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For an irreducible projective variety $Y\subset \mathbb{P}(V)$ over the complex numbers, its projective dual variety is the closure in the Zariski topology of the projective hyperplanes containing the embedded tangent space at some smooth point of $Y$. The projective biduality theorem recovers $Y$ as $(Y^{\ast }{)}^{\ast }$: the family of tangent projective hyperplanes determines the original projective variety.

For a nondegenerate smooth projective curve $C\subset {\mathbb{P}}^n$ with $n\ge 2$, the incidence pairs $(p,H)$ with $T_{p}C\subset H$ form a projective bundle over $C$ with fiber ${\mathbb{P}}^{n-2}$, hence dimension $n-1$. The general contact locus is a projective subspace of $C$; since $C$ is not a projective line, it is a single point. Thus $C^{\ast }$ is an irreducible projective hypersurface, and a general member represents one ordinary tangency. For a smooth plane conic, the tangent-line equation gives another smooth plane conic in the dual projective plane.

The ambient-dimension hypothesis excludes $C={\mathbb{P}}^1\subset {\mathbb{P}}^1$: no projective hyperplane contains its whole tangent space, so its dual is empty. The stated reflexivity and general tangency assertions also use characteristic zero; inseparable behavior in positive characteristic requires additional hypotheses.^[ref]

For $A=k[\epsilon ]/({\epsilon }^2)$ and $k=A/(\epsilon )$, the kernel and image of multiplication by $\epsilon$ both equal $(\epsilon )$. Thus the displayed sequence is a free resolution. Applying ${\mathrm{Hom}}_A(-,k)$ makes every differential zero, so ${\mathrm{Ext}}_A^i(k,k)\cong k$ for every $i\ge 0$. The module $k$ therefore has infinite projective dimension.

In bond percolation on the square lattice, the outer boundary of a finite percolation cluster, after filling its holes, consists of closed bonds. Their crossing edges in the planar dual graph form a connected enclosing contour, and hence belong to a single dual bond percolation cluster $K$. If $|K|=m$, both its coordinate widths are at most $m-1$, so the enclosed primal cluster contains at most $m^2$ vertices. A contour enclosing the origin also places every vertex of $K$ within distance $m+1$ of the origin. An exponential volume tail for the subcritical dual bond percolation clusters thus gives a bound $C{e}^{-c\sqrt{n}}$ for a finite primal cluster of size at least $n$.

A planar finite network with three boundary terminals induces one of five connection partitions: all three connected, all three separate, or one connected pair and an isolated third terminal. Put three dual terminals on the intervening boundary arcs. Passing to the planar dual graph interchanges the all-connected and all-separate partitions and permutes the three pair partitions. If the network is symmetric under all terminal permutations, equality $P_3=P_0$ of the first two probabilities makes its whole partition law dual-invariant. Independent cells on a self-dual triangular terminal arrangement then give identical primal and dual coarse connection laws. Infinite connectivity is preserved because cells are uniformly finite. For a martini lattice cell with six independent bonds of common parameter $p$, $P_3=3p^5-2p^6$ and $P_3-P_0=(2p^2-1)(p^4-3p^3+2p^2+1)$. The unique zero in $[0,1]$ is $1/\sqrt{2}$.

For a down-and-out European call with barrier equal to strike, zero interest and no dividends, hold one share and borrow the strike until barrier hitting or maturity. At continuous hitting the share price is exactly the debt, so liquidating leaves zero; on survival the portfolio pays $S_T-K$. Hence the contract has value $S-K$ and delta one before knock-out. The pathwise replication works beyond constant-volatility Black-Scholes model, but can fail through downward jumps or execution gaps, and nonzero carry, transaction costs or discrete monitoring require changes.

Suppose the generators of a binary linear code have Hamming weights divisible by four and have pairwise zero binary inner products. Their linear span is doubly even: for any two vectors,The binary inner product says the intersection size is even. Induction on the number of generators in a sum proves divisibility by four.

// Target: algebra.bigb

The motion of a star around the astronomical barycenter caused by an orbiting companion. In an isolated two-body problem with relative semi-major axis $a$, the stellar semi-major axis is $a_{\star }=M_{p}a/(M_{\star }+M_p)$. Doppler spectroscopy measures its line-of-sight velocity and astrometry measures its projected position on the sky.

For independent zero-mean square-integrable increments, their partial sums form a martingale, and their squares form a nonnegative submartingale. Apply the Doob maximal inequality to the squared process at threshold $x^2$. Independence and zero means make its terminal second moment the sum of the increment variances. Identical distributions are not required.

For a square-integrable martingale with $M_0=0$ and $V=\mathbb{E}{M}_n^2$, apply the Doob maximal inequality to $(M_k+c{)}^2$ for $c\ge 0$. An upper crossing of $x$ implies a square crossing of $(x+c{)}^2$, giving $(V+c^2)/(x+c{)}^2$. Minimization at $c=V/x$ gives the displayed bound. A one-step martingale taking $x$ with probability $V/(V+x^2)$ and $-V/x$ otherwise attains equality when $V>0$.

For a nonnegative submartingale with maximum $M_t$, the first-moment maximal estimate is $\lambda \mathbb{P}(M_t\ge \lambda )\le \mathbb{E}[X_t{1}_{\{M_t\ge \lambda \}}]$. Integrate $2\lambda$ times the tail probability only up to $R$, obtaining $\mathbb{E}(M_t\wedge R{)}^2\le 2\mathbb{E}[X_t(M_t\wedge R)]$. The Cauchy-Schwarz inequality gives $\Vert M_t\wedge R{\Vert }_2\le 2\Vert X_t{\Vert }_2$, and monotone convergence gives the result as $R\to \infty$. Truncation avoids assuming beforehand that the maximum has a finite second moment.

// Target: brownian-motion.bigb

For centered unit-variance independent and identically distributed increments with $m_4=\mathbb{E}{\xi }^4<\infty$, an interpolated increment is $N^{-1/2}{\sum }_j{c}_j{\xi }_j$. Here $c_j$ is the length of the overlap of $(Ns,Nt)$ with the $j$th unit cell, so $0\le c_j\le 1$ and $\sum c_j=N(t-s)$. Independence gives its fourth moment as $N^{-2}[3(\sum c_j^2{)}^2+(m_4-3)\sum c_j^4]$. This is at most $\max (3,m_4)N^{-2}(\sum c_j^2{)}^2$, and $\sum c_j^2\le N(t-s)$ gives the bound, including intervals crossing a mesh point. A uniform fourth-moment continuity criterion yields tightness of probability measures on $C_0[0,1]$.

For a Landau-Ginzburg theory with positive gradient stiffness $\kappa$, $r<0$ and quartic coupling $u>0$, the degenerate minima are $\pm M_0$, $M_0^2=|r|/u$. A planar domain wall solves $\kappa {\varphi }^{\prime \prime }=r\varphi +u{\varphi }^3$. Multiplying by ${\varphi }^{\prime }$ and using the limiting minima gives $\kappa ({\varphi }^{\prime }{)}^2/2=u({\varphi }^2-M_0^2{)}^2/4$. Separating this equation yields the displayed profile, and integrating its excess energy gives the positive surface tension $2\sqrt{2\kappa }|r{|}^{3/2}/(3u)$. Its area cost is subextensive compared with bulk volume, permitting phase-separated common-tangent construction states when the average order parameter is constrained.

If a continuous local martingale $M$ satisfiesthen its stochastic exponential is a true martingale through time $T$.

This is an exponential-moment criterion for the stochastic exponential to be a true martingale. In the infinite-horizon stopped-moment form, a zero-starting convergent continuous local martingale satisfying ${\sup }_T\mathbb{E}{e}^{M_T/2}<\infty$ has a uniformly integrable martingale $\mathcal{E}(M)$. The half-threshold for the exponential-martingale Hölder bound first controls strict scalings. The terminal scaling inequality for stochastic exponentials and terminal expectation criterion for a nonnegative local martingale then include the endpoint scaling.

The Itô formula gives $d\log Z=dM-\frac{1}{2}d[M]$ for a strictly positive solution. Therefore equal initial values give indistinguishable solutions. Conversely the displayed exponential satisfies the equation. This is uniqueness for a fixed continuous local martingale driver and does not require the exponential to be a true martingale.

The Newton polynomial expresses the interpolation polynomial in products that successively vanish at all previously used interpolation nodes. With $y_i=f(x_i)$, the coefficients are divided differences:A triangular divided difference table computes all its coefficients using exactly $n(n+1)/2$ divisions. The Newton polynomial and the Lagrange polynomial have the same polynomial degree bound and the same nodal values, hence are equal by uniqueness of polynomial interpolation.

For $f(0)=1$, $f(1)=2$, $f(2)=5$, the complete divided difference table isThe first divided differences are $1$ and $3$, and their divided difference is $(3-1)/(2-0)=1$. Reading the top entries givesThe initial term fits $(0,1)$; adding $x$ also fits $(1,2)$; adding $x(x-1)$ fits $(2,5)$ while leaving the first two data values unchanged. This is the practical beauty of the Newton polynomial: each new correction respects all previous interpolation constraints.

If a new distinct interpolation node $x_{n+1}$ with value $y_{n+1}$ arrives, append one term:For example, appending $(3,11)$ to the three-point data gives $c=(11-10)/(3\cdot 2\cdot 1)=1/6$, hence $p_3(x)=x^2+1+x(x-1)(x-2)/6$. Changing the order of the interpolation nodes changes the Newton polynomial coefficients and products, but not the final interpolation polynomial in exact arithmetic.

The products also permit nested evaluation analogous to Horner's method. If $a_k=f[x_0,\dots ,x_k]$, start with $b_n=a_n$ and work backwards by $b_k=a_k+(x-x_k)b_{k+1}$; the answer is $b_0$. The three-point example becomes $1+x[1+(x-1)]$, needing two multiplications for each evaluation. This organization is useful when adding observations incrementally or evaluating a stored Newton polynomial many times; the Runge phenomenon still depends on the interpolation nodes rather than on which representation is used.

To justify the update formula, its added product vanishes at every previous interpolation node. Choosing $c$ by the new residual forces agreement at $x_{n+1}$, so uniqueness of polynomial interpolation identifies the result. For $n\ge 0$, $p_n$ has polynomial degree at most $n$, and the added product has leading coefficient one. Thus the new coefficient $c$ is exactly $f[x_0,\dots ,x_{n+1}]$. The initial state is $p_0(x)=f(x_0)$, with the empty product interpreted as one.^[ref]

Appending $(4,57)$ to the four-point example in divided difference table leaves the Newton polynomial unchanged: $p_3(4)=57$, so the new fourth divided difference is zero. If the appended value is instead $58$, the new coefficient isThis polynomial still takes values $2,1,5,22$ at the original interpolation nodes, and now takes value $58$ at $4$. A zero update is a valid completed step: adding another exact sample of the original polynomial need not increase the polynomial degree.

For distinct interpolation nodes, write $D_i$ for the divided difference with $x_i$ omitted. Symmetry and the divided-difference recurrence give the displayed identity. Consequently, for $0<i<n$,The identity is algebraic and does not require the nodes to be ordered; the coefficients need not be nonnegative.

For a block code, separable distortion is the average of the individual symbol costs. With Hamming distortion it is the fraction of incorrectly reconstructed symbols, rather than the unnormalized number of mismatches. This normalization makes expected distortion comparable across different block lengths.^[ref]

Expected distortion is the expectation of the reconstruction cost under the joint probability distribution of $X$ and $\hat{X}$. It weights common source outcomes more strongly than rare ones. A bound on expected distortion is therefore different from a bound on the cost for every input, or on the probability of exceeding a specified cost.

The reconstruction alphabet is the set of values that the decoder may output. It need not equal the source alphabet. A restricted reconstruction alphabet can prevent zero expected distortion, even if the representation is allowed arbitrarily many bits.