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past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/ii/paper-3.bigb
= Paper 3
{scope}
https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2026/PaperII_3.pdf
= 1G
{parent=Paper 3}
{scope}
{title2=Number Theory}
= Solution
{parent=1G}
For $n=\prod p_i^{e_i}$ odd, define $(a/n)=\prod(a/p_i)^{e_i}$. Multiplying <quadratic reciprocity> for all prime-factor pairs gives
$$\left(\frac an\right)\left(\frac na\right)=(-1)^{((a-1)/2)((n-1)/2)}$$
for positive odd coprime $a,n$; the sign exponents add modulo two, proving Jacobi reciprocity from Legendre reciprocity. Since $959=7\cdot137$, $(5/7)=-1$ and $(5/137)=(2/5)=-1$, so $(5/959)=1$. Nevertheless $5$ is a nonresidue modulo both prime factors, so $x^2\equiv5\pmod{959}$ has no solution.
Solved by gpt-5.6-sol high.
= 2I
{parent=Paper 3}
{scope}
{title2=Topics in Analysis}
= Solution
{parent=2I}
A subset is dense when every nonempty <open set> meets it, equivalently its closure is the whole space. <Baire category theorem> says a complete metric space has dense intersection for every countable family of open dense sets. Here
$$S=\bigcap_{N\ge1}\bigcup_{j\ge N}A_j,$$
so Baire makes $S$ dense. It need not be open: enumerate the rationals $q_j$ and take $A_j=[0,1]\setminus\{q_1,\ldots,q_j\}$, giving the irrationals. Nor need $\cap_jA_j$ be nonempty: take $A_1=\varnothing$ and $A_j=[0,1]$ for $j\ge2$.
Solved by gpt-5.6-sol high.
= 3H
{parent=Paper 3}
{scope}
{title2=Coding \& Cryptography}
= Solution
{parent=3H}
A cyclic code is a linear subspace invariant under cyclic coordinate shift. Identifying words with $\mathbb F_2[X]/(X^n-1)$ makes such codes exactly <ideals>. Since this quotient is a principal <ideal> <ring>, every code is generated uniquely by a monic divisor of $X^n-1$. For $n=7$,
$$X^7-1=(X+1)(X^3+X+1)(X^3+X^2+1),$$
so its three distinct irreducible factors have $2^3=8$ monic divisors and there are eight cyclic codes.
Solved by gpt-5.6-sol high.
= 4I
{parent=Paper 3}
{scope}
{title2=Automata and Formal Languages}
= a
{parent=4i}
{scope}
= Solution
{parent=a}
A DFA consists of finite $Q$, alphabet $\Sigma$, transition $\delta$, initial $q_0$, and accepting set $F$. Extend by $\hat\delta(q,\epsilon)=q$ and $\hat\delta(q,wa)=\delta(\hat\delta(q,w),a)$. It accepts $\{w:\hat\delta(q_0,w)\in F\}$; a language is regular if some DFA accepts it.
Solved by gpt-5.6-sol high.
= b
{parent=4i}
{scope}
= Solution
{parent=b}
The <pumping lemma for regular languages> says that for some $N$, every accepted word $w$ of length at least $N$ factors $w=xyz$ with $|xy|\le N$, $|y|>0$, and $xy^kz$ accepted for every $k\ge0$.
Solved by gpt-5.6-sol high.
= c
{parent=4i}
{scope}
= i
{parent=c}
{scope}
= Solution
{parent=i}
Regular: a DFA need only remember the current run length of $b$'s, using states $0,\ldots,99$ and a rejecting sink at 100.
Solved by gpt-5.6-sol high.
= ii
{parent=c}
{scope}
= Solution
{parent=ii}
Since $w_i=(ab)^{2^{i-1}}$, the language is $\{(ab)^{2^j}:j\ge0\}$. If it were regular, pumping a sufficiently long member would alter its number of $ab$ blocks by a fixed positive amount, producing non-powers of two for some pumping exponent. Hence it is not regular.
Solved by gpt-5.6-sol high.
= 5K
{parent=Paper 3}
{scope}
{title2=Statistical Modelling}
= Solution
{parent=5K}
The residual-versus-fitted <smooth curve> bends downward, indicating nonlinearity, and the spread increases at high fitted values. The Q-Q plot is close centrally but departs in both tails, with observations 12, 17, and 67 notable. The scale-location trend confirms heteroscedasticity. The leverage plot identifies observations 1 and 4 as high leverage and 67 as a large residual, with potentially material Cook distance. The additive Gaussian linear model is therefore doubtful. Inspect those observations, transform sale price (often logarithmically), add nonlinear or interaction terms and omitted predictors such as lot size and age, and use weighted or robust regression if unequal variance remains.
Solved by gpt-5.6-sol high.
= 6C
{parent=Paper 3}
{scope}
{title2=Mathematical Biology}
= a
{parent=6c}
{scope}
= Solution
{parent=a}
The reaction is $F(n)=-n(n-A)(n-B)$. Its homogeneous equilibria are $0,A,B$; $F'(0)<0$ and $F'(B)<0$, while $F'(A)>0$. Thus the stable equilibria are $0$ and $B$.
Solved by gpt-5.6-sol high.
= b
{parent=6c}
{scope}
= Solution
{parent=b}
Substitution $n=f(\xi)$, $\xi=x-ct$, gives $f''+cf'+F(f)=0$ with $f(-\infty)=0$, $f(\infty)=B$. Multiplying by $f'$ gives $-c\int(f')^2d\xi=\int_0^BF(s)ds=B^4(1-2A/B)/12$. The populated state invades the empty state when $c<0$, equivalently $B>2A$; this is the condition preventing eventual extinction.
Solved by gpt-5.6-sol high.
= 7D
{parent=Paper 3}
{scope}
{title2=Further Complex Methods}
= Solution
{parent=7D}
A point $z_0$ is regular singular when $(z-z_0)p$ and $(z-z_0)^2q$ are analytic there. The Papperitz symbol records the three singular points and their two local exponents. Here the finite points are $0,1$ and the third is $\infty$. The indicial equations give exponents $(0,-1)$ at $0$ and $(0,0)$ at $1$. Comparing with the hypergeometric equation gives $c=2$, $a+b=2$, $ab=1$, hence $a=b=1$ and exponents $(1,1)$ at infinity. Thus
$$P\left\{\begin{matrix}0&1&\infty\\0&0&1\\-1&0&1\end{matrix};z\right\}.$$
Solved by gpt-5.6-sol high.
= 8A
{parent=Paper 3}
{scope}
{title2=Classical Dynamics}
= Solution
{parent=8A}
For the standard symplectic form, $D_f=\sum_a(f_{p_a}\partial_{q^a}-f_{q^a}\partial_{p_a})$; a transformation is canonical when it preserves the symplectic form, equivalently Poisson brackets. The Jacobi identity gives $[D_f,D_g]=D_{\{f,g\}}$ with the stated convention. For $Z^A=e^{-sD_f}z^A$, differentiate $\{Z^A,Z^B\}$ in $s$; Jacobi makes the <derivative> the Lie transport of the bracket, preserving the canonical constant brackets at $s=0$. Hence every map in the one-parameter flow is canonical.
Solved by gpt-5.6-sol high.
= 9D
{parent=Paper 3}
{scope}
{title2=Cosmology}
= a
{parent=9d}
{scope}
= Solution
{parent=a}
Substitution gives $\dot\phi=2M_{\rm Pl}/t$, $\ddot\phi=-2M_{\rm Pl}/t^2$ and $V=V_*e^{-\phi_*/M_{\rm Pl}}(t_*/t)^2=10M_{\rm Pl}^2/t^2$, exactly satisfying both equations when $\phi_*=M_{\rm Pl}\log(V_*t_*^2/(10M_{\rm Pl}^2))$. Since $a\propto t^2$, $\ddot a>0$: it inflates. $N=2\log(t_*/t)$ and $\Delta\phi=2M_{\rm Pl}\log(t_*/t)=NM_{\rm Pl}$, so 60 e-folds require a field excursion about $60M_{\rm Pl}$.
Solved by gpt-5.6-sol high.
= b
{parent=9d}
{scope}
= Solution
{parent=b}
In a decelerating universe the comoving Hubble radius grows, so regions now in causal contact extrapolate to many disconnected early Hubble patches. Inflation shrinks the comoving Hubble radius and stretches one causal patch beyond the present horizon. Radiation scaling gives a present-horizon physical size at $10^{-35}$ s of roughly $ct_0(a/a_0)\sim ct_0(t/t_0)^{1/2}\sim1$ m. Comparing this with the inflationary Hubble length (about $ct\sim10^{-27}$ m at that epoch) requires $e^N\sim10^{27}$, or $N\simeq62$ (order 60).
Solved by gpt-5.6-sol high.
= 10E
{parent=Paper 3}
{scope}
{title2=Quantum Information and Computation}
= i
{parent=10e}
{scope}
= Solution
{parent=i}
The all-zero amplitude is $2^{-n}\sum_x(-1)^{f(x)}=1-2k/2^n$, so its probability is $(1-2k/2^n)^2$. For a general nontrivial $k$ it lies strictly between zero and one.
Solved by gpt-5.6-sol high.
= ii
{parent=10e}
{scope}
= Solution
{parent=ii}
For constant $f$, $k=0$ or $2^n$, so the all-zero result occurs with probability one.
Solved by gpt-5.6-sol high.
= iii
{parent=10e}
{scope}
= Solution
{parent=iii}
For perfect balance, $k=2^{n-1}$ and the all-zero probability is zero. For other balances the one-shot outcomes overlap the constant case. Repeat <Deutsch-Jozsa algorithm> and declare nonconstant upon any nonzero result. Constants are never misclassified; a $k$-balanced <function> fails after $R$ trials with probability $q^R$, $q=(1-2k/2^n)^2$. It suffices that $R\ge\log\epsilon/\log q$, using that many oracle calls.
Solved by gpt-5.6-sol high.
= 11G
{parent=Paper 3}
{scope}
{title2=Number Theory}
= a
{parent=11g}
{scope}
= Solution
{parent=a}
With $p_{-1}=1,p_0=a_0$, $q_{-1}=0,q_0=1$ and $p_n=a_np_{n-1}+p_{n-2}$, $q_n=a_nq_{n-1}+q_{n-2}$, induction gives $p_nq_{n-1}-p_{n-1}q_n=(-1)^{n+1}$. <Matrix> multiplication gives $[a_0,\ldots,a_n,\beta]=(p_n\beta+p_{n-1})/(q_n\beta+q_{n-1})$; monotonicity in $\beta>0$ places it strictly between the adjacent convergents.
Solved by gpt-5.6-sol high.
= b
{parent=11g}
{scope}
= Solution
{parent=b}
Apply the formula to the complete quotient and use the <determinant> identity to express $b_n$. Eliminating the complete quotient from the consecutive expressions yields the stated quadratic for $q_{n+1}/q_n$. Taking the compatible signs for adjacent indices and using $c_n\le b_j$ gives
$$2c_na_{n+1}\le2b_na_{n+1}\le\sqrt{1-4b_nb_{n+1}}+\sqrt{1-4b_{n-1}b_n}\le2\sqrt{1-4c_n^2}.$$
Solved by gpt-5.6-sol high.
= c
{parent=11g}
{scope}
= Solution
{parent=c}
If all but finitely many convergents had $b_n>1/\sqrt5$, then eventually $c_n>1/\sqrt5$. Part (b) would imply $c_na_{n+1}\le\sqrt{1-4c_n^2}<c_n$, impossible since $a_{n+1}\ge1$. Thus infinitely many $b_n\le1/\sqrt5$, which is exactly $|\theta-p_n/q_n|\le1/(\sqrt5q_n^2)$.
Solved by gpt-5.6-sol high.
= 12J
{parent=Paper 3}
{scope}
{title2=Automata and Formal Languages}
= a
{parent=12j}
{scope}
= Solution
{parent=a}
A register machine has finitely many natural-number registers and labelled increment, conditional decrement/jump, and halt instructions. Starting with the input in designated registers, instructions determine a partial computation. Encode the finite instruction list by a standard effective Gödel coding, giving each machine an index in $\mathbb N$.
Solved by gpt-5.6-sol high.
= b
{parent=12j}
{scope}
= Solution
{parent=b}
For machine index $m$, $f_{m,k}$ is the partial $k$-ary <function> computed by that machine under the fixed coding, and $W_m$ is its halting domain (equivalently the r.e. set enumerated by machine $m$, under the course convention).
Solved by gpt-5.6-sol high.
= c
{parent=12j}
{scope}
= Solution
{parent=c}
Kleene's <recursion theorem> says that for every total recursive map $f$ on program indices there is an index $e$ with the same partial <function> as program $f(e)$.
Solved by gpt-5.6-sol high.
= d
{parent=12j}
{scope}
= Solution
{parent=d}
Take $i$ to index a machine halting on every input, so $W_i=\mathbb N$, and $j$ to index a machine halting on no input, so $W_j=\varnothing$.
Solved by gpt-5.6-sol high.
= e
{parent=12j}
{scope}
= Solution
{parent=e}
If $g_0$ were total recursive, its recursion-theorem fixed point $e$ would satisfy $W_e=W_j$ when $e\in\mathrm{Inf}$ and $W_e=W_i$ when $e\in\mathrm{Fin}$, a contradiction in either case. The same argument applies to $g_1$. A total recursive reduction $h$ with $n\in\mathrm{Inf}\iff h(n)\in\mathrm{Fin}$ would provide such a forbidden switch (and would collapse the known distinct arithmetical complexities), so none exists.
Solved by gpt-5.6-sol high.
= 13E
{parent=Paper 3}
{scope}
{title2=Mathematical Biology}
= a
{parent=13e}
{scope}
= Solution
{parent=a}
For jump rate $W(n,r)$ from $n$ to $n+r$, probability enters state $n$ from $n-r$ and leaves it from $n$. The resulting <kramers-Moyal master equation> is
$$
\partial_tP(n,t)=\sum_r
[P(n-r,t)W(n-r,r)-P(n,t)W(n,r)].
$$
Solved by gpt-5.6-sol high.
= b
{parent=13e}
{scope}
= Solution
{parent=b}
When jumps are small relative to a large population and moments exist, the <Kramers-Moyal expansion> through second order gives
$$
P_t=-\partial_x(uP)+\partial_x^2(DP),
\qquad
u=\sum_r rW(x,r),
\qquad
D=\frac12\sum_r r^2W(x,r).
$$
This <Fokker-Planck equation> is a conservation law with vanishing boundary flux; $u$ is the drift and $D$ the diffusivity. Equal constant rates for jumps $r=\mathord\pm1$ give pure diffusion.
Solved by gpt-5.6-sol high.
= c
{parent=13e}
{scope}
= Solution
{parent=c}
Integration by parts gives $d\langle x\rangle/dt=\langle u\rangle$ and $d\operatorname{var}x/dt=2\langle D\rangle+2\operatorname{Cov}(x,u)$.
Solved by gpt-5.6-sol high.
= d
{parent=13e}
{scope}
= Solution
{parent=d}
Here $u=-2\sinh(\lambda x)$ and $D=\cosh(\lambda x)$. Zero stationary flux gives $(DP)'=uP$, so $P\propto D^{-1}\exp\int u/D,dx\propto\cosh(\lambda x)^{-(1+2/\lambda)}$, a symmetric unimodal distribution.
Solved by gpt-5.6-sol high.
= e
{parent=13e}
{scope}
= i
{parent=e}
{scope}
= Solution
{parent=i}
For constants $u=a-b$ and $D=(a+b)/2$, $z=x-ut$ removes drift and gives $\widetilde P_t=D\widetilde P_{zz}$.
Solved by gpt-5.6-sol high.
= ii
{parent=e}
{scope}
= Solution
{parent=ii}
At the earliest negative minimum, $\widetilde P=0$, $\widetilde P_z=0$, $\widetilde P_{zz}\ge0$, so the diffusion equation gives $\widetilde P_t\ge0$, contradicting first passage below zero. This is the maximum principle.
Solved by gpt-5.6-sol high.
= iii
{parent=e}
{scope}
= Solution
{parent=iii}
The third <Kramers-Moyal expansion> coefficient is $F=\tfrac16\sum_rr^3W=(a-b)/6$. At a first minimum the third <derivative> has no fixed sign, so the maximum-principle argument fails; finite truncations beyond second order need not preserve positivity.
Solved by gpt-5.6-sol high.
= 14D
{parent=Paper 3}
{scope}
{title2=Cosmology}
= a
{parent=14d}
{scope}
= i
{parent=a}
{scope}
= Solution
{parent=i}
Writing $r=a(q+\psi)$ to first order gives $x''+\mathcal Hx'\simeq\psi''+\mathcal H\psi'=-\nabla_q\varphi$. The perturbation of Poisson's equation is $\nabla_q^2\varphi=(4\pi G/c^2)a^2\bar\rho_M\delta_M$.
Solved by gpt-5.6-sol high.
= ii
{parent=a}
{scope}
= Solution
{parent=ii}
The Jacobian is $a^3\det(I+\partial\psi/\partial q)\simeq a^3(1+\nabla\cdot\psi)$, so mass conservation gives $\delta_M=-\nabla\cdot\psi$. Taking the divergence of (i) and using Poisson yields $\delta_M''+\mathcal H\delta_M'-(4\pi G/c^2)\bar\rho_Ma^2\delta_M=0$.
Solved by gpt-5.6-sol high.
= b
{parent=14d}
{scope}
= Solution
{parent=b}
Set $\delta_R=e^{-S}u$ and neglect $S''$, $(S')^2$, and slow variation relative to $k^2$. Cancelling the first <derivative> gives $S'=\alpha k/(2\mathcal H)$, hence $S=\int^\tau\alpha k/(2\mathcal H)d\tau'$, and $u''+(c^2k^2/3)u\simeq0$. Thus acoustic oscillations are exponentially damped, unlike pressureless matter perturbations, which have a growing mode.
Solved by gpt-5.6-sol high.
= 15E
{parent=Paper 3}
{scope}
{title2=Quantum Information and Computation}
= a
{parent=15e}
{scope}
= Solution
{parent=a}
Across the cut $A|BC$, the two conditional $BC$ <vectors> $\alpha|00\rangle$ and $\gamma|00\rangle+\beta|11\rangle$ are linearly independent when $\alpha\beta\ne0$, so the Schmidt rank exceeds one. Hence the state cannot be a three-qubit product for any $\gamma$.
Solved by gpt-5.6-sol high.
= b
{parent=15e}
{scope}
= i
{parent=b}
{scope}
= Solution
{parent=i}
Charlie measures in the $X$ <basis> and sends one bit to Alice or Bob; the remaining pair becomes $|\Phi^+\rangle$ or $|\Phi^-\rangle$, corrected by a local $Z$. Alice then performs standard teleportation: CNOT from $|\xi\rangle$ to her GHZ qubit, Hadamard, two computational measurements, and sends the two outcomes to Bob, who applies the corresponding Pauli correction.
Solved by gpt-5.6-sol high.
= ii
{parent=b}
{scope}
= Solution
{parent=ii}
No. Without classical communication Bob's reduced state is independent of Alice's unknown input by no-signalling, whereas successful transfer would make it equal to that input.
Solved by gpt-5.6-sol high.
= c
{parent=15e}
{scope}
= Solution
{parent=c}
Reordering as $A_1A_2B_1B_2$ gives $\alpha^2|0000\rangle+\alpha\beta(|0011\rangle+|1100\rangle)+\beta^2|1111\rangle$. Each party applies CNOT $1\to2$ and measures qubit 2. When both obtain 1, qubits $A_1B_1$ are proportional to $|00\rangle+|11\rangle$. The success probability is $2|\alpha\beta|^2$.
Solved by gpt-5.6-sol high.
= d
{parent=15e}
{scope}
= Solution
{parent=d}
After local phases, take $\alpha,\beta\ge0$. Alice measures with Kraus operators $M_1=\operatorname{diag}(\alpha,\beta)$ and $M_2=\operatorname{diag}(\beta,\alpha)$, which satisfy $M_1^\dagger M_1+M_2^\dagger M_2=I$. Outcome 1 directly gives $\alpha|00\rangle+\beta|11\rangle$. Outcome 2 gives $\beta|00\rangle+\alpha|11\rangle$; Alice tells Bob the outcome and both apply $X$, swapping the two coefficients. Thus the conversion succeeds with certainty, in agreement with Nielsen majorization.
Solved by gpt-5.6-sol high.
= 16J
{parent=Paper 3}
{scope}
{title2=Logic \& Set Theory}
= a
{parent=16j}
{scope}
= Solution
{parent=a}
A class is axiomatisable if it is exactly the class of models of some set of first-order sentences in the given language.
Solved by gpt-5.6-sol high.
= b
{parent=16j}
{scope}
= Solution
{parent=b}
<Compactness theorem> says a set of first-order sentences has a model whenever each finite subset has a model.
Solved by gpt-5.6-sol high.
= c
{parent=16j}
{scope}
= Solution
{parent=c}
If a theory axiomatized precisely the Archimedean fields, add a constant $c$ and sentences $t_n<c$ for every $n$. Every finite subset has an Archimedean model with $c$ sufficiently large, so compactness gives a model of the whole set. Its field reduct satisfies the proposed theory but is non-Archimedean, a contradiction.
Solved by gpt-5.6-sol high.
= d
{parent=16j}
{scope}
= Solution
{parent=d}
Atomic one-variable formulas reduce to <polynomial> equalities and inequalities after clearing terms. Their solution sets are basic by the supplied fact. Basic sets are closed under finite union, intersection, and complement; structural induction through the Boolean connectives therefore proves the claim.
Solved by gpt-5.6-sol high.
= e
{parent=16j}
{scope}
= Solution
{parent=e}
Quantifier elimination makes $S=\{x:\exists y(y^2=x)\}$ basic. It contains every square and no negative element. If some $a>0$ were not a square, $S$ would have a boundary inside $(0,\infty)$; but multiplication by positive squares and the ordered-field inequalities force membership to be locally constant along positive multiplicative intervals, contradicting the finite-endpoint form of a basic set. Hence every positive element, and also zero, is a square.
Solved by gpt-5.6-sol high.
= 17J
{parent=Paper 3}
{scope}
{title2=Graph Theory}
= a
{parent=17j}
{scope}
= i
{parent=a}
{scope}
= Solution
{parent=i}
If $Av=\lambda v$, choose $i$ with $|v_i|$ maximal. Then $|\lambda||v_i|=|\sum_{j\sim i}v_j|\le d(i)|v_i|\le\Delta|v_i|$, so $|\lambda|\le\Delta$.
Solved by gpt-5.6-sol high.
= ii
{parent=a}
{scope}
= Solution
{parent=ii}
For a $\Delta$-regular graph, $A\mathbf1=\Delta\mathbf1$. Conversely equality in the preceding maximum-coordinate proof propagates equal absolute values and compatible signs through connectedness, forcing every vertex degree $\Delta$. Perron-Frobenius, or the same equality argument, makes the eigenspace one-dimensional.
Solved by gpt-5.6-sol high.
= b
{parent=17j}
{scope}
= Solution
{parent=b}
$(A^r)_{ij}$ counts length-$r$ walks from $i$ to $j$, so $\operatorname{tr}A^r$ counts length-$r$ closed walks. The given power sums identify the <spectrum> as $5$ once, $\sqrt5$ and $-\sqrt5$ three times each, and $-1$ five times. Thus $n=12$, regular degree $d=5$, and the size is $nd/2=30$.
Solved by gpt-5.6-sol high.
= c
{parent=17j}
{scope}
= Solution
{parent=c}
Strong regularity $(n,d,a,b)$ means degree $d$, with $a$ common neighbours for adjacent pairs and $b$ for nonadjacent pairs. Counting two-step walks gives $(n-d-1)b=d(d-a-1)$; <eigenvalue> multiplicities impose the standard rationality condition. Here $a=0,b=3$, so $n=1+d(d+2)/3$. Applying multiplicity integrality to roots of $x^2+3x+3-d$ leaves $d\in\{1,3,21\}$. For $d=3$, $K_{3,3}$ is the required example.
Solved by gpt-5.6-sol high.
= 18F
{parent=Paper 3}
{scope}
{title2=Galois Theory}
= a
{parent=18f}
{scope}
= Solution
{parent=a}
Finite means $[L:K]<\infty$; algebraic means every element satisfies a nonzero <polynomial> over $K$. Linear dependence among $1,x,\ldots,x^n$ proves finite implies algebraic. The algebraic closure $\overline{\mathbb Q}/\mathbb Q$ is algebraic but infinite.
Solved by gpt-5.6-sol high.
= b
{parent=18f}
{scope}
= Solution
{parent=b}
An algebraic $x$ is separable when its minimal <polynomial> has distinct roots. In characteristic zero its <derivative> cannot vanish. In characteristic $p$, repeatedly write an inseparable minimal <polynomial> as a <polynomial> in $X^p$; after finitely many steps the minimal <polynomial> of $x^{p^n}$ has nonzero <derivative> and is separable.
Solved by gpt-5.6-sol high.
= c
{parent=18f}
{scope}
= Solution
{parent=c}
Let $\alpha=2^{1/6}=\sqrt2/\sqrt[3]2$. Then $L=\mathbb Q(\alpha,\zeta_3)$ is the splitting field of $X^6-2$. Eisenstein gives $[\mathbb Q(\alpha):\mathbb Q]=6$, and adjoining nonreal $\zeta_3$ doubles it, so the degree is 12 and the extension is Galois. Valuation at a prime above 2 forces $n\mid6$ if $x^n=2$; conversely $x=\alpha^{6/n}$ works for $n=1,2,3,6$.
Solved by gpt-5.6-sol high.
= 19F
{parent=Paper 3}
{scope}
{title2=Representation Theory}
= a
{parent=19f}
{scope}
= Solution
{parent=a}
$SU(2)=\{A:A^*A=I,\det A=1\}$; every element is $\begin{pmatrix}a&b\\-\bar b&\bar a\end{pmatrix}$ with $|a|^2+|b|^2=1$, giving $S^3$. A maximal torus is $T=\{\operatorname{diag}(e^{it},e^{-it})\}$. A noncentral conjugacy class is determined by trace $2\cos t$, has stabilizer $T$, and is $SU(2)/T\simeq S^2$; it meets $T$ at the two inverse diagonal elements. Since $SU(2)$ has no nontrivial continuous characters, $\det\rho=1$.
Solved by gpt-5.6-sol high.
= b
{parent=19f}
{scope}
= Solution
{parent=b}
Map the quaternion generators to $i\sigma_x,i\sigma_y$ (and their products), obtaining the unique two-dimensional irreducible $V$ and an embedding $Q_8\subset SU(2)$. Clebsch-Gordan gives $V\otimes V\simeq\operatorname{Sym}^2V\oplus1$ and $V^{\otimes3}\simeq\operatorname{Sym}^3V\oplus V\oplus V$. On restriction, $\operatorname{Sym}^2V$ is the sum of the three nontrivial one-dimensional characters of $Q_8$, while the other summand is trivial.
Solved by gpt-5.6-sol high.
= 20H
{parent=Paper 3}
{scope}
{title2=Algebraic Topology}
= a
{parent=20h}
{scope}
= Solution
{parent=a}
Subdivide the torus so $p,q$ are vertices and add an edge path between them; identifying its endpoints yields a finite simplicial model, so $X$ is triangulable.
Solved by gpt-5.6-sol high.
= b
{parent=20h}
{scope}
= Solution
{parent=b}
Identifying two points in a path-connected space is homotopy equivalent to adjoining a circle, so $X\simeq T^2\vee S^1$. Hence $H_0=\mathbb Z$, $H_1=\mathbb Z^3$, $H_2=\mathbb Z$, and all higher homology vanishes.
Solved by gpt-5.6-sol high.
= c
{parent=20h}
{scope}
= Solution
{parent=c}
No. Collapse the one-skeleton of the torus summand to obtain a map $X\to S^2$ inducing degree one on $H_2$. A null-homotopic map induces zero on reduced homology.
Solved by gpt-5.6-sol high.
= 21I
{parent=Paper 3}
{scope}
{title2=Linear Analysis}
= a
{parent=21i}
{scope}
= Solution
{parent=a}
$\sigma(T)=\{\lambda:T-\lambda I\text{ is not invertible}\}$ and $r(T)=\sup_{\lambda\in\sigma(T)}|\lambda|$, well-defined because the <spectrum> is nonempty compact. If $|\lambda|>\|T^n\|^{1/n}$, factor $T^n-\lambda^nI$ and use a Neumann <series>; spectral mapping then excludes $\lambda$. Thus $r(T)\le\inf_n\|T^n\|^{1/n}$.
Solved by gpt-5.6-sol high.
= b
{parent=21i}
{scope}
= Solution
{parent=b}
Complex Stone-Weierstrass requires a self-conjugate unital subalgebra separating points; apply real Stone-Weierstrass to real and imaginary parts. Finite sums $f(x)g(y)$ form such a self-conjugate unital subalgebra of $C(K\times L)$ and separate points, so they are dense.
Solved by gpt-5.6-sol high.
= c
{parent=21i}
{scope}
= Solution
{parent=c}
$\|T_hf\|_\infty\le\|h\|_\infty\|f\|_\infty$, so $T_h$ is bounded, and $h\mapsto T_h$ is linear with operator norm at most one.
Solved by gpt-5.6-sol high.
= d
{parent=21i}
{scope}
= Solution
{parent=d}
Approximate $h$ uniformly by finite sums $\sum f_j(x)g_j(y)$; their <integral> operators have finite-dimensional range and converge in operator norm, so $T_h$ is compact. If $h(x,y)=0$ for $x\le y$, iterated integration over an ordered simplex gives $\|T_h^n\|\le\|h\|_\infty^n/n!$. Thus $r(T_h)=0$, and compact spectral theory gives $\sigma(T_h)=\{0\}$.
Solved by gpt-5.6-sol high.
= 22H
{parent=Paper 3}
{scope}
{title2=Analysis of Functions}
= a
{parent=22h}
{scope}
= Solution
{parent=a}
A point $x$ is a Lebesgue point of $f$ when $|B_r|^{-1}\int_{B_r(x)}|f(y)-f(x)|dy\to0$ as $r\downarrow0$.
Solved by gpt-5.6-sol high.
= b
{parent=22h}
{scope}
= Solution
{parent=b}
Approximate $f$ in $L^1$ by continuous $g$. The limsup local oscillation of $f$ is bounded by the maximal <function> of $f-g$ plus the vanishing oscillation of $g$. The weak $(1,1)$ <Hardy-Littlewood maximal inequality> makes the exceptional set have measure at most $C\|f-g\|_1/\epsilon$; let the approximation error tend to zero.
Solved by gpt-5.6-sol high.
= c
{parent=22h}
{scope}
= Solution
{parent=c}
At a Lebesgue point, $f*g_n(x)-f(x)=\int g(u)[f(x-u/n)-f(x)]du$. Compact support and boundedness of $g$ reduce this to a constant times the mean oscillation over a ball of radius $O(1/n)$, which tends to zero.
Solved by gpt-5.6-sol high.
= d
{parent=22h}
{scope}
= Solution
{parent=d}
Not for an arbitrary signed $L^1$ kernel. One can place very tall, very narrow spikes of $g$ at separated locations, with summable masses and total <integral> one, and matching still narrower spikes of an $L^1$ <function> accumulating at a Lebesgue point. Along a selected subsequence of dilations, a kernel spike lands on the matching <function> spike and contributes a fixed amount, while the total widths make the point a Lebesgue point. Thus compact support (or a suitable integrable radial majorant) cannot simply be dropped.
Solved by gpt-5.6-sol high.
= 23G
{parent=Paper 3}
{scope}
{title2=Riemann Surfaces}
= Solution
{parent=23G}
An elliptic <function> is meromorphic and $\Lambda$-periodic. The argument principle on opposite sides of a period parallelogram cancels, so zeros and poles have equal total multiplicity. The <Weierstrass elliptic function>
$$\wp(z)=z^{-2}+\sum_{0\ne\omega\in\Lambda}[(z-\omega)^{-2}-\omega^{-2}]$$
is elliptic. Its half-period values $e_i$ are distinct, and comparing poles and leading coefficients proves $\wp'^2=4\prod_i(\wp-e_i)$. The half-period addition formula gives
$$\wp(z+\omega_1/2)-e_1=\frac{(e_1-e_2)(e_1-e_3)}{\wp(z)-e_1}.$$
Differentiate and use the same formula at $z=\omega_1/4$ to identify $(e_1-e_2)(e_1-e_3)=(\wp(\omega_1/4)-e_1)^2$, yielding the required <derivative> ratio.
Solved by gpt-5.6-sol high.
= 24F
{parent=Paper 3}
{scope}
{title2=Algebraic Geometry}
= Solution
{parent=24F}
<Riemann-Roch theorem> is $\ell(D)-\ell(K-D)=\deg D+1-g$. Putting $D=0$ and using $\ell(0)=1$, $\ell(K)=g$ gives $\deg K=2g-2$. If an effective degree-two divisor has $\ell(D)=2$, its pencil defines a degree-two map to $\mathbb P^1$, so the curve is hyperelliptic. For $g=2$, Riemann-Roch applied to $K$ supplies such a degree-two pencil. For nonhyperelliptic genus 3, the canonical system has dimension 2, separates points and tangents, and hence embeds $C$ as a plane quartic. On a genus-3 hyperelliptic curve, $\ell(2Q)=2$ exactly at the $2g+2=8$ ramification points.
Solved by gpt-5.6-sol high.
= 25I
{parent=Paper 3}
{scope}
{title2=Differential Geometry}
= a
{parent=25i}
{scope}
= Solution
{parent=a}
An isometry is a smooth diffeomorphism with $I_x(v,w)=I_{\phi(x)}(d\phi_xv,d\phi_xw)$.
Solved by gpt-5.6-sol high.
= b
{parent=25i}
{scope}
= Solution
{parent=b}
If $f\circ\phi=f$, then $x$ and $\phi(x)$ have the same value, so every level set is preserved.
Solved by gpt-5.6-sol high.
= c
{parent=25i}
{scope}
= Solution
{parent=c}
Gauss's <Theorema Egregium> expresses Gaussian curvature intrinsically from the first fundamental form, hence $K\circ\phi=K$. Merely preserving curvature level sets is not sufficient: when $K$ is constant, every diffeomorphism does so, though most are not isometries.
Solved by gpt-5.6-sol high.
= d
{parent=25i}
{scope}
= Solution
{parent=d}
An isometry preserves both foliations. Independence makes their level curves transverse near $x$, so their pair of values gives local coordinates. The isometry fixes the intersection corresponding to every pair of nearby values, hence fixes every point in a sufficiently small neighbourhood.
Solved by gpt-5.6-sol high.
= e
{parent=25i}
{scope}
= Solution
{parent=e}
The one-dimensional kernel of $dK_x$ is the tangent to its level curve; the metric gives two <unit normals> and $dK_x(e_x)>0$ selects one. Isometries preserve the metric, $K$, and this orientation choice, so they carry $e_x$ to $e_{\phi(x)}$. Therefore $g(x)=dK_x(e_x)=|\nabla K(x)|$ is isometry invariant.
Solved by gpt-5.6-sol high.
= 26L
{parent=Paper 3}
{scope}
{title2=Probability and Measure}
= a
{parent=26l}
{scope}
= Solution
{parent=a}
Fatou says $\int\liminf f_n\le\liminf\int f_n$ for nonnegative <functions>. Apply it to $g+f_n$ and $g-f_n$ when $|f_n|\le g\in L^1$ to obtain both inequalities and hence dominated convergence.
Solved by gpt-5.6-sol high.
= b
{parent=26l}
{scope}
= Solution
{parent=b}
For every $\epsilon>0$, $1_{|X_n-X|>\epsilon}\to0$ almost surely and is dominated by 1, so its expectation, the corresponding probability, tends to zero.
Solved by gpt-5.6-sol high.
= c
{parent=26l}
{scope}
= Solution
{parent=c}
Yes. The $L^2$ bound makes the family uniformly integrable in $L^1$: $E[|X_n|1_{|X_n|>K}]\le E[X_n^2]/K\le1/K$. Together with convergence in probability, Vitali's theorem gives $E|X_n|\to0$.
Solved by gpt-5.6-sol high.
= d
{parent=26l}
{scope}
= Solution
{parent=d}
On disjoint events $A_n$ with $P(A_n)=c/n^2$, set $X_n=n1_{A_n}$. Then $X_n\to0$ almost surely and $E X_n^2=c$, but any common dominator has <integral> at least $\sum_n nP(A_n)=c\sum1/n=\infty$. Thus no integrable dominating variable exists.
Solved by gpt-5.6-sol high.
= 27K
{parent=Paper 3}
{scope}
{title2=Applied Probability}
= a
{parent=27k}
{scope}
= Solution
{parent=a}
This is Wald's identity. Expand the stopped sum and use $\{M\ge i\}$, which depends only on earlier interarrivals and is independent of $\xi_i$: $E\sum_{i=1}^M\xi_i=\sum_iE[1_{M\ge i}]E\xi_i=\lambda^{-1}EM$.
Solved by gpt-5.6-sol high.
= b
{parent=27k}
{scope}
= Solution
{parent=b}
Apply part (a) to the stopping time $N_t+1$ to obtain $E T_{N_t+1}=\lambda^{-1}E(N_t+1)$.
Solved by gpt-5.6-sol high.
= c
{parent=27k}
{scope}
= Solution
{parent=c}
The strong law gives $T_n/n\to1/\lambda$. Since $T_{N_t}\le t<T_{N_t+1}$ and $N_t\to\infty$, sandwiching yields $N_t/t\to\lambda$ almost surely.
Solved by gpt-5.6-sol high.
= d
{parent=27k}
{scope}
= Solution
{parent=d}
If $\xi_1\le C$, then $t<T_{N_t+1}\le t+C$. Taking expectations and using (b) yields $\lambda t-1<E N_t\le\lambda(t+C)-1$. Divide by $t$.
Solved by gpt-5.6-sol high.
= 28L
{parent=Paper 3}
{scope}
{title2=Principles of Statistics}
= a
{parent=28l}
{scope}
= Solution
{parent=a}
Let $q_u=G^{-1}(u)$. Since $\sqrt n(\hat\theta-\theta)\sim G$, an exact interval is $[\hat\theta-q_{1-\alpha/2}/\sqrt n,\hat\theta-q_{\alpha/2}/\sqrt n]$.
Solved by gpt-5.6-sol high.
= b
{parent=28l}
{scope}
= Solution
{parent=b}
A <bootstrap> sample consists of $n$ independent draws with replacement from the empirical distribution assigning mass $1/n$ to each observed $X_i$.
Solved by gpt-5.6-sol high.
= c
{parent=28l}
{scope}
= Solution
{parent=c}
Replace the two $G$ quantiles by conditional <bootstrap> quantiles $\hat q_u$ in the interval from (a). <Uniform convergence> implies $\hat q_u\to q_u$ in probability at continuity points, so Slutsky and continuity of $G$ give coverage tending to $1-\alpha$.
Solved by gpt-5.6-sol high.
= d
{parent=28l}
{scope}
= Solution
{parent=d}
Generate $B$ resamples, compute $\sqrt n(\hat\theta^{*j}-\hat\theta)$, and use their empirical $\alpha/2$ and $1-\alpha/2$ quantiles. Large $B$ approximates the otherwise unavailable conditional <bootstrap> distribution.
Solved by gpt-5.6-sol high.
= 29L
{parent=Paper 3}
{scope}
{title2=Stochastic Financial Models}
= a
{parent=29l}
{scope}
= Solution
{parent=a}
Let $V(n,x)$ be maximal expected utility from dates $n+1,\ldots,N$ plus terminal wealth, given $X_n=x$. Then $V(N,x)=U(x)$ and
$$V(n,x)=\sup_{c,h}\{U(c)+E[V(n+1,x-c+hR_{n+1})]\}.$$
Solved by gpt-5.6-sol high.
= b
{parent=29l}
{scope}
= Solution
{parent=b}
With $G=-U=e^{-x}$ and $F=-V$, maximization becomes
$$F(n,x)=\inf_{c,h}\{G(c)+E[F(n+1,x-c+hR)]\},\quad F(N,x)=e^{-x}.$$
Solved by gpt-5.6-sol high.
= c
{parent=29l}
{scope}
= Solution
{parent=c}
Substitution of $F(n+1,x)=B_{n+1}e^{-A_{n+1}x}$ separates the infima over investment and consumption. The investment contributes $\gamma=\inf_hE(e^{-hR})$ after rescaling, and the first-order condition preserves exponential form, with $A_n=A_{n+1}/(1+A_{n+1})$.
Solved by gpt-5.6-sol high.
= d
{parent=29l}
{scope}
= Solution
{parent=d}
Starting from $A_N=1$, recursion gives $A_n=1/(N-n+1)$. The coefficient recursion gives $B_n=(N-n+1)\gamma^{(N-n)/2}$, including $B_N=1$.
Solved by gpt-5.6-sol high.
= e
{parent=29l}
{scope}
= Solution
{parent=e}
Let $k=N-n+1$. The consumption first-order condition gives
$$C_n^*=\frac{X_{n-1}^*}{k+1}-\frac{k}{2}\log\gamma
=\frac{X_{n-1}^*}{N-n+2}-\frac{N-n+1}{2}\log\gamma.$$
Solved by gpt-5.6-sol high.
= 30D
{parent=Paper 3}
{scope}
{title2=Asymptotic Methods}
= a
{parent=30d}
{scope}
= Solution
{parent=a}
With phase $\phi(z)=(1+i)z-z^2/2$, the original ray is a contour <integral>. At the endpoint $z=-1$, the steepest direction solves $\operatorname{Im}(\phi(z)-\phi(-1))=0$ with decreasing real part; locally it is the direction for which $\phi'(-1)(z+1)$ is negative real.
Solved by gpt-5.6-sol high.
= b
{parent=30d}
{scope}
= Solution
{parent=b}
The saddle is $z_s=1+i$. Completing the square gives $\phi(z)=i-(z-1-i)^2/2$, so its steepest-descent line is the horizontal line $z=1+i+s$, $s\in\mathbb R$.
Solved by gpt-5.6-sol high.
= c
{parent=30d}
{scope}
= Solution
{parent=c}
Deform the real ray upward to the saddle line, joining it near the endpoint and at a large right cutoff. The integrand is entire. The far bridge vanishes by Gaussian decay; the endpoint bridge is $O(x^{-1}e^{-3x/2})$ because $\Re\phi(-1)=-3/2$.
Solved by gpt-5.6-sol high.
= d
{parent=30d}
{scope}
= Solution
{parent=d}
The saddle contributes the full Gaussian
$$e^{ix}\int_{-\infty}^{\infty}e^{-xs^2/2}ds=\sqrt{2\pi},x^{-1/2}e^{ix}.$$
Thus $\alpha=\sqrt{2\pi}$. Since the phase is exactly quadratic, there are no further algebraic saddle terms; the next contribution is the endpoint term $O(x^{-1}e^{-3x/2})$.
Solved by gpt-5.6-sol high.
= 31B
{parent=Paper 3}
{scope}
{title2=Dynamical Systems}
= a
{parent=31b}
{scope}
= Solution
{parent=a}
Set $y=\dot x$. Then $\dot x=y$, $\dot y=a+bx-y-x^2-y^2$.
Solved by gpt-5.6-sol high.
= b
{parent=31b}
{scope}
= Solution
{parent=b}
Fixed points have $y=0$ and $x^2-bx-a=0$. They coalesce on $a=-b^2/4$, the saddle-node curve. Away from it, $x_\pm=(b\pm\sqrt{b^2+4a})/2$; the Jacobian has trace $-1$ and <determinant> $2x-b$, so $x_+$ is stable and $x_-$ a saddle. At $(a,b)=(0,0)$ the branches $x=0$ and $x=b$ cross in a transcritical bifurcation.
Solved by gpt-5.6-sol high.
= c
{parent=31b}
{scope}
= Solution
{parent=c}
For $a=0$ and extended variable $\dot b=0$, the centre manifold has $y=bx-x^2+$ higher-order terms. Hence $\dot x=bx-x^2+\cdots$, the transcritical normal form: $x=0$ and $x=b$ exchange stability at $b=0$.
Solved by gpt-5.6-sol high.
= d
{parent=31b}
{scope}
= Solution
{parent=d}
For $b=1$, set $A=a+1/4$ and $X=x-1/2$. The centre manifold has $y=A-X^2+\cdots$, so $\dot X=A-X^2+\cdots$. This is a saddle-node at $a=-1/4$, with two branches for $a>-1/4$.
Solved by gpt-5.6-sol high.
= e
{parent=31b}
{scope}
= Solution
{parent=e}
For $a=-1,b=1$, $\dot y=-1+x-y-x^2-y^2\le-1/2<0$ after maximizing the two quadratic expressions. Thus no orbit is periodic. A proper forward bounding region or Lyapunov trapping <function> would, in a planar flow with no equilibria, force a periodic <limit> set by Poincare-Bendixson; therefore no such global bounding <function> exists.
Solved by gpt-5.6-sol high.
= 32B
{parent=Paper 3}
{scope}
{title2=Integrable Systems}
= Solution
{parent=32B}
The flow of $W=\phi u\partial_u+\sum\alpha_ix_i\partial_{x_i}$ is $(u,x_i)\mapsto(e^{\phi s}u,e^{\alpha_i s}x_i)$; invariants are exactly solutions of $Wf=0$. Two scaling symmetries of the PDE are
$$W_1=2u\partial_u+x\partial_x-t\partial_t,\qquad W_2=x\partial_x+y\partial_y+t\partial_t.$$
Invariants for $W_1$ include $y,xt,ut^2$; for $W_2$, $u,y/x,t/x$. Two common invariants are $ut/x$ and $y^2/(xt)$. Invariance under $u\partial_u+x\partial_x+\tfrac12y\partial_y$ and $\partial_y$ forces $u=xF(t)$. Substitution gives $F'=F^2$, hence $u=0$ or $u=-x/(t-C)$.
Solved by gpt-5.6-sol high.
= 33A
{parent=Paper 3}
{scope}
{title2=Principles of Quantum Mechanics}
= Solution
{parent=33A}
In the interaction picture,
$$a_f^{(1)}(t)=-\frac{i\lambda}{\hbar}\sum_{k=1}^2\alpha_kW_k\int_0^te^{(i\omega_{fi_k}-\gamma)t'}\cos(\omega t')dt'.$$
At infinity this is $-(i\lambda/\hbar)\sum_k\alpha_kW_kF_k$, where
$$F_k=\tfrac12\left[\frac1{\gamma-i(\omega_{fi_k}+\omega)}+\frac1{\gamma-i(\omega_{fi_k}-\omega)}\right],$$
so $b_1=1,b_2=1,b_3=-1$ and $\omega_{fi_k}=(E_f-E_{i_k})/\hbar$. The probability is $(\lambda^2/\hbar^2)|\sum_k\alpha_kW_kF_k|^2$, displaying interference between initial components. For $N$ equal response factors, Cauchy-Schwarz gives $|\sum_k\alpha_k|^2\le N$, attained by equal phases and $|\alpha_k|=N^{-1/2}$; the largest probability is $N\lambda^2|WF|^2/\hbar^2$ within perturbation theory.
Solved by gpt-5.6-sol high.
= 34E
{parent=Paper 3}
{scope}
{title2=Applications of Quantum Mechanics}
= a
{parent=34e}
{scope}
= Solution
{parent=a}
<Bloch theorem> follows because lattice translations commute with $H$ and with each other: simultaneous eigenstates obey $T_R\psi=e^{ik\cdot R}\psi$, so $\psi_k(r)=e^{ik\cdot r}u_k(r)$ with $u_k$ lattice-periodic. Conversely this form has the required translation character.
Solved by gpt-5.6-sol high.
= b
{parent=34e}
{scope}
= i
{parent=b}
{scope}
= Solution
{parent=i}
The listed points are precisely integer combinations of $a_1,a_2,a_3$ (the face-centred cubic lattice), as follows by sorting the parity classes of doubled Cartesian coordinates.
Solved by gpt-5.6-sol high.
= ii
{parent=b}
{scope}
= Solution
{parent=ii}
$|a_1\cdot(a_2\times a_3)|=a^3/2$.
Solved by gpt-5.6-sol high.
= iii
{parent=b}
{scope}
= Solution
{parent=iii}
Using $b_i=2\pi(a_j\times a_k)/V$ gives $b_1=(2\pi/a)(0,1,1)$, $b_2=(2\pi/a)(1,0,1)$, $b_3=(2\pi/a)(1,1,0)$. The reciprocal cell volume is $(2\pi)^3/V=16\pi^3/a^3$.
Solved by gpt-5.6-sol high.
= c
{parent=34e}
{scope}
= i
{parent=c}
{scope}
= Solution
{parent=i}
<Bloch theorem> $|k\rangle\propto\sum_Re^{ik\cdot R}|R\rangle$ are eigenstates with $E(k)=E_0-2\mu\sum_{i=1}^3\cos(k\cdot a_i)$.
Solved by gpt-5.6-sol high.
= ii
{parent=c}
{scope}
= Solution
{parent=ii}
The sum of three cosines ranges from $-3$ to $3$ on the Brillouin zone, so $\Delta E=12|\mu|$.
Solved by gpt-5.6-sol high.
= iii
{parent=c}
{scope}
= Solution
{parent=iii}
Near $k=0$, $E=E_0-6\mu+\mu\sum_i(k\cdot a_i)^2+O(k^4)$. The quadratic form has <eigenvalues> $a^2/4,a^2/4,a^2$ along its principal axes, so comparison with $\hbar^2k_j^2/(2m_j^*)$ gives $m^*=2\hbar^2/(\mu a^2)$ twice and $m^*=\hbar^2/(2\mu a^2)$ once (with signs interpreted at the selected band extremum).
Solved by gpt-5.6-sol high.
= 35E
{parent=Paper 3}
{scope}
{title2=Statistical Physics}
= a
{parent=35e}
{scope}
= Solution
{parent=a}
Mean field replaces edge and triangle products by $m^2$ and $m^3$. There are $3N$ edges and $2N$ elementary triangles, so
$$Z_N\simeq\sum_M\binom{N}{(N+M)/2}\exp\{\beta N(Bm+3Jm^2+2Km^3)\}.$$
Solved by gpt-5.6-sol high.
= b
{parent=35e}
{scope}
= Solution
{parent=b}
Stirling and replacing the magnetization sum by an <integral> gives the stated form with $P(m)=-\beta(Bm+3Jm^2+2Km^3)$, up to subexponential constants.
Solved by gpt-5.6-sol high.
= c
{parent=35e}
{scope}
= Solution
{parent=c}
Stationarity gives $\operatorname{artanh}m=\beta(B+6Jm+6Km^2)$, or $m=\tanh Q(m)$ with $Q(m)=\beta(B+6Jm+6Km^2)$.
Solved by gpt-5.6-sol high.
= d
{parent=35e}
{scope}
= Solution
{parent=d}
For $B=J=0$, $m=0$ remains stationary. At large $\beta K$, evaluating the free energy near a positive $m$ makes the negative cubic energy dominate the bounded <entropy> cost, so some $m>0$ has lower free energy than zero. At high <temperature> <entropy> uniquely favors zero. Continuity in <temperature> therefore forces a transition at an intermediate $\beta_*$.
Solved by gpt-5.6-sol high.
= 36D
{parent=Paper 3}
{scope}
{title2=Electrodynamics}
= Solution
{parent=36D}
Define $p(t)=\int x'\rho(t,x')d^3x'$. Expanding the retarded potential for source size $d\ll r$ and speed $v\ll c$, with radiation zone $r\gg d,c/\omega$, and using charge conservation gives $B=\mu_0\ddot p(t-r/c)\times\hat x/(4\pi rc)$. Then $S=\mu_0|\ddot p\times\hat x|^2\hat x/(16\pi^2cr^2)$ and $dP/d\Omega=\mu_0(|\ddot p|^2-(\hat x\cdot\ddot p)^2)/(16\pi^2c)$. For circular motion, $P=\mu_0q^2\omega^4R^2/(6\pi c)$ and $\langle dP/d\Omega\rangle=\mu_0q^2\omega^4R^2(1+\cos^2\theta)/(32\pi^2c)$. The stated axial and equatorial polarizations follow directly. Relativistically, perpendicular <acceleration> gives $P=\mu_0q^2\gamma^4\omega^4R^2/(6\pi c)$.
Solved by gpt-5.6-sol high.
= 37A
{parent=Paper 3}
{scope}
{title2=General Relativity}
= a
{parent=37a}
{scope}
= Solution
{parent=a}
Expansion of the Ricci identity gives $R^\mu{}_{\nu\alpha\beta}=\partial_\alpha\Gamma^\mu_{\beta\nu}-\partial_\beta\Gamma^\mu_{\alpha\nu}+\Gamma^\mu_{\alpha\lambda}\Gamma^\lambda_{\beta\nu}-\Gamma^\mu_{\beta\lambda}\Gamma^\lambda_{\alpha\nu}$.
Solved by gpt-5.6-sol high.
= b
{parent=37a}
{scope}
= Solution
{parent=b}
$R_{\mu\nu\alpha\beta}=-R_{\nu\mu\alpha\beta}=-R_{\mu\nu\beta\alpha}=R_{\alpha\beta\mu\nu}$ and $R_{\mu[\nu\alpha\beta]}=0$. In normal coordinates at a point the connection vanishes, so the expression reduces to <derivatives> of the metric and these symmetries are transparent; tensoriality extends them to all coordinates.
Solved by gpt-5.6-sol high.
= c
{parent=37a}
{scope}
= Solution
{parent=c}
Under a coordinate change, the inhomogeneous second-derivative terms arising from each <partial derivative> cancel between $V^\alpha\partial_\alpha W^\mu$ and $W^\alpha\partial_\alpha V^\mu$. The remainder transforms with one contravariant Jacobian, so the commutator is a <vector> field.
Solved by gpt-5.6-sol high.
= d
{parent=37a}
{scope}
= Solution
{parent=d}
Differentiate the Killing equation, permute indices, and combine using the Ricci identity to obtain $\nabla_\alpha\nabla_\beta V^\mu=R^\mu{}_{\beta\alpha\nu}V^\nu$ (up to the equivalent sign convention for $R$).
Solved by gpt-5.6-sol high.
= e
{parent=37a}
{scope}
= Solution
{parent=e}
The Lie <derivative> formulation is $\mathcal L_Vg=0$. Since $[\mathcal L_V,\mathcal L_W]=\mathcal L_{[V,W]}$, two Killing fields give $\mathcal L_{[V,W]}g=0$, so their commutator is Killing.
Solved by gpt-5.6-sol high.
= 38C
{parent=Paper 3}
{scope}
{title2=Fluid Dynamics}
= Solution
{parent=38C}
The axial equation is $\mu r^{-1}(ru')'=-G$. Regularity at zero, no slip $u_a(a)=0$, continuity of <velocity> and shear at $c$ give
$$u_a(r)=\frac{G(a^2-r^2)}{4\mu_a},\qquad u_c(r)=\frac{G(c^2-r^2)}{4\mu_c}+\frac{G(a^2-c^2)}{4\mu_a}.$$
The annulus is controlled only by its own <viscosity> and the outer no-slip condition, explaining the absence of $\mu_c$. For the annular control volume, kinetic-energy rates vanish; <pressure> work plus interfacial traction work equals $\int_V2\mu_ae_{ij}e_{ij}dV$. Substituting $u_a'=-Gr/(2\mu_a)$ evaluates both sides to the same boundary expression, verifying the mechanical-energy equation.
Solved by gpt-5.6-sol high.
= 39C
{parent=Paper 3}
{scope}
{title2=Waves}
= a
{parent=39c}
{scope}
= Solution
{parent=a}
<Stationary points> satisfy $x/t=\omega'(k_j)$. For <nondegenerate stationary points of a phase>[nondegenerate points],
$$\eta(x,t)\sim\frac1{\sqrt{2\pi t}}\sum_j\frac{F(k_j)}{\sqrt{|\omega''(k_j)|}}\exp\{i[k_jx-\omega(k_j)t-i\tfrac\pi4\operatorname{sgn}\omega''(k_j)]\}.$$
Solved by gpt-5.6-sol high.
= b
{parent=39c}
{scope}
= Solution
{parent=b}
Fourier modes of the harmonic potential decay as $e^{|k|y}$; the two surface conditions give $\eta_{tt}+|k|^3\eta=0$, hence $\omega^2=|k|^3$. The initial transform is $F(k)=-i\sqrt\pi,k e^{-k^2/4}/2$. At $x=Vt$, the contributing stationary wavenumber is $k_0=4V^2/9$, giving
$$\eta(Vt,t)\sim\frac{4V^{5/2}}{27\sqrt t}e^{-4V^4/81}\sin\left(\frac{4V^3t}{27}-\frac\pi4\right).$$
Different observer speeds select different group-velocity wavenumbers; fast observers see exponentially weaker, higher-frequency ripples.
Solved by gpt-5.6-sol high.
= 40B
{parent=Paper 3}
{scope}
{title2=Numerical Analysis}
= a
{parent=40b}
{scope}
= Solution
{parent=a}
Taylor expansion gives $\Gamma_9u=h^2\Delta u+(h^4/12)\Delta^2u+O(h^6)=h^2f+(h^4/12)\Delta f+O(h^6)$. Thus the local residual is $O(h^4)$.
Solved by gpt-5.6-sol high.
= b
{parent=40b}
{scope}
= Solution
{parent=b}
The error equation has right side $O(h^4)$. With $\|A_h^{-1}\|_2\le C/h^2$, it follows that $\|e\|_2=O(h^2)$.
Solved by gpt-5.6-sol high.
= c
{parent=40b}
{scope}
= Solution
{parent=c}
If $\Delta f=0$, the $h^4$ term vanishes, so the local residual is $O(h^6)$ and the global error is $O(h^4)$.
Solved by gpt-5.6-sol high.
= d
{parent=40b}
{scope}
= Solution
{parent=d}
The five-point expression satisfies $\Gamma_5f=h^2\Delta f+O(h^4)$. Therefore the modified right side is $h^2f+h^4\Delta f/12+O(h^6)$, cancelling the leading residual from part (a) and leaving the same $O(h^6)$ local error as in part (c).
Solved by gpt-5.6-sol high.