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In the usual nondegenerate convention, a strongly regular graph of order $n$ and parameters $(d,a,b)$ is a finite simple $d$-regular graph with $0<d<n-1$ in which adjacent vertices have $a$ common neighbours and distinct nonadjacent vertices have $b$. Its adjacency matrix of a graph $A$ obeysFor the usual nondegenerate parameters with $b>0$ the graph is connected, so eigenvalue $d$ has multiplicity one. On the perpendicular complement of the all-ones vector, $J=0$ and the possible eigenvalues areIf $r$ has multiplicity $m$, then zero trace gives $d+mr+(n-1-m)s=0$. Solving provesThe denominator is positive in this nondegenerate convention. Otherwise all eigenvalues on the all-ones complement would coincide, forcing $A=rI+(d-r)J/n$. All off-diagonal entries would then be the same, making the graph complete or edgeless. If complete graphs are admitted with arbitrary unused $b$, the printed claim needs qualification: $K_2$ with $(d,a,b)=(1,0,2)$ satisfies the common-neighbour conditions vacuously but makes the fraction $0/0$. For complete graphs with nonzero denominator, the spectral calculation still works with an unused root of multiplicity zero.

For the triangle-free graph claim, suppose first there are at least two vertices. Nonadjacent vertices have a common neighbour, so the graph is connected. If $u,v$ are adjacent, put $U=N(u)\setminus \{v\}$ and $V=N(v)\setminus \{u\}$. Every member of $U$ is nonadjacent to $v$, and exactly two of its three common neighbours with $v$ lie in $V$, the third being $u$. By symmetry every member of $V$ has exactly two neighbours in $U$. Counting these edges gives $2|U|=2|V|$, hence $d(u)=d(v)$. Connectivity makes the graph $d$-regular. Counting length-two paths from a vertex to non-neighbours givesHere $a=0$, $b=3$, and the larger-root multiplicity becomes $[n-1+d^2/\sqrt{4d-3}]/2$. Thus $t=\sqrt{4d-3}$ is rational and therefore an integer. It is odd, $d=(t^2+3)/4$, and integrality also gives $t\mid d^2$. Since $16d^2\equiv 9(\mathrm{mod}t)$, one obtains $t\mid 9$, henceThe complete bipartite graph $K_{3,3}$ realizes $d=3$: vertices in the same part have all three vertices of the other part as common neighbours, and vertices in opposite parts are adjacent. There are no triangles.

As literally worded, the one-vertex graph is a vacuous exception: it has no triangles or nonadjacent distinct pairs, but has $d=0$. The displayed positive-degree classification therefore requires at least two vertices. This qualification completes the triangle-free graphs with three common neighbours argument rather than silently excluding a counterexample.

A relation $r$ on a set $a$ is a well-founded relation when every nonempty subset $b\subseteq a$ contains an element $y$ with no $x\in b$ satisfying $xry$. The well-founded recursion theorem states that a rule assigning a value at $y$ from the already assigned values at its predecessors determines a unique function on $a$; the rule must give a unique set-valued output for every possible predecessor assignment. No choice or transitivity of $r$ is needed.

Apply recursion withThe empty supremum is zero. Well-founded induction makes every value an ordinal; the Axiom schema of replacement collects the values into a set. Choose an ordinal $\alpha$ larger than every value. Then $f:a\to \alpha$, and $xry$ implies $f(x)+1\le f(y)$, hence $f(x)<f(y)$.

Conversely, suppose such an ordinal-valued function exists. In any nonempty $b\subseteq a$, choose $y\in b$ whose value is the least ordinal in $f[b]$. There can be no $x\in b$ with $xry$, because it would have smaller value. Thus $r$ is well-founded, proving the equivalence.

Using the axiom of choice, fix a well-order $\prec$ of $a$. Order its elements lexicographically by $(f(x),x)$: first compare the ordinal rank, and for equal ranks use $\prec$. Every nonempty subset has a smallest rank and then a least element in that rank fibre, so this is a well-order. Since $xry$ strictly increases rank, it respects every original relation pair. Hence every well-founded relation extends to a well-order under choice.

The Gaussian log likelihood function, apart from a constant, isDifferentiating in $\beta$ gives $X^{\mathsf{T}}(Y-X\hat{\beta })=0$. Full column rank makes $X^{\mathsf{T}}X$ invertible, soThe distribution follows by applying this linear map to the Gaussian error vector. Differentiating the profiled likelihood function in ${\sigma }^2$ givesFor its distribution, $P=X(X^{\mathsf{T}}X{)}^{-1}{X}^{\mathsf{T}}$ is symmetric idempotent, and the residual is $(I-P)ϵ$. The needed form of Cochran's theorem says that quadratic forms in a standard Gaussian vector associated with orthogonal symmetric projection matrices summing to identity are independent random variables with chi-squared distributions with degrees equal to their ranks. Applying it to $P$ and $I-P$ givesThus the maximum-likelihood variance estimate uses divisor $n$ and has expectation $(n-p){\sigma }^2/n$; the unbiased residual estimate would use divisor $n-p$. The fitted coefficient and residual quadratic form are also independent.

The Lie group $PSL(2,\mathbb{C})$ is a second-countable space. For every point of its discrete subgroup, choose the first basic neighbourhood isolating that point; different elements require different such neighbourhoods. This injects $G$ into a countable set, proving countability.

A discrete subgroup is closed and meets compact sets in finitely many points. Indeed, distinct subgroup elements accumulating anywhere would give quotients converging to identity, contradicting an identity neighbourhood containing no other subgroup element. To apply this, use the positive Hermitian determinant-one model of hyperbolic space. At the base point $I$, the image under a determinant-one representative $g$ is $g{g}^{\ast }$ andThis distance formula follows by unitary diagonalization: eigenvalues are $e^t,e^{-t}$ and the axial hyperbolic displacement is $|t|$. Thus bounded displacement bounds all matrix entries. The determinant-one condition is closed, so the resulting set is compact; the quotient by $\{I,-I\}$ is compact too. Conjugation gives the same conclusion at any base point. These are the bounded displacement sets of hyperbolic isometries are compact facts.

Hence, for each $R$, only finitely many $g\in G$ satisfy $d(p,gp)\le R$. Any enumeration of an infinite group without repetitions therefore has $d(p,g_{n}p)\to \infty$. A finite discrete group is a harmless exception to the wording: it has no infinite enumeration of distinct elements, and proper discontinuity is immediate.

For a compact set $C$, put $R={\max }_{x\in C}d(p,x)$. If $gC\cap C\ne \varnothing$, choose $x,gx\in C$. The triangle inequality gives $d(p,gp)\le d(p,gx)+d(x,p)\le 2R$. Only finitely many such $g$ exist, proving properly discontinuous group action. Conversely, if the subgroup is not discrete, there are distinct $g_n\to e$. A closed ball about $p$ then meets its images under infinitely many $g_n$, violating this compact-set definition. Thus discreteness is equivalent to proper discontinuity. Here proper discontinuity allows finite point stabilizers; a stronger convention requiring freeness would exclude discrete elliptic subgroups and is not the equivalence intended.

In the nonrelativistic dilute regime, $E(p)=m{c}^2+p^2/(2m)+\cdots$ and the Fermi-Dirac distribution reduces to the Maxwell-Boltzmann distribution. HenceDifferentiating the Gaussian integral with respect to its coefficient gives ${\int }_0^{\infty }{p}^2{e}^{-a{p}^2}dp=\sqrt{\pi }/(4a^{3/2})$. ThereforeThe degeneracy factor here is the one in the supplied integral, namely one. Substituting $kT=m{c}^2/\alpha$, $\mu =0$ and dividing by the supplied photon density givesAt $\alpha =20$ this is about $4.8\times 10^{-8}$. The relic is already nonrelativistic at decoupling and becomes still colder as its momenta redshift, making it a possible cold dark matter component if it is stable and interacts sufficiently weakly with light.

For orientation, write ${\eta }_b=n_b/n_{\gamma }$. Ignoring subsequent entropy transfer, its density relative to baryons would be ${\rho }_X/{\rho }_b=(1/20)(n/n_{\gamma })/{\eta }_b$. With the cosmological baryon-to-photon ratio of order $6\times 10^{-10}$, this is of order four, demonstrating the right cosmological scale. The measured dark-to-baryonic density ratio is likewise of order five; an observational reference is arxiv.org/abs/1807.06209 . This is a candidate abundance argument, not an exact relic-density calculation: photon heating after decoupling changes $n/n_{\gamma }$, and stability and interaction properties were not specified. The supplied mass and freeze-out parameter alone therefore do not establish a precise fit.

For real $b\ne 0$, the integrand depends only on $b^2$, so first take $b>0$. Integrate $e^{iz}/(z^2-b^2)$ over the upper-half-plane semicircle, indenting above both real poles with small clockwise semicircles. The large arc tends to zero by Jordan lemma. Each indentation contributes $-i\pi$ times its residue, and there are no enclosed poles. Thus the Cauchy principal value on the whole real axis isThe imaginary part is odd and the real part even. Taking the real part and halving givesThe right-hand side is even in $b$, completing the negative-$b$ case as well.

Representative normal forms are $\dot{x}=\mu +x^2$ for a saddle-node bifurcation, $\dot{x}=\mu x-x^2$ for a transcritical bifurcation, and $\dot{x}=\mu x-x^3$ for a supercritical pitchfork bifurcation. Replacing the last cubic term by $+x^3$ gives the subcritical pitchfork bifurcation.

Treat $\mu$ as a variable with $\dot{\mu }=0$. The extended center manifold is tangent to $y=0$, so write $y=h(x,\mu )=A{x}^2+Bx\mu +C{\mu }^2+O((|x|+|\mu |{)}^3)$. Its invariance equation is $h_x\dot{x}=\dot{y}$. At quadratic order this readsMatching coefficients gives $A=2$, $B=-4$, $C=4$. Substitution into the first equation therefore givesHere $O(3)$ denotes total degree at least three in $(x,\mu )$. With $T=t$, $X=x-2\mu$ and $\tilde{\mu }=\mu$, this becomes $dX/dT=\tilde{\mu }+X^2+O(3)$. Thus $\alpha =1$, $\beta =2$, $\gamma (\mu )=\mu$, and the bifurcation is a saddle-node. For negative parameter, the negative-$X$ branch is stable and the positive-$X$ branch unstable along the center manifold; the transverse eigenvalue is near $-1$.

In a generalized linear model, the linear predictor is ${\eta }_i=x_i^T\beta$, a linear combination of the covariates with unknown coefficients. The link function relates this predictor to the mean: $g({\mu }_i)={\eta }_i$. The canonical link function chooses the natural parameter of the exponential family, namely $g(\mu )=\theta (\mu )$. Differentiating the normalization of the density shows that $\mu =K^{\prime }(\theta )$, so the canonical link function is $(K^{\prime }{)}^{-1}$ on the admissible parameter interval.

For a Bernoulli distribution, the probability mass function can be written asThus $\theta =\log (\mu /(1-\mu ))$ and $K(\theta )=\log (1+e^{\theta })$. The canonical link is the logit:

A binary block code is a subset $C\subseteq \{0,1{\}}^n$; its length is the common word length $n$, its size is $m=|C|$, and its minimum distance is the least Hamming distance between distinct words. There are $V(n,r)={\sum }_{j=0}^r\left(\genfrac{}{}{0ex}{}{n}{j}\right)$ words in a radius-$r$ Hamming ball, since one chooses the positions to change.

Put $t=\lfloor (d-1)/2\rfloor$. Two radius-$t$ Hamming balls about different codewords cannot meet: otherwise the triangle inequality would make their centres at distance at most $2t<d$. Comparing their combined size with $2^n$ gives the Hamming boundFor the other direction, start with one word and repeatedly add any word at distance at least $d$ from every word already selected. Finiteness ensures that this process stops. At that point the radius-$(d-1)$ Hamming balls cover the entire cube; an uncovered word could otherwise be added. These balls may overlap, but their total size is at least $2^n$. This gives the Gilbert–Varshamov bound

Extend the Möbius transformations to isometries of three-dimensional hyperbolic space. For any interior point $o$, the limit set of a Kleinian group is the set of boundary accumulation points of $Go$. Its definition is independent of $o$: two interior orbits stay at a bounded hyperbolic distance, so have the same boundary limits. The Kleinian limit set is closed and $G$-invariant. Iterating a loxodromic Möbius transformation in the positive and negative directions shows that its two fixed points belong to the Kleinian limit set.

The two fixed-point sets in this problem cannot have exactly one common point. Otherwise conjugate that point to infinity and the other fixed point of the first transformation to zero. The transformations then have the forms $f(z)=\alpha z$, with $|\alpha |>1$, and $g(z)=\beta z+b$, with $b\ne 0$. But $f^{-n}g{f}^n(z)=\beta z+b{\alpha }^{-n}$ is a sequence of distinct group elements converging in the Möbius group. Taking quotients of consecutive elements gives nonidentity elements converging to the identity, contrary to discrete subgroup. Consequently the Kleinian limit set contains at least three points.

Let $\xi$ be any point of the Kleinian limit set, and choose distinct $g_n$ with $g_{n}o\to \xi$. The rank-one convergence of divergent Möbius transformations supplies a subsequence converging to $\xi$ uniformly away from one exceptional boundary point $\eta$. To see the lemma directly, normalize determinant-one representing matrices by their norms. The norms tend to infinity, and a subsequential limit has rank one: its kernel is $\eta$ and its image is the attracting point. In the positive-Hermitian-matrix model of hyperbolic space, $g_{n}o$ has this same image limit, so the attracting point is indeed $\xi$.

Choose distinct ${\zeta }_1,{\zeta }_2$ in the Kleinian limit set other than $\eta$. Then $g_n{\zeta }_1$ and $g_n{\zeta }_2$ both tend to $\xi$, belong to the Kleinian limit set, and remain distinct because $g_n$ is injective. If $\xi$ were isolated, both would eventually equal $\xi$, an impossibility. The limit set has no isolated points.