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A prism polarizer associated with William Nicol. Its birefringent crystal arrangement selects one transmitted polarization; two such prisms can serve as the polarizer and analyser in a polariscope.

For $N\sim \mathrm{Pois}(\lambda )$ with $\lambda \ge 0$, insert its probability mass function into the expected value:The exponential series shows finiteness for every real $t$. Differentiating this moment-generating function gives mean and variance both $\lambda$. Taking the logarithm gives the cumulant-generating function $\lambda (e^t-1)$, so every positive-order cumulant is also $\lambda$.

When a Poisson distribution is observed in cells $0,1,\dots ,r-1,\ge r$, the last cell probability is a tail sum, not the probability of exactly $r$. With $p_i(\theta )=e^{-\theta }{\theta }^i/i!$ and $Q_r(\theta )=1-{\sum }_{i<r}{p}_i(\theta )$, telescoping derivatives give $Q_r^{\prime }=p_{r-1}$. Thus an interior maximum-likelihood estimate solves ${\sum }_{i<r}{n}_i(i/\theta -1)+n_r{p}_{r-1}(\theta )/Q_r(\theta )=0$. Substitution of the estimate into these cell probabilities supplies fitted expected counts for a likelihood-ratio test. With $r+1$ cells and one regular fitted parameter, the asymptotic chi-squared distribution has $r-1$ degrees of freedom. Replacing every grouped tail observation by $r$ changes the data model and generally does not give the grouped maximum likelihood estimate.

For planar Brownian motion started at $1$, let ${\Theta }_t$ be its continuous unwrapped angle, initialized at zero. Then $2{\Theta }_t/\log t$ converges in distribution to the standard symmetric Cauchy distribution, with density $1/[\pi (1+x^2)]$. The angle records all completed turns rather than a bounded principal argument. The assertion is a convergence in distribution at deterministic times; it does not assert convergence of sample paths. A precise statement appears as Theorem 7.31 in Mörters and Peres, Brownian Motion.

For a start $Z_0=r_0{e}^{i{\alpha }_0}$ in the punctured unit disc, the logarithmic lift $-i\log Z$ starts at ${\alpha }_0-i\log r_0$ in the upper half-plane. Local conformal invariance of planar Brownian motion changes its clock by $\int |Z_s{|}^{-2}ds$. That clock is finite at the disc exit because the stopped path is compact and avoids zero. Its terminal real coordinate is therefore the usual upper-half-plane Cauchy exit law from a Brownian half-plane. Subtracting ${\alpha }_0$ gives the displayed winding law. Its first absolute moment diverges logarithmically, even though the final point modulo angle lies on a compact circle.

For real constants $a,c$, apply Picard iteration to the initial value problem $y^{\prime }=ay/(1-x)$, $y(0)=c$, starting with $y_0(x)=c$. The recurrence isSince ${\int }_0^x{\log }^k(1-t)/(1-t)dt=-{\log }^{k+1}(1-x)/(k+1)$, mathematical induction gives the displayed partial sums. The exponential series converges at every finite argument, so $y_n(x)\to c(1-x{)}^{-a}$ for every real $x<1$, uniformly on each compact interval in that range. This is not a Taylor series in $x$, so its convergence is not restricted to $|x|<1$. The differential equation's coefficient is singular at $x=1$ even when a special solution has a removable singularity there.

A Rankine vortex models a finite rotating core joined continuously to an irrotational exterior. In cylindrical polar coordinates, its radial and vertical velocities vanish and its azimuthal velocity has the displayed form, where $a$ is the core radius and $\Omega$ is a signed angular velocity. The axial vorticity is $r^{-1}{\partial }_r(r{u}_{\theta })$, equal to $2\Omega$ inside and zero outside. The velocity is continuous at $r=a$, although its first radial derivative has different one-sided values. The circulation outside the core is $2\pi \Omega a^2$.

For every $m\ge 5$, let $A=A_m$ act by rooted permutations on the rooted tree of finite words over $m$ letters. Define $b(a)$ to have local permutation $a\in A$ at the pairwise incomparable vertices $1^{r}2$ for $r\ge 0$, and identity elsewhere. Then $b$ is an embedding of $A$ and satisfies $b(a)=(b(a),a,1,\dots ,1)$ with trivial root permutation. The group $H=⟨A,b(A)⟩$ is finitely generated and perfect because it is generated by two finite perfect subgroups.

The root permutation and sections define an injective homomorphism $H\to H\mathrm{wr}{A}_m$. To prove surjectivity, conjugate $b(c)$ by the rooted cycle $(234)$ and take its commutator with $b(a)$. In wreath coordinates this leaves only $b([a,c])$ in the first coordinate. Perfectness of $A_m$ therefore supplies the entire localized copy of $b(A)$ there. Canceling it from the recursion for $b(a)$ supplies a localized rooted $a$ in the second coordinate. Transitivity moves both kinds of generators to every coordinate, yielding the full base $H^m$ and hence the full wreath product. The group is infinite, since a nontrivial finite order $h$ could not satisfy $h=h^m|A_m|$.

For any perfect group $H$ and $m\ge 3$, the permutation wreath product $H\mathrm{wr}{A}_m$ is generated by elements of order three. The top $A_m$ is generated by 3-cycles. With $\sigma =(123)$, the product of the conjugate $h_1\sigma h_1^{-1}$ and ${\sigma }^{-1}$ is $h_1{h}_2^{-1}$; using ${\sigma }^{-1}$ gives $h_1{h}_3^{-1}$. The commutator of $h_1{h}_2^{-1}$ and $k_1{k}_3^{-1}$ is $[h,k{]}_1$, since distinct coordinate groups commute. Perfectness gives the entire first coordinate group, and transitivity gives all coordinates. No finite-generation assumption on $H$ is needed.

For a continuous map $f:X\to Y$ between Tychonoff spaces, extend it to $F:\beta X\to \beta Y$ by the Stone-Čech extension property. If $f$ is perfect, compactness of a fiber and closedness of $f$ show that no remainder point can map to a point of $Y$: separate such a point from the fiber by a neighborhood in $\beta X$, and use closedness to obtain a target neighborhood whose preimage stays inside it, contradicting density. Conversely, if $F^{-1}(Y)=X$, each fiber is compact, and for closed $A\subseteq X$ the equality $f(A)=F({\overline{A}}^{\beta X})\cap Y$ proves closedness. Equivalently $F(\beta X\setminus X)\subseteq \beta Y\setminus Y$.

Source paper: people.maths.ox.ac.uk/suli/MMSC_2020.pdf

This is the B1 paper for the MSc in mathematical modelling and scientific computing course of the University of Oxford. Questions 1–2 have original solutions below. The official revision page identifies public model solutions to questions 3–4, so those questions are skipped. The programme also advertises a student-only past-papers-and-solutions folder; its exact coverage of this paper has not been established.