past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-4.bigb

= Paper 4
{scope}

https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2025/paperii_4_2025.pdf

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

= a
{parent=1g}
{scope}

= Solution
{parent=a}

For an odd prime $p$, the <Legendre symbol> is
$$
\left(\frac ap\right)=
\begin{cases}
0,&p\mid a,\\
1,&a\not\equiv0\pmod p\text{ and }a\text{ is a square modulo }p,\\
-1,&a\text{ is a nonsquare modulo }p.
\end{cases}
$$
For distinct odd primes $p$ and $q$, <quadratic reciprocity> states that
$$
\boxed{
\left(\frac pq\right)\left(\frac qp\right)
=(-1)^{(p-1)(q-1)/4}.}
$$

Solved by gpt-5.6-sol high.

= b
{parent=1g}
{scope}

= Solution
{parent=b}

The discriminant is
$$
11^2-4\cdot31=-3.
$$
The <discriminant criterion for a quadratic congruence modulo an odd prime> therefore says that, for $p\ne3$, a solution exists exactly when
$$
\left(\frac{-3}{p}\right)=1.
$$
The <quadratic character of minus three> gives
$$
\left(\frac{-3}{p}\right)=\left(\frac p3\right),
$$
which equals $1$ precisely for $p\equiv1\pmod3$. When $p=3$, the polynomial is
$$
X^2+2X+1=(X+1)^2,
$$
so it also has a solution. Hence
$$
\boxed{p\equiv0\text{ or }1\pmod3}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=1g}
{scope}

= i
{parent=c}
{scope}

= Solution
{parent=i}

The class
$$
X=-1
$$
satisfies $X^3=-1$ for every prime. Thus a solution exists for every odd prime $p$, regardless of its residue class modulo $3$.

Solved by gpt-5.6-sol high.

= ii
{parent=c}
{scope}

= Solution
{parent=ii}

The <cube roots of minus one modulo an odd prime> are controlled by
$$
X^3+1=(X+1)(X^2-X+1).
$$
For $p\ne3$, the quadratic factor has roots exactly when its discriminant $-3$ is a square, namely when $p\equiv1\pmod3$. In that case there are three distinct roots; when $p\equiv2\pmod3$, only $X=-1$ remains. For $p=3$,
$$
X^3+1=(X+1)^3,
$$
so $X=-1$ is again the unique root. Therefore
$$
\boxed{X^3=-1\text{ has a unique solution exactly when }
p\equiv0\text{ or }2\pmod3}.
$$

Solved by gpt-5.6-sol high.

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

= Solution
{parent=2I}

Starting with $x_0=x>0$, the <continued-fraction algorithm> sets
$$
a_n=\lfloor x_n\rfloor,
\qquad
x_{n+1}=\frac1{x_n-a_n}
$$
unless $x_n$ is already an integer. This gives
$$
x=[a_0;a_1,a_2,\ldots].
$$

Every finite continued fraction is rational, since it is built from integers by finitely many additions and reciprocals. Conversely, if $x=p/q$ is rational in lowest terms, then
$$
\frac pq=a_0+\frac rq,
\qquad 0\leq r<q.
$$
When $r\ne0$, the next complete quotient is $q/r$. Thus each step is an application of the Euclidean algorithm and replaces the denominator by a smaller nonnegative remainder. The process must terminate. This proves the <termination criterion for a simple continued fraction>.

For $x=\sqrt3$,
$$
a_0=1,\qquad
x_1=\frac1{\sqrt3-1}=\frac{\sqrt3+1}{2},\qquad a_1=1,
$$
and
$$
x_2=\frac1{x_1-1}=\sqrt3+1,\qquad a_2=2.
$$
Finally,
$$
x_3=\frac1{x_2-2}=\frac1{\sqrt3-1}=x_1,
$$
so the complete quotients repeat. Hence the <continued fraction of the square root of three> is
$$
\boxed{\sqrt3=[1;\overline{1,2}]}.
$$

Solved by gpt-5.6-sol high.

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

= Solution
{parent=3K}

A binary <linear-feedback shift register> of degree $d$ has a state
$(s_n,\ldots,s_{n+d-1})\in\mathbb F_2^d$ and recurrence
$$
s_{n+d}=a_{d-1}s_{n+d-1}+\cdots+a_0s_n.
$$
Its feedback polynomial is
$$
f(X)=X^d+a_{d-1}X^{d-1}+\cdots+a_0.
$$

There are $2^d$ states, and the zero state is fixed. A nonzero periodic orbit therefore visits at most the other $2^d-1$ states, proving the <period bound for a linear-feedback shift register>.

For a maximal orbit with $d>1$, the all-one state cannot itself be fixed. From that state the incoming bit is $a_0+\cdots+a_{d-1}$ in $\mathbb F_2$. If an odd number of the $a_i$ were one, the incoming bit would be one and the all-one state would be fixed. Hence an even number of the $a_i$ are one; including the leading coefficient, the feedback polynomial has an odd number of nonzero coefficients. This is the <feedback-polynomial parity condition for maximal period>.

The given prefix has seven consecutive zeros followed by a one. Any register of degree at most seven would therefore enter the all-zero state before producing the final one, which is impossible. Degree eight is attained by
$$
s_{n+8}=s_n,
$$
with initial state $10000000$. It produces the next bit $1$ and has feedback polynomial
$$
\boxed{f(X)=X^8+1}.
$$
Thus this is the <minimal linear-feedback shift register for the prefix 100000001>.

Solved by gpt-5.6-sol high.

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

= a
{parent=4f}
{scope}

= Solution
{parent=a}

A <variable-based grammar> is a tuple
$$
G=(\Sigma,V,P,S),
$$
where $\Sigma$ and $V$ are disjoint finite sets of terminals and variables, $S\in V$, and each production $\alpha\to\beta$ has $\alpha,\beta\in(\Sigma\cup V)^*$ with at least one variable in $\alpha$.

Solved by gpt-5.6-sol high.

= b
{parent=4f}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

Choose a new start variable $S_0\notin V\cup V'$. The <concatenation grammar> is
$$
H=\left(\Sigma,V\cup V'\cup\{S_0\},
P\cup P'\cup\{S_0\to SS'\},S_0\right).
$$
Indeed, its derivations first produce a word from $G$ and a word from $G'$ in their original order, so $L(H)=L(G)L(G')$.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

For the <regular concatenation grammar>, let
$$
P_1=
\{\alpha\to\beta\in P:\beta\notin\Sigma^*\}
\cup
\{\alpha\to wS':(\alpha\to w)\in P,\ w\in\Sigma^*\}.
$$
Then define
$$
H^{\mathrm{reg}}=(\Sigma,V\cup V',P_1\cup P',S).
$$
If $G$ and $G'$ are regular, every nonterminal rule retained from $P$ is right-linear, and each replaced terminal rule is also right-linear. A regular derivation in $G$ uses exactly one terminal rule, at its final step; the replacement transfers control to $S'$, after which a word of $L(G')$ is generated. Hence
$$
L(H^{\mathrm{reg}})=L(G)L(G').
$$

Solved by gpt-5.6-sol high.

= c
{parent=4f}
{scope}

= Solution
{parent=c}

Let
$$
G=(\{a,b\},\{S\},\{S\to a\},S),
\qquad
G'=(\{a,b\},\{T\},\{T\to b\},T).
$$
Both are <regular grammar>s and their variable sets are disjoint. Their ordinary concatenation grammar adds a new rule
$$
S_0\to ST.
$$
This rule has two variables on its right-hand side, so it is not right-linear. Thus this is an explicit instance in which the <concatenation grammar need not be regular>.

Solved by gpt-5.6-sol high.

= d
{parent=4f}
{scope}

= Solution
{parent=d}

Take
$$
G=(\{a,b,c\},\{S,A,B\},
\{S\to AB,\ A\to a,\ B\to b\},S)
$$
and
$$
G'=(\{a,b,c\},\{T\},\{T\to c\},T).
$$
These are variable based and have disjoint variable sets. They satisfy
$$
L(G)L(G')=\{abc\}.
$$
In $H^{\mathrm{reg}}$, however, both terminal productions of $G$ are replaced:
$$
A\to aT,\qquad B\to bT.
$$
Consequently its only terminal derivation is
$$
S\Rightarrow AB\Rightarrow aTB\Rightarrow aTbT
\Rightarrow acbc,
$$
so
$$
L(H^{\mathrm{reg}})=\{acbc\}\ne\{abc\}.
$$
This realizes the <failure of the regular concatenation construction for a nonregular grammar>: two different terminal productions introduce two copies of $T$.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=5K}

For a model with $k$ fitted parameters, the <Akaike information criterion> is
$$
\operatorname{AIC}=-2\ell(\widehat\theta)+2k,
$$
where $\ell$ is the log-likelihood evaluated at the maximum-likelihood estimate.

In the normal linear model with known $\sigma^2$,
$$
-2\ell(\beta)
=n\log(2\pi\sigma^2)+\frac1{\sigma^2}\|Y-X\beta\|^2.
$$
The maximum-likelihood estimator is the ordinary least-squares estimator, with fitted mean $\widehat\mu=HY$, where
$$
H=X(X^TX)^{-1}X^T.
$$
Since only the $p$ components of $\beta$ are fitted,
$$
\operatorname{AIC}
=n\log(2\pi\sigma^2)
+\frac1{\sigma^2}\|Y-\widehat\mu\|^2+2p.
$$
After multiplying by $\sigma^2$ and discarding the model-independent constant, this is exactly <Mallows Cp>:
$$
\boxed{C_p=\|Y-\widehat\mu\|^2+2p\sigma^2}.
$$

To compare it with prediction error, write $Y=\mu+\varepsilon$ and $Y^*=\mu+\varepsilon^*$, where $\mu=X\beta$, the two errors are independent, and both have covariance $\sigma^2I$. Since $H\mu=\mu$, $H$ is an orthogonal projection of rank $p$, and
$$
Y^*-\widehat\mu=\varepsilon^*-H\varepsilon.
$$
The cross term has zero expectation, so
$$
\operatorname{MSPE}
=\mathbb E\|\varepsilon^*\|^2
+\mathbb E\|H\varepsilon\|^2
=n\sigma^2+p\sigma^2.
$$
On the other hand,
$$
\mathbb E\|Y-\widehat\mu\|^2
=\mathbb E\|(I-H)\varepsilon\|^2
=\sigma^2\operatorname{tr}(I-H)
=(n-p)\sigma^2.
$$
Therefore the <unbiased prediction-error identity for ordinary least squares> gives
$$
\mathbb E(C_p)=(n-p)\sigma^2+2p\sigma^2
=(n+p)\sigma^2
=\boxed{\operatorname{MSPE}}.
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=6a}
{scope}

= Solution
{parent=a}

The population map is
$$
x_{n+1}=g(x_n),
\qquad
g(x)=x\bigl(r+ke^{-\lambda x}\bigr).
$$
A positive equilibrium satisfies
$$
1=r+ke^{-\lambda x_*},
$$
and hence
$$
x_*=\frac1\lambda\log\left(\frac{k}{1-r}\right).
$$
It is strictly positive exactly when
$$
\boxed{k>1-r}.
$$
This is the nonzero equilibrium of the <survival-augmented Ricker map>.

Solved by gpt-5.6-sol high.

= b
{parent=6a}
{scope}

= Solution
{parent=b}

Differentiate the map:
$$
g'(x)=r+ke^{-\lambda x}(1-\lambda x).
$$
At the positive equilibrium, $ke^{-\lambda x_*}=1-r$, so
$$
g'(x_*)
=1-(1-r)\lambda x_*
=1-(1-r)\log\left(\frac{k}{1-r}\right).
$$
The <fixed point stability for an iteration> requires $|g'(x_*)|<1$. Existence already makes the logarithm positive, so the upper inequality is automatic. The lower inequality gives
$$
(1-r)\log\left(\frac{k}{1-r}\right)<2.
$$
Combining it with existence yields the <stability interval of the survival-augmented Ricker equilibrium>:
$$
\boxed{1-r<k<(1-r)\exp\left(\frac2{1-r}\right)}.
$$

Solved by gpt-5.6-sol high.

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

= Solution
{parent=7E}

The <Laplace integral method for a differential equation> seeks
$$
y(z)=\int_Ce^{zt}f(t)\,dt.
$$
Differentiating under the integral and substituting into
$$
zy''+y'-zy=0
$$
gives
$$
\int_Ce^{zt}\left[z(t^2-1)f(t)+tf(t)\right]dt=0.
$$
Since $ze^{zt}=\partial_te^{zt}$, integration by parts yields
$$
\left[e^{zt}(t^2-1)f(t)\right]_{\partial C}
-\int_Ce^{zt}\left[(t^2-1)f'(t)+tf(t)\right]dt=0.
$$
It is therefore enough to choose $f$ and $C$ so that
$$
(t^2-1)f'(t)+tf(t)=0
$$
and the boundary term vanishes. Solving the first-order equation on $-1<t<1$ gives
$$
f(t)=\frac{K}{\sqrt{1-t^2}}.
$$
Take $C$ to be the interval $[-1,1]$. Although $f$ has integrable endpoint singularities,
$$
(t^2-1)f(t)=-K\sqrt{1-t^2}
$$
vanishes at both endpoints, so the boundary term is zero. Thus
$$
y(z)=K\int_{-1}^1\frac{e^{zt}}{\sqrt{1-t^2}}\,dt.
$$
At $z=0$, substituting $t=\sin\theta$ gives
$$
\int_{-1}^1\frac{dt}{\sqrt{1-t^2}}=\pi.
$$
The normalization $y(0)=1$ therefore sets $K=1/\pi$. By the defining regularity and normalization of the <modified Bessel function>, this proves the <real integral representation of the modified Bessel function I0>:
$$
\boxed{I_0(z)=\frac1\pi\int_{-1}^1
\frac{e^{zs}}{\sqrt{1-s^2}}\,ds}.
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=8b}
{scope}

= Solution
{parent=a}

The <Jacobi identity for the Poisson bracket> is
$$
\boxed{\{F,\{G,H\}\}+\{G,\{H,F\}\}
+\{H,\{F,G\}\}=0}.
$$
For quantities with no explicit time dependence,
$$
\dot F=\{F,H\},
\qquad
\dot G=\{G,H\}.
$$
If $F$ and $G$ are conserved, the first two terms of the Jacobi identity vanish. Hence
$$
\{H,\{F,G\}\}=0,
$$
or, by antisymmetry,
$$
\boxed{\{\{F,G\},H\}=0}.
$$
Thus the <poisson bracket of conserved quantities> is itself conserved.

Solved by gpt-5.6-sol high.

= b
{parent=8b}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

For $Q=\lambda q$ and $P=\lambda p$,
$$
\{Q,P\}_{q,p}=\lambda^2.
$$
The one-dimensional criterion for a <canonical transformation> therefore gives
$$
\boxed{\lambda=1\text{ or }\lambda=-1}.
$$
For either value, the <type-two generating function for a canonical transformation>
$$
\boxed{S(q,P;\lambda)=\lambda qP}
$$
works, because $p=S_q=\lambda P$ and $Q=S_P=\lambda q$; when $\lambda^2=1$, the first equation is equivalent to $P=\lambda p$. For every other $\lambda$, the Poisson bracket is not one and the map is not canonical.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

For $Q=q$ and $P=\lambda p$,
$$
\{Q,P\}_{q,p}=\lambda.
$$
Thus the map is canonical only for
$$
\boxed{\lambda=1}.
$$
At that value it is the identity and is generated by
$$
\boxed{S(q,P)=qP}.
$$

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

For $Q=p$ and $P=\lambda q$,
$$
\{Q,P\}_{q,p}=-\lambda.
$$
The map is therefore canonical only for
$$
\boxed{\lambda=-1}.
$$
Then $Q=p$ and $P=-q$. The <type-one generating function for a canonical transformation>
$$
\boxed{\Phi(q,Q)=qQ}
$$
gives
$$
p=\Phi_q=Q,
\qquad
P=-\Phi_Q=-q,
$$
as required.

Solved by gpt-5.6-sol high.

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

= a
{parent=9e}
{scope}

= Solution
{parent=a}

The <potential slow-roll parameter> is
$$
\epsilon=\frac{M_{\rm Pl}^2}{2}\left(\frac{V'}V\right)^2.
$$
From the <slow-roll approximation>,
$$
\dot\phi=-\frac{V'}{3H},
\qquad
H^2=\frac{V}{3M_{\rm Pl}^2}.
$$
Therefore
$$
\frac{\dot\phi^2}{2V}
=\frac{V'^2}{18H^2V}
=\frac{M_{\rm Pl}^2V'^2}{6V^2}
=\frac\epsilon3.
$$
Thus $\epsilon\ll1$ implies $\dot\phi^2\ll V$, consistently with potential-energy domination.

The same equations give
$$
\frac{H}{\dot\phi}
=-\frac{3H^2}{V'}
=-\frac1{M_{\rm Pl}^2}\frac V{V'}.
$$
On the branch $V'/V>0$, the definition of $\epsilon$ turns this into
$$
\boxed{
\frac{H}{\dot\phi}
=-\frac1{M_{\rm Pl}^2}\frac V{V'}
=-\frac1{\sqrt2M_{\rm Pl}}\frac1{\sqrt\epsilon}}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=9e}
{scope}

= Solution
{parent=b}

By definition,
$$
N=\log\frac{a(t_f)}{a(t_i)}
=\int_{t_i}^{t_f}H\,dt.
$$
Using $dt=d\phi/\dot\phi$ and the result of part (a),
$$
\begin{aligned}
N
&=\int_{\phi_i}^{\phi_f}\frac{H}{\dot\phi}\,d\phi\\
&\simeq-\frac1{\sqrt2M_{\rm Pl}}
\int_{\phi_i}^{\phi_f}\frac{d\phi}{\sqrt{\epsilon(\phi)}}.
\end{aligned}
$$
Reversing the limits yields the <slow-roll e-fold count>
$$
\boxed{N\simeq\frac1{\sqrt2M_{\rm Pl}}
\int_{\phi_f}^{\phi_i}\frac{d\phi}{\sqrt{\epsilon(\phi)}}}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=9e}
{scope}

= Solution
{parent=c}

For the <natural inflation> potential, write $x=\phi/f$. Then
$$
V'= -\frac{V_0}{f}\sin x
$$
and the half-angle identity gives
$$
\left|\frac{V'}V\right|
=\frac1f\frac{\sin x}{1+\cos x}
=\frac1f\tan\frac x2.
$$
Therefore, on the stated rolling branch,
$$
\boxed{\sqrt{\epsilon(\phi)}
=\frac{M_{\rm Pl}}{\sqrt2f}
\tan\frac{\phi}{2f}}.
$$

Substitution into part (b) gives
$$
N=\frac f{M_{\rm Pl}^2}
\int_{\phi_f}^{\phi_i}\cot\frac{\phi}{2f}\,d\phi.
$$
Since
$$
\int\cot\frac{\phi}{2f}\,d\phi
=2f\log\sin\frac{\phi}{2f},
$$
the <e-fold count for the natural-inflation cosine potential> is
$$
\boxed{
N=\frac{2f^2}{M_{\rm Pl}^2}
\left[
\log\left(\sin\frac{\phi_i}{2f}\right)
-\log\left(\sin\frac{\phi_f}{2f}\right)
\right].}
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=10c}
{scope}

= Solution
{parent=a}

Apply a <Hadamard gate> to Alice's qubit, followed by a <controlled-NOT gate> from Alice to Bob and another from Alice to Charlie:
$$
|000\rangle
\longmapsto
\frac{|000\rangle+|100\rangle}{\sqrt2}
\longmapsto
\frac{|000\rangle+|110\rangle}{\sqrt2}
\longmapsto
\frac{|000\rangle+|111\rangle}{\sqrt2}.
$$
Thus this circuit gives the required <quantum circuit preparation of a three-qubit GHZ state>.

Solved by gpt-5.6-sol high.

= b
{parent=10c}
{scope}

= Solution
{parent=b}

Let Alice's two bits be $(x,z)$ and Bob's bit be $b$. Alice applies
$$
X_A^xZ_A^z,
$$
while Bob applies
$$
X_B^b.
$$
They then send their qubits to Charlie. Acting on
$$
|\chi_1^+\rangle
=\frac{|000\rangle+|111\rangle}{\sqrt2},
$$
these operations produce, up to an irrelevant global phase,
$$
\frac{|x,b,0\rangle+(-1)^z
|1\mathbin\oplus x,1\mathbin\oplus b,1\rangle}{\sqrt2}.
$$
The complete encoding table is
$$
\begin{array}{c|c|c}
(x,z)&b&\text{state measured by Charlie}\\ \hline
(0,0)&0&|\chi_1^+\rangle\\
(0,1)&0&|\chi_1^-\rangle\\
(0,0)&1&|\chi_2^+\rangle\\
(0,1)&1&|\chi_2^-\rangle\\
(1,0)&0&|\chi_4^+\rangle\\
(1,1)&0&|\chi_4^-\rangle\\
(1,0)&1&|\chi_3^+\rangle\\
(1,1)&1&|\chi_3^-\rangle
\end{array}
$$
where a possible overall minus sign in the two $x=z=1$ cases has no observable effect.

Charlie performs a projective measurement in the eight-state GHZ basis. The sign $+$ or $-$ reveals $z$. The basis index reveals $(x,b)$ according to
$$
1\leftrightarrow(0,0),
\quad
2\leftrightarrow(0,1),
\quad
3\leftrightarrow(1,1),
\quad
4\leftrightarrow(1,0).
$$
He therefore reconstructs Alice's two bits and Bob's one bit with certainty, which is the <three-party dense coding with a GHZ state> protocol.

Solved by gpt-5.6-sol high.

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

= a
{parent=11g}
{scope}

= Solution
{parent=a}

For a prime $p$ and a positive integer $N$, the <P-adic valuation> is
$$
v_p(N)=\max\{r\in\mathbb Z_{\geq0}:p^r\mid N\}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=11g}
{scope}

= Solution
{parent=b}

Because $p\geq k+2$, neither $(k+1)!$ nor $k!$ contains a factor $p$. On the other hand, $p\leq2k+1$, so $(2k+1)!$ contains a factor $p$; and
$$
2p\geq2k+4>2k+1,
$$
so it contains no second factor $p$. Therefore
$$
v_p\left(\binom{2k+1}{k+1}\right)
=v_p((2k+1)!)-v_p((k+1)!)-v_p(k!)
=1.
$$
This is the <valuation of a near-central binomial coefficient>.

Solved by gpt-5.6-sol high.

= c
{parent=11g}
{scope}

= Solution
{parent=c}

It suffices first to prove the <elementary primorial bound> for positive integers $n$, by strong induction. The case $n=1$ is immediate.

If $n=2k$, every prime $p$ with $k<p\leq2k$ divides $\binom{2k}{k}$: the numerator $(2k)!$ contains $p$, whereas the two copies of $k!$ do not. Since these primes are distinct,
$$
P(2k)\leq P(k)\binom{2k}{k}.
$$
The induction hypothesis and the binomial theorem give
$$
P(2k)\leq4^k\binom{2k}{k}
\leq4^k2^{2k}=4^{2k}.
$$

If $n=2k+1$, part (b) shows that every prime $k+2\leq p\leq2k+1$ divides $\binom{2k+1}{k+1}$. Hence
$$
P(2k+1)\leq P(k+1)\binom{2k+1}{k+1}.
$$
The two central coefficients of $(1+1)^{2k+1}$ are equal, so
$$
2\binom{2k+1}{k+1}\leq2^{2k+1},
$$
and therefore
$$
P(2k+1)\leq4^{k+1}4^k=4^{2k+1}.
$$
This completes the induction. For real $X\geq1$,
$$
P(X)=P(\lfloor X\rfloor)\leq4^{\lfloor X\rfloor}\leq4^X.
$$

Solved by gpt-5.6-sol high.

= d
{parent=11g}
{scope}

= Solution
{parent=d}

Every prime in $(\sqrt X,X]$ is at least $\sqrt X$. Consequently the <primorial> satisfies
$$
P(X)\geq(\sqrt X)^{\pi(X)-\pi(\sqrt X)}.
$$
Combining this with part (c) and taking logarithms gives
$$
\bigl(\pi(X)-\pi(\sqrt X)\bigr)\frac{\log X}{2}
\leq X\log4,
$$
so
$$
\pi(X)\leq\pi(\sqrt X)+\frac{2X\log4}{\log X}
\leq\sqrt X+\frac{2X\log4}{\log X}.
$$
For $X\geq2$, the inequality $\log X\leq\sqrt X$ implies
$$
\sqrt X\leq\frac{X}{\log X}.
$$
Thus the <prime-counting upper bound from a primorial estimate> yields
$$
\pi(X)\leq(1+2\log4)\frac{X}{\log X}.
$$
We may take $c=1+2\log4$.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=12i}

A subset $A$ of a metric space $X$ is a <nowhere dense set> when
$$
\operatorname{int}\overline A=\varnothing.
$$
Equivalently, every nonempty open set contains a nonempty open subset disjoint from $A$.

The <Baire category theorem> says that if $X$ is a complete metric space and $G_1,G_2,\ldots$ are open dense subsets of $X$, then $\bigcap_{n\geq1}G_n$ is dense in $X$. Equivalently, no nonempty open subset of $X$ is a countable union of nowhere-dense sets.

To prove it, take a nonempty open set $V$. Since $G_1$ is open and dense, there is a closed ball
$$
\overline B(x_1,r_1)\subset V\cap G_1
$$
with $0<r_1<2^{-1}$. Inductively, openness and density of $G_{n+1}$ allow us to choose
$$
\overline B(x_{n+1},r_{n+1})
\subset B(x_n,r_n)\cap G_{n+1},
\qquad 0<r_{n+1}<2^{-(n+1)}.
$$
The balls are nested, and for $m>n$ their centres satisfy
$$
d(x_m,x_n)<r_n<2^{-n}.
$$
Thus $(x_n)$ is Cauchy and has a limit $x\in X$. Every closed ball contains the tail of the sequence, so it contains $x$. Hence
$$
x\in V\cap\bigcap_{n\geq1}G_n.
$$
As $V$ was arbitrary, the intersection is dense.

Solved by gpt-5.6-sol high.

= i
{parent=12i}
{scope}

= Solution
{parent=i}

Let $C=C([0,1])$ with the uniform norm. For rational numbers $0\leq a<b\leq1$, let $M^+_{a,b}$ and $M^-_{a,b}$ be the sets of functions that are respectively nondecreasing and nonincreasing on $[a,b]$.

Both sets are closed. For example, if $f_j\in M^+_{a,b}$ and $f_j\to f$ uniformly, then for $x<y$ in $[a,b]$,
$$
f(x)=\lim_jf_j(x)\leq\lim_jf_j(y)=f(y).
$$
They also have empty interior. Given $f\in M^+_{a,b}$ and $\varepsilon>0$, continuity lets us choose $a<x<y<b$ sufficiently close that
$$
0\leq f(y)-f(x)<\frac{\varepsilon}{2}.
$$
Add a continuous triangular bump $h$ supported near $x$, with $h(x)=3\varepsilon/4$, $h(y)=0$, and $\lVert h\rVert_\infty<\varepsilon$. Then $(f+h)(x)>(f+h)(y)$, so the $\varepsilon$-ball about $f$ is not contained in $M^+_{a,b}$. The analogous upward bump at $y$ deals with $M^-_{a,b}$. Thus both sets are nowhere dense.

There are only countably many rational pairs $(a,b)$, so
$$
\bigcup_{a,b\in\mathbb Q,\ 0\leq a<b\leq1}
\left(M^+_{a,b}\cup M^-_{a,b}\right)
$$
is meagre. Completeness of $C$ and the <Baire category theorem> show that its complement is nonempty. Choose $f$ in that complement. If $f$ were monotone on an interval of positive length, that interval would contain a closed interval with rational endpoints, putting $f$ in one of the displayed sets. Hence $f$ is monotone on no interval of positive length, as described by the <generic nowhere-monotone continuous function> result.

Solved by gpt-5.6-sol high.

= ii
{parent=12i}
{scope}

= Solution
{parent=ii}

For $n\geq0$, put
$$
S_n=\{x\in\mathbb R:F^{(n)}(x)=0\},
\qquad E_n=\operatorname{int}S_n.
$$
Each $S_n$ is closed, the hypothesis gives $\mathbb R=\bigcup_nS_n$, and $E_n\subset E_{n+1}$ because a function that vanishes on an open set has derivative zero there.

The open set
$$
\Omega=\bigcup_{n\geq0}E_n
$$
is dense. Indeed, inside any nonempty <open interval> choose a <nondegenerate interval>[nondegenerate] closed subinterval $J$. The <complete metric space> $J$ is covered by the <closed sets> $J\cap S_n$, so the <Baire category theorem>[Baire's theorem] makes one of them contain a relative open interval, which lies in some $E_n$.

On each connected component $I$ of $\Omega$, the function $F$ is one polynomial. To see this, any compact subinterval $K\subset I$ is covered by the increasing family $(E_n)$. A finite subcover therefore gives $K\subset E_N$ for some $N$, and $F^{(N)}=0$ on $K$. Hence $F$ is polynomial on $K$; overlapping compact intervals force these polynomials to agree throughout $I$.

Let $D=\mathbb R\setminus\Omega$. This closed set has no isolated points. Otherwise, on the two sides of an isolated point $x$, $F$ would equal polynomials. Smoothness makes all their one-sided derivatives agree at $x$, so the two polynomials are identical and extend across $x$. Some derivative would then vanish on a neighborhood of $x$, contrary to $x\in D$.

Suppose that $D$ is nonempty. It is complete and
$$
D=\bigcup_{n\geq0}(D\cap S_n)
$$
is a countable closed cover. A second application of Baire's theorem gives an index $k$ and an open interval $U_0$ such that
$$
\varnothing\ne D\cap U_0\subset S_k.
$$
Choose $x_0\in D\cap U_0$ and a bounded open interval $U$ centred at $x_0$ whose closure lies in $U_0$.

Because $D$ has no isolated points, every $x\in D\cap U$ is approached by distinct points of $D\cap U$. Difference quotients first give $F^{(k+1)}(x)=0$, and induction gives
$$
F^{(m)}(x)=0\qquad(m\geq k, x\in D\cap U).
$$

No component of $\Omega$ meeting $U$ can cross $x_0$, so it has an endpoint in $D\cap U$. If its polynomial had degree $d\geq k$, then its $d$th derivative would approach a nonzero constant at that endpoint, contradicting the preceding display. Its degree is therefore less than $k$, so $F^{(k)}$ also vanishes on the part of that component in $U$. We conclude that $F^{(k)}=0$ throughout $U$, which says $U\subset E_k\subset\Omega$ and contradicts $D\cap U\ne\varnothing$.

Thus $D$ is empty. The whole line is the single component of $\Omega$, and the argument above shows that $F$ is one polynomial on $\mathbb R$. This is the <smooth function with a pointwise vanishing derivative> theorem.

Solved by gpt-5.6-sol high.

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

= a
{parent=13k}
{scope}

= Solution
{parent=a}

Let $Y_i$ be the number of survivors among the $n_i$ patients in row $i$. The fitted <grouped-binomial logistic regression> is
$$
Y_i\ \mathrel{\mathop\sim^{\mathrm{ind}}}\ \operatorname{Bin}(n_i,p_i),
$$
with
$$
\log\frac{p_i}{1-p_i}
=\beta_0+\beta_B\mathbf1\{\text{site}_i=B\}
+\beta_C\mathbf1\{\text{site}_i=C\}
+\beta_M\mathbf1\{\text{malignant}_i=\text{yes}\}.
$$
Site A and nonmalignant tumours are the reference levels. There is no site-by-malignancy interaction, so the malignancy log-odds effect is assumed to be the same at all three sites. In the R call, `survive/total` supplies the observed proportions and `weights = total` supplies the binomial denominators $n_i$.

Solved by gpt-5.6-sol high.

= b
{parent=13k}
{scope}

= Solution
{parent=b}

The estimates maximize the grouped-binomial likelihood, numerically obtained by <iteratively reweighted least squares>. At the fitted probabilities, the estimated covariance matrix is the inverse observed information, approximately
$$
\widehat{\operatorname{Var}}(\widehat\beta)
=(X^T\widehat W X)^{-1},
\qquad
\widehat W_{ii}=n_i\widehat p_i(1-\widehat p_i).
$$
The `Std. Error` column is the square root of its diagonal. Each `z value` is `Estimate/Std. Error`, and the two-sided <Wald test> p-value is
$$
2\Phi(-|z|)
$$
for the null hypothesis that the corresponding coefficient is zero.

Thus $\widehat\beta_0=1.6855$ is the fitted log odds for a nonmalignant patient at site A. The site coefficients compare B and C with A, while $\widehat\beta_M=-0.9048$ compares malignant with nonmalignant tumours after adjustment for site. In odds-ratio form,
$$
e^{0.8096}\approx2.25,
\qquad e^{-0.5423}\approx0.58,
\qquad e^{-0.9048}\approx0.405.
$$
At level $0.05$, the B-versus-A contrast is not significant ($p=0.0624$), the C-versus-A contrast is not significant ($p=0.2610$), and malignancy has a significant negative association with survival ($p=0.0175$). The intercept test merely compares the baseline survival probability with $1/2$.

The null deviance $11.693$ compares the intercept-only fit with the saturated model and has $6-1=5$ residual degrees of freedom. The residual deviance $0.85048$ compares the four-parameter fitted model with the saturated model and has $6-4=2$ degrees of freedom. Finally, the <Akaike information criterion> is
$$
-2\ell(\widehat\beta)+2\cdot4=29.003.
$$

Solved by gpt-5.6-sol high.

= c
{parent=13k}
{scope}

= Solution
{parent=c}

Changing the <reference level in a regression factor> from A to B does not change the statistical model, fitted probabilities, likelihood, or deviance. If the first parametrization is
$$
(\beta_0,\beta_B,\beta_C,\beta_M),
$$
then the second is
$$
(\beta'_0,\beta'_A,\beta'_C,\beta'_M)
=(\beta_0+\beta_B,-\beta_B,\beta_C-\beta_B,\beta_M).
$$
Numerically,
$$
\begin{aligned}
1.6855+0.8096&=2.4951,\\
-0.8096&=-0.8096,\\
-0.5423-0.8096&\approx-1.3520,\\
-0.9048&=-0.9048.
\end{aligned}
$$

The A-versus-B test is the same test as the former B-versus-A test, with its sign reversed, so its p-value remains $0.0624$. The malignancy coefficient is unchanged in this additive model, so its p-value remains $0.0175$. The intercept now represents site B rather than site A, so its test changes.

The `siteC` coefficient also changes meaning: in `fit1` it tests C versus A, whereas in `fit2` it tests C versus B. Its standard error and p-value therefore need not agree. The displayed tables show that C versus A is not significant ($p=0.2610$), while C versus B is significant ($p=0.0160$). There is no contradiction: these are different pairwise hypotheses.

Solved by gpt-5.6-sol high.

= d
{parent=13k}
{scope}

= Solution
{parent=d}

The reduced model `fit3` retains malignancy but omits both site indicators. The <analysis of deviance for nested generalized linear models> therefore tests
$$
H_0:\beta_B=\beta_C=0
$$
against the additive model with a site effect. The likelihood-ratio statistic is the reduction in deviance,
$$
7.4923-0.8505=6.6418.
$$
The full model adds two parameters, so under $H_0$ this is approximately $\chi^2_2$. Since
$$
6.6418>\chi^2_{2,0.95}=5.9915
$$
but is less than the $0.99$ quantile $9.2103$, the p-value is between $0.01$ and $0.05$; in fact, because $\chi^2_2$ has survival function $e^{-x/2}$,
$$
p=e^{-6.6418/2}\approx0.0361.
$$
We reject the no-site-effect null at the $5\%$ level and find evidence that survival differs by site after adjustment for malignancy.

Solved by gpt-5.6-sol high.

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

= a
{parent=14a}
{scope}

= Solution
{parent=a}

For a spatially homogeneous equilibrium of the <Brusselator>, the reaction terms obey
$$
0=\alpha-(\beta+1)u+u^2v,
\qquad
0=\beta u-u^2v.
$$
Positivity gives $uv=\beta$, and substitution into the first equation gives
$$
(u_*,v_*)=\left(\alpha,\frac\beta\alpha\right).
$$

The reaction Jacobian there is
$$
J=
\begin{pmatrix}
\beta-1&\alpha^2\\
-\beta&-\alpha^2
\end{pmatrix},
$$
so
$$
\operatorname{tr}J=\beta-1-\alpha^2<0,
\qquad
\det J=\alpha^2>0.
$$
The <linear stability analysis> therefore makes the equilibrium asymptotically stable. It is a stable node when
$$
(\beta-1-\alpha^2)^2\geq4\alpha^2
$$
and a stable focus when the reverse strict inequality holds. Since the upper-right entry of $J$ is positive and the lower-left entry is negative, focus trajectories rotate clockwise. Thus the local phase portrait consists of trajectories approaching the fixed point, either directly as a node or while spiralling clockwise as a focus.

Solved by gpt-5.6-sol high.

= b
{parent=14a}
{scope}

= Solution
{parent=b}

If $D=1$, both species have the same diffusion coefficient. For a perturbation proportional to $e^{\sigma t+ikx}$, diffusion replaces $J$ by $J-k^2I$, shifting both reaction eigenvalues to the left by $k^2$. Since the homogeneous equilibrium is already stable, every spatial mode remains stable.

For general $D$, put $q=k^2$. The mode matrix is
$$
A(q)=J-q\begin{pmatrix}D&0\\0&1\end{pmatrix}
=\begin{pmatrix}
\beta-1-Dq&\alpha^2\\
-\beta&-\alpha^2-q
\end{pmatrix}.
$$
Its trace is
$$
\operatorname{tr}A(q)=\beta-1-\alpha^2-(D+1)q<0,
$$
while its determinant is
$$
\Delta(q)=Dq^2+\bigl(D\alpha^2-(\beta-1)\bigr)q+\alpha^2.
$$
With negative trace, instability occurs exactly when $\Delta(q)<0$ for some $q>0$. The <two-species diffusion-driven instability criterion> says that this upward-opening quadratic has a negative minimum precisely when
$$
\beta-1-D\alpha^2>2\alpha\sqrt D.
$$
Equivalently, the condition is
$$
\boxed{\beta>(1+\alpha\sqrt D)^2}.
$$
This is the <Turing instability> region.

Solved by gpt-5.6-sol high.

= c
{parent=14a}
{scope}

= Solution
{parent=c}

At the first loss of stability, the minimum of $\Delta(q)$ just touches zero. The threshold condition from part (b) is
$$
\sqrt\beta=1+\alpha\sqrt{D_c},
$$
so
$$
D_c=\frac{(\sqrt\beta-1)^2}{\alpha^2}.
$$
The minimizing squared wavenumber is
$$
q_*=-\frac{D_c\alpha^2-(\beta-1)}{2D_c}
=\frac{\beta-1-D_c\alpha^2}{2D_c}.
$$
At threshold, $\beta-1-D_c\alpha^2=2\alpha\sqrt{D_c}$, and therefore
$$
k_*^2=q_*=\frac{\alpha}{\sqrt{D_c}}
=\frac{\alpha^2}{\sqrt\beta-1}.
$$
Thus the <Turing threshold of the Brusselator> selects
$$
\boxed{k_*=\frac{\alpha}{\sqrt{\sqrt\beta-1}}}.
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=15b}
{scope}

= Solution
{parent=a}

The rotational kinetic energy and squared angular momentum are the <torque-free rigid-body invariants>
$$
E=\frac12\left(I_1\omega_1^2+I_2\omega_2^2+I_3\omega_3^2\right),
$$
and
$$
L^2=I_1^2\omega_1^2+I_2^2\omega_2^2+I_3^2\omega_3^2.
$$
Differentiating the energy and inserting the Euler equations gives
$$
\begin{aligned}
\dot E
&=I_1\omega_1\dot\omega_1+I_2\omega_2\dot\omega_2+I_3\omega_3\dot\omega_3\\
&=\omega_1\omega_2\omega_3
\bigl[(I_2-I_3)+(I_3-I_1)+(I_1-I_2)\bigr]=0.
\end{aligned}
$$
Similarly,
$$
\begin{aligned}
\frac12\frac{dL^2}{dt}
&=I_1^2\omega_1\dot\omega_1+I_2^2\omega_2\dot\omega_2+I_3^2\omega_3\dot\omega_3\\
&=\omega_1\omega_2\omega_3
\bigl[I_1(I_2-I_3)+I_2(I_3-I_1)+I_3(I_1-I_2)\bigr]=0.
\end{aligned}
$$

Solved by gpt-5.6-sol high.

= b
{parent=15b}
{scope}

= Solution
{parent=b}

Setting $\omega_1=\omega_3=0$ makes all three Euler equations stationary. The invariants then give
$$
E=\frac12I_2\omega_2^2,
\qquad
L^2=I_2^2\omega_2^2=2EI_2,
$$
so the two rotations are
$$
\omega_2=\pm\sqrt{\frac{2E}{I_2}}
=\pm\frac{L}{I_2},
$$
where $L=\sqrt{L^2}$.

Linearize about $(0,\Omega,0)$, where $\Omega$ is either of these values. Writing the transverse perturbations as $(\xi,\zeta)=(\delta\omega_1,\delta\omega_3)$ gives
$$
I_1\dot\xi=(I_2-I_3)\Omega\zeta,
\qquad
I_3\dot\zeta=(I_1-I_2)\Omega\xi.
$$
Consequently
$$
\ddot\xi
=\Omega^2\frac{(I_3-I_2)(I_2-I_1)}{I_1I_3}\,\xi.
$$
The transverse eigenvalues are therefore
$$
s=\pm |\Omega|
\sqrt{\frac{(I_3-I_2)(I_2-I_1)}{I_1I_3}},
$$
one of which is positive. Both rotations are linearly unstable, which is the <intermediate axis theorem>.

Solved by gpt-5.6-sol high.

= c
{parent=15b}
{scope}

= Solution
{parent=c}

Subtracting $I_2$ times the energy identity from the angular-momentum identity, under $L^2=2EI_2$, gives
$$
I_1(I_2-I_1)\omega_1^2
=I_3(I_3-I_2)\omega_3^2.
$$
Combining this with
$$
I_1\omega_1^2+I_3\omega_3^2
=2E-I_2\omega_2^2
=\frac{L^2}{I_2}-I_2\omega_2^2
$$
yields
$$
\omega_1^2=
\frac{I_3-I_2}{I_1(I_3-I_1)}
\left(2E-I_2\omega_2^2\right)
$$
and
$$
\omega_3^2=
\frac{I_2-I_1}{I_3(I_3-I_1)}
\left(2E-I_2\omega_2^2\right).
$$

The second Euler equation now gives, after squaring or choosing one orientation of the orbit,
$$
\dot\omega_2
=\pm\sqrt{\frac{(I_3-I_2)(I_2-I_1)}{I_1I_3}}
\left(\mu^2-\omega_2^2\right),
\qquad
\mu=\sqrt{\frac{2E}{I_2}}.
$$
Indeed, the factor $I_3-I_1$ in $I_2\dot\omega_2=(I_3-I_1)\omega_3\omega_1$ cancels the denominator from the product of the two preceding expressions.

For the plus sign, separation gives
$$
\operatorname{artanh}\frac{\omega_2}{\mu}
=\mu\sqrt{\frac{(I_3-I_2)(I_2-I_1)}{I_1I_3}}\,(t-t_0).
$$
After choosing the time origin $t_0=0$,
$$
\boxed{\omega_2(t)=\mu\tanh(\lambda t)},
\qquad
\boxed{\lambda=\mu
\sqrt{\frac{(I_3-I_2)(I_2-I_1)}{I_1I_3}}}.
$$
For example, a compatible choice of the other components is
$$
\omega_1(t)=
\sqrt{\frac{2E(I_3-I_2)}{I_1(I_3-I_1)}}\operatorname{sech}(\lambda t),
$$
$$
\omega_3(t)=
\sqrt{\frac{2E(I_2-I_1)}{I_3(I_3-I_1)}}\operatorname{sech}(\lambda t).
$$
As $t$ runs from $-\infty$ to $+\infty$, the solution approaches the rotations $\omega_2=-\mu$ and $\omega_2=+\mu$. It is the <intermediate-axis separatrix>, a heteroclinic trajectory that exhibits the instability found in part (b).

Solved by gpt-5.6-sol high.

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

= a
{parent=16h}
{scope}

= Solution
{parent=a}

A <class in set theory> is a collection
$$
M=\{x\in V:V\models\psi(x,a_1,\ldots,a_r)\}
$$
defined by a formula $\psi$, with set parameters if these are allowed by the chosen convention. It is a set-class if there is an $m\in V$ whose elements are exactly the members of $M$. Otherwise it is a <proper class>.

Solved by gpt-5.6-sol high.

= b
{parent=16h}
{scope}

= Solution
{parent=b}

Let $\mathcal F$ be the class of all finite sets. It is proper. If $\mathcal F$ were a set, then for every set $x$ the singleton $\{x\}$ would belong to $\mathcal F$, so
$$
x\in\bigcup\mathcal F.
$$
Thus $\bigcup\mathcal F$ would be a universal set. Separation applied to the property $z\notin z$ would then give the usual Russell contradiction.

The class $\mathcal F$ is not transitive. For example, $\{\omega\}$ is finite, but its member $\omega$ is infinite. It is nevertheless $\varphi$-closed for every formula $\varphi$: if $x$ is finite, then every subset
$$
\{z\in x:\varphi(z)\}
$$
is finite.

Solved by gpt-5.6-sol high.

= c
{parent=16h}
{scope}

= Solution
{parent=c}

The collection of <hereditarily finite sets> is
$$
V_\omega=\bigcup_{n<\omega}V_n,
$$
where $V_0=\varnothing$ and $V_{n+1}=\mathcal P(V_n)$. Every $V_n$ is finite, and the axiom of infinity together with replacement and union makes $V_\omega$ a set. Hence this collection is not a proper class.

It is transitive: if $x\in V_n$, then $x\subset V_{n-1}$, so every $y\in x$ belongs to $V_{n-1}\subset V_\omega$. It is also $\varphi$-closed. If $x\in V_n$ and
$$
s=\{z\in x:\varphi(z)\},
$$
then $s\subset x\subset V_{n-1}$, so $s\in V_n\subset V_\omega$.

Solved by gpt-5.6-sol high.

= d
{parent=16h}
{scope}

= Solution
{parent=d}

Consider the structure $(M,\in)$.

For extensionality, let $x,y\in M$ have the same members in $M$. Since $M$ is transitive, every actual member of $x$ or $y$ lies in $M$. Thus $x$ and $y$ have the same members in $V$, and ambient extensionality gives $x=y$.

By hypothesis $\varnothing\in M$, and it has no members in the induced structure, so the empty-set axiom holds. If $x,y\in M$, then $\{x,y\}\in M$ by closure, and its internal members are exactly $x$ and $y$, proving pairing.

Finally, if $x\in M$, then $\bigcup x\in M$. For $z\in M$,
$$
z\in\bigcup x
\quad\Longleftrightarrow\quad
\exists y\in x\ (z\in y).
$$
Every such $y$ lies in $M$ by transitivity, so the same equivalence holds internally. Hence union is satisfied. This is the <basic set-theoretic axioms inherited by a transitive class> argument.

Solved by gpt-5.6-sol high.

= e
{parent=16h}
{scope}

= Solution
{parent=e}

The definition of $\varphi$-closure forms
$$
s_V=\{z\in x:V\models\varphi(z)\},
$$
whereas the <axiom schema of specification> inside $(M,\in)$ requires
$$
s_M=\{z\in x:M\models\varphi(z)\}.
$$
An arbitrary formula need not be absolute between a transitive class and the ambient universe, so $s_V$ and $s_M$ can differ. Closure under $s_V$ therefore supplies no reason for the required set $s_M$ to belong to $M$.

Solved by gpt-5.6-sol high.

= f
{parent=16h}
{scope}

= Solution
{parent=f}

Take
$$
\varphi^*=\varphi^M,
$$
the <formula relativization to a class>: replace each $\exists y\,\psi$ by
$$
\exists y\,(y\in M\land\psi^M)
$$
and each $\forall y\,\psi$ by
$$
\forall y\,(y\in M\mathbin\Rightarrow\psi^M),
$$
leaving atomic formulas unchanged. If $M$ is given by a defining formula, insert that formula wherever $y\in M$ occurs; this is why the map may depend on $M$.

For $z$ and any parameters in $M$, induction on formulas gives
$$
V\models\varphi^M(z)
\quad\Longleftrightarrow\quad
M\models\varphi(z).
$$
Therefore $\varphi^*$-closure puts
$$
\{z\in x:V\models\varphi^M(z)\}
=\{z\in x:M\models\varphi(z)\}
$$
in $M$ for every $x\in M$. This is exactly the <relativized closure criterion for separation>, so $(M,\in)$ satisfies the $\varphi$-instance of separation.

Solved by gpt-5.6-sol high.

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

= a
{parent=17f}
{scope}

= Solution
{parent=a}

The <diagonal Ramsey number> $R(t)$ is the least positive integer $n$ such that every red-blue colouring of the edges of $K_n$ contains a monochromatic $K_t$.

More generally, let $R(s,t)$ be the least $n$ forcing either a red $K_s$ or a blue $K_t$. We have
$$
R(1,t)=R(s,1)=1.
$$
Assuming the two smaller numbers exist, colour
$$
K_{R(s-1,t)+R(s,t-1)}
$$
and choose a vertex $v$. At least $R(s-1,t)$ of its incident edges are red, or at least $R(s,t-1)$ are blue. In the first case, the corresponding neighbourhood contains a red $K_{s-1}$, which extends with $v$ to a red $K_s$, or a blue $K_t$. The second case is symmetric. Thus
$$
R(s,t)\leq R(s-1,t)+R(s,t-1),
$$
which proves existence by induction.

Pascal's identity then gives the <binomial upper bound for a Ramsey number>
$$
R(s,t)\leq\binom{s+t-2}{s-1}.
$$
Consequently, for $t\geq2$,
$$
R(t)=R(t,t)
\leq\binom{2t-2}{t-1}
\leq2^{2t-2}<2^{2t}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=17f}
{scope}

= Solution
{parent=b}

The <graph Ramsey number> $R(G)$ is the least positive integer $n$ for which every red-blue colouring of $K_n$ contains a monochromatic copy of $G$. If $g=|V(G)|$, the Ramsey theorem gives a monochromatic $K_g$ in every colouring of $K_{R(g)}$, and that clique contains $G$. Hence
$$
R(G)\leq R(g)<\infty.
$$

Solved by gpt-5.6-sol high.

= i
{parent=b}
{scope}

= Solution
{parent=i}

Choose any vertex $v$ of $K_{2t}$. Its $2t-1$ incident edges have two colours, so at least
$$
\left\lceil\frac{2t-1}{2}\right\rceil=t
$$
have the same colour. Those edges, together with $v$, form a monochromatic $K_{1,t}$.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

Part (i) gives $R(K_{1,t})\leq2t$ for every $t$.

Suppose first that $t$ is odd. On $2t-1$ vertices, take a red $(t-1)$-regular graph; for example, label the vertices cyclically and join each vertex to the $(t-1)/2$ nearest vertices in each direction. Its blue complement is also $(t-1)$-regular. Neither colour contains a vertex of degree $t$, so there is no monochromatic $K_{1,t}$. Therefore
$$
R(K_{1,t})=2t
\qquad(t\text{ odd}).
$$

Now let $t$ be even. If a colouring of $K_{2t-1}$ had no monochromatic $K_{1,t}$, every vertex would have red and blue degree at most $t-1$. Since the two degrees sum to $2t-2$, each would equal $t-1$. The red graph would then be $(t-1)$-regular on $2t-1$ vertices, impossible because both numbers are odd and the sum of degrees must be even. Hence $R(K_{1,t})\leq2t-1$.

For the matching lower bound, colour a copy of $K_{t-1,t-1}$ red on $2t-2$ vertices and colour all remaining edges blue. Red degree is $t-1$ and the blue graph is two disjoint copies of $K_{t-1}$, of degree $t-2$. Again neither colour contains $K_{1,t}$. Thus the <Ramsey number of a star> is
$$
\boxed{R(K_{1,t})=
\begin{cases}
2t,&t\text{ odd},\\
2t-1,&t\text{ even}.
\end{cases}}
$$

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

Here $H$ is the <paw graph>, a triangle with a pendant edge.

For the lower bound, partition the vertices of $K_6$ into two triples. Colour the edges inside each triple red and all edges between the triples blue. Each red component is only a triangle, while the blue graph is the triangle-free graph $K_{3,3}$, so there is no monochromatic copy of $H$. Hence $R(H)>6$.

For the upper bound, every colouring of $K_7$ contains a monochromatic triangle because $R(3)=6$. Suppose that a triangle $T$ is red, and let $S$ be the other four vertices. If any edge from $T$ to $S$ were red, it would be a pendant edge extending $T$ to a red $H$. Thus all edges between $T$ and $S$ are blue.

If an edge $xy$ inside $S$ were blue, then for any $u\in T$ the vertices $u,x,y$ would form a blue triangle, and an edge from a second vertex of $T$ to $x$ would extend it to a blue $H$. Therefore every edge inside $S$ is red. The resulting red $K_4$ contains a red triangle and an additional incident edge, hence a red $H$. The blue-triangle case is symmetric, so the <Ramsey number of the paw graph> is
$$
\boxed{R(H)=7}.
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=18j}
{scope}

= Solution
{parent=a}

Fix a primitive $n$th root of unity $\zeta_n$. The $n$th <cyclotomic polynomial> is
$$
\Phi_n(X)=
\prod_{\substack{1\leq a\leq n\\(a,n)=1}}
(X-\zeta_n^a).
$$
Its roots are exactly the roots of unity of order $n$.

Partitioning all $n$th roots of unity by exact order gives the <cyclotomic factorization>
$$
X^n-1=\prod_{d\mid n}\Phi_d(X).
$$
We prove $\Phi_n\in\mathbb Z[X]$ by induction on $n$. The assertion is clear for $n=1$. If it holds for every proper divisor of $n$, then
$$
\Phi_n(X)=\frac{X^n-1}{\prod_{d\mid n,\ d<n}\Phi_d(X)}.
$$
The denominator is monic and belongs to $\mathbb Z[X]$. Exact polynomial division shows that the quotient lies in $\mathbb Q[X]$, and Gauss's lemma makes this monic quotient an element of $\mathbb Z[X]$.

Now let $K\subseteq\mathbb C$. The minimal polynomial of $\zeta_n$ over $K$ divides $\Phi_n$, so every $K$-conjugate of $\zeta_n$ is another primitive root $\zeta_n^a$ and lies in $K(\zeta_n)$. The extension is therefore normal. It is separable because the characteristic is zero, and hence it is Galois.

Every $\sigma\in\operatorname{Gal}(K(\zeta_n)/K)$ has
$$
\sigma(\zeta_n)=\zeta_n^a
$$
for some $a\in(\mathbb Z/n\mathbb Z)^\times$. Since $\zeta_n$ generates the extension, this gives an injective homomorphism
$$
\operatorname{Gal}(K(\zeta_n)/K)
\hookrightarrow(\mathbb Z/n\mathbb Z)^\times.
$$
The group on the right is abelian, so the Galois group is abelian, as summarized by the <cyclotomic field> construction.

Solved by gpt-5.6-sol high.

= b
{parent=18j}
{scope}

= Solution
{parent=b}

A real subfield is constructible when all its elements are straightedge-and-compass constructible; algebraically, it is contained in a tower of quadratic extensions of $\mathbb Q$.

Let $\zeta=e^{2\pi i/17}$ and
$$
F=\mathbb Q(\zeta+\zeta^{-1})
=\mathbb Q\left(2\cos\frac{2\pi}{17}\right).
$$
The cyclotomic extension $\mathbb Q(\zeta)/\mathbb Q$ has Galois group
$$
(\mathbb Z/17\mathbb Z)^\times,
$$
which is cyclic of order $16$. Complex conjugation corresponds to $-1$, so the maximal real subfield has
$$
\operatorname{Gal}(F/\mathbb Q)
\cong(\mathbb Z/17\mathbb Z)^\times/\{\mathord\pm1\},
$$
a cyclic group of order $8$.

This group has a chain
$$
G=G_0\supset G_1\supset G_2\supset G_3=\{1\}
$$
in which every index is two. By the Galois correspondence, the fixed fields form
$$
\mathbb Q=F^{G_0}\subset F^{G_1}\subset F^{G_2}\subset F,
$$
with every successive degree equal to two. Thus $F$, and in particular $\cos(2\pi/17)$, is constructible. This is the <constructibility of the real seventeenth cyclotomic field>.

Solved by gpt-5.6-sol high.

= c
{parent=18j}
{scope}

= Solution
{parent=c}

For $K=\mathbb R$, the answer is yes. A root of $\Phi_4(X)=X^2+1$ is $i$ or $-i$, so
$$
\mathbb R(\zeta_4)=\mathbb C=\overline{\mathbb R}.
$$

For $K=\mathbb Q$, the answer is no. Every $\mathbb Q(\zeta_n)/\mathbb Q$ is finite abelian Galois. If $\sqrt[3]{2}$ belonged to $\mathbb Q^{\mathrm{cyc}}$, then it would belong to some $\mathbb Q(\zeta_n)$. The <subextensions of an abelian Galois extension> result would make
$$
\mathbb Q(\sqrt[3]{2})/\mathbb Q
$$
Galois. This is false: $X^3-2$ has two nonreal roots absent from the real field $\mathbb Q(\sqrt[3]{2})$. Hence $\mathbb Q^{\mathrm{cyc}}$ is a proper subfield of $\overline{\mathbb Q}$.

For $K=\mathbb F_p$, the answer is yes. Given $m\geq1$, take
$$
n=p^m-1,
$$
which is coprime to $p$. Every root of $\Phi_n$ in characteristic $p$ has exact multiplicative order $n$. The degree
$$
[\mathbb F_p(\zeta_n):\mathbb F_p]
$$
is the least positive $d$ such that $n\mid p^d-1$. The value $d=m$ works, and no $d<m$ can work because
$$
0<p^d-1<p^m-1=n.
$$
Therefore
$$
\mathbb F_p(\zeta_{p^m-1})=\mathbb F_{p^m}.
$$
Every element algebraic over $\mathbb F_p$ lies in some finite field $\mathbb F_{p^m}$, so
$$
\mathbb F_p^{\mathrm{cyc}}=\overline{\mathbb F}_p.
$$
These are respectively the <maximal cyclotomic extension of the real numbers>, the <maximal cyclotomic extension of the rational numbers>, and the <maximal cyclotomic extension of a finite field>.

Solved by gpt-5.6-sol high.

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

= a
{parent=19j}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

The <Mackey restriction formula> says that for subgroups $H,K\leq G$ and a $K$-representation $W$,
$$
\operatorname{Res}_H^G\operatorname{Ind}_K^GW
\cong
\bigoplus_{x\in H\backslash G/K}
\operatorname{Ind}_{H\cap xKx^{-1}}^H
\operatorname{Res}_{H\cap xKx^{-1}}^{xKx^{-1}}({}^xW),
$$
where ${}^xW(xkx^{-1})=W(k)$.

For $H=K=B$, the <Bruhat decomposition of SL2 over a finite field> gives
$$
G=B\sqcup BwB,
\qquad
w=\begin{pmatrix}0&-1\\1&0\end{pmatrix}.
$$
Thus $B\backslash G/B$ has representatives $1,w$. The identity double coset contributes $\theta$, while
$$
B\cap wBw^{-1}=T
$$
contributes an induced representation from the diagonal subgroup. Therefore
$$
\boxed{
\operatorname{Res}_B^G\operatorname{Ind}_B^G\theta
\cong
\theta\oplus\operatorname{Ind}_T^B(\theta^w|_T),
}
$$
where
$$
\theta^w(t)=\theta(w^{-1}tw).
$$
For $t(a)=\operatorname{diag}(a,a^{-1})$, one has $w^{-1}t(a)w=t(a^{-1})$.

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

Apply <Frobenius reciprocity> and then the decomposition from part (i):
$$
\begin{aligned}
\left\langle\operatorname{Ind}_B^G\theta,
\operatorname{Ind}_B^G\varphi\right\rangle_G
&=\left\langle
\operatorname{Res}_B^G\operatorname{Ind}_B^G\theta,
\varphi\right\rangle_B\\
&=\langle\theta,\varphi\rangle_B
+\left\langle\operatorname{Ind}_T^B(\theta^w|_T),\varphi\right\rangle_B\\
&=\langle\theta,\varphi\rangle_B
+\langle\theta^w|_T,\varphi|_T\rangle_T.
\end{aligned}
$$
Since all the representations in the last line are one-dimensional,
$$
\boxed{
\left\langle\operatorname{Ind}_B^G\theta,
\operatorname{Ind}_B^G\varphi\right\rangle_G
=\mathbf1_{\theta=\varphi}
+\mathbf1_{\theta^w|_T=\varphi|_T}.
}
$$
It is therefore $2$ when both equalities hold, $1$ when exactly one holds, and $0$ when neither holds. This is the <mackey inner product for the finite SL2 principal series>.

Solved by gpt-5.6-sol high.

= b
{parent=19j}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

Write
$$
u(x)=\begin{pmatrix}1&x\\0&1\end{pmatrix}.
$$
Since $u(x)u(y)=u(x+y)$,
$$
\begin{aligned}
\chi(v\mathbin\cdot)(u(x+y))
&=\chi(u(vx+vy))\\
&=\chi(u(vx))\chi(u(vy))\\
&=\chi(v\mathbin\cdot)(u(x))\chi(v\mathbin\cdot)(u(y)).
\end{aligned}
$$
Thus $\chi(v\mathbin\cdot)$ is a one-dimensional representation of $U$; for $v=0$ it is the trivial representation.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

For $a\in\mathbb F_p^*$ put
$$
t(a)=\begin{pmatrix}a&0\\0&a^{-1}\end{pmatrix}.
$$
Conjugation gives
$$
t(a)u(x)t(a)^{-1}=u(a^2x).
$$
Suppose the $\chi$-weight space
$$
V_\chi=\{z\in V:u(x)z=\chi(u(x))z\text{ for all }x\}
$$
is nonzero. If $0\ne z\in V_\chi$, then $t(a)z$ lies in the $\chi(a^{-2}\mathbin\cdot)$-weight space, because
$$
u(x)t(a)z
=t(a)u(a^{-2}x)z
=\chi(u(a^{-2}x))t(a)z.
$$
Hence every character indexed by a nonzero square occurs in $\operatorname{Res}_U^BV$.

These characters are distinct. Indeed, if $\chi(v\mathbin\cdot)=\chi(v'\mathbin\cdot)$ with $v\ne v'$, multiplication by $v-v'$ would show that $\chi$ is trivial on all of $U$, contrary to hypothesis. There are $(p-1)/2$ nonzero squares, so
$$
\left\langle\operatorname{Res}_U^BV,\chi(v\mathbin\cdot)\right\rangle_U\ne0
$$
for at least $(p-1)/2$ values of $v\in\mathbb F_p^*$. This is the <square orbits of additive characters of a finite field> argument.

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

Put
$$
W=\operatorname{Ind}_T^B\theta.
$$
Since $B=U\rtimes T$ and $U\cap T=1$, restriction to $U$ gives its regular representation:
$$
\operatorname{Res}_U^BW\cong\mathbb C[U].
$$
As $U\cong(\mathbb F_p,+)$ is abelian, this is the direct sum of all $p$ one-dimensional characters of $U$, each with multiplicity one.

The diagonal subgroup $T$ permutes these one-dimensional weight spaces. It fixes the trivial character, while part (ii) shows that the nontrivial characters split into two orbits, indexed by the nonzero squares and nonsquares. Let $W_0,W_+,W_-$ be the sums of the weight spaces in these three orbits. Then
$$
W=W_0\oplus W_+\oplus W_-
$$
as $B$-representations, and
$$
\dim W_0=1,
\qquad
\dim W_+=\dim W_-=\frac{p-1}{2}.
$$

Each summand is irreducible. Any nonzero $B$-subrepresentation decomposes into $U$-weight spaces. If it contains one of the one-dimensional weight spaces in an orbit, transitivity of the $T$-action forces it to contain every weight space in that orbit, hence the whole corresponding $W_0$, $W_+$, or $W_-$. The three summands are pairwise nonisomorphic because their restrictions to $U$ have disjoint character supports. Therefore the <induction from the diagonal subgroup of upper-triangular SL2> decomposition has the required three irreducible constituents of dimensions
$$
\boxed{1,\quad\frac{p-1}{2},\quad\frac{p-1}{2}}.
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=20g}
{scope}

= Solution
{parent=a}

Let $K$ have $r$ real embeddings and $s$ conjugate pairs of complex embeddings. The <Dirichlet unit theorem> states that
$$
\mathcal O_K^\times\cong\mu(K)\times\mathbb Z^{r+s-1},
$$
where $\mu(K)$ is the finite cyclic group of roots of unity in $K$. Thus the unit rank is
$$
m=r+s-1.
$$
Equivalently, there are units $u_1,\ldots,u_m$ such that every unit has a unique expression
$$
u=\zeta u_1^{n_1}\cdots u_m^{n_m},
\qquad
\zeta\in\mu(K),\quad n_i\in\mathbb Z.
$$

Solved by gpt-5.6-sol high.

= b
{parent=20g}
{scope}

= Solution
{parent=b}

Choose the real embeddings $\sigma_1,\ldots,\sigma_r$ and one embedding $\tau_j$ from each pair of complex-conjugate embeddings. The <logarithmic embedding of number field units> is
$$
\operatorname{Log}:\mathcal O_K^\times\longrightarrow\mathbb R^{r+s},
$$
$$
\operatorname{Log}(u)=
\bigl(log|\sigma_1(u)|,\ldots,\log|\sigma_r(u)|,
2\log|\tau_1(u)|,\ldots,2\log|\tau_s(u)|\bigr).
$$
It is a group homomorphism from multiplication to addition.

If $u$ lies in its kernel, every conjugate of $u$ has modulus one. Under the Minkowski embedding, every power $u^n$ therefore lies in one fixed bounded subset. Each $u^n$ belongs to $\mathcal O_K$, whose Minkowski image is discrete, so that bounded subset contains only finitely many such lattice points. Hence $u^a=u^b$ for some $0\leq a<b$, and
$$
u^{b-a}=1.
$$
Thus $u$ is a root of unity. Conversely, every root of unity clearly has all conjugates of modulus one and lies in the kernel. This proves the <kernel of the logarithmic unit embedding> is precisely $\mu(K)$.

Solved by gpt-5.6-sol high.

= c
{parent=20g}
{scope}

= Solution
{parent=c}

The product formula places every unit logarithm in
$$
H=\left\{x\in\mathbb R^{r+s}:\sum_{j=1}^{r+s}x_j=0\right\}.
$$
For a fundamental system $u_1,\ldots,u_m$, where $m=r+s-1$, the vectors
$$
\operatorname{Log}(u_1),\ldots,\operatorname{Log}(u_m)
$$
are linearly independent, form a real basis of $H$, and generate the full lattice
$$
\operatorname{Log}(\mathcal O_K^\times)
=\bigoplus_{i=1}^m\mathbb Z\operatorname{Log}(u_i).
$$
This is the <Dirichlet unit lattice>.

Solved by gpt-5.6-sol high.

= d
{parent=20g}
{scope}

= Solution
{parent=d}

Let
$$
e=(1,\ldots,1,2,\ldots,2)\in\mathbb R^{r+s},
$$
with $r$ entries equal to one and $s$ entries equal to two. Its coordinate sum is
$$
r+2s=[K:\mathbb Q]\ne0,
$$
so $e\notin H$. Since the fundamental-unit logarithms form a basis of the codimension-one hyperplane $H$, the vectors
$$
e,\operatorname{Log}(u_1),\ldots,\operatorname{Log}(u_m)
$$
form a basis of $\mathbb R^{r+s}$. This gives unique real numbers $t,\lambda_i$ such that
$$
\operatorname{Log}(\beta)
=te+\sum_{i=1}^m\lambda_i\operatorname{Log}(u_i).
$$

The sum of the coordinates on the left is
$$
\log|N_{K/\mathbb Q}(\beta)|=\log N((\beta))=\log N(I).
$$
Every unit logarithm has coordinate sum zero, so summing the right side gives $t(r+2s)$. Therefore
$$
\boxed{t=\frac{\log N(I)}{[K:\mathbb Q]}}.
$$

Solved by gpt-5.6-sol high.

= e
{parent=20g}
{scope}

= Solution
{parent=e}

Start with any generator $I=(\beta)$ and use the decomposition from part (d). For each $i$, choose an integer $n_i$ such that
$$
-\frac12\leq\lambda_i+n_i\leq\frac12,
$$
and put
$$
u=u_1^{n_1}\cdots u_m^{n_m},
\qquad
\alpha=\beta u.
$$
Since $u$ is a unit, $(\alpha)=(\beta)=I$. Moreover,
$$
\operatorname{Log}(\alpha)
=te+h,
\qquad
h=\sum_{i=1}^m(\lambda_i+n_i)\operatorname{Log}(u_i).
$$
The vector $h$ lies in the fixed compact parallelepiped
$$
P=\left\{\sum_i c_i\operatorname{Log}(u_i):|c_i|\leq\frac12\right\}.
$$
Hence there is a constant $M_K$ such that every real coordinate of every $h\in P$ is at most $M_K$, and every complex coordinate is at most $2M_K$.

For a real embedding $\sigma_i$,
$$
\log|\sigma_i(\alpha)|=t+h_i
\leq\frac{\log N(I)}{[K:\mathbb Q]}+M_K.
$$
For a chosen complex embedding $\tau_j$, the weighted coordinate gives
$$
2\log|\tau_j(\alpha)|=2t+h_{r+j}
\leq2\frac{\log N(I)}{[K:\mathbb Q]}+2M_K.
$$
The conjugate embedding has the same modulus. Exponentiating and taking any $C>e^{M_K}$ proves the <archimedean balancing of a principal ideal generator>:
$$
\boxed{
|\sigma(\alpha)|<C N(I)^{1/[K:\mathbb Q]}
}
$$
for every embedding $\sigma:K\to\mathbb C$, with $C$ depending only on $K$.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=21f}

The <universal coefficient theorem for homology> and flatness of $\mathbb Q$ give
$$
H_i(X;\mathbb Q)
\cong H_i(X;\mathbb Z)\otimes_\mathbb Z\mathbb Q.
$$
Thus, if
$$
H_i(X;\mathbb Z)\cong\mathbb Z^{b_i}\oplus T_i
$$
with $T_i$ torsion, then $H_i(X;\mathbb Q)\cong\mathbb Q^{b_i}$.

For a finite triangulable space, its <Euler characteristic> is
$$
\chi(X)=\sum_i(-1)^i\dim_\mathbb QH_i(X;\mathbb Q).
$$
If $K$ is a finite triangulation and $f_i$ is its number of $i$-simplices, then the <Euler-Poincare formula> is
$$
\chi(X)=\sum_i(-1)^if_i.
$$
To prove it over $\mathbb Q$, write $Z_i=\ker\partial_i$ and $B_i=\operatorname{im}\partial_{i+1}$. Rank-nullity and $H_i=Z_i/B_i$ give
$$
\dim C_i
=\dim Z_i+\dim B_{i-1}
=\dim H_i+\dim B_i+\dim B_{i-1}.
$$
After multiplying by $(-1)^i$ and summing, the two boundary sums cancel, leaving the claimed equality.

The standard simplex $\Delta^3$ is contractible, so
$$
H_i(\Delta^3)\cong
\begin{cases}
\mathbb Z,&i=0,\\
0,&i>0.
\end{cases}
$$

Solved by gpt-5.6-sol high.

= i
{parent=21f}
{scope}

= Solution
{parent=i}

The simplices in a <barycentric subdivision> are strict chains of nonempty faces. A tetrahedron has $15$ nonempty faces, so $f_0=15$. Counting strict two-face chains gives
$$
f_1
=\sum_{j=1}^4\binom4j(2^j-2)
=50.
$$
A strict three-face chain amounts to choosing a terminal face and partitioning its vertices into three nonempty ordered blocks. Hence
$$
f_2
=\binom43\,3!\,S(3,3)+\binom44\,3!\,S(4,3)
=24+36=60,
$$
where $S(-,-)$ denotes a Stirling number of the second kind. Therefore
$$
\boxed{\chi(|M|)=f_0-f_1+f_2=15-50+60=25}.
$$
Equivalently, the full subdivision has $4!=24$ tetrahedra and Euler characteristic one, so deleting its three-dimensional simplices from the alternating count gives $1+24=25$. This is the f-vector computation for the <barycentric subdivision of a tetrahedron>.

Solved by gpt-5.6-sol high.

= ii
{parent=21f}
{scope}

= Solution
{parent=ii}

The complex $M$ is connected, so $H_0(M)\cong\mathbb Z$. Adding the 24 three-simplices to $M$ produces the contractible complex $(\Delta^3)'$. Adding three-simplices does not alter $H_1$, hence
$$
H_1(M)=H_1((\Delta^3)')=0.
$$
Also $H_i(M)=0$ for $i>2$ because $M$ is two-dimensional.

The group $H_2(M)=\ker(\partial_2)$ is a subgroup of the free group $C_2(M)$ and is therefore free abelian. If its rank is $b_2$, the Euler characteristic from part (i) gives
$$
25=1-0+b_2,
$$
so $b_2=24$. Consequently
$$
\boxed{
H_i(M)\cong
\begin{cases}
\mathbb Z,&i=0,\\
\mathbb Z^{24},&i=2,\\
0,&\text{otherwise}.
\end{cases}}
$$

Solved by gpt-5.6-sol high.

= iii
{parent=21f}
{scope}

= Solution
{parent=iii}

Yes. Let $b$ be the vertex of $(\Delta^3)'$ corresponding to the whole tetrahedron. Its link in $M$ is the 1-skeleton of the barycentric subdivision of $\partial\Delta^3$. This connected graph has one vertex for each of the 14 nonempty proper faces and 36 edges, so
$$
H_2(|M|,|M|\setminus\{b\})
\cong\widetilde H_1(\operatorname{link}(b))
\cong\mathbb Z^{36-14+1}
=\mathbb Z^{23}.
$$

No other point has this local homology rank. At the other vertices, corresponding respectively to faces of size $1,2,3$, the first Betti numbers of the links are $6,4,6$. A point in the interior of an edge is incident with at most six triangles, so its local $H_2$ has rank at most five; a point in a triangle interior has local $H_2\cong\mathbb Z$.

A homeomorphism preserves <local homology>, so it must send the unique point with local $H_2$ of rank $23$ to itself. Therefore every self-homeomorphism fixes $b$, which is the <fixed barycentre of the barycentric tetrahedral two-skeleton>.

Solved by gpt-5.6-sol high.

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

= a
{parent=22i}
{scope}

= Solution
{parent=a}

The <Riesz representation theorem> says that for every bounded linear functional $f$ on a Hilbert space $H$, there is a unique $z\in H$ such that
$$
f(x)=\langle x,z\rangle\qquad(x\in H),
$$
and $\lVert f\rVert=\lVert z\rVert$.

For $T\in L(H)$ and fixed $y\in H$, the map
$$
f_y(x)=\langle Tx,y\rangle
$$
is a bounded linear functional, with
$$
|f_y(x)|\leq\lVert T\rVert\lVert x\rVert\lVert y\rVert.
$$
Riesz therefore gives a unique vector, denoted $T^*y$, such that
$$
\langle Tx,y\rangle=\langle x,T^*y\rangle
$$
for every $x$. Uniqueness in the representation theorem shows that $y\mapsto T^*y$ is linear. Moreover,
$$
\lVert T^*y\rVert=\lVert f_y\rVert
\leq\lVert T\rVert\lVert y\rVert,
$$
so the <adjoint operator> belongs to $L(H)$ and $\lVert T^*\rVert\leq\lVert T\rVert$.

If $(e_j)$ is an orthonormal basis and $A=(a_{ij})$ is the matrix of $T$, then the matrix of $T^*$ is
$$
A^*=\overline A^{,T};
$$
its entries are $\overline{a_{ji}}$.

Solved by gpt-5.6-sol high.

= b
{parent=22i}
{scope}

= Solution
{parent=b}

The <spectral theorem for compact Hermitian operators> says that the nonzero spectrum of a compact Hermitian $T$ consists of real eigenvalues of finite multiplicity, with zero as the only possible accumulation point. Eigenvectors for distinct eigenvalues are orthogonal, and an orthonormal basis of $\ell^2$ can be chosen from eigenvectors together with a basis of $\ker T$. Thus
$$
Tx=\sum_{n\geq1}\lambda_n\langle x,e_n\rangle e_n,
$$
where the nonzero eigenvalues are repeated according to multiplicity and $\lambda_n\to0$ if there are infinitely many.

Define
$$
T_Nx=\sum_{n=1}^N\lambda_n\langle x,e_n\rangle e_n.
$$
Each $T_N$ is finite-rank and Hermitian, and
$$
\lVert T-T_N\rVert
=\sup_{n>N}|\lambda_n|\longrightarrow0.
$$
Hence every compact Hermitian operator is a norm limit of finite-rank Hermitian operators, by <finite-rank truncation of a compact Hermitian operator>.

Solved by gpt-5.6-sol high.

= c
{parent=22i}
{scope}

= Solution
{parent=c}

Let $P_N$ be the orthogonal projection of $\ell^2$ onto the span of its first $N$ standard basis vectors. Then $P_Ny\to y$ for every $y\in\ell^2$, and $\lVert I-P_N\rVert\leq1$.

Let $T$ be compact and let
$$
K=\overline{T(B)},
$$
where $B$ is the closed unit ball. The set $K$ is compact. Given $\varepsilon>0$, choose a finite $\varepsilon/3$-net $y_1,\ldots,y_m$ in $K$. Pointwise convergence lets us choose $N$ such that
$$
\lVert(I-P_N)y_j\rVert<\frac{\varepsilon}{3}
$$
for every $j$. If $y\in K$ and $\lVert y-y_j\rVert<\varepsilon/3$, then
$$
\lVert(I-P_N)y\rVert
\leq\lVert(I-P_N)(y-y_j)\rVert
+\lVert(I-P_N)y_j\rVert
<\frac{2\varepsilon}{3}.
$$
Thus $P_N\to I$ uniformly on $K$, and
$$
\lVert P_NT-T\rVert
=\sup_{\lVert x\rVert\leq1}\lVert(P_N-I)Tx\rVert
\longrightarrow0.
$$
Each $P_NT$ has image in an $N$-dimensional space, so it has finite rank. This <coordinate-projection approximation of a compact operator> proves the result.

Solved by gpt-5.6-sol high.

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

= a
{parent=23h}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

If $x=0$, the claim is immediate. Otherwise weak convergence gives
$$
\lVert x\rVert^2
=\langle x,x\rangle
=\lim_{n\to\infty}\langle x_n,x\rangle.
$$
By Cauchy-Schwarz,
$$
\langle x_n,x\rangle
\leq\lVert x_n\rVert\lVert x\rVert.
$$
Taking the lower limit and dividing by $\lVert x\rVert$ proves the <weak lower semicontinuity of the Hilbert norm>:
$$
\boxed{\lVert x\rVert\leq\liminf_n\lVert x_n\rVert}.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

Strong convergence immediately implies convergence of norms by the reverse triangle inequality.

Conversely, suppose $x_n\rightharpoonup x$ and $\lVert x_n\rVert\to\lVert x\rVert$. Then
$$
\begin{aligned}
\lVert x_n-x\rVert^2
&=\lVert x_n\rVert^2+\lVert x\rVert^2-2\langle x_n,x\rangle\\
&\longrightarrow
\lVert x\rVert^2+\lVert x\rVert^2-2\lVert x\rVert^2=0.
\end{aligned}
$$
Thus $x_n\to x$ strongly. This is the <Radon-Riesz theorem> for Hilbert spaces.

Solved by gpt-5.6-sol high.

= iii
{parent=a}
{scope}

= Solution
{parent=iii}

No. In $H=\ell^2$, let $x_n=e_n$ be the standard orthonormal sequence. For any $y=(y_j)\in\ell^2$,
$$
\langle e_n,y\rangle=y_n\longrightarrow0,
$$
so $e_n\rightharpoonup0$. However,
$$
\lVert e_n-e_m\rVert=\sqrt2
$$
for $n\ne m$. No subsequence is Cauchy, hence none converges strongly. This is a <weakly null orthonormal sequence>.

Solved by gpt-5.6-sol high.

= b
{parent=23h}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

Weak convergence $u_n\rightharpoonup u$ in $H_0^1(\Omega)$ implies
$$
Du_n\rightharpoonup Du
$$
in $L^2$. The $L^2$ norm is weakly lower semicontinuous, so
$$
\int_\Omega|Du|^2
\leq\liminf_n\int_\Omega|Du_n|^2.
$$

The <Rellich-Kondrashov compactness theorem for H01> upgrades the weak convergence to
$$
u_n\longrightarrow u
$$
strongly in $L^2(\Omega)$. Indeed, compactness gives this along every subsequence after passage to a further subsequence, and the weak limit uniquely identifies every such strong limit as $u$. Consequently,
$$
\begin{aligned}
\left|\int_\Omega V(u_n^2-u^2)\right|
&\leq\lVert V\rVert_\infty
\lVert u_n-u\rVert_2
\bigl(\lVert u_n\rVert_2+\lVert u\rVert_2\bigr)\\
&\longrightarrow0.
\end{aligned}
$$
Combining the two terms proves the <weak lower semicontinuity of a bounded-domain Schrodinger energy>:
$$
\boxed{E(u)\leq\liminf_nE(u_n)}.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

Choose a minimizing sequence $(u_n)$ with
$$
\lVert u_n\rVert_2=1,
\qquad
E(u_n)\longrightarrow\lambda.
$$
Since
$$
E(u_n)\geq\lVert Du_n\rVert_2^2-\lVert V\rVert_\infty,
$$
the sequence is bounded in $H_0^1(\Omega)$. Passing to a subsequence, weak compactness gives
$$
u_n\rightharpoonup w
$$
in $H_0^1(\Omega)$. Rellich compactness gives $u_n\to w$ strongly in $L^2$, and hence
$$
\lVert w\rVert_2=1.
$$
Part (i) now yields
$$
E(w)\leq\liminf_nE(u_n)=\lambda.
$$
The definition of $\lambda$ gives the reverse inequality because $w$ is admissible. Therefore
$$
\boxed{E(w)=\lambda,\qquad\lVert w\rVert_2=1}.
$$
This is the <constrained ground-state minimizer on a bounded domain> obtained by the direct method.

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

Choose $\phi\in C_c^\infty(\mathbb R)$ with $\lVert\phi\rVert_2=1$, and for $R>0$ set
$$
u_R(x)=R^{-1/2}\phi(x/R).
$$
A change of variables gives
$$
\lVert u_R\rVert_2=1,
\qquad
\int_\mathbb R|u_R'(x)|^2\,dx
=R^{-2}\int_\mathbb R|\phi'(x)|^2\,dx.
$$
The latter tends to zero as $R\to\infty$. Since the energy is nonnegative,
$$
\boxed{
\inf_{\substack{u\in H^1(\mathbb R)\\\lVert u\rVert_2=1}}
\int_\mathbb R|u'(x)|^2\,dx=0.
}
$$
This is <vanishing Dirichlet energy by dilation on the line>.

The infimum is not attained. If an admissible $u$ had zero energy, then its weak derivative would vanish almost everywhere, so $u$ would be constant almost everywhere. The only constant in $L^2(\mathbb R)$ is zero, contradicting $\lVert u\rVert_2=1$.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=24J}

For a divisor $D$ on a smooth projective curve $X$, define
$$
L(D)=\{h\in k(X)^\times:(h)+D\geq0\}\cup\{0\},
\qquad
\ell(D)=\dim_kL(D).
$$
The <Riemann-Roch theorem> states
$$
\boxed{\ell(D)-\ell(K_X-D)=\deg D+1-g},
$$
where $K_X$ is the divisor of a nonzero rational differential, $g$ is the genus of $X$, and $\deg D$ is the sum of the divisor coefficients because the ground field is algebraically closed.

Under the coordinate change $u=1/x$ and $v=y/x^n$, we have
$$
v^2=\frac{y^2}{x^{2n}}
=u^{2n}f(1/u).
$$
Hence
$$
\boxed{g(u)=u^{2n}f(1/u)}.
$$
This is a polynomial, and its value at zero is the nonzero leading coefficient of $f$. Thus the <even-degree hyperelliptic model> has two points $P_+,P_-$ above $u=0$, distinguished by the two values of $v$.

Consider the rational differential
$$
\omega=\frac{dx}{y}.
$$
At a finite branch point $x=a$, square-freeness gives a local parameter $y$, with $x-a$ a nonzero constant times $y^2$ to first order. Hence $dx/y$ is regular and nonzero there. It is also regular and nonzero at finite unramified points.

At infinity,
$$
dx=-u^{-2}du,
\qquad
y=vu^{-n},
$$
so
$$
\omega=-\frac{u^{n-2}}v\,du.
$$
Because $v(P_\pm)\ne0$ and $u$ is a local parameter at each point, $\omega$ has a zero of order $n-2$ at both. Therefore the <canonical divisor of an even-degree hyperelliptic curve> is
$$
\boxed{K_X=(\omega)=(n-2)(P_++P_-)},
$$
and
$$
\boxed{\deg K_X=2n-4}.
$$

Since $\deg K_X=2g-2$, it follows that
$$
2g-2=2n-4,
\qquad
\boxed{g=n-1}.
$$

The function $x$ has a simple pole at each of $P_+$ and $P_-$. Consequently
$$
1,x,x^2,\ldots,x^{n-2}\in L(K_X).
$$
These $n-1$ functions are linearly independent. Riemann-Roch with $D=K_X$ gives $\ell(K_X)=g=n-1$, so
$$
\boxed{L(K_X)=\operatorname{span}_k\{1,x,\ldots,x^{n-2}\}}.
$$

With respect to this basis, the <canonical map of a hyperelliptic curve> is
$$
(x,y)\longmapsto[1:x:x^2:\cdots:x^{n-2}].
$$
It identifies $(x,y)$ with $(x,-y)$ for general $x$, and hence is not an embedding.

If $X$ embedded in $\mathbb P^2$, its smooth plane image would have degree $d\geq4$: degrees one and two have genus zero, and degree three has genus one, whereas $g=n-1\geq2$. But the <canonical map of a smooth plane curve> of degree $d\geq4$ is an embedding by adjunction, contradicting the preceding calculation. Therefore
$$
\boxed{X\text{ cannot be embedded in }\mathbb P^2}.
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=25i}
{scope}

= Solution
{parent=a}

For $v$ in a sufficiently small neighbourhood of $0\in T_pS$, let $\gamma_v$ be the geodesic satisfying $\gamma_v(0)=p$ and $\dot\gamma_v(0)=v$. The <exponential map> at $p$ is
$$
\exp_p(v)=\gamma_v(1).
$$
Choose an oriented orthonormal basis of $T_pS$, write $e(\theta)=(\cos\theta,\sin\theta)$, and define <geodesic polar coordinates> by
$$
\phi(r,\theta)=\exp_p\bigl(r e(\theta)\bigr).
$$

The <Gauss lemma> says that radial and angular coordinate curves are orthogonal. Since $r\mapsto\phi(r,\theta)$ is a unit-speed geodesic,
$$
|\phi_r|=1.
$$
Moreover, torsion-freeness of the surface connection gives $D_r\phi_\theta=D_\theta\phi_r$, and therefore
$$
\begin{aligned}
\frac{\partial}{\partial r}\langle\phi_r,\phi_\theta\rangle
&=\langle D_r\phi_r,\phi_\theta\rangle
  +\langle\phi_r,D_r\phi_\theta\rangle\\
&=0+\langle\phi_r,D_\theta\phi_r\rangle
=\frac12\frac{\partial}{\partial\theta}|\phi_r|^2=0.
\end{aligned}
$$
At $r=0$, $\phi_\theta=0$, so $\langle\phi_r,\phi_\theta\rangle=0$ for every sufficiently small $r$. Thus the <first fundamental form> is
$$
\boxed{I=dr^2+G(r,\theta)\,d\theta^2},
\qquad G(r,\theta)=|\phi_\theta|^2,
$$
and $\sqrt G/r\to1$ as $r\downarrow0$ because $d(\exp_p)_0$ is the identity.

Solved by gpt-5.6-sol high.

= b
{parent=25i}
{scope}

= Solution
{parent=b}

An orientation of $S$ chooses a unit normal at every point. The <Gauss map> is
$$
N:S\longrightarrow S^2,
\qquad p\longmapsto N(p).
$$
For the unit tangent $T=\dot\alpha$, the <surface covariant derivative> is the tangential projection
$$
D_sT=(\ddot\alpha)^{\top}.
$$
The signed <geodesic curvature> is
$$
\boxed{\kappa_g=\langle D_sT,N\times T\rangle}.
$$
Because $|T|=1$, $D_sT$ is perpendicular to $T$ and hence
$$
D_sT=\kappa_g(N\times T).
$$
Consequently $\kappa_g=0$ means $D_sT=0$: the tangent is parallel along $\alpha$, so $\alpha$ is a <geodesic>.

Solved by gpt-5.6-sol high.

= c
{parent=25i}
{scope}

= Solution
{parent=c}

Let $M$ be a compact oriented surface with piecewise smooth, positively oriented boundary. If $\theta_j$ are the signed exterior turning angles at its corners, the global <Gauss-Bonnet theorem> with boundary is
$$
\boxed{
\int_M K\,dA
+\int_{\partial M}\kappa_g\,ds
+\sum_j\theta_j
=2\pi\chi(M)}.
$$
For a smooth boundary the corner sum is absent, and for a closed surface both boundary terms are absent.

Solved by gpt-5.6-sol high.

= d
{parent=25i}
{scope}

= Solution
{parent=d}

Suppose that two closed geodesics $\gamma_1$ and $\gamma_2$ were disjoint. Under the usual convention that a closed geodesic curve is simple, the <Jordan curve theorem> on the sphere shows that they bound an annulus $A\subset S$. Its Euler characteristic is zero. Give $A$ the induced orientation and apply the <Gauss-Bonnet theorem>:
$$
\int_AK\,dA+\int_{\partial A}\kappa_g\,ds=2\pi\chi(A)=0.
$$
Both boundary components are geodesics, so their <geodesic curvature> vanishes. It follows that
$$
\int_AK\,dA=0.
$$
This is impossible because $K>0$ everywhere and $A$ has positive area. Hence the two closed geodesics must intersect, as summarized by the <intersection of closed geodesics on a positively curved sphere>.

Solved by gpt-5.6-sol high.

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

= a
{parent=26h}
{scope}

= Solution
{parent=a}

Put $\mu=\mathbb E[X]$ and $\sigma^2=\operatorname{Var}(X)$. Independence of $X$ and the <uniform distribution> variable $U$ gives
$$
\mathbb E[Y]=\mathbb E[X]\mathbb E[U]=\boxed{\frac\mu2}.
$$
Also $\mathbb E[U^2]=1/3$, so
$$
\begin{aligned}
\operatorname{Var}(Y)
&=\mathbb E[X^2]\mathbb E[U^2]-\mathbb E[Y]^2\\
&=\frac{\sigma^2+\mu^2}{3}-\frac{\mu^2}{4}.
\end{aligned}
$$
Therefore
$$
\boxed{\operatorname{Var}(Y)=\frac13\operatorname{Var}(X)+\frac1{12}\mathbb E[X]^2}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=26h}
{scope}

= Solution
{parent=b}

The map $(x,u)\mapsto(x,y)=(x,xu)$ has inverse $u=y/x$ and Jacobian magnitude $1/x$. Hence the density of $(X,Y)$ is
$$
\boxed{f_{X,Y}(x,y)=\frac{g(x)}x
\mathbf 1_{\{x>0,\,0\leq y\leq x\}}}.
$$

For $(y,z)=(xu,x(1-u))$, the inverse is
$$
x=y+z,
\qquad
u=\frac y{y+z},
$$
and the Jacobian magnitude of $(y,z)\mapsto(x,u)$ is $1/(y+z)$. Thus the <uniform split of a random total> has joint density
$$
\boxed{f_{Y,Z}(y,z)=\frac{g(y+z)}{y+z}
\mathbf 1_{\{y,z\geq0\}}}.
$$
Integrating out the other piece yields
$$
\boxed{f_Y(y)=\int_0^\infty\frac{g(y+z)}{y+z}\,dz=h(y)},
\qquad y\geq0.
$$
The transformation $U\mapsto1-U$ preserves the uniform law, so $Z$ has the same density:
$$
\boxed{f_Z(z)=h(z)},
\qquad z\geq0.
$$

Solved by gpt-5.6-sol high.

= c
{parent=26h}
{scope}

= Solution
{parent=c}

Because $h'(x)=-g(x)/x$, the joint density from part (b) can be written
$$
f_{Y,Z}(y,z)=-h'(y+z).
$$
The pieces are therefore independent if and only if
$$
\boxed{-h'(y+z)=h(y)h(z)\quad(y,z\geq0)}.
$$
This condition is also sufficient because $h$ is the common marginal density.

Set $c=h(0)$. Taking $z=0$ gives the differential equation
$$
h'(y)=-c\,h(y).
$$
Since $h$ is a probability density, $c$ must be positive, and normalization gives
$$
\boxed{h(y)=ce^{-cy}\mathbf 1_{\{y\geq0\}}}.
$$
Conversely this function satisfies the displayed factorization. Thus the <exponential characterization by a uniform random split> shows that
$$
\boxed{Y,Z\text{ are independent }\operatorname{Exp}(c)\text{ variables}}.
$$
Equivalently, $g(x)=-xh'(x)=c^2xe^{-cx}$, so $X$ has the gamma distribution with shape two and rate $c$.

Solved by gpt-5.6-sol high.

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

= a
{parent=27l}
{scope}

= Solution
{parent=a}

Let $\Pi$ be a Poisson point process on $\mathbb R^d$ with intensity measure
$$
\mu(A)=\int_A\lambda(x)\,dx,
$$
where $\lambda$ is nonnegative and locally integrable. Let $f:\mathbb R^d\to\mathbb R^s$ be measurable and suppose its pushforward measure
$$
\nu(B)=\mu(f^{-1}(B))
=\int_{f^{-1}(B)}\lambda(x)\,dx
$$
is locally finite. The <mapping theorem for Poisson point processes> says that the image counting measure
$$
f(\Pi)=\sum_{x\in\Pi}\delta_{f(x)}
$$
is a Poisson random measure with intensity $\nu$.

For $f(\Pi)$ to be a spatial Poisson process in the usual simple sense, one also assumes that $\nu$ is diffuse:
$$
\nu(\{y\})=0\qquad(y\in\mathbb R^s).
$$
This prevents collisions with positive probability. If $\nu$ is absolutely continuous, its Radon--Nikodym derivative $\widetilde\lambda$ is the image intensity function, characterized by
$$
\int_B\widetilde\lambda(y)\,dy
=\int_{f^{-1}(B)}\lambda(x)\,dx.
$$

Solved by gpt-5.6-sol high.

= b
{parent=27l}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

Write $r=(x^2+y^2+z^2)^{1/2}$, so $f(x,y,z)=r^3$. For $t\geq0$,
$$
f^{-1}([0,t])
=\{(x,y,z):r\leq t^{1/3}\},
$$
whose volume is
$$
\frac{4\pi}{3}\bigl(t^{1/3}\bigr)^3
=\frac{4\pi}{3}t.
$$
Thus the pushforward intensity of $[0,t]$ is
$$
\lambda\,\operatorname{vol}(f^{-1}([0,t]))
=\frac{4\pi\lambda}{3}t.
$$
The pushforward is locally finite and diffuse, so the <radial volume transform of a homogeneous Poisson point process> and the mapping theorem show that
$$
\boxed{f(\Pi)\text{ is a homogeneous Poisson process of rate }
\rho=\frac{4\pi\lambda}{3}}.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

Put $T_k=R_k^3$. Part (i) identifies $T_k$ as the $k$th arrival time of a rate
$$
\rho=\frac{4\pi\lambda}{3}
$$
Poisson process. Hence $T_k$ has the gamma density
$$
f_{T_k}(t)=\frac{\rho^k}{(k-1)!}t^{k-1}e^{-\rho t},
\qquad t>0.
$$
The change of variables $t=r^3$, with $dt=3r^2\,dr$, gives the <kth-nearest-neighbour distance in a homogeneous Poisson point process>:
$$
\boxed{
f_{R_k}(r)
=\frac{3\rho^k}{(k-1)!}\,
r^{3k-1}e^{-\rho r^3},
\qquad r>0,
\quad
\rho=\frac{4\pi\lambda}{3}.}
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=28l}
{scope}

= Solution
{parent=a}

Let $Q=F^{-1}$ be the <quantile function>. For $p\in(0,1)$,
$$
Q(p)\leq t\quad\Longleftrightarrow\quad p\leq F(t).
$$
The reverse implication is immediate from the infimum defining $Q$. For the forward implication, if $Q(p)<t$ there is a point $s\leq t$ with $F(s)\geq p$. If $Q(p)=t$, choose $s_m<t+1/m$ with $F(s_m)\geq p$; monotonicity and right continuity give
$$
F(t)=\lim_{m\to\infty}F(t+1/m)\geq p.
$$
The endpoint values $p=0,1$ have zero probability under a continuous uniform variable.

Consequently
$$
\mathbb P(Q(U)\leq t)
=\mathbb P(U\leq F(t))
=F(t).
$$
Thus the <inverse transform sampling> identity is
$$
\boxed{F^{-1}(U)\sim F}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=28l}
{scope}

= Solution
{parent=b}

Choose an initial value $X_0$. One sweep of the <Gibbs sampler> consists of
$$
Y_m\sim f_{Y\mid X}(\,\cdot\mid X_{m-1}),
\qquad
X_m\sim f_{X\mid Y}(\,\cdot\mid Y_m).
$$
After a burn-in period, the pairs $(Y_m,X_m)$ are retained as approximate samples. Standard irreducibility and recurrence conditions are needed for convergence from the chosen starting point.

The transition density from $(y,x)$ to $(y',x')$ is
$$
k((y,x),(y',x'))
=f_{Y\mid X}(y'\mid x)f_{X\mid Y}(x'\mid y').
$$
Suppose $(Y,X)$ currently has density $f_{XY}$. The density after one sweep is
$$
\begin{aligned}
&\int_{\mathbb R^2}
f_{XY}(y,x)
f_{Y\mid X}(y'\mid x)
f_{X\mid Y}(x'\mid y')\,dy\,dx\\
&\quad=f_{X\mid Y}(x'\mid y')
\int_{\mathbb R}f_X(x)f_{Y\mid X}(y'\mid x)\,dx\\
&\quad=f_{X\mid Y}(x'\mid y')f_Y(y')
=f_{XY}(x',y').
\end{aligned}
$$
This proves the <stationarity of the two-coordinate Gibbs sampler>:
$$
\boxed{f_{XY}\text{ is stationary for the Gibbs transition kernel}.}
$$

Solved by gpt-5.6-sol high.

= c
{parent=28l}
{scope}

= Solution
{parent=c}

Write
$$
\bar z=\frac1n\sum_{i=1}^nz_i,
\qquad
S(\mu)=\sum_{i=1}^n(z_i-\mu)^2.
$$
Multiplying the likelihood by the <improper prior> gives the <Bayesian posterior> kernel
$$
\pi(\mu,\omega\mid z)
\propto
\omega^{n/2}
\exp\left\{-\omega\left(\lambda+\frac12S(\mu)\right)\right\},
\qquad \omega>0.
$$

Since
$$
S(\mu)=\sum_{i=1}^n(z_i-\bar z)^2+n(\mu-\bar z)^2,
$$
the first full conditional is
$$
\boxed{\mu\mid\omega,z\sim
N\left(\bar z,\frac1{n\omega}\right)}.
$$
Holding $\mu$ fixed and comparing powers and exponential rates gives
$$
\boxed{\omega\mid\mu,z\sim
\operatorname{Gamma}\left(
m,\lambda+\frac12S(\mu)\right)},
\qquad
m=\frac n2+1.
$$
The assumed parity of $n$ makes $m$ a positive integer.

Initialize $\mu_0$ arbitrarily. At iteration $j$, put
$$
r_j=\lambda+\frac12S(\mu_{j-1}).
$$
Draw $m$ independent uniform variables $U_{j1},\ldots,U_{jm}$ and set
$$
\omega_j=-\frac1{r_j}\sum_{\ell=1}^m\log U_{j\ell}.
$$
By <sampling an integer-shape gamma distribution>, this has the required $\operatorname{Gamma}(m,r_j)$ law. Next draw $G_j\sim N(0,1)$ independently and set
$$
\mu_j=\bar z+\frac{G_j}{\sqrt{n\omega_j}}.
$$
These are exactly the two full-conditional updates of the <normal mean-precision Gibbs sampler>. After discarding burn-in, the pairs $(\mu_j,\omega_j)$ are approximate samples from the posterior by <Markov chain Monte Carlo>.

Solved by gpt-5.6-sol high.

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

= a
{parent=29k}
{scope}

= Solution
{parent=a}

For $t>0$, differentiate
$$
u(t)=\mathbb E[f(\sqrt tZ)].
$$
The growth assumptions justify differentiation under the expectation, and <Gaussian integration by parts> gives
$$
\begin{aligned}
u'(t)
&=\frac1{2\sqrt t}\mathbb E[Zf'(\sqrt tZ)]\\
&=\frac1{2\sqrt t}\mathbb E\left[
\frac d{dZ}f'(\sqrt tZ)\right]\\
&=\frac12\mathbb E[f''(\sqrt tZ)].
\end{aligned}
$$
Since $u(0)=f(0)$, integration from $0$ to $t$ yields the <Brownian transition semigroup> identity
$$
\boxed{
\frac12\int_0^t\mathbb E[f''(\sqrt sZ)]\,ds
=\mathbb E[f(\sqrt tZ)]-f(0)}.
$$
Continuity at zero covers $t=0$.

Solved by gpt-5.6-sol high.

= b
{parent=29k}
{scope}

= Solution
{parent=b}

A process $(W_t)_{t\geq0}$ is a standard <Brownian motion> if:

* $W_0=0$ almost surely;
* its sample paths are almost surely continuous;
* it has independent increments; and
* for $0\leq s<t$,
  $$
  W_t-W_s\sim N(0,t-s).
  $$

Equivalently, every future increment is independent of the past filtration and has the stated centred normal law.

Solved by gpt-5.6-sol high.

= c
{parent=29k}
{scope}

= Solution
{parent=c}

Let $\mathcal F_s=\sigma(W_r:0\leq r\leq s)$. For $0\leq s<t$, the increment $W_t-W_s$ is independent of $\mathcal F_s$ and distributed as $\sqrt{t-s}Z$. Apply part (a) to the translated function $u\mapsto f(W_s+u)$, conditionally on $\mathcal F_s$:
$$
\begin{aligned}
\mathbb E[f(W_t)\mid\mathcal F_s]
&=f(W_s)
+\frac12\int_0^{t-s}
\mathbb E[f''(W_s+\sqrt rZ)\mid\mathcal F_s]\,dr\\
&=f(W_s)
+\frac12\mathbb E\left[
\int_s^t f''(W_r)\,dr\ \middle|\ \mathcal F_s
\right].
\end{aligned}
$$
Therefore
$$
\begin{aligned}
\mathbb E[M_t\mid\mathcal F_s]
&=\mathbb E[f(W_t)\mid\mathcal F_s]
-\frac12\int_0^sf''(W_r)\,dr\\
&\quad-\frac12\mathbb E\left[
\int_s^tf''(W_r)\,dr\ \middle|\ \mathcal F_s
\right]\\
&=f(W_s)-\frac12\int_0^sf''(W_r)\,dr
=M_s.
\end{aligned}
$$
The assumed growth conditions supply integrability, so $(M_t)$ is the <Brownian compensator martingale>.

Solved by gpt-5.6-sol high.

= d
{parent=29k}
{scope}

= Solution
{parent=d}

Let $\mathcal F_s=\sigma(W_r:0\leq r\leq s)$ and fix $c\in\mathbb R$. The martingale property between $s$ and $t$ gives
$$
\mathbb E[e^{cW_t}\mid\mathcal F_s]
=e^{cW_s}
+\frac{c^2}{2}\int_s^t
\mathbb E[e^{cW_r}\mid\mathcal F_s]\,dr.
$$
For fixed $s$, write the left-hand conditional expectation at time $t$ as $G(t)$. This integral equation has the unique solution
$$
G(t)=e^{cW_s}\exp\left(\frac{c^2}{2}(t-s)\right).
$$
Multiplication by the $\mathcal F_s$-measurable factor $e^{-cW_s}$ therefore yields the conditional moment-generating function
$$
\boxed{
\mathbb E[e^{c(W_t-W_s)}\mid\mathcal F_s]
=\exp\left(\frac{c^2}{2}(t-s)\right)}.
$$
This is the moment-generating function of $N(0,t-s)$ and is deterministic. Hence $W_t-W_s$ has that normal law conditionally on $\mathcal F_s$, and its conditional law does not depend on $\mathcal F_s$; the increment is therefore independent of the past.

The process starts at zero by assumption and has continuous paths. The independent centred Gaussian increments just obtained complete the definition of Brownian motion. This proves the <exponential test-function characterization of Brownian motion>.

Solved by gpt-5.6-sol high.

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

= a
{parent=30l}
{scope}

= Solution
{parent=a}

The candidate $\Pi$ is tacitly required to belong to $S$, as in the usual <variational characterization of convex projection>; this feasibility holds for the candidate constructed in part (b). Use the Frobenius inner product
$$
\langle A,B\rangle_F=\operatorname{Tr}(A^TB).
$$
For any $Z\in S$,
$$
M-Z=(M-\Pi)-(Z-\Pi),
$$
and hence
$$
\begin{aligned}
\|M-Z\|_F^2
&=\|M-\Pi\|_F^2+\|Z-\Pi\|_F^2
-2\langle M-\Pi,Z-\Pi\rangle_F\\
&\geq\|M-\Pi\|_F^2+\|Z-\Pi\|_F^2\\
&\geq\|M-\Pi\|_F^2.
\end{aligned}
$$
Thus $\Pi$ minimizes the squared Frobenius distance over $S$. Since that objective is strictly convex,
$$
\boxed{\Pi=\pi(M)}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=30l}
{scope}

= Solution
{parent=b}

Since $M\succeq0$ but $M\notin S$, its eigenvalues satisfy
$$
\sum_{i=1}^d\mu_i=\operatorname{Tr}M>s.
$$
The continuous decreasing function
$$
q(\rho)=\sum_{i=1}^d(\mu_i-\rho)_+
$$
falls from $q(0)>s$ to zero, so there is a unique $\rho>0$ for which $q(\rho)=s$.

Put
$$
p_i=(\mu_i-\rho)_+,
\qquad
\Pi=\sum_{i=1}^dp_iv_iv_i^T.
$$
Then $\Pi\succeq0$ and $\operatorname{Tr}\Pi=\sum_i p_i=s$, so $\Pi\in S$. Moreover
$$
A:=M-\Pi
=\sum_i\min(\mu_i,\rho)v_iv_i^T,
$$
and therefore
$$
0\preceq A\preceq\rho I.
$$
For any $Z\in S$,
$$
\operatorname{Tr}(AZ)\leq\rho\operatorname{Tr}Z\leq\rho s.
$$
Whenever $p_i>0$, the corresponding eigenvalue of $A$ is exactly $\rho$, while terms with $p_i=0$ contribute nothing. Consequently
$$
\operatorname{Tr}(A\Pi)=\rho\sum_i p_i=\rho s.
$$
It follows that
$$
\langle M-\Pi,Z-\Pi\rangle_F
=\operatorname{Tr}(AZ)-\operatorname{Tr}(A\Pi)\leq0.
$$
Part (a), equivalently the formula for <projection onto a positive semidefinite trace ball>, now gives
$$
\boxed{\pi(M)=\sum_{i=1}^d
\max(0,\mu_i-\rho)v_iv_i^T},
\qquad
\sum_i\max(0,\mu_i-\rho)=s.
$$

Solved by gpt-5.6-sol high.

= c
{parent=30l}
{scope}

= Solution
{parent=c}

The hypothesis class and margin loss are
$$
\boxed{\mathcal H
=\{h_M:x\mapsto x^TMx,\ M\succeq0,\ \operatorname{Tr}M\leq s\}}
$$
and
$$
\boxed{\phi(u)=\log(1+e^{-u})}.
$$
Thus this is a <positive semidefinite quadratic-form classifier> with <logistic loss>. Its empirical risk, viewed as a function of the matrix parameter, is
$$
F(M)=\widehat R_\phi(h_M)
=\frac1n\sum_{j=1}^n
\log\left(1+\exp(-Y_jX_j^TMX_j)\right).
$$
Differentiating under the sum shows that
$$
\nabla F(M)
=-\frac1n\sum_{j=1}^n
Y_jX_jX_j^T
\frac{\exp(-Y_jX_j^TMX_j)}
{1+\exp(-Y_jX_j^TMX_j)}.
$$
Hence the displayed $g_i$ is exactly $\nabla F(M^{(i)})$, and the algorithm is <projected gradient descent> on the <positive semidefinite trace ball>.

The scalar factor multiplying each $Y_jX_jX_j^T$ lies in $[0,1]$. Since
$$
\|xx^T\|_F
=\sqrt{\operatorname{Tr}(xx^Txx^T)}
=\|x\|_2^2,
$$
we have the uniform gradient bound
$$
\|g_i\|_F\leq\frac1n\sum_{j=1}^n\|X_j\|_2^2
\leq C^2.
$$
Let $\widehat M$ parametrize the empirical minimizer $\widehat h$. Positive semidefiniteness gives
$$
\|\widehat M\|_F
=\left(\sum_r\lambda_r(\widehat M)^2\right)^{1/2}
\leq\sum_r\lambda_r(\widehat M)
=\operatorname{Tr}\widehat M
\leq s.
$$
Thus the initial distance from $M^{(1)}=0$ to $\widehat M$ is at most $D=s$, while the gradient bound is $G=C^2$.

The <averaged projected-gradient bound>, in its slightly looser form
$$
F(\bar M)-F(\widehat M)
\leq\frac{D^2}{\eta k}+\eta G^2,
$$
applies because $F$ is convex. Choose
$$
\boxed{\eta=\frac{s}{C^2\sqrt k}}.
$$
Substitution gives
$$
\widehat R_\phi(\bar h)-\widehat R_\phi(\widehat h)
=F(\bar M)-F(\widehat M)
\leq\frac{s^2}{\eta k}+\eta C^4
=\boxed{\frac{2sC^2}{\sqrt k}}.
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=31d}
{scope}

= Solution
{parent=a}

Near the endpoint $t=0$,
$$
\frac1{1+t}\sim\sum_{n=0}^\infty(-1)^nt^n.
$$
The <Watson lemma> integrates this local expansion term by term, and
$$
\int_0^\infty e^{-xt}t^n\,dt
=\frac{\Gamma(n+1)}{x^{n+1}}
=\frac{n!}{x^{n+1}}.
$$
Therefore the <reciprocal-amplitude Laplace integral> has expansion
$$
\boxed{
U(x)\sim\sum_{n=0}^\infty
(-1)^nn!x^{-n-1}},
\qquad x\to+\infty,
$$
so
$$
\boxed{a_n=(-1)^nn!}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=31d}
{scope}

= Solution
{parent=b}

Differentiation under the integral sign gives
$$
U'(x)=-\int_0^\infty\frac{te^{-xt}}{1+t}\,dt,
\qquad
U''(x)=\int_0^\infty\frac{t^2e^{-xt}}{1+t}\,dt.
$$
Hence
$$
\begin{aligned}
xU''+(1-x)U'-U
&=\int_0^\infty e^{-xt}
\frac{xt^2-(1-x)t-1}{1+t}\,dt\\
&=\int_0^\infty e^{-xt}(xt-1)\,dt\\
&=x\frac1{x^2}-\frac1x=0.
\end{aligned}
$$
Thus
$$
\boxed{xU''(x)+(1-x)U'(x)-U(x)=0}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=31d}
{scope}

= Solution
{parent=c}

After division by $x$, the equation is
$$
y''+\left(\frac1x-1\right)y'-\frac1xy=0.
$$
The <removal of the first derivative from a second-order differential equation> uses
$$
y=e^{-\frac12\int(1/x-1)\,dx}v
=e^{x/2}x^{-1/2}v.
$$
With
$$
P=\frac1x-1,\qquad Q=-\frac1x,
$$
the normal-form coefficient is
$$
Q-\frac12P'-\frac14P^2
=-\frac14-\frac1{2x}+\frac1{4x^2}.
$$
Consequently
$$
\boxed{
v''-\frac14\left(1+\frac2x-\frac1{x^2}\right)v=0},
$$
so the constants in the PDF's expression
$1+Ax^{-1}+Bx^{-2}$ are
$$
\boxed{A=2,\qquad B=-1}.
$$

To classify infinity, put $z=1/x$. Since
$$
\frac{d^2v}{dx^2}=z^4v_{zz}+2z^3v_z,
$$
the coefficient of $v$ in the transformed equation has a pole of order four at $z=0$. This exceeds the order two allowed at a regular singular point. Hence $x=+\infty$ is an <irregular singular point at infinity>.

Solved by gpt-5.6-sol high.

= d
{parent=31d}
{scope}

= Solution
{parent=d}

Put $v=e^S$ and $p=S'$. The <Liouville-Green exponential ansatz> turns the normal-form equation into the <Riccati equation>
$$
p'+p^2
=\frac14\left(1+\frac2x-\frac1{x^2}\right).
$$
Write
$$
p\sim b_0+\frac{b_1}{x}+\frac{b_2}{x^2}+\cdots.
$$
Matching the constant, $x^{-1}$, and $x^{-2}$ coefficients gives
$$
b_0^2=\frac14,\qquad
2b_0b_1=\frac12,\qquad
b_1^2+2b_0b_2-b_1=-\frac14.
$$

For the recessive branch,
$$
b_0=-\frac12,\qquad b_1=-\frac12,\qquad b_2=1,
$$
so
$$
S_-(x)
=-\frac x2-\frac12\log x-\frac1x+O(x^{-2}).
$$
Since $y=e^{x/2}x^{-1/2}v$, this gives
$$
y_-(x)
=x^{-1}\exp\left(-\frac1x+O(x^{-2})\right)
=\frac1x-\frac1{x^2}+O(x^{-3}).
$$
This agrees with the first two terms of part (a),
$$
U(x)\sim\frac1x-\frac1{x^2}+\frac2{x^3}-\cdots.
$$
Continuing the Riccati recursion reproduces the factorial asymptotic series, so $U$ is proportional to this recessive branch.

For the dominant branch,
$$
b_0=\frac12,\qquad b_1=\frac12,\qquad b_2=0,
$$
and in fact
$$
S_+(x)=\frac x2+\frac12\log x+\text{constant}
$$
solves the phase equation exactly. It follows that
$$
v_+(x)\asymp e^{x/2}x^{1/2},
\qquad
\boxed{y_+(x)\asymp e^x}.
$$
Indeed, $e^x$ is an exact second solution of the original differential equation.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=32A}

A map is <Devaney chaos> when:

* every pair of nonempty open sets eventually overlap under an iterate, which is <topological transitivity>;
* it has <dense periodic points>; and
* nearby initial conditions can eventually separate by a fixed amount, which is <sensitive dependence on initial conditions>.

Write a point as a binary expansion
$$
x=0.b_1b_2b_3\ldots{}_2.
$$
Apart from the harmless choice of expansion at dyadic rationals, the <binary shift representation of the doubling map> is
$$
F(x)=0.b_2b_3b_4\ldots{}_2.
$$

Every open interval contains a binary cylinder specified by a finite initial word. Given cylinders $U$ and $V$, choose a binary sequence beginning with the word for $U$ and place the word for $V$ after it. A suitable iterate shifts the second word to the front, proving topological transitivity.

Given any cylinder, repeat its defining word forever. The resulting point is periodic and lies in that cylinder, so periodic points are dense.

Finally, given $x$ and any neighbourhood, choose $N$ so large that changing only digits after the first $N$ stays inside that neighbourhood. Choose the later tail so that after $N$ shifts it is either $0$ or $3/4$, whichever is farther from $F^N(x)$. The separation is at least $3/8$, so any smaller fixed constant, for example $\delta=1/4$, proves sensitivity. Hence the <doubling map> is chaotic in Devaney's sense.

Solved by gpt-5.6-sol high.

= a
{parent=32a}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

Concatenate every finite binary word, ordered first by length and then lexicographically:
$$
x=0.\,0\,1\,00\,01\,10\,11\,000\,001\,010\,011\cdots{}_2.
$$
Every finite binary block occurs in this expansion. Since iterating $F$ shifts digits to the left, its orbit enters every binary cylinder and is therefore dense. The digit string is not eventually periodic, so it gives a concrete nonperiodic chaotic orbit.

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

A fixed point satisfies
$$
2x\equiv x\pmod1,
$$
so $x$ is an integer. In $[0,1)$ the only possibility is
$$
\boxed{x=0}.
$$
Equivalently, a binary sequence fixed by the left shift is constant; $0.000\ldots{}_2=0$, while $0.111\ldots{}_2=1$ represents the excluded endpoint, which is the same circle point.

Solved by gpt-5.6-sol high.

= iii
{parent=a}
{scope}

= Solution
{parent=iii}

Points whose period divides two solve
$$
(2^2-1)x\in\mathbb Z,
$$
so, after removing the fixed point zero, the unique two-cycle is
$$
\boxed{\frac13\longmapsto\frac23\longmapsto\frac13}.
$$

Points whose period divides four are $j/15$ for $j=0,\ldots,14$. Removing the three points fixed by $F^2$, namely $0,5/15,10/15$, leaves twelve points of exact period four. They form the three orbits
$$
\boxed{
\frac1{15}\to\frac2{15}\to\frac4{15}\to\frac8{15}\to\frac1{15}},
$$
$$
\boxed{
\frac15\to\frac25\to\frac45\to\frac35\to\frac15},
$$
and
$$
\boxed{
\frac7{15}\to\frac{14}{15}\to\frac{13}{15}\to\frac{11}{15}\to\frac7{15}}.
$$
Starting at another point on one displayed orbit gives the corresponding cyclic rotation.

Solved by gpt-5.6-sol high.

= b
{parent=32a}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

The result follows by induction. It is true for $n=1$. If
$$
F^n(x)=2^nx\pmod1,
$$
then
$$
F^{n+1}(x)
=F(F^n(x))
=2(2^nx\pmod1)\pmod1
=2^{n+1}x\pmod1.
$$
Therefore
$$
\boxed{F^n(x)=2^nx\pmod1}.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

By part (i), $F^n(x)=x$ exactly when
$$
(2^n-1)x\in\mathbb Z.
$$
Thus the <periodic points of the doubling map> fixed by $F^n$ are
$$
x=\frac{j}{2^n-1},
\qquad j=0,\ldots,2^n-2,
$$
and their number is
$$
\boxed{2^n-1}.
$$
For $n=2$ this gives three points: the fixed point and the two points in the two-cycle. For $n=4$ it gives fifteen: those three points and the twelve points in the three four-cycles found above.

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

Take $n=2^k$. Every proper divisor of $n$ divides $n/2=2^{k-1}$, so the points fixed by $F^n$ that do not have exact period $n$ are precisely those fixed by $F^{n/2}$. Part (ii) therefore gives
$$
\begin{aligned}
\#\{x:\text{exact period }2^k\}
&=(2^{2^k}-1)-(2^{2^{k-1}}-1)\\
&=\boxed{2^{2^k}-2^{2^{k-1}}}.
\end{aligned}
$$
Because different starting points are counted separately, no division by the period is made. This is the <exact power-of-two periods of the doubling map>.

Solved by gpt-5.6-sol high.

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

= a
{parent=33c}
{scope}

= Solution
{parent=a}

The orbital angular momentum operator is
$$
L_i=\varepsilon_{ik\ell}X_kP_\ell.
$$
Using $[X_i,X_j]=[P_i,P_j]=0$ and $[X_i,P_j]=i\hbar\delta_{ij}$,
$$
\begin{aligned}
[L_i,X_j]
&=\varepsilon_{ik\ell}X_k[P_\ell,X_j]\\
&=-i\hbar\varepsilon_{ikj}X_k
=i\hbar\varepsilon_{ijk}X_k,
\end{aligned}
$$
and
$$
\begin{aligned}
[L_i,P_j]
&=\varepsilon_{ik\ell}[X_k,P_j]P_\ell\\
&=i\hbar\varepsilon_{ik\ell}\delta_{kj}P_\ell
=i\hbar\varepsilon_{ij\ell}P_\ell.
\end{aligned}
$$
Thus the <rotation commutators for orbital angular momentum> are
$$
\boxed{[L_i,X_j]=i\hbar\varepsilon_{ijk}X_k},
\qquad
\boxed{[L_i,P_j]=i\hbar\varepsilon_{ijk}P_k}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=33c}
{scope}

= Solution
{parent=b}

Contract the two Levi-Civita symbols:
$$
\begin{aligned}
L_iL_i
&=\varepsilon_{ijk}\varepsilon_{i\ell m}
X_jP_kX_\ell P_m\\
&=(\delta_{j\ell}\delta_{km}-\delta_{jm}\delta_{k\ell})
X_jP_kX_\ell P_m.
\end{aligned}
$$
Moving momenta past positions with
$P_kX_\ell=X_\ell P_k-i\hbar\delta_{k\ell}$ gives the <squared orbital angular momentum in position and momentum operators>:
$$
\boxed{
L^2=X^2P^2-(X\cdot P)^2+i\hbar X\cdot P}.
$$
Therefore
$$
\boxed{c_1=i\hbar,\qquad c_2=1,\qquad c_3=-1}.
$$

In position representation,
$$
P=-i\hbar\nabla,
\qquad
X\cdot P=-i\hbar r\partial_r,
$$
and
$$
\nabla^2=\partial_r^2+\frac2r\partial_r+\frac1{r^2}\nabla_{S^2}^2.
$$
Also $(r\partial_r)^2=r^2\partial_r^2+r\partial_r$. Substitution into the Cartesian identity gives
$$
\begin{aligned}
L^2
&=-\hbar^2r^2\nabla^2
+\hbar^2(r\partial_r)^2
+\hbar^2r\partial_r\\
&=-\hbar^2\nabla_{S^2}^2,
\end{aligned}
$$
because every radial derivative cancels. Hence
$$
\boxed{L^2=-\hbar^2\nabla_{S^2}^2},
$$
the <spherical Laplacian from orbital angular momentum>.

Solved by gpt-5.6-sol high.

= c
{parent=33c}
{scope}

= Solution
{parent=c}

The <first excited hydrogen eigenspace> is the four-dimensional $n=2$ space
$$
\boxed{
\{|2,0,0\rangle,\ |2,1,-1\rangle,\ |2,1,0\rangle,\ |2,1,1\rangle\}}.
$$
The first vector is the $2s$ state and the other three are the $2p$ states. All have the same unperturbed energy $E_2$.

Solved by gpt-5.6-sol high.

= i
{parent=c}
{scope}

= Solution
{parent=i}

The states $|n,\ell,m\rangle$ diagonalize $L^2$:
$$
L^2|n,\ell,m\rangle
=\hbar^2\ell(\ell+1)|n,\ell,m\rangle.
$$
The perturbation therefore introduces no mixing. The <orbital-angular-momentum-squared perturbation of hydrogen> leaves
$$
|2,0,0\rangle
$$
at energy
$$
\boxed{E_2},
$$
while
$$
|2,1,-1\rangle,\quad |2,1,0\rangle,\quad |2,1,1\rangle
$$
remain triply degenerate at
$$
\boxed{E_2+2g\hbar^2}.
$$
These statements are exact because the unperturbed Hamiltonian commutes with $L^2$.

Solved by gpt-5.6-sol high.

= ii
{parent=c}
{scope}

= Solution
{parent=ii}

Part (a) gives
$$
[L_3,X_3]=[L_3,P_3]=0.
$$
Hence both operators preserve the magnetic quantum number $m$. They are also odd under parity, while $|n,\ell,m\rangle$ has parity $(-1)^\ell$. Their matrix elements between states of equal parity vanish. The <position-momentum selection rules in the first excited hydrogen eigenspace> therefore leave only the coupling between
$$
|s\rangle:=|2,0,0\rangle,
\qquad
|p\rangle:=|2,1,0\rangle.
$$
The $m=\pm1$ states decouple, and all diagonal matrix elements vanish.

In the ordered basis $(|s\rangle,|p\rangle)$, remove the common energy $E_2$ and write
$$
\Delta H(t)=
\begin{pmatrix}
0&z(t)\\
z(t)^*&0
\end{pmatrix}
=
\begin{pmatrix}
0&ce^{i\omega t/\hbar}\\
ce^{-i\omega t/\hbar}&0
\end{pmatrix}.
$$
Let
$$
U(t)=
\begin{pmatrix}
e^{i\omega t/(2\hbar)}&0\\
0&e^{-i\omega t/(2\hbar)}
\end{pmatrix},
\qquad
\binom{a_0}{a_1}=U(t)\binom{b_0}{b_1}.
$$
The rotating-frame amplitudes obey a constant Schrödinger equation
$$
i\hbar\frac d{dt}\binom{b_0}{b_1}
=H_{\mathrm{eff}}\binom{b_0}{b_1},
\qquad
H_{\mathrm{eff}}=
\begin{pmatrix}
\omega/2&c\\
c&-\omega/2
\end{pmatrix}.
$$
Put
$$
q=\sqrt{c^2+\frac{\omega^2}{4}},
\qquad
\theta=\frac{qt}{\hbar}.
$$
Since $H_{\mathrm{eff}}^2=q^2I$,
$$
e^{-itH_{\mathrm{eff}}/\hbar}
=\cos\theta\,I-i\frac{\sin\theta}{q}H_{\mathrm{eff}}.
$$
For the initial amplitudes $(A_0,A_1)$, the exact coefficients are therefore
$$
\boxed{
a_0(t)=e^{i\omega t/(2\hbar)}
\left[
\left(\cos\theta-i\frac{\omega}{2q}\sin\theta\right)A_0
-i\frac cq\sin\theta\,A_1
\right]}
$$
and
$$
\boxed{
a_1(t)=e^{-i\omega t/(2\hbar)}
\left[
-i\frac cq\sin\theta\,A_0
+\left(\cos\theta+i\frac{\omega}{2q}\sin\theta\right)A_1
\right]}.
$$
Restoring the common unperturbed phase, the solution is
$$
\boxed{
|\psi(t)\rangle
=e^{-iE_2t/\hbar}
\bigl(a_0(t)|2,0,0\rangle+a_1(t)|2,1,0\rangle\bigr)}.
$$
This is the <rotating-frame solution for a harmonically driven two-level system>.

Solved by gpt-5.6-sol high.

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

= a
{parent=34b}
{scope}

= Solution
{parent=a}

In the <symmetric gauge>,
$$
\partial_iA_j+\partial_jA_i=0,
\qquad
\partial_iA_j-\partial_jA_i
=\varepsilon_{ijk}B_k.
$$
Together with $[p_i,f(x)]=-i\hbar\partial_i f$, this gives
$$
\begin{aligned}
[p_i+eA_i,p_j-eA_j]
&=-e[p_i,A_j]+e[A_i,p_j]\\
&=ie\hbar(\partial_iA_j+\partial_jA_i)=0,
\end{aligned}
$$
whereas
$$
\begin{aligned}
[p_i+eA_i,p_j+eA_j]
&=e[p_i,A_j]+e[A_i,p_j]\\
&=ie\hbar(\partial_jA_i-\partial_iA_j)\\
&=-ie\hbar\varepsilon_{ijk}B_k.
\end{aligned}
$$
Thus
$$
\boxed{[p_i+eA_i,p_j-eA_j]=0},
\qquad
\boxed{[p_i+eA_i,p_j+eA_j]
=-ie\hbar\varepsilon_{ijk}B_k}.
$$
These are the commutators of the <kinetic momentum and magnetic pseudomomentum>.

Solved by gpt-5.6-sol high.

= b
{parent=34b}
{scope}

= Solution
{parent=b}

With the convention in the question,
$$
T_r=\exp\left(\frac i\hbar r\cdot p\right)
$$
acts as
$$
(T_r\psi)(x)=\psi(x+r).
$$
Although $V(x+r)=V(x)$ for $r\in\Lambda$, the vector potential changes by
$$
A(x+r)=A(x)+\frac12B\times r.
$$
Consequently
$$
T_rHT_r^{-1}
=\frac1{2m}[p-eA(x+r)]^2+V(x)
\ne H
$$
for a nonzero magnetic field in general.

Now let $K=p+eA$. Since
$$
r\cdot K=-i\hbar r\cdot\nabla
+\frac e2r\cdot(B\times x)
$$
and the derivative in the $r$ direction annihilates
$r\cdot(B\times x)$, the two terms in its exponential commute. The <magnetic translation operator> therefore acts as
$$
\boxed{
(\mathcal T_r\psi)(x)
=\exp\left(\frac{ie}{2\hbar}
r\cdot(B\times x)\right)\psi(x+r)}.
$$

Part (a) gives $[K_i,p_j-eA_j]=0$, so $\mathcal T_r$ commutes with the kinetic term. Its phase commutes with $V$, while its translation sends $V(x)$ to $V(x+r)=V(x)$. Hence
$$
\boxed{[\mathcal T_r,H]=0\qquad(r\in\Lambda)}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=34b}
{scope}

= Solution
{parent=c}

Part (a) gives
$$
[K_i,K_j]=-ie\hbar\varepsilon_{ijk}B_k.
$$
For
$$
M=\frac i\hbar r\cdot K,
\qquad
N=\frac i\hbar r'\cdot K,
$$
their commutator is the scalar
$$
[M,N]
=\frac{ie}{\hbar}(r\times r')\cdot B.
$$
It therefore commutes with both $M$ and $N$. Applying the stated Baker--Campbell--Hausdorff identity in both orders yields the <magnetic translation algebra>
$$
\boxed{
\mathcal T_r\mathcal T_{r'}
=\exp\left[
\frac{ie}{\hbar}(r\times r')\cdot B
\right]
\mathcal T_{r'}\mathcal T_r}.
$$

Writing $\Phi=B\cdot(r\times r')$, the translations commute exactly when their phase is one:
$$
\frac{e\Phi}{\hbar}=2\pi n,
\qquad n\in\mathbb Z.
$$
Thus the cell flux must be an integral multiple of the <magnetic flux quantum>:
$$
\boxed{\Phi=\frac{2\pi\hbar}{e}n=\frac he n}.
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=35c}
{scope}

= Solution
{parent=a}

The latent heat for converting liquid to gas at coexistence is
$$
L=T(S_{\mathrm{gas}}-S_{\mathrm{liq}})
=H_{\mathrm{gas}}-H_{\mathrm{liq}},
$$
with all quantities taken per mole or per particle consistently. Along the <phase coexistence curve>, the two <Gibbs free energy> values agree:
$$
G_{\mathrm{gas}}(T,p)=G_{\mathrm{liq}}(T,p).
$$
Since
$$
dG=-S\,dT+V\,dp,
$$
differentiating the equality along the curve gives
$$
-S_{\mathrm{gas}}\,dT+V_{\mathrm{gas}}\,dp
=-S_{\mathrm{liq}}\,dT+V_{\mathrm{liq}}\,dp.
$$
Therefore the <Clausius-Clapeyron relation> is
$$
\boxed{
\frac{dp}{dT}
=\frac{S_{\mathrm{gas}}-S_{\mathrm{liq}}}
{V_{\mathrm{gas}}-V_{\mathrm{liq}}}
=\frac{L}
{T(V_{\mathrm{gas}}-V_{\mathrm{liq}})}}.
$$
At the critical point the liquid and gas become the same phase, so their entropy discontinuity disappears and
$$
\boxed{L\longrightarrow0}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=35c}
{scope}

= Solution
{parent=b}

Put
$$
x=\beta\mu B.
$$
The one-spin partition function is
$$
z_1=e^{-x}+1+e^x=1+2\cosh x.
$$
The spins are independent, so the <spin-1 paramagnet> has
$$
Z=z_1^N
$$
and free energy
$$
\boxed{
F=-\frac N\beta\log(1+2\cosh(\beta\mu B))}.
$$

The canonical heat capacity can be obtained from the energy variance:
$$
C=k_B\beta^2\frac{\partial^2}{\partial\beta^2}\log Z.
$$
Because
$$
\frac{d^2}{dx^2}\log(1+2\cosh x)
=\frac{2(\cosh x+2)}{(1+2\cosh x)^2},
$$
we obtain
$$
\boxed{
C=Nk_B(\beta\mu B)^2
\frac{2(\cosh(\beta\mu B)+2)}
{(1+2\cosh(\beta\mu B))^2}}.
$$

At high temperature, $x\to0$, and hence
$$
\boxed{-\beta F\longrightarrow N\log3},
\qquad
\boxed{C\sim\frac23Nk_B(\beta\mu B)^2\longrightarrow0}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=35c}
{scope}

= Solution
{parent=c}

Immediately after the reversal the state populations have not changed, but the new Hamiltonian satisfies
$$
H(-B)=-H(B).
$$
Thus
$$
e^{-\beta H(B)}
=e^{-(-\beta)H(-B)},
$$
so the same state is canonical for the reversed field with
$$
\boxed{\beta'=-\beta}.
$$
This is the <magnetic-field reversal and negative temperature> construction. Since
$$
\beta'\mu B'=(-\beta)\mu(-B)=\beta\mu B,
$$
the partition function is unchanged. Therefore
$$
\boxed{F'=-F},
\qquad
\boxed{C'=C}.
$$

A gas has kinetic energy unbounded above. Its partition function diverges for $\beta<0$, so it cannot have a normalizable <negative temperature> canonical state and cannot satisfy the zeroth-law equilibrium condition $\beta_{\mathrm{gas}}=\beta'<0$.

If an amount $\delta Q>0$ flows from the spins into a positive-temperature gas, the total entropy change is
$$
\delta S_{\mathrm{total}}
=\delta Q(\beta_{\mathrm{gas}}-\beta')
>0.
$$
The second law therefore selects heat flow
$$
\boxed{\text{from the negative-temperature spin system to the gas}}.
$$
This is the general ordering expressed by <heat flow from negative to positive temperature>.

Solved by gpt-5.6-sol high.

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

= a
{parent=36b}
{scope}

= Solution
{parent=a}

The <source-free Maxwell equations in a linear medium> are
$$
\nabla\cdot D=0,\qquad
\nabla\cdot B=0,
$$
$$
\nabla\times E=-\frac{\partial B}{\partial t},
\qquad
\nabla\times H=\frac{\partial D}{\partial t},
$$
with $D=\epsilon E$ and $B=\mu H$.

Take complex plane waves
$$
E=E_0e^{i(k\cdot x-\omega t)},
\qquad
B=B_0e^{i(k\cdot x-\omega t)}.
$$
The divergence equations give
$$
k\cdot E_0=k\cdot B_0=0,
$$
and <Faraday's law> gives
$$
\boxed{B_0=\frac1\omega k\times E_0}.
$$
Ampère's law gives
$$
k\times\frac{B_0}{\mu}=-\omega\epsilon E_0.
$$
Substituting the expression for $B_0$ and using
$k\times(k\times E_0)=-|k|^2E_0$ yields
$$
\frac{|k|^2}{\mu\omega}E_0
=\omega\epsilon E_0.
$$
Hence the <plane electromagnetic wave in a linear medium> has
$$
\boxed{\omega^2=\frac{|k|^2}{\epsilon\mu}},
\qquad
\boxed{v=\frac{\omega}{|k|}
=\frac1{\sqrt{\epsilon\mu}}},
$$
and equivalently
$$
\boxed{B_0=\sqrt{\epsilon\mu}\,\widehat k\times E_0}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=36b}
{scope}

= Solution
{parent=b}

Let $\widehat n$ be the unit normal pointing from the minus medium to the plus medium. Integrating the <Maxwell equations> across an infinitesimal pillbox or loop gives the <dielectric-interface boundary conditions>
$$
\boxed{\widehat n\cdot(D_+-D_-)=0},
\qquad
\boxed{\widehat n\cdot(B_+-B_-)=0},
$$
and
$$
\boxed{\widehat n\times(E_+-E_-)=0},
\qquad
\boxed{\widehat n\times(H_+-H_-)=0}.
$$
Thus tangential $E$ and $H$ are continuous, as are normal $D$ and $B$.

Because $D=\epsilon E$, the normal electric field generally jumps according to
$$
\epsilon_+E_{+,n}=\epsilon_-E_{-,n}.
$$
The tangential displacement field generally jumps because $\epsilon_+\ne\epsilon_-$. Since the permeability is common, continuity of tangential $H$ is continuity of tangential $B$; together with normal-$B$ continuity, every component of $B$ is continuous. The same common $\mu$ also makes the normal component of $H$ continuous.

Solved by gpt-5.6-sol high.

= c
{parent=36b}
{scope}

= Solution
{parent=c}

Choose the interface as $x=0$ and write the incident, reflected, and transmitted fields as plane waves with wavevectors $k_I,k_R,k_T$. The boundary conditions must hold for every tangential coordinate and every time. The <phase matching at a planar wave interface> therefore requires
$$
\omega_I=\omega_R=\omega_T
$$
and equality of tangential wavevectors:
$$
(k_I)_\parallel=(k_R)_\parallel=(k_T)_\parallel.
$$

The incident and reflected waves are in the same medium, so
$|k_I|=|k_R|$. Equality of their tangential components gives
$$
|k_I|\sin\theta_I=|k_R|\sin\theta_R,
$$
hence
$$
\boxed{\theta_R=\theta_I}.
$$

At fixed frequency, $|k_\pm|=\omega/v_\pm$ is proportional to the refractive index $n_\pm$. Tangential phase matching between incident and transmitted waves gives the <Snell law for electromagnetic waves>:
$$
|k_-|\sin\theta_I=|k_+|\sin\theta_T,
$$
or
$$
\boxed{n_-\sin\theta_I=n_+\sin\theta_T}.
$$
Here $n_\pm=c/v_\pm=c\sqrt{\epsilon_\pm\mu}$ if the vacuum speed $c$ is used to normalize the index.

Solved by gpt-5.6-sol high.

= d
{parent=36b}
{scope}

= Solution
{parent=d}

Take the plane of incidence to be the $xy$-plane, so all three electric polarizations point along $z$. Write their signed scalar amplitudes as $E_I,E_R,E_T$. Tangential-$E$ continuity gives
$$
E_I+E_R=E_T.
$$
For each plane wave,
$$
H=\frac1{\mu\omega}k\times E.
$$
Continuity of its tangential component gives
$$
n_-\cos\theta_I(E_I-E_R)
=n_+\cos\theta_T E_T,
$$
where the common proportionality between $|k|$ and $n$ cancels. Solving the two equations gives the <transverse-electric Fresnel reflection coefficient>
$$
\frac{E_R}{E_I}
=\frac{n_-\cos\theta_I-n_+\cos\theta_T}
{n_-\cos\theta_I+n_+\cos\theta_T}.
$$
Using Snell's law to eliminate $\theta_T$,
$$
n_+\cos\theta_T
=\sqrt{n_+^2-n_-^2\sin^2\theta_I}.
$$
Since $\epsilon_+>\epsilon_-$ and $\mu$ is common, $n_+>n_-$. Therefore
$$
(n_+\cos\theta_T)^2-(n_-\cos\theta_I)^2
=n_+^2-n_-^2>0.
$$
The numerator is consequently strictly negative throughout
$0\leq\theta_I\leq\pi/2$, and
$$
\boxed{
\frac{|E_R|}{|E_I|}
=
\frac{
\sqrt{n_+^2-n_-^2\sin^2\theta_I}
-n_-\cos\theta_I}
{
\sqrt{n_+^2-n_-^2\sin^2\theta_I}
+n_-\cos\theta_I}
>0}.
$$
Thus transverse-electric polarization has no Brewster angle in this setting.

Solved by gpt-5.6-sol high.

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

= a
{parent=37e}
{scope}

= Solution
{parent=a}

In the static limit, the supplied linearized Riemann tensor gives
$$
R_{00}=\delta^{ij}R_{i0j0}=-\frac12\nabla^2h_{00}.
$$
The trace of the <Einstein field equations> is $R=-\kappa T$. Here $T\simeq-\rho$, so $R=\kappa\rho$. The $00$ field equation is therefore
$$
R_{00}-\frac12\eta_{00}R=R_{00}+\frac12R=\kappa\rho,
$$
which yields $R_{00}=\kappa\rho/2$. Combining the two expressions for $R_{00}$ gives
$$
-\nabla^2h_{00}=\kappa\rho.
$$

Solved by gpt-5.6-sol high.

= b
{parent=37e}
{scope}

= Solution
{parent=b}

For a slowly moving particle in a static weak field, the spatial part of the <geodesic equation> reduces to
$$
\frac{d^2x^i}{dt^2}=-\Gamma^i{}_{00}.
$$
To first order in the perturbation,
$$
\Gamma^i{}_{00}
=-\frac12\delta^{ij}\partial_jh_{00},
$$
and hence
$$
\frac{dv^i}{dt}=\frac12\delta^{ij}\partial_jh_{00}.
$$
Newtonian acceleration is $-\delta^{ij}\partial_j\Phi$, so $h_{00}=-2\Phi$. Substitution into part (a), followed by the <Poisson equation> $\nabla^2\Phi=4\pi G\rho$, gives
$$
\kappa\rho=-\nabla^2h_{00}=2\nabla^2\Phi=8\pi G\rho.
$$
Thus the <Newtonian limit of general relativity> fixes
$$
\boxed{h_{00}=-2\Phi,\qquad \kappa=8\pi G}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=37e}
{scope}

= Solution
{parent=c}

For a point mass, $\Phi=-GM/r$, and part (b) gives
$$
h_{00}=\frac{2GM}{r}.
$$
The <linearized Ricci tensor and scalar> obey
$$
R=\partial_\mu\partial_\nu h^{\mu\nu}-\mathop{\Box}h.
$$
For a static perturbation with $h_{0i}=0$, this becomes
$$
R=\partial_i\partial_jh_{ij}+\nabla^2h_{00}-\nabla^2h_{ii}.
$$
Away from the source, $\nabla^2h_{00}=0$. For $h_{ij}=f(r)x_i x_j$, one has $h_{ii}=r^2f$, so the two supplied identities give
$$
R=
\left(r^2f''+8rf'+12f\right)
-\left(r^2f''+6rf'+6f\right)
=2rf'+6f.
$$
The vacuum trace equation $R=0$ therefore reduces to
$$
rf'+3f=0,
$$
whose general solution for $r>0$ is
$$
\boxed{f(r)=\frac{C}{r^3}}.
$$
The scalar equation alone leaves $C$ arbitrary. Imposing the remaining vacuum components $R_{ij}=0$ and matching the mass fixes $C=2GM$, as in the <linearized point-mass metric in radial Cartesian form>.

Solved by gpt-5.6-sol high.

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

= a
{parent=38d}
{scope}

= Solution
{parent=a}

The momentum equation is
$$
\rho\left(\partial_tu+(u\cdot\nabla)u\right)
=f+\nabla\cdot\sigma,
\qquad \nabla\cdot u=0.
$$
Taking its scalar product with $u$ gives
$$
\partial_t\left(\frac12\rho|u|^2\right)
+\nabla\cdot\left(\frac12\rho|u|^2u\right)
=u\cdot f+\nabla\cdot(u\cdot\sigma)-\sigma:\nabla u.
$$
For the <Newtonian fluid stress tensor> $\sigma=-pI+2\mu e$, its contraction with the velocity gradient is
$$
\sigma:\nabla u=2\mu e:e,
$$
because incompressibility removes the pressure term and the symmetric <rate-of-strain tensor> is orthogonal to the antisymmetric part of $\nabla u$. Integration over $\mathcal D$ and the divergence theorem therefore give the <kinetic-energy balance for an incompressible Newtonian fluid>:
$$
\frac{d}{dt}\int_{\mathcal D}\frac12\rho|u|^2\,dV
+\int_{\partial\mathcal D}\frac12\rho|u|^2u\cdot n\,dS
=\int_{\mathcal D}u\cdot f\,dV
+\int_{\partial\mathcal D}u\cdot\sigma\cdot n\,dS
-2\mu\int_{\mathcal D}e:e\,dV.
$$
The terms are respectively the rate of change of kinetic energy, outward advective flux of kinetic energy, power supplied by the body force, power supplied by surface traction, and irreversible viscous dissipation. Since $f=-\nabla\psi$, the body-force power may also be written $-\nabla\cdot(\psi u)$ and interpreted as exchange with potential energy.

Solved by gpt-5.6-sol high.

= b
{parent=38d}
{scope}

= Solution
{parent=b}

The linearized kinematic condition at the undisturbed surface is
$$
\partial_t\eta=\partial_z\phi\big|_{z=0}.
$$
For the given complex amplitudes this says
$$
-i\omega A=kB,
\qquad
\boxed{B=-\frac{i\omega}{k}A}.
$$

The velocity is $u=(\phi_x,0,\phi_z)$. Its nonzero complex strain amplitudes are
$$
e_{xx}=-k^2Be^{kz},\qquad
e_{zz}=k^2Be^{kz},\qquad
e_{xz}=e_{zx}=ik^2Be^{kz}.
$$
Using the stated rule for period averages,
$$
\langle e:e\rangle=2k^4|B|^2e^{2kz}.
$$
Thus the mean dissipation per unit horizontal area is
$$
\mathcal D
=2\mu\int_{-\infty}^0\langle e:e\rangle\,dz
=2\mu k^3|B|^2
=2\mu k\omega^2|A|^2
=\boxed{2\mu gk^2|A|^2}.
$$
The mean energy $E=\rho g|A|^2/2$ consequently obeys $dE/dt=-\mathcal D$. Writing $\nu=\mu/\rho$ gives the <viscous decay of a deep-water gravity wave>:
$$
\boxed{\frac{d|A|}{dt}=-2\nu k^2|A|},
\qquad
\boxed{|A(t)|=|A(0)|e^{-2\nu k^2t}}.
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=39a}
{scope}

= Solution
{parent=a}

In the frame in which the shock is stationary, the upstream gas approaches it with speed $V$, while the downstream gas recedes from it with speed $V-u_1$. The <Rankine-Hugoniot conditions for a perfect gas> are therefore
$$
\rho_0V=\rho_1(V-u_1),
$$
$$
p_0+\rho_0V^2
=p_1+\rho_1(V-u_1)^2,
$$
and
$$
\frac{\gamma p_0}{(\gamma-1)\rho_0}+\frac12V^2
=\frac{\gamma p_1}{(\gamma-1)\rho_1}
+\frac12(V-u_1)^2.
$$
These equations express conservation of mass flux, momentum flux, and total specific enthalpy respectively.

Solved by gpt-5.6-sol high.

= b
{parent=39a}
{scope}

= Solution
{parent=b}

When $u_1=V/3$, the downstream speed in the shock frame is $2V/3$. Mass conservation gives
$$
\rho_0V=\rho_1\frac{2V}{3},
$$
so
$$
\boxed{\frac{\rho_1}{\rho_0}=\frac32}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=39a}
{scope}

= Solution
{parent=c}

Momentum conservation and the density ratio from part (b) give
$$
p_1-p_0
=\rho_0V^2-\frac32\rho_0\left(\frac{2V}{3}\right)^2
=\frac13\rho_0V^2.
$$
For $\gamma=3$, energy conservation becomes
$$
\frac{3p_0}{2\rho_0}+\frac12V^2
=\frac{p_1}{\rho_0}+\frac{2}{9}V^2.
$$
Eliminating $\rho_0V^2$ between these equations yields $p_1=4p_0$. Hence
$$
\boxed{\frac{p_1}{p_0}=4}.
$$

Solved by gpt-5.6-sol high.

= d
{parent=39a}
{scope}

= Solution
{parent=d}

The momentum equation from part (c) is
$$
p_1-p_0=\frac13\rho_0V^2.
$$
Since $p_1-p_0=3p_0$, it follows that
$$
\rho_0V^2=9p_0,
\qquad
\boxed{V=3\sqrt{\frac{p_0}{\rho_0}}}.
$$
This completes the <piston-driven shock with specific-heat ratio three> calculation.

Solved by gpt-5.6-sol high.

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

= a
{parent=40d}
{scope}

= Solution
{parent=a}

The <Householder-John theorem> states the following. Let
$$
A=M-N
$$
be a splitting in which $A$ is Hermitian positive definite. If
$$
B=M^*+N
$$
is also Hermitian positive definite, then $M$ is nonsingular and every eigenvalue of $H=M^{-1}N$ has modulus less than one. Hence the associated stationary iteration converges.

First, $M$ must be nonsingular. If $Mx=0$ for some nonzero $x$, then $Nx=-Ax$, and therefore
$$
x^*Bx=x^*(M^*+N)x=-x^*Ax<0,
$$
contradicting the positive definiteness of $B$.

Now let $Hx=\lambda x$ with $x\ne0$. Then $Nx=\lambda Mx$, so
$$
Ax=(1-\lambda)Mx.
$$
Put $a=x^*Ax>0$ and $m=x^*Mx$. Since $\lambda\ne1$,
$$
m=\frac{a}{1-\lambda}.
$$
Consequently,
$$
0<x^*Bx
=\overline m+\lambda m
=a\left(\frac1{1-\overline\lambda}
+\frac{\lambda}{1-\lambda}\right)
=a\frac{1-|\lambda|^2}{|1-\lambda|^2}.
$$
It follows that $|\lambda|<1$. Thus $\rho(H)<1$, proving the theorem.

Solved by gpt-5.6-sol high.

= b
{parent=40d}
{scope}

= Solution
{parent=b}

Write
$$
A=D+L+U,
$$
where $D=\operatorname{diag}(a_1,\ldots,a_n)$ and $L,U$ are the strict lower and upper triangular parts. The <Jacobi method> is
$$
x^{(k+1)}=D^{-1}\left(b-(L+U)x^{(k)}\right),
$$
with iteration matrix
$$
H=-D^{-1}(L+U).
$$

Here $A$ is symmetric, so $U=L^T$. Regard the iteration as the splitting
$$
A=M-N,
\qquad
M=D,
\qquad
N=-(L+L^T).
$$
Let
$$
S=\operatorname{diag}(1,-1,1,-1,\ldots).
$$
Because $A$ is tridiagonal, changing alternating signs reverses the sign of every off-diagonal entry while preserving every diagonal entry. Therefore
$$
M^T+N=D-L-L^T=SAS.
$$
Since $S^T=S=S^{-1}$ and $A$ is positive definite, $SAS$ is positive definite. The <Householder-John theorem> now gives
$$
\rho(H)<1.
$$
The Jacobi iterates therefore converge to the unique solution $A^{-1}b$ for every starting vector, which is the <jacobi convergence for a symmetric positive-definite tridiagonal matrix>.

Solved by gpt-5.6-sol high.