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For $V=\lambda {\varphi }^4/4$ and $\varphi >0$, putThe slow-roll approximation gives $H\simeq B{\varphi }^2$ andTherefore, with $\gamma =c\sqrt{\lambda /(6\pi G)}$,Since $B/\gamma =2\pi G/c^2$, integration of the Hubble parameter gives

Under the slow-roll approximation,For $V=m^2{\varphi }^2/2$ and the branch $\varphi >0$, setThen $H\simeq Am\varphi$ andWith $\varphi (0)={\varphi }_0$ and $a(0)=a_0$,and integrating $\dot{a}/a=H$ gives

Along a solution, the chain rule and antisymmetry of $K$ givebecause $v^{T}Kv=0$ for every real vector $v$. ThereforeThe closed level curves around the strict minimum at the coexistence equilibrium are the periodic population cycles of the Lotka-Volterra predator-prey model.

Since $N_i=N_i^{\ast }{e}^{x_i}$, division of each population equation by $N_i$ givesDefine the Hamiltonian functionand the antisymmetric matrixThen

At a nonzero fixed point, both bracketed per-capita growth rates vanish:Hence the coexistence equilibrium of this Lotka-Volterra predator-prey model is

Let ${\epsilon }_i$ be independent standard logistic random variables and define the latent variableSet $Y_i=1$ exactly when $Z_i>0$. If $F$ is the logistic cumulative distribution function, its symmetry and the identity $F^{-1}(p)=\mathrm{logit}(p)$ givewhich is precisely the fitted logistic regression.

Two packs contain $40$ cigarettes. Holding height fixed, increasing average consumption from zero to two packs per day changes the log odds byThe corresponding odds ratio is thereforeThus the model estimates that a two-pack-per-day smoker has about $2.52$ times the odds of heart disease of an otherwise comparable nonsmoker. This is an association in the fitted observational model, rather than by itself a causal conclusion.

Let $Y_i$ be the indicator that man $i$ develops heart disease and let ${\pi }_i=\mathbb{P}(Y_i=1)$. The fitted logistic regression assumes independent Bernoulli responses with

The setis the diagonal halting set. It is recursively enumerable by simulating the coded machine, but it is not recursive by the halting problem diagonal argument.

By contrast, ${\mathbb{K}}_{100}$ is recursive: decode $n$, reject it if it is not a program code, otherwise simulate the program on input $n$ for exactly $100$ steps and accept precisely if it has halted. This bounded computation always terminates. Thus

A subset $E\subseteq \mathbb{N}$ is recursive when its indicator function is a total computable function. It is recursively enumerable when it is the domain of a partial computable function, equivalently when an algorithm can enumerate its members.

If $E$ is recursive, its indicator decides both $E$ and its complement, so both sets are recursively enumerable. Conversely, suppose programs enumerate $E$ and $\mathbb{N}\setminus E$. On input $n$, run the two enumerations in parallel by dovetailing. Exactly one must eventually print $n$; accept in the first case and reject in the second. This algorithm always halts and decides $E$. Hence

Because $A$ is a positive-definite matrix, it has a positive-definite inverse square root. Choose the exact inverse-square-root preconditionerThenwhich has one distinct eigenvalue, so preconditioned CG terminates in one step.

This construction is of little computational use because obtaining and applying $A^{-1/2}$ is generally at least as expensive in storage and arithmetic as directly factoring or solving the original system. A useful preconditioner must only approximate this transformation while remaining cheap to construct and apply.

Let $e_1=(1,0,\dots ,0{)}^T$. The displayed lower-bidiagonal matrix $Q$ satisfiesSince $P=Q^{-1}$, the Preconditioned conjugate gradient method usesSolving $Qz=e_1$ givesThe rank-one matrix $I+z{z}^T$ has eigenvalue $1+n$ in the direction $z$ and eigenvalue $1$ on the $(n-1)$-dimensional orthogonal complement of $z$. It therefore has only two distinct eigenvalues. Part (b) gives

For an iterate $x_m$, the Conjugate gradient method residual isTwo fundamental properties are the Conjugate-gradient residual orthogonality conditionsEquivalently, the residuals generated before convergence are mutually orthogonal and span the successive Krylov spaces.

Since $A$ has only $s$ distinct eigenvalues, its minimal polynomial has degree at most $s$. ConsequentlyThe two properties then put $r_s$ both inside and orthogonal to ${\mathcal{K}}_s(A,r_0)$, so $r_s=0$. Hence $A{x}_s=b$, and

The $m$th Krylov subspace isBecause $A$ has a complete eigenbasis, decomposewhere $v_j$ is the projection of $v$ onto the eigenspace belonging to the $j$th distinct eigenvalue ${\lambda }_j$; some $v_j$ may vanish. ThenEvery generator of ${\mathcal{K}}_m(A,v)$ therefore belongs to $\mathrm{span}\{v_1,\dots ,v_s\}$, and

For a discrete memoryless channel with input $X$ and output $Y$, its information capacity iswhere $I(X;Y)$ is the mutual information and the maximum is over all input distributions. Capacity is measured in bits when logarithms have base two.

Take a linear combination of the P- and SV-fields from part (b). The rigid boundary condition $\mathbf{u}=0$ at $z=H$ givesThe conditions at $z=-H$ are the same because the horizontal displacement is even and the vertical displacement is odd. A nonzero pair $(A,B)$ exists exactly when the determinant vanishes:Therefore the rigid elastic waveguide mode dispersion relation is

As $c$ increases, both $akH$ and $bkH$ increase without bound. The two sides consequently have infinitely many alternating tangent branches and poles. On successive continuity intervals their relative ordering reverses, so the intermediate value theorem gives infinitely many intersections, corresponding to infinitely many propagating waveguide modes.

Fix $\omega =kc$. For a P-wave,SetEqual-amplitude P-waves with vertical wavenumbers $\pm m_P$ then give, after an irrelevant common normalization,Thus $f_1$ is even and both scalar functions multiplying $1$ and $i$ are real.

Similarly, putUsing the in-plane transverse polarizations $(m_S,0,-k)$ and $(m_S,0,k)$ for the two vertical directions givesThis has the same required parity and reality properties.

Substitute the harmonic plane waveinto the elastic wave equation. The vector identitygivesIf $\mathbf{A}$ is parallel to $\mathbf{K}$, this becomesand gives a longitudinal P wave. If $\mathbf{A}\cdot \mathbf{K}=0$, it becomesand gives either polarization of an S wave.

For $\mathbf{K}=(k,0,m)$, convenient polarization vectors areThe SV-wave lies in the $xz$ propagation plane, whereas the SH-wave is perpendicular to that plane.