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The function $f$ is continuous at $y_0$ when for every $\epsilon >0$ there is a $\delta >0$ such thatSuppose $f$ has a local minimum at $c$ and is differentiable there. For sufficiently small $h>0$,Both one-sided limits equal $f^{\prime }(c)$, so it is both nonnegative and nonpositive. Therefore

A finite irreducible Markov chain is positive recurrent. Its mean recurrence time to state $i$, starting at $i$ and counting the first positive return, is $1/{\pi }_i$. This identity comes from the long-run fraction of time spent at that state: successive return cycles each contain one visit at their start, so the fraction is the reciprocal of the mean cycle length. It remains valid for a periodic chain.

Here ${\pi }_0=2^{-k}$, henceFor $k=1$ the return is deterministically after two minutes, which checks the formula.

If there are $i$ balls in box A, a ball currently in B is selected with probability $(k-i)/k$, increasing the count, and a ball in A is selected with probability $i/k$, decreasing it. For $k\ge 1$,with every other transition zero. This is the Ehrenfest urn model.

The stationary distribution isIt is normalized by the binomial theorem, and it obeys the detailed balance:All other pairs have zero flux in both directions. These identities prove stationarity and show that this is a reversible Markov chain in equilibrium. The finite chain is irreducible and has period two; reversibility does not require aperiodicity.

Expand the second stage by the Taylor theorem, writing $f=f(t_n,y_n)$ and $D_{y}f$ for the derivative in the state variable:The update therefore agrees with the exact solution's Taylor series:Thus the one-step local truncation error is $O(h^3)$, and the method has order at least two under the usual smoothness and Lipschitz continuity assumptions.

For $f=\lambda y$, the two-stage Runge-Kutta method givesOn the real axis, $R(z)=\frac{1}{2}(z+1{)}^2+\frac{1}{2}\ge \frac{1}{2}$, so $|R(z)|\le 1$ is equivalent to $(z+1{)}^2\le 1$. HenceThe decay convention gives $(-2,0)$. Since negative real values such as $z=-3$ lie outside the linear stability domain, the method is not A-stable. Its order is exactly two, since $R(z)$ lacks the cubic term of $e^z$.

Apply the numerical method to the scalar test equation $y^{\prime }=\lambda y$, with $z=h\lambda$. Its linear stability domain consists of those $z\in \mathbb{C}$ for which the discrete solution stays bounded as the step count tends to infinity. For a one-step method with stability function $R$, this isIf asymptotic decay is required, use $|R(z)|<1$ instead; this only changes the boundary conventions in part (b). A method is A-stable if its stability domain contains the left half-plane, so every decaying linear test equation is stable for every positive step size. For a multistep method, the corresponding definition uses the root condition for a multistep method for every characteristic root, including simplicity of unit-modulus roots.

In a charge-free and current-free vacuum, Maxwell equations giveTake the curl of Faraday's law and use the curl of the curl identity $\nabla \times (\nabla \times \mathbf{E})=\nabla (\nabla \cdot \mathbf{E})-{\nabla }^2\mathbf{E}$. The divergence term vanishes by Gauss's law, while the time derivative of the magnetic curl supplies $-{\mu }_0{ϵ}_0{\partial }_t^2\mathbf{E}$. ThusThis is the transverse electromagnetic wave equation.

Distinct eigenvalues of the self-adjoint angular momentum operator imply $⟨{\psi }_1,{\psi }_2⟩=0$. The given combination is not normalized: its squared norm is $({\cos }^2\omega t+3{\sin }^2\omega t)/4$, which is always positive. Therefore its properly normalized expectation value isIt vanishes when ${\tan }^2\omega t=1/3$. The first positive solution has $\omega t=\pi /6$, soNormalization changes the value of the expectation value away from zero, but not its zero times.

The product rule for a commutator gives $[A,B^2]=[A,B]B+B[A,B]$. Using the permitted cyclic angular momentum commutation relations,The first two expressions cancel, even though the factors need not commute. HenceThe squared angular momentum operator commutes with each of its components by the same cyclic calculation.

Use the canonical commutation relations $[{\hat{x}}_i,{\hat{x}}_j]=[{\hat{p}}_i,{\hat{p}}_j]=0$ and $[{\hat{x}}_i,{\hat{p}}_j]=i\hslash {\delta }_{ij}I$. For compactness write products without hats in the calculation, on a common domain such as smooth compactly supported wavefunctions. Expanding the commutator givesThe middle two commutators vanish because all the relevant indices differ. The first is $-i\hslash x_2{p}_1$, since $[p_3,x_3]=-i\hslash I$; the last is $i\hslash x_1{p}_2$, since $[x_3,p_3]=i\hslash I$. ThusThis is one of the orbital angular momentum commutation relations.

The orbital angular momentum linear operator is the cross product of the position operator and the momentum operator:In position space, $\hat{\mathbf{p}}=-i\hslash \nabla$, so $\hat{\mathbf{L}}=-i\hslash \mathbf{r}\times \nabla$. The Levi-Civita symbol gives ${\hat{L}}_1={\hat{x}}_2{\hat{p}}_3-{\hat{x}}_3{\hat{p}}_2$ and its cyclic analogues.

Define $F_i(x)=a_i+\frac{1}{6}{\sum }_{j\ne i}\sin x_j$. Solving the system is equivalent to finding a fixed point of $F:{\mathbb{R}}^3\to {\mathbb{R}}^3$. The mean value theorem implies $|\sin u-\sin v|\le |u-v|$. In the specified metric,Each coordinate difference occurs exactly twice. This metric makes ${\mathbb{R}}^3$ a nonempty complete metric space, because a Cauchy sequence in it is Cauchy in each real coordinate, and coordinate convergence implies convergence in the sum metric. Thus $F$ is a contraction mapping with constant $1/3$.

There is exactly one solution for every choice of $(a_1,a_2,a_3)$, by the contraction mapping theorem. Iteration of $F$ from any starting vector also constructs it.

The contraction mapping theorem states that a contraction mapping of a nonempty complete metric space into itself has a unique fixed point $x_{\ast }$. For any starting point $x_0$, the iterates $x_{n+1}=f(x_n)$ converge to $x_{\ast }$, with the quantitative estimateCompleteness and the self-map condition are essential: the geometric estimate makes the iterates a Cauchy sequence, completeness supplies their limit in $X$, and continuity of the contraction mapping makes that limit a fixed point. If two fixed points existed, their distance would be at most $q$ times itself, forcing equality of the points.

A contraction mapping on a metric space $(X,d)$ is a map $f:X\to X$ for which there is a single constant $q$ with $0\le q<1$ such thatThe same $q$ must work for every pair. Merely decreasing each nonzero distance strictly does not suffice for this definition.

An extreme point of a convex set cannot be written as a nontrivial convex combination of two distinct members. If a permutation matrix $P=\theta B+(1-\theta )C$ with $0<\theta <1$ and $B,C$ doubly stochastic, each zero entry of $P$ forces the corresponding entries of $B,C$ to be zero by nonnegativity. The unique possible nonzero entry in each row must then be one by the row sum. Thus $B=C=P$, proving that every permutation matrix is extreme.

Conversely, let $A$ be a doubly stochastic matrix that is not a permutation matrix. At least one entry lies strictly between zero and one. Form a bipartite graph whose two vertex classes are rows and columns, with edges precisely at these fractional entries. Every incident row has at least two fractional entries: a single fractional entry, with all other entries zero or one, could not give row sum one. The same holds for columns.

A finite nonempty graph whose incident vertices all have degree at least two contains a cycle. In a bipartite graph that cycle has even length. Assign alternating signs $+1,-1$ to its entries, giving a nonzero matrix $D$ with every row and column sum zero. Choose $\epsilon >0$ smaller than the distance of each cycle entry from both zero and one. Then $A+\epsilon D$ and $A-\epsilon D$ remain doubly stochastic matrices, are distinct, and satisfySo $A$ is not extreme. ThereforeThis alternating-cycle perturbation of a doubly stochastic matrix proves the extreme-point part of the Birkhoff-von Neumann theorem directly.

A maximum flow problem specifies a finite directed flow network, a source $s$, a sink $t$, and nonnegative arc capacities. A feasible flow assigns $0\le f_e\le c_e$ to each arc, with incoming flow equal to outgoing flow at each vertex other than $s,t$. The objective is to maximize the net flow leaving $s$, equivalently entering $t$.

The Ford-Fulkerson algorithm starts with zero flow. Its residual network has a forward arc of capacity $c_e-f_e$ and a reverse arc of capacity $f_e$ for each original arc. If an $s$–$t$ path has positive residual capacities, augment by their minimum: increase flow along forward arcs and decrease it along reverse arcs. This preserves capacity bounds and conservation, and increases the flow value by that positive bottleneck. Repeat until there is no augmenting path.

For rational capacities, choose a positive integer $D$ clearing every denominator. Initially all flows and residual capacities are integer multiples of $1/D$, and augmentation preserves that property. Every augmentation therefore increases the flow value by at least $1/D$. The flow value is bounded above by the sum of capacities of arcs leaving $s$. Thus only finitely many augmentations can occur, independently of how the augmenting paths are chosen.

To see why termination supplies a maximum, let $S$ consist of vertices reachable from $s$ in the final residual network. The sink is not in $S$. Each original arc from $S$ to its complement is saturated; each original arc entering $S$ has zero flow, since otherwise its reverse residual arc would leave $S$. The resulting flow value equals the capacity of this cut. No feasible flow exceeds a cut capacity. Hence the algorithm terminates at a maximum flow for rational capacities, proving the relevant max-flow min-cut theorem conclusion as well.

Put ${\mathrm{RSS}}_1={\sum }_i(Y_i-\hat{\alpha }-\hat{\beta }{x}_i{)}^2$ for the unrestricted fit. Under $H_0$, the best slope is still $\hat{\beta }$ because the design is centered. Orthogonality of the residuals to the constant vector givesAt either fitted mean, the maximum-likelihood estimator of variance is $\mathrm{RSS}/n$. The ratio used by the generalized likelihood-ratio test is consequentlySmall likelihood ratios correspond to large $|T_n|$. Under $H_0$, part (a) gives independent $\sqrt{n}\overline{Y}/\sigma \sim N(0,1)$ and ${\mathrm{RSS}}_1/{\sigma }^2\sim {\chi }_{n-2}^2$. Thus $T_n$ has the Student's t-distribution with $n-2$ degrees of freedom. The exact size-5% critical region iswhere $t_{\nu ,0.975}$ is the 97.5th percentile of that distribution. The denominator uses the unbiased residual variance estimate, not the likelihood variance denominator $n$.

Under an alternative, the same statistic has the representationThe joint law of $Z,W$ is parameter-free, and $\beta$ has disappeared. Thereforewhere $g(d)=\mathbb{P}(|(Z+\sqrt{n}d)/\sqrt{W/(n-2)}|>t_{n-2,0.975})$. This is also the noncentral t-distribution description of the power. As explained in part (a), an exact residual-based test requires $n>2$.

Let $Q={\sum }_i{x}_i^2>0$. Under the usual nondegenerate regression assumption $n>2$, maximizing the normal likelihood is equivalent to minimizing the residual sum of squares. The normal equations and ${\sum }_i{x}_i=0$ give the maximum-likelihood estimatorsWrite $A=\overline{\epsilon }$ and $B={\sum }_i{x}_i{\epsilon }_i/Q$. Then $\hat{\alpha }=\alpha +A$, $\hat{\beta }=\beta +B$, and the residual is $r_i={\epsilon }_i-A-x_{i}B$. Independence of the errors givesFor each $i$,Thus the three requested variables are pairwise uncorrelated.

They are linear functions of the jointly normal error vector. More strongly, the whole vector $(A,B,r_1,\dots ,r_n)$ has a multivariate normal distribution, and its block covariances show that $A$, $B$ and the residual vector are independent, by uncorrelated jointly normal variables are independent. Independence is preserved when applying a measurable function to the residual block. HenceThe residual projection is $I-11^{\mathsf{T}}/n-x{x}^{\mathsf{T}}/Q$, of rank $n-2$, so additionally ${\sum }_i{r}_i^2/{\sigma }^2\sim {\chi }_{n-2}^2$. This is the orthogonal projection of a Gaussian vector interpretation of the same result. The printed design conditions alone allow $n=2$; then the fit is exact, the positive-variance likelihood has no finite maximizer, and the residual test in part (b) is undefined. The standard testing question therefore needs $n>2$.