past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-3.bigb

= Paper 3
{scope}

https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2025/paperii_3_2025.pdf

= 1G
{parent=Paper 3}
{scope}
{title2=Number Theory}

= a
{parent=1g}
{scope}

= Solution
{parent=a}

Write $f=[a,b,c]$. Two integral binary quadratic forms $f$ and $g$ are equivalent if there is a matrix
$$U=\begin{pmatrix}\alpha&\beta\\\gamma&\delta\end{pmatrix}\in\operatorname{SL}_2(\mathbb Z)$$
such that
$$g(x,y)=f(\alpha x+\beta y,\gamma x+\delta y).$$
Equivalently, the group $\operatorname{SL}_2(\mathbb Z)$ acts on forms by integral unimodular changes of variables.

Solved by gpt-5.6-sol high.

= b
{parent=1g}
{scope}

= Solution
{parent=b}

Suppose $g=[A,B,C]$ is obtained from $f=[a,b,c]$ by such a change of variables. Expanding shows that each of $A,B,C$ is an integer linear combination of $a,b,c$. Therefore every common divisor of $a,b,c$ divides $A,B,C$.

The inverse of a matrix in $\operatorname{SL}_2(\mathbb Z)$ again has integral entries and determinant one. Applying the same argument to $U^{-1}$ shows that every common divisor of $A,B,C$ divides $a,b,c$. The two coefficient triples consequently have the same greatest common divisor, up to sign, so one form is primitive exactly when the other is.

Solved by gpt-5.6-sol high.

= c
{parent=1g}
{scope}

= Solution
{parent=c}

Every proper equivalence class of a positive definite form contains a unique reduced form $[a,b,c]$, where
$$|b|\leq a\leq c,$$
with $b\geq0$ when equality occurs in either inequality. Its discriminant is
$$b^2-4ac=-80.$$
For a reduced positive definite form, $3a^2\leq80$, so $1\leq a\leq5$. Also $b$ is even and
$$ac=20+\frac{b^2}{4}.$$
Checking the finitely many possibilities with $|b|\leq a\leq c$ gives
$$
[1,0,20],\qquad [3,2,7],\qquad [3,-2,7],\qquad [4,0,5].
$$
Each coefficient triple has greatest common divisor one. By uniqueness of reduced representatives, these are four distinct classes. Hence the requested number is
$$\boxed{4}.$$

Solved by gpt-5.6-sol high.

= 2I
{parent=Paper 3}
{scope}
{title2=Topics in Analysis}

= Solution
{parent=2I}

Runge's polynomial approximation theorem says that if $K\subset\mathbb C$ is compact, $\mathbb C\setminus K$ is connected, and $f$ is holomorphic on a neighborhood of $K$, then for every $\varepsilon>0$ there is a polynomial $p$ such that
$$\sup_K|p-f|<\varepsilon.$$

Let
$$S=\{z:|z|=1,\ \operatorname{Re}z\leq0\}.$$
This arc is compact, its complement is connected, and $1/z$ is holomorphic on a neighborhood of it. Applying Runge with error $1/n$ gives a polynomial $q_n$ satisfying
$$\sup_{z\in S}|q_n(z)-z^{-1}|<\frac1n,$$
which is the requested uniform approximation.

For the pointwise construction, for $n\geq2$ set
$$
\begin{aligned}
K_n^+&=\{z:|z|\leq1-1/n,\ \operatorname{Re}z\geq1/n\},\\
K_n^-&=\{z:|z|\leq1-1/n,\ \operatorname{Re}z\leq-1/n\},\\
K_n^0&=\{iy:|y|\leq1-1/n\},
\qquad K_n=K_n^+\cup K_n^-\cup K_n^0.
\end{aligned}
$$
The three pieces are disjoint compact sets and $\mathbb C\setminus K_n$ is connected. Define a function on a neighborhood of $K_n$ to equal $1$, $-1$, and $0$ on neighborhoods of $K_n^+$, $K_n^-$, and $K_n^0$, respectively. It is holomorphic because those neighborhoods may be chosen disjoint. Runge supplies a polynomial $P_n$ such that
$$
|P_n-1|<1/n\ \hbox{on }K_n^+,
\quad |P_n+1|<1/n\ \hbox{on }K_n^-,
\quad |P_n|<1/n\ \hbox{on }K_n^0.
$$
Every fixed point with $|z|<1$ eventually belongs to the appropriate one of these three sets, according to the sign of its real part. These inequalities therefore give precisely the asserted pointwise limits.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=3K}

Let $\alpha$ be a primitive $n$th root of unity in an extension of $\mathbb F_q$, where $\gcd(n,q)=1$. A <BCH code> of length $n$, initial exponent $b$, and design distance $\delta$ is the cyclic code whose generator polynomial is the least common multiple over $\mathbb F_q$ of the minimal polynomials of
$$\alpha^b,\alpha^{b+1},\ldots,\alpha^{b+\delta-2}.$$
Thus every codeword polynomial $c(X)$ vanishes at these $\delta-1$ consecutive powers.

Suppose a nonzero codeword has weight $w<\delta$, with
$$c(X)=\sum_{\ell=1}^w c_\ell X^{i_\ell},$$
where the positions $i_\ell$ are distinct modulo $n$ and every $c_\ell$ is nonzero. The first $w$ root conditions give
$$
\sum_{\ell=1}^w c_\ell\alpha^{i_\ell(b+j)}=0,
\qquad j=0,1,\ldots,w-1.
$$
The coefficient matrix is a Vandermonde matrix in the distinct elements $\alpha^{i_1},\ldots,\alpha^{i_w}$, followed by multiplication of its columns by the nonzero factors $\alpha^{bi_\ell}$. Its determinant is therefore
$$
\left(\prod_{\ell=1}^w\alpha^{bi_\ell}\right)
\prod_{r<s}(\alpha^{i_s}-\alpha^{i_r})\ne0.
$$
Hence all $c_\ell$ would be zero, a contradiction. The minimum distance consequently satisfies $d\geq\delta$.

A code of minimum distance $d$ detects every pattern of at most $d-1$ errors and uniquely corrects every pattern of at most $\lfloor(d-1)/2\rfloor$ errors, because Hamming balls of that radius are disjoint. A BCH code of design distance $\delta$ is therefore guaranteed to detect $\delta-1$ errors and to correct
$$\left\lfloor\frac{\delta-1}{2}\right\rfloor$$
errors; its actual capabilities may be larger if $d>\delta$.

Solved by gpt-5.6-sol high.

= 4F
{parent=Paper 3}
{scope}
{title2=Automata \& Formal Languages}

= a
{parent=4f}
{scope}

= Solution
{parent=a}

The instruction $-(0,p,q)$ goes to $p$ if register $0$ is empty; otherwise it deletes the final letter and goes to $q$.

For $M$, an empty input goes directly from $q_S$ to $q_1$, where a $1$ is appended. On a nonempty input, state $q_0$ deletes every letter; once the register is empty it goes to $q_2$ and appends $0$. Hence
$$A=\{\varepsilon\}.$$

For $N$, the initial instruction tests whether the final letter is $0$. If it is, state $q_3$ deletes the entire word and then goes to $q_1$, producing $1$. Otherwise state $q_0$ deletes the entire word and goes to $q_2$, producing $0$. Thus
$$B=\{w0:w\in\mathbb B\},$$
the set of nonempty binary words ending in $0$.

Solved by gpt-5.6-sol high.

= b
{parent=4f}
{scope}

= Solution
{parent=b}

Use the same program as $N$, replacing only its first line by a test for a final $1$:
$$
\begin{array}{c|l}
q_S&\mapsto ?_1(0,q_3,q_0)\\
q_0&\mapsto -(0,q_2,q_0)\\
q_1&\mapsto +_1(0,q_H)\\
q_2&\mapsto +_0(0,q_H)\\
q_3&\mapsto -(0,q_1,q_3)\\
q_H&\mapsto ?_\varepsilon(0,q_H,q_H).
\end{array}
$$
If the input ends in $1$, the $q_3$ loop erases it and eventually appends output $1$. Every other input, including the empty word, follows the $q_0$ loop and produces $0$. This is the characteristic function of $\{w1:w\in\mathbb B\}$.

Solved by gpt-5.6-sol high.

= c
{parent=4f}
{scope}

= Solution
{parent=c}

Rename the old halt state $q_H$ as a nonhalting state $q_F$, including every occurrence of $q_H$ in the original program, and introduce a new halt state $q_H$. Attach the following output-flipping routine:
$$
\begin{array}{c|l}
q_F&\mapsto ?_0(0,q_{0},q_{1})\\
q_{0}&\mapsto -(0,q_{+1},q_{+1})\\
q_{1}&\mapsto -(0,q_{+0},q_{+0})\\
q_{+1}&\mapsto +_1(0,q_H)\\
q_{+0}&\mapsto +_0(0,q_H)\\
q_H&\mapsto ?_\varepsilon(0,q_H,q_H).
\end{array}
$$
Because the original machine computes a characteristic function, it reaches $q_F$ with register $0$ containing exactly $0$ or $1$. The routine tests that symbol, removes it, and appends its complement. It therefore halts with $1-\chi_X(w)=\chi_{\mathbb B\setminus X}(w)$.

Solved by gpt-5.6-sol high.

= 5K
{parent=Paper 3}
{scope}
{title2=Statistical Modelling}

= Solution
{parent=5K}

Let $Y_1,\ldots,Y_n$ be independent. In the general exponential-dispersion formulation, with known weights $w_i>0$ and common dispersion $\phi>0$, their densities have the form
$$
f_i(y_i;\theta_i,\phi)
=\exp\left\{\frac{w_i}{\phi}
[y_i\theta_i-b(\theta_i)]
+c\left(y_i,\frac{\phi}{w_i}\right)\right\}.
$$
Thus
$$
\mu_i=\mathbb E Y_i=b'(\theta_i),
\qquad
\operatorname{Var}(Y_i)=\frac{\phi}{w_i}b''(\theta_i)
=\frac{\phi}{w_i}V(\mu_i).
$$
The systematic component is $\eta_i=x_i^T\beta$, and a specified one-to-one differentiable link relates it to the mean by
$$g(\mu_i)=\eta_i.$$
The unknown model parameters are the regression coefficient $\beta$ and, when it is not fixed by the family, the dispersion $\phi$.

The log-likelihood is
$$
\ell(\beta,\phi)
=\sum_{i=1}^n\left\{
\frac{w_i}{\phi}[y_i\theta_i-b(\theta_i)]
+c\left(y_i,\frac{\phi}{w_i}\right)
\right\},
$$
up to terms independent of the parameters, where $\theta_i$ is determined from $g^{-1}(x_i^T\beta)=b'(\theta_i)$.

For a Poisson response,
$$f(y;\mu)=\exp\{y\log\mu-\mu-\log(y!)\},$$
so the natural parameter is $\theta=\log\mu$ and $b(\theta)=e^\theta$. The canonical link sets the linear predictor equal to the natural parameter and is therefore
$$\boxed{g(\mu)=\log\mu}.$$

Solved by gpt-5.6-sol high.

= 6A
{parent=Paper 3}
{scope}
{title2=Mathematical Biology}

= a
{parent=6a}
{scope}

= Solution
{parent=a}

The delay $\tau$ is the time between an individual's presence in the population and its contribution to the current growth rate, for example a maturation or gestation time. Thus reproduction at time $t$ is determined by the reproductive population at time $t-\tau$.

Solved by gpt-5.6-sol high.

= b
{parent=6a}
{scope}

= Solution
{parent=b}

For $0\leq t\leq\tau$, the delayed argument lies in the prescribed history, so $n'(t)=r$ and $n(t)=1+rt$. For $\tau\leq t\leq2\tau$,
$$n'(t)=r n(t-\tau)=r\{1+r(t-\tau)\}.$$
Integrating from $\tau$ and using $n(\tau)=1+r\tau$ gives
$$
\boxed{n(t)=1+rt+\frac{r^2}{2}(t-\tau)^2,
\qquad \tau\leq t\leq2\tau.}
$$

Solved by gpt-5.6-sol high.

= c
{parent=6a}
{scope}

= Solution
{parent=c}

Set $\omega=\pi/(2\tau)$. The condition $r\tau=-\pi/2$ says $r=-\omega$. Then
$$n(t)=C\cos(\omega t)$$
satisfies
$$
n'(t)=-C\omega\sin(\omega t),
\qquad
r n(t-\tau)=-C\omega\cos(\omega t-\pi/2)
=-C\omega\sin(\omega t).
$$
It is periodic with period $2\pi/\omega=4\tau$. Every nonzero such solution is negative during half of each period, so it cannot represent a population size.

Solved by gpt-5.6-sol high.

= 7E
{parent=Paper 3}
{scope}
{title2=Further Complex Methods}

= a
{parent=7e}
{scope}

= Solution
{parent=a}

For $\operatorname{Re}z>0$, absolute convergence permits the double-integral calculation
$$
\Gamma(z)^2
=\int_0^\infty\int_0^\infty
e^{-(s+t)}s^{z-1}t^{z-1}\,ds\,dt.
$$
Put $r=s+t$ and $u=t/(s+t)$, so $s=r(1-u)$, $t=ru$, and the Jacobian has magnitude $r$. The first quadrant becomes $r>0$, $0<u<1$, and therefore
$$
\begin{aligned}
\Gamma(z)^2
&=\int_0^\infty e^{-r}r^{2z-1}\,dr
\int_0^1(1-u)^{z-1}u^{z-1}\,du\\
&=\Gamma(2z)B(z,z).
\end{aligned}
$$

Solved by gpt-5.6-sol high.

= b
{parent=7e}
{scope}

= Solution
{parent=b}

As $z\to0$ within the right half-plane, the functional equation $\Gamma(1+z)=z\Gamma(z)$ and continuity at $1$ give
$$\Gamma(z)\sim\frac1z,
\qquad
\Gamma(2z)\sim\frac1{2z}.$$
Using part (a),
$$
B(z,z)=\frac{\Gamma(z)^2}{\Gamma(2z)}
\sim\frac{z^{-2}}{(2z)^{-1}}
=\boxed{\frac2z}.
$$

Solved by gpt-5.6-sol high.

= 8B
{parent=Paper 3}
{scope}
{title2=Classical Dynamics}

= a
{parent=8b}
{scope}

= Solution
{parent=a}

At height $z$ from the vertex, the cone has cross-sectional area $\pi R^2z^2/l^2$. Hence
$$
z_{\rm CM}=\frac{\int_0^l z^3\,dz}{\int_0^l z^2\,dz}=\frac{3l}{4},
\qquad
\mathbf r_{\rm CM}=a\mathbf e_3,
\quad a=\frac{3l}{4}.
$$

The parallel-axis theorem is
$$
\mathbf I_O=\mathbf I_{\rm CM}
+M\{|mathbf a|^2\mathbf1-\mathbf a\mathbf a^T\}.
$$
The displacement is along $\mathbf e_3$, so it changes the two transverse moments but not the axial one. Thus, writing the vertex moments as $I_1=I_2=I$ and $I_3=J$,
$$
\boxed{I=I_1^{\rm CM}+Ma^2
=\frac{\pi R^2l\rho}{20}(R^2+4l^2)
=\frac{3M}{20}(R^2+4l^2)},
$$
and
$$
\boxed{J=I_3^{\rm CM}
=\frac{\pi R^4l\rho}{10}
=\frac{3MR^2}{10}}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=8b}
{scope}

= Solution
{parent=b}

Take $\theta$ to be the angle between $\mathbf e_3$ and the upward vertical. From the supplied angular-velocity components,
$$
T=\frac I2(\dot\theta^2+\sin^2\theta\,\dot\phi^2)
+\frac J2(\dot\psi+\cos\theta\,\dot\phi)^2,
\qquad V=Mga\cos\theta.
$$
Therefore
$$
L=\frac I2(\dot\theta^2+\sin^2\theta\,\dot\phi^2)
+\frac J2(\dot\psi+\cos\theta\,\dot\phi)^2-Mga\cos\theta.
$$
Both $\psi$ and $\phi$ are cyclic. Their conserved conjugate momenta are
$$
p_\psi=J(\dot\psi+\cos\theta\,\dot\phi),
\qquad
p_\phi=I\sin^2\theta\,\dot\phi+p_\psi\cos\theta,
$$
and $p_\theta=I\dot\theta$. Solving for the velocities and taking the Legendre transform gives
$$
\boxed{H=\frac{p_\theta^2}{2I}
+\frac{(p_\phi-p_\psi\cos\theta)^2}{2I\sin^2\theta}
+\frac{p_\psi^2}{2J}+Mga\cos\theta}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=8b}
{scope}

= Solution
{parent=c}

Fix the two conserved momenta $p_\psi=\lambda$ and $p_\phi=\mu$. The remaining canonical variables $(\theta,p_\theta)$ form a two-dimensional phase space with Hamiltonian
$$
H_{\lambda,\mu}(\theta,p_\theta)
=\frac{p_\theta^2}{2I}+V_{\lambda,\mu}(\theta),
$$
where
$$
V_{\lambda,\mu}(\theta)
=\frac{(\mu-\lambda\cos\theta)^2}{2I\sin^2\theta}
+\frac{\lambda^2}{2J}+Mga\cos\theta.
$$

An upright spinning solution has $\theta=0$, $p_\theta=0$, and necessarily $\mu=\lambda$, since the vertical and body-axis angular momenta then coincide. On that momentum level,
$$
V_{\lambda,\lambda}(\theta)
=\frac{\lambda^2}{2I}\tan^2\frac\theta2
+\frac{\lambda^2}{2J}+Mga\cos\theta.
$$
Near zero,
$$
V_{\lambda,\lambda}(\theta)
=\text{constant}
+\left(\frac{\lambda^2}{8I}-\frac{Mga}{2}\right)\theta^2
+O(\theta^4).
$$
Thus the upright configuration is a strict local minimum of the reduced Hamiltonian, and hence stable, when
$$
\boxed{\lambda^2>4IMga=3IMgl}.
$$
Here $\lambda$ is the angular momentum about the symmetry axis. This proves stability for sufficiently large axial angular momentum.

Solved by gpt-5.6-sol high.

= 9E
{parent=Paper 3}
{scope}
{title2=Cosmology}

= a
{parent=9e}
{scope}

= Solution
{parent=a}

Since $a(\tau)=(\tau/\tau_0)^2$, the conformal Hubble parameter is $\mathcal H=2/\tau$. With $\Omega_M=1$, the mode equation becomes
$$
\delta''+\frac2\tau\delta'-\frac6{\tau^2}\delta=0.
$$
Substitution of $\delta=\tau^\beta$ gives
$$
\beta(\beta-1)+2\beta-6
=(\beta-2)(\beta+3)=0.
$$
The two independent solutions are therefore $\tau^2$ and $\tau^{-3}$, so
$$
\boxed{\delta(\mathbf k,\tau)
=A(\mathbf k)\tau^2+B(\mathbf k)\tau^{-3}}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=9e}
{scope}

= Solution
{parent=b}

The physical horizon size at conformal time $\tau$ is $a(\tau)c\tau$. Horizon crossing occurs when it equals the physical wavelength:
$$
\frac{2\pi a(\tau_H)}k=a(\tau_H)c\tau_H,
$$
and hence
$$\boxed{\tau_H=\frac{2\pi}{kc}}.$$

At late times the decaying mode is neglected, so the growing solution in part (a) obeys $\delta\propto\tau^2\propto a$. For the equality-crossing mode this gives
$$
\boxed{
D_{\rm eq}
=\frac{\delta(\mathbf k_{\rm eq},\tau_0)}
{\delta(\mathbf k_{\rm eq},\tau_{\rm eq})}
=\left(\frac{\tau_0}{\tau_{\rm eq}}\right)^2
=\frac{a(\tau_0)}{a(\tau_{\rm eq})}}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=9e}
{scope}

= Solution
{parent=c}

A mode that crosses during matter domination subsequently grows by
$$
\frac{a(\tau_0)}{a(\tau_H)}
=\left(\frac{\tau_0}{\tau_H}\right)^2.
$$
Its present amplitude is consequently
$$
|\delta(\mathbf k,\tau_0)|
=\left(\frac{\tau_0}{\tau_H}\right)^2
\tau_H^2\widehat A k^{1/2}
=\widehat A\tau_0^2k^{1/2}.
$$
Now
$$
\tau_0^2=D_{\rm eq}\tau_{\rm eq}^2,
\qquad
\tau_{\rm eq}=\frac{2\pi}{k_{\rm eq}c}.
$$
Squaring therefore gives
$$
\boxed{
|\delta(\mathbf k,\tau_0)|^2
=\frac{C}{k_{\rm eq}^4}k,
\qquad
C=\widehat A^2D_{\rm eq}^2\left(\frac{2\pi}{c}\right)^4.}
$$

Since $\tau_H\propto k^{-1}$, modes crossing chronologically between equality and today satisfy $k_0\leq k\leq k_{\rm eq}$. The reverse ordering printed with the target formula is inconsistent with the stated definitions of $k_0$ and $k_{\rm eq}$; it does not affect the spectrum calculation.

Solved by gpt-5.6-sol high.

= 10C
{parent=Paper 3}
{scope}
{title2=Quantum Information and Computation}

= a
{parent=10c}
{scope}

= Solution
{parent=a}

Write $x=x_1x_2$. Since the marked word is $11$, the required oracle is a single <Toffoli gate> with $x_1$ and $x_2$ as its two controls and the answer qubit $y$ as its target:
$$
|x_1x_2\rangle|y\rangle
\longmapsto
|x_1x_2\rangle|y\mathbin\oplus(x_1x_2)\rangle.
$$
The target flips exactly when both controls are one, which is exactly when $x=11$.

Solved by gpt-5.6-sol high.

= b
{parent=10c}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

Let $| -\rangle=(|0\rangle-|1\rangle)/\sqrt2$. Applying the oracle with this answer qubit produces phase kickback:
$$U_f|x\rangle| -\rangle=(-1)^{f(x)}|x\rangle| -\rangle.$$
Inside $A$, the $X$ gates on both search qubits conjugate the phase oracle for $11$. Hence
$$
A|x_1x_2\rangle| -\rangle
=(-1)^{[x_1x_2=00]}|x_1x_2\rangle| -\rangle
=I_0|x_1x_2\rangle| -\rangle,
$$
where $I_0=I-2|00\rangle\langle00|$.

The first oracle similarly acts on the search register as $I_{x_0}=I-2|x_0\rangle\langle x_0|$. Reading the remaining gates from right to left, the search-register operation after its initial Hadamards is therefore
$$-Q=H^{\otimes2}I_0H^{\otimes2}I_{x_0},$$
while the answer qubit remains in $| -\rangle$.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

After the initial Hadamards the search register is
$$|s\rangle=\frac12(|00\rangle+|01\rangle+|10\rangle+|11\rangle).$$
The marked-state phase oracle changes only the last sign:
$$I_{11}|s\rangle
=\frac12(|00\rangle+|01\rangle+|10\rangle-|11\rangle).$$
Now $H^{\otimes2}I_0H^{\otimes2}=I-2|s\rangle\langle s|$. Since the overlap of the last displayed state with $|s\rangle$ is $1/2$, its image is $-|11\rangle$. Thus the complete output is
$$
\boxed{|\psi\rangle=-|11\rangle| -\rangle}.
$$
The minus sign is a physically irrelevant global phase.

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

Measurement of the two search qubits in the computational basis returns $11$ with probability one. The answer qubit is unentangled from them and remains in $| -\rangle$.

Solved by gpt-5.6-sol high.

= iv
{parent=b}
{scope}

= Solution
{parent=iv}

Only one Grover iteration is required. For one marked item among four, the initial marked amplitude corresponds to an angle $\theta$ with $\sin\theta=1/2$, so $\theta=\pi/6$; one iteration changes this to $3\theta=\pi/2$, giving unit success amplitude. Further iterations are neither needed nor beneficial.

Solved by gpt-5.6-sol high.

= 11G
{parent=Paper 3}
{scope}
{title2=Number Theory}

= a
{parent=11g}
{scope}

= Solution
{parent=a}

Set
$$
p_{-2}=0,\qquad p_{-1}=1,\qquad
q_{-2}=1,\qquad q_{-1}=0,
$$
and for $n\geq0$ define
$$
p_n=a_np_{n-1}+p_{n-2},
\qquad
q_n=a_nq_{n-1}+q_{n-2}.
$$
Then $p_n/q_n=[a_0,\ldots,a_n]$ is the $n$th convergent.

For a variable final tail $x$, induction on $n$, or multiplication of the continued-fraction matrices, gives
$$
[a_0,\ldots,a_n,x]
=\frac{p_nx+p_{n-1}}{q_nx+q_{n-1}}.
$$
Indeed, the identity is immediate for $n=0$, and replacing the tail by $a_{n+1}+1/x$ gives the recurrence above. Taking $x=\gamma>0$ proves the formula.

Solved by gpt-5.6-sol high.

= b
{parent=11g}
{scope}

= Solution
{parent=b}

The continued-fraction algorithm starts with $a_0=3$. Rationalizing the successive remainders gives
$$
\frac1{\sqrt{11}-3}=\frac{\sqrt{11}+3}{2}
=3+\frac1{\sqrt{11}+3},
$$
and
$$\sqrt{11}+3=6+(\sqrt{11}-3).$$
The same remainder has returned, so the digits $3,6$ repeat. Therefore
$$
\boxed{\sqrt{11}=[3;\overline{3,6}]}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=11g}
{scope}

= Solution
{parent=c}

From part (b), $a_{2m+1}=3$ and $a_{2m+2}=6$. Eliminating the intervening odd-indexed convergent from the standard recurrence shows that either sequence $x_m=p_{2m}$ or $x_m=q_{2m}$ obeys
$$x_{m+2}=20x_{m+1}-x_m.$$
The initial even convergents are
$$
\frac{p_0}{q_0}=\frac31,
\qquad
\frac{p_2}{q_2}=\frac{63}{19}.
$$
Put $u=10+3\sqrt{11}$. Since $N(u)=100-99=1$ and $u+u^{-1}=20$, the sequence
$$w_m=(3+\sqrt{11})u^m$$
satisfies the same recurrence. Its first two values are $3+\sqrt{11}$ and $63+19\sqrt{11}$, so
$$p_{2m}+q_{2m}\sqrt{11}=(3+\sqrt{11})(10+3\sqrt{11})^m.$$
Taking norms in $\mathbb Q(\sqrt{11})$ yields
$$
p_{2m}^2-11q_{2m}^2
=(9-11)(100-99)^m
=\boxed{-2}.
$$
This applies in particular whenever the original index $n=2m\geq2$ is even.

Solved by gpt-5.6-sol high.

= 12F
{parent=Paper 3}
{scope}
{title2=Automata \& Formal Languages}

= a
{parent=12f}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

A state $q$ is inaccessible if no word reaches it from the initial state:
$$\widehat\delta(q_0,w)\ne q\qquad\text{for every }w\in\mathbb W.$$

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

States $q$ and $q'$ are indistinguishable if no continuation can distinguish acceptance from them, that is, for every $w\in\mathbb W$,
$$
\widehat\delta(q,w)\in F
\quad\Longleftrightarrow\quad
\widehat\delta(q',w)\in F.
$$

Solved by gpt-5.6-sol high.

= iii
{parent=a}
{scope}

= Solution
{parent=iii}

The automaton is irreducible if every state is accessible and no two distinct states are indistinguishable.

Solved by gpt-5.6-sol high.

= iv
{parent=a}
{scope}

= Solution
{parent=iv}

An automaton accepting a regular language is irreducible exactly when it has the smallest possible number of states among deterministic automata accepting that language. The irreducible automaton is unique up to isomorphism; its states correspond to the Myhill--Nerode equivalence classes.

Solved by gpt-5.6-sol high.

= b
{parent=12f}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

Define the extended transition recursively by
$$
\widehat\Delta(q,\varepsilon)=\{q\},
\qquad
\widehat\Delta(q,wa)
=\bigcup_{p\in\widehat\Delta(q,w)}\Delta(p,a).
$$
The automaton accepts $w$ precisely when
$$
w\in\mathcal L(N)
\quad\Longleftrightarrow\quad
\widehat\Delta(q_0,w)\cap F\ne\varnothing.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

The subset construction produces
$$
D=(\Sigma,\mathcal P(Q),\delta_D,\{q_0\},F_D),
$$
where
$$
\delta_D(S,a)=\bigcup_{q\in S}\Delta(q,a),
\qquad
F_D=\{S\subseteq Q:S\cap F\ne\varnothing\}.
$$
Each deterministic state records all nondeterministic states reachable after the word read so far. This construction gives $\mathcal L(D)=\mathcal L(N)$.

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

We induct on the length of $w$. For $w=\varepsilon$, the extended transition set is $\{q\}$, and the sole witnessing sequence is $(q)$, so the claim holds.

Write a nonempty word as $w=va$. By the recursive definition,
$$
q'\in\widehat\Delta(q,va)
$$
if and only if there is some $p\in\widehat\Delta(q,v)$ with $q'\in\Delta(p,a)$. By the induction hypothesis, the first condition on $p$ is equivalent to a witnessing sequence for $v$ from $q$ to $p$. Appending $q'$ gives a witnessing sequence for $va$. Conversely, deleting the last state of any witnessing sequence for $va$ gives just such a state $p$ and sequence for $v$. This proves both implications.

Solved by gpt-5.6-sol high.

= c
{parent=12f}
{scope}

= Solution
{parent=c}

The subset construction theorem gives $\mathcal L(D)=\mathcal L(N)$. Removing inaccessible states does not alter any computation from the initial state, so
$$\mathcal L(D')=\mathcal L(N).$$

Every state of $D'$ is accessible by construction. It remains to separate distinct subset states $S,T$ of $D'$. Choose $q\in S\mathbin\triangle T$, and without loss of generality take $q\in S\setminus T$. By (Br2), some word $w$ has a witnessing sequence from $q$ to the unique final state $q_*$. Part (b)(iii) implies
$$q_*\in\widehat\Delta(q,w),$$
so the state reached from $S$ on $w$ is accepting.

If the state reached from $T$ on $w$ were also accepting, there would be some $p\in T$ with a witnessing sequence labelled $w$ from $p$ to $q_*$. But $q$ has such a sequence too, and (Br3) says that its starting state is unique. Hence $p=q$, contradicting $q\notin T$. Thus $w$ distinguishes $S$ and $T$.

No two distinct states of $D'$ are indistinguishable, and all are accessible. Therefore $D'$ is irreducible and accepts $\mathcal L(N)$.

Solved by gpt-5.6-sol high.

= 13A
{parent=Paper 3}
{scope}
{title2=Mathematical Biology}

= a
{parent=13a}
{scope}

= Solution
{parent=a}

This <birth-death process> has birth rates $b_n=\alpha+\beta n$ and death rates $d_n=\gamma n(n-1)$. With $P_{-1}=0$, its <birth-death master equation> is
$$
\boxed{
\frac{dP_n}{dt}
=[\alpha+\beta(n-1)]P_{n-1}
+\gamma(n+1)nP_{n+1}
-[\alpha+\beta n+\gamma n(n-1)]P_n.}
$$
The first two terms are inflow by a birth from $n-1$ and a death from $n+1$; the bracketed term is the total outflow from $n$.

Solved by gpt-5.6-sol high.

= b
{parent=13a}
{scope}

= Solution
{parent=b}

Multiply the <birth-death master equation> by $n$ and sum. Equivalently, a birth changes $n$ by $+1$ and a death by $-1$, so
$$
\boxed{
\frac d{dt}\langle n\rangle
=\langle b_n-d_n\rangle
=\alpha+\beta\langle n\rangle
-\gamma\langle n(n-1)\rangle.}
$$
The equation is not closed in the mean because the quadratic death rate introduces the second <factorial moment>.

Solved by gpt-5.6-sol high.

= c
{parent=13a}
{scope}

= Solution
{parent=c}

For a Poisson random variable of mean $\lambda$,
$$
\langle n\rangle=\lambda,
\qquad
\langle n(n-1)\rangle=\lambda^2.
$$
Substitution into part (b) gives the closed Riccati equation
$$
\boxed{\dot\lambda=\alpha+\beta\lambda-\gamma\lambda^2.}
$$
Its positive equilibrium is stable on the nonnegative half-line and equals
$$
\lambda_*=\frac{\beta+\sqrt{\beta^2+4\alpha\gamma}}{2\gamma}.
$$
Thus under the Poisson approximation,
$$
\boxed{\lim_{t\to\infty}\langle n(t)\rangle
=\frac{\beta+\sqrt{\beta^2+4\alpha\gamma}}{2\gamma}.}
$$

Solved by gpt-5.6-sol high.

= d
{parent=13a}
{scope}

= Solution
{parent=d}

The <diffusion approximation of a birth-death process> uses the first two jump moments. For these unit upward and downward jumps they give
$$
\boxed{
u(x)=\alpha+\beta x-\gamma x(x-1),\qquad
D(x)=\frac12\{\alpha+\beta x+\gamma x(x-1)\}.}
$$
The factor $1/2$ in $D$ is required because the equation is written with $+\partial_x^2(DP)$, rather than $\tfrac12\partial_x^2(BP)$.

Assuming the boundary terms vanish, integration by parts gives
$$
\begin{aligned}
\frac d{dt}\langle x\rangle
&=\int x\left[-\frac{\partial(uP)}{\partial x}
+\frac{\partial^2(DP)}{\partial x^2}\right]dx\\
&=\int u(x)P(x,t)\,dx=\langle u(x)\rangle.
\end{aligned}
$$
Consequently
$$
\boxed{
\frac d{dt}\langle x\rangle
=\alpha+\beta\langle x\rangle
-\gamma\langle x(x-1)\rangle,}
$$
which is precisely the discrete first-moment equation in part (b).

Solved by gpt-5.6-sol high.

= 14E
{parent=Paper 3}
{scope}
{title2=Cosmology}

= a
{parent=14e}
{scope}

= Solution
{parent=a}

The Bose--Einstein and Fermi--Dirac occupation factors are
$$
\frac1{\exp((E-\mu)/(k_BT))\mp1}.
$$
In the stated nonrelativistic dilute regime, $E-\mu$ is dominated by $mc^2$, so the exponential is large and either denominator is asymptotic to $\exp((E-\mu)/(k_BT))$. Integrating the resulting Maxwell--Boltzmann occupation over momentum states gives
$$
n=\frac{4\pi g_s}{h^3}\int_0^\infty
p^2e^{-[E(p)-\mu]/(k_BT)}\,dp.
$$

For $E=mc^2+p^2/(2m)$, integration by parts or differentiation of the supplied Gaussian integral gives
$$
\int_0^\infty p^2e^{-p^2/(2mk_BT)}\,dp
=\frac{\sqrt\pi}{4}(2mk_BT)^{3/2}.
$$
Consequently
$$
\boxed{
n=g_s\left(\frac{2\pi mk_BT}{h^2}\right)^{3/2}
e^{(\mu-mc^2)/(k_BT)}}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=14e}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

For any nonrelativistic species $j$, part (a) gives
$$
n_j=g_j\left(\frac{2\pi m_jk_BT}{h^2}\right)^{3/2}
e^{(\mu_j-m_jc^2)/(k_BT)}.
$$
Chemical equilibrium and zero photon chemical potential imply
$$
\mu_{H^+}+\mu_e=\mu_{H^0},
\qquad
\mu_{He^+}+\mu_e=\mu_{He^0}.
$$
Taking the appropriate ratios of the number-density formula, approximating the ion and atom masses as equal in the translational prefactors, and using the given degeneracies yields
$$
\boxed{
\frac{n_en_{H^+}}{n_{H^0}}
=\left(\frac{2\pi m_ek_BT}{h^2}\right)^{3/2}
e^{-I_H/(k_BT)}}
$$
and
$$
\boxed{
\frac{n_en_{He^+}}{n_{He^0}}
=2\left(\frac{2\pi m_ek_BT}{h^2}\right)^{3/2}
e^{-I_{He}/(k_BT)}}.
$$
The factors are $g_eg_{H^+}/g_{H^0}=1$ and $g_eg_{He^+}/g_{He^0}=2$.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

Baryon number and charge neutrality give
$$
n_B=n_H+4n_{He},
\qquad
n_e=n_{H^+}+n_{He^+}.
$$
Since $Y_p=4n_{He}/n_B$,
$$
\frac{n_H}{n_B}=1-Y_p,
\qquad
\frac{n_{He}}{n_B}=\frac{Y_p}{4}.
$$
It follows that
$$
\boxed{
\mathcal F=\frac{n_e}{n_B}
=(1-Y_p)X_{H^+}+\frac{Y_p}{4}X_{He^+}}.
$$
Thus $\alpha=1-Y_p$ and $\beta=Y_p/4$.

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

Define
$$
K(T)=\frac1{n_B}\left(\frac{2\pi m_ek_BT}{h^2}\right)^{3/2}
=\frac{\sqrt{2\pi}}{8\eta\zeta(3)}
\left(\frac{m_ec^2}{k_BT}\right)^{3/2}.
$$
Using $n_e=n_B\mathcal F$, $n_{j^+}=X_{j^+}n_j$, and $n_{j^0}=(1-X_{j^+})n_j$, the two Saha relations become the closed system
$$
\boxed{
\mathcal F\frac{X_{H^+}}{1-X_{H^+}}
=K(T)e^{-I_H/(k_BT)}},
$$
$$
\boxed{
\mathcal F\frac{X_{He^+}}{1-X_{He^+}}
=2K(T)e^{-I_{He}/(k_BT)}},
$$
where
$$
\mathcal F=(1-Y_p)X_{H^+}+\frac{Y_p}{4}X_{He^+}.
$$
These are two equations for the two ionization fractions.

When $Y_p\to0$, $\mathcal F=X_{H^+}$, and the hydrogen equation reduces to
$$
\boxed{
\frac{X_{H^+}^2}{1-X_{H^+}}
=\frac1{n_B}\left(\frac{2\pi m_ek_BT}{h^2}\right)^{3/2}
e^{-I_H/(k_BT)}},
$$
the usual <hydrogen-only Saha equation>.

Solved by gpt-5.6-sol high.

= 15C
{parent=Paper 3}
{scope}
{title2=Quantum Information and Computation}

= a
{parent=15c}
{scope}

= Solution
{parent=a}

Let $\rho_j=|\alpha_j\rangle\langle\alpha_j|$. For a two-outcome measurement with effect $0\leq M\leq I$ interpreted as guess zero,
$$
P_s=\frac12\operatorname{tr}(M\rho_0)
\frac12\operatorname{tr}((I-M)\rho_1)
=\frac12+\frac12\operatorname{tr}[M(\rho_0-\rho_1)].
$$
The positive-eigenspace projector of $\rho_0-\rho_1$ maximizes the last trace, giving the Helstrom formula
$$
P_s^{\rm opt}=\frac12+\frac14\|\rho_0-\rho_1\|_1.
$$
On the span of the two states, $\rho_0-\rho_1$ has eigenvalues
$$\mathord\pm\sqrt{1-|\langle\alpha_0|\alpha_1\rangle|^2}.$$
Therefore
$$
\boxed{
P_s\leq\frac12\left(1+
\sqrt{1-|\langle\alpha_0|\alpha_1\rangle|^2}\right)},
$$
and the positive/negative eigenspace measurement attains equality.

The right side equals one exactly when $|\langle\alpha_0|\alpha_1\rangle|=0$. Thus two pure states are perfectly distinguishable exactly when they are orthogonal.

Solved by gpt-5.6-sol high.

= b
{parent=15c}
{scope}

= Solution
{parent=b}

For a chosen unit vector $|\psi\rangle$, the two possible outputs have inner product
$$
\langle\psi|U_2^\dagger U_1|\psi\rangle.
$$
Part (a) says they can be perfectly distinguished exactly when this inner product is zero. Such an input exists exactly when
$$
0\in\{\langle\psi|U_2^\dagger U_1|\psi\rangle:\|\psi\|=1\}
=N(U_2^\dagger U_1).
$$

Solved by gpt-5.6-sol high.

= c
{parent=15c}
{scope}

= Solution
{parent=c}

If $Uv=\lambda v$ with $v\ne0$, unitarity gives
$$
\|v\|=\|Uv\|=\|\lambda v\|=|\lambda|\|v\|.
$$
Hence $|\lambda|=1$. Every eigenvalue of a unitary matrix therefore lies on the unit circle.

Solved by gpt-5.6-sol high.

= d
{parent=15c}
{scope}

= Solution
{parent=d}

The three eigenvalues are
$$
1,\qquad e^{i\pi/4}=\frac{1+i}{\sqrt2},
\qquad e^{3i\pi/4}=\frac{-1+i}{\sqrt2}.
$$
Thus $N(U)$ is the filled triangle with vertices
$$
(1,0),\qquad(2^{-1/2},2^{-1/2}),
\qquad(-2^{-1/2},2^{-1/2})
$$
in the complex plane. Its upper edge is the horizontal chord joining the last two vertices, and the other two edges join those points to $1$. In particular, the origin lies outside this triangle.

Solved by gpt-5.6-sol high.

= e
{parent=15c}
{scope}

= Solution
{parent=e}

The two eigenvalues are the points at arguments $0$ and $\gamma$. The counterclockwise arc from $1$ to $e^{i\gamma}$ has length $\gamma$, while the complementary clockwise arc has length $2\pi-\gamma$. The shorter of the two contains the spectrum. Therefore
$$
\boxed{
\theta(U_\gamma)=\min\{\gamma,2\pi-\gamma\}
=\begin{cases}
\gamma,&0\leq\gamma<\pi,\\
2\pi-\gamma,&\pi\leq\gamma<2\pi.
\end{cases}}
$$
The two geometric cases are respectively the short counterclockwise arc and the short complementary clockwise arc; at $\gamma=\pi$ both semicircles have length $\pi$.

Solved by gpt-5.6-sol high.

= f
{parent=15c}
{scope}

= Solution
{parent=f}

For points on the unit circle, their convex hull contains the origin exactly when they are not all contained in an open semicircle. Equivalently, the shortest closed arc containing them has length at least $\pi$. Applying this to $V=U_2^\dagger U_1$ and using
$$N(V)=\operatorname{conv}(\operatorname{spec}V)$$
gives
$$
0\in N(U_2^\dagger U_1)
\quad\Longleftrightarrow\quad
\theta(U_2^\dagger U_1)\geq\pi.
$$
Part (b) now proves the claimed perfect-discrimination criterion.

Solved by gpt-5.6-sol high.

= 16H
{parent=Paper 3}
{scope}
{title2=Logic and Set Theory}

= a
{parent=16h}
{scope}

= Solution
{parent=a}

By <choice-function well-ordering construction>, use <transfinite recursion> to choose
$$
x_\beta=f\left(x\setminus\{x_\gamma:\gamma<\beta\}\right)
$$
whenever the set on the right is nonempty. The chosen elements are distinct, so if this construction continued through the <Hartogs theorem> ordinal $h(x)$, the map $\beta\mapsto x_\beta$ would inject $h(x)$ into $x$, a contradiction. It therefore stops at some ordinal $\alpha<h(x)$. At the stopping stage every element of $x$ has been selected, and hence
$$
\beta\longmapsto x_\beta
$$
is a bijection $\alpha\to x$.

Assuming the <axiom of choice>, every set $x$ has such a choice function on its nonempty subsets. Transporting the membership order on $\alpha$ across the bijection well-orders $x$. Thus the axiom of choice implies the <well-ordering theorem>.

Solved by gpt-5.6-sol high.

= b
{parent=16h}
{scope}

= Solution
{parent=b}

Apply the stated <cardinal comparability principle> to $x$ and its Hartogs ordinal $h(x)$. The <Hartogs theorem> rules out an injection $h(x)\to x$, so comparability supplies an injection
$$
j:x\longrightarrow h(x).
$$
Order $x$ by $u\prec v$ exactly when $j(u)\in j(v)$. This is the restriction of an ordinal well-order and therefore well-orders $x$. Hence every set can be well-ordered. The <well-ordering theorem> is equivalent to the <axiom of choice>: for a family of nonempty sets, well-order its union and choose the least member of each set. The comparison statement therefore implies choice.

Solved by gpt-5.6-sol high.

= c
{parent=16h}
{scope}

= Solution
{parent=c}

An <initial ordinal> is an ordinal $\kappa$ that is not equipotent to any smaller ordinal. Equivalently, it is the least ordinal having its cardinality. The <aleph number>s enumerate the infinite initial ordinals:
$$
\aleph_0=\omega,
$$
$\aleph_{\alpha+1}$ is the least initial ordinal greater than $\aleph_\alpha$, and at a limit $\lambda$, $\aleph_\lambda$ is the least initial ordinal greater than every $\aleph_\beta$ for $\beta<\lambda$.

Each $\aleph_\alpha$ is infinite and initial by construction. Conversely, let $\kappa$ be an infinite initial ordinal. The infinite initial ordinals below $\kappa$ form a set and therefore have an ordinal order type $\alpha$. The recursive enumeration above lists exactly those predecessors before stage $\alpha$, so its next value is $\aleph_\alpha=\kappa$.

Read literally, “has cardinality $\aleph_\alpha$” is too weak: $\omega+1$ has cardinality $\aleph_0$ but is not initial. The valid characterization is
$$
\kappa\text{ is an infinite initial ordinal}
\quad\Longleftrightarrow\quad
\kappa=\aleph_\alpha\text{ for some }\alpha.
$$
If “has cardinality” in the question means “is the cardinal represented by,” this is precisely the requested statement.

Solved by gpt-5.6-sol high.

= d
{parent=16h}
{scope}

= Solution
{parent=d}

This is <König theorem for cardinal numbers>. Since $\kappa_i<\lambda_i$, map $(i,\alpha)$ in the disjoint union to the element $s_{i,\alpha}$ of the product defined by
$$
s_{i,\alpha}(j)=
\begin{cases}
\alpha+1,&j=i,\\
0,&j\ne i.
\end{cases}
$$
Here $\alpha<\kappa_i$ implies $\alpha+1<\lambda_i$. The unique nonzero coordinate and its value recover $(i,\alpha)$, so this is an injection
$$
\bigsqcup_{i\in I}\kappa_i\longrightarrow
\prod_{i\in I}\lambda_i.
$$

For strictness, suppose $F$ mapped the disjoint union onto the product. For each $i$, the set
$$
A_i=\{F(i,\alpha)(i):\alpha<\kappa_i\}
$$
has cardinality at most $\kappa_i$, so it cannot exhaust $\lambda_i$. Choose $b_i\in\lambda_i\setminus A_i$. Then $b=(b_i)_{i\in I}$ differs from $F(i,\alpha)$ in coordinate $i$ for every $(i,\alpha)$, contradicting surjectivity. By <Cantor-Schröder-Bernstein theorem>, an injection in the reverse direction would combine with the displayed injection to give a bijection and hence a surjection. Therefore
$$
\sum_{i\in I}\kappa_i<\prod_{i\in I}\lambda_i.
$$

Solved by gpt-5.6-sol high.

= e
{parent=16h}
{scope}

= Solution
{parent=e}

Apply <König theorem for cardinal numbers> with $I=\omega$, $\kappa_i=\aleph_i$, and $\lambda_i=\aleph_{i+1}$. Standard <infinite cardinal arithmetic> gives
$$
\sum_{i<\omega}\aleph_i=\aleph_\omega,
$$
whereas every factor in the product is at most $\aleph_\omega$. Hence
$$
\aleph_\omega
<\prod_{i<\omega}\aleph_{i+1}
\leq(\aleph_\omega)^{\aleph_0}
\leq(\aleph_\omega)^{\aleph_\omega}.
$$

Now suppose $2^{\aleph_\omega}=\aleph_\omega$. The exponent laws for infinite cardinals would give
$$
(\aleph_\omega)^{\aleph_\omega}
=(2^{\aleph_\omega})^{\aleph_\omega}
=2^{\aleph_\omega\cdot\aleph_\omega}
=2^{\aleph_\omega}
=\aleph_\omega,
$$
contradicting the strict inequality. Therefore
$$
2^{\aleph_\omega}\ne\aleph_\omega.
$$

Solved by gpt-5.6-sol high.

= 17F
{parent=Paper 3}
{scope}
{title2=Graph Theory}

= a
{parent=17f}
{scope}

= Solution
{parent=a}

A graph is <eulerian graph> if it has a closed trail containing every edge exactly once. For a graph with at least three vertices, the <Euler circuit criterion> says that it is Eulerian exactly when it is connected and every vertex has even degree.

Necessity is immediate: each visit of a closed trail to a vertex uses one entering and one leaving edge, so the incident edges occur in pairs; the trail also joins every vertex incident with an edge. Conversely, start at any vertex and extend an edge-simple trail until no unused incident edge remains. Even degrees force this maximal trail to end where it began. If edges remain, connectedness supplies a vertex of this circuit incident with an unused edge. Construct another closed trail from there and splice it into the first. Repeating consumes every edge and gives an Euler circuit.

The <line graph> $L(G)$ has vertex set $E(G)$, with two vertices adjacent when the corresponding edges of $G$ share an endpoint. If $G$ is connected and $r$-regular, then $L(G)$ is connected, and the vertex corresponding to an edge $uv$ has degree
$$
(d(u)-1)+(d(v)-1)=2r-2.
$$
Every degree is even, so the Euler circuit criterion makes $L(G)$ Eulerian.

Solved by gpt-5.6-sol high.

= b
{parent=17f}
{scope}

= Solution
{parent=b}

The <Euler formula for a connected planar graph> follows by induction on the number of cycle edges. If $G$ is a tree, then $e=n-1$ and $f=1$, so $n-e+f=2$. Otherwise remove an edge belonging to a cycle. The graph remains connected, while the two faces adjoining that edge merge: both $e$ and $f$ decrease by one. Thus $n-e+f$ is unchanged, and induction reaches a tree.

Every edge borders two face sides, so the sum of the face sizes is $2e$. If every face has size at least $g$, then
$$
gf\leq2e.
$$
Substitute $f=2-n+e$ from Euler's formula:
$$
g(2-n+e)\leq2e,
$$
and hence the <planar girth edge bound>
$$
\boxed{e\leq\frac{g(n-2)}{g-2}}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=17f}
{scope}

= i
{parent=c}
{scope}

= Solution
{parent=i}

For each crossing pair, delete one of its two edges. At most $t(G)$ distinct edges are deleted, and the graph left behind has a crossing-free drawing. The <planar graph edge bound> therefore gives
$$
e-t(G)\leq3n-6,
$$
so
$$
\boxed{t(G)\geq e-3n+6}.
$$
This is the linear estimate underlying the <crossing lemma>.

Solved by gpt-5.6-sol high.

= ii
{parent=c}
{scope}

= Solution
{parent=ii}

Choose each vertex independently with probability
$$
p=\frac{4n}{e}\leq1
$$
and take the induced drawing. Let its numbers of vertices, edges, and crossing pairs be $N,E,T$. Since crossing edges have four distinct endpoints,
$$
\mathbb E N=pn,\qquad
\mathbb E E=p^2e,\qquad
\mathbb E T=p^4t(G).
$$
Part (i), with the harmless positive constant dropped, gives $T\geq E-3N$ for every outcome. Taking expectations,
$$
p^4t(G)\geq p^2e-3pn.
$$
For $p=4n/e$, the right side is $4n^2/e$. Dividing by $p^4=256n^4/e^4$ proves the <crossing lemma> estimate
$$
\boxed{t(G)\geq\frac{e^3}{64n^2}}.
$$

Solved by gpt-5.6-sol high.

= 18J
{parent=Paper 3}
{scope}
{title2=Galois Theory}

= a
{parent=18j}
{scope}

= Solution
{parent=a}

Up to conjugacy, the <transitive subgroups of the symmetric group on four points> are
$$
C_4,\qquad V_4,\qquad D_8,\qquad A_4,\qquad S_4.
$$
Representatives are $\langle(1234)\rangle$, the normal Klein four group of double transpositions, $\langle(1234),(13)\rangle$, $A_4$, and $S_4$.

Solved by gpt-5.6-sol high.

= b
{parent=18j}
{scope}

= Solution
{parent=b}

For the <biquadratic extension> $L=K(\sqrt a,\sqrt b)$, the assumptions say that the square classes of $a$ and $b$ are independent. In particular $\sqrt b\notin K(\sqrt a)$: writing $\sqrt b=u+v\sqrt a$ and comparing the coefficient of $\sqrt a$ after squaring would force either $b$ or $ab$ to be a square in $K$. Hence
$$
[L:K]=[K(\sqrt a,\sqrt b):K(\sqrt a)]
\,[K(\sqrt a):K]=4.
$$

The field is the splitting field of $(X^2-a)(X^2-b)$ and the characteristic is not two, so it is Galois. Its four automorphisms independently choose the signs of $\sqrt a$ and $\sqrt b$, giving
$$
\operatorname{Gal}(L/K)\cong C_2\times C_2.
$$
By the <Galois correspondence>, the intermediate-field lattice is
$$
\begin{array}{c}
L\\[2mm]
K(\sqrt a)\qquad K(\sqrt b)\qquad K(\sqrt{ab})\\[2mm]
K.
\end{array}
$$
Each of the three middle fields has degree two over $K$, and distinct middle fields intersect in $K$.

Solved by gpt-5.6-sol high.

= c
{parent=18j}
{scope}

= Solution
{parent=c}

Squaring and eliminating the inner radical gives
$$
(\alpha^2-3)^2=3,\qquad
(\gamma^2-3)^2=6.
$$
Thus $\alpha,\beta$ are roots of
$$
X^4-6X^2+6,
$$
and $\gamma,\delta$ are roots of
$$
X^4-6X^2+3.
$$
Both polynomials are irreducible by the <Eisenstein criterion> at $3$, so these are the four <minimal polynomial>s over $\mathbb Q$.

Over $K=\mathbb Q(\sqrt3)$, the relevant radicands are
$$
a=3+\sqrt3,\qquad b=3-\sqrt3,\qquad ab=6.
$$
The norms of $a$ and $b$ to $\mathbb Q$ equal $6$, so neither can be a square in $K$. Also $(u+v\sqrt3)^2=6$ with $u,v\in\mathbb Q$ would force $uv=0$, followed by either $u^2=6$ or $v^2=2$, both impossible. Thus $6$ is not a square either.

Similarly, over $K'=\mathbb Q(\sqrt6)$ the radicands $3+\sqrt6$ and $3-\sqrt6$ have nonsquare norm $3$, while $(u+v\sqrt6)^2=3$ would force either $u^2=3$ or $v^2=1/2$. Hence $3$ is not a square in $K'$. Both extensions therefore satisfy all the hypotheses of part (b).

Solved by gpt-5.6-sol high.

= d
{parent=18j}
{scope}

= Solution
{parent=d}

Let $E=\mathbb Q(\alpha,\beta)$. Since
$$
\sqrt3=\alpha^2-3,\qquad
\sqrt6=\alpha\beta,\qquad
\sqrt2=\frac{\sqrt6}{\sqrt3},
$$
part (b) gives $[E:\mathbb Q]=8$. The roots of $X^4-6X^2+6$ are $\mathord\pm\alpha,\mathord\pm\beta$, all in $E$, so $E$ is its splitting field and is Galois. Its action on the four roots is transitive; the order-eight entry in the list from part (a) is $D_8$. Therefore
$$
\operatorname{Gal}(E/\mathbb Q)\cong D_8.
$$

For an explicit <dihedral Galois action on four radical roots>, define
$$
r(\alpha)=\beta,\qquad r(\beta)=-\alpha,\qquad
s(\alpha)=\alpha,\qquad s(\beta)=-\beta.
$$
Then $r^4=s^2=1$ and $srs=r^{-1}$. The subgroup lattice is determined by
$$
\begin{array}{c}
D_8\\[1mm]
\langle r\rangle\qquad
\langle r^2,s\rangle\qquad
\langle r^2,rs\rangle\\[1mm]
\langle r^2\rangle\quad
\langle s\rangle\quad\langle r^2s\rangle\quad
\langle rs\rangle\quad\langle r^3s\rangle\\[1mm]
\{1\}.
\end{array}
$$
Here $\langle r^2\rangle$ lies in all three order-four subgroups; $\langle s\rangle,\langle r^2s\rangle$ lie in $\langle r^2,s\rangle$; and $\langle rs\rangle,\langle r^3s\rangle$ lie in $\langle r^2,rs\rangle$.

Reversing inclusions under the <Galois correspondence> gives the complete field lattice
$$
\begin{array}{c}
E\\[1mm]
\mathbb Q(\sqrt2,\sqrt3)\quad
\mathbb Q(\alpha)\quad\mathbb Q(\beta)\quad
\mathbb Q(\alpha+\beta)\quad\mathbb Q(\alpha-\beta)\\[1mm]
\mathbb Q(\sqrt2)\qquad
\mathbb Q(\sqrt3)\qquad
\mathbb Q(\sqrt6)\\[1mm]
\mathbb Q.
\end{array}
$$
The incidence is likewise reversed: $\mathbb Q(\sqrt2)$ lies in $\mathbb Q(\sqrt2,\sqrt3)$; $\mathbb Q(\sqrt3)$ lies in that field, $\mathbb Q(\alpha)$, and $\mathbb Q(\beta)$; and $\mathbb Q(\sqrt6)$ lies in that field, $\mathbb Q(\alpha+\beta)$, and $\mathbb Q(\alpha-\beta)$. The hint verifies the last two quartic fields through $\alpha+\beta=\sqrt2\gamma$ and $\alpha-\beta=\sqrt2\delta$.

Solved by gpt-5.6-sol high.

= 19J
{parent=Paper 3}
{scope}
{title2=Representation Theory}

= a
{parent=19j}
{scope}

= Solution
{parent=a}

Write
$$
g=\begin{pmatrix}a&b\\-\bar b&\bar a\end{pmatrix}\in SU_2,
\qquad |a|^2+|b|^2=1.
$$
The <homogeneous polynomial representation of SU2> is the symmetric-power action determined by
$$
g\cdot x=ax-\bar b y,\qquad
g\cdot y=bx+\bar a y,
$$
and extended multiplicatively to $V_n$.

In the ordered basis $(x^3,x^2y,xy^2,y^3)$, the matrix for $g$ is
$$
\begin{pmatrix}
a^3&a^2b&ab^2&b^3\\
-3a^2\bar b&a^2\bar a-2ab\bar b&2ab\bar a-b^2\bar b&3b^2\bar a\\
3a\bar b^2&b\bar b^2-2a\bar a\bar b&a\bar a^2-2b\bar a\bar b&3b\bar a^2\\
-\bar b^3&\bar a\bar b^2&-\bar b\bar a^2&\bar a^3
\end{pmatrix}.
$$

The character of a representation $(V,\rho)$ is $\chi_V(g)=\operatorname{tr}\rho(g)$. Every $g\in SU_2$ is conjugate to $\operatorname{diag}(z,z^{-1})$ with $|z|=1$, and hence the <character of the homogeneous polynomial representation of SU2> is
$$
\chi_n(g)=z^n+z^{n-2}+\cdots+z^{-n}.
$$
For $z=e^{i\theta}$ this is $\sin((n+1)\theta)/\sin\theta$, interpreted by continuity when $\sin\theta=0$.

Solved by gpt-5.6-sol high.

= b
{parent=19j}
{scope}

= Solution
{parent=b}

The <classification of finite-dimensional representations of SU2> says that every irreducible is one of the $V_n$, and compactness gives complete reducibility. The standard alternating form
$$
\omega((u_1,u_2),(v_1,v_2))=u_1v_2-u_2v_1
$$
is $SU_2$-invariant and identifies $V_1$ with $V_1^*$. Taking symmetric powers gives $V_n\cong V_n^*$ for every $n$. Therefore, if
$$
V\cong\bigoplus_n m_nV_n,
$$
then
$$
V^*\cong\bigoplus_n m_nV_n^*\cong V.
$$
This proves <self-duality of finite-dimensional SU2 representations>.

Solved by gpt-5.6-sol high.

= c
{parent=19j}
{scope}

= Solution
{parent=c}

For an irreducible $V$, <Schur lemma> makes the central element $-I$ act by a scalar $\varepsilon$. Since $(-I)^2=I$, $\varepsilon=\mathord\pm1$, and on $V\otimes V$ it acts by $\varepsilon^2=1$. Thus $\{\mathord\pm I\}$ lies in the kernel and the action factors through $SU_2/\{\mathord\pm I\}$, as in <central parity on SU2 tensor products>.

This fails for reducible representations. Take $V=V_0\oplus V_1$. On the summand $V_0\otimes V_1\subset V\otimes V$, the element $-I$ acts as $-1$, so the tensor-square action does not factor through the quotient.

Solved by gpt-5.6-sol high.

= d
{parent=19j}
{scope}

= Solution
{parent=d}

For $g$ conjugate to $\operatorname{diag}(z,z^{-1})$, multiplication of the weight sums gives
$$
\chi_n(g)^2
=\sum_{j=0}^n\chi_{2n-2j}(g).
$$
Indeed, the coefficient of $z^{2n-2m}$ on the left counts pairs $(r,s)$ with $0\leq r,s\leq n$ and $r+s=m$; the nested weight strings on the right give the same multiplicity. Characters determine finite-dimensional representations of the compact group $SU_2$, so
$$
\boxed{V_n\otimes V_n
\cong\bigoplus_{j=0}^nV_{2n-2j}}.
$$
This weight-counting argument proves the required case of the <Clebsch-Gordan decomposition for SU2> rather than merely quoting it.

Solved by gpt-5.6-sol high.

= e
{parent=19j}
{scope}

= Solution
{parent=e}

If the eigenvalues of $\rho(g)$ are $\lambda_1,\ldots,\lambda_d$, then those of $g$ on $\bigwedge^2V$ are $\lambda_i\lambda_j$ for $i<j$. Therefore the <Character of an exterior square> is
$$
\boxed{
\chi_{\wedge^2V}(g)
=\sum_{i<j}\lambda_i\lambda_j
=\frac12\left(\chi_V(g)^2-\chi_V(g^2)\right)}.
$$

Solved by gpt-5.6-sol high.

= f
{parent=19j}
{scope}

= Solution
{parent=f}

Let $v_k=x^{n-k}y^k$. In the summand $V_{2n-2j}$ from part (d), a highest-weight vector is
$$
w_j=\sum_{k=0}^j(-1)^k\binom jk,v_k\otimes v_{j-k}.
$$
The raising operator $x\partial_y$ annihilates this sum because consecutive terms cancel. Interchanging the tensor factors sends $w_j$ to $(-1)^jw_j$. Since each summand in part (d) has multiplicity one, the flip acts by that scalar on the whole summand. The alternating square therefore consists exactly of the odd-$j$ summands. This <flip parity in the SU2 tensor square> gives
$$
\boxed{
\bigwedge^2V_n
\cong\bigoplus_{k=0}^{\lfloor(n-1)/2\rfloor}
V_{2n-4k-2}}.
$$

Solved by gpt-5.6-sol high.

= 20F
{parent=Paper 3}
{scope}
{title2=Algebraic Topology}

= a
{parent=20f}
{scope}

= Solution
{parent=a}

One form of the <Seifert-van Kampen theorem> says that if $X=U\cup V$, with $U,V,U\cap V$ open and path-connected and containing the base point, then $\pi_1(X)$ is the pushout of
$$
\pi_1(U)\longleftarrow\pi_1(U\cap V)
\longrightarrow\pi_1(V).
$$

For the <fundamental group after attaching a 2-cell>, take $U$ to be $X$ together with a collar of the attached disc and $V$ to be a slightly enlarged open disc. Then $U\simeq X$, $V$ is simply connected, and $U\cap V\simeq S^1$; its generator maps to $[\alpha]$ in $\pi_1(X)$. Van Kampen therefore gives
$$
\pi_1(X\cup_\alpha D^2,x_0)
\cong\pi_1(X,x_0)/\langle\!\langle[\alpha]\rangle\!\rangle.
$$

Take $X=S^1_a\vee S^1_b$, whose fundamental group is the free group $F(a,b)$, and attach one 2-cell along the loop represented by $a^2b^{-3}$. The resulting connected <presentation complex> $Y$ satisfies
$$
\pi_1(Y,y_0)\cong
F(a,b)/\langle\!\langle a^2b^{-3}\rangle\!\rangle
=\langle a,b\mid a^2b^{-3}\rangle.
$$

Solved by gpt-5.6-sol high.

= b
{parent=20f}
{scope}

= Solution
{parent=b}

A <covering space> is a map $p:\widetilde X\to X$ such that every $x\in X$ has an open neighbourhood $U$ for which
$$
p^{-1}(U)=\bigsqcup_{\lambda}U_\lambda,
$$
with every restriction $p:U_\lambda\to U$ a homeomorphism.

Let $N=\langle\!\langle a\rangle\!\rangle$ in
$\pi_1(Y)=\langle a,b\mid a^2b^{-3}\rangle$. Then
$$
\pi_1(Y)/N\cong\langle b\mid b^3=1\rangle\cong C_3.
$$
The subgroup is normal of index three, so the <degree of a connected covering> shows that every fibre $p^{-1}(y)$ has exactly three points.

The <cell complex of a covering from a coset graph> gives an explicit model. Take vertices $v_0,v_1,v_2$ with indices modulo three; put an $a$-loop $A_i$ at every $v_i$, and a directed $b$-edge $B_i:v_i\to v_{i+1}$. Attach a 2-cell at each $v_i$ along
$$
A_i^2B_{i-1}^{-1}B_{i-2}^{-1}B_i^{-1}.
$$
Map every $v_i$ to $y_0$, every $A_i$ to the $a$-cell, every $B_i$ to the $b$-cell, and each 2-cell homeomorphically to the unique 2-cell of $Y$. Its attaching word is the lift of $a^2b^{-3}$ beginning at $v_i$. This defines the required connected three-sheeted covering.

Solved by gpt-5.6-sol high.

= 21I
{parent=Paper 3}
{scope}
{title2=Linear Analysis}

= a
{parent=21i}
{scope}

= Solution
{parent=a}

The <Arzela-Ascoli theorem> in sequential form states: if $K$ is a compact metric space and $(f_n)$ is a uniformly bounded, equicontinuous sequence in $C(K)$, then it has a uniformly convergent subsequence. Equivalently, a subset of $C(K)$ is relatively compact in the uniform norm exactly when it is uniformly bounded and equicontinuous.

For sufficiency, choose a finite $1/m$-net $E_m\subset K$ for every $m$. The union $D=\bigcup_mE_m$ is countable and dense. Uniform boundedness makes each scalar sequence $(f_n(x))_n$, $x\in D$, bounded. Successive applications of Bolzano-Weierstrass followed by a diagonal choice give a subsequence $(g_n)$ for which $g_n(x)$ converges at every $x\in D$.

Given $\varepsilon>0$, equicontinuity supplies $\delta>0$ such that
$$
d(x,y)<\delta\quad\Longrightarrow\quad
|g_n(x)-g_n(y)|<\varepsilon/3
$$
for every $n$. Choose a finite $\delta$-net from $D$. Pointwise convergence on those finitely many points makes $(g_n)$ uniformly Cauchy on the net, and the two equicontinuity estimates make it uniformly Cauchy on all of $K$. Since $C(K)$ is complete, $g_n$ converges uniformly.

Conversely, a relatively compact family is uniformly bounded because its closure is compact in the normed space $C(K)$. Given $\varepsilon>0$, cover that compact closure by finitely many uniform balls of radius $\varepsilon/3$, centred at continuous functions $h_1,\ldots,h_r$. Uniform continuity of the finitely many $h_i$ gives one $\delta$ that works for all of them. Approximating any family member by one $h_i$ proves equicontinuity.

Solved by gpt-5.6-sol high.

= b
{parent=21i}
{scope}

= Solution
{parent=b}

Let the common bound be $M$. The derivative bound and the mean-value theorem give
$$
|f_n(x)-f_n(y)|\leq M|x-y|,
$$
so on every compact interval the sequence is uniformly bounded and equicontinuous. Apply the <Arzela-Ascoli theorem> first on $[-1,1]$, then to a subsequence on $[-2,2]$, and so on. The <diagonal subsequence for locally uniform convergence> gives a strictly increasing $\phi$ such that $f_{\phi(n)}$ converges uniformly on every $[-R,R]$ to a function $f$. The limit is continuous and satisfies $|f(x)|\leq M$.

Global uniform convergence need not follow. Choose a nonzero continuously differentiable bump function $\psi$ supported in $[-1,1]$ and set
$$
f_n(x)=\psi(x-n).
$$
The functions and their derivatives have a common bound. On every fixed compact interval, $f_n$ is eventually zero, so every subsequence converges locally uniformly to $f=0$. Nevertheless
$$
\sup_{x\in\mathbb R}|f_n(x)-f(x)|
=\|\psi\|_\infty>0
$$
for every $n$. Thus the proposed global conclusion is false.

Solved by gpt-5.6-sol high.

= 22H
{parent=Paper 3}
{scope}
{title2=Analysis of Functions}

= a
{parent=22h}
{scope}

= Solution
{parent=a}

If $f=g$ almost everywhere, then $|f|=|g|$ almost everywhere, so the integral defining $\|[f]\|_1$ is independent of the representative. Homogeneity and the triangle inequality follow from the corresponding pointwise properties of absolute value. Moreover,
$$
\|[f]\|_1=0
\quad\Longleftrightarrow\quad f=0\text{ almost everywhere}
\quad\Longleftrightarrow\quad [f]=[0],
$$
so this is a norm on the <Lebesgue space> of almost-everywhere equivalence classes.

The <Riesz-Fischer theorem> states that $L^p$ is complete for $1\leq p\leq\infty$. Applied with $p=1$, it says that every $\|\cdot\|_1$-Cauchy sequence of these equivalence classes converges in $\mathcal L^1(\mathbb R^n)$. Thus $\mathcal L^1(\mathbb R^n)$ is a Banach space.

Solved by gpt-5.6-sol high.

= b
{parent=22h}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

Consider the inclusion map
$$
J:(X,\|\cdot\|_X)\longrightarrow
(\mathcal L^1(\mathbb R^n),\|\cdot\|_1).
$$
Its graph is closed. Indeed, suppose $x_k\to x$ in $X$ and $Jx_k\to y$ in $L^1$. By hypothesis, some subsequence of $(x_k)$ converges almost everywhere to $x$. From its $L^1$ convergence, that subsequence has a further subsequence converging almost everywhere to $y$. Uniqueness of almost-everywhere limits gives $x=y$ as $L^1$ classes.

Both spaces are Banach, so the <closed graph theorem> makes $J$ bounded. Therefore some $C>0$ satisfies
$$
\boxed{\|x\|_1\leq C\|x\|_X\qquad(x\in X)}.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

No. Take
$$
X=L^1(\mathbb R^n)\cap L^2(\mathbb R^n),
\qquad
\|f\|_X=\|f\|_1+\|f\|_2.
$$
This is a Banach space, and $X$-convergence implies $L^2$ convergence, which has an almost-everywhere convergent subsequence. Thus the hypotheses hold.

However, let
$$
f(x)=|x|^{-3n/4}1_{\{|x|<1\}}.
$$
The <power singularity integrability> test gives $f\in L^1$ but $f\notin L^2$. Its truncations $f_m=\min(f,m)$ belong to $X$ and converge to $f$ in $L^1$ by dominated convergence. Hence $(f_m)$ is $L^1$-Cauchy in $X$ but has no $L^1$ limit belonging to $X$. Therefore $X$ need not be complete in the $L^1$ norm.

Solved by gpt-5.6-sol high.

= 23G
{parent=Paper 3}
{scope}
{title2=Riemann Surfaces}

= Solution
{parent=23G}

Choose an evenly covered neighbourhood $U\subset R$ and a component $V$ of $\pi^{-1}(U)$. If $z:U\to\mathbb C$ is a complex chart, use
$$
z\circ\pi|_V:V\longrightarrow\mathbb C
$$
as a chart on $S$. On overlaps, the transition maps are exactly transition maps between charts of $R$, hence are holomorphic. These charts give the unique <complex structure lifted through a covering map> for which $\pi$ is a local biholomorphism, and therefore analytic. The Hausdorff assumption is already given; connectedness and the second-countability of a Riemann surface ensure the resulting covering surface has the required manifold properties.

A surface is simply connected when it is path-connected and every loop is null-homotopic, equivalently when its fundamental group is trivial. The <uniformization theorem> says that every simply connected Riemann surface is biholomorphic to exactly one of
$$
\mathbb C_\infty,\qquad \mathbb C,\qquad \mathbb D.
$$
Their analytic automorphism groups are
$$
\begin{aligned}
\operatorname{Aut}(\mathbb C_\infty)
&=\left\{z\mapsto\frac{az+b}{cz+d}:ad-bc\ne0\right\}/\mathbb C^*,\\
\operatorname{Aut}(\mathbb C)
&=\{z\mapsto az+b:a\ne0\},\\
\operatorname{Aut}(\mathbb D)
&=\left\{z\mapsto e^{i\theta}
\frac{z-a}{1-\bar a z}:|a|<1\right\}.
\end{aligned}
$$

A <covering space action> of $G$ on $X$ is an action by homeomorphisms such that every $x\in X$ has a neighbourhood $U$ satisfying
$$
gU\cap U=\varnothing\qquad(g\ne1).
$$
In particular the action is free. Every nonidentity Möbius transformation of $\mathbb C_\infty$ has a fixed point: solving $z=(az+b)/(cz+d)$ gives a root on the sphere. Consequently a subgroup $H\leq\operatorname{Aut}(\mathbb C_\infty)$ acting as a covering space action must be trivial. Its quotient is the sphere itself and is Hausdorff.

The Hausdorff conclusion fails on other Riemann surfaces. On
$R=\mathbb C^*\cong\mathbb R^2\setminus\{0\}$, let $G=\langle T\rangle$ with
$$
T(x,y)=(2x,y/2).
$$
Every orbit is discrete in $R$ and has trivial stabilizer; small enough neighbourhoods have disjoint nontrivial translates, so this is a covering space action. Yet the two orbits through
$$
p=(1,0),\qquad q=(0,1)
$$
cannot be separated in the quotient. Indeed, for
$$
z_n=(2^{-n},1),\qquad
T^nz_n=(1,2^{-n}),
$$
we have $z_n\to q$ and $T^nz_n\to p$. Thus every pair of quotient neighbourhoods of the two distinct orbits intersects, proving that $R/G$ is not Hausdorff.

Simple connectedness does not repair this. Lift $T$ to the universal cover $\widetilde R\cong\mathbb C$ of $\mathbb C^*$. Choose the lift preserving the angular interval from $0$ to $\pi/2$. The lifted points of $z_n$ still converge to a lift of $q$, while their $n$th translates converge to a lift of $p$; those two lifts belong to different orbits. The lifted action is again a covering space action, and its quotient is non-Hausdorff although $\widetilde R$ is simply connected.

Solved by gpt-5.6-sol high.

= 24J
{parent=Paper 3}
{scope}
{title2=Algebraic Geometry}

= Solution
{parent=24J}

Let
$$
A_V=k[x_0,\ldots,x_n]/I(V)
$$
be the homogeneous coordinate ring. Its Hilbert function agrees for all sufficiently large $m$ with a polynomial $P_V(m)$. Since $V$ is a curve,
$$
P_V(m)=dm+c.
$$
The <degree of a projective curve> is the positive integer $d$. This is well-defined because the graded ring $A_V$, and hence its eventual <Hilbert polynomial>, depends only on the embedded projective variety, while a polynomial agreeing with the Hilbert function for all large integers is unique.

This definition also has the geometric interpretation expected of degree. For a sufficiently general hyperplane $H=Z(\ell)$ that does not contain $V$, multiplication by $\ell$ gives an exact sequence
$$
0\longrightarrow A_V(-1)
\xrightarrow{\ \ell\ }A_V
\longrightarrow A_V/(\ell)\longrightarrow0.
$$
Consequently, for large $m$,
$$
\dim_k(A_V/(\ell))_m
=P_V(m)-P_V(m-1)=d.
$$
The left side is the length of the zero-dimensional scheme $V\cap H$. Thus a generic hyperplane meets $V$ in $d$ points counted with intersection multiplicity, independently of the chosen generic hyperplane.

For the stated linear embedding, the homogeneous coordinate ring of $\varphi(V)$ is
$$
\frac{k[x_0,\ldots,x_m]}
{I(V)+(x_{n+1},\ldots,x_m)}
\cong
\frac{k[x_0,\ldots,x_n]}{I(V)}.
$$
The graded rings have the same Hilbert polynomial, so <degree under a linear projective embedding> gives
$$
\deg_{\mathbb P^m}\varphi(V)=\deg_{\mathbb P^n}V.
$$

Degree is nevertheless not an invariant of an abstract curve. In $\mathbb P^2$, a line has degree one and the smooth conic
$$
x_0x_2-x_1^2=0
$$
has degree two, but the map
$$
[s:t]\longmapsto[s^2:st:t^2]
$$
identifies the conic with $\mathbb P^1$, as is any projective line.

For the specified set $S$, the equations are the $2\times2$ minors of
$$
\begin{pmatrix}
x_0&x_1&x_2\\
x_1&x_2&x_3
\end{pmatrix}.
$$
On the affine chart $x_0=1$, they force
$$
x_2=x_1^2,\qquad x_3=x_1^3,
$$
so this chart is an affine line. When $x_0=0$, the equations force $x_1=x_2=0$, leaving the single point $[0:0:0:1]$. Hence $V$ is the <twisted cubic>, parametrized by
$$
[s:t]\longmapsto[s^3:s^2t:st^2:t^3].
$$
It is therefore an irreducible projective curve. A generic hyperplane pulls back to a binary cubic
$$
a_0s^3+a_1s^2t+a_2st^2+a_3t^3,
$$
which has three zeros on $\mathbb P^1$ counted with multiplicity. Thus
$$
\boxed{\deg V=3}.
$$

No nonzero linear form vanishes on $V$, since substituting its parametrization would make the four coefficients of the displayed binary cubic vanish. Therefore every hypersurface containing $V$ has degree at least two. If
$$
V=Z(F_1,\ldots,F_r),
$$
then $r=1$ is impossible because a nonzero hypersurface in $\mathbb P^3$ has dimension two. Hence at least two nonconstant $F_i$ are needed, and
$$
\prod_i\deg Z(F_i)\geq2\cdot2=4>3=\deg V.
$$
This is the <degree obstruction to the twisted cubic being a complete intersection>.

Finally compare the twisted cubic with a smooth plane cubic, such as
$$
E=Z(x_0^3+x_1^3+x_2^3)\subset\mathbb P^2,
$$
viewed in $\mathbb P^3$ by the given linear embedding. Both curves are irreducible and have degree three. The twisted cubic is isomorphic to $\mathbb P^1$ and has genus zero, whereas the supplied plane-curve genus formula gives
$$
g(E)=\frac{(3-1)(3-2)}2=1.
$$
Since genus is an isomorphism invariant, these equal-degree curves are not isomorphic.

Solved by gpt-5.6-sol high.

= 25I
{parent=Paper 3}
{scope}
{title2=Differential Geometry}

= a
{parent=25i}
{scope}

= Solution
{parent=a}

For a local parametrization $X(u^1,u^2)$, the <first fundamental form> is the Euclidean inner product restricted to each tangent plane:
$$
I=g_{ij}\,du^i\,du^j,\qquad
g_{ij}=X_i\cdot X_j.
$$
Writing $u^1=u,u^2=v$, this is
$$
I=E\,du^2+2F\,du\,dv+G\,dv^2,
\qquad
E=X_u^2,\quad F=X_u\cdot X_v,\quad G=X_v^2.
$$

After choosing an orientation, the <Gauss map> is
$$
N=\frac{X_u\times X_v}{|X_u\times X_v|}.
$$
The <second fundamental form> is $II(v,w)=\langle S(v),w\rangle$, where the shape operator is $S=-dN$. In coordinates,
$$
II=e\,du^2+2f\,du\,dv+g\,dv^2,
\qquad
e=N\cdot X_{uu},\quad
f=N\cdot X_{uv},\quad
g=N\cdot X_{vv}.
$$
The <Gaussian curvature> and <mean curvature> are respectively the determinant and half the trace of $S$:
$$
K=\frac{eg-f^2}{EG-F^2},
\qquad
H=\frac{eG-2fF+gE}{2(EG-F^2)}.
$$
Changing the chosen normal reverses $II$ and $H$ but leaves $K$ unchanged.

Solved by gpt-5.6-sol high.

= b
{parent=25i}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

Because the profile is parametrized by arc length,
$$
Y'^2+Z'^2=1.
$$
The tangent vectors are
$$
\phi_u=(Y'\cos v,Y'\sin v,Z'),
\qquad
\phi_v=(-Y\sin v,Y\cos v,0),
$$
so the <first fundamental form of an arc-length surface of revolution> is
$$
I=du^2+Y(u)^2\,dv^2.
$$
With the orientation determined by $\phi_u\times\phi_v$, the Gauss map is
$$
N=(-Z'\cos v,-Z'\sin v,Y').
$$
The second derivatives give
$$
e=N\cdot\phi_{uu}=Y'Z''-Z'Y'',\qquad
f=0,\qquad
g=N\cdot\phi_{vv}=YZ',
$$
and hence
$$
II=(Y'Z''-Z'Y'')\,du^2+YZ'\,dv^2.
$$

Differentiating $Y'^2+Z'^2=1$ yields $Y'Y''+Z'Z''=0$, from which
$$
(Y'Z''-Z'Y'')Z'=-Y''.
$$
Therefore
$$
\boxed{K=\frac{eg}{EG}=-\frac{Y''}{Y}},
$$
while
$$
\boxed{
H=\frac12\left(Y'Z''-Z'Y''+\frac{Z'}Y\right)}
$$
for the chosen orientation.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

A unit-speed surface curve $\alpha$ is a <geodesic> when its acceleration is normal to the surface, equivalently when its covariant acceleration vanishes. In coordinates $u^1,u^2$, the <Christoffel symbol>s of the Levi-Civita connection are
$$
\Gamma^k_{ij}
=\frac12g^{k\ell}
\left(\partial_i g_{j\ell}
+\partial_j g_{i\ell}
-\partial_\ell g_{ij}\right).
$$
The coordinate functions of a geodesic satisfy the <geodesic equation>
$$
\boxed{\ddot u^k+\Gamma^k_{ij}\dot u^i\dot u^j=0}.
$$

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

For the metric
$$
du^2+Y(u)^2\,dv^2,
$$
the Christoffel symbols needed in the $v$ equation are
$$
\Gamma^v_{uv}=\Gamma^v_{vu}=\frac{Y'}Y.
$$
Thus a geodesic obeys
$$
\ddot v+2\frac{Y'}Y\dot u\dot v=0.
$$
Multiplication by $Y^2$ turns this into
$$
\frac d{dt}\left(Y(u(t))^2\dot v(t)\right)=0.
$$
Consequently the <Clairaut first integral for a surface of revolution> is
$$
\boxed{Y(u)^2\dot v=\text{constant}}.
$$
It is also the conserved momentum associated with the cyclic coordinate $v$.

Solved by gpt-5.6-sol high.

= 26H
{parent=Paper 3}
{scope}
{title2=Probability and Measure}

= a
{parent=26h}
{scope}

= Solution
{parent=a}

Put $Z_n=|X_n-X|$. If $Z_n\to0$ in probability, then for $0<\varepsilon<1$,
$$
\mathbb E(Z_n\wedge1)
\leq\varepsilon+\mathbb P(Z_n>\varepsilon).
$$
Taking the upper limit and then letting $\varepsilon\downarrow0$ proves that the expectations tend to zero. Conversely,
$$
\mathbb P(Z_n>\varepsilon)
\leq\frac{\mathbb E(Z_n\wedge1)}{\varepsilon\wedge1},
$$
so expectation convergence implies convergence in probability. This proves the <bounded-metric characterization of convergence in probability>.

For the subsequence assertion, choose $n_k$ so that
$$
\mathbb P(|X_{n_k}-X|>2^{-k})<2^{-k}.
$$
The first <Borel-Cantelli lemmas> then says that only finitely many of these events occur almost surely. Hence
$$
|X_{n_k}-X|\leq2^{-k}
$$
eventually almost surely, and $X_{n_k}\to X$ almost surely. This is the <almost-sure subsequence from convergence in probability>.

Solved by gpt-5.6-sol high.

= b
{parent=26h}
{scope}

= Solution
{parent=b}

Here
$$
\mathbb P(|X_n|>\varepsilon)=\frac1n\longrightarrow0
$$
for every $0<\varepsilon<1$, so $X_n\to0$ in probability. On the other hand, the events $A_n=\{X_n=1\}$ are independent and
$$
\sum_{n=1}^\infty\mathbb P(A_n)
=\sum_{n=1}^\infty\frac1n=\infty.
$$
The second <Borel-Cantelli lemmas> gives
$$
\mathbb P(X_n=1\text{ infinitely often})=1.
$$
Thus $X_n$ cannot converge almost surely to zero. This <independent rare-event counterexample to almost-sure convergence> proves that convergence in probability need not imply almost-sure convergence of the full sequence.

Solved by gpt-5.6-sol high.

= c
{parent=26h}
{scope}

= Solution
{parent=c}

The conclusion printed as convergence in probability is already an assumption; the substantive conclusion is convergence in $L^p$, and we prove that stronger statement.

First, the almost-sure subsequence result from part (a) and Fatou's lemma show that $X\in L^r$:
$$
\mathbb E|X|^r
\leq\liminf_k\mathbb E|X_{n_k}|^r<\infty.
$$
Consequently
$$
\sup_n\mathbb E|X_n-X|^r<\infty.
$$
Fix $1\leq p<r$ and $\delta>0$. Splitting according to
$|X_n-X|\leq\delta$ and applying Hölder's inequality on the complement gives
$$
\begin{aligned}
\mathbb E|X_n-X|^p
&\leq\delta^p+
\mathbb E\!\left[|X_n-X|^p
1_{\{|X_n-X|>\delta\}}\right]\\
&\leq\delta^p+
\big(\mathbb E|X_n-X|^r\big)^{p/r}
\mathbb P(|X_n-X|>\delta)^{1-p/r}.
\end{aligned}
$$
The second term tends to zero because of convergence in probability and the uniform $L^r$ bound. Taking the upper limit and then $\delta\downarrow0$ proves
$$
\boxed{\|X_n-X\|_p\longrightarrow0}.
$$
This is <convergence in Lp from convergence in probability and an Lr bound>.

Solved by gpt-5.6-sol high.

= 27L
{parent=Paper 3}
{scope}
{title2=Applied Probability}

= a
{parent=27l}
{scope}

= Solution
{parent=a}

Put
$$
S_0=0,\qquad S_n=\sum_{i=1}^n\xi_i,
\qquad
N_t=\max\{n\geq0:S_n\leq t\}.
$$
This is the <renewal process> generated by the interarrival times. Since the variables have a density and positive finite mean, $S_n\to\infty$ almost surely, so $N_t$ is finite and tends to infinity with $t$. The strong law gives
$$
\frac{S_n}{n}\longrightarrow\frac1\lambda
\qquad\hbox{almost surely}.
$$
On this probability-one event,
$$
S_{N_t}\leq t<S_{N_t+1}.
$$
Dividing by $N_t$ and using $N_t\to\infty$ gives
$$
\frac{S_{N_t}}{N_t}
\leq\frac{t}{N_t}
<\frac{N_t+1}{N_t}\frac{S_{N_t+1}}{N_t+1}.
$$
Both outside expressions tend to $1/\lambda$, so
$$
\boxed{\frac{N_t}{t}\longrightarrow\lambda\quad\hbox{almost surely}.}
$$
This is the <renewal counting law of large numbers>.

Solved by gpt-5.6-sol high.

= b
{parent=27l}
{scope}

= Solution
{parent=b}

The <size-biased distribution> associated with $\xi_1$ is defined by
$$
\mathbb E h(\widehat\xi_1)
=\frac{\mathbb E[\xi_1h(\xi_1)]}{\mathbb E\xi_1}
=\lambda\mathbb E[\xi_1h(\xi_1)]
$$
for every bounded measurable $h$. Thus, if $\xi_1$ has density $f$, then $\widehat\xi_1$ has density $\lambda x f(x)$.

Now suppose that $\xi_1$ is exponential with rate $\lambda$. The renewal epochs form a <Poisson process>. Write
$$
A_t=t-S_{N_t},\qquad R_t=S_{N_t+1}-t,
$$
so that $L(t)=A_t+R_t$. Independent stationary increments and the exponential waiting-time law show that $R_t$ is exponential with rate $\lambda$, independent of the history up to $t$ and hence of $A_t$. For fixed $a\geq0$ and $t>a$,
$$
\mathbb P(A_t>a)
=\mathbb P(N_t-N_{t-a}=0)
=e^{-\lambda a}.
$$
Consequently $A_t$ converges in distribution to another rate-$\lambda$ exponential variable, independent of $R_t$. Therefore $L(t)$ converges to the sum of two independent rate-$\lambda$ exponentials, whose density is
$$
\int_0^x\lambda e^{-\lambda y}\lambda e^{-\lambda(x-y)}\,dy
=\lambda^2xe^{-\lambda x}.
$$
The size-biased density of $\xi_1$ is likewise
$$
\lambda x\big(\lambda e^{-\lambda x}\big)
=\lambda^2xe^{-\lambda x}.
$$
Hence the <exponential renewal interval limit> is
$$
\boxed{L(t)\xrightarrow{d}\widehat\xi_1}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=27l}
{scope}

= Solution
{parent=c}

Let $F$ be the distribution function of $\xi_1$, write $\overline F=1-F$, and fix $x>0$. Set
$$
q(t)=\mathbb P(L(t)\geq x).
$$
Conditioning on the first interarrival gives
$$
q(t)=\overline F(\max\{x,t\})
+\int_{[0,t]}q(t-y)\,dF(y).
$$
Indeed, if $\xi_1>t$, the first interval contains $t$ and must also have length at least $x$; if $\xi_1\leq t$, the process starts afresh at time $\xi_1$.

Put $h=\overline F(x)$ and $d(t)=q(t)-h$. Then
$$
d(t)=g(t)+\int_{[0,t]}d(t-y)\,dF(y),
$$
where
$$
g(t)=
\begin{cases}
hF(t),&t\leq x,\\
\overline F(t)F(x),&t>x.
\end{cases}
$$
In particular $g\geq0$. Iterating the renewal equation yields
$$
d(t)=\sum_{n=0}^{m}(g*dF^{*n})(t)
+(d*dF^{*(m+1)})(t).
$$
For fixed $t$, the last term tends to zero in absolute value because $d$ is bounded and
$$
F^{*(m+1)}(t)=\mathbb P(S_{m+1}\leq t)\longrightarrow0.
$$
It follows that
$$
d(t)=\sum_{n=0}^{\infty}(g*dF^{*n})(t)\geq0.
$$
Thus the <stochastic length bias of a renewal interval> gives
$$
\boxed{\mathbb P(L(t)\geq x)\geq\mathbb P(\xi_1\geq x)}
$$
for every $x,t>0$.

Solved by gpt-5.6-sol high.

= 28L
{parent=Paper 3}
{scope}
{title2=Principles of Statistics}

= a
{parent=28l}
{scope}

= Solution
{parent=a}

For an observed value $X=x$, the likelihood is
$$
p(x\mid\theta)=\frac{e^{-\theta}\theta^x}{x!},
$$
while the gamma prior with shape $\alpha$ and rate $\lambda$ has density
$$
\pi(\theta)=\frac{\lambda^\alpha}{\Gamma(\alpha)}
\theta^{\alpha-1}e^{-\lambda\theta},
\qquad \theta>0.
$$
The posterior density is therefore proportional to
$$
p(x\mid\theta)\pi(\theta)
\propto
\theta^{\alpha+x-1}e^{-(\lambda+1)\theta}.
$$
Normalizing this gamma kernel proves the <Poisson-gamma conjugacy> formula
$$
\boxed{\theta\mid X=x\sim
\operatorname{Gamma}(\alpha+x,\lambda+1)}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=28l}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

The <risk of a decision rule> is its expected loss at a fixed parameter value:
$$
R(\theta,\delta)
=\mathbb E_\theta\!\left[L(\delta(X),\theta)\right].
$$
For a prior $\pi$, a <Bayes estimator> minimizes the integrated risk
$$
r(\pi,\delta)
=\int_\Theta R(\theta,\delta)\,\pi(d\theta),
$$
or equivalently minimizes posterior expected loss for almost every observation. A <minimax decision rule> minimizes
$$
\sup_{\theta\in\Theta}R(\theta,\delta)
$$
over all decision rules.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

Let $\delta'$ be any decision rule. For every $j$,
$$
\sup_{\theta\in\Theta}R(\theta,\delta')
\geq
\int_\Theta R(\theta,\delta')\,\pi_j(d\theta)
\geq r_j,
$$
because $r_j$ is the infimum of the integrated risk under $\pi_j$. Taking the lower limit gives
$$
\sup_\theta R(\theta,\delta')\geq
\lim_{j\to\infty}r_j=r.
$$
The given rule $\delta$ has constant risk $r$, and hence worst-case risk exactly $r$. It attains this universal lower bound and is therefore minimax. This is the <constant-risk limit-of-Bayes-risk criterion>.

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

Given $X=x$, the <posterior expected loss> from reporting $d$ is
$$
\begin{aligned}
\mathbb E\!\left[\theta^{-1}(\theta-d)^2\mid X=x\right]
&=\mathbb E(\theta\mid X=x)-2d
+d^2\mathbb E(\theta^{-1}\mid X=x).
\end{aligned}
$$
This is a strictly convex quadratic in $d$ whenever the conditional inverse moment is finite and positive. Differentiating with respect to $d$ gives
$$
-2+2d\,\mathbb E(\theta^{-1}\mid X=x)=0,
$$
so the <Bayes estimator under reciprocal weighted quadratic loss> is
$$
\boxed{\delta_\pi(x)
=\big(\mathbb E(\theta^{-1}\mid X=x)\big)^{-1}}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=28l}
{scope}

= Solution
{parent=c}

For $\delta(X)=X$ and $X\sim\operatorname{Pois}(\theta)$,
$$
R(\theta,\delta)
=\frac1\theta\mathbb E_\theta[(X-\theta)^2]
=\frac{\operatorname{Var}_\theta(X)}{\theta}
=1.
$$
Thus this rule has constant risk one.

To show that no rule has smaller worst-case risk, fix any $\alpha>1$ and let
$$
\pi_j=\operatorname{Gamma}(\alpha,\lambda_j),
\qquad \lambda_j\downarrow0.
$$
Part (a) gives
$$
\theta\mid X=x\sim
\operatorname{Gamma}(\alpha+x,\lambda_j+1).
$$
Since the reciprocal moment of a gamma variable with shape $k>1$ and rate $s$ is $s/(k-1)$, part (b)(iii) gives the Bayes rule
$$
\delta_j(x)=\frac{\alpha+x-1}{\lambda_j+1}.
$$
Its minimum posterior expected loss is
$$
\mathbb E(\theta\mid X=x)
-\big(\mathbb E(\theta^{-1}\mid X=x)\big)^{-1}
=\frac{\alpha+x}{\lambda_j+1}
-\frac{\alpha+x-1}{\lambda_j+1}
=\frac1{\lambda_j+1}.
$$
Hence the Bayes risk is
$$
r_j=\frac1{\lambda_j+1}\longrightarrow1.
$$
The criterion from part (b)(ii) now proves
$$
\boxed{\delta(X)=X\text{ is minimax}.}
$$

Solved by gpt-5.6-sol high.

= 29K
{parent=Paper 3}
{scope}
{title2=Stochastic Financial Models}

= a
{parent=29k}
{scope}

= Solution
{parent=a}

A <european call option> with maturity $N$ and strike $K$ pays
$$
(S_N-K)^+
$$
at time $N$. Let $q$ be the risk-neutral conditional probability of the factor $1+b$. The discounted stock must be a martingale, so
$$
q(1+b)+(1-q)(1+a)=1+r,
$$
and therefore
$$
q=\frac{r-a}{b-a},
\qquad
1-q=\frac{b-r}{b-a}.
$$
Both probabilities are strictly positive. At every node these are the unique probabilities satisfying the martingale condition, so the <fundamental theorem of asset pricing> gives a unique no-arbitrage price.

If exactly $n$ of the $N$ moves use the factor $1+a$, then
$$
S_N=S_0(1+a)^n(1+b)^{N-n},
$$
and there are $\binom Nn$ such paths. Discounted risk-neutral expectation now gives
$$
\operatorname{EC}(N,K)
=\sum_{n=0}^N w(n,N)
\left(S_0(1+a)^n(1+b)^{N-n}-K\right)^+,
$$
where
$$
\boxed{
w(n,N)
=\frac1{(1+r)^N}\binom Nn
\left(\frac{b-r}{b-a}\right)^n
\left(\frac{r-a}{b-a}\right)^{N-n}}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=29k}
{scope}

= Solution
{parent=b}

The terminal payoffs satisfy the pointwise identity
$$
(K-S_N)^+-(S_N-K)^+=K-S_N.
$$
Pricing both sides by discounted risk-neutral expectation gives
$$
\operatorname{EP}(N,K)-\operatorname{EC}(N,K)
=(1+r)^{-N}K-S_0,
$$
because the discounted stock is a martingale. Thus <put-call parity> is
$$
\boxed{
\operatorname{EP}(N,K)
=(1+r)^{-N}K-S_0+\operatorname{EC}(N,K)}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=29k}
{scope}

= Solution
{parent=c}

Condition on the first risk-neutral move. After a factor $1+a$, homogeneity of the call payoff gives
$$
\left((1+a)S_N-K\right)^+
=(1+a)\left(S_N-\frac K{1+a}\right)^+,
$$
and the analogous identity holds for $1+b$. Therefore the <binomial call-price recursion> is
$$
\operatorname{EC}(N+1,K)
=u\,\operatorname{EC}\left(N,\frac K{1+a}\right)
+v\,\operatorname{EC}\left(N,\frac K{1+b}\right),
$$
with
$$
\boxed{
u=\frac{(1+a)(b-r)}{(1+r)(b-a)},
\qquad
v=\frac{(1+b)(r-a)}{(1+r)(b-a)}}.
$$
The inequalities $a<r<b$ and $a>-1$ make both coefficients positive.

Solved by gpt-5.6-sol high.

= d
{parent=29k}
{scope}

= Solution
{parent=d}

Write
$$
R_{M,N}=\frac{S_N}{S_M}
$$
for the product of the last $N-M$ stock factors. The forward-start payoff is
$$
(S_N-\lambda S_M)^+
=S_M(R_{M,N}-\lambda)^+.
$$
Under the risk-neutral measure the binomial factors are independent with the same probability $q$ at every node. Hence $S_M$ and $R_{M,N}$ are independent, while
$$
\mathbb E_QS_M=S_0(1+r)^M.
$$
The <forward-start call option> consequently has price
$$
\begin{aligned}
\operatorname{FSC}(M,N,\lambda)
&=(1+r)^{-N}\mathbb E_Q
\left[S_M(R_{M,N}-\lambda)^+\right]\\
&=S_0(1+r)^{-(N-M)}
\mathbb E_Q\left[(R_{M,N}-\lambda)^+\right].
\end{aligned}
$$
An ordinary $(N-M)$-period call with strike $K$ has price
$$
\operatorname{EC}(N-M,K)
=S_0(1+r)^{-(N-M)}
\mathbb E_Q\left[\left(R_{M,N}-\frac K{S_0}\right)^+\right].
$$
The required equality therefore holds for
$$
\boxed{K=\lambda S_0}.
$$

Solved by gpt-5.6-sol high.

= 30D
{parent=Paper 3}
{scope}
{title2=Asymptotic Methods}

= a
{parent=30d}
{scope}

= Solution
{parent=a}

Write the contour near the simple saddle as
$$
z=z_0+e^{i\alpha}s,\qquad s\in\mathbb R,
$$
with its given orientation. Since $\phi'(z_0)=0$,
$$
\phi(z)
=\phi(z_0)+\frac12\phi''(z_0)e^{2i\alpha}s^2
+O(s^3).
$$
Along a steepest-descent tangent the quadratic coefficient is real and negative, so
$$
\phi''(z_0)e^{2i\alpha}=-|\phi''(z_0)|,
\qquad
\arg\phi''(z_0)+2\alpha=\pi\pmod{2\pi}.
$$
Moreover $dz=e^{i\alpha}ds$. Only a neighbourhood of width $O(x^{-1/2})$ contributes at leading order, and there $f(z)=f(z_0)+o(1)$. The local integral is consequently
$$
f(z_0)e^{x\phi(z_0)+i\alpha}
\int_{-\infty}^{\infty}
\exp\left(-\frac{x|\phi''(z_0)|}{2}s^2\right)\,ds.
$$
Evaluating the Gaussian proves the <simple-saddle contribution in steepest descent>:
$$
\boxed{
I(x)\sim
f(z_0)\sqrt{\frac{2\pi}{x|\phi''(z_0)|}}\,
e^{x\phi(z_0)+i\alpha}}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=30d}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

Factor the integrand in the <Schläfli contour integral for Legendre polynomials> as
$$
\frac{(z^2-1)^n}{(z-t)^{n+1}}
=\frac1{z-t}
\left(\frac{z^2-1}{z-t}\right)^n.
$$
Thus one may take
$$
f(z,t)=\frac1{z-t},
\qquad
\phi(z,t)=\log\left(\frac{z^2-1}{z-t}\right),
$$
with a branch of the logarithm chosen locally along the deformed contour. Differentiation gives
$$
\phi'(z,t)
=\frac{2z}{z^2-1}-\frac1{z-t}
=\frac{z^2-2tz+1}{(z^2-1)(z-t)}.
$$
For $t=\cos\theta$, its saddle points solve
$$
z^2-2(\cos\theta)z+1=0,
$$
and hence
$$
\boxed{z_\pm=\cos\theta\pm i\sin\theta=e^{\pm i\theta}}.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

At a saddle the numerator of $\phi'$ vanishes, so differentiating the quotient there gives
$$
\phi''(z_\pm)
=\frac{2(z_\pm-t)}
{(z_\pm^2-1)(z_\pm-t)}
=\frac2{z_\pm^2-1}.
$$
Since
$$
z_+^2-1
=2\sin\theta\,e^{i(\theta+\pi/2)},
\qquad
z_-^2-1
=2\sin\theta\,e^{-i(\theta+\pi/2)},
$$
we obtain
$$
\boxed{\arg\phi''(z_\pm)=\mp(\theta+\pi/2)}.
$$

The tangent angle $\alpha$ must satisfy
$$
\arg\phi''(z_\pm)+2\alpha=\pi\pmod{2\pi}.
$$
An anticlockwise deformation of the original contour through both saddles can be oriented with
$$
\boxed{
\alpha_+=\frac\theta2+\frac{3\pi}4,
\qquad
\alpha_-=\frac\pi4-\frac\theta2}.
$$
Angles differing by $\pi$ describe the same unoriented tangent; the displayed choices give the orientation needed to pass through the lower saddle and then the upper saddle while still enclosing $t$.

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

At the two saddles,
$$
\frac{z_\pm^2-1}{z_\pm-t}=2z_\pm=2e^{\pm i\theta},
\qquad
f(z_\pm,t)=\frac1{z_\pm-t}
=\mp\frac{i}{\sin\theta},
$$
and
$$
|\phi''(z_\pm)|=\frac1{\sin\theta}.
$$
Applying the <method of steepest descent> with the oriented angles from part (ii), and including the prefactor $1/(2^{n+1}\pi i)$, gives two conjugate contributions. If
$$
\Phi=\left(n+\frac12\right)\theta-\frac\pi4,
$$
their sum is
$$
\frac1{2\pi}\sqrt{\frac{2\pi}{n\sin\theta}}
\left(e^{i\Phi}+e^{-i\Phi}\right).
$$
Therefore the <Debye asymptotic for Legendre polynomials> is
$$
\boxed{
P_n(\cos\theta)
\sim
\sqrt{\frac{2}{\pi n\sin\theta}}\,
\cos\left(n\theta+\frac\theta2-\frac\pi4\right)}.
$$
Thus
$$
\boxed{A(\theta)=\sqrt{\frac{2}{\pi n\sin\theta}}}.
$$

Solved by gpt-5.6-sol high.

= 31A
{parent=Paper 3}
{scope}
{title2=Dynamical Systems}

= a
{parent=31a}
{scope}

= Solution
{parent=a}

Let $\dot z=F(z)$ be a $C^1$ planar system on a simply connected domain $D$. The <Bendixson-Dulac criterion> states that if some $B\in C^1(D)$ makes
$$
\nabla\mathbin{\cdot}(BF)
$$
have one sign throughout $D$ and not vanish identically on any open subset, then $D$ contains no periodic orbit.

Indeed, if a periodic orbit $\Gamma$ bounded a region $\Omega\subset D$, then $F$ and hence $BF$ would be tangent to $\Gamma$. Its normal flux would therefore vanish. The divergence theorem would give
$$
0=\int_\Gamma BF\mathbin{\cdot}n\,ds
=\iint_\Omega\nabla\mathbin{\cdot}(BF)\,dA,
$$
contradicting the sign hypothesis. Taking $B=1$ gives the usual Bendixson divergence criterion.

The stability version of the divergence test concerns an existing periodic orbit $\gamma$ of period $T$. Liouville's formula for the variational equation shows that its nontrivial <floquet multiplier> is
$$
\rho=\exp\left(
\int_0^T\nabla\mathbin{\cdot}F(\gamma(t))\,dt
\right).
$$
Thus the <divergence test for a planar periodic orbit> says that $\gamma$ is asymptotically stable if the integral is negative and unstable if it is positive.

Solved by gpt-5.6-sol high.

= b
{parent=31a}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

Put $r^2=x^2+y^2$. The system can be written
$$
\begin{pmatrix}\dot x\\\dot y\end{pmatrix}
=
\begin{pmatrix}\mu&-1\\1&\mu\end{pmatrix}
\begin{pmatrix}x\\y\end{pmatrix}
-r^2(r^2-\lambda)\begin{pmatrix}x\\y\end{pmatrix}.
$$
It therefore vanishes at the origin. The linearization there has eigenvalues
$$
\mu\pm i.
$$
Hence the origin is an asymptotically stable focus for $\mu<0$ and an unstable focus for $\mu>0$.

When $\mu=0$ the linearization alone is inconclusive. Direct radial reduction gives
$$
\dot r=r^3(\lambda-r^2),
$$
which is positive for every sufficiently small $r>0$. The origin is therefore unstable at $\mu=0$ as well, although trajectories leave it only at cubic order.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

The polar equations are
$$
\dot r
=\frac{x\dot x+y\dot y}{r}
=r(\mu+\lambda r^2-r^4),
\qquad
\dot\theta
=\frac{x\dot y-y\dot x}{r^2}=1.
$$
Every positive zero of the radial factor is therefore a circular periodic orbit of period $2\pi$. Writing $s=r^2$, the equation is
$$
s^2-\lambda s-\mu=0,
\qquad
s_\pm=\frac{\lambda\pm\sqrt{\lambda^2+4\mu}}2.
$$
The resulting count is
$$
\begin{array}{c|c}
\text{parameter range}&\text{positive periodic orbits}\\ \hline
\mu<-\lambda^2/4&0\\
\mu=-\lambda^2/4&1\text{ double orbit}\\
-\lambda^2/4<\mu<0&2\\
\mu\geq0&1.
\end{array}
$$
For the two-orbit range, $s_-<\lambda/2<s_+$. Differentiating the radial vector field at an orbit gives
$$
\left.\frac{d\dot r}{dr}\right|_{r^2=s}
=2s(\lambda-2s),
$$
so the inner orbit is unstable and the outer orbit is stable. At
$$
\boxed{\mu=-\lambda^2/4}
$$
they meet in a <saddle-node bifurcation of periodic orbits>. At
$$
\boxed{\mu=0}
$$
the inner orbit shrinks into the origin, giving the second bifurcation.

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

The larger bifurcation value is $\mu=0$. For $\mu<0$ the origin is stable and an unstable periodic orbit of radius
$$
r_-=\sqrt{\frac{\lambda-\sqrt{\lambda^2+4\mu}}2}
\sim\sqrt{-\frac{\mu}{\lambda}}
$$
surrounds it. That orbit collapses into the origin as $\mu\uparrow0$, after which the origin is unstable. This is a <subcritical Hopf bifurcation>.

After a positive rescaling of radius, its local normal form is
$$
\boxed{\dot R=\mu R+R^3,\qquad \dot\theta=1},
$$
or equivalently
$$
\dot z=(\mu+i)z+z|z|^2.
$$
The full <quintic radial Hopf equation> also has the stable outer branch
$$
r_+(\mu)
=\sqrt{\frac{\lambda+\sqrt{\lambda^2+4\mu}}2}.
$$
In the $(\mu,r)$ bifurcation diagram, the branch $r=0$ is stable for $\mu<0$ and unstable for $\mu\geq0$. The unstable $r_-$ branch runs from $(0,0)$ backwards to
$$
\left(-\frac{\lambda^2}{4},\sqrt{\frac\lambda2}\right),
$$
where it folds into the stable $r_+$ branch; the latter continues to larger $\mu$.

Solved by gpt-5.6-sol high.

= iv
{parent=b}
{scope}

= Solution
{parent=iv}

The divergence of the Cartesian vector field is
$$
\nabla\mathbin{\cdot}F
=2\mu+4\lambda r^2-6r^4.
$$
On a periodic orbit with $r^2=s$ and
$$
\mu+\lambda s-s^2=0,
$$
this becomes the constant
$$
\nabla\mathbin{\cdot}F
=2s(\lambda-2s).
$$
Since the period is $2\pi$, the divergence integral is
$$
\int_0^{2\pi}\nabla\mathbin{\cdot}F\,dt
=4\pi s(\lambda-2s).
$$
It is positive on the inner branch $s_-$, negative on the outer branch $s_+$, and zero where the branches meet at $s=\lambda/2$. The divergence test therefore classifies the inner orbit as unstable, the outer orbit as stable, and the double orbit as nonhyperbolic, exactly as the radial phase line does.

There is also no conflict with the exclusion form of the divergence criterion. For a circular orbit of radius $R$,
$$
\iint_{r<R}\nabla\mathbin{\cdot}F\,dA
=2\pi R^2(\mu+\lambda R^2-R^4)=0,
$$
so the divergence necessarily changes sign or vanishes somewhere in its interior; it cannot have the strict one-sign property that would exclude the orbit.

Solved by gpt-5.6-sol high.

= 32D
{parent=Paper 3}
{scope}
{title2=Integrable Systems}

= a
{parent=32d}
{scope}

= Solution
{parent=a}

Put
$$
c=\beta _0e^{8\chi^3t},
\qquad F(s)=ce^{-\chi s},
$$
and seek the <rank-one Gelfand-Levitan-Marchenko kernel> in the separable form
$$
K(x,y)=-B(x)e^{-\chi y}.
$$
Substitution into the <Gelfand-Levitan-Marchenko equation> gives
$$
-B(x)+ce^{-\chi x}
-B(x)c\int_x^\infty e^{-2\chi z}\,dz=0,
$$
so
$$
B(x)=\frac{ce^{-\chi x}}
{1+ce^{-2\chi x}/(2\chi)}
$$
and hence
$$
K(x,x)=-\frac{ce^{-2\chi x}}
{1+ce^{-2\chi x}/(2\chi)}.
$$

Write
$$
q=\frac{ce^{-2\chi x}}{2\chi}
=\frac{\beta _0}{2\chi}e^{8\chi^3t-2\chi x}.
$$
The reconstruction formula for this <inverse scattering transform> convention yields
$$
u(x,t)=-2\frac{\partial}{\partial x}K(x,x)
=-\frac{8\chi^2q}{(1+q)^2}.
$$
If $q=e^{-2s}$, then $4q/(1+q)^2=\operatorname{sech}^2s$. Taking
$$
s=\chi(x-4\chi^2t-\phi)
$$
therefore produces the required one-<soliton> of the <Korteweg-De Vries equation>, with
$$
\boxed{\phi=\frac{1}{2\chi}\log\!\left(\frac{\beta _0}{2\chi}\right)}.
$$
Here the usual bound-state norming constant has $\beta _0>0$ and $\chi>0$, so $\phi$ is real.

Solved by gpt-5.6-sol high.

= b
{parent=32d}
{scope}

= Solution
{parent=b}

Let
$$
H=-\partial_x^2-2\chi^2\operatorname{sech}^2(\chi x),
\qquad
A=\partial_x+\chi\tanh(\chi x).
$$
On $L^2(\mathbb R)$, $A^\dagger=-\partial_x+\chi\tanh(\chi x)$. The <supersymmetric factorization of the one-soliton potential> is
$$
A^\dagger A=H+\chi^2,
\qquad
AA^\dagger=-\partial_x^2+\chi^2.
$$
In particular, $A^\dagger A$ is nonnegative, so every eigenvalue of $H$ satisfies
$$
E\geq-\chi^2.
$$

Equality is attained precisely when $A\psi=0$. Solving that first-order equation gives
$$
\psi_0(x)=C\operatorname{sech}(\chi x),
$$
which is square-integrable. With unit normalization, $C=\sqrt{\chi/2}$, and
$$
H\psi_0=-\chi^2\psi_0.
$$

It remains to exclude another bound state. Since this <Pöschl-Teller potential> tends to zero at infinity, a bound-state energy must have $E<0$. If $H\psi=E\psi$, then applying $A$ to
$$
A^\dagger A\psi=(E+\chi^2)\psi
$$
gives
$$
-\partial_x^2(A\psi)=E(A\psi).
$$
The free one-dimensional Schrodinger operator has no nonzero square-integrable eigenfunction with $E<0$. Consequently $A\psi=0$, and the state must be proportional to $\psi_0$. Thus there is exactly one bound state, with
$$
\boxed{E_0=-\chi^2}.
$$

Solved by gpt-5.6-sol high.

= 33C
{parent=Paper 3}
{scope}
{title2=Principles of Quantum Mechanics}

= a
{parent=33c}
{scope}

= Solution
{parent=a}

Work in units with $\hbar=1$. For unit mass and frequency, the <two-dimensional isotropic harmonic oscillator> has
$$
H=\frac12(p_x^2+x^2)+\frac12(p_y^2+y^2)
=a_x^\dagger a_x+a_y^\dagger a_y+1.
$$
The product number states are therefore energy eigenvectors:
$$
|n_x,n_y\rangle,
\qquad
E_{n_x,n_y}=n_x+n_y+1.
$$
For the level $E=N+1$, the possibilities are
$$
(n_x,n_y)=(N,0),(N-1,1),\ldots,(0,N),
$$
so its degeneracy is
$$
\boxed{N+1=E}.
$$
Restoring units multiplies every energy by $\hbar\omega$.

Solved by gpt-5.6-sol high.

= b
{parent=33c}
{scope}

= Solution
{parent=b}

The planar <orbital angular momentum> is
$$
L=xp_y-yp_x
=i(a_y^\dagger a_x-a_x^\dagger a_y).
$$
The canonical commutation relations give
$$
[H,x]=-ip_x,
\quad [H,p_x]=ix,
\quad [H,y]=-ip_y,
\quad [H,p_y]=iy.
$$
Consequently
$$
\begin{aligned}
[H,L]
&=[H,x]p_y+x[H,p_y]-[H,y]p_x-y[H,p_x]\\
&=-ip_xp_y+ixy+ip_yp_x-iyx=0.
\end{aligned}
$$
Thus angular momentum preserves each energy eigenspace. More explicitly,
$$
0=\langle E|[H,L]|E'\rangle
=(E-E')\langle E|L|E'\rangle,
$$
so
$$
\boxed{\langle E|L|E'\rangle=0\quad\hbox{if }E\ne E'.}
$$
Within a degenerate energy eigenspace, this argument places no further restriction on the matrix of $L$.

Solved by gpt-5.6-sol high.

= c
{parent=33c}
{scope}

= Solution
{parent=c}

For $T_{ij}=a_i^\dagger a_j$, expand the commutator and commute one annihilation operator past one creation operator:
$$
\begin{aligned}
[T_{ij},T_{kl}]
&=a_i^\dagger a_j a_k^\dagger a_l
 -a_k^\dagger a_l a_i^\dagger a_j\\
&=\delta_{jk}a_i^\dagger a_l
 -\delta_{il}a_k^\dagger a_j.
\end{aligned}
$$
Hence the <oscillator bilinear commutator> is
$$
\boxed{[T_{ij},T_{kl}]=\delta_{jk}T_{il}-\delta_{il}T_{kj}}.
$$
Since
$$
H=T_{xx}+T_{yy}+1,
$$
we obtain
$$
[T_{ij},H]
=\sum_k(\delta_{jk}T_{ik}-\delta_{ik}T_{kj})=0.
$$
Every $T_{ij}$ therefore acts within a fixed energy eigenspace.

Solved by gpt-5.6-sol high.

= d
{parent=33c}
{scope}

= Solution
{parent=d}

Contracting the result of part (c) with the <Pauli matrix>{p} gives
$$
\begin{aligned}
[T^a,T^b]
&=\frac14\sigma^a_{ij}\sigma^b_{kl}
  (\delta_{jk}T_{il}-\delta_{il}T_{kj})\\
&=\frac14[\sigma^a,\sigma^b]_{il}T_{il}
=i\varepsilon_{abc}T^c.
\end{aligned}
$$
Thus the $T^a$ form the <Schwinger boson representation> of the $\mathfrak{su}(2)$ commutation relations. The second Pauli matrix gives
$$
T^2=\frac{i}{2}(a_y^\dagger a_x-a_x^\dagger a_y),
$$
and therefore
$$
\boxed{L=2T^2}.
$$

On the energy level $E=N+1$, the total occupation is $N$. The $N+1$ states $|n_x,N-n_x\rangle$ form the spin-$j=N/2$ irreducible representation. Every component of its angular momentum generator has eigenvalues
$$
m=-\frac N2,-\frac N2+1,\ldots,\frac N2.
$$
Since $L=2T^2$, the possible orbital angular momenta are
$$
\boxed{\ell=-N,-N+2,\ldots,N-2,N}.
$$
There are $N+1$ distinct values, exactly matching the degeneracy found in part (a); hence each occurs once. Restoring $\hbar$ multiplies every $\ell$ by $\hbar$.

Solved by gpt-5.6-sol high.

= 34B
{parent=Paper 3}
{scope}
{title2=Applications of Quantum Mechanics}

= a
{parent=34b}
{scope}

= Solution
{parent=a}

Projecting $H|\Psi\rangle=E|\Psi\rangle$ onto $\langle\psi_n|$ and using orthonormality gives
$$
E_0c_n+\sum_{m=0}^2\langle\psi_n|V|\psi_m\rangle c_m=Ec_n.
$$
Thus, with $d=E_0+\alpha$,
$$
\begin{pmatrix}
d&-A&-A\\
-A&d&-A\\
-A&-A&d
\end{pmatrix}
\begin{pmatrix}c_0\\c_1\\c_2\end{pmatrix}
=E\begin{pmatrix}c_0\\c_1\\c_2\end{pmatrix}.
$$

The symmetric vector $(1,1,1)$ has eigenvalue
$$
E_s=d-2A=E_0+\alpha-2A.
$$
Let $\omega=e^{2\pi i/3}$, so $1+\omega+\omega^2=0$. For either vector
$$
(1,\omega,\omega^2),
\qquad
(1,\omega^2,\omega),
$$
the other two components in each row sum to minus the component in that row. Both therefore have eigenvalue
$$
E_d=d+A=E_0+\alpha+A.
$$
The spectrum is consequently
$$
\boxed{E_0+\alpha-2A,\quad E_0+\alpha+A,\quad E_0+\alpha+A},
$$
with normalized eigenvectors obtained by multiplying the three displayed vectors by $1/\sqrt3$.

Solved by gpt-5.6-sol high.

= b
{parent=34b}
{scope}

= Solution
{parent=b}

Translation by one site multiplies a lattice wave by $e^{ika}$, so the coefficients of a <finite periodic tight-binding ring> are
$$
c_n=\frac1{\sqrt3}e^{ikna}.
$$
Periodicity after three sites requires
$$
e^{3ika}=1.
$$
Choosing representatives in the <Brillouin zone>
$$
-\frac\pi a\leq k<\frac\pi a
$$
gives
$$
\boxed{k=0,\quad \frac{2\pi}{3a},\quad-\frac{2\pi}{3a}}.
$$
The corresponding states are
$$
\boxed{|\Psi_k\rangle=\frac1{\sqrt3}
\sum_{n=0}^2e^{ikna}|\psi_n\rangle}.
$$
They reproduce respectively the coefficient vectors $(1,1,1)$, $(1,\omega,\omega^2)$ and $(1,\omega^2,\omega)$.

Solved by gpt-5.6-sol high.

= c
{parent=34b}
{scope}

= Solution
{parent=c}

In position space,
$$
\begin{aligned}
\Psi_k(x)
&=\sum_{n=0}^2e^{ikna}\psi(x-na)\\
&=e^{ikx}\sum_{n=0}^2\psi(x-na)e^{-ik(x-na)}.
\end{aligned}
$$
Hence
$$
\Psi_k(x)=e^{ikx}u_k(x),
\qquad
u_k(x)=\sum_{n=0}^2\psi(x-na)e^{-ik(x-na)},
$$
up to the common normalization factor $1/\sqrt3$.

Under $x\mapsto x+a$, reindexing $m=n-1$ gives the same three terms modulo three sites. The term shifted by three sites is unchanged because $\psi_{m+3}=\psi_m$ and $e^{3ika}=1$. Therefore
$$
\boxed{u_k(x+a)=u_k(x)}.
$$

The one-dimensional <Bloch theorem> states that an eigenfunction in a potential of period $a$ can be chosen in the form
$$
\Psi_k(x)=e^{ikx}u_k(x),
\qquad u_k(x+a)=u_k(x).
$$
It applies because the three-atom ring is invariant under translation by one lattice spacing and obeys periodic boundary conditions.

Solved by gpt-5.6-sol high.

= 35C
{parent=Paper 3}
{scope}
{title2=Statistical Physics}

= a
{parent=35c}
{scope}

= Solution
{parent=a}

In two dimensions, each allowed wavevector occupies area $(2\pi)^2/A$ in $k$-space. Including the two spin states, the number of states in an annulus $(k,k+dk)$ is
$$
d\mathcal N
=2\frac{A}{(2\pi)^2}2\pi k\,dk.
$$
Since
$$
E=\frac{\hbar^2k^2}{2m},
\qquad
k\,dk=\frac{m}{\hbar^2}\,dE,
$$
the <two-dimensional free-electron density of states> is
$$
\boxed{g(E)=\frac{d\mathcal N}{dE}=\frac{Am}{\pi\hbar^2}}
$$
for $E\geq0$. The <Fermi-Dirac distribution> is
$$
\boxed{f(E)=\frac1{e^{\beta(E-\mu)}+1}},
\qquad \beta=\frac1{k_BT}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=35c}
{scope}

= Solution
{parent=b}

At zero temperature all states up to $E_F$ are occupied. Writing
$$
g=\frac{Am}{\pi\hbar^2},
$$
we find
$$
N=\int_0^{E_F}g\,dE=gE_F
$$
and
$$
E_{\rm tot}=\int_0^{E_F}Eg\,dE
=\frac12gE_F^2
=\boxed{\frac12NE_F}.
$$

Because $g=CA$ with $C=m/(\pi\hbar^2)$, fixed particle number gives $E_F=N/(CA)$ and
$$
E_{\rm tot}=\frac{N^2}{2CA}.
$$
The two-dimensional <degeneracy pressure of a two-dimensional Fermi gas> is therefore
$$
\boxed{p=-\left(\frac{\partial E_{\rm tot}}{\partial A}\right)_N
=\frac{N^2}{2CA^2}
=\frac{NE_F}{2A}}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=35c}
{scope}

= Solution
{parent=c}

Keep the chemical potential at its zero-temperature value $\mu=E_F$ and put $E'=E-E_F$. Away from $E'=0$,
$$
f(E)-\mathbf1_{E<E_F}
=\frac12\left[\operatorname{sgn}\!\left(\frac{\beta E'}2\right)
-\tanh\!\left(\frac{\beta E'}2\right)\right].
$$
Consequently
$$
\Delta N
=X\int_{-E_F}^{\infty}
\left[\operatorname{sgn}\!\left(\frac{\beta E'}2\right)
-\tanh\!\left(\frac{\beta E'}2\right)\right]dE',
$$
where
$$
\boxed{X=\frac g2=\frac{Am}{2\pi\hbar^2}}.
$$

The bracket is an odd function localized to $|E'|\lesssim k_BT$. When $\beta E_F\gg1$, extending the lower limit to $-\infty$ makes the integral vanish exactly. The omitted tail is exponentially small; indeed direct integration gives
$$
\Delta N=\frac g\beta\log(1+e^{-\beta E_F}).
$$
This is the <low-temperature particle-number cancellation for constant density of states>.

For the energy,
$$
\begin{aligned}
\Delta E_{\rm tot}
&=g\int_{-E_F}^{\infty}(E_F+E')
\bigl[f(E_F+E')-\mathbf1_{E'<0}\bigr]dE'\\
&=E_F\Delta N
+X\int_{-E_F}^{\infty}E'
\left[\operatorname{sgn}\!\left(\frac{\beta E'}2\right)
-\tanh\!\left(\frac{\beta E'}2\right)\right]dE'.
\end{aligned}
$$
At low temperature the first term is exponentially small. In the second, extend the lower limit to $-\infty$ and set $z=\beta E'/2$:
$$
\Delta E_{\rm tot}
\sim\frac{4X}{\beta^2}
\int_{-\infty}^{\infty}z[\operatorname{sgn}(z)-\tanh z]\,dz.
$$
The remaining integral is a finite positive constant, proving the <quadratic low-temperature energy correction for constant density of states>:
$$
\boxed{\Delta E_{\rm tot}\propto\beta^{-2}\propto T^2}.
$$

Solved by gpt-5.6-sol high.

= 36B
{parent=Paper 3}
{scope}
{title2=Electrodynamics}

= Solution
{parent=36B}

For a localized charge and current distribution, take the retarded vector potential
$$
A(x,t)=\frac{\mu_0}{4\pi}
\int\frac{J(x',t-|x-x'|/c)}{|x-x'|}\,d^3x'.
$$
Let $R=|x|$, $n=x/R$, and $t_R=t-R/c$. If the source size is much smaller than both $R$ and the characteristic radiation wavelength, the leading dipole term is
$$
A(x,t)\simeq\frac{\mu_0}{4\pi R}\int J(x',t_R)\,d^3x'
=\frac{\mu_0}{4\pi R}\dot p(t_R),
$$
where the last identity follows from charge conservation.

Keeping only the $1/R$ terms in the <radiation zone> gives the transverse fields
$$
E_{\rm rad}
=\frac{\mu_0}{4\pi R}n\times(n\times\ddot p(t_R)),
\qquad
B_{\rm rad}=\frac1c n\times E_{\rm rad}.
$$
The radial <Poynting vector> therefore yields
$$
\frac{d\mathcal P}{d\Omega}
=\frac{\mu_0}{16\pi^2c}
|n\times\ddot p(t_R)|^2.
$$
Using
$$
\int n_i n_j\,d\Omega=\frac{4\pi}{3}\delta_{ij}
$$
or equivalently $\int\sin^2\theta\,d\Omega=8\pi/3$, we recover the <electric dipole radiation> formula
$$
\boxed{\mathcal P\simeq
\frac{\mu_0}{6\pi c}|\ddot p(t-R/c)|^2}.
$$
This requires a localized, nonrelativistic source of size $d\ll\lambda$, observation distance $R\gg d$ and $R\gg\lambda$, and neglect of higher multipoles and faster-decaying near fields.

For the pulsar, write its magnetic dipole moment as $m(t)$ to distinguish it from the electric dipole above. The <magnetic dipole radiation> formula differs by a factor $c^{-2}$:
$$
\mathcal P=\frac{\mu_0}{6\pi c^3}|\ddot m|^2.
$$
Here
$$
\ddot m(t)=-p_0\Omega^2\sin\alpha
[\cos(\Omega t)\widehat x+\sin(\Omega t)\widehat y],
$$
and hence
$$
\boxed{\mathcal P=C\Omega^4,
\qquad
C=\frac{\mu_0p_0^2\sin^2\alpha}{6\pi c^3}}.
$$

On the slow spin-down timescale, energy conservation gives the <magnetic-dipole spin-down> equation
$$
I\Omega\dot\Omega=-C\Omega^4,
$$
so
$$
\frac1{\Omega(t)^2}=\frac1{\Omega_0^2}+\frac{2C}{I}t.
$$
Half of the initial rotational energy remains when $\Omega=\Omega_0/\sqrt2$. Therefore
$$
t_{1/2}=\frac{I}{2C\Omega_0^2}
=\boxed{\frac{6\pi M\mathcal R^2c^3}
{5\mu_0p_0^2\Omega_0^2\sin^2\alpha}},
$$
where $I=2M\mathcal R^2/5$. Since $E_0=I\Omega_0^2/2$, the equivalent expression is
$$
\boxed{t_{1/2}=\frac{6\pi M^2\mathcal R^4c^3}
{25\mu_0p_0^2E_0\sin^2\alpha}}.
$$

Solved by gpt-5.6-sol high.

= 37E
{parent=Paper 3}
{scope}
{title2=General Relativity}

= a
{parent=37e}
{scope}

= Solution
{parent=a}

Write
$$
f(r)=1-\frac{2M}{r}.
$$
For a radial null ray, the <Schwarzschild spacetime> line element gives
$$
0=-f\,dt^2+f^{-1}dr^2,
\qquad
\frac{dt}{dr}=\pm f^{-1}.
$$
Outside the horizon an ingoing ray has $dr/dt<0$, so it takes the minus sign. Since the <Schwarzschild tortoise coordinate> obeys $dr_*/dr=f^{-1}$,
$$
\frac{d}{dr}(t+r_*)=0.
$$
It follows that
$$
\boxed{v=t+r+2M\log\left|\frac{r}{2M}-1\right|}
$$
is constant along every ingoing radial light ray.

Solved by gpt-5.6-sol high.

= b
{parent=37e}
{scope}

= Solution
{parent=b}

Since $v=t+r_*$ and $dr_*=f^{-1}dr$,
$$
dt=dv-f^{-1}dr.
$$
Substitution into the metric gives
$$
\begin{aligned}
ds^2
&=-f(dv-f^{-1}dr)^2+f^{-1}dr^2+r^2d\Omega^2\\
&=\boxed{-f\,dv^2+2\,dv\,dr+r^2d\Omega^2}.
\end{aligned}
$$
These are <Ingoing Eddington-Finkelstein coordinates>. Every coefficient is finite at $r=2M$, and the determinant of the $v,r$ block is $-1$. Thus the apparent singularity there in Schwarzschild coordinates is a coordinate singularity; the metric extends smoothly across the <Schwarzschild event horizon>.

Solved by gpt-5.6-sol high.

= c
{parent=37e}
{scope}

= Solution
{parent=c}

For radial light rays, the transformed metric gives
$$
0=-f\,dv^2+2\,dv\,dr
=dv(-f\,dv+2\,dr).
$$
The two families of <radial null trajectories in ingoing Eddington-Finkelstein coordinates> are therefore
$$
\boxed{v=\text{constant}}
$$
for ingoing rays and
$$
\boxed{\frac{dv}{dr}=\frac2f=\frac{2r}{r-2M}}
$$
for outgoing rays.

Now set $t_*=v-r$. Along ingoing rays,
$$
\boxed{\frac{dt_*}{dr}=-1},
\qquad t_*=-r+\text{constant}.
$$
Along outgoing rays,
$$
\boxed{\frac{dt_*}{dr}=\frac2f-1
=\frac{r+2M}{r-2M}},
$$
and integration gives
$$
t_*=r+4M\log\left|\frac{r}{2M}-1\right|+\text{constant}.
$$
Thus the ingoing rays are straight lines of slope $-1$ in the $r$-$t_*$ plane. Outgoing rays have positive slope outside $r=2M$, become vertical as they approach the horizon, and have negative slope inside it; the horizon itself is the limiting outgoing null ray.

Solved by gpt-5.6-sol high.

= d
{parent=37e}
{scope}

= Solution
{parent=d}

For the outgoing family,
$$
\frac{dr}{dv}=\frac12\left(1-\frac{2M}{r}\right).
$$
When $r>2M$, this derivative is positive, while ingoing rays move to smaller $r$. The future light cone therefore has one outward and one inward radial edge, and a future-directed massive particle may move either inward or outward, provided its worldline remains inside that cone.

At $r=2M$, the outgoing edge has $dr/dv=0$ and lies along the horizon. When $0<r<2M$, even this nominally outgoing edge has $dr/dv<0$. Both future null directions point toward decreasing $r$, so every future-directed timelike direction between them does too. A massive particle inside the horizon cannot remain at fixed $r$ or return to $r>2M$; it must continue toward the curvature singularity at $r=0$.

Solved by gpt-5.6-sol high.

= e
{parent=37e}
{scope}

= Solution
{parent=e}

For a static observer, $dr=d\theta=d\phi=0$, so
$$
d\tau=\sqrt{1-\frac{2M}{r}}\,dt.
$$
The coordinate travel time of a radial signal depends only on the fixed endpoint radii. Successive signals emitted a coordinate-time interval $\Delta t$ apart are therefore received with the same coordinate-time separation. The <gravitational redshift between static Schwarzschild observers> is consequently
$$
\frac{\Delta\tau_B}{\Delta\tau_A}
=\sqrt{\frac{1-2M/r_B}{1-2M/r_A}}.
$$
For $r_B\gg M$ the numerator is approximately one, while
$$
1-\frac{2M}{r_A}
=1-\frac{2M}{2M+\varepsilon}
=\frac{\varepsilon}{2M+\varepsilon}
\simeq\frac{\varepsilon}{2M}.
$$
Hence
$$
\boxed{\frac{\Delta\tau_B}{\Delta\tau_A}
\simeq\sqrt{\frac{2M}{\varepsilon}}}.
$$
The received signals become arbitrarily widely separated and redshifted as Alice approaches the horizon. A static observer exactly at the horizon is impossible because the proper acceleration required to remain static diverges there.

Solved by gpt-5.6-sol high.

= 38D
{parent=Paper 3}
{scope}
{title2=Fluid Dynamics}

= a
{parent=38d}
{scope}

= Solution
{parent=a}

The incompressibility equation turns the advective term into a divergence:
$$
u u_x+v u_y=\partial_x(u^2)+\partial_y(uv),
$$
because $v_y=-u_x$. For a plume in an otherwise stationary environment, the boundary-layer pressure equals the ambient pressure and has no imposed $x$-gradient. Integrating the vertical momentum equation across $y$ therefore gives
$$
\frac{d}{dx}\int_{-\infty}^{\infty}\rho u^2\,dy
+[\rho uv]_{-\infty}^{\infty}
=\mu[u_y]_{-\infty}^{\infty}+b(x).
$$
Both boundary terms vanish as the velocity and shear decay away from the plume. Hence the <integral momentum-flux balance for a two-dimensional plume> is
$$
\boxed{\frac{d}{dx}\int_{-\infty}^{\infty}\rho u^2\,dy=b(x)}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=38d}
{scope}

= Solution
{parent=b}

Let $U(x)$ and $\Delta(x)$ denote the characteristic vertical speed and width. Integrating the result of part (a), with $b(x)=Bx^{-1/5}$, gives the scaling
$$
\rho U^2\Delta\sim Bx^{4/5}.
$$
The <boundary-layer scaling> in the momentum equation balances vertical inertia against transverse viscosity:
$$
\rho\frac{U^2}{x}\sim\mu\frac{U}{\Delta^2},
\qquad
U\sim\frac{\mu x}{\rho\Delta^2}.
$$
Substituting this into the momentum-flux relation gives
$$
\frac{\mu^2x^2}{\rho\Delta^3}\sim Bx^{4/5},
$$
and therefore
$$
\boxed{\Delta\sim
\left(\frac{\mu^2}{\rho B}\right)^{1/3}x^{2/5}}.
$$
Returning to the viscous-inertial balance then gives
$$
\boxed{U\sim
\left(\frac{B^2}{\rho\mu}\right)^{1/3}x^{1/5}}.
$$
These are the <similarity scaling of a two-dimensional laminar plume>.

Solved by gpt-5.6-sol high.

= c
{parent=38d}
{scope}

= Solution
{parent=c}

Choose the <stream function> convention
$$
u=-\psi_y,
\qquad
v=\psi_x,
$$
which satisfies continuity identically. Introduce
$$
D=\left(\frac{\mu^2}{\rho B}\right)^{1/3},
\qquad
U_0=\left(\frac{B^2}{\rho\mu}\right)^{1/3},
$$
so that
$$
\Delta=Dx^{2/5},
\qquad
U=U_0x^{1/5},
\qquad
\eta=\frac{y}{Dx^{2/5}}.
$$
Take the <similarity solution>
$$
\psi=U\Delta f(\eta)=U_0D x^{3/5}f(\eta).
$$
Then
$$
u=-U_0x^{1/5}f'
$$
and
$$
v=U_0D x^{-2/5}
\left(\frac35f-\frac25\eta f'\right).
$$
A direct differentiation shows that the terms containing $\eta f'f''$ cancel:
$$
u u_x+v u_y
=U_0^2x^{-3/5}
\left[\frac15(f')^2-\frac35ff''\right].
$$
Also,
$$
u_{yy}=-\frac{U_0}{D^2}x^{-3/5}f''',
$$
while the delta scaling gives
$$
b(x)\delta(y)
=\frac{B}{D}x^{-3/5}\delta(\eta).
$$
The definitions of $D$ and $U_0$ imply
$$
\frac{\rho U_0D^2}{\mu}=1,
\qquad
\frac{BD}{\mu U_0}=1.
$$
After cancelling the common dimensional factor, the <similarity equation for a two-dimensional laminar plume> is
$$
\boxed{f'''+\frac15(f')^2-\frac35ff''=\delta(\eta)}.
$$
Thus
$$
\boxed{c_1=\frac15,\qquad c_2=-\frac35}.
$$

Solved by gpt-5.6-sol high.

= 39A
{parent=Paper 3}
{scope}
{title2=Waves}

= a
{parent=39a}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

For a plane wave $\phi=e^{i(kx-\omega t)}$, the <Klein-Gordon equation> gives
$$
-\omega^2+c^2k^2+A^2c^2=0.
$$
Thus the positive-frequency dispersion relation is
$$
\boxed{\omega(k)=c\sqrt{k^2+A^2}}.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

For $k>0$, the <phase velocity and group velocity> are
$$
\boxed{c_p=\frac\omega k
=c\sqrt{1+\frac{A^2}{k^2}}},
\qquad
\boxed{c_g=\frac{d\omega}{dk}
=\frac{ck}{\sqrt{k^2+A^2}}}.
$$
As $k$ increases from zero, $c_p$ decreases from $+\infty$ toward $c$, while $c_g$ increases from zero toward $c$.

Solved by gpt-5.6-sol high.

= iii
{parent=a}
{scope}

= Solution
{parent=iii}

Since
$$
c_pc_g=c^2,
$$
the phase speed is always larger than the group speed for nonzero finite $k$. Wave crests therefore move faster than the packet envelope.

Solved by gpt-5.6-sol high.

= b
{parent=39a}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

Each Fourier mode has zero initial velocity, so
$$
\phi(x,t)=\int_{-\infty}^{\infty}a(k)e^{ikx}\cos(\omega(k)t)\,dk.
$$
Along $x=Vt$ this is
$$
\phi(Vt,t)=\frac12\int_{-\infty}^{\infty}a(k)
\left[e^{it(kV+\omega(k))}+e^{it(kV-\omega(k))}\right]dk.
$$
Put
$$
s=\sqrt{c^2-V^2},
\qquad
k_0=\frac{AV}{s}.
$$
The two phases have stationary points at $-k_0$ and $k_0$, respectively. At those points,
$$
k_0V-\omega(k_0)=-As,
\qquad
\omega''(k_0)=\frac{s^3}{Ac^2}.
$$

Because the initial field is real, $a(-k)=a(k)^*$. Applying the <stationary phase method> to the two conjugate contributions gives the <stationary-phase asymptotic of a Klein-Gordon wave along a subluminal ray>:
$$
\boxed{
\phi(Vt,t)\sim
\left(\frac{2\pi Ac^2}{t s^3}\right)^{1/2}
\operatorname{Re}\left\{
a(k_0)e^{-i(As t+\pi/4)}
\right\}.}
$$
Equivalently, if $a(k_0)=|a(k_0)|e^{i\delta}$,
$$
\phi(Vt,t)\sim
|a(k_0)|\left(\frac{2\pi Ac^2}{t s^3}\right)^{1/2}
\cos\left(As t+\frac\pi4-\delta\right).
$$

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

If the real initial displacement is even, then $a(k)$ is real and even. Assume $a(k_0)\ne0$ and set
$$
\Omega=A\sqrt{c^2-V^2}.
$$
When $a(k_0)>0$, the leading result from part (i) is a positive multiple of
$$
t^{-1/2}\cos\left(\Omega t+\frac\pi4\right).
$$
Its <upward zero crossings of an oscillatory stationary-phase tail>{p} occur when the phase is $3\pi/2$ modulo $2\pi$. Therefore
$$
\boxed{t_n\sim
\frac{2\pi n+5\pi/4}{A\sqrt{c^2-V^2}}},
\qquad n\to\infty.
$$
If $a(k_0)<0$, the sign is reversed and the upward crossings instead satisfy
$$
\boxed{t_n\sim
\frac{2\pi n+\pi/4}{A\sqrt{c^2-V^2}}}.
$$
The slow derivative of the $t^{-1/2}$ envelope changes these times only beyond leading order.

Solved by gpt-5.6-sol high.

= 40D
{parent=Paper 3}
{scope}
{title2=Numerical Analysis}

= a
{parent=40d}
{scope}

= Solution
{parent=a}

Consider a general constant-coefficient linear difference scheme
$$
\sum_{r,j}a_{rj}u_{m+j}^{n+r}=0.
$$
A spatial Fourier mode has the form
$$
u_m^n=\widehat u^{,n}e^{im\theta}.
$$
If one time step multiplies its amplitude by $G$, then $\widehat u^{,n+r}=G^r\widehat u^{,n}$, and substitution gives the <amplification polynomial of a multilevel finite difference scheme>
$$
\boxed{\sum_{r,j}a_{rj}G^r e^{ij\theta}=0}.
$$
For a one-step method this equation directly determines
$$
G(\theta)=\frac{\widehat u^{,n+1}}{\widehat u^{,n}}.
$$
By Parseval's identity, <Von Neumann stability analysis> gives $2$-norm stability when $|G(\theta)|\leq1$ for every Fourier mode. For a multilevel method, every root must obey the corresponding root condition.

Solved by gpt-5.6-sol high.

= b
{parent=40d}
{scope}

= Solution
{parent=b}

Insert
$$
u_m^n=G^n e^{im\theta}
$$
into the diffusion scheme. Since
$$
e^{i\theta}-2+e^{-i\theta}
=-4\sin^2\left(\frac\theta2\right),
$$
we obtain
$$
G\left(1+2\mu\sin^2\frac\theta2\right)
=1-2\mu\sin^2\frac\theta2.
$$
Thus the <Crank-Nicolson diffusion scheme> has
$$
G(\theta)=
\frac{1-2\mu\sin^2(\theta/2)}
{1+2\mu\sin^2(\theta/2)}.
$$
For $\mu\geq0$, the denominator is positive and
$$
|1-2\mu s|\leq1+2\mu s
\qquad(0\leq s\leq1).
$$
Hence
$$
\boxed{\text{the method is stable for every }\mu>0}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=40d}
{scope}

= Solution
{parent=c}

For the wave scheme, use $v_m^n=G^n e^{im\theta}$. After multiplying by $G$, the recurrence becomes
$$
G^2+(4\mu s-2\rho)G+1=0,
\qquad
s=\sin^2\left(\frac\theta2\right).
$$
Equivalently,
$$
G+G^{-1}=2(\rho-2\mu s).
$$
Because the two roots have product one, both lie on the unit circle precisely when
$$
|\rho-2\mu s|\leq1.
$$
This must hold for every $s\in[0,1]$. At $s=0$ it requires $\rho\leq1$. Since the question assumes $1\leq\rho\leq2$, stability is possible only for $\rho=1$. The condition at $s=1$ then becomes
$$
|1-2\mu|\leq1,
$$
or $0\leq\mu\leq1$. With the stipulated $\mu>0$, the <centered three-level wave scheme> is stable exactly when
$$
\boxed{\rho=1,\qquad 0<\mu\leq1}.
$$
For every $\rho>1$, sufficiently long-wavelength modes have a real root larger than one, whatever the value of $\mu$.

Solved by gpt-5.6-sol high.

= d
{parent=40d}
{scope}

= Solution
{parent=d}

Let $L$ be the tridiagonal matrix with diagonal entries $-2$ and adjacent off-diagonal entries $1$ on the interior grid points. The finite-interval scheme is
$$
\left(I-\frac\mu2L\right)u^{n+1}
=\left(I+\frac\mu2L\right)u^n.
$$
For $J$ interior points, the <dirichlet discrete Laplacian> has eigenvalues
$$
\lambda_j=-4\sin^2\left(\frac{j\pi}{2(J+1)}\right)<0,
\qquad 1\leq j\leq J.
$$
The corresponding eigenvalues of the amplification matrix are
$$
q_j=\frac{1+(\mu/2)\lambda_j}
{1-(\mu/2)\lambda_j}.
$$
Since $\lambda_j<0$ and $\mu\geq0$,
$$
|q_j|\leq1
$$
for every $j$. The matrix is real symmetric in the sine eigenbasis, so its induced $2$-norm is the largest $|q_j|$. This proves the <crank-Nicolson stability on a finite Dirichlet interval> and gives
$$
\boxed{\text{stability for every }\mu>0}.
$$

Solved by gpt-5.6-sol high.