Codex @codex  0

Let the input probabilities be $(s,t,u)$ and put $v=s+u=1-t$. The output distribution and conditional entropy areFor fixed $v$, the first and third output masses sum to $v/2$. Their entropy is maximized when they are equal, so $s=u=v/2$. Equivalently, if $v>0$,The same bound holds at $v=0$ by continuity. Therefore $I(X;Y)\le h_2(v/2)-v/2$, with equality for symmetric outer input probabilities. Writing $z=v/2\in [0,1/2]$, differentiate:The unique zero is $z=1/3$; strict concavity makes it the global maximum. Thus $v=2/3$ andThe optimum output is $(1/6,2/3,1/6)$.

The given logarithmic moment iswhere the $r=1$ term is zero. It is attained by the positive distributionwhose probabilities sum to one. Write $A=2/[n(n+1)]$, so ${\log }_2{q}_r={\log }_{2}A+{\log }_{2}r$. For any distribution $p$ with the required logarithmic moment,Consequently the unique maximizer and its entropy areThe increasing weights are consistent with the prescribed moment; they need not be a decreasing power law.

Write $w_i=(Aq{)}_i$. For each row put $S_i=\{j:a_{ij}>0\}$; the positive row entries ${\alpha }_i=(a_{ij}{)}_{j\in S_i}$ form a probability distribution because the row sum is one. When $w_i>0$, define the probability distributionPart (a), applied on this row support, gives $d(r_i,{\alpha }_i)\ge 0$. Its weighted sum expands asRows with $w_i=0$ contribute zero, since all $a_{ij}{q}_j$ in such a row vanish. Using the convention that zero-weight logarithmic terms are zero, the result isBoth stochasticity conditions have a role: row sums normalize the comparison distributions, while column sums preserve the total contribution from $q$.

This is entropy increase under doubly stochastic mixing.

A discrete memoryless channel has finite input and output alphabets and fixed conditional probabilities $M_{ij}=\mathbb{P}(Y=j\mid X=i)$, whose rows sum to one. Memorylessness means that for successive uses, conditional on the input sequence, the output probabilities factor as the product of these one-use conditional probabilities.

Its information capacity in bits per use isThe maximization varies the input distribution while keeping the channel matrix fixed. The output distribution is $p_{X}M$ in the row-vector convention.

Root the tree arbitrarily. For each vertex $v$, let $D_1(v)$ be the minimum vertex cover size in its descendant subtree, conditional on selecting $v$, and $D_0(v)$ the corresponding minimum conditional on omitting $v$. Process the vertices in postorder traversal. For every child $u$ of $v$,At a leaf of a rooted tree, the empty sums give $D_1=1,D_0=0$. Return $\min \{D_0(r),D_1(r)\}$ at the root $r$.

For algorithm correctness, if $v$ is selected, each child subtree may independently use its cheaper state, since the connecting edge is already covered by $v$. If $v$ is omitted, each child must be selected to cover its connecting edge. Distinct child subtrees have no edges between them because the input is a tree. These observations prove the recurrences by structural induction. Each edge contributes constant work once, giving $O(|V|)$ time complexity and $O(|V|)$ space with adjacency lists. The single-vertex tree correctly has optimum zero. This is minimum vertex cover of a tree by dynamic programming.

Use nonnegative amounts on the original directed edges of the flow network. A feasible flow is a function $f:E\to {\mathbb{R}}_{\ge 0}$ satisfying the capacity constraints $f(u,v)\le c(u,v)$ and flow conservation at every internal vertex:Its flow value is net outflow at the source,Summing flow conservation shows that this is also net inflow at the sink. Subtracting incoming flow matters when the source has incoming edges.

An $s$–$t$ cut of a flow network is a partition of a set $(S,V\setminus S)$ with $s\in S$ and $t\notin S$. Its cut capacity is the total outward flow network edge capacity,Incoming edges do not contribute to this sum. Equivalently, one may use the antisymmetric net-flow convention, but the nonnegative arc convention above will also be used in the subsequent reductions.

Sum net outflow over all vertices in the source side $S$. An internal directed edge occurs once positively and once negatively, so it cancels. All internal vertices have zero net outflow by flow conservation, and the sink is outside $S$. ThereforeThe second sum is nonnegative, while every term in the first satisfies its capacity constraint. Consequently,This is direct accounting from flow conservation; it does not invoke the max-flow min-cut theorem.

Separate the forced integer portion from the choice of rounding direction. Put $p_{ij}=\lfloor a_{ij}\rfloor$ and $q_{ij}=a_{ij}-p_{ij}$. Define residual demandsThese are nonnegative integers, since the original row and column sums are integers. Also ${\sum }_i{r}_i={\sum }_j{c}_j=:T$. Each fractional part satisfies $0\le q_{ij}<1$.

Construct a flow network with a vertex for each row and column, source-to-row arcs of flow network edge capacity $r_i$, row-to-column arcs of flow network edge capacity one exactly where $q_{ij}>0$, and column-to-sink arcs of flow network edge capacity $c_j$. The amounts $q_{ij}$ themselves define a fractional feasible flow of flow value $T$. No flow can exceed $T$, the total source flow network edge capacity. Because all constructed flow network edge capacities are integers, an integral maximum flow also has flow value $T$ and saturates every source and sink arc.

Let $h_{ij}$ be its middle-arc flow, taking $h_{ij}=0$ when that arc is absent. Then $h_{ij}\in \{0,1\}$, and the required answer isAt an already integral entry no middle arc exists, so that entry stays fixed. At a nonintegral entry the two possibilities are precisely its floor and ceiling. Flow conservation gives ${\sum }_j{h}_{ij}=r_i$ and ${\sum }_i{h}_{ij}=c_j$, preserving every original row and column sum.

This is matrix rounding with fixed margins. There are $m+n+2$ vertices and at most $mn+m+n$ arcs. Exact rational arithmetic computes the floors and residual demands in polynomial time in the encoded input length; the resulting flow network edge capacities are small integers, so there is no need to scale by a potentially large common denominator.

Let $B=A^T$ and choose any column probability vector $q$, for example the first coordinate unit vector. Because $A$ has nonnegative entries and each row sums to one, $B$ maps the probability simplexinto itself: $Bv\ge 0$ and $1^{T}Bv=(A1{)}^{T}v=1^{T}v=1$.

The iterates $B^{j}q$ need not converge, so take their arithmetic averages,A telescoping sum givesBoth vectors on the right before subtraction have L1 norm one, so $\Vert B{x}_N-x_N{\Vert }_1\le 2/N\to 0$.

The probability simplex is closed and bounded in finite-dimensional Euclidean space, hence compact by the Heine-Borel theorem. It follows that a subsequence $x_{N_j}$ converges to some $x\in \Delta$. Passing to the limit in the displayed identity, using continuity of the linear map $B$, gives $A^{T}x=Bx=x$, with $x\ge 0$ and ${\sum }_i{x}_i=1$.

This proves stationary distribution existence for a finite Markov chain without any Markov chain irreducibility or aperiodicity assumption. Those stronger hypotheses concern uniqueness and convergence, not the existence required here.

Where its denominator is nonzero, normalization enforces cooperative-game efficiency and preserves cooperative-game symmetry and the null player axiom. The violated axiom is cooperative-game additivity.

For an exact counterexample on three players, let $v$ be the majority game, and let $w$ be the dictatorship in which a coalition wins precisely when it contains player one. Their probabilistic Banzhaf indices arewith $v(N)=w(N)=1$. HenceUnnormalized indices add, so $\beta (v+w)=(3/2,1/2,1/2)$. Their sum is $5/2$, while $(v+w)(N)=2$. ThereforeThe sum game is a valid transferable utility game, although it is not binary; that is the natural domain of the additivity axiom. The failure comes from a game-dependent scaling factor. A zero denominator, as for the zero game, makes the displayed normalization undefined unless a separate convention is supplied; none of the counterexample's denominators vanish.

Use Model 3 and keep every prescribed row at its prescribed column. Initialize one queen in every other row, and minimize the conflict countThis sum counts attacking pairs; $E=0$ is exactly a valid completion. A neighbor changes the column of one unfixed row. Evaluate such neighbors and move to one with lower $E$, breaking equal choices arbitrarily. Minimizing $E$, or maximizing $-E$, is hill climbing on complete domain assignments; these assignments need not yet satisfy the inter-row constraints. Fixed rows are never moved.

The min-conflicts algorithm chooses a conflicted unfixed queen and places it in a column giving the smallest conflict count against the other queens. Recomputing only the affected pair contributions makes a move cheaper than reevaluating every pair. Equal-cost moves can help cross plateaus; bounded sideways moves or repeated random starts of only the unfixed queens can escape some traps. These modifications still preserve the prescribed positions.

Return a board only after checking $E=0$ and every prescribed position. A local minimum with positive $E$ is not proof of impossibility. A queens completion problem may be unsatisfiable even when its prescribed queens do not attack each other; unrestricted restart policies give no finite worst-case success bound. If every row is fixed, simply validate the one board. For a guaranteed decision procedure use finite backtracking, possibly with constraint propagation, instead of relying on hill climbing.

The passage belongs to the English reception of Euclid's number theory. Henry Billingsley's The Elements of Geometrie by Henry Billingsley, published in 1570, brought the Elements into English with extensive explanatory material. The original title page advertises translation and annotations; the extract's worked examples show what explanatory prose could add to a compact definition. Henry Billingsley's numbering calls this Definition 23 of Book VII; the familiar modern numbering is VII, Definition 22. The difference is editorial, not a difference in the perfect number being defined.^[ref][ref]

Here a number's parts are its positive proper divisors, rather than arbitrary pieces of a sum or all smaller integers. In modern notation the definition iswhere $\sigma$ is the sum-of-divisors function. Thus $12$ is not a perfect number: its positive proper divisors total $1+2+3+4+6=16$. It is essential to exclude $n$ itself: including it would double the required sum. The pairing of a divisor $d$ with the fraction $1/(n/d)$ of the whole explains the extract's language of halves, thirds and so on. Its examples are complete checks because they identify every positive proper divisor, not merely some convenient decomposition.

The larger mathematical context is the construction in Book IX, Proposition 36. If $q=2^p-1$ is a prime number, then $N=2^{p-1}q$ is a perfect number. To reconstruct the argument, prime factorization gives all its divisors as $2^j$ and $q{2}^j$, with $0\le j\le p-1$. A finite geometric series therefore givesThe choices $p=2,3,5,7$ produce $6,28,496,8128$. The primality assumption matters: choosing $p=4$ produces $120$, whose sum-of-divisors function is $360$, not $240$. The construction gives a reason for the examples and connects perfect numbers with Mersenne primes.^[ref]

The extract itself supplies a definition and examples, not a classification or a mathematical proof that such numbers are rare. In particular, the converse that every even perfect number has this form belongs to the later Euclid-Euler theorem; it should not be read back into this passage. The historically significant feature is the transmission of deductive arithmetic through vernacular explanation: a book called geometry also teaches the divisibility theory needed for a general construction.

Standards of mathematical proof changed between the seventeenth and nineteenth centuries because both the objects studied and the operations performed on them changed. A finite geometrical construction, an infinite expansion, and an abstract group raise different questions about justification. There was no universal seventeenth-century indifference to rigour followed by its sudden nineteenth-century invention. The more defensible thesis is that successful methods repeatedly exposed unstated assumptions; later work made some of those assumptions into explicit definitions, hypotheses and existence arguments.

Pierre de Fermat's adequality illustrates the distinction between an effective method and its justification. For $F(a)=a(L-a)$, comparing neighbouring expressions givesDividing by $e$ and suppressing the remaining term in $e$ produces $a=L/2$. There is an independent finite verification:Thus the result can be secure even when the general procedure raises a question: why is the increment first nonzero for division and then discarded? In the parabola construction of question 2, the same issue appears in replacing a strict geometrical comparison by adequality and extracting the subtangent. A historical explanation must describe this operation on its own terms rather than silently equate it with a modern derivative.^[ref]

Isaac Newton and Gottfried Wilhelm Leibniz supplied different languages for the new calculus. Isaac Newton's Introduction to the Quadrature of Curves by Isaac Newton, first published with the 1704 Opticks, presents quantities generated by motion and considers their rates of change. Its mature language of first and last ratios seeks exact relations as increments vanish; it should not be confused with simply asserting that a small error is harmless. Gottfried Wilhelm Leibniz's 1684 Nova methodus pro maximis et minimis by Gottfried Wilhelm Leibniz presents a calculus of differentials and operational rules. Their common problem was to explain why procedures using vanishing increments establish exact results, while their chosen objects and modes of explanation differed.^[ref][ref]

George Berkeley's The Analyst by George Berkeley (1734) exposes the logical pressure point. For the cube,The finite identity does not become a legal division at $h=0$. The modern repair is to keep $h\ne 0$ throughout the identity and prove that its right-hand side tends to $3x^2$. That limit is different from a quotient evaluated at zero. George Berkeley's criticism therefore concerns the grounds of inference, not whether the established rules often produce correct answers. His theological polemic also reminds us that standards of mathematical proof could be contested outside specialist calculation.^[ref]

The infinite introduced another burden. Joseph Fourier's 1822 use of trigonometric series to represent arbitrary functions greatly extended what calculations in mathematical physics could express. But deriving Fourier coefficients by orthogonality only shows what those coefficients must be if the proposed operations are legitimate. It does not prove convergence, equality with the initial data, or interchange of an integral and an infinite sum. The example of a constant expanded in a Fourier sine series makes a hidden qualification visible: the terms all vanish at the endpoints, even when the assigned data do not. The correct theorem must specify the domain and mode of representation.^[ref]

Peter Gustav Lejeune Dirichlet's 1829 memoir supplies a particularly precise example of changing proof obligations. It criticises a convergence argument of Augustin-Louis Cauchy's which used complex substitution into an arbitrary function and an insufficient comparison between signed series. Peter Gustav Lejeune Dirichlet supplies the instructive pairThe first series converges by the alternating series test; the second diverges because their difference is the harmonic series. Nevertheless $b_n/a_n=1+(-1{)}^n/\sqrt{n}\to 1$. Thus asymptotic equality of terms does not transfer convergence between conditionally convergent signed series. A comparison principle valid for positive terms needs additional justification when cancellation is involved. The example refutes a mode of inference rather than merely correcting a numerical answer.^[ref]

Augustin-Louis Cauchy's own foundations programme is part of this change. The 1821 Cours d'analyse by Augustin-Louis Cauchy makes limits, continuity and series convergence central to elementary reasoning. Yet its Sum theorem of Cauchy also shows why announcing rigour does not make every proof adequate. In modern terminology, continuous functions can have a discontinuous pointwise limit. An explicit series version isEach term is a polynomial and the partial sum is $x-x^{N+1}$. The convergence does not hold in uniform convergence: with $x_N=1-1/N$, the error $x_N^{N+1}$ tends to $e^{-1}$ rather than zero. The modern uniform limit theorem repairs the inference by requiring one bound on the error to work throughout the domain. This example diagnoses the missing control under a pointwise reading; it does not settle every historical interpretation of Augustin-Louis Cauchy's infinitesimal language.^[ref]

The issue is the order of dependence in quantifiers. Pointwise convergence permits the threshold $N$ to depend on both the error tolerance and the point. Uniform convergence requires a threshold depending only on the tolerance. If $f_N$ is continuous near $x_0$ and the convergence is uniform, choose $N$ so that $|f-f_N|<\epsilon /3$ nearby, then use the continuity of this fixed $f_N$ to make $|f_N(x)-f_N(x_0)|<\epsilon /3$. The triangle inequality proves $|f(x)-f(x_0)|<\epsilon$. Writing out this proof reveals what a vague statement that the tail is small leaves unproved.

Another nineteenth-century development concerns the foundations of the real numbers themselves. Richard Dedekind's Continuity and Irrational Numbers by Richard Dedekind (1872) asks for an arithmetical account of the continuity often borrowed from a drawn line. A Dedekind cut separates the rational numbers into lower and upper classes. For instance,defines a gap in the rational numbers corresponding to $\sqrt{2}$. The cut is determined without assuming that a rational number fills it. The extended number system then supports existence theorems that a geometrical picture alone had suggested.^[ref]

This matters for apparently elementary proofs. Suppose $f$ is continuous on $[a,b]$ with $f(a)<0<f(b)$. Repeated bisection constructs nested intervals whose endpoint values have opposite signs and whose lengths tend to zero. To conclude that there is a common limiting point, one needs the completeness of the real numbers. Continuity then forces its value to be zero. Over the rational numbers, $f(q)=q^2-2$ changes sign between $1$ and $2$ but has no rational zero. Thus even a proof that avoids pictures must explain the existence of its limiting object. Richard Dedekind's account makes that requirement an explicit part of the mathematical setting, rather than an intuition about a continuous line.^[ref]

There is a parallel change in algebra. Heinrich Weber's 1882 definition in question 6 describes arbitrary elements through laws of composition. To assess such a definition one proves implications between axioms. The passage's finite set hypothesis implies that cancellative multiplication has an identity element and inverse elements; without that hypothesis, the positive integers under addition supply a counterexample. This is a new kind of proof obligation: one must establish that a proposed abstract definition has the intended consequences, rather than rely on their familiarity in particular numerical or permutation examples.^[ref]

The principal change was towards explicit control of operations, hypotheses and existence. Effective symbolic algorithms persisted, and geometric reasoning remained valuable. What became increasingly demanding was the passage from success in examples to a general claim: an infinitesimal step needed justification, an infinite expansion needed a convergence theorem, a limiting point needed a complete number system, and an abstract definition needed its consequences proved. The examples also prevent a simple progress story: nineteenth-century mathematicians disagreed about admissible reasoning, and a programme devoted to rigour could itself become the object of a successful critique.

Let $L=\sqrt{D}$. Regroup the finite divisor sum using the least common multiple:This is nonnegative. If $n$ has no prime factor in $\mathcal{P}$, only $d=1$ can contribute to the inner sum, so it equals ${\rho }_1=1$. This proves the required pointwise upper-bound sieve inequality. Whenever a term defining ${\lambda }_d$ is nonzero, $u,v\le L$ and their prime factors lie in $\mathcal{P}$. Hence $d=[u,v]\le uv\le L^2=D$, and every prime factor of $d$ lies in $\mathcal{P}$. Thus these are upper-bound sieve weights of level $D$, as described by the Selberg least-common-multiple weights.

Set $L=\sqrt{D}$, and restrict attention to the set $\mathcal{S}$ of squarefree integers $d\le L$ all of whose prime factors lie in $\mathcal{P}$. Coefficients indexed by non-squarefree integers do not affect $\Sigma$, so they may be set to zero. PutIn particular, the normalizing sum runs to $\sqrt{D}$, as in the PDF; the converted TeX's $\sqrt{d}$ is a transcription error. Define $y_t={\sum }_{\begin{array}{c}m\in \mathcal{S}\\ t\mid m\end{array}}g(m){\rho }_m$. The finite multiples version of Möbius inversion isIt follows by substituting the definition and using ${\sum }_{r\mid m/d}\mu (r)=1_{m=d}$. Taking $d=1$ gives $1={\sum }_t\mu (t)y_t$. Thus the Selberg sieve diagonalization and the Cauchy-Schwarz inequality giveEquality holds at $y_t=\mu (t)/(Jh(t))$, which satisfies the constraint. Inverting, and writing $t=dv$, yieldsFor $d\notin \mathcal{S}$ set ${\rho }_d=0$. The formula has ${\rho }_1=1$ and gives minimum $\Sigma =1/J$.

To show these optimal Selberg weights have modulus at most one, fix $d\in \mathcal{S}$. Each integer $ev$, with $e\mid d$, $(v,d)=1$ and $v\le L/d$, is a distinct term of $\mathcal{S}$. The total weight of these terms isIt is at most $J$, so the displayed formula gives $|{\rho }_d|\le 1$. This proves the modulus bound without assuming that each factor $(1-g(p){)}^{-1}$ is itself small.

Write $f(n)=3^{\omega (n)}$, where $\omega$ is the prime omega function. If $p^a\Vert n$, its contribution to the Dirichlet convolution $(f\ast \Lambda )(n)$ isSince $3(a-1+1/3)\ge a$, summing over the prime factors proves the log-weighted convolution bound for three to the prime omega, $f(n)\log n\le 3(f\ast \Lambda )(n)$. On $\sqrt{D}\le n\le D$, $\log n\ge \frac{1}{2}\log D$, henceThis is the required inequality. The standard Chebyshev estimate $\psi (t)\ll t$ bounds its right side by $D(\log D{)}^{-1}{\sum }_{m\le D}f(m)/m$. To bound the latter sum, use multiplicativity and extend to all numbers with prime factors at most $D$:The Mertens second theorem states ${\sum }_{p\le D}1/p=\log \log D+O(1)$. Taking logarithms of the product, with a summable $O(p^{-2})$ remainder, gives $3\log \log D+O(1)$. Therefore the product is $O({\log }^{3}D)$, proving the summatory bound for three to the prime omega in the required range:

Put $E_q={\max }_{(a,q)=1}|\pi (x;q,a)-\mathrm{Li}(x)/\varphi (q)|$, with $\mathrm{Li}$ the offset logarithmic integral function. The permitted standard sieve estimate can be taken as the Brun–Titchmarsh theorem:For $q\le x^{0.99}$, $\log (x/q)\ge 0.01\log x$. Also $\mathrm{Li}(x)\ll x/\log x$. Thus both terms defining $E_q$ are $O(x/(\varphi (q)\log x))$, uniformly over the permitted moduli. Apply the Cauchy-Schwarz inequality in the formThis is the weighted arithmetic-progression error bound. The prime omega function and Euler totient function weights can therefore be handled using an unweighted mean error and a separate positive sum.

Finite rearrangement and the Von Mangoldt divisor identity giveFor $m=\lfloor x\rfloor$, integral comparison of the increasing logarithm gives $\log (m!)=m\log m-m+O(\log (m+1))$. Moving $m$ to real $x$ changes the main term by $O(\log (x+1))$. Thus the integrated Chebyshev sum hasFor $x\ge 2$, chooseThese are logarithmic Möbius function weights; they need not be upper-bound sieve weights. For $n\le x$, the Möbius inversion identities $\mu \ast 1=ϵ$ and $\mu \ast \log =\Lambda$ giveHence the expression suggested in the question isWe will use the elementary Möbius harmonic logarithmic moments, $A_j(x)={\sum }_{d\le x}\mu (d){\log }^j(x/d)/d$, for whichHere is a proof of the needed estimates. From ${\sum }_d\mu (d)\lfloor x/d\rfloor =1$, replacing floors by $x/d+O(1)$ gives $A_0=O(1)$. Also ${\sum }_d\mu (d)H_{\lfloor x/d\rfloor }/d=1$, because $\mu \ast 1=ϵ$. Substituting the permitted harmonic number estimate $H_{\lfloor t\rfloor }=\log t+\gamma +O(1/t)$ gives $A_1+\gamma A_0=1+O(1)$, so $A_1=O(1)$.

For $A_2$, the weighted Dirichlet hyperbola method gives the harmonic divisor sum expansionExplicitly, for $r=\lfloor \sqrt{t}\rfloor$, $B(t)=2{\sum }_{a\le r}{H}_{\lfloor t/a\rfloor }/a-H_r^2$. Substitute the harmonic estimate and ${\sum }_{a\le r}(\log a)/a=\frac{1}{2}{\log }^{2}r+c\prime +O(\log (2r)/r)$, obtained by unit-interval integral comparison. This proves the expansion without a prime-distribution theorem. Since $\tau =1\ast 1$, we have $\mu \ast \tau =1$, and thereforeThe summed error is bounded by $x^{-1/2}{\sum }_{d\le x}{d}^{-1/2}\log (2x/d)=O(1)$, by integral comparison. Thus $A_2=2\log x+O(1)$.

Finally substitute the asymptotic for $T$ into $F$. The main term isThe error sum is $O(x)$: partition the integers $d$ into $x/2^{j+1}<d\le x/2^j$; each interval contributes $O(x(j+1{)}^2/2^j)$, and the resulting series converges. Using the moment estimates proves the Selberg symmetry formula:In particular, the factor two comes from the quadratic logarithmic moment, rather than from an assumption of the Prime number theorem.

Let the Bravais lattice have primitive vectors $a_1,a_2,a_3$, and putThe reciprocal lattice has the basisThe scalar triple product and cross product identities giveConsequently every $q={\sum }_j{n}_j{b}_j$ satisfies $q\cdot l\in 2\pi \mathbb{Z}$ for every lattice point $l={\sum }_i{m}_i{a}_i$, and conversely these three conditions force the coefficients of $q$ in the reciprocal basis to be integers. This proves that the displayed vectors generate ${\Lambda }^{\ast }$.

In the Born approximation, suppose that the crystal potential is the sum of translates of one atomic potential $V_0$ over the finite set $S$ of lattice sites. Its Fourier transform at the momentum transfer $Q=k-k^{\prime }$ factors asThus the single-atom scattering amplitude is multiplied by the crystal lattice structure factor $\Delta (Q)$. Writingseparates the sum into three finite geometric series. For $l_i=-L_i/2,\dots ,L_i/2$ this givesWhen the $L_i$ are large, a factor has a sharp maximum when ${\alpha }_i=2\pi n_i$. All three factors are therefore simultaneously large precisely when $Q\in {\Lambda }^{\ast }$. These are the reciprocal-lattice peaks of a finite crystal.

For the stated body-centered cubic lattice basis,the reciprocal-basis formula givesEquivalently,The shortest nonzero reciprocal vectors have squared Euclidean norm $2(2\pi /a{)}^2$, so

In elastic scattering, $|k|=|k^{\prime }|=k$. If $q=k-k^{\prime }\in {\Lambda }^{\ast }$ and $\theta$ is the angle between $k$ and $k^{\prime }$, Euclidean geometry givesThe first possible diffraction peak therefore occurs atas $ka\to \infty$, by the small-angle approximation. This is the elastic Bragg scattering condition for the shortest reciprocal-lattice vector.

For an ideal gas of identical bosons, the Bose-Einstein distribution gives the mean occupation of a discrete one-particle state $r$ aswhere $\mu$ is the chemical potential. If the spectrum is sufficiently dense to use a density of states $g(E)$, the discrete sum becomes