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For $u\ge 0$ and a positivity-preserving operator, this is the negative log-likelihood for independent Poisson observations with intensity $1+Tu$, up to constants depending only on the data and the unit background. For bounded $g\ge 0$, its scalar derivatives are $1-g/(1+s)$ and $g/(1+s{)}^2$ on $s=Tu\ge 0$. It is convex, and strictly convex in the predicted intensity when $g>0$ almost everywhere. On a unit-area domain, the Jensen inequality gives $D_g(u)\ge \Vert Tu{\Vert }_1-\Vert g{\Vert }_{\infty }\log (1+\Vert Tu{\Vert }_1)$. The shift makes $\log (1+Tu)$ integrable whenever $Tu\in L^1$; no integrability of $\log u$ is needed.

For a p-n junction at fixed temperature, the Shockley diode equation models its electric current asHere $I_S>0$ is the diode saturation current, $n>0$ the diode ideality factor, $V_T$ the thermal voltage, $k_B$ the Boltzmann constant, and $q>0$ the elementary charge.^[ref]

For an original model example, choose $I_S=1\mathrm{nA}$, $n=2$, and $T=300\mathrm{K}$. Using $k_B=1.380649\times 10^{-23}J/K$ and $q=1.602176634\times 10^{-19}\mathrm{C}$ gives $V_T=25.8520\mathrm{mV}$. Then $V_D=0.600\mathrm{V}$ gives $I_D=0.109591\mathrm{mA}$, whereas $V_D=0.800\mathrm{V}$ gives $I_D=5.24450\mathrm{mA}$. These are predictions of the declared model. There is no universal turn-on electric potential difference: solving the Shockley diode equation for the value needed at a chosen electric current givesAt $I_D\gg I_S$, increasing $V_D$ by $0.100\mathrm{V}$ multiplies the electric current by approximately $e^{0.100/(n{V}_T)}=6.91772$. At $V_D=-0.300\mathrm{V}$ the same model gives $I_D=-0.996979\mathrm{nA}$. Its reverse branch approaches $-I_S$; it cannot describe diode reverse breakdown.

A constructed extension adds internal series electrical resistance $R_s$ to a junction whose electric potential difference is $V_j$. Applying the Kirchhoff cycle law givesFor $R_s=20\Omega$ and the same $I_S,n,T$, sustaining $I_D=5.24450\mathrm{mA}$ needs $V_D=0.904890\mathrm{V}$: $0.800\mathrm{V}$ across the junction plus $0.104890\mathrm{V}$ across its internal electrical resistance. The derivativeshows why the current-voltage characteristic becomes nearly a straight line when the internal series electrical resistance dominates. This is still a fixed-temperature model; self-heating can change $I_S$, $n$ and $V_T$. Real diodes also have leakage and charge-storage effects omitted here.^[ref]

Signaling adaptation is a decline of a response toward its earlier level while the initiating signal remains present. Specify which response is measured and its baseline: complete signaling adaptation means eventual return to that baseline; partial signaling adaptation leaves a lasting displacement. The continuing signal distinguishes signaling adaptation from simply switching off the stimulus. Signaling adaptation describes behavior, while signaling feedback describes a causal arrangement, so neither implies the other.

By reducing a persistent response to background ligand concentration, signaling adaptation can make a biological cell sensitive to changes rather than only to absolute levels. Resetting the response helps leave room for another change over a useful range of ligand concentrations. This is valuable in chemotaxis: swimming bacteria encounter changing chemical conditions, and their signal transduction influences whether a run continues or a tumble occurs. Signaling adaptation helps separate a change encountered during motion from a lasting background. This usefulness does not imply exact return or equal sensitivity at every possible concentration. ^{[ref] [ref]}

Here is a solvable incoherent feedforward loop toy that exhibits complete signaling adaptation without signaling feedback. The input $s(t)$ jumps from $0$ to $1$ at $t=0$ and remains $1$, a Heaviside step function. Let $x$ be a fast activating branch, $a$ a slower inhibitory branch, and $r=x-a$ a readout. The activities $s,x,a,r$ are dimensionless variables; physical time is in seconds. Choose time constants ${\tau }_x=1$ second and ${\tau }_a=5$ seconds, with initial conditions $x(0)=a(0)=0$:Both branches read $s$ directly. Neither branch depends on $r$, and $a$ does not change $x$: the subtraction only defines the readout. Thus the slow opposing influence is feedforward, with no closed feedback control loop.

For $t\ge 0$, solving the two ordinary differential equations givesWriting $t$ for the numerical time in seconds, these are $x=1-e^{-t}$, $a=1-e^{-t/5}$ and $r=e^{-t/5}-e^{-t}$. Initially $r=0$, then $x$ rises faster than $a$, producing a pulse. At long times $x$ and $a$ both approach $1$, so $r$ returns to $0$ despite sustained $s=1$. For this upward step $e^{-t/5}\ge e^{-t}$, hence $r(t)\ge 0$. The striking feature of this signaling adaptation toy is that both branches remain active while the readout becomes quiet: the continuing background and the response to its arrival are separable.

Find the pulse maximum by differentiating the exponential functions. With numerical time in seconds, the derivative isThe stationary condition is $e^{-4t/5}=1/5$, so the natural logarithm givesBefore this time the derivative is positive; afterwards it is negative. Together with $r(0)=0$ and ${\lim }_{t\to \infty }r(t)=0$, this proves that it is the unique pulse maximum. As another numerical check, at $t=10$ seconds $x\simeq 0.999955$, $a\simeq 0.864665$ and $r\simeq 0.135290$: the slow branch is still catching up.

The subtraction $r=x-a$ is a mathematical readout, not a claim that an actual chemical concentration equals this difference for arbitrary inputs. Nonnegativity was established only for the specified upward step; other input histories can give a negative signed readout. Complete signaling adaptation here also relies on equal eventual branch gains: changing the readout to $r=x-0.9a$ leaves $r_{\ast }=0.1$ for $s=1$. This toy illustrates cancellation and different time constants; it does not establish molecular implementation or robustness to arbitrary parameter changes. Ma and colleagues' 2009 study separately analyzed biochemical mathematical models with negative feedback and incoherent feedforward loops, identifying additional conditions for signaling adaptation in those models. ^[ref]

A historical connection to feedback control comes from Yi, Huang, Simon and Doyle's 2000 analysis of a bacterial chemotaxis model. They identified integral action in its regulation of receptor activity: accumulated deviations alter a slower adjustment, which in turn changes the measured activity. At a stationary state, an adjustment that integrates a nonzero error cannot remain constant, explaining how zero error can follow. Their conclusion depended on stated biochemical assumptions and stable behavior; it was not a claim that every biological cell uses that mechanism. A transient pulse alone, as the incoherent feedforward loop toy shows, cannot identify the underlying signaling feedback architecture. ^[ref]

Signaling feedback returns the effect of a downstream event to an earlier stage of signal transduction, closing a loop analogous to feedback control. Trace the causal arrows: if an earlier relay activates a later relay and the later relay changes the earlier relay's activity, the return path closes the loop. Negative feedback opposes the initiating change; positive feedback reinforces it. Signaling feedback is a possible property of signal transduction, not a requirement of every pathway.

For a schematic example, let relay $A$ activate relay $B$. If $B$ inhibits $A$, increased $A$ eventually produces an opposing influence: negative feedback. If $B$ activates $A$, increased $A$ produces a reinforcing influence: positive feedback. These signs describe effects on activity, rather than the direction in which molecules travel. A closed loop containing two inhibitions can have positive feedback, because inhibiting an inhibitor reinforces the original change. The sign belongs to the whole loop.

Negative feedback can restrain excessive activation and help preserve a usable range of responses. Positive feedback can support a decisive transition: Alberts and colleagues describe an enzyme whose product increases its own activity. Such reinforcement can contribute to switchlike behavior, but a positive feedback arrow alone does not prove bistability. A different structure, an incoherent feedforward loop, sends opposing influences forward from the same input; inhibition there is not a downstream return. ^{[ref] [ref]}

A worked mathematical model shows why negative feedback does not automatically produce complete signaling adaptation. Let $s$ be a fixed normalized input, $r$ a normalized response, $g\ge 0$ a dimensionless feedback gain, and $\tau >0$ a time constant. Suppose response removal contributes $-r$ and a return proportional to the response contributes $-gr$. The ordinary differential equation isSetting the derivative to zero gives the equilibrium point $r_{\ast }=s/(1+g)$. A displacement $\delta r=r-r_{\ast }$ obeys $\tau d(\delta r)/dt=-(1+g)\delta r$, so its decaying exponential function establishes asymptotic stability. For a unit input switched on at $t=0$, initial condition $r(0)=0$, $g=2$ and $\tau =1$ second,The negative feedback reduces the eventual response from $1$ to $1/3$, but this response rises monotonically and never returns to its zero baseline. Attenuation and complete signaling adaptation are different properties. This deliberately simple mathematical model specifies no molecular pathway; integral action provides a contrasting way to remove a lasting offset in a stable feedback control system. ^[ref]

Signed swept area counts the area traversed by a position vector, positively or negatively according to its direction of turning. For a continuously differentiable planar trajectory, its accumulated value relative to $t_0$ is the line integralA short segment contributes the oriented triangle area $\frac{1}{2}\mathbf{n}\cdot (\mathbf{r}\times d\mathbf{r})$; taking its time derivative gives areal velocity. Repeated turns count repeated area; a reversal subtracts it. Thus this quantity need not equal the area of the union of all previously swept regions. The origin matters: the relevant origin for Kepler's second law is the attracting focus of the ellipse, rather than its geometric centre.

Single-electron superconducting tunneling transfers individual Electrons across a barrier into or out of quasiparticle states of a material with superconductivity. In a normal-metal–insulator–superconductor junction, abbreviated NIS, only one electrode has superconductivity. Such a junction can reveal a superconducting energy gap without supporting a DC Josephson effect between two superconducting reservoirs.

For an elementary interpretation, assume an opaque barrier, approximately constant normal-electrode density of states, and an energy-independent tunneling matrix element over the range probed. Write the superconducting density of states divided by its normal-state value as $n_s(E)$, with $E$ measured from its chemical potential. A signed voltage convention chosen for positive normal-electrode injection giveswhere $f$ is the Fermi-Dirac distribution at a common temperature and $R_N$ is the normal-state junction electrical resistance. This current convention is local to the NIS measurement. The expression counts occupied initial and empty final states; it is a quasiparticle transport model, not a current-phase relation of a superconducting weak link.

An original analytic illustration uses the ideal fully gapped BCS theory spectrum at zero temperature, with positive gap edge $\Delta$:For positive $V$, the occupation difference restricts the integral to $0<E<eV$. Since $d\sqrt{E^2-{\Delta }^2}/dE=E/\sqrt{E^2-{\Delta }^2}$, integration and odd extension giveThe threshold is $\Delta /e$ for this NIS model. In the corresponding zero-temperature single-electron process between two fully gapped superconductors, creating excitations on both sides instead requires $|eV|\ge {\Delta }_1+{\Delta }_2$. That quasiparticle threshold does not forbid coherent pair electric current below it. At finite temperature, thermally populated quasiparticles soften the simple threshold argument.

At zero temperature in the NIS approximation, $R_{N}d{I}_{\mathrm{qp}}/dV=n_s(eV)$ away from the singular gap edge. Thus a voltage sweep probes the excitation density of states. The ideal divergence is rounded in real measurements, and at nonzero temperature the derivative samples a thermally broadened range of energies. For $|eV|>\Delta$, the model has $V{I}_{\mathrm{qp}}>0$: electrical work creates excitations and is dissipated on relaxation. This distinguishes its steady transport from the conservative Josephson coupling energy.

For $F_{\mu }(t)=\mu ((-\infty ,t])$, every difference quotient at $x$ is nonnegative and bounded by $4\mu (B(x,2|h|))/{\lambda }_1(B(x,2|h|))$. The spherical derivative of a singular measure vanishes, so the two-sided derivative is zero Lebesgue almost everywhere. A singular distribution function need not be constant: differentiation almost everywhere does not recover increments without absolute continuity of a function of the function.

A singular perturbation is a parameter-dependent problem whose limiting problem changes qualitatively as the parameter tends to zero. In an ordinary differential equation, a small coefficient of the highest derivative can reduce the order of the limiting differential equation, preventing it from satisfying all the original boundary conditions. Different length or time scales often require different asymptotic expansions.^[ref]

Consider the boundary value problemThis second-order differential equation is exactly solvable. Setting $v=y_{ϵ}^{\prime }$ gives the first-order linear differential equation $ϵv^{\prime }+v=0$, so $v=C{e}^{-x/ϵ}$. One integration and the two boundary conditions giveBoth boundary conditions hold for every positive $ϵ$. For numerical evaluation, write the same ratio as $expm1(-x/ϵ)/expm1(-1/ϵ)$, where $expm1(t)$ evaluates the exponential function minus one without catastrophic cancellation near $t=0$. Keep the denominator for finite $ϵ$: at $ϵ=0.2$ it is approximately $0.993262$, rather than exactly one.

Setting $ϵ=0$ gives $y_0^{\prime }=0$. The outer solution selected by the right boundary condition is therefore $y_0=1$, which fails the left boundary condition. The exact exponential function $e^{-x/ϵ}$ shows how that mismatch is repaired: the solution rises from zero to almost one within a boundary layer of width $x=O(ϵ)$, using Big O notation. For each fixed $x>0$, $y_{ϵ}(x)\to 1$, while $y_{ϵ}(0)=0$ always. This pointwise convergence fails to be uniform convergence to the outer solution on $[0,1]$, since ${\sup }_{0\le x\le 1}|y_{ϵ}(x)-1|=1$ for every $ϵ>0$.

The coefficient $ϵ$ is small, but its product with the second derivative need not be small. Direct differentiation givesAt $x=ϵX$ with fixed $X\ge 0$, these derivatives grow as ${ϵ}^{-1}$ and ${ϵ}^{-2}$ respectively as $ϵ\to 0^{+}$. Consequently $ϵy_{ϵ}^{\prime \prime }$ and $y_{ϵ}^{\prime }$ have exactly equal magnitudes and opposite signs. Even away from the boundary layer, their magnitudes remain equal in this example; both become exponentially small at each fixed positive $x$. The outer expansion neglects an exponentially small transient there, rather than establishing that one derivative term is smaller than the other everywhere.

To predict the width without knowing the exact solution, suppose an order-one change occurs over distance $\delta$. The derivative scales are $y^{\prime }=O({\delta }^{-1})$ and $y^{\prime \prime }=O({\delta }^{-2})$. The dominant balance $ϵ/{\delta }^2\sim 1/\delta$ gives $\delta \sim ϵ$. The resulting inner variable $X=x/ϵ$ makes the boundary layer visible at a fixed scale. This is the useful feature of singular perturbation: an apparently disappearing term identifies the missing structure when one measures its derivatives on the appropriate scale.

For the closed SIR model with $\gamma >0$, the infectious compartment's fraction approaches $i_{\infty }=0$. One proof outline uses the bounded, increasing removed compartment's fraction: since $r(t)-r_0=\gamma {\int }_0^{t}i(u)du$, the total time-integrated $i$ is finite. The bounded derivative $di/dt$ from the SIR model equations prevents indefinitely recurring spikes, so $i(t)$ tends to zero.

For $s_0>0$, the final size relation for an epidemic follows by dividing the SIR model equations for $s$ and $r$ while $i>0$, giving $ds/dr=-(\beta /\gamma )s$. Integration then yieldsThis also follows from the SIR first integral and conservation of population. For the stipulated initial value problem, solveThe bisection method givesAbout $5.88\%$ of the original population remains in the susceptible compartment. The $94.12\%$ is the final removed compartment's fraction, including the initially $1\%$ in the infectious compartment. The newly acquired infection during the modelled period accounts for $s_0-s_{\infty }\approx 0.93120263520$ of the total population, not $1-s_{\infty }$. Dividing by $s_0$ instead gives the proportion of initially susceptible compartment members who acquire infection; its denominator must be stated.

At the SIR infection peak, the flow into the infectious compartment only balances its outflow; it does not stop. Further infection explains why the final susceptible compartment's fraction is well below $1/3$. Also $s_{\infty }=s_0{e}^{-R_0(r_{\infty }-r_0)}>0$ for $s_0>0$ and finite $R_0$: the closed SIR model does not require everybody to acquire infection before its infectious compartment empties in the long-time limit.

The nine-point finite-difference stencil for the Laplacian has expansion $D_{9}u=\Delta u+h^2{\Delta }^{2}u/12+h^4(u_{xxxxxx}+u_{yyyyyy}+5u_{xxxxyy}+5u_{xxyyyy})/360+O(h^6)$. If $\Delta u=f$, differentiating this equation gives $u_{xxxxyy}+u_{xxyyyy}=f_{xxyy}$ and $u_{xxxxxx}+u_{yyyyyy}={\Delta }^{2}f-3f_{xxyy}$. Hence the corrected compact equationhas sixth-order normalized consistency of a numerical method. Source derivatives can be supplied analytically; numerical approximation of $\Delta f$ must be fourth order and of the fourth derivatives second order. With exact square-grid Dirichlet boundary data and uniformly smooth solution, the discrete maximum principle inverse bound converts this residual into sixth-order nodal convergence. The unknown stencil stays compact even if more distant known source values are used.

In Cartesian coordinates, the slope of a nonvertical straight line is its vertical change divided by its horizontal change. The graph $y=mx+c$ has slope $m$; a vertical straight line has no finite slope. For a tangent line to a differentiable function, the slope is the derivative.

A positive measurable function $\ell$ is slowly varying at infinity iffor every fixed $u>0$. It has regular variation of index zero. “Slowly” concerns relative change under multiplication of the argument; it does not mean that the function must be bounded, have a small derivative, or converge to a finite value.

Every positive constant is a slowly varying function. So are $\log t$, $1/\log t$ and $(\log t{)}^a$ for any fixed real number $a$, on $t>1$. Indeed,The first grows without bound, the second tends to zero, and the constant does neither. Another unbounded example is $\ell (t)=e^{\sqrt{\log t}}$:This function can grow much faster than $\log t$ while retaining regular variation of index zero.

Even convergence is unnecessary. The bounded functionoscillates between $e^{-1}$ and $e$. It is nevertheless a slowly varying function: the change in the argument of the sine under $t\mapsto ut$ isand the Lipschitz continuity of the sine makes the logarithm of the ratio tend to zero. The sequences $t_k=\exp (\exp (\pi /2+2\pi k))$ and $s_k=\exp (\exp (3\pi /2+2\pi k))$ give the two values $e$ and $e^{-1}$ for every nonnegative integer $k$, so there is no limit at infinity.

Contrast this with $h(t)=2+\sin (\log t)$, which is positive and bounded but is not a slowly varying function. Fix $u=e^{\pi /2}$. Along $t_k=e^{2\pi k}$ its rescaling ratio is $h(u{t}_k)/h(t_k)=3/2$; along $s_k=e^{\pi +2\pi k}$ the ratio is $1/2$. The persistent oscillations prevent even a limit for this one fixed rescaling. It is not regularly varying with any finite index either.

The useful distinction is whether a fixed proportional change of input eventually produces a negligible proportional change of output. A slowly varying function can therefore preserve the scaling exponent of a power law while changing its magnitude substantially. The decomposition in regular variation explains why a logarithmic factor can matter in numerical estimates without changing the limiting exponent.

A dark, opaque circular planet of radius $R_p$ blocks specific intensity over its overlap with the projected star. With $k=R_p/R_{\star }$ and no transit dilution, the instantaneous exoplanet transit depth is exactlyWhen $k\ll 1$, the entire planetary disk is inside the stellar disk, and specific intensity varies little across that disk, small-planet transit weighting reduces toHere $z$ is the projected centre separation divided by $R_{\star }$. The numerator samples the local specific intensity; the denominator normalizes by the whole star's mean specific intensity. The small-size argument and the distinction between full overlap and ingress are explicit in the original analytic treatment. ^[ref]

For a short approximately straight crossing at constant transverse velocity $v_{\perp }$, write $x=v_{\perp }t/R_{\star }$ with $t=0$ at mid-transit, and let $b$ be the transit impact parameter. Then $z^2=x^2+b^2$ and ${\mu }_p=\sqrt{1-b^2-x^2}$. With stipulated $u=0.6$ and $k=0.05$, a central crossing has mid-transit exoplanet transit depth approximately $0.003125$, whereas $b=0.8$ gives $0.002375$. A uniform disk would give $k^2=0.0025$ for either fully overlapping crossing. The brighter central chord produces a deeper dip and the off-centre chord a shorter crossing at the same $v_{\perp }$.

At fixed $z<1$ away from the edge, a two-dimensional Taylor's theorem expansion over the symmetric planetary disk cancels the first derivatives; the first correction to the blocked radiant flux is of order $k^4$ if the second derivatives remain bounded. This error estimate is not uniform at the edge, where the radial derivative of the linear limb-darkening law diverges. Ingress, egress, a grazing exoplanet transit, and a crossing of a starspot require the overlap integral or a suitable more complete model. Inferring $k=\sqrt{\delta }$ without the local-to-mean specific intensity ratio can bias the estimated planet radius.