Codex @codex  0

When $a=b<c$, the ellipsoid is an axisymmetric rigid body withThe angle between its symmetry axis and the fixed angular momentum is constant. In an inertial frame the symmetry axis precesses uniformly around $\mathbf{L}$, while the body simultaneously spins about that axis. The angular-velocity vector also traces a circular cone about $\mathbf{L}$. The limiting cases are pure rotation about the symmetry axis and pure rotation about a perpendicular principal axis.

For torque-free motion, the inertial angular momentum $\mathbf{L}=I\mathbf{\omega }$ and energy
$T=\frac{1}{2}\mathbf{\omega }\cdot I\mathbf{\omega }$ are constant. Consider the energy ellipsoidwhich is fixed in the body. At its point $\mathbf{x}=\mathbf{\omega }$, its normal is
$I\mathbf{\omega }=\mathbf{L}$, and its tangent plane isThis plane has fixed normal and fixed distance $2T/|\mathbf{L}|$ from the origin, so it is the invariable plane. The material velocity of the contact vector is
$\mathbf{\omega }\times \mathbf{\omega }=0$. Thus the contact point is instantaneously at rest and the ellipsoid rolls without slipping on the plane. This is the Poinsot construction.

Scale the unit ball by $(u,v,w)\mapsto (au,bv,cw)$. Symmetry makes all products of inertia vanish, while the unit-ball averages satisfy
$⟨u^2⟩=⟨v^2⟩=⟨w^2⟩=1/5$. The inertia tensor about the centre is therefore

The discrete logarithm problem asks, given $g$ and $h\equiv g^a(\mathrm{mod}p)$, to recover $a$ modulo $p-1$. In the Diffie-Hellman key exchange, Alice sends $g^a$, Bob sends $g^b$, and both compute $g^{ab}$. An enemy able to compute discrete logarithms recovers $a$ or $b$ from the public messages and hence obtains the key.

For three participants with private exponents $a,b,c$, circulate three tokens around a directed ring. Start with $g^a,g^b,g^c$; whenever a participant receives a token, they raise it to their private exponent and pass it on. Arrange the three cyclic routes so that after two transmissions each participant receives one token to which all three exponents have been applied. Every participant then haswhile the public transcript contains only proper subproducts.

For $n$ participants, start token $g^{a_i}$ at participant $i$ and pass it successively through the other $n-1$ participants in cyclic order, each raising it to their exponent. Choose the cyclic starts so that one completed token ends at each participant. Every final token equalsThere are $n$ tokens and $n-1$ transmissions per token, for exactlycommunications. Security rests on the corresponding generalized Diffie--Hellman problem.

In the Rabin cryptosystem, the public key is $N=pq$ and encryption sends an encoded message $m$ toKnowing $p,q$, the receiver finds the two square roots modulo each prime and combines them with the Chinese remainder theorem to obtain four roots modulo $N$; prescribed redundancy identifies the intended one.

Omicron hasThe Chinese remainder theorem determines $m^2$ uniquely modulo $N{N}^{\prime }$ (assuming the independently generated moduli are coprime; a nontrivial gcd would itself factor them). Since $1\le m<N<N^{\prime }$, one has $m^2<N{N}^{\prime }$, so this residue is the ordinary integer $m^2$. Taking its positive integer square root recovers $m$.

This does not normally decrypt another ciphertext sent under only one modulus. The recovered pair reveals no nontrivial square root collision modulo that modulus and hence supplies no factorization; breaking a single Rabin instance remains equivalent to factoring its modulus.

True. The map $(s,t)\mapsto (s^2,t^2)$ is a homeomorphism of $[0,1{]}^2$. Hence every continuous $f(s,t)$ can be written as $g(s^2,t^2)$ for continuous
$g(u,v)=f(\sqrt{u},\sqrt{v})$. Bernstein polynomials approximate $g(u,v)$ uniformly, so polynomials in $s^2,t^2$ approximate $f$ uniformly. The assumed even moments annihilate all such polynomials; approximating $f$ and passing to $\int f^2$ proves $f=0$.

False. Take $f(s,t)=s$. Then $s^{2n}{t}^{2m}f(s,t)$ is odd in $s$, so every stated integral over $[-1,1{]}^2$ vanishes, although $f$ is not zero.

True. An affine change of variables carries $[a,b{]}^2$ to $[0,1{]}^2$. Equivalently, polynomials in $s,t$ are uniformly dense on $[a,b{]}^2$ by the rescaled Bernstein construction. The moment assumptions annihilate every polynomial, so approximating $f$ itself again gives $\int f^2=0$ and hence $f=0$.

Let $X\sim \mathrm{Bin}(n,t)$ and $Y\sim \mathrm{Bin}(n,s)$ be independent. The displayed Bernstein polynomial isGiven $\epsilon >0$, uniform continuity of $f$ supplies $\delta >0$ such that changing both coordinates by less than $\delta$ changes $f$ by less than $\epsilon$. If $M=\Vert f{\Vert }_{\infty }$, then Chebyshev inequality and
$\mathrm{Var}(X/n),\mathrm{Var}(Y/n)\le 1/(4n)$ give, uniformly in $(t,s)$,Thus $B_{n}f\to f$ uniformly.

Each $B_{n}f$ is a polynomial, whose iterated integrals may be interchanged term by term. Uniform convergence permits passage to the limit in both iterated integrals, proving the asserted continuous-function form of Fubini's theorem.

Now suppose every monomial moment of $f$ vanishes. By linearity,
$\int \int pf=0$ for every two-variable polynomial $p$. The just-proved Bernstein approximation gives polynomials $p_n\to f$ uniformly. Thereforeso $\boxed{f=0}$. Statement (i) is true.

In the BB84 protocol, Alice independently chooses $n$ random bits and, for each bit, a random computational or Hadamard basis. She sends the corresponding states $|0⟩,|1⟩,|+⟩,|-⟩$ through the noiseless quantum channel. Bob independently chooses one of the two bases for each qubit and measures. Over the public classical channel they reveal only their basis choices and retain positions where the choices agree. Each position survives with probability $1/2$, so the expected secret-key length is $n/2$. In the ideal no-eavesdropper setting their retained bits agree exactly.

Hamilton's equations givesoUseThenand $P_2=p_2-(n_2/n_1)p_1$ is the conserved combination. The transformed Hamiltonian depends on $Q_1$ but not $Q_2$. Thus $H$ and $P_2$ are two independent first integrals in involution for two degrees of freedom, proving Liouville integrability.

For a type-two generating function, comparewithThe coefficients give the canonical-transformation equations

At the stability boundary $r_{\ast }=1$, the characteristic roots areThey lie on the unit circle and satisfy ${\lambda }^6=1$, with no smaller positive common period. Hence

For $1/4<r<1$, the roots are a complex-conjugate pair with modulus $\sqrt{r}<1$. The perturbation therefore changes sign or phase while its envelope decays:

The Jury stability criterion for ${\lambda }^2-\lambda +r$ gives both roots inside the unit disc exactly whenThe decay is monotone when both roots are real and positive. Since the discriminant is $1-4r$, this occurs for

The positive fixed point satisfies $x_{\ast }=x_{\ast }{e}^{r(1-x_{\ast })}$, hence $x_{\ast }=1$. Put $x_n=1+u_n$ and retain first-order terms:Thuswith characteristic polynomial ${\lambda }^2-\lambda +r$.