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A subset of $X$ is measurable relative to a sigma-algebra $\mathcal{F}$ when it belongs to $\mathcal{F}$. A measure defined on $\mathcal{F}$ assigns sizes to these sets. Measurability depends on the chosen sigma-algebra; for example, a Lebesgue measurable set need not be a Borel set.

In a Keplerian shearing sheet, an epicyclic guiding center $x_0$ carries rotating-frame energy $-3{\Omega }^2{x}_0^2/8$, plus nonnegative radial and vertical oscillation energies. Inelastic collisions dissipate the total Jacobi energy in a shearing sheet, while momentum conservation preserves $\sum x_0$. Thus $\sum x_0^2$ must grow: the ring spreads about its fixed mean radius. The effect is an angular momentum transport process, with some particles moving inward and others outward.

Expand a unit-mass particle's Lagrangian about an axisymmetric circular orbit of radius $r_0$ and frequency ${\Omega }_0$, using $r=r_0+x$ and $\phi ={\Omega }_{0}t+y/r_0$. Circular-orbit balance removes the linear radial term; total-time-derivative invariance of a Lagrangian removes $r_0{\Omega }_0\dot{y}$. To second order,Here $q$ is the orbital shear parameter and ${\Omega }_z$ is the vertical epicyclic frequency. The Euler-Lagrange equations retain the local Coriolis acceleration and tidal gravity.

The orbital shear parameter measures the fractional radial decrease in orbital angular frequency. A Kepler orbit has $q=3/2$, while rigid rotation has $q=0$. The radial epicyclic frequency obeys ${\kappa }_r^2=2(2-q){\Omega }^2$; this follows by differentiating the squared specific angular momentum $h^2=r^4{\Omega }^2$.

The shear velocity is the velocity scale $u_{\ast }=\sqrt{|\tau |/\rho }$ associated with a shear stress $\tau$ in a fluid of mass density $\rho$. It is a stress scale, not automatically the actual local fluid velocity at a grain.

A sheaf of abelian groups is a sheaf of sets whose sections on each open set form an abelian group and whose restriction maps are group homomorphisms. Its stalk of a sheaf is the corresponding group of germs. Kernels are computed on sections, whereas cokernels are obtained by sheafification of the sectionwise cokernel. An exact sequence can therefore be checked on stalks.

A morphism of sheaves assigns a map $\mathcal{F}(U)\to \mathcal{G}(U)$ on every open set, commuting with all restriction maps. For sheaves of abelian groups, sheaves of rings or sheaves of modules, the maps must also preserve the relevant algebraic structure. It induces maps on every stalk, and equality of sheaf morphisms can be checked there.

For a thin Newtonian fluid film on a conical slope at angle $\alpha$, neglecting the thickness-gradient contribution to the driving pressure gives volume flux per unit width $q=D{h}^3$, where $D=g\sin \alpha /(3\nu )$. Radial mass conservation isThe distance $x$ is measured along the slope; the horizontal radius is $x\cos \alpha$, so total volume is $2\pi \cos \alpha \int xhdx$.

An underdamped RLC response is a free series RLC circuit response with $0<R<2\sqrt{L/C}$, equivalently $0<\alpha <{\omega }_0$. Its characteristic roots are $-\alpha \pm \mathrm{i}{\omega }_d$, and its general electric charge isFor initial conditions $q(0)=q_0$, $i(0)=0$,In the dimensionless example ${\omega }_0=1$, $\alpha =0.5$, the ringing angular frequency is ${\omega }_d=\sqrt{3}/2\approx 0.866025$, below the driven electric current resonance at ${\omega }_0=1$. The electric charge first crosses zero at $t=4\pi /(3\sqrt{3})\approx 2.41840$ and reaches its first negative extremum at $t=2\pi /\sqrt{3}\approx 3.62760$, where $q/q_0=-e^{-\pi /\sqrt{3}}\approx -0.163034$.

The sinusoidal factor changes sign while its envelope decays as $e^{-\alpha t}$. The total energy follows the exact balance $E^{\prime }=-R{i}^2$, rather than a pure exponential decay at every instant; in particular, its rate of loss is zero whenever $i=0$. As $\alpha \to 0$, the displayed solution tends to the lossless harmonic oscillator $q=q_0\cos ({\omega }_{0}t)$. As $\alpha \to {\omega }_0$ from below, $\sin ({\omega }_{d}t)/{\omega }_d\to t$, recovering the critically damped RLC response continuously. The overdamped RLC response reaches the same limit through $\sinh (\beta t)/\beta \to t$.

A critically damped RLC response is the boundary between underdamped RLC response and overdamped RLC response. The free series RLC circuit has a repeated characteristic root $-{\omega }_0$, with general solutionFor initial conditions $q(0)=q_0$, $i(0)=0$,In the plotted normalization, $q(1)=2/e\approx 0.735759$ and $i(1)=-1/e\approx -0.367879$. The electric charge returns without ringing; the polynomial factor $1+{\omega }_{0}t$ matters, so a single exponential function does not give the whole response.

For fixed $L,C$ and these initial conditions, critical damping avoids ringing while giving a faster asymptotic return than any larger $R$: the overdamped RLC response has a root closer to zero. This makes critical damping a useful tuning target for settling after switching. An MIT laboratory exercise makes the distinction observable by asking students to compare all three cases with $L=47\mathrm{mH}$ and $C=47\mathrm{nF}$. Those values give $R_{\mathrm{crit}}=2\sqrt{L/C}=2000\Omega$ and ${\omega }_0\approx 21276.6{\mathrm{s}}^{-1}$. ^[ref].

An overdamped RLC response is a free series RLC circuit response with $R>2\sqrt{L/C}$, equivalently $\alpha >{\omega }_0$. Put $\beta =\sqrt{{\alpha }^2-{\omega }_0^2}$. The distinct negative characteristic roots are $r_{\pm }=-\alpha \pm \beta$, so the general electric charge isFor initial conditions $q(0)=q_0$, $i(0)=0$,For $q_0>0$, this electric charge decreases to zero without reversing sign. In the dimensionless example ${\omega }_0=1$, $\alpha =2$, the characteristic roots are $-2\pm \sqrt{3}\approx -0.268,-3.732$. The slow term dominates late times. Stronger damping can therefore make the final return slower: for $\alpha \gg {\omega }_0$, $r_{+}\sim -{\omega }_0^2/(2\alpha )=-1/(RC)$. Nonoscillation alone does not imply a fast response, and monotonicity here uses the specified initial conditions.

Drive a series RLC circuit with electric potential difference $V(t)=V_0\cos (\omega t)$, where $V_0>0$ and $\omega >0$. After the free transient decays, the electric current has the same angular frequency as the source:For fixed positive $R,L,C,V_0$, the electric current amplitude is largest exactly at $\omega ={\omega }_0=1/\sqrt{LC}$. The two electrical reactance contributions cancel there, leaving only the drop from electrical resistance. In particular,Every solution approaches this periodic response because its difference from the displayed solution is a decaying free response. The peak remains at ${\omega }_0$ even for an overdamped RLC response; free ringing is not required for a maximum of driven electric current. ^[ref].

For example, stipulate $L=10\mathrm{mH}$, $C=1\mu \mathrm{F}$, $R=20\Omega$, and $V_0=1\mathrm{V}$. Then ${\omega }_0=10^4{\mathrm{s}}^{-1}$, $f_0={\omega }_0/(2\pi )\approx 1591.55\mathrm{Hz}$, and the peak electric current amplitude is $50\mathrm{mA}$. Holding the source amplitude fixed is essential when comparing responses at different frequencies.

Three frequencies must be distinguished. The free underdamped RLC response rings at ${\omega }_d=\sqrt{{\omega }_0^2-{\alpha }^2}$. Driven electric current peaks at ${\omega }_0$. Driven electric charge has amplitudewhose peak occurs at $\sqrt{{\omega }_0^2-2{\alpha }^2}$ only if $\alpha <{\omega }_0/\sqrt{2}$. Otherwise it decreases from its zero-frequency limit $C{V}_0$, with no positive-frequency peak. The different peak locations follow because electric current is the time derivative of electric charge, multiplying a sinusoidal amplitude by $\omega$. The limiting responses are $I_{\mathrm{amp}}\sim V_{0}C\omega$ as $\omega \to 0^{+}$ and $I_{\mathrm{amp}}\sim V_0/(L\omega )$ as $\omega \to \infty$.

The Q factor measures selectivity:The average Joule heating power is $R{I}_{\mathrm{amp}}^2/2$. Its half-maximum angular frequencies and their separation are exactlyThese follow by setting $|\omega L-1/(\omega C)|=R$. In the numerical example, $Q=5$ and $\Delta \omega =2000{\mathrm{s}}^{-1}$. Reducing $R$ makes a taller, narrower electric current peak; varying $C$ tunes the selected frequency. Measuring the drop $v_R=Ri$ gives a band-pass filter that suppresses very low and very high frequencies while passing a range near resonance. This is useful for selecting a radio signal, though the range must be broad enough to pass its modulation. ^[ref].

A finite repetition-free sequence in a set $X$ is an injective function from an initial segment $\{0,\dots ,k-1\}$ of the natural numbers into $X$, including the empty sequence for $k=0$. For a finite set of size $m$, the number of these sequences isThe order of each list is part of the data, so it can be enumerated without choosing an order on its underlying support.

Use an orthogonal eigendecomposition $X=V\Lambda V^T$ and a vector $\xi$ of independent Rademacher random variables. Because $\Lambda$ is diagonal and ${\xi }_j^2=1$, every sign vector gives $\Vert \hat{x}{\Vert }_2^2=\mathrm{tr}X$, without taking an expectation. If the scaling denominator $M$ is positive, then $x$ satisfies every slab constraint and $\Vert x{\Vert }_2^2=\mathrm{tr}X/M^2$. Under spanning constraints and positive trace, $M>0$ automatically. A nonzero $\hat{x}$ with $M=0$ instead certifies an unbounded direction.

Suppose a graph with $n$ vertices and $m>0$ edges has a vector coloring with adjacent inner products at most $-1/2$. Take $r$ independent copies of random hyperplane rounding and use the $r$ signs as a color. Each edge remains monochromatic with probability at most $3^{-r}$, so the expected number $B$ of monochromatic edges is at most $m{3}^{-r}$ by linearity of expectation.

Choose $r=\max \{0,\lceil {\log }_3(4m/n)\rceil \}$. Then Markov inequality gives $\mathbb{P}(B>n/2)\le 1/2$. On success, delete one endpoint of every monochromatic edge. This random alteration method leaves at least $n/2$ vertices properly colored withThe removed vertices can retain their original colors, as the semicoloring only requires the retained induced subgraph to be proper. A failed draw can be detected and repeated, with at most two trials on average. The case $m=0$ needs just one color. This is the elementary independent-hyperplane construction in Karger, Motwani and Sudan's paper on approximate graph coloring.

For a spherical buoyant thermal fed only through the moving top of a Boussinesq point-source plume, put $\lambda =R/B$ and $B=(6\alpha /5)Z$. Relative inflow $\pi B^2(w_p-\dot{Z})$ and $\dot{V}=4\pi R^2\dot{R}$ implyThe ratio of Froude numbers is $5/[\sqrt{\lambda }(5+24\alpha {\lambda }^3)]$ when both use plume reduced gravity $g_p$. Normalizing the thermal with its own $g_t=9g_p/4$ multiplies this by $2/3$. These two definitions must not be interchanged when stating bounds.

For a self-adjoint operator $L$, solving $Lu=f$ requires $f$ to be orthogonal to every $v\in \ker L$, because $⟨v,Lu⟩=⟨Lv,u⟩=0$. When the range is closed, this condition is also sufficient: $(\mathrm{ran}L{)}^{\perp }=\ker L$ and closedness gives $\mathrm{ran}L=(\ker L{)}^{\perp }$. In finite dimensions the range is automatically closed. This Fredholm solvability condition determines amplitude equations by projecting perturbative forcing onto critical eigenfunctions.