= Paper 4
{scope}

https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2024/paperii_4_2024.pdf

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

= Solution
{parent=1F}

The Möbius <function> is
$$
\mu(n)=\begin{cases}
1,&n=1,\\
(-1)^r,&n\text{ is a product of }r\text{ distinct primes},\\
0,&n\text{ is divisible by a square}.
\end{cases}
$$
Möbius inversion says that
$$
g(n)=\sum_{d\mid n}f(d)
\quad\Longleftrightarrow\quad
f(n)=\sum_{d\mid n}\mu(d)g(n/d).
$$
Now
$$
\sum_{d^k\mid n}\mu(d)
$$
is the sum of $(-1)^{|S|}$ over subsets of the primes whose $k$th powers divide $n$. It is one if there are no such primes and zero otherwise, proving the power-free criterion.

Let
$$
c_n=\sum_{\substack{1\leq d\leq n\\(d,n)=1}}e^{2\pi i d/n}.
$$
Grouping all $n$th roots of unity by their exact order gives
$$
\sum_{m\mid n}c_m=\begin{cases}1,&n=1,\\0,&n>1.
\end{cases}
$$
Möbius inversion therefore yields $c_n=\mu(n)$.

Solved by gpt-5.6-sol high.

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

= a
{parent=2g}
{scope}

= Solution
{parent=a}

For each degree and coefficient bound there are finitely many integer <polynomials>. Hence $\mathbb Z[x]$ is countable. Every nonzero <polynomial> has finitely many roots, and every algebraic number is a root of one of these <polynomials>. A countable union of finite sets is countable.

Solved by gpt-5.6-sol high.

= b
{parent=2g}
{scope}

= Solution
{parent=b}

Choose a subsequence $a_{n_j}$ so sparse that
$$
\sum_{\ell>j}a_{n_\ell}<a_{n_j}
$$
for every $j$; this is possible because the tails of the convergent positive <series> tend to zero. Distinct binary <sequences> $(\varepsilon_j)$ then have distinct sums $\sum_j\varepsilon_ja_{n_j}$: at their first disagreement, the corresponding term exceeds the entire remaining tail. Thus these subsums form an uncountable set. The algebraic numbers are countable, so at least one subsum is transcendental. Set $\theta_{n_j}=\varepsilon_j$ and all other $\theta_n=0$.

Solved by gpt-5.6-sol high.

= c
{parent=2g}
{scope}

= Solution
{parent=c}

Let
$$
x=\sum_{j=1}^\infty\frac{\theta_j}{j!}
$$
and suppose $x=p/q$. Choose $n\geq\max(q,2)$ such that some $\theta_j=1$ for $j>n$. Then $n!x$ is an integer, while its first $n$ terms also sum to an integer. Their difference is strictly positive and satisfies
$$
0<n!\sum_{j>n}\frac{\theta_j}{j!}
\leq\sum_{r=1}^\infty\frac1{(n+1)^r}<1.
$$
This cannot be the difference of two integers, so $x$ is irrational.

Solved by gpt-5.6-sol high.

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

= a
{parent=3k}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

Because decryption determines $M$ from $(C,K)$,
$$
H(M\mid C,K)=0.
$$
The chain rule and monotonicity give
$$
H(M\mid C)\leq H(M,K\mid C)
=H(K\mid C)+H(M\mid C,K)=H(K\mid C).
$$

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

The unicity distance is the least ciphertext length for which the key is determined, or in the usual approximation the length at which the expected number of spurious keys falls to about zero. Assume equiprobable keys, a stationary plaintext source of <information entropy> rate $H_E$, and ciphertext symbols that are approximately uniform on $A$. The language redundancy per symbol is
$$
D=\log_2|A|-H_E.
$$
The key equivocation is then approximated by
$$
H(K\mid C_1,ldots,C_n)\simeq\log_2|K|-nD.
$$
Hence the classical closed-form estimate is
$$
n_0\simeq\frac{\log_2|K|}{\log_2|A|-H_E}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=3k}
{scope}

= Solution
{parent=b}

<Entropy> is nonnegative. Gibbs' inequality also gives
$$
H(X_1,ldots,X_n)\leq\log_2|A^n|=n\log_2 27.
$$
Divide by $n$ and pass to the assumed <limit> to obtain
$$
0\leq H_E\leq\log_2 27.
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=4j}
{scope}

= Solution
{parent=a}

A homomorphism $f:D\to D'$ is a map $f:Q\to Q'$ satisfying
$$
f(q_0)=q'_0,
\qquad f(\delta(q,a))=\delta'(f(q),a)
$$
for every state $q$ and symbol $a$, together with
$$
q\in F\quad\Longleftrightarrow\quad f(q)\in F'.
$$

Solved by gpt-5.6-sol high.

= b
{parent=4j}
{scope}

= Solution
{parent=b}

Induction on word length gives
$$
f(\delta^*(q_0,w))=\delta'^*(q'_0,w).
$$
The final-state condition then says that the state reached by $D$ is accepting exactly when the state reached by $D'$ is accepting. Therefore $w\in\mathcal L(D)$ exactly when $w\in\mathcal L(D')$, and the languages are equal.

Solved by gpt-5.6-sol high.

= c
{parent=4j}
{scope}

= i
{parent=c}
{scope}

= Solution
{parent=i}

The statement is false. Given any $D$, adjoin arbitrarily many unreachable states whose transitions remain among those new states, choosing their accepting status consistently away from the embedded copy. This produces infinitely many pairwise nonisomorphic finite deterministic automata $D'$ containing an injective homomorphic copy of $D$.

Solved by gpt-5.6-sol high.

= ii
{parent=c}
{scope}

= Solution
{parent=ii}

The statement is true; in fact any fixed finite $D$ works. If $D'\leq D$, then $D'$ has at most $|Q|$ states. For a fixed finite alphabet, there are only finitely many transition tables, initial states, and accepting subsets on at most $|Q|$ labelled states, hence only finitely many isomorphism classes.

Solved by gpt-5.6-sol high.

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

= a
{parent=5l}
{scope}

= Solution
{parent=a}

Conditionally on $x_i$,
$$
\mathbb P(Z_i=1\mid x_i)
=\Phi\left(\frac{\mu+x_i^T\beta-\tau}{\sigma}\right).
$$
Thus this is a probit regression with linear predictor
$$
\eta_i=\alpha+x_i^T\gamma,
\qquad \alpha=\frac{\mu-\tau}{\sigma},
\qquad \gamma=\frac\beta\sigma.
$$
Fit $(\alpha,\gamma)$ by maximum likelihood for Bernoulli observations, then recover
$$
\widehat\mu=\tau+\sigma\widehat\alpha,
\qquad \widehat\beta=\sigma\widehat\gamma.
$$

Solved by gpt-5.6-sol high.

= b
{parent=5l}
{scope}

= Solution
{parent=b}

Write $\widetilde x_*=(1,x_*^T)^T$, $\widehat\vartheta=(\widehat\alpha,\widehat\gamma^T)^T$, and let $\widehat V$ estimate the asymptotic covariance of $\widehat\vartheta$. Then
$$
\widehat\eta_* =\widetilde x_*^T\widehat\vartheta,
\qquad
\widehat s_*^2=\widetilde x_*^T\widehat V\widetilde x_*.
$$
An asymptotic confidence interval for the mean binary response $p_*=\Phi(\eta_*)$ is
$$
\left[
\Phi(\widehat\eta_*-z_{1-\alpha/2}\widehat s_*),
\Phi(\widehat\eta_*+z_{1-\alpha/2}\widehat s_*)
\right].
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=6a}
{scope}

= Solution
{parent=a}

Of the $a_t$ immature cells, a fraction $\lambda$ leaves by maturation and the rest remains immature. Of the $b_t$ mature cells, a fraction $\mu$ divides and is replaced by four immature cells, while a fraction $k$ dies; the remainder stays mature. These contributions give exactly
$$
a_{t+1}=(1-\lambda)a_t+4\mu b_t,
\qquad
b_{t+1}=\lambda a_t+(1-\mu-k)b_t.
$$

Solved by gpt-5.6-sol high.

= b
{parent=6a}
{scope}

= Solution
{parent=b}

Once a cell is mature, its eventual competing outcomes are division and death. Conditional on one of them occurring, division has probability $\mu/(\mu+k)$ and creates four immature offspring. Thus the expected lifetime offspring number is
$$
R_0=\frac{4\mu}{\mu+k},
$$
assuming maturation eventually occurs.

Solved by gpt-5.6-sol high.

= c
{parent=6a}
{scope}

= Solution
{parent=c}

Let $N_t=a_t+b_t$. Adding the two recurrences gives
$$
N_{t+1}-N_t=(3\mu-k)b_t.
$$
Since $b_t\geq0$, total population can increase only if $3\mu>k$. This is exactly
$$
R_0=\frac{4\mu}{\mu+k}>1,
$$
so the population-growth and offspring criteria agree.

Solved by gpt-5.6-sol high.

= d
{parent=6a}
{scope}

= Solution
{parent=d}

For $t\geq T$, set $\lambda=0$. Mature cells then decay geometrically:
$$
b_{T+j}=(1-\mu-k)^jb_T,
$$
while their divisions add $4\mu b_{T+j}$ to the permanently immature population. If $\mu+k>0$, mature cells disappear and
$$
N_\infty=a_T+4\mu b_T\sum_{j=0}^\infty(1-\mu-k)^j
=a_T+\frac{4\mu}{\mu+k}b_T.
$$
If $\mu=k=0$, no further change occurs and the <limit> is $a_T+b_T$.

Solved by gpt-5.6-sol high.

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

= a
{parent=7d}
{scope}

= Solution
{parent=a}

For $w''+P(z)w'+Q(z)w=0$, a finite point $z_0$ is regular singular when
$$
(z-z_0)P(z),
\qquad (z-z_0)^2Q(z)
$$
extend analytically to $z_0$. Inspection shows that $a,b,c$ satisfy these conditions. There are no other finite singularities; with the Fuchs relation on the six exponents, the point at infinity is ordinary in the corresponding sphere description.

Solved by gpt-5.6-sol high.

= b
{parent=7d}
{scope}

= Solution
{parent=b}

At $z=a$, insert $w\sim(z-a)^r$. The coefficient of the most singular power gives the indicial equation
$$
(r-\alpha)(r-\alpha')=0.
$$
Thus the two local exponents are $\alpha,\alpha'$. Cyclically, the exponents at $b$ are $\beta,\beta'$ and those at $c$ are $\gamma,\gamma'$. The assumed nonintegral exponent differences give two independent Frobenius solutions without logarithmic resonance at each point.

Solved by gpt-5.6-sol high.

= c
{parent=7d}
{scope}

= Solution
{parent=c}

The Papperitz symbol is
$$
P\left\{
\begin{matrix}
a&b&c&\\
\alpha&\beta&\gamma&z\\
\alpha'&\beta'&\gamma'&
\end{matrix}
\right\}.
$$
It records the three regular singular points and the two characteristic exponents at each.

Solved by gpt-5.6-sol high.

= d
{parent=7d}
{scope}

= Solution
{parent=d}

Set $a=0$, $b=1$, $c=\infty$, $\alpha=\beta=0$. Comparing the $w'$ coefficient gives
$$
\alpha'=1-C,
\qquad \beta'=C-A-B,
$$
and comparison of the $w$ coefficient gives the exponents $A,B$ at infinity. Hence
$$
P\left\{
\begin{matrix}
0&1&\infty&\\
0&0&A&z\\
1-C&C-A-B&B&
\end{matrix}
\right\}.
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=8e}
{scope}

= Solution
{parent=a}

The phase space of an $n$-degree-of-freedom mechanical system is the $2n$-dimensional space of canonical positions and <momenta> $(q_1,\ldots,q_n,p_1,\ldots,p_n)$, geometrically the cotangent bundle of configuration space. A point specifies a complete instantaneous state.

Solved by gpt-5.6-sol high.

= b
{parent=8e}
{scope}

= Solution
{parent=b}

For the canonical Poisson bracket,
$$
\{f,g\}=\sum_i\left(\frac{\partial f}{\partial q_i}\frac{\partial g}{\partial p_i}
-\frac{\partial f}{\partial p_i}\frac{\partial g}{\partial q_i}\right),
$$
Hamilton's equations are
$$
\dot q_i=\{q_i,H\}=\frac{\partial H}{\partial p_i},
\qquad
\dot p_i=\{p_i,H\}=-\frac{\partial H}{\partial q_i}.
$$
More generally $\dot f=\{f,H\}+\partial f/\partial t$.

Solved by gpt-5.6-sol high.

= c
{parent=8e}
{scope}

= Solution
{parent=c}

With $H=p_jp_j/(2m)$, the stated brackets give
$$
\dot q_i=\{q_i,H\}=\frac{p_i}{m},
$$
and
$$
\dot p_i=\{p_i,H\}
=\frac em\epsilon_{ijk}p_jB_k
=e(\dot{\mathbf q}\times\mathbf B)_i.
$$
Thus $m\ddot{\mathbf q}=e\dot{\mathbf q}\times\mathbf B$, the Lorentz-force equation with zero <electric field>. The magnetic interaction has been moved from the <Hamiltonian> into the noncanonical symplectic structure.

Solved by gpt-5.6-sol high.

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

= a
{parent=9d}
{scope}

= Solution
{parent=a}

From $E=\rho V$, $P=\rho/3$, and $dE=T,dS-P,dV$,
$$
T,dS=V,d\rho+\frac43\rho,dV.
$$
Using $\rho=4\sigma T^4/c$ gives
$$
dS=\frac{16\sigma}{3c}\left(T^3dV+3T^2VdT\right)
=d\left(\frac{16\sigma}{3c}VT^3\right).
$$
Therefore
$$
S=\frac{16\sigma}{3c}VT^3,
\qquad s=\frac SV=\frac{16\sigma}{3c}T^3,
$$
up to an irrelevant additive <entropy> constant.

Solved by gpt-5.6-sol high.

= b
{parent=9d}
{scope}

= Solution
{parent=b}

When $\Gamma\gg H$, reactions maintain thermal and chemical equilibrium, so the expansion is quasistatic and adiabatic. With no change in the equilibrium degrees of freedom, comoving <entropy> and photon number are conserved. Since
$$
s\propto T^3,
\qquad sa^3=\text{constant},
$$
one gets $T^3a^3=\text{constant}$ and hence $T\propto a^{-1}$.

Solved by gpt-5.6-sol high.

= c
{parent=9d}
{scope}

= Solution
{parent=c}

Before electron-positron annihilation, the electromagnetic plasma has
$$
g_{*s}=2+\frac78(2+2)=\frac{11}{2},
$$
whereas afterward only the two photon polarizations remain, so $g_{*s}=2$. <Entropy> conservation in the decoupled electromagnetic sector gives
$$
\frac{11}{2}T_{\rm before}^3a^3=2T_\gamma^3a^3.
$$
Neutrinos receive none of this <entropy> and continue cooling as $T_\nu\propto a^{-1}$. Therefore
$$
\frac{T_\nu}{T_\gamma}=\left(\frac4{11}\right)^{1/3}.
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=10e}
{scope}

= Solution
{parent=a}

Write $|\Phi^+\rangle=2^{-1/2}\sum_{j=0}^1|j\rangle|j\rangle$. Then
$$
(I\otimes A)|\Phi^+\rangle
=\frac1{\sqrt2}\sum_{i,j}A_{ij}|j\rangle|i\rangle
=\frac1{\sqrt2}\sum_{i,j}(A^T)_{ji}|j\rangle|i\rangle
=(A^T\otimes I)|\Phi^+\rangle.
$$

Solved by gpt-5.6-sol high.

= b
{parent=10e}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

Since
$$
|\psi_2\rangle=-\sin\theta|0\rangle+\cos\theta|1\rangle,
$$
application of $X$ gives $\cos\theta|0\rangle-\sin\theta|1\rangle$, and subsequent application of $Z$ gives
$$
\cos\theta|0\rangle+\sin\theta|1\rangle=|\psi_1\rangle.
$$
Thus $ZX|\psi_2\rangle=|\psi_1\rangle$.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

The <matrix> $U_\theta$ is real orthogonal. Part (a) gives
$$
(U_\theta\otimes U_\theta)|\Phi^+\rangle
=(I\otimes U_\theta U_\theta^T)|\Phi^+\rangle
=|\Phi^+\rangle.
$$
Expanding the left side in the computational <basis> yields
$$
|\Phi^+\rangle=\frac1{\sqrt2}
\left(|\psi_1\rangle|\psi_1\rangle+|\psi_2\rangle|\psi_2\rangle\right).
$$

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

Using the decomposition from part (ii), $U_\theta^{-1}$ maps Alice's $|\psi_1\rangle$ to $|0\rangle$ and $|\psi_2\rangle$ to $|1\rangle$. Alice obtains $0$ or $1$, each with probability $1/2$. Conditional on outcome $0$, Bob has $|\psi_1\rangle$; conditional on outcome $1$, Bob has $|\psi_2\rangle$.

Solved by gpt-5.6-sol high.

= iv
{parent=b}
{scope}

= Solution
{parent=iv}

Alice performs the operation and measurement from part (iii) and sends the one-bit outcome to Bob. If the bit is $0$, Bob does nothing. If it is $1$, Bob applies $ZX$, which maps $|\psi_2\rangle$ to $|\psi_1\rangle$ by part (i). In either case Bob finishes with the known target state $|\psi_1\rangle$, using one shared Bell pair and one classical bit.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=11F}

Set
$$
p_{-2}=0, p_{-1}=1,\qquad q_{-2}=1, q_{-1}=0,
$$
and recursively
$$
p_n=a_np_{n-1}+p_{n-2},\qquad q_n=a_nq_{n-1}+q_{n-2}.
$$
Then $p_n/q_n=[a_0,ldots,a_n]$. The <determinant> identity
$$
p_nq_{n-1}-p_{n-1}q_n=(-1)^{n-1}
$$
follows by induction. Writing the remaining complete quotient as $\theta_{n+1}>1$ gives
$$
\theta=\frac{p_n\theta_{n+1}+p_{n-1}}
{q_n\theta_{n+1}+q_{n-1}},
$$
so $\theta$ lies strictly between consecutive convergents and their order alternates. For odd $n$,
$$
\frac{p_{n-1}}{q_{n-1}}<\theta<\frac{p_n}{q_n}.
$$
Moreover $|\theta-p_n/q_n|<1/(q_nq_{n+1})$, and $q_n\to\infty$, proving convergence.

The continued-fraction algorithm gives
$$
\sqrt7=[2;\overline{1,1,1,4}].
$$

Solved by gpt-5.6-sol high.

= i
{parent=11f}
{scope}

= Solution
{parent=i}

The convergents begin
$$
2, 3, \frac52, \frac83, \frac{37}{14},\ldots.
$$
Since
$$
8^2-7\cdot3^2=1,
$$
a strictly positive solution is $(x,y)=(8,3)$.

Solved by gpt-5.6-sol high.

= ii
{parent=11f}
{scope}

= Solution
{parent=ii}

Reducing $x^2-7y^2=-1$ modulo seven gives $x^2\equiv-1\equiv6\pmod7$. The quadratic residues modulo seven are $0,1,2,4$, so no solution exists.

Solved by gpt-5.6-sol high.

= iii
{parent=11f}
{scope}

= Solution
{parent=iii}

Reducing $x^2-7y^2=5$ modulo seven gives $x^2\equiv5\pmod7$, again impossible because five is not a quadratic residue modulo seven.

Solved by gpt-5.6-sol high.

= iv
{parent=11f}
{scope}

= Solution
{parent=iv}

The denominators of the convergents are
$$
1,1,2,3,14,17,31,48,\ldots.
$$
They increase strictly from $q_2$ onward, and none equals $13$. Equivalently, $34/13=[2;1,1,1,1,2]$, which ceases to agree with $[2;\overline{1,1,1,4}]$ after the third partial quotient. Hence there is no positive index $x$ with $p_x/q_x=34/13$.

Solved by gpt-5.6-sol high.

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

= a
{parent=12g}
{scope}

= Solution
{parent=a}

Moving from $A_1$ to $A_n$, the colour changes an odd number of times because the endpoints have opposite colours: every change toggles the current colour, while every nonchange preserves it. Formally, encode red and green by zero and one; the sum modulo two of adjacent differences is the endpoint difference, which is one.

Solved by gpt-5.6-sol high.

= b
{parent=12g}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

Count red-green segments modulo two. On $DE$, part (a) says their number is odd; the other two outer sides contain no red-green segment. Every interior segment belongs to two small triangles and therefore contributes zero modulo two, while each boundary segment belongs to one. Thus the sum, over all small triangles, of their numbers of red-green edges is odd.

A triangle with all three colours has exactly one red-green edge. A triangle using only red and green has zero or two, and any other nontrichromatic triangle has zero red-green edges. Hence the parity sum counts precisely the trichromatic triangles modulo two, proving that their number is odd.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

Not necessarily, once the subdivision has at least two segments per side. Colour $D,E,F$ red, green, blue respectively and colour every other grid vertex red. No small triangle contains both exceptional vertices $E$ and $F$, and every small triangle therefore uses at most two colours. For the unsubdivided triangle the assertion is of course true.

Solved by gpt-5.6-sol high.

= c
{parent=12g}
{scope}

= i
{parent=c}
{scope}

= Solution
{parent=i}

The statement is false. The constant map to $E$ satisfies it: $E\in I$ and also $E\in J\subset J\cup K$.

Solved by gpt-5.6-sol high.

= ii
{parent=c}
{scope}

= Solution
{parent=ii}

The statement is true. At the vertices the side conditions force
$$
g(D)=D,\qquad g(E)=E,\qquad g(F)=F.
$$
On each side, $g$ remains in that same side, so the boundary restriction is homotopic within the boundary to the identity and has degree one. If it extended continuously over the filled triangle with image in the boundary circle, that degree-one boundary map would be null-homotopic, which is impossible. Equivalently, a sufficiently fine simplicial approximation would contradict the parity form of Sperner's lemma proved in part (b).

Solved by gpt-5.6-sol high.

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

= a
{parent=13l}
{scope}

= Solution
{parent=a}

With $p_i=\mathbb P(\mathrm{dis}_i=1)$, the fitted logistic model is
$$
\log\frac{p_i}{1-p_i}
=\beta_0+\beta_1\,mathrm{pollution}_i
+\beta_2\mathbf1_{\{\mathrm{risk}_i=2\}}.
$$
Risk category 1 is the reference level, so its effect is included in the intercept; adding a separate coefficient for both levels would make the design <matrix> linearly dependent.

Solved by gpt-5.6-sol high.

= b
{parent=13l}
{scope}

= Solution
{parent=b}

Holding risk category fixed, increasing pollution by one unit multiplies the disease odds by
$$
e^{0.18243}\simeq1.20.
$$
Thus the fitted odds increase by about twenty percent per unit of pollution.

Solved by gpt-5.6-sol high.

= c
{parent=13l}
{scope}

= Solution
{parent=c}

There are $53$ observations. The null model estimates one intercept, so its residual degrees of freedom are $53-1=52$. The fitted model estimates three coefficients, so its residual degrees of freedom are $53-3=50$.

Solved by gpt-5.6-sol high.

= d
{parent=13l}
{scope}

= Solution
{parent=d}

Fit the intercept-only model with
```
glm(dis ~ 1, family = binomial, data = disease)
```
Its deviance is $73.5$. Since it has one fitted coefficient, its AIC on the same convention is
$$
73.5+2=75.5.
$$

Solved by gpt-5.6-sol high.

= e
{parent=13l}
{scope}

= Solution
{parent=e}

The first likelihood-ratio test compares
$$
H_0:\beta_{\rm pollution}=\beta_{\rm risk2}=0
$$
with the alternative that at least one is nonzero. The deviance drop $73.5-55.3=18.2$ is compared with $\chi^2_2$, giving $p=0.000114$; there is strong evidence that the covariates improve the model.

The second test compares the additive model with
$$
H_0:\beta_{\rm pollution:risk}=0
$$
against a model with a pollution-by-risk interaction. Its $\chi^2_1$ likelihood-ratio p-value is $0.871$, so there is no evidence for an interaction.

Solved by gpt-5.6-sol high.

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

= a
{parent=14a}
{scope}

= Solution
{parent=a}

For a positive homogeneous equilibrium,
$$
v=1+u,\qquad \alpha u=\beta v,
$$
so
$$
u_* =\frac\beta{\alpha-\beta},\qquad
v_* =\frac\alpha{\alpha-\beta},
$$
requiring $\alpha>\beta$. The reaction Jacobian there is
$$
J=\frac1{\alpha-\beta}
\begin{pmatrix}
\beta&-\beta\\
\alpha\beta&-\beta^2
\end{pmatrix}.
$$
Its <determinant> is $\beta^2/(\alpha-\beta)>0$, while its trace is $\beta(1-\beta)/(\alpha-\beta)$. Stability therefore requires
$$
\beta>1,\qquad \alpha>\beta.
$$
This is the region above the line $\alpha=\beta$ and to the right of $\beta=1$ in the positive quadrant.

Solved by gpt-5.6-sol high.

= b
{parent=14a}
{scope}

= Solution
{parent=b}

For $q=k^2$, the mode <matrix> is $J-q\operatorname{diag}(D_1,D_2)$. Its trace remains negative. Its <determinant> is
$$
D_1D_2q^2-\frac{D_2\beta(1-\lambda\beta)}{\alpha-\beta}q
+\frac{\beta^2}{\alpha-\beta},
\qquad \lambda=\frac{D_1}{D_2}.
$$
It is negative for some $q>0$ exactly when
$$
1-\lambda\beta>0,
\qquad
(1-\lambda\beta)^2>4\lambda(\alpha-\beta).
$$
Together with homogeneous stability, the spatial-instability region is
$$
\beta>1,\qquad 0<\lambda<\frac1\beta,\qquad
\beta<\alpha<\beta+\frac{(1-\lambda\beta)^2}{4\lambda}.
$$
For fixed $\beta$, this is the region below the last curve, above $\alpha=\beta$, and left of $\lambda=1/\beta$.

Solved by gpt-5.6-sol high.

= c
{parent=14a}
{scope}

= Solution
{parent=c}

If $D_1=D_2=D$, the linearized <matrix> for wavenumber $k$ is
$$
J-Dk^2I.
$$
Its <eigenvalues> are those of the homogeneous Jacobian shifted left by $Dk^2$, so a stable homogeneous equilibrium remains stable for every spatial mode. Thus equal diffusivities cannot produce a Turing instability. In part (b), equal diffusivities mean $\lambda=1$; because homogeneous stability requires $\beta>1$, the necessary condition $1-\lambda\beta>0$ fails, consistently ruling out instability.

Solved by gpt-5.6-sol high.

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

= a
{parent=15e}
{scope}

= Solution
{parent=a}

The <kinetic energy> and squared angular-momentum magnitude are
$$
T=\frac12(I_1\omega_1^2+I_2\omega_2^2+I_3\omega_3^2),
\qquad
L^2=I_1^2\omega_1^2+I_2^2\omega_2^2+I_3^2\omega_3^2.
$$
Differentiate and substitute Euler's equations. In $\dot T$, the coefficient of $\omega_1\omega_2\omega_3$ is
$$
(I_2-I_3)+(I_3-I_1)+(I_1-I_2)=0.
$$
The analogous weighted sum in $dL^2/dt$ also cancels, so both quantities are conserved.

Solved by gpt-5.6-sol high.

= b
{parent=15e}
{scope}

= Solution
{parent=b}

The initial relation implies $L^2=2I_2T$, and conservation preserves it. With $I_3=I_1+I_2$, this identity reduces to
$$
\omega_3^2=\frac{I_2-I_1}{I_1+I_2}\omega_1^2.
$$
The energy then gives
$$
\omega_2^2=\frac{2T}{I_2}-\omega_1^2.
$$
The first Euler equation is $\dot\omega_1=-\omega_2\omega_3$. Squaring and substituting the preceding identities yields
$$
\dot\omega_1^2=\left(\frac{2T}{I_2}-\omega_1^2\right)
\frac{I_2-I_1}{I_2+I_1}\omega_1^2.
$$

Solved by gpt-5.6-sol high.

= c
{parent=15e}
{scope}

= Solution
{parent=c}

Set
$$
A=\sqrt{\frac{2T}{I_2}},
\qquad
\Omega=A\sqrt{\frac{I_2-I_1}{I_2+I_1}}
=\sqrt{\frac{2T(I_2-I_1)}{I_2(I_1+I_2)}}.
$$
Then $\omega_1=A\operatorname{sech}(\Omega t)$ satisfies the squared equation and the stated initial condition after choosing the origin of time at the maximum of $\omega_1$.

Solved by gpt-5.6-sol high.

= d
{parent=15e}
{scope}

= Solution
{parent=d}

With compatible signs,
$$
\omega_2=A\tanh(\Omega t),
\qquad
\omega_3=\Omega\operatorname{sech}(\Omega t).
$$
Thus as $t\to\infty$, $\omega_1$ and $\omega_3$ vanish while $\omega_2\to A$. The body approaches steady rotation about the intermediate principal axis; the full separatrix connects the two opposite intermediate-axis rotations.

Solved by gpt-5.6-sol high.

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

= a
{parent=16i}
{scope}

= Solution
{parent=a}

The rank is defined recursively by
$$
\operatorname{rank}(x)=\sup\{\operatorname{rank}(y)+1:y\in x\}.
$$
For a nonzero <limit> ordinal $\alpha$, every <function> $f:\alpha\to\alpha$, viewed as a set of ordered pairs, has rank $\alpha$, and the set of all such <functions> consequently has rank $\alpha+1$.

The von Neumann hierarchy is
$$
V_0=\varnothing,\qquad V_{\gamma+1}=\mathcal P(V_\gamma),
\qquad V_\lambda=\bigcup_{\gamma<\lambda}V_\gamma
$$
for <limit> $\lambda$. Foundation permits induction down the membership relation and shows that the ranks of all elements of a set form a set of ordinals. If $\rho=\operatorname{rank}(x)$, then every element of $x$ lies in $V_\rho$, so $x\subseteq V_\rho$ and hence $x\in V_{\rho+1}$. Thus every set occurs in the hierarchy.

Solved by gpt-5.6-sol high.

= b
{parent=16i}
{scope}

= Solution
{parent=b}

Define $\aleph_0=|\omega|$, let $\aleph_{\alpha+1}$ be the least cardinal greater than $\aleph_\alpha$, and at a <limit> take the least cardinal above all earlier $\aleph_\gamma$. In ZFC every set is well-orderable, so every infinite cardinal is an initial ordinal and equals a unique $\aleph_\alpha$.

By transfinite induction on infinite well-ordered cardinals $\kappa$, order $\kappa\times\kappa$ first by $\max(\xi,\eta)$ and then lexicographically. Every proper initial segment has cardinal below $\kappa$, using the inductive hypothesis for smaller infinite cardinals. Hence this well-order has cardinal at most $\kappa$, while the diagonal gives the reverse inequality, so $\kappa^2=\kappa$. Therefore, for $\aleph_\alpha\leq\aleph_\beta$,
$$
\aleph_\alpha+\aleph_\beta
=\aleph_\alpha\aleph_\beta=\aleph_\beta.
$$

Finally suppose nonempty $x$ had a set $y$ containing every set equinumerous with $x$. For every ordinal $\gamma$, replacing each $z\in x$ by a tagged ordered pair $(\gamma,z)$ gives a set $t_\gamma$ equinumerous with $x$ whose rank is at least $\gamma$. Then every $t_\gamma$ would belong to $y$, so the ranks of members of $y$ would be unbounded in the ordinals, contradicting the ordinal rank of $y$. This proves the displayed sentence.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=17I}

A strongly regular graph with parameters $(k,a,b)$ is $k$-regular, with every adjacent pair having $a$ common neighbours and every distinct nonadjacent pair having $b$ common neighbours. Its adjacency <matrix> satisfies
$$
A^2=(k-b)I+(a-b)A+bJ.
$$
On the orthogonal complement of the all-one <vector>, the two possible <eigenvalues> are
$$
r,s=\frac{a-b\pm\sqrt{(a-b)^2+4(k-b)}}2.
$$
Using $1+m_r+m_s=n$ and $k+m_rr+m_ss=0$ gives exactly the two displayed expressions for $m_r,m_s$. They are eigenspace dimensions and hence integers, proving the rationality condition.

The Petersen graph has parameters $(3,0,1)$, so its <spectrum> is
$$
3^{(1)},\qquad1^{(5)},\qquad(-2)^{(4)}.
$$
If three Petersen graphs partitioned $E(K_{10})$, their adjacency <matrices> would satisfy $A+B+C=J-I$. The five-dimensional eigenvalue-one spaces of $A$ and $B$ inside the nine-dimensional space $\mathbf1^\perp$ intersect nontrivially. For a nonzero common <vector> $v$, $Av=Bv=v$ and therefore
$$
Cv=(J-I-A-B)v=-3v,
$$
contradicting the Petersen <spectrum>. No such partition exists.

Solved by gpt-5.6-sol high.

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

= a
{parent=18h}
{scope}

= Solution
{parent=a}

Let
$$
f=(X^2-2)(X^2-3),
\qquad
g=X^4-10X^2+1.
$$
Both have splitting field $\mathbb Q(\sqrt2,\sqrt3)$. On the roots of $f$, the Galois <group> has two orbits of size two and embeds as
$$
\{1,(12),(34),(12)(34)\}.
$$
The roots of $g$, namely $\pm\sqrt2\pm\sqrt3$, form one orbit and give the regular Klein-four <subgroup>
$$
\{1,(12)(34),(13)(24),(14)(23)\}.
$$
The <subgroups> are not conjugate because their orbit decompositions differ.

Solved by gpt-5.6-sol high.

= b
{parent=18h}
{scope}

= Solution
{parent=b}

Let $F$ be the splitting field over $\mathbb F_p$ of $X^q-X$. Its <derivative> is $-1$, so it has $q$ distinct roots. The roots are closed under addition, subtraction, multiplication, and inversion, using $(x+y)^q=x^q+y^q$ and $(xy)^q=x^qy^q$. They therefore form a field with $q$ elements.

Every field with $q$ elements has multiplicative <group> of order $q-1$, so every element satisfies $x^q=x$; it is therefore a splitting field of the same <polynomial> and is unique up to isomorphism. If an irreducible factor over $\mathbb F_p$ has a root $\alpha\in\mathbb F_q$, its degree is $[\mathbb F_p(\alpha):\mathbb F_p]$, which divides $d$ by the tower law. In particular it is at most $d$.

Solved by gpt-5.6-sol high.

= c
{parent=18h}
{scope}

= Solution
{parent=c}

Dedekind's factorization/Frobenius theorem says that for a prime not dividing the discriminant, the degrees of the distinct irreducible factors modulo $p$ give the cycle lengths of an element of the Galois <group>.

The <polynomial> is irreducible over $\mathbb Q$ by the rational-root test followed by a comparison of possible monic quadratic factors. Its discriminant is $547^2$, a square, so its transitive Galois <group> lies in $A_4$. Modulo five,
$$
f(X)\equiv X(X^3-X-2),
$$
and the cubic has no root in $\mathbb F_5$, giving a three-cycle. A transitive <subgroup> of $A_4$ containing a three-cycle cannot be the Klein four <group>; hence
$$
\operatorname{Gal}(f/\mathbb Q)\cong A_4.
$$

Solved by gpt-5.6-sol high.

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

= Solution
{parent=19H}

The topological <group> $S^1=\{z\in\mathbb C:|z|=1\}$ has complex multiplication and the subspace topology. It is compact and abelian, so every finite-dimensional irreducible complex representation is one-dimensional. A continuous character lifts along $t\mapsto e^{it}$ to a continuous homomorphism $\mathbb R\to\mathbb C^\times$, hence has the form $t\mapsto e^{i\lambda t}$. Periodicity by $2\pi$ forces $\lambda=n\in\mathbb Z$, giving precisely $z\mapsto z^n$.

Solved by gpt-5.6-sol high.

= a
{parent=19h}
{scope}

= Solution
{parent=a}

Every element of $SU(2)$ is unitarily conjugate to an element of the maximal torus
$$
T=\{\operatorname{diag}(z,z^{-1}):z\in S^1\}.
$$
Since characters are class <functions>, $\chi_V$ is determined on $T$. The restriction decomposes into finitely many torus weights, so
$$
\chi_V(z)=\sum_{n\in\mathbb Z}a_nz^n,
\qquad a_n\in\mathbb Z_{\geq0},
\qquad\sum_na_n=\dim V.
$$
A Weyl-group element conjugates $\operatorname{diag}(z,z^{-1})$ to $\operatorname{diag}(z^{-1},z)$, so $\chi_V(z)=\chi_V(z^{-1})$ and $a_n=a_{-n}$.

Solved by gpt-5.6-sol high.

= b
{parent=19h}
{scope}

= Solution
{parent=b}

For the $(n+1)$-dimensional irreducible,
$$
\chi_{V_n}(z)=z^n+z^{n-2}+\cdots+z^{-n}.
$$
The Clebsch--Gordan rule gives
$$
V_4\otimes V_4\cong V_8\oplus V_6\oplus V_4\oplus V_2\oplus V_0.
$$
The even total-spin summands are symmetric and the odd ones alternating, hence
$$
S^2V_4\cong V_8\oplus V_4\oplus V_0,
\qquad
\Lambda^2V_4\cong V_6\oplus V_2.
$$

Solved by gpt-5.6-sol high.

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

= Solution
{parent=20F}

For a nonzero <ideal> $I\subset O_K$, its norm is
$$
N(I)=|O_K/I|.
$$
For $d=[K:\mathbb Q]$ and embeddings $\sigma_1,\ldots,\sigma_d$, define
$$
\operatorname{disc}(\alpha_1,\ldots,\alpha_d)
=\det(\sigma_i(\alpha_j))^2.
$$
The field discriminant is the discriminant of any <integral> <basis> of $O_K$. If $\alpha_1,\ldots,\alpha_d$ is a $\mathbb Z$-basis of $I$, then
$$
N(I)^2=\frac{|\operatorname{disc}(\alpha_1,\ldots,\alpha_d)|}{|D_K|}.
$$

Multiplication by $\alpha$ on an <integral> <basis> has integer <matrix> $M_\alpha$. Its image lattice is $\alpha O_K$, so
$$
N(\alpha O_K)=|\det M_\alpha|.
$$
The <determinant> of this $\mathbb Q$-linear map is the field norm $N_{K/\mathbb Q}(\alpha)$, proving
$$
N(\alpha O_K)=|N_{K/\mathbb Q}(\alpha)|.
$$

Solved by gpt-5.6-sol high.

= a
{parent=20f}
{scope}

= Solution
{parent=a}

If $Q\mid PO_L$, then $PO_L\subseteq Q$, so $P\subseteq Q\cap O_K$. The contraction is a nonzero prime <ideal> of $O_K$ containing the maximal <ideal> $P$, hence equals $P$. Conversely, if $P=Q\cap O_K$, every element of $P$ lies in $Q$, so the <ideal> it generates satisfies $PO_L\subseteq Q$, equivalently $Q\mid PO_L$.

Solved by gpt-5.6-sol high.

= b
{parent=20f}
{scope}

= Solution
{parent=b}

The equality of residue-field sizes says that the residue degree $f(Q/P)$ is one. Since $Q=\alpha O_L$, the principal <ideal> factorization of $\alpha$ has valuation one at $Q$ and zero at every other prime. Taking the relative <ideal> norm gives
$$
N_{L/K}(\alpha)O_K=N_{L/K}(\alpha O_L)
=N_{L/K}(Q)=P^{f(Q/P)}=P.
$$
Thus $P=N_{L/K}(\alpha)O_K$.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=21J}

For $K=M\cup N$, the simplicial Mayer--Vietoris <sequence> is
$$
\cdots\to H_i(M\cap N)\xrightarrow{(j_*,-k_*)}
H_i(M)\oplus H_i(N)\to H_i(K)
\xrightarrow{\partial}H_{i-1}(M\cap N)\to\cdots.
$$

Construct a simplicial mapping cone of a degree-$k$ map of a circle. Explicitly, let the target circle have vertices $v_0,v_1,v_2$, and let the source circle have vertices $w_0,\ldots,w_{3k-1}$. Map $w_j$ to $v_{j\bmod3}$, triangulate each quadrilateral of its mapping cylinder, and cone the source circle to one new vertex. The resulting finite simplicial complex is the mapping cone $K_k$.

Its reduced cellular, or equivalently simplicial, chain complex collapses to
$$
0\to\mathbb Z\xrightarrow{\times k}\mathbb Z\to0
$$
in degrees two and one. Consequently
$$
H_i(K_k)\cong
\begin{cases}
\mathbb Z,&i=0,\\
\mathbb Z/k,&i=1,\\
0,&\text{otherwise}.
\end{cases}
$$
The same computation follows from Mayer--Vietoris applied to the cone and the mapping-cylinder neighborhood.

Solved by gpt-5.6-sol high.

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

= a
{parent=22g}
{scope}

= Solution
{parent=a}

Suppose the <spectrum> were empty. The resolvent
$$
R(\lambda)=(\lambda I-T)^{-1}
$$
would then be an entire operator-valued <function>. For $|\lambda|>\|T\|$,
$$
R(\lambda)=\frac1\lambda\sum_{n=0}^\infty\left(\frac T\lambda\right)^n,
$$
because the <series> converges in operator norm and multiplication by $\lambda I-T$ telescopes to $I$. Hence $\|R(\lambda)\|\leq(|\lambda|-\|T\|)^{-1}$.

For fixed $x,y\in\ell^2$, the <scalar> <function> $\langle R(\lambda)x,y\rangle$ is entire, bounded outside a disc by the estimate and bounded inside by compactness. Liouville's theorem makes it constant, and its <limit> at infinity makes that constant zero. This for all $x,y$ would imply $R(\lambda)=0$, impossible for an inverse. Thus the <spectrum> is nonempty.

Solved by gpt-5.6-sol high.

= b
{parent=22g}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

This is the unilateral right shift. It is not compact because the orthonormal <sequence> $(Te_n)=(e_{n+1})$ has no convergent subsequence. Its norm is one. For $|\lambda|>1$ the Neumann <series> gives a resolvent; for $|\lambda|<1$, $\bar\lambda$ is an <eigenvalue> of the adjoint left shift, so $\lambda$ lies in the <spectrum> of $T$. Closedness supplies the boundary. Hence
$$
\sigma(T)=\{\lambda:|\lambda|\leq1\}.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

This is the unilateral left shift. It is not compact because $S e_{n+1}=e_n$. For every $|\lambda|<1$,
$$
(1,\lambda,\lambda^2,\ldots)\in\ell^2
$$
is an <eigenvector> with <eigenvalue> $\lambda$. The norm bound and closedness then give
$$
\sigma(S)=\{\lambda:|\lambda|\leq1\}.
$$

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

The weighted shift is compact because truncating after the first $N$ weights gives finite-rank operators converging in norm, as $1/n\to0$. Moreover
$$
\|R^m\|\leq\frac1{m!},
$$
so its spectral radius is zero. Since the <spectrum> is nonempty,
$$
\sigma(R)=\{0\}.
$$

Solved by gpt-5.6-sol high.

= iv
{parent=b}
{scope}

= Solution
{parent=iv}

Again the weights tend to zero, so finite-rank truncations prove compactness. Products of $m$ consecutive weights are bounded by $1/(m+1)!$, making the spectral radius zero. Therefore
$$
\sigma(Q)=\{0\}.
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=23g}
{scope}

= Solution
{parent=a}

Taking Fourier transforms gives
$$
(1+|\xi|^4)\widehat u(\xi)=\widehat f(\xi).
$$
Thus
$$
\widehat u(\xi)=\frac{\widehat f(\xi)}{1+|\xi|^4}.
$$
The multiplier and all its polynomially weighted <derivatives> have at most <polynomial> growth, so this defines a tempered distribution. In fact it maps $H^s$ continuously into $H^{s+4}$. Since the multiplier never vanishes, any homogeneous tempered solution has zero Fourier transform; uniqueness follows.

Solved by gpt-5.6-sol high.

= b
{parent=23g}
{scope}

= Solution
{parent=b}

A compactly supported smooth <function> belongs to $H^s(\mathbb R^n)$ for every real $s$. Part (a) therefore puts $u$ in $H^{s+4}$ for every $s$. Taking $s$ arbitrarily large and applying the Sobolev embedding theorem to every <derivative> shows that $u\in C^\infty(\mathbb R^n)$.

Solved by gpt-5.6-sol high.

= c
{parent=23g}
{scope}

= Solution
{parent=c}

In three dimensions, $\delta_0\in H^s$ for every $s<-3/2$. Hence the solution belongs to $H^{s+4}$ for every such $s$; choosing, for example, $s=-2$ gives $u\in H^2(\mathbb R^3)$. Since $2>3/2$, Sobolev embedding supplies a unique continuous representative. It solves the equation distributionally, and uniqueness follows from part (a), so this is the unique continuous solution.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=24F}

A morphism $\phi:\mathbb P^2\to\mathbb P^1$ is represented by homogeneous forms $F,G$ of the same degree with no common projective zero. If that degree were positive, the two plane curves $F=0$ and $G=0$ would intersect, producing a base point. Thus the degree is zero and $\phi$ is constant.

If a closed subvariety of $(\mathbb P^1)^3$ were isomorphic to $\mathbb P^2$, composing its embedding with each coordinate projection would give three constant maps. Their product would be constant, contradicting that it is an embedding.

The Riemann--Hurwitz theorem states
$$
2g_X-2=d(2g_Y-2)+\sum_{P\in X}(e_P-1)
$$
for a degree-$d$ nonconstant map of smooth projective curves.

Project $X$ to the second $\mathbb P^1$. This is a degree-two map. Its quadratic discriminant in the $x$ variables is homogeneous of degree six in $y$, so smoothness gives six branch points counted with multiplicity. Hence
$$
2g_X-2=2(-2)+6=2,
$$
and $g_X=2$. A smooth plane curve has genus $(d-1)(d-2)/2$, which is never two for an integer $d$, so $X$ is not isomorphic to a smooth plane curve.

Solved by gpt-5.6-sol high.

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

= a
{parent=25j}
{scope}

= Solution
{parent=a}

The homotopy lemma says that homotopic smooth maps between compact connected manifolds have the same mod-two degree, computed as the parity of the inverse image of a regular value. The homogeneity lemma says that any two points of a connected smooth manifold are related by a diffeomorphism isotopic to the identity.

Given regular values $p,q$ of $f$, choose such a diffeomorphism $h$ with $h(p)=q$. Then $h\circ f$ is homotopic to $f$, while $(h\circ f)^{-1}(q)=f^{-1}(p)$. The homotopy lemma proves that the two inverse-image counts have equal parity.

For smooth Brouwer, suppose a smooth self-map of a ball had no fixed point. Following the ray from $f(x)$ through $x$ to the boundary constructs a smooth retraction of the ball onto its sphere. Its restriction to the sphere is the identity, but it is also null-homotopic through the ball. The identity has odd mod-two degree and a constant map has even degree at a different regular value, contradicting the homotopy lemma. Thus a fixed point exists.

Solved by gpt-5.6-sol high.

= b
{parent=25j}
{scope}

= Solution
{parent=b}

Orthogonality and $\|\widetilde\phi(x)\|=\|x\|$ give
$$
\|x+\varepsilon\widetilde\phi(x)\|^2=(1+\varepsilon^2)\|x\|^2,
$$
so $\psi_\varepsilon$ maps each boundary sphere to itself. Also
$$
D\psi_\varepsilon=(1+\varepsilon^2)^{-1/2}
(I+\varepsilon D\widetilde\phi).
$$
On compact $A$, this <derivative> is uniformly close to the identity; for sufficiently small $\varepsilon$ its <determinant> is positive and it is locally invertible. The map is properly homotopic to the identity and has degree one, so the local diffeomorphism has one sheet and is a global diffeomorphism $A\to A$.

Solved by gpt-5.6-sol high.

= c
{parent=25j}
{scope}

= Solution
{parent=c}

Apply change of variables to the orientation-preserving diffeomorphism $\psi_\varepsilon:A\to A$:
$$
\operatorname{vol}(A)=\int_A\det D\psi_\varepsilon
=(1+\varepsilon^2)^{-(n+1)/2}
\int_A\det(I+\varepsilon D\widetilde\phi).
$$
Rearranging gives
$$
\int_A\det(I+\varepsilon D\widetilde\phi)
=\operatorname{vol}(A)(1+\varepsilon^2)^{(n+1)/2}.
$$

Solved by gpt-5.6-sol high.

= d
{parent=25j}
{scope}

= Solution
{parent=d}

The left side in part (c) is a <polynomial> in $\varepsilon$ of degree at most $n+1$. If $n$ is even, $(n+1)/2$ is a half-integer, so
$$
(1+\varepsilon^2)^{(n+1)/2}
$$
is not a <polynomial> and cannot agree with the left side on an interval. This contradiction shows that an everywhere nonzero tangent unit <vector> field $\phi$ cannot exist on an even-dimensional sphere.

Solved by gpt-5.6-sol high.

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

= a
{parent=26g}
{scope}

= Solution
{parent=a}

Uniform integrability means
$$
\lim_{K\to\infty}\sup_n
\mathbb E\bigl[|X_n|\mathbf1_{\{|X_n|>K\}}\bigr]=0.
$$
If $\sup_n\mathbb E|X_n|^p=C<\infty$ for $p>1$, then
$$
\mathbb E[|X_n|\mathbf1_{|X_n|>K}]
\leq K^{1-p}\mathbb E|X_n|^p\leq CK^{1-p},
$$
proving uniform integrability. For a counterexample, let $X_n=n$ with probability $1/n$ and zero otherwise. Then $\mathbb E|X_n|=1$, but for every $K$ and $n>K$ the tail expectation is one.

Solved by gpt-5.6-sol high.

= b
{parent=26g}
{scope}

= Solution
{parent=b}

Write $Y_n=n^{-1}\sum_{i=1}^n|X_i|$. Truncate each $|X_i|$ at a level $K$. The average of the truncated parts is at most $K$, while the expected average of the tails is
$$
\mathbb E[|X_1|\mathbf1_{|X_1|>K}],
$$
independently of $n$. Markov's inequality then shows that $(Y_n)$ is uniformly integrable as $K\to\infty$. Since $|S_n|/n\leq Y_n$, the <sequence> $(S_n/n)$ is uniformly integrable.

Convergence in probability to the constant $\mathbb EX_1$, together with uniform integrability, implies convergence of first absolute moments by the Vitali convergence theorem. Hence
$$
\mathbb E\left|\frac{S_n}{n}-\mathbb EX_1\right|\to0.
$$

Solved by gpt-5.6-sol high.

= c
{parent=26g}
{scope}

= i
{parent=c}
{scope}

= Solution
{parent=i}

By the union bound,
$$
\mathbb P(M_n>\varepsilon\sqrt n)
\leq n\mathbb P(|X_1|>\varepsilon\sqrt n).
$$
For an $L^2$ <random variable>, $t^2\mathbb P(|X_1|>t)\to0$. Taking $t=\varepsilon\sqrt n$ makes the right side tend to zero, proving $M_n/\sqrt n\to0$ in probability.

Solved by gpt-5.6-sol high.

= ii
{parent=c}
{scope}

= Solution
{parent=ii}

For any $\varepsilon>0$,
$$
M_n\leq\varepsilon\sqrt n+\sum_{i=1}^n
|X_i|\mathbf1_{\{|X_i|>\varepsilon\sqrt n\}}.
$$
Therefore
$$
\frac{\mathbb EM_n}{\sqrt n}
\leq\varepsilon+\sqrt n\,
\mathbb E[|X_1|\mathbf1_{|X_1|>\varepsilon\sqrt n}]
\leq\varepsilon+\frac1\varepsilon
\mathbb E[X_1^2\mathbf1_{|X_1|>\varepsilon\sqrt n}].
$$
The last term tends to zero. Taking the limsup and then $\varepsilon\downarrow0$ proves the claim.

Solved by gpt-5.6-sol high.

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

= a
{parent=27k}
{scope}

= Solution
{parent=a}

A dispatch cycle contains $N$ interarrival times and has mean length $N/\mu$. The expected total passenger waiting time in one cycle is
$$
\frac1\mu(1+2+\cdots+(N-1))=\frac{N(N-1)}{2\mu}.
$$
The renewal-reward average cost rate is therefore
$$
C(N)=\frac{K+cN(N-1)/(2\mu)}{N/\mu}
=\frac{\mu K}{N}+\frac c2(N-1).
$$
Thus the continuous optimum is $\sqrt{2\mu K/c}$, and the integer optimum is the positive integer minimizing $C(N)$; equivalently it is the smallest $N$ with
$$
N(N+1)\geq\frac{2\mu K}{c},
$$
with both adjacent values optimal in the equality case.

Solved by gpt-5.6-sol high.

= b
{parent=27k}
{scope}

= Solution
{parent=b}

The lifetime $L$ is uniform on $[0,3]$ with mean $3/2$. The equilibrium age and excess densities are both
$$
\frac{\mathbb P(L>x)}{\mathbb EL}=\frac23\left(1-\frac x3\right).
$$
Hence, for $0\leq x\leq3$,
$$
\mathbb P(A_t\leq x),\ \mathbb P(E_t\leq x)
\longrightarrow\frac{2x}{3}-\frac{x^2}{9}.
$$
The observed total lifetime is length-biased, so
$$
\mathbb P(A_t+E_t\leq x)
\longrightarrow\frac{\mathbb E[L\mathbf1_{L\leq x}]}{\mathbb EL}
=\frac{x^2}{9}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=27k}
{scope}

= Solution
{parent=c}

Campbell's formula for a <Poisson point process> gives
$$
\mathbb ES=\int_{\mathbb R^d}g(u)\lambda(u)du.
$$
Expanding $S^2$ separates equal and distinct Poisson points. The factorial-moment formula gives
$$
\mathbb ES^2
=\int g(u)^2\lambda(u)du
+\left(\int g(u)\lambda(u)du\right)^2.
$$
Subtracting the square of the mean yields
$$
\operatorname{var}S=\int_{\mathbb R^d}g(u)^2\lambda(u)du.
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=28l}
{scope}

= Solution
{parent=a}

The identity
$$
\frac1n\sum_{i=1}^n(X_i-\bar X_n)^2
=\frac1n\sum_{i=1}^nX_i^2-\bar X_n^2
$$
and the weak law applied to $X_i$ and $X_i^2$ give convergence in probability to
$$
\mathbb EX_1^2-\mu^2=\sigma^2.
$$

Solved by gpt-5.6-sol high.

= b
{parent=28l}
{scope}

= Solution
{parent=b}

The central <limit> theorem and Delta method give
$$
\sqrt n\{g(\bar X_n)-g(\mu)\}
\Rightarrow N(0,g'(\mu)^2\sigma^2).
$$
Thus one may take
$$
\sigma_n=\frac{|g'(\mu)|\sigma}{\sqrt n}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=28l}
{scope}

= Solution
{parent=c}

Put $d_i=\bar X_{n-1,i}-\bar X_n=(\bar X_n-X_i)/(n-1)$. Taylor's theorem gives
$$
T_{n-1,i}-T_n=g'(\bar X_n)d_i+R_{n,i},
\qquad |R_{n,i}|\leq M d_i^2.
$$
Since $\sum_i d_i=0$, centering the leave-one-out values and expanding the square gives
$$
v_{\rm JACK}=\frac{n-1}{n}(A_n+B_n+2C_n),
$$
with the stated $A_n,B_n$ and the cross term satisfying $|C_n|\leq\sqrt{A_nB_n}$ by Cauchy--Schwarz.

Now
$$
A_n=\frac{g'(\bar X_n)^2}{(n-1)^2}
\sum_i(X_i-\bar X_n)^2
\sim\frac{g'(\mu)^2\sigma^2}{n}.
$$
The fourth-moment assumption gives
$$
B_n\leq\frac{M^2}{(n-1)^4}\sum_i(X_i-\bar X_n)^4=O(n^{-3}),
$$
and hence $C_n=O(n^{-2})$. Since $\sigma_n^2=g'(\mu)^2\sigma^2/n$, it follows that
$$
\frac{v_{\rm JACK}}{\sigma_n^2}\to1.
$$

Solved by gpt-5.6-sol high.

= d
{parent=28l}
{scope}

= Solution
{parent=d}

By the Delta method and consistency of the jackknife variance, Slutsky's theorem gives
$$
\frac{T_n-\theta}{\sqrt{v_{\rm JACK}}}\Rightarrow N(0,1).
$$
Therefore an asymptotically valid interval is
$$
\left[T_n-z_{1-\alpha/2}\sqrt{v_{\rm JACK}},
T_n+z_{1-\alpha/2}\sqrt{v_{\rm JACK}}\right].
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=29l}
{scope}

= Solution
{parent=a}

For times $0\leq t_1<\cdots<t_m$, the <vector> $(W_{t_1},\ldots,W_{t_m})$ is a linear transformation of the independent Gaussian increments
$$
W_{t_1}, W_{t_2}-W_{t_1},\ldots,W_{t_m}-W_{t_{m-1}},
$$
so Brownian motion is a Gaussian process. Its mean is zero, and for $s\leq t$,
$$
\mathbb E(W_sW_t)=\mathbb E\{W_s(W_s+W_t-W_s)\}=s.
$$

Solved by gpt-5.6-sol high.

= b
{parent=29l}
{scope}

= Solution
{parent=b}

Fubini gives $\mathbb EI_t=0$. For $s\leq t$,
$$
\mathbb E(I_sI_t)=\int_0^s\int_0^t\min(u,v)dvdu
=\int_0^s\left(ut-\frac{u^2}{2}\right)du
=\frac{s^2(3t-s)}6.
$$

Solved by gpt-5.6-sol high.

= c
{parent=29l}
{scope}

= Solution
{parent=c}

For $s<t$,
$$
e^{-rt}S_t=e^{-rs}S_s
\exp\left(-\frac12\sigma^2(t-s)+\sigma(W_t-W_s)\right).
$$
The increment is independent of the Brownian filtration at time $s$, and its exponential factor has mean one by the Gaussian moment-generating <function>. Hence the conditional expectation of the discounted time-$t$ price is the discounted time-$s$ price.

Solved by gpt-5.6-sol high.

= d
{parent=29l}
{scope}

= Solution
{parent=d}

Risk-neutral valuation gives
$$
e^{-rT}\mathbb E[(S_T-K)^+]
=S_0\mathbb E\left[\left(e^{-\sigma^2T/2+\sigma\sqrt T Z}
-\frac{Ke^{-rT}}{S_0}\right)^+\right].
$$
With $v=\sigma^2T$ and $m=Ke^{-rT}/S_0$, this is exactly
$$
S_0F(v,m).
$$

Solved by gpt-5.6-sol high.

= e
{parent=29l}
{scope}

= Solution
{parent=e}

The average log price is
$$
\frac1T\int_0^T\log S_tdt
=\log S_0+\frac12\left(r-\frac12\sigma^2\right)T
+\frac\sigma T I_T.
$$
Part (b) gives $\operatorname{var}(I_T)=T^3/3$, so the Gaussian variance of the final term is
$$
v=\frac13\sigma^2T.
$$
Rewriting the lognormal factor in the normalized form $e^{-v/2+\sqrt vZ}$ and discounting gives
$$
S_0e^{-(r/2+\sigma^2/12)T}
F\left(\frac13\sigma^2T,
\frac{Ke^{-(r/2-\sigma^2/12)T}}{S_0}\right).
$$
Thus $\alpha=1/12$.

Solved by gpt-5.6-sol high.

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

= a
{parent=30k}
{scope}

= Solution
{parent=a}

The subdifferential is
$$
\partial f(\alpha)=\{s:f(\beta)\geq f(\alpha)+s^T(\beta-\alpha)
\text{ for every }\beta\}.
$$
If $0\in\partial f(\alpha)$, this inequality says $f(\beta)\geq f(\alpha)$, so $\alpha$ minimizes $f$. Conversely, the zero <vector> satisfies the subgradient inequality at any global minimizer.

Solved by gpt-5.6-sol high.

= b
{parent=30k}
{scope}

= Solution
{parent=b}

For a convex surrogate margin loss $\phi$, the empirical risk is
$$
\widehat R_\phi(h)=\frac1n\sum_{i=1}^n
\phi(y_ih(x_i)).
$$
It replaces the discontinuous classification error by a convex penalty on signed margins.

Solved by gpt-5.6-sol high.

= c
{parent=30k}
{scope}

= Solution
{parent=c}

The <matrix>
$$
P=X^T(XX^T)^{-1}X
$$
is the orthogonal projection onto the row space of $X$. Since $XP\beta=X\beta$, the empirical risks of $h_\beta$ and $h_{P\beta}$ are equal. Orthogonality gives
$$
\|\beta\|^2=\|P\beta\|^2+\|(I-P)\beta\|^2,
$$
so $q(\beta)\geq q(P\beta)$, with strict inequality unless $(I-P)\beta=0$. Therefore the minimizer lies in the row space:
$$
\widehat\beta=P\widehat\beta=X^T\widehat\alpha.
$$
Because $X^T$ is injective when $XX^T$ is invertible, minimizing over that row space is equivalent to minimizing $r(\alpha)=q(X^T\alpha)$.

Solved by gpt-5.6-sol high.

= d
{parent=30k}
{scope}

= Solution
{parent=d}

For hinge loss, choose
$$
t_i=y_is_i,
\qquad
s_i=\begin{cases}
1,&y_ix_i^T\widehat\beta<1,\\
0,&y_ix_i^T\widehat\beta>1,\\
\text{any value in }[0,1],&y_ix_i^T\widehat\beta=1.
\end{cases}
$$
The subgradient optimality condition is
$$
0=-\frac1nX^Tt+2\lambda X^T\widehat\alpha.
$$
Injectivity of $X^T$ gives $t/n=2\lambda\widehat\alpha$. Multiplying by $XX^T$ and writing $k_i$ for its $i$th column yields
$$
\frac1n\sum_i k_it_i=2\lambda XX^T\widehat\alpha.
$$
If $y_ix_i^T\widehat\beta>1$, then $t_i=0$, and $t/n=2\lambda\widehat\alpha$ implies $\widehat\alpha_i=0$.

Solved by gpt-5.6-sol high.

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

= a
{parent=31c}
{scope}

= Solution
{parent=a}

The coefficient changes sign where $1-\lambda^2/x^2=0$, so the single turning point is
$$
x_{\rm tp}=\lambda.
$$
Put $p(x)=(\lambda^2/x^2-1)^{1/2}$ and $q(x)=(1-\lambda^2/x^2)^{1/2}$. The <WKB approximation> satisfying the <boundary condition> at $x=1$ is
$$
y_1(x)\sim A p(x)^{-1/2}
\sin\left(\frac1\epsilon\int_1^x p(s)\,ds\right),
\qquad 1\leq x<\lambda,
$$
while decay at infinity selects
$$
y_2(x)\sim B q(x)^{-1/2}
\exp\left(-\frac1\epsilon\int_\lambda^xq(s)\,ds\right),
\qquad x>\lambda.
$$

Solved by gpt-5.6-sol high.

= b
{parent=31c}
{scope}

= Solution
{parent=b}

Near $x=\lambda$,
$$
1-\frac{\lambda^2}{x^2}=\frac{2(x-\lambda)}{\lambda}+O((x-\lambda)^2).
$$
With
$$
t=\left(\frac2\lambda\right)^{1/3}
\frac{x-\lambda}{\epsilon^{2/3}},
$$
the leading equation is $y_{tt}-ty=0$. Decay for $t>0$ selects $y=C\operatorname{Ai}(t)$. The <Airy turning-point connection formula> gives on the oscillatory side
$$
y\sim C' p(x)^{-1/2}
\sin\left(\frac1\epsilon\int_x^\lambda p(s)\,ds+\frac\pi4\right).
$$
For this to be proportional to the inner expression from part (a), the two phases must differ by an integer multiple of $\pi$. Thus the <WKB quantization condition> is
$$
\int_1^\lambda\left(\frac{\lambda^2}{x^2}-1\right)^{1/2}dx
=\epsilon\left(n+\frac34\right)\pi,
\qquad n=0,1,2,\ldots.
$$

Solved by gpt-5.6-sol high.

= c
{parent=31c}
{scope}

= Solution
{parent=c}

Write $\lambda=1+\delta$, where $\delta\downarrow0$, and put $x=\lambda(1-u)$. The integration range is $0\leq u\leq1-1/\lambda=\delta+O(\delta^2)$, and
$$
\left((1-u)^{-2}-1\right)^{1/2}
=(2u)^{1/2}+O(u^{3/2}).
$$
Consequently
$$
\int_1^\lambda\left(\frac{\lambda^2}{x^2}-1\right)^{1/2}dx
=\frac{2\sqrt2}{3}\delta^{3/2}+O(\delta^{5/2}).
$$
Substitution into the quantization condition gives
$$
\lambda=1+\alpha_n\epsilon^{2/3}+O(\epsilon^{4/3}),
\qquad
\alpha_n=\left[\frac{3\pi}{2\sqrt2}
\left(n+\frac34\right)\right]^{2/3}.
$$

Solved by gpt-5.6-sol high.

= d
{parent=31c}
{scope}

= Solution
{parent=d}

The action <integral> can be evaluated exactly:
$$
I(\lambda)=\lambda\log\left(\lambda+\sqrt{\lambda^2-1}\right)
-\sqrt{\lambda^2-1}.
$$
As $\lambda\to\infty$,
$$
I(\lambda)=\lambda\log(2\lambda)-\lambda+O(\lambda^{-1}).
$$
Its leading logarithmic term is $\lambda\log\lambda$. Hence the large <eigenvalues> satisfy, to leading order,
$$
\lambda\log\lambda
=\epsilon\left(n+\frac34\right)\pi.
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=32a}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

The <Poincare-Bendixson theorem> says that a nonempty compact $\omega$-limit set of a planar $C^1$ flow which contains only finitely many fixed points is either a fixed point, a periodic orbit, or a union of fixed points and connecting trajectories. In particular, if it contains no fixed point, it is a periodic orbit.

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

Along a periodic orbit the <vector> field is tangent and makes one full turn, so its index around the orbit is $+1$. If the enclosed region contained no fixed point, the normalized <vector> field would extend continuously across the disc and its boundary map would have degree zero. This contradicts the index $+1$. Thus every planar periodic orbit encloses a fixed point.

Solved by gpt-5.6-sol high.

= b
{parent=32a}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

Adding the two equations gives $\dot x+\dot y=b-x$, so a fixed point has
$$
(x_0,y_0)=\left(b,\frac{b}{a+b^2}\right).
$$
At this point the Jacobian has <determinant> $a+b^2>0$ and trace
$$
T=-1-a-b^2+\frac{2b^2}{a+b^2}.
$$
The fixed point is asymptotically stable exactly when $T<0$, or
$$
(a+b^2)(1+a+b^2)>2b^2.
$$
The equality curve exists only for $a\leq1/8$ and is
$$
b^2=\frac{1-2a\pm\sqrt{1-8a}}2.
$$
For $0<a<1/8$ the fixed point is unstable between these two curves and stable outside them; for $a>1/8$ it is stable everywhere.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

On $x=0$, $\dot x=ay\geq0$; on $y=0$, $\dot y=b>0$. On $x=x^*$, the identity $\dot x+\dot y=b-x$ gives a negative outward component because $x^*>b$. On $y=b/a$, one has $\dot y=-x^2b/a\leq0$. Finally, the sloping edge has outward normal proportional to $(1,1)$, and there
$$
(1,1)\cdot(\dot x,\dot y)=b-x\leq0.
$$
Thus the <vector> field points inward or is tangent on every edge, so trajectories cannot leave the closed polygon $D$.

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

The region $D$ is compact and positively invariant and contains the unique fixed point. If that fixed point is unstable, choose an initial point in $D$ which is not on its stable set. Its nonempty compact $\omega$-limit set cannot be the fixed point. The <Poincare-Bendixson theorem> therefore supplies a periodic orbit. Hence a sufficient parameter condition, and the one selected by the preceding analysis, is
$$
0<a<\frac18,
\qquad
\frac{1-2a-\sqrt{1-8a}}2<b^2<
\frac{1-2a+\sqrt{1-8a}}2.
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=33b}
{scope}

= Solution
{parent=a}

For unit mass,
$$
H_A=\frac12P_A^2+\frac12\omega^2X_A^2
=\hbar\omega\left(a_A^\dagger a_A+\frac12\right),
$$
where
$$
X_A=\sqrt{\frac{\hbar}{2\omega}}(a_A+a_A^\dagger),
\qquad
P_A=-i\sqrt{\frac{\hbar\omega}{2}}(a_A-a_A^\dagger).
$$
The Hilbert space is $L^2(\mathbb R)$, equivalently the Fock-space completion of the orthonormal number states $|n\rangle$, $n\geq0$.

Solved by gpt-5.6-sol high.

= b
{parent=33b}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

The unperturbed ground state is $|0,0\rangle$ with energy $\hbar\omega$. Since
$$
H_{\rm int}|0,0\rangle
=\frac{\lambda\hbar}{2\omega}|1,1\rangle,
$$
the <first-order energy correction> vanishes and <first-order nondegenerate perturbation theory> gives
$$
|\Omega\rangle=|0,0\rangle-\frac{\lambda}{4\omega^2}|1,1\rangle+O(\lambda^2),
\qquad
E_\Omega=\hbar\omega+O(\lambda^2).
$$

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

Put $g=\lambda/(4\omega^2)$. Normalization through order $g^2$ gives
$$
|\Omega\rangle=(1-\tfrac12g^2)|0,0\rangle-g|1,1\rangle+O(g^2).
$$
Thus
$$
\rho_{AB}=(1-g^2)|00\rangle\langle00|
-g(|00\rangle\langle11|+|11\rangle\langle00|)
+g^2|11\rangle\langle11|+O(g^3).
$$
Tracing over $B$ removes the cross terms. The <reduced density operator of a weakly coupled oscillator pair> is therefore
$$
\rho_A=(1-g^2)|0\rangle\langle0|+g^2|1\rangle\langle1|+O(g^3),
$$
whose trace is one to the required order.

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

If the <eigenvalues> of a <density operator> are $p_j\geq0$ with $\sum_jp_j=1$, then
$$
\gamma=\sum_jp_j^2\leq1,
$$
with equality exactly when one <eigenvalue> is one, namely for a pure state. In dimension $d$, Cauchy--Schwarz gives $\gamma\geq1/d$, with equality for the maximally mixed state. On an infinite-dimensional space the universal lower bound is zero as an infimum.

Solved by gpt-5.6-sol high.

= iv
{parent=b}
{scope}

= Solution
{parent=iv}

The <eigenvalues> found in part (ii) give the <purity of a density operator>
$$
\gamma_A=(1-g^2)^2+g^4+O(g^3)
=1-2g^2+O(g^3)
=1-\frac{\lambda^2}{8\omega^4}+O(\lambda^3).
$$
The displayed truncation lies between zero and one when $|\lambda|\leq2\sqrt2\,\omega^2$. Perturbation theory requires the stronger condition $|\lambda|\ll\omega^2$; the exact coupled oscillator is stable only for $|\lambda|<\omega^2$. The reduction in purity records entanglement between $A$ and $B$.

Solved by gpt-5.6-sol high.

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

= a
{parent=34e}
{scope}

= Solution
{parent=a}

Under the stated potential transformation, set
$$
\psi'(x,t)=e^{-ief(x,t)/\hbar}\psi(x,t).
$$
Then
$$
\left(\nabla+\frac{ie}{\hbar}(A+\nabla f)\right)\psi'
=e^{-ief/\hbar}\left(\nabla+\frac{ie}{\hbar}A\right)\psi
$$
and
$$
\left(\partial_t-\frac{ie}{\hbar}(\phi-\partial_tf)\right)\psi'
=e^{-ief/\hbar}\left(\partial_t-\frac{ie}{\hbar}\phi\right)\psi.
$$
Applying the spatial identity twice proves the <Gauge covariance of the Schrödinger equation>.

Solved by gpt-5.6-sol high.

= b
{parent=34e}
{scope}

= Solution
{parent=b}

Writing $D_A=\nabla+ieA/\hbar$, the current is
$$
J=-\frac{i\hbar}{2m}
(\psi^*D_A\psi-\psi(D_A\psi)^*).
$$
The two covariant factors acquire opposite phases under a gauge transformation, so $J$ is invariant.

Multiply the <Schrodinger equation> by $\psi^*$, subtract its complex conjugate multiplied by $\psi$, and use
$$
\psi^*D_A^2\psi-\psi(D_A^2\psi)^*
=\nabla\cdot\{\psi^*D_A\psi-\psi(D_A\psi)^*\}.
$$
The real scalar-potential terms cancel, leaving
$$
\partial_t|\psi|^2+\nabla\cdot J=0.
$$

Solved by gpt-5.6-sol high.

= c
{parent=34e}
{scope}

= Solution
{parent=c}

The proposed potentials give $\nabla\times A=(0,0,B)$ and $-\nabla\phi=(0,E,0)$. Substituting $\psi=e^{ikx}e^{-i\mathcal Et/\hbar}\varphi(y)$ gives
$$
\left[-\frac{\hbar^2}{2m}\frac{d^2}{dy^2}
+\frac{(\hbar k-eBy)^2}{2m}+eEy\right]\varphi
=\mathcal E\varphi.
$$
This is a harmonic oscillator with cyclotron frequency and centre
$$
\omega_1=\frac{|eB|}{m},
\qquad
y_0=\frac{\hbar k}{eB}-\frac{mE}{eB^2}
$$
for $eB>0$ as written. Completing the square yields
$$
\mathcal E=\hbar\omega_1\left(n+\frac12\right)+W,
\qquad
W=\frac{\hbar kE}{B}-\frac{mE^2}{2B^2}.
$$
Thus the <Landau levels in crossed electric and magnetic fields> retain their oscillator spacing but acquire a linear dependence on $k$, lifting their guiding-centre degeneracy.

Solved by gpt-5.6-sol high.

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

= a
{parent=35b}
{scope}

= Solution
{parent=a}

At a first-order transition the <Gibbs free energy> is continuous but a first <derivative> such as <entropy> or volume jumps; there is latent heat. At a second-order transition the first <derivatives> are continuous while a second <derivative>, such as <heat capacity> or susceptibility, is discontinuous or divergent, and there is no latent heat.

Solved by gpt-5.6-sol high.

= b
{parent=35b}
{scope}

= Solution
{parent=b}

Equality of <chemical potentials> gives $d\mu_{\rm l}=d\mu_{\rm g}$ along coexistence. Since
$$
d\mu=-s\,dT+v\,dp,
$$
one obtains
$$
(v_{\rm g}-v_{\rm l})\frac{dp}{dT}=s_{\rm g}-s_{\rm l}.
$$
Using $L=T(s_{\rm g}-s_{\rm l})$ gives the <Clausius-Clapeyron relation>
$$
\frac{dp}{dT}=\frac{L}{T(v_{\rm g}-v_{\rm l})}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=35b}
{scope}

= Solution
{parent=c}

With $v_{\rm g}\gg v_{\rm l}$ and $v_{\rm g}=k_BT/p$, the coexistence equation becomes
$$
\frac1p\frac{dp}{dT}=\frac{L}{k_BT^2}.
$$
For constant $L$ this integrates to
$$
p(T)=p_0\exp\left(-\frac{L}{k_BT}\right),
$$
where $p_0$ is constant. For $N$ gas particles, $V_{\rm g}=Nk_BT/p(T)$, and therefore
$$
\frac{dV_{\rm g}}{dT}
=V_{\rm g}\left(\frac1T-\frac{L}{k_BT^2}\right).
$$

Solved by gpt-5.6-sol high.

= d
{parent=35b}
{scope}

= i
{parent=d}
{scope}

= Solution
{parent=i}

In the <Dieterici equation>, $b$ is the excluded volume per particle caused by short-range repulsion. The positive constant $a$ measures cohesive attraction, which lowers the <pressure> through the Boltzmann factor.

Solved by gpt-5.6-sol high.

= ii
{parent=d}
{scope}

= Solution
{parent=ii}

At low density,
$$
\frac{p}{k_BT}
=\frac1{v-b}\exp\left(-\frac{a}{k_BTv}\right)
=\frac1v+\frac{b-a/(k_BT)}{v^2}+O(v^{-3}).
$$
Hence the second virial coefficient is
$$
B_2(T)=b-\frac{a}{k_BT}.
$$

Solved by gpt-5.6-sol high.

= iii
{parent=d}
{scope}

= Solution
{parent=iii}

At the critical point, $(\partial p/\partial v)_T=(\partial^2p/\partial v^2)_T=0$. The first logarithmic <derivative> gives
$$
\frac{a}{k_BT v^2}=\frac1{v-b}.
$$
At a stationary point the second-derivative condition reduces to
$$
\frac1{(v-b)^2}-\frac{2a}{k_BT v^3}=0.
$$
Combining the equations gives $v_c=2b$, and substitution then gives
$$
T_c=\frac{a}{4bk_B}.
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=36d}
{scope}

= Solution
{parent=a}

A dielectric is a medium whose constituent charges become polarized by an applied <electric field>; it may also acquire a magnetization. Spatial variation of $P$ leaves uncompensated dipole charge, giving $\rho_{\rm bd}=-\nabla\cdot P$. Time-dependent polarization transports charge and aligned microscopic current loops contribute magnetization current, giving the <bound charge and bound current>
$$
J_{\rm bd}=\partial_tP+\nabla\times M.
$$

Solved by gpt-5.6-sol high.

= b
{parent=36d}
{scope}

= Solution
{parent=b}

Split the microscopic sources into free and bound parts and define
$$
D=\epsilon_0E+P,
\qquad
H=\frac{B}{\mu_0}-M.
$$
Substitution of $\rho=\rho_{\rm f}-\nabla\cdot P$ into <Gauss's law> gives $\nabla\cdot D=\rho_{\rm f}$. Similarly, using $J=J_{\rm f}+\partial_tP+\nabla\times M$ in the <Ampère-Maxwell equation> gives $\nabla\times H=J_{\rm f}+\partial_tD$. The homogeneous <Maxwell equations> are unchanged. Thus
$$
\nabla\cdot D=\rho_{\rm f},\qquad
\nabla\times H=J_{\rm f}+\frac{\partial D}{\partial t},
$$
$$
\nabla\cdot B=0,\qquad
\nabla\times E=-\frac{\partial B}{\partial t}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=36d}
{scope}

= Solution
{parent=c}

Since there are no currents, $\nabla\times B=0$ in each constant-permeability region, while $\nabla\cdot B=0$ gives $\nabla^2\psi=0$. The displayed dipole forms are therefore harmonic. Regularity at the origin and the field at infinity give
$$
b_1=0,\qquad a_3=B_0.
$$
For $\psi=(ar+b/r^2)\cos\theta$,
$$
B_r=(a-2b/r^3)\cos\theta,
\qquad
B_\theta=-(a+b/r^3)\sin\theta.
$$
The <magnetic spherical-shell matching> conditions, continuity of $B_r$ and $H_\theta$, give the remaining four equations
$$
a_1-\frac{2b_1}{R_1^3}=a_2-\frac{2b_2}{R_1^3},
\qquad
\frac{a_1+b_1/R_1^3}{\mu_0}
=\frac{a_2+b_2/R_1^3}{\mu},
$$
$$
a_2-\frac{2b_2}{R_2^3}=a_3-\frac{2b_3}{R_2^3},
\qquad
\frac{a_2+b_2/R_2^3}{\mu}
=\frac{a_3+b_3/R_2^3}{\mu_0}.
$$
The two $R_1$ equations yield
$$
a_2=\frac{\mu_0+2\mu}{3\mu_0}a_1,
\qquad
b_2=\frac{\mu-\mu_0}{3\mu_0}R_1^3a_1.
$$
Together with $b_1=0$ and $a_3=B_0$, these are the requested expressions. Substituting them into the two $R_2$ equations gives a two-by-two linear system for $a_1$ and $b_3$.

Solved by gpt-5.6-sol high.

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

= a
{parent=37b}
{scope}

= Solution
{parent=a}

The <Levi-Civita connection> is
$$
\Gamma^\alpha{}_{\beta\gamma}
=\frac12g^{\alpha\delta}
(g_{\delta\beta,\gamma}+g_{\delta\gamma,\beta}-g_{\beta\gamma,\delta}).
$$
For $g_{\alpha\beta}=\eta_{\alpha\beta}+h_{\alpha\beta}$, retain only first-order terms in $h$. Substitution into the curvature formula and contraction gives the <linearized Ricci tensor and scalar>
$$
2R_{\alpha\beta}
=h_\alpha{}^\gamma{}_{,\gamma\beta}
+h_\beta{}^\gamma{}_{,\gamma\alpha}
-h_\gamma{}^\gamma{}_{,\alpha\beta}
-h_{\alpha\beta,\gamma\rho}\eta^{\gamma\rho}.
$$
Hence
$$
A=1,\qquad B=1,\qquad C=-1,\qquad D=-1.
$$

Solved by gpt-5.6-sol high.

= b
{parent=37b}
{scope}

= Solution
{parent=b}

For the given metric, direct substitution of the stated connection coefficients into the curvature formula leaves
$$
R_{uu}=-\frac12(H_{,xx}+H_{,yy}).
$$
The inverse metric has $g^{uu}=0$, and every other Ricci component vanishes, so
$$
R=g^{\alpha\beta}R_{\alpha\beta}=0.
$$
Thus the full, nonlinear vacuum Einstein equation is precisely the transverse Laplace equation from a <plane-fronted gravitational wave>:
$$
H_{,xx}+H_{,yy}=0.
$$
The dependence on $u$ is unrestricted.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=38C}

The interface is the material level set $F=r-R(\theta,t)=0$. The condition $DF/Dt=0$ gives
$$
-R_t+u-\frac vrR_\theta=0,
$$
which is the stated equation.

Let the perturbation potentials inside and outside be
$$
\phi_i=A r^k e^{ik\theta+\sigma t},
\qquad
\phi_o=B r^{-k}e^{ik\theta+\sigma t}.
$$
Linearizing the kinematic condition at $r=a$ gives
$$
(\sigma+ik\omega)\eta=kAa^{k-1},
\qquad
\sigma\eta=-kBa^{-k-1}.
$$
The linearized interior Euler equations give
$$
p_i'=-\rho[\sigma+i(k-2)\omega]\phi_i,
$$
whereas the stationary exterior gives $p_o'=-\rho\sigma\phi_o$. Since the basic interior <pressure> has $dp_i/dr=\rho\omega^2r$, <pressure> continuity on the displaced boundary is
$$
p_i'(a)+\rho\omega^2a\eta=p_o'(a).
$$
Eliminating $A$ and $B$ gives the <circular vortex-sheet mode> relation
$$
\sigma^2+i\omega(k-1)\sigma
-\frac12\omega^2k(k-1)=0,
$$
so
$$
\boxed{\displaystyle
\sigma=\frac\omega2\left[-i(k-1)\pm\sqrt{k^2-1}\right]}.
$$
The $k=1$ mode is a neutral displacement. Every $k>1$ mode has one exponentially growing branch, so the circular interface rolls up through a Kelvin--Helmholtz instability. Its pattern angular <velocity> is $\omega(k-1)/(2k)$, in the direction of the gyre but slower than the solid-body motion; the disturbances therefore propagate upstream relative to the rotating water.

Solved by gpt-5.6-sol high.

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

= a
{parent=39d}
{scope}

= Solution
{parent=a}

Mass and <momentum> conservation are
$$
\rho_t+(\rho u)_x=0,
\qquad
u_t+uu_x+\frac1\rho p_x=0.
$$
For homentropic ideal-gas flow, $c^2=dp/d\rho$ and
$$
\frac{dc}{c}=\frac{\gamma-1}{2}\frac{d\rho}{\rho}.
$$
Combining the two conservation equations then gives
$$
(\partial_t+(u\pm c)\partial_x)
\left(u\pm\frac{2c}{\gamma-1}\right)=0.
$$
Subtracting the harmless constant $\pm2c_0/(\gamma-1)$ proves that each <Riemann invariant>
$$
R_\pm=u\pm\frac{2(c-c_0)}{\gamma-1}
$$
is constant on $C_\pm$, where $dx/dt=u\pm c$.

Solved by gpt-5.6-sol high.

= b
{parent=39d}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

Before characteristics intersect, the right-running compression is a simple wave. The left-running invariant retains its undisturbed value $R_-=0$, so
$$
c=c_0+\frac{\gamma-1}{2}u.
$$
A $C_+$ characteristic emitted by the piston at time $\tau$ carries the constant value $u=\dot X(\tau)$ and has speed
$$
u+c=c_0+\frac{\gamma+1}{2}\dot X(\tau).
$$
It therefore has equation
$$
x=X(\tau)+\left[c_0+\frac{\gamma+1}{2}\dot X(\tau)\right](t-\tau),
$$
which gives the required parametrization. It applies between the piston and the leading undisturbed characteristic,
$$
X(t)\leq x\leq c_0t,
$$
for $0\leq\tau\leq t$, until the first characteristic intersection; for $x>c_0t$ the gas remains at rest.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

Put $A=(\gamma+1)/2$. A <shock formation by characteristic intersection> first occurs when the map from emission time to position loses monotonicity:
$$
0=\frac{\partial x}{\partial\tau}
=(1-A)\dot X-c_0+A\ddot X(t-\tau).
$$
Thus
$$
t=\tau+\frac{c_0+(A-1)\dot X(\tau)}{A\ddot X(\tau)}.
$$
For $X=c_0t^3/T^2$, write $s=\tau/T$ to obtain
$$
\frac tT=\frac{3A-1}{2A}s+\frac1{6As}.
$$
Minimizing over $s>0$ gives $s^2=[3(3A-1)]^{-1}$ and hence
$$
\boxed{\displaystyle
\frac{t_{\rm shock}}T=f(\gamma)
=\frac{\sqrt{2(3\gamma+1)}}{\sqrt3(\gamma+1)}}.
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=40a}
{scope}

= Solution
{parent=a}

The <power method> starts from $q_0\ne0$ and iterates
$$
q_{j+1}=\frac{Aq_j}{\|Aq_j\|}.
$$
If the <eigenvalue> of largest <modulus> is separated by a <spectral gap> and the initial <vector> is <nonorthogonal> to a corresponding <eigenvector>, the iterates converge in direction to that <eigenvector>.

<Inverse iteration> with shift $s$ instead solves
$$
(A-sI)y_{j+1}=q_j,
\qquad
q_{j+1}=\frac{y_{j+1}}{\|y_{j+1}\|}.
$$
It applies the <power method> to $(A-sI)^{-1}$ and therefore finds an <eigenvector> whose <eigenvalue> is closest to $s$.

Solved by gpt-5.6-sol high.

= b
{parent=40a}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

The <unshifted QR algorithm> computes the <eigenvalues> of $A$. For a real symmetric <matrix> it drives $A_k$ toward a diagonal or block-diagonal <matrix>; the accumulated orthogonal factors simultaneously approximate its <eigenvectors>.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

Since $A_k=Q_kR_k$,
$$
A_{k+1}=R_kQ_k=Q_k^TA_kQ_k.
$$
Thus every step is an orthogonal similarity, so it preserves <eigenvalues> and symmetry.

For a symmetric $r$-banded <matrix>, its subdiagonal entries can be eliminated by Givens rotations ordered along the band. Each rotation creates only the next local bulge; multiplication in reverse order chases that bulge out without creating entries beyond the original upper band. Symmetry supplies the corresponding lower band. This is <symmetric bandwidth preservation under QR iteration>, and proves that $A_{k+1}$ is again $r$-banded.

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

For $k=1$ the claim is just $A=Q_0R_0$. If
$$
A^k=\widetilde Q_k\widetilde R_k,
$$
then $A_k=\widetilde Q_k^TA\widetilde Q_k=Q_kR_k$, and hence
$$
A^{k+1}=A\widetilde Q_k\widetilde R_k
=\widetilde Q_kA_k\widetilde R_k
=(\widetilde Q_kQ_k)(R_k\widetilde R_k)
=\widetilde Q_{k+1}\widetilde R_{k+1}.
$$
Induction proves $A^k=\widetilde Q_k\widetilde R_k$ for all $k\geq1$.

Solved by gpt-5.6-sol high.

= iv
{parent=b}
{scope}

= Solution
{parent=iv}

Let $q_1$ and $q_n$ be the first and last columns of $\widetilde Q_k$. Since $\widetilde R_k$ is upper triangular,
$$
A^ke_1=(\widetilde R_k)_{11}q_1.
$$
Thus $q_1$ is the normalized $k$th <power method> iterate starting from $e_1$.

Taking the transpose of $A^k=\widetilde Q_k\widetilde R_k$ and using symmetry gives $A^k=\widetilde R_k^T\widetilde Q_k^T$. Therefore
$$
A^kq_n=(\widetilde R_k)_{nn}e_n,
\qquad
q_n=(\widetilde R_k)_{nn}A^{-k}e_n.
$$
So the last column is the normalized $k$th <inverse iteration> iterate with shift zero and starting <vector> $e_n$.

Solved by gpt-5.6-sol high.