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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.

For $r=-\lambda >0$ and vanishing solid-to-liquid diffusivity ratio, $H(r)=\sqrt{\pi }r{e}^{r^2}\mathrm{erfc}(-r)$ gives the interfacial liquid concentration. It increases from zero to $C_s$ as $r$ increases. Its derivative is $C_s{H}^{\prime }/(1+H{)}^2$, which decreases from $\sqrt{\pi }{C}_s$ to zero. For a direct proof, write $A=H/r=\sqrt{\pi }{e}^{r^2}\mathrm{erfc}(-r)$. Then $A^{\prime }=2rA+2$, so $A\ge \sqrt{\pi }+2r$. Differentiation gives $2H^{\prime 2}-(1+H)H^{\prime \prime }=2[(1+r^2+2r^4)A^2+r(4r^2-1)A+2r^2-2]>0$: the bracket increases with $A$, and substitution of its lower bound gives only positive polynomial coefficients, including constant $\pi -2$. Thus the concentration is strictly concave as a function of melting speed parameter $r$. Combining this with a linearized thermal balance produces one fold on the melting branch and, with the freezing branch, an interval of three possible interface speeds.

Smooth convergence on a compact interval requires uniform convergence of the functions and of their derivatives of every order separately, as measured by the derivative seminorms. The need for several scales is visible inFor $n\ge 2$, $p_0(f_n)=1/n$ because $t=\pi /(2n)$ lies in $[0,1]$. The derivative has supremum norm one for every $n$, already at $t=0$. Thus $f_n$ converges uniformly to zero, but fails the test for smooth convergence to zero. Andreas Buttenschoen, Applied Functional Analysis, spaces of continuous and differentiable functions discusses this example and the family of smoothness measurements.

More generally, for any fixed $K\ge 0$, $h_n(t)=\sin (nt)/n^{K+1}$ has $p_K(h_n)\le 1/n$ but $p_{K+1}(h_n)=1$ for $n\ge 2$. If a norm $N$ induced the smooth topology, its continuity would give $N(f)\le C{p}_K(f)$ for some finite $K$, while continuity of $p_{K+1}$ for $N$ would give $p_{K+1}(f)\le C^{\prime }N(f)$. Substituting $h_n$ contradicts these two bounds. Smooth convergence therefore cannot be reduced to one norm. Its beauty is that every finite order can be controlled without fixing a final order in advance.

For the first-order Cauchy-Riemann operator, the equation ${\varphi }_t-i{\varphi }_x=0$ admits the entire functions ${\varphi }_n(x,t)=e^{-\sqrt{n}}\cos (n(x+it))$. Every fixed-order derivative of the initial data converges uniformly to zero, since $n^r{e}^{-\sqrt{n}}\to 0$. Nevertheless ${\sup }_x|{\varphi }_n(x,t)|=e^{-\sqrt{n}}\cosh (n|t|)\to \infty$ for each $t\ne 0$. Thus even convergence of all initial derivatives does not give continuous dependence on initial data away from the initial line. Analytic existence and smooth-data stability are different properties.

A penalized function estimate balancing a residual loss against an integrated derivative penalty. For squared-error loss and the second derivative roughness penalty, the estimate is a cubic smoothing spline. Higher derivative orders give other spline degrees.