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Assume $a=c/r$ decreases outward, tends to zero at $R$, and define $h(a)=-d\ln r/da$. Differentiation of the Duvall law gives $F^{\prime }(w)={\int }_0^{w}h(b)b/[w^2\sqrt{w^2-b^2}]db$. Interchanging integrals and using ${\int }_b^{a}dw/[w\sqrt{w^2-b^2}\sqrt{a^2-w^2}]=\pi /(2ab)$ proves the displayed Abel transform inversion. For a finite outer reference value $w_s$, both lower limits become $w_s$, after removing the known outer-layer contribution to the travel time. Inversion requires a monotone branch and amplifies errors in differentiated frequency data.

For saturated Darcy flow through a long dam occupying a channel of width $\alpha z^2$, let hydraulic conductivity be $K=k\rho g/\mu$ and free-surface height $h(x)$. The Dupuit approximation gives $Q=-K\alpha h^3{h}_x/3$. Fixed heads $h(0)=H$ and $h(L)=\eta$ give $h^4=H^4-(H^4-{\eta }^4)x/L$ and the displayed steady discharge. Porosity relates pore velocity to superficial Darcy velocity but does not independently multiply this discharge when $k$ is intrinsic permeability. For depth-dependent permeability, integrate the horizontal conductivity over the saturated cross-section first.

For $n\ge 1$, any distinct node set in $[-1,1]$ different from the $n+1$ Chebyshev polynomial extrema admits one degree-$n$ polynomial $q$, bounded by one at all those nodes, whose every nonconstant derivative violates the corresponding sharp endpoint bound. In decreasing order set $q(t_i)=(-1{)}^i$. The endpoint derivative signs of Lagrange cardinal polynomials give $q^{(k)}(1)={\sum }_i|L_i^{(k)}(1)|$. Interpolating $T_n$ on the same nodes showsEach term is nonnegative and at least one is positive, because the other node set includes a nonextremal point. Uniqueness is a claim about node sets: permuting the same nodes does not change their norming property.

Start with cash one, switch to $X/a$ when a nonnegative supermartingale first reaches at most $a$, and switch to cash $q=b/a$ when it next reaches at least $b$. Repeat with coefficients multiplied by $q$ after every completed upcrossing. At each switch the new value is at most the old one, so pasting supermartingales with downward jumps gives a supermartingale when stopped after any fixed number of switches. Its initial value is $\min (X_0/a,1)$, and its value at the $k$th completed crossing is exactly $q^k$. Applying its expected value bound proves the Dubins upcrossing inequality.

For an irreducible projective variety $Y\subset \mathbb{P}(V)$, its projective dual is the closure of hyperplanes containing the embedded tangent space at some smooth point of $Y$. Over the complex numbers, biduality gives $(Y^{\ast }{)}^{\ast }=Y$. For a nondegenerate smooth curve in characteristic zero, the dual is an irreducible hypersurface; a general point corresponds to one ordinary tangency. Its defining incidence space consists of a point on the curve and a hyperplane containing its tangent line.

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.