past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ii/paper-2.bigb

= Paper 2
{scope}

https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2024/paperii_2_2024.pdf

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

= Solution
{parent=1F}

A <reduced positive definite binary quadratic form> $[a,b,c]$ satisfies
$$
|b|\leq a\leq c,
$$
with $b\geq0$ when either $|b|=a$ or $a=c$. For $d\equiv0$ or $3\pmod4$, the <class number of a negative discriminant> $h(-d)$ is the number of proper equivalence classes of primitive positive definite <integral> forms of discriminant $-d$, equivalently the number of their reduced representatives.

Write
$$
d=\prod_{j=1}^k p_j^{e_j}.
$$
Partitioning the $k$ complete prime-power factors between $r$ and $s$ gives $2^k$ ordered factorizations $d=rs$ with $\gcd(r,s)=1$. After identifying $(r,s)$ with $(s,r)$, there are $2^{k-1}$ choices with $r\leq s$. Each gives the primitive reduced form
$$
[r,0,s],
$$
whose discriminant is $-4rs=-4d$. The uniqueness of reduced representatives makes these classes distinct, proving the <coprime-factorization lower bound for a quadratic-form class number>
$$
h(-4d)\geq2^{k-1}.
$$

The inequality can be strict. For $d=5$, there is one distinct prime factor, while the reduced primitive forms of discriminant $-20$ are
$$
[1,0,5]\quad\hbox{and}\quad[2,2,3].
$$
Thus $h(-20)=2>1=2^{1-1}$.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=2G}

The <Baire category theorem> says that a complete metric space cannot be a countable union of nowhere-dense closed sets. Equivalently, every countable intersection of open dense sets is dense.

A point $x$ of a metric space is an <isolated point> when some $\varepsilon>0$ satisfies $B(x,\varepsilon)=\{x\}$.

Solved by gpt-5.6-sol high.

= i
{parent=2g}
{scope}

= Solution
{parent=i}

False. The rational numbers with the Euclidean metric are countable and have no <isolated point>.

Solved by gpt-5.6-sol high.

= ii
{parent=2g}
{scope}

= Solution
{parent=ii}

False. Every discrete metric space is complete and every one of its points is isolated; a singleton is the smallest example.

Solved by gpt-5.6-sol high.

= iii
{parent=2g}
{scope}

= Solution
{parent=iii}

False. The interval $(0,1)$ with the Euclidean metric is uncountable and has no isolated points, but the Cauchy <sequence> $1/n$ has no <limit> in the space.

Solved by gpt-5.6-sol high.

= iv
{parent=2g}
{scope}

= Solution
{parent=iv}

False. A singleton, or $\mathbb Z$ with the discrete metric, is complete and countable.

Solved by gpt-5.6-sol high.

= v
{parent=2g}
{scope}

= Solution
{parent=v}

True. If a nonempty countable complete metric space had no isolated point, every singleton would be closed and nowhere dense. Their countable union would be the whole space, contradicting the <Baire category theorem>. This is the basic <isolated points in a countable complete metric space> argument.

Solved by gpt-5.6-sol high.

= vi
{parent=2g}
{scope}

= Solution
{parent=vi}

False. The subspace
$$
X=\{0\}\cup\{1/n:n\geq1\}\subset\mathbb R
$$
is closed and hence complete, and it is countable; however, $0$ is not isolated.

Solved by gpt-5.6-sol high.

= vii
{parent=2g}
{scope}

= Solution
{parent=vii}

True. Part (v) supplies one isolated point $p$. Since $\{p\}$ is open, $X\setminus\{p\}$ is closed and therefore complete. It is nonempty when $X$ has at least two points, so part (v) applied to this subspace supplies an isolated point $q$. Because $d(p,q)>0$, a sufficiently small ball showing that $q$ is isolated in the subspace also avoids $p$; hence $q$ is isolated in $X$.

Solved by gpt-5.6-sol high.

= viii
{parent=2g}
{scope}

= Solution
{parent=viii}

True. If there were only finitely many isolated points, delete all of them. The remainder would be a nonempty closed, countable, complete subspace, so part (v) would give it an isolated point. Its positive distance from the finite deleted set would make it isolated in the original space as well, a contradiction. Thus the stronger <isolated points in a countable complete metric space> result applies.

Solved by gpt-5.6-sol high.

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

= a
{parent=3k}
{scope}

= Solution
{parent=a}

The original binary <Hamming code> has parameters $[7,4,3]$, so it has $2^4=16$ codewords. Its minimum distance three makes the radius-one balls about its codewords disjoint. Each ball has
$$
\binom70+\binom71=8
$$
<vectors>, and $16\cdot8=128=2^7$, the size of the ambient space. Equality in the <Hamming bound> therefore shows that these balls partition $\mathbb F_2^7$: the code is perfect.

Solved by gpt-5.6-sol high.

= b
{parent=3k}
{scope}

= Solution
{parent=b}

The added bit is the overall parity bit. The original minimum-weight words have weight three, so they acquire a parity bit equal to one and have extended weight four. Every original even-weight difference keeps its weight, while every odd-weight difference gains one; hence the new minimum distance is four.

The <minimum-distance error-detection and correction guarantee> therefore says that the code detects three errors and corrects one error.

Solved by gpt-5.6-sol high.

= c
{parent=3k}
{scope}

= Solution
{parent=c}

Yes, for a binary code. Correcting $e$ errors requires minimum distance at least $2e+1$. Apply the <parity extension>: append to each word the bit that makes its total weight even. Every odd distance increases by one and every even distance is unchanged, so the extended minimum distance is at least $2e+2$. It therefore detects every pattern of at most $2e+1$ errors.

Solved by gpt-5.6-sol high.

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

= a
{parent=4j}
{scope}

= Solution
{parent=a}

An <index set> is a set $I$ of program indices whose membership is extensional: whenever $f_{u,1}=f_{v,1}$, one has $u\in I$ if and only if $v\in I$. It is nontrivial when it is neither empty nor the set of all indices.

Solved by gpt-5.6-sol high.

= b
{parent=4j}
{scope}

= Solution
{parent=b}

For the fixed effective enumeration of unary partial computable <functions>, the <infinite-domain index set> is
$$
\mathbf{Inf}=\{w:\operatorname{dom}(f_{w,1})\text{ is infinite}\}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=4j}
{scope}

= Solution
{parent=c}

The <Rice theorem> states that every nontrivial index set of partial computable <functions> is not computable.

Solved by gpt-5.6-sol high.

= d
{parent=4j}
{scope}

= Solution
{parent=d}

Fix $x_0\in X$. The property defining $\mathbf{Inf}_X$ depends only on the partial <function> computed, so it is an index set. It is nontrivial: an index of the nowhere-defined <function> is outside it, whereas an index of the total <constant function> $n\mapsto x_0$ has infinite domain and range $\{x_0\}\subseteq X$, and hence lies in it. The <Rice theorem> now shows that this <infinite-domain range-restriction index set> is not computable.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=5L}

Put $K=nY$. Apart from a factor independent of $p$, the likelihood is
$$
L(p)=p^{nY}(1-p)^{n(1-Y)}.
$$
Thus the <score function> and <Fisher information matrix> are
$$
S(p)=\ell'(p)=n\left(\frac Yp-\frac{1-Y}{1-p}\right)
=\frac{n(Y-p)}{p(1-p)},
\qquad
I(p)=\frac n{p(1-p)}.
$$

The <Newton method> and <Fisher scoring> updates are respectively
$$
p^{(r+1)}=p^{(r)}-\frac{\ell'(p^{(r)})}{\ell''(p^{(r)})},
\qquad
p^{(r+1)}=p^{(r)}+I(p^{(r)})^{-1}S(p^{(r)}).
$$
Since $\widehat p=Y$, initialization at $p^{(0)}=Y$ requires zero steps for either algorithm. Moreover,
$$
p+I(p)^{-1}S(p)=p+\frac{p(1-p)}n\frac{n(Y-p)}{p(1-p)}=Y,
$$
so <binomial-proportion Fisher scoring> reaches the MLE in one step from every interior $p^{(0)}\ne Y$.

Solved by gpt-5.6-sol high.

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

= a
{parent=6a}
{scope}

= Solution
{parent=a}

In the <SIR model with demography and permanent immunity>, $\mu$ is the common per-capita death rate and, through the term $\mu N$, the replacement birth rate. The coefficient $\beta$ is the mass-action transmission rate per susceptible-infective pair, and $\nu$ is the per-capita recovery rate from infection into permanent immunity.

Solved by gpt-5.6-sol high.

= b
{parent=6a}
{scope}

= Solution
{parent=b}

The infective equation is
$$
Y'=Y\{\beta X-(\mu+\nu)\}.
$$
Since $X\leq N$, the threshold population is
$$
N_c=\frac{\mu+\nu}{\beta}.
$$
If $N<N_c$ and $\delta=\mu+\nu-\beta N>0$, then
$$
Y'\leq-\delta Y,
\qquad Y(t)\leq Y(0)e^{-\delta t}\longrightarrow0.
$$
Also
$$
Z(t)=e^{-\mu t}Z(0)+\nu\int_0^t e^{-\mu(t-s)}Y(s)\,ds\longrightarrow0.
$$
Any steady state with $Y>0$ would require $X=N_c>N$, which is impossible. Thus the infection cannot be maintained, and both infected and immune populations tend to zero from every admissible initial condition.

Solved by gpt-5.6-sol high.

= c
{parent=6a}
{scope}

= Solution
{parent=c}

For a positive equilibrium, $Y'=0$ gives $X^*=N_c$. The equations $Z'=0$ and $X^*+Y^*+Z^*=N$ then give the <endemic equilibrium of the SIR model with demography>
$$
X^*=N_c,
\qquad
Y^*=\frac{\mu}{\mu+\nu}(N-N_c),
\qquad
Z^*=\frac{\nu}{\mu+\nu}(N-N_c),
$$
which is strictly positive when $N>N_c$.

Use $Z=N-X-Y$ to reduce to the $(X,Y)$ system. At the endemic equilibrium its Jacobian is
$$
J=\begin{pmatrix}
-\mu-\beta Y^*&-(\mu+\nu)\\
\beta Y^*&0
\end{pmatrix}.
$$
Its trace is negative and its <determinant> is $\beta Y^*(\mu+\nu)>0$. The two <eigenvalues> therefore have negative real part, so the endemic equilibrium is locally asymptotically stable.

Solved by gpt-5.6-sol high.

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

= a
{parent=7d}
{scope}

= Solution
{parent=a}

Substituting
$$
y(z)=\int_\gamma e^{zt}f(t)\,dt,
\qquad y^{(3)}(z)=\int_\gamma e^{zt}t^3f(t)\,dt,
$$
and using $ze^{zt}=\partial_te^{zt}$ gives
$$
zy^{(3)}+2y
=[e^{zt}t^3f(t)]_{\partial\gamma}
+\int_\gamma e^{zt}\{2f(t)-(t^3f(t))'\}\,dt.
$$
It is therefore enough to require
$$
(t^3f)'=2f
$$
and choose $\gamma$ so that the <integral> converges and $e^{zt}t^3f(t)$ vanishes at its endpoints. Solving the amplitude equation,
$$
\frac{f'}f=\frac2{t^3}-\frac3t,
\qquad
f(t)=Ct^{-3}e^{-1/t^2}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=7d}
{scope}

= Solution
{parent=b}

For $x>0$, take the real contour $\gamma=(-\infty,0)$. At $t\to0^-$ the factor $e^{-1/t^2}$ kills every power, while at $t\to-\infty$ the factor $e^{xt}$ makes the integrand and endpoint term vanish. The <laplace integral solution of a singular third-order equation> is therefore
$$
y(x)=C\int_{-\infty}^0 e^{xt-1/t^2}t^{-3}\,dt.
$$
With $u=-1/t$, and absorbing a minus sign into the arbitrary constant, this becomes
$$
\boxed{y(x)=C_1\int_0^\infty u\exp\left(-u^2-\frac xu\right)du}.
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=8e}
{scope}

= Solution
{parent=a}

The <generalized momentum> conjugate to $q$ is
$$
p_q=\frac{\partial L}{\partial\dot q}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=8e}
{scope}

= Solution
{parent=b}

A coordinate $q$ is an <ignorable coordinate> when $L$ has no explicit dependence on it, so $\partial L/\partial q=0$. Its Euler--Lagrange equation is
$$
\frac d{dt}\frac{\partial L}{\partial\dot q}
=\frac{\partial L}{\partial q}=0,
$$
which proves that $p_q$ is conserved.

Solved by gpt-5.6-sol high.

= c
{parent=8e}
{scope}

= Solution
{parent=c}

Let $q=(q_1,q_2,q_3)$ and $n=(1,1,1)$. Since
$$
q_1q_2+q_2q_3+q_3q_1
=\frac12\{(q\cdot n)^2-|q|^2\},
$$
the <Lagrangian> depends on $q$ only through $|q|^2$ and $(q\cdot n)^2$, and on the <velocity> through $|\dot q|^2$. Rotations about the $n$-axis preserve all three quantities, proving the invariance.

The <axially symmetric isotropic-mass Lagrangian> therefore conserves the axial <angular momentum>
$$
J_n=n\cdot(q\times\dot q)
=(q_2\dot q_3-q_3\dot q_2)
 +(q_3\dot q_1-q_1\dot q_3)
 +(q_1\dot q_2-q_2\dot q_1),
$$
up to an irrelevant normalization of $n$. Since $L$ has no explicit time dependence, <conservation of energy from time-translation invariance> gives the second conserved quantity
$$
E=\frac12|\dot q|^2+\frac12|q|^2
+\alpha(q_1q_2+q_2q_3+q_3q_1).
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=9d}
{scope}

= Solution
{parent=a}

Chemical equilibrium gives $\mu_D=\mu_p+\mu_n$. Substitution of the three Maxwell--Boltzmann densities therefore cancels the <chemical potentials> and leaves the rest-mass factor
$$
\exp\left(\frac{(m_p+m_n-m_D)c^2}{k_BT}\right)
=e^{B_D/(k_BT)}.
$$
Using $g_p=g_n=2$, $g_D=4$, $m_n\simeq m_p$, and $m_D\simeq2m_p$, the translational prefactor is
$$
\frac{4(2\pi m_Dk_BT/h^2)^{3/2}}
 {4(2\pi m_pk_BT/h^2)^{3/2}(2\pi m_nk_BT/h^2)^{3/2}}
\simeq\left(\frac{\pi m_pk_BT}{h^2}\right)^{-3/2}.
$$
Hence
$$
\boxed{\frac{n_D}{n_pn_n}\simeq
\left(\frac{\pi m_pk_BT}{h^2}\right)^{-3/2}
e^{B_D/(k_BT)}}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=9d}
{scope}

= Solution
{parent=b}

Because $X_i=n_i/n_B$,
$$
\frac{X_D}{X_pX_n}=n_B\frac{n_D}{n_pn_n}.
$$
Using $n_B=\eta n_\gamma$ and the given photon density in the <deuterium equilibrium abundance> formula gives
$$
\boxed{
\frac{X_D}{X_pX_n}
\simeq\frac{16\zeta(3)}{\sqrt\pi}\eta
\left(\frac{k_BT}{m_pc^2}\right)^{3/2}
e^{B_D/(k_BT)}}.
$$
Although the exponential favors deuterium once $k_BT<B_D$, the very small factor $\eta$ expresses the enormous excess of photons. The high-energy tail therefore photodissociates newly formed deuterium until the <temperature> is well below $B_D/k_B$; this is the <deuterium bottleneck>.

Solved by gpt-5.6-sol high.

= c
{parent=9d}
{scope}

= Solution
{parent=c}

A larger $\eta$ increases the deuterium fraction at fixed <temperature>, so the bottleneck ends at a higher <temperature> and earlier time. With weak decoupling unchanged, fewer free neutrons then beta-decay before nuclear reactions bind them. Since primordial helium production is neutron-limited, the <baryon-density effect on primordial helium> makes $Y_{\rm He}$ larger than the standard value.

Solved by gpt-5.6-sol high.

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

= a
{parent=10e}
{scope}

= Solution
{parent=a}

The powers are
$$
4^0\equiv1,
\qquad4^1\equiv4,
\qquad4^2\equiv7,
\qquad4^3\equiv1\pmod9.
$$
Thus the least positive period is $\boxed{r=3}$.

Solved by gpt-5.6-sol high.

= b
{parent=10e}
{scope}

= Solution
{parent=b}

The possible values are $y\in\{1,4,7\}$. Each occurs for four of the twelve equally weighted values of $x$, so
$$
\mathbb P(y=1)=\mathbb P(y=4)=\mathbb P(y=7)=\frac13.
$$

Solved by gpt-5.6-sol high.

= c
{parent=10e}
{scope}

= Solution
{parent=c}

The equality $f(x)=4$ holds exactly for $x\equiv1\pmod3$. Projecting onto the measured second-register value and normalizing gives
$$
\boxed{|\alpha\rangle
=\frac12\bigl(|1\rangle+|4\rangle+|7\rangle+|10\rangle\bigr)}.
$$

Solved by gpt-5.6-sol high.

= d
{parent=10e}
{scope}

= Solution
{parent=d}

With $\omega=e^{2\pi i/12}$, the <quantum Fourier transform> acts by
$$
\boxed{\operatorname{QFT}_{12}|x\rangle
=\frac1{\sqrt{12}}\sum_{c=0}^{11}\omega^{xc}|c\rangle}.
$$
(The opposite phase convention gives the same measurement probabilities.)

Solved by gpt-5.6-sol high.

= e
{parent=10e}
{scope}

= Solution
{parent=e}

Applying the transform to the coset state gives amplitude
$$
\frac1{2\sqrt{12}}\omega^c
\sum_{j=0}^3\omega^{3jc}
$$
for $|c\rangle$. The geometric sum vanishes unless $\omega^{3c}=1$, which means $c\equiv0\pmod4$. For $c=0,4,8$ its magnitude is $4/(2\sqrt{12})=1/\sqrt3$. Thus the <quantum Fourier transform of a periodic coset state> gives
$$
\boxed{c\in\{0,4,8\},\qquad \mathbb P(c)=\frac13}.
$$

Solved by gpt-5.6-sol high.

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

= Solution
{parent=11G}

Put $A_j=\left(\begin{smallmatrix}a_j&b_j\\1&0\end{smallmatrix}\right)$. The identity
$$
A_0\cdots A_n=
\begin{pmatrix}p_n&b_np_{n-1}\\q_n&b_nq_{n-1}\end{pmatrix}
$$
follows by induction: multiplying the formula for $n-1$ by $A_n$ gives the two stated recurrences in the first column and $b_n(p_{n-1},q_{n-1})^T$ in the second. The associated Möbius transformations $z\mapsto a_j+b_j/z$, applied from the right to $a_n$, show that the first-column ratio is the displayed <positive generalized continued fraction>.

Now set every $b_j=1$. Taking <determinants> in the <matrix> identity gives
$$
p_nq_{n-1}-q_np_{n-1}=(-1)^{n+1}.
$$
Consequently adjacent convergents differ by
$$
\frac{p_{n+1}}{q_{n+1}}-\frac{p_n}{q_n}
=\frac{(-1)^{n+2}}{q_nq_{n+1}}.
$$
The recurrence and $a_j\geq1$ imply $q_{n+1}\geq q_n+q_{n-1}$, so $q_n\to\infty$. The <determinant> signs show that the even convergents increase, the odd convergents decrease, and every even one is below every odd one. Their adjacent separation tends to zero, so both subsequences have a common <limit> $x$. Since $x$ lies between each adjacent pair, the <convergence of a positive simple continued fraction> gives exactly
$$
\left|\frac{p_n}{q_n}-x\right|
+\left|\frac{p_{n+1}}{q_{n+1}}-x\right|
=\frac1{q_nq_{n+1}}.
$$

If all $a_j=1$ as well, the recurrence and initial values give
$$
p_n=F_{n+2},\qquad q_n=F_{n+1}.
$$
The characteristic roots of $F_{n+2}=F_{n+1}+F_n$ are
$\phi=(1+\sqrt5)/2$ and $\psi=(1-\sqrt5)/2=-\phi^{-1}$, hence
$$
F_n=\frac{\phi^n-\psi^n}{\sqrt5}.
$$
Therefore $F_{n+2}/F_{n+1}\to\phi$. Moreover,
$$
F_{n+1}-\phi F_n=\psi^n,\qquad
F_{n+2}-\phi F_{n+1}=\psi^{n+1}.
$$
Using $F_n\phi^{-n}\to1/\sqrt5$ yields
$$
F_nF_{n+1}\left|\frac{F_{n+1}}{F_n}-\phi\right|
=F_{n+1}\phi^{-n}\longrightarrow\frac{\phi}{\sqrt5},
$$
and
$$
F_nF_{n+1}\left|\frac{F_{n+2}}{F_{n+1}}-\phi\right|
=F_n\phi^{-(n+1)}\longrightarrow\frac1{\phi\sqrt5}.
$$
This is the <all-one continued fraction and Fibonacci ratios> case.

Solved by gpt-5.6-sol high.

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

= a
{parent=12k}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

In the <Rabin cryptosystem>, the public key is $N=pq$, usually with distinct secret primes $p,q\equiv3\pmod4$. A message $m$ is encrypted as
$$
c\equiv m^2\pmod N.
$$
Knowing $p$ and $q$, the receiver finds the two square roots of $c$ modulo each prime and combines them by the Chinese remainder theorem, obtaining the four square roots modulo $N$; redundancy identifies the intended message.

Factoring $N$ plainly enables this decryption. Conversely, suppose an algorithm returns a square root of a chosen quadratic residue. Choose random invertible $x$, submit $x^2\bmod N$, and receive a root $y$. With probability at least $1/2$, $y\not\equiv\pm x\pmod N$; then
$$
\gcd(x-y,N)\quad\hbox{or}\quad\gcd(x+y,N)
$$
is a nontrivial factor. Thus decryption and factoring are equivalent up to a randomized polynomial-time reduction.

Solved by gpt-5.6-sol high.

= b
{parent=12k}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

For the <RSA cryptosystem>, choose $N=pq$ and a public exponent $e$ coprime to $\phi(N)$. The public key is $(N,e)$, while the private exponent satisfies
$$
ed\equiv1\pmod{\phi(N)}.
$$
Encryption sends $m$ to $c\equiv m^e\pmod N$, and decryption computes $c^d\equiv m\pmod N$.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

Let the intercepted ciphertexts be
$$
c_i\equiv m^{e_i},\qquad c_j\equiv m^{e_j}\pmod N.
$$
Since $\gcd(e_i,e_j)=1$, the extended Euclidean algorithm gives integers $a,b$ with $ae_i+be_j=1$. The <common-modulus RSA attack> recovers
$$
\boxed{m\equiv c_i^ac_j^b\pmod N}.
$$
Negative powers are evaluated using modular inverses. If an inverse does not exist, its greatest common divisor with $N$ already factors the <modulus>.

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

Yes. Your public and private exponents reveal $ed-1$, a nonzero multiple of $\lambda(N)$. The standard <RSA private exponent reveals the factorization> algorithm uses repeated squaring of random residues to obtain a nontrivial square root of one and hence factors $N$ with a greatest common divisor. Once $p$ and $q$ are known, compute $\phi(N)$ and invert every other customer's public exponent to obtain that customer's private exponent.

Solved by gpt-5.6-sol high.

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

= a
{parent=13d}
{scope}

= Solution
{parent=a}

An <elliptic function> is a meromorphic $f:\mathbb C\to\widehat{\mathbb C}$ with two real-linearly independent periods $\omega_1,\omega_2$. A fundamental cell may be taken as the half-open parallelogram
$$
z_0+\{s\omega_1+t\omega_2:0\leq s,t<1\},
$$
with $z_0$ chosen so that its boundary contains no zero or pole under discussion. The argument principle and cancellation on opposite sides show that a nonconstant elliptic <function> has the same number of zeros and poles in a cell, counting multiplicity.

Solved by gpt-5.6-sol high.

= b
{parent=13d}
{scope}

= Solution
{parent=b}

If an elliptic <function> has no poles, it is entire. It is bounded on the closure of a fundamental parallelogram, and periodicity transfers that bound to the whole plane. Liouville's theorem therefore makes it constant.

Solved by gpt-5.6-sol high.

= c
{parent=13d}
{scope}

= Solution
{parent=c}

Choose a fundamental parallelogram with no pole on its boundary. Opposite sides are translates traversed in opposite directions, so periodicity gives
$$
\oint_{\partial P}f(z)\,dz=0.
$$
The residue theorem then gives
$$
\sum_j\operatorname{Res}(f;z_j)=0.
$$
A lone simple pole has a nonzero residue by definition of its order, so an elliptic <function> cannot have exactly one simple pole in a fundamental cell.

Solved by gpt-5.6-sol high.

= d
{parent=13d}
{scope}

= Solution
{parent=d}

Fix $w\in\mathbb C$ and apply the stated clockwise argument principle to $h=f-w$. Its logarithmic <derivative> is periodic, so the <integrals> over opposite sides of a fundamental cell cancel. Hence $P-Z=0$. The poles of $f-w$ are precisely the poles of $f$, with unchanged multiplicities, while its zeros are the solutions of $f(z)=w$. Thus the <value multiplicity of an elliptic function> is the same fixed pole count for every finite $w$, counting multiplicity.

Solved by gpt-5.6-sol high.

= e
{parent=13d}
{scope}

= Solution
{parent=e}

At every lattice point $\omega=m\omega_1+n\omega_2$, the translated summand supplies a principal part
$$
\frac1{(z-\omega)^2}.
$$
All other terms are holomorphic nearby. Hence the <Weierstrass elliptic function> has a double pole at every lattice point, with zero residue, and no other singularities.

Solved by gpt-5.6-sol high.

= f
{parent=13d}
{scope}

= Solution
{parent=f}

For a nonzero lattice point $\omega$ and $|z|<|\omega|$,
$$
\frac1{(z-\omega)^2}-\frac1{\omega^2}
=\sum_{r=1}^{\infty}(r+1)\frac{z^r}{\omega^{r+2}}.
$$
Pairing $\omega$ with $-\omega$ cancels odd powers. There is no constant term, so $a_0=0$, and the <laurent coefficients of the Weierstrass elliptic function> are
$$
\boxed{a_{2k}=(2k+1)
\sum_{(m,n)\ne(0,0)}
\frac1{(m\omega_1+n\omega_2)^{2k+2}},
\qquad k\geq1.}
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=14e}
{scope}

= Solution
{parent=a}

Write $r_1=(x_1,y_1)$, $r_2=(x_2,y_2)$,
$\rho_1=|r_1|$, and $\rho_2=|r_2-r_1|$. Then
$$
L=\frac{m_1}2|\dot r_1|^2+\frac{m_2}2|\dot r_2|^2
-\frac{k_1}2(\rho_1-\ell)^2-\frac{k_2}2(\rho_2-\ell)^2
-m_1gy_1-m_2gy_2.
$$
With $e_y=(0,1)$, the Euler--Lagrange equations are
$$
m_1\ddot r_1
=-k_1\left(1-\frac{\ell}{\rho_1}\right)r_1
+k_2\left(1-\frac{\ell}{\rho_2}\right)(r_2-r_1)-m_1ge_y,
$$
$$
m_2\ddot r_2
=-k_2\left(1-\frac{\ell}{\rho_2}\right)(r_2-r_1)-m_2ge_y.
$$

Solved by gpt-5.6-sol high.

= b
{parent=14e}
{scope}

= Solution
{parent=b}

At the downward vertical equilibrium, the lower spring supports $m_2g$ and the upper spring supports $(m_1+m_2)g$. Thus the <two vertically suspended springs> have lengths
$$
L_2=\ell+\frac{m_2g}{k_2},\qquad
L_1=\ell+\frac{(m_1+m_2)g}{k_1},
$$
and positions
$$
\boxed{(x_1,y_1)=(0,-L_1),\qquad
(x_2,y_2)=(0,-L_1-L_2).}
$$

Solved by gpt-5.6-sol high.

= c
{parent=14e}
{scope}

= Solution
{parent=c}

For equal masses and springs,
$$
L_1=\ell+\frac{2mg}{k},\qquad L_2=\ell+\frac{mg}{k}.
$$
A small horizontal displacement $x_1$ changes each spring direction to first order. The equilibrium tensions $2mg$ and $mg$ give transverse restoring stiffness
$$
K_x=\frac{2mg}{L_1}+\frac{mg}{L_2}.
$$
Therefore
$$
\omega_1^2=\frac{K_x}{m}
=\frac{k}{m}\left(
2-\frac1{1+2mg/(k\ell)}
-\frac1{1+mg/(k\ell)}
\right),
$$
as required.

The other motions are vertical. If $\eta_1,\eta_2$ are downward displacements from equilibrium, their linearized equations are
$$
m\begin{pmatrix}\ddot\eta_1\\\ddot\eta_2\end{pmatrix}
=-k\begin{pmatrix}2&-1\\-1&1\end{pmatrix}
\begin{pmatrix}\eta_1\\\eta_2\end{pmatrix}.
$$
The <vertical normal modes of two equal suspended masses> are
$$
\boxed{\omega_\pm^2=\frac{k}{2m}(3\pm\sqrt5)},
\qquad
\boxed{(\eta_1,\eta_2)\propto
\left(1,\frac{1\mp\sqrt5}{2}\right)}.
$$
Neither frequency contains $g$: gravity only shifted the equilibrium lengths.

Solved by gpt-5.6-sol high.

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

= a
{parent=15e}
{scope}

= Solution
{parent=a}

The <eigenphase Hadamard-test probability> is
$$
p_0=\frac{1+\operatorname{Re}e^{2\pi i\theta}}2
=\frac{1+\cos(2\pi\theta)}2
=\cos^2(\pi\theta).
$$
Thus
$$
\cos(2\pi\theta)=2p_0-1,
$$
or $\theta=\pm(2\pi)^{-1}\arccos(2p_0-1)\pmod1$. This test alone cannot distinguish $\theta$ from $1-\theta$.

Solved by gpt-5.6-sol high.

= b
{parent=15e}
{scope}

= Solution
{parent=b}

The state entering the controlled swap has lower-register states
$|\phi_1\rangle$ and $V|\phi_2\rangle$. The <swap test> therefore gives
$$
\boxed{p_0=\frac12\left(1+
|\langle\phi_1|V|\phi_2\rangle|^2\right)}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=15e}
{scope}

= i
{parent=c}
{scope}

= Solution
{parent=i}

Writing $N=2^m$, the Hadamards first give
$$
(H^{\otimes m}\otimes I)|0\rangle^{\otimes m}|\psi\rangle
=\frac1{\sqrt N}\sum_{k=0}^{N-1}|k\rangle|\psi\rangle.
$$
Since $U^k|\psi\rangle=e^{2\pi ik\theta}|\psi\rangle$ and $\theta=j/N$,
$$
\boxed{|\Phi\rangle=
\frac1{\sqrt N}\sum_{k=0}^{N-1}
e^{2\pi ijk/N}|k\rangle|\psi\rangle.}
$$

Solved by gpt-5.6-sol high.

= ii
{parent=c}
{scope}

= Solution
{parent=ii}

The first register is
$$
\boxed{|\phi_j\rangle=\frac1{\sqrt N}
\sum_{k=0}^{N-1}e^{2\pi ijk/N}|k\rangle.}
$$
For $j,j'$,
$$
\langle\phi_{j'}|\phi_j\rangle
=\frac1N\sum_{k=0}^{N-1}e^{2\pi i(j-j')k/N}.
$$
The geometric sum is one when $j=j'$ and zero otherwise. There are $N$ such orthonormal <vectors> in an $N$-dimensional space, so they form an orthonormal <basis>.

Solved by gpt-5.6-sol high.

= iii
{parent=c}
{scope}

= Solution
{parent=iii}

The operator $F$ maps the computational orthonormal <basis> $\{|j\rangle\}$ to the orthonormal <basis> $\{|\phi_j\rangle\}$. It therefore preserves <inner products> and is surjective, so $F^\dagger F=FF^\dagger=I$ and $F$ is unitary.

Solved by gpt-5.6-sol high.

= iv
{parent=c}
{scope}

= Solution
{parent=iv}

First apply $F^\dagger$, which sends $|\phi_j\rangle$ to $|j\rangle$. Then measure in the computational <basis>. Because the phase has an exact $m$-bit representation, <exact quantum phase estimation> returns
$$
\boxed{p_j=1}.
$$

Solved by gpt-5.6-sol high.

= d
{parent=15e}
{scope}

= Solution
{parent=d}

The overlap is
$$
|\langle0|-\rangle|=\frac1{\sqrt2}.
$$
The <Helstrom-Holevo bound> for two equiprobable pure states gives
$$
p_{\rm succ}^{\max}
=\frac12\left(1+\sqrt{1-|\langle0|-\rangle|^2}\right)
=\boxed{\frac12\left(1+\frac1{\sqrt2}\right)}
=\cos^2\frac\pi8.
$$
The <Breidbart basis> is the eigenbasis of the difference of the two weighted state projectors, so its decision rule is precisely the Helstrom measurement and attains this bound.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=16I}

The <Hartogs theorem> says that for every set $X$ there is an ordinal that does not inject into $X$. Indeed, the well-orderings of subsets of $X$ form a set, since each is a relation in $\mathcal P(X\times X)$. By replacement, their order types form a set of ordinals. Its supremum plus one cannot be the order type of a well-ordered subset of $X$, and hence cannot inject into $X$. The least such ordinal is the Hartogs ordinal $h(X)$.

<Ordinal exponentiation> is defined by
$$
\alpha^0=1,\qquad
\alpha^{\beta+1}=\alpha^\beta\alpha,\qquad
\alpha^\lambda=\sup_{\xi<\lambda}\alpha^\xi
$$
for nonzero <limit> $\lambda$. Fix $\alpha,\beta$ and induct on $\gamma$. The zero case is immediate. If the identity holds at $\gamma$, then
$$
\alpha^{\beta+(\gamma+1)}
=\alpha^{(\beta+\gamma)+1}
=\alpha^{\beta+\gamma}\alpha
=\alpha^\beta\alpha^{\gamma+1}.
$$
At a <limit> $\lambda$, continuity of exponentiation and right continuity of ordinal multiplication give
$$
\alpha^{\beta+\lambda}
=\sup_{\gamma<\lambda}\alpha^{\beta+\gamma}
=\sup_{\gamma<\lambda}\alpha^\beta\alpha^\gamma
=\alpha^\beta\alpha^\lambda.
$$

Put $\rho=\omega^\alpha$. The set
$$
B=\{\eta:\rho\eta\leq\gamma\}
$$
is nonempty and bounded, so let $\beta=\sup B$. Continuity of multiplication in its right argument gives $\rho\beta\leq\gamma$. There is a unique tail $\delta$ with $\gamma=\rho\beta+\delta$. If $\delta\geq\rho$, then
$\rho(\beta+1)=\rho\beta+\rho\leq\gamma$, contrary to maximality. Hence $\delta<\rho$. If
$\rho\beta+\delta=\rho\beta'+\delta'$ with both remainders below $\rho$ and $\beta<\beta'$, then
$$
\rho\beta+\delta<\rho(\beta+1)\leq\rho\beta',
$$
a contradiction. This proves existence and uniqueness in the <division by an additively indecomposable ordinal>.

Now let $X$ be nonempty and well ordered. Its least element is not a <limit> point, so $X'\ne X$. Every <derivative> is itself a well-ordered subset. If $X^{(\alpha)}$ were nonempty for every $\alpha<h(X)$, then each difference
$$
X^{(\alpha)}\setminus X^{(\alpha+1)}
$$
would be nonempty. Choosing its least member would inject $h(X)$ into $X$, since the nested successive differences are disjoint. This contradicts Hartogs' lemma. Thus some <derivative> is empty, as asserted by the <derived-set iteration of a well-order>.

For an ordinal $\xi$, a point $\gamma<\xi$ is a <limit> point exactly when it is a nonzero <limit> ordinal. Division by $\omega$ writes $\gamma=\omega\beta+n$ with $n<\omega$; this is a nonzero <limit> exactly when $n=0$ and $\beta>0$. Hence
$$
\boxed{\xi'=\{\omega\beta<\xi:\beta>0\}}.
$$
Within this derived set, $\omega\beta$ is a <limit> point exactly when $\beta$ is a nonzero <limit> ordinal, equivalently $\beta=\omega\eta$ for some $\eta>0$. Therefore the <derived sets of an ordinal> satisfy
$$
\boxed{\xi''=\{\omega^2\eta<\xi:\eta>0\}}.
$$
Finally $\omega'=\varnothing$, so the index of $\omega$ is $1$, while
$(\omega^2)'=\{\omega,2\omega,3\omega,\ldots\}$ and
$(\omega^2)''=\varnothing$, so the index of $\omega^2$ is $2$.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=17I}

The <Turan theorem> states that every $n$-vertex graph with no $K_{r+1}$ has at most $e(T_r(n))$ edges, where $T_r(n)$ is the complete $r$-partite graph with part sizes differing by at most one.

Here is an induction on $n$ and $r$. If a $K_{r+1}$-free graph $G$ contains no $K_r$, induction on $r$ gives
$$
e(G)\leq e(T_{r-1}(n))\leq e(T_r(n)).
$$
Otherwise choose a copy $S$ of $K_r$. Every vertex outside $S$ has at most $r-1$ neighbors in $S$, and induction on $n$ gives
$$
e(G)\leq e(T_r(n-r))+\binom r2+(r-1)(n-r)=e(T_r(n)).
$$
The last identity is obtained by removing one vertex from every part of $T_r(n)$. This proves the theorem; the standard equality analysis forces the balanced complete $r$-partite graph.

Now suppose $G$ is rhombus-free. If it is triangle-free, the $r=2$ case just proved gives
$e(G)\leq e(T_2(n))$. Otherwise remove the three vertices of a triangle. No remaining vertex can be adjacent to two vertices of that triangle, since those two triangle vertices and the outside vertex would form a second triangle sharing an edge with the first. Induction therefore gives
$$
e(G)\leq
\left\lfloor\frac{(n-3)^2}{4}\right\rfloor+3+(n-3)
\leq\left\lfloor\frac{n^2}{4}\right\rfloor
=e(T_2(n)).
$$
This proves the <rhombus-free edge bound> and hence the requested strict contrapositive.

For equality at $n=6$, take the <triangular prism graph>: two disjoint triangles joined by a matching. It has $6+3=9=e(T_2(6))$ edges. Every edge belongs to at most one triangle, so there is no rhombus, while the presence of triangles proves that it is not isomorphic to $T_2(6)=K_{3,3}$.

Solved by gpt-5.6-sol high.

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

= a
{parent=18h}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

The extension $L/K$ is finite when $L$ is finite-dimensional as a <vector space> over $K$, and its degree is $[L:K]=\dim_KL$.

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

The extension $L/K$ is separable when it is algebraic and the minimal <polynomial> over $K$ of every element of $L$ has no repeated root in an algebraic closure.

Solved by gpt-5.6-sol high.

= iii
{parent=a}
{scope}

= Solution
{parent=iii}

The extension $L/K$ is simple when there is an element $\alpha\in L$ such that $L=K(\alpha)$.

Solved by gpt-5.6-sol high.

= Solution
{parent=a}

Finite and separable imply simple by the primitive element theorem. Separable and simple imply finite: if $L=K(\alpha)$ and the extension is separable, then $\alpha$ is algebraic, so $[L:K]=\deg m_{\alpha,K}<\infty$.

Finite and simple do not imply separable. In characteristic $p$, take
$$
K=\mathbb F_p(t),\qquad L=K(t^{1/p}).
$$
Then $L/K$ has degree $p$ and is simple, but the generator has inseparable minimal <polynomial> $X^p-t$. These are the <finite separable simple extension implications>.

Solved by gpt-5.6-sol high.

= b
{parent=18h}
{scope}

= Solution
{parent=b}

The rational-root test shows that $f=X^3-X-1$ is irreducible. Its discriminant is
$$
\Delta=-4(-1)^3-27(-1)^2=-23,
$$
which is not a square in $\mathbb Q$. The <Galois group of an irreducible cubic> therefore gives
$$
\operatorname{Gal}(L/\mathbb Q)\cong S_3.
$$

Quadratic subfields correspond to order-three <subgroups>. The unique such <subgroup> is $A_3$, and its fixed field is generated by the alternating product of root differences, whose square is $\Delta$. Thus the unique quadratic subfield is
$$
\boxed{\mathbb Q(\sqrt{-23})}.
$$

Let the roots be $\alpha,\beta,\gamma$ and choose
$\delta=(\alpha-\beta)(\alpha-\gamma)(\beta-\gamma)$, so $\delta^2=-23$. Put $\theta=\alpha+\delta$. A nonidentity even permutation moves $\alpha$ and fixes $\delta$, so it cannot fix $\theta$. An odd permutation negates $\delta$. If it fixes $\alpha$, it plainly does not fix $\theta$. If it sends $\alpha$ to $\beta$ and fixed $\theta$, then $\beta-\alpha=2\delta$. Cancelling the factor $\beta-\alpha$ in the expression for $\delta$ would give
$$
1=\mathord\pm2(\alpha-\gamma)(\beta-\gamma)
=\mathord\pm2f'(\gamma).
$$
But $f'(\gamma)=3\gamma^2-1$, so this would make $\gamma^2$ equal to $1/6$ or $1/2$, contradicting the irreducibility of its cubic minimal <polynomial>. The other transpositions are identical. Hence $\theta$ has trivial stabilizer, so
$$
[\mathbb Q(\theta):\mathbb Q]=6=[L:\mathbb Q],
\qquad
\boxed{L=\mathbb Q(\alpha+\sqrt{-23})}.
$$

Solved by gpt-5.6-sol high.

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

= Solution
{parent=19H}

The character of $(\rho,V)$ is
$$
\chi_V(g)=\operatorname{tr}\rho(g).
$$
The stated action is a representation because successive action by $(g_2,h_2)$ and $(g_1,h_1)$ gives
$$
\sigma(h_1h_2)\,\alpha\,\rho(g_2^{-1}g_1^{-1})
=\sigma(h_1h_2)\,\alpha\,\rho((g_1g_2)^{-1}).
$$

Identifying $\operatorname{Hom}_{\mathbb C}(V,W)$ with $V^*\otimes W$ gives the <character of a Hom representation>
$$
\boxed{\chi_{\operatorname{Hom}(V,W)}(g,h)
=\chi_V(g^{-1})\chi_W(h)
=\overline{\chi_V(g)}\,\chi_W(h)}.
$$
Here the final equality follows after unitarizing the finite-group representation.

The permutation character of $\mathbb CG$ counts fixed points:
$$
\chi_{\mathbb CG}(g,h)
=\#\{x\in G:gxh^{-1}=x\}
=\#\{x:x^{-1}gx=h\}.
$$
Thus
$$
\chi_{\mathbb CG}(g,h)=
\begin{cases}
|C_G(g)|,&g\text{ and }h\text{ are conjugate},\\
0,&\text{otherwise}.
\end{cases}
$$

On the other hand, the character of
$\bigoplus_i\operatorname{Hom}(V_i,V_i)$ is
$$
\sum_i\overline{\chi_i(g)}\chi_i(h).
$$
The column form of <character orthogonality> says that this has exactly the same two values above. Complex representations of a finite <group> are semisimple, and semisimple representations with the same character are isomorphic. Hence the <two-sided regular representation decomposition> is
$$
\boxed{\mathbb CG\cong
\bigoplus_{i=1}^r\operatorname{Hom}_{\mathbb C}(V_i,V_i)}
$$
as a $G\times G$ representation.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=20F}

The <Dirichlet unit theorem> states that for a number field of signature $(r_1,r_2)$,
$$
\mathcal O_K^\times\cong\mu(K)\times\mathbb Z^{r_1+r_2-1}.
$$

For $K=\mathbb Q(\sqrt5)$,
$$
\mathcal O_K=\mathbb Z[\phi],
\qquad \phi=\frac{1+\sqrt5}{2},
\qquad N(\phi)=-1.
$$
The field is real quadratic, so the unit rank is one and its only roots of unity are $\pm1$. To see that $\phi$ is fundamental, suppose a positive unit satisfies $1<u<\phi$. If its norm is $1$, its <integral> trace $u+u^{-1}$ lies strictly between $2$ and $\sqrt5$; if its norm is $-1$, its trace $u-u^{-1}$ lies strictly between $0$ and $1$. Neither interval contains an integer. Reducing any positive unit by a suitable power of $\phi$ now proves the <units of the quadratic field Q square root of five> formula
$$
\boxed{\mathcal O_K^\times=\{\mathord\pm\phi^n:n\in\mathbb Z\}}.
$$

If $\mathcal O_L^\times/\mathcal O_K^\times$ is finite, the two unit <groups> have the same rank, namely one. Put $d=[L:K]$ and let $(r_1,r_2)$ be the signature of $L$. Then
$$
r_1+r_2-1=1,
\qquad
r_1+2r_2=[L:\mathbb Q]=2d.
$$
Therefore $r_1=4-2d$ and $r_2=2d-2$. For a proper extension, nonnegativity forces
$$
\boxed{d=2},
$$
with signature $(0,2)$.

This degree occurs: take $L=K(i)$. It is a totally imaginary quadratic extension of $K$, so the unit ranks agree and the quotient is finite. It is nontrivial because $i\in\mathcal O_L^\times$ but $i\notin\mathcal O_K^\times$ (and its coset has order two). This is the <finite relative unit quotient over a real quadratic field> example.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=21J}

The <Seifert-van Kampen theorem> says that if $X=U\cup V$, where $U,V,U\cap V$ are path connected <open sets> containing $x_0$, then
$$
\pi_1(X,x_0)\cong
\bigl(\pi_1(U,x_0)*\pi_1(V,x_0)\bigr)\big/
\langle\!\langle i_{U*}(\gamma)i_{V*}(\gamma)^{-1}:
\gamma\in\pi_1(U\cap V,x_0)\rangle\!\rangle.
$$

The cell attachment is the quotient
$$
X\cup_fD^n=(X\sqcup D^n)/(u\sim f(u)\text{ for }u\in\partial D^n),
$$
with the image of $x_0=f(*)$ as base point. For $n=2$, take one <open set> that deformation retracts onto $X$ together with a boundary collar, and another that consists of the interior of the disc with a collar and is contractible. Their intersection deformation retracts onto $S^1$. Its generator maps to $[f]$ in the first set and to the identity in the second. Van Kampen therefore proves the <fundamental group after attaching a 2-cell> formula
$$
\boxed{\pi_1(X\cup_fD^2,x_0)
\cong\pi_1(X,x_0)/\langle\!\langle[f]\rangle\!\rangle}.
$$

Applying van Kampen to the two circles gives
$$
\pi_1(S^1\vee S^1,*)\cong F(a,b).
$$
Attach three discs along loops representing $a^2$, $b^3$, and $(ab)^2$. The resulting <presentation complex for the symmetric group on three letters> has <group>
$$
\langle a,b\mid a^2=b^3=(ab)^2=1\rangle.
$$
Sending $a$ to $(12)$ and $b$ to $(123)$ gives a surjection onto $S_3$. The relations imply $aba=b^{-1}$, so every word reduces to one of
$$
1,b,b^2,a,ab,ab^2.
$$
The presented <group> has at most six elements and surjects onto the six-element <group> $S_3$, so this map is an isomorphism.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=22G}

The <Baire category theorem> states that a complete metric space is not a countable union of closed sets with empty interior. To prove it, let $G_1,G_2,\ldots$ be open dense sets and start with any nonempty <open set> $V$. Inductively choose closed balls
$$
\overline B(x_n,r_n)\subset
B(x_{n-1},r_{n-1})\cap G_n,
\qquad 0<r_n<2^{-n},
$$
with the first ball inside $V\cap G_1$. The centres are Cauchy. Completeness supplies a <limit> lying in every ball, hence in $V\cap\bigcap_nG_n$. Thus the intersection is dense, which is the equivalent form of the theorem.

Now each $nS$ is closed and $\bigcup_{n\geq1}nS=X$. Baire gives
$$
B(x,r)\subset nS
$$
for some $x,r,n$. Symmetry gives $B(-x,r)\subset nS$. If $\|z\|<r$, then $x+z$ and $-x+z$ lie in these two balls; convexity makes their midpoint $z$ belong to $nS$. Hence
$$
B(0,r/n)\subset S,
$$
which proves the <closed convex absorbing set has an origin neighbourhood> result.

Convexity cannot be dropped. In $X=\mathbb R$, set
$$
S=\{0\}\cup\bigcup_{k\geq0}
\{x:4^{-k}\leq|x|\leq2\cdot4^{-k}\}.
$$
This is closed and symmetric but has gaps arbitrarily close to zero. For any $x\ne0$, choose $k$ so that the interval
$$
\left[\frac{|x|}{2\cdot4^{-k}},\frac{|x|}{4^{-k}}\right]
$$
has length at least one, and choose an integer $n$ in it. Then $x/n\in S$, so $\bigcup_n nS=\mathbb R$, while $S$ is not a neighbourhood of zero. This is a <closed symmetric absorbing set without an origin neighbourhood>.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=23G}

The <Hahn-Banach theorem> says that if $Y$ is a linear subspace of a real normed space $X$ and $g:Y\to\mathbb R$ is bounded and linear, then there is a bounded linear extension $G:X\to\mathbb R$ with $\|G\|=\|g\|$.

For $x\in X$, the map $\widehat x(f)=f(x)$ is linear on $X'$ and
$$
|\widehat x(f)|\leq\|f\|\,\|x\|,
$$
so $\widehat x\in X''$ and $\|\widehat x\|\leq\|x\|$. The assignment $Jx=\widehat x$ is visibly linear. If $x\ne0$, define $g$ on $\operatorname{span}\{x\}$ by
$$
g(\lambda x)=\lambda\|x\|.
$$
It has norm one, and Hahn--Banach extends it to an $f\in X'$ of norm one with $f(x)=\|x\|$. Therefore
$$
\|\widehat x\|\geq|\widehat x(f)|=\|x\|.
$$
Thus the <canonical embedding into the bidual> is an isometry and
$$
\boxed{\|x\|=\sup_{\|f\|\leq1}|f(x)|}.
$$

On $C([0,1])$, point evaluation $\ell(f)=f(0)$ has norm one for the $L^\infty$ norm, because a <continuous function>'s essential supremum equals its supremum. Hahn--Banach extends it to $\Lambda\in(L^\infty)'$. If $\Lambda$ were represented by some $g\in L^1$, then
$$
f(0)=\int_0^1f(t)g(t)\,dt
$$
for every continuous $f$. Choose continuous $0\leq f_n\leq1$ with $f_n(0)=1$ and support in $[0,1/n]$. Absolute continuity of the <integral> makes the right side tend to zero, while the left side is always one, a contradiction. This <singular functional on L infinity> proves
$$
(L^\infty)'\ne L^1(\mu).
$$

Solved by gpt-5.6-sol high.

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

= Solution
{parent=24H}

The <valency theorem> states that for a nonconstant analytic map between compact connected Riemann surfaces,
$$
\sum_{z\in f^{-1}(w)}m_f(z)=\deg f
$$
for every target point $w$.

For a rational map $f=p/q$ on the Riemann sphere, first cancel common factors; then the <degree of a rational map of the Riemann sphere> is
$$
\deg f=\max(\deg p,\deg q).
$$
The analytic isomorphisms are exactly the degree-one maps, namely the <Möbius transformations>.

The required transformation is
$$
\boxed{h(z)=\frac{z+1}{z-1}}.
$$
It sends $\infty\leftrightarrow1$ and $0\leftrightarrow-1$. Its fixed points satisfy
$$
\boxed{z^2-2z-1=0}.
$$

The <octahedral rotation orbits on the Riemann sphere> have possible sizes
$$
\boxed{6,\ 8,\ 12,\ 24},
$$
corresponding respectively to vertices, face centres, edge centres, and generic points.

In the displayed $F$, no numerator factor vanishes at $0$ or at a fourth root of one. Hence $0,\pm1,\pm i$ are poles of order four. The numerator and denominator have degrees $24$ and $20$, so $F(z)\sim z^4$ at infinity; infinity is also a pole of order four. Thus
$$
F^{-1}(\infty)=\{0,\infty,\pm1,\pm i\},
\qquad m_F=4\text{ at each point},
$$
and
$$
\boxed{\deg F=24}.
$$

Finally let $z$ have stabilizer of order $e$. A local coordinate turns its stabilizer action into rotation by $e$th roots of unity. Since $F$ is invariant, the first nonconstant term of its local expansion has exponent divisible by $e$, so $m_F(z)\geq e$. The orbit has $24/e$ points and therefore contributes at least $24$ to the fibre through $F(z)$. The valency theorem and $\deg F=24$ leave no room for any further point. The <degree-sized invariant separates finite-group orbits>, so $F(z)=F(w)$ implies that $z$ and $w$ lie in the same orbit.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=25F}

For $X\subseteq\mathbb A^n$ and $P\in X$, the <Zariski tangent space> is
$$
T_PX=\{v\in k^n:df_P(v)=0\text{ for every }f\in I(X)\}.
$$
The dimension can be defined as $\min_{P\in X}\dim T_PX$. Equivalently, the <krull dimension of an affine variety> is the supremum of lengths of strict chains of irreducible closed subsets, or the Krull dimension of $k[X]$. A point is singular when $\dim T_PX>\dim X$.

For
$$
X=Z(x_1^2-x_2^2,x_3^2-x_4^2),
$$
the differentials give the <tangent spaces of a product of two nodal line pairs>
$$
\boxed{T_PX=\{v:x_1v_1-x_2v_2=0,\quad
x_3v_3-x_4v_4=0\}}.
$$
The variety is a union of four two-dimensional linear spaces. If both pairs $(x_1,x_2)$ and $(x_3,x_4)$ are nonzero, the two displayed equations are independent and $\dim T_PX=2$. If exactly one pair is zero, the dimension is three; at the origin it is four. Thus the singular locus is the union of the loci where either coordinate pair vanishes.

For $Y$, rank at most one is equivalent to vanishing of all $2\times2$ minors:
$$
y_0y_3-y_1y_2=0,\qquad
y_0y_4-y_2^2=0,\qquad
y_1y_4-y_2y_3=0.
$$
These homogeneous equations prove that $Y$ is projectively Zariski closed.

The following affine charts show both its dimension and smoothness. On $y_0=1$,
$$
y_3=y_1y_2,\qquad y_4=y_2^2,
$$
so $(y_1,y_2)$ are free. On $y_1=1$,
$$
y_2=y_0y_3,\qquad y_4=y_0y_3^2,
$$
so $(y_0,y_3)$ are free. On $y_3=1$,
$$
y_2=y_1y_4,\qquad y_0=y_1^2y_4,
$$
and on $y_4=1$,
$$
y_1=y_2y_3,\qquad y_0=y_2^2.
$$
These four charts cover $Y$, since the point with only $y_2$ nonzero violates the second minor. Every chart is isomorphic to $\mathbb A^2$. Hence the <smooth projective surface from overlapping rank-one coordinates> satisfies
$$
\boxed{\dim Y=2}
$$
and is nonsingular everywhere.

Solved by gpt-5.6-sol high.

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

= a
{parent=26j}
{scope}

= Solution
{parent=a}

For a unit-speed space curve, set
$$
t=\dot\alpha,\qquad
\kappa=|\dot t|,\qquad
n=\frac{\dot t}{\kappa},\qquad b=t\times n.
$$
These form the Frenet trihedron. With the signed torsion convention compatible with the question, the <Frenet-Serret formulas> are
$$
\dot t=\kappa n,\qquad
\dot n=-\kappa t-\tau b,\qquad
\dot b=\tau n.
$$
For an oriented surface with <unit normal> $N$, the signed <geodesic curvature> is
$$
\kappa_g=\langle\dot t,N\times t\rangle.
$$

On the unit sphere $N=\alpha$. Since $\alpha\cdot t=0$, <differentiation> gives
$$
1+\kappa\,\alpha\cdot n=0.
$$
Write $\alpha=An+Bb$. Then $A=-1/\kappa$, and differentiating this expression and comparing the $n$ coefficient with $\dot\alpha=t$ gives
$$
\dot A+\tau B=0,\qquad
B=-\frac{\dot\kappa}{\tau\kappa^2}.
$$
Thus, wherever the displayed quantities are defined,
$$
\boxed{\alpha=-\kappa^{-1}n-\tau^{-1}\kappa^{-2}\dot\kappa\,b}.
$$
Finally,
$$
\kappa_g
=\kappa n\cdot(\alpha\times t)
=-\frac{\dot\kappa}{\tau\kappa},
$$
which is the stated <curvature decomposition for a spherical curve>.

Solved by gpt-5.6-sol high.

= b
{parent=26j}
{scope}

= Solution
{parent=b}

For a positively oriented simple closed curve bounding a disc $D$ on the unit sphere, the local <Gauss-Bonnet theorem> gives
$$
\int_\alpha\kappa_g\,ds+\int_DK\,dA=2\pi.
$$
Here $K=1$, so
$$
\boxed{\int_0^L\kappa_g\,ds+A=2\pi}.
$$
At fixed length $L$, maximizing the enclosed area $A$ is therefore exactly the same as minimizing $\int_0^L\kappa_g\,ds$.

Solved by gpt-5.6-sol high.

= c
{parent=26j}
{scope}

= Solution
{parent=c}

From part (a),
$$
\alpha=-\frac1\kappa n+\frac{\kappa_g}{\kappa}b.
$$
As $\alpha$ is orthogonal to $t$,
$$
\begin{aligned}
\alpha\cdot\beta
&=\left(-\frac1\kappa n+\frac{\kappa_g}{\kappa}b\right)
\cdot\left(-\varphi t-\frac{\dot\varphi}{\kappa}n
-\frac{\dot\varphi}{\kappa\kappa_g}b\right)\\
&=\frac{\dot\varphi}{\kappa^2}
-\frac{\dot\varphi}{\kappa^2}=0.
\end{aligned}
$$
To compute $\dot\beta\cdot\dot\alpha=\dot\beta\cdot t$, only its $t$ component matters. Differentiating $-\varphi t$ contributes $-\dot\varphi t$, while differentiating $-\kappa^{-1}\dot\varphi n$ contributes $+\dot\varphi t$ through $\dot n=-\kappa t-\tau b$. Every other term is normal to $t$. Hence
$$
\boxed{\beta\perp\alpha,\qquad\dot\beta\perp\dot\alpha}.
$$

Solved by gpt-5.6-sol high.

= d
{parent=26j}
{scope}

= Solution
{parent=d}

At an area critical point the assumed first-variation identity holds for every smooth $\varphi$ compactly supported where $\kappa_g\ne0$:
$$
0=\int_0^L\kappa_g^{-1}\dot\varphi\,ds
=-\int_0^L(\kappa_g^{-1})'\varphi\,ds.
$$
The <fundamental lemma of the calculus of variations> gives
$(\kappa_g^{-1})'=0$ on each component of $\{\kappa_g\ne0\}$. Thus $\kappa_g$ is constant there. A nonzero constant component cannot have a boundary point where continuity makes $\kappa_g=0$. Since the closed curve is connected, either $\kappa_g\equiv0$ or it is one nonzero constant everywhere. Therefore every area-maximizing curve has constant geodesic curvature.

Solved by gpt-5.6-sol high.

= e
{parent=26j}
{scope}

= Solution
{parent=e}

For a unit-speed spherical curve,
$$
\dot t=-\alpha+\kappa_g(\alpha\times t):
$$
the normal component is $-\alpha$ because
$\alpha\cdot\dot t=-1$, and the tangential normal component is the definition of $\kappa_g$. If $\kappa_g$ is constant, then
$$
\begin{aligned}
\dot e
&=\frac d{ds}(\alpha\times t+\kappa_g\alpha)\\
&=\alpha\times\dot t+\kappa_g t\\
&=\kappa_g\alpha\times(\alpha\times t)+\kappa_g t=0.
\end{aligned}
$$
Moreover $e\cdot\alpha=\kappa_g$. Hence the whole curve lies in the fixed plane
$$
\boxed{e\cdot x=\kappa_g},
$$
so it is planar.

Solved by gpt-5.6-sol high.

= f
{parent=26j}
{scope}

= Solution
{parent=f}

The preceding parts show that an area maximizer among curves of fixed length is a plane section of the sphere, hence a circle. If its smaller cap has angular radius $0\leq\theta\leq\pi/2$, then
$$
L=2\pi\sin\theta,\qquad
A=2\pi(1-\cos\theta).
$$
For this circle,
$$
A(4\pi-A)
=4\pi^2(1-\cos^2\theta)
=4\pi^2\sin^2\theta=L^2.
$$
No curve of the same length encloses more area than the maximizing circle. Equivalently every curve enclosing the smaller area $A$ satisfies the <spherical isoperimetric inequality>
$$
\boxed{L^2\geq A(4\pi-A)},
$$
with equality precisely for circles.

Solved by gpt-5.6-sol high.

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

= a
{parent=27g}
{scope}

= Solution
{parent=a}

The <Borel-Cantelli lemmas> are:

* If $\sum_n\mathbb P(A_n)<\infty$, then
  $\mathbb P(A_n\ \text{infinitely often})=0$.
* If the $A_n$ are independent and
  $\sum_n\mathbb P(A_n)=\infty$, then
  $\mathbb P(A_n\ \text{infinitely often})=1$.

Solved by gpt-5.6-sol high.

= b
{parent=27g}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

Fix $\varepsilon>0$. Since a rate-one exponential variable has
$\mathbb P(X_n>x)=e^{-x}$,
$$
\sum_n\mathbb P\{X_n>(1+\varepsilon)\log n\}
=\sum_n n^{-(1+\varepsilon)}<\infty.
$$
The first Borel--Cantelli lemma gives the upper bound
$\limsup X_n/\log n\leq1$. Conversely,
$$
\sum_n\mathbb P\{X_n>(1-\varepsilon)\log n\}
=\sum_n n^{-(1-\varepsilon)}=\infty.
$$
These events are independent, so the second lemma makes them occur infinitely often. Hence the lower bound is $1-\varepsilon$. Letting rational $\varepsilon\downarrow0$ proves
$$
\boxed{\limsup_{n\to\infty}\frac{X_n}{\log n}=1}
\quad\hbox{almost surely}.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

For $\varepsilon>0$,
$$
\begin{aligned}
\mathbb P\{M_n\leq(1-\varepsilon)\log n\}
&=\left(1-n^{-(1-\varepsilon)}\right)^n\\
&\leq\exp(-n^\varepsilon),
\end{aligned}
$$
using $1-x\leq e^{-x}$. The final probabilities are summable, so Borel--Cantelli implies that eventually
$$
M_n>(1-\varepsilon)\log n.
$$
Intersecting over positive rational $\varepsilon$ proves the second <extremes of independent exponential variables> assertion
$$
\boxed{\liminf_{n\to\infty}\frac{M_n}{\log n}\geq1}
\quad\hbox{almost surely}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=27g}
{scope}

= Solution
{parent=c}

First assume the stated $\varepsilon$--$\delta$ condition. If $\nu(A)=0$, then $\nu(A)<\delta$ for the $\delta$ belonging to every $\varepsilon>0$, so $\mu(A)<\varepsilon$ for every $\varepsilon$ and $\mu(A)=0$. Hence $\mu\ll\nu$.

Conversely, suppose $\mu\ll\nu$ but the uniform condition fails. Then for some $\varepsilon>0$ there are sets $A_n$ with
$$
\nu(A_n)<2^{-n},\qquad \mu(A_n)\geq\varepsilon.
$$
Put $B_m=\bigcup_{n\geq m}A_n$. Then $\nu(B_m)\leq2^{1-m}$ and $B_m\downarrow B=\limsup A_n$, so $\nu(B)=0$. Absolute continuity gives $\mu(B)=0$. But $\mu(B_m)\geq\varepsilon$, and the finiteness of $\mu(\Omega)$ permits continuity from above:
$$
\mu(B)=\lim_m\mu(B_m)\geq\varepsilon,
$$
a contradiction. This proves <uniform absolute continuity for a finite measure>.

Solved by gpt-5.6-sol high.

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

= a
{parent=28k}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

Let $E_n$ be the holding time in state $n$. The variables are independent with $E_n\sim\operatorname{Exp}(n\lambda)$, and the explosion time would be
$$
T_\infty=\sum_{n\geq1}E_n.
$$
For $s>0$,
$$
\mathbb E e^{-sT_\infty}
=\prod_{n\geq1}\frac{n\lambda}{n\lambda+s}.
$$
This product is zero because
$\sum_n\log(1+s/(n\lambda))=\infty$. If $T_\infty$ were finite with positive probability, the nonnegative variable $e^{-sT_\infty}$ would have positive expectation. Therefore $T_\infty=\infty$ almost surely, and the <Yule process> is nonexplosive.

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

For ordered birth times $0<s_1<\cdots<s_n<t$, with $s_0=0$, the joint density of births at those times followed by no further birth before $t$ is
$$
\left[\prod_{k=1}^n k\lambda
e^{-k\lambda(s_k-s_{k-1})}\right]
e^{-(n+1)\lambda(t-s_n)}
=n!\lambda^n e^{\lambda\sum_ks_k-(n+1)\lambda t}.
$$
Also
$$
\mathbb P(X_t=n+1)
=e^{-\lambda t}(1-e^{-\lambda t})^n
=e^{-(n+1)\lambda t}(e^{\lambda t}-1)^n.
$$
Division gives
$$
n!\prod_{k=1}^n
\frac{\lambda e^{\lambda s_k}}{e^{\lambda t}-1},
$$
which is exactly the joint density of the order statistics of $n$ independent variables with the stated density. This proves the <conditional Yule birth times> result.

Solved by gpt-5.6-sol high.

= b
{parent=28k}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

Starting with one individual, let $S$ be the first event time, of density
$(\lambda+\mu)e^{-(\lambda+\mu)s}$. At that event there is a death with conditional probability $\mu/(\lambda+\mu)$, causing extinction immediately, or a birth with probability $\lambda/(\lambda+\mu)$, leaving two independent descendant populations. The branching property therefore gives
$$
\boxed{h(t)=\int_0^t e^{-(\lambda+\mu)s}
\{\mu+\lambda h(t-s)^2\}\,ds}.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

Changing variables in the <integral> equation,
$$
h(t)=e^{-(\lambda+\mu)t}
\int_0^t e^{(\lambda+\mu)u}
\{\mu+\lambda h(u)^2\}\,du.
$$
<Differentiation> gives
$$
h'=-(\lambda+\mu)h+\mu+\lambda h^2
=\boxed{(h-1)(\lambda h-\mu)}.
$$

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

Separating variables in
$h'=(h-1)(\lambda h-\mu)$ with $h(0)=0$ gives the <extinction probability of a linear birth-death process>
$$
\boxed{
h(t)=\frac{\mu\left(1-e^{-(\lambda-\mu)t}\right)}
{\lambda-\mu e^{-(\lambda-\mu)t}},
\qquad\lambda\ne\mu.}
$$
For $\lambda>\mu$ this tends to $\mu/\lambda$; for $\lambda<\mu$ it tends to one, as expected.

Solved by gpt-5.6-sol high.

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

= a
{parent=29l}
{scope}

= Solution
{parent=a}

Let $\widehat\theta_n$ be the MLE and let $I(\theta)$ be the one-observation Fisher information <matrix>. One standard form of the Wald statistic is
$$
\boxed{W_n(\theta)
=n(\widehat\theta_n-\theta)^T
I(\widehat\theta_n)(\widehat\theta_n-\theta)}.
$$
Any consistent information estimator gives the same asymptotics. Under $H_0$,
$W_n(\theta_0)\Rightarrow\chi_p^2$. If
$c_{p,\alpha}$ is the $(1-\alpha)$ quantile of $\chi_p^2$, reject when
$$
W_n(\theta_0)>c_{p,\alpha}.
$$
The asymptotic type-I error is $\alpha$.

Solved by gpt-5.6-sol high.

= b
{parent=29l}
{scope}

= Solution
{parent=b}

With $\ell_n(\theta)=\sum_i\log f(X_i,\theta)$, define the likelihood-ratio statistic
$$
\boxed{\Lambda_n
=2\{\ell_n(\widehat\theta_n)-\ell_n(\theta_0)\}}.
$$
Wilks' theorem gives $\Lambda_n\Rightarrow\chi_p^2$ under $H_0$, so reject when
$$
\Lambda_n>c_{p,\alpha}.
$$
This again has asymptotic type-I error $\alpha$.

Solved by gpt-5.6-sol high.

= c
{parent=29l}
{scope}

= Solution
{parent=c}

For <scalar> $\theta$, Taylor's theorem about the MLE, where
$\ell_n'(\widehat\theta_n)=0$, gives some $\widetilde\theta_n$ between $\theta_0$ and $\widehat\theta_n$ such that
$$
\Lambda_n
=-\ell_n''(\widetilde\theta_n)
(\widehat\theta_n-\theta_0)^2.
$$
The Wald statistic is
$$
W_n(\theta_0)
=nI(\widehat\theta_n)(\widehat\theta_n-\theta_0)^2.
$$
Under the null, consistency gives
$\widehat\theta_n,\widetilde\theta_n\to\theta_0$ in probability. A uniform law of large numbers for the observed information on a neighborhood of $\theta_0$ gives
$$
-\frac1n\ell_n''(\widetilde\theta_n)\xrightarrow P I(\theta_0),
\qquad
I(\widehat\theta_n)\xrightarrow P I(\theta_0)>0.
$$
Cancellation of the common squared displacement and Slutsky's theorem prove the <Wald and likelihood-ratio asymptotic equivalence>
$$
\boxed{\frac{\Lambda_n}{W_n(\theta_0)}\xrightarrow P1}.
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=30l}
{scope}

= Solution
{parent=a}

The increment of the stopped process is
$$
M_{(n+1)\wedge T}-M_{n\wedge T}
=\mathbf1_{\{T>n\}}(M_{n+1}-M_n).
$$
Because $T$ is a stopping time, $\{T>n\}\in\mathcal F_n$. Taking conditional expectation therefore gives zero. The stopped variables are adapted and integrable, so the <stopped martingale in discrete time> is a martingale.

Solved by gpt-5.6-sol high.

= b
{parent=30l}
{scope}

= Solution
{parent=b}

The martingale property gives
$$
\mathbb E M_{n\wedge T}=\mathbb E M_0=M_0
$$
because $\mathcal F_0$ is trivial. Since $T<\infty$ almost surely,
$M_{n\wedge T}\to M_T$ almost surely. The assumed uniform bound permits bounded convergence, so
$$
\boxed{\mathbb E M_T=M_0}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=30l}
{scope}

= Solution
{parent=c}

For the symmetric increment $\xi_{n+1}$,
$$
\mathbb E(w^{X_{n+1}}\mid\mathcal F_n)
=w^{X_n}\frac{w+w^{-1}}2.
$$
The same multiplier holds for $w^{-X_n}$. Thus choose
$$
\boxed{z=\frac2{w+w^{-1}}>0}.
$$
Then each of $z^nw^{X_n}$ and $z^nw^{-X_n}$ is a martingale, and every linear combination
$$
(Aw^{X_n}+Bw^{-X_n})z^n
$$
is a martingale.

Solved by gpt-5.6-sol high.

= d
{parent=30l}
{scope}

= Solution
{parent=d}

Given $0<z<1$, take
$$
w=\frac{1+\sqrt{1-z^2}}z>1,
\qquad z=\frac2{w+w^{-1}}.
$$
Choose $A,B$ so that
$$
Aw^{-a}+Bw^a=1,\qquad
Aw^b+Bw^{-b}=1.
$$
Solving,
$$
A=\frac{w^a-w^{-b}}{w^{a+b}-w^{-(a+b)}},
\qquad
B=\frac{w^b-w^{-a}}{w^{a+b}-w^{-(a+b)}}.
$$
The stopped discounted martingale is bounded, and $T$ is finite, so part (b) gives
$$
A+B=\mathbb E\!\left[
z^T(Aw^{X_T}+Bw^{-X_T})\right]=\mathbb E(z^T).
$$
Therefore the <discounted symmetric random-walk exit transform> is
$$
\boxed{
\mathbb E(z^T)=
\frac{w^a+w^b-w^{-a}-w^{-b}}
{w^{a+b}-w^{-(a+b)}}},
\qquad
w=\frac{1+\sqrt{1-z^2}}z.
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=31k}
{scope}

= Solution
{parent=a}

Start with the root region $\mathbb R^p$. For any current terminal region $R$, coordinate $k$, and threshold $s$, form
$$
R_-=\{x\in R:x_k\leq s\},\qquad
R_+=\{x\in R:x_k>s\},
$$
discarding splits that leave an empty child. For each child use its training-response mean
$$
\widehat\gamma_\pm=
\frac{\sum_iY_i'\mathbf1_{R_\pm}(X_i')}
{\sum_i\mathbf1_{R_\pm}(X_i')}.
$$
Compare candidate splits by the resulting residual sum of squares
$$
\sum_{X_i'\in R_-}(Y_i'-\widehat\gamma_-)^2+
\sum_{X_i'\in R_+}(Y_i'-\widehat\gamma_+)^2,
$$
together with the unchanged residual sums in the other leaves. Choose the leaf, coordinate, and threshold minimizing the total. Replace that leaf by its two children and repeat until the stopping rule or the prescribed $J$ leaves is reached. The resulting terminal regions are $\widehat R_1,\ldots,\widehat R_J$, and
$$
\widehat\gamma_j=
\frac{\sum_iY_i'\mathbf1_{\widehat R_j}(X_i')}
{\sum_i\mathbf1_{\widehat R_j}(X_i')}.
$$
This is the standard <regression tree> construction.

Solved by gpt-5.6-sol high.

= b
{parent=31k}
{scope}

= Solution
{parent=b}

Let $m_i=\mathbb E(Y_i\mid X_i)$. Conditional on $X_{1:n}$,
$$
\gamma_j=\frac1{N_j+1}\sum_i
m_i\mathbf1_{\widehat R_j}(X_i).
$$
The responses are conditionally independent, so cross-covariances vanish and
$$
\boxed{
\mathbb E[(\widetilde\gamma_j-\gamma_j)^2\mid X_{1:n}]
=\frac1{(N_j+1)^2}\sum_i
\operatorname{Var}(Y_i\mid X_i)
\mathbf1_{\widehat R_j}(X_i)}.
$$

Because the regions partition predictor space,
$$
\widetilde T(X)-\mathbb E(\widetilde T(X)\mid X,X_{1:n})
=\sum_j\mathbf1_{\widehat R_j}(X)
(\widetilde\gamma_j-\gamma_j).
$$
Only one summand is nonzero. Put
$q_j=\mathbb P(X\in\widehat R_j)$; the partition is deterministic because $D'$ is fixed, and $N_j\sim\operatorname{Bin}(n,q_j)$. The variance bound and
$N_j/(N_j+1)^2\leq1/(N_j+1)$ give
$$
\begin{aligned}
\mathbb E[\{\widetilde T(X)-\mathbb E(\widetilde T(X)\mid X,X_{1:n})\}^2]
&\leq\sigma^2\sum_jq_j\,
\mathbb E\frac1{N_j+1}\\
&\leq\sigma^2\sum_{j:q_j>0}\frac1n
\leq\boxed{\frac{\sigma^2J}{n}}.
\end{aligned}
$$
This is the <conditional variance of a fixed regression-tree partition> bound.

Solved by gpt-5.6-sol high.

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

= a
{parent=32c}
{scope}

= Solution
{parent=a}

An <asymptotic sequence> satisfies
$$
\phi_{n+1}(x)=o(\phi_n(x))
\qquad(x\to x_0).
$$
The expansion
$f\sim\sum_{n\geq0}a_n\phi_n$ means that for every $N\geq0$,
$$
f(x)-\sum_{n=0}^Na_n\phi_n(x)=o(\phi_N(x)).
$$
For $N=0$, division by $\phi_0$ gives
$$
a_0=\lim_{x\to x_0}\frac f{\phi_0}.
$$
For $n\geq1$, the definition with $N=n$ says
$$
f-\sum_{k=0}^{n-1}a_k\phi_k
=a_n\phi_n+o(\phi_n).
$$
Division by $\phi_n$ gives the second coefficient formula. In particular, asymptotic-expansion coefficients are unique.

Solved by gpt-5.6-sol high.

= b
{parent=32c}
{scope}

= Solution
{parent=b}

Since
$$
\frac{x}{x-1}=\frac1{1-x^{-1}}
\sim\sum_{j=0}^\infty x^{-j},
$$
we have
$$
\boxed{
f(x)\sim x^{-1}+\sum_{n=1}^\infty x^{-n+1}e^{-x}
=\sum_{n=0}^\infty\phi_n(x)}.
$$
Thus $a_n=1$ for every $n\geq0$.

Explicitly,
$$
\frac f{\phi_0}
=1+\frac{x^2e^{-x}}{x-1}\longrightarrow1.
$$
For $n\geq1$, the remainder after terms $0,\ldots,n-1$ is
$$
e^{-x}\left(\frac1{1-x^{-1}}
-\sum_{j=0}^{n-2}x^{-j}\right)
=\frac{x^{-n+1}e^{-x}}{1-x^{-1}}.
$$
Its ratio to $\phi_n$ tends to one, verifying every coefficient formula.

Because $e^{-x}=o(x^{-N})$ for every $N$, the exponentially small part is invisible to the power <sequence> $\psi_n=x^{-n}$. Hence
$$
\boxed{f(x)\sim x^{-1}}
$$
with coefficient one at $\psi_1$ and all other power coefficients zero.

Solved by gpt-5.6-sol high.

= c
{parent=32c}
{scope}

= Solution
{parent=c}

Repeated integration by parts starts with
$$
\operatorname{si}(x)
=\frac{\cos x}{x}
-\frac1x\int_1^\infty\frac{\cos(xt)}{t^2}\,dt
=\frac{\cos x}{x}+\frac{\sin x}{x^2}
-\frac2{x^2}\int_1^\infty\frac{\sin(xt)}{t^3}\,dt.
$$
Continuing alternates sine and cosine and multiplies by successive integers. After finitely many steps the remainder is bounded by a constant times the next inverse power, so the <fixed-lower-limit sine-integral expansion> is
$$
\operatorname{si}(x)\sim
\cos x\sum_{n=0}^\infty(-1)^n(2n)!x^{-2n-1}
+\sin x\sum_{n=0}^\infty(-1)^n(2n+1)!x^{-2n-2}.
$$
Therefore
$$
\boxed{a_n=(-1)^n(2n)!,\qquad
b_n=(-1)^n(2n+1)!}.
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=33a}
{scope}

= Solution
{parent=a}

A <transcritical bifurcation> has normal form
$$
\dot x=\mu x-x^2.
$$
Its equilibrium branches are $x=0$ and $x=\mu$. Linearization gives <derivative> $\mu$ on the first and $-\mu$ on the second, so the branches cross at $(\mu,x)=(0,0)$ and exchange stability: for $\mu<0$, $x=0$ is stable and $x=\mu$ unstable; for $\mu>0$ the labels reverse.

A small constant perturbation
$$
\dot x=\mu x-x^2+\varepsilon
$$
has equilibria $x=(\mu\pm\sqrt{\mu^2+4\varepsilon})/2$. For $\varepsilon>0$ the two branches avoid the crossing; for $\varepsilon<0$ there is a parameter interval with no equilibrium, bounded by two saddle-node points. Thus arbitrarily small perturbations change the bifurcation diagram, so the crossing is not structurally stable.

Solved by gpt-5.6-sol high.

= b
{parent=33a}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

At $r=1$, the Jacobian at the origin is
$$
J=\begin{pmatrix}
-1&1&0\\
1&-1&0\\
0&0&-1
\end{pmatrix}.
$$
Its <eigenvalues> are $0,-2,-1$. The zero <eigenvalue> makes the equilibrium nonhyperbolic.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

<Eigenvectors> for $0,-2,-1$ are respectively
$$
(1,1,0),\qquad(1,-1,0),\qquad(0,0,1).
$$
Hence
$$
\boxed{E^c=\operatorname{span}\{(1,1,0)\},\qquad
E^s=\operatorname{span}\{(1,-1,0),(0,0,1)\},\qquad
E^u=\{0\}}.
$$

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

With $x=v+w$ and $y=v-w$, the equations at $r=1$ become
$$
\dot v=\frac12\{a(v+w)^3-z(v-w)\},
$$
$$
\dot w=-2w+\frac12\{a(v+w)^3+z(v-w)\},
\qquad
\dot z=-z+v^2-w^2.
$$
The symmetry $(v,w,z)\mapsto(-v,-w,z)$ suggests
$$
w_c(v)=c_3v^3+O(v^5),\qquad
z_c(v)=d_2v^2+d_4v^4+O(v^6).
$$
Substitute these into the invariance equations
$$
w_c'(v)\dot v=\dot w,\qquad
z_c'(v)\dot v=\dot z.
$$
Comparison at orders $v^2,v^3,v^4$ gives
$$
d_2=1,\qquad c_3=\frac{a+1}{4},\qquad d_4=1-a.
$$
Thus the <center manifold of the 2024 Cambridge cubic system> is
$$
\boxed{w_c(v)=\frac{a+1}{4}v^3+O(v^5),\qquad
z_c(v)=v^2+(1-a)v^4+O(v^6)}.
$$

Solved by gpt-5.6-sol high.

= iv
{parent=b}
{scope}

= Solution
{parent=iv}

Substitution of the center-manifold <series> into the $v$ equation gives at cubic order
$$
\frac12(av^3-v^3)=\frac{a-1}{2}v^3.
$$
Keeping the $v^5$ terms from $3av^2w$, $z w$, and the quartic term in $z$ gives
$$
\boxed{
\dot v=\frac{a-1}{2}v^3+
\frac{3a^2+8a-3}{8}v^5+O(v^7)
=\frac{a-1}{2}v^3+
\frac{(3a-1)(a+3)}8v^5+O(v^7)}.
$$

Solved by gpt-5.6-sol high.

= v
{parent=b}
{scope}

= Solution
{parent=v}

The transverse <eigenvalues> $-2$ and $-1$ are stable, so the <center manifold> reduction determines local asymptotic stability. If $a<1$, the leading center term has negative coefficient and $v$ flows toward zero from both sides. If $a>1$, it flows away. At $a=1$ the cubic term vanishes but the quintic coefficient is
$$
\frac{(3-1)(1+3)}8=1>0,
$$
so the origin is again unstable. Therefore
$$
\boxed{\text{the origin is asymptotically stable exactly when }a<1}.
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=34c}
{scope}

= Solution
{parent=a}

The stated operators form a <Lax pair> when the KdV equation is equivalent to
$$
\boxed{\partial_tL=[L,A]=LA-AL}.
$$
Indeed, if $L\phi=k^2\phi$, <differentiation> and this identity show
$$
(L-k^2)(A\phi+\partial_t\phi)
=([L,A]-L_t)\phi=0.
$$
Thus $\partial_t+A$ preserves every spectral eigenspace, which is the <isospectral Lax equation> formulation.

Solved by gpt-5.6-sol high.

= b
{parent=34c}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

Translation by $2\pi$ preserves the scattering equation, so it acts linearly on its two-dimensional solution space. Because $u$ is real and the initial conditions are conjugate,
$\varphi_-=\overline{\varphi_+}$, forcing the transfer <matrix> to have the form
$$
\widehat T=\begin{pmatrix}a&b\\\bar b&\bar a\end{pmatrix}.
$$
The Wronskian of two solutions of
$-\phi''+u\phi=k^2\phi$ is independent of $x$. Translation therefore preserves it, so
$$
\det\widehat T=1.
$$
Consequently
$$
\boxed{|a|^2-|b|^2=1}.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

By part (a), each $A\varphi_\pm+\partial_t\varphi_\pm$ lies in the same two-dimensional eigenspace. Reality and conjugacy again force the coefficient <matrix> to have the form
$$
\Lambda=\begin{pmatrix}\lambda&\mu\\\bar\mu&\bar\lambda\end{pmatrix},
\qquad
(\partial_t+A)\Psi=\Lambda\Psi.
$$
Because $u$ and hence $A$ are periodic, $\partial_t+A$ commutes with translation by $2\pi$. Applying it to
$\Psi(x+2\pi)=\widehat T\Psi(x)$ gives
$$
\Lambda\widehat T\Psi
=(\partial_t\widehat T)\Psi+\widehat T\Lambda\Psi.
$$
The <basis> $\Psi$ is independent, so the <periodic KdV transfer matrix> obeys
$$
\boxed{\partial_t\widehat T=[\Lambda,\widehat T]}.
$$
Taking traces,
$$
\frac d{dt}(a+\bar a)=\operatorname{tr}[\Lambda,\widehat T]=0.
$$
Hence $\boxed{\operatorname{Re}a}$ is independent of time.

Solved by gpt-5.6-sol high.

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

= a
{parent=35b}
{scope}

= Solution
{parent=a}

A <density operator> satisfies
$$
\rho=\rho^\dagger,\qquad \rho\succeq0,\qquad\operatorname{tr}\rho=1.
$$
An $N\times N$ Hermitian <matrix> has $N^2$ real parameters, and the trace condition removes one, leaving
$$
\boxed{N^2-1}
$$
real free parameters.

If $A_H(t)=U(t)^\dagger A_SU(t)$, equality of <expectation values> requires
$$
\operatorname{tr}(\rho_HU^\dagger A_SU)
=\operatorname{tr}(U\rho_HU^\dagger A_S).
$$
Therefore
$$
\boxed{\rho_S(t)=U(t)\rho_HU(t)^\dagger}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=35b}
{scope}

= Solution
{parent=b}

The spin commutators are
$$
[S_i,S_j]=i\hbar\varepsilon_{ijk}S_k,
\qquad [S^2,S_i]=0.
$$
Set
$$
r_i=\frac{2\langle S_i\rangle_\psi}{\hbar}.
$$
Purity gives $r_x^2+r_y^2+r_z^2=1$, so the supplied sign determines
$$
r_y=\operatorname{sgn}\langle S_y\rangle_\psi\,
\sqrt{1-r_x^2-r_z^2}.
$$
For $r_z\ne-1$, choose the overall phase so that the upper component is nonnegative. The normalized state is
$$
\boxed{
|\psi\rangle=
\sqrt{\frac{1+r_z}{2}}\,|+\rangle_z+
\frac{r_x+ir_y}{\sqrt{2(1+r_z)}}\,|-\rangle_z}.
$$
If $r_z=-1$, it is simply $|-\rangle_z$ up to phase.

Solved by gpt-5.6-sol high.

= c
{parent=35b}
{scope}

= Solution
{parent=c}

For a mixed spin-half state, the <Bloch vector> formula gives
$$
\boxed{
\rho=\frac12\left(
I+\frac2\hbar\langle S_x\rangle\sigma_x
+\frac2\hbar\langle S_y\rangle\sigma_y
+\frac2\hbar\langle S_z\rangle\sigma_z
\right)}.
$$
Physical consistency is equivalent to the resulting Bloch <vector> having length at most one.

Three linearly independent Hermitian operators do not always suffice. The operators
$$
I,\quad\sigma_x,\quad\sigma_z
$$
are linearly independent, but the first expectation is always one and none detects the $y$ component. States with Bloch <vectors> $(0,c,0)$ and $(0,-c,0)$ have identical expectations for all three. The correct criterion for <informationally complete three-observable qubit tomography> is that the three traceless parts span the Pauli space.

Solved by gpt-5.6-sol high.

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

= a
{parent=36e}
{scope}

= Solution
{parent=a}

For any normalized trial state $|\psi\rangle$, the spectral decomposition gives the <variational method> bound
$$
E_0\leq\langle\psi|H|\psi\rangle.
$$
Choose a family $|\psi(\alpha)\rangle$, form its Rayleigh quotient, and minimize over the variational parameters. Every resulting value is an upper bound, and the minimum is the best bound within that family.

Solved by gpt-5.6-sol high.

= b
{parent=36e}
{scope}

= Solution
{parent=b}

For $\psi=\sum_n\alpha_n\phi_n$ and an orthonormal trial set,
$$
\frac{\langle\psi|H|\psi\rangle}{\langle\psi|\psi\rangle}
=\frac{\alpha^\dagger\mathcal H\alpha}{\alpha^\dagger\alpha},
\qquad
\mathcal H_{nm}=\langle\phi_n|H|\phi_m\rangle.
$$
Stationarity subject to $\alpha^\dagger\alpha=1$ gives
$$
\mathcal H\alpha=E\alpha.
$$
The <finite-subspace variational method>, equivalently the <Rayleigh-Ritz variational principle>, says that the minimum quotient is the smallest <eigenvalue> of $\mathcal H$. It is therefore the optimal upper bound on $E_0$ available from this span.

Solved by gpt-5.6-sol high.

= c
{parent=36e}
{scope}

= Solution
{parent=c}

The kinetic <matrix> is diagonal:
$$
\langle\phi_n|T|\phi_m\rangle
=\delta_{nm}\frac{\hbar^2\pi^2n^2}{2ma^2}.
$$
For the linear potential,
$$
\langle\phi_n|V|\phi_n\rangle=\frac{V_0}{2}
=\frac{9\hbar^2}{2ma^2},
$$
and
$$
\left\langle\phi_1\left|\frac xa\right|\phi_2\right\rangle
=2\int_0^1y\sin(\pi y)\sin(2\pi y)\,dy
=-\frac{16}{9\pi^2}.
$$
Thus, in units $\hbar^2/(ma^2)$,
$$
\mathcal H=
\begin{pmatrix}
(\pi^2+9)/2&-16/\pi^2\\
-16/\pi^2&(4\pi^2+9)/2
\end{pmatrix}.
$$
The lower <eigenvalue> gives the <two-mode variational bound for a linearly tilted square well>
$$
\boxed{
E_0\leq\frac{\hbar^2}{ma^2}
\left[
\frac{5\pi^2+18}{4}
-\sqrt{\frac{9\pi^4}{16}+\frac{256}{\pi^4}}
\right]}
\approx9.25936\,\frac{\hbar^2}{ma^2}.
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=37b}
{scope}

= Solution
{parent=a}

An intensive quantity is unchanged when the size of a homogeneous system is scaled; <temperature> and <pressure> are examples. An extensive quantity scales in proportion to system size; volume and particle number are examples. Energy and <entropy> are also extensive, while <chemical potential> is intensive. These are <intensive and extensive thermodynamic quantities>.

Solved by gpt-5.6-sol high.

= b
{parent=37b}
{scope}

= Solution
{parent=b}

Differentiate
$$
f(\lambda x_1,\ldots,\lambda x_n)=\lambda^kf(x_1,\ldots,x_n)
$$
with respect to $\lambda$. The chain rule gives
$$
\sum_i x_i\frac{\partial f}{\partial x_i}
(\lambda x_1,\ldots,\lambda x_n)
=k\lambda^{k-1}f(x_1,\ldots,x_n).
$$
Setting $\lambda=1$ proves the <Euler theorem for homogeneous functions>
$$
\boxed{\sum_i x_i\frac{\partial f}{\partial x_i}=kf}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=37b}
{scope}

= Solution
{parent=c}

Replicating a macroscopic equilibrium system scales $S,V,N$, and $E$ by the same factor, so $E(S,V,N)$ is homogeneous of degree one. Euler's theorem and the first law
$$
dE=T\,dS-p\,dV+\mu\,dN
$$
give
$$
\boxed{E=TS-pV+\mu N}.
$$
Differentiating this identity and comparing with the first law cancels the $T\,dS$, $-p\,dV$, and $\mu\,dN$ terms, leaving the <Gibbs-Duhem equation>
$$
S\,dT-V\,dp+N\,d\mu=0.
$$
Therefore
$$
\boxed{d\mu=\frac{V\,dp-S\,dT}{N}}.
$$

Solved by gpt-5.6-sol high.

= d
{parent=37b}
{scope}

= Solution
{parent=d}

Let $\xi$ be the forward reaction extent. Then
$$
dN_A=-d\xi,\qquad dN_B=2d\xi,\qquad dN_C=d\xi.
$$
At fixed $T,p$,
$$
dG=\mu_A\,dN_A+\mu_B\,dN_B+\mu_C\,dN_C
=(-\mu_A+2\mu_B+\mu_C)d\xi.
$$
Minimization with respect to $\xi$ gives the <chemical-potential balance for a reaction>
$$
\boxed{\mu_A=2\mu_B+\mu_C}.
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=38b}
{scope}

= Solution
{parent=a}

Contract the Riemann tensor to obtain
$$
R_{\mu\nu}=R^\alpha{}_{\mu\alpha\nu},
\qquad R=g^{\mu\nu}R_{\mu\nu},
$$
and define the <Einstein tensor>
$$
G_{\mu\nu}=R_{\mu\nu}-\frac12Rg_{\mu\nu}.
$$
Contracting the differential Bianchi identity twice, using the curvature symmetries and metric compatibility, gives the <contracted Bianchi identity>
$$
\nabla_\nu R_\mu{}^\nu=\frac12\nabla_\mu R.
$$
It follows immediately that
$$
\boxed{\nabla_\nu G_\mu{}^\nu=0}.
$$
The Einstein equations with cosmological constant are
$$
G_{\mu\nu}+\Lambda g_{\mu\nu}=\kappa T_{\mu\nu}.
$$
Since $\nabla g=0$, the left side is divergence-free, so consistency requires and implies the local conservation law $\nabla^\nu T_{\mu\nu}=0$.

Solved by gpt-5.6-sol high.

= b
{parent=38b}
{scope}

= Solution
{parent=b}

Write $H=\dot a/a$. The nonzero Christoffel symbols for the flat FLRW metric are
$$
\Gamma^t_{ij}=a\dot a\,\delta_{ij},
\qquad
\Gamma^i_{tj}=\Gamma^i_{jt}=H\delta^i_j.
$$
The mixed Einstein tensor is
$$
G_t{}^t=-3H^2,\qquad
G_i{}^j=-\left(2\frac{\ddot a}{a}+H^2\right)\delta_i^j.
$$
For a mixed tensor,
$$
\nabla_\nu G_\mu{}^\nu
=\partial_\nu G_\mu{}^\nu
+\Gamma^\nu_{\nu\lambda}G_\mu{}^\lambda
-\Gamma^\lambda_{\nu\mu}G_\lambda{}^\nu.
$$
For $\mu=t$, direct substitution gives
$$
\begin{aligned}
\nabla_\nu G_t{}^\nu
&=\frac d{dt}(-3H^2)+3H(-3H^2)
-3H\left[-\left(2\frac{\ddot a}{a}+H^2\right)\right]\\
&=-6H\left(\dot H+H^2-\frac{\ddot a}{a}\right)=0.
\end{aligned}
$$
For each spatial $\mu$, homogeneity makes the <partial derivatives> vanish and the remaining diagonal connection terms cancel pairwise. This verifies the <flat FLRW Einstein-tensor divergence> directly.

In vacuum,
$$
G_\mu{}^\nu+\Lambda\delta_\mu^\nu=0.
$$
The $tt$ equation is
$$
H^2=\frac{\Lambda}{3}.
$$
Choosing the positive root gives the expanding <de Sitter scale factor in flat slicing>
$$
\boxed{a(t)=a_0\exp\!\left(\sqrt{\frac{\Lambda}{3}}\,t\right)},
$$
and the spatial equation is then automatically satisfied.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=39C}

The incompressible Stokes equations are
$$
-\nabla p+\mu\nabla^2u=0,\qquad\nabla\cdot u=0.
$$
Taking the curl gives $\nabla^2\omega=0$. For a planar <stream function>, $\omega=-\nabla^2\psi$, so the <biharmonic stream function for planar Stokes flow> satisfies
$$
\boxed{\nabla^4\psi=0}.
$$

For $\psi=r^2f(\theta)$,
$$
u_r=rf',\qquad u_\theta=-2rf.
$$
The polar rate-of-strain components are
$$
e_{rr}=\partial_ru_r=f',\qquad
e_{\theta\theta}=\frac1r\partial_\theta u_\theta+\frac{u_r}{r}=-f',
$$
$$
e_{r\theta}=\frac12\left(
\frac1r\partial_\theta u_r+\partial_ru_\theta-\frac{u_\theta}{r}
\right)=\frac12f''.
$$
Hence
$$
\boxed{
e=\begin{pmatrix}f'&f''/2\\f''/2&-f'\end{pmatrix},
\qquad
\tau=2\mu e
=\mu\begin{pmatrix}2f'&f''\\f''&-2f'\end{pmatrix}.}
$$

Since
$$
\nabla^2(r^2f)=4f+f'',
$$
biharmonicity gives
$$
f^{(4)}+4f''=0,
\qquad
f=A+B\theta+C\cos2\theta+D\sin2\theta.
$$
No slip at $\theta=-\alpha$, no penetration at $\theta=0$, and the imposed tangential traction give
$$
f(-\alpha)=f'(-\alpha)=f(0)=0,\qquad
\mu f''(0)=S.
$$
Put $\Delta=\sin2\alpha-2\alpha\cos2\alpha$. Solving,
$$
f(\theta)=\frac{S}{4\mu}\left[
1-\cos2\theta+
\frac{2(1-\cos2\alpha)\theta+
(1-\cos2\alpha-2\alpha\sin2\alpha)\sin2\theta}
{\Delta}\right].
$$
The upper-surface <velocity> is radial and equals $rf'(0)$. Therefore the <similarity solution for tangentially forced Stokes wedge> gives
$$
\boxed{
U(r)=\frac{Sr}{\mu}
\frac{1-\cos2\alpha-\alpha\sin2\alpha}
{\sin2\alpha-2\alpha\cos2\alpha}}.
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=40d}
{scope}

= Solution
{parent=a}

Taking the divergence of the elastic equation gives
$$
\rho\,\vartheta_{tt}=(\lambda+2\mu)\nabla^2\vartheta.
$$
Taking its curl gives
$$
\rho\,\Omega_{tt}=\mu\nabla^2\Omega.
$$
Thus the <Helmholtz separation of elastic waves> produces longitudinal P-waves and transverse S-waves with
$$
\boxed{c_p=\sqrt{\frac{\lambda+2\mu}{\rho}},
\qquad c_s=\sqrt{\frac{\mu}{\rho}}}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=40d}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

A P-wave is longitudinal. For
$\widehat k=(\sin\theta,0,\cos\theta)$, its amplitude is
$$
\boxed{A=A_0(\sin\theta,0,\cos\theta)},
$$
parallel to the wavevector. Substitution gives $\omega=c_p\kappa$.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

An <SV-wave> is transverse but polarized in the vertical plane containing the wavevector:
$$
\boxed{B_V=B_0(\cos\theta,0,-\sin\theta)}.
$$
It is perpendicular to $\widehat k$ and has $\omega=c_s\kappa$.

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

An <SH-wave> is polarized horizontally, normal to the vertical propagation plane:
$$
\boxed{B_H=B_0(0,1,0)}.
$$
It is also perpendicular to $\widehat k$ and has $\omega=c_s\kappa$. A horizontal boundary distinguishes these two otherwise degenerate S polarizations.

Solved by gpt-5.6-sol high.

= c
{parent=40d}
{scope}

= i
{parent=c}
{scope}

= Solution
{parent=i}

Seek an SH displacement
$$
u=(0,Y(z),0)e^{i(\kappa x-\omega t)}.
$$
The S-wave equation gives
$$
Y''+q^2Y=0,\qquad
q^2=\frac{\omega^2}{c_s^2}-\kappa^2.
$$
Rigidity at $z=0$ gives $Y(0)=0$, while zero tangential traction at the free surface gives
$\mu Y'(h)=0$. Thus
$$
Y(z)=C\sin(qz),\qquad \cos(qh)=0,
$$
so
$$
q=q_n=\frac{(2n+1)\pi}{2h},\qquad
\boxed{\omega^2=c_s^2(\kappa^2+q_n^2)}.
$$
These are the <guided SH modes between rigid and free planes>.

Solved by gpt-5.6-sol high.

= ii
{parent=c}
{scope}

= Solution
{parent=ii}

At zero horizontal wavenumber, the dispersion relation gives the vertical-mode cutoff
$$
\boxed{\omega_n=c_sq_n
=\frac{(2n+1)\pi}{2h}c_s},
\qquad n=0,1,2,\ldots.
$$

Solved by gpt-5.6-sol high.

= iii
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= Solution
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Since
$$
\omega=\sqrt{c_s^2\kappa^2+\omega_n^2},
$$
the phase and <group> <velocities> are
$$
\boxed{
c=\frac{\omega}{\kappa}
=\sqrt{c_s^2+\frac{\omega_n^2}{\kappa^2}},
\qquad
c_g=\frac{d\omega}{d\kappa}
=\frac{c_s^2\kappa}{\sqrt{c_s^2\kappa^2+\omega_n^2}}
=\frac{c_s^2}{c}.}
$$

Solved by gpt-5.6-sol high.

= iv
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= Solution
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The preceding formulas give
$$
\frac c{c_g}=\frac{c^2}{c_s^2}
=1+\frac{\omega_n^2}{c_s^2\kappa^2}.
$$
Therefore
$$
\boxed{c=mc_g,\qquad
m=1+\left(\frac{(2n+1)\pi}{2\kappa h}\right)^2>1}.
$$

Solved by gpt-5.6-sol high.

= v
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= Solution
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A horizontal wavelength $h$ means $\kappa=2\pi/h$. For $n=1$,
$$
\frac{(2n+1)\pi}{2\kappa h}
=\frac{3\pi}{4\pi}=\frac34.
$$
Hence
$$
\boxed{m=1+\frac9{16}=\frac{25}{16}}.
$$

Solved by gpt-5.6-sol high.

= 41A
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{title2=Numerical Analysis}

= a
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= Solution
{parent=a}

The exact average is $I(h)=\widehat h_0$. For one Fourier mode,
$$
\frac1{2N}\sum_{k=-N+1}^Ne^{i\pi nk/N}
=\begin{cases}
1,&2N\mid n,\\
0,&2N\nmid n.
\end{cases}
$$
Thus the <periodic trapezoidal Fourier aliasing> identity gives
$$
I_N(h)=\sum_{j\in\mathbb Z}\widehat h_{2Nj},
\qquad
\boxed{|I_N(h)-I(h)|
=\left|\sum_{j\ne0}\widehat h_{2Nj}\right|}.
$$
Under the coefficient bound,
$$
|I_N-I|
\leq2M\sum_{j=1}^\infty c^{2Nj}
=\boxed{\frac{2Mc^{2N}}{1-c^{2N}}}
\leq\frac{2M}{1-c^2}c^{2N}.
$$
This decays exponentially in $N$.

Solved by gpt-5.6-sol high.

= b
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= Solution
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Write
$$
w'=\sum_l i\pi l\,\widehat w_l e^{i\pi lx},
\qquad
u_x=\sum_m i\pi m\,\widehat u_m e^{i\pi mx}.
$$
Projecting the equation onto modes $|n|\leq D$ gives the <Fourier-Galerkin matrix for a drift-diffusion equation>
$$
\boxed{\dot{\widehat u}_n
=\sum_{|m|\leq D}B_{nm}\widehat u_m,\qquad
B_{nm}=-\pi^2n^2\delta_{nm}
+\pi^2(n-m)m\,\widehat w_{n-m},}
$$
where $\widehat w_l=0$ for $|l|>d$.

For $w=\cos\pi x$, only
$\widehat w_{\pm1}=1/2$ are nonzero. Hence
$$
B_{nm}=-\pi^2n^2\delta_{nm}
+\frac{\pi^2}{2}(n-m)m\,\mathbf1_{\{|n-m|=1\}}.
$$
The column belonging to the constant mode $m=0$ is zero, so
$$
\boxed{B\text{ is not invertible}}.
$$
Remove that zero mode. In every remaining row $n$, the sum of the magnitudes of the off-diagonal entries is at most $\pi^2|n|$; for $|n|=1$, the would-be coupling to the removed zero mode vanishes. The diagonal entry is $-\pi^2n^2$. Gershgorin's theorem therefore places every <eigenvalue> of the nonconstant block in
$$
\operatorname{Re}z\leq-\pi^2(n^2-|n|)\leq0.
$$
Together with the zero <eigenvalue>, all <eigenvalues> of $B$ have nonpositive real part.

Solved by gpt-5.6-sol high.