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.

Dissociation of an ionic compound separates its constituent ions. It differs from creating charged species by ionization of neutral molecules. Dissolved ions can also associate into ion pairs, so a dissociated description does not exclude all subsequent association.

With weak kinematic viscosity $\nu$, zero buoyancy diffusion and locally unchanged plane internal gravity wave polarization, the displacement pseudomomentum of an internal gravity wave obeys $P_t+{\partial }_z(c_{gz}P)=-\nu K^{2}P$. For constant wavenumbers and a stationary wave envelope, $P\propto |\hat{\sigma }{|}^2$ proves the displayed amplitude decay. Decay is along the group velocity, so $c_{gz}<0$ describes a source above the receiving level. In a slowly varying shear flow, use wave action and the changing intrinsic frequency rather than treating $P/|\hat{\sigma }{|}^2$ as constant.

Plumbing identifies a base disk times a fiber disk in one oriented rank-two disk bundle with such a neighborhood in another, exchanging base and fiber coordinates. Their zero sections then intersect transversely once. For disk bundles over $S^2$ arranged along a chain with Euler numbers $-a_i$, the intersection form has diagonal entries $-a_i$, adjacent entries one and other entries zero. The same manifold has a Kirby diagram consisting of adjacent Hopf-linked unknots with framings $-a_i$. Its boundary is described by the corresponding negative continued fraction, with the orientation fixed by the surgery convention.

Let $R$ be a one-dimensional normal Noetherian local domain. Choose $0\ne x$ in its maximal ideal $\mathfrak{m}$ and a minimal $n$ with ${\mathfrak{m}}^n\subseteq (x)$. If $n>1$, choose $y\in {\mathfrak{m}}^{n-1}\setminus (x)$ and put $z=y/x$. Then $z\notin R$ but $z\mathfrak{m}\subseteq R$. If $z\mathfrak{m}\subseteq \mathfrak{m}$, the determinant trick makes $z$ integral, contrary to normality. Thus $zu$ is a unit for some $u\in \mathfrak{m}$, and $v=u(zv)/(zu)$ for every $v\in \mathfrak{m}$ proves $\mathfrak{m}=(u)$. The case $n=1$ is already principal. Such a local domain is a discrete valuation ring.

For a normalized discrete valuation, a uniformizer $\pi$, and a chosen set $A$ of representatives of the residue field, every existing nonzero element has a unique displayed expansion, with its first digit not in the maximal ideal. Starting from $y_0={\pi }^{-v(x)}x$, choose $a_j$ representing $y_j$ modulo the maximal ideal and put $y_{j+1}=(y_j-a_j)/\pi$. The error after $r$ digits has valuation at least $v(x)+r$, so the sums converge to $x$ even without completeness. Conversely arbitrary digit sequences need not converge in an incomplete field: in $\mathbb{Q}$ with its three-adic valuation, the digits of $\sqrt{10}\in {\mathbb{Q}}_3$ give a counterexample. Hensel lemma gives this square root, but it is not rational.

A field with a surjective discrete valuation, complete for the associated non-Archimedean metric. Its valuation ring is a discrete valuation ring, and its maximal ideal is generated by a uniformizer. Completeness allows successive congruence solutions to converge to actual roots, which underlies the Hensel lemma. Non-Archimedean local fields are examples; completeness alone does not require the residue field to be finite.

Let the numerical support of a probability distribution be $x_1<\cdots <x_n$. Its cumulative distribution function counts how many values have been passed:It is zero below $x_1$, equals $j/n$ on $x_j\le t<x_{j+1}$ for $1\le j<n$, and is one at and above $x_n$. Every jump is $1/n$, even when the horizontal gaps differ.

For inverse transform sampling, take an ideal continuous uniform distribution for $U$ on $[0,1)$. Partition that interval into $n$ intervals of width $1/n$ and outputEach interval has probability $1/n$, proving the desired discrete uniform distribution. Using the floor function, the index is $j=1+\lfloor nU\rfloor$; for consecutive integer values the output is $X=a+\lfloor nU\rfloor$. With the usual quantile function convention $Q(u)=\inf \{t:F_X(t)\ge u\}$, one instead writes $Q(u)=x_{\lceil nu\rceil }$ for $0<u\le 1$, using the ceiling function. The endpoint conventions differ only at a finite set of points with probability zero under the ideal continuous uniform distribution.

For $S=\{1,2,4,8\}$, the four intervals $[0,1/4)$, $[1/4,1/2)$, $[1/2,3/4)$, and $[3/4,1)$ select $1,2,4,8$, respectively. Thus $U=3/10$ selects $2$, while $U=9/10$ selects $8$. The upper half of the interval selects values at least four, recovering probability $1/2$. Equal widths refer to the sampling probability coordinate, not to distances between output values. This is an exact mathematical construction assuming an ideal continuous uniform distribution; a finite computer random-number source needs its own justification of equal output chances.

Suppose the support of a probability distribution consists of all consecutive integers from $a$ to $b$, including both endpoints, with $a\le b$. Its cardinality is $n=b-a+1$, and a discrete uniform distribution on it satisfiesFor a short derivation, set $K=X-a$, whose possible values are $0,\dots ,n-1$. The finite sums giveand henceAdding $a$ shifts the expected value by $a$ and leaves the variance unchanged. For completeness, the first sum follows by pairing $k$ with $n-1-k$. For the second, sum $(k+1{)}^3-k^3=3k^2+3k+1$ from $k=0$ to $n-1$: the left side is $n^3$, so inserting the first sum gives the stated result.

The edge case $a=b$ has $n=1$: $X=a$ with probability one, expected value $a$, and variance zero. On an arithmetic progression $\{a,a+h,\dots ,a+(n-1)h\}$ with $h>0$, the same argument instead gives expected value $a+h(n-1)/2$ and variance $h^2(n^2-1)/12$. Equal spacing is the extra hypothesis behind these formulas; the discrete uniform distribution on $\{1,2,4,8\}$ requires the general sums above.

For $U_t=-cDU$ with $c>0$, a positive discrete norm matrix $H$ satisfying the displayed identity gives $d(U^{T}HU)/dt=-c(U_N^2-U_0^2)$. Homogeneous inflow data $U_0=0$ therefore leave only the nonpositive outflow flux. Inhomogeneous inflow contributes a controlled boundary forcing term. This estimate includes the numerical boundary closure, unlike an interior-only Fourier stability analysis. For any dissipative semidiscrete operator in the $H$ norm, Backward Euler method steps preserve the contraction by the identity $\Vert U^{n+1}{\Vert }_H^2-\Vert U^n{\Vert }_H^2=2k⟨U^{n+1},A{U}^{n+1}{⟩}_H-\Vert U^{n+1}-U^n{\Vert }_H^2$.

For discrete opfibrations $A\to D\leftarrow B$ with weakly filtered categories, the map taking a pair to its connected components of a category is a bijection. For surjectivity, send two base objects in the same component to a common target and lift both arrows. For injectivity, choose common targets upstairs for two representatives; a cocone over the resulting finite connected diagram downstairs makes their lifted composites agree. Hence the representatives have a common target in the pullback in a category. Over a filtered category, the component functor on the slice of discrete opfibrations preserves all finite categorical limits: it preserves pullbacks and takes the terminal object to a singleton.

An integrable predictable strategy whose discounted gains ${\varphi }_n\Delta {\tilde{S}}_n$ are integrable preserves the martingale property of discounted prices in finite discrete time. Bounded holdings suffice. Without integrability the claim is false: let ${\tilde{S}}_0=3/2$, ${\tilde{S}}_1=1+U$, ${\tilde{S}}_2=1+U+\epsilon /2$, where $U$ is uniform on $(0,1)$ and $\epsilon$ is an independent fair sign. Holding zero stock first and $1/U$ shares in the second period, funded by the bank, gives zero-initial-cost terminal wealth $\epsilon /(2U)$, which is not integrable. Replication pricing requires the corresponding true-martingale condition, not merely formal self-financing.