Write $s=e^{-\lambda t}$ for individual survival probability. Conditional on a clump of size $n$, its death probability is $(1-s{)}^n$. Averaging with the geometric distribution givesThusHere $0\le \theta <1$ and $t>0$; different clumps and experimental repetitions are understood independent for the sampling model.

The live-clump counts have independent binomial distributions, all with the same $p$. Up to a constant, their combined likelihood function is $p^R(1-p{)}^{N-R}$, maximized at $\hat{p}=R/N$ when $0<R<N$. Solving $p=s/(1-\theta +\theta s)$ for $s$ gives $s=p(1-\theta )/(1-\theta p)$. ThereforeMonotonicity of $p$ in $\lambda$ makes this the maximum-likelihood estimator. At the sampling boundaries, $R=N$ gives $\hat{\lambda }=0$, while $R=0$ gives only a supremum as $\lambda \to \infty$, with no finite maximizer. These are the limiting interpretations of the formula.

Up to a parameter-independent multinomial factor, the likelihood function is $L={\prod }_{i,j}{\theta }_{ij}^{n_{ij}}$. A Lagrange multiplier for $\sum {\theta }_{ij}=1$ gives $n_{ij}/{\theta }_{ij}=\lambda$, whence $\lambda =n$ andSince all counts are positive, the log-likelihood function is strictly concave on the positive simplex and tends to minus infinity at its boundary, proving this is the unique maximum.

Under independence of random variables, ${\theta }_{ij}={\varphi }_i{\rho }_j$, so the log-likelihood function separates as ${\sum }_i{n}_{i\cdot }\log {\varphi }_i+{\sum }_j{n}_{\cdot j}\log {\rho }_j$. Applying the same constrained maximization twice gives ${\hat{\varphi }}_i=n_{i\cdot }/n$, ${\hat{\rho }}_j=n_{\cdot j}/n$, andTwice the difference between the unrestricted and null maximized log-likelihood functions isthe required likelihood-ratio test statistic. Under the null hypothesis, its asymptotic chi-squared distribution has $(r-1)(c-1)$ statistical degrees of freedom, the parameter-count difference $(rc-1)-(r+c-2)$. Here that is four. Since $25.192>q_4=18.467$, reject independence at the 0.1 percent significance level. For four statistical degrees of freedom, the asymptotic upper-tail probability is $e^{-25.192/2}(1+25.192/2)\approx 4.60\times 10^{-5}$. This establishes an association in these data; it is not a causal conclusion.

The independent identically distributed central limit theorem states that for independent identically distributed random variables $U_k$ with finite mean $\mu$ and positive finite variance ${\sigma }^2$,For $0<u<1$, continuity, strict monotonicity and the endpoint limits of the cumulative distribution function give a unique $x=F^{-1}(u)$. Then $P(F(X)\le u)=P(X\le x)=F(x)=u$. Thus $Y=F(X)$ has the uniform distribution on $(0,1)$, by the probability integral transform.

For $W=-\log (1-Y)$, $P(W\le w)=1-e^{-w}$ when $w\ge 0$, and it is zero otherwise. Therefore its probability density function is $e^{-w}{\chi }_{(0,\infty )}(w)$: the exponential distribution of rate one. By integration by parts, $E[W]=1$ and $E[W^2]=2$, so $\mathrm{Var}(W)=1$.

The variables $W_k=-\log (1-F_k(X_k))$ are independent because they are separate measurable functions of independent random variables. The probability integral transform makes all of them identically distributed as $\mathrm{Exp}(1)$, even though the $X_k$ need not be identically distributed. The displayed $Z_n$ is exactly $n^{-1/2}{\sum }_{k=1}^n(W_k-1)$. The central limit theorem therefore gives

For potential flow, each velocity potential satisfies Laplace's equation in its fluid. The exact kinematic boundary condition on $y=\eta$ is ${\eta }_t+{\varphi }_{j,x}{\eta }_x={\varphi }_{j,y}$. Keeping only first-order terms givesThe linearized Bernoulli equation is $p_j=p_0-{\rho }_j({\varphi }_{j,t}+g\eta )$ at the interface, so pressure continuity givesFor a decaying disturbance away from the interface, put $\psi =kx-\omega t$ and takeThey satisfy Laplace's equation and the kinematic boundary conditions for $\eta =A\cos \psi$. Substituting in the dynamic condition gives the dispersion relationIf the heavier fluid is on top, ${\omega }^2<0$. Instead of oscillation, a Fourier disturbance has an exponentially growing component with growth rate $\sqrt{gk({\rho }_2-{\rho }_1)/({\rho }_1+{\rho }_2)}$. This is the Rayleigh-Taylor instability in the absence of surface tension.

Orient $C$ counterclockwise around the solid body and let $\mathbf{n}$ point from the solid into the fluid. The pressure force is $-{\int }_{C}p\mathbf{n}ds$. Since $\mathbf{n}ds=(dy,-dx)$,For steady irrotational flow, Bernoulli's equation gives $p=p_0-\rho |w^{\prime }{|}^2/2$. The constant term integrates to zero. The boundary is a streamline, so $dw=w^{\prime }dz$ is real there, giving $w^{\prime }dz=\overline{w^{\prime }}d\overline{z}$. Thereforewhich is Blasius force theorem with the specified orientation.

The prescribed stream function is the imaginary part of $w_0(z)=\alpha (z-c{)}^3$. The zero-circulation solution from the Milne-Thomson circle theorem isOn $|z|=a$ the two terms are conjugates, so the boundary stream function is constant. The added term has no singularities in the exterior and does not change the prescribed growing far-field part. Additive constants are immaterial.

Differentiating givesThe coefficient of $z^{-1}$ in $(w^{\prime }{)}^2$ is $36{\alpha }^{2}c{a}^2(a^2+c^2)$. The residue theorem and Blasius force theorem giveThe zero-circulation convention is relevant: if an independently imposed circulation $\Gamma$ is allowed, adding $\Gamma \log z/(2\pi i)$ changes $F_y$ to $-3\rho \alpha \Gamma c^2$. Thus the claimed zero transverse force belongs to the standard circle-theorem construction, or to a single-valued exterior complex potential of a two-dimensional potential flow, which excludes that extra circulation.

In vacuum, the source-free Maxwell equations arePut $\psi =\omega (t-z/c)$ and denote the constant electric amplitude by $E_0$. A corresponding traveling-wave magnetic field isBoth divergences vanish because both the electric field and the magnetic field depend only on $z,t$ and have zero $z$ component. Direct differentiation gives $\nabla \times \mathbf{E}=(\omega E_0/c)(\cos \psi ,\sin \psi ,0)=-{\partial }_t\mathbf{B}$ and $\nabla \times \mathbf{B}=(\omega E_0/c^2)(-\sin \psi ,\cos \psi ,0)=c^{-2}{\partial }_t\mathbf{E}$. A time-independent background magnetic field satisfying the homogeneous equations could be added; the formula is the wave component.

Introduce the magnetic vector potential and electric potential by $\mathbf{B}=\nabla \times \mathbf{A}$ and $\mathbf{E}=-\nabla \varphi -{\partial }_t\mathbf{A}$. These identities automatically give the divergence equation for $\mathbf{B}$ and Faraday's law. For this wave choose $\varphi =0$ and $\mathbf{A}=(c/\omega )\mathbf{B}$; the derivatives just calculated verify both potential equations.

The electromagnetic force is the Lorentz force, $Q(\mathbf{E}+\mathbf{v}\times \mathbf{B})$. On the smooth constraint plane $z=z_0$, the constraint force is normal to the plane. Both $\mathbf{v}$ and $\mathbf{B}$ are tangential, so $\mathbf{v}\times \mathbf{B}$ is normal and is balanced by that constraint force; it does not affect the tangential motion. ThusA particular motion is ${\mathbf{r}}_{\parallel }={\mathbf{r}}_0-[Q{E}_0/(m{\omega }^2)](\cos \psi ,\sin \psi )$, a circle of radiusThe general planar motion also includes a constant drift velocity; the particle moves on a circle when that drift is zero. The constraint force can do no work because it is perpendicular to the velocity.

Write $q=(x,X{)}^{\mathsf{T}}$, $M=\left(\begin{array}{cc}a& b\\ b& c\end{array}\right)$ and $K=\left(\begin{array}{cc}\alpha & \beta \\ \beta & \gamma \end{array}\right)$. The Euler-Lagrange equations give $M\ddot{q}+Kq=0$. A normal mode is a motion $q=v\cos (\omega t+\phi )$ with fixed nonzero vector $v$; its normal frequency is $\omega >0$. Substitution gives $(K-{\omega }^{2}M)v=0$. Thus the squared normal frequencies are the assumed positive distinct roots ofFor either root, solve the associated two linear equations for the mode ratio.

For the rods, the small angles are $\theta =x/a$ and $\varphi =(X-x)/(2a)$. The first rod has kinetic energy $m{\dot{x}}^2/6$. The second rod's center has horizontal velocity $(\dot{x}+\dot{X})/2$ and its angular velocity is $(\dot{X}-\dot{x})/(2a)$. Its center-of-mass moment of inertia is $2m{a}^2/3$, givingAdding these kinetic energies gives $T=m(3{\dot{x}}^2+2\dot{x}\dot{X}+2{\dot{X}}^2)/6$. The first center rises by $a{\theta }^2/4$, and the second by $a{\theta }^2/2+a{\varphi }^2/2$. Multiplying by the respective masses and $g$ givesTherefore the stated Lagrangian is $T-V$.

Set $\mu =2a{\omega }^2/(3g)$. The mode equation isHenceThe mode ratio is $r=X/x=(6-3\mu )/(1+\mu )$, so the ratio of the rods' angles is $\varphi /\theta =(r-1)/2=(5-4\mu )/[2(1+\mu )]$. It is positive for ${\mu }_{-}=(17-3\sqrt{21})/10$ and negative for ${\mu }_{+}=(17+3\sqrt{21})/10$. Thus the lower-frequency normal mode has the rods in phase, and the higher-frequency normal mode has them out of phase.