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past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/ii/paper-2.bigb
= Paper 2
{scope}
https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2026/PaperII_2.pdf
= 1G
{parent=Paper 2}
{scope}
{title2=Number Theory}
= Solution
{parent=1G}
The <möbius function> is $mu(1)=1$, $mu(n)=0$ if a prime square divides $n$, and $mu(n)=(-1)^r$ when $n$ is a product of $r$ distinct primes. The <Riemann zeta function> is $zeta(s)=\sum_{n\ge1}n^{-s}$ for $\Re s>1$.
Both sides of the proposed identity are multiplicative. At $n=p^a$, the right side is $1$ for $a=0,1$ and $1+\mu(p)=0$ for $a\ge2$, exactly $mu(p^a)^2$. Absolute convergence permits rearrangement, so
$$
\sum_n\frac{\mu(n)^2}{n^s}=\sum_d\frac{\mu(d)}{d^{2s}}\sum_m\frac1{m^s}=\frac{\zeta(s)}{\zeta(2s)}.
$$
Solved by gpt-5.6-sol high.
= 2I
{parent=Paper 2}
{scope}
{title2=Topics in Analysis}
= Solution
{parent=2I}
<Runge theorem> says that if $K\subset\mathbb C$ is compact and $\mathbb C\setminus K$ is connected, every <function> holomorphic near $K$ is uniformly approximable on $K$ by <polynomials>.
Choose a compact exhaustion $K_j$ of the slit disc $\Delta\setminus L^+$, rounding the two sides of the slit so that $0\notin K_j$ and $\mathbb C\setminus K_j$ is connected. Runge applied to $1/z$ gives a <polynomial> $p_j$ with $\sup_{K_j}|p_j-1/z|<1/j$. Every compact subset of the slit disc lies in all sufficiently large $K_j$, proving locally <uniform convergence>.
Solved by gpt-5.6-sol high.
= 3H
{parent=Paper 2}
{scope}
{title2=Coding \& Cryptography}
= a
{parent=3h}
{scope}
= Solution
{parent=a}
The source is Bernoulli when the $X_n$ are independent and identically distributed with a common law $p$ on the finite alphabet.
Solved by gpt-5.6-sol high.
= b
{parent=3h}
{scope}
= Solution
{parent=b}
It is reliably encodable at rate $r$ if there are block encoders into at most $2^{nr}$ codewords and decoders whose block error probability tends to zero as $n\to\infty$.
Solved by gpt-5.6-sol high.
= c
{parent=3h}
{scope}
= Solution
{parent=c}
The <information rate> is $\lim_{n\to\infty}n^{-1}H(X_1,\ldots,X_n)$ when the <limit> exists.
Solved by gpt-5.6-sol high.
= d
{parent=3h}
{scope}
= Solution
{parent=d}
The <asymptotic equipartition property> is $-n^{-1}\log_2p(X_1,\ldots,X_n)\to H$ in probability. For an i.i.d. source this is the weak law applied to $-\log_2p(X_i)$, and independence gives $H(X_1,\ldots,X_n)=nH(X)$. Thus its <information rate> and AEP constant are both $H(X)$.
Solved by gpt-5.6-sol high.
= 4I
{parent=Paper 2}
{scope}
{title2=Automata and Formal Languages}
= a
{parent=4i}
{scope}
= Solution
{parent=a}
A state $q$ is inaccessible when no word $w\in\Sigma^*$ has $\widehat\delta(q_0,w)=q$.
Solved by gpt-5.6-sol high.
= b
{parent=4i}
{scope}
= Solution
{parent=b}
Set $p\sim q$ when $\widehat\delta(p,w)\in F\iff\widehat\delta(q,w)\in F$ for every word $w$. Put $\delta'([q],a)=[\delta(q,a)]$ and $F'=\{[q]:q\in F\}$. If $p\sim q$, then testing continuations $aw$ shows $\delta(p,a)\sim\delta(q,a)$, so $\delta'$ is well-defined.
Solved by gpt-5.6-sol high.
= c
{parent=4i}
{scope}
= Solution
{parent=c}
The <Myhill-Nerode theorem> says this quotient is, up to isomorphism, the unique accessible DFA with the least possible number of states accepting the language.
Solved by gpt-5.6-sol high.
= d
{parent=4i}
{scope}
= Solution
{parent=d}
Take states $0,\ldots,2n-1$, transition $j\mapsto j+1\pmod{2n}$, initial state $0$, and make every even state accepting. Every state is reached by some $a^j$, and two states are equivalent exactly when they have the same parity. The quotient therefore has two states and accepts the even-length words.
Solved by gpt-5.6-sol high.
= 5K
{parent=Paper 2}
{scope}
{title2=Statistical Modelling}
= Solution
{parent=5K}
<Logistic regression> assumes independent $Y_i\mid X_i\sim\operatorname{Bernoulli}(\pi_i)$ with
$$\log\frac{\pi_i}{1-\pi_i}=\beta_0+X_i^T\beta.$$
For the binomial exponential family, $\ell=y\theta-m\log(1+e^\theta)+\text{constant}$, so the mean is $m e^\theta/(1+e^\theta)$ and the canonical parameter is its logit. It may be fitted by <iteratively reweighted least squares>.
Under LDA, Bayes' rule gives
$$
\log\frac{P(Y=1\mid x)}{P(Y=0\mid x)}=log\frac{\pi_1}{\pi_0}+(\mu_1-\mu_0)^T\Sigma^{-1}x- frac12(\mu_1^T\Sigma^{-1}\mu_1-\mu_0^T\Sigma^{-1}\mu_0),
$$
which is a logistic model with an intercept.
Solved by gpt-5.6-sol high.
= 6C
{parent=Paper 2}
{scope}
{title2=Mathematical Biology}
= a
{parent=6c}
{scope}
= Solution
{parent=a}
Conservation requires zero flux: $n_x(0,t)=n_x(L,t)=0$. Indeed $d\int_0^L n,dx/dt=D[n_x]_0^L=0$.
Solved by gpt-5.6-sol high.
= b
{parent=6c}
{scope}
= Solution
{parent=b}
Add exponential growth: $n_t=Dn_{xx}+\alpha n$.
Solved by gpt-5.6-sol high.
= c
{parent=6c}
{scope}
= Solution
{parent=c}
With hostile Dirichlet boundaries, expand in $\sin(k\pi x/L)$. The $k$th amplitude grows at rate $\alpha-D(k\pi/L)^2$. Persistence requires the principal rate to be nonnegative, hence $\alpha\ge D\pi^2/L^2$; strict inequality gives growth.
Solved by gpt-5.6-sol high.
= 7D
{parent=Paper 2}
{scope}
{title2=Further Complex Methods}
= Solution
{parent=7D}
The <laplace transform> is $\widehat f(p)=\int_0^\infty e^{-pt}f(t),dt$, and inversion is
$$f(t)=\frac1{2\pi i}\int_{c-i\infty}^{c+i\infty}e^{pt}\widehat f(p),dp,$$
where the vertical <bromwich contour> lies to the right of all singularities. For $t>0$ it is closed leftwards, and for $t<0$ rightwards, where the exponential decays.
Transforming the PDE and imposing boundedness as $x\to\infty$ gives
$$
U(x,p)=\frac1{p(p+1)}+\frac{p-1}{p(p+1)}e^{-x\sqrt{p/D}},
$$
with the square root cut conventionally along $(-\infty,0]$. Therefore
$$u(x,t)=\frac1{2\pi i}\int_{c-i\infty}^{c+i\infty}e^{pt}U(x,p),dp.$$
There are poles at $p=0,-1$ (with cancellation assessed in the complete expression) and a <branch point> at $0$; close left for $t>0$ and wrap the cut while including the relevant residues.
Solved by gpt-5.6-sol high.
= 8A
{parent=Paper 2}
{scope}
{title2=Classical Dynamics}
= a
{parent=8a}
{scope}
= Solution
{parent=a}
With $x=\ell\sin\theta$ and $y=-at^2/2-\ell\cos\theta$, discard a total <derivative> from the <kinetic energy> to obtain
$$L=\tfrac12m\ell^2\dot\theta^2+ m\ell(g-a)\cos\theta.$$
Thus $\ddot\theta+(g-a)\sin\theta/\ell=0$. If $a=g$, then $\ddot\theta=0$: the pendulum is weightless in the lift and rotates uniformly (or remains fixed).
Solved by gpt-5.6-sol high.
= b
{parent=8a}
{scope}
= Solution
{parent=b}
The equilibria are $\theta=0,\pi$ modulo $2\pi$ unless $a=g$, when every angle is an equilibrium. Linearisation shows $ heta=0$ stable for $a<g$ and $\theta=\pi$ stable for $a>g$; the other is unstable.
Solved by gpt-5.6-sol high.
= 9D
{parent=Paper 2}
{scope}
{title2=Cosmology}
= a
{parent=9d}
{scope}
= Solution
{parent=a}
Since $p\propto L^{-1}=V^{-1/3}$, $dp/dV=-p/(3V)$. At fixed occupations, $dE=\int \mathcal E'(p)(dp/dV)\bar n(p),dp,dV$. Equating this to $-P,dV$ yields $P=(3V)^{-1}\int p\mathcal E'(p)\bar n(p),dp$.
Solved by gpt-5.6-sol high.
= b
{parent=9d}
{scope}
= Solution
{parent=b}
For $\mathcal E=pc$, $p\mathcal E'=\mathcal E$, so $P=E/(3V)=\rho/3$. Adiabatic energy conservation gives $\rho\propto a^{-4}$, equivalently conserved particle number gives $n\propto a^{-3}$; with $\rho\propto T^4$ or $n\propto T^3$, both imply $T\propto a^{-1}$.
Solved by gpt-5.6-sol high.
= c
{parent=9d}
{scope}
= Solution
{parent=c}
Here $\mathcal E=mc^2+p^2/(2m)+\cdots$, so $p\mathcal E'=p^2/m$. Consequently $P=(3V)^{-1}\int p^2/m,\bar n,dp=2U/(3V)$. For an adiabatic monatomic gas, $u\propto a^{-5}$ and $n\propto a^{-3}$, hence $T\propto u/n\propto a^{-2}$.
Solved by gpt-5.6-sol high.
= 10E
{parent=Paper 2}
{scope}
{title2=Quantum Information and Computation}
= a
{parent=10e}
{scope}
= Solution
{parent=a}
The reflection is $I_{|\psi\rangle}=I-2|\psi\rangle\langle\psi|$. Hence $I_{|x_0\rangle}|x\rangle=-|x_0\rangle$ for $x=x_0$ and equals $|x\rangle$ otherwise.
Solved by gpt-5.6-sol high.
= b
{parent=10e}
{scope}
= i
{parent=b}
{scope}
= Solution
{parent=i}
The marked-subspace reflection is $I_G=I-2\sum_{j=1}^M|g_j\rangle\langle g_j|$.
Solved by gpt-5.6-sol high.
= ii
{parent=b}
{scope}
= Solution
{parent=ii}
Write $|G\rangle=M^{-1/2}\sum_g|g\rangle$ and $|B\rangle=(N-M)^{-1/2}\sum_{x\notin G}|x\rangle$. Then $|\psi_0\rangle=\sin\alpha|G\rangle+\cos\alpha|B\rangle$. Both reflections preserve $\operatorname{span}\{|G\rangle,|B\rangle\}$, hence so does their product.
Solved by gpt-5.6-sol high.
= iii
{parent=b}
{scope}
= Solution
{parent=iii}
In the orthonormal <basis> $(|G\rangle,|B\rangle)$, direct multiplication gives
$$Q_M=\begin{pmatrix}\cos2\alpha&\sin2\alpha\\-\sin2\alpha&\cos2\alpha\end{pmatrix}$$
(up to the equivalent opposite orientation convention).
Solved by gpt-5.6-sol high.
= iv
{parent=b}
{scope}
= Solution
{parent=iv}
The displayed orthogonal <matrix> is a rotation through $2\alpha$, where $\sin^2\alpha=M/N$.
Solved by gpt-5.6-sol high.
= v
{parent=b}
{scope}
= Solution
{parent=v}
After $r$ iterations the marked amplitude is $\sin((2r+1)\alpha)$. Choose $r$ nearest to $\pi/(4\alpha)-1/2$; for $M\ll N$, this is about $(\pi/4)\sqrt{N/M}$ oracle iterations.
Solved by gpt-5.6-sol high.
= 11I
{parent=Paper 2}
{scope}
{title2=Topics in Analysis}
= a
{parent=11i}
{scope}
= Solution
{parent=a}
For a closed path avoiding zero, choose a continuous lift $\gamma(t)/|\gamma(t)|=e^{i\theta(t)}$ and set $w(\gamma;0)=(\theta(1)-\theta(0))/(2\pi)$. Conjugation replaces $\theta$ by $-\theta$, as does reversing the path, so both <winding number>s are $-w(\gamma;0)$.
Solved by gpt-5.6-sol high.
= b
{parent=11i}
{scope}
= Solution
{parent=b}
<Homotopy invariance of winding number> states that a homotopy $H(s,t)$ through closed paths in $\mathbb C^*$ preserves <winding number>. Uniform continuity divides the $s$ interval so successive paths differ pointwise by less than the minimum <modulus> of the earlier path; the supplied perturbation lemma makes their <winding number>s equal. Chaining the subdivisions proves the result.
Solved by gpt-5.6-sol high.
= c
{parent=11i}
{scope}
= Solution
{parent=c}
If no $z\in S^1$ obeyed $z^n=-g(z)$, the loops $z^n$ and $-g(z)$ could be joined without crossing zero by the straight-line homotopy, so they would have equal <winding number>. The first has winding $n$. Since $g|_{S^1}$ extends over the disc, it is null-homotopic and has winding $0$; multiplication by $-1$ does not change it. This contradicts $n\ne0$.
Solved by gpt-5.6-sol high.
= 12H
{parent=Paper 2}
{scope}
{title2=Coding \& Cryptography}
= a
{parent=12h}
{scope}
= Solution
{parent=a}
<Shannon noiseless coding theorem> says the minimum expected binary prefix-code length $L$ satisfies $H(X)\le L<H(X)+1$. Kraft and Gibbs give $H\le L$ for every prefix code. Taking lengths $\ell_i=\lceil-\log_2p_i\rceil$ satisfies Kraft because $\sum2^{-\ell_i}\le1$, and gives $L<\sum p_i(-\log_2p_i+1)=H+1$.
Solved by gpt-5.6-sol high.
= b
{parent=12h}
{scope}
= i
{parent=b}
{scope}
= Solution
{parent=i}
Equiprobability gives $H=\log_2m$.
Solved by gpt-5.6-sol high.
= ii
{parent=b}
{scope}
= Solution
{parent=ii}
For $m=2^b$, assign every outcome one of the $b$-bit words. This complete fixed-length prefix code has expected length $b$ and meets the <information entropy> lower bound.
Solved by gpt-5.6-sol high.
= iii
{parent=b}
{scope}
= Solution
{parent=iii}
If two lengths differed by at least two, replacing two deepest sibling leaves by their parent and splitting a shallower leaf lowers total length, contradicting optimality. Thus lengths are $b=\lfloor\log_2m\rfloor$ and $b+1$. Kraft equality gives $n_b=2^{b+1}-m$ and $n_{b+1}=2m-2^{b+1}$, hence
$$L=b+2-\frac{2^{b+1}}m.$$
Solved by gpt-5.6-sol high.
= 13D
{parent=Paper 2}
{scope}
{title2=Further Complex Methods}
= Solution
{parent=13D}
A <branch point> is a point around which analytic continuation changes a value; a branch cut removes curves so continuation becomes single-valued, and a branch is one such single-valued analytic choice. Define
$$\operatorname{Arcsin}z=\int_0^z(1-t^2)^{-1/2}dt$$
on $\mathbb C\setminus((-∞,-1]\cup[1,∞))$, choosing the square root equal to $1$ at zero. Monodromy around $\pm1$ explains multivaluedness.
On the upper lip $0+$ of the cut convention, the branch remains positive on $(0,1)$, so $\int_{0+}^1dt/\sqrt{1-t^2}=\pi/2$. Define $\operatorname{Arccos}z=\pi/2-\operatorname{Arcsin}z$ on the same slit domain.
Also
$$\arctan z=\frac1{2i}[\log(1+iz)-\log(1-iz)].$$
Changing logarithm branches changes the value by multiples of $\pi$ (already visible at $z=1$). Principal logarithms give a single branch on the plane cut from $\pm i$ outward. Differentiating both sides and checking the value at zero proves
$$\operatorname{Arctan}z=\operatorname{Arcsin}\frac z{\sqrt{1+z^2}}$$
where the compatible principal branches are defined.
Solved by gpt-5.6-sol high.
= 14A
{parent=Paper 2}
{scope}
{title2=Classical Dynamics}
= a
{parent=14a}
{scope}
= Solution
{parent=a}
Let $r$ be the cylindrical radius of the point of contact and $\phi$ its azimuth. Rolling without slip or twist gives $|\omega|=v/a$ and therefore
$$L=\frac{7m}{10}(\dot r^2\csc^2\alpha+r^2\dot\phi^2)-mgr\cot\alpha,$$
up to an irrelevant additive constant. The factor $7/5$ combines translational and rotational inertia.
Solved by gpt-5.6-sol high.
= b
{parent=14a}
{scope}
= Solution
{parent=b}
Time and azimuthal invariance give energy
$$E=\frac{7m}{10}(\dot r^2\csc^2\alpha+r^2\dot\phi^2)+mgr\cot\alpha$$
and <angular momentum> $J=(7m/5)r^2\dot\phi$. They represent mechanical energy and vertical <angular momentum>.
Solved by gpt-5.6-sol high.
= c
{parent=14a}
{scope}
= Solution
{parent=c}
The radial <effective potential> is $V_{\rm eff}=mgr\cot\alpha+5J^2/(14mr^2)$. Initially $r=8a$, $J=(7m/5)(8a)v_0$. Comparing the initial energy with $V_{\rm eff}(2a)$ gives
$$\frac7{10}mv_0^2+8mga\cot\alpha\le \frac{56}{5}mv_0^2+2mga\cot\alpha,$$
which is exactly $v_0^2\ge(4/7)ga\cot\alpha$. Larger speed raises the centrifugal barrier at the neck faster than it raises the initial <kinetic energy>.
Solved by gpt-5.6-sol high.
= d
{parent=14a}
{scope}
= Solution
{parent=d}
A holonomic constraint is $F(q,t)=0$. In the multiplier formulation its generalized constraint force is $Q_i=\lambda\partial_iF$. Along an allowed motion, its power is $\sum_iQ_i\dot q_i=\lambda(\dot F-\partial_tF)$, which vanishes when the constraint has no explicit time dependence.
Solved by gpt-5.6-sol high.
= e
{parent=14a}
{scope}
= Solution
{parent=e}
At the circular minimum, $v^2=(5/7)gr\cot\alpha$ and $r=8a$. Projecting the centre-of-mass equation on the cone normal gives
$$N=mg\sin\alpha+\frac{mv^2}{r}\cos\alpha
=\frac{mg(5+2\sin^2\alpha)}{7\sin\alpha}.$$
This is the magnitude of the normal constraint force.
Solved by gpt-5.6-sol high.
= 15E
{parent=Paper 2}
{scope}
{title2=Quantum Information and Computation}
= a
{parent=15e}
{scope}
= Solution
{parent=a}
After transmission Alice and Bob announce <bases>, retain matching-basis bits, reveal a random sample, and abort if its error rate is too large; privacy amplification and error correction produce the key. A Breidbart intercept-resend causes error $1/4$ conditional on matching Alice and Bob <bases>. Since matching occurs with probability $1/2$, one transmitted qubit detects Eve with probability $1/8$.
Solved by gpt-5.6-sol high.
= b
{parent=15e}
{scope}
= Solution
{parent=b}
Conditional on Alice and Bob using the same <basis>, Eve chooses the wrong <basis> with probability $2/3$, then Bob disagrees with probability $1/2$, so the conditional error is $1/3$. Alice and Bob match with probability $1/3$, hence the per-transmission detection probability is $1/9$.
Solved by gpt-5.6-sol high.
= c
{parent=15e}
{scope}
= Solution
{parent=c}
The channel is $\rho\mapsto(1-p)\rho+pI/2$, so
$$P(0)=(1-p)|\langle0|\psi\rangle|^2+p/2.$$
Without Eve the sifted error is $p/2$. If Eve alone produces sifted error $e$ ($1/4$ for BB84, $1/3$ for six-state), the combined error is $(1-p)e+p/2$, strictly above the noise baseline exactly when $p<1$. Thus with known stationary $p$ and enough test bits, Eve is statistically detectable for $0\le p<1$ in either protocol; at $p=1$ the channel is completely mixed and erases the distinction.
Solved by gpt-5.6-sol high.
= 16J
{parent=Paper 2}
{scope}
{title2=Logic \& Set Theory}
= a
{parent=16j}
{scope}
= Solution
{parent=a}
Using the definition twice gives $P(a,b)=1$ when $a\le b$; when $a>b$, $a\to b=b$, then $b\to a=1$, and $1\to a=a$.
Solved by gpt-5.6-sol high.
= b
{parent=16j}
{scope}
= Solution
{parent=b}
If $b\le c$, then $b\to c=1$ and $T=1\to a=a$. If $b>c$, then $b\to c=c$, so $T=c\to a$, which is $1$ exactly when $c\le a$ and is $a$ otherwise.
Solved by gpt-5.6-sol high.
= c
{parent=16j}
{scope}
= Solution
{parent=c}
The implicational Hilbert axioms are $A\to(B\to A)$, $(A\to(B\to C))\to((A\to B)\to(A\to C))$, and Peirce's law $((A\to B)\to A)\to A$.
Solved by gpt-5.6-sol high.
= d
{parent=16j}
{scope}
= Solution
{parent=d}
A direct case split in the linearly ordered chain verifies types 1 and 2. Peirce's value is $P(a,b)$ from part (a); for $n\ge3$, choose $0<a>b$, obtaining $P(a,b)=a<1$. Thus type 3 need not be $n$-valid.
Solved by gpt-5.6-sol high.
= e
{parent=16j}
{scope}
= Solution
{parent=e}
One example is
$$\varphi=((p\to q)\to p)\to(((q\to0)\to p)\to p).$$
Applying the two case formulas from (a) and (b) checks the three possible values in $H_3$ and always gives $1$. In $H_4$, the valuation $p=1/2,q=1/3$ gives $v(\varphi)=1/2$, so it is not 4-valid.
Solved by gpt-5.6-sol high.
= 17J
{parent=Paper 2}
{scope}
{title2=Graph Theory}
= a
{parent=17j}
{scope}
= Solution
{parent=a}
<Ramsey theorem> follows from $R(s,t)\le R(s-1,t)+R(s,t-1)$: at a vertex, either at least $R(s-1,t)$ neighbours or at least $R(s,t-1)$ non-neighbours produce the desired clique or independent set. Induction with $R(2,t)=R(s,2)=t$ gives $R(s,t)\le\binom{s+t-2}{s-1}$, hence $R(t)\le\binom{2t-2}{t-1}<2^{2t}$.
Solved by gpt-5.6-sol high.
= b
{parent=17j}
{scope}
= Solution
{parent=b}
Apply Ramsey's theorem to a sufficiently large complete graph on integer vertices, colouring edge $\{i,j\}$ by the colour of $|i-j|$. Choose a monochromatic clique $y_0<y_1<\cdots<y_{n-1}$. Set $x_i=y_i-y_{i-1}$ for $1\le i\le n-1$ and $x_n=y_{n-1}-y_0$. Every $x_i$ is the difference along an edge of the clique, so all have one colour, and the differences telescope to $x_1+\cdots+x_{n-1}=x_n$.
Solved by gpt-5.6-sol high.
= c
{parent=17j}
{scope}
= Solution
{parent=c}
Alternating suitable half-open vertical strips of width $\sqrt3/2$ gives a two-colouring in which the horizontal projections forced by a unit equilateral triangle cannot all occupy strips of one parity; boundary strips are assigned consistently. For three colours, embed the <Moser spindle>, a finite unit-distance graph of chromatic number four, in the plane. Any three-colouring therefore has a monochromatic unit edge.
Solved by gpt-5.6-sol high.
= 18F
{parent=Paper 2}
{scope}
{title2=Galois Theory}
= a
{parent=18f}
{scope}
= Solution
{parent=a}
Define $\Phi_n(X)=\prod_{1\le k\le n,(k,n)=1}(X-e^{2\pi ik/n})$. From $X^n-1=\prod_{d\mid n}\Phi_d(X)$, induction and monic division show $\Phi_n\in\mathbb Z[X]$. Every $\Phi_n$ is irreducible over $\mathbb Q$. For prime $p$, $\Phi_p(X+1)=((X+1)^p-1)/X$ is Eisenstein at $p$, proving the assertion.
Solved by gpt-5.6-sol high.
= b
{parent=18f}
{scope}
= Solution
{parent=b}
For odd $n>1$, the primitive $2n$th roots are exactly the negatives of primitive $n$th roots. Since $\varphi(2n)=\varphi(n)$, their monic root <polynomials> satisfy $\Phi_{2n}(X)=\Phi_n(-X)$.
Solved by gpt-5.6-sol high.
= c
{parent=18f}
{scope}
= Solution
{parent=c}
Identify $G\simeq(\mathbb Z/15\mathbb Z)^\times\simeq C_4\times C_2$, with $\sigma(\zeta)=\zeta^2$ and $\tau(\zeta)=\zeta^{-1}$. Its <subgroups> are
$$1;\ \langle\sigma^2\rangle,\langle\tau\rangle,\langle\sigma^2\tau\rangle;\ \langle\sigma\rangle,\langle\sigma\tau\rangle,\langle\sigma^2,\tau\rangle;\ G.$$
Their fixed fields, in the same layers, are
$$L;\ \mathbb Q(\sqrt{-3},\sqrt5),\ L^+=\mathbb Q(\zeta+\zeta^{-1}),\ \mathbb Q(\zeta_5);\ \mathbb Q(\sqrt{-15}),\mathbb Q(\sqrt{-3}),\mathbb Q(\sqrt5);\ \mathbb Q.$$
The real fields are $\mathbb Q$, $\mathbb Q(\sqrt5)$, and $L^+$. This is the <Galois correspondence>.
Solved by gpt-5.6-sol high.
= 19F
{parent=Paper 2}
{scope}
{title2=Representation Theory}
= a
{parent=19f}
{scope}
= Solution
{parent=a}
The kernel is $\{g:\rho(g)=I\}$ and is normal because $\rho(hgh^{-1})=\rho(h)\rho(g)\rho(h)^{-1}$. A one-dimensional representation kills every commutator. Conversely, irreducibles killing $G'$ factor through the finite abelian <group> $G/G'$ and are one-dimensional.
Solved by gpt-5.6-sol high.
= b
{parent=19f}
{scope}
= Solution
{parent=b}
By <Schur lemma>, $z\in Z$ acts on irreducible $V$ by a <scalar>. <Induced representation> shows that $V$ occurs in $\operatorname{Ind}_Z^G\Theta$ iff that <scalar> character is $\Theta$, and then with multiplicity $\dim V$.
Solved by gpt-5.6-sol high.
= c
{parent=19f}
{scope}
= Solution
{parent=c}
Here $|G|=p^3$, $Z=G'=\{(x,y,c):x=y=0\}\simeq C_p$. Degree divisibility and $\sum d_i^2=p^3$ exclude degrees strictly between $1$ and $p$. There are $p^2$ linear characters, corresponding to $G/G'$. For trivial $\Theta$, induction is their direct sum. For each of the $p-1$ nontrivial central characters, induction is $p$ copies of one degree-$p$ irreducible $V_\Theta$. Thus the complete list is $p^2$ degree-one and $p-1$ degree-$p$ representations; their squared degrees sum to $p^3$.
Solved by gpt-5.6-sol high.
= 20G
{parent=Paper 2}
{scope}
{title2=Number Fields}
= a
{parent=20g}
{scope}
= Solution
{parent=a}
For $x\in\mathcal O_K$, all $\operatorname{Tr}(\alpha^{i-1}x)$ are integers, and duality gives $x=\sum_i\operatorname{Tr}(\alpha^{i-1}x)y_i$. Hence $\mathcal O_K\subset\sum\mathbb Zy_i$. Every <ideal> of $R$ is therefore a <subgroup> of a finitely generated free abelian <group> and is finitely generated over $\mathbb Z$, hence over $R$; so $R$ is Noetherian.
Solved by gpt-5.6-sol high.
= b
{parent=20g}
{scope}
= i
{parent=b}
{scope}
= Solution
{parent=i}
If $cI\subset R$, then $cI$ is an <ideal> of the Noetherian <ring> $R$, hence finitely generated; scaling back shows $I$ is finitely generated.
Solved by gpt-5.6-sol high.
= ii
{parent=b}
{scope}
= Solution
{parent=ii}
If $I=\sum R x_j$ with $x_j\in K$, clear the finitely many denominators relative to the $\mathbb Q$-basis $1,\alpha,\ldots,\alpha^{n-1}$ to obtain $0\ne c\in\mathbb Z$ with $cx_j\in\mathbb Z[\alpha]\subset R$. Thus $cI\subset R$.
Solved by gpt-5.6-sol high.
= c
{parent=20g}
{scope}
= i
{parent=c}
{scope}
= Solution
{parent=i}
Given a nonzero fractional $I$, scale it to an <integral> <ideal> $aI$. By (i), $(aI)J=(r)$ for some nonzero <ideal> $J$. Then $I^{-1}=ar^{-1}J$ is fractional and $II^{-1}=R$.
Solved by gpt-5.6-sol high.
= ii
{parent=c}
{scope}
= Solution
{parent=ii}
If every fractional <ideal> is invertible and $I\subset R$ is nonzero, choose $I^{-1}$ and clear denominators: $0\ne cI^{-1}\subset R$. Then $I(cI^{-1})=(c)$ is principal.
Solved by gpt-5.6-sol high.
= d
{parent=20g}
{scope}
= Solution
{parent=d}
Since $\beta$ is <integral>, $R[\beta]$ is generated over $R$ by finitely many powers, hence is fractional. It is invertible by (c). Because it is a <ring>, $I^2=I$; multiplying by $I^{-1}$ gives $I=R$. Thus every $\beta\in\mathcal O_K$ lies in $R$, so $R=\mathcal O_K$.
Solved by gpt-5.6-sol high.
= e
{parent=20g}
{scope}
= Solution
{parent=e}
In $R=\mathbb Z[\sqrt5]$, take $I=J_1=(2,1+\sqrt5)$ and $J_2=(2)$. Direct multiplication gives $I^2=(4,2+2\sqrt5,6+2\sqrt5)=2I=IJ_2$, but $I\ne(2)$.
Solved by gpt-5.6-sol high.
= 21H
{parent=Paper 2}
{scope}
{title2=Algebraic Topology}
= Solution
{parent=21H}
Homotopy equivalence means maps $f:X\to Y$, $g:Y\to X$ with $gf\simeq\mathrm{id}_X$ and $fg\simeq\mathrm{id}_Y$. Radial retraction of $\mathbb C^*$ and contraction of $\mathbb C$ give $\mathbb C^*\times\mathbb C\simeq S^1$. Coordinates $(a,b)\mapsto(a-b,b)$ identify $X\simeq\mathbb C^*\times\mathbb C$, so $\pi_1(X)=\mathbb Z$. The map to unordered roots is a two-sheeted covering. Its induced <subgroup> is $2\mathbb Z$ inside $\pi_1(Y)$, while exchanging roots is a generator; hence $\pi_1(Y,p_0)\simeq\mathbb Z$.
Solved by gpt-5.6-sol high.
= 22I
{parent=Paper 2}
{scope}
{title2=Linear Analysis}
= a
{parent=22i}
{scope}
= Solution
{parent=a}
The <Riesz representation theorem> says every bounded linear functional $f$ on a Hilbert space has a unique $z$ with $f(x)=\langle x,z\rangle$ and $\|f\|=\|z\|$. For $f\ne0$, decompose relative to $\ker f$; its one-dimensional orthogonal complement supplies $z$, and Cauchy-Schwarz proves the norm statement and uniqueness.
Solved by gpt-5.6-sol high.
= b
{parent=22i}
{scope}
= Solution
{parent=b}
For fixed $y$, Riesz gives a unique $Sy$ with $\theta(x,y)=\langle x,Sy\rangle$. Conjugate <linearity> in $y$ and the inner-product convention make $S$ linear. Moreover $\|Sy\|=\sup_{\|x\|\le1}|\theta(x,y)|\le\|\theta\|\|y\|$, and taking both suprema proves $\|S\|=\|\theta\|$.
Solved by gpt-5.6-sol high.
= c
{parent=22i}
{scope}
= Solution
{parent=c}
The adjoint is the unique $T^*$ satisfying $\langle Tx,y\rangle=\langle x,T^*y\rangle$. Apply Riesz to $x\mapsto\langle Tx,y\rangle$. The preceding norm calculation gives $\|T^*\|\le\|T\|$, while $\|Tx\|=\sup_{\|y\|=1}|\langle x,T^*y\rangle|$ gives the reverse inequality.
Solved by gpt-5.6-sol high.
= d
{parent=22i}
{scope}
= Solution
{parent=d}
Hermitian means $T=T^*$, unitary $T^*T=TT^*=I$, and normal $T^*T=TT^*$. If $\|(T-\lambda)x_n\|\to0$ for unit $x_n$, then $\langle Tx_n,x_n\rangle\to\lambda$ and each term is real, so $\lambda\in\mathbb R$. For self-adjoint $T$, a spectral value that is not approximate would make $T-\lambda$ bounded below; its closed range has orthogonal complement $\ker(T-\bar\lambda)=0$, hence is onto, contradicting spectrality.
Solved by gpt-5.6-sol high.
= 23H
{parent=Paper 2}
{scope}
{title2=Analysis of Functions}
= a
{parent=23h}
{scope}
= Solution
{parent=a}
$H^s(\mathbb R^d)=\{u\in\mathcal S':\int(1+|\xi|^2)^s|\widehat u(\xi)|^2d\xi<\infty\}$. For integer $s\ge0$, an equivalent norm is $\sum_{|\alpha|\le s}\|D^\alpha u\|_2^2$.
Solved by gpt-5.6-sol high.
= b
{parent=23h}
{scope}
= Solution
{parent=b}
If $s>d/2$, then $H^s(\mathbb R^d)\hookrightarrow L^\infty$. Indeed Fourier inversion and Cauchy-Schwarz give $|u(x)|\le C\|(1+|\xi|^2)^{s/2}\hat u\|_2\|(1+|\xi|^2)^{-s/2}\|_2$, and the second factor is finite exactly when $s>d/2$.
Solved by gpt-5.6-sol high.
= c
{parent=23h}
{scope}
= Solution
{parent=c}
$H_0^1(\Omega)$ is the closure of $C_c^\infty(\Omega)$ in $H^1(\Omega)$.
Solved by gpt-5.6-sol high.
= d
{parent=23h}
{scope}
= Solution
{parent=d}
For smooth compactly supported $u$, multiply by $1-\varphi(nx)$. The removed term has $L^2$ norm $O(n^{-3/2})$, while its <gradient> terms are $O(n^{-1/2})$ in dimension three. Thus it tends to zero in $H^1$. Density then proves $H_0^1(\mathbb R^3\setminus\{0\})=H^1(\mathbb R^3)$.
Solved by gpt-5.6-sol high.
= e
{parent=23h}
{scope}
= Solution
{parent=e}
No. Point evaluation is continuous on $H^1(\mathbb R)$, and every <limit> of <functions> supported away from zero has trace $u(0)=0$. A smooth <function> with nonzero value at zero belongs to $H^1$ but not to this $H_0^1$.
Solved by gpt-5.6-sol high.
= 24G
{parent=Paper 2}
{scope}
{title2=Riemann Surfaces}
= a
{parent=24g}
{scope}
= Solution
{parent=a}
For a nonconstant holomorphic map $f:X\to Y$ of degree $d$ between compact <Riemann surfaces>, <Riemann-Hurwitz formula> is $2g_X-2=d(2g_Y-2)+\sum_p(e_p-1)$. Triangulate $Y$ with branch values as vertices and lift the triangulation: faces and edges lift $d$ times, while the vertex deficit is exactly the ramification sum. Comparing Euler characteristics proves the formula.
Solved by gpt-5.6-sol high.
= b
{parent=24g}
{scope}
= Solution
{parent=b}
Compactify $y^d=g(z)$ by adding the points over infinity. Since $d\mid n$, there are $d$ such points and they are unramified. Each of the $n$ simple finite roots is totally ramified with index $d$. Riemann-Hurwitz for the degree-$d$ map to $\mathbb P^1$ gives
$$2g-2=-2d+n(d-1),\qquad g=\frac{(d-1)(n-2)}2.$$
Solved by gpt-5.6-sol high.
= 25F
{parent=Paper 2}
{scope}
{title2=Algebraic Geometry}
= Solution
{parent=25F}
For $V=Z(f_1,\ldots,f_r)$, $T_{V,p}$ is the common kernel of $df_i(p)$; $\dim V$ is the transcendence degree of its <function> field, and $p$ is smooth when $\dim T_{V,p}=\dim V$. A nonzero maximal Jacobian minor defines a nonempty open smooth locus (generic rank gives nonemptiness).
A line $ux+vy+wz=0$ corresponds to $[u:v:w]\in(\mathbb P^2)^*$. Lines through $P=[a:b:c]$ satisfy $au+bv+cw=0$, a projective line. For a <smooth curve> $F=0$, the dual morphism is $P\mapsto[F_x(P):F_y(P):F_z(P)]$.
For $F=x^3+y^3+z^3$, eliminating $x,y,z$ from $[u:v:w]=[x^2:y^2:z^2]$ gives
$$u^6+v^6+w^6-2u^3v^3-2u^3w^3-2v^3w^3=0.$$
The dual sextic is singular: its nine cusps are the tangent lines at the nine flexes of the cubic. Thus points of $X$ mapping to singularities are precisely inflection points, where the tangent has higher contact.
Solved by gpt-5.6-sol high.
= 26I
{parent=Paper 2}
{scope}
{title2=Differential Geometry}
= a
{parent=26i}
{scope}
= Solution
{parent=a}
For $v\in T_p\Sigma$, let $\gamma_v$ be the maximal geodesic with $\gamma_v(0)=p$, $\dot\gamma_v(0)=v$. Then $\exp_p(v)=\gamma_v(1)$ on the open star-shaped set of $v$ for which time $1$ exists.
Solved by gpt-5.6-sol high.
= b
{parent=26i}
{scope}
= Solution
{parent=b}
At the north pole, for $v=(v_1,v_2,0)$ and $r=|v|$, $\exp_p(v)=\cos r\,p+(\sin r/r)v$. Polar coordinates give $(\sin r\cos\theta,\sin r\sin\theta,\cos r)$. They are one-to-one for $0<r<\pi$, covering the sphere minus the two poles as a coordinate chart, with the north pole added as the polar origin; the south pole is the cut locus.
Solved by gpt-5.6-sol high.
= c
{parent=26i}
{scope}
= Solution
{parent=c}
A great-circle arc of angle $\arccos(p\cdot q)\le\pi$ joins the points. The spherical distance inequality follows by integrating the polar metric $ds^2=dr^2+\sin^2r,d\theta^2$, so every curve has length at least the radial change. If $q\ne-p$, the shorter great-circle arc is unique; equality forces constant angular coordinate and monotone radial motion, hence only a reparametrisation.
Solved by gpt-5.6-sol high.
= d
{parent=26i}
{scope}
= Solution
{parent=d}
A geodesic still joins most pairs, but if the unique minimizing great-circle arc passes through the deleted north pole, that arc is unavailable; the complementary great-circle arc remains a geodesic, so a geodesic does exist. A length minimizer need not: for points whose shorter arc passes through the missing pole, curves detouring arbitrarily close to it approach the spherical distance but never attain it.
Solved by gpt-5.6-sol high.
= 27L
{parent=Paper 2}
{scope}
{title2=Probability and Measure}
= a
{parent=27l}
{scope}
= Solution
{parent=a}
A measure is a map $\mu:\mathcal A\to[0,\infty]$ with $\mu(\varnothing)=0$ and countable additivity on disjoint families.
Solved by gpt-5.6-sol high.
= b
{parent=27l}
{scope}
= Solution
{parent=b}
The map $A\mapsto r^{-d}\lambda(rA)$ is a translation-invariant Borel measure assigning the usual volume to the unit cube. Uniqueness of Lebesgue measure makes it equal to $\lambda$, hence $\lambda(rA)=r^d\lambda(A)$.
Solved by gpt-5.6-sol high.
= c
{parent=27l}
{scope}
= Solution
{parent=c}
The disjoint sets $E_k$ contain at most one applicable summand at each $x$. Thus $F_n$ stabilizes to $F=2^k$ when $x\in E_k$ and to $0$ when $f(x)=0$ or $\infty$. By definition $F\le f\le2F$ (with the extended-value interpretation).
Solved by gpt-5.6-sol high.
= d
{parent=27l}
{scope}
= Solution
{parent=d}
Monotone convergence gives $\int F=\sum_k2^k\lambda(E_k)$. Since $F\le f\le2F$, $f$ is integrable exactly when this <series> is finite.
Solved by gpt-5.6-sol high.
= e
{parent=27l}
{scope}
= Solution
{parent=e}
The annulus where $2^k<|x|^{-b}\le2^{k+1}$ has volume comparable to $2^{-kd/b}$ for large positive $k$. The <series> is therefore comparable to $\sum2^{k(1-d/b)}$, which converges exactly when $b<d$.
Solved by gpt-5.6-sol high.
= 28K
{parent=Paper 2}
{scope}
{title2=Applied Probability}
= a
{parent=28k}
{scope}
= Solution
{parent=a}
A simple birth process has rates $q_{n,n+1}=n\lambda$. Its pgf solves $G_t=\lambda z(z-1)G_z$, giving $G(z,t)=e^{-\lambda t}z/[1-(1-e^{-\lambda t})z]$. Thus from one ancestor $X_t$ is geometric with parameter $e^{-\lambda t}$ and mean $e^{\lambda t}$. From $i$ ancestors it is a sum of $i$ such variables, hence negative binomial with mean $ie^{\lambda t}$.
Solved by gpt-5.6-sol high.
= b
{parent=28k}
{scope}
= Solution
{parent=b}
The jump chain rises from $i$ with probability $(i+1)/(i+2)$ and resets to $0$ otherwise. The probability of reaching level $i$ in an excursion is $2/(i+1)$, so reset occurs almost surely but the expected excursion length diverges: recurrence is null. Holding means grow like $i+1$, so there is no explosion and the expected return time is infinite. Hence neither the chain nor its jump chain is positive recurrent, and the continuous-time chain has no invariant probability distribution.
Solved by gpt-5.6-sol high.
= c
{parent=28k}
{scope}
= Solution
{parent=c}
For an irreducible nonexplosive countable-state chain, existence of a stationary probability $\pi P(t)=\pi$ implies positive recurrence, hence recurrence, and $\pi$ is the unique invariant distribution. Thus all three questions have affirmative answers (the last is also exactly the stated identity).
Solved by gpt-5.6-sol high.
= 29L
{parent=Paper 2}
{scope}
{title2=Principles of Statistics}
= a
{parent=29l}
{scope}
= Solution
{parent=a}
The MLE is $\hat\theta=X_{(n)}$. Since $Y=X_{(n)}/\theta\sim\operatorname{Beta}(n,1)$, its risk is $E(Y-1)^2=2/[(n+1)(n+2)]$.
Solved by gpt-5.6-sol high.
= b
{parent=29l}
{scope}
= Solution
{parent=b}
It is inadmissible. For $cX_{(n)}$, the risk is $c^2n/(n+2)-2cn/(n+1)+1$, minimized by $c=(n+2)/(n+1)$ with risk $1/(n+1)^2$, strictly smaller for every $\theta$.
Solved by gpt-5.6-sol high.
= c
{parent=29l}
{scope}
= i
{parent=c}
{scope}
= Solution
{parent=i}
The Bayes risk is $r(\pi,\delta)=\int R(\delta,\theta)\,\pi(d\theta)$.
Solved by gpt-5.6-sol high.
= ii
{parent=c}
{scope}
= Solution
{parent=ii}
A Bayes estimator minimizes Bayes risk, equivalently minimizes posterior expected loss almost surely.
Solved by gpt-5.6-sol high.
= d
{parent=29l}
{scope}
= Solution
{parent=d}
The posterior is Pareto with shape $\alpha+n$ and lower endpoint $m=\max\{\ell,X_{(n)}\}$. Minimizing $E[(\delta-\theta)^2/\theta^2\mid X]$ gives $\delta=E(\theta^{-1}\mid X)/E(\theta^{-2}\mid X)$, hence
$$\hat\theta_{\alpha,\ell}=m\frac{\alpha+n+2}{\alpha+n+1}.$$
Solved by gpt-5.6-sol high.
= 30L
{parent=Paper 2}
{scope}
{title2=Stochastic Financial Models}
= a
{parent=30l}
{scope}
= Solution
{parent=a}
An adapted integrable process is a supermartingale when $E[X_n\mid\mathcal F_{n-1}]\le X_{n-1}$ almost surely for every $n$.
Solved by gpt-5.6-sol high.
= b
{parent=30l}
{scope}
= Solution
{parent=b}
$E[X_{n+1}\mid\mathcal F_n]=X_n[pc+(1-p)c^{-1}]$. Thus the condition is $pc+(1-p)/c\le1$, or $pc^2-c+1-p\le0$. Since the roots are $(1-p)/p$ and $1$, the range is $(1-p)/p\le c\le1$.
Solved by gpt-5.6-sol high.
= c
{parent=30l}
{scope}
= Solution
{parent=c}
Conditional Jensen and monotonicity give $E[U(X_n)\mid\mathcal F_{n-1}]\le U(E[X_n\mid\mathcal F_{n-1}])\le U(X_{n-1})$.
Solved by gpt-5.6-sol high.
= d
{parent=30l}
{scope}
= Solution
{parent=d}
Predictability gives $E[H_k(X_k-X_{k-1})]=E[H_kE(X_k-X_{k-1}\mid\mathcal F_{k-1})]\le0$. Sum over $k\le n$.
Solved by gpt-5.6-sol high.
= e
{parent=30l}
{scope}
= Solution
{parent=e}
Fix $A\in\mathcal F_{n-1}$ and choose the nonnegative predictable strategy with $H_n=1_A$ and all other holdings zero. The assumed inequality yields $E[1_A(X_n-X_{n-1})]\le0$ for every $A$, equivalent to the supermartingale conditional inequality.
Solved by gpt-5.6-sol high.
= 31K
{parent=Paper 2}
{scope}
{title2=Mathematics of Machine Learning}
= a
{parent=31k}
{scope}
= Solution
{parent=a}
$\widehat R_\phi(h)=n^{-1}\sum_{i=1}^n\phi(Y_ih(X_i))$.
Solved by gpt-5.6-sol high.
= b
{parent=31k}
{scope}
= Solution
{parent=b}
$\mathcal R_n(\mathcal H)=E\sup_{h\in\mathcal H}|n^{-1}\sum_i\varepsilon_i h(X_i)|$, with the expectation over data and independent Rademacher signs.
Solved by gpt-5.6-sol high.
= c
{parent=31k}
{scope}
= Solution
{parent=c}
Because $\widehat R(\hat h)\le\widehat R(h^*)$, adding and subtracting empirical risks and taking expectations leaves at most $E\sup_h(R(h)-\widehat R(h))$. Symmetrization and the unit-Lipschitz contraction lemma for hinge loss bound this by $2\mathcal R_n(\mathcal H)$.
Solved by gpt-5.6-sol high.
= d
{parent=31k}
{scope}
= Solution
{parent=d}
Each $Z_j=n^{-1}\sum_i\varepsilon_iX_{ij}$ is $1/\sqrt n$-sub-Gaussian. Applying the exponential-moment bound to the $2p$ variables $\pm Z_j$ and optimizing the parameter gives $E\max_j|Z_j|\le\sqrt{2\log(2p)/n}$.
Solved by gpt-5.6-sol high.
= e
{parent=31k}
{scope}
= Solution
{parent=e}
For fixed hidden units, the supremum over $\|\alpha\|_1\le\lambda$ is $\lambda$ times the largest absolute unit correlation. Taking the further supremum over admissible $\beta_k$ gives exactly the displayed bound.
Solved by gpt-5.6-sol high.
= f
{parent=31k}
{scope}
= Solution
{parent=f}
The Rademacher contraction lemma for ReLU (which is 1-Lipschitz and vanishes at zero), duality between $\ell_1$ and $\ell_\infty$, and part (d) give $\mathcal R_n(\mathcal H)\le2\lambda\sqrt{2\log(2p)/n}$. Combining with part (c) gives the claimed $4\lambda\sqrt{2\log(2p)}/\sqrt n$.
Solved by gpt-5.6-sol high.
= 32D
{parent=Paper 2}
{scope}
{title2=Asymptotic Methods}
= a
{parent=32d}
{scope}
= i
{parent=a}
{scope}
= Solution
{parent=i}
Repeated integration by parts gives
$$E(x)\sim e^{-x}x^{-1}\sum_{n\ge0}(-1)^n n!x^{-n},$$
so $a_n=(-1)^nn!$.
Solved by gpt-5.6-sol high.
= ii
{parent=a}
{scope}
= Solution
{parent=ii}
Writing $t=x+u$ gives $E(x)=e^{-x}x^{-1}\int_0^\infty e^{-u}(1+u/x)^{-1}du$. Expanding the last factor and applying <Watson lemma> yields the same coefficients $(-1)^n\Gamma(n+1)$.
Solved by gpt-5.6-sol high.
= iii
{parent=a}
{scope}
= Solution
{parent=iii}
Successive term magnitudes have ratio $(n+1)/x$, so truncate near the least term, $n\simeq x-1$. Stirling gives least relative term $\sqrt{2\pi x}e^{-x}$, hence an exponentially small relative remainder of that order.
Solved by gpt-5.6-sol high.
= b
{parent=32d}
{scope}
= Solution
{parent=b}
Substitute $u=nt$ and use the $\pi$-periodicity of $\cos(n\cos u)$ to get $J_n=\int_0^\pi\cos(n\cos u)du=\pi J_0(n)$. Stationary phase at $0,\pi$ gives
$$J_n\sim\sqrt{\frac{2\pi}{n}}\cos(n-\pi/4).$$
Thus $A=\sqrt{2\pi/n}$, $\alpha=n-\pi/4$, and the next term is $O(n^{-3/2})$.
Solved by gpt-5.6-sol high.
= 33B
{parent=Paper 2}
{scope}
{title2=Dynamical Systems}
= a
{parent=33b}
{scope}
= Solution
{parent=a}
For $H=x^2+y^2$, $\dot H=2\mu x^2(1-ax^2-y^2)$. It points outward on a sufficiently small circle $r<1/\sqrt a$ and inward on a circle $r>1$. The intervening annulus is compact, forward invariant, and contains no equilibrium. <Poincare-Bendixson theorem> therefore gives a periodic orbit.
Solved by gpt-5.6-sol high.
= b
{parent=33b}
{scope}
= Solution
{parent=b}
The nontrivial <floquet multiplier> is the factor by which a transverse perturbation changes after one period. Liouville's formula makes its logarithm the <integral> of the divergence. Here $\nabla\cdot f=\mu(1-3ax^2-y^2)$, yielding the stated formula.
Solved by gpt-5.6-sol high.
= c
{parent=33b}
{scope}
= Solution
{parent=c}
For small $\mu$, average $\dot H$ around $x=r\cos2t$, $y=-r\sin2t$. This gives $\langle\dot H\rangle=\mu[r^2-(3a+1)r^4/4]$. The unique positive balance is $r=2/\sqrt{3a+1}$, which lies between $1/\sqrt a$ and $1$ for $a>1$.
Solved by gpt-5.6-sol high.
= d
{parent=33b}
{scope}
= Solution
{parent=d}
On that leading orbit, averaging the divergence gives $\sigma=\mu[1-(3a+1)r^2/2]+O(\mu^2)=-\mu+O(\mu^2)<0$, so the cycle is stable.
Solved by gpt-5.6-sol high.
= e
{parent=33b}
{scope}
= Solution
{parent=e}
The averaged energy drift is $F(r)=\mu[r^2-(3a+1)r^4/4]$. Linearizing radial energy balance at its nonzero zero gives decay rate $-\mu$ after converting $H=r^2$, exactly the Floquet exponent found from divergence.
Solved by gpt-5.6-sol high.
= 34B
{parent=Paper 2}
{scope}
{title2=Integrable Systems}
= Solution
{parent=34B}
If $Lf=\lambda f$ and $f_t+Af=0$, <differentiation> and the <isospectral Lax equation>[Lax equation] give $(L-\lambda)f_t=-A(L-\lambda)f$, so $\lambda_t=0$ after projection onto the normalized <eigenfunction> for a <simple eigenvalue>[nondegenerate eigenvalue].
For $\Psi=(f,f_x)^T$,
$$M_L=\begin{pmatrix}0&1\\u-\lambda&0\end{pmatrix},$$
$$M_A=\begin{pmatrix}-u_x-b&-(u-\lambda+a)\\-u_{xx}-b_x-(u-\lambda+a)(u-\lambda)&-2u_x-b-a_x\end{pmatrix}.$$
The compatibility $\Psi_{xt}=\Psi_{tx}$ is precisely $\partial_tM_L-\partial_xM_A-[M_A,M_L]=0$. If all coefficients are $x$-independent, this becomes $\dot M_L=[M_A,M_L]$, and cyclicity of trace gives $d\operatorname{Tr}(M_L^k)/dt=0$.
Solved by gpt-5.6-sol high.
= 35A
{parent=Paper 2}
{scope}
{title2=Principles of Quantum Mechanics}
= a
{parent=35a}
{scope}
= Solution
{parent=a}
Orthonormality gives $\langle\mathcal B_1|\mathcal B_i\rangle=\delta_{1i}$.
Solved by gpt-5.6-sol high.
= b
{parent=35a}
{scope}
= Solution
{parent=b}
Tracing out $B$ gives $\rho_A=I/2$. Its <Von Neumann entropy> is $-\operatorname{tr}\rho_A\log_2\rho_A=1$, and $\operatorname{tr}\rho_A^2=1/2$, so it is mixed.
Solved by gpt-5.6-sol high.
= c
{parent=35a}
{scope}
= Solution
{parent=c}
No. A pure product state has a pure reduced density <matrix>, whereas part (b) gives $I/2$.
Solved by gpt-5.6-sol high.
= d
{parent=35a}
{scope}
= Solution
{parent=d}
The values are $+1,+1,-1,-1$ for $\mathcal B_1,\ldots,\mathcal B_4$. Thus $z$-spin outcomes are perfectly correlated for the first pair and perfectly anticorrelated for the second.
Solved by gpt-5.6-sol high.
= e
{parent=35a}
{scope}
= Solution
{parent=e}
Using $T=2\,\mathrm{SWAP}-I$, the symmetric triplet <Bell state> have <eigenvalue> $1$, while the antisymmetric singlet has <eigenvalue> $-3$. With the ordering shown, $\mathcal B_1,\mathcal B_2,\mathcal B_3$ are triplet combinations and $\mathcal B_4$ is the singlet.
Solved by gpt-5.6-sol high.
= f
{parent=35a}
{scope}
= Solution
{parent=f}
Applying $I\otimes I,X,Z,$ or $XZ$ to one qubit permutes the Bell <basis> up to phases. Composing two such Pauli operations maps any chosen Bell state to any other, proving local equivalence.
Solved by gpt-5.6-sol high.
= 36E
{parent=Paper 2}
{scope}
{title2=Applications of Quantum Mechanics}
= a
{parent=36e}
{scope}
= Solution
{parent=a}
For any normalized trial state $\psi$, the spectral expansion gives $\langle\psi|H|\psi\rangle\ge E_0$. Minimize this Rayleigh quotient over a trial family to obtain the best upper bound.
Solved by gpt-5.6-sol high.
= b
{parent=36e}
{scope}
= Solution
{parent=b}
Write the normalized trial state as $\sqrt{1-\epsilon^2}|0\rangle+\epsilon|\eta\rangle$ with $\eta\perp0$. The cross term vanishes because $H|0\rangle=E_0|0\rangle$, so the energy error is $\epsilon^2(\langle\eta|H|\eta\rangle-E_0)=O(\epsilon^2)$.
Solved by gpt-5.6-sol high.
= c
{parent=36e}
{scope}
= Solution
{parent=c}
The unperturbed ground state is even about $a/2$, while $V_{\mathcal E}=q\mathcal E(x-a/2)$ is odd. Its first-order expectation therefore vanishes.
Solved by gpt-5.6-sol high.
= d
{parent=36e}
{scope}
= Solution
{parent=d}
The correction is odd, obeys the well <boundary conditions>, and is linear in the field, exactly the parity and order of the first-order <wavefunction> correction.
Solved by gpt-5.6-sol high.
= e
{parent=36e}
{scope}
= Solution
{parent=e}
Put $y=x-a/2$ and $\epsilon=q\mathcal E$. To quadratic order the Rayleigh quotient contains $\epsilon^2(A\alpha^2+2B\alpha)$, where $A=\hbar^2/(2m)$ and $B=\langle y^2\rangle=a^2(\pi^2-6)/(12\pi^2)$. Hence
$$\alpha_{\rm opt}=-\frac BA=-\frac{ma^2(\pi^2-6)}{6\pi^2\hbar^2}.$$
Solved by gpt-5.6-sol high.
= 37E
{parent=Paper 2}
{scope}
{title2=Statistical Physics}
= a
{parent=37e}
{scope}
= Solution
{parent=a}
The first law and Gibbs-Duhem relation give $d\Gamma=-S,dT+V,dp-N,d\mu$. Extensivity gives Euler's relation $E=TS-pV+\mu N$, hence $\Gamma=E-TS+pV-\mu N=0$.
Solved by gpt-5.6-sol high.
= b
{parent=37e}
{scope}
= i
{parent=b}
{scope}
= Solution
{parent=i}
Two polarizations and periodic wavevectors give $g(\omega)=A\omega/(\pi c_s^2)$.
Solved by gpt-5.6-sol high.
= ii
{parent=b}
{scope}
= Solution
{parent=ii}
Requiring $\int_0^{\omega_D}g(\omega)d\omega=2N$ gives $\omega_D=2c_s\sqrt{\pi N/A}$.
Solved by gpt-5.6-sol high.
= iii
{parent=b}
{scope}
= Solution
{parent=iii}
For $k_BT\ll\hbar\omega_D$, extend the <integral> to infinity:
$$\log Z=-\int_0^\infty g(\omega)\log(1-e^{-\beta\hbar\omega})d\omega
=\frac{A k_B^2T^2}{\pi c_s^2\hbar^2}\sum_{n\ge1}\frac1{n^3}.$$
Thus $X=\zeta(3)/\pi$.
Solved by gpt-5.6-sol high.
= iv
{parent=b}
{scope}
= Solution
{parent=iv}
Since $F=-k_BT\log Z$, $S=-\partial F/\partial T=3\zeta(3)A k_B^3T^2/(\pi c_s^2\hbar^2)$.
Solved by gpt-5.6-sol high.
= 38A
{parent=Paper 2}
{scope}
{title2=General Relativity}
= a
{parent=38a}
{scope}
= Solution
{parent=a}
$G_{\mu\nu}=R_{\mu\nu}-\tfrac12Rg_{\mu\nu}$. The contracted Bianchi identity $\nabla^\mu G_{\mu\nu}=0$ is required to match $\nabla^\mu T_{\mu\nu}=0$.
Solved by gpt-5.6-sol high.
= b
{parent=38a}
{scope}
= Solution
{parent=b}
Computing $R=6(\ddot a/a+(\dot a^2+k)/a^2)$ gives
$$\dot a^2=\frac{8\pi\rho}{3}a^2-k,\qquad \ddot a=-\frac{4\pi}{3}(\rho+3p)a.$$
Thus $\alpha=8\pi\rho/3$, $\beta=-k$, and $\gamma=-4\pi(\rho+3p)/3$ in the stated notation.
Solved by gpt-5.6-sol high.
= c
{parent=38a}
{scope}
= Solution
{parent=c}
Staticity requires $\rho_0+3p_0=0$, so $\rho_m=\Lambda/(4\pi)$. The first Friedmann equation then requires $k=1$ and $a_0=\Lambda^{-1/2}$.
Solved by gpt-5.6-sol high.
= d
{parent=38a}
{scope}
= Solution
{parent=d}
Linearizing the constraint gives $\delta\rho=-2\rho_0\delta a/a_0$. Since $\delta p=\delta\rho/3$, the <acceleration> equation gives $\delta\ddot a=2\Lambda\delta a$. Hence perturbations contain $e^{\sqrt{2\Lambda}t}$: the Einstein static universe is unstable.
Solved by gpt-5.6-sol high.
= 39C
{parent=Paper 2}
{scope}
{title2=Fluid Dynamics II}
= a
{parent=39c}
{scope}
= Solution
{parent=a}
<Stokes flow> obeys $-\nabla p+\mu\nabla^2u=0$, $\nabla\cdot u=0$. Reversing all applied forces reverses $u$ while leaving <streamlines> unchanged. Reflection in the vertical plane containing the cylinder axis rules out rotation normal to that plane; reflection combined with reversibility rules out rotation within it and spin about the axis. Thus its orientation remains fixed.
Solved by gpt-5.6-sol high.
= b
{parent=39c}
{scope}
= Solution
{parent=b}
Lubrication gives $u_r=(p_r/2\mu)z(z-h)$ and continuity gives $\int_0^h u_rdz=-Vr/2$. Hence $p_r=6\mu Vr/h^3$ and $p-p_0=3\mu V(r^2-a^2)/h^3$. Balancing the upward <pressure> force with $W$ yields
$$\dot h=-\frac{2Wh^3}{3\pi\mu a^4}.$$
Therefore $h^{-2}=h_0^{-2}+4Wt/(3\pi\mu a^4)$ and $h\sim(3\pi\mu a^4/4Wt)^{1/2}$.
Solved by gpt-5.6-sol high.
= 40C
{parent=Paper 2}
{scope}
{title2=Waves}
= a
{parent=40c}
{scope}
= Solution
{parent=a}
Substitution gives the displayed relation with $c_P^2=(\lambda+2\mu)/\rho$ and $c_S^2=\mu/\rho$. Resolving $A$ parallel and perpendicular to $k$ gives a longitudinal P mode of speed $c_P$ and two transverse polarizations of speed $c_S$, conventionally SV in and SH normal to the propagation plane.
Solved by gpt-5.6-sol high.
= b
{parent=40c}
{scope}
= Solution
{parent=b}
In the lower solid $l^2+m^2=\omega^2/c_S^2$. Put $m'^2=\omega^2/c_S'^2-l^2$ and $Z=\mu'm'\cot(m'h)$. Matching displacement and shear stress at $y=0$, with zero displacement at $h$, gives
$$R=-\frac{Z+i\mu m}{Z-i\mu m}.$$
For real propagating parameters $|R|=1$: the rigidly backed lossless layer transmits no mean energy, so all incident energy is reflected (though with a phase shift).
Solved by gpt-5.6-sol high.
= 41B
{parent=Paper 2}
{scope}
{title2=Numerical Analysis}
= a
{parent=41b}
{scope}
= Solution
{parent=a}
With $x=\cos\theta$, $T_k(x)=\cos k\theta$, proving orthogonality. In the convention $f\sim c_0+\sum_{k\ge1}c_kT_k$, $c_0=\pi^{-1}\int_0^\pi f(\cos\theta)d\theta$ and $c_k=2\pi^{-1}\int_0^\pi f(\cos\theta)\cos k\theta d\theta$. For $f=\sqrt{1-x^2}$, $c_0=2/\pi$, odd coefficients vanish, and $c_{2m}=-4/[\pi(4m^2-1)]$.
Solved by gpt-5.6-sol high.
= b
{parent=41b}
{scope}
= Solution
{parent=b}
The substitution $x=\cos\theta$ identifies the weighted norm with the $L^2(0,\pi)$ norm of $f(\cos\theta)$. Completeness of cosine Fourier <series> therefore makes the Chebyshev partial sums converge in $\|\cdot\|_w$.
Solved by gpt-5.6-sol high.
= c
{parent=41b}
{scope}
= Solution
{parent=c}
Parseval's identity for the cosine <series>, with $\|T_0\|_w^2=\pi$ and $\|T_k\|_w^2=\pi/2$ for $k\ge1$, gives the claimed equality.
Solved by gpt-5.6-sol high.
= d
{parent=41b}
{scope}
= Solution
{parent=d}
Repeated integration by parts in the cosine-coefficient <integral> shows $|c_k|\le C_rk^{-r}$ for every $r$ when $f$ is smooth (endpoint terms vanish after using the even periodic extension). Parseval then bounds the squared tail by $C_r\sum_{k>N}k^{-2r}$, which is $O(N^{-2r+1})$. Choosing $r$ arbitrarily large proves spectral convergence.
Solved by gpt-5.6-sol high.