Past exam of the mathematics course of the University of Cambridge / 2011 / ii / Paper 4

Suppose the prime numbers were the finite list $p_1,\dots ,p_r$. The integer $N=p_1\cdots p_r+1>1$ has a prime factor, by repeatedly factoring any composite factor until a prime is reached. But division by any $p_j$ leaves remainder one, so that prime factor is absent from the list. This contradiction proves there are infinitely many prime numbers. The argument does not require $N$ itself to be prime.

For any positive integer $m$, put $N=(m+1)!$. Each of $N+2,N+3,\dots ,N+m+1$ is composite, since $N+j$ is divisible by $j$ and larger than $j$. Let $p$ be the largest prime number at most $N+1$, and let $q$ be the smallest prime number at least $N+m+2$. Such $q$ exists by part (i), and no prime lies strictly between $p$ and $q$. Therefore these are consecutive prime numbers withSince $m$ is arbitrary, gaps between consecutive prime numbers are unbounded.

Normalize the nonvanishing loop to $\gamma (t)/|\gamma (t)|$. A continuous argument lift is a continuous function $\theta :[0,1]\to \mathbb{R}$ such that $\gamma (t)/|\gamma (t)|=e^{i\theta (t)}$. Such a lift exists on an interval; any two differ by a constant integer multiple of $2\pi$. The winding number isIts value is independent of the lift. This definition applies to continuous loops without assuming a differentiable contour.

The homotopy invariance of winding number states: if $H:[0,1{]}^2\to \mathbb{C}\setminus \{0\}$ is continuous and $H(s,0)=H(s,1)$ for every $s$, then $w(H(s,\cdot );0)$ is independent of $s$. Neither the loops nor their moving basepoints may pass through zero. One may justify invariance by uniformly subdividing the homotopy into nearby loops whose normalized values have a continuous argument difference of magnitude less than $\pi$; that argument difference has equal endpoint values, so their winding numbers agree.

Put $z(t)=e^{2\pi it}$ and use the straight homotopy $H(s,t)=z(t)+s(\gamma (t)-z(t))$. For $0\le s<1$, the triangle inequality givesAt $s=1$ it is nonzero by the hypothesis. Thus this is a homotopy of closed loops avoiding the origin, even though the given distance bound is non-strict. The homotopy invariance of winding number gives

If the boundary loop meets zero, the assertion already holds. Otherwise suppose, for contradiction, that $f$ has no zero anywhere in the disc. Then $K(s,t)=f((1-s)e^{2\pi it})$ is a homotopy through nonvanishing closed loops from the boundary loop to the constant loop $f(0)$. Constant loops have winding number zero, whereas part (i) gives winding number one. This contradicts homotopy invariance of winding number. Hence $\boxed{f(z)=0\text{ for some }z\in B}$.

An inversion in a circle on the Riemann sphere is the anti-Möbius involution fixing that generalized circle pointwise. For a Euclidean circle with centre $a$ and radius $R>0$, it isIt reverses the argument relative to the centre and sends radius $r$ to $R^2/r$, so exactly the points of the circle are fixed. For a straight line, regarded as a circle through infinity, inversion is the ordinary reflection in that line. These formulas prove existence. For uniqueness, the composition of two anti-Möbius transformations fixing the same circle pointwise is a Möbius transformation fixing at least three distinct points, hence the identity. Therefore the two inversions agree.

Every inversion has the form $(a\overline{z}+b)/(c\overline{z}+d)$ with $ad-bc\ne 0$. Composing two conjugate-linear fractional expressions cancels the complex conjugations and gives $(Az+B)/(Cz+D)$ with nonzero determinant. Induction proves that an even number of inversions gives a Möbius transformation.

Conversely translations are compositions of reflections in parallel lines, rotations are compositions of reflections in two lines through their centre, and positive dilations are compositions of two concentric circle inversions: $J_R{J}_1(z)=R^{2}z$. Thus every nonconstant affine map $z\mapsto az+b$ is an even composition. The map $z\mapsto 1/z$ is $J_{|z|=1}$ composed with reflection in the real axis, also an even composition. If $c\ne 0$, factor a general Möbius transformation asThis is a composition of affine maps and $1/z$. If $c=0$, it is already affine. Every Möbius transformation is therefore a composition of an even number of inversions, with the identity represented by zero inversions or by any inversion repeated twice.

Use the BB84 protocol to establish a secret random key, then encrypt the actual message with a one-time pad. Alice chooses independent random bits and independently chooses between the computational orthonormal basis $\{|0⟩,|1⟩\}$ and the diagonal basis $\{|+⟩,|-⟩\}$, where $|\pm ⟩=(|0⟩\pm |1⟩)/\sqrt{2}$. She sends one photon or other qubit in each corresponding state. Bob independently chooses one of the two bases for each measurement.

Over an authenticated public classical channel, they announce their bases but not their bit values. They retain only positions where their bases match; absent disturbance these sifted bits agree. They disclose and discard a randomly selected test sample to estimate the error rate. If it is too large they abort. Otherwise they perform information reconciliation and privacy amplification to produce a shared key about which an eavesdropper has negligible information. Authentication is essential: without it, an active attacker can impersonate the two parties. It may use a previously shared short authentication key, subsequently replenished from the new secret key.

The no-cloning theorem prevents an attacker from copying an unknown nonorthogonal state perfectly: preservation of inner products would require $⟨\psi |\varphi ⟩=⟨\psi |\varphi {⟩}^2$, impossible when its modulus is strictly between zero and one. In a simple intercept-and-resend attack, Eve chooses the wrong basis with probability $1/2$, and then introduces a wrong sifted bit with probability $1/2$. Thus the test error rate is $1/4$, and the probability that $m$ independent tested bits show no error is $(3/4{)}^m$. More general attacks are handled by bounding Eve's information from the observed statistics and shortening the reconciled key through privacy amplification.

Finally Alice sends $C=M\oplus K$, using a uniform independent key of the message's length exactly once. The one-time pad makes the ciphertext independent of the message. The security claim is secrecy with arbitrarily small failure probability after testing and privacy amplification, not that a finite quantum transmission can never be disturbed or intercepted. The scheme assumes trusted devices and an authenticated classical channel; it does not rely on a computational hardness assumption.

Model each person's infections over the fixed exposure period as an independent Poisson random variable with mean ${\mu }_j$, conditional on the explanatory variables. The sum over $n_j$ people is then Poisson distributed with mean $n_j{\mu }_j$, because independent Poisson counts add. This is reasonable for repeatable events occurring at an approximately constant rate without substantial clustering; the count need not be at most $n_j$, so a binomial model would not describe repeated infections.

The Poisson regression uses a log link and the known offset $\log n_j$. The omitted factor levels are the reference categories under treatment contrasts: their effects are absorbed into the intercept. Including all levels together with an intercept would make the design matrix linearly dependent. For the proposed new group, the relevant predictor uses the intercept, the infrequent-swimmer coefficient and the age-20–24 coefficient; beach and female are reference levels. HenceTo add the age-by-sex interaction, one possible R command iswhere swimmers denotes the data frame. The product expands to both main effects and their interaction. There are two additional interaction coefficients because age has three levels and sex has two.

The nested-model comparison is a likelihood-ratio test. The reduction in Poisson deviance is $51.714-44.319\simeq 7.3948$, asymptotically ${\chi }_2^2$ under the null hypothesis that both interaction coefficients are zero. Its reported $p$-value $0.02479$ gives evidence for an age-by-sex interaction at the five-percent level, though not at the one-percent level.

Neither fitted Poisson model fits well in an absolute sense. Comparing residual deviances with ${\chi }_{18}^2$ and ${\chi }_{16}^2$ respectively gives small tail probabilities: the deviance-to-degree-of-freedom ratios are about $2.87$ and $2.77$. Improvement relative to the simpler model does not establish goodness of fit. Omitted structure, dependence or overdispersion could account for the remaining lack of fit; if the Poisson variance assumption is false, the nominal interaction-test calibration also needs reconsideration. The age label in the introductory legend is inconsistent with the actual model levels; the prediction uses the unambiguous age-20–24 row.

The preceding summer supplies $\gamma M_n$ new seeds. After one winter, $\alpha \gamma M_n$ survive and the fraction $\mu$ germinates. Seeds from summer $n-1$ must survive two winters and fail to germinate after the first, so their germinating contribution is $\nu {\alpha }^2(1-\mu )\gamma M_{n-1}$. Older seeds cannot contribute. Applying the density-dependent maturation function givesThis is a seed-bank population recurrence. Put $A=a+b$. Its fixed points satisfy $M_{\ast }=N(A{M}_{\ast })$. The zero solution always exists. If $A>1$, the graph of $N(AM)$ leaves the origin with slope $A>1$, is concave because $N^{\prime }$ decreases, and is bounded above by $N_{\max }$; it therefore crosses the straight line $M$ again at a positive value. Concavity makes this positive crossing unique: the ratio $N(S)/S$ is nonincreasing, starts at one, tends to zero, and is strictly decreasing at any level below one that it attains after the initial slope has dropped. Thus there are exactly two nonnegative steady states when $A>1$.

Linearization at extinction uses $N^{\prime }(0)=1$:The positive root is $r_{+}=(a+\sqrt{a^2+4b})/2$. Since the polynomial at $r=1$ is $1-A<0$, and tends to positive infinity, $r_{+}>1$. Its associated two-component perturbation can be chosen nonnegative, so the growing mode is physically admissible. Extinction is unstable whenever $\boxed{a+b>1}$.

For a nearly Hamiltonian system, the unperturbed Hamiltonian is conserved and its level curves typically contain a family of periodic orbits. A weak perturbation changes the energy slowly. The energy balance method integrates $\dot{H}$ over one unperturbed period to obtain the leading energy change $\Delta H(H)$. A zero of this change selects a possible persistent periodic orbit. A simple zero persists for sufficiently small perturbation by a Poincaré return map and the implicit function theorem. A negative slope of the averaged energy drift gives attraction; a positive slope gives repulsion. A multiple zero requires higher-order analysis, and a zero-free leading drift rules out periodic motion in the corresponding perturbative energy range.

Set $H=({\dot{x}}^2+x^2)/2$. ThenOn an unperturbed circle write $x=r\cos t$, $\dot{x}=-r\sin t$. The supplied integrals giveEquivalently, the leading averaged amplitude equation isWith $s=r^2$, its nonzero stationary amplitudes solve $\beta s^2-2\alpha s+8=0$. If ${\alpha }^2>8\beta$, both roots are positive and simple:The averaged drift is positive below $r_{-}$, negative between $r_{-}$ and $r_{+}$, and positive above $r_{+}$. Thus the inner orbit is attracting and the outer orbit repelling, with leading displacement amplitudes $r_{-}$ and $r_{+}$ and period $2\pi +O(ϵ)$. The simple zeros of the Poincaré return map persist for sufficiently small $ϵ$.

If ${\alpha }^2<8\beta$, the leading quadratic is strictly positive for all $s\ge 0$; the energy increases on every nonzero unperturbed orbit, so the energy balance method yields no periodic orbits in any fixed compact range of nonzero amplitudes for sufficiently small $ϵ$. This perturbative calculation does not itself establish a uniform global exclusion at amplitudes growing as $ϵ$ tends to zero. At equality, the leading drift has a double zero at $r^2=\alpha /\beta$. It is positive on both sides, with a semistable candidate at that amplitude. First-order averaging alone cannot decide whether the exact system has zero, one, or two nearby periodic orbits at equality: higher-order drift determines the unfolding of this degenerate zero. The original PDF has ${\alpha }^2=8\beta$; the extra exponent on $\beta$ in the converted TeX is an OCR error.

In the common sector $0<\arg z<\pi /2$, rotate the positive real contour clockwise to the negative imaginary axis. The large quarter-circle contributes zero because $\mathrm{Re}(z\zeta )>0$ there. The rotated ray passes through the simple pole $\zeta =-i$, whose residue isTake a small semicircular indentation into the right half-plane, excluding the pole from the clockwise contour. The downward ray traverses this semicircle clockwise, contributing $-i\pi R$ in the limit. Thus the deformed downward integral is $\tilde{F}-i\pi R$, and it equals the original integral. ConsequentlyThis half-residue, rather than a full $2\pi i$ residue, results from a pole lying on the rotated integration ray.

Parametrizing $\zeta =-iu$ givesFor $\mathrm{Im}z>0$ its tail decays exponentially. Near $u=1$, subtract the pole coefficient before integrating; the remaining integrand is regular, and the principal-value pole term is explicit. This proves holomorphic dependence on $z$ throughout the upper half-plane, locally uniformly on compact subsets. Agreement in the overlap and the identity theorem therefore give the requested analytic continuation:

The Poisson bracket in canonical coordinates isUsing Hamilton's equations ${\dot{q}}_i=H_{p_i}$ and ${\dot{p}}_i=-H_{q_i}$ in the chain rule yieldsThe partial time derivative holds phase-space coordinates fixed; the total derivative follows the actual trajectory.

The autonomous Hamiltonian is constant and positive on every trajectory, since $q\ne 0$. Hamilton's equations give $\dot{q}=2p$, $\dot{p}=2/q^3$, and henceThus $u=pq-2Ht$ is conserved. Put $s=u+2Ht$. Multiplying the energy equation by $q^2$ gives $H{q}^2=s^2+1$. The full classical path is thereforeThe sign $\eta$ labels the two disconnected half-lines and is fixed by initial data. Differentiating verifies both Hamilton equations. The inverse-square potential prevents the path from crossing zero.

Expand the relativistic energy as $ϵ(p)=m{c}^2+p^2/(2m)+O(p^4/(m^3{c}^2))$. In the dilute nonrelativistic regime the occupation denominator is dominated by its exponential, so the Fermi-Dirac distribution reduces to the Maxwell-Boltzmann distribution. The number density is thenThe assumed low temperature makes the contributing momenta nonrelativistic; the large positive $(m{c}^2-\mu )/(kT)$ makes occupation numbers small.

For chemical equilibrium of hydrogen ionization, ${\mu }_p+{\mu }_e={\mu }_H+{\mu }_{\gamma }$, and thermal photons have ${\mu }_{\gamma }=0$. Apply the same dilute-gas expression to protons, electrons and ground-state hydrogen. Their density ratio isAssume charge neutrality with no other important ionic species, so $n_p=n_e$; neglect the binding-energy mass correction in $m_p/m_H$, so this ratio is approximately one. Including proton and electron spin gives $g_p=g_e=2$ and ground-state $g_H=4$, hence $g_p{g}_e/g_H=1$. Equivalently one may consistently omit nuclear spin from both proton and hydrogen degeneracies. Thus the Saha equation isThe assumptions are thermal and chemical equilibrium at a common temperature, dilute nonrelativistic species, neutral total charge, predominantly ground-state hydrogen, and negligible interaction/partition-function corrections. The Maxwell-Boltzmann density formula applies to dilute hydrogen whether its composite spin makes it a boson or a fermion.

For odd coprime positive integers $a={\prod }_i{p}_i^{e_i}$ and $b={\prod }_j{q}_j^{f_j}$, define the Jacobi symbol by multiplying Legendre symbols over the prime factorization of the denominator. Multiplicativity in the numerator givesBy quadratic reciprocity for the Legendre symbol, this is $(-1{)}^{UV}$, where $U={\sum }_i{e}_i(p_i-1)/2$ and $V={\sum }_j{f}_j(q_j-1)/2$. For odd $x,y$, $(xy-1)/2\equiv (x-1)/2+(y-1)/2(\mathrm{mod}2)$, since $(x-1)(y-1)/2$ is even. Thus $U\equiv (a-1)/2$ and $V\equiv (b-1)/2(\mathrm{mod}2)$. We obtainIf $a,b$ are not coprime, both symbols are zero and the equivalent identity $(a/b)=(-1{)}^{(a-1)(b-1)/4}(b/a)$ still holds. The convention $(a/1)=1$ handles denominator one.

Factor $n={\prod }_j{q}_j^{e_j}$. Since it is not a square, one exponent, say $e_1$, is odd. Choose a nonzero quadratic nonresidue $a$ modulo $q_1$; one exists because the squaring map on the nonzero residues is two-to-one and hence has only $(q_1-1)/2$ images. The Chinese remainder theorem supplies a positive odd integer $m>1$ satisfyingThen $\gcd (m,n)=1$ and $(m/n)=(-1{)}^{e_1}=-1$. The Jacobi reciprocity law proved in part (i), together with $m\equiv 1(\mathrm{mod}4)$, gives $(n/m)=-1$. Factoring $m={\prod }_{\ell }{p}_{\ell }^{r_{\ell }}$ givesAll $p_{\ell }$ are odd and none divides $n$. At least one factor must be negative, yielding an odd prime number $\boxed{p\text{ with }(n/p)=-1}$. No theorem about primes in arithmetic progressions is needed.

A Euclidean lattice in ${\mathbb{R}}^2$ is a discrete additive subgroup, equivalently the integer span of zero, one or two real-linearly independent vectors. Its rank is the number of these generators, or the dimension of its real span. Discreteness makes its intersection with any bounded set finite, so shortest nonzero vectors exist.

First $\Lambda \cap \mathbb{R}{w}_1=\mathbb{Z}{w}_1$. Otherwise an element $t{w}_1$ with noninteger $t$ could be reduced by its nearest integer multiple of $w_1$ to a nonzero vector of length at most $|w_1|/2$, contradicting minimality. Therefore $w_2\notin \mathbb{Z}{w}_1$ is not parallel to $w_1$.

Write any $v\in \Lambda$ as $v=a{w}_1+b{w}_2$ with real coefficients. Subtract nearest integer multiples to obtain $r=s{w}_1+t{w}_2\in \Lambda$, with $|s|,|t|\le 1/2$. If $t\ne 0$, it is not on the first axis, yetwith strict inequality because the independent vectors cannot achieve equality in the triangle inequality; if one coefficient vanishes the inequality is also strict. This contradicts minimality of $w_2$. If $t=0$ and $r\ne 0$, then $|r|\le |w_1|/2$, contradicting minimality of $w_1$. Thus $r=0$ and both original coefficients are integers. Hence $\boxed{\Lambda =\mathbb{Z}{w}_1+\mathbb{Z}{w}_2}$.

For $w_1=(1,0)$ and $w_2=(x,y)$, minimality implies $x^2+y^2\ge 1$ and $|w_2|\le |w_2\pm w_1|$, hence $|x|\le 1/2$. The exact region isThese conditions are sufficient: in the generated lattice, vectors with second coefficient $\pm 1$ have smallest length when the first coefficient is zero, allowing ties at $|x|=1/2$. A vector with second coefficient of magnitude at least two has length at least $2|y|>\sqrt{x^2+y^2}$, since $|y{|}^2\ge 3/4$ and $x^2\le 1/4$. Vectors on the first axis have length at least one. Thus the shaded region includes exactly all allowable choices, including ties on its boundary.

For a second lattice basis, write $v_1=a{w}_1+b{w}_2$, $v_2=c{w}_1+d{w}_2$ with integer coefficients, as just proved. Conversely each $w_i$ is an integer combination of the $v_j$. The change-of-basis matrix $\left(\begin{array}{cc}a& c\\ b& d\end{array}\right)$ and its inverse therefore both have integer entries. Their determinants are integers with product one, giving $\boxed{ad-bc=\pm 1}$. Thus lattice bases are related by a unimodular matrix.

For ${\sigma }^2>0$, the Gaussian linear model has likelihood functionMaximizing over $\beta$ minimizes the residual sum of squares. Full column rank makes $X^{T}X$ positive definite, and differentiating gives the unique least-squares estimatorPut $P=X(X^{T}X{)}^{-1}{X}^T$ and $R=\Vert (I-P)Y{\Vert }^2$. If $R>0$, differentiating the profile log-likelihood in ${\sigma }^2$ givesThe denominator is $n$, not the residual degrees of freedom $n-p$; the latter gives the unbiased estimator.

A linear transformation of the multivariate normal distribution remains normal, soIts covariance with the residual vector is ${\sigma }^2(X^{T}X{)}^{-1}{X}^T(I-P)=0$. Jointly Gaussian vectors with zero cross-covariance are independent, hence $\hat{\beta }$ is independent of $(I-P)Y$, and therefore of ${\hat{\sigma }}^2$, a function of that residual. Orthogonal coordinates also give $R/{\sigma }^2\sim {\chi }_{n-p}^2$.

The allowed boundary case $p=n$ needs qualification. Then $P=I$, $R=0$ for every observation, and the likelihood increases without bound as ${\sigma }^2\downarrow 0$ after fitting $Y$ exactly. There is no maximum in the positive-variance model; $R/n=0$ is only a formal boundary estimate. When $n>p$, $R>0$ almost surely under a genuinely positive variance, and the displayed MLEs apply. A zero residual in any exceptional exact-fit sample has the same nonexistence issue. The estimator $\hat{\beta }$ and its normal distribution remain well-defined in the square full-rank case.

The First Lyapunov theorem states that for a locally Lipschitz autonomous system $\dot{x}=f(x)$ with equilibrium zero, if a $C^1$ function $V$ is positive definite near zero and $\dot{V}=\nabla V\cdot f\le 0$, then the equilibrium is Lyapunov stable. To prove it, choose a small sphere $|x|=\epsilon$ inside that neighborhood. Compactness and positivity give $m={\min }_{|x|=\epsilon }V(x)>0$. Continuity at zero gives a smaller ball $|x|<\delta$ where $V<m$. Since $V$ cannot increase along a trajectory, a trajectory starting there cannot first cross the sphere where $V\ge m$. It stays within the sphere for all future time, with continuation justified by confinement in a compact subset of the vector field's domain. This proves stability for every sufficiently small $\epsilon$.

The LaSalle invariance principle states that on a compact positively invariant set on which $\dot{V}\le 0$, every trajectory approaches the largest invariant subset of $\{\dot{V}=0\}$. If that subset is just the equilibrium, the trapping and stability argument proves asymptotic stability.

For example take $\dot{x}=y$, $\dot{y}=-x-y$ and $V=(x^2+y^2)/2$. Then $\dot{V}=-y^2$, which is not strictly negative on the punctured plane. Nevertheless its sublevel discs are compact and invariant, and a trajectory confined to $y=0$ must have $\dot{y}=-x=0$, hence $x=0$. LaSalle gives asymptotic stability despite the non-strict orbital derivative of this particular energy function.

The printed phrase about nonexistence of a strict Lyapunov function must be understood as a limitation of the chosen candidate. In this example a strict function actually exists:More generally the converse Lyapunov theorem supplies local strict functions for asymptotically stable sufficiently smooth systems. LaSalle avoids having to find such a function; it does not show that none exists.

Choose the ellipsoidal Lyapunov function $V=x^2+2y^2$. Direct substitution of the printed vector field givesThe weighted Cauchy-Schwarz inequality gives $(x-y{)}^2\le \frac{3}{2}(x^2+2y^2)=3V/2$, soFor any initial point with $V(0)<2/3$, choose $c$ with $V(0)<c<2/3$. The compact ellipse $V\le c$ is positively invariant, since $\dot{V}<0$ on its boundary. Inside it,This establishes global forward existence for those trajectories and convergence to the origin, while the First Lyapunov theorem establishes stability. The origin is asymptotically stable and its basin contains $\boxed{x^2+2y^2<2/3}$. The PDF coefficient in the second equation is $x^{2}y/2$, with the entire linear term $-x$; placing the factor $1/2$ on the linear term would destroy this factorization.

A canonical transformation is a differentiable invertible change of phase-space coordinates preserving the symplectic form, equivalently the canonical Poisson brackets. At each fixed time, let $D$ be its Jacobian in the ordering $(q,p)\mapsto (Q,P)$ and put $J=\left(\begin{array}{cc}0& I\\ -I& 0\end{array}\right)$. The condition iswhich is equivalent to $D^{T}JD=J$. Time dependence changes the Hamiltonian by a generating-function time derivative, but not this fixed-time symplectic condition.

The Hamilton-Jacobi equation is $S_t+\frac{1}{2}[(S_q{)}^2+{\omega }^2{q}^2]=0$. Separating $S=W(q,E)-Et$ gives $(W_q{)}^2=2E-{\omega }^2{q}^2$. On the positive-momentum branch, therefore,Choose the integration constant so that the given angle relation is obtained. For $E>0$, $\beta =S_E$ yields $q=\sqrt{2E}\sin [\omega (t+\beta )]/\omega$, while $p=S_q=\sqrt{2E-{\omega }^2{q}^2}$ on this branch. Continuing the angle across the turning points gives the full orbitThe square-root expression is locally the positive branch; the cosine formula correctly changes sign on the rest of the orbit.

The generating function has differential $dS=pdq+\beta d\alpha -Hdt$. At fixed time, taking an exterior derivative gives $dq\wedge dp=d\beta \wedge d\alpha$. Thus the canonical new coordinate is $Q=\beta$ and its conjugate momentum is $P=\alpha$. Directly, $\alpha =(p^2+{\omega }^2{q}^2)/2$ and $\beta =atan2(\omega q,p)/\omega -t$ locally implyThe angle is a local chart or a periodic variable, and is undefined at the zero-energy point. The source's phrase “variables $(\alpha ,\beta )$” lists the new momentum before its coordinate: if interpreted instead as the conventional coordinate-then-momentum ordering, that ordered pair would have $\{\alpha ,\beta \}=-1$ and be anti-canonical.

The cumulative hierarchy is defined by transfinite recursion:Prove transitivity and nesting together by transfinite induction. The empty set is transitive. If $V_{\alpha }$ is transitive, each $x\in V_{\alpha }$ satisfies $x\subseteq V_{\alpha }$, so $V_{\alpha }\subseteq \mathcal{P}(V_{\alpha })$. If $y\in x\in \mathcal{P}(V_{\alpha })$, then $y\in V_{\alpha }\subseteq \mathcal{P}(V_{\alpha })$; thus the successor stage is transitive. At a limit, if $y\in x\in V_{\lambda }$, choose an earlier stage containing $x$ and apply its transitivity. The union is transitive and contains every earlier stage. These observations also give $V_{\alpha }\subseteq V_{\beta }$ for all $\alpha \le \beta$.

By Axiom of foundation and well-founded recursion, define the rank of a set by $\mathrm{rank}(x)=\sup \{\mathrm{rank}(y)+1:y\in x\}$. Axiom schema of replacement ensures this is an ordinal. Induction shows every element of $x$ lies in $V_{\mathrm{rank}(x)}$, soThus every set appears in the hierarchy; the argument uses foundation, not the existence of a set containing all sets.

If $\mathrm{rank}(x)=\beta$, every subset of $x$ has rank at most $\beta$, while $x$ itself belongs to $\mathcal{P}(x)$. ConsequentlyOnly successor ordinals occur. Conversely $\mathrm{rank}(V_{\beta })=\beta$, so choosing $x=V_{\beta }$ realizes every successor $\beta +1$. The exact answer is all successor ordinals, excluding zero and nonzero limit ordinals.

With one edge colour, $f(1)=3$. Suppose the assertion is known for $k-1$ colours, and take $n=k(f(k-1)-1)+2$. At any vertex, among its $n-1$ incident edges some colour occurs on at least $f(k-1)$ edges. Let $S$ be their other endpoints. If an edge within $S$ has that colour, it completes a monochromatic triangle with the chosen vertex. Otherwise $S$ uses only $k-1$ colours internally, and the inductive hypothesis gives a monochromatic triangle there. ThusInduction now yieldsTogether with the base case, $\boxed{f(k)\le 3k!}$ for every positive $k$. This is a constructive multicolour Ramsey bound.

In any two-colouring of $K_7$, there is a monochromatic triangle because $f(2)=6$. Call its colour blue. If any edge from one of its vertices to an outside vertex is blue, the triangle and that edge give the required copy of $H$. Otherwise all edges from that triangle to the four outside vertices are yellow. If any edge among those four is yellow, it makes a yellow triangle with a vertex of the original triangle, and another incident yellow edge supplies the pendant edge. If none is yellow, the four outside vertices form a blue $K_4$, which also contains $H$. Hence $g(2)\le 7$.

For the lower bound, divide six vertices into two triples. Colour edges inside each triple blue and all cross-edges yellow. Each blue component is only a triangle, and the yellow graph is bipartite, so neither colour contains $H$. Therefore $\boxed{g(2)=7}$. Copies are not required to be induced: an additional edge in a monochromatic $K_4$ does not invalidate the copy of $H$.

For three colours, $f(3)=17$ supplies a triangle in $K_{17}$. If it has no pendant edge of the same colour, each of its vertices has fourteen edges to outside vertices in the other two colours. Choose one vertex $v$ and a colour occurring on at least seven of those edges, with endpoint set $S$. If an internal edge of $S$ has this colour, it creates a triangle with $v$ and a fourth neighbour supplies the pendant edge. Otherwise $S$ is internally two-coloured, and $g(2)=7$ supplies $H$. Thus $g(3)\le 17$. Conversely the definition $f(3)=17$ ensures a three-colouring of $K_{16}$ without any monochromatic triangle, and therefore without $H$. Hence $\boxed{g(3)=17}$.