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For a coordinate curve $v=v_0$, the length between $u_1$ and $u_2$ iswhere $E$ is a coefficient of the first fundamental form. Thus $\partial E/\partial v=0$ makes the two $u$-directed sides of every coordinate rectangle equally long. Similarly,so $\partial G/\partial u=0$ makes the two $v$-directed sides equally long.

Conversely, suppose every pair of opposite sides has equal length. Fix $v_0,v_1$. Equality of the first pair for every interval $[u_1,u_2]$ givesThe fundamental theorem of calculus implies that the continuous integrand vanishes, so $E$ is independent of $v$. The same argument shows that $G$ is independent of $u$. Hence the stated equal-length property is equivalent to $E_v=G_u=0$.

The Cayley-Hamilton theorem states that every square matrix satisfies its characteristic polynomial: if ${\chi }_A(t)=\det (tI-A)$, then ${\chi }_A(A)=0$.

For the proof, the adjugate identity givesWrite $\mathrm{adj}(tI-A)=B_0+B_{1}t+\cdots +B_{n-1}{t}^{n-1}$ and compare coefficients. Multiplying the resulting recurrence by successive powers of $A$ and summing telescopically yields ${\chi }_A(A)=0$.

By polynomial division, $p=q{\chi }_A+r$ with $\mathrm{deg}r<n$. Therefore $p(A)=r(A)$ by Cayley-Hamilton, while $p(\lambda )=r(\lambda )$ for every eigenvalue $\lambda$, since ${\chi }_A(\lambda )=0$.

For the displayed matrix,The remainder of $t^{1000}$ modulo this polynomial has the form $a{t}^2+c$, because its values at $i$ and $-i$ are both $1$. The equations $-a+c=1$ and $4a+c=2^{1000}$ giveSince $(A^2{)}_{11}=3$, the requested entry is

A function $f:{\mathbb{R}}^n\to \mathbb{R}$ is a convex function whenfor every $x_0,x_1$ and $t\in [0,1]$.

Fix $b_0,b_1\in \mathbb{R}$, $t\in [0,1]$, and $\epsilon >0$. Since the value function is finite, choose $x_i$ withFor the convex combination $x_t=(1-t)x_0+t{x}_1$, convexity of $g$ givesso $x_t$ is feasible for the intermediate right-hand side. Convexity of $f$ then givesLetting $\epsilon \downarrow 0$ provesso $\varphi$ is convex.

A simple hypothesis specifies a single probability distribution. For a test with rejection region $R$, its size of a statistical test is the rejection probability under the simple null hypothesis, while its statistical power against the simple alternative is the rejection probability under that alternative. The Neyman-Pearson lemma states that among tests of size at most $\alpha$, a test rejecting for the largest values of the likelihood ratio $f_1/f_0$ has greatest power, with boundary randomization if needed.

Herewhich decreases with $|x|$. The best test therefore rejects for $|X|\le c$. Under $H_0$, ${\mathbb{P}}_0(|X|\le c)=c$, so size $0.05$ requires $c=0.05=1/20$. Its power isThus reject $H_0$ exactly when $|X|\le 0.05$.

Write $\mathbf{g}=-g{\mathbf{e}}_z$. The vertical component of the balance gives hydrostatic pressure,Its horizontal gradient is ${\nabla }_{h}p=\rho g{\nabla }_{h}h$. The horizontal balance is consequentlyThus $\mathbf{u}\cdot {\nabla }_{h}h=0$: the velocity is tangent to every level set of $h$, so those contours are streamlines of the geostrophic flow.

Across a vertical section joining the contours $h_0$ and $h_0+\Delta h$, an element of horizontal distance normal to the contours is $dn=dh/|{\nabla }_{h}h|$, while $|\mathbf{u}|=(g/f)|{\nabla }_{h}h|$. The volumetric flow rate is thereforeFor dimensional analysis, $[g]=L{T}^{-2}$, $[f]=T^{-1}$ and $[h_0]=[\Delta h]=L$, so the result has dimension $L^3{T}^{-1}$, exactly that of a volume flux.

For a time-independent current, the Maxwell equations reduce to magnetostatics:Write $\mathbf{B}=\nabla \times \mathbf{A}$ using a vector potential and choose the Coulomb gauge $\nabla \cdot \mathbf{A}=0$. The vector identity for the curl of a curl gives the Poisson equationThe free-space Green function therefore givesTaking the curl with respect to $\mathbf{r}$, moving it under the integral, and using $\nabla (1/|\mathbf{r}-{\mathbf{r}}^{\prime }|)=-(\mathbf{r}-{\mathbf{r}}^{\prime })/|\mathbf{r}-{\mathbf{r}}^{\prime }{|}^3$ gives the Biot-Savart law

Along a curve $s\mapsto (x(s),y(s))$, the chain rule givesTogether with the second-order partial differential equation, these form a linear system for $(u_{xx},u_{xy},u_{yy})$ whose coefficient matrix isA characteristic curve is precisely one along which this system is singular, equivalently where the principal symbol vanishes on a normal covector. Taking the determinant yields