past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ib/paper-2.bigb

= Paper 2
{scope}

https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2025/paperib_2_2025.pdf

= 1E
{parent=Paper 2}
{scope}
{title2=Groups, Rings and Modules}

= a
{parent=1e}
{scope}

= Solution
{parent=a}

An <integral domain> is a nonzero commutative <ring> with identity and no zero divisors. A <ring> is Noetherian when every <ideal> is finitely generated, equivalently every ascending <ideal> chain stabilizes. A <principal ideal domain> is an <integral> domain in which every <ideal> is generated by one element.

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= b
{parent=1e}
{scope}

= Solution
{parent=b}

Equality of constant terms is preserved by componentwise addition, subtraction, and multiplication, and $(1,1)$ belongs to the set, so $R$ is a subring. The projection is surjective and
$$\ker p=\{(0,yg(y)):g\in\mathbb C[y]\}.$$
The nonzero elements $(x,0)$ and $(0,y)$ have product zero, so $R$ is not an <integral> domain and hence not a PID. In fact $R\cong\mathbb C[x,y]/(xy)$, which also shows that it is Noetherian.

Solved by gpt-5.6-sol high.

= 2G
{parent=Paper 2}
{scope}
{title2=Analysis and Topology}

= i
{parent=2g}
{scope}

= Solution
{parent=i}

Pointwise, $\tanh(nx)$ tends to $-1,0,1$ according as $x<0,x=0,x>0$. This <limit> is discontinuous, whereas a <uniform limit> of <continuous functions> is continuous. Convergence is therefore not uniform.

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= ii
{parent=2g}
{scope}

= Solution
{parent=ii}

The <limit> is $\sin x$, and
$$\sup_x|f_n(x)-\sin x|=1/n\to0,$$
so convergence is uniform.

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= iii
{parent=2g}
{scope}

= Solution
{parent=iii}

The pointwise <limit> is $f(x)=e^x/\cosh(ax)$ and
$$|f_n-f|=(e^{1/n}-1)\frac{e^x}{\cosh(ax)}.$$
The final factor is bounded on $\mathbb R$ exactly when $|a|\ge1$; when $|a|<1$ it diverges as $x\to+\infty$. Thus convergence is uniform exactly for $|a|\ge1$.

Solved by gpt-5.6-sol high.

= iv
{parent=2g}
{scope}

= Solution
{parent=iv}

The <fundamental theorem of calculus> gives
$$f_n(x)=2n\int_{x-1/n}^{x+1/n}h'(t)\,dt.$$
Thus $f_n(x)\to4h'(x)$. Uniform continuity of $h'$ makes the difference from $4h'(x)$ uniformly small over every interval of radius $1/n$, so convergence is uniform on all of $\mathbb R$.

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= 3B
{parent=Paper 2}
{scope}
{title2=Methods}

= Solution
{parent=3b}

Let $r_1,r_2$ be the roots of $\alpha r^2+\beta r+\gamma=0$. The <causal Green function> is $G(t,\tau)=H(t-\tau)g(t-\tau)$, where
$$g(s)=\frac{e^{r_1s}-e^{r_2s}}{\alpha(r_1-r_2)}$$
for distinct roots, and $g(s)=se^{rs}/\alpha$ for a repeated root. These formulas enforce $g(0)=0$ and $\alpha g'(0)=1$, hence the required delta jump.

For the oscillator,
$$y(t)=\int_0^t\frac{\sin(\omega(t-\tau))}{\omega}\sin(\lambda\tau)\,d\tau.$$
If $\lambda\ne\omega$, this is
$$y(t)=\frac{\sin(\lambda t)-(\lambda/\omega)\sin(\omega t)}{\omega^2-\lambda^2}.$$
At resonance $\lambda=\omega$, the limiting expression is
$$y(t)=\frac{\sin(\omega t)-\omega t\cos(\omega t)}{2\omega^2}.$$

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= 4B
{parent=Paper 2}
{scope}
{title2=Electromagnetism}

= Solution
{parent=4b}

<Maxwell equations> are
$$\nabla\cdot E=\rho/\varepsilon_0,\quad \nabla\cdot B=0,
\quad\nabla\times E=-\partial_tB,
\quad\nabla\times B=\mu_0J+\mu_0\varepsilon_0\partial_tE.$$
Taking the divergence of the <Ampère-Maxwell equation> and using <Gauss's law> gives
$$\partial_t\rho+\nabla\cdot J=0.$$
Therefore
$$\dot Q=-\int_{\partial D}J\cdot n\,dS=0$$
because $J$ vanishes on the boundary. Componentwise integration by parts also gives
$$\frac d{dt}\int_Dx\rho\,d^3x=-\int_Dx\nabla\cdot J\,d^3x=\int_DJ\,d^3x,$$
with the boundary term again zero.

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= 5D
{parent=Paper 2}
{scope}
{title2=Fluid Dynamics}

= Solution
{parent=5d}

For incompressible inviscid flow without body force,
$$\nabla\cdot u=0,\qquad \partial_tu+(u\cdot\nabla)u=-\rho^{-1}\nabla p.$$
Taking curl gives
$$\partial_t\omega+(u\cdot\nabla)\omega=(\omega\cdot\nabla)u,\qquad \omega=\nabla\times u,$$
equivalently $\partial_t\omega=\nabla\times(u\times\omega)$.

Direct <differentiation> of the stated field gives $\omega=u$. Hence $u\times\omega=0$, so the <vorticity equation> gives $\partial_t\omega=0$ and the field is steady. Finally
$$(u\cdot\nabla)u=\nabla(|u|^2/2)-u\times\omega=\nabla(|u|^2/2).$$
Euler's equation therefore implies
$$\nabla\left(p+\frac12\rho|u|^2\right)=0,$$
so $H$ is spatially uniform.

Solved by gpt-5.6-sol high.

= 6H
{parent=Paper 2}
{scope}
{title2=Statistics}

= a
{parent=6h}
{scope}

= Solution
{parent=a}

The likelihood is $\theta e^{-\theta X}$. Multiplication by the prior gives a density proportional to $\theta^m e^{-(\lambda+X)\theta}$, so
$$\theta\mid X\sim\Gamma(m+1,\lambda+X)$$
in shape-rate parametrisation.

Solved by gpt-5.6-sol high.

= b
{parent=6h}
{scope}

= Solution
{parent=b}

Squared-error posterior risk is minimized by the posterior mean. Hence
$$\hat\theta_{\rm Bayes}=E[\theta\mid X]=\frac{m+1}{\lambda+X}.$$

Solved by gpt-5.6-sol high.

= c
{parent=6h}
{scope}

= Solution
{parent=c}

Put $k=m+1$ and $L=\lambda+X$. The posterior risk is
$$\tfrac12\{e^{-ra}M(r)+e^{ra}M(-r)\},$$
where $M(s)=(L/(L-s))^k$. Differentiating in $a$ gives
$$e^{2ra}=\frac{M(r)}{M(-r)}=\left(\frac{L+r}{L-r}\right)^k.$$
Thus
$$\hat\theta=\frac{m+1}{2r}\log\frac{\lambda+X+r}{\lambda+X-r}.$$

Solved by gpt-5.6-sol high.

= 7H
{parent=Paper 2}
{scope}
{title2=Optimisation}

= a
{parent=7h}
{scope}

= Solution
{parent=a}

A set is convex when it contains every segment joining two of its points. A <function> on a <convex set> is convex when
$$f(tx+(1-t)y)\le tf(x)+(1-t)f(y)$$
for all $x,y$ and $0\le t\le1$.

Solved by gpt-5.6-sol high.

= b
{parent=7h}
{scope}

= Solution
{parent=b}

Take $(t,x),(s,y)$ and $0\le\alpha\le1$, and put $T=\alpha t+(1-\alpha)s$. Then
$$\frac{\alpha x+(1-\alpha)y}{T}
=\frac{\alpha t}{T}\frac xt+\frac{(1-\alpha)s}{T}\frac ys.$$
Convexity of $f$, followed by multiplication by $T$, proves convexity of its perspective $g(t,x)=tf(x/t)$.

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= c
{parent=7h}
{scope}

= Solution
{parent=c}

Let $z=tx+(1-t)y$. Apply the hypothesis with first argument $z$ and second argument successively $x,y$:
$$f(z)-f(x)\le(1-t)\lambda(z)^T(y-x),$$
$$f(z)-f(y)\le t\lambda(z)^T(x-y).$$
Multiply these inequalities by $t$ and $1-t$ respectively and add. The right sides cancel, leaving $f(z)\le tf(x)+(1-t)f(y)$.

Solved by gpt-5.6-sol high.

= 8F
{parent=Paper 2}
{scope}
{title2=Linear Algebra}

= Solution
{parent=8f}

A <Jordan block> is $J_m(\alpha)=\alpha I+N$, where $N$ has ones on the superdiagonal and zeros elsewhere. Since $N^2=0$ for $m=2$, Taylor expansion gives
$$p(J_2(\alpha))=p(\alpha)I+p'(\alpha)N
=\begin{pmatrix}p(\alpha)&p'(\alpha)\\0&p(\alpha)\end{pmatrix}.$$
The <Jordan normal form> is the block-diagonal <matrix>, unique up to block order, to which $A$ is similar and whose blocks are Jordan blocks.

A block of size at least two for <eigenvalue> $\lambda$ contributes a <generalized eigenvector> in $\ker(A-\lambda I)^2\setminus\ker(A-\lambda I)$. Thus the displayed equality for every $\lambda$ is equivalent to every block having size one, which is diagonalizability.

The transpose of each Jordan block is similar to it by reversing its <basis>. Hence $A^T$ has the same Jordan form as $A$.

Write $A=PJP^{-1}$. Let $R$ be block diagonal with the reversal <matrix> on each Jordan block. Both $R$ and $RJ$ are symmetric and $R$ is invertible. Therefore
$$A=(PRP^T)\{P^{-T}(RJ)P^{-1}\}=CD,$$
where $C,D$ are symmetric and $C$ is nonsingular.

Solved by gpt-5.6-sol high.

= 9E
{parent=Paper 2}
{scope}
{title2=Groups, Rings and Modules}

= Solution
{parent=9e}

A <euclidean domain> is an <integral> domain with a <function> $d$ on nonzero elements such that for $a,b\ne0$ there are $q,r$ with $a=bq+r$ and either $r=0$ or $d(r)<d(b)$.

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= a
{parent=9e}
{scope}

= Solution
{parent=a}

For a nonzero <ideal> $I$, choose $a\in I$ of minimal Euclidean value. Divide any $x\in I$ by $a$: $x=qa+r$. Then $r\in I$, and minimality forces $r=0$. Hence $I=(a)$; the zero <ideal> is also principal.

Solved by gpt-5.6-sol high.

= b
{parent=9e}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

The class $a$ is a unit modulo $p^e$ exactly when $\phi(a)\ne0$. This says a representative is not divisible by $p$, so it is coprime to $p^e$ and Bézout gives an inverse.

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= ii
{parent=b}
{scope}

= Solution
{parent=ii}

The class $a$ is nilpotent exactly when $\phi(a)=0$. Divisibility by $p$ makes $a^e$ zero modulo $p^e$. Conversely, the image of a nilpotent in the field $R/(p)$ is nilpotent and therefore zero.

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= c
{parent=9e}
{scope}

= Solution
{parent=c}

The <Chinese remainder theorem> identifies
$$R/I\cong\prod_iR/(p_i^{e_i}).$$
Thus a class is a unit exactly when its image modulo every $p_i$ is nonzero, equivalently when no $p_i$ divides a representative.

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= d
{parent=9e}
{scope}

= Solution
{parent=d}

A unit modulo $d$ has nonzero image modulo every $p_i$. Any representative therefore satisfies the criterion in part (c), so it is already a unit modulo $I$. Its image is the given unit modulo $d$, proving surjectivity on unit <groups>.

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= 10G
{parent=Paper 2}
{scope}
{title2=Analysis and Topology}

= Solution
{parent=10g}

The <product topology> has <basis> $U\times V$ with $U$ open in $X$ and $V$ open in $Y$. Since $\Pi_X^{-1}(U)=U\times Y$ and similarly for $\Pi_Y$, both projections are continuous.

If $Y$ is Hausdorff, a point $(y_1,y_2)$ off the diagonal has disjoint product neighborhoods, so the diagonal is closed. Conversely, if the diagonal is closed, its complement supplies disjoint neighborhoods of any two distinct points.

The graph is the inverse image of the diagonal under $(x,y)\mapsto(f(x),y)$, so a continuous map into a <Hausdorff space> has closed graph. Conversely, if $X,Y$ are compact Hausdorff and $C\subseteq Y$ is closed, then $\Gamma_f\cap(X\times C)$ is compact. Its projection onto $X$ is $f^{-1}(C)$, compact and hence closed. Thus $f$ is continuous.

The <function> $f(0)=0$, $f(x)=1/x$ for $x\ne0$ is discontinuous but has closed graph: a convergent graph <sequence> approaching $x=0$ cannot have $x\ne0$ and bounded second coordinate unless it eventually reaches the isolated graph point $(0,0)$.

Solved by gpt-5.6-sol high.

= 11F
{parent=Paper 2}
{scope}
{title2=Geometry}

= a
{parent=11f}
{scope}

= Solution
{parent=a}

A <topological surface> is a Hausdorff second-countable space locally homeomorphic to $\mathbb R^2$. Pairing opposite sides of the hexagon gives neighborhoods modelled on discs, including at edge and vertex classes; compactness follows from the compact polygon quotient. Cutting and rearranging the polygon gives the usual opposite-side square, hence a <torus>.

The full double cone is not a surface at the origin: deleting the vertex from a small neighborhood leaves two components, unlike a punctured disc. The upper cone with $z>0$ excludes that vertex and has ordinary annular charts, so it is a surface.

Solved by gpt-5.6-sol high.

= b
{parent=11f}
{scope}

= Solution
{parent=b}

Writing the upper cone as $(r\cos\theta,r\sin\theta,r)$ gives metric
$$ds_C^2=2\,dr^2+r^2d\theta^2.$$
On any polar branch, the map
$$\pi(s,\varphi)=\left(\frac{s}{\sqrt2}\cos(\sqrt2\varphi),
\frac{s}{\sqrt2}\sin(\sqrt2\varphi),\frac{s}{\sqrt2}\right)$$
pulls this back to $ds^2+s^2d\varphi^2$, so it is a local isometry. Strictly, because $\sqrt2\varphi$ is not $2\pi$-periodic, this formula is globally single-valued on the universal polar cover and locally on $\mathbb R^2\setminus\{0\}$; that is the <developing map> used here.

An isometry fixing $p$ lifts locally to a Euclidean isometry fixing a lift of $p$. Compatibility with the cone's angular identification leaves only the identity and reflection across a radial line. Descending to the cone, these are the identity and reflection in the plane through the cone axis and $p$.

Solved by gpt-5.6-sol high.

= 12
{parent=Paper 2}
{scope}
{title2=Complex Analysis OR Complex Methods}

= 12.1G
{parent=12}
{scope}

= Solution
{parent=12.1g}

Apply the residue theorem to $f(z)/(z-w)^{n+1}$; its residue at $w$ is $f^{(n)}(w)/n!$, which proves Cauchy's <derivative> formula.

For a compact subdisc choose a slightly larger contour still inside $D(a,R)$. Cauchy's <derivative> formula bounds the supremum of $g_k^{(n)}-g^{(n)}$ on the smaller disc by a fixed constant times the supremum of $g_k-g$ on the contour. Local <uniform convergence> therefore passes to every <derivative>.

On $|z-a|=\varepsilon$, $g$ has a positive minimum <modulus>. For large $k$, $|g_k-g|<|g|$ there. Rouché's theorem then says $g_k$ and $g$ have the same number of zeros inside, namely $m$, counting multiplicity.

Solved by gpt-5.6-sol high.

= 12.2A
{parent=12}
{scope}

= a
{parent=12.2a}
{scope}

= Solution
{parent=a}

If $f$ had a pole at one of the finitely many candidate points, choose a <polynomial> vanishing to sufficiently high order at all the other candidates. <Hermite interpolation> lets its remaining Taylor coefficients be chosen so that the residue of $p^2f$ at the selected pole is nonzero. A small contour around that pole would then have nonzero <integral> by the <residue theorem>, contradicting the hypothesis. Hence none of the candidate points is a pole and $f$ is entire.

Solved by gpt-5.6-sol high.

= b
{parent=12.2a}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

Schwarz reflection shows that the Taylor coefficients along the real axis are real. Since $v(x,0)=0$ and $v(x,y)>0$ for $y>0$, one has $v_y(x,0)\ge0$. The Cauchy–Riemann equations give $h'(x)=u_x(x,0)=v_y(x,0)\ge0$.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

If $h'(0)=0$, let $a_kz^k$ be the first nonzero Taylor term. Then $k\ge2$ and $a_k$ is real. But $\operatorname{Im}(a_kr^ke^{ik\theta})$ changes sign as $0<\theta<\pi$, and for small $r$ the leading term controls the sign. This contradicts positivity in the upper half-plane. Hence $h'(0)\ne0$.

Solved by gpt-5.6-sol high.

= 13C
{parent=Paper 2}
{scope}
{title2=Variational Principles}

= Solution
{parent=13c}

Put $W=1+\rho'^2+\rho^2\phi'^2$. The Euler–Lagrange equations are
$$\frac d{dz}\left(\frac{n\rho'}{\sqrt W}\right)-n_\rho\sqrt W-\frac{n\rho\phi'^2}{\sqrt W}=0,$$
$$\frac d{dz}\left(\frac{n\rho^2\phi'}{\sqrt W}\right)-n_\phi\sqrt W=0.$$
When $n=n(\rho)$, the helical ansatz $\rho=R$, $\phi=\phi_0+\omega z$ automatically satisfies the second equation. The first becomes
$$n(R)R\omega^2+(1+R^2\omega^2)n'(R)=0,$$
so $a=1$.

The endpoints have the same cylindrical angle exactly when $\omega L=2\pi k$ for some integer $k$. Thus a positive $L$ with $L=2\pi k/\omega(R)$ admits the two special solutions of opposite handedness, using $\omega$ and $-\omega$.

Solved by gpt-5.6-sol high.

= 14B
{parent=Paper 2}
{scope}
{title2=Methods}

= Solution
{parent=14b}

Transverse force balance on a short element gives
$$\rho y_{tt}=\tau y_{xx},\qquad c=\sqrt{\tau/\rho}.$$
With fixed ends,
$$y(x,t)=\sum_{n\ge1}[A_n\cos(\omega_nt)+B_n\sin(\omega_nt)]\sin\frac{n\pi x}{L},
\quad\omega_n=\frac{n\pi c}{L}.$$
Here $A_n=0$. If $b_n$ are the sine coefficients of initial <velocity>, then
$$b_n=\frac{4v}{n\pi\sqrt\varepsilon}\sin\frac{n\pi l}{L}\sin\frac{n\pi\varepsilon}{2L},
\qquad B_n=\frac{b_n}{\omega_n}.$$
The total deposited energy is $E=\rho v^2/2$, while
$$\frac{E_n}{E}=\frac{8L}{n^2\pi^2\varepsilon}
\sin^2\frac{n\pi l}{L}\sin^2\frac{n\pi\varepsilon}{2L}.$$
The seventh mode is eliminated by striking at a node $l=jL/7$, $j=1,\ldots,6$. For a narrow hammer, each fixed-mode fraction is $O(\varepsilon)$; beyond the inverse-width cutoff the envelope falls as $n^{-2}$, modulated by the two sine factors.

Solved by gpt-5.6-sol high.

= 15C
{parent=Paper 2}
{scope}
{title2=Quantum Mechanics}

= Solution
{parent=15c}

The commutator follows directly on a test <function>:
$$[f,p]=i\hbar f'.$$
Expansion gives
$$V_+=\frac{f^2-\hbar f'}{2m},\qquad V_-=\frac{f^2+\hbar f'}{2m}.$$
To make $V_-=2m$, take
$$f(x)=2m\tanh(2mx/\hbar).$$
Then
$$V_+(x)=2m-4m\operatorname{sech}^2(2mx/\hbar),$$
which tends to $2m$ at both ends.

Applying $p+if$ to $e^{ikx}$ gives a partner scattering state with asymptotic amplitudes $\hbar k-2mi$ at $-\infty$ and $\hbar k+2mi$ at $+\infty$, and no reflected wave. Their <moduli> agree. Subtracting the common constant $2m$ therefore shows that the displayed $-4m\operatorname{sech}^2$ potential is reflectionless:
$$\mathcal R=0,\qquad \mathcal T=1.$$

Solved by gpt-5.6-sol high.

= 16B
{parent=Paper 2}
{scope}
{title2=Electromagnetism}

= Solution
{parent=16b}

Taking curls gives
$$B_1=(0,0,2b_1),\qquad B_2=(-b_2x,-b_2y,2b_2z).$$
Both have zero divergence and curl, so they satisfy the static vacuum equations. Their upward fluxes through the loop are
$$\Phi_1=2\pi b_1r^2,\qquad \Phi_2=2\pi b_2r^2z.$$
For clockwise current $I$, the net force is zero in $B_1$, while in $B_2$
$$F_z=-2\pi b_2Ir^2.$$

Put $K=2\pi b_2r^2$. Motion in $B_1$ changes no flux, so no current is induced and $z=z_0-gt^2/2$. In $B_2$, using counterclockwise current as positive,
$$I=-\frac{K\dot z}{R},\qquad F_z=-\frac{K^2}{R}\dot z.$$
Thus
$$m\ddot z=-mg-\frac{K^2}{R}\dot z.$$
With $\Gamma=K^2/R$, the <velocity> tends to $-mg/\Gamma$ and
$$z=z_0-\frac{mg}{\Gamma}\left[t-\frac m\Gamma(1-e^{-\Gamma t/m})\right].$$
At terminal speed, $I^2R=mg|\dot z|$, so Joule loss equals gravitational-energy loss. For $R=0$, Ohm's quasistatic formula is singular: flux is conserved, persistent current stores magnetic energy, and <ideal> motion has no resistive terminal state (<inductance> must be retained).

Solved by gpt-5.6-sol high.

= 17A
{parent=Paper 2}
{scope}
{title2=Numerical Analysis}

= a
{parent=17a}
{scope}

= Solution
{parent=a}

Write the internal stage as
$$k_2=f(y_n+h[(1-\theta)k_1+\theta k_2]).$$
The two methods have $\theta=1/2$ and $3/4$. Expansion gives stability <function>
$$R_\theta(z)=\frac{1+(1-\theta)z+(1/2-\theta)z^2}{1-\theta z}
=1+z+\frac{z^2}{2}+\frac\theta2z^3+O(z^4).$$
Both therefore have order two; order three would require $\theta=1/3$.

Solved by gpt-5.6-sol high.

= b
{parent=17a}
{scope}

= Solution
{parent=b}

The <linear stability domain> is $\{z\in\mathbb C:|R(z)|\le1\}$ for the test equation $y'=\lambda y$, $z=h\lambda$. A method is A-stable when this domain contains the entire closed left half-plane.

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= c
{parent=17a}
{scope}

= Solution
{parent=c}

For $\theta=1/2$,
$$R(z)=\frac{1+z/2}{1-z/2},$$
so $|R(z)|\le1$ exactly when $\operatorname{Re}z\le0$. Thus the first method is A-stable with stability domain the closed left half-plane. The $\theta=3/4$ method is not A-stable: along the negative real axis its quadratic numerator makes $|R(z)|$ grow without bound as $z\to-\infty$.

Solved by gpt-5.6-sol high.

= 18H
{parent=Paper 2}
{scope}
{title2=Markov Chains}

= a
{parent=18h}
{scope}

= Solution
{parent=a}

If $S_k<n\le S_{k+1}$, the construction gives $X_n=S_{k+1}-n$. This is exactly the stated maximum over the next renewal endpoint.

Solved by gpt-5.6-sol high.

= b
{parent=18h}
{scope}

= Solution
{parent=b}

From state $i>0$ the chain moves deterministically to $i-1$. From zero it jumps to $j$ when the next lifetime is $j+1$. Thus
$$p_{i,i-1}=1\ (i>0),\qquad p_{0,j}=q_{j+1}\ (0\le j<N),$$
with all other probabilities zero.

Solved by gpt-5.6-sol high.

= c
{parent=18h}
{scope}

= Solution
{parent=c}

The balance recurrence is
$$\pi_i=\pi_{i+1}+\pi_0q_{i+1}.$$
Working down from $N-1$ gives $\pi_i=\pi_0P(T_1>i)$. Since the sum of these tails is $E[T_1]$, the unique invariant law is
$$\pi_i=\frac{P(T_1>i)}{E[T_1]}=
\frac{\sum_{j=i+1}^Nq_j}{\sum_{j=1}^Njq_j}.$$

Solved by gpt-5.6-sol high.

= d
{parent=18h}
{scope}

= Solution
{parent=d}

Take $N\ge2$ and $q_2=1$. The chain alternates deterministically between zero and one, so $P(X_n=0)$ alternates between one and zero and does not converge.

Solved by gpt-5.6-sol high.

= e
{parent=18h}
{scope}

= Solution
{parent=e}

Adding $\varepsilon I$ leaves every invariant distribution unchanged. On the unique closed communicating class it also supplies a self-loop, making the chain aperiodic; transient states do not affect the <limit>. Therefore
$$P(X_n^{(\varepsilon)}=0)\longrightarrow\pi_0
=\frac1{E[T_1]}=\frac1{\sum_{j=1}^Njq_j},$$
independently of $\varepsilon\in(0,1)$.

Solved by gpt-5.6-sol high.