past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-1.bigb

= Paper 1
{scope}

https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2025/paperii_1_2025.pdf

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

= a
{parent=1g}
{scope}

= Solution
{parent=a}

Starting at $r=\lceil\sqrt{14351}\rceil=120$, one immediately finds
$$r^2-N=120^2-14351=49=7^2.$$
Therefore
$$14351=(120-7)(120+7)=113\cdot127.$$

Solved by gpt-5.6-sol high.

= b
{parent=1g}
{scope}

= Solution
{parent=b}

Since $N=3p$, its nontrivial difference-of-squares representation is
$$3p=\left(\frac{p+3}{2}\right)^2-\left(\frac{p-3}{2}\right)^2.$$
As $p$ is prime, every earlier square difference would give another nontrivial factorization, which is impossible. Thus the first successful value is $r=(p+3)/2$, and
$$k=\frac{p+3}{2}-\lfloor\sqrt{3p}\rfloor.$$

Solved by gpt-5.6-sol high.

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

= Solution
{parent=2i}

<Liouville approximation theorem> states that if an irrational algebraic number $\alpha$ has degree $d\geq2$, then some $C(\alpha)>0$ satisfies
$$\left|\alpha-\frac pq\right|>\frac{C(\alpha)}{q^d}$$
for every rational $p/q$ with $q>0$.

The omitted $n=0$ term is rational and does not affect transcendence. Let
$$\xi=\sum_{n\geq1}10^{-n^n},\qquad
\xi_N=\sum_{n=1}^N10^{-n^n}=\frac{p_N}{q_N},\quad q_N=10^{N^N}.$$
The decimal expansion has ones at the increasingly separated positions $n^n$ and zeros elsewhere, so it is not eventually periodic and $\xi$ is irrational. Moreover,
$$0<\xi-\xi_N<2\,10^{-(N+1)^{N+1}}.$$
For every fixed $d$, the exponent $(N+1)^{N+1}$ eventually exceeds $dN^N$ by an arbitrarily large amount, so this upper bound is smaller than $Cq_N^{-d}$. Liouville's theorem therefore rules out every finite algebraic degree, proving that $\xi$ is transcendental.

There are only countably many integer <polynomials> and each has finitely many roots, so the algebraic numbers are countable. Since $\mathbb R$ is uncountable, its complement, the set of transcendental numbers, is uncountable.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=3k}

For a binary <prefix code> with word lengths $l_1,\ldots,l_m$, choose $L\geq\max l_i$. Each codeword is the prefix of exactly $2^{L-l_i}$ words of length $L$, and these descendant sets are disjoint. Hence
$$\sum_i2^{L-l_i}\leq2^L,\qquad\text{so}\qquad\sum_i2^{-l_i}\leq1.$$
This is <Kraft inequality>. Conversely, if integer lengths satisfy this inequality, place words greedily as leaves of the binary tree; the unused capacity ensures that the requested leaves can all be chosen, giving a prefix code.

For Shannon--Fano coding, order symbols by probability and assign
$$l_i=\lceil-\log_2p_i\rceil.$$
Kraft's inequality applies because $2^{-l_i}\leq p_i$ and therefore $\sum_i2^{-l_i}\leq1$. Thus codewords of those lengths exist. Since
$$-\log_2p_i\leq l_i< -\log_2p_i+1,$$
the expected length obeys
$$H(X)\leq\mathbb E L<H(X)+1.$$

Solved by gpt-5.6-sol high.

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

= Solution
{parent=4f}

The closure table is

$$
\begin{array}{c|ccc}
&\text{union}&\text{intersection}&\text{complement}\\ \hline
\text{regular}&\text{Yes}&\text{Yes}&\text{Yes}\\
\text{context-free}&\text{Yes}&\text{No}&\text{No}\\
\text{computable}&\text{Yes}&\text{Yes}&\text{Yes}\\
\text{computably enumerable}&\text{Yes}&\text{Yes}&\text{No}
\end{array}
$$

For the context-free intersection failure, take
$$L_1=\{a^nb^nc^k:n,k\geq0\},\qquad L_2=\{a^kb^nc^n:n,k\geq0\}.$$
Both are context-free, but $L_1\cap L_2=\{a^nb^nc^n:n\geq0\}$ is not. If context-free languages were also closed under complement, their closure under union and De Morgan's law would imply closure under intersection, so complement closure also fails. Finally, the halting set is computably enumerable but its complement is not, disproving complement closure for computably enumerable languages.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=5k}

The change of variable $x=(y/\lambda)^k$ has <derivative> $kx/y$, so
$$f_\lambda(y)=e^{-(y/\lambda)^k}\frac{k}{\lambda}\left(\frac y\lambda\right)^{k-1},\qquad y>0.$$
For fixed $k$ this is
$$k y^{k-1}\exp\left\{\eta y^k-A(\eta)\right\},\qquad
\eta=-\lambda^{-k}<0,\quad A(\eta)=-\log(-\eta).$$
Thus the sample's <sufficient statistic> is $T=\sum_iY_i^k$.

Since $(Y/\lambda)^k\sim\operatorname{Exp}(1)$,
$$\mathbb E(Y^k)=\lambda^k.$$
The log likelihood, up to constants, is $-kn\log\lambda-T/\lambda^k$. Differentiating gives the unique maximum
$$\widehat\lambda=\left(\frac1n\sum_{i=1}^nY_i^k\right)^{1/k}.$$

Solved by gpt-5.6-sol high.

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

= a
{parent=6a}
{scope}

= Solution
{parent=a}

The $\beta YN$ interaction converts supporters into opponents. The $\alpha YN$ interaction converts opponents into undecided people, while $\zeta U$ converts undecided people into opponents. Summing the three equations gives $(Y+N+U)'=0$, as required for a fixed population.

Solved by gpt-5.6-sol high.

= b
{parent=6a}
{scope}

= Solution
{parent=b}

Let the conserved population be $S>0$. At equilibrium $Y'=-\beta YN=0$, so $Y$ and $N$ cannot both be positive. The other equations then force $U=0$. The only equilibria are
$$E_Y=(S,0,0),\qquad E_N=(0,S,0).$$

Near $E_Y$, using $(N,U)$ as coordinates gives the linearization
$$\binom{N'}{U'}=
\begin{pmatrix}(\beta-\alpha)S&\zeta\\ \alpha S&-\zeta\end{pmatrix}
\binom NU.$$
Its <determinant> is $-\beta S\zeta<0$, so $E_Y$ is a saddle and is unstable. Near $E_N$, coordinates $(Y,U)$ give the triangular linearization with <eigenvalues> $-\beta S$ and $-\zeta$. Hence $E_N$ is asymptotically stable.

Solved by gpt-5.6-sol high.

= c
{parent=6a}
{scope}

= Solution
{parent=c}

No positive parameter choices can make everyone eventually favour the proposition for every initial condition. The all-opposed state $E_N$ is itself an equilibrium, and part (b) shows that an entire neighborhood is attracted to it. Thus initial conditions in that neighborhood cannot converge to $E_Y$.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=7e}

The <Cauchy principal value> deletes symmetric intervals around the real poles $-1$ and $1$ and takes a symmetric <limit> at infinity:
$$\operatorname{PV}\int_{-\infty}^{\infty}f(x)dx
=\lim_{R\to\infty,\ \varepsilon\downarrow0}
\left(\int_{-R}^{-1-\varepsilon}+\int_{-1+\varepsilon}^{1-\varepsilon}+\int_{1+\varepsilon}^{R}\right)f(x)dx.$$
The integrand is even. The standard <principal-value beta integral> gives
$$\operatorname{PV}\int_0^\infty\frac{x^{s-1}}{1-x^a}dx=\frac\pi a\cot\frac{\pi s}{a}.$$
Taking $s=1$, $a=6$, and reversing the denominator,
$$\operatorname{PV}\int_{-\infty}^{\infty}\frac{dx}{x^6-1}
=-\frac{2\pi}{6}\cot\frac\pi6=-\frac\pi{\sqrt3}.$$

Solved by gpt-5.6-sol high.

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

= a
{parent=8b}
{scope}

= Solution
{parent=a}

The <Lagrangian> is
$$L=\frac m2(\dot y_1^2+\dot y_3^2)+\frac M2\dot y_2^2
-\frac k2\left[(y_1-y_2)^2+(y_2-y_3)^2\right]
-g(my_1+M y_2+my_3).$$
Writing $T=2m+M$, the center of mass is
$$Y=\frac{my_1+M y_2+my_3}{T}.$$
All internal forces cancel, so $\ddot Y=-g$ and
$$Y(t)=A+\dot Y(0)t-\frac12gt^2.$$
If the molecule is released from rest, the middle term vanishes.

Solved by gpt-5.6-sol high.

= b
{parent=8b}
{scope}

= Solution
{parent=b}

Since $y_1=(Q_s+Q_a)/2$, $y_3=(Q_s-Q_a)/2$, elimination of the middle coordinate by the center of mass gives
$$y_2=\frac{TY-mQ_s}{M}.$$
After substitution,
$$\widehat L=\frac m4(\dot Q_s^2+\dot Q_a^2)
+\frac{(T\dot Y-m\dot Q_s)^2}{2M}
-\frac{kT^2}{M^2}\left(Y-\frac{Q_s}{2}\right)^2
-\frac k4Q_a^2-TgY.$$
The antisymmetric coordinate is decoupled. Its Euler--Lagrange equation is
$$\frac m2\ddot Q_a+\frac k2Q_a=0,\qquad\text{or}\qquad
\ddot Q_a+\frac kmQ_a=0.$$

Solved by gpt-5.6-sol high.

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

= a
{parent=9e}
{scope}

= Solution
{parent=a}

Successive radial wave crests travel the same fixed comoving distance, so
$$\int_{t_e}^{t_0}\frac{dt}{a(t)}
=\int_{t_e+\delta t_e}^{t_0+\delta t_0}\frac{dt}{a(t)}.$$
To first order, $\delta t_0/a(t_0)=\delta t_e/a(t_e)$. Since wavelength is proportional to the period,
$$1+z=\frac{\lambda_0}{\lambda_e}=\frac{\delta t_0}{\delta t_e}
=\frac{a(t_0)}{a(t_e)}.$$

Solved by gpt-5.6-sol high.

= b
{parent=9e}
{scope}

= Solution
{parent=b}

Each photon loses a factor $(1+z)^{-1}$ in energy, and <cosmological time dilation> reduces the photon arrival rate by another factor $(1+z)^{-1}$. Hence
$$L_{\rm received}=\frac{L}{(1+z)^2},$$
so $n=2$. The physical area of the sphere today is $4\pi a(t_0)^2x^2$, giving flux
$$F=\frac{L}{4\pi a(t_0)^2x^2(1+z)^2}.$$

Solved by gpt-5.6-sol high.

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

= a
{parent=10c}
{scope}

= Solution
{parent=a}

Immediately before the final Hadamard, the state is
$$\frac1{\sqrt2}\left(|0\rangle|\psi\rangle-i|1\rangle U|\psi\rangle\right).$$
The outcome-one component after that Hadamard is $(|\psi\rangle+iU|\psi\rangle)/2$. Thus, with $z=\langle\psi|U|\psi\rangle$,
$$p(1)=\frac14\lVert|\psi\rangle+iU|\psi\rangle\rVert^2
=\frac12(1-\operatorname{Im}z).$$

Solved by gpt-5.6-sol high.

= b
{parent=10c}
{scope}

= Solution
{parent=b}

For every product <vector>,
$$ (A\otimes B)(|x\rangle\otimes|y\rangle)=A|x\rangle\otimes B|y\rangle.$$
Product <vectors> span the two-register state space, so the single $A\otimes B$ gate equals parallel application of $A$ to the first register and $B$ to the second. Unitarity guarantees that both descriptions are valid quantum circuits.

Solved by gpt-5.6-sol high.

= c
{parent=10c}
{scope}

= Solution
{parent=c}

Both $Z\otimes Z$ and $X\otimes I$ are unitary and Hermitian, so their expectations are real. Run the real-part <Hadamard test> twice, once with controlled $Z\otimes Z$ and once with controlled $X\otimes I$. If the respective ancilla outcome-one probabilities are $p_{ZZ}$ and $p_{XI}$, then
$$\langle\psi|Z\otimes Z|\psi\rangle=1-2p_{ZZ},\qquad
\langle\psi|X\otimes I|\psi\rangle=1-2p_{XI}.$$
The identity in part (b) implements each controlled tensor product by gates on its two target wires with the common control. By <linearity>,
$$\langle\psi|W|\psi\rangle=2-2(p_{ZZ}+p_{XI}).$$

Solved by gpt-5.6-sol high.

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

= a
{parent=11k}
{scope}

= Solution
{parent=a}

Huffman's algorithm repeatedly merges the two least probable current symbols, labels the two new edges $0,1$, and replaces the pair by a compound symbol whose probability is their sum. Reading paths from the final root gives a prefix code.

For optimality, first observe that in some optimal full binary tree the two least probable symbols are sibling leaves at maximum depth: exchange them with any deepest sibling pair, putting the smaller probabilities at no smaller depths, without increasing expected length. Contract that sibling pair to one symbol of combined probability. The original expected length equals the contracted tree's expected length plus the pair's combined probability. Induction on the alphabet size now proves that choosing the two least probabilities and recursing is optimal, which is precisely Huffman's algorithm.

If a symbol has length one, it must be $a_1$. At the three-weight stage, the other two weights $x,y$ must be merged while $p_1$ survives, so $x,y\leq p_1$. Thus $1-p_1=x+y\leq2p_1$, proving $p_1\geq1/3$. Therefore $p_1<1/3$ forces every length to be at least two.

For the other bound, suppose $p_1$ is first merged at the three-weight stage, with weights $p_1,q,r$, where $q\leq p_1\leq r$. The preceding merge created $r=x+y$ from two weights no larger than the surviving $q$, so $r\leq2q$. Since also $r\geq p_1$, we have $q\geq p_1/2$, and hence
$$1=p_1+q+r\geq p_1+\frac{p_1}{2}+p_1=\frac52p_1.$$
Thus this can happen only when $p_1\leq2/5$. If $p_1>2/5$, it survives until the final merge and receives a length-one word. At equality, ties can support either tree shape.

For code (a), take
$$ (p_1,p_2,p_3,p_4)=\left(\frac25,\frac15,\frac15,\frac15\right).$$
Huffman merging produces lengths $(1,2,3,3)$ and expected length $2$.

Solved by gpt-5.6-sol high.

= b
{parent=11k}
{scope}

= Solution
{parent=b}

For the same distribution $(2/5,1/5,1/5,1/5)$, the balanced code has every length equal to two and hence expected length $2$. Part (a) exhibited a Huffman code with that same expected length, so the balanced code is also optimal. Thus both listed codes are optimal for this distribution, after assigning their words in nondecreasing length order to nonincreasing probabilities.

Solved by gpt-5.6-sol high.

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

= a
{parent=12f}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

For $C,D\subseteq\mathbb B$, one writes $C\leq_mD$ if there is a total computable <function> $f:\mathbb B\to\mathbb B$ such that
$$x\in C\iff f(x)\in D$$
for every $x$. This is a computable <many-one reduction>.

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

One writes $C\equiv_mD$ when both $C\leq_mD$ and $D\leq_mC$.

Solved by gpt-5.6-sol high.

= iii
{parent=a}
{scope}

= Solution
{parent=iii}

Fix an effective coding of machines and write $W_w$ for the computably enumerable language recognized by code $w$. A set $I$ of codes is an <index set> when membership depends only on the recognized language:
$$W_u=W_v\implies(u\in I\iff v\in I).$$
It is nontrivial when it is neither empty nor all of $\mathbb B$.

Solved by gpt-5.6-sol high.

= b
{parent=12f}
{scope}

= Solution
{parent=b}

Let $K=\{w:w\in W_w\}$ and let $e_\varnothing$ be any index for the empty language. The proof of <Rice theorem> gives
$$e_\varnothing\notin I\implies K\leq_m I,
\qquad e_\varnothing\in I\implies\mathbb B\setminus K\leq_m I.$$
In either case choose a language on the opposite side of the nontrivial property and switch from the empty enumerator to its enumerator if the simulated diagonal computation halts.

Solved by gpt-5.6-sol high.

= c
{parent=12f}
{scope}

= Solution
{parent=c}

The empty language belongs to $\mathbf{Emp}$, so the second case applies:
$$\mathbb B\setminus K\leq_m\mathbf{Emp}.$$
It does not belong to $\mathbf{Inf}$, so the first case applies:
$$K\leq_m\mathbf{Inf}.$$

Solved by gpt-5.6-sol high.

= d
{parent=12f}
{scope}

= Solution
{parent=d}

The set of codes of even length is not an <index set> because code length is syntactic rather than a property of $W_w$. By the <padding lemma>, every computably enumerable language has equivalent machine descriptions of arbitrarily many lengths, in particular descriptions of both parities. Hence equal recognized languages need not give equal membership.

Solved by gpt-5.6-sol high.

= e
{parent=12f}
{scope}

= Solution
{parent=e}

Since the empty language does not lie in
$$\mathbf{Two}=\{w:|W_w|\geq2\},$$
Rice's construction gives $K\leq_m\mathbf{Two}$. Conversely, $\mathbf{Two}$ is computably enumerable: dovetail the enumerator for $W_w$ until two distinct words appear, then accept. The diagonal halting set $K$ is many-one complete for computably enumerable sets, so $\mathbf{Two}\leq_mK$. Therefore
$$K\equiv_m\mathbf{Two}.$$

Solved by gpt-5.6-sol high.

= f
{parent=12f}
{scope}

= Solution
{parent=f}

For $K\leq_m\mathbf{Cof}$, map $x$ to a machine that initially enumerates nothing and, if the simulation establishing $x\in K$ halts, enumerates every word. Its language is cofinite exactly when $x\in K$.

For $\mathbb B\setminus K\leq_m\mathbf{Cof}$, map $x$ to a machine that enumerates successively longer finite initial segments of $\mathbb B$ while simulating $x$. If the simulation never halts, every word is eventually enumerated; if it halts, enumeration stops with a finite language and therefore an infinite complement. Thus the constructed language is cofinite exactly when $x\notin K$.

The parameter theorem makes both code transformations total and computable.

Solved by gpt-5.6-sol high.

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

= a
{parent=13k}
{scope}

= Solution
{parent=a}

Conditionally on $X$, the log likelihood differs from
$$-\frac1{2\sigma^2}(Y-X\beta)^T\Sigma^{-1}(Y-X\beta)$$
only by terms independent of $\beta$. <Differentiation> gives the weighted normal equations
$$X^T\Sigma^{-1}X\widehat\beta=X^T\Sigma^{-1}Y,$$
and therefore
$$\widehat\beta(\Sigma)=(X^T\Sigma^{-1}X)^{-1}X^T\Sigma^{-1}Y.$$

Solved by gpt-5.6-sol high.

= b
{parent=13k}
{scope}

= Solution
{parent=b}

Dividing row $i$ of both $Y$ and $X$ by $\sqrt{v(X_i)}$ transforms the model to one with covariance $\sigma^2I$. Ordinary least squares without an intercept on these transformed data is exactly the estimator in part (a). In R, the unweighted estimator is returned by

```R
fit2 <- lm(Y ~ X - 1)
fit2$coefficients
```

Solved by gpt-5.6-sol high.

= c
{parent=13k}
{scope}

= Solution
{parent=c}

Write $Y_i=X_i^T\beta+\varepsilon_i$, where conditionally $\mathbb E(\varepsilon_i|X_i)=0$ and $\operatorname{Var}(\varepsilon_i|X_i)=\sigma^2v(X_i)$. Then
$$\widehat\beta(\Sigma)-\beta=
\left(\frac1n\sum_i\frac{X_iX_i^T}{v(X_i)}\right)^{-1}
\frac1n\sum_i\frac{X_i\varepsilon_i}{v(X_i)}.$$
The law of large numbers sends the <matrix> to
$$A=\mathbb E\left[\frac{XX^T}{v(X)}\right],$$
while the <vector> tends to zero, proving consistency. The central <limit> theorem and Slutsky's theorem give
$$\sqrt n(\widehat\beta(\Sigma)-\beta)\Rightarrow N(0,\sigma^2A^{-1}).$$

Similarly, with $B=\mathbb E(XX^T)$ and $C=\mathbb E(v(X)XX^T)$,
$$\sqrt n(\widehat\beta(I)-\beta)\Rightarrow
N(0,\sigma^2B^{-1}CB^{-1}).$$
These conclusions require the displayed <matrices> to be finite and nonsingular.

Solved by gpt-5.6-sol high.

= d
{parent=13k}
{scope}

= Solution
{parent=d}

For $p=1$, part (c) gives
$$\operatorname{Var}(\widehat\beta(\Sigma))\sim
\frac{\sigma^2}{n\mathbb E[X_1^2/v(X_1)]},$$
$$\operatorname{Var}(\widehat\beta(I))\sim
\frac{\sigma^2\mathbb E[v(X_1)X_1^2]}{n(\mathbb E[X_1^2])^2}.$$
Consequently
$$\rho=\frac{(\mathbb E[X_1^2])^2}
{\mathbb E[X_1^2/v(X_1)]\,\mathbb E[v(X_1)X_1^2]}.$$

Solved by gpt-5.6-sol high.

= e
{parent=13k}
{scope}

= Solution
{parent=e}

Weighted least squares uses the known conditional variances and should be at least as efficient, so $\rho\leq1$. Indeed, apply Cauchy--Schwarz to
$$\frac{|X_1|}{\sqrt{v(X_1)}}\quad\text{and}\quad |X_1|\sqrt{v(X_1)}.$$
It gives
$$\bigl(\mathbb E[X_1^2]\bigr)^2
\leq\mathbb E[X_1^2/v(X_1)]\,\mathbb E[v(X_1)X_1^2],$$
which is exactly $\rho\leq1$. Equality holds precisely when $v(X_1)$ is constant on the part of the support where $X_1\ne0$, up to null sets.

Solved by gpt-5.6-sol high.

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

= a
{parent=14e}
{scope}

= Solution
{parent=a}

Use the <Hankel contour> about the negative real axis: it starts at $-\infty$ below the cut, circles the origin counterclockwise, and returns to $-\infty$ above the cut. Take $-\pi<\arg t<\pi$. For $\operatorname{Re}s>0$, the small circle vanishes and the two banks give
$$\int_{-\infty}^{(0+)}\frac{t^{s-1}}{e^t+e^{-t}}dt
=2i\sin(\pi s)\int_0^\infty\frac{x^{s-1}}{e^x+e^{-x}}dx.$$
The <gamma reflection formula> then gives
$$\frac{\Gamma(1-s)}{2\pi i}(2i\sin\pi s)\int_0^\infty\frac{x^{s-1}}{e^x+e^{-x}}dx
=\frac1{\Gamma(s)}\int_0^\infty\frac{x^{s-1}}{e^x+e^{-x}}dx=\beta(s).$$

Near zero,
$$\frac1{e^t+e^{-t}}=\frac12-\frac14t^2+\frac5{48}t^4-\cdots.$$
Subtracting any desired number of these terms makes the local contour <integral> converge in successively larger left half-planes. The resulting apparent singularities are removable after multiplication by $\Gamma(1-s)$, so the Hankel formula analytically continues $\beta$ to all $s\in\mathbb C$.

Solved by gpt-5.6-sol high.

= b
{parent=14e}
{scope}

= Solution
{parent=b}

At a nonpositive integer, the branch disappears and the collapsed Hankel contour extracts the residue at zero. For $s=0$,
$$\beta(0)=\operatorname{Res}_{t=0}\frac1{t(e^t+e^{-t})}=\frac12.$$
For $s=-2$, the coefficient of $t^{-1}$ in
$$t^{-3}\left(\frac12-\frac14t^2+O(t^4)\right)$$
is $-1/4$; multiplication by $\Gamma(3)=2$ gives
$$\beta(-2)=-\frac12.$$
When $s=-2n-1$, the integrand is $t^{-2n-2}$ times an even power <series>, so it has no $t^{-1}$ coefficient. Hence
$$\beta(-2n-1)=0\qquad(n\geq0).$$

Solved by gpt-5.6-sol high.

= c
{parent=14e}
{scope}

= Solution
{parent=c}

The nonzero poles are
$$t=\pm ia_n,\qquad a_n=\frac\pi2(2n+1),\qquad n\geq0.$$
Their paired residues sum to
$$-(-1)^na_n^{s-1}\cos\frac{\pi s}{2}.$$
The paired <series> converges for $\operatorname{Re}s<1$ by the alternating-series test, and absolutely for $\operatorname{Re}s<0$. Therefore
$$\sum\operatorname{Res}=-\cos\frac{\pi s}{2}
\left(\frac\pi2\right)^{s-1}\beta(1-s).$$
Using the stated contour closure, the Hankel <integral> is $-2\pi i$ times this sum. Substitution into part (a) gives
$$\beta(s)=\Gamma(1-s)\cos\frac{\pi s}{2}
\left(\frac\pi2\right)^{s-1}\beta(1-s).$$
Now $1/\Gamma(1-s)=\Gamma(s)\sin(\pi s)/\pi$ and $\sin(\pi s)=2\sin(\pi s/2)\cos(\pi s/2)$, so
$$\boxed{\beta(1-s)=\Gamma(s)\left(\frac\pi2\right)^{-s}
\sin\frac{\pi s}{2}\,\beta(s)}.$$
It was derived on a nonempty half-plane. Both sides have analytic continuations, with the apparent gamma poles cancelled by the sine factor or the trivial zeros from part (b); the <identity theorem> therefore extends it to every $s\in\mathbb C$.

Solved by gpt-5.6-sol high.

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

= a
{parent=15e}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

At fixed time and angles, proper radial distance is
$$D=a(t)\int_0^{\Delta r}\frac{dr}{\sqrt{1-kr^2}}.$$
Thus
$$D=\begin{cases}
a\Delta r,&k=0,\\
\dfrac a{\sqrt k}\sin^{-1}(\sqrt k\,\Delta r),&k>0,\\
\dfrac a{\sqrt{-k}}\sinh^{-1}(\sqrt{-k}\,\Delta r),&k<0.
\end{cases}$$
These are respectively flat Euclidean, positively curved spherical, and negatively curved hyperbolic spatial geometries.

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

Define <conformal time> by $d\tau=dt/a(t)$. The radial coordinate and the angular-radius <function> are
$$
\begin{array}{c|c|c}
k&\chi(r)&f(\chi)\\ \hline
0&r&\chi\\
k>0&k^{-1/2}\sin^{-1}(\sqrt k\,r)&k^{-1/2}\sin(\sqrt k\,\chi)\\
k<0&(-k)^{-1/2}\sinh^{-1}(\sqrt{-k}\,r)&(-k)^{-1/2}\sinh(\sqrt{-k}\,\chi).
\end{array}
$$
Substitution yields the stated conformal metric. A radial null ray satisfies
$$-c^2d\tau^2+d\chi^2=0,$$
so $d\chi/d\tau=\pm c$ and $\chi=\chi_0\pm c(\tau-\tau_0)$.

Solved by gpt-5.6-sol high.

= b
{parent=15e}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

With $P=-2\rho/3$, the continuity equation becomes
$$\dot\rho=-\frac{\dot a}{a}\rho,$$
hence $\rho a$ is constant. Since $a(t_0)=1$, one has $\rho=\rho_0/a$. The Friedmann equation is therefore
$$\left(\frac{\dot a}{a}\right)^2=\frac\gamma a+\frac\beta{a^2},$$
where
$$\gamma=\frac{8\pi G\rho_0}{3c^2},\qquad \beta=-kc^2>0.$$

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

The expanding branch obeys
$$\dot a=\sqrt{\gamma a+\beta}.$$
Integrating from $a=0$ at $t=0$ gives
$$\frac2\gamma\left(\sqrt{\gamma a+\beta}-\sqrt\beta\right)=t.$$
Solving for $a$,
$$a(t)=\sqrt\beta\,t+\frac\gamma4t^2
=t\left(\sqrt\beta+\frac{\gamma t}{4}\right).$$

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

Setting $a(t_0)=1$ gives
$$t_0=\frac{2(\sqrt{\beta+\gamma}-\sqrt\beta)}\gamma
=\frac2{\sqrt{\beta+\gamma}+\sqrt\beta}.$$
At the present epoch,
$$H_0=\dot a(t_0)=\sqrt{\beta+\gamma},$$
so
$$t_0H_0=\frac{2\sqrt{\beta+\gamma}}
{\sqrt{\beta+\gamma}+\sqrt\beta}.$$
For $\beta\gg\gamma$, this tends to $1$, giving $t_0\sim H_0^{-1}$. For $\gamma\gg\beta$, it tends to $2$, giving $t_0\sim2H_0^{-1}$.

Solved by gpt-5.6-sol high.

= iv
{parent=b}
{scope}

= Solution
{parent=iv}

From the explicit solution, $\ddot a=\gamma/2$. Since $a(t_0)=1$ and $\dot a(t_0)^2=\beta+\gamma$,
$$q_0=-\frac{\gamma}{2(\beta+\gamma)}.$$
Because an open universe has $\beta>0$, this is strictly greater than $-1/2$. It approaches $-1/2$ only in the zero-curvature <limit>, so this model cannot attain $q_0\leq-0.5$ for the permitted parameters and therefore cannot reproduce $q_0\approx-0.55$.

Solved by gpt-5.6-sol high.

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

= a
{parent=16h}
{scope}

= Solution
{parent=a}

For a well-order $(A,r)$, define recursively
$$F(x)=\{F(y):y\mathrel r x\}.$$
<Transfinite recursion> makes $F$ an isomorphism from $(A,r)$ onto a transitive set of transitive sets, hence onto an ordinal. If two ordinals were isomorphic, the first point where the isomorphism differed from the identity would contradict preservation of their initial segments. Thus this ordinal is unique.

Solved by gpt-5.6-sol high.

= b
{parent=16h}
{scope}

= Solution
{parent=b}

The ordinal $\beta$ is the ordered concatenation of its initial segment $\alpha$ and its tail $\beta\setminus\alpha$. If the tail has order type $\gamma$, this concatenation has order type $\alpha+\gamma$. Hence $\alpha+\gamma=\beta$.

Solved by gpt-5.6-sol high.

= c
{parent=16h}
{scope}

= Solution
{parent=c}

Every proper initial segment of $\omega_1\setminus\omega$ is countable, while the whole tail is uncountable. Its order type is therefore the least uncountable ordinal itself:
$$\operatorname{otp}(\omega_1\setminus\omega)=\omega_1.$$
Equivalently, part (b) gives $\omega+\gamma=\omega_1$, and the unique possible uncountable initial ordinal $\gamma$ is $\omega_1$.

Solved by gpt-5.6-sol high.

= d
{parent=16h}
{scope}

= Solution
{parent=d}

Write the nonzero ordinal in <Cantor normal form>
$$\alpha=\omega^{\delta_1}n_1+\omega^{\delta_2}n_2+\cdots+\omega^{\delta_r}n_r,$$
where $\delta_1>\cdots>\delta_r$ and $0<n_i<\omega$. Taking $\delta=\delta_1$, $n=n_1$, and
$$\eta=\omega^{\delta_2}n_2+\cdots+\omega^{\delta_r}n_r$$
gives $\alpha=\omega^\delta n+\eta$ with $\eta<\omega^\delta$.

Solved by gpt-5.6-sol high.

= e
{parent=16h}
{scope}

= Solution
{parent=e}

Use the shuffle bound: if a well-order is partitioned into subsets of order types $\xi$ and $\eta$, its order type is at most the <Hessenberg natural sum> $\xi\mathbin\#\eta$. This follows by merging the two induced well-orders and comparing their Cantor-normal-form terms.

If both $X$ and $Y$ had order type strictly below $\omega^\delta$, each would have leading Cantor exponent below $\delta$. Their natural sum would still be below $\omega^\delta$. The shuffle bound would then give
$$\omega^\delta\leq\operatorname{otp}(X)\mathbin\#\operatorname{otp}(Y)<\omega^\delta,$$
a contradiction. Thus at least one part has order type $\omega^\delta$.

Solved by gpt-5.6-sol high.

= f
{parent=16h}
{scope}

= Solution
{parent=f}

The shuffle bound gives $\alpha\leq\beta\mathbin\#\beta$. By part (d), write
$$\beta=\omega^\delta n+\eta,\qquad 1\leq n<\omega,\qquad\eta<\omega^\delta.$$
Natural addition doubles corresponding Cantor coefficients, so
$$\beta\mathbin\#\beta=\omega^\delta(2n)+(\eta\mathbin\#\eta)
<\omega^\delta(2n+1).$$
On the other hand, ordinary ordinal addition absorbs each intermediate lower-order remainder, giving
$$\beta+\beta+\beta=\omega^\delta(3n)+\eta.$$
Consequently $\alpha<\beta+\beta+\beta$.

The stronger inequality with two copies is false. Take $\beta=\omega$, $\alpha=\omega+\omega$, let $X$ be the first block and $Y$ the second. Both have order type $\omega$, but $\alpha=\beta+\beta$.

Solved by gpt-5.6-sol high.

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

= a
{parent=17f}
{scope}

= Solution
{parent=a}

<Menger theorem> says that the maximum number of pairwise vertex-disjoint $A$--$B$ paths equals the minimum size of an $A$--$B$ separating vertex set, with the standard convention that the path endpoints lie in $A\cup B$. The connectivity $\kappa(G)$ is the minimum number of vertices whose deletion disconnects $G$ or leaves a single vertex; $\kappa(K_n)=n-1$.

The vertex form says that for distinct nonadjacent vertices $x,y$, the maximum number of internally vertex-disjoint $x$--$y$ paths equals the minimum size of an $x$--$y$ separator disjoint from $\{x,y\}$. It follows from the set form by splitting off the endpoints, or by applying it to their neighbor sets after deleting $x,y$.

Now let $C$ be a longest cycle. If $|C|<2k\leq|G|$, some component $H$ of $G-C$ exists. Its neighbor set $N_C(H)$ is a vertex separator, so it contains at least $k$ vertices. No two of these attachment vertices can be consecutive on $C$: a path through the connected component $H$ between consecutive attachments would replace their edge by a path of at least two edges and create a longer cycle. Thus $C$ contains at least one nonattachment between each of at least $k$ attachments, and $|C|\geq2k$, a contradiction.

Solved by gpt-5.6-sol high.

= b
{parent=17f}
{scope}

= Solution
{parent=b}

The <edge connectivity> $\lambda(G)$ is the minimum number of edges whose deletion disconnects $G$. Deleting all edges incident with a minimum-degree vertex proves $\lambda(G)\leq\delta(G)$. For a minimum edge cut separating vertex sets $A,B$, its endpoints on either suitable side give a vertex separator of size at most the number of cut edges; the complete-graph convention gives the same conclusion. Hence
$$\delta(G)\geq\lambda(G)\geq\kappa(G).$$

To realize prescribed $d\geq\ell\geq k\geq1$, take two disjoint copies $A,B$ of $K_{d+1}$. Add a bipartite set of exactly $\ell$ cross edges whose maximum matching has size $k$: include $a_ib_i$ for $1\leq i\leq k$, and, when $\ell>k$, add $a_1b_j$ for $k<j\leq\ell$. This is simple because $\ell\leq d$, has a matching of size $k$, and all its edges are covered by $\{a_1,\ldots,a_k\}$.

Vertices untouched by cross edges have degree $d$, so $\delta(G)=d$. The cross edges form an edge cut of size $\ell$; every cut that splits either clique uses at least $d\geq\ell$ clique edges, hence $\lambda(G)=\ell$. By König's theorem, the cross graph has a vertex cover of size $k$, whose deletion separates the two surviving clique pieces. Deleting fewer than $k$ vertices leaves each clique connected and at least one cross edge, so $\kappa(G)=k$.

Solved by gpt-5.6-sol high.

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

= a
{parent=18j}
{scope}

= Solution
{parent=a}

For a monic <polynomial> with roots $r_1,\ldots,r_n$, its discriminant is
$$\operatorname{Disc}(f)=\prod_{i<j}(r_i-r_j)^2.$$
Here $\operatorname{Disc}(g)=(u^3-v^3)^2$. Both $u^3-v^3$ and the Vandermonde product in the $\alpha_i$ are alternating homogeneous cubics. Evaluation at $(\alpha_1,\alpha_2,\alpha_3)=(0,1,-1)$ fixes the constant and gives
$$ (u^3-v^3)^2=-27\prod_{i<j}(\alpha_i-\alpha_j)^2=-27\operatorname{Disc}(f).$$

Using $\alpha_1+\alpha_2+\alpha_3=0$, $\sum_{i<j}\alpha_i\alpha_j=a$, and $\alpha_1\alpha_2\alpha_3=-b$, direct expansion gives
$$uv=-3a,\qquad u^3+v^3=-27b.$$
Consequently
$$g(X)=X^2+27bX-27a^3,$$
and
$$\operatorname{Disc}(g)=729b^2+108a^3=-27(-4a^3-27b^2).$$
Thus
$$\boxed{\operatorname{Disc}(f)=-4a^3-27b^2}.$$

Solved by gpt-5.6-sol high.

= b
{parent=18j}
{scope}

= Solution
{parent=b}

Solve the quadratic resolvent:
$$u^3,v^3=\frac{-27b\pm\sqrt{-27\operatorname{Disc}(f)}}2.$$
Choose cube roots with $uv=-3a$. Inverting the discrete Fourier transform gives
$$\alpha_1=\frac{u+v}{3},\qquad
\alpha_2=\frac{\omega^2u+\omega v}{3},\qquad
\alpha_3=\frac{\omega u+\omega^2v}{3}.$$
These expressions use only square and cube roots. A general cubic is first translated to remove its quadratic term, proving the existence of a formula by radicals.

Solved by gpt-5.6-sol high.

= c
{parent=18j}
{scope}

= Solution
{parent=c}

For an irreducible cubic over $\mathbb Q$, the Galois <group> is $A_3$ when its discriminant is a rational square and $S_3$ otherwise. The rational-root test shows that the first two cubics are irreducible.

For $X^3-21X-22$,
$$\Delta=23976=2^3\,3^4\,37$$
is not a square, so the Galois <group> is $S_3$.

For $X^3-21X-28$,
$$\Delta=15876=126^2,$$
so the Galois <group> is $A_3\cong C_3$.

Finally,
$$X^3-21X-34=(X+2)(X^2-2X-17),$$
whose remaining roots are $1\pm3\sqrt2$. Its splitting field is $\mathbb Q(\sqrt2)$ and its Galois <group> is $C_2$.

Solved by gpt-5.6-sol high.

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

= a
{parent=19j}
{scope}

= Solution
{parent=a}

Over $\mathbb C$, the operator representing $r$ has an <eigenvector> $v$ with <eigenvalue> $\lambda\ne0$. The relation $trt^{-1}=r^{-1}$ gives
$$r(tv)=t(r^{-1}v)=\lambda^{-1}tv.$$
Therefore $\operatorname{span}\{v,tv\}$ is a nonzero $G$-invariant subspace of dimension at most two. If $V$ is irreducible it equals this subspace, so $\dim V\leq2$.

Solved by gpt-5.6-sol high.

= b
{parent=19j}
{scope}

= Solution
{parent=b}

In one dimension the defining relations force
$$r\mapsto\varepsilon,\qquad t\mapsto\delta,\qquad
\varepsilon,\delta\in\{1,-1\},$$
giving four characters.

Every irreducible two-dimensional representation is, after a choice of <basis>,
$$r\mapsto\begin{pmatrix}\lambda&0\\0&\lambda^{-1}\end{pmatrix},
\qquad
t\mapsto\begin{pmatrix}0&1\\1&0\end{pmatrix},$$
where $\lambda\in\mathbb C^\times\setminus\{1,-1\}$. It is irreducible because the two distinct $r$-eigenlines are interchanged by $t$. Conversely, part (a) gives this <basis> from $v,tv$. If the only <eigenvalue> of $r$ were $\pm1$, its eigenspace would be $t$-invariant and contain a $t$-eigenvector, contradicting irreducibility. Parameters $\lambda$ and $\lambda^{-1}$ give isomorphic representations, and there are no other identifications.

Solved by gpt-5.6-sol high.

= c
{parent=19j}
{scope}

= Solution
{parent=c}

The quotient $G\to D_{2n}$ adds the relation $r^n=1$. All four one-dimensional characters factor through some such quotient: choose even $n$ when $r$ acts as $-1$. A two-dimensional representation from part (b) factors through $D_{2n}$ exactly when $\lambda^n=1$. Hence the irreducible representations that never arise from a finite dihedral quotient are precisely the two-dimensional ones for which $\lambda$ is not a <root of unity>.

Solved by gpt-5.6-sol high.

= d
{parent=19j}
{scope}

= Solution
{parent=d}

Take $t\mapsto A$ and $r\mapsto B$. Then $A^2=I$ and
$$ABA^{-1}=\begin{pmatrix}1&-1\\0&1\end{pmatrix}=B^{-1},$$
so these <matrices> define a representation. The line $\mathbb Ce_1$ is fixed by both <matrices> and is a one-dimensional subrepresentation.

Any complementary line has a generator $e_2+ce_1$. But
$$B(e_2+ce_1)=e_2+(c+1)e_1,$$
which cannot lie on the same line. Thus no complement is $B$-invariant, and therefore no $G$-invariant complement exists.

Solved by gpt-5.6-sol high.

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

= a
{parent=20g}
{scope}

= Solution
{parent=a}

Use the module criterion: an element $x$ is an algebraic integer if and only if it lies in a nonzero finitely generated $\mathbb Z$-module $M$ with $xM\subseteq M$.

Let the coefficients of the monic <polynomial> be $c_0,\ldots,c_{n-1}$ and put $A=\mathbb Z[c_0,\ldots,c_{n-1}]$. A <ring> generated by finitely many algebraic integers is a finitely generated $\mathbb Z$-module. If $\alpha$ is a root, then
$$M=A+A\alpha+\cdots+A\alpha^{n-1}$$
is finitely generated, and the monic relation
$$\alpha^n=-c_{n-1}\alpha^{n-1}-\cdots-c_0$$
shows that $\alpha M\subseteq M$. The criterion proves that every root is an algebraic integer.

Solved by gpt-5.6-sol high.

= b
{parent=20g}
{scope}

= Solution
{parent=b}

Since $17$ is square-free and $17\equiv1\pmod4$, the standard quadratic-field integral-basis theorem gives
$$\mathcal O_K=\mathbb Z\left[\frac{1+\sqrt{17}}2\right].$$
Indeed, the displayed generator has monic <polynomial> $x^2-x-4$, and the discriminant of this <basis> is $17$, the <field discriminant>; no larger order is possible.

Solved by gpt-5.6-sol high.

= c
{parent=20g}
{scope}

= Solution
{parent=c}

One has $\alpha^2=4+\sqrt{17}\in K$ and
$$N_{K/\mathbb Q}(\alpha^2)=(4+\sqrt{17})(4-\sqrt{17})=-1.$$
If $\alpha$ belonged to $K$, multiplicativity would give
$$N_{K/\mathbb Q}(\alpha)^2=-1,$$
which is impossible because $N_{K/\mathbb Q}(\alpha)\in\mathbb Q$. Therefore $\alpha\notin K$ and $[L:K]=2$.

Solved by gpt-5.6-sol high.

= d
{parent=20g}
{scope}

= Solution
{parent=d}

If $\beta\in\mathcal O_L$, all of its $K$-conjugates are algebraic integers, so their sum and product give
$$\operatorname{Tr}_{L/K}(\beta),\ N_{L/K}(\beta)\in\mathcal O_K.$$

Conversely, if these two quantities lie in $\mathcal O_K$, then $\beta$ is a root of
$$X^2-\operatorname{Tr}_{L/K}(\beta)X+N_{L/K}(\beta),$$
a monic <polynomial> whose coefficients are algebraic integers. Part (a), or transitivity of <integral> dependence, implies $\beta\in\mathcal O_L$.

Solved by gpt-5.6-sol high.

= e
{parent=20g}
{scope}

= Solution
{parent=e}

Let $\beta=a+b\alpha\in\mathcal O_L$. The nontrivial $K$-automorphism sends $\alpha$ to $-\alpha$, so
$$\operatorname{Tr}_{L/K}(\beta)=2a\in\mathcal O_K.$$
Also $\alpha$ is <integral>, hence $\alpha\beta$ is <integral>, and
$$\operatorname{Tr}_{L/K}(\alpha\beta)=2b\alpha^2
=2b(4+\sqrt{17})\in\mathcal O_K.$$
But $4+\sqrt{17}$ has norm $-1$, so it is a unit of $\mathcal O_K$. Multiplying by its inverse proves $2b\in\mathcal O_K$.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=21f}

The simplicial <Mayer-Vietoris theorem> says that if a simplicial complex $X=M\cup N$ for subcomplexes $M,N$, there is a natural long exact <sequence>
$$\cdots\to H_n(M\cap N)\xrightarrow{(i_*,-j_*)}
H_n(M)\oplus H_n(N)\to H_n(X)\to H_{n-1}(M\cap N)\to\cdots.$$
The reduced version extends through dimension zero.

For every simplex $\sigma$ of $K$, all faces of $\sigma\cup\{c_i\}$ are either faces of $\sigma$ or have the form $\tau\cup\{c_i\}$ with $\tau$ a face of $\sigma$, so they lie in $L$. Each $c_i*K$ is a cone and hence a simplicial complex. For distinct $i,j$, the rays in the last two coordinates through $c_i$ and $c_j$ are not positive multiples. Thus the two cones meet exactly in $K$, so their simplices have common faces only in $K$. Hence $L$ is a simplicial complex.

Topologically, $L$ is the union of three cones on $K$ along their common base. The first two cones form the suspension $\Sigma K$. Attaching the third contractible cone along the equatorial copy of $K$ gives
$$L\simeq\Sigma K\vee\Sigma K.$$
This also follows by applying reduced Mayer--Vietoris twice: every cone has zero reduced homology, and the two inclusion maps from the common base are null-homotopic inside the cones. Therefore
$$\boxed{\widetilde H_n(L)\cong
\widetilde H_{n-1}(K)\oplus\widetilde H_{n-1}(K)}$$
for every $n$, with the convention that negative reduced homology vanishes. Since $K$ is nonempty, $L$ is connected, so explicitly
$$H_0(L)\cong\mathbb Z,\qquad
H_1(L)\cong\widetilde H_0(K)^{\oplus2},$$
$$H_n(L)\cong H_{n-1}(K)^{\oplus2}\quad(n\geq2).$$

Solved by gpt-5.6-sol high.

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

= a
{parent=22i}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

The dual $X^*$ is the <vector space> of bounded <linear maps> $f:X\to\mathbb F$, with norm
$$\lVert f\rVert=\sup_{\lVert x\rVert\leq1}|f(x)|.$$
If $(f_n)$ is Cauchy in this norm, then $(f_n(x))$ is Cauchy in the complete <scalar> field for every $x$. Define $f(x)=\lim_nf_n(x)$. Pointwise passage to the <limit> preserves <linearity>. Given $\varepsilon>0$, choose $N$ such that $\lVert f_n-f_m\rVert<\varepsilon$ for $m,n\geq N$; sending $m\to\infty$ gives
$$|f_n(x)-f(x)|\leq\varepsilon\lVert x\rVert.$$
Thus $f$ is bounded and $\lVert f_n-f\rVert\leq\varepsilon$. Hence $X^*$ is Banach, even if $X$ was only normed.

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

Let $1<p<\infty$ and $q=p/(p-1)$. Every $y\in\ell^q$ defines
$$f_y(x)=\sum_{n\geq1}x_ny_n.$$
Hölder's inequality gives $\lVert f_y\rVert\leq\lVert y\rVert_q$, and equality follows by using a normalized <sequence> proportional to $\operatorname{sgn}(y_n)|y_n|^{q-1}$.

Conversely, for $f\in(\ell^p)^*$ put $y_n=f(e_n)$. On the first $N$ coordinates, choose the same norming <vector> to obtain
$$\left(\sum_{n=1}^N|y_n|^q\right)^{1/q}\leq\lVert f\rVert.$$
Letting $N\to\infty$ shows $y\in\ell^q$. On finitely supported <sequences>, <linearity> gives $f=f_y$; density of those <sequences> in $\ell^p$ extends this identity to all $x$. Therefore
$$\boxed{(\ell^p)^*\cong\ell^q}$$
isometrically.

Solved by gpt-5.6-sol high.

= iii
{parent=a}
{scope}

= Solution
{parent=iii}

The endpoint duals are
$$ (\ell^1)^*\cong\ell^\infty,\qquad (c_0)^*\cong\ell^1,$$
under the same coordinate pairing $\langle x,y\rangle=\sum_nx_ny_n$.

Solved by gpt-5.6-sol high.

= b
{parent=22i}
{scope}

= Solution
{parent=b}

Yes. For $x=(x_n)\in c$, let $L=\lim_nx_n$ and define
$$T(x)=(L,x_1-L,x_2-L,\ldots).$$
The tail tends to zero, so $T(x)\in c_0$, and $\lVert T(x)\rVert_\infty\leq2\lVert x\rVert_\infty$. Conversely, for $y=(y_n)\in c_0$, define
$$T^{-1}(y)_n=y_1+y_{n+1}.$$
This <sequence> converges to $y_1$ and has norm at most $2\lVert y\rVert_\infty$. The formulas are inverse bounded <linear maps>, so $c$ and $c_0$ are isomorphic Banach spaces.

Solved by gpt-5.6-sol high.

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

= a
{parent=23h}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

Since $\nu\ll\mu$ and both measures are finite, the Radon--Nikodym theorem supplies a measurable $w:E\to[0,\infty]$ such that
$$\nu(A)=\int_Aw\,d\mu$$
for every measurable $A$. Moreover,
$$\int_Ew\,d\mu=\nu(E)<\infty,$$
so $w$ is finite $\mu$-almost everywhere and is $\mu$-integrable. The reverse absolute continuity implies $w>0$ $\mu$-almost everywhere: if $w=0$ on $A$, then $\nu(A)=0$, hence $\mu(A)=0$.

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

For $0<p\leq1$, the elementary bound $t^p\leq1+t$ gives
$$\int_Ew^p,d\mu\leq\mu(E)+\int_Ew,d\mu<\infty.$$
No exponent $p>1$ is forced. On $E=(0,1)$ with Lebesgue measure, take $w(x)=x^{-1/p}$. Then $w$ is positive and integrable, so $d\nu=w,d\mu$ defines a finite measure mutually absolutely continuous with $\mu$, but
$$\int_0^1w(x)^pdx=\int_0^1\frac{dx}{x}=\infty.$$
Thus the required range is exactly $0<p\leq1$.

Solved by gpt-5.6-sol high.

= b
{parent=23h}
{scope}

= Solution
{parent=b}

Yes. Define the probability measure
$$\mu=\sum_{n=1}^\infty2^{-n}\nu_n.$$
Countable additivity follows by Tonelli's theorem and $\mu(E)=\sum_n2^{-n}=1$. If $\mu(A)=0$, every nonnegative term $2^{-n}\nu_n(A)$ is zero, so $\nu_n(A)=0$ for every $n$. Hence every $\nu_n\ll\mu$.

Solved by gpt-5.6-sol high.

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

= i
{parent=24g}
{scope}

= Solution
{parent=i}

A Riemann surface is a connected, Hausdorff, second-countable topological space with an atlas to open subsets of $\mathbb C$ whose transition maps are holomorphic.

Every open connected subset inherits the restricted charts and the other topological properties, so it is a Riemann surface. Removing finitely many points gives an open subset because points are closed. It remains connected: a Riemann surface is path connected, and a path meeting finitely many deleted points can be perturbed inside coordinate discs around those points. Countably many points may also be removed successfully; for example, $\mathbb C\setminus\mathbb Z$ is open and connected and hence is a Riemann surface.

The sphere $S^2$ is the <Riemann sphere>. <Stereographic projection> from the north and south poles gives two complex charts, and their overlap map is $z\mapsto1/z$ up to the chosen conjugation convention, which is holomorphic after choosing compatible orientations.

Solved by gpt-5.6-sol high.

= ii
{parent=24g}
{scope}

= Solution
{parent=ii}

Projection onto the first two coordinates is a homeomorphism
$$X\longrightarrow\mathbb R^2\setminus\{0\},$$
because the last two coordinates are uniquely recovered as $(x/r,y/r)$. Identifying $(x,y)$ with $z=x+iy$ transports the standard Riemann-surface structure of $\mathbb C^*$ to $X$. Topologically this is a cylinder.

Solved by gpt-5.6-sol high.

= iii
{parent=24g}
{scope}

= Solution
{parent=iii}

The equation factors as
$$zw-2iw-iz-2=(z-2i)(w-i).$$
Thus $Y$ is the union of the two complex lines $z=2i$ and $w=i$, meeting at $(2i,i)$. A sufficiently small neighborhood of the intersection, with that point removed, has two connected components, whereas a punctured disc is connected. The intersection therefore has no neighborhood homeomorphic to an open subset of $\mathbb C$. Hence $Y$ cannot be a Riemann surface.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=25j}

A rational map $X\dashrightarrow Y$ between irreducible projective varieties is an equivalence class of morphisms from nonempty open subsets of $X$ to $Y$, with representatives identified when they agree on a nonempty open subset. It is regular at $p$ if some representative is defined on a neighborhood of $p$.

For
$$\phi(x:y:z)=(xy:xz:z^2),$$
the coordinates vanish together exactly at $(0:1:0)$ and $(1:0:0)$. Near the first point, the paths $z=0$ and $x=0$ have limiting images $(1:0:0)$ and $(0:0:1)$; near the second, the paths $z=0$ and $y=0$ have limiting images $(1:0:0)$ and $(0:1:0)$. No continuous extension exists there. Elsewhere the coordinates do not vanish together, so the map is regular.

Define
$$\psi(u:v:w)=(v^2:uw:vw).$$
Where all coordinates are nonzero,
$$\psi\phi(x:y:z)=(x^2z^2:xyz^2:xz^3)=(x:y:z),$$
and similarly $\phi\psi$ is the identity. Thus $\phi$ is birational and is an isomorphism on $\mathbb P^2\setminus Z(xyz)$.

For irreducibility of $P=x^2z^4-x^3y^3+z^6$, dehomogenize at $z=1$. Over $k(x)$ this is, up to a unit,
$$y^3-\frac{x^2+1}{x^3}.$$
The rational <function> on the right is not a cube, since its valuation at either root of $x^2+1$ is one. The cubic is therefore irreducible over $k(x)$; Gauss's lemma gives irreducibility in $k[x,y]$. Since $z$ does not divide $P$, homogenization preserves irreducibility.

Writing $(u:v:w)=(xy:xz:z^2)$ transforms the equation into
$$C:\quad v^2w-u^3+w^3=0.$$
Its <partial derivatives> are $-3u^2$, $2vw$, and $v^2+3w^2$ up to common signs. Simultaneous vanishing forces $u=v=w=0$, impossible projectively, so $C$ is nonsingular. The inverse $\psi$ restricts on dense open subsets, proving that $V$ and $C$ are birational.

Solved by gpt-5.6-sol high.

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

= a
{parent=26i}
{scope}

= Solution
{parent=a}

A <critical point> of a smooth map is a point where its <derivative> is not surjective; its image is a critical value. A <regular value> is a target value all of whose preimages have surjective <derivative>.

For $f(X)=X^TAX$ with $A=A^T$,
$$Df_X(H)=2X^TAH.$$
Thus $X$ is critical exactly when $AX=0$, and every critical point has value zero. Hence zero is the only possible critical value; it is a critical value precisely when $A$ is singular. If $A$ is singular, every point of its kernel is critical, so there can be infinitely many critical points; if $A$ is invertible, only the origin is critical.

Solved by gpt-5.6-sol high.

= b
{parent=26i}
{scope}

= Solution
{parent=b}

Let $\operatorname{Skew}_{2n}$ be the skew-symmetric $2n$ by $2n$ <matrices> and define
$$F:M_{2n}(\mathbb R)\to\operatorname{Skew}_{2n},\qquad F(A)=A^TJA.$$
Then $Sp_n(\mathbb R)=F^{-1}(J)$, and
$$DF_A(H)=H^TJA+A^TJH.$$
At a symplectic $A$, put $H=AX$. The <derivative> becomes $X^TJ+JX$. Given skew $K$, choosing $X=-\tfrac12JK$ gives $X^TJ+JX=K$, so the <derivative> is surjective.

The <preimage theorem> makes the level set a submanifold. Since
$$\dim M_{2n}=4n^2,\qquad \dim\operatorname{Skew}_{2n}=\frac{(2n)(2n-1)}2=2n^2-n,$$
its dimension is $2n^2+n$.

Solved by gpt-5.6-sol high.

= c
{parent=26i}
{scope}

= i
{parent=c}
{scope}

= Solution
{parent=i}

If $P=P^T=P^2$ and $\operatorname{tr}P=1$, the spectral theorem gives <eigenvalues> $1,0,0$. Thus $P$ is orthogonal projection onto a line and
$$P=XX^T$$
for a unit <vector> $X$; the only choices are $X$ and $-X$.

The smooth map $\pi:S^2\to S_3(\mathbb R)$, $X\mapsto XX^T$, has image $Gr_{1,3}(\mathbb R)$ and fibers $\{X,-X\}$. Near $X_0$, restrict to the hemisphere $X\cdot X_0>0$; there $\pi$ is one-to-one and has smooth inverse obtained by choosing the positive local unit <eigenvector>. Combining this with a chart of $S^2$ supplies two-dimensional local parametrizations. Hence $Gr_{1,3}(\mathbb R)$ is a two-dimensional submanifold, diffeomorphic to $\mathbb RP^2$.

Solved by gpt-5.6-sol high.

= ii
{parent=c}
{scope}

= Solution
{parent=ii}

No. The set is closed and bounded in the finite-dimensional space $S_3(\mathbb R)$, hence compact and nonempty. A global parametrization would be a homeomorphism from a nonempty open subset of $\mathbb R^2$ onto this compact surface, but no nonempty open subset of $\mathbb R^2$ is compact.

Solved by gpt-5.6-sol high.

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

= a
{parent=27h}
{scope}

= Solution
{parent=a}

Lévy's continuity theorem states that if characteristic <functions> $\phi_n$ converge pointwise to a <function> $\phi$ continuous at zero, then $\phi$ is a characteristic <function> and the corresponding laws converge weakly to its law. Conversely, convergence in distribution implies <pointwise convergence> of characteristic <functions>.

Solved by gpt-5.6-sol high.

= b
{parent=27h}
{scope}

= Solution
{parent=b}

For $X_j\sim\operatorname{Unif}[-j,j]$,
$$\phi_{X_j}(t)=\frac1{2j}\int_{-j}^je^{itx}dx=\frac{\sin(jt)}{jt}.$$
Independence therefore gives
$$\phi_{n^{-3/2}S_n}(\xi)
=\prod_{j=1}^n\frac{\sin(j\xi/n^{3/2})}{j\xi/n^{3/2}}
=\frac{n^{3n/2}}{\xi^n n!}\prod_{j=1}^n\sin\left(\frac{j\xi}{n^{3/2}}\right).$$

Solved by gpt-5.6-sol high.

= c
{parent=27h}
{scope}

= Solution
{parent=c}

Uniformly for the arguments involved,
$$\log\frac{\sin x}{x}=-\frac{x^2}{6}+O(x^4).$$
Hence
$$\log\phi_{n^{-3/2}S_n}(\xi)
=-\frac{\xi^2}{6n^3}\sum_{j=1}^nj^2
+O\left(\frac{\xi^4}{n^6}\sum_{j=1}^nj^4\right)
\longrightarrow-\frac{\xi^2}{18},$$
because $n^{-3}\sum j^2\to1/3$ and the error is $O(n^{-1})$. The limiting characteristic <function> is that of $N(0,1/9)$. Lévy's theorem gives
$$n^{-3/2}S_n\Rightarrow N(0,1/9).$$

Solved by gpt-5.6-sol high.

= d
{parent=27h}
{scope}

= Solution
{parent=d}

Since $\operatorname{Var}(X_j)=j^2/3$,
$$V_n:=\sum_{j=1}^n\sigma_j^2=\frac13\sum_{j=1}^nj^2\sim\frac{n^3}{9}.$$
Thus $V_n^{-1/2}S_n$ differs by a factor tending to $3$ from $n^{-3/2}S_n$. Slutsky's theorem and part (c) yield
$$V_n^{-1/2}S_n\Rightarrow N(0,1).$$

Solved by gpt-5.6-sol high.

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

= a
{parent=28l}
{scope}

= Solution
{parent=a}

Let $p_n(t)=\mathbb P(X_t=n)$. The infinitesimal definition gives the forward equations
$$p_0'=-\lambda p_0,\qquad p_n'=\lambda p_{n-1}-\lambda p_n\quad(n\geq1),$$
with $p_0(0)=1$. Solving recursively gives
$$\mathbb P(X_t=n)=e^{-\lambda t}\frac{(\lambda t)^n}{n!},$$
so $X_t\sim\operatorname{Poisson}(\lambda t)$.

Solved by gpt-5.6-sol high.

= b
{parent=28l}
{scope}

= Solution
{parent=b}

Conditionally on $X_t=n$, the $n$ arrival times are the order statistics of independent $\operatorname{Unif}[0,t]$ variables. For $n\geq1$, the last jump occurs before $3t/4$ exactly when every point lies before that time, so the probability is
$$\left(\frac34\right)^n.$$
The number falling in $(t/4,3t/4)$ is therefore
$$\operatorname{Binomial}\left(n,\frac12\right).$$

Solved by gpt-5.6-sol high.

= c
{parent=28l}
{scope}

= Solution
{parent=c}

A particle arriving at time $s\leq t$ survives until $t$ with probability $e^{-\mu(t-s)}$. Independently retaining each Poisson arrival with this time-dependent probability is <Poisson thinning>. Hence the number alive at time $t$ is Poisson with mean
$$\lambda\int_0^te^{-\mu(t-s)}ds
=\frac\lambda\mu(1-e^{-\mu t}).$$
Thus
$$N_t^{\rm alive}\sim\operatorname{Poisson}\left(\frac\lambda\mu(1-e^{-\mu t})\right).$$

Solved by gpt-5.6-sol high.

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

= a
{parent=29l}
{scope}

= Solution
{parent=a}

The <delta method> gives
$$\sqrt n\bigl(\varphi(\widehat\psi_n)-\varphi(\psi)\bigr)
\Rightarrow N\left(0,\,[\varphi'(\psi)]^2v\right).$$
If $\varphi'(\psi)=0$, this denotes the degenerate distribution at zero.

Solved by gpt-5.6-sol high.

= b
{parent=29l}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

The log likelihood is
$$\ell(\theta)=n\log\theta-\theta\sum_{i=1}^nX_i.$$
Its strictly concave maximizer is
$$\widehat\theta_n=\frac{n}{\sum_iX_i}=\frac1{\overline X}.$$

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

Since $\mathbb EX_i=1/\theta$ and $\operatorname{Var}(X_i)=1/\theta^2$, the central <limit> theorem gives
$$\sqrt n(\overline X-1/\theta)\Rightarrow N(0,1/\theta^2).$$
Applying the <delta method> to $g(x)=1/x$ yields
$$\sqrt n(\widehat\theta_n-\theta)\Rightarrow N(0,\theta^2).$$
Thus, with $z_{1-\alpha/2}=\Phi^{-1}(1-\alpha/2)$, an asymptotic interval centered at the MLE is
$$C_n=\left[\widehat\theta_n-z_{1-\alpha/2}\frac{\widehat\theta_n}{\sqrt n},
\widehat\theta_n+z_{1-\alpha/2}\frac{\widehat\theta_n}{\sqrt n}\right],$$
optionally intersected with $(0,\infty)$. Slutsky's theorem gives coverage tending to $1-\alpha$.

Solved by gpt-5.6-sol high.

= c
{parent=29l}
{scope}

= i
{parent=c}
{scope}

= Solution
{parent=i}

For $y=0,1,\ldots$,
$$\mathbb P_\theta(Y_i=y)=e^{-\theta y}(1-e^{-\theta}),$$
so $Y_i$ is geometric with success probability $1-e^{-\theta}$. The log likelihood is
$$n\log(1-e^{-\theta})-n\theta\overline Y.$$
Its score equation gives
$$e^{-\theta}=\frac{\overline Y}{1+\overline Y},$$
and hence
$$\widetilde\theta_n=\log\left(\frac{1+\overline Y}{\overline Y}\right).$$
When $\overline Y=0$, the extended MLE is $+\infty$; this event has probability tending to zero.

Solved by gpt-5.6-sol high.

= ii
{parent=c}
{scope}

= Solution
{parent=ii}

Put $r=e^\theta$. The geometric moments are
$$m:=\mathbb EY_i=\frac1{r-1},\qquad
\operatorname{Var}(Y_i)=\frac r{(r-1)^2}.$$
For $g(y)=\log((1+y)/y)$, one has $g(m)=\theta$ and
$$g'(m)=-\frac1{m(1+m)}=-\frac{(r-1)^2}{r}.$$
The central <limit> theorem for $\overline Y$ followed by the <delta method> therefore gives
$$\sqrt n(\widetilde\theta_n-\theta)Rightarrow
N\left(0,\frac{(e^\theta-1)^2}{e^\theta}\right).$$

Solved by gpt-5.6-sol high.

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

= a
{parent=30k}
{scope}

= Solution
{parent=a}

For $X=\theta^TZ$,
$$\mathbb EX=\theta^Tb,\qquad \operatorname{Var}(X)=\theta^TV\theta.$$
The objective has <gradient> $b-V\theta$ and Hessian $-V$, which is negative definite. Its unique maximizer is therefore
$$\boxed{\theta_M=V^{-1}b}.$$

Solved by gpt-5.6-sol high.

= b
{parent=30k}
{scope}

= Solution
{parent=b}

Put $D=b^TV^{-1}b$ and decompose any $\theta$ in the $V$-inner product as
$$\theta=\lambda\theta_M+\eta,\qquad \eta^TV\theta_M=\eta^Tb=0.$$
Then
$$\mathbb EX=\lambda D,\qquad
\operatorname{Var}(X)=\lambda^2D+\eta^TV\eta.$$
For fixed $\lambda$, a nonzero $\eta$ leaves the mean unchanged and strictly raises variance, so strict monotonicity of $F$ excludes it from a maximizer. If $D>0$ and $\lambda<0$, the zero portfolio has a larger mean and smaller variance, so this is also impossible. Hence every maximizer is
$$\theta^*=\lambda\theta_M\qquad(\lambda\geq0).$$
If $b=0$, the unique variance-minimizing maximizer is $\theta^*=0$, which is the same conclusion with $\lambda=0$.

Solved by gpt-5.6-sol high.

= c
{parent=30k}
{scope}

= Solution
{parent=c}

Assume first that $D=b^TV^{-1}b>0$. Since $X_M=\theta_M^TZ$,
$$\mathbb EX_M=D,\qquad \operatorname{Var}(X_M)=D,$$
and for $Y=\phi^TZ$,
$$\mathbb EY=\phi^Tb,\qquad
\operatorname{Cov}(X_M,Y)=\theta_M^TV\phi=b^T\phi.$$
The least-squares affine-regression coefficients are therefore
$$\beta=\frac{b^T\phi}{D},\qquad
\alpha=\mathbb EY-\beta\mathbb EX_M=0.$$
The minimum is
$$\operatorname{Var}(Y)-\frac{\operatorname{Cov}(X_M,Y)^2}{\operatorname{Var}(X_M)}
=\phi^TQ\phi,$$
where
$$\boxed{Q=V-\frac{bb^T}{b^TV^{-1}b}}.$$
This <matrix> is symmetric and positive semidefinite. If $b=0$, then $X_M=0$, one may take $\alpha=0$ and arbitrary $\beta$, and the minimum is $\phi^TV\phi$, corresponding to $Q=V$.

Solved by gpt-5.6-sol high.

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

= a
{parent=31l}
{scope}

= Solution
{parent=a}

For a fixed sample $x_{1:n}$ and independent Rademacher signs $\sigma_i$,
$$\widehat{\mathcal R}(H(x_{1:n}))
=\mathbb E_\sigma\sup_{h\in H}\frac1n\sum_{i=1}^n\sigma_i h(x_i).$$
The <Rademacher complexity> is
$$\mathcal R_n(H)=\mathbb E_{X_{1:n}}\widehat{\mathcal R}(H(X_{1:n})).$$
The contraction lemma says that if each $\psi_i$ is $L$-Lipschitz and $\psi_i(0)=0$, then
$$\mathbb E_\sigma\sup_{h\in H}\frac1n\sum_i\sigma_i\psi_i(h(x_i))
\leq L\widehat{\mathcal R}(H(x_{1:n})).$$
Subtracting constants handles maps not vanishing at zero.

Solved by gpt-5.6-sol high.

= b
{parent=31l}
{scope}

= Solution
{parent=b}

For a positive semidefinite <matrix>, the sum of the <eigenvalues> is its trace. Thus
$$S=\{M=M^T:M\succeq0,\ \operatorname{tr}M\leq s\}.$$
If $M,N\in S$ and $0\leq t\leq1$, then $tM+(1-t)N\succeq0$ and
$$\operatorname{tr}(tM+(1-t)N)
=t\operatorname{tr}M+(1-t)\operatorname{tr}N\leq s.$$
Hence $S$ is convex.

Solved by gpt-5.6-sol high.

= c
{parent=31l}
{scope}

= Solution
{parent=c}

For fixed $x_{1:n}$ put $A_\sigma=\sum_i\sigma_i x_ix_i^T$. Then
$$\widehat{\mathcal R}(H(x_{1:n}))
=\frac1n\mathbb E_\sigma\sup_{M\succeq0,\operatorname{tr}M\leq s}
\operatorname{tr}(MA_\sigma)
\leq\frac s n\mathbb E_\sigma\lVert A_\sigma\rVert_{\rm op}
\leq\frac s n\mathbb E_\sigma\lVert A_\sigma\rVert_F.$$
Jensen's inequality and cancellation of cross terms between independent signs give
$$\mathbb E_\sigma\lVert A_\sigma\rVert_F
\leq\left(\sum_i\operatorname{tr}[(x_ix_i^T)^2]\right)^{1/2}
=\left(\sum_i\lVert x_i\rVert_2^4\right)^{1/2}
\leq C^2\sqrt n.$$
Therefore
$$\boxed{\mathcal R_n(H)\leq\frac{C^2s}{\sqrt n}}.$$

Solved by gpt-5.6-sol high.

= d
{parent=31l}
{scope}

= Solution
{parent=d}

The <hinge loss> $\phi(t)=\max(0,1-t)$ is one-Lipschitz. Centering it at zero and applying the contraction lemma gives
$$\mathcal R_n(\phi\circ H)\leq\mathcal R_n(H).$$
For an empirical risk minimizer and a population minimizer, the standard expected ERM inequality with the $1/n$ Rademacher convention used in part (a) is
$$\mathbb E R_\phi(\widehat h)-R_\phi(h^*)
\leq2\mathcal R_n(\phi\circ H).$$
Part (c) consequently yields
$$\mathbb E R_\phi(\widehat h)-R_\phi(h^*)
\leq\frac{2C^2s}{\sqrt n}.$$
Thus $K=2C^2s$ for this normalization of Rademacher complexity.

Solved by gpt-5.6-sol high.

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

= a
{parent=32a}
{scope}

= Solution
{parent=a}

A $C^1$ <function> $V$ near an equilibrium at the origin is a <Lyapunov function> if $V(0)=0$, $V(x)>0$ for $x\ne0$, and
$$\dot V(x)=\nabla V(x)\cdot f(x)\leq0.$$
The <first Lyapunov theorem> says these conditions imply stability. The second says that if the last inequality is strict away from the origin, then the origin is asymptotically stable.

To prove the first, fix a sufficiently small ball of radius $\varepsilon$. Positive definiteness and compactness give
$$m=\min_{|x|=\varepsilon}V(x)>0.$$
Continuity at zero gives $\delta>0$ such that $|x_0|<\delta$ implies $V(x_0)<m$. Since $V$ cannot increase along a trajectory, that trajectory cannot meet the sphere $|x|=\varepsilon$, where $V\geq m$. It therefore remains in the $\varepsilon$-ball for all forward time, which is stability.

Solved by gpt-5.6-sol high.

= b
{parent=32a}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

The Jacobian at the origin is $-I$, whose two <eigenvalues> are $-1$. The linearization theorem for a <hyperbolic equilibrium> therefore implies that the origin is asymptotically stable.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

Set
$$Q(x,y)=12x^2+6y^2.$$
Direct substitution of the system gives the exact factorization
$$\dot Q=-2Q+24(x+y)(2x^2+y^2)
=2Q(2x+2y-1).$$
Cauchy--Schwarz in the ellipsoidal norm gives
$$x+y\leq\sqrt{(12x^2+6y^2)\left(\frac1{12}+\frac16\right)}
=\frac{\sqrt Q}{2}.$$
Thus if $0<Q<1$, then $x+y<1/2$ and $\dot Q<0$. Every trajectory beginning in $Q<1$ remains in a compact smaller sublevel set and tends to the only point where $Q=0$, namely the origin. Hence the basin contains the stated region.

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

Yes. On $Q\leq1$ the same inequality gives $\dot Q\leq0$, so the closed ellipse is positively invariant. Equality in $\dot Q=0$ occurs only at the origin and at the boundary point
$$p=(1/6,1/3).$$
At that point the <vector> field is
$$f(p)=(1/54,-1/54)\ne0,$$
so $\{p\}$ is not invariant. The largest invariant subset of $\{\dot Q=0\}$ is therefore the origin. LaSalle's invariance principle shows that every trajectory in the closed ellipse converges to the origin. The inequality may consequently be extended to $Q\leq1$.

Solved by gpt-5.6-sol high.

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

= a
{parent=33d}
{scope}

= Solution
{parent=a}

Equality of the mixed <derivatives> gives
$$-U_y\Phi+UV\Phi=-V_x\Phi+VU\Phi.$$
For a fundamental <matrix> solution this is equivalent to the <zero-curvature condition>
$$\boxed{V_x-U_y+[U,V]=0}.$$

Solved by gpt-5.6-sol high.

= b
{parent=33d}
{scope}

= Solution
{parent=b}

Substituting the displayed <matrices> into $V_x-U_y+[U,V]$ makes every off-diagonal entry vanish. The two nonzero diagonal entries are opposites, the first being
$$-u_{xy}+e^u-e^{-2u}.$$
Thus compatibility is equivalent to the <Tzitzeica equation>
$$\boxed{u_{xy}=F(u)=e^u-e^{-2u}}.$$

Solved by gpt-5.6-sol high.

= c
{parent=33d}
{scope}

= Solution
{parent=c}

The flow of $x\partial_x-\alpha y\partial_y$ is
$$G_\varepsilon:(x,y,u)\longmapsto(e^\varepsilon x,e^{-\alpha\varepsilon}y,u).$$
Under the induced action on solutions, $u_{xy}$ acquires the factor $e^{(\alpha-1)\varepsilon}$, while $e^u-e^{-2u}$ is unchanged. The equation is invariant for every $\varepsilon$ exactly when
$$\alpha=1.$$

Solved by gpt-5.6-sol high.

= d
{parent=33d}
{scope}

= Solution
{parent=d}

For $\alpha=1$, the invariant of the characteristic equations is $z=xy$, so an invariant solution has $u(x,y)=f(z)$. Then
$$u_{xy}=f'(z)+zf''(z).$$
The invariant solutions are therefore characterized by
$$\boxed{zf''+f'=e^f-e^{-2f}}.$$

Solved by gpt-5.6-sol high.

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

= a
{parent=34c}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

Using $B^2=0$ and $BB^\dagger=1-B^\dagger B$,
$$H^2=B^\dagger BB^\dagger B
=B^\dagger(1-B^\dagger B)B=H.$$
Thus $H$ is a projection and its <eigenvalues> are $0$ and $1$.

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

The state $B|1\rangle$ has norm
$$\lVert B|1\rangle\rVert^2=\langle1|B^\dagger B|1\rangle=1$$
and satisfies $H B|1\rangle=B^\dagger B^2|1\rangle=0$. Hence it is the normalized ground state, denoted $|0\rangle$. Also
$$\lVert B^\dagger|1\rangle\rVert^2
=\langle1|BB^\dagger|1\rangle
=\langle1|(1-H)|1\rangle=0,$$
so
$$B|1\rangle=|0\rangle,\qquad B^\dagger|1\rangle=0.$$

Solved by gpt-5.6-sol high.

= iii
{parent=a}
{scope}

= Solution
{parent=iii}

In the ordered <basis> $(|0\rangle,|1\rangle)$,
$$B=\begin{pmatrix}0&1\\0&0\end{pmatrix},\qquad
B^\dagger=\begin{pmatrix}0&0\\1&0\end{pmatrix},\qquad
H=\begin{pmatrix}0&0\\0&1\end{pmatrix}.$$

Solved by gpt-5.6-sol high.

= b
{parent=34c}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

The product states $|n_1n_2\rangle$, $n_a\in\{0,1\}$, are exact <eigenvectors>, with
$$H_{\rm tot}|n_1n_2\rangle=(E_1n_1+E_2n_2)|n_1n_2\rangle.$$
Thus the energies attached to $|00\rangle,|01\rangle,|10\rangle,|11\rangle$ are respectively
$$0,\quad E_2,\quad E_1,\quad E_1+E_2.$$

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

In the ordered <basis> $(|00\rangle,|01\rangle,|10\rangle,|11\rangle)$,
$$B_1=\begin{pmatrix}0&0&1&0\\0&0&0&1\\0&0&0&0\\0&0&0&0\end{pmatrix},\qquad
B_2=\begin{pmatrix}0&1&0&0\\0&0&0&0\\0&0&0&1\\0&0&0&0\end{pmatrix},$$
$B_1^\dagger$ and $B_2^\dagger$ are their transposes, and
$$H_{\rm tot}=\operatorname{diag}(0,E_2,E_1,E_1+E_2).$$

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

For $H_0=E_2B_2^\dagger B_2$, the eigenspaces of energies $0$ and $E_2$ are each two-dimensional because $n_1$ is free. <Degenerate perturbation theory> requires diagonalizing $E_1B_1^\dagger B_1$ within each space. It is already diagonal in the product <basis>, with <eigenvalues> $0,E_1$. Therefore the first-order <eigenvectors> remain
$$|00\rangle,|10\rangle,|01\rangle,|11\rangle,$$
and their energies are $0,E_1,E_2,E_2+E_1$, respectively. These are the exact answers, so all higher corrections vanish.

Solved by gpt-5.6-sol high.

= c
{parent=34c}
{scope}

= i
{parent=c}
{scope}

= Solution
{parent=i}

Define $|\pm\rangle_1=(|0\rangle_1\pm|1\rangle_1)/\sqrt2$. Since $B_1+B_1^\dagger$ is the Pauli $X$ operator on the first oscillator, the exact <eigenvectors> are
$$|\pm\rangle_1\otimes|n_2\rangle_2,$$
with energies
$$E_{\pm,n_2}=\pm g+E_2n_2,\qquad n_2=0,1.$$

Solved by gpt-5.6-sol high.

= ii
{parent=c}
{scope}

= Solution
{parent=ii}

Within each degenerate unperturbed eigenspace of fixed $n_2$, the perturbation has <matrix>
$$g\begin{pmatrix}0&1\\1&0\end{pmatrix}.$$
Its <eigenvectors> are $|+\rangle_1$ and $|-\rangle_1$, with first-order shifts $+g$ and $-g$. Thus <degenerate perturbation theory> gives exactly the states and energies found in part (i); again, all higher corrections vanish because the perturbation commutes with the second-oscillator <Hamiltonian> and is exactly diagonalized within each degenerate block.

Solved by gpt-5.6-sol high.

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

= a
{parent=35b}
{scope}

= Solution
{parent=a}

Put
$$G_k(x-x')=e^{ik|x-x'|}.$$
Away from $x=x'$, this satisfies $(d^2/dx^2+k^2)G_k=0$. Its first <derivative> has jump $2ik$, so, distributionally,
$$
\left(\frac{d^2}{dx^2}+k^2\right)G_k(x-x')=2ik\delta(x-x').
$$
Applying $d^2/dx^2+k^2$ to the proposed <integral> equation therefore gives
$$
\left(\frac{d^2}{dx^2}+k^2\right)\psi(x)
=\frac{2m}{\hbar^2}V(x)\psi(x).
$$
Since $k^2=2mE/\hbar^2$, this is precisely the stated <Schrodinger equation>. The term $e^{ikx}$ is the incident wave and the choice $e^{ik|x-x'|}$ gives outgoing waves on both sides of each source point.

Solved by gpt-5.6-sol high.

= b
{parent=35b}
{scope}

= Solution
{parent=b}

Set
$$
q=\frac{im\lambda}{\hbar^2k},\qquad r=e^{ika},\qquad
\boldsymbol\Psi=\begin{pmatrix}\psi(-a)\\\psi(0)\\\psi(a)\end{pmatrix}.
$$
The <integral> equation becomes
$$
\psi(x)=e^{ikx}+q\left\{e^{ik|x+a|}\psi(-a)+e^{ik|x|}\psi(0)+e^{ik|x-a|}\psi(a)\right\}.
$$
Evaluating it at the three delta <functions> gives
$$
\begin{aligned}
\psi(-a)&=r^{-1}+q\{\psi(-a)+r\psi(0)+r^2\psi(a)\},\\
\psi(0)&=1+q\{r\psi(-a)+\psi(0)+r\psi(a)\},\\
\psi(a)&=r+q\{r^2\psi(-a)+r\psi(0)+\psi(a)\}.
\end{aligned}
$$
Thus
$$
\left[I-q\begin{pmatrix}1&r&r^2\\r&1&r\\r^2&r&1\end{pmatrix}\right]
\boldsymbol\Psi
=\begin{pmatrix}r^{-1}\\1\\r\end{pmatrix}.
$$

For $x>a$, all three Green-function terms are proportional to $e^{ikx}$. Hence
$$
S_{++}(k)=1+q\left\{r\psi(-a)+\psi(0)+r^{-1}\psi(a)\right\}.
$$

Now define $\gamma=ik\hbar^2/(\lambda m)$, so that $q=-1/\gamma$. If $M$ denotes the <matrix> multiplying $\boldsymbol\Psi$, direct evaluation gives
$$
\det M=\frac{r^4}{\gamma^3}\left[1-\gamma-2r^{-2}(1+\gamma)+r^{-4}(1+\gamma)^3\right].
$$
The displayed algebraic equation is therefore exactly the condition that the finite-dimensional scattering system become singular. Its solutions are poles of the analytically continued <scattering amplitude>.

To see the imaginary-axis poles directly, write $k=i\kappa$ with $\kappa>0$. When $a\to0$, the equation reduces to
$$\gamma^2(\gamma+3)=0,$$
and its nonzero root is $\gamma=-3$, or
$$\kappa=\frac{3m\lambda}{\hbar^2}.$$
This is the <bound state> of the coincident potential $-3\lambda\delta(x)$. When $a\to\infty$, the three centres decouple and the roots approach
$$\gamma=-1,\qquad \kappa=\frac{m\lambda}{\hbar^2},$$
with exponentially small splitting. A pole at $k=i\kappa$ has energy $E=-\hbar^2\kappa^2/(2m)$ and an exponentially decaying <wavefunction>, so these upper imaginary-axis singularities represent <bound states>.

Solved by gpt-5.6-sol high.

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

= a
{parent=36c}
{scope}

= Solution
{parent=a}

For discrete states $s$ of energy $E_s$, the <canonical partition function> is
$$Z(\beta)=\sum_s e^{-\beta E_s},\qquad \beta=(k_BT)^{-1}.$$
For a classical continuous system the sum is replaced by the appropriately normalized phase-space <integral>, including the Gibbs factor $1/N!$ for identical particles.

It packages the equilibrium probabilities $P_s=e^{-\beta E_s}/Z$ and hence all canonical <thermodynamics>. In particular,
$$
F=-k_BT\log Z,\qquad
\langle E\rangle=-\frac{\partial\log Z}{\partial\beta},\qquad
(\Delta E)^2=\frac{\partial^2\log Z}{\partial\beta^2},
$$
and $S=k_B(\log Z+\beta\langle E\rangle)$.

Solved by gpt-5.6-sol high.

= b
{parent=36c}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

The one-particle partition <function> is
$$
z_1=\frac{A}{h^3}\int_{\mathbb R^3}e^{-\beta p^2/(2m)}\,d^3p
\int_0^\infty e^{-\beta mgz}\,dz
=\frac{A}{\lambda^3\beta mg},
$$
where $\lambda=(2\pi\hbar^2\beta/m)^{1/2}$. Thus
$$
Z_N=\frac{z_1^N}{N!},\qquad
\log z_1=\text{constant}-\frac52\log\beta.
$$
Canonical <differentiation> now gives
$$
\langle E\rangle=-\partial_\beta\log Z_N=\frac{5N}{2\beta}
=\frac52Nk_BT
$$
and
$$
(\Delta E)^2=\partial_\beta^2\log Z_N
=\frac{5N}{2\beta^2}.
$$
Therefore
$$
\frac{\Delta E}{\langle E\rangle}=\sqrt{\frac{2}{5N}}.
$$
The factor $5/2$ consists of $3/2$ from <kinetic energy> and $1$ from gravitational <potential energy>.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

The normalized one-particle height density is exponential:
$$
f_Z(z)=\beta mg e^{-\beta mgz},\qquad z>0.
$$
Consequently
$$
\langle z\rangle=\frac{1}{\beta mg}=\frac{k_BT}{mg}.
$$
The number density and local ideal-gas <pressure> are
$$
n(z)=\frac{N}{A\langle z\rangle}e^{-z/\langle z\rangle},
\qquad
p(z)=n(z)k_BT
=\frac{Nmg}{A}e^{-z/\langle z\rangle}.
$$
This also obeys hydrostatic balance $p'(z)=-mgn(z)$.

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

Since $\langle z\rangle=(\beta mg)^{-1}$,
$$
Z_N=\frac1{N!}\left(\frac{A\langle z\rangle}{\lambda^3}\right)^N.
$$
Using $\log N!\simeq N\log N-N$ and $S=k_B(\log Z_N+\beta\langle E\rangle)$ gives
$$
\boxed{
S=Nk_B\left[\log\left(\frac{A\langle z\rangle}{N\lambda^3}\right)+\frac72\right].
}
$$
This has the Sackur-Tetrode form with the effective volume $A\langle z\rangle$, but its additive constant is $7/2$ rather than $5/2$. The extra unit arises because the atmosphere's vertical extent grows with <temperature> and its mean gravitational energy per atom is $k_BT$.

Solved by gpt-5.6-sol high.

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

= i
{parent=37b}
{scope}

= Solution
{parent=i}

Assume $qE>0$ and put
$$\alpha=\frac{qE}{mc}.$$
The nonzero four-velocity equations are
$$
\dot u^0=\alpha u^x,\qquad \dot u^x=\alpha u^0,
$$
with $u^0(0)=c$ and $u^x(0)=0$. Hence
$$
u^0=c\cosh(\alpha\tau),\qquad
u^x=c\sinh(\alpha\tau).
$$
Integrating from the origin gives the uniformly accelerated trajectory
$$
\boxed{
ct=\frac c\alpha\sinh(\alpha\tau),\qquad
x=\frac c\alpha\{\cosh(\alpha\tau)-1\},\qquad y=z=0.
}
$$

The light ray has $x_L=-h+ct$. At interception, $x_L=x$, so
$$
h=ct-x=\frac c\alpha\left(1-e^{-\alpha\tau}\right).
$$
A finite solution exists exactly when
$$h<h_c=\frac c\alpha=\frac{mc^2}{qE}.$$
Then
$$
\tau_c=-\frac1\alpha\log\left(1-\frac{\alpha h}{c}\right),
$$
and, writing $H=\alpha h/c$,
$$
t_c=\frac1\alpha\sinh(\alpha\tau_c)
=\frac{H(2-H)}{2\alpha(1-H)}.
$$
At $h=h_c$ interception is approached only as $t\to\infty$, and for $h>h_c$ it never occurs. The limiting backward light ray is the <Rindler horizon> of the accelerated particle.

Solved by gpt-5.6-sol high.

= ii
{parent=37b}
{scope}

= Solution
{parent=ii}

Again let $\alpha=qE/(mc)$. For $\mathbf B=(0,0,E/c)$, the Lorentz-force equations are
$$
\dot u^0=\alpha u^x,\qquad
\dot u^x=\alpha(u^0+u^y),\qquad
\dot u^y=-\alpha u^x.
$$
The sum $u^0+u^y$ is constant and initially equals $c$. Therefore
$$
\dot u^x=\alpha c,\qquad
u^x=\alpha c\tau,\qquad
u^0=c\left(1+\frac{\alpha^2\tau^2}{2}\right),\qquad
u^y=-\frac{c\alpha^2\tau^2}{2}.
$$
A further integration gives
$$
\boxed{
ct=c\tau+\frac{c\alpha^2\tau^3}{6},\qquad
x=\frac{c\alpha\tau^2}{2},\qquad
y=-\frac{c\alpha^2\tau^3}{6},\qquad z=0.
}
$$
Eliminating $\tau$ between $x$ and $y$ yields
$$
\boxed{9c\,y^2=2\alpha x^3,}
$$
a <semicubical parabola>, with the physical branch selected by $y\leq0$ for the signs shown.

Solved by gpt-5.6-sol high.

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

= a
{parent=38e}
{scope}

= Solution
{parent=a}

Write $f(r)=1+r^2/a^2$. The geodesic <Lagrangian> is
$$
L=\frac12\left[-f\dot t^2+f^{-1}\dot r^2+r^2\dot\theta^2+r^2\sin^2\theta\dot\phi^2\right].
$$
Its $\theta$ equation is
$$
\frac d{d\tau}(r^2\dot\theta)-r^2\sin\theta\cos\theta\dot\phi^2=0.
$$
Spherical symmetry lets us rotate the conserved angular-momentum plane into $\theta=\pi/2$. Equivalently, initial data $\theta=\pi/2$, $\dot\theta=0$ solve this equation for all $\tau$.

The cyclic coordinates give
$$E=f\dot t,\qquad h=r^2\dot\phi.$$
Using the timelike normalization $g_{\mu\nu}\dot x^\mu\dot x^\nu=-1$ gives
$$
\dot r^2+f\left(1+\frac{h^2}{r^2}\right)=E^2.
$$
After separating the constant terms,
$$
\boxed{
\frac12\dot r^2+V(r)=\frac12\left(E^2-1-\frac{h^2}{a^2}\right),
\qquad
V(r)=\frac12\left(\frac{r^2}{a^2}+\frac{h^2}{r^2}\right).
}
$$

Solved by gpt-5.6-sol high.

= b
{parent=38e}
{scope}

= Solution
{parent=b}

A geodesic passing through $r=0$ must have $h=0$. Its radial equation is then
$$
\dot r^2+\frac{r^2}{a^2}=E^2-1.
$$
On the outgoing and returning portion it is the harmonic solution
$$
r(\tau)=a\sqrt{E^2-1}\sin(\tau/a),\qquad 0\leq\tau\leq\pi a.
$$
It reaches
$$
\boxed{r_{\max}=a\sqrt{E^2-1}}
$$
when $\tau=\pi a/2$ and returns to the origin at
$$
\boxed{\Delta\tau=\pi a.}
$$
The return time is independent of the launch energy, a characteristic focusing property of global anti-de Sitter spacetime.

Solved by gpt-5.6-sol high.

= c
{parent=38e}
{scope}

= Solution
{parent=c}

A circular orbit at $r=r_0$ requires $V'(r_0)=0$, hence
$$
\frac{r_0}{a^2}-\frac{h^2}{r_0^3}=0,
\qquad
h^2=\frac{r_0^4}{a^2}.
$$
This can be satisfied for every $r_0>0$. Moreover,
$$
V''(r_0)=\frac1{a^2}+\frac{3h^2}{r_0^4}=\frac4{a^2}>0,
$$
so every such circular orbit is stable. The radial energy equation gives
$$E^2=\left(1+\frac{r_0^2}{a^2}\right)^2,$$
and future direction selects $E=1+r_0^2/a^2=f(r_0)$. Therefore
$$
\dot t=\frac E{f(r_0)}=1,
$$
so <proper time> and coordinate time differ only by an additive constant on the orbit. Also $\dot\phi=\pm1/a$.

Solved by gpt-5.6-sol high.

= 39D
{parent=Paper 1}
{scope}
{title2=Fluid Dynamics}

= a
{parent=39d}
{scope}

= Solution
{parent=a}

The steady Stokes equations are linear and contain no inertial term. If every imposed boundary <velocity> and body force is reversed, then the <velocity> and <pressure> departure from any uniform reference <pressure> reverse sign. Thus every fluid particle retraces its path in the reversed experiment. This is the principle of kinematic reversibility.

Solved by gpt-5.6-sol high.

= b
{parent=39d}
{scope}

= Solution
{parent=b}

Reflect the configuration in the horizontal plane through the cylinder axis. The wall and cylinder geometry are unchanged. Downward translation becomes upward translation, a possible wall-normal migration <velocity> is unchanged, and the axial angular <velocity> changes sign because it is an axial <vector>.

On the other hand, reversing the original <Stokes flow> makes all three <velocities> change sign: translation becomes upward, wall-normal migration reverses, and rotation reverses. The reflected and reversed problems therefore have the same forcing and geometry except that their wall-normal <velocities> have opposite signs. Uniqueness of <Stokes flow> forces that <velocity> to equal its own negative, so it is zero. The rotational <velocities> already agree under the two operations, so this argument places no restriction on rotation.

Solved by gpt-5.6-sol high.

= c
{parent=39d}
{scope}

= i
{parent=c}
{scope}

= Solution
{parent=i}

At vertical displacement $x$ from the axis, the gap is
$$
h(x)=h_0+a-\sqrt{a^2-x^2}.
$$
For $|x|\ll a$,
$$
\sqrt{a^2-x^2}=a-\frac{x^2}{2a}+O(x^4/a^3),
$$
so
$$
\boxed{h(x)\simeq h_0+\frac{x^2}{2a}
=h_0\left(1+\frac{x^2}{2ah_0}\right).}
$$
The narrow-gap length scale is therefore $(ah_0)^{1/2}$.

Solved by gpt-5.6-sol high.

= ii
{parent=c}
{scope}

= Solution
{parent=ii}

Take $x$ upward, let $y$ measure distance from the wall, and work in the cylinder frame. The wall at $y=0$ moves upward with speed $V$. Choose the sign of $\Omega$ so that the near-cylinder surface at $y=h$ moves upward with speed $a\Omega$. Lubrication theory gives
$$
\mu\frac{\partial^2u}{\partial y^2}=\frac{dp}{dx},
\qquad u(0)=V,\qquad u(h)=a\Omega.
$$
Thus
$$
\boxed{
u(y)=V+\frac{a\Omega-V}{h}y
+\frac1{2\mu}\frac{dp}{dx}y(y-h).}
$$
The vertical volume flux per unit cylinder length is independent of $x$ and equals
$$
Q=\int_0^h u\,dy
=\frac h2(V+a\Omega)-\frac{h^3}{12\mu}\frac{dp}{dx}.
$$
Therefore
$$
\frac{dp}{dx}=\frac{6\mu(V+a\Omega)}{h^2}-\frac{12\mu Q}{h^3}.
$$
Because $p(+\infty)=p(-\infty)$, its <integral> over $x$ vanishes. With
$$t=\frac{x}{\sqrt{2ah_0}},\qquad h=h_0(1+t^2),$$
the quoted <integrals> give
$$
\frac{\int_{-\infty}^{\infty}h^{-2}\,dx}
{\int_{-\infty}^{\infty}h^{-3}\,dx}
=\frac{4h_0}{3}.
$$
It follows that
$$
\boxed{Q=\frac{2h_0}{3}(V+a\Omega).}
$$

Solved by gpt-5.6-sol high.

= iii
{parent=c}
{scope}

= Solution
{parent=iii}

<Pressure> acts normally to the circular cylinder and so has no moment about its axis. To leading lubrication order, the tangential shear at the cylinder is
$$
\tau_c=\mu\left.\frac{\partial u}{\partial y}\right|_{y=h}
=\frac{\mu(a\Omega-V)}h+\frac h2\frac{dp}{dx}.
$$
Substitution of the <pressure> <gradient> gives
$$
\tau_c=\mu\left[\frac{4a\Omega+2V}{h}-\frac{6Q}{h^2}\right].
$$
A torque-free cylinder requires $\int_{-\infty}^{\infty}\tau_c\,dx=0$. Also,
$$
\int h^{-1}dx=\pi\sqrt{2ah_0}\,h_0^{-1},
\qquad
\int h^{-2}dx=\frac{\pi}{2}\sqrt{2ah_0}\,h_0^{-2}.
$$
Using $Q=2h_0(V+a\Omega)/3$, the <torque> condition reduces to
$$
(4a\Omega+2V)-2(V+a\Omega)=2a\Omega=0.
$$
Hence
$$\boxed{\Omega=0.}$$

Solved by gpt-5.6-sol high.

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

= a
{parent=40a}
{scope}

= Solution
{parent=a}

Let $\rho'$ and $p'$ be the density and <pressure> perturbations. Homentropy gives $p'=c_0^2\rho'$. The linearized mass and <momentum> equations are
$$
\rho'_t+\rho_0\nabla\cdot\mathbf u=0,
\qquad
\rho_0\mathbf u_t=-\nabla p'.
$$
For <potential flow>, $\mathbf u=\nabla\phi$, and the <momentum> equation integrates to
$$p'=-\rho_0\phi_t,$$
where a <function> of time has been absorbed into $\phi$. Substitution into mass conservation yields
$$
-\frac{\rho_0}{c_0^2}\phi_{tt}+\rho_0\nabla^2\phi=0,
$$
so
$$\boxed{\phi_{tt}-c_0^2\nabla^2\phi=0.}$$

Solved by gpt-5.6-sol high.

= b
{parent=40a}
{scope}

= Solution
{parent=b}

Multiply the wave equation by $\rho_0\phi_t/c_0^2$ and use product rules. This gives
$$
\frac{\partial}{\partial t}
\left[\frac{\rho_0}{2}|\nabla\phi|^2
+\frac{\rho_0}{2c_0^2}\phi_t^2\right]
+\nabla\cdot(-\rho_0\phi_t\nabla\phi)=0.
$$
Therefore
$$
\boxed{
\mathcal E=\frac{\rho_0}{2}|\nabla\phi|^2
+\frac{\rho_0}{2c_0^2}\phi_t^2
=\frac{\rho_0}{2}|\mathbf u|^2+\frac{p'^2}{2\rho_0c_0^2},
\qquad
\mathbf I=-\rho_0\phi_t\nabla\phi=p'\mathbf u.
}
$$
These are respectively the kinetic-plus-compressional acoustic energy density and its flux.

Solved by gpt-5.6-sol high.

= c
{parent=40a}
{scope}

= i
{parent=c}
{scope}

= Solution
{parent=i}

Choose membrane displacement $\eta$ positive into the fluid. The linear <boundary conditions> at $z=0$ are
$$
\phi_z=\eta_t,
\qquad
p'=-\mu\eta.
$$
A wave localized near the membrane has complex form
$$
\phi=\operatorname{Re}\left\{\Phi e^{ik(x-ct)}e^{-\alpha z}\right\}.
$$
The wave equation requires
$$
\alpha=k\sqrt{1-\frac{c^2}{c_0^2}},
$$
so localization requires $0<c<c_0$. If $\eta=\operatorname{Re}\{H e^{ik(x-ct)}\}$, the kinematic condition and $p'=-\rho_0\phi_t$ give
$$
-\alpha\Phi=-ikcH,
\qquad
i\rho_0kc\Phi=-\mu H.
$$
Eliminating $H$ yields
$$\rho_0k^2c^2=\mu\alpha.$$
On squaring and writing $s=c/c_0$, we obtain
$$
\boxed{As^4+s^2-1=0,
\qquad
A=\left(\frac{\rho_0kc_0^2}{\mu}\right)^2.}
$$

Solved by gpt-5.6-sol high.

= ii
{parent=c}
{scope}

= Solution
{parent=ii}

For real physical fields of period $T=2\pi/(kc)$, define
$$
\langle I_z\rangle(x,z)=\frac1T\int_{t_0}^{t_0+T}p'(x,z,t)u_z(x,z,t)\,dt.
$$
Equivalently, for complex amplitudes,
$$
\langle I_z\rangle=\frac12\operatorname{Re}(\widehat p\,\widehat u_z^*).
$$
Here
$$
\widehat p=i\rho_0kc\Phi e^{-\alpha z},
\qquad
\widehat u_z=-\alpha\Phi e^{-\alpha z}.
$$
Their product is purely imaginary, so
$$\boxed{\langle I_z\rangle=0.}$$
<Pressure> and normal <velocity> are in temporal quadrature: energy is stored and returned by the evanescent field, but none is transported away from the membrane on average.

Solved by gpt-5.6-sol high.

= iii
{parent=c}
{scope}

= Solution
{parent=iii}

The physical root of the dispersion relation is
$$
s^2=\frac{\sqrt{1+4A}-1}{2A}.
$$
For $A\ll1$,
$$
\boxed{c=c_0\left(1-\frac A2+O(A^2)\right).}
$$
This is a weakly localized, nearly grazing acoustic wave with speed close to the bulk sound speed.

For $A\gg1$,
$$
\boxed{
c=c_0A^{-1/4}\left(1-\frac1{4\sqrt A}+O(A^{-1})\right)
\sim\sqrt{\frac{\mu}{\rho_0k}}.
}
$$
Now $c\ll c_0$, the decay rate is $\alpha\simeq k$, and the fluid is effectively incompressible. A layer of depth $1/k$ supplies <added mass> per area of order $\rho_0/k$, whose balance with the spring stiffness gives the final scaling.

Solved by gpt-5.6-sol high.

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

= a
{parent=41d}
{scope}

= Solution
{parent=a}

Starting with $A_0=A$, suppose columns $1,\ldots,j-1$ have already been reduced to tridiagonal form. Let $x$ be the part of column $j$ in rows $j+1,\ldots,n$. A <Householder reflection> $P_j$ on these coordinates can map $x$ to $\pm\lVert x\rVert e_1$. Extend it by the identity on the first $j$ coordinates and call the resulting orthogonal <matrix> $U_j$.

The update
$$A_j=U_j^TA_{j-1}U_j$$
zeros every entry below row $j+1$ in column $j$. Because it is an <orthogonal similarity>, it preserves <eigenvalues>; because the same transformation is applied on both sides, it preserves symmetry and zeros the corresponding row entries without disturbing earlier columns. After $n-2$ such steps,
$$
H=U^TAU,\qquad U=U_1U_2\cdots U_{n-2},
$$
is symmetric and tridiagonal and has the same <eigenvalues> as $A$. Every reflector is obtained from finitely many <matrix> entries using finitely many arithmetic operations and a square root, so this is a finite construction.

Solved by gpt-5.6-sol high.

= b
{parent=41d}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

The spectral expansion gives
$$
\left(\frac H{\lambda_n}\right)^ke_1
=\sum_{j=1}^n b_j\left(\frac{\lambda_j}{\lambda_n}\right)^kw_j.
$$
For $j\leq n-2$, the coefficient tends to zero because $|\lambda_j|<|\lambda_n|$. Thus every possible <limit> belongs to
$$W=\operatorname{span}\{w_{n-1},w_n\}.$$
The identical argument with the coefficients $c_j$ proves
$$\boxed{v_1,v_2\in W.}$$
The statement is conditional because, when $\lambda_{n-1}=-\lambda_n$, the component along $w_{n-1}$ generally alternates rather than converges.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

Let $E_2=(e_1,e_2)$ and
$$S_k=\operatorname{range}(H^{k+1}E_2).$$
Because $Q_kR_k$ is the QR factorization of $H^{k+1}$, $S_k$ is the span of the first two columns $q_1^{(k)},q_2^{(k)}$ of $Q_k$.

The first row of the tridiagonal <eigenvalue> equation is
$$
h_{11}b_j+h_{12}c_j=\lambda_jb_j.
$$
In the generic case $\lambda_{n-1}\ne\lambda_n$, the two dominant coefficient <vectors> are independent, since
$$
\det\begin{pmatrix}b_{n-1}&c_{n-1}\\b_n&c_n\end{pmatrix}
=\frac{b_{n-1}b_n}{h_{12}}(\lambda_n-\lambda_{n-1})\ne0.
$$
Here $h_{12}\ne0$, since otherwise the displayed <eigenvalue> equation and the nonzero $b_j$ would force all <eigenvalues> to equal $h_{11}$. The strict <spectral gap> below the dominant pair now implies
$$S_k\longrightarrow W=\operatorname{span}\{w_{n-1},w_n\}.$$
This is two-dimensional <subspace iteration>. If the <dominant eigenvalue> is repeated, $H$ acts as a <scalar> on $W$; the same QR argument either captures all of $W$ or separates its missing direction from the strictly smaller eigenspaces, and the conclusion below is unchanged.

Since $q_2^{(k)}\in S_k$ and $q_3^{(k)}\perp S_k$,
$$
\operatorname{dist}(q_2^{(k)},W)\to0,
\qquad
\lVert P_Wq_3^{(k)}\rVert\to0.
$$
The space $W$ is invariant under $H$, so
$$
 h_{3,2}^{(k+1)}
 =(q_3^{(k)})^THq_2^{(k)}\longrightarrow0.
$$
Shifting the index does not affect the <limit>, and therefore
$$\boxed{h_{3,2}^{(k)}\to0.}$$
Thus the unshifted QR algorithm asymptotically deflates a $2\times2$ block associated with the equal-modulus dominant pair.

Solved by gpt-5.6-sol high.