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Putand take the branch cut $[-1,1]$. On $\mathbb{C}\setminus [-1,1]$, choose the analytic branch of a square root whose upper boundary value isIts lower boundary value is $s(x-i0)=-\sqrt{1-x^2}$. This fixes the integrand to be $+1$ at $0^{+}$ as required.

For $0\le \arg z\le \pi$, start at $0^{+}$ and use any path to $z$ lying above the cut; path deformations within that region do not alter the integral by the Cauchy integral theorem. For $\pi <\arg z<2\pi$, start in the same way, continue counterclockwise around the branch point $-1$ from the upper bank to the lower bank, and then continue to $z$ through the lower half-plane. This second prescription is the continuation obtained as $argz$ increases through $\pi$.

Again put $x=1+u$. ThenThe linearized equation is thereforewhose sole characteristic exponent is $-\alpha <0$ and is independent of $T$. By delay-independent stability from an undelayed linearization,

Put $x(t)=1+u(t)$. Retaining terms linear in $u$ givesThe exponential ansatz $u=e^{\lambda t}$ gives the characteristic equation of a delay differential equationAt $T=0$ its root is $\lambda =-\alpha$, and the remaining delay roots lie far in the left half-plane, so the equilibrium is stable for sufficiently small $T$.

A stability change can first occur when $\lambda =i\omega$. Separating real and imaginary parts givesFor the first positive crossing, $\omega T=\pi /3$ and $\omega =\sqrt{3}\alpha$. ThusAll positive-frequency crossings have $\omega T=\pi /3+2\pi k$, so there is no earlier imaginary-axis crossing. Implicit differentiation ofgives, with $q=\alpha T_c$,so the conjugate pair crosses into the right half-plane as $T$ increases. Therefore the stability threshold for delayed quadratic crowding is

Let $K\subseteq \mathbb{N}$ be the undecidable set denoted $\mathbb{K}$ in the question. The language satisfies the pumping lemma for regular languages with pumping length one. For any nonempty $s\in L$, write $s=xyz$ with $x=\epsilon$, $y$ its first symbol, and $z$ the remaining suffix.

If $s\in b^{\ast }$, then $x{y}^{i}z\in b^{\ast }$ for every $i\ge 0$. Otherwise $s=wa{b}^n$ for some $n\in K$. If the chosen first symbol lies in $w$, pumping it preserves the same final suffix $a{b}^n$. If $w=\epsilon$, then $s=a{b}^n$: pumping down gives $b^n\in b^{\ast }$, while every positive pump gives $a^i{b}^n$ and all but its last $a$ can be absorbed into $w$. Thus every sufficiently long word has a valid pumping decomposition.

Nevertheless, if $L$ were regular language, closure under intersection would makeregular. A finite automaton deciding this unary language would decide whether $n\in K$ by running it on $a{b}^n$, contradicting the halting problem. ThereforeThis is a nonregular language satisfying the pumping lemma.

No. The procedure exposes a square-root oracle for the Rabin cryptosystem. The verifier chooses $m$, sends $r=m^2\mathrm{mod}N$, and receives another square root $m_1$. Ifthen the factorization from distinct square roots modulo a semiprime givesFor a product of two odd primes, a square normally has four roots, and a returned root lies outside the pair $\{\pm m\}$ with probability about $1/2$. Repeating chosen-square challenges therefore factors $N$ with overwhelming probability. The attacker can then compute square roots and impersonate Alice. This is the chosen-square attack on Rabin authentication, so the proposed protocol is insecure.

In Diffie-Hellman key exchange, Alice and Bob publicly agree on a cyclic group $G=⟨g⟩$ in which the discrete logarithm problem is believed hard. Alice chooses a secret $a$ and sends $A=g^a$; Bob chooses a secret $b$ and sends $B=g^b$. They independently computeThusis their shared key. A passive eavesdropper sees $g,g^a,g^b$ but is believed unable to compute $g^{ab}$ efficiently; this is the computational Diffie--Hellman assumption. The basic protocol does not authenticate the participants and is therefore vulnerable to a man-in-the-middle attack unless the exchanged values are authenticated.

Insert the Fourier seriesinto the rectangle rule. The finite geometric sum obeysConsequently the quadrature value iswhereasThus the exact Periodic N-point rectangle-rule error is

By the Fourier coefficient decay of an analytic periodic function, there are constants $C>0$ and $0<\rho <1$ such thatIt follows thatThereforewhich is exponential, and hence spectral, convergence.

Expand the unknown in Fourier modes:By Termwise differentiation of a Fourier series,Also,Therefore the coefficient of $e^{i\pi mx}$ on the left-hand side isThe right-hand side has coefficientsbecause the coefficient of each of $e^{\pm i\pi nx}$ is $1/(n^2+1)$ and the constant coefficient is one. Hence the Fourier spectral system for a cosine-modulated second derivative isIts coefficient array is an infinite tridiagonal matrix.

For a spectral truncation, retain the modes $-M\le n\le M$, setand impose the displayed coefficient equation for every $-M\le m\le M$. This gives an explicit $(2M+1)$-dimensional tridiagonal system. The resulting finite Fourier series automatically satisfies the periodic boundary conditions.

The obstacle and wave pattern are stationary, so $\omega =\Omega =0$. Withthe dispersion relation becomesHence $\cos \varphi >0$, so $-\pi /2<\varphi <\pi /2$, and

Differentiating the dispersion relation with respect to the wavevector giveswhere the zero-frequency relation
$\alpha /\sqrt{\kappa }=V\cos \varphi$ was used. Its first component is always negative, so the ray angle $\psi$ lies inMoreover,

Put $q=|\tan \varphi |$. The magnitude of the slope relative to the negative $x_1$-axis isIts derivative isso its global maximum occurs at $q=1/\sqrt{2}$ and equalsTherefore all group-velocity rays lie in the stationary square-root-dispersion wakea wedge of semi-angle ${\tan }^{-1}(2^{-3/2})$ extending in the negative $x_1$-direction.

In a homogeneous, time-independent medium, $\partial \Omega /\partial x_i=0$ and $\partial \Omega /\partial t=0$. Therefore $\mathbf{k}$ and $\omega$ are constant along each ray. The group velocity ${\mathbf{c}}_g={\nabla }_{\mathbf{k}}\Omega$ is consequently constant, andso every ray is a straight line.

For $\omega =0$, let $s=|\mathbf{x}|$ denote distance along a ray from the origin and letThe phase equation givesBoth vectors on the right are constant along the ray, soThe final formula in the question denotes the physical phase $\Theta =\theta /\epsilon$ simply by $\theta$, giving its stated expression.

The physical rapid phase isDefine the local wavevector, angular frequency, and group velocity byThe equality of mixed partial derivatives gives the phase-compatibility equationsA ray is defined to move with the local group velocity:

Differentiate the local dispersion relationwith respect to $x_i$. Along a ray,The last two terms cancel because $\mathbf{k}$ is a gradient. Similarly,Finally,Thus the Hamiltonian ray-tracing equations are

Let the sphere have radius $a$; set $a=1$ for the unit sphere implicit in the question. Sinceone findsOn $r=a$, the no-slip boundary condition requires $\mathbf{u}=\mathbf{U}$ for every surface normal. The coefficient of
$(\mathbf{U}\cdot \mathbf{n})\mathbf{n}$ must vanish, while the coefficient of $\mathbf{U}$ must equal one:Solving,For a unit sphere,Substitution reproduces the standard translating sphere in Stokes flow.

Let $\mathbf{u}$ be the Stokes flow and let $\mathbf{v}$ be any other admissible incompressible flow with the same no-slip boundary condition on the sphere and the same decay at infinity. PutThen $\nabla \cdot \mathbf{w}=0$ and $\mathbf{w}=0$ on the sphere. The viscous dissipation satisfiesFor the Stokes stressincompressibility of $\mathbf{w}$ givesThe divergence theorem, the homogeneous boundary data, decay at infinity, and $\nabla \cdot \mathbf{\sigma }=0$ giveThereforeThis proves the Minimum-dissipation theorem for Stokes flow.

For the constant density ${\rho }_0$,The definition of $m$ then givesSubstitution in the Tolman–Oppenheimer–Volkoff equation yieldsThe minus sign follows both from the stated TOV equation and from the outward decrease of pressure; a version of the requested intermediate formula without that sign is a typographical error.

SetSeparating variables givesSinceintegration and the surface condition $P(R)=0$ giveWritingone obtains the pressure of the Interior Schwarzschild metric:In particular, the central pressure of a constant-density relativistic star isIt diverges whenor equivalentlyThis is the limiting equality case of Buchdahl's theorem.

Use the geodesic Lagrangianwhere here the overdot on a coordinate means differentiation with respect to the affine parameter. Applying the Euler-Lagrange equation to each coordinate and comparing with the geodesic equationdetermines the Christoffel symbols. To display derivatives of the metric coefficients unambiguously, write ${\lambda }_t={\partial }_t\lambda$, ${\lambda }_r={\partial }_r\lambda$, and similarly for $\mu$. The nonzero symbols, together with those obtained by symmetry in the two lower indices, areThis is the Christoffel symbols of a diagonal spherical spacetime metric.