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For a free particle, the Hamiltonian operator isSince ${\chi }^{\prime \prime }=-k^2\chi$,Thuswhere the momentum follows because $-i\hslash d\chi /dx=\hslash k\chi$.

If $4a^3+27b^2=0$ and $(a,b)\ne (0,0)$, there is a nonzero real number $r$ such thatThen $(0,0,r)$ is singular andFor $r>0$, the singular point is locally isolated because all terms inare nonnegative nearby. For $r<0$, the surface is locally a double cone with vertex at that point. Neither neighbourhood is homeomorphic to an open disc, so

LetBy the regular level set theorem, $F^{-1}(0)$ is a smooth surface wherever $\nabla F\ne 0$. A critical point on the level set would satisfyThese equations imply $4a^3+27b^2=0$. Therefore

The stated plan isIts six positive cells form a spanning tree of the supplier--consumer bipartite graph, so it is basic and feasible. Taking $u_1=0$, the basic-cell equations $u_i+v_j=c_{ij}$ giveThe complete matrix of reduced costs ${\bar{c}}_{ij}=c_{ij}-u_i-v_j$ isThe negative entry ${\bar{c}}_{12}=-2$ proves that the plan is not optimal.

Enter cell $(1,2)$. The alternating cycle isand the step is $\theta =\min (3,1)=1$. The new plan iswhose cost is $26$, down from $28$. New potentials areand the reduced-cost matrix isEvery reduced cost is nonnegative, so the transportation optimality criterion shows thatThe zero reduced cost in cell $(2,3)$ also indicates an alternative optimum.

A transportation problem chooses nonnegative shipments $x_{ij}$ from $n$ suppliers to $m$ consumers so that row sums equal supplies and column sums equal demands, while minimizing $\sum c_{ij}{x}_{ij}$.

The north-west corner rule givesIt has $3+4-1=6$ positive cells and its support contains no cycle, so it is a nondegenerate basic feasible solution.

A degenerate basic feasible solution isIts five positive cells form a forest with two balanced components; adding one zero cell that joins the components completes a basis of six cells. Thus at least one basic variable is zero.

Let $\nu =n/2-p$. Under $H_0$, the two residual sums of squares are independent and satisfyHenceUnder $H_1$, the numerator is scaled by the smaller variance, so small values provide evidence against the null. If $F_{\nu ,\nu }^{-1}(\alpha )$ denotes the lower $\alpha$ quantile, a size-$\alpha$ test rejects when

Write $\hat{\beta }=AY$, whereAny other linear unbiased estimator is $CY$ with $CX=I$, so $C=A+D$ and $DX=0$. The cross covariance vanishes becauseThereforeBy the Gauss-Markov theorem,

The estimator is an affine transformation of the multivariate normal distribution. Its mean is $\beta$, and direct covariance calculation gives

The log likelihood, up to constants, isDifferentiation gives the generalized least squares estimatorand maximizing over the scale givesThe divisor $n$ is appropriate for maximum likelihood rather than unbiased estimation.

The Dahlquist equivalence theorem says that a consistent linear multistep method is convergent exactly when it is zero-stable. HereThe consistency conditions hold for every $a$. The two quadratic roots have product one. They are distinct and on the unit circle exactly when $0<a<2$; at $a=2$ the root $-1$ is repeated, while outside this interval one root has modulus greater than one. The root condition for a multistep method therefore givesThe polynomial exactness conditions hold through degree two, while the degree-three residual is $6-a$. Every convergent member thus has(The exceptional value $a=6$ has formal order four but is not zero-stable.)

Insert the exact data $y(t)=P(t)$ into the linear multistep method. Its local residual is the linear functionalTaylor expansion about $t_n$ shows that a method has order $p$ exactly when this functional annihilates the monomials $1,t,\dots ,t^p$; by linearity this is equivalent to annihilating every polynomial of degree at most $p$. This is precisely

Set the similarity variableThenThe factors of $t$ cancel, leaving the self-similar ordinary differential equation

The vorticity is normal to the plane, so the stretching term vanishes. Axisymmetry also makes angular advection vanish, and part (a) gives no radial advection. Hence the scalar vorticity equation is the radial diffusion equation

Take the curl of the Navier-Stokes equation. The pressure term disappears, curl commutes with the Laplacian, and the incompressible vector identitygivesTherefore the material derivative form is

Using the polar-coordinate curl with $u_{\theta }=r\Omega (r)$ gives

Axisymmetry and incompressibility giveThus $r{u}_r=C$. Finiteness at the origin forces $C=0$, so

Take the wire along the $x$-axis and let $r$ be perpendicular distance from it. Gauss's law and cylindrical symmetry giveFor an observer boosted parallel to the wire,The moving observer sees length contraction and hence line density ${\lambda }^{\prime }=\gamma \lambda$, together with current $I^{\prime }=-\gamma \lambda v$. The transformed magnetic field is exactlywith the direction prescribed by the current. This is Ampère's law for the current seen in the moving frame.

Transforming the electromagnetic field tensor by $F^{\prime }=\Lambda F{\Lambda }^T$ and reading off its components givesThese are the parallel and transverse field transformation laws for a boost in the $x$ direction.

For $X^{\mu }=(ct,x,y,z)$, the stated Lorentz transformation gives