Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2007 / Paper 1 / Section B

Because $\mathbf{b}\cdot \mathbf{x}$ is a scalar, applying the linear operator again givesThis is the first nontrivial case of the powers of a rank-one operator; its matrix is the outer product $\mathbf{a}{\mathbf{b}}^T$.

The vector triple product identity and $\mathbf{n}\cdot \mathbf{n}=1$ giveApply $H$ once more. The term parallel to $\mathbf{n}$ vanishes under its cross product with $\mathbf{n}$, leaving $H^3\mathbf{x}=-{\alpha }^3\mathbf{n}\times \mathbf{x}=-{\alpha }^{2}H\mathbf{x}$. The result includes $\alpha =0$.

The displayed power series is the matrix exponential $G=e^H$. From $H^3=-{\alpha }^{2}H$, its odd powers and nonconstant even powers sum separately:The scalar power series are $\sin \alpha /\alpha$ and $(1-\cos \alpha )/{\alpha }^2$, with continuous values $1$ and $1/2$ at zero. Substituting $H\mathbf{x}=\alpha \mathbf{n}\times \mathbf{x}$ and the preceding formula for $H^2\mathbf{x}$ yieldsThis is the Rodrigues rotation formula: the component parallel to $\mathbf{n}$ is fixed, while the perpendicular component undergoes a rotation through $\alpha$. For $\alpha =0$ the linear operator is the identity operator.

Using the principal complex logarithm, $\mathrm{Log}i=i\pi /2$ and $\mathrm{Log}(-e)=1+i\pi$. Thus the values and their real parts and imaginary parts areFor the final row, first calculate $i^i=e^{-\pi /2}$ and only then raise this positive real number to the power $i$. Treating nested powers as ordinary real powers would ignore the branching of complex exponentiation.

If all branches are requested, $\sqrt{i}$ also has the negative of its principal value, and the complex logarithm of $-e$ is $1+i(\pi +2\pi k)$. Allowing nonprincipal logarithms of the positive base $2$ gives $e^{-2\pi k}{e}^{i\ln 2}$. Finally, all values of $(i^i{)}^i$ are $-i{e}^{-2\pi m}$, for integers $k,m$ as appropriate. These alternatives do not alter the principal-value table.

Euler's formula givesCubing the expression for the sine, and grouping the $e^{3ix}$ and $e^{-3ix}$ terms, givesThe required equation therefore becomes $\sin x(1-4{\sin }^{2}x)=0$. The complete real solution set is

The quadratic formula gives $z=-b\pm \sqrt{b^2-1}$. For $-1\le b\le 1$, write this asEach root of a polynomial has modulus $1$, since $b^2+(1-b^2)=1$. As $b$ increases from $-1$ to $1$, the upper root traverses the upper unit semicircle from $1$ to $-1$, and the lower root traverses the lower semicircle in the opposite angular direction. At $b=-1$ the roots coincide at $1$; at $b=1$ they coincide at $-1$. For intermediate $b$ they are complex conjugates.

Since the derivative of the arctangent is $1/(1+x^2)=1-x^2+O(x^4)$ near zero, termwise integration givesThe integration constant is zero because $\arctan 0=0$.

The expression in the PDF is $1/a^x-1$, with the subtraction outside the fraction. Using the exponential series with $a^{-x}=e^{-x\ln a}$ givesFor $a>0$, $a\ne 1$, these are the first two nonzero terms. If $a=1$, the function is identically zero, so there are no nonvanishing terms.

The square is on the cosine: near zero, $\ln ({\cos }^{2}x)=2\ln (\cos x)$. Substituting $\cos x=1-x^2/2+x^4/24+O(x^6)$ into the logarithmic series givesTherefore

Since $g(0)=0$, only the first three positive powers of $g$ contribute through degree three. The composition coefficients of power series follow by using $g^2=g_1^2{x}^2+2g_1{g}_2{x}^3+O(x^4)$ and $g^3=g_1^3{x}^3+O(x^4)$ in $f_0+f_{1}g+f_2{g}^2+f_3{g}^3+\cdots$. HenceHere $f_m$ and $g_m$ are coefficients of the power series, rather than raw derivatives; their definitions already contain any factorial factors.

The exponential series proves normalization of a probability distribution:Let $M$ and $F$ count the two labels. Conditional on $N=M+F=k+l$, independent labeling gives a binomial distribution, so the joint probability mass function isEach bracket is a normalized Poisson distribution. Summing over $l\ge 0$ therefore givesThe factorization also proves that $M$ and $F$ are independent random variables. This is Poisson thinning; the phrase about ignoring the other count refers to taking a marginal distribution, rather than fixing that count.

Let $D$, $C$, and $V_0$ denote the three disjoint categories, and $+$ or $-$ the test result. The law of total probability givesTo calculate the requested conditional probability, first combine the two categories with the virus. Their joint negative-test probability is $0.1(0.05)+0.2(0.1)=0.025$, whereas $\mathbb{P}(-)=1-0.31=0.69$. Bayes' theorem therefore givesThese numbers use the specified population proportions and test probabilities; the negative result reduces the initial combined proportion $0.3$ to about $3.62\%$ within this model.

Writing the differential one-form as $Mdx+Ndy$, its coefficients satisfy $M_y=e^y\cos x=N_x$. It is an exact differential form, with potential $F=(1+e^y)\sin x$, becauseThe implicit solution is therefore

On the real domain $y>0$, $M=y(x^2+\ln y)$ and $N=x$ have $M_y=x^2+\ln y+1$ and $N_x=1$, so this differential one-form is not exact. An integrating factor for a differential one-form is $\mu =1/y$, which is nonzero on this domain. Multiplication givesHence the implicit solution is

On the interval $x>-1$ containing the initial condition, division by $1+x$ gives a linear ordinary differential equation with integrating factor $(1+x{)}^{-3}$. ConsequentlyThe initial value gives $C=1$. ThusThis polynomial also satisfies the original equation at $x=-1$ and gives an extension to all real $x$. The initial value problem is uniquely determined on $x>-1$; the vanishing leading coefficient at $-1$ does not ensure unique continuation to its left.

Put $S={\theta }_1+{\theta }_2$ and $D={\theta }_1-{\theta }_2$. The gradient has components $3\sin S+\sin D$ and $3\sin S-\sin D$, so a stationary point requires $\sin S=\sin D=0$. In the closed square, $S\in [0,2\pi ]$ and $D\in [-\pi ,\pi ]$, and the compatible possibilities are precisely the four corners and $(\pi /2,\pi /2)$.

The Hessian matrix isIts eigenvectors $(1,1)$ and $(1,-1)$ have eigenvalues $6\cos S$ and $2\cos D$. The second-derivative test gives the following classification:

The corner classifications also hold for the smooth extension beyond the square, and hence for the restricted region. The global bounds $-4\le U\le 4$ show that the stated minima and maxima are global within it. The central point has opposite curvature along the two diagonal directions.

The region is $0\le y\le 1$, $2y\le x\le 2$. Reversing the order describes the same triangle as $0\le x\le 2$, $0\le y\le x/2$. Since the integrand is continuous on this compact region, Fubini's theorem permitsThe new inner integral introduces the factor $x$ needed for the change of variables $u=x^2$.

In cylindrical coordinates the spherical cap has $1<z<\sqrt{2}$, $0<r<\sqrt{2-z^2}$ and $0<\theta <2\pi$. Its mass density is $kz$ and the volume element is $rdrd\theta dz$. ThereforeThis is the mass of a spherical cap with height-proportional density with $R=\sqrt{2}$ and $c=1$.

For its curved surface area, write $z=\sqrt{2-r^2}$ with $0\le r\le 1$. The surface integral of one becomesIt agrees with the curved area of a spherical cap, $2\pi Rh$, for $h=\sqrt{2}-1$. The planar closing disk is not part of the requested curved surface.

False. Take the diagonal matrix $A=\mathrm{diag}(1,2)$ and its eigenvectors $\mathbf{x}=(1,0{)}^T$, $\mathbf{y}=(0,1{)}^T$. Their sum $(1,1{)}^T$ is mapped to $(1,2{)}^T$, which is not a scalar multiple of that sum. More generally, $A(\mathbf{x}+\mathbf{y})=a\mathbf{x}+c\mathbf{y}$ need not have the form $\lambda (\mathbf{x}+\mathbf{y})$ when $a\ne c$.

True. By linearity, $(A+B)\mathbf{x}=A\mathbf{x}+B\mathbf{x}=(a+b)\mathbf{x}$. Since $\mathbf{x}\ne 0$, it is an eigenvector of $A+B$ with eigenvalue

True. Applying the matrices in order gives $AB\mathbf{x}=A(b\mathbf{x})=ab\mathbf{x}$. The eigenvalue isThis conclusion only uses the shared eigenvector; it does not require $A$ and $B$ to commute on other vectors.

If $A\mathbf{v}=\lambda \mathbf{v}$ for a nonzero eigenvector, then $A^2\mathbf{v}={\lambda }^2\mathbf{v}=A\mathbf{v}=\lambda \mathbf{v}$. Thus $\lambda (\lambda -1)=0$, soBoth values need not occur: the zero matrix has only $0$, and the identity matrix only $1$. More precisely, an idempotent linear map whose rank of a linear map is $r$ has $r$ copies of $1$ and $n-r$ copies of $0$. Indeed, its image of a linear map and kernel of a linear map form a direct sum, giving an eigenbasis without assuming an orthogonal projection.

On an eigenvector of $A$ with eigenvalue $\lambda$, the linear operator $B=I-A$ has eigenvalue $1-\lambda$. ThereforeThe multiplicities are exchanged: if the matrix rank of $A$ is $r$, then $B$ has $n-r$ copies of $1$ and $r$ copies of $0$. Again, a value with zero multiplicity is absent.

Computing the characteristic polynomial with the convention $\det (\lambda I-H)$ givesThus the three eigenvalues areThey can also be checked directly on the respective eigenvectors $(1,0,-1{)}^T$, $(1,-1,1{)}^T$, and $(1,2,1{)}^T$.

For the lowest eigenvalue, $-1$, solving $(H+I)\mathbf{v}=0$ gives $v_2=0$ and $v_3=-v_1$. A unit eigenvector is thereforeAny nonzero scalar multiple is an equally valid eigenvector.

For $x\in (n,n+1)$, strict decrease gives $f(x)<f(n)$, while for $x\in (n-1,n)$ it gives $f(x)>f(n)$. Integrating over these unit intervals provesSumming the rectangle bounds givesThe partial sums and the improper integral are nondecreasing. These bounds show that one is bounded exactly when the other is bounded, proving the integral test for convergence. In the convergent case, sum the same interval inequalities for $n=N+1,N+2,\dots$ to obtainFor $f(x)=1/(x\ln x)$, the antiderivative is $\ln (\ln x)$, which grows without bound, so the first series diverges. For $f(x)=1/[x(\ln x{)}^2]$, the antiderivative is $-1/\ln x$, whose tail integral is finite, so the second series converges, withBoth functions are positive and strictly decreasing on $[2,\infty )$, as required.

The ratio test uses $L={\lim }_{n\to \infty }|a_{n+1}/a_n|$, when it exists for eventually nonzero terms: $L<1$ gives absolute convergence, $L>1$ gives divergence, and $L=1$ is inconclusive.

For $a_n=n/(2^n-1)$,Thus the first series converges absolutely. For the alternating second series,so it diverges. Alternating signs cannot rescue terms whose absolute values fail to tend to zero.

For $x>0$ write $f(x)=x^{-a}I(x)$, where $I(x)={\int }_0^x{e}^{-t}{t}^{a-1}dt$. Because $a>0$, the lower endpoint is integrable. At each positive upper endpoint the fundamental theorem of calculus gives $I^{\prime }(x)=e^{-x}{x}^{a-1}$. The product rule therefore yieldsDifferentiating once more and substituting this expression for $f^{\prime }$ givesSubstitution now cancels the $e^{-x}$ terms and the terms proportional to $f$, proving $x{f}^{\prime \prime }+(1+a+x)f^{\prime }+af=0$.

The change of variables $t=xu$ identifies $f$ as the scaled lower incomplete gamma integral ${\int }_0^1{e}^{-xu}{u}^{a-1}du$. It has a smooth extension at zero, with $f(0)=1/a$, $f^{\prime }(0)=-1/(a+1)$ and $f^{\prime \prime }(0)=1/(a+2)$. Thus the apparent singularities in the derivative formulas cancel there; those formulas with denominators $x$ were derived for $x>0$.

The kernel has an integrable singularity at the moving endpoint $t=x$. Direct substitution into the ordinary Leibniz integral rule would produce the undefined boundary value $h(x)/\sqrt{x-x}$ and the nonintegrable differentiated kernel $-h(t)/[2(x-t{)}^{3/2}]$. Those separately divergent expressions cannot be treated as ordinary finite terms. The derivative of an Abel integral is obtained by rewriting the integral before differentiating.

Assume $h$ is continuously differentiable on the interval concerned. For fixed $x>0$, take $u=h(t)$ and $dv=(x-t{)}^{-1/2}dt$, so $v=-2\sqrt{x-t}$. Integration by parts, interpreted first with an endpoint cutoff and then by a limit, givesThe upper endpoint term vanishes, and the lower endpoint supplies the explicit square-root term.

Differentiate the rewritten expression. Its new kernel vanishes at the moving upper endpoint, and its partial derivative is an integrable inverse square root. The derivative of an Abel integral isFor a rigorous route, rewrite $2{\int }_0^x{h}^{\prime }(t)\sqrt{x-t}dt$ as ${\int }_0^x{\int }_0^s{h}^{\prime }(t)/\sqrt{s-t}dtds$ using Fubini's theorem, and apply the fundamental theorem of calculus to the continuous inner integral. The assumption $h(0)\ne 0$ implies $g(x)=2h(0)\sqrt{x}+O(x^{3/2})$, so the extension $g(0)=0$ has no finite right derivative at zero.