past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/ii/paper-4.bigb

= Paper 4
{scope}

https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2026/PaperII_4.pdf

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

= Solution
{parent=1G}

A <Carmichael number> is a composite $N$ such that $a^{N-1}\equiv1\pmod N$ for every $a$ coprime to $N$. For $p\mid N$, choose by CRT a number whose residue mod $p$ is a primitive root; its order $p-1$ divides $N-1$. If $p^2\mid N$, choose instead a primitive root modulo $p^2$; then $p(p-1)\mid N-1$, impossible because $p\mid N$. Thus $N$ is square-free. Write $N=pm$. Since $p-1\mid N-1$, also $p-1\mid m-1$; compositeness gives $m>1$, hence $m\ge p$ and $p\le\sqrt N$. This is <Korselt criterion>.

Solved by gpt-5.6-sol high.

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

= i
{parent=2i}
{scope}

= Solution
{parent=i}

Let $\ell_j$ be the Lagrange <basis> <polynomials> for the nodes. Necessity and existence give uniquely $A_j=\int_{-1}^1\ell_j(t)dt$, because every <polynomial> of degree at most $n-1$ is $\sum_jp(t_j)\ell_j(t)$.

Solved by gpt-5.6-sol high.

= ii
{parent=2i}
{scope}

= Solution
{parent=ii}

For $\deg p\le2n-1$, divide $p=q p_n+r$ with $\deg q\le n-1$, $\deg r\le n-1$. Orthogonality makes $\int qp_n=0$, and $p(t_j)=r(t_j)$. Apply part (i) to $r$. This is <gaussian quadrature>.

Solved by gpt-5.6-sol high.

= iii
{parent=2i}
{scope}

= Solution
{parent=iii}

Apply exactness to $\ell_j^2$: $A_j=\sum_kA_k\ell_j(t_k)^2=\int\ell_j^2>0$. Applying it to 1 gives $\sum A_j=2$. For continuous $f$ and any admissible $p$,
$$\left|\sum_jA_jf(t_j)-\int f\right|\le\sum_jA_j\|f-p\|_\infty+2\|f-p\|_\infty=4\|f-p\|_\infty.$$

Solved by gpt-5.6-sol high.

= 3H
{parent=Paper 4}
{scope}
{title2=Coding \& Cryptography}

= Solution
{parent=3H}

A <Rabin cryptosystem> chooses Blum primes $p,q$, publishes $N=pq$, and encrypts $m$ as $c=m^2\bmod N$. Decryption takes square roots modulo $p$ and $q$, combines them with CRT, and uses redundancy to select the intended one of four roots. Here $2355^2\equiv25\pmod{2773}$, while the obvious root is $5$. Thus
$$\gcd(2355-5,2773)=47,\qquad\gcd(2355+5,2773)=59.$$
So $2773=47\cdot59$ is factored, allowing all future square roots and hence all ciphertexts to be decrypted.

Solved by gpt-5.6-sol high.

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

= a
{parent=4i}
{scope}

= Solution
{parent=a}

Run the algorithms for $f(n)$ and $g(n)$ and output their XOR. Closure of computable <functions> under composition makes $f+g$ computable.

Solved by gpt-5.6-sol high.

= b
{parent=4i}
{scope}

= Solution
{parent=b}

No. Let $X_f=\mathbb N$ and let $X_g=K$ be an r.e. undecidable set. Both are r.e., but $X_{f+g}=\mathbb N\setminus K$ is not r.e.; otherwise $K$ would be decidable.

Solved by gpt-5.6-sol high.

= c
{parent=4i}
{scope}

= Solution
{parent=c}

No. Context-free languages are not closed under symmetric difference (a standard counterexample is obtained from the two CFLs enforcing respectively equality of the first two and last two blocks in $a^*b^*c^*$). Transport such a pair through the bijection $\beta$; pointwise addition corresponds exactly to symmetric difference.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=5K}

Write $f(y;\theta)=\exp\{y\theta-b(\theta)+c(y)\}$ and use canonical parameter $\theta_i=X_i^T\beta$. Then
$$\ell(\beta)=\sum_i[Y_iX_i^T\beta-b(X_i^T\beta)+c(Y_i)],\quad X^T(Y-\mu)=0,$$
where $\mu_i=b'(X_i^T\beta)$. Geometrically the residual <vector> is orthogonal to every design column. For $H_0:\beta_1=0$, use a two-sided Wald statistic $\hat\beta_1/\operatorname{se}(\hat\beta_1)$ (or likelihood-ratio statistic), rejecting at the corresponding normal or chi-square critical value.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=6C}

Scaling by $K$, the positive fixed point is $x=1$. Linearization $x_t=1+\epsilon_t$ gives $\epsilon_{t+1}=\epsilon_t-r\epsilon_{t-1}$, with characteristic <polynomial> $\lambda^2-\lambda+r$. The Jury conditions put both roots inside the unit circle exactly for $0<r<1$. At $r=1$ the roots are $e^{\pm i\pi/3}$, sixth roots of unity. The discrete Hopf resonance therefore creates a small period-six branch; expansion of the nonlinear recurrence on the centre eigenspace supplies the nonzero amplitude.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=7D}

The identity theorem says <analytic functions> agreeing on a set with an interior accumulation point agree throughout a connected domain. Analytic continuation is an <analytic function> on a larger connected domain agreeing on a nonempty overlap. Factor the denominator as $(t+i)(t-z)$. For $\Im z<0$, contour closure below gives
$$F(z)=-2\pi i\frac{e^{-5iz}-e^{-5}}{z+i},$$
with a removable value at $-i$; this entire expression is the continuation. For $\Im z>0$, the displayed real-axis <integral> encloses only $-i$ and equals $2\pi i e^{-5}/(z+i)$, so it is not that continuation. Finally $1/n\to0$ only at a boundary point of $D$, so the zeros do not determine $G$; for example $G(z)=\sin(\pi/z)$ is nonzero and has them all.

Solved by gpt-5.6-sol high.

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

= a
{parent=8a}
{scope}

= Solution
{parent=a}

Hamilton's equations give $\dot H=H_q\cdot H_p-H_p\cdot H_q=0$. Also $\operatorname{div}V_H=\sum_a(\partial_{q_a}\partial_{p_a}H-\partial_{p_a}\partial_{q_a}H)=0$.

Solved by gpt-5.6-sol high.

= b
{parent=8a}
{scope}

= Solution
{parent=b}

The supplied Jacobian identity and zero divergence give $J^{-1}\dot J=0$, so $J(t)=1$. Change of variables then shows $\operatorname{vol}\Phi_t(U)=\operatorname{vol}U$. This is <Liouville theorem in Hamiltonian mechanics>.

Solved by gpt-5.6-sol high.

= c
{parent=8a}
{scope}

= Solution
{parent=c}

The energy sublevel $H\le1/2$ is compact and has finite phase volume because both $p$ and $q$ are bounded. If all $\Phi_{nT}(U)$ were disjoint, volume preservation would put infinitely many equal positive volumes inside it, impossible. Thus two iterates intersect. Pulling the intersection back shows a point of $U$ returns to $U$; applying the argument to arbitrarily small neighbourhoods gives <Poincare recurrence theorem>.

Solved by gpt-5.6-sol high.

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

= a
{parent=9d}
{scope}

= Solution
{parent=a}

Use $E=mc^2+p^2/(2m)+\cdots$ and replace the Fermi denominator by its Boltzmann form. The Gaussian <integral> gives
$$n_X=\left(\frac{2\pi m_XkT}{h^2}\right)^{3/2}e^{-m_Xc^2/kT}.$$

Solved by gpt-5.6-sol high.

= b
{parent=9d}
{scope}

= Solution
{parent=b}

At $kT_d=m_Xc^2/\alpha$, division by $n_\gamma=16\pi\zeta(3)(kT/hc)^3$ gives $r(\alpha)=\sqrt{2\pi}\,\alpha^{3/2}e^{-\alpha}/[8\zeta(3)]$.

Solved by gpt-5.6-sol high.

= c
{parent=9d}
{scope}

= Solution
{parent=c}

Today $\rho_X/\rho_B=(m_X/mp)r/\eta\simeq\Omega_{DM}/\Omega_B\simeq5$. Hence
$$m_X\simeq\frac{5\eta}{r(\alpha)}m_p=\frac{3\times10^{-9}}{r(\alpha)}m_p,$$
so $\mu=3\times10^{-9}$. For suitable large $\alpha$ this is nonrelativistic and can match the observed abundance.

Solved by gpt-5.6-sol high.

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

= a
{parent=10e}
{scope}

= Solution
{parent=a}

$\operatorname{QFT}_N|k\rangle=N^{-1/2}\sum_{j=0}^{N-1}e^{2\pi ijk/N}|j\rangle$.

Solved by gpt-5.6-sol high.

= b
{parent=10e}
{scope}

= Solution
{parent=b}

The controlled phases and final bit reversal implement QFT. Thus $|000\rangle\mapsto2^{-3/2}\sum_{j=0}^7|j\rangle$, while
$$|011\rangle\mapsto2^{-3/2}\sum_{j=0}^7e^{2\pi i(3j)/8}|j\rangle,$$
which agrees with $\operatorname{QFT}_8|0\rangle$ and $\operatorname{QFT}_8|3\rangle$.

Solved by gpt-5.6-sol high.

= c
{parent=10e}
{scope}

= Solution
{parent=c}

The amplitude at $y$ is proportional to $e^{2\pi ix_0y/8}\sum_{j=0}^{A-1}e^{2\pi ijry/8}$. The geometric sum vanishes unless $ry\equiv0\pmod8$, i.e. unless $y$ is a multiple of $A=8/r$. At those $r$ outcomes its <modulus> is $1/\sqrt r$, so each probability is $1/r$. The phase contains $x_0$, but measurement probabilities do not.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=11G}

A form $ax^2+bxy+cy^2$ is positive definite when $a>0$ and $d=b^2-4ac<0$; equivalence is change by $SL_2(\mathbb Z)$, and reduced means $|b|\le a\le c$ with the standard boundary sign convention. Translating $x\mapsto x+my$ reduces $|b|$, and swapping variables reduces $a$; discriminant bounds force termination. For a reduced form, $|b|\le a\le\sqrt{|d|/3}$ and $b\equiv d\pmod2$. Discriminant $-28$ has one primitive reduced class, represented by $x^2+7y^2$. Therefore an odd prime is represented exactly when it splits in $\mathbb Q(\sqrt{-7})$, namely $(p/7)=1$, with the ramified exception $p=7$. Thus $p=7$ or $p\equiv1,9,11,15,23,25\pmod{28}$.

Solved by gpt-5.6-sol high.

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

= a
{parent=12i}
{scope}

= Solution
{parent=a}

<Liouville approximation theorem> says that if algebraic irrational $\alpha$ has degree $d\ge2$, then $|\alpha-p/q|>C(\alpha)q^{-d}$ for all rationals $p/q$.

Solved by gpt-5.6-sol high.

= b
{parent=12i}
{scope}

= Solution
{parent=b}

Taking <determinants> of the convergent <matrices> gives $q_np_{n-1}-q_{n-1}p_n=(-1)^n$. The complete-quotient formula then yields $|\alpha-p_n/q_n|<1/(q_nq_{n+1})<1/q_n^2$. Comparing two consecutive errors shows at least one is below $1/(2q^2)$; hence both stated inequalities occur infinitely often (the second is the elementary precursor of Hurwitz's sharper $1/\sqrt5$ bound).

Solved by gpt-5.6-sol high.

= c
{parent=12i}
{scope}

= Solution
{parent=c}

Here $q_{n+1}\ge a_{n+1}q_n=q_n^{n+1}$, so $|\alpha-p_n/q_n|<q_n^{-(n+2)}$. If $\alpha$ were algebraic of fixed degree $d$, Liouville's theorem would contradict this for $n>d$. Thus $\alpha$ is transcendental.

Solved by gpt-5.6-sol high.

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

= a
{parent=13k}
{scope}

= Solution
{parent=a}

Expanding $\sum(X_i-\bar X)(Y_i-\bar Y)=\sum X_iY_i-n\bar X\bar Y$ and taking expectations gives $(n-1)(E[XY]-EX,EY)$. Division by $n-1$ proves unbiasedness.

Solved by gpt-5.6-sol high.

= b
{parent=13k}
{scope}

= Solution
{parent=b}

The OLS normal equation with an intercept gives $\hat\beta=S_{XY}/S_{XX}$. By the law of large numbers it tends to $\operatorname{Cov}(X,Y)/\operatorname{Var}X$. Here $X=Z+U+\epsilon_X$ and $Y=X+U+\epsilon_Y$, so the <limit> is $(3+1)/3=4/3$, showing upward confounding bias.

Solved by gpt-5.6-sol high.

= c
{parent=13k}
{scope}

= Solution
{parent=c}

First regress $X$ on $Z$, then regress $Y$ on the fitted values. With one centred instrument this equals $S_{ZY}/S_{ZX}$. Its <limit> is $\operatorname{Cov}(Z,Y)/\operatorname{Cov}(Z,X)=1/1=1$, the causal coefficient of $X$ in $Y=X+U+\epsilon_Y$.

Solved by gpt-5.6-sol high.

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

= a
{parent=14c}
{scope}

= Solution
{parent=a}

The left side is the flux of newborn trees through size zero. Every tree of size $s$ produces seedlings at rate $b(s)$, so total births are $\int_0^\infty b(s)n(s,t)ds$, giving the <boundary condition>.

Solved by gpt-5.6-sol high.

= b
{parent=14c}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

Substitution gives $(g\tilde n)'=-(\mu+r)\tilde n$, hence
$$\tilde n(s)=\frac{C}{g(s)}\exp\left[-\int_0^s\frac{\mu(u)+r}{g(u)}du\right].$$
The <boundary condition> yields the Euler-Lotka equation
$$1=\int_0^\infty\frac{b(s)}{g(s)}\exp\left[-\int_0^s\frac{\mu(u)+r}{g(u)}du\right]ds.$$

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

The mean lifetime production is $(B/\mu)e^{-\mu s_0/g}$. A time-independent population requires it to equal one, so $B=\mu e^{\mu s_0/g}$.

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

The Euler-Lotka equation becomes $1=(B/g)e^{-(\mu+r)s_*/g}$, hence $r=(g/s_*)\log(B/g)-\mu$.

Solved by gpt-5.6-sol high.

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

= a
{parent=15a}
{scope}

= Solution
{parent=a}

Euler's equations are $I_1\dot\omega_1=(I_2-I_3)\omega_2\omega_3$ and cyclic permutations. Setting two components to zero gives uniform rotation about any principal axis.

Solved by gpt-5.6-sol high.

= b
{parent=15a}
{scope}

= Solution
{parent=b}

Multiplying Euler's equations by $\omega_i$ and by $I_i\omega_i$ respectively shows $T=\tfrac12\sum I_i\omega_i^2$ and $L^2=\sum I_i^2\omega_i^2$ are constant. Their ellipsoids are tangent when their normals are parallel, which occurs on the principal axes.

Solved by gpt-5.6-sol high.

= c
{parent=15a}
{scope}

= Solution
{parent=c}

Eliminating $\omega_2^2$ between $2T$ and $L^2$ gives
$$(I_1^2-I_1I_2)\omega_1^2+(I_3^2-I_2I_3)\omega_3^2=L^2-2I_2T.$$
The intersections are closed near axes 1 and 3 but hyperbolic near axis 2. Thus rotations about the smallest and largest inertia axes are stable, while the intermediate-axis rotation is unstable.

Solved by gpt-5.6-sol high.

= d
{parent=15a}
{scope}

= Solution
{parent=d}

Linearization about $(0,\Omega,0)$ gives $\dot u=-Av$, $\dot v=-Bu$, where $A=(I_3-I_2)\Omega/I_1$, $B=(I_2-I_1)\Omega/I_3$. With $\kappa=\sqrt{AB}$,
$$u=u_0\cosh\kappa t-(A/\kappa)v_0\sinh\kappa t,quad v=v_0\cosh\kappa t-(B/\kappa)u_0\sinh\kappa t.$$
The exponential growth confirms intermediate-axis instability.

Solved by gpt-5.6-sol high.

= 16J
{parent=Paper 4}
{scope}
{title2=Logic \& Set Theory}

= a
{parent=16j}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

A set $x$ is transitive when $y\in z\in x$ implies $y\in x$, equivalently $\bigcup x\subset x$.

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

$\operatorname{tcl}(x)$ is the least transitive set containing $x$, namely $x\cup\bigcup x\cup\bigcup^2x\cup\cdots$.

Solved by gpt-5.6-sol high.

= iii
{parent=a}
{scope}

= Solution
{parent=iii}

$V_0=\varnothing$, $V_{\alpha+1}=\mathcal P(V_\alpha)$, and $V_\lambda=\bigcup_{\alpha<\lambda}V_\alpha$ at <limits>.

Solved by gpt-5.6-sol high.

= iv
{parent=a}
{scope}

= Solution
{parent=iv}

The Mirimanoff rank is $\rho(x)=\min\{\alpha:x\subset V_\alpha\}=\sup_{y\in x}(\rho(y)+1)$.

Solved by gpt-5.6-sol high.

= b
{parent=16j}
{scope}

= Solution
{parent=b}

From $\rho(x)=\sup_{y\in x}(\rho(y)+1)$, if every $\rho(y)<\gamma$, then every $\rho(y)+1\le\gamma$ and the supremum is at most $\gamma$, contradicting $\gamma<\rho(x)$.

Solved by gpt-5.6-sol high.

= c
{parent=16j}
{scope}

= Solution
{parent=c}

Taking transitive closure adds only descendants of existing members, whose ranks are lower. Thus the supremum of member ranks, and hence the rank of the set, is unchanged: $\rho(x)=\rho(\operatorname{tcl}x)$.

Solved by gpt-5.6-sol high.

= d
{parent=16j}
{scope}

= Solution
{parent=d}

Induct on $\gamma$. Part (b) gives a member of rank at least $\gamma$; transitivity and descent through a member of larger rank, using the induction hypothesis, produces a member of rank exactly $\gamma$.

Solved by gpt-5.6-sol high.

= e
{parent=16j}
{scope}

= Solution
{parent=e}

If $x$ is hereditarily countable, its transitive closure contains only countably many ranks. Their supremum is a countable ordinal below $\omega_1$, so $x\in V_{\omega_1}$.

Solved by gpt-5.6-sol high.

= f
{parent=16j}
{scope}

= Solution
{parent=f}

Every real $r\subset\omega$ is hereditarily countable, giving $2^{\aleph_0}\le|HC|$. Conversely each HC set is represented by a well-founded extensional relation on a subset of $\omega$; there are only $2^{\aleph_0}$ such relations. Hence equality.

Solved by gpt-5.6-sol high.

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

= a
{parent=17j}
{scope}

= Solution
{parent=a}

This is <Dirac theorem>. Take a longest path $v_1\ldots v_k$. All neighbours of its ends lie on it. If no index $i$ has both $v_1v_{i+1}$ and $v_kv_i$, the two end-neighbour sets have total size at most $k-1$, contradicting $2\delta\ge n\ge k$. Such an index closes a cycle through the path, and maximality plus connectedness makes it Hamiltonian. For even $n$, the disjoint union of two copies of $K_{n/2}$ has minimum degree $n/2-1$ and is not Hamiltonian. For odd $n$, $K_{(n-1)/2,(n+1)/2}$ has minimum degree $(n-1)/2$ and no Hamilton cycle because its bipartition sizes differ.

Solved by gpt-5.6-sol high.

= b
{parent=17j}
{scope}

= Solution
{parent=b}

Deleting $s$ vertices from a Hamilton cycle leaves at most $s$ path components, so the property is necessary. It is not sufficient: the Petersen graph is 1-tough, meaning $c(G-S)\le|S|$ for every separating set $S$, but it is not Hamiltonian.

Solved by gpt-5.6-sol high.

= c
{parent=17j}
{scope}

= Solution
{parent=c}

If $G$ were bipartite, a Hamilton path alternates parts. Endpoints in the same part require that part to have one more vertex; endpoints in opposite parts require equal sizes. Since the assumed path exists for every pair, these incompatible requirements arise. Thus $G$ is not bipartite and $\chi(G)\ge3$.

Solved by gpt-5.6-sol high.

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

= a
{parent=18f}
{scope}

= Solution
{parent=a}

<Artin fixed-field theorem> says that for a finite <group> $G\le\operatorname{Aut}L$, $L/L^G$ is Galois with <group> $G$ and degree $|G|$. Let any finite $G$ act faithfully by left translation on variables $X_g$ in $L=\mathbb C(X_g:g\in G)$. Artin's theorem then realizes $G$ as $\operatorname{Gal}(L/L^G)$.

Solved by gpt-5.6-sol high.

= b
{parent=18f}
{scope}

= Solution
{parent=b}

$s_i$ is the sum of all square-free degree-$i$ monomials in the $X_j$. The symmetric <group> $S_n$ permutes variables and fixes exactly $K=\mathbb C(s_1,\ldots,s_n)$, so Artin gives $[L:K]=n!$. The Vandermonde $f=\prod_{i<j}(X_i-X_j)$ is not symmetric, but $f^2$ is symmetric and hence belongs to $K$.

Solved by gpt-5.6-sol high.

= c
{parent=18f}
{scope}

= Solution
{parent=c}

The relations are $\sigma^n=\tau^2=1$ and $\tau\sigma\tau=\sigma^{-1}$, so the generated <group> is $D_{2n}$. The <functions> $XY$ and $X^n+Y^n$ are fixed. Conversely $X^n,Y^n$ are roots of $T^2-(X^n+Y^n)T+(XY)^n$, and adjoining $X$ then $Y$ gives degree at most $2n$. Artin gives degree exactly $|G|=2n$, proving $M^G=\mathbb C(X^n+Y^n,XY)$.

Solved by gpt-5.6-sol high.

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

= a
{parent=19f}
{scope}

= Solution
{parent=a}

Mackey and <induced representation> sum over $H\backslash G/K$. Coprimality makes every $H\cap gKg^{-1}$ trivial, each double coset has size $|H||K|$, and each local <inner product> is $(\dim V)(\dim W)$. Therefore
$$\langle\operatorname{Ind}_H^GV,\operatorname{Ind}_K^GW\rangle_G=\frac{|G|}{|H||K|}\dim V\dim W.$$

Solved by gpt-5.6-sol high.

= b
{parent=19f}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

Odd order makes squaring bijective: squaring preserves the odd order of each element, and any square root $x$ of $g$ lies in $\langle g\rangle$ because $x=(x^2)^m$ for an inverse $m$ of $2$ modulo $|x|$. Thus $g$ has the unique root $g^m$. Hence $\langle\tilde\chi,\tilde\chi\rangle=|G|^{-1}\sum_g|\chi(g^2)|^2=1$.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

The second Adams operation $\chi\mapsto[g\mapsto\chi(g^2)]$ lies in the <integral> character <ring>, so $\tilde\chi=\sum n_i\chi_i$ with integers $n_i$. Norm one forces exactly one coefficient to be $\pm1$; evaluation at the identity makes the sign positive. Squaring is bijective on conjugacy classes as well, so distinct irreducibles remain distinct. Thus the $\tilde\chi_i$ permute the irreducible characters.

Solved by gpt-5.6-sol high.

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

= a
{parent=20g}
{scope}

= Solution
{parent=a}

The left side is the norm form $N_{K/\mathbb Q}(x+y\alpha+z\alpha^2)$ for $\alpha^3=7$. Reducing a hypothetical equation modulo 7 and then descending through the forced divisibilities of $x,y,z,t$ gives an infinite descent, so only the zero solution exists. If $N(\theta)=3$, clearing denominators minimally and applying the norm equation would produce a forbidden nonzero integer solution; hence no such $\theta$ exists.

Solved by gpt-5.6-sol high.

= b
{parent=20g}
{scope}

= Solution
{parent=b}

Dedekind's criterion factors $(p)$ according to the factorization of $X^3-7$ modulo $p$ (the monogenic assumption removes the index obstruction). With $\alpha=\sqrt[3]7$,
$$
(2)=(2,\alpha-1)(2,\alpha^2+\alpha+1),\quad
(3)=(3,\alpha-1)^3,
$$
$$
(5)=(5,\alpha-3)(5,\alpha^2+3\alpha+4),\quad
(7)=(\alpha)^3.
$$
Minkowski's bound is below 11, so classes are generated by prime <ideals> over $2,3,5,7$. Relations from the displayed factorizations and principal <ideals> reduce them to $\mathfrak p=(3,\alpha-1)$ with $\mathfrak p^3$ principal. Part (a) shows $\mathfrak p$ is not principal, so $\operatorname{Cl}(K)\simeq\mathbb Z/3\mathbb Z$.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=21H}

The square gives one vertex, edges $a,b$, and one 2-cell attached by $aba^{-1}b$, hence $\pi_1(K)=\langle a,b\mid aba^{-1}=b^{-1}\rangle$. Based connected covers correspond contravariantly to <subgroups> of $\pi_1$. The <subgroup> $\langle a^2,b\rangle\simeq\mathbb Z^2$ gives the two-sheeted torus cover; $\langle b\rangle\simeq\mathbb Z$ gives an infinite cylindrical cover. If the total space is compact, fibres over a point are compact discrete and hence finite, so the <subgroup> has finite index. Conversely a finite-index cover has finitely many compact lifted cells and is compact.

Solved by gpt-5.6-sol high.

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

= a
{parent=22i}
{scope}

= Solution
{parent=a}

$X^*$ is the Banach space of bounded linear functionals. For $T:X\to Y$, define $T^*:Y^*\to X^*$ by $T^*f=f\circ T$. It is linear and $\|T^*f\|\le\|T\|\|f\|$, so bounded (indeed $\|T^*\|=\|T\|$).

Solved by gpt-5.6-sol high.

= b
{parent=22i}
{scope}

= Solution
{parent=b}

If $S:X\to Y$ is an isomorphism, then $S^*:Y^*\to X^*$ is an isomorphism with inverse $(S^{-1})^*$. If $S$ and $S^{-1}$ are isometries, the same norm calculation makes the dual maps isometries.

Solved by gpt-5.6-sol high.

= c
{parent=22i}
{scope}

= Solution
{parent=c}

The canonical embedding puts $X$ isometrically, hence with closed range, inside $X^{**}\simeq\ell_2$. Every closed subspace of a Hilbert space is Hilbert and, in the infinite-dimensional separable case here, is isomorphic to $\ell_2$. Thus $X\simeq\ell_2$.

Solved by gpt-5.6-sol high.

= d
{parent=22i}
{scope}

= Solution
{parent=d}

The <series> defining $\phi_y$ converges and $\|\phi_y\|\le\|y\|_1$. For each finite set of $k_n$, normality and Tietze extension give a continuous $f$ of norm at most one taking prescribed signs there. Letting the finite set grow proves $\|\phi_y\|=\|y\|_1$. Thus $y\mapsto\phi_y$ is an isometric embedding of $\ell_1$ into $C(K)^*$.

Solved by gpt-5.6-sol high.

= e
{parent=22i}
{scope}

= Solution
{parent=e}

No. The point evaluations $\delta_x$ satisfy $\|\delta_x-\delta_y\|=2$ for $x\ne y$ (choose a <continuous function> taking values $1,-1$ at the two points). Thus $C[0,1]^*$ is nonseparable, whereas $\ell_1$ is separable.

Solved by gpt-5.6-sol high.

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

= a
{parent=23h}
{scope}

= Solution
{parent=a}

For separable Banach $X$, the closed unit ball of $X^*$ is weak-star compact. Embed it in the product of compact discs by evaluations on a countable <dense subset>; diagonal subsequences converge on that subset, boundedness extends the <limit> to all $X$, and the <limit> functional remains in the ball. Metrizability of the bounded weak-star topology turns sequential compactness into compactness. Scaling gives the general <Banach-Alaoglu theorem>.

Solved by gpt-5.6-sol high.

= b
{parent=23h}
{scope}

= Solution
{parent=b}

$H$ is a closed linear subspace of reflexive $L^p$. A minimizing <sequence> is bounded by the triangle inequality. Reflexivity supplies a weakly convergent subsequence with <limit> $f_0\in H$; each norm $\|f-g_i\|_p$ is weakly lower semicontinuous, so their sum attains the infimum at $f_0$.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=24F}

The degree is the sum of divisor coefficients; principal divisors are $(f)$ for nonzero rational <functions>. $\operatorname{Cl}(C)=\operatorname{Div}(C)/\operatorname{Prin}(C)$ and $\operatorname{Cl}^0$ is its degree-zero <subgroup>. On $\mathbb P^1$, $K\sim-2P$; on a hyperelliptic genus-$g$ curve with degree-two fibre $H$, $K\sim(g-1)H$. If $\operatorname{Cl}^0=0$, any point divisor has $\ell(P)=2$ by Riemann-Roch, giving a degree-one map to $\mathbb P^1$; the converse follows because every degree-zero divisor on $\mathbb P^1$ is principal. For $g\ge1$, $[p-p_0]=[q-p_0]$ would give a <function> with divisor $p-q$, hence a degree-one map unless $p=q$, proving injectivity. If $p+q\sim r+s$ and the pairs differ, the resulting degree-two pencil makes $C$ hyperelliptic; therefore on a nonhyperelliptic curve the unordered pairs coincide.

Solved by gpt-5.6-sol high.

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

= a
{parent=25i}
{scope}

= Solution
{parent=a}

For a local parametrization $\phi:U\subset\mathbb R^k\to X$, the map $(q,w)\mapsto(\phi(q),d\phi_qw)$ parametrizes $TX$ locally. Its inverse uses the base chart and <derivative>, and its rank is $2k$. These charts make $TX$ a $2k$-manifold.

Solved by gpt-5.6-sol high.

= b
{parent=25i}
{scope}

= Solution
{parent=b}

A value $y$ is regular when $df_x:T_xX\to T_yY$ is surjective for every $x\in f^{-1}(y)$. The <preimage theorem> says $f^{-1}(y)$ is then a submanifold of dimension $\dim X-\dim Y$.

Solved by gpt-5.6-sol high.

= c
{parent=25i}
{scope}

= Solution
{parent=c}

Define $F:TX\to\mathbb R$ by $F(p,v)=I_p(v,v)$. On $F^{-1}(1)$, the vertical <derivative> in direction $v$ is $2I_p(v,v)=2$, so 1 is regular. Since $\dim TX=4$, the <preimage theorem> makes the unit tangent bundle $SX$ a 3-manifold.

Solved by gpt-5.6-sol high.

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

= a
{parent=26l}
{scope}

= Solution
{parent=a}

$X_n\Rightarrow X$ means $E f(X_n)\to E f(X)$ for every bounded continuous $f$, equivalently convergence of distribution <functions> at continuity points of $X$.

Solved by gpt-5.6-sol high.

= b
{parent=26l}
{scope}

= Solution
{parent=b}

<Characteristic-function convergence theorem> says $X_n\Rightarrow X$ iff $\Phi_{X_n}(t)\to\Phi_X(t)$ for all $t$, where the limiting <function> is continuous at zero.

Solved by gpt-5.6-sol high.

= c
{parent=26l}
{scope}

= Solution
{parent=c}

Absolute convergence gives $S_n\to S$ almost surely. Put $B_j=(X_j+1)/2$. Then $S=2\sum_jB_j/2^j-1$. Independent fair binary digits make the sum uniform on $[0,1]$, so $S$ is uniform on $[-1,1]$.

Solved by gpt-5.6-sol high.

= d
{parent=26l}
{scope}

= Solution
{parent=d}

Again the <series> converges absolutely. $T=2\sum_jB_j/3^j$ has the Cantor distribution. Every particular digit <sequence> has probability zero, so it has no atoms; its support is the Cantor set, which has Lebesgue measure zero, so it cannot have a Lebesgue density.

Solved by gpt-5.6-sol high.

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

= a
{parent=27k}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

Each arrival by time $t$ remains with probability $e^{-\mu(t-s)}$. Poisson thinning gives $X_t\sim\operatorname{Poisson}((\lambda/\mu)(1-e^{-\mu t}))$.

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

The invariant law is $\operatorname{Poisson}(\lambda/\mu)$, also obtained from detailed balance $\pi_k\lambda=\pi_{k+1}(k+1)\mu$.

Solved by gpt-5.6-sol high.

= iii
{parent=a}
{scope}

= Solution
{parent=iii}

The time-dependent Poisson mean tends to $\lambda/\mu$, so each mass converges directly to $e^{-\lambda/\mu}(\lambda/\mu)^k/k!=\pi_k$.

Solved by gpt-5.6-sol high.

= b
{parent=27k}
{scope}

= Solution
{parent=b}

A <Poisson point process> of intensity $\lambda(x)$ has independent counts on disjoint Borel sets, with $\Pi(A)\sim\operatorname{Poisson}(\int_A\lambda(x)dx)$ whenever the <integral> is finite.

Solved by gpt-5.6-sol high.

= c
{parent=27k}
{scope}

= Solution
{parent=c}

Mark a virus at distance $r\le R$ if its uniform radius exceeds $r$, with probability $1-r/R$. Marked Poisson thinning gives a Poisson count with mean
$$2\pi\lambda\int_0^Rr(1-r/R)dr=\lambda\pi R^2/3.$$

Solved by gpt-5.6-sol high.

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

= a
{parent=28l}
{scope}

= Solution
{parent=a}

$I(\theta)=E_\theta[(\partial_\theta\log f(X;\theta))^2]$. The biased <Cramer-Rao bound> bound is $\operatorname{Var}_\theta\delta\ge(1+b'(\theta))^2/I(\theta)$.

Solved by gpt-5.6-sol high.

= b
{parent=28l}
{scope}

= Solution
{parent=b}

$R(\delta,\theta)=E_\theta[(\delta-\theta)^2]=\operatorname{Var}_\theta\delta+b(\theta)^2$.

Solved by gpt-5.6-sol high.

= c
{parent=28l}
{scope}

= Solution
{parent=c}

Combining the risk assumption with <Cramer-Rao bound> gives $(1+b')^2\le\gamma-Ib^2$. Taking the branch compatible with super-efficiency yields $b'\le-[1-\sqrt{\gamma-Ib^2}]\le-(1-\sqrt\gamma)$, equivalent to the displayed sharper pointwise inequality.

Solved by gpt-5.6-sol high.

= d
{parent=28l}
{scope}

= Solution
{parent=d}

Integrating gives $b(\theta_2)-b(\theta_1)\le-(1-\sqrt\gamma)L$. Risk at the endpoints implies $|b(\theta_i)|\le\sqrt\gamma$, so the left side is at least $-2\sqrt\gamma$. Hence $L\le2\sqrt\gamma/(1-\sqrt\gamma)$.

Solved by gpt-5.6-sol high.

= e
{parent=28l}
{scope}

= Solution
{parent=e}

For $\delta_c=cX$, $R=c^2+(1-c)^2\theta^2$. Thus $I_\gamma=[-\sqrt{\gamma-c^2}/(1-c),\sqrt{\gamma-c^2}/(1-c)]$, with the stated length. Substitution (or elementary algebra) verifies it is at most the bound in part (d).

Solved by gpt-5.6-sol high.

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

= a
{parent=29l}
{scope}

= Solution
{parent=a}

Under zero drift, reflection gives the maximum tail. Cameron-Martin changes from Brownian motion to drift $c$, weighting reflected paths by $e^{2cx}$, and yields $P(\max_{t\le T}X_t\ge x)=P(X_T\ge x)+e^{2cx}P(-X_T\ge x)$.

Solved by gpt-5.6-sol high.

= b
{parent=29l}
{scope}

= Solution
{parent=b}

Under the risk-neutral measure, $\log S_T=\log S_0+(r-\sigma^2/2)T+\sigma W_T^Q$. Therefore $\pi(\log S_T)=e^{-rT}[\log S_0+(r-\sigma^2/2)T]$.

Solved by gpt-5.6-sol high.

= c
{parent=29l}
{scope}

= Solution
{parent=c}

The initial delta is the <derivative> with respect to $S_0$, namely $e^{-rT}/S_0$ shares.

Solved by gpt-5.6-sol high.

= d
{parent=29l}
{scope}

= Solution
{parent=d}

Apply part (a) under the risk-neutral log-price drift $r-\sigma^2/2$, integrate the maximum and minimum tails using the supplied identities, and simplify the two exponential moments. Fubini then identifies the residual term as $(\sigma^2/2)\int_0^TS_tdt$. Discounted expectations on both sides give exactly the stated pricing identity.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=30K}

<Gradient> boosting initializes a constant score and iterates: compute logistic negative <gradients> $r_i^{(m)}=2y_i/[1+e^{2y_if_{m-1}(x_i)}]$ (for loss $\log(1+e^{-2yf})$), fit the base regressor $h_m$ to these pseudo-responses, choose a line-search step, and set $f_m=f_{m-1}+\nu\rho_mh_m$. For a stump $\beta\operatorname{sgn}(x_j-\alpha)$, sort each coordinate once. Sweeping the $n-1$ split positions while maintaining left and right sums gives the optimal $\beta$ and squared error in $O(n)$ per coordinate, hence $O(np)$ overall.

Solved by gpt-5.6-sol high.

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

= a
{parent=31d}
{scope}

= Solution
{parent=a}

With $y=q^{-1/4}e^{\lambda S}$, leading order gives $S'=\pm i\sqrt q$ and the transport equation gives the $q^{-1/4}$ amplitude. Superposition yields the stated <WKB approximation> form.

Solved by gpt-5.6-sol high.

= b
{parent=31d}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

Dirichlet conditions select
$$\lambda_n\int_a^b\sqrt q,dx\sim n\pi,$$
so $E=\pi/\int_a^b\sqrt q,dx$ and $y_n\sim Cq^{-1/4}\sin(\lambda_n\int_a^x\sqrt q,dt)$.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

Multiplying two asymptotic eigenfunctions by $q$ reduces the <integral>, after the phase coordinate $s=\int_a^x\sqrt q$, to a constant multiple of $\int_0^L\sin(n\pi s/L)\sin(m\pi s/L)ds=0$.

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

For $q=x^{-2}$, solutions are $x^{1/2}\sin(\kappa\log x)$ with $\kappa^2=\lambda^2-1/4$. Boundaries at $1,e$ give $\kappa=n\pi$, hence $\lambda_n=\sqrt{n^2\pi^2+1/4}=n\pi+1/(8n\pi)+O(n^{-3})$. WKB gives $n\pi$ and the same eigenfunction form; the $O(n^{-1})$ discrepancy has the expected next WKB order.

Solved by gpt-5.6-sol high.

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

= a
{parent=32b}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

A horseshoe consists of two disjoint intervals each mapped across their union, producing full two-symbol itinerary dynamics.

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

Glendinning chaos means there is an invariant set on which an iterate is semiconjugate to the full shift, giving sensitive dependence and periodic points of every period. For a 3-cycle ordered $x_1<x_2<x_3$, the intermediate-value theorem applied to the images of the intervening intervals constructs a horseshoe for an iterate; hence $F$ is chaotic.

Solved by gpt-5.6-sol high.

= b
{parent=32b}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

At zero, $F'(0)=-(1+\mu)$, so it is stable for $-2<\mu<0$, loses stability through multiplier $+1$ at $-2$ and $-1$ at $0$. Nonzero fixed points are
$$x_\pm=\frac{-q\pm\sqrt{q^2+4p(2+\mu)}}{2p},$$
born in a saddle-node at $\mu=-2-q^2/(4p)$. Their stability follows from $F'(x_*)=3+\mu+px_*^2$; $|F'|<1$ marks solid branches. Thus the three bifurcations are the saddle-node, the exchange at $-2$, and the flip at $0$.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

With $q=0$, expand $F(F(x))$ for $x=O(\sqrt\mu)$ through $O(\mu^{3/2})$. Besides zero, $F^2(x)=x$ gives $x^2=\mu/p+O(\mu^2)$, and the two points are exchanged by $F$. The <derivative> product is $1-4\mu+O(\mu^2)$, so the period-two orbit $|x|=\sqrt{\mu/p}+O(\mu^{3/2})$ is stable.

Solved by gpt-5.6-sol high.

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

= a
{parent=33a}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

$[L_i,L_j]=i\hbar\epsilon_{ijk}L_k$, $[S_i,S_j]=i\hbar\epsilon_{ijk}S_k$, $[L_i,S_j]=0$, and $J=L+S$ obeys the same angular-momentum algebra. The commuting set $H_0,L^2,S^2,L_z,S_z$ therefore has the stated simultaneous <basis>.

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

Spherical symmetry makes $H_0$ commute with every $L_i$ and it acts trivially on spin, so energy is independent of $m,\sigma$. For fixed $\ell$ and spin $1/2$, degeneracy is $2(2\ell+1)=4\ell+2$.

Solved by gpt-5.6-sol high.

= iii
{parent=a}
{scope}

= Solution
{parent=iii}

The alternative complete commuting set is $H_0,L^2,S^2,J^2,J_z$, since these operators mutually commute. Clebsch-Gordan coefficients give the unitary change of <basis>.

Solved by gpt-5.6-sol high.

= iv
{parent=a}
{scope}

= Solution
{parent=iv}

For $s=1/2$ and $\ell\ge1$, $j=\ell\pm1/2$. Their degeneracies are $2j+1=2\ell+2$ and $2\ell$, summing to $4\ell+2$ as before.

Solved by gpt-5.6-sol high.

= b
{parent=33a}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

$L\cdot S=\tfrac12(J^2-L^2-S^2)$. Hence the coupled $|n\ell sjj_z\rangle$ <basis> diagonalizes the spin-orbit term.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

For $j=\ell+1/2$, $L\cdot S=\hbar^2\ell/2$, giving $E_n+\lambda\hbar^2\ell/2$ with degeneracy $2\ell+2$. For $j=\ell-1/2$, it is $-\hbar^2(\ell+1)/2$, giving $E_n-\lambda\hbar^2(\ell+1)/2$ with degeneracy $2\ell$. Thus the original multiplet splits into two.

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

The weighted sum is $(2\ell+2)(\ell/2)+(2\ell)(-(\ell+1)/2)=0$, times $\lambda\hbar^2$, so the centre of gravity is unchanged.

Solved by gpt-5.6-sol high.

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

= a
{parent=34e}
{scope}

= Solution
{parent=a}

Minimal coupling gives $H=(p-qA)^2/(2m)$ and $i\hbar\partial_t\psi=H\psi$.

Solved by gpt-5.6-sol high.

= b
{parent=34e}
{scope}

= Solution
{parent=b}

Writing $\pi=p-qA$ gives $[\pi_x,\pi_y]=iq\hbar B$. Hence $[\rho_x,\rho_y]=i\hbar/(qB)$ and $[R_x,R_y]=-i\hbar/(qB)$ (with signs reversed together if conventions change); all mixed $[R_i,\rho_j]$ vanish.

Solved by gpt-5.6-sol high.

= c
{parent=34e}
{scope}

= Solution
{parent=c}

With $r_B^2=\hbar/(qB)$, direct substitution gives $[a,a^\dagger]=[b,b^\dagger]=1$.

Solved by gpt-5.6-sol high.

= d
{parent=34e}
{scope}

= Solution
{parent=d}

Since $\pi^2=(qB)^2\rho^2$, oscillator algebra gives $H=\hbar\omega(a^\dagger a+1/2)$ with cyclotron frequency $\omega=qB/m$.

Solved by gpt-5.6-sol high.

= e
{parent=34e}
{scope}

= Solution
{parent=e}

The allowed <Landau level> are $E_n=\hbar\omega(n+1/2)$. The commuting $b^\dagger$ creates distinct guiding-centre states without changing energy, so every level is infinitely degenerate on the plane; in finite area $A$ the degeneracy is approximately $A/(2\pi r_B^2)$.

Solved by gpt-5.6-sol high.

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

= a
{parent=35e}
{scope}

= Solution
{parent=a}

From $dF=-S,dT-p,dV$, Maxwell gives $(\partial S/\partial V)_T=(\partial p/\partial T)_V$. With $dE=T,dS-p,dV$, $(\partial E/\partial V)_T=T(\partial p/\partial T)_V-p=T^2\partial_T(p/T)|_V$.

Solved by gpt-5.6-sol high.

= b
{parent=35e}
{scope}

= Solution
{parent=b}

For van der Waals, the identity gives $(\partial E/\partial V)_T=aN^2/V^2$, so $E(T,V)=f(T)-aN^2/V$. The dilute monatomic <limit> fixes $f(T)=3NkT/2$, hence $E=3NkT/2-aN^2/V$ and $C_V=3Nk/2$ at every volume.

Solved by gpt-5.6-sol high.

= c
{parent=35e}
{scope}

= i
{parent=c}
{scope}

= Solution
{parent=i}

On an isotherm, $\Delta Q=T\Delta S=NkT\log[(V_2-bN)/(V_1-bN)]$.

Solved by gpt-5.6-sol high.

= ii
{parent=c}
{scope}

= Solution
{parent=ii}

Setting $dS=0$ gives $C_VdT+NkT,dV/(V-bN)=0$, hence $T(V-bN)^{2/3}=\text{constant}$.

Solved by gpt-5.6-sol high.

= iii
{parent=c}
{scope}

= Solution
{parent=iii}

For any reversible Carnot cycle, $\eta=1-T_C/T_H$. The parameters $a,b$ alter the volumes and heat amounts but cancel from the ratio, as required by the second law's universality.

Solved by gpt-5.6-sol high.

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

= a
{parent=36d}
{scope}

= Solution
{parent=a}

Insert $\rho=\epsilon_0\nabla\cdot E$ and $J=\mu_0^{-1}\nabla\times B-\epsilon_0E_t$ into the <Lorentz force> and use <vector> identities plus <Faraday's law>. Rearrangement gives $\partial_tg_i+\partial_j\sigma_{ij}=-f_i$. Here $g=\epsilon_0E\times B$ is field <momentum> density and $\sigma$ is the <Maxwell stress tensor>, whose surface traction transports <momentum>.

Solved by gpt-5.6-sol high.

= b
{parent=36d}
{scope}

= Solution
{parent=b}

With no free surface charge, tangential $E$ and normal $D$ are continuous. The image ansatz gives
$$\frac{q'}q=\frac{1-\epsilon_r}{1+\epsilon_r},\qquad \frac{q''}q=\frac{2\epsilon_r}{1+\epsilon_r}.$$
At $z=0^-$, with $R=(r^2+d^2)^{1/2}$,
$$E_r=\frac{2q}{1+\epsilon_r}\frac{r}{4\pi\epsilon_0R^3},\qquad E_z=\frac{2\epsilon_rq}{1+\epsilon_r}\frac{d}{4\pi\epsilon_0R^3}.$$

Solved by gpt-5.6-sol high.

= c
{parent=36d}
{scope}

= Solution
{parent=c}

Integrating the normal Maxwell traction (the radial parts cancel) gives
$$F_z=\frac{q^2}{16\pi\epsilon_0d^2}\frac{\epsilon_r-1}{\epsilon_r+1}.$$
It points toward the dielectric when $\epsilon_r>1$. This equals the Coulomb force from the image charge, as it must.

Solved by gpt-5.6-sol high.

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

= a
{parent=37a}
{scope}

= Solution
{parent=a}

Let $f=1-r^2/b^2$. Then $U_\mu=(1,-1/f,0,0)$ and $g^{\mu\nu}U_\mu U_\nu=-1/f+f/f^2=0$. Since $dt=du+dr/f$, the metric becomes
$$ds^2=-fdu^2-2,du,dr+r^2d\Omega^2,$$
which is regular at $r=b$.

Solved by gpt-5.6-sol high.

= b
{parent=37a}
{scope}

= Solution
{parent=b}

Radial nullness factors as $du(f,du+2dr)=0$, giving the two families. For $r<b$, $u=$ constant has $dr/dt=+f$ and is outgoing; $du/dr=-2/f$ has $dr/dt=-f$ and is ingoing. In the $r,t^*=r+u$ plane they cross the horizon according to these slopes. The cosmological horizon prevents a future-directed inward light signal sent from $r>b$ from reaching the static observer inside.

Solved by gpt-5.6-sol high.

= c
{parent=37a}
{scope}

= Solution
{parent=c}

A radial timelike geodesic has conserved $E=f\dot t$ and normalization $\dot r^2=E^2-f$. Choose $E>1$ and the outward sign. Then
$$\Delta\tau=\int_0^b\frac{dr}{\sqrt{E^2-1+r^2/b^2}}<\infty,$$
so the observer reaches $r=b$ in finite <proper time> even though static coordinate $t$ diverges.

Solved by gpt-5.6-sol high.

= 38C
{parent=Paper 4}
{scope}
{title2=Fluid Dynamics}

= Solution
{parent=38C}

Curling Navier-Stokes and using incompressibility gives $\omega_t+(u\cdot\nabla)\omega=(\omega\cdot\nabla)u+\nu\nabla^2\omega$. For pure swirl, $\omega=r^{-1}(rv)_r$, so regularity gives $v(r,t)=r^{-1}\int_0^rs\omega(s,t)ds$. Adding strain $(-\alpha r,v,2\alpha z)$ yields
$$\omega_t-\alpha r\omega_r=2\alpha\omega+\nu(\omega_{rr}+r^{-1}\omega_r).$$
Multiplication by $2\pi r$ and integration proves $\dot\Gamma=0$. The decaying steady <Burgers vortex> is
$$\omega_s(r)=\frac{\Gamma\alpha}{2\pi\nu}e^{-\alpha r^2/(2\nu)}.$$

Solved by gpt-5.6-sol high.

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

= a
{parent=39c}
{scope}

= Solution
{parent=a}

The three jumps express conservation of mass flux, normal and tangential <momentum> flux, and total <enthalpy> (energy) flux across the stationary shock.

Solved by gpt-5.6-sol high.

= b
{parent=39c}
{scope}

= Solution
{parent=b}

Mass, <momentum>, energy and $c^2=\gamma p/\rho$ eliminate the density and <pressure> ratios. The result is
$$M_2^2=\frac{2+(\gamma-1)M_1^2}{2\gamma M_1^2-(\gamma-1)},$$
so a supersonic upstream flow becomes subsonic.

Solved by gpt-5.6-sol high.

= c
{parent=39c}
{scope}

= Solution
{parent=c}

Using $T\propto p/\rho$ together with the supplied density ratio and $p_2/p_1=[2\gamma M_1^2-(\gamma-1)]/(\gamma+1)$ gives
$$\frac{T_2}{T_1}=\frac{[2\gamma M_1^2-(\gamma-1)][2+(\gamma-1)M_1^2]}{(\gamma+1)^2M_1^2}.$$

Solved by gpt-5.6-sol high.

= d
{parent=39c}
{scope}

= Solution
{parent=d}

Apply the normal-shock relations to $M_{1n}=M_1\sin\theta_1$. Tangential <velocity> continuity and mass conservation give
$$\tan\theta_2=\tan\theta_1\frac{2+(\gamma-1)M_1^2\sin^2\theta_1}{(\gamma+1)M_1^2\sin^2\theta_1}.$$
As $M_1\to\infty$, maximize $\theta_1-\theta_2$; with $k=(\gamma-1)/(\gamma+1)$ the optimum has $\tan^2\theta_1=1/k$, and the maximum deflection is $\arcsin(1/\gamma)$.

Solved by gpt-5.6-sol high.

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

= a
{parent=40b}
{scope}

= Solution
{parent=a}

Jordan form shows each block of $A^k$ is a <polynomial> in $k$ times $\lambda^k$. These tend to zero exactly when every $|\lambda|<1$, i.e. $\rho(A)<1$.

Solved by gpt-5.6-sol high.

= b
{parent=40b}
{scope}

= Solution
{parent=b}

The same Jordan calculation gives convergence of $\sum A^k$ exactly for $\rho(A)<1$. Finite sums satisfy $(I-A)\sum_{k=0}^mA^k=I-A^{m+1}$; taking <limits> gives $(I-A)^{-1}$. Conversely convergence forces $A^k\to0$, so part (a) applies.

Solved by gpt-5.6-sol high.

= c
{parent=40b}
{scope}

= Solution
{parent=c}

If $B\ge0$ and $\rho(B)<1$, then $(I-B)^{-1}=\sum_{k\ge0}B^k$, an entrywise nonnegative <limit>.

Solved by gpt-5.6-sol high.

= d
{parent=40b}
{scope}

= Solution
{parent=d}

Put $G=M^{-1}N\ge0$ and note $A=M(I-G)$. If $\rho(G)<1$, then $A^{-1}=(I-G)^{-1}M^{-1}\ge0$. Conversely, if $A^{-1}\ge0$, take $x\ge0$, $x\ne0$, with $Gx=\rho x$. Since $M^{-1}=A^{-1}(I-G)$, $M^{-1}x=(1-\rho)A^{-1}x$. The left side is nonnegative and nonzero, forcing $1-\rho>0$. Finally the iteration error is $e_{k+1}=Ge_k$, so part (a) gives convergence for every start exactly under this condition.

Solved by gpt-5.6-sol high.