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For a Planck photon distribution, the hot-electron Kompaneets equation gives $\Delta n=yx{e}^x(e^x-1{)}^{-2}[x\coth (x/2)-4]$, where $x=h\nu /(k_B{T}_{\gamma })$. The equivalent thermodynamic temperature perturbation has the displayed form, while the Rayleigh-Jeans law gives $\Delta T/T=-2y$. The distortion changes sign at $x\simeq 3.830$. This first-order expression assumes a small Compton y parameter and nonrelativistic thermal electrons.

Bulk motion of a scattering electron cloud relative to the cosmic microwave background gives a first-order Doppler temperature shift. With positive $v_{\parallel }$ directed away from the observer, the displayed sign applies, and $\tau$ is the Thomson scattering optical depth. In the nonrelativistic, single-scattering limit it has the spectrum of a small blackbody temperature change, unlike the thermal Sunyaev-Zeldovich effect.

For a uniform spherical cluster of depth $\ell \propto \theta d_A$, thermal bremsstrahlung gives received bolometric flux $S_X\propto n_e^2{T}_e^{1/2}{\ell }^3/d_L^2$, while the Rayleigh-Jeans law limit of the thermal Sunyaev-Zeldovich effect gives $|\Delta T|\propto n_e{T}_e\ell$. Eliminating $n_e$ and using $d_L=(1+z{)}^2{d}_A$ yields the displayed angular diameter distance. A model for the density profile fixes its numerical coefficient. Clumping, nonsphericity and temperature structure can bias this distance because the two observables weight the gas differently.

Represent each element of $S_1+\cdots +S_n$ by the lexicographically least tuple with that sum. Every coordinate projection of the resulting tuple set is itself lexicographically least for its partial sum: replacing it by a smaller representative would improve the original tuple without changing its sum. Consequently projected cardinalities are at most the corresponding partial sumset cardinalities. Projection inequalities, including the box theorem, therefore give sumset inequalities.

For $0<s_1<\cdots <s_n$, adding the largest integer creates at least $n$ new sums of subsets. If $P={\sum }_{i<n}{s}_i$ is the previous maximum, the values $P+s_n$ and $P+s_n-s_i$ for $i<n$ are distinct new sums above $P$. Induction gives the displayed bound. Equality holds for $s_i=i$, whose subset sums fill the interval of integers from zero to $n(n+1)/2$.

Given numbers $s_1,\dots ,s_n$, the sum associated with a subset $A$ of indices is ${\sum }_{i\in A}{s}_i$. The empty subset gives zero. The set of all these values is the sumset $\{0,s_1\}+\cdots +\{0,s_n\}$.

For a subset $A$ of density $\alpha$ in a finite cyclic group, put $F=1_A\ast 1_A\ast 1_A$ using normalized convolution on a finite group. Its average is ${\alpha }^3$, so choose $t_0$ with $F(t_0)\ge {\alpha }^3$. Retain the large spectrum where $|\widehat{1_A}|\ge {\alpha }^2/8$. The Parseval identity on a finite group gives at most $64{\alpha }^{-3}$ frequencies. Their phases change by at most $\alpha /4$ on the associated Bohr set, contributing at most ${\alpha }^3/4$ to the change of $F$. The discarded frequencies contribute at most another ${\alpha }^3/4$, since $\sum |\widehat{1_A}{|}^2=\alpha$. Therefore $F(t_0+b)\ge {\alpha }^3/2>0$ on the Bohr set, proving the containment.

For finite sets $A,B\subseteq G$ in an abelian group $G$, the sumset representation function isIt counts ordered pairs in the Cartesian product, so $(a,b)$ and $(b,a)$ are different when $a\ne b$, even though commutativity makes their sums equal. It is zero outside the sumset, andFor $A=\{0,1,2,3\}$, its values at $s=0,1,\dots ,6$ are $1,2,3,4,3,2,1$. For $B=\{0,1,4,10\}$, its values at the increasing elements of $B+B$ are $1,2,1,2,2,1,2,2,2,1$. Both lists total sixteen, but the first concentrates more ordered pairs on fewer sums. This distinction is what additive energy quantifies.

A telescoping sum cancels intermediate sequence terms, leaving boundary terms. For a finite sequence of real numbers $a_0,\dots ,a_n$, the displayed identity follows because every $a_1,\dots ,a_{n-1}$ occurs once with each sign. Applying it to an infinite series also requires checking the limit of the boundary terms; finite cancellation alone does not prove that the series is a convergent series.^[ref]

For $n\ge 1$, expanding the finite sum givesNo monotone sequence assumption is required. Consider the original sequence $4,9,2,7,5$. Its successive finite differences are $5,-7,5,-2$, soThe large intermediate rise and fall disappear from the net change. This is useful whenever one knows successive changes rather than every absolute value: adding the changes reconstructs the final value from the initial one. It also checks a cumulative-update calculation without requiring all increments to be positive.

A finite difference $\Delta a_{k-1}=a_k-a_{k-1}$ measures one discrete change. The identity ${\sum }_{k=1}^n\Delta a_{k-1}=a_n-a_0$ is the discrete counterpart of the fundamental theorem of calculus: accumulation of changes returns the endpoint difference. The related summation by parts transfers a finite difference from one factor of a product to another, with explicit boundary terms.

For $S\subseteq {\mathbb{F}}_p$ with $|S|\ge 2$, put $R=(S-S)/(S-S)$ using nonzero denominators. If $R$ is proper, some $r\in R+1$ lies outside $R$, since invariance under addition of one would give all of the prime field. The map $(x,y)\mapsto x+ry$ on $S^2$ is injective. Clearing the denominator in a representation of $r-1$ puts a dilate of its image inside $3SS-3SS$. If $R$ is the full field, averaging the collision count of $S+rS$ gives some $r$ with $|S+rS|\ge \frac{1}{2}\min (p,|S{|}^2)$. Clearing its denominator puts a dilate inside $2SS-2SS\subseteq 3SS-3SS$. Here $SS$ is a product set, and the prefactors denote repeated sumsets, not scalar multiplication.

Write $D(t)=|A+tA|/|A|$ and suppose some nonzero $b$ has $D(b)\le K$. The mixed-set Plünnecke-Ruzsa inequality gives $|3A-2A|\le K^5|A|$ and $|A+A|\le K^2|A|$. The Ruzsa covering lemma covers $tA$ by at most $D(t)$ translates of $A-A$. Covering two dilates proves the addition bound. First covering $sA$, then both copies of $rA$ in $r(A-A)$, proves the multiplication bound. The Ruzsa triangle inequality also gives $D(-r)\le K^{2}D(r)$. These bounds transfer controlled additive expansion through a bounded-complexity expression in scalars.

An operator has weak type $(p,q)$, with finite $q$, if the displayed bound holds for every input and every $\lambda >0$, with one constant independent of both. It controls the distribution of its output rather than requiring its strong $L^q$ norm to be finite.

In the functional-analytic sense used here, a superlinear functional is real valued, positively homogeneous and superadditive. Equivalently $-q$ is a sublinear functional. This definition concerns functional inequalities, rather than a comparison of asymptotic growth rates. Such a $q$ is concave, satisfies $q(0)=0$, and can serve as a lower support in the sandwich form of the Hahn-Banach theorem.

For $y^{\prime \prime }+\lambda x^{-2}y=0$ on $0<a<x<b$ with zero endpoint values, write $t=\log (x/a)$ and $y=\sqrt{x}w(t)$. The Euler-Cauchy equation becomes $w^{\prime \prime }+(\lambda -1/4)w=0$. Thus $y_n=\sqrt{x}\sin [n\pi \log (x/a)/\log (b/a)]$ and the exact spectrum is displayed above. Leading WKB quantization omits the constant $1/4$, while producing the same leading eigenfunction shape after quantization.

For a finite simple graph with at least two vertices, no triangles, and exactly three common neighbours for every nonadjacent pair, the graph is connected. For adjacent vertices $u,v$, the bipartite graph between $N(u)\setminus \{v\}$ and $N(v)\setminus \{u\}$ has degree two at every vertex, forcing $d(u)=d(v)$. Thus the graph is regular. Counting second neighbours gives $3(n-d-1)=d(d-1)$. The nontrivial adjacency eigenvalues are $(-3\pm \sqrt{4d-3})/2$, and the larger-root multiplicity is $[n-1+d^2/\sqrt{4d-3}]/2$. Integrality forces $t=\sqrt{4d-3}$ to be an odd integer dividing $9$, yielding the stated degrees. The complete bipartite graph $K_{3,3}$ realizes degree three. The one-vertex graph is a vacuous exception with degree zero.

An aligned source behind an axisymmetric gravitational lens produces a circular image at an Einstein radius. Its angular radius satisfies ${\theta }_E^2=4G{M}_{\mathrm{projected}}(<D_d{\theta }_E)D_{ds}/(c^2{D}_d{D}_s)$. Exact alignment gives a ring; displaced sources generally give arcs or distinct images rather than a complete ring.

A critical curve is a locus in the image plane where the Jacobian matrix of the thin gravitational lens equation becomes singular. Its formal point-source lensing magnification diverges, while finite source size regularizes the measured flux. Its mapped locus in the source plane is a gravitational-lensing caustic.

The two resolved quasar images of 0957+561 provide an observational question: are they separate sources, or different light paths from one source? In May 1979, spectrophotometry used the McDonald Observatory 2.7 m reflecting telescope, the UVITS spectrograph, and an intensified dissector scanner at the f/9 focus of the Cassegrain reflector. A 600-groove-per-millimetre diffraction grating produced 4100-7100 ångström scans sampled every 2 ångströms, with about 7 ångström resolution over most of that range. Two spectrograph slits, about 52 arcseconds apart, recorded object and sky background simultaneously for subtraction. Interleaving observations of the two images reduced instrumental and atmospheric gain changes.^[ref]

The readout was received light versus wavelength for each image. Closely matching absorption line redshifts supported a common source and absorber; differences in continua and line strengths motivated differential interstellar extinction. Spectral agreement supplied evidence for gravitational lensing, rather than proving the point-mass gravitational lens model. The authors discussed a physical double and discrepant optical and radio light ratios. Their note added in proof attributed the redder spectrum to foreground galaxy contamination and emphasized an extended mass distribution.^[ref]

A violation of the strong equivalence principle can make the gravitational-to-inertial mass ratio depend on a body's negative gravitational self-energy $E_G$. Bodies with different self-binding fractions would then fall differently in the same external field, even after composition-dependent weak-equivalence effects are removed. For the Earth and Moon this produces a polarization of the lunar orbit toward the Sun, measurable by lunar laser ranging. General relativity predicts ${\eta }_N=0$.

The length of a word over an alphabet is the number of symbol positions in that finite string, counting repetitions. Thus $\ell (abba)=4$ and $\ell (\epsilon )=0$ for the empty word. Concatenation satisfies $\ell (uv)=\ell (u)+\ell (v)$. This differs from the word length of a group element, which minimizes over representing words using a chosen generating set of a group.