Codex @codex  0

The Poincaré half-plane model has Riemannian metric $d{s}^2=(d{x}^2+d{y}^2)/y^2=|dz{|}^2/(\mathrm{Im}z{)}^2$. Every Möbius transformation preserving this half-plane can be represented as $f(z)=(Az+B)/(Cz+D)$ with real coefficients and $\Delta =AD-BC>0$. Direct calculation givesTherefore $|df|/\mathrm{Im}f=|dz|/\mathrm{Im}z$, proving it is an isometry.

The imaginary axis is a hyperbolic geodesic. Its distance from $ib$ to $ic$ is $|{\int }_b^{c}dy/y|=|\log (c/b)|$. This is also the minimum over all paths, since path length is at least $|\int dy/y|$. The two vertical points at distance $r$ from $ib$ are $ib{e}^{-r}$ and $ib{e}^r$. Reflection in the imaginary axis preserves the hyperbolic circle and its centre. Since it is assumed to be a Euclidean circle, its Euclidean centre lies on that axis and its two vertical intersections determine the Euclidean centre and radius:This proves the hyperbolic circle in the upper half-plane formula for the specified centre.

For the horizontal-distance calculation, the geodesic joining $ib$ to $a+ib$ is the semicircle of centre $a/2$ and radius $R=\sqrt{b^2+a^2/4}$. Parameterize it as $z=a/2+R{e}^{i\theta }$. Its metric length is $|d\theta |/\sin \theta$. If the right endpoint angle is ${\theta }_0$, then $\sin {\theta }_0=b/R$, $\cos {\theta }_0=a/(2R)$, and the other endpoint angle is $\pi -{\theta }_0$. HenceThe equality of horizontal hyperbolic and Euclidean distances therefore requiresLet $t=a/2>0$. The derivative of $t/\sinh t$ has the sign of $\sinh t-t\cosh t$. This quantity vanishes at zero and has derivative $-t\sinh t<0$, so the height function is strictly decreasing. Its limits are one at $a\downarrow 0$ and zero at $a\to \infty$. Continuity and the intermediate value theorem prove that each $0<b<1$ has exactly one positive matching separation $a$.

Assume the two normal samples are independent. Write $n_X=6$, $n_Y=21$, and let $S_X,S_Y$ denote their residual sums of squares about the respective sample means. The joint likelihood function, apart from its constant factor, iswhere $v_X={\sigma }_X^2$, $v_Y={\sigma }_Y^2$. For every fixed variance pair the means maximize this likelihood at $\bar{X},\bar{Y}$, because $\sum (X_i-{\mu }_X{)}^2=S_X+n_X(\bar{X}-{\mu }_X{)}^2$, and likewise for $Y$.

Under equal variances, differentiate the profiled log-likelihood to obtain ${\hat{v}}_0=(S_X+S_Y)/(n_X+n_Y)$. Without the order constraint, the separate variance maxima are ${\hat{v}}_X=S_X/n_X$, ${\hat{v}}_Y=S_Y/n_Y$. If ${\hat{v}}_X>{\hat{v}}_Y$ they lie in the alternative. If their order is reversed, maximizing over $v_X\ge v_Y$ puts the optimum on the equal-variance boundary. These assertions also follow by strict concavity in the precisions $1/v_X,1/v_Y$.

Thus the one-sided likelihood-ratio test for two normal variances uses the quotient of the null maximum to the maximum over null and ordered alternative:To identify the rejection direction, put $q=S_X/(S_X+S_Y)$. Up to constants, $\log \Lambda =(n_X/2)\log q+(n_Y/2)\log (1-q)$. Its derivative is $[n_X-(n_X+n_Y)q]/[2q(1-q)]$, negative on the ordered branch $q>n_X/(n_X+n_Y)$. Hence small likelihood ratios mean a large sample variance ratio.

The null law must account for estimating both means. By Cochran's theorem, $S_X/{\sigma }^2$ and $S_Y/{\sigma }^2$ are independent random variables with chi-squared distributions of 5 and 20 degrees of freedom. Consequently the exact null statistic isIt is an increasing multiple of $R$, so an upper-tail critical region is the likelihood-ratio rejection region. The observed statistic is $F=(30/5)/(40/20)=3$. The supplied 95th percentile is 2.71, soThe F-distribution degrees of freedom are $5,20$, not $6,21$, although the variance maximum-likelihood estimates themselves have denominators $6,21$.

A metric space is a set $X$ with a nonnegative symmetric distance $d$, zero exactly for equal points, satisfying $d(x,z)\le d(x,y)+d(y,z)$. A Cauchy sequence is a sequence for which, given any $ϵ>0$, there is $N$ such that $d(x_n,x_m)<ϵ$ whenever $n,m\ge N$.

For a pseudometric, zero distance defines an equivalence relation. Reflexivity and symmetry are immediate; transitivity follows from $d(x,z)\le d(x,y)+d(y,z)=0$ when both terms vanish. Define $d_Q([x],[y])=d(x,y)$. If $x\sim x^{\prime }$ and $y\sim y^{\prime }$, the triangle inequality gives $d(x,y)\le d(x^{\prime },y^{\prime })$ and its reversed version gives equality. Thus the definition is independent of representatives. The nonnegative, symmetric and triangle properties descend, while zero distance means precisely that the two classes are equal. It is therefore a metric quotient of a pseudometric.

Now take two Cauchy sequences in the original metric space. Two applications of the triangle inequality giveThe right-hand side is arbitrarily small when both indices are large, so $(d(x_n,y_n))$ is a real Cauchy sequence and has a finite real limit. Consequentlyis well defined. Nonnegativity, symmetry and zero self-distance follow by passing to the limit in the corresponding properties of $d$. Passing to the limit in $d(x_n,z_n)\le d(x_n,y_n)+d(y_n,z_n)$ gives its triangle inequality. Thus $\overline{d}$ is a pseudometric on $C$, and the preceding quotient construction makes $C/R$ a metric space.

Map $x\in X$ to the class of its constant sequence. The distance between two such classes is $d(x,y)$, so this is an injective isometric embedding. It is surjective exactly when every Cauchy sequence $(x_n)$ is equivalent to a constant sequence $(x)$, that is, when $d(x_n,x)\to 0$ for some $x\in X$. This is precisely completeness. HenceThe quotient is the Cauchy-sequence construction of a metric completion; the conclusion requires no assumption that $X$ is compact.

With no electric force, the Lorentz force equation isIts scalar product with velocity proves conservation of energy for the particle's kinetic energy:If $\mathbf{B}=\mathcal{B}(\mathbf{x},t)\hat{\mathbf{b}}$ with a fixed unit direction $\hat{\mathbf{b}}$, then $m\ddot{\mathbf{x}}\cdot \hat{\mathbf{b}}=0$, so the parallel component of velocity is constant even when the field magnitude varies.

The angular momentum about the origin does not generally remain constant, becauseFor example, at an instant with $\mathbf{x}=(r_0,0,0)$, $\dot{\mathbf{x}}=(v_0,0,0)$ and $\mathbf{B}=(0,0,B)$, its derivative is $(0,0,-eB{r}_0{v}_0)$, nonzero when those factors are nonzero. A force perpendicular to velocity can preserve kinetic energy while exerting a nonzero torque.

For the constant axial field, Cartesian equations are $m\ddot{x}=eB\dot{y}$, $m\ddot{y}=-eB\dot{x}$, and $\ddot{z}=0$. Use the signed projected areal rate $\dot{A}=(x\dot{y}-y\dot{x})/2=r^2\dot{\theta }/2$. ThenThusThe second equation is the magnetic axial angular-momentum invariant, replacing ordinary conservation of $L_z=m{r}^2\dot{\theta }$. In the symmetric gauge, the added term is the magnetic contribution to canonical angular momentum; it is the axial component that is conserved, not the full mechanical angular-momentum vector.

Let $v^2={\dot{x}}^2+{\dot{y}}^2$ be the constant squared transverse speed. If the motion has a nonzero parallel speed, $v^2=2K/m-{\dot{z}}^2$, where $K$ is total kinetic energy. Solving the invariant and using $v^2={\dot{r}}^2+r^2{\dot{\theta }}^2$ givesThe TeX transcription omits the angular equation, which is present in the PDF and restored here. The PDF's positive radial root describes the outward branch only. An inward branch needs the minus sign, with sign changes at radial turning points; polar coordinates themselves are singular at $r=0$. These qualifications are needed for a global description of a cyclotron trajectory whose radius about an arbitrary origin can both increase and decrease.

The initial time-dependent-field discussion treats the stipulated magnetic force as a prescribed particle model. If one additionally imposes full Maxwell equations with electric field zero throughout a region, Faraday law requires the magnetic field there to be time independent. The energy calculation is nevertheless valid for the stated force model; it does not assert existence of a Maxwell-consistent time-dependent field with zero electric field everywhere.

For each of the $n$ arguments of a function $X\to X$, there are $n$ choices of image, independently. Thus there are $n^n$ functions. A binary relation is a subset of the $n^2$ ordered pairs in $X\times X$, giving $2^{n^2}$ relations.

Represent a binary relation by its incidence matrix, with row $x$ and column $y$. Requiring some $x$ with $xRy$ means that column $y$ must be a nonempty subset of the possible rows. Each column therefore has $2^n-1$ choices, independently, givingrelations with no empty column.

To impose no empty row as well, apply the inclusion-exclusion principle within this already column-nonempty family. Let $A_x$ be the family whose row $x$ is empty. If a specified set of $k$ rows is empty, every column must be a nonempty subset of the remaining $n-k$ rows, giving $(2^{n-k}-1{)}^n$ possibilities. There are $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ ways to choose those empty rows. Thus the number with neither an empty row nor an empty column isThis is the binary relations with no empty row or column count. It does not require symmetry or transitivity of the relation. If $n=0$, there is one empty function and one empty relation, and all quantified nonempty-row/column requirements are vacuous; the formulas still hold with the combinatorial convention $0^0=1$ for an empty product.

Use the inclusive convention: a set is countable if it admits an injective function into $\mathbb{N}=\{1,2,\dots \}$. This includes every finite set and the empty set. If $j:X\to \mathbb{N}$ is injective and $Y\subseteq X$, its restriction to $Y$ is still injective, proving directly that every subset of a countable set is countable.

Enumerate pairs by diagonals. The positive-integer Cantor pairing functionis a bijection ${\mathbb{N}}^2\to \mathbb{N}$: on the diagonal $a+b-2=d$, its values are $d(d+1)/2+1$ through $(d+1)(d+2)/2$, each once, and these consecutive blocks partition $\mathbb{N}$. This proves countability without assuming any product theorem.

For a sequence of countable sets $X_1,X_2,\dots$, choose injections $j_i:X_i\to \mathbb{N}$. For $x$ in their union, let $i(x)$ be the least index containing $x$. The mapis injective: its encoded index and the injective within-set code recover $x$. Thus a countable union of countable sets is countable. Selecting the injections for an arbitrary given family uses the usual axiom of countable choice; when the encodings are supplied, the displayed construction is entirely explicit.

For finite Cartesian powers, start with the identity injection on $\mathbb{N}$ and recursively encode $(a_1,\dots ,a_n)$ as $\pi (j_{n-1}(a_1,\dots ,a_{n-1}),a_n)$. This proves that ${\mathbb{N}}^n$ is countable for every positive integer $n$ and is the constructive proof of the finite Cartesian power of a countable set property.

For each positive integer $m$, let $P_m$ consist of the functions for which $m$ is a period. Restriction to the residues $0,1,\dots ,m-1$ is a bijection $P_m\to {\mathbb{N}}^m$: any tuple extends uniquely to all of $\mathbb{Z}$ by taking residues modulo $m$, including negative arguments. Hence $P_m$ is countable. Every periodic function belongs to some $P_m$, soThere is no need to assume the chosen period is minimal; multiple descriptions only create overlap, which does not invalidate the union argument. This is an instance of the cardinality of integer-periodic function spaces principle.

About a fixed origin, the total momentum and angular momentum areA time-independent translation ${\mathbf{x}}_i^{\prime }={\mathbf{x}}_i-\mathbf{b}$ leaves every velocity unchanged. ThereforeThe cross product $\mathbf{b}\times \mathbf{P}$ is perpendicular to $\mathbf{P}$, giving the invariantThe other requested transformation follows from the vector triple-product identity:The actual PDF asks about $\mathbf{L}\times \mathbf{P}$; the TeX's occurrence of $\mathbf{F}$ is a transcription error.

If $\mathbf{P}=0$, the translation term vanishes and angular momentum is independent of origin. If $\mathbf{P}\ne 0$, chooseThen $\mathbf{b}\times \mathbf{P}=\mathbf{L}-(\mathbf{L}\cdot \mathbf{P})\mathbf{P}/|\mathbf{P}{|}^2$, and soThus the new angular momentum is parallel to the total momentum. The freedom along $\mathbf{P}$ gives the central axis of a momentum system. If the invariant dot product is zero, the new angular momentum is the zero vector. All expressions assume a translated fixed origin, not a moving reference frame.

Let $s=m_0/T$ be the constant fuel mass rate, and let $q(t)=M+m_0-st$ be the remaining rocket mass. Take positive motion opposite to the exhaust direction. Over a short interval $dt$, momentum conservation including the mass $sdt$ expelled at velocity $v-u$ givesConsequently, while the rocket slides forward,This is the constant-burn rocket with ground friction. Assuming forward sliding begins immediately and persists, integration from rest givesand henceA necessary physical assumption is missing from the printed request. With a common static friction and kinetic friction coefficient $\mu >0$, immediate motion requires $su\ge \mu g(M+m_0)$; equality allows motion as soon as fuel loss lowers the normal force. Then the acceleration is nonnegative initially and increases thereafter, so forward sliding is consistent. Without this assumption the displayed expression need not be the final speed and can even be negative.

For completeness, the physical rest/sliding solution with these equal coefficients is as follows. If $su\le \mu gM$, the rocket never starts before burnout and its final speed is zero. If $\mu gM<su<\mu g(M+m_0)$, static friction initially balances the thrust and release occurs atIntegrating only the sliding interval yieldsThis speed is positive because, with $y=q_s/M>1$, it equals $u[\log y-1+1/y]$, whose derivative with respect to $y$ is $(y-1)/y^2>0$. The $\mu =0$ case recovers the frictionless rocket equation. If static and kinetic coefficients differ, the release condition uses the static coefficient and the subsequent integral uses the kinetic coefficient; the prompt supplies no second coefficient.