Past exam of the mathematics course of the University of Cambridge / 2009 / ib / Paper 1

Let $r=\dim \mathrm{im}T$. Since the kernel of a linear map and its image agree, the rank-nullity theorem gives $\dim V=r+r$. Thus $\dim V=2r$ is even. Every $Tv$ lies in $\ker T$, so $T^2=0$. The minimal polynomial therefore divides $X^2$. It cannot be $1$, and it cannot be $X$ since $T\ne 0$; hence the minimal polynomial is $X^2$.

Choose $0\ne v\in A_3$ and write $v=a_1+a_2$ using the direct sum of vector spaces $V=A_1\oplus A_2$. Neither component is zero: otherwise $v$ would lie in one of the zero intersections $A_3\cap A_1$ or $A_3\cap A_2$. They are linearly independent, so $W=\mathrm{span}\{a_1,a_2\}$ is a two-dimensional vector subspace. Uniqueness of the direct-sum decomposition gives $W\cap A_1=\mathrm{span}\{a_1\}$ and $W\cap A_2=\mathrm{span}\{a_2\}$. Also $v\in W\cap A_3$, but $W$ cannot be contained in $A_3$ because $a_1\in A_1\setminus \{0\}$. Hence $W\cap A_3$ is one-dimensional. This constructs the required common plane for three pairwise complementary subspaces.

An ideal hyperbolic triangle has its three vertices at infinity, with sides given by geodesics. In curvature $-1$ its angles are zero and its area is $\pi$, using the hyperbolic triangle area formula $\pi -(\alpha +\beta +\gamma )$.

To calculate the area of a hyperbolic disc, use the Poincare disc model centered at the origin. Hyperbolic radial distance is $\rho ={\int }_0^{R}2dr/(1-r^2)=2\mathrm{artanh}R$, so its Euclidean radius is $R=\tanh (\rho /2)$. The metric area density is $4rdrd\theta /(1-r^2{)}^2$, givingEvery hyperbolic triangle, including ones with ideal vertices, has area at most $\pi$. But $A(2)=2\pi (\cosh 2-1)>\pi$. A hyperbolic triangle is geodesically convex, so if it contains a complete hyperbolic circle, it also contains the enclosed hyperbolic disc, its geodesic convex hull. This would violate the area comparison. Therefore no hyperbolic triangle contains a complete circle of radius two.

For a holomorphic function, compute its complex derivative along the real and imaginary directions. The real direction gives $f^{\prime }=u_x+i{v}_x$, while division by an imaginary increment gives $f^{\prime }=v_y-i{u}_y$. Equating real and imaginary parts proves the Cauchy-Riemann equations:For the specified real part, $u_x=-e^{-x}\cos y$ and $u_y=-e^{-x}\sin y$. Integrating $v_y=u_x$ gives $v=-e^{-x}\sin y+C(x)$; the other relation in the Cauchy-Riemann equations then forces $C^{\prime }(x)=0$. Thus the harmonic conjugate and the holomorphic function are

Use $x^{\mu }=(ct,x,y,z)$ and metric signature $(+,-,-,-)$. Since $d\tau =dt/\gamma$, the four-velocity and four-momentum areThese use the proper time and Lorentz factor; the mass shell identity is $E^2-c^2|\mathbf{p}{|}^2=M^2{c}^4$.

Initially the parent four-momentum is $(Mc,0)$. By four-momentum conservation, the remaining single particle has energy $M{c}^2-E$ and three-momentum $-\mathbf{p}$. Applying its mass shell identity gives $m^2{c}^4=(M{c}^2-E{)}^2-c^2|\mathbf{p}{|}^2$, henceThe positive root is selected for the rest mass; physically admissible decay data also give nonnegative remaining energy.

The divergence of this axisymmetric velocity field is $r^{-1}{\partial }_r(-\alpha r^2)+{\partial }_z(2\alpha z)=-2\alpha +2\alpha =0$. Thus it satisfies incompressibility. The apparent axis singularity in the azimuthal component is removable, since $u_{\theta }\sim \gamma \beta r$ as $r\to 0$.

Only the axial vorticity component is nonzero:The cross product is $\mathbf{u}\times \mathbf{\omega }=(u_{\theta }w,\alpha rw,0)$. Its components are independent of $z$ and $\theta$, so its curl also has only an axial component:Meanwhile $w^{\prime }=-2\beta rw$ and $w^{\prime \prime }=(-2\beta +4{\beta }^2{r}^2)w$, givingThus the required steady vorticity equation holds with $\nu =\alpha /(2\beta )$. This is the balance between vortex stretching and viscous diffusion in a Burgers vortex.

Subtract the exact solution from the Jacobi method to get $e_n=H{e}_{n-1}$, hence $e_n=H^n{e}_0$. Convergence from every initial vector is equivalent to $H^n\to 0$. If every eigenvalue has modulus below one, a Jordan block $\lambda I+N$ has powers that are finite sums $\left(\genfrac{}{}{0ex}{}{n}{k}\right){\lambda }^{n-k}{N}^k$. The polynomial factor in $n$ is dominated by geometric decay; for $\lambda =0$, the block is nilpotent. Consequently $H^n\to 0$.

Conversely, for a complex eigenvector $v$ with eigenvalue $\lambda$, $H^{n}v={\lambda }^{n}v$. Convergence for real errors also implies convergence for their complexification, so this tends to zero only if $|\lambda |<1$. Therefore the iteration converges from every starting vector if and only if $\rho (H)<1$, where $\rho$ is the spectral radius.

For the specified matrix,Its eigenvalues are $0,\mu ,-\mu$, so $\rho (H)=|\mu |$. Hence $-1<\mu <1$ is the convergence range. The original coefficient matrix has determinant $12(1-{\mu }^2)$, so the endpoints also violate its assumed nonsingularity.

An unbiased estimator $\hat{\theta }$ satisfies ${\mathbb{E}}_{\theta }\hat{\theta }=\theta$ for every parameter value. Under the standard full-column-rank assumption on the design matrix, the Gaussian log-likelihood for the normal linear model, up to terms independent of $\beta$, is $-\Vert Y-X\beta {\Vert }^2/(2{\sigma }^2)$. Its maximum is the ordinary least squares minimum. Differentiating gives $X^{T}X\hat{\beta }=X^{T}Y$, and full column rank makes $X^{T}X$ positive definite. ThusSince $\mathbb{E}Y=X\beta$, its expectation is $(X^{T}X{)}^{-1}{X}^{T}X\beta =\beta$, proving unbiasedness. Estimating an unknown ${\sigma }^2$ as well does not change the maximizing coefficient vector.

The printed hypotheses give $p<n$ but do not explicitly require full column rank. Without that condition the maximum-likelihood estimators are not unique: they are $X^{+}Y+z$ with $z\in \ker X$, as in rank-deficient ordinary least squares. If $0\ne z\in \ker X$, the observation distributions at $\beta$ and $\beta +z$ are identical. Every estimator then has the same expectation at both parameters, so no estimator can be unbiased for both entire coefficient vectors. This proves nonidentifiability prevents unbiased coefficient estimation. The stated estimator and unbiasedness claim require the usual full-rank condition.

The point $(x_1,x_2,x_3)=(4,2,2)$ is nonnegative and makes all three inequalities equalities, with objective value $20$. To certify global optimality, combine the first upper bound with coefficient $1/2$, the second with coefficient one, and the lower bound with coefficient $-3/2$:This is a weak duality certificate for the linear program. The feasible point attains the bound, so the optimal solution is $(4,2,2)$ and the maximum is $20$. Equality forces each of the three constraints to be active; their simultaneous equations have this unique solution.

The dual space is $V^{\ast }={\mathrm{Hom}}_F(V,F)$, the vector space of all linear functionals with pointwise operations. For finite $d=\dim V$, choose a basis $v_1,\dots ,v_d$ and define its dual basis by ${\epsilon }_j(v_i)={\delta }_{ij}$. Every linear functional is $\ell ={\sum }_j\ell (v_j){\epsilon }_j$, and evaluation on each $v_i$ proves that the ${\epsilon }_j$ are linearly independent. Thus $\dim V^{\ast }=\dim V=d$. This dimension formula is a finite-dimensional assertion.

On the space of polynomials of degree at most $n$, define the evaluation map $E(p)=(p(a_0),\dots ,p(a_n))$. Its kernel is zero: a nonzero degree-at-most-$n$ polynomial cannot have $n+1$ distinct roots. Since domain and codomain both have dimension $n+1$, $E$ is an isomorphism. The evaluations therefore form a basis of the dual space. The linear functional $p\mapsto p^{\prime }(0)$ has a unique expansion in that basis, proving the existence and uniqueness of the requested weights.

Explicitly, use the Lagrange interpolation polynomial basisThe differentiation weights from nodal evaluations are ${\lambda }_j={\ell }_j^{\prime }(0)$. If a node is zero, no singular reciprocal formula is needed: differentiate the finite product directly. For $n=0$ the sole weight is zero.

In a principal ideal domain, an ascending chain of ideals stabilizes: its union is an ideal $(d)$, and its generator belongs to one member, which then contains the union. This proves the needed ascending chain condition. If a nonzero nonunit had no finite irreducible factorization, split it into two nonunits and select a factor that still has no such factorization. Continuing would give a strictly ascending chain of principal ideals, because $a=bc$ with $c$ a nonunit implies $(a)⊊(b)$. This contradicts the chain condition, so factorization exists.

An irreducible element $p$ generates a maximal ideal: any intermediate ideal is $(d)$ with $p=dc$, and irreducibility forces $d$ or $c$ to be a unit. Therefore $p$ is a prime element. In any two irreducible factorizations, the first prime factor divides a factor of the other, hence is associate to it. Cancel and continue. This proves every principal ideal domain is a unique factorization domain.

In $\mathbb{Z}[\sqrt{-3}]$, the multiplicative field norm is $N(a+b\sqrt{-3})=a^2+3b^2$. The only units are $\pm 1$. There is no element of norm two, so every element of norm four is irreducible: a factorization into nonunits would require norms two and two. Consequentlyis a pair of genuinely different irreducible factorizations. Neither $1+\sqrt{-3}$ nor $1-\sqrt{-3}$ is associate to $2$.

Take the Eisenstein integers $R=\mathbb{Z}[\omega ]$, where $\omega =(-1+\sqrt{-3})/2$ and ${\omega }^2+\omega +1=0$. Since $\sqrt{-3}=1+2\omega$, the embedded subring is $\mathbb{Z}+2\mathbb{Z}\omega$, whose additive index in $R$ is two. Its larger ring has Eisenstein-integer norm $N(a+b\omega )=a^2-ab+b^2$ and is Euclidean. Indeed, write a complex quotient in the real basis $1,\omega$ and round both coordinates to integers. The coordinate errors $u,v$ satisfy $|u|,|v|\le 1/2$, giving $|u+v\omega {|}^2=u^2-uv+v^2\le 3/4<1$. Thus division leaves a remainder of smaller norm. A Euclidean domain is a principal ideal domain, because division by a nonzero element of least norm in an ideal leaves zero remainder. This gives the requested index-two embedding into a principal ideal domain.

Let $\varphi (t)=\mathrm{dist}(t,\mathbb{Z})$. It is continuous and takes values in $[0,1/2]$. The finite sums $f_N(x)={\sum }_{n=1}^N{2}^{-n}\varphi (2^{n}x)$ are continuous and their tails obeyfor every real $x$. To prove continuity without invoking an unproved uniform-limit theorem, fix $x$ and $\epsilon >0$. Choose $N$ with $2\cdot 2^{-N-1}<\epsilon /2$, then use continuity of $f_N$ to make $|f_N(y)-f_N(x)|<\epsilon /2$ near $x$. The triangle inequality then gives $|f(y)-f(x)|<\epsilon$. Thus $f$ is continuous; it is the scaled Takagi function $T(2x)/2$.

For the general secant claim, differentiability gives $g(x+h)=g(x)+g^{\prime }(x)h+r(h)$, where $|r(h)|\le \epsilon |h|$ for sufficiently small $h$, with $r(0)=0$. Since the two endpoints straddle $x$,Hence the secant limit is $g^{\prime }(x)$, including cases where an endpoint equals $x$. This proves secants straddling a differentiability point.

Now take the nested dyadic intervals $u_N=2^{-N}\lfloor 2^{N}x\rfloor$, $v_N=u_N+2^{-N}$. They straddle $x$ and shrink to it. For every $n\ge N$, $2^n{u}_N$ and $2^n{v}_N$ are integers, so the corresponding summand vanishes at both endpoints. For $n<N$, all corners of that summand lie on the dyadic grid of mesh $2^{-N}$; on the chosen open interval it is linear with slope ${\epsilon }_n\in \{-1,1\}$. Therefore its total secant slope isRefining to the nested interval preserves the slopes of all old terms and adds exactly one new term of slope $+1$ or $-1$. Thus $|S_{N+1}-S_N|=1$ for every $N$, so $S_N$ cannot converge. This contradicts the secant consequence of differentiability at any $x$. Therefore $f$ is nowhere differentiable. At dyadic $x$ the intervals chosen on its right satisfy the same argument. The printed reference to a denominator $2^{-n}$ is naturally read as mesh $2^{-n}$, or denominator $2^n$, as used here.

Suppose $(x_n,f(x_n))$ converges to $(x,y)$. Continuity gives $f(x_n)\to f(x)$, and uniqueness of limits in a metric space gives $y=f(x)$. Hence every convergent sequence in the graph has its limit in the graph, proving ${\Gamma }_f$ is closed.

If $f$ were not continuous at $x$, there would be $\epsilon >0$ and $x_n\to x$ with $d_Y(f(x_n),f(x))\ge \epsilon$. A compact metric space is sequentially compact, so some subsequence $f(x_{n_k})$ converges to $y$. The closed graph then contains its product limit $(x,y)$, forcing $y=f(x)$, contrary to the lower bound. Thus $f$ is continuous, the compact-codomain form of the closed graph theorem for compact spaces.

Take $f(0)=0$ and $f(x)=1/x$ for $x\ne 0$. Its graph is $\{(x,y):xy=1\}\cup \{(0,0)\}$, a union of two closed sets, since multiplication is continuous. But $f(1/n)=n$ does not tend to $f(0)$, so the function is not continuous.

Choose the principal branch of a multivalued function, $-\pi <\arg s<\pi$, for $s^{1/2}$. The integrand $G(s)=e^{st}/((s^2+1)\sqrt{s})$ has simple poles at $\pm i$ and a branch cut along the negative real axis. Its residues arewhose sum is $\sin (t-\pi /4)$.

For $t>0$, close the Bromwich contour to the left using a rectangle with outer real part $-R$ and horizontal edges at imaginary parts $\pm R$, indenting the cut and zero. On every distant edge, $|s^2+1|\ge |s{|}^2-1$, $|\sqrt{s}|=|s{|}^{1/2}$ and $|e^{st}|\le e^{\gamma t}$. Each edge has length $O(R)$ and $|s|\ge R$, so its integral is $O(R^{-3/2})$. The small circle of radius $\epsilon$ about zero contributes $O({\epsilon }^{1/2})$. The tails of the original vertical integral are absolutely bounded by $O(R^{-3/2})$ as well.

On the upper and lower banks at $s=-r$, the square roots are $i\sqrt{r}$ and $-i\sqrt{r}$, respectively. With the upper bank traversed toward zero and the lower bank away from zero, their combined integral isThe cut integral converges near zero since $r^{-1/2}$ is integrable, and at infinity by exponential decay. The residue theorem, followed by $R\to \infty$ and $\epsilon \to 0$, therefore givesFor $t<0$, close the Bromwich contour to the right. There are no enclosed poles or cuts, and $|e^{st}|\le e^{\gamma t}$ there. The same distant-edge estimates show $f(t)=0$ for $t<0$.

For the large-$t$ expansion, $1/(1+r^2)=1-r^2+r^4/(1+r^2)$. Scaling $r=u/t$ and using the given integral yields ${\int }_0^{\infty }{e}^{-rt}{r}^{-1/2}dr=\sqrt{\pi }{t}^{-1/2}$. Integration by parts twice gives the next coefficient $\Gamma (5/2)=3\sqrt{\pi }/4$. The remainder is bounded by ${\int }_0^{\infty }{e}^{-rt}{r}^{7/2}dr=\Gamma (9/2)t^{-9/2}$. Thus, without needing a globally convergent termwise integration of the local power series,In particular, the requested two leading terms are $\sin (t-\pi /4)+(\pi t{)}^{-1/2}$, interpreted as an additive asymptotic expansion. This is Bromwich inversion with a square-root branch cut.

Write $y={\sum }_{k\ge 0}{c}_k{x}^k$ with $c_0=1$. Substitution into the ordinary differential equation givesUnless the series terminates, its coefficient ratio is $O(1/k)$, so it converges for every finite $x$. A coefficient first vanishes exactly when $\lambda$ is a nonnegative integer $n$, and then $c_n=(-1{)}^n/n!\ne 0$, $c_{n+1}=0$. Thus $y$ is a polynomial precisely for $\lambda =n\in \{0,1,2,\dots \}$, in which caseis the normalized Laguerre polynomial.

Multiplying the differential equation by $e^{-x}$ gives $(x{e}^{-x}{y}_n^{\prime }{)}^{\prime }+n{e}^{-x}{y}_n=0$. Multiply by $y_m$, subtract the equation with the indices exchanged, and integrate over $[0,\infty )$. The boundary term $x{e}^{-x}(y_m{y}_n^{\prime }-y_n{y}_m^{\prime })$ vanishes both at zero and infinity. ThereforeThis is the weighted orthogonality of the Laguerre polynomials.

Their leading coefficients show $\mathrm{deg}(y_m{y}_n)=m+n$. Since $y_0,\dots ,y_{m+n}$ have distinct degrees, they form a basis of that polynomial space. Hence $y_m{y}_n={\sum }_{p=0}^{m+n}{a}_p{y}_p$, and comparison of leading coefficients gives $a_{m+n}=(m+n)!/(m!n!)\ne 0$. Likewise the degrees of $y_m$ and $y_m{y}_n$ for $0\le m<n$ run through $0,\dots ,n-1$ and $n,\dots ,2n-1$, respectively. A nontrivial linear combination cannot cancel its highest-degree term, so these $2n$ functions are linearly independent.

For $n=2$, this gives the basis $y_0,y_1,y_2,y_1{y}_2$ of the cubic polynomial space, establishing the specified expansion. Its weighted integral is $a_0$: $y_0=1$ integrates to one, $y_1$ and $y_2$ are orthogonal to $y_0$, and $y_1{y}_2$ integrates to zero. At the two roots ${\alpha }_1,{\alpha }_2$ of $y_2$, the expansion reduces to $f({\alpha }_i)=a_0+a_1{y}_1({\alpha }_i)$. Solving these two equations for $a_0$ givesFor explicit verification, the recurrence gives $y_1=1-x$, $y_2=1-2x+x^2/2$. Thus ${\alpha }_1=2-\sqrt{2}$, ${\alpha }_2=2+\sqrt{2}$ are distinct positive roots and $w_1=(2+\sqrt{2})/4$, $w_2=(2-\sqrt{2})/4$. This is the two-node Gauss-Laguerre quadrature rule, exact for every cubic.

The parity operator acts by $(P\psi )(x)=\psi (-x)$. For an even potential, differentiating twice and using $V(-x)=V(x)$ proves $HP=PH$ on the parity-invariant domain of the Hamiltonian operator. Therefore each energy eigenspace is invariant under $P$. The projectors $(I+P)/2$ and $(I-P)/2$ split it into even and odd subspaces, giving a basis of simultaneous eigenstates. An odd state obeys $\psi (0)=-\psi (0)$, hence ${\psi }_{\mathrm{odd}}(0)=0$.

The infinite walls impose $\psi (-a)=\psi (a)=0$, with the wavefunction zero outside. It is continuous at the central delta potential. Integrate the stationary Schrödinger equation over $[-\epsilon ,\epsilon ]$; the energy integral tends to zero and the Dirac delta function contributes $\kappa \psi (0)$. This givesFor the given piecewise positive-energy form, $E={\hslash }^2{\lambda }^2/(2m)$. Replacing $x$ by $-x$ swaps the two pieces and leaves their values unchanged, proving $P\psi =\psi$. The value at zero is $A$, and the one-sided derivatives are $-B\lambda$ and $B\lambda$. The jump condition gives $B=-m\kappa A/({\hslash }^2\lambda )$. The wall at $a$ gives $A\cos (\lambda a)-B\sin (\lambda a)=0$, henceFor $\kappa =0$ the undivided wall equation instead gives the usual free even roots. For $\kappa >0$, set $c=m\kappa /{\hslash }^2>0$ and $k_n=(n+1/2)\pi /a$. In every interval $k_n<\lambda <(n+1)\pi /a$ there is exactly one root: writing $\delta =a(\lambda -k_n)\in (0,\pi /2)$ gives $\cot \delta =\lambda /c$, whose two sides are respectively strictly decreasing and increasing. These are all positive even roots; the quadratic form ${\hslash }^2\int |{\psi }^{\prime }{|}^2/(2m)+\kappa |\psi (0){|}^2$ is nonnegative, so there are no negative eigenvalues. Thus ${\lambda }_n>k_n$ and ${\eta }_n={\hslash }^2({\lambda }_n^2-k_n^2)/(2m)>0$.

The printed limit of this absolute energy difference is false. The exact quantization equation gives ${\delta }_n=\arctan (c/{\lambda }_n)$, so ${\delta }_n\to 0$ but ${\lambda }_n{\delta }_n\to c$. ConsequentlyThus the high-energy shift from a central delta barrier is a finite positive constant, not zero. The corrected vanishing statement is the relative shift ${\eta }_n/E_n^{(0)}\to 0$, or the wave-number difference ${\lambda }_n-k_n\to 0$. Physically, highly energetic particles are weakly affected in relative terms by a fixed point barrier. The free even wavefunction normalized in $[-a,a]$ has density $1/a$ at the origin, so the barrier's first-order expectation is $\kappa /a$, matching the exact limit.

Odd states vanish at zero, so the delta term and the derivative jump both vanish. Their normalized wavefunctions and energies arewith zero wavefunction outside. They do not depend on $\kappa$. These even and odd sectors describe the central delta barrier in a symmetric infinite well.

In electrostatic equilibrium, the electric field inside a perfect conductor is zero. Apply the Faraday's law $\oint \mathbf{E}\cdot d\mathbf{l}=0$ to a thin rectangular loop crossing the surface, with its long sides tangential and its short sides shrinking to zero. It yields continuity of the tangential electric field. Since its interior value is zero, $E_x(0+)=E_y(0+)=0$. Earthing fixes the electric potential at the plane to zero. For completeness, a thin Gauss law pillbox gives $E_z(0+)=\sigma /{\epsilon }_0$; the normal component can be nonzero because of induced surface charge. These are the electrostatic boundary conditions at a conductor.

The charge and its proposed image charge give the exterior electric potentialAt $z=0$ the terms cancel, so $\Phi =0$ everywhere on the plane and its tangential derivatives vanish. Thus the image construction satisfies the earthed-conductor boundary conditions. It has the correct point-charge singularity, is harmonic elsewhere in $z>0$, and decays at infinity; electrostatic uniqueness identifies it with the exterior solution.

At instantaneous height $z$, the image charge is a distance $2z$ below the real charge, giving force $m\ddot{z}=-K/z^2$ with $K=q^2/(16\pi {\epsilon }_0)$. Multiplying by $\dot{z}$ and using rest at $z=d$ givesThe induced-conductor potential energy is $-K/z$, not the full energy of a pair of independently movable charges. On the falling branch, integrate the reciprocal speed:The last integral is $\pi /2$ by $s={\sin }^2\theta$. Hence the fall time of a charge toward a grounded plane in the electrostatic Newtonian model isFor $q=0$ there is no force and the particle remains at rest.

Use the steady inviscid, hydrostatic, slowly varying channel approximation, without a hydraulic jump. Conservation of volume flux per unit width gives $Q=u(x)h(x)=u_1{h}_1$. The free-surface Bernoulli equation isDividing by $h_1$ yieldsThe parameter $F$ here is the squared upstream Froude number. For nonzero discharge, regarding $D$ as a function of $h$, its derivative is $dD/dh=-1+Q^2/(g{h}^3)$. It has its unique maximum at $h_c=(Q^2/g{)}^{1/3}=h_1{F}^{1/3}$, where the local Froude number is one. Substitution gives $D^{\ast }=h_1(1-3F^{1/3}/2+F/2)$. This maximum is the hydraulic control over a smooth hump.

The upstream flow is subcritical flow, with $h_1>h_c$. On this branch $dD/dh<0$, so the depth decreases as the bed rises; the velocity $Q/h$ increases. Since the crest height is below $D^{\ast }$, the depth remains above $h_c$ and never becomes critical. Over the falling side of the hump the depth increases again to $h_1$ downstream. The free surface falls over the hump: $d(h+D)/dD=-{\mathrm{Fr}}^2/(1-{\mathrm{Fr}}^2)<0$ on the subcritical branch.

The upstream flow is supercritical flow, with $h_1<h_c$. Now $dD/dh>0$, so the depth increases over the rising hump and the velocity decreases, while the free surface rises. Since $D_{\max }<D^{\ast }$, the depth stays below $h_c$. On the descending bed it decreases back to $h_1$ downstream, remaining supercritical throughout.

At the crest the two depth branches meet at $h=h_c=h_1{F}^{1/3}$. For a smooth nondegenerate maximum of the bed, $D^{\ast }-D$ is proportional to $(x-x_c{)}^2$, while expansion of the depth relation gives $D^{\ast }-D\sim 3(h-h_c{)}^2/(2h_c)$. A differentiable flow through the crest therefore switches branches rather than following the cusp obtained by staying on the same branch. Subcritical upstream flow becomes supercritical downstream; supercritical upstream flow becomes subcritical downstream.

When $F<1$ and the bed returns to zero, put $y=h_{\mathrm{down}}/h_1$. The head equation becomes $2y^3-(2+F)y^2+F=0$, orThe root $y=1$ is the original subcritical branch. The other positive root is smaller and lies on the supercritical branch, givingThis is the smooth transcritical, shock-free solution. A downstream hydraulic jump would require further downstream conditions and energy loss. A degenerate flat crest can also require a branch-selection condition; the standard critical-control interpretation uses a nondegenerate smooth crest. For $F=1$, $D^{\ast }=0$, so a positive hump is incompatible with the prescribed upstream critical head. If $F=0$, the head equation is the still-water relation $h+D=h_1$; a crest of height $h_1$ dries out, so the wet transcritical-flow interpretation requires $F>0$.

A critical region $C$ is the set of sample outcomes on which a test rejects its null hypothesis. Its power function is $\beta (\theta )={\mathbb{P}}_{\theta }(X\in C)$; its size of a statistical test is ${\sup }_{\theta \in {\Theta }_0}\beta (\theta )$. Randomized tests replace the indicator of $C$ by a rejection probability $\varphi (X)\in [0,1]$.

The Neyman-Pearson lemma states that for two simple hypotheses with densities $p_0,p_1$, the test ${\varphi }^{\ast }$ that is one where $p_1>k{p}_0$, zero where $p_1<k{p}_0$, and randomized on equality to have null rejection probability $\alpha$, maximizes power among tests of size at most $\alpha$. Here $k\ge 0$. To prove it, let $\varphi$ be any competing test. Pointwise, $({\varphi }^{\ast }-\varphi )(p_1-k{p}_0)\ge 0$, since both factors have the same sign off the equality set. Integrating givesThis proves the lemma, including equality-set randomization and points where $p_0=0$.

Take the exponential parameter to be its rate: $p_{\lambda }(x)=\lambda e^{-\lambda x}$ for $x\ge 0$. For $T={\sum }_i{X}_i$, the joint likelihood ratio is $({\lambda }_1/{\lambda }_0{)}^n{e}^{-({\lambda }_1-{\lambda }_0)T}$, decreasing in $T$. Hence for $0<\alpha <1$ the most powerful test isThere is no boundary randomization because $T$ has a continuous gamma distribution. Its density follows by repeated convolution: $p_T(t)={\lambda }^n{t}^{n-1}{e}^{-\lambda t}/(n-1)!$. Integrating by parts gives the exponential-rate likelihood-ratio test powerThe threshold is the unique positive solution of this formula at $\lambda ={\lambda }_0$ with value $\alpha$. Differentiating the finite sum yieldsThus the power function is strictly increasing, and its supremum over $0<\lambda \le {\lambda }_0$ is $\beta ({\lambda }_0)=\alpha$. Therefore the same test has size $\alpha$ for the composite null $\lambda \le {\lambda }_0$. Since the threshold does not depend on the particular larger rate, it is also uniformly most powerful against all $\lambda >{\lambda }_0$. The endpoint sizes zero and one have their trivial constant tests.