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The vector sum of two vectors $v,w$ is the vector $v+w$ obtained by vector addition.

The Von Mangoldt function is nonnegative and satisfies ${\sum }_{d\mid m}\Lambda (d)=\log m$. Therefore the middle short-divisor coefficient in the Vaughan identity has the displayed pointwise bound. Grouping its long-divisor term as ${\mu }_{>U}(m)b(n)$, with $b(n)={\sum }_{d\mid n,d>V}\Lambda (d)$, gives $0\le b(n)\le \log n$. This grouping permits bounded-coefficient Type II sums after division by $\log N$.

For reduced $a/q$ with $|\alpha -a/q|\le q^{-2}$ and $1\le q\le N$, the displayed bound follows by splitting Vaughan identity with parameters $U=V=N^{2/5}$. Type I sums use short geometric-sum bounds; type II sums use the Cauchy-Schwarz inequality and rational spacing to control bilinear cancellation. A badly approximable number has a Dirichlet approximant with $q\asymp N^{1/2}$, yielding $O(N^{4/5}{\log }^{4}N)$ and hence $O(N^{9/10})$.

// Target: analytic-number-theory.bigb

If a random variable $X$ takes values in $[m,M]$, then

For a nonnegative count with positive mean, the Fano factor compares its variance to its mean. A Poisson distribution has factor one. Correlated production bursts or weak restoring feedback can give factors larger than one even at large molecular copy number.

// Target: mathematical-biology.bigb

Use the Fourier transform convention $\hat{\psi }(\omega )=\int \psi (x)e^{-i\omega x}dx$. If $\psi$ and $x^r\psi$ for $r<q$ are integrable functions, moment differentiation of the Fourier transform yields a $C^{q-1}$ Fourier transform andThe dominated convergence theorem justifies these derivatives. Under these assumptions, $q$ vanishing moments are equivalent to vanishing derivatives through order $q-1$ at zero. For $q>1$ this gives a Peano zero $\hat{\psi }(\omega )=o(|\omega {|}^{q-1})$; for $q=1$ it gives $\hat{\psi }(\omega )=o(1)$. It does not by itself justify a $q$th derivative or an $O(|\omega {|}^q)$ estimate.

If also $\int |x{|}^q|\psi (x)|dx<\infty$, subtract the degree-$(q-1)$ Taylor polynomial of $e^{-i\omega x}$ inside the integral. Its Taylor remainder gives the stronger boundFor an integrable function with compact support, every weighted integral exists and its Fourier transform extends to an entire function. The language “zero of order at least $q$” is then literal analytic function terminology. If the next integral is nonzero, the order is exactly $q$. The smooth-mask vanishing-moment criterion relates these cancellations to a refinement mask under its additional hypotheses.^[ref]

Let $\psi$ be an integrable function with at least one vanishing moment. For a unit jump discontinuity $f(x)=H(x-x_0)$, where $H$ is the Heaviside step function, substitution into the coefficient givesThe signed integral of the whole analyzing function is zero, but the jump discontinuity cuts through its support of a function and usually leaves an unbalanced partial integral. At fixed relative position $\theta$ with nonzero partial integral, the coefficient decays as $\sqrt{a}$, slower than the $a^{q+1/2}$ Taylor remainder bound on a smooth region. “Edges survive” means this relative contrast: the normalized coefficients still tend to zero as $a\to 0$.

For an original exact comparison, use the smooth polynomial $f(x)=x+x^2$ at $b=0$. The Haar wavelet $h$ and the second-difference analyzing function $\eta$ giveThe linear term cancels for $\eta$; it remains for $h$. For the fixed unit jump discontinuity $H(x)$, choose $b=-a/4$ so that $\theta =1/4$ at every scale. Then

Position matters. If an analyzing function's support of a function lies entirely on one constant side of the jump discontinuity, its coefficient is zero. Even a crossing can cancel: $\eta$ has zero partial integral at $\theta =3/2$. A jump discontinuity on a dyadic interval boundary can likewise escape finer Haar wavelets whose supports do not cross it. These examples describe particular functions and normalizations, not a universal ranking of wavelets for signal processing.

The Haar wavelet provides an explicit one-vanishing moment comparison:It cancels a constant polynomial, but not a nonconstant affine function. To obtain two vanishing moments by an equally transparent construction, take three adjacent unit intervals with heights proportional to $1,-2,1$:The L2 norm normalization follows from $1^2+(-2{)}^2+1^2=6$. Direct integration givesThus $\eta$ has exactly two vanishing moments. In fact its coefficient is a finite difference of adjacent local integrals:The weights are those of the second-order central difference. The first moment balances the centres of the three intervals; it is not enough merely to balance their signed areas.

This $\eta$ is a normalized analyzing function, with no orthonormal basis assertion. Indeed, its unit translate has nonzero L2 inner product:Hence these translates already fail orthogonality. The Daubechies wavelet construction imposes additional refinement and orthogonality requirements; two vanishing moments and unit L2 norm by themselves do not supply those requirements.

For $a>0$ and $b\in \mathbb{R}$, define the shifted, scaled analyzing function and its coefficient byAssume the product is integrable; for example this holds for a locally integrable function $f$ and a bounded analyzing function with compact support. The bar is complex conjugation. This normalization preserves the L2 norm of the analyzing function. If $p$ is a polynomial of degree of a polynomial less than $q$, then $p(b+au)$ is a sum of monomials $u^r$ with $r<q$. Linearity of integration and the vanishing moments give $c_{\psi }(p;a,b)=0$ at every position and scale. Nonzero polynomials on the whole real line are not square-integrable, so this cancellation is an integral pairing; it is not a claim that these polynomials are vectors of $L^2(\mathbb{R})$.

Here is a quantitative reason that smooth regions yield small details. Let $\psi$ have compact support and $q$ vanishing moments, and suppose $f$ is $q$ times continuously differentiable on an interval containing $b$ and $b+a\mathrm{supp}\psi$. Subtract its degree-$(q-1)$ Taylor polynomial at $b$. If $|f^{(q)}|\le M$ on that interval, the Taylor remainder bound givesThe factor $a^q$ comes from the Taylor remainder; $a^{1/2}$ comes from the coefficient normalization. Thus the analyzing function measures the departure from the local polynomial trend, rather than the height of $f$ itself. The wavelet coefficient decay for Hölder functions uses the same argument with a Hölder-Taylor remainder bound. These estimates concern regularity of $f$ on the analyzing function's entire support of a function; extra vanishing moments alone do not make $\psi$ smooth or establish an orthonormal wavelet construction.^[ref]

A valuation ring for a nontrivial real-valued rank-one valuation is a principal ideal domain exactly when its value group is discrete. A least positive value makes the value group cyclic and lets every nonzero ideal be generated by an element of smallest valuation. Conversely, if the maximal ideal is principal, its generator has the least positive value. Nontriviality matters: a field with the trivial valuation is a principal ideal domain, but not a discrete valuation ring in the usual nontrivial convention.

For a stable free Klein-Gordon field with a complete timelike Killing flow and a positive self-adjoint one-particle energy generator, its positive spectral subspace selects positive-frequency solutions of positive Klein-Gordon inner product. Their annihilation operators define the ground-state Fock vacuum. Degenerate modes can be mixed by a unitary matrix without changing this splitting or vacuum. Merely being stationary near infinity is insufficient if the Killing field becomes spacelike in an ergoregion of a stationary spacetime; zero modes, instability or additional boundary conditions also require separate treatment. The choice of a ground state does not prevent other states, such as thermal states, from using the same particle operators.

// Target: statistical-physics.bigb

The symmetric similarity solution for a localized fixed volume $V$ in a V-shaped channel gravity current has $h(x,t)=[3(X^2-x^2)/(7Ct){]}_{+}^{1/3}$. Integrating the similarity conservation of mass equation once gives $C{h}^4{h}_x=-2x{h}^2/(7t)$, and vanishing thickness at the two noses gives the profile. Its volume fixes $X=[V/(I\tan \alpha ){]}^{3/7}(7Ct/3{)}^{2/7}$, where $I={\int }_{-1}^1(1-s^2{)}^{2/3}ds$. At any fixed nonzero position, the thickness first rises after arrival and then falls; its maximum occurs when $X^2=7x^2/3$.

For a shallow V-channel with floor $z=\alpha |y|$, $0<\alpha \ll 1$, let the free surface be $z=h(x,t)$. Hydrostatic pressure gives $p_x=\rho g{h}_x$. A vertical strip has thickness $H_y=h-\alpha |y|$ and, with no slip below and zero shear above, flux $-\rho g{H}_y^3{h}_x/(3\mu )$. Integrating across $|y|<h/\alpha$ givesConservation of volume is $(h^2{)}_t=C(h^4{h}_x{)}_x$, $C=\rho g/(6\mu )$. Here $\alpha$ is the wall slope to the horizontal, not the opening half-angle measured from the vertical in a general V-section. Vertical shear dominates transverse shear by the shallow-slope approximation.

Combinatory logic describes computation by application and fixed combinator constants, eliminating bound variables through bracket abstraction. Its intensional weak reduction does not identify every pair of terms that becomes beta-equivalent after translation into lambda calculus.

Eta conversion expresses extensionality for a lambda abstraction whose fresh argument is only passed to a function. The side condition prevents removal of a genuine dependency. In a categorical model with $D\cong D^D$, the inverse laws for currying and evaluation validate this conversion as well as beta reduction.

In a homogeneous ambient fluid, the displayed nonsteady power-law ansatz solves the unsteady top-hat line-plume balances for $\tau =t+t_{\ast }>0$. It gives $Q=R\alpha z^2/(3\tau )$, $M=2R\alpha z^3/(9{\tau }^2)$ and $F=2R\alpha z^3/(9{\tau }^3)$. The solution is singular at zero age and carries zero source flux at its unshifted ideal origin. It describes a separable branch away from a source; a finite-width boundary, virtual-origin shift or matching region is needed to supply nonzero physical source flux. It is distinct from the steady constant-flux line plume, whose top-hat half-width grows as $\alpha z$.

For an unramified extension $L/K$ of non-Archimedean local fields and any finite extension $M/K$, reduction is a bijection between $K$-field homomorphisms $L\to M$ and $k_K$-field homomorphisms $k_L\to k_M$. Choose an integral generator with separable irreducible reduction. Each specified residue image lifts uniquely to a root in $M$ by Hensel lemma. Conversely a $K$-embedding preserves the unique extended valuation and therefore induces the residue embedding. The same argument applies when $M$ is any non-Archimedean local extension with the given compatible embedding of $K$.

For non-Archimedean local fields, let $L/K$ be an unramified extension of degree $f$, and let $E/K$ be finite. The degree-$f$ unramified extension of $E$ has a residue field containing the degree-$f$ residue extension of $K$. Lift an integral residue generator of $L$ into that field by Hensel lemma. This embeds $L$ into an unramified extension of $E$, so $LE/E$ is an unramified subextension. Applying this argument to each completion of a number field at a prime ideal proves the finite-prime assertion for number fields, even when $E/K$ is ramified.

The local principal series of GL2 induces a smooth linear character of its diagonal torus, inflated to its upper-triangular subgroup. Its unnormalized Jacquet module has two one-dimensional composition factors, $\chi$ and ${\chi }^w{\delta }^{-1}$, where $\delta (\mathrm{diag}(a,d))=|d/a|$. When these coincide, the module is a nonsplit self-extension, so one must not count the linear character twice as two independent quotients.

Evaluation of an intertwiner at the identity gives a $B$-equivariant functional. Conversely a functional $\ell$ gives $A(v)(g)=\ell (\pi (g)v)$, which has the required covariance and is fixed on the right by any compact-open subgroup fixing $v$. These inverse constructions prove the displayed adjunction. If $\xi$ is trivial on the unipotent radical, it is equivalent to a character-valued functional on the Jacquet module.