A subset $A\subseteq S$ is closed under a binary operation on $S$ when the result of combining any two elements of $A$ is again in $A$. The even integers are closed under addition, but the odd integers are not: $1+3=4$ is even. Closure permits restricting the binary operation to $A\times A\to A$. It is different from being a closed set in topology.

An algebraic magma is a set equipped with a binary operation. No associative property, commutativity, or identity element is required. A semigroup adds the associative property; a monoid also adds an identity element.

An error exponent is the limiting exponential decay rate of an error probability $p_n$ as sample size $n$ grows. Different constraints on the type I error and type II error give different optimal rates. The Chernoff information describes the optimal average-error rate for simple Bayesian hypotheses; Stein's lemma gives the optimal type-II rate when type-I error is held below a fixed nontrivial bound.

A coordination game rewards compatible choices. Players can have a common interest in coordinating while having different preferred coordinated outcomes, so their interests need not be opposite as in a zero-sum game.

Both players choose heads $H$ or tails $T$ without seeing the other's choice. The row player receives $1$ when the choices match and $-1$ when they differ; the column player receives the opposite payoff. Thus matching pennies is a zero-sum game, with ordered payoff pairs:

At $(H,H)$ the column player improves by switching to $T$; at $(H,T)$ the row player improves by switching to $T$; at $(T,H)$ the row player improves by switching to $H$; at $(T,T)$ the column player improves by switching to $H$. Every pure strategy strategy profile therefore has a profitable unilateral deviation, so none is a Nash equilibrium.

Let $p$ be the row player's probability of heads and $q$ the column player's probability of heads. Their random choices are independent. Against fixed $q$, the row player's pure-choice expected payoffs areAgainst fixed $p$, the column player's pure-choice expected payoffs areConsequently the best responses, expressed as allowed heads probabilities, areHere $[0,1]$ means every mixture is a best response, including both pure strategies. If $q>1/2$, the row player must choose $p=1$, which makes the column player's best response $q=0$, a contradiction. If $q<1/2$, the row player must choose $p=0$, which forces $q=1$, also a contradiction. Hence $q=1/2$; for the column player to mix at this $q$, the two column payoffs must tie, forcing $p=1/2$. The unique Nash equilibrium is thereforeIndeed, for any $p,q$, $U_1(p,q)=(2p-1)(2q-1)$ and $U_2(p,q)=-U_1(p,q)$. Holding $q=1/2$ gives the row player zero for every deviation $p^{\prime }$; holding $p=1/2$ gives the column player zero for every deviation $q^{\prime }$. A player is indifferent against the equilibrium mixture, but choosing an arbitrary best response to it need not preserve the opponent's best response. This is why that indifference does not produce additional Nash equilibria.

For a teaching treatment with the same payoff convention, see MIT 17.810, lecture 3, slides 29–38, on mixed strategies and matching pennies.

If $I_m$ is an integer interval inside $[1,N]$, discrete Fourier inversion gives the displayed identity. The exponential geometric sum bound gives a common envelope $H_N(\theta )\ll \min (N,\Vert \theta {\Vert }^{-1})$, whose integral is $O(\log (2N))$. At each frequency, $\widehat{1_{I_m}}(\theta )/H_N(\theta )$ is a bounded coefficient depending only on $m$, while $e(n\theta )$ depends only on $n$. A uniform bounded-coefficient Type II sum estimate therefore survives an $m$-dependent interval cutoff at a cost of one logarithm. This is particularly useful for the condition $mn\le N$.

This method uses cancellation in Type I sums and Type II sums to control a sum weighted by the Von Mangoldt function. The Vaughan identity separates the short-factor contributions from bilinear ones. Fourier separation of interval cutoffs permits a hyperbolic product cutoff. Ben Green's 2007 notes, Sections 2.3–2.4 give a version adapted to binary digit sums.

A linear bijection between spaces carrying compatible bilinear forms, Hermitian forms or quadratic forms is an isometry of a space with a form when it preserves the specified form. Preserving the polar form alone need not preserve a quadratic form in characteristic two. Subspaces may have degenerate restricted forms even when the ambient form is nonsingular.

For a bilinear form $B$ on a finite-dimensional vector space $V$, the right bilinear orthogonal complement of a vector subspace $W$ isIt is a vector subspace by linearity in the second argument. For a general nonsymmetric bilinear form, left and right complements must be distinguished.

If $B$ is a nondegenerate bilinear form, the map $v\mapsto B(\cdot ,v)$ is a linear isomorphism from $V$ to its dual space. It identifies $W^{\perp }$ with the annihilator of a vector subspace $W$, givingIf $B$ is also a symmetric bilinear form or an alternating bilinear form, reversing a zero pairing preserves zero. Hence $W\subseteq (W^{\perp }{)}^{\perp }$, and the dimension formula gives equality. Nondegeneracy of the ambient form is required for this double-complement conclusion. For the zero bilinear form, every complement is all of $V$, so a proper $W$ cannot be recovered that way.

Even in a nondegenerate ambient space, the restricted form can be degenerate. For $B(x,y)=x_1{y}_1-x_2{y}_2$ on ${\mathbb{R}}^2$, the line $W=\mathrm{span}\{(1,1)\}$ equals $W^{\perp }$. In general, for a symmetric bilinear form or an alternating bilinear form, the restricted radical of a bilinear form is $W\cap W^{\perp }$; quotienting $W$ by this radical of a bilinear form gives a nondegenerate bilinear form.

Over a field whose characteristic is not two, a bilinear form has a unique decomposition into a symmetric bilinear form $S$ and an alternating bilinear form $K$:Adding and subtracting the equations $B(v,w)=S(v,w)+K(v,w)$ and $B(w,v)=S(v,w)-K(v,w)$ both proves uniqueness and gives these formulas. In matrix notation, $S$ has matrix $(A+A^T)/2$ and $K$ has skew-symmetric matrix $(A-A^T)/2$.^[ref]

The diagonal values $q(v)=B(v,v)$ alone do not determine a general bilinear form, because $K(v,v)=0$ for every vector. For the matrix of a bilinear form in the example above,Thus two different bilinear forms, represented by $A$ and its symmetric matrix part, have exactly the same quadratic form. To recover a bilinear form from its diagonal, one needs extra information such as the hypothesis that it is a symmetric bilinear form.

Choose an ordered basis $(e_1,\dots ,e_n)$ of a finite-dimensional vector space. The matrix of a bilinear form records all pairings of basis vectors, with $A_{ij}=B(e_i,e_j)$. Writing $v={\sum }_i{v}_i{e}_i$ and $w={\sum }_j{w}_j{e}_j$, separate linearity giveswhere the last expression uses the coordinate vectors of $v,w$. Conversely, every matrix $A$ defines a bilinear form by this formula.^[ref]

For a concrete bilinear form on ${\mathbb{R}}^2$, takeThen $B((1,0),(0,1))=2$ but $B((0,1),(1,0))=0$: a bilinear form need not be a symmetric bilinear form. For $v=(2,1{)}^T$ and $w=(-1,3{)}^T$, first compute $Aw=(5,3{)}^T$, then $v^{T}Aw=2\cdot 5+1\cdot 3=13$. Reversing the inputs gives $Av=(4,1{)}^T$ and $w^{T}Av=-4+3=-1$.

Under change of basis, let the columns of an invertible matrix $P$ be the new basis vectors in old coordinates. Thus $v=P{v}^{\prime }$ and $w=P{w}^{\prime }$, soThis is matrix congruence, rather than the matrix similarity $P^{-1}AP$ for an endomorphism. Both inputs have changed coordinates, which explains the two factors of $P$. For the displayed $A$ and $P=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$, direct multiplication gives $A^{\prime }=\left(\begin{array}{cc}1& 3\\ 1& 4\end{array}\right)$; its $(1,2)$ entry is $B(e_1,e_1+e_2)=3$.^[ref]

A uniform big O bound for a family of functions uses one multiplier and one threshold for every allowed parameter:The order of the quantifiers matters. For $f_{\alpha }(n)=\alpha n$, $0\le \alpha \le 10$ admits $C=10$ and $N=1$, while the unbounded family $\alpha >0$ admits no such constant. Each individual member nevertheless satisfies $f_{\alpha }=O(n)$. A uniform bound is useful when an approximation must work across a whole parameter range, rather than separately for each fixed parameter.

At the specified limit, big theta notation means that two positive constants give both-sided bounds:sufficiently near the limit. Equivalently, $f=O(g)$ and $g=O(f)$ in big O notation. In operation-count analysis, $3n^2+7n+4=\Theta (n^2)$ for $n\ge 1$, using $c=3$ and $C=14$. This supplies a tight scale without asserting that the normalized ratio tends to one.

For an example with no limiting ratio, let $f(n)=n(2+(-1{)}^n)$ for positive integers $n$. Then $n\le f(n)\le 3n$, so $f(n)=\Theta (n)$, but $f(n)/n$ alternates between one and three. A tight growth order is therefore weaker than asymptotic equivalence. The constant bounds also do not fix an exact factor by which a function changes when its argument doubles.

Donald E. Knuth's 1976 paper proposes using $\Theta$ for two-sided constant bounds in algorithm analysis. He credits Bob Tarjan and Mike Paterson with independently suggesting the notation. His discussion gives a concrete motivation for distinguishing an eventual upper estimate from a tight comparison. ^[ref]

When $g$ is nonzero sufficiently near the specified limit, little o notation meansThus the numerator becomes negligible relative to the denominator's scale. As $n\to \infty$, $n=o(n^2)$ because $n/n^2=1/n\to 0$; as $h\to 0$, $h^2=o(h)$ because $h^2/h=h\to 0$. The direction of the limit changes the comparison. ^[ref]

Equivalently, for every $\epsilon >0$, the bound $|f|\le \epsilon |g|$ holds sufficiently near that limit, with a threshold allowed to depend on $\epsilon$. Big O notation needs only one fixed multiplier, making it weaker. For example, $2n=O(n)$ but $2n$ is not $o(n)$. The useful relation $f\sim g⟺f-g=o(g)$ turns asymptotic equivalence into a statement about a negligible remainder.

The Bhattacharyya distance is the negative natural logarithm of the Bhattacharyya coefficient, with $-\ln 0=\infty$. It is zero for identical probability distributions and infinite for disjoint probability supports. Despite the name, it need not satisfy the triangle inequality required of a metric.

An antiparallel beta sheet has neighboring beta strands running in opposite N→C sequence directions. This specifies strand orientation rather than whether the beta strands belong to one polypeptide chain or several.

A parallel beta sheet has neighboring beta strands running in the same direction along their amino acid residue sequences, from the N-terminus toward the C-terminus.

A beta strand is an extended segment of protein backbone in the geometry characteristic of a beta sheet. Several beta strands can associate through backbone hydrogen bonds. A beta strand need not be an entire polypeptide chain.

For a causal autoregressive process with mean $m$ and innovations orthogonal to the observed past, define ${\hat{X}}_{T,h}=X_{T+h}$ for $h\le 0$ and replace future innovations by zero. The displayed recursion is its best linear prediction from a finite past when at least the last $p$ observations are available. Future innovation sums are orthogonal to those observations, which proves the projection property. Mere white noise does not guarantee that a linear forecast equals a conditional mean: that stronger interpretation requires martingale-difference or independent innovations.

The regular power series has $J_0(\rho )=1-{\rho }^2/4+{\rho }^4/64-\cdots$. Its quartic truncation has a double zero at $\rho =\sqrt{8}$, giving a rough disc-well energy estimate $4{\hslash }^2/(\mu a^2)$. A truncation zero is not an exact Bessel zero or a rigorous eigenvalue bound. The double-root behavior particularly warns against treating a low-order local expansion as accurate far from its center.