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An independent set with the largest possible cardinality in the given finite graph. Its size is the independence number. This differs from a maximal independent set, which only cannot be enlarged by adding vertices: another independent set can still be larger.

For target $p=N(0,{\sigma }^2)$ and proposal $q=N(0,1)$, this is the importance weight. Assuming only $E_p{f}^2<\infty$, the weighted estimator has guaranteed finite variance for $0<\sigma \le 1$, since $E_q(w_{\sigma }^2{f}^2)=E_p(w_{\sigma }{f}^2)\le {\sigma }^{-1}{E}_p{f}^2$. For every $\sigma >1$, choosing $f^2=e^{a{x}^2}$ with $\max (0,{\sigma }^{-2}-1/2)<a<(2{\sigma }^2{)}^{-1}$ gives finite target second moment but infinite weighted second moment. With the additional assumption that $f$ is bounded, the sufficient range is instead $\sigma <\sqrt{2}$. The integrand hypothesis therefore matters.

To estimate $\mathbb{P}(Z>z)$ for a standard normal variable using proposal $N(\mu ,1)$, the weighted indicator has second moment $e^{{\mu }^2}\bar{\Phi }(z+\mu )$. Differentiating it at a minimum gives the displayed equation. Since $\phi (a)/\bar{\Phi }(a)=E[Z\mid Z>a]$, the minimizing shift is half the target conditional mean above the shifted threshold. The resulting estimator remains unbiased with the exact likelihood-ratio weight.

Put $\theta =\sqrt{-30}$. The ring of integers is $\mathbb{Z}[\theta ]$ and its discriminant is $-120$. The Minkowski bound for ideal classes $4\sqrt{30}/\pi <7$ shows that the ideal class group is generated by the classes of the ramified prime ideals $P_2=(2,\theta )$, $P_3=(3,\theta )$ and $P_5=(5,\theta )$. There is no additional prime ideal of norm four because two ramifies with residue degree one. The relations $P_i^2=(i)$ and $(\theta )=P_2{P}_3{P}_5$ show that these classes have order dividing two and their product is one. None is a principal ideal: the element field norm $a^2+30b^2$ cannot equal two, three or five. Thus there are exactly four classes, with the displayed group structure. This also supplies the class-number input for the Hilbert class field of Q of square root minus thirty.

Write $\eta =\sqrt{-34}$. The ring of integers is $\mathbb{Z}[\eta ]$ with discriminant $-136$. The Minkowski bound for ideal classes is less than eight, so classes of prime ideals of norm at most seven generate the ideal class group. These are the ramified prime $P_2=(2,\eta )$ and conjugate pairs at five and seven; three is inert with prime norm nine. Set $P_5=(5,\eta -1)$ and $P_7=(7,\eta +1)$. Conjugate prime classes are inverse, so these three classes suffice.

The field norm identities $N(4+\eta )=50$ and $N(1+\eta )=35$ give $(4+\eta )=P_2{P}_5^2$ and $(1+\eta )={\bar{P}}_5{P}_7$, by reduction modulo the indicated primes and equality of norms. Thus $[P_5{]}^2=[P_2]$ and $[P_7]=[P_5]$. Since $P_2^2=(2)$ but $a^2+34b^2=2$ has no integer solution, $[P_2]$ has order two. Therefore $[P_5]$ has order four and generates the entire group, proving the displayed isomorphism.

Let $d>1$ be a positive square-free integer congruent to one modulo four, with distinct prime factors $p_1,\dots ,p_k$. The ring of integers is $\mathbb{Z}[\sqrt{-d}]$. Each ramified prime ideal $P_i=(p_i,\sqrt{-d})$ satisfies $P_i^2=(p_i)$, so its ideal class has order dividing two. Their product is the principal ideal $(\sqrt{-d})$.

There are no further subset relations. For a nonempty proper subset $S$, put $m={\prod }_{i\in S}{p}_i$. If ${\prod }_{i\in S}{P}_i=(u+v\sqrt{-d})$, the ideal norm gives $m=u^2+d{v}^2$. But $1<m<d$ forces $v=0$ and makes the square-free integer $m$ a square, a contradiction. Thus the homomorphism ${\mathbb{F}}_2^k\to \mathrm{Cl}(K)$ defined by the ramified classes has exactly the all-ones relation, and its image is $C_2^{k-1}$. This identifies a subgroup, without asserting that these classes exhaust the ideal class group.

Let $d$ be a positive square-free integer congruent to two modulo four and $p\ge 3$ a prime. If $p$ does not divide $h(\mathbb{Q}(\sqrt{-d}))$ and $p\mid d$, then $x^2+d=y^p$ has no positive integer solution. More generally, if $\gcd (x,a)=1$ and $x^2+a^{2}d=y^p$, the two conjugate principal ideals are coprime. Parity excludes a shared prime above two; for an odd common prime, divisibility of both conjugates would force the underlying rational prime to divide both $x$ and $d$, giving valuation one on the left side, incompatible with a $p$th power.

Unique prime ideal factorization consequently gives $(x+a\sqrt{-d})=A^p$. The class-number condition forces $A$ principal; the only units are $\pm 1$, both odd powers. Hence $x+a\sqrt{-d}=(u+v\sqrt{-d}{)}^p$. When $a=1$ and $p\mid d$, the coefficient of $\sqrt{-d}$ on the right is divisible by $p$: its linear term has coefficient $p$, and every higher odd term contains $d$. It cannot equal one. Conversely, an actual solution under these conditions forces divisibility of the class number by $p$. For example $13^2+174=7^3$ proves $3\mid h(\mathbb{Q}(\sqrt{-174}))$.

Tissue swelling classically associated with severe hypothyroidism, often relatively non-pitting; distinguish it from fluid-overload anasarca.

The cat map is the area-preserving map of the real torus given byIts integer matrix $A=\left(\begin{array}{cc}2& 1\\ 1& 1\end{array}\right)$ has determinant one and inverse $\left(\begin{array}{cc}1& -1\\ -1& 2\end{array}\right)$, so reducing coordinates modulo one retains an invertible transformation of the torus.

Geometrically, $A$ stretches a unit square into a parallelogram. Cutting that parallelogram along the integer grid and translating its pieces by integer vectors brings it back into a unit square. Edges are identified on the torus, which is why cutting the drawing does not make the torus map discontinuous.

The eigenvalues $(3+\sqrt{5})/2$ and $(3-\sqrt{5})/2$ give one expanding and one contracting eigenvector. Their product one agrees with area preservation, while the absence of eigenvalues on the unit circle makes this a hyperbolic toral automorphism. The Fibonacci powers of the cat-map matrix give its iterates explicitly. Thus an integer matrix that is easy to multiply provides a concrete model of simultaneous uniform stretching, contraction and global wrapping.

An instrument comparing dry-bulb and wet-bulb temperature to investigate atmospheric humidity.

Alternating right and left Householder transformations reduce a real square matrix to a bidiagonal matrix. For a lower-bidiagonal result, first clear row $j$ beyond column $j$ using a right transformation on trailing columns. Then clear column $j$ below row $j+1$ using a left transformation on trailing rows. Previously completed rows and columns are preserved. Since $B^{T}B=Q^T{A}^{T}AQ$ is a symmetric tridiagonal matrix, its eigenvalues are the squares of the singular values of $A$. This explains the reduction structurally; explicitly forming $A^{T}A$ is unnecessary and can worsen numerical conditioning.

Every real orthogonal matrix of dimension $n$ is a product of at most $n$ Householder matrices. For a first-column unit vector $q\ne e_1$, the reflection with normal $q-e_1$ sends $q$ to $e_1$. Orthogonality then makes the transformed matrix $\mathrm{diag}(1,Q^{\prime })$, reducing the problem by one dimension. If $q=e_1$, omit that reflection. The dimension-one cases are $1$, the empty product, and $-1$, a single reflection. Induction and reversal of the transformations prove the factorization.

For $\dot{r}=r(a-g(r))$, $\dot{\theta }=1$ and $g(r)=1+[1+(r-1{)}^2{]}^{-1}$, the origin loses stability at $a=3/2$ and a stable small circular orbit appears for larger $a$. Its radius is $1-\sqrt{(2-a)/(a-1)}\sim 2(a-3/2)$. The Cartesian vector field is continuously differentiable but not generally twice differentiable at the origin, so the usual smooth Hopf normal form and its square-root amplitude law do not apply directly. The explicit radial equation proves the transition.

A real cubic $p(\lambda )={\lambda }^3+A{\lambda }^2+B\lambda +C$ has roots $-A,\pm i\sqrt{B}$ when $B>0$ and $AB=C$. For $A>0$ this gives a stable transverse third direction. To conclude a generic Hopf bifurcation as a parameter varies, also require a transverse crossing of the imaginary pair and a nonzero first Lyapunov coefficient. The algebraic imaginary-root condition alone is not sufficient.

The first Lyapunov coefficient is the cubic radial coefficient in the center-manifold normal form of a smooth Hopf bifurcation, after an appropriate normalization. In $\dot{r}=\beta r+l_1{r}^3+\cdots$, its nonzero sign distinguishes stable and unstable bifurcating small cycles relative to the eigenvalue-crossing direction. If $l_1=0$, higher-order terms are required and the generic cubic Hopf theorem does not apply.

The eigenvalue crossing speed is the derivative of the real part of a parameter-dependent eigenvalue at a stability boundary. A nonzero speed for an imaginary pair is the transversality condition in a generic Hopf bifurcation.

A transverse eigenvalue crossing occurs when the real part of an eigenvalue passes through zero with nonzero parameter derivative. This spectral condition is separate from the nonlinear nondegeneracy conditions in the Hopf theorem.

The convolution estimate gives the upper bound. Apply the operator to the Fejér kernel, whose $L^1$ norm is one. The summability kernel on the circle property implies $f\ast K_n\to f$ in $B$, giving the reverse bound. This identifies the operator norm even when no single input attains it.

A homeomorphism $g:\partial Q\to \partial Q$ of the boundary of a closed square $Q$ extends to a homeomorphism of $Q$ by fixing its centre $c$ and setting $H(c+r(p-c))=c+r(g(p)-c)$ for $p\in \partial Q$ and $0<r\le 1$. Each noncentral point lies on a unique ray from $c$ meeting the boundary once, so the formula is well defined and its inverse is obtained from $g^{-1}$. Both maps are continuous off the centre by the radial parametrization, and at the centre because their distances from $c$ are bounded by $r$ times the radius of the square. It follows that complements of any two closed boundary arcs are homeomorphic: first map the arcs to one another by a boundary homeomorphism, then extend radially.

A map from an open subset of $\mathbb{C}$ into a complex Banach space is weakly holomorphic when every composition with a bounded complex-linear functional is a scalar holomorphic function. Norm holomorphy requires its difference quotients to converge in the Banach norm. Weak holomorphy is initially a condition on scalar observations; the weak holomorphy implies norm holomorphy argument recovers an actual vector derivative. A closed target subspace containing values on a sequence with an interior accumulation point contains every value on the connected domain: annihilate the subspace, apply the scalar identity theorem, and use Hahn-Banach separation.