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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.

For the sextic even Landau potentialput $q={\varphi }^2$. The stationary-point equation isso the nonzero stationary points haveAt such a point, $\epsilon =-q^2-gq$, and henceA first-order transition occurs when a nonzero minimum has the same free energy as the disordered minimum $G(0)=0$. ThereforeAt $g=0,\epsilon =0$, the first-order line meets the continuous line and forms the tricritical point. For $g>0$, the transition at $\epsilon =0$ is continuous.

For $g<0$, the disordered state $\varphi =0$ is a local minimum when $\epsilon >0$ and loses stability at the disordered spinodal point $\epsilon =0$. The ordered minima exist whileand merge with the intervening maxima at the ordered spinodal $\epsilon =g^2/4$. Consequently the two spinodals enclose a region of metastability. If increasing $\epsilon$ corresponds to heating, the ordered phase can persist above the coexistence curve throughoutwhile the disordered phase can persist below it throughout

Landau theory represents the appropriate thermodynamic potential as an analytic function of a small order parameter, retaining every low-order term allowed by the system's symmetry. At fixed control parameters, the equilibrium phase is the global minimum of the Landau free energy; local minima describe metastable phases. A continuous phase transition occurs when a minimum changes continuously and loses stability, whereas a first-order phase transition occurs when distinct minima exchange global stability.

Forthe stationary points satisfyIf $\epsilon >0$, $\varphi =0$ is the unique minimum. If $\epsilon <0$, $\varphi =0$ is a maximum and the two degenerate minima areThus the reflection symmetry $\varphi \mapsto -\varphi$ is spontaneously broken below $\epsilon =0$, and

Write $u=kr-l\pi /2$. Sincethe outgoing-to-incoming coefficient ratio is $-S_l$. Hence the Partial-wave S-matrix element and scattering phase shift areThe scattering length from a partial-wave S-matrix is

For the stated S-wave scattering solution, $\tanh (\lambda r)\to 1$ as $r\to \infty$, soUsing the same incoming/outgoing convention,Therefore

The analytic continuation of $S_0$ has a bound-state pole of the scattering amplitude at $k=i\lambda$ in the upper half-plane. Since $E={\hslash }^2{k}^2/(2m)$, this givesTo extract the bound-state wavefunction, multiply the scattering solution by $k-i\lambda$ and take the pole limit:Thus the unnormalized exterior solution isThese data are collected in the Hyperbolic-tangent S-wave exterior solution.

In the classical experiment, particles whose impact parameter lies between $b$ and $b+db$ occupy an incident annulus of areaFor a spherically symmetric potential, azimuthal symmetry scatters this annulus into the solid-angle stripThus the central-potential classical differential cross-section isIf the relation between $b$ and $\theta$ has several branches, one sums this expression over every branch. The corresponding diagram consists of an incoming line offset by $b$ from the centre, an outgoing asymptote making angle $\theta$ with the incident direction, and the annulus and solid-angle strip just described.

For quantum scattering by a central potential, the far-field wavefunction iswhere $f(\theta )$ is the scattering amplitude. The incident plane wave has probability currentTo leading order in $1/r$, the scattered spherical wave has radial currentThe scattered rate through area $r^{2}d\Omega$ divided by the incident flux is therefore the quantum differential cross-section from a scattering amplitude:

Write $r=e^x$, $ϵ=\hslash /\sqrt{2m}$, $L=l+1/2$, and $\lambda =-|\lambda |$. Since $V(r)=-V_0/r$, the transformed equation hasThe two turning radii $a<b$ are the positive roots ofsoInside the allowed interval,The semiclassical action integral is thereforeEquating this with $(n+1/2)\pi$ from part (a), and using $L=l+1/2$, givesSince ${ϵ}^2={\hslash }^2/(2m)$,Thus the Coulomb WKB quantization with the Langer correction yields

Between the classical turning points, $q<0$, while boundedness selects exponentially decaying WKB approximation solutions in the forbidden regions $x<a$ and $x>b$. The Airy turning-point connection formula at the left turning point matches the left decaying solution toin the allowed region. Matching the right decaying solution across $b$ givesFor these expressions to be proportional throughout $(a,b)$, their phases must be compatible. Each simple turning point contributes a phase shift $\pi /4$, giving the two-turning-point WKB quantization condition

Given the present state $(S_t^1,M_t)=(x,m)$, take an independent fresh copy of the stock path and denote its initial value, terminal value after $T-t$ steps, and running maximum by $(y,z,w)=(S_0^1,S_{T-t}^1,M_{T-t})$. Scaling the fresh path to start at $x$ makes its terminal value $xz/y$ and its future running maximum $xw/y$. Including the maximum already attained givesHence defineIndependent increments then give the Running-maximum state reduction in the Cox--Ross--Rubinstein modelwith