Write the top row as $(u,v,5-u-v)$. The bottom row is then $(3-u,3-v,u+v-1)$, and nonnegativity becomesThe feasible set is a hexagon in this plane. Its six vertices, hence all basic feasible solutions of this transportation problem, areThere are four independent constraints; each displayed support consists of four cells forming a transportation spanning tree, so each is a nondegenerate basic solution. Conversely, a basic feasible solution is a vertex, so the hexagon enumeration is exhaustive.

For any feasible $x^{\prime }$, the proposed dual inequalities giveIf equality holds on every cell with $x_{ij}>0$, this last quantity equals $f(x)$. Hence $f(x)\le f(x^{\prime })$ for every feasible $x^{\prime }$. This is weak duality with complementary slackness.

The Northwest corner method first allocates three units to $(1,1)$, then two to $(1,2)$, one to $(2,2)$ and four to $(2,3)$:Choose ${\lambda }_1=0$, ${\lambda }_2=2$, $({\mu }_1,{\mu }_2,{\mu }_3)=(a,2,-1)$. Equality holds on its four occupied cells. The other inequalities are $-1\le 3$ and $a+2\le b$. Thus $a+2\le b$ implies that $x^{\ast }$ minimizes $f$.

The last printed clause changes from minimization to maximization. To answer that wording, use the transportation simplex algorithm for maximization, equivalently minimize the negative costs. There are only two possible entering cells at $x^{\ast }$:

Put $d=b-a$. On the feasible hexagon,If $d\le -2$, this expression is maximized by taking $u=3,v=0$, so $E$ is a global maximizer. If $d>-2$, the feasible vertex $(u,v)=(1,0)$ has objective $2(d+2)$ larger than $E$, so $E$ is not a maximizer. Nor can $C$ ever be one: comparison with $x^{\ast }$ requires $d\ge 2$, while comparison with the vertex $(0,3)$ requires $d\le -2$. Therefore one pivot to a maximizer is possible exactly when $a\ge b+2$, by entering $(1,3)$. Its reduced cost for maximization is $4>0$; the other reduced cost is $d-2\le -4$, so the improving pivot is unambiguous.

If “transportation algorithm” is required to retain the preceding minimization objective, no such step can reach a maximizer: every minimizing step is nonincreasing, whereas the feasible point $E$ always has $f(E)=f(x^{\ast })+8$. If the final word was intended to be “minimizes”, the corresponding genuinely improving one-pivot condition would instead be $a-2\le b<a+2$; at $b=a+2$, $x^{\ast }$ is already optimal and a zero-cost pivot is possible. These distinctions preserve the printed maximization request without silently altering it.

A quadratic form on a finite-dimensional real vector space is a function $q(v)=B(v,v)$ for a symmetric bilinear form $B$. Conversely $B(u,v)=\frac{1}{2}(q(u+v)-q(u)-q(v))$ recovers that bilinear form. Two quadratic forms are congruent if a linear isomorphism $T$ satisfies $q_2(v)=q_1(Tv)$; in coordinates this is matrix congruence, $A_2=P^T{A}_{1}P$, with $P$ invertible.

Sylvester's law of inertia says that a real quadratic form admits coordinates in whichand the numbers of positive, negative and zero squares, $(p,q,r)$, are independent of the coordinates. Here $p+q+r=\dim V$; $p$ and $q$ are the maximal dimensions of subspaces on which the quadratic form is positive definite and negative definite, while $r$ is the dimension of its radical of a bilinear form. These numbers are the inertia of a bilinear form. The rank of a quadratic form is $p+q$, and the signature of a quadratic form, in the signed-difference convention, is $p-q$.

Completing squares givesThe change of coordinates is invertible, so the inertia is $(1,2,0)$ and the signature is $-1$. If signature is recorded as an ordered pair instead, that pair is $(1,2)$.

A sufficient statistic is one for which the conditional distribution of the full sample, given its value, is independent of the parameter. For independent observations the likelihood is ${\prod }_{i=1}^6{p}_i^{n_i}$, depending on the data only through the counts. The factorization criterion for sufficiency therefore makes $(n_1,\dots ,n_6)$ sufficient. Directly, all ordered samples with given counts have the same probability, so their conditional distribution is uniform over the $n!/\prod n_i!$ possible arrangements, independently of the $p_i$.

The unrestricted maximum-likelihood estimator is ${\hat{p}}_i=n_i/n$. Thus the generalized likelihood-ratio test useswith zero-count terms interpreted by continuity. Reject for small $\Lambda$, equivalently large $D$.

Write $n_i=n/6+{\delta }_i$, so $\sum {\delta }_i=0$. A Taylor series expansion of $(n/6+{\delta }_i)\log (1+6{\delta }_i/n)$ gives ${\delta }_i+3{\delta }_i^2/n+O(|{\delta }_i{|}^3/n^2)$. Summing cancels the linear terms, and yieldsUnder the null hypothesis, the multinomial central limit theorem places the standardized count deviations in the five-dimensional subspace with coordinate sum zero, with identity covariance on that subspace. Their squared norm consequently tends to the chi-squared distribution with five degrees of freedom. This gives the Pearson chi-squared goodness-of-fit test.

For $T=8.12$, the asymptotic p-value is about $0.150$. The 5% critical value of ${\chi }_5^2$ is approximately $11.07$. Do not reject at the usual 5% level, or at 10%. A significance level is not specified in the paper, so an unconditional yes-or-no decision is not determined: using this approximation, rejection would require a chosen level at least about $15\%$.

Irrotational flow permits $\mathbf{u}=\nabla \varphi$, and incompressible flow imposes $\nabla \cdot \mathbf{u}=0$, hence $\Delta \varphi =0$. In the laboratory frame, the no-penetration condition at the translating sphere is ${\partial }_r\varphi (a,\theta )=U\cos \theta$.

For separation of variables, take the angular factor $g(\theta )=\cos \theta$, which satisfies $\frac{1}{\sin \theta }(\sin \theta g^{\prime }{)}^{\prime }=-2g$. The radial equation iswith solutions $r$ and $r^{-2}$. Rest at infinity excludes the $r$ term. The normal boundary condition fixes the remaining coefficient, giving the potential flow around a translating sphereThere is no azimuthal velocity. With fluid density ${\rho }_f$, the kinetic energy isThe angular integral is $2\pi$, and the radial integral is $1/(3a^3)$, so $K_f={\rho }_f\pi a^3{U}^2/3$. Since $M=(4\pi a^3/3){\rho }_s$ and $k={\rho }_f/{\rho }_s$, $K_f=kM{U}^2/4$. Thus the added mass of a sphere is $kM/2$.

During a fall through $h$, gravity does work $Mgh$ and buoyancy does work $-kMgh$. Starting from rest, conservation of energy givesConsequently $U=\sqrt{4(1-k)gh/(2+k)}$. The mass cancels: the reduced speed comes from both buoyancy and the fluid's added mass.

Use the Fourier transform convention $\hat{F}(k)={\int }_{\mathbb{R}}F(x)e^{-ikx}dx$, with inverse factor $1/(2\pi )$. Splitting the inverse integral at zero givesThis is the Poisson kernel for the upper half-plane at height $a$.

Take the Fourier transform in $x$ of the Laplace equation. The transformed boundary-value problem isFor $k\ne 0$ the general solution is a linear combination of $e^{|k|y}$ and $e^{-|k|y}$; the two conditions set their coefficients to zero and one respectively. At $k=0$, the solution is the constant one. Inverting therefore givesThe upper boundary holds pointwise. At the lower boundary the Dirac delta function must be interpreted distributionally: the kernel has integral one and concentrates at zero, so for every continuous compactly supported test function $h$, $\int \psi (x,y)h(x)dx\to h(0)$. This also specifies the solution in the assumed class admitting the stated Fourier transforms.

For a variation $y+\epsilon \eta$ preserving both endpoint values and slopes, $\eta ={\eta }^{\prime }=0$ at the endpoints. The first variation isIntegrate the last term twice and the middle term once. All boundary terms vanish, givingThe fundamental lemma of the calculus of variations yields the higher-order Euler-Lagrange equationFor the present functional this becomes $1+y-y^{(4)}=0$. Its general solution is $y=-1+A\cosh x+B\sinh x+C\cos x+D\sin x$. The left endpoint gives $C=-A$ and $D=-B$. The right endpoint then givesThe coefficient determinant is $2(1+\cosh \pi )>0$, so $A=1$, $B=0$. The stationary function is $y_{\ast }(x)=-1+\cosh x-\cos x$.

For any admissible perturbation $\eta$, exact quadratic expansion, followed by the stationarity equation and integration by parts, givesApply the supplied Poincaré inequality first to $\eta$, then to ${\eta }^{\prime }$, whose endpoint values also vanish:Thus $J[y_{\ast }+\eta ]\le J[y_{\ast }]$ for every admissible perturbation, proving a global maximum through the clamped second-variation inequality.

Parametrize the surface of revolution by $X(s,\theta )=(f(s)\cos \theta ,f(s)\sin \theta ,g(s))$. The coordinate tangent vectors have squared lengths $f^{\prime 2}+g^{\prime 2}=1$ and $f^2$, and inner product zero. Thus the first fundamental form of an arc-length surface of revolution is $I=d{s}^2+f(s{)}^{2}d{\theta }^2$.

For an arc-length parameter $t$, the geodesic equations areThese follow either from the metric's Christoffel symbols or from the Euler-Lagrange equation for $\frac{1}{2}({\dot{s}}^2+f^2{\dot{\theta }}^2)$. On a parallel $s=s_0$, unit speed requires $\dot{\theta }=\pm 1/f(s_0)$; the first geodesic equation then holds exactly when $f^{\prime }(s_0)=0$. Thus geodesic parallels of a surface of revolution are precisely the critical points of the radius function.

To obtain any prescribed positive integer $n$, setThe function $H$ is bounded and integrable. Choose $\epsilon >0$ so $\epsilon \sup |H|<1$, and choose $R>\epsilon {\int }_{\mathbb{R}}|H|$. Then $f>0$, $g$ is smooth, and $f^{\prime 2}+g^{\prime 2}=1$. The derivative $f^{\prime }=\epsilon H$ vanishes at exactly $1,\dots ,n$. Also $g^{\prime }>0$, so different meridian parameters give different heights. The resulting surface has exactly $n$ geodesic parallels.

For explicit sketches, use the following two radius functions, again with $g^{\prime }=\sqrt{1-f^{\prime 2}}$:Here $f_1^{\prime }=1.2s{e}^{-s^2}$ has only the zero $0$, while $f_2^{\prime }=0.6(s^2-1)e^{-s^2}$ has exactly the zeros $-1,1$. Both radii stay positive and both derivatives have absolute value below one. The highlighted rings are the geodesic parallels; the surfaces continue beyond the displayed window.

If every parallel is a geodesic, then $f^{\prime }=0$ everywhere, so $f=R$ is a positive constant. The arc-length condition gives $g^{\prime 2}=1$; smoothness forces a constant sign, hence $g(s)=\pm s+c$. The surface is a circular cylinder of radius $R$. It is a circular cylinder because the meridian is a straight line parallel to the axis.

The inner circle has centre $2/5$ and radius $2/5$. The pole $z=1/2$ of the Möbius transformation lies inside that removed disc, so the transformation is holomorphic on $D$, with derivative $3/(2z-1{)}^2\ne 0$.

For $z=x+iy$,On the outer boundary this is one. On the inner boundary, $|z{|}^2=4x/5$, and the ratio is four. More directly, inside the outer circle the numerator exceeds the denominator by $3(1-|z{|}^2)>0$, whereas outside the inner circle four times the denominator exceeds the numerator by $3(5|z{|}^2-4x)>0$. Thus the image is $1<|w|<2$. The inverse transformation is $z=(w-2)/(2w-1)$, so this is a bijective conformal map onto that annulus.

A radial harmonic function on an annulus has form $A+B\log |w|$. The two boundary values give $A=1$, $B=1/\log 2$. By conformal invariance of harmonicity,It has the required boundary values, and the maximum principle for harmonic functions proves uniqueness among solutions continuous on the closed region.

The contraction mapping theorem states that a map $f$ of a nonempty complete metric space into itself, satisfying $d(fx,fy)\le qd(x,y)$ for some $0\le q<1$, has a unique fixed point. Iteration from any point converges to it; for example $d(f^{n}x,\tau (f))\le q^{n}d(x,fx)/(1-q)$.

Assume $X$ is nonempty. Its boundedness makes $\delta$ finite. Symmetry, positivity and separation follow pointwise from $d$, and taking the supremum of $d(fx,hx)\le d(fx,gx)+d(gx,hx)$ proves the triangle inequality. Hence $\delta$ is a uniform metric.

If $(f_n)$ is a Cauchy sequence in this uniform metric, each $(f_n(x))$ converges by completeness of $X$; denote its limit by $f(x)$. Given $ϵ>0$, choose $N$ so that $\delta (f_n,f_m)<ϵ$ for $n,m\ge N$. Letting $m\to \infty$ gives $d(f_n(x),f(x))\le ϵ$ for all $x$, so $f_n\to f$ uniformly. To prove continuity, fix $x$ and useChoose $N$ so the outside terms are small, then use continuity of $f_N$ for the middle term. Thus $f\in \mathrm{Maps}(X,X)$, proving completeness of the continuous-map space in the uniform metric.

The contraction subspace is not necessarily complete. On $X=[0,1]$, $f_n(x)=(1-1/n)x$, $n\ge 2$, are contraction mappings and converge to the identity by uniform convergence. Their only possible uniform limit is not a contraction, so this Cauchy sequence has no limit in $\mathcal{C}$.

Nevertheless the fixed point assignment is continuous. Fix $f\in \mathcal{C}$ with contraction constant $q_f<1$, and put $u=\tau (f)$, $v=\tau (g)$. ThenConsequently $d(\tau (f),\tau (g))\le \delta (f,g)/(1-q_f)$. This proves continuity of the fixed-point assignment at $f$ without requiring one contraction constant valid for every $g$.

A matrix is in Jordan normal form if it is block diagonal with Jordan blocks $J_r(\lambda )=\lambda I+N$, where $N$ has ones immediately above the diagonal and zeros elsewhere. For a Jordan normal form matrix $J$, number its diagonal positions $1,\dots ,n$ and putChoose positive ${\epsilon }_m\to 0$ avoiding the finitely many values solving ${\lambda }_i+i\epsilon ={\lambda }_j+j\epsilon$ for $i\ne j$. The matrix remains upper triangular and now has distinct eigenvalues, so it is diagonalizable. Its entries converge to those of $J$. For arbitrary $A=SJ{S}^{-1}$, use $A_m=S{J}_m{S}^{-1}$. This proves density of diagonalizable complex matrices.

If $A$ is diagonalizable with eigenvalues ${\lambda }_1,{\lambda }_2$, conjugating $B$ by the same basis change conjugates $T_A$ to $T_{\mathrm{diag}({\lambda }_1,{\lambda }_2)}$. On the four matrix units $E_{ij}$, this acts by $({\lambda }_i+{\lambda }_j)E_{ij}$. Thus, writing $s=\mathrm{tr}A$ and $d=\det A$, its characteristic polynomial isThe entries of $T_A$ depend linearly on those of $A$, and the coefficients of its characteristic polynomial depend polynomially, hence continuously, on those entries. Apply the formula to the approximating $A_m$ and pass to the limit. The boxed formula therefore holds for every $A$, including non-diagonalizable matrices. This is the two-by-two anticommutator characteristic polynomial.

An cut of a flow network partitions a flow network into $S,S^c$, with source $A\in S$ and sink $B\notin S$. Its cut capacity is the sum of capacities of arcs directed from $S$ to $S^c$; a minimum cut minimizes this sum.

Use the suggested nodes. Give $A\to a_e$ capacity $1$, each $a_e\to b_v$ capacity $M=m+kn+1$, and each $b_v\to B$ capacity $k$. By the max-flow min-cut theorem, a flow of value $kn$ exists once every cut capacity is at least $kn$.

Consider a cut of a flow network containing no crossing arc of capacity $M$, and let $U$ be the intersections on the sink side. Every street incident to $U$ must also be on that side, or its arc to an endpoint in $U$ would cross the cut of a flow network. Hence at least $d(U)$ source-to-street arcs cross, as do the $n-|U|$ intersection-to-sink arcs from the other intersections. Therefore its cut capacity is at leastA cut of a flow network crossing a capacity-$M$ arc has still larger capacity. The cut of a flow network just before the sink has capacity $kn$, proving the maximum flow value is exactly $kn$.

The integral max-flow theorem supplies an integral maximum flow. Every intersection sends $k$ units to the sink; each street can supply at most one unit, assigned to one endpoint. Orient each assigned street away from its assigned endpoint. Orient unassigned streets arbitrarily. Then each vertex has at least $k$ outgoing streets. An Ford-Fulkerson algorithm with integer capacities constructs this flow and therefore the street directions. Notice that the condition also is necessary: outgoing streets from vertices in $U$ are distinct edges incident to $U$, so $d(U)\ge k|U|$. This is an outdegree orientation criterion.

The Rao-Blackwell theorem says that if $S$ is a sufficient statistic and $\delta$ a finite-variance estimator, then ${\delta }^{\ast }=\mathbb{E}(\delta \mid S)$ is an estimator with the same expectation and no larger variance. Sufficiency ensures the conditional distribution, and hence this formula as a function of the data, does not involve the unknown parameter. The tower property of conditional expectation gives $\mathbb{E}{\delta }^{\ast }=\mathbb{E}\delta$, while the law of total variance givesThis proves the theorem and shows that equality holds exactly when $\delta$ is already determined by $S$, almost surely. In particular, unbiasedness is preserved.

Assume $n\ge 3$. Independence gives $\mathbb{E}\hat{\theta }=p_0{p}_1{p}_2=\theta$, so the proposed indicator is unbiased and has variance $\theta (1-\theta )$. The type counts $S=(n_0,n_1,n_2)$ are a sufficient statistic: conditional on them, every ordering of the labels is equally likely. Consequently,For example, the conditional probability of the ordered first three labels is $(n_0/n)(n_1/(n-1))(n_2/(n-2))$. This is Rao-Blackwellization of a multinomial probability product.

If all three $p_i$ are positive, there is positive probability that all three counts are positive. On such an event the conditional indicator has probability strictly between zero and one, so its conditional variance is positive. Thus $\mathrm{Var}({\theta }^{\ast })<\theta (1-\theta )$ for interior parameter values. If any $p_i=0$, then $\theta =0$ and both estimators have zero variance; the printed strict inequality is impossible at those boundary parameters. The valid unrestricted statement is the non-strict inequality.

The free surface is a material surface, so its kinematic boundary condition isWith constant atmospheric pressure $p_a$ and no surface tension, the dynamic boundary condition for an inviscid interface follows from Unsteady Bernoulli equation:Write $\varphi =Ux+\phi$, where $\phi$ and $\eta$ are small. Choose the time-dependent potential gauge so the undisturbed right-hand side is $U^2/2$. Retaining first-order terms, evaluated at $y=0$, givesThe incompressibility condition also gives Laplace's equation for $\phi$.

Set $\Omega =\omega -kU$. Substitution of the specified normal mode into the two linearized boundary conditions yields $-i\Omega a=ikb$ and $\Omega b+ga=0$. Eliminating $b$ gives the dispersion relationThe decay factor $e^{ky}$ in a fluid extending to $y=-\infty$ uses $k>0$. With signed wavenumber the decaying factor is $e^{|k|y}$ and the right-hand side is $g|k|$. The frequency shift is the advection of a deep-water gravity wave by the uniform current.

Take the divergence of the Ampère-Maxwell equation, $\nabla \times \mathbf{B}={\mu }_0\mathbf{J}+{\mu }_0{ϵ}_0{\partial }_t\mathbf{E}$. Since the divergence of a curl is zero vanishes, and Gauss's law gives $\nabla \cdot \mathbf{E}=\rho /{ϵ}_0$, the result isThis is the charge continuity equation.

At an internal point of a homogeneous Ohmic conductor, Ohm's law gives $\mathbf{J}=\sigma \mathbf{E}$. Since $\sigma$ is constant, the charge continuity equation becomes ${\partial }_t\rho =-(\sigma /{ϵ}_0)\rho$. ThereforeThis is charge relaxation. Here $\rho$ is the total charge appearing in the vacuum form of Gauss's law; in a homogeneous dielectric description of free charge the corresponding constant permittivity replaces ${ϵ}_0$.

For a finite isolated body, charge travels to the surface. The bulk formula applies at internal points, not across the interface where electrical conductivity jumps and surface charge density accumulates. Integrating the charge continuity equation over the body, including its surface charge, gives constant total electric charge because no electric current crosses the external boundary. Bulk decay and conservation of electric charge are therefore compatible.

Parametrize the characteristics by $s$, starting from $(x,y,u)=(1,r,r)$. Their ordinary differential equations areThey give $x=e^s$, $y=e^s(r+s)$ and $u=r+s$. Thus $s=\log x$ and $r=y/x-\log x$, soIndeed, $u_x=-y/x^2$, $u_y=1/x$ give $x{u}_x+(x+y)u_y=1$, and $u(1,y)=y$. The initial line is noncharacteristic, and its characteristics cover the half-plane $x>0$. The data do not determine a solution on $x<0$; a smooth solution through the origin is impossible because the differential equation there would read $0=1$.