past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ii/paper-2.bigb

= Paper 2
{scope}

https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2023/PaperII_2.pdf

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

= Solution
{parent=1G}

The <Lagrange root bound over a field> says that a nonzero <polynomial> of <degree of a polynomial>[degree] $r$ over a <field> has at most $r$ roots. In particular, a polynomial <modular congruence> of degree $r$ modulo a <prime number> has at most $r$ incongruent solutions unless all its coefficients vanish modulo that prime.

Suppose that $d$ is good and that the positive <integer divisor> $e$ divides $d$. The $d$ roots of $X^d-1$ form a finite multiplicative <subgroup> $H$ of $\mathbb F_p^\times$. By the fact that every <finite multiplicative subgroup of a field is cyclic>, $H=\langle g\rangle$ is a <cyclic group> of order $d$. The solutions in $H$ of $x^e=1$ are
$$
1,g^{d/e},g^{2d/e},\ldots,g^{(e-1)d/e},
$$
so there are at least $e$ of them. The <Lagrange root bound over a field> gives at most $e$ roots in all of $\mathbb F_p$, hence exactly $e$. Thus every divisor of a good number is good.

Now put $n=pq$. By the <Chinese remainder theorem for unit groups>, a base $b$ is a <Fermat pseudoprime>[Fermat-pseudoprime] base precisely when its two components satisfy
$$
b^{n-1}=1\quad\hbox{in }\mathbb F_p^\times
\qquad\hbox{and}\qquad
b^{n-1}=1\quad\hbox{in }\mathbb F_q^\times.
$$
The <power roots in a finite field> and
$$
\gcd(n-1,p-1)=\gcd(q-1,p-1)=10,
\qquad
\gcd(n-1,q-1)=\gcd(p-1,q-1)=10
$$
show that each component has ten choices. Therefore there are
$$
10\cdot10=100
$$
Fermat-pseudoprime bases.

To impose the <strong pseudoprime> condition, write $n-1=2^s m$ with $m$ odd. Since $10\mid n-1$, we have $5\mid m$. In each cyclic group of ten Fermat components, raising to the $m$th power sends five elements to $1$ and five elements to $-1$. A pair of components passes the strong test exactly when their signs agree: the pair $(1,1)$ satisfies the first alternative, and $(-1,-1)$ satisfies the second at $j=0$. Components of opposite sign become $(1,1)$ after squaring and can never jointly equal $-1$. Hence the number of strong-pseudoprime bases is
$$
5^2+5^2=50.
$$

Solved by gpt-5.6-sol high.

= 2F
{parent=Paper 2}
{scope}
{title2=Topics In Analysis}

= Solution
{parent=2F}

For a continuous closed path $\gamma:[0,1]\to\mathbb C\setminus\{0\}$, choose a continuous argument lift
$$
\frac{\gamma(t)}{|\gamma(t)|}=e^{2\pi i\theta(t)}.
$$
Its <winding number of a continuous closed path> about zero is
$$
w(\gamma,0)=\theta(1)-\theta(0)\in\mathbb Z.
$$
For a piecewise smooth path this equals $(2\pi i)^{-1}\int_\gamma dz/z$.

If $|\gamma(t)|>|\phi(t)|$, then
$$
H(s,t)=\gamma(t)+s\phi(t),\qquad 0\leq s\leq1,
$$
never vanishes, since $|H(s,t)|\geq|\gamma(t)|-s|\phi(t)|>0$. Thus $H$ is a homotopy through closed paths avoiding zero. By <homotopy invariance of winding number>, or directly by the <dominated perturbation preserves winding number>[dominated-perturbation lemma],
$$
w(\gamma+\phi,0)=w(\gamma,0).
$$

More generally, $\gamma_0$ and $\gamma_1$ are homotopic by paths in $\Gamma$ when there is a <continuous function>
$$
H:[0,1]^2\longrightarrow\mathbb C\setminus\{0\}
$$
with $H(0,t)=\gamma_0(t)$, $H(1,t)=\gamma_1(t)$, and $H(s,0)=H(s,1)$ for every $s$. The winding-number theorem states that such a homotopy implies
$$
w(\gamma_0,0)=w(\gamma_1,0).
$$

For the <Fundamental theorem of algebra>, let $P(z)=a_nz^n+\cdots+a_0$ with $n\geq1$. For sufficiently large $R$,
$$
|a_nR^ne^{2\pi int}|>
|a_{n-1}R^{n-1}e^{2\pi i(n-1)t}+\cdots+a_0|
$$
for every $t$. The dominated-perturbation lemma shows that $t\mapsto P(Re^{2\pi it})$ has the same winding number as $t\mapsto a_nR^ne^{2\pi int}$, namely $n$. If $P$ had no zero, however,
$$
H(s,t)=P(sRe^{2\pi it})
$$
would be a homotopy in $\Gamma$ from that loop to the constant loop $P(0)$, whose winding number is zero. This contradiction proves that $P$ has a complex root. This is the <winding-number proof of the fundamental theorem of algebra>.

Finally suppose that a continuous retraction $r:D^2\to S^1$ existed. The boundary loop $\eta(t)=e^{2\pi it}$ has winding number one, while $(1-s)\eta(t)$ contracts it to zero inside the <disc>. Composing this contraction with $r$ gives a homotopy through loops in $S^1$ from $r\circ\eta=\eta$ to the constant loop $r(0)$. Their winding numbers are respectively one and zero, contradicting homotopy invariance. Hence there is no such retraction, as in the <winding-number proof of the no-retraction theorem>.

Solved by gpt-5.6-sol high.

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

= a
{parent=3i}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

A <Bernoulli source> is a sequence $(X_n)_{n\geq1}$ of <independent and identically distributed random variables> with a common <probability mass function> on the finite <alphabet> $\mathcal A$.

The source is <reliable source encoding at a rate>[reliably encodable at rate] $r$ if, for each block length $n$, there are an encoder with at most $2^{nr}$ outputs and a decoder such that the block error probability
$$
P\bigl(\widehat X_1^n\ne X_1^n\bigr)
$$
tends to zero as $n\to\infty$.

Its <information rate> is the limiting <information entropy> per source symbol,
$$
h=\lim_{n\to\infty}\frac1nH(X_1,\ldots,X_n),
$$
when this <limit of a function>[limit] exists.

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

For a <Bernoulli source>, independence makes entropy additive, so
$$
H(X_1,\ldots,X_n)=\sum_{j=1}^nH(X_j)=nH(X_1).
$$
Its <information rate> is therefore $H(X_1)$. If $c$ is an <optimal prefix code> and $L(c)=\mathbb E|c(X_1)|$ is its <expected codeword length>, the <Shannon noiseless coding theorem> gives
$$
H(X_1)\leq L(c)<H(X_1)+1.
$$
Consequently the information rate is at most the expected word length of the optimal code.

Solved by gpt-5.6-sol high.

= b
{parent=3i}
{scope}

= Solution
{parent=b}

The first $m$ letters of the blocked source contain exactly the first $mN$ letters of the original source, so
$$
H(Y_1,\ldots,Y_m)=H(X_1,\ldots,X_{mN}).
$$
If the original <information rate> is $h$, then
$$
\lim_{m\to\infty}\frac1mH(Y_1,\ldots,Y_m)
=N\lim_{m\to\infty}\frac1{mN}H(X_1,\ldots,X_{mN})
=Nh.
$$
Thus fixed-length blocking multiplies information rate by $N$, exactly as recorded by <information rate under fixed-length blocking>.

Solved by gpt-5.6-sol high.

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

= a
{parent=4i}
{scope}

= Solution
{parent=a}

A subset $A\subseteq\Sigma^*$ is a <computably enumerable set> if there is a program, equivalently a <register machine>, that halts on input $w$ exactly when $w\in A$. It need not halt when $w\notin A$.

Solved by gpt-5.6-sol high.

= b
{parent=4i}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

Statement (i) implies (ii) directly from the definition: if $A$ is <computably enumerable set>[computably enumerable], its semidecision program computes a <partial computable function> whose domain is exactly $A$; for example, return the empty word whenever the program halts.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

Suppose $A=\operatorname{dom}f$ for a <partial computable function> $f$. Define
$$
g(w)=w\quad\hbox{if }f(w)\mathbin\downarrow,
$$
and leave $g(w)$ undefined otherwise. A program computes $g$ by running $f(w)$ and returning $w$ if that computation halts. Hence $g$ is partial computable and
$$
\operatorname{ran}g=A,
$$
which proves (ii)$\Rightarrow$(iii).

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

Suppose $A=\operatorname{ran}f$ for a <partial computable function> $f$. Since $A\ne\varnothing$, fix $a_0\in A$. Decode each input $w$ as a pair $((w)_0,(w)_1)=(u,t)$. Using the <truncated computation function>, define
$$
h(w)=
\begin{cases}
f(u),&\text{if the computation of $f(u)$ halts within $t$ steps},\\
a_0,&\text{otherwise}.
\end{cases}
$$
This is a <total computable function>, and every output lies in $A$. Conversely, if $a=f(u)$, choosing $t$ at least the halting time gives an input $w$ with $h(w)=a$. Thus $\operatorname{ran}h=A$, proving (iii)$\Rightarrow$(iv).

Solved by gpt-5.6-sol high.

= iv
{parent=b}
{scope}

= Solution
{parent=iv}

Suppose $A=\operatorname{ran}h$ for a <total computable function> $h$. Effectively enumerate the words as $w_0,w_1,\ldots$. On input $x$, compute
$$
h(w_0),h(w_1),\ldots
$$
in order and halt as soon as one equals $x$. Every individual computation terminates because $h$ is total. The search halts exactly for $x\in\operatorname{ran}h=A$, so $A$ is <computably enumerable set>[computably enumerable]. This proves (iv)$\Rightarrow$(i) and completes the equivalence recorded by the <domain and range characterizations of a nonempty computably enumerable set>.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=5J}

The commands choose $n=100$ observations and $p=2$ predictors; generate a $100\times2$ matrix $X$ of <independent random variables>[independent] <standard normal distribution>[standard-normal] values; generate <independent random variables>[independent binary random variables] $Y_i$ with success probability $1/2$; and report the observed number of successes, $\sum_iY_i=48$. The first `glm` call then fits the <Bernoulli logistic-regression model>
$$
Y_i\sim\operatorname{Bernoulli}(p_i),
\qquad
\log\frac{p_i}{1-p_i}
=\beta_0+\beta_1X_{i1}+\beta_2X_{i2}.
$$
The prediction command returns the fitted response probabilities $\widehat p_i$ and sums them.

The likelihood is
$$
L(\beta)=\prod_{i=1}^{100}
 p_i^{Y_i}(1-p_i)^{1-Y_i},
\qquad
p_i=\frac{e^{\eta_i}}{1+e^{\eta_i}},
\quad
\eta_i=\beta_0+\beta_1X_{i1}+\beta_2X_{i2}.
$$
At the <maximum-likelihood estimator>, the intercept component of the <score function> is
$$
0=\frac{\partial\log L}{\partial\beta_0}
 =\sum_{i=1}^{100}(Y_i-\widehat p_i).
$$
Therefore
$$
\sum_i\widehat p_i=\sum_iY_i=48
$$
exactly, by the <fitted-mean balance for logistic regression with an intercept>.

The second fit uses <probit regression>,
$$
p_i=\Phi(\eta_i),
$$
where $\Phi$ is the <standard normal distribution> function. Its intercept score is
$$
\sum_i
\frac{\phi(\eta_i)}{p_i(1-p_i)}(Y_i-p_i)=0,
$$
a weighted residual equation. It does not imply $\sum_i\widehat p_i=48$, so exact equality is not expected. The output should nevertheless be close to $48$: the data were generated with constant success probability $1/2$ independently of $X$, so the fitted slopes should be small, the linear predictors should cluster near a common intercept, and the score weights should be nearly constant.

Solved by gpt-5.6-sol high.

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

= a
{parent=6c}
{scope}

= Solution
{parent=a}

This is the <SIR model with waning immunity>. The <mass-action interaction> term $\beta IS$ transfers people from the susceptible compartment to the infected compartment, $\nu I$ transfers infectives to the recovered compartment, and $fR$ returns recovered people to susceptibility as immunity wanes. Adding the equations gives
$$
\frac d{dt}(S+I+R)=0,
$$
so the total population $N$ is conserved.

When $f=0$ and infection is rare, $S\simeq N$ and
$$
I'=(\beta S-\nu)I\simeq(\beta N-\nu)I.
$$
Thus $I$ initially decays if $\beta N<\nu$. Equivalently, the <basic reproduction number> is $\mathcal R_0=\beta N/\nu<1$, so the <epidemic invasion threshold> is not crossed and no epidemic occurs.

Solved by gpt-5.6-sol high.

= b
{parent=6c}
{scope}

= Solution
{parent=b}

At a nontrivial equilibrium $I^*>0$, the equation $I'=0$ gives
$$
S^*=\frac\nu\beta.
$$
The equation $R'=0$ gives $R^*=\nu I^*/f$, and conservation of population then yields
$$
I^*=\frac{f(\beta N-\nu)}{\beta(f+\nu)},
\qquad
R^*=\frac{\nu(\beta N-\nu)}{\beta(f+\nu)}.
$$
These values are positive precisely in the epidemic regime $\beta N>\nu$.

Eliminate $R=N-S-I$. The two-dimensional system is
$$
S'=f(N-S-I)-\beta IS,
\qquad
I'=(\beta S-\nu)I.
$$
Set $a=\beta I^*=f(\beta N-\nu)/(f+\nu)>0$. The <jacobian matrix> at the endemic equilibrium is
$$
J=
\begin{pmatrix}
-(f+a)&-(f+\nu)\\
a&0
\end{pmatrix},
$$
so its eigenvalues satisfy
$$
\lambda^2+(f+a)\lambda+a(f+\nu)=0.
$$
Their sum is $-(f+a)<0$ and their product is $a(f+\nu)>0$, proving <linear stability of a planar equilibrium>[local asymptotic stability].

The discriminant is
$$
\Delta=(f+a)^2-4a(f+\nu).
$$
Writing $r=\beta N-\nu>0$, we have $a=fr/(f+\nu)$. As $f\to0^+$,
$$
\Delta=-4fr+O(f^2)<0,
$$
whereas as $f\to\infty$,
$$
\Delta=f^2-2rf+O(1)>0.
$$
Hence sufficiently slow immunity loss gives a stable focus: $I(t)$ approaches $I^*$ through damped oscillations, corresponding to successively smaller epidemic waves. Sufficiently rapid immunity loss gives a stable node: small disturbances are sums of two decaying real modes and show no forced oscillation. These conclusions and the equilibrium formulas are collected in the <Endemic equilibrium of the SIR model with waning immunity>.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=7E}

For $\Re s>1$, absolute convergence permits separation into odd and even terms:
$$
\begin{aligned}
\sum_{n=1}^\infty(-1)^{n-1}n^{-s}
&=\sum_{n=1}^\infty n^{-s}-2\sum_{n=1}^\infty(2n)^{-s}\\
&=(1-2^{1-s})\zeta(s).
\end{aligned}
$$
Using the defining integral of the <Gamma function> and the substitution $u=nt$ gives
$$
\Gamma(s)n^{-s}=\int_0^\infty t^{s-1}e^{-nt}\,dt.
$$
Termwise integration is justified by absolute convergence when $\Re s>1$, and the geometric sum is
$$
\sum_{n=1}^\infty(-1)^{n-1}e^{-nt}=\frac1{1+e^t}.
$$
Consequently the <Dirichlet eta function> satisfies
$$
(1-2^{1-s})\zeta(s)=\eta(s)
=\frac1{\Gamma(s)}\int_0^\infty\frac{t^{s-1}}{1+e^t}\,dt.
$$

Near zero the integrand is $O(t^{\Re s-1})$, while at infinity it decays exponentially. The integral therefore defines a holomorphic function for $\Re s>0$. Thus
$$
\boxed{\displaystyle
\zeta(s)=\frac1{(1-2^{1-s})\Gamma(s)}
\int_0^\infty\frac{t^{s-1}}{1+e^t}\,dt}
$$
provides the desired continuation wherever the displayed denominator is nonzero. At a nonreal zero of $1-2^{1-s}$, use instead
$$
(1-k^{1-s})\zeta(s)
=\sum_{n\geq1}a_n n^{-s},
\qquad
a_n=\begin{cases}1,&k\nmid n,\\1-k,&k\mid n,
\end{cases}
$$
with an integer $k$ for which $1-k^{1-s}\ne0$; the bounded partial sums of $a_n$ give convergence for $\Re s>0$. This shows that those apparent singularities are removable and yields the <Analytic continuation of the Riemann zeta function to the right half-plane>.

At $s=1$,
$$
\eta(1)=\int_0^\infty\frac{dt}{1+e^t}=\log2,
$$
whereas
$$
1-2^{1-s}=1-e^{-(s-1)\log2}
=(s-1)\log2+O((s-1)^2).
$$
Hence
$$
\zeta(s)=\frac1{s-1}+O(1),
$$
so $s=1$ is a simple pole and its <residue> is
$$
\boxed{\operatorname{Res}_{s=1}\zeta(s)=1}.
$$

Solved by gpt-5.6-sol high.

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

= Solution
{parent=8D}

A <fixed space frame> is an inertial orthonormal frame whose axes remain fixed in space. A <principal body frame> is attached to the rigid body and aligned with the principal axes of its inertia tensor, so that the tensor is diagonal with entries $I_1,I_2,I_3$.

If $e_i(t)$ are the body axes and $\omega$ is the <angular velocity>, then the <derivative of a body-fixed basis vector> is
$$
\dot e_i=\omega\times e_i.
$$
In the principal frame,
$$
L=\sum_{i=1}^3I_i\omega_i e_i.
$$
For torque-free motion, the <angular momentum> has zero space derivative. Therefore
$$
0=\frac{dL}{dt}
=\sum_iI_i\dot\omega_i e_i+\omega\times L.
$$
Taking body-frame components gives the <Euler equations for a torque-free rigid body>:
$$
\begin{aligned}
I_1\dot\omega_1&=(I_2-I_3)\omega_2\omega_3,\\
I_2\dot\omega_2&=(I_3-I_1)\omega_3\omega_1,\\
I_3\dot\omega_3&=(I_1-I_2)\omega_1\omega_2.
\end{aligned}
$$

For an axisymmetric body, put $I_1=I_2=I$ and let $e_3$ be its symmetry axis. Decompose $\omega=\omega_1e_1+\omega_2e_2+\omega_3e_3$. Then
$$
\begin{aligned}
L
&=I\omega_1e_1+I\omega_2e_2+I_3\omega_3e_3\\
&=I\omega+(I_3-I)\omega_3e_3.
\end{aligned}
$$
Thus $L$ is a linear combination of $\omega$ and $e_3$, proving that the angular momentum, angular velocity, and symmetry axis are always coplanar. This is the <coplanarity in a torque-free axisymmetric rigid body>.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=9B}

The given formula is the integral of the <Planck photon distribution>. Set
$$
x=\frac{h\nu}{k_BT},
\qquad d\nu=\frac{k_BT}{h}\,dx.
$$
Then
$$
n=\frac{8\pi}{c^3}\left(\frac{k_B}{h}\right)^3T^3
\int_0^\infty\frac{x^2}{e^x-1}\,dx
=\alpha T^3,
$$
where
$$
\alpha=\frac{8\pi}{c^3}\left(\frac{k_B}{h}\right)^3
\int_0^\infty\frac{x^2}{e^x-1}\,dx.
$$
Each photon of frequency $\nu$ has energy $h\nu$, so the energy density is
$$
\begin{aligned}
\rho
&=\frac{8\pi h}{c^3}\int_0^\infty
\frac{\nu^3}{e^{h\nu/(k_BT)}-1}\,d\nu\\
&=\frac{8\pi k_B^4}{c^3h^3}T^4
\int_0^\infty\frac{x^3}{e^x-1}\,dx
=\xi T^4.
\end{aligned}
$$
This is the <temperature scaling of thermal photon number and energy densities>.

After decoupling, <cosmological redshift> gives
$$
\nu(t)=\frac{a(t_{\rm dec})}{a(t)}\nu_{\rm dec},
\qquad
\nu_{\rm dec}=\frac{a(t)}{a(t_{\rm dec})}\nu(t).
$$
The occupation number is conserved along the freely propagating photon trajectory. Its exponential argument at decoupling becomes
$$
\frac{h\nu_{\rm dec}}{k_BT_{\rm dec}}
=\frac{h\nu(t)}{k_B\bigl(a(t_{\rm dec})T_{\rm dec}/a(t)\bigr)}.
$$
Thus the spectrum still has the Planck form if
$$
\boxed{T(t)=\frac{a(t_{\rm dec})}{a(t)}T_{\rm dec}}.
$$
Frequency and temperature redshift by the same factor, which is the <redshift preservation of a thermal photon spectrum>.

Solved by gpt-5.6-sol high.

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

= a
{parent=10d}
{scope}

= Solution
{parent=a}

Expanding either <Bell state> and using $\langle0|1\rangle_B=0$ removes the cross terms:
$$
\begin{aligned}
\langle\Phi^\pm|Q\otimes I|\Phi^\pm\rangle
&=\frac12(\langle00|\pm\langle11|)(Q\otimes I)(|00\rangle\pm|11\rangle)\\
&=\frac12\bigl(\langle0|Q|0\rangle+\langle1|Q|1\rangle\bigr)\\
&=\frac12\operatorname{tr}Q.
\end{aligned}
$$
Thus the two expectation values agree. Equivalently, both states have the same <reduced density matrix> $I/2$ on qubit $A$; the positivity of $Q$ makes it a possible measurement effect but is not needed for the algebraic equality.

Solved by gpt-5.6-sol high.

= b
{parent=10d}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

Yes. The states $|\Phi^+\rangle$ and $|\Phi^-\rangle$ are orthogonal, since
$$
\langle\Phi^+|\Phi^-\rangle=\frac12(1-1)=0.
$$
A joint projective measurement in the Bell basis therefore distinguishes them with certainty. Concretely, a controlled-NOT followed by a <Hadamard gate> converts the Bell basis into the computational basis before measurement.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

No. Taking the partial trace over $B$ gives
$$
\rho_A^+=\rho_A^-=\frac12(|0\rangle\langle0|+|1\rangle\langle1|)=\frac I2.
$$
Every measurement on $A$ alone therefore has identical outcome probabilities in the two cases. This is the <local indistinguishability of Bell-state phase>.

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

Yes. Alice and Bob both measure in the Hadamard basis $|+\rangle,|-\rangle$. The identities
$$
|\Phi^+\rangle=\frac{|++\rangle+|--\rangle}{\sqrt2},
\qquad
|\Phi^-\rangle=\frac{|+-\rangle+|-+\rangle}{\sqrt2}
$$
show that their outcomes agree for $|\Phi^+\rangle$ and differ for $|\Phi^-\rangle$. After exchanging one classical bit they know which state they had. This is the <LOCC discrimination of two Bell states>.

Solved by gpt-5.6-sol high.

= c
{parent=10d}
{scope}

= Solution
{parent=c}

No. In <quantum dense coding>, Alice's four encodings produce four different joint Bell states, but the qubit she transmits has reduced state
$$
\rho_A=\frac I2
$$
in every case. Charlie intercepts only that qubit, so every measurement he can perform has the same probability distribution for messages $00,01,10,11$. He gains no information about the two-bit string.

Solved by gpt-5.6-sol high.

= 11F
{parent=Paper 2}
{scope}
{title2=Topics In Analysis}

= a
{parent=11f}
{scope}

= Solution
{parent=a}

The <Baire category theorem> states that every countable intersection of open dense subsets of a <complete metric space> is dense. Equivalently, no nonempty open subset of a complete metric space is a countable union of <nowhere dense sets>.

To prove the first form, let $G_1,G_2,\ldots$ be open dense subsets of a complete metric space $X$, and let $V\subseteq X$ be nonempty and open. Choose a closed ball
$$
\overline B(x_1,r_1)\subseteq V\cap G_1,
\qquad 0<r_1<2^{-1}.
$$
Inductively, density and openness of $G_{n+1}$ allow a closed ball
$$
\overline B(x_{n+1},r_{n+1})
\subseteq B(x_n,r_n)\cap G_{n+1},
\qquad 0<r_{n+1}<2^{-(n+1)}.
$$
The balls are nested and $d(x_m,x_n)<2^{-n}$ for $m>n$, so $(x_n)$ is Cauchy. Completeness gives $x_n\to x\in X$. For every $n$, all later centres lie in $\overline B(x_n,r_n)$; closedness gives $x\in\overline B(x_n,r_n)\subseteq G_n$. The first ball also lies in $V$, hence
$$
x\in V\cap\bigcap_{n=1}^\infty G_n.
$$
Since every nonempty open $V$ meets the intersection, that intersection is dense.

Solved by gpt-5.6-sol high.

= b
{parent=11f}
{scope}

= Solution
{parent=b}

Every summand is at most $2^{-r}$, and $\sum_{r\geq0}2^{-r}=2$, so $d(f,g)$ is finite and well defined. Nonnegativity and symmetry follow from the uniform norm. If $d(f,g)=0$, its $r=0$ summand gives $\|f-g\|_\infty=0$, hence $f=g$. Finally,
$$
\min\{1,a+b\}\leq\min\{1,a\}+\min\{1,b\}
$$
and the triangle inequality for each uniform norm give the triangle inequality for $d$. Thus $d$ is a metric.

Let $(f_n)$ be $d$-Cauchy. For fixed $r$ and $0<\varepsilon<1$, sufficiently large $m,n$ satisfy
$$
d(f_m,f_n)<2^{-r}\varepsilon,
$$
which forces $\|f_m^{(r)}-f_n^{(r)}\|_\infty<\varepsilon$. Hence, for every $r$, the continuous functions $f_n^{(r)}$ converge uniformly to some continuous $g_r$. The fundamental theorem of calculus gives
$$
f_n^{(r)}(x)-f_n^{(r)}(0)
=\int_0^x f_n^{(r+1)}(t)\,dt.
$$
Uniform convergence permits passage to the limit, yielding
$$
g_r(x)-g_r(0)=\int_0^xg_{r+1}(t)\,dt.
$$
Therefore $g_r'=g_{r+1}$ for every $r$, so $g_0\in C^\infty([0,1])$ and $g_0^{(r)}=g_r$.

To prove convergence in $d$, first choose $R$ so that $\sum_{r\geq R}2^{-r}$ is small, then use uniform convergence for the finitely many derivatives $r<R$. Thus $d(f_n,g_0)\to0$. This proves the <completeness of the smooth-function metric>.

Solved by gpt-5.6-sol high.

= c
{parent=11f}
{scope}

= Solution
{parent=c}

For rational $q\in(0,1)$ and positive integer $M$, define
$$
U_{q,M}=\bigcup_{m\geq M}
\left\{f\in C^\infty([0,1]):|f^{(m)}(q)|>m!m^m\right\}.
$$
Evaluation of the $m$th derivative at $q$ is continuous in the smooth-function metric, so $U_{q,M}$ is open.

It is also dense. Given $f$ and a metric tolerance $\varepsilon>0$, choose $R\geq1$ so that the contribution of all derivatives of order at least $R$ is below $\varepsilon/2$, and choose $m\geq\max\{M,R\}$. For large $K$, perturb $f$ by
$$
h_K(x)=\delta_K\cos\left(K(x-q)-\frac{m\pi}{2}\right),
\qquad
\delta_K=\frac{\varepsilon}{4R K^{R-1}}.
$$
For every $r<R$, $\|h_K^{(r)}\|_\infty\leq\varepsilon/(4R)$, so $d(f+h_K,f)<\varepsilon$. On the other hand,
$$
h_K^{(m)}(q)=\delta_KK^m
=\frac{\varepsilon}{4R}K^{m-R+1}\longrightarrow\infty.
$$
For sufficiently large $K$, $|f^{(m)}(q)+h_K^{(m)}(q)|>m!m^m$, proving density.

The space is complete by part (b). The <Baire category theorem> therefore makes
$$
G=\bigcap_{q\in\mathbb Q\cap(0,1)}
\bigcap_{M=1}^\infty U_{q,M}
$$
dense. Put $E=C^\infty([0,1])\setminus G$. Each complement $U_{q,M}^c$ is closed and nowhere dense, so $E$ is a <meagre set>, or a set of first category. Every $f\notin E$ has the required derivative growth. This is the <generic superfactorial derivative growth at rational points>.

Solved by gpt-5.6-sol high.

= d
{parent=11f}
{scope}

= Solution
{parent=d}

For $f\notin E$ and rational $q\in(0,1)$, every $M$ admits $m\geq M$ with
$$
\left|\frac{f^{(m)}(q)}{m!}\right|>m^m.
$$
Consequently
$$
\limsup_{m\to\infty}
\left|\frac{f^{(m)}(q)}{m!}\right|^{1/m}=\infty.
$$
By the <Cauchy-Hadamard theorem>, the <Taylor series> of $f$ at $q$ has <radius of convergence> zero.

Suppose some Taylor series represented $f$ throughout a neighborhood of a point $x\in[0,1]$. That neighborhood contains a rational $q\in(0,1)$. A function represented locally by a convergent power series is a <real analytic function>, and re-expanding that series about $q$ gives the Taylor series of $f$ at $q$ a positive radius of convergence. This contradicts the preceding conclusion. Thus $f$ agrees with no Taylor series on a neighborhood of any point; it is a <nowhere-analytic generic smooth function>.

Solved by gpt-5.6-sol high.

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

= a
{parent=12i}
{scope}

= Solution
{parent=a}

A <one-time pad> over the <finite additive group> $\mathbb Z_n$ uses a key $K$ that is <uniform distribution>[uniform] on $\mathbb Z_n$, <independent random variables>[independent] of the plaintext $X$, as long as the plaintext, and never reused. Encryption and decryption are
$$
C=X+K\pmod n,
\qquad
X=C-K\pmod n.
$$

For <independent random variables> $X,Y$, <conditioning reduces entropy> and translation by a known element of the <finite additive group> preserves entropy, so
$$
H(X+Y)\geq H(X+Y\mid Y)=H(X\mid Y)=H(X).
$$
Interchanging $X$ and $Y$ gives $H(X+Y)\geq H(Y)$, hence
$$
\boxed{H(X+Y)\geq\max\{H(X),H(Y)\}}.
$$
This is the <entropy of a sum of independent finite-group variables>.

The result explains why adding <independent random variables>[independent pad symbols] cannot reduce uncertainty. For a <uniform distribution>[uniform] pad, $X+K$ is itself uniform and is <independent random variables>[independent] of $X$, which is the <perfect secrecy>[perfect-secrecy property]. Independence is necessary: if $X$ is nonconstant and $Y=-X\pmod n$, then $X+Y=0$ is constant, so
$$
H(X+Y)=0<H(X)=H(Y).
$$

Solved by gpt-5.6-sol high.

= b
{parent=12i}
{scope}

= Solution
{parent=b}

Write the intercepted ciphertext, known plaintext, and desired plaintext without spaces:
$$
C=\text{LRPFOJQLCUD},\qquad
P=\text{FLYXATXONCE},\qquad
P'=\text{REMAINXHERE}.
$$
Because the pad is additive, Eve applies the <malleability of an additive one-time pad> and sends
$$
C'=C-P+P'\pmod{26}.
$$
Letter-by-letter, using $A=1,\ldots,Z=26$, this gives
$$
\begin{array}{c|ccccccccccc}
C&L&R&P&F&O&J&Q&L&C&U&D\\
P&F&L&Y&X&A&T&X&O&N&C&E\\
P'&R&E&M&A&I&N&X&H&E&R&E\\ \hline
C'&X&K&D&I&W&D&Q&E&T&J&D
\end{array}
$$
Thus Eve should deliver
$$
\boxed{\text{XKDIWDQETJD}.}
$$
Indeed, if $C=P+K$, then $C'=C-P+P'=P'+K$, so Ollie decrypts the desired message.

Solved by gpt-5.6-sol high.

= c
{parent=12i}
{scope}

= i
{parent=c}
{scope}

= Solution
{parent=i}

Since $\alpha$ generates $K^\times$, it cannot lie in a proper subfield, so its <minimal polynomial> over $\mathbb F_2$ has degree $d$. Write it as
$$
P(X)=X^d+c_{d-1}X^{d-1}+\cdots+c_1X+c_0.
$$
The identity $P(\alpha)=0$ and characteristic two give
$$
\alpha^d=\sum_{j=0}^{d-1}c_j\alpha^j.
$$
Multiplying by $\alpha^n$ and applying the linear map $T$ yields
$$
x_{n+d}=T(\alpha^{n+d})
=\sum_{j=0}^{d-1}c_jT(\alpha^{n+j})
=\sum_{j=0}^{d-1}c_jx_{n+j}.
$$
This is a binary linear recurrence of order at most $d$, so $(x_n)$ is the output of a <linear-feedback shift register> of length at most $d$.

Solved by gpt-5.6-sol high.

= ii
{parent=c}
{scope}

= Solution
{parent=ii}

Because $\alpha^{2^d-1}=1$,
$$
x_{n+2^d-1}=T(\alpha^{n+2^d-1})=T(\alpha^n)=x_n,
$$
so the period divides $2^d-1$.

Conversely, suppose $r$ is an eventual period. Then for every sufficiently large $n$,
$$
0=x_{n+r}-x_n
=T\bigl(\alpha^n(\alpha^r-1)\bigr).
$$
Any block of $2^d-1$ consecutive powers of $\alpha$ runs through every element of $K^\times$. Hence
$$
T\bigl((\alpha^r-1)y\bigr)=0
$$
for every $y\in K$, including $y=0$. The assumed <nondegenerate bilinear form> forces $\alpha^r-1=0$. Since $\alpha$ generates the cyclic group $K^\times$ of order $2^d-1$, this means $2^d-1\mid r$. The least positive period is therefore
$$
\boxed{2^d-1},
$$
as in a <trace-generated maximal-period linear-feedback sequence>.

Solved by gpt-5.6-sol high.

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

= a
{parent=13e}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

Put
$$
D=\mathbb C\setminus\bigl((-\infty,-1]\cup[1,\infty)\bigr).
$$
This slit <complex plane> is a <simply connected domain>. On $D$ there is one branch of $\sqrt{1-z^2}$ that equals $1$ at zero, and its reciprocal is <holomorphic function>[holomorphic]. The <primitive of a holomorphic function on a simply connected domain> therefore makes the integral from zero path-independent, so $G$ is single-valued.

For unrestricted paths, going once around either <branch point> changes the sign of the square root. The resulting <analytic continuation> therefore need not return to the original germ, and the value of $g(z)$ can depend on the path.

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

The endpoint values of the principal branch are
$$
G(-1)=-\frac\pi2,
\qquad
G(1)=\frac\pi2.
$$
By the <monodromy reflection at a square-root branch point>, continuation around $-1$ acts on a value $w$ as
$$
R_-(w)=-\pi-w,
$$
while continuation around $1$ acts as
$$
R_+(w)=\pi-w.
$$

The contour $C_1$ makes the first continuation and then follows the principal path to $z$, so
$$
\boxed{g_{C_1}(z)=-\pi-G(z)}.
$$
The contour $C_2$ continues successively around $-1$ and $1$. Hence
$$
\boxed{g_{C_2}(z)=R_+R_-(G(z))=2\pi+G(z)}.
$$

Solved by gpt-5.6-sol high.

= iii
{parent=a}
{scope}

= Solution
{parent=iii}

Every path can change sheets only by applying the two reflections
$$
R_-(w)=-\pi-w,
\qquad
R_+(w)=\pi-w.
$$
Their products satisfy
$$
R_+R_-(w)=w+2\pi,
\qquad
R_-R_+(w)=w-2\pi.
$$
Thus an even number of reflections gives a translation by an integral multiple of $2\pi$, while an odd number gives a translated reflection. All possible values are therefore
$$
\boxed{2n\pi+G(z)
\quad\hbox{and}\quad
(2n+1)\pi-G(z),\qquad n\in\mathbb Z,}
$$
in agreement with the <branches of the complex inverse sine>.

Solved by gpt-5.6-sol high.

= b
{parent=13e}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

The additional <branch points> $\pm1/k$ lie on the two deleted rays because $0<k<1$. On the same <simply connected domain>
$$
D=\mathbb C\setminus\bigl((-\infty,-1]\cup[1,\infty)\bigr),
$$
choose both square roots to equal $1$ at zero. Their product has a single-valued holomorphic reciprocal on $D$, so its integral from zero defines the single-valued branch $H$ of the <elliptic integral of the first kind>.

An unrestricted contour may wind around any of $-1,1,-1/k,1/k$. Continuation around one such point flips one square root, so the value can change by <monodromy reflection at a square-root branch point>[square-root monodromy]. Thus $h$ is multivalued.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

At $z=1$, the branch value is the <complete elliptic integral of the first kind>
$$
K=H(1).
$$
A contour that goes once around $1$ and then to $z$ applies the reflection about $K$, giving
$$
R_1(H(z))=2K-H(z).
$$

Similarly, the branch value at $-1$ is $-K$, so a loop around $-1$ acts as $R_{-1}(w)=-2K-w$. A contour that loops first around $-1$ and then around $1$ therefore gives
$$
R_1R_{-1}(H(z))=2K-(-2K-H(z))
=\boxed{4K+H(z)}.
$$

Finally approach the interval $(1,1/k)$ through the upper half-plane. There
$$
\sqrt{1-t^2}=-i\sqrt{t^2-1},
$$
so the branch value at $1/k$ is
$$
H(1/k)=K+i\int_1^{1/k}
\frac{dt}{\sqrt{(t^2-1)(1-k^2t^2)}}
=K+iL.
$$
It follows that a loop around $1/k$ acts as
$$
R_{1/k}(w)=2K+2iL-w.
$$
A contour that loops first around $1$ and then around $1/k$ consequently gives
$$
R_{1/k}R_1(H(z))
=2K+2iL-(2K-H(z))
=\boxed{2iL+H(z)}.
$$
Here $L=K'(k)$ is the <complementary complete elliptic integral of the first kind>. These contours exhibit all three requested values.

Solved by gpt-5.6-sol high.

= c
{parent=13e}
{scope}

= Solution
{parent=c}

Part (b) gives paths with the same endpoint $z$ but integral values
$$
H(z),\qquad H(z)+4K,\qquad H(z)+2iL.
$$
Applying the local inverse and then continuing it shows that
$$
\mathcal H(w+4K)=\mathcal H(w),
\qquad
\mathcal H(w+2iL)=\mathcal H(w).
$$
Since $K,L>0$, the first <period of a function>[period] is real and the second is purely imaginary, so they are <linear independence>[linearly independent] over $\mathbb R$. Hence $\mathcal H$ is an <elliptic function> with two periods
$$
\boxed{4K\quad\hbox{and}\quad2iL}.
$$
Indeed, it is the <Jacobi elliptic sine>, and these generators give its <period lattice of the Jacobi elliptic sine>[period lattice].

Solved by gpt-5.6-sol high.

= d
{parent=13e}
{scope}

= Solution
{parent=d}

The inverse is nonconstant; locally at zero,
$$
\mathcal H(0)=0,
\qquad
\mathcal H'(0)=\frac1{H'(0)}=1.
$$
If the assumed <meromorphic function> $\mathcal H$ had no <poles>, it would be <entire function>[entire]. It is bounded on the compact closure of a <fundamental parallelogram>, and its two <periods> then make it bounded throughout the <complex plane>. The <Liouville theorem> would force it to be constant, a contradiction. Therefore $\mathcal H$ has at least one pole, as stated by the general result that a <nonconstant elliptic function has a pole>.

Solved by gpt-5.6-sol high.

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

= a
{parent=14d}
{scope}

= Solution
{parent=a}

Because the three positive gaps make one circuit,
$$
\boxed{\alpha+\beta+\gamma=2\pi}.
$$
They describe only relative positions. The missing <degree of freedom> is the collective rigid rotation of all three particles around the circle, represented by the mean angular coordinate $\Phi$ in <collective rotation and gap coordinates on a circle>.

Solved by gpt-5.6-sol high.

= b
{parent=14d}
{scope}

= Solution
{parent=b}

Choose ordered particle angles $q_1,q_2,q_3$ with
$$
\alpha=q_2-q_1,\qquad
\beta=q_3-q_2,\qquad
\gamma=2\pi-\alpha-\beta,
$$
and put $\Phi=(q_1+q_2+q_3)/3$. Solving for the individual angles gives
$$
q_1=\Phi-\frac{2\alpha+\beta}{3},\qquad
q_2=\Phi+\frac{\alpha-\beta}{3},\qquad
q_3=\Phi+\frac{\alpha+2\beta}{3}.
$$
The <kinetic energy> is therefore
$$
T=\frac{mr^2}{2}\sum_{j=1}^3\dot q_j^2
=\frac{3mr^2}{2}\dot\Phi^2
+\frac{mr^2}{3}
 \left(\dot\alpha^2+\dot\alpha\dot\beta+\dot\beta^2\right).
$$
The collective angle is an <ignorable coordinate> and decouples from the gaps, so its term may be omitted from the relative <Lagrangian mechanics>[Lagrangian]. Substituting $\gamma=2\pi-\alpha-\beta$ into the <potential energy> gives
$$
\boxed{
L_{\rm rel}
=\frac{mr^2}{3}
 \left(\dot\alpha^2+\dot\alpha\dot\beta+\dot\beta^2\right)
-V_0\left(
e^{-2\alpha}+e^{-2\beta}+e^{-4\pi+2\alpha+2\beta}
\right).
}
$$

Solved by gpt-5.6-sol high.

= c
{parent=14d}
{scope}

= Solution
{parent=c}

The function $U(s)=V_0e^{-2s}$ is strictly <convex function>[convex]. Subject to $\alpha+\beta+\gamma=2\pi$, the <Jensen inequality> gives
$$
U(\alpha)+U(\beta)+U(\gamma)
\geq3U\left(\frac{\alpha+\beta+\gamma}{3}\right),
$$
with equality only at
$$
\boxed{\alpha=\beta=\gamma=\frac{2\pi}{3}}.
$$
This is the minimum of the relative potential, so it is a <stable equilibrium in Lagrangian mechanics>[stable relative equilibrium].

Let $q_j=q_j^{(0)}+x_j$ near an equally spaced configuration, and put
$$
A=V_0e^{-4\pi/3}.
$$
The gap perturbations are
$$
\delta\alpha=x_2-x_1,\qquad
\delta\beta=x_3-x_2,\qquad
\delta\gamma=x_1-x_3,
$$
whose sum vanishes. A <Taylor series>[Taylor expansion] gives the quadratic energies
$$
T_2=\frac{mr^2}{2}\sum_j\dot x_j^2,
\qquad
V_2=2A\left[
(x_2-x_1)^2+(x_3-x_2)^2+(x_1-x_3)^2
\right].
$$
Thus the linearized equations are
$$
mr^2\ddot x_i+4A(2x_i-x_j-x_k)=0,
\qquad \{i,j,k\}=\{1,2,3\}.
$$
The matrix in parentheses is the <graph Laplacian> of $K_3$. Its constant eigenvector has eigenvalue zero, while every vector whose components sum to zero has eigenvalue three. One orthonormal set of three <normal modes> and their <angular frequencies> is therefore
$$
\begin{array}{c|c}
\text{mode}&\omega\\ \hline
\displaystyle \frac1{\sqrt3}(1,1,1)&0\\[2mm]
\displaystyle \frac1{\sqrt2}(1,-1,0)&
\displaystyle\sqrt{\frac{12V_0e^{-4\pi/3}}{mr^2}}\\[3mm]
\displaystyle \frac1{\sqrt6}(1,1,-2)&
\displaystyle\sqrt{\frac{12V_0e^{-4\pi/3}}{mr^2}}
\end{array}
$$
The zero mode is rigid rotation. The positive and equal squared frequencies on the two-dimensional relative subspace prove stability and agree with the general <normal modes of three equal masses with a symmetric gap potential>.

Solved by gpt-5.6-sol high.

= d
{parent=14d}
{scope}

= Solution
{parent=d}

Permuting the gaps, or equivalently relabelling the particles, leaves both quadratic forms invariant. The three-dimensional <permutation representation> splits as the rigid-rotation line spanned by $(1,1,1)$ and the two-dimensional sum-zero <standard representation of the symmetric group>.

The two displayed vibrational vectors merely choose a basis of that sum-zero plane. A permutation generally mixes them, but their common frequency leaves the whole plane invariant. This is precisely the <degenerate normal modes from permutation symmetry> required by the physical symmetry.

Solved by gpt-5.6-sol high.

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

= a
{parent=15d}
{scope}

= Solution
{parent=a}

Writing $\omega_N=e^{2\pi i/N}$, the <quantum Fourier transform> acts on a computational-basis state as
$$
\boxed{
\operatorname{QFT}_N|x\rangle
=\frac1{\sqrt N}\sum_{y=0}^{N-1}\omega_N^{xy}|y\rangle.
}
$$

Solved by gpt-5.6-sol high.

= b
{parent=15d}
{scope}

= Solution
{parent=b}

Applying the transform twice gives
$$
\operatorname{QFT}_N^2|x\rangle
=\frac1N\sum_{z=0}^{N-1}
 \left(\sum_{y=0}^{N-1}\omega_N^{y(x+z)}\right)|z\rangle.
$$
The <root-of-unity filter> makes the inner sum equal to $N$ exactly when $z\equiv-x\pmod N$ and zero otherwise. Hence
$$
\boxed{\operatorname{QFT}_N^2|x\rangle=|-x\bmod N\rangle}.
$$
Applying modular negation twice gives
$$
\boxed{\operatorname{QFT}_N^4=I}.
$$
Therefore $\operatorname{QFT}_N^2$ is an involution. Every one of its <eigenvalues> $\lambda$ satisfies $\lambda^2=1$, so its spectrum is contained in $\{1,-1\}$, as summarized by the <square of the quantum Fourier transform>.

Solved by gpt-5.6-sol high.

= c
{parent=15d}
{scope}

= Solution
{parent=c}

The two registers start in $|0\rangle|0\rangle$. A <quantum Fourier transform> on the first and evaluation of $f$ give
$$
|0\rangle|0\rangle
\longmapsto
\frac14\sum_{x=0}^{15}|x\rangle|0\rangle
\longmapsto
\frac14\sum_{x=0}^{15}|x\rangle|f(x)\rangle.
$$
The first <quantum measurement in the computational basis>[measurement], on the function register, returns $1$. The compatible inputs are $1,5,9,13$, so the normalized state becomes
$$
\frac12\bigl(|1\rangle+|5\rangle+|9\rangle+|13\rangle\bigr)|1\rangle.
$$

Apply $\operatorname{QFT}_{16}$ to the first register and put $\omega=e^{2\pi i/16}$. Its state is
$$
\frac18\sum_{y=0}^{15}
\omega^y\left(1+\omega^{4y}+\omega^{8y}+\omega^{12y}\right)|y\rangle.
$$
The geometric sum vanishes unless $4\mid y$. Thus this is
$$
\frac12\left(
|0\rangle+i|4\rangle-|8\rangle-i|12\rangle
\right),
$$
up to the unchanged second register. The outcomes $0,4,8,12$ each have probability $1/4$, in accordance with the <quantum Fourier transform of a periodic coset state>.

For the stated second outcome,
$$
\frac{12}{16}=\frac34.
$$
The <continued-fraction algorithm> therefore returns denominator $4$. Since the numerator $3$ is coprime to the true period, <exact period recovery from a Fourier sample> succeeds, and direct evaluation confirms that the least period is
$$
\boxed{r=4}.
$$

Solved by gpt-5.6-sol high.

= d
{parent=15d}
{scope}

= Solution
{parent=d}

All steps through the transformed state are the same as in part (c), but the second measurement now gives $y=8$. The classical fraction is
$$
\frac8{16}=\frac12,
$$
so reduction or the <continued-fraction algorithm> proposes denominator $2$. This is not a period, since for example
$$
f(0)=2\ne3=f(2).
$$
Equivalently, the sample has $s=2$ and true period $r=4$, with $\gcd(s,r)=2$, so it reveals only $r/\gcd(s,r)=2$. The algorithm does not succeed from this sample and must be repeated.

Solved by gpt-5.6-sol high.

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

= a
{parent=16h}
{scope}

= Solution
{parent=a}

For a nonzero <ordinal> $\alpha$, let
$$
S=\{\xi:\omega^\xi\leq\alpha\}.
$$
The set is nonempty because $0\in S$, and it is an initial segment because <ordinal exponentiation> by $\omega$ is strictly increasing. Put $\beta=\sup S$.

If $\beta$ is a successor, the definition of the <supremum> forces $\beta\in S$. If $\beta$ is a <limit ordinal>, then continuity of ordinal exponentiation gives
$$
\omega^\beta
=\sup_{\xi<\beta}\omega^\xi
\leq\alpha,
$$
so again $\beta\in S$. Thus $\omega^\beta\leq\alpha$, while $\omega^{\beta+1}>\alpha$ by the definition of the supremum. Hence $\beta$ is the greatest required exponent.

Apply <division by an additively indecomposable ordinal> with $\lambda=\omega^\beta$:
$$
\alpha=\omega^\beta q+\gamma,
\qquad
\gamma<\omega^\beta.
$$
Since $\alpha\geq\omega^\beta$, the quotient $q$ is nonzero. It must be finite: if $q\geq\omega$, then
$$
\omega^{\beta+1}
=\omega^\beta\omega
\leq\omega^\beta q
\leq\alpha,
$$
contradicting the maximality of $\beta$. Writing $q=n$ gives
$$
\boxed{\alpha=\omega^\beta n+\gamma},
\qquad
0<n<\omega,\quad \gamma<\omega^\beta.
$$
This is the <greatest power of omega below an ordinal>[leading-term decomposition].

Solved by gpt-5.6-sol high.

= b
{parent=16h}
{scope}

= Solution
{parent=b}

First let $\delta=\omega^\alpha$. If $\beta,\gamma<\delta$, every exponent appearing in their <Cantor normal forms> is strictly below $\alpha$. The rules for <ordinal addition> can delete lower terms but cannot introduce an exponent at least $\alpha$. Therefore
$$
\beta+\gamma<\omega^\alpha=\delta,
$$
so $\delta$ is <additively closed ordinal>[additively closed].

Conversely, suppose nonzero $\delta$ is additively closed. Part (a) gives
$$
\delta=\omega^\alpha n+\rho,
\qquad
0<n<\omega,\quad \rho<\omega^\alpha.
$$
If $\rho>0$, then both $\omega^\alpha n$ and $\rho$ are smaller than $\delta$, but their sum is $\delta$, contradicting closure. Hence $\rho=0$. If now $n>1$, then
$$
\omega^\alpha(n-1)<\delta,
\qquad
\omega^\alpha<\delta,
$$
while
$$
\omega^\alpha(n-1)+\omega^\alpha
=\omega^\alpha n=\delta,
$$
another contradiction. Thus $n=1$, and
$$
\boxed{\delta=\omega^\alpha}.
$$
This proves the classification of <additively closed ordinals>.

Solved by gpt-5.6-sol high.

= c
{parent=16h}
{scope}

= Solution
{parent=c}

Suppose first that $\delta>2$ is <multiplicatively closed ordinal>[multiplicatively closed]. It is also additively closed. Indeed, for $\beta,\gamma<\delta$, let $\mu=\max\{\beta,\gamma\}$. Monotonicity of <ordinal addition> gives
$$
\beta+\gamma\leq\mu+\mu=\mu\,2<\delta,
$$
where the last inequality uses $\mu<\delta$, $2<\delta$, and multiplicative closure. Part (b) therefore gives
$$
\delta=\omega^\lambda
$$
for some nonzero ordinal $\lambda$.

For any $\rho,\sigma<\lambda$, strict monotonicity gives
$$
\omega^\rho,\omega^\sigma<\omega^\lambda=\delta.
$$
Multiplicative closure and the exponent law now imply
$$
\omega^{\rho+\sigma}
=\omega^\rho\omega^\sigma
<\omega^\lambda,
$$
hence $\rho+\sigma<\lambda$. Thus $\lambda$ is additively closed, and part (b) gives $\lambda=\omega^\alpha$. Consequently
$$
\boxed{\delta=\omega^{\omega^\alpha}}.
$$

Conversely, let $\delta=\omega^\lambda$ with $\lambda=\omega^\alpha$. By part (b), $\lambda$ is additively closed. Take nonzero $\beta,\gamma<\delta$, with leading exponents $\rho,\sigma<\lambda$ in <Cantor normal form>. If $\gamma$ is finite, the product $\beta\gamma$ has leading exponent $\rho<\lambda$. If $\gamma$ is infinite, the <leading exponent of an ordinal product> is
$$
\rho+\sigma<\lambda
$$
by additive closure of $\lambda$. In either case
$$
\beta\gamma<\omega^\lambda=\delta.
$$
Products involving zero are immediate, so $\delta$ is multiplicatively closed. This is the <multiplicative closure criterion for a power of omega> and completes both directions.

Solved by gpt-5.6-sol high.

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

= a
{parent=17h}
{scope}

= Solution
{parent=a}

For a positive integer $t$, the <chromatic polynomial> $P_G(t)$ is defined to be the number of proper <graph colourings>[vertex colourings] of $G$ using a fixed palette of $t$ colours.

We prove that this counting function is a polynomial by induction on the number of <edges>. If $G$ has $n$ vertices and no edges, every assignment of colours is proper, so
$$
P_G(t)=t^n.
$$
Otherwise choose an edge $e=uv$. Every proper colouring of $G-e$ either gives $u,v$ different colours, in which case it is a colouring of $G$, or gives them the same colour, in which case it corresponds to a proper colouring of the contracted graph $G/e$. Hence the <deletion-contraction recurrence for the chromatic polynomial> is
$$
P_G(t)=P_{G-e}(t)-P_{G/e}(t).
$$
Both terms on the right are polynomials by induction, so $P_G$ is a polynomial.

Solved by gpt-5.6-sol high.

= b
{parent=17h}
{scope}

= Solution
{parent=b}

Factor the proposed polynomial:
$$
f(t)=t(t-1)^2(t-2).
$$
In particular,
$$
f(2)=0.
$$
By the <two-colourability criterion for bipartite graphs>, every finite <bipartite graph> has at least one proper two-colouring, so its chromatic polynomial is positive at $2$. Therefore $f$ cannot be the chromatic polynomial of a bipartite graph.

It is nevertheless a chromatic polynomial. Start with the <complete graph> $K_3$, whose vertices may be coloured in
$$
t(t-1)(t-2)
$$
ways, and attach one <leaf of a graph>[leaf] to any vertex. After the triangle is coloured, the leaf has $t-1$ available colours. The <chromatic polynomial after attaching a leaf> therefore gives
$$
P_G(t)=t(t-1)(t-2)(t-1)=f(t).
$$

Solved by gpt-5.6-sol high.

= c
{parent=17h}
{scope}

= Solution
{parent=c}

Let $d=d(G)$ and suppose
$$
c_0I+c_1A+\cdots+c_dA^d=0.
$$
If some coefficient is nonzero, let $k$ be the largest index with $c_k\ne0$. Choose a path
$$
v_0,v_1,\ldots,v_d
$$
whose endpoints realize the <graph diameter>. Every initial segment is a shortest path, since a shorter route from $v_0$ to $v_k$ could be followed by the remaining segment to shorten the path from $v_0$ to $v_d$. Thus
$$
d(v_0,v_k)=k.
$$

By the <walk count from powers of an adjacency matrix>,
$$
(A^j)_{v_0v_k}=0\quad(j<k),
\qquad
(A^k)_{v_0v_k}>0.
$$
Taking the $(v_0,v_k)$ entry of the assumed relation and using maximality of $k$ gives
$$
c_k(A^k)_{v_0v_k}=0,
$$
a contradiction. Every coefficient is therefore zero, proving the <linear independence of adjacency powers up to the diameter>:
$$
\boxed{I,A,A^2,\ldots,A^{d(G)}\ \text{are linearly independent}.}
$$

Solved by gpt-5.6-sol high.

= d
{parent=17h}
{scope}

= Solution
{parent=d}

Take the infinite family
$$
G_n=K_n,\qquad n\geq2.
$$
Each <complete graph> is connected and has <graph diameter> one. If $A_n$ is its <adjacency matrix of a graph>[adjacency matrix], then
$$
A_n=J-I
$$
and direct multiplication gives the <adjacency-matrix quadratic relation for a complete graph>
$$
A_n^2=(n-2)A_n+(n-1)I.
$$
Consequently
$$
A_n^2-(n-2)A_n-(n-1)I=0,
$$
so $I,A_n,A_n^2$, which are precisely the powers through $d(G_n)+1=2$, are linearly dependent for every $n\geq2$.

Solved by gpt-5.6-sol high.

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

= a
{parent=18i}
{scope}

= Solution
{parent=a}

An element $x\in L$ is <separable algebraic element>[separable over] $K$ when its <minimal polynomial> $m_x(T)\in K[T]$ has distinct roots in a <splitting field>.

Because $m_x$ is irreducible, it is separable exactly when
$$
\gcd(m_x,m_x')=1.
$$
If $m_x'\ne0$, then $\deg m_x'<\deg m_x$, so irreducibility makes this greatest common divisor equal to one. If $m_x'=0$, the <polynomial with zero formal derivative in characteristic p>[zero-derivative criterion] gives
$$
m_x(T)=g(T^p)
$$
for some $g\in K[T]$, and $m_x$ is inseparable. Therefore
$$
\boxed{x\text{ is separable over }K
\iff m_x(T)\ne g(T^p)\text{ for every }g\in K[T].}
$$

Solved by gpt-5.6-sol high.

= b
{parent=18i}
{scope}

= Solution
{parent=b}

Suppose first that $x$ is a <purely inseparable algebraic element> over $K$, so
$$
x^{p^N}=y\in K
$$
for some $N\geq0$. Then $m_x$ divides $T^{p^N}-y$. Over an algebraic closure the <Frobenius endomorphism> is injective, so this polynomial has only one distinct root:
$$
T^{p^N}-y=(T-x)^{p^N}.
$$

Repeatedly use the zero-derivative criterion to write
$$
m_x(T)=q(T^{p^n})
$$
with $q'\ne0$. The polynomial $q$ is irreducible and therefore separable by part (a), but all its roots are powers of the single root $x$. Hence $q$ has degree one. Since $m_x$ is monic,
$$
m_x(T)=T^{p^n}-a
$$
for some $a\in K$.

Conversely, if the minimal polynomial has this form, then
$$
x^{p^n}=a\in K,
$$
so $x$ is purely inseparable. Thus the <minimal polynomial of a purely inseparable element> is exactly
$$
\boxed{m_x(T)=T^{p^n}-a,\qquad n\geq0,\ a\in K.}
$$

Solved by gpt-5.6-sol high.

= c
{parent=18i}
{scope}

= Solution
{parent=c}

Let $y_1,\ldots,y_r$ be the distinct roots of $g$ in its splitting field $L$. For each $i$, let $x_i$ be the unique $p$th root of $y_i$ in an algebraic closure. Then
$$
T^p-y_i=(T-x_i)^p,
$$
so the roots of
$$
f(T)=g(T^p)
$$
are precisely the $x_i$. Hence
$$
M=L(x_1,\ldots,x_r).
$$
But $y_i=x_i^p$, so $L=K(y_1,\ldots,y_r)\subseteq K(x_1,\ldots,x_r)$. It follows that
$$
M=K(x_1,\ldots,x_r),
$$
which proves that $M$ is also the <splitting field> of $f$ over $K$, as in the <splitting field of a polynomial obtained by Frobenius substitution>.

Moreover $x_i^p=y_i\in L$, so every root of $f$ is purely inseparable over $L$. In fact $M/L$ is purely inseparable.

Now take $\sigma\in\operatorname{Aut}(L/K)$. The <extension count for field embeddings>[extension theorem for field embeddings] extends $\sigma$ to a $K$-embedding of $M$ into an algebraic closure. Since $M$ is a splitting field of $f$ over $K$, its image is again $M$, so this extension is an automorphism $\tau$ of $M$. It is unique: for every root $x_i$,
$$
\tau(x_i)^p=\sigma(y_i),
$$
and the <Frobenius endomorphism> gives only one possible $p$th root. Since the $x_i$ generate $M$ over $L$, $\tau$ is uniquely determined. This is the <unique embedding extension through a purely inseparable extension>.

Solved by gpt-5.6-sol high.

= d
{parent=18i}
{scope}

= Solution
{parent=d}

Assume now that $g$ is irreducible and separable. Its splitting field $L/K$ is <Finite Galois extension>[Galois], and its Galois group acts transitively on the roots $y_i$ by the <Galois group of a polynomial>. Part (c) extends every such automorphism uniquely to $M$. If $x_i^p=y_i$, the extension sends $x_i$ to the unique $p$th root above the image of $y_i$. The <transitivity lifted through unique pth roots> therefore shows that
$$
\operatorname{Aut}(M/K)
$$
acts transitively on the roots of $f$.

Every automorphism of $M/K$ preserves $L$, because $L$ is the splitting field of $g$ over $K$. Transitivity therefore implies that either all roots $x_i$ lie in $L$ or none does. Since $x_i^p\in L$, the <minimal polynomial of a purely inseparable element> has degree either one or $p$. Consequently,
$$
\boxed{\text{either every root lies in }L,\text{ or every root has degree }p\text{ over }L.}
$$

Let $h\in K[T]$ be a monic irreducible factor of $f$. If $h$ is inseparable, part (a) gives
$$
h(T)=q(T^p)
$$
for a nonconstant monic $q\in K[T]$. Since $q(T^p)$ divides $g(T^p)$, polynomial division in $K[U]$ and substitution $U=T^p$ show that $q$ divides $g$. Irreducibility of $g$ forces $q=g$, hence $h=f$. Therefore
$$
\boxed{h=f\quad\text{or}\quad h\text{ is separable},}
$$
which is the <irreducible factors after Frobenius substitution> dichotomy.

If every coefficient of $g$ is a $p$th power, say
$$
g(T)=\sum_i a_i^pT^i,
$$
then
$$
f(T)=g(T^p)
=\left(\sum_i a_iT^i\right)^p,
$$
so $f$ is reducible.

Conversely, suppose $f$ is reducible and factor it into distinct monic irreducibles:
$$
f=\prod_i h_i^{e_i}.
$$
Every $h_i$ is then a proper factor, hence separable by the preceding dichotomy. Since $f'=0$, unique factorization and $h_i'\ne0$ force $p\mid e_i$ for every $i$. Thus $f=q^p$ for some monic $q\in K[T]$. The <Frobenius endomorphism> raises coefficients to their $p$th powers, and the coefficient of $T^{ip}$ in $f=g(T^p)$ is the coefficient of $T^i$ in $g$. Hence every coefficient of $g$ is a $p$th power in $K$. We have proved the <reducibility criterion after Frobenius substitution>:
$$
\boxed{g(T^p)\text{ is reducible over }K
\iff\text{ every coefficient of }g\text{ is a }p\text{th power in }K.}
$$

Solved by gpt-5.6-sol high.

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

= Solution
{parent=19H}

Write a matrix as $(r,b)$, where $r=a^2$ is a nonzero <quadratic residue> modulo $11$. The square subgroup is
$$
Q=\langle4\rangle=\{1,4,5,9,3\}\cong C_5.
$$
Matrix multiplication becomes
$$
(r,b)(s,c)=(rs,b+rc),
\qquad
(r,b)^{-1}=(r^{-1},-r^{-1}b).
$$
Thus $H$ is the <affine semidirect product of cyclic groups of orders eleven and five>
$$
H\cong(\mathbb F_{11},+)\rtimes Q\cong C_{11}\rtimes C_5.
$$
There are five choices for $r$ and eleven for $b$, so $|H|=55$. It is nonabelian, since
$$
(4,0)(1,1)=(4,4)\ne(4,1)=(1,1)(4,0).
$$

Let
$$
N=\{(1,b):b\in\mathbb F_{11}\}\cong C_{11}.
$$
Conjugation by the complement gives
$$
(r,0)(1,b)(r,0)^{-1}=(1,rb).
$$
Hence the ten nonidentity translations split into the two orbits
$$
T_Q=\{(1,b):b\in Q\},
\qquad
T_N=\{(1,b):b\in2Q\},
$$
each of size five. For $r\ne1$,
$$
(1,t)(r,b)(1,t)^{-1}
=(r,b+(1-r)t).
$$
Since $1-r\ne0$, varying $t$ gives every $b\in\mathbb F_{11}$. Conjugation cannot change $r$ in the abelian quotient $H/N$. Therefore, with
$$
C_k=\{(4^k,b):b\in\mathbb F_{11}\},
\qquad 1\leq k\leq4,
$$
the seven conjugacy classes are
$$
\{1\},\quad T_Q,\quad T_N,\quad C_1,\quad C_2,\quad C_3,\quad C_4,
$$
of sizes $1,5,5,11,11,11,11$, agreeing with the <conjugacy classes in the affine semidirect product of orders eleven and five>.

The commutators with $(4,0)$ generate every translation because multiplication by $4-1=3$ is invertible in $\mathbb F_{11}$. Hence the <commutator subgroup> is $N$, and the <abelianization> is
$$
H^{\mathrm{ab}}\cong H/N\cong C_5.
$$
Put $\zeta=e^{2\pi i/5}$. The <one-dimensional characters factor through the abelianization>, giving five characters
$$
\chi_j(4^k,b)=\zeta^{jk},
\qquad 0\leq j\leq4.
$$

For the remaining characters, put $\omega=e^{2\pi i/11}$ and define a character of $N$ by
$$
\theta_m(1,b)=\omega^{mb}.
$$
The complement has two free orbits on the nontrivial $\theta_m$, indexed by $m\in Q$ and $m\in2Q$. By <induction from an abelian normal subgroup with a free character orbit>, the characters
$$
\psi_Q=\operatorname{Ind}_N^H\theta_1,
\qquad
\psi_N=\operatorname{Ind}_N^H\theta_2
$$
are irreducible of degree five and vanish outside $N$.

Set
$$
\eta=\sum_{q\in Q}\omega^q
=\frac{-1+i\sqrt{11}}2.
$$
The <quadratic periods modulo eleven> give the nonsquare sum $\overline\eta=(-1-i\sqrt{11})/2$. Multiplication by a square preserves $Q$ and multiplication by a nonsquare exchanges the two square classes, so the complete <character table> is
$$
\begin{array}{c|ccc|cccc}
 &1&T_Q&T_N&C_1&C_2&C_3&C_4\\
\text{class size}&1&5&5&11&11&11&11\\ \hline
\chi_0&1&1&1&1&1&1&1\\
\chi_1&1&1&1&\zeta&\zeta^2&\zeta^3&\zeta^4\\
\chi_2&1&1&1&\zeta^2&\zeta^4&\zeta&\zeta^3\\
\chi_3&1&1&1&\zeta^3&\zeta&\zeta^4&\zeta^2\\
\chi_4&1&1&1&\zeta^4&\zeta^3&\zeta^2&\zeta\\
\psi_Q&5&\eta&\overline\eta&0&0&0&0\\
\psi_N&5&\overline\eta&\eta&0&0&0&0
\end{array}.
$$
Finally,
$$
5\cdot1^2+2\cdot5^2=55=|H|,
$$
and there are seven rows for the seven conjugacy classes. Thus these are all irreducible characters, as described by the <irreducible characters of the affine semidirect product of orders eleven and five>.

Solved by gpt-5.6-sol high.

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

= a
{parent=20h}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

The <trace form of a number field> is
$$
B(x,y)=\operatorname{Tr}_{K/\mathbb Q}(xy).
$$
It is bilinear because the <field trace> is linear. If $x\ne0$, choose $y=x^{-1}$. Then
$$
B(x,y)=\operatorname{Tr}_{K/\mathbb Q}(1)
=[K:\mathbb Q]\ne0.
$$
Thus no nonzero $x$ lies in the radical, and $B$ is a <nondegenerate bilinear form>. This is the <nondegeneracy of the trace form of a number field>.

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

Let $\omega_1,\ldots,\omega_n$ be an <integral basis> of $\mathcal O_K$. Since each $\alpha_j$ is an <algebraic integer>, there is an integer matrix $A=(a_{ij})$ such that
$$
\alpha_j=\sum_i a_{ij}\omega_i.
$$
The $\alpha_j$ form a $\mathbb Q$-basis, so $\det A\ne0$.

Products of algebraic integers are algebraic integers, and their traces are rational algebraic integers, hence ordinary integers. Therefore every entry
$$
\operatorname{Tr}_{K/\mathbb Q}(\alpha_i\alpha_j)
$$
is integral. Part (i) makes the trace Gram matrix nonsingular, so $\Delta(\alpha_1,\ldots,\alpha_n)$ is a nonzero integer.

Let $G_\omega$ and $G_\alpha$ be the two trace Gram matrices. A change of basis gives
$$
G_\alpha=A^TG_\omega A.
$$
Taking determinants yields the <discriminant-index formula for an integral lattice>
$$
\Delta(\alpha_1,\ldots,\alpha_n)
=(\det A)^2d_K,
$$
where $d_K=\det G_\omega$ is the <field discriminant>. Since $\det A$ is a nonzero integer,
$$
|\Delta(\alpha_1,\ldots,\alpha_n)|\geq|d_K|>0.
$$
Equality holds exactly when $|\det A|=1$, which is equivalent to $A$ being unimodular and to $\alpha_1,\ldots,\alpha_n$ being a $\mathbb Z$-basis of $\mathcal O_K$. Thus the minimum is the positive integer
$$
\boxed{|d_K|}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=20h}
{scope}

= Solution
{parent=b}

Put $\theta=\sqrt[3]{3}$. The given ring of integers has integral basis $1,\theta,\theta^2$. The discriminant of the power basis, equivalently of $T^3-3$, is
$$
d_K=-27\cdot3^2=-243.
$$
The field has signature $(1,1)$: one real embedding and one conjugate pair of complex embeddings. The <Minkowski bound for ideal classes> is therefore
$$
\left(\frac4\pi\right)\frac{3!}{3^3}\sqrt{243}
=\frac{8\sqrt3}{\pi}<5.
$$
Every ideal class consequently has an integral representative of norm at most four.

It remains to inspect prime ideals above $2$ and $3$. Modulo $2$,
$$
T^3-3\equiv T^3+1
=(T+1)(T^2+T+1).
$$
The <Dedekind factorization theorem> gives
$$
(2)=\mathfrak p_2\mathfrak q_2,
$$
where
$$
\mathfrak p_2=(2,\theta-1),\quad N(\mathfrak p_2)=2,
\qquad
\mathfrak q_2=(2,\theta^2+\theta+1),\quad N(\mathfrak q_2)=4.
$$
But
$$
N_{K/\mathbb Q}(\theta-1)=3-1=2,
$$
so the principal ideal $(\theta-1)$ is contained in $\mathfrak p_2$ and has the same norm; hence
$$
\mathfrak p_2=(\theta-1).
$$
Also
$$
(\theta-1)(\theta^2+\theta+1)=\theta^3-1=2,
$$
so
$$
\mathfrak q_2=(\theta^2+\theta+1).
$$
Both primes above $2$ are principal.

Finally,
$$
(\theta)^3=(3),
\qquad
N((\theta))=|N_{K/\mathbb Q}(\theta)|=3.
$$
Thus the unique prime above $3$ is the principal ideal $(\theta)$. By <unique factorization of ideals in a number field>, every integral ideal of norm at most four is built from these principal prime ideals. Every ideal class is therefore trivial, and the <Class group of Q of cube root of three> is
$$
\boxed{\operatorname{Cl}\bigl(\mathbb Q(\sqrt[3]{3})\bigr)=1.}
$$

Solved by gpt-5.6-sol high.

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

= Solution
{parent=21G}

Choose a path $\gamma$ in $X$ from $x_0$ to $x_1$. For every $\widehat x\in p^{-1}(x_0)$, the <path lifting theorem> gives a unique lift of $\gamma$ starting at $\widehat x$. Send $\widehat x$ to the endpoint of this lift. Lifting the reversed path gives the inverse map, so this is the <fibre bijection by path lifting>
$$
p^{-1}(x_0)\cong p^{-1}(x_1).
$$
In particular, all fibres have the same cardinality.

A connected covering is a <normal covering map> when its deck transformations act transitively on a fibre. Equivalently, for a choice of $\widehat x_0$ above $x_0$,
$$
p_*\pi_1(\widehat X,\widehat x_0)
\triangleleft
\pi_1(X,x_0).
$$
The relevant <lifting criterion for a covering space> says that a based map $f:(Y,y_0)\to(X,x_0)$ lifts through $p$ exactly when
$$
f_*\pi_1(Y,y_0)
\subseteq
p_*\pi_1(\widehat X,\widehat x_0).
$$
For a universal covering, $\widehat X$ is simply connected, so the displayed covering subgroup is trivial and hence normal. Thus a <universal covering map is normal>.

Now consider connected finite covers of the closed orientable surface $\Sigma_g$. By the <classification of connected covering spaces>, degree-$n$ connected covers correspond to index-$n$ subgroups of the <fundamental group of a closed orientable surface>, and normal covers correspond to normal subgroups.

The cases in which normality is forced are:

* $n=1$, because the subgroup is the whole fundamental group.
* $n=2$, because every <index-two subgroup is normal>.
* $g=1$, because $\pi_1(\Sigma_1)\cong\mathbb Z^2$ is abelian, so all its subgroups are normal.
* For $g=0$, the sphere is simply connected, so a connected cover necessarily has $n=1$.

It remains to show that these are the only forced cases. Let $g\geq2$ and $n\geq3$. Write
$$
\pi_1(\Sigma_g)=
\left\langle a_1,b_1,\ldots,a_g,b_g
\mathrel{\Big|}
\prod_{i=1}^g[a_i,b_i]=1
\right\rangle.
$$
In the <symmetric group> $S_n$, put
$$
\sigma=(1\,2\,\cdots\,n),
\qquad
\tau=(1\,2).
$$
Define a homomorphism by
$$
a_1\mapsto\sigma,\quad b_1\mapsto\tau,\quad
a_2\mapsto\tau,\quad b_2\mapsto\sigma,
$$
and send all remaining generators to the identity. The relation is respected because
$$
[\sigma,\tau][\tau,\sigma]=1.
$$
The cycle $\sigma$ and transposition $\tau$ generate $S_n$, so the homomorphism is surjective.

Let
$$
H=\phi^{-1}\bigl(\operatorname{Stab}_{S_n}(1)\bigr).
$$
The natural action of $S_n$ is transitive, so $H$ has index $n$. Its image is the point stabilizer $S_{n-1}$, which is not normal in $S_n$ for $n\geq3$; hence $H$ is not normal. The connected covering corresponding to $H$ is therefore an explicit degree-$n$ nonnormal cover. This is the <nonnormal finite cover of a higher-genus orientable surface>.

Consequently, among connected covers that exist, normality is forced exactly when
$$
\boxed{n\leq2\quad\text{or}\quad g\leq1,}
$$
with the qualification that $g=0$ permits only $n=1$, as summarized by the <forced normality of finite connected covers of orientable surfaces>.

Solved by gpt-5.6-sol high.

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

= a
{parent=22f}
{scope}

= Solution
{parent=a}

Let
$$
m=\frac{v+w}{2},
\qquad
R(z)=v+w-z=2m-z.
$$
The map $R$ is reflection about $m$ and is an isometry. It interchanges $v,w$, so it preserves $S_1^{vw}$. Inductively, if it preserves $S_{n-1}^{vw}$, then it preserves its <diameter of a metric space>[diameter] and all distances in the defining condition for $S_n^{vw}$; hence it preserves every $S_n^{vw}$.

The midpoint belongs to $S_1^{vw}$ because
$$
\|m-v\|=\|m-w\|=\frac12\|v-w\|.
$$
Suppose $m\in S_{n-1}^{vw}$. For every $z\in S_{n-1}^{vw}$, reflection invariance gives $R(z)\in S_{n-1}^{vw}$ and
$$
\|m-z\|
=\frac12\|R(z)-z\|
\leq\frac12\operatorname{diam}(S_{n-1}^{vw}).
$$
Thus $m\in S_n^{vw}$, so all the sets are nonempty and contain $m$.

Put $D_n=\operatorname{diam}(S_n^{vw})$. If $x,y\in S_n^{vw}$ with $n\geq2$, then $y\in S_{n-1}^{vw}$ and the defining property of $x$ gives
$$
\|x-y\|\leq\frac12D_{n-1}.
$$
Therefore
$$
D_n\leq\frac12D_{n-1},
\qquad
D_n\leq2^{1-n}D_1\longrightarrow0.
$$
If $z$ lies in every $S_n^{vw}$, then both $z$ and $m$ lie in $S_n^{vw}$, so
$$
\|z-m\|\leq D_n\longrightarrow0.
$$
Hence $z=m$. We have proved the <metric extraction of a midpoint by shrinking diameters>:
$$
\boxed{\bigcap_{n\geq1}S_n^{vw}
=\left\{\frac{v+w}{2}\right\}.}
$$

Solved by gpt-5.6-sol high.

= b
{parent=22f}
{scope}

= Solution
{parent=b}

Here “isometry” is used in the standard <Mazur-Ulam theorem> sense of a surjective distance-preserving map. Surjectivity is needed; a merely distance-preserving embedding need not preserve midpoints, as shown by a <nonsurjective isometry need not preserve midpoints>.

Distance preservation and surjectivity give
$$
u(S_1^{vw})=S_1^{u(v)u(w)}.
$$
Assume inductively that
$$
u(S_{n-1}^{vw})=S_{n-1}^{u(v)u(w)}.
$$
Then the two sets have the same diameter, and the universal distance condition defining the next set transfers through the bijection $u$. Hence
$$
u(S_n^{vw})=S_n^{u(v)u(w)}
$$
for every $n$.

Because $u$ is injective, it also preserves the intersection of this nested family. Part (a) therefore gives
$$
\left\{
u\left(\frac{v+w}{2}\right)
\right\}
=u\left(\bigcap_{n\geq1}S_n^{vw}\right)
=\bigcap_{n\geq1}S_n^{u(v)u(w)}
=\left\{
\frac{u(v)+u(w)}2
\right\}.
$$
Thus
$$
\boxed{
u\left(\frac{v+w}{2}\right)
=\frac{u(v)+u(w)}2.
}
$$

Solved by gpt-5.6-sol high.

= c
{parent=22f}
{scope}

= Solution
{parent=c}

Assume $u(0)=0$. Applying midpoint preservation to $2v$ and $0$ gives
$$
u(v)=\frac{u(2v)+u(0)}2=\frac12u(2v),
$$
so $u(2v)=2u(v)$. Consequently,
$$
u(v+w)
=u\left(2\frac{v+w}{2}\right)
=2u\left(\frac{v+w}{2}\right)
=u(v)+u(w).
$$
Thus $u$ is additive. It follows successively that
$$
u(nv)=nu(v)
\quad(n\in\mathbb Z),
\qquad
u(qv)=qu(v)
\quad(q\in\mathbb Q).
$$

An isometry is continuous. For any $\lambda\in\mathbb R$, choose rationals $q_j\to\lambda$. Then
$$
u(\lambda v)
=\lim_{j\to\infty}u(q_jv)
=\lim_{j\to\infty}q_ju(v)
=\lambda u(v).
$$
Together with additivity, this proves that $u$ is real-linear. Hence the origin-fixing case of the <Mazur-Ulam theorem> gives
$$
\boxed{u(av+bw)=au(v)+bu(w)}
$$
for all $a,b\in\mathbb R$ and $v,w\in V$.

Solved by gpt-5.6-sol high.

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

= a
{parent=23f}
{scope}

= Solution
{parent=a}

For a measurable set $E\subseteq U$, define
$$
\nu(E)=\Lambda(\mathbf1_E),
$$
where $\mathbf1_E$ is its <indicator function>. This is finite because $U$ has <finite measure> and hence $\mathbf1_E\in L^p(U)$. If the sets $E_k$ are pairwise disjoint and $E=\bigcup_{k\geq1}E_k$, then
$$
\left\|\mathbf1_E-
\sum_{k=1}^N\mathbf1_{E_k}\right\|_p^p
=|E\setminus\bigcup_{k=1}^NE_k|\longrightarrow0.
$$
The continuity of the <positive linear functional> $\Lambda$ therefore gives
$$
\nu(E)=\sum_{k=1}^{\infty}\nu(E_k),
$$
so $\nu$ is a <measure> by <countable additivity>. Moreover, a set of zero <Lebesgue measure> has zero indicator in $L^p$, so $\nu$ is <absolute continuity of measures>[absolutely continuous] with respect to Lebesgue measure.

The <Radon-Nikodym theorem> now supplies a measurable function $\omega\geq0$ such that
$$
\nu(E)=\int_E\omega.
$$
Consequently, first for nonnegative simple functions and then, by <approximation by nonnegative simple functions>, for every nonnegative $f\in L^p(U)$,
$$
\int_U f\omega=\Lambda(f)
\leq\|\Lambda\|\,\|f\|_p.
$$
Applying the stated characterization of the <Lp norm> to $\omega$ shows that $\omega\in L^q(U)$ and $\|\omega\|_q\leq\|\Lambda\|$. The <Holder inequality> makes $f\mapsto\int_Uf\omega$ continuous on $L^p(U)$, while bounded simple functions form a <dense subset>; hence
$$
\boxed{\Lambda(f)=\int_U f\omega\quad(f\in L^p(U)).}
$$
This is the <positive functional representation on Lp>.

Solved by gpt-5.6-sol high.

= b
{parent=23f}
{scope}

= Solution
{parent=b}

For $f\in L^1(\mathbb R^n)$, use the angular-frequency convention
$$
\boxed{\widehat f(\xi)=
\int_{\mathbb R^n}f(x)e^{-ix\mathbin\cdot\xi}\,dx.}
$$
This is the <Fourier transform> on $L^1(\mathbb R^n)$.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

Choose a nonzero $f\in L^p(\mathbb R^n)\cap L^1(\mathbb R^n)$ whose transform belongs to $L^q$, for example a Gaussian by the <Fourier transform of a Gaussian>. For $\lambda>0$, set $f_\lambda(x)=f(\lambda x)$. The <scaling property of the Fourier transform> gives
$$
\|f_\lambda\|_p
=\lambda^{-n/p}\|f\|_p,
\qquad
\|\widehat {f_\lambda}\|_q
=\lambda^{-n(1-1/q)}\|\widehat f\|_q.
$$
The assumed estimate, applied for every $\lambda>0$, becomes
$$
\lambda^{-n(1-1/q)}\|\widehat f\|_q
\leq C\lambda^{-n/p}\|f\|_p.
$$
Since both norms are nonzero, this can hold as $\lambda$ tends both to zero and to infinity only if the powers agree. Thus the <scaling necessity for an Lp to Lq Fourier bound> yields
$$
1-\frac1q=\frac1p,
\qquad
\boxed{q=\frac p{p-1}}.
$$
Hence $q$ is uniquely the <conjugate exponents>[conjugate exponent] of $p$.

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

Although $x\mapsto|x|^{-1}$ is not integrable on all of $\mathbb R^3$, it defines a <tempered distribution>. For $x\ne0$, the <Gamma function> integral gives
$$
\frac1{|x|}
=\frac1{\sqrt\pi}\int_0^\infty
t^{-1/2}e^{-t|x|^2}\,dt.
$$
The three-dimensional form of the <Fourier transform of a Gaussian> is
$$
\mathcal F(e^{-t|x|^2})(\xi)
=\left(\frac\pi t\right)^{3/2}
e^{-|\xi|^2/(4t)}.
$$
Therefore, up to a nonzero constant depending on the Fourier-transform convention,
$$
\widehat{|x|^{-1}}(\xi)
=C\int_0^\infty t^{-2}e^{-|\xi|^2/(4t)}\,dt.
$$
For $\xi\ne0$, the substitution $u=|\xi|^2/(4t)$ turns the last integral into
$$
\frac4{|\xi|^2}\int_0^\infty e^{-u}\,du.
$$
Thus, as asserted by the <Fourier transform of inverse distance in three dimensions>,
$$
\boxed{\widehat{|x|^{-1}}(\xi)=C'|\xi|^{-2}}
$$
as a tempered distribution, where $C'\ne0$.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=24F}

If $F:X\to Y$ is a nonconstant <holomorphic map> between compact connected <Riemann surfaces>, its <local degree of a holomorphic map>[local degree] at $p$ is the integer $m_F(p)\geq1$ for which suitable local coordinates give
$$
F(z)-F(p)=az^{m_F(p)}+O(z^{m_F(p)+1}),
\qquad a\ne0.
$$
The <valency theorem> states that
$$
\sum_{p\in F^{-1}(y)}m_F(p)
$$
is independent of $y\in Y$. This common value is the <degree of a holomorphic map>, denoted $\deg F$.

Now let $f$ be a nonconstant <rational function> of degree $d$. If its distinct finite <poles> have orders $m_1,\ldots,m_r$, and its pole order at infinity is $m_\infty\geq0$, then
$$
d=m_1+\cdots+m_r+m_\infty.
$$
The <derivative> $f'$ has a pole of order $m_j+1$ at each finite pole. When $m_\infty>0$, the expansion $f(z)\sim cz^{m_\infty}$ shows that $f'(z)\sim cm_\infty z^{m_\infty-1}$ at infinity. Hence the <degree of the derivative of a rational function> is
$$
\deg f'=
\begin{cases}
d+r-1,&m_\infty>0,\\
d+r,&m_\infty=0.
\end{cases}
$$
In the first case $r\geq0$, while in the second $1\leq r\leq d$; therefore
$$
\boxed{d-1\leq\deg f'\leq2d.}
$$
For every $d\geq1$, the lower bound is attained by $f(z)=z^d$, whose derivative has degree $d-1$ (with a constant assigned degree zero). For distinct $a_1,\ldots,a_d$, the function
$$
f(z)=\sum_{j=1}^d\frac1{z-a_j}
$$
has $d$ simple poles, degree $d$, and a derivative with $d$ double poles, so $\deg f'=2d$. Thus both rational bounds are sharp for every $d$.

Let next $g$ be a nonconstant <elliptic function> for the <period lattice> $\Lambda$. Its <degree of an elliptic function>[degree] is the total order of its poles in a <fundamental parallelogram>; by the valency theorem, this is also the degree of the induced map $\mathbb C/\Lambda\to\widehat{\mathbb C}$. If these poles have orders $n_1,\ldots,n_s$, then $d=\deg g=\sum_jn_j$, and $g'$ has poles of orders $n_j+1$. The <degree of the derivative of an elliptic function> is consequently
$$
\deg g'=d+s.
$$
Since every nonconstant elliptic function has at least one pole and $1\leq s\leq d$,
$$
\boxed{d+1\leq\deg g'\leq2d.}
$$

Let $d\geq3$ be odd. The <Weierstrass elliptic function> supplies the lower-bound example
$$
g(z)=\wp(z)^{(d-3)/2}\wp'(z).
$$
It has one pole modulo $\Lambda$, of order $d$, so its derivative has one pole of order $d+1$. For the upper bound, choose distinct points $a_1,\ldots,a_d$ modulo $\Lambda$ and nonzero constants $c_j$ with $\sum_jc_j=0$. The quasi-periodicity of the <Weierstrass zeta function> makes
$$
g(z)=\sum_{j=1}^dc_j\zeta(z-a_j)
$$
elliptic. It has exactly $d$ simple poles, while $g'$ has $d$ double poles. Therefore the two bounds are attained for every required odd degree.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=25G}

For a <divisor on an algebraic curve> $D$ on a smooth projective curve $X$ of genus $g$, the <Riemann-Roch theorem> states
$$
\ell(D)-\ell(K_X-D)=\deg D+1-g,
$$
where $K_X$ is a <canonical divisor>. Taking $D=0$ gives $1-\ell(K_X)=1-g$, so $\ell(K_X)=g$. Taking $D=K_X$ then gives
$$
g-1=\deg K_X+1-g,
\qquad
\boxed{\deg K_X=2g-2}.
$$

To obtain a uniform projective embedding, choose a divisor $D$ of degree $2g+1$. Since every divisor appearing below has degree greater than $2g-2$, Riemann--Roch gives
$$
\ell(D)=g+2,
\quad
\ell(D-P)=g+1,
\quad
\ell(D-P-Q)=g,
\quad
\ell(D-2P)=g.
$$
The first two equalities show that the <complete linear system of a divisor> $|D|$ has no base point. The strict drops in the last two comparisons show respectively that its sections separate distinct points $P,Q$ and tangent directions at $P$. Thus $D$ is a <very ample divisor>, in accordance with the general fact that a <high-degree divisor is very ample on a smooth projective curve>, and its sections define a closed embedding
$$
\boxed{X\hookrightarrow\mathbb P^{\ell(D)-1}=\mathbb P^{g+1}.}
$$
The ambient dimension therefore depends only on $g$.

The <Riemann-Hurwitz formula> for a nonconstant morphism $h:C_1\to C_2$ of degree $n$ is
$$
2g(C_1)-2=n\bigl(2g(C_2)-2\bigr)
+\sum_{P\in C_1}(e_P-1).
$$
Choose a <smooth plane quartic> $C$, so $g(C)=3$, and form the <product of projective varieties> $S=C\times C$. This is a smooth projective variety of dimension two. If $B\subset S$ is an irreducible curve, pass to its <normalization of an algebraic curve>[normalization] $\widetilde B$. At least one coordinate projection $\widetilde B\to C$ is nonconstant, since otherwise $B$ would be a point. For that projection, Riemann--Hurwitz gives
$$
2g(\widetilde B)-2
\geq n(2\cdot3-2)\geq4,
$$
so the <geometric genus> of $B$ is at least three. Hence $S$ is the required surface; this is the <product surface without low-genus curves> construction.

Finally let $X\subset\mathbb P^2$ be a smooth plane curve of degree $d$ and let $p\notin X$. After a projective change of coordinates, take $p=[0:0:1]$. Projection away from $p$ is
$$
[x:y:z]\longmapsto[x:y].
$$
Its two homogeneous coordinate functions cannot vanish simultaneously on $X$, because their common zero in $\mathbb P^2$ is $p$. The criterion for a <morphism of algebraic varieties> therefore shows that the restriction
$$
\pi:X\longrightarrow\mathbb P^1
$$
is a morphism. A fibre is the intersection with a line through $p$, and a general such line meets $X$ in $d$ points counted with multiplicity. Thus the <projection of a plane curve from an exterior point> has degree $d$.

By the <genus of a smooth plane curve>, $g(X)=(d-1)(d-2)/2$. Applying Riemann--Hurwitz to $\pi$ and its <ramification divisor> $R_\pi$ gives
$$
\deg R_\pi
=2g(X)-2+2d
=(d-1)(d-2)-2+2d
=d(d-1).
$$
Every ramification point contributes at least one to this degree, so
$$
\boxed{\#\{P\in X:\pi\text{ ramifies at }P\}\leq d(d-1)}.
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=26g}
{scope}

= Solution
{parent=a}

Let $\gamma$ be a $C^3$ <regular curve> parametrized by <arc length>, so $|\gamma'(s)|=1$. Define the <Frenet frame>
$$
T=\gamma',
\qquad
\kappa=|T'|,
\qquad
N=\frac{T'}{\kappa},
\qquad
B=T\times N,
$$
and define the torsion by
$$
\tau=-B'\mathbin\cdot N.
$$
Thus curvature requires two derivatives, while $N,B,$ and torsion require $\kappa>0$; the displayed classical torsion requires three derivatives. At a point where $\kappa=0$, $N,B,$ and $\tau$ are not defined by this construction.

Since $(T,N,B)$ is an orthonormal frame, differentiating its inner products shows that its derivative matrix is skew-symmetric. The definition gives $T'=\kappa N$. Next,
$$
N'\mathbin\cdot T=-N\mathbin\cdot T'=-\kappa,
\qquad
N'\mathbin\cdot N=0,
$$
so $N'=-\kappa T+cB$. As $B'=T\times N'$ and $T\times B=-N$, we have $B'=-cN$; hence $c=-B'\cdot N=\tau$. The <Frenet-Serret formulas> are therefore
$$
\boxed{
T'=\kappa N,
\qquad
N'=-\kappa T+\tau B,
\qquad
B'=-\tau N.}
$$

Solved by gpt-5.6-sol high.

= b
{parent=26g}
{scope}

= Solution
{parent=b}

If $\kappa\equiv0$, then $T'=0$, so the unit tangent is constant and
$$
\gamma(s)=p+sT.
$$
Thus both curves are straight line segments. Equal arc length makes their parameter intervals equally long, and a rotation in $SO(3)$ followed by a translation maps one to the other. Hence they are related by a <proper Euclidean motion of Euclidean three-space>.

The conclusion fails for $\kappa\equiv1$, because curvature alone does not determine a space curve. A unit circle has curvature one and torsion zero. The <circular helix> with $a=b=1/2$,
$$
\eta(s)=\left(\frac12\cos(\sqrt2s),
\frac12\sin(\sqrt2s),
\frac{s}{\sqrt2}\right),
$$
is unit speed and has curvature one but torsion one. Proper Euclidean motions preserve torsion, so equal-length pieces of these two curves cannot be related by one.

Solved by gpt-5.6-sol high.

= c
{parent=26g}
{scope}

= Solution
{parent=c}

Write a <proper Euclidean motion of Euclidean three-space> as $E(x)=Ax+b$, where $A\in SO(3)$. Then
$$
(E\circ\gamma)^{(j)}=A\gamma^{(j)}
\qquad(j\geq1).
$$
Because $A$ preserves norms, dot products, and cross products, the formulas
$$
\kappa=|T'|,
\qquad
\tau=\frac{\det(\gamma',\gamma'',\gamma''')}
{|\gamma'\times\gamma''|^2}
$$
show directly that curvature and torsion are unchanged by $E$.

For the shifted curve $\gamma_{s_0}(s)=\gamma(s-s_0)$, every derivative at $s$ equals the corresponding derivative of $\gamma$ at $s-s_0$. Consequently
$$
\kappa_{\gamma_{s_0}}(s)=\kappa_\gamma(s-s_0),
\qquad
\tau_{\gamma_{s_0}}(s)=\tau_\gamma(s-s_0).
$$
Both are therefore <pointwise Euclidean invariant of a curve>[pointwise Euclidean invariants] in the stated sense.

Solved by gpt-5.6-sol high.

= d
{parent=26g}
{scope}

= Solution
{parent=d}

No. For sufficiently smooth curves,
$$
Q(\gamma,s)=\kappa_\gamma'(s)
$$
is unchanged by proper Euclidean motions and transforms correctly under translations of the arc-length parameter, so it is a <pointwise Euclidean invariant of a curve>. It is not determined by the two numbers $\kappa(s)$ and $\tau(s)$.

For a concrete comparison near $s=0$, use the <plane curve reconstructed from curvature> construction with
$$
\kappa_1(s)=1,
\qquad
\kappa_2(s)=1+s.
$$
The resulting unit-speed planar curves both have $\kappa(0)=1$ and $\tau(0)=0$, but their curvature derivatives at zero are respectively $0$ and $1$. Thus no single function $f$ can satisfy $Q(\gamma,s)=f(\kappa(s),\tau(s))$ for all curves.

Indeed, the covariance conditions alone even permit nonlocal examples such as $Q(\gamma,s)=\kappa_\gamma(s+1)$. This is the distinction recorded by <pointwise Euclidean invariants need not depend only on current curvature and torsion>.

Solved by gpt-5.6-sol high.

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

= a
{parent=27k}
{scope}

= Solution
{parent=a}

Use the convention
$$
\widehat f(u)=\int_{\mathbb R^d}e^{iu\cdot x}f(x)\,dx.
$$
The <Fourier inversion theorem> says that, when $f,\widehat f\in L^1(\mathbb R^d)$,
$$
f(x)=\frac1{(2\pi)^d}
\int_{\mathbb R^d}e^{-iu\cdot x}\widehat f(u)\,du
$$
for almost every $x$.

Since
$$
\overline{\widehat f(u)}
=\int_{\mathbb R^d}e^{-iu\cdot x}\overline{f(x)}\,dx,
$$
the <Fubini's theorem> and Fourier inversion give
$$
\begin{aligned}
\int_{\mathbb R^d}|\widehat f(u)|^2\,du
&=\int_{\mathbb R^d}\widehat f(u)
 \int_{\mathbb R^d}e^{-iu\cdot x}\overline{f(x)}\,dx\,du\\
&=\int_{\mathbb R^d}\overline{f(x)}
 \int_{\mathbb R^d}e^{-iu\cdot x}\widehat f(u)\,du\,dx\\
&=(2\pi)^d\int_{\mathbb R^d}|f(x)|^2\,dx.
\end{aligned}
$$
Thus the <Plancherel theorem> identity in this convention is
$$
\boxed{\|\widehat f\|_2^2=(2\pi)^d\|f\|_2^2.}
$$

The inverse integral is a continuous function of $x$, since $\widehat f\in L^1$ permits dominated convergence. If $f$ is continuous, it and the inverse integral are continuous and agree almost everywhere. They must agree everywhere: otherwise their continuous difference would be nonzero on an open set of positive measure. This is the <continuous version of Fourier inversion>.

Solved by gpt-5.6-sol high.

= b
{parent=27k}
{scope}

= Solution
{parent=b}

Put
$$
H(u)=\frac{4\sin^2(u/2)}{u^2},
\qquad H(0)=1.
$$
This function is bounded near zero and is $O(u^{-2})$ at infinity, so $H\in L^1(\mathbb R)$ and the defining integral for $g$ converges absolutely.

For the <triangular function>
$$
h(x)=(1-|x|)\mathbf1_{[-1,1]}(x),
$$
direct integration gives
$$
\begin{aligned}
\widehat h(u)
&=\int_{-1}^1(1-|x|)e^{iux}\,dx\\
&=2\int_0^1(1-x)\cos(ux)\,dx\\
&=\frac{2(1-\cos u)}{u^2}
=\frac{4\sin^2(u/2)}{u^2}=H(u).
\end{aligned}
$$
The <Fourier inversion theorem>, or equivalently the <Fourier transform of a triangular function>, therefore yields
$$
\boxed{g(x)=(1-|x|)\mathbf1_{[-1,1]}(x).}
$$
In particular $g(x)=0$ whenever $|x|>1$. Finally,
$$
\boxed{\|g\|_2^2
=2\int_0^1(1-x)^2\,dx
=\frac23.}
$$
This also agrees with the <Plancherel theorem> applied to $h$ and $H$.

Solved by gpt-5.6-sol high.

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

= a
{parent=28j}
{scope}

= Solution
{parent=a}

A simple birth process with parameter $\lambda$ starting from one individual is the <Yule process>: when its population is $n$, it waits an exponential time of rate $n\lambda$ and then moves to $n+1$.

For $f(n)=n$, the <Markov jump-process generator> satisfies
$$
(Lf)(n)=n\lambda\bigl(f(n+1)-f(n)\bigr)=n\lambda.
$$
Therefore $\mu(t)=\mathbb E X_t$ obeys
$$
\mu'(t)=\lambda\mu(t),
\qquad \mu(0)=1.
$$
It follows, as recorded by the <mean population of a Yule process>, that
$$
\boxed{\mathbb E X_t=e^{\lambda t}.}
$$

Solved by gpt-5.6-sol high.

= b
{parent=28j}
{scope}

= Solution
{parent=b}

For $i\geq1$, the total holding rate is $(\lambda+1)2^i$, and the <jump chain> moves upward and downward with probabilities
$$
p=\frac{\lambda}{\lambda+1},
\qquad
q=\frac1{\lambda+1},
$$
while it moves from zero to one with probability one. Thus it is a <reflected nearest-neighbour random walk>. Recurrence is unaffected by the holding times, so it is recurrent exactly when $p\leq q$. In the given range,
$$
\boxed{X\text{ is recurrent exactly when }\lambda=1.}
$$
For $\lambda>1$ the jump chain is transient.

The <detailed balance for a birth-death process>[detailed-balance equations] are
$$
\pi_i\lambda2^i=\pi_{i+1}2^{i+1}
\qquad(i\geq0),
$$
where the formula also covers the boundary $i=0$. Hence every candidate invariant measure has
$$
\pi_i=\pi_0\left(\frac\lambda2\right)^i.
$$
At $\lambda=1$ this is summable and the chain is nonexplosive, giving
$$
\pi_i=2^{-i-1},
\qquad i\geq0.
$$
For $\lambda=2$ it is not summable. When $1<\lambda<2$, it is a summable formal solution of $\pi Q=0$, but the chain explodes, so it is not an invariant probability for the minimal process; this is precisely the caveat in <invariant distribution of an explosive chain>. Consequently
$$
\boxed{X\text{ has an invariant distribution exactly when }\lambda=1.}
$$

It remains to classify explosion. At $\lambda=1$, the recurrent jump chain visits zero infinitely often. The holding time at zero has rate one, so <nonexplosion from recurrent visits to a slow state> shows that the accumulated time is infinite almost surely.

For $\lambda>1$, let $V_i$ be the total number of visits of the transient jump chain to $i$. Using the stated uniform visit bound and writing $q_i=(\lambda+1)2^i$ for $i\geq1$,
$$
\mathbb E\!\left[\sum_{n\geq0}\frac1{q_{Y_n}}\right]
=\sum_{i\geq0}\frac{\mathbb E V_i}{q_i}
<\infty,
$$
because $\sum_i2^{-i}<\infty$. Conditional on the jump path, this is the expected sum of all holding times, so the total lifetime is finite almost surely. This is <explosion of an upward-biased walk with geometrically increasing rates>. Therefore
$$
\boxed{X\text{ is explosive exactly when }1<\lambda\leq2.}
$$

Solved by gpt-5.6-sol high.

= c
{parent=28j}
{scope}

= Solution
{parent=c}

The assertion is false. Take the chain from part (b) with $\lambda=1$. It is irreducible and nonexplosive, and it has the invariant distribution
$$
\pi_i=2^{-i-1}.
$$
Hence every state is a <positive recurrent state> for the continuous-time chain.

Its jump chain is the reflected simple symmetric random walk on $\mathbb N$. This chain is recurrent but has only the infinite invariant measure
$$
\mu_0=1,
\qquad
\mu_i=2\quad(i\geq1),
$$
so every state is a <null recurrent state>. Equivalently, the <invariant-measure transfer between a jump chain and a CTMC> divides this infinite jump-chain measure by the rapidly growing holding rates and produces the summable continuous-time invariant measure. Thus <fast holding rates create continuous-time positive recurrence>, and positive recurrence need not pass from a continuous-time chain to its jump chain.

Solved by gpt-5.6-sol high.

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

= a
{parent=29k}
{scope}

= Solution
{parent=a}

Provided $X$ and $Y$ have finite positive variances, their <correlation coefficient> is
$$
\boxed{
\rho_{X,Y}=
\frac{\operatorname{Cov}(X,Y)}
{\sqrt{\operatorname{Var}(X)\operatorname{Var}(Y)}}.}
$$
It lies in $[-1,1]$ by the Cauchy--Schwarz inequality.

Solved by gpt-5.6-sol high.

= b
{parent=29k}
{scope}

= Solution
{parent=b}

Write $Z_i=(X_i,Y_i)^T$ for the independent observations and put $\Omega=\Sigma^{-1}$. Apart from constants, the log-likelihood is
$$
\ell(\Omega)=\frac n2\log|\Omega|
-\frac12\sum_{i=1}^n Z_i^T\Omega Z_i.
$$
Using the stated matrix derivatives, its <score function> is
$$
\frac{\partial\ell}{\partial\Omega}
=\frac n2\Omega^{-1}
-\frac12\sum_{i=1}^nZ_iZ_i^T.
$$
At a root of the score, $\Omega^{-1}=n^{-1}\sum_iZ_iZ_i^T$. Hence the <maximum-likelihood covariance estimator for centered Gaussian data> is
$$
\boxed{
\widehat\Sigma=\frac1n\sum_{i=1}^nZ_iZ_i^T
=\begin{pmatrix}
 n^{-1}\sum_iX_i^2&n^{-1}\sum_iX_iY_i\\
 n^{-1}\sum_iX_iY_i&n^{-1}\sum_iY_i^2
\end{pmatrix}.}
$$

Solved by gpt-5.6-sol high.

= c
{parent=29k}
{scope}

= Solution
{parent=c}

Set $a=\Sigma_{11}$, $c=\Sigma_{12}$, and $b=\Sigma_{22}$. The estimator is the sample mean of the independent random vectors
$$
W_i=(X_i^2,X_iY_i,Y_i^2)^T,
\qquad
\mathbb EW_i=(a,c,b)^T.
$$
The supplied fourth moments give
$$
\begin{aligned}
\operatorname{Var}(X^2)&=2a^2,&
\operatorname{Cov}(X^2,XY)&=2ac,&
\operatorname{Cov}(X^2,Y^2)&=2c^2,\\
\operatorname{Var}(XY)&=ab+c^2,&
\operatorname{Cov}(XY,Y^2)&=2bc,&
\operatorname{Var}(Y^2)&=2b^2.
\end{aligned}
$$
Therefore the <multivariate central limit theorem> gives
$$
\boxed{
\sqrt n
\left[
\begin{pmatrix}
\widehat\Sigma_{11}\\
\widehat\Sigma_{12}\\
\widehat\Sigma_{22}
\end{pmatrix}
-
\begin{pmatrix}a\\c\\b\end{pmatrix}
\right]
\xrightarrow d
N_3\!\left(0,
\begin{pmatrix}
2a^2&2ac&2c^2\\
2ac&ab+c^2&2bc\\
2c^2&2bc&2b^2
\end{pmatrix}\right).}
$$
This is the <asymptotic covariance of the bivariate Gaussian covariance estimator>.

Solved by gpt-5.6-sol high.

= d
{parent=29k}
{scope}

= Solution
{parent=d}

Let
$$
g(a,c,b)=\frac{c}{\sqrt{ab}}=\rho.
$$
Its gradient, in the coordinate order $(a,c,b)$, is
$$
\nabla g=
\left(-\frac{\rho}{2a},\frac1{\sqrt{ab}},-\frac{\rho}{2b}\right)^T.
$$
Apply the <Delta method> to the limit in part (c). Multiplying the displayed covariance matrix there on both sides by this gradient gives
$$
(\nabla g)^TV\nabla g=(1-\rho^2)^2.
$$
Consequently the <asymptotic distribution of the Gaussian sample correlation> is
$$
\boxed{
\sqrt n(\widehat\rho-\rho_{X,Y})
\xrightarrow d
N\!\left(0,
\left(1-\frac{\Sigma_{12}^2}{\Sigma_{11}\Sigma_{22}}\right)^2
\right).}
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=30k}
{scope}

= Solution
{parent=a}

The process is adapted because $X_0$ is $\mathcal F_0$-measurable, $A_k$ is $\mathcal F_{k-1}$-measurable, and $M_k-M_{k-1}$ is $\mathcal F_k$-measurable. Using the <predictable process>[predictability] of $A_n$ and the martingale increment property,
$$
\begin{aligned}
\mathbb E[X_n\mid\mathcal F_{n-1}]
&=X_{n-1}+A_n\mathbb E[M_n-M_{n-1}\mid\mathcal F_{n-1}]\\
&=X_{n-1}.
\end{aligned}
$$
The assumed integrability therefore proves that $X$ is the <martingale transform> of $M$ by $A$ and is a martingale.

Solved by gpt-5.6-sol high.

= b
{parent=30k}
{scope}

= Solution
{parent=b}

The increments of $Y_n=M_{n\wedge T}$ satisfy
$$
Y_n-Y_{n-1}
=\mathbf1_{\{T\geq n\}}(M_n-M_{n-1}).
$$
Since $\{T\geq n\}=\{T>n-1\}\in\mathcal F_{n-1}$, the indicator is predictable. Part (a), applied to this bounded integrand, shows that $Y$ is a martingale. Equivalently, this is the <stopped martingale> theorem.

Solved by gpt-5.6-sol high.

= c
{parent=30k}
{scope}

= Solution
{parent=c}

Write $\Delta M_n=M_n-M_{n-1}$. Conditional on $\mathcal F_{n-1}$, the variables $X_{n-1}$, $M_{n-1}$, and $A_n$ are known, while $\mathbb E[\Delta M_n\mid\mathcal F_{n-1}]=0$. Hence
$$
\begin{aligned}
\operatorname{Cov}(X_n,M_n\mid\mathcal F_{n-1})
&=\operatorname{Cov}(A_n\Delta M_n,\Delta M_n\mid\mathcal F_{n-1})\\
&=A_n\operatorname{Var}(\Delta M_n\mid\mathcal F_{n-1})\\
&=A_n\operatorname{Var}(M_n\mid\mathcal F_{n-1}).
\end{aligned}
$$
The denominator is positive almost surely, so the <recovery of a martingale-transform integrand by conditional covariance> gives
$$
\boxed{A_n=
\frac{\operatorname{Cov}(X_n,M_n\mid\mathcal F_{n-1})}
{\operatorname{Var}(M_n\mid\mathcal F_{n-1})}.}
$$

Solved by gpt-5.6-sol high.

= d
{parent=30k}
{scope}

= Solution
{parent=d}

Let $D_n=M_n-M_{n-1}$. Given $\mathcal F_{n-1}$, the new information at time $n$ is only the independent symmetric sign $\xi_n$. Thus
$$
D_n=C_n+B_n\xi_n
$$
for some $\mathcal F_{n-1}$-measurable $C_n,B_n$. Taking conditional expectation and using the martingale property gives $C_n=0$. More explicitly, one can take
$$
B_n=\mathbb E[D_n\xi_n\mid\mathcal F_{n-1}]
=\mathbb E[M_n\xi_n\mid\mathcal F_{n-1}],
$$
which is predictable. Therefore
$$
\boxed{M_n=M_0+\sum_{k=1}^nB_k\xi_k.}
$$
This is the <predictable representation in a Rademacher filtration>.

Solved by gpt-5.6-sol high.

= e
{parent=30k}
{scope}

= Solution
{parent=e}

Choose $N$ with $T\leq N$. By part (d),
$$
M_T=M_0+\sum_{k=1}^NB_k\xi_k\mathbf1_{\{T\geq k\}}.
$$
Each $B_k\mathbf1_{\{T\geq k\}}$ is $\mathcal F_{k-1}$-measurable. Distinct summands are orthogonal: if $j<k$, conditioning on $\mathcal F_{k-1}$ makes every factor except $\xi_k$ known and $\mathbb E[\xi_k\mid\mathcal F_{k-1}]=0$. The cross terms with $M_0$ vanish in the same way, while $\xi_k^2=1$. Expanding the square therefore gives
$$
\begin{aligned}
\mathbb E[M_T^2]
&=M_0^2+\sum_{k=1}^N\mathbb E[B_k^2\mathbf1_{\{T\geq k\}}]\\
&=\boxed{M_0^2+\mathbb E\left[\sum_{k=1}^TB_k^2\right]}.
\end{aligned}
$$
This is the <stopped martingale isometry in a Rademacher filtration>.

Solved by gpt-5.6-sol high.

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

= a
{parent=31j}
{scope}

= Solution
{parent=a}

For fixed sample points $x_{1:n}=(x_1,\ldots,x_n)$, write
$$
\mathcal H(x_{1:n})
=\{(h(x_1),\ldots,h(x_n)):h\in\mathcal H\}.
$$
The <shattering coefficient> is
$$
\boxed{s(\mathcal H,n)=\sup_{x_{1:n}\in\mathcal X^n}|\mathcal H(x_{1:n})|.}
$$
Thus it is the largest number of distinct binary labelings that the <hypothesis class> can realize on $n$ points.

A set of $n$ points is shattered when all $2^n$ labelings are realized. The <VC dimension> is therefore
$$
\boxed{\operatorname{VC}(\mathcal H)
=\sup\{n:s(\mathcal H,n)=2^n\},}
$$
with value infinity if arbitrarily large finite sets are shattered.

Solved by gpt-5.6-sol high.

= b
{parent=31j}
{scope}

= Solution
{parent=b}

For every fixed $x_{1:n}$,
$$
(\mathcal H_1\cup\mathcal H_2)(x_{1:n})
=\mathcal H_1(x_{1:n})\cup\mathcal H_2(x_{1:n}).
$$
The cardinality of a union is at most the sum of the two cardinalities, so
$$
|(\mathcal H_1\cup\mathcal H_2)(x_{1:n})|
\leq s(\mathcal H_1,n)+s(\mathcal H_2,n).
$$
Taking the supremum over samples proves the <shattering coefficient of a union> bound
$$
\boxed{s(\mathcal H_1\cup\mathcal H_2,n)
\leq s(\mathcal H_1,n)+s(\mathcal H_2,n).}
$$

Solved by gpt-5.6-sol high.

= c
{parent=31j}
{scope}

= Solution
{parent=c}

The functions $x\mapsto x^T\beta$ form a real vector space of dimension $p$. Hence the <VC dimension of a vector space> gives
$$
\operatorname{VC}(\mathcal H_{\mathcal F_1})\leq p.
$$
Applying the <Sauer-Shelah lemma> and then the <Sauer-Shelah growth bound>,
$$
s(\mathcal H_{\mathcal F_1},n)
\leq\sum_{j=0}^p\binom nj
\leq\boxed{(n+1)^p}.
$$
This is the <growth bound for homogeneous linear classifiers>.

Solved by gpt-5.6-sol high.

= d
{parent=31j}
{scope}

= Solution
{parent=d}

First fix an ordered tuple $(g_1,\ldots,g_m)\in\mathcal G^m$. On writing
$$
z(x)=(g_1(x),\ldots,g_m(x))\in\mathbb R^m,
$$
the resulting classifiers are $x\mapsto\operatorname{sgn}(\alpha^Tz(x))$. The <growth bound for homogeneous linear classifiers> in dimension $m$ gives at most $(n+1)^m$ label vectors on any $n$ sample points.

If $\mathcal G$ is finite, there are $|\mathcal G|^m$ ordered tuples of hidden functions. The <shattering coefficient of a union> therefore gives
$$
\boxed{s(\mathcal H_{\mathcal F_2},n)
\leq(n+1)^m|\mathcal G|^m.}
$$

For an arbitrary $\mathcal G$, fix $x_{1:n}$. Choose one representative for every distinct vector in $\mathcal G(x_{1:n})$, obtaining a finite class $\mathcal G'$ with
$$
|\mathcal G'|=|\mathcal G(x_{1:n})|
\leq s(\mathcal G,n).
$$
Every $m$-tuple from $\mathcal G$ agrees on the sample with an $m$-tuple from $\mathcal G'$. The finite-class argument now gives
$$
|\mathcal H_{\mathcal F_2}(x_{1:n})|
\leq(n+1)^m s(\mathcal G,n)^m.
$$
Taking the supremum over $x_{1:n}$ proves the general <growth bound for signs of m-term linear combinations>:
$$
\boxed{s(\mathcal H_{\mathcal F_2},n)
\leq(n+1)^m s(\mathcal G,n)^m.}
$$

Solved by gpt-5.6-sol high.

= e
{parent=31j}
{scope}

= Solution
{parent=e}

A function in the stated network class has the form
$$
f(x)=\sum_{j=1}^m\alpha_j\operatorname{sgn}(x^T\beta_j).
$$
Thus this is the class $\mathcal F_2$ from part (d), with the hidden-unit class
$$
\mathcal G=\mathcal H_{\mathcal F_1}.
$$
Parts (c) and (d) give
$$
\begin{aligned}
s(\mathcal H_{\mathcal F_3},n)
&\leq(n+1)^m s(\mathcal G,n)^m\\
&\leq(n+1)^m\bigl((n+1)^p\bigr)^m\\
&=\boxed{(n+1)^{(p+1)m}}.
\end{aligned}
$$
This is the <growth bound for a single-hidden-layer sign network>.

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}

For $n\geq1$, put $r_n=\phi_n/\phi_{n-1}$. Since $(\phi_n)$ is an <asymptotic sequence>, $r_n\to0$. We can rewrite
$$
\psi_n=\frac{\phi_n}{1+r_n},
$$
so
$$
\frac{\psi_n}{\phi_n}=\frac1{1+r_n}\longrightarrow1.
$$
Together with $\psi_0=\phi_0$, this proves
$$
\boxed{\psi_n\sim\phi_n\quad\text{for every }n\geq0.}
$$
Moreover,
$$
\frac{\psi_{n+1}}{\psi_n}
=\frac{\psi_{n+1}}{\phi_{n+1}}
 \frac{\phi_{n+1}}{\phi_n}
 \frac{\phi_n}{\psi_n}
\longrightarrow1\cdot0\cdot1=0.
$$
Hence $(\psi_n)$ is an asymptotic sequence. This is the <harmonic-mean refinement of an asymptotic sequence>.

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

First,
$$
\frac{\chi_1}{\chi_0}
=\sqrt{\frac{\phi_1}{\phi_0}}\longrightarrow0.
$$
For $n\geq1$,
$$
\frac{\chi_{n+1}}{\chi_n}
=\sqrt{\frac{\phi_{n+1}}{\phi_{n-1}}}
=\sqrt{\frac{\phi_{n+1}}{\phi_n}
       \frac{\phi_n}{\phi_{n-1}}}
\longrightarrow0.
$$
Thus $(\chi_n)$ is an asymptotic sequence.

Termwise <asymptotic equivalence> fails for every $n\geq1$, because
$$
\frac{\chi_n}{\phi_n}
=\sqrt{\frac{\phi_{n-1}}{\phi_n}}
\longrightarrow\infty.
$$
For example, as $x\to0$, the positive scale $\phi_n(x)=|x|^n$ gives $\chi_n(x)=|x|^{n-1/2}$ for $n\geq1$. This illustrates the <geometric-mean refinement of an asymptotic sequence>.

Solved by gpt-5.6-sol high.

= b
{parent=32e}
{scope}

= Solution
{parent=b}

No. Let $x_0=0$ and, for $0<|x|<1$, define
$$
\phi_n(x)=|x|^n,
$$
while
$$
\psi_0(x)=1+|x|^{1/2},
\qquad
\psi_n(x)=|x|^n\quad(n\geq1).
$$
Both families are asymptotic sequences, and $\psi_n\sim\phi_n$ for every $n$.

Take $f(x)=1$. It has the exact expansion
$$
f(x)\sim1\cdot\phi_0(x)+\sum_{n=1}^{\infty}0\cdot\phi_n(x).
$$
Suppose it had an expansion $f\sim\sum_{n\geq0}b_n\psi_n$. The order-zero condition forces $b_0=1$. At order one we would then require some constant $b_1$ such that
$$
1-\psi_0-b_1\psi_1
=-|x|^{1/2}-b_1|x|=o(|x|).
$$
After division by $|x|$, the left side is $-|x|^{-1/2}-b_1$, which cannot tend to zero. This contradiction proves that no such coefficients exist. Thus <termwise equivalent asymptotic scales need not preserve expansions>.

Solved by gpt-5.6-sol high.

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

= a
{parent=33a}
{scope}

= Solution
{parent=a}

For $\dot x=f(x)$, a fixed point $x^*$ satisfies $f(x^*)=0$. It is <Lyapunov stability>[Lyapunov stable] when, for every $\varepsilon>0$, there is $\delta>0$ such that
$$
|x(0)-x^*|<\delta
\quad\Longrightarrow\quad
|x(t)-x^*|<\varepsilon\quad(t\geq0).
$$

A <Lyapunov function> on a neighbourhood of $x^*$ is a continuously differentiable function $V$ such that
$$
V(x^*)=0,
\qquad V(x)>0\quad(x\ne x^*),
\qquad
\dot V(x)=\nabla V(x)\cdot f(x)\leq0.
$$
The <First Lyapunov theorem> states that such a function makes $x^*$ Lyapunov stable. If $\dot V<0$ away from $x^*$, then $x^*$ is locally asymptotically stable.

To prove stability, choose a closed ball contained in the domain of $V$. On its boundary sphere $|x-x^*|=\varepsilon$, compactness and positive definiteness give
$$
m=\min_{|x-x^*|=\varepsilon}V(x)>0.
$$
Continuity at $x^*$ supplies $\delta<\varepsilon$ such that $|x-x^*|<\delta$ implies $V(x)<m$. Along a trajectory, $V$ cannot increase. Such a trajectory therefore cannot first reach the boundary sphere, where its value would be at least $m$. This proves Lyapunov stability.

If $\dot V<0$ away from $x^*$, take a sufficiently small compact sublevel set of $V$. The <LaSalle invariance principle> says that every trajectory in it approaches the largest invariant subset of $\{\dot V=0\}$, which is just $\{x^*\}$. Hence the equilibrium is also <asymptotic stability>[asymptotically stable].

Solved by gpt-5.6-sol high.

= b
{parent=33a}
{scope}

= Solution
{parent=b}

Put $y=\dot x$. The first-order system is
$$
\dot x=y,
\qquad
\dot y=-2x(1-x^2)^2-\mu y.
$$
At an <equilibrium point of a dynamical system>, $y=0$ and $x(1-x^2)^2=0$, so the three fixed points are
$$
\boxed{(0,0),\ (1,0),\ (-1,0).}
$$

Define
$$
U(x)=x^2-x^4+\frac{x^6}{3},
\qquad
E(x,y)=\frac12y^2+U(x).
$$
Then $U'(x)=2x(1-x^2)^2=-F(x)$ and
$$
\dot E
=y\dot y+U'(x)\dot x
=-\mu y^2\leq0.
$$
Moreover,
$$
U(x)=x^2\left(1-x^2+\frac{x^4}{3}\right)>0
\quad(x\ne0),
$$
so $E$ is positive definite at the origin. It is therefore a <damped mechanical energy as a Lyapunov function>, and the <First Lyapunov theorem> proves that the origin is Lyapunov stable.

Since $U(\pm1)=1/3$, choose a compact energy sublevel $\{E\leq c\}$ with $0<c<1/3$. In this set, $\dot E=0$ means $y=0$. A trajectory can remain in $y=0$ only when $F(x)=0$, and the chosen sublevel excludes $x=\pm1$. Thus the largest invariant subset of $\{\dot E=0\}$ is the origin. The <LaSalle invariance principle> proves that the origin is asymptotically stable.

Finally, $E$ is radially unbounded because its leading terms are $x^6/3+y^2/2$. Every forward trajectory is therefore bounded. LaSalle's principle puts its <omega-limit set> inside
$$
\{(x,0):F(x)=0\}
=\{(-1,0),(0,0),(1,0)\}.
$$
An omega-limit set of a bounded continuous trajectory is nonempty and connected. A connected subset of this three-point set is a singleton, so every trajectory has precisely one of the three fixed points as its omega-limit set.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=34E}

The spectral equation $(S)$ is
$$
-\phi_{xx}+u\phi=\lambda\phi,
\qquad\text{equivalently}\qquad
\phi_{xx}=(u-\lambda)\phi.
$$
This is the <KdV Schrodinger spectral problem>. Differentiating it with respect to $t$ gives
$$
\phi_{xxt}=(u_t-\dot\lambda)\phi+(u-\lambda)\phi_t.
$$
For
$$
Q=\phi_t+u_x\phi-2(u+2\lambda)\phi_x,
$$
a direct differentiation, followed by substitution of the two preceding identities, yields
$$
Q_{xx}=(u-\lambda)Q+
\bigl(u_t+u_{xxx}-6uu_x-\dot\lambda\bigr)\phi.
$$
Therefore
$$
\begin{aligned}
\partial_x(\phi_xQ-\phi Q_x)
&=\phi_{xx}Q-\phi Q_{xx}\\
&=\phi^2\bigl(\dot\lambda-u_t-u_{xxx}+6uu_x\bigr).
\end{aligned}
$$
Comparison with $(*)$ gives
$$
\boxed{\alpha=-1.}
$$
This is the reusable <KdV Schrodinger spectral problem> <Wronskian>[Wronskian identity].

Now take $u$ to satisfy the <Korteweg-De Vries equation> and $\phi=\varphi_n$, where $\lambda_n=-\kappa_n^2$. Both $\varphi_n$ and its derivative decay at infinity, as does $u$, so integration of $(*)$ over the real line gives
$$
0=\dot\lambda_n\int_{-\infty}^{\infty}\varphi_n^2\,dx
=\dot\lambda_n.
$$
The normalization was used in the last equality. Hence
$$
\boxed{\lambda_n(t)=\lambda_n(0)},
$$
which is the <Isospectrality of the KdV discrete spectrum>.

With the KdV equation and $\dot\lambda_n=0$, the Wronskian identity says
$$
\partial_x(\varphi_{n,x}Q-\varphi_nQ_x)=0.
$$
Decay at infinity makes the constant zero. Thus $(Q/\varphi_n)_x=0$ between zeros, and continuation across the isolated zeros gives
$$
Q(x,t)=h_n(t)\varphi_n(x,t).
$$
Multiplying by $\varphi_n$ and integrating,
$$
h_n=\int\varphi_n\varphi_{n,t}\,dx
+\int u_x\varphi_n^2\,dx
-2\int(u+2\lambda_n)\varphi_n\varphi_{n,x}\,dx.
$$
The first integral is zero by differentiating the normalization. Integration by parts turns the last integral into $\int u_x\varphi_n^2dx$, so
$$
h_n=2\int u_x\varphi_n^2\,dx.
$$
To evaluate this, differentiate $(S)$ in $x$ and use it to write
$$
u_x\varphi_n^2
=\varphi_n\varphi_{n,xxx}-\varphi_{n,x}\varphi_{n,xx}.
$$
Its integral is $-2\int\varphi_{n,x}\varphi_{n,xx}dx=0$ by decay. Hence
$$
\boxed{h_n=0,\qquad Q=0.}
$$

As $x\to+\infty$, rapid decay of $u$ and $u_x$, together with
$$
\varphi_n\sim c_n(t)e^{-\kappa_nx},
\qquad
\varphi_{n,x}\sim-\kappa_nc_n(t)e^{-\kappa_nx},
$$
reduces $Q=0$ to
$$
c_n'-4\kappa_n^3c_n=0.
$$
Since $\kappa_n$ is constant,
$$
\boxed{c_n(t)=c_n(0)e^{4\kappa_n^3t},}
$$
as stated by the <Evolution of a KdV discrete norming constant>.

Solved by gpt-5.6-sol high.

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

= i
{parent=35b}
{scope}

= Solution
{parent=i}

Let $\Delta=E_+-E_->0$. Hermiticity gives
$$
\langle-|\Delta H|+\rangle=-i\lambda,
$$
so in the ordered basis $(| -\rangle,|+\rangle)$,
$$
H_0+\Delta H=
\begin{pmatrix}
E_-&-i\lambda\\
i\lambda&E_+
\end{pmatrix}.
$$
For a <nondegenerate energy eigenvalue>[nondegenerate unperturbed level], the <first-order nondegenerate perturbation theory> formulas are
$$
E_n^{(1)}=\langle n|\Delta H|n\rangle,
\qquad
|n^{(1)}\rangle
=\sum_{m\ne n}|m\rangle
\frac{\langle m|\Delta H|n\rangle}{E_n-E_m}.
$$
The diagonal matrix elements vanish, so both linear energy corrections are zero. The state corrections are
$$
| -^{(1)}\rangle
=|+\rangle\frac{i\lambda}{E_--E_+}
=-\frac{i\lambda}{\Delta}|+\rangle,
$$
$$
|+^{(1)}\rangle
=|-\rangle\frac{-i\lambda}{E_+-E_-}
=-\frac{i\lambda}{\Delta}|-\rangle.
$$
Thus, through linear order,
$$
\boxed{
\begin{aligned}
E_-^{\rm pert}&=E_-+O(\lambda^2),&
|-\rangle_{\rm pert}&=|-\rangle-\frac{i\lambda}{\Delta}|+\rangle+O(\lambda^2),\\
E_+^{\rm pert}&=E_++O(\lambda^2),&
|+\rangle_{\rm pert}&=|+\rangle-\frac{i\lambda}{\Delta}|-\rangle+O(\lambda^2).
\end{aligned}}
$$
Normalization changes only at quadratic order.

Solved by gpt-5.6-sol high.

= ii
{parent=35b}
{scope}

= Solution
{parent=ii}

The characteristic equation is
$$
(E_--E)(E_+-E)-\lambda^2=0.
$$
Hence the <imaginary-coupled two-level Hamiltonian> has exact energies
$$
\boxed{
E_{\rm L,H}=\frac{E_-+E_+}{2}
\mp\sqrt{\frac{\Delta^2}{4}+\lambda^2}.}
$$
Choose a real angle $\theta$ satisfying
$$
\tan(2\theta)=\frac{2\lambda}{\Delta},
\qquad \theta\to0\quad\text{as }\lambda\to0.
$$
Normalized exact eigenstates are
$$
\boxed{
|L\rangle=\cos\theta|-\rangle-i\sin\theta|+\rangle,
\qquad
|H\rangle=-i\sin\theta|-\rangle+\cos\theta|+\rangle.}
$$
Since
$$
\theta=\frac{\lambda}{\Delta}+O(\lambda^3),
\qquad
\sqrt{\frac{\Delta^2}{4}+\lambda^2}
=\frac\Delta2+O(\lambda^2),
$$
these states and energies reproduce exactly the linear-order perturbative results from part (i).

Solved by gpt-5.6-sol high.

= iii
{parent=35b}
{scope}

= Solution
{parent=iii}

The exact energies and mixing coefficients contain
$$
\sqrt{\frac{\Delta^2}{4}+\lambda^2}.
$$
As a function of complex $\lambda$, its nearest <branch points> to the origin occur when its argument vanishes:
$$
\lambda=\pm\frac{i\Delta}{2}.
$$
Their distance from the expansion point $\lambda=0$ determines the <radius of convergence>. Therefore
$$
\boxed{R=\frac{\Delta}{2}=\frac{E_+-E_-}{2}.}
$$
The real-axis perturbation series converges for $|\lambda|<(E_+-E_-)/2$.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=36D}

For any nonzero state, the <Rayleigh quotient> is
$$
R[\psi]=\frac{\langle\psi|\widehat H|\psi\rangle}
{\langle\psi|\psi\rangle}.
$$
Expand $|\psi\rangle=\sum_nc_n|\psi_n\rangle$ in normalized energy eigenstates. Then
$$
R[\psi]=\frac{\sum_n|c_n|^2E_n}{\sum_n|c_n|^2}
\geq E_0.
$$
Because the ground state is unique, equality holds exactly when all coefficients except $c_0$ vanish. Thus the <Rayleigh-Ritz variational principle> gives its minimum at the ray of $|\psi_0\rangle$.

For the proposed exact state, write $\widetilde\psi=e^{-\beta x^n}$. Its logarithmic derivatives give
$$
\frac{\widetilde\psi''}{\widetilde\psi}
=\beta^2n^2x^{2n-2}-\beta n(n-1)x^{n-2}.
$$
The stationary Schrödinger equation, after multiplication by $2m/\hbar^2$, is
$$
\left[-\frac{d^2}{dx^2}+x^6-3x^2+2\right]\widetilde\psi
=\varepsilon\widetilde\psi,
\qquad E=\frac{\hbar^2}{2m}\varepsilon.
$$
Matching the highest power requires $2n-2=6$, so $n=4$, and cancellation of $x^6$ requires $16\beta^2=1$. Normalizability selects $\beta=1/4$. Then $12\beta=3$ also cancels the quadratic term, leaving $\varepsilon=2$. Therefore
$$
\boxed{n=4,\qquad\beta=\frac14,\qquad E=\frac{\hbar^2}{m}.}
$$
This is the <exact ground state of a solvable sextic potential> candidate $e^{-x^4/4}$.

For the Gaussian trial state $\psi_\alpha=e^{-\alpha x^2/2}$, normalization cancels from the quotient. With respect to the probability density proportional to $e^{-\alpha x^2}$,
$$
\langle x^2\rangle=\frac1{2\alpha},
\qquad
\langle x^6\rangle=\frac{15}{8\alpha^3},
\qquad
\left\langle-\frac{d^2}{dx^2}\right\rangle=\frac\alpha2.
$$
Hence
$$
R[\psi_\alpha]
=\frac{\hbar^2}{2m}\varepsilon(\alpha),
\qquad
\varepsilon(\alpha)
=2+\frac\alpha2-\frac3{2\alpha}+\frac{15}{8\alpha^3}.
$$
The stationary equation is
$$
\varepsilon'(\alpha)=0
\quad\Longleftrightarrow\quad
4\alpha^4+12\alpha^2-45=0.
$$
Writing $z=\alpha^2$, the quadratic has exactly one positive root,
$$
z=\frac{-3+3\sqrt6}{2}
=\frac{-3+\sqrt{54}}2.
$$
Since the quotient tends to infinity as $\alpha\downarrow0$ or $\alpha\to\infty$, this is the unique global minimizer. Thus
$$
\boxed{p=-3,\qquad q=54,\qquad
\alpha_*=\sqrt{\frac{-3+\sqrt{54}}2}.}
$$
Using the stationary equation to replace $15/(8\alpha_*^3)$ by $\alpha_*/6+1/(2\alpha_*)$ gives the best estimate
$$
\boxed{
E_0^*=\frac{\hbar^2}{2m}
\left(2+\frac{2\alpha_*}{3}-\frac1{\alpha_*}\right).}
$$
This is the <Gaussian variational estimate for a solvable sextic potential>.

The exact eigenfunction $e^{-x^4/4}$ is positive and has no nodes. The <nodeless theorem for a one-dimensional ground state> therefore identifies it as the true ground state, with dimensionless energy $2$. The variational value is consistent: since $\alpha_*^2>3/2$,
$$
\frac{2\alpha_*}{3}-\frac1{\alpha_*}>0,
$$
so $E_0^*>\hbar^2/m=E_0$, as every trial-state upper bound must satisfy.

Solved by gpt-5.6-sol high.

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

= a
{parent=37a}
{scope}

= Solution
{parent=a}

The extension of the <one-dimensional freely jointed chain> is
$$
l=(n_+-n_-)a=(2n_+-n)a.
$$
As the <integer> $n_+$ runs from $0$ to $n$, the possible extensions are therefore
$$
\boxed{l=-na,-(n-2)a,\ldots,(n-2)a,na.}
$$
For fixed $n_+$ and $n_-=n-n_+$, choosing which $n_+$ of the $n$ links point in the positive direction determines the <microstate>. The <state degeneracy> is consequently the <binomial coefficient>
$$
\boxed{g(l)=\binom n{n_+}=\frac{n!}{n_+!n_-!}.}
$$

Solved by gpt-5.6-sol high.

= b
{parent=37a}
{scope}

= Solution
{parent=b}

The <expected value>[mean] extension in an ensemble of $A$ molecules is
$$
L=\frac1A\sum_i a_i l_i,
$$
with constraints $\sum_i a_i=A$ and $\sum_i a_i l_i=AL$. For occupancies $\{a_i\}$, the number of assignments of the $A$ distinguishable ensemble members to the states is the <multinomial coefficient>
$$
W=\frac{A!}{\prod_i a_i!}.
$$
Thus maximizing the probability at fixed $A$ and $L$ is equivalent to maximizing the stated <Lagrange multiplier> expression.

Because $A\gg1$, the <Stirling formula> gives $\log(a_i!)\simeq a_i\log a_i-a_i$. Differentiating with respect to each $a_i$ gives
$$
-\log a_i+\tau l_i-\alpha=0,
\qquad
a_i=e^{-\alpha}e^{\tau l_i}.
$$
Normalization by $\sum_i a_i=A$ therefore yields the <probability mass function>
$$
\boxed{p_i=\frac{a_i}{A}=\frac{e^{\tau l_i}}{Z}},
\qquad
\boxed{Z=\sum_i e^{\tau l_i}}.
$$
Here $Z$ is the fixed-tension <partition function> that normalizes the probabilities.

Solved by gpt-5.6-sol high.

= c
{parent=37a}
{scope}

= Solution
{parent=c}

Several <microstates> can have the same extension. If $g(l)$ denotes their <state degeneracy>, grouping equal terms in the state sum gives
$$
Z=\sum_i e^{\tau l_i}
=\boxed{\sum_l g(l)e^{\tau l}}.
$$
For the <one-dimensional freely jointed chain>, a <macrostate> with $n_+$ positive links and $n_-=n-n_+$ negative links has
$$
l=(n_+-n_-)a,
\qquad
g(l)=\frac{n!}{n_+!n_-!}.
$$
Hence
$$
Z=\sum_{n_+=0}^n
\frac{n!}{n_+!n_-!}
(e^{\tau a})^{n_+}(e^{-\tau a})^{n_-}.
$$
The <binomial theorem> now gives the <fixed-tension partition function of a one-dimensional chain>
$$
\boxed{Z=(e^{\tau a}+e^{-\tau a})^n
=(2\cosh(\tau a))^n.}
$$

Solved by gpt-5.6-sol high.

= d
{parent=37a}
{scope}

= Solution
{parent=d}

Substituting the <fixed-tension partition function of a one-dimensional chain> into the given free-energy definition gives
$$
\boxed{G=-k_BT\log Z=-nk_BT\log(2\cosh(\tau a)).}
$$
Differentiation with respect to the parameter conjugate to extension gives
$$
\frac{\partial G}{\partial\tau}
=-nk_BT\,a\tanh(\tau a).
$$
The same result follows directly from the <expected value> under $p_i=e^{\tau l_i}/Z$:
$$
L=\sum_i p_i l_i
=\frac1Z\frac{\partial Z}{\partial\tau}
=\frac{\partial\log Z}{\partial\tau}.
$$
Consequently
$$
\boxed{L=-\frac1{k_BT}\frac{\partial G}{\partial\tau}},
\qquad
\boxed{\tanh(\tau a)=\frac{L}{na}}.
$$

Solved by gpt-5.6-sol high.

= e
{parent=37a}
{scope}

= Solution
{parent=e}

If a physical tension $f$ acts on a state of extension $l_i$, its mechanical energy is $-fl_i$. The <canonical ensemble> therefore assigns that state the <Boltzmann factor>
$$
e^{-(-fl_i)/(k_BT)}=e^{fl_i/(k_BT)}.
$$
Comparison with the statistical weight $e^{\tau l_i}$ gives
$$
\boxed{\tau=\frac{f}{k_BT}},
\qquad
\boxed{f=k_BT\tau}.
$$
Equivalently, $f$ is the <thermodynamic conjugate variable> to $L$ in the <Gibbs free energy>, just as pressure is conjugate to volume.

Solved by gpt-5.6-sol high.

= f
{parent=37a}
{scope}

= Solution
{parent=f}

The condition $L\ll na$ implies $|\tau a|\ll1$. The <small-argument expansion of the hyperbolic tangent> gives $\tanh x=x+O(x^3)$, so
$$
\frac{L}{na}\simeq\tau a.
$$
Using $f=k_BT\tau$ yields the <entropic Hooke law for a one-dimensional chain>
$$
\boxed{f\simeq\frac{k_BT}{na^2}L}.
$$
Thus the effective spring constant is $k_BT/(na^2)$.

At fixed tension, the exact extension is
$$
L=na\tanh\!\left(\frac{fa}{k_BT}\right).
$$
Increasing the <temperature> decreases the positive argument of the <hyperbolic tangent>, so the chain contracts. In the small-extension regime this becomes
$$
\boxed{L\simeq\frac{na^2f}{k_BT}},
$$
which is inversely proportional to $T$.

Solved by gpt-5.6-sol high.

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

= a
{parent=38b}
{scope}

= Solution
{parent=a}

For coordinates $x^\alpha$, the <Christoffel symbol>[Christoffel symbols] of the <Levi-Civita connection> are
$$
\Gamma^\alpha_{\beta\gamma}
=\frac12g^{\alpha\delta}
\left(
\partial_\beta g_{\delta\gamma}
+\partial_\gamma g_{\delta\beta}
-\partial_\delta g_{\beta\gamma}
\right),
$$
where $g^{\alpha\delta}$ is the <inverse metric>.

For the <two-dimensional hyperbolic metric in polar coordinates>,
$$
g_{rr}=a^2,
\qquad
g_{\phi\phi}=a^2\sinh^2r,
\qquad
g^{rr}=a^{-2},
\qquad
g^{\phi\phi}=\frac1{a^2\sinh^2r}.
$$
Only $g_{\phi\phi}$ has a nonzero derivative, namely
$$
\partial_r g_{\phi\phi}=2a^2\sinh r\cosh r.
$$
It follows that the nonzero connection coefficients are
$$
\boxed{\Gamma^r_{\phi\phi}=-\sinh r\cosh r},
\qquad
\boxed{\Gamma^\phi_{r\phi}=\Gamma^\phi_{\phi r}=\coth r}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=38b}
{scope}

= Solution
{parent=b}

For an <affine parameter> $\lambda$, the <geodesic equation> and the connection coefficients from part (a) give
$$
\ddot r-\sinh r\cosh r\,\dot\phi^2=0,
\qquad
\ddot\phi+2\coth r\,\dot r\dot\phi=0,
$$
where dots denote derivatives with respect to $\lambda$.

On a line of constant $\phi$, one has $\dot\phi=0$. The equations reduce to $\ddot r=0$, so $r$ itself can be chosen as an affine parameter. Every such radial line is therefore a <geodesic>.

For a circle $r=r_0>0$, the first equation instead requires
$$
\sinh r_0\cosh r_0\,\dot\phi^2=0.
$$
Both hyperbolic factors are positive when $r_0>0$, so $\dot\phi=0$. This gives only a constant point, not a parametrized circle. Thus no circle of constant positive $r$ is a geodesic, in agreement with the <coordinate geodesics of the hyperbolic polar metric>.

Solved by gpt-5.6-sol high.

= c
{parent=38b}
{scope}

= Solution
{parent=c}

Using the stated convention for the <Riemann curvature tensor>, the first independent component is
$$
\begin{aligned}
R^r{}_{\phi r\phi}
&=\partial_r\Gamma^r_{\phi\phi}
-\Gamma^\phi_{\phi r}\Gamma^r_{\phi\phi}\\
&=-(\cosh^2r+\sinh^2r)+\cosh^2r\\
&=\boxed{-\sinh^2r}.
\end{aligned}
$$
The other is
$$
\begin{aligned}
R^\phi{}_{r\phi r}
&=-\partial_r\Gamma^\phi_{r\phi}
-\Gamma^\phi_{r\phi}\Gamma^\phi_{\phi r}\\
&=\operatorname{csch}^2r-\coth^2r\\
&=\boxed{-1},
\end{aligned}
$$
using the <hyperbolic Pythagorean identity> $\coth^2r-\operatorname{csch}^2r=1$. Antisymmetry in the final two indices supplies the components with those indices reversed. These are the <curvature of the two-dimensional hyperbolic polar metric> components.

Solved by gpt-5.6-sol high.

= d
{parent=38b}
{scope}

= Solution
{parent=d}

Contracting the <Riemann curvature tensor> gives the nonzero components of the <Ricci tensor>:
$$
R_{rr}=R^\phi{}_{r\phi r}=-1,
\qquad
R_{\phi\phi}=R^r{}_{\phi r\phi}=-\sinh^2r.
$$
The <Ricci scalar> is the contraction with the <inverse metric>. Therefore
$$
\begin{aligned}
R
&=g^{rr}R_{rr}+g^{\phi\phi}R_{\phi\phi}\\
&=\frac1{a^2}(-1)
+\frac1{a^2\sinh^2r}(-\sinh^2r)\\
&=\boxed{-\frac2{a^2}}.
\end{aligned}
$$
It is independent of both coordinates, so the <hyperbolic plane> has constant negative scalar curvature.

Solved by gpt-5.6-sol high.

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

= a
{parent=39c}
{scope}

= Solution
{parent=a}

The <Navier-Stokes equation> for constant <mass density> $\rho$, <dynamic viscosity> $\mu$, and <body force> per unit mass $\mathbf f$ is
$$
\rho\left(\partial_t\mathbf u+\mathbf u\mathbin{\cdot}\nabla\mathbf u\right)
=\nabla\mathbin{\cdot}\boldsymbol\sigma+\rho\mathbf f,
\qquad
\nabla\mathbin{\cdot}\mathbf u=0,
$$
where the <Newtonian fluid stress tensor> is
$$
\boldsymbol\sigma=-p\mathbf I+2\mu\mathbf e,
\qquad
\mathbf e=\frac12\left(\nabla\mathbf u+(\nabla\mathbf u)^T\right).
$$
Taking the <inner product>[dot product] with $\mathbf u$ gives
$$
\rho\mathbf u\mathbin{\cdot}
\left(\partial_t\mathbf u+\mathbf u\mathbin{\cdot}\nabla\mathbf u\right)
=\nabla\mathbin{\cdot}(\boldsymbol\sigma\mathbf u)
-\boldsymbol\sigma:\nabla\mathbf u
+\rho\mathbf f\mathbin{\cdot}\mathbf u.
$$
Incompressibility implies
$$
\boldsymbol\sigma:\nabla\mathbf u=2\mu\mathbf e:\mathbf e,
$$
and converts the left side into a local time derivative plus the divergence of kinetic-energy flux. The <divergence theorem> therefore gives the <kinetic-energy balance for an incompressible Newtonian fluid>
$$
\boxed{
\begin{aligned}
\frac d{dt}\int_\Omega\frac12\rho|\mathbf u|^2\,dV
={}&-\int_{\partial\Omega}\frac12\rho|\mathbf u|^2
\mathbf u\mathbin{\cdot}\mathbf n\,dS\\
&+\int_{\partial\Omega}
\mathbf u\mathbin{\cdot}\boldsymbol\sigma\mathbf n\,dS
+\int_\Omega\rho\mathbf f\mathbin{\cdot}\mathbf u\,dV
-2\mu\int_\Omega\mathbf e:\mathbf e\,dV.
\end{aligned}}
$$
The four terms are respectively outward advective transport of kinetic energy, mechanical power supplied by surface traction, power supplied by the body force, and irreversible viscous dissipation into heat.

Solved by gpt-5.6-sol high.

= b
{parent=39c}
{scope}

= Solution
{parent=b}

Let $a$ denote the bubble size, $U$ its speed, and $\operatorname{Re}=Ua/\nu\gg1$ the <Reynolds number>. Outside a thin <boundary layer>, inertia dominates and the flow is an <irrotational flow> with velocity scale $U$ and strain scale $U/a$. Its viscous dissipation has order
$$
\mathcal D_{\rm outer}
\sim \mu\left(\frac Ua\right)^2a^3
\sim\mu U^2a
$$
in three dimensions.

The boundary-layer thickness is
$$
\delta\sim\left(\frac{\nu a}{U}\right)^{1/2}
=a\operatorname{Re}^{-1/2}.
$$
For a clean bubble the <stress-free boundary condition> prescribes tangential stress rather than tangential velocity. The outer flow already has tangential strain of order $U/a$, so cancelling that stress requires a velocity correction of only
$$
\Delta u\sim\frac Ua\delta,
$$
not a correction of order $U$. Its gradient remains $O(U/a)$, and its boundary-layer dissipation is
$$
\mathcal D_{\rm layer}
\sim\mu\left(\frac Ua\right)^2a^2\delta
\sim\mathcal D_{\rm outer}\frac\delta a,
$$
which is asymptotically smaller.

In steady translation, the drag power $DU$ balances viscous dissipation. The leading estimate therefore follows from the known outer flow alone:
$$
\boxed{D\sim\frac{\mathcal D_{\rm outer}}U\sim\mu Ua}
$$
in three dimensions. This is the <dissipation estimate for high-Reynolds-number bubble drag>.

Solved by gpt-5.6-sol high.

= c
{parent=39c}
{scope}

= i
{parent=c}
{scope}

= Solution
{parent=i}

Work in the frame translating with the bubble. The bubble is the fixed circle $r=a$, while the velocity tends to $-U\mathbf e_x$ at infinity. The <potential flow around a circular cylinder> has <velocity potential>
$$
\boxed{
\phi=-U\left(r+\frac{a^2}{r}\right)\cos\theta
}.
$$
It is <harmonic function>[harmonic] for $r>a$ and gives
$$
\boxed{
u_r=-U\left(1-\frac{a^2}{r^2}\right)\cos\theta,
\qquad
u_\theta=U\left(1+\frac{a^2}{r^2}\right)\sin\theta.}
$$
Thus $u_r=0$ on $r=a$, satisfying the <kinematic boundary condition>, and $\mathbf u\to-U\mathbf e_x$ as $r\to\infty$. The laboratory-frame flow is obtained by adding $U\mathbf e_x$.

Solved by gpt-5.6-sol high.

= ii
{parent=c}
{scope}

= Solution
{parent=ii}

For the outer flow from part (i), the <rate-of-strain tensor> in polar coordinates has components
$$
e_{rr}=-\frac{2Ua^2}{r^3}\cos\theta,
\qquad
e_{\theta\theta}=\frac{2Ua^2}{r^3}\cos\theta,
\qquad
e_{r\theta}=-\frac{2Ua^2}{r^3}\sin\theta.
$$
Therefore
$$
\mathbf e:\mathbf e
=e_{rr}^2+e_{\theta\theta}^2+2e_{r\theta}^2
=\frac{8U^2a^4}{r^6}.
$$
The leading dissipation per unit axial length is
$$
\begin{aligned}
\mathcal D
&=2\mu\int_a^\infty\int_0^{2\pi}
\mathbf e:\mathbf e\,r\,d\theta\,dr\\
&=16\mu U^2a^4(2\pi)
\int_a^\infty r^{-5}\,dr\\
&=8\pi\mu U^2.
\end{aligned}
$$
Balancing this with drag power $DU$ gives the <drag on a two-dimensional circular bubble from outer-flow dissipation>
$$
\boxed{D=8\pi\mu U}
$$
per unit axial length, directed opposite to the bubble velocity.

Solved by gpt-5.6-sol high.

= iii
{parent=c}
{scope}

= Solution
{parent=iii}

A rigid body obeys the <no-slip boundary condition>. Its outer <potential flow> generally has an $O(U)$ tangential slip velocity at the surface, so the boundary layer must make an $O(U)$ velocity correction across thickness $\delta$. Its strain is therefore $O(U/\delta)$, and in two dimensions its dissipation per unit axial length is
$$
\mathcal D_{\rm layer}
\sim\mu\left(\frac U\delta\right)^2a\delta
=\mu U^2\frac a\delta,
$$
which exceeds the $O(\mu U^2)$ outer dissipation by $a/\delta\sim\operatorname{Re}^{1/2}$.

The dominant drag consequently depends on the detailed boundary-layer velocity field and possibly on <boundary-layer separation>. It cannot be obtained by integrating the outer irrotational strain alone.

Solved by gpt-5.6-sol high.

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

= a
{parent=40c}
{scope}

= Solution
{parent=a}

Let $k_S=\omega/c_S$ and $k_P=\omega/c_P$. <Snell law for elastic waves> fixes the reflected waves' common tangential wavenumber:
$$
k_S\sin\theta=k_P\sin\phi,
\qquad
\sin\phi=\frac{c_P}{c_S}\sin\theta.
$$
Choose the reflected <SV-wave> and converted reflected <P-wave> as
$$
\mathbf u_{RS}
=\operatorname{Re}\left\{
R(\cos\theta,0,\sin\theta)
 e^{ik_S(x\sin\theta-z\cos\theta)-i\omega t}
\right\},
$$
$$
\mathbf u_{RP}
=\operatorname{Re}\left\{
T(\sin\phi,0,-\cos\phi)
 e^{ik_P(x\sin\phi-z\cos\phi)-i\omega t}
\right\}.
$$
The polarizations are respectively perpendicular and parallel to their <wavevector>[wavevectors]. Phase matching makes all three exponential factors equal at $z=0$. The <rigid boundary condition for an elastic wave> $\mathbf u_I+\mathbf u_{RS}+\mathbf u_{RP}=0$ there gives
$$
\cos\theta(1+R)+T\sin\phi=0,
$$
$$
-\sin\theta+R\sin\theta-T\cos\phi=0.
$$
Solving this two-by-two <linear system> yields
$$
\boxed{R=-\frac{\cos(\theta+\phi)}{\cos(\theta-\phi)}},
\qquad
\boxed{T=-\frac{\sin2\theta}{\cos(\theta-\phi)}}.
$$
These fields give the <reflection of an SV-wave from a rigid plane>.

Solved by gpt-5.6-sol high.

= b
{parent=40c}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

The reflected P-wave is propagating only while its normal wavenumber is real, equivalently while $|\sin\phi|\leq1$. Since
$$
\sin\phi=\frac{c_P}{c_S}\sin\theta,
$$
it is <evanescent wave>[evanescent] when
$$
\boxed{\sin\theta>\frac{c_S}{c_P}}.
$$
Put
$$
s=\frac{c_P}{c_S}\sin\theta>1,
\qquad
q=\sqrt{s^2-1},
\qquad
\cos\phi=iq.
$$
Then its spatial factor is
$$
e^{ik_Psx+k_Pqz},
$$
which decays as $z\to-\infty$. The reflected SV amplitude becomes
$$
R=-\frac{iq\cos\theta-s\sin\theta}
{iq\cos\theta+s\sin\theta}.
$$
The numerator and denominator have equal <modulus>[complex modulus], so
$$
\boxed{|R|=1}.
$$
Thus evanescent mode conversion changes the reflected SV phase but not its amplitude, as described by <unit-modulus SV reflection with evanescent P conversion>.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

Average the instantaneous <elastic-wave energy flux> over one temporal period at fixed position. For complex amplitudes,
$$
\langle P_z\rangle
=-\frac12\operatorname{Re}
\left(\widehat\sigma_{zj}\widehat{\dot u}_j^*\right).
$$
Write the evanescent P displacement amplitude as
$$
\widehat{\mathbf u}
=T\mathbf p\,e^{ik_Psx+k_Pqz},
\qquad
\mathbf p=(s,0,-iq),
\qquad
\mathbf p\mathbin{\cdot}\mathbf p=s^2-q^2=1.
$$
For an <linear elasticity>[isotropic linear elastic material] with <Lamé parameters> $\lambda,\mu$, its stress and velocity amplitudes are
$$
\widehat{\boldsymbol\sigma}
=ik_PT\left(\lambda\mathbf I+2\mu\mathbf p\mathbf p\right)
 e^{ik_Psx+k_Pqz},
\qquad
\widehat{\dot{\mathbf u}}=-i\omega\widehat{\mathbf u}.
$$
Apart from a real common factor,
$$
\widehat\sigma_{zj}\widehat{\dot u}_j^*
\propto
\lambda p_z^*+2\mu p_z(\mathbf p\mathbin{\cdot}\mathbf p^*)
=iq\left[\lambda-2\mu(s^2+q^2)\right],
$$
which is purely imaginary. Its real part vanishes, and hence
$$
\boxed{\langle P_z\rangle=0}.
$$
The converted P-wave is a reactive near-boundary field and transports no time-averaged acoustic energy in the normal direction.

Solved by gpt-5.6-sol high.

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

= a
{parent=41c}
{scope}

= Solution
{parent=a}

For the two-periodic <Fourier series>, orthogonality of the exponential basis gives the <Fourier coefficient>
$$
\boxed{\widehat c_n=\frac12\int_{-1}^{1}c(x)e^{-i\pi nx}\,dx},
$$
where any interval of length two could replace $[-1,1]$.

Let $C_{nm}=\widehat c_{n-m}$ for $-d\leq n,m\leq d$. Since $c$ is real,
$$
\widehat c_{-k}=\overline{\widehat c_k},
$$
so $C_{mn}=\overline{C_{nm}}$ and $C$ is a <Hermitian positive-definite matrix>[Hermitian matrix]. For any nonzero $z=(z_{-d},\ldots,z_d)^T$,
$$
\begin{aligned}
z^*Cz
&=\sum_{n,m=-d}^d\overline{z_n}\widehat c_{n-m}z_m\\
&=\frac12\int_{-1}^{1}c(x)
\left|\sum_{m=-d}^dz_me^{i\pi mx}\right|^2dx.
\end{aligned}
$$
The <trigonometric polynomial> in the modulus cannot vanish identically unless every $z_m$ vanishes. Since $c(x)>0$, the integral is strictly positive. Hence $C$ is <Hermitian positive-definite matrix>[Hermitian positive definite], which is the <positive Fourier-symbol Toeplitz matrix> result.

Solved by gpt-5.6-sol high.

= b
{parent=41c}
{scope}

= Solution
{parent=b}

Differentiating the truncated Fourier expansion gives
$$
u_x=\sum_{m=-d}^di\pi m\widehat u_m e^{i\pi mx}.
$$
The $n$th Fourier coefficient of the product $cu_x$ is the <discrete convolution>
$$
i\pi\sum_{m=-d}^dm\widehat c_{n-m}\widehat u_m.
$$
Projecting the advection equation onto modes $|n|\leq d$ therefore gives
$$
\frac{d\widehat u_n}{dt}
=-i\pi\sum_{m=-d}^dm\widehat c_{n-m}\widehat u_m.
$$
Thus
$$
\boxed{\frac{d\widehat{\mathbf u}}{dt}
=i\pi B\widehat{\mathbf u}},
\qquad
\boxed{B_{nm}=-m\widehat c_{n-m}}.
$$
Equivalently, if
$$
C_{nm}=\widehat c_{n-m},
\qquad
D=\operatorname{diag}(-d,-d+1,\ldots,d),
$$
then
$$
\boxed{B=-CD}.
$$
This is the <Fourier spectral method for variable-coefficient advection>.

Solved by gpt-5.6-sol high.

= c
{parent=41c}
{scope}

= Solution
{parent=c}

The matrix $C$ from part (a) is <Hermitian positive-definite matrix>[Hermitian positive definite], while the real <diagonal matrix> $D$ is Hermitian. The stated product theorem, or similarity
$$
CD=C^{1/2}(C^{1/2}DC^{1/2})C^{-1/2},
$$
shows that every <eigenvalue> of $CD$ is real. Hence every eigenvalue of $B=-CD$ is real; this is the <real spectrum of a positive-Hermitian times Hermitian product>.

Moreover, $C$ is invertible and $D$ has rank $2d$, so for $d\geq1$ the matrix $B$ has nonzero real eigenvalues. If $\lambda\ne0$ is one of them, the corresponding mode of the semidiscrete equation has eigenvalue $i\pi\lambda$. The <explicit Euler method> has amplification factor
$$
G=1+i\pi\lambda\Delta t,
\qquad
|G|=\sqrt{1+\pi^2\lambda^2\Delta t^2}>1
$$
for every $\Delta t>0$. Therefore the explicit Euler discretization is unstable, as in <explicit Euler instability on a nonzero imaginary eigenvalue>.

Solved by gpt-5.6-sol high.

= d
{parent=41c}
{scope}

= Solution
{parent=d}

Using <Euler's formula>,
$$
\cos(\pi x)=\frac12e^{i\pi x}+\frac12e^{-i\pi x},
$$
$$
-\frac12\sin(\pi x)
=\frac i4e^{i\pi x}-\frac i4e^{-i\pi x}.
$$
Thus
$$
\widehat c_0=2,
\qquad
\widehat c_1=\frac12+\frac i4,
\qquad
\widehat c_{-1}=\frac12-\frac i4.
$$
Ordering the modes as $(-1,0,1)$, let $a=1/2+i/4$. Then
$$
C=\begin{pmatrix}
2&\overline a&0\\
a&2&\overline a\\
0&a&2
\end{pmatrix},
\qquad
D=\begin{pmatrix}-1&0&0\\0&0&0\\0&0&1\end{pmatrix}.
$$
Consequently
$$
\boxed{
B=-CD=
\begin{pmatrix}
2&0&0\\
\frac12+\frac i4&0&-\frac12+\frac i4\\
0&0&-2
\end{pmatrix}.}
$$
This matrix is <triangular matrix>[triangular], so its eigenvalues are its diagonal entries:
$$
\boxed{\operatorname{spec}(B)=\{-2,0,2\}.}
$$

Solved by gpt-5.6-sol high.