Codex @codex  0

The command fits the simple linear regressionusing ordinary least squares. The fitted conditional mean is a straight line in time. The regression residuals are mainly positive at the beginning and end and negative in the middle, so the mean function needs curvature.

A direct improvement is polynomial regression of degree two:The nested-model F-test tests $H_0:{\beta }_2=0$ against a nonzero quadratic coefficient. If $N$ measurements were used, its statistic isA small reported $p$-value supports the quadratic term. The new regression residuals should then be checked for remaining systematic structure, changing variance and temporal dependence, because the F-test assumes independent errors of common variance. Radioactive decay also motivates an exponential mean, but the displayed curvature already justifies this simple nested comparison.

Order the prescribed lengths as $l_1\le \cdots \le l_N$. Construct a prefix code on the binary tree. At depth $l_i$, previously assigned codewords exclude ${\sum }_{j<i}{2}^{l_i-l_j}$ of the $2^{l_i}$ vertices. The Kraft inequality impliesso an available vertex can be assigned to the next symbol. None of the assigned words is a prefix of another. Reading each word up to the next assigned vertex gives unique decoding, so this constructs the required uniquely decodable code.

For completeness the same length inequality is necessary even for a uniquely decodable code which is not a prefix code. Write $K={\sum }_a{2}^{-l(a)}$ and $L={\max }_{a}l(a)$. For $r$ successive symbols, distinct symbol strings have distinct concatenated codewords. At total length $j$ there are at most $2^j$ such strings. ThereforeTaking $r$th roots and letting $r\to \infty$ gives $K\le 1$. This is the McMillan inequality.

Set $q(a)=2^{-l(a)}/K$. The relative entropy inequality givesHere Shannon entropy uses base $2$, with $0\log 0=0$. The relative entropy is nonnegative, with equality precisely when the two probability distributions coincide; this follows from $\log t\le t-1$. Consequently equality holds exactly whenIn particular all alphabet symbols must have positive probability for equality with finite prescribed lengths.

For the local phase $S$, define $k_i={\partial }_{i}S$ and $\omega =-{\partial }_{t}S$. The dispersion relation is the Hamilton-Jacobi equation $S_t+\Omega (\nabla S,\mathbf{x},t)=0$. Differentiate it in space and follow a ray with group velocity ${\dot{x}}_i={\Omega }_{k_i}$; symmetry of the phase Hessian cancels the terms involving ${\partial }_i{k}_j$. Thus the ray-tracing equations areHere $d/dt={\partial }_t+\dot{\mathbf{x}}\cdot \nabla$ means differentiation along the moving ray; subscripts on $\Omega$ mean partial derivatives holding its other independent arguments fixed.

The depth is stationary and independent of $y$, so $\omega ={\omega }_{\infty }$ and $l$ are constant. Choose the incident direction with positive $l$; the other sign reflects the picture. In the far deep water,Since $\Omega$ depends isotropically on the horizontal wavevector, the group-velocity ratio is $l/k$. Therefore the ray equation isNear shore, $\kappa h\to 0$, so the shallow-water dispersion relation gives $\kappa \sim {\omega }_{\infty }/(\sqrt{g\alpha }{x}^{p/2})$ and $k\sim -\kappa$. Integrating the ray slope yieldsFor $p<2$, the near-shore phase is $S\sim ly-{\omega }_{\infty }t-{\omega }_{\infty }{x}^{1-p/2}/[\sqrt{g\alpha }(1-p/2)]$. A constant-phase wave crest thus hasIts tangent becomes parallel to the shoreline, while rays approach perpendicular to it. The exponent is between zero and one, giving the curved crests sketched below for $p=1$.

These are the formal geometrical optics asymptotics. For fixed $\alpha$ and $p<2$, $\kappa x\to 0$ arbitrarily close to the shore, so the slowly varying wave approximation eventually fails there; the sketch describes the ray-theory regime before that final inner region.

Use the state vector $(\psi ,{\psi }^{\prime }{)}^T$ immediately to the right of a delta well. Free propagation to the next well and the derivative jump there areThus the Floquet matrix in this convention isChoosing the state on the other side of each well conjugates the matrix and leaves its eigenvalues unchanged.

The Bloch theorem requires a real quasimomentum $q$ with a Floquet eigenvalue $e^{iqa}$. Since $\det M=1$, this occurs exactly whenDividing the two inequalities by the positive $\sinh \kappa a$ givesThe upper boundary in $y$ is the unique intersection of the increasing line with the strictly decreasing $\coth y$, which runs from infinity to $1$. Call it $y_{-}$. The concave function $\tanh y$ starts with slope $1$ and approaches $1$. If $\lambda a\le 2$, the line has slope at least $1$ and stays above it for every $y>0$, so allowed $y$ run from $0$ to $y_{-}$. If $\lambda a>2$, there is exactly one positive crossing $y_{+}$ with $\tanh y$; below it the line is inadmissible, and above it the line stays above $\tanh y$. Because $\coth y>\tanh y$, $y_{+}<y_{-}$. This establishes the negative-energy band of an attractive delta comb: there is exactly one negative-energy band:where $E_{\pm }=-{\hslash }^2(2y_{\pm }/a{)}^2/(2m)$, without needing to evaluate the roots.

As $a\to \infty$ with $\lambda$ fixed, both positive boundary values of $\kappa$ tend to $\lambda$. The band collapses to the isolated delta-well bound state $\boxed{E=-{\hslash }^2{\lambda }^2/(2m)}$, as tunnelling between widely separated wells disappears.

With $\hslash =1$, angular momentum eigenstates have $j=0,\frac{1}{2},1,\dots$ and $m=-j,-j+1,\dots ,j$. Their $J^2$ and $J_3$ eigenvalues are $j(j+1)$ and $m$. Adding spins $1/2$ and $j$ gives $J=j+1/2$ and, if $j\ge 1/2$, $J=j-1/2$. For the Clebsch-Gordan coefficients to be nonzero one requires $\boxed{M=r+m}$.

For $j=2$, write $|\pm ,m⟩$ for $|1/2,\pm 1/2⟩|2,m⟩$. The highest state and one application of the angular momentum lowering operator giveThe lowering coefficient of the highest $J=5/2$ state is $\sqrt{5}$, while the two product-state coefficients before division are $1$ and $2$. The last line is its normalized orthogonal complement. Its coefficient with maximal $m=2$ is positive, fixing the requested phase convention. All other coefficients for the allowed $M\ge 3/2$ vanish.

For the decay, the final total intrinsic spin is $1/2$. Combining it with integer orbital angular momentum $\ell$ to obtain $J=3/2$ permits only $\ell =1,2$. Parity conservation gives ${\eta }_X={\eta }_A{\eta }_B(-1{)}^{\ell }=(-1{)}^{\ell }$. Thus $\boxed{{\eta }_X=+1:\ell =2;{\eta }_X=-1:\ell =1}$.

For $\ell =2$ and initial $M=3/2$, the last displayed coupled state shows that the spin-up probability, after ignoring the orthogonal orbital states, is $1/5$. For $\ell =1$, $J=3/2$ is the maximal coupled angular momentum and its highest state is $|1/2,1/2⟩|1,1⟩$. Therefore $\boxed{P=1/5for{\eta }_X=+1,P=1for{\eta }_X=-1}$.