThe rule is not exact for every polynomial of degree eight: its quadrature value on $q_4^2$ is zero because every node is a root of $q_4$, whereas

For fixed interpolation nodes $c_0,\dots ,c_{s-1}$, interpolate any $p\in {\mathbb{P}}_{s-1}[x]$ asIntegrating shows that the unique weights making the quadrature rule exact through degree $s-1$ are

Now let the nodes be the zeros of the degree-$s$ orthogonal polynomial $q_s$. Given any $p\in {\mathbb{P}}_{2s-1}[x]$, polynomial division givesOrthogonality gives ${\int }_a^b{q}_{s}h=0$, while $q_s(c_k)=0$ makes the quadrature sum for $q_{s}h$ vanish. The rule is already exact on $r$, so it is exact on $p$. This is the defining precision property of Gaussian quadrature.

Define the cardinal polynomialsThey satisfy ${\ell }_k(x_i)={\delta }_{ik}$. Hence evaluating the proposed polynomial interpolation formula at $x_i$ gives $p_n(x_i)=a_i$, so the required coefficients areThe polynomial ${\sum }_k{f}_k{\ell }_k$ has degree at most $n$ and takes every prescribed value. Its uniqueness follows because the difference of two such polynomials would have $n+1$ distinct roots while having degree at most $n$.

Using $\gamma =\lambda$ and the Pythagorean trigonometric identity,so the speed is independent of $\theta$. Since $\gamma =\pi /\alpha >1$,and the speed tends to zero at the vertex. The steady irrotational Bernoulli equation givesThus the fluid pressure is also independent of $\theta$, withIn particular, the pressure difference between the vertex and $r_0$ is

A streamline satisfiesIntegration givesThe radius is smallest on the wedge bisector $\theta =\alpha /2$, where $\sin (\gamma \theta )=1$. Labelling that radius by $r_{\min }$ therefore givesEvery streamline arrives from infinity along one wall, reaches its closest point on the bisector, and returns to infinity along the other wall:

For a velocity potential, $\mathbf{u}=\nabla \varphi$. The incompressible flow condition $\nabla \cdot \mathbf{u}=0$ therefore becomes Laplace equation ${\nabla }^2\varphi =0$. The Laplace equation in polar coordinates givesso positivity of $\gamma$ and $\lambda$ requiresThe velocity components areThe no-penetration boundary condition on both walls says $u_{\theta }=0$ at $\theta =0$ and $\theta =\alpha$, hence $\lambda \alpha =n\pi$. If $n\ge 2$, $\sin (n\pi \theta /\alpha )$ changes sign inside the wedge. The stated sign condition therefore selects $n=1$, and

Stokes theorem givesFor the upward orientation, the outer circle $r=2,z=1$ is counterclockwise from above and the inner circle $r=1,z=4$ is clockwise. On a counterclockwise circle of radius $R$ and height $z$,The outer circle contributes $-16\pi$. The inner circle would contribute $-7\pi$ counterclockwise and hence contributes $+7\pi$ with its induced clockwise orientation. Thusagreeing with the direct calculation.

The surface is an annular band of the downward-opening paraboloid, bounded by the circles $r=1,z=4$ and $r=2,z=1$. Choose the upward orientation.

The curl of $\mathbf{A}=(3y,-xz,y{z}^2)$ isFor the graph $z=5-x^2-y^2$, the upward vector surface integral element isOver the annulus $1<r<2$, the odd term $2x{z}^2$ integrates to zero. Therefore