Past exam of the mathematics course of the University of Oxford / 2023 / Prelims / Mathematics II (A10139W1)

Source: released examination PDF, pages 185–190.

Solutions to all seven questions, including every optional question, of the examination sat on 20 June 2023.

For a nonempty set $A\subseteq \mathbb{R}$, a supremum is a real number $s$ such that $a\le s$ for every $a\in A$, and every upper bound $u$ of $A$ satisfies $s\le u$. Equivalently, $s$ is an upper bound and, for every $\epsilon >0$, some $a\in A$ satisfies $a>s-\epsilon$. By completeness of the real numbers, it exists whenever $A$ is nonempty and bounded above.

An increasing monotone sequence satisfies $a_{n+1}\ge a_n$ for every $n$; a bounded sequence satisfies $|a_n|\le M$ for some finite $M$ and every $n$. Put $s=\sup \{a_n:n\ge 1\}$. Given $\epsilon >0$, the defining property of the supremum supplies $N$ with $a_N>s-\epsilon$. For every $n\ge N$, monotonicity givesThis is exactly the definition of convergent sequence, so the bounded monotone sequence theorem gives $\boxed{a_n\to s}$.

Choose $M$ with $|b_n|\le M$. Every tail of a sequence is nonempty and bounded, so its supremum $c_n$ exists and lies in $[-M,M]$. Removing one term cannot increase a supremum, hence $c_{n+1}\le c_n$. The bounded monotone sequence theorem, applied to $-c_n$, provesThus $L$ is the limit superior of $(b_n)$.

For $\epsilon >0$, eventually $c_n<L+\epsilon$, and then $b_n\le c_n<L+\epsilon$. Only finitely many indices can satisfy $b_n\ge L+\epsilon$. If only finitely many satisfied $b_n>L-\epsilon$, some entire tail of a sequence would have $b_n\le L-\epsilon$. Its supremum would also be at most $L-\epsilon$, contradicting $c_n\ge L$.

Inductively choose $n_k>n_{k-1}$ so late that $c_{n_k}<L+1/k$ and $b_{n_k}>L-1/k$. There are infinitely many choices for the latter inequality, so both requirements can be met. Then $|b_{n_k}-L|<1/k$, proving $\boxed{b_{n_k}\to L}$. This constructs a subsequence attaining the limit superior.

At even integers $n=2k$, the terms are $1+1/(2k)\to 1$. At odd integers they are $-1-1/(2k-1)\to -1$. The largest eventual values therefore come from the subsequence at even integers, andExplicitly, every sufficiently late term is below $1+\epsilon$, while infinitely many even integers give terms greater than $1-\epsilon$.

The sine has values $1,0,-1,0$ on the four residue classes of $n$ modulo four. For $n=4k+1$, the exponential function gives $b_n=e^{1-1/n}\to e$; the other residue classes tend to $1,e^{-1},1$. All terms satisfy $b_n\le e$, and the first class approaches $e$. Hence

A series ${\sum }_{n=1}^{\infty }{a}_n$ converges when its partial sums $s_N={\sum }_{n=1}^N{a}_n$ have a finite limit of a sequence.

The comparison test says that if $0\le a_n\le b_n$ eventually and $\sum b_n$ converges, then $\sum a_n$ converges. Indeed, the partial sums of the nonnegative terms $a_n$ are increasing, and are bounded above by a fixed initial contribution plus the sum of the comparison series. The bounded monotone sequence theorem proves their convergence. Finite initial terms do not affect convergence. Applying this result to $|a_n|$ also proves absolute convergence whenever $|a_n|\le b_n$ eventually with $\sum b_n<\infty$.

The implication is true, with absolute convergence. Since $n^{3/2}{a}_n\to 0$, there is $N$ such that $|a_n|\le n^{-3/2}$ for $n\ge N$. The p-series $\sum n^{-3/2}$ converges because its exponent exceeds one. The comparison test proves $\sum |a_n|<\infty$, and absolute convergence implies convergence of $\sum a_n$.

The implication is false. SetThen $n{a}_n=1/\log (n+1)\to 0$. For $n\ge 2$, $\log (n+1)\le \log (2n)\le 2\log n$, so $a_n\ge 1/(2n\log n)$. The integral test givesThus the comparison test shows that $\sum a_n$ diverges.

The binomial theorem gives $\left(\genfrac{}{}{0ex}{}{2n}{n}\right)\le {\sum }_{k=0}^{2n}\left(\genfrac{}{}{0ex}{}{2n}{k}\right)=4^n$. ThereforeThe geometric series with ratio $4/9$ converges. By the comparison test, the given series is absolutely convergent.

For $t\ge 0$, $|\sin t|\le t$, soThe integral test applies to the decreasing positive comparison function, andThe comparison test proves that the series converges; since all terms are nonnegative, this is also absolute convergence.

Let $u(x)=(\log x)x^{-2/3}$. Its derivative iswhich is negative for $x>e^{3/2}$. Also $u(x)\to 0$, since a positive power eventually dominates the natural logarithm. Thus the alternating series test proves convergence of $\sum (-1{)}^{n}u(n)$.

For $n\ge 3$, $u(n)\ge n^{-2/3}$, whose p-series diverges. The comparison test therefore rules out absolute convergence. The answer is conditional convergence.

For a power series $\sum a_n{z}^n$, its radius of convergence $R\in [0,\infty ]$ separates absolute convergence for $|z|<R$ from divergence for $|z|>R$. One may define $R=\sup \{r\ge 0:\sum |a_n|r^n<\infty \}$, allowing $R=\infty$.

To justify convergence inside the radius, choose $|z|<r<R$ with $\sum |a_n|r^n<\infty$. In particular $|a_n|r^n\le M$ for some $M$, soThe comparison test with a geometric series proves absolute convergence. Conversely, convergence at a point of modulus $\rho$ makes $a_n{z}^n$ bounded there and gives absolute convergence at all smaller moduli by the same estimate. Consequently a convergent point of modulus greater than $R$ would contradict the definition of $R$.

For $0<R<\infty$, the real domain of convergence is one of $(-R,R)$, $[-R,R)$, $(-R,R]$, or $[-R,R]$. At $R=1$ these are realized respectively byThe endpoint assertions follow from the harmonic series, alternating series test, and p-series. Replacing $z$ by $z/R$ realizes every finite positive radius. The degenerate domain $\{0\}$ occurs for $\sum n!z^n$, while the entire real line occurs for $\sum z^n/n!$. Their radii are zero and infinity by the ratio test.

The Cauchy-Hadamard theorem states $R^{-1}=\mathrm{limsup}|a_n{|}^{1/n}$. At the prime numbers $n=p$, this root equals $p^{2/p}\to 1$. There are infinitely many prime numbers, so these indices form an unbounded subsequence. For every other positive index the root is zero. The limit superior is therefore one, and $\boxed{R=1}$.

By Stirling's approximation,Here $(2\pi n{)}^{1/(2n)}\to 1$ and $n^{1/n^2}\to 1$, as is seen by taking the natural logarithm. The Cauchy-Hadamard theorem gives $\boxed{R=e^{-1}}$.

At the nonzero indices $n=m!$ with $m\ge 2$,since $m/(m!)=1/(m-1)!\to 0$. At all other sufficiently large indices the coefficient is zero. The initial coefficients do not affect the limit superior, which equals one. The Cauchy-Hadamard theorem therefore gives $\boxed{R=1}$.

A limit point $p$ of $E\subseteq \mathbb{R}$ is a point such that, for every $\epsilon >0$, the punctured neighbourhood $(p-\epsilon ,p+\epsilon )\setminus \{p\}$ meets $E$. A closed set contains all of its limit points.

Every nonempty open interval contains a rational number; applying this to $(p,p+\epsilon )$ shows that every $p\in \mathbb{R}$ is a limit point of $\mathbb{Q}$. Thus the limit points of $\mathbb{Q}$ form $\mathbb{R}$, and $\mathbb{Q}$ is not closed.

If $p\in \mathbb{Z}$, radius $1/2$ excludes all other integers. If $p\notin \mathbb{Z}$, its distances to the two neighbouring integers have a positive minimum, and a smaller radius excludes every integer. Thus $\mathbb{Z}$ has no limit points in $\mathbb{R}$ and is closed.

Suppose $E$ is a closed set and $p\notin E$. Then $p$ is not a limit point of $E$, so some punctured neighbourhood of $p$ misses $E$. As $p\notin E$, the entire neighbourhood misses $E$. Thus $\mathbb{R}\setminus E$ is an open set.

Conversely, suppose every point $p\notin E$ has a neighbourhood disjoint from $E$. No such point can be a limit point of $E$, so every limit point lies in $E$. This proves the equivalence between containing all limit points and having an complement that is an open set.

A point is in the closure $\overline{E}$ exactly when every neighbourhood meets $E$. This is equivalent to belonging to $E$ or being a limit point of $E$.

If $p\notin \overline{E}$, choose $\epsilon >0$ with $(p-\epsilon ,p+\epsilon )\cap E=\varnothing$. Every $q$ with $|q-p|<\epsilon /2$ has a smaller neighbourhood contained in this interval and disjoint from $E$. Thus $(p-\epsilon /2,p+\epsilon /2)$ misses $\overline{E}$, and the complement of $\overline{E}$ is open. Hence $\overline{E}$ is closed.

Every closed set $F$ containing $E$ contains all limit points of $E$: a limit point outside $F$ would have a neighbourhood missing $F$, and hence missing $E$. Therefore $\overline{E}\subseteq F$. Since $\overline{E}$ is itself one of these closed sets,

The limit of a function $f:E\to \mathbb{R}$ at a limit point $p$ of $E$ is $\ell$ whenThe function converges as $x\to p$ if such a finite $\ell$ exists. The puncture excludes any value assigned at $p$.

For $p\in E$, continuity at $p$ meansWhen $p$ is also a limit point, this says that the limit of a function is $f(p)$. At an isolated point of $E$, every function is continuous relative to $E$, even though the punctured limit does not have a unique forced value.

Take $y=f(p)$ with $p\in \overline{E}$. Given $\epsilon >0$, continuity gives $\delta >0$ such that $|x-p|<\delta$ implies $|f(x)-f(p)|<\epsilon$. Since $p\in \overline{E}$, choose $x\in E$ with $|x-p|<\delta$. Thus every neighbourhood of $y$ meets $f(E)$, provingEquality need not hold. Let $f(x)=e^x$ and $E=\mathbb{R}$. The exponential function is continuous, butThe additional point zero comes from arguments escaping to $-\infty$, not from a finite point of $\overline{E}$.

The sequence of functions $(f_n)$ converges uniformly to $f$ on $E$ ifThe same $N$ must work for every $x$. Equivalently, the supremum norm of $f_n-f$ tends to zero. This is stronger than pointwise convergence, where $N$ may depend on $x$.

The Uniform Cauchy criterion says that $(f_n:E\to \mathbb{R})$ converges uniformly on $E$ if and only ifFor necessity, use the triangle inequality through the uniform limit. For sufficiency, the condition makes $(f_n(x){)}_n$ a Cauchy sequence for each $x$, so completeness of the real numbers defines a pointwise limit $f(x)$. Apply the condition with $\epsilon /2$ and let $m\to \infty$ to obtain $|f_n(x)-f(x)|\le \epsilon /2<\epsilon$ uniformly in $x$.

Use uniform convergence to choose $N$ with $|f_n(x)-f_N(x)|\le 1$ for all $n\ge N$ and $x\in E$; this follows, for example, from the Uniform Cauchy criterion. Since each $f_j$ is a bounded function, put $M_j={\sup }_E|f_j|<\infty$ for $1\le j\le N$. Thenfor every $n$ and $x$. Thus $(f_n)$ is a uniformly bounded family of functions. The finite initial segment is essential: uniform convergence directly controls only a tail of a sequence.

Write $f_n\to f$ and $g_n\to g$ with uniform convergence. The triangle inequality givesHence $f_n+g_n\to f+g$ with uniform convergence.

If every $f_n,g_n$ is a bounded function, part iii gives constants $M_f,M_g$ bounding the two families. Their pointwise limits satisfy $|f|\le M_f$ and $|g|\le M_g$. ConsequentlyThus both sums and products have uniform convergence under the stated hypotheses. The product estimate is the uniform convergence product rule for bounded functions.

On $E=\mathbb{R}$, take $f_n(x)=x$ and $g_n(x)=1/n$. Then $f_n\to f(x)=x$ with uniform convergence, because there is no error, and $g_n\to 0$ with uniform convergence, with error $1/n$. Their products $f_n{g}_n=x/n$ converge pointwise to zero, but for every $n$ the choice $x=n$ gives $|f_n(x)g_n(x)|=1$. Thus the products do not converge uniformly. Here the first family consists of unbounded functions, exactly the missing hypothesis in the preceding estimate.

The mean value theorem states that if $f:[a,b]\to \mathbb{R}$ is continuous on $[a,b]$ and differentiable on $(a,b)$, where $a<b$, there is $c\in (a,b)$ such thatNo continuity assumption on the derivative is needed.

The constancy theorem states that a differentiable function on an interval whose derivative vanishes everywhere is constant. At any included endpoints, continuity there is understood.

Indeed, take arbitrary $x<y$ in the interval. The mean value theorem applies on $[x,y]$ and gives $f(y)-f(x)=f^{\prime }(c)(y-x)=0$. Every pair of values is equal, proving the assertion. The interval hypothesis ensures that the whole segment $[x,y]$ lies in the domain.

The derivative of $h(x)=f(x)-\sin x$ vanishes. By the constancy theorem on the interval $I$,This need not hold on a disconnected domain. On $I=(-\infty ,0)\cup (0,\infty )$, define $f(x)=\sin x$ for $x<0$ and $f(x)=\sin x+1$ for $x>0$. The function is differentiable throughout $I$ with $f^{\prime }(x)=\cos x$, but no single $C$ works on both components.

Suppose distinct fixed points $x<y$ existed. Then $h(t)=f(t)-t$ satisfies $h(x)=h(y)=0$. Rolle's theorem supplies $c\in (x,y)$ with $h^{\prime }(c)=0$, hence $f^{\prime }(c)=1$, contrary to the hypothesis. Thus there is at most one fixed point. The argument does not require $f^{\prime }$ to be continuous.

For every real $t$, $f(t)-t=1/(1+e^t)>0$, so there is no fixed point. Direct differentiation givesWriting $u=e^t>0$, the inequality $(1+u{)}^2\ge 4u$ showsIn particular $0<f^{\prime }(t)<1$, but $f^{\prime }(t)\to 1$ as $t\to \pm \infty$. There is no common contraction constant smaller than one. Strict pointwise bounds on the derivative therefore do not suffice for the contraction mapping theorem.

By the mean value theorem, $|f(x)-f(y)|\le A|x-y|$ for all real $x,y$. Thus $f$ is a contraction mapping on the complete metric space $\mathbb{R}$. Starting from arbitrary $x_1$, define $x_{n+1}=f(x_n)$. Induction gives $|x_{n+1}-x_n|\le A^{n-1}|x_2-x_1|$. For $m>n$, the triangle inequality and a geometric series yieldHence $(x_n)$ is a Cauchy sequence and completeness of the real numbers supplies a limit $x$. The Lipschitz continuity of $f$ implies $f(x)=\lim f(x_n)=\lim x_{n+1}=x$. Thus the limit is a fixed point, independently of the initial value. If $y$ were another, $|x-y|\le A|x-y|$ would force $x=y$. This proves the contraction mapping theorem in this setting, with error estimate

Assume, as required for the displayed integrals, that the $f_n$ are Riemann integrable on $[a,b]$. Their uniform limit $f$ is also Riemann integrable: if $\sup |f-f_N|<\epsilon$, the upper and lower Darboux sums for a partition of an interval differ for $f$ by at most the corresponding difference for $f_N$ plus $2\epsilon (b-a)$. Choosing $N$ and then a sufficiently fine partition of an interval proves the Riemann integrability criterion for $f$.

The integral estimate then givesThis proves that uniform convergence preserves integrals on a compact interval. Some integrability hypothesis is necessary: uniform convergence alone cannot make the integral of an arbitrary starting function exist.

Let $f_n$ be the continuous triangular bumpand set $f_1=0$. At $x=0$ every term is zero, and each fixed $x>0$ eventually lies outside the support, so $f_n\to 0$ pointwise. The triangle has base $2/n$ and height $n$, henceThis concentrating triangular bump shows that pointwise convergence does not preserve integrals, even for continuous functions on a compact interval.

Take $f_n(x)=x^n$ on $[0,1]$. Each function is continuous, and the pointwise limit is $f(x)=0$ for $0\le x<1$, while $f(1)=1$. NeverthelessThe limiting function is Riemann integrable, since changing a value at one point does not change its Riemann integral. There is no uniform convergence: for every $n$, ${\sup }_{[0,1]}|f_n-f|=1$, with the supremum approached as $x\uparrow 1$ from below. Equivalently, the uniform limit theorem would require the limit to be continuous.

No: the unbounded length of the domain matters. Let $\varphi (t)=\max \{1-|t|,0\}$ and define $f_n(x)=n^{-1}\varphi (x/n)$ on $\mathbb{R}$. These are continuous functions with compact support, and ${\sup }_{\mathbb{R}}|f_n|=1/n\to 0$, so the sequence has uniform convergence to zero. However, the change of variables $t=x/n$ givesThe integral of the limit is zero. The supports widen as the heights decrease; a uniform bound on the heights gives no bound on the total area over an infinite interval.

The functions $g_n$ are nonnegative. The change of variables $u=nx$ and the given Gaussian integral implyFor every fixed $\delta >0$,Thus the mass concentrates at the endpoint zero. This is a one-sided approximate identity.

Because $f$ is continuous on the compact interval $[0,1]$, it is a bounded function; put $B={\sup }_{[0,1]}|f|$. Given $\epsilon >0$, choose $\delta \in (0,1)$ with $|f(x)-f(0)|<\epsilon$ for $0\le x<\delta$. Splitting the integral near and away from zero yieldsTaking the limit superior gives a bound of $\epsilon$, and $\epsilon$ was arbitrary. ConsequentlyOnly continuity at zero and boundedness on $[0,1]$ were needed for this final estimate; the assumed continuity everywhere also guarantees the ordinary integrals exist.