past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-2.bigb

= Paper 2
{scope}

https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2025/paperii_2_2025.pdf

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

= a
{parent=1g}
{scope}

= Solution
{parent=a}

Since $136=8\cdot17$ and the factors are coprime,
$$
(\mathbb Z/136\mathbb Z)^\times\cong
(\mathbb Z/8\mathbb Z)^\times\times(\mathbb Z/17\mathbb Z)^\times.
$$
Hence
$$\boxed{|G|=\phi(136)=\phi(8)\phi(17)=4\cdot16=64.}$$

Solved by gpt-5.6-sol high.

= b
{parent=1g}
{scope}

= Solution
{parent=b}

The required integer is the exponent of the unit <group>. Since $\lambda(8)=2$ and $\lambda(17)=16$,
$$\boxed{m=\lambda(136)=\operatorname{lcm}(2,16)=16.}$$

Solved by gpt-5.6-sol high.

= c
{parent=1g}
{scope}

= Solution
{parent=c}

The residue $3$ is a <primitive root> modulo $17$. Choose
$$g\equiv1\pmod8,\qquad g\equiv3\pmod{17}.$$
Writing $g=1+8k$ gives $k\equiv13\pmod{17}$, so one may take
$$\boxed{g=105.}$$
Its Chinese-remainder components have orders $1$ and $16$, hence its order modulo $136$ is $16$.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=2I}

Chebyshev's equal-ripple criterion says that a <polynomial> $q$ of degree at most $m$ is a best uniform approximation to a <continuous function> $F$ if and only if the error $F-q$ attains alternating extrema of maximum magnitude at $m+2$ ordered points.

Since $T_n$ has leading coefficient $2^{n-1}$, the <polynomial> $2^{1-n}T_n$ is monic and alternates between $\pm2^{1-n}$ at $t_j=\cos(j\pi/n)$. Therefore
$$
\boxed{q_*(t)=t^n-2^{1-n}T_n(t),\qquad
\min_{\deg q<n}\lVert t^n-q\rVert_\infty=2^{1-n}.}
$$

At each $t_j$, both $T_n-f$ and $T_n+f$ have sign $(-1)^j$, because $|f(t_j)|<1$. Each has a root in every interval between consecutive extrema, hence all its at most $n$ roots lie in $(-1,1)$. For $t>1$ both <polynomials> remain positive, giving $|f(t)|<T_n(t)$. For $t<-1$ both have sign $(-1)^n$, giving $|f(t)|<|T_n(t)|$. On $[-1,1]$ the hypothesis gives $|f|<1$. Thus
$$\boxed{|f(t)|<\max\{1,|T_n(t)|\}\quad(t\in\mathbb R).}$$

Solved by gpt-5.6-sol high.

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

= Solution
{parent=3K}

The received word $001$ has Hamming distances $1$ from $000$ and $2$ from $111$, so
$$
P(001\mid000)=\frac9{64},\qquad P(001\mid111)=\frac3{64}.
$$
After multiplying by the priors, the unnormalized posterior probabilities are $9/320$ and $12/320$. Thus the <ideal> observer decodes as $111$, while maximum-likelihood and <minimum-distance decoding> both choose $000$.

The <ideal> observer minimizes average error but needs priors and channel statistics. Maximum likelihood needs the channel law but not priors, and can be suboptimal for unequal codeword probabilities. Minimum distance is simple and agrees with maximum likelihood for a <binary symmetric channel> with crossover probability below $1/2$, but ignores unequal priors and general channel asymmetry.

Solved by gpt-5.6-sol high.

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

= a
{parent=4f}
{scope}

= Solution
{parent=a}

A language $L$ satisfies the context-free pumping lemma if there is an integer $p\geq1$ such that every $z\in L$ with $|z|\geq p$ has a decomposition
$$z=uvwxy$$
with
$$|vwx|\leq p,\qquad |vx|\geq1,$$
and
$$uv^iwx^iy\in L\quad\hbox{for every }i\geq0.$$
Every context-free language has this property, although the property by itself is not sufficient for context-freeness.

Solved by gpt-5.6-sol high.

= b
{parent=4f}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

This language is not context-free. If $p$ were a pumping length, apply the lemma to
$$z=0^p1^p0^p.$$
Because $|vwx|\leq p$, the pumped region meets at most two adjacent blocks and cannot meet both zero blocks. Pumping with $i=0$ or $i=2$ changes at least one symbol count. If it changes either zero block, that block no longer agrees with the untouched zero block; if it changes only the middle block, the number of ones no longer agrees with the two zero counts. In every case the pumped word leaves the language, contradicting the lemma.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

The language is context-free. A grammar with start symbol $S$ is
$$
S\longrightarrow0S0\mid1S1\mid00\mid11.
$$
Every derivation adds matching symbols to the two ends and starts with a matching central pair, so it produces exactly the nonempty even palindromes $ww^R$.

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

This language is not context-free. Intersect it with the regular language
$$R=0^+1^+0^+.$$
An even palindrome in $R$ has the form $0^r1^{2s}0^r$, and its first half is $w=0^r1^s$. The required equality of the numbers of zeros and ones in $w$ forces $r=s$. Hence the intersection is
$$
L\cap R=\{0^n1^{2n}0^n:n>0\}.
$$
This is not context-free: applying the pumping lemma to $0^p1^{2p}0^p$, a pumped substring of length at most $p$ cannot meet both zero blocks. Pumping therefore either destroys equality of the two zero counts or changes the middle count away from twice their common value.

Context-free languages are closed under intersection with regular languages, so a context-free $L$ would make $L\cap R$ context-free. This contradiction proves the claim.

Solved by gpt-5.6-sol high.

= iv
{parent=b}
{scope}

= Solution
{parent=iv}

This language is context-free. Let
$$
S\longrightarrow0S0\mid0T0,
\qquad
T\longrightarrow1T0\mid10.
$$
The nonterminal $T$ produces exactly $1^m0^m$ with $m>0$. The base production for $S$ supplies one leading and one trailing zero, and each recursive production supplies one more of each. Thus $S$ produces exactly
$$0^n1^m0^{n+m},\qquad n,m>0.$$

Solved by gpt-5.6-sol high.

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

= Solution
{parent=5K}

Write $Y_{gj}$ for the yield in replicate $j$ of genotype $g\in\{aa,Aa,AA\}$. Model 1 is the one-way cell-means model
$$
Y_{gj}=\mu_g+\varepsilon_{gj},
\qquad
\varepsilon_{gj}\mathrel{\overset{\mathrm{iid}}\sim}N(0,\sigma^2).
$$
Thus its three coefficients estimate the three genotype means directly.

Put $D_i=\mathbf1\{\text{count}_i\geq1\}$. Model 2 is
$$
Y_i=\beta_0+\beta_1D_i+\varepsilon_i,
\qquad
\varepsilon_i\mathrel{\overset{\mathrm{iid}}\sim}N(0,\sigma^2).
$$
It imposes $\mu_{Aa}=\mu_{AA}=\beta_0+\beta_1$ and $\mu_{aa}=\beta_0$. The reported finite-sample p-values assume independent normal errors, a common unknown variance in every genotype, fixed full-rank design <matrices>, and correctness of the relevant mean model. Random assignment of genotypes supports interpreting the fitted differences as treatment effects.

Solved by gpt-5.6-sol high.

= 1
{parent=5k}
{scope}

= Solution
{parent=1}

The displayed analysis of variance compares Model 2 with the unrestricted three-mean Model 1. It tests
$$
H_0:\mu_{Aa}=\mu_{AA},
$$
which is precisely full dominance: once an $A$ allele is present, a second copy has no further effect. Under $H_0$,
$$
F=\frac{(54.148-48.487)/1}{48.487/57}=6.6551
\sim F_{1,57}.
$$
The p-value $0.01249$ rejects full dominance at the $5\%$ level, providing evidence that the heterozygote and $AA$ means differ.

Solved by gpt-5.6-sol high.

= 2
{parent=5k}
{scope}

= Solution
{parent=2}

No dominance says that mean yield is affine in the number of $A$ alleles:
$$
\mu_{aa}=\alpha,
\qquad
\mu_{Aa}=\alpha+\delta,
\qquad
\mu_{AA}=\alpha+2\delta.
$$
Equivalently, $2\mu_{Aa}=\mu_{aa}+\mu_{AA}$. Fit that reduced model and compare it with Model 1 using

```r
model3 <- lm(yield ~ count, data = potato)
anova(model3, model1)
```

The resulting one-degree-of-freedom F-test tests the no-dominance constraint against unrestricted genotype means.

Solved by gpt-5.6-sol high.

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

= a
{parent=6a}
{scope}

= Solution
{parent=a}

The term $-IS$ removes healthy foxes through mass-action contact with infectives. The Laplacian $\nabla^2I$ models random migration of infected foxes. The matching term $+IS$ creates newly infected animals, while $-\mu I$ removes infectives at per-capita rate $\mu$. The equation $R_t=\mu I$ records those removed from infectious <circulation>, for example foxes that have died from rabies; they neither migrate nor return to susceptibility in this model.

Solved by gpt-5.6-sol high.

= b
{parent=6a}
{scope}

= Solution
{parent=b}

In a homogeneous population,
$$
\dot I=(S-\mu)I.
$$
An infected fox remains infectious for mean time $1/\mu$ and creates new infections at initial rate $S_0$, so the basic reproductive ratio is
$$
\boxed{\mathcal R_0=\frac{S_0}{\mu}.}
$$
Initially $I$ grows exactly when $S_0-\mu>0$, equivalently $\mathcal R_0>1$. Below that threshold each infective produces fewer than one replacement on average and the infection decays.

Solved by gpt-5.6-sol high.

= c
{parent=6a}
{scope}

= Solution
{parent=c}

With $\xi=x-ct$, time <derivatives> become $-c$ times $\xi$ <derivatives> and $I_{xx}=I''$. Hence
$$
\boxed{-cS'=-IS,}
$$
$$
\boxed{-cI'=I''+IS-\mu I,}
$$
or equivalently
$$cS'=IS,
\qquad
I''+cI'+(S-\mu)I=0.$$

Solved by gpt-5.6-sol high.

= d
{parent=6a}
{scope}

= Solution
{parent=d}

At the leading edge $S\simeq1$ and $I$ is small, so
$$I''+cI'+(1-\mu)I=0.$$
Seek a decaying mode $I\sim e^{-\lambda\xi}$ with $\lambda>0$. Its characteristic equation is
$$
\lambda^2-c\lambda+(1-\mu)=0.
$$
A nonoscillatory positive leading edge requires real positive roots, hence
$$c^2\geq4(1-\mu).$$
For an invasive infection, $mu<1$, and the selected minimum speed is therefore
$$
\boxed{c_{\min}=2\sqrt{1-\mu}.}
$$
If $\mu\geq1$, the state ahead of the wave is below the epidemic threshold and no growing infection front is selected.

Solved by gpt-5.6-sol high.

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

= a
{parent=7e}
{scope}

= Solution
{parent=a}

The partial-fraction decomposition is
$$
\frac1{1+t^2}=\frac1{2i}\left(\frac1{t-i}-\frac1{t+i}\right).
$$
The residues at $i$ and $-i$ are respectively $1/(2i)$ and $-1/(2i)$. If two paths from $0$ to $z$ have winding-number differences $n_+$ and $n_-$ about those poles, their <integrals> differ by
$$
2\pi i\left(\frac{n_+}{2i}-\frac{n_-}{2i}\right)
=\pi(n_+-n_-).
$$
Loops can wind around either pole independently, so the value changes by arbitrary integer multiples of $\pi$ and $F$ is multivalued.

Along the straight real segment,
$$\int_0^1\frac{dt}{1+t^2}=\frac\pi4.$$
Consequently the complete set of values is
$$
\boxed{F(1)=\frac\pi4+k\pi,\qquad k\in\mathbb Z.}
$$

Solved by gpt-5.6-sol high.

= b
{parent=7e}
{scope}

= Solution
{parent=b}

Let $\Gamma$ be any closed curve in $\mathbb C\setminus B$. Since the arc $B$ joins $-i$ to $i$ without meeting $\Gamma$, those two points lie in the same component of $\mathbb C\setminus\Gamma$. Their winding numbers about $\Gamma$ are therefore equal. The residue theorem gives
$$
\oint_\Gamma\frac{dt}{1+t^2}
=\pi\{\operatorname{wind}(\Gamma,i)-\operatorname{wind}(\Gamma,-i)\}=0.
$$
Thus every closed-path period vanishes. <Integrals> from $0$ to $z$ along any two paths avoiding $B$ agree, so $F_B$ is single-valued. Notice that this argument uses cancellation of the two residues; simple connectedness of the slit domain is not required.

Solved by gpt-5.6-sol high.

= c
{parent=7e}
{scope}

= Solution
{parent=c}

On the positive real axis, whose path does not cross the left-displaced cut,
$$
F_B(x)=\int_0^x\frac{dt}{1+t^2}=\arctan x
=\frac\pi2+O(x^{-1}).
$$
Since $F_B'(z)=1/(1+z^2)=O(|z|^{-2})$ near infinity, integrating this <derivative> along large arcs extends the same expansion uniformly there:
$$
F_B(z)=\frac\pi2+O(|z|^{-1}).
$$
Therefore
$$
\oint_{\gamma_R}\frac{F_B(t)}t\,dt
=\frac\pi2\oint_{\gamma_R}\frac{dt}{t}+O(R^{-1})
=i\pi^2+O(R^{-1}),
$$
and hence
$$
\boxed{\lim_{R\to\infty}\oint_{\gamma_R}\frac{F_B(t)}t\,dt=i\pi^2.}
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=8b}
{scope}

= Solution
{parent=a}

For a one-dimensional closed energy orbit, define
$$
I(E)=\frac1{2\pi}\oint p(q,E)\,dq.
$$
After locally inverting this relation to write $E=H(I)$, use the Hamilton--Jacobi generating <function>
$$
W(q,I)=\int^q p(q',I)\,dq'
$$
and define the angle $\phi=\partial W/\partial I$ modulo $2\pi$. The transformation is canonical, the <Hamiltonian> depends only on $I$, and
$$
\dot I=0,
\qquad
\dot\phi=\frac{dH}{dI}=\omega(I).
$$

The adiabatic-invariance principle says that if an external parameter changes on a time scale much longer than the orbital period, and no separatrix is crossed, the action $I$ remains approximately constant. Its small within-cycle changes do not accumulate at leading adiabatic order.

Solved by gpt-5.6-sol high.

= b
{parent=8b}
{scope}

= Solution
{parent=b}

Put $\epsilon=|E|>0$. On the energy surface,
$$
\frac{p^2}{2m}=\frac1{|q|}-\epsilon\geq0,
$$
so
$$
\boxed{|q|\leq q_{\max}=\frac1\epsilon.}
$$
Equality occurs at a turning point, so this is the smallest possible bound.

Using the natural collision continuation through the singular point, a full orbit runs between both turning points. Hence
$$
I=\frac1{2\pi}\oint p\,dq
=\frac2\pi\int_0^{1/\epsilon}
\sqrt{2m\left(\frac1q-\epsilon\right)}\,dq.
$$
Set $q=x^2/\epsilon$. Then
$$
\int_0^{1/\epsilon}\sqrt{2m\left(\frac1q-\epsilon\right)}\,dq
=2\sqrt{\frac{2m}{\epsilon}}
\int_0^1\sqrt{1-x^2}\,dx
=\frac\pi2\sqrt{\frac{2m}{\epsilon}}.
$$
Therefore
$$
\boxed{I=\sqrt{\frac{2m}{|E|}}.}
$$
If instead one identifies the singularity as a reflecting endpoint and regards one half-line as the orbit, the action is half this value; the following scaling is identical.

Adiabatic invariance keeps $m/|E|$ constant. Thus doubling $m$ doubles $|E|$:
$$
\boxed{|E|_{\rm final}=2|E|_{\rm initial},
\qquad E_{\rm final}=2E_{\rm initial}.}
$$
The energy becomes twice as negative.

Solved by gpt-5.6-sol high.

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

= a
{parent=9e}
{scope}

= Solution
{parent=a}

Chemical equilibrium in $n+\nu_e\leftrightarrow p+e^-$ gives
$$\mu_n+\mu_{\nu_e}=\mu_p+\mu_e.$$
The negligible lepton <chemical potentials> therefore imply $\mu_n\simeq\mu_p$. Nonrelativistic Maxwell--Boltzmann number densities satisfy
$$
n_i\simeq g_i\left(\frac{m_ik_BT}{2\pi\hbar^2}\right)^{3/2}
\exp\left(\frac{\mu_i-m_ic^2}{k_BT}\right).
$$
Neutrons and protons have equal spin degeneracy and nearly equal masses, so their prefactor ratio is approximately one. Hence
$$
\boxed{\frac{n_n}{n_p}\simeq
\exp\left[-\frac{(m_n-m_p)c^2}{k_BT}\right]
=e^{-Q/(k_BT)}.}
$$

Solved by gpt-5.6-sol high.

= b
{parent=9e}
{scope}

= Solution
{parent=b}

The ratio
$$\frac{\Gamma_W}{H}\simeq\left(\frac{k_BT}{\kappa}\right)^3$$
falls rapidly as the universe cools. Below $k_BT\simeq0.7\,\mathrm{MeV}$, weak reactions occur less than once per expansion time and can no longer track the falling equilibrium ratio. The neutron-to-proton ratio therefore freezes at approximately $1/6$.

Before nuclei can form efficiently, free neutrons continue to beta-decay,
$$n\to p+e^-+\bar\nu_e.$$
During the <deuterium bottleneck> this lowers the ratio to about $1/7$ by $k_BT\simeq0.07\,\mathrm{MeV}$.

Solved by gpt-5.6-sol high.

= c
{parent=9e}
{scope}

= Solution
{parent=c}

Once deuterium survives photodissociation, nuclear reactions rapidly assemble nearly every available neutron into tightly bound helium-4. Protons are more abundant, so neutrons are the limiting constituent. If $r=n_n/n_p=1/7$, then
$$n_{\rm He}\simeq\frac{n_n}{2}.$$
Neglecting the neutron--proton mass difference,
$$
Y_p=\frac{\rho_{\rm He}}{\rho_B}
\simeq\frac{4m_p(n_n/2)}{m_p(n_n+n_p)}
=\frac{2r}{1+r}.
$$
Thus
$$\boxed{Y_p\simeq\frac14.}$$

Solved by gpt-5.6-sol high.

= d
{parent=9e}
{scope}

= Solution
{parent=d}

Increasing $\kappa$ makes $\Gamma_W/H$ smaller at every fixed <temperature>, so freeze-out occurs earlier, at a higher $T$. The equilibrium value $e^{-Q/(k_BT)}$ is then closer to one. Although some neutrons subsequently decay, the final neutron fraction is larger, and
$$Y_p\simeq\frac{2(n_n/n_p)}{1+n_n/n_p}$$
increases, approaching unity if the surviving neutron and proton abundances become comparable.

Such a universe begins with less hydrogen and more helium. It has less fuel for long-lived hydrogen-burning stars; stellar evolution would tend to proceed through hotter, faster helium-burning stages, reducing typical stellar lifetimes. The shorter stable-energy window, altered heavy-element production, and lower primordial hydrogen abundance would make familiar water-rich, slowly evolving habitats less likely, although a detailed conclusion would require a stellar-evolution model.

Solved by gpt-5.6-sol high.

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

= a
{parent=10c}
{scope}

= Solution
{parent=a}

By the Born rule,
$$
P_\psi(A,B)=|\langle a_0b_0|\psi\rangle|^2
+|\langle a_1b_1|\psi\rangle|^2,
$$
$$
P_\psi(B,C)=|\langle b_0c_0|\psi\rangle|^2
+|\langle b_1c_1|\psi\rangle|^2,
$$
and
$$
P_\psi(C,A)=|\langle c_0a_0|\psi\rangle|^2
+|\langle c_1a_1|\psi\rangle|^2.
$$
Each is the expectation of the projector onto the span of the two displayed matching-label product states.

Solved by gpt-5.6-sol high.

= b
{parent=10c}
{scope}

= Solution
{parent=b}

In the $A$ <basis>, the transformations whose columns are the new <basis> <vectors> are
$$
U_B=\begin{pmatrix}1/2&\sqrt3/2\\\sqrt3/2&-1/2\end{pmatrix},
\qquad
U_C=\begin{pmatrix}1/2&\sqrt3/2\\-\sqrt3/2&1/2\end{pmatrix}.
$$
Thus $U_C$ is a rotation through $-\pi/3$, while $U_B$ is a real orthogonal reflection, equivalently a rotation with one basis-vector phase reversed.

For any real orthogonal <matrix> $U$ with columns $|u_j\rangle$,
$$
\sum_j|u_ju_j\rangle
=\sum_{j,k,l}U_{kj}U_{lj}|a_ka_l\rangle
=\sum_{k,l}(UU^T)_{kl}|a_ka_l\rangle
=\sum_k|a_ka_k\rangle.
$$
Applying this to $U_B$ and $U_C$ and normalizing proves
$$
\boxed{
\frac{|a_0a_0\rangle+|a_1a_1\rangle}{\sqrt2}
=\frac{|b_0b_0\rangle+|b_1b_1\rangle}{\sqrt2}
=\frac{|c_0c_0\rangle+|c_1c_1\rangle}{\sqrt2}.}
$$

Solved by gpt-5.6-sol high.

= c
{parent=10c}
{scope}

= Solution
{parent=c}

For $|\psi\rangle=|00\rangle$, independence of the product state gives
$$
P_\psi(A,B)=\frac14,
\qquad
P_\psi(C,A)=\frac14.
$$
For the remaining pair,
$$
P_\psi(B,C)
=\left(\frac14\right)\left(\frac14\right)
+\left(\frac34\right)\left(\frac34\right)
=\frac58.
$$
Therefore
$$
\boxed{P_\psi(A,B)+P_\psi(B,C)+P_\psi(C,A)=\frac98\geq1.}
$$

Solved by gpt-5.6-sol high.

= d
{parent=10c}
{scope}

= Solution
{parent=d}

For the real Bell state $|\phi^+\rangle$, real one-qubit <vectors> satisfy
$$
\langle x_jy_j|\phi^+\rangle
=\frac1{\sqrt2}\langle x_j|y_j\rangle.
$$
For each adjacent pair $(A,B)$, $(B,C)$, and $(C,A)$, the two same-label overlaps have magnitude $1/2$. Hence every matching-label probability is
$$
P_{\phi^+}(X,Y)=\frac12\left(\frac14+\frac14\right)=\frac14.
$$
It follows that
$$
\boxed{P_{\phi^+}(A,B)+P_{\phi^+}(B,C)+P_{\phi^+}(C,A)=\frac34.}
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=11i}
{scope}

= Solution
{parent=a}

Assume (i), and suppose a map $f$ as in (ii) existed. Then
$$A=f^{-1}(I),\qquad B=f^{-1}(J),\qquad C=f^{-1}(K)$$
are closed, cover $T$, and contain $I,J,K$, respectively. But
$$A\cap B\cap C=f^{-1}(I\cap J\cap K)=\varnothing,$$
since the three sides have no common point. This contradicts (i), so (i) implies (ii).

Conversely, suppose (i) fails. Let $A,B,C$ be a counterexample and put
$$
d_A(x)=\operatorname{dist}(x,A),\quad
d_B(x)=\operatorname{dist}(x,B),\quad
d_C(x)=\operatorname{dist}(x,C).
$$
No point lies in all three closed sets, so $D=d_A+d_B+d_C>0$. Identify $T$ affinely with the standard simplex so that side $I$ is the side on which the first barycentric coordinate is zero, and similarly for $J,K$. Define $f(x)$ to be the point with barycentric coordinates
$$
\frac1D(d_A(x),d_B(x),d_C(x)).
$$
Because $A\cup B\cup C$ covers $T$, at least one distance is zero, so $f(x)\in\partial T$. If $x\in I\subset A$, then $d_A(x)=0$, so $f(x)\in I$; likewise for $J,K$. Thus $f$ is a forbidden map from (ii). This proves (ii) implies (i).

Solved by gpt-5.6-sol high.

= b
{parent=11i}
{scope}

= Solution
{parent=b}

Brouwer's fixed-point theorem in the plane states that every continuous map from a closed disk to itself has a fixed point. Equivalently, the same holds for every nonempty compact convex subset of $\mathbb R^2$.

On the closed unit disk define
$$
F(z)=-\frac{z^6-2z^5+4z^2+2}{9}.
$$
For $|z|\leq1$,
$$
|F(z)|\leq\frac{1+2+4+2}{9}=1,
$$
so $F$ maps the disk continuously into itself. Brouwer supplies $z$ with $F(z)=z$, which is equivalent to
$$
\boxed{z^6-2z^5+4z^2+9z+2=0.}
$$
Thus this <polynomial> has a root with $|z|\leq1$.

Solved by gpt-5.6-sol high.

= c
{parent=11i}
{scope}

= Solution
{parent=c}

Write $\beta(s)=(\beta_1(s),\beta_2(s))$, $\gamma(t)=(\gamma_1(t),\gamma_2(t))$, and
$$h(s,t)=\beta(s)-\gamma(t).$$
The endpoint conditions imply
$$
\begin{array}{ll}
h_1(s,-1)\geq0,&h_1(s,1)\leq0,\\
h_2(-1,t)\leq0,&h_2(1,t)\geq0.
\end{array}
$$
Let $P(x)=\max(-1,\min(1,x))$ and define the continuous self-map of $I^2$
$$
F(s,t)=\bigl(P(s-h_2(s,t)),\ P(t+h_1(s,t))\bigr).
$$
By Brouwer, $F$ has a fixed point $(s,t)$. If $s$ is interior, its fixed-point equation gives $h_2=0$. At $s=-1$ or $s=1$, the projection equation and the corresponding displayed boundary sign again force $h_2=0$. The identical argument in the second coordinate gives $h_1=0$. Hence
$$h(s,t)=0,$$
so
$$\boxed{\beta(s)=\gamma(t).}$$
The two paths intersect.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=12K}

The <unicity distance> is the ciphertext length at which the key is uniquely determined, or equivalently at which <key equivocation> vanishes:
$$
U=\min\{n:H(K\mid C^n)=0\}.
$$
For Shannon's idealized calculation, assume a uniformly chosen key independent of an iid plaintext source, deterministic invertible encryption for each key, and uniformly distributed ciphertext over an alphabet $\Sigma$. Since $M^n$ is determined by $(C^n,K)$,
$$
\begin{aligned}
H(K\mid C^n)
&=H(M^n,K\mid C^n)\\
&=H(M^n,K)-H(C^n)\\
&=\log|K|+nH-n\log|\Sigma|.
\end{aligned}
$$
Setting this to zero gives
$$
\boxed{U=\frac{\log|K|}{\log|\Sigma|-H}.}
$$
Equivalently, each ciphertext symbol supplies the source redundancy $\log|\Sigma|-H$ bits toward identifying the key.

For the stated cipher, the original formula uses $k_{i\bmod3}$, so $|K|=2^3=8$, while $|\Sigma|=3$. Split first according to whether $m=0$. This event and its complement each have probability $1/2$, and conditional on $m\ne0$ the probabilities are $(2p,1-2p)$. Hence
$$
H(m)=1+\frac12H(2p,1-2p).
$$
If $U\geq20$, then
$$
\log_2(3)-H(m)\leq\frac3{20},
$$
so, using $\log_2(3)=1.585$,
$$
1+\frac12H(2p,1-2p)\geq1.435.
$$
Therefore
$$\boxed{H(2p,1-2p)\geq0.87.}$$

<Binary entropy> is symmetric about probability $1/2$ and increases up to that point. Since equality holds at $2p=0.30$, it also holds at $2p=0.70$. Thus
$$
H(2p,1-2p)\geq0.87
\quad\Longleftrightarrow\quad
0.30\leq2p\leq0.70,
$$
and all requested values are
$$\boxed{0.15\leq p\leq0.35.}$$

Finally let $p=0$, so the source is uniform on $\{0,2\}$. Use one uniform key bit $k$ and encrypt each symbol by
$$
c_i=\begin{cases}m_i,&k=0,\\2-m_i,&k=1.
\end{cases}
$$
Thus the second key swaps $0$ and $2$. Every ciphertext has two possible plaintexts, a binary string and its complement, with equal source probability. No amount of ciphertext distinguishes the keys: $H(K\mid C^n)=1$ for every $n$. This cipher has infinite unicity distance.

Solved by gpt-5.6-sol high.

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

= a
{parent=13e}
{scope}

= Solution
{parent=a}

With $z=\cos x$,
$$
\frac{dz}{dx}=-\sin x,
\qquad
\frac{d^2z}{dx^2}=-\cos x=-z.
$$
The chain rule gives
$$
\frac{d^2w}{dx^2}
=\left(\frac{dz}{dx}\right)^2\frac{d^2w}{dz^2}
+\frac{d^2z}{dx^2}\frac{dw}{dz}
=(1-z^2)w_{zz}-zw_z.
$$
Hence $w_{xx}+n^2w=0$ becomes
$$
\boxed{(1-z^2)w_{zz}-zw_z+n^2w=0.}
$$

Solved by gpt-5.6-sol high.

= b
{parent=13e}
{scope}

= Solution
{parent=b}

A <Papperitz symbol> records the three regular singular points of a second-order Fuchsian equation and the two characteristic exponents at each point. Near $z=1$, substituting $w\sim(z-1)^\rho$ into the leading terms gives
$$\rho(2\rho-1)=0,$$
so the exponents are $0$ and $1/2$. The same calculation at $z=-1$ gives $0$ and $1/2$. At infinity, $w\sim z^s$ gives
$$-s^2+n^2=0,$$
so $s=\pm n$. Since an exponent $\rho$ at infinity denotes behavior $z^{-\rho}$, these are recorded as $-n,n$. Thus the equation has P-symbol
$$
\boxed{
P\left\{\begin{matrix}
1&-1&\infty&\\
0&0&-n&z\\
\tfrac12&\tfrac12&n&
\end{matrix}\right\}.}
$$
The exponent sum is $1$, as required by the <Fuchs relation> for three singular points.

Solved by gpt-5.6-sol high.

= c
{parent=13e}
{scope}

= Solution
{parent=c}

Set
$$u=\frac{1-z}{2}.$$
This Möbius change sends $z=1,-1,\infty$ to $u=0,1,\infty$, respectively, without changing the corresponding exponents. The hypergeometric P-symbol has exponents
$$
\begin{array}{c|ccc}
&0&1&\infty\\ \hline
&0&0&A\\
&1-C&C-A-B&B.
\end{array}
$$
Matching it with $(0,1/2)$ at both finite singularities and $(-n,n)$ at infinity gives
$$C=\frac12,\qquad A=n,\qquad B=-n.$$
The exponent-zero solution normalized to one at $u=0$ is therefore
$$
\boxed{w_1(z)=F\left(n,-n;\frac12;\frac{1-z}{2}\right).}
$$

The second standard local solution at $u=0$ is
$$
u^{1-C}F(A-C+1,B-C+1;2-C;u).$$
After absorbing the constant $2^{-1/2}$ into its normalization, this becomes
$$
\boxed{
w_2(z)=(1-z)^{1/2}
F\left(-n+\frac12,n+\frac12;\frac32;\frac{1-z}{2}\right).}
$$
Their exponents at $z=1$ are $0$ and $1/2$, so they are linearly independent.

Solved by gpt-5.6-sol high.

= d
{parent=13e}
{scope}

= Solution
{parent=d}

Take $n=1$ in part (c). The original equation $w_{xx}+w=0$ has the solution $\sin x=\sqrt{1-z^2}$. The solution $w_2$ is normalized by $w_2/(1-z)^{1/2}\to1$ as $z\to1$, so
$$
w_2(z)=\frac1{\sqrt2}\sqrt{1-z^2}
=(1-z)^{1/2}\sqrt{\frac{1+z}{2}}.
$$
Comparing this with the formula for $w_2$ and using $u=(1-z)/2$ gives
$$
\boxed{F\left(-\frac12,\frac32;\frac32;u\right)
=\sqrt{1-u}.}
$$
This also agrees with the general identity $F(A,C;C;u)=(1-u)^{-A}$.

Solved by gpt-5.6-sol high.

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

= a
{parent=14b}
{scope}

= Solution
{parent=a}

With the pivot as origin and upward vertical coordinate,
$$
\mathbf r_1=l(\sin\theta_1,-\cos\theta_1),
\qquad
\mathbf r_2=\mathbf r_1+l(\sin\theta_2,-\cos\theta_2).
$$
Therefore
$$
|\dot{\mathbf r}_1|^2=l^2\dot\theta_1^2,
$$
$$
|\dot{\mathbf r}_2|^2=l^2\left[\dot\theta_1^2+\dot\theta_2^2
+2\cos(\theta_1-\theta_2)\dot\theta_1\dot\theta_2\right].
$$
The potential energy is
$$V=-(M+m)gl\cos\theta_1-mgl\cos\theta_2.$$
Thus $L=T-V$ is
$$
\boxed{
L=\frac12Ml^2\dot\theta_1^2
+\frac12ml^2\left[\dot\theta_1^2+\dot\theta_2^2
+2\cos(\theta_1-\theta_2)\dot\theta_1\dot\theta_2\right]
+Mgl\cos\theta_1+mg l(\cos\theta_1+\cos\theta_2).}
$$

Solved by gpt-5.6-sol high.

= b
{parent=14b}
{scope}

= Solution
{parent=b}

Put $\Delta=\theta_1-\theta_2$ and $\omega_0^2=g/l$. The Euler--Lagrange equations, after division by the appropriate powers of $l$, are
$$
\boxed{(M+m)\ddot\theta_1+m\cos\Delta\,\ddot\theta_2
+m\sin\Delta\,\dot\theta_2^2+(M+m)\omega_0^2\sin\theta_1=0,}
$$
$$
\boxed{\ddot\theta_2+\cos\Delta\,\ddot\theta_1
-\sin\Delta\,\dot\theta_1^2+\omega_0^2\sin\theta_2=0.}
$$
Time-translation invariance conserves the total energy
$$
E=\frac12(M+m)l^2\dot\theta_1^2+\frac12ml^2\dot\theta_2^2
+ml^2\cos\Delta\,\dot\theta_1\dot\theta_2
-(M+m)gl\cos\theta_1-mgl\cos\theta_2.
$$

Setting both angles and velocities to zero satisfies the equations, so the downward configuration is an equilibrium. Writing $\theta_i=z_i$ and retaining linear terms gives
$$
\boxed{(M+m)\ddot z_1+m\ddot z_2+(M+m)\omega_0^2z_1=0,}
$$
$$
\boxed{\ddot z_1+\ddot z_2+\omega_0^2z_2=0.}
$$
Equivalently, the mass and stiffness matrices are
$$
\mathsf M=l^2\begin{pmatrix}M+m&m\\m&m\end{pmatrix},
\qquad
\mathsf K=gl\begin{pmatrix}M+m&0\\0&m\end{pmatrix}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=14b}
{scope}

= Solution
{parent=c}

Seek $z=v e^{i\Omega t}$ and put $\mu=m/M$. The generalized characteristic equation is
$$
\det(\mathsf K-\Omega^2\mathsf M)=0,
$$
or
$$
(1+\mu)(\omega_0^2-\Omega^2)^2-\mu\Omega^4=0.
$$
As a quadratic in $X=\Omega^2$, this is
$$X^2-2(1+\mu)\omega_0^2X+(1+\mu)\omega_0^4=0,$$
so
$$
\boxed{
\Omega_\pm^2=\omega_0^2\left(1+\mu\pm\sqrt{\mu(1+\mu)}\right).}
$$
A corresponding displacement vector is
$$
v_\pm=\begin{pmatrix}\omega_0^2-\Omega_\pm^2\\\Omega_\pm^2\end{pmatrix}.
$$
Thus the four normal modes are $v_+e^{\pm i\Omega_+t}$ and $v_-e^{\pm i\Omega_-t}$. Both squared frequencies are positive for every $\mu>0$, proving linear stability.

For $\mu\ll1$,
$$
\Omega_\pm
=\omega_0\left(1\pm\frac12\sqrt\mu+O(\mu)\right).
$$
Therefore, labeling the larger and smaller positive frequencies by $\omega$ and $\omega'$,
$$
\boxed{\omega-\omega'=\omega_0\sqrt\mu+O(\mu),}
$$
so $\alpha=\omega_0$.

Solved by gpt-5.6-sol high.

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

= a
{parent=15c}
{scope}

= Solution
{parent=a}

The order of $a$ modulo $N$ is the least positive integer $r$ such that
$$
\boxed{a^r\equiv1\pmod N.}
$$
It exists because $a$ is a unit in the finite group $(\mathbb Z/N\mathbb Z)^\times$.

Solved by gpt-5.6-sol high.

= b
{parent=15c}
{scope}

= Solution
{parent=b}

Define
$$
\boxed{f(x)=a^x\pmod N,\qquad x\in\mathbb Z,}
$$
using the inverse of $a$ for negative $x$. Since $a^r\equiv1$, $f(x+r)=f(x)$. Conversely, if $f(x)=f(y)$ then $a^{x-y}\equiv1$, so the definition of order gives $r\mid x-y$. Two integers in the same block of $r$ consecutive integers cannot differ by a nonzero multiple of $r$. Hence $f$ is one-to-one within each period.

Solved by gpt-5.6-sol high.

= c
{parent=15c}
{scope}

= Solution
{parent=c}

Since $r$ is even,
$$
(a^{r/2}-1)(a^{r/2}+1)=a^r-1\equiv0\pmod N.
$$
The first factor is not divisible by $N$, since that would contradict minimality of $r$, and the second is not divisible by $N$ by hypothesis. Nevertheless their product is divisible by $N$. Therefore
$$
1<\gcd(a^{r/2}-1,N)<N,
$$
and Euclid's algorithm computes a nontrivial factor. One can also compute $\gcd(a^{r/2}+1,N)$; together the two gcds expose factors lying on opposite sides of the congruence $a^{r/2}\equiv\pm1$ modulo the prime-power divisors of $N$.

Solved by gpt-5.6-sol high.

= d
{parent=15c}
{scope}

= i
{parent=d}
{scope}

= Solution
{parent=i}

With the positive-exponent convention,
$$
\boxed{
\operatorname{QFT}_M|x\rangle
=\frac1{\sqrt M}\sum_{u=0}^{M-1}e^{2\pi i xu/M}|u\rangle.}
$$
The opposite sign convention is equivalent after conjugating all Fourier phases.

Solved by gpt-5.6-sol high.

= ii
{parent=d}
{scope}

= Solution
{parent=ii}

Put $Q=2^m$. Applying the transform term by term gives
$$
|\phi_2\rangle
=\frac1{\sqrt{AQ}}\sum_{u=0}^{Q-1}
\sum_{j=0}^{A-1}e^{2\pi i(x_0+jr)u/Q}|u\rangle
=\sum_{u=0}^{Q-1}g(u)|u\rangle,
$$
where
$$
\boxed{
g(u)=\frac{e^{2\pi i x_0u/Q}}{\sqrt{AQ}}
\sum_{j=0}^{A-1}e^{2\pi i jru/Q}.}
$$
Thus, when $e^{2\pi iru/Q}\ne1$,
$$
\boxed{
g(u)=\frac{e^{2\pi i x_0u/Q}}{\sqrt{AQ}}
\frac{1-e^{2\pi iAru/Q}}{1-e^{2\pi iru/Q}}.}
$$
If the denominator vanishes, the continuous limiting value is
$$g(u)=\sqrt{A/Q}\,e^{2\pi ix_0u/Q}.$$
The magnitude is a Dirichlet-kernel peak near integers $u$ for which $ru/Q$ is close to an integer.

Solved by gpt-5.6-sol high.

= iii
{parent=d}
{scope}

= Solution
{parent=iii}

The assumed measurement estimate gives
$$
\left|\frac cQ-\frac kr\right|<\frac1{2Q}<\frac1{2N^2},
$$
because $Q=2^m>N^2$. Since $(k,r)=1$, the fraction $k/r$ is reduced.

Suppose two distinct reduced fractions $k/r$ and $k'/r'$ with $r,r'<N$ both obeyed $(*)$. Then
$$
\left|\frac kr-\frac{k'}{r'}\right|
<\frac1{N^2}.
$$
But distinct reduced fractions satisfy
$$
\left|\frac kr-\frac{k'}{r'}\right|
=\frac{|kr'-k'r|}{rr'}\geq\frac1{rr'}>\frac1{N^2},
$$
a contradiction. Hence at most one such fraction exists.

Solved by gpt-5.6-sol high.

= iv
{parent=d}
{scope}

= Solution
{parent=iv}

Here $N^2=441$, so the least power of two exceeding it is $Q=2^9=512$. The continued fraction is
$$
\frac{427}{512}=[0;1,5,42,2],
$$
whose convergents begin
$$0,\quad1,\quad\frac56,\quad\frac{211}{253},\quad\frac{427}{512}.$$
The only nontrivial convergent with denominator below $N=21$ is $5/6$, and indeed
$$
\left|\frac{427}{512}-\frac56\right|
=\frac1{1536}<\frac1{2\cdot21^2}.
$$
The uniqueness result therefore gives
$$\boxed{r=6.}$$
As a check, $10^6\equiv1\pmod{21}$, while $10^2\not\equiv1$ and $10^3\not\equiv1$, so the order is exactly six.

Solved by gpt-5.6-sol high.

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

= a
{parent=16h}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

A sentence is a formula with no free variables. Thus its truth in an $L$-structure does not depend on a variable assignment.

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

A theory in $L$ is a set of $L$-sentences. Here a theory need not already contain every sentence derivable from it.

Solved by gpt-5.6-sol high.

= iii
{parent=a}
{scope}

= Solution
{parent=iii}

An $L$-structure $M$ is a model of $T$ when
$$M\models\sigma\qquad\text{for every }\sigma\in T.$$

Solved by gpt-5.6-sol high.

= iv
{parent=a}
{scope}

= Solution
{parent=iv}

A theory $T$ is consistent when no contradiction is derivable from it, equivalently when $T\nvdash\bot$. In a classical proof system this is also equivalent to there being no sentence $\sigma$ for which both $T\vdash\sigma$ and $T\vdash\neg\sigma$.

Solved by gpt-5.6-sol high.

= b
{parent=16h}
{scope}

= Solution
{parent=b}

Suppose first that $M\models T$. By the <soundness theorem for first-order logic>, every theorem of $T$ is true in $M$. A contradiction cannot be true in a structure, so $T$ is consistent.

Conversely, the <Godel completeness theorem> says that every semantically valid consequence is derivable. Its equivalent model-existence form says directly that every consistent first-order theory has a model. Hence
$$
\boxed{T\text{ is consistent}\quad\Longleftrightarrow\quad T\text{ has a model}.}
$$

Solved by gpt-5.6-sol high.

= c
{parent=16h}
{scope}

= Solution
{parent=c}

The <compactness theorem> states that a set $T$ of first-order sentences has a model if and only if every finite subset of $T$ has a model.

The forward implication follows by using the same model for every finite subset. Conversely, suppose every finite subset of $T$ has a model. If $T$ were inconsistent, a formal derivation of $\bot$ from $T$ would use only finitely many assumptions, say those in $T_0\subseteq T$. Then $T_0$ would be inconsistent. By part (b), $T_0$ would have no model, contrary to the hypothesis. Thus $T$ is consistent, and part (b) gives a model of $T$.

Solved by gpt-5.6-sol high.

= d
{parent=16h}
{scope}

= Solution
{parent=d}

Take $T_f$ to consist of $T$ together with the following sentences.

* $f$ is injective and surjective:
$$
\forall x\,\forall y\,(f(x)=f(y)\Rightarrow x=y),
\qquad
\forall y\,\exists x\,f(x)=y.
$$
* For every $n$-ary operation symbol $\omega$ of $L$,
$$
\forall x_1\cdots\forall x_n\,
f(\omega(x_1,\ldots,x_n))
=\omega(f(x_1),\ldots,f(x_n)).
$$
This includes $f(c)=c$ for each constant symbol $c$.
* For every $n$-ary relation symbol $R$ of $L$,
$$
\forall x_1\cdots\forall x_n\,
\bigl(R(x_1,\ldots,x_n)\Longleftrightarrow
R(f(x_1),\ldots,f(x_n))\bigr).
$$

The first pair of axioms makes the interpretation $\vartheta$ a bijection. The remaining schemes say exactly that it preserves every operation and relation, so $(M,\vartheta)\models T_f$ exactly when $M\models T$ and $\vartheta\in\operatorname{Aut}(M)$.

Finally, $T$ is consistent and hence has a model $M$ by part (b). Expanding $M$ by interpreting $f$ as the identity automorphism gives a model of $T_f$, so $T_f$ is consistent.

Solved by gpt-5.6-sol high.

= e
{parent=16h}
{scope}

= i
{parent=e}
{scope}

= Solution
{parent=i}

A model of $T_G$ consists of a model $M\models T$ and automorphisms
$$
\vartheta_g=(f_g)^M\in\operatorname{Aut}(M)
$$
such that $\vartheta_g\vartheta_h=\vartheta_{gh}$ and $\vartheta_g\ne\vartheta_h$ whenever $g\ne h$. Thus $g\mapsto\vartheta_g$ is an injective homomorphism $G\hookrightarrow\operatorname{Aut}(M)$. Conversely, any such embedding supplies interpretations of the $f_g$ and hence a model of $T_G$.

Therefore $G$ is $T$-good exactly when $T_G$ has a model. By part (b), this is equivalent to consistency of $T_G$, and hence
$$
G\text{ is }T\text{-bad}\quad\Longleftrightarrow\quad T_G\text{ is inconsistent}.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=e}
{scope}

= Solution
{parent=ii}

Suppose $T_G$ is inconsistent. By the <compactness theorem>, some finite subset $\Delta\subseteq T_G$ is inconsistent. Only finitely many added symbols $f_g$ occur in $\Delta$; let $X\subseteq G$ be the finite set of their indices. Every sentence of $\Delta$ belongs to the sublanguage $L_X$, so
$$
\Delta\subseteq T_G\cap L_X.
$$
Consequently $T_G\cap L_X$ is inconsistent.

Solved by gpt-5.6-sol high.

= iii
{parent=e}
{scope}

= Solution
{parent=iii}

If $T_G\cap L_X$ is inconsistent for some finite $X\subseteq G$, then its superset $T_G$ is inconsistent. Together with parts (i) and (ii), this proves the equivalence of all three conditions.

Solved by gpt-5.6-sol high.

= f
{parent=16h}
{scope}

= Solution
{parent=f}

We describe a collection of finite multiplication patterns. For finite sets
$$
E\subseteq\{1,\ldots,n\}^3,
\qquad
D\subseteq\{1,\ldots,n\}^2,
$$
write
$$
P_{E,D}(x_1,\ldots,x_n)=
\bigwedge_{(i,j,k)\in E}x_ix_j=x_k
\ \wedge\!
\bigwedge_{(i,j)\in D}x_i\ne x_j.
$$
Call $(E,D)$ $T$-forbidden if the following theory, in $L$ expanded by unary symbols $f_1,\ldots,f_n$, is inconsistent:

* $T$, together with the assertion that every $f_i$ is an automorphism;
* $\forall z\,f_i(f_j(z))=f_k(z)$ for every $(i,j,k)\in E$;
* $\exists z\,f_i(z)\ne f_j(z)$ for every $(i,j)\in D$.

Let $T^*$ contain the group axioms and, for every $T$-forbidden finite pattern, the sentence
$$
\forall x_1\cdots\forall x_n\,\neg P_{E,D}(x_1,\ldots,x_n).
$$
This is a first-order theory in the language of groups.

Suppose $G$ is $T$-good. Choose $M\models T$ and an embedding $\rho:G\hookrightarrow\operatorname{Aut}(M)$. If a tuple $(g_1,\ldots,g_n)$ in $G$ realized a forbidden pattern, interpreting $f_i$ as $\rho(g_i)$ would give a model of its supposedly inconsistent automorphism theory. Thus no such tuple exists, and $G\models T^*$.

Conversely, suppose $G$ is $T$-bad. By part (e), there is a finite set $X=\{g_1,\ldots,g_n\}\subseteq G$ for which $T_G\cap L_X$ is inconsistent. Let
$$
E=\{(i,j,k):g_ig_j=g_k\},
\qquad
D=\{(i,j):g_i\ne g_j\}.
$$
The associated automorphism theory is precisely $T_G\cap L_X$, up to renaming its function symbols, so $(E,D)$ is $T$-forbidden. But $(g_1,\ldots,g_n)$ realizes $P_{E,D}$ in $G$, and therefore $G$ violates the corresponding forbidding axiom. Hence $G\not\models T^*$.

We have proved
$$
\boxed{G\models T^*\quad\Longleftrightarrow\quad G\text{ is }T\text{-good}.}
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=17f}
{scope}

= Solution
{parent=a}

A matching from $X$ to $Y$ is a set of pairwise vertex-disjoint edges that covers every vertex of $X$. Equivalently, it chooses for each $x\in X$ a distinct neighbour in $Y$.

<Hall marriage theorem> says that such a matching exists if and only if
$$
|N(S)|\geq |S|\qquad\text{for every }S\subseteq X.
$$
Necessity is immediate: the distinct partners of the vertices in $S$ all belong to $N(S)$.

For sufficiency, induct on $|X|$. The claim is clear when $|X|=1$. First suppose every nonempty proper $S\subset X$ satisfies the strict inequality $|N(S)|\geq |S|+1$. Choose an edge $xy$ and delete $x$ and $y$. For $S\subseteq X\setminus\{x\}$, deletion removes at most one neighbour, so
$$
|N_{G-\{x,y\}}(S)|\geq |N_G(S)|-1\geq |S|.
$$
Induction gives a matching of $X\setminus\{x\}$, and adding $xy$ completes it.

Otherwise there is a nonempty proper $S\subset X$ with $|N(S)|=|S|$. Hall's condition holds in the bipartite graph induced by $S\cup N(S)$, so induction matches $S$ onto $N(S)$. For $A\subseteq X\setminus S$,
$$
|N(A\cup S)|\geq |A|+|S|,
$$
and hence $A$ has at least $|A|$ neighbours outside $N(S)$. Induction therefore matches $X\setminus S$ into $Y\setminus N(S)$. The two matchings are disjoint and together cover $X$.

Solved by gpt-5.6-sol high.

= i
{parent=a}
{scope}

= Solution
{parent=i}

Let the bipartition be $X\sqcup Y$. Counting edges from each side gives
$$
k|X|=|E(G)|=k|Y|,
$$
so $|X|=|Y|=n/2$. For $S\subseteq X$, the $k|S|$ edges leaving $S$ all end in $N(S)$, while each vertex of $N(S)$ receives at most $k$ of them. Therefore
$$k|S|\leq k|N(S)|,$$
and Hall's condition holds. There is a matching covering $X$, of size $n/2$. No matching can use more than one edge at each vertex, so this is maximal and
$$
\boxed{\nu(G)=\frac n2.}
$$

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

Let $M$ be a maximum matching and put $m=|M|$. The $n-2m$ unmatched vertices form an independent set, since an edge between two of them could be added to $M$. Because $G$ is $k$-regular, exactly $k(n-2m)$ edges run from unmatched vertices to vertices covered by $M$.

Each endpoint of a matched edge has one incident edge in $M$, and hence at most $k-1$ incident edges from unmatched vertices. The two endpoints of each of the $m$ matched edges therefore receive at most $2k-2$ such edges. Thus
$$
k(n-2m)\leq(2k-2)m.
$$
Rearranging gives
$$
\boxed{\nu(G)=m\geq\frac{k}{4k-2}n.}
$$

Solved by gpt-5.6-sol high.

= iii
{parent=a}
{scope}

= Solution
{parent=iii}

For every $r\geq1$, take the disjoint union
$$G_r=rK_3$$
of $r$ triangles. This graph is $2$-regular, has $n=3r$ vertices, and each triangle contributes exactly one edge to a maximum matching. Hence
$$
\nu(G_r)=r=\frac n3=\frac{2}{4\cdot2-2}n,
$$
so equality holds for this infinite family.

Solved by gpt-5.6-sol high.

= b
{parent=17f}
{scope}

= Solution
{parent=b}

The eigenvalues of a graph are the eigenvalues of its adjacency matrix. Order the vertices with those of $X$ first and those of $Y$ second. The adjacency matrix then has block form
$$
A=\begin{pmatrix}0&B\\B^T&0\end{pmatrix},
$$
where $B$ is the $|X|\times|Y|$ <bipartite adjacency matrix>.

Since zero is not an eigenvalue, $A$ is invertible. If $|X|\ne|Y|$, then
$$
\operatorname{rank}A\leq2\min(|X|,|Y|)<|X|+|Y|,
$$
a contradiction. Thus $|X|=|Y|=r$. Moreover, if $B$ were singular, a nonzero $v$ with $Bv=0$ would give $A(0,v)^T=0$. Hence $B$ is invertible.

In the determinant expansion
$$
\det B=\sum_{\pi\in S_r}\operatorname{sgn}(\pi)
\prod_{i=1}^r B_{i,\pi(i)},
$$
at least one product is nonzero. For its permutation $\pi$, all entries $B_{i,\pi(i)}$ equal one, so the corresponding $r$ edges pair every vertex of $X$ with a distinct vertex of $Y$. They form the required matching.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=18j}

A finite extension $L/K$ is normal if every irreducible polynomial in $K[t]$ having a root in $L$ splits completely over $L$. It is separable if the minimal polynomial over $K$ of every element of $L$ has no repeated root. It is Galois if
$$
|\operatorname{Aut}(L/K)|=[L:K].
$$

Solved by gpt-5.6-sol high.

= a
{parent=18j}
{scope}

= Solution
{parent=a}

Put $L_i=K(\alpha_1,\ldots,\alpha_i)$, with $L_0=K$, and let
$$d_i=[L_i:L_{i-1}].$$
Fix a $K$-embedding $\sigma:L_{i-1}\to\overline K$. If $m_i(t)$ is the minimal polynomial of $\alpha_i$ over $L_{i-1}$, an extension of $\sigma$ to $L_i$ is determined by the image of $\alpha_i$, which must be a root of the polynomial obtained by applying $\sigma$ to the coefficients of $m_i$. Conversely, each distinct root gives one extension. Since $\overline K$ is algebraically closed, there is at least one such root, and there are at most $d_i$ distinct roots.

Starting from the unique embedding of $K$, induction therefore gives
$$
1\leq |\operatorname{Hom}_K(L_i,\overline K)|
\leq d_1\cdots d_i=[L_i:K].
$$
If $L/K$ is separable, each $m_i$ is separable, so every transformed polynomial has exactly $d_i$ distinct roots. Every inequality is then an equality. Taking $i=n$ proves
$$
1\leq |\operatorname{Hom}_K(L,\overline K)|\leq[L:K],
$$
with equality in the separable case.

Solved by gpt-5.6-sol high.

= b
{parent=18j}
{scope}

= Solution
{parent=b}

After fixing $L\subseteq\overline K$, every $K$-automorphism of $L$ is a $K$-embedding $L\to\overline K$. Hence
$$
|\operatorname{Aut}(L/K)|\leq
|\operatorname{Hom}_K(L,\overline K)|.
$$
If $L/K$ is normal, every $K$-embedding of $L$ into $\overline K$ has image $L$. It is consequently a $K$-automorphism of $L$, so the two sets, and therefore their cardinalities, are equal.

Solved by gpt-5.6-sol high.

= c
{parent=18j}
{scope}

= Solution
{parent=c}

If $L/K$ is separable, part (a) gives
$$|\operatorname{Hom}_K(L,\overline K)|=[L:K].$$
If it is also normal, part (b) gives
$$|\operatorname{Aut}(L/K)|=|\operatorname{Hom}_K(L,\overline K)|.$$
Thus
$$
|\operatorname{Aut}(L/K)|=[L:K],
$$
which is exactly the defining condition for $L/K$ to be Galois.

Solved by gpt-5.6-sol high.

= d
{parent=18j}
{scope}

= Solution
{parent=d}

Take $p=11$. Write $Y=X^5$, so the extension is
$$
\mathbb F_{11}(X)/\mathbb F_{11}(Y).
$$
The polynomial $t^5-Y$ is irreducible over $\mathbb F_{11}(Y)$, for example by Eisenstein's criterion at $Y$ in $\mathbb F_{11}[Y]$, and has $X$ as a root. Thus the extension has degree five.

The element $3\in\mathbb F_{11}^{\times}$ has order five. Hence all roots of $t^5-Y$ are
$$
X,\ 3X,\ 3^2X,\ 3^3X,\ 3^4X,
$$
and already lie in $\mathbb F_{11}(X)$. The extension is therefore the splitting field of $t^5-Y$ and is normal. Its derivative is $5t^4\ne0$ in characteristic eleven, so the polynomial is separable. By part (c),
$$
\boxed{\mathbb F_{11}(X)/\mathbb F_{11}(X^5)\text{ is Galois}.}
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=19j}
{scope}

= Solution
{parent=a}

Start with any positive-definite Hermitian inner product $\langle-,-\rangle$ on $\mathbb C^n$ and average it over the finite group:
$$
\langle v,w\rangle_G=
\frac1{|G|}\sum_{g\in G}
\langle\rho(g)v,\rho(g)w\rangle.
$$
This remains positive definite. For $h\in G$, reindexing $g\mapsto gh$ gives
$$
\langle\rho(h)v,\rho(h)w\rangle_G
=\langle v,w\rangle_G,
$$
so the form is $G$-invariant.

Choose an orthonormal basis for $\langle-,-\rangle_G$. In this basis every $\rho(g)$ preserves the standard Hermitian form and is therefore unitary. If $P$ is the change-of-basis matrix, then
$$
\rho'(g)=P^{-1}\rho(g)P
$$
defines an isomorphic representation with $\rho'(G)\leq U_n$.

Solved by gpt-5.6-sol high.

= b
{parent=19j}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

An invariant bilinear form $B$ defines a linear map
$$
\Phi_B:V\longrightarrow V^*,
\qquad
\Phi_B(v)(w)=B(v,w).
$$
Invariance gives
$$
\Phi_B(gv)(w)=B(gv,w)=B(v,g^{-1}w)
=(g\Phi_B(v))(w),
$$
so $\Phi_B$ is an <intertwining operator>[$G$-homomorphism]. Its <kernel> is a $G$-subrepresentation of <irreducible representation>[irreducible] $V$. Hence either $\Phi_B=0$, so $B=0$, or $\Phi_B$ is <injective function>[injective] and therefore a <linear isomorphism>[linear isomorphism] $V\to V^*$. In the latter case $B$ is a <nondegenerate bilinear form>[nondegenerate].

If $B$ and $C$ are two nonzero invariant forms, then
$$
\Phi_C^{-1}\Phi_B\in\operatorname{End}_G(V).
$$
By <Schur lemma>, this endomorphism is scalar, so $B=\lambda C$. Cases involving the zero form are immediate. Thus any two invariant bilinear forms are proportional.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

Given a nonzero invariant form $B$, its transpose
$$B^T(v,w)=B(w,v)$$
is again nonzero and $G$-invariant. By part (i), $B^T=\lambda B$ for some nonzero scalar $\lambda$. Transposing once more gives
$$
B=(B^T)^T=\lambda B^T=\lambda^2B.
$$
Since $B\ne0$, $\lambda^2=1$, and hence $\lambda\in\{1,-1\}$. Therefore
$$
\boxed{B(w,v)=\lambda B(v,w),\qquad \lambda=\pm1.}
$$
Thus every nonzero invariant bilinear form on an irreducible complex representation is either symmetric or alternating.

Solved by gpt-5.6-sol high.

= c
{parent=19j}
{scope}

= Solution
{parent=c}

The induced representation is
$$
\operatorname{Ind}_H^G(V)=\mathbb C[G]\otimes_{\mathbb C[H]}V,
$$
with $G$ acting by left multiplication on the first tensor factor. If $R$ is a set of representatives for the left cosets $G/H$, then as a vector space it is $\bigoplus_{r\in R}r\otimes V$.

The operator $g$ permutes these summands. A summand indexed by $xH$ contributes to the trace exactly when $gxH=xH$, equivalently when $x^{-1}gx\in H$, and its contribution is $\chi_V(x^{-1}gx)$. Accounting for the $|H|$ representatives of each coset gives
$$
\boxed{
\chi_{\operatorname{Ind}_H^G V}(g)
=\frac1{|H|}
\sum_{\substack{x\in G\\x^{-1}gx\in H}}
\chi_V(x^{-1}gx).}
$$

For a $G$-representation $W$, restrict $W$ to $H$ in the left-hand induced representation. Define
$$
\Phi:\mathbb C[G]\otimes_{\mathbb C[H]}(W\otimes V)
\longrightarrow
W\otimes\bigl(\mathbb C[G]\otimes_{\mathbb C[H]}V\bigr)
$$
by
$$
\Phi\bigl(x\otimes(w\otimes v)\bigr)
=xw\otimes(x\otimes v).
$$
It is well-defined because both representatives of the balancing relation give
$$
xhw\otimes(x\otimes hv).
$$
It is $G$-equivariant, and its inverse is
$$
w\otimes(x\otimes v)\longmapsto
x\otimes(x^{-1}w\otimes v).
$$
Consequently
$$
\boxed{
\operatorname{Ind}_H^G(\operatorname{Res}_H^G W\otimes V)
\cong W\otimes\operatorname{Ind}_H^G(V).}
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=20g}
{scope}

= Solution
{parent=a}

Let $K$ be a number field. The class group is
$$
\operatorname{Cl}(K)=
\{\text{nonzero fractional <ideals> of }\mathcal O_K\}/
\{\text{principal fractional <ideals>}\},
$$
with multiplication induced by ideal multiplication. Equivalently, two fractional ideals $I,J$ represent the same class when $I=(\alpha)J$ for some $\alpha\in K^\times$.

Solved by gpt-5.6-sol high.

= b
{parent=20g}
{scope}

= Solution
{parent=b}

Let $I$ be a nonzero integral ideal. By the stated estimate, choose $0\ne\alpha\in I$ with
$$|N_{K/\mathbb Q}(\alpha)|\leq C N(I).$$
Since $(\alpha)\subseteq I$, the fractional ideal
$$J=(\alpha)I^{-1}$$
is integral. Its class is $[I]^{-1}$ and
$$
N(J)=\frac{|N_{K/\mathbb Q}(\alpha)|}{N(I)}\leq C.
$$
Every ideal class therefore has, after taking inverses, an integral representative of norm at most $C$.

There are only finitely many integral ideals of bounded norm. Indeed, if $N(J)=m$, then the finite group $\mathcal O_K/J$ has order $m$, so multiplication by $m$ kills it and
$$m\mathcal O_K\subseteq J.$$
For each of the finitely many integers $m\leq C$, such ideals correspond to ideals of the finite ring $\mathcal O_K/m\mathcal O_K$, of which there are only finitely many. Hence $\operatorname{Cl}(K)$ is finite.

Solved by gpt-5.6-sol high.

= c
{parent=20g}
{scope}

= Solution
{parent=c}

If $I=0$, take $n=1$. Otherwise, by part (b), the element $[I]$ of the finite group $\operatorname{Cl}(K)$ has finite order, say $n\geq1$. Therefore
$$
[I^n]=[I]^n=1,
$$
which says precisely that $I^n$ is principal. Since $I^n$ is integral, if $I^n=(\alpha)$ then necessarily $\alpha\in\mathcal O_K$.

Solved by gpt-5.6-sol high.

= d
{parent=20g}
{scope}

= Solution
{parent=d}

First let $I=\mathfrak p^e$ for a prime ideal $\mathfrak p$ and $e\geq1$. If $\alpha\beta\in\mathfrak p^e$ but $\alpha\notin\mathfrak p^e$, then in the prime-ideal factorization valuations
$$
v_{\mathfrak p}(\alpha)+v_{\mathfrak p}(\beta)\geq e,
\qquad
v_{\mathfrak p}(\alpha)<e.
$$
Thus $v_{\mathfrak p}(\beta)>0$, so $\beta\in\mathfrak p$ and $\beta^e\in\mathfrak p^e=I$. Hence every prime-ideal power is primary.

Conversely, the radical of any <primary ideal> is prime. To see this, suppose $ab\in\sqrt I$ and $a\notin\sqrt I$. For some $m$, $a^mb^m\in I$, while $a^m\notin I$. Primaryness gives $(b^m)^k\in I$ for some $k$, and hence $b\in\sqrt I$.

For nonzero $I$, unique factorization of ideals gives
$$I=\mathfrak p_1^{e_1}\cdots\mathfrak p_r^{e_r},$$
so
$$\sqrt I=\mathfrak p_1\cdots\mathfrak p_r.$$
This radical is prime only when $r=1$. Therefore a nonzero primary ideal is $\mathfrak p^e$. The zero ideal is itself prime because $\mathcal O_K$ is a domain, so it is its own first power. Consequently
$$
\boxed{I\text{ is primary}\quad\Longleftrightarrow\quad
I\text{ is a power of a prime <ideal>}.}
$$

Solved by gpt-5.6-sol high.

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

= Solution
{parent=21f}

A based loop is a continuous map $\alpha:[0,1]\to X$ with $\alpha(0)=\alpha(1)=x_0$. Two loops are equivalent when they are joined by a homotopy that keeps both endpoints at $x_0$. The <fundamental group> $\pi_1(X,x_0)$ is the set of these based-homotopy classes.

For loops $\alpha$ and $\beta$, define concatenation by
$$
(\alpha*\beta)(t)=
\begin{cases}
\alpha(2t),&0\leq t\leq\tfrac12,\\
\beta(2t-1),&\tfrac12\leq t\leq1.
\end{cases}
$$
If $H$ and $K$ are based homotopies from $\alpha$ to $\alpha'$ and from $\beta$ to $\beta'$, concatenating $H(s,-)$ and $K(s,-)$ gives a based homotopy from $\alpha*\beta$ to $\alpha'*\beta'$. Thus multiplication $[\alpha][\beta]=[\alpha*\beta]$ is well-defined.

Concatenating three paths with different breakpoints changes only the speed of traversal. Linear interpolation between the two increasing piecewise-linear parametrizations gives a based homotopy, proving associativity on classes. The constant loop $c(t)=x_0$ is an identity, since deleting its stationary half is another endpoint-fixing reparametrization. The inverse of $[\alpha]$ is represented by $\bar\alpha(t)=\alpha(1-t)$. Indeed, $\alpha*\bar\alpha$ contracts through
$$
H(s,t)=
\begin{cases}
\alpha(2t(1-s)),&0\leq t\leq\tfrac12,\\
\alpha(2(1-t)(1-s)),&\tfrac12\leq t\leq1,
\end{cases}
$$
and the analogous contraction handles $\bar\alpha*\alpha$. Hence the operation satisfies all group axioms.

Solved by gpt-5.6-sol high.

= a
{parent=21f}
{scope}

= Solution
{parent=a}

Use the continuous determinant map
$$
\det:U(2)\longrightarrow S^1.
$$
It induces a homomorphism on fundamental groups. Under the standard identification
$$
\pi_1(S^1,1)\cong\mathbb Z,
$$
the class of $t\mapsto e^{2\pi it}$ corresponds to its <winding number> $1$. Since
$$
(\det\circ\gamma)(t)=e^{2\pi it},
$$
we obtain
$$
\det_*([\gamma]^k)=k\in\mathbb Z.
$$
For $k\ne0$ this is not the identity, so $[\gamma]^k$ could not have been the identity in $\pi_1(U(2),I)$.

Solved by gpt-5.6-sol high.

= b
{parent=21f}
{scope}

= Solution
{parent=b}

Besides concatenation $*$, loops in $U(2)$ have pointwise multiplication
$$
(\alpha\mathbin{\cdot}\beta)(t)=\alpha(t)\beta(t).
$$
Continuity of matrix multiplication makes this a loop, and pointwise multiplication of homotopies shows that it is well-defined on homotopy classes. Both operations have the constant loop $e(t)=I$ as identity, and they satisfy the interchange law
$$
(\alpha*\beta)\mathbin{\cdot}(\gamma*\delta)
=(\alpha\mathbin{\cdot}\gamma)*(\beta\mathbin{\cdot}\delta).
$$

For homotopy classes $a,b$, the interchange law gives
$$
a\mathbin{\cdot}b
=(a*e)\mathbin{\cdot}(e*b)
=(a\mathbin{\cdot}e)*(e\mathbin{\cdot}b)
=a*b.
$$
It then gives
$$
a*b=a\mathbin{\cdot}b
=(e*a)\mathbin{\cdot}(b*e)
=(e\mathbin{\cdot}b)*(a\mathbin{\cdot}e)
=b*a.
$$
Thus loop concatenation is commutative on homotopy classes, and
$$
\boxed{\pi_1(U(2),I)\text{ is abelian}.}
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=22i}
{scope}

= Solution
{parent=a}

The <bounded inverse theorem> states that a bounded bijective linear map between Banach spaces has a bounded inverse.

The <closed graph theorem> states that if $X$ and $Y$ are Banach spaces and $T:X\to Y$ is linear, then $T$ is continuous if and only if its graph
$$
\Gamma(T)=\{(x,Tx):x\in X\}
$$
is closed in the Banach space $X\times Y$.

If $T$ is continuous and $(x_n,Tx_n)\to(x,y)$, then $Tx_n\to Tx$, so uniqueness of limits gives $y=Tx$. Hence the graph is closed.

Conversely, suppose $\Gamma(T)$ is closed. It is then a Banach space. The coordinate projection
$$
P_X:\Gamma(T)\longrightarrow X,
\qquad (x,Tx)\longmapsto x,
$$
is bounded and bijective. By the bounded inverse theorem, $P_X^{-1}:x\mapsto(x,Tx)$ is bounded. Composing it with the bounded second-coordinate projection gives
$$
T=P_Y\circ P_X^{-1},
$$
so $T$ is continuous.

Solved by gpt-5.6-sol high.

= b
{parent=22i}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

Take $X=Y=\ell^2$ and define
$$
S(x_1,x_2,\ldots)=
\left(x_1,\frac{x_2}{2},\frac{x_3}{3},\ldots\right).
$$
This is bounded and injective. Its inverse on $\operatorname{im}S$ is not continuous: for the standard unit vectors $e_n$,
$$
\|Se_n\|_2=\frac1n\longrightarrow0,
\qquad
\|S^{-1}(Se_n)\|_2=\|e_n\|_2=1.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

Set $R=S^{-1}\circ T:X\to X$, which is well-defined and linear because $\operatorname{im}T\subseteq\operatorname{im}S$. We show that its graph is closed.

Suppose $x_n\to x$ and $Rx_n\to z$ in $X$. Continuity of $S$ and $T$ gives
$$
Sz=\lim_n S(Rx_n)=\lim_n Tx_n=Tx=S(Rx).
$$
Since $S$ is injective, $z=Rx$. Thus $\Gamma(R)$ is closed. The domain and codomain $X$ are Banach spaces, so the closed graph theorem implies
$$
\boxed{S^{-1}\circ T\in L(X,X).}
$$

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

Let $X=c_{00}$, the finitely supported sequences, with the $\ell^2$ norm. This normed space is not complete. Let $Y=\ell^2$ and define
$$
S(x_1,x_2,\ldots)=
\left(x_1,\frac{x_2}{2},\frac{x_3}{3},\ldots\right),
\qquad
T(x)=x.
$$
Both maps $X\to Y$ are bounded, and $S$ is injective. Moreover
$$
\operatorname{im}S=c_{00}=\operatorname{im}T,
$$
because multiplication of a finitely supported sequence by $(1,2,3,\ldots)$ preserves finite support. But
$$
(S^{-1}T)e_n=ne_n,
$$
so $\|(S^{-1}T)e_n\|_2=n$ while $\|e_n\|_2=1$. Hence $S^{-1}T$ is not continuous, showing that completeness of $X$ is essential.

Solved by gpt-5.6-sol high.

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

= a
{parent=23h}
{scope}

= Solution
{parent=a}

The <Rellich-Kondrashov compactness theorem for H01> states that if $\Omega\subset\mathbb R^d$ is bounded and open, then the inclusion
$$
H_0^1(\Omega)\hookrightarrow L^2(\Omega)
$$
is compact: every bounded sequence in $H_0^1(\Omega)$ has a subsequence converging strongly in $L^2(\Omega)$.

Let $(u_n)$ be bounded in $H_0^1(\Omega)$. Since this is a Hilbert space, Banach--Alaoglu and reflexivity give a subsequence, still denoted $(u_n)$, and $u\in H_0^1(\Omega)$ such that $u_n\rightharpoonup u$ weakly in $H_0^1(\Omega)$ and hence in $L^2(\Omega)$. Extend all these functions by zero to $\mathbb R^d$. The zero extensions belong to $H^1(\mathbb R^d)$ and remain uniformly bounded there.

For each $\xi\in\mathbb R^d$,
$$
\widehat u_n(\xi)=\int_\Omega u_n(x)e^{-ix\cdot\xi}\,dx
\longrightarrow
\int_\Omega u(x)e^{-ix\cdot\xi}\,dx=\widehat u(\xi),
$$
because $e^{ix\cdot\xi}\mathbf1_\Omega\in L^2(\Omega)$. On each ball $|\xi|\leq R$, Cauchy--Schwarz gives a uniform bound on $|\widehat u_n-\widehat u|$, so dominated convergence yields
$$
\int_{|\xi|\leq R}|\widehat u_n-\widehat u|^2\,d\xi\longrightarrow0.
$$

The gradient bound controls high frequencies. By Plancherel,
$$
\int_{|\xi|>R}|\widehat u_n-\widehat u|^2\,d\xi
\leq \frac1{R^2}
\int_{\mathbb R^d}|\xi|^2|\widehat u_n-\widehat u|^2\,d\xi
\leq\frac C{R^2},
$$
uniformly in $n$. First choose $R$ large and then $n$ large. The low- and high-frequency estimates show $\widehat u_n\to\widehat u$ in $L^2(\mathbb R^d)$, and a final application of Plancherel gives $u_n\to u$ strongly in $L^2(\Omega)$.

Solved by gpt-5.6-sol high.

= b
{parent=23h}
{scope}

= Solution
{parent=b}

No. Choose a nonzero $\varphi\in C_c^\infty(\mathbb R)$ and a number $L$ larger than the diameter of its support, and set
$$
u_n(x)=\varphi(x-nL).
$$
Translation preserves both the $L^2$ norm and the derivative norm, so $(u_n)$ is bounded in $H^1(\mathbb R)$. Distinct $u_n$ have disjoint supports, and hence
$$
\|u_n-u_m\|_2^2
=\|u_n\|_2^2+\|u_m\|_2^2
=2\|\varphi\|_2^2
\qquad(n\ne m).
$$
Thus the sequence has no $L^2$-Cauchy subsequence and therefore no strongly convergent subsequence. The embedding $H^1(\mathbb R)\hookrightarrow L^2(\mathbb R)$ is not compact.

Solved by gpt-5.6-sol high.

= c
{parent=23h}
{scope}

= Solution
{parent=c}

Fix $R>0$ and choose $\chi_R\in C_c^\infty((-R-1,R+1))$ with $\chi_R=1$ on $[-R,R]$. The sequence $(\chi_Rf_n)$ is bounded in $H_0^1((-R-1,R+1))$. By part (a), each subsequence has a further subsequence converging strongly in $L^2$ on this bounded interval. The original weak convergence in $L^2(\mathbb R)$ forces every such strong limit to be zero. It follows that the whole sequence satisfies
$$
\|f_n\|_{L^2([-R,R])}\longrightarrow0;
$$
otherwise a subsequence bounded away from zero would contradict the preceding compactness argument.

The pointwise hypothesis gives a uniform tail estimate:
$$
\int_{|x|>R}|f_n(x)|^2\,dx
\leq C^2\int_{|x|>R}(1+x^2)^{-2}\,dx,
$$
whose right-hand side tends to zero as $R\to\infty$, independently of $n$. Given $\varepsilon>0$, first choose $R$ so that this tail is below $\varepsilon/2$, and then choose $n$ so that the integral on $[-R,R]$ is below $\varepsilon/2$. Therefore
$$
\boxed{\|f_n\|_{L^2(\mathbb R)}\longrightarrow0.}
$$

Solved by gpt-5.6-sol high.

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

= Solution
{parent=24g}

The identity theorem says that if two holomorphic functions on a connected Riemann surface agree on a set having an accumulation point, then they agree everywhere.

Apply local coordinates to their difference $F$. The power-series proof in the plane shows that near any point, either $F$ vanishes identically or its zeros are isolated: if the first nonzero Taylor coefficient has order $m$, then $F(z)=(z-z_0)^mG(z)$ with $G(z_0)\ne0$. An accumulation point of the zero set therefore has a neighbourhood on which $F=0$. The set of points having such a neighbourhood is nonempty and open. It is also closed: near a limit point, a coordinate neighbourhood contains an open set on which $F=0$, and the planar identity theorem makes $F$ vanish throughout that coordinate neighbourhood. Connectedness now makes this set the whole surface.

For a nonempty connected open set $U\subset\mathbb R^2$, a function $h:U\to\mathbb R$ is harmonic when $h\in C^2(U)$ and
$$
\Delta h=h_{xx}+h_{yy}=0.
$$
On a sufficiently small disc, the one-form
$$-h_y\,dx+h_x\,dy$$
is closed and hence equals $dk$ for some $k$. The Cauchy--Riemann equations then make $h+ik$ holomorphic. Holomorphic functions are smooth, so $h\in C^\infty(U)$.

A real-valued function $H$ on a Riemann surface is harmonic when, in every holomorphic chart $z$, its coordinate expression is harmonic. This definition is chart-independent. If $w=w(z)$ is a holomorphic transition map, the chain rule and Cauchy--Riemann equations give
$$
\Delta_z(H\circ w)=|w'(z)|^2(\Delta_w H)\circ w.
$$
Thus vanishing of the Laplacian is preserved by every change of holomorphic coordinate.

Solved by gpt-5.6-sol high.

= i
{parent=24g}
{scope}

= Solution
{parent=i}

No. For example, let
$$
f(z)=\max(1-|z|,0),
\qquad
g(z)=\max(1-|z-3|,0).
$$
Both functions are continuous and nonzero, but their supports lie in disjoint closed unit discs. Hence $f(z)g(z)=0$ for every $z\in\mathbb C$.

Solved by gpt-5.6-sol high.

= ii
{parent=24g}
{scope}

= Solution
{parent=ii}

Yes. The sets
$$
U=\{z:f(z)\ne0\},
\qquad
V=\{z:g(z)\ne0\}
$$
are open. The equation $fg=0$ makes them disjoint, while the assumption that $f$ and $g$ never vanish simultaneously gives $U\cup V=\mathbb C$. Since $\mathbb C$ is connected, one of $U,V$ must be empty. Therefore one of $f,g$ is identically zero.

Solved by gpt-5.6-sol high.

= iii
{parent=24g}
{scope}

= Solution
{parent=iii}

No. Define the standard smooth bump
$$
\eta(t)=
\begin{cases}
e^{-1/(1-t)},&t<1,\\
0,&t\geq1,
\end{cases}
$$
and set, for $x\in\mathbb R^2$,
$$
f(x)=\eta(|x|^2),
\qquad
g(x)=\eta(|x-(3,0)|^2).
$$
These are nonzero smooth functions supported in disjoint unit discs, so $fg=0$ identically.

Solved by gpt-5.6-sol high.

= iv
{parent=24g}
{scope}

= Solution
{parent=iv}

Yes. Suppose $f$ is not identically zero. Then
$$U=\{x\in\mathbb R^2:f(x)\ne0\}$$
is a nonempty open set, and $fg=0$ forces $g=0$ on $U$. Harmonic functions satisfy unique continuation: locally they are real parts of holomorphic functions and hence are real analytic, so a harmonic function vanishing on a nonempty open subset of a connected domain vanishes everywhere. Thus $g\equiv0$. Interchanging $f$ and $g$ proves that one factor must be identically zero.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=25J}

For $X\subseteq\mathbb A^n$ and $P\in X$, the <Zariski tangent space> is
$$
T_{X,P}=\{v\in k^n:
\sum_{i=1}^n\frac{\partial h}{\partial x_i}(P)v_i=0
\text{ for every }h\in I(X)\}.
$$
Equivalently, it is the kernel of the Jacobian of generators of $I(X)$ at $P$. With the tangent-space definition used here,
$$
\dim X=\min_{P\in X}\dim T_{X,P},
$$
and $P$ is smooth when $\dim T_{X,P}=\dim X$.

Now let $X=Z(f)$. Write $g$ for the square-free product of the distinct irreducible factors of $f$. Hilbert's Nullstellensatz gives
$$
I(X)=\sqrt{(f)}=(g).
$$
Thus $T_{X,P}=\ker(dg_P)$, whose dimension is at least $n-1$. There is some $P\in X$ with $dg_P\ne0$. Otherwise every partial derivative of $g$ would vanish on $X$, hence belong to $(g)$ by the Nullstellensatz. Its degree is smaller than $\deg g$, so every partial derivative would be zero; in characteristic zero this would make $g$ constant, a contradiction. At such a point the tangent space has dimension $n-1$, and therefore
$$
\boxed{\dim Z(f)=n-1.}
$$

Next suppose $f$ is irreducible of degree at least two and $P$ is smooth. Translate $P$ to the origin and write
$$
\ell=df_P.
$$
Then $H=T_{X,P}=Z(\ell)$ is a hyperplane. The restriction $f|_H$ is nonzero: otherwise $\ell$ would divide $f$, contradicting irreducibility and $\deg f\geq2$. Since $f(P)=0$, the restriction is nonconstant, so the first part gives
$$
\dim(X\cap H)=n-2.
$$
For the natural intersection cut out by $(f,\ell)$, the two differentials at $P$ are $df_P=\ell$ and $d\ell=\ell$. Hence they impose only one independent tangent equation and
$$
T_{X\cap H,P}=H,
\qquad
\dim T_{X\cap H,P}=n-1>n-2.
$$
Therefore $X\cap T_{X,P}$ is singular at $P$.

Finally, represent $\varphi:\mathbb A^2\to\mathbb A^3$ by
$$
A=\begin{pmatrix}
x_1&x_2\\x_3&x_4\\x_5&x_6
\end{pmatrix}.
$$
It is not injective exactly when $\operatorname{rank}A\leq1$, equivalently when all its $2\times2$ minors vanish. Thus
$$
Y=Z(I),
\qquad
I=(x_1x_4-x_2x_3,\ x_1x_6-x_2x_5,\ x_3x_6-x_4x_5),
$$
an ideal generated by three quadrics. By the stated assumption this is the full radical ideal of $Y$.

Every nonzero point of $Y$ has rank one. The action of $GL_3(k)\times GL_2(k)$ by row and column operations is transitive on rank-one matrices, so it suffices to compute at
$$
A_0=\begin{pmatrix}1&0\\0&0\\0&0\end{pmatrix}.
$$
At $A_0$, the differentials of the three minors are respectively
$$dx_4,\qquad dx_6,\qquad0.$$
Hence $\dim T_{Y,A_0}=6-2=4$, and the same holds at every rank-one point. At the zero matrix all three quadrics have zero differential, so $T_{Y,0}=\mathbb A^6$ has dimension six. It follows that
$$
\boxed{\dim Y=4,\qquad \operatorname{Sing}(Y)=\{0\}.}
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=26i}
{scope}

= Solution
{parent=a}

A smooth curve $\alpha:I\to\mathbb R^3$ is regular when $\alpha'(t)\ne0$ for every $t\in I$. Its arc-length between parameters $t_0$ and $t_1$ is
$$
L(\alpha)=\int_{t_0}^{t_1}|\alpha'(t)|\,dt.
$$

Fix $t_0$ and define
$$
s(t)=\int_{t_0}^t|\alpha'(r)|\,dr.
$$
Regularity gives $s'(t)=|\alpha'(t)|>0$, so $s$ is strictly increasing and has a smooth inverse $t=t(s)$ on its image. The reparametrized curve $\beta(s)=\alpha(t(s))$ satisfies
$$
|\beta'(s)|=|\alpha'(t(s))|\frac{dt}{ds}=1.
$$
Thus every <regular curve> has an <arc-length parametrization>.

For a unit-speed curve with nonzero curvature, let
$$
T=\beta',\qquad
\kappa=|T'|,\qquad
N=\frac{T'}{\kappa},\qquad
B=T\times N.
$$
Its torsion is
$$
\tau=-B'\cdot N,
$$
equivalently
$$
\tau=\frac{\det(\beta',\beta'',\beta''')}{|\beta'\times\beta''|^2}.
$$
If the curve lies in an affine plane, all three derivative vectors lie in the parallel two-dimensional vector plane, so their determinant is zero. Hence $\tau=0$ wherever torsion is defined.

Solved by gpt-5.6-sol high.

= b
{parent=26i}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

Put
$$h(x)=u_+(x)-u_-(x).$$
Vertical slicing and Fubini's theorem give
$$
\mathcal A(\Omega)=\int_{x_0}^{x_1}h(x)\,dx.
$$
The vertical slice of $S_L(\Omega)$ is $(-h(x)/2,h(x)/2)$ and has the same length $h(x)$. Therefore
$$
\boxed{\mathcal A(S_L(\Omega))
=\int_{x_0}^{x_1}h(x)\,dx
=\mathcal A(\Omega).}
$$

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

The original perimeter is
$$
P(\Omega)=\int_{x_0}^{x_1}
\left(\sqrt{1+u_+'(x)^2}+\sqrt{1+u_-'(x)^2}\right)dx.
$$
The two boundary graphs of the symmetrized domain are $\pm h/2$, so
$$
P(S_L(\Omega))
=\int_{x_0}^{x_1}\sqrt{4+(u_+'-u_-')^2}\,dx.
$$
Apply the triangle inequality in $\mathbb R^2$ to the vectors
$$
(1,u_+'),\qquad(1,-u_-').
$$
It gives pointwise
$$
\sqrt{1+u_+'{}^2}+\sqrt{1+u_-'{}^2}
\geq\sqrt{4+(u_+'-u_-')^2}.
$$
After integration,
$$
\boxed{P(S_L(\Omega))\leq P(\Omega).}
$$

Equality in the Euclidean triangle inequality holds exactly when the two displayed vectors are nonnegative scalar multiples. Their first coordinates are both one, so this means
$$u_+'=-u_-'.$$
Equality of perimeters therefore holds exactly when $u_+(x)+u_-(x)=c$ is constant. In that case the midpoint of every vertical chord lies on $y=c/2$, and $\Omega$ has the horizontal line $y=c/2$, parallel to $L$, as an axis of symmetry. The converse is immediate.

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

Fix any direction and choose a line $L$ in that direction lying outside the bounded domain. Steiner symmetrization in that direction preserves convexity and, by part (i), preserves area. Since $\Omega$ minimizes perimeter among admissible convex domains of that area,
$$
P(\Omega)\leq P(S_L(\Omega)).
$$
Part (ii) gives the reverse inequality, so equality holds. Its equality characterization says that $\Omega$ has an axis of symmetry parallel to $L$. Since the direction was arbitrary, $\Omega$ admits axes of symmetry in every direction.

Solved by gpt-5.6-sol high.

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

= a
{parent=27h}
{scope}

= Solution
{parent=a}

The <monotone convergence theorem> states that if $f_n$ are nonnegative measurable functions with $f_n(x)\uparrow f(x)$ almost everywhere, then
$$
\int f_n\,d\mu\uparrow\int f\,d\mu,
$$
where either side may be infinite.

The integrals increase and are bounded above by $\int f$, so let their limit be $L\leq\int f$. Let $s$ be a nonnegative simple function with $s\leq f$, and fix $0<c<1$. The sets
$$
E_n=\{x:f_n(x)\geq c s(x)\}
$$
increase and cover $\{s>0\}$ up to a null set. Hence continuity of measure from below gives
$$
\int f_n\,d\mu
\geq c\int_{E_n}s\,d\mu
\longrightarrow c\int s\,d\mu.
$$
Thus $L\geq c\int s$. Taking the supremum over all simple $s\leq f$ and then letting $c\uparrow1$ gives $L\geq\int f$. Therefore equality holds.

Solved by gpt-5.6-sol high.

= b
{parent=27h}
{scope}

= Solution
{parent=b}

Since $\theta_n$ is even,
$$
\widehat\theta_n(\xi)
=2\int_0^n\left(1-\frac xn\right)\cos(x\xi)\,dx.
$$
For $\xi\ne0$, integration by parts gives
$$
\widehat\theta_n(\xi)
=\frac{2(1-\cos(n\xi))}{n\xi^2}
=n\left(\frac{\sin(n\xi/2)}{n\xi/2}\right)^2.
$$
At $\xi=0$ the continuous limiting value is the area of the triangle,
$$
\widehat\theta_n(0)=\int_{-n}^n\left(1-\frac{|x|}{n}\right)dx=n.
$$
In particular, $\widehat\theta_n\geq0$.

Solved by gpt-5.6-sol high.

= c
{parent=27h}
{scope}

= Solution
{parent=c}

Regard the same triangular functions as cutoffs in the frequency variable. They satisfy
$$0\leq\theta_n(\xi)\uparrow1.$$
Since $\widehat f\geq0$, monotone convergence gives
$$
\|\widehat f\|_1
=\lim_{n\to\infty}\int_{\mathbb R}\theta_n(\xi)\widehat f(\xi)\,d\xi.
$$
Fubini's theorem and evenness of $\theta_n$ give
$$
\int\theta_n(\xi)\widehat f(\xi)\,d\xi
=\int f(x)\widehat\theta_n(x)\,dx.
$$
Part (b) shows that $\widehat\theta_n\geq0$. Moreover, Fourier inversion at zero yields
$$
\int_{\mathbb R}\widehat\theta_n(x)\,dx
=2\pi\theta_n(0)=2\pi.
$$
Consequently
$$
0\leq\int\theta_n\widehat f
=\left|\int f\widehat\theta_n\right|
\leq\|f\|_\infty\int\widehat\theta_n
=2\pi\|f\|_\infty.
$$
Passing to the limit proves
$$
\boxed{\|\widehat f\|_{L^1}\leq2\pi\|f\|_{L^\infty}.}
$$
Thus one may take the universal constant $\alpha=2\pi$.

Solved by gpt-5.6-sol high.

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

= a
{parent=28l}
{scope}

= Solution
{parent=a}

The jump chain moves away from zero with probability $2/3$ and towards zero with probability $1/3$ on either half-line. Starting from $1$, the probability that this biased walk ever hits zero is $(1/3)/(2/3)=1/2$, and the same holds starting from $-1$. Thus the return probability to zero is $1/2<1$. The jump chain, and hence $X$, is transient.

It is also explosive. The transient jump chain visits zero only finitely often. After its last visit it stays on one half-line, and the strong law for its increments gives
$$
\frac{|Y_n|}{n}\longrightarrow\frac13
$$
almost surely. In particular, eventually $|Y_n|\geq n/6$. Conditional on the jump-chain path, the mean total remaining holding time is
$$
\sum_n\frac1{q_{Y_n}}
=\sum_n3^{-|Y_n|}<\infty.
$$
The sum of the actual nonnegative holding times is therefore finite almost surely, so infinitely many jumps occur in finite time.

Despite this, the chain has an invariant distribution in the continuous-time balance-equation sense $\pi Q=0$. Detailed balance across the edge from zero to one gives $\pi_1=\pi_0/2$, and on each half-line it gives
$$
\frac{\pi_{i+1}}{\pi_i}
=\frac{q_{i,i+1}}{q_{i+1,i}}=\frac23.
$$
By symmetry and normalization,
$$
\boxed{
\pi_0=\frac14,
\qquad
\pi_i=\pi_{-i}=\frac18\left(\frac23\right)^{i-1}quad(i\geq1).}
$$
This example also shows why, for an explosive chain, existence of an invariant distribution need not imply positive recurrence.

Solved by gpt-5.6-sol high.

= b
{parent=28l}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

The statement is false. Let the jump-chain state space be $\mathbb Z_+$, with
$$
p_{0i}=2^{-i}\quad(i\geq1),
\qquad
p_{i0}=1\quad(i\geq1).
$$
It is irreducible and has invariant distribution
$$
\mu_0=\frac12,
\qquad
\mu_i=2^{-(i+1)}\quad(i\geq1),
$$
so it is positive recurrent.

Give the continuous-time chain holding rates
$$q_0=1,\qquad q_i=2^{-i}quad(i\geq1),$$
and set $q_{ij}=q_ip_{ij}$. The rates are bounded, so the chain is nonexplosive. A return cycle from zero jumps to $i$ with probability $2^{-i}$ and then has mean holding time $2^i$ there. Consequently its mean return time is at least
$$
\sum_{i\geq1}2^{-i}2^i=\infty.
$$
Thus the continuous-time chain is not positive recurrent.

Equivalently, the jump-chain invariant distribution would induce the continuous-time invariant measure $\mu_i/q_i$, but this has infinite total mass.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

This statement is also false. Let the jump chain be simple symmetric random walk on $\mathbb Z$, which is null recurrent, and choose holding rates
$$q_i=(1+|i|)^2.$$
Thus $q_{i,i+1}=q_{i,i-1}=q_i/2$. The counting measure $\mu_i=1$ is invariant for the jump chain, so
$$
\pi_i=\frac{C}{(1+|i|)^2}
$$
is an invariant distribution for the continuous-time chain, where $C$ is the normalizing constant.

The chain is nonexplosive: its jump chain returns to zero infinitely often, and the independent holding times on those visits have rate $q_0=1$, so their sum diverges almost surely. For an irreducible nonexplosive continuous-time chain, existence of an invariant distribution is equivalent to positive recurrence. Hence $X$ is positive recurrent although its jump chain is null recurrent.

Solved by gpt-5.6-sol high.

= c
{parent=28l}
{scope}

= i
{parent=c}
{scope}

= Solution
{parent=i}

While the queue is nonempty, service completions occur at rate $\mu$. A completion reduces the queue length only when the customer leaves, which has probability $1-p$. Thinning therefore makes $L$ a birth--death chain with birth rate
$$\lambda$$
and effective death rate
$$\delta=\mu(1-p).$$
The standard birth--death classification gives
$$
\boxed{
L\text{ is transient if }\lambda>\mu(1-p),
\quad
L\text{ is recurrent if }\lambda\leq\mu(1-p).}
$$
At equality it is null recurrent.

Solved by gpt-5.6-sol high.

= ii
{parent=c}
{scope}

= Solution
{parent=ii}

The recurrent chain is positive recurrent precisely when the drift is strictly towards zero:
$$
\boxed{\lambda<\mu(1-p).}
$$
The equality case is null recurrent.

Solved by gpt-5.6-sol high.

= iii
{parent=c}
{scope}

= Solution
{parent=iii}

Assume $\rho=\lambda/[\mu(1-p)]<1$. Detailed balance gives
$$
\pi_{n+1}\mu(1-p)=\pi_n\lambda,
$$
so normalization yields
$$
\boxed{\pi_n=(1-\rho)\rho^n,\qquad n\geq0.}
$$
This is a geometric distribution on $\mathbb Z_+$, and therefore
$$
\boxed{\mathbb E_\pi L=\frac{\rho}{1-\rho}
=\frac{\lambda}{\mu(1-p)-\lambda}.}
$$

Solved by gpt-5.6-sol high.

= iv
{parent=c}
{scope}

= Solution
{parent=iv}

In the stable regime, $L$ is exactly the queue-length process of an $M/M/1$ queue with arrival rate $\lambda$ and effective service rate $\mu(1-p)$. By <Burke theorem>, when this queue is in equilibrium its external departure process is a Poisson process of rate
$$
\boxed{\lambda.}
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=29l}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

The log-likelihood, score, and Fisher information are
$$
\ell_n(\theta)=\sum_{r=1}^n\log f(X_r,\theta),
\qquad
S_n(\theta)=\nabla_\theta\ell_n(\theta),
$$
and
$$
I_n(\theta)=\mathbb E_\theta
\bigl[S_n(\theta)S_n(\theta)^T\bigr]
=-\mathbb E_\theta\bigl[\nabla_\theta^2\ell_n(\theta)\bigr].
$$
For independent identically distributed observations,
$$I_n(\theta)=nI_1(\theta).$$

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

Writing the score as a column vector and differentiating under the integral,
$$
\mathbb E_\theta S_1(\theta)
=\int \nabla_\theta\log f(x,\theta)f(x,\theta)\,dx
=\int\nabla_\theta f(x,\theta)\,dx
=\nabla_\theta\int f(x,\theta)\,dx=0.
$$

Solved by gpt-5.6-sol high.

= iii
{parent=a}
{scope}

= Solution
{parent=iii}

For the simple null, the restricted maximum-likelihood estimator is $\widetilde\theta=\theta_0$, and hence
$$
T_n=(n^{-1/2}S_n(\theta_0))^T
I_1(\theta_0)^{-1}
(n^{-1/2}S_n(\theta_0)).
$$
The individual scores are independent, have mean zero by part (ii), and have covariance $I_1(\theta_0)$. The multivariate central limit theorem gives
$$
n^{-1/2}S_n(\theta_0)
\xrightarrow{d}N_p(0,I_1(\theta_0)).
$$
After multiplying by $I_1(\theta_0)^{-1/2}$, the limit is standard normal in $\mathbb R^p$. The continuous mapping theorem therefore yields
$$
\boxed{T_n\xrightarrow{d}\chi_p^2.}
$$

Solved by gpt-5.6-sol high.

= b
{parent=29l}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

Write $v=\sigma^2$. For one observation, the score is
$$
s_\mu(x;\mu,v)=\frac{x-\mu}{v},
\qquad
s_v(x;\mu,v)=-\frac1{2v}+\frac{(x-\mu)^2}{2v^2},
$$
and
$$
I_1(\mu,v)=
\begin{pmatrix}v^{-1}&0\\0&(2v^2)^{-1}\end{pmatrix}.
$$
Under the restriction $v=1$, the <restricted maximum-likelihood estimator> is $\widetilde\mu=\bar X$. At $(\bar X,1)$ the first score component is zero, while the second is
$$
\frac12\sum_{i=1}^n\bigl((X_i-\bar X)^2-1\bigr).
$$
Since $I_1(\bar X,1)^{-1}=\operatorname{diag}(1,2)$, substitution into the score statistic gives
$$
\boxed{
T_n=\left[\frac1{\sqrt{2n}}
\sum_{i=1}^n\bigl((X_i-\bar X)^2-1\bigr)\right]^2.}
$$

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

Under $H_0$, regardless of the unknown value of $\mu$,
$$
Q_n=\sum_{i=1}^n(X_i-\bar X)^2\sim\chi_{n-1}^2.
$$
Write $Q_n$ as a sum of $n-1$ independent squared standard normals. Since such a square has mean one and variance two, the central limit theorem gives
$$
\frac{Q_n-(n-1)}{\sqrt{2(n-1)}}\xrightarrow{d}N(0,1).
$$
Now
$$
\frac{Q_n-n}{\sqrt{2n}}
=\sqrt{\frac{n-1}{n}}
\frac{Q_n-(n-1)}{\sqrt{2(n-1)}}
-\frac1{\sqrt{2n}},
$$
so Slutsky's theorem makes this converge to $N(0,1)$. Squaring and using the continuous mapping theorem yields
$$
\boxed{T_n\xrightarrow{d}\chi_1^2.}
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=30k}
{scope}

= Solution
{parent=a}

The <optional sampling theorem for a supermartingale> at a bounded stopping time gives
$$
\mathbb E[V_\tau]\leq\mathbb E[V_0]=V_0,
$$
where $V_0$ is constant because $\mathcal F_0$ is trivial. Since $Z_\tau\leq V_\tau$ almost surely,
$$
\boxed{\mathbb E[Z_\tau]\leq V_0.}
$$

Solved by gpt-5.6-sol high.

= b
{parent=30k}
{scope}

= Solution
{parent=b}

For $0\leq k<N$, put
$$
\Delta A_{k+1}=A_{k+1}-A_k
=V_k-\mathbb E[V_{k+1}\mid\mathcal F_k].
$$
This is $\mathcal F_k$-measurable, and it is nonnegative because $V$ is a supermartingale. Hence $A_n$, being the sum of increments known by time $n-1$, is $\mathcal F_{n-1}$-measurable for $n\geq1$, so $A$ is previsible; and its nonnegative increments make it nondecreasing.

Solved by gpt-5.6-sol high.

= c
{parent=30k}
{scope}

= Solution
{parent=c}

The process $M$ is adapted and integrable. Since $A_{n+1}$ is $\mathcal F_n$-measurable,
$$
\begin{aligned}
\mathbb E[M_{n+1}\mid\mathcal F_n]
&=\mathbb E[V_{n+1}\mid\mathcal F_n]+A_{n+1}\\
&=\mathbb E[V_{n+1}\mid\mathcal F_n]+A_n
+V_n-\mathbb E[V_{n+1}\mid\mathcal F_n]\\
&=V_n+A_n=M_n.
\end{aligned}
$$
Thus $M$ is a martingale. This is the finite-horizon Doob decomposition $V=M-A$.

Solved by gpt-5.6-sol high.

= d
{parent=30k}
{scope}

= Solution
{parent=d}

For $n<N$,
$$
A_{n+1}-A_n
=V_n-\mathbb E[V_{n+1}\mid\mathcal F_n]\geq0.
$$
The recursion
$$
V_n=\max\{Z_n,\mathbb E[V_{n+1}\mid\mathcal F_n]\}
$$
shows that if $V_n>Z_n$, then $V_n=\mathbb E[V_{n+1}\mid\mathcal F_n]$ and the increment of $A$ is zero. Conversely, if that increment is positive, the maximum must be attained by $Z_n$, so $V_n=Z_n$. The two nonnegative quantities therefore satisfy
$$
\min\{V_n-Z_n,A_{n+1}-A_n\}=0.
$$
For $n=N$, this also holds because $V_N=Z_N$, regardless of the convention $A_{N+1}=\infty$.

Solved by gpt-5.6-sol high.

= e
{parent=30k}
{scope}

= Solution
{parent=e}

Because $A$ is nondecreasing and $A_{n+1}$ is $\mathcal F_n$-measurable,
$$
\{\tau^*\leq n\}=\{A_{n+1}>0\}\in\mathcal F_n.
$$
The convention $A_{N+1}=\infty$ ensures $\tau^*\leq N$, so $\tau^*$ is a bounded stopping time.

By minimality, $A_{\tau^*}=0$ and $A_{\tau^*+1}-A_{\tau^*}>0$. Part (d) then gives
$$
V_{\tau^*}=Z_{\tau^*}.
$$
Moreover $M_{\tau^*}=V_{\tau^*}+A_{\tau^*}=Z_{\tau^*}$. Optional sampling for the martingale $M$ now yields
$$
\boxed{
\mathbb E[Z_{\tau^*}]
=\mathbb E[M_{\tau^*}]
=M_0=V_0.}
$$
Together with part (a), this proves that $\tau^*$ is optimal and $V_0$ is the optimal stopping value.

Solved by gpt-5.6-sol high.

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

= a
{parent=31l}
{scope}

= Solution
{parent=a}

A set of points $x_1,\ldots,x_n$ is shattered by $H$ when every one of the $2^n$ binary label vectors is realized by some $h\in H$. The <shattering coefficient> is
$$
s(H,n)=\max_{x_1,\ldots,x_n}
\left|\{(h(x_1),\ldots,h(x_n)):h\in H\}\right|,
$$
and the <VC dimension> is
$$
VC(H)=\sup\{n:s(H,n)=2^n\}.
$$

The Sauer--Shelah lemma states that if $VC(H)=D<\infty$, then
$$
s(H,n)\leq\sum_{j=0}^{D}\binom nj
$$
for every $n$; in particular $s(H,n)\leq(n+1)^D$.

Solved by gpt-5.6-sol high.

= b
{parent=31l}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

Two ordered points $x_1<x_2$ are shattered by intervals: an interval can contain neither, either one alone, or both. Three ordered points cannot be shattered, because an interval containing the first and third must also contain the middle one. Hence
$$
\boxed{VC(H_1)=2.}
$$

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

The choice $\delta=1$ labels an interval positively and its complement negatively, while $\delta=-1$ reverses those labels. On three ordered points, every subset is either a consecutive block or the complement of a consecutive block, so all eight labelings occur. Thus three points are shattered.

Four ordered points are not shattered: the alternating labeling $+,-,+,-$ has neither its positive set nor its negative set consecutive, so it cannot arise from an interval or its complement. Therefore
$$
\boxed{VC(H_2)=3.}
$$

Solved by gpt-5.6-sol high.

= c
{parent=31l}
{scope}

= Solution
{parent=c}

Only the symmetric part of $M$ contributes to $x^TMx$. The functions
$$
x\longmapsto x^TMx
$$
therefore form a real vector space of dimension
$$
D=\frac{d(d+1)}2=\binom{d+1}{2},
$$
with coordinates given by the quadratic monomials $x_i^2$ and $x_ix_j$ for $i<j$. The theorem on signs of a finite-dimensional function space says that the sign class of a $D$-dimensional real vector space has VC dimension at most $D$. Consequently
$$
\boxed{VC(H_3)\leq\binom{d+1}{2}.}
$$

Solved by gpt-5.6-sol high.

= d
{parent=31l}
{scope}

= Solution
{parent=d}

Use the empirical Rademacher convention
$$
\widehat{\mathcal R}(A)
=\frac1n\mathbb E_\sigma
\sup_{a\in A}\sum_{i=1}^n\sigma_i a_i.
$$
Because $H_3$ is symmetric under $h\mapsto-h$, optimizing a linear functional over the positive $\ell^1$ hull defining $\mathcal F$ gives
$$
\widehat{\mathcal R}(\mathcal F(x_{1:n}))
\leq\widehat{\mathcal R}(H_3(x_{1:n})).
$$

By part (c), $VC(H_3)\leq D=d(d+1)/2$. Sauer--Shelah therefore bounds the number $N$ of distinct label vectors on the sample by
$$
N\leq s(H_3,n)\leq(n+1)^D.
$$
Massart's finite-class lemma states that for $A\subseteq\{-1,1\}^n$,
$$
\widehat{\mathcal R}(A)
\leq\sqrt{\frac{2\log|A|}{n}}.
$$
Applying it to the distinct label vectors of $H_3$ gives
$$
\widehat{\mathcal R}(\mathcal F(x_{1:n}))
\leq\sqrt{\frac{2D\log(n+1)}n}
=\boxed{\sqrt{\frac{(d^2+d)\log(n+1)}n}}.
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=32d}
{scope}

= Solution
{parent=a}

Integration by parts gives the recurrence
$$
\gamma(x,y)=y^{x-1}e^{-y}+(x-1)\gamma(x-1,y).
$$
Iterating $N$ times yields
$$
\gamma(x,y)=y^{x-1}e^{-y}
\sum_{n=0}^{N-1}(x-1)(x-2)\cdots(x-n)y^{-n}
+R_N,
$$
where the empty product for $n=0$ is one and
$$
R_N=(x-1)(x-2)\cdots(x-N)\gamma(x-N,y).
$$
For fixed $x$, the remainder has the order of the first omitted term as $y\to\infty$. Therefore
$$
\boxed{a_0(x)=1,\qquad
a_n(x)=\prod_{j=1}^n(x-j)\quad(n\geq1).}
$$

Solved by gpt-5.6-sol high.

= b
{parent=32d}
{scope}

= Solution
{parent=b}

For fixed $y$,
$$
\gamma(x,y)=\Gamma(x)-\int_0^y t^{x-1}e^{-t}\,dt.
$$
The omitted integral is at most $y^x/x$, which is negligible compared with $\Gamma(x)$ as $x\to\infty$. Stirling's formula therefore gives
$$
\boxed{
\gamma(x,y)\sim\Gamma(x)
\sim\sqrt{2\pi}\,x^{x-1/2}e^{-x}.}
$$

Solved by gpt-5.6-sol high.

= c
{parent=32d}
{scope}

= Solution
{parent=c}

Put $y=\lambda x$ and $t=y+s$. Factoring out the endpoint value gives
$$
\gamma(x,y)=y^{x-1}e^{-y}
\int_0^\infty
\exp\left((x-1)\log\left(1+\frac{s}{\lambda x}\right)-s\right)ds.
$$
For fixed $s$,
$$
(x-1)\log\left(1+\frac{s}{\lambda x}\right)-s
=-\frac{\lambda-1}{\lambda}s
-\frac1x\left(\frac{s}{\lambda}+\frac{s^2}{2\lambda^2}\right)
+O(x^{-2}).
$$
Watson's endpoint lemma justifies termwise integration, so with $\alpha=(\lambda-1)/\lambda$,
$$
\begin{aligned}
\frac{\gamma(x,y)}{y^{x-1}e^{-y}}
&\sim\int_0^\infty e^{-\alpha s}
\left[1-\frac1x\left(\frac{s}{\lambda}
+\frac{s^2}{2\lambda^2}\right)\right]ds\\
&=\frac1\alpha
-\frac1x\left(\frac1{\lambda\alpha^2}
+\frac1{\lambda^2\alpha^3}\right).
\end{aligned}
$$
Consequently
$$
\boxed{
f(\lambda)=\frac{\lambda}{\lambda-1},
\qquad
g(\lambda)=-\frac{\lambda^2}{(\lambda-1)^3}.}
$$

Solved by gpt-5.6-sol high.

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

= Solution
{parent=33A}

The <centre manifold theorem> states that near an equilibrium whose linearization has centre, stable, and unstable spectral subspaces $E^c,E^s,E^u$, there is a local invariant manifold $W^c$ tangent to $E^c$ at the equilibrium. It can be written locally as a graph over $E^c$, and the dynamics on it govern the local nonhyperbolic behaviour. The manifold need not be unique, but its finite Taylor expansion is determined to the required order by the invariance equation.

For a parameter-dependent system, the key step in forming the extended centre manifold is to promote the parameter to a dynamical variable:
$$
\dot\mu=0.
$$
The parameter direction then has zero eigenvalue and is included in the extended centre subspace.

Solved by gpt-5.6-sol high.

= a
{parent=33a}
{scope}

= Solution
{parent=a}

The Jacobian is
$$
J(x,y)=
\begin{pmatrix}
\mu-y^2-6x^2&-2xy\\
-2xy&1-x^2-3y^2
\end{pmatrix}.
$$
At $(0,1)$,
$$J(0,1)=\operatorname{diag}(\mu-1,-2),$$
so one eigenvalue crosses zero at
$$\boxed{\mu=1}$$
while the other remains negative. Thus a bifurcation occurs there.

On the $x$-axis the nonzero fixed point is
$$
(x,y)=\left(\sqrt{\frac\mu2},0\right).
$$
Its eigenvalues are
$$
-2\mu,\qquad1-\frac\mu2.
$$
The second crosses zero at
$$\boxed{\mu_c=2},$$
which is the second bifurcation value.

Solved by gpt-5.6-sol high.

= b
{parent=33a}
{scope}

= Solution
{parent=b}

Set
$$
\varepsilon=\mu-1,
\qquad
v=y-1,
\qquad
\dot\varepsilon=0.
$$
Then
$$
\begin{aligned}
\dot x&=x(\varepsilon-2v-v^2-2x^2),\\
\dot v&=-2v-x^2-vx^2-3v^2-v^3.
\end{aligned}
$$
The extended centre variables are $(x,\varepsilon)$ and the stable variable is $v$. Write the centre manifold as $v=h(x,\varepsilon)$. At quadratic order, the invariance equation
$$
h_x\dot x+h_\varepsilon\dot\varepsilon
=-2h-x^2-vx^2-3v^2-v^3
$$
gives
$$
h(x,\varepsilon)=-\frac{x^2}{2}+\text{higher-order terms}.
$$
Substitution into the $x$ equation gives the leading normal form
$$
\boxed{\dot x=\varepsilon x-x^3+\text{higher-order terms}.}
$$
This is a supercritical pitchfork: the $x=0$ branch loses stability as $\varepsilon$ becomes positive and stable branches $x=\pm\sqrt\varepsilon$ emerge. The original system is invariant under $x\mapsto-x$, so a pitchfork is exactly the symmetry-forced bifurcation expected. The physical quadrant $x\geq0$ displays its positive half.

Solved by gpt-5.6-sol high.

= c
{parent=33a}
{scope}

= Solution
{parent=c}

At a fixed point with $x,y>0$, the two bracketed factors must vanish:
$$
\mu-y^2-2x^2=0,
\qquad
1-x^2-y^2=0.
$$
Subtracting gives
$$
x^2=\mu-1,
\qquad
y^2=2-\mu.
$$
Hence the interior branch is
$$
\boxed{(x,y)=(\sqrt{\mu-1},\sqrt{2-\mu}),\qquad1<\mu<2.}
$$
Near $\mu=1$, if $\varepsilon=\mu-1$, then
$$x=\sqrt\varepsilon,\qquad
y-1=\sqrt{1-\varepsilon}-1=-\frac\varepsilon2+O(\varepsilon^2)
=-\frac{x^2}{2}+O(x^4),$$
which agrees with both the centre manifold and the positive branch of the pitchfork normal form.

Solved by gpt-5.6-sol high.

= d
{parent=33a}
{scope}

= Solution
{parent=d}

The fixed-point branches and their stability are as follows.

* $(0,0)$ has eigenvalues $\mu$ and $1$, so it is unstable for every $\mu>0$.
* $(0,1)$ has $x=0$ and is stable for $0<\mu<1$, then a saddle for $\mu>1$.
* $(\sqrt{\mu/2},0)$ exists for all $\mu>0$; it is a saddle for $0<\mu<2$ and stable for $\mu>2$.
* $(\sqrt{\mu-1},\sqrt{2-\mu})$ exists for $1<\mu<2$. At it,
$$
J=-2\begin{pmatrix}2x^2&xy\\xy&y^2\end{pmatrix},
$$
which is negative definite, so this branch is stable.

Thus the $x$-versus-$\mu$ diagram has the stable $x=0$ branch up to $\mu=1$, a stable branch $x=\sqrt{\mu-1}$ joining the bifurcation points $(1,0)$ and $(2,1)$, and the branch $x=\sqrt{\mu/2}$ changing from unstable to stable at $\mu=2$. The origin branch also lies at $x=0$ but remains unstable throughout.

Solved by gpt-5.6-sol high.

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

= a
{parent=34d}
{scope}

= Solution
{parent=a}

A Hamiltonian system on a $2n$-dimensional symplectic manifold is completely integrable when it has $n$ first integrals $F_1=H,F_2,\ldots,F_n$ whose differentials are independent on the regular set and which are in involution:
$$
\{F_i,F_j\}=0\qquad\text{for all }i,j.
$$

The Arnold--Liouville theorem states that every compact connected regular common level set of these integrals is an $n$-torus. In a neighbourhood of it there are action--angle coordinates $(I_1,\ldots,I_n,\theta_1,\ldots,\theta_n)$ in which
$$
\omega=\sum_{i=1}^n dI_i\wedge d\theta_i,
\qquad
H=H(I),
$$
and Hamilton's equations become
$$
\dot I_i=0,
\qquad
\dot\theta_i=\frac{\partial H}{\partial I_i}.
$$
Thus the motion on each invariant torus is linear in the angles.

Solved by gpt-5.6-sol high.

= b
{parent=34d}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

Set
$$
F_i=\frac12\left(p_i^2+W_i^2q_i^2+a_iq_i\right),
\qquad i=1,\ldots,n.
$$
Then $\sum_iF_i=H$. Each $F_i$ involves only $(p_i,q_i)$, so
$$
\{F_i,F_j\}=0
$$
for all $i,j$. Their differentials have disjoint coordinate supports and are independent wherever none of them vanishes, which is an open dense regular set.

Hamilton's equations are
$$
\dot q_i=p_i,
\qquad
\dot p_i=-W_i^2q_i-\frac{a_i}{2}.
$$
After the shift
$$
Q_i=q_i+\frac{a_i}{2W_i^2},
$$
they become $n$ independent harmonic oscillators,
$$
\ddot Q_i+W_i^2Q_i=0.
$$
Thus the $n$ independent commuting integrals establish complete integrability.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

Completing the square gives
$$
F_i=\frac12\left(p_i^2+W_i^2Q_i^2\right)
-\frac{a_i^2}{8W_i^2}.
$$
Hence the positive oscillator energy is
$$
E_i=F_i+\frac{a_i^2}{8W_i^2}.
$$
Using the convention
$$
I_i=\frac1{2\pi}\oint p_i\,dq_i,
$$
the integral is the phase-plane ellipse area divided by $2\pi$. With oscillator frequency $\omega_i=|W_i|$, that area is $2\pi E_i/\omega_i$. Therefore
$$
\boxed{
I_i=\frac1{|W_i|}
\left(F_i+\frac{a_i^2}{8W_i^2}\right).}
$$
Equivalently,
$$
H=\sum_{i=1}^n\left(|W_i|I_i-\frac{a_i^2}{8W_i^2}\right),
$$
so the angle velocities are the constant frequencies $|W_i|$.

Solved by gpt-5.6-sol high.

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

= a
{parent=35c}
{scope}

= Solution
{parent=a}

The spin components satisfy
$$
[S_i,S_j]=i\hbar\varepsilon_{ijk}S_k,
\qquad
[S^2,S_i]=0.
$$
The irreducible representations are indexed by
$$s=0,\frac12,1,\frac32,\ldots$$
and have orthonormal basis $|s,\sigma\rangle$, where
$$
\sigma=-s,-s+1,\ldots,s.
$$
On this basis,
$$
S^2|s,\sigma\rangle=\hbar^2s(s+1)|s,\sigma\rangle,
\qquad
S_z|s,\sigma\rangle=\hbar\sigma|s,\sigma\rangle.
$$
Each Cartesian component is Hermitian,
$$S_i^\dagger=S_i,$$
and $S_+^\dagger=S_-$. Also
$$
\operatorname{tr}S_z=\hbar\sum_{\sigma=-s}^s\sigma=0.
$$
Rotational equivalence, or the off-diagonal form of $S_x,S_y$ in this basis, gives
$$
\boxed{\operatorname{tr}S_x=operatorname{tr}S_y=operatorname{tr}S_z=0.}
$$

Solved by gpt-5.6-sol high.

= b
{parent=35c}
{scope}

= Solution
{parent=b}

Use the ordered basis
$$
\left(|3/2\rangle,|1/2\rangle,|-1/2\rangle,|-3/2\rangle\right).
$$
The ladder coefficients are $\sqrt3,2,\sqrt3$, so
$$
S_+=\hbar\begin{pmatrix}
0&\sqrt3&0&0\\
0&0&2&0\\
0&0&0&\sqrt3\\
0&0&0&0
\end{pmatrix},
\qquad
S_-=S_+^\dagger.
$$
Therefore
$$
\boxed{
S_x=\frac\hbar2\begin{pmatrix}
0&\sqrt3&0&0\\
\sqrt3&0&2&0\\
0&2&0&\sqrt3\\
0&0&\sqrt3&0
\end{pmatrix}}
$$
and
$$
\boxed{
S_y=\frac\hbar2\begin{pmatrix}
0&-i\sqrt3&0&0\\
i\sqrt3&0&-2i&0\\
0&2i&0&-i\sqrt3\\
0&0&i\sqrt3&0
\end{pmatrix}.}
$$

Solved by gpt-5.6-sol high.

= c
{parent=35c}
{scope}

= Solution
{parent=c}

The required state is the maximal-<spin coherent state> pointing at azimuth $\varphi$ in the equatorial plane. In the same ordered $S_z$ basis it is
$$
\boxed{
|\mathbf n,3/2\rangle
=\frac1{2\sqrt2}
\begin{pmatrix}
1\\
\sqrt3e^{i\varphi}\\
\sqrt3e^{2i\varphi}\\
e^{3i\varphi}
\end{pmatrix}.}
$$
Its norm is $(1+3+3+1)/8=1$. Direct multiplication using the matrices in part (b), or rotation of $|\hat{\mathbf z},3/2\rangle$, gives
$$
(S_x\cos\varphi+S_y\sin\varphi)|\mathbf n,3/2\rangle
=\frac32\hbar|\mathbf n,3/2\rangle.
$$
It therefore yields the value $3\hbar/2$ with probability one when spin is measured along $\mathbf n$.

Solved by gpt-5.6-sol high.

= d
{parent=35c}
{scope}

= Solution
{parent=d}

Write $\omega=\gamma B$. Since
$$H=-\omega S_z,$$
the time-evolution operator is
$$
U(t)=e^{i\omega tS_z/\hbar}.
$$

For the initial state $|\hat{\mathbf z},3/2\rangle=|3/2\rangle$,
$$
\boxed{|\psi(t)\rangle=e^{3i\omega t/2}|3/2\rangle.}
$$
Only an overall phase changes, so every spin measurement has time-independent probabilities.

For the initial state along $\hat{\mathbf x}$, part (c) with $\varphi=0$ gives
$$
\boxed{
|\psi(t)\rangle=\frac1{2\sqrt2}
\begin{pmatrix}
e^{3i\omega t/2}\\
\sqrt3e^{i\omega t/2}\\
\sqrt3e^{-i\omega t/2}\\
e^{-3i\omega t/2}
\end{pmatrix}.}
$$
Up to the overall phase $e^{3i\omega t/2}$, this is the coherent state of part (c) with $\varphi=-\omega t$. Thus the spin direction undergoes Larmor precession about the $z$-axis with angular velocity $-\gamma B$.

Solved by gpt-5.6-sol high.

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

= a
{parent=36b}
{scope}

= Solution
{parent=a}

Let $\psi_1$ be the normalized ground state for
$$H_1=T+V_1.$$
Using it as a trial state for $H_2=T+V_2$, the variational principle gives
$$
E_2\leq\langle\psi_1,H_2\psi_1\rangle
=E_1+\langle\psi_1,(V_2-V_1)\psi_1\rangle.
$$
The pointwise assumption $V_2\leq V_1$ makes the last expectation nonpositive. Hence
$$
\boxed{E_2\leq E_1.}
$$

Solved by gpt-5.6-sol high.

= b
{parent=36b}
{scope}

= Solution
{parent=b}

Use the normalized scaled trial state
$$
\psi_\lambda(x)=\lambda^{1/2}\psi_0(\lambda x),
\qquad \lambda>0.
$$
If
$$T_0=\langle\psi_0,T\psi_0\rangle>0,
\qquad
V_0=\langle\psi_0,V\psi_0\rangle<0,$$
then scaling derivatives and changing variables give
$$
E(\lambda)=\langle\psi_\lambda,H\psi_\lambda\rangle
=\lambda^2T_0+\lambda^nV_0.
$$
If a finite ground state existed, $\lambda=1$ would minimize this expression, and hence
$$
0=E'(1)=2T_0+nV_0.
$$
It would follow that
$$
E_0=T_0+V_0=\left(1-\frac2n\right)T_0>0
$$
for $n>2$. But a bound state of an attractive potential tending to zero at infinity must have negative energy. This is a contradiction.

Indeed, the scaling gives the stronger collapse statement
$$
E(\lambda)=\lambda^2T_0+\lambda^nV_0\longrightarrow-\infty
\qquad(\lambda\to\infty)
$$
when $n>2$. Thus the ideal singular Hamiltonian is not bounded below and has no finite ground bound state.

Solved by gpt-5.6-sol high.

= c
{parent=36b}
{scope}

= Solution
{parent=c}

In the basis $(|\psi_1\rangle,|\psi_2\rangle)$,
$$
H=\begin{pmatrix}E_1&h\\h&E_2\end{pmatrix}.
$$
Writing $\Delta=E_2-E_1>0$, its exact eigenvalues are
$$
\boxed{
E_\pm=\frac{E_1+E_2}{2}
\pm\sqrt{\frac{\Delta^2}{4}+h^2}.}
$$

For the normalized trial state,
$$
\begin{aligned}
E(\beta)
&=E_1\sin^2\beta+E_2\cos^2\beta
+2h\sin\beta\cos\beta\\
&=\frac{E_1+E_2}{2}
+\frac\Delta2\cos2\beta+h\sin2\beta.
\end{aligned}
$$
The minimum of $A\cos2\beta+h\sin2\beta$ is $-\sqrt{A^2+h^2}$ with $A=\Delta/2$. Therefore
$$
\boxed{
\min_\beta E(\beta)
=\frac{E_1+E_2}{2}
-\sqrt{\frac{\Delta^2}{4}+h^2}=E_-.}
$$
The variational estimate is exact because the trial family ranges over all real normalized states in the two-dimensional Hilbert space, and the real symmetric Hamiltonian has a real ground eigenvector.

Solved by gpt-5.6-sol high.

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

= a
{parent=37c}
{scope}

= Solution
{parent=a}

For a simple compressible system with fixed particle number, the first law is
$$
dE=T\,dS-p\,dV.
$$
Define the <enthalpy> by $H=E+pV$. Then
$$
dH=dE+p\,dV+V\,dp
=T\,dS+V\,dp.
$$
Thus $H$ has natural variables $(S,p)$ and
$$
T=\left(\frac{\partial H}{\partial S}\right)_p,
\qquad
V=\left(\frac{\partial H}{\partial p}\right)_S.
$$
Equality of mixed partial derivatives gives the Maxwell relation
$$
\boxed{
\left(\frac{\partial T}{\partial p}\right)_S
=\left(\frac{\partial V}{\partial S}\right)_p.}
$$

Solved by gpt-5.6-sol high.

= b
{parent=37c}
{scope}

= Solution
{parent=b}

With vibrations frozen out, each diatomic molecule has three translational and two rotational quadratic degrees of freedom. Equipartition therefore gives
$$
E=\frac52Nk_BT.
$$
Using the ideal-gas law $pV=Nk_BT$,
$$
\boxed{H=E+pV=\frac72Nk_BT.}
$$

Solved by gpt-5.6-sol high.

= c
{parent=37c}
{scope}

= i
{parent=c}
{scope}

= Solution
{parent=i}

Let $(T_1,V_1)$ be the final equilibrium state. The suddenly added weight supplies the constant external pressure $p_1$ during the displacement. Since the process is adiabatic, the first law gives
$$
\frac52Nk_B(T_1-T_0)=p_1(V_0-V_1).
$$
The initial and final ideal-gas equations are
$$
p_0V_0=Nk_BT_0,
\qquad
p_1V_1=Nk_BT_1.
$$
Substitution yields
$$
\frac52(T_1-T_0)=\frac{p_1}{p_0}T_0-T_1,
$$
and hence
$$
T_1=T_0\frac{5+2p_1/p_0}{7}.
$$
Therefore
$$
\boxed{\Delta T=T_1-T_0
=\frac{2T_0}{7}\left(\frac{p_1}{p_0}-1\right).}
$$

Solved by gpt-5.6-sol high.

= ii
{parent=c}
{scope}

= Solution
{parent=ii}

At fixed pressure, the heat supplied equals the enthalpy change:
$$
Q=\Delta H=\frac72Nk_B\Delta T.
$$
Thus
$$
\boxed{\Delta T=\frac{2Q}{7Nk_B}.}
$$
Equivalently, the gas does work
$$
W=p\Delta V=Nk_B\Delta T=\frac{2Q}{7},
$$
while its internal energy increases by $5Q/7$.

Solved by gpt-5.6-sol high.

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

= a
{parent=38e}
{scope}

= Solution
{parent=a}

For a covariant rank-two tensor,
$$
\nabla_\alpha g_{\mu\nu}
=\partial_\alpha g_{\mu\nu}
-\Gamma^\rho_{\alpha\mu}g_{\rho\nu}
-\Gamma^\rho_{\alpha\nu}g_{\mu\rho}.
$$
Impose <metric compatibility> $\nabla g=0$, write this equation and its two cyclic permutations in $\alpha,\mu,\nu$, and use the assumed torsion-free symmetry of the connection. Solving gives the unique Levi--Civita connection
$$
\boxed{
\Gamma^\gamma_{\alpha\beta}
=\frac12g^{\gamma\delta}
\left(\partial_\alpha g_{\beta\delta}
+\partial_\beta g_{\alpha\delta}
-\partial_\delta g_{\alpha\beta}\right).}
$$

Solved by gpt-5.6-sol high.

= b
{parent=38e}
{scope}

= Solution
{parent=b}

Set $U_\mu=\partial_\mu\phi$. Torsion-freeness and commutation of ordinary partial derivatives imply that its covariant Hessian is symmetric:
$$
\nabla_\alpha U_\beta=\nabla_\beta U_\alpha.
$$
Take the cyclic sum of the three commutators acting respectively on $U_\gamma,U_\alpha,U_\beta$. Their left-hand sides cancel pairwise by this Hessian symmetry, leaving
$$
\left(R^\nu{}_{\gamma\alpha\beta}
+R^\nu{}_{\alpha\beta\gamma}
+R^\nu{}_{\beta\gamma\alpha}\right)U_\nu=0.
$$
At a chosen point, a scalar field can be selected whose gradient has any prescribed covector value. The coefficient of $U_\nu$ must therefore vanish. Relabelling indices gives the <first Bianchi identity>
$$
\boxed{
R^\mu{}_{\alpha\beta\gamma}
+R^\mu{}_{\beta\gamma\alpha}
+R^\mu{}_{\gamma\alpha\beta}=0.}
$$

Solved by gpt-5.6-sol high.

= c
{parent=38e}
{scope}

= Solution
{parent=c}

Using the Leibniz rule for the connection, or first checking decomposable tensors and extending linearly, the commutator acts on each covariant index:
$$
[\nabla_\alpha,\nabla_\beta]W_{\mu\nu}
=-R^\rho{}_{\mu\alpha\beta}W_{\rho\nu}
-R^\rho{}_{\nu\alpha\beta}W_{\mu\rho}.
$$
Choose $W_{\mu\nu}=g_{\mu\nu}$. Metric compatibility makes the left side zero. Applying the formula with commutator indices $\beta,\gamma$ gives
$$
0=-R^\rho{}_{\mu\beta\gamma}g_{\rho\alpha}
-R^\rho{}_{\alpha\beta\gamma}g_{\mu\rho}
=-R_{\alpha\mu\beta\gamma}-R_{\mu\alpha\beta\gamma}.
$$
Hence
$$
\boxed{R_{\mu\alpha\beta\gamma}
=-R_{\alpha\mu\beta\gamma}.}
$$

Solved by gpt-5.6-sol high.

= d
{parent=38e}
{scope}

= Solution
{parent=d}

Define the <Ricci tensor> by contracting the first and third Riemann indices:
$$
R_{\alpha\gamma}=R^\mu{}_{\alpha\mu\gamma}.
$$
Contract the first Bianchi identity in its upper index and third lower index. The middle term vanishes by antisymmetry in the first pair, while antisymmetry in the final pair turns the last term into $-R_{\gamma\alpha}$. Thus
$$R_{\alpha\gamma}-R_{\gamma\alpha}=0,$$
so the Ricci tensor is symmetric.

For the stated constant-curvature form,
$$
\begin{aligned}
R_{\alpha\gamma}
&=g^{\mu\beta}R_{\mu\alpha\beta\gamma}
=(n-1)K g_{\alpha\gamma},\\
R&=g^{\alpha\gamma}R_{\alpha\gamma}
=n(n-1)K.
\end{aligned}
$$
The <contracted Bianchi identity> is
$$
\nabla^\alpha R_{\alpha\beta}=\frac12\nabla_\beta R.
$$
Using $\nabla g=0$ gives
$$
(n-1)\partial_\beta K
=\frac12n(n-1)\partial_\beta K,
$$
or
$$
\frac{(n-1)(n-2)}2\,\partial_\beta K=0.
$$
In four dimensions,
$$
\boxed{R_{\alpha\beta}=3Kg_{\alpha\beta},
\qquad R=12K,
\qquad K\text{ is constant}.}
$$
More generally, $K$ is constant for every $n>2$. In dimension $n=2$, the contracted Bianchi identity gives no such conclusion, and the Gaussian curvature may vary from point to point.

Solved by gpt-5.6-sol high.

= 39D
{parent=Paper 2}
{scope}
{title2=Fluid Dynamics}

= Solution
{parent=39d}

At zero Reynolds number, an incompressible Newtonian fluid obeys the Stokes equations
$$
-\nabla p+\mu\nabla^2\mathbf u=0,
\qquad
\nabla\cdot\mathbf u=0.
$$
Taking the divergence gives $\nabla^2p=0$. Taking the curl and writing $\boldsymbol\omega=\nabla\times\mathbf u$ gives
$$
\mu\nabla^2\boldsymbol\omega=0.
$$
Thus both pressure and vorticity are harmonic in the fluid.

In the laboratory frame, the boundary conditions are
$$
\mathbf u\to0\quad(r\to\infty),
$$
and, on $r=a$,
$$
\mathbf u\cdot\mathbf n=\mathbf U\cdot\mathbf n,
\qquad
(\mathbf I-\mathbf n\mathbf n^T)\mathbf u
=(\mathbf I-\mathbf n\mathbf n^T)\mathbf U.
$$
The first is impermeability of the solid surface and the second is viscous no slip. Together they say $\mathbf u=\mathbf U$ on the sphere.

For the stated solution, put
$$
A(r)=\frac{3a}{4r}+\frac{a^3}{4r^3},
\qquad
B(r)=\frac{3a}{4r^3}-\frac{3a^3}{4r^5},
\qquad
s=\mathbf U\cdot\mathbf x.
$$
Then $u_j=AU_j+Bs x_j$, and
$$
\boxed{
\frac{\partial u_j}{\partial x_i}
=\frac{A'}r x_iU_j
+\frac{B'}r s x_ix_j
+B U_ix_j+B s\delta_{ij},}
$$
where
$$
A'=-\frac{3a}{4r^2}-\frac{3a^3}{4r^4},
\qquad
B'=-\frac{9a}{4r^4}+\frac{15a^3}{4r^6}.
$$

Solved by gpt-5.6-sol high.

= i
{parent=39d}
{scope}

= Solution
{parent=i}

Taking the curl of $A\mathbf U$ gives $\nabla A\times\mathbf U$, while
$$
\nabla\times(Bs\mathbf x)=B\mathbf U\times\mathbf x.
$$
Therefore
$$
\boldsymbol\omega
=\left(B-\frac{A'}r\right)\mathbf U\times\mathbf x
=\boxed{\frac{3a}{2r^3}\,\mathbf U\times\mathbf x}.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=39d}
{scope}

= Solution
{parent=ii}

For $r>0$,
$$
\boxed{\nabla\left(\frac1r\right)=-\frac{\mathbf x}{r^3}},
$$
and
$$
\boxed{\nabla^2\left(\frac1r\right)
=-\frac3{r^3}+\frac{3|\mathbf x|^2}{r^5}=0.}
$$
The pressure and vorticity can be written
$$
p=-\frac32\mu a\,\mathbf U\cdot\nabla\left(\frac1r\right),
\qquad
\boldsymbol\omega=-\frac{3a}{2}\mathbf U\times
\nabla\left(\frac1r\right).
$$
Constant-coefficient derivatives commute with the Laplacian, so both fields are harmonic away from the sphere.

Solved by gpt-5.6-sol high.

= iii
{parent=39d}
{scope}

= Solution
{parent=iii}

Taking the trace of the velocity gradient gives
$$
\nabla\cdot\mathbf u
=s\left(\frac{A'}r+rB'+4B\right).
$$
Direct substitution yields
$$
\frac{A'}r+rB'+4B=0,
$$
so
$$
\boxed{\nabla\cdot\mathbf u=0.}
$$
This verifies incompressibility directly.

Solved by gpt-5.6-sol high.

= iv
{parent=39d}
{scope}

= Solution
{parent=iv}

For a Newtonian fluid,
$$
\sigma_{ij}=-p\delta_{ij}
+\mu\left(\frac{\partial u_i}{\partial x_j}
+\frac{\partial u_j}{\partial x_i}\right).
$$
On $r=a$, with $\mathbf n=\mathbf x/a$,
$$
A'=-\frac{3}{2a},
\qquad B=0,
\qquad B'=\frac{3}{2a^3},
\qquad
p=\frac{3\mu}{2a}(\mathbf U\cdot\mathbf n).
$$
The surface velocity gradient is therefore
$$
\frac{\partial u_j}{\partial x_i}
=\frac{3}{2a}n_i\bigl((\mathbf U\cdot\mathbf n)n_j-U_j\bigr).
$$
Contracting the stress with $n_j$, the normal part of the viscous stress cancels the pressure contribution, leaving the uniform traction
$$
\boxed{(\boldsymbol\sigma\cdot\mathbf n)_i
=-\frac{3\mu}{2a}U_i.}
$$

Solved by gpt-5.6-sol high.

= v
{parent=39d}
{scope}

= Solution
{parent=v}

Integrating the traction over the sphere gives
$$
\mathbf F
=\int_{r=a}\boldsymbol\sigma\cdot\mathbf n\,dS
=-\frac{3\mu}{2a}\mathbf U\,(4\pi a^2).
$$
Hence the force exerted by the fluid on the sphere is
$$
\boxed{\mathbf F=-6\pi\mu a\mathbf U.}
$$
This is Stokes' drag law; the sign is opposite to the sphere's velocity.

Solved by gpt-5.6-sol high.

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

= a
{parent=40a}
{scope}

= Solution
{parent=a}

Take the divergence of the elastic equation. With $\theta=\nabla\cdot\mathbf u$,
$$
\rho\theta_{tt}=(\lambda+2\mu)\nabla^2\theta.
$$
Thus dilatational P-waves have speed
$$
\boxed{c_P=\sqrt{\frac{\lambda+2\mu}{\rho}}.}
$$

Taking the curl and writing $\boldsymbol\omega=\nabla\times\mathbf u$ eliminates the gradient term and gives
$$
\rho\boldsymbol\omega_{tt}=\mu\nabla^2\boldsymbol\omega.
$$
Thus rotational S-waves have speed
$$
\boxed{c_S=\sqrt{\frac\mu\rho}.}
$$

Solved by gpt-5.6-sol high.

= b
{parent=40a}
{scope}

= Solution
{parent=b}

Let $\eta=\hat{\mathbf k}\cdot\mathbf x-ct$. Then
$$
\nabla^2\mathbf u=\mathbf f''(\eta),
\qquad
\nabla(\nabla\cdot\mathbf u)
=\hat{\mathbf k}(\hat{\mathbf k}\cdot\mathbf f''(\eta)),
$$
and substitution gives
$$
\rho c^2\mathbf f''
=(\lambda+\mu)\hat{\mathbf k}
(\hat{\mathbf k}\cdot\mathbf f'')+\mu\mathbf f''.
$$

For a P-wave, $c=c_P$ and the varying part of $\mathbf f$ is longitudinal:
$$
\mathbf f(\eta)=\hat{\mathbf k}F(\eta),
\qquad
\theta=F'(\eta),
\qquad
\boldsymbol\omega=0.
$$
For an S-wave, $c=c_S$ and $\mathbf f$ is transverse:
$$
\hat{\mathbf k}\cdot\mathbf f=0,
\qquad
\theta=0,
\qquad
\boldsymbol\omega=\hat{\mathbf k}\times\mathbf f'(\eta).
$$

Solved by gpt-5.6-sol high.

= c
{parent=40a}
{scope}

= Solution
{parent=c}

All wavevectors lie in the $xz$ plane. The incident P-wave travels upward toward the interface. In the lower medium there are a reflected P-wave, polarized parallel to its reflected wavevector, and a reflected SV-wave, polarized in the $xz$ plane perpendicular to its wavevector. In the upper medium there are transmitted P and SV waves with the corresponding longitudinal and transverse polarizations. No SH wave is generated by this in-plane incident field.

Frequency and tangential phase matching give Snell's law: every wave has the same $\omega$ and the same tangential wavenumber $k_x$, while its normal wavenumber has the sign appropriate to reflection or transmission.

For a perfectly bonded interface, displacement and traction are continuous:
$$
[u_x]=[u_z]=0,
\qquad
[\sigma_{xz}]=[\sigma_{zz}]=0
$$
in the two-dimensional problem. In three dimensions one also includes $[u_y]=[\sigma_{yz}]=0$. Here
$$
\sigma_{ij}=\lambda(\nabla\cdot\mathbf u)\delta_{ij}
+\mu(\partial_i u_j+\partial_j u_i).
$$
These four in-plane conditions determine the four reflected and transmitted amplitudes.

Solved by gpt-5.6-sol high.

= d
{parent=40a}
{scope}

= Solution
{parent=d}

Let $\omega=kc_P$, and let $\psi$ be the transmission angle. Since $\hat c_P=2c_P$, tangential phase matching gives
$$
k\sin\phi=\hat k\sin\psi,
\qquad
\hat k=\frac k2,
\qquad
\boxed{\sin\psi=2\sin\phi.}
$$
The assumption of no evanescence requires $\sin\phi\leq1/2$.

With a common suppressed factor $e^{-i\omega t}$, take
$$
\begin{aligned}
\mathbf u_I&=(\sin\phi,0,\cos\phi)
e^{ik(x\sin\phi+z\cos\phi)},\\
\mathbf u_R&=R(\sin\phi,0,-\cos\phi)
e^{ik(x\sin\phi-z\cos\phi)},\\
\mathbf u_T&=T(\sin\psi,0,\cos\psi)
e^{i\hat k(x\sin\psi+z\cos\psi)}.
\end{aligned}
$$
For an inviscid liquid, continuity of normal displacement and normal stress gives
$$
\cos\phi(1-R)=T\cos\psi,
\qquad
Z(1+R)=\hat ZT,
$$
where $Z=\lambda/c_P$ and $\hat Z=\hat\lambda/\hat c_P$. Solving,
$$
\boxed{
R=\frac{\hat Z\cos\phi-Z\cos\psi}
{\hat Z\cos\phi+Z\cos\psi},
\qquad
T=\frac{2Z\cos\phi}
{\hat Z\cos\phi+Z\cos\psi}.}
$$

The condition $|R|=|T|$ gives
$$
|\hat Z\cos\phi-Z\cos\psi|=2Z\cos\phi.
$$
The sign that could instead produce $Z\cos\psi=(\hat Z+2Z)\cos\phi$ leads to $\sin^2\phi>1$ and is inadmissible. Hence
$$
Z\cos\psi=(\hat Z-2Z)\cos\phi.
$$
Using $\cos^2\psi=1-4\sin^2\phi$ and squaring gives
$$
\boxed{
\sin^2\phi
=\frac{3Z^2-4Z\hat Z+\hat Z^2}
{\hat Z(\hat Z-4Z)}.}
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=41d}
{scope}

= Solution
{parent=a}

Put $\omega_{2N}=e^{-\pi i/N}$ and split the transform into its even and odd input samples. If $E_k$ and $O_k$ are the two length-$N$ transforms, then
$$
Y_k=E_k+\omega_{2N}^kO_k,
\qquad
Y_{k+N}=E_k-\omega_{2N}^kO_k,
\qquad 0\leq k<N.
$$
Thus a length-$2N$ transform requires two length-$N$ transforms and $O(N)$ further multiplications. Its multiplication count satisfies
$$M(2N)\leq2M(N)+CN.$$
Since $N$ is a power of two, iteration through $\log_2(2N)$ levels gives $M(2N)=O(N\log N)$.

Solved by gpt-5.6-sol high.

= b
{parent=41d}
{scope}

= Solution
{parent=b}

Pair the two entries of the even-reflected sequence that contain $x_n$. For every $k$,
$$
\begin{aligned}
Y_k
&=\sum_{n=0}^{N-1}x_n
\left(e^{-\pi ink/N}+e^{-\pi i(2N-1-n)k/N}\right)\\
&=\sum_{n=0}^{N-1}x_n
\left(e^{-\pi ink/N}+e^{\pi i(n+1)k/N}\right)\\
&=2e^{\pi ik/(2N)}
\sum_{n=0}^{N-1}x_n
\cos\left(\frac{\pi}{N}\left(n+\frac12\right)k\right).
\end{aligned}
$$
Consequently
$$
\boxed{\frac12e^{-\pi ik/(2N)}Y_k
=\sum_{n=0}^{N-1}x_n
\cos\left(\frac{\pi}{N}\left(n+\frac12\right)k\right)}
$$
for $0\leq k<2N$, as required.

Solved by gpt-5.6-sol high.

= c
{parent=41d}
{scope}

= Solution
{parent=c}

Constructing the reflected vector $y$ costs $O(N)$ assignments. Part (a) computes all of its Fourier coefficients in $O(N\log N)$ multiplications, and part (b) recovers the required coefficients $Z_0,\ldots,Z_{N-1}$ with one phase multiplication per coefficient. Hence the whole <discrete cosine transform> costs $O(N\log N)$ multiplications.

Solved by gpt-5.6-sol high.

= d
{parent=41d}
{scope}

= Solution
{parent=d}

Let $C_j(\xi)$ denote the $j$th discrete cosine coefficient of $\xi_n=(-1)^nx_n$. For $0\leq k<N$,
$$
\begin{aligned}
C_{N-1-k}(\xi)
&=\sum_{n=0}^{N-1}(-1)^nx_n
\cos\left(\pi\left(n+\frac12\right)
-\frac{\pi}{N}\left(n+\frac12\right)(k+1)\right)\\
&=\sum_{n=0}^{N-1}x_n
\sin\left(\frac{\pi}{N}\left(n+\frac12\right)(k+1)\right)
=\widetilde Z_k.
\end{aligned}
$$
The identity uses $\cos(\pi(n+1/2)-u)=(-1)^n\sin u$. Sign-modulating the input and reversing the $N$ cosine-transform outputs each cost $O(N)$ operations, so part (c) gives the sine transform in $O(N\log N)$ multiplications.

Solved by gpt-5.6-sol high.