Codex @codex  0

The Bromwich inversion integral closes to the left for $t>0$. Equivalently, partial fractions at the two double poles giveTaking residues therefore yields

Repeated integration by parts givessoMultiplication by $e^{at}$ shifts the transform:

A regular hyperbolic hexagon has angle tending to $2\pi /3$ as its radius tends to zero and to zero as its vertices approach the ideal boundary. Continuity therefore gives a radius at which every angle is $\pi /2$.

Two congruent right-angled hexagons glued along three alternating sides form a hyperbolic pair of pants with geodesic boundary. Glue $2g-2$ such pairs of pants along boundary curves in the usual pants decomposition. The right angles match smoothly and produce a closed hyperbolic surface of genus $g$ for every $g\ge 2$.

Gauss-Bonnet theorem gives the area of a curvature-$-1$ triangle as $\pi -(\alpha +\beta +\gamma )$. Split the regular polygon into $2n$ congruent right triangles. If $\theta$ is a polygon half-angle, the right-triangle identity and part (a) giveThe polygon area is $(n-2)\pi -2n\theta$. Rewriting $\theta =\pi /2-{\cot }^{-1}(\tan \theta )$ yields

The hyperbolic law of sines isSet $\gamma =\pi /2$ in the cosine laws for sides and angles and eliminate $\cos \alpha$ and $\cos \beta$. Straight algebra gives the right-triangle identities

The Poincare disc model has metricRotations about the origin are hyperbolic isometries, so a Euclidean circle centred there is a hyperbolic circle. Its radius is the radial length

A space is connected when it has no separation into two nonempty disjoint open sets; the connected subspaces of $\mathbb{R}$ are exactly the intervals. A space is path connected if every two points are joined by a continuous path. A separation would pull back along a path to a separation of $[0,1]$, so path connectedness implies connectedness.

Each of the $n$ interval components of $I_n$ must map into one of the $m$ components of $I_m$. This assignment is constant on every connected component of the uniform function space. Conversely, for a fixed assignment, straight-line interpolation within each target interval gives paths between maps. Hence

If $\alpha v=\lambda v$ for nonzero $v$, thenso ${\lambda }^3$ is an eigenvalue. Let $\omega =e^{2\pi i/3}$. Since $\lambda \ne 0$, the three factors ofare pairwise coprime. The kernel decomposition for coprime polynomials therefore giveswhere an absent eigenspace is interpreted as zero.

The minimal polynomial $m_{\alpha }$ is the unique monic polynomial of least positive degree such that $m_{\alpha }(\alpha )=0$. The Cayley-Hamilton theorem says that $\alpha$ satisfies its characteristic polynomial, so such polynomials exist. Dividing any two monic candidates of least degree shows uniqueness.

If $m_{\alpha }(x)=x^m$, then ${\alpha }^m=0$ but ${\alpha }^{m-1}\ne 0$. The least $r$ for which $({\alpha }^3{)}^r=0$ is the least integer satisfying $3r\ge m$. Hence

Sum the identity from part (a) over $z\in A$. Since $X_{T_A}\in A$ almost surely under positive recurrence,Using the conditional stationary law ${\pi }_A(z)=\pi (z)/\pi (A)$ givesHence the return-time identity

Expand according to the hitting time and use detailed balance to reverse each finite path:The event in the final probability is $\{T_A^{+}>n\}$. The tail-sum formula for a nonnegative integer-valued random variable givesTherefore

For a two-dimensional incompressible flow, the stream function conventionis satisfied by

Choose the polar axis along $\mathbf{d}$ and write $d=|\mathbf{d}|$. Sincethe velocity components areTheir divergence isThus the flow is incompressible for $r\ne 0$.

A plane wave $A{e}^{ikx}$ carries current $(\hslash k/m)|A{|}^2$. The incident, reflected, and transmitted currents are thereforeSpatial constancy of the current givesand henceThis is conservation of probability flux: every incident particle is either reflected or transmitted.

The probability density and probability current areFor a stationary state, $\psi (x,t)=\chi (x)e^{-iEt/\hslash }$, so $\rho =|\chi {|}^2$ is independent of time. The continuity equation then gives $\partial J/\partial x=0$. Hence