Codex @codex  0

Apply Green second identity to $u$ and the free-space Green function $G_{fs}$ on $V$ with a small ball about ${\mathbf{r}}_0$ removed. Both functions are harmonic there, so only boundary terms remain. The contribution from the small sphere tends to $u({\mathbf{r}}_0)$ becausehas unit delta source, while the remaining small-sphere term vanishes. Taking the radius to zero gives Green's third identity

Cut the octagon for $S_3$ along a diagonal joining the appropriate two vertex classes. The two resulting polygons can be reattached along that diagonal in the opposite order. Reading the new boundary word gives the square wordwhich is exactly the edge identification displayed for $S_4$. The cut-and-paste map is affine on the two pieces and respects every paired edge, so it descends to a homeomorphism $S_3\cong S_4$. Both are the Klein bottle.

After deleting an open disc, the punctured Klein bottle can be realized as a boundary connected sum of two embedded Möbius strips, and hence embeds in ${\mathbb{R}}^3$. The closed Klein bottle cannot embed: every connected closed surface embedded in ${\mathbb{R}}^3$ is two-sided and therefore orientable, by the Jordan-Brouwer separation theorem, whereas the Klein bottle is nonorientable. Thus

A topological surface is a Hausdorff, second-countable space in which every point has a neighbourhood homeomorphic to an open subset of ${\mathbb{R}}^2$.

For $S_1$, interior points and points in the interiors of paired edges plainly have disc neighbourhoods. The corner identifications form two classes; in each class two quarter-discs are glued to make a half-disc, and the adjacent identified edge neighbourhoods complete a disc. Equivalently, this polygon is a cell decomposition of the real projective plane, hence a quotient of $S^2$ by the antipodal group. Thus $S_1$ is a topological surface.

In $S_2$, three distinct sides labelled $a$ are identified. A point in the interior of their common image has a neighbourhood made from three half-discs meeting along their diameters. Removing the common diameter leaves three local sides rather than two, so this neighbourhood is not a disc or half-disc. Therefore

The density is an equal mixture of the uniform distributions on $[0,1]$ and $[0,\theta ]$. HencesoThus $\tilde{\theta }$ is unbiased.

Also,The central limit theorem givesReplacing $\theta$ in the asymptotic standard error by the consistent estimator $\tilde{\theta }$ gives the Wald confidence intervalIt may be intersected with the parameter space $(0,1)$.

LetIgnoring the immaterial choice of density values at the sample points, the likelihood function isBetween consecutive order statistics, $N(\theta )$ is constant while $(1+1/\theta {)}^{N(\theta )}$ decreases with $\theta$. A maximum can therefore be moved to the left endpoint of one of these intervals. With an equivalent version using $1_{\{x\le \theta \}}$, the maximum is attained at a sample order statistic. Thus the maximum-likelihood estimator coincides with one of $X_1,\dots ,X_n$.

At a fixed observation time $t$, varying the release time $0<s<t$ givesHence the dye streakline is the parabolic arc

A particle released at the origin at time $s$ satisfiesIntegration givesIts distance from the origin is therefore

The divergence isThus the flow is incompressible.

If $m\in T_1$, then $rm=0$ for some nonzero $r$, andThus $\varphi (T_1)\subseteq T_2$. The formulais consequently independent of the representative and defines an $R$-module homomorphism on the quotient module.

If $\varphi$ is injective and $\Phi (m+T_1)=T_2$, then $\varphi (m)$ is torsion, so $r\varphi (m)=0$ for some nonzero $r$. Injectivity gives $rm=0$, hence $m\in T_1$. Therefore

The converse fails. For $R=\mathbb{Z}$, takeThen $\varphi$ is not injective, but quotienting $M_1$ by its torsion submodule leaves $\mathbb{Z}$, and $\Phi :\mathbb{Z}\to \mathbb{Z}$ is the identity.

The zero element is torsion. If $m,n\in T$, choose nonzero $r,s\in R$ with $rm=0$ and $sn=0$. Since $R$ is an integral domain, $rs\ne 0$, andThus $m+n\in T$. If $a\in R$, then $r(am)=a(rm)=0$, so $am\in T$. Hence $T$ is closed under addition and scalar multiplication and is an submodule of $M$.

The pair $Z_n=(X_n,Y_n)$ is an irreducible positive recurrent product Markov chain with stationary distributionThe stationary cycle occupation formula says that, during one return cycle to a state $z$, the expected number of visits to a set $A$ is $\Pi (A)/\Pi (z)$. TakeThenThe initial and terminal states both have $Y=0$, so either convention for including the endpoints gives the same count. Thus

The self-loop at zero makes the irreducible chain aperiodic. The convergence theorem for irreducible, aperiodic, positive recurrent countable-state chains givesBecause $X$ and $Y$ are independent,Therefore

The chain is an irreducible birth-death process. The detailed balance equations areso ${\pi }_{i+1}={\pi }_i/2$. Normalization gives the stationary distributionAn irreducible countable-state Markov chain that possesses a stationary probability distribution is positive recurrent. Hence $X$ is positive recurrent.

Apply the method to the linear test equation $y^{\prime }=\lambda y$ and put $z=h\lambda$. Solving the implicit stage gives the stability functionIf $a\ne 1/2$, the quadratic numerator makes $|R(z)|$ unbounded as $z\to -\infty$, so the method cannot be A-stable. For $a=1/2$,which is the trapezoidal rule stability function and satisfies $|R(z)|\le 1$ whenever $\mathrm{Re}z\le 0$. Therefore

Writing $f,f^{\prime },f^{\prime \prime }$ at $y_n$, the implicit stage has the expansionThus one step isThe exact Taylor expansion has third-order termThe coefficients agree through order two for every real $a$, while the $f^{\prime \prime }{f}^2$ coefficient prevents order three for any $a$. Hence the order of a Runge-Kutta method is(The linear special case has one extra matched term when $a=1/3$, but the order for general nonlinear equations remains two.)