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past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ib/paper-3.bigb
= Paper 3
{scope}
https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2021/paperib_3_2021.pdf
= 1G
{parent=Paper 3}
{scope}
{title2=Groups, Rings and Modules}
= Solution
{parent=1G}
Let $G$ act by left multiplication on the set $G/H$ of $n$ left <coset>[cosets]. This gives a <group action> homomorphism
$$
\varphi:G\longrightarrow S_n.
$$
Its kernel
$$
K=\bigcap_{g\in G}gHg^{-1}
$$
is a <normal subgroup> of $G$, and the <first isomorphism theorem> gives $G/K\cong\operatorname{im}\varphi$. By <Lagrange's theorem>, $|G/K|$ divides $n!$. The image acts transitively on the $n$ cosets, so the <orbit-stabilizer theorem> shows that $n$ divides $|G/K|$; in particular,
$$
|G/K|\geq n.
$$
Now suppose $G$ is nonabelian and <simple group>[simple]. Since $H$ is proper, the coset action is nontrivial, so $K\ne G$. Simplicity gives $K=\{e\}$, and $\varphi$ embeds $G$ into $S_n$. Composing with the <sign homomorphism> gives
$$
G\longrightarrow S_n\longrightarrow\{\pm1\}.
$$
Its kernel is normal. A nontrivial map would embed the simple group $G$ into the abelian group of order two, which is impossible because $G$ is nonabelian. The sign is therefore always $+1$, so
$$
\boxed{G\cong\varphi(G)\leq A_n}.
$$
Solved by gpt-5.6-sol high.
= 2E
{parent=Paper 3}
{scope}
{title2=Geometry}
= Solution
{parent=2E}
For a <geodesic triangle> with interior angles $\alpha,\beta,\gamma$, the local <Gauss-Bonnet theorem> is
$$
\boxed{\int_TK\,dA=\alpha+\beta+\gamma-\pi},
$$
because its geodesic sides have zero <geodesic curvature>.
Triangulate a closed oriented surface into $F$ geodesic triangles, with $E$ edges and $V$ vertices. Summing the local formula, the angles around each vertex total $2\pi$, so
$$
\int_SK\,dA=2\pi V-\pi F.
$$
Since every triangular face has three edges and every edge belongs to two faces, $3F=2E$. Hence
$$
\int_SK\,dA=2\pi(V-E+F)
=\boxed{2\pi\chi(S)},
$$
which is the global Gauss-Bonnet theorem.
For the sphere $S_r$, the unit normal is $N(p)=p/r$. Its <shape operator> is, up to the conventional sign, $(1/r)I$ on each tangent plane. Both <principal curvature>[principal curvatures] therefore have magnitude $1/r$, and the <Gaussian curvature> is
$$
\boxed{K=\frac1{r^2}}.
$$
An octant has one eighth of the sphere's area:
$$
\operatorname{area}(T)=\frac18(4\pi r^2)=\frac{\pi r^2}{2}.
$$
Thus $\int_TK\,dA=\pi/2$. Its three great-circle sides meet at three right angles, so
$$
\alpha+\beta+\gamma-\pi
=3\frac\pi2-\pi=\frac\pi2.
$$
The two sides of the local Gauss-Bonnet formula agree directly.
Solved by gpt-5.6-sol high.
= 3B
{parent=Paper 3}
{scope}
{title2=Complex Methods}
= Solution
{parent=3B}
Computing the <Laplacian> gives
$$
\Delta\!\left(x\cosh y\sin x\right)=2\cos x\cosh y
$$
and
$$
\Delta\!\left(Ay\sinh y\cos x\right)=2A\cos x\cosh y.
$$
Thus <Laplace's equation> holds exactly when
$$
\boxed{A=-1}.
$$
For $z=x+iy$,
$$
\sin z=\sin x\cosh y+i\cos x\sinh y,
$$
so
$$
\operatorname{Re}(z\sin z)
=x\sin x\cosh y-y\cos x\sinh y
=\phi(x,y).
$$
One suitable <analytic function> is therefore
$$
\boxed{f(z)=z\sin z}.
$$
Solved by gpt-5.6-sol high.
= 4D
{parent=Paper 3}
{scope}
{title2=Variational Principles}
= Solution
{parent=4D}
For
$$
F(x,y,y')=y'^2+yy'+y'+y^2+yx^2,
$$
the <Euler-Lagrange equation> is
$$
\frac d{dx}(2y'+y+1)-(y'+2y+x^2)=0,
$$
or
$$
y''-y=\frac{x^2}{2}.
$$
The general solution of this <inhomogeneous linear differential equation> is
$$
y(x)=Ce^x+De^{-x}-\frac{x^2}{2}-1.
$$
The condition $y(0)=-1$ gives $C+D=0$, and the condition at $x=1$ gives $C=1$. Therefore
$$
\boxed{y(x)=e^x-e^{-x}-\frac{x^2}{2}-1}.
$$
Solved by gpt-5.6-sol high.
= 5A
{parent=Paper 3}
{scope}
{title2=Methods}
= Solution
{parent=5A}
The function $f$ has zero mean and its cosine coefficients vanish:
$$
a_0=0,\qquad a_n=0.
$$
Its sine coefficients are
$$
b_n=\frac1\pi\left(\int_0^\pi\sin(n\theta)\,d\theta
-\int_\pi^{2\pi}\sin(n\theta)\,d\theta\right),
$$
so
$$
\boxed{
b_n=\begin{cases}
\dfrac4{\pi n},&n\text{ odd},\\
0,&n\text{ even}.
\end{cases}}
$$
Away from the corners, $F'=f$. Differentiating the <Fourier series> shows that $nB_n=a_n$ and $-nA_n=b_n$. Also,
$$
A_0=\frac1\pi\int_0^{2\pi}F(\theta)\,d\theta=\pi.
$$
Consequently
$$
\boxed{B_n=0,\qquad
A_n=\begin{cases}
-\dfrac4{\pi n^2},&n\text{ odd},\\
0,&n\text{ even},
\end{cases}}
$$
and
$$
F(\theta)=\frac\pi2-\frac4\pi
\sum_{r=0}^\infty\frac{\cos((2r+1)\theta)}{(2r+1)^2}.
$$
Evaluating the series for $f$ at $\theta=\pi/2$ gives the <Leibniz formula for pi>,
$$
\boxed{\sum_{r=0}^\infty\frac{(-1)^r}{2r+1}=\frac\pi4}.
$$
Evaluating the continuous Fourier series for $F$ at $\theta=0$ gives
$$
0=\frac\pi2-\frac4\pi\sum_{r=0}^\infty\frac1{(2r+1)^2},
$$
and hence
$$
\boxed{\sum_{r=0}^\infty\frac1{(2r+1)^2}=\frac{\pi^2}{8}}.
$$
Solved by gpt-5.6-sol high.
= 6C
{parent=Paper 3}
{scope}
{title2=Quantum Mechanics}
= i
{parent=6c}
{scope}
= Solution
{parent=i}
The states $\chi_1$ and $\chi_2$ are orthonormal <energy eigenstate>[energy eigenstates] with hydrogenic energies
$$
E_n=-\frac{mK^2}{2\hbar^2n^2}=-\frac{K}{2an^2}.
$$
The <Born rule> therefore gives
$$
\langle E\rangle
=\frac14E_1+\frac34E_2
=\frac7{16}E_1
=\boxed{-\frac{7mK^2}{32\hbar^2}}
=\boxed{-\frac{7K}{32a}}.
$$
Solved by gpt-5.6-sol high.
= ii
{parent=6c}
{scope}
= Solution
{parent=ii}
At time $t$ the state is
$$
\chi(t)=\frac12e^{-iE_1t/\hbar}\chi_1
+\frac{\sqrt3}{2}e^{-iE_2t/\hbar}\chi_2.
$$
Direct evaluation of the radial <integral> gives the <matrix element>[matrix elements]
$$
\langle\chi_1|r|\chi_1\rangle=\frac{3a}{2},
\qquad
\langle\chi_2|r|\chi_2\rangle=6a,
$$
and
$$
\begin{aligned}
\langle\chi_1|r|\chi_2\rangle
&=4\pi\int_0^\infty r^3\chi_1(r)\chi_2(r)\,dr\\
&=-\frac{64a}{81\sqrt2}.
\end{aligned}
$$
It follows that
$$
\boxed{
R(t)=\frac{39a}{8}
-\frac{32a}{81}\sqrt{\frac32}
\cos\!\left(\frac{E_2-E_1}{\hbar}t\right)
}.
$$
Thus $R(t)$ oscillates sinusoidally about $39a/8$. Its <angular frequency> is
$$
\boxed{\omega=\frac{E_2-E_1}{\hbar}
=\frac{3mK^2}{8\hbar^3}
=\frac{3K}{8a\hbar}}.
$$
Solved by gpt-5.6-sol high.
= 7A
{parent=Paper 3}
{scope}
{title2=Fluid Dynamics}
= a
{parent=7a}
{scope}
= Solution
{parent=a}
Writing $R=x^2+y^2$, direct differentiation gives
$$
\frac{\partial u}{\partial x}
=\frac{-2x(x^2-3y^2)}{R^3},
\qquad
\frac{\partial v}{\partial y}
=\frac{2x(x^2-3y^2)}{R^3},
$$
so $\nabla\mathbin{\cdot}\mathbf u=0$. Likewise,
$$
\frac{\partial v}{\partial x}
=\frac{-2y(3x^2-y^2)}{R^3}
=\frac{\partial u}{\partial y},
$$
so the two-dimensional <vorticity> $\partial_xv-\partial_yu$ vanishes. The flow is therefore <incompressible flow>[incompressible] and <irrotational flow>[irrotational] away from the origin.
Solved by gpt-5.6-sol high.
= b
{parent=7a}
{scope}
= Solution
{parent=b}
With the convention
$$
u=\frac{\partial\psi}{\partial y},
\qquad
v=-\frac{\partial\psi}{\partial x},
$$
a <stream function> is
$$
\boxed{\psi(x,y)=\frac{y}{x^2+y^2}},
$$
up to an additive constant. Differentiation recovers both velocity components.
Solved by gpt-5.6-sol high.
= c
{parent=7a}
{scope}
= Solution
{parent=c}
For a <velocity potential> $\phi$, one requires $u=\partial_x\phi$ and $v=\partial_y\phi$. Integration gives
$$
\boxed{\phi(x,y)=-\frac{x}{x^2+y^2}},
$$
again up to an additive constant.
Solved by gpt-5.6-sol high.
= 8H
{parent=Paper 3}
{scope}
{title2=Markov Chains}
= a
{parent=8h}
{scope}
= Solution
{parent=a}
States $i$ and $j$ <communicating state>[communicate] when each is accessible from the other: there exist $m,n\geq0$ with
$$
(P^m)_{ij}>0,\qquad(P^n)_{ji}>0.
$$
This is an <equivalence relation>, and its equivalence classes are the <communicating class>[communicating classes]. A class is <closed communicating class>[closed] when no positive-probability transition leaves it.
Solved by gpt-5.6-sol high.
= b
{parent=8h}
{scope}
= Solution
{parent=b}
Reading the directed edges with positive transition probability, the communicating classes are
$$
\boxed{\{3\},\qquad\{4\},\qquad\{1,2,5,6\}}.
$$
State $3$ can move to $2$ or $4$, and state $4$ can move to $2$ or $5$, so their singleton classes are not closed. Every transition from $1,2,5,$ or $6$ remains within $\{1,2,5,6\}$, so
$$
\boxed{\{1,2,5,6\}\text{ is the unique closed class}}.
$$
Solved by gpt-5.6-sol high.
= c
{parent=8h}
{scope}
= Solution
{parent=c}
Let $h_i=\mathbb E_iT_6$. Then $h_6=0$, and <first-step analysis> on the closed class gives
$$
\begin{aligned}
h_1&=1+\frac14h_5,\\
h_2&=1+\frac14h_1+\frac12h_5,\\
h_5&=1+\frac14h_1+\frac12h_2.
\end{aligned}
$$
Solving these <linear equation>[linear equations] gives
$$
h_5=\frac{20}{7},
\qquad
\boxed{h_1=\frac{12}{7}}.
$$
Solved by gpt-5.6-sol high.
= 9E
{parent=Paper 3}
{scope}
{title2=Linear Algebra}
= a
{parent=9e}
{scope}
= i
{parent=a}
{scope}
= Solution
{parent=i}
The <rank-nullity theorem> says that for a linear map $T:V\to Z$ with finite-dimensional domain,
$$
\dim V=\dim\ker T+\dim\operatorname{im}T.
$$
The <direct sum> $U\oplus W$ consists of pairs $(u,w)$ with componentwise vector-space operations. The canonical inclusions are
$$
i(u)=(u,0),
\qquad
j(w)=(0,w).
$$
For subspaces $U,W\leq V$, define
$$
q:U\oplus W\longrightarrow U+W,
\qquad
q(u,w)=u+w.
$$
This is surjective, and
$$
\ker q=\{(x,-x):x\in U\cap W\}
=\operatorname{im}l.
$$
The <first isomorphism theorem for vector spaces> therefore gives
$$
\boxed{(U\oplus W)/\operatorname{im}l\cong U+W\leq V}.
$$
Solved by gpt-5.6-sol high.
= ii
{parent=a}
{scope}
= Solution
{parent=ii}
Row reduction of the matrices whose columns are the displayed spanning vectors gives
$$
\dim U=3,\qquad\dim W=3.
$$
Row-reducing the matrix formed from all six columns gives
$$
\dim(U+W)=4.
$$
The <dimension formula for a sum of subspaces> now yields
$$
\boxed{\dim(U\cap W)
=\dim U+\dim W-\dim(U+W)
=3+3-4=2}.
$$
Solved by gpt-5.6-sol high.
= b
{parent=9e}
{scope}
= Solution
{parent=b}
Expand $\det(A+tB)$ by <multilinearity of the determinant> in its columns. The coefficient of $t^j$ is a sum of determinants obtained by choosing $j$ columns from $B$ and the remaining columns from $A$. Since the columns of $B$ span a space of dimension $k$, every choice of more than $k$ columns from $B$ is <linear dependence>[linearly dependent]. All coefficients of $t^j$ for $j>k$ therefore vanish, so
$$
\deg\det(A+tB)\leq k.
$$
If $k=n$, then $B$ is invertible and the coefficient of $t^n$ is $\det B\ne0$. Hence the degree is exactly $n$.
For a zero-polynomial example with $k=1$, take
$$
A=\begin{pmatrix}0&0\\0&0\end{pmatrix},
\qquad
B=\begin{pmatrix}1&0\\0&0\end{pmatrix}.
$$
Then $\operatorname{rank}B=1$, while $\det(A+tB)=0$ for every $t$.
Solved by gpt-5.6-sol high.
= 10G
{parent=Paper 3}
{scope}
{title2=Groups, Rings and Modules}
= i
{parent=10g}
{scope}
= Solution
{parent=i}
Assume that $p$ is a <prime element>. If $p=ab$, then $p$ divides $ab$, so $p$ divides $a$ or $b$. If $p\mid a$, write $a=pc$; cancellation in the <integral domain> gives $1=bc$, so $b$ is a <unit>. The other case is symmetric. Hence every prime element is <irreducible element>[irreducible].
Solved by gpt-5.6-sol high.
= ii
{parent=10g}
{scope}
= Solution
{parent=ii}
Suppose $p$ is irreducible and
$$
(p)\subseteq(a)\subseteq R.
$$
Then $p=ab$ for some $b\in R$. Irreducibility says that either $a$ is a unit, in which case $(a)=R$, or $b$ is a unit, in which case $(a)=(p)$. Thus no proper ideal lies strictly between $(p)$ and $R$, so
$$
\boxed{(p)\text{ is a maximal ideal}}.
$$
Solved by gpt-5.6-sol high.
= iii
{parent=10g}
{scope}
= Solution
{parent=iii}
For any commutative ring with identity, the <maximal ideal quotient criterion> states that an ideal $I$ is maximal exactly when $R/I$ is a field. Applying it to $I=(p)$ proves
$$
(p)\text{ maximal}\quad\Longleftrightarrow\quad R/(p)\text{ is a field}.
$$
Solved by gpt-5.6-sol high.
= iv
{parent=10g}
{scope}
= Solution
{parent=iv}
Every <field> is an <integral domain>, so if $R/(p)$ is a field then it is an integral domain.
Solved by gpt-5.6-sol high.
= v
{parent=10g}
{scope}
= Solution
{parent=v}
By the <prime ideal quotient criterion>,
$$
R/(p)\text{ is an integral domain}
\quad\Longleftrightarrow\quad
(p)\text{ is a prime ideal}.
$$
For a principal ideal, this is precisely the assertion that $p$ is a prime element. This closes the cycle and proves that (i)--(v) are equivalent.
Now let $\varphi:R\to S$ be a surjective ring homomorphism, where $R$ is a <principal ideal domain> and $S$ is an integral domain. Write
$$
\ker\varphi=(a).
$$
If $a=0$, the <first isomorphism theorem for rings> makes $\varphi$ an isomorphism. Otherwise
$$
S\cong R/(a)
$$
is an integral domain, so the equivalences just proved show that $(a)$ is maximal and $S$ is a field.
Next suppose that the <polynomial ring> $R[X]$ is a principal ideal domain. It follows first that $R$ is an integral domain. For nonzero $a\in R$, the ideal
$$
(a,X)=(f)
$$
is principal. Since $f$ divides the nonzero constant $a$, it is constant; since it also divides $X$, that constant divides the coefficient $1$, so $f$ is a unit. Therefore $(a,X)=R[X]$, and there are polynomials $g,h$ such that
$$
1=ag(X)+Xh(X).
$$
Setting $X=0$ gives $1=ag(0)$, so every nonzero $a\in R$ is a unit. Hence
$$
\boxed{R\text{ is a field}}.
$$
Finally, let $R$ be an integral domain in which every two nonzero elements have a <greatest common divisor>. If an irreducible $p$ divides $ab$, then $\gcd(p,a)$ is either a unit or an associate of $p$. In the second case $p\mid a$. In the first case, the <Euclid lemma in a greatest-common-divisor domain> gives $p\mid b$. Thus every irreducible element of $R$ is prime.
Solved by gpt-5.6-sol high.
= 11F
{parent=Paper 3}
{scope}
{title2=Analysis and Topology}
= Solution
{parent=11F}
A <topological space> is <connected space>[connected] when it is not the union of two disjoint nonempty open subsets. It is <path-connected space>[path-connected] when every two points $x,y$ can be joined by a <path>, meaning a continuous map $\gamma:[0,1]\to X$ with $\gamma(0)=x$ and $\gamma(1)=y$.
To prove that $[0,1]$ is connected, suppose it had a separation $[0,1]=U\cup V$ with $0\in U$. Let
$$
c=\sup\{x\in[0,1]:[0,x]\subseteq U\}.
$$
Whichever of $U$ or $V$ contains $c$, its relative openness supplies an interval about $c$ that contradicts either the definition of the <supremum> or the fact that all points immediately below $c$ lie in $U$. Thus no separation exists.
If $X$ is path-connected and $X=U\cup V$ were a separation, choose $u\in U$ and $v\in V$. A path from $u$ to $v$ would make $[0,1]$ the union of the disjoint nonempty open sets $\gamma^{-1}(U)$ and $\gamma^{-1}(V)$, contradicting the connectedness just proved. Hence
$$
\boxed{\text{path-connected}\Longrightarrow\text{connected}}.
$$
Now let $X\subseteq\mathbb R^n$ be open. Every point $x\in X$ lies in an open ball contained in $X$, and an <open ball> is path-connected by straight line segments. It follows that every <path component> of $X$ is open. If there were more than one path component, one component and the union of all the others would separate $X$. Consequently a connected open subset of Euclidean space is path-connected. The converse follows from the preceding implication.
The same argument answers the locally Euclidean question affirmatively. Every point has a neighbourhood homeomorphic to an open subset of $\mathbb R^n$, hence a path-connected open neighbourhood after restricting to a sufficiently small ball. Thus $X$ is <locally path-connected space>[locally path-connected], its path components are open, and connectedness forces there to be only one.
For the final example, put
$$
Y=A\cup\bigcup_{n\geq1}C_n.
$$
The set $Y$ is path-connected: each vertical segment $C_n$ meets the horizontal segment $A$. The segment $B$ is also path-connected and lies in the <closure> of $Y$, because $(1/n,y)\to(0,y)$ for every $y\in[1/2,1]$. A connected set together with a connected subset of its closure is connected, so
$$
\boxed{X=Y\cup B\text{ is connected}}.
$$
It is not path-connected. The image of the first coordinate of any path in $X$ lies in
$$
\{0\}\cup\{1/n:n\geq1\}\cup(0,1]
$$
but, while the path has positive height, its first coordinate lies in the totally disconnected set $\{0\}\cup\{1/n:n\geq1\}$. A path beginning on $B$ cannot leave $x=0$ before reaching height zero, and $X$ contains no point $(0,y)$ with $0\leq y<1/2$. Hence no path joins $B$ to $Y$, and
$$
\boxed{X\text{ is not path-connected}}.
$$
Solved by gpt-5.6-sol high.
= 12E
{parent=Paper 3}
{scope}
{title2=Geometry}
= Solution
{parent=12E}
For a smooth curve $\gamma(t)$ on a <Riemannian surface>, its <energy of a curve> is
$$
\boxed{\mathcal E(\gamma)=\frac12\int_0^1
\langle\dot\gamma,\dot\gamma\rangle\,dt}.
$$
Choose a local parameterization $\mathbf X(u,v)$ and write $\gamma(t)=\mathbf X(u(t),v(t))$. With coefficients of the <first fundamental form>
$$
E=\mathbf X_u\mathbin{\cdot}\mathbf X_u,\qquad
F=\mathbf X_u\mathbin{\cdot}\mathbf X_v,\qquad
G=\mathbf X_v\mathbin{\cdot}\mathbf X_v,
$$
the Lagrangian is
$$
L=\frac12(E\dot u^2+2F\dot u\dot v+G\dot v^2).
$$
The <Euler-Lagrange equations> are equivalently
$$
\begin{aligned}
0={}&E\ddot u+F\ddot v
+(\mathbf X_{uu}\mathbin{\cdot}\mathbf X_u)\dot u^2
+2(\mathbf X_{uv}\mathbin{\cdot}\mathbf X_u)\dot u\dot v
+(\mathbf X_{vv}\mathbin{\cdot}\mathbf X_u)\dot v^2,\\
0={}&F\ddot u+G\ddot v
+(\mathbf X_{uu}\mathbin{\cdot}\mathbf X_v)\dot u^2
+2(\mathbf X_{uv}\mathbin{\cdot}\mathbf X_v)\dot u\dot v
+(\mathbf X_{vv}\mathbin{\cdot}\mathbf X_v)\dot v^2.
\end{aligned}
$$
After multiplying by the inverse metric these become the <geodesic equation>
$$
\ddot q^k+\Gamma^k_{ij}\dot q^i\dot q^j=0,
$$
where the <Christoffel symbols> are determined by $E,F,G$.
If a straight line segment lies in the surface, parameterize it by
$$
\gamma(t)=P+t(Q-P).
$$
Then $\ddot\gamma=0$, so its acceleration has zero tangential component. The two displayed equations hold directly, and the segment is a geodesic.
For the one-sheeted <hyperboloid>
$$
H=\{(x,y,z):x^2+y^2-z^2=1\},
$$
put
$$
\rho_0=\sqrt{x_0^2+y_0^2}=\sqrt{1+z_0^2},\qquad
e_r=\frac1{\rho_0}(x_0,y_0,0),\qquad
e_\theta=\frac1{\rho_0}(-y_0,x_0,0).
$$
Two distinct ruling lines through $P$ are
$$
\boxed{\gamma_\pm(t)=P+t(z_0e_r\pm e_\theta+\rho_0e_z)}.
$$
Indeed, the direction $d_\pm$ satisfies
$$
x_0d_x+y_0d_y-z_0d_z=0,
\qquad
d_x^2+d_y^2-d_z^2=0,
$$
so substitution shows that every point of the line lies in $H$. These are geodesics by the straight-line argument.
A third geodesic is the meridian through $P$. Choose $s_0$ with $\sinh s_0=z_0$; then
$$
\boxed{\gamma_m(t)=\cosh(t+s_0)e_r+\sinh(t+s_0)e_z}.
$$
Its acceleration is $\gamma_m$, which is normal to $H$, so it is a geodesic. If $z_0\ne0$, these give the required three distinct subsets.
If $z_0=0$, there is also the equatorial circle. Writing $P=(\cos\theta_0,\sin\theta_0,0)$,
$$
\boxed{\gamma_e(t)=(\cos(t+\theta_0),\sin(t+\theta_0),0)}.
$$
Its acceleration is normal to $H$ along $z=0$, so it is a fourth geodesic distinct from the meridian and the two rulings.
Finally, write $\rho=\sqrt{x^2+y^2}=\sqrt{1+z^2}$. <Clairaut's relation> gives the conserved quantity
$$
c=\rho\sin\psi.
$$
Choose initial data in $z>0$ with
$$
1<c<\rho(0).
$$
Since $|\sin\psi|\leq1$, every point of the resulting geodesic satisfies $\rho\geq c$, and therefore
$$
z^2=\rho^2-1\geq c^2-1>0.
$$
Continuity keeps the geodesic in the component $z>0$, and the stated completeness assumption defines it for every real time. The continuum of choices of $c$ supplies infinitely many such geodesics.
Solved by gpt-5.6-sol high.
= 13G
{parent=Paper 3}
{scope}
{title2=Complex Analysis}
= Solution
{parent=13G}
Fix $a\notin[\gamma]$ and let
$$
d=\operatorname{dist}(a,[\gamma])>0.
$$
For $|z-a|<d$ and $\lambda\in[\gamma]$, the <geometric series> gives
$$
\frac1{\lambda-z}
=\frac1{\lambda-a}\frac1{1-(z-a)/(\lambda-a)}
=\sum_{n=0}^{\infty}\frac{(z-a)^n}{(\lambda-a)^{n+1}}.
$$
The series converges uniformly on every smaller closed disc, so it may be integrated term by term along the curve. Hence
$$
\boxed{
f(z)=\sum_{n=0}^{\infty}c_n(z-a)^n,
\qquad
c_n=\int_\gamma\frac{\phi(\lambda)}
{(\lambda-a)^{n+1}}\,d\lambda
},
$$
which is a <power series> about $a$.
If $f$ is holomorphic on a neighbourhood of the closed disc $\overline D(a,r)$, the <Cauchy integral formula> says
$$
f(z)=\frac1{2\pi i}\int_{|\zeta-a|=r}
\frac{f(\zeta)}{\zeta-z}\,d\zeta,
\qquad |z-a|<r.
$$
Differentiation under the integral is valid uniformly on smaller discs and gives the <Cauchy derivative formula>
$$
\boxed{
f^{(n)}(z)=\frac{n!}{2\pi i}
\int_{|\zeta-a|=r}
\frac{f(\zeta)}{(\zeta-z)^{n+1}}\,d\zeta
}.
$$
This proves inductively that every holomorphic function has complex derivatives of every order. In particular,
$$
\boxed{
f'(z)=\frac1{2\pi i}
\int_{|\zeta-a|=r}\frac{f(\zeta)}{(\zeta-z)^2}\,d\zeta
}.
$$
Now suppose $f_n\to f$ <locally uniform convergence>[locally uniformly] on $U$. Given a compact set $K\subset U$, choose finitely many closed discs whose slightly larger concentric discs remain in $U$ and whose interiors cover $K$. Applying the derivative formula to $f_n-f$ on the larger boundary circles gives a uniform <Cauchy estimate>
$$
\sup_K|f_n'-f'|
\leq C\sup_L|f_n-f|,
$$
where $L\subset U$ is the compact union of those circles. The right-hand side tends to zero, proving that
$$
\boxed{f_n'\to f'\text{ locally uniformly}}.
$$
Finally, choose open neighbourhoods $U_j$ of the closed discs $D_j$ so small that
$$
U_1\cap U_2
$$
lies in the given neighbourhood on which $f$ is holomorphic. Inside that neighbourhood choose a positively oriented piecewise smooth contour $\Gamma$ surrounding $D_1\cap D_2$. It may be chosen as the boundary of a slightly enlarged lens and split into arcs
$$
\Gamma=\Gamma_1+\Gamma_2
$$
so that $\Gamma_1$ stays away from $D_1$ and $\Gamma_2$ stays away from $D_2$. Define
$$
f_1(z)=\frac1{2\pi i}\int_{\Gamma_1}
\frac{f(\zeta)}{\zeta-z}\,d\zeta
\quad\text{near }D_1,
$$
and
$$
f_2(z)=\frac1{2\pi i}\int_{\Gamma_2}
\frac{f(\zeta)}{\zeta-z}\,d\zeta
\quad\text{near }D_2.
$$
Because each arc avoids the corresponding disc, these formulas define holomorphic functions on possibly smaller neighbourhoods $U_1,U_2$. On their overlap, the <Cauchy integral formula> for the full contour gives
$$
\boxed{f=f_1+f_2}.
$$
Solved by gpt-5.6-sol high.
= 14A
{parent=Paper 3}
{scope}
{title2=Methods}
= a
{parent=14a}
{scope}
= Solution
{parent=a}
Let $Q_k=P^{(k)}$. Differentiating <Legendre's differential equation> $k$ times and using the <Leibniz rule> gives
$$
(1-x^2)Q_k''-2(k+1)xQ_k'
+\bigl[\lambda-k(k+1)\bigr]Q_k=0.
$$
The induction step follows by differentiating this equation once: the derivative of the coefficient $-2(k+1)x$ contributes the additional term $-2(k+1)Q_k'$. Thus
$$
\boxed{\lambda_k=\lambda-k(k+1)}.
$$
For fixed $\lambda$, this is negative for every sufficiently large $k$.
Solved by gpt-5.6-sol high.
= b
{parent=14a}
{scope}
= Solution
{parent=b}
Multiplication by $(1-x^2)^k$ puts the equation in <self-adjoint differential equation>[self-adjoint form]:
$$
\boxed{
\frac d{dx}\left[(1-x^2)^{k+1}Q_k'\right]
+\lambda_k(1-x^2)^kQ_k=0
}.
$$
Multiply by $Q_k$ and integrate over $[-1,1]$. Regularity and the vanishing factor at both endpoints remove the boundary term in <integration by parts>, leaving
$$
\int_{-1}^1(1-x^2)^{k+1}(Q_k')^2\,dx
=\lambda_k\int_{-1}^1(1-x^2)^kQ_k^2\,dx.
$$
If $Q_k$ is not identically zero, the integral on the right without $\lambda_k$ is positive, whereas the left side is nonnegative. Therefore
$$
\boxed{\lambda_k\geq0}.
$$
Solved by gpt-5.6-sol high.
= c
{parent=14a}
{scope}
= Solution
{parent=c}
Part (a) makes $\lambda_k<0$ for all sufficiently large $k$, while part (b) would make $\lambda_k\geq0$ whenever $P^{(k)}$ were nonzero. Hence some derivative of $P$ vanishes identically, so $P$ is a <polynomial>.
Let its degree be $n$. Then $P^{(n)}$ is a nonzero constant. Substituting $k=n$ into the differentiated equation annihilates both derivative terms and gives
$$
\lambda_nP^{(n)}=0.
$$
Thus $\lambda_n=0$, and
$$
\boxed{\lambda=n(n+1)},\qquad n=0,1,2,\ldots.
$$
The nonzero regular solutions are therefore constant multiples of the <Legendre polynomial> of degree $n$.
Solved by gpt-5.6-sol high.
= 15D
{parent=Paper 3}
{scope}
{title2=Electromagnetism}
= a
{parent=15d}
{scope}
= Solution
{parent=a}
In a region with no current, the <Ampere-Maxwell equation> and <Faraday's law> are
$$
\nabla\times\mathbf B=\mu_0\epsilon_0\frac{\partial\mathbf E}{\partial t},
\qquad
\nabla\times\mathbf E=-\frac{\partial\mathbf B}{\partial t}.
$$
Therefore
$$
\begin{aligned}
\frac{\partial w}{\partial t}
&=\epsilon_0\mathbf E\mathbin{\cdot}\partial_t\mathbf E
+\frac1{\mu_0}\mathbf B\mathbin{\cdot}\partial_t\mathbf B\\
&=\frac1{\mu_0}\left[
\mathbf E\mathbin{\cdot}(\nabla\times\mathbf B)
-\mathbf B\mathbin{\cdot}(\nabla\times\mathbf E)\right]\\
&=-\frac1{\mu_0}\nabla\mathbin{\cdot}(\mathbf E\times\mathbf B),
\end{aligned}
$$
where the last line uses the <divergence and curl of a cross product>. Thus <Poynting's theorem> takes the form
$$
\boxed{\frac{\partial w}{\partial t}+\nabla\mathbin{\cdot}\mathbf S=0},
\qquad
\boxed{\mathbf S=\frac1{\mu_0}\mathbf E\times\mathbf B}.
$$
The vector $\mathbf S$ is the <Poynting vector>.
Solved by gpt-5.6-sol high.
= b
{parent=15d}
{scope}
= Solution
{parent=b}
The inverse <Lorentz transformation of electromagnetic fields>, from the primed rest frame to the unprimed frame, gives
$$
\boxed{
E_x=E_x',\qquad E_y=\gamma E_y',\qquad E_z=\gamma E_z'
}
$$
and
$$
\boxed{
B_x=0,\qquad
B_y=-\frac{\gamma v}{c^2}E_z',\qquad
B_z=\frac{\gamma v}{c^2}E_y'
}.
$$
Equivalently, $\mathbf B=\gamma\,\mathbf v\times\mathbf E'/c^2$ for $\mathbf v=v\mathbf e_x$.
Using $1/\mu_0=\epsilon_0c^2$ and $\gamma^2=(1-v^2/c^2)^{-1}$,
$$
\begin{aligned}
w
&=\frac{\epsilon_0}{2}\left[
E_x'^2+\gamma^2(E_y'^2+E_z'^2)
+\gamma^2\frac{v^2}{c^2}(E_y'^2+E_z'^2)
\right]\\
&=\boxed{\frac{\epsilon_0}{2}\left[
E_x'^2+\frac{c^2+v^2}{c^2-v^2}
(E_y'^2+E_z'^2)\right]}.
\end{aligned}
$$
The primed field is static, so $w$ depends on $t,x$ through
$$
x'=\gamma(x-vt)
$$
and has no explicit $t'$ dependence. The <chain rule> gives
$$
\partial_tw=-\gamma v\,\partial_{x'}w,
\qquad
\partial_xw=\gamma\,\partial_{x'}w.
$$
Consequently
$$
\boxed{
\frac{\partial w}{\partial t}
+\nabla\mathbin{\cdot}(wv\mathbf e_x)
=\partial_tw+v\partial_xw=0
}.
$$
Thus the field-energy profile is transported rigidly with velocity $v\mathbf e_x$.
Solved by gpt-5.6-sol high.
= 16A
{parent=Paper 3}
{scope}
{title2=Fluid Dynamics}
= a
{parent=16a}
{scope}
= Solution
{parent=a}
For the parallel flow $\mathbf u=(u(y,t),0)$ with no pressure gradient, the <Navier-Stokes equations> reduce to the <diffusion equation>
$$
\partial_tu=\nu\,\partial_y^2u.
$$
Substitution of
$$
u=\operatorname{Re}\!\left[U_0f(y)e^{i\omega t}\right]
$$
gives
$$
\nu f''=i\omega f.
$$
The boundary velocities require $f(L_0)=1$ and $f(-L_0)=-1$. The problem is antisymmetric under $y\mapsto-y$, so $f$ is odd. With
$$
k=\sqrt{\frac{i\omega}{\nu}}
=(1+i)\sqrt{\frac{\omega}{2\nu}}
=\frac{(1+i)\Delta}{L_0},
$$
the unique odd solution is
$$
\boxed{
f(y)=\frac{\sinh[(1+i)\Delta\widehat y]}
{\sinh[(1+i)\Delta]},
\qquad
\widehat y=\frac y{L_0},
\qquad
\Delta=\left(\frac{\omega L_0^2}{2\nu}\right)^{1/2}
}.
$$
Solved by gpt-5.6-sol high.
= b
{parent=16a}
{scope}
= Solution
{parent=b}
Using
$$
\sinh(a+ib)=\sinh a\cos b+i\cosh a\sin b
$$
in the quotient from part (a), multiplying numerator and denominator by the complex conjugate of the denominator, and taking the real part after multiplication by $e^{i\omega t}$ gives
$$
\boxed{
\begin{aligned}
\frac{u(y,t)}{U_0}
={}&\frac{\cos\omega t\,
[\cosh\Delta_+\cos\Delta_-
-\cosh\Delta_-\cos\Delta_+]}
{\cosh(2\Delta)-\cos(2\Delta)}\\
&+\frac{\sin\omega t\,
[\sinh\Delta_+\sin\Delta_-
-\sinh\Delta_-\sin\Delta_+]}
{\cosh(2\Delta)-\cos(2\Delta)},
\end{aligned}}
$$
where
$$
\Delta_\pm=\Delta(1\pm\widehat y).
$$
The denominator follows from
$$
2|\sinh((1+i)\Delta)|^2
=\cosh(2\Delta)-\cos(2\Delta).
$$
Solved by gpt-5.6-sol high.
= c
{parent=16a}
{scope}
= Solution
{parent=c}
When $\Delta\ll1$, the small-argument expansion $\sinh z=z+O(z^3)$ gives
$$
f(y)=\frac{(1+i)\Delta\widehat y+O(\Delta^3)}
{(1+i)\Delta+O(\Delta^3)}
=\widehat y+O(\Delta^2).
$$
Therefore
$$
\boxed{
u(y,t)\simeq\frac{U_0y}{L_0}\cos\omega t
}.
$$
The <viscous diffusion time> $L_0^2/\nu$ is then short compared with the oscillation period. Viscosity communicates the wall motion across the entire layer almost instantaneously, producing the linear profile of quasi-steady <Couette flow>.
Solved by gpt-5.6-sol high.
= 17B
{parent=Paper 3}
{scope}
{title2=Numerical Analysis}
= a
{parent=17b}
{scope}
= Solution
{parent=a}
By the <Leibniz rule>,
$$
\frac{d^n}{dx^n}\left(x^{n+1/2}e^{-x}\right)
=e^{-x}\sum_{j=0}^n\binom nj(-1)^{n-j}
\frac{d^j}{dx^j}x^{n+1/2}.
$$
After multiplication by $(-1)^nx^{-1/2}e^x$, the term $j=0$ is $x^n$, and every other term has lower degree. Thus $p_n$ is a <monic polynomial> of degree $n$. Direct calculation gives
$$
\boxed{
p_0(x)=1,\qquad
p_1(x)=x-\frac32,\qquad
p_2(x)=x^2-5x+\frac{15}{4}
}.
$$
These are the monic normalization of the <generalized Laguerre polynomial>[generalized Laguerre polynomials] with parameter $1/2$.
Solved by gpt-5.6-sol high.
= b
{parent=17b}
{scope}
= Solution
{parent=b}
Suppose $m<n$. The <Rodrigues formula> and $n$ integrations by parts give
$$
\begin{aligned}
\langle p_n,p_m\rangle
&=(-1)^n\int_0^\infty
p_m(x)\frac{d^n}{dx^n}
\left(x^{n+1/2}e^{-x}\right)dx\\
&=\int_0^\infty
p_m^{(n)}(x)x^{n+1/2}e^{-x}\,dx=0,
\end{aligned}
$$
because $p_m^{(n)}=0$. Every boundary term vanishes: the exponential controls infinity, while the remaining power of $x$ has positive exponent at zero. Symmetry of the <inner product> gives orthogonality whenever $m\ne n$.
For the norm, monicity gives $p_n^{(n)}=n!$, so the same calculation yields
$$
\boxed{
\langle p_n,p_n\rangle
=n!\int_0^\infty x^{n+1/2}e^{-x}\,dx
=n!\,\Gamma\!\left(n+\frac32\right)
}.
$$
Solved by gpt-5.6-sol high.
= c
{parent=17b}
{scope}
= Solution
{parent=c}
For <monic orthogonal polynomials>, taking the inner product of the recurrence with $p_{n-1}$ gives
$$
\beta_n
=\frac{\langle xp_n,p_{n-1}\rangle}
{\langle p_{n-1},p_{n-1}\rangle}
=\frac{\langle p_n,xp_{n-1}\rangle}
{\langle p_{n-1},p_{n-1}\rangle}
=\frac{\langle p_n,p_n\rangle}
{\langle p_{n-1},p_{n-1}\rangle}.
$$
Using part (b) and the <Gamma function recurrence>,
$$
\boxed{\beta_n=n\left(n+\frac12\right)}.
$$
The Rodrigues expansion also shows that the coefficient of $x^{n-1}$ in $p_n$ is
$$
-n\left(n+\frac12\right).
$$
Comparing the coefficients of $x^n$ in
$$
p_{n+1}=(x-\alpha_n)p_n-\beta_np_{n-1}
$$
therefore gives
$$
-(n+1)\left(n+\frac32\right)
=-n\left(n+\frac12\right)-\alpha_n,
$$
and hence
$$
\boxed{\alpha_n=2n+\frac32}.
$$
Solved by gpt-5.6-sol high.
= 18H
{parent=Paper 3}
{scope}
{title2=Statistics}
= i
{parent=18h}
{scope}
= Solution
{parent=i}
For the <normal linear model>, the <maximum likelihood estimators> are
$$
\boxed{\widehat\beta=(X^TX)^{-1}X^TY}
$$
and
$$
\boxed{
\widehat\sigma^2
=\frac1n\|Y-X\widehat\beta\|^2
}.
$$
The divisor is $n$ for maximum likelihood, rather than the degrees-of-freedom divisor used by the unbiased residual-variance estimator.
Solved by gpt-5.6-sol high.
= ii
{parent=18h}
{scope}
= Solution
{parent=ii}
Let
$$
H=X(X^TX)^{-1}X^T
$$
be the <hat matrix>. Then
$$
\widehat\beta=(X^TX)^{-1}X^T(HY)
$$
depends only on the orthogonal projection $HY$, while
$$
n\widehat\sigma^2=\|(I-H)Y\|^2
$$
depends only on the orthogonal residual projection $(I-H)Y$.
The random vectors $HY$ and $(I-H)Y$ are jointly <multivariate normal distribution>[multivariate normal], and their cross-covariance is
$$
\sigma^2H(I-H)=0.
$$
Uncorrelated jointly Gaussian vectors are <independent random variables>[independent]. Therefore
$$
\boxed{\widehat\beta\ \text{and}\ \widehat\sigma^2
\text{ are independent}}.
$$
Solved by gpt-5.6-sol high.
= iii
{parent=18h}
{scope}
= Solution
{parent=iii}
Since $\widehat\beta$ is an affine transformation of $Y$,
$$
\boxed{
\widehat\beta\sim
N_p\!\left(\beta,\sigma^2(X^TX)^{-1}\right)
}.
$$
The residual projection $I-H$ has rank $n-p$. <Cochran's theorem> therefore gives
$$
\boxed{
\frac{n\widehat\sigma^2}{\sigma^2}
=\frac{\|(I-H)Y\|^2}{\sigma^2}
\sim\chi^2_{n-p}
}.
$$
Solved by gpt-5.6-sol high.
= iv
{parent=18h}
{scope}
= Solution
{parent=iv}
Under $H_0$, write the <singular value decomposition>
$$
X=UDV^T,\qquad D_{jj}=\sqrt{\lambda_j}.
$$
Then
$$
\widehat\beta
=VD^{-1}U^T\varepsilon.
$$
The coordinates of $U^T\varepsilon/\sigma$ are independent standard normal variables. Hence
$$
\frac{\|\widehat\beta\|^2}{\sigma^2}
\mathrel{\overset d=}
\sum_{j=1}^p\lambda_j^{-1}W_j,
\qquad
W_j\sim\chi^2_1
$$
independently. By parts (ii) and (iii),
$$
Z=\frac{n\widehat\sigma^2}{\sigma^2}\sim\chi^2_{n-p}
$$
is independent of all the $W_j$. Consequently
$$
\boxed{
T\mathrel{\overset d=}
\frac{\sum_{j=1}^p\lambda_j^{-1}W_j}{Z}
}.
$$
Solved by gpt-5.6-sol high.
= v
{parent=18h}
{scope}
= Solution
{parent=v}
For arbitrary $\beta$,
$$
\mathbb E\|\widehat\beta\|^2
=\|\beta\|^2
+\operatorname{tr}\!\left(\sigma^2(X^TX)^{-1}\right)
=\|\beta\|^2+\sigma^2\sum_{j=1}^p\lambda_j^{-1}.
$$
Independence from part (ii) and $Z=n\widehat\sigma^2/\sigma^2\sim\chi^2_{n-p}$ give
$$
\mathbb ET
=\frac{\mathbb E\|\widehat\beta\|^2}{\sigma^2}
\mathbb E\!\left(\frac1Z\right).
$$
Because $n-p>2$,
$$
\mathbb E(Z^{-1})=\frac1{n-p-2}.
$$
Thus
$$
\boxed{
\mathbb ET
=\frac{\|\beta\|^2/\sigma^2
+\sum_{j=1}^p\lambda_j^{-1}}
{n-p-2}
}.
$$
Solved by gpt-5.6-sol high.
= 19H
{parent=Paper 3}
{scope}
{title2=Optimisation}
= Solution
{parent=19H}
An $m\times n$ <matrix game> with payoff matrix $A$ is a two-player <zero-sum game>: player I chooses a row and receives $A_{ij}$, while player II chooses a column and loses the same amount. For <mixed strategy>[mixed strategies] $p$ and $q$, the expected payoff to player I is
$$
p^TAq.
$$
Player I's <optimal mixed strategy> maximizes the payoff guaranteed against every $q$, while player II's minimizes the largest payoff obtainable by any $p$. Thus
$$
\max_p\min_qp^TAq
=v
=\min_q\max_pp^TAq
$$
by the <minimax theorem>. Equivalently, optimal strategies satisfy
$$
p^TA\geq v\mathbf1^T,
\qquad
Aq\leq v\mathbf1.
$$
Here $A^T=-A$, so this is an <antisymmetric zero-sum game>. For every probability vector $r$,
$$
r^TAr=0.
$$
It follows that the row player's guaranteed payoff cannot exceed zero and the column player's worst loss cannot be below zero. Minimax therefore gives
$$
\boxed{v=0}.
$$
If $p$ is optimal for player I, then
$$
p^TA\geq0.
$$
Transposing and using $A^T=-A$ gives
$$
Ap\leq0,
$$
which is precisely the optimality condition for player II. Thus every optimal strategy for player I is also optimal for player II.
The condition $Ap\leq0$ explicitly reads
$$
\begin{aligned}
p_2+p_3-4p_4&\leq0,\\
-p_1+2p_3+2p_4&\leq0,\\
-p_1-2p_2+3p_4&\leq0,\\
4p_1-2p_2-3p_3&\leq0.
\end{aligned}
$$
The probability vector
$$
\boxed{p=\frac17(2,4,0,1)^T}
$$
satisfies
$$
Ap=(0,0,-1,0)^T\leq0,
$$
so it is optimal for both players.
To prove uniqueness, let $q$ be any optimal strategy. Since $p$ is optimal for player II and $q$ is optimal for player I,
$$
0\leq q^TAp=-q_3,
$$
and hence $q_3=0$. Writing $q=(a,b,0,d)^T$, the first, second, and fourth inequalities in $Aq\leq0$ give
$$
b\leq4d,\qquad a\geq2d,\qquad b\geq2a.
$$
Consequently
$$
4d\geq b\geq2a\geq4d,
$$
so equality holds throughout: $a=2d$ and $b=4d$. The normalization $a+b+d=1$ yields $d=1/7$. Therefore the displayed $p$ is the unique optimal strategy.
Solved by gpt-5.6-sol high.