An inguinal hernia entering through the deep inguinal ring, lateral to the inferior epigastric vessels. It can follow the inguinal canal toward the scrotum.

For a finite probability distribution with largest atom $\Pi$, its base-two Shannon entropy satisfies $H\ge 2(1-\Pi )$. If $\Pi \ge 1/2$, coarsen the distribution to the indicator of a largest atom: Shannon entropy cannot increase under a deterministic function, so $H\ge h_2(\Pi )$. The concavity of binary entropy places it above the chord from $(1/2,1)$ to $(1,0)$, giving $h_2(\Pi )\ge 2(1-\Pi )$. If $\Pi \le 1/2$, every nonzero probability $p$ has $-{\log }_{2}p\ge -{\log }_2\Pi$, so $H\ge -{\log }_2\Pi$. The function $-{\log }_2\Pi -2(1-\Pi )$ vanishes at $1/2$ and has negative derivative on $(0,1/2]$, proving the same bound. Equality in the combined bound occurs for a point mass or a distribution uniform on two points.

For a random finite binary string $M$ of length $L$, conditioning on $L=\ell$ leaves at most $2^{\ell }$ possible strings. Thus its conditional entropy is at most $\ell$. Since the length is a function of the string, the chain rule for information entropy givesThe term $H(L)$ cannot generally be removed. For concatenations of $m$ words from a fixed finite decipherable code with maximum length $d$, it is at most ${\log }_2(md+1)$. Decipherability preserves the Shannon entropy of the source block. For an independent source block, dividing the resulting inequality by $m$ and taking $m\to \infty$ proves the expected codeword length is at least the one-symbol Shannon entropy.

An inflammatory exudate rich in fibrin.

An inode-based file system uses direct pointers for small files and successive levels of indirect block pointers for larger ones. If an address block contains $q$ pointers, a depth-$k$ region addresses $q^k$ file data blocks. With the inode resident, an uncached byte access in that region requires $k$ address-block reads and one file data block read.

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.