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Choose a right inverse $s:A\to B$ of the surjection $f$, so $f(s(a))=a$, and defineFor each $a$, the element $b=s(a)$ then witnesses both $f(b)=a$ and $g(b)=h(a)$. As in part (a), arbitrary simultaneous choices use the axiom of choice. If $A$ is empty, surjectivity and the existence of $f:B\to A$ force $B$ to be empty, and the unique empty function $h:A\to C$ already meets the requirement.

Suppose $g$ is constant on each fiber of a function of $f$. Since every fiber is nonempty, $h(a)$ must be its common $g$-value: any witness $b$ in that fiber gives that value. This defines exactly one $h$ and yields $g=h\circ f$, an instance of factorization through a surjection.

Conversely, suppose a fiber contains $b,b^{\prime }$ with $f(b)=f(b^{\prime })=a_0$ but $g(b)\ne g(b^{\prime })$. Starting with any admissible $h$, define $h_1,h_2$ to agree with it away from $a_0$, but put $h_1(a_0)=g(b)$ and $h_2(a_0)=g(b^{\prime })$. Both remain admissible, witnessed at $a_0$ by $b$ and $b^{\prime }$ respectively, yet they are distinct. HenceThe distinction is important: the original existence property asks only for one witness in each fiber. Only when $g$ is fiberwise constant does that property force $g=h\circ f$ for every $b$.

A left inverse of $f$ is a function $\ell :B\to A$ satisfying $\ell \circ f={\mathrm{id}}_A$. If such $\ell$ exists and $f(a)=f(a^{\prime })$, applying $\ell$ gives $a=a^{\prime }$, so $f$ is an injection. Conversely, suppose $f$ is an injection and fix $a_0\in A$, possible because $A$ is nonempty. DefineThe first value is unique by injectivity, and $\ell (f(a))=a$ for every $a$. ThusNo choice from a family of nonempty sets is needed here: each preimage in the first case is unique, and the second case uses one fixed element.

A right inverse is a function $r:B\to A$ with $f\circ r={\mathrm{id}}_B$. Its existence makes every $b$ the image of $r(b)$, proving that $f$ is a surjection. Conversely, if $f$ is a surjection, every fiber of a function $f^{-1}(\{b\})$ is nonempty. Using the usual axiom of choice for arbitrary sets, choose $r(b)\in f^{-1}(\{b\})$ for each $b$. Then $f(r(b))=b$, givingThe converse in this second equivalence is the right-inverse characterization of the axiom of choice; it is not an unrestricted choice-free theorem about arbitrary sets.

With the Sun fixed, its gravitational force on the planet is a central force. In the fixed orbital plane, the transverse acceleration in polar coordinates therefore vanishes:During $dt$, the position vector sweeps an infinitesimal sector of area $dA=\frac{1}{2}{r}^{2}d\theta$. Hence the areal velocity isIntegrating over any two equal time intervals gives equal swept areas, proving Kepler's second law. Here $L$ is the signed component of angular momentum normal to the orbital plane. The argument uses conservation of angular momentum; an inverse-square dependence is not needed for this particular law, only a central force.

The particle's angular momentum about the force centre is $\mathbf{L}=m\mathbf{r}\times \dot{\mathbf{r}}$. A central force exerts zero torque about that centre, soBy conservation of angular momentum, $\mathbf{L}$ is a constant vector. If it is nonzero, the identity $\mathbf{r}\cdot \mathbf{L}=0$ places every point of the orbit in the fixed plane through the centre perpendicular to $\mathbf{L}$:If $\mathbf{L}=0$, position and velocity are parallel. Away from the force centre, ${\dot{\mathbf{e}}}_r=(\dot{\mathbf{r}}-\dot{r}{\mathbf{e}}_r)/r=0$, so the motion is radial along a fixed line, which lies in any plane containing that line. Thus the zero-angular-momentum case also gives planar motion. As usual, these assertions apply while the force law supplies a regular orbit; continuation through a singular force centre is a separate issue.

A central force acts along the line joining the particle to a fixed centre. Taking that centre as the origin, its usual position-dependent form isPositive $f$ gives an outward force and negative $f$ an inward one. For a conventional central force the magnitude depends only on distance from the centre, not on direction. The angular-momentum conclusions below require only that the force remain radial about that fixed centre.

The integer $137$ is prime: none of the primes $2,3,5,7,11$ below $\sqrt{137}$ divides it. Since $43\not\equiv 0(\mathrm{mod}137)$, Fermat's little theorem gives $43^{136}\equiv 1$. Therefore $43^{135}$ is the modular inverse of $43$. The extended Euclidean algorithm yieldsand back-substitution gives $1=51(43)-16(137)$. ConsequentlyIndeed, $43\cdot 51=2193=16\cdot 137+1$, directly checking the required modular inverse.

Eliminate $x$ by twice the first modular congruence minus three times the second. This givesThe modular inverse of $3$ is $16$, because $3\cdot 16=48\equiv 1$. Hence $y\equiv 16\cdot 13\equiv 20$. Substitution into the second modular congruence gives $6x\equiv 14-7(20)\equiv 15$, and $6^{-1}\equiv 8$, so $x\equiv 8\cdot 15\equiv 26$. ThusIn particular, $(x,y)=(26,20)$ works: $9(26)+12(20)=474=10(47)+4$ and $6(26)+7(20)=296=6(47)+14$. The elimination is reversible because $3$ and $6$ have modular inverses, so the displayed classes contain all solutions.

Place the four fixed charges at $(sa,ta)$ with $s,t\in \{-1,1\}$, and write the moving particle's position as $(x,y)$. Introduce its mass $m>0$, which is needed for an oscillation frequency but is not named in the question. Let $C=Qq/(4\pi {ϵ}_0)>0$. By Coulomb's law and superposition, the moving particle's potential energy isThe electrostatic potential itself is $V/q$. Put $r^2=x^2+y^2$ and ${\delta }_{st}=-(sx+ty)/a+r^2/(2a^2)$. Near the centre, a Taylor expansion givesOn summing over the four corners, linear terms cancel. To quadratic order, $\sum {\delta }_{st}=2r^2/a^2$ and $\sum {\delta }_{st}^2=4r^2/a^2+O(r^4/a^4)$, since $\sum st=0$. Reflection symmetry removes all odd-degree terms. HencewhereThus the centre has zero planar force and a positive-definite planar Hessian matrix $\kappa I_2$. It is a strict local minimum of the constrained potential energy, so it is a stable equilibrium. More explicitly, conservation of the total energy confines sufficiently small-energy motions within a small neighbourhood of this isolated local minimum.

Keeping only the quadratic potential, the two coordinates obey independent small oscillations:Both planar modes have the same angular frequency, and the ordinary frequency isThe planar constraint is essential to this constrained electrostatic equilibrium. Along the perpendicular axis, $V(0,0,z)=4C/(2a^2+z^2{)}^{1/2}=V(0,0)-C{z}^2/(\sqrt{2}{a}^3)+O(z^4)$, so the same stationary point is unstable to unconstrained normal displacements. That also checks the cancellation of the three-dimensional Hessian trace required by Laplace's equation away from charges; it does not contradict the positive-definite restriction to the given plane.

With no electric field, the Lorentz force isFor nonrelativistic motion, the work-energy theorem givesConsequently $\boxed{K(t)=K(0)}$. The reason is geometric: magnetic forces do no work because they are perpendicular to the instantaneous velocity. They can change its direction but not its magnitude. The same energy conclusion holds relativistically, using $dE/dt=\mathbf{F}\cdot \mathbf{v}$ and subtracting the constant rest energy to obtain kinetic energy.

Let the incoming projectile momentum be ${\mathbf{p}}_1$. The stationary target has four-momentum $(m_{2}c,0)$, so the total incoming norm isHere the relativistic energy-momentum relation eliminated $|{\mathbf{p}}_1{|}^2$. Write $W$ for the total energy in the centre-of-momentum frame; then $W^2=c^2{P}_{\mathrm{in}}^2$. If $M_f={\sum }_i{\mu }_i$, the relativistic collision mass threshold gives $W\ge M_f{c}^2$. Squaring and rearranging with $m_2>0$ yields the fixed-target production thresholdThe actual incident total energy also obeys $E_1\ge m_1{c}^2$. Thus a negative algebraic reaction bound would not allow a negative-energy projectile.

For a photon, $m_1=0$ and $E_1=h\nu$, where $h$ is the Planck constant. The necessary bound becomesIn the usual massive multi-product channel with $M_f>m_2$, conservation alone permits the threshold configuration in which all products are at rest relative to one another in the centre-of-momentum frame. Boosting that configuration supplies the required laboratory momentum. Hence the photon fixed-target production threshold isFor a multi-product channel with $M_f\le m_2$, there is no positive mass threshold: the lower limit of allowed positive frequencies is zero. The compact expression for that lower limit is $\boxed{{\nu }_{\inf }=\max \{0,M_f^2-m_2^2\}{c}^2/(2m_{2}h)}$; it need not correspond to a zero-frequency incident photon.

For completeness, if the only product is one particle of mass ${\mu }_1$, its invariant mass must match exactly. A positive incident frequency is possible only when ${\mu }_1>m_2$, and is then fixed at $\nu =({\mu }_1^2-m_2^2)c^2/(2m_{2}h)$; frequencies above that value do not produce the same single particle while conserving four-momentum. If a multi-product channel includes massless particles, an all-products-at-rest configuration is impossible for those particles with nonzero energy, so the nominal massive threshold may be only an infimum. These qualifications distinguish the requested necessary inequalities from unrestricted claims of sufficiency.

The two incoming particles have equal energies and opposite momenta, so their total four-momentum in this frame isIt is already a centre-of-momentum frame. By four-momentum conservation, the final spatial momenta sum to zero and their energies sum to $2E$. Each product has energySumming gives the necessary relativistic collision mass thresholdFor massive products the equality threshold has all of them at rest in this frame. Above it, multiple products can carry balancing momenta and relative kinetic energy. The displayed condition is necessary, as requested; a single final particle must instead have exactly the incoming invariant mass, and additional physical conservation laws could further restrict a real reaction.

For $|\mathbf{v}|<c$, define the Lorentz factor $\gamma =(1-|\mathbf{v}{|}^2/c^2{)}^{-1/2}$. A particle of rest mass $m$ hasThese expressions give the relativistic energy-momentum relation directly:so $\boxed{E^2=c^2\mathbf{p}\cdot \mathbf{p}+m^2{c}^4}$.

The four-momentum iswhere $U^{\mu }$ is the four-velocity, the proper-time derivative of $(ct,\mathbf{x})$. It is a four-vector, so its Minkowski metric norm is invariant under Lorentz transformations. Choose metric signature $(+,-,-,-)$; in the particle's rest frame, $P^{\mu }=(mc,0)$ and $P_{\mu }{P}^{\mu }=m^2{c}^2$. In any frame the same norm is $E^2/c^2-|\mathbf{p}{|}^2$. Equating the two yields the same relativistic energy-momentum relation. With the opposite metric convention both norm expressions change sign, leaving the energy relation unchanged.

Again assume independent die scores. Their sum has the discrete convolution of the two score distributions, soAll terms are nonnegative. The arithmetic-geometric mean inequality applied to the two extreme-score contributions givesIf all eleven probabilities were equal, normalization would make each $1/11$, but the inequality would require $1/11\ge 2/11$. This contradiction proves the sum cannot be uniform on $\{2,\dots ,12\}$, whatever the individual die probabilities.

Use the usual assumption that the two die scores are independent random variables. A fair score has expected value $7/2$ and variance $35/12$. Linearity of expectation and the variance of a sum therefore giveCount the ordered pairs of scores to obtain the probability mass functionsOn these attainable probability supports, the maxima occur at $x=7$ and $y=0$, each with probability $1/6$; the minima occur at $x=2,12$ and $y=-5,5$, each with probability $1/36$. Outside the respective probability support, the mass is zero.

They are not independent: $Y=0$ implies equal die scores and hence $X$ is an even number, so $P(X=7,Y=0)=0$, whereas $P(X=7)P(Y=0)=1/36>0$. In fact $\mathrm{Cov}(X,Y)=\mathrm{Var}(S_1)-\mathrm{Var}(S_2)=0$: this is an example of uncorrelated random variables that are dependent.

Apply differentiation under the integral sign on the moving drop:The moving-boundary term vanishes because $h(R,t)=0$. At the origin, symmetry and regularity give $h_r(0,t)=0$. At the front, the stated $h\propto (R-r{)}^{1/3}$ gives $h^3{h}_r=O((R-r{)}^{1/3})\to 0$. Thus $\boxed{dM/dt=0}$, expressing volume conservation for this axisymmetric viscous gravity current.

With $\eta =r/t^{\alpha }$ and $R={\eta }_0{t}^{\alpha }$, substitute the self-similar solution into the conserved integral:For a nonzero finite drop volume, the power of $t$ must vanish, so $\boxed{\alpha =1/8}$. Balancing powers in the evolution equation gives the same result: its left side scales as $t^{-5/4}$ and its right side as $t^{-1-2\alpha }$. The front-edge exponent alone would not determine this spreading exponent without the conserved volume.

The chain rule gives ${\partial }_x={\partial }_u+{\partial }_v$ and ${\partial }_t={\partial }_u-{\partial }_v$. Hence ${\partial }_x^2-{\partial }_t^2=4{\partial }_u{\partial }_v$, and the forced wave equation becomes $y_{uv}=1$. Integrating twice gives $y=uv+F(u)+G(v)$. At $t=0$ the initial conditions areThus $F-G$ is constant, which cancels from the final expression. Combining the remaining sum yieldsDifferentiation checks the forcing and both initial conditions.

If $\rho =0$, the bivariate normal density in part (a) factorizes asIntegrating over any product of measurable sets proves independence of random variables. Conversely, independent variables with finite second moments have $\mathbb{E}[X_1{X}_2]=\mathbb{E}[X_1]\mathbb{E}[X_2]$, so their covariance is zero and $\rho =\mathrm{Cov}(X_1,X_2)/({\sigma }_1{\sigma }_2)=0$. This proves independence of uncorrelated jointly normal variables; zero correlation alone would not imply independence without the joint normal assumption.

With ${\sigma }_1,{\sigma }_2>0$ and $|\rho |<1$, the bivariate normal distribution has probability density functionThe strict correlation bound makes its covariance matrix positive definite, so this is an ordinary two-dimensional density rather than a distribution supported on a line.

The Galton-Watson extinction fixed point equation isIts smallest root in $[0,1]$ is $q=(3\sqrt{3}-5)/4$, not the other allowed fixed point $1$. Consequently the large-depth escape probability isFor finite $N$, the exact escape probability is $1-q_N$ and is slightly larger. Indeed ${\varphi }^{\prime }(q)=2-\sqrt{3}<1$, and the mean value theorem on $[q_N,q]$ gives $0\le q-q_N\le q(2-\sqrt{3}{)}^N$. This also quantifies how quickly the finite-depth answer approaches the displayed limit.

View the accessible rooms as the open cluster of bond percolation on a rooted ternary tree. Each accessible room has three forward corridors, independently open with probability $2/3$, so its number of accessible children has the binomial distribution $\mathrm{Bin}(3,2/3)$. The counts at successive levels form a binomial branching process with offspring probability generating functionLet $q_N$ be the probability that the root has no open path to level $N$. Then $q_0=0$, and independence of the three descendant subtrees givesThe sequence increases to the smallest fixed point $q=\varphi (q)$ in $[0,1]$, the branching extinction probability. Thus $q_N$ is close to that solution for large $N$.

Connectivity is the relevant escape event: if no window is reachable, escape is impossible; if one is reachable, ordinary random wandering reaches a window eventually with probability one. Before reaching level $N$ the guest moves within a finite connected graph, so an accessible exit cannot be avoided forever. This interpretation does not require the guest to move monotonically outward.