= Paper 2
{scope}

https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2021/paperii_2_2021.pdf

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

= Solution
{parent=1I}

The <Möbius function> is defined by $\mu(1)=1$ and, for $n>1$,
$$
\mu(n)=
\begin{cases}
0,&p^2\mid n\text{ for some prime }p,\\
(-1)^r,&n\text{ is a product of }r\text{ distinct primes}.
\end{cases}
$$
It is a <multiplicative arithmetic function>: $\mu(mn)=\mu(m)\mu(n)$ whenever $m$ and $n$ are <coprime integers>[coprime].

The summand $\mu(d)^2/\phi(d)$ is multiplicative, so its <divisor sum> is multiplicative. For a <prime power> $p^a$ with $a\geq1$, only $d=1,p$ contribute and
$$
\sum_{d\mid p^a}\frac{\mu(d)^2}{\phi(d)}
=1+\frac1{p-1}=\frac p{p-1}
=\frac{p^a}{\phi(p^a)}.
$$
Multiplying these local identities over the prime divisors of $n$ gives
$$
\boxed{\sum_{d\mid n}\frac{\mu(d)^2}{\phi(d)}=\frac n{\phi(n)}}.
$$

For the final claim, choose distinct primes $p_0,\ldots,p_k$. The moduli $p_j^2$ are pairwise coprime, so the <Chinese remainder theorem> gives an integer $a$ satisfying
$$
a\equiv-j\pmod{p_j^2},\qquad 0\leq j\leq k.
$$
Every positive integer $n\equiv a\pmod{\prod_jp_j^2}$ then has $p_j^2\mid n+j$. Hence
$$
\boxed{\mu(n)=\mu(n+1)=\cdots=\mu(n+k)=0}.
$$
There are infinitely many positive representatives of this congruence class.

Solved by gpt-5.6-sol high.

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

= i
{parent=2h}
{scope}

= Solution
{parent=i}

The assertion is true. Continuity and compactness give a maximum of $\phi$ on $\overline\Omega$. At an interior maximum the <Hessian matrix> is negative semidefinite, so its trace satisfies $\Delta\phi\leq0$, contradicting $\Delta\phi>0$. The maximum must therefore occur on $\partial\Omega$. This is the strict form of the <maximum principle for subharmonic functions>.

Solved by gpt-5.6-sol high.

= ii
{parent=2h}
{scope}

= Solution
{parent=ii}

The assertion is false. On the <unit disk>[open unit disk], let
$$
\phi(x,y)=-(x^2+y^2).
$$
Then $\Delta\phi=-4<0$, but the unique maximum is $phi(0,0)=0$, whereas $phi=-1$ on the boundary.

Solved by gpt-5.6-sol high.

= iii
{parent=2h}
{scope}

= Solution
{parent=iii}

The assertion is false. On the open unit disk take, for $0<\varepsilon<1$,
$$
\phi(x,y)=-(x^2+y^2)+\varepsilon(x^4+y^4).
$$
Its relevant fourth derivatives satisfy
$$
\phi_{xxxx}+\phi_{yyyy}=48\varepsilon>0.
$$
Nevertheless, since $x^4+y^4\leq(x^2+y^2)^2$, every point with $0<x^2+y^2\leq1$ has
$$
\phi(x,y)\leq-r^2+\varepsilon r^4<0=\phi(0,0).
$$
Thus the maximum lies strictly inside $\Omega$.

Solved by gpt-5.6-sol high.

= iv
{parent=2h}
{scope}

= Solution
{parent=iv}

The assertion is true by the <maximum principle for harmonic functions>. A direct reduction to part (i) is also possible: for $\varepsilon>0$, set
$$
\phi_\varepsilon(x)=\phi(x)+\varepsilon|x|^2.
$$
Then $\Delta\phi_\varepsilon=4\varepsilon>0$, so
$$
\max_{\overline\Omega}\phi_\varepsilon
=\max_{\partial\Omega}\phi_\varepsilon.
$$
Because $\Omega$ is bounded, $\phi_\varepsilon\to\phi$ uniformly as $\varepsilon\downarrow0$. Taking the limit yields
$$
\max_{\overline\Omega}\phi=\max_{\partial\Omega}\phi.
$$

Solved by gpt-5.6-sol high.

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

= Solution
{parent=3K}

For a <binary symmetric channel> with crossover probability $p$, the capacity is
$$
C=1-H_2(p),
$$
where $H_2$ is the <binary entropy>. <Shannon second coding theorem> says that for every transmission rate $R<C$ and every $\epsilon>0$, sufficiently long block codes exist with rate at least $R$ and decoding-error probability below $\epsilon$; conversely, a sequence of codes whose error tends to zero cannot have limiting rate above $C$.

For the general channel let $X$ and $Y$ be its input and output. Every conditional output distribution is a permutation of $(p_1,\ldots,p_n)$, hence
$$
H(Y\mid X)=-\sum_{i=1}^np_i\log p_i
$$
for every input distribution. Since $H(Y)\leq\log n$,
$$
I(X;Y)\leq\log n+\sum_i p_i\log p_i.
$$
The uniform input makes every output probability equal to $1/n$: the column-permutation assumption and the total sum imply that all column sums are equal to one. It therefore attains $H(Y)=\log n$, proving
$$
\boxed{C=\log n+\sum_i p_i\log p_i}.
$$

In the displayed four-output channel each row is a permutation of
$$
\left(\frac13,\frac13,\frac16,\frac16\right).
$$
With base-two logarithms,
$$
\begin{aligned}
C
&=2+\frac23\log\frac13+\frac13\log\frac16\\
&=2-\log3-\frac13
=\boxed{\frac53-\log3}.
\end{aligned}
$$

Solved by gpt-5.6-sol high.

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

= Solution
{parent=4F}

For a <deterministic finite automaton> the <extended transition function of a deterministic finite automaton> is recursively
$$
\widehat\delta(q,\epsilon)=q,
\qquad
\widehat\delta(q,wa)=\delta(\widehat\delta(q,w),a).
$$
The accepted language is
$$
L(D)=\{w\in\Sigma^*: \widehat\delta(q_0,w)\in F\}.
$$
For a <nondeterministic finite automaton>, $\widehat\delta(q,w)$ is the set of all states reachable from $q$ while reading $w$; recursively,
$$
\widehat\delta(q,\epsilon)=\{q\},
\qquad
\widehat\delta(q,wa)=\bigcup_{r\in\widehat\delta(q,w)}\delta(r,a).
$$
Thus
$$
L(N)=\{w\in\Sigma^*: \widehat\delta(q_0,w)\cap F\ne\varnothing\}.
$$

The <powerset construction> has state set $\mathcal P(Q)$, initial state $\{q_0\}$, transition
$$
\overline\delta(S,a)=\bigcup_{q\in S}\delta(q,a),
$$
and accepting states $\{S\subseteq Q:S\cap F\ne\varnothing\}$. Induction on $|w|$ gives
$$
\widehat{\overline\delta}(\{q_0\},w)=\widehat\delta(q_0,w),
$$
so the two automata accept exactly the same language. If $|Q|=m$ and $F$ has one state, exactly $2^{m-1}$ of the $2^m$ subset states contain it and are accepting, including states that may be unreachable.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=5J}

A <generalized linear model> assumes that the independent responses $Y_i$ have distributions in an <exponential family>, with means $\mu_i$, and that
$$
g(\mu_i)=\eta_i=x_i^T\beta.
$$
The invertible function $g$ is the <link function>; it connects the mean to the <linear predictor>.

For <binomial regression>, write $Y_i\sim\operatorname{Bin}(m_i,p_i)$ independently, so $\mu_i=m_ip_i$. The <logistic regression> link is
$$
\log\frac{p_i}{1-p_i}=x_i^T\beta,
\qquad
p_i=\frac{e^{x_i^T\beta}}{1+e^{x_i^T\beta}},
$$
while <probit regression> uses
$$
\Phi^{-1}(p_i)=x_i^T\beta,
\qquad
p_i=\Phi(x_i^T\beta),
$$
where $\Phi$ is the standard-normal distribution function. The logit is the <canonical link function> for the binomial family.

Solved by gpt-5.6-sol high.

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

= a
{parent=6e}
{scope}

= Solution
{parent=a}

The <birth-death master equation> must include jumps of size two on death. For $n\geq2$,
$$
\boxed{\dot p_n=bp_{n-1}+d(n+2)p_{n+2}-(b+dn)p_n}.
$$
At the boundary states it is
$$
\boxed{\dot p_0=2dp_2+Dp_1-bp_0},
\qquad
\boxed{\dot p_1=bp_0+3dp_3-(b+D)p_1}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=6e}
{scope}

= Solution
{parent=b}

Differentiate $\langle f(n)\rangle=\sum_{n\geq0}f_np_n$ and substitute the three master equations. Reindexing every gain term by its source state gives one contribution per possible transition:
$$
\begin{aligned}
\frac d{dt}\langle f(n)\rangle
={}&b\sum_{n=0}^\infty(f_{n+1}-f_n)p_n\\
&-d\sum_{n=2}^\infty(f_n-f_{n-2})np_n
-D(f_1-f_0)p_1.
\end{aligned}
$$
This is the <Markov jump-process generator>[generator identity for a continuous-time Markov chain] applied to the test function $f$.

Solved by gpt-5.6-sol high.

= c
{parent=6e}
{scope}

= Solution
{parent=c}

Set $f_n=n$. A birth changes $f$ by $1$, an ordinary pair death by $-2$, and singleton death by $-1$. Hence
$$
\dot\mu=b-2d\sum_{n\geq2}np_n-Dp_1
=b-2d\mu+(2d-D)p_1.
$$
When $D=2d$ this closes to
$$
\boxed{\dot\mu=b-2d\mu},
$$
so the stationary mean is $\mu_*=b/(2d)$.

Solved by gpt-5.6-sol high.

= d
{parent=6e}
{scope}

= Solution
{parent=d}

For $f_n=n^2$, the birth increment is $2n+1$, while
$$
f_n-f_{n-2}=n^2-(n-2)^2=4n-4.
$$
With $D=2d$, the generator identity gives
$$
\boxed{
\frac d{dt}\langle n^2\rangle
=b(2\mu+1)-4d(\langle n^2\rangle-\mu)-2dp_1}.
$$
At stationarity $\mu=b/(2d)$, and solving this equation for the <variance> yields
$$
\sigma^2=\langle n^2\rangle-\mu^2
=\frac{3b}{4d}-\frac{p_1}{2}.
$$
For $b,d>0$ the chain is irreducible on the nonnegative integers, so its stationary distribution has $0<p_1<1$. Therefore
$$
\boxed{\frac{3b}{4d}-\frac12<\sigma^2<\frac{3b}{4d}}.
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=7e}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

The point $z_0$ is an <ordinary point criterion for a second-order equation>[ordinary point] when both $p$ and $q$ are <holomorphic function>[holomorphic] at $z_0$.

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

The point $z_0$ is a <regular singular point criterion for a second-order equation>[regular singular point] when it is singular but
$$
(z-z_0)p(z),
\qquad
(z-z_0)^2q(z)
$$
extend holomorphically to $z_0$.

Solved by gpt-5.6-sol high.

= iii
{parent=a}
{scope}

= Solution
{parent=iii}

A singular point which is not regular singular is an <irregular singular point>: at least one of $(z-z_0)p(z)$ and $(z-z_0)^2q(z)$ fails to extend holomorphically to $z_0$.

Solved by gpt-5.6-sol high.

= b
{parent=7e}
{scope}

Put $t=1/z$ and $y(t)=w(1/t)$. Since
$$
w'=-t^2y',
\qquad
w''=t^4y''+2t^3y',
$$
the transformed equation is
$$
y''+\left(\frac2t-\frac{p(1/t)}{t^2}\right)y'
+\frac{q(1/t)}{t^4}y=0.
$$

= i
{parent=b}
{scope}

= Solution
{parent=i}

By the transformed equation, infinity is ordinary precisely when
$$
\frac2t-\frac{p(1/t)}{t^2},
\qquad
\frac{q(1/t)}{t^4}
$$
are holomorphic at $t=0$. Equivalently, as convergent Laurent expansions near infinity,
$$
p(z)=\frac2z+O(z^{-2}),
\qquad
q(z)=O(z^{-4}).
$$

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

By the <regular singular point at infinity> criterion, infinity is regular singular precisely when
$$
2-\frac{p(1/t)}t,
\qquad
\frac{q(1/t)}{t^2}
$$
are holomorphic at $t=0$, but the ordinary-point conditions fail. Equivalently,
$$
p(z)=O(z^{-1}),
\qquad
q(z)=O(z^{-2})
$$
with convergent expansions in $1/z$, excluding the ordinary case.

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

Infinity is irregular singular when at least one of
$$
2-zp(z),
\qquad
z^2q(z)
$$
fails to be holomorphic as a function of $1/z$ at $1/z=0$.

Solved by gpt-5.6-sol high.

= c
{parent=7e}
{scope}

= Solution
{parent=c}

Suppose there were no singular point in the <Riemann sphere>. Then every finite point would be ordinary, so $p$ and $q$ would be <entire functions>. Ordinary behavior at infinity would force $p(z)=2/z+O(z^{-2})$ and $q(z)=O(z^{-4})$. In particular $p(z)\to0$, so <Liouville theorem> would give $p\equiv0$, contradicting the required leading term $2/z$. Thus infinity cannot also be ordinary, and every such differential equation has at least one singular point in $\overline{\mathbb C}$.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=8D}

Let $R$ be the <center of mass> and $g$ the uniform gravitational acceleration. The gravitational <torque> about $R$ is
$$
\sum_i(r_i-R)\times m_i g
=\left(\sum_i m_i(r_i-R)\right)\times g=0.
$$

Choose body-fixed <principal body frame>[principal axes of inertia]. The inertial derivative of the <angular momentum> is
$$
N=\left(\frac{dL}{dt}\right)_{\rm body}+\omega\times L.
$$
For torque-free motion $N=0$ and $L_i=I_i\omega_i$, giving the <Euler equations for a torque-free rigid body>
$$
\begin{aligned}
I_1\dot\omega_1+(I_3-I_2)\omega_2\omega_3&=0,\\
I_2\dot\omega_2+(I_1-I_3)\omega_3\omega_1&=0,\\
I_3\dot\omega_3+(I_2-I_1)\omega_1\omega_2&=0.
\end{aligned}
$$

For a <symmetric top>, let $I_1=I_2=I$ and let $I_3$ be the axial moment. Then $\omega_3$ is constant and
$$
\dot\omega_1=-\Omega\omega_2,
\qquad
\dot\omega_2=\Omega\omega_1,
\qquad
\Omega=\frac{I_3-I}{I}\omega_3.
$$
Thus $\omega_1^2+\omega_2^2$ is constant and the transverse angular-velocity vector rotates uniformly about the body symmetry axis. Its rate is $|\Omega|$; it precesses in the positive axial sense when $(I_3-I)\omega_3>0$ and in the opposite sense when this product is negative.

Solved by gpt-5.6-sol high.

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

= a
{parent=9b}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

For bosons the occupation number is $n_{\mathbf p}=0,1,2,\ldots$. Summing the <geometric series> for the single-mode <grand canonical ensemble> gives the <Bose-Einstein distribution>
$$
\boxed{\langle N(\mathbf p)\rangle
=\frac1{e^{\beta(E_{\mathbf p}-\mu)}-1}}.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

For fermions the <Pauli exclusion principle> permits only $n_{\mathbf p}=0,1$. The two-term partition sum gives the <Fermi-Dirac distribution>
$$
\boxed{\langle N(\mathbf p)\rangle
=\frac1{e^{\beta(E_{\mathbf p}-\mu)}+1}}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=9b}
{scope}

= Solution
{parent=b}

Write $n_b=n_e+n_H\simeq\eta n_\gamma$. Charge neutrality gives $n_p=n_e$, and therefore
$$
n_e=X_en_b,
\qquad
n_H=(1-X_e)n_b.
$$
Substitution into the <Saha ionization equation> yields
$$
\frac{1-X_e}{X_e^2}
=n_b\left(\frac{2\pi\hbar^2}{m_ek_BT}\right)^{3/2}
e^{I/(k_BT)}.
$$
Using the given photon density,
$$
\boxed{
\frac{1-X_e}{X_e^2}
=\eta\frac{2\zeta(3)(2\pi)^{3/2}}{\pi^2}
\left(\frac{k_BT}{m_ec^2}\right)^{3/2}
e^{I/(k_BT)}}.
$$

The <baryon-to-photon ratio> is extremely small. Even when $k_BT\ll I$, the high-energy tail of the enormous photon population contains enough ionizing photons per baryon to delay <cosmological recombination>. The exponential factor must become large enough to overcome both $\eta\sim10^{-9}$ and the phase-space factor, which occurs near $k_BT\simeq0.3\,\mathrm{eV}$ rather than near $I$.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=10D}

Apply a <Hadamard gate> to every qubit. The amplitude of $|0^n\rangle$ becomes
$$
\langle0^n|H^{\otimes n}|\xi\rangle
=\frac1{2^n}\sum_{x\in\mathcal B_n}(-1)^{f(x)}.
$$
It equals $1$ in case (I) and $0$ in case (II). A <quantum measurement in the computational basis>[computational-basis measurement] therefore returns $0^n$ with certainty in case (I) and never returns it in case (II). This is the <Deutsch-Jozsa algorithm>.

For the distributed problem, Alice prepares
$$
|\xi_A\rangle=2^{-n/2}\sum_x(-1)^{f_A(x)}|x\rangle
$$
using one query to her <Boolean quantum oracle>[quantum oracle], and sends these $n$ qubits to Bob. Bob applies his phase oracle, producing
$$
2^{-n/2}\sum_x(-1)^{f_A(x)\oplus f_B(x)}|x\rangle.
$$
The function $h=f_A\oplus f_B$ is constant zero in case (1) and balanced in case (2). Bob applies $H^{\otimes n}$ and measures as above: outcome $0^n$ identifies case (1), while every other outcome identifies case (2), with certainty.

Solved by gpt-5.6-sol high.

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

= a
{parent=11h}
{scope}

= Solution
{parent=a}

Bilinearity and symmetry are immediate. Since $r\geq0$, $\langle u,u\rangle\geq0$. If this is zero, continuity of $u^2r$ makes it identically zero. The points where $r>0$ are dense because only finitely many are excluded, so $u$ vanishes on a dense set and hence everywhere. Thus the form is an <inner product>.

Solved by gpt-5.6-sol high.

= b
{parent=11h}
{scope}

= Solution
{parent=b}

Apply the <Gram-Schmidt process> to $1,x,\ldots,x^n$. The last resulting vector $P_n$ is nonzero, has degree exactly $n$, and is orthogonal to every polynomial in $\operatorname{span}(1,x,\ldots,x^{n-1})$.

Solved by gpt-5.6-sol high.

= c
{parent=11h}
{scope}

= Solution
{parent=c}

Let $x_1,\ldots,x_k$ be the distinct zeros in $(-1,1)$ at which $P_n$ changes sign, and put $Q(x)=\prod_{j=1}^k(x-x_j)$. Then $P_nQ$ has constant sign and is nonzero except at finitely many points, so
$$
\int_{-1}^1P_nQr\ne0.
$$
If $k<n$, this contradicts the orthogonality of $P_n$ to the lower-degree polynomial $Q$. Hence $k\geq n$; degree $n$ forces exactly $n$ distinct, simple roots, all in $(-1,1)$.

Solved by gpt-5.6-sol high.

= d
{parent=11h}
{scope}

= Solution
{parent=d}

Let $\ell_j$ be the <Lagrange interpolation polynomial> for the roots $\omega_j$, and define
$$
A_j=\int_{-1}^1\ell_j(x)r(x)\,dx.
$$
For $\deg Q\leq2n-1$, <polynomial division> gives $Q=SP_n+R$ with $\deg S,\deg R<n$. Orthogonality kills the first term, while interpolation gives $R=\sum_jR(\omega_j)\ell_j$. Therefore
$$
\int Qr=\sum_jA_jQ(\omega_j).
$$
Taking $Q=\ell_j$ proves uniqueness. Taking $Q=\ell_j^2$ also shows
$$
A_j=\int\ell_j^2r>0,
\qquad
\sum_jA_j=\int r.
$$
This is <Gaussian quadrature>.

Solved by gpt-5.6-sol high.

= e
{parent=11h}
{scope}

= Solution
{parent=e}

Given $\epsilon>0$, the <Weierstrass approximation theorem> supplies a polynomial $Q$ with $\|f-Q\|_\infty<\epsilon$. For all sufficiently large $n$, the quadrature is exact on $Q$. Positivity of the weights and $\sum_jA_j=\int r$ then give
$$
\left|\int fr-\sum_jA_jf(\omega_j)\right|
\leq2\epsilon\int_{-1}^1r.
$$
Letting $\epsilon\downarrow0$ proves convergence.

Solved by gpt-5.6-sol high.

= f
{parent=11h}
{scope}

= Solution
{parent=f}

Write $f=Q_{2n-1}+E_n$, where $Q_{2n-1}$ contains the terms through degree $2n-1$. On $[-1,1]$,
$$
|E_n(x)|\leq K\sum_{m=2n}^\infty R^{-m}
=\frac{KR^{-2n+1}}{R-1}.
$$
The quadrature error of $Q_{2n-1}$ is zero. Bounding the integral and quadrature errors of $E_n$ separately, using positive weights with total $\int r$, gives exactly
$$
\boxed{\left|\int fr-\sum_jA_jf(\omega_j)\right|
\leq\frac{2KR^{-2n+1}}{R-1}\int r}.
$$

Solved by gpt-5.6-sol high.

= g
{parent=11h}
{scope}

= Solution
{parent=g}

With $x=\cos\theta$, the weight becomes $\sin^2\theta\,d\theta$. Since
$$
U_n(\cos\theta)=\frac{\sin((n+1)\theta)}{\sin\theta},
$$
the sine orthogonality relations show that $P_n$ is a scalar multiple of the <Chebyshev polynomial of the second kind> $U_n$. Its roots are
$$
\boxed{\omega_j(n)=\cos\frac{j\pi}{n+1}},
\qquad 1\leq j\leq n.
$$
The printed normalization $P_n(0)=1$ is possible only for even $n$; then $U_n(0)=(-1)^{n/2}$ and
$$
P_n=(-1)^{n/2}U_n.
$$
For odd $n$, every such orthogonal polynomial is odd and vanishes at zero, so that normalization is inconsistent.

Solved by gpt-5.6-sol high.

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

= a
{parent=12k}
{scope}

= Solution
{parent=a}

A <cyclic code>[binary cyclic code] is a linear subspace $C\leq\mathbb F_2^n$ invariant under cyclic coordinate shift. Identify a word $(c_0,\ldots,c_{n-1})$ with $c_0+c_1X+\cdots+c_{n-1}X^{n-1}$ in
$$
\mathbb F_2[X]/(X^n-1).
$$
Cyclic shift is multiplication by $X$, so cyclic codes are exactly ideals of this quotient. Since $\mathbb F_2[X]$ is a <principal ideal domain>, each is generated by a unique monic divisor $g\mid X^n-1$, giving the required bijection.

Solved by gpt-5.6-sol high.

= b
{parent=12k}
{scope}

= Solution
{parent=b}

A <linear feedback shift register> is a linear map
$$
M(x_1,\ldots,x_d)=(x_2,\ldots,x_d,\sum_ia_ix_i).
$$
For the stated matrix $H$, a word $c=(c_0,\ldots,c_{k-1})$ lies in $C$ exactly when $\sum_jc_jM^jx=0$. If $\sigma c$ is its cyclic shift, periodicity $M^kx=x$ gives
$$
H(\sigma c)=M(Hc)=0.
$$
Thus $\ker H$ is shift-invariant and is a binary cyclic code.

Solved by gpt-5.6-sol high.

= c
{parent=12k}
{scope}

= Solution
{parent=c}

If $c\in C$ had odd <Hamming weight>, add all $n$ cyclic shifts of $c$. Each coordinate of the sum is the sum of all coordinates of $c$, namely one in $\mathbb F_2$. Linearity and cyclicity would therefore put $(1,\ldots,1)$ in $C$, contrary to the hypothesis. Hence every codeword has even weight.

Solved by gpt-5.6-sol high.

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

= a
{parent=13e}
{scope}

= Solution
{parent=a}

The zero initial condition turns the time <Laplace transform> of the heat equation into
$$
p\widehat T=\kappa\widehat T_{xx}.
$$
Decay as $x\to\infty$ selects
$$
\widehat T(x,p)=\widehat f(p)e^{-x\sqrt{p/\kappa}}.
$$
The boundary value gives
$$
\widehat f(p)=a\left(\frac1p-\frac1{p+b}\right)=\frac{ab}{p(p+b)},
$$
so
$$
\boxed{\widehat T(x,p)=\frac{ab}{p(p+b)}e^{-x\sqrt{p/\kappa}}}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=13e}
{scope}

= Solution
{parent=b}

Use the principal square root and deform the <Bromwich contour> around the negative real-axis branch cut. The pole at $p=0$ contributes $a$. The two boundary values at $p=-b$ combine to
$$
-ae^{-bt}\cos\left(x\sqrt{b/\kappa}\right).
$$
On the cut put $p=-v^2$; the jump of $e^{-x\sqrt{p/\kappa}}$ is $-2i\sin(xv/\sqrt\kappa)$ and $dp=-2v,dv$. Taking the principal value at $v=\sqrt b$ gives
$$
\boxed{T(x,t)=a\left[1-e^{-bt}\cos\left(\sqrt{b/\kappa}\,x\right)\right]
+\frac{2ab}{\pi}\mathcal P\int_0^\infty
\frac{e^{-v^2t}\sin(xv/\sqrt\kappa)}{v(v^2-b)}\,dv}.
$$
The expression has the prescribed boundary value and decays into the bar.

Solved by gpt-5.6-sol high.

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

= a
{parent=14d}
{scope}

= Solution
{parent=a}

<Hamilton's equations> give $\dot q=p$ and $\dot p=-\omega^2q$, hence $\ddot q+\omega^2q=0$. The phase-space energy curve is an ellipse. Its <action variable> is
$$
I=\frac1{2\pi}\oint p\,dq
=\frac1{2\pi}\times\pi\sqrt{2E}\frac{\sqrt{2E}}\omega
=\frac E\omega.
$$
Thus $\boxed{E=\omega I}$.

Solved by gpt-5.6-sol high.

= b
{parent=14d}
{scope}

= Solution
{parent=b}

Substituting $x=t^{1/2}y(\log t)$ gives
$$
y''+\left(\epsilon^{-2}-\frac14\right)y=0.
$$
Thus
$$
k=\sqrt{\epsilon^{-2}-\frac14}
$$
and the initial conditions select
$$
\boxed{x(t)=\frac{\sqrt t}{k}\sin(k\log t)}.
$$
Direct differentiation verifies both the equation and $x(1)=0$, $\dot x(1)=1$.

Solved by gpt-5.6-sol high.

= c
{parent=14d}
{scope}

= Solution
{parent=c}

Put $\theta=k\log t$. Since
$$
\dot x=\frac{\sin\theta+2k\cos\theta}{2k\sqrt t},
\qquad
\epsilon^{-2}=k^2+\frac14,
$$
the instantaneous oscillator energy is
$$
E(t)=\frac12\dot x^2+\frac{x^2}{2(\epsilon t)^2}
=\frac{1-\cos2\theta+2k\sin2\theta+4k^2}{8k^2t}.
$$
The instantaneous frequency is $\omega(t)=1/(\epsilon t)$, so
$$
I(t)=\frac{E(t)}{\omega(t)}
=\frac{\epsilon}{8k^2}\left(4k^2+1-\cos2\theta+2k\sin2\theta\right).
$$
Its oscillatory part divided by its mean is $O(k^{-1})=O(\epsilon)$. It remains bounded and periodic in $\log t$, so the variations do not accumulate. This is the expected behavior of an <adiabatic invariance of the action>[adiabatic invariant]: slow parameter variation produces small bounded oscillations rather than secular drift.

Solved by gpt-5.6-sol high.

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

= a
{parent=15d}
{scope}

= Solution
{parent=a}

Alice and Bob need a uniformly random secret key $K\in\{0,1\}^m$ known only to them. Alice publicly sends the <one-time pad> ciphertext $C=M\oplus K$, and Bob recovers $M=C\oplus K$. For every observed $c$ and possible message $m$ there is exactly one equally likely key $k=c\oplus m$, so $C$ is independent of $M$ and Eve learns nothing.

Solved by gpt-5.6-sol high.

= b
{parent=15d}
{scope}

= Solution
{parent=b}

Alice adjoins $|0\rangle_{A'}$ and applies a local unitary whose action on the relevant inputs is
$$
|0\rangle|0\rangle\mapsto
\frac st|0\rangle|0\rangle+sqrt{1-\frac{s^2}{t^2}}|1\rangle|0\rangle,
\qquad
|0\rangle|1\rangle\mapsto|0\rangle|1\rangle.
$$
Measuring $A'$ and obtaining zero leaves the unnormalized state
$$
s(|00\rangle+|11\rangle),
$$
so the normalized state is the <Bell state> $|\phi^+\rangle$. The success probability is $2s^2>0$.

Applying this <entanglement concentration> independently to $n$ pairs yields an expected $2s^2n$ Bell pairs. Alice and Bob measure successful pairs in the computational basis to obtain identical secret random bits; public communication identifies successful positions without revealing outcomes. These bits form a secret key for a one-time pad of expected length $2s^2n$.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=16G}

<Ordinal exponentiation> is defined by
$$
\omega^0=1,
\qquad
\omega^{\beta+1}=\omega^\beta\omega,
\qquad
\omega^\lambda=\sup_{\beta<\lambda}\omega^\beta
$$
for limit $\lambda$. <Transfinite induction> gives $\omega^\alpha\geq\alpha$. Since $\omega^{\alpha+1}>\alpha$, there is a least $\alpha^*$ with $\omega^{\alpha^*}>\alpha$. If nonzero $\alpha^*$ were a limit, the defining supremum would imply $\omega^\beta>\alpha$ for some $\beta<\alpha^*$, a contradiction. Hence $\alpha^*$ is a successor.

For $\alpha>0$, write $\alpha^*=\beta+1$. Then $\omega^\beta\leq\alpha<\omega^{\beta+1}$. There is a largest positive integer $n$ with $\omega^\beta n\leq\alpha$, and ordinal division gives
$$
\alpha=\omega^\beta n+\gamma,
\qquad
\gamma<\omega^\beta.
$$
Repeating on $\gamma$ terminates because there is no infinite strictly decreasing sequence of ordinals, producing the <Cantor normal form>
$$
\alpha=\omega^{\beta_1}n_1+\cdots+\omega^{\beta_k}n_k,
\qquad
\beta_1>\cdots>\beta_k.
$$

For two monomials,
$$
\omega^{\delta_1}m_1+\omega^{\delta_2}m_2=
\begin{cases}
\omega^{\delta_2}m_2,&\delta_1<\delta_2,\\
\omega^{\delta_1}(m_1+m_2),&\delta_1=\delta_2,\\
\omega^{\delta_1}m_1+\omega^{\delta_2}m_2,&\delta_1>\delta_2.
\end{cases}
$$
For general $\alpha+\alpha'$, compare the leading exponent $\beta'_1$ with the exponents of $\alpha$: discard every trailing term of $\alpha$ having exponent below $\beta'_1$, combine coefficients if the last retained exponent equals $\beta'_1$, and then append all remaining terms of $\alpha'$. This is its Cantor normal form.

Solved by gpt-5.6-sol high.

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

= a
{parent=17g}
{scope}

= Solution
{parent=a}

A <tree> is a connected acyclic graph; acyclic means containing no cycle. Every component of an acyclic graph is a tree. If the component orders are $n_i$, each has $n_i-1$ edges, so with $c\geq1$ components,
$$
e(G)=\sum_i(n_i-1)=n-c\leq n-1.
$$

Solved by gpt-5.6-sol high.

= b
{parent=17g}
{scope}

= Solution
{parent=b}

If a cubic graph has no cycle of length at most $2r+1$, the <breadth-first search> ball of radius $r$ about a vertex is a tree and contains
$$
1+3(1+2+\cdots+2^{r-1})=3\cdot2^r-2
$$
vertices. Taking $r=\lfloor\log_2 n\rfloor$ gives a contradiction for a suitable nearby radius, and in particular yields a cycle of length at most $100\log n$.

Choose such a cycle $C$ of length $m\leq100\log n$. Removing its vertices deletes at most $3m$ edges. The remaining graph has at least $3n/2-3m$ edges on $n-m$ vertices. For all sufficiently large $n$, this exceeds $n-m-1$, so part (a) says the remainder contains a cycle. It is vertex-disjoint from $C$.

Solved by gpt-5.6-sol high.

= c
{parent=17g}
{scope}

= Solution
{parent=c}

Choose a bipartition maximizing the number of crossing edges. If the graph has an edge, a maximum cut is nontrivial. Moving a vertex with more neighbors on its own side than across would strictly increase the cut, which is impossible. Thus every vertex has at least as many neighbors across as on its own side. If the graph has no edges, every nontrivial partition works. Hence every graph on at least two vertices has an unfriendly partition.

Solved by gpt-5.6-sol high.

= d
{parent=17g}
{scope}

= Solution
{parent=d}

The complete graph $K_4$ is cubic and has no friendly partition: a $1+3$ split fails at the singleton, while in a $2+2$ split every vertex has one neighbor on its own side and two across.

For a sufficiently large cubic graph, part (b) supplies disjoint cycles $C_1,C_2$. Put $C_1$ initially in $S$ and $C_2$ in $T$, and assign all remaining vertices to maximize the number of edges lying within a part. Every cycle vertex already has two same-side neighbors. Any other vertex with more opposite-side than same-side neighbors could be moved to increase the objective. Thus the final nonempty partition is friendly.

Solved by gpt-5.6-sol high.

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

= a
{parent=18i}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

An irreducible degree-$n$ polynomial over $\mathbb F_q$ is separable and its roots lie in $\mathbb F_{q^n}$, with any one root generating that field. Hence its splitting field is $\mathbb F_{q^n}$ and
$$
\boxed{\operatorname{Gal}(f)\cong C_n},
$$
generated by the <Frobenius endomorphism>[Frobenius automorphism] $x\mapsto x^q$.

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

If $n_i=\deg h_i$ and $N=\operatorname{lcm}(n_1,\ldots,n_r)$, the splitting field is the compositum of the fields $\mathbb F_{q^{n_i}}$, namely $\mathbb F_{q^N}$. Therefore
$$
\boxed{\operatorname{Gal}(f)\cong C_N},
$$
again generated by Frobenius.

Solved by gpt-5.6-sol high.

= iii
{parent=a}
{scope}

= Solution
{parent=iii}

The multiplicative group of $L=\mathbb F_{q^N}$ is cyclic of order $m=q^N-1$. For a primitive element $\zeta$, $L=\mathbb F_q(\zeta)$, so this is a <cyclotomic field>[cyclotomic extension]. The cyclotomic character is
$$
\operatorname{Gal}(L/\mathbb F_q)\hookrightarrow(\mathbb Z/m\mathbb Z)^\times,
\qquad
\sigma\mapsto a_\sigma,quad \sigma(\zeta)=\zeta^{a_\sigma}.
$$
Its image is the cyclic subgroup $\langle q\rangle$, of order $N$.

Solved by gpt-5.6-sol high.

= b
{parent=18i}
{scope}

= Solution
{parent=b}

Irreducibility makes the quartic Galois group a transitive subgroup of $S_4$. Its square discriminant puts it inside $A_4$, leaving $V_4$ or $A_4$. The cubic resolvent is
$$
y^3-48y-64.
$$
The <rational root theorem> shows it is irreducible over $\mathbb Q$. Thus the Galois group acts transitively on the three pairings of four roots, excluding $V_4$. Consequently
$$
\boxed{\operatorname{Gal}(x^4+8x+12/\mathbb Q)\cong A_4}.
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=19i}
{scope}

= Solution
{parent=a}

Let the distinct values of the faithful character $\chi$ be $a_1,\ldots,a_r$. If an irreducible character $\psi$ occurred in none of $1,\chi,\ldots,\chi^{r-1}$, then
$$
0=|G|\langle\chi^n,\psi\rangle
=\sum_{j=1}^ra_j^nS_j,
\qquad
S_j=\sum_{\chi(g)=a_j}\overline{\psi(g)}
$$
for $0\leq n<r$. The <Vandermonde determinant>[Vandermonde matrix] is invertible, so every $S_j=0$. But faithfulness implies that $\chi(g)=\chi(1)$ only for $g=1$: in a unitary realization, equality of the trace with the dimension forces every eigenvalue to be one. For that fiber $S_j=\psi(1)>0$, a contradiction.

Solved by gpt-5.6-sol high.

= b
{parent=19i}
{scope}

= Solution
{parent=b}

The irreducible characters are orthonormal under the <character orthogonality>[character inner product]. The dimension of the space of class functions is the number of conjugacy classes, which by hypothesis equals their number, so they form an orthonormal basis.

Conjugate elements have equal values under every character. Conversely, if $g,h$ have equal values under all characters, they have equal values under every class function by the basis result. Applying the indicator of the conjugacy class of $g$ shows that $h$ lies in that class.

Solved by gpt-5.6-sol high.

= c
{parent=19i}
{scope}

= Solution
{parent=c}

Let $K=\ker\chi$, which is nontrivial. Every tensor power affording $\chi^n$ is trivial on $K$, so each of its irreducible constituents has $K$ in its kernel. Some irreducible representation is nontrivial on $K$; otherwise the regular representation, which contains every irreducible, would also be trivial on $K$. Choose its character $\psi$. Then $\psi$ is a constituent of no $\chi^n$, and
$$
\boxed{\langle\chi^n,\psi\rangle=0\quad(n\geq0)}.
$$

Solved by gpt-5.6-sol high.

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

= Solution
{parent=20G}

An element $\alpha$ is <algebraic element>[algebraic] over $\mathbb Q$ when it is a root of a nonzero polynomial in $\mathbb Q[X]$. Clearing denominators gives
$$
a_0+a_1\alpha+\cdots+a_d\alpha^d=0,
\qquad a_i\in\mathbb Z,quad a_0\ne0.
$$
For $\alpha\ne0$, take $\beta=-(a_1+a_2\alpha+\cdots+a_d\alpha^{d-1})\in\mathbb Z[\alpha]$; then $\alpha\beta=a_0\ne0$.

Since $K=\operatorname{Frac}R$, write $x=a/b$ with $a,b\in R$ and $b\ne0$. Applying the first result to $b$ gives $b\beta=m\in\mathbb Z\setminus\{0\}$, whence $x=a\beta/m$. Changing signs makes $m>0$.

Choose a $\mathbb Z$-basis $e_1,\ldots,e_d$ of $\mathcal O_K$. Write each $e_i=r_i/m_i$ as above and take a common multiple $m$. Then
$$
m\mathcal O_K\subseteq R\subseteq\mathcal O_K.
$$
Thus $R$ is a free abelian group of rank $d=[K:\mathbb Q]$ and has finite index in $\mathcal O_K$. If $0\ne a\in I$, then $aR\subseteq I$ and multiplication by $a$ has nonzero determinant on the rank-$d$ lattice $R$, so $(R:aR)$, and hence $(R:I)$, is finite. Also $m\mathcal O_K\subseteq R$ is an ideal of $R$.

Put $h=(\mathcal O_K:R)$ and $I=m\mathcal O_K$. Then
$$
(R:I)=\frac{m^d}{h},
\qquad
(R:I^2)=\frac{m^{2d}}h.
$$
The assumed multiplicativity of indices gives
$$
\frac{m^{2d}}h=\left(\frac{m^d}h\right)^2,
$$
so $h=1$. Therefore $\boxed{R=\mathcal O_K}$.

Solved by gpt-5.6-sol high.

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

= a
{parent=21f}
{scope}

= Solution
{parent=a}

The <Seifert-van Kampen theorem> says that if a path-connected space is the union of path-connected open sets $U,V$ with path-connected intersection, then
$$
\pi_1(U\cup V)=\pi_1(U)*\pi_1(V)/\langle\!\langle i_{U*}(w)i_{V*}(w)^{-1}:w\in\pi_1(U\cap V)\rangle\!\rangle.
$$
Using the usual polygonal cell decompositions, the torus attaches a two-cell to $S^1\vee S^1$ along $aba^{-1}b^{-1}$, while the projective plane attaches one along $a^2$. Hence
$$
\boxed{\pi_1(T^2)=\langle a,b\mid aba^{-1}b^{-1}\rangle\cong\mathbb Z^2},
\qquad
\boxed{\pi_1(\mathbb {RP}^2)=\langle a\mid a^2\rangle\cong\mathbb Z/2}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=21f}
{scope}

= Solution
{parent=b}

A <covering space>[covering map] induces an injection on fundamental groups. There is no injection $\mathbb Z^2\hookrightarrow\mathbb Z/2$, and there is no injection $\mathbb Z/2\hookrightarrow\mathbb Z^2$ because $\mathbb Z^2$ is torsion-free. Thus neither proposed covering exists.

Solved by gpt-5.6-sol high.

= c
{parent=21f}
{scope}

= Solution
{parent=c}

Label the four vertices cyclically by $\mathbb Z/4$. Reading the arrows in the diagram, an $a$-edge moves one step in one direction and a $b$-edge one step in the other. Thus the monodromy homomorphism is
$$
\rho:F(a,b)\longrightarrow\mathbb Z/4,
\qquad
\rho(a)=-1,quad \rho(b)=1
$$
(up to simultaneously reversing the chosen numbering). By the <classification of connected covering spaces>[Galois correspondence for covering spaces], the subgroup associated with the chosen vertex above $x_0$ is
$$
\boxed{\ker\rho=\{w\in F(a,b):\#_b(w)-\#_a(w)\equiv0\pmod4\}}.
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=22h}
{scope}

= Solution
{parent=a}

If a subspace $W\subset V$ contains a ball about some $w\in W$, translation gives a ball about zero in $W$, and scaling then gives every vector of $V$. Thus every proper subspace has empty interior.

If a complete infinite-dimensional $V$ had a countable <basis>[Hamel basis] $(e_n)$, then
$$
V=\bigcup_{n\geq1}\operatorname{span}(e_1,\ldots,e_n).
$$
Each finite-dimensional subspace is closed and, being proper, nowhere dense. This contradicts the <Baire category theorem>. The polynomials have the countable Hamel basis $1,x,x^2,\ldots$ and are infinite-dimensional, so no norm can make that whole vector space complete.

Solved by gpt-5.6-sol high.

= b
{parent=22h}
{scope}

= Solution
{parent=b}

The identity map
$$
I:(V,\|\cdot\|_1)\to(V,\|\cdot\|_2)
$$
is a continuous linear bijection. Since both spaces are Banach, the <bounded inverse theorem> makes $I^{-1}$ continuous, giving $\|x\|_1\leq C_2\|x\|_2$.

Completeness is essential. On the space $c_{00}$ of finitely supported sequences, $\|x\|_2\leq\|x\|_1$, but for $x=(1,\ldots,1,0,\ldots)$ the ratio $\|x\|_1/\|x\|_2=\sqrt n$ is unbounded. Both normed spaces are incomplete.

Solved by gpt-5.6-sol high.

= c
{parent=22h}
{scope}

= Solution
{parent=c}

Define
$$
T:V^*\to\ell^1,
\qquad
Tf=(f(x_n))_{n\geq1}.
$$
The hypothesis makes $T$ everywhere defined. The dual $V^*$ is Banach even when $V$ is incomplete. If $f_j\to f$ in $V^*$ and $Tf_j\to a$ in $\ell^1$, then coordinatewise convergence gives $a_n=f(x_n)$, so the graph of $T$ is closed. The <closed graph theorem> therefore gives
$$
\|Tf\|_1\leq C\|f\|_{V^*}.
$$
Taking the supremum over the unit ball proves the required finite bound.

Solved by gpt-5.6-sol high.

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

The <Schwartz space> $\mathcal S(\mathbb R^n)$ consists of smooth functions for which every seminorm
$$
p_{N,k}(\phi)=\sup_{x,\,|\alpha|\leq k}(1+|x|^2)^{N/2}|D^\alpha\phi(x)|
$$
is finite. Convergence means convergence in every such seminorm. A <tempered distribution> is a continuous linear functional on this Fréchet space; $u_j\to u$ in $\mathcal S'$ means $u_j[\phi]\to u[\phi]$ for every test function $\phi$.

= a
{parent=23h}
{scope}

= Solution
{parent=a}

Continuity of $u$ at zero supplies finitely many Schwartz seminorms and a constant $C$ bounding $|u[\phi]|$. Increasing their two indices to common maxima $N,k$ gives
$$
|u[\phi]|\leq C\|\phi\|_{N,k}.
$$
Thus $u$ is bounded for the $X_{N,k}$ norm. Since $\mathcal S$ is dense in the Banach space $X_{N,k}$, it has a unique continuous extension $\widetilde u\in X_{N,k}'$.

Solved by gpt-5.6-sol high.

= b
{parent=23h}
{scope}

= Solution
{parent=b}

Take the extension from part (a), and choose $M>N+n+1$. The positive function
$$
w(x)=(1+|x|^2)^{-M/2}
$$
belongs to $X_{N,k}$. Positivity extends from nonnegative Schwartz functions to nonnegative elements obtained as $X_{N,k}$ limits. Since
$$
-\|\phi\|_{M,0}w\leq\phi\leq\|\phi\|_{M,0}w,
$$
positivity gives
$$
|v[\phi]|\leq \widetilde v[w]\,\|\phi\|_{M,0}.
$$
Taking $K=\widetilde v[w]$ proves that every <positive distribution> has order zero.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=24F}

An analytic continuation along $\alpha$ is a chain of function elements $(f_t,U_t)$ covering the path, beginning with the germ of $(f,U)$ at $\alpha(0)$, such that neighboring elements agree on the component of overlap met by the path. The terminal element is $(g,V)$.

The <monodromy theorem> says that if continuation is possible along every path in $D$, then continuations along endpoint-fixed homotopic paths have the same terminal germ. It follows by subdividing a homotopy square into small rectangles on which the <identity theorem> identifies neighboring germs. Hence
$$
g(\alpha(1))=h(\beta(1)).
$$

The requested final example with homotopic paths and different terminal values cannot exist under these hypotheses; it contradicts the theorem just proved. With “non-homotopic” in place of “homotopic”, take the germ of $\log z$ near $1$ in $\mathbb C^*$. Continuation along the constant path gives $0$, whereas continuation once counterclockwise around zero gives $2\pi i$.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=25I}

Inside $\mathbb A^{n^2}$,
$$
GL_n(k)=D(\det)=\{A:\det A\ne0\},
$$
a nonempty <Zariski-open set>. Affine space is smooth and irreducible; smoothness passes to open subsets, and every nonempty open subset of an irreducible space is irreducible.

The map
$$
A\longmapsto(A,(\det A)^{-1})
$$
identifies $GL_n$ with the closed hypersurface
$$
\{(A,t)\in\mathbb A^{n^2+1}:t\det A-1=0\}.
$$
In these coordinates multiplication is $(A,s)(B,t)=(AB,st)$, whose entries are polynomial functions, so it is a morphism.

A morphism $\mathbb A^n\to\mathbb A^1\setminus\{0\}$ is a polynomial with no zero. Over an algebraically closed field every nonconstant polynomial in $n$ variables has a zero: specialize all but one variable so that a nonconstant coefficient remains, then use algebraic closure. The morphism is therefore constant.

A morphism $\mathbb P^n\to\mathbb P^1$ is represented by homogeneous polynomials $F,G$ of the same degree with no common projective zero. If the degree is positive and $n\geq2$, the <projective dimension theorem> forces the two hypersurfaces $V(F)$ and $V(G)$ to meet. Thus the degree is zero and the morphism is constant.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=26F}

The map $\phi$ is an <immersion> when $d\phi$ has rank two, equivalently $\phi_u$ and $\phi_v$ are linearly independent. It is <isothermal coordinates>[isothermal] when its first fundamental form has
$$
E=G=\lambda^2>0,
\qquad F=0.
$$
In general the <mean curvature> is
$$
H=\frac{eG-2fF+gE}{2(EG-F^2)}.
$$
The surface is <minimal surface>[minimal] when $H=0$. In isothermal coordinates the <Laplace-Beltrami operator> identity is
$$
\phi_{uu}+\phi_{vv}=2\lambda^2H N.
$$
Thus $H=0$ exactly when each coordinate function $x,y,z$ is <harmonic function>[harmonic].

For
$$
\phi(u,v)=(f(u)\cos v,f(u)\sin v,g(u)),
$$
the isothermal conditions are
$$
f'^2+g'^2=f^2,
$$
and harmonicity gives
$$
f''-f=0,
\qquad g''=0.
$$
Consequently
$$
f=Ae^u+Be^{-u},
\qquad g=cu+d,
$$
and the isothermal equation reduces to
$$
c^2=4AB.
$$
These are all the solutions for which $f$ never vanishes. When $AB>0$ they parameterize a <catenoid>, up to translations, reflection and scaling. The limiting cases $AB=0$, $c=0$, with exactly one of $A,B$ nonzero, parameterize a punctured plane.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=27H}

For a nonnegative measurable $f$, define
$$
f_n(x)=2^{-n}\left\lfloor2^n\min(f(x),n)\right\rfloor.
$$
Each $f_n$ takes only finitely many values, its level sets are measurable, and $f_n(x)\to f(x)$ pointwise. Thus measurable nonnegative functions are pointwise limits of <approximation by nonnegative simple functions>[simple functions].

If $f$ is integrable, the same functions satisfy $0\leq f_n\leq f$ and converge pointwise. The <dominated convergence theorem> gives
$$
\int_E|f_n-f|\,d\mu\longrightarrow0.
$$
They are integrable because they are bounded above by $f$.

If $0\leq f\leq M$, instead set
$$
g_n=2^{-n}\lfloor2^nf\rfloor.
$$
This has finitely many values and
$$
0\leq f-g_n<2^{-n}
$$
everywhere, so $g_n\to f$ uniformly.

Solved by gpt-5.6-sol high.

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

= a
{parent=28k}
{scope}

= Solution
{parent=a}

Put $q_x=-q_{x,x}=\sum_{y\ne x}q_{x,y}$. The <jump chain> records the successive states visited by $X$ and has transition probabilities
$$
P(x,y)=\frac{q_{x,y}}{q_x},\qquad y\ne x.
$$
By the exponential holding-time construction, conditional on the current state $x$, the destination of the next jump has this distribution independently of the earlier path. Hence the jump chain has the <Markov property>.

Solved by gpt-5.6-sol high.

= b
{parent=28k}
{scope}

= Solution
{parent=b}

An irreducible continuous-time chain is recurrent when, after leaving any state, it returns there almost surely. Non-explosion ensures that returns of $X$ correspond exactly to returns of its embedded jump chain, so one is recurrent precisely when the other is.

The expected total occupation time of zero is
$$
\mathbb E_0\int_0^\infty\mathbf1_{\{X_t=0\}}dt
=\int_0^\infty p_{0,0}(t)dt.
$$
Each visit contributes an independent holding time of mean $1/q_0$. Thus this expectation is $(1/q_0)$ times the expected number of visits. If the integral is infinite, the jump chain has infinitely many expected visits; for an irreducible chain this is equivalent to recurrence.

Solved by gpt-5.6-sol high.

= c
{parent=28k}
{scope}

= Solution
{parent=c}

For the first proposed generator,
$$
q_{x,y}=(1+|x|)^{-2}e^{-|x-y|^2},
$$
the state-dependent factor cancels when the jump probabilities are normalized. Hence
$$
P(x,y)=\frac{e^{-|x-y|^2}}{\sum_{z\ne0}e^{-z^2}}
$$
depends only on $y-x$ and is symmetric. The jump chain is therefore a one-dimensional symmetric random walk with finite variance, and is recurrent. By part (b), the corresponding continuous-time process is recurrent. This proves that at least one of the two generators has the required property.

Solved by gpt-5.6-sol high.

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

= a
{parent=29j}
{scope}

= Solution
{parent=a}

The <Kolmogorov-Smirnov theorem> states that, for continuous $F$,
$$
\sqrt n\sup_x|\widehat F_n(x)-F(x)|
\xrightarrow{d}\sup_{0\leq t\leq1}|B(t)|,
$$
where $B$ is a <Brownian bridge>; the finite-sample law is also independent of continuous $F$. To test $H_0:F=F_0$, compute
$$
D_n=\sup_x|\widehat F_n(x)-F_0(x)|
$$
and reject when $\sqrt nD_n$ exceeds the appropriate null quantile.

Solved by gpt-5.6-sol high.

= b
{parent=29j}
{scope}

= Solution
{parent=b}

Split $Z_1,\ldots,Z_{mn}$ into $m$ blocks of size $n$. Compute the same Kolmogorov-Smirnov statistic against $F_0$ for the observed $X$ sample and for every $Z$ block. Under $H_0$ the $m+1$ statistics are exchangeable and, because the distribution is continuous, ties have probability zero. Reject exactly when the statistic from the $X$ sample is the largest. Its null rejection probability is exactly
$$
\boxed{\frac1{m+1}}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=29j}
{scope}

= Solution
{parent=c}

Let $h=n^{-1/3}$. Away from the boundary, the estimator is
$$
\widehat f_n(x)=\frac1{nh}\sum_{i=1}^n
\mathbf1_{\{|X_i-x|\leq h/2\}}.
$$
Its variance is at most
$$
\frac1{nh^2}\mathbb P(|X_1-x|\leq h/2)
\leq\frac1{nh}=n^{-2/3},
$$
and the <bias of a kernel density estimator> is bounded by $h\|f'\|_\infty/2$, so its squared bias is at most $h^2/4$. This gives the claimed bound, with room to spare, whenever the full window lies in the support.

At $x=0$, however, the assertion is false as printed. For example
$$
f(x)=\frac12e^{-x/2},\qquad x\geq0,
$$
satisfies $\sup(|f|+|f'|)=3/4<1$, but
$$
\mathbb E\widehat f_n(0)longrightarrow\frac{f(0)}2=\frac14,
$$
whereas $f(0)=1/2$. The mean-square error therefore tends to $1/16$, contradicting a bound tending to zero. The intended statement needs $x\geq h/2$, a boundary correction, or a condition such as $f(0)=0$.

Solved by gpt-5.6-sol high.

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

= a
{parent=30k}
{scope}

= Solution
{parent=a}

A risky-asset portfolio $\varphi$ financed through the risk-free asset has zero-cost discounted gain
$$
G_\varphi=\varphi^T(S_1-(1+r)S_0).
$$
It is an <arbitrage> when $G_\varphi\geq0$ almost surely and $\mathbb P(G_\varphi>0)>0$.

Solved by gpt-5.6-sol high.

= b
{parent=30k}
{scope}

= Solution
{parent=b}

If $G\geq0$ were an arbitrage gain, then $X_1^*+tG$ would be attainable from the same initial wealth for every $t>0$. Since $U$ is increasing,
$$
U(X_1^*+tG)\geq U(X_1^*)
$$
almost surely, with strict inequality on a set of positive probability under the usual strictly increasing meaning of utility. This contradicts optimality of $X_1^*$. Hence existence of an optimizer implies <arbitrage>[no-arbitrage].

Solved by gpt-5.6-sol high.

= c
{parent=30k}
{scope}

= Solution
{parent=c}

Let $u(x)$ be maximal expected utility without the claim. The <utility indifference price> $\pi(Y)$ is determined by
$$
\sup_\varphi\mathbb E\,U\!\left((x-\pi(Y))(1+r)+G_\varphi+Y\right)=u(x).
$$
For claims $Y_1,Y_2$, mix optimal portfolios for their indifference problems with weights $\lambda$ and $1-\lambda$. Concavity of $U$ shows that buying $\lambda Y_1+(1-\lambda)Y_2$ at price
$$
\lambda\pi(Y_1)+(1-\lambda)\pi(Y_2)
$$
gives expected utility at least $u(x)$. Since maximal utility decreases with purchase price,
$$
\pi(\lambda Y_1+(1-\lambda)Y_2)
\geq\lambda\pi(Y_1)+(1-\lambda)\pi(Y_2).
$$
Thus $\pi$ is concave.

Solved by gpt-5.6-sol high.

= d
{parent=30k}
{scope}

= Solution
{parent=d}

At the optimal terminal wealth, normalized marginal utility defines an <equivalent martingale measure> $Q$ and hence a marginal utility price
$$
p_Q(Y)=\frac1{1+r}\mathbb E_QY.
$$
This is an arbitrage-free price, so $p_Q(Y)\in I$. Concavity of utility implies that the finite-quantity indifference bid does not exceed its marginal price:
$$
\pi(Y)\leq p_Q(Y).
$$
Consequently
$$
\boxed{\pi(Y)\leq\sup\{p:p\in I\}}.
$$

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 a <convex function> $f$, its <subdifferential> is
$$
\partial f(x)=\{g:f(y)\geq f(x)+g^T(y-x)\text{ for every }y\}.
$$
For $f(x)=\gamma|x|$,
$$
\partial f(x)=
\begin{cases}
\{\gamma\},&x>0,\\
[-\gamma,\gamma],&x=0,\\
\{-\gamma\},&x<0.
\end{cases}
$$
The defining inequality shows directly that $0\in\partial f(x)$ exactly when $x$ is a global minimizer.

Strict convexity means
$$
f(\lambda x+(1-\lambda)y)<\lambda f(x)+(1-\lambda)f(y)
$$
for distinct $x,y$ and $0<\lambda<1$. Two distinct minimizers would make the midpoint have a strictly smaller value, so a minimizer is unique.

Solved by gpt-5.6-sol high.

= b
{parent=31j}
{scope}

= Solution
{parent=b}

With the other coordinates fixed, the smooth part of the objective has derivative at $\beta_1=0$
$$
\frac1n\left(\kappa_1-(\kappa_2-\kappa_1)\right)
=\frac{2\kappa_1-\kappa_2}{n}.
$$
The subdifferential of the $\ell^1$ penalty at zero is $\gamma[-1,1]$. Therefore
$$
\partial_{\beta_1}f(0,\beta_2,\ldots,\beta_p)
=\frac{2\kappa_1-\kappa_2}{n}+\gamma[-1,1].
$$
The stated condition $|2\kappa_1-\kappa_2|\leq\gamma$ is stronger than the needed condition $|2\kappa_1-\kappa_2|\leq n\gamma$, and hence places zero in this subdifferential. The one-variable exponential loss is strictly convex because the mismatch set is nonempty, so the minimizer is unique:
$$
\boxed{\operatorname{argmin}_{\beta_1\in\mathbb R}f(\beta_1,\beta_2,\ldots,\beta_p)=\{0\}}.
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=32a}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

The functions $(\phi_n)$ form an <asymptotic sequence> as $t\to\infty$ when
$$
\phi_{n+1}(t)=o(\phi_n(t))
$$
for every $n$.

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

The notation
$$
x(t)\sim\sum_{n=0}^\infty a_n\phi_n(t)
$$
means that for every $N\geq0$,
$$
x(t)-\sum_{n=0}^Na_n\phi_n(t)=o(\phi_N(t))
\qquad(t\to\infty).
$$
The series need not converge for any fixed $t$.

Solved by gpt-5.6-sol high.

= b
{parent=32a}
{scope}

= Solution
{parent=b}

Let $\Phi(t)=2t^4-t^2$. Its stationary points on $[0,1]$ are the endpoint $0$ and the interior point $1/2$, with
$$
\Phi(0)=0,quad \Phi''(0)=-2,\qquad
\Phi(1/2)=-\frac18,quad \Phi''(1/2)=4.
$$
The endpoint stationary-phase contribution to $\int_0^1e^{ix\Phi(t)}dt$ is
$$
\frac{\sqrt\pi}{2\sqrt x}e^{-i\pi/4},
$$
and the interior contribution is
$$
\sqrt{\frac\pi{2x}}e^{i(\pi/4-x/8)}.
$$
The nonstationary endpoint contributes only $O(x^{-1})$. Taking imaginary parts gives
$$
\boxed{I(x)\sim\sqrt{\frac\pi{2x}}
\left[\sin\left(\frac\pi4-\frac x8\right)-\frac12\right]}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=32a}
{scope}

= Solution
{parent=c}

Write $\sinh(x\Phi)=(e^{x\Phi}-e^{-x\Phi})/2$. The global maximum of $\Phi$ is the nonstationary endpoint $t=1$, where
$$
\Phi(1)=1,\qquad \Phi'(1)=6.
$$
Endpoint <Laplace's method> therefore gives
$$
\frac12\int_0^1e^{x\Phi(t)}dt\sim\frac{e^x}{12x}.
$$
The second exponential has its maximum at $t=1/2$ and is only of order $x^{-1/2}e^{x/8}$, exponentially smaller. Hence
$$
\boxed{J(x)\sim\frac{e^x}{12x}}.
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=33a}
{scope}

= Solution
{parent=a}

On each side of the stated parallelogram, substitution into the outward normal component of
$$
(\dot x,\dot y)=\left(y-\mu(x^3/3-x),-x+F\right)
$$
shows that the leading $L^3$ terms point inward; the remaining terms are $O(L)$, so the region is positively invariant for sufficiently large $L$.

There is one equilibrium,
$$
(x_*,y_*)=(F,\mu(F^3/3-F)).
$$
Its Jacobian has determinant one and trace $\mu(1-F^2)$. If $F^2<1$ the equilibrium is unstable. A trajectory starting nearby remains in the compact trapping region, and its omega-limit set contains no stable equilibrium. The <Poincare-Bendixson theorem> therefore supplies a periodic orbit.

Solved by gpt-5.6-sol high.

= b
{parent=33a}
{scope}

= Solution
{parent=b}

Eliminating $y$ gives
$$
\ddot x+x-F=\mu(1-x^2)\dot x.
$$
For $\mu\ll1$, take $x=F+A\cos(t+\delta)$ over one fast period. The oscillator energy satisfies
$$
\dot E=\mu(1-x^2)\dot x^2,
$$
and averaging gives
$$
\langle\dot E\rangle
=\frac{\mu A^2}{2}\left(1-F^2-\frac{A^2}{4}\right).
$$
Thus, if $F^2<1$, amplitudes approach
$$
A=2\sqrt{1-F^2},
$$
producing a stable weakly nonlinear limit cycle. If $F^2>1$, every nonzero small amplitude loses energy and trajectories approach the stable equilibrium.

Solved by gpt-5.6-sol high.

= c
{parent=33a}
{scope}

= Solution
{parent=c}

For $\mu\gg1$, trajectories rapidly approach the cubic nullcline
$$
y=\mu(x^3/3-x)
$$
and then move slowly along its attracting outer branches $|x|>1$. For $0<F<1$, the equilibrium lies on the repelling middle branch. Slow motion to a fold, a rapid jump to the opposite outer branch, and repetition produce a <relaxation oscillation>. For $F>1$, the equilibrium lies on the attracting right branch; trajectories undergo any necessary fast jump and then drift to that equilibrium, so no relaxation cycle remains.

Solved by gpt-5.6-sol high.

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

= a
{parent=34d}
{scope}

= Solution
{parent=a}

The <isospectral Lax equation>[Lax equation] $L_t=[A,L]$ with
$$
L=-\partial_x^2+u,\qquad
A=4\partial_x^3-3u\partial_x-3\partial_xu
$$
reduces, after expanding the commutator, to the KdV equation
$$
u_t+u_{xxx}-6uu_x=0.
$$
Thus the evolution is isospectral: the eigenvalues of $L$ are constant. The fourth derivative shown in the converted question is a transcription error.

Solved by gpt-5.6-sol high.

= b
{parent=34d}
{scope}

= Solution
{parent=b}

The <discrete scattering data>[scattering data for the Schrodinger operator] consist of the reflection coefficient $R(k,t)$ on the continuous spectrum, the negative discrete eigenvalues $-\kappa_n^2$, and their norming constants $c_n$. Isospectrality gives
$$
\kappa_n(t)=\kappa_n(0).
$$
Comparing the asymptotic plane waves in $\psi_t=A\psi$ gives
$$
R(k,t)=R(k,0)e^{8ik^3t},
\qquad
c_n(t)=c_n(0)e^{4\kappa_n^3t}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=34d}
{scope}

= Solution
{parent=c}

Each discrete term satisfies
$$
\partial_t(c_n^2e^{-\kappa_nx})=8\kappa_n^3c_n^2e^{-\kappa_nx}
=-8\partial_x^3(c_n^2e^{-\kappa_nx}),
$$
and the same identity follows for the Fourier integral from $R_t=8ik^3R$. Hence
$$
F_t+8F_{xxx}=0.
$$
Apply $\partial_x^2-\partial_y^2-u(x,t)$ to the <Gelfand-Levitan-Marchenko equation>. Differentiating under the integral and integrating the $z$ derivatives by parts, the boundary terms combine to $u=-2\partial_xK(x,x,t)$. The terms containing $F$ cancel, leaving
$$
\boxed{G(x,y,t)+\int_x^\infty G(x,z,t)F(z+y,t)\,dz=0}.
$$
Uniqueness of the Marchenko equation then also gives $G=0$.

Solved by gpt-5.6-sol high.

= d
{parent=34d}
{scope}

= Solution
{parent=d}

Use
$$
(\partial_x+\partial_y)^3
=\partial_x^3+\partial_y^3+3\partial_x\partial_y(\partial_x+\partial_y)
$$
and differentiate the spatial equation
$$
(\partial_x^2-\partial_y^2)K=uK
$$
to eliminate the mixed derivative in the time equation. After collecting the terms involving $u$ and $u_x$, one obtains
$$
\boxed{K_t+(\partial_x+\partial_y)^3K
-3u(\partial_x+\partial_y)K=0}.
$$
Set $y=x$, use $u=-2\partial_xK(x,x,t)$ with the derivative along the diagonal, and differentiate once more in $x$. The resulting compatibility condition is
$$
\boxed{u_t+u_{xxx}-6uu_x=0}.
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=35b}
{scope}

= Solution
{parent=a}

For a nondegenerate level, <time-independent perturbation theory> gives
$$
\boxed{E_n(\lambda)=E_n+\lambda\langle n|\Delta H|n\rangle
+\lambda^2\sum_{m\ne n}
\frac{|\langle m|\Delta H|n\rangle|^2}{E_n-E_m}+O(\lambda^3)}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=35b}
{scope}

= Solution
{parent=b}

Since $X=L(a+a^\dagger)$, the perturbation is
$$
\lambda\hbar\omega(a+a^\dagger)^3.
$$
Every diagonal matrix element vanishes by <parity>, so all first-order shifts are zero. For the ground state,
$$
(a+a^\dagger)^3|0\rangle=3|1\rangle+\sqrt6|3\rangle.
$$
The second-order shift is therefore
$$
\lambda^2(\hbar\omega)^2
\left(\frac9{-\hbar\omega}+\frac6{-3\hbar\omega}\right)
=-11\lambda^2\hbar\omega.
$$
Thus
$$
\boxed{E_0=\hbar\omega\left(\frac12-11\lambda^2\right)+O(\lambda^3)}.
$$
For every nonzero real $\lambda$, the cubic potential is unbounded below on one side, so there are no genuine bound-state eigenvalues continuously connected to the harmonic levels. The formal perturbation series does not converge; it is at best asymptotic to resonance energies.

Solved by gpt-5.6-sol high.

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

= a
{parent=36b}
{scope}

= Solution
{parent=a}

The s-wave phase shift is defined by
$$
\psi_0(r)\sim\frac{C}{r}\sin(kr+\delta_0).
$$
Expanding the sine and comparing coefficients with the given expression gives
$$
C\cos\delta_0=A,\qquad C\sin\delta_0=Ag(k),
$$
so $\boxed{\tan\delta_0=g(k)}$.

Solved by gpt-5.6-sol high.

= b
{parent=36b}
{scope}

= Solution
{parent=b}

The <scattering amplitude> is defined by
$$
\psi\sim e^{ikz}+f(\theta)\frac{e^{ikr}}r.
$$
For a central potential its <partial-wave S-matrix>[partial-wave expansion] is
$$
f(\theta)=\frac1k\sum_{l=0}^\infty(2l+1)e^{i\delta_l}\sin\delta_lP_l(\cos\theta).
$$
Using orthogonality of the <Legendre polynomials> in $\sigma_T=\int|f|^2d\Omega$ gives
$$
\boxed{\sigma_T=\frac{4\pi}{k^2}\sum_{l=0}^\infty(2l+1)\sin^2\delta_l}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=36b}
{scope}

= Solution
{parent=c}

Here $k\cot\delta_0=-K$, so the s-wave amplitude is
$$
f_0(k)=\frac1{-K-ik}.
$$
It has a pole at $k=iK$, corresponding to the bound-state energy
$$
\boxed{E=-\frac{\hbar^2K^2}{2m}}.
$$
Moreover,
$$
\boxed{\sigma_T\simeq4\pi|f_0|^2=\frac{4\pi}{K^2+k^2}}.
$$

Solved by gpt-5.6-sol high.

= d
{parent=36b}
{scope}

= Solution
{parent=d}

Since $\tan\delta_0=\gamma/(K_0-k)$,
$$
\boxed{\sigma_0(k)=\frac{4\pi}{k^2}
\frac{\gamma^2}{(K_0-k)^2+\gamma^2}}.
$$
The exact nearby stationary point satisfies $K_0-k=\gamma^2/K_0+O(\gamma^4/K_0^3)$, so to leading order in $\gamma/K_0$ the local maximum is at $k=K_0$. This is a <scattering resonance> with
$$
E_R=\frac{\hbar^2K_0^2}{2m},
\qquad
\boxed{\Gamma=\frac{2\hbar^2K_0\gamma}{m}},
$$
where $\Gamma$ is the full energy width of the Breit-Wigner peak.

Solved by gpt-5.6-sol high.

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

= a
{parent=37c}
{scope}

= Solution
{parent=a}

The <microcanonical ensemble> fixes energy, volume and particle number; the <canonical ensemble> fixes temperature, volume and particle number; and the <grand canonical ensemble> fixes temperature, volume and chemical potential while allowing particle exchange. For macroscopic systems with short-range interactions, away from phase coexistence and other nonconcave regimes, relative fluctuations vanish in the thermodynamic limit and the ensembles give equivalent local thermodynamics.

Solved by gpt-5.6-sol high.

= b
{parent=37c}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

The center-of-mass and relative terms in the Hamiltonian are independent, so the phase-space integral factorizes as stated. Gaussian integration gives
$$
\boxed{Z_{\rm trans}=\frac{V}{(2\pi\hbar)^3}
\left(\frac{2\pi M}{\beta}\right)^{3/2}}.
$$
In the large-volume limit the relative coordinate may be integrated over all space. Since
$$
4\pi\int_0^\infty r^2e^{-\beta\alpha r}dr
=\frac{8\pi}{(\beta\alpha)^3},
$$
we obtain
$$
\boxed{Z_{\rm int}=\frac1{(2\pi\hbar)^3}
\left(\frac{2\pi\mu}{\beta}\right)^{3/2}
\frac{8\pi}{(\beta\alpha)^3}}.
$$
The radial probability is proportional to $r^2e^{-\beta\alpha r}$, hence
$$
\boxed{\langle r\rangle=\frac{3}{\beta\alpha}=\frac{3k_BT}{\alpha}}.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

For an ideal gas of indistinguishable classical mesons,
$$
\boxed{Z_N=\frac{Z_1^N}{N!}}.
$$
Apart from temperature-independent factors, $Z_{\rm trans}\propto T^{3/2}$ and $Z_{\rm int}\propto T^{9/2}$, so $Z_1\propto T^6$. Therefore
$$
U=-\frac{\partial}{\partial\beta}\log Z_N=6Nk_BT,
\qquad
\boxed{C_V=6Nk_B}.
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=38c}
{scope}

= Solution
{parent=a}

An affine geodesic follows from
$$
\mathcal L=\frac12\left[r^{-2}(-\dot t^2+\dot r^2)+r^2\dot\theta^2\right].
$$
Its constant norm may be chosen as
$$
2\mathcal L=\varepsilon,
\qquad
\varepsilon=-1,0,+1
$$
for timelike, null and spacelike geodesics respectively.

Solved by gpt-5.6-sol high.

= b
{parent=38c}
{scope}

= Solution
{parent=b}

The Killing coordinates $t$ and $\theta$ give
$$
E=\frac{\dot t}{r^2},
\qquad
J=r^2\dot\theta.
$$
Together with the norm $\varepsilon=2\mathcal L$, these are the three constants of motion.

Solved by gpt-5.6-sol high.

= c
{parent=38c}
{scope}

= i
{parent=c}
{scope}

= Solution
{parent=i}

For a radial null ray, $d\theta=0$ and the metric gives $dt=dr$ for the outgoing branch. Starting at $r=t=0$, it reaches $r=R$ at
$$
\boxed{t_1=R}.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=c}
{scope}

= Solution
{parent=ii}

A stationary observer has four-velocity $u=r\partial_t$. If $E=-k_t$ is the conserved photon energy, the measured frequency is proportional to
$$
-u\cdot k=rE.
$$
Hence
$$
\boxed{\nu_2=2\nu_1}.
$$
The photon is blueshifted as it moves outward.

Solved by gpt-5.6-sol high.

= d
{parent=38c}
{scope}

= Solution
{parent=d}

Parametrize the spacelike geodesic by unit length. With $J=r^2\dot\theta=b$, its norm equation gives
$$
\dot r^2=r^2-b^2.
$$
The total angular change is therefore
$$
|\Delta\theta|=2\int_b^\infty
\frac{b\,dr}{r^2\sqrt{r^2-b^2}}.
$$
Putting $r=b\sec u$ evaluates the half-integral as $1/b$, and hence
$$
\boxed{|\Delta\theta|=\frac2b}.
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=39a}
{scope}

= Solution
{parent=a}

The <lubrication theory>[lubrication approximation] requires $H/L\ll1$, small wall slope, and negligible inertia, for example $\rho UH/\mu\ll L/H$. To leading order,
$$
p_z=0,\qquad
\mu\mathbf u_{zz}=\nabla p,\qquad
\nabla\cdot\mathbf u+w_z=0.
$$
With no slip at $z=0,h$,
$$
\mathbf u=\mathbf U\left(1-\frac zh\right)
+\frac{z(z-h)}{2\mu}\nabla p.
$$
Integration gives
$$
\boxed{\mathbf q=\frac{\mathbf Uh}{2}-\frac{h^3}{12\mu}\nabla p}.
$$
Integrated incompressibility gives $\nabla\cdot\mathbf q=0$, so locally $\mathbf q=\nabla\wedge(0,0,\psi)$. Solving the flux law for $\nabla p$ and taking its curl yields
$$
\boxed{\nabla\cdot\left(h^{-3}\nabla\psi\right)
=-Uh^{-3}h_y}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=39a}
{scope}

= Solution
{parent=b}

At infinity,
$$
\mathbf q=-\frac{h_0^3}{12\mu}\nabla p
=\frac{h_0^3G}{12\mu}\mathbf e_x,
$$
so
$$
\psi\sim Ar\sin\theta,\qquad
\boxed{A=\frac{h_0^3G}{12\mu}}.
$$
Continuity of $q_r=r^{-1}\psi_\theta$ gives $[\psi_\theta]^+_-=0$. Since $\nabla p=-12\mu h^{-3}\mathbf q$, continuity of tangential pressure derivative gives $[h^{-3}\psi_r]^+_-=0$.

The harmonic solutions regular at zero and with the prescribed far field are
$$
\psi_>=\left(Ar+\frac Br\right)\sin\theta,
\qquad
\psi_<=Cr\sin\theta.
$$
The interface conditions give
$$
B=Aa^2\frac{h_1^3-h_0^3}{h_0^3+h_1^3},
\qquad
C=\frac{2Ah_1^3}{h_0^3+h_1^3}.
$$
Integrating the outward flux across the appropriate semicircle gives
$$
\boxed{2aC=\frac{4Aah_1^3}{h_0^3+h_1^3}}.
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=40a}
{scope}

= Solution
{parent=a}

Write the SH displacement as $u=U(z)e^{i(kx-\omega t)}$ with $c=\omega/k$. A trapped mode requires
$$
\bar c_s<c<c_s.
$$
In the substrate $U=Ae^{qz}$, where
$$
q=k\sqrt{1-c^2/c_s^2},
$$
and the stress-free condition at $z=h$ makes the layer solution $U=B\cos(m(z-h))$, where
$$
m=k\sqrt{c^2/\bar c_s^2-1}.
$$
Continuity of displacement and shear stress at $z=0$ eliminates $A,B$ and gives
$$
\boxed{\tan(mh)=\frac{\mu}{\bar\mu}\frac qm
=\frac{\mu}{\bar\mu}
\frac{\sqrt{1-c^2/c_s^2}}{\sqrt{c^2/\bar c_s^2-1}}}.
$$
The bars lost from the converted statement are required dimensionally and by the stated ordering of wave speeds.

Solved by gpt-5.6-sol high.

= b
{parent=40a}
{scope}

= Solution
{parent=b}

Both sides of the dispersion relation are positive on an interval beginning at zero phase, so a zeroth branch exists for every $k>0$. Higher branches first appear when $c=c_s$, for which the right side vanishes and $mh=n\pi$. Thus
$$
k_c^{(n)}=\frac{n\pi\bar c_s}{h\sqrt{c_s^2-\bar c_s^2}},
$$
and
$$
\boxed{\omega_c^{(n)}=c_sk_c^{(n)}
=\frac{n\pi\bar c_sc_s}{h\sqrt{c_s^2-\bar c_s^2}}}.
$$
The $n=1$ dispersion curve begins on the line $\omega=c_sk$ at $(k_c^{(1)},\omega_c^{(1)})$, bends below it, and approaches the layer shear-wave line $\omega=\bar c_sk$ from above as $k\to\infty$.

Solved by gpt-5.6-sol high.

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

= a
{parent=41e}
{scope}

= Solution
{parent=a}

Reverse the second half of the DFT sum by putting $m=2N-n-1$. Since $\omega_{2N}^{2Nk}=1$,
$$
\begin{aligned}
Y_k
&=\sum_{n=0}^{N-1}x_n
\left(\omega_{2N}^{nk}+\omega_{2N}^{-(n+1)k}\right)\\
&=2\omega_{2N}^{-k/2}\sum_{n=0}^{N-1}x_n
\cos\left[\frac\pi N\left(n+\frac12\right)k\right].
\end{aligned}
$$
This is the required identity.

Solved by gpt-5.6-sol high.

= b
{parent=41e}
{scope}

= Solution
{parent=b}

For a length-$M$ DFT with $M$ even, split the input into its even and odd entries. Two length-$M/2$ DFTs determine the result, followed by at most $M$ multiplications by twiddle factors. Hence the multiplication count satisfies
$$
T(M)\leq2T(M/2)+M,
$$
so induction gives $T(M)\leq CM\log_2M$ for powers of two. This proves the needed <Cooley--Tukey fast Fourier transform>[fast Fourier transform] bound rather than assuming it.

Part (a) computes the length-$2N$ DFT of the reflected vector $y$ and then recovers
$$
z_k=\frac12\omega_{2N}^{k/2}Y_k,\qquad0\leq k<N.
$$
Forming $y$ uses no multiplications and the final recovery uses only $N$. Therefore the <discrete cosine transform> costs at most
$$
C(2N)\log(2N)+N=O(N\log N).
$$

Solved by gpt-5.6-sol high.