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The poles inside $C_n$ are zero and $\pm k\pi$ for $1\le k\le n$. At a nonzero pole,and the residue at $-k\pi$ is its negative. At zero, the Laurent series contains only even powers because the function is even, so its residue is zero. All residues cancel and therefore

A map $f:{\mathbb{R}}^n\to {\mathbb{R}}^m$ is differentiable at $x$ if there is a linear map $L$ such thatthen $Df{|}_x=L$. The inverse function theorem says that if $f$ is continuously differentiable near $x$ and $Df{|}_x$ is invertible, then $f$ restricts to a $C^1$ diffeomorphism between neighborhoods of $x$ and $f(x)$.

Since $F(A)=A^{T}A$ is polynomial,Thus $\ker DF{|}_I=T$, the space of skew-symmetric matrices.

DefineThen $D\Phi {|}_I(H)=2H$, so the inverse function theorem supplies open neighborhoods $I\in U$, $0\in V$ on which $\Phi :U\to V$ is a $C^1$ diffeomorphism. Its symmetric part is $A^{T}A-I$ and its skew part is $A-A^T$. Consequentlyand therefore $\Phi (U\cap \mathcal{O})=V\cap T$.

For any $R\in \mathcal{O}$, left multiplication $A\mapsto R^{T}A$ is a homeomorphism preserving $\mathcal{O}$ and carrying $R$ to $I$. Transporting the preceding chart gives a neighborhood of $R$ in $\mathcal{O}$ homeomorphic to an open subset of $T$.

Letwhere $\phi$ is the golden ratio. If the union-closed family consists of one nonempty set, any element of that set has frequency one, so assume its cardinality exceeds one. Choose independent uniform members $A,B$ and let $X,Y\in \{0,1{\}}^n$ be their characteristic vectors of sets. Then $H(X)=H(Y)=\log |\mathcal{A}|>0$.

Suppose for a contradiction that every element has frequency $p_i<c$. Put $q_i=1-p_i>1/\phi$, and let $Z=X\mathrm{OR}Y$, the characteristic vector of a set of $A\cup B$. The chain rule for information entropy and conditioning reduces entropy givebecause $Z_{<i}$ is a function of $(X_{<i},Y_{<i})$.

Fix the two prefixes and setThe two conditioned bits are independent random variables, and $Z_i=0$ exactly when both are zero. The supplied binary entropy product inequality therefore givesAveraging over the independent prefixes yieldswhenever either conditional entropy is positive. Summing and using $H(X)>0$ gives $H(Z)>H(X)$.

But set union keeps $A\cup B$ inside the union-closed family, so $Z$ is supported on $\mathcal{A}$. The maximum entropy distribution on a finite set gives $H(Z)\le \log |\mathcal{A}|=H(X)$, a contradiction. Some element must therefore occur in at least $c|\mathcal{A}|$ members, proving the entropy bound for a union-closed family.

For $C,D\subseteq \mathbb{B}$, one writes $C{\le }_{m}D$ if there is a total computable function $f:\mathbb{B}\to \mathbb{B}$ such thatfor every $x$. This is a computable many-one reduction.

Define conformal time by $d\tau =dt/a(t)$. The radial coordinate and the angular-radius function areSubstitution yields the stated conformal metric. A radial null ray satisfiesso $d\chi /d\tau =\pm c$ and $\chi ={\chi }_0\pm c(\tau -{\tau }_0)$.

A rational map $X⇢Y$ between irreducible projective varieties is an equivalence class of morphisms from nonempty open subsets of $X$ to $Y$, with representatives identified when they agree on a nonempty open subset. It is regular at $p$ if some representative is defined on a neighborhood of $p$.

Forthe coordinates vanish together exactly at $(0:1:0)$ and $(1:0:0)$. Near the first point, the paths $z=0$ and $x=0$ have limiting images $(1:0:0)$ and $(0:0:1)$; near the second, the paths $z=0$ and $y=0$ have limiting images $(1:0:0)$ and $(0:1:0)$. No continuous extension exists there. Elsewhere the coordinates do not vanish together, so the map is regular.

DefineWhere all coordinates are nonzero,and similarly $\varphi \psi$ is the identity. Thus $\varphi$ is birational and is an isomorphism on ${\mathbb{P}}^2\setminus Z(xyz)$.

For irreducibility of $P=x^2{z}^4-x^3{y}^3+z^6$, dehomogenize at $z=1$. Over $k(x)$ this is, up to a unit,The rational function on the right is not a cube, since its valuation at either root of $x^2+1$ is one. The cubic is therefore irreducible over $k(x)$; Gauss's lemma gives irreducibility in $k[x,y]$. Since $z$ does not divide $P$, homogenization preserves irreducibility.

Writing $(u:v:w)=(xy:xz:z^2)$ transforms the equation intoIts partial derivatives are $-3u^2$, $2vw$, and $v^2+3w^2$ up to common signs. Simultaneous vanishing forces $u=v=w=0$, impossible projectively, so $C$ is nonsingular. The inverse $\psi$ restricts on dense open subsets, proving that $V$ and $C$ are birational.

Lévy's continuity theorem states that if characteristic functions ${\varphi }_n$ converge pointwise to a function $\varphi$ continuous at zero, then $\varphi$ is a characteristic function and the corresponding laws converge weakly to its law. Conversely, convergence in distribution implies pointwise convergence of characteristic functions.

Uniformly for the arguments involved,Hencebecause $n^{-3}\sum j^2\to 1/3$ and the error is $O(n^{-1})$. The limiting characteristic function is that of $N(0,1/9)$. Lévy's theorem gives

A $C^1$ function $V$ near an equilibrium at the origin is a Lyapunov function if $V(0)=0$, $V(x)>0$ for $x\ne 0$, andThe First Lyapunov theorem says these conditions imply stability. The second says that if the last inequality is strict away from the origin, then the origin is asymptotically stable.

To prove the first, fix a sufficiently small ball of radius $\epsilon$. Positive definiteness and compactness giveContinuity at zero gives $\delta >0$ such that $|x_0|<\delta$ implies $V(x_0)<m$. Since $V$ cannot increase along a trajectory, that trajectory cannot meet the sphere $|x|=\epsilon$, where $V\ge m$. It therefore remains in the $\epsilon$-ball for all forward time, which is stability.

SetThe integral equation becomesEvaluating it at the three delta functions givesThus

For $x>a$, all three Green-function terms are proportional to $e^{ikx}$. Hence

Now define $\gamma =ik{\hslash }^2/(\lambda m)$, so that $q=-1/\gamma$. If $M$ denotes the matrix multiplying $\mathbf{\Psi }$, direct evaluation givesThe displayed algebraic equation is therefore exactly the condition that the finite-dimensional scattering system become singular. Its solutions are poles of the analytically continued scattering amplitude.

To see the imaginary-axis poles directly, write $k=i\kappa$ with $\kappa >0$. When $a\to 0$, the equation reduces toand its nonzero root is $\gamma =-3$, orThis is the bound state of the coincident potential $-3\lambda \delta (x)$. When $a\to \infty$, the three centres decouple and the roots approachwith exponentially small splitting. A pole at $k=i\kappa$ has energy $E=-{\hslash }^2{\kappa }^2/(2m)$ and an exponentially decaying wavefunction, so these upper imaginary-axis singularities represent bound states.

The one-particle partition function iswhere $\lambda =(2\pi {\hslash }^2\beta /m{)}^{1/2}$. ThusCanonical differentiation now givesandThereforeThe factor $5/2$ consists of $3/2$ from kinetic energy and $1$ from gravitational potential energy.

Let ${\rho }^{\prime }$ and $p^{\prime }$ be the density and pressure perturbations. Homentropy gives $p^{\prime }=c_0^2{\rho }^{\prime }$. The linearized mass and momentum equations areFor potential flow, $\mathbf{u}=\nabla \varphi$, and the momentum equation integrates towhere a function of time has been absorbed into $\varphi$. Substitution into mass conservation yieldsso

For a one-dimensional closed energy orbit, defineAfter locally inverting this relation to write $E=H(I)$, use the Hamilton--Jacobi generating functionand define the angle $\varphi =\partial W/\partial I$ modulo $2\pi$. The transformation is canonical, the Hamiltonian depends only on $I$, and

The adiabatic-invariance principle says that if an external parameter changes on a time scale much longer than the orbital period, and no separatrix is crossed, the action $I$ remains approximately constant. Its small within-cycle changes do not accumulate at leading adiabatic order.

False. Thomae’s function is Riemann integrable and discontinuous at every rational, an infinite countable set.

False. Continuous functions can converge pointwise to the non-Riemann-integrable Dirichlet function; for example enumerate the rationals and use continuous narrow bumps whose first $n$ peaks cover the first $n$ rationals while widths shrink suitably pointwise off $\mathbb{Q}$.

False without continuity. The function equal to $1$ at one point and $0$ elsewhere is nonnegative, Riemann integrable with integral $0$, but is not identically zero.

False. The Cantor function has derivative $0$ outside the Cantor set, hence its derivative extends to the Riemann-integrable zero function, while $F(1)-F(0)=1$.