Past exam of the mathematics course of the University of Cambridge / 2019 / ia

The powers of the imaginary unit have period four. HencewhileThus $x+iy=1-i$, so

Since $(1+i)/(1-i)=i$ and $(1+i{)}^2=2i$,

Use the principal complex logarithm, for which $\mathrm{Log}i=i\pi /2$, and setThen the definition of complex exponentiation givesand thereforeOther choices of branch of the complex logarithm produce further valid answers.

Write $z=r{e}^{i\theta }$ with $r>0$. On a fixed branch of the complex logarithm on which $\log z=\log r+i\theta$ and $\log \bar{z}=\log r-i\theta$, the equation becomesIts real and imaginary parts give the same condition $\log r=\theta$, orThis is a logarithmic spiral. For the principal complex logarithm, it is the portion parametrized by $-\pi <\theta <\pi$ that avoids the branch cut.

The Leibniz formula for determinants isEquivalently, the determinant is the unique alternating function of the rows that is multilinear and takes value one on the identity matrix. Multiplying one row by $\lambda$ therefore gives

The scalar multiplication $\lambda A$ multiplies every one of the $n$ rows by $\lambda$. Applying multilinearity of the determinant once to each row gives

Interchanging two rows reverses the sign of an alternating multilinear map, soIn the Leibniz formula for determinants, the same result follows by composing every permutation with the corresponding transposition, which reverses its sign.

Let the columns be $C_1,\dots ,C_n$, and replace $C_k$ by $C_k+\lambda C_l$, where $k\ne l$. By multilinearity of the determinant,The second determinant vanishes because it has two equal columns. Hence

The Bolzano-Weierstrass theorem states that every bounded sequence of real numbers has a convergent subsequence.

Condition (i) is insufficient because the allowed limit may be zero. For example, $a_n=1/n$ is nonzero and converges to zero, whereas $1/a_n=n$ does not converge in $\mathbb{R}$.

Condition (ii) is sufficient. If $a_n\to \ell \ne 0$, then eventually $|a_n|\ge |\ell |/2$. Thereforeso

Condition (iii) is also sufficient. If $|a_n|$ did not tend to infinity, some $M>0$ would contain infinitely many terms with $|a_n|\le M$. Those terms form a bounded subsequence, and the Bolzano-Weierstrass theorem would give it a convergent subsequence, contrary to the hypothesis. Hence $|a_n|\to \infty$ and

LetThe set $S$ contains zero. If $r\in S$ and $0\le s<r$, then $s\in S$ by comparison. Moreover, if the original power series diverges at $x_0$, no $r>|x_0|$ can lie in $S$, because absolute convergence at $r$ would imply absolute convergence at $x_0$. Thus $R=\sup S$ is finite and nonnegative.

If $|x|<R$, choose $r\in S$ with $|x|<r$; comparison gives absolute convergence at $x$. If $|x|>R$ and the series converged at $x$, then its terms $a_n{x}^n$ would be bounded. For any $y$ with $R<|y|<|x|$,so the geometric series comparison would put $|y|$ in $S$, contradicting the definition of $R$. This proves the radius of convergence property.

For ${\sum }_{n\ge 1}{x}^n/3^n$, the ratio is $x/3$, so the geometric series criterion gives

The inequality $(1+1/n{)}^n\le e$ gives, for $|x|<1$,For $|x|>1$, the terms have magnitude at least $|x{|}^n$ and therefore do not tend to zero, so the series diverges by the term test for divergence. Hence

Using the Levi-Civita symbol and the epsilon-delta identity,This proves the vector triple product identityApplying it inside the scalar triple product gives Lagrange's identity

The vector triple product identity givesTaking the dot product with $\mathbf{a}\times \mathbf{b}$ therefore yields

WriteThe first equation is then automatic, while the second becomesThus the general solution isAs $\mathbf{u}$ varies, $\mathbf{x}$ and $\mathbf{y}$ are antipodal points of the sphere with center $\mathbf{a}$ and radius $\sqrt{\mathbf{a}\cdot \mathbf{a}-c}$, so they are opposite ends of a diameter.

Let the vertex position vectors be $\mathbf{a},\mathbf{b},\mathbf{c},\mathbf{d}$. SupposeSubtracting the second scalar product from the first givesso the remaining pair of opposite edges of a tetrahedron is also perpendicular.

PutDirect expansion gives, cyclically,with the other two differences equal to minus twice the scalar products of the other opposite-edge pairs. All three scalar products vanish, hence

The columns of $P$ are the coordinates of the vectors ${\mathbf{f}}_j$, so they form a basis exactly whenFor a linear transformation $\alpha$, define its matrix representation in the standard basis byThe change-of-basis matrix satisfies $[\mathbf{v}{]}_{\mathbf{e}}=P[\mathbf{v}{]}_{\mathbf{f}}$. Thereforeand multiplication by $P^{-1}$ gives the similar matrix relation

The characteristic polynomial of the displayed $A$ isEigenvectors $(2,-1,2{)}^T$, $(1,0,0{)}^T$, and $(0,-1,3{)}^T$ giveIndeed, $\det P=1$ andso $P^{-1}AP=B$. Finally, the diagonalization of a matrix gives

ChooseThe Leibniz formula for determinants shows that its leading term is $(-z{)}^n$, as required. A scalar $\lambda$ is an eigenvalue exactly when $A-\lambda I$ has a nonzero kernel, equivalently when it is singular, soThe fundamental theorem of algebra gives a complex root of the degree-$n$ polynomial ${\chi }_A$, so every nonempty complex square matrix has at least one complex eigenvalue.

Eigenvectors belonging to distinct eigenvalues are linearly independent. Under the hypothesis they therefore form a basis $v_1,\dots ,v_n$. For every $j$,because ${\chi }_A({\lambda }_j)=0$. A linear map that vanishes on a basis is zero, henceThis proves the Cayley-Hamilton theorem in the distinct-eigenvalue case.

Over $\mathbb{C}$, triangularize $B$ and denote its eigenvalues, with algebraic multiplicity, by ${\beta }_1,\dots ,{\beta }_n$. ThenThe numbers ${\omega }_l=e^{2\pi il/m}$ are all the root of unity solutions of $w^m=1$, soConsequentlyIf $\lambda$ is an eigenvalue of $B^m$, choose $z$ with $z^m=\lambda$. The displayed identity makes at least one factor ${\chi }_B({\omega }_{l}z)$ zero. Thus $\mu ={\omega }_{l}z$ is an eigenvalue of $B$ and

LetSince $J_{-}^2=-I$ and $J_{+}^2=I$, separating the even and odd terms of the matrix exponential givesand

Direct multiplication and the identities ${\cos }^2\theta +{\sin }^2\theta =1$ and ${\cosh }^2\theta -{\sinh }^2\theta =1$ giveThus $R$ is an orthogonal matrix, while $S$ preserves the indefinite quadratic form $x_1^2-x_2^2$.

The matrix $A$ fixes the first coordinate and restricts to one half of the rotation generator on the last two coordinates. Hence

The matrix $B$ fixes the second coordinate and satisfies the hyperbolic-generator relation on the first and third coordinates. Therefore

The rotation matrix $e^{xA}$ commutes with $C$ and is orthogonal, soSince $C^n=\mathrm{diag}(1,(-1{)}^n,(-1{)}^n)$,

On the first and third coordinates, the identity $SJS=J$ from part (b) gives $e^{xB}C{e}^{xB}=C$; the second coordinate is also unchanged. Thus

The Cauchy functional equation first gives $g(0)=0$ and $g(-x)=-g(x)$. If $h\to 0$, then continuity at the given point $z$ givesso $g$ is continuous at zero. For every $x$,so $g$ is continuous everywhere.

Put $c=g(1)$. Additivity gives $g(n)=nc$ for integers $n$ and then $g(m/n)=(m/n)c$ for rational numbers. For any real $x$, choose rationals $q_j\to x$; continuity gives

For the multiplicative equation, $h(0)=h(0{)}^2$. If $h(0)=0$, then $h(x)=h(x)h(0)=0$ for every $x$. Otherwise $h(0)=1$. In that caseand $h(x)$ cannot vanish because $1=h(0)=h(x)h(-x)$. Hence $h$ is everywhere positive. The continuous function $g=\log h$ satisfies the Cauchy functional equation, so $g(x)=cx$. Therefore the complete family is

The intermediate value theorem states that if $f:[a,b]\to \mathbb{R}$ is continuous and $y$ lies between $f(a)$ and $f(b)$, then some $c\in [a,b]$ satisfies $f(c)=y$.

Assume $f(a)<y<f(b)$ and letThe least-upper-bound property makes $c$ well defined. If $f(c)<y$, continuity gives points immediately to the right of $c$ in $S$, contradicting that $c$ is an upper bound. If $f(c)>y$, continuity gives a left neighborhood of $c$ disjoint from $S$, contradicting the definition of the supremum. Thus $f(c)=y$. The endpoint and reversed-order cases follow directly or by replacing $f$ with $-f$.

The mean value theorem states that if $g$ is continuous on $[\alpha ,\beta ]$ and differentiable on $(\alpha ,\beta )$, then some $c\in (\alpha ,\beta )$ satisfiesFor the functions in the question, the definition of the derivative makes $h$ continuous at $a$ and $f$ continuous at $b$, withIf the middle secant slope equals $k$, take $[\alpha ,\beta ]=[a,b]$. If it exceeds $k$, the intermediate value theorem applied to $h$ gives $\beta \in (a,b)$ with $h(\beta )=k$, and we take $\alpha =a$. If it is below $k$, apply the theorem to $f$ to obtain $\alpha \in (a,b)$ with $f(\alpha )=k$, and take $\beta =b$. In every caseThe mean value theorem on this subinterval then produces $c\in (\alpha ,\beta )\subset (a,b)$ withThis is the Darboux theorem for derivatives for the stated pair of derivative values.

For positive $a_n,b_n$,The comparison test for series therefore gives

The arithmetic-geometric mean inequality givesThus convergence of ${\sum }_n{a}_n$ implies convergence of ${\sum }_n\sqrt{a_n{a}_{n+1}}$ by the comparison test for series.

The converse is false. DefineThe even terms make ${\sum }_n{a}_n$ diverge, whileso the neighboring geometric-mean series converges.

Suppose ${\sum }_n{a}_n$ diverges. Positivity implies $s_n\to \infty$. Put $t_n=a_n/s_n\in (0,1)$ for $n\ge 2$. Since $s_{n-1}=s_n-a_n$,If ${\sum }_n{t}_n$ converged, then eventually $t_n\le 1/2$ andSumming would bound the telescoping quantity $\log (s_n/s_1)$, contradicting $s_n\to \infty$. ThereforeThe converse is true. If ${\sum }_n{a}_n<\infty$, then $s_n\ge s_1>0$ andTaking the contrapositive shows that divergence of the ratio series forces divergence of ${\sum }_n{a}_n$.