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The map $x\mapsto -x$ is an involution, soIf $\lambda$ is an eigenvalue of ${\mathrm{QFT}}_N$, then ${\lambda }^4=1$. Hence every eigenvalue belongs toBecause $X^4-1$ has distinct roots, the transform is diagonalizable; which fourth roots occur and their multiplicities depend on $N$.

Applying the transform twice givesBy the finite geometric series, the inner sum over $y$ is $N$ when $x+z\equiv 0(\mathrm{mod}N)$ and zero otherwise. Thereforewhich is the stated $|N-x⟩$ with the label interpreted modulo $N$.

For $N=4$, the binary label $10$ is the integer $2$, sowhere $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$ and $|-⟩=(|0⟩-|1⟩)/\sqrt{2}$. It is a product state and hence has no quantum entanglement.

For $\omega =e^{2\pi i/N}$, the quantum Fourier transform isIts matrix is unitary because character orthogonality for the cyclic group gives orthogonal rows.

SetThis is an elliptic function. Write $a=g_2/20$ and $b=g_3/28$ in the expansion from part (i). At a lattice point, the coefficient of $z^{-2}$ in $F$ is $g_2-20a=0$, and its constant coefficient is $g_3-28b=0$; there are no other singular terms. Every apparent pole is therefore a removable singularity, and $F$ extends to an entire elliptic function with value zero at the lattice points.

An entire elliptic function is bounded on a fundamental parallelogram and hence everywhere. By Liouville theorem it is constant, and its value at zero is zero. Thus $F\equiv 0$, proving the Weierstrass elliptic differential equation

For each nonzero period lattice point $\omega$, the binomial series givesPairing $\omega$ with $-\omega$ cancels every odd power. Hence the Laurent series of the Weierstrass elliptic function iswhich is the stated coefficient formula.

Eliminate $R=N-S-I$ to obtain the two-dimensional autonomous differential equationIts Jacobian matrix at the endemic equilibrium isSince $I^{\ast }>0$,The trace-determinant stability criterion shows that both eigenvalues have negative real part. The Endemic equilibrium of the SIR model with waning immunity is therefore locally asymptotically stable.

At an endemic equilibrium, $I^{\ast }>0$, so $0=\dot{I}=I^{\ast }(\beta S^{\ast }-\gamma )$ forcesThe equation $\dot{R}=0$ gives $R^{\ast }=\gamma I^{\ast }/\alpha$. The conservation law $S^{\ast }+I^{\ast }+R^{\ast }=N$ therefore yieldsA positive endemic state exists exactly when $\beta N>\gamma$, equivalently when the basic reproduction number ${\mathcal{R}}_0=\beta N/\gamma$ exceeds one.

Adding the three equations of the SIR model with waning immunity givesHence conservation of population holds: $N=S+I+R$ is an integral of motion.

The hat matrix $P$ is an orthogonal projection of rank $p$, so $I-P$ is an orthogonal projection of rank $n-p$. Because $\epsilon /\sigma$ is a standard multivariate normal distribution vector, Cochran's theorem givesThus ${\hat{\sigma }}^2\stackrel{d}{=}({\sigma }^2/n){\chi }_{n-p}^2$. Substituting this into the preceding formula gives the Distribution of the Akaike information criterion in a normal linear model:

For the normal linear model, the Gaussian likelihood givesMaximization over $\beta$ gives the ordinary least squares estimator $\hat{\beta }$, and maximization over the variance givesAt this value the quadratic term in $-2\ell$ is $n$. There are $p$ regression coefficients and one variance parameter, so substitution into the Akaike information criterion yields

Conditional independence makes the log-likelihoodA maximum-likelihood estimator is any $\hat{\theta }\in {\mathrm{argmax}}_{\theta \in \Theta }\ell (\theta )$. If the model has $q$ fitted real parameters, its Akaike information criterion is

The language $L=\{a^{m^2}{b}^{m^2}:m\ge 0\}$ is not context-free. Suppose it had pumping length $p$ in the pumping lemma for context-free languages, and apply the lemma to $w=a^{p^2}{b}^{p^2}$. Write $w=uvxyz$ with $|vxy|\le p$, $|vy|>0$, and $u{v}^{i}x{y}^{i}z\in L$ for every $i\ge 0$.

Let $\alpha$ and $\beta$ be the respective numbers of $a$s and $b$s in $vy$. If $\alpha \ne \beta$, pumping with $i=0$ or $i=2$ destroys equality of the two block lengths. Hence a valid decomposition would require $\alpha =\beta =d>0$. Pumping with $i=2$ would then produce $a^{p^2+d}{b}^{p^2+d}$, where $1\le d\le p$. Butso $p^2+d$ is not a square number. This contradiction proves the square-count language is not context-free.

A context-free grammar is a quadruple $G=(V,\Sigma ,P,S)$ consisting of a finite set $V$ of nonterminal symbols, a disjoint finite alphabet $\Sigma$ of terminal symbols, a start symbol $S\in V$, and a finite set $P$ of productions $A\to \alpha$ with $A\in V$ and $\alpha \in (V\cup \Sigma {)}^{\ast }$. A sentence is a terminal word over an alphabet $w\in {\Sigma }^{\ast }$ with $S{\Rightarrow }^{\ast }w$, and the generated formal language is $\mathcal{L}(G)=\{w\in {\Sigma }^{\ast }:S{\Rightarrow }^{\ast }w\}$.

The first language is context-free. For example, the productionsgenerate exactly $\{a^{2m}{b}^{2m}:m\ge 0\}$: each recursive production adds two $a$s to the left and two $b$s to the right.

Insert the Fourier modeinto the leapfrog advection scheme. This givesand hence the amplification polynomialIts roots areIf $0<\mu <1$, the square root is real for every $\theta$, andIf $\mu >1$, choosing $|\sin \theta |>1/\mu$ makes the roots non-real multiples of $i$ with reciprocal moduli, one of which exceeds one, so the method is unstable.

The endpoint $\mu =1$ requires care. At $\theta =\pi /2$ the polynomial isso a generic mode has the form $(c_1+c_{2}n)i^n$ and grows linearly. It satisfies the weaker test $|{\zeta }_{\pm }|\le 1$ often quoted in elementary Fourier analysis, but it violates the root condition for a multistep method because the unit-modulus root is repeated. Therefore the rigorous uniform stability range for arbitrary two-level initial data isIf “stable” is being used only for the non-amplification condition on root moduli, the conventionally quoted range is $0<\mu \le 1$, with $\mu =1$ a defective marginal endpoint rather than a uniformly stable case.

Let $D\in {\mathbb{R}}^{M\times M}$ be the Toeplitz antisymmetric tridiagonal matrix with $D_{m,m+1}=1$, $D_{m,m-1}=-1$, and zero diagonal. Fixed homogeneous endpoint values give the homogeneous updateThe hint gives the eigenvaluesBecause each $d_j$ is purely imaginary, $1-(\mu /4)d_j\ne 0$ for every real $\mu$, so the left matrix is invertible. The two matrices are polynomials in $D$ and therefore have the same orthonormal eigenvectors. The amplification matrixhas eigenvaluesThe numerator and denominator are complex conjugates, so $|q_j|=1$. Moreover, the common orthonormal eigenbasis makes $Q$ a normal matrix. By the matrix 2-norm of a normal matrix,for every $n$. Thus the Crank-Nicolson centered-advection scheme on a finite interval is unconditionally stable:

A complex square matrix $A$ is a normal matrix if it commutes with its conjugate transpose:By unitary diagonalization of a normal matrix, there are a unitary matrix $U$ and a diagonal matrix $\Lambda =\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_N)$ such thatThe matrix 2-norm is invariant under multiplication by unitary matrices, soFor a diagonal matrix,with equality on a coordinate vector belonging to an eigenvalue of greatest modulus. Hencethe matrix 2-norm of a normal matrix.

The stationary contribution travels with group velocityso $k_0=4V^2/(9S^2)$. The conditionis therefore equivalent, up to a constant factor, to $k_0ϵ\ll 1$: the dominant wavelength is much longer than the impacted interval. Its Fourier transform can then be replaced bywhich is the total initial vertical velocity integrated over space. The far field cannot resolve $W$ and $ϵ$ separately and depends only on their product $Wϵ$.

Put $x=Vt$ with fixed $V>0$. The relevant stationary point will have $k=O(V^2/S^2)$, so the assumption $ϵV^2/S^2\ll 1$ allows the long-wave approximationUsing $\sin A\cos B=[\sin (A+B)+\sin (A-B)]/2$, the phase $S{k}^{3/2}+Vk$ has no stationary point for $k>0$. The other phase iswithandThe stationary part of the integral is thereforeBy the one-dimensional stationary-phase formula, this is asymptotic toSincewe obtainUsing $V=x/t$, this has the requested formwhere

Use the Fourier transform conventionThe deep-water capillary-wave dispersion relation makes each Fourier mode satisfyThe transformed initial data areThereforeThe integrand is even, so Fourier inversion gives the surface response to a localized impulsive velocity