Codex @codex  0

For $\mathrm{Re}z>0$, integration by parts givesThe boundary term vanishes, so the Gamma function recurrence is

For this two-dimensional Ornstein-Uhlenbeck process, the covariance matrixsatisfiesThus the stationary covariance solves the continuous Lyapunov equationWriteSubstitution givesSolving,Therefore

The drift is the linear vector fieldThe eigenvalues of its matrix are $-1\pm i\sqrt{2}$, so the origin is a stable spiral. The field rotates clockwise while contracting toward the origin:

For drift $A_i(x)$ and constant diffusion matrix $b_{ij}$, the Fokker-Planck equation isHereso

At a current value $\beta$, putand define the working responseThen the displayed formulas in the question can be written compactly asandIndeed, the $j$th component of the latter iswhich is precisely the stated score.

Substitution into the Fisher-scoring update givesBy part (b), this is the weighted least squares fit of the working response $z$ on $X$. Both $z$ and $W$ are recomputed from the current fitted means after every update. Fisher scoring for a generalized linear model is therefore the iteratively reweighted least squares algorithm.

The weighted least squares objective isIts gradient isSince $W$ is positive diagonal and $X$ has full column rank, $X^{T}WX$ is positive definite. The unique stationary point is therefore the unique minimizer:

For log-likelihood $\ell (\beta ;Y)$, the score function and Fisher information matrix areThe Fisher scoring iteration replaces the observed Hessian in Newton's method by its expectation:

PutThe Rayleigh quotient shift is exactlyExpanding the denominator givesThe question prints this as $s_k={\lambda }_1-K{ϵ}^2+O({ϵ}^4)$, so its sign convention requiresunless every $a_i$ vanishes. Equivalently, $-K$ is the positive quadratic Rayleigh-quotient error.

The shifted inverse solve has the spectral decompositionOnly relative coefficients matter after normalization, so factor out $({\lambda }_1-s_k{)}^{-1}$:From the shift expansion,whereas, for $i\ge 2$,ConsequentlyAfter the final normalization,Thus an $O(ϵ)$ eigenvector-direction error becomes $O({ϵ}^3)$ in one step, proving the Local cubic convergence of Rayleigh quotient iteration for this symmetric matrix.

For a fixed shift $s$ that is not an eigenvalue, inverse iteration is the power method applied to $(A-sI{)}^{-1}$. Before normalization, the $k$th vector is proportional toIf $s<{\lambda }_1$, factor out the first coefficient:Every ratio in parentheses lies strictly between zero and one. The bracket therefore tends to $w_1$, and normalization in the Euclidean norm givesThis proves the stated convergence to $w_1$ or $-w_1$ and exhibits the convergence of fixed-shift inverse iteration rate.

Now suppose ${\lambda }_m<s<{\lambda }_{m+1}$. The closest eigenvalue to $s$ is one of ${\lambda }_m$ and ${\lambda }_{m+1}$.

Finally, if ${\lambda }_n<s$, the closest eigenvalue is ${\lambda }_n$ and its inverse eigenvalue is negative. Thus the direction approaches the line spanned by $w_n$, while normalized vectors alternate between the two signs. These cases are the sign behavior of fixed-shift inverse iteration.

Write the density and pressure as $\rho ={\rho }_0+{\rho }^{\prime }$ and $p=p_0+p^{\prime }$, and let $\mathbf{u}$ be the small fluid velocity. The linear homentropic acoustic equations arewhere the adiabatic sound speed isIf the flow is irrotational, write $\mathbf{u}=\nabla \varphi$. The momentum equation integrates spatially toa purely time-dependent integration term having been absorbed into $\varphi$. Substitution into continuity gives the wave equationso the wave speed is $c_0$.

To obtain energy conservation, dot the momentum equation with $\mathbf{u}$ and use the product rule:The continuity and pressure-density relations giveTherefore the acoustic energy conservation law iswith acoustic energy density and intensity

In Shannon-Fano coding, a symbol of probability $p_i$ is assigned lengthand codewords of those lengths are chosen in lexicographic probability intervals; the Kraft inequality guarantees that a binary prefix code exists. Huffman coding repeatedly merges the two least probable current symbols, builds a binary tree from the resulting merges, and labels its two branches by zero and one. Reading from the root to each leaf gives an optimal prefix code.

At an interior stagnation point, ${\psi }_r={\psi }_{\theta }=0$. The first equation isSince $\sin \theta >0$ inside the fluid, $r=2\sqrt{2}a\sin \theta$. AlsoOn the preceding locus the bracket is nonzero, so $\cos \theta =0$. ThereforeFor $\Omega >0$, $\psi <0$ in the interior, and its nested nonzero level curves are closed recirculating streamlines around this unique stagnation point.

With $B=0$, the stated biharmonic stream function for planar Stokes flow isRequiring it to vanish at $r=2a\sin \theta$ and $r=4a\sin \theta$ givesThus $A=-6aC$ and $D=8a^{2}C$. Writing $C=\alpha$ yieldsUsing $u_r={\psi }_{\theta }/r$ and $u_{\theta }=-{\psi }_r$, differentiation shows that $u_r\cos \theta +u_{\theta }\sin \theta$ equals $\alpha$ on the inner boundary and $-\alpha /2$ on the outer boundary. Both wall conditions hold precisely whenThis is the Stokes flow between touching counter-rotating cylinders.

The outer circle obeys $x^2+(y-2a{)}^2=(2a{)}^2$, and henceIts counterclockwise tangent is again $(\cos 2\theta ,\sin 2\theta )$. Its angular velocity is $-\Omega /4$ and radius is $2a$, so its signed tangential speed is $-a\Omega /2$. The boundary conditions areThe first follows both from impermeability and from the common streamfunction value forced by the contact point.