past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ii/paper-4.bigb

= Paper 4
{scope}

https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2021/paperii_4_2021.pdf

= 1I
{parent=Paper 4}
{scope}
{title2=Number Theory}

= Solution
{parent=1I}

<Legendre formula> gives
$$v_p{2n\choose n}=\sum_{j\geq1}(\lfloor2n/p^j\rfloor-2\lfloor n/p^j\rfloor),$$
whose summands are zero or one. If the sum is at least $k$, some nonzero summand has index $j\geq k$, whence $p^k\leq p^j\leq2n$.

Grouping the von Mangoldt function by prime gives
$$\psi(x)=\sum_{p\leq x}\lfloor\log x/\log p\rfloor\log p.$$
Every prime-power contribution to $\log {2n\choose n}$ occurs in $\psi(2n)$ by the first part. Finally ${2n\choose n}=\prod_{j=1}^n(n+j)/j\geq2^n$, so $\psi(2n)\geq n\log2$.

Solved by gpt-5.6-sol high.

= 2H
{parent=Paper 4}
{scope}
{title2=Topics in Analysis}

= a
{parent=2h}
{scope}

= Solution
{parent=a}

The <Brouwer fixed-point theorem> says that every continuous map from a closed two-dimensional disk to itself has a fixed point.

Solved by gpt-5.6-sol high.

= b
{parent=2h}
{scope}

= Solution
{parent=b}

The equivalent <no-retraction theorem> says there is no continuous retraction of the closed disk onto its boundary circle. A fixed-point-free map gives a retraction by projecting from $f(x)$ through $x$ to the boundary; conversely a retraction composed with the antipodal boundary map gives a fixed-point-free self-map.

Solved by gpt-5.6-sol high.

= c
{parent=2h}
{scope}

= Solution
{parent=c}

On the simplex define $f(x)=Ax/(\mathbf1^TAx)$. Strict positivity makes the denominator positive and maps the simplex continuously into its interior. By <Brouwer fixed-point theorem>, $f(x)=x$ for some $x$. Thus $Ax=λx$, where $λ=\mathbf1^TAx>0$, and because $Ax$ is strictly positive, so is $x$.

Solved by gpt-5.6-sol high.

= 3K
{parent=Paper 4}
{scope}
{title2=Coding and Cryptography}

= Solution
{parent=3K}

For <Rabin cryptosystem>, choose primes $p,q\equiv3\pmod4$, publish $N=pq$, and encrypt $x$ as $x^2\bmod N$. The factors let the receiver take square roots modulo $p$ and $q$ and combine them using the <Chinese remainder theorem>; redundancy identifies the intended one of four roots.

For <RSA cryptosystem>, choose $N=pq$, select $e$ coprime to $\varphi(N)$ and $d$ with $ed\equiv1\pmod{\varphi(N)}$. Encrypt $x$ as $x^e$ and decrypt by raising to $d$ modulo $N$. Rabin inversion is provably equivalent to factoring, while ordinary RSA lacks that reduction; Rabin's disadvantage is its fourfold decryption ambiguity.

Solved by gpt-5.6-sol high.

= 4F
{parent=Paper 4}
{scope}
{title2=Automata and Formal Languages}

= i
{parent=4f}
{scope}

= Solution
{parent=i}

The <pumping lemma for regular languages> says that sufficiently long accepted words split as $xyz$ with $|xy|\leq P$, $|y|>0$, and $xy^kz$ accepted for all $k\geq0$. Applying it to $0^P1^P01$ pumps only zeros, so language (i) is not regular.

Solved by gpt-5.6-sol high.

= ii
{parent=4f}
{scope}

= Solution
{parent=ii}

This language is not regular. Intersecting with the regular language $0^*11 0^*$ isolates strings whose two zero blocks must have equal length; the pumping lemma then gives a contradiction.

Solved by gpt-5.6-sol high.

= iii
{parent=4f}
{scope}

= Solution
{parent=iii}

A word containing neither $01$ nor $10$ cannot change symbol. The language is therefore $0^*\cup1^*$, hence is regular.

Solved by gpt-5.6-sol high.

= 5J
{parent=Paper 4}
{scope}
{title2=Statistical Modelling}

= Solution
{parent=5J}

This is <Poisson regression> with independent counts and log mean $\log μ_d=1.5624+0.1658d$. The residual deviance $9.90$ on $12$ degrees of freedom has $p=0.625$, and the plotted curve tracks the data, so there is no evidence of poor fit.

The fitted doubling time is $\log2/0.1658=4.16$ days. A 95% interval for the growth coefficient is $0.1658±1.96(0.0166)=(0.133,0.198)$, corresponding to doubling times $(3.50,5.21)$ days. Seven days, whose rate is about $0.099$, is unsupported.

Solved by gpt-5.6-sol high.

= 6E
{parent=Paper 4}
{scope}
{title2=Mathematical Biology}

= a
{parent=6e}
{scope}

= Solution
{parent=a}

The terms are logistic birth and crowding, $rn-rn^2/K$, linear predation mortality $-δn$, and diffusion $Dn_{xx}$. The coast is reflecting, so $n_x(0)=0$; lethal habitat beyond $L$ gives the absorbing condition $n(L)=0$.

Solved by gpt-5.6-sol high.

= b
{parent=6e}
{scope}

= Solution
{parent=b}

Linearizing at extinction gives $n_t=(r-δ)n+Dn_{xx}$. The principal eigenfunction for Neumann data at $0$ and Dirichlet data at $L$ is $\cos(πx/(2L))$, with growth rate
$$r-δ-Dπ^2/(4L^2).$$
A population is viable exactly when this is positive, requiring $r>δ$ and $L>(π/2)\sqrt{D/(r-δ)}$. Thus intrinsic growth must beat predation and the refuge must exceed the diffusion-dependent critical size.

Solved by gpt-5.6-sol high.

= 7E
{parent=Paper 4}
{scope}
{title2=Further Complex Methods}

= a
{parent=7e}
{scope}

= Solution
{parent=a}

A Riemann P-symbol specifies a second-order Fuchsian equation with regular singular points $a,b,c$ and the two local exponents shown beneath each point. The Fuchs relation is $α+α\prime+β+β\prime+γ+γ\prime=1$.

Solved by gpt-5.6-sol high.

= b
{parent=7e}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

Set $z\prime=(1-z)/2$, sending $1,-1,∞$ to $0,1,∞$. Then rescale the dependent variable by
$$y=z\prime^{-m/2}(1-z\prime)^{m/2}u.$$
This shifts the exponent pairs to $(0,m)$ at zero and $(0,-m)$ at one, while leaving the exponents at infinity unchanged.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

The hypergeometric exponent differences give $1-c=m$, $c-a-b=-m$, and exponents $a,b$ at infinity. Hence one may take
$$a=n,\qquad b=1-n,\qquad c=1-m,\qquad z\prime=(1-z)/2,$$
so a solution is ${}_2F_1(n,1-n;1-m;(1-z)/2)$, with the prefactor from part (i) when returning to the original dependent variable.

Solved by gpt-5.6-sol high.

= 8D
{parent=Paper 4}
{scope}
{title2=Classical Dynamics}

= Solution
{parent=8D}

A Lagrange top is an axially symmetric rigid body with one point fixed, centre of mass a distance $l$ along its symmetry axis, principal moments $I_1=I_2,I_3$, mass $M$, and Euler angles $θ,φ,ψ$ in gravity $g$.

The cyclic coordinates give $p_ψ=I_3(\dot ψ+\dot φ\cosθ)$ and $p_φ=I_1\dot φ\sin^2θ+p_ψ\cosθ$; the energy $E$ is the third integral. Eliminating $\dot φ,\dot ψ$ from $E$ and setting $u=\cosθ$, with $\dot u=-\dot θ\sinθ$, yields $\dot u^2=f(u)$ after multiplication by $1-u^2$; all denominators cancel and $f$ is cubic.

Solved by gpt-5.6-sol high.

= 9B
{parent=Paper 4}
{scope}
{title2=Cosmology}

= a
{parent=9b}
{scope}

= Solution
{parent=a}

For $I=\frac12\sum m_i|x_i|^2$, differentiation gives
$$\ddot I=2T+\sum_i x_i\mathbin\cdot F_i=2T+V,$$
because the Newtonian potential is homogeneous of degree $-1$. Bounded positions and velocities imply $\overline{\ddot I}=0$, so the <virial theorem> gives $\overline T=-\overline V/2$.

Solved by gpt-5.6-sol high.

= b
{parent=9b}
{scope}

= Solution
{parent=b}

Observed stellar velocities estimate $T$, while visible mass and size predict $V$. Galaxies and clusters often have far more kinetic energy than luminous matter can bind under the <virial theorem>. The additional gravitational potential is evidence for unseen dark matter.

Solved by gpt-5.6-sol high.

= 10D
{parent=Paper 4}
{scope}
{title2=Quantum Information and Computation}

= a
{parent=10d}
{scope}

= Solution
{parent=a}

Writing $ω=e^{2πi/N}$, $|χ_k\rangle=N^{-1/2}\sum_jω^{-jk}|j\rangle$. Therefore $S|χ_k\rangle=ω^k|χ_k\rangle$.

Solved by gpt-5.6-sol high.

= b
{parent=10d}
{scope}

= Solution
{parent=b}

If $|v
angle=\sum_kc_k|χ_k
angle$, then QFT measurement has probabilities $|c_k|^2$. Applying $S$ multiplies $c_k$ by the phase $ω^k$, so the probabilities are unchanged.

Solved by gpt-5.6-sol high.

= c
{parent=10d}
{scope}

= Solution
{parent=c}

The offset state is $S^{x_0}$ applied to the offset-zero periodic state. Part (b) therefore makes its QFT output distribution independent of $x_0$. Explicitly, only outcomes divisible by $A=N/r$ occur, each with probability $1/r$; $x_0$ contributes only a phase.

Solved by gpt-5.6-sol high.

= 11I
{parent=Paper 4}
{scope}
{title2=Number Theory}

= a
{parent=11i}
{scope}

= Solution
{parent=a}

$N$ is a <Fermat pseudoprime> to base $b$ when it is composite, coprime to $b$, and $b^{N-1}\equiv1\pmod N$. Here $N\mid b^k-1$ gives coprimality and $k\mid N-1$, so $b^{N-1}=(b^k)^{(N-1)/k}\equiv1\pmod N$.

Solved by gpt-5.6-sol high.

= b
{parent=11i}
{scope}

= Solution
{parent=b}

Both $n$ and $m$ divide $b^{2p}-1$, hence so does $N=nm$ after checking $\gcd(n,m)=1$ under the hypothesis. Also $n\equiv p\pmod{b-1}$ and the elementary factorization gives $N\equiv1\pmod{2p}$; part (a) with $k=2p$ applies. As odd primes $p$ grow, $N$ grows, yielding infinitely many examples.

Solved by gpt-5.6-sol high.

= c
{parent=11i}
{scope}

= Solution
{parent=c}

By the <Chinese remainder theorem>, $b^{n-1}\equiv1\pmod n$ iff it holds modulo every $p_j$. Since $n-1=p_jr_j-1\equiv r_j-1\pmod{p_j-1}$ and the order of $b$ modulo $p_j$ divides $p_j-1$, this is equivalent to $\operatorname{ord}_{p_j}(b)\mid r_j-1$ for every $j$.

Solved by gpt-5.6-sol high.

= d
{parent=11i}
{scope}

= Solution
{parent=d}

For each prime $k\geq5$, choose a prime divisor $p$ of $2^k-1$ and a prime divisor $q\ne3$ of $2^k+1$. Their orders are $k$ and $2k$, respectively, so $p\equiv1\pmod k$, $q\equiv1\pmod{2k}$, and the criterion in part (c) shows $pq$ is a base-two pseudoprime. The supplied gcd fact makes examples from distinct prime $k$ distinct, so there are infinitely many.

Solved by gpt-5.6-sol high.

= 12H
{parent=Paper 4}
{scope}
{title2=Topics in Analysis}

= Solution
{parent=12H}

Multiplying the matrices inductively gives columns $(p_n,q_n)^T$ and $(p_{n-1},q_{n-1})^T$. Their determinant is $(-1)^{n-1}$, and writing $x=(p_nx_{n+1}+p_{n-1})/(q_nx_{n+1}+q_{n-1})$ gives
$$|x-p_n/q_n|={1\over q_n(q_nx_{n+1}+q_{n-1})}\leq{1\over q_nq_{n+1}}.$$
If $Ay^2+By+C=0$, then for rationals $p/q$ near $y$, factorization against the conjugate root gives $|p/q-y|\geq M/q^2$; rationals away from $y$ are handled by reducing $M$. Finally $q_{n+1}\geq a_{n+1}q_n$, so the upper bound is at most $1/(a_{n+1}q_n^2)$. Unbounded partial quotients contradict the fixed lower bound for a quadratic irrational.

Solved by gpt-5.6-sol high.

= 13J
{parent=Paper 4}
{scope}
{title2=Statistical Modelling}

= a
{parent=13j}
{scope}

= Solution
{parent=a}

Differentiating the ridge objective gives $(X^TX+λI)\widehat β=X^TY$, hence $\widehat μ=X(X^TX+λI)^{-1}X^TY=H_λY$.

Solved by gpt-5.6-sol high.

= b
{parent=13j}
{scope}

= Solution
{parent=b}

Since $Y-HY=(I-H)μ+(I-H)ε$, the cross term has mean zero and $H$ is symmetric. Thus the expectation is $\|(I-H)μ\|^2+σ^2\operatorname{tr}(I-H)^2$, which expands to the stated expression.

Solved by gpt-5.6-sol high.

= c
{parent=13j}
{scope}

= Solution
{parent=c}

Independence gives $\mathbb E\|Y^*-HY\|^2=\|(I-H)μ\|^2+nσ^2+σ^2\operatorname{tr}H^2$. Comparing with part (b) shows the difference is $2σ^2\operatorname{tr}H$, proving unbiasedness.

Solved by gpt-5.6-sol high.

= d
{parent=13j}
{scope}

= Solution
{parent=d}

Here $H=hI$, $h=(1+λ)^{-1}$, and prediction risk is
$$R(h)=nσ^2+(1-h)^2\|μ\|^2+nσ^2h^2.$$
It decreases then increases, with $h^*=\|μ\|^2/(\|μ\|^2+nσ^2)$, equivalently $λ^*=nσ^2/\|μ\|^2$. Its minimum is $nσ^2+nσ^2\|μ\|^2/(\|μ\|^2+nσ^2)$; the endpoint limits are $2nσ^2$ and $nσ^2+\|μ\|^2$.

Solved by gpt-5.6-sol high.

= 14E
{parent=Paper 4}
{scope}
{title2=Mathematical Biology}

= a
{parent=14e}
{scope}

= Solution
{parent=a}

This bistable reaction-diffusion equation models an invading population with an Allee effect: densities below $r$ decline, while those above it grow toward carrying capacity one.

Solved by gpt-5.6-sol high.

= b
{parent=14e}
{scope}

= Solution
{parent=b}

The homogeneous equilibria are $0,r,1$. Since $f\prime(0)=-r<0$, $f\prime(1)=r-1<0$, and $f\prime(r)=r(1-r)>0$, the first and last are linearly stable and the middle one unstable.

Solved by gpt-5.6-sol high.

= c
{parent=14e}
{scope}

= Solution
{parent=c}

Multiplying $Dn\prime\prime+f(n)=0$ by $n\prime$ and using the endpoint conditions gives the logistic front
$$n_0(x)={1\over1+e^{x/\sqrt{2D}}}.$$
It has $n_0(0)=1/2$ and direct differentiation verifies the equation for $r=1/2$.

Solved by gpt-5.6-sol high.

= d
{parent=14e}
{scope}

= Solution
{parent=d}

With $ξ=x-ct$, the equation is $Du\prime\prime+cu\prime+f(u)=0$. Since $f_r(u)-f_{1/2}(u)=u(1-u)(1/2-r)$ and $n_0\prime=-n_0(1-n_0)/\sqrt{2D}$, the same profile works with
$$c=\sqrt{D/2}(1-2r).$$

Solved by gpt-5.6-sol high.

= e
{parent=14e}
{scope}

= Solution
{parent=e}

The speed is positive for $r<1/2$, zero at $1/2$, and negative for $r>1/2$. The state with the larger basin-weighted reaction potential invades the other; raising the Allee threshold favours extinction.

Solved by gpt-5.6-sol high.

= 15D
{parent=Paper 4}
{scope}
{title2=Classical Dynamics}

= a
{parent=15d}
{scope}

= Solution
{parent=a}

The <Poisson bracket> is $\{f,g\}=\sum_i(f_{q_i}g_{p_i}-f_{p_i}g_{q_i})$. Canonical variables satisfy $\{Q_i,Q_j\}=\{P_i,P_j\}=0$ and $\{Q_i,P_j\}=δ_{ij}$.

Solved by gpt-5.6-sol high.

= b
{parent=15d}
{scope}

= Solution
{parent=b}

To first order the conditions are $\partial_{p_j}δq_i-\partial_{p_i}δq_j=0$, $\partial_{q_j}δp_i-\partial_{q_i}δp_j=0$, and $\partial_{q_j}δq_i+\partial_{p_i}δp_j=0$. The choices $δq_i=εF_{p_i}$ and $δp_i=-εF_{q_i}$ satisfy them by equality of mixed partials.

Solved by gpt-5.6-sol high.

= c
{parent=15d}
{scope}

= Solution
{parent=c}

Taylor expansion gives $δH=H_q\mathbin\cdotδq+H_p\mathbin\cdotδp=-ε\{F,H\}$. Hamilton's equation gives $\dot F=\{F,H\}$, so $F$ is conserved exactly when the generated infinitesimal canonical transformation leaves $H$ invariant.

Solved by gpt-5.6-sol high.

= d
{parent=15d}
{scope}

= Solution
{parent=d}

For a constant translation vector $a$, take $F=a\mathbin\cdot\sum_i p_i$. Then $δr_i=εa$ and $δp_i=0$. The potential depends only on position differences and the kinetic energy only on momenta, so this is a symmetry and total momentum is conserved.

Solved by gpt-5.6-sol high.

= 16G
{parent=Paper 4}
{scope}
{title2=Logic and Set Theory}

= Solution
{parent=16G}

<Axiom of foundation> says every nonempty set $A$ contains $a$ with $a\cap A=\emptyset$. Define $T_0=x$, $T_{n+1}=\bigcup T_n$, and $\operatorname{TC}(x)=\bigcup_{n<ω}T_n$ (including $x$ if that convention is desired). Replacement and union form this set, and it is transitive; induction shows every transitive set containing $x$ contains it.

The principle of membership induction says that if $(\forall y\in x\ P(y))\Rightarrow P(x)$ for every $x$, then $P$ holds for every set. Otherwise Foundation applied to the set of counterexamples in a suitable transitive closure gives a minimal counterexample.

Apply induction to $P(x):F(x)=x$. If it holds for all $y\in x$, then extensionality and preservation plus surjectivity give $z\in F(x)$ iff $z=F(y)$ for some $y\in x$, iff $z\in x$. Hence $F(x)=x$.

Solved by gpt-5.6-sol high.

= 17G
{parent=Paper 4}
{scope}
{title2=Graph Theory}

= Solution
{parent=17G}

<Hall marriage theorem> states that a bipartite graph with parts $A,B$ has a matching saturating $A$ iff $|N(S)|\geq|S|$ for every $S\subseteq A$. Starting with a maximum matching, an unmatched vertex and its alternating reachable set would violate Hall unless an augmenting path exists; flipping along that path increases the matching, proving sufficiency.

Every vertex cover meets each edge of a matching, so $γ\leqβ$. Endpoints of a maximal matching cover every edge, so $β\leq2γ$. A disjoint union of $k$ triangles has $(γ,β)=(k,2k)$. For $T_3(30)$, $γ=15$ and $β=30-10=20$. In bipartite graphs the alternating-path proof constructs a cover of size equal to a maximum matching, giving König theorem.

The chromatic index is the minimum number of matchings partitioning the edges. Label vertices of $K_{2^r}$ by the vector space over $\mathbb F_2$; for each nonzero $a$, pair $x$ with $x+a$. These $2^r-1$ perfect matchings partition all edges, while degree gives the matching lower bound, so $χ\prime=n-1$.

Solved by gpt-5.6-sol high.

= 18I
{parent=Paper 4}
{scope}
{title2=Galois Theory}

= a
{parent=18i}
{scope}

= Solution
{parent=a}

An action by field automorphisms is a homomorphism $G	o\operatorname{Aut}(L)$. Elements fixed by every group element are closed under addition, multiplication, additive inverses, and inversion of nonzero elements, so $L^G$ is a subfield.

Solved by gpt-5.6-sol high.

= b
{parent=18i}
{scope}

= Solution
{parent=b}

The orbit polynomial $\prod_{β\in Gα}(X-β)$ has coefficients fixed by $G$, lies in $K[X]$, and has distinct roots. Thus $α$ is separable and its degree is the orbit size $|G:G_α|$, which divides $|G|$. If $L=K(α)$ and the action is faithful, only the identity fixes $α$, so the degree is $|G|$.

Solved by gpt-5.6-sol high.

= c
{parent=18i}
{scope}

= Solution
{parent=c}

The relations $σ^n=τ^2=1$ and $τσ τ=σ^{-1}$ give the dihedral group; its $2n$ substitutions $z\mapstoζ_n^jz$ and $z\mapstoζ_n^j/z$ are distinct. The element $w=z^n+z^{-n}$ is fixed. Since $z$ satisfies $X^{2n}-wX^n+1=0$, the extension has degree at most $2n$, while the automorphism group gives degree at least $2n$. Hence $L^G=\mathbb C(w)$.

Solved by gpt-5.6-sol high.

= 19I
{parent=Paper 4}
{scope}
{title2=Representation Theory}

= a
{parent=19i}
{scope}

= Solution
{parent=a}

$S^1=\{z\in\mathbb C:|z|=1\}$. A continuous finite-dimensional representation is unitary after averaging an inner product over Haar measure, and commuting unitary matrices diagonalize simultaneously. Its irreducibles are therefore $z\mapsto z^n$, $n\in\mathbb Z$. Their characters obey $\int_{S^1}z^n\overline{z^m}\,dz=δ_{nm}$.

Solved by gpt-5.6-sol high.

= b
{parent=19i}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

$SU(2)$ consists of unitary $2\times2$ matrices of determinant one. Every element is unitarily diagonalizable with eigenvalues $e^{±iθ}$, and two are conjugate exactly when they have the same trace. Thus $t(g)=\operatorname{tr}(g)/2=\cosθ$ bijects conjugacy classes with $[-1,1]$.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

For $-1<x<1$, the centralizer of the diagonal representative is the maximal torus $T$. Hence the orbit is $SU(2)/T$. Identifying $SU(2)$ with $S^3$ makes this quotient the Hopf quotient $S^3/S^1\cong S^2$.

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

If $V$ is the defining representation, $χ_V=2t$. The character of $\operatorname{Sym}^3V$ is $(2t)^3-2(2t)=8t^3-4t$. Therefore $8t^3$ is the character of $\operatorname{Sym}^3V\oplus V\oplus V$.

Solved by gpt-5.6-sol high.

= 20G
{parent=Paper 4}
{scope}
{title2=Number Fields √}

= a
{parent=20g}
{scope}

= Solution
{parent=a}

$\mathcal O_K=\mathbb Z[\sqrt{30}]$ and $d_K=120$, so every class has an ideal of norm at most $\sqrt{30}$. Factoring the primes $2,3,5$ and checking principal norms shows that the ramified prime ideals represent one nontrivial class of order two; hence $\operatorname{Cl}(K)\cong\mathbb Z/2$. The continued fraction of $\sqrt{30}$ gives the least unit greater than one as $11+2\sqrt{30}$, of norm $1$.

Solved by gpt-5.6-sol high.

= b
{parent=20g}
{scope}

= Solution
{parent=b}

The two-sign map $K^*\to\{±1\}^2$ is surjective, with kernel the totally positive elements, so their index is four. Narrow equivalence is reflexive, symmetric, and transitive because totally positive elements form a group; ideal multiplication makes the classes an abelian group.

The map from the narrow class group to the ordinary class group is onto. Its kernel records sign patterns of principal generators modulo signs realized by units. A norm-$-1$ fundamental unit realizes the two mixed signs, making the kernel trivial; otherwise only equal signs occur and the kernel has order two. Thus the narrow class number is $h_K$ in the first case and $2h_K$ otherwise. Here it is $4$.

Solved by gpt-5.6-sol high.

= 21F
{parent=Paper 4}
{scope}
{title2=Algebraic Topology}

= a
{parent=21f}
{scope}

= Solution
{parent=a}

For a finite triangulation, the <Euler characteristic> is $χ(X)=V-E+F$, independent of the triangulation.

Solved by gpt-5.6-sol high.

= b
{parent=21f}
{scope}

= Solution
{parent=b}

Choose a triangulation of the sphere having every branch point as a vertex. Its lift has twice every edge and face, and twice every nonbranch vertex, but only one vertex above each of the $n$ branch points. Hence
$$χ(Σ_g)=2χ(S^2)-n=4-n.$$
Since $χ(Σ_g)=2-2g$, this gives $n=2g+2$, the degree-two <Riemann-Hurwitz formula>.

Solved by gpt-5.6-sol high.

= 22H
{parent=Paper 4}
{scope}
{title2=Linear Analysis}

= a
{parent=22h}
{scope}

= Solution
{parent=a}

For fixed $y$, the map $x\mapsto\langle Tx,y\rangle$ is a bounded functional. The <Riesz representation theorem> gives a unique vector $T^*y$ representing it. Linearity, uniqueness, and $\|T^*y\|\leq\|T\|\|y\|$ make $T^*$ the unique bounded <adjoint operator>.

Solved by gpt-5.6-sol high.

= b
{parent=22h}
{scope}

= Solution
{parent=b}

A countable orthonormal basis is a frame with $A=B=1$. For any frame, a vector orthogonal to every $e_i$ makes the lower frame bound force its norm to vanish. Therefore the span is dense.

Solved by gpt-5.6-sol high.

= c
{parent=22h}
{scope}

= Solution
{parent=c}

The upper frame bound gives $\|Ux\|_{\ell^2}\leq\sqrt B\|x\|$. For $c\in\ell^2$, the adjoint is the norm-convergent synthesis series
$$U^*c=\sum_i c_i e_i,$$
because its inner product with $x$ is $\sum_i c_i\langle e_i,x\rangle$.

Solved by gpt-5.6-sol high.

= d
{parent=22h}
{scope}

= Solution
{parent=d}

Given $ε>0$, choose $N$ with $\sum_{i>N}|\langle a,e_i\rangle|^2<ε^2$. Every cube point has uniformly bounded tail by this sum, while its first $N$ coordinates lie in a compact finite-dimensional box. A finite net for that box is therefore an $ε$-net for the cube. The coordinate inequalities define a closed set, so the cube is complete and totally bounded, hence <compact set>[compact].

Solved by gpt-5.6-sol high.

= 23H
{parent=Paper 4}
{scope}
{title2=Analysis of Functions}

= a
{parent=23h}
{scope}

= Solution
{parent=a}

Strong convergence means $\|f_j-f\|_p\to0$; weak convergence means $\int f_jh\to\int fh$ for every $h\in L^q$. By duality choose $h$ of norm one with $\int fh$ arbitrarily close to $\|f\|_p$. <Holder inequality> and passage to the limit give $\|f\|_p\leq\liminf\|f_j\|_p$.

Solved by gpt-5.6-sol high.

= b
{parent=23h}
{scope}

= Solution
{parent=b}

Because $1<p<∞$, $L^p$ is reflexive, so a bounded sequence has a weakly convergent subsequence $g_{j_k}\rightharpoonup g$. Then
$$|\int g_{j_k}(h_k-h)|\leq K\|h_k-h\|_q\to0,$$
and weak convergence handles $\int g_{j_k}h$. Thus the displayed limit holds (with the evident OCR correction $\int gh$ on the right).

Strong convergence of the tests is essential: in $L^2(0,2π)$ take the same orthonormal sequence for $g_j$ and $h_j$. Both converge weakly to zero, but $\int g_jh_j=1$.

Solved by gpt-5.6-sol high.

= 24I
{parent=Paper 4}
{scope}
{title2=Algebraic Geometry}

= Solution
{parent=24I}

Choose a nonconstant rational function $t$ on $C$. Functions in $L(D)$ are integral over a finite-dimensional bounded-pole space over $k(t)$; equivalently, evaluation of sufficiently many principal parts embeds $L(D)$ into a finite-dimensional vector space. Hence $L(D)$ is finite dimensional.

If $E=D+\operatorname{div}(h)$, multiplication by $h$ gives the isomorphism $L(E)\to L(D)$. A <canonical divisor> is the divisor of a nonzero rational differential. <Riemann-Roch theorem> says $\ell(D)-\ell(K-D)=\deg D+1-g$. Taking $D=0$ gives $1-g=1+g-\deg K$, hence $\deg K=2g-2$.

For a smooth plane curve of degree $d$, the adjunction formula gives $K_C\sim(d-3)H|_C$. A cubic has $d=3$, so $K_C\sim0$.

Solved by gpt-5.6-sol high.

= 25F
{parent=Paper 4}
{scope}
{title2=Differential Geometry}

= i
{parent=25f}
{scope}

= Solution
{parent=i}

The geodesic curvature is the tangential component of $α\prime\prime$, namely $k_g=\langle α\prime\prime,N\times α\prime\rangle$ up to orientation; a curve is geodesic when $k_g=0$. <Gauss-Bonnet theorem> for a compact region says $\int K+\int_{\partial}k_g+\sum\text{exterior angles}=2πχ$.

For $K=0$, a flat cylinder has infinitely many parallel closed geodesics.

Solved by gpt-5.6-sol high.

= ii
{parent=25f}
{scope}

= Solution
{parent=ii}

For $K>0$ the number can again be infinite: remove two points away from an infinite family of great circles on the round sphere. The result is a cylinder of positive curvature containing those intersecting simple closed geodesics.

Solved by gpt-5.6-sol high.

= iii
{parent=25f}
{scope}

= Solution
{parent=iii}

For $K<0$ there is at most one. Two essential simple closed geodesics cannot form a geodesic bigon by <Gauss-Bonnet theorem>, so they are disjoint; the annulus between them would have $χ=0$ and geodesic boundary, forcing $\int K=0$, a contradiction. Thus multiple geodesics in the flat case are disjoint, while in the positive example they need not be.

Solved by gpt-5.6-sol high.

= 26H
{parent=Paper 4}
{scope}
{title2=Probability and Measure}

= Solution
{parent=26H}

The union bound gives $\mathbb P(\bigcup_{n\geq m}A_n)\leq\sum_{n\geq m}\mathbb P(A_n)\to0$. Taking the decreasing intersection proves the first <Borel-Cantelli lemmas>[Borel-Cantelli lemma].

If $X_n\to X$ almost surely, then $1_{\{|X_n-X|>ε\}}\to0$ almost surely; <dominated convergence theorem> gives convergence of its expectation, hence convergence in probability.

If convergence in probability holds, any subsequence has a further one with $\mathbb P(|X_{n(k_r)}-X|>2^{-r})<2^{-r}$. Borel-Cantelli then gives almost-sure convergence. Conversely, failure in probability supplies a subsequence with probabilities bounded below by some positive constant, and no further subsequence can converge almost surely because that would imply convergence in probability.

Solved by gpt-5.6-sol high.

= 27K
{parent=Paper 4}
{scope}
{title2=Applied Probability}

= a
{parent=27k}
{scope}

= Solution
{parent=a}

Condition on the first short time interval. A jump contributes the generator term $Qf$, while local time grows at unit speed in the current coordinate and contributes $\partial_{\ell_i}f$. The <Markov property> and division by the interval length therefore give $\partial_tf_t=Mf_t$.

Solved by gpt-5.6-sol high.

= b
{parent=27k}
{scope}

= Solution
{parent=b}

Symmetry of $Q$ gives $\partial_{y_i}e^{y^TQy/2}=e^{y^TQy/2}\sum_jq_{ij}y_j$. Also $\partial_{y_i}f(i,y^2/2)=y_i\partial_{\ell_i}f$. Substitute these identities into the left side and integrate the resulting total $y_i$ derivative by parts. Compact support removes the boundary term and leaves exactly the right side.

Solved by gpt-5.6-sol high.

= c
{parent=27k}
{scope}

= Solution
{parent=c}

For $f_t(i,\ell)=\mathbb E_i[g(\ell+L(t))1_{X(t)=j}]$, part (a) gives $Mf_t=\partial_tf_t$. Apply part (b), integrate in $t$ from zero to infinity, and integrate by parts in time. Compact support makes the terminal term vanish, while $f_0(i,\ell)=g(\ell)1_{i=j}$. Moving the resulting Gaussian derivatives once more by parts yields the stated identity with $y_iy_j$.

Solved by gpt-5.6-sol high.

= 28J
{parent=Paper 4}
{scope}
{title2=Principles of Statistics}

= a
{parent=28j}
{scope}

= Solution
{parent=a}

The posterior is $\operatorname{Gamma}(α+X,β+1)$, so under quadratic loss the Bayes estimator is $δ(X)=(α+X)/(β+1)$. At fixed $θ$ its risk is
$$R(θ,δ)={θ+(α-βθ)^2\over(β+1)^2}.$$

Solved by gpt-5.6-sol high.

= b
{parent=28j}
{scope}

= Solution
{parent=b}

Put $c=(β+1)/(β+2)$. The posterior mean of $e^{-θ}$ is $δ(X)=c^{α+X}$. Using the Poisson probability-generating function, its risk is
$$c^{2α}e^{θ(c^2-1)}-2c^αe^{θ(c-2)}+e^{-2θ}.$$

Solved by gpt-5.6-sol high.

= c
{parent=28j}
{scope}

= Solution
{parent=c}

An admissible Bayes estimator whose Bayes risk equals its maximum risk is minimax. Indeed every estimator has maximum risk at least its prior-average risk, which is at least the Bayes risk; equality for the stated estimator attains this lower bound. Admissibility excludes a distinct estimator improving it everywhere.

Solved by gpt-5.6-sol high.

= d
{parent=28j}
{scope}

= Solution
{parent=d}

Neither proper gamma prior yields the condition in part (c). The risk in part (a) is unbounded as $θ\to∞$ because $β>0$. The risk in part (b) is analytic, tends to zero at infinity, and is not identically zero, so its prior average is strictly below its supremum for a gamma prior with positive density everywhere. Thus this sufficient condition proves neither estimator minimax for any $α,β>0$.

Solved by gpt-5.6-sol high.

= 29K
{parent=Paper 4}
{scope}
{title2=Stochastic Financial Models}

= a
{parent=29k}
{scope}

= Solution
{parent=a}

A <Brownian motion> starts at zero, has continuous paths, and has independent increments with $W_t-W_s\sim N(0,t-s)$. The stated covariance makes every finite collection jointly Gaussian; disjoint increments have zero covariance and hence are independent. Their variances are interval lengths, proving all defining properties.

Solved by gpt-5.6-sol high.

= b
{parent=29k}
{scope}

= Solution
{parent=b}

The transformed process is centred Gaussian and
$$\mathbb E[\widetilde W_s\widetilde W_t]=st\min(1/s,1/t)=\min(s,t).$$
The Brownian strong law makes $tW_{1/t}\to0$ almost surely as $t\downarrow0$, giving continuity at zero; elsewhere continuity is immediate. Part (a) now applies.

Solved by gpt-5.6-sol high.

= c
{parent=29k}
{scope}

= Solution
{parent=c}

The joint Gaussian density of the independent increments $ΔW_i$ changes under shifts $cΔt_i$ by
$$\exp\{c\sum_iΔW_i-c^2\sum_iΔt_i/2\}=\exp(cW_T-c^2T/2).$$
Multiplying the density and integrating the bounded cylinder function $g$ proves the identity.

Solved by gpt-5.6-sol high.

= d
{parent=29k}
{scope}

= Solution
{parent=d}

Apply <Cameron-Martin theorem for a linear drift>[Cameron-Martin theorem] to the absolutely continuous path $h(t)=xt/T$ up to $T$ and $h(t)=x$ thereafter. Its derivative is $x/T$ on $[0,T]$ and zero later, so the Radon-Nikodym factor is $\exp((x/T)W_T-x^2/(2T))$, yielding the formula for every bounded measurable functional of the path after $T$.

Solved by gpt-5.6-sol high.

= 30J
{parent=Paper 4}
{scope}
{title2=Mathematics of Machine Learning}

= Solution
{parent=30J}

A random forest independently bootstrap-resamples the data for each tree, and at each split considers a fresh random subset of features. It averages the resulting regression trees: $f_{\rm rf}=B^{-1}\sum_b\widehat T^{(b)}$. Each leaf prediction is an average of responses, so it lies in $[-M,M]$.

The <bounded differences inequality> says that if changing coordinate $i$ changes $G$ by at most $c_i$, then $\mathbb P(G-\mathbb EG\geq t)\leq\exp(-2t^2/\sum c_i^2)$. Replacing one tree changes the forest pointwise by at most $2M/B$, and hence changes the supremum $G$ by at most that amount. Therefore
$$G\leq\mathbb EG+M\sqrt{2\log(1/δ)/B}$$
with probability at least $1-δ$.

Solved by gpt-5.6-sol high.

= 31A
{parent=Paper 4}
{scope}
{title2=Asymptotic Methods}

= a
{parent=31a}
{scope}

= Solution
{parent=a}

After setting $z=1/x$, the transformed coefficients have a singularity beyond the regular-singular growth bounds at $z=0$. Thus infinity is an <irregular singular point>.

Solved by gpt-5.6-sol high.

= b
{parent=31a}
{scope}

= Solution
{parent=b}

The standard removal of the first derivative, $y=u/x$, gives $u\prime\prime+q(x)u=0$ with $q(x)=x^{-1}-x^{-2}$.

Solved by gpt-5.6-sol high.

= c
{parent=31a}
{scope}

= Solution
{parent=c}

The <Liouville-Green exponential ansatz>[Liouville-Green approximation] gives
$$u_±\sim q^{-1/4}\exp\left(±i\int^x\sqrt q\,ds\right).$$
Here
$$q^{-1/4}=x^{1/4}(1+1/(4x)+5/(32x^2)+\cdots),$$
$$\int^x\sqrt q\,ds=2\sqrt x+x^{-1/2}+x^{-3/2}/12+\cdots.$$
The control ratio $|q\prime|q^{-3/2}=O(x^{-1/2})$ tends to zero, verifying consistency.

Solved by gpt-5.6-sol high.

= d
{parent=31a}
{scope}

= Solution
{parent=d}

Since $y=u/x$, two real independent approximations are
$$y_1\sim x^{-3/4}(1+1/(4x)+5/(32x^2))\cos(2\sqrt x+x^{-1/2}+x^{-3/2}/12),$$
$$y_2\sim x^{-3/4}(1+1/(4x)+5/(32x^2))\sin(2\sqrt x+x^{-1/2}+x^{-3/2}/12).$$

Solved by gpt-5.6-sol high.

= 32A
{parent=Paper 4}
{scope}
{title2=Dynamical Systems}

= a
{parent=32a}
{scope}

= Solution
{parent=a}

Order a period-three orbit as $a<b<c$. Its images produce two subintervals $I_0,I_1$ with covering relations $I_0\to I_0,I_1$ and $I_1\to I_0$. Any closed walk in this directed graph gives, by the interval-covering lemma and the <intermediate value theorem>, a fixed point of the corresponding iterate. Primitive closed walks of every length exist, giving an orbit of every positive period; this is the period-three case of <Sharkovsky theorem>.

Solved by gpt-5.6-sol high.

= b
{parent=32a}
{scope}

= Solution
{parent=b}

The covering graph has at least one primitive cycle of lengths two and four and two inequivalent primitive cycles of length five. Thus the guaranteed minimum numbers of distinct orbits are respectively $1,1,2$.

Solved by gpt-5.6-sol high.

= c
{parent=32a}
{scope}

= i
{parent=c}
{scope}

= Solution
{parent=i}

For the stated map, the graph cycles give the period-two orbit
$$1/3\mapsto5/6\mapsto1/3,$$
and the period-four orbit
$$2/9\mapsto13/18\mapsto5/9\mapsto8/9\mapsto2/9.$$

Solved by gpt-5.6-sol high.

= ii
{parent=c}
{scope}

= Solution
{parent=ii}

The two period-five orbits are
$$1/9\to11/18\to7/9\to4/9\to17/18\to1/9,$$
$$2/15\to19/30\to11/15\to8/15\to14/15\to2/15.$$
Direct use of the two affine branches verifies every arrow.

Solved by gpt-5.6-sol high.

= 33B
{parent=Paper 4}
{scope}
{title2=Principles of Quantum Mechanics}

= a
{parent=33b}
{scope}

= Solution
{parent=a}

First-order <time-dependent perturbation theory> gives
$$c_1(t)=-{i\over\hbar}\int_0^t e^{i(ω_1-ω_0)t\prime}\langle1|V(t\prime)|0\rangle\,dt\prime.$$
The lowest nontrivial transition probability is $|c_1(t)|^2$, with errors of higher order in $V$.

Solved by gpt-5.6-sol high.

= b
{parent=33b}
{scope}

= Solution
{parent=b}

Move to a rotating frame. With detuning $Δ=ω_1-ω_0+ω$ (changing sign with the phase convention), the constant two-level Hamiltonian has generalized Rabi frequency $Ω=\sqrt{Δ^2+4v^2}$. Starting in state zero,
$$P_1(t)={4v^2\over Ω^2}\sin^2(Ωt/2).$$
The transition envelope is maximized on resonance, $Δ=0$, namely $ω=ω_0-ω_1$ for the matrix phases printed in the question.

Solved by gpt-5.6-sol high.

= 34B
{parent=Paper 4}
{scope}
{title2=Applications of Quantum Mechanics}

= a
{parent=34b}
{scope}

= Solution
{parent=a}

$E_0(k)=\hbar^2k^2/(2m)$. Periodic boundary conditions give $k=2πn/L$. Orthogonality of plane waves gives $\langle k|U|k\prime\rangle=U_l$ when $k\prime-k=2πl/a$, and zero otherwise.

Solved by gpt-5.6-sol high.

= b
{parent=34b}
{scope}

= Solution
{parent=b}

Nondegenerate perturbation theory fails where free energies related by a reciprocal vector coincide, at $k=lπ/a$. Diagonalizing the two-state block places the gap centre at $\hbar^2(lπ/a)^2/(2m)+λU_0$ and gives width $2λ|U_l|+O(λ^2)$.

Solved by gpt-5.6-sol high.

= c
{parent=34b}
{scope}

= i
{parent=c}
{scope}

= Solution
{parent=i}

For $V=2λ\cos(2πx/a)$, only $U_{±1}=1$ are nonzero. To first order there is one gap of width $2λ$ at the first Brillouin-zone boundary $k=±π/a$; higher gaps vanish at this order. The allowed bands are the folded free-electron parabolas with that interval removed.

Solved by gpt-5.6-sol high.

= ii
{parent=c}
{scope}

= Solution
{parent=ii}

For the Dirac comb, Poisson summation gives $U_l=1$ for every integer $l$. Hence every crossing at $k=lπ/a$ opens a gap of width $2λ$ to first order, producing an infinite sequence of allowed bands between consecutive gaps.

Solved by gpt-5.6-sol high.

= d
{parent=34b}
{scope}

= Solution
{parent=d}

Filled bands separated by gaps explain the distinction between electrical insulators and conductors: a completely filled band cannot change its net crystal momentum under a weak field, whereas a partially filled band supports current.

Solved by gpt-5.6-sol high.

= 35C
{parent=Paper 4}
{scope}
{title2=Statistical Physics}

= a
{parent=35c}
{scope}

= Solution
{parent=a}

A first-order phase transition has a discontinuous first derivative of the equilibrium free energy, such as entropy or volume, and latent heat. A second-order transition has continuous first derivatives but a discontinuous or divergent second derivative.

Solved by gpt-5.6-sol high.

= b
{parent=35c}
{scope}

= Solution
{parent=b}

At fixed temperature, volume, and particle number, spontaneous changes satisfy $ΔF\leq0$ because $F=E-TS$ accounts for heat exchanged with the reservoir. Equilibrium therefore minimizes the Helmholtz free energy.

Solved by gpt-5.6-sol high.

= c
{parent=35c}
{scope}

= Solution
{parent=c}

Minimization gives $m=0$ for $T>T_c$ and $m=±\sqrt{a(T_c-T)/b}$ below. Thus
$$F_{eq}=F_0\quad(T\geq T_c),\qquad F_{eq}=F_0-{a^2(T_c-T)^2\over4b}\quad(T<T_c).$$
Using $S=-F_T$ and $C=-TF_{TT}$, entropy is continuous but the low-temperature heat capacity has the additional $Ta^2/(2b)$. The jump is $T_ca^2/(2b)$, so the transition is second order.

Solved by gpt-5.6-sol high.

= d
{parent=35c}
{scope}

= Solution
{parent=d}

Writing $x=m^2$, nonzero equilibria satisfy $a(T-T_c)+bx+cx^2=0$. For $b<0$, coexistence occurs at $x=-3b/(4c)$ and $T-T_c=3b^2/(16ac)$, so magnetization jumps and the transition is first order.

For $b=0$, $m=0$ above $T_c$ and $m^4=a(T_c-T)/c$ below: this is the tricritical continuous case. Below $T_c$,
$$F_{eq}-F_0=-{a^{3/2}\over3\sqrt c}(T_c-T)^{3/2},$$
so the extra heat capacity is $a^{3/2}T/[4\sqrt{c(T_c-T)}]$ and diverges on approach from below; above it is absent.

Solved by gpt-5.6-sol high.

= 36C
{parent=Paper 4}
{scope}
{title2=Electrodynamics}

= a
{parent=36c}
{scope}

= Solution
{parent=a}

For a linear dielectric, $D=εE$ (equivalently $D=ε_0E+P$). With no free charge, Gauss law gives $\nabla\mathbin\cdot D=0$. Across an interface without free surface charge, $D_n$ is continuous; electrostatic Faraday law makes $E_t$ continuous.

Solved by gpt-5.6-sol high.

= b
{parent=36c}
{scope}

= Solution
{parent=b}

Imposing those two boundary conditions on the ansatz gives, with $κ=(ε-ε_0)/(ε+2ε_0)$,
$$α={3ε_0\over ε+2ε_0},\qquad β=3a^3κ,\qquad δ=-a^3κ.$$
Thus the internal field is uniform and the external correction has dipolar angular form.

Solved by gpt-5.6-sol high.

= c
{parent=36c}
{scope}

= Solution
{parent=c}

The far field of a dipole is $[3(p\cdot\widehat x)\widehat x-p]/(4πε_0r^3)$. Comparison yields the induced <electric dipole moment>
$$p=4πε_0a^3{ε-ε_0\over ε+2ε_0}E_0.$$

Solved by gpt-5.6-sol high.

= 37C
{parent=Paper 4}
{scope}
{title2=General Relativity}

= a
{parent=37c}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

Separate conservation gives $ρ_{ph}=ρ_0a^{-3(1+w)}$, which grows because $w<-1$. At late times $\dot a/a=C a^{-3(1+w)/2}$. Integration gives
$$a(t)\sim A(t^*-t)^{2/[3(1+w)]}.$$
The exponent is negative, so $a\to∞$ at finite $t^*$: the Big Rip.

Solved by gpt-5.6-sol high.

Direct use of the <Christoffel symbol> formula gives $Γ^0_{ii}=a\dot a$ and $Γ^i_{0i}=Γ^i_{i0}=\dot a/a$. Substitution in the Ricci formula gives $R_{00}=-3\ddot a/a$ and $R_{ii}=a\ddot a+2\dot a^2$.

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

The sketch begins at $a=0$ at the Big Bang, follows the concave-down dust law $a\propto t^{2/3}$, crosses into accelerated phantom domination, and rises with a vertical asymptote at the finite time $t^*$.

Solved by gpt-5.6-sol high.

The scalar curvature is $R=6(\ddot a/a+H^2)$. Therefore $G_{00}=3H^2$, and the $00$ Einstein equation gives the flat <Friedmann equation>
$$H^2={8π\over3}ρ+{Λ\over3}.$$

Solved by gpt-5.6-sol high.

= b
{parent=37c}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

For a null vector, the perfect-fluid tensor gives $T_{αβ}V^αV^β=(ρ+P)(u\cdot V)^2$. Phantom energy has $ρ+P=(1+w)ρ<0$, so it violates the null energy condition.

Solved by gpt-5.6-sol high.

= 38A
{parent=Paper 4}
{scope}
{title2=Fluid Dynamics II}

= a
{parent=38a}
{scope}

= Solution
{parent=a}

The radial strain draws fluid inward, axial strain expels it away from the midplane, and $v(r)$ supplies swirl. In cylindrical coordinates $r^{-1}\partial_r(r(-αr))+\partial_z(2αz)=-2α+2α=0$, so the flow is incompressible.

Solved by gpt-5.6-sol high.

= b
{parent=38a}
{scope}

= Solution
{parent=b}

The only vorticity component is $ω=(1/r)(rv)\prime$. Its steady equation is $-αrω\prime=2αω+ν(ω\prime\prime+r^{-1}ω\prime)$, solved by $ω=ω_0e^{-αr^2/(2ν)}$. Regularity at zero gives
$$v(r)={ω_0ν\over αr}(1-e^{-αr^2/(2ν)}).$$
Its maximum satisfies $e^s=1+2s$, $s=αr^{*2}/(2ν)$, so $r^*=O(\sqrt{ν/α})$.

Solved by gpt-5.6-sol high.

= c
{parent=38a}
{scope}

= Solution
{parent=c}

Adding inward side flux and outward end-cap flux gives the net inward angular-momentum advection
$$\mathcal A={8πρz_0ω_0ν^2\over α}[1-(1+s_0)e^{-s_0}],\qquad s_0={αr_0^2\over2ν}.$$
It is positive because $e^{s_0}>1+s_0$, and tends to the stated value.

Solved by gpt-5.6-sol high.

= d
{parent=38a}
{scope}

= Solution
{parent=d}

Integrating the viscous traction $2ρνe_{rθ}$ over the side and end surfaces gives
$$\mathcal T=-{8πρz_0ω_0ν^2\over α}[1-(1+s_0)e^{-s_0}]=-\mathcal A.$$
It is always negative. In steady state the cylinder stores no angular momentum, so the divergence theorem for angular-momentum balance requires viscous torque to cancel net advective influx.

Solved by gpt-5.6-sol high.

= 39A
{parent=Paper 4}
{scope}
{title2=Waves}

= a
{parent=39a}
{scope}

= Solution
{parent=a}

In the shock frame, conservation of mass, normal momentum, and energy gives
$$ρ_1U=ρ_2(U-u_2),$$
$$p_1+ρ_1U^2=p_2+ρ_2(U-u_2)^2,$$
$$h_1+U^2/2=h_2+(U-u_2)^2/2.$$
These are the <Rankine-Hugoniot conditions for a perfect gas>[Rankine-Hugoniot relations].

Solved by gpt-5.6-sol high.

= b
{parent=39a}
{scope}

= Solution
{parent=b}

Insert $h=γp/[ρ(γ-1)]$, eliminate $p_2$ and the downstream normal speed using mass conservation, and solve for the density ratio. This gives
$$ {ρ_1\overρ_2}={2c_1^2+(γ-1)U^2\over(γ+1)U^2},\qquad c_1^2={γp_1\overρ_1}.$$

Solved by gpt-5.6-sol high.

= c
{parent=39a}
{scope}

= Solution
{parent=c}

Tangential velocity is unchanged, while the downstream normal velocity is $Uρ_1/ρ_2$. Thus the incident and outgoing angles to the shock have tangents $1$ and $ρ_1/ρ_2$. Using the tangent-difference identity, the ratio from part (b), and $M^2=2U^2/c_1^2$ gives
$$\tanδ={M^2-2\over γM^2+2}.$$

Solved by gpt-5.6-sol high.

= 40E
{parent=Paper 4}
{scope}
{title2=Numerical Analysis}

= a
{parent=40e}
{scope}

= Solution
{parent=a}

Let $Mx=(A-B)x$ and suppose $Hx=λx$, so $-Bx=λMx$. Put $a=x^*Ax>0$ and $m=x^*Mx$. Since $A=M-(-B)$, $a=(1-λ)m$. Also
$$x^*(A-B-B^T)x=2\operatorname{Re}m-a
=a{1-|λ|^2\over|1-λ|^2}.$$
Positive definiteness makes this positive, hence $|λ|<1$ for every eigenvalue and $ρ(H)<1$.

Solved by gpt-5.6-sol high.

= b
{parent=40e}
{scope}

= Solution
{parent=b}

After multiplying the five-point equations by $-1$ if necessary, their matrix $A$ is symmetric positive definite. For any ordering write $A=D+L+L^T$, where $D=4I$. Gauss-Seidel method uses $M=D+L$ and has iteration matrix $-M^{-1}L^T$. In part (a) take $B=L^T$; then $A-B=M$ and
$$A-B-B^T=D,$$
which is positive definite. Therefore the iteration matrix has spectral radius below one for every ordering, proving convergence.

Solved by gpt-5.6-sol high.