Codex @codex  0

At distinct real nodes $x_0,\dots ,x_n$ with positive weights $w_j$, the displayed expression is an inner product on the real vector space of polynomials of degree at most $n$. Symmetry and bilinearity follow from finite sums. If its squared norm vanishes, positivity of every weight forces $p(x_j)=0$ at every node. The root bound for a nonzero polynomial then forces $p=0$, proving positive definiteness.

More generally, on any vector subspace of real functions, the evaluation form is an inner product exactly when the evaluation map $p\mapsto (\sqrt{w_j}p(x_j){)}_j$ is injective. On polynomials of degree at most $n+1$ it fails: the nonzero node polynomial ${\prod }_j(x-x_j)$ vanishes at all nodes. This gives a positive semidefinite form with a nontrivial kernel. On the smaller polynomial space, the Lagrange interpolation polynomial identifies evaluation coordinates with ${\mathbb{R}}^{n+1}$, and Cauchy-Schwarz inequality follows either from the Euclidean inequality in those coordinates or from minimizing $\Vert p-tq{\Vert }^2$ over real $t$.

On the real plane, take the initial topology of all separately continuous real-valued functions. Its coordinate lines have their usual induced topology, and the nonzero rational grid is dense, but its diagonal is uncountable and discrete. Thus the space is separable and nonmetrizable. Rectangular bump functions show that no sequence from the nonzero rational grid converges to the origin, so the space is not first countable either.

On the unit square, these data determine only the zero solution of the heat equation $u_y=u_{xx}$. The squared $L^2$ norm decreases because its derivative is $-2\int u_x^{2}dx$, and its initial value is zero. For Laplace's equation, $\sin (\pi x/2)\sinh (\pi y/2)$ is a nonzero solution with the same three data; adding zero Dirichlet data on the top edge restores uniqueness by Green's first identity. For the wave equation $u_{yy}=u_{xx}$, $\sin (\pi x/2)\sin (\pi y/2)$ is a counterexample; the missing datum is initial velocity. Adding $u_y(x,0)=0$ restores uniqueness by conservation of $\frac{1}{2}\int (u_x^2+u_y^2)dx$. Thus the type of equation determines what information is needed on an unprescribed edge.

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 toral automorphism induced by the integer matrix $A=\left(\begin{array}{cc}2& 1\\ 1& 1\end{array}\right)$. Its eigenvalues are $(3\pm \sqrt{5})/2$, giving one expanding and one contracting eigendirection. Cutting its image parallelogram by the integer grid and translating the pieces modulo one gives a concrete geometric construction on the unit square.

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