Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics II (A10139W1)

Source: released examination PDF, bundle pages 58–62.

The infimum of $A$ is its greatest lower bound: $\alpha =\inf A$ means that $\alpha \le x$ for every $x\in A$, and that every lower bound $l$ of $A$ satisfies $l\le \alpha$. The supremum is its least upper bound: $\beta =\sup A$ means that $x\le \beta$ for every $x\in A$, and every upper bound $u$ satisfies $\beta \le u$. The completeness of the real numbers ensures that both exist for a nonempty bounded set of real numbers.

Put $s=\sup (-A)$. For every $x\in A$, $-x\le s$, hence $-s\le x$, so $-s$ is a lower bound of $A$. If $l$ is any lower bound of $A$, then $-x\le -l$ for every $x\in A$, making $-l$ an upper bound of $-A$. The definition of supremum gives $s\le -l$, or $l\le -s$. Thus $-s$ is the greatest lower bound, and

Let $M=\max \{\sup A,\sup B\}$. Every element of the union $A\cup B$ is at most $M$, so $M$ is an upper bound. Conversely, every upper bound of $A\cup B$ is an upper bound of each of $A$ and $B$, hence is at least both suprema and therefore at least $M$. Consequently

True. Write $s=\sup A$ and $u=\sup \{\sup R_m:m\ge 1\}$. All these suprema exist in $[0,1]$. Since $R_m\subseteq A$, we have $\sup R_m\le s$ for every $m$, hence $u\le s$. Conversely, every $x_{m,n}\le \sup R_m\le u$, so $u$ is an upper bound of $A$ and $s\le u$. Thus the iterated extrema of an array satisfy

True. Set $l=\inf A$ and $v=\inf \{\inf R_m:m\ge 1\}$. Because $R_m\subseteq A$, every row infimum satisfies $\inf R_m\ge l$, so $v\ge l$. But $x_{m,n}\ge \inf R_m\ge v$ for all $m,n$, making $v$ a lower bound of $A$. Therefore $v\le l$, and

True. For every $m,n$,Fix $m$. The number $\sup R_m$ is an upper bound for every column infimum, so ${\sup }_n\inf C_n\le \sup R_m$. This holds for every $m$, making ${\sup }_n\inf C_n$ a lower bound of the row suprema. Taking their infimum gives the weak minimax inequality

False. Take $x_{1,n}=0$ for every $n$, and $x_{m,n}=1$ for every $m\ge 2$ and every $n$. Each row is constant, so its infimum and supremum coincide. Nevertheless,The two outer extrema can select different rows.

A sequence $(a_n)$ has limit of a sequence $L$ ifFor the reciprocal rule, the required condition is $\boxed{L\ne 0}$. Indeed, eventually $|a_n-L|<|L|/2$, so $|a_n|\ge |L|/2>0$. The reciprocals are then defined on this tail, andA finite number of zero initial terms does not affect this tail limit of a sequence; if a reciprocal sequence is required at every index, those terms must be excluded or assigned arbitrary values separately.

Rewrite the recurrence relation asSince $a_0=1>0$ and $c>0$, a positive $a_n$ gives a positive $a_{n+1}$. Mathematical induction therefore proves $\boxed{a_n>0\text{ for all }n\ge 0}$, and also shows that every division in the recurrence relation is defined.

For every $n\ge 0$, direct algebra givesThus $\boxed{a_n^2\ge c\text{ for every }n\ge 1}$. This identity is the key estimate in Newton iteration for a square root.

For $n\ge 1$, the preceding parts give a positive denominator and a nonpositive numerator inHence $\boxed{a_{n+1}\le a_n\text{ for every }n\ge 1}$. The restriction $n\ge 1$ matters: if $c>1$, then $a_1=(1+c)/2>a_0$.

The tail $(a_n{)}_{n\ge 1}$ is a monotone sequence bounded below by zero. The bounded monotone sequence theorem therefore gives $a_n\to L\ge 0$. Taking the limit of a sequence in $a_n^2\ge c$ yields $L^2\ge c>0$, so $L>0$. The reciprocal rule from part (a) permits passage to the limit of a sequence in the recurrence relation:Multiplication by $2L$ givesThis proves existence of the positive square root without assuming that existence when constructing the sequence.

Only (ii) is necessarily true. Every real $a_n$ satisfies $a_n^2\ge 0>c$, so (ii) holds automatically. For a single counterexample to (i) and (iii), choose $c=-3$ and $a_0=1$. The recurrence relation then gives the nonzero periodic sequenceHere $a_1<0$, disproving (i), and $a_2>a_1$, disproving (iii). This counterexample satisfies the requirement that no iterate is zero.

Write $d=-c>0$. Suppose, for a contradiction, that only finitely many sign changes occur. Since every $a_n\ne 0$, the sign is then constant for $n\ge N$. Set $b_n=|a_n|>0$ on that tail. Multiplying the recurrence relation by this constant sign givesConsequently $b_{N+j}<2^{-j}{b}_N$ for $j\ge 1$, and $b_n\to 0$ by comparison with a geometric sequence. But positivity of $b_{n+1}$ requires $b_n^2>d$ at every step, which is incompatible with $b_n\to 0$. Thus the negative-parameter square-root iteration changes sign infinitely often.

A real or complex sequence $(a_n)$ is a Cauchy sequence ifFor $z_n=x_n+i{y}_n$, the complex modulus givesThus the same $N$ works for both coordinate Cauchy sequences. Conversely, if both real coordinate Cauchy sequences converge to $x$ and $y$, then $|z_n-(x+iy)|\le |x_n-x|+|y_n-y|\to 0$. This proves the needed completeness for complex numbers from the assumed completeness of the real numbers.

Now suppose ${\sum }_{n\ge 0}|z_n|$ converges. Its real partial sums form a Cauchy sequence. Given $\epsilon >0$, sufficiently late tails therefore satisfy ${\sum }_{k=n+1}^m|z_k|<\epsilon$. For $s_m={\sum }_{k=0}^m{z}_k$, the triangle inequality yieldsThe complex partial sums are a Cauchy sequence, hence converge by the coordinate argument. Therefore absolute convergence implies convergence for a complex series.

Define the radius of convergence of the power series bywith $R=\infty$ when this set is unbounded. It is nonempty because the power series converges at zero.

If $|z|<R$, the definition of supremum supplies a point $w$ of convergence with $|z|<|w|$. The terms $a_n{w}^n$ of a convergent series tend to zero and are bounded, say $|a_n{w}^n|\le M$. ThereforeThe right-hand side has a convergent geometric series, since $|z|/|w|<1$. The comparison test proves absolute convergence at $z$.

If $|z|>R$ and the power series converged at $z$, then $|z|$ would belong to the set whose supremum is $R$, an immediate contradiction. ThusNo assertion about the boundary $|z|=R$ follows from this argument. If $R=0$ the first assertion is vacuous; if $R=\infty$ the second is vacuous.

For fixed $z\ne 0$, the ratio of successive absolute values isThe ratio test says that a series converges absolutely when this ratio has a limit of a sequence strictly below one. Thus this power series converges absolutely for every complex $z$; at zero the conclusion is immediate. Its radius of convergence is $\boxed{R=\infty }$.

Split the power series into even and odd indices:whenever the two series converge absolutely. If $|z|<1$, the even part is a convergent geometric series, while the odd part is bounded termwise in absolute value by ${\sum }_j|z{|}^{2j+1}$, also a convergent geometric series. Hence there is absolute convergence throughout $|z|<1$.

If $|z|>1$, the odd terms have absolute values $|z{|}^{2j+1}/(2j+1{)}^2$. Their successive ratio tends to $|z{|}^2>1$, so these terms grow eventually and do not tend to zero. The term test for divergence therefore forces divergence of the original power series. Consequently its radius of convergence is

If $x\ne y$, one of $x<y$ and $y<x$ holds. Since $f$ is strictly increasing, the corresponding values are strictly ordered, so $f(x)\ne f(y)$. Thus $f$ is an injection. Every element of the image of a function $f([a,b])$ equals $f(x)$ for some $x\in [a,b]$, which is precisely surjectivity onto that image of a function. Hence $f:[a,b]\to f([a,b])$ is a bijection.

Let $u<v$ lie in $f([a,b])$, and put $x=g(u)$ and $y=g(v)$. Equality $x=y$ would imply $u=v$. If $x>y$, strict monotonicity of $f$ would give $u=f(x)>f(y)=v$. Both are impossible, so $x<y$. Therefore the inverse function $g$ is strictly increasing.

For $x_0>a$, defineThis nonempty set is bounded above by $f(x_0)$, so $l\le f(x_0)$. Given $\epsilon >0$, the supremum property supplies $s<x_0$ with $l-\epsilon <f(s)\le l$. Whenever $s<t<x_0$, monotonicity givesThis proves the left-hand one-sided limit $f^{-}(x_0)=l$.

For $x_0<b$, defineHere $r\ge f(x_0)$. Choose $u>x_0$ with $r\le f(u)<r+\epsilon$, using the infimum property. For $x_0<t<u$, monotonicity gives $r\le f(t)\le f(u)<r+\epsilon$, proving the right-hand one-sided limit $f^{+}(x_0)=r$. Thuswhere the relevant one-sided limits exist, including the indicated single one-sided limit at each endpoint.

At an interior point, continuity implies both one-sided limits equal $f(x_0)$. Conversely, equality $f^{-}(x_0)=f^{+}(x_0)$ and the displayed inequalities force both to equal $f(x_0)$, giving continuity. Therefore

If $f$ is a continuous function, the intermediate value theorem ensures that every value between $f(a)$ and $f(b)$ is attained. Monotonicity excludes values outside that range, so $f([a,b])=[f(a),f(b)]$.

Conversely, suppose this equality of sets holds. At an interior point, if $f^{-}(x_0)<f(x_0)$, any $y$ strictly between these two numbers lies in $[f(a),f(b)]$. Yet inputs below $x_0$ have values at most $f^{-}(x_0)$ and inputs at or above $x_0$ have values at least $f(x_0)$, so no input attains $y$. This contradicts surjectivity. The same argument excludes $f(x_0)<f^{+}(x_0)$, so part (b) gives continuity at every interior point.

At $a$, a strict gap $f(a)<f^{+}(a)$ likewise leaves every value between them unattained; hence $f^{+}(a)=f(a)$, giving right continuity. At $b$, the corresponding argument gives $f^{-}(b)=f(b)$ and left continuity. These are the appropriate endpoint notions on $[a,b]$. Hence

With the strict monotonicity assumed at the start of the question, parts (a) and (c) give a strictly increasing inverse functionIts image of a function is exactly the interval between its endpoint values. Apply part (c) to $g$, with domain $[f(a),f(b)]$, to conclude that $g$ is a continuous function, including at both endpoints.

If the word “increasing” in this subpart were interpreted as merely nondecreasing, an inverse function need not exist: a constant function is a counterexample. The preceding strict hypothesis supplies the required injectivity.

The sequence $(f_n)$ converges by uniform convergence to $f$ on $E$ ifThe index $M$ must work simultaneously for every $x\in E$.

Given $\epsilon >0$, choose $m\ge N$ so large that $|f_m(x)-f(x)|<\epsilon /3$ for every $x\in E$. The continuity of this one fixed $f_m$ at $x_0$ gives $\delta >0$ such that $x\in E$ and $|x-x_0|<\delta$ imply $|f_m(x)-f_m(x_0)|<\epsilon /3$. The triangle inequality then givesThus $f$ is continuous at $x_0$, with continuity understood relative to $E$. This is the pointwise form of the uniform limit theorem.

No. Let $E=\mathbb{R}$, $x_0=0$ andThe supremum norm satisfies $\Vert f_n-f{\Vert }_{\infty }=1/n\to 0$, so there is uniform convergence to the continuous function $f$. But every $f_n$ is discontinuous at zero: $f_n(0)=1/n$ whereas $f_n(x)=0$ for every $x\ne 0$. No tail consists of functions continuous at $x_0$.

At $x=0$, $f_n(x)=0$. For fixed $x>0$,Thus the pointwise limit is zero everywhere on $[0,1]$. However, with $t=nx$, the inequality $1+t^2\ge 2t$ gives $f_n(x)\le 1/2$, and equality occurs at $x=1/n\in [0,1]$. Henceso there is no uniform convergence. The moving evaluation point $1/n$ captures a peak missed by each fixed positive input.

The finite geometric series identity gives, for each $|x|<1$,For every fixed $n$, the absolute value of this error tends to infinity as $x\to 1^{-}$. Thusand there is no uniform convergence on $(-1,1)$.

Yes. By uniform convergence, choose $N$ so that $|p_n(x)-f(x)|<1$ for every $n\ge N$ and every $x\in \mathbb{R}$. ThenThe polynomial $p_n-p_N$ is bounded on the whole real line. The supplied growth property of nonconstant polynomials implies that this difference must be constant; write $p_n=p_N+c_n$. Evaluating at zero gives $c_n=p_n(0)-p_N(0)\to c=f(0)-p_N(0)$. For each $x$, take the limit of a sequence in $p_n(x)=p_N(x)+c_n$ to obtainThis is a polynomial. The uniform limits of polynomials on the real line therefore have a tail in which all nonconstant coefficients have already stabilized; only the constant coefficients may continue to change.

The function $f$ is differentiable at $c$ if the finite limit of a functionexists. If it exists, put $L=f^{\prime }(c)$ and defineThe definition of derivative gives $ϵ(x)\to 0$ as $x\to c$. For $x\ne c$, rearrangement gives the desired local linear approximationAt $x=c$, this identity holds because both terms containing $x-c$ vanish. Conversely, if such an identity holds, then for $x\ne c$ its difference quotient is $L+ϵ(x)\to L$. Hence $f$ is differentiable at $c$, and necessarily $\boxed{L=f^{\prime }(c)}$.

Use the local linear approximation from part (a), with $L=f^{\prime }(c)$. Since $x_n\le c\le y_n$ and $x_n\ne y_n$, we have $y_n-x_n>0$, and subtraction givesThe coefficients $(y_n-c)/(y_n-x_n)$ and $(c-x_n)/(y_n-x_n)$ are nonnegative and sum to one. ThereforeHere $ϵ(c)=0$ handles any endpoint of a secant that happens to equal $c$. Thus secants straddling a differentiability point satisfy

No. On $(-1,1)$ take $c=0$ andAt zero, its difference quotient has absolute value $|x\sin (1/x)|\le |x|\to 0$, so $f^{\prime }(0)=0$. Away from zero, the product rule and chain rule apply. DefineThese are distinct positive inputs tending to zero. Since $\sin A_n=0$ and $\sin (A_n+B_n)=\sin B_n$,The even-indexed subsequence tends to $-2/\pi$, whereas the odd-indexed subsequence tends to $2/(3\pi )$. Thus the limit of a sequence does not exist. The two points lie on the same side of $c$, so the nonnegative-weight estimate in part (b) is unavailable. This is a concrete instance of differentiability does not imply continuity of the derivative.

By the Heine-Cantor theorem, a continuous function on the compact interval $[a,b]$ is uniformly continuous. Given $\epsilon >0$, choose $\delta >0$ such that $|f(s)-f(t)|<\epsilon$ whenever $s,t\in [a,b]$ and $|s-t|<\delta$.

Choose a finite partition of an interval $a=t_0<t_1<\cdots <t_m=b$ with every width below $\delta$. Define the step functionFor each $t<b$, its distance from the chosen left endpoint is below $\delta$, so $|f(t)-\varphi (t)|<\epsilon$; at $b$ the error is zero. HenceA step function may have separately assigned values at its finitely many partition points; these values do not affect its Riemann integral. If $a=b$, simply take $\varphi =f$ on that singleton.

Put $E_n={\sup }_{t\in [a,b]}|f(t)-{\varphi }_n(t)|\to 0$, so the hypothesis is uniform convergence. Every step function is bounded and Riemann integrable: away from its finitely many partition points it is constant, and those exceptional points can be covered by intervals of arbitrarily small total length. A partition of an interval containing their boundaries makes the difference of its upper Darboux sum and lower Darboux sum arbitrarily small.

Choose one $n$ with $E_n<\infty$. Boundedness of ${\varphi }_n$ and $|f|\le |{\varphi }_n|+E_n$ show that $f$ is a bounded function. Assume $a<b$, since for $a=b$ all the integrals are zero. Given $\eta >0$, choose $n$ so that $2E_n(b-a)<\eta /2$, and then a partition of an interval $P$ such that $U({\varphi }_n,P)-L({\varphi }_n,P)<\eta /2$. On every partition subinterval, $|f-{\varphi }_n|\le E_n$ givesThusThe Riemann integrability criterion proves that $f$ is Riemann integrable. Finally, linearity of the integral and monotonicity of the Riemann integral giveThereforeThis proves the relevant case of uniform convergence preserves integrals on a compact interval.

The printed assertion needs an endpoint qualification: the enumeration includes zero, but $1_{[0,0)}$ is identically zero. In fact $f$ is continuous at zero; the claimed discontinuities occur at every $r_j>0$.

Define the finite step function sumsThe assumed convergence of ${\sum }_{k}1/k^2$ and the bound on each indicator function giveThus the defining series converges everywhere and has uniform convergence. Part (b) proves that $f$ is Riemann integrable; it also gives, if desired,For $r_j>0$ and $0\le t<r_j$, every indicator function is nonincreasing, and the $j$th drops by one between $t$ and $r_j$. HenceTaking inputs from the left tending to $r_j$ proves a discontinuity there.

For completeness, continuity at zero follows directly from a finite-head and tail argument. Given $\epsilon >0$, choose $N$ with ${\sum }_{k>N}1/k^2<\epsilon$. Choose $\delta >0$ below every positive $r_k$ with $k\le N$. For $0\le t<\delta$, the first $N$ indicator functions have exactly their values at zero; the one with $r_k=0$, if present, is always zero. Thus $|f(t)-f(0)|\le {\sum }_{k>N}1/k^2<\epsilon$. This confirms that the endpoint exception is real. Replacing the printed enumeration of $[0,1)$ by one of $(0,1)$ would remove it.

Finally, the density of the rational numbers supplies a positive rational $r_j$ in every nonempty open subinterval of $[0,1]$. Thus a Riemann integrable function can have a discontinuity in every open subinterval: dense jumps do not prevent Riemann integrability.