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The cosmological Lagrangian displacement sends an initially labelled matter element from its Lagrangian coordinate $\mathbf{q}$ to $\mathbf{x}(\mathbf{q},t)=\mathbf{q}+\mathbf{\psi }(\mathbf{q},t)$. Its time derivative determines the peculiar velocity through $\mathbf{v}=a\dot{\mathbf{\psi }}$.

A covalent bond joins chemical atoms through shared Electrons. In the familiar two-centre description, one bond pair contributes to a connection between two atomic nuclei. Sharing means that the Electrons belong to the combined arrangement and interact with both atomic nuclei; it does not mean that identifiable dots shuttle along drawn paths. The ^[ref] emphasizes the concentration of Electrons between atomic nuclei and the resulting attraction and characteristic separation. The pair picture is especially useful for simple molecules, but is not a universal description of every covalent bond.

A Lewis structure turns this picture into practical bookkeeping. Count the valence electrons, indicate which chemical atoms are connected, and allocate two Electrons to each line. A bond pair is shared; a lone pair is assigned to one chemical atom without connecting it to another. A double line represents two bond pairs. Dots and lines record allocations, rather than positions, trajectories, or a measured three-dimensional arrangement. ^[ref]

For molecular hydrogen, ${\mathrm{H}}_2$, each neutral hydrogen chemical atom supplies one valence electron. Thus $1+1=2$ valence electrons make one bond pair, represented by $\mathrm{H}-\mathrm{H}$, with no lone pairs. Each hydrogen counts the same shared pair when checking its two-electron arrangement. The molecule still contains two Electrons, not four: counting a pair around both endpoints does not create extra Electrons.

For water, ${\mathrm{H}}_2\mathrm{O}$, neutral oxygen supplies six valence electrons and the two hydrogen chemical atoms supply one each: $6+1+1=8$. Two O–H connections consume four Electrons as two bond pairs. The remaining $8-4=4$ Electrons form two lone pairs on oxygen. Around oxygen, the two bond pairs and two lone pairs count as eight Electrons; each hydrogen counts two. The whole water molecule has ten Electrons, because two additional oxygen Electrons are omitted from this valence electron drawing.

For ammonia, ${\mathrm{NH}}_3$, the total is $5+3\times 1=8$ valence electrons: five from nitrogen and three from hydrogen. Three N–H connections use six Electrons as three bond pairs, leaving two Electrons as one lone pair on nitrogen. Thus water and ammonia have the same total of eight valence electrons but different allocations: two plus two pairs versus three plus one. This small calculation already distinguishes their connectivity without a detailed quantum mechanics calculation.

For molecular oxygen, ${\mathrm{O}}_2$, $6+6=12$ valence electrons give the conventional $\mathrm{O}=\mathrm{O}$ Lewis structure. Its two bond pairs use four Electrons, leaving eight as four lone pairs, two on each oxygen. Each oxygen consequently counts eight Electrons locally. However, actual molecular oxygen has two unpaired Electrons and exhibits paramagnetism: it is attracted into a stronger magnetic field. The all-paired Lewis structure cannot capture this observation. Molecular orbital theory describes the unpaired Electrons; the successful counting diagram alone is insufficient. ^[ref]

Why does sharing produce a stable separation? For molecular hydrogen, the electronic arrangement responds as the atomic nuclei approach. Their interactions can lower the energy, but pushing them too close raises it. A curve of potential energy against separation therefore has a local minimum, rather than falling forever. Attraction and repulsion balance there. This is a statement about the combined system, not a claim that stationary Electrons sit between otherwise motionless atomic nuclei. ^[ref]

The following Morse potential makes those features calculable. Set $x=r/r_e$, where $r$ is separation and $r_e$ its equilibrium value, and let $D$ be the well depth. With the self-chosen parameter $a=2$, define the dimensionless quantitiesHere $F=-dV/dr$ is the net separating force, obtained from the derivative of the potential energy. Positive $f$ pushes the chemical atoms apart; negative $f$ draws them together. These are illustrative curves with $r_e=D=1$ in chosen units, not measured molecular hydrogen data or fitted parameters.

At $x=1$, $u=-1$ and $f=0$. The second derivative is $u^{\prime \prime }(1)=8>0$, confirming a local minimum. At $x=0.5$, $u=(1-e{)}^2-1\approx 1.9525$; at $x=1.5$, $u=(1-e^{-1}{)}^2-1\approx -0.6004$. As $x$ increases without bound, $u$ approaches zero from below. Thus moving from the well bottom to separated chemical atoms requires an increase of $D$ in this model's potential energy. An actual quantum mechanics dissociation threshold also depends on the initial motion of the atomic nuclei, so the well depth alone is not every measured dissociation energy.

The force panel separates $f$ into a positive term $4e^{-4(x-1)}$ and a negative term $-4e^{-2(x-1)}$. At $x=1$ they are $+4$ and $-4$, cancelling exactly; nearer separations favour the positive term. These are algebraic components of the chosen Morse potential, not separately measured microscopic forces. Their usefulness is to show how a balance arises from the slope of one energy curve.

The beauty of the pair picture is its economy: a few marks summarize connectivity and constrain what a proposed molecule can contain. Its historical anchor is Gilbert N. Lewis's 1916 paper ^[ref], also available as an ^[ref]. Lewis structures remain useful because they answer counting questions quickly; quantum mechanics supplies the richer account when energy, shape, or unpaired Electrons matter.

For $0<k<1$, a neutral mode of the equal-width Hazel model behaves as $w\sim z^{1-k}$ near $z=0$. Its derivative diverges as $z^{-k}$, so it is not a classical continuously differentiable mode across the critical level of an internal gravity wave. The horizontal velocity is proportional to $w^{\prime }/k$; its local kinetic energy is finite only for $k<1/2$, because ${\int }_0^{ϵ}{z}^{-2k}dz$ converges precisely then. One possible neutral-mode convention is the boundary value from $U-c$ with $c_i\downarrow 0$, which fixes the phase of the power on the negative-$z$ side. The Miles–Howard theorem concerns growing modes with nonreal $c$ and does not rule out such singular neutral limits.

For a smooth function $f:U\to \mathbb{R}$ on an open set $U\subseteq {\mathbb{R}}^n$, a critical point is an input $x_{\ast }$ where the gradient vanishes: $\nabla f(x_{\ast })=0$. Equivalently, every first partial derivative is zero. Such a critical point is a stationary point; in one variable the condition is $f^{\prime }(x_{\ast })=0$.

There is a broader convention in elementary one-variable calculus: a "critical number" is an interior input where the derivative is zero or does not exist, provided the function itself is defined there. Thus a corner can be a critical number without being a stationary point. Here the smooth meaning of critical point is used unless the broader convention is explicitly stated. ^[ref]

For the polynomial $f(x)=x^4-2x^2$ on the real line,The stationary points at $x=\pm 1$ are strict local minima and attain the global minimum; the stationary point at $x=0$ is a strict local maximum. One can check the local classifications from $f^{\prime \prime }(x)=12x^2-4$, which is positive at $\pm 1$ and negative at zero. The identity $f(x)=(x^2-1{)}^2-1\ge -1$ proves the global minimum directly. Since $f(x)\to +\infty$ as $|x|\to \infty$, there is no global maximum on $\mathbb{R}$.

A zero derivative alone does not guarantee an extremum. The polynomial $g(x)=x^3$ has $g^{\prime }(0)=0$ but is strictly increasing through zero: every neighbourhood of zero contains both positive and negative values, so zero is neither a local minimum nor a local maximum. Conversely, the absolute value function $h(x)=|x|$ has a strict local minimum and a unique global minimum at zero, but its left and right derivatives are $-1$ and $+1$. It is not differentiable there, so zero is a critical number under the broader convention and is not a stationary point.

The usefulness of critical points is that they reduce the search for smooth interior extrema to equations. Every local minimum or local maximum of a differentiable function at an interior point has zero gradient; this condition is necessary, and $x^3$ shows that it is not sufficient. The striking feature is that all the first directional derivatives must vanish together, even though the surrounding values can form a valley, a peak, or a saddle point of a scalar function. ^[ref]

Proof outline: at an interior local minimum $x_{\ast }$ in one variable, the difference quotient $(f(x_{\ast }+h)-f(x_{\ast }))/h$ is nonnegative for small $h>0$ and nonpositive for small $h<0$. If the derivative exists, the two sides have the same limit, forcing $f^{\prime }(x_{\ast })=0$. Reverse the inequalities for a local maximum. In several variables, apply this argument along each coordinate line to obtain zero partial derivatives and hence zero gradient. At a domain boundary, both directions may not be available: $f(x)=x$ on $[0,1]$ attains its global minimum at zero although $f^{\prime }(0)=1$ for its smooth extension.

A critical point of a smooth map is an input; a critical value is an output reached from at least one such input. For the fold $F(x,y)=(x,y^2)$, all inputs $(a,0)$ are critical points of a smooth map, and all outputs $(a,0)$ are critical values. The identical coordinate pairs live in different copies of ${\mathbb{R}}^2$. To classify an output $(a,b)$, inspect its entire fiber of a function:For $b>0$ both inputs have matrix rank two, so the output is a regular value. For $b=0$ its input has matrix rank one, so the output is a critical value. For $b<0$ the fiber of a function is the empty set, and the output is a regular value despite being outside the image of a function.

In general, for a smooth map $f:X\to Y$, write $C_f$ for its set of critical points of a smooth map. Its set of critical values is $f(C_f)\subseteq Y$. Equivalently, $q\in Y$ is a critical value if some $p\in f^{-1}(q)$ has a differential that is not surjective. A regular value requires a surjective differential at every input in that fiber of a function. The universal condition holds vacuously for an empty fiber of a function: every output outside $f(X)$ is a regular value. These quantifiers are explicit in Loring W. Tu, An Introduction to Manifolds, Definition 8.22 and its two following remarks.

The distinction is useful when solving equations. For a smooth map $f:X^m\to Y^n$ between smooth manifolds without boundary and a regular value $q$ with nonempty fiber of a function, the regular level set theorem makes $f^{-1}(q)$ an embedded submanifold of dimension $m-n$, with tangent spaceThe implicit function theorem explains the freedom count: locally, $n$ independent equations fix $n$ coordinates and leave $m-n$ free. The rank-nullity theorem gives the same count for the displayed kernel of a linear map. See Eckhard Meinrenken, Differentiable Manifolds, Theorems 1.11 and 1.15 for the local projection description and its consequence for level sets.

For example, $H:{\mathbb{R}}^2\to \mathbb{R}$, $H(x,y)=x^2+y^2$, has gradient $(2x,2y)$ and just one critical point, the origin, with critical value zero. For $c>0$, its fiber of a function is the circle of radius $\sqrt{c}$, a one-dimensional embedded submanifold; negative $c$ are regular values with empty fibers. At $c=0$, the fiber of a function is a single point instead of a circle. A critical value does not force its fiber of a function to be nonsmooth: that singleton is itself a zero-dimensional embedded submanifold, but the regular dimension conclusion $2-1=1$ does not apply.

Nor does a single regular input certify a regular value. The scalar smooth function $h(t)=t^2(t-1)$ has $h^{-1}(0)=\{0,1\}$ and derivative $h^{\prime }(t)=3t^2-2t$. The input zero is a critical point, while the input one is regular because $h^{\prime }(1)=1$. Their shared output zero is nevertheless a critical value. Conversely, different critical points of a smooth map can share one critical value: $f(t)=(t^2-1{)}^2$ sends both critical inputs $-1$ and $1$ to zero. Counting inputs and counting critical outputs are different tasks.

Finally, for $G(t)=(t,t^2):\mathbb{R}\to {\mathbb{R}}^2$, every output on the parabola is a critical value, and every output off it is a regular value with empty fiber of a function. There is no nonempty regular fiber of a function here, so the negative expression $1-2$ never becomes a claimed manifold dimension. Both diagrams show exact mathematical examples, not measured data.

The integral $I(x)={\int }_0^{\infty }{e}^{xt-t^3/3}dt$ is a rescaled Scorer Hi function. It has the three useful limitsFor the negative-argument limit, scale $t=s/M$ and use the dominated convergence theorem. At zero, substitute $s=t^3/3$ and use the Gamma integral. For the positive-argument limit, the exponent has its maximum at $t=\sqrt{M}$, with second derivative $-2\sqrt{M}$; Laplace's method gives the displayed factor. These estimates describe a cubic endpoint-to-saddle transition.

If the first two phase derivatives vanish at an endpoint but the third does not, an oscillatory contribution is concentrated on a width proportional to $N^{-1/3}$. For a smooth nonzero endpoint oscillatory integral amplitude, rescaling reduces the leading contribution to a cubic oscillatory integral. Its constant follows from the Gamma function Fourier integral. This is a degenerate form of the stationary phase method, and a quadratic formula is insufficient.

In a smooth interior region with locally constant shear ${\Omega }^{\prime }$, a wound azimuthal Fourier mode has radial wavenumber $k_s\simeq -m{\Omega }^{\prime }t$. Integrating its resistive damping rate $\eta k_s^2$ gives the exponential decay in the title. The characteristic time is ${\tau }_c=[3/(\eta m^2{\Omega }^{\prime 2}){]}^{1/3}$. For $m=1$ and ${\Omega }^{\prime }=-{\Omega }_0/d$, ${\Omega }_0{\tau }_c=(3\mathrm{Rm}{)}^{1/3}$, where $\mathrm{Rm}={\Omega }_0{d}^2/\eta$. This is a high-magnetic Reynolds number interior transient on a time long compared with the rotation time but short compared with $d^2/\eta$. Physical magnetic field components may have additional algebraic factors from taking derivatives of the wound magnetic vector potential. A maintained exterior field, an axial core and a circulation-edge boundary layer prevent interpreting this estimate as an exact, uniform, infinite-time solution.

Suppose $T\ge 0$ has an absolutely continuous probability distribution, with survival function $S(0)=1$. On an interval where $S(t)>0$, its cumulative hazard isHere $h$ is the hazard function; its integral is dimensionless because the units of rate and time cancel. The identity follows by integrating the derivative of $\log S$, so mere continuity of the survival function is not enough: the absolute continuity assumption supplies a probability density function. If $S$ reaches zero, interpret $-\log S$ as $+\infty$. The normalization $S(0)=1$ also matters; an initial probability mass at zero would require an initial term $-\log S(0)$.

Accumulating hazard functions adds on the $H$ scale and multiplies survival factors on the $S$ scale. For $S(a)>0$ and $b>a$,For instance, a cumulative hazard increment of $2$ gives event probability $1-e^{-2}\approx 0.864665$, rather than a probability of $2$. With $h(t)=0.2$ per year, $H(5)=1$ and $S(5)=e^{-1}\approx 0.367879$. With the increasing hazard function $h(t)=0.04t$ per year when $t$ is expressed in years, $H(t)=0.02t^2$, so $H(5)=0.5$ gives $S(5)=e^{-0.5}\approx 0.606531$. Its coefficient has units of inverse years squared.

The constant hazard function produces an exponential distribution; the increasing example is a Weibull distribution. Their survival functions both decrease, although one hazard function stays constant and the other increases. At year ten both have $H(10)=2$ and $S(10)=e^{-2}$ despite different earlier histories. The accumulation identity is described in Lu Tian's Stanford notes, Section 1.1.

For a discrete hazard $q_j$ at each possible event time $t_j$, the exact counterpart is $S(t)={\prod }_{t_j\le t}(1-q_j)$. Taking its negative logarithm sums $-\log (1-q_j)$, rather than $q_j$. Therefore exponentiating a Nelson–Aalen estimator does not give exactly the Kaplan–Meier estimator; small-jump comparison of cumulative hazard estimators quantifies the difference.

For a twice continuously differentiable regular curve $r(t)=(x(t),y(t))$ in the plane, unsigned curvature is the absolute value of its signed curvature:The nonvanishing velocity vector makes the denominator positive. For a graph of a function $(t,f(t))$, this reduces to $|f^{\prime \prime }(t)|/(1+f^{\prime }(t{)}^2{)}^{3/2}$; omitting the absolute value gives the signed curvature of a plane graph for increasing $t$.

For the parabola $r(t)=(t,t^2)$, the derivatives are $r^{\prime }=(1,2t)$ and $r^{\prime \prime }=(0,2)$. ThereforeThe curvature is greatest at the vertex $t=0$, where $\kappa =2$ and the radius of curvature is $1/2$. At $t=\pm 1/2$ the curvature is $1/\sqrt{2}$, so the parabola bends less sharply despite its larger speed $\sqrt{2}$.

Its unit tangent vector and left unit normal areSince $k>0$, $N_L$ is also its principal normal vector. At $t=0$ the osculating circle consequently has center $(0,0)+(1/2)(0,1)=(0,1/2)$ and equationThe lower branch $y=1/2-\sqrt{1/4-x^2}$ has value $0$, derivative $0$, and second derivative $2$ at $x=0$, matching the parabola $y=x^2$ through second order. Away from the vertex the two curves separate: an osculating circle describes a local bend, not the whole curve.

Vanishing curvature at one point does not imply a straight segment. The graph of a function $y=x^3$ has signed curvature $6x/(1+9x^4{)}^{3/2}$, which vanishes at its inflection point $x=0$ and changes sign across it. By contrast, if curvature vanishes throughout a connected regular interval, $dT/ds=0$ throughout that interval: the constant unit tangent vector forces the image to lie on a straight line.

Definitions of the Riemann curvature tensor can differ by an overall sign, independent of the metric signature. Reversing that definition reverses the Ricci tensor, Ricci scalar and Einstein tensor, as well as the curvature side of the geodesic deviation identity and the Palatini identity. A derivation must choose the contraction and derivative ordering explicitly before combining formulas from different conventions. Changing a metric signature alone is a separate operation.

Damping dissipates the energy of an oscillation, reducing its amplitude. A linearly damped string applies this mechanism to each normal mode.

Why can stronger damping make a return to equilibrium slower? A weak brake lets a spring exchange potential energy with kinetic energy many times; a very strong brake makes the final motion creep. Critical damping marks the boundary between these behaviours. The distinction depends on the force law: the classification below concerns a single damped harmonic oscillator with viscous damping, rather than every mechanism that removes energy.

Let $x(t)$ be displacement from equilibrium, $m>0$ the moving mass, $k>0$ the spring constant, and $b\ge 0$ the constant coefficient of viscous damping. With no driving force, Newton's second law givesThe two initial conditions specify displacement and velocity. Define the undamped eigenfrequency as an angular frequency ${\omega }_0=\sqrt{k/m}$ and the damping ratio as $\zeta =b/(2\sqrt{mk})$. The ordinary differential equation becomes $\ddot{x}+2\zeta {\omega }_0\dot{x}+{\omega }_0^{2}x=0$. A constant weight in a vertical spring merely shifts equilibrium; it is already absorbed in the definition of $x$.^[ref]

For example, the ideal parameters $m=1\mathrm{kg}$ and $k=4\mathrm{N}{\mathrm{m}}^{-1}$ give ${\omega }_0=2{\mathrm{s}}^{-1}$ and critical damping at $b=4\mathrm{kg}{\mathrm{s}}^{-1}$. Values $b=1,4,8\mathrm{kg}{\mathrm{s}}^{-1}$ give damping ratios $0.25,1,2$, respectively. These are model choices, not measured values. For release from rest with $x_0>0$, set $X=x/x_0$ and $\tau ={\omega }_{0}t$; then $X^{\prime \prime }+2\zeta X^{\prime }+X=0$, where primes now denote derivatives with respect to $\tau$.

The left panel uses identical initial conditions $X(0)=1$, $X^{\prime }(0)=0$. Underdamping produces repeated crossings of equilibrium; critical damping and overdamping give positive decreasing responses. The right panel keeps critical damping but changes initial velocity: $X^{\prime }(0)=-2$ gives one crossing, although there is no oscillation. Both panels are exact model predictions.

This linear functional extracts one coefficient of a B-spline expansion using local polynomial data. With ${\psi }_i(x)={\prod }_{\ell =1}^{k-1}(x-t_{i+\ell })/(k-1)!$ and a point $\xi$ inside a knot cell in the support, it is ${\sum }_{r=0}^{k-1}(-1{)}^r{s}^{(r)}(\xi ){\psi }_i^{(k-1-r)}(\xi )$. The local polynomial identity for the Marsden dual functional gives the desired coefficient. Its bounded restriction to the local spline space can be extended to continuous data on $[t_i,t_{i+k}]$ by the Hahn-Banach theorem, with the same bound. Such extensions supply spline quasi-interpolation coefficients without requiring derivatives of the data.

De Casteljau's algorithm evaluates a Bézier curve by repeated linear interpolation between neighboring control points. Fix $t\in [0,1]$, start with $P_i^{(0)}(t)=P_i$, and computeEach new row has one fewer point. The final point is $P_0^{(n)}(t)=C(t)$. Use the same $t$ at every stage: it describes the fraction of each line segment traversed from its first endpoint toward its second. A cubic Bézier curve needs three levels containing $3$, $2$, and $1$ linear interpolations, respectively.^[ref]

For the Bézier curve with control points $P_0=(0,0)$, $P_1=(1,2)$, $P_2=(3,2)$, $P_3=(4,0)$, evaluation at $t=1/2$ takes the midpoint of every neighboring pair. The first level givesThe second level givesand the third level givesThe intermediate control polygons shrink toward the evaluated point. Each step is a convex combination, giving a geometric explanation of the convex hull property; every step also commutes with an affine map.

The same construction splits the Bézier curve exactly, with no approximation. The outer edges of the triangular array give the new control points: for a cubic, $(P_0,Q_0,R_0,S)$ specify the left piece and $(S,R_1,Q_2,P_3)$ specify the right piece.^[ref] At $t=1/2$ these two lists areWrite $C_L$ and $C_H$ for the Bézier curves defined by these lists, each with its own parameter $u\in [0,1]$. ThenTheir images reproduce the original curve, while their derivatives account for the shorter parameter intervals. At the join,The chain rule gives $C_L^{\prime }(1)=C_H^{\prime }(0)=C^{\prime }(1/2)/2$; in the original parameter, both pieces have derivative $(4.5,0)$. This is the interval scaling in the parametric first-derivative join of Bézier curves.

Repeated midpoint splitting is useful for drawing vector graphics paths: each piece has a smaller control polygon and its own convex hull enclosure. A renderer can refine pieces until they are sufficiently close to line segments at the chosen display scale. Evaluating $C(t)$ and splitting it are thus two results of the same short algorithm.

For a proof outline, expanding the recursion givesPascal's identity proves this by induction; at $r=n$, $i=0$ it is precisely the Bernstein basis formula for the Bézier curve. The geometric construction remains useful even when expanding the polynomial by hand would be cumbersome.

// Target: analysis.bigb

This initial value problem uses a square root and absolute value to define a continuous function on all of ${\mathbb{R}}^2$. For every finite waiting time $c\ge 0$, a forward solution of a differential equation on $t\ge 0$ isThe constant solution of a differential equation $y_{\infty }(t)=0$ is another choice. For example, at $t=3$, waits $c=0,1,2,3$ give values $9,4,1,0$. These are dimensionless mathematical trajectories; no measured physical waiting time is being supplied.

Verification uses the original ordinary differential equation, including the join. Before departure, both $y_c^{\prime }$ and $2\sqrt{|y_c|}$ vanish. For $t>c$, their common value is $2(t-c)$. For $c>0$, the difference quotient at the join isThus the derivative exists at $c$, equals the rate law there, and is continuous across the join. For $c=0$ the same computation gives the right derivative at the starting endpoint. For $c>0$, the left and right second derivatives at departure are $0$ and $2$: these forward solutions of a differential equation are continuously differentiable functions, without being twice differentiable at departure. The local two-sided statement of the Peano existence theorem is separate from this explicitly forward family.

The rate law tends to zero as $y\to 0$, but its secant slopes from zero are unbounded:No finite local Lipschitz constant exists at zero. At $\epsilon =10^{-2},10^{-4},10^{-6}$ the quotient is $20,200,2000$, even though the rates are shrinking to zero. Continuity controls the rate itself; Lipschitz continuity would control its change relative to the state displacement.

There is also a numerical lesson. The Forward Euler method starting exactly at zero gives $Y_{k+1}=Y_k+2\Delta t\sqrt{|Y_k|}=0$ at every step. It reproduces the constant solution of a differential equation for every positive step size, while the moving trajectories remain equally valid. A computation producing one trajectory cannot establish uniqueness. Likewise, the separable differential equation calculation that divides by $\sqrt{|y|}$ only describes the moving pieces: checking the original equation recovers the zero pieces and their joins.