= Paper 1
{scope}

https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2024/paperii_1_2024.pdf

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

= Solution
{parent=1F}

For an odd composite $N$ and a unit $b$ modulo $N$, $N$ is a <Fermat pseudoprime> to base $b$ when
$$
b^{N-1}\equiv1\pmod N.
$$
It is an <Euler pseudoprime> to base $b$ when
$$
b^{(N-1)/2}\equiv\left(\frac bN\right)\pmod N,
$$
where the right side is the Jacobi symbol.

By the Chinese remainder theorem,
$$
(\mathbb Z/105\mathbb Z)^\times
\cong(\mathbb Z/3\mathbb Z)^\times
\times(\mathbb Z/5\mathbb Z)^\times
\times(\mathbb Z/7\mathbb Z)^\times,
$$
and $\varphi(105)=2\cdot4\cdot6=48$. The condition $b^{104}=1$ is automatic modulo $3$ and $5$. Modulo $7$ it has
$$
\gcd(104,6)=2
$$
solutions. Thus there are $2\cdot4\cdot2=16$ Fermat <bases>, giving proportion
$$
\boxed{\frac{16}{48}=\frac13}.
$$

For the Euler condition, $b^{52}=1$ modulo $3$, so the Jacobi symbol must be $+1$. The power condition is automatic modulo $5$, while modulo $7$ it again restricts $b$ to the two solutions of $b^2=1$. For each of those two residues, exactly half of the $2\cdot4$ choices modulo $3$ and $5$ have Jacobi symbol $+1$. Hence there are $2\cdot4=8$ Euler <bases> and proportion
$$
\boxed{\frac8{48}=\frac16}.
$$

Solved by gpt-5.6-sol high.

= 2G
{parent=Paper 1}
{scope}
{title2=Topics in Analysis}

= a
{parent=2g}
{scope}

= Solution
{parent=a}

The $n$th <Chebyshev polynomial> is defined on $[-1,1]$ by
$$
T_n(x)=\cos(n\arccos x).
$$
The cosine addition formula gives
$$
T_0(x)=1,
\qquad T_1(x)=x,
\qquad T_{n+1}(x)=2xT_n(x)-T_{n-1}(x).
$$
Induction therefore shows that every $T_n$ is a <polynomial>. Its defining cosine formula also gives
$$
\boxed{-1\leq T_n(x)\leq1\quad(-1\leq x\leq1)}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=2g}
{scope}

= Solution
{parent=b}

The assertion is clear for $T_1=x$. If $T_n$ has degree $n$ and leading coefficient $2^{n-1}$, the recurrence shows that the leading term of $T_{n+1}$ is
$$
2x\cdot2^{n-1}x^n=2^nx^{n+1};
$$
the lower-degree <polynomial> $T_{n-1}$ cannot cancel it. Hence
$$
\boxed{\operatorname{lc}(T_n)=2^{n-1}\quad(n\geq1)}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=2g}
{scope}

= Solution
{parent=c}

For $x=\cos\theta\in[-1,1]$,
$$
T_n(-x)=\cos(n(\pi-\theta))=(-1)^n\cos(n\theta)=(-1)^nT_n(x).
$$
Both sides are <polynomials> in $x$. Since they agree on an interval, the <polynomial> identity theorem extends the equality to every real $x$. Thus $T_n$ is even for even $n$ and odd for odd $n$.

Solved by gpt-5.6-sol high.

= d
{parent=2g}
{scope}

= Solution
{parent=d}

The $n$ roots of $T_n$ are distinct and lie in $(-1,1)$. Repeated use of Rolle's theorem shows that every root of $T_n^{(k)}$, for $0\leq k<n$, also lies in $(-1,1)$. Its leading coefficient is positive, so
$$
T_n^{(k)}(x)>0\qquad(x\geq1).
$$
Taking $k=r+1$ shows that
$$
\frac d{dx}T_n^{(r)}(x)=T_n^{(r+1)}(x)>0
$$
for $0\leq r\leq n-1$; when $r=n-1$, the <derivative> is the positive constant $T_n^{(n)}$. This is the <positivity of Chebyshev derivatives beyond the unit interval>.

Solved by gpt-5.6-sol high.

= e
{parent=2g}
{scope}

= Solution
{parent=e}

Writing $x=\cos\theta$ and taking the <limit> as $\theta\to0$ gives
$$
T_n'(1)=\lim_{\theta\to0}
\frac{n\sin(n\theta)}{\sin\theta}=n^2.
$$
Part (d), with $r=1$, shows that $T_n'$ is increasing on $[1,\infty)$, so
$$
T_n(x)-T_n(1)=\int_1^xT_n'(t)\,dt
\geq n^2(x-1).
$$
Since $T_n(1)=1$ and $n^2\geq n$,
$$
\boxed{T_n(x)\geq1+n(x-1)\quad(x\geq1)}.
$$

Solved by gpt-5.6-sol high.

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

= Solution
{parent=3K}

Binary <Huffman coding> repeatedly merges the two least probable current symbols, assigns opposite bits to the two children of every merge, and reads each original symbol's codeword from the root. The resulting prefix code minimizes expected word length.

The merge weights here may be chosen as
$$
.02+.03=.05,\qquad .04+.04=.08,\qquad
.05+.08=.13,\qquad .12+.13=.25,\qquad
.25+.26=.51,\qquad .49+.51=1.
$$
One corresponding code is
$$
\begin{array}{c|ccccccc}
 &x_1&x_2&x_3&x_4&x_5&x_6&x_7\\ \hline
\text{code}&0&10&110&11110&11111&11101&11100.
\end{array}
$$
Its expected word length is
$$
\boxed{.49+2(.26)+3(.12)+5(.04+.04+.03+.02)=2.02}.
$$

Solved by gpt-5.6-sol high.

= 4J
{parent=Paper 1}
{scope}
{title2=Automata and Formal Languages}

= a
{parent=4j}
{scope}

= Solution
{parent=a}

Take the productions
$$
S\to AA,
\qquad A\to a,
$$
and the word $w=aa$. There is only one parse tree: its root uses $S\to AA$ and each child uses $A\to a$. There are nevertheless exactly two derivations,
$$
S\Rightarrow AA\Rightarrow aA\Rightarrow aa
$$
and
$$
S\Rightarrow AA\Rightarrow Aa\Rightarrow aa,
$$
according as the left or right occurrence of $A$ is replaced first. This illustrates that derivations record replacement order while a <context-free grammar> parse tree does not.

Solved by gpt-5.6-sol high.

= b
{parent=4j}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

The first displayed tree supplies the productions $S\Rightarrow AS$ and $S\Rightarrow AA$, as well as an $A$-parse tree producing $ab$. By assumption there is also an $A$-parse tree producing $baba$. Substitute these trees as follows:
$$
S\Rightarrow AS\Rightarrow A(AA)
\Rightarrow ab\,baba\,ab.
$$
The resulting yield is
$$
\boxed{abbabaab\in L(G)}.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

The displayed trees provide $A$-parse trees producing $ab$, $bb$, and $aa$, together with the same root productions as above. Therefore
$$
S\Rightarrow AS\Rightarrow A(AA)
\Rightarrow ab\,bb\,aa,
$$
whose yield is
$$
\boxed{abbbaa\in L(G)}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=4j}
{scope}

= Solution
{parent=c}

Assume that $L$ is context-free and let $p$ be its pumping length. Apply the <pumping lemma for context-free languages> to
$$
w=a^pb^pa^p.
$$
Write $w=uvxyz$, with $|vxy|\leq p$ and $|vy|>0$. The substring $vxy$ meets at most two of the three constant-letter blocks.

If $v$ and $y$ affect only the first block, only the middle block, or only the last block, pumping down immediately makes the final block length differ from the minimum of the first two lengths. If they meet the first and second blocks, pumping down decreases at least one of those block lengths while leaving the final length $p$, again violating the defining minimum.

It remains that they meet the second and third blocks. Let pumping change their lengths by $s$ and $t$, respectively. Pumping down fails unless $s=t$. If $s=t>0$, pumping up gives lengths
$$
p,\qquad p+s,\qquad p+s,
$$
whose last entry is not $\min(p,p+s)=p$. Thus some pumping exponent always leaves $L$, contradicting the lemma. Hence
$$
\boxed{L\text{ is not context-free}}.
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=5l}
{scope}

= Solution
{parent=a}

A <generalized linear model> has three components:

* a random component in which independent responses belong to an exponential dispersion family;
* a systematic component $\eta_i=x_i^T\beta$;
* a link <function> satisfying $g(\mu_i)=\eta_i$, where $\mu_i=\mathbb E Y_i$.

Solved by gpt-5.6-sol high.

= b
{parent=5l}
{scope}

= Solution
{parent=b}

Write the exponential-dispersion density as
$$
f(y_i;\theta_i,\phi)
=\exp\left\{\frac{y_i\theta_i-b(\theta_i)}{\phi}+c(y_i,\phi)\right\}.
$$
With the canonical link, $\theta_i=\eta_i=x_i^T\beta$. Therefore
$$
\boxed{
\mathbb E Y_i=b'(x_i^T\beta),
\qquad
\operatorname{Var}(Y_i)=\phi b''(x_i^T\beta)}.
$$
Known prior weights replace $\phi$ by $\phi/w_i$.

Solved by gpt-5.6-sol high.

= c
{parent=5l}
{scope}

= Solution
{parent=c}

In the Poisson GLM the responses are independent and
$$
Y_i\sim\operatorname{Poisson}(\mu_i).
$$
The <Poisson canonical link> is logarithmic, so
$$
\boxed{\log\mu_i=x_i^T\beta,
\qquad \mu_i=e^{x_i^T\beta},
\qquad \operatorname{Var}(Y_i)=\mu_i}.
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=6a}
{scope}

= Solution
{parent=a}

For $F(n)=\alpha n-\beta n^3$, the nonnegative equilibria are
$$
\boxed{n=0,
\qquad n=K:=\sqrt{\alpha/\beta}}.
$$
Since
$$
F'(0)=\alpha>0,
\qquad F'(K)=\alpha-3\beta K^2=-2\alpha<0,
$$
the zero equilibrium is unstable and $K$ is asymptotically stable.

Solved by gpt-5.6-sol high.

= b
{parent=6a}
{scope}

= Solution
{parent=b}

For $0<n<K$, $F(n)>0$, so solutions increase monotonically toward $K$. For $n>K$, $F(n)<0$, so they decrease monotonically toward $K$. The solution $n=0$ remains at zero. Thus a phase-line sketch has arrows away from $0$ and toward $K$; a time-series sketch has all positive curves tending to the same horizontal asymptote $K$. This is the <cubic saturation population model>.

Solved by gpt-5.6-sol high.

= c
{parent=6a}
{scope}

= Solution
{parent=c}

With the same low-density rate and carrying capacity, the two per-capita growth rates are
$$
\frac{\dot n}{n}=\alpha\left(1-\frac{n^2}{K^2}\right)
$$
for the cubic model and
$$
\frac{\dot n}{n}=\alpha\left(1-\frac nK\right)
$$
for logistic growth. The logistic correction is linear in a small population, whereas the cubic correction is quadratic. Consequently the cubic trajectory initially follows pure exponential growth more closely and lies above the logistic trajectory before both approach $K$. Plotting measured per-capita growth against $n$ distinguishes a linear decline from a curve with zero slope at the origin.

Solved by gpt-5.6-sol high.

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

= a
{parent=7d}
{scope}

= Solution
{parent=a}

For singularities $a<c_1<\cdots<c_N<b$, the Cauchy principal value is
$$
\boxed{
\operatorname{PV}\int_{-\infty}^{\infty}f(x)\,dx
=\lim_{\varepsilon\downarrow0}
\left[
\int_{-\infty}^{c_1-\varepsilon}f(x)\,dx+
\sum_{j=1}^{N-1}\int_{c_j+\varepsilon}^{c_{j+1}-\varepsilon}f(x)\,dx+
\int_{c_N+\varepsilon}^{\infty}f(x)\,dx
\right]},
$$
where the assumed tail <integrals> make the two unbounded pieces meaningful. Equivalently, remove a symmetric interval of radius $\varepsilon$ about each $c_j$ and take the <limit> of the remaining <integral>.

Solved by gpt-5.6-sol high.

= b
{parent=7d}
{scope}

= Solution
{parent=b}

Using the convention adopted here, the <Hilbert transform> is
$$
\boxed{
Hf(y)=\frac1\pi\operatorname{PV}
\int_{-\infty}^{\infty}\frac{f(x)}{y-x}\,dx}.
$$
The convention with denominator $x-y$ differs by a minus sign.

Solved by gpt-5.6-sol high.

= c
{parent=7d}
{scope}

= Solution
{parent=c}

For $f(x)=1/(x^2+1)$, contour integration or partial fractions gives
$$
Hf(y)=\frac{y}{1+y^2}
$$
under the convention in part (b). Therefore
$$
\boxed{Hf(-1)=-\frac12}.
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=8e}
{scope}

= Solution
{parent=a}

Let $\theta$ be the angle from the downward vertical to the hoop's centre, measured at the suspension point, and let $\phi$ be the absolute angle from the downward vertical to the bead's radius from the hoop centre. The bead position is the sum of two <vectors> of length $a$. Including the hoop's moment of inertia about the pivot gives
$$
T=\frac{ma^2}{2}
\left(3\dot\theta^2+\dot\phi^2
+2\dot\theta\dot\phi\cos(\theta-\phi)\right),
$$
while
$$
V=-mga(2\cos\theta+\cos\phi).
$$
Thus one suitable <Lagrangian> is
$$
\boxed{
L=\frac{ma^2}{2}
\left(3\dot\theta^2+\dot\phi^2
+2\dot\theta\dot\phi\cos(\theta-\phi)\right)
+mga(2\cos\theta+\cos\phi)}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=8e}
{scope}

= Solution
{parent=b}

To quadratic order about $\theta=\phi=0$,
$$
T_2=\frac{ma^2}{2}
\begin{pmatrix}\dot\theta&\dot\phi\end{pmatrix}
\begin{pmatrix}3&1\\1&1\end{pmatrix}
\binom{\dot\theta}{\dot\phi},
$$
and
$$
V_2=\frac{mga}{2}
\begin{pmatrix}\theta&\phi\end{pmatrix}
\begin{pmatrix}2&0\\0&1\end{pmatrix}
\binom{\theta}{\phi}.
$$
For a normal mode, put $c=\omega^2a/g$. The characteristic equation is
$$
\det\begin{pmatrix}2-3c&-c\\-c&1-c\end{pmatrix}
=2c^2-5c+2=0,
$$
so $c=1/2$ or $2$. The <small oscillations of a hoop with a sliding bead> therefore have
$$
\boxed{\omega_1=\frac1{\sqrt2}\sqrt{\frac ga},
\qquad
\omega_2=\sqrt2\sqrt{\frac ga}}.
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=9d}
{scope}

= Solution
{parent=a}

The slow-roll equations imply
$$
\frac1a\frac{da}{d\phi}
=\frac{H}{\dot\phi}
=-\frac{3H^2}{c^2V'(\phi)}
=-\frac{8\pi G}{c^4}\frac{V}{V'}.
$$
For $V=\lambda\phi^n/n$, $V/V'=\phi/n$. Integrating from $\phi_i$ and using $a(\phi_i)=1$ gives
$$
\boxed{
a(\phi)=\exp\left[
\frac{4\pi G}{c^4n}(\phi_i^2-\phi^2)
\right]}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=9d}
{scope}

= Solution
{parent=b}

From $3H\dot\phi=-c^2V'$ and $H^2=8\pi GV/(3c^2)$,
$$
\frac{\dot\phi^2}{2c^2}
=\frac{c^2V'^2}{18H^2}
=\boxed{\frac{c^4n^2V(\phi)}{48\pi G\phi^2}}.
$$
The <slow-roll approximation> fails when this kinetic contribution is comparable with $V$, giving
$$
\boxed{\phi_{\rm end}\simeq
\frac{nc^2}{\sqrt{48\pi G}}}.
$$

The number of e-folds is
$$
N_e=\log\frac{a_{\rm end}}{a_i}
=\frac{4\pi G}{c^4n}
(\phi_i^2-\phi_{\rm end}^2).
$$
Using the estimate above,
$$
N_e\simeq\frac n{12}
\left[\left(\frac{\phi_i}{\phi_{\rm end}}\right)^2-1\right].
$$
For $n=6$ and the roughly $60$ e-folds needed for the flatness problem,
$$
\boxed{\phi_i\simeq11\phi_{\rm end}}.
$$
This is the <monomial slow-roll e-fold count>.

Solved by gpt-5.6-sol high.

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

= a
{parent=10e}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

Let $| -\rangle=(|0\rangle-|1\rangle)/\sqrt2$. After the Hadamards and the oracle, <Bernstein-Vazirani phase kickback> gives
$$
\boxed{
|\Phi_1\rangle
=\frac1{2^{n/2}}
\sum_{x\in\{0,1\}^n}
(-1)^{a\cdot x\oplus b}|x\rangle| -\rangle}.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

The Walsh-Hadamard identity
$$
H^{\otimes n}
\frac1{2^{n/2}}\sum_x(-1)^{a\cdot x}|x\rangle=|a\rangle
$$
and $H| -\rangle=|1\rangle$ give
$$
\boxed{|\Phi_2\rangle=(-1)^b|a\rangle|1\rangle}.
$$

Solved by gpt-5.6-sol high.

= iii
{parent=a}
{scope}

= Solution
{parent=iii}

A final oracle call evaluates $f$ on the definite input $a$:
$$
\boxed{
|\Phi_3\rangle=(-1)^b|a\rangle
|1\oplus f(a)\rangle
=(-1)^b|a\rangle
|1\oplus a\cdot a\oplus b\rangle}.
$$
Since $a\cdot a$ modulo two is the parity of the Hamming weight of $a$, this form also determines the last-qubit measurement.

Solved by gpt-5.6-sol high.

= b
{parent=10e}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

The first register of $|\Phi_3\rangle$ is exactly $|a\rangle$, so the probability of observing the string $a$ is
$$
\boxed{1}.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

Measuring the first register does not disturb the product-state last register. Its state is
$$
\boxed{|1\oplus a\cdot a\oplus b\rangle}.
$$

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

If $a$ has odd Hamming weight, then $a\cdot a=1$ and the final bit is $b$, so the probability is $1$. If $a$ has even Hamming weight, then $a\cdot a=0$ and the final bit is $1\oplus b$, so the probability is $0$. Thus
$$
\boxed{
\mathbb P(\text{last outcome}=b)=
\begin{cases}1,&|a|\text{ odd},\\0,&|a|\text{ even}.
\end{cases}}
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=11k}
{scope}

= Solution
{parent=a}

For a received word $y$, the ideal-observer rule chooses a message $i$ maximizing the posterior probability
$$
\mathbb P(M=i\mid Y=y).
$$
Maximum-likelihood decoding chooses $i$ maximizing
$$
\mathbb P(Y=y\mid M=i),
$$
and minimum-distance decoding chooses a codeword minimizing its Hamming distance from $y$.

Bayes' formula gives
$$
\mathbb P(M=i\mid Y=y)
\propto \mathbb P(Y=y\mid M=i)\mathbb P(M=i).
$$
Equal message priors therefore make ideal-observer and maximum-likelihood decoding identical. On a <binary symmetric channel>, if $d=d_H(y,c_i)$, then
$$
\mathbb P(Y=y\mid M=i)=p^d(1-p)^{n-d}.
$$
For $p<1/2$, this strictly decreases with $d$, so maximum likelihood and minimum distance agree.

Solved by gpt-5.6-sol high.

= b
{parent=11k}
{scope}

= Solution
{parent=b}

Choose $\delta$ with
$$
p<\delta<\frac14.
$$
A greedy packing construction gives binary length-$n$ codes of minimum distance greater than $2\delta n$ and size at least
$$
\frac{2^n}{\sum_{j\leq2\delta n}\binom nj}.
$$
Their asymptotic rate is at least
$$
1-h_2(2\delta)>0,
$$
where $h_2$ is binary <Binary entropy>. Minimum-distance decoding corrects every error pattern of weight at most $\delta n$. Since a channel error count is $\operatorname{Bin}(n,p)$, the law of large numbers gives
$$
\mathbb P\{\operatorname{Bin}(n,p)>\delta n\}\longrightarrow0.
$$
Thus a fixed positive rate is achievable with error tending to zero, proving that the operational capacity is nonzero. This is the <positive-rate coding bound below one-quarter crossover>.

Solved by gpt-5.6-sol high.

= c
{parent=11k}
{scope}

= Solution
{parent=c}

<Shannon second coding theorem> states that the operational capacity of a discrete memoryless channel equals
$$
C=\max_{P_X}I(X;Y):
$$
every rate below $C$ is achievable with error probability tending to zero, while rates above $C$ are not.

Put $d=b-a$. If $d=0$, then $Y=X+a$, so $X$ is recovered exactly and $C=1$ bit. If $|d|\geq2$, the two output supports
$$
\{a,b\},\qquad\{a+1,b+1\}
$$
are disjoint, so again $Y$ determines $X$ and $C=1$.

If $|d|=1$, one output value is common to both inputs and occurs with probability $1/2$ independently of the input, while either of the other two values reveals the input. The channel is therefore a binary erasure channel with erasure probability $1/2$, whose capacity is $1/2$ bit. Hence
$$
\boxed{
C=\begin{cases}
1/2,&|b-a|=1,\\
1,&|b-a|\ne1.
\end{cases}}
$$

Solved by gpt-5.6-sol high.

= 12J
{parent=Paper 1}
{scope}
{title2=Automata and Formal Languages}

= a
{parent=12j}
{scope}

= Solution
{parent=a}

The upper register index is the largest register number occurring in an instruction of $P$, with value zero if only register $0$ is used. A configuration records the current state and the contents of every register up to that index.

For input $\vec w$, let $C(0,M,\vec w)$ be the initial configuration: the designated initial state, the words of $\vec w$ in the input registers, and the empty word in every remaining register. Recursively, if $C(t,M,\vec w)$ is halting, keep it fixed; otherwise let $C(t+1,M,\vec w)$ be the unique configuration into which $M$ transforms it. The resulting <sequence>
$$
\boxed{\{C(t,M,\vec w):t\in\mathbb N\}}
$$
is the computation <sequence>.

Solved by gpt-5.6-sol high.

= b
{parent=12j}
{scope}

= Solution
{parent=b}

Choose an input word $w$ of length at least $n+1$. The final value $f_{M,1}(w)=\chi_L(w)$ has length at most one. Every symbol initially in register $0$ must either be deleted or transferred elsewhere, and either action begins with a remove instruction
$$
-(0,q,q').
$$
No other instruction can lower the length of register $0$. Consequently at least
$$
|w|-1\geq n
$$
such remove instructions occur in this computation. Since words of arbitrary length exist, the assertion follows.

Solved by gpt-5.6-sol high.

= c
{parent=12j}
{scope}

= Solution
{parent=c}

Two registers suffice. Use register $0$ for the input and register $1$ as a stack of markers.

The finite control first rejects the empty word. While the next input symbol is $a$, remove it from register $0$ and append one marker to register $1$. On seeing the first $b$, enter a second phase. For every $b$ removed from register $0$, remove one marker from register $1$; reject if a marker is unavailable or if an $a$ is encountered in this phase. Accept exactly when both registers become empty simultaneously.

The first phase stores precisely the number of $a$'s, and the second compares it with the number of $b$'s. Register $1$ is the only scratch register, so the construction is a one-register-machine computation in the question's terminology. Therefore
$$
\boxed{\{a^nb^n:n>0\}\text{ is $1$-computable}}.
$$

Solved by gpt-5.6-sol high.

= d
{parent=12j}
{scope}

= Solution
{parent=d}

Consider computations while a terminal block of $b$'s is removed from register $0$. Until that block has disappeared, the available instruction and next state depend on the current state and on the visible $b$, not on the untouched prefix $x$.

Take a block longer than $|Q|$. Record the state after each successive $b$ has been removed. Two records have the same state $q$, by the pigeonhole principle. If the corresponding removed tail lengths are $k\ne\ell$, the same finite instruction <sequences> work in front of every prefix $x$. Thus there are fixed times $t,t'$ such that
$$
\boxed{
C(t,M,xb^k)=(q,x)=C(t',M,xb^\ell)
}
$$
for every word $x$.

Solved by gpt-5.6-sol high.

= e
{parent=12j}
{scope}

= Solution
{parent=e}

Suppose a zero-register machine decided $L=\{a^nb^n:n>0\}$, and choose $k\ne\ell$ as in part (d). Take $m\geq\max\{k,\ell\}$ and set
$$
x=a^mb^{m-k}.
$$
Then
$$
xb^k=a^mb^m\in L,
$$
whereas
$$
xb^\ell=a^mb^{m-k+\ell}\notin L.
$$
Part (d) puts both computations into the identical configuration $(q,x)$. Determinism forces their subsequent computations and outputs to agree, a contradiction. Therefore
$$
\boxed{L\text{ is not $0$-computable}}.
$$
This is the <one-register separation for equal block lengths>.

Solved by gpt-5.6-sol high.

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

= a
{parent=13l}
{scope}

= Solution
{parent=a}

The fitted <normal linear model> is
$$
Y_i=\beta_0+\beta_1\,\mathrm{weight}_i
+\beta_2\,\mathrm{width}_i+\varepsilon_i,
\qquad
\varepsilon_i\mathrel{\mathrm{iid}}{\sim}N(0,\sigma^2),
$$
for $i=1,\ldots,173$.

Solved by gpt-5.6-sol high.

= b
{parent=13l}
{scope}

= Solution
{parent=b}

Let $X$ be the $173\times3$ design <matrix> and write $p=3$. The displayed estimate in row $j$ is
$$
\widehat\beta_j,\qquad
\widehat\beta=(X^TX)^{-1}X^TY.
$$
With
$$
s^2=\frac{\operatorname{RSS}}{n-p}
=\frac{\|Y-X\widehat\beta\|^2}{170},
$$
its standard error is
$$
\operatorname{se}(\widehat\beta_j)
=s\sqrt{[(X^TX)^{-1}]_{jj}}.
$$
The $t$-value and two-sided $p$-value are
$$
t_j=\frac{\widehat\beta_j}{\operatorname{se}(\widehat\beta_j)},
\qquad
2\mathbb P\{|T_{170}|\geq|t_j|\},
$$
respectively.

Solved by gpt-5.6-sol high.

= c
{parent=13l}
{scope}

= Solution
{parent=c}

The nested-model $F$-test has hypotheses
$$
H_0:\beta_1=\beta_2=0,
\qquad
H_1:(\beta_1,\beta_2)\ne(0,0).
$$
Its statistic is
$$
F=\frac{(1704.9-1477.7)/2}{1477.7/170}=13.069.
$$
The reported $p$-value $5.252\times10^{-6}$ gives strong evidence against $H_0$: weight and width are jointly useful.

Neither individual coefficient is significant at $5\%$. This is consistent with the predictor correlation $0.877$: strong <multicollinearity> makes the two partial effects difficult to distinguish and inflates their individual standard errors, even though their joint contribution is clear.

Solved by gpt-5.6-sol high.

= d
{parent=13l}
{scope}

= Solution
{parent=d}

With a well-fitting Gaussian linear model, the residual-versus-fitted plot should be an unstructured horizontal band around zero with roughly constant width. The normal Q-Q plot should lie close to a straight line.

Here the residual spread grows with the fitted value and the response's discreteness produces pronounced bands, indicating nonconstant variance. The Q-Q plot bends strongly upward in the upper tail, indicating right-skewness or heavy positive tails and failure of Gaussian residuals. Thus both homoscedasticity and normality are doubtful.

Solved by gpt-5.6-sol high.

= e
{parent=13l}
{scope}

= Solution
{parent=e}

Two natural improvements are:

* fit a count-response GLM, initially a Poisson model with log link, or a negative-binomial model if the data are overdispersed;
* allow nonlinear predictor effects, for example quadratic terms or smooth <functions> of weight and width, with model selection or validation used to control complexity.

The first addresses discreteness and mean-dependent variance; the second addresses systematic lack of fit in the conditional mean.

Solved by gpt-5.6-sol high.

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

= a
{parent=14d}
{scope}

= Solution
{parent=a}

A system is causal if outputs up to time $t$ depend only on inputs up to time $t$. For a time-invariant convolution system this is equivalent to an impulse response $g(t)$ that vanishes for $t<0$.

It is BIBO stable if every bounded input produces a bounded output. For a causal convolution system, absolute integrability
$$
\int_0^\infty|g(t)|\,dt<\infty
$$
is the standard stability criterion.

Solved by gpt-5.6-sol high.

= b
{parent=14d}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

Taking Laplace transforms and using $y(0)=0$ gives
$$
(\alpha s+1)Y(s)=F(s).
$$
Hence the transfer <function> is
$$
\boxed{G(s)=\frac{Y(s)}{F(s)}=\frac1{\alpha s+1}}.
$$
Its pole is at $-1/\alpha$, and its causal impulse response is
$$
g(t)=\frac1\alpha e^{-t/\alpha}H(t).
$$
It is absolutely integrable exactly when
$$
\boxed{\alpha>0}.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

For $\alpha>0$,
$$
G_{\rm CL}(s)
=\frac1{\alpha s+1+k}.
$$
Its only pole is $-(1+k)/\alpha$, so direct inspection gives stability exactly when
$$
\boxed{k>-1}
$$
apart from the excluded boundary.

For the <Nyquist stability criterion>, let $L(s)=kG(s)$. The open loop has no right-half-plane poles, so $P=0$. On the clockwise right-half-plane contour, the winding number $N$ of $1+L$ about zero, equivalently of $L$ about $-1$, satisfies
$$
N=P-Z,
$$
where $Z$ is the number of unstable closed-loop poles. The Nyquist locus
$$
L(i\omega)=\frac{k}{1+i\alpha\omega}
$$
is the circle with diameter joining $0$ and $k$ on the real axis, completed by its conjugate half.

If $k>-1$, this circle does not wind around $-1$, so $N=0$, $Z=0$, and the loop is stable. If $k<-1$, it winds once with $N=-1$, so $Z=1$, and the loop is unstable. Nyquist therefore gives precisely the same condition $k>-1$.

Solved by gpt-5.6-sol high.

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

= a
{parent=15d}
{scope}

= Solution
{parent=a}

The radial equation is
$$
\ddot r=-\frac{GM}{r^2}+\frac{\Lambda c^2}{3}r.
$$
Multiplying by $\dot r$ and integrating gives
$$
\frac12\dot r^2-\frac{GM}{r}
-\frac{\Lambda c^2}{6}r^2=E.
$$
Since $\rho$ is an energy density,
$$
M=\frac{4\pi r^3}{3c^2}\rho.
$$
Set $r=ax_0$, divide the energy equation by $r^2/2$, and define
$$
\frac{2E}{x_0^2}=-kc^2.
$$
Then
$$
\boxed{
\left(\frac{\dot a}{a}\right)^2
=\frac{8\pi G}{3c^2}\rho-\frac{kc^2}{a^2}
+\frac{\Lambda c^2}{3}}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=15d}
{scope}

= Solution
{parent=b}

Differentiate the Friedmann equation and use
$$
\dot\rho=-3H(\rho+P),
\qquad
\dot H=\frac{\ddot a}{a}-H^2.
$$
After cancelling the curvature term with the original Friedmann equation, one obtains the <Friedmann acceleration equation>
$$
\boxed{
\frac{\ddot a}{a}
=-\frac{4\pi G}{3c^2}(\rho+3P)
+\frac{\Lambda c^2}{3}}.
$$

The Newtonian derivation does not explain why <pressure> gravitates through the $3P$ term; that input enters only through a relativistic-looking continuity law. It also treats space and time as Euclidean and absolute, so it cannot supply the spacetime meaning of curvature $k$, horizons, or the global geometry of the cosmological model.

Solved by gpt-5.6-sol high.

= c
{parent=15d}
{scope}

= Solution
{parent=c}

Radiation conservation gives
$$
\rho_R=\rho_{R0}a^{-4}.
$$
Consequently
$$
H^2=H_0^2\Omega_{R0}a^{-4}
+\frac{\Lambda c^2}{3}.
$$
Evaluating at $t_0$, where $a=1$, shows that
$$
\frac{\Lambda c^2}{3}=H_0^2(1-\Omega_{R0}).
$$
Put
$$
A=H_0^2\Omega_{R0},
\qquad
B=H_0^2(1-\Omega_{R0}),
\qquad
b=a^2.
$$
Then $\dot b=2\sqrt{A+Bb^2}$. Choosing the big bang as $t=0$ and integrating gives
$$
b=\sqrt{\frac AB}\sinh(2\sqrt B\,t).
$$
Therefore
$$
\boxed{
a(t)=\alpha[\sinh(\beta t)]^{1/2}},
$$
where
$$
\boxed{
\alpha=\left(\frac{\Omega_{R0}}{1-\Omega_{R0}}\right)^{1/4},
\qquad
\beta=2H_0\sqrt{1-\Omega_{R0}}}.
$$

At early times, $\sinh(\beta t)\sim\beta t$, so $a\propto t^{1/2}$, as in a radiation-dominated universe. At late times,
$$
a\sim\frac{\alpha}{\sqrt2}e^{\beta t/2},
$$
the expected de Sitter expansion with $H\to\sqrt{\Lambda c^2/3}$.

<Acceleration> begins when the radiation deceleration and cosmological-constant <acceleration> balance:
$$
H_0^2\Omega_{R0}a^{-4}
=H_0^2(1-\Omega_{R0}).
$$
Thus $a=\alpha$, which in the exact solution means $\sinh(\beta t_\Lambda)=1$. Hence
$$
\boxed{
t_\Lambda=\frac{\operatorname{arsinh}1}
{2H_0\sqrt{1-\Omega_{R0}}}}.
$$
This is the <radiation-to-cosmological-constant transition>.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=16I}

The <soundness theorem for propositional logic> says that if $\Gamma\vdash t$, then every valuation satisfying every member of $\Gamma$ also satisfies $t$.

The proposed <function> need not respect the connectives. For example, if a primitive proposition $p$ is independent of $S$, then neither $p$ nor $\neg p$ is provable from $S$, so the definition gives
$$
v(p)=v(\neg p)=0.
$$

Order the consistent supersets of $S$ by inclusion. The union of a chain is consistent, since a finite proof of a contradiction would already use assumptions from one member of the chain. <Zorn lemma> therefore gives a maximal consistent extension $T$. It is deductively closed: if $T\vdash t$, adjoining $t$ preserves consistency, so maximality forces $t\in T$. Moreover, for every $t$, exactly one of $t$ and $\neg t$ belongs to $T$. They cannot both belong by consistency; if neither belonged, the inconsistency of both proper extensions would give $T\vdash\neg t$ and $T\vdash\neg\neg t$, again a contradiction. The usual induction on formulae now shows that
$$
v(t)=1\quad\Longleftrightarrow\quad t\in T
$$
defines a valuation satisfying $T$, and hence $S$.

Now suppose every finite subset of $S$ has a model. By soundness every finite subset is consistent. Any proof of a contradiction from $S$ uses only finitely many assumptions, so $S$ itself is consistent. Applying the preceding maximal-consistent-extension construction gives a model of $S$. This proves the <propositional compactness theorem>.

Solved by gpt-5.6-sol high.

= i
{parent=16i}
{scope}

= Solution
{parent=i}

For distinct $x,x'\in X$ and distinct $y,y'\in Y$, take all clauses
$$
\neg(p_{x,y}\wedge p_{x,y'}),
\qquad
\neg(p_{x,y}\wedge p_{x',y}).
$$
The first family says that each $x$ is paired with at most one $y$; the second says that each $y$ is paired with at most one $x$. There is no existence clause, so the domain may be any subset of $X$. Thus the relations $f_v$ are exactly the injective partial <functions> from $X$ to $Y$.

Solved by gpt-5.6-sol high.

= ii
{parent=16i}
{scope}

= Solution
{parent=ii}

Start with the two families of uniqueness and injectivity clauses from part (i), and for every $x\in X$ add the finite clause
$$
\bigvee_{y\in A_x}p_{x,y}.
$$
This clause is a legitimate proposition because $A_x$ is finite and nonempty. It makes every $x$ have a value in $A_x$, while the earlier clauses make that value unique and make the resulting <function> injective. Conversely, every injection $g:X\to Y$ with $g(x)\in A_x$ satisfies all these clauses.

Solved by gpt-5.6-sol high.

= iii
{parent=16i}
{scope}

= Solution
{parent=iii}

No such set of propositions exists. Suppose that $S$ had precisely the total injections as its models, fix $x_0\in X$, and adjoin
$$
\neg p_{x_0,y}\qquad(y\in Y)
$$
to $S$. Every finite subset of the enlarged theory has a model. Indeed, it excludes only finitely many possible values $F\subset Y$ for $x_0$. Starting with any injection $h:X\to Y$, either $h(x_0)\notin F$, or one may choose $x_1\ne x_0$ with $h(x_1)\notin F$ and swap the values of $x_0$ and $x_1$.

The <propositional compactness theorem> would therefore give a model of the whole enlarged theory. Its relation $f_v$ is a model of $S$ but has no value at $x_0$, contradicting the assumed description of the models of $S$. This is the <compactness obstruction to expressing totality over an infinite codomain>.

Solved by gpt-5.6-sol high.

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

= a
{parent=17i}
{scope}

= Solution
{parent=a}

Let
$$
P=v_1v_2\cdots v_m
$$
be a longest path. Every neighbour of either endpoint lies on $P$. Among the $m-1$ possible cuts between consecutive vertices, mark a cut $i$ when $v_m v_i$ is an edge and also mark it when $v_1v_{i+1}$ is an edge. There are at least
$$
d(v_m)+d(v_1)\geq n>m-1
$$
marks, so some cut receives both marks. The path edges together with $v_mv_i$ and $v_{i+1}v_1$ form a cycle through all vertices of $P$.

The degree assumption makes $G$ connected: two components would each contain at least $n/2+1$ vertices. If $m<n$, connectedness gives an edge from a vertex outside the cycle to a vertex on it; breaking the cycle there produces a path with $m+1$ vertices, contrary to maximality. Hence $m=n$, so $G$ is Hamiltonian. This proves the <Dirac theorem> in this case.

Solved by gpt-5.6-sol high.

= b
{parent=17i}
{scope}

= Solution
{parent=b}

Let $P=v_0\ldots v_\ell$ be a longest path. All neighbours of $v_0$ lie on $P$, in the colour class opposite to $v_0$, so $\ell\geq2k-1$. If $\ell\geq2k$, the first $2k$ edges give the required path. Otherwise $\ell=2k-1$, and all $k$ vertices of the opposite colour on $P$ must be neighbours of $v_0$; in particular $v_0v_\ell$ is an edge, giving a cycle of length $2k$. The graph $K_{k,k}$ has minimum degree $k$ but only $2k$ vertices, so it has no path of length $2k$.

For the stronger assertion, assume there is no $4$-cycle. Rotate a longest path about each edge from $v_0$ to obtain $k$ possible endpoints in the colour class of $v_0$. Every neighbour of each rotated endpoint lies on $P$, since otherwise the corresponding path could be extended. In a $4$-cycle-free bipartite graph, two vertices in one colour class have at most one common neighbour. The <neighbourhood bound for a square-free bipartite graph> therefore shows that the colour class opposite to $v_0$ contains at least $2k-1$ vertices of $P$. If the endpoints have opposite colours, repeating the argument from $v_\ell$ gives the same bound for the other class. If they have the same colour, their class has one more vertex on $P$ than the opposite class. In either case $P$ has at least $4k-2$ vertices and length at least $4k-3$. Consequently $G$ contains either such a path or a $4$-cycle.

Solved by gpt-5.6-sol high.

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

= a
{parent=18h}
{scope}

= Solution
{parent=a}

Let $f$ be the common minimal <polynomial>. Evaluation gives
$$
\mathbb Q[X]/(f)\cong\mathbb Q(\alpha),
\qquad
\mathbb Q[X]/(f)\cong\mathbb Q(\beta),
$$
so composing the two isomorphisms gives $\mathbb Q(\alpha)\cong\mathbb Q(\beta)$ and sends $\alpha$ to $\beta$.

Take $\alpha=\beta=\sqrt[4]{2}$. The field is real and contains the two roots $\pm\alpha$ of $X^4-2$, but not $\pm i\alpha$. Hence exactly two $\mathbb Q$-automorphisms are possible, while the degree is four.

In general, after choosing one isomorphism $\phi:\mathbb Q(\alpha)\to\mathbb Q(\beta)$, every other one is uniquely $\phi\circ\sigma$ for a $\mathbb Q$-automorphism $\sigma$ of $\mathbb Q(\alpha)$. Thus
$$
k=|\operatorname{Aut}_{\mathbb Q}(\mathbb Q(\alpha))|.
$$
By the <automorphism-count divisibility theorem>, this order divides $[\mathbb Q(\alpha):\mathbb Q]$. Therefore $k$ always divides the degree.

Solved by gpt-5.6-sol high.

= b
{parent=18h}
{scope}

= Solution
{parent=b}

Put $d=[L:K]$. If $\alpha=0$ the result is immediate, so suppose $\alpha=x^n\ne0$ with $x\in L$. The <field norm> is multiplicative, sends an element of $K$ to its $d$th power, and commutes with powers. Therefore
$$
\alpha^d=N_{L/K}(\alpha)=N_{L/K}(x)^n.
$$
Choose integers $r,s$ with $rd+sn=1$. Then
$$
\alpha=(\alpha^d)^r(\alpha^n)^s
      =\bigl(N_{L/K}(x)^r\alpha^s\bigr)^n.
$$
The expression in parentheses belongs to $K$, so $\alpha$ is an $n$th power in $K$. This is <coprime-degree descent for powers>.

Solved by gpt-5.6-sol high.

= c
{parent=18h}
{scope}

= Solution
{parent=c}

Let $\beta$ satisfy $\beta^p=\alpha$. Since $K=\mathbb Q(\alpha)\subseteq\mathbb Q(\beta)$,
$$
[\mathbb Q(\beta):\mathbb Q]=[K(\beta):K][K:\mathbb Q].
$$
The <polynomial> $f(X^p)$ has degree $p[K:\mathbb Q]$ and has root $\beta$. It is irreducible exactly when $[K(\beta):K]=p$.

If $\alpha=\gamma^p$ for some $\gamma\in K$, then $\gamma$ is a root of $f(X^p)$ of degree at most $[K:\mathbb Q]$, so $f(X^p)$ is reducible. Conversely, if it is reducible, then
$$
d=[K(\beta):K]<p.
$$
Since $p$ is prime, $1\leq d<p$ implies $\gcd(d,p)=1$. Applying <coprime-degree descent for powers> to $K(\beta)/K$ and $\beta^p=\alpha$ shows that $\alpha$ is a $p$th power in $K$. Hence
$$
f(X^p)\text{ is irreducible}\quad\Longleftrightarrow\quad
\alpha\notin K^p.
$$

Solved by gpt-5.6-sol high.

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

= Solution
{parent=19H}

A complex representation of $G$ is a complex <vector space> $V$ together with a homomorphism
$$
\rho:G\longrightarrow\operatorname{GL}(V).
$$
It is faithful when $\ker\rho=\{e\}$, equivalently when $\rho$ is injective. Every finite <group> has a faithful complex representation: in the <regular representation>, $G$ permutes the <basis> $\{e_h:h\in G\}$ by
$$
\rho(g)e_h=e_{gh},
$$
and a <group> element fixing every <basis> <vector> must be the identity.

If $g$ is conjugate to $g^k$, then $\rho(g)$ is similar to $\rho(g^k)=\rho(g)^k$. Thus the two <matrices> have the same <spectrum>, while an <eigenvalue> $\lambda$ of $\rho(g)$ gives the <eigenvalue> $\lambda^k$ of $\rho(g)^k$. Hence
$$
\lambda\in S(g)\quad\Longrightarrow\quad\lambda^k\in S(g).
$$

Now take a $p$-cycle $g\in S_p$. It is conjugate to $g^k$ for every $1\leq k<p$. Faithfulness makes $\rho(g)$ have order $p$, so its <spectrum> contains a nontrivial $p$th root of unity $\lambda$. The preceding implication puts all the $p-1$ distinct values
$$
\lambda,\lambda^2,\ldots,\lambda^{p-1}
$$
in the <spectrum>. Therefore $\dim V\geq p-1$. This is the <spectrum orbit bound for a faithful symmetric-group representation>.

Finally let $G$ be the <group> of all permutations of $\mathbb N$. It has no faithful finite-dimensional complex representation. For every $n$, $G$ contains a <subgroup> $(C_2)^n$ generated by $n$ disjoint transpositions. In a $d$-dimensional complex representation, commuting involutions are simultaneously diagonalizable and hence map into the <group> of diagonal sign <matrices>, which has order $2^d$. A faithful restriction would require $2^n\leq2^d$ for every $n$, an impossibility.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=20F}

An algebraic number is an algebraic integer when it is a root of a monic <polynomial> in $\mathbb Z[X]$.

If $\alpha$ and $\beta$ are algebraic integers, the <ring> $\mathbb Z[\alpha,\beta]$ is a finitely generated $\mathbb Z$-module. Multiplication by $\alpha\beta$ preserves this module, so the <algebraic integer module criterion> shows that $\alpha\beta$ is an algebraic integer.

The <ring of integers of a quadratic field> gives
$$
\mathcal O_{\mathbb Q(\sqrt2)}=\mathbb Z[\sqrt2],
$$
since $2\equiv2\pmod4$. For a direct verification, if $x=a+b\sqrt2$ with $a,b\in\mathbb Q$ is <integral>, then its trace $2a$ and norm $a^2-2b^2$ are integers. Writing $2a=m\in\mathbb Z$ and reducing the norm condition in lowest terms shows first that $b$ can have denominator at most $2$; a denominator $2$ with odd numerator would require $m^2\equiv2\pmod4$, which is impossible. Thus $b\in\mathbb Z$, and then the norm condition forces $m$ even, so $a\in\mathbb Z$. Conversely, every $a+b\sqrt2$ with $a,b\in\mathbb Z$ is <integral> because $\sqrt2$ is <integral> and algebraic integers form a <ring>.

Now let $f$ have degree $d$ and roots $\alpha_1,\ldots,\alpha_d$. Since
$$
2=M(f)=|a_d|\prod_j\max\{1,|\alpha_j|\}
$$
and $|\alpha_1|>1$, the integer $|a_d|$ cannot be $2$; hence $|a_d|=1$. Thus $f$ is monic up to sign and $\alpha_1$ is an algebraic integer.

Let $B$ be the product of all roots with <modulus> greater than one, counting multiplicity. Complex roots occur in conjugate pairs, so $B$ is real up to the signs of the real roots and
$$
|B|=M(f)=2.
$$
As a product of algebraic integers, $B$ is an algebraic integer; the equality forces $B=\pm2$. Therefore
$$
\frac{|N_{\mathbb Q(\alpha_1)/\mathbb Q}(\alpha_1)|}{2}
$$
is the absolute value of the product of the remaining conjugates, is rational, and is an algebraic integer. It is consequently a nonzero rational integer. Every remaining factor has <modulus> at most one, so this integer is at most one. It must equal one, and hence
$$
\boxed{|N_{\mathbb Q(\alpha_1)/\mathbb Q}(\alpha_1)|=2}.
$$
This is the <mahler-measure norm argument at measure two>.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=21J}

A <covering space> is a map $p:\widetilde X\to X$ such that every $x\in X$ has an open neighbourhood $U$ for which
$$
p^{-1}(U)=\coprod_{\alpha}U_\alpha
$$
and every restriction $p:U_\alpha\to U$ is a homeomorphism.

For path lifting, cover the compact image of $\gamma$ by evenly covered sets. The <Lebesgue number lemma> supplies a subdivision
$$
0=t_0<t_1<\cdots<t_m=1
$$
for which each $\gamma([t_{j-1},t_j])$ lies in one such set. Starting in the sheet containing $\widetilde x_0$, lift the first segment with the inverse of that sheet. Its endpoint selects the sheet for the next segment, and induction constructs a continuous lift. Two lifts starting at the same point agree on the first segment because the relevant sheet map is injective, and the same argument at successive endpoints proves uniqueness.

Write $q:Y\to Y/\phi$ for the quotient map. The identity $p\circ Q=q\circ\operatorname{pr}_Y$ on $Y\times\mathbb Z$, where $Q$ is the quotient defining $\widetilde{Y/\phi}$, proves continuity of $p$ by the quotient property. Away from the two gluing regions, a small <open set> lifts to one copy in every level. Near a glued point choose $U\Subset A$ and use the matched pair
$$
U\times\{i\}\ \cup\ \phi(U)\times\{i-1\}.
$$
After the prescribed identifications, each such pair maps homeomorphically to $q(U\cup\phi(U))$, and the pairs are disjoint for different $i$. Thus $p$ is a covering map.

Each copy of $Y$ is path-connected, and adjacent copies meet through the identified copies of the nonempty set $A$, so $\widetilde{Y/\phi}$ is path-connected. Translation
$$
\tau[(y,i)]=[(y,i+1)]
$$
is a deck transformation. Its powers act freely and transitively on every fibre, so the covering is regular. The <covering-space subgroup and deck-group quotient> then gives
$$
G\trianglelefteq\pi_1(Y/\phi,[a_0]),
\qquad
\pi_1(Y/\phi,[a_0])/G\cong\langle\tau\rangle\cong\mathbb Z.
$$

Solved by gpt-5.6-sol high.

= i
{parent=21j}
{scope}

= Solution
{parent=i}

At the first subdivision interval, choose the unique sheet containing $\widetilde x_0$ and use its inverse to define the lift. This makes $\widetilde\gamma(0)=\widetilde x_0$ by construction; every later segment begins at the endpoint already obtained.

Solved by gpt-5.6-sol high.

= ii
{parent=21j}
{scope}

= Solution
{parent=ii}

On each subdivision interval the lift is the inverse of one restriction $p:U_\alpha\to U$, so $p\circ\widetilde\gamma=\gamma$ there and therefore on all of $[0,1]$. If two such lifts agree at the start of an interval, injectivity on the selected sheet makes them agree throughout it. Induction proves uniqueness.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=22G}

The <closest point theorem in a Hilbert space> states that every nonempty closed convex subset $C$ of a Hilbert space has a unique point nearest to any given $x$.

Let $d=\inf_{y\in C}\|x-y\|$ and choose $y_n\in C$ with $\|x-y_n\|\to d$. The parallelogram identity and convexity give
$$
\|y_n-y_m\|^2
=2\|x-y_n\|^2+2\|x-y_m\|^2
 -4\left\|x-\frac{y_n+y_m}{2}\right\|^2
\leq2\|x-y_n\|^2+2\|x-y_m\|^2-4d^2,
$$
so $(y_n)$ is Cauchy. Completeness and closedness give a <limit> $y\in C$ with $\|x-y\|=d$. If $y,z$ were both minimizers, the same identity with their midpoint would force $\|y-z\|=0$.

Apply the theorem to a closed subspace $F$. For $x\in H$, let $y\in F$ be closest and put $z=x-y$. For every $u\in F$, minimality of $\|z-tu\|^2$ at $t=0$ for real and imaginary $t$ gives $\langle z,u\rangle=0$. Hence
$$
x=y+z\in F+F^\perp.
$$
The intersection is zero, so $H=F\oplus F^\perp$.

If $T$ is a shift with orthonormal <basis> $(e_n)$, it is an isometry,
$$
(\operatorname{Im}T)^\perp=\operatorname{span}\{e_1\},
\qquad
\bigcap_{n\geq1}\operatorname{Im}(T^n)=\{0\}.
$$
Conversely, suppose these three properties hold. Choose a unit <vector> $e_1$ spanning $(\operatorname{Im}T)^\perp$ and set $e_n=T^{n-1}e_1$. Isometry makes this <sequence> orthonormal. Iterating the orthogonal decomposition
$$
H=\operatorname{span}\{e_1\}\oplus\operatorname{Im}T
$$
gives
$$
H=\operatorname{span}\{e_1,\ldots,e_n\}\oplus\operatorname{Im}(T^n).
$$
A <vector> orthogonal to every $e_n$ lies in every $\operatorname{Im}(T^n)$ and is therefore zero. Thus $(e_n)$ is an orthonormal <basis> and $Te_n=e_{n+1}$, so $T$ is a shift. This is the <wandering-vector characterization of a unilateral shift>.

Solved by gpt-5.6-sol high.

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

= a
{parent=23g}
{scope}

= Solution
{parent=a}

Tonelli's theorem and the triangle inequality give
$$
\begin{aligned}
\int_0^\infty |G(y)g(y)|\,dy
&\leq\int_0^\infty\int_0^\infty
 |F(x,y)g(y)|\,dx\,dy\\
&=\int_0^\infty\left(\int_0^\infty
 |F(x,y)g(y)|\,dy\right)dx.
\end{aligned}
$$
Applying <Holder inequality> in the $y$ variable for each fixed $x$ yields
$$
\int_0^\infty |G(y)g(y)|\,dy
\leq
\int_0^\infty
\left(\int_0^\infty |F(x,y)|^p\,dy\right)^{1/p}
\|g\|_q\,dx,
$$
and $\|g\|_q\leq1$ gives the claimed inequality. This is the duality proof of the <Minkowski integral inequality>.

Solved by gpt-5.6-sol high.

= b
{parent=23g}
{scope}

= Solution
{parent=b}

For $f_z(y)=f(yz)$, the substitution $u=yz$ gives
$$
\|f_z\|_p^p=z^{-1}\|f\|_p^p,
\qquad
\|f_z\|_p=z^{-1/p}\|f\|_p.
$$
Homogeneity of $K$ and the substitution $x=yz$ give
$$
Tf(y)=\int_0^\infty K(yz,y)f(yz)y\,dz
     =\int_0^\infty K(z,1)f_z(y)\,dz.
$$
Apply part (a), or equivalently the <Minkowski integral inequality>, to this <integral> and use the preceding scaling identity:
$$
\begin{aligned}
\|Tf\|_p
&\leq\int_0^\infty |K(z,1)|\,\|f_z\|_p\,dz\\
&=\|f\|_p\int_0^\infty |K(z,1)|z^{-1/p}\,dz\\
&=\|f\|_p.
\end{aligned}
$$

Solved by gpt-5.6-sol high.

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

= Solution
{parent=24H}

A covering map is locally a disjoint union of homeomorphisms onto the base. A space is simply connected when it is path-connected and every loop is null-homotopic. By the <uniformization theorem>, the simply connected Riemann surfaces are
$$
\mathbb C_\infty,\qquad\mathbb C,\qquad\mathbb D
$$
up to analytic isomorphism.

A lattice is a discrete <subgroup>
$$
L=\mathbb Z\omega_1\oplus\mathbb Z\omega_2\subset\mathbb C
$$
with $\omega_1/\omega_2\notin\mathbb R$. The <Weierstrass elliptic function>
$$
\wp_L(z)=\frac1{z^2}+
\sum_{0\ne\omega\in L}
\left(\frac1{(z-\omega)^2}-\frac1{\omega^2}\right)
$$
converges normally on compact subsets of $\mathbb C\setminus L$, defines a meromorphic map $\mathbb C\to\mathbb C_\infty$, and is nonconstant because it has a double pole at every lattice point. Reindexing the normally convergent <derivative> <series> shows that $\wp_L'$ is $L$-periodic; evenness fixes the integration constants, so $\wp_L(z+\omega)=\wp_L(z)$.

The invariance makes
$$
\overline\wp([z])=\wp_L(z)
$$
well-defined and unique on $\mathbb C/L$. It is analytic because the quotient projection is locally biholomorphic.

Neither map is a covering. For any nonzero $\omega\in L$, periodicity and oddness of $\wp_L'$ give
$$
\wp_L'(\omega/2)=\wp_L'(-\omega/2)=-\wp_L'(\omega/2),
$$
so $\wp_L'(\omega/2)=0$. The map is not a local homeomorphism there; since the quotient projection is locally biholomorphic, the descended map fails for the same reason.

There is no covering $\mathbb C/L\to\mathbb C_\infty$. The torus is connected, while every connected covering of the simply connected sphere is a homeomorphism. A torus is not homeomorphic to a sphere, for example because their fundamental <groups> are $\mathbb Z^2$ and $0$.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=25F}

A nonempty topological space is irreducible when it is not the union of two proper closed subsets, equivalently when any two nonempty open subsets intersect.

The Zariski topology on $\mathbb A^n$ is Noetherian because $k[x_1,\ldots,x_n]$ satisfies the ascending chain condition on <ideals>. Suppose a closed set $X$ that is not a finite union of irreducible closed sets were minimal among such counterexamples. It is reducible, say $X=X_1\cup X_2$ with both $X_i$ proper closed subsets. Minimality expresses each $X_i$ as a finite union of irreducible closed sets, and combining those decompositions contradicts the choice of $X$. Thus every closed $X$ has a finite irreducible decomposition.

Using $x_3=x_2^2$, the second defining <polynomial> reduces to
$$
x_1^2-x_2^2-x_2^4+x_3^2=x_1^2-x_2^2.
$$
Consequently
$$
Z(x_3-x_2^2,x_1^2-x_2^2-x_2^4+x_3^2)
=
Z(x_3-x_2^2,x_1-x_2)
\cup
Z(x_3-x_2^2,x_1+x_2).
$$
Each component is the image of $t\mapsto(\pm t,t,t^2)$ and is isomorphic to $\mathbb A^1$, hence irreducible. In characteristic two the two displayed components coincide.

Suppose a <polynomial> $P(x,y)=\sum_{j=0}^m p_j(x)y^j$ vanishes on every $(x,e^x)$. Then
$$
\sum_{j=0}^m p_j(x)e^{jx}=0
$$
as an <entire function>. These exponential-polynomial <functions> are linearly independent: applying $(D-m)^{\deg p_m+1}$ kills the term with largest exponent, while it acts injectively on $p_j(x)e^{jx}$ for $j<m$; induction on the number of terms finishes the proof. Hence every $p_j$ is zero, so no nonzero <polynomial> vanishes on the graph. Its Zariski closure is all of $\mathbb A^2$.

Finally refine an open cover of an affine variety $X$ by distinguished opens $D(f_j)$, choosing one inside an original member around each point. Since these distinguished opens cover,
$$
V((f_j)_j)=\varnothing.
$$
The <Hilbert Nullstellensatz> implies that the $f_j$ generate the unit <ideal>. Thus
$$
1=g_1f_{j_1}+\cdots+g_rf_{j_r}
$$
for finitely many of them, so $D(f_{j_1}),\ldots,D(f_{j_r})$ cover $X$. The corresponding original <open sets> form a finite subcover. This is <quasi-compactness of an affine variety>.

Solved by gpt-5.6-sol high.

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

= a
{parent=26j}
{scope}

= Solution
{parent=a}

Choose a smooth <unit normal> $N$. For $v,w\in T_p\Sigma$, the <first fundamental form> and <second fundamental form> are
$$
I_p(v,w)=\langle v,w\rangle,
\qquad
II_p(v,w)=\langle S_pv,w\rangle,
$$
where the shape operator is $S_p=-dN_p$. The principal curvatures are the <eigenvalues> $\kappa_1,\kappa_2$ of $S_p$; the <mean curvature> and <Gaussian curvature> are
$$
H=\frac12\operatorname{tr}S_p=\frac{\kappa_1+\kappa_2}{2},
\qquad
K=\det S_p=\kappa_1\kappa_2.
$$
The surface is minimal when $H=0$ everywhere.

Solved by gpt-5.6-sol high.

= b
{parent=26j}
{scope}

= Solution
{parent=b}

Write $h_x=\partial_xh$ and $h_y=\partial_yh$. Since
$$
\phi_x=(1,0,h_x),\qquad \phi_y=(0,1,h_y),
$$
the first fundamental form in the coordinate <basis> is
$$
\boxed{
I^S=
\begin{pmatrix}
1+h_x^2&h_xh_y\\
h_xh_y&1+h_y^2
\end{pmatrix}.}
$$

Solved by gpt-5.6-sol high.

= c
{parent=26j}
{scope}

= Solution
{parent=c}

Put $W=\sqrt{1+h_x^2+h_y^2}$. An upward-pointing Gauss map is
$$
N=\frac{(-h_x,-h_y,1)}{W}.
$$
For <tangent vectors> $v,w$, <differentiation> of
$$
d\phi_t=d\phi+t\,dN
$$
gives
$$
\left.\frac d{dt}\right|_{t=0}I^{S_t}(v,w)
=\langle dN(v),d\phi(w)\rangle
 +\langle d\phi(v),dN(w)\rangle.
$$
The Weingarten map is self-adjoint and $S=-dN$, so the right side is $-2II^S(v,w)$. Hence
$$
II^S=-\frac12\left.\frac d{dt}\right|_{t=0}I^{S_t}.
$$
Alternatively differentiating $\langle N,\phi_i\rangle=0$ gives
$$
II_{ij}=-\langle N_i,\phi_j\rangle
       =\langle N,\phi_{ij}\rangle.
$$
Therefore
$$
\boxed{
II^S=\frac1W
\begin{pmatrix}
h_{xx}&h_{xy}\\
h_{xy}&h_{yy}
\end{pmatrix}.}
$$
This is the <fundamental forms of a graph surface>.

Solved by gpt-5.6-sol high.

= d
{parent=26j}
{scope}

= Solution
{parent=d}

The inverse first-fundamental-form <matrix> is
$$
(I^S)^{-1}=\frac1{W^2}
\begin{pmatrix}
1+h_y^2&-h_xh_y\\
-h_xh_y&1+h_x^2
\end{pmatrix}.
$$
Thus $2H=\operatorname{tr}((I^S)^{-1}II^S)$ vanishes exactly when
$$
\boxed{
(1+h_y^2)h_{xx}-2h_xh_yh_{xy}+(1+h_x^2)h_{yy}=0.}
$$
This is the <minimal surface equation for a graph>.

Solved by gpt-5.6-sol high.

= e
{parent=26j}
{scope}

= Solution
{parent=e}

The area element is $\sqrt{\det I^S}\,dx\,dy=W\,dx\,dy$, so
$$
\boxed{A(h)=\int_\Omega\sqrt{1+h_x^2+h_y^2}\,dx\,dy.}
$$
For the <Lagrangian> $L(p,q)=\sqrt{1+p^2+q^2}$, stationarity under every compactly supported variation gives the <Euler-Lagrange equation>
$$
\partial_x\!\left(\frac{h_x}{W}\right)
+\partial_y\!\left(\frac{h_y}{W}\right)=0.
$$
Multiplication by $W^3$ expands this into
$$
(1+h_y^2)h_{xx}-2h_xh_yh_{xy}+(1+h_x^2)h_{yy}=0,
$$
recovering part (d).

Solved by gpt-5.6-sol high.

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

= a
{parent=27g}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

A measure $\mu$ on $(\Omega,\mathcal F)$ is $\sigma$-finite when
$$
\Omega=\bigcup_{n\geq1}E_n
\qquad\text{with}\qquad
\mu(E_n)<\infty.
$$
The $\sigma$-finite uniqueness theorem says that two measures agreeing on a generating $\pi$-system agree on the generated $\sigma$-algebra, provided the space is covered by countably many sets in that system having finite common measure. Equivalently, a $\sigma$-finite premeasure has at most one extension to the generated $\sigma$-algebra.

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

Fix $x\in\mathbb R$ and define $\mu_x(B)=\lambda(B+x)$. This is a $\sigma$-finite measure on the Borel sets. For every half-open interval,
$$
\mu_x((a,b])=\lambda((a+x,b+x])=b-a=\lambda((a,b]).
$$
The half-open intervals form a generating $\pi$-system, so the $\sigma$-finite uniqueness theorem gives $\mu_x=\lambda$. Thus <Lebesgue measure> is translation invariant.

Solved by gpt-5.6-sol high.

= iii
{parent=a}
{scope}

= Solution
{parent=iii}

Let $\mu$ have the stated properties. Translation invariance and the disjoint partition of $(0,1]$ into $n$ intervals of length $1/n$ give
$$
\mu((0,1/n])=\frac1n.
$$
Finite additivity then gives $\mu((0,r])=r$ for every positive rational $r$. Continuity from below and above extends this to every real $r>0$, and translation invariance yields
$$
\mu((a,b])=b-a.
$$
Hence $\mu$ and $\lambda$ agree on the half-open intervals. Both are $\sigma$-finite, so uniqueness of extension gives $\mu=\lambda$ on the Borel $\sigma$-algebra.

Solved by gpt-5.6-sol high.

= b
{parent=27g}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

The convergence $X_n\to X$ in distribution means
$$
F_{X_n}(x)\longrightarrow F_X(x)
$$
for every continuity point $x$ of $F_X$.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

Let $F^{-1}(u)=\inf\{x:u\leq F(x)\}$. The generalized-inverse identity
$$
F^{-1}(u)\leq x\quad\Longleftrightarrow\quad u\leq F(x)
$$
gives
$$
\mathbb P(\widetilde X\leq x)
=\lambda\{u:u\leq F_X(x)\}=F_X(x),
$$
and similarly $\mathbb P(\widetilde X_n\leq x)=F_{X_n}(x)$. Thus the coupled variables have the required marginal distributions.

At every continuity point $u$ of the nondecreasing <function> $F_X^{-1}$, convergence of the distribution <functions> at continuity points of $F_X$ implies
$$
F_{X_n}^{-1}(u)\longrightarrow F_X^{-1}(u).
$$
Indeed, values just below and above $F_X^{-1}(u)$ can be chosen at continuity points of $F_X$ and trap the two generalized inverses. A monotone <function> has at most countably many discontinuities, so this convergence holds for Lebesgue-almost every $u\in(0,1)$. Therefore $\widetilde X_n\to\widetilde X$ almost surely. This is the <quantile coupling for convergence in distribution>.

Solved by gpt-5.6-sol high.

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

= a
{parent=28k}
{scope}

= Solution
{parent=a}

Conditioned on the jump-chain states, the successive holding times are independent exponentials with rates $q_{i_0},\ldots,q_{i_n}$. Hence
$$
q_{i_n}\mathbb P(J_n\leq t<J_{n+1}\mid Y_0=i_0,\ldots,Y_n=i_n)
$$
equals the <integral> over $s_0,\ldots,s_{n-1}\geq0$ with $\sum_{r<n}s_r\leq t$ of
$$
\left(\prod_{r=0}^{n-1}q_{i_r}e^{-q_{i_r}s_r}\right)
q_{i_n}e^{-q_{i_n}(t-\sum_{r<n}s_r)}.
$$
This simplex <integral> is unchanged by reversing the $n+1$ time portions and the rates attached to them. It is therefore
$$
q_{i_0}\mathbb P(J_n\leq t<J_{n+1}\mid Y_0=i_n,\ldots,Y_n=i_0),
$$
as required. This is <holding-time reversal along a fixed CTMC jump path>.

Solved by gpt-5.6-sol high.

= b
{parent=28k}
{scope}

= Solution
{parent=b}

The Markov property at deterministic times gives
$$
\mathbb P(Z_{n+1}=j\mid Z_0,\ldots,Z_n=i)
=\mathbb P_i(X_h=j).
$$
Thus $(Z_n)$ is a discrete-time Markov chain with transition <matrix>
$$
P(h)=(p_{ij}(h))_{i,j\in I};
$$
when the minimal chain is nonexplosive, $P(h)=e^{hQ}$.

For an irreducible chain, state $i$ is recurrent in continuous time exactly when
$$
\int_0^\infty p_{ii}(t)\,dt=\infty,
$$
and it is recurrent for the skeleton exactly when $\sum_{n\geq0}p_{ii}(nh)=\infty$. If $t\in[nh,(n+1)h]$, the event of staying at $i$ supplies
$$
e^{-q_i h}p_{ii}(nh)\leq p_{ii}(t)
\leq e^{q_i h}p_{ii}((n+1)h).
$$
Integrating over each interval shows that the <integral> diverges exactly when the <series> does. Irreducibility then makes recurrence equivalent for the two chains. This is <recurrence equivalence for a CTMC and its fixed-time skeleton>.

Solved by gpt-5.6-sol high.

= c
{parent=28k}
{scope}

= Solution
{parent=c}

On the finite state space, $P_t=e^{tQ}$ and
$$
\mathbb E_xf(X_t)=(P_tf)(x).
$$
Therefore
$$
\lim_{t\downarrow0}\frac{\mathbb E_xf(X_t)-f(x)}t
=\left(\lim_{t\downarrow0}\frac{P_t-I}{t}f\right)(x)
=Qf(x).
$$
The <matrix> semigroup satisfies $\frac d{dt}P_tf=P_tQf$. Integrating from $0$ to $t$ gives the backward <integral> equation
$$
\boxed{
\mathbb E_xf(X_t)=f(x)+\int_0^t\mathbb E_x[Qf(X_s)]\,ds.}
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=29l}
{scope}

= Solution
{parent=a}

Under the standard differentiability, identifiability, moment, and nonsingularity assumptions,
$$
\boxed{
\sqrt n(\widehat\theta-\theta_0)
\xrightarrow{d}N(0,I(\theta_0)^{-1}).}
$$
Here the one-observation score is
$$
S_\theta(X)=\frac{\partial}{\partial\theta}\log f(X,\theta),
$$
and the <Fisher information matrix> is
$$
I(\theta)=\mathbb E_\theta[S_\theta(X)S_\theta(X)^T]
=-\mathbb E_\theta\left[\frac{\partial^2}{\partial\theta\partial\theta^T}log f(X,\theta)\right].
$$

Solved by gpt-5.6-sol high.

= b
{parent=29l}
{scope}

= Solution
{parent=b}

Put $M_n=\max_{1\leq i\leq n}|X_i|$. The likelihood is
$$
L(\theta)=\frac1{(2\theta)^n}\mathbf1_{\{\theta\geq M_n\}}.
$$
It decreases on its nonzero range, so
$$
\boxed{\widehat\theta=M_n.}
$$

Solved by gpt-5.6-sol high.

= c
{parent=29l}
{scope}

= Solution
{parent=c}

For $0\leq m\leq\theta_0$,
$$
\mathbb P_{\theta_0}(M_n\leq m)
=\left(\frac m{\theta_0}\right)^n.
$$
Hence, for $z\geq0$,
$$
\mathbb P\{n(\theta_0-M_n)>z\}
=\left(1-\frac z{n\theta_0}\right)^n
\longrightarrow e^{-z/\theta_0}.
$$
Thus
$$
\boxed{n(\theta_0-\widehat\theta)\xrightarrow d Z,
\qquad Z\sim\operatorname{Exp}(1/\theta_0).}
$$
This is the <endpoint asymptotics of the symmetric-uniform maximum likelihood estimator>.

Solved by gpt-5.6-sol high.

= d
{parent=29l}
{scope}

= Solution
{parent=d}

Factorizing and using part (c),
$$
n(\widehat\theta^2-\theta_0^2)
=-n(\theta_0-\widehat\theta)(\widehat\theta+\theta_0)
\xrightarrow d -2\theta_0Z
$$
by the <Slutsky theorem>, since $\widehat\theta\to\theta_0$ in probability. Thus the limiting variable is a negative scaled exponential, with $Z\sim\operatorname{Exp}(1/\theta_0)$.

Solved by gpt-5.6-sol high.

= e
{parent=29l}
{scope}

= Solution
{parent=e}

No. Part (c) shows that $n(\widehat\theta-\theta_0)=O_{\mathbb P}(1)$, and therefore
$$
\sqrt n(\widehat\theta-\theta_0)
=\frac{n(\widehat\theta-\theta_0)}{\sqrt n}
\xrightarrow{\mathbb P}0.
$$
Every distributional <limit> is the point mass at zero and has variance zero. The regular maximum-likelihood central <limit> theorem fails because the support depends on $\theta$.

Solved by gpt-5.6-sol high.

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

= a
{parent=30l}
{scope}

= Solution
{parent=a}

In discounted terms, an arbitrage is a portfolio with zero initial value whose terminal payoff is nonnegative in every state and strictly positive with positive probability. A risk-neutral measure is a probability measure equivalent to the physical measure under which every discounted asset price is a martingale. The one-period <fundamental theorem of asset pricing> says that a finite one-period market is arbitrage-free exactly when it has a risk-neutral measure.

Solved by gpt-5.6-sol high.

= b
{parent=30l}
{scope}

= Solution
{parent=b}

If $q\in\mathcal Q$ and $\phi\in\mathcal A$, then
$$
q^TP\phi=(P^Tq)^T\phi=0,
$$
but strict positivity of $q$ and the nonzero <vector> $P\phi\geq0$ make the left side positive. Thus the two sets cannot both be nonempty.

Conversely, suppose $\mathcal A=\varnothing$. The subspace $L=\operatorname{Im}P$ is disjoint from the compact simplex
$$
\Delta=\{x\in\mathbb R^n:x_i\geq0,\ \sum_i x_i=1\}.
$$
Strict separation gives a <vector> $q$ that vanishes on $L$ and is positive on all of $\Delta$. The first property says $P^Tq=0$; testing the vertices of $\Delta$ shows every $q_i>0$. After normalization $\sum_iq_i=1$, this gives $q\in\mathcal Q$. This is <Stiemke theorem> in normalized form. Therefore
$$
\boxed{\mathcal Q=\varnothing\quad\Longleftrightarrow\quad\mathcal A\ne\varnothing.}
$$

Solved by gpt-5.6-sol high.

= c
{parent=30l}
{scope}

= Solution
{parent=c}

Let $P$ be the state-by-asset <matrix> of discounted gains and identify $Y$ with its state <vector>. Fix $q^0\in\mathcal Q$. If $q\geq0$, $P^Tq=0$, and $q\ne0$, then
$$
q_\varepsilon=\frac{q+\varepsilon q^0}{\sum_i(q_i+\varepsilon q_i^0)}
$$
belongs to $\mathcal Q$. The assumption and passage to the <limit> imply $q^TY\geq0$ for every nonnegative $q\in\ker P^T$.

The <finite-state superhedging alternative> now gives $\theta$ with
$$
Y-P\theta\geq0.
$$
This residual cannot vanish identically, since then every $Q\in\mathcal Q$ would have $E_QY=q^TP\theta=0$. Hence it is strictly positive in at least one state, which has positive physical probability. This is the required inequality.

Solved by gpt-5.6-sol high.

= d
{parent=30l}
{scope}

= Solution
{parent=d}

The same mixing argument with a fixed $q^0\in\mathcal Q$ shows that
$$
q^TZ\geq0
$$
for every $q\geq0$ in $\ker P^T$, including boundary <vectors>. By the <finite-state superhedging alternative>, there is $\phi\in\mathbb R^d$ such that
$$
\boxed{Z-P\phi\geq0,}
$$
which is the asserted almost-sure inequality. No strict residual is required here.

Solved by gpt-5.6-sol high.

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

= a
{parent=31k}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

Let $\varepsilon_1,\ldots,\varepsilon_n$ be independent Rademacher signs. For the set of evaluation <vectors>
$$
\mathcal F(z_{1:n})=\left\{(f(z_1),\ldots,f(z_n)):f\in\mathcal F\right\},
$$
the empirical <Rademacher complexity> is
$$
\widehat R(\mathcal F(z_{1:n}))
=\mathbb E_\varepsilon\left[
\sup_{f\in\mathcal F}\frac1n\sum_{i=1}^n\varepsilon_i f(z_i)
\right].
$$
For an independent sample $Z_1,\ldots,Z_n$,
$$
R_n(\mathcal F)=\mathbb E_{Z_{1:n}}
\widehat R(\mathcal F(Z_{1:n})).
$$

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

For points $z_1,ldots,z_n$, let
$$
H(z_{1:n})=\{(h(z_1),\ldots,h(z_n)):h\in H\}.
$$
The shattering coefficient and <VC dimension> are
$$
s(H,n)=\sup_{z_{1:n}}|H(z_{1:n})|,
\qquad
VC(H)=\sup\{n:s(H,n)=2^n\}.
$$
Thus a set of $n$ points is shattered when every binary labeling occurs on it.

Solved by gpt-5.6-sol high.

= iii
{parent=a}
{scope}

= Solution
{parent=iii}

Take any $p+1$ points $z_1,ldots,z_{p+1}\in\mathbb R^p$. For each coordinate $j$, choose a point attaining the largest $j$th coordinate. At most $p$ points are chosen, so some point $z_k$ is not among them.

Consider the labeling that assigns $1$ to every chosen coordinate maximizer and $0$ to $z_k$. Any lower orthant
$$
\prod_{j=1}^p(-\infty,a_j]
$$
containing all the chosen points must have $a_j\geq\max_i(z_i)_j$ for every $j$, and hence contains $z_k$ as well. This labeling is impossible. No $p+1$ points are shattered, so
$$
\boxed{VC(H)\leq p.}
$$
This is the <VC dimension bound for lower orthants>.

Solved by gpt-5.6-sol high.

= b
{parent=31k}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

Put $D=y^{(1)}-y^{(0)}$ and define the loss class
$$
\mathcal F=\{f_h(y^{(0)},y^{(1)},x)=D\,h(x):h\in H\}.
$$
Since $\widehat h$ minimizes $\widehat Q$,
$$
\begin{aligned}
Q(\widehat h)-Q(h^*)
&\leq (Q-\widehat Q)(\widehat h)
 +(\widehat Q-Q)(h^*).
\end{aligned}
$$
The expectation of the second term is zero. Symmetrization of the first gives
$$
\mathbb E\sup_{h\in H}(Q-\widehat Q)(h)
\leq2R_n(\mathcal F).
$$
Therefore the <expected excess-risk bound for empirical risk minimization> gives
$$
\boxed{\mathbb E Q(\widehat h)\leq Q(h^*)+2R_n(\mathcal F).}
$$

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

Condition on the sample. The set of <vectors>
$$
\bigl(D_i h(X_i)\bigr)_{i=1}^n,qquad h\in H,
$$
has at most $s(H,n)$ distinct members, each with Euclidean norm at most $M\sqrt n$. The <Massart finite-class lemma>, the <Sauer-Shelah growth bound>, and $VC(H)\leq p$ give
$$
\widehat R(\mathcal F)
\leq M\sqrt{\frac{2\log s(H,n)}n}
\leq M\sqrt{\frac{2p\log(n+1)}n}.
$$
Taking expectations and substituting this into part (i) yields
$$
\boxed{
\mathbb E Q(\widehat h)
\leq Q(h^*)+2M\sqrt{\frac{2p\log(n+1)}n}.}
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=32a}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

In Glendinning's definition, a continuous interval map $F$ is chaotic when $F^m$ has a horseshoe for some $m\geq1$. Thus there are two disjoint subintervals whose images under $F^m$ each cover their union, producing full two-symbol itinerary dynamics.

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

A map is chaotic in Devaney's sense when it is topologically transitive, its periodic points are dense, and it has sensitive dependence on initial conditions. Explicitly, transitivity means that for every nonempty open $U,V$ some $n$ has $F^n(U)\cap V\ne\varnothing$, while sensitivity means that one $\delta>0$ works so that arbitrarily close initial points eventually separate by more than $\delta$.

Solved by gpt-5.6-sol high.

= b
{parent=32a}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

Write the period-three points in increasing order and take the two intervals between consecutive points. For either possible cyclic ordering, the interval-covering transition graph is, after interchanging its vertices,
$$
A=\begin{pmatrix}1&1\\1&0\end{pmatrix}.
$$
The loop at the first vertex gives a fixed point. For every $n\geq2$, the closed itinerary
$$
0^{n-1}1
$$
has least period $n$. The interval-covering periodic-orbit theorem supplies a point with that itinerary and therefore a cycle of least period $n$. Thus $F$ has periodic orbits of every positive period. This is the period-three case of the <Sharkovsky theorem>.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

For the Fibonacci transition <matrix> above,
$$
\operatorname{tr}(A^n)=L_n,
$$
where $L_n$ is the $n$th Lucas number. It counts closed symbolic itineraries of length $n$. Since $7$ is prime, the itineraries fixed by the seventh shift have least period either $1$ or $7$. Hence the number of cyclic classes of primitive length-seven itineraries is
$$
\frac{\operatorname{tr}(A^7)-\operatorname{tr}(A)}7
=\frac{29-1}{7}=4.
$$
Distinct primitive cyclic itineraries give distinct cycles, so $F$ has at least four distinct $7$-cycles. This is the <fibonacci transition-graph cycle count>.

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

An itinerary fixed by the eighth shift has least period dividing $8$. Those with period dividing $4$ are exactly the itineraries fixed by the fourth shift. Thus the primitive length-eight itineraries form
$$
\operatorname{tr}(A^8)-\operatorname{tr}(A^4)=47-7=40
$$
pointed words. Each cycle has eight cyclic shifts, so they give
$$
\boxed{40/8=5}
$$
distinct $8$-cycles. Therefore at least five such cycles are forced.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=33C}

Differentiate the spatial equation with respect to $t$ and the temporal equation with respect to $x$. Substitution of
$$
\psi_x=ik\psi+u,
\qquad
\psi_t=-ik^2\psi+iu_x-ku
$$
into $\psi_{xt}-ik\psi_t$ cancels every term involving $\psi$, $u$, and $u_x$, leaving
$$
\psi_{xt}-ik\psi_t=iu_{xx}.
$$
The left side is $u_t$ by differentiating $\psi_x-ik\psi=u$, so consistency is exactly $u_t=iu_{xx}$. This is a <scalar> <Lax pair> formulation.

Taking the half-line Fourier transform and integrating twice by parts gives
$$
\partial_t\widehat u(k,t)+ik^2\widehat u(k,t)
=kh(t)-iu_x(0,t).
$$
Multiplication by $e^{ik^2t}$ and integration in time yields
$$
\boxed{
\widehat u(k,t)e^{ik^2t}-\widehat u_0(k)
=\int_0^t e^{ik^2\tau}
 [kh(\tau)-iu_x(0,\tau)]\,d\tau.}
$$

Replace $k$ by $-k$ and subtract. The unknown Neumann boundary value cancels:
$$
e^{ik^2t}\bigl[\widehat u(k,t)-\widehat u(-k,t)\bigr]
=\widehat u_0(k)-\widehat u_0(-k)
 +2k\int_0^t e^{ik^2\tau}h(\tau)\,d\tau.
$$
For $x>0$, Fourier inversion of the zero extension gives
$$
\int_{\mathbb R}e^{ikx}\widehat u(-k,t)\,dk=0.
$$
Inverting the preceding identity therefore gives
$$
\begin{aligned}
u(x,t)
={}&\frac1{2\pi}\int_{-\infty}^{\infty}
 e^{-ik^2t+ikx}[\widehat u_0(k)-\widehat u_0(-k)]\,dk\\
&+\frac1\pi\int_{-\infty}^{\infty}\int_0^t
 e^{-ik^2(t-\tau)+ikx}\,k h(\tau)\,d\tau\,dk.
\end{aligned}
$$
Thus
$$
\boxed{G(k,\tau)=k h(\tau).}
$$
This elimination is the <global relation for the half-line free Schrodinger equation>.

Solved by gpt-5.6-sol high.

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

= a
{parent=34b}
{scope}

= Solution
{parent=a}

Set $c=\langle\psi|\phi\rangle$. The off-diagonal Hermiticity identity holds automatically, while
$$
\langle\psi|H|\psi\rangle=gc
$$
must be real. Thus
$$
\boxed{g\langle\psi|\phi\rangle\in\mathbb R}
$$
is necessary and sufficient. For $g\ne0$, put $r=gc/|g|\in\mathbb R$. Since the two normalized spanning <vectors> are linearly independent, $|r|<1$.

The normalized <eigenvectors> and <eigenvalues> are
$$
\boxed{
|\pm\rangle=
\frac{(g^*/|g|)|\psi\rangle\pm|\phi\rangle}
 {\sqrt{2(1\pm r)}},
\qquad E_\pm=\pm|g|.}
$$
Direct application of $H$ verifies the <eigenvalue> equations. Their unnormalized <inner product> is
$$
-\frac g{|g|}c+\frac{g^*}{|g|}c^*=-r+r=0,
$$
so the distinct <eigenvectors> are orthogonal, as required for a <Hermitian operator>. If $g=0$, $H=0$ and every <vector> is an <eigenvector>.

Solved by gpt-5.6-sol high.

= b
{parent=34b}
{scope}

= Solution
{parent=b}

Now $g=1$, so $c=\langle\psi|\phi\rangle$ is real and
$$
|\pm\rangle=\frac{|\psi\rangle\pm|\phi\rangle}
 {\sqrt{2(1\pm c)}}.
$$
The perturbation is diagonal in this <basis>:
$$
\Delta(t)|+\rangle=2V(t)|+\rangle,
\qquad
\Delta(t)|-\rangle=0.
$$
Hence the instantaneous <eigenvalues> are $1+2V(t)$ and $-1$ with time-independent <eigenvectors>, so <Hamiltonians> at different times commute. In units $\hbar=1$, put
$$
W(t)=\int_0^tV(s)\,ds.
$$
Expanding the initial state and evolving its two components gives
$$
|\psi(t)\rangle
=\frac{\sqrt{2(1+c)}}2e^{-i[t+2W(t)]}|+\rangle
 +\frac{\sqrt{2(1-c)}}2e^{it}|-\rangle.
$$
Taking its overlap with $|\phi\rangle$ and simplifying gives the exact transition probability
$$
\boxed{
\mathbb P_{\psi\to\phi}(t)
=c^2+(1-c^2)\sin^2\!\bigl(t+W(t)\bigr).}
$$
With explicit $\hbar$, each phase in the final sine is divided by $\hbar$.

Solved by gpt-5.6-sol high.

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

= a
{parent=35e}
{scope}

= Solution
{parent=a}

For real $k$, conservation of the radial <probability current>, equivalently constancy of the radial Wronskian, equates the incoming and outgoing fluxes and gives
$$
S_l(k)^*S_l(k)=1.
$$
For a real potential, the regular radial solution depends on $k$ through $k^2$. Comparing its asymptotic form at $k$ and $-k$ therefore gives
$$
S_l(k)S_l(-k)=1.
$$
The first identity places $S_l(k)$ on the unit circle, so
$$
S_l(k)=e^{2i\delta_l(k)}
$$
for a real phase shift. Choosing the phase continuously, the second identity gives
$$
\boxed{\delta_l(-k)=-\delta_l(k).}
$$
These are the <unitarity and reflection identities for a partial-wave S-matrix>.

Solved by gpt-5.6-sol high.

= b
{parent=35e}
{scope}

= Solution
{parent=b}

At small $k$,
$$
\frac d{dk}\log S_0(k)\bigg|_{k=0}
=-2i\left(\frac1{3\lambda}+\frac1{2\lambda}\right)
=-\frac{5i}{3\lambda}.
$$
Since $S_0=e^{2i\delta_0}$,
$$
\delta_0(k)=-\frac{5k}{6\lambda}+O(k^3).
$$
The scattering length and low-energy total cross-section are therefore
$$
\boxed{a_s=-\lim_{k\to0}\frac{\tan\delta_0(k)}k
=\frac5{6\lambda},
\qquad
\sigma_T\sim4\pi a_s^2=\frac{25\pi}{9\lambda^2}.}
$$

The poles at $k=2i\lambda$ and $k=3i\lambda$ lie on the positive imaginary axis. They represent two <bound states>, with energies
$$
E=-\frac{\hbar^2(2\lambda)^2}{2m},
\qquad
E=-\frac{\hbar^2(3\lambda)^2}{2m}.
$$
There are no resonance poles away from the imaginary axis in the lower half-plane. Positivity of $\lambda$ puts the poles on the physical bound-state axis, where $e^{ikr}$ decays; changing its sign would move them to the unphysical half-plane. This illustrates the <bound-state poles of a partial-wave S-matrix>.

Solved by gpt-5.6-sol high.

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

= a
{parent=36b}
{scope}

= Solution
{parent=a}

A <canonical ensemble> describes a system with fixed particle number and external parameters that exchanges energy with a heat bath at fixed <temperature> $T$. With $\beta=(k_BT)^{-1}$,
$$
\boxed{Z(\beta)=\sum_{i=0}^\infty e^{-\beta E_i},
\qquad p_i=\frac{e^{-\beta E_i}}Z.}
$$

Solved by gpt-5.6-sol high.

= b
{parent=36b}
{scope}

= Solution
{parent=b}

The classical one-particle phase-space <integral> is
$$
Z_1=\frac1{h^3}\int_{\mathbb R^3}\!d^3x
\int_{\mathbb R^3}\!d^3p\,
 e^{-\beta(pc+U(x))}.
$$
The <momentum> <integral> is
$$
4\pi\int_0^\infty p^2e^{-\beta cp}\,dp
=\frac{8\pi}{\beta^3c^3}.
$$
Hence
$$
\boxed{
Z_1=\frac{8\pi(k_BT)^3}{h^3c^3}
\int_{\mathbb R^3}e^{-U(x)/(k_BT)}\,d^3x
=\frac{(k_BT)^3}{\pi^2(\hbar c)^3}
\int e^{-U/(k_BT)}\,d^3x.}
$$

Solved by gpt-5.6-sol high.

= c
{parent=36b}
{scope}

= i
{parent=c}
{scope}

= Solution
{parent=i}

With $r=|x|$, set
$$
r=V^{1/3}(k_BT)^{1/(2n)}u.
$$
Then
$$
\int e^{-\beta U(x)}d^3x
=4\pi V(k_BT)^{3/(2n)}I_n.
$$
Part (b) gives
$$
Z_1=\frac{4}{\pi(\hbar c)^3}
V(k_BT)^{3+3/(2n)}I_n.
$$
For $N$ identical classical non-interacting particles,
$$
Z_N=\frac{Z_1^N}{N!}.
$$
Using $\log N!=N\log N-N+o(N)$,
$$
F=-Nk_BT\left[\log V+A\log(k_BT)+\log I_n+B\right]+o(N),
$$
where
$$
\boxed{
A=3+\frac{3}{2n},
\qquad
B=1+\log\!\left(\frac{4}{\pi N(\hbar c)^3}\right).}
$$
This is the <canonical partition function for an ultrarelativistic gas in a power-law trap>.

Solved by gpt-5.6-sol high.

= ii
{parent=c}
{scope}

= Solution
{parent=ii}

Only the $\log V$ term depends on $V$, so
$$
p=-\left(\frac{\partial F}{\partial V}\right)_{T,N}
=\frac{Nk_BT}{V}.
$$
Thus the equation of state is
$$
\boxed{pV=Nk_BT.}
$$

Solved by gpt-5.6-sol high.

= iii
{parent=c}
{scope}

= Solution
{parent=iii}

Write $A=3+3/(2n)$. Since $\log Z_N$ contains $-NA\log\beta$,
$$
\boxed{
E=-\partial_\beta\log Z_N=NAk_BT,}
$$
$$
\boxed{
(\Delta E)^2=\partial_\beta^2\log Z_N
=NA(k_BT)^2,
\qquad C_V=NAk_B.}
$$
Consequently
$$
\frac{\Delta E}{E}=\frac1{\sqrt{NA}}\longrightarrow0
$$
in the thermodynamic <limit>: relative energy fluctuations vanish.

Solved by gpt-5.6-sol high.

= iv
{parent=c}
{scope}

= Solution
{parent=iv}

The normalized one-particle spatial density gives the local number density
$$
\boxed{
n(x)=\frac{N}{4\pi V(k_BT)^{3/(2n)}I_n}
\exp\!\left[-\frac{|x|^{2n}}
 {V^{2n/3}k_BT}\right].}
$$
The radial probability density is proportional to
$$
r^2\exp\!\left[-\frac{r^{2n}}{V^{2n/3}k_BT}\right].
$$
Its logarithmic <derivative> vanishes when
$$
\frac2r-\frac{2n r^{2n-1}}{V^{2n/3}k_BT}=0,
$$
so the most likely radius is
$$
\boxed{r_{\rm mode}=V^{1/3}\left(\frac{k_BT}{n}\right)^{1/(2n)}.}
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=37d}
{scope}

= Solution
{parent=a}

Varying the potential gives
$$
\delta F_{\mu\nu}=\partial_\mu\delta A_\nu-
\partial_\nu\delta A_\mu.
$$
Antisymmetry implies
$$
\frac12F^{\mu\nu}\delta F_{\mu\nu}
=F^{\mu\nu}\partial_\mu\delta A_\nu.
$$
After integration by parts and omission of the boundary term,
$$
\delta S=-\frac1{\mu_0c}\int
\left[-\partial_\mu F^{\mu\nu}+m^2A^\nu-\mu_0J^\nu\right]
\delta A_\nu\,d^4x.
$$
Stationarity for arbitrary compactly supported $\delta A_\nu$ therefore gives the <Proca equation>
$$
\boxed{\partial_\mu F^{\mu\nu}-m^2A^\nu=-\mu_0J^\nu.}
$$

Solved by gpt-5.6-sol high.

= b
{parent=37d}
{scope}

= Solution
{parent=b}

Lorentz invariance requires $A^\mu$ and $J^\mu$ to transform as four-vectors, $\chi$ as a <scalar>, $m$ and $\mu_0$ as <scalars>, and the integration domain and <boundary conditions> to respect the transformation. Then every contracted term and $d^4x$ are Lorentz invariant.

Under $A_\mu\mapsto A_\mu+\partial_\mu\chi$, the field strength is unchanged. The source term changes by a boundary term plus
$$
-\int\chi\,\partial_\mu J^\mu,d^4x,
$$
which vanishes for a conserved current and suitable boundary behaviour of $\chi$. The mass term is not gauge invariant. Thus the action has the usual local gauge invariance when
$$
\boxed{m=0,\qquad \partial_\mu J^\mu=0,}
$$
with the gauge <function> or its boundary contribution suitably controlled. For $m>0$, the mass term breaks this gauge symmetry.

Solved by gpt-5.6-sol high.

= c
{parent=37d}
{scope}

= Solution
{parent=c}

Take the divergence of the field equation. Commuting <derivatives> and antisymmetry of $F^{\mu\nu}$ give
$$
\partial_\nu\partial_\mu F^{\mu\nu}=0.
$$
Current conservation then yields
$$
-m^2\partial_\nu A^\nu=-\mu_0\partial_\nu J^\nu=0.
$$
For $m>0$,
$$
\boxed{\partial_\mu A^\mu=0.}
$$
This is the <Lorenz constraint in Proca theory>; it follows dynamically rather than being a gauge choice.

Solved by gpt-5.6-sol high.

= d
{parent=37d}
{scope}

= Solution
{parent=d}

Using
$$
\partial_\mu F^{\mu\nu}
=\Box A^\nu-\partial^\nu(\partial_\mu A^\mu)
$$
and writing
$$
\alpha=\frac1{c^2}\frac{\partial\phi}{\partial t}+\nabla\cdot A,
$$
the temporal and spatial components give, with the stated metric conventions,
$$
\boxed{
\Box\phi+\frac{\partial\alpha}{\partial t}-m^2\phi
=-c\mu_0J^0,}
$$
$$
\boxed{
\Box A-\nabla\alpha-m^2A=-\mu_0J.}
$$
For $m>0$, part (c) sets $\alpha=0$. For $m=0$, $\alpha=0$ is the Lorenz gauge condition.

Solved by gpt-5.6-sol high.

= e
{parent=37d}
{scope}

= Solution
{parent=e}

For the static point source, take $A=0$ and $\alpha=0$. The <scalar> equation becomes
$$
(\nabla^2-m^2)\phi=-c^2\mu_0q\,\delta(x).
$$
Its spatial Fourier transform satisfies
$$
(|k|^2+m^2)\widehat\phi(k)=c^2\mu_0q.
$$
The stated inverse transform therefore gives
$$
\boxed{\phi(r)=\lambda\frac{e^{-mr}}r,\qquad A=0,}
$$
where $\lambda$ absorbs the transform normalization and source strength. This is the <Yukawa potential>: a nonzero vector-field mass screens the interaction beyond the range $m^{-1}$. As $m\to0$, it tends to the long-range Coulomb potential $\lambda/r$. The same <limit> removes the mass term that broke gauge invariance in part (b).

Solved by gpt-5.6-sol high.

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

= a
{parent=38b}
{scope}

= Solution
{parent=a}

The Killing <vectors> $\partial_t$ and $\partial_\phi$ give conserved specific energy and <angular momentum>,
$$
E=\left(1-\frac{2m}{r}\right)\dot t,
\qquad
h=r^2\dot\phi.
$$
The normalization $g_{\mu\nu}\dot x^\mu\dot x^\nu=-1$ then rearranges to
$$
\frac12\dot r^2-\frac{m}{r}+\frac{h^2}{2r^2}-\frac{mh^2}{r^3}
=\frac{E^2-1}{2},
$$
so the displayed radial energy is also constant. This is the <effective potential for timelike Schwarzschild geodesics>.

Solved by gpt-5.6-sol high.

= b
{parent=38b}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

For a circular orbit, the <derivative> of
$$
V_{\rm eff}(r)=-\frac mr+\frac{h^2}{2r^2}-\frac{mh^2}{r^3}
$$
vanishes. Thus
$$
\frac m{r^2}-\frac{h^2}{r^3}+\frac{3mh^2}{r^4}=0,
\qquad
\boxed{h^2=\frac{mr^2}{r-3m}.}
$$
A real finite <angular momentum> therefore requires
$$
\boxed{r>3m.}
$$

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

For a circular orbit, normalization and the value of $h$ give
$$
E^2=\frac{(r-2m)^2}{r(r-3m)},
\qquad
\dot t^2=\frac r{r-3m},
\qquad
\dot\phi^2=\frac{m}{r^2(r-3m)}.
$$
Consequently
$$
\boxed{
\left(\frac{d\phi}{dt}\right)^2
=\frac{\dot\phi^2}{\dot t^2}=\frac m{r^3}.}
$$
This is the relativistic form of the circular-orbit <Kepler third law>.

Solved by gpt-5.6-sol high.

= c
{parent=38b}
{scope}

= Solution
{parent=c}

Substitute $r=1/u$ and $\dot r=-hu'$ into the conserved radial energy. <Differentiation> with respect to $\phi$ gives
$$
h^2u'u''-mu'+h^2uu'-3mh^2u^2u'=0.
$$
After division by $h^2u'$ and continuity through turning points,
$$
\boxed{u''+u=\frac m{h^2}+3mu^2.}
$$

Put $q=m/h^2$ and substitute
$$
u=q(1+\alpha)+A\cos[(1+\beta)\phi].
$$
Keeping the leading nontrivial powers of $m/h$ and terms through first order in $A$, the constant and oscillatory terms give
$$
q(1+\alpha)=q+3mq^2,
\qquad
-2\beta A=6mqA.
$$
Therefore
$$
\boxed{
\alpha=\frac{3m^2}{h^2},
\qquad
\beta=-\frac{3m^2}{h^2}.}
$$
The radial oscillation has angular period $2\pi/(1+\beta)$, so successive periapses advance by
$$
\boxed{\Delta\phi=-2\pi\beta
=\frac{6\pi m^2}{h^2}}
$$
to leading order. This is the <Schwarzschild perihelion precession>.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=39C}

Take downward <velocity> as positive. Lubrication theory reduces the axial <momentum> equation to a constant <pressure> <gradient>, and the no-slip conditions are $u(0)=0$, $u(h)=U$. Twice integrating gives
$$
\boxed{u(y)=-\frac{\Delta p}{2\mu L}y(h-y)+U\frac yh.}
$$
The flux per unit circumference and the total gap flux are
$$
q=\int_0^hu(y)dy
=-\frac{\Delta p\,h^3}{12\mu L}+\frac{Uh}{2},
\qquad
Q\simeq2\pi a q.
$$
The shear at the cylinder is
$$
\boxed{
\tau_h=\mu u_y(h)=\frac{\Delta p\,h}{2L}+\frac{\mu U}{h},}
$$
with the fluid traction on the falling cylinder opposing its motion.

The descending solid displaces volume at rate $\pi a^2U$, so closed-container mass conservation gives
$$
2\pi a\left(-\frac{\Delta p\,h^3}{12\mu L}+\frac{Uh}{2}\right)
=-\pi a^2U.
$$
Since $h\ll a$, the Couette-flux term is smaller, and
$$
\boxed{\Delta p\simeq\frac{6\mu aLU}{h^3}.}
$$
The <pressure> or form drag is
$$
F_p\simeq\pi a^2\Delta p
=\frac{6\pi\mu a^3LU}{h^3}.
$$
Using the pressure-dominated part of $\tau_h$, the side shear force is of order
$$
F_\mu\simeq2\pi aL\frac{\Delta p h}{2L}
=\frac{6\pi\mu a^2LU}{h^2},
$$
so $F_p/F_\mu\simeq a/h\gg1$. Balancing form drag against the excess weight $\Delta\rho g\pi a^2L$ gives
$$
\boxed{U\simeq\frac{\Delta\rho\,g h^3}{6\mu a}.}
$$
This is <annular lubrication drag on a settling cylinder>.

Solved by gpt-5.6-sol high.

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

= a
{parent=40d}
{scope}

= Solution
{parent=a}

Let $\widetilde\rho$, $\widetilde p$, and $u$ be first-order perturbations. Linearized mass and <momentum> conservation and the homentropic relation are
$$
\partial_t\widetilde\rho+\rho_0\nabla\cdot u=0,
\qquad
\rho_0\partial_tu=-\nabla\widetilde p,
\qquad
\widetilde p=c_0^2\widetilde\rho.
$$
Differentiate the first equation in time and use the divergence of the second to obtain
$$
\boxed{\partial_t^2\widetilde p-c_0^2\nabla^2\widetilde p=0.}
$$
The perturbation is irrotational, so write $u=\nabla\phi$. Integrating the <momentum> equation spatially, with a purely time-dependent constant absorbed into $\phi$, gives
$$
\boxed{\widetilde p=-\rho_0\partial_t\phi.}
$$

Solved by gpt-5.6-sol high.

= b
{parent=40d}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

Use complex fields with time factor $e^{i\omega t}$ and put $k_\pm=\omega/c_\pm$. The outgoing potential in $x>0$ is
$$
\Phi_+(x)=T e^{-ik_+x}.
$$
Continuity of <pressure> and <velocity> at $x=0$ gives
$$
\Phi_-(0)=\frac{\rho_+}{\rho_-}T,
\qquad
\Phi_-'(0)=-ik_+T.
$$
Hence
$$
\Phi_-(x)=T\left[
\frac{\rho_+}{\rho_-}\cos(k_-x)
-i\frac{c_-}{c_+}\sin(k_-x)
\right].
$$
The piston condition $\Phi_-'(-L)=i\omega\epsilon$, with $\lambda=k_-L$, gives
$$
\boxed{
T=\epsilon c_-
\frac{i(\rho_+/\rho_-)\sin\lambda-(c_-/c_+)\cos\lambda}
 { (\rho_+/\rho_-)^2\sin^2\lambda+(c_-/c_+)^2\cos^2\lambda}.}
$$
This is the <transmitted velocity potential from a piston through an acoustic interface>.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

For the outgoing wave, the <pressure> and <velocity> amplitudes are $-i\omega\rho_+T$ and $-ik_+T$. Its time-averaged acoustic energy flux is therefore
$$
\langle\mathcal I\rangle
=\frac12\operatorname{Re}(p u^*)
=\frac{\rho_+\omega^2}{2c_+}|T|^2.
$$
If $\lambda=n\pi$, then $|T|=\epsilon c_+$ and
$$
\boxed{
\langle\mathcal I\rangle_{n\pi}
=\frac12\rho_+c_+\epsilon^2\omega^2.}
$$
If $\lambda=(n+\tfrac12)\pi$, then
$$
|T|=\epsilon c_-\frac{\rho_-}{\rho_+},
$$
and
$$
\boxed{
\langle\mathcal I\rangle_{(n+1/2)\pi}
=\frac{\epsilon^2\omega^2\rho_-^2c_-^2}
 {2\rho_+c_+}.}
$$
When $\rho_+\ll\rho_-$ and $c_+\simeq c_-$, the half-integer family is strongly resonantly enhanced relative to the integer family by approximately $(\rho_-/\rho_+)^2$ at equal $\omega$.

Solved by gpt-5.6-sol high.

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

= a
{parent=41a}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

Expanding the square and using orthogonality of Fourier modes,
$$
\begin{aligned}
\frac1{2\pi}\int_{-\pi}^{\pi}|\widehat v(\theta)|^2d\theta
&=\sum_{m,j}v_m\overline{v_j}
\frac1{2\pi}\int_{-\pi}^{\pi}e^{-i(m-j)\theta}d\theta\\
&=\sum_m|v_m|^2.
\end{aligned}
$$
Finite truncations justify the calculation directly, and passage to the $\ell^2$ <limit> proves <Parseval identity> for every square-summable <sequence>.

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

Fourier transformation in $m$ turns the recurrence into
$$
(1+\mu)\widehat u^{,n+1}
=2\mu\cos\theta\,\widehat u^{,n}
 +(1-\mu)\widehat u^{,n-1}.
$$
An amplification factor $z$ therefore satisfies
$$
(1+\mu)z^2-2\mu\cos\theta\,z-(1-\mu)=0.
$$
For $\mu>0$, the quadratic <Schur stability criterion> gives
$$
|a_2|<1,qquad
1+a_1+a_2=\frac{2\mu(1-\cos\theta)}{1+\mu}\geq0,
$$
$$
1-a_1+a_2=\frac{2\mu(1+\cos\theta)}{1+\mu}\geq0,
$$
where $a_1=-2\mu\cos\theta/(1+\mu)$ and $a_2=(\mu-1)/(\mu+1)$. Thus both roots lie in the closed unit disk; roots on it at $\theta=0,\pi$ are simple. At $\mu=0$, the roots are $\pm1$. Parseval's identity then gives stability precisely for
$$
\boxed{\mu\geq0,}
$$
which is the full stated parameter range.

Solved by gpt-5.6-sol high.

= b
{parent=41a}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

The <matrix> exponential is the absolutely convergent <series>
$$
e^X=\sum_{j=0}^\infty\frac{X^j}{j!}.
$$
Multiplying the two expansions gives
$$
e^{kB}e^{kC}
=I+k(B+C)+k^2\left(\frac12B^2+BC+\frac12C^2\right)+O(k^3).
$$
On the other hand,
$$
e^{k(B+C)}
=I+k(B+C)+\frac{k^2}{2}(B^2+BC+CB+C^2)+O(k^3).
$$
Subtracting yields
$$
\boxed{
e^{kB}e^{kC}=e^{k(B+C)}
+\frac{k^2}{2}(BC-CB)+O(k^3).}
$$
This is the leading <Lie-Trotter splitting commutator error>.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

Expanding the numerical propagation <matrix>,
$$
(I-kB)^{-1}(I-kC)^{-1}
=I+k(B+C)+k^2(B^2+BC+C^2)+O(k^3).
$$
Its difference from $e^{kA}$ is
$$
\frac{k^2}{2}(B^2+C^2+BC-CB)+O(k^3).
$$
Therefore the one-step local truncation error is
$$
\boxed{O(k^2),}
$$
and the method is first-order accurate globally.

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

If $BC=CB$, the splitting commutator vanishes, but the backward-Euler resolvents still contribute
$$
\frac{k^2}{2}(B^2+C^2)+O(k^3).
$$
Thus commutativity alone does not change the order: the local error remains $O(k^2)$ and the global method remains first order, barring an additional special cancellation.

Solved by gpt-5.6-sol high.