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Choose an optimal feasible vector $x$ having as few positive coordinates as possible, and letIf the columns $(A_i{)}_{i\in S}$ were linearly dependent, there would be a nonzero vector $h$, supported on $S$, such that $Ah=0$. For all sufficiently small positive $\epsilon$, bothwould remain feasible.

If $c^{T}h\ne 0$, one of these two perturbations would increase the objective, contradicting optimality. Hence $c^{T}h=0$. We may then increase $\epsilon$ in one of the two directions until at least one positive coordinate first becomes zero. The resulting vector is still feasible and optimal but has smaller positive support, contradicting the choice of $x$.

A vector $x$ is a basic feasible solution whenand the columns $A_i$ for which $x_i>0$ are linearly independent. Equivalently, after extending those columns to a basis of the column space of $A$, one sets all nonbasic variables to zero and solves for the basic variables, obtaining a nonnegative vector.

Set ${\theta }_0=1/2$ and $z=(\theta -1/2{)}^2$. Sincethe risk is affine in $z$:Its supremum is therefore controlled by the midpoint $z=0$ and the endpoint limit $z\to 1/4$. Requiring both endpoint values to be at most the MLE's maximal risk $1/(4n)$ givesEquivalently,Their intersection isFor precisely this range, the shrinkage estimator has maximal mean squared error no greater than that of $\hat{\theta }$.

Forthe bias and variance areThe bias-variance decomposition of mean squared error yieldsThis is the risk formula for an affine shrinkage estimator for a binomial proportion.

For the weighted loss, the Bayes estimator under parameter-weighted squared error is the mean of the posterior after multiplication bySince the original posterior density is proportional tothe reweighted density is $\mathrm{Beta}(X+\alpha ,n-X+\beta )$. Its mean givesWriting this in the required form,

The uniform prior is the Beta distribution $\mathrm{Beta}(1,1)$. By Beta-binomial conjugacy, after observing $X$ heads the posterior isThe Bayes estimator under squared error loss is the posterior mean, hence

The estimator is unbiased because ${\mathbb{E}}_{\theta }X=n\theta$, andThe bias-variance decomposition of mean squared error therefore givesThe quadratic $\theta (1-\theta )$ is maximized at $\theta =1/2$, so

Here $X$ has the binomial distribution $\mathrm{Bin}(n,\theta )$, so, up to a factor independent of $\theta$, the likelihood function isFor $0<X<n$, differentiating the log-likelihood givesand the boundary cases give the same formula. Thus the binomial proportion maximum-likelihood estimator is

Since $dt=da/\dot{a}$ on the collapsing branch, the time for $a$ to decrease from $a_0$ to zero isEquivalently, scaling $a=a_{0}x$ gives

At the moving boundary,Substitution into the result of part (b) gives the Rayleigh collapse equationTreat $v=\dot{a}$ as a function of $a$, so that $\ddot{a}=vdv/da$. With $w=v^2$, the equation becomesMultiplication by the integrating factor $a^3$ after division by $a$ givesUsing $w(a_0)=0$,The collapsing branch has negative radial velocity, so

The radial velocity and velocity potential areChoose the gauge in the Unsteady Bernoulli equation by evaluating it at infinity, where $u=0$, $\varphi =0$, and $p=p_0$:At the cavity surface the vacuum pressure is zero, and the kinematic boundary condition gives $u(a,t)=\dot{a}$. SinceBernoulli's equation becomesTherefore

In spherical symmetry, the velocity is radial and constant over each sphere. HenceApply the divergence theorem to the fluid shell between radii $R_1$ and $R_2$. Since the flow is incompressible,Thus $Q$ is independent of $R$ and is a function of time alone. This is the flux law for spherically symmetric incompressible radial flow.

The Hamiltonian is the sum of three commuting one-dimensional harmonic-oscillator Hamiltonians. Its product eigenstates arewith energiesEquivalently, the level with $N=n_x+n_y+n_z$ has energy $E_N=\hslash \omega (N+3/2)$, as in the three-dimensional isotropic harmonic oscillator.

Up to normalization, the ground-state wavefunction isIt is radial, so $\nabla {\Psi }_{000}$ is parallel to the position vector. Since the orbital angular momentum is $L=-i\hslash r\times \nabla$, every component of $L$ annihilates this state. Thereforeand the ground state has $\ell =m=0$.

The first excited level has $N=1$ and is spanned byThe statesatisfies $L_z{\Phi }_0=0$. Applying the ladder operators givesHence convenient $m=\pm 1$ eigenstates areup to normalization and irrelevant overall phases.

Finally, the isotropic Hamiltonian is rotationally invariant. The rotational invariance of a central-potential Hamiltonian givesThese commuting self-adjoint operators can be simultaneously diagonalized within each energy eigenspace, which is why joint eigenstates of $H,L^2,L_z$ must exist.

The orbital angular momentum commutation relations giveConsequentlyThe same commutation relations imply $[L^2,L_x]=[L^2,L_y]=0$, and henceThese are the defining identities for the angular momentum ladder operators.

Ifthen, whenever $L_{\pm }\varphi \ne 0$,Thus $L_{\pm }\varphi$ has quantum numbers $(\ell ,m\pm 1)$.

Here the integrandhas no explicit $x$-dependence. The Beltrami identity therefore givesfor a positive constant $R$. Thuswhose nonvertical solutions are the semicircles

The endpoint $(a,a)$ and the prescribed value at $x=b$ both lie onbecauseThe required positive arc is consequentlyThe condition $b<2a$ keeps the endpoint above the $x$-axis. The sketch is the descending part of the upper semicircle of radius $a$ centred at $(a,0)$, from $(a,a)$ toThis is precisely a Geodesic in the Poincare half-plane model.

Forone has $L_y=0$ andA circle of radius $R$ centred at $(0,c)$ on the $y$-axis satisfiesOn any arc that does not cross $y=c$,which is constant. Hence $d{L}_{y^{\prime }}/dx=0=L_y$, so the Euler-Lagrange equation holds and $\delta I=0$. These circles are the coordinate-swapped form of a Geodesic in the Poincare half-plane model.

Let $y_{\epsilon }=y+\epsilon \eta$, where the variation $\eta$ satisfies $\eta (a)=\eta (b)=0$. Differentiating the functional at $\epsilon =0$ gives its first variationUsing integration by parts and the fixed endpoint conditions,If this vanishes for every admissible $\eta$, the fundamental lemma of the calculus of variations gives the Euler-Lagrange equation

For fixed $(x_1,x_2)$, apply the given one-variable identity to the smooth functionThe chain rule givesThereforewhereEvery derivative of the integrand is continuous. Repeated differentiation under the integral sign therefore shows that $h$ is smooth. This is the Second-order Hadamard lemma.