past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ib/paper-1.bigb

= Paper 1
{scope}

https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2025/paperib_1_2025.pdf

= 1F
{parent=Paper 1}
{scope}
{title2=Linear Algebra}

= Solution
{parent=1f}

The Leibniz definition is
$$\det A=\sum_{\pi\in S_n}\operatorname{sgn}(\pi)\prod_{i=1}^nA_{i,\pi(i)}.$$
The adjugate is the transpose of the cofactor <matrix>. Cofactor expansion gives
$$A\operatorname{adj}(A)=\operatorname{adj}(A)A=(\det A)I.$$
For the displayed tridiagonal <matrix>, expansion along the last row gives $D_n=2D_{n-1}-D_{n-2}$, with $D_1=2,D_2=3$. Induction yields
$$\det A_n=D_n=n+1.$$

Solved by gpt-5.6-sol high.

= 2F
{parent=Paper 1}
{scope}
{title2=Geometry}

= Solution
{parent=2f}

In local coordinates $u=(u^1,u^2)$ with metric $g_{ij}=\partial_i\sigma\cdot\partial_j\sigma$, the energy is
$$E(\gamma)=\frac12\int_a^b g_{ij}(u)\dot u^i\dot u^j\,dt.$$
The Euler–Lagrange equations are equivalently
$$\ddot u^k+\Gamma^k_{ij}\dot u^i\dot u^j=0,$$
where $\Gamma^k_{ij}=\tfrac12g^{k\ell}(\partial_i g_{j\ell}+\partial_jg_{i\ell}-\partial_\ell g_{ij})$.

For the plane-section curve $\eta$, constant speed gives $\eta''\perp\eta'$. Both $\eta''$ and the surface normal lie in $P$, while $P$ is spanned by $\eta'$ and that normal along the intersection. Hence $\eta''$ is normal to the surface. Its tangential <acceleration> vanishes, which is the <geodesic equation>.

Solved by gpt-5.6-sol high.

= 3
{parent=Paper 1}
{scope}
{title2=Complex Analysis OR Complex Methods}

= 3.1G
{parent=3}
{scope}

= Solution
{parent=3.1g}

Jordan's lemma states that if $f$ is holomorphic in the upper half-plane apart from finitely many poles and $|f(z)|\le M/|z|$ on sufficiently large upper semicircles, then for $a>0$ the <integral> of $e^{iaz}f(z)$ over those arcs tends to zero. Indeed, split the arc away from its endpoints, where exponential decay is uniform, and bound the two short endpoint arcs using $|e^{iaRe^{i\theta}}|=e^{-aR\sin\theta}$ and $\sin\theta\ge2\theta/\pi$ on $[0,\pi/2]$.

Apply the upper semicircle to $ze^{iz}/(1+z^2)$. The sole enclosed pole is $i$, with residue $e^{-1}/2$. Therefore the contour <integral> is $\pi i/e$; taking imaginary parts gives
$$\int_{-\infty}^{\infty}\frac{x\sin x}{1+x^2}\,dx=\frac\pi e.$$

Solved by gpt-5.6-sol high.

= 3.2A
{parent=3}
{scope}

= a
{parent=3.2a}
{scope}

= Solution
{parent=a}

If $f$ is meromorphic inside and on a positively oriented simple closed contour $C$, with no pole on $C$, then
$$\oint_Cf(z)\,dz=2\pi i\sum_{a\text{ inside }C}\operatorname{Res}(f,a).$$

Solved by gpt-5.6-sol high.

= b
{parent=3.2a}
{scope}

= Solution
{parent=b}

Integrate $e^{inz}/(z^4+1)$ over the upper semicircle. Jordan's lemma removes the arc and the upper poles are $e^{i\pi/4}$ and $e^{3i\pi/4}$. Summing their residues and taking real parts gives
$$\int_{-\infty}^{\infty}\frac{\cos(nx)}{x^4+1}\,dx
=\frac\pi{\sqrt2}e^{-n/\sqrt2}\left(\cos\frac n{\sqrt2}+\sin\frac n{\sqrt2}\right).$$

Solved by gpt-5.6-sol high.

= 4C
{parent=Paper 1}
{scope}
{title2=Variational Principles}

= Solution
{parent=4c}

For every <tangent vector> $v\perp x_0$, stationarity of $Q$ on the sphere gives $2v^TAx_0=0$. Hence $Ax_0$ is parallel to $x_0$, say $Ax_0=Ex_0$. Taking the <inner product> with $x_0$ gives $E=Q(x_0)$. The spectral theorem shows that this is the largest <eigenvalue>.

For $A=\begin{pmatrix}1&t\\t&1\end{pmatrix}$ the <eigenvalues> are $1\pm t$, so
$$E(t)=1+|t|.$$
Its graph is a V translated upward and is convex.

Solved by gpt-5.6-sol high.

= 5A
{parent=Paper 1}
{scope}
{title2=Numerical Analysis}

= a
{parent=5a}
{scope}

= Solution
{parent=a}

The <Dahlquist equivalence theorem> says that a consistent linear multistep method is convergent exactly when it is zero-stable.

Solved by gpt-5.6-sol high.

= b
{parent=5a}
{scope}

= Solution
{parent=b}

The first characteristic <polynomial> factors as
$$\rho(z)=z^3+(2\alpha-3)(z^2-z)-1
=(z-1)\{z^2+(2\alpha-2)z+1\}.$$
The root condition holds precisely for $0<\alpha<2$: in this range the two reciprocal roots are distinct and on the unit circle; at either endpoint a unit root is repeated, and outside it one root has <modulus> greater than one. Consistency is given, so these and only these values are convergent. The exceptional order-three value $\alpha=6$ is outside this interval; consequently every convergent case has order two.

Solved by gpt-5.6-sol high.

= 6H
{parent=Paper 1}
{scope}
{title2=Statistics}

= Solution
{parent=6h}

The Neyman–Pearson lemma says that among tests of size at most $\alpha$ for two simple hypotheses, a <likelihood-ratio test> rejecting where $p(x;\theta_1)/p(x;\theta_0)>k$ is most powerful, with boundary randomisation if required. If $\varphi$ is that test and $\psi$ any competing test, choose $k$ so the sizes agree. Pointwise,
$$(\varphi-\psi)(p_1-kp_0)\ge0.$$
Integration and the size inequality yield $E_1\varphi\ge E_1\psi$.

Here
$$\frac{p(x;\theta_1)}{p(x;\theta_0)}=\frac{\theta_1}{\theta_0}e^{-(\theta_1-\theta_0)|x|},$$
which decreases with $|x|$. Thus reject for $|X|\le c$, where
$$\alpha=P_{\theta_0}(|X|\le c)=1-e^{-\theta_0c},\qquad
c=-\frac1{\theta_0}\log(1-\alpha).$$

Solved by gpt-5.6-sol high.

= 7H
{parent=Paper 1}
{scope}
{title2=Optimisation}

= a
{parent=7h}
{scope}

= Solution
{parent=a}

Multiplying $Ax\le b$ by $y\ge0$ gives the upper bound $c^Tx\le y^TAx\le y^Tb$ whenever $A^Ty\ge c$. Minimising the bound yields the dual
$$\text{minimise }b^Ty\quad\text{subject to }A^Ty\ge c, y\ge0.$$

Solved by gpt-5.6-sol high.

= b
{parent=7h}
{scope}

= Solution
{parent=b}

The primal point $x=(1/2,1/2,0)$ is feasible and has value $5/2$. The dual point $y=(1/2,3/2)$ is feasible because
$$A^Ty=(2,3,13/2)^T\ge(2,3,4)^T,$$
and has value $2(1/2)+3/2=5/2$. Weak duality proves both are optimal. Thus the requested optimal primal solution is
$$x_1=x_2=\frac12,\qquad x_3=0.$$

Solved by gpt-5.6-sol high.

= 8F
{parent=Paper 1}
{scope}
{title2=Linear Algebra}

= Solution
{parent=8f}

For a <linear map> $T:V\to W$ with $V$ finite-dimensional,
$$\dim V=\operatorname{rk}T+\dim\ker T.$$
Now
$$\operatorname{rk}(\alpha\beta)=\dim\operatorname{im}\beta-dim(\ker\alpha\cap\operatorname{im}\beta)
\ge\operatorname{rk}\beta-\dim\ker\alpha,$$
which is the required inequality after rank-nullity for $\alpha$.

If $X$ and $Y$ represent the same map, then $Y=C^{-1}XB$, where $B$ changes new domain coordinates to old ones and $C$ does the same in the codomain.

Invertible block row and column operations reduce
$$\begin{pmatrix}P&Q\\R&S\end{pmatrix}$$
to $\operatorname{diag}(P,S-RP^{-1}Q)$, proving the rank formula. Apply it to $\begin{pmatrix}I_n&Q\\R&I_m\end{pmatrix}$ first with the upper-left block and then with the lower-right block. Equating the results gives
$$\operatorname{rk}(I_n-QR)=\operatorname{rk}(I_m-RQ)+n-m.$$

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= 9E
{parent=Paper 1}
{scope}
{title2=Groups, Rings and Modules}

= a
{parent=9e}
{scope}

= Solution
{parent=a}

Let $k=|G/H|$. The <coset action> gives a nontrivial homomorphism $G\to S_k$ whose kernel is normal. Simplicity makes it injective. The sign map is trivial on the nonabelian simple image, so $G$ embeds in $A_k$. For $k\le4$, $A_k$ is solvable, as are its <subgroups>, whereas a nonabelian simple <group> is not. Hence $k\ge5$.

Solved by gpt-5.6-sol high.

= b
{parent=9e}
{scope}

= Solution
{parent=b}

Let $S$ act by conjugation on the Sylow $p$-subgroups. The only fixed point is $S$: if $S$ normalizes another $T$, then $ST$ is a $p$-subgroup, forcing $S=T$. For $T=gSg^{-1}\ne S$, a stabilizer element in $S$ normalizes $T$, hence lies in $T$ by the same argument; the hypothesis then makes it trivial. Every nontrivial orbit therefore has size $|S|$, so
$$n_p\equiv1\pmod{|S|}.$$

Solved by gpt-5.6-sol high.

= c
{parent=9e}
{scope}

= i
{parent=c}
{scope}

= Solution
{parent=i}

Sylow gives $n_7\mid24$ and $n_7\equiv1\pmod7$. Simplicity excludes $n_7=1$, so $n_7=8$. Distinct order-seven <subgroups> intersect trivially, hence there are
$$8(7-1)=48$$
elements of order seven.

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= ii
{parent=c}
{scope}

= Solution
{parent=ii}

If all distinct Sylow 2-subgroups met trivially, part (b) would give $n_2\equiv1\pmod8$. But Sylow gives $n_2\mid21$ and $n_2$ odd, so $n_2\in\{3,7,21\}$ after excluding normality, none congruent to one modulo eight. Thus some distinct pair has nontrivial intersection. That intersection is a nontrivial finite 2-group, so Cauchy's theorem supplies an element of order two.

Solved by gpt-5.6-sol high.

= 10G
{parent=Paper 1}
{scope}
{title2=Analysis and Topology}

= Solution
{parent=10g}

Uniform continuity means that for every $\varepsilon>0$ there is a single $\delta>0$ such that $d(x,y)<\delta$ implies $|f(x)-f(y)|<\varepsilon$ for all $x,y$. A $d'$-Cauchy <sequence> converges uniformly pointwise to a <bounded function>; passing a uniform-continuity estimate through a sufficiently close member proves that the <limit> is uniformly continuous. Thus $C_{b,u}(X)$ is complete.

If $f\in C_0(\mathbb R^n)$, decay at infinity and compactness of a ball make it bounded. Uniform continuity on a sufficiently large compact ball, together with small values outside it, proves global uniform continuity. Hence $C_0\subset C_{b,u}$.

A <uniform limit> of <functions> vanishing at infinity also vanishes at infinity, so $C_0$ is closed. It is not compact: translate a fixed compactly supported bump of height one so that the supports are disjoint. The resulting <sequence> has pairwise sup distance one and no convergent subsequence.

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= 11F
{parent=Paper 1}
{scope}
{title2=Geometry}

= Solution
{parent=11f}

An allowable parametrisation is a smooth homeomorphism from an open subset of $\mathbb R^2$ onto an open subset of $S$, with <derivative> of rank two. Away from the axis, the rotation orbit has nonzero tangent. A transverse curve supplied by the submanifold theorem, followed by the rotation action, gives
$$\sigma(u,v)=(f(u)\cos v,f(u)\sin v,g(u)),$$
with $|v|<\pi$ and $(f',g')\ne(0,0)$.

For a ruled parametrisation, regularity is exactly
$$\psi_s\times\psi_t=(a'+tb')\times b\ne0.$$

Rotate and translate along the axis so the specified ruling is
$$L(t)=(d,st,ct),\qquad d>0.$$
Here $s\ne0$ because the line is not parallel to the axis, and $c\ne0$: if $c=0$, rotating the horizontal tangent line makes the ruled parametrisation singular at its closest point. Rotating $L$ gives
$$x^2+y^2=d^2+\frac{s^2}{c^2}z^2,$$
a <one-sheet hyperboloid>. Rotation invariance puts this whole surface in $\Sigma$. Connectedness and the fact that a complete embedded hyperboloid cannot be a proper subset of another connected embedded surface force equality. Rescaling radial and axial coordinates gives a diffeomorphism with $x^2+y^2=1+z^2$.

Solved by gpt-5.6-sol high.

= 12
{parent=Paper 1}
{scope}
{title2=Complex Analysis OR Complex Methods}

= 12.1G
{parent=12}
{scope}

= Solution
{parent=12.1g}

For any circle $|\zeta|=\rho$ with $r<\rho<R$, Cauchy's formula applied inside and outside the circle gives the locally uniformly convergent Laurent expansion
$$f(z)=\sum_{n=-\infty}^{\infty}a_nz^n,\qquad
a_n=\frac1{2\pi i}\oint_{|\zeta|=\rho}\frac{f(\zeta)}{\zeta^{n+1}}\,d\zeta.$$
At zero, the singularity is removable exactly when all $a_n$ with $n<0$ vanish; it is a pole of order $k$ when $a_{-k}\ne0$ and $a_n=0$ for $n<-k$; it is essential when infinitely many negative coefficients are nonzero.

Solved by gpt-5.6-sol high.

= i
{parent=12.1g}
{scope}

= Solution
{parent=i}

Since $\csc^2z=z^{-2}+1/3+O(z^2)$, the principal parts cancel and $f_1=-1/3+O(z^2)$. The singularity is removable.

Solved by gpt-5.6-sol high.

= ii
{parent=12.1g}
{scope}

= Solution
{parent=ii}

Termwise integration of the exponential <series> gives
$$f_2(z)=2\sum_{n=0}^{\infty}\frac{(-1)^n}{n!(2n+1)}z^{-2n}.$$
There are infinitely many negative Laurent coefficients, so zero is an <essential singularity>.

Solved by gpt-5.6-sol high.

= 12.2A
{parent=12}
{scope}

= a
{parent=12.2a}
{scope}

= Solution
{parent=a}

For distinct $z,w$ in the convex disc, integrate along the segment:
$$f(z)-f(w)=f'(z_0)(z-w)+\int_w^z(f'(\zeta)-f'(z_0))\,d\zeta.$$
The second term has <modulus> strictly below $|f'(z_0)||z-w|$, so the sum cannot vanish. Thus $f$ is one-to-one.

Solved by gpt-5.6-sol high.

= b
{parent=12.2a}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

Harmonic means twice continuously <differentiable> with $u_{xx}+u_{yy}=0$. On the simply connected plane, choose an entire <harmonic conjugate> so $F=u+iv$ is entire. If $u\ge0$, then $e^{-F}$ is bounded; Liouville's theorem makes it, and hence $u$, constant.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

Choose an entire $F$ with real part $u$. The bound implies that $e^F$ has at most <polynomial> growth, so Cauchy's estimates make it a <polynomial>. It has no zeros, hence that <polynomial> is constant. Therefore $F$, and in particular $u$, is constant.

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= 13B
{parent=Paper 1}
{scope}
{title2=Methods}

= a
{parent=13b}
{scope}

= Solution
{parent=a}

Legendre's equation is
$$-\frac d{dx}\left((1-x^2)y'\right)=\lambda y.$$
With $\langle f,g\rangle=\int_{-1}^1f\bar g\,dx$, integration by parts has no endpoint term because $1-x^2=0$ there, so the operator is self-adjoint. Sturm–Liouville <eigenvalues> are real, can be ordered increasingly, and their eigenfunctions are orthogonal and complete under standard regularity assumptions.

Substitution of $y=\sum a_nx^n$ gives
$$\frac{a_{n+2}}{a_n}=\frac{n(n+1)-\lambda}{(n+1)(n+2)}.$$
The <series> terminates at degree $\ell$ when $\lambda=\ell(\ell+1)$. Normalizing at one gives
$$P_1(x)=x,\qquad P_3(x)=\frac12(5x^3-3x).$$

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= b
{parent=13b}
{scope}

= Solution
{parent=b}

The separated axisymmetric solution is
$$\Phi(r,x)=\sum_{\ell=0}^{\infty}(A_\ell r^\ell+B_\ell r^{-\ell-1})P_\ell(x).$$
Since $x(1-x^2)=\tfrac25(P_1-P_3)$, regularity at the origin and the <boundary condition> give
$$\Phi(r,x)=\frac25\left[\frac rR P_1(x)-\left(\frac rR\right)^3P_3(x)\right].$$

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= 14C
{parent=Paper 1}
{scope}
{title2=Quantum Mechanics}

= Solution
{parent=14c}

Write $E=-\hbar^2\kappa^2/(2m)$. An even <bound state> is proportional to $\cos(kx)$ inside the well and to $e^{-\kappa|x|}$ outside, where
$$k^2=\frac{m}{a\hbar^2}-\kappa^2.$$
Continuity of the logarithmic <derivative> at $a$ gives
$$\kappa=k\tan(ka).$$
For sufficiently small $a$, $ka$ lies in the first monotone branch, so this equation has exactly one positive root and the even state is unique up to scale.

As $a\downarrow0$, $\kappa\sim k^2a\to m/\hbar^2$. Hence
$$E_0=-\frac{m}{2\hbar^2}.$$
The wells converge distributionally to $V_0(x)=-\delta(x)$, whose normalized even <bound state> is
$$\psi_0(x)=\sqrt{\frac m{\hbar^2}}e^{-m|x|/\hbar^2}.$$
Its <derivative> jump is precisely the one imposed by the <delta potential>.

Solved by gpt-5.6-sol high.

= 15B
{parent=Paper 1}
{scope}
{title2=Electromagnetism}

= Solution
{parent=15b}

Using $\rho=\varepsilon_0\nabla\cdot E$, $E=-\nabla\phi$, integration by parts, and $\phi=0$ on the boundary gives
$$U=\frac{\varepsilon_0}{2}\int_V|E|^2\,d^3x.$$
Put $C=q/(4\pi\varepsilon_0)$. <Gauss's law> gives the radial field
$$E_r=\begin{cases}0,&r<R,\\C/r^2,&R<r<2R,\\-C/r^2,&2R<r<3R,\\0,&r>3R.\end{cases}$$
Taking zero potential outside,
$$\phi=\begin{cases}C/(3R),&r<R,\\C(1/r-2/(3R)),&R<r<2R,\\C(1/(3R)-1/r),&2R<r<3R,\\0,&r>3R.\end{cases}$$
The charge formula gives
$$U=\frac12\sum_iQ_i\phi(r_i)=\frac{q^2}{12\pi\varepsilon_0R}.$$
The field formula gives the same result after integrating $4\pi r^2E_r^2$ over the two nonzero annuli.

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= 16D
{parent=Paper 1}
{scope}
{title2=Fluid Dynamics}

= Solution
{parent=16d}

For steady inviscid flow, Bernoulli's equation along a <streamline> is
$$p+\frac12\rho|u|^2+\chi=\text{constant}.$$
Writing steady Euler flow in divergence form and integrating over $V$ gives
$$\int_{\partial V}(\rho(u\cdot n)u+pn+\chi n)\,dS=0$$
by incompressibility and the divergence theorem.

Solved by gpt-5.6-sol high.

= i
{parent=16d}
{scope}

= Solution
{parent=i}

The speeds are $U=q/A$ upstream and $v=q/(2a)$ in either daughter vessel. Bernoulli therefore gives
$$p_{\rm up}-p_{\rm down}=\frac\rho2(v^2-U^2)
=\frac{\rho q^2}{2}\left(\frac1{4a^2}-\frac1{A^2}\right).$$

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= ii
{parent=16d}
{scope}

= Solution
{parent=ii}

Take the downstream tissue <pressure> as gauge zero. The two transverse <momentum> fluxes cancel. The force of the fluid on the junction is axial and equals
$$F=\left[A(p_{\rm up}-p_{\rm down})+\rho q(U-v\cos\alpha)\right]e_x,$$
where $U=q/A$, $v=q/(2a)$ and the <pressure> difference is that in part (i).

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= 17A
{parent=Paper 1}
{scope}
{title2=Numerical Analysis}

= a
{parent=17a}
{scope}

= Solution
{parent=a}

For a unit <vector> $v$, a <Householder reflection> is $H=I-2vv^T$. Clearly $H^T=H$, and $H^TH=(I-2vv^T)^2=I$.

Solved by gpt-5.6-sol high.

= b
{parent=17a}
{scope}

= Solution
{parent=b}

The <vector> $v$ has <eigenvalue> $-1$, while every <vector> in $v^\perp$ has <eigenvalue> $1$. Their multiplicities are one and $n-1$, respectively.

Solved by gpt-5.6-sol high.

= c
{parent=17a}
{scope}

= Solution
{parent=c}

At step $k$, choose a Householder reflection acting only on coordinates $k,\ldots,n$ that maps the trailing part of column $k$ to a multiple of the first coordinate in that block. It zeros every entry below the diagonal without changing earlier columns. Induction produces $H_n\cdots H_1A=R$ upper triangular.

Solved by gpt-5.6-sol high.

= d
{parent=17a}
{scope}

= Solution
{parent=d}

For symmetric $A$, apply at stage $k$ the same reflection on both sides, choosing it to zero entries below the first subdiagonal in column $k$. <Orthogonal similarity> preserves symmetry, so the corresponding row entries vanish too and previous zeros remain. After finitely many stages $Q A Q^T$ is symmetric tridiagonal. Constructing each reflector uses only arithmetic and one square root.

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= 18H
{parent=Paper 1}
{scope}
{title2=Statistics}

= a
{parent=18h}
{scope}

= Solution
{parent=a}

Let $S_{xx}=\sum X_i^2$. The likelihood equations give
$$\hat\beta=\frac{\sum X_iY_i}{S_{xx}},\qquad
\hat\sigma^2=\frac1n\sum(Y_i-X_i\hat\beta)^2.$$
These are orthogonal projections of a Gaussian <vector> onto the span of $X$ and its orthogonal complement, hence are independent.

Solved by gpt-5.6-sol high.

= b
{parent=18h}
{scope}

= Solution
{parent=b}

With $s^2=\sum(Y_i-X_i\hat\beta)^2/(n-1)$,
$$\frac{\hat\beta-\beta}{s/\sqrt{S_{xx}}}\sim t_{n-1}.$$
Thus the interval is
$$\hat\beta\ \pm\ t_{n-1,1-\alpha/2}\frac{s}{\sqrt{S_{xx}}}.$$

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= c
{parent=18h}
{scope}

= Solution
{parent=c}

Unbiasedness requires $\sum c_iX_i=1$. Cauchy–Schwarz gives $1\le(\sum c_i^2)S_{xx}$, so
$$\operatorname{Var}(\tilde\beta)=\sigma^2\sum c_i^2\ge\frac{\sigma^2}{S_{xx}}.$$
Equality holds exactly for $c_i=X_i/S_{xx}$.

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= d
{parent=18h}
{scope}

= Solution
{parent=d}

The reverse-regression estimate is $\hat b=\sum X_iY_i/\sum Y_i^2$. Therefore
$$\hat b\hat\beta=\frac{(\sum X_iY_i)^2}{(\sum X_i^2)(\sum Y_i^2)}\le1$$
by Cauchy–Schwarz. Equality means the two data <vectors> are proportional, which is exactly when both fitted residual sums, and hence both variance MLEs, vanish.

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= 19H
{parent=Paper 1}
{scope}
{title2=Markov Chains}

= a
{parent=19h}
{scope}

= Solution
{parent=a}

A chain is reversible with respect to $\pi$ when $\pi_iP_{ij}=\pi_jP_{ji}$ for all states. For random walk on a finite connected undirected graph, $\pi_i=\deg(i)/(2|E|)$ satisfies this because both sides equal $1/(2|E|)$ on an edge and zero otherwise.

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= b
{parent=19h}
{scope}

= Solution
{parent=b}

The graph has six edges, and $A$ has degree one. Hence $\pi_A=1/12$. Kac's return-time formula gives
$$\mathbb E_A T_A^+=\frac1{\pi_A}=12.$$

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= c
{parent=19h}
{scope}

= Solution
{parent=c}

After the forced first step $A\to B$, let $h_i=P_i(T_A<T_F)$. Symmetry gives $h_C=h_D=x$, while
$$h_B=(1+2x)/3,\quad x=(h_B+h_E)/2,\quad h_E=2x/3.$$
Solving gives $h_B=2/3$. Thus the requested probability is $2/3$.

Solved by gpt-5.6-sol high.

= d
{parent=19h}
{scope}

= Solution
{parent=d}

Let $g_i=E_i[T_A\mathbf1_{\{T_A<T_F\}}]$ for the chain absorbed at $A$ or $F$. With $g_C=g_D=y$, <First-step analysis> gives
$$g_B=(2+2y)/3,\qquad g_E=(1+2y)/3,\qquad y=(1+g_B+g_E)/2.$$
Thus $y=3$ and $g_B=8/3$. Including the initial step from $A$, the conditional mean is
$$1+\frac{g_B}{h_B}=1+\frac{8/3}{2/3}=5.$$

Solved by gpt-5.6-sol high.