past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ii/paper-1.bigb

= Paper 1
{scope}

https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2023/paperii_1_2023.pdf

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

= i
{parent=1g}
{scope}

= Solution
{parent=i}

The <linear Diophantine equation solvability criterion> says that $ax+by=c$ has an integer solution exactly when $\gcd(a,b)$ divides $c$. The <Euclidean algorithm> gives
$$
\begin{aligned}
2314&=2(1007)+300,\\
1007&=3(300)+107,\\
300&=2(107)+86,\\
107&=86+21,\\
86&=4(21)+2,\\
21&=10(2)+1.
\end{aligned}
$$
Thus $\gcd(1007,2314)=1$, which divides $37$. Therefore
$$
\boxed{1007x+2314y=37\text{ has integer solutions}.}
$$

Solved by gpt-5.6-sol high.

= ii
{parent=1g}
{scope}

= Solution
{parent=ii}

Applying the <Euclidean algorithm>,
$$
\begin{aligned}
3211&=2508+703,\\
2508&=3(703)+399,\\
703&=399+304,\\
399&=304+95,\\
304&=3(95)+19,\\
95&=5(19).
\end{aligned}
$$
Hence $\gcd(2508,3211)=19$. Since $19$ does not divide $55$, the <linear Diophantine equation solvability criterion> gives
$$
\boxed{2508x+3211y=55\text{ has no integer solution}.}
$$

Solved by gpt-5.6-sol high.

= iii
{parent=1g}
{scope}

= Solution
{parent=iii}

The quadratic form factors as a sum of two squares:
$$
5x^2+16xy+13y^2
=(x+2y)^2+(2x+3y)^2.
$$
The change of variables
$$
u=x+2y,
\qquad
v=2x+3y
$$
has determinant $-1$, so its matrix is a <unimodular matrix> and it maps $\mathbb Z^2$ bijectively to itself. The original equation is therefore equivalent to
$$
u^2+v^2=365.
$$
The <sum of two squares theorem> also decides this systematically: $365=5\cdot73$, and every prime factor congruent to $3$ modulo $4$ occurs to an even power, vacuously here. Thus $365$ is a sum of two integer squares. Explicitly,
$$
365=2^2+19^2.
$$
Inverting the change of variables gives, for example, $(x,y)=(32,-15)$. Consequently
$$
\boxed{5x^2+16xy+13y^2=365\text{ has integer solutions}.}
$$
This is the <unimodular reduction of a discriminant-minus-four form to two squares>.

Solved by gpt-5.6-sol high.

= 2F
{parent=Paper 1}
{scope}
{title2=Topics In Analysis}

= a
{parent=2f}
{scope}

= Solution
{parent=a}

The <Liouville approximation theorem> states that if $\alpha$ is an irrational algebraic number of degree $d\geq2$, then there is a constant $c(\alpha)>0$ such that every reduced rational $p/q$ satisfies
$$
\boxed{\left|\alpha-\frac pq\right|>
\frac{c(\alpha)}{q^d}}.
$$

Let $P\in\mathbb Z[X]$ be the <minimal polynomial> of $\alpha$, of degree $d$. Since $P$ is irreducible of degree greater than one, $P(p/q)\ne0$. Moreover,
$$
q^dP(p/q)\in\mathbb Z\setminus\{0\},
$$
and hence
$$
|P(p/q)|\geq q^{-d}.
$$
On the compact interval $I=[\alpha-1,\alpha+1]$, put
$$
M=\max_{x\in I}|P'(x)|.
$$
If $p/q\in I$, the <mean value theorem> gives a point $\xi$ between $\alpha$ and $p/q$ such that
$$
|P(p/q)|=|P'(\xi)|\left|\frac pq-\alpha\right|
\leq M\left|\frac pq-\alpha\right|.
$$
Thus $|\alpha-p/q|\geq M^{-1}q^{-d}$. If $p/q\notin I$, then $|\alpha-p/q|>1\geq q^{-d}$. Taking any
$$
0<c(\alpha)<\min\{1,M^{-1}\}
$$
proves the stated strict inequality.

Solved by gpt-5.6-sol high.

= b
{parent=2f}
{scope}

= Solution
{parent=b}

Because both fractions are reduced and $q>v$, they are distinct. Their difference is
$$
\left|\frac pq-\frac uv\right|
=\frac{|pv-qu|}{qv}.
$$
The numerator is a nonzero integer, so its absolute value is at least one. Therefore
$$
\left|\frac pq-\frac uv\right|
\geq\frac1{qv}>\boxed{\frac1{q^2}},
$$
where the strict inequality uses $v<q$. This is the <separation of distinct reduced fractions>.

Solved by gpt-5.6-sol high.

= c
{parent=2f}
{scope}

= Solution
{parent=c}

Since the positive series converges, $a_n\to0$. We construct the subsequence recursively. Choose $n(1)$ arbitrarily. Once $n(1)<\cdots<n(k-1)$ have been chosen, write the reduced partial sums as
$$
s_j=\sum_{r=1}^j a_{n(r)}=\frac{p_j}{q_j}
\qquad(1\leq j<k).
$$
Choose $n(k)>n(k-1)$ so far out that
$$
0<a_{n(k)}<2^{-k}min_{1\leq j<k}q_j^{-j}.
$$
This is possible because $a_n\to0$.

Let
$$
\alpha=\sum_{k=1}^{\infty}a_{n(k)}.
$$
For every fixed $j$, each later choice includes $q_j^{-j}$ in its minimum, so
$$
0<\alpha-s_j
<q_j^{-j}\sum_{k=j+1}^{\infty}2^{-k}
=2^{-j}q_j^{-j}.
$$

First, $\alpha$ is irrational. If $\alpha=u/v$ were reduced and rational, the strictly increasing rational sequence $s_j\to\alpha$ would have unbounded denominators $q_j$; only finitely many reduced fractions in a bounded interval have bounded denominator. Choose $j>2$ with $q_j>v$. Part (b) would give
$$
|\alpha-s_j|>q_j^{-2},
$$
contrary to $|\alpha-s_j|<2^{-j}q_j^{-j}<q_j^{-2}$.

If $\alpha$ were algebraic of degree $d$, part (a) would give a constant $c>0$ with
$$
|\alpha-s_j|>c q_j^{-d}.
$$
For $j>d$ our construction instead gives
$$
|\alpha-s_j|<2^{-j}q_j^{-j}
\leq2^{-j}q_j^{-d},
$$
which contradicts the lower bound once $2^{-j}<c$. Hence $\alpha$ is transcendental. This proves the <transcendental subseries of a positive rational convergent series> construction.

Solved by gpt-5.6-sol high.

= 3I
{parent=Paper 1}
{scope}
{title2=Coding and Cryptography}

= a
{parent=3i}
{scope}

= Solution
{parent=a}

A binary code $c:\mathcal A\to\{0,1\}^*$ is a <prefix code> when no codeword is a proper prefix of another. If the source letters have probabilities $p_1,\ldots,p_m$ and codeword lengths $l_1,\ldots,l_m$, the <expected value> of its codeword length is
$$
L=\sum_{r=1}^m p_r l_r.
$$
It is an <optimal prefix code> when it minimizes $L$ among all binary prefix codes for that source.

Suppose $p_i>p_j$ but $l_i>l_j$. Exchanging the codewords assigned to $\mu_i$ and $\mu_j$ leaves the set of words, and hence prefix-freeness, unchanged. The change in expected length is
$$
p_i l_j+p_j l_i-(p_i l_i+p_j l_j)
=(p_i-p_j)(l_j-l_i)<0,
$$
contradicting optimality. Thus
$$
\boxed{p_i>p_j\ \Longrightarrow\ l_i\leq l_j},
$$
which is the <probability monotonicity of optimal prefix-code lengths>.

Represent the code by its binary prefix tree and choose a codeword $wb$ of maximal length, where $b\in\{0,1\}$. Its sibling is $w(1-b)$. If that sibling were not a codeword, it could not have a codeword below it either, since such a word would be longer than $wb$. Replacing $wb$ by $w$ would then preserve prefix-freeness and strictly reduce $L$. This is impossible for an optimal code, so $w(1-b)$ is also a maximal-length codeword. The pair differs only in its last digit, proving the <deepest sibling property of an optimal prefix code>.

Solved by gpt-5.6-sol high.

= b
{parent=3i}
{scope}

= Solution
{parent=b}

The <Huffman coding> merges may be chosen as
$$
0.1+0.1=0.2,\qquad
0.2+0.2=0.4,\qquad
0.2+0.4=0.6,\qquad
0.4+0.6=1.
$$
Resolving the tied weights so that the original $0.4$ symbol is merged only at the final step gives the code
$$
\begin{array}{c|ccccc}
\text{probability}&0.4&0.2&0.2&0.1&0.1\\ \hline
\text{codeword}&0&100&101&110&111.
\end{array}
$$
Its lengths are $(1,3,3,3,3)$ and its expected length is
$$
0.4+3(0.2+0.2+0.1+0.1)=\boxed{2.2}.
$$
The <Huffman coding> theorem proves optimality, so an optimal coding with all but one word of the same length does exist. This is an instance of a <nonunique optimal prefix code>: a different resolution of the ties also gives the optimal profile $(2,2,2,3,3)$.

An optimal code cannot have five distinct lengths. By the <deepest sibling property of an optimal prefix code>, at least two maximal-length codewords have the same length. Thus the answers are respectively
$$
\boxed{\text{(i) yes},\qquad\text{(ii) no}}.
$$

Solved by gpt-5.6-sol high.

= 4I
{parent=Paper 1}
{scope}
{title2=Automata & Formal Languages}

= a
{parent=4i}
{scope}

= Solution
{parent=a}

A <context-free grammar> is in <Chomsky normal form> when every production is of one of the forms
$$
A\to BC,
\qquad
A\to a,
$$
where $A,B,C$ are nonterminal symbols and $a$ is a terminal symbol. The conventional exceptional production $S\to\epsilon$ may be allowed when the language contains the empty word, with $S$ excluded from right-hand sides.

Solved by gpt-5.6-sol high.

= b
{parent=4i}
{scope}

= Solution
{parent=b}

The length is
$$
\boxed{2n-1}.
$$
Indeed, the <parse tree> has $n$ terminal-production nodes and, as a full binary tree with $n$ leaves, $n-1$ binary-production nodes. This is the <Chomsky-normal-form derivation length>.

Solved by gpt-5.6-sol high.

= c
{parent=4i}
{scope}

= Solution
{parent=c}

One <parse tree> can be recorded unambiguously by the nested expression
$$
S\bigl(
 A(A(a),B(b)),
 C(B(b),A(A(A(a),B(b)),B(B(b),A(a))))
\bigr).
$$
Every internal node uses one of the given productions, and the leaves read from left to right are
$$
a,b,b,a,b,b,a.
$$
Equivalently, the corresponding leftmost <grammar derivation> is
$$
\begin{aligned}
S&\Rightarrow AC\Rightarrow ABC\Rightarrow aBC\Rightarrow abC
\Rightarrow abBA\Rightarrow abbA\\
&\Rightarrow abbAB\Rightarrow abbABB\Rightarrow abbaBB
\Rightarrow abbabB\Rightarrow abbabBA\\
&\Rightarrow abbabbA\Rightarrow\boxed{abbabba}.
\end{aligned}
$$
Hence the required word is generated by the grammar.

Solved by gpt-5.6-sol high.

= d
{parent=4i}
{scope}

= i
{parent=d}
{scope}

= Solution
{parent=i}

Take
$$
G'=(\{a,b\},\{S,A,B\},P',S),
$$
where
$$
P'=\{S\to ABA,\ A\to AB,\ B\to BA,\ A\to a,\ B\to b\}.
$$
This grammar is in <Weak Chomsky normal form>. In the original grammar, $C$ occurs only in $S\to AC$ and has the sole production $C\to BA$. Thus every use of
$$
S\to AC\to ABA
$$
is replaced in $G'$ by $S\to ABA$, while every later production is unchanged. Conversely, the ternary production can always be expanded into those two original steps. Therefore $G'$ is equivalent to $G$ in the sense of an <equivalent grammar>.

Solved by gpt-5.6-sol high.

= ii
{parent=d}
{scope}

= Solution
{parent=ii}

The compressed initial production gives the leftmost derivation
$$
\begin{aligned}
S&\Rightarrow ABA\Rightarrow ABBA\Rightarrow aBBA
\Rightarrow abBA\Rightarrow abbA\Rightarrow abbAB\\
&\Rightarrow abbABB\Rightarrow abbaBB\Rightarrow abbabB
\Rightarrow abbabBA\Rightarrow abbabbA\Rightarrow\boxed{abbabba}.
\end{aligned}
$$
It has $12$ steps, whereas part (b) gives $2\cdot7-1=13$ for every derivation in <Chomsky normal form>. The <Weak Chomsky normal form> ternary rule has compressed exactly one pair of binary productions, so the derivation is strictly shorter as required.

Solved by gpt-5.6-sol high.

= 5J
{parent=Paper 1}
{scope}
{title2=Statistical Modelling}

= Solution
{parent=5J}

For $X=x$, the last flip must be the $r$th head. Among the preceding $x+r-1$ flips, exactly $x$ are tails and $r-1$ are heads. There are $\binom{x+r-1}{x}$ choices for their positions, and every resulting sequence has probability $(1-p)^x p^r$. The <negative binomial stopping argument> therefore gives
$$
\boxed{\mathbb P(X=x)=\binom{x+r-1}{x}(1-p)^x p^r},
\qquad x=0,1,\ldots,
$$
so $X$ has the failures-before-the-$r$th-success <negative binomial distribution>.

To put the mass function into <exponential family> form, set
$$
\theta=\log(1-p),
\qquad -\infty<\theta<0.
$$
Since $p=1-e^\theta$,
$$
\mathbb P_\theta(X=x)
=\binom{x+r-1}{x}
 \exp\!\left\{\theta x+r\log(1-e^\theta)\right\}
=h(x)\exp\{\theta T(x)-A(\theta)\},
$$
where
$$
\boxed{T(x)=x},
\qquad
\boxed{A(\theta)=-r\log(1-e^\theta)},
\qquad
h(x)=\binom{x+r-1}{x}.
$$
Thus the <natural parameter of an exponential family> is $\theta$, and the <Fisher-Neyman factorization theorem> shows that $T(X)=X$ is a <sufficient statistic>. This is the <negative binomial exponential family>.

The <exponential-family derivative identities> now give
$$
\mathbb E[X]=A'(\theta)
=\frac{r e^\theta}{1-e^\theta}
=\boxed{\frac{r(1-p)}p}
$$
and
$$
\operatorname{var}(X)=A''(\theta)
=\frac{r e^\theta}{(1-e^\theta)^2}
=\boxed{\frac{r(1-p)}{p^2}}.
$$

Solved by gpt-5.6-sol high.

= 6C
{parent=Paper 1}
{scope}
{title2=Mathematical Biology}

= Solution
{parent=6C}

With population $n$, the transition diagram has the two outgoing arrows
$$
n\xrightarrow{\lambda}n+3,
\qquad
n\xrightarrow{\beta n}n-1.
$$
Thus this is a <batch-birth linear-death process>. If $P_n(t)=\mathbb P(N_t=n)$ and $P_j=0$ for $j<0$, probability enters state $n$ from $n-3$ by a birth and from $n+1$ by a death. Hence the master equation is
$$
\boxed{
\dot P_n
=\lambda P_{n-3}+\beta(n+1)P_{n+1}
 -(\lambda+\beta n)P_n
}.
$$

The <Markov jump-process generator> is
$$
(Lf)(n)=\lambda\bigl(f(n+3)-f(n)\bigr)
       +\beta n\bigl(f(n-1)-f(n)\bigr).
$$
For $f(n)=n$, this gives $Ln=3\lambda-\beta n$. Therefore the <expected value> $m(t)=\mathbb E[N_t]$ satisfies
$$
m'=3\lambda-\beta m,
\qquad m(0)=n_0,
$$
so
$$
\boxed{
m(t)=\frac{3\lambda}{\beta}(1-e^{-\beta t})+n_0e^{-\beta t}
}.
$$

For $f(n)=n^2$,
$$
Ln^2
=\lambda(6n+9)+\beta n(-2n+1).
$$
Writing $s=\mathbb E[N_t^2]$ and using the generator identity gives
$$
s'=6\lambda m+9\lambda-2\beta s+\beta m.
$$
The <variance> $v=s-m^2$ consequently obeys
$$
\begin{aligned}
v'&=s'-2mm'\\
  &=9\lambda+\beta m-2\beta v.
\end{aligned}
$$
If $v(0)=v_0<\infty$, substitution of the formula for $m$ yields
$$
v(t)=v_0e^{-2\beta t}
 +\frac{6\lambda}{\beta}(1-e^{-2\beta t})
 +\left(n_0-\frac{3\lambda}{\beta}\right)
  (e^{-\beta t}-e^{-2\beta t}).
$$
All transient terms vanish, and the <moments of a batch-birth linear-death process> therefore give
$$
\boxed{\operatorname{var}(N_t)\longrightarrow\frac{6\lambda}{\beta}}.
$$

Solved by gpt-5.6-sol high.

= 7E
{parent=Paper 1}
{scope}
{title2=Further Complex Methods}

= Solution
{parent=7E}

Taking reciprocals in the finite product defining the <gamma function> gives
$$
\frac1{\Gamma(z)}
=z\lim_{n\to\infty}n^{-z}
 \prod_{k=1}^n\left(1+\frac zk\right).
$$
Let $H_n=\sum_{k=1}^n k^{-1}$. Since
$$
n^{-z}
=e^{z(H_n-\log n)}e^{-zH_n}
=e^{z(H_n-\log n)}\prod_{k=1}^n e^{-z/k}
$$
and $H_n-\log n\to\gamma$, where $\gamma$ is the <Euler--Mascheroni constant>, passage to the limit gives the <Weierstrass product for the reciprocal gamma function>
$$
\boxed{
\frac1{\Gamma(z)}
=ze^{\gamma z}\prod_{k=1}^{\infty}
 \left(1+\frac zk\right)e^{-z/k}
}.
$$

The logarithmic derivative of this identity is
$$
-\psi(z)
=\frac1z+\gamma
 +\sum_{k=1}^{\infty}
  \left(\frac1{z+k}-\frac1k\right).
$$
Since $k^{-1}-(z+k)^{-1}=z/[k(z+k)]$, the <digamma function> therefore satisfies
$$
\boxed{
\psi(z)=-\gamma-\frac1z
 +z\sum_{k=1}^{\infty}\frac1{k(z+k)}
}.
$$

For real $z>0$, termwise differentiation gives the <trigamma function>
$$
\psi'(z)=\frac1{z^2}
 +\sum_{k=1}^{\infty}\frac1{(z+k)^2}>0.
$$
Thus $\psi$ is strictly increasing on the positive real axis. At the two positive integers needed here, telescoping gives
$$
\psi(1)=-\gamma<0,
\qquad
\psi(2)=1-\gamma>0.
$$
The <intermediate value theorem> supplies a zero in $(1,2)$, and strict increase makes it unique. This is the <positive zero of the digamma function>.

Solved by gpt-5.6-sol high.

= 8D
{parent=Paper 1}
{scope}
{title2=Classical Dynamics}

= a
{parent=8d}
{scope}

= Solution
{parent=a}

Under the stated <gauge transformation>, the <charged-particle electromagnetic Lagrangian> becomes
$$
\begin{aligned}
L'&=\frac12m\dot r^2
-q\left(\phi-\partial_tf
 -\dot r\cdot(A+\nabla f)\right)\\
&=L+q\left(\partial_tf+\dot r\cdot\nabla f\right)
 =L+q\frac{df(r(t),t)}{dt}.
\end{aligned}
$$
Thus
$$
\boxed{L'-L=q\,df/dt}.
$$
The action changes only by $q[f(t_2)-f(t_1)]$, whose variation vanishes for fixed endpoints. By <total-time-derivative invariance of a Lagrangian>, the <Euler-Lagrange equation> is unchanged.

Solved by gpt-5.6-sol high.

= b
{parent=8d}
{scope}

= Solution
{parent=b}

In Cartesian components,
$$
\frac{\partial L}{\partial\dot r_i}=m\dot r_i+qA_i
$$
and
$$
\frac d{dt}\frac{\partial L}{\partial\dot r_i}
=m\ddot r_i+q\partial_tA_i+q\dot r_j\partial_jA_i,
\qquad
\frac{\partial L}{\partial r_i}
=-q\partial_i\phi+q\dot r_j\partial_iA_j.
$$
The <Euler-Lagrange equation> therefore gives
$$
m\ddot r_i
=q\left[-\partial_i\phi-\partial_tA_i
 +\dot r_j(\partial_iA_j-\partial_jA_i)\right].
$$
Using
$$
E=-\nabla\phi-\partial_tA,
\qquad B=\nabla\times A,
$$
and $[\dot r\times B]_i=\dot r_j(\partial_iA_j-\partial_jA_i)$, this becomes the <Lorentz force>
$$
\boxed{m\ddot r=q(E+\dot r\times B)}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=8d}
{scope}

= Solution
{parent=c}

Let $\widehat b=B/|B|$. For a constant uniform <magnetic field>, choose the symmetric gauge
$$
A=\frac12B\times r,
$$
which has $\nabla\times A=B$ and $A\cdot\widehat b=0$. With $E=0$, one may take $\phi=0$. The coordinate $r_\parallel=r\cdot\widehat b$ is then absent from the Lagrangian, so it is an <ignorable coordinate>. Its <generalized momentum>
$$
p_\parallel
=\frac{\partial L}{\partial\dot r_\parallel}
=(m\dot r+qA)\cdot\widehat b
=m\dot r\cdot\widehat b
$$
is conserved. Hence
$$
\boxed{\frac d{dt}(p\cdot\widehat b)=0},
$$
which is the <canonical momentum parallel to a uniform magnetic field> in this gauge.

Solved by gpt-5.6-sol high.

= 9B
{parent=Paper 1}
{scope}
{title2=Cosmology}

= a
{parent=9b}
{scope}

= Solution
{parent=a}

For a comoving region, $E=\rho V$ and $V=a^3V_0$. The work law gives
$$
d(\rho V)=-P\,dV,
$$
so differentiation with respect to time yields
$$
V\dot\rho+(\rho+P)\dot V=0.
$$
Since $\dot V/V=3\dot a/a$, division by $V$ gives the <cosmological perfect-fluid continuity equation>
$$
\boxed{\dot\rho+3\frac{\dot a}{a}(\rho+P)=0}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=9b}
{scope}

= Solution
{parent=b}

Put $K=8\pi G/(3c^2)$. Differentiating $Q=K\rho a^2-\dot a^2$ gives
$$
\dot Q
=K(a^2\dot\rho+2a\dot a\rho)-2\dot a\ddot a.
$$
The continuity equation and the <Friedmann acceleration equation> imply
$$
a^2\dot\rho+2a\dot a\rho
=-a\dot a(\rho+3P)
$$
and
$$
-2\dot a\ddot a
=K a\dot a(\rho+3P).
$$
These terms cancel, so
$$
\boxed{\dot Q=0}.
$$
This is the <First integral of the Friedmann acceleration equation>.

Solved by gpt-5.6-sol high.

= c
{parent=9b}
{scope}

= Solution
{parent=c}

For the <barotropic equation of state> $P=w\rho$, the continuity equation separates as
$$
\frac{d\rho}{\rho}=-3(1+w)\frac{da}{a}.
$$
Since $a(t_0)=1$,
$$
\rho=\rho_0a^{-3(1+w)},
\qquad \rho_0=\rho(t_0),
$$
which is the <constant-equation-of-state density scaling>.

When $Q=0$, the expanding branch satisfies
$$
\dot a=\sqrt{\frac{8\pi G\rho_0}{3c^2}}
 a^{-(1+3w)/2}=H_0a^{-(1+3w)/2},
$$
where $H_0=\dot a(t_0)>0$. Because $w>-1$, integration gives
$$
a(t)^{3(1+w)/2}
=1+\frac32(1+w)H_0(t-t_0),
$$
and hence the <flat constant-equation-of-state scale factor>
$$
\boxed{
a(t)=\left[1+\frac32(1+w)H_0(t-t_0)\right]^{2/[3(1+w)]}
}.
$$
It vanishes at
$$
\boxed{t_*=t_0-\frac{2}{3(1+w)H_0}<t_0},
$$
as $|w|<1$ ensures $1+w>0$.

Solved by gpt-5.6-sol high.

= 10D
{parent=Paper 1}
{scope}
{title2=Quantum Information and Computation}

= a
{parent=10d}
{scope}

= Solution
{parent=a}

Apply the hypothetical <quantum cloning> device repeatedly to obtain $N$ copies of the unknown state. If $c=|\langle\phi|\psi\rangle|<1$, the two possibilities are
$$
|\phi\rangle^{\otimes N},
\qquad |\psi\rangle^{\otimes N},
$$
whose overlap has modulus $c^N$. For equal prior probabilities, the <Helstrom-Holevo bound> gives optimal success probability
$$
P_N=\frac12\left(1+\sqrt{1-c^{2N}}\right)
\longrightarrow1.
$$
Thus, for any $\varepsilon>0$, sufficiently many clones permit success probability above $1-\varepsilon$. Orthogonal states already have $c=0$ and can be distinguished perfectly. This is <clone-assisted asymptotic state discrimination>.

Here distinct states means distinct rays. Vectors differing only by a global phase represent the same <quantum state> and cannot be distinguished.

Solved by gpt-5.6-sol high.

= b
{parent=10d}
{scope}

= Solution
{parent=b}

Run the perfect discriminator on the supplied state and record whether the result is $i=0$ or $i=1$. For each $i$, choose a unitary $V_i$ satisfying
$$
V_i|0\rangle=|\phi_i\rangle,
$$
which is possible by extending $|\phi_i\rangle$ to an orthonormal basis. Prepare two fresh systems in $|0\rangle|0\rangle$ and, controlled by the classical outcome $i$, apply $V_i\otimes V_i$. The output is
$$
|\phi_i\rangle|\phi_i\rangle.
$$
The discriminator may destroy its input; the two freshly prepared outputs still implement cloning on this known pair. This is the construction that <perfect discrimination implies cloning for a known state family>.

Solved by gpt-5.6-sol high.

= c
{parent=10d}
{scope}

= Solution
{parent=c}

Suppose first that the stated unitary and environment states exist. Preservation of the <inner product> gives
$$
\langle\phi_0|\phi_1\rangle
=\langle\psi_0|\psi_1\rangle
 \langle e_0|e_1\rangle.
$$
The environment states are normalized, so the <Cauchy-Schwarz inequality> gives $|\langle e_0|e_1\rangle|\leq1$. Therefore
$$
|\langle\phi_0|\phi_1\rangle|
\leq|\langle\psi_0|\psi_1\rangle|.
$$

Conversely, write
$$
a=\langle\phi_0|\phi_1\rangle,
\qquad b=\langle\psi_0|\psi_1\rangle
$$
and assume $|a|\leq|b|$. If $b\ne0$, put $c=a/b$, so $|c|\leq1$, and choose
$$
|e_0\rangle=|0\rangle,
\qquad
|e_1\rangle=c|0\rangle+\sqrt{1-|c|^2}\,|1\rangle.
$$
Then $\langle e_0|e_1\rangle=c$ and the desired output vectors have inner product $bc=a$, exactly matching the input vectors. If $b=0$, the inequality forces $a=0$, and one may take $|e_0\rangle=|e_1\rangle=|0\rangle$.

In either case the input pair and output pair have the same <Gram matrix>. The isometry taking one pair to the other extends to a <unitary matrix>, so the required $U$ exists. Hence
$$
\boxed{
U\text{ exists}
\quad\Longleftrightarrow\quad
|\langle\phi_0|\phi_1\rangle|
\leq|\langle\psi_0|\psi_1\rangle|
},
$$
which is the <environment-assisted two-state pure-state transformation> criterion.

Solved by gpt-5.6-sol high.

= 11I
{parent=Paper 1}
{scope}
{title2=Coding and Cryptography}

= a
{parent=11i}
{scope}

= Solution
{parent=a}

A <binary symmetric channel> independently flips each input bit with probability $p$, so, with rows indexed by the input and columns by the output, its channel matrix is
$$
\boxed{
\begin{pmatrix}
1-p&p\\
p&1-p
\end{pmatrix}}
.$$
If $p>1/2$, complementing every received bit converts the channel into one with crossover probability $1-p<1/2$ without changing its information-carrying ability. The <completely noisy binary symmetric channel>[case $p=1/2$] has identical rows and <binary symmetric channel capacity>[capacity] zero, so it suffices to study $p<1/2$.

The <Shannon second coding theorem> states that for every rate below the <channel capacity> there are arbitrarily long codes whose decoding error tends to zero, whereas no sequence of codes with rate above capacity can have vanishing error. It also identifies
$$
C=\max_{P_X}I(X;Y).
$$
For the present channel,
$$
I(X;Y)=H(Y)-H(Y\mid X)=H(Y)-h_2(p).
$$
The output entropy is at most one bit, with equality for a uniform input. Hence the <binary symmetric channel capacity> is
$$
\boxed{C=1-h_2(p)}
$$
bits per channel use.

Solved by gpt-5.6-sol high.

= b
{parent=11i}
{scope}

= Solution
{parent=b}

Let $X\in\{0,1\}$ denote whether $00$ or $11$ was sent, and write
$$
Y_j=X\mathbin\oplus E_j,
$$
where $E_1,E_2$ are independent Bernoulli variables of parameter $p$. Because $X$ is uniform, $Y_1$ is uniform, while $H(Y_1\mid X)=h_2(p)$. Thus the <mutual information> in the first received digit is
$$
\boxed{I(X;Y_1)=1-h_2(p)}.
$$

By the <chain rule for mutual information>, the extra information in the second digit is $I(X;Y_2\mid Y_1)$. Conditional on $X$, the second channel error is independent of the first output, so
$$
H(Y_2\mid X,Y_1)=H(E_2)=h_2(p).
$$
Moreover,
$$
\mathbb P(Y_2\ne Y_1)
=\mathbb P(E_2\ne E_1)=2p(1-p).
$$
Given either value of $Y_1$, the second output therefore differs from it with probability $2p(1-p)$, and
$$
H(Y_2\mid Y_1)=h_2(2p(1-p)).
$$
Consequently the <incremental information in a twofold binary repetition code> is
$$
\boxed{
I(X;Y_2\mid Y_1)
=h_2(2p(1-p))-h_2(p)
}
$$
bits.

Solved by gpt-5.6-sol high.

= c
{parent=11i}
{scope}

= Solution
{parent=c}

Every row of the channel matrix is a permutation of $(1-2\alpha,\alpha,\alpha)$, and every column sum is one. It is therefore a weakly symmetric channel. The uniform input makes the output uniform, with entropy $\log_2 3$, while the conditional output entropy is the entropy of one row:
$$
H(Y\mid X)
=-(1-2\alpha)\log_2(1-2\alpha)
 -2\alpha\log_2\alpha.
$$
The <weakly symmetric channel capacity> formula now gives the <ternary symmetric channel capacity>
$$
\boxed{
C=\log_2 3
 +(1-2\alpha)\log_2(1-2\alpha)
 +2\alpha\log_2\alpha
}
$$
bits per channel use, for $0\leq\alpha\leq1/2$, with the convention $0\log_2 0=0$.

Solved by gpt-5.6-sol high.

= 12I
{parent=Paper 1}
{scope}
{title2=Automata & Formal Languages}

= i
{parent=12i}
{scope}

= Solution
{parent=i}

The relation $A\leq_mB$ means that there is a total computable function $g:\mathbb W\to\mathbb W$ such that
$$
\boxed{w\in A\quad\Longleftrightarrow\quad g(w)\in B}
$$
for every word $w$. Such a $g$ is a <many-one reduction>.

Solved by gpt-5.6-sol high.

= ii
{parent=12i}
{scope}

= Solution
{parent=ii}

A set $A$ is $\mathcal C$-complete under many-one reductions when
$$
\boxed{A\in\mathcal C
\quad\text{and}\quad
(\forall B\in\mathcal C)\ B\leq_mA}.
$$
The first condition is membership in the class; the second is $\mathcal C$-hardness. This is a <many-one complete set>.

Solved by gpt-5.6-sol high.

= iii
{parent=12i}
{scope}

= Solution
{parent=iii}

A set $A\subseteq\mathbb W$ is a <Sigma-1 set> when it is the domain of a <partial computable function>: there is a partial computable $f$ such that
$$
\boxed{w\in A\quad\Longleftrightarrow\quad f(w)\mathbin\downarrow}.
$$
Equivalently, $w\in A$ iff $\exists t\,R(w,t)$ for some computable predicate $R$; one may take $R(w,t)$ to say that the <truncated computation function> observes a halt within $t$ steps.

Solved by gpt-5.6-sol high.

= iv
{parent=12i}
{scope}

= Solution
{parent=iv}

For an effective enumeration $(W_w)_{w\in\mathbb W}$ of the computably enumerable languages, the diagonal <halting problem> is
$$
\boxed{\mathbf K=\{w\in\mathbb W:w\in W_w\}}.
$$
Equivalently, if $f_{w,1}$ is the unary partial function coded by $w$, then
$$
w\in\mathbf K
\quad\Longleftrightarrow\quad
f_{w,1}(w)\mathbin\downarrow.
$$
Thus $\mathbf K$ is the <diagonal halting set>.

Solved by gpt-5.6-sol high.

= v
{parent=12i}
{scope}

= Solution
{parent=v}

First, $\mathbf K\in\Sigma_1$: the partial procedure on input $w$ simulates $f_{w,1}(w)$ and halts exactly when that computation halts.

For hardness, let $A\in\Sigma_1$. Choose a partial computable function $g$ with
$$
w\in A\quad\Longleftrightarrow\quad g(w)\mathbin\downarrow.
$$
For each fixed $w$, define a unary program $P_w$ as follows: on any input $x$, simulate $g(w)$; if that simulation halts, halt and output $x$, and otherwise run forever. The <S-m-n theorem> gives a total computable function $h$ that maps $w$ to a code for $P_w$. Its accepted language is
$$
W_{h(w)}=
\begin{cases}
\mathbb W,&g(w)\mathbin\downarrow,\\
\varnothing,&g(w)\mathbin\uparrow.
\end{cases}
$$
In particular,
$$
w\in A
\quad\Longleftrightarrow\quad
h(w)\in W_{h(w)}
\quad\Longleftrightarrow\quad
h(w)\in\mathbf K.
$$
Thus $h$ is a <many-one reduction> $A\leq_m\mathbf K$. Since $A$ was arbitrary, $\mathbf K$ is $\Sigma_1$-hard; together with membership this proves the <many-one completeness of the halting problem>.

Solved by gpt-5.6-sol high.

= vi
{parent=12i}
{scope}

= Solution
{parent=vi}

Membership is immediate from the supplied characterization of $\Pi_2$. Define the partial computable function
$$
F(v,x)=f_{v,1}(x).
$$
Then
$$
v\in\mathbf{Tot}
\quad\Longleftrightarrow\quad
W_v=\mathbb W
\quad\Longleftrightarrow\quad
(\forall x\in\mathbb W)\ F(v,x)\mathbin\downarrow,
$$
so $\mathbf{Tot}\in\Pi_2$.

Now let $P\in\Pi_2$. There is a partial computable $f$ such that
$$
w\in P
\quad\Longleftrightarrow\quad
(\forall x\in\mathbb W)\ f(w,x)\mathbin\downarrow.
$$
For each $w$, define a unary partial function
$$
g_w(x)=
\begin{cases}
x,&f(w,x)\mathbin\downarrow,\\
\text{undefined},&f(w,x)\mathbin\uparrow.
\end{cases}
$$
Operationally, its program simulates $f(w,x)$ and returns $x$ if that computation halts. By the <S-m-n theorem>, a total computable map $h$ produces an index $h(w)$ for $g_w$. Therefore
$$
\begin{aligned}
w\in P
&\Longleftrightarrow
(\forall x)\ g_w(x)\mathbin\downarrow\\
&\Longleftrightarrow W_{h(w)}=\mathbb W
\Longleftrightarrow h(w)\in\mathbf{Tot}.
\end{aligned}
$$
Thus every <Pi-2 set> many-one reduces to $\mathbf{Tot}$. Combined with membership, this proves that the <totality problem> is $\Pi_2$-complete.

Solved by gpt-5.6-sol high.

= 13J
{parent=Paper 1}
{scope}
{title2=Statistical Modelling}

= i
{parent=13j}
{scope}

= Solution
{parent=i}

Write
$$
Y=\mu+\varepsilon,
\qquad
Y^*=\mu+\varepsilon^*,
$$
where $\varepsilon$ and $\varepsilon^*$ are independent, centered, and have covariance matrix $\sigma^2I_n$. Then
$$
HY-Y^*=-(I-H)\mu+H\varepsilon-\varepsilon^*.
$$
The deterministic term is orthogonal in expectation to the two centered random terms, and the random terms are independent. Since the <hat matrix> is a symmetric idempotent projection of rank $p$,
$$
\mathbb E\|H\varepsilon\|^2
=\sigma^2\operatorname{tr}(H^TH)
=\sigma^2\operatorname{tr}H=p\sigma^2,
$$
while $\mathbb E\|\varepsilon^*\|^2=n\sigma^2$. Therefore
$$
\boxed{
\mathbb E\|HY-Y^*\|^2
=\|(I-H)\mu\|^2+(n+p)\sigma^2
}.
$$
In this <bias-variance decomposition for linear prediction>, $\|(I-H)\mu\|^2$ is squared model bias, $p\sigma^2$ is variance from fitting $p$ coefficients, and $n\sigma^2$ is irreducible noise in the future response. Enlarging the model tends to reduce the first term while increasing the fitted-model variance, which is the bias-variance tradeoff.

Solved by gpt-5.6-sol high.

= ii
{parent=13j}
{scope}

= Solution
{parent=ii}

Because $I-H$ is an orthogonal projection of rank $n-p$,
$$
\begin{aligned}
\mathbb E\|Y-HY\|^2
&=\mathbb E\|(I-H)(\mu+\varepsilon)\|^2\\
&=\|(I-H)\mu\|^2
 +\sigma^2\operatorname{tr}(I-H)\\
&=\|(I-H)\mu\|^2+(n-p)\sigma^2.
\end{aligned}
$$
It follows that <Mallows Cp> satisfies
$$
\begin{aligned}
\mathbb E[C_p]
&=\mathbb E\|Y-HY\|^2+2p\sigma^2\\
&=\|(I-H)\mu\|^2+(n+p)\sigma^2\\
&=\mathbb E\|HY-Y^*\|^2,
\end{aligned}
$$
where the last equality is part (i). Thus $C_p$ is an <unbiased estimator> of the independent-copy prediction error, as summarized by the <unbiased prediction-error identity for ordinary least squares>.

Solved by gpt-5.6-sol high.

= iii
{parent=13j}
{scope}

= Solution
{parent=iii}

Let
$$
\widehat\beta=(X^TX)^{-1}X^TY,
\qquad
\widehat\sigma^2
=\frac{\|Y-X\widehat\beta\|^2}{n-p}.
$$
The pivotal quadratic form has an $F_{p,n-p}$ distribution, so the $(1-\alpha)$ <normal linear-model confidence ellipsoid> is
$$
\boxed{
\left\{b\in\mathbb R^p:
\frac{(b-\widehat\beta)^TX^TX(b-\widehat\beta)}
 {p\widehat\sigma^2}
\leq F_{p,n-p}(1-\alpha)
\right\}
}.
$$
Here $F_{p,n-p}(1-\alpha)$ is the $(1-\alpha)$ quantile of the indicated $F$ distribution.

Solved by gpt-5.6-sol high.

= iv
{parent=13j}
{scope}

= Solution
{parent=iv}

Let $\widehat\beta_{(i)}$ be the least-squares estimate after deleting observation $i$, and put $\widehat Y=X\widehat\beta$ and $\widehat Y_{(i)}=X\widehat\beta_{(i)}$. <Cook's distance> is
$$
\boxed{
D_i
=\frac{\|\widehat Y_{(i)}-\widehat Y\|^2}
 {p\widehat\sigma^2}
=\frac{(\widehat\beta_{(i)}-\widehat\beta)^TX^TX
 (\widehat\beta_{(i)}-\widehat\beta)}
 {p\widehat\sigma^2}
}.
$$
Thus $D_i$ is the squared displacement caused by deleting observation $i$, measured in the same $X^TX$ metric and scale as the confidence ellipsoid. In particular, $\widehat\beta_{(i)}$ lies outside the $(1-\alpha)$ ellipsoid centered at $\widehat\beta$ exactly when
$$
D_i>F_{p,n-p}(1-\alpha).
$$

Solved by gpt-5.6-sol high.

= v
{parent=13j}
{scope}

= Solution
{parent=v}

Here $p=4$ and $n-p=96$. The supplied quantile gives
$$
F_{4,96}(0.9)=2.00.
$$
Since
$$
D_i=3.1>2.00,
$$
the coefficient estimate after deleting this observation lies outside the $90\%$ confidence ellipsoid centered at the full-data estimate. It also greatly exceeds the common rough reference value $D_i=1$. The observation is therefore highly influential and merits investigation:
$$
\boxed{\text{yes, its influence is concerning}.}
$$

Solved by gpt-5.6-sol high.

= 14E
{parent=Paper 1}
{scope}
{title2=Further Complex Methods}

= i
{parent=14e}
{scope}

= Solution
{parent=i}

After division by $z$, the coefficient functions are
$$
P(z)=\frac1z-1,
\qquad Q(z)=\frac\lambda z.
$$
Both $zP(z)=1-z$ and $z^2Q(z)=\lambda z$ are analytic at zero, so the <regular singular point criterion for a second-order equation> shows that $z=0$ is regular singular. Substitution of $y=z^r$ into the leading terms gives the indicial equation
$$
r(r-1)+r=r^2=0.
$$
The exponent is repeated. Directly substituting $y=\sum_{n\geq0}a_nz^n$ gives
$$
(n+1)^2a_{n+1}+(\lambda-n)a_n=0,
\qquad
\boxed{a_{n+1}=\frac{n-\lambda}{(n+1)^2}a_n}.
$$
Thus $a_0$ determines the entire power series, so there is only one such solution $y_1$ up to scale. The <logarithmic solution from a repeated Frobenius exponent> predicts
$$
\boxed{y_2(z)=y_1(z)\log z+\text{a power series}},
$$
so its leading nonanalytic term is proportional to $\log z$.

For the contour ansatz, differentiation under the integral and one integration by parts give
$$
\begin{aligned}
0={}&\int_\gamma e^{zt}
 \{z(t^2-t)+(t+\lambda)\}f(t)\,dt\\
={}&\left[e^{zt}t(t-1)f(t)\right]_{\partial\gamma}
 +\int_\gamma e^{zt}
 \left\{(t+\lambda)f-[t(t-1)f]'\right\}dt.
\end{aligned}
$$
The integral vanishes when
$$
t(t-1)f'+(t-1-\lambda)f=0,
$$
whose solution is the <Laguerre contour-integral amplitude>
$$
\boxed{f(t)=t^{-\lambda-1}(t-1)^\lambda}.
$$
The contour and branches must make $f$ single-valued along the traversed path and must kill the endpoint term
$$
\left[e^{zt}t^{-\lambda}(t-1)^{\lambda+1}\right]_{\partial\gamma}.
$$

Now suppose $\lambda<0$ is nonintegral. Near $t=0$, $f(t)=O(t^{-\lambda-1})$ is integrable and the endpoint factor is $O(t^{-\lambda})\to0$. We may therefore choose
$$
\gamma_1:quad 0\longrightarrow
\text{one counterclockwise loop around }1
\longrightarrow0,
$$
a finite loop based at the branch point $0$ and avoiding $1$ except by encirclement. A second choice is
$$
\gamma_2:quad 0\longrightarrow-\infty
$$
along one bank of the negative real axis, with a consistent branch. At $-\infty$, $e^{zt}$ kills the algebraic endpoint factor because $\operatorname{Re}z>0$.

The $\gamma_1$ integral is analytic in $z$ because its contour is finite, and it is a nonzero solution. By uniqueness of the analytic local solution, it is a constant multiple of $y_1$. This is the <finite Laguerre contour solution> construction.

Solved by gpt-5.6-sol high.

= ii
{parent=14e}
{scope}

= Solution
{parent=ii}

Let $\lambda>0$ be nonintegral. This time $f(t)=O((t-1)^\lambda)$ is integrable at $t=1$, and the endpoint factor behaves as $(t-1)^{\lambda+1}\to0$. Suitable contours are therefore
$$
\gamma_1:quad 1\longrightarrow
\text{one counterclockwise loop around }0
\longrightarrow1
$$
and
$$
\gamma_2:quad 1\longrightarrow-\infty,
$$
where the second path makes a fixed detour above or below the branch point $0$ and the branch of $t^{-\lambda-1}(t-1)^\lambda$ is continued along it. The exponential again controls the endpoint at $-\infty$ for $\operatorname{Re}z>0$.

As in part (i), the finite $\gamma_1$ integral is analytic near zero and hence is a constant multiple of the unique power-series solution $y_1$.

Solved by gpt-5.6-sol high.

= iii
{parent=14e}
{scope}

= Solution
{parent=iii}

Write $\lambda=-m$ with $m\in\mathbb N$. The amplitude is now the rational function
$$
f(t)=\frac{t^{m-1}}{(t-1)^m}.
$$
Choose $\gamma_1$ to be a small positively oriented circle around the pole $t=1$, and choose $\gamma_2$ from $t=0$ along the negative real axis to $-\infty$. At zero the endpoint factor behaves as $t^m$ and vanishes; at negative infinity the exponential decays.

By the <residue theorem>, the finite-contour solution is
$$
y_{\gamma_1}(z)
=\frac{2\pi i}{(m-1)!}
 \left.\frac{d^{m-1}}{dt^{m-1}}
 \left(e^{zt}t^{m-1}\right)\right|_{t=1}.
$$
Every derivative term contains $e^z$ times a power of $z$ of degree at most $m-1$, and the highest-degree term is nonzero. Hence
$$
\boxed{y_1(z)=e^zP_{m-1}(z)}
$$
up to a constant, where $P_{m-1}$ is a polynomial of degree $m-1$. This is the negative-integer case of the <integer-parameter Laguerre contour residues>.

Solved by gpt-5.6-sol high.

= iv
{parent=14e}
{scope}

= Solution
{parent=iv}

Write $\lambda=N$ with $N\in\mathbb Z_{\geq0}$. Then
$$
f(t)=\frac{(t-1)^N}{t^{N+1}}.
$$
Take $\gamma_1$ to be a small positively oriented circle around $t=0$. For $\gamma_2$, start at $t=1$, travel to $-\infty$, and avoid the pole at zero by a fixed upper or lower detour. The endpoint factor vanishes at $t=1$, and the exponential controls $-\infty$.

The finite contour extracts the residue at zero:
$$
y_{\gamma_1}(z)
=2\pi i\,[t^N]\left(e^{zt}(t-1)^N\right).
$$
Only powers $z^0,\ldots,z^N$ occur, and the coefficient of $z^N$ is nonzero. Thus this solution is a polynomial of degree $N$ and, up to normalization, is the <Laguerre polynomial> $L_N(z)$:
$$
\boxed{y_1(z)\propto L_N(z)}.
$$
This also follows from the recurrence in part (i), because $a_{N+1}=0$ and every later coefficient vanishes.

Solved by gpt-5.6-sol high.

= 15B
{parent=Paper 1}
{scope}
{title2=Cosmology}

= a
{parent=15b}
{scope}

= Solution
{parent=a}

Write $A=8\pi G/(3c^2)$ and $H=\dot a/a$. Differentiating the <Friedmann equation> gives
$$
2H\dot H-2\frac{Kc^2}{a^2}H=A\dot\rho.
$$
The <cosmological perfect-fluid continuity equation> says $\dot\rho=-3H(\rho+P)$, so, first where $H\ne0$ and then everywhere by continuity,
$$
\dot H-\frac{Kc^2}{a^2}
=-\frac{4\pi G}{c^2}(\rho+P).
$$
Using $Kc^2/a^2=A\rho-H^2$ from the original equation gives
$$
\boxed{
\dot H+H^2
=-\frac{4\pi G}{3c^2}(\rho+3P)
}.
$$
Since $\dot H+H^2=\ddot a/a$ for the <Hubble parameter>, this is the <Friedmann acceleration equation>.

Solved by gpt-5.6-sol high.

= b
{parent=15b}
{scope}

= Solution
{parent=b}

Separate continuity gives
$$
\rho_r=\rho_{r0}a^{-4}
=\beta\frac{3c^2H_0^2}{8\pi G}a^{-4},
\qquad
\rho_\Lambda=\rho_{\Lambda0}
=\frac{3c^2H_0^2}{8\pi G},
$$
using the scalings for <radiation in cosmology> and <cosmological constant energy>. At $t=t_0$, where $a=1$, the Friedmann equation becomes
$$
H_0^2+Kc^2=(\beta+1)H_0^2,
$$
so
$$
\boxed{Kc^2=\beta H_0^2}.
$$
At general $a$,
$$
H^2+\frac{\beta H_0^2}{a^2}
=H_0^2\left(\frac\beta{a^4}+1\right),
$$
which rearranges to
$$
\boxed{
H^2=\frac{H_0^2}{a^4}(\beta-\beta a^2+a^4)
}.
$$

Part (a), applied to $\rho=\rho_r+\rho_\Lambda$ and $P=\rho_r/3-\rho_\Lambda$, gives
$$
\dot H+H^2
=-H_0^2\left(\frac\beta{a^4}-1\right).
$$
Subtracting the displayed formula for $H^2$ yields
$$
\boxed{
\dot H=-\beta\frac{H_0^2}{a^4}(2-a^2)
}.
$$
These relations define the <closed radiation--cosmological-constant turning polynomial>.

Solved by gpt-5.6-sol high.

= c
{parent=15b}
{scope}

= Solution
{parent=c}

Put $x=a^2$. Apart from the positive factor $H_0^2/a^4$,
$$
H^2\quad\hbox{has the sign of}\quad
F(x)=x^2-\beta x+\beta.
$$
For $\beta>4$, its discriminant $\beta(\beta-4)$ is positive and its two roots are
$$
x_\pm=\frac{\beta\pm\sqrt{\beta^2-4\beta}}2.
$$
Both are positive. Moreover $F(0)>0$ and $F(2)=4-\beta<0$, so
$$
0<x_-<2<x_+.
$$
A universe expanding from $a=0$ follows the allowed region $0\leq a^2\leq x_-$ and reaches $a^2=x_-$ with $H=0$. There,
$$
\dot H
=-\beta\frac{H_0^2}{a^4}(2-a^2)<0,
$$
so $H$ crosses from positive to negative: expansion reverses into contraction.

As $a\to0$ on the contracting branch,
$$
H\sim-\frac{H_0\sqrt\beta}{a^2},
\qquad
\dot a=aH\sim-\frac{H_0\sqrt\beta}{a}.
$$
Hence $a\,da\sim-H_0\sqrt\beta\,dt$, so zero scale factor is reached in finite time. The universe therefore undergoes the <supercritical closed radiation--cosmological-constant recollapse> to a <Big Crunch>.

Solved by gpt-5.6-sol high.

= d
{parent=15b}
{scope}

= Solution
{parent=d}

For $\beta=4$,
$$
H^2=\frac{H_0^2}{a^4}(a^2-2)^2.
$$
The branch expanding from $a=0$ has
$$
H=\frac{H_0(2-a^2)}{a^2}
$$
until it approaches $a^2=2$. With $x=a^2$,
$$
\dot x=2a\dot a=2a^2H=2H_0(2-x),
\qquad x(0)=0.
$$
Solving this linear equation gives the <critical closed radiation--cosmological-constant solution>
$$
\boxed{a(t)=\sqrt{2(1-e^{-2H_0t})}}.
$$

For $H_0t\ll1$,
$$
1-e^{-2H_0t}\sim2H_0t,
\qquad
\boxed{a(t)\sim2\sqrt{H_0t}}.
$$
For $H_0t\gg1$,
$$
a(t)=\sqrt2\left(1-\frac12e^{-2H_0t}
+O(e^{-4H_0t})\right),
$$
so
$$
\boxed{a(t)\longrightarrow\sqrt2}
$$
exponentially from below.

Solved by gpt-5.6-sol high.

= 16H
{parent=Paper 1}
{scope}
{title2=Logic and Set Theory}

= a
{parent=16h}
{scope}

= Solution
{parent=a}

The <compactness theorem> says that a set $T$ of first-order sentences has a model if and only if every finite subset of $T$ has a model. One direction is immediate: a model of $T$ models every subset. Conversely, suppose every finite subset is satisfiable. If $T$ had no model, then $T\models\bot$. The <Godel completeness theorem> would give a formal proof $T\vdash\bot$. Every proof is finite, so it would use only a finite subset $T_0\subseteq T$, making $T_0$ inconsistent and hence unsatisfiable. This contradiction proves compactness.

The <Upward Lowenheim-Skolem theorem> says that a first-order theory with an infinite model has models of arbitrarily large cardinality. Fix a cardinal $\kappa$ and expand the language by constants $c_\alpha$ for $\alpha<\kappa$. Add the sentences
$$
c_\alpha\ne c_\beta
\qquad(\alpha\ne\beta)
$$
to $T$. Any finite subset mentions only finitely many new constants, which can be interpreted as distinct elements of the given infinite model. Compactness therefore gives a model of the expanded theory in which the $c_\alpha$ are all distinct. Its reduct is a model of $T$ of cardinality at least $\kappa$. Together with the downward theorem, this gives a model of exactly $\kappa$ whenever $\kappa\geq|L|+\aleph_0$.

The <Downward Lowenheim-Skolem theorem> says that if $A$ is an infinite $L$-structure and
$$
|L|+\aleph_0\leq\kappa\leq|A|,
$$
then $A$ has an elementary substructure of cardinality $\kappa$. To see why, first choose $\kappa$ elements. For every existential formula, add a Skolem function selecting a witness whenever one exists, and repeatedly close the chosen set under all language operations and these witness functions. There are at most $\kappa$ functions of finite arity, so the closure still has size $\kappa$. Every existential statement true in $A$ with parameters from the closure has a witness in the closure; the Tarski--Vaught test therefore says that the resulting substructure is elementary.

Solved by gpt-5.6-sol high.

= b
{parent=16h}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

Expand $L$ by a unary function symbol $f$. Let $T^*$ contain:

* every sentence of $T$;
* the <first-order axiomatization of an automorphism> saying that $f$ is bijective and preserves every function and relation of $L$;
* the sentence $\exists x\,[f(x)\ne x]$;
* for each $n\geq1$, a sentence $\theta_n$ asserting the existence of at least $n$ distinct elements.

Every finite subset $S\subseteq T^*$ contains only finitely many size requirements, say no requirement beyond $\theta_N$. By hypothesis there is a finite nonrigid model $A\models T$ with at least $N$ elements. Interpret $f$ as a nonidentity automorphism of $A$. This expanded structure satisfies $S$; indeed it satisfies every automorphism axiom, including any finitely many that occur in $S$.

Thus every finite subset of $T^*$ is satisfiable. By the <compactness theorem>, $T^*$ has a model $M^*$. The sentences $\theta_n$ make its domain infinite, while $\exists x[f(x)\ne x]$ and the automorphism axioms make $f^{M^*}$ a nonidentity automorphism. The $L$-reduct $M$ is therefore an infinite nonrigid model of $T$. This is the <compactness transfer of finite nonrigidity>.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

No. Take $L=\{<\}$ and let $T$ be the theory of linear orders. It has finite models of every positive size. Every finite linear order is a <rigid first-order structure>: an order automorphism fixes the least element, then the next element, and inductively every element.

However, $(\mathbb Z,<)$ is an infinite model of $T$ and is not rigid. For example,
$$
n\longmapsto n+1
$$
is a nonidentity order automorphism. Hence arbitrarily large rigid finite models do not force all infinite models to be rigid.

Solved by gpt-5.6-sol high.

= 17H
{parent=Paper 1}
{scope}
{title2=Graph Theory}

= a
{parent=17h}
{scope}

= Solution
{parent=a}

Fix $k\geq1$ and sample $G$ from the <Erdős-Rényi model> with $p=\tfrac12n^{-2/3}$. Let $X$ count copies of the <complete bipartite graph> $K_{3,3}$. By the <expected subgraph count in the Erdős-Rényi model>,
$$
\mathbb EX
=\frac12\binom n3\binom{n-3}3p^9
\longrightarrow \frac{1}{72\,2^9}<1.
$$

Put $m=\lceil n/k\rceil$ and let $Z$ count <independent sets>[independent set] of size $m$. If the <chromatic number> satisfies $\chi(G)\leq k$, some colour class has at least $m$ vertices, so $Z\geq1$. Moreover,
$$
\begin{aligned}
\mathbb EZ
&=\binom nm(1-p)^{\binom m2}\\
&\leq\left(\frac{en}{m}\right)^m
   \exp\left(-p\binom m2\right)
\longrightarrow0,
\end{aligned}
$$
because the positive part of the exponent is $O(n)$ whereas $p\binom m2$ has order $n^{4/3}$. The <first moment method> therefore gives $\mathbb P(\chi(G)\leq k)\leq\mathbb EZ=o(1)$.

Using <Markov inequality> for $X$ and the <union bound>,
$$
\begin{aligned}
\mathbb P(X=0\text{ and }\chi(G)>k)
&\geq1-\mathbb P(X\geq1)-\mathbb P(\chi(G)\leq k)\\
&\geq1-\mathbb EX-o(1)>0
\end{aligned}
$$
for all sufficiently large $n$. Hence at least one such $G$ contains no $K_{3,3}$ and has $\chi(G)>k$.

Solved by gpt-5.6-sol high.

= b
{parent=17h}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

Let $X$ count copies of the <complete graph> $K_4$ in $G$. Its <expected value> is
$$
\mu=\mathbb EX=\binom n4p^6
\sim\frac{(\log n)^6}{24}\longrightarrow\infty.
$$

Write $X$ as a sum of <indicator random variables>[indicator random variable]. Indicators for two distinct copies are independent when the copies share at most one vertex, since their edge sets are then disjoint. Pairs sharing two vertices have eleven edges in their union, while pairs sharing three have nine. It follows that
$$
\operatorname{var}(X)
\leq \mu+O(n^6p^{11})+O(n^5p^9)
=O((\log n)^6)+o(1).
$$
Thus $\operatorname{var}(X)/\mu^2\to0$. By the <second moment method>, $X/\mu\to1$ in probability, so $\mathbb P(X\geq100)\to1$. This is the <sparse clique-count concentration> estimate.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

Let $Y$ count unordered pairs of distinct $K_4$ copies that have a common vertex. Classifying such a pair by whether it shares one, two, or three vertices gives
$$
\mathbb EY
=O(n^7p^{12})+O(n^6p^{11})+O(n^5p^9)
=o(1).
$$
Indeed, after substituting $p=n^{-2/3}\log n$, these terms are respectively
$$
O(n^{-1}(\log n)^{12}),\qquad
O(n^{-4/3}(\log n)^{11}),\qquad
O(n^{-1}(\log n)^9).
$$

The <first moment method> gives $\mathbb P(Y>0)\leq\mathbb EY\to0$. Part (i) says that $X\geq100$ with probability tending to one, so the <union bound> shows that, with probability tending to one, both $X\geq100$ and $Y=0$. On that event any one hundred of the $K_4$ copies are pairwise vertex-disjoint, proving <vertex-disjoint sparse clique copies> in this case.

Solved by gpt-5.6-sol high.

= 18I
{parent=Paper 1}
{scope}
{title2=Galois Theory}

= a
{parent=18i}
{scope}

= Solution
{parent=a}

A finite extension $L/K$ is <normal field extension>[normal] when every <irreducible polynomial> in $K[T]$ having one root in $L$ splits completely over $L$.

Suppose first that $L/K$ is normal. Choose $\alpha_1,\ldots,\alpha_m$ with $L=K(\alpha_1,\ldots,\alpha_m)$, and let $f_j$ be the <minimal polynomial> of $\alpha_j$ over $K$. Every $f_j$ splits in $L$ by normality, and $L$ is generated by their roots because it is already generated by the particular roots $\alpha_j$. Thus $L$ is the <splitting field> of $f_1\cdots f_m$.

Conversely, suppose that $L$ is the splitting field of $f\in K[T]$. Every $K$-embedding $\sigma:L\to\overline K$ permutes the roots of $f$, so $\sigma(L)=L$. By the embedding characterization of <normal field extension>[normality], $L/K$ is normal. This proves the <finite normal extension as a splitting field> criterion.

Solved by gpt-5.6-sol high.

= b
{parent=18i}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

This is true. If $M/K$ is normal, the <finite normal extension as a splitting field> criterion writes $M$ as the splitting field over $K$ of some $f\in K[T]$. Since $K\subseteq L\subseteq M$, the same field $M$ is the splitting field of $f$, now regarded as a polynomial over $L$. Hence $M/L$ is normal.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

This is false. Let
$$
K=\mathbb Q,qquad L=\mathbb Q(\sqrt[3]2),qquad
M=\mathbb Q(\sqrt[3]2,\omega),
$$
where $\omega$ is a primitive cube root of unity. Then $M$ is the <splitting field> of $T^3-2$ over $\mathbb Q$, so $M/K$ is normal. The <irreducible polynomial> $T^3-2$ has a root in $L$ but does not split there, so $L/K$ is not normal.

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

This is false. Take
$$
K=\mathbb Q,qquad L=\mathbb Q(\sqrt2),qquad
M=\mathbb Q(\sqrt[4]2).
$$
Each of $L/K$ and $M/L$ is a <quadratic extension> and hence normal. However, $T^4-2$ has the root $\sqrt[4]2$ in $M$ while its roots $i\sqrt[4]2$ and $-i\sqrt[4]2$ are absent. Thus $M/K$ is not normal.

Solved by gpt-5.6-sol high.

= c
{parent=18i}
{scope}

= Solution
{parent=c}

Put $\alpha=\sqrt[4]7$. The roots of $T^4-7$ are $\alpha,i\alpha,-\alpha,-i\alpha$, so its <splitting field> is
$$
L=\mathbb Q(\alpha,i).
$$
The polynomial is <irreducible polynomial>[irreducible] over $\mathbb Q$ by the Eisenstein criterion at $7$, and $\mathbb Q(\alpha)\subset\mathbb R$ does not contain $i$. The <tower law for field extensions> therefore gives $[L:\mathbb Q]=8$.

Define automorphisms
$$
r(\alpha)=i\alpha,quad r(i)=i,qquad
s(\alpha)=\alpha,quad s(i)=-i.
$$
They satisfy $r^4=s^2=1$ and $srs=r^{-1}$. The eight maps $r^j$ and $r^js$ are distinct, so they exhaust the Galois group. Hence
$$
\operatorname{Gal}(L/\mathbb Q)\cong D_8,
$$
the <dihedral group> of order eight.

The <Galois correspondence> gives all intermediate fields. The full list, with a single generator for each field, is
$$
\begin{array}{c|c|c}
\text{subgroup}&\text{fixed field}&\text{normal over }\mathbb Q\\ \hline
D_8&\mathbb Q=\mathbb Q(0)&\text{yes}\\
\langle r\rangle&\mathbb Q(i)&\text{yes}\\
\langle r^2,s\rangle&\mathbb Q(\sqrt7)&\text{yes}\\
\langle r^2,rs\rangle&\mathbb Q(\sqrt{-7})&\text{yes}\\
\langle r^2\rangle&\mathbb Q(\sqrt7+i)&\text{yes}\\
\langle s\rangle&\mathbb Q(\alpha)&\text{no}\\
\langle r^2s\rangle&\mathbb Q(i\alpha)&\text{no}\\
\langle rs\rangle&\mathbb Q((1+i)\alpha)&\text{no}\\
\langle r^3s\rangle&\mathbb Q((1-i)\alpha)&\text{no}\\
\{1\}&L=\mathbb Q(\alpha+i)&\text{yes}.
\end{array}
$$
For example, $\sqrt7+i$ has stabilizer $\langle r^2\rangle$, and $(1+i)\alpha$ has stabilizer $\langle rs\rangle$; the other entries follow similarly. Also $\alpha+i$ has trivial stabilizer, so its orbit has eight elements and the <primitive element theorem> conclusion $L=\mathbb Q(\alpha+i)$ holds explicitly.

Finally, the <normal subextension criterion> says that normal intermediate fields correspond exactly to normal subgroups. In $D_8$ these are $D_8$, $\langle r\rangle$, the two index-two Klein four-groups, $\langle r^2\rangle$, and $\{1\}$, which gives exactly the “yes” rows above. This is the <subfields of the splitting field of x to the fourth minus seven> lattice.

Solved by gpt-5.6-sol high.

= 19H
{parent=Paper 1}
{scope}
{title2=Representation Theory}

= Solution
{parent=19H}

A <complex representation> of a finite group $G$ is a homomorphism
$$
\rho:G\longrightarrow\operatorname{GL}(V)
$$
for a finite-dimensional complex vector space $V$. It is an <irreducible representation> when its only invariant subspaces are $0$ and $V$, and its <degree of a representation>[degree] is $\dim V$. Two representations are <isomorphic representations>[isomorphic] when there is an invertible linear map intertwining their $G$-actions.

Write the <dihedral group> as
$$
D_{2n}=\langle r,s:r^n=s^2=1,\ srs=r^{-1}\rangle.
$$
In an irreducible representation choose an <eigenvector> $0\ne v\in V$ of $\rho(r)$, which is possible because $r^n=1$. If its <eigenvalue> is $\lambda$, then
$$
\rho(r)\rho(s)v
=\rho(s)\rho(r)^{-1}v
=\lambda^{-1}\rho(s)v.
$$
Thus $W=\operatorname{span}\{v,\rho(s)v\}$ is invariant under both $r$ and $s$. Irreducibility forces $W=V$, proving $\dim V\leq2$ as in <irreducible complex representations of a finite dihedral group>.

Let $\omega=e^{2\pi i/n}$. Besides the one-dimensional representations described below, define for each indicated $j$
$$
\rho_j(r)=
\begin{pmatrix}\omega^j&0\\0&\omega^{-j}\end{pmatrix},
\qquad
\rho_j(s)=
\begin{pmatrix}0&1\\1&0\end{pmatrix}.
$$
These matrices satisfy the defining relations. Since $\omega^j\ne\omega^{-j}$, the only one-dimensional $r$-invariant subspaces are the two coordinate axes, and $s$ interchanges them. Hence $\rho_j$ is irreducible. Their <character of a representation>[characters] satisfy
$$
\chi_j(r)=2\cos\frac{2\pi j}{n},
$$
whose values are distinct in the ranges below, so the $\rho_j$ are pairwise nonisomorphic. This is the family of <two-dimensional representations of a finite dihedral group>.

If $n$ is odd, a <one-dimensional character> must send $r$ to a scalar $\varepsilon$ satisfying $\varepsilon^n=1$ and $\varepsilon=\varepsilon^{-1}$. Thus $\varepsilon=1$, while $s$ may independently map to $1$ or $-1$. These give two one-dimensional irreducibles. Taking
$$
j=1,\ldots,\frac{n-1}{2}
$$
gives $(n-1)/2$ two-dimensional irreducibles, for a total of
$$
2+\frac{n-1}{2}=\frac{n+3}{2}.
$$

If $n$ is even, $r$ may map to either $1$ or $-1$, and again $s$ may map independently to either sign. This gives four pairwise nonisomorphic one-dimensional irreducibles. Taking
$$
j=1,\ldots,\frac n2-1
$$
gives $n/2-1$ two-dimensional irreducibles, for a total of
$$
4+\left(\frac n2-1\right)=\frac{n+6}{2}.
$$
Different dimensions distinguish the one- and two-dimensional families, completing the required construction and justification.

Solved by gpt-5.6-sol high.

= 20H
{parent=Paper 1}
{scope}
{title2=Number Fields}

= a
{parent=20h}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

By <unique factorization of ideals in a number field>, each nonzero integral ideal has a unique expression
$$
\mathfrak a=\prod_{\mathfrak p}\mathfrak p^{e_{\mathfrak p}},
$$
with only finitely many nonzero exponents. Since the <ideal norm> is multiplicative, formally expanding one geometric series for each prime ideal gives
$$
\begin{aligned}
\prod_{\mathfrak p}
\left(1-N(\mathfrak p)^{-s}\right)^{-1}
&=\prod_{\mathfrak p}\sum_{e\geq0}N(\mathfrak p)^{-es}\\
&=\sum_{0\ne\mathfrak a\subseteq\mathcal O_K}N(\mathfrak a)^{-s}\\
&=\zeta_K(s).
\end{aligned}
$$
Uniqueness of ideal factorization says that every term occurs exactly once. This proves the <Euler product for the Dedekind zeta function> as a formal identity.

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

For each rational prime $p$, define
$$
\chi(p)=
\begin{cases}
1,&(p)=\mathfrak p_1\mathfrak p_2\text{ with }\mathfrak p_1\ne\mathfrak p_2,\\
-1,&(p)\text{ remains prime in }\mathcal O_K,\\
0,&(p)=\mathfrak p^2\text{ is ramified}.
\end{cases}
$$
These are the three cases in <splitting of rational primes in a quadratic field>.

Set $x=p^{-s}$. In the split case the local factor of the <Dedekind zeta function> is $(1-x)^{-2}$; in the inert case it is $(1-x^2)^{-1}$ because the prime ideal has norm $p^2$; and in the ramified case it is $(1-x)^{-1}$. On the other hand, the product of the local factors of $\zeta_{\mathbb Q}(s)$ and $L(\chi,s)$ is
$$
(1-x)^{-1}(1-\chi(p)x)^{-1},
$$
which gives exactly those three expressions when $\chi(p)=1,-1,0$. Multiplying over all rational primes proves the <quadratic Dedekind zeta factorization>
$$
\zeta_K(s)=\zeta_{\mathbb Q}(s)L(\chi,s)
$$
formally.

Solved by gpt-5.6-sol high.

= b
{parent=20h}
{scope}

= Solution
{parent=b}

Because $p\equiv11\pmod{12}$, we have $p\equiv3\pmod4$ and
$$
\mathcal O_K=\mathbb Z\left[\frac{1+\sqrt{-p}}2\right].
$$
For $\theta=(1+\sqrt{-p})/2$, the <minimal polynomial> is
$$
f(T)=T^2-T+\frac{p+1}{4}.
$$
Since $p\equiv2\pmod3$, reduction modulo $3$ gives $f(T)\equiv T(T-1)$. Its two roots are distinct, so
$$
(3)=\mathfrak p_1\mathfrak p_2
$$
with distinct prime ideals of norm $3$. Thus $3$ splits completely, as recorded in <splitting of three in Q of square root minus p>.

The <ideal class group> $\operatorname{Cl}_K$ is the group of nonzero fractional ideals of $\mathcal O_K$ modulo the subgroup of principal fractional ideals. Saying that $\mathfrak p_1$ has order $n$ means that $n$ is the least positive integer for which $\mathfrak p_1^n$ is principal.

Suppose for contradiction that $n$ is odd and $\mathfrak p_1$ has order $n$. Then $\mathfrak p_1^n=(\alpha)$ for some $\alpha\in\mathcal O_K$, and taking <ideal norm>[norms] gives
$$
|N_{K/\mathbb Q}(\alpha)|=N(\mathfrak p_1)^n=3^n.
$$
Write $\alpha=(a+b\sqrt{-p})/2$, where $a,b\in\mathbb Z$ have the same parity. Then
$$
4\cdot3^n=a^2+pb^2.
$$
If $b\ne0$, the right-hand side is at least $p$, contradicting $p>3^{n+2}>4\cdot3^n$. Hence $b=0$, so $\alpha$ is a rational algebraic integer and therefore an integer. But then $3^n=N(\alpha)=\alpha^2$, impossible when $n$ is odd. This proves the <odd-order obstruction for a split prime in an imaginary quadratic field> and shows that the order cannot equal $n$.

Solved by gpt-5.6-sol high.

= 21G
{parent=Paper 1}
{scope}
{title2=Algebraic Topology}

= Solution
{parent=21G}

Given <group homomorphisms> $i_A:C\to A$ and $i_B:C\to B$, their <amalgamated free product> is a group $A*_CB$ with maps $j_A:A\to A*_CB$ and $j_B:B\to A*_CB$ satisfying $j_Ai_A=j_Bi_B$, such that every pair $f_A:A\to H$, $f_B:B\to H$ satisfying $f_Ai_A=f_Bi_B$ factors through a unique homomorphism $A*_CB\to H$.

The <Seifert-van Kampen theorem> says that if $X=U_1\cup U_2$, where $U_1,U_2$, and $U_1\cap U_2$ are open and path-connected and $x_0\in U_1\cap U_2$, then the inclusion maps induce
$$
\pi_1(X,x_0)\cong
\pi_1(U_1,x_0)*_{\pi_1(U_1\cap U_2,x_0)}
\pi_1(U_2,x_0).
$$

Here is a direct proof of the requested generation statement. For a <based loop> $\gamma$ in $X$, the <Lebesgue number lemma> supplies a subdivision
$$
0=t_0<t_1<\cdots<t_m=1
$$
such that every arc $\gamma|_{[t_{j-1},t_j]}$ lies in one $U_{\varepsilon_j}$. Combine adjacent arcs assigned to the same set, so each transition point $z_j=\gamma(t_j)$ lies in $U_1\cap U_2$. Since the intersection is <path-connected space>[path-connected], choose a path $q_j$ in it from $x_0$ to $z_j$, taking $q_0=q_m$ constant. Then
$$
q_{j-1}*\gamma|_{[t_{j-1},t_j]}*\overline{q_j}
$$
is a loop in $U_{\varepsilon_j}$. In the product of their classes, each $\overline{q_j}*q_j$ cancels by <path reversal>, leaving $[\gamma]$. Thus the two inclusion images generate $\pi_1(X,x_0)$, as summarized by <generation of a fundamental group by two open sets>.

Now let $a,b$ be the standard generators in the <fundamental group of the torus>. A <Möbius band> retracts to a core circle, while its boundary traverses that core twice. If $x$ and $y$ denote the core classes of the two attached bands, the <Seifert-van Kampen theorem> gives
$$
\pi_1(Y)
\cong
\langle a,b,x,y\mid [a,b]=1,\ x^2=ab,\ y^2=a^2b^3\rangle.
$$

The two attaching classes $A=ab$ and $B=a^2b^3$ commute. Their exponent matrix has determinant
$$
\det\begin{pmatrix}1&1\\2&3\end{pmatrix}=1,
$$
so they form a basis of $\langle a,b\rangle\cong\mathbb Z^2$, with
$$
a=A^3B^{-1},\qquad b=A^{-2}B.
$$
Substituting $A=x^2$ and $B=y^2$ eliminates $a,b$ and leaves the two-generator one-relator presentation
$$
\boxed{\pi_1(Y)\cong\langle x,y\mid[x^2,y^2]=1\rangle.}
$$
This is the calculation in <two Möbius bands attached to a torus>.

Finally send $x$ to $(12)$ and $y$ to $(23)$ in the <symmetric group> $S_3$. Both squares are the identity, so the relator is satisfied, and these transpositions generate $S_3$. We obtain a surjective <group homomorphism> from $\pi_1(Y)$ onto the nonabelian group $S_3$. Therefore $\pi_1(Y)$ is nonabelian.

Solved by gpt-5.6-sol high.

= 22F
{parent=Paper 1}
{scope}
{title2=Linear Analysis}

= a
{parent=22f}
{scope}

= Solution
{parent=a}

The <open mapping theorem> states that every surjective bounded linear map $S:X\to Y$ between <Banach spaces> maps open sets to open sets. Equivalently, there is $c>0$ such that
$$
B_Y(0,c)\subseteq S(B_X(0,1)).
$$

The <closed graph theorem> states that a linear map $T:X\to Y$ between Banach spaces is continuous if and only if its <graph of a linear operator>[graph]
$$
\Gamma(T)=\{(x,Tx):x\in X\}
$$
is closed in $X\times Y$.

Assume the open mapping theorem and suppose $\Gamma(T)$ is closed. As a closed subspace of the Banach space $X\times Y$, the graph is itself Banach. The coordinate projection
$$
P:\Gamma(T)\longrightarrow X,\qquad (x,Tx)\longmapsto x
$$
is a bounded linear bijection. By the open mapping theorem, $P^{-1}$ is bounded. The other coordinate projection $Q:\Gamma(T)\to Y$ is bounded, and
$$
T=Q\circ P^{-1}.
$$
Thus $T$ is bounded and hence continuous. This is the <closed graph theorem from the bounded inverse theorem> argument.

Solved by gpt-5.6-sol high.

= b
{parent=22f}
{scope}

= Solution
{parent=b}

The <continuous dual space> of a normed vector space $V$ is
$$
V^*=\{f:V\to\mathbb F:f\text{ is linear and continuous}\},
$$
with the <operator norm>
$$
\lVert f\rVert=\sup_{\lVert v\rVert\leq1}|f(v)|.
$$

Let $(f_n)$ be a <cauchy sequence> in this norm. For each $v\in V$,
$$
|f_n(v)-f_m(v)|\leq\lVert f_n-f_m\rVert\,\lVert v\rVert,
$$
so $(f_n(v))$ is Cauchy in the scalar field. Define $f(v)=\lim_n f_n(v)$. Passing to limits in the linearity identities shows that $f$ is linear.

A norm-Cauchy sequence is bounded, say $\lVert f_n\rVert\leq M$. Hence $|f(v)|\leq M\lVert v\rVert$, so $f\in V^*$. Given $\varepsilon>0$, choose $N$ such that $\lVert f_n-f_m\rVert<\varepsilon$ for $m,n\geq N$. Letting $m\to\infty$ gives
$$
|f_n(v)-f(v)|\leq\varepsilon\lVert v\rVert
$$
for every $v$, and therefore $\lVert f_n-f\rVert\leq\varepsilon$. Thus $f_n\to f$ in operator norm, proving <completeness of the dual space>: $V^*$ is Banach.

Solved by gpt-5.6-sol high.

= c
{parent=22f}
{scope}

= Solution
{parent=c}

Write $T_v=T(v)$. We prove that $T$ has a closed graph. Suppose
$$
v_n\longrightarrow v\quad\text{in }V,
\qquad
T_{v_n}\longrightarrow f\quad\text{in }V^*.
$$
Put $u_n=v_n-v$ and $g=f-T_v$. Then $u_n\to0$ and $T_{u_n}\to g$ in $V^*$.

Fix $w\in V$ and $t\in\mathbb R$. Positivity and linearity give
$$
\begin{aligned}
0&\leq T_{u_n+tw}(u_n+tw)\\
 &=T_{u_n}(u_n)
 +t\bigl(T_{u_n}(w)+T_w(u_n)\bigr)
 +t^2T_w(w).
\end{aligned}
$$
Because $(T_{u_n})$ converges in the <continuous dual space>, its norms are bounded, so
$$
|T_{u_n}(u_n)|\leq\lVert T_{u_n}\rVert\lVert u_n\rVert\longrightarrow0.
$$
Also $T_{u_n}(w)\to g(w)$, while $T_w(u_n)\to0$ because the fixed functional $T_w$ is continuous. Passing to the limit yields
$$
0\leq t g(w)+t^2T_w(w)
$$
for every real $t$. If $g(w)\ne0$, a sufficiently small $t$ of the opposite sign makes the right-hand side negative. Hence $g(w)=0$ for every $w$, so $g=0$ and $f=T_v$.

The graph is therefore closed. The <closed graph theorem> now proves that $T:V\to V^*$ is continuous. This is precisely <continuity of a positive linear map into a dual space>.

Solved by gpt-5.6-sol high.

= 23F
{parent=Paper 1}
{scope}
{title2=Analysis of Functions}

= a
{parent=23f}
{scope}

= Solution
{parent=a}

For $s>\tfrac12$, the <Sobolev trace theorem> states that the restriction map initially defined on smooth functions extends uniquely to a bounded linear operator
$$
\gamma:H^s(\mathbb R^n)
\longrightarrow H^{s-1/2}(\mathbb R^{n-1}),
\qquad
(\gamma u)(x')=u(x',0).
$$

Take first $u$ in the <Schwartz space>. Up to the harmless constant determined by the <Fourier transform> convention,
$$
\widehat{\gamma u}(\xi')
=\int_{\mathbb R}\widehat u(\xi',\xi_n)\,d\xi_n.
$$
The <Cauchy-Schwarz inequality>, with weights $1+|\xi'|^2+\xi_n^2$, gives
$$
\begin{aligned}
|\widehat{\gamma u}(\xi')|^2
&\leq
\left(\int_{\mathbb R}
(1+|\xi'|^2+\xi_n^2)^s
|\widehat u(\xi',\xi_n)|^2\,d\xi_n\right)\\
&\quad\times
\left(\int_{\mathbb R}
(1+|\xi'|^2+\xi_n^2)^{-s}\,d\xi_n\right).
\end{aligned}
$$
After the substitution $\xi_n=(1+|\xi'|^2)^{1/2}t$, the second factor is
$$
C_s(1+|\xi'|^2)^{1/2-s},
\qquad
C_s=\int_{\mathbb R}(1+t^2)^{-s}\,dt<\infty,
$$
where finiteness is exactly the condition $s>1/2$. Multiplying by $(1+|\xi'|^2)^{s-1/2}$ and integrating in $\xi'$ yields
$$
\lVert\gamma u\rVert_{H^{s-1/2}(\mathbb R^{n-1})}
\leq C_s^{1/2}\lVert u\rVert_{H^s(\mathbb R^n)}.
$$
Density of the Schwartz space in the <Sobolev space> completes the unique bounded extension.

Solved by gpt-5.6-sol high.

= b
{parent=23f}
{scope}

= Solution
{parent=b}

For $1\leq p<\infty$, choose nonzero $\phi\in C_c(\mathbb R^{n-1})$ and $\psi\in C_c(\mathbb R)$ with $\psi(0)=1$. Define
$$
u_k(x',x_n)=\phi(x')\psi(kx_n).
$$
These functions are continuous and belong to the <Lebesgue space> $L^p(\mathbb R^n)$, while their restrictions to the hyperplane are all the same:
$$
u_k(x',0)=\phi(x').
$$
Changing variables in the normal coordinate gives
$$
\lVert u_k\rVert_{L^p(\mathbb R^n)}^p
=\frac1k\lVert\phi\rVert_{L^p(\mathbb R^{n-1})}^p
\lVert\psi\rVert_{L^p(\mathbb R)}^p
\longrightarrow0.
$$
If the claimed bounded $T$ existed, then
$$
0<\lVert\phi\rVert_p
=\lVert Tu_k\rVert_p
\leq\lVert T\rVert\,\lVert u_k\rVert_p
\longrightarrow0,
$$
a contradiction. This scaling argument is the <failure of an Lp hyperplane trace>.

Solved by gpt-5.6-sol high.

= c
{parent=23f}
{scope}

= Solution
{parent=c}

Compact support and the <fundamental theorem of calculus> give, for every $(x,y)$,
$$
\begin{aligned}
|u(x,y)|^2
&=\left|\int_{-\infty}^y\partial_t|u(x,t)|^2\,dt\right|\\
&\leq2\int_{\mathbb R}|u(x,t)|\,|\partial_2u(x,t)|\,dt
\leq2A(x),
\end{aligned}
$$
where
$$
A(x)=\int_{\mathbb R}|u(x,t)|\,|\nabla u(x,t)|\,dt.
$$
Applying the same argument in the other coordinate gives
$$
|u(x,y)|^2\leq2B(y),
\qquad
B(y)=\int_{\mathbb R}|u(t,y)|\,|\nabla u(t,y)|\,dt.
$$

Multiplication and <Fubini's theorem> now yield
$$
\begin{aligned}
\int_{\mathbb R^2}|u(x,y)|^4\,dx\,dy
&\leq4\int_{\mathbb R^2}A(x)B(y)\,dx\,dy\\
&=4\left(\int_{\mathbb R^2}|u|\,|\nabla u|\right)^2.
\end{aligned}
$$
Finally, the <Cauchy-Schwarz inequality> gives
$$
\left(\int_{\mathbb R^2}|u|\,|\nabla u|\right)^2
\leq
\left(\int_{\mathbb R^2}|u|^2\right)
\left(\int_{\mathbb R^2}|\nabla u|^2\right).
$$
Thus the required estimate holds with $C=4$; it is the <Ladyzhenskaya inequality in two dimensions>.

Solved by gpt-5.6-sol high.

= 24F
{parent=Paper 1}
{scope}
{title2=Riemann Surfaces}

= Solution
{parent=24F}

A <germ of a holomorphic function> at $z\in D$ is an equivalence class $[f]_z$ of pairs consisting of a neighbourhood $U$ of $z$ and a holomorphic function $f:U\to\mathbb C$. Two pairs represent the same germ if their functions agree on some neighbourhood of $z$.

The <space of germs of holomorphic functions> is
$$
\mathcal G=\{[f]_z:z\in D\}.
$$
For every holomorphic $f$ on an open $U\subseteq D$, the set
$$
\widetilde U_f=\{[f]_z:z\in U\}
$$
is declared open; these sets form a basis. The forgetful map
$$
\pi:\mathcal G\to D,
\qquad \pi([f]_z)=z,
$$
maps each $\widetilde U_f$ homeomorphically onto $U$. The inverses of these restrictions give the charts defining the complex structure, so $\pi$ is a local biholomorphism.

The evaluation map is
$$
\mathcal E:\mathcal G\to\mathbb C,
\qquad \mathcal E([f]_z)=f(z).
$$
It is well-defined by the germ equivalence relation. On the chart $\widetilde U_f$, its coordinate expression is
$$
\mathcal E\circ(\pi|_{\widetilde U_f})^{-1}(z)=f(z),
$$
which is holomorphic. Hence $\mathcal E$ is analytic, as stated in <evaluation map on a space of germs>.

Now put
$$
D=\mathbb C\setminus\{\zeta:\zeta^8=1\},
\qquad
R=\{(z,w)\in D\times\mathbb C:w^2=z^8-1\}.
$$
This gives the <germ surface of the square root of z to the eighth minus one>. An explicit gluing description is obtained by pairing the eight roots into four adjacent pairs and cutting the plane along four disjoint arcs joining the members of each pair. On the complement of the cuts choose one branch $q(z)$ of $\sqrt{z^8-1}$. Take two copies, labelled by $q$ and $-q$, and glue the upper bank of each cut in one copy to the lower bank in the other, and conversely. The cut interiors are restored by the gluing, while the eight endpoints remain absent because they are not in $D$.

On this surface define
$$
\pi(z,w)=z,
\qquad
\mathcal E(z,w)=w.
$$
For each $(z,w)\in R$, the <holomorphic function> $z^8-1$ is nonzero near $z$, so it has a unique local square root $q$ with $q(z)=w$. Define
$$
\Phi:R\longrightarrow\mathcal G,
\qquad
\Phi(z,w)=[q]_z.
$$
This is well-defined, injective, and analytic in the displayed local sheets; its inverse on its image is $[q]_z\mapsto(z,q(z))$. Thus $\Phi$ is the required analytic embedding, and it intertwines both the forgetful and evaluation maps.

To compactify, first add one point above each of the eight roots of unity. These are simple branch points of the double cover. Since the polynomial has even degree eight, the <even-degree hyperelliptic model> has two distinct, unbranched points above infinity. Consequently
$$
|\overline R\setminus R|=8+2=10.
$$

Finally apply the <Riemann-Hurwitz formula> to the degree-two meromorphic map $\bar\pi:\overline R\to\mathbb C_\infty$. Its only ramification consists of the eight simple finite branch points, so
$$
2g(\overline R)-2
=2(2g(\mathbb C_\infty)-2)+8
=2(-2)+8=4.
$$
Therefore
$$
\boxed{g(\overline R)=3.}
$$
This is the <compactification of y squared equals x to the eighth minus one>.

Solved by gpt-5.6-sol high.

= 25G
{parent=Paper 1}
{scope}
{title2=Algebraic Geometry}

= Solution
{parent=25G}

Let $X\subseteq\mathbb A^n$ have ideal $I(X)$, and let $p\in X$. Its <Zariski tangent space> is
$$
T_pX
=\left\{v\in k^n:
\sum_{j=1}^n\frac{\partial f}{\partial x_j}(p)v_j=0
\text{ for every }f\in I(X)\right\}.
$$
If $X$ is irreducible, $p$ is smooth when
$$
\dim T_pX=\dim X;
$$
otherwise it is singular. Equivalently, for generators of $I(X)$, the <Jacobian matrix>[Jacobian] has the maximum rank $n-\dim X$ at $p$.

Now let the irreducible affine plane cubic be $X=V(f)\subseteq\mathbb A^2$. Suppose distinct points $p,q\in X$ were both singular. Parametrize their joining line by
$$
\ell(t)=p+t(q-p)
$$
and put $h(t)=f(\ell(t))$. Then $\deg h\leq3$. At $t=0$,
$$
h(0)=0,
\qquad
h'(0)=\nabla f(p)\mathbin\cdot(q-p)=0,
$$
so zero is a root of multiplicity at least two. The same argument at $t=1$ gives another root of multiplicity at least two. Therefore $h$ has at least four roots counted with multiplicity and must vanish identically. The line through $p,q$ is then contained in $X$, so its linear equation divides $f$, contradicting irreducibility. This proves <singular points of an irreducible affine plane cubic> are unique when they exist.

For the density statement, embed an irreducible affine variety $X$ of dimension $d$ in $\mathbb A^n$ and choose generators $f_1,\ldots,f_r$ of its ideal. By <dimension from minimum tangent dimension>, some point has tangent dimension $d$, so the Jacobian has rank $n-d$ there. Some $(n-d)\times(n-d)$ minor $\Delta$ is consequently nonzero on $X$. The distinguished open set
$$
X_\Delta=\{p\in X:\Delta(p)\ne0\}
$$
is nonempty, and at each of its points the Jacobian rank is at least $n-d$. Since every tangent space has dimension at least $d$, the rank is also at most $n-d$. Thus every point of $X_\Delta$ is smooth. A nonempty open subset of an <irreducible topological space> is dense, proving <density of the smooth locus>.

Finally write the smooth irreducible <projective hypersurface> as
$$
X=V_+(F)\subseteq\mathbb P^n
$$
for an irreducible homogeneous polynomial $F$. The closure of $\pi^{-1}(X)$ is its <affine cone over a projective hypersurface>
$$
Y=V(F)\subseteq\mathbb A^{n+1}.
$$
At a nonzero point $y\in Y$, simultaneous vanishing of all partial derivatives of $F$ would make the projective point $[y]\in X$ singular. Since $X$ is smooth, this cannot happen. Hence every nonzero point of $Y$ is smooth, and the origin is its only possible singular point.

Both possibilities occur. If $X$ is a projective hyperplane, then $F$ is linear and $Y$ is an affine linear subspace, hence smooth even at the origin. For a singular cone, take the smooth conic
$$
X=V_+(x_0x_2-x_1^2)\subseteq\mathbb P^2.
$$
Its affine cone is
$$
Y=V(x_0x_2-x_1^2)\subseteq\mathbb A^3.
$$
All first derivatives vanish at the origin, so the vertex is singular, while the preceding argument shows that it is the only singular point.

Solved by gpt-5.6-sol high.

= 26G
{parent=Paper 1}
{scope}
{title2=Differential Geometry}

= a
{parent=26g}
{scope}

= Solution
{parent=a}

A point $y_0\in Y$ is a <regular value> of the smooth map $f:X\to Y$ if, for every $x\in f^{-1}(y_0)$, the derivative
$$
D f_x:T_xX\longrightarrow T_{y_0}Y
$$
is surjective. A value outside the image is regular vacuously.

For a nonvacuous example, take
$$
f:\mathbb R^2\to\mathbb R,
\qquad f(x,y)=x^2+y^2.
$$
The value $1$ is regular: its fibre is the unit circle, and at every $(x,y)$ on that circle,
$$
D f_{(x,y)}(u,v)=2xu+2yv
$$
is a nonzero linear functional and hence surjective.

<Sard theorem> states that the set of critical values of a smooth map between smooth manifolds has measure zero in the target. Equivalently, almost every target point is a regular value.

Solved by gpt-5.6-sol high.

= b
{parent=26g}
{scope}

= Solution
{parent=b}

Assume first that $Y$ is connected; otherwise the definition is made separately on each connected component. Choose a <regular value> $y\in Y$ and define the <degree modulo two> by
$$
\deg_2(f)=|f^{-1}(y)|\pmod2.
$$

The following course results make this well-defined.

* By <Sard theorem>, regular values are dense, so one can choose $y$.
* By the <preimage theorem>, $f^{-1}(y)$ is a zero-dimensional submanifold because $\dim X=\dim Y$. It is discrete and closed in compact $X$, hence finite.
* The parity is independent of the regular value. Indeed, after choosing a path between two regular values transverse to $f$, its inverse image is a compact one-dimensional manifold whose boundary is the disjoint union of the two fibres. Every compact one-manifold has an even number of boundary points.

The same transverse-preimage argument applied to a smooth homotopy proves homotopy invariance. Consequently the definition depends only on $f$, rather than on the chosen regular value.

Solved by gpt-5.6-sol high.

= c
{parent=26g}
{scope}

= Solution
{parent=c}

For $v\in T_pS$, let $\gamma_v$ be the maximal <geodesic> satisfying
$$
\gamma_v(0)=p,
\qquad
\dot\gamma_v(0)=v.
$$
The <exponential map> is
$$
\exp_p(v)=\gamma_v(1)
$$
on the <domain of the exponential map>
$$
\mathcal D_p
=\{v\in T_pS:\gamma_v\text{ is defined throughout }[0,1]\}.
$$
This is an open, star-shaped neighbourhood of zero. In local coordinates the geodesic equation is a smooth ordinary differential equation, so smooth dependence on its initial position and velocity proves that $(p,v)\mapsto\exp_p(v)$ is smooth on its domain.

For both requested phenomena consider the embedded surface
$$
S=\{(x,y,0):(x,y)\ne(0,0)\}
$$
with $p=(1,0,0)$. Its geodesics are Euclidean straight lines for as long as they remain in $S$. The initial vector $v=(-2,0,0)$ would give
$$
\gamma_v(t)=(1-2t,0,0),
$$
which reaches the missing origin at $t=1/2$. Thus $v\notin\mathcal D_p$, and the domain is not all of $T_pS$.

The point $q=(-1,0,0)$ is not in the image of $\exp_p$. Any geodesic from $p$ to $q$ would have to be their unique Euclidean straight line, which passes through the omitted origin. Hence this exponential map is also not surjective, as summarized by <exponential map of the punctured plane>.

Solved by gpt-5.6-sol high.

= d
{parent=26g}
{scope}

= Solution
{parent=d}

The induced metric makes the compact surface $S$ a compact Riemannian manifold. By the <Hopf-Rinow theorem> it is geodesically complete, so every $\exp_p$ is defined on all of $T_pS$. In particular, $\exp_p(V(p))$ is defined for every $p$.

The global exponential map
$$
\operatorname{Exp}:TS\to S,
\qquad (p,v)\mapsto\exp_p(v),
$$
is smooth by smooth dependence of geodesics on initial data. Since $p\mapsto(p,V(p))$ is a smooth section of $TS$, their composition
$$
\phi(p)=\operatorname{Exp}(p,V(p))
$$
is smooth.

For $0\leq t\leq1$, define
$$
\phi_t(p)=\exp_p(tV(p)).
$$
Completeness makes this a well-defined smooth homotopy, with $\phi_0=\operatorname{id}_S$ and $\phi_1=\phi$. By homotopy invariance of <degree modulo two>,
$$
\deg_2(\phi)=\deg_2(\operatorname{id}_S)=1.
$$
This is the <exponential displacement map on a compact surface>.

Solved by gpt-5.6-sol high.

= 27K
{parent=Paper 1}
{scope}
{title2=Probability and Measure}

= a
{parent=27k}
{scope}

= Solution
{parent=a}

<Dynkin lemma>, also called the pi-lambda theorem, states that if a <Dynkin system> $\mathcal D$ contains a <pi-system> $\mathcal P$, then
$$
\sigma(\mathcal P)\subseteq\mathcal D.
$$

Let $\lambda(\mathcal P)$ be the smallest Dynkin system containing $\mathcal P$. It is enough to prove that $\lambda(\mathcal P)$ is a <sigma-algebra>, because then
$$
\sigma(\mathcal P)\subseteq\lambda(\mathcal P)\subseteq\mathcal D.
$$

First fix $A\in\mathcal P$ and define
$$
\mathcal G_A
=\{B\in\lambda(\mathcal P):A\cap B\in\lambda(\mathcal P)\}.
$$
This is a Dynkin system. It contains $\Omega$ because $A\in\lambda(\mathcal P)$; closure under relative complements follows from
$$
A\cap B^c=A\setminus(A\cap B),
$$
and closure under disjoint unions follows by distributing $A\cap-$ over the union. Since $\mathcal P$ is closed under intersections, $\mathcal P\subseteq\mathcal G_A$. Minimality therefore gives
$$
\lambda(\mathcal P)\subseteq\mathcal G_A.
$$
Thus $A\cap B\in\lambda(\mathcal P)$ whenever $A\in\mathcal P$ and $B\in\lambda(\mathcal P)$.

Now fix $B\in\lambda(\mathcal P)$ and define
$$
\mathcal H_B
=\{A\in\lambda(\mathcal P):A\cap B\in\lambda(\mathcal P)\}.
$$
The same argument shows that $\mathcal H_B$ is a Dynkin system, and the preceding paragraph shows that it contains $\mathcal P$. Hence $\mathcal H_B=\lambda(\mathcal P)$. We have proved that $\lambda(\mathcal P)$ is itself a pi-system.

A Dynkin system that is also a pi-system is a sigma-algebra: it is closed under arbitrary finite intersections, hence finite unions by complements, and any countable union can be disjointified before using closure under disjoint unions. Therefore $\lambda(\mathcal P)=\sigma(\mathcal P)$, proving the lemma.

Solved by gpt-5.6-sol high.

= b
{parent=27k}
{scope}

= Solution
{parent=b}

Fix $A_1\in\mathcal A_1$ and let
$$
\mathcal D_{A_1}
=\left\{B\in\mathcal F:
\mathbb P(A_1\cap B)=\mathbb P(A_1)\mathbb P(B)\right\}.
$$
This is a <Dynkin system>. It contains $\Omega$. If $B\in\mathcal D_{A_1}$, then
$$
\begin{aligned}
\mathbb P(A_1\cap B^c)
&=\mathbb P(A_1)-\mathbb P(A_1\cap B)\\
&=\mathbb P(A_1)(1-\mathbb P(B))\\
&=\mathbb P(A_1)\mathbb P(B^c),
\end{aligned}
$$
so it is closed under complements. Countable additivity gives closure under countable disjoint unions.

The hypothesis says $\mathcal A_2\subseteq\mathcal D_{A_1}$. Since $\mathcal A_2$ is a <pi-system>, <Dynkin lemma> yields
$$
\sigma(\mathcal A_2)\subseteq\mathcal D_{A_1}.
$$
Thus the factorization holds for every $A_1\in\mathcal A_1$ and every $B\in\sigma(\mathcal A_2)$.

Now fix such a $B$ and define
$$
\mathcal E_B
=\left\{A\in\mathcal F:
\mathbb P(A\cap B)=\mathbb P(A)\mathbb P(B)\right\}.
$$
The identical calculation makes $\mathcal E_B$ a Dynkin system. The first step gives $\mathcal A_1\subseteq\mathcal E_B$, so a second application of Dynkin lemma gives
$$
\sigma(\mathcal A_1)\subseteq\mathcal E_B.
$$
Therefore the probability factorization holds for every
$$
A\in\sigma(\mathcal A_1),
\qquad B\in\sigma(\mathcal A_2),
$$
which is precisely independence of the two <independent sigma-algebras>[sigma-algebras]. This is <independence extended from generating pi-systems>.

Solved by gpt-5.6-sol high.

= 28J
{parent=Paper 1}
{scope}
{title2=Applied Probability}

= a
{parent=28j}
{scope}

= Solution
{parent=a}

Superpose the two independent <Poisson processes>. The combined process has rate $9$, and each arrival is independently a Number 1 bus with probability $1/9$ and a Number 8 bus with probability $8/9$. Therefore the number of Number 8 buses before the first Number 1 has the <geometric distribution> on $\{0,1,2,\ldots\}$, and
$$
\mathbb P(\text{exactly three Number 8 buses})
=\left(\frac89\right)^3\frac19
=\boxed{\frac{512}{6561}}.
$$
This is <number of one Poisson type before another> with rates $1$ and $8$.

Solved by gpt-5.6-sol high.

= b
{parent=28j}
{scope}

= Solution
{parent=b}

For $0<j_1<\cdots<j_n<t$, the probability of one jump in each infinitesimal interval around $j_1,\ldots,j_n$ and no other jump by time $t$ gives the joint density
$$
\lambda^n e^{-\lambda t}
$$
for $(J_1,ldots,J_n)$ together with the event $N_t=n$. Since
$$
\mathbb P(N_t=n)=e^{-\lambda t}\frac{(\lambda t)^n}{n!},
$$
the conditional density is
$$
\frac{\lambda^ne^{-\lambda t}}
{e^{-\lambda t}(\lambda t)^n/n!}
=\frac{n!}{t^n}
$$
on that ordered simplex.

Now $n$ independent <uniform distribution>[uniform] points on $[0,t]$ have joint density $t^{-n}$ on the cube. Each ordered vector has $n!$ permutations, so their <order statistics> have density $n!/t^n$ on the same simplex. The densities agree, proving <Poisson process conditional arrival times>.

Solved by gpt-5.6-sol high.

= c
{parent=28j}
{scope}

= Solution
{parent=c}

Condition on $N_t=n$. By <Poisson process conditional arrival times>, the jump times are the ordered version of independent $U_1,\ldots,U_n$ uniformly distributed on $[0,t]$. The sum is symmetric in the marked pairs, and the marks are independent, so
$$
\begin{aligned}
&\mathbb E\left[
\left.
\exp\left\{\theta\sum_{i=1}^{N_t}g(J_i,X_i)\right\}
\right|N_t=n\right]\\
&\qquad=
\left[
\frac1t\int_0^t
\mathbb E\left(e^{\theta g(s,X_1)}\right)ds
\right]^n.
\end{aligned}
$$
Write the bracket as $a$. Averaging over the <Poisson distribution> of $N_t$ gives
$$
\begin{aligned}
\mathbb E\exp\left\{\theta\sum_{i=1}^{N_t}g(J_i,X_i)\right\}
&=e^{-\lambda t}\sum_{n=0}^\infty
\frac{(\lambda t a)^n}{n!}\\
&=\exp\{\lambda t(a-1)\}\\
&=\exp\left\{
\lambda\int_0^t
\left(\mathbb E(e^{\theta g(s,X_1)})-1\right)ds
\right\}.
\end{aligned}
$$
This is the <exponential formula for a marked Poisson sum>.

Solved by gpt-5.6-sol high.

= d
{parent=28j}
{scope}

= Solution
{parent=d}

Use the convention $J_0=0$, so $A(t)=t$ when $N_t=0$. Conditional on $N_t=n\geq1$, the last jump $J_n$ is the largest of $n$ independent uniform points on $[0,t]$. The stated <order statistic> formula gives
$$
\mathbb E(J_n\mid N_t=n)=\frac{nt}{n+1},
$$
and hence, for every $n\geq0$,
$$
\mathbb E(A(t)\mid N_t=n)=\frac{t}{n+1}.
$$
Putting $\mu=\lambda t$ and averaging over $N_t$,
$$
\begin{aligned}
\mathbb EA(t)
&=t e^{-\mu}\sum_{n=0}^\infty
\frac{\mu^n}{n!(n+1)}\\
&=t e^{-\mu}\frac{e^\mu-1}{\mu}\\
&=\boxed{\frac{1-e^{-\lambda t}}{\lambda}}.
\end{aligned}
$$
This agrees with <age of a Poisson process>.

Solved by gpt-5.6-sol high.

= 29K
{parent=Paper 1}
{scope}
{title2=Principles of Statistics}

= a
{parent=29k}
{scope}

= Solution
{parent=a}

Under the usual regularity conditions, consistency and <asymptotic normality of a maximum likelihood estimator> give
$$
\boxed{
\sqrt n(\widehat\theta_n-\theta_0)
\xrightarrow d
N_p\bigl(0,I(\theta_0)^{-1}\bigr),
}
$$
where $I(\theta_0)$ is the one-observation <Fisher information matrix>. In a scalar model this reads $N(0,I(\theta_0)^{-1})$.

Solved by gpt-5.6-sol high.

= b
{parent=29k}
{scope}

= Solution
{parent=b}

The <multivariate Wald statistic> for a candidate parameter $\theta\in\mathbb R^p$ is
$$
W_n(\theta)
=n(\widehat\theta_n-\theta)^T
I(\widehat\theta_n)
(\widehat\theta_n-\theta).
$$
Replacing $I(\widehat\theta_n)$ by any consistent estimator of $I(\theta_0)$ gives the same limit.

Under $P_{\theta_0}$, part (a) and consistency of the information matrix imply, by <Slutsky theorem>, that
$$
I(\widehat\theta_n)^{1/2}
\sqrt n(\widehat\theta_n-\theta_0)
\xrightarrow d N_p(0,I_p).
$$
Consequently
$$
\boxed{W_n(\theta_0)\xrightarrow d\chi_p^2,}
$$
using the characterization of the <chi-squared distribution> as the squared norm of a standard normal vector.

If $q_{p,1-\alpha}$ is the $(1-\alpha)$ quantile of $\chi_p^2$, an asymptotic $(1-\alpha)$ confidence region is
$$
\boxed{
C_n=\{\theta\in\mathbb R^p:W_n(\theta)\leq q_{p,1-\alpha}\}.
}
$$

Solved by gpt-5.6-sol high.

= c
{parent=29k}
{scope}

= Solution
{parent=c}

Let
$$
R=\begin{pmatrix}I_k&0\end{pmatrix}\in\mathbb R^{k\times p},
\qquad
r=R\theta^*.
$$
Then the null hypothesis is $R\theta_0=r$. Define the <Wald statistic for linear restrictions>
$$
W_{n,R}
=n(R\widehat\theta_n-r)^T
\left[R I(\widehat\theta_n)^{-1}R^T\right]^{-1}
(R\widehat\theta_n-r).
$$

Under $H_0$, part (a) and the continuous mapping theorem give
$$
\sqrt n(R\widehat\theta_n-r)
=\sqrt nR(\widehat\theta_n-\theta_0)
\xrightarrow d N_k(0,V_0),
$$
where
$$
V_0=R I(\theta_0)^{-1}R^T.
$$
The matrix $V_0$ is positive definite because $I(\theta_0)$ is positive definite and $R$ has full row rank. Consistency gives
$$
R I(\widehat\theta_n)^{-1}R^T\xrightarrow p V_0.
$$
Therefore <Slutsky theorem> yields
$$
V_0^{-1/2}\sqrt n(R\widehat\theta_n-r)
\xrightarrow dN_k(0,I_k),
$$
and applying the continuous squared-norm map proves rigorously that
$$
\boxed{W_{n,R}\xrightarrow d\chi_k^2.}
$$

Let $q_{k,1-\alpha}$ be the $(1-\alpha)$ quantile of $\chi_k^2$. Rejecting $H_0$ when
$$
W_{n,R}>q_{k,1-\alpha}
$$
has rejection probability tending to $\alpha$ under every fixed parameter satisfying $H_0$, and is therefore an asymptotically valid level-$\alpha$ test.

Solved by gpt-5.6-sol high.

= 30K
{parent=Paper 1}
{scope}
{title2=Stochastic Financial Models}

= a
{parent=30k}
{scope}

= Solution
{parent=a}

If $y_1\leq y_2$, then the <utility function> is increasing, so for every $X\in\mathcal X$,
$$
\mathbb E[U(X+y_1)]\leq\mathbb E[U(X+y_2)].
$$
Taking <suprema> gives $F(y_1)\leq F(y_2)$.

For $0\leq t\leq1$, let $X_1$ and $X_2$ attain the suprema at $y_1$ and $y_2$. Because $\mathcal X$ is a <vector space>,
$$
X_t=tX_1+(1-t)X_2\in\mathcal X.
$$
The <concavity> of $U$ now gives
$$
\begin{aligned}
F(ty_1+(1-t)y_2)
&\geq\mathbb E\left[U\left(X_t+ty_1+(1-t)y_2\right)\right]\\
&\geq t\mathbb E[U(X_1+y_1)]
 +(1-t)\mathbb E[U(X_2+y_2)]\\
&=tF(y_1)+(1-t)F(y_2).
\end{aligned}
$$
Thus $F$ is increasing and concave, as stated by <optimized affine shift of concave utility>.

Solved by gpt-5.6-sol high.

= b
{parent=30k}
{scope}

= Solution
{parent=b}

For $0\leq t\leq1$, <concavity> of $U$ implies
$$
\begin{aligned}
G(ts_1+(1-t)s_2)
&=\mathbb E\left[U\left(t(m+s_1Z)+(1-t)(m+s_2Z)\right)\right]\\
&\geq tG(s_1)+(1-t)G(s_2),
\end{aligned}
$$
so $G$ is concave.

Since $\mathbb E Z=0$, <Jensen inequality> gives
$$
G(s)=\mathbb E[U(m+sZ)]
\leq U(m+s\mathbb EZ)=U(m)=G(0).
$$
If $0\leq s<t$, concavity and $s=(1-s/t)0+(s/t)t$ yield
$$
G(s)\geq\left(1-\frac st\right)G(0)+\frac stG(t)\geq G(t).
$$
Hence $G$ is decreasing on $[0,\infty)$; this is <scaled centered risk under concave utility>.

Solved by gpt-5.6-sol high.

= c
{parent=30k}
{scope}

= Solution
{parent=c}

Let the <value function> at time $n$ be the greatest conditional expected terminal utility attainable from wealth $x$. Independence of the innovations gives the <Bellman equation>
$$
\boxed{
V(N,x)=U(x),\qquad
V(n,x)=\sup_{\theta\in\mathbb R^d}
\mathbb E\left[V\left(n+1,(1+r)x+\theta^T\xi_{n+1}\right)\right]
}
$$
for $0\leq n<N$. A policy obtained from the maximizing choices is admissible because each time-$n$ holding is a <predictable process>, hence is known before $\xi_n$ is observed.

We prove the claimed properties by <backward induction>. They hold at $n=N$ because $V(N,\cdot)=U$ is increasing and concave. Suppose they hold at time $n+1$. For fixed $\theta$, the map
$$
x\longmapsto
\mathbb E\left[V\left(n+1,(1+r)x+\theta^T\xi_{n+1}\right)\right]
$$
is increasing because $1+r>0$, and taking a supremum preserves this inequality.

For concavity, let $\theta_x$ and $\theta_y$ be optimal at wealths $x$ and $y$. For $0\leq t\leq1$, use the admissible portfolio
$$
\theta_t=t\theta_x+(1-t)\theta_y.
$$
The next wealth is the same <convex combination> of the next wealths generated from $(x,\theta_x)$ and $(y,\theta_y)$. Therefore
$$
\begin{aligned}
V(n,tx+(1-t)y)
&\geq\mathbb E\left[V\left(n+1,(1+r)(tx+(1-t)y)+\theta_t^T\xi_{n+1}\right)\right]\\
&\geq tV(n,x)+(1-t)V(n,y).
\end{aligned}
$$
This completes the induction and proves <monotonicity and concavity of a portfolio value function>.

Solved by gpt-5.6-sol high.

= d
{parent=30k}
{scope}

= Solution
{parent=d}

Condition on the information available before time $n$. The current wealth is then fixed, while $\xi_n\sim N(b,\Sigma)$ is independent of that information. Put
$$
B=b^T\Sigma^{-1}b>0.
$$
For any portfolio $\theta$, use the <inner product> induced by the <positive-definite matrix> $\Sigma$ to write
$$
\theta=\lambda\Sigma^{-1}b+\eta,
\qquad
\lambda=\frac{\theta^Tb}{B},
\qquad
\eta^Tb=0.
$$
The <linear image of a multivariate normal vector> shows that the two portfolio returns are jointly normal, and
$$
\operatorname{cov}\left((\lambda\Sigma^{-1}b)^T\xi_n,\eta^T\xi_n\right)
=\lambda\eta^T\Sigma\Sigma^{-1}b
=\lambda\eta^Tb=0.
$$
Thus <independence of uncorrelated jointly normal variables> makes the residual return $\eta^T\xi_n$ independent of the aligned return; it also has <expected value> zero.

The continuation value $V(n,\cdot)$ is concave by part (c). Conditional <Jensen inequality> therefore shows that adding the independent centered residual cannot improve the objective:
$$
\mathbb E\left[V\left(n,m+(\lambda\Sigma^{-1}b)^T\xi_n+\eta^T\xi_n\right)\right]
\leq
\mathbb E\left[V\left(n,m+(\lambda\Sigma^{-1}b)^T\xi_n\right)\right],
$$
where $m=(1+r)X_{n-1}$. Since the optimal portfolio is unique, its residual must be zero.

It remains to determine the sign. If $\lambda<0$, the portfolios $\lambda\Sigma^{-1}b$ and $-\lambda\Sigma^{-1}b$ have the same return <variance>, namely $\lambda^2B$, while their means are $\lambda B$ and $-\lambda B$. The latter return has the distribution of the former plus the positive constant $-2\lambda B$. Since $V(n,\cdot)$ is increasing, replacing $\lambda$ by $-\lambda$ cannot reduce the objective, contradicting uniqueness. Hence $\lambda\geq0$.

Applying this conditional argument at every time gives nonnegative, past-measurable random variables $\lambda_n$ such that
$$
\boxed{\theta_n^*=\lambda_n\Sigma^{-1}b},
\qquad 1\leq n\leq N.
$$
This is the <Gaussian one-fund theorem>.

Solved by gpt-5.6-sol high.

= 31J
{parent=Paper 1}
{scope}
{title2=Mathematics of Machine Learning}

= a
{parent=31j}
{scope}

= Solution
{parent=a}

A set $C\subseteq\mathbb R^d$ is <convex> when every <line segment> between two of its points remains in the set: for every $x,y\in C$ and $0\leq t\leq1$,
$$
\boxed{tx+(1-t)y\in C}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=31j}
{scope}

= Solution
{parent=b}

A function $f:C\to\mathbb R$ is <strictly convex> when, for distinct $x,y\in C$ and $0<t<1$,
$$
\boxed{f(tx+(1-t)y)<tf(x)+(1-t)f(y)}.
$$

Suppose distinct points $x$ and $y$ were both minimizers, with minimum value $m$. Since $C$ is a <convex set>, their midpoint belongs to $C$, while strict convexity gives
$$
f\left(\frac{x+y}{2}\right)
<\frac{f(x)+f(y)}2=m,
$$
a contradiction. Thus any minimizer is unique, which is the <uniqueness of a minimizer of a strictly convex function>.

Solved by gpt-5.6-sol high.

= c
{parent=31j}
{scope}

= Solution
{parent=c}

For $x\in\mathbb R^d$, the <euclidean projection onto a convex set> $C$ is
$$
\boxed{
\pi_C(x)=\operatorname*{argmin}_{z\in C}\|x-z\|_2
}.
$$
The squared distance $z\mapsto\|x-z\|_2^2$ is a <strictly convex function>. Its restriction to the <convex set> $C$ can therefore have at most one minimizer by part (b). Closedness, together with finite dimensionality, supplies existence, although existence was not required to be proved.

Solved by gpt-5.6-sol high.

= d
{parent=31j}
{scope}

= Solution
{parent=d}

Assume $\pi\in C$ and
$$
(x-\pi)^T(z-\pi)\leq0
$$
for every $z\in C$. Expanding the squared <euclidean distance> gives
$$
\begin{aligned}
\|x-z\|_2^2
&=\|(x-\pi)-(z-\pi)\|_2^2\\
&=\|x-\pi\|_2^2+\|z-\pi\|_2^2
  -2(x-\pi)^T(z-\pi)\\
&\geq\|x-\pi\|_2^2.
\end{aligned}
$$
Hence $\pi$ minimizes the distance from $x$ over $C$, so $\pi=\pi_C(x)$. This proves the sufficient direction of the <variational characterization of convex projection>.

Solved by gpt-5.6-sol high.

= e
{parent=31j}
{scope}

= Solution
{parent=e}

The set $C$ is the <second-order cone>. Put
$$
\rho=\|u\|_2,
\qquad
\alpha=\frac12\left(1+\frac t\rho\right).
$$
When $\rho\geq|t|$ and $\rho>0$, one has $0\leq\alpha\leq1$, and the proposed point
$$
\pi=\alpha(u,\rho)
$$
lies on the boundary of $C$ because $\|\alpha u\|_2=\alpha\rho$.

Set $\delta=(\rho-t)/2\geq0$. Since
$$
(u,t)-\pi=\delta\left(\frac u\rho,-1\right),
$$
for any $(v,s)\in C$ we obtain
$$
\begin{aligned}
\bigl((u,t)-\pi\bigr)^T\bigl((v,s)-\pi\bigr)
&=\delta\left(\frac{u^Tv}{\rho}-s\right)\\
&\leq\delta(\|v\|_2-s)\\
&\leq0,
\end{aligned}
$$
where the first inequality is the <Cauchy-Schwarz inequality> and the second uses $\|v\|_2\leq s$. Part (d) therefore gives
$$
\boxed{
\pi_C(u,t)
=\frac12\left(1+\frac t{\|u\|_2}\right)
\bigl(u,\|u\|_2\bigr)
}
$$
whenever $\|u\|_2\geq|t|$ and $u\ne0$. At $(u,t)=(0,0)$ the projection is plainly $(0,0)$, consistent with the limiting formula.

If $\rho\leq-t$, then for every $(v,s)\in C$,
$$
(u,t)^T(v,s)
\leq\rho\|v\|_2+ts
\leq(\rho+t)s
\leq0.
$$
Applying part (d) with $\pi=0$ gives
$$
\boxed{\pi_C(u,t)=(0,0)}.
$$
Together with the trivial case $\rho\leq t$, this is the full <projection onto the second-order cone>.

Solved by gpt-5.6-sol high.

= f
{parent=31j}
{scope}

= Solution
{parent=f}

Using the harmless normalization $1/(2n)$, the <squared-loss empirical risk for linear prediction> is
$$
R_n(\beta)
=\frac1{2n}\sum_{i=1}^n(\beta^TX_i-Y_i)^2,
$$
with explicit <gradient>
$$
\boxed{
\nabla R_n(\beta)
=\frac1n\sum_{i=1}^n(\beta^TX_i-Y_i)X_i
}.
$$
Starting from any $\beta_0\in C$, <projected gradient descent> with step sizes $\eta_k>0$ is
$$
\widetilde\beta_{k+1}
=\beta_k-\eta_k\nabla R_n(\beta_k),
\qquad
\boxed{\beta_{k+1}=\pi_C(\widetilde\beta_{k+1})}.
$$

For a fully explicit update, write $\widetilde\beta_{k+1}=(u_k,t_k)$ and $\rho_k=\|u_k\|_2$. The projection from part (e) is
$$
\pi_C(u_k,t_k)=
\begin{cases}
(u_k,t_k),&\rho_k\leq t_k,\\
(0,0),&\rho_k\leq-t_k,\\
\displaystyle
\frac12\left(1+\frac{t_k}{\rho_k}\right)(u_k,\rho_k),
&\rho_k>|t_k|.
\end{cases}
$$
Every iterate therefore lies in the prescribed hypothesis class, and any convergent run under the standard convex-optimization step-size conditions targets its empirical-risk minimizer.

Solved by gpt-5.6-sol high.

= 32A
{parent=Paper 1}
{scope}
{title2=Dynamical Systems}

= a
{parent=32a}
{scope}

= Solution
{parent=a}

<Dulac theorem> states the following. Let $D$ be a simply connected region for the planar system $\dot z=f(z)$, and suppose there is a continuously differentiable function $B:D\to\mathbb R$ for which
$$
\nabla\mathbin{\cdot}(Bf)
$$
has one strict sign throughout $D$. Then there is no <periodic orbit> contained in $D$.

To prove it, suppose a periodic orbit $\Gamma$ existed and let $A$ be its interior. Simple connectedness ensures $A\subset D$. By the planar <divergence theorem>,
$$
\iint_A\nabla\mathbin{\cdot}(Bf)\,dA
=\int_\Gamma Bf\mathbin{\cdot}n\,ds.
$$
The vector field $f$ is tangent to its trajectory $\Gamma$, so its scalar product with the outward normal $n$ vanishes. The right-hand side is therefore zero. The left-hand side cannot be zero because its integrand has one strict sign, a contradiction.

The <Poincare-Bendixson theorem> states that if a forward trajectory of a smooth planar system remains in a compact set and its nonempty <omega-limit set> contains no equilibrium point, then that omega-limit set is a periodic orbit.

Solved by gpt-5.6-sol high.

= b
{parent=32a}
{scope}

= Solution
{parent=b}

At an <equilibrium point of a dynamical system>, the second equation gives
$$
x(s-xy)=0.
$$
The choice $x=0$ is impossible because then $\dot x=r>0$. Hence $y=s/x$, and substitution into the first equation gives
$$
0=r-(1+s)x+sx=r-x.
$$
Thus the unique fixed point is
$$
\boxed{(x_*,y_*)=\left(r,\frac sr\right)}.
$$

Choose
$$
\boxed{\alpha=\frac r{1+s}},
$$
which satisfies $0<\alpha<r$. Let
$$
R=\left\{(x,y):x\geq\alpha,\quad 0\leq y\leq\frac s\alpha,
\quad x+y\leq r+\frac s\alpha\right\}.
$$
We check the direction of the vector field on its four edges.

On $x=\alpha$,
$$
\dot x=r-(1+s)\alpha+\alpha^2y=\alpha^2y\geq0.
$$
On $y=0$,
$$
\dot y=sx>0.
$$
On $y=s/\alpha$, where $x\geq\alpha$,
$$
\dot y=sx\left(1-\frac x\alpha\right)\leq0.
$$
Finally, throughout the system,
$$
\dot x+\dot y=r-x.
$$
On the sloping edge $x+y=r+s/\alpha$, the bound $y\leq s/\alpha$ implies $x\geq r$, and hence $\dot x+\dot y\leq0$. Every boundary edge therefore points inward. Violations of the corresponding four inequalities are driven toward the boundary in the same order, so positive trajectories enter this compact <trapping region> and then remain there. This is the <Brusselator trapping region>.

The <jacobian matrix> at the unique fixed point is
$$
J_*=
\begin{pmatrix}
-(1+s)+2x_*y_*&x_*^2\\
 s-2x_*y_*&-x_*^2
\end{pmatrix}
=
\begin{pmatrix}
 s-1&r^2\\
 -s&-r^2
\end{pmatrix}.
$$
Its determinant and trace are
$$
\det J_*=r^2>0,
\qquad
\operatorname{tr}J_*=s-1-r^2.
$$
When $s-1>r^2$, the trace is positive, so <linear stability of a planar equilibrium> shows that the fixed point is a repeller. It lies in the interior of $R$.

Choose a trajectory in $R$ other than the equilibrium. Compactness of $R$ gives a nonempty compact omega-limit set, and the repelling equilibrium cannot belong to that set. Since there are no other equilibria, the <Poincare-Bendixson theorem> gives a <periodic orbit>. Hence
$$
\boxed{s-1>r^2\quad\Longrightarrow\quad\text{the system has a periodic orbit}.}
$$
This is the <Brusselator periodic-orbit criterion>.

Solved by gpt-5.6-sol high.

= 33E
{parent=Paper 1}
{scope}
{title2=Integrable Systems}

= Solution
{parent=33E}

Equality of the mixed derivatives of the auxiliary vector $\Psi$ gives
$$
\Psi_{xt}=U_t\Psi+UV\Psi,
\qquad
\Psi_{tx}=V_x\Psi+VU\Psi.
$$
Thus the <zero-curvature condition> for this convention is
$$
\boxed{U_t-V_x+[U,V]=0}.
$$
Substituting the displayed matrices and collecting powers of the spectral parameter $\lambda$ makes every diagonal entry and every $\lambda$-dependent term cancel. The remaining matrix is
$$
U_t-V_x+[U,V]
=
\begin{pmatrix}
0&iq_t-q_{xx}-2rq^2\\
 ir_t+r_{xx}+2qr^2&0
\end{pmatrix}.
$$
Consequently compatibility is equivalent to
$$
\boxed{
 ir_t+r_{xx}+2qr^2=0,
\qquad
 iq_t-q_{xx}-2rq^2=0
},
$$
so the requested constant is
$$
\boxed{a=2}.
$$
This is the <AKNS Lax pair for the nonlinear Schrodinger equation>.

The reduction
$$
q=\overline r
$$
is preserved by the two equations, which become complex conjugates and reduce to the <Focusing nonlinear Schrodinger equation>
$$
\boxed{ir_t+r_{xx}+2|r|^2r=0}.
$$
The analogous reduction
$$
q=-\overline r
$$
gives the <Defocusing nonlinear Schrödinger equation>
$$
\boxed{ir_t+r_{xx}-2|r|^2r=0}.
$$

For a complex field $r$ and rapidly decreasing boundary conditions, define
$$
H_{\mathrm{foc}}[r,\overline r]
=\int_{\mathbb R}\left(|r_x|^2-|r|^4\right)dx,
\qquad
H_{\mathrm{def}}[r,\overline r]
=\int_{\mathbb R}\left(|r_x|^2+|r|^4\right)dx.
$$
Their <variational derivatives>, after <integration by parts>, are
$$
\frac{\delta H_{\rm foc}}{\delta\overline r}
=-r_{xx}-2|r|^2r,
\qquad
\frac{\delta H_{\rm def}}{\delta\overline r}
=-r_{xx}+2|r|^2r.
$$
Both equations therefore have the <Hamiltonian field equation>
$$
\boxed{
ir_t=\frac{\delta H}{\delta\overline r}
},
$$
or, including the conjugate equation,
$$
\begin{pmatrix}r_t\\ \overline r_t\end{pmatrix}
=
\begin{pmatrix}0&-i\\ i&0\end{pmatrix}
\begin{pmatrix}\delta H/\delta r\\ \delta H/\delta\overline r\end{pmatrix}.
$$
This is the <Hamiltonian form of the cubic nonlinear Schrodinger equations>.

Now put
$$
r(x,t)=e^{-iEt}f(x),
$$
where $E$ and $f$ are real and $f$ is smooth and rapidly decreasing. For the focusing equation,
$$
f''+Ef+2f^3=0.
$$
Multiplication by $f'$ and use of the decay at infinity gives the first integral
$$
(f')^2=-Ef^2-f^4.
$$
A nonzero solution requires $E<0$. Writing $E=-\kappa^2$ with $\kappa>0$, separation of variables, or direct substitution, gives
$$
\boxed{
f(x)=\pm\kappa\operatorname{sech}\bigl(\kappa(x-x_0)\bigr)
}.
$$
Hence
$$
\boxed{
r(x,t)=\pm\kappa e^{i\kappa^2t}
\operatorname{sech}\bigl(\kappa(x-x_0)\bigr)
}
$$
is the <Bright standing soliton of the focusing nonlinear Schrodinger equation>.

For the defocusing equation the profile instead satisfies
$$
f''+Ef-2f^3=0,
$$
with first integral
$$
(f')^2=f^4-Ef^2=f^2(f^2-E).
$$
If a nonzero rapidly decreasing profile existed, $|f|$ would attain a positive maximum $A$. At that point $f'=0$, so the identity forces $E=A^2>0$. But along either tail, where $0<|f|<A$, the right-hand side $f^2(f^2-E)$ is negative, which is impossible. Thus there is no nonzero solution of the prescribed form, as recorded by <No rapidly decaying standing wave for the defocusing cubic nonlinear Schrodinger equation>.

Solved by gpt-5.6-sol high.

= 34B
{parent=Paper 1}
{scope}
{title2=Principles of Quantum Mechanics}

= a
{parent=34b}
{scope}

= Solution
{parent=a}

In units with $\hbar=1$, the <quantum harmonic oscillator> Hamiltonian and <number operator> are
$$
\boxed{H=\omega\left(N+\frac12\right)},
\qquad
\boxed{N=A^\dagger A}.
$$
The nontrivial commutators among the <creation and annihilation operators> and $N$ are
$$
\boxed{
[A,A^\dagger]=1,
\qquad
[N,A]=-A,
\qquad
[N,A^\dagger]=A^\dagger
}.
$$
Their reversed commutators have the opposite signs, while the commutator of any operator with itself is zero.

The operator $N$ is <self-adjoint> because
$$
(A^\dagger A)^\dagger=A^\dagger A.
$$
Its eigenvalues are therefore real. Moreover, for every state $|\psi\rangle$,
$$
\langle\psi|N|\psi\rangle
=\langle A\psi|A\psi\rangle
=\|A|\psi\rangle\|^2\geq0,
$$
so every eigenvalue is nonnegative.

If $N|\nu\rangle=\nu|\nu\rangle$, the commutator with $A$ gives
$$
N(A|\nu\rangle)=(\nu-1)A|\nu\rangle,
$$
and
$$
\|A|\nu\rangle\|^2=\nu\|\,|\nu\rangle\|^2.
$$
Repeated application of $A$ lowers the eigenvalue by one. It must terminate before producing a negative eigenvalue. If $m$ is the last occupied rung, then $A|\nu-m\rangle=0$, and the norm identity forces $\nu-m=0$. Hence
$$
\boxed{\nu\in\{0,1,2,\ldots\}}.
$$
This is the <integer spectrum of the number operator>.

Solved by gpt-5.6-sol high.

= b
{parent=34b}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

For occupation numbers $n,m\geq0$, the Hamiltonian of the <two commensurate quantum harmonic oscillators> is
$$
H_0=N_A+2N_B+\frac32,
\qquad
E_{n,m}=n+2m+\frac32.
$$
The five lowest distinct levels and their product-state bases are
$$
\begin{array}{c|c|c}
E&\text{states }|n,m\rangle&\text{degeneracy}\\ \hline
\frac32&|0,0\rangle&1\\
\frac52&|1,0\rangle&1\\
\frac72&|2,0\rangle,|0,1\rangle&2\\
\frac92&|3,0\rangle,|1,1\rangle&2\\
\frac{11}2&|4,0\rangle,|2,1\rangle,|0,2\rangle&3
\end{array}
$$
The repeated levels exhibit <quantum degeneracy>.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

The degenerate $E_0=9/2$ subspace has ordered basis
$$
|3,0\rangle,\qquad |1,1\rangle.
$$
Using the ladder actions,
$$
\begin{aligned}
H'|3,0\rangle
&=A^2B^\dagger|3,0\rangle
=\sqrt6\,|1,1\rangle,\\
H'|1,1\rangle
&=A^{\dagger2}B|1,1\rangle
=\sqrt6\,|3,0\rangle.
\end{aligned}
$$
Thus <degenerate perturbation theory> asks us to diagonalize
$$
H'\big|_{E_0}
=\begin{pmatrix}0&\sqrt6\\\sqrt6&0\end{pmatrix}.
$$
Its normalized eigenvectors and first-order energies are
$$
\boxed{
|\psi_+\rangle
=\frac{|3,0\rangle+|1,1\rangle}{\sqrt2},
\qquad
E_+=\frac92+\sqrt6\lambda
},
$$
$$
\boxed{
|\psi_-\rangle
=\frac{|3,0\rangle-|1,1\rangle}{\sqrt2},
\qquad
E_-=\frac92-\sqrt6\lambda
}.
$$
There is no order-$\lambda$ admixture from other unperturbed levels.

In fact, $H'$ preserves $n+2m$, so this two-dimensional eigenspace of $H_0$ is invariant. Directly,
$$
H'|\psi_\pm\rangle=\pm\sqrt6|\psi_\pm\rangle,
\qquad
H_0|\psi_\pm\rangle=\frac92|\psi_\pm\rangle.
$$
Therefore
$$
\boxed{
H|\psi_\pm\rangle
=\left(\frac92\pm\sqrt6\lambda\right)|\psi_\pm\rangle
}
$$
for every real $\lambda$, with no omitted higher-order terms. This is <exact diagonalization within a degenerate subspace>, specifically the <resonant two-to-one oscillator coupling>.

Solved by gpt-5.6-sol high.

= 35D
{parent=Paper 1}
{scope}
{title2=Applications of Quantum Mechanics}

= a
{parent=35d}
{scope}

= Solution
{parent=a}

For a wave incident from the left, define the reflection and transmission amplitudes by
$$
\psi(x)\sim
\begin{cases}
e^{ikx}+r e^{-ikx},&x\to-\infty,\\
t e^{ikx},&x\to+\infty,
\end{cases}
\qquad
E=\frac{\hbar^2k^2}{2m}.
$$
Reflection invariance gives the same amplitudes for incidence from the right. If the incoming amplitudes from the left and right are ordered first, and the outgoing amplitudes toward the left and right second, the <One-dimensional S-matrix> is
$$
\boxed{
S=\begin{pmatrix}r&t\\t&r\end{pmatrix}
}.
$$

For $V(x)=V_0\delta(x)$, the <wavefunction> is continuous and its derivative has the jump
$$
\psi'(0^+)-\psi'(0^-)
=\frac{2mV_0}{\hbar^2}\psi(0).
$$
The scattering ansatz gives
$$
1+r=t,
\qquad
2ik r=\frac{2mV_0}{\hbar^2}t.
$$
With
$$
\gamma=\frac{mV_0}{\hbar^2k},
$$
we obtain the <scattering by a delta potential> amplitudes
$$
\boxed{
t=\frac1{1+i\gamma},
\qquad
r=\frac{-i\gamma}{1+i\gamma}
}.
$$
They satisfy
$$
|r|^2+|t|^2=1,
\qquad
r^*t+t^*r=0.
$$
Therefore
$$
S^\dagger S
=\begin{pmatrix}
|r|^2+|t|^2&r^*t+t^*r\\
t^*r+r^*t&|r|^2+|t|^2
\end{pmatrix}
=I,
$$
so the scattering matrix is unitary.

Solved by gpt-5.6-sol high.

= b
{parent=35d}
{scope}

= Solution
{parent=b}

For a potential of period $a$, define the <Floquet matrix for a one-dimensional periodic potential> by
$$
\begin{pmatrix}\psi(x+a)\\\psi'(x+a)\end{pmatrix}
=M(E)\begin{pmatrix}\psi(x)\\\psi'(x)\end{pmatrix}.
$$
The Schrodinger equation has no first-derivative term, so the Wronskian is constant and $\det M=1$. Its multipliers are therefore reciprocal. They lie on the unit circle precisely when
$$
\boxed{|\operatorname{tr}M(E)|\leq2}.
$$
Those energies form continuous allowed bands of bounded <Bloch theorem> solutions. Values with $|\operatorname{tr}M|>2$ have a growing and a decaying multiplier and form forbidden gaps. The band edges have $\operatorname{tr}M=\pm2$, as in <Floquet discriminant and energy bands>.

For the delta-comb <Kronig-Penney model>, start immediately to the right of one delta function. Free propagation to the next delta and the derivative jump there are represented by
$$
P=
\begin{pmatrix}
\cos ka&\dfrac{\sin ka}{k}\\
-k\sin ka&\cos ka
\end{pmatrix},
\qquad
D=
\begin{pmatrix}
1&0\\
\dfrac{2mV_0}{\hbar^2}&1
\end{pmatrix}.
$$
Thus
$$
\boxed{
M(E)=DP
=
\begin{pmatrix}
\cos ka&\dfrac{\sin ka}{k}\\
\dfrac{2mV_0}{\hbar^2}\cos ka-k\sin ka&
\cos ka+\dfrac{2mV_0}{\hbar^2k}\sin ka
\end{pmatrix}
}.
$$
Consequently the <Floquet discriminant of the delta-comb Kronig-Penney model> is
$$
\boxed{
\frac12\operatorname{tr}M(E)
=\cos(ka)+\gamma\sin(ka)
},
\qquad
\gamma=\frac{mV_0}{\hbar^2k}.
$$
All band edges are therefore determined by
$$
\cos(ka)+\gamma\sin(ka)=\sigma,
\qquad \sigma=\pm1.
$$
In factorized form, the periodic edges $\sigma=1$ obey
$$
\boxed{
\sin\frac{ka}{2}=0
\quad\text{or}\quad
\tan\frac{ka}{2}=\gamma
},
$$
while the antiperiodic edges $\sigma=-1$ obey
$$
\boxed{
\cos\frac{ka}{2}=0
\quad\text{or}\quad
\tan\frac{ka}{2}=-\frac1\gamma
}.
$$

To express the same edges through the single-barrier scattering data, the <One-dimensional transfer matrix from scattering amplitudes> gives
$$
\operatorname{tr}M
=\frac{(t^2-r^2)e^{ika}+e^{-ika}}t.
$$
At an edge put $z=e^{-ika}$ and $\operatorname{tr}M=2\sigma$. Then
$$
z^2-2\sigma tz+t^2-r^2=0,
$$
so the <scattering-amplitude equations for one-dimensional band edges> are
$$
\boxed{e^{-ika}=\sigma t\pm r}.
$$
Explicitly,
$$
\begin{array}{c|c}
\operatorname{tr}M=2&e^{-ika}=t+r\quad\text{or}\quad t-r\\
\operatorname{tr}M=-2&e^{-ika}=-t+r\quad\text{or}\quad -t-r.
\end{array}
$$
For the delta barrier, $t-r=1$ and
$$
t+r=\frac{1-i\gamma}{1+i\gamma},
$$
which reproduces the four factorized edge equations above.

Solved by gpt-5.6-sol high.

= 36A
{parent=Paper 1}
{scope}
{title2=Statistical Physics}

= a
{parent=36a}
{scope}

= Solution
{parent=a}

The <microcanonical ensemble> describes an isolated system at fixed energy $E$, volume $V$, and particle number $N$, assigning equal weight to accessible states in a narrow energy shell.

The <canonical ensemble> describes a system that can exchange energy with a heat reservoir at fixed temperature $T$, while $V$ and $N$ remain fixed. A microstate $s$ has probability
$$
p_s=\frac{e^{-\beta E_s}}{Z},
\qquad
Z=\sum_s e^{-\beta E_s},
\qquad
\beta=\frac1{k_BT}.
$$

The <grand canonical ensemble> allows exchange of both energy and particles at fixed $T$, $V$, and <chemical potential> $\mu$. Its state weights are
$$
p_s=\frac{e^{-\beta(E_s-\mu N_s)}}{\Xi}.
$$

The choice is irrelevant for bulk thermodynamic observables in the thermodynamic limit when the system is additive, has sufficiently short-range interactions, and lies away from singular phase-coexistence regimes. Then relative energy and particle-number fluctuations vanish as the system size grows. These are the standard conditions for <equivalence of statistical ensembles>.

Solved by gpt-5.6-sol high.

= b
{parent=36a}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

The classical <canonical partition function> uses one phase-space cell $h^2$ per state. In <polar coordinates>,
$$
\begin{aligned}
Z_1
&=\frac1{h^2}
\int_{\mathbb R^2}e^{-\beta p^2/(2m)}\,d^2p
\int_{r<R}e^{-\beta\kappa r^2/2}\,d^2x\\
&=\frac1{h^2}
\left(\frac{2\pi m}{\beta}\right)
\left(2\pi\int_0^Rr e^{-\beta\kappa r^2/2}\,dr\right)\\
&=\boxed{
\frac{4\pi^2m}{h^2\beta^2\kappa}
\left(1-e^{-\beta\kappa R^2/2}\right)
}.
\end{aligned}
$$
Equivalently, using $h=2\pi\hbar$,
$$
Z_1=\frac{m}{\hbar^2\beta^2\kappa}
\left(1-e^{-\beta\kappa R^2/2}\right).
$$
This is the <particle in a finite two-dimensional harmonic trap> partition function.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

Put
$$
a=\frac12\kappa R^2.
$$
The <thermodynamic derivatives of the canonical partition function> give
$$
\begin{aligned}
\langle E\rangle
&=-\frac{\partial}{\partial\beta}\log Z_1\\
&=-\frac{\partial}{\partial\beta}
\left[-2\log\beta+\log(1-e^{-\beta a})+\text{constant}\right]\\
&=\boxed{
\frac2\beta-\frac a{e^{\beta a}-1}
}.
\end{aligned}
$$
The two momentum coordinates are unrestricted quadratic degrees of freedom, so the <equipartition theorem> gives mean kinetic energy $1/\beta$. Hence
$$
\boxed{
\langle V\rangle
=\langle E\rangle-\frac1\beta
=\frac1\beta-\frac a{e^{\beta a}-1}
}.
$$
The same result follows from $\langle V\rangle=-(\kappa/\beta)\,\partial_\kappa\log Z_1$.

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

If
$$
a=\frac12\kappa R^2\gg k_BT,
$$
then $e^{\beta a}$ is large and the wall correction is exponentially small:
$$
\boxed{\langle E\rangle\sim2k_BT}.
$$
This is exactly the <equipartition theorem> result from the two kinetic and two harmonic-potential quadratic terms.

If $a\ll k_BT$, use
$$
\frac a{e^{\beta a}-1}
=\frac1\beta-\frac a2+O(\beta a^2).
$$
Then
$$
\boxed{
\langle E\rangle
=k_BT+\frac{\kappa R^2}{4}
+O\left(\frac{\kappa^2R^4}{k_BT}\right)
\sim k_BT
}.
$$
The leading term is the kinetic energy. The particle is nearly uniform in the disk, so $\langle r^2\rangle\to R^2/2$ and $\langle V\rangle\to\kappa R^2/4$, much smaller than $k_BT$. The hard wall therefore suppresses the positional equipartition contribution, as in the <hard-wall correction to harmonic equipartition>.

Solved by gpt-5.6-sol high.

= iv
{parent=b}
{scope}

= Solution
{parent=iv}

For $N$ identical, noninteracting classical particles, the one-particle phase-space integrals factorize and indistinguishability supplies the Gibbs factor. Thus the <classical ideal-gas partition function> rule gives
$$
\boxed{
Z_N=\frac{Z_1^N}{N!}
=\frac1{N!}
\left[
\frac{4\pi^2m}{h^2\beta^2\kappa}
\left(1-e^{-\beta\kappa R^2/2}\right)
\right]^N
}.
$$

Solved by gpt-5.6-sol high.

= 37A
{parent=Paper 1}
{scope}
{title2=Electrodynamics}

= a
{parent=37a}
{scope}

= Solution
{parent=a}

For a timelike worldline, $-\eta_{\mu\nu}\dot x^\mu\dot x^\nu>0$. Its square root is a Lorentz scalar, and
$$
\sqrt{-\eta_{\mu\nu}\dot x^\mu\dot x^\nu}\,d\lambda
=\sqrt{-\eta_{\mu\nu}dx^\mu dx^\nu}
=c\,d\tau.
$$
Thus the first term is the Lorentz-invariant free-particle action $-mc^2\int d\tau$. Written as a line element, it is also invariant under a change of the worldline parameter $\lambda$.

The interaction is the scalar line integral
$$
q\int A_\mu dx^\mu.
$$
Under an electromagnetic <gauge transformation>
$$
A_\mu\longmapsto A_\mu+\partial_\mu\chi,
$$
it changes only by
$$
q\int\partial_\mu\chi\,dx^\mu
=q\,[\chi]_{\rm endpoints}.
$$
This endpoint term does not affect a fixed-endpoint variational problem. Hence both Lorentz invariance and gauge invariance select the displayed <relativistic charged-particle action>, with the expected free-particle and minimal-coupling terms.

Solved by gpt-5.6-sol high.

= b
{parent=37a}
{scope}

= Solution
{parent=b}

Write
$$
L=-mc\sqrt{-\dot x^2}+qA_\nu(x)\dot x^\nu.
$$
Since
$$
\sqrt{-\dot x^2}=c\frac{d\tau}{d\lambda},
\qquad
\dot x^\mu=\frac{d\tau}{d\lambda}u^\mu,
$$
its derivatives are
$$
\frac{\partial L}{\partial\dot x^\mu}
=m u_\mu+qA_\mu,
\qquad
\frac{\partial L}{\partial x^\mu}
=q(\partial_\mu A_\nu)\dot x^\nu.
$$
The <Euler-Lagrange equation> therefore gives
$$
\begin{aligned}
0
&=\frac d{d\lambda}(m u_\mu+qA_\mu)
-q(\partial_\mu A_\nu)\dot x^\nu\\
&=m\frac{du_\mu}{d\lambda}
+q(\partial_\nu A_\mu-\partial_\mu A_\nu)\dot x^\nu.
\end{aligned}
$$
Using the <electromagnetic field tensor>
$$
F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu
$$
and dividing by $d\tau/d\lambda$ gives
$$
\boxed{
m\frac{du_\mu}{d\tau}=qF_{\mu\nu}u^\nu
}.
$$
Raising the first index yields the equivalent requested form
$$
\boxed{
m\frac{du^\mu}{d\tau}=qF^{\mu\nu}u_\nu
}.
$$
This is the <Lorentz-force equation from the worldline action>.

Solved by gpt-5.6-sol high.

= c
{parent=37a}
{scope}

= Solution
{parent=c}

The <four-velocity> and particle energy are
$$
u^\mu=\gamma(c,\mathbf v),
\qquad
\mathcal E=\gamma mc^2,
\qquad
p^0=mu^0=\frac{\mathcal E}{c}.
$$
With the conventions for the given metric, $F^{0i}=E_i/c$. The temporal component of the covariant <Lorentz force> is therefore
$$
\frac{d(\mathcal E/c)}{d\tau}
=qF^{0i}u_i
=\frac q c\gamma\mathbf E\mathbin{\cdot}\mathbf v.
$$
Since $dt/d\tau=\gamma$, cancellation of $\gamma/c$ gives the <relativistic electromagnetic work-energy theorem>
$$
\boxed{
\frac{d\mathcal E}{dt}=q\mathbf E\mathbin{\cdot}\mathbf v
}.
$$
The absence of $\mathbf B$ reflects that magnetic force is perpendicular to the velocity and does no work.

Solved by gpt-5.6-sol high.

= d
{parent=37a}
{scope}

= Solution
{parent=d}

The spatial part of the <Lorentz force> is
$$
\frac d{dt}(\gamma m\mathbf v)
=q(\mathbf E+\mathbf v\times\mathbf B).
$$
Part (c), together with $\mathcal E=\gamma mc^2$, gives
$$
\frac{d\gamma}{dt}
=\frac q{mc^2}\mathbf E\mathbin{\cdot}\mathbf v.
$$
Writing $\mathbf a=d\mathbf v/dt$ and expanding the momentum derivative,
$$
m\gamma\mathbf a+m\mathbf v\frac{d\gamma}{dt}
=q(\mathbf E+\mathbf v\times\mathbf B).
$$
Substitution of $d\gamma/dt$ gives the <coordinate acceleration of a relativistic charged particle>:
$$
\boxed{
\frac{d\mathbf v}{dt}
=\frac q{m\gamma}
\left[
\mathbf E+\mathbf v\times\mathbf B
-\frac1{c^2}\mathbf v(\mathbf v\mathbin{\cdot}\mathbf E)
\right]
}.
$$

In the nonrelativistic limit $|\mathbf v|/c\to0$, one has $\gamma\to1$ and the final term is smaller by order $v^2/c^2$. Hence
$$
\boxed{
m\frac{d\mathbf v}{dt}
=q(\mathbf E+\mathbf v\times\mathbf B)
},
$$
which is precisely Newton's second law with the ordinary Lorentz force.

Solved by gpt-5.6-sol high.

= 38B
{parent=Paper 1}
{scope}
{title2=General Relativity}

= a
{parent=38b}
{scope}

= Solution
{parent=a}

Put
$$
Y=\nabla_\rho\phi\nabla^\rho\phi.
$$
Using metric compatibility and the <symmetric Hessian of a scalar field>, the divergence of the proposed tensor is
$$
\begin{aligned}
2\nabla^\mu T_{\mu\nu}
&=(\nabla^\mu\nabla_\mu\phi)\nabla_\nu\phi
+\nabla^\mu\phi\,\nabla_\mu\nabla_\nu\phi
+A\nabla_\nu Y+2B\phi\nabla_\nu\phi\\
&=m^2\phi\nabla_\nu\phi
+\frac12\nabla_\nu Y
+A\nabla_\nu Y+2B\phi\nabla_\nu\phi\\
&=\left(A+\frac12\right)\nabla_\nu Y
+(m^2+2B)\phi\nabla_\nu\phi.
\end{aligned}
$$
For this to vanish for arbitrary solutions and local field data, the two coefficients must vanish. Hence
$$
\boxed{A=-\frac12,\qquad B=-\frac{m^2}{2}}.
$$
The resulting tensor is the <Conserved Klein-Gordon scalar stress-energy ansatz>.

Solved by gpt-5.6-sol high.

= b
{parent=38b}
{scope}

= Solution
{parent=b}

Write the <Einstein field equations> in $D$ dimensions as
$$
R_{\mu\nu}-\frac12g_{\mu\nu}R=\kappa T_{\mu\nu}.
$$
Taking the trace gives
$$
\left(1-\frac D2\right)R=\kappa T,
\qquad
R=-\frac{2\kappa}{D-2}T.
$$
Therefore the <Trace-reversed Einstein field equations in D dimensions> are
$$
R_{\mu\nu}
=\kappa\left(T_{\mu\nu}-\frac1{D-2}g_{\mu\nu}T\right).
$$

For the values from part (a), let $Y=\nabla_\rho\phi\nabla^\rho\phi$. Then
$$
T_{\mu\nu}
=\frac12\nabla_\mu\phi\nabla_\nu\phi
-\frac14g_{\mu\nu}(Y+m^2\phi^2),
$$
and
$$
T=\frac{2-D}{4}Y-\frac D4m^2\phi^2.
$$
Substitution cancels the $Yg_{\mu\nu}$ terms and yields
$$
\boxed{
R_{\mu\nu}
=\frac\kappa2\left[
\nabla_\mu\phi\nabla_\nu\phi
+\frac{m^2}{D-2}g_{\mu\nu}\phi^2
\right]
}.
$$
For the convention $\kappa=8\pi G$, the prefactor is $4\pi G$. This is the <Ricci curvature sourced by a Klein-Gordon scalar field>.

Solved by gpt-5.6-sol high.

= c
{parent=38b}
{scope}

= Solution
{parent=c}

Because the condition is local and covariant, choose <normal coordinates> at the point and a Lorentz frame in which the timelike observer is $X^\mu=\delta^\mu_0$. Rescaling a timelike vector by a positive factor does not affect the sign, so it suffices to use this unit vector. With signature $(-,+,\ldots,+)$,
$$
\nabla_\rho\phi\nabla^\rho\phi
=-(\partial_0\phi)^2+\sum_{i=1}^{D-1}(\partial_i\phi)^2.
$$
The measured energy density is therefore
$$
\begin{aligned}
2T_{\mu\nu}X^\mu X^\nu
=2T_{00}
&=(\partial_0\phi)^2
-A\left[-(\partial_0\phi)^2+\sum_i(\partial_i\phi)^2\right]
-B\phi^2\\
&=(1+A)(\partial_0\phi)^2
-A\sum_i(\partial_i\phi)^2-B\phi^2.
\end{aligned}
$$
The value of $\phi$, its time derivative, and its spatial first derivatives can be varied independently at a point. Nonnegativity for all such data is therefore equivalent to nonnegativity of all three coefficients:
$$
1+A\geq0,
\qquad
-A\geq0,
\qquad
-B\geq0.
$$
Thus the most general constraints are
$$
\boxed{-1\leq A\leq0,
\qquad B\leq0}.
$$
These conditions are also sufficient in every timelike frame, because any timelike vector can be brought to the chosen rest frame. This is the <weak energy condition for a quadratic scalar stress-energy ansatz>.

The conserved values from part (a),
$$
A=-\frac12,
\qquad
B=-\frac{m^2}{2},
$$
satisfy these inequalities for real $m$, as required.

Solved by gpt-5.6-sol high.

= 39C
{parent=Paper 1}
{scope}
{title2=Fluid Dynamics II}

= a
{parent=39c}
{scope}

= Solution
{parent=a}

The steady incompressible <Stokes flow> equations are
$$
-\nabla p+\mu\nabla^2\mathbf u=0,
\qquad
\nabla\mathbin{\cdot}\mathbf u=0.
$$
Taking the <curl> eliminates the pressure gradient and commutes with the Laplacian, so for the <vorticity> $\boldsymbol\omega=\nabla\times\mathbf u$,
$$
\boxed{\nabla^2\boldsymbol\omega=0}.
$$
This is the vorticity part of <Harmonic pressure and vorticity in Stokes flow>.

For a two-dimensional incompressible flow, choose the <stream function> convention
$$
u_x=\psi_y,
\qquad
u_y=-\psi_x.
$$
Then
$$
\omega_z=\partial_xu_y-\partial_yu_x=-\nabla^2\psi.
$$
Applying the harmonic-vorticity equation gives
$$
\boxed{\nabla^4\psi=0},
$$
the <Biharmonic stream function for planar Stokes flow> equation.

Solved by gpt-5.6-sol high.

= b
{parent=39c}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

In the exterior domain $r>a$, the streamfunction obeys
$$
\boxed{\nabla^4\psi=0}.
$$
Using the <Streamfunction in polar coordinates> convention
$$
u_r=\frac1r\psi_\theta,
\qquad
u_\theta=-\psi_r,
$$
the no-slip condition on the stationary disk is
$$
u_r(a,\theta)=u_\theta(a,\theta)=0.
$$
Thus $\psi$ is constant on the disk and $\psi_r=0$ there. Choose the arbitrary constant to be zero:
$$
\boxed{\psi(a,\theta)=0,
\qquad
\psi_r(a,\theta)=0}.
$$

The imposed shear $\mathbf u_\infty=\gamma y\,\mathbf e_x$ has streamfunction
$$
\psi_\infty=\frac\gamma2y^2
=\frac{\gamma r^2}{4}(1-\cos2\theta).
$$
The far-field condition is therefore
$$
\boxed{\nabla\psi-\nabla\psi_\infty\longrightarrow0
\quad(r\to\infty)}.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

Only the axisymmetric and $\cos2\theta$ modes are needed. The exterior biharmonic forms compatible with the far-field velocity are
$$
\psi_0(r)=\frac\gamma4r^2+C_0\log r+D_0,
$$
$$
\psi_2(r,\theta)
=\left(-\frac\gamma4r^2+C_2+D_2r^{-2}\right)\cos2\theta.
$$
Terms $r^2\log r$ and $r^4\cos2\theta$ are excluded because they would dominate the imposed shear.

The two conditions at $r=a$ applied separately to each Fourier mode give
$$
C_0=-\frac{\gamma a^2}{2},
\qquad
D_0=-\frac{\gamma a^2}{4}+\frac{\gamma a^2}{2}\log a,
$$
$$
C_2=\frac{\gamma a^2}{2},
\qquad
D_2=-\frac{\gamma a^4}{4}.
$$
Hence
$$
\boxed{
\psi(r,\theta)
=\frac\gamma4\left[
 r^2-a^2-2a^2\log\frac ra
-\left(r^2-2a^2+\frac{a^4}{r^2}\right)\cos2\theta
\right]
}.
$$
Direct differentiation gives $\psi=\psi_r=0$ at $r=a$. At infinity the remaining disturbance velocity is $O(r^{-1})$, so the imposed shear is recovered. This is the <Biharmonic stream function for a fixed disk in planar shear>.

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

The <polar shear stress in a Newtonian fluid> is
$$
\sigma_{r\theta}
=\mu\left[
 r\frac\partial{\partial r}\left(\frac{u_\theta}{r}\right)
+\frac1r\frac{\partial u_r}{\partial\theta}
\right].
$$
On $r=a$, no slip holds for every $\theta$, so $u_\theta=0$ and $\partial_\theta u_r=0$. Therefore
$$
\sigma_{r\theta}(a,\theta)
=\mu\,\partial_ru_\theta(a,\theta)
=-\mu\psi_{rr}(a,\theta).
$$
Differentiating the solution from part (ii) gives
$$
\psi_{rr}(a,\theta)=\gamma-2\gamma\cos2\theta,
$$
and hence
$$
\boxed{
\sigma_{r\theta}(a,\theta)
=-\mu\gamma+2\mu\gamma\cos2\theta
}.
$$

The moment arm and line element are both $a$, so the torque exerted by the fluid on the disk per unit axial length is
$$
\begin{aligned}
\mathcal T_z
&=a^2\int_0^{2\pi}\sigma_{r\theta}(a,\theta)\,d\theta\\
&=\boxed{-2\pi\mu\gamma a^2}.
\end{aligned}
$$
For $\gamma>0$ the negative sign denotes the clockwise sense of the ambient shear rotation. This agrees with the <hydrodynamic torque on a fixed disk in planar shear>.

Solved by gpt-5.6-sol high.

= 40C
{parent=Paper 1}
{scope}
{title2=Waves}

= a
{parent=40c}
{scope}

= Solution
{parent=a}

Let $\rho'$ and $p'$ be the density and pressure perturbations and let $\mathbf u$ be the velocity perturbation. The <linear acoustics> equations about a uniform quiescent state are
$$
\rho'_t+\rho_0\nabla\mathbin{\cdot}\mathbf u=0,
\qquad
\rho_0\mathbf u_t=-\nabla p',
\qquad
p'=c_0^2\rho'.
$$
The last relation is the <homentropic pressure perturbation> law.

Taking the scalar product of the momentum equation with $\mathbf u$ gives
$$
\frac{\partial}{\partial t}\left(\frac12\rho_0|\mathbf u|^2\right)
=-\mathbf u\mathbin{\cdot}\nabla p'
=-\nabla\mathbin{\cdot}(p'\mathbf u)+p'\nabla\mathbin{\cdot}\mathbf u.
$$
The continuity and pressure relations imply
$$
p'\nabla\mathbin{\cdot}\mathbf u
=-\frac{p'p'_t}{\rho_0c_0^2}
=-\frac\partial{\partial t}
\left(\frac{p'^2}{2\rho_0c_0^2}\right).
$$
Therefore the <acoustic energy conservation> law is
$$
\boxed{
\frac\partial{\partial t}(K+W)+\nabla\mathbin{\cdot}\mathbf I=0
},
$$
with
$$
\boxed{
K=\frac12\rho_0|\mathbf u|^2,
\qquad
W=\frac{p'^2}{2\rho_0c_0^2},
\qquad
\mathbf I=p'\mathbf u
}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=40c}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

Use the <acoustic velocity potential> convention
$$
\mathbf u=\nabla\phi,
\qquad
p'=-\rho_0\phi_t.
$$
For a spherically symmetric outgoing wave with time dependence $e^{i\omega t}$ and $k=\omega/c_0$, write
$$
\widehat\phi(r)=C\frac{e^{-ik(r-a)}}r.
$$
Linearizing the boundary condition at the mean surface $r=a$ gives
$$
\widehat\phi'(a)=i\omega\epsilon.
$$
Since
$$
\widehat\phi'(a)=-\frac{C}{a^2}(1+ika),
$$
we find
$$
C=-\frac{i\omega\epsilon a^2}{1+ika}.
$$
Thus the <outgoing acoustic field of a pulsating sphere> is
$$
\boxed{
\phi(r,t)
=\operatorname{Re}\left[
-\frac{i\omega\epsilon a^2}{1+i\omega a/c_0}
\frac{e^{i\omega t-i(\omega/c_0)(r-a)}}r
\right]
}.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

For real harmonic quantities
$$
f(t)=\operatorname{Re}(\widehat f e^{i\omega t}),
\qquad
g(t)=\operatorname{Re}(\widehat g e^{i\omega t}),
$$
define the period average by
$$
\langle fg\rangle
=\frac\omega{2\pi}\int_0^{2\pi/\omega}f(t)g(t)\,dt
=\frac12\operatorname{Re}(\widehat f\widehat g^*).
$$
This is the <time average of harmonic power>.

At $r=a$, put $x=\omega a/c_0$. The velocity and pressure amplitudes are
$$
\widehat u_r(a)=i\omega\epsilon,
\qquad
\widehat p(a)=-i\rho_0\omega\widehat\phi(a)
=-\frac{\rho_0\omega^2\epsilon a}{1+ix}.
$$
Hence the mean outward work rate per unit area is
$$
\langle p'u_r\rangle
=\frac12\operatorname{Re}
\left(\widehat p\widehat u_r^*\right)
=\frac{\rho_0\omega^4\epsilon^2a^2}
{2c_0(1+x^2)}.
$$
Multiplication by the area $4\pi a^2$ yields
$$
\begin{aligned}
\langle\mathcal P\rangle
&=\frac{2\pi\rho_0\omega^4\epsilon^2a^4/c_0}
{1+\omega^2a^2/c_0^2}\\
&=\boxed{
2\pi a^2\rho_0\omega^2\epsilon^2c_0
\frac{\omega^2a^2}{c_0^2+\omega^2a^2}
}.
\end{aligned}
$$
This is the <mean acoustic power of a pulsating sphere>.

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

For a harmonic field, the time-averaged energy densities at $r=a$ are
$$
\langle K\rangle
=\frac{\rho_0}{4}|\widehat u_r|^2
=\frac{\rho_0\omega^2\epsilon^2}{4},
$$
$$
\langle W\rangle
=\frac{|\widehat p|^2}{4\rho_0c_0^2}
=\frac{\rho_0\omega^2\epsilon^2}{4}
\frac{x^2}{1+x^2}.
$$
From part (ii),
$$
|\langle I_r\rangle|
=\frac{\rho_0\omega^2\epsilon^2c_0}{2}
\frac{x^2}{1+x^2}.
$$
Consequently
$$
\boxed{
\frac{c_0\langle K+W\rangle}{|\langle I_r\rangle|}
=\frac{1+2x^2}{2x^2}
=1+\frac{c_0^2}{2\omega^2a^2}
}.
$$
This is the <surface acoustic energy-to-flux ratio of a pulsating sphere>.

Solved by gpt-5.6-sol high.

= iv
{parent=b}
{scope}

= Solution
{parent=iv}

If $c_0\ll\omega a$, then $x\gg1$. The sphere is acoustically large, and
$$
\langle K\rangle\sim\langle W\rangle,
\qquad
|\langle I_r\rangle|
\sim c_0\langle K+W\rangle.
$$
Pressure and velocity are nearly in phase, so the near-surface field behaves locally like a progressive radiating wave. The power approaches
$$
2\pi a^2\rho_0\omega^2\epsilon^2c_0.
$$

If $c_0\gg\omega a$, then $x\ll1$. The sphere is acoustically compact, and
$$
\frac{\langle W\rangle}{\langle K\rangle}\sim x^2\ll1,
\qquad
\frac{c_0\langle K+W\rangle}{|\langle I_r\rangle|}
\sim\frac1{2x^2}\gg1.
$$
The surface field is mainly reactive kinetic energy: pressure and velocity are nearly in quadrature, energy is stored and returned during each cycle, and only a small mean <acoustic energy flux> is radiated. The mean power scales as
$$
\langle\mathcal P\rangle
\sim\frac{2\pi\rho_0\omega^4\epsilon^2a^4}{c_0}.
$$

Solved by gpt-5.6-sol high.

= 41C
{parent=Paper 1}
{scope}
{title2=Numerical Analysis}

= a
{parent=41c}
{scope}

= Solution
{parent=a}

At an interior grid point, the five-point <finite difference method> gives
$$
\nabla^2u(ih,jh,t)
\approx\frac1{h^2}
\left(u_{i+1,j}+u_{i-1,j}+u_{i,j+1}+u_{i,j-1}-4u_{ij}\right).
$$
Let
$$
T=
\begin{pmatrix}
-2&1&&\\
1&-2&\ddots&\\
&\ddots&\ddots&1\\
&&1&-2
\end{pmatrix}\in\mathbb R^{m\times m},
$$
where the zero Dirichlet boundary values are already incorporated. Stack the unknowns with the $y$ index varying fastest. Using the <Kronecker product>, define
$$
\boxed{A_x=T\otimes I_m,
\qquad A_y=I_m\otimes T}.
$$
Then the semidiscrete equation is
$$
\boxed{
\frac{d\mathbf u}{dt}
=\frac1{h^2}(A_x+A_y)\mathbf u,
\qquad
\mathbf u(t)\in\mathbb R^{m^2}
}.
$$
Moreover,
$$
A_xA_y
=(T\otimes I)(I\otimes T)
=T\otimes T
=(I\otimes T)(T\otimes I)
=A_yA_x.
$$
Thus the directional matrices commute, as in the <five-point Dirichlet Laplacian as a Kronecker sum>.

Solved by gpt-5.6-sol high.

= b
{parent=41c}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

The first stage is
$$
(I-\mu A_y)\mathbf u^{n+1/2}=\mathbf u^n.
$$
Because $A_y=I\otimes T$, this consists of $m$ independent tridiagonal systems of dimension $m$. The <Thomas algorithm> solves each in $O(m)$ operations, for a total of $O(m^2)$.

The second stage,
$$
\mathbf u^{n+1}=(I+\mu A_x)\mathbf u^{n+1/2},
$$
is multiplication by a matrix with only $O(m^2)$ nonzero entries, so it also costs $O(m^2)$. Hence one complete step uses at most
$$
\boxed{O(m^2)}
$$
arithmetic operations.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

Eliminating the intermediate vector gives the amplification matrix
$$
\boxed{
\mathbf u^{n+1}=C\mathbf u^n,
\qquad
C=(I+\mu A_x)(I-\mu A_y)^{-1}
}.
$$
The commuting real symmetric matrices admit <simultaneous diagonalization> in the orthonormal basis $v^{(p,q)}$ from the question. On this vector,
$$
\boxed{
c_{pq}=\frac{1+\mu\lambda_p}{1-\mu\lambda_q}
},
\qquad
\lambda_j=-4\sin^2\left(\frac{j\pi h}{2}\right).
$$
Since the basis is orthonormal, stability in the discrete Euclidean norm is equivalent to $|c_{pq}|\leq1$ for every $p,q$.

Put
$$
\alpha=4\sin^2\frac{\pi h}{2},
\qquad
\beta=4\cos^2\frac{\pi h}{2}.
$$
Then $-\beta\leq\lambda_j\leq-\alpha$. The smallest denominator is $1+\mu\alpha$, while the largest numerator modulus occurs at an endpoint of this interval. The condition involving $1-\mu\alpha$ is automatic, and the other is
$$
|1-\mu\beta|\leq1+\mu\alpha.
$$
For $m>1$, this is equivalent to
$$
\mu(\beta-\alpha)\leq2.
$$
Since $\beta-\alpha=4\cos(\pi h)$, the exact finite-grid stability condition is
$$
\boxed{
0<\mu\leq\frac1{2\cos(\pi h)}
}.
$$
For $m=1$ there is only one mode and the method is stable for every $\mu>0$. A simple grid-independent condition, and the limiting condition as $h\to0$, is
$$
\boxed{0<\mu\leq\frac12}.
$$
This is the <stability limit of one-implicit-direction diffusion splitting>.

Solved by gpt-5.6-sol high.

= c
{parent=41c}
{scope}

= Solution
{parent=c}

The first two modified stages give
$$
\widetilde{\mathbf u}^{n+1}
=(I+\mu A_x)(I-\mu A_y)^{-1}\mathbf u^n.
$$
The correction stage is
$$
(I-\mu A_x)\mathbf u^{n+1}
=\widetilde{\mathbf u}^{n+1}-\mu A_x\mathbf u^n.
$$
Using $A_xA_y=A_yA_x$,
$$
\begin{aligned}
\widetilde{\mathbf u}^{n+1}-\mu A_x\mathbf u^n
&=\left[(I+\mu A_x)(I-\mu A_y)^{-1}-\mu A_x\right]\mathbf u^n\\
&=(I+\mu^2A_xA_y)(I-\mu A_y)^{-1}\mathbf u^n.
\end{aligned}
$$
Therefore
$$
\boxed{
\mathbf u^{n+1}=D\mathbf u^n,
\qquad
D=(I-\mu A_x)^{-1}
(I+\mu^2A_xA_y)
(I-\mu A_y)^{-1}
}.
$$

On the common eigenvector $v^{(p,q)}$, the eigenvalue of $D$ is
$$
\boxed{
d_{pq}
=\frac{1+\mu^2\lambda_p\lambda_q}
{(1-\mu\lambda_p)(1-\mu\lambda_q)}
}.
$$
Set $a=-\mu\lambda_p>0$ and $b=-\mu\lambda_q>0$. Then
$$
d_{pq}=\frac{1+ab}{(1+a)(1+b)}.
$$
Both numerator and denominator are positive, and
$$
(1+a)(1+b)-(1+ab)=a+b>0.
$$
Hence
$$
0<d_{pq}<1
$$
for every mode and every $\mu>0$. Orthogonal diagonalization now gives $\|D\|_2<1$, so the method is stable without a time-step restriction. This is the <unconditionally stable corrected directional diffusion splitting>.

Solved by gpt-5.6-sol high.