Define the reflected function $\varphi (x)=\psi (-x)$. The chain rule gives ${\varphi }^{\prime }(x)=-{\psi }^{\prime }(-x)$ and ${\varphi }^{\prime \prime }(x)={\psi }^{\prime \prime }(-x)$. Evaluating the time-independent Schrödinger equation for $\psi$ at $-x$ and using the even function identity $V(-x)=V(x)$ givesThus reflection preserves the energy eigenvalue. The same substitution $y=-x$ in the norm givesThis is the action of the parity operator $R\psi (x)=\psi (-x)$, which is a unitary operator. For a reflection-symmetric physical problem it preserves the Hamiltonian operator's operator domain as well as its differential expression.

The time-independent Schrödinger equation is the energy eigenvalue equation for the Hamiltonian operator:Here $\hslash$ is the reduced Planck constant. A corresponding stationary wavefunction evolves as $\Psi (x,t)=\psi (x)e^{-iEt/\hslash }$, which follows by substitution in the Time-dependent Schrödinger equation. Its position probability density is time independent because the time factor has unit modulus. Physical energy eigenstates must also satisfy the system's boundary conditions; for a bound state, wavefunction normalization requires ${\int }_{-\infty }^{\infty }|\psi (x){|}^{2}dx=1$.

Let $a={\mathbb{E}}_{\psi }(A)$, which is real. The dispersion, or quantum uncertainty, is the nonnegative standard deviationIndeed, self-adjointness and normalization giveA squared norm is nonnegative, so the square root is defined. It vanishes precisely when $(A-aI)\psi =0$, namely when $\psi$ is an eigenvector of $A$. Conversely, an eigenvector with eigenvalue $\lambda$ has $a=\lambda$ and hence zero dispersion. Thus zero dispersion is equivalent to being an eigenstate.

A normalized quantum state satisfies $⟨\psi ,\psi ⟩=1$. Its expectation value is $\boxed{{\mathbb{E}}_{\psi }(A)=⟨\psi ,A\psi ⟩}$. The conjugate symmetry of the inner product and self-adjointness giveso the expectation value is real. If $A\varphi =\lambda \varphi$ with $\varphi \ne 0$, normalize $\varphi$ by dividing by its norm. ThenEvery eigenvalue of a self-adjoint operator is real.

The commutator of two linear operators is $\boxed{[A,B]=AB-BA}$. Write $T^{†}$ for the adjoint operator, using the inner product convention linear in its second argument. Since $A^{†}=A$ and $B^{†}=B$,Thus the commutator is anti-self-adjoint. With $C=-i[A,B]$, conjugation of the scalar givesAll products are understood on the stated common operator domain; for physical unbounded linear operators, the state must lie in the relevant product domains and have the finite moments used below.

For identical fermions the full wave function is antisymmetric:In particular it vanishes when the two complete one-particle arguments coincide. Applying this to the position-only model assumes that no independent spin symmetry changes the required spatial parity.

For identical bosons the full wave function is symmetric under the particle exchange operator:In the given position-only model this is the spatial parity under the particle exchange operator. If additional spin variables are present, the symmetry applies to the complete state including them.

Compare every complete row with the given vector, returning as soon as a match is found. The length check prevents recycling of the vector; seq_len also handles an empty matrix correctly. This implementation treats a missing comparison as a nonmatch and always returns a Boolean.

row.match <- function(X, y) {
  stopifnot(is.matrix(X), length(y) == ncol(X))
  for (i in seq_len(nrow(X))) {
    eq <- X[i, ] == y
    if (all(!is.na(eq) & eq)) return(TRUE)
  }
  FALSE
}

For the integer-coordinate matrices used below there are no missing values, so this is ordinary exact row equality.

In the $(u,v)$ plane the feasible set lies in the first quadrant above the three lines $v=3/2$, $v=5/2-2u$ and $v=2-(2/3)u$. Its lower boundary has verticesand the region is unbounded upward and to the right. The objective at these vertices is respectively $21$, $43/2$ and $55/2$, so $A$ is optimal for the original supply.

Let the available amount of R be $r\ge 0$. Comparing the three dual vertex objectives gives the optimal valueThus the marginal shadow price of R is $3/4$ below $22/3$, then $3/8$ between $22/3$ and $22$, and zero beyond $22$. At the two breakpoints, optimal dual prices range between the adjacent one-sided prices: $[3/8,3/4]$ and $[0,3/8]$ respectively. At zero supply the right marginal value is $3/4$; dual multipliers on the zero resource can be nonunique. These slope changes describe why the first extra $4/3$ units from the original supply six have the original price, but subsequent units do not.

The dual prices $u,v\ge 0$ solveThe optimal tableau gives $(u,v)=c_B^T{B}^{-1}=(6,3)B^{-1}=(3/4,3/2)$. This vector is dual feasible and has objective $21$, equal to the primal value, proving optimality.

Thus an extra unit of R is worth $3/4$ while the current basis stays feasible. Increasing its supply by $\delta$ changes the basic values toFor increases, feasibility is retained exactly up to $\delta =4/3$. Therefore the firm can buy up to $4/3$ extra units at a marginal value of $3/4$ per unit. The reduced profits do not change during this right-hand-side variation.

For the proposed fourth good, the reduced profit is $8-(4u+4v)=8-9=-1$. Its profit is below the resource opportunity cost, so it is not worth producing at the current optimum. The old dual solution also satisfies the new good's inequality, certifying that introducing it cannot improve the optimum.

With nonnegative production amounts $x_1,x_2,x_3$, the primal linear programme isIntroduce slacks $s_R,s_S\ge 0$. The initial basic variables are the slacks, and the reduced profits of the goods are $5,6,3$. The greatest is six, so $x_2$ enters. The ratio test compares $6/2=3$ and $11/3$, so $s_R$ leaves. The resulting dictionary iswith profit $18-7x_1+3x_3-3s_R$. The greatest positive reduced profit is now three, so $x_3$ enters; its only positive pivot coefficient is two in the second row, and $s_S$ leaves. The final tableau, expressed as equations, isAll nonbasic reduced profits are nonpositive, proving optimality atThe ordered optimal basis is $(x_2,x_3)$, henceThe inverse is read from the two slack columns of the final basic equations.

Let $\varphi$ denote the standard normal density. Under the shifted sampling density $q(y)=\varphi (y-c)$, the likelihood ratio is $\varphi (y)/q(y)=e^{-cy+c^2/2}$. Simulate independent $Y_k\sim N(c,1)$ and useMultiplying the weight by $q$ restores the original density, so each summand $H$ has expectation ${\int }_c^{\infty }\varphi (y)dy=z_c$. The estimator is unbiased. For the second moment,because $y\ge c$. Since $n{v}_c={\mathbb{E}}_q{H}^2-z_c^2$,For $c>0$, $e^{-c^2/2}<1$ and $z_c>0$, so $n{v}_c<z_c-z_c^2=n{v}_0$, proving $v_c<v_0$. Nonnegativity of variance also implies $z_c^2\le e^{-c^2/2}{z}_c$; division by the positive $z_c$ gives

For the independent standard normals stipulated in this part, simulate $n$ values and average the exceedance indicators:Each indicator is Bernoulli with probability $z_c$, so expectation and independence giveThus the Monte Carlo estimator is unbiased.

The assertion is false for the algorithm as printed. Its second output depends only on $V$. For $0<V\le 1/2$ its sine factor is nonnegative. For $1/2<V<1$, its radius is at most $\sqrt{2\log 2}$, so in every case $Y\ge -\sqrt{2\log 2}$. Therefore $\mathrm{Pr}(Y<-2)=0$, whereas a standard normal variable has positive probability below $-2$. The printed output cannot be a pair of independent standard normals.

The minimal correction is to use $U$ in both logarithms. Let $R=\sqrt{-2\log U}$, $\Theta =2\pi V$, and set $X=R\cos \Theta$, $Y=R\sin \Theta$. Independence of $U,V$ gives independent $R,\Theta$ with densities $r{e}^{-r^2/2}$ for $r>0$ and $1/(2\pi )$ for the angle. The polar-coordinate Jacobian is $r$, so the joint Cartesian density isThis proves independent standard normal outputs for the corrected Box-Muller transform. The private transcription retains the source's original $\log V$ and records this correction separately.

Let $k=\chi (G)$. For $k\le 1$ the bound is immediate. Otherwise choose a vertex-minimal subgraph $H$ with chromatic number $k$. Each vertex has degree at least $k-1$ in $H$: if one had degree at most $k-2$, a $(k-1)$-colouring of the smaller graph obtained by deleting it could be extended to it using an absent neighbour colour, contradicting $\chi (H)=k$.

The graph $H$ has at least $k$ vertices. By the degree-sum identity,Solving the quadratic inequality gives