Writing the log-likelihood as $\ell (\beta )=\log L(\beta )$, the Fisher information matrix is the expected negative Hessian matrixUnder the usual regularity conditions it is equivalently the expected outer product of the score function.

Introduce nonterminals for the terminals and one for the pair $bb$:Because the original start symbol occurs on a right-hand side, take a fresh start symbol $S_0$. After eliminating the unit production $S\to T$, one grammar in Chomsky normal form isThe original grammar generates exactlyThe displayed grammar generates the same set: $AS$ adds an $a$ on the left, $SU$ adds one $bb$ pair on the right, and $CC$ terminates the derivation with $cc$. Since the original grammar does not generate $ϵ$,

Use the CYK algorithm. For a nonempty input word $v=v_1\cdots v_n$, let $T_{i,j}$ be the set of nonterminals that derive the substring $v_i\cdots v_j$. InitializeFor increasing substring lengths, computeAccept exactly when the start symbol belongs to $T_{1,n}$. The proof is a mathematical induction on substring length: a one-letter derivation uses a terminal production, while every longer parse tree has a root production $A\to BC$ whose two subtrees yield two adjacent nonempty substrings. The recurrence considers every possible split point and therefore finds exactly all possible roots. For a fixed grammar the algorithm uses $O(n^3)$ time. Under the strict form in part (a), it rejects $ϵ$ immediately.

A context-free grammar is in Chomsky normal form when every production is of one of the formswhere $A,B,C$ are nonterminal symbols and $a$ is a terminal symbol. Under this strict convention there is no epsilon production, so such a grammar cannot generate the empty word. Conversion of an arbitrary grammar therefore givesIf $ϵ\notin \mathcal{L}(G)$, the two languages are equal. An alternative extended convention permits the exceptional start production $S_0\to ϵ$; under that convention conversion preserves the whole language.

Let $L=[1,-2,1]$ be the symmetric Dirichlet second-difference matrix on the $M$ interior points. The method isThe orthogonal sine vectors diagonalize $L$, with eigenvaluesBecause the two time-level matrices are real symmetric polynomials in the same symmetric matrix, they share this orthonormal eigenbasis. Whenever the left matrix is invertible, the amplification matrix is therefore normal, so its Euclidean operator norm is the largest modulus of its eigenvalues. The $j$th amplification factor isFor the physical Courant number $\mu >0$,Hence $|G_j|\le 1$ exactly when $a-c{s}_j\ge 0$. Requiring this uniformly for every mesh, whose values $s_j$ become dense in $(0,1)$, givesEquivalently, $a\ge 0$ and $c\le a$. Under these inequalities the denominator is nonzero for every $0<s<1$: if $c\le \mu$ this is immediate, while if $c>\mu$ then $a\ge c$ gives $a+(\mu -c)s\ge a+\mu -c\ge \mu >0$. Thus the necessary and sufficient mesh-uniform stability conditions areThis is the stability of a two-parameter implicit-explicit diffusion scheme. If consistency with the stated diffusion equation is also required, comparison of the leading Taylor terms gives $ak{u}_t=(k/2)u_{xx}+o(k)$, so $a=1/2$ and the stable consistent family has $c\le 1/2$.

Let $U_m^n=u(mh,nk)$ be the sampled exact solution and let $A_h$ denote the update matrix of the Forward Euler diffusion scheme. For $0\le \mu \le 1/2$,is a convex combination, with the homogeneous boundary values included. ThereforeThis is the parabolic discrete maximum principle and proves max-norm stability directly.

The Taylor theorem and the second-order central difference give the exact-grid residualfor a sufficiently smooth solution on a fixed interval $0\le t\le T$. If $e^n=U^n-u^n$, thenIteration for $nk\le T$ yieldsThus $\mu \le 1/2$ implies convergence when the initial grid values converge to the initial data.

If $p>1/2$, complementing every received bit converts the channel into a binary symmetric channel with crossover probability $1-p<1/2$. This deterministic and invertible output relabelling does not change the channel capacity. Since $h_2(p)=h_2(1-p)$,Thus the capacity rises again from zero as $p$ increases beyond $1/2$, reaching one at $p=1$, where every bit is flipped deterministically and can be recovered by complementation.

At $p=1/2$, the output of the binary symmetric channel is independent of its input. It conveys no mutual information, and therefore

A reliable transmission rate $R$ means that length-$n$ codes can carry $M_n$ messages with $n^{-1}{\log }_2{M}_n$ tending to at least $R$ while the probability measure of the decoding-error event tends to zero. By the Shannon second coding theorem, the supremum of reliable rates is the binary symmetric channel capacityFor small positive $p$, the binary entropy function has the asymptotic expansionso $C(p)$ starts at one bit per channel use and decreases rapidly from one as errors are introduced.

Let $Z_j={\rho }_j{c}_j$, $k_j=\omega /c_j$, and $\varphi =k_{1}L$. Use the harmonic convention $e^{-i\omega t}$ and write the pressure and velocity amplitudes asContinuity of pressure and normal velocity at both interfaces is equivalently the layer transfer relationSubstituting $p(0)=1+R$, $u(0)=(1-R)/Z_0$, $p(L)=T$, and $u(L)=T/Z_0$, then adding the two resulting equations, givesHence the acoustic transmission through a uniform layer isFor either $\lambda \ll 1$ or $\lambda \gg 1$, the strong acoustic impedance mismatch makes transmission small except near $k_{1}L=n\pi$, where $|T|=1$. Away from those resonances,At the antiresonant points $k_{1}L=(n+\frac{1}{2})\pi$, the minimum is $2/(\lambda +{\lambda }^{-1})$.

For a progressive plane wave, writeThe momentum equation gives ${\rho }_0\omega U=kP$, while the dispersion relation is $\omega =c_{0}k$. Therefore the pressure and longitudinal velocity amplitudes obeyThe constant $Z$ is the acoustic impedance.

The linear homentropic acoustic equations areDifferentiate the first equation in time, take the divergence of the second, and use the third. This eliminates ${\rho }^{\prime }$ and $\mathbf{u}$ and gives the acoustic pressure wave equation

Use the axisymmetric streamfunction conventionFor the similarity solution $\psi =\nu zg(\eta )$, $\eta =r/z$, regularity on the axis requires $g(0)=g^{\prime }(0)=0$, while matching to still ambient fluid requires $g^{\prime }(\eta )/\eta \to 0$ as $\eta \to \infty$. The given equation has the first integralThe axis conditions give $C=0$, and separation yieldsfor some $a>0$.

The axial velocity isNormalizing by the conserved momentum flux givesso $a=3M/(32{\nu }^2)$. The similarity solution for a laminar round jet is therefore

Use the continuity equation to write the axial momentum equation in conservative form:Integrating from the axis to infinity, regularity at $r=0$ and decay of $u,w,w_r$ as $r\to \infty$ eliminate the boundary terms. Therefore the conserved momentum flux of a laminar round jet satisfies

Let $W(z)$ be the centre-line velocity scale and $\delta (z)$ the jet width. Conservation gives $M\sim W^2{\delta }^2$, hence $W\delta \sim M^{1/2}$. The axial inertial and viscous terms scale asso $W{\delta }^2\sim \nu z$. Combining the two relations givesThus the width grows linearly and the centre-line speed decays as $z^{-1}$. The volume flux scales asso it increases downstream. By fluid entrainment, the extra flux must be ambient fluid drawn radially into the jet.

The states $|2⟩$ and $|3⟩$ are degenerate and couple to $|1⟩$ with amplitudes $\epsilon V_0$ and ${\epsilon }^2{V}_0$. The bright and dark state reduction of a star-coupled Hamiltonian gives one exact dark eigenstate proportional to $-\epsilon |2⟩+|3⟩$, with energy $E$. The orthogonal bright state couples to $|1⟩$ with magnitudeDiagonalizing the remaining two-dimensional Hermitian matrix gives the exact spectrumWriting $\Delta =E_1-E\ne 0$ and expanding the branches according to the eigenvalue to which they converge,These are exactly the three perturbative energies from part b through order ${\epsilon }^2$.