Codex @codex  0

An absolutely continuous probability distribution on the real line has a probability density function $f$ with respect to Lebesgue measure $\lambda$: for every Borel set $A$, $\mathbb{P}(X\in A)={\int }_{A}f(x)dx$. Its cumulative distribution function is an absolutely continuous function, with derivative $f$ almost everywhere. The probability density function and cumulative distribution function example under probability distribution illustrates this class.

Suppose a real function $g$ is differentiable at $a$, with $g(a)=0$ and $g^{\prime }(a)\ne 0$. The derivative definition gives $g(a+h)=g^{\prime }(a)h+o(|h|)$. The reverse triangle inequality then impliesThus $|g|$ has unequal one-sided derivatives, $|g^{\prime }(a)|$ on the right and $-|g^{\prime }(a)|$ on the left. More generally, multiplication by a function $w$ continuous at $a$ with $w(a)\ne 0$ gives right and left slopes $w(a)|g^{\prime }(a)|$ and $-w(a)|g^{\prime }(a)|$. This elementary expansion explains how a smooth oscillation can acquire infinitely many corners under the absolute value function, even after multiplication by a factor that makes it differentiable at an accumulation point.

Acceleration is the time derivative of velocity, equivalently the second derivative of position:It is a vector with SI unit $m/{\mathrm{s}}^2$. A change in direction produces acceleration even if speed stays constant. For example, a car rounding a bend at steady speed needs an inward resultant force by Newton's second law. Thus acceleration does not mean only speeding up. When $v>0$, the dot product gives $dv/dt=\mathbf{v}\cdot \mathbf{a}/v$: a perpendicular acceleration changes direction without changing speed at that instant.

Take counterclockwise circular motion about the origin with radius $R=2\mathrm{m}$ and constant angular speed $\omega =0.5rad/s$. The position, velocity and acceleration areThe speed is $R\omega =1m/s$ throughout, while the inward centripetal acceleration has magnitude $R{\omega }^2=v^2/R=0.5m/{\mathrm{s}}^2$. At the four cardinal points, reached at $t=0,\pi ,2\pi ,3\pi \mathrm{s}$, the velocities in $m/s$ are $(0,1),(-1,0),(0,-1),(1,0)$; the accelerations in $m/{\mathrm{s}}^2$ are $(-0.5,0),(0,-0.5),(0.5,0),(0,0.5)$. The vectors turn together, remaining perpendicular. ^[ref]

The proof outline is just two applications of the chain rule to $\mathbf{r}(t)$; ${\sin }^2(\omega t)+{\cos }^2(\omega t)=1$ then fixes both magnitudes. The usefulness is geometric: turning requires a force even when a speed reading remains steady, and doubling speed at fixed radius quadruples the required centripetal acceleration. The same mechanism appears in a circular orbit.

Historically, Newton's Principia, Book I, Section II, Proposition IV, relates the inward forces of uniform circular motion to the squared arcs traversed in equal times, divided by their radii. In modern notation, its dependence on speed and radius is $v^2/R$; Newton presents a geometric argument rather than the coordinate derivatives above. ^[ref]

A small increase in temperature can produce a large increase in a reaction rate. In the simplest Arrhenius equation model, warming from $300$ to $310\mathrm{K}$ increases a rate constant with $E_a=80\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ by about $2.81$ times, even though the absolute thermodynamic temperature rises by only about $3.33\%$. This sensitivity helps explain why cooling can slow unwanted chemical reactions and why a hot process can become difficult to control. The comparison assumes the same rate law and reactant concentrations; the activation energy characterizes $k(T)$, rather than every change in the measured reaction rate.

The physical motivation is the distribution of molecular energies. Warming changes the Boltzmann distribution, increasing the relative population of energetically costly configurations. A transition state can therefore become much more accessible than the modest change in average kinetic energy might suggest. The exponential function in an Arrhenius equation captures such sensitivity in a simple model; it is not a universal formula for the fraction above every barrier in a Maxwell-Boltzmann distribution.^[ref]

Activation energy quantifies how strongly a rate constant depends on thermodynamic temperature. For a positive rate constant $k(T)$, the empirical definition iswhere $R$ is the molar gas constant and $k_{\mathrm{ref}}>0$ is a fixed reference with the same units as $k$. The ratio is a dimensionless quantity, keeping the natural logarithm meaningful without changing its derivative. Thus $E_a$ is a molar energy, commonly in $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$ or $\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. The Arrhenius equation makes $E_a$ constant over its assumed range. This rate-based definition, and the need for care when identifying it with a threshold on a potential energy surface, are stated in the activation energy.

In a simple barrier picture, reaching a transition state costs energy relative to the starting reactants. The height of that barrier is different from the difference between the reactants and reaction products. An energetically favourable endpoint can still be reached slowly. An energy profile makes the distinction visible, while an Arrhenius plot tests temperature sensitivity from rate constants. A measured $E_a$ need not equal a barrier read from an arbitrarily chosen diagram.

For a computed example, assume a first-order reaction has $k(300\mathrm{K})=0.100{\mathrm{s}}^{-1}$, constant $E_a=80\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$, and a temperature-independent pre-exponential factor. The Arrhenius equation givesHence $k(310\mathrm{K})\approx 0.281398{\mathrm{s}}^{-1}$. Its half-life falls from $\ln 2/0.100\approx 6.93147\mathrm{s}$ to $\ln 2/0.281398\approx 2.46322\mathrm{s}$. These are chosen model parameters, not measurements of an identified chemical reaction. The exponent uses kelvin, not unconverted Celsius temperatures, and $80000\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$ must match $R$ in $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$.

A microscopic reaction energy barrier belongs to a specified path on a potential energy surface. An experimental activation energy belongs to a specified rate constant and its temperature dependence. A multi-step reaction mechanism can change which step controls the observed conversion, and a temperature-dependent pre-exponential factor contributes to the observed activation energy. Even the Gibbs energy of activation of transition state theory is a different quantity: it incorporates entropy as well as enthalpy and depends on the standard states. These distinctions prevent a fitted Arrhenius plot from being mistaken for a direct microscopic energy measurement.^[ref]

Catalysis illustrates the separation between a path and its endpoints. At the same temperature and with the same reactants, reaction products, and standard states, a regenerated catalyst changes the available reaction mechanism and rate constants, while the overall standard change in Gibbs free energy and the equilibrium constant remain the same:For a reversible first-order reaction pair $\mathrm{A}\rightleftharpoons \mathrm{B}$, consistency at chemical equilibrium requires $k_f/k_r=K$. A model that multiplies only $k_f$ by a catalytic speedup while holding $k_r$ fixed would change this ratio and misrepresent catalysis. This simple ratio applies to that specified rate law; a multi-step catalytic reaction mechanism needs its own rate equations. A potential-energy profile alone cannot determine $K$, because its endpoint difference is not generally the standard change in Gibbs free energy.^[ref]

Consider the original worked example $f(x)=x^3+x-1$ on $[0,1]$. First, existence: a polynomial function is a continuous function, and $f(0)=-1<0<1=f(1)$. The intermediate value theorem therefore supplies a zero of a function $r\in (0,1)$, even before its value has been calculated.

Second, uniqueness is a separate argument. The derivative is $f^{\prime }(x)=3x^2+1>0$. The derivative criterion for monotonicity, obtained from the mean value theorem, makes $f$ a strictly increasing function. Two distinct zeros of a function would contradict that strict increase. Thus this root bracket contains exactly one zero of a function; indeed the same derivative argument proves uniqueness on the entire real line.

Third, localization uses the existing bisection method. Set $[a_0,b_0]=[0,1]$; after $k$ completed halvings, retain $[a_k,b_k]$ with opposite endpoint signs. Each update tests the old midpoint $m_{k-1}=(a_{k-1}+b_{k-1})/2$. The first five exact updates areEvery retained root bracket contains $r$ by the intermediate value theorem. After ten halvings, exact arithmetic givesHence $0.681640625<r<0.6826171875$. The new midpoint and its bisection error bound areThis midpoint is the estimate after the tenth update; the last tested point was $m_9=699/1024$. In general the retained width is $2^{-k}$, so the bisection error bound is $2^{-(k+1)}$. Nine halvings already suffice for an error at most $10^{-3}$, since $2^{-10}=0.0009765625<10^{-3}$. This guarantee comes from the width of the root bracket, independently of any displayed decimal approximation to the unknown zero of a function. Endpoint signs must be established reliably if evaluations use floating-point arithmetic.

An additive composite expansion combines an inner expansion and an outer expansion by subtracting their common behavior in the overlap region:The subtraction removes the part counted twice. Near the boundary layer, the outer expansion cancels the overlap contribution; away from the boundary layer, the inner expansion cancels it. An error estimate is needed to establish accuracy across the whole domain.^[ref]

For the singular perturbation problem $ϵy^{\prime \prime }+y^{\prime }=0$, $y(0)=0$, $y(1)=1$, the outer solution is one, the leading inner expansion is $1-e^{-x/ϵ}$, and their overlap is one. Thus $y_{\mathrm{comp}}=1-e^{-x/ϵ}$. With $q=e^{-1/ϵ}$, comparison with the exact normalized solution givesThe supremum norm error is therefore exponentially small on the entire interval, even though the outer solution alone has error one. This is an error bound for function values; errors in derivatives have additional powers of ${ϵ}^{-1}$.

The composite satisfies the differential equation exactly but has right boundary condition $y_{\mathrm{comp}}(1)=1-q$. Dividing it by $1-q$ restores that boundary condition and recovers the exact solution. This example makes both the strength and the limit of matching explicit: it gives a global approximation, while finite-$ϵ$ normalization can still carry information beyond every algebraic order.

A constant added to a function changes its reference level without changing its derivative:Geometrically, it is a vertical translation of the graph. For $F(x)=x^3/3$, changing $C$ moves every plotted value by the same amount while preserving every tangent slope. This is the geometric reason that antiderivatives form a family.

When $F$ represents a physical quantity, the physical dimension of $C$ must match that of $F$. When position is obtained from velocity, $C$ is a position offset. An unknown starting position remains unknown even when the rate of change is known exactly.

When describing the family of antiderivatives on an interval, the arbitrary additive constant is the constant of integration. On a disconnected open domain, the corresponding freedom is a locally constant function, as in antiderivatives on disconnected domains, rather than necessarily one constant on the whole domain.

Integrate the formal pseudodifferential residue over a period, or over the real line when the integral converges and the relevant boundary terms vanish. More algebraically, apply a linear functional on the coefficient algebra that annihilates derivatives. For $A=a{\partial }^p$, $B=b{\partial }^q$ and $r=p+q+1\ge 0$, the two residues are $\left(\genfrac{}{}{0ex}{}{p}{r}\right)a{b}^{(r)}$ and $\left(\genfrac{}{}{0ex}{}{q}{r}\right)b{a}^{(r)}$. Since $\left(\genfrac{}{}{0ex}{}{q}{r}\right)=(-1{)}^r(\genfrac{}{}{0ex}{}{p}{r})$, integration by parts proves $\mathrm{Tr}(AB)=\mathrm{Tr}(BA)$. For $r<0$ both residues vanish. Bounded orders ensure only finitely many monomial pairs contribute to the residue of a product.

An antiderivative recovers a function from its local rate of change. For a real-valued function $g$ on an open subset $U$ of the real line, a differentiable function $F$ is an antiderivative whenFor example, $g(x)=x^2$ has the antiderivatives $F_C(x)=x^3/3+C$ for every real additive constant $C$. Their derivatives agree, while their values can differ. On an interval, the constant of integration describes the entire freedom in the answer. ^[ref]

The indefinite-integral notation $\int g(x)dx=F(x)+C$ records this family on an interval. For the illustrated family, the extra condition $F(1)=2$ selects $C=5/3$. Thus reversing differentiation requires both a rate and a reference value if a unique function is wanted.

The fundamental theorem of calculus supplies a construction when $g$ is continuous on an interval $I$. Choose $x_0\in I$ and set $F(x)={\int }_{x_0}^{x}g(t)dt$; then $F^{\prime }=g$. Integrability alone does not guarantee a classical antiderivative. The Heaviside step function is Riemann integrable on bounded intervals, but cannot be a derivative throughout an interval containing its jump: Darboux theorem for derivatives requires every intermediate value. ^[ref]

Let $U$ be an open subset of the real line, and let $F,G$ be antiderivatives of the same function on $U$. Each connected component $I_j$ is an open interval. On that component, $(G-F{)}^{\prime }=0$, so the mean value theorem gives $G-F=C_j$. The constants on different components are independent: there is no subinterval inside $U$ joining them. Conversely, adding any chosen constant on each component preserves the derivative everywhere on $U$.

For example, the domain of $g(x)=1/x$ is $U=\mathbb{R}\setminus \{0\}$. Every antiderivative on this domain has the formwith arbitrary real $C_{-},C_{+}$. The logarithm of the absolute value differentiates to $1/x$ on each side. The function with $C_{-}=2$ and $C_{+}=-1$ is therefore a valid antiderivative, although it cannot be written as $\log |x|+C$ with one global constant.

An initial condition at $x=1$ fixes $C_{+}$ and leaves $C_{-}$ free. Fixing both constants requires one reference value in each connected component, or additional conditions that couple the components. The locally constant function $F(x)-\log |x|$ captures this remaining freedom.

The Arrhenius equation models the temperature dependence of a rate constant byHere $A$ is the pre-exponential factor, $E_a$ the activation energy, $R$ the molar gas constant, and $T$ the absolute thermodynamic temperature. In its original form, $A$ and $E_a$ are held constant over the range being modeled. It is an empirical relationship, not a universal rule for every reaction mechanism. See the Arrhenius equation.

Eliminating $A$ between two positive temperatures gives the useful predictionFor $E_a>0$, warming increases $k$. The logarithmic sensitivity is $d\ln (k/k_{\mathrm{ref}})/dT=E_a/(R{T}^2)$, so there is no universal doubling per ten degrees. The same $10\mathrm{K}$ rise from $300$ to $310\mathrm{K}$ gives a factor $1.67749$ for $E_a=40\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$, versus $2.81398$ for $80\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$.

The useful simplification is that a strong change in reaction rate becomes a straight-line relation after changing variables. Measurements at several temperatures can estimate the slope and test whether one pair of constants describes the chosen range. Curvature can indicate a failure of that approximation; fitting a line does not uniquely establish a microscopic reaction mechanism.

Svante Arrhenius's 1889 paper on sucrose conversion through catalysis by a chemical acid compared the strong observed temperature dependence with the much smaller increase in collision frequency. He discussed van't Hoff's earlier temperature relations and proposed a small population of hypothetical active sucrose, replenished from ordinary sucrose, to interpret the dependence. The inspected translated excerpt derives an inverse-temperature relation in its equations (6)–(7) and then applies it to the reaction rate. That historical hypothesis is distinct from a modern transition state description. See the On the Reaction Velocity of the Inversion of Cane Sugar by Acids. The excerpt omits parts of the original paper; it does not provide a complete original apparatus description or dataset.

The pre-exponential factor must have the same units as the rate constant, since the exponential function supplies a dimensionless quantity. For a rate law $v=k[\mathrm{A}{]}^m$, both $A$ and $k$ have units $(\mathrm{mol}{\mathrm{L}}^{-1}{)}^{1-m}{\mathrm{s}}^{-1}$. Thus a first-order reaction uses ${\mathrm{s}}^{-1}$, whereas a second-order reaction uses $\mathrm{L}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{s}}^{-1}$. Calling $A$ a frequency factor does not give it frequency units for every order of reaction. Its value includes the physical and geometrical factors allowed by the chosen reaction mechanism; equal activation energies need not mean equal reaction rates.

To read the blue Arrhenius plot, define the dimensionless quantity $X=1000\mathrm{K}/T$ and $y=\ln [k/k(300\mathrm{K})]$. Its numerical slope is $m_X=-60000/(1000R)\approx -7.21634$, giving $E_a=-(1000\mathrm{K})R{m}_X=60\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. The slope against $1/T$, by contrast, is $-7216.34\mathrm{K}$. The orange tangent at $300\mathrm{K}$ gives $E_{a,\mathrm{app}}\approx 64.9887\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$; its finite-interval slope between $280$ and $340\mathrm{K}$ gives about $65.1227\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. The distinction between a local derivative and a fitted interval matters even with perfectly noise-free synthetic evaluations.

Two measured positive rate constants give a two-point estimateSeveral temperatures let linear regression test the constant-parameter approximation: inspect the regression residuals, keep the same rate law and composition convention, and account for uncertainty in both $k$ and $T$. A straight line over a short range supports a useful local approximation; it does not uniquely identify a reaction mechanism. A fixed change of units or of $k_{\mathrm{ref}}$ shifts the intercept without changing the inferred activation energy.^[ref]

An asymptotic power series uses powers as its asymptotic sequence. Near a finite point $a$, the condition in the title applies along the specified domain for every fixed $N$; at infinity, inverse powers are convenient. It need not be a convergent power series. Its coefficients record the successive limiting responses, as the integral example below demonstrates.

The finite-order Taylor theorem supplies an asymptotic power series for a smooth function near a point, but smoothness does not imply that its Taylor series converges to the function. Define $g(x)=e^{-1/x}$ for $x>0$ and $g(x)=0$ for $x\le 0$. Every right-hand derivative is an exponential function times a polynomial in $1/x$, tending to zero as $x\downarrow 0$. Consequently $g$ is a smooth function with every derivative zero at the origin. Its all-zero Taylor series misses $g(x)>0$ to the right. Zero and $g$ also have the same all-zero asymptotic power series on that side, despite being different functions. An asymptotic expansion can determine all its own coefficients without determining the represented function uniquely.

For a square well of width $L$, exterior wavenumber $k$ and interior wavenumber $q>k$, matching the wavefunction and its derivative gives the reflection probability $R=C{\sin }^2(qL)/(1+C{\sin }^2(qL))$, where $C=(q^2-k^2{)}^2/(4k^2{q}^2)$. For example $q=3k$ and $kL=\pi /2$ give $R=16/25$, whereas zero potential gives $R=0$. Moreover deeper wells at $qL\in \pi \mathbb{Z}$ have zero reflection by resonant transmission through a square well. Reflection therefore cannot be ordered solely by pointwise ordering of attractive potentials.

For ${\mathcal{L}}_{\mathrm{int}}=-g\varphi \bar{\psi }{\gamma }^5\psi$, the variational Dirac equation and adjoint Dirac equation are $(i/\partial -m-g\varphi {\gamma }^5)\psi =0$ and $i({\partial }_{\mu }\bar{\psi }){\gamma }^{\mu }+m\bar{\psi }+g\varphi \bar{\psi }{\gamma }^5=0$. Substitute these into the derivative of $j_5^{\mu }=\bar{\psi }{\gamma }^{\mu }{\gamma }^5\psi$. Anticommuting the chirality matrix past ${\gamma }^{\mu }$ gives two identical mass terms and two identical interaction terms, proving the displayed identity. These are classical field equations; the formula does not assert an unrenormalized quantum composite-operator identity. With Hermitian coupling $g=i{g}_P$, the second term is $-2g_P\varphi \bar{\psi }\psi$.

For a posterior distribution with finite first moment, the posterior risk derivative at continuity points is $(\gamma +\delta )F(a)-\gamma$. Convexity therefore makes a posterior $\gamma /(\gamma +\delta )$-quantile a Bayes estimator. With atoms, the condition is $F(a^{-})\le \gamma /(\gamma +\delta )\le F(a)$. Increasing the cost of underestimation moves the optimal estimate upward.