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Take a singly quantized vortex of positive circulation. The superfluid velocity is $\mathbf{u}=(\hslash /m)\nabla S$, where $S$ is the complex argument of the condensate. Since $S=\theta$, quantized circulation gives $u_{\theta }=\hslash /(mr)$. Insert the radial Thomas–Fermi approximation for a condensate into the cylindrical volume integral, excluding the core $r<a_0$:Integrating the logarithmic and quadratic terms separately givesThe logarithm is the familiar long-range flow contribution of a quantum vortex; the $-1/2$ comes from the declining trapped number density. If the bucket truncates the cloud, put $R=R_0<R_r$ and the same integration gives $E_{\mathrm{kin}}=\pi H_0{\hslash }^2{n}_0[\log (R/a_0)-(R^2-a_0^2)/(2R_r^2)]/m$, provided $a_0<R$. A vortex of integer winding number $q$ has $u_{\theta }=q\hslash /(mr)$ and multiplies this flow energy by $q^2$. The core radius acts as a cutoff of order the local healing length; its internal gradient and interaction energies are not computed by this shell estimate.

With the sign convention of the displayed operator, the Lax equation is $L_t=[L,A]$. Differentiating $L\phi =k^2\phi$ shows that ${\phi }_t+A\phi$ is another solution at the same spectral parameter. The normalization at $-\infty$ fixessince $A{e}^{-ikx}=4i{k}^3{e}^{-ikx}$ there. At $+\infty$, comparison of the two plane-wave coefficients givesThus the transmission denominator $a$ is time independent. This is the Time evolution of KdV scattering data for the stated normalization.

The logarithmic derivative $S={\phi }_x/\phi +ik$ obeys the Riccati equationIts formal large-$k$ expansion givesThere is an important qualification to the integral identity in the question. For real $k$ with nonzero reflection, $e^{ikx}\phi \sim a+b{e}^{2ikx}$ has no limit at $+\infty$, so the literal improper integral of the exact $S$ need not exist. Continue to $\mathrm{Im}k>0$, where the reflected exponential decays relative to $e^{-ikx}$, and take sufficiently large $k$ away from zeros and with a consistent logarithm. Then the normalization givesThe second equality is an asymptotic expansion, not an asserted convergent series. Time independence of $a$ fixes every coefficient, so each $\int S_{n}dx$ is a conserved quantity of the Korteweg-De Vries equation. The first two nontrivial examples are $-\int udx$ and $\int u^{2}dx$.

For the usual real-valued KdV potential and real formal $k$, write $S=X+iY$ with real coefficient functions. Its imaginary part of the Riccati equation is $Y^{\prime }-2kX+2XY=0$, henceIn the formal expansion, $X$ contains precisely the even-indexed $S_n$ and $Y$ the odd-indexed ones. Rapid decay of $u$ and its derivatives makes every coefficient of the logarithm vanish at both spatial ends. Therefore $\boxed{{\int }_{\mathbb{R}}{S}_{2j}dx=0(j\ge 1)}$. This is a coefficientwise formal identity; it does not claim that the exact real-axis reflected solution has a convergent integral of $X$. The analytically continued scattering identity and the formal expansion are the conventions needed to make the requested argument valid.

After division by $z$, the coefficients are $P=2/z-1$ and $Q=-1/z$. At $0$, $zP$ and $z^{2}Q$ are analytic functions, so $0$ is a regular singular point. Its indicial equation is $r(r-1)+2r=r(r+1)=0$. There are no other finite singular points of a differential equation. At infinity the regular singular point criterion for a second-order equation would require $P=O(z^{-1})$, but $P\to -1$; hence infinity is an irregular singular point.

Substituting $y=f/z$ reduces the equation to $f^{\prime \prime }-f^{\prime }=0$, soA linearly independent pair is $1/z$ and $(e^z-1)/z$. Their Wronskian is $e^z/z^2\ne 0$. The second has a removable singularity at $0$ and value $1$, whereas the first has a simple pole, agreeing with the two characteristic exponents at a regular singular point, $0,-1$. The integer difference does not require a logarithm. At infinity $e^z/z$ has an essential singularity, consistent with the irregular singular point; the existence of the simpler solution $1/z$ does not make infinity regular.

Expand the finite Euler product into geometric series. By unique prime factorization, it is the sum of $1/n$ over positive integers whose prime factors do not exceed $x$, and hence contains every term up to $m=\lfloor x\rfloor$. The harmonic sum satisfiesThe last inequality uses $m+1>x$, including when $x$ is an integer. For each prime number $p$,The telescoping series over all integers $n\ge 2$ bounds the sum of these remainders by $\frac{1}{2}{\sum }_{n\ge 2}[n(n-1){]}^{-1}=1/2$. Taking logarithms of the product bound therefore givesThis explicit constant suffices; no asymptotic theorem about the distribution of prime numbers is needed.

Let $g=a{U}_0\gg 1$ and $z=ka$. A denominator zero away from the removable zero at $z=0$ obeysNear $z=\pi$, write $z=\pi +\delta$ and choose the logarithm branch with $\delta \to 0$. Expanding givesConsequentlyThe lower-half-plane pole is a scattering resonance: an odd standing wave is nearly confined between the two high repulsive barriers but leaks out slowly. If $E=E_R-i\Gamma /2={\hslash }^2{k}^2/(2m)$, then $\Gamma \simeq 2{\hslash }^2\pi \gamma /(m{a}^2)$ and its probability lifetime is $\hslash /\Gamma$. It is not a genuine bound state of the repulsive potential.

All entropies and information rates below use base-two logarithms. The quantum fidelity uses the unsquared convention.

Use base-two logarithms. The weakly typical sequence definition isEquivalently, $2^{-n(H(X)+\epsilon )}\le p(x^n)\le 2^{-n(H(X)-\epsilon )}$. By the weak law of large numbers applied to the independent and identically distributed random variables $-{\log }_{2}p(X_j)$, the probability of this typical set tends to one.

This captures the usual probability scale of a long sample, but it need not make every symbol frequency representative. For a fair binary source, every word has probability $2^{-n}$ and $H(X)=1$, so even the all-zero word is weakly typical. The strongly typical sequence definition additionally controls empirical symbol frequencies and excludes this word for small tolerance. Thus weak typicality agrees with the high-probability-set intuition, while allowing individually unrepresentative members.

There are two positive roots for $u>1$: the derivative of $x{e}^{1/x}$ is $e^{1/x}(1-1/x)$, so the function decreases to its minimum $e$ at $x=1$ and then increases. Negative $x$ cannot solve the equation. For the larger root, expansion of the exponential function in $1/x$ givesso series reversion yieldsFor the smaller root, set $y=1/x$ and take a logarithm. Then $y-\log y=u$, so successive substitution gives $y=u+\log u+O(\log u/u)$ andThese are the first two nonzero terms of the two asymptotic expansions. Equivalently, the roots are $-1/W_0(-e^{-u})$ and $-1/W_{-1}(-e^{-u})$, from the real branches of Lambert W.

Apply the Jensen inequality to the convex exponential function under the trial equilibrium measure:The given identity therefore implies $e^{-F}\ge e^{-F_0-⟨H-H_0{⟩}_0}$. Taking minus the logarithm reverses the inequality and proves the Feynman-Bogoliubov inequalityThe units $\beta =1$ make $H$ and $F$ dimensionless here. Equality holds when $H-H_0$ is constant on the support of the trial measure. Optimizing the trial parameters tightens an upper bound, which is the basis of the Gaussian variational approximation.

On either open half-line, part (i) says $\hat{u}=-v^{\prime }$. Differentiating the formula in part (iii) makes all elementary terms cancel except the logarithm and the bracket. ThereforewhereandwhereBoth functions are smooth on their respective half-lines, proving the required assertion.

Put $L=L^{\ast }(X)$ and $\mu =\mathbb{E}L$. Since $L$ is a function of $X$, the chain rule for information entropy and part (c) givePart (a), applied to the nonnegative integer-valued random variable $L$, yieldsThe logarithm inequality $\log (1+t)\le t\log e$ implies $\mu \log (1+1/\mu )\le \log e$. ThusPart (c) also gives $\mu \le H(X)$, so monotonicity of the logarithm lets us replace $\log (1+\mu )$ by $\log (H(X)+1)$. Rearranging proves

The pH of a chemical solution is the negative base-ten logarithm of its hydrogen ion thermodynamic activity:The thermodynamic activity is a dimensionless quantity, so the logarithm has a dimensionless argument. The definition uses thermodynamic activity, rather than total chemical acid concentration. Single-ion thermodynamic activities require a convention; practical pH measurements are tied to calibrated standards. See IUPAC Gold Book, pH.

In a dilute ideal chemical solution in water, a useful approximation isSquare brackets denote concentration in moles per litre. In water, $[{\mathrm{H}}^{+}]$ is shorthand for the hydronium concentration, often written $[{\mathrm{H}}_3{\mathrm{O}}^{+}]$; it does not describe isolated bare protons. Departures from ideal behavior need activity coefficients. The concentration formula approximates the thermodynamic activity definition.

For an ideal strong acid supplying one proton per initial unit, stipulate a total concentration of $0.001\mathrm{mol}{\mathrm{L}}^{-1}$ and negligible self-ionization of water. Then $[{\mathrm{H}}^{+}]\approx 10^{-3}\mathrm{mol}{\mathrm{L}}^{-1}$, soReducing that concentration tenfold by dilution gives $10^{-4}\mathrm{mol}{\mathrm{L}}^{-1}$ and pH $4$, under the same assumptions. More generally, dividing the hydrogen ion thermodynamic activity by ten raises pH by one. This is why the scale compresses an enormous range: equal steps of pH represent multiplicative changes, rather than equal differences in concentration. These are model calculations, rather than measured samples.

Neutrality depends on temperature. In the dilute ideal water model, define the dimensionless ion product $K_w=([{\mathrm{H}}^{+}]/c^{\circ })([{\mathrm{OH}}^{-}]/c^{\circ })$ and $\mathrm{p}{K}_w=-{\log }_{10}{K}_w$. Equal hydronium and hydroxide concentrations implyAt $25{}^{\circ }\mathrm{C}$, the textbook approximation $K_w\approx 10^{-14}$ gives neutral pH $7$. Since $K_w$ varies with temperature, neutrality need not mean pH $7$. Nor is pH restricted to $0$–$14$: the activity definition permits negative pH when $a_{{\mathrm{H}}^{+}}>1$, and pH above $14$ when $a_{{\mathrm{H}}^{+}}<10^{-14}$. High concentrations can invalidate the dilute ideal approximation. See Flowers et al., Chemistry 2e, pH and pOH.

S. P. L. Sørensen introduced pH at the Carlsberg Laboratory in 1909, as recorded in the Carlsberg Foundation historical timeline.

Phase folding of a light curve maps timestamps onto a trial cycle of duration $P>0$. Given a reference time $t_0$, define the phase in cycles bywhere $\mathrm{frac}(x)=x-\lfloor x\rfloor$ is the fractional part, using the floor function also for negative $x$. Plot brightness against ${\varphi }_i$ rather than $t_i$; measurements separated by an integer multiple of $P$ occupy the same phase. For a periodic function $F(t+P)=F(t)$, this reveals the shared shape of many cycles. Folding changes coordinates; it neither averages samples nor chooses the correct $P$ by itself. Welsh et al.'s folded Kepler plot separates the exoplanet transit and orbital brightness modulation. ^[ref].

The second figure uses a completely synthetic trapezoidal dip: $P=3$ days, $t_0=1$ day, and the same $1\%$ depth and four-hour duration as the first figure. Different colours identify four successive cycles. Correct folding superimposes the dips; a wrong trial orbital period spreads them out. If $P^{\prime }=P+ϵ$, the centre of the $n$th repeated event shifts by approximately $-nϵ/P$ cycles when folded on $P^{\prime }$, for $|ϵ|\ll P$. With $P^{\prime }=3.03$ days, the fourth plotted event is displaced by $-3(0.03)/3.03\simeq -0.0297$ cycles. Transit-timing variations can also smear a constant-period fold, so retain the original timestamps and inspect individual events.

One may combine nearby phases with a weighted mean, but the uncertainty reduction requires assumptions. For $N$ independent measurements with common standard deviation $\sigma$ and effectively the same expected brightness within the bin, their arithmetic mean has standard deviation $\sigma /\sqrt{N}$. Thus sixteen independent $0.0006$-precision samples give $0.00015$ precision on that phase-bin mean. Shared calibration errors, changing starspots, and bins wide enough to mix ingress with out-of-transit brightness violate these assumptions. A folded plot can conceal changes between cycles. Gaussian process models can represent correlated variability and interpolate irregular samples, but their predictions and uncertainties depend on their assumed covariance function.

The bottom panels show the same model as relative brightness $f=F/F_0$ and as magnitude difference in one fixed photometric passband with a fixed magnitude zero point:A dip to $f=0.99$ gives $\Delta m\simeq +0.010912$ magnitudes: fainter means a larger apparent magnitude. The figure places larger magnitude differences lower on the page so the two dip shapes are easy to compare; reversing that plotting convention does not change the formula. For $f=1-\delta$ and small $\delta$, expansion of the logarithm gives $\Delta m\simeq (2.5/\ln 10)\delta \simeq 1.085736\delta$. The exact transform is nonlinear: transforming an exposure's mean brightness is generally different from averaging its instantaneous apparent magnitudes. The ordinary logarithm cannot be applied to a nonpositive background-subtracted measurement; keep such measurements and their uncertainties in brightness units for an appropriate statistical model.

Choose the following original phenomenological mathematical model solely to calculate a paired-pulse readout. It is not experimental data, a fit to the teaching experiment, or a simulation of the Hodgkin-Huxley model. The conditioning action potential defines time zero. Test pulses have fixed width $0.2\mathrm{ms}$; $I$ denotes their inward electric current amplitude. Impose an absolute neuronal refractory period $t_{\mathrm{abs}}=2\mathrm{ms}$. No finite test amplitude generates an event for $0<\Delta <t_{\mathrm{abs}}$. After that hold, stipulate a required amplitudewith $I_{\infty }=0.2\mathrm{nA}$, dimensionless $A=2$ and recovery time constant ${\tau }_r=3\mathrm{ms}$. Emit a test event exactly when $I\ge I_{\mathrm{req}}(\Delta )$, including equality. For this ideal event readout, assign events to pulse onset and assume the conditioning event is successfully evoked. There is no calculated membrane potential, spike waveform, response latency or stochastic success probability.

The exponential decay represents a combined recovery deficit; it is neither the Hodgkin-Huxley model's $h$ nor a measured potassium electrical conductance. One such curve cannot separately identify sodium recovery and persistent potassium electrical conductance. A constant pulse-width amplitude criterion also does not establish an equivalent rule for other pulse widths or voltage trajectories.

At the hold boundary, $I_{\mathrm{req}}(2\mathrm{ms})=0.2(1+2)=0.6\mathrm{nA}$. At $\Delta =4\mathrm{ms}$,at $\Delta =6\mathrm{ms}$,Thus a $0.35\mathrm{nA}$ test fails at $1\mathrm{ms}$ because of the imposed hold, fails at $4\mathrm{ms}$ because recovery is insufficient for that amplitude, and succeeds at $6\mathrm{ms}$. At $4\mathrm{ms}$, a $0.5\mathrm{nA}$ pulse would instead succeed: failure of the weaker pulse was relative refractoriness.

For $I>I_{\infty }$, solve $I\ge I_{\mathrm{req}}(\Delta )$ using the logarithm. The earliest allowed delay isFor $I=0.35\mathrm{nA}$ this is $2+3\ln (8/3)\approx 4.942488\mathrm{ms}$; for $I=0.5\mathrm{nA}$ it is $2+3\ln (4/3)\approx 2.863046\mathrm{ms}$. If $I\ge 0.6\mathrm{nA}$, the imposed $2\mathrm{ms}$ hold is the only delay restriction. If $I\le I_{\infty }$, no finite delay passes the rule: the required amplitude stays strictly above its asymptotic baseline. A finite experimental “fully recovered” time therefore also needs a tolerance and a readout definition.

The pKa is the negative base-ten logarithm of the dimensionless acid dissociation constant. A lower pKa means a larger acid dissociation constant and a stronger tendency to donate a proton under the same standard state, solvent, and temperature conventions. The Henderson-Hasselbalch equation places equal weak acid and conjugate base thermodynamic activities at pH equal to pKa. See Flowers et al., Chemistry 2e, the Henderson-Hasselbalch equation.

For a sequence of functions $f_n:X\to \mathbb{R}$, pointwise convergence to a function $f:X\to \mathbb{R}$ meansIt is the ordinary limit of a sequence at each fixed input. The threshold $N=N(x,\epsilon )$ may depend on the input. Uniform convergence instead requires one threshold for the entire domain of a function.^[ref]

Consider the polynomial functions $f_n(x)=x^n$, for positive integers $n$, on the interval $[0,1]$. For $0\le x<1$ the geometric sequence $(x^n)$ tends to zero, while $f_n(1)=1$. ThusEach $f_n$ is a continuous function, but its pointwise limit is discontinuous at one. Holding an input still hides what happens near that endpoint. The analogous even-power example on $[-1,1]$ appears in the source.^[ref]

For $0<x<1$ and $0<\epsilon <1$, taking the logarithm of $x^n<\epsilon$ and dividing by the negative real number $\log x$ reverses the inequality:The floor function followed by adding one handles the strict inequality even when the ratio is an integer. For tolerance $\epsilon =0.01$, the first successful indices at $x=0.5$, $0.8$, and $0.99$ are respectively $7$, $21$, and $459$. At $x=0.5$, $2^{-6}=0.015625$ fails and $2^{-7}=0.0078125$ succeeds. At $x=0$ and $x=1$, the error from this pointwise limit is already zero, so $N=1$ works. Nevertheless, $N(x,\epsilon )$ grows without bound as $x\to 1$ from below. A finite list of accurate test inputs cannot establish uniform convergence.

An even sharper warning concerns integration. On $[0,1]$, define the continuous functionsTheir supports are $[0,1/n]$, with height $n$ at $1/(2n)$ and value zero at both ends. Every fixed $x>0$ eventually lies beyond the support, and $g_n(0)=0$ always, so pointwise convergence gives the zero function. Nevertheless, each integral is the area of a triangle: ${\int }_0^1{g}_n(x)dx=\frac{1}{2}(1/n)n=\frac{1}{2}$. Shrinking width compensates increasing height. Hence the limit of a sequence of these integrals differs from the integral of the pointwise limit; interchanging them requires additional conditions, such as uniform convergence preserves integrals on a compact interval.