Past exam of the mathematics course of the University of Cambridge / 2018 / ia / Paper 1

For $z\ne 0$, the principal value of complex exponentiation is $z^w=\exp (w\mathrm{Log}z)$, where the principal complex logarithm is $\mathrm{Log}z=\log |z|+i\mathrm{Arg}z$ with $-\pi <\mathrm{Arg}z\le \pi$. De Moivre's theorem states that $(\cos \theta +i\sin \theta {)}^n=\cos (n\theta )+i\sin (n\theta )$ for every integer $n$.

Since $\sqrt{3}+i=2e^{i\pi /6}$, the six roots are

Write $z=r{e}^{i\theta }$ with $-\pi <\theta \le \pi$. Its principal sixth root is $r^{1/6}{e}^{i\theta /6}$, whose complex argument lies in $(-\pi /6,\pi /6]$. Equality with $2e^{i\pi /6}$ forces $r=64$ and $\theta =\pi$, so $z=-64$.

For $z=x+iy$,Matching its modulus and complex argument with $2e^{i\pi /6}$ gives

The base $e^{5i\pi /2}$ is the complex number $i$. Principal exponentiation uses the principal logarithm of that base, so this is the same equation as in part iii:

The vector triple product identity gives $\mathbf{n}\times (\mathbf{n}\times \mathbf{x})=\mathbf{n}(\mathbf{n}\cdot \mathbf{x})-\mathbf{x}$, and henceThus $\Phi$ is the identity on the orthogonal complement ${\mathbf{n}}^{\perp }$ and multiplies the component parallel to $\mathbf{n}$ by $\alpha$. It is invertible exactly when $\alpha \ne 0$, and then

The only noninvertible case is $\alpha =0$. Then $\Phi$ is the orthogonal projection onto ${\mathbf{n}}^{\perp }$, soThese dimensions agree with the rank-nullity theorem.

In Einstein notation, $(\mathbf{n}\cdot \mathbf{x})n_i=n_i{n}_j{x}_j$, so the matrix and, for $\alpha \ne 0$, its matrix inverse are

Let $(u_n)$ be increasing and bounded above, and let $L=\sup \{u_n:n\in \mathbb{N}\}$. For every $\epsilon >0$, the definition of the supremum supplies $N$ with $L-\epsilon <u_N\le L$. Monotonicity gives $L-\epsilon <u_n\le L$ for every $n\ge N$, proving $u_n\to L$. This is the monotone bounded sequence theorem.

For the recurrence, $f(x_n)>0$ makes $(x_n)$ increasing. If it were bounded above, it would converge to some $L$. Since $x_n\le L$ and $f$ is decreasing, $f(x_n)\ge f(L)>0$, whencea contradiction. Therefore $x_n\to \infty$.

The radius of convergence of the power series ${\sum }_{n\ge 0}{a}_n{z}^n$ isIf $|z|<R$, choose $w$ with $|z|<|w|<R$. Since $a_n{w}^n\to 0$, those terms are bounded, say $|a_n{w}^n|\le M$. Thereforeand comparison with a geometric series proves absolute convergence. If $|z|>R$, convergence would imply absolute convergence at every smaller modulus by the same argument, contradicting the definition of $R$.

If $|a_n|\le |b_n|$, convergence of $\sum b_n{w}^n$ implies absolute convergence of $\sum a_n{z}^n$ whenever $|z|<|w|$. Letting $|w|$ approach the radius of the $b$-series shows that $R_a\ge R_b$.

The dot product and Euclidean norm are $\mathbf{x}\cdot \mathbf{y}={\sum }_i{x}_i{y}_i$ and $|\mathbf{x}|=\sqrt{\mathbf{x}\cdot \mathbf{x}}$. For every real $t$,Its quadratic discriminant is nonpositive, proving the Cauchy-Schwarz inequality $|\mathbf{x}\cdot \mathbf{y}|\le |\mathbf{x}||\mathbf{y}|$. In the inequality asked for, equality holds exactly when $\mathbf{x}$ is a positive scalar multiple of $\mathbf{y}$, and

Using the Levi-Civita symbol and its contraction identity,

Put $T=\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})$, the scalar triple product. Cyclic symmetry gives $\mathbf{m}\cdot \mathbf{a}=\mathbf{m}\cdot \mathbf{b}=\mathbf{m}\cdot \mathbf{c}=T$, so each requested angle $\varphi$ satisfies $\cos \varphi =T/|\mathbf{m}|$. If every pairwise angle is $\theta$ and $t=\cos \theta$, the preceding identity givesand therefore $|\mathbf{m}|=\sqrt{3}(1-\cos \theta )$. This is $0$ when $\theta =0$; for a right-handed orthonormal basis and $\theta =\pi /2$, $\mathbf{m}=\mathbf{a}+\mathbf{b}+\mathbf{c}$ has norm $\sqrt{3}$.

The three planes have equations $\mathbf{a}\cdot \mathbf{x}=\mathbf{a}\cdot \mathbf{p}$, $\mathbf{b}\cdot \mathbf{x}=\mathbf{b}\cdot \mathbf{q}$, and $\mathbf{c}\cdot \mathbf{x}=\mathbf{c}\cdot \mathbf{r}$. They meet at one point exactly when $\Delta =\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})\ne 0$, meaning their normals are linearly independent vectors. The point is

The characteristic polynomial of the displayed rotation matrix is $(1-\lambda )({\lambda }^2-2\lambda \cos \theta +1)$, so its eigenvalues are $1,e^{i\theta },e^{-i\theta }$, all of modulus one. The real eigenvector $(0,0,1{)}^T$ spans the rotation axis, while the arguments $\pm \theta$ of the conjugate eigenvalues record the rotation angle on its perpendicular plane.

Applying first the $z$-rotation and then the $x$-rotation givesIts $1$-eigenspace is spanned by $(1,-1,1{)}^T$, so $\mathbf{n}=(1,-1,1{)}^T/\sqrt{3}$. The matrix trace identity $\mathrm{tr}R=1+2\cos \varphi$ gives $0=1+2\cos \varphi$. Thus $\varphi =2\pi /3$.

The quadratic form has symmetric matrixwhose characteristic polynomial is $(\lambda -2)(\lambda -3)(\lambda -8)$. All its eigenvalues are positive, so $A$ is positive definite and ${\mathbf{x}}^{T}A\mathbf{x}=1$ is an ellipsoid. The spectral theorem for real symmetric matrices gives semi-axis lengthsThe nearest points lie in the eigendirection of the largest eigenvalue. Since $(5,2,1{)}^T$ is an $8$-eigenvector, they are

Suppose $A\mathbf{v}=\lambda \mathbf{v}$ for a nonzero complex vector. Because a real symmetric matrix is Hermitian, ${\mathbf{v}}^{\ast }A\mathbf{v}$ is real. But ${\mathbf{v}}^{\ast }A\mathbf{v}=\lambda {\mathbf{v}}^{\ast }\mathbf{v}$, so $\lambda$ is real. The real matrix $A-\lambda I$ is singular; the real or imaginary part of a nonzero complex null vector supplies a nonzero real eigenvector. Thus every eigenvalue is real and has a real eigenvector.

If $A\mathbf{u}=\lambda \mathbf{u}$ and $A\mathbf{v}=\mu \mathbf{v}$, symmetry givesFor $\lambda \ne \mu$, ${\mathbf{u}}^T\mathbf{v}=0$, so the eigenvectors are orthogonal.

Eigenvectors belonging to distinct eigenvalues are linearly independent. With $n$ distinct eigenvalues there are $n$ eigenvectors forming a basis, and the matrix is diagonalizable. In this eigenbasis, the Rayleigh quotient is a weighted average of the eigenvalues, so it is maximized by choosing $\mathbf{v}$ in the eigenspace of the largest eigenvalue.

From $A\mathbf{v}=\lambda B\mathbf{v}$ we have $(A-\lambda B)\mathbf{v}=0$ with $\mathbf{v}\ne 0$. Thus $A-\lambda B$ is singular and $\det (A-\lambda B)=0$.

Multiplication by ${\mathbf{v}}^{\ast }$ givesThe numerator is real because $A$ is symmetric, while the denominator is strictly positive because $B$ is positive definite. Hence $\lambda$ is real. As $A-\lambda B$ is real and singular, its kernel contains a nonzero real generalized eigenvector.

For $A\mathbf{u}=\lambda B\mathbf{u}$ and $A\mathbf{v}=\mu B\mathbf{v}$, symmetry givesIf $\lambda \ne \mu$, ${\mathbf{u}}^{T}B\mathbf{v}=0$: the eigenvectors are orthogonal in the inner product induced by $B$.

The generalized characteristic polynomial isThe generalized Rayleigh quotient is maximized by the largest generalized eigenvalue, $4$. Solving $(A-4B)\mathbf{v}=0$ gives

A real square matrix $M$ is an orthogonal matrix when $M^{T}M=I$. The planar rotationobeys $R^{T}R=I$ by the Pythagorean trigonometric identity.

For $A=\left(\begin{array}{cc}a& b\\ b& c\end{array}\right)$, the off-diagonal entry of $RA{R}^T$ is $\frac{1}{2}(a-c)\sin 2\theta +b\cos 2\theta$. ThusIf $(a-c,b)\ne (0,0)$, equivalently $\theta =\frac{1}{2}atan2(-2b,a-c)+k\pi /2$; if $A$ is scalar, every angle works.

Because $RA{R}^T$ is an orthogonal similarity, its diagonal entries are the eigenvaluesThe matrix $C=\left(\begin{array}{cc}1& 2\\ 2& 1\end{array}\right)$ has eigenvalues $3,-1$ along $(1,1{)}^T,(1,-1{)}^T$. Raising its orthogonal diagonalization to the even power gives

The function is continuous at $x$ when, for every $\epsilon >0$, there is $\delta >0$ such that $|y-x|<\delta$ implies $|f(y)-f(x)|<\epsilon$.

If this holds and $x_n\to x$, eventually $|x_n-x|<\delta$, so $f(x_n)\to f(x)$. Conversely, if continuity fails, some ${\epsilon }_0>0$ allows, for every $n$, a point $x_n$ with $|x_n-x|<1/n$ but $|f(x_n)-f(x)|\ge {\epsilon }_0$. Then $x_n\to x$ while $f(x_n)\nrightarrow f(x)$, a contradiction. This is the sequential characterization of continuity in metric spaces.

Let the nonconstant polynomial have degree $d$ and leading coefficient $c$. If $d$ is odd, its limits at $+\infty$ and $-\infty$ have opposite signs, so the intermediate value theorem shows that its image is $\mathbb{R}$. If $d$ is even and $c>0$, it tends to $+\infty$ at both ends. Outside a sufficiently large compact interval it exceeds $f(0)$, while on that interval the extreme value theorem gives a global minimum $a$. Continuity and the intermediate value theorem then give image $[a,\infty )$. For $c<0$, applying this argument to $-f$ gives $(-\infty ,a]$. These are exactly the three possible images.

Applying the functional equation at $x^{{\alpha }^{-1}}$ repeatedly givesSince $x^{{\alpha }^{-n}}\to 1$, continuity at $1$ gives $f(x)=f(1)$. Thus $f$ is constant.

Without continuity the claim is false. The function $f(1)=1$ and $f(x)=0$ for $x\ne 1$ satisfies $f(x)=f(x^{\alpha })$ because $x^{\alpha }=1$ for positive $x$ exactly when $x=1$, but it is not constant.

Differentiability at $x_0$ meansHence $f(x)-f(x_0)=(x-x_0)(f^{\prime }(x_0)+o(1))\to 0$ as $x\to x_0$. Therefore every differentiable function is continuous at each point where it is differentiable.

The mean value theorem says that if $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then $f(b)-f(a)=f^{\prime }(c)(b-a)$ for some $c\in (a,b)$.

For $F(t)=\cos (e^{-t})$ and $t\ge 0$,The mean value theorem gives $|\cos (e^{-x})-\cos (e^{-y})|\le |x-y|$.

For the second inequality put $t=\sqrt{1+x}\ge 1$. It becomes $2\log t\le t-t^{-1}$. The difference $G(t)=t-t^{-1}-2\log t$ satisfies $G(1)=0$ andThus $\log (1+x)\le x/\sqrt{1+x}$.

For $x\ne 0$, $f(x)=|x{|}^x=e^{x\log |x|}$, so $f^{\prime }(x)=|x{|}^x(\log |x|+1)$. Although $x\log |x|\to 0$ makes $f$ continuous at zero, $(f(x)-1)/x\sim \log |x|\to -\infty$, so it is not differentiable there.

Since the cosine function is even, $g(x)=\cos |x|=\cos x$, so it is differentiable everywhere with $g^{\prime }(x)=-\sin x$. Finally, $h(x)=x|x|$ is differentiable everywhere, including at zero, and $h^{\prime }(x)=2|x|$. Therefore

At a nonzero rational $p/q$ in lowest terms, irrational points approach $p/q$ with function value zero, whereas $f(p/q)=1/q>0$, so $f$ is discontinuous. At zero, $|p/q|\ge 1/q$ for nonzero $p$, hence $0\le f(x)\le |x|$ and continuity follows.

At an irrational $x$, given $\epsilon >0$, only finitely many rationals in a bounded neighbourhood have reduced denominator $q\le 1/\epsilon$. Choose a neighbourhood avoiding those points; throughout it, $f<\epsilon$. Consequently the continuity set is $(\mathbb{R}\setminus \mathbb{Q})\cup \{0\}$. This is a variant of the Thomae function.

The comparison test for series states that if $0\le a_n\le b_n$ eventually and $\sum b_n$ converges, then $\sum a_n$ converges. The partial sums of $\sum a_n$ are increasing and bounded above by a constant plus the bounded partial sums of $\sum b_n$, so the monotone bounded sequence theorem applies.

If $\sum x_n$ converges, then $x_n^2\le x_n$ proves convergence of $\sum x_n^2$. Also $x_n\to 0$, so eventually $x_n\le 1/2$ and $x_n/(1-x_n)\le 2x_n$, proving convergence of $\sum x_n/(1-x_n)$. The converse for the square series is false: $x_n=1/(n+1)$ has convergent square series but divergent original series. The converse for the fraction series is true because $x_n\le x_n/(1-x_n)$.

If $(x_n)$ is positive and decreasing and $\sum x_n$ converges, put $m=\lfloor n/2\rfloor$. Thenso $n{x}_n\to 0$. The converse fails for $x_n=1/(n\log n)$ for $n\ge 2$: it is decreasing and $n{x}_n\to 0$, but the integral test for convergence shows that its series diverges.

Even convergence of $\sum x_n$ need not imply $(n\log n)x_n\to 0$. Choose integers $N_j=\lceil e^{2^j}\rceil$ and set $x_n=(N_j\log N_j{)}^{-1}$ for $N_{j-1}<n\le N_j$. This is decreasing, and the $j$th block contributes at most $1/\log N_j\le 2^{-j}$, so the series converges. At $n=N_j$, however, $(n\log n)x_n=1$.

Assume $a_n\to 0$. The function is bounded, and every interval has lower Darboux sum zero because it contains an irrational. Given $\epsilon >0$, choose $N$ so that $a_n<\epsilon /2$ for $n>N$. Put small intervals around the finitely many points $q_1,\dots ,q_N$ whose total length makes their contribution to the upper sum less than $\epsilon /2$, and refine their endpoints to a partition. On every remaining subinterval the function is at most $\epsilon /2$, so the upper sum is less than $\epsilon$. The Riemann integrability criterion proves that $f$ is Riemann integrable with integral zero.

The condition $a_n\to 0$ is sufficient but not necessary. Assign $a_n=1$ when $q_n=1/k$ for some positive integer $k$, and $a_n=1/n$ otherwise. Then $(a_n)$ does not tend to zero. Outside a short interval near zero only finitely many unit spikes occur and can be enclosed in intervals of arbitrarily small total length; all remaining values can be made uniformly small except for finitely many further points. The same Darboux-sum argument proves that this $f$ is Riemann integrable with integral zero.

The fundamental theorem of calculus says first that if $h$ is continuous on $[a,b]$ and $H(x)={\int }_a^{x}h(t)dt$, then $H^{\prime }(x)=h(x)$. Indeed,whose absolute value tends to zero by continuity. Consequently, if $F$ is differentiable with continuous derivative, applying the first part to $h=F^{\prime }$ shows that $F(x)-{\int }_a^x{F}^{\prime }(t)dt$ has zero derivative and is constant. Hence

A derivative need not be Riemann integrable. Define $f(0)=0$ and $f(x)=x^2\sin (x^{-2})$ for $x\ne 0$. Then $f$ is differentiable at zero with $f^{\prime }(0)=0$, while for $x\ne 0$,which is unbounded near zero. Since every Riemann-integrable function on a compact interval is bounded, $g=f^{\prime }{|}_{[0,1]}$ need not be Riemann integrable.