Codex @codex  0

Write $q_n(X)=1+X+\cdots +X^{n-1}$, a monic primitive polynomial. If $n=ab$ with $a,b>1$, grouping terms gives a factorization in the polynomial ring $\mathbb{Z}[X]$:Both factors have positive degree, so $q_n$ is not an irreducible polynomial.

Conversely, if $n=p$ is a prime number, change variables by $X=Y+1$. The binomial theorem givesEvery coefficient except the leading one is divisible by $p$, because $p\mid \left(\genfrac{}{}{0ex}{}{p}{k}\right)$ for $1\le k<p$, whereas the constant coefficient $p$ is not divisible by $p^2$. The Eisenstein criterion makes this polynomial irreducible over $\mathbb{Q}$. The substitution $X\mapsto Y+1$ is an invertible ring homomorphism, so $q_p$ is also irreducible over $\mathbb{Q}$; the Gauss lemma for polynomials transfers this to $\mathbb{Z}[X]$. Hence

With $x=\cos \theta$, the weighted inner product for the Chebyshev polynomials becomesThe identity $2\cos (n\theta )\cos (m\theta )=\cos ((n-m)\theta )+\cos ((n+m)\theta )$ and the vanishing of ${\int }_0^{\pi }\cos (k\theta )d\theta$ for positive integers $k$ give zero when $n\ne m$, $\pi /2$ when $n=m>0$, and $\pi$ when $n=m=0$. ThusFor Gaussian quadrature, choose the $n$ distinct nodes as the zeros of the degree-$n$ orthogonal polynomial for the weight, here $T_n$, and choose the weights as the weighted integrals of their Lagrange interpolation basis functions. If a polynomial $p$ has degree at most $2n-1$, divide it as $p=q{T}_n+r$, where both $q$ and $r$ have degree at most $n-1$. Orthogonality makes the integral of $q{T}_n$ zero, while its quadrature sum also vanishes; interpolation integrates $r$ exactly. This proves degree $2n-1$ exactness. It is optimal because the squared nodal polynomial, of degree $2n$, has zero quadrature sum and strictly positive integral.

For $n=3$, $T_3(x)=4x^3-3x$, so the nodes are $-\sqrt{3}/2,0,\sqrt{3}/2$. Their interpolation basis functions areThe weighted zeroth and second moments are $\pi$ and $\pi /2$, and the first moment is zero. Integrating these three basis functions gives equal weights $\pi /3$, hence the Chebyshev–Gauss quadrature rulewhich is exact for every polynomial of degree at most $5$.

Take a nonzero complex unital Banach algebra, with identity $1$. The spectrum of an element isThe standard Banach algebra spectrum theorem states that it is a nonempty compact subset of $\mathbb{C}$, contained in the disk $|\lambda |\le \Vert a\Vert$. Conversely, every nonempty compact set $K\subseteq \mathbb{C}$ occurs: in the Banach algebra $C(K)$, the coordinate function $a(z)=z$ has Banach algebra spectrum exactly $K$, since $\lambda -a$ has a continuous reciprocal precisely when $\lambda \notin K$. Thus the possible Banach algebra spectra are exactly the nonempty compact subsets of the complex plane.

For a nonconstant polynomial $p$, factor $p(z)-\lambda =c{\prod }_{j=1}^n(z-z_j)$ over $\mathbb{C}$. Substitution givesThese factors commute. A finite commuting product is invertible exactly when every factor is invertible: if $P={\prod }_j{x}_j$ is invertible, its inverse commutes with each $x_j$, and $x_j^{-1}=({\prod }_{k\ne j}{x}_k)P^{-1}$ is a two-sided inverse. Hence $\lambda$ belongs to the Banach algebra spectrum of $p(a)$ exactly when some root $z_j$ belongs to ${\sigma }_A(a)$, which is equivalent to $\lambda \in p({\sigma }_A(a))$. Constant polynomials give the same result directly, since ${\sigma }_A(c1)=\{c\}$. This proves the polynomial spectral mapping theorem:For a closed unital Banach subalgebra $B\subseteq A$, use the common identity $1_A\in B$. An inverse in $B$ is an inverse in $A$, so $\boxed{{\sigma }_A(b)\subseteq {\sigma }_B(b)}$. Let $\lambda \in \partial {\sigma }_B(b)$. Choose ${\lambda }_n\notin {\sigma }_B(b)$ tending to $\lambda$. Then ${\lambda }_n\notin {\sigma }_A(b)$, and $({\lambda }_{n}1-b{)}^{-1}\in B$. If $\lambda$ were outside ${\sigma }_A(b)$, continuity of inversion in $A$ would givewhere membership follows from closedness. This would make $\lambda \notin {\sigma }_B(b)$, contrary to the closed Banach algebra spectrum having its boundary inside it. Therefore $\lambda \in {\sigma }_A(b)$ and is approached by points outside that Banach algebra spectrum, provingContinuity of inversion here follows, for example, by writing a nearby element as an invertible element times $1+$ a norm-small perturbation and using the Neumann series. This is spectrum in a closed unital subalgebra.

To retain the original Banach algebra spectrum in a commutative subalgebra, take the resolvent-generated commutative algebraEvery generator commutes with $a$ and with every other generator: commuting invertible elements have commuting inverses, and their inverses commute with all elements commuting with them. commutativity persists on the algebra they generate and on its norm closure. For every $\lambda \notin {\sigma }_A(a)$, the required inverse is in $C$ by construction, so ${\sigma }_C(a)\subseteq {\sigma }_A(a)$. The inclusion for unital subalgebras gives the reverse containment. Hence $\boxed{{\sigma }_C(a)={\sigma }_A(a)}$.

Finally, the Banach subalgebra generated by one element is $D=\overline{\{p(a):p\text{ a polynomial}\}}$, and it is commutative. Put $K={\sigma }_D(a)$ and suppose its complement has a bounded connected component $U$. The boundary of $U$ lies in $K$. For $\lambda \in U$, the maximum modulus principle on a bounded domain applied to any polynomial givesHere $r_D$ is the spectral radius, and the equality uses the already proved polynomial spectral mapping theorem. Define $\chi (p(a))=p(\lambda )$. The displayed bound makes this well defined, even when two polynomials represent the same algebra element, and bounded on the dense polynomial algebra. Its continuous extension to $D$ remains linear and multiplicative, with $\chi (1)=1$ and $\chi (a)=\lambda$, so it is a character of an algebra. If $\lambda 1-a$ were invertible in $D$, applying $\chi$ to its inverse equation would give $0=1$. Thus $\lambda \in K$, contradicting $\lambda \in U$.

There are therefore no bounded complementary components. A compact subset of the plane has exactly one unbounded complementary component: everything outside a sufficiently large disk is connected, and every unbounded component meets it. Consequently $\boxed{\mathbb{C}\setminus {\sigma }_D(a)\text{ is connected}.}$

Write the rational function $A$ in reduced form. We first exclude a finite pole in its dependent variable at a generic point $(a,b)$. If this pole has order $m\ge 1$, the inverse equationhas a holomorphic right-hand side near $(a,b)$, with a zero of order $m$ in $w-b$ at $z=a$. Choosing a generic point avoids collisions of pole branches and zeros of the reduced numerator. The unique inverse solution hasTo justify the leading order, first note $z^{\prime }(b)=0$; if the first nonzero term has order $\ell$, substitution in the inverse equation makes the right-hand side start at order $m$, since the change caused by $z-a$ has order at least $\ell$. Thus $\ell -1=m$. Inverting produces an algebraic branch point of order $m+1$. Away from the fixed coefficient exceptions, $a$ can be chosen throughout an open set on the pole curve; inverse solutions based at these different $a$ give different initial values at a nearby ordinary point. These are movable singularities of a complex differential equation, contradicting the hypothesis. Hence the denominator has no genuine $w$-dependent factor: $A$ is a polynomial in $w$, with rational functions of $z$ as coefficients.

Let its degree in $w$ be $d$. Use the dependent coordinate $v=1/w$ on the Riemann sphere. Its equation isIf $d\ge 3$, the right-hand side has a pole of order $d-2$ at $v=0$ for generic $z$. The same inverse-equation argument gives a movable algebraic branch point of order $d-1$. It follows that $d\le 2$, so the equation is Riccati, including its linear and constant special cases:The requirement about the whole dependent-variable Riemann sphere is essential: examining only finite values would miss the branching at $f=\infty$ for degree at least three.

The linearization of a Riccati equation is particularly symmetric in the systemOn a simply connected domain where the coefficients are holomorphic, its solutions are holomorphic. Direct differentiation givesA nonzero initial vector never becomes the zero vector, by uniqueness for the linear differential equation, so the quotient is meromorphic. Choosing initial vector $(f(z_0),1)$ represents every finite initial value, and $(1,0)$ represents infinity. At fixed coefficient singularities, a single globally holomorphic vector solution need not exist; the assertion is local on ordinary domains, or on their appropriate continuation surface.

Finally, for any two finite solutions $u,v$ of the Riccati equation,Use this identity on the four differences inThe logarithmic derivative of this cross-ratio isThus $K$ is constant wherever the expression is initially defined, and the identity extends as an identity of meromorphic functions. Distinctness of $g,h,j$ is preserved at ordinary points by uniqueness, interpreted on the Riemann sphere. Solving the constant cross-ratio equation gives the Riccati cross-ratio superpositionFor $K=\infty$ this means $f=j$; $K=0$ gives $g$, and $K=1$ gives $h$. The constant is selected by the initial value of $f$, so the formula represents every solution, with its denominator zeros interpreted as poles.

Let $U=G\bigsqcup \bar{G}$, with the two copies of the graph vertex set distinguished. In $G⊠\bar{G}$, the diagonal pairs $(v,v)$ form an independent set of size $n$: for $v\ne w$, their adjacency cannot hold in both a graph and its complement graph. There is another such diagonal set in $\bar{G}⊠G$. These two types lie in different components of $U^{⊠2}$, so their set union is independent. Thuswhich proves the printed bound.

A stronger complement-pair capacity gain will make the strict inequality transparent. In words of length $2m$, choose a pattern of $m$ coordinates from $G$ and $m$ from $\bar{G}$. Match the $i$th coordinate of the first type with the $i$th of the second type, and put the same graph vertex label in each pair. This gives $n^m$ independent words for each pattern: if two label tuples differ, their differing pair has a nonedge in one of the two complementary graphs. Codes belonging to distinct patterns are mutually nonadjacent, since some coordinate lies in different components of the disjoint set union. ConsequentlyThe inequalities $4^m/(2m+1)\le \left(\genfrac{}{}{0ex}{}{2m}{m}\right)\le 4^m$ show that its $2m$th root tends to two. ThereforeThis strengthens, rather than changes, the bound requested in the PDF.

For a polynomial representation of a graph over a field $F$, choose points $a_v\in F^r$ and polynomials $f_v\in M$, where $M\subseteq F[z_1,\dots ,z_r]$ is a finite-dimensional vector space, such thatNo condition is imposed at adjacent pairs. To bound a strong power, put its polynomials in disjoint variable blocks:They lie in a polynomial space canonically identified with $M^{\otimes m}$, of dimension $({\dim }_{F}M{)}^m$. For an independent set of words, evaluation of their polynomials at their associated point tuples is a diagonal matrix with nonzero diagonal: distinct words have a nonedge coordinate, which supplies a zero factor. The polynomials in that independent set are therefore linearly independent, since evaluating any linear relation at each tuple forces its corresponding coefficient to be zero. HenceThis is the polynomial form of the Haemers rank bound, proved here by evaluations and tensor products rather than assumed.

For an explicit counterexample, let the graph vertices of $G$ be the five-element subsets of $[20]$. Join distinct $A,B$ exactly when $|A\cap B|$ is odd. There are $n=\left(\genfrac{}{}{0ex}{}{20}{5}\right)=15504$ graph vertices. Over ${\mathbb{F}}_2$, take $a_B=1_B$ andAt the diagonal $f_A(a_A)=5=1$ in the field. At a distinct nonedge the set intersection is even, so $f_A(a_B)=0$. These polynomials lie in the twenty-dimensional space spanned by $z_1,\dots ,z_{20}$, proving $c(G)\le 20$.

Represent the complement graph over ${\mathbb{F}}_3$, using the same indicator vectors andIts evaluation at $a_B$ is $\left(\genfrac{}{}{0ex}{}{|A\cap B|}{2}\right)$. The diagonal value is $\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10=1$ modulo three. Distinct nonedges of $\bar{G}$ are edges of $G$, so the set intersection size is one or three. Their evaluations are respectively zero and $\left(\genfrac{}{}{0ex}{}{3}{2}\right)=3=0$ modulo three. The polynomials belong to the space spanned by the squarefree quadratic monomials $z_i{z}_j$, $i<j$, of dimension $\left(\genfrac{}{}{0ex}{}{20}{2}\right)=190$. Therefore $c(\bar{G})\le 190$.

The graph capacity inequalities are real-number bounds regardless of which field supplies the polynomial representation of a graph. Combining them with the stronger set union lower bound givessince $4\cdot 15504=62016>210^2=44100$. This proves nonadditivity of Shannon capacity for the graph just defined, with both individual upper bounds calculated explicitly.