For a smooth flow map ${\phi }_t$ generated by a vector field $F$, the chain rule and $\det (I+tDF)=1+t\mathrm{tr}(DF)+o(t)$ give ${\partial }_{t}J[{\phi }_t](x)=(\nabla \cdot F)({\phi }_t(x))J[{\phi }_t](x)$. With initial value one, its solution is the exponential of the time integral of this divergence. Thus a divergence-free generator preserves Euclidean volume and orientation. This connects the Jacobian determinant with a volume-preserving vector field.

For a zero-potential interval of width $a$, an infinite wall at its left end, and constant exterior potential $U_0$ to its right, the $j$th bound state branch exists precisely when $U_0>{\mathcal{U}}_j$, for $j=0,1,\dots$. Matching the wavefunction and its derivative gives $z\cot z=-\sqrt{z_0^2-z^2}$ with $z_0=a\sqrt{2m{U}_0}/\hslash$. At equality the new threshold solution is not normalizable, so exactly one bound state corresponds to $U_0\in ({\mathcal{U}}_0,{\mathcal{U}}_1]$.

Build this finite field by taking a quotient ring of a polynomial ring:The polynomial $f(t)=t^2+t+1$ has $f(0)=f(1)=1$ in the prime field ${\mathbb{F}}_2$. A polynomial of polynomial degree two is an irreducible polynomial exactly when it has no polynomial root in the coefficient field. Thus the quotient by an irreducible polynomial is a field construction applies. Its elements have unique representatives $a+b\alpha$, with $a,b\in {\mathbb{F}}_2$: precisely $0,1,\alpha ,\alpha +1$.

The defining relation is ${\alpha }^2+\alpha +1=0$. Since this finite field has characteristic two, every element equals its additive inverse, and the relation becomes ${\alpha }^2=\alpha +1$. Addition acts separately on the two coefficients:All coefficient calculations use modular arithmetic modulo two. For example, $(\alpha +1)+\alpha =1$ and $(\alpha +1)+(\alpha +1)=0$. Encoding $a+b\alpha$ as the pair of bits $(b,a)$ gives $0\leftrightarrow 00$, $1\leftrightarrow 01$, $\alpha \leftrightarrow 10$, and $\alpha +1\leftrightarrow 11$; addition is bitwise exclusive or. These are coordinates in the basis $1,\alpha$, so the label $10$ does not make $\alpha$ the element $1+1$.

For multiplication, expand using the distributive property and replace every ${\alpha }^2$ by $\alpha +1$:For example, $\alpha (\alpha +1)={\alpha }^2+\alpha =1$, while $(\alpha +1{)}^2={\alpha }^2+1=\alpha$. The following complete tables are computed from the coefficient rules. Colors identify results, rather than their numerical sizes. Every row of the addition table lists all four elements, and every nonzero row of the multiplication table does too: adding an element, or multiplying by a nonzero element, can be undone.

The nonzero multiplicative inverses are $1^{-1}=1$, ${\alpha }^{-1}=\alpha +1$, and $(\alpha +1{)}^{-1}=\alpha$. Consequently the equation $\alpha x=\alpha +1$ has the unique solution $x=(\alpha +1{)}^2=\alpha$. Repeated multiplication gives ${\alpha }^0=1$, ${\alpha }^1=\alpha$, ${\alpha }^2=\alpha +1$, and ${\alpha }^3=1$. They cycle through every element of the multiplicative group of a finite field.

Compare the ring $\mathbb{Z}/4\mathbb{Z}$, also with four elements. Its addition satisfies $1+1=2\ne 0$, and its multiplication satisfies $2^2=0$. Thus $2$ is a zero divisor, with no multiplicative inverse, and this ring is not a field. Multiplying by $2$ collapses different inputs to the same output: both $0$ and $2$ give $0$, while both $1$ and $3$ give $2$. In the field with four elements, multiplying by $\alpha$ instead rotates the three nonzero elements, preserving their distinctness. The figure makes visible why counting four elements alone does not specify their arithmetic.

The finite field ${\mathbb{F}}_9$ can be represented as ${\mathbb{F}}_3[X]/(X^2+1)$, since the quadratic has no root modulo three. Its elements are $a+b\beta$ for $a,b\in {\mathbb{F}}_3$, with ${\beta }^2=-1$. Multiplication reduces to $(a+b\beta )(c+d\beta )=(ac-bd)+(ad+bc)\beta$ modulo three.

Neglecting metric perturbations, a free massive scalar field fluctuation obeys $\delta {\chi }^{\prime \prime }+2\mathcal{H}\delta {\chi }^{\prime }-{\nabla }^2\delta \chi +a^2{m}^2\delta \chi =0$. Rescaling it by the scale factor removes the conformal-time friction and gives $v^{\prime \prime }-{\nabla }^{2}v+(a^2{m}^2-a^{\prime \prime }/a)v=0$. Growth of $v$ need not mean growth of the physical fluctuation $v/a$.

A field is formally real when $-1$ is not a finite sum of squares. The Artin-Schreier ordering criterion says this is equivalent to the existence of an ordering compatible with the field operations. Such a field has characteristic zero and embeds into a real closed field.

The double factorial multiplies the positive integers of the same parity as $n$ up to $n$: $(2m-1)!!=1\cdot 3\cdots (2m-1)$ and $(2m)!!=2\cdot 4\cdots 2m$. Set $(-1)!!=0!!=1$. In particular, $(2m-1)!!=(2m)!/(2^{m}m!)$, the number of pairings of $2m$ labelled objects.

For disjoint pairs $A_i,B_i$ such that $A_i\cap B_j\ne \varnothing$ whenever $i\ne j$, the displayed sum is at most one. In a uniformly random permutation, the events that all of $A_i$ precedes all of $B_i$ are disjoint events, and their probabilities are the summands. Taking $B_i$ to be the complement of the members $A_i$ of an antichain recovers the LYM inequality.

Here two words must both have symbols greater than one in some common coordinate, without requiring the symbols to agree. Their supports $S(x)$ therefore form an intersecting family. The fibre of a support of size $r$ contains $(k-1{)}^r$ words, relating this problem to a weighted set family problem.

If every cross-intersection of two set families on $[n]$ has even size, the spans of their indicator vectors are orthogonal in ${\mathbb{F}}_2^n$. The orthogonal complement over the binary field dimension formula bounds the sum of their dimensions by $n$, giving the displayed cardinality bound.

Appending a coordinate one to each indicator vector changes every odd cross-intersection dot product into an even one in ${\mathbb{F}}_2^{n+1}$. The argument for the even cross-intersection bound then gives the displayed bound. It is a valid bound without a claim that it is sharp.

For $i<j$, replace a member $A$ containing $j$ but not $i$ by $(A\setminus \{j\})\cup \{i\}$ only when the replacement is absent. An elementary set shift preserves family size and the intersecting family property. The inclusion $\partial (S_{ij}\mathcal{F})\subseteq S_{ij}(\partial \mathcal{F})$ shows that it cannot increase the lower shadow.

If distinct members of a set family on $[n]$ have intersection size $\lambda$, then its size is at most $n$ for $\lambda >0$, and at most $n+1$ for $\lambda =0$, assuming $n\ge 1$. In the positive case, use the Gram matrix of the characteristic vectors of sets; a member of size $\lambda$ is handled by disjoint remainders. In the zero case nonempty members are disjoint.

In a four-dimensional theory with unbroken $\mathcal{N}\ge 2$ supersymmetry, a full charged hypermultiplet contains $\mathcal{N}=1$ chiral superfields in $R$ and $\overline{R}$. Their net complex chiral gauge spectrum vanishes. The vector multiplet is in the real adjoint gauge group representation. Half-hypermultiplets in pseudoreal representations do not supply genuinely complex gauge chirality. Thus the chiral matter of the Standard Model can have at most unbroken $\mathcal{N}=1$ supersymmetry in four dimensions.

Modular exponentiation computes the remainder $a^n\mathrm{mod}m$ for an integer base $a$, a nonnegative integer exponent $n$, and, here, an integer modulus $m\ge 2$. It applies exponentiation by squaring while reducing after each multiplication, including each squaring. The result is the unique representative in $\{0,\dots ,m-1\}$. There is no requirement that the base be coprime to the modulus when the exponent is nonnegative.^[ref]

Take $a=2$, $n=13=(1101{)}_2$, and $m=17$. Initialize $r=2$; the remaining digits giveHence $2^{13}\mathrm{mod}17=15$. The final operation multiplies $16$ by $2$, since the preceding square has already been reduced. The computation never needs to construct $2^{13}=8192$.

In general reduce the base first, $x=a\mathrm{mod}m$. For $n>0$, initialize $r=x$ and process the remaining binary digits with $r\leftarrow r^2\mathrm{mod}m$, followed, for a digit one, by $r\leftarrow rx\mathrm{mod}m$. The loop invariant iswhere $p$ is the processed prefix value. The congruence is preserved because modular arithmetic respects multiplication: if $u\equiv v(\mathrm{mod}m)$, then $u^2\equiv v^2$ and $ux\equiv va(\mathrm{mod}m)$. Thus changing representatives never changes the final congruence class.

Both factors in every product are less than $m$, so the unreduced product is less than $m^2$, and the stored result is again less than $m$. The same bounds hold for a gigantic exponent, such as $n=2^{1000}$: there are $1000$ squarings, each on bounded-size residues. For base two, the corresponding ordinary power would have $2^{1000}+1$ binary digits. The benefit is avoiding the huge ordinary power altogether, rather than constructing it and taking one final remainder. Initial reduction is essential when the supplied base is much larger than the modulus.

For $n=0$, return $1$ under the same empty-product convention as exponentiation by squaring. For $n=-k<0$, require $\gcd (a,m)=1$, find the modular multiplicative inverse $a^{-1}\mathrm{mod}m$, and compute $(a^{-1}{)}^k\mathrm{mod}m$; the extended Euclidean algorithm can find that inverse. Its cost is separate from the positive-exponent binary powering multiplication count. For example, $2^{-1}\equiv 9(\mathrm{mod}17)$, while $2^{-1}\mathrm{mod}4$ is undefined because $2$ and $4$ are not coprime.

A reversible computation gives the coherent quantum modular exponentiation used in quantum order finding and discrete-logarithm Fourier sampling.

The time constant $\tau >0$ in exponential decay is the time over which the decaying quantity is multiplied by $e^{-1}$. A response approaching equilibrium as $q(t)=q_{\infty }+(q(0)-q_{\infty })e^{-t/\tau }$ has the same time constant. For a resistor-inductor circuit it is $L/R$.

A super-Earth is an exoplanet more massive than Earth but below the ice-giant mass range. The label does not guarantee a rocky exoplanet interior, a specific radius, or habitability. Radius-based usage can overlap the sub-Neptune population.

The Lagrangian pressure perturbation is the change in pressure experienced by a displaced fluid element: $\Delta p=\delta p+\mathbf{\xi }\cdot \nabla p_0$. For a free surface against fixed external pressure, $\Delta p=0$, rather than $\delta p=0$ at the original surface.

Moving a uniform-density boundary produces a delta-function sheet in the Eulerian mass density perturbation. For a slab displaced vertically by ${\eta }_{\pm }$, ${\rho }^{\prime }={\rho }_0{\eta }_{+}\delta (z-H)-{\rho }_0{\eta }_{-}\delta (z+H)$. Poisson equation then gives a continuous gravitational potential with derivative jumps fixed by those sheets.

An Eulerian fluid perturbation compares a perturbed field with its equilibrium value at one fixed spatial point. It differs from the perturbation following a displaced material element: $\Delta f=f_1+\mathbf{\xi }\cdot \nabla f$. Linearizing a static fluid velocity gives ${\mathbf{v}}_1={\partial }_t\mathbf{\xi }$.