Colour produced when wavelength-dependent interference selectively transmits or reflects part of white light. Between crossed polarizers, a uniform birefringent plate gives a factor ${\sin }^2(\pi \Gamma /\lambda )$, where $\Gamma$ is its optical path difference. The observed colour also depends on illumination and the spectral response of the observer.

For a lossless uniaxial dielectric, $n_{\mathrm{ex}}^{-2}={\cos }^2\vartheta /n_o^2+{\sin }^2\vartheta /n_e^2$, where $\vartheta$ is the angle between the wave vector and optic axis. Along the axis $n_{\mathrm{ex}}=n_o$, while perpendicular propagation gives $n_{\mathrm{ex}}=n_e$.

A known retarder inserted in a polarization microscope to compare phase delay with an unknown specimen. Parallel slow directions add signed retardations; a slow direction paired with a fast one subtracts them. General relative angles require composing polarization transformations rather than scalar addition.

For independent random variables $X,Y$ with the same binomial distribution $\mathrm{Bin}(n,p)$, partition the tie event into the mutually exclusive events $X=Y=j$. independence makes each probability the square of its binomial probability, proving the formula. At $p=1/2$, Vandermonde's identity gives $\mathrm{Pr}(X=Y)=4^{-n}\left(\genfrac{}{}{0ex}{}{2n}{n}\right)$. Symmetry under exchange of the two variables then gives $\mathrm{Pr}(X>Y)=\mathrm{Pr}(Y>X)=(1-\mathrm{Pr}(X=Y))/2$.

An upper binomial tail probability is $\mathbb{P}(K\ge k)={\sum }_{j=k}^n\left(\genfrac{}{}{0ex}{}{n}{j}\right)p^j(1-p{)}^{n-j}$ for a binomial count $K$. A strict upper tail starts at $k+1$ instead; this distinction matters for a discrete law. For four fair Bernoulli trials, $\mathbb{P}(K\ge 3)=5/16$, while $\mathbb{P}(K>3)=1/16$. Lower and upper tails satisfy $\mathbb{P}(K\ge k)=1-\mathbb{P}(K\le k-1)$. These sums calibrate one-sided hypothesis tests and the order statistic distribution formula.

The binomial probability mass at $k$ isfor $K\sim \mathrm{Bin}(n,p)$. Each selected set of $k$ successes has probability $p^k(1-p{)}^{n-k}$ by independence, and the binomial coefficient counts such subsets. For $n=4,p=1/2$, the masses are $(1,4,6,4,1)/16$. They sum to one by the binomial theorem. Endpoint laws $p=0$ and $p=1$ are concentrated at zero and $n$, respectively.

A binomial count $K$ records the number of successes in $n$ independent Bernoulli trials with common success probability $p$. It is the sum of the success indicator variables, so $K\sim \mathrm{Bin}(n,p)$, with expected value $np$ and variance $np(1-p)$. Ten independent coin flips give a binomial count of heads; ten dependent outcomes generally do not, even if each marginal success probability is $p$.

If $N$ independent trials give a specified digit with probability $p$, its observed fraction is $f=K/N$ with binomial count $K$. Since $\mathbb{E}[f]=p$ and $\mathrm{Var}(K)=Np(1-p)$, the mean squared discrepancy is the displayed inverse-sample-size law. Dividing by $p^2$ converts it to a relative squared discrepancy without changing the log-log slope. Nested prefixes are correlated; a single deterministic expansion's plotted discrepancies do not prove that its digits are independent or normally distributed in the number-theoretic sense.

The least significant bit is the binary digit in the position of weight $2^0$. For a nonnegative integer $n$, its value is $n\mathrm{mod}2$, so it records parity. In $13=(1101{)}_2$ it is one; in $12=(1100{)}_2$ it is zero.

The phrase must be distinguished from the least significant set bit, the lowest position containing a one. For $12$, that set bit has weight four, even though the least significant bit is zero. The distinction matters in bit masks, divisibility by powers of two, and binary arithmetic.

The most significant bit is the binary digit with greatest positional weight in a specified binary representation. For a positive integer written without leading zeros, it is the leading one: in $13=(1101{)}_2$, it has weight $2^3$.

For a fixed-width word, the term instead denotes the leftmost position even when that bit is zero. In two's-complement signed representation, that position is the sign bit. Stating the width and whether the value is signed prevents confusing the highest position of a word with the highest nonzero bit of an integer.

Two separated short-range scatterers are coupled because the outgoing wave from either centre supplies part of the regular incoming field at the other. Applying the Bethe-Peierls boundary condition at both centres gives the displayed two-by-two linear system, where $s_j$ is the incident plane-wave phase. matrix inversion sums repeated scattering between the centres. Near a small determinant, effective-range corrections may matter even if the nominal microscopic range is short.

For independent geometric random variables with success probability $p$, observing $d$ terminal events and $F$ preceding failed Bernoulli trials gives likelihood function proportional to $p^d(1-p{)}^F$. Multiplication by a beta distribution prior $\mathrm{Beta}(a,b)$ gives the posterior distribution $\mathrm{Beta}(a+d,b+F)$. With complete waiting times $x_1,\dots ,x_n$, take $d=n$ and $F={\sum }_i(x_i-1)$.

A waiting time that is right-censored after $c$ failed trials contributes $(1-p{)}^c$: increase $F$ by $c$ without increasing $d$. This requires the censoring mechanism to be ignorable, as with an externally fixed observation horizon. Counting a censored waiting time as an observed event changes the likelihood function and biases the update.

A Bernoulli string is a finite sequence of independent binary digits, each taking value one with a specified probability. Unless stated otherwise, all positions use the same probability.

Number the leads left 1,top 2,right 3,bottom 4. Let narrow leads 1,2 have $M$ spin-resolved modes and wide leads 3,4 have $N>M$. Straight-through injection sends $M$ modes between 1 and 3 and between 2 and 4; each excess $N-M$ mode connects 3 and 4. With current 1 to 4, $V_4=0$ and zero current at 2,3, $V_2=0$, $V_3=M{V}_1/N$ and $I=(e^2/h)M(N-M)V_1/N$. Thus $R_{14,23}=-h/[e^2(N-M)]$. At $N=M$ this ideal network disconnects the current source from its drain.

For left-deflected injection in the left 1,top 2,right 3,bottom 4 geometry, use 1 to 2 with $M$ modes, 2 to 3 with $M$, 3 to 4 with $N$, and 4 to 1 with $M$ plus 4 to 3 with $N-M$. The zero-current probes give $V_2=V_1$ and $V_3=M{V}_1/N$, hence $R_{14,23}=(h/e^2)(1/M-1/N)$. Reversing the transmission matrix gives equal probe potentials and zero bend resistance. Fixed-lead resistance need not be even in field; reciprocity also exchanges current and voltage pairs.

For a balanced nonnegative transmission matrix and $I_i=g_0(N_i{V}_i-{\sum }_j{T}_{ij}{V}_j)$, electrical power obeyswith diagonal reflection included consistently or removed from both sides. A negative four-terminal voltage-to-current ratio is compatible with this positive total power because the measured voltage pair is not the current source's pair.

If $\alpha ,\beta >1$ are irrational numbers with $1/\alpha +1/\beta =1$, the Beatty sequences $\lfloor n\alpha \rfloor$ and $\lfloor n\beta \rfloor$ partition the positive integers.

Let $\alpha >1$ be irrational and $\beta =\alpha /(\alpha -1)$, so $1/\alpha +1/\beta =1$. The Beatty sequences $\lfloor n\alpha \rfloor$ and $\lfloor n\beta \rfloor$, $n\ge 1$, partition the positive integers.

Indeed, the numbers of terms at most $m$ are $\lfloor (m+1)/\alpha \rfloor$ and $\lfloor (m+1)/\beta \rfloor$. Their arguments sum to the integer $m+1$, and neither is integral, so the counts sum to $m$. Subtracting the count at $m-1$ shows that every integer occurs exactly once in the two lists together. This proof interprets reciprocity as an exact complementarity of two densities.