past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ii/paper-2.bigb

= Paper 2
{scope}

https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2022/paperii_2_2022.pdf

= 1I
{parent=Paper 2}
{scope}
{title2=Number Theory}

= Solution
{parent=1I}

Write an integral <binary quadratic form> as $[a,b,c]=ax^2+bxy+cy^2$. It is positive definite when $a>0$ and its <discriminant of a binary quadratic form>[discriminant] $D=b^2-4ac$ is negative. Two forms are <proper equivalence of binary quadratic forms>[properly equivalent] when one is obtained from the other by a change of variables in $\operatorname{SL}_2(\mathbb Z)$. A <reduced positive definite binary quadratic form> satisfies
$$
|b|\leq a\leq c,
$$
with $b\geq0$ when $|b|=a$ or $a=c$.

To prove reduction, choose in a proper equivalence class a form whose positive leading coefficient $a$ is minimal. Replacing $x$ by $x+ny$ changes $b$ by $2an$, so choose $n$ to arrange $|b|\leq a$. If the resulting $c<a$, the determinant-one substitution $(x,y)\mapsto(-y,x)$ would put $c$ in the leading position, contradicting minimality. Thus $a\leq c$. Sign changes on the boundary give the stated convention. Hence every positive definite form is properly equivalent to a reduced one.

A unimodular integral substitution is a bijection of $\mathbb Z^2$, so properly equivalent forms represent exactly the same integers.

Both given forms have discriminant $-23$. Reduction gives
$$
[6,5,2]\sim[2,1,3],
\qquad
[9,25,18]\sim[2,-1,3].
$$
The two reduced forms represent the same integers because
$$
[2,-1,3](x,y)=[2,1,3](x,-y),
$$
an integral change of variables of determinant $-1$. They are not properly equivalent: the uniqueness theorem for reduced positive definite forms says that each proper class has one reduced representative, and $[2,1,3]\ne[2,-1,3]$. Therefore the original forms represent the same integers but are not equivalent in the required proper sense.

Solved by gpt-5.6-sol high.

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

= i
{parent=2g}
{scope}

= Solution
{parent=i}

A set is of first Baire category, or <meagre set>[meagre], when it is a countable union of <nowhere dense set>[nowhere dense sets]. The statement is true: a countable union of countable unions of nowhere dense sets is again a countable union of nowhere dense sets.

The relevant <Baire category theorem> says that a nonempty complete metric space is not meagre in itself.

Solved by gpt-5.6-sol high.

= ii
{parent=2g}
{scope}

= Solution
{parent=ii}

False. Fix $y_0\in\mathbb R$ and take
$$
A=\mathbb R\times\{y_0\}.
$$
This closed set has empty interior in $\mathbb R^2$, so it is nowhere dense and hence meagre. Its section at $y_0$ is $C_{y_0}=\mathbb R$, which is not meagre in itself by the <Baire category theorem>.

Solved by gpt-5.6-sol high.

= iii
{parent=2g}
{scope}

= Solution
{parent=iii}

True. Write $C=\bigcup_{n\geq1}C_n$ with every $C_n$ nowhere dense in $\mathbb R$. Then
$$
A=C\times\mathbb R=\bigcup_{n\geq1}(C_n\times\mathbb R).
$$
The closure of $C_n\times\mathbb R$ is
$\overline{C_n}\times\mathbb R$, which has empty interior because $\overline{C_n}$ has empty interior. Each product is therefore nowhere dense, so $A$ is meagre in $\mathbb R^2$.

Solved by gpt-5.6-sol high.

= iv
{parent=2g}
{scope}

= Solution
{parent=iv}

False. The coordinate axes
$$
A=\mathbb R\times\{0\},\qquad B=\{0\}\times\mathbb R
$$
are both closed nowhere dense subsets of $\mathbb R^2$, but every $(x,y)$ is $(x,0)+(0,y)$, so
$$
A+B=\mathbb R^2,
$$
which is not meagre by the <Baire category theorem>.

Solved by gpt-5.6-sol high.

= 3K
{parent=Paper 2}
{scope}
{title2=Coding and Cryptography}

= Solution
{parent=3K}

A <discrete memoryless channel> has finite input alphabet $\mathcal A$, finite output alphabet $\mathcal B$, and transition probabilities $P(y\mid x)$; successive outputs are conditionally independent given the corresponding inputs. The <Shannon second coding theorem> states that rates below
$$
C=\max_{P_X}I(X;Y)
$$
admit block codes with error probability tending to zero, whereas rates above $C$ cannot have vanishing error.

For any joint input law of $(X_1,X_2)$, conditional independence of the product channel gives
$$
\begin{aligned}
I(X_1,X_2;Y_1,Y_2)
&=H(Y_1,Y_2)-H(Y_1,Y_2\mid X_1,X_2)\\
&\leq H(Y_1)+H(Y_2)-H(Y_1\mid X_1)-H(Y_2\mid X_2)\\
&=I(X_1;Y_1)+I(X_2;Y_2)
\leq C_1+C_2.
\end{aligned}
$$
Choose $X_1$ and $X_2$ independently with capacity-achieving input laws for their respective channels. Then the inequality becomes equality, so the product-channel capacity is
$$
\boxed{C_1+C_2}.
$$

Solved by gpt-5.6-sol high.

= 4I
{parent=Paper 2}
{scope}
{title2=Automata and Formal Languages}

= i
{parent=4i}
{scope}

= Solution
{parent=i}

The <pumping lemma for regular languages> states that if $L$ is regular, then some $p\geq1$ has the following property: every $w\in L$ with $|w|\geq p$ can be written $w=xyz$ with $|xy|\leq p$, $|y|\geq1$, and $xy^iz\in L$ for every $i\geq0$. Indeed, while a deterministic finite automaton with $p$ states reads the first $p$ symbols, two of the first $p+1$ visited states coincide; the intervening nonempty loop may be traversed any number of times.

The language $\{0^{n^2}1:n\geq0\}$ is not regular. If its pumping length were $p$, apply the lemma to $w=0^{p^2}1$. The pumped block is $y=0^r$ for some $1\leq r\leq p$. Pumping once more gives $p^2+r$ zeros, but
$$
p^2<p^2+r<(p+1)^2,
$$
so their number is not a square.

Solved by gpt-5.6-sol high.

= ii
{parent=4i}
{scope}

= Solution
{parent=ii}

The language $\{0^n1^n:n\geq0\}$ is not regular. For pumping length $p$, take $0^p1^p$. Any permitted pumped block lies entirely among the initial zeros. Pumping it down changes the number of zeros without changing the number of ones, producing a word outside the language and contradicting the pumping lemma.

Solved by gpt-5.6-sol high.

= iii
{parent=4i}
{scope}

= Solution
{parent=iii}

This language is not regular. If it were regular, its intersection with the <regular language> $0^*1^*$ would be regular because regular languages are closed under intersection. That intersection is exactly
$$
\{0^n1^n:n\geq0\},
$$
which part (ii) proved nonregular.

Solved by gpt-5.6-sol high.

= 5J
{parent=Paper 2}
{scope}
{title2=Statistical Modelling}

= a
{parent=5j}
{scope}

= Solution
{parent=a}

A one-parameter <exponential family> has densities
$$
f(y;\theta)=h(y)\exp\{\theta T(y)-A(\theta)\},
$$
with support independent of the <natural parameter of an exponential family>[natural parameter] $\theta$. In the present canonical setting the natural statistic is $T(y)=y$.

Solved by gpt-5.6-sol high.

= b
{parent=5j}
{scope}

= Solution
{parent=b}

Independence gives the joint density
$$
\prod_{i=1}^nh(y_i)
\exp\left\{\theta\sum_{i=1}^ny_i-nA(\theta)\right\}.
$$
By the factorization criterion, its dependence on the sample relevant to $\theta$ is only through $\sum_iY_i$, equivalently through $\overline Y$. Thus $\overline Y$ is a <sufficient statistic> for $\theta$.

Solved by gpt-5.6-sol high.

= c
{parent=5j}
{scope}

= Solution
{parent=c}

The <log-likelihood> has derivative
$$
\ell'(\theta)=\sum_iY_i-nA'(\theta).
$$
Differentiating the normalizing identity for the family gives
$$
A'(\theta)=\mathbb E_\theta Y_1=\mu(\theta),
\qquad
A''(\theta)=\operatorname{Var}_\theta(Y_1)\geq0.
$$
Hence an interior maximum satisfies, and under nondegeneracy uniquely satisfies,
$$
\boxed{\mu(\widehat\theta_{\rm MLE})=\overline Y}.
$$

Solved by gpt-5.6-sol high.

= 6C
{parent=Paper 2}
{scope}
{title2=Mathematical Biology}

= a
{parent=6c}
{scope}

= Solution
{parent=a}

Set
$$
x=\frac{N_1}{K_1},\qquad y=\frac{N_2}{K_2},\qquad \tau=r_2t.
$$
Then
$$
x'=\rho x(1-x-\alpha y),\qquad
y'=y(1-\beta x).
$$
The steady states are
$$
(0,0),\qquad(1,0),\qquad
\left(\frac1\beta,\frac{1-\beta^{-1}}{\alpha}\right),
$$
where the coexistence state is biologically relevant exactly when $\beta>1$.

Solved by gpt-5.6-sol high.

= b
{parent=6c}
{scope}

= Solution
{parent=b}

The Jacobian is
$$
J=\begin{pmatrix}
\rho(1-2x-\alpha y)&-\rho\alpha x\\
-\beta y&1-\beta x
\end{pmatrix}.
$$
For $\beta>1$, the origin has eigenvalues $\rho,1$ and is an unstable node. At $(1,0)$ the eigenvalues are $-\rho$ and $1-\beta$, so it is a stable node. At the coexistence state,
$$
\det J=-\frac{\rho(\beta-1)}{\beta}<0,
$$
so it is a saddle.

The nullclines are $x=0$, $x+\alpha y=1$, $y=0$, and $x=1/\beta$. The stable manifold of the coexistence saddle separates trajectories tending to $(1,0)$ from trajectories on which $x\to0$ and $y$ grows without bound. The coordinate axes are invariant.

Solved by gpt-5.6-sol high.

= c
{parent=6c}
{scope}

= Solution
{parent=c}

There is no stable positive coexistence state: the only one is a saddle for every $\rho,\alpha>0$ when $\beta>1$. Moreover, the <Bendixson-Dulac criterion> with Dulac function $B(x,y)=1/(xy)$ applies in the positive quadrant, since
$$
\nabla\cdot(BF)
=\frac{\partial}{\partial x}\left(\frac{\rho(1-x-\alpha y)}y\right)
+\frac{\partial}{\partial y}\left(\frac{1-\beta x}{x}\right)
=-\frac{\rho}{y}<0.
$$
Thus no periodic orbit can support long-term coexistence. Except on the saddle's stable manifold, one species is excluded: species 2 becomes extinct for initial data in the basin of $(1,0)$, ecologically when species 1 is sufficiently abundant and species 2 sufficiently scarce. On the other side, species 1 becomes extinct and the unregulated second species grows.

Solved by gpt-5.6-sol high.

= 7E
{parent=Paper 2}
{scope}
{title2=Further Complex Methods}

= Solution
{parent=7E}

Choose paths from $0$ to $z$ in the cut plane and deform them continuously with the endpoint, never crossing the specified cut. With $0\leq\arg z<2\pi$, use the sheet reached from the positive real axis; when an endpoint is moved into the left half-plane, the path passes around the upper endpoint $i$ of the cut. The cut makes all such admissible paths homotopic with fixed endpoints, so the integral is single valued and its endpoint derivative is $(1+z^2)^{-1/2}$; hence it is <holomorphic function>[analytic].

For $z=-\sinh u$, continue the endpoint counterclockwise from the positive real axis through the upper half-plane. In
$$
\operatorname{Arcsinh}z
=\log\!\left(z+\sqrt{1+z^2}\right),
$$
the continued square root equals $-\cosh u$ at $z=-\sinh u$. The logarithm is reached with argument $\pi$, and therefore
$$
\boxed{\operatorname{Arcsinh}(-\sinh u)=u+i\pi}.
$$

The identities
$$
\sinh(w+2\pi i)=\sinh w,\qquad
\sinh((2k+1)\pi i-w)=\sinh w
$$
show directly why continuation without an argument restriction is multivalued. The complete set of values of the <multivalued inverse hyperbolic sine> at $\sinh u$ is
$$
\boxed{u+2\pi ik\quad\hbox{and}\quad
-u+(2k+1)\pi i,\qquad k\in\mathbb Z}.
$$

Solved by gpt-5.6-sol high.

= 8B
{parent=Paper 2}
{scope}
{title2=Classical Dynamics}

= Solution
{parent=8B}

Order the phase-space coordinates as
$$
x=(q_1,\ldots,q_n,p_1,\ldots,p_n),
\qquad
\Omega=\begin{pmatrix}0&I\\-I&0\end{pmatrix}.
$$
Then <Hamilton's equations> are
$$
\dot x_a=\Omega_{ab}\frac{\partial H}{\partial x_b}.
$$
The <Poisson bracket> is
$$
\{f,g\}=\frac{\partial f}{\partial x_a}
\Omega_{ab}\frac{\partial g}{\partial x_b},
\qquad
\boxed{\{x_a,x_b\}=\Omega_{ab}}.
$$

For $X=X(x)$ with Jacobian $J_{ab}=\partial X_a/\partial x_b$, the chain rule gives
$$
\dot X=J\Omega J^T\nabla_XH.
$$
This has Hamiltonian form with the same canonical matrix for every $H$ exactly when
$$
\boxed{J\Omega J^T=\Omega}.
$$
Taking determinants gives $(\det J)^2=1$; the symplectic condition fixes the orientation, so $\det J=1$. Hence the change-of-variables formula shows that the phase-space volume element is invariant. This is <Liouville theorem in Hamiltonian mechanics>.

Solved by gpt-5.6-sol high.

= 9A
{parent=Paper 2}
{scope}
{title2=Cosmology}

= Solution
{parent=9A}

By the shell theorem, only the mass inside the particle's radius contributes:
$$
M(r)=\frac{4\pi}{3}\rho r^3.
$$
The conserved Newtonian energy per unit particle mass is
$$
\frac12\dot r^2-\frac{GM}{r}=E.
$$
Writing $r=a(t)r_0$ and using conservation of enclosed mass, so that $\rho a^3$ is constant, gives
$$
\boxed{\left(\frac{\dot a}{a}\right)^2
=\frac{8\pi G}{3}\rho-\frac{k}{a^2}},
\qquad
k=-\frac{2E}{r_0^2}.
$$
This is the first <Friedmann equation> without a cosmological constant.

The derivation has two central defects. It uses instantaneous Newtonian gravity and omits the relativistic gravitational effect of pressure, so it is not valid for general cosmological matter. It also starts from a finite ball with a preferred centre and boundary, contrary to exact homogeneity and isotropy; the assumption $r\ll L$ only hides that defect locally.

Solved by gpt-5.6-sol high.

= 10D
{parent=Paper 2}
{scope}
{title2=Quantum Information and Computation}

= a
{parent=10d}
{scope}

= Solution
{parent=a}

In the <BB84 protocol>, Alice independently chooses $n$ random bits and, for each bit, a random computational or Hadamard basis. She sends the corresponding states $|0\rangle,|1\rangle,|+\rangle,|-\rangle$ through the noiseless quantum channel. Bob independently chooses one of the two bases for each qubit and measures. Over the public classical channel they reveal only their basis choices and retain positions where the choices agree. Each position survives with probability $1/2$, so the expected secret-key length is $n/2$. In the ideal no-eavesdropper setting their retained bits agree exactly.

Solved by gpt-5.6-sol high.

= b
{parent=10d}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

The controlled-NOT leaves computational-basis signals unentangled:
$$
|0\rangle_A|0\rangle_E\mapsto|00\rangle,\qquad
|1\rangle_A|0\rangle_E\mapsto|11\rangle.
$$
For the two Hadamard-basis choices it gives
$$
|+\rangle_A|0\rangle_E\mapsto
\frac{|00\rangle+|11\rangle}{\sqrt2}=|\phi^+\rangle,
$$
$$
|-\rangle_A|0\rangle_E\mapsto
\frac{|00\rangle-|11\rangle}{\sqrt2}=|\phi^-\rangle.
$$
Both outputs are <Bell state>[maximally entangled Bell states].

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

Conditioned on a sifted-key position, Alice and Bob used the computational basis with probability $1/2$ and the Hadamard basis with probability $1/2$. Eve causes no error in the computational basis. In the Hadamard basis Bob's reduced state is maximally mixed, so his bit is wrong with probability $1/2$. The final bit-error rate is therefore
$$
\boxed{\frac12\cdot0+\frac12\cdot\frac12=\frac14}.
$$

Solved by gpt-5.6-sol high.

= 11G
{parent=Paper 2}
{scope}
{title2=Topics in Analysis}

= i
{parent=11g}
{scope}

= Solution
{parent=i}

Let $X\sim\operatorname{Bin}(n,t)$ and $Y\sim\operatorname{Bin}(n,s)$ be independent. The displayed <Bernstein polynomial> is
$$
B_nf(t,s)=\mathbb E f(X/n,Y/n).
$$
Given $\varepsilon>0$, <uniform continuity> of $f$ supplies $\delta>0$ such that changing both coordinates by less than $\delta$ changes $f$ by less than $\varepsilon$. If $M=\lVert f\rVert_\infty$, then <Chebyshev inequality> and
$\operatorname{Var}(X/n),\operatorname{Var}(Y/n)\leq1/(4n)$ give, uniformly in $(t,s)$,
$$
|B_nf(t,s)-f(t,s)|
\leq\varepsilon+2M\,
\mathbb P\!\left(|X/n-t|\geq\delta
\ \hbox{or}\ |Y/n-s|\geq\delta\right)
\leq\varepsilon+\frac{M}{n\delta^2}.
$$
Thus $B_nf\to f$ uniformly.

Each $B_nf$ is a polynomial, whose iterated integrals may be interchanged term by term. Uniform convergence permits passage to the limit in both iterated integrals, proving the asserted continuous-function form of <Fubini's theorem>.

Now suppose every monomial moment of $f$ vanishes. By linearity,
$\int\!\!\int p f=0$ for every two-variable polynomial $p$. The just-proved Bernstein approximation gives polynomials $p_n\to f$ uniformly. Therefore
$$
\int_{[0,1]^2}f^2
=\lim_n\int_{[0,1]^2}f p_n=0,
$$
so $\boxed{f=0}$. Statement (i) is true.

Solved by gpt-5.6-sol high.

= ii
{parent=11g}
{scope}

= Solution
{parent=ii}

True. An affine change of variables carries $[a,b]^2$ to $[0,1]^2$. Equivalently, polynomials in $s,t$ are uniformly dense on $[a,b]^2$ by the rescaled Bernstein construction. The moment assumptions annihilate every polynomial, so approximating $f$ itself again gives $\int f^2=0$ and hence $f=0$.

Solved by gpt-5.6-sol high.

= iii
{parent=11g}
{scope}

= Solution
{parent=iii}

False. Take $f(s,t)=s$. Then $s^{2n}t^{2m}f(s,t)$ is odd in $s$, so every stated integral over $[-1,1]^2$ vanishes, although $f$ is not zero.

Solved by gpt-5.6-sol high.

= iv
{parent=11g}
{scope}

= Solution
{parent=iv}

True. The map $(s,t)\mapsto(s^2,t^2)$ is a homeomorphism of $[0,1]^2$. Hence every continuous $f(s,t)$ can be written as $g(s^2,t^2)$ for continuous
$g(u,v)=f(\sqrt u,\sqrt v)$. Bernstein polynomials approximate $g(u,v)$ uniformly, so polynomials in $s^2,t^2$ approximate $f$ uniformly. The assumed even moments annihilate all such polynomials; approximating $f$ and passing to $\int f^2$ proves $f=0$.

Solved by gpt-5.6-sol high.

= 12K
{parent=Paper 2}
{scope}
{title2=Coding and Cryptography}

= a
{parent=12k}
{scope}

= Solution
{parent=a}

In the <Rabin cryptosystem>, the public key is $N=pq$ and encryption sends an encoded message $m$ to
$$
c\equiv m^2\pmod N.
$$
Knowing $p,q$, the receiver finds the two square roots modulo each prime and combines them with the <Chinese remainder theorem> to obtain four roots modulo $N$; prescribed redundancy identifies the intended one.

Omicron has
$$
m^2\equiv c\pmod N,\qquad m^2\equiv c'\pmod{N'}.
$$
The Chinese remainder theorem determines $m^2$ uniquely modulo $NN'$ (assuming the independently generated moduli are coprime; a nontrivial gcd would itself factor them). Since $1\leq m<N<N'$, one has $m^2<NN'$, so this residue is the ordinary integer $m^2$. Taking its positive integer square root recovers $m$.

This does not normally decrypt another ciphertext sent under only one modulus. The recovered pair reveals no nontrivial square root collision modulo that modulus and hence supplies no factorization; breaking a single Rabin instance remains equivalent to factoring its modulus.

Solved by gpt-5.6-sol high.

= b
{parent=12k}
{scope}

= Solution
{parent=b}

The <discrete logarithm problem> asks, given $g$ and $h\equiv g^a\pmod p$, to recover $a$ modulo $p-1$. In the <Diffie-Hellman key exchange>, Alice sends $g^a$, Bob sends $g^b$, and both compute $g^{ab}$. An enemy able to compute discrete logarithms recovers $a$ or $b$ from the public messages and hence obtains the key.

For three participants with private exponents $a,b,c$, circulate three tokens around a directed ring. Start with $g^a,g^b,g^c$; whenever a participant receives a token, they raise it to their private exponent and pass it on. Arrange the three cyclic routes so that after two transmissions each participant receives one token to which all three exponents have been applied. Every participant then has
$$
g^{abc},
$$
while the public transcript contains only proper subproducts.

For $n$ participants, start token $g^{a_i}$ at participant $i$ and pass it successively through the other $n-1$ participants in cyclic order, each raising it to their exponent. Choose the cyclic starts so that one completed token ends at each participant. Every final token equals
$$
g^{a_1a_2\cdots a_n}.
$$
There are $n$ tokens and $n-1$ transmissions per token, for exactly
$$
\boxed{n(n-1)=n^2-n}
$$
communications. Security rests on the corresponding generalized Diffie--Hellman problem.

Solved by gpt-5.6-sol high.

= 13E
{parent=Paper 2}
{scope}
{title2=Further Complex Methods}

= Solution
{parent=13E}

Substitution of
$$
w(z)=\int_\gamma e^{zt}f(t)\,dt
$$
and integration by parts give
$$
w'''-zw
=\int_\gamma e^{zt}(t^3f+f')\,dt
-\left[e^{zt}f(t)\right]_{\partial\gamma}.
$$
Thus take
$$
\boxed{f(t)=e^{-t^4/4}},
$$
and require the endpoint term to vanish. Infinite contour ends must lie in sectors where $\operatorname{Re}(t^4)>0$, namely sectors centred on the positive and negative real and imaginary axes, each of angular width $\pi/4$.

Let the common final segment be the positive real ray from $0$ to $+\infty$, and choose the initial rays from $-\infty$, $+i\infty$, and $-i\infty$ to $0$. Call the resulting contours $\gamma_1,\gamma_2,\gamma_3$. Put
$$
A_r=\int_0^\infty t^re^{-t^4/4}\,dt
=4^{(r-3)/4}\Gamma\!\left(\frac{r+1}{4}\right).
$$
Since $w_i^{(r)}(0)=\int_{\gamma_i}t^re^{-t^4/4}\,dt$, for $r=0,1,2$ the three rows are
$$
\begin{pmatrix}
2A_0&0&2A_2\\
(1-i)A_0&2A_1&(1+i)A_2\\
(1+i)A_0&2A_1&(1-i)A_2
\end{pmatrix}.
$$
Consequently
$$
\det\begin{pmatrix}
w_1(0)&w_1'(0)&w_1''(0)\\
w_2(0)&w_2'(0)&w_2''(0)\\
w_3(0)&w_3'(0)&w_3''(0)
\end{pmatrix}
=-16iA_0A_1A_2.
$$
Using the <Gamma function> reflection identity,
$$
\boxed{\det=-2\sqrt2\,i\,\pi^{3/2}\ne0}.
$$
The initial-data vectors are therefore <linear independence>[linearly independent], so the three contour integrals are linearly independent solutions.

Solved by gpt-5.6-sol high.

= 14B
{parent=Paper 2}
{scope}
{title2=Classical Dynamics}

= a
{parent=14b}
{scope}

= Solution
{parent=a}

Scale the unit ball by $(u,v,w)\mapsto(au,bv,cw)$. Symmetry makes all products of inertia vanish, while the unit-ball averages satisfy
$\langle u^2\rangle=\langle v^2\rangle=\langle w^2\rangle=1/5$. The <inertia tensor> about the centre is therefore
$$
\boxed{I=\frac M5
\begin{pmatrix}
b^2+c^2&0&0\\
0&a^2+c^2&0\\
0&0&a^2+b^2
\end{pmatrix}}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=14b}
{scope}

= Solution
{parent=b}

For torque-free motion, the inertial angular momentum $\mathbf L=I\boldsymbol\omega$ and energy
$T=\tfrac12\boldsymbol\omega\cdot I\boldsymbol\omega$ are constant. Consider the energy ellipsoid
$$
\mathbf x\cdot I\mathbf x=2T,
$$
which is fixed in the body. At its point $\mathbf x=\boldsymbol\omega$, its normal is
$I\boldsymbol\omega=\mathbf L$, and its tangent plane is
$$
\mathbf L\cdot\mathbf x=2T.
$$
This plane has fixed normal and fixed distance $2T/|\mathbf L|$ from the origin, so it is the invariable plane. The material velocity of the contact vector is
$\boldsymbol\omega\times\boldsymbol\omega=0$. Thus the contact point is instantaneously at rest and the ellipsoid rolls without slipping on the plane. This is the <Poinsot construction>.

Solved by gpt-5.6-sol high.

= c
{parent=14b}
{scope}

= Solution
{parent=c}

When $a=b<c$, the ellipsoid is an axisymmetric rigid body with
$$
I_1=I_2=\frac M5(a^2+c^2)>I_3=\frac{2M}{5}a^2.
$$
The angle between its symmetry axis and the fixed angular momentum is constant. In an inertial frame the symmetry axis precesses uniformly around $\mathbf L$, while the body simultaneously spins about that axis. The angular-velocity vector also traces a circular cone about $\mathbf L$. The limiting cases are pure rotation about the symmetry axis and pure rotation about a perpendicular principal axis.

Solved by gpt-5.6-sol high.

= 15D
{parent=Paper 2}
{scope}
{title2=Quantum Information and Computation}

= a
{parent=15d}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

A Bell measurement is the projective measurement of two qubits in the orthonormal <Bell state> basis
$$
|\phi^\pm\rangle=\frac{|00\rangle\pm|11\rangle}{\sqrt2},
\qquad
|\psi^\pm\rangle=\frac{|01\rangle\pm|10\rangle}{\sqrt2}.
$$
Its four classical outcomes identify the corresponding rank-one projector.

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

Alice holds the unknown input $|\alpha\rangle$ and one qubit of a shared $|\phi^+\rangle$ pair; Bob holds the other. Alice performs a Bell measurement on her two qubits and sends its two-bit outcome to Bob. For outcomes
$\phi^+,\phi^-,\psi^+,\psi^-$, Bob's qubit is respectively
$$
|\alpha\rangle,\quad Z|\alpha\rangle,\quad
X|\alpha\rangle,\quad XZ|\alpha\rangle
$$
up to global phase. Bob applies respectively $I,Z,X,ZX$ and obtains $|\alpha\rangle$. This is <quantum teleportation>.

Solved by gpt-5.6-sol high.

= iii
{parent=a}
{scope}

= Solution
{parent=iii}

Replacing the resource by $(I\otimes U)|\phi^+\rangle$ applies $U$ to Bob's uncorrected output. If $P\in\{I,Z,X,XZ\}$ labels the Bell outcome, the uncorrected state is $UP|\alpha\rangle$, and the unchanged correction produces
$$
\boxed{P^{-1}UP|\alpha\rangle}.
$$
Thus the four outputs are $U|\alpha\rangle$, $ZUZ|\alpha\rangle$, $XUX|\alpha\rangle$, and $(XZ)^{-1}U(XZ)|\alpha\rangle$, up to global phases.

Solved by gpt-5.6-sol high.

= b
{parent=15d}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

For two <nonorthogonal vectors>[nonorthogonal] input states $|\alpha\rangle,|\beta\rangle$, preservation of <inner products> by $G$ gives
$$
\langle\alpha|\beta\rangle
=\langle\alpha|\beta\rangle
\langle\widetilde P_U(\alpha)|\widetilde P_U(\beta)\rangle.
$$
Hence the two output-program states agree up to <quantum phase>[phase]. Any two <quantum state>[qubit states] can be joined through a state <nonorthogonal vectors>[nonorthogonal] to both, so after absorbing phases
$|\widetilde P_U\rangle$ is independent of the input.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

Inner-product preservation between programs for $U$ and $V$ gives
$$
\langle\alpha|\beta\rangle\langle P_U|P_V\rangle
=\langle\alpha|U^\dagger V|\beta\rangle
\langle\widetilde P_U|\widetilde P_V\rangle
$$
for all $|\alpha\rangle,|\beta\rangle$. If $\langle P_U|P_V\rangle\ne0$, this identity forces every <matrix element> of $U^\dagger V$ to be the same <scalar multiple> of the corresponding <identity matrix> element. Hence $U^\dagger V=e^{i\theta}I$, contrary to physical distinctness. Therefore
$$
\boxed{\langle P_U|P_V\rangle=0}.
$$

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

There are <uncountable set>[uncountably] many physically distinct one-qubit <unitary operators>[unitaries]. Part (ii) would require an <orthogonal vectors>[orthogonal] program state for every one, whereas a $K$-qubit register has <dimension of a vector space>[dimension] only $2^K$. This contradiction is the <no-programming theorem>, so no deterministic gate satisfying (PROG) exists.

Solved by gpt-5.6-sol high.

= iv
{parent=b}
{scope}

= Solution
{parent=iv}

Yes. Store $U$ in the two-qubit program
$$
|P_U\rangle=(I\otimes U)|\phi^+\rangle.
$$
Apply the teleportation Bell measurement to the input and the first program qubit. On the $\phi^+$ outcome, which has probability $1/4$ independently of $U$ and $|\alpha\rangle$, the remaining qubit is exactly $U|\alpha\rangle$. Accept that heralded outcome and declare failure otherwise. This gives a probabilistic universal programmable gate with constant success probability $p=1/4$.

Solved by gpt-5.6-sol high.

= 16F
{parent=Paper 2}
{scope}
{title2=Logic and Set Theory}

= a
{parent=16f}
{scope}

= Solution
{parent=a}

The inductive definition of <ordinal addition> is
$$
\alpha+0=\alpha,\qquad
\alpha+(\beta+1)=(\alpha+\beta)+1,\qquad
\alpha+\lambda=\sup_{\beta<\lambda}(\alpha+\beta)
$$
for a <limit ordinal> $\lambda$.

The synthetic definition takes the <order type> of the <disjoint union> of a copy of $\alpha$ followed by a copy of $\beta$: every point of the first copy precedes every point of the second, and each copy retains its original <total order>.

Fix $\alpha$ and use <transfinite induction> on $\beta$. The synthetic sum with the <empty order> is $\alpha$. Appending a <greatest element> produces the <successor ordinal> rule. At a <limit ordinal> $\lambda$, the second copy is the union of its <initial segments> of types $\beta<\lambda$, so the whole order has type
$\sup_{\beta<\lambda}(\alpha+\beta)$. Thus the synthetic operation satisfies the inductive recursion; uniqueness in <transfinite recursion> proves the definitions equivalent.

Solved by gpt-5.6-sol high.

= b
{parent=16f}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

False. With $\alpha=\beta=1$ and $\gamma=\omega$,
$$
(\alpha+\beta)\gamma=2\omega=\omega,
\qquad
\alpha\gamma+\beta\gamma=\omega+\omega=\omega\cdot2.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

True. Left distributivity over a sum in the right argument,
$$
\boxed{\alpha(\beta+\gamma)=\alpha\beta+\alpha\gamma},
$$
follows by transfinite induction on $\gamma$ from the recursive definition of <ordinal multiplication>; the successor and limit steps are exactly its defining clauses.

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

False. The <ordinal number>[ordinal] $\alpha=\omega\cdot2$ is a <limit ordinal>, but
$$
\alpha\omega=(\omega\cdot2)\omega=\omega^2,
\qquad
\omega\alpha=\omega(\omega\cdot2)=\omega^2\cdot2.
$$

Solved by gpt-5.6-sol high.

= iv
{parent=b}
{scope}

= Solution
{parent=iv}

True. Every $\beta<\omega_1$ is a <countable ordinal>, and
$$
\beta+\omega_1=\sup_{\xi<\omega_1}(\beta+\xi)=\omega_1.
$$
Writing any $\alpha\geq\omega_1$ as $\omega_1+\delta$ and using <associativity of ordinal addition>[associativity] gives
$$
\beta+\alpha=(\beta+\omega_1)+\delta
=\omega_1+\delta=\alpha.
$$

Solved by gpt-5.6-sol high.

= v
{parent=b}
{scope}

= Solution
{parent=v}

True. The hypothesis says that $\alpha+\alpha$ and $\beta$ commute under <ordinal addition>. By the <commuting ordinal addition> classification, comparison of their <Cantor normal forms> implies that $\alpha$ and $\beta$ themselves commute. Therefore
$$
\alpha+\alpha+\beta
=\alpha+\beta+\alpha
=\beta+\alpha+\alpha,
$$
so both outer expressions equal the middle one.

Solved by gpt-5.6-sol high.

= 17F
{parent=Paper 2}
{scope}
{title2=Graph Theory}

= a
{parent=17f}
{scope}

= Solution
{parent=a}

The <extremal number> $\operatorname{ex}(n,H)$ is the maximum number of <edges> in an $n$-vertex <graph> containing no <subgraph> <isomorphic graphs>[isomorphic] to $H$.

Let $G$ be <triangle-free graph>[triangle-free] with $e$ edges. For every edge $uv$, the <neighbourhoods of a vertex>[neighbourhoods] of $u$ and $v$ are disjoint apart from their endpoints, so
$d(u)+d(v)\leq n$. Summing over edges and applying <Cauchy-Schwarz inequality> gives
$$
\sum_vd(v)^2=\sum_{uv\in E}(d(u)+d(v))\leq ne,
$$
while
$$
\sum_vd(v)^2\geq\frac{(\sum_vd(v))^2}{n}
=\frac{4e^2}{n}.
$$
Thus $e\leq n^2/4$, proving the required <Mantel theorem> bound.

Solved by gpt-5.6-sol high.

= b
{parent=17f}
{scope}

= Solution
{parent=b}

Let $T$ be the number of <triangles in a graph>[triangles]. Each edge $uv$ has at least
$d(u)+d(v)-n$ <common neighbours>, and summing common-neighbour counts over <edges> counts every triangle three times. Hence
$$
\begin{aligned}
3T
&\geq\sum_{uv\in E}(d(u)+d(v)-n)\\
&=\sum_vd(v)^2-ne
\geq\frac{4e^2}{n}-ne
=e\left(\frac{4e}{n}-n\right).
\end{aligned}
$$
If $e>(1+\delta)n^2/4$, this is greater than
$\delta en>\delta n^3/4$. Therefore
$$
\boxed{T>\frac{\delta}{12}n^3},
$$
so one may take $\varepsilon=\delta/12$.

Solved by gpt-5.6-sol high.

= c
{parent=17f}
{scope}

= Solution
{parent=c}

If $G$ contains no $K_{3,2}$, every <pair> of <vertices> has at most two <common neighbours>. <Double counting>[Double-counting] a vertex together with an <unordered pair> of its neighbours gives
$$
\sum_v\binom{d(v)}2\leq2\binom n2.
$$
Consequently
$$
\sum_vd(v)^2\leq2n(n-1)+2e.
$$
By Cauchy--Schwarz,
$$
\frac{4e^2}{n}\leq2n(n-1)+2e.
$$
This quadratic inequality implies $e<cn^{3/2}$ for an absolute constant $c$; for example $c=2$ works for every $n\geq1$. Thus
$$
\boxed{\operatorname{ex}(n,K_{3,2})<2n^{3/2}}.
$$

Solved by gpt-5.6-sol high.

= d
{parent=17f}
{scope}

= Solution
{parent=d}

Form the <graph> whose <vertices> are the points $x_i$, joining two when their <Euclidean distance> is one. Two distinct points have at most two common unit-distance neighbours, because two <unit circles> intersect in at most two points. The graph is therefore $K_{3,2}$-free. Part (c) bounds its <edges>[unordered edges] by $2n^{3/2}$, so the number of <ordered pairs>[ordered] unit-distance pairs is at most
$$
\boxed{4n^{3/2}}.
$$

Solved by gpt-5.6-sol high.

= 18H
{parent=Paper 2}
{scope}
{title2=Galois Theory}

= a
{parent=18h}
{scope}

= Solution
{parent=a}

If the <minimal polynomial> of $\gamma$ has <degree of a polynomial>[degree] $d$, then
$\mathbb F_p(\gamma)$ has $p^d$ elements and is an <intermediate field> of
$L/\mathbb F_p$. The <tower law for field extensions> gives
$$
n=[L:\mathbb F_p]
=[L:\mathbb F_p(\gamma)]d,
$$
so $d\mid n$.

The <multiplicative group of a finite field is cyclic>. Choose a generator
$\gamma$ of $L^\times$, of <order of a group element>[order] $p^n-1$. If $\gamma$ lay in a proper <intermediate field> of degree $d<n$, its order would <divisibility>[divide] $p^d-1<p^n-1$, a contradiction. Thus
$\mathbb F_p(\gamma)=L$ and its <minimal polynomial> has degree $n$.

For arbitrary $r\geq1$, let $E$ be the <splitting field> of
$X^{p^r}-X$ over $\mathbb F_p$. Its <roots of a polynomial>[roots] form a <field>: the <Frobenius endomorphism> shows they are closed under <addition>, <subtraction>, <multiplication>, and <multiplicative inverse>[inversion]. The <derivative> is $-1$, so there are exactly $p^r$ distinct roots. Thus this root field has order $p^r$. Applying the preceding generator argument supplies an element whose <minimal polynomial> over $\mathbb F_p$ has degree $r$, proving that an <irreducible polynomial> of every positive degree exists.

Solved by gpt-5.6-sol high.

= b
{parent=18h}
{scope}

= Solution
{parent=b}

An element $\alpha$ is <separable algebraic element>[separable] over $K$ when its <minimal polynomial> has <distinct elements>[distinct] roots in a <splitting field>. A <polynomial> has a <multiple root>[repeated root] exactly when it has a <common root> with its <formal derivative in positive characteristic>[formal derivative]. Hence
$\gcd(f,f')=1$ implies that $f$ has no repeated root and $\alpha$ is separable.

Let $f$ be the <minimal polynomial> of $\beta$. Repeatedly factor through the <Frobenius endomorphism>[Frobenius] until
$$
f(X)=g(X^{p^h}),
$$
where $g'\ne0$. Irreducibility of $f$ makes $g$ irreducible, and
$\alpha=\beta^{p^h}$ has separable minimal polynomial $g$. Put
$$
M=K(\alpha)=K(\beta^{p^h}).
$$
Then $M/K$ is separable and
$$
[L:M]=p^h.
$$
Every $\gamma\in L=K(\beta)$ is a polynomial in $\beta$ of degree below $[L:K]$. The freshman's-dream identity gives
$$
\gamma^{p^h}\in K(\beta^{p^h})=M,
$$
so $L/M$ is <purely inseparable field extension>[purely inseparable].

Finally, any intermediate field satisfying the stated conditions consists of elements separable over $K$, so it lies in the maximal separable subextension just constructed. Conversely, the condition that all relevant $p^h$th powers lie in that field puts $\beta^{p^h}$, and hence $M$, inside it. The two inclusions prove uniqueness.

Solved by gpt-5.6-sol high.

= 19H
{parent=Paper 2}
{scope}
{title2=Representation Theory}

= Solution
{parent=19H}

<Irreducible character degrees>[Irreducible degrees] divide the <order of a finite group>[order] of a finite $2$-group and satisfy
$$
\sum_i d_i^2=|G|=16.
$$
The number of <linear characters> is $|G/G'|$. For a <nontrivial group>[nontrivial] finite $2$-group this is at least $4$ unless the group is <abelian group>[abelian], and every <nonlinear irreducible character>[nonlinear] degree is at least $2$. A degree $4$ constituent is therefore impossible in the nonabelian case. If there are $\ell$ linear and $m$ degree-two characters, then
$$
\ell+4m=16,\qquad \ell\in\{4,8,16\}.
$$
Thus the only degree collections are
$$
\boxed{1^{16}},\qquad
\boxed{1^8,2^2},\qquad
\boxed{1^4,2^3},
$$
with respectively $r=16,10,7$.

For $1^{16}$ take the <cyclic group> $G=C_{16}=\langle g\rangle$. Every <conjugacy class> is a <singleton set>[singleton], and its <character table> is
$$
\chi_j(g^k)=\zeta_{16}^{\,jk},
\qquad 0\leq j,k<16,
$$
where $\zeta_{16}=e^{2\pi i/16}$.

For $1^8,2^2$, take $G=D_8\times C_2$, where
$D_8=\langle r,s:r^4=s^2=1,\ srs=r^{-1}\rangle$ and
$C_2=\langle z\rangle$. On the five classes
$$
1,\quad r^2,\quad\{r,r^3\},\quad
\{s,r^2s\},\quad\{rs,r^3s\},
$$
the four linear characters of $D_8$, indexed by
$\varepsilon,\delta\in\{\pm1\}$, and its degree-two character are
$$
\chi_{\varepsilon,\delta}=(1,1,\varepsilon,\delta,\varepsilon\delta),
\qquad
\psi=(2,-2,0,0,0).
$$
The ten classes of $G$ are $(C,1)$ and $(C,z)$. Its full table consists of
$$
(\chi\otimes\eta)(C,z^e)=\chi(C)\eta^e,
$$
where $\chi$ runs over the above five rows and
$\eta\in\{1,-1\}$ runs over the two characters of $C_2$. This explicitly gives eight linear and two degree-two rows.

For $1^4,2^3$, take the <dihedral group>
$$
D_{16}=\langle r,s:r^8=s^2=1,\ srs=r^{-1}\rangle.
$$
Order its seven classes as
$$
1,\ r^4,\ \{r,r^7\},\ \{r^2,r^6\},\
\{r^3,r^5\},\
\{r^{2j}s\}_{j=0}^3,\ \{r^{2j+1}s\}_{j=0}^3.
$$
The four linear rows are
$$
\chi_{\varepsilon,\delta}
=(1,1,\varepsilon,1,\varepsilon,\delta,\varepsilon\delta),
\qquad \varepsilon,\delta=\pm1.
$$
The three degree-two rows, for $k=1,2,3$, are
$$
\psi_k=
\left(
2,\ 2(-1)^k,\
2\cos\frac{k\pi}{4},\
2\cos\frac{k\pi}{2},\
2\cos\frac{3k\pi}{4},\
0,\ 0
\right).
$$
The row norms and mutual inner products, weighted by the displayed class sizes, verify irreducibility and completeness.

Solved by gpt-5.6-sol high.

= 20H
{parent=Paper 2}
{scope}
{title2=Number Fields}

= a
{parent=20h}
{scope}

= Solution
{parent=a}

Distinct <prime-power ideals> are pairwise <comaximal ideals>[comaximal]. For comaximal ideals, <product of ideals>[product] equals <intersection of ideals>[intersection], so <mathematical induction>[induction] gives
$$
I=P_1^{m_1}\cdots P_k^{m_k}
=P_1^{m_1}\cap\cdots\cap P_k^{m_k}.
$$
The <Chinese remainder theorem> therefore gives the ring isomorphism
$$
\mathcal O_K/I\longrightarrow
\prod_i\mathcal O_K/P_i^{m_i}.
$$

The <ideal approximation theorem>, proved by applying this <Chinese remainder theorem>[Chinese-remainder] map one <prime power> deeper, supplies
$$
\alpha\in I,\qquad
\alpha\notin P_iI\quad(1\leq i\leq k).
$$
Equivalently, $v_{P_i}(\alpha)=m_i$ at every prime dividing $I$.

Since $(\alpha)\subseteq I$, the <fractional ideal>
$$
I'=(\alpha)I^{-1}
$$
is an <integral ideal> and $II'=(\alpha)$ is a <principal ideal>. At every $P_i\mid I$,
$v_{P_i}(I')=v_{P_i}(\alpha)-m_i=0$, so $I+I'=\mathcal O_K$.

Choose $\beta\in I$ and $\gamma\in I'$ with $\beta+\gamma=1$. Then
$\alpha,\beta\in I$. For any $x\in I$,
$$
x=x\beta+x\gamma\in(\beta)+II'=(\beta)+(\alpha).
$$
Thus
$$
\boxed{I=(\alpha,\beta)},
$$
so every <ideal> of $\mathcal O_K$ has a <generating set of an ideal>[generating set] of two elements.

Solved by gpt-5.6-sol high.

= b
{parent=20h}
{scope}

= Solution
{parent=b}

The <Dedekind factorization theorem> says that, when a rational prime $p$ does not divide the index of an order $\mathbb Z[\theta]\subseteq\mathcal O_K$, a factorization
$$
\overline f=\prod_i\overline g_i^{e_i}
$$
of the <minimal polynomial> modulo $p$ gives
$$
p\mathcal O_K=\prod_i(p,g_i(\theta))^{e_i}.
$$

Let $K=\mathbb Q(\sqrt d)$ with square-free $d$, and let $p$ be odd. The resulting <splitting of rational primes in a quadratic field> is:
$$
\begin{array}{c|c}
p\mid d & (p)=P^2\quad\text{(ramified)},\\
\left(\frac dp\right)=1 &(p)=(p,\sqrt d-a)(p,\sqrt d+a),
\quad a^2\equiv d\pmod p,\\
\left(\frac dp\right)=-1 &(p)\ \text{is prime}.
\end{array}
$$

Finally suppose $d>0$ and
$\alpha=A/C+(B/C)\sqrt d$ has norm $-1$, with integers
$A,B,C$ chosen <coprime integers>[coprime]. Then
$$
A^2-dB^2=-C^2.
$$
If an <odd integer>[odd] $p\equiv3\pmod4$ <ramification of a prime>[ramified], then $p\mid d$. <Reduction modulo an ideal>[Reduction] modulo $p$ gives
$A^2\equiv-C^2\pmod p$. Since $-1$ is not a <quadratic residue>[square] modulo such a <prime number>[prime], $p\mid A,C$. Dividing the equation's <divisibility> shows $p\mid B$ as well (use that <square-free integer>[square-free] $d$ has <P-adic valuation>[$p$-adic valuation] one), contradicting <coprime integers>[coprimality]. Hence no prime $p\equiv3\pmod4$ ramifies.

Solved by gpt-5.6-sol high.

= 21I
{parent=Paper 2}
{scope}
{title2=Algebraic Topology}

= Solution
{parent=21I}

The <snake lemma> associates to a <commutative diagram> with <exact sequences>[exact] rows an <exact sequence>
$$
\ker f_1\to\ker f_2\to\ker f_3
\xrightarrow{\delta}\operatorname{coker}f_1
\to\operatorname{coker}f_2\to\operatorname{coker}f_3.
$$
For $K=L_1\cup L_2$, apply it degree by degree to the short exact sequence of chain complexes
$$
0\to C_*(L_1\cap L_2)
\xrightarrow{c\mapsto(c,-c)}
C_*(L_1)\oplus C_*(L_2)
\xrightarrow{(a,b)\mapsto a+b}
C_*(K)\to0.
$$
The resulting <connecting homomorphisms>[connecting maps] splice the <kernels> modulo <chain boundaries>[boundaries] into the <Mayer-Vietoris theorem>[Mayer--Vietoris sequence]
$$
\cdots\to H_n(L_1\cap L_2)\to
H_n(L_1)\oplus H_n(L_2)\to H_n(K)
\xrightarrow{\partial}H_{n-1}(L_1\cap L_2)\to\cdots.
$$

Now let $K$ satisfy the stated <simplicial pseudomanifold> conditions. In an $n$-<chain cycle>[cycle], cancellation at a common $(n-1)$-<face of a simplex>[face] determines the <chain coefficient>[coefficient] of either incident $n$-<simplex> from the other, up to the <orientation> sign. Connectivity through such faces therefore determines every top-simplex coefficient from one <integer>. If the signs are globally compatible, their oriented sum is a cycle and <generator of a group>[generates] $H_n(K)\cong\mathbb Z$; if they are inconsistent, that integer must be zero and $H_n(K)=0$.

Let $x$ be the resulting <fundamental class of an orientable simplicial pseudomanifold>[fundamental class], represented by $z$. Since the intersection has dimension below $n$, split uniquely
$z=z_1+z_2$ into top chains in $L_1$ and $L_2$. Then
$$
\partial z_1=-\partial z_2\in C_{n-1}(L_1\cap L_2),
$$
and the Mayer--Vietoris boundary is
$$
\boxed{\partial x=[\partial z_1]=-[\partial z_2]}.
$$
It is nonzero exactly when both $L_1$ and $L_2$ contain top-dimensional simplices; if one side contains none, the fundamental cycle already lies in the other side, while if both do, connectedness forces a nonempty interface and its oriented boundary represents a nonzero class.

Finally take $K\cong S^3$, $L_1\cong S^1\times D^2$, and
$L_1\cap L_2\cong T^2$. The preceding boundary
$$
H_3(S^3)\longrightarrow H_2(T^2)
$$
is a nonzero map between copies of $\mathbb Z$ and sends the fundamental class to the oriented boundary torus, hence is an isomorphism. Exactness then gives
$$
H_2(L_2)=0.
$$
The next part of the sequence is
$$
0\to H_1(T^2)\cong\mathbb Z^2
\to H_1(L_1)\oplus H_1(L_2)
\to0.
$$
Since $H_1(L_1)\cong\mathbb Z$, it follows that
$H_1(L_2)\cong\mathbb Z$. The degree-zero sequence makes $L_2$ connected. Therefore
$$
\boxed{H_0(L_2)\cong\mathbb Z,\quad
H_1(L_2)\cong\mathbb Z,\quad
H_i(L_2)=0\ (i\geq2)}.
$$

Solved by gpt-5.6-sol high.

= 22G
{parent=Paper 2}
{scope}
{title2=Linear Analysis}

= a
{parent=22g}
{scope}

= Solution
{parent=a}

Let $(T_m)$ be a <Cauchy sequence> in the <operator norm>. For each $v\in V$,
$$
\|T_mv-T_nv\|\leq\|T_m-T_n\|\,\|v\|,
$$
so $(T_mv)$ is a Cauchy sequence. <Completeness> of the <Banach space> $V$ permits the definition
$Tv=\lim_mT_mv$. Pointwise passage to the <limit> proves that $T$ is a <linear map>[linear]. A norm-Cauchy sequence is a <bounded sequence>, say $\|T_m\|\leq C$, and therefore
$\|Tv\|\leq C\|v\|$, so $T$ is bounded. Finally, letting $m\to\infty$ in
$\|(T_n-T_m)v\|\leq\varepsilon\|v\|$ gives
$\|T_n-T\|\leq\varepsilon$. Thus $T_n\to T$ in operator norm and
$\mathcal B(V,V)$ is Banach.

Solved by gpt-5.6-sol high.

= b
{parent=22g}
{scope}

= Solution
{parent=b}

A <compact operator> maps every bounded sequence to a sequence with a norm-convergent subsequence. A <Hilbertian basis> is a complete orthonormal sequence.

If $T$ is compact and $\lambda_n$ does not tend to zero, some subsequence has
$|\lambda_{n_j}|\geq\varepsilon$. The vectors
$Te_{n_j}=\lambda_{n_j}e_{n_j}$ are pairwise separated by at least $\sqrt2\varepsilon$, so they have no convergent subsequence, a contradiction.

Conversely, if $\lambda_n\to0$, define the <finite-rank operator>
$$
T_Ne_n=\begin{cases}\lambda_ne_n,&n\leq N,\\0,&n>N.\end{cases}
$$
Then
$$
\|T-T_N\|=\sup_{n>N}|\lambda_n|\longrightarrow0.
$$
An operator-norm limit of compact operators is compact, so $T$ is compact.

Solved by gpt-5.6-sol high.

= c
{parent=22g}
{scope}

= Solution
{parent=c}

For $x=\sum_nx_ne_n$, <Cauchy-Schwarz inequality> gives
$$
\|Tx\|
\leq\sum_n|x_n|\,\|Te_n\|
\leq\|x\|\left(\sum_n\|Te_n\|^2\right)^{1/2}.
$$
Thus every $T\in H(V,V)$ is bounded and
$\|T\|\leq\|T\|_*$. The truncations
$$
T_Nx=\sum_{n\leq N}x_nTe_n
$$
have finite rank and satisfy
$$
\|T-T_N\|\leq
\left(\sum_{n>N}\|Te_n\|^2\right)^{1/2}\to0.
$$
Hence every such <Hilbert-Schmidt operator> is compact.

Define
$$
\langle S,T\rangle_*=\sum_n\langle Se_n,Te_n\rangle.
$$
This is an inner product inducing $\|\cdot\|_*$. If $(T_j)$ is Cauchy in this norm, then each $T_je_n$ converges to some $y_n$, and the usual Fatou/tail argument gives
$\sum_n\|y_n\|^2<\infty$ and
$\sum_n\|T_je_n-y_n\|^2\to0$. Defining
$Tx=\sum_nx_ny_n$ produces a bounded operator by the preceding estimate and gives
$T_j\to T$ in $\|\cdot\|_*$. Thus $(H(V,V),\|\cdot\|_*)$ is a Hilbert space.

The norms are not equivalent in infinite dimension. The rank-$N$ orthogonal projection has
$$
\|P_N\|=1,\qquad \|P_N\|_*=\sqrt N.
$$
Although $\|T\|\leq\|T\|_*$, no uniform reverse inequality can hold.

Solved by gpt-5.6-sol high.

= 23G
{parent=Paper 2}
{scope}
{title2=Analysis of Functions}

= Solution
{parent=23G}

A functional $p:X\to\mathbb R$ is <sublinear functional>[sublinear] when
$$
p(u+v)\leq p(u)+p(v),\qquad p(tu)=tp(u)\quad(t\geq0).
$$

Any extension to $\widetilde M=M+\mathbb Rx$ must have the form
$$
\widetilde\ell(y+tx)=\ell(y)+tc.
$$
Choose $c$ satisfying
$$
\sup_{y\in M}\{\ell(y)-p(y-x)\}
\leq c\leq
\inf_{y\in M}\{p(y+x)-\ell(y)\}.
$$
This interval is nonempty: for $y,z\in M$,
$$
\ell(y)+\ell(z)=\ell(y+z)
\leq p(y+z)
\leq p(y-x)+p(z+x),
$$
which rearranges to the required lower bound being at most the upper bound. For $t>0$, positive homogeneity reduces
$\widetilde\ell(y+tx)\leq p(y+tx)$ to the upper inequality after replacing $y$ by $y/t$; for $t<0$ it reduces to the lower inequality. The case $t=0$ is the hypothesis on $\ell$. Thus this choice gives the required dominated linear extension.

The dominated form of the <Hahn-Banach theorem> states that a linear functional on a subspace, bounded above by a sublinear functional, extends linearly to the whole real vector space while retaining that bound.

Let $M=\operatorname{span}\{z_1,\ldots,z_n\}$. Its coordinate maps
$$
\lambda_j\!\left(\sum_ka_kz_k\right)=a_j
$$
are continuous because every linear map on a finite-dimensional normed space is continuous. Hahn--Banach extends each to
$\ell_j\in Z'$ without increasing its norm, and then
$$
\ell_j(z_k)=\delta_{jk}.
$$

For an arbitrary finite-dimensional subspace $M$, choose a basis $z_1,\ldots,z_n$ and these extended coordinate functionals. Then
$$
N=\bigcap_{j=1}^n\ker\ell_j
$$
is closed. Every $z\in Z$ has the decomposition
$$
z=\sum_j\ell_j(z)z_j+
\left(z-\sum_j\ell_j(z)z_j\right)\in M+N.
$$
Applying every $\ell_j$ shows $M\cap N=\{0\}$, so
$$
\boxed{Z=M\oplus N}.
$$

Solved by gpt-5.6-sol high.

= 24F
{parent=Paper 2}
{scope}
{title2=Riemann Surfaces}

= a
{parent=24f}
{scope}

= Solution
{parent=a}

The sum of two <meromorphic function>[meromorphic functions] has no poles away from the union of their pole sets, and at each $p_i$ its pole order is at most the larger of the two orders, hence at most $m_i$. Scalar multiplication cannot increase a pole order. The zero function belongs to the set, so $V$ is a complex vector space.

Solved by gpt-5.6-sol high.

= b
{parent=24f}
{scope}

= Solution
{parent=b}

Choose a local coordinate $z_i$ at each $p_i$ and map $f\in V$ to all coefficients of the negative-power terms in its Laurent expansions:
$$
f\longmapsto
(a_{i,-1},\ldots,a_{i,-m_i})_{i=1}^n
\in\mathbb C^{\sum_i m_i}.
$$
The kernel consists of globally holomorphic functions on the compact Riemann surface $R$. By the <maximum modulus principle>, every such function is constant, so the kernel has dimension one. Rank--nullity gives
$$
\boxed{\dim_{\mathbb C}V\leq1+\sum_i m_i}.
$$
For $n=0$, this simply says $\dim V=1$.

Solved by gpt-5.6-sol high.

= c
{parent=24f}
{scope}

= Solution
{parent=c}

Lift $f$ to a doubly periodic meromorphic function on $\mathbb C$. Integrate it around a fundamental parallelogram whose boundary avoids all poles. Integrals over opposite edges cancel by periodicity, while the <residue theorem> gives
$$
0=\frac1{2\pi i}\int_{\partial P}f(z)\,dz
=\sum_{i=1}^n\operatorname{res}_{p_i}f.
$$

Thus the principal-part map from part (b) takes values in the codimension-one hyperplane on which the sum of residue coefficients is zero. The Mittag--Leffler existence criterion on a compact Riemann surface says that prescribed principal parts occur precisely when their residues pair trivially with every holomorphic one-form. On a complex torus the holomorphic one-forms are the scalar multiples of $dz$, so the single condition is exactly the displayed residue sum. Hence the image has dimension
$\sum_i m_i-1$, and the kernel of constants has dimension one. For $n\geq1$,
$$
\boxed{\dim_{\mathbb C}V=\sum_{i=1}^n m_i}.
$$
Equivalently, this is the genus-one case of the <Riemann-Roch theorem> for a positive divisor.

Solved by gpt-5.6-sol high.

= 25H
{parent=Paper 2}
{scope}
{title2=Algebraic Geometry}

= Solution
{parent=25H}

The <Riemann-Hurwitz formula> for a nonconstant degree-$n$ morphism
$f:C\to C'$ of smooth projective connected curves in characteristic zero is
$$
2g(C)-2=n(2g(C')-2)+\deg R_f,
$$
where $R_f$ is the effective ramification divisor. If $g(C')\geq2$ and
$g(C)<g(C')$, the right side is at least $2g(C')-2>2g(C)-2$, impossible. If $g(C')=1$, then $g(C)=0$ and the left side is $-2$ while the right side is nonnegative. Thus every morphism is constant.

Choose $p\in C_d$ and project from $p$ to the pencil of lines through it. Bézout's theorem says that a general such line meets $C_d$ in $p$ plus $d-1$ further points, so the resolved projection
$$
\phi:C_d\to\mathbb P^1
$$
has degree $d-1$. Since the <genus of a smooth plane curve> is
$g=(d-1)(d-2)/2$, Riemann--Hurwitz gives
$$
\deg R_\phi
=2g-2+2(d-1)
=(d-2)(d+1).
$$
Every branch point receives at least one ramification point, hence
$$
\boxed{|B|\leq(d-2)(d+1)}.
$$

The adjunction formula gives
$$
K_{C_d}\sim(d-3)D.
$$
Since $\deg D=d$ and $\deg K_{C_d}=d(d-3)$, linear equivalence
$D\sim K_{C_d}$ would force $d=d(d-3)$, hence $d=4$. It is therefore impossible for $d\geq5$.

For a smooth plane quartic, projection from a point of the curve gives a degree-three map, so its <gonality> is at most three. Its genus is three and
$K_C\sim D$, so its plane embedding is the <canonical map>. A genus-three curve with a degree-two map to $\mathbb P^1$ would be <hyperelliptic curve>[hyperelliptic], and its canonical map would factor through that double cover and map onto a conic rather than embed the curve. Thus no degree-two map exists, and a degree-one map is excluded by the genus. The quartic's gonality is
$$
\boxed{3}.
$$
For a genus-one curve, Riemann--Roch supplies a degree-two map to $\mathbb P^1$, while degree one would make it isomorphic to $\mathbb P^1$. Its gonality is therefore
$$
\boxed{2}.
$$

Solved by gpt-5.6-sol high.

= 26I
{parent=Paper 2}
{scope}
{title2=Differential Geometry}

= Solution
{parent=26I}

A <smooth manifold> of dimension $k$ is a Hausdorff, second-countable space with an atlas of smoothly compatible charts to open subsets of $\mathbb R^k$. A value $y\in Y$ is a <regular value> of a smooth map $f:X\to Y$ when every $x\in f^{-1}(y)$ has surjective derivative $df_x$.

The <inverse function theorem> says that a smooth map between equal-dimensional manifolds whose derivative is invertible at a point is a diffeomorphism between neighbourhoods of that point and its image. If $y$ is a regular value of a map from an $m$-manifold to an $n$-manifold, choose coordinates corresponding to an invertible $n$ by $n$ minor of $df$. Applying the inverse theorem to $f$ together with the remaining $m-n$ coordinates makes $f$ the coordinate projection onto the first $n$ coordinates. Its level set is therefore locally $\mathbb R^{m-n}$. This proves the <preimage theorem>.

The critical-point set is closed because failure of full rank is the simultaneous vanishing of all maximal minors of $df$. If $X$ is compact, this set is compact, so its image under $f$ is compact and hence closed in the manifold $Y$. Its complement, the set of regular values, is open.

For the printed equations put
$$
F_1=x+y-z^2-w^2,\qquad
F_2=x^2+y^2-\frac{z^4}{2}.
$$
A rank calculation shows that $(a,0)$ is a regular value when $a>0$: if
$\nabla F_2=\lambda\nabla F_1$, then
$$
x=y=\lambda/2,\qquad z(z^2-\lambda)=0,\qquad
\lambda w=0,\qquad \lambda^2=z^4.
$$
At a point of the level set these relations force $a\leq0$. Hence for $a>0$, the preimage theorem makes $X_a$ a two-dimensional manifold.

As printed, however, the assertion for every $a\ne0$ is false. If $a<0$, the points
$$
(0,0,0,\pm\sqrt{-a})
$$
belong to $X_a$. Near either point the first equation solves smoothly for $w$, while the second equation is
$$
x^2+y^2=z^4/2.
$$
Its positive-$z$ and negative-$z$ sheets meet only at the origin, so deleting the meeting point disconnects every sufficiently small neighbourhood. A punctured neighbourhood in a two-manifold is connected. Thus $X_a$ is not a manifold there.

For $a=0$, Cauchy--Schwarz gives
$$
x+y\leq\sqrt{2(x^2+y^2)}=z^2.
$$
The first equation requires $w^2=x+y-z^2\geq0$, so equality must hold. Therefore
$$
X_0=\{(z^2/2,z^2/2,z,0):z\in\mathbb R\}.
$$
This is the image of a smooth embedding with nonzero derivative, and hence
$\boxed{X_0\text{ is a one-dimensional manifold}}$.

Solved by gpt-5.6-sol high.

= 27G
{parent=Paper 2}
{scope}
{title2=Probability and Measure}

= a
{parent=27g}
{scope}

= Solution
{parent=a}

The <monotone convergence theorem> states that if
$0\leq f_n\uparrow f$ pointwise, then
$$
\int f_n\,d\mu\uparrow\int f\,d\mu,
$$
allowing the value $+\infty$.

Monotonicity of the integral shows that the increasing limit $L$ of the left side is at most $\int f$. Conversely, let $s$ be a nonnegative simple function with $s\leq f$, and fix $0<c<1$. The sets
$$
E_n=\{f_n\geq cs\}
$$
increase to the support of $s$. Hence
$$
\int f_n\,d\mu\geq c\int_{E_n}s\,d\mu
\longrightarrow c\int s\,d\mu.
$$
Let $c\uparrow1$ and take the supremum over all simple $s\leq f$. This gives
$L\geq\int f$, proving equality.

Solved by gpt-5.6-sol high.

= b
{parent=27g}
{scope}

= Solution
{parent=b}

Since $f_n\uparrow f$, the functions
$$
g_n=f_n-f_1
$$
are nonnegative and increase to $g=f-f_1$. The monotone convergence theorem gives
$\mu(g_n)\uparrow\mu(g)$. Because $f_1$ is integrable, adding the finite number
$\mu(f_1)$ yields
$$
\boxed{\mu(f_n)\uparrow\mu(f)}.
$$

Integrability of $f_1$ is essential. On $\mathbb R$ with Lebesgue measure, let
$$
f_n=-\mathbf1_{[n,\infty)}.
$$
Then $f_n\uparrow0$ pointwise, but every integral is $-\infty$, so
$\mu(f_n)$ does not increase to $\mu(0)=0$.

Solved by gpt-5.6-sol high.

= 28J
{parent=Paper 2}
{scope}
{title2=Applied Probability}

= a
{parent=28j}
{scope}

= Solution
{parent=a}

The first condition is that $X$ is Markov with transition semigroup
$P(t)=e^{tQ}$. The second is that, for every function
$f:S\to\mathbb R$,
$$
M_t^f=f(X_t)-f(X_0)-\int_0^t(Qf)(X_s)\,ds
$$
is a martingale.

The first condition implies the second by conditioning over a short time interval, using
$P(h)f=f+hQf+o(h)$, summing increments, and passing to the limit. This is Dynkin's formula.

Conversely assume the <martingale problem for a continuous-time Markov chain>. Fix $T>0$ and a function $f$, and put
$$
u(t,x)=(e^{(T-t)Q}f)(x).
$$
The backward equation gives $\partial_tu+Qu=0$. Applying the martingale identity to this time-dependent function shows that $u(t,X_t)$ is a martingale. Therefore, for $s\leq T$,
$$
\mathbb E[f(X_T)\mid\mathcal F_s]
=u(s,X_s)=(e^{(T-s)Q}f)(X_s).
$$
Taking indicator functions $f$ proves both the Markov property and the transition semigroup $e^{tQ}$.

Solved by gpt-5.6-sol high.

= b
{parent=28j}
{scope}

= Solution
{parent=b}

Write $d_G(x)$ for the degree of $x$ in $G$ and $d_A(x)$ for the number of its neighbours lying in $A$. Each trial succeeds with probability
$d_A(x)/d_G(x)$. The stated geometric-sum fact shows that the total holding time at $x$ is exponential with rate $d_A(x)/d_G(x)$. Conditional on success, each neighbour in $A$ is chosen uniformly. Hence
$$
\boxed{q_{xy}=
\begin{cases}
1/d_G(x),&xy\in E(A),\\
0,&x\ne y,\ xy\notin E(A),
\end{cases}
\qquad
q_{xx}=-d_A(x)/d_G(x)}.
$$

Let
$$
\pi(x)=\frac{d_G(x)}{\sum_{z\in A}d_G(z)}.
$$
For every edge $xy$ of $A$,
$$
\pi(x)q_{xy}
=\frac1{\sum_{z\in A}d_G(z)}
=\pi(y)q_{yx}.
$$
Thus <detailed balance> holds, so $\pi$ is invariant. Connectedness of $A$ makes the chain irreducible and this invariant distribution unique.

Solved by gpt-5.6-sol high.

= 29K
{parent=Paper 2}
{scope}
{title2=Principles of Statistics}

= a
{parent=29k}
{scope}

= Solution
{parent=a}

Since $\overline X_n\sim N(\theta,1/n)$, the estimator has variance
$a^2/n$ and bias $(a-1)\theta+b$. Its quadratic <risk function> is
$$
\boxed{R(\theta,\widehat\theta_{a,b})
=\frac{a^2}{n}+\bigl((a-1)\theta+b\bigr)^2}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=29k}
{scope}

= Solution
{parent=b}

If $a>1$, compare with
$$
\widetilde\theta=(2-a)\overline X_n-b.
$$
Its bias is the negative of the original bias, so the squared-bias terms agree, while
$$
(2-a)^2<a^2.
$$
Its risk is therefore strictly smaller for every $\theta$. Thus
$\widehat\theta_{a,b}$ is not an <admissible estimator>.

Solved by gpt-5.6-sol high.

= c
{parent=29k}
{scope}

= Solution
{parent=c}

Put $\theta_0=-b/(a-1)$ and compare with the constant estimator
$\delta=\theta_0$. Its risk is $(\theta-\theta_0)^2$. For $a<0$,
$$
R(\theta,\widehat\theta_{a,b})
=\frac{a^2}{n}+(a-1)^2(\theta-\theta_0)^2
>(\theta-\theta_0)^2
$$
for every $\theta$, since $|a-1|>1$. Hence the original estimator is inadmissible.

Solved by gpt-5.6-sol high.

= d
{parent=29k}
{scope}

= Solution
{parent=d}

The risk of $\overline X_n$ is constantly $1/n$. Suppose an estimator $\delta$ dominates it, and put
$m(\theta)=\mathbb E_\theta\delta$ and $b(\theta)=m(\theta)-\theta$. The biased <Cramer-Rao bound>, using total information $n$, gives
$$
R(\theta,\delta)\geq
\frac{m'(\theta)^2}{n}+b(\theta)^2.
$$
Domination would imply
$$
(1+b'(\theta))^2+n b(\theta)^2\leq1
$$
for every real $\theta$. The standard differential-inequality argument shows that the only globally defined differentiable solution is $b\equiv0$: a nonzero value forces $b'$ to retain a sign and magnitude that makes $b$ leave the permitted bounded interval in one time direction.

Thus $m(\theta)=\theta$. The Cramer--Rao inequality now gives
$\operatorname{Var}_\theta\delta\geq1/n$, so domination forces equality everywhere. Equality in Cramer--Rao makes the centred estimator proportional to the Gaussian score:
$$
\delta-\theta=\frac1n\sum_i(X_i-\theta),
$$
and hence $\delta=\overline X_n$ almost surely. No strict improvement exists, so $\overline X_n$ is admissible.

Solved by gpt-5.6-sol high.

= e
{parent=29k}
{scope}

= Solution
{parent=e}

No. Every estimator in parts (b) and (c) has $a\ne1$, so its risk contains the nonconstant quadratic term
$((a-1)\theta+b)^2$ and has infinite supremum over $\theta\in\mathbb R$. The estimator $\overline X_n$ has constant risk $1/n$, so those estimators cannot be <minimax estimator>[minimax].

Solved by gpt-5.6-sol high.

= 30K
{parent=Paper 2}
{scope}
{title2=Stochastic Financial Models}

= a
{parent=30k}
{scope}

= Solution
{parent=a}

Put
$$
q=\mu-(1+r)S_0,\qquad h=m-(1+r)x.
$$
For a risky holding vector $\theta$ and the remaining wealth in the bank,
$$
X_1=(1+r)x+\theta^T(S_1-(1+r)S_0),
$$
so
$$
\mathbb E X_1=(1+r)x+\theta^Tq,\qquad
\operatorname{Var}(X_1)=\theta^TV\theta.
$$
When $V$ is nonsingular it is positive definite. Cauchy--Schwarz in the
$V$ inner product gives
$$
(\theta^Tq)^2
\leq(\theta^TV\theta)(q^TV^{-1}q),
$$
with equality exactly when $\theta=\lambda V^{-1}q$. Under
$\theta^Tq=h$,
$$
\boxed{\theta^*=
\frac{h}{q^TV^{-1}q}V^{-1}q,\qquad
\min\operatorname{Var}(X_1)
=\frac{h^2}{q^TV^{-1}q}}.
$$

For the constraint $\mathbb E X_1\geq m$, if $h\leq0$ the zero risky portfolio is feasible and the minimum variance is zero. If $h>0$, the inequality binds and the preceding optimizer and minimum apply.

Solved by gpt-5.6-sol high.

= b
{parent=30k}
{scope}

= Solution
{parent=b}

For every nonzero-risk portfolio,
$$
\frac{\mathbb E X_1-(1+r)X_0}
{\sqrt{\operatorname{Var}(X_1)}}
=\frac{\theta^Tq}{\sqrt{\theta^TV\theta}}
\leq\sqrt{q^TV^{-1}q}.
$$
Equality holds precisely for
$$
\boxed{\theta=\lambda V^{-1}q,\qquad\lambda>0}.
$$
The maximum is therefore $\sqrt{q^TV^{-1}q}$. For each resulting mean, part (a) shows that this portfolio has the least possible variance, so every maximizer lies on the <mean-variance efficient ray>.

Solved by gpt-5.6-sol high.

= c
{parent=30k}
{scope}

= Solution
{parent=c}

For a symmetric matrix,
$\operatorname{range}V=(\ker V)^\perp$. Since $q\notin\operatorname{range}V$, there is
$\eta\in\ker V$ with $\eta^Tq\ne0$. For any target excess mean $h$, set
$$
\theta=\frac{h}{\eta^Tq}\eta.
$$
Then $\theta^Tq=h$ and
$$
\operatorname{Var}(X_1)=\theta^TV\theta=0.
$$
Thus the required minimum is zero for every $m,x$.

Choose the sign of $\eta$ so that $\eta^Tq>0$, buy the risky portfolio $\eta$, and finance it by borrowing its time-zero cost in the bank. Its initial wealth is zero, while its terminal excess payoff has variance zero and positive mean $\eta^Tq$, so it is a strictly positive constant almost surely. This is an <arbitrage>.

Solved by gpt-5.6-sol high.

= 31J
{parent=Paper 2}
{scope}
{title2=Mathematics of Machine Learning}

= a
{parent=31j}
{scope}

= Solution
{parent=a}

A function has the bounded-differences property with constants
$L_1,\ldots,L_n$ when replacing only coordinate $i$ changes its value by at most $L_i$. The <bounded differences inequality> states that, for <independent random variables>[independent] inputs,
$$
\mathbb P(F-\mathbb EF\geq t)
\leq\exp\left(-\frac{2t^2}{\sum_iL_i^2}\right),
$$
and the analogous lower-tail bound also holds.

Solved by gpt-5.6-sol high.

= b
{parent=31j}
{scope}

= Solution
{parent=b}

Replace the $i$th example of $D$ to obtain $D'$. The population-loss term changes by at most $\beta$ by the assumed <uniform stability of a learning algorithm>. In the empirical average, each of the $n-1$ unchanged examples contributes a change at most $\beta$. For the replaced term, first change the hypothesis, costing at most $\beta$, and then change the evaluated example; boundedness of the loss costs at most $M$. Thus the empirical term changes by at most
$$
\frac{(n-1)\beta+\beta+M}{n}
=\beta+\frac Mn.
$$
Combining the two terms gives
$$
\boxed{|F(D)-F(D')|\leq2\beta+\frac Mn}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=31j}
{scope}

= Solution
{parent=c}

Let $Z=(X,Y)$ be an independent test example and write $Z_i=(X_i,Y_i)$. Exchangeability of $Z$ and $Z_i$ gives
$$
\mathbb E\,\ell(H_D,Z)
=\frac1n\sum_{i=1}^n
\mathbb E\,\ell(H_{D_i(Z)},Z_i).
$$
The datasets $D_i(Z)$ and $D$ differ in one coordinate, so stability gives
$$
\mathbb E\,\ell(H_{D_i(Z)},Z_i)
\leq\mathbb E\,\ell(H_D,Z_i)+\beta.
$$
Averaging over $i$ and subtracting the empirical loss proves
$$
\boxed{\mathbb EF(D)\leq\beta}.
$$

Solved by gpt-5.6-sol high.

= d
{parent=31j}
{scope}

= Solution
{parent=d}

Apply the bounded-differences inequality with every
$L_i=2\beta+M/n$. With probability at least $1-\delta$,
$$
F(D)\leq\mathbb EF(D)
+\left(2\beta+\frac Mn\right)
\sqrt{\frac{n\log(1/\delta)}2}.
$$
Using part (c) and substituting the definition of $F$ gives
$$
\boxed{
\frac1n\sum_{i=1}^n\ell(H_D(X_i),Y_i)
+\beta+(2n\beta+M)\sqrt{\frac{\log(1/\delta)}{2n}}
\geq\mathbb E\ell(H_D(X),Y)}.
$$

Solved by gpt-5.6-sol high.

= 32E
{parent=Paper 2}
{scope}
{title2=Asymptotic Methods}

= a
{parent=32e}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

This is an <asymptotic sequence>. Since
$$
\log(\cos u)\sim-\frac{u^2}{2}\qquad(u\to0),
$$
one has
$$
\frac{\phi_{n+1}(x)}{\phi_n(x)}
\sim x^2\longrightarrow0.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

This is not an asymptotic sequence, because
$$
\frac{\psi_{n+1}(x)}{\psi_n(x)}
=\left(\frac{n+1}{n}\right)^{1/x}
\longrightarrow1
$$
as $x\to\infty$, rather than zero.

Solved by gpt-5.6-sol high.

= iii
{parent=a}
{scope}

= Solution
{parent=iii}

This is not an asymptotic sequence. The functions $\sin(x^n)$ oscillate and have arbitrarily large zeros; the ratio
$\sin(x^{n+1})/\sin(x^n)$ has no limit, let alone limit zero.

Solved by gpt-5.6-sol high.

= b
{parent=32e}
{scope}

= Solution
{parent=b}

Recall that positivity and asymptoticity mean
$$
\frac{\phi_{j+1}}{\phi_j}\to0,\qquad
\frac{\psi_{j+1}}{\psi_j}\to0
$$
for each fixed $j$. For $0\leq k\leq n$, compare the $k$th term of
$\chi_{n+1}$ with the $k$th term of $\chi_n$:
$$
\frac{\phi_k\psi_{n+1-k}}{\phi_k\psi_{n-k}}
=\frac{\psi_{n+1-k}}{\psi_{n-k}}\to0.
$$
For the remaining term, compare with the final term of $\chi_n$:
$$
\frac{\phi_{n+1}\psi_0}{\phi_n\psi_0}
=\frac{\phi_{n+1}}{\phi_n}\to0.
$$
Since $\chi_n$ is positive and dominates every comparison term, summing these finitely many bounds gives
$$
\boxed{\frac{\chi_{n+1}(x)}{\chi_n(x)}\to0}.
$$
Thus $(\chi_n)$ is an asymptotic sequence.

Solved by gpt-5.6-sol high.

= 33B
{parent=Paper 2}
{scope}
{title2=Dynamical Systems}

= a
{parent=33b}
{scope}

= Solution
{parent=a}

An <interval map> $F:I\to I$ has a <horseshoe for an interval map>[horseshoe] if there are two <closed real intervals>[closed subintervals] $J_0,J_1\subseteq I$ with disjoint interiors such that
$$
F(J_i)\supseteq J_0\cup J_1
\qquad(i=0,1).
$$
It is <Glendinning chaos>[chaotic in Glendinning's sense] if some positive <iteration of a map>[iterate] $F^n$ has a horseshoe.

Suppose $a<b<c$ are the points of a <cycle of an interval map>[3-cycle] and put $J_0=[a,b]$, $J_1=[b,c]$. There are two possible cyclic orders. If
$$
a\mapsto b\mapsto c\mapsto a,
$$
then the <intermediate value theorem> gives
$$
J_0\longrightarrow J_1,
\qquad
J_1\longrightarrow J_0\cup J_1.
$$
Consequently both $F^2(J_0)$ and $F^2(J_1)$ contain $J_0\cup J_1$. If instead
$$
a\mapsto c\mapsto b\mapsto a,
$$
then
$$
J_0\longrightarrow J_0\cup J_1,
\qquad
J_1\longrightarrow J_0,
$$
and again both second images contain $J_0\cup J_1$. Thus in either cyclic order $F^2$ has a horseshoe.

Solved by gpt-5.6-sol high.

= b
{parent=33b}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

Let
$$
I_0=[x_0,x_1],\qquad I_1=[x_1,x_2],\qquad I_2=[x_2,x_3].
$$
The <interval covering relation>[covering relations] are
$$
I_0\longrightarrow I_1,\qquad
I_1\longrightarrow I_2,\qquad
I_2\longrightarrow I_0,I_1,I_2.
$$
Thus the <directed covering graph of an interval map>[covering graph] has <adjacency matrix of a directed graph>[adjacency matrix]
$$
A=\begin{pmatrix}0&1&0\\0&0&1\\1&1&1\end{pmatrix}.
$$
The two closed length-three walks
$$
I_2\to I_2\to I_2\to I_2,
\qquad
I_2\to I_0\to I_1\to I_2
$$
give a <horseshoe from two closed covering walks> for $F^3$, so $F$ must be <Glendinning chaos>[chaotic]. Moreover,
$$
\operatorname{tr}A=1,
\qquad
\operatorname{tr}(A^3)=7.
$$
By <counting cycles in an interval covering graph>, the number of primitive length-three cyclic itineraries is
$$
\frac{7-1}{3}=2.
$$
Hence $F$ must have at least, and the connect-the-dots realization shows that it need have only, $\boxed{2}$ distinct 3-cycles.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

Set
$$
I_0=[x_0,x_1],\qquad I_1=[x_1,x_3],\qquad I_2=[x_3,x_2].
$$
Their covering graph is
$$
I_0\longrightarrow I_1,I_2,\qquad
I_1\longrightarrow I_0,I_1,I_2,\qquad
I_2\longrightarrow I_0,I_1,
$$
with <adjacency matrix of a directed graph>[adjacency matrix]
$$
A=\begin{pmatrix}0&1&1\\1&1&1\\1&1&0\end{pmatrix}.
$$
The closed walks $I_1\to I_1\to I_1$ and $I_1\to I_0\to I_1$ give a <horseshoe from two closed covering walks> for $F^2$, so $F$ must be <Glendinning chaos>[chaotic]. Since
$$
\operatorname{tr}A=1,
\qquad
\operatorname{tr}(A^3)=13,
$$
<counting cycles in an interval covering graph> gives
$$
\frac{13-1}{3}=\boxed4
$$
distinct 3-cycles in the minimum case, attained by the corresponding <connect-the-dots interval map>.

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

Take
$$
I_0=[x_0,x_2],\qquad I_1=[x_2,x_1],\qquad I_2=[x_1,x_3].
$$
Now
$$
I_0\longrightarrow I_2,\qquad
I_1\longrightarrow I_1,I_2,\qquad
I_2\longrightarrow I_0,
$$
so
$$
A=\begin{pmatrix}0&0&1\\0&1&1\\1&0&0\end{pmatrix}.
$$
The recurrent pieces of this <directed covering graph of an interval map>[covering graph] are the two-cycle $I_0\leftrightarrow I_2$ and the loop at $I_1$; neither contains two competing closed routes. Also
$$
\operatorname{tr}(A^3)=\operatorname{tr}A=1,
$$
so this ordering forces no 3-cycle.

The bound is attained by the <piecewise-linear function>
$$
F(x)=
\begin{cases}
x+\dfrac23,&0\leq x\leq\dfrac13,\\[3pt]
\dfrac53-2x,&\dfrac13\leq x\leq\dfrac23,\\[3pt]
1-x,&\dfrac23\leq x\leq1.
\end{cases}
$$
Indeed,
$$
0\mapsto\frac23\mapsto\frac13\mapsto1\mapsto0
$$
is a 4-cycle with the required spatial order. On $I_0\cup I_2$, the fourth iterate is the <identity function>. Within $I_1$, the fixed point is $5/9$ and
$$
F(x)-\frac59=-2\left(x-\frac59\right),
$$
so every other point eventually leaves $I_1$ and enters $I_0\cup I_2$. The map therefore has no horseshoe in any iterate and is not <Glendinning chaos>[chaotic]. The minimum number of distinct 3-cycles is consequently $\boxed0$.

Solved by gpt-5.6-sol high.

= 34E
{parent=Paper 2}
{scope}
{title2=Integrable Systems}

= Solution
{parent=34E}

For $N=1$, write
$$
q=\exp\left(-2lX-\frac{T}{2l}\right).
$$
The given evolution of the <discrete scattering data> gives
$$
\lambda_1=il,
\qquad
c_1(T)=2l\exp\left(-\frac{T}{2l}\right).
$$
Solving the resulting two scalar linear equations for the components of $\psi_1$ and substituting in the reconstruction formula gives
$$
u_X=-\frac{8lq}{1+q^2}.
$$
Choosing the additive multiple of $2\pi$ so that $u\to0$ as $X\to+\infty$ and <integral>[integrating] in $X$ yields the <one-soliton solution of the sine-Gordon equation in light-cone coordinates>
$$
\boxed{u(X,T)=4\arctan\exp\left(-2lX-\frac{T}{2l}\right)}.
$$
It depends only on $X+T$ exactly when the two positive coefficients in the exponent agree:
$$
2l=\frac1{2l}.
$$
The unique positive solution is $l=1/2$, and then
$$
u(X,T)=F(X+T),
\qquad
\boxed{F(z)=4\arctan(e^{-z})}.
$$

The transformations satisfy $g^s g^r=g^{s+r}$, $g^0$ is the <identity map>, and $(g^s)^{-1}=g^{-s}$, so they form a <group>[one-parameter group]. If
$$
U(X,T)=u(e^{-s}X,e^sT),
$$
then the <chain rule> gives
$$
U_{XT}=e^{-s}e^s u_{XT}=\sin U.
$$
Thus $g^s$ is a <Lie point symmetry>. Applied to the one-soliton family, it replaces $l$ by $l'=e^{-s}l$. Taking $e^s=2l$ gives $l'=1/2$, so every member is transformed to the function $F(X+T)$ found above.

For the stated $N=2$ solution, set $x=X+T$ and $t=T-X$. When $l^2+m^2=1/4$, its two arguments reduce to
$$
2mX-\frac{2mT}{4(l^2+m^2)}=-2mt,
\qquad
\frac{2lT}{4(l^2+m^2)}+2lX=2lx.
$$
At fixed $x$, only $\sin(2mt)$ varies, so the <Sine-Gordon breather> has fundamental period
$$
\boxed{\mathcal T=\frac{\pi}{m}}.
$$

Finally put $r^2=l^2+m^2$. Under the same symmetry, the parameters become
$$
l'=e^{-s}l,
\qquad
m'=e^{-s}m,
$$
and hence $(l')^2+(m')^2=e^{-2s}r^2$. Choosing $e^s=2r$ makes this sum $1/4$, proving that every solution in the family is symmetry-equivalent to the normalized breather.

Solved by gpt-5.6-sol high.

= 35A
{parent=Paper 2}
{scope}
{title2=Principles of Quantum Mechanics}

= a
{parent=35a}
{scope}

= Solution
{parent=a}

Since $\mathbf S=(\hbar/2)\boldsymbol\sigma$,
$$
\mathbf n\mathbin{\cdot}\mathbf S
=\frac\hbar2
\begin{pmatrix}
\cos\theta&\sin\theta\\
\sin\theta&-\cos\theta
\end{pmatrix}.
$$
Its normalized <spin one-half along an axis>[eigenstates] are
$$
\boxed{|\uparrow_\theta\rangle
=\cos\frac\theta2|\uparrow\rangle
+\sin\frac\theta2|\downarrow\rangle},
\qquad E_\uparrow=\frac\hbar2,
$$
and
$$
\boxed{|\downarrow_\theta\rangle
=-\sin\frac\theta2|\uparrow\rangle
+\cos\frac\theta2|\downarrow\rangle},
\qquad E_\downarrow=-\frac\hbar2.
$$

Solved by gpt-5.6-sol high.

= b
{parent=35a}
{scope}

= Solution
{parent=b}

Write $c=\cos(\theta/2)$ and $s=\sin(\theta/2)$. Then
$$
|\uparrow_\theta\downarrow_\theta\rangle
-|\downarrow_\theta\uparrow_\theta\rangle
=(c^2+s^2)
\bigl(|\uparrow\downarrow\rangle-|\downarrow\uparrow\rangle\bigr).
$$
Thus $|\psi\rangle$ is unchanged. It is the <spin-one-half singlet state>, so joint rotations leave it invariant. Equivalently, direct use of the <Pauli matrix>[Pauli matrices] shows
$$
(S_i^{(1)}+S_i^{(2)})|\psi\rangle=0
\qquad(i=x,y,z).
$$
It is therefore an <eigenstate> of every Cartesian component of the combined spin, with eigenvalue zero.

Solved by gpt-5.6-sol high.

= c
{parent=35a}
{scope}

= Solution
{parent=c}

Using the states from part (a),
$$
|\chi\rangle=\frac1{\sqrt2}
\left[
-s|\uparrow\uparrow\rangle
+c|\uparrow\downarrow\rangle
-c|\downarrow\uparrow\rangle
-s|\downarrow\downarrow\rangle
\right].
$$
The <Born rule> therefore gives
$$
\begin{array}{c|c}
\text{outcome}&\text{probability}\\ \hline
(+\hbar/2,+\hbar/2)&\frac12\sin^2(\theta/2)\\
(+\hbar/2,-\hbar/2)&\frac12\cos^2(\theta/2)\\
(-\hbar/2,+\hbar/2)&\frac12\cos^2(\theta/2)\\
(-\hbar/2,-\hbar/2)&\frac12\sin^2(\theta/2).
\end{array}
$$
After the equal-sign outcomes the state is respectively $|\uparrow\uparrow\rangle$ or $|\downarrow\downarrow\rangle$. These are <spin-one-half triplet state>[triplet states], hence <eigenstates> of $\mathbf S^2$ with eigenvalue
$$
\boxed{2\hbar^2}.
$$
The opposite-sign product states are each a nontrivial <linear combination> of the singlet and the $m=0$ triplet, so they are not eigenstates of $\mathbf S^2$.

Solved by gpt-5.6-sol high.

= 36D
{parent=Paper 2}
{scope}
{title2=Applications of Quantum Mechanics}

= Solution
{parent=36D}

Let $G_n=2\pi n/a$. The free states $|k\rangle$ and $|k-G_n\rangle$ are degenerate at the Bragg point $k=G_n/2=n\pi/a$. Since
$$
\langle k|V|k-G_n\rangle=V_n,
$$
<degenerate perturbation theory> restricts the Hamiltonian near this crossing to
$$
\begin{pmatrix}
\hbar^2k^2/(2m)+V_0&V_n\\
V_n^*&\hbar^2(k-G_n)^2/(2m)+V_0
\end{pmatrix}.
$$
Writing $k=n\pi/a+\kappa$ and diagonalizing gives the <nearly-free electron dispersion near a one-dimensional band gap>
$$
\boxed{
E_\pm(k)=V_0+\frac{\hbar^2}{2m}
\left[\left(\frac{n\pi}{a}\right)^2+\kappa^2\right]
\pm\sqrt{
\left(\frac{\hbar^2n\pi\kappa}{ma}\right)^2+|V_n|^2}}.
$$
Thus the two continuous <energy bands>[bands] avoid crossing, and at $\kappa=0$ their separation is
$$
\boxed{\Delta E_n=2|V_n|}.
$$
This is the <nearly-free electron model>. The <dispersion relation> is the relation $E=E(k)$ between energy and Bloch wavevector.

For the specified potential,
$$
\frac83V_0\cos^4\left(\frac{2\pi x}{a}\right)
=V_0+\frac43V_0\cos\left(\frac{4\pi x}{a}\right)
+\frac13V_0\cos\left(\frac{8\pi x}{a}\right).
$$
Hence
$$
V_{\pm2}=\frac23V_0,\qquad
V_{\pm4}=\frac16V_0,
$$
and all other nonconstant Fourier coefficients vanish. The gaps are therefore
$$
\begin{array}{c|c|c}
n&\text{crossing wavevectors}&\text{gap width}\\ \hline
2&k=\pm 2\pi/a&4|V_0|/3\\
4&k=\pm 4\pi/a&|V_0|/3.
\end{array}
$$
Their centre energies are respectively
$$
V_0+\frac{2\hbar^2\pi^2}{ma^2},
\qquad
V_0+\frac{8\hbar^2\pi^2}{ma^2}.
$$
The extended-zone sketch consists of the free-particle parabolas shifted upward by $V_0$, with avoided crossings of these two widths at the listed Bragg points; elsewhere they meet to this order because the corresponding Fourier coefficients vanish.

Solved by gpt-5.6-sol high.

= 37A
{parent=Paper 2}
{scope}
{title2=Statistical Physics}

= a
{parent=37a}
{scope}

= Solution
{parent=a}

The <microcanonical ensemble> describes an isolated system with fixed energy, volume, and particle number. It assigns equal probabilities to the accessible microstates in the chosen narrow energy shell. The <canonical ensemble> describes a system in thermal contact with a large heat reservoir: its temperature, volume, and particle number are fixed, while its energy can fluctuate.

Solved by gpt-5.6-sol high.

= b
{parent=37a}
{scope}

= Solution
{parent=b}

For the <Gibbs entropy>
$$
S=-k_B\sum_n p_n\log p_n,
$$
first maximize subject only to $\sum_np_n=1$ over $\Omega$ accessible microstates. A <Lagrange multiplier> gives
$$
-(\log p_n+1)+\alpha=0,
$$
so every $p_n$ is equal. Normalization yields the <microcanonical ensemble>
$$
\boxed{p_n=\frac1\Omega}.
$$

For the canonical ensemble, impose both normalization and the mean-energy constraint $\sum_np_nE_n=E$. Stationarity of
$$
-\sum_np_n\log p_n-\alpha\sum_np_n-\beta\sum_np_nE_n
$$
gives $p_n=C e^{-\beta E_n}$. The <canonical partition function> fixes $C$, so
$$
\boxed{p_n=\frac{e^{-\beta E_n}}{Z}},
\qquad
Z=\sum_ne^{-\beta E_n}.
$$
Thermodynamic consistency identifies $\beta=(k_BT)^{-1}$. This is the <maximum-entropy derivation of equilibrium ensembles>.

Solved by gpt-5.6-sol high.

= c
{parent=37a}
{scope}

= Solution
{parent=c}

Substitution of the canonical probabilities into the <Gibbs entropy> gives
$$
S=-k_B\sum_np_n(-\beta E_n-\log Z)
=k_B\log Z+k_B\beta E.
$$
Hence
$$
\boxed{S=k_B\log Z+\frac ET}.
$$

Solved by gpt-5.6-sol high.

= d
{parent=37a}
{scope}

= Solution
{parent=d}

Put $x=\beta\epsilon$. The one-particle <canonical partition function> is
$$
z=e^x+1+e^{-x}=1+2\cosh x,
$$
and independence gives $Z=z^N$. Therefore the <independent symmetric three-level system> has
$$
\boxed{E=-\frac{\partial\log Z}{\partial\beta}
=-\frac{2N\epsilon\sinh x}{1+2\cosh x}}
$$
and, using part (c),
$$
\boxed{
S=Nk_B\left[
\log(1+2\cosh x)
-\frac{2x\sinh x}{1+2\cosh x}
\right]}.
$$
As $T\to\infty$, $x\to0$, all three levels become equally likely and
$$
E\to0,\qquad S\to Nk_B\log3.
$$
As $T\to0^+$, every particle occupies the nondegenerate ground level $-\epsilon$, and
$$
E\to-N\epsilon,\qquad S\to0.
$$

Solved by gpt-5.6-sol high.

= e
{parent=37a}
{scope}

= Solution
{parent=e}

Because the one-particle spectrum is bounded above, a <negative temperature> is possible. A population-inverted configuration with more particles in the $+\epsilon$ level than in the $-\epsilon$ level has
$$
\frac{N_+}{N_-}=e^{-2\beta\epsilon}>1,
$$
which requires $\beta<0$. Equivalently, above the maximum-entropy energy $E=0$, adding energy reduces the number of compatible microstates, so $\partial S/\partial E=1/T<0$. The limiting configuration with every particle at $+\epsilon$ has $E=N\epsilon$, zero entropy, and $T\to0^-$.

Solved by gpt-5.6-sol high.

= 38D
{parent=Paper 2}
{scope}
{title2=General Relativity}

= a
{parent=38d}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

For
$$
g_{\theta\theta}=1,\qquad g_{\phi\phi}=\sin^2\theta,
$$
the nonzero <Christoffel symbols> needed here are
$$
\Gamma^\theta_{\phi\phi}=-\sin\theta\cos\theta,
\qquad
\Gamma^\phi_{\theta\phi}
=\Gamma^\phi_{\phi\theta}=\cot\theta.
$$
On a line of constant longitude, $\dot\phi=0$, and the <geodesic equation> reduces to $\ddot\theta=0$ after choosing an <affine parameter>. Thus every meridian is a geodesic.

For a nonconstant line of constant latitude, $\dot\theta=0$ and the $\theta$ equation requires
$$
-\sin\theta\cos\theta\,\dot\phi^2=0.
$$
Away from the coordinate-degenerate poles, this holds only when
$$
\boxed{\theta=\pi/2},
$$
so only the equator is a constant-latitude geodesic.

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

Along the constant latitude, use $\phi$ as parameter. The <parallel transport> equations are
$$
\frac{dV^\theta}{d\phi}
-\sin\theta\cos\theta\,V^\phi=0,
\qquad
\frac{dV^\phi}{d\phi}
+\cot\theta\,V^\theta=0.
$$
With $W=\sin\theta\,V^\phi$, these become
$$
\frac{dV^\theta}{d\phi}=\cos\theta\,W,
\qquad
\frac{dW}{d\phi}=-\cos\theta\,V^\theta.
$$
The initial components $(1,0)$ therefore evolve as
$$
V^\theta(\phi)=\cos(\phi\cos\theta),
\qquad
V^\phi(\phi)=-\frac{\sin(\phi\cos\theta)}{\sin\theta}.
$$
After one circuit, the <parallel transport around a latitude of the unit sphere> gives
$$
\boxed{
(V^\theta,V^\phi)
=\left(
\cos(2\pi\cos\theta),
-\frac{\sin(2\pi\cos\theta)}{\sin\theta}
\right)}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=38d}
{scope}

= Solution
{parent=b}

Contracting the stated <Einstein field equations> with $g^{\alpha\beta}$ in $n$ dimensions gives
$$
T=R-\frac n2R=-\frac{n-2}{2}R.
$$
Thus $R=-2T/(n-2)$, and substitution back yields the <trace-reversed Einstein field equations in D dimensions>
$$
\boxed{
R_{\alpha\beta}
=T_{\alpha\beta}-\frac1{n-2}g_{\alpha\beta}T
},
\qquad n>2.
$$

Solved by gpt-5.6-sol high.

= c
{parent=38d}
{scope}

= Solution
{parent=c}

At any point choose orthonormal coordinates. The antisymmetries and pair symmetry of the <Riemann curvature tensor> leave only $R_{1212}$ independent. In these coordinates,
$$
R_{11}=R_{1212},\qquad
R_{22}=R_{1212},\qquad
R_{12}=0,
$$
while $R=2R_{1212}$. Since this is a tensor identity, it follows in every coordinate system that the <Ricci tensor in two dimensions> satisfies
$$
\boxed{R_{\alpha\beta}=\frac12g_{\alpha\beta}R}.
$$
The <Einstein tensor> consequently vanishes identically. Hence the two-dimensional vacuum field equation contains no local gravitational dynamics, while coupling the unmodified equation to matter would require $T_{\alpha\beta}=0$. This is the basic degeneracy of <two-dimensional general relativity>.

Solved by gpt-5.6-sol high.

= 39C
{parent=Paper 2}
{scope}
{title2=Fluid Dynamics II}

= a
{parent=39c}
{scope}

= Solution
{parent=a}

The <Reynolds number> is
$$
\boxed{\operatorname{Re}=\frac{\mathcal U\mathcal L}{\nu}}.
$$
When it is large, viscosity is negligible in most of the flow, but the inviscid outer solution cannot generally satisfy the <no-slip boundary condition> at a rigid wall. A thin <boundary layer> resolves this mismatch: tangential gradients remain of outer scale $\mathcal L$, normal gradients have the much shorter scale $\delta$, and the small viscous coefficient is offset by $u_{yy}=O(\mathcal U/\delta^2)$. Matching a viscous inner solution to an inviscid outer solution retains the leading viscous effects without solving the full equations uniformly everywhere.

Solved by gpt-5.6-sol high.

= b
{parent=39c}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

The wall speed sets
$$
\boxed{U(x)=\alpha x}.
$$
Using $x$ as the local streamwise length gives
$$
\operatorname{Re}_x=\frac{U(x)x}{\nu}
=\frac{\alpha x^2}{\nu},
$$
which is large when $x\gg\sqrt{\nu/\alpha}$. The <boundary-layer scaling>
$$
\frac{U^2}{x}\sim\frac{\nu U}{\delta^2}
$$
then gives
$$
\delta(x)\sim\sqrt{\frac{\nu x}{U(x)}}
=\boxed{\sqrt{\frac{\nu}{\alpha}}},
$$
so the stretching-sheet layer has constant thickness to leading order.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

With zero imposed pressure gradient, the <two-dimensional boundary-layer equations> are
$$
u_x+v_y=0,
\qquad
uu_x+vu_y=\nu u_{yy}.
$$
Introduce the <stream function> by
$$
u=\psi_y,\qquad v=-\psi_x.
$$
The boundary conditions are
$$
\psi_y(x,0)=\alpha x,\qquad
\psi_x(x,0)=0,\qquad
\psi_y(x,y)\to0\quad(y\to\infty).
$$
The additive constant in $\psi$ may be chosen so that $\psi(x,0)=0$.

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

Put
$$
\delta=\sqrt{\frac{\nu}{\alpha}},
\qquad
\eta=\frac y\delta,
\qquad
\psi=\alpha x\delta f(\eta).
$$
Then
$$
u=\alpha x f'(\eta),
\qquad
v=-\alpha\delta f(\eta).
$$
Substitution in the boundary-layer momentum equation, using $\nu/\delta^2=\alpha$, gives
$$
\boxed{f'''+ff''-(f')^2=0}.
$$
The wall and far-field conditions become
$$
\boxed{f(0)=0,\qquad f'(0)=1,\qquad f'(\infty)=0}.
$$
These are the similarity equations for the <boundary layer over a linearly stretching sheet>.

Solved by gpt-5.6-sol high.

= iv
{parent=b}
{scope}

= Solution
{parent=iv}

Set $f=A+Be^{-C\eta}$ with $C>0$. The differential equation reduces to
$$
BC^2(A-C)e^{-C\eta}=0,
$$
so $A=C$. The two wall conditions give
$$
A+B=0,\qquad -BC=1.
$$
Hence $C^2=1$, and positivity of $C$ yields
$$
\boxed{A=1,\qquad B=-1,\qquad C=1},
$$
or
$$
\boxed{f(\eta)=1-e^{-\eta}}.
$$

Solved by gpt-5.6-sol high.

= i
{disambiguate=final}
{parent=b}
{scope}

= Solution
{parent=i-final}

As $\eta\to\infty$,
$$
f(\eta)\to1,\qquad f'(\eta)\to0.
$$
Thus the tangential velocity matches the quiescent outer fluid, but the normal velocity tends to
$$
\boxed{v\to-\alpha\delta=-\sqrt{\alpha\nu}}.
$$
The stretching sheet entrains fluid toward itself at a constant leading-order speed. Consequently a globally stationary exterior flow is impossible: the outer flow must provide this normal influx and complete the mass balance away from the local boundary-layer approximation.

Solved by gpt-5.6-sol high.

= 40C
{parent=Paper 2}
{scope}
{title2=Waves}

= a
{parent=40c}
{scope}

= Solution
{parent=a}

Taking the divergence of the <elastic wave in an isotropic solid>[Cauchy momentum equation] and commuting constant-coefficient derivatives gives
$$
\rho_0\frac{\partial^2}{\partial t^2}(\nabla\cdot\mathbf u)
=(\lambda+2\mu)\nabla^2(\nabla\cdot\mathbf u).
$$
Thus the dilatation is a wave of speed
$$
\boxed{c_P=\sqrt{\frac{\lambda+2\mu}{\rho_0}}}.
$$
Taking the curl, and using $\nabla\times\nabla f=0$, gives
$$
\rho_0\frac{\partial^2}{\partial t^2}(\nabla\times\mathbf u)
=\mu\nabla^2(\nabla\times\mathbf u),
$$
so the rotation propagates at
$$
\boxed{c_S=\sqrt{\frac{\mu}{\rho_0}}}.
$$
These are respectively the speeds of <P-wave>[P-waves] and <S-wave>[S-waves].

Solved by gpt-5.6-sol high.

= b
{parent=40c}
{scope}

= Solution
{parent=b}

All wavevectors lie in the $xz$ plane and have the same tangential component. The incident P-wave travels upward toward $z=0$. In the lower solid there are a downward reflected P-wave, polarized parallel to its wavevector, and a downward reflected SV-wave, polarized in the $xz$ plane perpendicular to its wavevector. In the upper solid there are corresponding upward transmitted P- and SV-waves. The SH polarization decouples from this in-plane incident field. Their angles obey the <Snell law for elastic and acoustic waves>.

For a welded interface, the two in-plane displacement components are continuous:
$$
[u_x]=[u_z]=0.
$$
The two in-plane traction components are also continuous:
$$
[\sigma_{xz}]=[\sigma_{zz}]=0.
$$
These four independent scalar conditions determine the four reflected and transmitted amplitudes. This is <mode conversion at a planar elastic interface>.

Solved by gpt-5.6-sol high.

= c
{parent=40c}
{scope}

= i
{parent=c}
{scope}

= Solution
{parent=i}

Tangential phase matching gives
$$
k\sin\theta=k'\sin\theta',
\qquad
\boxed{\frac{\sin\theta}{c_P}
=\frac{\sin\theta'}{c'_P}}.
$$
A transmitted propagating wave exists when its normal wavenumber is real, equivalently
$$
\boxed{\left|\frac{c'_P}{c_P}\sin\theta\right|\leq1}.
$$
For incidence angles between zero and $\pi/2$, this is automatic if $c'_P\leq c_P$ and imposes the corresponding critical-angle bound if $c'_P>c_P$.

Solved by gpt-5.6-sol high.

= ii
{parent=c}
{scope}

= Solution
{parent=ii}

Suppressing the common tangential phase, suitable displacement fields are
$$
\mathbf u_I=
(\sin\theta,0,\cos\theta)
e^{i[k(\sin\theta\,x+\cos\theta\,z)-\omega t]},
$$
$$
\mathbf u_R=
R(\sin\theta,0,-\cos\theta)
e^{i[k(\sin\theta\,x-\cos\theta\,z)-\omega t]},
$$
and
$$
\mathbf u_T=
T(\sin\theta',0,\cos\theta')
e^{i[k'(\sin\theta'\,x+\cos\theta'\,z)-\omega t]}.
$$
For an inviscid elastic liquid, $\mu=0$, $c_P^2=\lambda/\rho$, and $\boldsymbol\sigma=\lambda(\nabla\cdot\mathbf u)I$. Continuity of normal displacement and normal traction at $z=0$ gives
$$
\cos\theta(1-R)=T\cos\theta',
\qquad
\lambda k(1+R)=\lambda'k'T.
$$
Together with the phase-matching relation from part (i), these determine $R$ and $T$.

Solved by gpt-5.6-sol high.

= iii
{parent=c}
{scope}

= Solution
{parent=iii}

Eliminating $T$ from the two boundary conditions gives
$$
\frac{1+R}{1-R}
=\frac{\lambda'k'\cos\theta}
{\lambda k\cos\theta'}.
$$
The <Snell law for elastic and acoustic waves> implies
$$
\frac{k'}k=\frac{\sin\theta}{\sin\theta'}.
$$
Therefore the <displacement reflection from an interface between two inviscid elastic liquids> is
$$
\boxed{
R=\frac{\lambda'\sin2\theta-\lambda\sin2\theta'}
{\lambda'\sin2\theta+\lambda\sin2\theta'}}.
$$

Solved by gpt-5.6-sol high.

= iv
{parent=c}
{scope}

= Solution
{parent=iv}

No reflection requires
$$
\lambda'\sin2\theta=\lambda\sin2\theta'.
$$
Using $\lambda=\rho c_P^2$, $\lambda'=\rho'(c'_P)^2$, and
$\sin\theta'=(c'_P/c_P)\sin\theta$, this reduces to the normal <acoustic impedance> matching condition
$$
\rho'c'_P\cos\theta=\rho c_P\cos\theta'.
$$
After squaring and eliminating $\theta'$, the required incidence angle satisfies
$$
\boxed{
\sin^2\theta=
\frac{\rho'^2(c'_P)^2-\rho^2c_P^2}
{(c'_P)^2(\rho'^2-\rho^2)}
}.
$$
Such a no-reflection angle exists only when the right-hand side lies in $[0,1]$ and the transmitted wave is propagating.

Solved by gpt-5.6-sol high.

= 41C
{parent=Paper 2}
{scope}
{title2=Numerical Analysis}

= a
{parent=41c}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

Multiplication by $e^{-im\theta}$ and summation over $m\in\mathbb Z$ give
$$
\sum_m e^{-im\theta}u^n_{m+k}
=e^{ik\theta}\widehat u^{\,n}(\theta).
$$
Thus the recurrence becomes
$$
A(\theta)\widehat u^{\,n+1}(\theta)
=B(\theta)\widehat u^{\,n}(\theta),
$$
where
$$
A(\theta)=\sum_{k=r}^sa_ke^{ik\theta},
\qquad
B(\theta)=\sum_{k=r}^sb_ke^{ik\theta}.
$$
Assuming $A(\theta)\ne0$, the <amplification factor of a two-sided one-step stencil> is
$$
\boxed{
H(\theta)=\frac{B(\theta)}{A(\theta)}
=\frac{\sum_{k=r}^sb_ke^{ik\theta}}
{\sum_{k=r}^sa_ke^{ik\theta}}
}.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

Iteration gives
$$
\widehat u^{\,n}(\theta)=H(\theta)^n\widehat u^{\,0}(\theta).
$$
If $|H(\theta)|\leq1$ everywhere, the <Parseval identity> gives
$$
\|u^n\|_{\ell^2}^2
=\frac1{2\pi}\int_{-\pi}^{\pi}
|H(\theta)|^{2n}|\widehat u^{\,0}(\theta)|^2\,d\theta
\leq\|u^0\|_{\ell^2}^2.
$$
Conversely, if $|H(\theta_0)|>1$, continuity supplies a neighbourhood on which $|H|\geq1+\epsilon$. Choose nonzero square-integrable Fourier data supported there. Its norm then grows at least as $(1+\epsilon)^n$, contradicting boundedness. Hence
$$
\boxed{\{u^n\}\text{ is bounded for every }u^0
\iff |H(\theta)|\leq1\text{ for all }\theta}.
$$
This is the one-step <Von Neumann stability analysis> criterion.

Solved by gpt-5.6-sol high.

= b
{parent=41c}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

The first scheme has
$$
H(\theta)=1+\mu(1-e^{-i\theta}).
$$
At $\theta=\pi$,
$$
|H(\pi)|=1+2\mu.
$$
Since the <Courant number> obeys $\mu\geq0$, stability therefore holds exactly for
$$
\boxed{\mu=0}.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

For the second scheme,
$$
\boxed{
H(\theta)
=1-\mu^2(1-\cos\theta)-i\mu\sin\theta
}.
$$
A direct calculation gives
$$
|H(\theta)|^2
=1-4\mu^2(1-\mu^2)\sin^4\frac\theta2.
$$
This is at most one for every $\theta$ exactly when
$$
\boxed{0\leq\mu\leq1}.
$$

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

The centred second difference has Fourier symbol
$$
e^{-i\theta}-2+e^{i\theta}
=-4\sin^2\frac\theta2.
$$
Writing $q=\sin^2(\theta/2)$ gives
$$
\boxed{
H(\theta)=
\frac{1-2(\mu+\alpha)q}
{1+2(\mu-\alpha)q}
}.
$$
For the denominator to stay positive and $|H|\leq1$, one needs
$$
-\bigl[1+2(\mu-\alpha)q\bigr]
\leq1-2(\mu+\alpha)q
\leq1+2(\mu-\alpha)q.
$$
The right inequality follows from $\mu\geq0$, while the left is equivalent to $2-4\alpha q\geq0$. Hence stability for every Fourier mode is
$$
\boxed{0\leq\alpha\leq\frac12,\qquad \mu\geq0},
$$
apart from the singular endpoint $(\mu,\alpha)=(0,1/2)$, where both sides annihilate the $\theta=\pi$ mode and the recurrence does not determine its next value.

Solved by gpt-5.6-sol high.