past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ii/paper-1.bigb

= Paper 1
{scope}

https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2021/paperii_1_2021.pdf

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

= Solution
{parent=1I}

<Euler's criterion> states that for an odd <prime number> $p$ and $p\nmid a$,
$$
\boxed{
a^{(p-1)/2}\equiv\left(\frac ap\right)\pmod p
},
$$
where the right-hand side is the <Legendre symbol>.

Let $g$ be a <primitive root> modulo $p$. Its <multiplicative order> is $p-1$, so $g^{(p-1)/2}\ne1$. This power has square $g^{p-1}=1$, and the only roots of $X^2-1$ modulo the odd prime $p$ are $\pm1$. Therefore
$$
g^{(p-1)/2}\equiv-1\pmod p.
$$
Euler's criterion now gives $(g/p)=-1$, so every primitive root is a <quadratic nonresidue>.

Now let $p=2^{2^k}+1$ be a <Fermat prime>, with $k\geq1$. The number of primitive roots modulo $p$ is
$$
\varphi(p-1)
=\varphi\left(2^{2^k}\right)
=2^{2^k-1}
=\frac{p-1}{2}.
$$
There are also exactly $(p-1)/2$ quadratic nonresidues. Since every primitive root is a nonresidue, these equally large sets coincide. Thus every quadratic nonresidue modulo $p$ is a primitive root, as summarized by <quadratic nonresidues modulo a Fermat prime>.

Finally, $p\equiv1\pmod4$ and $p\equiv2\pmod3$. <Quadratic reciprocity> therefore gives
$$
\left(\frac3p\right)
=\left(\frac p3\right)
=\left(\frac23\right)
=-1.
$$
Hence $3$ is a quadratic nonresidue modulo $p$, and consequently
$$
\boxed{3\text{ is a primitive root modulo every Fermat prime }p
\text{ with }k\geq1}.
$$

Solved by gpt-5.6-sol high.

= 2H
{parent=Paper 1}
{scope}
{title2=Topics in Analysis}

= Solution
{parent=2H}

The set $K\cap P$ is compact because $K$ is closed and bounded, and the product
$$
G(x)=\prod_{j=1}^n x_j
$$
is continuous. By the <extreme value theorem>, $G$ has a maximizer. Since $K$ meets the interior of the <nonnegative orthant> $P$, some feasible product is positive, so every maximizer lies in $\operatorname{Int}P$.

On $\operatorname{Int}P$, maximizing $G$ is equivalent to maximizing
$$
F(x)=\log G(x)=\sum_{j=1}^n\log x_j.
$$
The Hessian of $F$ is
$$
D^2F(x)=\operatorname{diag}\left(-\frac1{x_1^2},\ldots,-\frac1{x_n^2}\right),
$$
which is negative definite. Thus $F$ is a <strictly concave function>, and it has at most one maximizer on the <convex set> $K\cap P$. Denote this unique point by $x^*$.

For any $x\in K\cap P$, the line segment
$$
x(t)=(1-t)x^*+tx,\qquad 0\leq t<1,
$$
lies in $K\cap\operatorname{Int}P$. Since $F(x(t))$ is maximal at $t=0$, its right derivative there is nonpositive:
$$
0\geq\left.\frac d{dt}F(x(t))\right|_{t=0}
=\sum_{j=1}^n\frac{x_j-x_j^*}{x_j^*}.
$$
Therefore
$$
\boxed{\sum_{j=1}^n\frac{x_j}{x_j^*}\leq n}.
$$
This is the supporting inequality for the <product maximizer on a compact convex subset of the positive orthant>.

Suppose now that $K$ is invariant under cyclic coordinate permutation. That permutation preserves the product, so uniqueness forces it to fix $x^*$. Hence
$$
x^*=(r,\ldots,r)
$$
for some $r>0$. More explicitly,
$$
\boxed{
r=\max_{x\in K\cap P}\frac1n\sum_{j=1}^n x_j
}.
$$
Indeed, <cyclic symmetry averaging> puts the diagonal point with coordinate equal to the mean of any $x\in K$ back in $K$, while the preceding inequality with $x_j^*=r$ gives $\sum_jx_j\leq nr$.

Finally, given $a=(a_1,\ldots,a_n)\in\operatorname{Int}P$, define
$$
K=\left\{x\in P:
\sum_{j=1}^n\frac{x_j}{a_j}\leq n
\right\}.
$$
This set is nonempty, closed, convex and bounded, and it contains $a$. By the <arithmetic-geometric mean inequality>,
$$
\frac{\prod_jx_j}{\prod_ja_j}
=\prod_{j=1}^n\frac{x_j}{a_j}
\leq
\left(\frac1n\sum_{j=1}^n\frac{x_j}{a_j}\right)^n
\leq1.
$$
Equality holds only when all $x_j/a_j$ are equal and their sum is $n$, namely only at $x=a$. Thus this $K$ has
$$
\boxed{x^*=a}.
$$

Solved by gpt-5.6-sol high.

= 3K
{parent=Paper 1}
{scope}
{title2=Coding and Cryptography}

= i
{parent=3k}
{scope}

= Solution
{parent=i}

For a <binary block code>, $n$ is the common length of its codewords, $m=|C|$ is the number of codewords, and
$$
d=\min_{\substack{x,y\in C\\x\ne y}}d_H(x,y)
$$
is its minimum <Hamming distance>.

The <parity extension> is
$$
C^+=\left\{
\left(x_1,\ldots,x_n,\sum_{j=1}^nx_j\right):x\in C
\right\}.
$$
Every extended word has even <Hamming weight>. The distance between two words increases by one exactly when their original distance is odd, so
$$
\boxed{
C^+\text{ has parameters }[n+1,m,d^+],
\qquad
d^+=
\begin{cases}
d,&d\text{ even},\\
d+1,&d\text{ odd}.
\end{cases}}
$$

Solved by gpt-5.6-sol high.

= ii
{parent=3k}
{scope}

= Solution
{parent=ii}

Fix a coordinate, say the last one. The <punctured code> is
$$
C^-=\{(x_1,\ldots,x_{n-1}):(x_1,\ldots,x_n)\in C\}.
$$
Deleting one coordinate lowers each pairwise distance by at most one. Thus, when $d\geq2$, no two words merge and
$$
\boxed{
C^-\text{ has parameters }[n-1,m,d^-],
\qquad d-1\leq d^-\leq d
}.
$$
In guaranteed-distance notation this is recorded as an $[n-1,m,d-1]$ code. If $d=1$, puncturing a coordinate on which a distance-one pair differs can merge words, so its exact size must then be stated separately.

Solved by gpt-5.6-sol high.

= iii
{parent=3k}
{scope}

= Solution
{parent=iii}

For a chosen coordinate and symbol $\varepsilon\in\{0,1\}$, the <shortened code> retains the words having that symbol in the chosen coordinate and then deletes that coordinate. For the last coordinate,
$$
\overline C_\varepsilon
=\{(x_1,\ldots,x_{n-1}):(x_1,\ldots,x_{n-1},\varepsilon)\in C\}.
$$
At least one of the two coordinate fibres contains at least $\lceil m/2\rceil$ words. Choose that fibre and, if necessary, discard surplus words. Because all retained words agree in the deleted coordinate, their mutual distances do not change. Hence one can always obtain
$$
\boxed{\overline C\text{ with guaranteed parameters }
[n-1,\lceil m/2\rceil,d]}.
$$
Without discarding words, its size is that of the chosen fibre and its minimum distance is at least $d$.

For the final calculation, the given code is the whole space $\mathbb F_2^3$, so its parity extension is the length-four <even-weight binary code>. In a <binary symmetric channel>, parity fails to notice a nonzero error exactly when an even number of bits flips. The possible error weights are therefore two and four. Their total probability is
$$
\boxed{
\binom42p^2(1-p)^2+p^4
=6p^2(1-p)^2+p^4
}.
$$
This excludes the weight-zero event, since it is not an <undetected error>.

Solved by gpt-5.6-sol high.

= 4F
{parent=Paper 1}
{scope}
{title2=Automata and Formal Languages}

= Solution
{parent=4F}

For the given effective enumeration of partial computable functions, the <diagonal halting set> is
$$
\boxed{K=\{n\in\mathbb N:f_{n,1}(n)\text{ is defined}\}}.
$$

A <many-one reduction> $A\leq_mB$ is a <total computable function> $g:\mathbb N\to\mathbb N$ such that
$$
\boxed{x\in A\Longleftrightarrow g(x)\in B}.
$$

The <S-m-n theorem> states that for every $m,k\geq1$ there is a total computable function $s_k^m$ satisfying
$$
f_{s_k^m(e,x_1,\ldots,x_m),k}(y_1,\ldots,y_k)
=f_{e,m+k}(x_1,\ldots,x_m,y_1,\ldots,y_k)
$$
whenever either side is defined.

Suppose first that $X$ is <recursively enumerable set>[recursively enumerable]. Choose a program that halts exactly on inputs in $X$, and let $e$ index a two-variable program that, on $(x,y)$, ignores $y$ and runs that program on $x$. By the S-m-n theorem,
$$
g(x)=s_1^1(e,x)
$$
is total computable and indexes the unary function
$$
f_{g(x),1}(y)=f_{e,2}(x,y).
$$
Therefore
$$
x\in X
\Longleftrightarrow f_{g(x),1}(g(x))\text{ is defined}
\Longleftrightarrow g(x)\in K.
$$
Thus $X\leq_mK$.

Conversely, suppose $X\leq_mK$ through a total computable $g$. On input $x$, compute $g(x)$ and simulate machine $g(x)$ on its own code. This procedure halts exactly when $g(x)\in K$, equivalently exactly when $x\in X$. Hence $X$ is recursively enumerable. We have proved the <many-one completeness of the halting problem>:
$$
\boxed{X\text{ is recursively enumerable}\Longleftrightarrow X\leq_mK}.
$$

Finally, use the stated fact that $0\notin K$ and define
$$
S=\{0\}\cup\{2n+1:n\in K\}.
$$
Then $0\in S\setminus K$, so $S\ne K$. The total computable map $n\mapsto2n+1$ satisfies
$$
n\in K\Longleftrightarrow2n+1\in S,
$$
and therefore $K\leq_mS$. This is a <distinct representative of the halting many-one degree>.

Solved by gpt-5.6-sol high.

= 5J
{parent=Paper 1}
{scope}
{title2=Statistical Modelling}

= Solution
{parent=5J}

Expand the exponent in the <inverse Gaussian distribution> with unit <shape parameter>:
$$
-\frac{(x-\mu)^2}{2\mu^2x}
=-\frac{x}{2\mu^2}+\frac1\mu-\frac1{2x}.
$$
Consequently its <probability density function> can be written as
$$
f(x;\mu)=h(x)\exp\{\eta x-A(\eta)\},
$$
where
$$
h(x)=\frac1{\sqrt{2\pi x^3}}\exp\left(-\frac1{2x}\right),
\qquad
\eta=-\frac1{2\mu^2}<0,
\qquad
A(\eta)=-\sqrt{-2\eta}.
$$
Indeed, $-A(\eta)=\sqrt{-2\eta}=1/\mu$. This is the canonical form of a one-parameter <exponential family>, and its <natural parameter of an exponential family>[natural parameter] is therefore
$$
\boxed{\eta=-\frac1{2\mu^2}}.
$$

Here the natural statistic is $T(X)=X$. The <exponential-family derivative identities> give
$$
\mathbb E_\eta X=A'(\eta)=(-2\eta)^{-1/2}=\mu
$$
and
$$
\operatorname{var}_\eta X=A''(\eta)=(-2\eta)^{-3/2}=\mu^3.
$$
Thus
$$
\boxed{\mathbb E X=\mu,
\qquad \operatorname{var}X=\mu^3}.
$$

Solved by gpt-5.6-sol high.

= 6E
{parent=Paper 1}
{scope}
{title2=Mathematical Biology}

= a
{parent=6e}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

The <per capita>[per-capita] net growth rate is the birth rate minus the death rate,
$$
be^{-aN}-d.
$$
Multiplying by the population size gives the <ordinary differential equation>
$$
\boxed{\frac{dN}{dt}=N\left(be^{-aN}-d\right)}.
$$
This is the <exponential density-dependent birth model>.

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

At a positive <equilibrium of an autonomous differential equation>[equilibrium], division by $N$ gives
$$
be^{-aN}=d.
$$
Since $b>d$, this has the positive solution
$$
\boxed{N^*=\frac1a\log\frac bd}.
$$

Write the vector field as $F(N)=N(be^{-aN}-d)$. Its <derivative> is
$$
F'(N)=be^{-aN}-d-abNe^{-aN}.
$$
At $N=N^*$, the first two terms cancel and $be^{-aN^*}=d$, so
$$
F'(N^*)=-adN^*<0.
$$
The one-dimensional <linear stability analysis> therefore shows that $N^*$ is locally <asymptotic stability>[asymptotically stable].

Solved by gpt-5.6-sol high.

= b
{parent=6e}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

The susceptible equation may be written
$$
\dot S
=\underbrace{be^{-aS}S}_{\text{birth of susceptibles}}
-\underbrace{\beta SI}_{\text{infection}}
-\underbrace{dS}_{\text{natural death}}.
$$
The factor $be^{-aS}$ is a density-dependent per-capita birth rate. The <law of mass action> gives the infection incidence $\beta SI$, which transfers individuals out of the susceptible compartment.

Likewise,
$$
\dot I
=\underbrace{\beta SI}_{\text{new infections}}
-\underbrace{dI}_{\text{natural death}}
-\underbrace{\delta I}_{\text{disease-induced death}}.
$$
There is no birth term in the infective equation, so only susceptible individuals reproduce in this model. Infectives die at total per-capita rate $d+\delta$. These terms define the <susceptible-infective model with exponential density-dependent birth>.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

The <disease-free equilibrium> is $(S,I)=(N^*,0)$. For
$$
F(S,I)=S(be^{-aS}-\beta I-d),
\qquad
G(S,I)=I(\beta S-d-\delta),
$$
the <Jacobian matrix> there is
$$
J(N^*,0)=
\begin{pmatrix}
-adN^*&-\beta N^*\\
0&\beta N^*-(d+\delta)
\end{pmatrix}.
$$
Because this matrix is triangular, its <eigenvalues>[eigenvalue] are its diagonal entries:
$$
-adN^*<0,
\qquad
\beta N^*-(d+\delta).
$$
The second eigenvalue is positive exactly when
$$
\boxed{N^*>\frac{d+\delta}{\beta}}.
$$
Under this condition the <linear stability analysis> gives an unstable disease-free equilibrium, agreeing with the <epidemic invasion threshold>.

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

At an <endemic equilibrium>, $I^*>0$. The equation $\dot I=0$ then gives
$$
\boxed{S^*=\frac{d+\delta}{\beta}}.
$$
Since also $S^*>0$, the equation $\dot S=0$ gives
$$
\boxed{\beta I^*+d=be^{-aS^*}},
\qquad
I^*=\frac{be^{-aS^*}-d}{\beta}.
$$
This infective population is positive precisely when
$$
S^*<\frac1a\log\frac bd=N^*,
$$
which is exactly the assumed instability condition for the disease-free equilibrium. Thus the <endemic equilibrium of the susceptible-infective model with exponential birth> exists.

At this equilibrium the relations above reduce the <Jacobian matrix> to
$$
J(S^*,I^*)=
\begin{pmatrix}
-aS^*be^{-aS^*}&-\beta S^*\\
\beta I^*&0
\end{pmatrix}.
$$
Hence
$$
\operatorname{tr}J=-aS^*be^{-aS^*}<0,
\qquad
\det J=\beta^2S^*I^*>0.
$$
By the <trace-determinant stability criterion>, both eigenvalues have negative real part. The <linearization stability theorem> therefore proves that the endemic equilibrium is locally <asymptotic stability>[asymptotically stable].

Solved by gpt-5.6-sol high.

= 7E
{parent=Paper 1}
{scope}
{title2=Further Complex Methods}

= Solution
{parent=7E}

Consider the <meromorphic function>
$$
f(z)=\frac{e^{iz}}{z(z^2-1)}.
$$
Close the <contour integration>[contour] with a large semicircle in the upper half-plane, indenting above the three real <poles>[pole] $-1,0,1$. The large-arc integral vanishes by <Jordan lemma>[Jordan's lemma], while the <upper-half-plane indentation rule> gives
$$
\operatorname{PV}\int_{-\infty}^{\infty}
\frac{e^{ix}}{x(x^2-1)}\,dx
=i\pi\sum_{a\in\{-1,0,1\}}\operatorname{Res}(f,a).
$$
The three <residues>[residue] are
$$
\operatorname{Res}(f,-1)=\frac{e^{-i}}2,
\qquad
\operatorname{Res}(f,0)=-1,
\qquad
\operatorname{Res}(f,1)=\frac{e^i}2.
$$
Their sum is $\cos1-1$, so the <residue theorem> yields
$$
\operatorname{PV}\int_{-\infty}^{\infty}
\frac{e^{ix}}{x(x^2-1)}\,dx
=i\pi(\cos1-1).
$$

The real part of the integrand is an <odd function>, so its full-line principal value is zero. Its imaginary part,
$$
\frac{\sin x}{x(x^2-1)},
$$
is an <even function>. Taking imaginary parts and halving the full-line integral therefore gives
$$
\boxed{
\operatorname{PV}\int_0^\infty\frac{\sin x}{x(x^2-1)}\,dx
=\frac\pi2(\cos1-1)
}.
$$

Solved by gpt-5.6-sol high.

= 8D
{parent=Paper 1}
{scope}
{title2=Classical Dynamics}

= a
{parent=8d}
{scope}

= Solution
{parent=a}

The <kinetic energy> of the two equal masses is
$$
T=\frac m2(\dot x_1^2+\dot x_2^2).
$$
The extensions of the left, coupling, and right springs are respectively $x_1$, $x_2-x_1$, and $x_2$. Their <potential energy> is therefore
$$
U=\frac12k_1x_1^2+\frac12k_3(x_2-x_1)^2+\frac12k_2x_2^2.
$$
Thus the mechanical <Lagrangian> $L=T-U$ has the requested quadratic form with the symmetric <small-oscillation mass and stiffness matrices>
$$
\boxed{
(T_{ij})=
\begin{pmatrix}m&0\\0&m\end{pmatrix},
\qquad
(V_{ij})=
\begin{pmatrix}
k_1+k_3&-k_3\\
-k_3&k_2+k_3
\end{pmatrix}
}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=8d}
{scope}

= Solution
{parent=b}

The <Euler-Lagrange equation>[Euler--Lagrange equations] for the quadratic Lagrangian are
$$
m\ddot x+\mathsf Vx=0.
$$
For a <normal mode> $x(t)=a e^{i\omega t}$, they become the <generalized eigenvalue problem for small oscillations>
$$
\mathsf Va=m\omega^2a.
$$
Put $\lambda=m\omega^2/k$. With the stated spring constants,
$$
\frac{\mathsf V}{k}
=
\begin{pmatrix}
1+\varepsilon(1+\delta)&-\varepsilon\\
-\varepsilon&1+\varepsilon(1-\delta)
\end{pmatrix}
=(1+\varepsilon)I
+\varepsilon
\begin{pmatrix}
\delta&-1\\
-1&-\delta
\end{pmatrix}.
$$
The final matrix has <characteristic polynomial>
$$
\nu^2-(1+\delta^2),
$$
so its <eigenvalues>[eigenvalue] are $\nu_\pm=\pm\sqrt{1+\delta^2}$. Hence
$$
\lambda_\pm
=1+\varepsilon\pm\varepsilon\sqrt{1+\delta^2}
=1+\varepsilon\left(1\pm\sqrt{1+\delta^2}\right).
$$
The two <angular frequencies> therefore satisfy
$$
\boxed{
\omega_\pm^2=\lambda_\pm\frac{k}{m},
\qquad
\lambda_\pm=1+\varepsilon\left(1\pm\sqrt{1+\delta^2}\right)
}.
$$
This is the spectrum recorded by <normal modes of two equal masses between three springs>.

Solved by gpt-5.6-sol high.

= 9B
{parent=Paper 1}
{scope}
{title2=Cosmology}

= a
{parent=9b}
{scope}

= Solution
{parent=a}

In the homogeneous background, $\mathbf v=0$ and the physical velocity is the <Hubble flow>
$$
\mathbf u=aH\mathbf x=\dot a\,\mathbf x.
$$
The <comoving coordinate> $\mathbf x$ is held fixed by the time derivative in the fluid equation, so
$$
\frac{\partial\mathbf u}{\partial t}=\ddot a\,\mathbf x.
$$
Moreover, spatial homogeneity gives $\nabla\bar P=0$. The <Euler equations for an inviscid fluid>[Euler equation] therefore reduces to
$$
\bar\rho a\ddot a\,\mathbf x=-\bar\rho\nabla\bar\Phi,
\qquad
\nabla\bar\Phi=-a\ddot a\,\mathbf x.
$$
Taking the <divergence> in three spatial dimensions gives
$$
\nabla^2\bar\Phi=-3a\ddot a.
$$
Consistency with the <Poisson equation>,
$$
\nabla^2\bar\Phi=\frac{4\pi G}{c^2}\bar\rho a^2,
$$
then requires
$$
\boxed{\frac{\ddot a}{a}
=-\frac{4\pi G}{3c^2}\bar\rho}.
$$
This is the pressureless, zero-cosmological-constant <Raychaudhuri equation>, obtained by the <Newtonian fluid derivation of the Raychaudhuri equation>.

Solved by gpt-5.6-sol high.

= b
{parent=9b}
{scope}

= Solution
{parent=b}

The elementary Newtonian derivation of the <Friedmann equation> follows a test particle on the boundary of a finite expanding sphere. It therefore introduces a distinguished centre and can misleadingly suggest that all matter expands away from one origin.

The fluid derivation is local: the <Euler equations for an inviscid fluid>[Euler equation] and <Poisson equation> must be compatible in every comoving neighbourhood. It consequently describes a genuinely homogeneous infinite universe, in which every comoving observer sees the same <Hubble flow>, without selecting a physical centre. This is the improvement captured by the <Newtonian fluid derivation of the Raychaudhuri equation>.

Solved by gpt-5.6-sol high.

= c
{parent=9b}
{scope}

= Solution
{parent=c}

Insert
$$
\rho=\bar\rho+\epsilon\,\delta\rho,
\qquad
\mathbf v=\epsilon\,\delta\mathbf v
$$
into the continuity equation. Its zeroth-order part is the homogeneous <cosmological continuity equation>
$$
\dot{\bar\rho}+3H\bar\rho=0.
$$
Since $\bar\rho$ is spatially homogeneous, the terms of first order in $\epsilon$ give
$$
\frac{\partial\delta\rho}{\partial t}
+3H\delta\rho
+\frac{\bar\rho}{a}\nabla\cdot\delta\mathbf v=0.
$$
On the other hand, the <quotient rule> and the background equation imply
$$
\frac{\partial}{\partial t}
\left(\frac{\delta\rho}{\bar\rho}\right)
=\frac1{\bar\rho}\frac{\partial\delta\rho}{\partial t}
-\frac{\delta\rho}{\bar\rho^2}\dot{\bar\rho}
=\frac1{\bar\rho}\frac{\partial\delta\rho}{\partial t}
+3H\frac{\delta\rho}{\bar\rho}.
$$
Substitution of the first-order equation yields the <linearized cosmological continuity equation>
$$
\boxed{
\frac{\partial}{\partial t}
\left(\frac{\delta\rho}{\bar\rho}\right)
=-\frac1a\nabla\cdot\delta\mathbf v
}.
$$
The ratio $\delta\rho/\bar\rho$ is the <density contrast>.

Solved by gpt-5.6-sol high.

= 10D
{parent=Paper 1}
{scope}
{title2=Quantum Information and Computation}

= a
{parent=10d}
{scope}

= Solution
{parent=a}

With equal <prior probability>[priors], the probability of a correct result from this <quantum measurement> is
$$
P_S
=\frac12|\langle b_0|a_0\rangle|^2
+\frac12|\langle b_1|a_1\rangle|^2.
$$
The two required amplitudes are
$$
\langle b_0|a_0\rangle=\cos\beta,
\qquad
\langle b_1|a_1\rangle
=\frac{\cos\beta-\sin\beta}{\sqrt2}.
$$
Therefore
$$
\begin{aligned}
P_S
&=\frac12\cos^2\beta
+\frac14(\cos\beta-\sin\beta)^2\\
&=\frac12+\frac14(\cos2\beta-\sin2\beta)\\
&=\frac12+\frac{\sqrt2}{4}
\cos\left(2\beta+\frac\pi4\right).
\end{aligned}
$$
Its maximum occurs at $\beta=-\pi/8$ and is
$$
\boxed{
P_S^{\max}
=\frac12+\frac1{2\sqrt2}
=\frac12\left(1+\frac1{\sqrt2}\right)
}.
$$

Here $|\langle a_0|a_1\rangle|=1/\sqrt2$, so $\theta=\pi/4$ and $\sin\theta=1/\sqrt2$. The stated <Helstrom-Holevo bound> is consequently exactly the same value, and the basis above is the <Helstrom measurement for the zero and plus states>.

Solved by gpt-5.6-sol high.

= b
{parent=10d}
{scope}

= Solution
{parent=b}

The <no-cloning theorem for two pure states> states that, for distinct nonorthogonal states $|c_0\rangle$ and $|c_1\rangle$, there is no <unitary operator> $U$ and fixed blank state $|0\rangle$ such that
$$
U\bigl(|c_j\rangle|0\rangle\bigr)
=|c_j\rangle|c_j\rangle,
\qquad j=0,1.
$$

To derive this from state discrimination, let
$$
s=|\langle c_0|c_1\rangle|,
\qquad 0<s<1,
$$
and suppose such a cloner existed. Applying it repeatedly would produce $N$ copies. The <inner product> of the two possible $N$-copy states has magnitude $s^N$, so their optimal success probability under the <Helstrom-Holevo bound> would be
$$
P_N=\frac12\left(1+\sqrt{1-s^{2N}}\right).
$$
Already for $N=2$,
$$
P_2>P_1
=\frac12\left(1+\sqrt{1-s^2}\right),
$$
because $0<s<1$ implies $s^4<s^2$. But cloning followed by the two-copy <Helstrom measurement for two pure states> would itself be a quantum procedure applied to the original single state, contradicting the optimal one-copy bound $P_1$. Hence the assumed unitary cloner cannot exist.

Equivalently, <clone-assisted asymptotic state discrimination> would give $P_N\to1$, although two nonorthogonal states are not <perfect distinguishability of pure states>[perfectly distinguishable]. Thus the Helstrom--Holevo theorem implies the no-cloning theorem.

Solved by gpt-5.6-sol high.

= 11K
{parent=Paper 1}
{scope}
{title2=Coding and Cryptography}

= Solution
{parent=11K}

Take every logarithm in base two. The <information entropy> of $X$ is
$$
\boxed{H(X)=-\sum_{i=1}^Np_i\log_2p_i},
$$
with $0\log_20=0$. If the codeword $c(\mu_i)$ has length $s_i$, then its <expected codeword length> is
$$
\boxed{\mathbb E(S)=\sum_{i=1}^Np_is_i}.
$$

For any <decipherable code>, the <Kraft inequality> gives
$$
K=\sum_{i=1}^N2^{-s_i}\leq1.
$$
Define the <probability distribution> $q_i=2^{-s_i}/K$. By <Gibbs inequality>,
$$
0\leq\sum_i p_i\log_2\frac{p_i}{q_i}
=-H(X)+\mathbb E(S)+\log_2K.
$$
Since $\log_2K\leq0$,
$$
\mathbb E(S)\geq H(X)-\log_2K\geq H(X).
$$

Conversely, choose the integer lengths
$$
s_i=\lceil-\log_2p_i\rceil.
$$
Then $2^{-s_i}\leq p_i$, so their Kraft sum is at most one and the converse part of Kraft's inequality supplies a binary <prefix code>. Moreover,
$$
\mathbb E(S)
=\sum_i p_i\lceil-\log_2p_i\rceil
<\sum_i p_i(-\log_2p_i+1)
=H(X)+1.
$$
Thus the minimum expected length $S^*$ satisfies the <Shannon noiseless coding theorem>
$$
\boxed{H(X)\leq S^*<H(X)+1}.
$$

For a decipherable code with arbitrary lengths $s_1,\ldots,s_N$, apply the lower bound to the uniform source $p_i=1/N$. Its entropy is $\log_2N$ and its expected length is $(s_1+\cdots+s_N)/N$. Hence
$$
\boxed{N\log_2N\leq s_1+\cdots+s_N},
$$
the <total length lower bound for a decipherable binary code>.

Now consider the cumulative construction. Since
$$
b_j-b_i=\sum_{k=i}^{j-1}p_k\geq p_i\geq2^{-s_i}
\qquad(j>i),
$$
$b_i$ and $b_j$ cannot have the same first $s_i$ binary digits: numbers sharing those digits lie in one half-open dyadic interval of length $2^{-s_i}$. Also, $p_i\geq p_j$ implies $s_i\leq s_j$. Therefore no earlier codeword $b_i^*$ is a prefix of a later one, and no later, weakly longer codeword can be a prefix of an earlier one. The <cumulative Shannon code> is thus a <prefix code>, hence decipherable.

An optimal code minimizes <expected codeword length> over all decipherable codes for the specified source probabilities. The cumulative construction need not be optimal. For example, if
$$
(p_1,p_2)=(0.6,0.4),
$$
then $(s_1,s_2)=(1,2)$ and the construction gives codewords $0$ and $10$, with expected length $1.4$. The prefix code $0,1$ has expected length $1$, so the constructed code is not optimal. In general, <Huffman coding> produces an optimal prefix code.

Solved by gpt-5.6-sol high.

= 12F
{parent=Paper 1}
{scope}
{title2=Automata and Formal Languages}

= i
{parent=12f}
{scope}

= Solution
{parent=i}

A <partial recursive function> is obtained from the three kinds of <initial function of recursion theory>—
the <zero function>, <successor function>, and <projection functions>[projection function]—by finitely many applications of <function composition in recursion theory>, <primitive recursion>, and <unbounded minimization>. Explicitly, primitive recursion has the form
$$
f(\mathbf x,0)=g(\mathbf x),
\qquad
f(\mathbf x,n+1)=h(\mathbf x,n,f(\mathbf x,n)),
$$
while minimization takes the least $y$ for which $g(\mathbf x,y)=0$ and is undefined if no such $y$ is found.

For the function in part (i), define
$$
s(0)=1,
\qquad
s(n+1)=0.
$$
The initial value $1=S(Z(0))$ and the identically zero recursion step are <primitive recursive function>[primitive recursive], so this is a primitive-recursive definition of the required zero test. It uses no minimization.

Solved by gpt-5.6-sol high.

= ii
{parent=12f}
{scope}

= Solution
{parent=ii}

Using the zero test $s$ from part (i), define by <primitive recursion>
$$
r(0)=0,
\qquad
r(n+1)=s(r(n)).
$$
Since $s(0)=1$ and $s(1)=0$, the values alternate:
$$
r(2k)=0,
\qquad
r(2k+1)=1.
$$
Thus $r$ is the required <parity> function and, like $s$, is primitive recursive without minimization.

Solved by gpt-5.6-sol high.

= iii
{parent=12f}
{scope}

= Solution
{parent=iii}

Let $g(n,k)=r(n)$, where $r$ is the total primitive-recursive parity function from part (ii), and apply <unbounded minimization> in $k$:
$$
p(n)=\mu k\,[g(n,k)=0].
$$
If $n$ is even, then $g(n,0)=0$ and $p(n)=0$. If $n$ is odd, then $g(n,k)=1$ for every $k$, so the search never terminates and $p(n)$ is undefined. Hence $p$ is partial recursive. Because every function built without minimization is total, $p$ cannot be defined without minimization; the functions $s$ and $r$ can.

It remains to characterize the functions built using only the initial functions and composition. They are exactly the <composition-only recursive functions>[functions]
$$
\boxed{f(x_1,\ldots,x_k)=c
\quad\hbox{or}\quad
f(x_1,\ldots,x_k)=x_i+c}
$$
for some $c\in\mathbb N$ and some input coordinate $i$.

Indeed, the <zero function> is the first form, the <successor function> and each <projection function> have the second form, and composing functions of these forms preserves the classification: a constant outer function remains constant, while an outer function $y_j+c$ selects one inner function and adds $c$. This proves necessity by <structural induction>. Conversely, applying the successor function $c$ times to zero produces the constant $c$, and applying it $c$ times to $\pi_i^k$ produces $x_i+c$.

Every function in this class is directly <total computable function>[computable]: an algorithm either writes the fixed constant $c$, or copies the $i$th input and performs $c$ successor steps.

Solved by gpt-5.6-sol high.

= 13J
{parent=Paper 1}
{scope}
{title2=Statistical Modelling}

= a
{parent=13j}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

Let $Y_{go}$ be the count for treatment
$$
g\in\{C,L,S\}
$$
(Control, LD, or SD) and outcome
$$
o\in\{B,W\}
$$
(Better or Worse), and write $\mu_{go}=\mathbb E Y_{go}$. Both fits take the six cell counts to be independent <Poisson distributions>[Poisson] variables.

The additive <log-linear model> fitted by `fit1` is
$$
Y_{go}\sim\operatorname{Pois}(\mu_{go}),
\qquad
\log\mu_{go}=\lambda+\alpha_g+\beta_o,
$$
with reference constraints $\alpha_C=0$ and $\beta_B=0$. Thus treatment changes the overall group count but not the relative frequencies of the two outcomes.

The interaction model fitted by `fit2` is
$$
Y_{go}\sim\operatorname{Pois}(\mu_{go}),
\qquad
\log\mu_{go}
=\lambda+\alpha_g+\beta_o+\gamma_{go},
$$
where, under reference coding,
$$
\gamma_{CB}=\gamma_{CW}=\gamma_{LB}=\gamma_{SB}=0
$$
and the two free interactions are $\gamma_{LW}$ and $\gamma_{SW}$. This is the <saturated log-linear model> for the $3\times2$ table.

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

Let
$$
N_g=Y_{gB}+Y_{gW}
$$
and condition on $N_g=n_g$. The corresponding model for each treatment group is
$$
(Y_{gB},Y_{gW})\mid N_g=n_g
\sim\operatorname{Mult}(n_g;\pi_{gB},\pi_{gW}).
$$
For `fit1`, the additive Poisson means imply
$$
\pi_{go}
=\frac{e^{\beta_o}}{e^{\beta_B}+e^{\beta_W}},
$$
so the outcome probabilities are common to all three treatments. For `fit2`,
$$
\pi_{go}
=\frac{e^{\beta_o+\gamma_{go}}}
{\sum_{r\in\{B,W\}}e^{\beta_r+\gamma_{gr}}},
$$
so each treatment may have its own outcome probabilities.

The <Poisson trick> is <Poisson-multinomial conditioning>: independent Poisson cell counts, conditional on their total, are multinomial with probabilities proportional to their means. With free treatment main effects, profiling the Poisson likelihood over the group totals gives the multinomial likelihood up to a parameter-independent factor. The Poisson regressions therefore fit these multinomial models and give the same likelihood-ratio comparisons for the outcome parameters.

Solved by gpt-5.6-sol high.

= iii
{parent=a}
{scope}

= Solution
{parent=iii}

The additive model has four free <regression coefficients>[coefficients]: an intercept, two treatment contrasts, and one outcome contrast. The interaction model adds
$$
(3-1)(2-1)=2
$$
treatment-by-outcome <interaction terms>[interaction term], making six parameters for six cells. It is saturated, so the residual degrees of freedom fall from two to zero and the <likelihood-ratio test> has
$$
\boxed{6-4=2\text{ degrees of freedom}}.
$$

The deviance reduction is $44.48$, with p-value $2.194\times10^{-10}$. The <analysis of deviance for nested generalized linear models> therefore gives overwhelming evidence against a common outcome distribution across the three groups. The observed proportions and the negative treatment--Worse interactions show that both doses reduce the chance of worsening relative to Control, with a larger reduction for LD.

Solved by gpt-5.6-sol high.

= b
{parent=13j}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

Because the second model is saturated, its fitted worsening probabilities are the observed proportions:
$$
\widehat p_C=\frac{101}{5728+101}=0.01733,
\qquad
\widehat p_L=\frac3{1364+3}=0.002195,
\qquad
\widehat p_S=\frac{27}{4413+27}=0.006081.
$$
Thus the <drug efficacy as a risk reduction> is
$$
\widehat E_L
=1-\frac{\widehat p_L}{\widehat p_C}
=1-0.1266
=\boxed{87.3\%},
$$
and
$$
\widehat E_S
=1-\frac{\widehat p_S}{\widehat p_C}
=1-0.3509
=\boxed{64.9\%}.
$$
These are reasonably similar to the published estimates of $90.0\%$ and $62.1\%$.

The interaction coefficients provide the same quick comparison on the odds scale:
$$
e^{-2.08156}=0.125
\quad\hbox{and}\quad
e^{-1.05847}=0.347.
$$
They are treatment-versus-control <odds ratios>[odds ratio]. Since worsening is rare in every group, each odds ratio is close to its <risk ratio>, giving the approximate efficacies $87.5\%$ and $65.3\%$.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

Equal LD and SD efficacy is equivalent to
$$
p_L=p_S,
$$
because both efficacy definitions use the same control probability. In this two-outcome table that is also equivalent to equality of the LD and SD log-odds interactions,
$$
H_0:\gamma_{LW}=\gamma_{SW}.
$$
The summary gives the two estimates and their individual <standard errors>[standard error], but a test of their difference needs
$$
\operatorname{Var}(\widehat\gamma_{LW}-\widehat\gamma_{SW})
=\operatorname{Var}(\widehat\gamma_{LW})
+\operatorname{Var}(\widehat\gamma_{SW})
-2\operatorname{Cov}(\widehat\gamma_{LW},\widehat\gamma_{SW}).
$$
The required <covariance> is absent from the displayed table, so the two separate coefficient p-values cannot test equality.

Instead, fit the <equal-efficacy Poisson log-linear model>. For example, create one indicator for a Worse outcome in either treated group:
```r
data$treated_worse <- with(data, treatment != "Control" & outcome == "Worse")
fit_equal <- glm(count ~ treatment + outcome + treated_worse,
                 family = poisson, data = data)
anova(fit_equal, fit2, test = "LRT")
```
The reduced model has one common treated-versus-control outcome interaction, while retaining separate LD and SD main effects for their different group totals. The full model has two such interactions. Their <analysis of deviance for nested generalized linear models> is therefore a one-degree-of-freedom likelihood-ratio test of equal efficacy.

Solved by gpt-5.6-sol high.

= 14E
{parent=Paper 1}
{scope}
{title2=Further Complex Methods}

= a
{parent=14e}
{scope}

= Solution
{parent=a}

The <identity theorem> states that if two <holomorphic functions>[holomorphic functions] on a connected <open set> $\mathcal D$ agree on a subset having an accumulation point in $\mathcal D$, then they agree throughout $\mathcal D$. In particular, agreement on any nonempty open subset is sufficient.

Solved by gpt-5.6-sol high.

= b
{parent=14e}
{scope}

= Solution
{parent=b}

An <analytic continuation> of $f_1$ from $\mathcal D_1$ to $\mathcal D_2$ is a holomorphic function $f_2$ on $\mathcal D_2$ whose restriction to the nonempty overlap $\mathcal D_1\cap\mathcal D_2$ agrees with $f_1$.

Suppose $f_2$ and $\widetilde f_2$ are two such continuations. Both equal $f_1$ on $\mathcal D_1\cap\mathcal D_2$, so
$$
f_2=\widetilde f_2
$$
on that nonempty open set. Since $\mathcal D_2$ is <connected space>[connected], the <identity theorem> gives
$$
\boxed{f_2=\widetilde f_2\text{ throughout }\mathcal D_2}.
$$
Thus analytic continuation to $\mathcal D_2$ is unique.

Solved by gpt-5.6-sol high.

= c
{parent=14e}
{scope}

= Solution
{parent=c}

For $\operatorname{Im}z>0$, close the $t$-contour in the upper half-plane. <Jordan lemma> removes the large semicircle, and the only enclosed <pole> is the order-$n$ pole at $t=z$. By the <residue at a pole of order n>,
$$
\operatorname{Res}_{t=z}\frac{e^{it}}{(t-z)^n}
=\frac1{(n-1)!}
\left.\frac{d^{\,n-1}}{dt^{\,n-1}}e^{it}\right|_{t=z}
=\frac{i^{n-1}e^{iz}}{(n-1)!}.
$$
The <residue theorem> therefore gives
$$
F(z)=\frac{2\pi i^ne^{iz}}{(n-1)!}.
$$

To continue the integral itself, move the $t$-contour downward locally as $z$ approaches and crosses the real axis, always keeping the pole above the contour. This <analytic continuation by contour deformation> defines
$$
\widetilde F(z)
=\int_{\mathcal C_z}\frac{e^{it}}{(t-z)^n}\,dt,
$$
where $\mathcal C_z$ passes below $z$. Closing $\mathcal C_z$ upward continues to enclose the pole, even when $\operatorname{Im}z\leq0$, so
$$
\boxed{
\widetilde F(z)=\frac{2\pi i^ne^{iz}}{(n-1)!}
}
$$
there as well. The right-hand side is an <entire function>, so it is the unique analytic continuation of $F$ to the whole <complex plane>.

Solved by gpt-5.6-sol high.

= d
{parent=14e}
{scope}

= Solution
{parent=d}

If $\operatorname{Im}z>0$, the same upper-half-plane contour calculation gives
$$
G(z)=\frac{2\pi i^ne^{iz}}{(n-1)!}.
$$
If $\operatorname{Im}z<0$, the pole lies below the real integration contour. Closing upward encloses no pole, so <Jordan lemma> and the <Cauchy integral theorem> give
$$
G(z)=0.
$$
Hence at a real point $x$ the upper and lower boundary values satisfy
$$
\boxed{
G(x+i0)-G(x-i0)
=\frac{2\pi i^ne^{ix}}{(n-1)!}
}.
$$
Thus crossing from the upper half-plane to the lower half-plane produces the negative of this jump.

An <analytic continuation> must be holomorphic, hence <continuous function>[continuous], across every point where it is defined. The undeformed real-axis formula for $G$ has the nonzero jump above, while the continuation constructed in part (c) retains $2\pi i^ne^{iz}/(n-1)!$ below the axis. Therefore $G$ cannot be the analytic continuation of $F$.

Solved by gpt-5.6-sol high.

= 15B
{parent=Paper 1}
{scope}
{title2=Cosmology}

= a
{parent=15b}
{scope}

= Solution
{parent=a}

Write the action density as
$$
\mathcal L
=a^3\left[
\frac12\dot\phi^2
-\frac{c^2}{2a^2}\nabla\phi\cdot\nabla\phi
-V(\phi)
\right].
$$
Under a compactly supported <variation> $\phi\mapsto\phi+\epsilon\eta$,
$$
\delta S
=\int d^3x\,dt
\left[
a^3\dot\phi\,\dot\eta
-ac^2\nabla\phi\cdot\nabla\eta
-a^3V'(\phi)\eta
\right].
$$
Integrating the first term by parts in time and the second in space, and discarding the <boundary condition>[boundary terms], gives
$$
\delta S
=\int d^3x\,dt
\left[
-\frac{\partial}{\partial t}(a^3\dot\phi)
+ac^2\nabla^2\phi
-a^3V'(\phi)
\right]\eta.
$$
The <principle of stationary action> requires this to vanish for every $\eta$. Since
$$
\frac{\partial}{\partial t}(a^3\dot\phi)
=a^3(\ddot\phi+3H\dot\phi),
$$
division by $-a^3$ yields the field <Euler-Lagrange equation>
$$
\boxed{
\ddot\phi+3H\dot\phi
-\frac{c^2}{a^2}\nabla^2\phi
+\frac{dV}{d\phi}=0
}.
$$
This is the equation generated by the <inflaton action in an expanding universe>.

Solved by gpt-5.6-sol high.

= b
{parent=15b}
{scope}

= Solution
{parent=b}

When $V$ is approximately constant, $dV/d\phi\simeq0$, and exponential expansion gives the constant <Hubble parameter> $H=H_{\rm inf}$. Take the spatial <Fourier transform>
$$
\phi(\mathbf x,t)
=\int\frac{d^3k}{(2\pi)^3}
\widehat\phi_{\mathbf k}(t)e^{i\mathbf k\cdot\mathbf x}.
$$
Since
$$
\nabla^2e^{i\mathbf k\cdot\mathbf x}
=-k^2e^{i\mathbf k\cdot\mathbf x},
$$
each independent <inflationary scalar Fourier mode> satisfies
$$
\boxed{
\ddot{\widehat\phi}_{\mathbf k}
+3H_{\rm inf}\dot{\widehat\phi}_{\mathbf k}
+\frac{c^2k^2}{a^2}\widehat\phi_{\mathbf k}=0
}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=15b}
{scope}

= Solution
{parent=c}

<Conformal time> is defined by
$$
d\tau=\frac{dt}{a(t)}.
$$
For $a=a_0e^{H_{\rm inf}t}$, choose $\tau\to0$ as $t\to+\infty$. Integration gives
$$
\boxed{
\tau=-\frac{e^{-H_{\rm inf}t}}{a_0H_{\rm inf}}
=-\frac1{aH_{\rm inf}},
\qquad -\infty<\tau<0
},
$$
as recorded by <conformal time during de Sitter expansion>.

Writing a prime for $d/d\tau$, the relations
$$
\frac d{dt}=\frac1a\frac d{d\tau},
\qquad
\frac{a'}a=-\frac1\tau
$$
turn the Fourier-mode equation into
$$
\widehat\phi_{\mathbf k}''
+2\frac{a'}a\widehat\phi_{\mathbf k}'
+c^2k^2\widehat\phi_{\mathbf k}=0.
$$
Set
$$
\widetilde\phi_{\mathbf k}
=a\widehat\phi_{\mathbf k}
=-\frac1{H_{\rm inf}\tau}\widehat\phi_{\mathbf k}.
$$
A direct substitution gives
$$
\widetilde\phi_{\mathbf k}''
+\left(c^2k^2-\frac{a''}a\right)
\widetilde\phi_{\mathbf k}=0.
$$
Here $a''/a=2/\tau^2$, so the <canonically rescaled de Sitter scalar mode> obeys
$$
\boxed{
\frac{d^2\widetilde\phi_{\mathbf k}}{d\tau^2}
+\left(c^2k^2-\frac2{\tau^2}\right)
\widetilde\phi_{\mathbf k}=0
}.
$$

Solved by gpt-5.6-sol high.

= d
{parent=15b}
{scope}

= Solution
{parent=d}

In the far past, $|ck\tau|\gg1$, the term $2/\tau^2$ is negligible and the rescaled mode is a <quantum harmonic oscillator> of angular frequency $ck$. The stated choice of its ground state is the <Bunch-Davies vacuum>.

At late times, after the mode is far outside the Hubble scale, $|ck\tau|\ll1$. Its <variance> becomes
$$
P_k
=\frac{\hbar H_{\rm inf}^2}{2c^3k^3}
\left(1+c^2k^2\tau^2\right)
\longrightarrow
\frac{\hbar H_{\rm inf}^2}{2c^3k^3}.
$$
Since $P_k$ has dimensions of length cubed, the dimensionless power is proportional to $k^3P_k$. More precisely, with the standard isotropic Fourier convention,
$$
\Delta^2(k)=\frac{k^3P_k}{2\pi^2}
\longrightarrow
\frac{\hbar H_{\rm inf}^2}{4\pi^2c^3},
$$
which is independent of the <wavenumber> $k$. Thus equal logarithmic intervals in $k$ carry equal late-time power. This is the sense in which inflation naturally generates a <scale-invariant inflationary power spectrum>.

Solved by gpt-5.6-sol high.

= 16G
{parent=Paper 1}
{scope}
{title2=Logic and Set Theory}

= a
{parent=16g}
{scope}

= Solution
{parent=a}

A set $S$ is a <deductively closed set of formulae> when every consequence of $S$ already belongs to it:
$$
S\vdash\varphi\Longrightarrow\varphi\in S.
$$
It is a <consistent set of formulae> when it does not prove a contradiction,
$$
S\nvdash\bot.
$$

If $S$ is inconsistent, there is a formal derivation of $\bot$ from assumptions in $S$. By the <finite character of formal proofs>, that derivation mentions only finitely many assumptions $s_1,\ldots,s_m\in S$. Hence
$$
\{s_1,\ldots,s_m\}\vdash\bot,
$$
so this finite subset of $S$ is already inconsistent.

Solved by gpt-5.6-sol high.

= b
{parent=16g}
{scope}

= Solution
{parent=b}

Suppose the <finitary set of formulae> $S$ is equivalent to the finite set
$$
F=\{f_1,\ldots,f_m\}.
$$
For each $f_j$, the relation $S\vdash f_j$ has a finite proof, so it uses a finite set $R_j\subseteq S$ of assumptions. Put
$$
R=\bigcup_{j=1}^mR_j.
$$
This is a finite subset of $S$, and $R\vdash f_j$ for every $j$, hence $R\vdash F$. Since equivalence also gives $F\vdash S$, transitivity of deduction yields
$$
\boxed{R\vdash S}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=16g}
{scope}

= Solution
{parent=c}

If some $T_i$ were inconsistent, it would prove every formula by the classical principle of explosion. Because $T_i$ is a <deductively closed set of formulae>, it would then contain every formula and could have no proper extension. This contradicts $T_i\subsetneq T_{i+1}$. Thus every $T_i$ is consistent.

Let
$$
T=\bigcup_{i=0}^{\infty}T_i.
$$
If $T$ were inconsistent, the <finite character of formal proofs> would give a finite inconsistent subset $R\subseteq T$. The chain is increasing, so all finitely many members of $R$ lie in some common $T_N$. Then $T_N$ would be inconsistent, a contradiction. Hence $T$ is consistent.

If $T\vdash\varphi$, a proof again uses only finitely many assumptions from $T$, all lying in some $T_N$. Thus
$$
T_N\vdash\varphi.
$$
Since $T_N$ is deductively closed, $\varphi\in T_N\subseteq T$. Therefore $T$ is deductively closed.

Finally, suppose that $T$ were finitary. Part (b) would provide a finite $R\subseteq T$ such that $R\vdash T$. Choose $N$ with $R\subseteq T_N$. Then $T_N\vdash T$; since $T_N$ is deductively closed, this implies $T\subseteq T_N$. But
$$
T_N\subsetneq T_{N+1}\subseteq T,
$$
a contradiction. Consequently $T$ is not finitary, proving the <increasing union of deductively closed sets>.

Solved by gpt-5.6-sol high.

= 17G
{parent=Paper 1}
{scope}
{title2=Graph Theory}

= a
{parent=17g}
{scope}

= Solution
{parent=a}

The <binomial random graph> $G(n,p)$ has vertex set $[n]=\{1,\ldots,n\}$, and each unordered pair of distinct vertices is included as an <edge of a graph>[edge] independently with probability $p$.

Let $X_t$ count copies of the <complete graph> $K_t$. Each $t$-element vertex set forms a copy with probability $p^{\binom t2}$, so <linearity of expectation> gives
$$
\mathbb EX_t=\binom ntp^{\binom t2}.
$$
The event $E_t$ is $\{X_t\geq1\}$. By the <first moment method>,
$$
\begin{aligned}
\mathbb P(E_t)
&\leq\mathbb EX_t\\
&\leq\frac{n^t}{t!}p^{t(t-1)/2}\\
&=\frac1{t!}
\left(pn^{2/(t-1)}\right)^{t(t-1)/2}
\longrightarrow0.
\end{aligned}
$$
This is the <clique count in a binomial random graph>.

Solved by gpt-5.6-sol high.

= b
{parent=17g}
{scope}

= Solution
{parent=b}

<Chebyshev inequality> states that, for a random variable $Y$ of finite <variance> and any $a>0$,
$$
\mathbb P(|Y-\mathbb EY|\geq a)
\leq\frac{\operatorname{var}Y}{a^2}.
$$

Let $X=\sum_T I_T$ count <triangles in a graph>[triangles], where $T$ ranges over the three-element vertex sets and $I_T$ is the corresponding <indicator random variable>. Then
$$
\mu=\mathbb EX=\binom n3p^3.
$$
Two distinct triangle indicators are independent unless their triangles share an edge. There are
$$
\binom n2\binom{n-2}2
$$
unordered pairs sharing an edge, and each covariance is
$$
p^5-p^6\leq p^5.
$$
Therefore
$$
\operatorname{var}X
\leq\binom n3p^3
+2\binom n2\binom{n-2}2p^5.
$$
It follows that
$$
\frac{\operatorname{var}X}{\mu^2}
=O\left(\frac1{n^3p^3}+\frac1{n^2p}\right)
\longrightarrow0
$$
when $np\to\infty$. Chebyshev's inequality now gives
$$
\mathbb P(X=0)
\leq\mathbb P(|X-\mu|\geq\mu)
\leq\frac{\operatorname{var}X}{\mu^2}
\longrightarrow0.
$$
Hence
$$
\boxed{\mathbb P(E_3)\longrightarrow1},
$$
as summarized by the <triangle count in a binomial random graph>.

Solved by gpt-5.6-sol high.

= c
{parent=17g}
{scope}

= Solution
{parent=c}

Partition the vertex set into disjoint sets $A$ and $B$ of sizes $\lfloor n/2\rfloor$ and $\lceil n/2\rceil$. The <induced subgraph> on $A$ has distribution
$$
G(\lfloor n/2\rfloor,p).
$$
For $p=n^{-0.9}$,
$$
p|A|\sim\frac12n^{0.1}\longrightarrow\infty,
$$
so part (b) shows that $A$ contains a triangle with probability tending to one.

On that event, choose a triangle using only the edges internal to $A$. The edges from its three vertices to $B$ remain independent of this choice. The probability that all $3|B|$ such edges are absent is
$$
(1-p)^{3|B|}
\leq e^{-3p|B|}
\longrightarrow0,
$$
because $p|B|\sim n^{0.1}/2$. Thus, with probability tending to one, some vertex of $B$ is adjacent to a vertex of the triangle. Those four vertices and the four required edges form the given graph $H$, a <triangle with an attached leaf in a binomial random graph>. Therefore
$$
\boxed{\mathbb P(F)\longrightarrow1}.
$$

Solved by gpt-5.6-sol high.

= 18I
{parent=Paper 1}
{scope}
{title2=Galois Theory}

= a
{parent=18i}
{scope}

= Solution
{parent=a}

A field extension $L/K$ is a <splitting field> for $f\in K[X]$ when $f$ splits into linear factors over $L$ and $L$ is generated over $K$ by the roots of $f$. Equivalently, it is the smallest extension inside an algebraic closure over which $f$ splits.

To prove existence, choose an irreducible factor of $f$, adjoin one of its roots, and factor $f$ over the resulting finite extension. If a nonlinear factor remains, adjoin one of its roots and repeat. Each step lowers the total degree of the unsplit factors, so after finitely many finite extensions the polynomial splits. The field generated by the roots obtained is therefore a splitting field.

The precise uniqueness theorem says that if $L$ and $L'$ are splitting fields of the same polynomial $f$ over $K$, then there is a <field embedding> isomorphism
$$
\varphi:L\longrightarrow L'
$$
whose restriction to $K$ is the identity. More generally, an isomorphism of base fields carrying one polynomial to another extends to an isomorphism of their splitting fields. Thus splitting fields are unique up to base-field isomorphism; inside a fixed algebraic closure, the field generated by all the roots is literally unique. These facts are summarized by <existence and uniqueness of splitting fields>.

Let
$$
\alpha=\sqrt[3]{2}>0,
\qquad
\omega=e^{2\pi i/3}.
$$
The roots of $X^3-2$ are $\alpha,\omega\alpha,\omega^2\alpha$, so
$$
\boxed{L=\mathbb Q(\alpha,\omega)}
$$
is the <splitting field of x cubed minus two>. As a subfield of $\mathbb C$ it is unique, because any splitting field in $\mathbb C$ must contain all three roots and hence must contain both $\alpha$ and
$$
\omega=\frac{\omega\alpha}{\alpha};
$$
minimality then forces it to equal $\mathbb Q(\alpha,\omega)$.

Solved by gpt-5.6-sol high.

= b
{parent=18i}
{scope}

= Solution
{parent=b}

The seventh <cyclotomic polynomial> is
$$
\Phi_7(X)=X^6+X^5+X^4+X^3+X^2+X+1.
$$
It is irreducible over $\mathbb Q$, and all its roots $\zeta_7^a$, $a=1,\ldots,6$, lie in $L=\mathbb Q(\zeta_7)$. Hence $L/\mathbb Q$ is a degree-six <finite Galois extension>, with
$$
G=\operatorname{Gal}(L/\mathbb Q)
=\{\sigma_a:a\in(\mathbb Z/7\mathbb Z)^\times\},
\qquad
\sigma_a(\zeta_7)=\zeta_7^a.
$$
The group $(\mathbb Z/7\mathbb Z)^\times$ is cyclic of order six; for example,
$$
G=\langle\sigma_3\rangle\cong C_6.
$$

By the <Galois correspondence>, the subfields are the fixed fields of the four subgroups of $C_6$:
$$
\begin{array}{c|c|c}
\operatorname{Gal}(L/M)&[M:\mathbb Q]&M\\ \hline
G&1&\mathbb Q\\
\{\sigma_1,\sigma_2,\sigma_4\}&2&
\mathbb Q(\eta),\quad
\eta=\zeta_7+\zeta_7^2+\zeta_7^4\\
\{\sigma_1,\sigma_6\}&3&
\mathbb Q(\theta),\quad
\theta=\zeta_7+\zeta_7^{-1}\\
\{\sigma_1\}&6&L.
\end{array}
$$
Thus these are all the subfields.

For the <quadratic Gaussian period in the seventh cyclotomic field>, put
$$
\bar\eta=\zeta_7^3+\zeta_7^5+\zeta_7^6.
$$
The cyclotomic relation gives $\eta+\bar\eta=-1$, while direct multiplication gives $\eta\bar\eta=2$. Hence
$$
\eta^2+\eta+2=0.
$$
Its discriminant is $-7$, so
$$
\eta=\frac{-1+\sqrt{-7}}2,
\qquad
m_{\eta,\mathbb Q}(X)=X^2+X+2.
$$

For the <real cubic subfield of the seventh cyclotomic field>, divide
$$
1+\zeta_7+\cdots+\zeta_7^6=0
$$
by $\zeta_7^3$. With $\theta=\zeta_7+\zeta_7^{-1}$,
$$
\zeta_7^2+\zeta_7^{-2}=\theta^2-2,
\qquad
\zeta_7^3+\zeta_7^{-3}=\theta^3-3\theta,
$$
so
$$
\theta^3+\theta^2-2\theta-1=0.
$$
This cubic has no rational root, and therefore
$$
m_{\theta,\mathbb Q}(X)=X^3+X^2-2X-1.
$$

The requested <primitive element of a field extension>[primitive elements], <minimal polynomial of an algebraic element>[minimal polynomials], and automorphism groups are consequently
$$
\begin{array}{c|c|c|c|c}
M&\text{primitive element}&m(X)&\operatorname{Aut}(M/\mathbb Q)&
\operatorname{Aut}(L/M)\\ \hline
\mathbb Q&0&X&1&C_6\\
\mathbb Q(\eta)&\eta&X^2+X+2&C_2&
\{\sigma_1,\sigma_2,\sigma_4\}\cong C_3\\
\mathbb Q(\theta)&\theta&X^3+X^2-2X-1&C_3&
\{\sigma_1,\sigma_6\}\cong C_2\\
L&\zeta_7&\Phi_7(X)&C_6&1.
\end{array}
$$
More explicitly, the nontrivial automorphism of $\mathbb Q(\eta)$ sends
$$
\eta\longmapsto\bar\eta=-1-\eta,
$$
and the three automorphisms of $\mathbb Q(\theta)$ cyclically permute
$$
\zeta_7+\zeta_7^{-1},\qquad
\zeta_7^2+\zeta_7^{-2},\qquad
\zeta_7^4+\zeta_7^{-4}.
$$

Finally, $G$ is abelian, so every subgroup is normal. The <subextensions of an abelian Galois extension> theorem shows that
$$
\boxed{\text{all four subfields }M\text{ are Galois over }\mathbb Q}.
$$

Solved by gpt-5.6-sol high.

= 19I
{parent=Paper 1}
{scope}
{title2=Representation Theory}

= a
{parent=19i}
{scope}

= Solution
{parent=a}

A representation is completely reducible, or <semisimple representation>[semisimple], when it is a <direct sum> of <irreducible representations>[irreducible representation]. Equivalently, every invariant subspace has an invariant complement.

<Maschke's theorem> says that every finite-dimensional representation of a finite group $G$ over a field of characteristic zero is completely reducible. More generally, it is enough that the field characteristic does not divide $|G|$.

<Schur lemma> states the following.

* If $V$ and $W$ are irreducible $G$-representations, every <intertwining operator> $T:V\to W$ is either zero or an isomorphism.
* If $V$ is finite-dimensional over an algebraically closed field, every $G$-endomorphism of $V$ is a scalar multiple of the identity.

Indeed, $\ker T\subseteq V$ and $\operatorname{im}T\subseteq W$ are invariant subspaces. If $T\ne0$, irreducibility forces
$$
\ker T=0,
\qquad
\operatorname{im}T=W,
$$
so $T$ is an isomorphism. For an endomorphism of a finite-dimensional complex representation, choose an <eigenvalue> $\lambda$ of $T$. The intertwiner $T-\lambda I$ has nonzero kernel, so the first part forces
$$
T-\lambda I=0.
$$
This proves the <proof of Schur lemma>.

Now let $\rho:G\to\operatorname{GL}(V)$ be faithful and irreducible over $\mathbb C$. For every $z\in Z(G)$, the operator $\rho(z)$ commutes with every $\rho(g)$. Schur's lemma therefore gives
$$
\rho(z)=\lambda_zI
$$
for some $\lambda_z\in\mathbb C^\times$. Faithfulness makes $z\mapsto\lambda_z$ an embedding
$$
Z(G)\hookrightarrow\mathbb C^\times.
$$
Since every <finite subgroup of the multiplicative complex numbers> is cyclic,
$$
\boxed{Z(G)\text{ is cyclic}}.
$$
This is the <cyclic-center obstruction to a faithful irreducible representation>.

Solved by gpt-5.6-sol high.

= b
{parent=19i}
{scope}

= Solution
{parent=b}

The complex <regular representation> has basis $(e_h)_{h\in G}$ and left action
$$
\rho(g)e_h=e_{gh}.
$$
If $\rho(g)$ is the identity, then
$$
e_g=\rho(g)e_1=e_1,
$$
so $g=1$. Thus $\mathbb CG$ is a <faithful representation>.

By <Maschke's theorem>, the regular representation decomposes as a direct sum of irreducibles,
$$
\mathbb CG=\bigoplus_jV_j.
$$
Its kernel is the intersection of the kernels of the constituent representations:
$$
\ker\rho=\bigcap_j\ker(\rho|_{V_j})=\{1\}.
$$
Each constituent kernel is a <normal subgroup> of $G$. If $G$ is simple, every such kernel is either $\{1\}$ or $G$. They cannot all be $G$, because then their intersection would be $G$, contradicting faithfulness of the regular representation. Hence some $V_j$ has trivial kernel. Therefore every finite simple group has a <faithful irreducible representation of a finite simple group>.

Solved by gpt-5.6-sol high.

= c
{parent=19i}
{scope}

= i
{parent=c}
{scope}

= Solution
{parent=i}

Yes. If $g$ generates the <cyclic group> $C_n$, define the one-dimensional <complex representation>
$$
\rho(g)=e^{2\pi i/n}.
$$
It is irreducible because it is one-dimensional, and it is faithful because the displayed scalar has multiplicative order $n$. Thus every $C_n$, including the trivial case $n=1$, has a faithful irreducible representation.

Solved by gpt-5.6-sol high.

= ii
{parent=c}
{scope}

= Solution
{parent=ii}

Yes. Write
$$
D_8=\langle r,s:r^4=s^2=1,\ srs=r^{-1}\rangle.
$$
The <faithful irreducible representation of D8> is
$$
\rho(r)=
\begin{pmatrix}i&0\\0&-i\end{pmatrix},
\qquad
\rho(s)=
\begin{pmatrix}0&1\\1&0\end{pmatrix}.
$$
These matrices satisfy the defining relations. The two eigenspaces of $\rho(r)$ are exchanged by $\rho(s)$, so there is no common one-dimensional invariant subspace; the representation is irreducible. Its eight matrices $\rho(r)^j$ and $\rho(r)^j\rho(s)$ are distinct, so it is faithful.

Solved by gpt-5.6-sol high.

= iii
{parent=c}
{scope}

= Solution
{parent=iii}

No. The <center of a group> satisfies
$$
Z(C_2\times D_8)
=C_2\times Z(D_8)
\cong C_2\times C_2,
$$
because $Z(D_8)=\{1,r^2\}\cong C_2$. This centre is not <cyclic group>[cyclic]. The <cyclic-center obstruction to a faithful irreducible representation> therefore rules out a faithful irreducible complex representation of $C_2\times D_8$.

Solved by gpt-5.6-sol high.

= 20G
{parent=Paper 1}
{scope}
{title2=Number Fields}

= a
{parent=20g}
{scope}

= Solution
{parent=a}

The element $\alpha$ is a root of
$$
f(T)=T^3-5T+8.
$$
By the <rational root theorem>, any rational root of this monic polynomial must be one of
$$
\mathord\pm1,\mathord\pm2,\mathord\pm4,\mathord\pm8.
$$
Direct substitution shows that none is a root. A reducible cubic over a field has a linear factor, so $f$ is an <irreducible polynomial> over $\mathbb Q$. It is therefore the <minimal polynomial of an algebraic element>[minimal polynomial] of $\alpha$, and
$$
\boxed{[K:\mathbb Q]=\deg f=3}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=20g}
{scope}

= Solution
{parent=b}

Put
$$
\beta=\frac{\alpha+\alpha^2}{2}.
$$
Using $\alpha^3=5\alpha-8$, multiplication by $\beta$ on the ordered basis $(1,\alpha,\beta)$ is determined by
$$
\beta\cdot1=\beta,
\qquad
\beta\alpha=-4+2\alpha+\beta,
\qquad
\beta^2=-4-\alpha+3\beta.
$$
Its matrix, with these coordinate vectors as columns, is
$$
M_\beta=
\begin{pmatrix}
0&-4&-4\\
0&2&-1\\
1&1&3
\end{pmatrix}.
$$
The <characteristic polynomial> is
$$
\det(TI-M_\beta)=T^3-5T^2+11T-12.
$$
By the <Cayley-Hamilton theorem>, this monic integer polynomial annihilates $\beta$, so $\beta$ is an <algebraic integer>.

Let
$$
A=\mathbb Z\langle1,\alpha,\beta\rangle.
$$
The element $\alpha$ is also integral because it satisfies the monic polynomial $f$. Hence $A\subseteq\mathcal O_K$. Since
$$
\alpha^2=2\beta-\alpha,
$$
the lattice $\mathbb Z[\alpha]$ has index two in $A$. The given power-basis discriminant is
$$
\Delta(1,\alpha,\alpha^2)=-4\cdot307,
$$
so the <discriminant-index formula for an integral lattice> gives
$$
\Delta(1,\alpha,\beta)
=\frac{-4\cdot307}{2^2}
=-307.
$$

If $m=[\mathcal O_K:A]$, then
$$
-307=m^2d_K.
$$
Because $307$ is prime, the square $m^2$ can divide $307$ only when $m=1$. Thus
$$
\boxed{\mathcal O_K=A
=\mathbb Z\langle1,\alpha,\beta\rangle},
$$
so $(1,\alpha,\beta)$ is an <integral basis of the cubic field of discriminant minus 307>.

Finally, the relation $\beta^2=-4-\alpha+3\beta$ gives
$$
\alpha=-4+3\beta-\beta^2.
$$
Consequently $\mathcal O_K=\mathbb Z[\beta]$, which will allow direct use of Dedekind's theorem.

Solved by gpt-5.6-sol high.

= c
{parent=20g}
{scope}

= Solution
{parent=c}

The <minimal polynomial of an algebraic element>[minimal polynomial] of $\beta$ is
$$
g(T)=T^3-5T^2+11T-12,
$$
and part (b) showed that $\mathcal O_K=\mathbb Z[\beta]$. Modulo $3$,
$$
\overline g(T)
=T(T^2+T-1).
$$
The quadratic has no root in $\mathbb F_3$, so it is irreducible. The <Dedekind factorization theorem> therefore gives
$$
\boxed{
(3)=(3,\beta)(3,\beta^2+\beta-1)
}.
$$
The two factors are distinct <prime ideals>[prime ideal], with norms $3$ and $3^2=9$, respectively.

Modulo $2$,
$$
\overline g(T)
=T(T^2+T+1),
$$
and the quadratic is irreducible over $\mathbb F_2$. Hence
$$
\boxed{
(2)=(2,\beta)(2,\beta^2+\beta+1)
}.
$$
This is a product of two distinct proper prime ideals, so $(2)$ is not prime. These decompositions are collected in <prime ideals above two and three in the cubic field of discriminant minus 307>.

Solved by gpt-5.6-sol high.

= 21F
{parent=Paper 1}
{scope}
{title2=Algebraic Topology}

= a
{parent=21f}
{scope}

= Solution
{parent=a}

Spaces $X$ and $Y$ are <homotopy equivalence>[homotopy equivalent] if there are continuous maps
$$
f:X\to Y,
\qquad
g:Y\to X
$$
such that
$$
g\circ f\simeq\operatorname{id}_X,
\qquad
f\circ g\simeq\operatorname{id}_Y.
$$
The maps $f$ and $g$ are called homotopy inverses.

Solved by gpt-5.6-sol high.

= b
{parent=21f}
{scope}

= Solution
{parent=b}

A subspace $Y\subseteq X$ is a <retract> of $X$ if there is a continuous retraction
$$
r:X\to Y
$$
such that $r(y)=y$ for every $y\in Y$. A space is a <contractible space> if its identity map is <homotopy>[homotopic] to a constant map.

Let $i:Y\hookrightarrow X$ be the inclusion and let
$$
H:X\times I\to X
$$
be a contraction, with $H(x,0)=x$ and $H(x,1)=x_0$. Then
$$
K(y,t)=r(H(i(y),t))
$$
is continuous and satisfies
$$
K(y,0)=r(i(y))=y,
\qquad
K(y,1)=r(x_0).
$$
Thus $K$ contracts $Y$, proving that every <retract of a contractible space> is contractible.

Solved by gpt-5.6-sol high.

= c
{parent=21f}
{scope}

= Solution
{parent=c}

Choose a contraction
$$
h:A\times I\to A,
\qquad
h(a,0)=a,
\qquad
h(a,1)=a_0.
$$
Apply the <homotopy extension property> to the initial map $\operatorname{id}_X:X\to X$ and to $h$, regarded as a homotopy into $X$. It supplies
$$
H:X\times I\to X
$$
such that
$$
H(x,0)=x,
\qquad
H(a,t)=h(a,t).
$$

At $t=1$, the map $H_1$ sends every point of $A$ to $a_0$. It is therefore constant on the equivalence classes of the <quotient topology> $X/A$. By the <universal property of the quotient topology>, there is a continuous map
$$
g:X/A\to X
$$
such that
$$
g\circ q=H_1,
$$
where $q:X\to X/A$ is the <quotient map>. The homotopy $H$ immediately gives
$$
g\circ q=H_1\simeq H_0=\operatorname{id}_X.
$$

For every $t$, the map $q\circ H_t$ is constant on $A$, because $H(a,t)=h(a,t)\in A$. It therefore descends to
$$
\overline H:(X/A)\times I\to X/A,
\qquad
\overline H([x],t)=q(H(x,t)).
$$
This descended map is continuous: the product $q\times\operatorname{id}_I$ is a quotient map because $I$ is compact Hausdorff, and $q\circ H$ is constant on its fibres.
At the endpoints,
$$
\overline H([x],0)=[x],
\qquad
\overline H([x],1)=q(g([x])).
$$
Thus
$$
q\circ g\simeq\operatorname{id}_{X/A}.
$$
The maps $q$ and $g$ are homotopy inverses, so
$$
\boxed{X\simeq X/A}.
$$
This proves the theorem on <collapsing a contractible cofibration>.

Solved by gpt-5.6-sol high.

= 22H
{parent=Paper 1}
{scope}
{title2=Linear Analysis}

= a
{parent=22h}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

Assume $x_n\rightharpoonup x_\infty$. For each $n$, define the bounded functional
$$
T_n(h)=\langle h,x_n\rangle.
$$
Weak convergence makes $(T_n(h))_n$ convergent, hence bounded, for every $h\in H$. The <uniform boundedness principle> gives
$$
\sup_n\|T_n\|<\infty.
$$
By the <Riesz representation theorem>, $\|T_n\|=\|x_n\|$, so $(x_n)$ is bounded.

Taking $h=e_i$ in the definition of <weak convergence in a Hilbert space> gives
$$
\langle x_n,e_i\rangle
\longrightarrow
\langle x_\infty,e_i\rangle
$$
for every $i$. Thus (i) implies (ii).

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

Conversely, suppose
$$
\|x_n\|\leq C
$$
and define
$$
a_i=\lim_{n\to\infty}\langle x_n,e_i\rangle.
$$
For every $N$, <Bessel inequality> and passage to the limit give
$$
\sum_{i=1}^N|a_i|^2
=\lim_{n\to\infty}
\sum_{i=1}^N|\langle x_n,e_i\rangle|^2
\leq C^2.
$$
Hence $\sum_i|a_i|^2\leq C^2$. The partial sums $\sum_{i=1}^Na_ie_i$ are therefore Cauchy; completeness of the <Hilbert space> gives an element
$$
x_\infty=\sum_{i=1}^{\infty}a_ie_i,
\qquad
\|x_\infty\|\leq C.
$$

Let $P_Nh=\sum_{i=1}^N\langle h,e_i\rangle e_i$. For fixed $N$, coordinate convergence gives
$$
\langle x_n-x_\infty,P_Nh\rangle\longrightarrow0.
$$
The <Hilbertian basis> property gives $P_Nh\to h$, and
$$
|\langle x_n-x_\infty,h-P_Nh\rangle|
\leq(\|x_n\|+\|x_\infty\|)\|h-P_Nh\|
\leq2C\|h-P_Nh\|.
$$
First choose $N$ large and then $n$ large. This proves
$$
\langle x_n,h\rangle\longrightarrow\langle x_\infty,h\rangle
$$
for every $h\in H$, so $x_n\rightharpoonup x_\infty$. This establishes the <coordinate criterion for weak convergence in a separable Hilbert space>.

Solved by gpt-5.6-sol high.

= b
{parent=22h}
{scope}

= Solution
{parent=b}

Let $\|x_n\|\leq C$. For each $i$, <Bessel inequality> gives
$$
|\langle x_n,e_i\rangle|\leq C,
$$
so the $i$th coordinate sequence is bounded. By the <Bolzano-Weierstrass theorem>, choose a subsequence on which the first coordinate converges. From it choose a further subsequence on which the second coordinate converges, and continue.

Take the diagonal subsequence $(x_{\phi(n)})$. For every fixed $i$, its $i$th coordinate converges, and it remains bounded by $C$. The <coordinate criterion for weak convergence in a separable Hilbert space> supplies an $x_\infty\in H$ such that
$$
\boxed{x_{\phi(n)}\rightharpoonup x_\infty}.
$$
This proves the theorem on a <weak subsequence of a bounded Hilbert-space sequence>.

Solved by gpt-5.6-sol high.

= c
{parent=22h}
{scope}

= i
{parent=c}
{scope}

= Solution
{parent=i}

Suppose first that $\|x_n-x_\infty\|\to0$. The reverse <triangle inequality> gives
$$
\big|\|x_n\|-\|x_\infty\|\big|
\leq\|x_n-x_\infty\|\longrightarrow0,
$$
so (i) implies (ii).

It also implies (iii). Given $\epsilon>0$, choose $N$ so that
$$
\|x_n-x_\infty\|^2<\frac\epsilon8
\qquad(n\geq N).
$$
By the <Parseval identity for a Hilbertian basis>, choose $I_0$ so that the basis tail of $x_\infty$ beyond $I_0$ is less than $\epsilon/8$. For $n\geq N$ and $I\geq I_0$,
$$
\sum_{i\geq I}|\langle x_n,e_i\rangle|^2
\leq
2\|x_n-x_\infty\|^2
+2\sum_{i\geq I}|\langle x_\infty,e_i\rangle|^2
<\frac\epsilon2.
$$
There are only finitely many $n<N$, so increasing $I$ makes each of their tails less than $\epsilon$. Thus one $I$ works for every $n$, proving (iii).

Solved by gpt-5.6-sol high.

= ii
{parent=c}
{scope}

= Solution
{parent=ii}

Assume (ii). Weak convergence gives
$$
\langle x_n,x_\infty\rangle
\longrightarrow
\langle x_\infty,x_\infty\rangle
=\|x_\infty\|^2.
$$
Using the <inner product> identity,
$$
\begin{aligned}
\|x_n-x_\infty\|^2
&=\|x_n\|^2+\|x_\infty\|^2
-2\operatorname{Re}\langle x_n,x_\infty\rangle\\
&\longrightarrow
\|x_\infty\|^2+\|x_\infty\|^2-2\|x_\infty\|^2=0.
\end{aligned}
$$
Hence (ii) implies (i); this is the <Radon-Riesz theorem> in a Hilbert space.

Solved by gpt-5.6-sol high.

= iii
{parent=c}
{scope}

= Solution
{parent=iii}

Assume (iii), and let $\delta>0$. Choose $I$ such that
$$
\sum_{i\geq I}|\langle x_n,e_i\rangle|^2<\delta
$$
for every $n$. Coordinate convergence implies, for every finite $J\geq I$,
$$
\sum_{i=I}^J|\langle x_\infty,e_i\rangle|^2
=\lim_{n\to\infty}
\sum_{i=I}^J|\langle x_n,e_i\rangle|^2
\leq\delta.
$$
Letting $J\to\infty$ shows that the corresponding tail of $x_\infty$ is also at most $\delta$.

By the <Parseval identity for a Hilbertian basis>,
$$
\begin{aligned}
\|x_n-x_\infty\|^2
&=\sum_{i<I}
|\langle x_n-x_\infty,e_i\rangle|^2\\
&\quad+\sum_{i\geq I}
|\langle x_n-x_\infty,e_i\rangle|^2.
\end{aligned}
$$
The finite first sum tends to zero by weak coordinate convergence, while the second is at most
$$
2\sum_{i\geq I}|\langle x_n,e_i\rangle|^2
+2\sum_{i\geq I}|\langle x_\infty,e_i\rangle|^2
\leq4\delta.
$$
Since $\delta$ is arbitrary, $\|x_n-x_\infty\|\to0$. Thus (iii) implies (i), completing the equivalence and proving the <uniform basis-tail criterion for strong convergence>.

Solved by gpt-5.6-sol high.

= 23H
{parent=Paper 1}
{scope}
{title2=Analysis of Functions}

= a
{parent=23h}
{scope}

= Solution
{parent=a}

For an <Lebesgue integrable function>[integrable] function $f:\mathbb R^n\to\mathbb C$, the <Lebesgue differentiation theorem> states that
$$
\lim_{r\downarrow0}\frac1{\lambda(B(x,r))}
\int_{B(x,r)}|f(y)-f(x)|\,d\lambda(y)=0
$$
for <almost everywhere> $x\in\mathbb R^n$. In particular,
$$
\lim_{r\downarrow0}\frac1{\lambda(B(x,r))}
\int_{B(x,r)}f(y)\,d\lambda(y)=f(x)
$$
at every <Lebesgue point> of $f$.

Almost every $x\in\mathbb R$ is a Lebesgue point of $g$. At such an $x$, for $t>0$,
$$
\left|\frac{G(x+t)-G(x)}t-g(x)\right|
\leq\frac1t\int_x^{x+t}|g(y)-g(x)|\,d\lambda(y)\longrightarrow0.
$$
For $t<0$, the same estimate over $[x+t,x]$ gives the identical <limit>. Therefore
$$
\boxed{G'(x)=g(x)}
$$
at every Lebesgue point of $g$, so $G$ is <differentiable function>[differentiable] $\lambda$-almost everywhere. This is the <differentiation of an indefinite Lebesgue integral>.

Solved by gpt-5.6-sol high.

= b
{parent=23h}
{scope}

= Solution
{parent=b}

Every <Lebesgue measurable set> $A$ differs from a <Borel set> $B$ by a <Lebesgue measure>[Lebesgue-null] set $N$. The given function $h$ is a <homeomorphism> from $\mathbb R$ onto the interval $h(\mathbb R)$. Hence $h(B)$ is Borel relative to $h(\mathbb R)$ and therefore Lebesgue measurable in $\mathbb R$. Moreover,
$$
h(A)\mathbin{\triangle}h(B)\subseteq h(A\mathbin{\triangle}B)\subseteq h(N),
$$
and $h(N)$ is null by <Lusin condition N>. The <completeness of Lebesgue measure> now shows that $h(A)$ is measurable, so
$$
\nu(A):=\lambda(h(A))
$$
is defined for every $A\in\mathcal M$.

Because $h$ is <injective function>[injective], it maps pairwise disjoint sets to pairwise disjoint sets and commutes with arbitrary unions. The <countable additivity> of $\lambda$ therefore gives
$$
\nu\left(\bigcup_{j=1}^{\infty}A_j\right)
=\lambda\left(\bigcup_{j=1}^{\infty}h(A_j)\right)
=\sum_{j=1}^{\infty}\lambda(h(A_j))
=\sum_{j=1}^{\infty}\nu(A_j).
$$
Thus $\nu$ is the <image-length measure of a strictly increasing continuous function>. If $\lambda(A)=0$, the null-set hypothesis gives $\nu(A)=\lambda(h(A))=0$, so
$$
\boxed{\nu\ll\lambda}.
$$

The measure is <sigma-finite measure>[sigma-finite], since
$$
\nu([-m,m])=h(m)-h(-m)<\infty.
$$
The <Radon-Nikodym theorem> consequently supplies a nonnegative locally integrable function $f$ such that
$$
\nu(A)=\int_Af\,d\lambda.
$$
For $a<x$, strict increase and continuity give
$$
h([a,x])=[h(a),h(x)]
$$
and hence
$$
h(x)-h(a)=\nu([a,x])=\int_a^x f(t)\,dt.
$$
The analogous oriented identity holds for $x<a$. By <differentiation of an indefinite Lebesgue integral>,
$$
\boxed{h'(x)=f(x)\quad\text{for almost every }x}.
$$

Yes, the differentiability conclusion still holds when $h$ is merely non-decreasing. In fact, the <Lebesgue theorem on differentiability of monotone functions> says that every <monotone function> is differentiable almost everywhere; neither continuity nor the null-set hypothesis is needed for that conclusion. Flat intervals mean that the preceding image-set argument no longer gives disjoint images, so its natural replacement is the <Lebesgue-Stieltjes measure> determined by
$$
\mu_h((a,b])=h(b)-h(a).
$$

Solved by gpt-5.6-sol high.

= 24F
{parent=Paper 1}
{scope}
{title2=Riemann Surfaces}

= a
{parent=24f}
{scope}

= Solution
{parent=a}

Suppose first that $u$ is <harmonic function>[harmonic]. Since
$$
\frac{\partial}{\partial y}u_x
=\frac{\partial}{\partial x}u_y
\quad\text{and}\quad
\frac{\partial}{\partial x}u_x
=-\frac{\partial}{\partial y}u_y,
$$
the <Cauchy-Riemann equations> show that
$$
g:=u_x-iu_y
$$
is <holomorphic function>[holomorphic] on $D$. An open disc is a <simply connected domain>, so $g$ has a <primitive of a holomorphic function on a simply connected domain>[holomorphic primitive] $f$. If $f=p+iq$, then
$$
p_x=\operatorname{Re}f'=u_x,
\qquad
p_y=-\operatorname{Im}f'=u_y.
$$
Thus $p-u$ has zero <gradient> and is constant on the connected disc. Subtracting that real constant from $f$ gives
$$
\boxed{u=\operatorname{Re}f}.
$$

Conversely, if $u=\operatorname{Re}f$ for a holomorphic $f=u+iv$, differentiating the Cauchy-Riemann equations gives
$$
\Delta u=u_{xx}+u_{yy}=v_{yx}-v_{xy}=0.
$$
Hence $u$ is harmonic. This proves the <harmonic function as the real part of a holomorphic function> criterion on $D$.

Solved by gpt-5.6-sol high.

= b
{parent=24f}
{scope}

= Solution
{parent=b}

Take the <punctured complex plane>
$$
D=\mathbb C^*=\mathbb C\setminus\{0\}
$$
and
$$
u(z)=\log|z|.
$$
In <polar coordinates>, $u(r,\theta)=\log r$, so
$$
\Delta u=u_{rr}+\frac1r u_r+\frac1{r^2}u_{\theta\theta}
=-\frac1{r^2}+\frac1{r^2}=0.
$$
Thus $u$ is harmonic on $D$.

If $u=\operatorname{Re}f$ for a holomorphic function $f$, the calculation from part (a) would give
$$
f'(z)=u_x-iu_y=\frac1z.
$$
The integral of a derivative around a closed curve is zero, whereas the <residue theorem> gives
$$
\int_{|z|=1}\frac{dz}{z}=2\pi i.
$$
This contradiction proves that no such $f$ exists. It is precisely the <log modulus has no global harmonic conjugate on the punctured plane> obstruction.

Solved by gpt-5.6-sol high.

= c
{parent=24f}
{scope}

= Solution
{parent=c}

Lift $u$ through the <exponential function>[complex exponential] by setting
$$
v(w)=u(e^w),\qquad w\in\mathbb C.
$$
The <conformal invariance of harmonicity> makes $v$ harmonic. Since $e^{w+2\pi i}=e^w$ and $u(2z)=u(z)$,
$$
v(w+2\pi i)=v(w),
\qquad
v(w+\log2)=v(w).
$$
Thus $v$ is bounded on the compact rectangle
$$
\{s\log2+2\pi it:0\leq s,t\leq1\},
$$
and its two <period lattice>[periods] make it bounded on the whole plane. The <harmonic Liouville theorem> makes $v$ constant. Since the exponential map is surjective onto $\mathbb C^*$, $u$ is constant.

For the requested counterexample after deleting a countable set, take
$$
S=\{2^m:m\in\mathbb Z\}
$$
and let $\wp_\Lambda$ be the <Weierstrass elliptic function> for the lattice
$$
\Lambda=(\log2)\mathbb Z+2\pi i\mathbb Z.
$$
Define
$$
u(z)=\operatorname{Re}\wp_\Lambda(\log z),
\qquad z\in\mathbb C^*\setminus S.
$$
Changing a branch of the logarithm adds $2\pi i$, a period of $\wp_\Lambda$, so $u$ is well defined. Its poles project precisely to $S$, and the <conformal invariance of harmonicity> makes $u$ harmonic away from them. The other period gives
$$
u(2z)=\operatorname{Re}\wp_\Lambda(\log z+\log2)=u(z).
$$
Moreover, $z\notin S$ implies $2z\notin S$. Finally, $u$ is nonconstant because as positive real $z\to1$,
$$
\wp_\Lambda(\log z)\sim\frac1{(\log z)^2},
$$
which is real and unbounded. This is a <scale-periodic harmonic function from an elliptic function>.

Solved by gpt-5.6-sol high.

= d
{parent=24f}
{scope}

= Solution
{parent=d}

Since $\mathbb C$ is <simply connected domain>[simply connected], write
$$
u=\operatorname{Re}f
$$
for an <entire function> $f$. If $u$ were bounded above, then
$$
|e^f|=e^u
$$
would be bounded, so the <Liouville theorem> would make $e^f$ constant. Its derivative $f'e^f$ would then vanish, whence $f$ and $u$ would be constant. Applying the same argument to $e^{-f}$ rules out a lower bound. Thus a nonconstant $u$ is unbounded both above and below.

Given any $c\in\mathbb R$, choose $z_-,z_+\in\mathbb C$ with
$$
u(z_-)<c<u(z_+).
$$
The restriction of $u$ to the line segment from $z_-$ to $z_+$ is <continuous function>[continuous], so the <intermediate value theorem> supplies a point on that segment where $u=c$. Therefore
$$
\boxed{u(\mathbb C)=\mathbb R}.
$$
This proves the <surjectivity of a nonconstant entire harmonic function>.

Solved by gpt-5.6-sol high.

= 25I
{parent=Paper 1}
{scope}
{title2=Algebraic Geometry}

= Solution
{parent=25I}

The <Hilbert basis theorem> makes $k[x_1,\ldots,x_n]$ a <Noetherian ring>. Reversing the inclusion of ideals shows that its <Zariski-closed set>[Zariski-closed subsets] satisfy the descending-chain condition, so every affine variety is a <Noetherian Zariski topology>[Noetherian topological space].

Suppose some nonempty closed subset had no finite decomposition into <irreducible topological space>[irreducible] closed subsets. It would be reducible, so it could be written
$$
V=V_1\cup V_2
$$
with $V_1,V_2$ proper closed subsets. At least one of these would again admit no finite irreducible decomposition. Repeating that choice would produce an infinite strictly descending chain of closed subsets, contradicting Noetherianity. Therefore
$$
\boxed{V\text{ is a finite union of irreducible subvarieties}.}
$$
Removing any member contained in another leaves exactly the <irreducible component>[irreducible components].

Solved by gpt-5.6-sol high.

= i
{parent=25i}
{scope}

= Solution
{parent=i}

A point $p$ belongs to $V(I_1+I_2)$ exactly when every polynomial $f_1+f_2$, with $f_j\in I_j$, vanishes at $p$. This implies that every member of each ideal vanishes by taking the other summand to be zero, and the converse follows by addition. Hence
$$
\boxed{V(I_1+I_2)=V(I_1)\cap V(I_2)}.
$$
This is the first <vanishing loci of ideal sums and intersections> identity.

Solved by gpt-5.6-sol high.

= ii
{parent=25i}
{scope}

= Solution
{parent=ii}

Since $I_1\cap I_2\subseteq I_j$, every point of either $V(I_j)$ lies in $V(I_1\cap I_2)$. Conversely, suppose
$$
p\notin V(I_1)\cup V(I_2).
$$
Choose $f_j\in I_j$ with $f_j(p)\ne0$. Then
$$
f_1f_2\in I_1I_2\subseteq I_1\cap I_2
$$
but $(f_1f_2)(p)\ne0$, so $p\notin V(I_1\cap I_2)$. Therefore
$$
\boxed{V(I_1\cap I_2)=V(I_1)\cup V(I_2)}.
$$

Solved by gpt-5.6-sol high.

= Decomposition of the given vanishing locus
{parent=25I}
{scope}

= Solution
{parent=Decomposition of the given vanishing locus}

Subtracting the two defining equations gives
$$
y^2+z^2=(y-iz)(y+iz)=0.
$$
Consequently the variety is
$$
V(y-iz,x^2-z^2-1)
\ \cup\
V(y+iz,x^2-z^2-1).
$$
Each member is irreducible. Indeed, after imposing its linear equation, its <coordinate ring> is
$$
\mathbb C[x,z]/(x^2-z^2-1).
$$
The invertible linear substitution
$$
u=x-z,\qquad v=x+z
$$
turns this into
$$
\mathbb C[u,v]/(uv-1)\cong\mathbb C[u,u^{-1}],
$$
which is an <integral domain>. The two curves are distinct and neither contains the other, so they are precisely the <irreducible components of two intersecting complex hyperbolas>.

Solved by gpt-5.6-sol high.

= Non-isomorphism of the two unions of lines
{parent=25I}
{scope}

= Solution
{parent=Non-isomorphism of the two unions of lines}

The three coordinate axes are the three <irreducible components> of
$$
V=V(xy,xz,yz)\subseteq\mathbb A_k^3.
$$
Their common intersection is the origin. Likewise, if $\ell_1,\ell_2,\ell_3$ are distinct linear forms defining the three lines in $\mathbb A_k^2$, then
$$
W=V(\ell_1\ell_2\ell_3),
$$
and its three irreducible components have only the origin in common. Any <isomorphism of algebraic varieties> would permute irreducible components and therefore map the origin of $V$ to the origin of $W$.

Every defining polynomial $xy,xz,yz$ has zero differential at the origin, so
$$
T_0V=k^3
\quad\text{and}\quad
\dim T_0V=3.
$$
The defining polynomial of $W$ is homogeneous of degree three, so its differential also vanishes at the origin, giving
$$
T_0W=k^2
\quad\text{and}\quad
\dim T_0W=2.
$$
An isomorphism induces an isomorphism of <Zariski tangent space>[Zariski tangent spaces], which is impossible here. Hence $V$ and $W$ are not isomorphic. This is the <tangent-space obstruction between two unions of three lines>.

Solved by gpt-5.6-sol high.

= 26F
{parent=Paper 1}
{scope}
{title2=Differential Geometry}

= a
{parent=26f}
{scope}

= Solution
{parent=a}

For each $p\in S$, the <first fundamental form> is defined without coordinates by
$$
I_p(v,w)=v\mathbin{\cdot}w,
\qquad v,w\in T_pS,
$$
the restriction of the Euclidean <inner product> to the <tangent space> of $S$.

If $\phi(u,v)$ is a local parametrization and
$$
\xi=\xi^1\partial_u+\xi^2\partial_v,
\qquad
\eta=\eta^1\partial_u+\eta^2\partial_v,
$$
then
$$
d\phi(\xi)=\xi^1\phi_u+\xi^2\phi_v,
\qquad
d\phi(\eta)=\eta^1\phi_u+\eta^2\phi_v.
$$
Taking their dot product gives
$$
I=E\,du^2+2F\,du\,dv+G\,dv^2,
$$
where
$$
\boxed{E=\phi_u\cdot\phi_u,\qquad
F=\phi_u\cdot\phi_v,\qquad
G=\phi_v\cdot\phi_v}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=26f}
{scope}

= Solution
{parent=b}

Since $z>0$, the cone has the parametrization
$$
\phi(r,\theta)
=(r\cos\theta,r\sin\theta,\sqrt a\,r),
\qquad r>0.
$$
Its coordinate derivatives satisfy
$$
\phi_r\cdot\phi_r=1+a,
\qquad
\phi_r\cdot\phi_\theta=0,
\qquad
\phi_\theta\cdot\phi_\theta=r^2,
$$
so its <first fundamental form> is
$$
I=(1+a)\,dr^2+r^2\,d\theta^2.
$$

On a sufficiently narrow angular patch, define
$$
R=\sqrt{1+a}\,r,
\qquad
\Theta=\frac{\theta}{\sqrt{1+a}},
$$
and map the cone to the plane point with <polar coordinates> $(R,\Theta)$. The Euclidean metric pulls back as
$$
dR^2+R^2d\Theta^2
=(1+a)\,dr^2+r^2\,d\theta^2=I.
$$
The map is therefore a <local isometry from a circular cone to the plane>, proving that $\Sigma$ is locally isometric to the Euclidean plane.

Solved by gpt-5.6-sol high.

= c
{parent=26f}
{scope}

= Solution
{parent=c}

For a chosen <unit normal> field $N$, the <second fundamental form> is
$$
II_p(v,w)=\langle-dN_p(v),w\rangle.
$$
In a local parametrization $\phi$, its coefficients, up to the equivalent simultaneous sign convention, are
$$
e=N\cdot\phi_{uu},
\qquad
f=N\cdot\phi_{uv},
\qquad
g=N\cdot\phi_{vv}.
$$
The <Gaussian curvature> is the <determinant> of the <shape operator>, hence
$$
K=\det(-dN)=\frac{eg-f^2}{EG-F^2}.
$$
<Theorema Egregium> states that $K$ is determined entirely by the first fundamental form. In particular, it is intrinsic and is preserved by every <local isometry>.

Solved by gpt-5.6-sol high.

= i
{parent=c}
{scope}

= Solution
{parent=i}

Let
$$
F(x,y,z)=x^2+y^2+z^2-2xy-2yz.
$$
Its <gradient> is
$$
\nabla F=2(x-y,\ y-x-z,\ z-y).
$$
If this vanishes, the first and third coordinates give $x=y=z$, and the middle coordinate then gives $y=0$. Thus the origin is the only <critical point> of $F$. It has been removed from $V$, so $0$ is a <regular value> of the restriction under consideration. The <regular level set theorem> shows that
$$
V=F^{-1}(0)\setminus\{0\}
$$
is a smooth surface.

Solved by gpt-5.6-sol high.

= ii
{parent=c}
{scope}

= Solution
{parent=ii}

Completing the square gives
$$
F=(x-y)^2+(z-y)^2-y^2.
$$
Thus $V$ is a two-napped cone. On the nappe where $\varepsilon y>0$, with $\varepsilon\in\{1,-1\}$, use
$$
\phi_\varepsilon(r,\theta)
=\bigl(\varepsilon r+r\cos\theta,\ \varepsilon r,\ 
\varepsilon r+r\sin\theta\bigr),
\qquad r>0.
$$
This has the conical form
$$
\phi_\varepsilon(r,\theta)=r\,c_\varepsilon(\theta).
$$
Its second derivatives obey
$$
\phi_{rr}=0,
\qquad
\phi_{r\theta}=c_\varepsilon'(\theta)=\frac1r\phi_\theta.
$$
The latter vector is tangent, so for either choice of unit normal,
$$
e=N\cdot\phi_{rr}=0,
\qquad
f=N\cdot\phi_{r\theta}=0.
$$
Since the parametrization is regular, $EG-F^2>0$, and therefore
$$
\boxed{K=\frac{eg-f^2}{EG-F^2}=0}
$$
at every point of $V$. This is the <Gaussian curvature of a cone away from its vertex>.

Solved by gpt-5.6-sol high.

= 27H
{parent=Paper 1}
{scope}
{title2=Probability and Measure}

= a
{parent=27h}
{scope}

= Solution
{parent=a}

<Fatou lemma> states that for any sequence $(f_n)$ of nonnegative measurable functions on a <measure space>,
$$
\boxed{\int\liminf_{n\to\infty}f_n\,d\mu
\leq
\liminf_{n\to\infty}\int f_n\,d\mu}.
$$

For the proof, define
$$
g_n=\inf_{k\geq n}f_k.
$$
Then $(g_n)$ is a nonnegative increasing sequence and
$$
g_n\uparrow\liminf_{k\to\infty}f_k.
$$
The <monotone convergence theorem> therefore gives
$$
\int\liminf_kf_k\,d\mu
=\lim_{n\to\infty}\int g_n\,d\mu.
$$
For each $k\geq n$, we have $g_n\leq f_k$, so <monotonicity of the Lebesgue integral> yields
$$
\int g_n\,d\mu
\leq\inf_{k\geq n}\int f_k\,d\mu.
$$
Taking the limit in $n$ proves the asserted inequality and the <proof of Fatou lemma>.

Solved by gpt-5.6-sol high.

= b
{parent=27h}
{scope}

= Solution
{parent=b}

On the <probability space> $([0,1],\mathcal B,\lambda)$, let
$$
f_n=n\mathbf1_{(0,1/n)}.
$$
For every $x\in[0,1]$, eventually $x\notin(0,1/n)$, so $f_n(x)\to0$. Nevertheless,
$$
\int_0^1f_n\,d\lambda=1
$$
for every $n$. Consequently
$$
\int\liminf_nf_n\,d\lambda=0
<1=\liminf_n\int f_n\,d\lambda.
$$
This gives <strict inequality in Fatou lemma>.

Solved by gpt-5.6-sol high.

= c
{parent=27h}
{scope}

= Solution
{parent=c}

Yes. Almost-sure convergence gives
$$
X=\liminf_{n\to\infty}X_n
$$
almost surely. Since the <random variable>[random variables] are nonnegative, <Fatou lemma> applies and gives
$$
\mathbb E X
\leq\liminf_{n\to\infty}\mathbb E X_n
\leq\sup_{n\in\mathbb N}\mathbb E X_n.
$$
The inequalities remain valid in the extended nonnegative reals, so no integrability assumption beyond nonnegativity is required.

Solved by gpt-5.6-sol high.

= 28K
{parent=Paper 1}
{scope}
{title2=Applied Probability}

= a
{parent=28k}
{scope}

= Solution
{parent=a}

Let $N_j(A)$ be the number of particles of gas $j$ in a bounded <Lebesgue measurable set> $A$. Because Lebesgue measure is diffuse, two independent spatial <Poisson point process>[Poisson point processes] have no common particle almost surely. Hence the mixture count is
$$
N(A)=N_1(A)+N_2(A).
$$
Writing $\lambda_j=z_j|A|$, independence gives the <probability generating function>
$$
\begin{aligned}
\mathbb E[s^{N(A)}]
&=\mathbb E[s^{N_1(A)}]\mathbb E[s^{N_2(A)}]\\
&=\exp(\lambda_1(s-1))\exp(\lambda_2(s-1))\\
&=\exp\bigl((z_1+z_2)|A|(s-1)\bigr).
\end{aligned}
$$
Thus
$$
N(A)\sim\operatorname{Pois}\bigl((z_1+z_2)|A|\bigr).
$$

If $A_1,\ldots,A_m$ are pairwise disjoint, the family
$$
\{N_j(A_r):j=1,2,\ 1\leq r\leq m\}
$$
is independent: counts in disjoint sets are independent within each gas, and the two gases are independent. Therefore the sums $N(A_r)$ are independent. These two facts prove directly that the mixture is a Poisson point process. By the <superposition theorem for Poisson point processes>, its activity is
$$
\boxed{z_1+z_2}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=28k}
{scope}

= Solution
{parent=b}

Put $\lambda_i=z|V_i|$. Then
$$
N(V_i)\sim\operatorname{Pois}(\lambda_i),
\qquad
\mathbb E N(V_i)=\lambda_i,
$$
and $\lambda_i\to\infty$. The <Poisson central limit theorem> gives
$$
\frac{N(V_i)-\lambda_i}{\sqrt{\lambda_i}}
\xrightarrow{\mathrm d}N(0,1).
$$
Since $\sqrt{\lambda_i}=\sqrt z\,\sqrt{|V_i|}$,
$$
\frac{N(V_i)-\mathbb E N(V_i)}{\sqrt{|V_i|}}
=\sqrt z\,
\frac{N(V_i)-\lambda_i}{\sqrt{\lambda_i}}
\xrightarrow{\mathrm d}\boxed{N(0,z)}.
$$

Equivalently, the <characteristic function> of the left-hand side is
$$
\exp\left\{
\lambda_i\left(
e^{it/\sqrt{|V_i|}}-1-\frac{it}{\sqrt{|V_i|}}
\right)\right\}
\longrightarrow e^{-zt^2/2},
$$
which is the characteristic function of that centered <normal distribution>.

Solved by gpt-5.6-sol high.

= c
{parent=28k}
{scope}

= Solution
{parent=c}

Let
$$
S_g=\sum_{x\in\Pi}g(x).
$$
The compact support of $g$ ensures that this sum has only finitely many nonzero terms almost surely. The intensity measure of the ideal gas is
$$
\mu(dx)=z\,dx.
$$
Applying the first-moment identity in <Campbell theorem> gives
$$
\boxed{\mathbb E S_g=z\int_{\mathbb R^3}g(x)\,dx}.
$$

For the second moment, separate equal and distinct particles:
$$
\begin{aligned}
\mathbb E S_g^2
&=\mathbb E\sum_{x\in\Pi}g(x)^2
+\mathbb E\sum_{\substack{x,y\in\Pi\\x\ne y}}g(x)g(y)\\
&=z\int_{\mathbb R^3}g(x)^2\,dx
+z^2\left(\int_{\mathbb R^3}g(x)\,dx\right)^2.
\end{aligned}
$$
Subtracting $(\mathbb ES_g)^2$ yields
$$
\boxed{\operatorname{Var}(S_g)
=z\int_{\mathbb R^3}g(x)^2\,dx}.
$$
These are the mean and variance formulas in Campbell's theorem for a homogeneous Poisson point process.

Solved by gpt-5.6-sol high.

= 29J
{parent=Paper 1}
{scope}
{title2=Principles of Statistics}

= a
{parent=29j}
{scope}

= Solution
{parent=a}

Let
$$
\ell_n(\theta)=\log f_\theta(X_1,\ldots,X_n)
$$
be the <log-likelihood>. The <score function> and scalar <Fisher information> are
$$
S_n(\theta)=\frac{\partial}{\partial\theta}\ell_n(\theta),
\qquad
\boxed{I_n(\theta)=\mathbb E_\theta[S_n(\theta)^2]}.
$$
Under the usual regularity conditions the score has mean zero, so this is also its <variance>.

The information tensorizes when it is additive over observations. For identically distributed observations,
$$
\boxed{I_n(\theta)=nI_1(\theta)}.
$$
For an independent sample the joint score is the sum of the individual scores, and the <mean-zero score identity> makes all cross covariances vanish. This is the <tensorization of Fisher information>.

Solved by gpt-5.6-sol high.

= b
{parent=29j}
{scope}

= Solution
{parent=b}

Write $f_\theta(x)$ for the joint density and assume regularity permits differentiation under the <integral>. Since
$$
\int f_\theta(x)\,dx=1,
$$
one differentiation gives
$$
\mathbb E_\theta S_n(\theta)
=\int \frac{\partial_\theta f_\theta}{f_\theta}f_\theta\,dx
=\partial_\theta\int f_\theta\,dx=0.
$$
Differentiating the score itself,
$$
\partial_\theta S_n
=\frac{\partial_\theta^2f_\theta}{f_\theta}
-\left(\frac{\partial_\theta f_\theta}{f_\theta}\right)^2.
$$
Taking expectation gives
$$
\begin{aligned}
\mathbb E_\theta[\partial_\theta S_n(\theta)]
&=\int\partial_\theta^2f_\theta(x)\,dx
-\mathbb E_\theta[S_n(\theta)^2]\\
&=-I_n(\theta).
\end{aligned}
$$
Thus the <information identity> is
$$
\boxed{I_n(\theta)
=-\mathbb E_\theta\!\left[
\frac{\partial}{\partial\theta}S_n(\theta)
\right]
=-\mathbb E_\theta[\ell_n''(\theta)]}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=29j}
{scope}

= Solution
{parent=c}

The model factors into the <normal distribution> densities
$$
X_1\sim N(\theta,1)
$$
and, conditionally on $X_{i-1}$,
$$
X_i\mid X_{i-1}
\sim N\!\left(
\theta(1-\sqrt\gamma)+\sqrt\gamma X_{i-1},
1-\gamma
\right).
$$
Ignoring constants independent of $\theta$, the log likelihood is
$$
\ell_n(\theta)
=-\frac12(X_1-\theta)^2
-\frac1{2(1-\gamma)}
\sum_{i=2}^n
\left[
X_i-\sqrt\gamma X_{i-1}
-\theta(1-\sqrt\gamma)
\right]^2.
$$
Its score is
$$
S_n(\theta)
=(X_1-\theta)
+\frac{1-\sqrt\gamma}{1-\gamma}
\sum_{i=2}^n
\left[
X_i-\sqrt\gamma X_{i-1}
-\theta(1-\sqrt\gamma)
\right].
$$
In terms of the independent innovations,
$$
S_n(\theta)
=\varepsilon_1
+\frac{1-\sqrt\gamma}{\sqrt{1-\gamma}}
\sum_{i=2}^n\varepsilon_i.
$$
<Variance additivity for independent random variables> therefore yields
$$
\boxed{
I_n(\theta)
=1+(n-1)\frac{(1-\sqrt\gamma)^2}{1-\gamma}
=1+(n-1)\frac{1-\sqrt\gamma}{1+\sqrt\gamma}}.
$$

Since $I_1(\theta)=1$, the equality $I_n=nI_1$ holds for all $n$ exactly when
$$
\frac{1-\sqrt\gamma}{1+\sqrt\gamma}=1,
$$
namely when
$$
\boxed{\gamma=0}.
$$
Thus the information tensorizes only in the independent case.

Finally, the <Cramer-Rao bound> gives every unbiased estimator $\widehat\theta$ the lower bound
$$
\boxed{
\operatorname{Var}_\theta(\widehat\theta)
\geq
\frac1{I_n(\theta)}
=\left[
1+(n-1)\frac{1-\sqrt\gamma}{1+\sqrt\gamma}
\right]^{-1}}.
$$
This is the <Fisher information in a stationary Gaussian autoregressive location model>.

Solved by gpt-5.6-sol high.

= 30K
{parent=Paper 1}
{scope}
{title2=Stochastic Financial Models}

= a
{parent=30k}
{scope}

= Solution
{parent=a}

A process $(X_n)_{n\geq0}$ is a <martingale> with respect to a filtration $(\mathcal F_n)_{n\geq0}$ when:

* $X_n$ is $\mathcal F_n$-measurable for every $n$, so the process is adapted;
* $\mathbb E|X_n|<\infty$ for every $n$;
* for every $n\geq0$,
  $$
  \boxed{\mathbb E[X_{n+1}\mid\mathcal F_n]=X_n\quad\text{almost surely}.}
  $$

Equivalently, every increment has conditional mean zero given the past.

Solved by gpt-5.6-sol high.

= b
{parent=30k}
{scope}

= Solution
{parent=b}

Because $X$ is a martingale,
$$
\mathbb E[\xi_n\mid\mathcal F_{n-1}]=0.
$$
Writing
$$
p_n=\mathbb P(\xi_n=1\mid\mathcal F_{n-1}),
$$
and using $\xi_n\in\{-1,1\}$ gives
$$
0=p_n-(1-p_n)=2p_n-1.
$$
Thus
$$
\mathbb P(\xi_n=1\mid\mathcal F_{n-1})
=\mathbb P(\xi_n=-1\mid\mathcal F_{n-1})
=\frac12.
$$

For signs $e_1,\ldots,e_m\in\{-1,1\}$, repeated use of the <tower property of conditional expectation> gives
$$
\begin{aligned}
\mathbb P(\xi_1=e_1,\ldots,\xi_m=e_m)
&=\mathbb E\left[
\mathbf1_{\{\xi_1=e_1,\ldots,\xi_{m-1}=e_{m-1}\}}
\mathbb P(\xi_m=e_m\mid\mathcal F_{m-1})
\right]\\
&=\frac12\mathbb P(\xi_1=e_1,\ldots,\xi_{m-1}=e_{m-1})\\
&=2^{-m}.
\end{aligned}
$$
Hence the increments are <independent and identically distributed random variables>[IID] symmetric signs, proving the <symmetric signs forced by the martingale property> result. Thus $X_n-X_0$ is a <simple symmetric random walk>.

Solved by gpt-5.6-sol high.

= c
{parent=30k}
{scope}

= Solution
{parent=c}

Because $H_k$ is <predictable process>[previsible], $H_k$ is $\mathcal F_{k-1}$-measurable. The finite sum $\widehat X_n$ is therefore $\mathcal F_n$-measurable. Boundedness of $H$ and integrability of the martingale increments imply
$$
\mathbb E|\widehat X_n|<\infty.
$$
Finally,
$$
\begin{aligned}
\mathbb E[\widehat X_n-\widehat X_{n-1}\mid\mathcal F_{n-1}]
&=\mathbb E[H_n(X_n-X_{n-1})\mid\mathcal F_{n-1}]\\
&=H_n\mathbb E[X_n-X_{n-1}\mid\mathcal F_{n-1}]\\
&=0.
\end{aligned}
$$
Thus
$$
\mathbb E[\widehat X_n\mid\mathcal F_{n-1}]
=\widehat X_{n-1},
$$
so $\widehat X$ is a martingale. It is the <martingale transform> of $X$ by $H$.

Solved by gpt-5.6-sol high.

= d
{parent=30k}
{scope}

= Solution
{parent=d}

Define
$$
H_n=
\begin{cases}
1,&n\leq T_a,\\
-1,&n>T_a.
\end{cases}
$$
The event $\{n\leq T_a\}$ means that level $a$ has not been reached by time $n-1$, so it belongs to $\mathcal F_{n-1}$. Hence $H$ is a bounded <predictable process>. A direct check across the hitting time gives
$$
\widehat X_n-\widehat X_{n-1}
=H_n(X_n-X_{n-1}),
$$
and therefore
$$
\widehat X_n=\sum_{k=1}^nH_k(X_k-X_{k-1}).
$$
Part (c) shows that $\widehat X$ is a martingale. Its increments still take values in $\{-1,1\}$, so part (b) shows that they are IID symmetric signs. Consequently $\widehat X$ is a simple symmetric random walk. This is the path transformation behind the <reflection principle for simple symmetric random walk>.

Solved by gpt-5.6-sol high.

= e
{parent=30k}
{scope}

= Solution
{parent=e}

The reflection map from part (d) bijects paths that hit $a$ and finish below $a$ with paths whose reflected endpoint lies strictly above $a$. Since the reflected walk has the same distribution,
$$
\mathbb P(M_n\geq a)
=\mathbb P(X_n\geq a)+\mathbb P(X_n>a)
=\mathbb P(X_n\geq a)+\mathbb P(X_n\geq a+1).
$$
Subtracting the corresponding identity at level $a+1$ gives
$$
\boxed{\mathbb P(M_n=a)
=\mathbb P(X_n=a)+\mathbb P(X_n=a+1)}.
$$

For a simple symmetric random walk,
$$
\mathbb P(X_n=k)
=2^{-n}\binom{n}{(n+k)/2}
$$
when $|k|\leq n$ and $n+k$ is even, and it is zero otherwise. Therefore the explicit answer is
$$
\boxed{
\mathbb P(M_n=a)
=2^{-n}\left[
\binom{n}{(n+a)/2}
+\binom{n}{(n+a+1)/2}
\right]},
$$
where a binomial coefficient is interpreted as zero when its lower argument is not an integer in $\{0,\ldots,n\}$. This is the <point probability for the maximum of simple symmetric random walk>.

Solved by gpt-5.6-sol high.

= 31J
{parent=Paper 1}
{scope}
{title2=Mathematics of Machine Learning}

= Solution
{parent=31J}

For points $x_1,\ldots,x_n\in\mathcal X$, the restriction of $\mathcal H$ to those points is the set of binary vectors
$$
\mathcal H|_{x_{1:n}}
=\{(h(x_1),\ldots,h(x_n)):h\in\mathcal H\}.
$$
The <shattering coefficient> is
$$
\boxed{
s(\mathcal H,n)
=\sup_{x_1,\ldots,x_n\in\mathcal X}
|\mathcal H|_{x_{1:n}}|}.
$$
The <VC dimension> is
$$
\boxed{
\operatorname{VC}(\mathcal H)
=\sup\{n:s(\mathcal H,n)=2^n\}},
$$
with value infinity when arbitrarily large finite sets are shattered.

If $\mathcal H'\subseteq\mathcal H$, then for every choice of points,
$$
\mathcal H'|_{x_{1:n}}\subseteq\mathcal H|_{x_{1:n}}.
$$
Thus every set shattered by $\mathcal H'$ is also shattered by $\mathcal H$, and
$$
\operatorname{VC}(\mathcal H')\leq\operatorname{VC}(\mathcal H).
$$

Now put $d=\dim\mathcal F$ and suppose that $x_1,\ldots,x_n$ are shattered by
$$
\mathcal H
=\{\mathbf1_{\{u:f(u)\leq0\}}:f\in\mathcal F'\}.
$$
Consider the <linear map>
$$
T:\mathcal F\longrightarrow\mathbb R^n,
\qquad
Tf=(f(x_1),\ldots,f(x_n)).
$$
If $n>d$, the <rank-nullity theorem> shows that $\operatorname{im}T$ is a proper subspace of $\mathbb R^n$. Choose a nonzero
$$
\alpha\in(\operatorname{im}T)^\perp
$$
and replace $\alpha$ by $-\alpha$ if necessary so that at least one coordinate is positive.

Ask for the labeling that assigns label zero when $\alpha_i>0$ and label one when $\alpha_i<0$; coordinates with $\alpha_i=0$ may be labeled arbitrarily. Shattering would supply $f\in\mathcal F'$ such that
$$
\alpha_i>0\Longrightarrow f(x_i)>0,
\qquad
\alpha_i<0\Longrightarrow f(x_i)\leq0.
$$
Every product $\alpha_i f(x_i)$ is then nonnegative, and at least one is strictly positive. Hence
$$
\alpha\cdot Tf>0,
$$
contradicting $\alpha\perp\operatorname{im}T$. Therefore $n\leq d$, proving
$$
\boxed{\operatorname{VC}(\mathcal H)\leq\dim\mathcal F}.
$$
This is the <VC dimension of a vector space> argument.

Finally, a closed <Euclidean ball> with center $c$ and radius $r$ is
$$
\{x:\lVert x-c\rVert_2^2\leq r^2\}
=\{x:f_{c,r}(x)\leq0\},
$$
where
$$
f_{c,r}(x)
=\lVert x\rVert_2^2-2c^Tx+\lVert c\rVert_2^2-r^2.
$$
Every such function lies in
$$
\mathcal F
=\operatorname{span}
\{\lVert x\rVert_2^2,x_1,\ldots,x_d,1\},
$$
whose dimension is at most $d+2$. Applying the result just proved gives the <VC dimension upper bound for Euclidean balls>
$$
\boxed{\operatorname{VC}(\mathcal H)\leq d+2}.
$$

Solved by gpt-5.6-sol high.

= 32A
{parent=Paper 1}
{scope}
{title2=Dynamical Systems}

= a
{parent=32a}
{scope}

= Solution
{parent=a}

Let $x_*$ be an <equilibrium point of a dynamical system>. A <Lyapunov function> on a neighbourhood of $x_*$ is a continuously differentiable function $V$ such that
$$
V(x_*)=0,\qquad V(x)>0\quad(x\ne x_*),
$$
and whose derivative along trajectories satisfies
$$
\dot V(x)=\nabla V(x)\cdot f(x)\leq0.
$$

The <First Lyapunov theorem> says that the existence of such a positive-definite $V$ proves <Lyapunov stability> of $x_*$. If $\dot V<0$ away from $x_*$, the <Second Lyapunov theorem> strengthens this to <asymptotic stability>.

<LaSalle invariance principle> says that if a trajectory remains in a compact positively invariant set $\Omega$ on which $\dot V\leq0$, then it approaches the largest invariant subset of
$$
\{x\in\Omega:\dot V(x)=0\}.
$$
In particular, if that largest invariant subset consists only of $x_*$, every trajectory in $\Omega$ approaches $x_*$.

Solved by gpt-5.6-sol high.

= b
{parent=32a}
{scope}

= Solution
{parent=b}

Set
$$
U(x)=\frac{x^2}{(1+x^2)^2},
\qquad
E(x,y)=\frac12y^2+U(x).
$$
The stated identity gives
$$
U'(x)=\frac{2x(1-x^2)}{(1+x^2)^3}.
$$
Hence, along the system,
$$
\begin{aligned}
\dot E
&=y\dot y+U'(x)\dot x\\
&=y[-U'(x)-ky]+U'(x)y\\
&=-ky^2\leq0.
\end{aligned}
$$
Near the origin, $E$ is positive definite because $U(x)>0$ for $x\ne0$. The <First Lyapunov theorem> therefore gives stability.

Choose a sufficiently small compact <invariant sublevel set>
$$
\Omega_c=\{(x,y):E(x,y)\leq c\},
\qquad 0<c<\frac14.
$$
On $\{\dot E=0\}$ we have $y=0$. A trajectory can remain there only if
$$
\dot y=-U'(x)=0,
$$
which occurs at $x=0,\pm1$. The choice $c<1/4=U(\pm1)$ excludes the two latter points, so the largest invariant subset of $\{\dot E=0\}\cap\Omega_c$ is the origin. <LaSalle invariance principle> now gives convergence to the origin. Thus, for every $k>0$,
$$
\boxed{(0,0)\text{ is asymptotically stable}.}
$$
This is an instance of <damped mechanical energy as a Lyapunov function>.

Solved by gpt-5.6-sol high.

= c
{parent=32a}
{scope}

= i
{parent=c}
{scope}

= Solution
{parent=i}

When $k=0$, the system is conservative and
$$
E(x,y)=\frac12y^2+\frac{x^2}{(1+x^2)^2}
$$
is constant. The fixed points are
$$
(0,0),\qquad(1,0),\qquad(-1,0).
$$
The potential $U$ has a strict minimum $U(0)=0$, strict maxima
$$
U(\pm1)=\frac14,
$$
and tends to zero as $|x|\to\infty$. The <conservative planar phase portrait> therefore has:

* a center at the origin, surrounded by closed periodic energy curves for $0<E<1/4$;
* saddles at $(\pm1,0)$;
* the level $E=1/4$ as the separatrix, including the two heteroclinic branches joining the saddles around the central periodic region;
* unbounded outer trajectories on either side of the barriers, and crossing trajectories when $E>1/4$.

The portrait is symmetric under $(x,y)\mapsto(-x,-y)$ and under time reversal $(x,y,t)\mapsto(x,-y,-t)$.

Solved by gpt-5.6-sol high.

= ii
{parent=c}
{scope}

= Solution
{parent=ii}

For $0<k\ll1$, the same three fixed points remain. At the origin the Jacobian is
$$
\begin{pmatrix}0&1\\-2&-k\end{pmatrix}.
$$
Its eigenvalues have negative real part; for small $k$ they are complex, so the origin is a <stable spiral>. At $x=\pm1$, $U''(\pm1)<0$, making the Jacobian determinant negative, so both points remain <saddle equilibrium>[saddles].

Every nonstationary trajectory moves across the former energy contours in the direction of decreasing
$$
E=\frac12y^2+U(x),
\qquad
\dot E=-ky^2<0.
$$
The unstable branches of the two saddles directed into the central region spiral into the origin, while their outward branches escape toward $x=\pm\infty$. The stable manifolds of $(\pm1,0)$ are the separatrices that form the boundary of the <basin of attraction> of the origin. Initial points between these stable separatrices lose enough energy to remain between the barriers and spiral to the origin; points outside escape to one of the two infinities.

This gives the weakly damped <damped rational double-barrier phase portrait> and identifies the requested domain of stability as the open region bounded by those stable manifolds.

Solved by gpt-5.6-sol high.

= d
{parent=32a}
{scope}

= Solution
{parent=d}

At $(1,y_0)$ the trajectory initially moves into $x>1$. Suppose $y$ had a first zero at time $t_*>0$. Before that time $y>0$, hence $x(t_*)>1$. Since $U'(x)<0$ for $x>1$,
$$
\dot y(t_*)=-U'(x(t_*))>0,
$$
which is incompatible with crossing from positive values to zero. Thus
$$
y(t)>0,\qquad x(t)>1
$$
for all $t>0$.

The energy from part (b) decreases strictly because
$$
\dot E=-ky^2<0.
$$
Since $U\geq0$,
$$
\frac12y(t)^2
\leq E(t)
<E(0)
=\frac12y_0^2+\frac14.
$$
Therefore
$$
\boxed{0<y(t)<\sqrt{y_0^2+\frac12}}
\qquad(t>0).
$$

Finally, fix $\varepsilon>0$. If the trajectory never entered $0<y<\varepsilon$, positivity would force $y(t)\geq\varepsilon$ for all $t>0$. Then
$$
\dot E(t)=-ky(t)^2\leq-k\varepsilon^2,
$$
so
$$
E(t)\leq E(0)-k\varepsilon^2t,
$$
which becomes negative for large $t$, contradicting $E\geq0$. Hence the trajectory must enter every such strip. This proves <outward escape from the rational potential barrier>.

Solved by gpt-5.6-sol high.

= 33D
{parent=Paper 1}
{scope}
{title2=Integrable Systems}

= a
{parent=33d}
{scope}

= Solution
{parent=a}

The system prescribes both first derivatives of the same vector $\psi$, so it is overdetermined: the two mixed derivatives must agree. Differentiating gives
$$
\begin{aligned}
\partial_{\bar z}\partial_z\psi
&=(\partial_{\bar z}U)\psi+U\partial_{\bar z}\psi
=(\partial_{\bar z}U+UV)\psi,\\
\partial_z\partial_{\bar z}\psi
&=(\partial_zV)\psi+V\partial_z\psi
=(\partial_zV+VU)\psi.
\end{aligned}
$$
Thus every solution satisfies
$$
\left(
\partial_{\bar z}U-\partial_zV+[U,V]
\right)\psi=0.
$$
For the system to possess arbitrary nontrivial local solution data, the coefficient matrix must vanish:
$$
\boxed{\partial_{\bar z}U-\partial_zV+[U,V]=0}.
$$
This is the <zero-curvature condition>, or the <compatibility condition for an overdetermined linear system>.

Solved by gpt-5.6-sol high.

= b
{parent=33d}
{scope}

= Solution
{parent=b}

Substitute
$$
U=\frac12
\begin{pmatrix}
u_z&\lambda^{-1}e^{-u}\\
\lambda^{-1}e^u&-u_z
\end{pmatrix},
\qquad
V=\frac12
\begin{pmatrix}
-u_{\bar z}&\lambda e^u\\
\lambda e^{-u}&u_{\bar z}
\end{pmatrix}.
$$
The off-diagonal entries of
$$
U_{\bar z}-V_z+[U,V]
$$
cancel identically. Its upper-left entry is
$$
u_{z\bar z}
+\frac14(e^{-2u}-e^{2u})
=u_{z\bar z}-\frac12\sinh(2u),
$$
and its lower-right entry is the negative of this. Hence compatibility is equivalent to
$$
\boxed{u_{z\bar z}=\frac12\sinh(2u)}.
$$
This is the <Sinh-Gordon equation> in complex light-cone coordinates.

Solved by gpt-5.6-sol high.

= c
{parent=33d}
{scope}

= Solution
{parent=c}

When $u$ is independent of $y$ and $z=x+iy$,
$$
\partial_z=\partial_{\bar z}=\frac12\frac d{dx}.
$$
Put
$$
L=U-V,\qquad M=U+V.
$$
The zero-curvature equation becomes
$$
\frac12L_x+[U,V]=0.
$$
Since
$$
[M,L]=[U+V,U-V]=-2[U,V],
$$
we obtain the <isospectral Lax equation>
$$
\boxed{L_x=[M,L]}.
$$
Therefore, for every positive integer $n$, the cyclic property of the <matrix trace> gives
$$
\begin{aligned}
\frac d{dx}\operatorname{tr}(L^n)
&=n\operatorname{tr}(L^{n-1}[M,L])\\
&=n\operatorname{tr}(L^{n-1}ML-ML^n)=0.
\end{aligned}
$$
These are the <trace invariants of a Lax equation>.

The PDE from part (b) reduces to
$$
\frac14u_{xx}=\frac12\sinh(2u).
$$
With $\phi=2u$, this is
$$
\phi''=4\sinh\phi.
$$
To extract a nontrivial invariant, use $n=2$. Writing $p=u_x$, direct multiplication gives
$$
\operatorname{tr}(L^2)
=\frac12\left[
p^2+\lambda^2+\lambda^{-2}-2\cosh(2u)
\right].
$$
The terms involving $\lambda$ are constant, and $p=\phi'/2$. Hence
$$
\boxed{
\frac12(\phi')^2-4\cosh\phi=C
}
$$
is a <first integral>. Direct differentiation verifies it:
$$
\frac d{dx}
\left[\frac12(\phi')^2-4\cosh\phi\right]
=\phi'(\phi''-4\sinh\phi)=0.
$$

Solved by gpt-5.6-sol high.

= 34B
{parent=Paper 1}
{scope}
{title2=Principles of Quantum Mechanics}

= a
{parent=34b}
{scope}

= Solution
{parent=a}

The <Born rule> depends only on transition probabilities
$$
|\langle\phi|\psi\rangle|^2
$$
between rays. A physical transformation must preserve these probabilities. By <Wigner theorem>, such a ray transformation is induced by a unitary or antiunitary operator. For ordinary continuously connected transformations one chooses the unitary branch, so
$$
\boxed{U(g)^\dagger U(g)=I_{\mathcal H}}.
$$

The states $U(g_1)U(g_2)|\psi\rangle$ and $U(g_1g_2)|\psi\rangle$ represent the same ray. They may therefore differ by a state-independent <quantum phase>:
$$
\boxed{
U(g_1)U(g_2)
=e^{i\phi(g_1,g_2)}U(g_1g_2)}.
$$
Thus quantum transformations form a <projective unitary representation>.

If $g$ is a symmetry of the <Hamiltonian operator> $H$, it also obeys
$$
\boxed{U(g)^\dagger H U(g)=H},
$$
equivalently $[U(g),H]=0$. For a differentiable one-parameter subgroup, write
$$
U(s)=e^{-isQ/\hbar}
$$
with a self-adjoint generator $Q$. Differentiating the symmetry equation at $s=0$ yields
$$
[Q,H]=0.
$$
For a time-independent $Q$, the <Heisenberg picture> equation is
$$
\frac{dQ}{dt}=\frac i\hbar[H,Q]=0.
$$
Hence the generator is a conserved observable. Conversely, a self-adjoint conserved $Q$ commutes with $H$ and its unitary group generates symmetries. This is the <conserved generator of a continuous quantum symmetry>.

Solved by gpt-5.6-sol high.

= b
{parent=34b}
{scope}

= Solution
{parent=b}

The <unitary time evolution> is
$$
U(t)=e^{-iHt/\hbar}.
$$
First suppose that time reversal $T$ is linear and unitary. Differentiating
$$
U(t)T=TU(-t)
$$
at $t=0$ gives
$$
-\frac i\hbar HT
=T\frac i\hbar H,
$$
and linearity permits the scalar $i$ to pass through $T$. Hence
$$
\boxed{HT=-TH}.
$$
If $H|E\rangle=E|E\rangle$, then
$$
H(T|E\rangle)=-E(T|E\rangle).
$$
Thus every energy has its negative as an energy. The Coulomb Hamiltonian has arbitrarily large positive energies in its continuum, so this symmetry would force energies arbitrarily far below zero. It would have no lowest <energy eigenvalue> and therefore no stable <ground state>. This is the <unitary time reversal reverses the energy spectrum> obstruction.

Now let $T$ be antiunitary. It is conjugate linear, so
$$
T(i|\psi\rangle)=-iT|\psi\rangle.
$$
Differentiating the same intertwining equation gives
$$
-\frac i\hbar HT|\psi\rangle
=T\left(\frac i\hbar H|\psi\rangle\right)
=-\frac i\hbar TH|\psi\rangle.
$$
Therefore
$$
\boxed{HT=TH}.
$$
Time reversal now maps an energy eigenstate to another state with the same energy:
$$
H(T|E\rangle)=E(T|E\rangle).
$$
The spectrum is preserved rather than reflected through zero, so the lower-bounded Coulomb spectrum and its stable ground state are compatible with time-reversal symmetry. This is why <antiunitary time reversal preserves the energy spectrum>.

Solved by gpt-5.6-sol high.

= 35B
{parent=Paper 1}
{scope}
{title2=Application of Quantum Mechanics}

= a
{parent=35b}
{scope}

= Solution
{parent=a}

Let $H$ be the one-dimensional <Hamiltonian operator> and let
$$
E_0\leq E_1\leq E_2\leq\cdots
$$
be its discrete <energy eigenvalue>[energy eigenvalues]. Expanding a normalized trial wavefunction in an orthonormal energy basis,
$$
|\psi\rangle=\sum_nc_n|n\rangle,
$$
gives
$$
\langle\psi|H|\psi\rangle
=\sum_n|c_n|^2E_n
\geq E_0\sum_n|c_n|^2
=E_0.
$$
Thus the <Rayleigh-Ritz variational principle> gives
$$
\boxed{
E_0\leq
\frac{\int_{\mathbb R}\psi^*(x)H\psi(x)\,dx}
{\int_{\mathbb R}|\psi(x)|^2\,dx}}
$$
for every nonzero admissible trial wavefunction. One chooses a parameterized family and minimizes the quotient.

When $V(x)=V(-x)$, parity commutes with $H$. In one dimension the nondegenerate ground state is even and the first excited state is odd. Restricting the trial family to odd wavefunctions makes every trial state orthogonal to the ground state, so
$$
\boxed{
E_1\leq
\frac{\langle\psi_{\rm odd}|H|\psi_{\rm odd}\rangle}
{\langle\psi_{\rm odd}|\psi_{\rm odd}\rangle}}.
$$
This is the <odd-state variational principle for an even potential>.

Solved by gpt-5.6-sol high.

= b
{parent=35b}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

The condition $2m=\hbar^2$ makes
$$
H=-\frac{d^2}{dx^2}-V_0e^{-x^2}.
$$
For
$$
\psi_a(x)=e^{-ax^2/2},
\qquad a>0,
$$
the squared norm is the <Gaussian integral>
$$
\langle\psi_a|\psi_a\rangle
=\int_{\mathbb R}e^{-ax^2}\,dx
=\sqrt{\frac\pi a}.
$$
Since $\psi_a'=-ax\psi_a$, integration by parts gives the kinetic quotient
$$
\frac{\int\psi_a(-\psi_a'')\,dx}
{\int\psi_a^2\,dx}
=\frac{\int|\psi_a'|^2\,dx}{\int\psi_a^2\,dx}
=a^2\frac{\int x^2e^{-ax^2}\,dx}
{\int e^{-ax^2}\,dx}
=\frac a2.
$$
The potential quotient is
$$
-V_0
\frac{\int e^{-(a+1)x^2}\,dx}
{\int e^{-ax^2}\,dx}
=-V_0\sqrt{\frac a{1+a}}.
$$
The variational principle therefore yields the <Gaussian variational bound for an attractive Gaussian well>
$$
\boxed{
E_0\leq E(a)
=\frac a2-V_0\frac{\sqrt a}{\sqrt{1+a}}}.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

The positive zero of $E(a)$ satisfies
$$
\frac a2=V_0\sqrt{\frac a{1+a}}.
$$
Squaring and using $a>0$ gives
$$
a(1+a)=4V_0^2,
$$
so
$$
\boxed{
a_0=\frac{\sqrt{1+16V_0^2}-1}{2}}.
$$

Furthermore,
$$
E'(a)
=\frac12-\frac{V_0}{2\sqrt a(1+a)^{3/2}}.
$$
At an extremum this vanishes, and hence
$$
V_0=\sqrt a(1+a)^{3/2}.
$$
Squaring gives the required equation
$$
\boxed{(1+a)^3=\frac{V_0^2}{a}}.
$$

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

As $a\downarrow0$,
$$
E(a)=-V_0\sqrt a+O(a),
$$
so $E(a)<0$ for all sufficiently small positive $a$. On the other hand,
$$
E(a)\longrightarrow+\infty
\qquad(a\to\infty).
$$
The derivative satisfies
$$
E'(a)\to-\infty\quad(a\downarrow0),
\qquad
E'(a)\to\frac12\quad(a\to\infty).
$$
The extremum equation
$$
a(1+a)^3=V_0^2
$$
has exactly one positive solution because its left side is strictly increasing. Thus $E$ has one global minimum, and that minimum is negative.

The potential tends to zero at spatial infinity, so zero is the threshold of the continuous spectrum. A trial state with negative energy expectation forces the spectrum below zero; the corresponding lowest spectral value is a normalizable <bound state>. Hence every $V_0>0$ produces at least one bound state.

Solved by gpt-5.6-sol high.

= iv
{parent=b}
{scope}

= Solution
{parent=iv}

Since
$$
-V_0<V(x)\leq0
$$
except at the single point $x=0$, and the kinetic-energy expectation is nonnegative, every normalized wavefunction satisfies
$$
\langle H\rangle>-V_0.
$$
Therefore
$$
E_0>-V_0.
$$

For the upper bound, use the admissible choice
$$
a=V_0^2.
$$
Then
$$
\begin{aligned}
E(V_0^2)
&=\frac12V_0^2
-\frac{V_0^2}{\sqrt{1+V_0^2}}\\
&=-\frac12V_0^2+\frac12V_0^4+O(V_0^6).
\end{aligned}
$$
The <Rayleigh-Ritz variational principle> now gives
$$
\boxed{
-V_0<E_0
\leq-\frac12V_0^2+O(V_0^3)}.
$$
Indeed, the displayed calculation provides the sharper remainder $O(V_0^4)$.

Solved by gpt-5.6-sol high.

= 36C
{parent=Paper 1}
{scope}
{title2=Statistical Physics}

= a
{parent=36c}
{scope}

= Solution
{parent=a}

For a fixed number $N$ of particles, the <ideal gas> equation of state is
$$
\boxed{pV=Nk_BT}.
$$
Its <internal energy of an ideal gas>[internal energy] depends only on temperature, and
$$
dE=C_V\,dT.
$$
When the <heat capacity at constant volume> is constant, a choice of the energy zero gives
$$
\boxed{E=C_VT}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=36c}
{scope}

= Solution
{parent=b}

At constant volume, the <First law of thermodynamics> gives
$$
\delta Q=dE=C_V\,dT.
$$
At constant pressure,
$$
\delta Q=C_p\,dT=dE+p\,dV.
$$
Differentiating $pV=Nk_BT$ at fixed $p$ gives
$$
p\,dV=Nk_B\,dT.
$$
Consequently,
$$
C_p\,dT=(C_V+Nk_B)dT,
$$
and the <Mayer relation> is
$$
\boxed{C_p-C_V=Nk_B}.
$$
The <specific-heat ratio> is therefore
$$
\boxed{
\gamma=\frac{C_p}{C_V}
=1+\frac{Nk_B}{C_V}}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=36c}
{scope}

= Solution
{parent=c}

An <adiabatic process> exchanges no heat with its surroundings, so $\delta Q=0$. For a reversible process it is also <entropy>[isentropic]. The first law then gives
$$
C_V\,dT=dE=-p\,dV.
$$
Using $p=Nk_BT/V$ and the Mayer relation,
$$
\frac{dT}{T}
=-\frac{Nk_B}{C_V}\frac{dV}{V}
=-(\gamma-1)\frac{dV}{V}.
$$
Integration yields
$$
TV^{\gamma-1}=\text{constant}.
$$
Substituting $T=pV/(Nk_B)$ gives the <reversible ideal-gas adiabat>
$$
\boxed{pV^\gamma=\text{constant}}.
$$

Solved by gpt-5.6-sol high.

= d
{parent=36c}
{scope}

= Solution
{parent=d}

In the $(p,V)$ plane, the <Otto cycle> consists of:

* $A\to B$: an adiabatic compression along $pV^\gamma=\text{constant}$ from $V_1$ to $V_2$;
* $B\to C$: a vertical constant-volume rise at $V_2$;
* $C\to D$: an adiabatic expansion back to $V_1$;
* $D\to A$: a vertical constant-volume fall at $V_1$.

It is a clockwise loop, so its enclosed area is the positive net work output. In the $(T,S)$ plane, the two reversible adiabats $A\to B$ and $C\to D$ are vertical constant-entropy segments. Constant-volume heating $B\to C$ moves upward and to the right, while constant-volume cooling $D\to A$ moves downward and to the left.

Let
$$
r=\frac{V_1}{V_2}>1.
$$
The two adiabatic relations give
$$
T_B=T_A r^{\gamma-1},
\qquad
T_C=T_D r^{\gamma-1}.
$$
Because heat is exchanged at constant volume,
$$
Q_1=C_V(T_C-T_B),
\qquad
Q_2=C_V(T_D-T_A),
$$
where $Q_2$ denotes the positive amount rejected. Hence
$$
\begin{aligned}
\frac{Q_2}{Q_1}
&=\frac{T_C/r^{\gamma-1}-T_A}
{T_C-T_A r^{\gamma-1}}\\
&=\frac1{r^{\gamma-1}}.
\end{aligned}
$$
The <thermal efficiency> is therefore
$$
\boxed{
\eta=\frac W{Q_1}
=1-\frac{Q_2}{Q_1}
=1-\frac1{r^{\gamma-1}}}.
$$
For a fixed gas, $\gamma$ is fixed and the idealized efficiency increases monotonically with the compression ratio $r$, so it is maximized by making $r$ as large as the physical constraints allow.

Solved by gpt-5.6-sol high.

= 37C
{parent=Paper 1}
{scope}
{title2=Electrodynamics}

= a
{parent=37c}
{scope}

= Solution
{parent=a}

With
$$
\partial_\mu=\left(\frac1c\frac{\partial}{\partial t},\nabla\right),
$$
the <electromagnetic field tensor> is the antisymmetric tensor
$$
\boxed{
F^{\mu\nu}
=\partial^\mu A^\nu-\partial^\nu A^\mu}.
$$
Under a Lorentz transformation $x'^\mu=\Lambda^\mu{}_\nu x^\nu$, the four-potential transforms as a four-vector and therefore
$$
\boxed{
F'^{\mu\nu}(x')
=\Lambda^\mu{}_\rho\Lambda^\nu{}_\sigma
F^{\rho\sigma}(x)}.
$$
Thus $F$ is an antisymmetric rank-two <tensor> under <Lorentz transformation>[Lorentz transformations].

Solved by gpt-5.6-sol high.

= b
{parent=37c}
{scope}

= Solution
{parent=b}

Define the dual tensor by
$$
\widetilde F^{\mu\nu}
=\frac12\epsilon^{\mu\nu\rho\sigma}F_{\rho\sigma}.
$$
For metric signature $(-,+,+,+)$ and $\epsilon^{0123}=+1$, two independent quadratic <electromagnetic field invariants> are
$$
\boxed{
F_{\mu\nu}F^{\mu\nu}
=2\left(B^2-\frac{E^2}{c^2}\right)}
$$
and
$$
\boxed{
F_{\mu\nu}\widetilde F^{\mu\nu}
=-\frac4c\,\mathbf E\cdot\mathbf B}.
$$
Changing orientation or the sign convention for $F$ can change the displayed overall signs but not their vanishing.

For a vacuum <plane electromagnetic wave> with wavevector $\mathbf k$,
$$
\mathbf k\cdot\mathbf E=0,
\qquad
\mathbf B=\frac{\mathbf k\times\mathbf E}{\omega},
\qquad
\omega=c|\mathbf k|.
$$
Hence
$$
\mathbf E\cdot\mathbf B=0,
\qquad
B^2=\frac{E^2}{c^2},
$$
for every transverse polarization. Both Lorentz scalars therefore vanish.

Solved by gpt-5.6-sol high.

= c
{parent=37c}
{scope}

= Solution
{parent=c}

Choose, for example,
$$
\boxed{
\mathbf B_0=(0,0,B_0),
\qquad
\mathbf E_0=(cB_0,0,0)},
\qquad B_0\ne0.
$$
The fields are perpendicular and satisfy $E_0^2=c^2B_0^2$, so both invariants vanish. Every nonzero constant background with vanishing invariants can be brought to this form by a spatial rotation, up to signs.

In this frame, and with a convention consistent with the <relativistic Lorentz force>, the mixed tensor is
$$
F^\mu{}_\nu
=B_0
\begin{pmatrix}
0&1&0&0\\
1&0&1&0\\
0&-1&0&0\\
0&0&0&0
\end{pmatrix}.
$$
Calling the displayed dimensionless matrix $N$, direct multiplication gives
$$
N^2=
\begin{pmatrix}
1&0&1&0\\
0&0&0&0\\
-1&0&-1&0\\
0&0&0&0
\end{pmatrix},
\qquad
N^3=0.
$$
Consequently
$$
\boxed{
F^\mu{}_\rho F^\rho{}_\sigma F^\sigma{}_\nu=0}.
$$
This is a tensor equation. Since it holds in the canonical frame, it holds in every Lorentz frame for any <null electromagnetic field>.

Solved by gpt-5.6-sol high.

= d
{parent=37c}
{scope}

= Solution
{parent=d}

Let $X^\mu(\tau)$ be the worldline, $u^\mu=dX^\mu/d\tau$ its <four-velocity>, and
$$
\Omega=\frac{qB_0}{m}.
$$
The <relativistic Lorentz force> in the constant background is
$$
\frac{du^\mu}{d\tau}
=\frac qm F^\mu{}_\nu u^\nu.
$$
Because $F^3=0$, its exponential terminates:
$$
u(\tau)
=\exp\left(\frac qmF\tau\right)u(0)
=\left[
I+\frac qmF\tau
+\frac12\left(\frac qmF\tau\right)^2
\right]u(0).
$$
The particle starts from rest, so
$$
u(0)=(c,0,0,0).
$$
Using the matrix from part (c) gives
$$
u^0=c\left(1+\frac12\Omega^2\tau^2\right),
\qquad
u^1=c\Omega\tau,
\qquad
u^2=-\frac c2\Omega^2\tau^2,
\qquad
u^3=0.
$$
Integrating from the origin and using $X^0=ct$ yields the parametric trajectory
$$
\boxed{
\begin{aligned}
t(\tau)&=\tau+\frac{\Omega^2\tau^3}{6},\\
x(\tau)&=\frac{c\Omega\tau^2}{2},\\
y(\tau)&=-\frac{c\Omega^2\tau^3}{6},\\
z(\tau)&=0.
\end{aligned}}
$$
This is the <relativistic trajectory in a constant null crossed field>. The real cubic relation for $t(\tau)$ is strictly increasing, so it determines $\tau$ uniquely as a function of coordinate time if an explicit $\mathbf x(t)$ is desired.

Solved by gpt-5.6-sol high.

= 38C
{parent=Paper 1}
{scope}
{title2=General Relativity}

= a
{parent=38c}
{scope}

= Solution
{parent=a}

Contract the defining formula in its first and third indices. In four dimensions,
$$
\begin{aligned}
C^\alpha{}_{\beta\alpha\delta}
={}&R_{\beta\delta}\\
&-\frac12\left(
4R_{\beta\delta}
+g_{\beta\delta}R
-R_{\beta\delta}
-R_{\beta\delta}
\right)\\
&+\frac16(4g_{\beta\delta}-g_{\beta\delta})R\\
={}&0.
\end{aligned}
$$
Thus
$$
\boxed{C^\alpha{}_{\beta\alpha\delta}=0}.
$$

The <Weyl tensor> has the same algebraic symmetries as the <Riemann curvature tensor>: it is antisymmetric within each index pair, symmetric under exchanging the pairs, and obeys the first Bianchi identity. A contraction within an antisymmetric pair vanishes against the symmetric metric. Every contraction between the two pairs can be converted by the pair symmetries to the one just calculated, possibly with a sign. Hence every contraction vanishes, proving the <trace-free property of the Weyl tensor>.

Solved by gpt-5.6-sol high.

= b
{parent=38c}
{scope}

= Solution
{parent=b}

Write
$$
H_{\alpha\beta}
=\partial_\alpha\partial_\beta\omega\big|_p,
\qquad
H=\eta^{\alpha\beta}H_{\alpha\beta}.
$$
The <Levi-Civita connection> of
$$
g_{\alpha\beta}=e^{2\omega}\eta_{\alpha\beta}
$$
is
$$
\Gamma^\rho_{\alpha\beta}
=\delta^\rho_\alpha\partial_\beta\omega
+\delta^\rho_\beta\partial_\alpha\omega
-\eta_{\alpha\beta}\partial^\rho\omega.
$$
The assumed first-derivative condition makes $\Gamma^\rho_{\alpha\beta}(p)=0$. Differentiating once more gives, at $p$,
$$
\begin{aligned}
R_{\alpha\beta\gamma\delta}
=e^{2\omega}\bigl(
&\eta_{\alpha\delta}H_{\beta\gamma}
-\eta_{\alpha\gamma}H_{\beta\delta}\\
&-\eta_{\beta\delta}H_{\alpha\gamma}
+\eta_{\beta\gamma}H_{\alpha\delta}
\bigr).
\end{aligned}
$$
Its contractions are
$$
R_{\beta\delta}
=-2H_{\beta\delta}-\eta_{\beta\delta}H,
\qquad
R=-6e^{-2\omega}H.
$$
Substitution into the defining expression for $C_{\alpha\beta\gamma\delta}$ cancels every Hessian term, leaving
$$
\boxed{C_{\alpha\beta\gamma\delta}(p)=0}.
$$
Since $p$ was arbitrary, this is the <Weyl tensor of a conformally flat metric>.

Solved by gpt-5.6-sol high.

= c
{parent=38c}
{scope}

= i
{parent=c}
{scope}

= Solution
{parent=i}

The Schwarzschild exterior is vacuum, so
$$
R_{\mu\nu}=0,\qquad R=0,
$$
and consequently $C_{\alpha\beta\gamma\delta}=R_{\alpha\beta\gamma\delta}$. Put
$$
f(r)=1-\frac{2M}{r}.
$$
The relevant <Christoffel symbol>[Christoffel symbols] are
$$
\Gamma^t{}_{tr}=\frac{f'}{2f},
\qquad
\Gamma^r{}_{rr}=-\frac{f'}{2f},
\qquad
\Gamma^r{}_{tt}=\frac12ff'.
$$
With the convention
$$
R^\alpha{}_{\beta\gamma\delta}
=\partial_\gamma\Gamma^\alpha_{\beta\delta}
-\partial_\delta\Gamma^\alpha_{\beta\gamma}+\cdots,
$$
one finds
$$
R_{trtr}=\frac12f''(r).
$$
Since $f''=-4M/r^3$,
$$
\boxed{
C_{trtr}=-\frac{2M}{r^3}}
\qquad(r\gg2M).
$$
This is the leading <radial Weyl curvature of Schwarzschild spacetime>. The opposite convention for the Riemann tensor reverses the sign but not the physical tidal magnitude.

Solved by gpt-5.6-sol high.

= ii
{parent=c}
{scope}

= Solution
{parent=ii}

This curvature component produces radial <tidal gravity and the equivalence principle>[tidal acceleration]: nearby freely falling particles separated radially are stretched relative to one another, while transverse separations are compressed. At the Earth, after removing the Earth's own field, the Sun and especially the Moon produce the familiar ocean tides known since antiquity.

There is no conflict with the equivalence principle. A freely falling local frame can remove the uniform gravitational acceleration at one event by setting the connection to zero there. It cannot remove the <Riemann curvature tensor>, which controls relative acceleration through geodesic deviation and is detectable across an extended body such as the Earth. Tides measure this spatial variation of gravity rather than gravity at a single point.

Solved by gpt-5.6-sol high.

= 39A
{parent=Paper 1}
{scope}
{title2=Fluid Dynamics II}

= a
{parent=39a}
{scope}

= Solution
{parent=a}

For an incompressible Newtonian fluid with negligible inertia and no body force, the <Stokes flow>[Stokes equations] are
$$
\boxed{
-\nabla p+\mu\nabla^2\mathbf u=0,
\qquad
\nabla\cdot\mathbf u=0}.
$$
They are linear because any linear combination of solutions corresponds to the same linear combination of boundary data and applied forces. They are reversible because replacing all imposed velocities and forces by their negatives sends
$$
(\mathbf u,p)\longmapsto(-\mathbf u,-p).
$$
Fluid particles therefore retrace their paths under reversal. This is <kinematic reversibility of Stokes flow>.

Solved by gpt-5.6-sol high.

= b
{parent=39a}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

The geometry and boundary velocities are axisymmetric and linear in the axial pseudovector $\boldsymbol\Omega$. The pressure is a true scalar. No nonzero parity-even scalar can be formed linearly from $\boldsymbol\Omega$ and the radial vector $\mathbf x$: $\boldsymbol\Omega\cdot\mathbf x$ is a pseudoscalar. Hence the flow-induced pressure is spatially constant.

Equivalently, taking the divergence of the Stokes momentum equation gives the <harmonic pressure and vorticity in Stokes flow> result
$$
\nabla^2p=0,
$$
and symmetry plus the homogeneous boundary data excludes every nonconstant harmonic pressure mode. Since pressure is defined up to an additive constant, we choose
$$
\boxed{p=0}.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

Both candidate fields are divergence-free. For solid-body rotation,
$$
\mathbf u=\boldsymbol\Omega\times\mathbf x,
$$
each component is linear in $\mathbf x$, so $\nabla^2\mathbf u=0$. Also,
$$
\mathbf u=\boldsymbol\Omega\times\nabla\left(\frac1r\right)
$$
is harmonic for $r>0$, because $\nabla^2(1/r)=0$ there and constant-coefficient derivatives commute with the Laplacian. Thus both solve the Stokes equations with $p=0$.

The rotationally symmetric solution is consequently
$$
\mathbf u
=\left(A+\frac B{r^3}\right)
\boldsymbol\Omega\times\mathbf x.
$$
The <no-slip boundary condition> gives
$$
A+\frac B{a^3}=1,
\qquad
A+\frac B{b^3}=0.
$$
Therefore
$$
A=-\frac{a^3}{b^3-a^3},
\qquad
B=\frac{a^3b^3}{b^3-a^3},
$$
and
$$
\boxed{
\mathbf u(\mathbf x)
=\frac{a^3}{b^3-a^3}
\left(\frac{b^3}{r^3}-1\right)
\boldsymbol\Omega\times\mathbf x}.
$$
This is the <rotational Stokes flow between concentric spheres>.

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

Write
$$
f(r)=A+\frac B{r^3},
\qquad
\mathbf w=\boldsymbol\Omega\times\mathbf x,
\qquad
\mathbf u=f\mathbf w.
$$
The velocity gradient of the solid-rotation part is antisymmetric, so it drops out of the <Newtonian fluid stress tensor>. Since
$$
\partial_i f=-\frac{3B}{r^5}x_i,
$$
and $p=0$, the stress is
$$
\boxed{
\sigma_{ij}
=-\frac{3\mu B}{r^5}
\left[
x_i(\boldsymbol\Omega\times\mathbf x)_j
+x_j(\boldsymbol\Omega\times\mathbf x)_i
\right]}.
$$
Equivalently, with $\mathbf n=\mathbf x/r$,
$$
\sigma_{ij}
=-\frac{3\mu B}{r^3}
\left[
n_i(\boldsymbol\Omega\times\mathbf n)_j
+n_j(\boldsymbol\Omega\times\mathbf n)_i
\right].
$$

Solved by gpt-5.6-sol high.

= iv
{parent=b}
{scope}

= Solution
{parent=iv}

On $r=c$, the traction $\boldsymbol\sigma\mathbf n$ exerted by the outer fluid on the inner fluid is
$$
\boldsymbol\sigma\mathbf n
=-\frac{3\mu B}{c^3}
(\boldsymbol\Omega\times\mathbf n).
$$
The inner fluid exerts the opposite traction on the outer fluid. Its couple is therefore
$$
\begin{aligned}
\mathbf G
&=-\int_{r=c}\mathbf x\times
(\boldsymbol\sigma\mathbf n)\,dS\\
&=3\mu Bc^{-2}
\int_{r=c}\mathbf n\times
(\boldsymbol\Omega\times\mathbf n)\,dS.
\end{aligned}
$$
Using
$$
\mathbf n\times(\boldsymbol\Omega\times\mathbf n)
=\boldsymbol\Omega-\mathbf n(\mathbf n\cdot\boldsymbol\Omega)
$$
and the supplied spherical integral,
$$
\int_{r=c}n_in_j\,dS
=\frac{4\pi c^2}{3}\delta_{ij},
$$
we obtain
$$
\int_{r=c}\mathbf n\times
(\boldsymbol\Omega\times\mathbf n)\,dS
=\frac{8\pi c^2}{3}\boldsymbol\Omega.
$$
Thus the radius cancels:
$$
\mathbf G=8\pi\mu B\boldsymbol\Omega
=\boxed{
\frac{8\pi\mu a^3b^3}{b^3-a^3}
\boldsymbol\Omega}.
$$
This is the <torque in rotational Stokes flow between concentric spheres>.

When $a\ll b$,
$$
\mathbf G\sim8\pi\mu a^3\boldsymbol\Omega,
$$
the rotational drag torque for a sphere in an effectively unbounded fluid. If $h=b-a\ll a$, then
$$
b^3-a^3\sim3a^2h,
\qquad
a^3b^3\sim a^6,
$$
so
$$
\mathbf G\sim
\frac{8\pi\mu a^4}{3h}\boldsymbol\Omega,
$$
the inverse-gap growth expected from a thin <Couette flow>.

Solved by gpt-5.6-sol high.

= 40A
{parent=Paper 1}
{scope}
{title2=Waves}

= a
{parent=40a}
{scope}

= Solution
{parent=a}

Use the <acoustic velocity potential> convention
$$
\mathbf u=\nabla\phi,
\qquad
\widetilde p=-\rho_0\frac{\partial\phi}{\partial t}.
$$
Put
$$
k=\frac{\omega}{c_0},
\qquad
\alpha=kR.
$$
For a spherically symmetric harmonic solution, write
$$
\phi(r,t)=F(r)\sin\omega t.
$$
The wave equation and
$$
\nabla^2F=\frac1r\frac{d^2}{dr^2}(rF)
$$
give the radial Helmholtz equation
$$
\frac{d^2}{dr^2}(rF)+k^2(rF)=0.
$$

The rigid inner sphere requires
$$
u_r(R,t)=F'(R)\sin\omega t=0.
$$
A form satisfying this condition automatically is
$$
F(r)=\frac C r
\left[
\cos k(r-R)+\frac1{kR}\sin k(r-R)
\right].
$$
Indeed, if the bracket is $h(r)$, then $h(R)=1$ and $h'(R)=1/R$, so $(h/r)'=0$ at $R$.

At the outer sphere,
$$
\varepsilon p_0\cos\omega t
=-\rho_0\omega F(2R)\cos\omega t,
$$
so
$$
F(2R)=-\frac{\varepsilon p_0}{\rho_0\omega}.
$$
Since
$$
F(2R)=\frac C{2R}
\left(\cos\alpha+\frac{\sin\alpha}{\alpha}\right),
$$
we obtain
$$
C=-\frac{2R\varepsilon p_0}
{\rho_0\omega(\cos\alpha+\alpha^{-1}\sin\alpha)}.
$$
Therefore
$$
\boxed{
\phi(r,t)
=-\frac{2R\varepsilon p_0}{\rho_0\omega r}
\frac{\cos k(r-R)+(kR)^{-1}\sin k(r-R)}
{\cos\alpha+\alpha^{-1}\sin\alpha}
\sin\omega t}.
$$
The assumption $\alpha\leq\pi/2$ keeps the displayed denominator positive.

Solved by gpt-5.6-sol high.

= b
{parent=40a}
{scope}

= Solution
{parent=b}

At every radius, the pressure perturbation has the form
$$
\widetilde p(r,t)
=-\rho_0\omega F(r)\cos\omega t,
$$
whereas the radial velocity is
$$
u_r(r,t)=F'(r)\sin\omega t.
$$
Define the average over one forcing period $T=2\pi/\omega$ by
$$
\langle I_r\rangle
=\frac1T\int_0^T
\widetilde p(r,t)u_r(r,t)\,dt.
$$
Then
$$
\langle I_r\rangle
=-\rho_0\omega F(r)F'(r)
\frac1T\int_0^T\cos\omega t\sin\omega t\,dt
=\boxed{0}.
$$
The same is true for the net flux through every concentric sphere.

Pressure and velocity are in temporal quadrature, so this is a <reactive acoustic energy flux>. The forcing establishes a standing wave in the closed, lossless annulus: compressional and kinetic energy exchange during each cycle, but no energy is transported outward on average. This is the <pressure-forced standing acoustic wave in a spherical annulus>.

Solved by gpt-5.6-sol high.

= c
{parent=40a}
{scope}

= Solution
{parent=c}

At larger frequencies the denominator
$$
D(\alpha)=\cos\alpha+\frac{\sin\alpha}{\alpha}
$$
can vanish. Its zeros satisfy
$$
\tan\alpha=-\alpha
$$
and are the normal-mode frequencies for a pressure node at the outer sphere and zero radial velocity at the inner sphere. Near one of these <radial acoustic resonance in a spherical annulus>[radial acoustic resonances], the ideal undamped harmonic response becomes very large; exactly at resonance the linear lossless initial-value problem develops secular growth rather than a bounded periodic solution.

In a physical system, fluid viscosity, thermal losses, acoustic radiation, and damping in the elastic shell keep the amplitude finite and shift its phase. A nonzero pressure-velocity phase component then supplies positive mean power. At sufficiently large amplitude the linear acoustic approximation can also fail, leading to nonlinear frequency shifts or shock formation.

Solved by gpt-5.6-sol high.

= 41E
{parent=Paper 1}
{scope}
{title2=Numerical Analysis}

= a
{parent=41e}
{scope}

= Solution
{parent=a}

Each <QR decomposition> $A_k=Q_kR_k$ gives
$$
A_{k+1}=R_kQ_k=Q_k^TA_kQ_k.
$$
Induction therefore yields the first <accumulated QR factorization identity>:
$$
\boxed{A_{k+1}=\overline Q_k^TA\overline Q_k},
\qquad
\overline Q_k=Q_0Q_1\cdots Q_k.
$$

We next prove the corresponding factorization of a <matrix power>. The case $k=0$ is just $A=Q_0R_0$. If
$$
A^{k+1}=\overline Q_k\overline R_k,
\qquad
\overline R_k=R_kR_{k-1}\cdots R_0,
$$
then the similarity formula above implies
$$
A\overline Q_k=\overline Q_kA_{k+1}.
$$
Consequently,
$$
\begin{aligned}
A^{k+2}
&=A\overline Q_k\overline R_k\\
&=\overline Q_kA_{k+1}\overline R_k\\
&=\overline Q_kQ_{k+1}R_{k+1}\overline R_k\\
&=\overline Q_{k+1}\overline R_{k+1}.
\end{aligned}
$$
The product $\overline Q_k$ is <orthogonal matrix>[orthogonal], and the product $\overline R_k$ is <triangular matrix>[upper triangular]. Hence
$$
\boxed{A^{k+1}=\overline Q_k\overline R_k}
$$
is a QR decomposition.

Solved by gpt-5.6-sol high.

= b
{parent=41e}
{scope}

= Solution
{parent=b}

The assertion is false under the hypotheses as printed: nonzero coefficients do not ensure the <two-column dominant-subspace condition>.

For a counterexample, take $n=3$, eigenvalues
$$
(\lambda_1,\lambda_2,\lambda_3)=(1,2,-2),
$$
and an orthonormal eigenbasis $w_1,w_2,w_3$ for which
$$
e_1=\frac{-2w_1+w_2+w_3}{\sqrt6},
\qquad
e_2=\frac{w_1+w_2+w_3}{\sqrt3}.
$$
The two displayed coefficient vectors are orthonormal and can be completed to an <orthogonal matrix>, so such a real <symmetric matrix> $A$ exists. Every $b_i$ and $c_i$ is nonzero, but
$$
b_2c_3-b_3c_2=0.
$$
Indeed, with $t=1/\sqrt2$,
$$
A^ke_1-tA^ke_2
$$
is a nonzero multiple of $w_1$, while the remaining direction in
$\operatorname{span}(A^ke_1,A^ke_2)$ is
$$
v_k=\frac{w_2+(-1)^kw_3}{\sqrt2}.
$$
Thus
$$
\operatorname{span}(A^ke_1,A^ke_2)
=\operatorname{span}(w_1,v_k).
$$
By the <simultaneous iteration interpretation of the QR algorithm>, the leading two columns of the accumulated $Q$ factor span this same space. The leading block $B_k$ is therefore an orthogonal-coordinate representation of the <compression of a linear operator>[compression] of $A$ to this space. Since
$$
w_1^TAw_1=1,
\qquad
v_k^TAv_k=\frac{2+(-2)}2=0,
\qquad
w_1^TAv_k=0,
$$
we have
$$
\operatorname{Spec}(B_k)=\{0,1\}
$$
for every relevant $k$, rather than $\{-2,2\}$. Hence the requested <Hausdorff distance>[Hausdorff convergence] does not hold.

For completeness, the intended statement becomes true if one adds
$$
\Delta=b_{n-1}c_n-b_nc_{n-1}\ne0.
$$
Let $Z_k$ consist of the first two columns of $\overline Q_{k-1}$. The <accumulated QR factorization identity> gives
$$
\operatorname{col}(Z_k)
=\operatorname{span}(A^ke_1,A^ke_2).
$$
After division by $|\lambda_n|^k$, every component along
$w_1,\ldots,w_{n-2}$ tends to zero because of the strict <spectral gap>, while $\Delta\ne0$ makes the two surviving dominant components independent. Therefore these two-dimensional subspaces converge to the <dominant invariant subspace>
$$
E=\operatorname{span}(w_{n-1},w_n).
$$
There are two-by-two orthogonal matrices $U_k$ such that
$$
Z_kU_k\longrightarrow W_*=(w_{n-1},w_n).
$$
Because
$$
B_k=Z_k^TAZ_k,
$$
up to the harmless index convention at $k=0$, we obtain
$$
U_k^TB_kU_k
\longrightarrow
W_*^TAW_*
=\operatorname{diag}(\lambda_{n-1},\lambda_n)
$$
in the <matrix 2-norm>. Orthogonal similarity preserves the <spectrum>, and the stated spectral perturbation bound now gives
$$
d_H\!\left(\operatorname{Spec}(B_k),
\{\lambda_{n-1},\lambda_n\}\right)\longrightarrow0.
$$

Solved by gpt-5.6-sol high.