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past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ib/paper-2.bigb
= Paper 2
{scope}
https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2021/paperib_2_2021.pdf
= 1G
{parent=Paper 2}
{scope}
{title2=Groups, Rings and Modules}
= Solution
{parent=1G}
Suppose first that $M$ is finitely generated. Its quotient $M/N$ is generated by the images of any finite generating set of $M$. A <principal ideal domain> is <Noetherian ring>[Noetherian], so every submodule of the finitely generated $R$-module $M$ is finitely generated; in particular, $N$ is finitely generated.
Conversely, suppose
$$
N=(n_1,\ldots,n_r)
$$
and $M/N$ is generated by the cosets $\overline m_1,\ldots,\overline m_s$. For any $m\in M$, its coset is an $R$-linear combination of the $\overline m_j$, so
$$
m-\sum_{j=1}^sa_jm_j\in N.
$$
Expressing this remainder in terms of the $n_i$ shows that
$$
M=(n_1,\ldots,n_r,m_1,\ldots,m_s).
$$
Thus $M$ is finitely generated if and only if both the <submodule> $N$ and the <quotient module> $M/N$ are finitely generated.
Solved by gpt-5.6-sol high.
= 2F
{parent=Paper 2}
{scope}
{title2=Analysis and Topology}
= a
{parent=2f}
{scope}
= Solution
{parent=a}
Let $\|K\|_\infty=\max_{[0,1]^2}|K|$. For $f,g\in C([0,1])$,
$$
|Tf(x)-Tg(x)|
\leq\int_0^1|K(x,y)|\,|f(y)-g(y)|\,dy
\leq\|K\|_\infty\|f-g\|_\infty.
$$
Taking the supremum over $x$ proves that the <integral operator> $T$ is <Lipschitz continuous>, hence continuous, in the <uniform norm>.
It remains to see that $Tf$ is continuous. Since $K$ is <uniform continuity>[uniformly continuous] on the compact square and $f$ is bounded,
$$
|Tf(x)-Tf(x')|
\leq\|f\|_\infty\int_0^1|K(x,y)-K(x',y)|\,dy\to0
$$
as $x'\to x$. Therefore $T:C([0,1])\to C([0,1])$ is well defined and continuous.
Solved by gpt-5.6-sol high.
= b
{parent=2f}
{scope}
= Solution
{parent=b}
Yes. <Fubini's theorem> and the triangle inequality give
$$
\begin{aligned}
d_1(Tf,Tg)
&\leq\int_0^1\int_0^1
|K(x,y)|\,|f(y)-g(y)|\,dy\,dx\\
&\leq\|K\|_\infty d_1(f,g).
\end{aligned}
$$
Thus $T$ is Lipschitz continuous with respect to $d_1$ as well.
Solved by gpt-5.6-sol high.
= 3C
{parent=Paper 2}
{scope}
{title2=Methods}
= i
{parent=3c}
{scope}
= Solution
{parent=i}
Multiplication by the <integrating factor> $e^{2x}$ gives
$$
e^{2x}Ly=(e^{2x}y')'.
$$
Hence the <Sturm-Liouville problem> is
$$
\widetilde Ly=-\lambda wy,
\qquad
\boxed{\widetilde Ly=(e^{2x}y')',\quad w(x)=e^{2x}}.
$$
Solved by gpt-5.6-sol high.
= ii
{parent=3c}
{scope}
= Solution
{parent=ii}
Set $y=e^{-x}u$. Then
$$
y''+2y'=e^{-x}(u''-u),
$$
so the eigenvalue equation becomes
$$
u''+(\lambda-1)u=0,
\qquad u(0)=u(1)=0.
$$
The <Dirichlet eigenvalue>[Dirichlet eigenvalues] and corresponding eigenfunctions are therefore
$$
\boxed{\lambda_n=1+n^2\pi^2,\qquad
y_n(x)=e^{-x}\sin(n\pi x),\quad n\geq1}.
$$
They form an infinite discrete increasing set. Their weighted inner products satisfy
$$
\int_0^1w\,y_ny_m\,dx
=\int_0^1\sin(n\pi x)\sin(m\pi x)\,dx
=\frac12\delta_{nm}.
$$
Thus the <Sturm-Liouville eigenfunction expansion> coefficient is
$$
\boxed{
A_n=
\frac{\int_0^1e^{2x}(x-x^2)y_n(x)\,dx}
{\int_0^1e^{2x}y_n(x)^2\,dx}
=2\int_0^1e^x(x-x^2)\sin(n\pi x)\,dx }.
$$
Solved by gpt-5.6-sol high.
= 4D
{parent=Paper 2}
{scope}
{title2=Electromagnetism}
= Solution
{parent=4D}
<Gauss's law> in electrostatics states
$$
\oint_{\partial V}E\mathbin{\cdot}dS
=\frac{Q_{\rm enclosed}}{\varepsilon_0}.
$$
Cylindrical symmetry and a coaxial Gaussian cylinder give
$$
E(r)=
\begin{cases}
0,&0<r<a,\\[2pt]
\dfrac{Q}{2\pi\varepsilon_0r}\,e_r,&a<r<b,\\[6pt]
0,&r>b.
\end{cases}
$$
The field vanishes inside each perfect conductor, and outside the cable because the total enclosed charge per unit length is zero.
Choose the outer conductor's potential to be zero. Since $E=-\nabla V$,
$$
V(r)=
\begin{cases}
\dfrac{Q}{2\pi\varepsilon_0}\log(b/a),&0<r\leq a,\\[6pt]
\dfrac{Q}{2\pi\varepsilon_0}\log(b/r),&a<r<b,\\[6pt]
0,&r\geq b.
\end{cases}
$$
The <capacitance per unit length> is therefore
$$
\boxed{C=\frac{Q}{V(a)-V(b)}
=\frac{2\pi\varepsilon_0}{\log(b/a)}}.
$$
The <electrostatic energy> per unit length is
$$
U=\frac{\varepsilon_0}{2}
\int_a^b|E|^2\,2\pi r\,dr
=\boxed{\frac{Q^2}{4\pi\varepsilon_0}\log\frac ba}.
$$
Substitution of $C$ verifies $U=Q^2/(2C)$.
Solved by gpt-5.6-sol high.
= 5A
{parent=Paper 2}
{scope}
{title2=Fluid Dynamics}
= a
{parent=5a}
{scope}
= Solution
{parent=a}
If $z(t)$ is the water depth and $A(z)=\pi r(z)^2$ is the horizontal cross-sectional area, <Torricelli's law> gives the volume flux through the hole as
$$
q_0=\pi r_0^2\sqrt{2gz}.
$$
Conservation of volume gives
$$
A(z)\dot z=-\pi r_0^2\sqrt{2gz}.
$$
For a prescribed constant fall rate $\dot z=-\alpha$, $\alpha>0$, this requires
$$
A(z)=\frac{\pi r_0^2\sqrt{2g}}{\alpha}\sqrt z.
$$
Consequently
$$
\boxed{r(z)=
r_0\left(\frac{\sqrt{2g}}{\alpha}\right)^{1/2}z^{1/4}},
$$
so the container radius must be proportional to the fourth root of height.
Solved by gpt-5.6-sol high.
= b
{parent=5a}
{scope}
= Solution
{parent=b}
Since $z(t)=h_I-\alpha t$, the free-surface area is
$$
\boxed{
A(t)=\frac{\pi r_0^2\sqrt{2g}}{\alpha}
\sqrt{h_I-\alpha t}}
$$
until the emptying time $t_e=h_I/\alpha$. Equivalently,
$$
\boxed{\frac{A(t)}{A(0)}
=\sqrt{1-\frac{t}{t_e}}}.
$$
Solved by gpt-5.6-sol high.
= 6H
{parent=Paper 2}
{scope}
{title2=Statistics}
= Solution
{parent=6H}
Under the null hypothesis of independence between treatment and outcome, the expected counts in each treatment row are one half of the column totals:
$$
(10,20,10,10).
$$
The <Pearson chi-squared test of independence> statistic is
$$
\begin{aligned}
X^2
&=2\left(
\frac{(14-10)^2}{10}
+\frac{(21-20)^2}{20}
+\frac{(10-10)^2}{10}
+\frac{(5-10)^2}{10}
\right)\\
&=\boxed{8.30}.
\end{aligned}
$$
The number of <degrees of freedom> is
$$
(2-1)(4-1)=3.
$$
At the 5% level the critical value is $7.81$. Since $8.30>7.81$, we reject the null hypothesis and find statistically significant evidence that the drug and placebo have different effects.
Solved by gpt-5.6-sol high.
= 7H
{parent=Paper 2}
{scope}
{title2=Optimisation}
= Solution
{parent=7H}
Introduce slack variables $s_1,s_2,s_3$. In the initial <simplex algorithm> dictionary, $x_2$ has the largest positive objective coefficient. The ratio test gives
$$
\min\left\{\frac73,\frac52,\frac21\right\}=2,
$$
so $x_2$ enters and $s_3$ leaves. Solving the third constraint for $x_2$ gives
$$
x_2=2-x_1-2x_3-s_3.
$$
The objective becomes
$$
z=3x_1+6x_2+4x_3
=12-3x_1-8x_3-6s_3.
$$
Every reduced cost is now nonpositive, so the simplex optimality criterion gives
$$
\boxed{(x_1,x_2,x_3)=(0,2,0),\qquad z_{\max}=12}.
$$
The other two slacks both equal one.
The <dual linear program> is
$$
\begin{aligned}
\text{minimize}\quad&7y_1+5y_2+2y_3,\\
\text{subject to}\quad
&2y_1+4y_2+y_3\geq3,\\
&3y_1+2y_2+y_3\geq6,\\
&y_1+2y_2+2y_3\geq4,\\
&y_1,y_2,y_3\geq0.
\end{aligned}
$$
The vector
$$
\boxed{(y_1,y_2,y_3)=(0,0,6)}
$$
is dual feasible and has objective value $12$. By <weak duality>, it and the displayed primal point are optimal; they also satisfy <complementary slackness>.
Solved by gpt-5.6-sol high.
= 8E
{parent=Paper 2}
{scope}
{title2=Linear Algebra}
= a
{parent=8e}
{scope}
= Solution
{parent=a}
A direct determinant calculation gives
$$
\chi_A(t)=\det(tI-A)=(t-1)^3.
$$
Moreover,
$$
A-I=
\begin{pmatrix}
-3&-6&-9\\
3&6&9\\
-1&-2&-3
\end{pmatrix}
$$
has rank one, so its kernel has dimension two, while $(A-I)^2=0$ and $A\ne I$. Hence the <minimal polynomial> is
$$
\boxed{m_A(t)=(t-1)^2}.
$$
There are two <Jordan block>[Jordan blocks], because the eigenspace has dimension two, and the largest has size two, because the minimal polynomial has exponent two. Thus
$$
\boxed{\operatorname{JNF}(A)=J_2(1)\oplus J_1(1)}.
$$
Solved by gpt-5.6-sol high.
= b
{parent=8e}
{scope}
= i
{parent=b}
{scope}
= Solution
{parent=i}
Choose the least $k\geq1$ such that
$$
(f-\alpha I)^kv=0.
$$
Then
$$
w=(f-\alpha I)^{k-1}v
$$
is nonzero by minimality, and
$$
(f-\alpha I)w=0.
$$
Therefore $fw=\alpha w$, so $w$ is the required nonzero <eigenvector>.
Solved by gpt-5.6-sol high.
= ii
{parent=b}
{scope}
= Solution
{parent=ii}
We use induction on $d$. Suppose
$$
\sum_{i=1}^dc_iw_i=0.
$$
Applying $f-\alpha_dI$ gives
$$
\sum_{i=1}^{d-1}c_i(\alpha_i-\alpha_d)w_i=0.
$$
By induction, every $c_i(\alpha_i-\alpha_d)=0$ for $i<d$. The eigenvalues are distinct, so $c_i=0$ for $i<d$, and the original relation then gives $c_d=0$. Thus eigenvectors with distinct eigenvalues are <linear independence>[linearly independent].
Solved by gpt-5.6-sol high.
= iii
{parent=b}
{scope}
= Solution
{parent=iii}
If $\alpha\ne\beta$, the operator $f-\beta I$ is invertible on $W_{\alpha,n}$. Indeed, there it equals
$$
(\alpha-\beta)I+N,
\qquad N=f-\alpha I,\qquad N^n=0,
$$
whose inverse is the finite <geometric series>
$$
\frac1{\alpha-\beta}
\sum_{k=0}^{n-1}
\left(-\frac{N}{\alpha-\beta}\right)^k.
$$
Now use induction on $d$. If $\sum_{i=1}^dv_i=0$, apply $(f-\alpha_dI)^n$. The $v_d$ term vanishes, while every transformed vector
$$
(f-\alpha_dI)^nv_i,\qquad i<d,
$$
is nonzero by the invertibility just proved and still lies in $W_{\alpha_i,n}$. The induction hypothesis rules out the resulting relation. Allowing scalar coefficients, and omitting zero terms, gives the same argument. Hence the <generalized eigenspaces for distinct eigenvalues form a direct sum>, so $v_1,\ldots,v_d$ are linearly independent.
Solved by gpt-5.6-sol high.
= 9G
{parent=Paper 2}
{scope}
{title2=Groups, Rings and Modules}
= Solution
{parent=9G}
A subset $S\subseteq M$ freely generates $M$ when every $m\in M$ has a unique expression
$$
m=\sum_{s\in S}r_ss
$$
with $r_s\in R$ and only finitely many nonzero coefficients. Equivalently, $S$ is a <basis of a module> and $M$ is the <free module> on $S$.
If $S$ freely generates $M$ and $f:S\to N$ is any function, define
$$
\phi\left(\sum_sr_ss\right)=\sum_sr_sf(s).
$$
Unique coordinates make $\phi$ well defined; it is an $R$-module homomorphism and is the only extension of $f$. Conversely, apply the proposed universal property to the free module $F=R^{(S)}$ and the inclusion of $S$ into $M$. It gives maps $F\to M$ and $M\to F$ extending the corresponding functions on $S$. Uniqueness makes both composites identity maps, so $M\cong F$ and $S$ freely generates $M$. This is the <universal property of a free module>.
Now let $T$ generate the free module $M$ and $|T|=m$. Choose a <maximal ideal> $\mathfrak m$ of the nontrivial ring $R$. Then $k=R/\mathfrak m$ is a field, and
$$
M/\mathfrak mM
$$
is a $k$-vector space. The images of a basis $S$ of $M$ form a vector-space basis, while the images of $T$ span it. Therefore
$$
|S|=\dim_k(M/\mathfrak mM)\leq m.
$$
In particular $S$ is finite. Applying this result in both directions to two finite bases shows that they have equal cardinality, the <rank of a free module> $\operatorname{rk}M$.
A Euclidean domain is a <principal ideal domain>. By the <submodule theorem for free modules over a principal ideal domain>, every submodule $N$ of the finite-rank free module $M$ is free. Since a basis of $M$ generates it, the preceding inequality applied in the standard proof gives
$$
\boxed{\operatorname{rk}N\leq\operatorname{rk}M}.
$$
The <primary decomposition theorem for finitely generated modules over a principal ideal domain> states that
$$
M\cong R^r\oplus
\bigoplus_{p}\bigoplus_j R/(p^{\,e_{p,j}}),
$$
where $p$ ranges over finitely many nonassociate irreducibles and the positive exponents are uniquely determined up to order. Equivalently, the torsion part decomposes into its primary cyclic summands.
Let $H$ be a finite subgroup of the multiplicative group of a field. It is a finite abelian group, hence the theorem over $\mathbb Z$ gives an invariant-factor decomposition
$$
H\cong C_{d_1}\times\cdots\times C_{d_r},
\qquad d_1\mid\cdots\mid d_r.
$$
Its exponent is $d_r$, so every element of $H$ is a root of $X^{d_r}-1$. A degree-$d_r$ polynomial over a field has at most $d_r$ roots, whence
$$
|H|\leq d_r.
$$
But $d_r\leq|H|$, and equality forces all earlier factors to be trivial. Thus
$$
\boxed{H\text{ is cyclic}}.
$$
Solved by gpt-5.6-sol high.
= 10F
{parent=Paper 2}
{scope}
{title2=Analysis and Topology}
= Solution
{parent=10F}
Let $C\subset\mathbb R$ be compact and let $M=\|f\|_\infty$. Since every $k_n$ is supported in $[-R,R]$, the function $f$ is <uniform continuity>[uniformly continuous] on the compact set
$$
C_R=\{x-t:x\in C,\ |t|\leq R\}.
$$
Given $\varepsilon>0$, choose $\delta>0$ so that
$$
|f(x-t)-f(x)|<\frac{\varepsilon}{2}
$$
whenever $x\in C$ and $|t|<\delta$. Since $\int k_n=1$ and $k_n\geq0$,
$$
\begin{aligned}
|f_n(x)-f(x)|
&\leq\int_{\mathbb R}k_n(t)|f(x-t)-f(x)|\,dt\\
&\leq\frac{\varepsilon}{2}
+2M\int_{|t|\geq\delta}k_n(t)\,dt.
\end{aligned}
$$
Property 3 makes the final term smaller than $\varepsilon/2$ for all sufficiently large $n$, uniformly in $x\in C$. Hence $f_n\to f$ uniformly on every compact set. The sequence $(k_n)$ is an <approximate identity>.
For the second part, extend $g$ to a continuous function $\widetilde g$ on $\mathbb R$ by setting it equal to zero outside $[0,1]$; continuity at the endpoints uses $g(0)=g(1)=0$. Define
$$
c_n=\int_{-1}^1(1-t^2)^n\,dt,
\qquad
k_n(t)=\frac{(1-t^2)^n}{c_n}\mathbf1_{[-1,1]}(t).
$$
These nonnegative kernels have integral one. For every $\delta>0$, their mass outside $[-\delta,\delta]$ tends to zero exponentially relative to the mass near zero, so they satisfy property 3.
For $x\in[0,1]$, the convolution is
$$
p_n(x)
=\int_{\mathbb R}k_n(t)\widetilde g(x-t)\,dt
=\frac1{c_n}\int_0^1
\bigl(1-(x-y)^2\bigr)^ng(y)\,dy.
$$
Because $|x-y|\leq1$ on the square $[0,1]^2$, no cutoff remains in this formula. Expanding the $n$th power shows that $p_n$ is a <polynomial> in $x$. The first part, applied to the compact interval $[0,1]$, gives
$$
\boxed{\|p_n-g\|_\infty\to0}.
$$
This proves the <Weierstrass approximation theorem> for functions with the stated endpoint values.
Solved by gpt-5.6-sol high.
= 11E
{parent=Paper 2}
{scope}
{title2=Geometry}
= Solution
{parent=11E}
The <Poincare half-plane model> is
$$
\mathbb H=\{z=x+iy:y>0\},
\qquad
ds^2=\frac{dx^2+dy^2}{y^2},
$$
with the orientation inherited from the complex plane. For
$$
\gamma(z)=\frac{az+b}{cz+d},
\qquad
\begin{pmatrix}a&b\\c&d\end{pmatrix}\in SL_2(\mathbb R),
$$
one has
$$
\gamma'(z)=\frac1{(cz+d)^2},
\qquad
\operatorname{Im}\gamma(z)=\frac{\operatorname{Im}z}{|cz+d|^2}.
$$
These identities show directly that $\gamma^*ds^2=ds^2$. The map is holomorphic with nonzero derivative, so it preserves orientation. The matrices $I$ and $-I$ induce the same map, giving the action of <PSL2(R)>.
Conversely, let $F$ be an orientation-preserving isometry. The PSL2(R) action is transitive on $\mathbb H$, so compose $F$ with an element taking $F(i)$ back to $i$. The resulting isometry fixes $i$ and acts on $T_i\mathbb H$ by an orientation-preserving orthogonal map, hence a rotation. The stabilizer of $i$ in PSL2(R),
$$
\left\{
\begin{bmatrix}
\cos\alpha&\sin\alpha\\
-\sin\alpha&\cos\alpha
\end{bmatrix}
\right\},
$$
realizes every such tangent rotation. An isometry is determined by its value and differential at one point because it preserves geodesics and the <exponential map>. Thus the composed isometry belongs to PSL2(R), and so does $F$.
The map $\tau(z)=-\overline z$ is an orientation-reversing isometry. Composing any orientation-reversing isometry with $\tau$ gives an orientation-preserving one, so
$$
\operatorname{Isom}(\mathbb H)
=PSL_2(\mathbb R)\sqcup PSL_2(\mathbb R)\tau.
$$
A hyperbolic line is a vertical Euclidean line or a semicircle orthogonal to the real axis. Its <hyperbolic reflection> $\sigma_\ell$ is the unique orientation-reversing isometry that fixes every point of $\ell$. If $\ell,\ell'$ meet at angle $\theta$, then
$$
\rho=\sigma_\ell\sigma_{\ell'}
$$
is the hyperbolic rotation about $A$ through angle $2\theta$. The generators satisfy
$$
\sigma_\ell^2=\sigma_{\ell'}^2=1,
\qquad
\sigma_\ell\rho\sigma_\ell=\rho^{-1}.
$$
If $\theta/\pi=p/q$ in lowest terms, then $\rho$ has order $q$ and the generated group is the finite <dihedral group> of order $2q$. If $\theta/\pi$ is irrational, $\rho$ has infinite order and the generated group is the <infinite dihedral group>. In the degenerate case $\ell=\ell'$, the group has order two.
Solved by gpt-5.6-sol high.
= 12B
{parent=Paper 2}
{scope}
{title2=Complex Analysis or Complex Methods}
= a
{parent=12b}
{scope}
= Solution
{parent=a}
Write the <Taylor series> of the entire function as
$$
f(z)=\sum_{k=0}^{\infty}c_kz^k.
$$
The <Cauchy estimate> on the circle $|z|=R$ gives
$$
|c_k|\leq\frac{\max_{|z|=R}|f(z)|}{R^k}
\leq aR^{n/2-k}+bR^{-k}.
$$
If $k>n/2$, letting $R\to\infty$ gives $c_k=0$. Since $n$ is odd,
$$
\boxed{\deg f\leq\lfloor n/2\rfloor}.
$$
For the second question, suppose such an $f$ existed. It never vanishes, so $g=1/f$ is analytic on $\mathbb C\setminus\{0\}$ and
$$
|g(z)|\leq\sqrt{|z|}.
$$
The <Riemann removable singularity theorem> extends $g$ analytically across zero with $g(0)=0$. Applying the result just proved with $n=1$, $a=1$, and $b=0$ makes $g$ a polynomial of degree at most zero. It must then be identically zero, contradicting $g=1/f$ away from zero. Therefore no such function exists.
Solved by gpt-5.6-sol high.
= b
{parent=12b}
{scope}
= i
{parent=b}
{scope}
= Solution
{parent=i}
<Liouville theorem> states that every bounded entire function is constant.
Because $\mathbb C$ is simply connected, the harmonic function $u$ has a global <harmonic conjugate> $v$, so
$$
F=u+iv
$$
is entire. Positivity gives
$$
|e^{-F}|=e^{-u}<1.
$$
Liouville's theorem makes $e^{-F}$ constant. Differentiating this nonzero constant gives $F'=0$, so $F$, and hence $u$, is constant.
Solved by gpt-5.6-sol high.
= ii
{parent=b}
{scope}
= Solution
{parent=ii}
Set
$$
g(z)=\frac{f(z)-b}{a}.
$$
Then $g$ is entire and never takes a real value. The continuous function $\operatorname{Im}g$ is nowhere zero; because $\mathbb C$ is connected, it has one sign everywhere. Thus either $\operatorname{Im}g$ or $-\operatorname{Im}g$ is a positive <harmonic function>. Part (i) makes it constant. The <Cauchy-Riemann equations> then force the real part of $g$ to be constant as well. Hence $g$ and therefore $f$ are constant.
Solved by gpt-5.6-sol high.
= 13D
{parent=Paper 2}
{scope}
{title2=Variational Principles}
= Solution
{parent=13D}
Take a <variation> $x+\varepsilon\xi$ with $\xi(a)=\xi(b)=0$. Expanding the <action> gives
$$
S[x+\varepsilon\xi]
=S[x]+\varepsilon\int_a^b\bigl(\dot x\dot\xi-V'(x)\xi\bigr)\,dt
+\frac{\varepsilon^2}{2}\int_a^b\bigl(\dot\xi^2-V''(x)\xi^2\bigr)\,dt
+O(\varepsilon^3).
$$
An <integration by parts> and the fixed endpoint conditions give the <first variation>
$$
\delta S=-\int_a^b\bigl(\ddot x+V'(x)\bigr)\xi\,dt.
$$
The <fundamental lemma of the calculus of variations> therefore yields the <Euler-Lagrange equation>
$$
\boxed{\ddot x+V'(x)=0},
$$
and the <second variation> is
$$
\boxed{\delta^2S=\frac12\int_a^b\bigl(\dot\xi^2-V''(x)\xi^2\bigr)\,dt}.
$$
Linearizing the equation of motion about $x$ gives
$$
0=\ddot x+V'(x)+\varepsilon\bigl(\ddot u+V''(x)u\bigr)+O(\varepsilon^2),
$$
so the <Jacobi equation> is
$$
\ddot u+V''(x)u=0.
$$
When $u$ has no zero on $[a,b]$,
$$
\dot\xi^2-V''(x)\xi^2
=\left(\dot\xi-\frac{\dot u}{u}\xi\right)^2
+\frac d{dt}\left(\frac{\dot u}{u}\xi^2\right).
$$
The total derivative integrates to zero because $\xi(a)=\xi(b)=0$, hence
$$
\boxed{\delta^2S=\frac12\int_a^b
\left(\dot\xi-\frac{\dot u}{u}\xi\right)^2dt\geq0}.
$$
For the <simple harmonic oscillator>, the Jacobi equation is $\ddot u+\omega^2u=0$. Put $t_0=(a+b)/2$ and choose
$$
u(t)=\cos\bigl(\omega(t-t_0)\bigr).
$$
If $b-a<\pi/\omega$, then $|\omega(t-t_0)|<\pi/2$ throughout $[a,b]$, so $u$ is positive there. The preceding square identity proves that the classical path is a local minimum of the action whenever the elapsed time is less than half an <oscillation period>.
Solved by gpt-5.6-sol high.
= 14A
{parent=Paper 2}
{scope}
{title2=Methods}
= a
{parent=14a}
{scope}
= Solution
{parent=a}
Direct evaluation of the <integral> gives
$$
\begin{aligned}
\widetilde f(k)
&=\int_0^1e^{-ikx}\,dx-\int_{-1}^0e^{-ikx}\,dx\\
&=\frac{2-e^{ik}-e^{-ik}}{ik}
=-\frac{2i(1-\cos k)}{k}.
\end{aligned}
$$
The continuous value at $k=0$ is zero, in agreement with the vanishing <integral> of the <odd function> $f$. Thus
$$
\boxed{\widetilde f(k)=-\frac{2i(1-\cos k)}k}.
$$
Solved by gpt-5.6-sol high.
= b
{parent=14a}
{scope}
= Solution
{parent=b}
Because $e^{-\lambda|k|}$ is an <even function>, <Fourier inversion> reduces to a cosine integral:
$$
\begin{aligned}
g(x)
&=\frac1{2\pi}\int_{-\infty}^{\infty}e^{-\lambda|k|}e^{ikx}\,dk\\
&=\frac1\pi\int_0^\infty e^{-\lambda k}\cos(kx)\,dk
=\frac1\pi\operatorname{Re}\frac1{\lambda-ix}.
\end{aligned}
$$
Therefore
$$
\boxed{g(x)=\frac{\lambda}{\pi(x^2+\lambda^2)}}.
$$
Solved by gpt-5.6-sol high.
= c
{parent=14a}
{scope}
= Solution
{parent=c}
Extend the boundary data oddly to the whole real axis. The extended function is precisely the function $f$ from part (a). Taking the <Fourier transform> in $x$, the <Laplace equation> becomes
$$
\partial_y^2\widetilde u-k^2\widetilde u=0.
$$
Decay as $y\to\infty$ selects
$$
\widetilde u(k,y)=\widetilde f(k)e^{-|k|y}.
$$
Part (b), together with the <convolution theorem>, says that the inverse transform is convolution with the <Poisson kernel>
$$
P_y(s)=\frac{y}{\pi(s^2+y^2)}.
$$
For $x,y>0$ this gives
$$
\begin{aligned}
u(x,y)
&=\frac y\pi\int_0^1\left[
\frac1{(x-v)^2+y^2}-\frac1{(x+v)^2+y^2}
\right]dv\\
&=\boxed{\frac{4xy}{\pi}\int_0^1
\frac{v\,dv}{[(x-v)^2+y^2][(x+v)^2+y^2]}}.
\end{aligned}
$$
The odd extension enforces $u(0,y)=0$, while the <Poisson integral> takes the prescribed boundary values at every continuity point and decays at infinity.
Solved by gpt-5.6-sol high.
= d
{parent=14a}
{scope}
= Solution
{parent=d}
By the <superposition principle>, interchange $x$ and $y$ in the solution from part (c) and add the two solutions:
$$
\boxed{w(x,y)=u(x,y)+u(y,x)}.
$$
The first term supplies the required data on the positive $x$-axis and vanishes on the positive $y$-axis; the second does the reverse. Explicitly,
$$
\begin{aligned}
w(x,y)=\frac{4xy}{\pi}\bigg(&
\int_0^1\frac{v\,dv}{[(x-v)^2+y^2][(x+v)^2+y^2]}\\
&+\int_0^1\frac{v\,dv}{[(y-v)^2+x^2][(y+v)^2+x^2]}
\bigg).
\end{aligned}
$$
It is <harmonic> in the quarter-plane, has the stated boundary values, and decays in both unbounded directions.
Solved by gpt-5.6-sol high.
= 15C
{parent=Paper 2}
{scope}
{title2=Quantum Mechanics}
= a
{parent=15c}
{scope}
= Solution
{parent=a}
For a one-dimensional <wavefunction>,
$$
\boxed{\rho(x,t)=|\psi(x,t)|^2},
\qquad
\boxed{j(x,t)=\frac{\hbar}{2mi}
\left(\psi^*\frac{\partial\psi}{\partial x}
-\psi\frac{\partial\psi^*}{\partial x}\right)}
=\frac{\hbar}{m}\operatorname{Im}\left(\psi^*\partial_x\psi\right).
$$
The <probability continuity equation> is
$$
\partial_t\rho+\partial_xj=0.
$$
For a <stationary state> $\psi(x,t)=\phi(x)e^{-iEt/\hbar}$, the <probability density> is independent of time, so $\partial_xj=0$ and the <probability current> is constant in space.
For the <plane wave>
$$
\psi=Ae^{ikx-iEt/\hbar},
$$
one obtains
$$
\boxed{\rho=|A|^2},
\qquad
\boxed{j=\frac{\hbar k}{m}|A|^2}.
$$
This is a <momentum eigenstate> with momentum $p=\hbar k$ and, when it satisfies the free <Schrodinger equation>, energy $E=\hbar^2k^2/(2m)$. Its constant density and current describe a spatially uniform beam carrying probability in the sign of $k$. It is not a <normalizable wavefunction>, so it represents an idealized state rather than a localized particle.
Solved by gpt-5.6-sol high.
= b
{parent=15c}
{scope}
= Solution
{parent=b}
Put
$$
k=\frac{\sqrt{2mE}}{\hbar},
\qquad
K=\frac{\sqrt{2m(E+V_0)}}{\hbar}.
$$
The <time-independent Schrodinger equation> has the forms
$$
\psi(x)=
\begin{cases}
Ae^{ikx}+Be^{-ikx},&x<0,\\
Ce^{iKx}+De^{-iKx},&0\leq x\leq a,\\
Fe^{ikx},&x>a.
\end{cases}
$$
The first region contains the incident and <reflected wave>[reflected waves], while the final region contains only the transmitted wave.
Continuity of $\psi$ and $\psi'$ at both edges of the <finite square well> gives the standard transmission coefficient
$$
T=\frac{|F|^2}{|A|^2}
=\left[
1+\frac{(K^2-k^2)^2}{4k^2K^2}\sin^2(Ka)
\right]^{-1}.
$$
When $V_0=3E$, one has $K=2k$, and therefore
$$
\frac{(K^2-k^2)^2}{4k^2K^2}=\frac9{16},
\qquad
Ka=\frac{a\sqrt{8mE}}{\hbar}.
$$
Hence the <transmission probability> is
$$
\boxed{
T=\frac{16}{16+9\sin^2\!\left(a\sqrt{8mE}/\hbar\right)}
}.
$$
Solved by gpt-5.6-sol high.
= 16D
{parent=Paper 2}
{scope}
{title2=Electromagnetism}
= a
{parent=16d}
{scope}
= Solution
{parent=a}
Write
$$
\frac1{|\mathbf x-\mathbf y|}
=\frac1{|\mathbf x|}
\left(1-2\frac{\mathbf x\mathbin{\cdot}\mathbf y}{|\mathbf x|^2}
+\frac{|\mathbf y|^2}{|\mathbf x|^2}\right)^{-1/2}.
$$
Using the <Taylor series>
$$
(1+s)^{-1/2}=1-\frac12s+\frac38s^2+O(s^3)
$$
and retaining terms through second order in $|\mathbf y|/|\mathbf x|$ gives
$$
\boxed{
\frac1{|\mathbf x-\mathbf y|}
=\frac1{|\mathbf x|}
\left[
1+\frac{\mathbf x\mathbin{\cdot}\mathbf y}{|\mathbf x|^2}
+\frac{3(\mathbf x\mathbin{\cdot}\mathbf y)^2-|\mathbf x|^2|\mathbf y|^2}
{2|\mathbf x|^4}
+O\!\left(\frac{|\mathbf y|^3}{|\mathbf x|^3}\right)
\right]}.
$$
This is the beginning of the <multipole expansion>.
Solved by gpt-5.6-sol high.
= b
{parent=16d}
{scope}
= Solution
{parent=b}
The <method of images> replaces the earthed plane by an image charge $-q$ at $(-a,0,0)$. The two Coulomb potentials cancel at $x=0$, so uniqueness for the <Dirichlet problem> makes the resulting field the physical field in $x>0$. With $\mathbf r=(x,y,z)$ and $\mathbf e_x=(1,0,0)$,
$$
\boxed{\Phi(\mathbf r)=\frac{q}{4\pi\epsilon_0}
\left(\frac1{|\mathbf r-a\mathbf e_x|}
-\frac1{|\mathbf r+a\mathbf e_x|}\right)}
$$
and
$$
\boxed{\mathbf E(\mathbf r)=\frac{q}{4\pi\epsilon_0}
\left(
\frac{\mathbf r-a\mathbf e_x}{|\mathbf r-a\mathbf e_x|^3}
-\frac{\mathbf r+a\mathbf e_x}{|\mathbf r+a\mathbf e_x|^3}
\right)}.
$$
The <electric multipole expansion> from part (a) gives, for $r\gg a$,
$$
\Phi(\mathbf r)
=\frac{q}{4\pi\epsilon_0}\frac{2ax}{r^3}+O(r^{-4})
=\frac{\mathbf p\mathbin{\cdot}\mathbf r}{4\pi\epsilon_0r^3}+O(r^{-4}),
\qquad
\boxed{\mathbf p=2qa\,\mathbf e_x}.
$$
Thus the leading field is that of an <electric dipole>.
On the plane,
$$
E_x(0,y,z)
=-\frac{2qa}{4\pi\epsilon_0(a^2+y^2+z^2)^{3/2}}.
$$
Taking the plane normal to be $+\mathbf e_x$ and using <polar coordinates>,
$$
\int_{\mathbb R^2}E_x\,dy\,dz
=-\frac{qa}{\epsilon_0}\int_0^\infty
\frac{\rho\,d\rho}{(a^2+\rho^2)^{3/2}}
=\boxed{-\frac q{\epsilon_0}}.
$$
With the outward normal of the region $x>0$, the sign is reversed. This is consistent with <Gauss's law>: all electric flux from the real charge terminates on the grounded conductor.
The <electrostatic boundary condition> gives the induced <surface charge density>
$$
\boxed{\sigma(y,z)=\epsilon_0E_x(0,y,z)
=-\frac{qa}{2\pi(a^2+y^2+z^2)^{3/2}}},
$$
and its integral is
$$
\boxed{Q_{\rm induced}=-q}.
$$
In the plane $z=0$, the field lines leave the positive charge, meet the conductor normally, and are the right-half-plane portions of the field lines joining the real charge to its negative image.
Solved by gpt-5.6-sol high.
= c
{parent=16d}
{scope}
= Solution
{parent=c}
Successive reflections in the two grounded planes require four charges:
$$
\begin{array}{c|c}
\text{position}&\text{charge}\\ \hline
(a,b,0)&q\\
(-a,b,0)&-q\\
(a,-b,0)&-q\\
(-a,-b,0)&q.
\end{array}
$$
Their total charge and <electric dipole moment> vanish. Applying the second-order expansion from part (a), the terms proportional to $a^2x^2$, $b^2y^2$, and $a^2+b^2$ also cancel, while the mixed terms add. Thus
$$
\boxed{
\Phi(\mathbf r)\sim
\frac{q}{4\pi\epsilon_0}\frac{12abxy}{r^5}
=\frac{3qabxy}{\pi\epsilon_0r^5}
},
\qquad r\gg a,b.
$$
This is an <electric quadrupole> potential.
Solved by gpt-5.6-sol high.
= 17B
{parent=Paper 2}
{scope}
{title2=Numerical Analysis}
= a
{parent=17b}
{scope}
= Solution
{parent=a}
For a nonzero vector $v\in\mathbb R^n$, a <Householder reflection> is
$$
H=I-2\frac{vv^T}{v^Tv}.
$$
It is immediately <symmetric matrix>[symmetric]. If $P=vv^T/(v^Tv)$, then $P^2=P$, so
$$
H^TH=H^2=(I-2P)^2=I.
$$
Hence $H$ is also an <orthogonal matrix>, with $H^{-1}=H$.
The <similarity transformation> can be expanded as
$$
HAH
=A-2P A-2A P+4PAP.
$$
Computing $v^TA$, $Av$, and the scalar $v^TAv$ costs $O(n^2)$ <arithmetic operation>[arithmetic operations], after which the remaining updates are <outer product>[outer products] and scalar multiples, also costing $O(n^2)$. Thus $HAH^{-1}$ can be formed in $O(n^2)$ operations rather than by two general $O(n^3)$ <matrix multiplication>[matrix multiplications].
Solved by gpt-5.6-sol high.
= b
{parent=17b}
{scope}
= Solution
{parent=b}
Starting with $A_1=A$, for $k=1,\ldots,n-2$ choose a <Householder reflection> $H_k$ that acts only on coordinates $k+1,\ldots,n$ and maps the tail of column $k$,
$$
(A_k)_{k+1:n,k},
$$
to a multiple of its first coordinate vector. Set
$$
A_{k+1}=H_kA_kH_k.
$$
This zeros all entries in column $k$ below its first subdiagonal. Because $H_k$ fixes the first $k$ coordinates, it preserves the zeros created in earlier columns. After $n-2$ steps,
$$
T=A_{n-1}
$$
is an <upper Hessenberg matrix>.
Each $H_k$ is <orthogonal matrix>[orthogonal] and symmetric. If
$$
Q=H_{n-2}\cdots H_1,
$$
then $Q$ is orthogonal and
$$
\boxed{T=QAQ^T}.
$$
Part (a) shows that each similarity update costs $O(n^2)$ arithmetic operations, and there are $O(n)$ updates. The total cost is therefore
$$
\boxed{O(n^3)}.
$$
Solved by gpt-5.6-sol high.
= 18H
{parent=Paper 2}
{scope}
{title2=Markov Chains}
= a
{parent=18h}
{scope}
= i
{parent=a}
{scope}
= Solution
{parent=i}
A distribution $\pi$ is an <invariant distribution> when
$$
\pi_j=\sum_{i\in I}\pi_i p_{ij}
$$
for every state $j$. The pair $(\pi,P)$ satisfies <detailed balance> when
$$
\pi_i p_{ij}=\pi_jp_{ji}
$$
for every $i,j$. Summing this identity over $i$ gives
$$
\sum_i\pi_i p_{ij}
=\sum_i\pi_jp_{ji}
=\pi_j\sum_i p_{ji}
=\pi_j,
$$
so detailed balance implies invariance.
For an irreducible positive recurrent <Markov chain>, <Kac's lemma> relates the invariant mass to the <mean recurrence time>:
$$
\boxed{\mathbb E_iT_i^+=\frac1{\pi_i}}.
$$
Solved by gpt-5.6-sol high.
= ii
{parent=a}
{scope}
= Solution
{parent=ii}
The renewal-reward form of <Kac's lemma> says that the expected number of visits to state $i$ during one return cycle from $k$ to $k$ is
$$
\boxed{\frac{\pi_i}{\pi_k}}.
$$
Here time spent means the number of discrete time instants at which the chain occupies $i$ between consecutive visits to $k$.
Solved by gpt-5.6-sol high.
= b
{parent=18h}
{scope}
= i
{parent=b}
{scope}
= Solution
{parent=i}
The chain is a <birth-death chain>. Detailed balance between $0$ and $1$ gives
$$
\pi_0p=\pi_1q^{-1},
\qquad\text{so}\qquad
\pi_1=qp\,\pi_0.
$$
For $i\geq1$, detailed balance between $i$ and $i+1$ gives
$$
\pi_iq^{-(i+2)}=\pi_{i+1}q^{-(i+1)},
\qquad
\pi_{i+1}=\frac{\pi_i}{q}.
$$
Consequently
$$
\pi_i=qp\,\pi_0q^{-(i-1)},\qquad i\geq1.
$$
The expected occupation time of the positive even states during a return cycle to state $1$ is therefore
$$
\frac{\pi_2+\pi_4+\cdots}{\pi_1}
=\sum_{r=1}^{\infty}q^{-(2r-1)}
=\boxed{\frac{q}{q^2-1}}.
$$
Solved by gpt-5.6-sol high.
= ii
{parent=b}
{scope}
= Solution
{parent=ii}
Normalizing the invariant distribution found in part (i) gives
$$
1=\pi_0\left(1+\frac{q^2p}{q-1}\right),
\qquad
\pi_1=\frac{qp(q-1)}{q-1+q^2p}.
$$
The <mean recurrence time> formula now yields
$$
\boxed{
\mathbb E_1T_1^+
=\frac1{\pi_1}
=\frac{q-1+q^2p}{qp(q-1)}
}.
$$
Solved by gpt-5.6-sol high.
= iii
{parent=b}
{scope}
= Solution
{parent=iii}
Let $h_i=\mathbb E_iT_0$ be the <expected hitting time> of state $0$. For $i\geq1$, <first-step analysis> gives
$$
q^{-(i+2)}(h_{i+1}-h_i)
+q^{-i}(h_{i-1}-h_i)=-1.
$$
With $d_i=h_i-h_{i-1}$ this becomes
$$
d_{i+1}-q^2d_i=-q^{i+2}.
$$
The minimal nonnegative solution has
$$
d_i=\frac{q^{i+1}}{q-1},
$$
so
$$
\boxed{\mathbb E_1T_0=h_1=d_1=\frac{q^2}{q-1}}.
$$
Equivalently, <Kac's lemma> gives $\mathbb E_0T_0^+=1/\pi_0$. A first step from $0$ either returns immediately with probability $1-p$ or moves to $1$ with probability $p$, so
$$
\frac1{\pi_0}=1+p\,\mathbb E_1T_0.
$$
Substituting $\pi_0=(q-1)/(q-1+q^2p)$ gives the same result.
Solved by gpt-5.6-sol high.