past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ib/paper-3.bigb

= Paper 3
{scope}

https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2024/paperib_3_2024.pdf

= 1E
{parent=Paper 3}
{scope}
{title2=Groups, Rings and Modules}

= i
{parent=1e}
{scope}

= Solution
{parent=i}

The <Smith normal form> of an integer $m\times n$ <matrix> $A$ is a diagonal <matrix>
$$
D=\operatorname{diag}(d_1,\ldots,d_r,0,\ldots,0),
\qquad d_i>0,
\qquad d_i\mid d_{i+1},
$$
for which
$$
UAV=D
$$
with $U\in GL_m(\mathbb Z)$ and $V\in GL_n(\mathbb Z)$. It exists and is unique.

The <structure theorem for finitely generated modules over a principal ideal domain>, specialized to $\mathbb Z$, says that every finitely generated abelian <group> has a unique invariant-factor decomposition
$$
M\cong\mathbb Z^s\oplus
\mathbb Z/d_1\mathbb Z\oplus\cdots\oplus
\mathbb Z/d_r\mathbb Z,
\qquad
1<d_1\mid\cdots\mid d_r.
$$
Equivalently, its finite part is a direct sum of cyclic <groups> of prime-power order.

Solved by gpt-5.6-sol high.

= ii
{parent=1e}
{scope}

= Solution
{parent=ii}

Integer row and column operations give
$$
\begin{pmatrix}-4&-6\\2&2\end{pmatrix}
\sim
\begin{pmatrix}2&2\\-4&-6\end{pmatrix}
\sim
\begin{pmatrix}2&2\\0&-2\end{pmatrix}
\sim
\begin{pmatrix}2&0\\0&2\end{pmatrix}.
$$
Thus the Smith normal form is
$$
\boxed{\operatorname{diag}(2,2)}.
$$
The two relation <vectors> are the columns of the displayed <matrix>, so its cokernel is the presented module. Consequently
$$
\boxed{M\cong\mathbb Z/2\mathbb Z\oplus\mathbb Z/2\mathbb Z}.
$$

Solved by gpt-5.6-sol high.

= iii
{parent=1e}
{scope}

= Solution
{parent=iii}

By the finite case of the structure theorem, a finite abelian <group> is a direct sum of cyclic <groups> of prime-power order. If it is indecomposable, this decomposition can have only one nonzero summand, so the <group> is cyclic of order $p^n$ for some prime $p$.

Conversely, every nontrivial <subgroup> of the cyclic <group> $C_{p^n}$ contains its unique <subgroup> of order $p$. Hence two nontrivial <subgroups> cannot have trivial intersection. They therefore cannot be the two summands of an internal direct sum. Thus the <indecomposable finite abelian groups> are precisely
$$
\boxed{C_{p^n}\quad(n\geq1)}.
$$

Solved by gpt-5.6-sol high.

= 2G
{parent=Paper 3}
{scope}
{title2=Geometry}

= Solution
{parent=2G}

Use the global parametrization
$$
X(u,v)=(u,v,h(u,v)).
$$
Its <tangent vectors> are
$$
X_u=(1,0,h_u),
\qquad
X_v=(0,1,h_v),
$$
and
$$
X_u\times X_v=(-h_u,-h_v,1)\ne0.
$$
Thus $X$ is a regular parametrization and the graph is a smooth surface.

The <first fundamental form> is
$$
\boxed{
I=(1+h_u^2)\,du^2+2h_uh_v\,du\,dv+(1+h_v^2)\,dv^2}.
$$
Choosing the upward orientation, the <Gauss map> is
$$
\boxed{
N(u,v)=\frac{(-h_u,-h_v,1)}
{\sqrt{1+h_u^2+h_v^2}}}.
$$

Solved by gpt-5.6-sol high.

= 3B
{parent=Paper 3}
{scope}
{title2=Complex Methods}

= Solution
{parent=3B}

The <Cauchy integral theorem> states that if $f$ is holomorphic on a simply connected domain, then its <integral> around every closed piecewise smooth contour in that domain is zero.

Put $q=b-k$. Completing the square gives
$$
-ax^2+iqx
=-a\left(x-\frac{iq}{2a}\right)^2-
\frac{q^2}{4a}.
$$
After the change of variable $z=x-iq/(2a)$, the <integral> runs along the horizontal line $\operatorname{Im}z=-q/(2a)$. The integrand $e^{-az^2}$ is entire. Apply Cauchy's theorem to a rectangle joining this line to the real axis; the two vertical <integrals> tend to zero as their real parts tend to $\pm\infty$, because $a>0$. The contour may therefore be shifted to the real line. Hence the <Fourier transform of a Gaussian> gives
$$
\begin{aligned}
\widehat f_{a,b}(k)
&=e^{-q^2/(4a)}\int_{-\infty}^{\infty}e^{-az^2}\,dz\\
&=\boxed{\sqrt{\frac\pi a}
\exp\left[-\frac{(k-b)^2}{4a}\right]}.
\end{aligned}
$$

Solved by gpt-5.6-sol high.

= 4C
{parent=Paper 3}
{scope}
{title2=Variational Principles}

= Solution
{parent=4C}

For a general <Lagrangian>,
$$
p=\frac{\partial L}{\partial\dot x},
\qquad
H=p\cdot\dot x-L.
$$
Along an Euler-Lagrange trajectory,
$$
\frac{dH}{dt}=-\frac{\partial L}{\partial t},
$$
so energy is conserved when $L$ has no explicit time dependence.

For the given <Lagrangian>, write $v=|\dot x|$. At points where $v\ne0$,
$$
\boxed{p=mc\frac{\dot x}{v}}.
$$
The Euler-Lagrange equation is therefore
$$
\boxed{
\frac d{dt}\left(mc\frac{\dot x}{|\dot x|}\right)
=-\nabla V(x)}.
$$
In expanded form its left-hand side is
$$
mc\left(
\frac{\ddot x}{v}
-\frac{\dot x(\dot x\cdot\ddot x)}{v^3}
\right).
$$
The <momentum> has fixed magnitude,
$$
\boxed{p\cdot p=m^2c^2},
$$
so this is constant without needing to solve the equation of motion. The <Hamiltonian> is
$$
\boxed{H=p\cdot\dot x-L=V(x)}.
$$

The <singular Legendre transform of a degree-one velocity Lagrangian> applies here: $p$ determines only the direction of $\dot x$, not its magnitude, and the <velocity> Hessian is not invertible. Thus the usual inverse Legendre transform cannot reconstruct the <Lagrangian> from this <Hamiltonian>.

Solved by gpt-5.6-sol high.

= 5B
{parent=Paper 3}
{scope}
{title2=Methods}

= Solution
{parent=5B}

Set $u(r,\theta)=R(r)\Theta(\theta)$. Periodicity and <separation of variables> give the angular modes $1$, $\cos n\theta$, and $\sin n\theta$. The radial equation is
$$
r^2R''+rR'-n^2R=0,
$$
whose solutions are $r^{\pm n}$ for $n\geq1$; the zero mode has solutions $1$ and $\log r$. Thus a general real harmonic <function> on the annulus is
$$
\begin{aligned}
u(r,\theta)={}&A_0+B_0\log r\\
&+\sum_{n=1}^{\infty}
\left(A_nr^n+B_nr^{-n}\right)\cos n\theta\\
&+\sum_{n=1}^{\infty}
\left(C_nr^n+D_nr^{-n}\right)\sin n\theta.
\end{aligned}
$$

The boundary data contain only the $n=2$ cosine mode, so write
$$
u=(Ar^2+Br^{-2})\cos2\theta.
$$
The condition at $r=a$ gives $B=-Aa^4$, and the condition at $r=b$ fixes $A$. The <Dirichlet problem on an annulus for one Fourier mode> therefore has solution
$$
\boxed{
u(r,\theta)=
\frac{b^2(r^4-a^4)}{r^2(b^4-a^4)}\cos2\theta}.
$$

Solved by gpt-5.6-sol high.

= 6A
{parent=Paper 3}
{scope}
{title2=Quantum Mechanics}

= i
{parent=6a}
{scope}

= Solution
{parent=i}

The <canonical commutation relation> is
$$
\boxed{[x,p]=i\hbar I}.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=6a}
{scope}

= Solution
{parent=ii}

For a normalized state, the uncertainty of a Hermitian <observable> $O$ is
$$
\boxed{
\Delta O=\sqrt{\langle O^2\rangle-\langle O\rangle^2}}.
$$

Solved by gpt-5.6-sol high.

= iii
{parent=6a}
{scope}

= Solution
{parent=iii}

Positivity of the squared norm of the supplied state gives, for every real $s$,
$$
\begin{aligned}
0&\leq\lVert(p-isx)\psi\rVert^2\\
&=\langle(p+isx)(p-isx)\rangle\\
&=\langle p^2\rangle+s^2\langle x^2\rangle
+is\langle[x,p]\rangle\\
&=(\Delta p)^2+s^2(\Delta x)^2-s\hbar,
\end{aligned}
$$
where the stated zero means were used in the last line. This quadratic in $s$ is nonnegative for every real $s$, so its discriminant is nonpositive:
$$
\hbar^2-4(\Delta x)^2(\Delta p)^2\leq0.
$$
This is the <quadratic-norm proof of the Heisenberg uncertainty relation>, and proves
$$
\boxed{\Delta x\,\Delta p\geq\frac\hbar2}.
$$

Solved by gpt-5.6-sol high.

= 7D
{parent=Paper 3}
{scope}
{title2=Fluid Dynamics}

= Solution
{parent=7D}

The steady <Euler equations for an inviscid fluid> are
$$
\rho(u\cdot\nabla)u=-\nabla p-\nabla\chi.
$$
Taking the <scalar> product with $u$ and using
$$
u\cdot(u\cdot\nabla u)
=u\cdot\nabla\left(\frac12|u|^2\right)
$$
gives
$$
u\cdot\nabla
\left(\frac12\rho|u|^2+p+\chi\right)=0.
$$
Thus
$$
\boxed{u\cdot\nabla H=0}.
$$
The <Bernoulli function> $H$ is constant along each <streamline>: fluid particles in steady flow exchange <pressure>, kinetic, and <potential energy> without changing their total mechanical energy density.

Let $y(t)$ be the downward displacement of the free surface from its initial level, and let $U$ be the speed in the tube. Conservation of volume gives
$$
aU=A\dot y.
$$
The free surface and the outlet are both at atmospheric <pressure>, and their vertical separation is $H+h_0-y$. Applying the <Bernoulli equation> between them, while retaining the small free-surface speed, gives
$$
\frac12\left(U^2-\dot y^2\right)
=g(H+h_0-y).
$$
Therefore
$$
\dot y=
\sqrt{\frac{2g(H+h_0-y)}{A^2/a^2-1}}.
$$
The surface reaches the upper tube end when $y=h_0$. Hence the <draining time of a uniform tank through a siphon> is
$$
\begin{aligned}
t
&=\sqrt{\frac{A^2/a^2-1}{2g}}
\int_0^{h_0}\frac{dy}{\sqrt{H+h_0-y}}\\
&=\boxed{
\sqrt{2}\left(\frac{A^2}{a^2}-1\right)^{1/2}
\frac{\sqrt{H+h_0}-\sqrt H}{\sqrt g}}.
\end{aligned}
$$

Solved by gpt-5.6-sol high.

= 8H
{parent=Paper 3}
{scope}
{title2=Markov Chains}

= i
{parent=8h}
{scope}

= Solution
{parent=i}

The independent identically distributed <sequence> $(X_n)$ is itself a <Markov chain>, since the conditional law of $X_{n+1}$ is its common marginal law and does not depend on the past.

Also,
$$
S_{n+1}=S_n+X_{n+1}.
$$
Because $X_{n+1}$ is independent of the past, the conditional law of $S_{n+1}$ depends on the history only through $S_n$. Thus the partial sums $(S_n)$ are a Markov chain.

Solved by gpt-5.6-sol high.

= ii
{parent=8h}
{scope}

= Solution
{parent=ii}

The running minima satisfy
$$
L_{n+1}=\min\{L_n,X_{n+1}\}.
$$
Since $X_{n+1}$ is independent of the past, the conditional law of $L_{n+1}$ depends only on $L_n$. Hence $(L_n)$ is a Markov chain. This is the <running minimum of an independent sequence is Markov> property.

Solved by gpt-5.6-sol high.

= iii
{parent=8h}
{scope}

= Solution
{parent=iii}

The moving sums $(K_n)$ are not necessarily Markov. For a counterexample, let the $X_n$ be independent Bernoulli variables with parameter $1/2$. On the event
$$
K_n=1,
\qquad K_{n-1}=0,
$$
we must have $X_{n-1}=0$ and $X_n=1$, so
$$
\mathbb P(K_{n+1}=2\mid K_n=1,K_{n-1}=0)=\frac12.
$$
On the other hand, $K_n=1$ and $K_{n-1}=2$ force $X_{n-1}=1$ and $X_n=0$, whence
$$
\mathbb P(K_{n+1}=2\mid K_n=1,K_{n-1}=2)=0.
$$
Both conditioning events have positive probability. Knowledge of $K_n$ alone therefore does not determine the next-step law. The <overlapping moving sum need not be Markov>.

Solved by gpt-5.6-sol high.

= 9G
{parent=Paper 3}
{scope}
{title2=Linear Algebra}

= Solution
{parent=9G}

Fix $y\in V$. The map $x\mapsto\theta(x,y)$ is a linear functional, so the finite-dimensional <Riesz representation theorem> gives a unique <vector> $\beta(y)$ such that
$$
\theta(x,y)=\langle x,\beta(y)\rangle
\qquad(x\in V).
$$
For <scalars> $\lambda,\mu$,
$$
\begin{aligned}
\langle x,\beta(\lambda y+\mu z)\rangle
&=\theta(x,\lambda y+\mu z)\\
&=\overline\lambda\,\theta(x,y)
 +\overline\mu\,\theta(x,z)\\
&=\langle x,\lambda\beta(y)+\mu\beta(z)\rangle.
\end{aligned}
$$
Uniqueness gives
$$
\boxed{\beta(\lambda y+\mu z)=\lambda\beta(y)+\mu\beta(z)},
$$
so $\beta$ is linear.

For $\alpha\in\operatorname{End}(V)$, apply this result to
$$
\theta(x,y)=\langle\alpha x,y\rangle.
$$
It produces a unique <linear map> $\alpha^*$ satisfying
$$
\boxed{\langle\alpha x,y\rangle=\langle x,\alpha^*y\rangle},
$$
which proves existence and uniqueness of the adjoint.

Solved by gpt-5.6-sol high.

= a
{parent=9g}
{scope}

= Solution
{parent=a}

Suppose first that $\alpha(U)\subseteq U$. If $y\in U^\perp$ and $u\in U$, then
$$
\langle u,\alpha^*y\rangle
=\langle\alpha u,y\rangle=0.
$$
Thus $\alpha^*(U^\perp)\subseteq U^\perp$.

Conversely, assume that latter inclusion. For $u\in U$ and $y\in U^\perp$,
$$
\langle\alpha u,y\rangle
=\langle u,\alpha^*y\rangle=0.
$$
Hence $\alpha u\in(U^\perp)^\perp=U$. This proves the <adjoint criterion for an invariant orthogonal complement>:
$$
\boxed{\alpha(U)\subseteq U
\iff \alpha^*(U^\perp)\subseteq U^\perp}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=9g}
{scope}

= Solution
{parent=b}

Write $B(x,y)=\langle\alpha x,y\rangle$. The hypothesis says $B(x,x)=0$. Expanding at $x+y$ gives
$$
B(x,y)+B(y,x)=0,
$$
whereas expanding at $x+iy$ gives
$$
-iB(x,y)+iB(y,x)=0.
$$
Therefore $B(x,y)=0$ for every $x,y$, so
$$
\boxed{\alpha=0}.
$$
This is the complex <polarization argument for a vanishing quadratic form>.

The conclusion is false over a real <inner product> space. On $\mathbb R^2$, the nonzero operator
$$
J=\begin{pmatrix}0&-1\\1&0\end{pmatrix}
$$
satisfies $\langle Jx,x\rangle=0$ for every $x$.

Solved by gpt-5.6-sol high.

= c
{parent=9g}
{scope}

= Solution
{parent=c}

For every $x$,
$$
\|\alpha x\|^2=\langle x,\alpha^*\alpha x\rangle,
\qquad
\|\alpha^*x\|^2=\langle x,\alpha\alpha^*x\rangle.
$$
If $\alpha$ is normal, these quantities are equal. Conversely, equality of the norms for all $x$ gives
$$
\langle x,(\alpha^*\alpha-\alpha\alpha^*)x\rangle=0.
$$
The operator in parentheses is self-adjoint, so polarization makes it zero. Therefore
$$
\boxed{\alpha\alpha^*=\alpha^*\alpha
\iff \|\alpha x\|=\|\alpha^*x\|\quad\hbox{for all }x}.
$$
The same equivalence holds over a real <inner product> space: for a self-adjoint operator $T$, the real polarization identity
$$
4\langle Tx,y\rangle
=\langle T(x+y),x+y\rangle
-\langle T(x-y),x-y\rangle
$$
shows that a vanishing quadratic form forces $T=0$.

Solved by gpt-5.6-sol high.

= d
{parent=9g}
{scope}

= Solution
{parent=d}

Proceed by induction on $\dim V$. Over $\mathbb C$, $\alpha$ has an <eigenvector> $v$, say $\alpha v=\lambda v$. The normal operator $\alpha-\lambda I$ satisfies the norm equality from part (c), so
$$
\|(\alpha^*-\overline\lambda I)v\|
=\|(\alpha-\lambda I)v\|=0.
$$
Thus $\alpha^*v=\overline\lambda v$. Part (a) now shows that $v^\perp$ is invariant under both $\alpha$ and $\alpha^*$. The restriction of $\alpha$ to $v^\perp$ is normal. By induction it has an orthonormal eigenbasis, and adjoining the normalized <vector> $v$ proves the finite-dimensional <spectral theorem for normal operators>.

Hence
$$
\boxed{V\text{ has an orthonormal basis of eigenvectors of }\alpha}.
$$

Solved by gpt-5.6-sol high.

= 10E
{parent=Paper 3}
{scope}
{title2=Groups, Rings and Modules}

= a
{parent=10e}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

Suppose $I$ is prime. If
$$
(a+I)(b+I)=0+I,
$$
then $ab\in I$, so $a\in I$ or $b\in I$. Thus one factor is zero and $R/I$ is an <integral> domain.

Conversely, if $R/I$ is an <integral> domain and $ab\in I$, then
$$
(a+I)(b+I)=0+I,
$$
so $a+I=0+I$ or $b+I=0+I$. Therefore $a\in I$ or $b\in I$, and $I$ is prime. This proves the <prime ideal quotient criterion>.

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

In a Boolean <ring>, $(r+1)^2=r+1$. Using $r^2=r$ and $1^2=1$ gives
$$
r+2r+1=r+1,
$$
so
$$
\boxed{2r=0\quad\text{for every }r\in R}.
$$

If $R$ is also a nonzero <integral> domain, then
$$
r(r-1)=r^2-r=0
$$
forces every $r$ to equal $0$ or $1$. Hence $R\cong\mathbb F_2$.

If $I$ is prime in a Boolean <ring>, then $R/I$ is a nonzero Boolean <integral> domain by part (i), and hence is $\mathbb F_2$. Since the quotient is a field,
$$
\boxed{I\text{ is maximal}}.
$$
This is the <prime ideals of a Boolean ring are maximal> property.

Solved by gpt-5.6-sol high.

= b
{parent=10e}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

Reduce coefficients modulo $I$:
$$
\Phi:R[X]\longrightarrow(R/I)[X],
\qquad
\sum_ka_kX^k\longmapsto\sum_k(a_k+I)X^k.
$$
This is a surjective <ring homomorphism> and its kernel consists exactly of <polynomials> all of whose coefficients lie in $I$, namely $I[X]$. The first isomorphism theorem gives the <coefficientwise quotient of a polynomial ring>
$$
\boxed{R[X]/I[X]\cong(R/I)[X]}.
$$
If $I$ is prime, then $R/I$ is an <integral> domain, so $(R/I)[X]$ is an <integral> domain. The quotient criterion therefore shows that $I[X]$ is prime in $R[X]$.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

Take $R=\mathbb Z$ and $I=(2)$. The <ideal> $I$ is maximal, but
$$
\mathbb Z[X]/I[X]\cong\mathbb F_2[X]
$$
is not a field, since $X$ is nonzero and not invertible. Therefore
$$
\boxed{I[X]\text{ need not be maximal}}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=10e}
{scope}

= Solution
{parent=c}

Let $G=\mathbb F_p^\times$, and let $m$ be the least common multiple of the orders of its elements. For every prime power $q^a$ dividing $m$, some element of $G$ has order divisible by $q^a$, and a suitable power of it has order exactly $q^a$. Multiplying these elements over the distinct primes produces, because their orders are coprime, an element of order $m$.

Every element of $G$ is a root of $X^m-1$. A nonzero <polynomial> of degree $m$ over a field has at most $m$ roots, so $p-1\leq m$. On the other hand, every element order divides $p-1$ by Lagrange's theorem, so $m\leq p-1$. Hence $m=p-1$, and
$$
\boxed{\mathbb F_p^\times\text{ is cyclic}}.
$$

The squaring homomorphism has kernel $\{1,-1\}$ because $p$ is odd. Its image $H$ therefore has order $(p-1)/2$ and index two. If $p=3$, then $3=0$ is already a square. Otherwise $2,3\in G$. In the two-element quotient $G/H$, either $2H=H$, or $3H=H$, or both are the nontrivial coset, in which case $6H=H$. Thus one of $2,3,6$ is a square modulo $p$.

Finally,
$$
f(x)=(x^2-2)(x^2-3)(x^2-6).
$$
Whichever of $2,3,6$ is a square supplies a root, so
$$
\boxed{f\text{ has a root in }\mathbb F_p}.
$$
This is the <index-two square-class argument for three related residues>.

Solved by gpt-5.6-sol high.

= 11F
{parent=Paper 3}
{scope}
{title2=Analysis and Topology}

= a
{parent=11f}
{scope}

= Solution
{parent=a}

A <function> $f:(X,d_X)\to(Y,d_Y)$ is uniformly continuous if for every $\varepsilon>0$ there is $\delta>0$ such that
$$
d_X(x,x')<\delta
\implies d_Y(f(x),f(x'))<\varepsilon
$$
for all $x,x'\in X$.

Suppose $f_n\to f$ uniformly and every $f_n$ is uniformly continuous. Given $\varepsilon>0$, choose $N$ such that
$$
d_Y(f_N(x),f(x))<\varepsilon/3
$$
for every $x$. Uniform continuity of $f_N$ supplies $\delta>0$ such that $d_X(x,x')<\delta$ implies
$$
d_Y(f_N(x),f_N(x'))<\varepsilon/3.
$$
The triangle inequality then gives $d_Y(f(x),f(x'))<\varepsilon$. Thus the <uniform limit theorem for uniformly continuous functions> proves that $f$ is uniformly continuous.

<Pointwise convergence> is insufficient. On $[0,1]$, the uniformly <continuous functions> $f_n(x)=x^n$ converge pointwise to
$$
f(x)=\begin{cases}0,&0\leq x<1,\\1,&x=1,\end{cases}
$$
which is discontinuous and therefore not uniformly continuous.

Solved by gpt-5.6-sol high.

= b
{parent=11f}
{scope}

= Solution
{parent=b}

The inequalities imply
$$
B_{d_1}(x,r/\beta)\subseteq B_{d_2}(x,r)
\quad\text{and}\quad
B_{d_2}(x,\alpha r)\subseteq B_{d_1}(x,r).
$$
Thus the two metrics induce the same <open sets> and are equivalent.

The converse fails. On $\mathbb R$, let
$$
d_1(x,y)=|x-y|,
\qquad
d_2(x,y)=|\arctan x-\arctan y|.
$$
The metrics are equivalent because $\arctan:\mathbb R\to(-\pi/2,\pi/2)$ is a homeomorphism, but
$$
\frac{d_2(0,n)}{d_1(0,n)}\longrightarrow0,
$$
so no positive lower comparison constant exists. This is an <equivalent metrics need not be bi-Lipschitz equivalent> example.

Equivalent metrics on the codomain also need not give the same <uniform convergence>. Take $X=\mathbb N$, $Y=\mathbb R$, and use the two metrics above. Define
$$
f(k)=k,
\qquad
f_n(k)=\begin{cases}2n,&k=n,\\k,&k\ne n.
\end{cases}
$$
Then
$$
\sup_kd_2(f_n(k),f(k))
=\arctan(2n)-\arctan n\longrightarrow0,
$$
so $f_n\to f$ uniformly for $d_2$, whereas
$$
\sup_kd_1(f_n(k),f(k))=n,
$$
so convergence is not uniform for $d_1$. This is the <equivalent codomain metrics need not preserve uniform convergence> phenomenon.

Solved by gpt-5.6-sol high.

= 12E
{parent=Paper 3}
{scope}
{title2=Geometry}

= Solution
{parent=12E}

In polar coordinates the metric is
$$
ds^2=\frac{4(dr^2+r^2d\theta^2)}{(1-r^2)^2}.
$$
Along a radius, the hyperbolic distance from the origin to Euclidean radius $r$ is therefore
$$
R=\int_0^r\frac{2\,dt}{1-t^2}
=\log\frac{1+r}{1-r},
$$
so $r=\tanh(R/2)$. The Riemannian area element is
$$
dA=\frac{4r}{(1-r^2)^2}\,dr\,d\theta.
$$
Consequently a hyperbolic disc of radius $R$ has area
$$
\begin{aligned}
A(R)
&=\int_0^{2\pi}\int_0^{\tanh(R/2)}
\frac{4r}{(1-r^2)^2}\,dr\,d\theta\\
&=4\pi\sinh^2(R/2)
=2\pi(\cosh R-1).
\end{aligned}
$$
This proves the <area of a hyperbolic disc> formula from the metric.

For area $\pi/2$ we get $\cosh R=5/4$. Hence $\sinh R=3/4$ and $e^R=2$, so $R=\log2$. Tangent equal discs have centers at distance $2R$. Since the centers lie successively on the same radial geodesic,
$$
d(O,c_n)=2nR=n\log4.
$$
If $r_n$ is the Euclidean coordinate of $c_n$, the radial distance formula gives
$$
\frac{1+r_n}{1-r_n}=4^n.
$$
Therefore the <radial chain of equal hyperbolic discs> has
$$
\boxed{r_n=\frac{4^n-1}{4^n+1}}.
$$

No such isometry to the stated upper-half-plane configuration exists for $n\geq3$. The centers $c_0,\ldots,c_n$ lie on one hyperbolic geodesic, so their images under an isometry would also lie on one geodesic. In the upper-half-plane model, geodesics are vertical lines or semicircles orthogonal to the real axis. Neither type can contain three distinct points of the horizontal line $y=1$, while the centers of the distinct discs $D'_0,\ldots,D'_n$ all lie on that line. This is the <geodesic obstruction for a horizontal chain of hyperbolic discs>.

Solved by gpt-5.6-sol high.

= 13F
{parent=Paper 3}
{scope}
{title2=Complex Analysis}

= Solution
{parent=13F}

For a closed piecewise <smooth curve> $\gamma$ avoiding $w$, its <winding number> about $w$ is
$$
\boxed{
\operatorname{wind}(\gamma,w)
=\frac1{2\pi i}\int_\gamma\frac{dz}{z-w}}.
$$

The <argument principle> says that if a positively oriented closed curve $\gamma$ bounds a domain $U$, and a meromorphic <function> $F$ has no zeros or poles on $\gamma$, then
$$
\boxed{
\frac1{2\pi i}\int_\gamma\frac{F'(z)}{F(z)}\,dz
=N_U(F)-P_U(F)},
$$
where zeros and poles are counted with multiplicity.

Suppose $F$ and $G$ are holomorphic on a neighbourhood of $\overline U$ and
$$
|G(z)|<|F(z)|
\qquad(z\in\gamma).
$$
For $0\leq t\leq1$, $F+tG$ has no boundary zero, because a zero would imply $|F|=t|G|<|F|$. Its argument-principle count is integer-valued and continuous in $t$, hence constant. Thus <Rouché's theorem> states that
$$
\boxed{F\text{ and }F+G\text{ have the same number of zeros in }U}.
$$

Now let $w_0=f(z_0)$. Since $|w_0|<1\leq|f|$ on the unit circle, Rouché's theorem shows that $f$ and $f-w_0$ have the same number of zeros in the unit disc. The latter has the zero $z_0$, so $f$ has at least one zero there. For any $w$ with $|w|<1$, the same boundary inequality shows that $f-w$ has the same positive number of zeros as $f$. Therefore
$$
\boxed{\mathbb D\subseteq f(\mathbb D)}.
$$
This is the <unit-disc image from a boundary modulus lower bound>.

Finally, take a sufficiently small positively oriented circle $C$ around zero. The residue
$$
k=\operatorname{res}_{0}\frac{g'}g
=\frac1{2\pi i}\int_C\frac{g'}g\,dz
$$
is the winding number of $g(C)$ around zero, so $k\in\mathbb Z$. The pole is simple, hence its residue is nonzero and $k\ne0$. For
$$
h(z)=z^{-k}g(z)
$$
we have
$$
\frac{h'}h=\frac{g'}g-\frac{k}{z}.
$$
The second term cancels the complete principal part at zero, so the <integer residue of a logarithmic derivative> proves that $h'/h$ has a removable singularity there.

Solved by gpt-5.6-sol high.

= 14B
{parent=Paper 3}
{scope}
{title2=Methods}

= Solution
{parent=14B}

Separation $y=T(t)X(x)$ and the fixed-end conditions give
$$
X_n(x)=\sin(n\pi x),
\qquad
T_n''+T_n'+n^2\pi^2T_n=0.
$$
Put
$$
\omega_n=\sqrt{n^2\pi^2-\frac14}.
$$
The initial displacement and zero initial <velocity> then give the <separated solution of the damped string equation>
$$
\boxed{
y(t,x)=\sum_{n=1}^{\infty}a_ne^{-t/2}
\left(
\cos(\omega_nt)+\frac{\sin(\omega_nt)}{2\omega_n}
\right)\sin(n\pi x)}.
$$

For the triangular initial displacement,
$$
\begin{aligned}
a_n
&=2\int_0^1y(0,x)\sin(n\pi x)\,dx\\
&=\boxed{\frac{4\sin(n\pi/2)}{n^2\pi^2}}.
\end{aligned}
$$
Thus the even coefficients vanish and the odd ones alternate in sign.

Let
$$
T_n(t)=a_ne^{-t/2}
\left(\cos(\omega_nt)+\frac{\sin(\omega_nt)}{2\omega_n}\right).
$$
Then
$$
T_n'(t)=-a_ne^{-t/2}
\frac{n^2\pi^2}{\omega_n}\sin(\omega_nt).
$$
Orthogonality of the sine and cosine modes, equivalently the <Parseval identity>, gives
$$
\boxed{
E(t)=\frac14\sum_{n=1}^{\infty}a_n^2e^{-t}
\left[
\frac{n^4\pi^4}{\omega_n^2}\sin^2(\omega_nt)
+n^2\pi^2
\left(\cos(\omega_nt)+
\frac{\sin(\omega_nt)}{2\omega_n}\right)^2
\right]}.
$$

The sign of its <derivative> is clearest directly from the equation. Integration by parts, using $y_t=0$ at the fixed endpoints, yields
$$
\begin{aligned}
E'(t)
&=\int_0^1(y_ty_{tt}+y_xy_{xt})\,dx\\
&=\int_0^1y_t(y_{tt}-y_{xx})\,dx
=-\int_0^1y_t^2\,dx\leq0.
\end{aligned}
$$
This is the <energy dissipation identity for a linearly damped string>.

Solved by gpt-5.6-sol high.

= 15C
{parent=Paper 3}
{scope}
{title2=Electromagnetism}

= Solution
{parent=15C}

In electrostatic equilibrium, a nonzero <electric field> inside a conductor would move its free charges. Hence $E=0$ in the conducting material and the potential is constant throughout each connected conductor. Immediately outside its surface the tangential <electric field> is zero. A Gaussian pillbox across the surface gives
$$
(E_{\rm out}-E_{\rm in})\cdot n=\frac\sigma{\epsilon_0}.
$$
Since $E_{\rm in}=0$, the <electrostatic boundary conditions at a conductor> give
$$
\boxed{\sigma=\epsilon_0E_{\rm out}\cdot n}.
$$

For the widely separated shells, let their charges be $Q_1,Q_2$. Their common potential requires
$$
\frac{Q_1}{4\pi\epsilon_0R_1}
=\frac{Q_2}{4\pi\epsilon_0R_2},
\qquad Q_1+Q_2=Q.
$$
Thus the <charge sharing between distant connected spheres> is
$$
\boxed{
Q_1=\frac{R_1}{R_1+R_2}Q,
\qquad
Q_2=\frac{R_2}{R_1+R_2}Q}.
$$

For the <charge on connected concentric spherical shells>, the potentials at the two radii are
$$
\Phi(R_1)=\frac1{4\pi\epsilon_0}
\left(\frac{Q_1}{R_1}+\frac{Q_2}{R_2}\right),
\qquad
\Phi(R_2)=\frac{Q_1+Q_2}{4\pi\epsilon_0R_2}.
$$
Equality forces $Q_1=0$. Hence
$$
\boxed{Q_1=0,
\qquad Q_2=Q},
$$
so all charge lies on the exterior of the outer shell.

For the neutral sphere in the uniform field, the far-field condition gives $\alpha=-E$. Constancy of the potential on $r=R$ gives $\beta=ER^3$. Therefore
$$
\boxed{
\Phi(r,\theta)=-E\left(r-\frac{R^3}{r^2}\right)\cos\theta}.
$$
The outward normal field at the surface is
$$
E_r(R,\theta)
=-\left.\frac{\partial\Phi}{\partial r}\right|_{r=R}
=3E\cos\theta.
$$
The <induced charge on a conducting sphere in a uniform electric field> is consequently
$$
\boxed{\sigma(\theta)=3\epsilon_0E\cos\theta}.
$$
Finally,
$$
\int_{S^2}\sigma\,dA
=6\pi\epsilon_0ER^2\int_0^\pi
\cos\theta\sin\theta\,d\theta=0,
$$
confirming neutrality.

Solved by gpt-5.6-sol high.

= 16D
{parent=Paper 3}
{scope}
{title2=Fluid Dynamics}

= Solution
{parent=16D}

Take the sphere to move in the positive $z$-direction and use spherical coordinates centred on it. An axisymmetric harmonic potential that decays at infinity has the dipole form $A\cos\theta/r^2$. The no-penetration condition in the laboratory frame is
$$
\left.\frac{\partial\phi}{\partial r}\right|_{r=a}
=U\cos\theta.
$$
It fixes $A=-Ua^3/2$, so the <potential flow around a translating sphere> is
$$
\boxed{
\phi(r,\theta)=-\frac{Ua^3}{2r^2}\cos\theta}.
$$
The laboratory-frame <velocity> components are
$$
\boxed{
u_r=\frac{Ua^3}{r^3}\cos\theta,
\qquad
u_\theta=\frac{Ua^3}{2r^3}\sin\theta,
\qquad
u_\varphi=0}.
$$

The fluid <kinetic energy> is
$$
\begin{aligned}
K_f
&=\frac\rho2\int_a^\infty\int_{S^2}
\frac{U^2a^6}{r^6}
\left(\cos^2\theta+\frac14\sin^2\theta\right)
r^2\,d\Omega\,dr\\
&=\boxed{\frac{\pi}{3}\rho a^3U^2}
=\frac12\left(\frac12\rho V\right)U^2,
\end{aligned}
$$
where $V=4\pi a^3/3$. Thus the <added mass of a sphere> is $\rho V/2$.

When the sphere falls, replacing fluid of density $\rho$ by material of density $\rho_s$ lowers the gravitational <potential energy> at rate
$$
\boxed{\dot P=-(\rho_s-\rho)VgU}.
$$
The total <kinetic energy> of sphere and fluid is
$$
K=\frac12\left(\rho_s+\frac\rho2\right)VU^2.
$$
Conservation of total energy gives
$$
\left(\rho_s+\frac\rho2\right)VU\frac{dU}{dt}
-(\rho_s-\rho)VgU=0.
$$
For $U\ne0$, the <acceleration of a freely falling sphere with added mass> is
$$
\boxed{
\frac{dU}{dt}=\frac{\rho_s-\rho}{\rho_s+\rho/2}\,g}.
$$
The same formula holds at release by continuity.

Solved by gpt-5.6-sol high.

= 17A
{parent=Paper 3}
{scope}
{title2=Numerical Analysis}

= Solution
{parent=17A}

If the bound is to hold with finite $c$, the error functional must vanish on every <polynomial> whose fourth <derivative> is zero. Thus the scheme must be exact for degrees zero through three. Applying it to powers of $t=x+1$ gives
$$
\begin{aligned}
a_{-1}+a_0+a_1+a_2&=0,\\
a_0+2a_1+3a_2&=0,\\
a_0+4a_1+9a_2&=2,\\
a_0+8a_1+27a_2&=0.
\end{aligned}
$$
Solving,
$$
\boxed{a_{-1}=2,
\qquad a_0=-5,
\qquad a_1=4,
\qquad a_2=-1}.
$$
This is the <four-point one-sided second-derivative formula>.

The <Peano kernel theorem> says that if a linear functional $L$ annihilates all <polynomials> of degree below $r$, then for $f\in C^r[a,b]$,
$$
L(f)=\int_a^bK(t)f^{(r)}(t)\,dt,
\qquad
K(t)=L\left(\frac{(x-t)_+^{r-1}}{(r-1)!}\right).
$$

Here $r=4$ and $L=e$. Under the stated nonnegativity assumption,
$$
|e(f)|\leq
\left(\int_{-1}^2K(t)\,dt\right)
\max_{[-1,2]}|f^{(4)}|.
$$
The <integral> equals $e(q)$ for $q(x)=(x+1)^4/24$, since $q^{(4)}=1$. Now $q''(-1)=0$ and
$$
\eta(q)=\frac{-5+4\cdot16-81}{24}=-\frac{11}{12}.
$$
Therefore
$$
\int_{-1}^2K(t)\,dt=e(q)=\frac{11}{12}.
$$
Equality is attained by $q$, so the <sharp Peano-kernel constant for the four-point endpoint second derivative> is
$$
\boxed{c=\frac{11}{12}}.
$$

Solved by gpt-5.6-sol high.

= 18H
{parent=Paper 3}
{scope}
{title2=Statistics}

= a
{parent=18h}
{scope}

= Solution
{parent=a}

Under $H_0$,
$$
\sqrt n\,\overline X\sim N(0,1).
$$
Let $z_{1-\alpha/2}=\Phi^{-1}(1-\alpha/2)$. The standard two-sided level-$\alpha$ test rejects exactly when
$$
\boxed{|\sqrt n\,\overline X|>z_{1-\alpha/2}}.
$$
For the observed value $\overline X=\overline x$, its <two-sided Gaussian p-value> is
$$
\boxed{
p(\overline x)=2\left[1-\Phi(\sqrt n\,|\overline x|)\right]}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=18h}
{scope}

= Solution
{parent=b}

Conditionally on $\mu$,
$$
\overline X\mid\mu\sim N\left(\mu,\frac1n\right),
$$
so
$$
\boxed{
f(\overline x\mid\mu)
=\sqrt{\frac n{2\pi}}
\exp\left[-\frac n2(\overline x-\mu)^2\right]}.
$$

Under the continuous half of the prior, write $\overline X=\mu+Z$ with independent
$$
\mu\sim N(0,\tau^2),
\qquad
Z\sim N(0,1/n).
$$
Thus $\overline X\sim N(0,\tau^2+1/n)$ there. Mixing this density with the point-null density gives the <marginal likelihood for a Gaussian point-null mixture>
$$
\boxed{
\begin{aligned}
m(\overline x)={}&\frac12\sqrt{\frac n{2\pi}}
 e^{-n\overline x^2/2}\\
&+\frac1{2\sqrt{2\pi(\tau^2+1/n)}}
 \exp\left[-\frac{\overline x^2}{2(\tau^2+1/n)}\right].
\end{aligned}}
$$

Solved by gpt-5.6-sol high.

= c
{parent=18h}
{scope}

= Solution
{parent=c}

Bayes' formula divides the point-null contribution to the mixture density by the full marginal density:
$$
q(\overline x)
=\frac{\phi_{1/n}(\overline x)}
{\phi_{1/n}(\overline x)+
 \phi_{\tau^2+1/n}(\overline x)}.
$$
Since
$$
\frac{\phi_{\tau^2+1/n}(\overline x)}
{\phi_{1/n}(\overline x)}
=\frac1{\sqrt{1+n\tau^2}}
\exp\left[
\frac{n^2\tau^2\overline x^2}{2(1+n\tau^2)}
\right],
$$
the <posterior probability of a Gaussian point null> is
$$
\boxed{
q(\overline x)=
\left\{
1+\frac1{\sqrt{1+n\tau^2}}
\exp\left[
\frac{n^2\tau^2\overline x^2}{2(1+n\tau^2)}
\right]
\right\}^{-1}}.
$$

Solved by gpt-5.6-sol high.

= d
{parent=18h}
{scope}

= i
{parent=d}
{scope}

= Solution
{parent=i}

For $n=100$ and $\tau=1$,
$$
p(0)=1,
\qquad
q(0)=\frac1{1+1/\sqrt{101}}<1.
$$
Both are continuous, so when $|\overline x|$ is sufficiently small,
$$
\boxed{p(\overline x)>q(\overline x)}.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=d}
{scope}

= Solution
{parent=ii}

For large $|\overline x|$, the supplied normal-tail approximation gives
$$
p(\overline x)
\sim\frac1{5\sqrt{2\pi}\,|\overline x|}
 e^{-50\overline x^2}.
$$
Meanwhile,
$$
q(\overline x)
\sim\sqrt{101}
\exp\left(-\frac{5000}{101}\overline x^2\right).
$$
Since $5000/101<50$, the posterior point-null probability decays more slowly. Therefore, for sufficiently large $|\overline x|$,
$$
\boxed{q(\overline x)>p(\overline x)}.
$$
This reversal is the <Jeffreys-Lindley paradox for a Gaussian point null>.

Solved by gpt-5.6-sol high.

= 19H
{parent=Paper 3}
{scope}
{title2=Optimisation}

= a
{parent=19h}
{scope}

= Solution
{parent=a}

A feasible flow assigns $f_{ij}$ to each directed edge so that
$$
0\leq f_{ij}\leq C_{ij}
$$
and inflow equals outflow at every vertex other than the source and sink. Its value $|f|$ is the net outflow from the source. For a set $S$ containing the source but not the sink, the associated cut has capacity
$$
C(S,V\setminus S)
=\sum_{i\in S,\,j\notin S}C_{ij}.
$$
The <max-flow min-cut theorem> states
$$
\boxed{
\max_f|f|=\min_{S\ni s,\,t\notin S}C(S,V\setminus S)}.
$$

For every flow and cut, conservation at vertices inside $S$ gives
$$
|f|=f(S,V\setminus S)-f(V\setminus S,S)
\leq C(S,V\setminus S).
$$
This proves the weak inequality.

A maximum flow exists because the feasible-flow polytope is nonempty, closed, and bounded. Form its residual graph: a forward edge has residual capacity $C_{ij}-f_{ij}$, and a reverse edge has residual capacity $f_{ij}$. If the residual graph contained a source-to-sink path, augmenting by the smallest positive residual capacity on that path would increase the flow, contradicting maximality.

Let $S$ be the vertices reachable from the source in the residual graph. The sink is not in $S$. Every original edge from $S$ to its complement is saturated, and every original edge from the complement into $S$ carries zero flow; otherwise the appropriate residual edge would make its other endpoint reachable. Hence
$$
|f|=f(S,V\setminus S)-f(V\setminus S,S)
=C(S,V\setminus S).
$$
The maximum flow therefore equals the capacity of this cut, completing the proof.

Solved by gpt-5.6-sol high.

= b
{parent=19h}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

Label the seven black intermediate vertices, from left to right and top to bottom within each column, by $a,b,c,d,e,f,g$: thus $a$ is directly below $r$, $b$ is the lower-left vertex, $c,d$ are the next upper and lower vertices, $e$ is the middle-right vertex, and $f,g$ are the upper-right and lower-right vertices.

For $x=4$, the following nonzero edge flows are feasible:
$$
\begin{array}{c|cccccccccc}
\text{edge}&sr&rc&cf&ft&sb&bd&de&et&dg&gt\\ \hline
\text{flow}&4&4&4&4&5&5&3&3&2&2.
\end{array}
$$
Their value is $9$. The cut with source side $\{s\}$ has capacity
$$
C(\{s\},V\setminus\{s\})=x+5=9.
$$
By the max-flow min-cut theorem,
$$
\boxed{\delta^*(4)=9}.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

Two cuts give the upper bounds
$$
\delta^*(x)\leq x+5
$$
from the source cut, and
$$
\delta^*(x)\leq14
$$
from the cut whose source side is
$$
S=\{s,r,a,b,c,d,f\}.
$$
The latter cut crosses the edges $ce,de,dg,ft$, of respective capacities $3,3,2,6$.

At $x=0$, a flow of value $5$ is
$$
sb=bd=5,
\qquad de=et=3,
\qquad dg=gt=2.
$$
At $x=9$, a flow of value $14$ is given by
$$
\begin{array}{c|rrrrrrrrrrrrr}
\text{edge}&sr&sb&rc&ra&ac&bd&cf&ce&de&dg&et&ft&gt\\ \hline
\text{flow}&9&5&6&3&3&5&6&3&3&2&6&6&2.
\end{array}
$$
For $0\leq x\leq9$, take the convex combination of these two flows with weights $1-x/9$ and $x/9$. It is feasible at capacity $x$ and has value $5+x$. For $x\geq9$, the second flow remains feasible and has value $14$. The two cut bounds are therefore attained, and the <parametric maximum flow with one source capacity> is
$$
\boxed{
\delta^*(x)=\min\{x+5,14\}
=\begin{cases}
x+5,&0\leq x\leq9,\\
14,&x\geq9.
\end{cases}}
$$

Solved by gpt-5.6-sol high.