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A directed walk is closed when its first and last vertices agree.

For every subspace $U$ of a finite-dimensional vector space $V$, extending a basis of $U$ to a basis of $V$ produces a subspace $W$ such that $V=U\oplus W$.

For finite-dimensional subspaces $U,W$ of a common vector space,

If $x$ is a non-zero-divisor in a finite-dimensional Noetherian ring $A$, thenEvery prime chain above $(x)$ can be extended downward by a minimal prime of $A$, because a non-zero-divisor belongs to no minimal prime.

The digamma function is strictly increasing on the positive real axis, whileIt therefore has exactly one positive zero, and that zero lies in $(1,2)$.

The trigamma function is ${\psi }^{\prime }(z)$. For real $z>0$,

When the permeabilities agree, the signed reflected-to-incident electric-field ratio for polarization normal to the plane of incidence isFor incidence from lower to higher refractive index it never vanishes.

Equality of the boundary fields for every tangential position and time forces equal frequency and equal tangential wavevector for incident, reflected, and transmitted plane waves.

The off-diagonal recursionwith $R(1,t)=R(s,1)=1$ givesIn particular, $R(t)\le \left(\genfrac{}{}{0ex}{}{2t-2}{t-1}\right)<2^{2t}$.

Increasing the baryon-to-photon ratio raises the equilibrium deuterium abundance at fixed temperature, so the deuterium bottleneck ends earlier. Fewer neutrons decay before nuclear burning, and neutron-limited helium synthesis produces a larger primordial helium mass fraction if weak freeze-out is unchanged.

For diagonal entries $a_i$, superdiagonal entries $b_i$, and subdiagonal entries $c_i$, the leading principal determinants satisfy

For an $n$ by $n$ matrix,

The determinant is linear in each row and in each column when all the others are held fixed.

For an ordinal $\xi$,Thus the derivative index of $\omega$ is one and that of ${\omega }^2$ is two.

The chain rule calculates the derivative of a function composition by combining local rates. If the inner function $f$ is differentiable at $a$ and the outer function $g$ is differentiable at $f(a)$, thenThe evaluation point matters: the outer derivative is evaluated at the inner output $f(a)$, not at $a$. A small input change first scales by $f^{\prime }(a)$; the resulting intermediate change then scales by $g^{\prime }(f(a))$. In terms of linear approximations, composing those two local descriptions multiplies their scale factors. ^[ref].

For example, $h(x)=\sin (x^2)$ applies a quadratic function and then the sine. With $f(x)=x^2$ and $g(u)=\sin u$, the chain rule givesHere all angles are in radians. Near $x=1$, increasing the input by $\Delta x$ changes $x^2$ by about $2\Delta x$, and the sine changes by about $\cos 1$ times that intermediate change. For $\Delta x=0.01$, the linear approximation predicts an output increase of $0.0108060$; the exact increase is $\sin (1.0201)-\sin 1\approx 0.0106894$.

The same rule handles several nested stages. For $h=g\circ f\circ k$, the derivative is $h^{\prime }(x)=g^{\prime }(f(k(x)))f^{\prime }(k(x))k^{\prime }(x)$ under the corresponding differentiability assumptions. Writing each intermediate output prevents evaluating a derivative at the wrong stage. It also makes the chain rule useful for propagating sensitivities through a mathematical model, where one measured quantity influences another through several transformations.

For a quick proof idea, replace $f$ near $a$ by its linear approximation, then replace $g$ near $f(a)$ by its linear approximation. The composite leading term is their product; differentiability ensures the remaining error is smaller than the original input change. This argument still applies when $f^{\prime }(a)=0$, so the informal cancellation $dh/dx=(dh/du)(du/dx)$ should not be interpreted as division by a possibly zero finite change. The hypotheses are sufficient, but are not necessary for every composite to be differentiable, as the nondifferentiable inner function with a differentiable nondegenerate composite example shows.

If $f$ and $g$ are differentiable and $g\ne 0$, then

If $f$ and $g$ are differentiable at $a$, then

The completion form of dependent random choice starts with $t=k-O(\log k)$ vertices having at least $c{2}^{-t}n$ common neighbours in one colour. Reapplying common-neighbourhood sampling inside the remaining polynomial-size set either completes these vertices to a monochromatic $K_{k,k}$ in that colour or produces a $K_{k,k}$ in the other colour. It yields

If a bipartite graph with parts $A,B$ has density at least $p$, then averaging ordered distinct $t$-tuples gives a common neighbourhood of size at leastIn particular this is at least $p^t|B|/2$ when $|A|$ is sufficiently large compared with $t^2/p$.

If a bipartite graph $H$ has $k$ vertices and maximum degree $d$, thenA dependent random choice argument finds, in one colour, enough vertices whose every subset of at most $d$ vertices has a large common neighbourhood; a greedy embedding then places $H$.