past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/ii/paper-1.bigb

= Paper 1
{scope}

https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2026/PaperII_1%20updated.pdf

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

= Solution
{parent=1G}

A <primitive root> modulo $N$ is a unit of order $\varphi(N)$; a nonzero <quadratic residue> modulo an odd prime $p$ is a square.

If a primitive root $g$ modulo $p$ were $x^2$, then its order would divide $(p-1)/2$, the order of the <subgroup> of squares, contradicting $\operatorname{ord}_p(g)=p-1$.

Modulo $11$, $2$ has order $10$, so the primitive roots are $2^k$ for $\gcd(k,10)=1$:
$$
2,\quad 8,\quad 7,\quad 6.
$$
Finally $2^{10}=1024\equiv56\not\equiv1\pmod{121}$, so the lifting criterion makes $2$ primitive modulo $121$. It is plainly the smallest possible positive primitive root.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=2I}

The planar <Brouwer fixed-point theorem> says every continuous $F:D^2\to D^2$ has a fixed point. The equivalent <no-retraction theorem> says no continuous $r:D^2\to S^1$ restricts to the identity on $S^1$. A fixed-point-free map gives a retraction by extending the ray from $F(x)$ through $x$ to the boundary; a retraction followed by the antipodal map gives a fixed-point-free self-map.

Suppose the stated sphere map is not onto and omits $p$. Compactness gives $\varepsilon>0$ such that its image lies in $S^2\setminus B_\varepsilon(p)$. That set is homeomorphic to a closed disc. Restricting $f$ to it gives a continuous self-map of a disc, so Brouwer supplies a fixed point, contrary to the hypothesis. Hence $f$ is surjective.

Solved by gpt-5.6-sol high.

= 3H
{parent=Paper 1}
{scope}
{title2=Coding \& Cryptography}

= a
{parent=3h}
{scope}

= Solution
{parent=a}

Apply <Huffman coding>. Combining weights
$$
.02+.03=.05,\quad .04+.04=.08,\quad .05+.08=.13,\quad
.13+.14=.27,\quad .27+.28=.55
$$
gives, for example,
$$
x_1:0,\ x_2:10,\ x_3:110,\ x_4:11110,\ x_5:11111,\ x_6:11101,\ x_7:11100.
$$
Any consistent interchange of siblings is equivalent.

Solved by gpt-5.6-sol high.

= b
{parent=3h}
{scope}

= Solution
{parent=b}

For the binary code above, the lengths are $1,2,3,5,5,5,5$. Thus
$$
\bar\ell=.45+2(.28)+3(.14)+5(.04+.04+.03+.02)=2.08.
$$
This is optimal by <Huffman coding>.

Solved by gpt-5.6-sol high.

= c
{parent=3h}
{scope}

= Solution
{parent=c}

The ternary merges are $(.02,.03,.04)\mapsto.09$, then $(.04,.09,.14)\mapsto.27$, then $(.27,.28,.45)\mapsto1$. One code is
$$
x_1:0,\ x_2:1,\ x_3:20,\ x_4:21,\ x_5:220,\ x_6:221,\ x_7:222.
$$
Its expected length is
$$
.45+.28+2(.14+.04)+3(.04+.03+.02)=1.36.
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=4j}
{scope}

= Solution
{parent=a}

The basic <functions> are zero $Z(\mathbf x)=0$, successor $S(n)=n+1$, and projections $P_i^k(\mathbf x)=x_i$. Close these under composition and primitive recursion:
$$
f(0,\mathbf x)=g(\mathbf x),\qquad
f(n+1,\mathbf x)=h(n,f(n,\mathbf x),\mathbf x).
$$

Solved by gpt-5.6-sol high.

= b
{parent=4j}
{scope}

= Solution
{parent=b}

Define $A(0,m)=m$ and $A(n+1,m)=S(A(n,m))$. This is primitive recursion from a projection and successor, so $A(n,m)=n+m$ is <primitive recursive function>.

Solved by gpt-5.6-sol high.

= c
{parent=4j}
{scope}

= Solution
{parent=c}

If $f$ is primitive recursive, then $d(n)=f(P_1^1(n),P_1^1(n))$ is obtained by composition with projections. Hence the diagonal is primitive recursive.

Solved by gpt-5.6-sol high.

= d
{parent=4j}
{scope}

= Solution
{parent=d}

Assume a complete list $f$ were primitive recursive. Closure under diagonal composition and successor makes
$$
d(k)=f(k,k)+1
$$
primitive recursive. Completeness gives a row $n$ with $f(n,k)=d(k)$ for every $k$, but at $k=n$ this says $f(n,n)=f(n,n)+1$. This is the <diagonal argument>.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=5K}

The <normal linear model> is $Y=X\beta+\varepsilon$ with $\varepsilon\sim N(0,\sigma^2I_n)$. Full column rank gives
$$
\widehat\beta=(X^TX)^{-1}X^TY.
$$
With hat <matrix> $H=X(X^TX)^{-1}X^T$, the $i$th <regression leverage> is $h_{ii}$. Since $e=(I-H)Y=(I-H)\varepsilon$ and $I-H$ is an orthogonal projection,
$$
\operatorname{Var}(e_i)=\sigma^2(1-h_{ii}).
$$
High-leverage observations can move their own fitted values strongly and may exert disproportionate influence; residual size alone can therefore conceal them.

Solved by gpt-5.6-sol high.

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

= a
{parent=6c}
{scope}

= Solution
{parent=a}

The cross-infection terms transfer susceptible individuals into the infected class: $\beta S\widehat I$ is male incidence caused by infected females and $\widehat\beta\widehat S I$ is female incidence caused by infected males. The terms $\nu I$ and $\widehat\nu\widehat I$ are recovery flows back to susceptibility. Adding each susceptible-infected pair shows that $N$ and $\widehat N$ are conserved.

Solved by gpt-5.6-sol high.

= b
{parent=6c}
{scope}

= Solution
{parent=b}

Substitute $S=N-I$ and $\widehat S=\widehat N-\widehat I$:
$$
\dot I=\beta(N-I)\widehat I-\nu I,\qquad
\dot{\widehat I}=\widehat\beta(\widehat N-\widehat I)I-\widehat\nu\widehat I.
$$
Besides the origin, solve both nullcline equations. Eliminating one infected population gives
$$
I_*=\frac{\widehat NN-\widehat\nu\nu/(\widehat\beta\beta)}
{\nu/\beta+\widehat N},\qquad
\widehat I_*=\frac{\widehat NN-\widehat\nu\nu/(\widehat\beta\beta)}
{\widehat\nu/\widehat\beta+N}.
$$
The assumed <epidemic threshold> inequality makes both positive.

Solved by gpt-5.6-sol high.

= c
{parent=6c}
{scope}

= Solution
{parent=c}

At the origin the Jacobian is
$$
J_0=\begin{pmatrix}-\nu&\beta N\\ \widehat\beta\widehat N&-\widehat\nu\end{pmatrix},
\qquad
\det J_0=\nu\widehat\nu-\beta\widehat\beta N\widehat N<0.
$$
Thus it is a saddle and unstable. At the positive equilibrium the trace is negative. The product of the off-diagonal entries is $\nu\widehat\nu$ by the nullcline identities, while the product of the magnitudes of the diagonal entries is strictly larger, so the <determinant> is positive. Both <eigenvalues> therefore have negative real part and the endemic equilibrium is stable.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=7D}

The ordinary <integral> fails because the simple real poles at $-2$ and $3$ produce nonintegrable logarithmic singularities. Partial fractions give
$$
\frac1{x^2-x-6}=\frac15\left(\frac1{x-3}-\frac1{x+2}\right).
$$
Closing above for $\omega>0$ and below for $\omega<0$, with symmetric semicircular indentations, yields
$$
I(\omega)=\frac{i\pi}{5}\operatorname{sgn}(\omega)
\left(e^{3i\omega}-e^{-2i\omega}\right).
$$
Taking real and imaginary parts,
$$
\operatorname{PV}\int\frac{\cos\omega x}{x^2-x-6}\,dx
=-\frac{\pi}{5}\operatorname{sgn}(\omega)(\sin3\omega+\sin2\omega),
$$
$$
\operatorname{PV}\int\frac{\sin\omega x}{x^2-x-6}\,dx
=\frac{\pi}{5}\operatorname{sgn}(\omega)(\cos3\omega-\cos2\omega).
$$
All <integrals> here run over $\mathbb R$.

Solved by gpt-5.6-sol high.

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

= a
{parent=8a}
{scope}

= Solution
{parent=a}

Stationarity of the action under variations with fixed endpoints gives the Euler-Lagrange equation. The <generalized momentum> is $p=\partial L/\partial\dot q$. The <Legendre transform in mechanics> is
$$
H(q,p,t)=p\dot q-L(q,\dot q,t),
$$
after solving $p(q,\dot q,t)$ for $\dot q$. This requires at least local invertibility, $\partial^2L/\partial\dot q^2\ne0$; global single-valuedness requires global invertibility.

Solved by gpt-5.6-sol high.

= b
{parent=8a}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

By <Hamiltonian of a charged particle>,
$$
\mathbf p=m\dot{\mathbf r}+e\mathbf A,\qquad
H=\frac{|\mathbf p-e\mathbf A|^2}{2m}+e\phi.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

Here
$$
H=\frac{p_x^2+p_y^2+(p_z+eBx)^2}{2m}.
$$
Hence
$$
\dot x=p_x/m,\quad\dot y=p_y/m,\quad\dot z=(p_z+eBx)/m,
$$
$$
\dot p_x=-\frac{eB}{m}(p_z+eBx)=-eB\dot z,\qquad
\dot p_y=\dot p_z=0.
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=9d}
{scope}

= Solution
{parent=a}

Differentiate $\mathcal G=\sum_i\mathbf p_i\cdot\mathbf r_i$:
$$
\dot{\mathcal G}=\sum_i(\mathbf F_i\cdot\mathbf r_i+\mathbf p_i\cdot\dot{\mathbf r}_i)
=\sum_i\mathbf F_i\cdot\mathbf r_i+2T.
$$

Solved by gpt-5.6-sol high.

= b
{parent=9d}
{scope}

= Solution
{parent=b}

Pair the $ij$ and $ji$ terms and use $\mathbf f_{ji}=-\mathbf f_{ij}$:
$$
\sum_i\mathbf F_i\cdot\mathbf r_i
=\sum_i\sum_{j<i}\mathbf f_{ij}\cdot(\mathbf r_i-\mathbf r_j).
$$
For gravity, $\mathbf f_{ij}=-Gm^2(\mathbf r_i-\mathbf r_j)/|\mathbf r_i-\mathbf r_j|^3$, so each summand is $-Gm^2/|\mathbf r_i-\mathbf r_j|$. Their sum is $V$.

Solved by gpt-5.6-sol high.

= c
{parent=9d}
{scope}

= Solution
{parent=c}

For a confined system, $\mathcal G(t)$ stays bounded, so its long-time averaged <derivative> is zero. The preceding identities give the <virial theorem>
$$
2\langle T\rangle=-\langle V\rangle.
$$

Solved by gpt-5.6-sol high.

= d
{parent=9d}
{scope}

= Solution
{parent=d}

Estimate $T\sim M\langle v^2\rangle/2$ and $V\sim-GM^2/\langle R\rangle$. The <virial theorem> gives
$$
M\sim\frac{\langle R\rangle\langle v^2\rangle}{G}
$$
up to a profile-dependent factor. If this exceeds the stellar/gas luminous mass, the difference is attributed primarily to dark matter (with modelling uncertainties and non-equilibrium effects checked).

Solved by gpt-5.6-sol high.

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

= a
{parent=10e}
{scope}

= Solution
{parent=a}

The <quantum no-signalling> principle says Alice's uncommunicated local choice cannot alter Bob's reduced state. Expanding the two tensor products cancels the cross terms and gives
$$
|\mu\mu\rangle+|\nu\nu\rangle=|00\rangle+|11\rangle.
$$
Alice obtains either result with probability $1/2$; conditionally Bob has probability $\cos^2\theta$ or $\sin^2\theta$ of $0$, respectively. Averaging,
$$
P_B(0)=\tfrac12(\cos^2\theta+\sin^2\theta)=\tfrac12,\qquad P_B(1)=\tfrac12,
$$
independent of $\theta$, exactly as no-signalling requires.

Solved by gpt-5.6-sol high.

= b
{parent=10e}
{scope}

= Solution
{parent=b}

The <Helstrom-Holevo bound> for equal priors is
$$
P_{\max}=\frac12\left(1+\sqrt{1-|\langle\alpha_0|\alpha_1\rangle|^2}\right).
$$
Perfect cloning twice changes overlap $c$ to $c^2$, so discrimination would succeed with $\frac12(1+\sqrt{1-c^4})$, strictly exceeding $\frac12(1+\sqrt{1-c^2})$ for $0<c<1$. For $|0\rangle,|+\rangle$, $c=1/\sqrt2$, producing $\frac12(1+\sqrt3/2)>\frac12(1+1/\sqrt2)$, a contradiction.

Solved by gpt-5.6-sol high.

= 11H
{parent=Paper 1}
{scope}
{title2=Coding \& Cryptography}

= a
{parent=11h}
{scope}

= Solution
{parent=a}

The parity-check <matrix> has rank $3$, so its kernel has dimension $7-3=4$ and size $16$; see <linear code>. Its columns are precisely the seven distinct nonzero <vectors> of $\mathbb F_2^3$. Thus no weight-one or weight-two word is in the kernel, while three dependent columns give a weight-three word. Hence $d=3$.

Solved by gpt-5.6-sol high.

= b
{parent=11h}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

By <parity extension>, each code keeps size $16$, gains length $8$, and has distance $4$. Extension is injective. A common extended word punctures to a word in $H\cap K$, so only the zero and all-one words occur; both extend to $00000000$ and $11111111$.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

Write the three blocks as $u=a+x$, $v=b+x$, $w=a+b+x$. Over $\mathbb F_2$,
$$
x=u+v+w,\qquad a=u+x,\qquad b=v+x.
$$
Thus the parameters are unique and $|G|=|H'|^2|K'|=16^3=2^{12}$.

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

Every nonzero word has weight divisible by $4$. Weight $4$ is impossible: using $u+v=a+b$, $u+w=b$, $v+w=a$, and $u+v+w=x$, the distance-four properties force any nonzero one of these sums to have weight $4$; coordinate counting in total weight $4$ then forces two blocks to vanish, which would put a nontrivial word in $H'\cap K'$. That intersection contains only $0$ and $\mathbf1$, and $\mathbf1$ gives block weight at least $8$. Hence $d(G)\ge8$, and taking $a\ne0,b=x=0$ gives weight $2\operatorname{wt}(a)=8$, so $d(G)=8$.

Solved by gpt-5.6-sol high.

= iv
{parent=b}
{scope}

= Solution
{parent=iv}

Puncturing gives a binary code of length $23$, size $2^{12}$, and distance at least $7$. Radius-three balls are disjoint and
$$
2^{12}\sum_{j=0}^3\binom{23}{j}
=2^{12}(1+23+253+1771)=2^{23}.
$$
The <Hamming bound> is an equality, so $G^-$ is perfect.

Solved by gpt-5.6-sol high.

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

= a
{parent=12i}
{scope}

= Solution
{parent=a}

A variable-based grammar is $(N,\Sigma,P,S)$ with disjoint nonterminals and terminals, productions between words over $N\cup\Sigma$, and start variable $S$. Its language is $\{w\in\Sigma^*:S\Rightarrow^*w\}$.

Solved by gpt-5.6-sol high.

= b
{parent=12i}
{scope}

= Solution
{parent=b}

A type-2 rule is $A\to\alpha$ with one nonterminal $A$ on the left; a type-2 grammar and its language are <context-free language>s. The required repetition criterion is the <pumping lemma for context-free languages>.

Solved by gpt-5.6-sol high.

= c
{parent=12i}
{scope}

= i
{parent=c}
{scope}

= Solution
{parent=i}

This language is context-free. One grammar is
$$
S\to aSc\mid aTc,\qquad T\to bT\mid b,
$$
which independently chooses at least one outer $a,c$ pair and at least one $b$.

Solved by gpt-5.6-sol high.

= ii
{parent=c}
{scope}

= Solution
{parent=ii}

It is not context-free. Intersecting it with the regular language $a^*b^*c^*$ would give $\{a^nb^nc^n:n\ge0\}$, which violates the <pumping lemma for context-free languages>. Context-free languages are closed under intersection with regular languages.

Solved by gpt-5.6-sol high.

= d
{parent=12i}
{scope}

= Solution
{parent=d}

No production introduces $A$, so the two rules containing $A$ are unreachable. The effective derivations are
$$
S\Rightarrow a^{n}bC(BC)^{n-1}\Rightarrow a^nb^nC^n\Rightarrow a^nb^nc^n.
$$
Thus the language is $\{a^nb^nc^n:n\ge1\}$, which is not <context-free language>, although the displayed grammar itself also visibly has non-type-2 left sides.

Solved by gpt-5.6-sol high.

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

= a
{parent=13k}
{scope}

= Solution
{parent=a}

For one observation,
$$
P_\lambda(Y=y)=\frac1{y!}\exp\{y\log\lambda-\lambda\}.
$$
Thus this is an <exponential family> with natural parameter $\theta=\log\lambda$, statistic $T(y)=y$, base measure $1/y!$, and cumulant $A(\theta)=e^\theta$.

Solved by gpt-5.6-sol high.

= b
{parent=13k}
{scope}

= Solution
{parent=b}

For nonnegative $y_i$ summing to $m$, divide the joint Poisson mass by $P(M=m)=e^{-\lambda_+}\lambda_+^m/m!$:
$$
P(\mathbf Y=\mathbf y\mid M=m)=\frac{m!}{\prod y_i!}\prod_i
\left(\frac{\lambda_i}{\lambda_+}\right)^{y_i}.
$$
This is <independent Poisson conditioning>.

Solved by gpt-5.6-sol high.

= c
{parent=13k}
{scope}

= Solution
{parent=c}

In the <log-linear model> fit1 has only three main effects, so its fitted mean factors into a species factor, a diameter factor, and a height factor: mutual independence. Fit2 adds the species-diameter interaction, allowing those variables to associate, while asserting $\text{Height}\perp(\text{Species},\text{Diameter})$.

The likelihood-ratio deviance drop $12.606$ on one degree of freedom has $p=0.0003845$, strong evidence against species-diameter independence. Test fit2's remaining independence claim by comparing it with a model adding both Species:Height and Diameter:Height interactions (equivalently compare fit2 with the saturated model); its residual deviance $12.431$ is assessed against $\chi^2_3$.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=14D}

Put $\Gamma(z)=I(z)$ for $\Re z>0$. Integration by parts gives $\Gamma(z+1)=z\Gamma(z)$, both proving $\Gamma(n)=(n-1)!$ and defining the meromorphic continuation. A keyhole-contour evaluation of $\int_0^\infty t^{z-1}/(1+t)\,dt$ gives
$$
\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin\pi z}.
$$

For
$$
\Gamma(z)\Gamma(z+\tfrac12)=e^{g(z)}\Gamma(2z),
$$
the reflection formula at $z$ and $z+\tfrac12$ gives $g(z+1)-g(z)=-2\log2$. Thus $h(z)=g(z)+2z\log2$ is entire and $1$-periodic; poles and zeros have cancelled in the quotient. Its Fourier expansion and the exponential growth bound make it a finite Laurent <polynomial> in $e^{2\pi iz}$; the same growth bound for $e^h$ excludes every nonconstant mode. Hence $h=b$. At $z=1/2$, $e^b=2\sqrt\pi$, yielding the duplication formula.

Set $z=1/6$ and use reflection at $1/3$:
$$
\Gamma(\tfrac16)\Gamma(\tfrac23)=2^{2/3}\sqrt\pi\,\Gamma(\tfrac13),\qquad
\Gamma(\tfrac13)\Gamma(\tfrac23)=\frac{2\pi}{\sqrt3}.
$$
Therefore
$$
\Gamma(\tfrac16)=2^{-1/3}\sqrt{\frac3\pi}\,\Gamma(\tfrac13)^2.
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=15d}
{scope}

= Solution
{parent=a}

The conservation equation with $P=w\rho$ gives $d\rho/\rho=-3(1+w)\,da/a$, hence
$$
\rho=\rho_0a^{-3(1+w)}.
$$
The critical density and density parameter are
$$
\rho_{\rm crit}=\frac{3c^2H^2}{8\pi G},\qquad \Omega_i=\frac{\rho_i}{\rho_{\rm crit}}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=15d}
{scope}

= Solution
{parent=b}

Radiation has $w=1/3$, matter $w=0$, and a cosmological constant has constant density. With negligible curvature and present-day normalization,
$$
H^2=H_0^2\left(\frac{\Omega_{R0}}{a^4}+
\frac{\Omega_{M0}}{a^3}+\Omega_{\Lambda0}\right),
$$
where the present density fractions sum to one.

Solved by gpt-5.6-sol high.

= c
{parent=15d}
{scope}

= Solution
{parent=c}

<Phantom energy> with $w=-4/3$ has $\rho_P\propto a$, so flatness gives
$$
\dot a=H_0\sqrt\Omega\,a\sqrt{1+\beta^2a},\qquad
\beta^2=\frac{1-\Omega}{\Omega}.
$$
For $b^2=1+\beta^2a$, integration yields
$$
\log\frac{b-1}{b+1}=H_0\sqrt\Omega(t-t_0)+\log\kappa,\qquad
\kappa=\frac{1-\sqrt\Omega}{1+\sqrt\Omega}.
$$
Solving gives the displayed scale factor
$$
a(t)=\frac{4\Omega e^{H_0\sqrt\Omega(t-t_0)}}
{\left[(1+\sqrt\Omega)-(1-\sqrt\Omega)e^{H_0\sqrt\Omega(t-t_0)}\right]^2}.
$$
For $1-\Omega\ll1$ and times near $t_0$, this approaches de Sitter growth $e^{H_0\sqrt\Omega(t-t_0)}$. The denominator vanishes at
$$
t_{\rm BR}-t_0=\frac1{H_0\sqrt\Omega}
\log\frac{1+\sqrt\Omega}{1-\sqrt\Omega}.
$$
Even a tiny phantom fraction therefore ends expansion in finite time; its divergent density ultimately disrupts loosely bound systems.

Solved by gpt-5.6-sol high.

= 16J
{parent=Paper 1}
{scope}
{title2=Logic \& Set Theory}

= a
{parent=16j}
{scope}

= Solution
{parent=a}

A choice <function> on a family $\mathcal A$ of nonempty sets satisfies $c(A)\in A$. Zermelo's theorem says every set has a total order under which every nonempty subset has a least element.

Solved by gpt-5.6-sol high.

= b
{parent=16j}
{scope}

= Solution
{parent=b}

Choose from every nonempty subset of $X$. By transfinite recursion set
$$
x_\alpha=c\!\left(X\setminus\{x_\beta:\beta<\alpha\}\right)
$$
while the remainder is nonempty. This cannot continue through <Hartogs theorem>'s ordinal $h(X)$, since that would inject $h(X)$ into $X$. It therefore exhausts $X$ at some ordinal, and the selection order well-orders $X$.

Solved by gpt-5.6-sol high.

= c
{parent=16j}
{scope}

= Solution
{parent=c}

Hessenberg's theorem in ZF states that every infinite well-orderable cardinal satisfies $\kappa\times\kappa=\kappa$; equivalently $|\alpha\times\alpha|=|\alpha|$ for every infinite ordinal $\alpha$.

Solved by gpt-5.6-sol high.

= d
{parent=16j}
{scope}

= Solution
{parent=d}

Choice well-orders every infinite $X$, so Hessenberg gives $X^2\cong X$. Conversely assume this square identity for every infinite set. For arbitrary $X$, form the disjoint union $Y=X\sqcup h(X)$. A pairing $Y^2\cong Y$, together with the well-order on $h(X)$, yields by the standard tracing/comparison argument an injection either $h(X)\to X$ or $X\to h(X)$. The first is forbidden by <Hartogs theorem>, so the second well-orders $X$. This is <Tarski cardinal-square theorem>, and hence choice follows.

Solved by gpt-5.6-sol high.

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

= a
{parent=17j}
{scope}

= Solution
{parent=a}

A matching from $X$ to $Y$ consists of disjoint edges saturating every vertex of $X$. The criterion is <Hall marriage theorem>.

For sufficiency, induct on $|X|$. If a nonempty proper $S\subset X$ is tight, $|N(S)|=|S|$, apply induction to $S,N(S)$ and to the graph left after deleting them. If no proper set is tight, choose any edge $xy$, delete its endpoints, and Hall still holds because every nonempty proper subset formerly had at least one spare neighbour. Induction completes the matching. Necessity is immediate.

Solved by gpt-5.6-sol high.

= b
{parent=17j}
{scope}

= Solution
{parent=b}

For $S\subseteq X$, count with reciprocal degrees. On every edge $xy$, $d(x)\ge d(y)$, so
$$
|S|=\sum_{x\in S}\sum_{y\sim x}\frac1{d(x)}
\le\sum_{y\in N(S)}\sum_{\substack{x\in S\\x\sim y}}\frac1{d(y)}
\le|N(S)|.
$$
Hall's condition holds, so <Hall marriage theorem> supplies the matching.

Solved by gpt-5.6-sol high.

= c
{parent=17j}
{scope}

= Solution
{parent=c}

Assume no deletion works. For each $x\in A_i$, some $A_j$ contains $A_i\setminus\{x\}$; otherwise deleting $x$ preserves the <antichain>. For different deleted elements these witnesses are distinct, since one witness containing both complements would contain $A_i$. Thus every element of $A_i$ belongs to at least $|A_i|$ sets.

Use the incidence graph with left class $\bigcup_iA_i$ and right class $\{A_i\}$. Along every incidence edge, the degree of the element is at least the degree $|A_i|$ of the set vertex. Part (b) gives a matching from the union into the $k$ sets, forcing $|\bigcup_iA_i|\le k$, a contradiction.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=18F}

Orbits of the root action are precisely roots belonging to the same irreducible factor, proving transitivity iff irreducibility.

For a same-degree counterexample, let $L=\mathbb Q(\sqrt2,\sqrt3)$,
$$
f=(X^2-2)(X^2-3),\qquad
g=\min_{\mathbb Q}(\sqrt2+\sqrt3)=X^4-10X^2+1.
$$
Both split in $L$, but the first action has two orbits of size two while the second is transitive, so the <subgroups> are not conjugate in $S_4$.

Over a finite field, <Frobenius cycle type> proves the assertion. The given quartic reduces over $\mathbb F_2$ to $X^4+X^3+1$, which is irreducible, so its <group> there is cyclic of order four generated by a $4$-cycle.

Over $\mathbb Q$ the <polynomial> is irreducible by the same reduction. Modulo $3$ it factors as linear times irreducible cubic, so <Dedekind factorization cycle test> gives a $3$-cycle; modulo $2$ it gives a $4$-cycle. Its discriminant is $229$, and modulo $23$ its factor degrees are $1,1,2$, giving a transposition. These cycle types force the transitive <group> to be $S_4$.

Solved by gpt-5.6-sol high.

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

= a
{parent=19f}
{scope}

= Solution
{parent=a}

Let $\phi=(1+\sqrt5)/2$ and $\bar\phi=(1-\sqrt5)/2$. In class order $(1,2A,3A,5A,5B)$, <character orthogonality> gives
$$
\begin{array}{c|rrrrr}
&1&2A&3A&5A&5B\\
1&1&1&1&1&1\\
3&3&-1&0&\phi&\bar\phi\\
3'&3&-1&0&\bar\phi&\phi\\
4&4&0&1&-1&-1\\
5&5&1&-1&0&0
\end{array}
$$
with class sizes $1,15,20,12,12$. The rows are orthonormal and their squared degrees sum to $60$, so the table is complete.

A three-dimensional representation of $S_5$ would decompose into irreducibles. Since $S_5$ has irreducible degrees $1,1,4,4,5,5,6$, only one-dimensional constituents could occur, whose restrictions to $A_5$ are trivial. It cannot restrict to either irreducible $3$.

Solved by gpt-5.6-sol high.

= b
{parent=19f}
{scope}

= Solution
{parent=b}

Its character is the number of fixed points, and averaging it gives the number of orbits. Complementation $\Sigma\mapsto X\setminus\Sigma$ is a $G$-equivariant bijection $\Gamma_n\to\Gamma_{d-n}$.

The exterior-power assertion is false. For $n=2$ and $g=(12)(34)\in A_5$, $g$ fixes two $2$-subsets, so $\chi_{\mathbb C[\Gamma_2]}(g)=2$. The natural permutation character has $\chi(g)=1$ and $\chi(g^2)=5$, hence
$$
\chi_{\wedge^2\mathbb C[X]}(g)=\tfrac12(\chi(g)^2-\chi(g^2))=-2.
$$
The representations are not isomorphic.

Solved by gpt-5.6-sol high.

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

= a
{parent=20g}
{scope}

= Solution
{parent=a}

The answer is <ring of integers of a quadratic field>:
$$
\{1,(1+\sqrt d)/2\},\ D_K=d\quad(d\equiv1\pmod4),
$$
and otherwise $\{1,\sqrt d\},\ D_K=4d$.

Solved by gpt-5.6-sol high.

= b
{parent=20g}
{scope}

= Solution
{parent=b}

The trace/norm integrality conditions give the stated congruences for
$\theta=(a+bi+c\sqrt m+di\sqrt m)/2$. Solving them yields these <integral> <bases>:
$$
m\equiv1\pmod4:\quad
1,\ i,\ \frac{1+\sqrt m}{2},\ \frac{i+i\sqrt m}{2},
$$
$$
m\equiv3\pmod4:\quad
1,\ i,\ \frac{i+\sqrt m}{2},\ \frac{1+i\sqrt m}{2}.
$$
The three quadratic subfields have fundamental discriminants $-4,D_m,D_{-m}$. By <discriminant of a biquadratic field>,
$$
D_L=(-4)D_mD_{-m}=16m^2
$$
in both residue classes.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=21H}

Its homology and chain formulas are
$$
L(f)=\sum_i(-1)^i\operatorname{tr}(f_*|H_i(K;\mathbb Q))
=\sum_i(-1)^i\operatorname{tr}(f_\#|C_i(K;\mathbb Q)).
$$
For <Lefschetz fixed-point theorem>, argue contrapositively: after barycentric subdivision, a fixed-point-free map has a simplicial approximation such that no simplex meets its image. Every diagonal coefficient on chains is then zero, so $L(f)=0$.

The antipodal map on $S^n$ has degree $(-1)^{n+1}$. For even $n$ this is $-1$, whereas the identity has degree $1$, so they are not homotopic.

If the tangent field $v$ were nowhere zero, then
$$
H(x,t)=\cos(\pi t)x+\sin(\pi t)\frac{v(x)}{|v(x)|}
$$
would stay on the sphere because $x\cdot v(x)=0$, and would homotope the identity to the antipodal map. This is impossible for even $n$, so $v$ has a zero.

Solved by gpt-5.6-sol high.

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

= a
{parent=22i}
{scope}

= Solution
{parent=a}

The open mapping lemma says: if a bounded <linear map> between Banach spaces satisfies $\overline{T(B_X)}\supset rB_Y$, then $T(B_X)$ contains a (possibly smaller) ball. Given $y$, repeatedly choose corrections $x_n$ whose residuals shrink geometrically; completeness makes $\sum x_n$ converge and maps it to $y$. Baire category applied to $Y=\bigcup_nT(nB_X)$ supplies the closure hypothesis for a surjective map. This proves the <Open mapping theorem (functional analysis)>.

Solved by gpt-5.6-sol high.

= b
{parent=22i}
{scope}

= Solution
{parent=b}

If $\|Tx\|\ge\delta\|x\|$ and $Tx_n\to y$, then $x_n$ is Cauchy, so $x_n\to x$ and $y=Tx$; the range is closed. Conversely a closed range is Banach, and $T:X\to T(X)$ has bounded inverse by the <Open mapping theorem (functional analysis)>, giving the lower bound. This is the <closed-range criterion for an injection>.

Solved by gpt-5.6-sol high.

= c
{parent=22i}
{scope}

= i
{parent=c}
{scope}

= Solution
{parent=i}

If $\|Tx\|\ge\delta\|x\|$ and $\|S-T\|<\delta$, then
$\|Sx\|\ge(\delta-\|S-T\|)\|x\|$. Thus bounded-below maps form an <open set>.

Solved by gpt-5.6-sol high.

= ii
{parent=c}
{scope}

= Solution
{parent=ii}

Surjectivity and <Open mapping theorem (functional analysis)> give $rB_Y\subset T(B_X)$. If $\|S-T\|<r$, iteratively solve with $T$ and correct the error $(S-T)x$; the errors shrink geometrically. Completeness yields an exact preimage under $S$.

Solved by gpt-5.6-sol high.

= iii
{parent=c}
{scope}

= Solution
{parent=iii}

For an isomorphism $T$, if $\|T^{-1}(S-T)\|<1$, then
$$
S=T[I+T^{-1}(S-T)]
$$
is invertible by <neumann-series perturbation>.

Solved by gpt-5.6-sol high.

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

= a
{parent=23h}
{scope}

= Solution
{parent=a}

With the convention in <convolution theorem>, absolute integrability permits Fubini:
$$
\widehat{f*g}(\xi)=\int\!\int f(y)g(x-y)e^{-2\pi ix\cdot\xi}\,dy\,dx
=\widehat f(\xi)\widehat g(\xi).
$$

Solved by gpt-5.6-sol high.

= b
{parent=23h}
{scope}

= Solution
{parent=b}

For Schwartz $f,g$, Fourier inversion and Fubini give
$$
\int\widehat f(\xi)\overline{\widehat g(\xi)}\,d\xi
=\int f(x)\overline{g(x)}\,dx.
$$
Taking $g=f$ proves <plancherel theorem>; polarization gives the full identity.

Solved by gpt-5.6-sol high.

= c
{parent=23h}
{scope}

= Solution
{parent=c}

Approximate $g\in L^2$ by Schwartz $g_n$. Young's inequality gives
$\|f*(g_n-g)\|_2\le\|f\|_1\|g_n-g\|_2$, while multiplication by bounded $\widehat f$ is continuous in $L^2$. Pass to the <limit> in
$\widehat{f*g_n}=\widehat f\,\widehat g_n$.

Solved by gpt-5.6-sol high.

= d
{parent=23h}
{scope}

= i
{parent=d}
{scope}

= Solution
{parent=i}

If $g\in L^2$ is orthogonal to every translate of $f$, then the correlation $g*\widetilde f$ vanishes. By <convolution theorem>,
$\widehat g\,\overline{\widehat f}=0$ almost everywhere. Since $\widehat f$ vanishes only at one point, a null set, $\widehat g=0$ almost everywhere. By <plancherel theorem>, $g=0$. Orthogonal-complement characterization proves density in $L^2$.

Solved by gpt-5.6-sol high.

= ii
{parent=d}
{scope}

= Solution
{parent=ii}

The nonzero bounded functional
$$
\Lambda(h)=\widehat h(1)=\int h(x)e^{-2\pi ix}\,dx
$$
is continuous on $L^1$. For every translate, $\Lambda(f(\cdot-t))=e^{-2\pi it}\widehat f(1)=0$. Thus it annihilates $Y$, and <density by annihilators> shows that $Y$ is not dense in $L^1$.

Solved by gpt-5.6-sol high.

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

= a
{parent=24g}
{scope}

= Solution
{parent=a}

A covering evenly covers every base point. It is <regular covering map> when deck transformations act transitively on fibres. For a concrete nonregular example, restrict $z\mapsto z^3-3z$ to
$$
\mathbb C\setminus p^{-1}\{-2,2\}\longrightarrow\mathbb C\setminus\{-2,2\}.
$$
It is a surjective three-sheeted unbranched cover with monodromy $S_3$, so its deck <group> is not transitive.

Solved by gpt-5.6-sol high.

= b
{parent=24g}
{scope}

= Solution
{parent=b}

$(g,V)$ continues $(f,U)$ when they agree on a nonempty connected component reached by an overlapping chain of <function> elements. A <complete analytic function> contains every continuation. Its Riemann surface is the set of its germs; germ neighbourhoods are charts, projection records the base point, and germ evaluation gives a holomorphic map to $\mathbb C$.

Solved by gpt-5.6-sol high.

= c
{parent=24g}
{scope}

= i
{parent=c}
{scope}

= Solution
{parent=i}

Since $V$ is simply connected, $g$ has a primitive $G_0$. On connected $U\cap V$, $G_0-F$ has zero <derivative> and is constant. Adjust that constant to obtain $G=F$ on the overlap, hence the required continuation.

Solved by gpt-5.6-sol high.

= ii
{parent=c}
{scope}

= Solution
{parent=ii}

Differentiate a germ of a continuation of $F$ to obtain a germ of the continuation of $f$. This defines a surjective cover $R_{\widetilde F}\to R_F$. Two primitive germs above the same <derivative> germ differ by an additive period. Translation by periods gives all deck transformations and acts transitively on each fibre, so the cover is regular.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=25F}

In the affine chart $z=1$, the equation is $y^2=x^2(x+1)$, irreducible because $x+1$ is not a square in $k(x)$. The <partial derivatives> show the unique singular point is $P=[0:0:1]$; the point at infinity is smooth.

A line $y=tx$ through $P$ meets the curve again at
$$
x=t^2-1,\qquad y=t(t^2-1).
$$
Thus $t\mapsto[t^2-1:t(t^2-1):1]$ is birational, with inverse $t=y/x$ away from $P$. The incidence closure has one point over every smooth $Q$ and two points over $P$, corresponding to $t=\pm1$. This is the <normalization of a nodal curve> and separates the two branches.

Dominated by projective space does not imply being projective space. A smooth quadric surface $Y\simeq\mathbb P^1\times\mathbb P^1$ is rational and hence dominated by $\mathbb P^2$, but is not any $\mathbb P^k$; for example its Picard rank is $2$, not $1$.

Solved by gpt-5.6-sol high.

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

= a
{parent=26i}
{scope}

= Solution
{parent=a}

Curvature requires a $C^2$ regular curve; torsion requires $C^3$ and nonzero curvature. The parameter-independent formulas are
$$
\kappa=\frac{|\gamma'\times\gamma''|}{|\gamma'|^3},\qquad
\tau=\frac{\det(\gamma',\gamma'',d^3\gamma/dt^3)}{|\gamma'\times\gamma''|^2}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=26i}
{scope}

= Solution
{parent=b}

For unit speed on an oriented surface,
$$
T'=\kappa_n n+\kappa_g(n\times T).
$$
These coefficients are normal and <geodesic curvature>; the curve is geodesic iff $\kappa_g=0$.

Solved by gpt-5.6-sol high.

= c
{parent=26i}
{scope}

= Solution
{parent=c}

A <line segment> has zero normal curvature in its tangent direction, impossible when Gaussian curvature is positive because both principal curvatures have the same nonzero sign. Two disjoint closed geodesics would bound an annulus. Applying <Gauss-Bonnet theorem>, its geodesic boundary contributes zero and $\chi=0$, but $\int K\,dA>0$, a contradiction.

Solved by gpt-5.6-sol high.

= d
{parent=26i}
{scope}

= Solution
{parent=d}

If a regular $C^3$ extension existed at either finite endpoint, reparametrization by arc length would extend the Frenet data. Because $\kappa\ge\varepsilon$, the denominator in the torsion formula stays away from zero; compactness of a short closed parameter interval would then bound both $\kappa$ and $|\tau|$. This contradicts the stated limsup at that endpoint. Applying the argument at both ends proves nonextendibility.

Solved by gpt-5.6-sol high.

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

= a
{parent=27l}
{scope}

= Solution
{parent=a}

The statements and proof are in <Borel-Cantelli lemmas>. For the first,
$\mathbb P(\bigcup_{n\ge m}A_n)\le\sum_{n\ge m}\mathbb P(A_n)\to0$. For independent events in the divergent case,
$$
\mathbb P(\text{none of }A_m,\ldots,A_n)
=\prod_{j=m}^n(1-p_j)\le e^{-\sum_{j=m}^np_j}\to0,
$$
and then let $m\to\infty$.

Solved by gpt-5.6-sol high.

= b
{parent=27l}
{scope}

= Solution
{parent=b}

The characteristic <function> of $S_n/n$ is
$$
\Phi(\xi/n)^n=e^{-n|\xi|/n}=e^{-|\xi|}.
$$
Thus every $S_n/n$ already has the standard Cauchy law, and convergence in law follows from <characteristic-function convergence theorem>.

Solved by gpt-5.6-sol high.

= c
{parent=27l}
{scope}

= Solution
{parent=c}

If $S_n/n$ converged almost surely to a finite <limit>, then
$$
\frac{X_n}{n}=\frac{S_n}{n}-\frac{n-1}{n}\frac{S_{n-1}}{n-1}\longrightarrow0.
$$
But $\mathbb P(X_n/n>1)=\mathbb P(X_1>n)\sim1/(\pi n)$. These independent-event probabilities have divergent sum, so the second <Borel-Cantelli lemmas> says $X_n/n>1$ infinitely often almost surely, a contradiction.

Solved by gpt-5.6-sol high.

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

= a
{parent=28k}
{scope}

= Solution
{parent=a}

For ordered times $0<t_1<\cdots<t_n<t$, the joint density together with $N_t=n$ is $e^{-\lambda t}\lambda^n$. Dividing by $P(N_t=n)=e^{-\lambda t}(\lambda t)^n/n!$ gives $n!/t^n$, exactly the density of uniform order statistics. See <Poisson process>.

Solved by gpt-5.6-sol high.

= b
{parent=28k}
{scope}

= Solution
{parent=b}

Conditionally on $T=t$, $N_T$ is Poisson with mean $\lambda t$. Mixing over the exponential density gives
$$
P(N_T=n)=\int_0^\infty e^{-\lambda t}\frac{(\lambda t)^n}{n!}\nu e^{-\nu t}\,dt
=\frac{\nu\lambda^n}{(\lambda+\nu)^{n+1}}.
$$
Thus it is geometric on $\{0,1,\ldots\}$ with success probability $\nu/(\lambda+\nu)$.

Solved by gpt-5.6-sol high.

= c
{parent=28k}
{scope}

= Solution
{parent=c}

An arrival at $s\le t$ remains active with probability $1-(t-s)$ when $0\le t-s\le1$, and zero later. By <Poisson thinning>, the detector count is Poisson with mean
$$
m(t)=\lambda\int_{\max(0,t-1)}^t[1-(t-s)]\,ds
=\begin{cases}\lambda(t-t^2/2),&0\le t\le1,\\ \lambda/2,&t\ge1.\end{cases}
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=29l}
{scope}

= Solution
{parent=a}

The <maximum-likelihood estimator> is any measurable maximizer
$\widehat\theta_n\in\arg\max_\theta\prod_{i=1}^nf(X_i;\theta)$.

Solved by gpt-5.6-sol high.

= b
{parent=29l}
{scope}

= Solution
{parent=b}

Under regularity,
$$
\sqrt n(\widehat\theta_n-\theta)\Rightarrow N(0,I(\theta)^{-1}),
\qquad I(\theta)=E_\theta[(\partial_\theta\log f(X;\theta))^2].
$$

Solved by gpt-5.6-sol high.

= c
{parent=29l}
{scope}

= Solution
{parent=c}

The likelihood is
$$
e^{-\sum_iX_i+n\theta}\mathbf1_{\{\theta\le X_{(1)}\}},
$$
so $\widehat\theta_n=X_{(1)}$. Regular asymptotics fail because the support $[\theta,\infty)$ depends on $\theta$ and the likelihood maximum lies on its boundary.

Solved by gpt-5.6-sol high.

= d
{parent=29l}
{scope}

= Solution
{parent=d}

Write $X_i=\theta+E_i$ with independent rate-one exponentials. Their minimum is exponential of rate $n$, so
$$
n(\widehat\theta_n-\theta)\sim\operatorname{Exp}(1)
$$
for every $n$. Thus $\alpha=1$ and the nondegenerate <limit> is rate-one exponential.

Solved by gpt-5.6-sol high.

= e
{parent=29l}
{scope}

= Solution
{parent=e}

The MLE error is rate-$n$ exponential, hence
$$
\operatorname{MSE}(\widehat\theta_n)=E(\widehat\theta_n-\theta)^2=\frac2{n^2}.
$$
Since $\widetilde\theta_n=\bar X_n-1$ is unbiased with variance $1/n$,
$$
\frac{\operatorname{MSE}(\widehat\theta_n)}
{\operatorname{MSE}(\widetilde\theta_n)}=\frac2n\to0.
$$
The boundary-based MLE is asymptotically much more efficient.

Solved by gpt-5.6-sol high.

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

= a
{parent=30l}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

Use the optimizer for $Z_0$ also with $Z_1$. Strict increase of $U$ and a strict inequality on a positive-probability event give $u(Z_1)>u(Z_0)$.

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

If $X_i$ optimizes $Z_i$, then $pX_1+(1-p)X_0$ is attainable because the wealth set is affine. Concavity gives
$$
u(pZ_1+(1-p)Z_0)\ge E\,U(p(X_1+Z_1)+(1-p)(X_0+Z_0))
\ge pu(Z_1)+(1-p)u(Z_0).
$$

Solved by gpt-5.6-sol high.

= b
{parent=30l}
{scope}

= Solution
{parent=b}

The time-zero indifference price is the unique $\pi(Y)$ satisfying
$$
u(Y-(1+r)\pi(Y))=u(0).
$$
The left side is continuous and strictly decreasing in the cash amount by part (a), and its limiting values bracket $u(0)$; existence follows by the intermediate-value theorem and uniqueness by strict monotonicity.

Solved by gpt-5.6-sol high.

= c
{parent=30l}
{scope}

= Solution
{parent=c}

Differentiate expected utility at the optimum $X_0$ in each portfolio direction:
$$
0=E[U'(X_0)(S_1-(1+r)S_0)].
$$
Rearranging gives the claimed <vector> identity.

Solved by gpt-5.6-sol high.

= d
{parent=30l}
{scope}

= Solution
{parent=d}

The marginal utility price is
$$
\pi_0(Y)=\frac{E[U'(X_0)Y]}{(1+r)E[U'(X_0)]}.
$$
The value <function> is concave; its tangent at zero lies above its graph. Evaluating that supporting inequality at the utility-indifference displacement $Y-(1+r)\pi(Y)$ gives
$E[U'(X_0)(Y-(1+r)\pi(Y))]\ge0$, hence $\pi_0(Y)\ge\pi(Y)$.

Solved by gpt-5.6-sol high.

= e
{parent=30l}
{scope}

= Solution
{parent=e}

By <Gaussian conditional expectation>,
$$
E[Y\mid S_1]=E(Y)+\operatorname{cov}(Y,S_1)V^{-1}(S_1-\mu).
$$
Multiply by $U'(X_0)$, take expectations, and use part (c):
$$
\frac{E[U'(X_0)S_1]}{E[U'(X_0)]}=(1+r)S_0.
$$
Substitution into part (d) gives
$$
\pi_0(Y)=\frac1{1+r}\left[E(Y)-(\mu-(1+r)S_0)^TV^{-1}
\operatorname{cov}(S_1,Y)\right].
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=31k}
{scope}

= Solution
{parent=a}

For points $x_1,\ldots,x_n$, count the distinct <vectors> $(h(x_1),\ldots,h(x_n))$; maximize over samples to get $s(H,n)$. The <shattering coefficient> equals $2^n$ exactly when some sample is shattered, and the largest such $n$ is $\operatorname{VC}(H)$.

Solved by gpt-5.6-sol high.

= b
{parent=31k}
{scope}

= Solution
{parent=b}

Order the $n$ sample points. Each of the $m-1$ cut points can occupy one of at most $n+1$ gaps. Once those gaps are chosen the fixed $\beta_k$ determine every sign, so
$$
s(H_{F_1},n)\le(n+1)^{m-1}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=31k}
{scope}

= Solution
{parent=c}

For each coordinate, the $m$ interval indicators sum to one, so the span of all displayed indicators has dimension at most
$$
1+p(m-1).
$$
Therefore <VC dimension of a vector space> gives
$\operatorname{VC}(H_{F_2})\le1+p(m-1)$.

Solved by gpt-5.6-sol high.

= d
{parent=31k}
{scope}

= Solution
{parent=d}

On an $n$-sample, each coordinate partition has at most $(n+1)^{m-1}$ placements, hence all $p$ partitions have at most $(n+1)^{p(m-1)}$ configurations. For each fixed configuration, the sign class from a <vector space> of dimension $1+p(m-1)$ has at most $(n+1)^{1+p(m-1)}$ patterns. Multiplication gives
$$
s(H_{F_3},n)\le(n+1)^{1+2p(m-1)}.
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=32b}
{scope}

= Solution
{parent=a}

The required properties are those of a <strict Lyapunov function>; the domain of stability is the set of initial states whose forward trajectories converge to the fixed point.

For $V=(x^2+y^2)/2$,
$$
\dot V=(x^2+y^2)(-2+x^2+4y^2).
$$
Thus $\dot V<0$ when $V<1/4$, proving that disc lies in the basin. On $V=1$, $\dot V=6y^2\ge0$ (with the tangency points checked from the <vector> field), so trajectories outside cannot enter the basin through that boundary. Hence the basin is contained in $V\le1$.

Solved by gpt-5.6-sol high.

= b
{parent=32b}
{scope}

= Solution
{parent=b}

The fixed points are $(0,0)$ and $(\pm|\mu|,\mu^2)$. At the origin the <eigenvalues> are $\mu^2$ and $-1$, so it is a saddle. At either nonzero point the characteristic <polynomial> is
$$
\lambda^2+\lambda+2\mu^2.
$$
They are stable nodes for $0<|\mu|<1/(2\sqrt2)$, degenerate nodes at equality, and stable spirals above it.

The stable manifold of the origin is exactly $x=0$. Write the <stable manifold> in the unstable direction as $y=ax^2+bx^4+O(x^6)$. Graph invariance gives
$$
a=\frac1{1+2\mu^2},\qquad
b=\frac{2a^2}{1+4\mu^2}.
$$
Thus $W^u$ has
$$
y=\frac{x^2}{1+2\mu^2}
+\frac{2x^4}{(1+2\mu^2)^2(1+4\mu^2)}+O(x^6).
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=33b}
{scope}

= Solution
{parent=a}

Hamilton's equations are $\dot x_i=p_i$ and $\dot p_i=-\partial_iV$. A first <integral> $F$ satisfies $\{F,H\}=0$. See <integrable Hamiltonian system>: in three degrees of freedom, integrability requires three independent commuting first <integrals>.

Solved by gpt-5.6-sol high.

= b
{parent=33b}
{scope}

= Solution
{parent=b}

The conclusion is <Arnold-Liouville theorem>: a compact connected regular common level is a three-torus, and nearby action-angle coordinates make the flow linear in the angles.

Solved by gpt-5.6-sol high.

= c
{parent=33b}
{scope}

= Solution
{parent=c}

Rotational invariance gives $\{m_y,H\}=\{m_z,H\}=0$. Direct calculation gives
$$
m_x=\{m_y,m_z\}=yp_z-zp_y.
$$
Take $G=m_x^2+m_y^2+m_z^2$; angular-momentum brackets imply $\{G,m_x\}=0$, and central symmetry gives $\{G,H\}=0$. Hence $(H,G,m_x)$ are in involution.

At $\mathbf x=(1,0,0)$, $m_y=-p_z$, $m_z=p_y$. The differentials of $G$ and $m_x$ have independent components in the $p_y,p_z$ and transverse position directions when $p_y,p_z\ne0$; $dH$ supplies an independent radial/<momentum> component. Thus they are locally independent and the system is Liouville integrable.

Solved by gpt-5.6-sol high.

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

= a
{parent=34a}
{scope}

= Solution
{parent=a}

$\Delta_\psi Q=\sqrt{\langle Q^2\rangle-\langle Q\rangle^2}$. The <Robertson uncertainty principle> is
$$
\Delta A\,\Delta B\ge\frac12|\langle[A,B]\rangle|.
$$

Solved by gpt-5.6-sol high.

= b
{parent=34a}
{scope}

= Solution
{parent=b}

The displacement operator is unitary, so $\||\alpha\rangle\|=1$. By <coherent state>, $A|\alpha\rangle=\alpha|\alpha\rangle$, hence $\|A|\alpha\rangle\|=|\alpha|$.

Solved by gpt-5.6-sol high.

= c
{parent=34a}
{scope}

= Solution
{parent=c}

Using $X=(A+A^\dagger)/\sqrt{2\omega}$ and $H=\omega(A^\dagger A+1/2)$,
$$
\langle X\rangle=\sqrt{\frac2\omega}\Re\alpha,\qquad
\Delta X=\frac1{\sqrt{2\omega}},
$$
$$
\langle H\rangle=\omega(|\alpha|^2+\tfrac12),\qquad
\Delta H=\omega|\alpha|.
$$

Solved by gpt-5.6-sol high.

= d
{parent=34a}
{scope}

= Solution
{parent=d}

Since $[X,H]=iP$ and $\langle P\rangle=\sqrt{2\omega}\Im\alpha$, the bound is
$\Delta X\Delta H\ge\sqrt{\omega/2}|\Im\alpha|$. The left side is $\sqrt{\omega/2}|\alpha|$, so equality holds exactly when $\Re\alpha=0$.

Solved by gpt-5.6-sol high.

= e
{parent=34a}
{scope}

= Solution
{parent=e}

<Coherent state> evolution gives
$$
|\alpha(t)\rangle=e^{-i\omega t/2}|\alpha e^{-i\omega t}\rangle.
$$
The condition from part (d) becomes
$$
\Re(\alpha e^{-i\omega t})=0.
$$

Solved by gpt-5.6-sol high.

= 35E
{parent=Paper 1}
{scope}
{title2=Applications of Quantum Mechanics}

= a
{parent=35e}
{scope}

= Solution
{parent=a}

Insert the incident plane wave for $\psi$ in the <integral>. At large $r$, $|\mathbf r-\mathbf r'|\simeq r-\widehat{\mathbf r}\cdot\mathbf r'$, so the outgoing amplitude is
$$
f(\mathbf k',\mathbf k)=-\frac{m}{2\pi\hbar^2}\widetilde V(\mathbf q),
\quad
\widetilde V(\mathbf q)=\int e^{-i\mathbf q\cdot\mathbf r}V(\mathbf r)\,d^3r,
\quad\mathbf q=\mathbf k'-\mathbf k.
$$
Thus <Born approximation> gives the stated differential cross section.

Solved by gpt-5.6-sol high.

= b
{parent=35e}
{scope}

= Solution
{parent=b}

For $V(r)\sim r^{-n}$, the small-$q$ transform scales as $q^{n-3}$. Since $d\Omega$ contributes $q\,dq$ near forward scattering, the total Born cross section behaves as $\int_0 q^{2n-5}\,dq$, finite exactly for $n>2$. Coulomb decay $n=1$ gives the familiar divergent forward Rutherford cross section and requires long-range treatment.

Solved by gpt-5.6-sol high.

= c
{parent=35e}
{scope}

= Solution
{parent=c}

The Gaussian transform is
$$
\widetilde V(\mathbf q)=(2\pi)^{3/2}V_0r_0^3e^{-q^2r_0^2/2}.
$$
Therefore
$$
\frac{d\sigma}{d\Omega}
=\frac{2\pi m^2V_0^2r_0^6}{\hbar^4}e^{-q^2r_0^2},
\qquad q=2k\sin(\vartheta/2).
$$
A sufficient weak-scattering regime is <kinetic energy> much larger than $|V_0|$ (more intrinsically $m|V_0|r_0^2/\hbar^2\ll1$).

Solved by gpt-5.6-sol high.

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

= a
{parent=36e}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

At equal <pressure> and <temperature>, <ideal> gases have equal molecular number density. The mean air molecular mass is $.8(28)+.2(32)=28.8$, while helium has mass $4$, so
$$
\rho_{\rm He}/\rho_{\rm air}=4/28.8=5/36.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

Monatomic helium has molar $C_p=5R/2$ and diatomic air has $7R/2$. Dividing by molar masses,
$$
\frac{c_p^{\rm He}}{c_p^{\rm air}}
=\frac{(5/2)R/4}{(7/2)R/28.8}=\frac{36}{7}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=36e}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

The <ideal gas free energy> contributes $-Nk_BT\log V$. With $V=4\pi r^3/3$ and surface energy $4\pi tr^2$, the radius-dependent free energy is therefore
$$
F(r)=-Nk_BT\log V+4\pi tr^2+\text{constant}.
$$
For equilibrium against external <pressure> $P_0$, minimize $F+P_0V$.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

<Differentiation> gives
$$
-\frac{3Nk_BT}{r}+8\pi tr+4\pi P_0r^2=0,
$$
equivalently the Laplace-pressure condition
$$
P_{\rm in}=\frac{Nk_BT}{V}=P_0+\frac{2t}{r}.
$$

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

Let $r_0=(3Nk_BT/(4\pi P_0))^{1/3}$ be the zero-tension radius. From
$P_0r^3+2tr^2=P_0r_0^3$, put $r=r_0+\delta$ and retain first order:
$$
3P_0r_0^2\delta+2tr_0^2=0,\qquad
r=r_0-\frac{2t}{3P_0}+O(t^2).
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=37d}
{scope}

= Solution
{parent=a}

With $u^\mu=\gamma(c,\mathbf v)$ and $d/d\tau=\gamma d/dt$, the spatial components of the <relativistic Lorentz force> give
$$
\frac{d(\gamma m\mathbf v)}{dt}=q(\mathbf E+\mathbf v\times\mathbf B).
$$
The time component gives $mc^2\,d\gamma/dt=q\mathbf E\cdot\mathbf v$. It says <magnetic fields> do no work and electric power changes relativistic energy.

Solved by gpt-5.6-sol high.

= b
{parent=37d}
{scope}

= Solution
{parent=b}

For a wave travelling in $+x$,
$$
\mathbf B=\frac1c\widehat{\mathbf x}\times\mathbf E
=\frac{E_0}{c}(0,-\sin\omega\xi,\cos\omega\xi).
$$
At $x=0$, choose
$$
\mathbf v=\frac{qE_0}{\gamma m\omega}(0,\sin\omega t,-\cos\omega t).
$$
It is parallel to $-\mathbf B$, so the magnetic force vanishes, while its <derivative> supplies $q\mathbf E/(\gamma m)$. Constant speed then gives
$$
\gamma^2=1+\left(\frac{qE_0}{m\omega c}\right)^2.
$$
The circular radius is
$$
R=\frac{|q|E_0}{\gamma m\omega^2}.
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=38a}
{scope}

= Solution
{parent=a}

Spherical symmetry puts any geodesic in a plane through the origin, chosen as $\theta=\pi/2$. The cyclic coordinates give
$$
E=(1-r_s/r)\dot t,\qquad L=r^2\dot\phi.
$$
The null condition gives
$$
\dot r^2=F(r)=E^2-(1-r_s/r)\frac{L^2}{r^2}.
$$
For $u=1/r$, $(du/d\phi)^2=\dot r^2/L^2$, hence
$$
\left(\frac{du}{d\phi}\right)^2=\frac1{b^2}-u^2+r_su^3,\qquad b=L/E.
$$

Solved by gpt-5.6-sol high.

= b
{parent=38a}
{scope}

= Solution
{parent=b}

<Differentiation> gives $u''+u=\frac32r_su^2$. With $u=b^{-1}(v_0+\varepsilon v_1+\cdots)$,
$$
v_0''+v_0=0,\qquad v_1''+v_1=\frac32v_0^2.
$$
The symmetry selects $v_0=\sin\phi$, the straight line at closest distance $b$. Since
$\frac32\sin^2\phi=\frac34(1-\cos2\phi)$,
$$
v_1=\frac34+\frac14\cos2\phi.
$$
The two asymptotic zeros shift from $0,\pi$ to $-\varepsilon,\pi+\varepsilon$, so the deflection is $2\varepsilon=2r_s/b=4GM/(bc^2)$, the <Schwarzschild light deflection>. It is tested by stellar deflection near the Sun and gravitational lensing.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=39C}

Boundary-layer theory assumes steady incompressible high-Reynolds-number flow, $\delta/x\ll1$, weak streamwise viscous diffusion, nearly constant <pressure> across the layer, and outer <pressure> <gradient> $-\rho^{-1}p_x=U\,U'$.

For $\phi=-A r^k\cos(k\theta)/k$,
$$
u_r=-Ar^{k-1}\cos k\theta,\qquad u_\theta=Ar^{k-1}\sin k\theta.
$$
At $\theta=\pm\pi/k$ the normal <velocity> vanishes; along the upper wall $U(x)=Ax^{k-1}$. <Boundary-layer scaling> gives
$$
\delta=\sqrt{\frac{\nu x}{U}}=(\nu x^{2-k}/A)^{1/2}.
$$
Set $m=k-1$. The <Falkner-Skan equation> becomes
$$
f^{(3)}=(k-1)f'^2-\frac{k}{2}ff''+(1-k),
$$
so $\alpha=k-1$, $\beta=-k/2$, $\gamma=1-k$, with
$f(0)=f'(0)=0$ and $f'(\infty)=1$. Finally
$$
\frac{\delta}{x}=(\nu/A)^{1/2}x^{-k/2}\ll1
$$
requires $x\gg(\nu/A)^{1/k}$; thus $a=1/k$ and the tip region violates the approximation.

Solved by gpt-5.6-sol high.

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

= a
{parent=40c}
{scope}

= Solution
{parent=a}

Use a <velocity potential> proportional to $e^{ik(x-Vt)+ik\beta_1y}$, where
$\beta_1=\sqrt{(V/c_1)^2-1}$. The wall condition gives amplitude $-V\epsilon/\beta_1$. With $p'=-\rho_1\partial_t\phi$, <acoustic energy flux> gives
$$
\langle I_y\rangle=\frac{\rho_1k^2V^3\epsilon^2}{2\beta_1}.
$$
For $V<c_1$ the vertical wave number is imaginary: the disturbance is evanescent and carries no mean vertical energy flux.

Solved by gpt-5.6-sol high.

= b
{parent=40c}
{scope}

= Solution
{parent=b}

Let $H=k\beta_0h$, $\beta_0=\sqrt{1-V^2/c_0^2}$, and
$\beta_1=\sqrt{V^2/c_1^2-1}$. Match <pressure> and normal <velocity> between an evanescent lower potential and an outgoing upper wave. The transmitted amplitude has
$$
|A_1|^2=\frac{V^2\epsilon^2}
{\beta_1^2\cosh^2H+(\rho_1/\rho_0)^2\beta_0^2\sinh^2H}.
$$
Therefore
$$
\langle I_y\rangle=
\frac{\rho_1k^2V^3\epsilon^2\beta_1}
{2[\beta_1^2\cosh^2H+(\rho_1/\rho_0)^2\beta_0^2\sinh^2H]}.
$$
For $H\gg1$ this is
$$
\frac{2\rho_1k^2V^3\epsilon^2\beta_1e^{-2H}}
{\beta_1^2+(\rho_1/\rho_0)^2\beta_0^2},
$$
showing tunnelling suppression. For $H\ll1$ it tends to
$\rho_1k^2V^3\epsilon^2/(2\beta_1)$, as if the upper medium met the wall.

Solved by gpt-5.6-sol high.

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

= a
{parent=41b}
{scope}

= Solution
{parent=a}

The <power method> repeatedly normalizes $Aq^{(k)}$ and estimates the <eigenvalue> with the Rayleigh quotient $(q^{(k)})^TAq^{(k)}$. It converges when the <dominant eigenvalue> is unique in <modulus> and the initial <vector> has a nonzero component in its eigendirection.

Solved by gpt-5.6-sol high.

= b
{parent=41b}
{scope}

= Solution
{parent=b}

The <real spectral theorem> gives real <eigenvalues> and orthogonal distinct eigenspaces. Constructively, a <Householder reflection> sends an <eigenvector> to $e_1$; symmetry makes the remaining block symmetric. Induction diagonalizes that block, producing orthogonal $Q$ with $Q^TAQ=\Lambda$. Multiplying by $Q$ gives $AQe_k=\lambda_kQe_k$.

Solved by gpt-5.6-sol high.

= c
{parent=41b}
{scope}

= Solution
{parent=c}

Write $q^{(0)}=\sum_i c_iq_i$ with $c_1\ne0$. After $k$ multiplications, the tangent of the angle to $q_1$ is at most
$$
\frac{\sqrt{\sum_{i\ge2}c_i^2|\lambda_i|^{2k}}}
{|c_1||\lambda_1|^k}
\le\left|\frac{\lambda_2}{\lambda_1}\right|^k\tan\theta_0.
$$
Since $|\sin\theta_k|\le\tan\theta_k$, the first bound follows. Also
$$
\lambda^{(k)}-\lambda_1
=\sum_{i\ge2}(\lambda_i-\lambda_1)|q_i^Tq^{(k)}|^2.
$$
Bound the coefficient by $\max_{i\ge2}|\lambda_i-\lambda_1|$ and the sum by $\sin^2\theta_k$, then use the squared first estimate to obtain the second bound.

Solved by gpt-5.6-sol high.