Past exam of the mathematics course of the University of Cambridge / 2025 / ia / Paper 1

Translate and scale the circle to $|z|=1$. For distinct points $u,v,w$ on it, the angle at $w$ is right exactly whenis purely imaginary. Since $\bar{u}=u^{-1}$ and similarly for $v,w$, direct simplification gives $\bar{R}=(v/u)R$. Thus $\bar{R}=-R$ exactly when $v=-u$, which says that the opposite side joins antipodal points and is a diameter. The converse is the same calculation in reverse.

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The vertices satisfyAfter removing the root $z=0$, the product of the other roots has modulus equal to the constant term of $((z+1{)}^N-1)/z$, namely $N$. Hence the product of the $N-1$ chord lengths from one vertex of a unit regular polygon is $N$. Multiplying this identity over all $N$ vertices counts every chord twice, so the product of all chord lengths is $N^{N/2}$. Scaling the circle by $R$ scales each of its $N(N-1)/2$ chords by $R$, giving

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Expansion gives $\det A=(a-3)(a+2)$. Thus uniqueness fails precisely for $a=-2$ and $a=3$.

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Solvability requires $b$ to be orthogonal to the left nullspace. For $a=-2$ one may take $n=(-3,-2,2{)}^T$; for $a=3$ one may take $n=(1,-1,1{)}^T$.

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For $b=(2,b,0{)}^T$, compatibility gives $b=-3$ when $a=-2$ and $b=2$ when $a=3$. The respective solution families areandwhere $t\in \mathbb{R}$.

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For every $z\in \mathbb{C}$,

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Termwise differentiation givesThe differentiated series has the same radius of convergence $R$.

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Fix $a$ and define $g(b)=e^{a+b}{e}^{-b}$ using only the power series. Termwise differentiation gives $(e^z)\prime =e^z$, and the product rule gives $g\prime (b)=0$. Hence $g$ is constant, so $g(b)=g(0)=e^a$. Multiplication by $e^b$ yields $e^{a+b}=e^a{e}^b$.

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The removable definition gives the entire seriesTherefore

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The Darboux–Riemann criterion says that a bounded function on $[a,b]$ is Riemann integrable exactly when, for every $\epsilon >0$, some partition $P$ satisfies $U(f,P)-L(f,P)<\epsilon$. Indeed, every lower sum is at most every upper sum. Taking the supremum of lower sums and infimum of upper sums, the criterion makes their difference smaller than every positive $\epsilon$, so they are equal; this common value is the Riemann integral.

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Given $\epsilon >0$, choose a partition of $[a,d]$ whose upper-minus-lower sum is below $\epsilon$, and refine it by inserting $b,c$. The contribution from subintervals lying in $[b,c]$ is nonnegative and no larger than the total difference. Restricting the refined partition to $[b,c]$ therefore gives upper-minus-lower sum below $\epsilon$, so the criterion proves integrability there.

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Taking the parallelogram spanned by $e_2,e_3$ as base gives base area $|e_2\times e_3|$ and height $|e_1\cdot (e_2\times e_3)|/|e_2\times e_3|$. A tetrahedron has one third of the corresponding pyramid volume and the triangle has half the parallelogram base, henceThe reciprocal vectors arewhich directly satisfy $f_i\cdot e_j={\delta }_{ij}$.

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The final face contains $v_i=v+e_i$, so $c\cdot (v+e_i)+d=0$. Therefore $e_i\cdot c=-(c\cdot v+d)$.

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The $i$th face through $v$ contains the two edge directions $e_j$ with $j\ne i$, so $a_i\cdot e_j=0$ for those $j$. The one-dimensional space with these two orthogonality conditions is spanned by $f_i$, hence $a_i={\lambda }_i{f}_i$ for some nonzero real ${\lambda }_i$.

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Dotting $c={\sum }_j{\gamma }_j{a}_j$ with $e_i$ gives $e_i\cdot c={\gamma }_i{\lambda }_i$, proving ${\gamma }_i=(e_i\cdot c)/{\lambda }_i$. AlsoCombining these with $V=|\Delta |/6$ gives the displayed face formula, with the orientation chosen so its signed right-hand side is positive.

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The distinguished unit vector is $n=(1,1,1)/\sqrt{3}$. Comparing diagonal, symmetric off-diagonal, and antisymmetric parts givesso $A_{ij}=\alpha {\delta }_{ij}+\beta n_i{n}_j+\gamma {\epsilon }_{ijk}{n}_k$.

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On the line spanned by $n$, $A$ multiplies by $L=\alpha +\beta =a+b+c$. On $n^{\perp }$, it is the composition of a rotation with a dilation by $r=\sqrt{{\alpha }^2+{\gamma }^2}$. Consequently every area in that plane is multiplied by $r^2={\alpha }^2+{\gamma }^2$.

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For a plane reflection, the $n$ direction is reversed while $n^{\perp }$ is fixed. Thus $L=-1$, $\alpha =1$, and $\gamma =0$, equivalently

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The map is a rotation about the $n$ axis exactly whenor explicitlyThese conditions make the axis fixed and the perpendicular-plane action length preserving; they are also necessary.

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Let $N=n{n}^T$ and $K_{ij}={\epsilon }_{ijk}{n}_k$. Since $K^2=N-I$, inversion separately on $\mathbb{R}n$ and $n^{\perp }$ givesprovided $(\alpha +\beta )({\alpha }^2+{\gamma }^2)\ne 0$.

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The line/plane decomposition givesSubstituting the values in part (a) and simplifying yields

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An eigenvalue $\lambda$ is a scalar for which $\ker (A-\lambda I)$ contains a nonzero vector; that kernel is its eigenspace. The characteristic polynomial has degree $n$, so the fundamental theorem of algebra supplies at least one complex root and hence an eigenvalue. Its eigenspace dimension can be any integer from $1$ to $n$.

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With ${\chi }_A(t)=\det (tI-A)$, direct expansion givesThus the eigenvalues are $3,1,-1,-2$. Kernel calculation gives eigenspace dimensions $1,2,1,1$, respectively.

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The eigenvalue equation is $u_{i+1}=\lambda u_i$, soThis is a nonzero square-summable vector exactly when $c\ne 0$ and $|\lambda |<1$. Hence every point of the open unit disc is an eigenvalue, with the displayed one-dimensional eigenspace.

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For the transpose shift, $(Cu{)}_1=0$ and $(Cu{)}_i=u_{i-1}$ for $i\ge 2$. If $Cu=\lambda u$, the first equation and the subsequent recurrence force every component to vanish, both for $\lambda =0$ and for $\lambda \ne 0$. Thus $C$ has no eigenvalues, illustrating that a bounded infinite-dimensional operator need not have one.

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A complex matrix is diagonalisable when it is similar to a diagonal matrix, equivalently when the vector space has a basis of its eigenvectors.

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If $Av=\lambda v$, then $p(A)v=p(\lambda )v$, proving one inclusion. Conversely, if $\mu$ is an eigenvalue of $p(A)$, factor $p(z)-\mu =c{\prod }_j(z-{\lambda }_j)$. If none of the ${\lambda }_j$ were eigenvalues of $A$, every $A-{\lambda }_{j}I$ would be invertible, making $p(A)-\mu I$ invertible, a contradiction. Thus $\mu =p(\lambda )$ for some eigenvalue $\lambda$ of $A$.

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Write $A=S\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_n)S^{-1}$. The power series gives $e^A=S\mathrm{diag}(e^{{\lambda }_1},\dots ,e^{{\lambda }_n})S^{-1}$. Therefore

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The vectors $(1,0,1{)}^T,(0,1,0{)}^T,(1,0,-1{)}^T$ are eigenvectors of $B$ with eigenvalues $2,1,0$. Applying the exponential to these eigenspaces gives

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A sequence $(a_n)$ is Cauchy when, for every $\epsilon >0$, some $N$ satisfies $|a_m-a_n|<\epsilon$ whenever $m,n\ge N$.

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The general principle of convergence says that a real sequence converges exactly when it is Cauchy. Convergent sequences are Cauchy by the triangle inequality. Conversely, a Cauchy sequence is bounded, so Bolzano–Weierstrass gives a convergent subsequence $a_{n_k}\to a$; the Cauchy property then forces the entire sequence to converge to $a$.

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True. For $m>n$, $|a_m-a_n|\le {\sum }_{j=n}^{m-1}{d}_j$. Convergence of $\sum d_j$ makes this tail arbitrarily small, so $(a_n)$ is Cauchy and hence convergent.

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False. The sequence $a_n=(-1{)}^n/n$ converges to zero, but $|a_{n+1}-a_n|=1/n+1/(n+1)$, whose series diverges.

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Writeuniformly for $j\ge 1$. The accumulated error from $j=n+1$ to $2n$ tends to zero, while the corresponding harmonic sum tends to $\log 2$. Hence $b_n\to \log 2$.

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The statement ${\lim }_{t\to x}f(t)=\ell$ means that for every $\epsilon >0$ there is $\delta >0$ such that $t\in [0,1]$ and $0<|t-x|<\delta$ imply $|f(t)-\ell |<\epsilon$.

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The function is continuous at $x$ exactly when ${\lim }_{t\to x}f(t)=f(x)$.

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If $x_n\to x$, continuity of $f$ gives $f(x_n)\to f(x)$, and continuity of $g$ at $f(x)$ then gives $g(f(x_n))\to g(f(x))$. The sequential criterion proves continuity of $g\circ f$ at $x$.

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Both functions are continuous at every $x>0$. At zero, $f_1(1/(\pi /2+2\pi n))=1$ while $f_1(1/(3\pi /2+2\pi n))=-1$, so $f_1$ is discontinuous. Since $|x\sin (1/x)|\le x\to 0$, $f_2$ is continuous at zero and hence everywhere.

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Let $S=\{0\}\cup \{1/n:n\ge 1\}$ and let $D$ be the Dirichlet function, equal to $1$ on rationals and $0$ on irrationals. Then $f(x)=d(x,S)D(x)$ is continuous exactly on $S$: the distance factor squeezes it to zero on $S$, while away from $S$ it is a positive continuous factor times an everywhere-discontinuous function.

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Using the preceding $f$, define $g(0)=f(0)+1$ and $g(x)=f(x)$ for $x>0$. Changing the value only at zero makes zero discontinuous while preserving continuity at every $1/n$ and discontinuity everywhere else. Thus the continuity set is precisely $\{1/n:n\ge 1\}$.

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Differentiability at $x$ means that the limitexists as a finite real number.

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If $f$ is continuous on $[x,y]$ and differentiable on $(x,y)$, then some $c\in (x,y)$ satisfies $f(y)-f(x)=f\prime (c)(y-x)$.

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If $f\prime \ge 0$, the mean value theorem gives $f(y)-f(x)=f\prime (c)(y-x)\ge 0$. Conversely, if $f$ is increasing, every difference quotient with positive or negative increment is nonnegative; taking its limit gives $f\prime (x)\ge 0$.

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For $x>0$, apply the mean value theorem to $[0,x]$: $f(x)/x=f\prime (c)$ for some $c<x$. Since $f\prime$ is increasing, $f(x)/x\le f\prime (x)$. Thereforeand part (c) proves that the quotient is increasing.

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For continuous $f$, the function $F(x)={\int }_a^{x}f(t)dt$ is differentiable with $F\prime =f$. Conversely, if $F\prime$ is continuous, then ${\int }_a^{b}F\prime (t)dt=F(b)-F(a)$.

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The fundamental theorem giveswhich is the desired inequality.

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The assumptions imply $f\ge 0$. Since $(f^2)\prime =2ff\prime \le 2f$ and both sides vanish at zero, integration gives $f(x{)}^2\le 2F(x)$ where $F(x)={\int }_0^{x}f$. NowThe bracket vanishes at zero, proving ${\int }_0^x{f}^3\le ({\int }_0^{x}f{)}^2$.

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No. Take the constant function $f(t)=c>0$, which has $f\prime =0$. The claimed inequality becomes $c^{3}x\le c^2{x}^2$, false whenever $0<x<c$.

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