past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ii/paper-3.bigb

= Paper 3
{scope}

https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2022/paperii_3_2022.pdf

= 1I
{parent=Paper 3}
{scope}
{title2=Number Theory}

= Solution
{parent=1I}

The <Lagrange theorem for polynomial congruences> says that if $p$ is prime and $f\in\mathbb Z[X]$ has degree $d$ with at least one coefficient not divisible by $p$, then
$$
f(x)\equiv0\pmod p
$$
has at most $d$ incongruent solutions modulo $p$.

The <Chinese remainder theorem> states that for pairwise coprime $m_1,\ldots,m_r$, every system
$$
x\equiv a_i\pmod{m_i}
\qquad(1\leq i\leq r)
$$
has exactly one solution modulo $M=\prod_i m_i$. For two moduli, choose $u,v$ with $um+vn=1$ by <Bezout identity>. Then
$$
x=a\,vn+b\,um
$$
is congruent to $a$ modulo $m$ and to $b$ modulo $n$. If $x,y$ are two solutions, both $m$ and $n$ divide $x-y$; coprimality makes $mn$ divide $x-y$. This proves existence and uniqueness for two factors, and induction proves the general statement.

Now
$$
1729=7\cdot13\cdot19
$$
and $12^3+1=1729$, so $x=12$ is a solution. For $1\leq x<12$, one has $0<x^3+1<1729$, so no such positive integer can satisfy the congruence. Hence the smallest is
$$
\boxed{x=12}.
$$
Modulo $7,13,19$, the roots are respectively
$$
\{3,5,6\},\qquad
\{4,10,12\},\qquad
\{8,12,18\}.
$$
Each list has three elements, and the Chinese remainder theorem combines the choices independently. Therefore the number of solutions with $1\leq x\leq1729$ is
$$
\boxed{3^3=27}.
$$

Solved by gpt-5.6-sol high.

= 2G
{parent=Paper 3}
{scope}
{title2=Topics in Analysis}

= i
{parent=2g}
{scope}

= Solution
{parent=i}

Because $\overline\Omega$ is compact, $\phi$ attains a maximum. If a maximizer lay in $\Omega$, its <Hessian matrix> would be negative semidefinite, so its trace would satisfy $\Delta\phi\leq0$, contradicting $\Delta\phi>0$. Hence every strictly <subharmonic function> under the stated hypotheses attains a maximum on $\partial\Omega$.

Statement (i) is true. For $\epsilon>0$, the function
$$
\phi_\epsilon(x)=\phi(x)+\epsilon|x|^2
$$
has $\Delta\phi_\epsilon=4\epsilon>0$, so
$$
\max_{\overline\Omega}\phi_\epsilon
=\max_{\partial\Omega}\phi_\epsilon.
$$
Letting $\epsilon\downarrow0$ gives
$$
\boxed{\max_{\overline\Omega}\phi
=\max_{\partial\Omega}\phi}.
$$
This is the <maximum principle for harmonic functions>.

Solved by gpt-5.6-sol high.

= ii
{parent=2g}
{scope}

= Solution
{parent=ii}

Statement (ii) is true. Since $\Delta(-\phi)=0$, applying the maximum principle to $-\phi$ gives
$$
\boxed{\min_{\overline\Omega}\phi
=\min_{\partial\Omega}\phi}.
$$

Solved by gpt-5.6-sol high.

= iii
{parent=2g}
{scope}

= Solution
{parent=iii}

Statement (iii) is false. On the unit disc, take
$$
\phi(x,y)=1-x^2-y^2,
\qquad
f=\Delta\phi=-4.
$$
The function $f$ is continuous on the closed disc, but $\phi$ has its strict maximum $1$ at the interior point $(0,0)$ and equals zero on the boundary.

Solved by gpt-5.6-sol high.

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

= a
{parent=3k}
{scope}

= Solution
{parent=a}

For binary length-$n$ <linear codes> $C_2\subseteq C_1$, their <bar product of binary linear codes> is
$$
\boxed{
C_1|C_2
=\{(u,u+v):u\in C_1,\ v\in C_2\}
}.
$$
It is a binary linear code of length $2n$ and dimension $\dim C_1+\dim C_2$.

Solved by gpt-5.6-sol high.

= b
{parent=3k}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

The code $\operatorname{RM}(d,0)$ is the binary repetition code of length $2^d$, while
$$
\operatorname{RM}(d,d)=\mathbb F_2^{\,2^d}
$$
is the full binary code. For $0<r<d$, the <Reed-Muller bar-product recursion> is
$$
\operatorname{RM}(d,r)
=\operatorname{RM}(d-1,r)
\mid\operatorname{RM}(d-1,r-1).
$$
Thus its rank $k(d,r)$ satisfies Pascal's recursion
$$
k(d,r)=k(d-1,r)+k(d-1,r-1),
$$
with the stated boundary values. Therefore
$$
\boxed{k(d,r)=\sum_{j=0}^r\binom dj}.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

A square-free monomial in $d$ variables of degree $j\leq d-1$ evaluates to one at exactly $2^{d-j}$ points of $\mathbb F_2^d$, an even number. Since parity of <Hamming weight> is a linear functional over $\mathbb F_2$, every sum of such evaluation words also has even weight. Hence every word in $\operatorname{RM}(d,d-1)$ has even weight.

If $f$ and $g$ have degrees at most $r$ and $d-r-1$, then $fg$ has degree at most $d-1$. The binary inner product of their evaluation words is the parity of the weight of the evaluation of $fg$, and is therefore zero. Thus
$$
\operatorname{RM}(d,d-r-1)
\subseteq\operatorname{RM}(d,r)^\perp.
$$
Finally, binomial symmetry gives
$$
\dim\operatorname{RM}(d,r)
+\dim\operatorname{RM}(d,d-r-1)=2^d,
$$
so the inclusion has equal dimensions and
$$
\boxed{
\operatorname{RM}(d,r)^\perp
=\operatorname{RM}(d,d-r-1)
}.
$$
This is the <dual of a Reed-Muller code>.

Solved by gpt-5.6-sol high.

= 4I
{parent=Paper 3}
{scope}
{title2=Automata and Formal Languages}

= i
{parent=4i}
{scope}

= Solution
{parent=i}

A <context-free grammar> is a quadruple $(V,\Sigma,P,S)$ consisting of finite nonterminal and terminal alphabets, a set of productions $A\to\alpha$ with one nonterminal $A$ on the left, and a start symbol $S$. A <context-free language> is a language generated by such a grammar.

The <pumping lemma for context-free languages> says that there is a pumping length $p$ such that every $w$ in the language with $|w|\geq p$ can be written
$$
w=uvxyz
$$
with $|vxy|\leq p$, $|vy|>0$, and $uv^ixy^iz$ in the language for every integer $i\geq0$.

The language in (i) is not context-free. If it were, pump
$$
w=a^p b^{2p}c^p.
$$
The substring $vxy$, having length at most $p$, cannot meet both the $a$- and $c$-blocks. If pumping changes the number of $a$'s or $c$'s, those two counts cease to agree. If it changes neither, it changes the number of $b$'s while the outer counts remain $p$, so the middle count ceases to be $2p$. Every permitted decomposition therefore fails, contradicting the pumping lemma.

Solved by gpt-5.6-sol high.

= ii
{parent=4i}
{scope}

= Solution
{parent=ii}

The language in (ii) is context-free. The grammar
$$
S\to aSc\mid B,
\qquad
B\to bbB\mid\epsilon
$$
generates exactly
$$
\{a^n b^{2i}c^n:n,i\geq0\}.
$$
The recursive uses of the first rule match the numbers of $a$'s and $c$'s, while $B$ independently generates any even number of $b$'s.

Solved by gpt-5.6-sol high.

= 5J
{parent=Paper 3}
{scope}
{title2=Statistical Modelling}

= Solution
{parent=5J}

The log density is
$$
-\log(2\sigma)-\frac{|y-\mu|}{\sigma}.
$$
Its kink occurs at the unknown location $\mu$, so its parameter dependence cannot be separated into a fixed finite-dimensional statistic of $y$ and the smooth canonical form required of an exponential dispersion family. Thus the two-parameter location-scale <Laplace distribution> is not such a family.

For independent observations, the <log-likelihood> is
$$
\ell(\beta,\sigma)
=-n\log(2\sigma)-\frac1\sigma
\sum_{i=1}^n|Y_i-X_i^T\beta|
=-n\log(2\sigma)-\frac{S(\beta)}{\sigma}.
$$
For every fixed $\sigma>0$, maximizing this expression in $\beta$ is equivalent to minimizing $S(\beta)$. Hence
$$
\boxed{
\widehat\beta\in\operatorname*{arg\,min}_\beta
\sum_{i=1}^n|Y_i-X_i^T\beta|
},
$$
the <least absolute deviations> estimator. At this value,
$$
\frac{\partial\ell}{\partial\sigma}
=-\frac n\sigma+\frac{S(\widehat\beta)}{\sigma^2},
$$
so, in the nondegenerate case $S(\widehat\beta)>0$,
$$
\boxed{\widehat\sigma=\frac{S(\widehat\beta)}n}.
$$

Solved by gpt-5.6-sol high.

= 6C
{parent=Paper 3}
{scope}
{title2=Mathematical Biology}

= a
{parent=6c}
{scope}

= Solution
{parent=a}

Probability enters state $n$ through a birth from $n-1$ or a pair-death event from $n+2$, and leaves through either event at state $n$. Thus the <Kramers-Moyal master equation> is
$$
\boxed{
\frac{\partial P(n,t)}{\partial t}
=\lambda P(n-1,t)+\beta(n+2)^2P(n+2,t)
-(\lambda+\beta n^2)P(n,t)
},
$$
with probabilities and impossible transition rates taken as zero outside the nonnegative states.

A birth changes $n$ by $+1$, while a death event changes it by $-2$. Applying these increments to the first moment gives
$$
\boxed{
\frac d{dt}\langle n\rangle
=\lambda-2\beta\langle n^2\rangle
}.
$$
Consequently every steady state must satisfy
$$
\boxed{\langle n^2\rangle=\frac{\lambda}{2\beta}}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=6c}
{scope}

= Solution
{parent=b}

Taylor-expand the gain terms:
$$
\lambda P(n-1)
=\lambda\left(P-P_n+\frac12P_{nn}+\cdots\right),
$$
and, writing $Q(n)=\beta n^2P(n)$,
$$
Q(n+2)=Q+2Q_n+2Q_{nn}+\cdots.
$$
After cancellation of the loss terms, the second-order <Kramers-Moyal expansion> is
$$
\frac{\partial P}{\partial t}
=-\frac{\partial}{\partial n}
\left[(\lambda-2\beta n^2)P\right]
+\frac12\frac{\partial^2}{\partial n^2}
\left[(\lambda+4\beta n^2)P\right].
$$
Hence the <constant-birth pair-annihilation process> has
$$
\boxed{A(n)=\lambda-2\beta n^2},
\qquad
\boxed{B(n)=\lambda+4\beta n^2}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=6c}
{scope}

= Solution
{parent=c}

The deterministic drift vanishes at the positive stable point
$$
n_*=\sqrt{\frac{\lambda}{2\beta}},
\qquad
A'(n_*)=-4\beta n_*<0.
$$
When $n_*$ is large, fluctuations sample only a narrow range around it, so the <linear noise approximation> replaces
$$
A(n)\simeq A'(n_*)(n-n_*),
\qquad
B(n)\simeq B(n_*)=3\lambda.
$$
The resulting Ornstein-Uhlenbeck stationary density is the stated normal distribution with
$$
\boxed{\mu=n_*=\sqrt{\frac{\lambda}{2\beta}}},
$$
and
$$
\boxed{
\sigma^2=-\frac{B(n_*)}{2A'(n_*)}
=\frac{3\lambda}{8\beta n_*}
=\frac34\sqrt{\frac{\lambda}{2\beta}}
=\frac{3\mu}{4}
}.
$$

Solved by gpt-5.6-sol high.

= d
{parent=6c}
{scope}

= Solution
{parent=d}

For the Gaussian approximation,
$$
\langle n^2\rangle\simeq\mu^2+\sigma^2
=\frac{\lambda}{2\beta}+\frac{3\mu}{4}.
$$
The extra term reflects the subleading error made by centring the linear approximation at the deterministic fixed point. It is smaller than $\mu^2$ by relative order $1/\mu$, so the exact steady-state condition from part (a) is recovered at leading order. A higher-order expansion shifts the mean and restores the moment balance more accurately.

The approximation is reliable when
$$
\mu=\sqrt{\frac{\lambda}{2\beta}}\gg1,
\qquad
\frac{\sigma}{\mu}=O(\mu^{-1/2})\ll1.
$$
Thus one needs $\lambda/\beta\gg1$: births must support a large typical population, with jumps of one or two individuals small compared with the population scale.

Solved by gpt-5.6-sol high.

= 7E
{parent=Paper 3}
{scope}
{title2=Further Complex Methods}

= Solution
{parent=7E}

Taking the <Laplace transform> in $t$ and using the zero initial condition gives
$$
p\widetilde T=\kappa\widetilde T_{xx}.
$$
The solution bounded as $x\to\infty$ is
$$
\widetilde T(x,p)=\widetilde T_0(p)
\exp\left(-x\sqrt{\frac p\kappa}\right).
$$
The transformed boundary condition is
$$
\boxed{\widetilde T_0(p)=\frac{\omega}{p^2+\omega^2}},
$$
so this is the <sinusoidally forced heat equation on a half-line>. Integrating in $x$ gives
$$
\boxed{
\widetilde I(p)
=\int_0^\infty\widetilde T(x,p)\,dx
=\frac{\omega\sqrt\kappa}
{\sqrt p\,(p^2+\omega^2)}
}.
$$

Use the principal square root, with a <branch cut> along the negative real axis. The <Bromwich contour> is a vertical line to the right of $p=\pm i\omega$ and the branch point $p=0$. For $t>0$, close it in the left half-plane, indent around the cut, and let the large semicircle recede. The simple poles at $p=\pm i\omega$ contribute
$$
\sqrt{\frac{\kappa}{\omega}}
\cos\left(\omega t-\frac{3\pi}{4}\right)
=-\sqrt{\frac{\kappa}{2\omega}}\cos\omega t
+\sqrt{\frac{\kappa}{2\omega}}\sin\omega t.
$$
Hence
$$
\boxed{A=-\sqrt{\frac{\kappa}{2\omega}}},
\qquad
\boxed{B=\sqrt{\frac{\kappa}{2\omega}}}.
$$

On the upper and lower sides of the cut, put $p=-s$; the two values of $\sqrt p$ are $i\sqrt s$ and $-i\sqrt s$. Their jump and the opposite contour orientations combine to give
$$
\frac1\pi\int_0^\infty
\frac{\omega\sqrt\kappa\,e^{-st}}
{(s^2+\omega^2)\sqrt s}\,ds.
$$
This is precisely <Bromwich inversion with a square-root branch cut>: the residues produce the permanent periodic response, while the cut integral is the transient that decays with time.

Solved by gpt-5.6-sol high.

= 8B
{parent=Paper 3}
{scope}
{title2=Classical Dynamics}

= Solution
{parent=8B}

The <generalized momentum>[generalized momenta] conjugate to the two <ignorable coordinates> are
$$
\boxed{
p_\phi
=I_1\dot\phi\sin^2\theta
+I_3(\dot\psi+\dot\phi\cos\theta)\cos\theta
}
$$
and
$$
\boxed{
p_\psi=I_3(\dot\psi+\dot\phi\cos\theta)
}.
$$
Since the Lagrangian contains neither $\phi$ nor $\psi$, both momenta are constant in time.

Eliminating the two angular velocities gives
$$
\dot\phi=\frac{p_\phi-p_\psi\cos\theta}
{I_1\sin^2\theta},
\qquad
\dot\psi+\dot\phi\cos\theta=\frac{p_\psi}{I_3}.
$$
Conservation of energy then yields the <heavy symmetric top reduction>
$$
\frac12I_1\dot\theta^2+V_{\rm eff}(\theta)=E,
$$
where
$$
\boxed{
V_{\rm eff}(\theta)
=\frac{(p_\phi-p_\psi\cos\theta)^2}
{2I_1\sin^2\theta}
+\frac{p_\psi^2}{2I_3}
+Mgl\cos\theta
}.
$$

As $\theta\to0$, the first term diverges unless
$$
\boxed{p_\phi=p_\psi}.
$$
Writing their common value as $p$, the small-angle expansion is
$$
V_{\rm eff}(\theta)
=\frac{p^2}{2I_3}+Mgl
+\left(\frac{p^2}{8I_1}-\frac{Mgl}{2}\right)\theta^2
+O(\theta^4).
$$
By the <effective potential stability criterion>, the vertical state is stable exactly when
$$
\boxed{p^2>4I_1Mgl}.
$$
At the vertical state $p=I_3(\dot\psi+\dot\phi)$, so this is the <gyroscopic stabilization of an inverted symmetric top> by sufficiently rapid spin.

Solved by gpt-5.6-sol high.

= 9A
{parent=Paper 3}
{scope}
{title2=Cosmology}

= Solution
{parent=9A}

Differentiate the <Friedmann equation>
$$
H^2=\frac{8\pi G}{3c^2}\rho-\frac{kc^2}{R^2a^2}
$$
and use the <cosmological perfect-fluid continuity equation>
$$
\dot\rho=-3H(\rho+P).
$$
After dividing by $2H$,
$$
\dot H=-\frac{4\pi G}{c^2}(\rho+P)
+\frac{kc^2}{R^2a^2}.
$$
Since $\ddot a/a=\dot H+H^2$, eliminating the curvature term gives the <Friedmann acceleration equation>
$$
\boxed{
\frac{\ddot a}{a}
=-\frac{4\pi G}{3c^2}(\rho+3P)
}.
$$

Under the <strong energy condition in a Friedmann universe>, $\rho+3P\geq0$, so $\ddot a/a\leq0$. Therefore
$$
\dot H=\frac{\ddot a}{a}-H^2\leq-H^2,
$$
and wherever $H\ne0$,
$$
\boxed{
\frac d{dt}H^{-1}
=-\frac{\dot H}{H^2}\geq1
}.
$$

If $H(t_0)>0$, integrating the inequality backward shows that $H^{-1}$ reaches zero no later than $t_0-H(t_0)^{-1}$. Thus $H\to+\infty$ and $a\to0$ at a finite time in the past. If $H(t_0)<0$, the same argument forward in time makes $H\to-\infty$ and $a\to0$ at a finite future time. The sign of $H$ distinguishes an expanding universe with a past Big Bang from a contracting universe ending in a <Big Crunch>. Hence a Friedmann universe obeying the strong energy condition cannot remain nonsingular for unlimited proper time in both directions.

Solved by gpt-5.6-sol high.

= 10D
{parent=Paper 3}
{scope}
{title2=Quantum Information and Computation}

= a
{parent=10d}
{scope}

= Solution
{parent=a}

Use the available unitaries to prepare
$$
|s\rangle=\frac1{\sqrt N}\sum_{i\in\mathbb Z_N}|i\rangle
$$
in the first register and
$$
|-\rangle=\frac{|0\rangle-|1\rangle}{\sqrt2}
$$
in the answer qubit. The <quantum phase kickback> identity gives
$$
O_x|s\rangle|-\rangle
=\frac1{\sqrt N}\sum_i(-1)^{x_i}|i\rangle|-\rangle.
$$
Choose a unitary $F$ with $F|0\rangle=|s\rangle$, apply $F^{-1}$ to the first register, and measure it. Its amplitude at $|0\rangle$ is
$$
\langle s|
\frac1{\sqrt N}\sum_i(-1)^{x_i}|i\rangle
=\frac1N\sum_i(-1)^{x_i}.
$$
This equals $+1$ or $-1$ for a constant string and zero for a balanced string. Thus outcome $0$ means constant, while every other outcome means balanced, with certainty after one oracle query. This is the <Deutsch-Jozsa test with an arbitrary uniform-state unitary>.

Solved by gpt-5.6-sol high.

= b
{parent=10d}
{scope}

= Solution
{parent=b}

With the restricted operations, begin in $|0\rangle|0\rangle$ and apply $F$ to the first register:
$$
\frac1{\sqrt N}\sum_i|i\rangle|0\rangle.
$$
Now apply
$$
O_x,\qquad Z,\qquad O_x.
$$
For every basis state this <compute-phase-uncompute construction> acts as
$$
|i\rangle|0\rangle
\mapsto |i\rangle|x_i\rangle
\mapsto(-1)^{x_i}|i\rangle|x_i\rangle
\mapsto(-1)^{x_i}|i\rangle|0\rangle.
$$
The first register is therefore the same phase state used in part (a). Apply $F^{-1}$ and measure. Again, outcome $0$ occurs with probability one for a constant string and probability zero for a balanced string. The construction uses exactly two queries to $O_x$ and only the permitted operations.

Solved by gpt-5.6-sol high.

= 11I
{parent=Paper 3}
{scope}
{title2=Number Theory}

= a
{parent=11i}
{scope}

= Solution
{parent=a}

An integer $b$ is a <primitive root> modulo $n$ when it is coprime to $n$ and its residue class generates the <multiplicative group of integers modulo n>. Equivalently,
$$
\boxed{\operatorname{ord}_n(b)=\varphi(n)}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=11i}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

We prove that (i) implies (ii). Since $b$ is primitive modulo $p$, write
$$
b^{p-1}=1+cp.
$$
Being primitive modulo $p^2$ is equivalent to $p\nmid c$. The binomial theorem then gives inductively
$$
v_p\left(b^{p^{j}(p-1)}-1\right)=j+1.
$$
It follows that the order modulo $p^m$ is exactly
$$
p^{m-1}(p-1)=\varphi(p^m),
$$
so $b$ is primitive modulo every $p^m$. This is <lifting a primitive root to odd prime powers>. The implication (ii)$\Rightarrow$(i) is immediate by taking $m=2$.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

This implication is already contained in part (i): being primitive modulo every $p^m$ plainly includes being primitive modulo $p^2$, while primitivity modulo $p^2$ lifts to every higher power by the valuation calculation there.

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

Suppose first that $b$ is not primitive modulo $p^2$. Since it is primitive modulo $p$, its order modulo $p^2$ is then $p-1$, so
$$
b^{p-1}\equiv1\pmod{p^2}.
$$
Because $p-1$ divides $p^2-1$, the composite number $p^2$ satisfies
$$
b^{p^2-1}\equiv1\pmod{p^2};
$$
it is a <Fermat pseudoprime> to base $b$ and is divisible by $p^2$. Thus (iii) fails.

Conversely, suppose $b$ is primitive modulo $p^2$ and a base-$b$ pseudoprime $N$ is divisible by $p^a$ with $a\geq2$. By part (i),
$$
\operatorname{ord}_{p^a}(b)=p^{a-1}(p-1).
$$
The pseudoprime congruence forces this order to divide $N-1$, so in particular $p\mid N-1$. But $p\mid N$, a contradiction. Hence no such pseudoprime exists. This proves (i)$\Leftrightarrow$(iii), and all three statements are equivalent.

Solved by gpt-5.6-sol high.

= c
{parent=11i}
{scope}

= Solution
{parent=c}

By <lifting a primitive root to odd prime powers>, it is enough to test primitivity modulo $25$. The positive integers below eight either fail to be units or have orders
$$
\operatorname{ord}_{25}(1)=1,\quad
\operatorname{ord}_{25}(4)=10,\quad
\operatorname{ord}_{25}(6)=5,\quad
\operatorname{ord}_{25}(7)=4,
$$
whereas
$$
\operatorname{ord}_{25}(2)
=\operatorname{ord}_{25}(3)
=\operatorname{ord}_{25}(8)=20.
$$
Therefore the three smallest required integers are
$$
\boxed{2,\ 3,\ 8}.
$$

Solved by gpt-5.6-sol high.

= d
{parent=11i}
{scope}

= Solution
{parent=d}

By the <primitive-root modulus classification>, $P(n)>0$ only for
$$
n=1,2,4,p^a,\ \text{or }2p^a
$$
with $p$ an odd prime, and then
$$
P(n)=\varphi(\varphi(n)).
$$
Fix $k\geq1$. The <finite fibres of the Euler totient function> imply that there are only finitely many possible values $m=\varphi(n)$ satisfying $\varphi(m)=k$. For each such $m$, the same result gives only finitely many $n$ satisfying $\varphi(n)=m$. The union of these finitely many finite sets is finite, proving the claim.

Solved by gpt-5.6-sol high.

= 12I
{parent=Paper 3}
{scope}
{title2=Automata and Formal Languages}

= Solution
{parent=12I}

A state $q$ is <accessible state of a deterministic finite automaton>[accessible] when
$$
\widehat\delta(q_0,w)=q
$$
for some word $w\in\Sigma^*$. States $p$ and $q$ are equivalent, or <indistinguishable states of a deterministic finite automaton>[indistinguishable], when
$$
\widehat\delta(p,w)\in F
\quad\Longleftrightarrow\quad
\widehat\delta(q,w)\in F
$$
for every continuation $w$.

This equivalence is preserved by transitions: if $p\sim q$, then $\delta(p,a)\sim\delta(q,a)$. Hence define $D/{\sim}$ with states $[q]$, initial state $[q_0]$, transition
$$
\overline\delta([q],a)=[\delta(q,a)],
$$
and accepting classes $[q]$ with $q\in F$. Induction on word length gives
$$
\widehat{\overline\delta}([q_0],w)
=[\widehat\delta(q_0,w)],
$$
so $D$ and its <quotient deterministic finite automaton by indistinguishable states> accept the same language. If two quotient states were equivalent, their representatives would be equivalent in $D$, so the classes would be equal. Thus no two distinct quotient states are equivalent.

For the unary alphabet, accessibility means that all states occur on the orbit
$$
q_0,\ q_1=\delta(q_0,1),\ q_2,\ldots.
$$
Finiteness makes this orbit a directed tail entering a directed cycle. The quotient retains exactly one accepting state and merges precisely those positions having the same future acceptance pattern.

More explicitly, if the unique accepting state lies before the cycle, every state after it can never reach acceptance and these states collapse to one rejecting sink. The minimal diagram is then a directed chain from the initial state through the unique accepting state and onward to that sink, which has a self-loop.

If the accepting state lies on the cycle, all cycle states are distinct because their next visits to the accepting state occur in different residue classes modulo the cycle length. Any remaining tail is a chain feeding the cycle. In a minimal diagram with a nonempty tail, the accepting state is the cycle vertex immediately preceding the entry vertex; otherwise the final tail state has the same future acceptance sequence as a cycle state and would be merged. A pure cycle with one accepting vertex is also possible. These are exactly the minimal <accessible unary deterministic finite automaton>[accessible unary DFAs] with one accepting state.

Solved by gpt-5.6-sol high.

= 13C
{parent=Paper 3}
{scope}
{title2=Mathematical Biology}

= Solution
{parent=13C}

<Fick's first law> gives the diffusive flux
$$
\boxed{\mathbf J=-D(C)\nabla C}.
$$
Local <conservation law>[conservation] says $C_t+\nabla\cdot\mathbf J=0$, and therefore
$$
\boxed{C_t=\nabla\cdot(D(C)\nabla C)}.
$$

For $D(C)=kC$, conservation of the deposited amount requires
$$
2\pi\int_0^\infty rC(r,t)\,dr=2\pi M.
$$
The dimensions satisfy $[M]=[C]L^2$ and $[k]=L^2/([C]T)$. Thus $Mkt$ has dimension $L^4$, so the similarity length is $(Mkt)^{1/4}$. Requiring the concentration scale to have dimension $[C]$ gives
$$
\boxed{\alpha=\beta=\frac12,\qquad\gamma=\frac14}.
$$
Hence
$$
C(r,t)=\sqrt{\frac{M}{kt}}\,F(\xi),
\qquad
\xi=\frac{r}{(Mkt)^{1/4}},
\qquad
\int_0^\infty\xi F(\xi)\,d\xi=1.
$$

In polar coordinates, substitution into
$C_t=k r^{-1}(rCC_r)_r$ gives
$$
-\frac12F-\frac14\xi F'
=\frac1\xi(\xi FF')'.
$$
Multiplying by $\xi$ and integrating once, with regularity and zero radial flux at the origin, yields
$$
\xi FF'=-\frac14\xi^2F.
$$
Where $F>0$, this reduces to $F'=-\xi/4$, so
$$
F(\xi)=A-\frac{\xi^2}{8}.
$$
Nonnegativity and zero flux at the moving front give the compactly supported profile
$$
F(\xi)=\left(A-\frac{\xi^2}{8}\right)_+,
\qquad
\xi_0^2=8A.
$$
The normalization gives
$$
1=\int_0^{\xi_0}\xi\left(A-\frac{\xi^2}{8}\right)d\xi
=2A^2,
$$
and hence $A=1/\sqrt2$ and $\xi_0^4=32$. This is the <two-dimensional Barenblatt solution with diffusivity proportional to concentration>, and its support radius is
$$
\boxed{r_0(t)=(32Mkt)^{1/4}},
$$
so $N=32$.

Now add the linear reaction term and write
$$
C(r,t)=e^{at}G(r,\tau(t)).
$$
Substitution cancels the terms $aC$ and leaves
$$
e^{at}\tau'G_\tau
=ke^{2at}\nabla\cdot(G\nabla G).
$$
Choosing
$$
\boxed{\tau'=e^{at},\qquad
\tau(t)=
\begin{cases}
(e^{at}-1)/a,&a\ne0,\\
t,&a=0
\end{cases}}
$$
produces $G_\tau=k\nabla\cdot(G\nabla G)$, as in the <linear-reaction time change for quadratic nonlinear diffusion>. Therefore
$$
\boxed{
C(r,t)=e^{at}\sqrt{\frac{M}{k\tau(t)}}
\left(
\frac1{\sqrt2}
-\frac{r^2}{8\sqrt{Mk\tau(t)}}
\right)_+
}.
$$

For $a=0$, the front grows like $t^{1/4}$ while the central concentration decays like $t^{-1/2}$. For $a>0$, $\tau\sim e^{at}/a$, so the front grows like $e^{at/4}$ and the central concentration like $e^{at/2}$. For $a<0$, $\tau\to1/|a|$: the front approaches a finite limiting radius while the concentration decays exponentially to zero.

Solved by gpt-5.6-sol high.

= 14A
{parent=Paper 3}
{scope}
{title2=Cosmology}

= a
{parent=14a}
{scope}

= Solution
{parent=a}

The <flatness problem> is that $|\Omega-1|$ grows during ordinary decelerating expansion, so its small present value requires the early density to have been extraordinarily close to critical. The <horizon problem> is the near uniformity of regions whose past light cones do not meet in a noninflationary hot Big Bang model.

Inflation requires accelerated expansion,
$$
\boxed{\ddot a>0}.
$$
For a constant equation of state $P=w\rho$, this means $w<-1/3$; representative scale factors are $a\propto t^p$ with $p>1$ and $a\propto e^{Ht}$. Equivalently, the comoving Hubble radius $(aH)^{-1}$ decreases. Since
$$
|\Omega-1|\propto(aH)^{-2},
$$
many e-folds drive the universe toward spatial flatness. A small region that was initially in causal contact is stretched to a size far larger than the later Hubble radius, allowing the observed universe to originate from one equilibrated patch.

Solved by gpt-5.6-sol high.

= b
{parent=14a}
{scope}

= Solution
{parent=b}

For the <inflaton>,
$$
\dot\rho=\dot\phi\ddot\phi+V'(\phi)\dot\phi,
\qquad
\rho+P=\dot\phi^2.
$$
The continuity equation therefore becomes
$$
\dot\phi\left[\ddot\phi+3H\dot\phi+V'(\phi)\right]=0.
$$
Away from an isolated turning point division by $\dot\phi$ gives, and continuity extends through such points,
$$
\boxed{\ddot\phi+3H\dot\phi+\frac{dV}{d\phi}=0}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=14a}
{scope}

= Solution
{parent=c}

The <Friedmann acceleration equation> gives
$$
\frac{\ddot a}{a}
=-\frac{4\pi G}{3c^2}(\rho+3P).
$$
For the scalar field,
$$
\rho+3P=2\dot\phi^2-2V(\phi),
$$
and hence
$$
\boxed{
\frac{\ddot a}{a}
=\frac{8\pi G}{3c^2}
\left[V(\phi)-\dot\phi^2\right]
}.
$$
An inflationary phase occurs precisely when potential energy dominates enough that
$$
\boxed{V(\phi)>\dot\phi^2}.
$$

Solved by gpt-5.6-sol high.

= d
{parent=14a}
{scope}

= Solution
{parent=d}

The <slow-roll approximation> assumes
$$
\dot\phi^2\ll V(\phi),
\qquad
|\ddot\phi|\ll3H|\dot\phi|.
$$
The first condition makes the energy density potential dominated, and the second neglects inertial acceleration relative to Hubble friction. The scalar and Friedmann equations reduce to
$$
\boxed{3H\dot\phi\simeq-V'(\phi)},
\qquad
\boxed{H^2\simeq\frac{8\pi G}{3c^2}V(\phi)}.
$$

Solved by gpt-5.6-sol high.

= e
{parent=14a}
{scope}

= Solution
{parent=e}

For $V=\lambda\phi^4/4$ and positive $\phi$, the slow-roll Friedmann equation gives
$$
H\simeq\phi^2\sqrt{\frac{2\pi G\lambda}{3c^2}}.
$$
The scalar equation then reduces to
$$
\dot\phi\simeq
-c\sqrt{\frac{\lambda}{6\pi G}}\,\phi.
$$
With $\phi(t_0)=\phi_0$,
$$
\boxed{
\phi(t)=\phi_0
\exp\left[-c\sqrt{\frac{\lambda}{6\pi G}}(t-t_0)\right]
}.
$$
Furthermore,
$$
\frac{d\log a}{d\phi}
=\frac H{\dot\phi}
\simeq-\frac{3H^2}{V'}
=-\frac{2\pi G}{c^2}\phi.
$$
Thus the <quartic-potential slow-roll solution> is
$$
\boxed{
a(\phi)=a_0
\exp\left[
\frac{\pi G}{c^2}(\phi_0^2-\phi^2)
\right]
}.
$$

Solved by gpt-5.6-sol high.

= 15D
{parent=Paper 3}
{scope}
{title2=Quantum Information and Computation}

= a
{parent=15d}
{scope}

= Solution
{parent=a}

Start two registers in $|0\rangle|0\rangle$. Apply $\operatorname{QFT}_N$ to the first and then reversibly evaluate $f$ into the second, obtaining
$$
\frac1{\sqrt N}\sum_{x\in\mathbb Z_N}|x\rangle|a^x\bmod N\rangle.
$$
The evaluation can be implemented by <repeated squaring> and reversible modular multiplication using a number of elementary arithmetic operations polynomial in $\log N$.

Measure the second register. Because $f$ is injective within one period and $r\mid N$, the first register becomes a uniformly weighted periodic coset
$$
\sqrt{\frac rN}\sum_{j=0}^{N/r-1}|x_0+jr\rangle.
$$
The <quantum Fourier transform of a periodic coset state> followed by measurement returns
$$
\boxed{c=s\frac Nr},
\qquad s\ \hbox{uniform in }\mathbb Z_r.
$$
Reduce the <rational number> $c/N$ to lowest terms. Since
$$
\frac cN=\frac sr,
$$
its denominator is $q=r/\gcd(s,r)$. Test whether $a^q\equiv1\pmod N$ by repeated squaring. The promised injectivity means that $r$ is the least positive period, and $q\mid r$, so this congruence holds exactly when $q=r$. The test therefore certifies whether the run succeeded; this is <heralded exact quantum period finding when the period divides the register size>.

One run succeeds with probability
$$
\frac{\varphi(r)}r,
$$
because success is equivalent to $s$ being <coprime integers>[coprime] to $r$. A standard estimate for the <Euler totient function> gives $\varphi(r)/r\geq C/\log\log r$ for all sufficiently large $r$, with the finitely many smaller cases absorbed by changing the positive constant $C$. Repeating independently $O(\log\log N)$ times therefore makes the probability that every run fails at most $1/2$. Every step, including the repetitions and the classical verification, takes time polynomial in $\log N$.

Solved by gpt-5.6-sol high.

= b
{parent=15d}
{scope}

= Solution
{parent=b}

Use the convention
$$
\operatorname{QFT}_N|k\rangle
=\frac1{\sqrt N}\sum_{j=0}^{N-1}e^{2\pi ijk/N}|j\rangle.
$$
Changing the summation variable from $j$ to $j-1$ gives
$$
S\operatorname{QFT}_N|k\rangle
=e^{-2\pi ik/N}\operatorname{QFT}_N|k\rangle.
$$
Thus the Fourier-basis states are <eigenvector>[eigenvectors] of the cyclic shift, with respective <eigenvalue>[eigenvalues] $e^{-2\pi ik/N}$, and the <cyclic shift diagonalization by the quantum Fourier transform> is
$$
S=\operatorname{QFT}_N D\operatorname{QFT}_N^{-1},
\qquad
D|k\rangle=e^{-2\pi ik/N}|k\rangle.
$$

For $N=4$, write $k=2x+y$ with $x,y\in\{0,1\}$. Then
$$
e^{-2\pi ik/4}=e^{-i\pi x}e^{-i\pi y/2},
$$
so
$$
D=P(-\pi)\otimes P(-\pi/2),
\qquad
P(\theta)=\begin{pmatrix}1&0\\0&e^{i\theta}\end{pmatrix}.
$$
Hence the required <quantum circuit> applies, from input to output,
$$
\boxed{
\operatorname{QFT}_4^{-1}
\;\longrightarrow\;
P(-\pi)\otimes P(-\pi/2)
\;\longrightarrow\;
\operatorname{QFT}_4
}.
$$
The first phase gate acts on the most significant qubit $x$ and the second on the least significant qubit $y$.

Solved by gpt-5.6-sol high.

= 16F
{parent=Paper 3}
{scope}
{title2=Logic and Set Theory}

= Solution
{parent=16F}

The <compactness theorem> says that a set $T$ of <first-order sentence>[first-order sentences] has a <model of a first-order theory>[model] if and only if every finite subset of $T$ has a model. One implication follows by taking the same model. Conversely, if $T$ had no model, then by the <Godel completeness theorem> it would prove a contradiction. A formal proof uses only finitely many assumptions, so some finite subset of $T$ would already have no model. This contradiction proves compactness.

The <Upward Lowenheim-Skolem theorem> says that if an $L$-theory $T$ has an infinite model, then it has models of arbitrarily large cardinality; more precisely, it has a model of cardinality at least $\kappa$ for every cardinal $\kappa$. Add new constants $c_\alpha$ for $\alpha<\kappa$ and the sentences
$$
c_\alpha\ne c_\beta\qquad(\alpha\ne\beta).
$$
Every finite subset of the enlarged theory can be interpreted in the given infinite model, since it mentions only finitely many constants. Compactness supplies a model of the whole enlarged theory, in which the $c_\alpha$ are pairwise distinct. Its reduct to $L$ is a model of $T$ having at least $\kappa$ elements. If $\kappa\geq|L|+\aleph_0$, the <Downward Lowenheim-Skolem theorem> gives a model of cardinality exactly $\kappa$.

Solved by gpt-5.6-sol high.

= i
{parent=16f}
{scope}

= Solution
{parent=i}

This class is not <first-order axiomatizable class>[axiomatisable]. Suppose that a theory $T$ axiomatized it. Add constants $m_0,m_1,\ldots$, require each $m_i$ to be a <maximal element of a partially ordered set>[maximal element], and require $m_i\ne m_j$ for $i\ne j$. Every finite part of this enlarged theory has a model: choose a finite <partially ordered set> with enough maximal elements. By the <compactness theorem> the whole theory has a model, but its reduct is a model of $T$ with infinitely many maximal elements, a contradiction. This is the <compactness obstruction to axiomatizing finitely many maximal elements>.

Solved by gpt-5.6-sol high.

= ii
{parent=16f}
{scope}

= Solution
{parent=ii}

This class is axiomatisable when an uncountable set of constants is allowed. Add constants $(c_\alpha)_{\alpha<\omega_1}$ and the axioms
$$
\operatorname{Max}(c_\alpha),
\qquad
c_\alpha\ne c_\beta\quad(\alpha\ne\beta),
$$
along with the <partially ordered set> axioms. Every reduct of a model has at least $\aleph_1$ maximal elements. Conversely, in every poset with uncountably many maximal elements the constants can be interpreted as distinct maximal elements. Thus the expanded theory axiomatizes precisely the desired reducts.

Solved by gpt-5.6-sol high.

= iii
{parent=16f}
{scope}

= Solution
{parent=iii}

This class is axiomatisable in the original language. For every positive integer $n$, let $\sigma_n$ say
$$
\text{``there are at least $n$ maximal elements''}
\quad\lor\quad
\text{``there are at least $n$ minimal elements''.}
$$
Both clauses are <first-order sentence>[first-order]. If one of the two sets is infinite, every $\sigma_n$ holds. Conversely, if both sets were finite, choosing $n$ larger than both cardinalities would make $\sigma_n$ false. Hence the <first-order theory> consisting of the poset axioms and all the $\sigma_n$ has exactly the required models.

Solved by gpt-5.6-sol high.

= iv
{parent=16f}
{scope}

= Solution
{parent=iv}

This class is not axiomatisable. If $T$ axiomatized it, add constants $m_i$ and $n_i$ for $i\in\mathbb N$, requiring the $m_i$ to be distinct maximal elements and the $n_i$ to be distinct <minimal element of a partially ordered set>[minimal elements]. Every finite fragment has a model of $T$: make one of the two sets infinite and make the other finite but large enough to interpret the finitely many constants mentioned. Compactness would then give a model of $T$ having both infinitely many maximal and infinitely many minimal elements, contrary to the exclusive condition. This is the <compactness obstruction to an exclusive disjunction of infinitude>.

Solved by gpt-5.6-sol high.

= v
{parent=16f}
{scope}

= Solution
{parent=v}

This class is not axiomatisable. It contains an infinite model, for example the <total order> $\mathbb Q$. If a theory $T$ in any fixed expanded language axiomatized it, choose a cardinal
$$
\kappa>\max\{|L|,|\mathbb R|\}.
$$
The <Upward Lowenheim-Skolem theorem> would give a model of $T$ of cardinality at least $\kappa$. Its underlying order cannot be isomorphic to a subset of $\mathbb R$, because every such subset has cardinality at most $|\mathbb R|$. This contradiction proves the claim.

Solved by gpt-5.6-sol high.

= 17F
{parent=Paper 3}
{scope}
{title2=Graph Theory}

= a
{parent=17f}
{scope}

= Solution
{parent=a}

Colour every <vertex of a graph> independently and uniformly with three colours, and retain precisely the <edges> whose endpoints have different colours. Each edge is retained with probability $2/3$, so <linearity of expectation> gives expected retained edge count
$$
\frac23e(G).
$$
Some colouring therefore retains at least $2e(G)/3$ edges. Its retained <subgraph> is three-colourable, and deleting edges if necessary leaves exactly
$$
\boxed{\left\lfloor\frac23e(G)\right\rfloor}
$$
edges without increasing its <chromatic number>. This is the <three-colourable two-thirds subgraph lemma>.

To prove sharpness, take $G=K_n$. If $H\subseteq K_n$ is three-colourable and its colour classes have sizes $n_1,n_2,n_3$, then they are <independent set>[independent sets], so
$$
e(H)\leq n_1n_2+n_1n_3+n_2n_3
=\frac12\left(n^2-\sum_i n_i^2\right)
\leq\frac{n^2}{3},
$$
where the final inequality follows from <Cauchy-Schwarz inequality>. Since $e(K_n)=n(n-1)/2$,
$$
\frac{e(H)}{e(K_n)}\leq\frac{2n}{3(n-1)}
=\frac23+\frac{2}{3(n-1)}.
$$
Given $\varepsilon>0$, choose $n$ with $2/[3(n-1)]\leq\varepsilon$. Then every three-colourable subgraph has at most $(2/3+\varepsilon)e(G)$ edges.

Solved by gpt-5.6-sol high.

= b
{parent=17f}
{scope}

= Solution
{parent=b}

Choose a <cut of a graph>[partition] $V=A\sqcup B$ that maximizes the number of edges crossing between its two parts. If some $v\in A$ had
$$
|N(v)\cap B|<|N(v)\cap A|,
$$
moving $v$ to $B$ would replace the former crossing edges incident with $v$ by the latter and strictly increase the cut size. This contradicts maximality. The same argument applies to every $v\in B$. Hence every vertex has at least as many neighbours in the opposite part as in its own part, so the maximum cut is an <unfriendly partition of a graph>.

Solved by gpt-5.6-sol high.

= c
{parent=17f}
{scope}

= Solution
{parent=c}

Enumerate the vertices as $v_1,v_2,\ldots$. Since every vertex has finite degree, choose increasing finite vertex sets $W_n$ whose union is $V$ and such that the closed neighbourhood of $v_i$ lies in $W_n$ whenever $i\leq n$. By part (b), each finite induced graph $G[W_n]$ has an unfriendly two-colouring.

There are only two colours. Pass successively to an infinite subsequence on which the colour of $v_1$ is constant, then one on which the colour of $v_2$ is constant, and so on. The <diagonal argument> gives a limiting colouring in which, for every fixed finite set of vertices, all its colours agree with those in infinitely many of the finite colourings.

Fix $v_i$. Its entire finite neighbourhood lies in every sufficiently large $W_n$, and along the diagonal subsequence the colours of $v_i$ and all its neighbours eventually stabilize. The unfriendly inequality for $v_i$ in $G[W_n]$ therefore passes unchanged to the limit. This holds for every vertex, proving the <unfriendly partition theorem for a countable locally finite graph>.

Solved by gpt-5.6-sol high.

= d
{parent=17f}
{scope}

= Solution
{parent=d}

Colour every vertex independently red or blue with probability $1/2$. Fix a vertex $v$ and enumerate infinitely many distinct neighbours $w_1,w_2,\ldots$. The probability that all neighbours after $w_n$ are red is
$$
\lim_{m\to\infty}2^{-(m-n)}=0.
$$
Taking the countable union over $n$, the probability that $v$ has only finitely many blue neighbours is zero. The same argument with the colours exchanged shows that the probability of only finitely many red neighbours is zero.

The vertex set is countable, so the union of these two null events over all vertices still has probability zero. Thus with probability one every vertex has infinitely many neighbours of each colour. Taking $A$ and $B$ to be the two colour classes gives an unfriendly partition. This is the <random unfriendly partition of a countable infinite-degree graph>.

Solved by gpt-5.6-sol high.

= 18H
{parent=Paper 3}
{scope}
{title2=Galois Theory}

= a
{parent=18h}
{scope}

= Solution
{parent=a}

The automorphisms $1,\sigma,\ldots,\sigma^{n-1}$ are distinct and hence linearly independent as maps $L\to L$ by the <linear independence of distinct field embeddings>. Therefore the operator
$$
T=\sum_{j=0}^{n-1}\zeta^{-j}\sigma^j
$$
is not the zero map. Choose $\beta\in L$ for which $\alpha=T(\beta)\ne0$. Reindexing the sum and using $\sigma^n=1$ gives
$$
\sigma(\alpha)
=\sum_{j=0}^{n-1}\zeta^{-j}\sigma^{j+1}(\beta)
=\zeta\alpha.
$$
This is the <Lagrange resolvent eigenvector for a cyclic field automorphism>.

Let $F=L^\sigma$ be the <fixed field>. Since $\sigma(\alpha^n)=\alpha^n$, the element $\alpha$ is a root of
$$
X^n-\alpha^n\in F[X].
$$
Its orbit under $\langle\sigma\rangle$ is
$$
\alpha,\zeta\alpha,\ldots,\zeta^{n-1}\alpha,
$$
and these $n$ elements are distinct because $\alpha\ne0$ and $\zeta$ is a <primitive root of unity>. Hence the <minimal polynomial> has at least $n$ distinct roots. It also divides the displayed degree-$n$ polynomial, so
$$
\boxed{m_{\alpha,F}(X)=X^n-\alpha^n}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=18h}
{scope}

= Solution
{parent=b}

An <algebraic closure> of a field $K$ is an <algebraically closed field> that is an algebraic extension of $K$.

By definition, every element of
$$
\overline{\mathbb Q}
=\{z\in\mathbb C:z\text{ is algebraic over }\mathbb Q\}
$$
is algebraic over $\mathbb Q$, and the question allows us to use that this set is a field. It remains to prove that it is algebraically closed.

Let $f\in\overline{\mathbb Q}[X]$ be nonconstant. Its finitely many coefficients generate a finite extension $E/\mathbb Q$. By the <fundamental theorem of algebra>, $f$ has a root $z\in\mathbb C$. Since $z$ is a root of a polynomial over $E$, it is algebraic over $E$; since $E/\mathbb Q$ is algebraic, <transitivity of algebraic extensions> makes $z$ algebraic over $\mathbb Q$. Thus $z\in\overline{\mathbb Q}$. Repeating after division by $X-z$ shows that $f$ splits there, so $\overline{\mathbb Q}$ is algebraically closed and is an algebraic closure of $\mathbb Q$.

Solved by gpt-5.6-sol high.

= c
{parent=18h}
{scope}

= Solution
{parent=c}

Because $\operatorname{char}K=0$, the finite extension $M/K$ is separable. By the <primitive element theorem>, write $M=K(\theta)$. The <minimal polynomial> of $\theta$ over $K$ has a root in the <algebraically closed field> $L$, so sending $\theta$ to that root defines a $K$-<field embedding> $\iota:M\hookrightarrow L$.

The cyclic group $\langle\sigma\rangle$ acts faithfully on $L$ after replacing $d$ by the order of $\sigma$. The <Artin fixed-field theorem> gives
$$
[L:K]=|\langle\sigma\rangle|
$$
and says that $L/K$ is a finite Galois extension with cyclic Galois group $\langle\sigma\rangle$. The image $E=\iota(M)$ is an <intermediate field>. Every subgroup of a cyclic group is normal, so the <normal subextension criterion> shows that $E/K$ is Galois, and its Galois group is a quotient of $\langle\sigma\rangle$, hence cyclic. Transporting this structure through $\iota$ proves that $M/K$ is Galois with cyclic Galois group. This is the <finite extensions of the fixed field of a finite-order automorphism of an algebraically closed field> theorem.

Algebraic closedness is necessary. Take $L=\mathbb Q$ and $\sigma=\operatorname{id}$, so $K=L^\sigma=\mathbb Q$, and let $M=\mathbb Q(\sqrt[3]{2})$. The extension has degree three but is not normal, since the two nonreal roots of $X^3-2$ do not belong to $M$. It is therefore not Galois.

Solved by gpt-5.6-sol high.

= 19H
{parent=Paper 3}
{scope}
{title2=Representation Theory}

= Solution
{parent=19H}

Let $V_n$ be the space of homogeneous degree-$n$ polynomials in $x,y$. For $g\in SU(2)$ and a column vector $v=(x,y)^T$, define
$$
(\rho_n(g)P)(v)=P(g^{-1}v).
$$
Substitution preserves degree, and
$$
\rho_n(gh)P(v)=P(h^{-1}g^{-1}v)
=\rho_n(g)\rho_n(h)P(v),
$$
so $\rho_n:SU(2)\to\operatorname{GL}(V_n)$ is a <continuous representation of a topological group>. This is the <homogeneous polynomial representation of SU2>.

To prove <irreducible representation>[irreducibility], restrict to the diagonal circle
$$
t_\theta=\operatorname{diag}(e^{i\theta},e^{-i\theta}).
$$
The monomials $x^{n-k}y^k$ are its one-dimensional weight spaces, with distinct weights up to our harmless inverse-action convention. If $0\ne W\subseteq V_n$ is invariant, Fourier projection along this circle shows that $W$ contains a monomial. Differentiating the action of $SU(2)$ and complexifying gives the operators
$$
E=x\frac{\partial}{\partial y},
\qquad
F=y\frac{\partial}{\partial x}.
$$
Repeated applications of $E$ and $F$ connect every monomial $x^{n-k}y^k$ to every other one. Hence $W$ contains the entire monomial basis, so $W=V_n$.

Every $g\in SU(2)$ is conjugate to $\operatorname{diag}(z,z^{-1})$ with $|z|=1$. Reading the eigenvalues on the monomial basis gives the <character of the homogeneous polynomial representation of SU2>
$$
\boxed{
\chi_n(g)=z^n+z^{n-2}+\cdots+z^{-n}
}.
$$
For $z=e^{i\theta}$ this is
$$
\chi_n(g)=\frac{\sin((n+1)\theta)}{\sin\theta},
$$
with the values at $\sin\theta=0$ obtained by continuity.

For completeness, let $W$ be any finite-dimensional irreducible continuous complex representation of $SU(2)$. Its restriction to the diagonal circle splits into integral weight spaces. Choose a vector of largest weight $n$. The raising operator kills it, and the $\mathfrak{sl}_2$ commutation relations show that $n$ is a nonnegative integer and that successive applications of the lowering operator form a string of weights
$$
n,n-2,\ldots,-n.
$$
Their span is an invariant copy of $V_n$; irreducibility forces it to equal $W$. Thus the <classification of finite-dimensional representations of SU2> says
$$
\boxed{W\cong V_n\quad\text{for a unique }n\geq0}.
$$

If the eigenvalues of $\rho(g)$ are $\lambda_1,\ldots,\lambda_d$, then the eigenvalues on $\bigwedge^2V$ are $\lambda_i\lambda_j$ for $i<j$. Therefore the <character of an exterior square> is
$$
\boxed{
\chi_{\wedge^2V}(g)
=\frac12\bigl(\chi(g)^2-\chi(g^2)\bigr)
}.
$$
The <exterior square of an SU2 irreducible representation> or the <Clebsch-Gordan decomposition for SU2> with flip parity gives
$$
\boxed{\bigwedge^2V_4\cong V_6\oplus V_2}.
$$
The dimensions $7+3=10=\binom52$ check the result.

The third <elementary symmetric polynomial> in the $\lambda_i$, together with <Newton identities>, gives
$$
\boxed{
\chi_{\wedge^3V}(g)
=\frac16\left(
\chi(g)^3-3\chi(g)\chi(g^2)+2\chi(g^3)
\right)
}.
$$
For the five-dimensional representation $V_4$, the canonical duality
$$
\bigwedge^3V_4
\cong(\bigwedge^2V_4)^*\otimes\det V_4
$$
finishes the decomposition. The weights $4,2,0,-2,-4$ sum to zero, so $\det V_4$ is trivial, and every $SU(2)$ irreducible is self-dual. Consequently
$$
\boxed{\bigwedge^3V_4\cong V_6\oplus V_2}.
$$

Solved by gpt-5.6-sol high.

= 20I
{parent=Paper 3}
{scope}
{title2=Algebraic Topology}

= Solution
{parent=20I}

Suppose first that $F:D^n\to X$ extends $f$. Regarding $S^{n-1}$ as the boundary of the unit disc,
$$
H(x,t)=F((1-t)x)
$$
is a <homotopy> from $f$ to the constant map with value $F(0)$. Thus $f$ is <null-homotopic map>[null-homotopic]. Conversely, if $H:S^{n-1}\times[0,1]\to X$ is a homotopy from $f$ to a constant, then $H$ is constant on $S^{n-1}\times\{1\}$ and hence descends to the quotient
$$
S^{n-1}\times[0,1]/(S^{n-1}\times\{1\}),
$$
which is the cone on $S^{n-1}$ and is homeomorphic to $D^n$. The descended map extends $f$. This proves the <extension-null-homotopy criterion for a sphere>.

A <universal cover> of $X$ is a covering map $p:\widetilde X\to X$ whose total space is path-connected and simply connected. Let $p_1:(\widetilde X_1,\widetilde x_1)\to(X,x)$ and $p_2:(\widetilde X_2,\widetilde x_2)\to(X,x)$ be two universal covers. The <lifting criterion for a covering space> applies because
$$
(p_1)_*\pi_1(\widetilde X_1)=0
\subseteq(p_2)_*\pi_1(\widetilde X_2)=0,
$$
so $p_1$ has a based lift $h:\widetilde X_1\to\widetilde X_2$ satisfying $p_2h=p_1$. Similarly there is a based lift $k:\widetilde X_2\to\widetilde X_1$ of $p_2$. Both $kh$ and the identity are lifts of $p_1$ that agree at the base point, so uniqueness of lifts gives $kh=1$. Likewise $hk=1$. Hence $h$ is a homeomorphism. This is the <uniqueness of a universal covering space>.

Now let $p:\widetilde X\to X$ be universal with $\widetilde X$ contractible. If an extension $F:|K|\to X$ exists, functoriality of the <fundamental group> gives
$$
f_*=F_*i_*.
$$
Thus the required factorization holds with $\Phi=F_*$.

Conversely, suppose $f_*=\Phi i_*$. We extend over the simplices of $K$ by dimension. The boundary of every 2-simplex is a loop in $|K^1|$ that becomes null-homotopic in $|K|$, so its class lies in $\ker i_*$. The factorization gives
$$
f_*[\partial\sigma]=\Phi i_*[\partial\sigma]=1.
$$
Hence $f|_{\partial\sigma}$ is null-homotopic in $X$ and extends over the disc $|\sigma|$ by the first part. Since distinct 2-simplices meet along the already fixed 1-skeleton, these extensions combine to a map on $|K^2|$.

Inductively suppose the map is defined on $|K^{m-1}|$ for $m\geq3$. For an $m$-simplex $\sigma$, its boundary is $S^{m-1}$, which is simply connected. The boundary map therefore lifts through $p$ by the covering-space lifting criterion. Its lift into the <contractible space> $\widetilde X$ is null-homotopic, so the boundary map itself is null-homotopic and extends over $|\sigma|$. Extending over every $m$-simplex and then every skeleton constructs $F:|K|\to X$. The weak topology on a simplicial complex makes the cellwise map continuous. This proves the <extension criterion into a space with contractible universal cover>.

Solved by gpt-5.6-sol high.

= 21G
{parent=Paper 3}
{scope}
{title2=Linear Analysis}

= a
{parent=21g}
{scope}

= Solution
{parent=a}

Let $A,B$ be disjoint closed subsets of the <metric space> $(X,d)$; the cases where one is empty are immediate. The distance functions
$$
d(x,A)=\inf_{a\in A}d(x,a),
\qquad
d(x,B)=\inf_{b\in B}d(x,b)
$$
are continuous, because each is 1-Lipschitz. Their sum is strictly positive: if both distances vanished, closedness would put $x$ in $A\cap B$. Therefore
$$
u(x)=\frac{d(x,A)}{d(x,A)+d(x,B)}
$$
is continuous, equals zero on $A$, and equals one on $B$. The sets
$$
u^{-1}([0,1/3))
\quad\text{and}\quad
u^{-1}((2/3,1])
$$
are disjoint open neighbourhoods of $A$ and $B$. Thus every metric space is a <normal topological space>; this is the <normality of every metric space>.

Solved by gpt-5.6-sol high.

= b
{parent=21g}
{scope}

= Solution
{parent=b}

The <Urysohn lemma> states that if $A$ and $B$ are disjoint closed subsets of a normal space $X$, there is a continuous $u:X\to[0,1]$ with $u=0$ on $A$ and $u=1$ on $B$.

The <Tietze extension theorem> states that if $A$ is a closed subset of a normal space $X$, then every continuous $f:A\to\mathbb R$ has a continuous extension $F:X\to\mathbb R$. In the bounded form, if $f(A)\subseteq[a,b]$, the extension can be chosen with $F(X)\subseteq[a,b]$.

Solved by gpt-5.6-sol high.

= c
{parent=21g}
{scope}

= Solution
{parent=c}

If $X$ is compact and $f:X\to\mathbb R$ is continuous, then $f(X)$ is compact by the <continuous image of a compact space> theorem. Every compact subset of $\mathbb R$ is bounded, so $f$ is bounded.

Conversely, suppose the metric space $X$ is not compact. For metric spaces, compactness is equivalent to <sequentially compact space>[sequential compactness], so there is a sequence of distinct points $(x_n)$ having no convergent subsequence. The set
$$
A=\{x_n:n\geq1\}
$$
is closed: an accumulation point would supply a convergent subsequence. It is also discrete for the same reason. Consequently the function
$$
g:A\to\mathbb R,
\qquad g(x_n)=n,
$$
is continuous. By part (a), $X$ is normal, and the <Tietze extension theorem> extends $g$ to a continuous $G:X\to\mathbb R$. Since $G(x_n)=n$, this extension is unbounded. Therefore, if every continuous real-valued function on $X$ is bounded, $X$ must be compact. This is the <bounded-continuous-function characterization of compact metric spaces>.

Solved by gpt-5.6-sol high.

= 22G
{parent=Paper 3}
{scope}
{title2=Analysis of Functions}

= Solution
{parent=22G}

Use the convention
$$
\widehat f(\xi)=\int_{\mathbb R^n}f(x)e^{-ix\cdot\xi}\,dx.
$$
The <Riemann-Lebesgue lemma> states that if $f\in L^1(\mathbb R^n)$, then $\widehat f$ is continuous and
$$
\widehat f(\xi)\longrightarrow0
\qquad\text{as }|\xi|\to\infty.
$$
Continuity follows from the <dominated convergence theorem>, since $e^{-ix\cdot\xi_j}\to e^{-ix\cdot\xi}$ and the integrands are dominated by $|f(x)|$.

For the decay, first take $g\in C_c^1(\mathbb R^n)$. Choose a coordinate $j$ for which $|\xi_j|\geq|\xi|/\sqrt n$. Integration by parts gives
$$
\widehat g(\xi)
=\frac1{i\xi_j}
\int_{\mathbb R^n}\partial_jg(x)e^{-ix\cdot\xi}\,dx,
$$
and hence
$$
|\widehat g(\xi)|
\leq\frac{\sqrt n}{|\xi|}\|\partial_jg\|_1
\longrightarrow0.
$$
Since $C_c^1(\mathbb R^n)$ is dense in $L^1(\mathbb R^n)$, choose $g$ with $\|f-g\|_1<\varepsilon$. The elementary Fourier bound gives
$$
|\widehat f(\xi)-\widehat g(\xi)|\leq\|f-g\|_1<\varepsilon,
$$
so the decay for $g$ implies the decay for $f$.

The <Parseval identity> says that for $f,g\in L^2(\mathbb R^n)$, with their Fourier transforms defined by the <Plancherel theorem>,
$$
\boxed{
\int_{\mathbb R^n}f(x)\overline{g(x)}\,dx
=\frac1{(2\pi)^n}
\int_{\mathbb R^n}\widehat f(\xi)\overline{\widehat g(\xi)}\,d\xi
}.
$$
In particular,
$$
\|\widehat f\|_2=(2\pi)^{n/2}\|f\|_2.
$$

For the given radial function, as $|x|\to0$,
$$
|f(x)|\sim |x|^a,
$$
while as $|x|\to\infty$,
$$
|f(x)|\sim |x|^{-b}.
$$
Using <polar coordinates>, local integrability of $|f|^q$ is therefore determined by
$$
\int_0^1r^{aq+n-1}\,dr<\infty
\quad\Longleftrightarrow\quad aq>-n,
$$
and integrability at infinity by
$$
\int_1^\infty r^{-bq+n-1}\,dr<\infty
\quad\Longleftrightarrow\quad bq>n.
$$
The assumptions $2a>-n$ and $b>n$ imply both $f\in L^2(\mathbb R^n)$ and $f\in L^1(\mathbb R^n)$. The Riemann--Lebesgue lemma and the direct estimate $|\widehat f|\leq\|f\|_1$ give $\widehat f\in L^\infty$, while Parseval gives $\widehat f\in L^2$. Finally, for every $2\leq p<\infty$,
$$
\int_{\mathbb R^n}|\widehat f|^p
\leq
\|\widehat f\|_\infty^{p-2}
\|\widehat f\|_2^2<\infty.
$$
Thus
$$
\boxed{\widehat f\in L^p(\mathbb R^n)\quad\text{for every }2\leq p\leq\infty}.
$$

Solved by gpt-5.6-sol high.

= 23F
{parent=Paper 3}
{scope}
{title2=Riemann Surfaces}

= a
{parent=23f}
{scope}

= Solution
{parent=a}

Choose a local coordinate $w$ with $w(p)=0$. Differentiation at $p$ gives a homomorphism
$$
\lambda:H\to\mathbb C^\times,
\qquad \lambda(h)=h'(p).
$$
It is injective. Indeed, a nonidentity <Möbius transformation> fixing $p$ and having derivative one there is parabolic, hence conjugate to a nonzero translation and of infinite order; no such element can belong to the finite group $H$. Every finite subgroup of $\mathbb C^\times$ is a cyclic group of roots of unity, so $H$ is cyclic. Write $n=|H|$.

The averaged coordinate
$$
z(q)=\frac1n\sum_{h\in H}\lambda(h)^{-1}w(hq)
$$
has derivative one at $p$, so it is a coordinate on a smaller neighbourhood. For $g\in H$, reindexing by $k=hg$ gives
$$
z(gq)=\lambda(g)z(q).
$$
After shrinking to an $H$-invariant disc $U$, the group therefore acts as all rotations $z\mapsto\zeta^jz$, where $\zeta$ is a primitive $n$th root of unity. The invariant coordinate
$$
u=z^n
$$
identifies $H\backslash U$ with a disc and gives it a <Riemann surfaces>[Riemann-surface] chart. In these coordinates the quotient map is exactly
$$
\boxed{z\longmapsto z^n}.
$$
This is the <local cyclic quotient of a Riemann surface>.

Solved by gpt-5.6-sol high.

= b
{parent=23f}
{scope}

= Solution
{parent=b}

Let $q\in\mathbb C_\infty$ and let $G_q$ be its <stabilizer subgroup>. Choose a sufficiently small disc $U$ about $q$ such that
$$
gU\cap U=\varnothing
\qquad\text{whenever }g\notin G_q.
$$
Part (a) says that $G_q$ is cyclic and supplies a coordinate in which $U\to G_q\backslash U$ is $z\mapsto z^{|G_q|}$. Since the global orbit map identifies no additional points within $U$, this is also a chart on a neighbourhood of the orbit $Gq$ in
$$
R=G\backslash\mathbb C_\infty.
$$

Use these quotient charts at every orbit. On overlaps, a local inverse branch of one power map followed by a group translate and the other power map gives the transition function; it is holomorphic. The charts therefore define a conformal structure on $R$, and the local expressions $z\mapsto z^{|G_q|}$ show directly that the quotient map $\mathbb C_\infty\to R$ is holomorphic. This is the <finite conformal quotient of a Riemann surface> construction.

Solved by gpt-5.6-sol high.

= c
{parent=23f}
{scope}

= Solution
{parent=c}

Let $\zeta=e^{2\pi i/n}$ and define Möbius transformations
$$
r(z)=\zeta z,
\qquad
s(z)=\frac1z.
$$
They satisfy
$$
r^n=s^2=1,
\qquad srs=r^{-1},
$$
and the $2n$ maps $r^k$ and $sr^k$ are distinct. They therefore give a faithful action of the <dihedral group> $D_{2n}$ on the <Riemann sphere>.

The <rational function>
$$
\boxed{
f(z)=z^n+z^{-n}=\frac{z^{2n}+1}{z^n}
}
$$
is invariant under both $r$ and $s$, so it is constant on every orbit. Conversely, for nonzero finite $z_1,z_2$, put $u=z_1^n$ and $v=z_2^n$. Equality of the function values gives
$$
u+u^{-1}=v+v^{-1}
\quad\Longleftrightarrow\quad
(u-v)(uv-1)=0.
$$
If $u=v$, then $z_2=\zeta^kz_1=r^kz_1$ for some $k$. If $uv=1$, then $z_2=\zeta^k/z_1=r^ksz_1$ for some $k$. Finally, zero and infinity both map to infinity and are interchanged by $s$. Hence
$$
\boxed{f(z_1)=f(z_2)\quad\Longleftrightarrow\quad z_1,z_2\text{ lie in the same }D_{2n}\text{-orbit}}.
$$
This is the <orbit-separating invariant for the standard dihedral action on the Riemann sphere>.

Solved by gpt-5.6-sol high.

= 24H
{parent=Paper 3}
{scope}
{title2=Algebraic Geometry}

= Solution
{parent=24H}

Let $X$ be an irreducible variety of dimension $d$. A point $p\in X$ is nonsingular, or smooth, when
$$
\dim T_pX=d,
$$
where $T_pX$ is the <Zariski tangent space>; it is a <singular point of an algebraic variety>[singular point] when $\dim T_pX>d$. Equivalently, the local ring $\mathcal O_{X,p}$ is regular exactly at a nonsingular point.

For an irreducible affine variety $X\subseteq\mathbb A^N$ over a perfect field, the <Jacobian criterion> expresses the singular locus by the vanishing of the relevant Jacobian minors, so it is <Zariski-closed set>[Zariski closed]. At a point where the tangent-space dimension is minimal it equals $d$; equivalently, some $(N-d)$-rowed Jacobian minor is nonzero. Its nonvanishing locus is therefore a nonempty <Zariski-open set> contained in the smooth locus. Every nonempty open subset of an <irreducible topological space> is dense, proving the <density of the smooth locus>.

Assume the ground field has characteristic other than two. For
$$
Q=V(X_0^2+\cdots+X_{n-1}^2)\subseteq\mathbb P^n,
$$
the partial derivatives are $2X_0,\ldots,2X_{n-1},0$. They vanish simultaneously at the unique projective point
$$
\boxed{[0:\cdots:0:1]}.
$$
Thus this point is the singular locus of the projective quadric cone.

For varieties $X,Y$,
$$
T_{(x,y)}(X\times Y)\cong T_xX\oplus T_yY,
\qquad
\dim(X\times Y)=\dim X+\dim Y.
$$
If $Y$ is smooth of dimension $m$, the product point $(x,y)$ is singular exactly when $x$ is singular. Hence
$$
\operatorname{Sing}(X\times Y)=Z\times Y
$$
and
$$
\boxed{\dim\operatorname{Sing}(X\times Y)=k+m}.
$$
This is the <singular locus of a product with a smooth variety>.

To construct the requested examples, put $r=n-k>0$. If $r\geq2$, let $Q_r\subseteq\mathbb P^{r+1}$ be an irreducible quadric cone of dimension $r$ with one singular vertex. If $r=1$, use instead the irreducible cuspidal cubic
$$
C=V(Y^2Z-X^3)\subseteq\mathbb P^2,
$$
whose only singular point is $[0:0:1]$. Let $W$ denote the chosen $r$-dimensional variety and embed
$$
W\times\mathbb P^k
$$
in projective space by the <Segre embedding>. It is irreducible and has dimension $r+k=n$, while its singular locus is the vertex or cusp point times $\mathbb P^k$, which is nonempty and has dimension exactly $k$. This is the <projective variety with a prescribed-dimensional singular locus> construction.

Finally, suppose that the irreducible plane curve $C\subseteq\mathbb P^2$ were smooth of degree $d$. Since it is birational to a smooth projective curve of genus two, its <geometric genus> would be two. But the <genus of a smooth plane curve> is
$$
g(C)=\frac{(d-1)(d-2)}2,
$$
and no integer $d$ makes this number equal to two. Therefore $C$ cannot be smooth and must contain a singular point.

Solved by gpt-5.6-sol high.

= 25I
{parent=Paper 3}
{scope}
{title2=Differential Geometry}

= a
{parent=25i}
{scope}

= Solution
{parent=a}

The <first fundamental form> of an embedded surface $S\subseteq\mathbb R^3$ is the inner product induced on each tangent plane:
$$
I_p(u,v)=u\cdot v,
\qquad u,v\in T_pS.
$$
For a tangent vector field $V$ along a curve $\alpha$, its <surface covariant derivative> is
$$
\frac{DV}{dt}=\left(\frac{dV}{dt}\right)^\top,
$$
the tangential projection of its ordinary derivative. A curve is a <geodesic> when
$$
\frac{D\dot\alpha}{dt}=0.
$$

A <local isometry> preserves the first fundamental form, so its Christoffel symbols and Levi--Civita covariant derivatives satisfy
$$
\nabla^R_{(\phi\circ\alpha)'}(D\phi V)
=D\phi(\nabla^S_{\alpha'}V).
$$
Taking $V=\alpha'$ shows that if $\alpha$ is geodesic, then $\phi\circ\alpha$ is geodesic.

The converse is false. The dilation
$$
\phi:\mathbb R^2\to\mathbb R^2,
\qquad \phi(x)=2x,
$$
maps every affinely parametrized straight-line geodesic to another such geodesic, but
$$
|D\phi(v)|=2|v|,
$$
so it does not preserve the first fundamental form. This is the <geodesic-preserving homothety that is not a local isometry>.

Solved by gpt-5.6-sol high.

= b
{parent=25i}
{scope}

= Solution
{parent=b}

The <Gaussian curvature> $K$ is the determinant of the shape operator, equivalently the product of the two principal curvatures. Yes,
$$
\boxed{K_R\circ\phi=K_S}.
$$
This is the <Theorema Egregium>: Gaussian curvature depends only on the first fundamental form, which a local isometry preserves.

Solved by gpt-5.6-sol high.

= c
{parent=25i}
{scope}

= i
{parent=c}
{scope}

= Solution
{parent=i}

Use angular coordinate $v$. The two surfaces have parametrizations
$$
R(u,v)=(e^u\cos v,e^u\sin v,u),
$$
and
$$
S(s,v)=(\cosh s\cos v,\cosh s\sin v,s).
$$
The supplied surface-of-revolution formula gives
$$
K_R(u)=-\frac1{(1+e^{2u})^2},
\qquad
K_S(s)=-\frac1{\cosh^4s}.
$$
Since $s>0$, the change of coordinate
$$
u=\log(\sinh s)
$$
is a diffeomorphism from $(0,\infty)$ to $\mathbb R$ and satisfies $1+e^{2u}=\cosh^2s$. Therefore
$$
\boxed{
\phi(S(s,v))=R(\log(\sinh s),v)
}
$$
is a diffeomorphism and obeys $K_R\circ\phi=K_S$. This is the <curvature-matching diffeomorphism between two surfaces of revolution>.

Solved by gpt-5.6-sol high.

= ii
{parent=c}
{scope}

= Solution
{parent=ii}

No such local isometry exists. The curvatures above are strictly monotone functions of $u$ and $s$, so curvature preservation would force the $R$-coordinate of any local isometry to be
$$
u(s,v)=\log(\sinh s).
$$
The first fundamental forms are
$$
I_R=(1+e^{2u})\,du^2+e^{2u}\,d\theta^2,
\qquad
I_S=\cosh^2s\,(ds^2+dv^2).
$$
Write the putative map as $(s,v)\mapsto(u(s),\theta(s,v))$. The mixed coefficient in its pullback is
$$
2\sinh^2s\,\theta_s\theta_v.
$$
A local isometry is locally nonsingular, so $\theta_v\ne0$; matching the zero mixed coefficient of $I_S$ therefore gives $\theta_s=0$. The $dv^2$ coefficients would then require
$$
\sinh^2s\,\theta_v^2=\cosh^2s,
\qquad
\theta_v^2=\coth^2s.
$$
The left side depends only on $v$, while the right side varies with $s$, an impossibility. Equal pointwise Gaussian curvature under a diffeomorphism therefore does not suffice for local isometry.

Solved by gpt-5.6-sol high.

= 26G
{parent=Paper 3}
{scope}
{title2=Probability and Measure}

= Solution
{parent=26G}

Convergence in distribution makes $(X_n)$ tight. Hence for every $\eta>0$ there is $M$ such that $\mathbb P(|X_n|>M)<\eta$ for all sufficiently large $n$. Since $Y_n-c\xrightarrow{P}0$,
$$
\mathbb P(|X_n(Y_n-c)|>\varepsilon)
\leq
\mathbb P(|X_n|>M)
+\mathbb P(|Y_n-c|>\varepsilon/M)
\longrightarrow\leq\eta.
$$
Letting $\eta\downarrow0$ gives $X_n(Y_n-c)\xrightarrow{P}0$. Also, the continuous-mapping theorem gives $cX_n\xrightarrow{d}cX$. Since
$$
X_nY_n=cX_n+X_n(Y_n-c),
$$
the <Slutsky theorem> yields
$$
\boxed{X_nY_n\xrightarrow{d}cX}.
$$

Now write the requested statistic as
$$
\frac{n^{-1/2}\sum_{i=1}^nZ_i}
{n^{-1}\sum_{i=1}^nZ_i^2}.
$$
The <central limit theorem> gives
$$
n^{-1/2}\sum_{i=1}^nZ_i\xrightarrow{d}N(0,1).
$$
Since $\mathbb E[Z_i^2]=1<\infty$, the <strong law of large numbers> gives
$$
\frac1n\sum_{i=1}^nZ_i^2\longrightarrow1
$$
almost surely, and therefore its reciprocal converges in probability to one. Applying the product result proves
$$
\boxed{
\frac{\sqrt n\sum_{i=1}^nZ_i}{\sum_{i=1}^nZ_i^2}
\xrightarrow{d}N(0,1)
}.
$$
This is a <self-normalized central limit theorem with a second-moment denominator>.

Solved by gpt-5.6-sol high.

= 27J
{parent=Paper 3}
{scope}
{title2=Applied Probability}

= a
{parent=27j}
{scope}

= Solution
{parent=a}

For independent identically distributed nonnegative random variables $\xi_1,\xi_2,\ldots$, define the renewal epochs
$$
S_0=0,
\qquad
S_n=\sum_{i=1}^n\xi_i.
$$
The associated <renewal process> is
$$
\boxed{N(t)=\max\{n:S_n\leq t\}},
$$
the number of renewals by time $t$.

Solved by gpt-5.6-sol high.

= b
{parent=27j}
{scope}

= Solution
{parent=b}

If $0<\mu=\mathbb E\xi_1<\infty$, the <size-biased distribution> of $\xi_1$ is the law of $\widehat\xi$ defined by
$$
\mathbb E h(\widehat\xi)
=\frac{\mathbb E[\xi_1h(\xi_1)]}{\mu}
$$
for bounded measurable $h$. Equivalently,
$$
\mathbb P(\widehat\xi\in dx)=\frac{x}{\mu}\,\mathbb P(\xi_1\in dx).
$$

Solved by gpt-5.6-sol high.

= c
{parent=27j}
{scope}

= Solution
{parent=c}

The <residual lifetime process>, or excess process, is
$$
\boxed{E(t)=S_{N(t)+1}-t}.
$$
If the inter-renewal distribution is non-arithmetic and $0<\mu=\mathbb E\xi_1<\infty$, then the <renewal excess limit theorem> says
$$
E(t)\xrightarrow{d}E_\infty,
\qquad
\boxed{
\mathbb P(E_\infty>x)
=\frac1\mu\int_x^\infty\mathbb P(\xi_1>y)\,dy
}.
$$
Equivalently, $E_\infty$ has the law of $U\widehat\xi$, where $U$ is uniform on $(0,1)$ and independent of the size-biased inter-renewal time $\widehat\xi$.

Solved by gpt-5.6-sol high.

= d
{parent=27j}
{scope}

= Solution
{parent=d}

Let the <Poisson process> $X$ have rate $\lambda$, let $\overline F(t)=\mathbb P(\xi_1>t)$ be the survival function of an inter-renewal time of $N$, and let $\overline H$ be the survival function of the first event time of $Y$. Independence gives
$$
\boxed{\overline H(t)=e^{-\lambda t}\overline F(t)}.
$$
Put
$$
\mu_F=\int_0^\infty\overline F(t)\,dt,
\qquad
\mu_H=\int_0^\infty\overline H(t)\,dt.
$$

At a large time, the excess of $Y$ is the minimum of the excesses of $X$ and $N$. The Poisson excess is an independent rate-$\lambda$ exponential random variable. Applying the renewal excess limit theorem to $N$ and to the assumed renewal process $Y$ gives
$$
\boxed{
\frac1{\mu_H}\int_x^\infty\overline H(y)\,dy
=\frac{e^{-\lambda x}}{\mu_F}
\int_x^\infty\overline F(y)\,dy
}.
$$
Substituting $\overline F(y)=e^{\lambda y}\overline H(y)$ yields the integral equation
$$
\frac1{\mu_H}\int_x^\infty\overline H(y)\,dy
=\frac{e^{-\lambda x}}{\mu_F}
\int_x^\infty e^{\lambda y}\overline H(y)\,dy.
$$

Set $A(x)=\int_x^\infty\overline H(y)dy$ and $c=\mu_H/\mu_F<1$. Differentiating the equation almost everywhere gives
$$
(1-c)\overline H(x)=\lambda A(x).
$$
Since $A'=-\overline H$, it follows that
$$
A'(x)=-\frac{\lambda}{1-c}A(x).
$$
Using $\overline H(0)=1$ shows that $\overline H(x)=e^{-x/\mu_H}$. Thus the first event time of $Y$ is exponential. This is the <exponential first interarrival in a renewal superposition containing a Poisson process> theorem.

Solved by gpt-5.6-sol high.

= 28K
{parent=Paper 3}
{scope}
{title2=Principles of Statistics}

= a
{parent=28k}
{scope}

= Solution
{parent=a}

Each leave-one-out statistic has the same distribution as $T_{n-1}$, so
$$
\mathbb E\widehat B_n
=(n-1)(B_{n-1}(\theta)-B_n(\theta)).
$$
Consequently the bias of the <jackknife bias correction> is
$$
\mathbb E\widetilde T_{\rm JACK}-\theta
=nB_n-(n-1)B_{n-1}.
$$
The assumed expansion gives
$$
nB_n=a+\frac bn+O(n^{-2})
$$
and
$$
(n-1)B_{n-1}
=a+\frac b{n-1}+O(n^{-2})
=a+\frac bn+O(n^{-2}).
$$
Their difference is therefore $O(n^{-2})$, as required.

Solved by gpt-5.6-sol high.

= b
{parent=28k}
{scope}

= Solution
{parent=b}

Since $\overline X_n\sim N(\mu,1/n)$,
$$
\mathbb E T_n=\mathbb E\overline X_n^2
=\mu^2+\frac1n,
$$
so
$$
\boxed{\operatorname{Bias}(T_n)=\frac1n}.
$$
Every leave-one-out mean has variance $1/(n-1)$, and hence
$$
\mathbb E\widehat B_n
=(n-1)\left(\frac1{n-1}-\frac1n\right)
=\frac1n.
$$
Therefore
$$
\boxed{\operatorname{Bias}(\widetilde T_{\rm JACK})=0}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=28k}
{scope}

= Solution
{parent=c}

The <central limit theorem> gives
$$
\sqrt n(\overline X_n-\mu)\xrightarrow{d}N(0,1).
$$
Applying the <delta method> to $g(x)=x^2$ yields
$$
\boxed{
\sqrt n(T_n-\mu^2)
\xrightarrow{d}N(0,4\mu^2)
}.
$$
For $\mu=0$ this denotes the degenerate distribution at zero, which is the correct limit at the $\sqrt n$ scale.

Solved by gpt-5.6-sol high.

= d
{parent=28k}
{scope}

= Solution
{parent=d}

The given identities imply
$$
\frac1n\sum_{i=1}^n
\left(\overline X_{n-1,i}^2-\overline X_n^2\right)
=\frac1{n(n-1)^2}
\sum_{i=1}^n(X_i-\overline X_n)^2.
$$
Thus, writing
$$
s_n^2=\frac1{n-1}\sum_{i=1}^n(X_i-\overline X_n)^2,
$$
we have the exact identity
$$
\widetilde T_{\rm JACK}
=T_n-\frac{s_n^2}{n}.
$$
The <sample variance> satisfies $s_n^2\xrightarrow{P}1$, so
$$
\sqrt n(\widetilde T_{\rm JACK}-T_n)
=-\frac{s_n^2}{\sqrt n}\xrightarrow{P}0.
$$
The <Slutsky theorem> and part (c) now give
$$
\boxed{
\sqrt n(\widetilde T_{\rm JACK}-\mu^2)
\xrightarrow{d}N(0,4\mu^2)
}.
$$

Solved by gpt-5.6-sol high.

= 29K
{parent=Paper 3}
{scope}
{title2=Stochastic Financial Models}

= a
{parent=29k}
{scope}

= Solution
{parent=a}

For $0\leq s\leq t$, the increment $W_t-W_s$ is independent of $\mathcal F_s$ and is $N(0,t-s)$. Hence
$$
\mathbb E(M_t\mid\mathcal F_s)
=M_s\,
\mathbb E\exp\left(c(W_t-W_s)-\frac{c^2}{2}(t-s)\right)
=M_s.
$$
Also $\mathbb E M_t=1$, so $M$ is the <exponential Brownian martingale>.

Solved by gpt-5.6-sol high.

= b
{parent=29k}
{scope}

= Solution
{parent=b}

Set
$$
\theta=\frac{\mu+\frac12\sigma^2-r}{\sigma}
$$
and define $Q$ on $\mathcal F_T$ by
$$
\frac{dQ}{dP}
=\exp\left(-\theta W_T-\frac12\theta^2T\right).
$$
The <Cameron-Martin theorem for a linear drift> says that
$$
W_t^Q=W_t+\theta t
$$
is a <Brownian motion> under $Q$. Therefore
$$
S_t=S_0
\exp\left(\left(r-\frac12\sigma^2\right)t+\sigma W_t^Q\right).
$$
The discounted stock is an exponential Brownian martingale, so $e^{-rt}S_t$ is a $Q$-martingale. Thus $Q$ is a <risk-neutral measure for the Black-Scholes model>.

Solved by gpt-5.6-sol high.

= c
{parent=29k}
{scope}

= Solution
{parent=c}

Risk-neutral valuation gives
$$
V_0=e^{-rT}\mathbb E_Q[S_T^p].
$$
Since $W_T^Q\sim N(0,T)$,
$$
\mathbb E_Q[S_T^p]
=S_0^p
\exp\left[p\left(r-\frac12\sigma^2\right)T
+\frac12p^2\sigma^2T\right].
$$
Hence the <power payoff in the Black-Scholes model> has price
$$
\boxed{
V_0=S_0^p
\exp\left(\left((p-1)r+\frac12p(p-1)\sigma^2\right)T\right)
}.
$$

Solved by gpt-5.6-sol high.

= d
{parent=29k}
{scope}

= Solution
{parent=d}

Under the risk-neutral measure, write
$$
S_t=S_0e^{X_t},
\qquad
X_t=\left(r-\frac12\sigma^2\right)t+\sigma W_t^Q.
$$
Let $M_T=\max_{t\leq T}X_t$ and $m_T=\min_{t\leq T}X_t$. Then
$$
Y_1=S_0e^{M_T},
\qquad
Y_2=S_0e^{pX_T-m_T}.
$$
The time-reversed increment process
$$
\widetilde X_t=X_T-X_{T-t},
\qquad0\leq t\leq T,
$$
has the same finite-dimensional distributions as $X$, because $X$ has stationary independent increments. Moreover,
$$
\max_{t\leq T}\widetilde X_t=X_T-m_T.
$$
Choosing
$$
\boxed{p=1}
$$
therefore makes $Y_2=S_0e^{X_T-m_T}$ equal in distribution to $Y_1$. Their discounted expectations, and hence their time-zero Black--Scholes prices, are equal. This is <Brownian time reversal for fixed-strike lookback extrema>.

Solved by gpt-5.6-sol high.

= 30E
{parent=Paper 3}
{scope}
{title2=Asymptotic Methods}

= a
{parent=30e}
{scope}

= Solution
{parent=a}

Put
$$
\phi(t)=t^3-2t^2+t=t(t-1)^2.
$$
On $[0,2]$, its global maximum is the endpoint $t=2$, where
$$
\phi(2)=2,
\qquad
\phi'(2)=5,
$$
and the amplitude $\log t$ has value $\log2$. The endpoint form of <Laplace's method> therefore gives
$$
\boxed{
I(x)\sim\frac{\log2}{5x}e^{2x}
}.
$$
Indeed, with $u=2-t$, Taylor expansion gives $\phi(2-u)=2-5u+O(u^2)$ and $\log(2-u)=\log2+O(u)$; the scale contributing to the integral is $u=O(x^{-1})$. Outside any fixed neighbourhood of $2$, $\phi$ is bounded above by $2-\delta$, so that part is exponentially smaller. This also controls the integrable logarithmic singularity at zero.

Solved by gpt-5.6-sol high.

= b
{parent=30e}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

Near zero, the absolute value of the integrand is comparable to
$$
t^{z-1}|\log t|,
$$
which is integrable for $z>0$. At infinity, the factor $e^{-t}$ dominates every power of $t$ and $\log t$, so the integral also converges there.

Direct use of Laplace's method is inconvenient because the phase
$$
z\log t-t
$$
does not have the form $z\phi(t)$ with a fixed phase: its stationary point $t=z$ moves to infinity as $z\to\infty$. Rescaling is needed to fix the maximum.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

Set $t=zs$. Then
$$
\Gamma'(z)
=e^{z\log z}
\int_0^\infty
\frac{\log z+\log s}{s}
e^{z(\log s-s)}\,ds.
$$
The fixed phase
$$
\phi(s)=\log s-s
$$
has its unique maximum at $s=1$, with
$$
\phi(1)=-1,
\qquad
\phi''(1)=-1.
$$
The amplitude at the maximum is $\log z$. The interior-maximum form of <Laplace's method> gives
$$
\Gamma'(z)
\sim e^{z\log z}e^{-z}\log z
\sqrt{\frac{2\pi}{z}}.
$$
Thus
$$
\boxed{
\Gamma'(z)\sim
\sqrt{\frac{2\pi}{z}},e^{z\log z-z}\log z
},
$$
so $\boxed{a=2\pi}$. This is the <Laplace asymptotic for the derivative of the Gamma function>.

Solved by gpt-5.6-sol high.

= 31B
{parent=Paper 3}
{scope}
{title2=Dynamical Systems}

= a
{parent=31b}
{scope}

= Solution
{parent=a}

The Jacobian at the origin is
$$
A=\begin{pmatrix}-a&3\\-1&-a\end{pmatrix},
$$
whose eigenvalues are
$$
-a\pm i\sqrt3.
$$
Their real parts are negative because $a>0$. The <linearization stability theorem> therefore makes the origin a locally asymptotically stable hyperbolic sink.

Solved by gpt-5.6-sol high.

= b
{parent=31b}
{scope}

= Solution
{parent=b}

A <Lyapunov function> for an equilibrium is a continuously differentiable function that is positive definite there and whose derivative along nonconstant trajectories is negative definite in a neighbourhood.

For $V=x^2+ky^2$, positivity requires $k>0$, and
$$
\dot V
=-2ax^2-2aky^2+2(3-k)xy
+2(x^2+y^2)(x^2+ky^2).
$$
The quartic term is negligible sufficiently close to the origin, so $V$ is a local Lyapunov function exactly when
$$
ax^2+aky^2-(3-k)xy
$$
is positive definite. Its determinant condition is
$$
4a^2k>(3-k)^2.
$$
The two roots of equality are
$$
k_{1,2}(a)=3+2a^2\mp2a\sqrt{a^2+3}
=\left(\sqrt{a^2+3}\mp a\right)^2.
$$
Therefore
$$
\boxed{
(\sqrt{a^2+3}-a)^2<k<(\sqrt{a^2+3}+a)^2
}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=31b}
{scope}

= Solution
{parent=c}

For $k=1$, the interval in part (b) contains one exactly when $a>1$. With $V=x^2+y^2=r^2$,
$$
\dot V=-2ar^2+4xy+2r^4
\leq2r^2(r^2-a+1),
$$
because $2xy\leq r^2$. Hence $\dot V<0$ throughout
$$
0<r^2<a-1.
$$
Every trajectory starting in the disc $x^2+y^2<a-1$ remains there and approaches the origin. Thus this disc is contained in the <basin of attraction> of the origin.

Solved by gpt-5.6-sol high.

= d
{parent=31b}
{scope}

= Solution
{parent=d}

The <Poincare-Bendixson theorem> states that a nonempty compact $\omega$-limit set of a planar $C^1$ flow which contains no equilibrium is a periodic orbit.

The origin is the only equilibrium: at a nonzero equilibrium, putting $q=x^2+y^2-a$ would require
$$
\begin{pmatrix}q&3\\-1&q\end{pmatrix}
\binom xy=0,
$$
but its determinant $q^2+3$ is positive. By local asymptotic stability, choose a small simple closed Lyapunov level curve on which the original vector field points inward. On a sufficiently large circle,
$$
\frac12\frac d{dt}(x^2+y^2)
=(x^2+y^2)^2-a(x^2+y^2)+2xy>0,
$$
so the original vector field points outward.

Reverse time. The annulus between these two curves is then a compact positively invariant <trapping region>: the reversed field points into it at both boundaries. It contains no equilibrium. The Poincaré--Bendixson theorem applied to any reversed trajectory in the annulus produces a periodic orbit. Reversing time does not change its image, so the original system also has a periodic orbit.

Solved by gpt-5.6-sol high.

= 32E
{parent=Paper 3}
{scope}
{title2=Integrable Systems}

= Solution
{parent=32E}

Write $\xi=V_1$ and $\eta=\phi$. The vector field
$$
V=\xi(x,u)\partial_x+\eta(x,u)\partial_u
$$
generates a <Lie point symmetry of an ordinary differential equation> if its local flow maps solution graphs to solution graphs. Its $n$th <prolongation of a vector field> is
$$
\operatorname{pr}^{(n)}V
=V+\sum_{j=1}^n\eta^{(j)}\partial_{u^{(j)}},
$$
where
$$
\eta^{(0)}=\eta,
\qquad
\eta^{(j)}=D_x\eta^{(j-1)}-u^{(j)}D_x\xi
$$
and
$$
D_x=\partial_x+u'\partial_u+u''\partial_{u'}+\cdots
$$
is the <total derivative operator>. The infinitesimal invariance criterion is
$$
\boxed{
\operatorname{pr}^{(n)}V(\Delta)=0
\quad\text{whenever }\Delta=0
}.
$$

Put $p=u'$ and $q=u''$. Direct calculation gives
$$
\eta^{(1)}
=\eta_x+(\eta_u-\xi_x)p-\xi_up^2
$$
and
$$
\eta^{(2)}
=\eta_{xx}+(2\eta_{xu}-\xi_{xx})p
+(\eta_{uu}-2\xi_{xu})p^2-\xi_{uu}p^3
+(\eta_u-2\xi_x-3\xi_up)q.
$$
For
$$
\Delta=q-\frac{p^2}{u}+u^2,
$$
the invariance condition is
$$
\eta^{(2)}
-\frac{2p}{u}\eta^{(1)}
+\frac{p^2}{u^2}\eta
+2u\eta=0
$$
after imposing $q=p^2/u-u^2$. The coefficient of $p^3$ is
$$
-\xi_{uu}-\frac1u\xi_u.
$$
It must vanish identically, so
$$
\partial_u(u\xi_u)=0.
$$
Since $u>0$, integration gives
$$
\boxed{\xi(x,u)=F(x)\log u+G(x)}
$$
for functions $F,G$.

Now take
$$
\xi=cx+d,
\qquad
\eta=-2cu.
$$
Then
$$
\eta^{(1)}=-3cp,
\qquad
\eta^{(2)}=-4cq,
$$
and substitution gives
$$
\operatorname{pr}^{(2)}V(\Delta)
=-4c\left(q-\frac{p^2}{u}+u^2\right)
=-4c\Delta.
$$
Thus the infinitesimal invariance criterion holds.

The generated one-parameter transformation is, for $c\ne0$,
$$
\boxed{
\widetilde x=e^{c\varepsilon}\left(x+\frac dc\right)-\frac dc,
\qquad
\widetilde u=e^{-2c\varepsilon}u
},
$$
while for $c=0$ it is $\widetilde x=x+d\varepsilon$, $\widetilde u=u$. Combining these transformations gives the two-parameter symmetry group
$$
\boxed{
(x,u)\longmapsto(\lambda x+a,\lambda^{-2}u),
\qquad \lambda>0,\ a\in\mathbb R
}.
$$
This is the <affine-scaling symmetry of u double prime equals u prime squared over u minus u squared>.

Solved by gpt-5.6-sol high.

= 33A
{parent=Paper 3}
{scope}
{title2=Principles of Quantum Mechanics}

= a
{parent=33a}
{scope}

= Solution
{parent=a}

In position representation,
$$
L_z=-i\hbar\left(x\partial_y-y\partial_x\right),
$$
which differentiates only with respect to $x$ and $y$. Multiplication by $z$ therefore commutes with it:
$$
\boxed{[L_z,z]=0}.
$$
Taking the matrix element of this commutator between angular-momentum eigenstates gives
$$
0=\hbar(m'-m)
\langle n',\ell',m'|z|n,\ell,m\rangle.
$$
Thus the matrix element can be nonzero only when
$$
\boxed{m'=m}.
$$
This is the magnetic part of the <electric-dipole selection rules for hydrogen>.

Solved by gpt-5.6-sol high.

= b
{parent=33a}
{scope}

= Solution
{parent=b}

Write
$$
\Lambda_\ell=\ell(\ell+1).
$$
Taking the stated double-commutator identity between the two eigenstates gives, if the matrix element is nonzero,
$$
\left(\Lambda_{\ell'}-\Lambda_\ell\right)^2
=2\left(\Lambda_{\ell'}+\Lambda_\ell\right).
$$
For nonnegative integers $\ell,\ell'$, this equation has precisely the possibilities
$$
|\ell'-\ell|=1
\qquad\text{or}\qquad
\ell'=\ell=0.
$$
Indeed, $\ell'=\ell$ forces $\ell=0$, while for unequal values the difference
$(\ell'-\ell)(\ell'+\ell+1)$ satisfies the equation only when its first factor has magnitude one.

Under parity, $z\mapsto-z$, while a hydrogen state of orbital angular momentum $\ell$ has parity $(-1)^\ell$. Therefore a nonzero matrix element requires
$$
(-1)^{\ell'+\ell+1}=1,
$$
so $\ell'$ and $\ell$ must have opposite parity. This excludes every case $\ell'=\ell$, including $\ell'=\ell=0$. Combining the results gives
$$
\boxed{\Delta\ell=\pm1,\qquad\Delta m=0}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=33a}
{scope}

= Solution
{parent=c}

The hydrogen ground-state charge cloud is spherically symmetric and has even parity, so it has no permanent <electric dipole moment>. A uniform electric field couples through a perturbation proportional to $\mathbf E\cdot\mathbf r$, which is odd under parity. Its expectation in the ground state therefore vanishes, giving no first-order energy correction.

The field can nevertheless polarize the atom: it admixes odd-parity excited states and creates an induced dipole. Such mixing contributes from second order onward, producing the quadratic <Stark effect of the hydrogen ground state>.

Solved by gpt-5.6-sol high.

= d
{parent=33a}
{scope}

= Solution
{parent=d}

Choose the $z$-axis along $\mathbf E$, so the perturbation is
$$
\Delta H=e|\mathbf E|z
$$
up to an irrelevant sign convention for the electron charge. By <second-order nondegenerate perturbation theory>,
$$
E_{100}^{(2)}
=e^2|\mathbf E|^2
\sum_{\alpha\ne100}
\frac{|\langle\alpha|z|1,0,0\rangle|^2}
{E_1-E_\alpha}.
$$
Parts (a) and (b) restrict the contributing bound states to
$$
|n,1,0\rangle,\qquad n\geq2.
$$
For hydrogen,
$$
E_n=-\frac R{n^2},
$$
and hence
$$
\frac1{E_1-E_n}
=\frac1{-R+R/n^2}
=\frac1R\frac{n^2}{1-n^2}.
$$
Therefore
$$
\boxed{
E_{100}^{(2)}
=\frac{e^2|\mathbf E|^2}{R}
\sum_{n=2}^{\infty}
\frac{n^2}{1-n^2}
|\langle n,1,0|z|1,0,0\rangle|^2
}.
$$
Every denominator is negative, so the discrete-state contribution lowers the ground-state energy.

Solved by gpt-5.6-sol high.

= 34D
{parent=Paper 3}
{scope}
{title2=Applications of Quantum Mechanics}

= Solution
{parent=34D}

Every primitive basis of the <Bravais lattice> has the form
$$
\boxed{
a_1'=p a_1+q a_2,
\qquad
a_2'=r a_1+s a_2
},
$$
where $p,q,r,s\in\mathbb Z$ and
$$
\boxed{ps-qr=\pm1}.
$$
The unimodular determinant condition is exactly what makes the integer change of basis invertible over $\mathbb Z$.

Let $b_i$ be the primitive vectors of the <reciprocal lattice>, with
$$
a_i\cdot b_j=2\pi\delta_{ij}.
$$
Solving these four equations gives
$$
\boxed{
b_1=\left(2\pi,\frac{2\pi}{\sqrt3}\right),
\qquad
b_2=\left(0,\frac{4\pi}{\sqrt3}\right)
}.
$$
The six reciprocal-lattice points nearest the origin are
$$
\pm b_1,\qquad
\pm b_2,\qquad
\pm(b_2-b_1),
$$
all at distance $4\pi/\sqrt3$.

The <Wigner-Seitz cell> is bounded by the perpendicular bisectors of the segments joining the origin to those six points. It is the regular hexagon with vertices
$$
\boxed{
\left(\frac{4\pi}{3},0\right),
\left(\frac{2\pi}{3},\frac{2\pi}{\sqrt3}\right),
\left(-\frac{2\pi}{3},\frac{2\pi}{\sqrt3}\right),
\left(-\frac{4\pi}{3},0\right),
\left(-\frac{2\pi}{3},-\frac{2\pi}{\sqrt3}\right),
\left(\frac{2\pi}{3},-\frac{2\pi}{\sqrt3}\right)
}.
$$
Its area equals the reciprocal primitive-cell area:
$$
\boxed{
|b_1\times b_2|=\frac{8\pi^2}{\sqrt3}
}.
$$
This is the <reciprocal lattice and Wigner-Seitz cell of the unit triangular lattice>.

For a periodic potential, <Bloch theorem> identifies wavevectors differing by a reciprocal-lattice vector. The first <Brillouin zone> is the Wigner--Seitz cell just found. More generally, the $n$th Brillouin zone consists of wavevectors reached from the origin after crossing exactly $n-1$ reciprocal-lattice Bragg planes; its boundaries are the perpendicular bisectors
$$
k\cdot G=\frac{|G|^2}{2},
\qquad G\in\Lambda^*\setminus\{0\}.
$$
For this triangular reciprocal lattice, the first zone is the central regular hexagon. The second is the sixfold-symmetric collection of regions immediately outside its six sides, bounded next by the bisectors associated with the next reciprocal points. These zones organize free-particle states into bands: Bragg coupling is strongest at their boundaries and opens energy gaps when the periodic potential is introduced.

Solved by gpt-5.6-sol high.

= 35A
{parent=Paper 3}
{scope}
{title2=Statistical Physics}

= a
{parent=35a}
{scope}

= Solution
{parent=a}

<Bosons> have symmetric many-particle states, so any number may occupy the same one-particle state. At low <temperature>, an ideal Bose gas therefore places a macroscopic population in its lowest one-particle state; below the critical temperature this is <Bose-Einstein condensation>.

<Fermions> have antisymmetric many-particle states and obey the <Pauli exclusion principle>, so each one-particle state can hold at most one fermion of each available internal label. At zero temperature, a noninteracting Fermi gas fills distinct states in increasing order of <energy> up to the <Fermi energy>. Its many-particle ground state is thus the filled <Fermi sea>.

Solved by gpt-5.6-sol high.

= b
{parent=35a}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

For an <ultrarelativistic particle>, $\varepsilon=pc$. Including the two <spin angular momentum>[spin] states of an electron, counting wavevectors in a sphere gives the <density of states>
$$
g(\varepsilon)
=\frac{V\varepsilon^2}{\pi^2\hbar^3c^3}.
$$
The <grand canonical partition function> of the ideal <Fermi gas> is therefore
$$
\log\mathcal Z
=\int_0^\infty g(\varepsilon)
\log\!\left(1+e^{-\beta(\varepsilon-\mu)}\right)d\varepsilon.
$$
Since the <grand potential> satisfies $pV=k_BT\log\mathcal Z$, <integration by parts> gives
$$
pV
=\frac13\int_0^\infty
\varepsilon g(\varepsilon)
\frac{d\varepsilon}{e^{\beta(\varepsilon-\mu)}+1}
=\frac E3.
$$
Thus
$$
\boxed{pV=AE,\qquad A=\frac13}.
$$
This is the <equation of state of an ideal ultrarelativistic gas>.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

At zero temperature the <Fermi-Dirac distribution> is the indicator of $0\leq\varepsilon<E_F$. Hence
$$
N=\int_0^{E_F}g(\varepsilon)\,d\varepsilon
=\frac{VE_F^3}{3\pi^2\hbar^3c^3}.
$$
Solving for the <Fermi energy> gives
$$
\boxed{
E_F=D\left(\frac NV\right)^{1/3},
\qquad
D=\hbar c(3\pi^2)^{1/3}
}.
$$

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

The zero-temperature <internal energy> is
$$
E=\int_0^{E_F}\varepsilon g(\varepsilon)\,d\varepsilon
=\frac{VE_F^4}{4\pi^2\hbar^3c^3}
=\frac34NE_F.
$$
Using part (i),
$$
p=\frac{NE_F}{4V}
=\frac D4N^{4/3}V^{-4/3}.
$$
Consequently the <adiabatic equation of state> is
$$
\boxed{
pV^a=K,
\qquad
a=\frac43,
\qquad
K=\frac{\hbar c(3\pi^2)^{1/3}}4N^{4/3}
}.
$$
The pressure is <degeneracy pressure>: it remains nonzero at zero temperature because of the <Pauli exclusion principle>. By comparison, a classical ultrarelativistic <ideal gas> at fixed temperature obeys $pV=Nk_BT$, so its corresponding volume exponent is one and its pressure vanishes as $T\to0$.

Solved by gpt-5.6-sol high.

= c
{parent=35a}
{scope}

= Solution
{parent=c}

Let $g(E)$ denote the total <density of states>, including both <spin angular momentum>[spin] directions. The field lowers one spin energy by $\mu_BB$ and raises the other by the same amount. To first order in $B$, each spin population changes by the one-spin density $g(E_F)/2$ times the displacement of its Fermi edge. The population imbalance is therefore
$$
N_+-N_-
=2\mu_BB\frac{g(E_F)}2
=\mu_BB\,g(E_F).
$$
Multiplying by the <Bohr magneton> carried by each excess aligned electron gives
$$
\boxed{
M=\mu_B^2B\,g(E_F)
}.
$$
Thus, with this total-density convention,
$$
\boxed{\alpha=1,\qquad\gamma=2,\qquad\delta=1}.
$$
The zero-temperature <magnetic susceptibility> is
$$
\boxed{\chi=\frac{\partial M}{\partial B}
=\mu_B^2g(E_F)>0}.
$$
This weak positive response is <Pauli paramagnetism>. If susceptibility per unit volume is required, both $M$ and $g$ are divided by $V$.

Solved by gpt-5.6-sol high.

= 36B
{parent=Paper 3}
{scope}
{title2=Electrodynamics}

= a
{parent=36b}
{scope}

= Solution
{parent=a}

The <Lorenz gauge> is $\partial_\mu A^\mu=0$. Substituting the <electromagnetic four-potential> into the covariant <Maxwell equations> then removes the mixed derivative and gives
$$
\boxed{\Box A^\mu=\mu_0J^\mu},
\qquad
\Box=\frac1{c^2}\partial_t^2-\nabla^2.
$$
Equivalently,
$$
\left(\nabla^2-\frac1{c^2}\partial_t^2\right)A^\mu=-\mu_0J^\mu.
$$

Solved by gpt-5.6-sol high.

= b
{parent=36b}
{scope}

= Solution
{parent=b}

The outgoing Green function of the <electromagnetic wave equation> has support on the past light cone. Convolution with that <causal Green function>[retarded Green function] yields the <retarded electromagnetic potential>
$$
\boxed{
A^\mu(\mathbf x,t)
=\frac{\mu_0}{4\pi}
\int_{\mathbb R^3}
\frac{J^\mu(\mathbf x',t')}{|\mathbf x-\mathbf x'|}
\,d^3x'
},
$$
where the source is evaluated at the <retarded time>
$$
\boxed{
t'=t-\frac{|\mathbf x-\mathbf x'|}{c}
}.
$$
This is the causal solution with no incoming radiation.

Solved by gpt-5.6-sol high.

= c
{parent=36b}
{scope}

= Solution
{parent=c}

The <electric dipole moment> of a localized charge density is
$$
\mathbf p(t)=\int_{\mathbb R^3}\mathbf x\,\rho(\mathbf x,t)\,d^3x.
$$
Differentiate and use the <continuity equation> $\dot\rho+\nabla\cdot\mathbf J=0$:
$$
\dot p_i
=-\int x_i\partial_jJ_j\,d^3x
=\int J_i\,d^3x
-\int_{\partial\mathbb R^3}x_iJ_jn_j\,dS.
$$
Localization makes the boundary integral vanish. Component by component,
$$
\boxed{\dot{\mathbf p}(t)=\int_{\mathbb R^3}\mathbf J(\mathbf x,t)\,d^3x}.
$$
This is the <time derivative of the electric dipole moment>.

Solved by gpt-5.6-sol high.

= d
{parent=36b}
{scope}

= Solution
{parent=d}

Let $R=|\mathbf x|$, $\widehat{\mathbf R}=\mathbf x/R$, and $t_R=t-R/c$. To leading nonzero <electromagnetic multipole expansion>[multipole expansion] order, the loop is a time-dependent <magnetic dipole moment>[magnetic dipole] with
$$
\mathbf m(t)=\pi r^2I(t)\mathbf n.
$$
Expanding the <retarded electromagnetic potential> across the small loop gives
$$
\mathbf A(\mathbf x,t)
=\frac{\mu_0}{4\pi}
\left[
\frac{\mathbf m(t_R)\times\widehat{\mathbf R}}{R^2}
+\frac{\dot{\mathbf m}(t_R)\times\widehat{\mathbf R}}{cR}
\right]
+O\!\left(\frac{r^3}{R^3}\right).
$$
The $R^{-2}$ term is the near magnetic-dipole field and the $R^{-1}$ term is the <magnetic dipole radiation> field. Substituting the Fourier series for the current gives
$$
\boxed{
\mathbf A(\mathbf x,t)
=\frac{\mu_0r^2}{4}
(\mathbf n\times\widehat{\mathbf R})
\sum_{n=0}^{\infty}I_n
\left[
\frac{\sin(n\omega t_R)}{R^2}
+\frac{n\omega}{cR}\cos(n\omega t_R)
\right]
}.
$$
The $n=0$ summand vanishes for the stated sine series.

Solved by gpt-5.6-sol high.

= 37D
{parent=Paper 3}
{scope}
{title2=General Relativity}

= a
{parent=37d}
{scope}

= Solution
{parent=a}

Write
$$
f(r)=1-\frac{2M}{r}.
$$
For a static observer, $d\tau=\sqrt{f(r)}\,dt$. Conservation of the photon's Killing energy therefore gives the <gravitational redshift between static Schwarzschild observers>
$$
\frac{\lambda_{\rm obs}}{\lambda}
=\sqrt{\frac{f(r)}{f(r_{\rm st})}}.
$$
For the stated distant observer, $r\gg r_{\rm st}$ and $f(r)\simeq1$, so
$$
\boxed{
\lambda_{\rm obs}
\simeq\frac{\lambda}{\sqrt{1-2M/r_{\rm st}}}
}.
$$
As $r_{\rm st}\to2M^+$, the observed wavelength tends to infinity: light emitted from a static source arbitrarily near the <Schwarzschild event horizon> is infinitely redshifted at infinity.

Solved by gpt-5.6-sol high.

= b
{parent=37d}
{scope}

= Solution
{parent=b}

A static worldline has <four-velocity>
$$
u^\mu=(f^{-1/2},0,0,0).
$$
Its <four-acceleration> is $a^\mu=u^\nu\nabla_\nu u^\mu$. The only nonzero component is
$$
a^r=\Gamma^r_{tt}(u^t)^2=\frac{M}{r^2}.
$$
Taking its norm with $g_{rr}=f^{-1}$ gives the <proper acceleration of a static Schwarzschild observer>
$$
\boxed{
|a|=\sqrt{a_\mu a^\mu}
=\frac{M}{r_{\rm st}^2\sqrt{1-2M/r_{\rm st}}}
}.
$$
It points radially outward, because that is the acceleration needed to prevent the station from falling. As $r_{\rm st}\to\infty$ the square-root factor tends to one, and $|a|\sim M/r_{\rm st}^2$, the <Newtonian gravitational field> in units $G=1$.

Solved by gpt-5.6-sol high.

= c
{parent=37d}
{scope}

= Solution
{parent=c}

The station's proper rest mass is
$$
M_{\rm st}=4\pi r_{\rm st}^2\rho.
$$
A static local energy $dE_{\rm loc}$ at radius $r_{\rm st}$ contributes the redshifted energy
$$
dE_\infty=\sqrt{1-\frac{2M}{r_{\rm st}}}\,dE_{\rm loc}
$$
to the mass measured at infinity. To first order in $M_{\rm st}/M$, the exterior mass parameter is therefore
$$
\boxed{
M_{\rm out}
=M+4\pi r_{\rm st}^2\rho
\sqrt{1-\frac{2M}{r_{\rm st}}}
}.
$$
Terms quadratic in $\rho$ describe the shell's self-gravity and are neglected.

The exterior of the shell is vacuum and spherically symmetric. The <Birkhoff theorem> therefore makes it a <Schwarzschild spacetime> with mass $M_{\rm out}$:
$$
\boxed{
ds^2
=-\left(1-\frac{2M_{\rm out}}r\right)dt^2
+\left(1-\frac{2M_{\rm out}}r\right)^{-1}dr^2
+r^2d\Omega^2,
\qquad r>r_{\rm st}
}.
$$
This is the <redshifted gravitational mass of a static thin shell>.

Solved by gpt-5.6-sol high.

= 38C
{parent=Paper 3}
{scope}
{title2=Fluid Dynamics II}

= a
{parent=38c}
{scope}

= Solution
{parent=a}

Choose the vertical plane through the rod's axis. Reflection in that plane leaves the horizontal rod, the vertical gravitational force, and the <Stokes flow>[Stokes equations] unchanged, but reverses any velocity normal to the plane. A second reflection in the vertical plane perpendicular to the rod reverses velocity along the rod. By uniqueness of <Stokes flow>, both horizontal velocity components vanish. The rod therefore translates vertically.

Equivalently, the <hydrodynamic resistance matrix> of an axisymmetric rod has one coefficient parallel to its axis and another perpendicular to it. A vertical force on a horizontal rod lies entirely in the perpendicular eigenspace, so its velocity is vertical.

Solved by gpt-5.6-sol high.

= b
{parent=38c}
{scope}

= Solution
{parent=b}

The uniform rod is invariant under inversion through its centre. This centrosymmetry makes the translation-rotation blocks of its generalized <hydrodynamic resistance matrix> vanish: a polar translation velocity changes sign under inversion, whereas an axial angular velocity does not, so an invariant linear coupling between them is impossible.

Gravity acts through the centre of mass and supplies no <torque>. The remaining rotational resistance is positive definite, so zero applied torque implies zero <angular velocity>. Thus an initially inclined rod keeps its orientation while settling.

Solved by gpt-5.6-sol high.

= c
{parent=38c}
{scope}

= Solution
{parent=c}

At negligible <Reynolds number>, the <Stokes flow>[Stokes equations] and their no-slip boundary conditions are linear in velocity. Scaling or superposing a prescribed rigid translation therefore scales or superposes the resulting stress and total hydrodynamic force. Hence, for a fixed rod orientation,
$$
\boxed{\mathbf F=-\mathbf R\cdot\mathbf U}.
$$
The minus sign records that viscous drag opposes motion, and $\mathbf R$ is the translational <hydrodynamic resistance matrix>.

Solved by gpt-5.6-sol high.

= d
{parent=38c}
{scope}

= Solution
{parent=d}

The <Lorentz reciprocal theorem for Stokes flow> states that two Stokes solutions $(\mathbf u^{(1)},\boldsymbol\sigma^{(1)})$ and $(\mathbf u^{(2)},\boldsymbol\sigma^{(2)})$ in the same fluid domain satisfy
$$
\int_{\partial\mathcal D}
\mathbf u^{(1)}\cdot\boldsymbol\sigma^{(2)}\mathbf n\,dS
=
\int_{\partial\mathcal D}
\mathbf u^{(2)}\cdot\boldsymbol\sigma^{(1)}\mathbf n\,dS.
$$
Apply it to the same rod translating with arbitrary velocities $\mathbf U_1$ and $\mathbf U_2$. The integral at infinity vanishes, and the rod-surface integrals give
$$
\mathbf U_1\cdot\mathbf F_2
=\mathbf U_2\cdot\mathbf F_1.
$$
Since $\mathbf F_i=-\mathbf R\mathbf U_i$,
$$
\mathbf U_1\cdot\mathbf R\mathbf U_2
=\mathbf U_2\cdot\mathbf R\mathbf U_1
$$
for every pair of vectors. Therefore
$$
\boxed{\mathbf R=\mathbf R^T}.
$$

Solved by gpt-5.6-sol high.

= e
{parent=38c}
{scope}

= Solution
{parent=e}

The power supplied to overcome drag equals the <viscous dissipation>:
$$
-\mathbf F\cdot\mathbf U
=2\mu\int_{\mathcal D}\mathbf e:\mathbf e\,dV,
\qquad
\mathbf e=\frac12\left(\nabla\mathbf u+\nabla\mathbf u^T\right).
$$
For a nonzero rigid translation the strain cannot vanish throughout a fluid at rest at infinity, so the integral is strictly positive. Since $\mathbf F=-\mathbf R\mathbf U$,
$$
\boxed{\mathbf U\cdot\mathbf R\mathbf U>0
\quad\text{for every }\mathbf U\ne0}.
$$
Thus the symmetric resistance matrix is <positive-definite matrix>[positive definite].

Solved by gpt-5.6-sol high.

= f
{parent=38c}
{scope}

= Solution
{parent=f}

Let $\mathbf e$ be the downward vertical unit vector and let the rod's weight after buoyancy be $W\mathbf e$. Force balance gives
$$
\mathbf R\mathbf U=W\mathbf e.
$$
Because
$$
\mathbf R
=c_1\mathbf t\mathbf t
+c_2(\mathbf 1-\mathbf t\mathbf t),
$$
its <matrix inverse> is
$$
\mathbf R^{-1}
=\frac1{c_1}\mathbf t\mathbf t
+\frac1{c_2}(\mathbf 1-\mathbf t\mathbf t).
$$
Using $\mathbf e\cdot\mathbf t=\cos\theta$ gives
$$
\mathbf U
=W\left[
\frac{\cos\theta}{c_1}\mathbf t
+\frac1{c_2}\left(\mathbf e-\cos\theta\,\mathbf t\right)
\right].
$$
Therefore
$$
\mathbf e\cdot\mathbf U
=W\left(\frac{\cos^2\theta}{c_1}
+\frac{\sin^2\theta}{c_2}\right),
$$
and
$$
|\mathbf U|
=W\sqrt{
\frac{\cos^2\theta}{c_1^2}
+\frac{\sin^2\theta}{c_2^2}
}.
$$
The required angle is consequently
$$
\boxed{
\cos\alpha
=
\frac{\cos^2\theta/c_1+\sin^2\theta/c_2}
{\sqrt{\cos^2\theta/c_1^2+\sin^2\theta/c_2^2}}
}.
$$
If $c_1=c_2$, then $\cos\alpha=1$, so the rod falls vertically. For an anisotropic rod with $c_2>c_1$, the component parallel to its long axis encounters less drag, and the settling direction is generally deflected toward that axis.

Solved by gpt-5.6-sol high.

= 39C
{parent=Paper 3}
{scope}
{title2=Waves}

= a
{parent=39c}
{scope}

= Solution
{parent=a}

Assume the medium varies on scales much longer than the wavelength and period, so the leading-order <WKB method> gives the local <eikonal equation>. For the phase $\theta$ in the stated ansatz, define
$$
\mathbf k=\nabla\theta,
\qquad
\omega=-\partial_t\theta.
$$
The local <dispersion relation> becomes the Hamilton--Jacobi equation
$$
-\partial_t\theta
=\Omega(\nabla\theta;\mathbf x,t).
$$
Differentiating it with respect to $x_i$, using equality of mixed derivatives, and choosing a ray whose velocity is
$$
\dot x_i=\frac{\partial\Omega}{\partial k_i}
$$
gives
$$
\dot k_i
=\partial_tk_i+\dot x_j\partial_{x_j}k_i
=-\frac{\partial\Omega}{\partial x_i}.
$$
Here the dot, or $d/dt$, is the <total derivative along a ray>,
$$
\frac d{dt}
=\frac{\partial}{\partial t}
+\dot{\mathbf x}\cdot\nabla_{\mathbf x},
$$
and not merely a partial time derivative at fixed position.

Finally, differentiate $\omega=\Omega(\mathbf k;\mathbf x,t)$ along the same trajectory:
$$
\frac{d\omega}{dt}
=\frac{\partial\Omega}{\partial t}
+\frac{\partial\Omega}{\partial x_i}\dot x_i
+\frac{\partial\Omega}{\partial k_i}\dot k_i
=\frac{\partial\Omega}{\partial t},
$$
because the last two terms cancel. The <Hamiltonian ray-tracing equations> are therefore
$$
\boxed{
\frac{dx_i}{dt}=\frac{\partial\Omega}{\partial k_i},
\qquad
\frac{dk_i}{dt}=-\frac{\partial\Omega}{\partial x_i},
\qquad
\frac{d\omega}{dt}=\frac{\partial\Omega}{\partial t}
}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=39c}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

For the given <internal gravity wave>,
$$
\Omega(k,m;x,z)
=\frac{Nk}{\sqrt{k^2+m^2}}
+\gamma(kx-mz).
$$
The wavevector part of the <Hamiltonian ray-tracing equations> is
$$
\dot k=-\frac{\partial\Omega}{\partial x}=-\gamma k,
\qquad
\dot m=-\frac{\partial\Omega}{\partial z}=\gamma m.
$$
Applying the initial data gives
$$
\boxed{
k(t)=k_0e^{-\gamma t},
\qquad
m(t)=m_0e^{\gamma t}
}.
$$
In particular, $k(t)m(t)=k_0m_0$ is conserved.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

The vertical <phase velocity and group velocity>[group velocity] is the $m$ derivative of the local dispersion relation:
$$
\dot z
=\frac{\partial\Omega}{\partial m}
=-\frac{Nkm}{(k^2+m^2)^{3/2}}-\gamma z.
$$
At release,
$$
\dot z(0)
=-\frac{Nk_0m_0}{(k_0^2+m_0^2)^{3/2}}
-\gamma z_0.
$$
Initial upward propagation is possible only when $m_0<0$. In that case $\dot z(0)>0$ precisely when
$$
\boxed{
z_0<z_m,
\qquad
z_m
=-\frac{Nk_0m_0}
{\gamma(k_0^2+m_0^2)^{3/2}}
}.
$$

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

Assume the condition in part (ii), so $m_0<0$. Substituting the wavevector solution into the vertical ray equation gives the linear <ordinary differential equation>
$$
\dot z+\gamma z=F(t),
$$
where
$$
F(t)
=\frac{-Nk_0m_0}
{\left(k_0^2e^{-2\gamma t}+m_0^2e^{2\gamma t}\right)^{3/2}}
>0.
$$
The <integrating factor> $e^{\gamma t}$ gives
$$
z(t)
=e^{-\gamma t}
\left[
z_0+\int_0^te^{\gamma s}F(s)\,ds
\right].
$$
The bracket is strictly positive, so $z(t)>0$ at every finite time. Moreover,
$$
F(t)=O(e^{-3\gamma t}),
\qquad
e^{\gamma t}F(t)=O(e^{-2\gamma t}),
$$
so the improper integral converges to a finite positive value. It follows that
$$
\boxed{z(t)\sim Ce^{-\gamma t}\longrightarrow0
\quad(t\to\infty)}
$$
for some $C>0$. The packet therefore approaches $z=0$ asymptotically and never reaches it at finite time. This is the <vertical trapping of an internal gravity wave by planar strain>.

Solved by gpt-5.6-sol high.

= 40C
{parent=Paper 3}
{scope}
{title2=Numerical Analysis}

= a
{parent=40c}
{scope}

= Solution
{parent=a}

The <power method> starts with a unit vector $\mathbf x_0$ and iterates
$$
\mathbf x_{k+1}
=\frac{A\mathbf x_k}{\|A\mathbf x_k\|}.
$$
Its eigenvalue estimate is the <Rayleigh quotient>
$$
r(\mathbf x_k)
=\frac{\mathbf x_k^TA\mathbf x_k}
{\mathbf x_k^T\mathbf x_k}.
$$

By the <real spectral theorem>, choose an orthonormal eigenbasis $\mathbf v_1,\ldots,\mathbf v_n$ and write
$$
\mathbf x_0=\sum_{j=1}^nc_j\mathbf v_j.
$$
Assume $|\lambda_1|>|\lambda_2|$, $\lambda_1\ne0$, and $c_1\ne0$. Before normalization,
$$
A^k\mathbf x_0
=c_1\lambda_1^k
\left[
\mathbf v_1
+\sum_{j=2}^n
\frac{c_j}{c_1}
\left(\frac{\lambda_j}{\lambda_1}\right)^k
\mathbf v_j
\right].
$$
Thus the direction error is $O(|\lambda_2/\lambda_1|^k)$. Orthogonality makes the linear terms vanish from the Rayleigh quotient:
$$
r(\mathbf x_k)-\lambda_1
=
\frac{
\sum_{j=2}^nc_j^2\lambda_j^{2k}(\lambda_j-\lambda_1)
}{
\sum_{j=1}^nc_j^2\lambda_j^{2k}
}.
$$
Consequently
$$
\boxed{
r(\mathbf x_k)-\lambda_1
=O\!\left(\left|\frac{\lambda_2}{\lambda_1}\right|^{2k}\right)
}.
$$
This is the <quadratic Rayleigh-quotient improvement for the power method>.

Solved by gpt-5.6-sol high.

= b
{parent=40c}
{scope}

= Solution
{parent=b}

An orthonormal eigenbasis is
$$
\mathbf v_9=\frac1{\sqrt3}(1,1,1)^T,
\qquad
\mathbf v_6=\frac1{\sqrt6}(1,-2,1)^T,
\qquad
\mathbf v_2=\frac1{\sqrt2}(1,0,-1)^T.
$$
The starting vector has the <orthogonal decomposition by a closed subspace>[orthogonal decomposition]
$$
\mathbf x_0
=0\mathbf v_9+\frac{\sqrt3}{2}\mathbf v_6+\frac12\mathbf v_2.
$$
Hence its component in the dominant eigenspace is exactly zero. Normalizing $A^k\mathbf x_0$ gives
$$
\boxed{
\mathbf x_k
=
\frac{
6^k(1,-2,1)^T+2^k(1,0,-1)^T
}{
\sqrt2\sqrt{3\cdot36^k+4^k}
}
}.
$$
Since the two remaining eigenvectors are orthogonal,
$$
\boxed{
r(\mathbf x_k)
=\frac{18\cdot36^k+2\cdot4^k}
{3\cdot36^k+4^k}
}.
$$
Therefore
$$
\boxed{\lim_{k\to\infty}r(\mathbf x_k)=6}.
$$
The limit is not the leading eigenvalue $9$ because the nonorthogonality assumption in part (a) fails. On the <active spectrum of the power method>, the leading eigenvalue is $6$ and the next is $2$; indeed
$$
r(\mathbf x_k)-6
=-\frac{4\cdot4^k}{3\cdot36^k+4^k}
=O(3^{-2k}),
$$
in agreement with the squared spectral-ratio estimate.

Solved by gpt-5.6-sol high.

= c
{parent=40c}
{scope}

= Solution
{parent=c}

Fix a shift $s$ that is not an <eigenvalue> of $A$. <Inverse iteration> computes
$$
(A-sI)\mathbf y_{k+1}=\mathbf x_k,
\qquad
\mathbf x_{k+1}
=\frac{\mathbf y_{k+1}}{\|\mathbf y_{k+1}\|},
$$
and may use $r(\mathbf x_k)$ as the eigenvalue estimate.

For a real symmetric matrix, suppose that one eigenvalue $\lambda_*$ is uniquely closest to $s$ and that $\mathbf x_0$ has a nonzero component in its eigenspace. Since the iteration is the <power method> applied to $(A-sI)^{-1}$, the normalized vectors converge, up to sign, to an eigenvector for $\lambda_*$. If $\lambda_{\rm next}$ is the second-closest eigenvalue to the shift, the asymptotic direction-error ratio is
$$
\boxed{
\frac{|\lambda_*-s|}
{|\lambda_{\rm next}-s|}
}.
$$
Thus a shift close to the desired eigenvalue gives rapid convergence. The Rayleigh quotients converge to $\lambda_*$ and, for a simple eigenvalue, their errors are of second order in the vector error. This is <convergence of fixed-shift inverse iteration>.

Solved by gpt-5.6-sol high.