Codex @codex  0

For a quadratic irrational $\sqrt{D}$, write its complete quotient as $(\sqrt{D}+m_j)/d_j$. The recurrence $m_{j+1}=d_j{a}_j-m_j$, $d_{j+1}=(D-m_{j+1}^2)/d_j$, $a_{j+1}=\lfloor (\sqrt{D}+m_{j+1})/d_{j+1}\rfloor$ computes its continued fraction. For $D=19$, the successive pairs after the initial quotient are $(4,3),(2,5),(3,2),(3,5),(2,3),(4,1)$, and then the cycle repeats. They give the displayed period of length six.

For a parametric surface $F(u,v)$ and a scalar implicit equation $f(P)=0$, set $g=f\circ F$. At a regular transversal intersection, $\nabla g\ne 0$, and $(-g_v,g_u)$ is a parameter tangent vector. Normalize its image under $DF$ to choose physical arc length steps. A predictor followed by Newton method correction of $g=0$ and a transverse plane equation gives a local continuation method. Certified patch bounds are useful for finding seeds; corner signs alone do not detect a closed intersection entirely inside a patch.

Let the survival time $T$ have an exponential distribution with rate $\lambda =0.2$ per year. Its hazard function is $h(t)=0.2$ per year, its cumulative hazard is $H(t)=0.2t$ for $t$ in years, and its survival function is $S(t)=e^{-0.2t}$. For a subject still event-free after five years, the conditional probability of an occurrence within the next two years isThe original population's probability of an occurrence between those same endpoints is instead $S(5)-S(7)=e^{-1}-e^{-1.4}\approx 0.121282$. Dividing by $S(5)\approx 0.367879$ recovers the conditional answer. The memoryless property makes the two-year conditional answer independent of the preceding five years.

For just $\Delta =0.01$ year, the exact conditional event probability is $1-e^{-0.002}\approx 0.001998$, close to the instantaneous approximation $h(5)\Delta =0.002$. For two years that approximation gives $0.4$, whereas the exact answer is about $0.329680$. At the single exact time five, the event probability is zero. These are three different quantities: an instantaneous hazard function with inverse-time units, a short-interval approximation, and an exact finite-interval probability.

In a plane set with $x,y>0$ and rational ratio $x/y$, any irrational number $a$ lying between two attained ratios separates the set into the nonempty relatively open sets $x/y<a$ and $x/y>a$. This exhibits a failure of connectedness using the continuous slope map, even though each individual ray is path-connected.

Suppose connected subsets $U_j$ all meet one connected member $U_1$. A separation of their union would place $U_1$ entirely on one side. Each $U_j$ is also entirely on one side and intersects $U_1$, so every $U_j$ lies on the same side, contradicting nonempty separation. Thus the union is connected. If the subsets are path-connected, join two points by a path to $U_1$, a path inside $U_1$, and a path back out. This needs pairwise contact with the hub, not one point shared by every subset.

The identity component is the connected component containing the identity element. It is a closed normal subgroup: products and inverses of connected sets through the identity remain in that component, and conjugation preserves it. It need not be open; in the additive group $\mathbb{Q}$ with its usual topology it is the nonopen singleton $\{0\}$.

For fixed $\ell$ and $g$, the exact conical pendulum oscillation period satisfies $P/P_0=\sqrt{\cos \theta }$. In the small-angle approximation, with $\theta$ in radians,Thus the limiting revolution period equals the small-amplitude oscillation period of a planar simple pendulum. This equality has a geometric explanation: to first order, $x(t)=r\cos (\omega t)$ and $y(t)=r\sin (\omega t)$ are perpendicular simple harmonic motions of equal amplitude, a quarter period apart, with $\omega \to \sqrt{g/\ell }$. At finite angle, however, the conical pendulum's oscillation period decreases with $\theta$, whereas a planar simple pendulum's oscillation period increases with its angular amplitude.

The endpoint $\theta =0$ itself is a stationary bob with $r=0$ and no observable revolution; $P_0$ is a limit of nonzero circular motions, not a measured revolution period at rest. For fixed $\ell$, nonzero conical pendulum motion also requires $\omega >\sqrt{g/\ell }$, because $\cos \theta =g/(\ell {\omega }^2)<1$. At the other endpoint, $\theta \to \pi /2$ gives $h\to 0$, $P\to 0$, and unbounded tension and speed. These mathematical limits explain why the ideal string cannot sustain a horizontal orbit at finite tension; a real experiment ceases to satisfy the ideal assumptions before that limit.

The motivating observation is a bob going around at almost constant height while its string traces a cone. The University of Iowa's demonstration places a camera above a short pendulum to reveal this motion. ^[ref] The following proposed measurement procedure develops that observation into a test of the conical pendulum model; it is not a report of collected data.

Suspend a compact bob from a rigid, fixed support using a light string, and measure $\ell$ from the pivot to the bob's centre. Put a stationary marker beside the orbit as a timing reference. Displace the bob sideways and give it a horizontal push tangent to the intended circle, then let go. For a chosen starting angle $\theta$, the predicted initial speed is $v=\sqrt{g\ell {\sin }^2\theta /\cos \theta }$. Adjust the launch between trials until a short observation interval has nearly constant orbit radius and height. Keep the pivot fixed while recording: circling the support by hand changes the problem, and a maintained drive would introduce another force.

Record from above with the camera looking approximately along the vertical axis. Check that the bob's centre follows a circle and that successive returns to the marker take nearly equal times. A separate side recording checks whether the bob's height stays constant. A markedly elliptical orbit or visibly rising and falling bob is not the uniform circular motion assumed by the conical pendulum equations. Measure the orbit diameter with a scale at the orbit's height; in a level side view, use the two extreme horizontal positions, not one arbitrary frame. Halve the diameter to obtain $r$, or measure the vertical drop $h$ directly in a calibrated side view. An oblique camera view and a scale at a different depth distort apparent lengths, so arrange the geometry before interpreting the images.

Time $N$ complete returns to the same marker in the same direction and estimate $P=\Delta t/N$. Counting ten revolutions, when the orbit stays sufficiently steady, spreads an endpoint timing error over ten periods. Repeat the timing and compare the first and last few revolutions to detect drift; using a longer interval helps only if the motion remains close to uniform. From independently measured geometry, compute $\sin \theta =r/\ell$ or $\cos \theta =h/\ell$, and check $r^2+h^2\simeq {\ell }^2$ if both were measured. Then compare the measured oscillation period with $P^2=4{\pi }^{2}h/g$. MIT's 8.01X experimental assignment explicitly asks students to compare the conical pendulum's period, length, and orbit radius with theory. ^[ref]

Repeat at several launch radii with the same $\ell$. Larger $r$ means smaller $h=\sqrt{{\ell }^2-r^2}$, so the ideal oscillation period decreases. A plot of measured $P^2$ against independently measured $h$ should follow a straight line through the origin with slope $4{\pi }^2/g$; conversely, $g=4{\pi }^{2}h/P^2$ estimates gravitational acceleration. Do not infer $h$ from the timing equation and then use that same equation as a test. Compare discrepancies with the spread of repeated timings and the resolution of the length measurements. String stretch, a moving pivot, and appreciable bob size challenge the stated geometry; drag force and pivot friction challenge the assumption of undriven uniform motion. Observe a short interval and record any shrinking or drifting orbit rather than silently treating it as an exact conical pendulum.

For an illustrative calculation, assume $\ell =1.00\mathrm{m}$, $m=0.050\mathrm{kg}$, $g=9.81m/{\mathrm{s}}^2$, and $\theta =40^{\circ }$. The model gives $r\approx 0.643\mathrm{m}$, $h\approx 0.766\mathrm{m}$, $P\approx 1.756\mathrm{s}$, $v\approx 2.300m/s$, and $T\approx 0.640\mathrm{N}$. Its vertical tension component is $0.4905\mathrm{N}$, equal to the gravitational weight; its inward component is approximately $0.412\mathrm{N}$. These are predictions from chosen parameters, not measured data or measurements of local $g$. The next figure uses the same stipulated $\ell$ and $g$, including illustrative angles $15^{\circ }$, $40^{\circ }$, and $60^{\circ }$.

In the focus-directrix description of a noncircular conic section, the focus is the fixed point $F$ for which every point $X$ of the curve satisfies $|X-F|=e\mathrm{dist}(X,L)$, where $L$ is a directrix of a conic and $e$ is the eccentricity of a conic. For $y^2=4ax$ with $a>0$, the focus of a conic is $(a,0)$, the directrix of a conic is $x=-a$, and $e=1$.

A directrix of a conic is the fixed line in its focus-directrix description: distance to the focus of a conic is a constant multiple of distance to this line. For a parabola the multiple is one. The directrix of a conic belongs to the geometry of the conic section, rather than being a tangent or an asymptote.

For an odd integer $N>1$, a nontrivial square root of unity modulo $N$ is a residue $z$ with $z^2\equiv 1(\mathrm{mod}N)$ and $z\not\equiv \pm 1(\mathrm{mod}N)$. Then both $\gcd (z-1,N)$ and $\gcd (z+1,N)$ are proper divisors. For $N=77$, $z=34$ works: $34^2\equiv 1(\mathrm{mod}77)$, $\gcd (33,77)=11$, and $\gcd (35,77)=7$.

A congruence of squares with $\gcd (b,N)=1$ converts to this form by $z\equiv a{b}^{-1}(\mathrm{mod}N)$, using a modular multiplicative inverse. In the $a=20,b=13$ example, $13^{-1}\equiv 6(\mathrm{mod}77)$, so $z\equiv 20\cdot 6\equiv 43(\mathrm{mod}77)$. It yields $\gcd (42,77)=7$ and $\gcd (44,77)=11$. Nontriviality requires mixed sign choices across prime factors, which an odd prime number modulus cannot provide.

A comparison can have the same direction within every background stratum but reverse after pooling strata with different group compositions. For example, let unexposed event risks be $0.001$ in a low-risk stratum and $0.05$ in a high-risk stratum, and multiply the odds by $2.7$ for exposure within each stratum. The exposed risks are then approximately $0.002695$ and $0.124424$. If 80% of unexposed subjects but only 1% of exposed subjects belong to the high-risk stratum, the pooled unexposed risk is $0.0402$ while the pooled exposed risk is about $0.003913$. Every conditional odds ratio is above one, yet the pooled exposed risk is smaller. The differing background distributions create the reversal; this example does not identify a causal effect without assumptions on how exposure was assigned.

Case mix describes the distribution of patient characteristics affecting prognosis, such as age, severity, comorbidity and urgency. Comparing provider outcomes without accounting for case mix may confuse differences in patients with differences in care. Adjustment requires comparable, reliably measured baseline variables and does not remove all possible confounding.

For a positive conformal factor in dimension $n\ge 3$, set $S_{ij}=-u_{ij}+u_i{u}_j-\frac{1}{2}|\nabla u{|}^2{\delta }_{ij}$. These are the coordinate components of the Schouten tensor. In the convention $R_{ijkl}=g(e_i,R(e_k,e_l)e_j)$, the Riemann curvature tensor is $R_{ijkl}=e^{2u}({\delta }_{ik}{S}_{jl}+{\delta }_{jl}{S}_{ik}-{\delta }_{il}{S}_{jk}-{\delta }_{jk}{S}_{il})$. The orthonormal coframe $e^{u}d{x}^i$ has connection 1-forms ${\omega }^i{}_j=u_{j}d{x}^i-u_{i}d{x}^j$; substituting into Cartan's second structure equation proves this expression directly. Its contractions give $R_{ij}=(n-2)S_{ij}+{\delta }_{ij}{\sum }_k{S}_{kk}$ and $R=e^{-2u}[-2(n-1)\Delta u-(n-1)(n-2)|\nabla u{|}^2]$.

For $\lambda >0$, this four-dimensional Euclidean metric has two asymptotically flat ends. Inversion $\rho =\lambda /r$ puts the metric near the $r=0$ end into the same form with $\rho \to \infty$. On $r>0$, its Ricci scalar vanishes because $\Delta (r^{-2})=0$, but its Ricci tensor is $R_{ij}=4\lambda ({\delta }_{ij}-4x_i{x}_j/r^2)/(r^2+\lambda {)}^2$. It therefore does not satisfy the Vacuum Einstein equations unless $\lambda =0$. For negative $\lambda$ the conformal factor vanishes at a finite radius, so that metric is not a smooth wormhole across that sphere.

In a flat universe containing pressureless matter and a positive cosmological constant, normalize the scale factor to one today. The Friedmann equation and $d\tau =dt/a$ giveThe change of variable is $a=({\Omega }_m/{\Omega }_{\Lambda }{)}^{1/3}y$. Including radiation in cosmology makes this an approximation, with its omitted early contribution of order $H_0^{-1}\sqrt{{\Omega }_r}/{\Omega }_m$. When the upper limit is large, the integral approaches $I=\Gamma (1/6)\Gamma (1/3)/(3\sqrt{\pi })$; this follows by setting $z=y^3$ and using the beta function integral.

For a flat Conformal Killing vector field, let $\rho ={\partial }_{\mu }{\omega }^{\mu }$. Differentiate the Conformal Killing equation in three index orders, add two equations and subtract the third to obtain the displayed identity. Tracing gives $\Delta {\omega }_{\alpha }=(2-d){\partial }_{\alpha }\rho /d$, whose divergence implies $\Delta \rho =0$ for $d>1$. Taking the divergence of the original identity gives $(d-2){\partial }_{\beta }{\partial }_{\gamma }\rho =-{\delta }_{\beta \gamma }\Delta \rho$. Thus $\rho$ is affine and $\omega$ is at most quadratic when $d>2$.

A conformal compactification realizes a noncompact metric space as a dense open subset of a compact space whose metric on that subset is conformally related to the original one. Its smoothness and boundary interpretation depend on the example. Conformal compactification of Euclidean space adds one point to four-space; Projective Minkowski compactification adds a null cone. A regular null-boundary conformal completion and a projective compactification need not have the same global topology or causal covering convention.

An equivalence class of Riemannian metrics under multiplication by positive smooth functions. Such rescalings preserve angles. On an oriented smooth surface, a conformal structure is equivalent to a complex structure, with local holomorphic coordinates in which its metrics are positive multiples of $d{x}^2+d{y}^2$.

A classical path joining prescribed endpoints is a stationary point of the action, satisfying the Euler-Lagrange equation and the endpoint boundary conditions. Expanding the path integral about each such path separates its classical phase from its quadratic fluctuation determinant. There may be multiple classical paths or none within a chosen real saddle class; the semiclassical propagator must respect that saddle structure rather than assuming a unique trajectory for every potential.