The pooled CC count is $83+50=133$, the case total is 1939, and the grand total is 3874. Under the null, the common estimated CC proportion is $133/3874$. Multiplication by the fixed case total givesThis is the fitted expected count, rather than a separate observation.

For observed counts $O_{ij}$ in an $r$ by $c$ contingency table, Pearson's chi-squared test statistic for independence isUnder the null model and the usual large-expected-count conditions, its reference distribution is ${\chi }^2$ with $(r-1)(c-1)$ degrees of freedom.

For $k>0$, phase velocity is the speed of a fixed phase, $c_p=\omega /k$. The group velocity $c_g=d\omega /dk$ transports a narrow-band wave envelope. With $q=kH$,As $H\to 0$ with $k$ fixed, $\tanh (kH)/(kH)\to 1$. ThereforeBoth velocities tend to the shallow-water speed $\sqrt{gH}$ at leading order.

Take the undisturbed surface to be $y=0$ and treat $\eta$, $\varphi$, and their induced velocities as first order in wave amplitude. Products of first-order quantities are second order and are discarded. Evaluating surface conditions at $y=\eta$ differs from evaluation at zero only by second-order terms. Thus the linearized surface gravity wave problem isHere $u={\varphi }_x$, $v={\varphi }_y$. The linear free-surface conditions can also be combined into ${\varphi }_{tt}+g{\varphi }_y=0$ at zero.

Condition (1b) is the kinematic boundary condition: surface particles remain on the moving free-surface graph. Condition (1c) is the dynamic boundary condition for an inviscid interface from unsteady Bernoulli's equation, with the surface pressure fixed at the atmospheric value and absorbed into the potential's time gauge. Condition (1d) is impermeability of the fixed horizontal bed, so its normal velocity vanishes.

Direct differentiation gives the convective accelerationThe steady Euler equations therefore require $p_x=-\rho \sin x\cos x$ and $p_y=\rho \sin y\cos y$. Integrating both derivatives consistently givesThe additive constant is arbitrary. With a corresponding redefinition of this constant, the Bernoulli function is $p/\rho +|\mathbf{u}{|}^2/2=C_0-{\psi }^2$, making its constancy along streamlines explicit.

For steady inviscid flow of constant density without a body-force potential, Bernoulli's theorem givesThe constant can differ between streamlines, because this flow has nonzero vorticity. Dotting the steady Euler equation with $\mathbf{u}$ gives $\mathbf{u}\cdot \nabla (p/\rho +|\mathbf{u}{|}^2/2)=0$, which proves precisely the along-streamline form.

The vorticity is perpendicular to the plane:It is generally nonzero, so this is a rotational flow even though it has a streamfunction.

Use the two-dimensional streamfunction convention $u={\psi }_y$, $v=-{\psi }_x$. HereThe divergence is $u_x+v_y=-\cos x\sin y+\cos x\sin y=0$, so the flow is incompressible.

The inverse function theorem states that if $f$ is continuously differentiable near $a\in {\mathbb{R}}^n$ and $Df(a)$ is invertible, then neighborhoods of $a$ and $f(a)$ exist on which $f$ is a bijection with a continuously differentiable inverse, satisfying $D(f^{-1})(f(a))=Df(a{)}^{-1}$.

Here $a=(\pi /2,\pi )$, $f(a)=(0,0)$, andThe hypotheses hold because both components are smooth. Inverting the matrix yields

For a differentiable map $f:\Omega \subset {\mathbb{R}}^n\to {\mathbb{R}}^k$, the Jacobian matrix represents its derivative and has entries $(Df{)}_{ij}=\partial f_i/\partial x_j$. Differentiating the two components gives

The unit circle is path-connected, since it is the continuous image of the interval $[0,2\pi ]$ under $\theta \mapsto (\cos \theta ,\sin \theta )$. It is therefore connected, and part (ii) makes $f(C)$ connected. A connected subset of $\mathbb{R}$ is an interval: omission of a point lying between two of its points would separate it by the two open rays about that point.

Moreover, the circle is compact, so the continuous image of a compact space is compact. Hence here the image is a closed bounded interval, possibly a singleton.

If $f(A)$ had a separation into two nonempty relatively open disjoint pieces, their inverse images under the restriction $f{|}_A$ would be two nonempty relatively open disjoint subsets covering $A$. This would contradict connectedness of $A$. Thus a continuous image of a connected set is connected.

A connected set is one whose relative topology admits no decomposition into two disjoint nonempty open sets. A path-connected set permits, between any two points, a continuous map $\gamma :[0,1]\to A$ with the prescribed endpoints.

Path connectedness implies connectedness, because a continuous image of an interval is connected. The converse can fail, for example for the topologist's sine curve. On open subsets of Euclidean spaces ${\mathbb{R}}^n$, the two properties are equivalent; more generally this holds in locally path-connected spaces.

A basis of a topology is a collection $\mathcal{B}$ of open sets such that every open set is a union of members of $\mathcal{B}$. Equivalently, for every open $U$ and $y\in U$ there is $B\in \mathcal{B}$ with $y\in B\subseteq U$.

A map $f:X\to Y$ is continuous exactly when $f^{-1}(B)$ is open in $X$ for every basic set $B$ of $Y$. Necessity follows from the definition of a continuous map. For sufficiency, express an arbitrary open set as a union of basic sets and use preservation of unions by inverse images.