Past exam of the mathematics course of the University of Cambridge / 2021 / ia

By complex exponentiation, a value of $z^i$ has the formFor this to equal one, its exponent must be $2\pi in$ for some $n\in \mathbb{Z}$. Hence a logarithm of $z$ must equal $2\pi n$, so $z$ is positive real andWith the principal branch this is the same list; in the multivalued convention these are precisely the $z$ for which one value of $z^i$ equals one.

The circle isThe direction from its center $c=5i$ to $z_0=15/4$ isso the corresponding unit direction is $3/5-4i/5$. Moving a distance five from the center in the two directions givesThus the intersections are

Let $P_B,P_{B^{\prime }},P_C,P_{C^{\prime }}$ have the corresponding basis vectors as columns in standard coordinates. From the PDF,The matrix representation of a linear map in standard coordinates is $P_{C}A{P}_B^{-1}$. The change of basis formula therefore givesMultiplication yields

The alternating series test says that if $a_n\ge 0$, $a_{n+1}\le a_n$, and $a_n\to 0$, then ${\sum }_{n\ge 1}(-1{)}^{n+1}{a}_n$ converges.

To prove it, let $S_N$ be the partial sums. The even sums satisfyso $(S_{2m})$ is increasing. The odd sums satisfyso $(S_{2m+1})$ is decreasing. Also $S_{2m}\le S_{2m+1}$, so both are bounded and converge. Their difference is $a_{2m+1}\to 0$, hence their limits agree and the whole sequence $(S_N)$ converges.

Taking $a_n=1/n$ proves convergence of the alternating harmonic series. Its even partial sums lie below its limit $S$, while its odd partial sums lie above it. Sinceandwe obtain

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

For a proof, place all terms in a closed bounded interval $I_1$. Bisect it and choose a closed half $I_2$ containing infinitely many terms. Continue inductively, choosing nested closed intervalswith infinitely many sequence terms and with lengths tending to zero. Choose $n_k>n_{k-1}$ such that $x_{n_k}\in I_k$. The nested interval theorem gives a unique point $x$ in every $I_k$. Because both $x_{n_k}$ and $x$ lie in $I_k$,Thus $(x_{n_k})$ is a convergent subsequence.

Now suppose every convergent subsequence of the bounded sequence $(x_n)$ converges to $L$. If $x_n$ did not converge to $L$, there would be an $ϵ>0$ and a subsequence $(x_{n_k})$ satisfyingfor every $k$. This subsequence is bounded, so Bolzano--Weierstrass gives a convergent subsubsequence. By hypothesis its limit is $L$, contradicting the displayed inequality. Hence the unique subsequential limit of a bounded sequence principle gives

Using Einstein notation andwe obtainThus

The scalar triple product isPut $\Delta =[a,b,c]$. When $\Delta \ne 0$, the three vectorsform the reciprocal basis to $a,b,c$, so $e_i^{\prime }\cdot e_j={\delta }_{ij}$. Their scalar triple product is $1/\Delta$. Thereforesince the cyclic permutation preserves orientation. The identity also holds when $\Delta =0$, either by continuity or directly because the cross products are then linearly dependent.

For the given basis $e_1,e_2,e_3$, the same identities giveIf ${\sum }_i{\alpha }_i{e}_i^{\prime }=0$, dotting with $e_j$ gives ${\alpha }_j=0$, so the $e_i^{\prime }$ are linearly independent and hence form a basis. Moreover, $e_i\cdot e_j^{\prime }={\delta }_{ij}$, so the original basis is reciprocal to the primed basis. Uniqueness of a reciprocal basis gives

Every vector $K$ has a unique expansion $K={\sum }_i{\kappa }_i{e}_i^{\prime }$, and thenThis is an integer for every integer triple $(n_1,n_2,n_3)$ exactly when every ${\kappa }_i$ is an integer. Hence all such points areThey form the reciprocal lattice in the convention without a factor of $2\pi$.

Using the convention of the question, the characteristic polynomial and characteristic equation areForone hasDirect multiplication givesentry by entry. Thus ${\chi }_A(A)=0$, verifying the Cayley-Hamilton theorem for $2\times 2$ matrices.

If $A_{ji}$ denotes the matrix obtained by deleting row $j$ and column $i$, then the adjugate matrix isFor nonsingular $A$,Therefore, when $A$ and $B$ are nonsingular,

The determinant $\det (M-tI)$ is a polynomial in $t$ with leading term $(-t{)}^n$, so it is not the zero polynomial and has only finitely many roots. Hence one may choose $\alpha >0$ smaller than every positive root, if any. Then

Each entry of $\mathrm{adj}(M-tI)$ is, up to sign, the determinant of an $(n-1)\times (n-1)$ minor whose entries are affine polynomials in $t$. The Leibniz formula for determinants therefore makes each adjugate entry a polynomial in $t$ of degree at most $n-1$.

For arbitrary $A,B$, choose positive sequences $s_j,t_j\to 0$ for which $A-s_{j}I$ and $B-t_{j}I$ are nonsingular. The nonsingular identity from part (b) applies to their product:Every entry is a polynomial, hence continuous, in the matrix entries. Letting $j\to \infty$ provesfor all square matrices.

DefineThe Cayley-Hamilton theorem forgivesIf $A$ is nonsingular, multiplication by $A^{-1}$ shows $P(A)=\det (A)A^{-1}=\mathrm{adj}(A)$.

For arbitrary $A$, apply the nonsingular result to $A-tI$ for a sequence of nonzero $t\to 0$ avoiding the finitely many singular values. Both the characteristic coefficients and the adjugate entries depend polynomially on the matrix entries, so the limit gives

By the real spectral theorem, a real symmetric matrix has an orthogonal diagonalizationWriting $y=Q^{T}x$ givesThis is nonnegative for every $x$ exactly when every ${\lambda }_i\ge 0$. Thus the quadratic-form and eigenvalue definitions of a positive semidefinite matrix agree.

When $A$ is positive semidefinite, defineThis matrix is symmetric and positive semidefinite, and its square is $A$, so it is the principal square root of a positive semidefinite matrix.

For nonsingular $M$, the matrix $M^{T}M$ is symmetric andfor every nonzero $x$. Thus $M^{T}M$ is positive definite, andexists and is nonsingular. Let $R=M{P}^{-1}$. Since $P$ is symmetric,Hence $R$ is an orthogonal matrix andthe polar decomposition of an invertible real matrix.

In three dimensions, $P$ stretches or contracts along three mutually perpendicular eigenvector directions by its positive eigenvalues. The orthogonal map $R$ is then applied: it is a rotation when $\det M>0$, and a rotation combined with a reflection when $\det M<0$.

The determinant factors asThus for $\mu \ne \pm 1$, the kernel is $\{0\}$ and the inhomogeneous system has exactly one solution.

At $\mu =-1$, row reduction givesand the augmented matrix has the same rank two as $A$, so there are infinitely many solutions. At $\mu =1$,but the augmented matrix has rank three while $A$ has rank two, so there is no solution. Therefore the number of solutions is

Writing $n$ as a column vector, the reflection isso its reflection matrix is

Choose the $z$-axis along the line of intersection of the two planes. Their unit normals lie in the $xy$-plane; after choosing the $x$-axis, takewhere $\theta$ is the oriented angle between the planes. The two reflection matrices restrict toon the $xy$-plane and both fix the $z$-axis. Their product isThus the composition of two plane reflections is a rotation about the intersection line through angle $2\theta$.

A plane reflection has determinant $-1$, while a rotation matrix has determinant $+1$. The determinant of their product, in either order, is therefore $-1$. Since every rotation has determinant $+1$, such a product can never be another rotation.

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 $f(c)=y$ for some $c\in [a,b]$.

Let $x_1<x_2<x_3$. If, for example, $f(x_2)$ exceeded both $f(x_1)$ and $f(x_3)$, choose $y$ strictly between $f(x_2)$ and $\max \{f(x_1),f(x_3)\}$. Applying the theorem on both $[x_1,x_2]$ and $[x_2,x_3]$ would give two distinct preimages of $y$, contradicting injectivity. The analogous argument excludes $f(x_2)$ below both endpoint values. Since the three values are distinct,

Fix $a<b$. If $f(a)<f(b)$, applying the displayed betweenness property to triples containing $a,b$ forces the same increasing order for every pair $x<y$; an order reversal would create a triple whose middle value is not between the other two. Thus $f$ is strictly increasing. If $f(a)>f(b)$, the same argument shows that it is strictly decreasing. This proves that every continuous bijection of the real line is monotone.

True. If $x_n\to x$, continuity of $g$ gives $g(x_n)\to g(x)$, and continuity of $f$ then givesEquivalently, this is the composition of continuous functions theorem.

This can be false because a strictly increasing $g$ need not be surjective. TakeThen $g$ is strictly increasing and $f\circ g=0$ is continuous, while $f$ is discontinuous at zero.

True. Part (a) shows that $f$ is strictly monotone. Suppose it is increasing and let $y_0=f(x_0)$. Given $ϵ>0$,Choose $\delta >0$ smaller than both distances from $y_0$ to the two outer values. If $|y-y_0|<\delta$, monotonicity forcesThus $f^{-1}$ is continuous at $y_0$. The decreasing case is identical with inequalities reversed.

This can be false. The functionis a differentiable bijection of $\mathbb{R}$, buthas an unbounded difference quotient at zero and is not differentiable there. The inverse function theorem requires the additional local condition $f^{\prime }(x)\ne 0$.

Pointwise on $[a,b]$,because $g(x)>0$. To see the implication directly from the definition, every Riemann sum of the nonnegative continuous function $(f-m)g$ is nonnegative, so its limit is nonnegative:Applying the same argument to $(M-f)g$ gives ${\int }_a^b(M-f)g\ge 0$. By linearity, these are exactly the monotonicity of the Riemann integral bounds

LetContinuity at zero gives $m_n,M_n\to f(0)$. Part (a), with the positive weight $n{e}^{-nx}$, givesThe squeeze theorem therefore yields

Since $f$ is bounded on $[0,1]$, say $|f|\le K$, the omitted tail satisfiesAdding the tail proves

The Taylor theorem with Lagrange remainder states that if $f$ is $n$ times differentiable between $a$ and $x$, then some $\xi$ between $a$ and $x$ satisfies

Let $T_{n-1}$ denote the displayed polynomial and choose $K$ so thatsatisfies $F(x)=0$. By construction,Starting with the two zeros $a,x$ and applying Rolle theorem repeatedly, there is a $\xi$ between them with $F^{(n)}(\xi )=0$. Sincewe have $K=f^{(n)}(\xi )/n!$, proving the theorem.

Repeatedly differentiating $f^{\prime \prime }=-f$ and using $f(0)=1$, $f^{\prime }(0)=0$ givesAlso,so $f^2+(f^{\prime }{)}^2=1$. Every derivative is one of $\pm f,\pm f^{\prime }$, and hence has absolute value at most one.