Past exam of the mathematics course of the University of Cambridge / 2022 / ia / Paper 1

The triangle inequality givesHence the left-hand side of the proposed equation is always at least $|a-b|$. There are no solutions when $c<|a-b|$.

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Equality in the triangle inequality holds exactly when the two displacement vectors $a-z$ and $z-b$ point in the same direction. Thus $z$ lies on the closed line segment joining $a$ and $b$. Conversely, every point of that segment satisfiesThe solution set is therefore precisely that segment in the complex plane.

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By the focal definition of an ellipse, the solution set is the ellipse with foci $a$ and $b$. Its centre is $(a+b)/2$, its major axis lies along the line through the foci, and its semiaxes areThe sketch is consequently a nondegenerate ellipse symmetric about both the focal line and its perpendicular bisector.

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The required root of unity issoTherefore

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The logarithms of $1+\omega =e^{i\pi /3}$ areThe definition of complex exponentiation therefore gives all values asSince $1+\omega =\frac{1}{2}+i\frac{\sqrt{3}}{2}$, these can be written

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Taking determinants of $M^{T}JM=J$ givesSince $\det J=-1\ne 0$, one has $\det (M{)}^2=1$, and henceBoth values occur, for example for $M=I$ and $M=\mathrm{diag}(1,-1)$.

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If $M_1^{T}J{M}_1=J$ and $M_2^{T}J{M}_2=J$, then matrix multiplication and the transpose rule giveThus the product also satisfies the equation.

The equation and part (i) imply that $M_1$ is invertible. Multiplying
$M_1^{T}J{M}_1=J$ on the left by $(M_1^{-1}{)}^T$ and on the right by $M_1^{-1}$ givesHence the set of solutions is closed under products and matrix inverses.

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WriteThe first column has unit length for the indefinite quadratic form:Because $p=M_{11}>0$, there is a unique $u\in \mathbb{R}$ such thatThe two columns are orthogonal for the same form and the second has squared length $-1$:The vectors with these properties areUsing the hyperbolic cosine and hyperbolic sine, the two possible families are therefore

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The alternating series test says that if $a_n\ge 0$ decreases to $0$, thenconverges. Taking $a_n=n^{-1/2}$ proves convergence of the given series.

It is not absolute convergence, because the series of absolute values is the p-serieswhich diverges since $p=1/2\le 1$. Thus the original series has conditional convergence.

For an explicit divergent rearrangement of a series, take unused positive terms, which are the even-indexed terms, until the partial sum exceeds $1$, then take the first unused negative term. Next take positive terms until the sum exceeds $2$, then the next unused negative term, and continue. Both the positive and negative subseries have infinite total magnitude, so this procedure uses every term. The negative term inserted at stage $m$ tends to zero, while the preceding partial sum exceeds $m$; consequently these rearranged partial sums tend to $+\infty$. This is a divergent series with exactly the prescribed terms.

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For the sum $F=f+g$:

For the product $G=fg$, the answer is **no** in both cases:

Every displayed $f$ and $g$ is a continuous function and is nonzero at points arbitrarily close to $a$, so the extra condition in the question is satisfied.

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For $A=(A_1,A_2,A_3)$ and $B=(B_1,B_2,B_3)$, the scalar product is the inner productThe cross product isIt is perpendicular to both vectors and has magnitude $|A||B|\sin \theta$.

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The vector triple product identity givesApplying the same identity after reversing the outer cross product givesThe formulas differ because the cross product is not associative.

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LetLinear independence implies that the scalar triple product $Delta$ is nonzero. Applying the vector triple-product formulas to $(A\times B)\times (C\times D)$ recovers the coefficients of $D$ in the basis $A,B,C$. Equivalently, the reciprocal basis to $A,B,C$ isTaking scalar products with $D$ therefore yields

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Because the sphere passes through the origin and has centre $P$, its radius is $|P|$. A point $X$ lies on it exactly whenor equivalentlyWith $P=\alpha A+\beta B+\gamma C$, the three required equations are the Gram matrix systemFor the specified vectors this becomesThusand the centre is

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An anticlockwise rotation matrix and reflection in the $x$-axis areThe columns of $R$ are the images of the standard coordinate vectors, while $M(x,y{)}^T=(x,-y{)}^T$.

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Direct multiplication givesHenceThis is the defining conjugation relation between a rotation and reflection in a dihedral group.

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No. Every power $R^a$ has determinant $1$, whereasMatrices with different determinants cannot be equal.

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The relationsimply $RM=M{R}^{-1}$. Moving every occurrence of $M$ to the left and reducing exponents therefore puts every word into one of the normal formsExplicitly, with $\theta =2\pi /n$,andThe $n$ rotations are distinct. The $n$ reflected matrices are also distinct, and no reflected matrix equals a rotation because their determinants are $-1$ and $1$, respectively. Hence there are exactlymatrices, forming the dihedral group.

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Let $Av=\lambda v$ for a nonzero eigenvector $v$ of the Hermitian matrix $A$. Hermitian symmetry of the inner product givesSince $lVertv{\Vert }^2>0$, $lambda=\overline{\lambda }$, so every eigenvalue is real.

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For an orthogonal projection matrix $P=P^{†}=P^2$,Thus $P$ is a positive semidefinite matrix.

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SetThen $v=v_0+v_1$, and idempotence givesMoreover,because both $P$ and $I-P$ are Hermitian. This is the orthogonal decomposition into the kernel and image of the projection.

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Put $H=A-B$. This is a nonzero Hermitian matrix. By the finite-dimensional spectral theorem, it has a unit eigenvector $v$ with a nonzero real eigenvalue $lambda$. Define the rank-one orthogonal projection matrixThe cyclic property of the matrix trace givesHence this $P$ distinguishes $A$ and $B$ through the required traces.

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No. Consider the two orthogonal projectionsTheir product isFor $x=(-1,2{)}^T$,Thus $PQ$ is not a positive semidefinite matrix. It is also not Hermitian; products of orthogonal projections need not remain orthogonal projections unless the factors commute.

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For the standard basis vector $e_j$, matrix multiplication selects the $j$th column, soConsequentlyand the columns of $PM$ are $P{c}_1,\dots ,P{c}_n$.

Suppose first that $v,Av,\dots ,A^{n-1}v$ are linearly independent, and defineThen $S$ is invertible. The first $n-1$ columns of $AS$ are the last $n-1$ columns of $S$. By the Cayley-Hamilton theorem,This is exactly the last-column rule for the companion matrix $C$, so

Conversely, suppose $S^{-1}AS=C$, or $AS=SC$, and write the columns of $S$ as $s_1,\dots ,s_n$. Comparing the first $n-1$ columns givesHenceSince $S$ is invertible, its columns are linearly independent. Taking $v=s_1$ therefore makeslinearly independent. Thus $A$ is similar to $C$ exactly when it has the stated cyclic vector.

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Suppose $a_n\to L$. Given $\epsilon >0$, choose $N$ such that $|a_k-L|<\epsilon /2$ for $k>N$. ThenThe first term tends to zero because its numerator is fixed, and the second is at most $\epsilon /2$. Thus the Cesaro mean tends to $L$.

The converse is false. For $a_n=(-1{)}^n$, the Cesaro means tend to $0$, while the original sequence alternates between $-1$ and $1$ and does not converge.

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First suppose $x_n\to L>0$. Continuity of the natural logarithm gives $\log x_n\to \log L$. Part (a), applied to this sequence, yieldsApplying the continuous exponential function,If $L=0$, then for every $\epsilon >0$ all sufficiently late $x_k$ are below $\epsilon$. Splitting off the fixed initial product shows that the limsup of the geometric means is at most $\epsilon$; hence it is zero. This proves the geometric mean of a sequence result in all cases.

Now supposeThe telescoping product isThereforeand so

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A Cauchy sequence is a sequence $(x_n)$ such that for every $\epsilon >0$ there is $N$ for whichThe general principle of convergence, or completeness of the real numbers, states that a real sequence converges if and only if it is Cauchy.

For the final claim, let $m=\lfloor n/2\rfloor$. Since $(a_n)$ is decreasing and positive,Because the series $\sum a_k$ converges, its tails tend to zero. Also $n-m\ge n/2$, soThe squeeze theorem gives

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Let $f:[a,b]\to \mathbb{R}$ be continuous. If it were unbounded, one could choose $x_n\in [a,b]$ with $|f(x_n)|>n$. The Bolzano-Weierstrass theorem gives a subsequence $x_{n_j}\to x\in [a,b]$, but continuity would then give $f(x_{n_j})\to f(x)$, contradicting unboundedness.

Now let $M=\sup f([a,b])$. Choose $x_n$ with $f(x_n)>M-1/n$. A convergent subsequence and continuity give $f(x)=M$ at its limit. Applying the same argument to $-f$ shows that the infimum is attained. This proves the extreme value theorem on a closed bounded interval.

The functionis continuous and bounded, but attains neither its infimum $0$ nor its supremum $1$. For the second example, enumerate the rationals in $[0,1]$ as $(q_n)$ and setEvery nondegenerate interval contains infinitely many rational numbers, hence some $q_n$ with arbitrarily large $n$; therefore $\psi$ is unbounded on every such interval.

For the running extrema, compactness lets us writeThe function $f$ is uniformly continuous on $[a,b]$. Given $\epsilon >0$, choose $\delta >0$ such that $|f(u)-f(v)|<\epsilon$ whenever $|u-v|<\delta$. If $x<y$ and $y-x<\delta$, every new value $f(\xi )$ with $x\le \xi \le y$ is at most $f(x)+\epsilon \le M(x)+\epsilon$. Since $M(y)\ge M(x)$,Interchanging $x,y$ handles the other order. Applying this argument to $-f$ proves continuity of $m$ as well.

Finally fix $T>0$ and put $h(x)=g(x+T)-g(x)$. For each positive integer $n$, there must be some $x_n>n$ with $|h(x_n)|<1/n$. Otherwise, for some $n$ the continuous function $h$ would satisfy $|h(x)|\ge 1/n$ on the interval $(n,\infty )$. By the intermediate value theorem, $h$ would have a constant sign there. The valueswould then increase or decrease by at least $1/n$ at every step, contradicting boundedness of $g$. Thus $x_n\to \infty$ and

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Rolle theorem: if $f$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a)=f(b)$, then $f^{\prime }(c)=0$ for some $c\in (a,b)$. Indeed, the extreme value theorem gives a maximum and minimum. If both occur only at the endpoints then $f$ is constant; otherwise an interior extremum $c$ satisfies $f^{\prime }(c)=0$ by comparing the two-sided difference quotients.

Mean value theorem: if $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then some $c\in (a,b)$ satisfiesApply Rolle's theorem towhich has $h(a)=h(b)=0$.

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LetThen $h(a)=h(b)=0$. Since $f$ is not linear, $h(c)\ne 0$ for some $c\in (a,b)$. If $h(c)>0$, the mean value theorem on $[a,c]$ gives a point $\xi$ withIf $h(c)<0$, apply it on $[c,b]$ to obtainIn either case $h^{\prime }=f^{\prime }-m$, and therefore

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No. On $[-1,1]$, takeThen $f^{\prime }(0)=0$. For every $x_1<0<x_2$,This quadratic expression is strictly positive: it equalsThus no chord with endpoints on opposite sides of $\xi$ has slope $f^{\prime }(\xi )=0$.

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SetIts endpoint values are equal. Rolle theorem therefore supplies $\xi \in (a,b)$ withIf $g^{\prime }(\xi )=0$, then (1) and $g(b)\ne g(a)$ force $f^{\prime }(\xi )=0$, contrary to the hypothesis. Hence $g^{\prime }(\xi )\ne 0$, and division in (1) gives

The condition is necessary. On $[0,2\pi ]$, letHere $g(2\pi )-g(0)=\pi /2$ while $f(2\pi )-f(0)=0$, so the endpoint ratio is zero. ButWhere $g^{\prime }(x)\ne 0$, their ratio is $1/(1+\frac{1}{2}\cos x)$ and is never zero. At the remaining points both derivatives vanish, so the derivative ratio is undefined. Thus the conclusion fails when simultaneous zeros are allowed.

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It is enough to treat an increasing function; replacing $f$ by $-f$ handles a decreasing one. Let $\mathcal{R}$ be the common refinement of dissections $\mathcal{D}$ and ${\mathcal{D}}^{\prime }$. Refinement raises lower Darboux sums and lowers upper ones, so

For the uniform dissection ${\mathcal{D}}_n=\{k/n:0\le k\le n\}$, monotonicity gives the telescoping differenceThis can be made smaller than any $\epsilon >0$, so the Riemann integrability criterion proves that $f$ is integrable.

The integral lies between the lower and upper sums, and the displayed sum in the question is the right-endpoint upper sum. Hence the generally valid sharp estimate isThe strict inequality printed in the question is false for arbitrary monotone functions: if $f(x)=0$ for $x<1$ and $f(1)=1$, the two sides of (1) are both $1/n$.

For the final claim, writeSince $F^{\prime }$ is continuous on a compact interval, it is uniformly continuous. Uniformly for $0\le t\le 1$,After the substitution $x=(k-1+t)/n$, summing the uniform errors givesThe right-hand sum is a Riemann integral, so the fundamental theorem of calculus yields

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