Past exam of the mathematics course of the University of Cambridge / 2001 / ia / Paper 1

Using the Levi-Civita symbol and the contraction of two Levi-Civita symbols, the $i$th component of the vector triple product isThus the triple-product identity is $a\times (b\times c)=(a\cdot c)b-(a\cdot b)c$.

Apply this with the unit vector $n$ to obtain $\Pi (x)=x-(n\cdot x)n$. Dot products and scalar multiplication are linear in $x$, so $\Pi (sx+ty)=s\Pi (x)+t\Pi (y)$. Its matrix representation isThe fixed-vector equation is equivalent to $(n\cdot x)n=0$. Since $n\ne 0$, all its solutions are $x\in n^{\perp }$. Also $\boxed{\Pi (n)=0}$. Every vector decomposes as $x=\Pi (x)+(n\cdot x)n$, with $\Pi (x)$ orthogonal to $n$. Hence $\Pi$ is the orthogonal projection onto the plane through the origin normal to $n$: it fixes that plane and removes the perpendicular component.

A family $x_1,\dots ,x_n$ is linearly independent if ${\sum }_{i=1}^n{a}_i{x}_i=0$ forces $a_1=\cdots =a_n=0$. For the three specified vectors, direct substitution givesThe coefficients are not all zero, so the vectors are linearly dependent.

For the implication question, using the usual interpretation that a basis contains the listed family as independent entries, the answers in order (i)–(x) areThe key distinctions are that the unrestricted zero combination always exists, whereas linear dependence requires a nontrivial combination; and dependence does not necessarily let the specified third vector be solved in terms of the first two. For example $x=y=e_1$, $z=e_2$ makes (iii) and (iv) false, while $x=e_1,y=e_2,z=e_3$ disproves (vi).

There is a set-versus-family qualification in (v). If “contains” is interpreted as literal set containment, repeated nonzero vectors need only occur once in the basis. Then $x=y=e_1,z=e_2$ is dependent as a family but all three belong to the basis $\{e_1,e_2\}$, so (v) becomes false. With three distinct vectors, or with family-containment as above, basis extension gives (v) true. Implication (x) remains true on either reading.

The definition of a convergent sequence isTo prove the product rule, convergence first gives an eventual bound $|u_n|\le |l|+1=:M$. ThenChoose $n$ sufficiently large that $|v_n-k|<ϵ/(2M)$ and $|u_n-l|<ϵ/[2(|k|+1)]$, as well as the bound on $u_n$. The right side is less than $ϵ$, proving $u_n{v}_n\to lk$ directly from the limit of a sequence definition.

If $l\ne 0$, eventually $|u_n-l|<|l|/2$, so the reverse triangle inequality gives $|u_n|>|l|/2$. ConsequentlyThus $1/u_n\to 1/l$. The condition $u_n\ne 0$ ensures the reciprocal sequence is defined at every index; the nonzero limit supplies the eventual uniform bound away from zero. Finally $\boxed{u_n=1/(n+1)\to 0}$ has every term nonzero, showing that this termwise property alone does not prevent a zero limit.

Suppose $f$ is continuous on $[a,b]$, differentiable on $(a,b)$ and satisfies $f(a)=f(b)$, with $a<b$. Apply the extreme value theorem to both $f$ and $-f$ to obtain a maximum and a minimum. If $f$ is constant, every interior derivative is zero. Otherwise some value differs from the common endpoint value, so either a maximum or a minimum occurs at an interior point $c$. At an interior maximum, difference quotients are nonpositive from the right and nonnegative from the left; since the derivative exists, it must be zero. The minimum case is identical with the inequalities reversed. This proves Rolle's theorem: $\boxed{f^{\prime }(c)=0}$ for some $c\in (a,b)$.

For the mean value theorem, subtract the endpoint chord:Then $g(a)=g(b)=0$, so Rolle's theorem supplies $c\in (a,b)$ with $g^{\prime }(c)=0$. HenceIf $f^{\prime }(t)>0$ everywhere, applying the mean value theorem to any $x<y$ gives $f(y)-f(x)=f^{\prime }(c)(y-x)>0$. Thus $f$ is strictly increasing. The converse is false: $\boxed{f(t)=t^3}$ is strictly increasing but $f^{\prime }(0)=0$.

To find the kernel of a linear map, perform the row operations $R_2\leftarrow R_2-R_1$ and $R_3\leftarrow R_3-(1+\alpha )R_1$, followed by $R_3\leftarrow R_3-R_2$. The equivalent homogeneous equations areThe resulting determinant is ${\alpha }^2(3-\alpha )$. If $\alpha \notin \{0,3\}$ all coordinates vanish. At $\alpha =0$ only $x-y+z=0$ remains; at $\alpha =3$ the equations give $y=2z$, $x=-5z$. Therefore kernel bases areThe empty family is the basis of the zero vector subspace.

For the change of basis, let $B_0,C_0$ be the matrices whose columns are the two ordered bases in standard coordinates. Then $v=B_0[v{]}_B=C_0[v{]}_C$, so $[v{]}_B=S[v{]}_C$ with $S=B_0^{-1}{C}_0$. If $A$ and $A^{\prime }$ represent the same linear map in the two bases, $[\Phi (v){]}_B=A[v{]}_B$ and also $[\Phi (v){]}_B=S{A}^{\prime }[v{]}_C$. Therefore $S{A}^{\prime }=AS$, andFor the concrete bases, their vectors satisfy $c_1=b_1+b_2$, $c_2=b_2+b_3$, $c_3=b_1+b_2+b_3$. These coordinate columns giveAt $\alpha =0$, every row of $A_0$ is $(1,-1,1)$, so every row of $A_{0}S$ is $(0,0,1)$. Multiplying on the left by the displayed inverse givesThe original PDF asks for this full product; its final factor $S$ is missing from the converted TeX.

If $A{x}_p=b$, then $Ax=b$ is equivalent to $A(x-x_p)=0$. Thus the solution set is the affine solution space of a linear equation $\boxed{x_p+\ker A}$, and every vector of that form solves the equation.

For the symmetric matrix case, let $A{u}_i={\lambda }_i{u}_i$. Symmetry givesDistinct eigenvalues therefore imply $u_i\cdot u_j=0$. Normalizing the three vectors gives an orthonormal basis. Expand $x={\sum }_i{t}_i{u}_i$ and $b={\sum }_i{b}_i{u}_i$, with $b_i=b\cdot u_i$. The equation $(A-{\lambda }_{k}I)x=b$ becomes $({\lambda }_i-{\lambda }_k)t_i=b_i$ for every $i$. Thus solvability is equivalent to $b_k=0$, and all solutions areThis proves the claimed orthogonality condition and the full solution family rather than only a particular solution.

For the numerical symmetric matrix, direct multiplication gives $A{u}_1=0$ for $u_1=(1,1,1{)}^T/\sqrt{3}$, and $A{u}_2=-\sqrt{2}{u}_2$ for $u_2=(1,-1,0{)}^T/\sqrt{2}$. Their orthogonal complement is spanned by $u_3=(1,1,-2{)}^T/\sqrt{6}$, and multiplication gives $A{u}_3=\sqrt{3}{u}_3$. The supplied right side has $b\cdot u_1=0$, so the linear system is solvable. Its other coefficients are $b\cdot u_2=2$ and $b\cdot u_3=3\sqrt{2}$. HenceEquivalently, all solutions are

Complete the square in the real-valued complex plane equation:Dividing by $\alpha$ gives center and radiusIf the radicand is zero, the locus is the single point $c$, a degenerate circle. If it is negative, the locus is empty. The absolute value of $\alpha$ is essential even when $\alpha <0$.

For the requested sketch, the five centers and radii are respectively $(-1/2,\sqrt{3}/2)$, $(-1,\sqrt{2})$, $(-2,\sqrt{6})$, $(2,\sqrt{2})$ and $(1,0)$, all centers on the real axis.

The original PDF's map is $f(z)=(\beta /\bar{\beta })\bar{z}$; the converted TeX drops the conjugate on its denominator. Write $\beta =|\beta |e^{i\theta }$, so $f(z)=e^{2i\theta }\bar{z}$. This is the reflection symmetry of a complex circle equation. It fixes $c=\beta /\alpha$ and satisfiesThus it maps every point of the circle into the same circle. Conversely, for any point $w$ on that circle, $z=f(w)$ also lies on it and $f(z)=w$. This proves $f(C)=C$, including the surjectivity required by the question.

For any three nonempty circles, fix $u\in C_3$. The triangle inequality gives, for every $z\in C_1,w\in C_2$,Taking the maximum over $z,w$ proves the requested inequality.

For centers $c_1,c_2$ and radii $r_1,r_2$, the maximum separation of two circles is bounded above by $|c_1-c_2|+r_1+r_2$. If the centers differ, choose $z=c_1+r_{1}v$ and $w=c_2-r_{2}v$, where $v=(c_1-c_2)/|c_1-c_2|$, to attain the bound. If the centers coincide, any unit $v$ gives opposite points attaining $r_1+r_2$. ThereforeThe function $m$ is not a metric on positive-radius circles, since $m(C,C)=2r$, even though it satisfies this triangle-type inequality.

The line through the origin is $l_0=\mathrm{span}(u)$, containing zero and closed under addition and scalar multiplication, so it is a vector subspace. If $l_{x_1}=l_{x_2}$, then $x_1\in l_{x_2}$ and $x_1=x_2+\lambda u$ for some real $\lambda$. Conversely that equality gives the same set after shifting the line parameter. ThusChanging representatives to $x^{\prime }=x+su$, $y^{\prime }=y+tu$ changes their sum by $(s+t)u$ and a scalar multiple by $\alpha su$. The line classes therefore do not change, proving both operations well-defined. Associativity follows from $l_{(x+y)+z}=l_{x+(y+z)}$. The zero vector of this quotient vector space is the entire line $l_0$, because $l_x+l_0=l_x$.

For a basis $(u,b_1,b_2)$ of ${\mathbb{R}}^3$, every vector is $x=su+t{b}_1+v{b}_2$, so $l_x=t{l}_{b_1}+v{l}_{b_2}$. If this combination equals $l_0$, then $t{b}_1+v{b}_2=su$; independence of the three original basis vectors forces $t=v=s=0$. Hence $(l_{b_1},l_{b_2})$ is a basis of the space of parallel lines as a quotient vector space.

For the specified direction, add $zu$ to $(x,y,z{)}^T$ to obtain the representative $(x+z,y+3z,0{)}^T$. Thus its coordinates in the two standard line classes are $(x+z,y+3z{)}^T$. The two supplied image classes have coordinates $(1,1{)}^T$ and $(-3,1{)}^T$, while their input coordinates are $(1,-1{)}^T$ and $(1,1{)}^T$. If $a_1,a_2$ are the matrix columns, then $a_1-a_2=(1,1{)}^T$ and $a_1+a_2=(-3,1{)}^T$. Solving gives

Let $M$ bound all $|a_n{w}^n|$ and put $q=|z|/|w|<1$. ThenThe geometric series ${\sum }_{n=0}^{\infty }M{q}^n$ converges, so the comparison test for series proves absolute convergence of $\sum a_n{z}^n$. This is the bounded power-series terms force interior absolute convergence principle.

Define the radius of convergence byallowing $\rho =0$ and $\rho =\infty$. If $|z|<\rho$, the supremum supplies a point of convergence $w$ with $|w|>|z|$; convergence makes its terms bounded, so the argument above gives absolute convergence at $z$. If $|z|>\rho$, convergence is impossible by definition. At $|z|=\rho$, no general conclusion follows. Thus the convergence region has a circular interior centered at zero, with boundary behavior depending on the power series.

For $z\ne 0$,If $R|z|<1$, choose $q$ strictly between that limit and $1$. Eventually $|a_n{z}^n|\le q^n$, proving absolute convergence by the geometric series comparison. If $R|z|>1$, choose $q>1$ below the limit; eventually $|a_n{z}^n|\ge q^n$, so the terms do not tend to zero and the series diverges. The point $z=0$ is harmless. This proves the root test in the needed case and gives

All three examples below have radius of convergence one, because the $n$th roots of their coefficient moduli tend to one.

For convergence everywhere on the boundary, take ${\sum }_{n=1}^{\infty }{z}^n/n^2$. On $|z|=1$ it converges absolutely: for $n\ge 2$, $1/n^2\le 1/[n(n-1)]=1/(n-1)-1/n$, whose sum telescopes to a finite value.

For divergence everywhere on the boundary, take ${\sum }_{n=0}^{\infty }{z}^n$. If $|z|=1$, every term has modulus one, so the term test for divergence applies.

For mixed boundary behavior, take ${\sum }_{n=1}^{\infty }{z}^n/n$. At $z=1$ it is the divergent harmonic series; each successive dyadic block contributes at least $1/2$. At every other point $|z|=1$, summation by parts proves convergence. Indeed put $B_k={\sum }_{j=m}^k{z}^j$, so the finite geometric series gives $|B_k|\le 2/|1-z|=:C_z$. ThenThus the partial sums form a Cauchy sequence and converge. This derives the relevant Dirichlet test rather than only naming it.

To prove the intermediate value theorem in this setting, let $A=\{t\in [a,b]:f(t)<v\}$ and $c=\sup A$. The endpoint inequalities and continuity give an interval immediately to the right of $a$ contained in $A$ and an interval immediately to the left of $b$ disjoint from $A$. Thus $a<c<b$. Choose points of $A$ tending to $c$; continuity gives $f(c)\le v$. If $f(c)<v$, continuity would place some point larger than $c$ in $A$, contradicting the supremum. Hence $\boxed{f(c)=v}$ at an interior point.

For a continuous self-map of $[a,b]$, define $g(t)=f(t)-t$. The range condition implies $g(a)\ge 0$ and $g(b)\le 0$. If either is zero, its endpoint is a fixed point. Otherwise apply the just-proved intermediate value theorem to $-g$ with target zero to obtain $d\in (a,b)$ and $g(d)=0$. ThereforeThis proves the fixed-point property of a closed interval without requiring $f$ itself to be monotone.

The fixed-point property fails on an open interval. The continuous functionmaps $(a,b)$ into itself, because $a<t<f(t)<b$. Its only possible fixed point satisfies $t=(t+b)/2$, hence $t=b$, which is outside the domain. It is therefore a counterexample even with the self-map requirement retained.

The fixed-point property also fails on the real line. The continuous function $\boxed{f(t)=t+1}$ maps $\mathbb{R}$ into itself but has no fixed point, since $t+1=t$ is impossible.

By continuity of $g^{\prime \prime }$ at $\alpha$, for any $ϵ>0$ choose $\delta (ϵ)>0$ such that $|g^{\prime \prime }(s)|\le |g^{\prime \prime }(\alpha )|+ϵ=:M$ whenever $|s-\alpha |<\delta (ϵ)$. Twice using the fundamental theorem of calculus gives the remainder asThe inner difference is ${\int }_{\alpha }^s{g}^{\prime \prime }(r)dr$ and has modulus at most $M|s-\alpha |$. Integrating this bound over the interval between $\alpha$ and $t$ proves, for either sign of $t-\alpha$,This is the local quadratic Taylor remainder estimate. The given $ϵ$ may enter the bound on the second derivative; no stronger differentiability assumption is needed.

The intermediate value theorem gives a root $r\in (a,b)$, unique because $f^{\prime }>0$ makes $f$ strictly increasing. We show that Newton root-finding iteration, started at $b$, always stays to the right of $r$. If $x>r$, the mean value theorem gives $f(x)=f^{\prime }(\xi )(x-r)$ for some $\xi \in (r,x)$. Since $f^{\prime \prime }>0$, $f^{\prime }$ is strictly increasing, so $0<f^{\prime }(\xi )<f^{\prime }(x)$. It follows thatInduction gives $r<x_{n+1}<x_n\le b$, so the bounded monotone sequence theorem supplies a limit $\alpha \ge r$. The continuous positive $f^{\prime }$ is bounded away from zero on $[a,b]$, making $F$ continuous. Passing to the limit in $x_{n+1}=F(x_n)$ gives $\alpha =F(\alpha )$ and hence $f(\alpha )=0$. Uniqueness implies $\boxed{\alpha =r}$.

For the rate, apply the Taylor theorem with Lagrange remainder to $f$ between $r$ and $x_n$:With $e_n=x_n-r>0$, rearrangement givesThus the monotone Newton convergence for a strictly convex function has quadratic convergence. More concretely, if $K={\max }_{[a,b]}{f}^{\prime \prime }/[2{\min }_{[a,b]}{f}^{\prime }]$, then $e_{n+1}\le K{e}_n^2$. Once $K{e}_N<1$, induction gives $e_{N+j}\le K^{-1}(K{e}_N{)}^{2^j}$, exhibiting the repeated squaring of small errors.

For a bounded function on a nondegenerate compact interval $[a,b]$, a partition $P$ has Darboux sumsThe Riemann integrability criterion isEquivalently the supremum of lower sums and infimum of upper sums agree; their common value is the Riemann integral. A bounded nonintegrable example is the Dirichlet function $\boxed{f=1_{\mathbb{Q}}}$. Every positive-length interval contains rational and irrational points, so every lower sum is zero and every upper sum is $b-a>0$.

Suppose the continuous nonnegative function has $f(t_0)>0$ somewhere. Continuity supplies a positive-length interval within $[a,b]$ on which $f\ge f(t_0)/2$; at an endpoint use a one-sided interval. By monotonicity of the Riemann integral, its contribution to the integral is strictly positive, a contradiction. Therefore $\boxed{f\equiv 0}$ under the stated continuous-function assumptions.

For the general integrable case, choose $c\in (a,b)$ and let $f(c)=1$ while $f(t)=0$ for $t\ne c$. This nonnegative indicator function is Riemann integrable with integral zero: isolate $c$ in partition intervals of arbitrarily small total length, making the upper sum arbitrarily small while the lower sum stays zero. Yet $f(c)=1$. This is the single-point changes preserve Riemann integrals phenomenon and supplies the counterexample.

Put $G(t)={\int }_a^{t}f(x)dx$. At an interior point and for $h\ne 0$ small enough,By continuity at $t$, the modulus is bounded by the supremum of $|f(x)-f(t)|$ between $t$ and $t+h$, which tends to zero. Thus the fundamental theorem of calculus givesAt the interval endpoints the same argument supplies the corresponding one-sided derivatives.

For an integrable discontinuous function, the single-point spike from part (i) has $G\equiv 0$, so $G^{\prime }(c)=0\ne f(c)=1$; this disproves the derivative identity even when the primitive is differentiable. To see that differentiability itself may fail, take the step function $f(x)=0$ for $x<c$ and $f(x)=1$ for $x\ge c$. It is Riemann integrable, but its primitive is $G(t)=\max (0,t-c)$ and has left derivative zero and right derivative one at $c$. Hence the full assertion is false for general Riemann-integrable functions.