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The Chebyshev differential equation $-(1-x^2)y^{\prime \prime }+x{y}^{\prime }=\lambda y$ on $-1<x<1$, with zero endpoint values, has the Sturm-Liouville formThus $p=\sqrt{1-x^2}$ and the weight defining the weighted inner product is $w=(1-x^2{)}^{-1/2}$. The substitution $x=\cos \theta$ maps the problem to $-Y^{\prime \prime }=\lambda Y$ on $0<\theta <\pi$ with homogeneous Dirichlet boundary conditions. It also transforms $w(x)dx$ into $-d\theta$. Consequently the normalized eigenfunctions areTheir weighted inner product is the ordinary sine inner product on $[0,\pi ]$. Equivalently, $y_n=\sqrt{2/\pi }\sqrt{1-x^2}{U}_{n-1}(x)$, where $U_{n-1}$ is a Chebyshev polynomial of the second kind. These are not the cosine solutions $T_n$, which have different endpoint values. The original endpoints are singular because $p$ vanishes and $w$ diverges there, but the transformed Dirichlet problem specifies the self-adjoint realization. The eigenfunctions vanish at both endpoints and their quasi-derivatives $p{y}_n^{\prime }$ have finite limits, so the boundary form vanishes.

George Elwood Smith reports fabricating MOS capacitors, thermally generating the first packet, and clocking it to an output that pushed it into the substrate for electric current detection. A later integrated eight-bit device demonstrated imaging. Dark or reversed-clock control traces are not specified; feasibility alone does not calibrate charge-transfer efficiency.^[ref]

A discriminating repeat would compare identical cycles with and without initial carrier injection, then suppress or reverse the transfer sequence while retaining output switching. An electric current transient synchronized with gate switching can include capacitor feedthrough; only the difference tied to initial electric charge and the expected transfer delay tests packet transport. These are proposed controls, not controls documented in the lecture. For an optical version, a blocked exposure estimates dark current and read noise; a spatially isolated illuminated optical detector pixel exposes any deferred charge in CCD readout trail. Photon shot noise persists even with ideal transfer.

Deferred charge in CCD readout reaches later output samples than its originating packet. George Elwood Smith describes surface states at the silicon–silicon dioxide interface capturing electrons and releasing them later, smearing the image. A buried channel keeps the stored packet away from that interface.^[ref]

An original comparison separates two interpretations of the same main-packet deficit. If a single transfer defers fraction $ϵ$ by exactly one sample and loses nothing, an isolated input packet $N$ produces samples ${\eta }_{\mathrm{tr}}N$ followed by $ϵN$: summing both recovers $N$. A truly lossy transfer yields only ${\eta }_{\mathrm{tr}}N$, with no later recovery. For $N=10000$ and $ϵ=0.01$, both give $9900$ in the intended sample, but only the deferred case gives a $100$-electron trail. Such a trail redistributes position information even when total electric charge is conserved. Repeated delays require a time-response model; $N_{\mathrm{in}}{\eta }_{\mathrm{tr}}^n$ alone cannot predict trail shape. Electron-hole recombination, escape to another contact, or release beyond the recorded window can instead remove signal from the measured sum.

The charge-transfer efficiency is the fraction of a packet delivered into its intended next site or time slot in one transfer, under stated conditions. It is distinct from quantum efficiency, which concerns conversion of incident photons into collected charge carriers. Define the complementary inefficiency as $ϵ=1-{\eta }_{\mathrm{tr}}$.

In an original model with constant $0\le {\eta }_{\mathrm{tr}}\le 1$, let $N_j$ denote the mean number of original packet electrons still delivered on time after $j$ transfers. Assuming no reinjection into that packet,Repeated substitution proves the result; for different efficiencies the fraction is ${\prod }_{j=1}^n{\eta }_j$. For $ϵ\ll 1$, a Taylor expansion of $\ln (1-ϵ)$ gives ${\eta }_{\mathrm{tr}}^n\simeq e^{-nϵ}$, while the simpler $1-nϵ$ requires $nϵ\ll 1$.

Choose $N_{\mathrm{in}}=10000$ and $n=2000$. For ${\eta }_{\mathrm{tr}}=0.9999$, the expected on-time output is $8187.23$ electrons; for ${\eta }_{\mathrm{tr}}=0.99999$, it is $9801.99$. These are model means, so fractional electron counts do not describe an individual packet. Even a small per-transfer imperfection can imprint distance-dependent image errors.

If every original electron independently stays on time with probability $p={\eta }_{\mathrm{tr}}^n$, a fixed initial count gives a binomial distribution with variance $N_{\mathrm{in}}p(1-p)$. An initial Poisson distribution with mean $\mu$ instead gives an on-time Poisson distribution with mean and variance $\mu p$, by Poisson thinning. A deterministic attenuation formula alone does not specify noise. Real trapping can correlate packets and depends on timing, illumination and temperature; the independent constant-efficiency model then fails.

For an electron packet, a sufficiently positive gate electric potential attracts electrons beneath the insulating layer of a MOS capacitor. The corresponding potential energy is $U=-e\varphi$, where $e>0$ is the magnitude of elementary charge: a larger electric potential gives a deeper potential well for electrons. A drawing of potential energy must therefore reverse the vertical sense of a drawing of electric potential.

A simplified three-phase transfer raises the adjacent gate's electric potential before lowering the old holding gate. This overlaps the accessible regions, then leaves the packet under the new gate. Repeat with the next gate to move it forward. Isolated wells with an intervening barrier would store the packet without transferring it.^[ref]

Let $X_1,X_2,\dots$ be independent and identically distributed random variables taking $-1$ and $1$ with equal probabilities, the Rademacher distribution. The two opposite complex exponentials give ${\phi }_X(t)=(e^{it}+e^{-it})/2=\cos t$. Their expected value is zero and their variance is one. For $S_n={\sum }_{j=1}^n{X}_j$, the characteristic functions of its sample mean and its variance normalization are respectivelyFor each fixed $t$, the bases are positive for all sufficiently large $n$. The Taylor expansion $\log \cos u=-u^2/2+O(u^4)$ at zero then givesExponentiating proves the two displayed pointwise limits. The first limit is the characteristic function of the constant zero; the second is that of the standard normal distribution. The Lévy continuity theorem gives the corresponding convergence in distribution statements. Averaging has variance $1/n$, whereas dividing the sum by $\sqrt{n}$ preserves variance one. This sign calculation illustrates why the law of large numbers and the central limit theorem use different scales; it is not a proof for arbitrary probability distributions.

Let $B_1,B_2,B_3$ be independent random variables, each with the Bernoulli distribution of parameter $1/3$, and let $S=B_1+B_2+B_3$. Multiplying their characteristic functions and expanding by the binomial theorem givesThe coefficients are exactly the probability mass function of the binomial distribution with parameters $3,1/3$: the sum takes $0,1,2,3$ with probabilities $8/27,12/27,6/27,1/27$. Direct outcome counting verifies them: an outcome with $k$ ones has probability $(1/3{)}^k(2/3{)}^{3-k}$, and there are $\left(\genfrac{}{}{0ex}{}{3}{k}\right)$ such outcomes. The coefficients sum to one, recovering ${\phi }_S(0)=1$.

At $t=\pi$, ${\phi }_S(\pi )=(1/3{)}^3=1/27$. This is also $\mathbb{P}(S\text{ even})-\mathbb{P}(S\text{ odd})=(8+6-12-1)/27$: the complex exponential now records parity. The moments agree by two routes: linearity of expectation gives $\mathbb{E}[S]=1$, while the variance of a sum of independent random variables gives $\mathrm{Var}(S)=3(2/9)=2/3$. Summing $k^2\mathbb{P}(S=k)$ gives the second moment $5/3$, also recovered as $-{\phi }_S^{\prime \prime }(0)$.

Let $A$ be a fair bit, let $N$ be an independent Bernoulli random variable with $\mathbb{P}(N=1)=p$, and set $B=A\oplus N$, where $\oplus$ is exclusive or. Thus $N=1$ flips the first bit. This constructs a binary symmetric channel with a fair input, for $0\le p\le 1$. The joint probability mass function isThe column sums show that $B$ is also a fair bit, whatever $p$ is. Both marginal distributions therefore have information entropy one bit. Given $A$, only the flip decision remains unknown, giving conditional entropy $H(B\mid A)=h_2(p)$, whereis the binary entropy, with $h_2(0)=h_2(1)=0$. The chain rule for information entropy and the definition of mutual information now give

Take $p=1/10$. The four joint probabilities, in row order, are $(0.45,0.05,0.05,0.45)$. Directly from the definition of joint entropy,Conditioning on either value of $A$ gives probabilities $0.9$ and $0.1$ for the two values of $B$, henceAdding $H(A)+H(B)=2$ would overcount by exactly $I(A;B)$. These are average uncertainties under the stipulated probability distribution; the four possible pairs are still represented by four distinct outcomes.

There is another way to see the decomposition: $(A,B)\mapsto (A,A\oplus B)=(A,N)$ is a bijection. It merely relabels the four outcomes, so it preserves joint entropy. The independence of $A$ and $N$ then gives $H(A,B)=H(A,N)=H(A)+H(N)=1+h_2(p)$. Describing the initial bit and the flip decision separates shared uncertainty from new uncertainty.

At $p=0$, the bits are identical: $B=A$, so $H(B\mid A)=0$ and $I(A;B)=1$. At $p=1$, they are complementary: $B=1-A$, with exactly the same conditional entropy and mutual information. In either case, observing one determines the other. At $p=1/2$, all four joint probabilities are $1/4$; the bits are independent, $H(A,B)=2$, and $I(A;B)=0$. Perfect disagreement is therefore just as informative as perfect agreement.

For $p>1/2$, guessing $A=1-B$ succeeds with probability $p$, better than guessing $A=B$. Changing $B$ to $1-B$ replaces $p$ by $1-p$ and preserves mutual information, since the relabeling is invertible. What matters is how well the observation distinguishes the input, rather than how often its symbol matches the input. The same symmetry makes the binary entropy curve peak at independence, while the mutual information curve reaches its minimum there.

If $U$ is a unitary matrix with no eigenvalue equal to one, the displayed inverse Cayley transform of a Hermitian matrix is Hermitian. Since $U^{†}=U^{-1}$, taking the Hermitian conjugate gives $B^{†}=-i(I-U^{-1}{)}^{-1}(I+U^{-1})=i(I-U{)}^{-1}(I+U)=B$, using commutativity of rational functions of $U$. An eigenvalue $e^{i\theta }\ne 1$ is mapped to the real number $-\cot (\theta /2)$. Conversely $U=(B-iI)(B+iI{)}^{-1}$, so this correspondence identifies Hermitian matrices with unitary matrices whose spectrum avoids one.

Dispersive atom-cavity coupling is the off-resonant regime where real excitation exchange is suppressed but virtual mixing shifts the energies. For a two-level system describing an chemical atom coupled to one mode, the scale $\chi \simeq g^2/\Delta$ appears when $|\Delta |$ is large compared with the relevant occupation-enhanced coupling $g\sqrt{n+1}$.

An effective term proportional to $\hslash \chi a^{†}a{\sigma }_z$ then makes the atomic phase depend on photon number and the cavity frequency depend on atomic state. Measuring that phase can infer field occupation without resonantly absorbing a photon in the ideal approximation. Finite detuning, losses and probe backaction still matter; dispersive does not mean completely uncoupled or universally disturbance-free. A model must state detuning sign and the retained perturbative order.

If $s>0$ and $\Theta$ has the continuous uniform distribution on $(-\pi /2,\pi /2)$, then $Y=s\tan \Theta$ has the Cauchy distribution with location zero and scale $s$. The tangent function is strictly increasing on this interval, so its cumulative distribution function isDifferentiating gives the probability density function $s/[\pi (s^2+y^2)]$. The probability of either endpoint is zero, so including endpoints in the uniform interval does not alter this probability distribution. The slow quadratic decay makes the expected value undefined: its positive and negative parts both have infinite integrals. Its second moment is also infinite, so it has no finite standard deviation.

For speed $c(y)=1/y$ in $y>0$, Fermat principle gives the elapsed-time integrand $y\sqrt{1+(y^{\prime }{)}^2}$. Its symmetric stationary light rays through $(\pm d,H)$ are the displayed catenaries. Set $t=d/a$. The endpoint equation is $\cosh t/t=H/d$. This function decreases and then increases, with unique minimum where $t\tanh t=1$. Thus above the minimum there are two stationary rays, at the minimum one, and below it none. The travel time on either ray is $T=ad+(a^2/2)\sinh (2d/a)$, and its minimum height is $a$.

For $d=1$ and $H=\cosh 1$, one solution is $a=1$, with $T=1+\sinh 2/2$. A second has $a\simeq 0.7021026952$, with $T\simeq 2.8224491447$. Hence these endpoints do not uniquely specify the ray. Stationarity is distinct from global minimization: paths descending almost to $y=0$ have limiting travel time $H^2$, below either stationary value in this example.

Given injections $f:X\to Y$ and $g:Y\to X$, start with the set difference $A_0=X\setminus g(Y)$: these are exactly the inputs for which $g^{-1}$ is unavailable. Follow both functions repeatedly, using the natural-number recursion theorem to defineThus $A$ contains all the $X$ members reachable from any starting point in $A_0$. The notation $f(A_k)$ and $g(f(A_k))$ denotes images of a function. If $A_0$ is the empty set, then every $A_k$ is empty and $g^{-1}$ alone gives the desired bijection.

Define a function on all of $X$ byHere $g^{-1}$ is the inverse function from $g(Y)$ to $Y$, obtained by restricting the codomain of $g$ to its image. It is well defined because $g$ is injective. Every $x\notin A$ belongs to $g(Y)$, since $A_0\subseteq A$, so the second rule is always available where it is used.

The decisive set identity isThe last equality holds because every $A_k$ with $k\ge 1$ lies in $g(Y)$, whereas $A_0$ is disjoint from $g(Y)$. Consequently, for any $y\in Y$,Indeed, $g(y)$ cannot belong to $A_0$; if it lies in $A$, the identity gives $g(y)=g(f(a))$ for some $a\in A$, and injectivity gives $y=f(a)$. The converse follows directly by applying $g$.

Now $f$ restricts to a bijection from $A$ onto $f(A)$. The displayed equivalence says that $g$ restricts to a bijection from $Y\setminus f(A)$ onto $X\setminus A$, so its inverse function bijects the remaining inputs with the remaining outputs. The two input sets partition $X$ and the two output sets partition $Y$. Their combined function $h$ is therefore both injective and surjective.

No choice function is needed: $A$ is fixed by the given injections, and each required $g^{-1}(x)$ is unique. We neither choose a representative from each chain nor enumerate $X$. This is a proof in ordinary classical ZF, even when no well-ordering of the sets exists there.^[ref]

Take $X=\{0,1,2,\dots \}$, the nonnegative integers, and $Y=\{1,2,3,\dots \}$, the positive integers. Choose the injectionsHere $g$ is an inclusion map, while $f$ reaches only the positive even integers. In particular, $f$ is not a bijection: it misses every positive odd integer. Simply using the inverse function of $g$ on its image misses input zero.

The remedy is to follow the alternating injections starting at the missing input zero:The subscripts distinguish an input in $X$ from an output in $Y$, even when their numerical values agree. The resulting subset of $X$ is $A=\{0,2,6,14,30,\dots \}$. Its sequence of members satisfies $a_0=0$ and $a_{k+1}=2a_k+2$, giving $a_k=2^{k+1}-2$ by mathematical induction. Define the function

The image $h(A)=\{2,6,14,30,\dots \}=A\setminus \{0\}$ covers all the positive members of $A$ exactly once. The other inputs cover $Y\setminus A$ exactly once by staying fixed. These output sets are disjoint and their union is $Y$, proving that $h$ is a bijection. In particular, $h(0)=2$, $h(2)=6$, $h(6)=14$, while $h(1)=1$ and $h(4)=4$.

For completeness, its inverse function isThe familiar bijection $n\mapsto n+1$ also works, but this example shows something more specific: the two chosen injections can repair the missing zero by moving only a sparse sequence of inputs. The repair never has to shift the entire set.

BABAR used asymmetric electron-Positron collisions near the Upsilon meson $4S$ resonance to prepare neutral B meson pairs. It reconstructed one decay into charmonium plus a kaon, and used flavour tagging on the other B meson with charged leptons or kaons. Silicon tracking reconstructed decay vertices for the decay-time difference. Self-tagging decays calibrated mistags and time resolution; charged decays provided a control. The tagged time distributions established CP violation, limited by backgrounds, mistags and vertex alignment.^[ref]

The Lorentz boost converts vertex separation to $\Delta t\simeq \Delta z/({\beta }_{\mathrm{boost}}{\gamma }_{\mathrm{boost}}c)$ at leading order. Unequal beam energies thus make a time difference measurable as a distance.^[ref]

For a gas of fixed composition in hydrostatic equilibrium, with $z$ upward and downward gravitational acceleration of magnitude $g$, an adiabatic process gives the displaced element density change $\delta {\rho }_{\mathrm{element}}=\rho {\gamma }^{-1}(d\ln p/dz)\xi$. Comparing with the ambient change $(d\rho /dz)\xi$ yieldsThe ideal gas law $p=\rho \mathcal{R}T$ and $dp/dz=-\rho g$ give the title expression, where $c_p=\gamma \mathcal{R}/(\gamma -1)$ is the constant-pressure specific heat capacity. A temperature gradient less negative than the adiabatic lapse rate gives positive $N^2$; a steeper decline gives convective instability.

A colloidal grain in a liquid is large enough to see with a microscope, yet small enough for thermal motion to compete with its tendency to settle. In the Perrin sedimentation experiment, the equilibrium number density falls with height: a visible miniature atmosphere. Counting grains at different heights measures an energy scale; tracking their Brownian motion measures a transport rate. Connecting both to the same Avogadro constant makes the experiment a quantitative test of kinetic theory. The Wiener measure describes an idealized path law; these measurements instead test a physical model and its material parameters.

Jean Baptiste Perrin's 1909 memoir traces his sedimentation experiments to May 1908. It reports relative counts $100$, $47$, $22.6$, $12$ at heights $5$, $35$, $65$, $95\mu \mathrm{m}$, and compares radius estimates from Stokes settling velocity, counted grain mass, and rows of touching grains.^[ref]

The 1926 lecture describes resin precipitated from alcohol into water, washed and fractionated by centrifugation to obtain nearly monodisperse gamboge or mastic grains. Diameter and density were measured separately. A sealed slide-and-cover-glass chamber prevented evaporation; a vertical microscope resolved roughly $2\mu \mathrm{m}$ layers through a roughly $0.1\mathrm{mm}$ depth. Refocusing and photographing compared layer populations after equilibrium; changing temperature reversibly changed the distribution.^[ref]

In Les Atomes (1913), Jean Baptiste Perrin restricts the eyepiece view to fewer than five or six grains and alternates observations between heights. Focus displacement requires a refractive index correction to recover actual layer separation. Horizontal positions were marked with a camera lucida, usually every $30\mathrm{s}$, and distances calibrated with a micrometer; the polygon joins sampled positions rather than recording the intervening trajectory.^[ref][ref]

A photon-recoil lag correlation asks how often a photon record occurs at a specified time displacement from an electron record. Let $r_e,r_{\gamma }$ be observed separate-channel rates, $S$ the rate of detected genuine pairs before a timing cut, and $f(\ell )$ their normalized timing probability density function. For an original dilute, stationary model with observation time $T$, neglecting end effects and detector dead time, the expected pair density isThis counts all pairs of records, rather than at most one detector trigger per electron. It uses independent unrelated detections for the flat term and an added same-event contribution. A normal distribution for timing spread is a chosen approximation, not a claim about the historical instrument. Intrinsic emission delay, counter response and readout uncertainty can all affect the observed probability density function.

Define the accepted lag interval by $|\ell |<w/2$: $w$ is its full width. Integrating the flat baseline gives the existing accidental coincidence expectation $T{r}_e{r}_{\gamma }w$. The additional area is $TS{\int }_{-w/2}^{w/2}f(\ell )d\ell$. Moving a window of the same full width away from the peak tests the baseline under the model's stationarity assumption. It does not remove correlated instrumental backgrounds or changing rates automatically.

For the lag model, choose $T=1000\mathrm{s}$, $r_e=2{\mathrm{s}}^{-1}$, $r_{\gamma }=0.5{\mathrm{s}}^{-1}$, $S=0.1{\mathrm{s}}^{-1}$ and timing standard deviation $\sigma =1\mathrm{ms}$. A full width $w=2\mathrm{ms}$ gives $2$ expected accidental coincidences and $100\mathrm{erf}(1/\sqrt{2})\simeq 68.27$ accepted genuine pairs, using the error function. The total is approximately $70.27$, whereas the displaced interval from $4$ to $6\mathrm{ms}$ has approximately $2.003$ expected pairs. A finite sample fluctuates around these expectations; these values are not measured counts or a significance calculation.

The right panel isolates the information lost by retaining singles totals alone. In a separate idealized model, stipulate $M=10^6$ independent $1\mathrm{ms}$ bins, with Bernoulli distribution indicators $X,Y$ for electron and photon detection in a bin. Set $p_e=\mathrm{Pr}(X=1)=0.002$ and $p_{\gamma }=\mathrm{Pr}(Y=1)=0.0005$. Independence predicts $\mathrm{Pr}(X=1,Y=1)=p_e{p}_{\gamma }=10^{-6}$, hence one expected double detection. Alternatively chooseThese nonnegative probabilities sum to one and preserve both marginal probabilities, yet predict $101$ double detections in expectation. Their covariance is $0.000101-p_e{p}_{\gamma }=0.000100$. This is an exact joint probability distribution illustrating pairing; it does not model the timing spread in the left panel or reconstruct either historical hypothesis quantitatively. Its lesson is why two singles totals cannot decide whether the observed events share a cause.