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Let g {\displaystyle {\mathfrak {g}}} be any semisimple Lie algebra. To specify a Lie bialgebra structure we thus need to specify a compatible Lie algebra structure on the dual vector space. Choose a Cartan subalgebra t ⊂ g {\displaystyle {\mathfrak {t}}\subset {\mathfrak {g}}} and a choice of positive roots. Let b ± ⊂...
Wikipedia - Lie bialgebra
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Then define a Lie algebra g ′ := { ( X − , X + ) ∈ b − × b + | π ( X − ) + π ( X + ) = 0 } {\displaystyle {\mathfrak {g'}}:=\{(X_{-},X_{+})\in {\mathfrak {b}}_{-}\times {\mathfrak {b}}_{+}\ {\bigl \vert }\ \pi (X_{-})+\pi (X_{+})=0\}} which is a subalgebra of the product b − × b + {\displaystyle {\mathfrak {b}}_{-}\tim...
Wikipedia - Lie bialgebra
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Let g {\displaystyle {\mathfrak {g}}} be the set of matrices on the form X = ( 0 θ x − θ 0 y 0 0 0 ) , θ , x , y ∈ R . {\displaystyle X=\left({\begin{matrix}0&\theta &x\\-\theta &0&y\\0&0&0\end{matrix}}\right),\quad \theta ,x,y\in \mathbb {R} .} Then g {\displaystyle {\mathfrak {g}}} is solvable, but not split solvable...
Wikipedia - Solvable Lie group
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Let h 0 ∗ {\displaystyle {\mathfrak {h}}_{0}^{*}} be the real subspace of h ∗ {\displaystyle {\mathfrak {h}}^{*}} generated by the roots of g {\displaystyle {\mathfrak {g}}} , where h ∗ {\displaystyle {\mathfrak {h}}^{*}} is the space of linear functionals λ: h → C {\displaystyle \lambda :{\mathfrak {h}}\to \mathbb {C}...
Wikipedia - Highest-weight module
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The motivation for these definitions is simple: The weights of finite-dimensional representations of g {\displaystyle {\mathfrak {g}}} satisfy the first integrality condition, while if G is a group with Lie algebra g {\displaystyle {\mathfrak {g}}} , the weights of finite-dimensional representations of G satisfy the se...
Wikipedia - Highest-weight module
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By elementary results for s l ( 2 , C ) {\displaystyle sl(2,\mathbb {C} )} , the eigenvalues of H α {\displaystyle H_{\alpha }} in any finite-dimensional representation must be an integer. We conclude that, as stated above, the weight of any finite-dimensional representation of g {\displaystyle {\mathfrak {g}}} is alge...
Wikipedia - Highest-weight module
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An element λ {\displaystyle \lambda } is then algebraically integral if and only if it is an integral combination of the fundamental weights. The set of all g {\displaystyle {\mathfrak {g}}} -integral weights is a lattice in h 0 {\displaystyle {\mathfrak {h}}_{0}} called the weight lattice for g {\displaystyle {\mathfr...
Wikipedia - Highest-weight module
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There are two simple roots, γ 1 {\displaystyle \gamma _{1}} and γ 2 {\displaystyle \gamma _{2}} . The first fundamental weight, ω 1 {\displaystyle \omega _{1}} , should be orthogonal to γ 2 {\displaystyle \gamma _{2}} and should project orthogonally to half of γ 1 {\displaystyle \gamma _{1}} , and similarly for ω 2 {\d...
Wikipedia - Highest-weight module
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The weight lattice is then the triangular lattice. Suppose now that the Lie algebra g {\displaystyle {\mathfrak {g}}} is the Lie algebra of a Lie group G. Then we say that λ ∈ h 0 {\displaystyle \lambda \in {\mathfrak {h}}_{0}} is analytically integral (G-integral) if for each t in h {\displaystyle {\mathfrak {h}}} suc...
Wikipedia - Highest-weight module
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For G semisimple, the set of all G-integral weights is a sublattice P(G) ⊂ P( g {\displaystyle {\mathfrak {g}}} ). If G is simply connected, then P(G) = P( g {\displaystyle {\mathfrak {g}}} ). If G is not simply connected, then the lattice P(G) is smaller than P( g {\displaystyle {\mathfrak {g}}} ) and their quotient i...
Wikipedia - Highest-weight module
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Let i: A → A {\displaystyle i:A\to A} be an anti-automorphism of R {\displaystyle R} -algebras with i 2 = id {\displaystyle i^{2}=\operatorname {id} } (just called "involution" from now on). A cell ideal of A {\displaystyle A} w.r.t. i {\displaystyle i} is a two-sided ideal J ⊆ A {\displaystyle J\subseteq A} such that ...
Wikipedia - Cellular algebra
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i {\displaystyle i} is defined as a direct decomposition A = ⨁ k = 1 m U k {\displaystyle A=\bigoplus _{k=1}^{m}U_{k}} into free R {\displaystyle R} -submodules such that i ( U k ) = U k {\displaystyle i(U_{k})=U_{k}} J k := ⨁ j = 1 k U j {\displaystyle J_{k}:=\bigoplus _{j=1}^{k}U_{j}} is a two-sided ideal of A {\disp...
Wikipedia - Cellular algebra
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Let i: X → Y be a (closed) regular embedding of codimension d, Y' → Y a morphism and i': X' = X ×Y Y' → Y' the induced map. Let N be the pullback of the normal bundle of i to X'. Then the refined Gysin homomorphism i! refers to the composition i !
Wikipedia - Gysin homomorphism
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: A k ( Y ′ ) ⟶ σ A k ( N ) ⟶ Gysin A k − d ( X ′ ) {\displaystyle i^{! }:A_{k}(Y'){\overset {\sigma }{\longrightarrow }}A_{k}(N){\overset {\text{Gysin}}{\longrightarrow }}A_{k-d}(X')} where σ is the specialization homomorphism; which sends a k-dimensional subvariety V to the normal cone to the intersection of V and X'...
Wikipedia - Gysin homomorphism
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encodes intersection product in intersection theory in that one either shows the intersection product of X and V to be given by the formula X ⋅ V = i ! , {\displaystyle X\cdot V=i^{!
Wikipedia - Gysin homomorphism
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},} or takes this formula as a definition.Example: Given a vector bundle E, let s: X → E be a section of E. Then, when s is a regular section, s ! {\displaystyle s^{!}} is the class of the zero-locus of s, where is the fundamental class of X.
Wikipedia - Gysin homomorphism
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Let k = C, and A2 be the two-dimensional affine space over C. Polynomials in the ring C can be viewed as complex valued functions on A2 by evaluating at the points in A2. Let subset S of C contain a single element f (x, y): f ( x , y ) = x + y − 1. {\displaystyle f(x,y)=x+y-1.} The zero-locus of f (x, y) is the set of ...
Wikipedia - Abstract algebraic variety
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(In the classical topology coming from the topology on the complex numbers, a complex line is a real manifold of dimension two.) This is the set Z( f ): Z ( f ) = { ( x , 1 − x ) ∈ C 2 } . {\displaystyle Z(f)=\{(x,1-x)\in \mathbf {C} ^{2}\}.}
Wikipedia - Abstract algebraic variety
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Thus the subset V = Z( f ) of A2 is an algebraic set. The set V is not empty. It is irreducible, as it cannot be written as the union of two proper algebraic subsets. Thus it is an affine algebraic variety.
Wikipedia - Abstract algebraic variety
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Let k = C, and A2 be the two-dimensional affine space over C. Polynomials in the ring C can be viewed as complex valued functions on A2 by evaluating at the points in A2. Let subset S of C contain a single element g(x, y): g ( x , y ) = x 2 + y 2 − 1. {\displaystyle g(x,y)=x^{2}+y^{2}-1.} The zero-locus of g(x, y) is t...
Wikipedia - Differential algebraic variety
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Let k > 1 {\displaystyle k>1} be an integer parameter. Geometric grouping proceeds in two steps: Partition the instance J {\displaystyle J} into several instances J 0 , J 1 , … {\displaystyle J_{0},J_{1},\ldots } such that, in each instance J r {\displaystyle J_{r}} , all sizes are in the interval [ B / 2 r + 1 , B / 2...
Wikipedia - Karmarkar–Karp bin packing algorithms
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Let K r , K r ′ {\displaystyle K_{r},K'_{r}} be the resulting instances. Let K := ∪ r K r {\displaystyle K:=\cup _{r}K_{r}} and K ′ := ∪ r K r ′ {\displaystyle K':=\cup _{r}K'_{r}} .Then, the number of different sizes is bounded as follows: For all r, m ( K r ′ ) = 1 {\displaystyle m(K'_{r})=1} and m ( K r ) ≤ n ( J r ...
Wikipedia - Karmarkar–Karp bin packing algorithms
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Summing over all r gives m ( K ) ≤ 2 F O P T ( J ) / k + log 2 ⁡ ( 1 / g ) {\displaystyle m(K)\leq 2FOPT(J)/k+\log _{2}(1/g)} .The number of bins is bounded as follows: For all r, O P T ( K r ′ ) ≤ k {\displaystyle OPT(K'_{r})\leq k} - since K r ′ {\displaystyle K'_{r}} has k ⋅ 2 r {\displaystyle k\cdot 2^{r}} items, a...
Wikipedia - Karmarkar–Karp bin packing algorithms
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Let k > 1 {\displaystyle k>1} be an integer parameter. Order the items by descending size. Partition them into groups such that the total size in each group is at least k ⋅ B {\displaystyle k\cdot B} . Since the size of each item is less than B, The number of items in each group is at least k + 1 {\displaystyle k+1} .
Wikipedia - Karmarkar–Karp bin packing algorithms
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The number of items in each group is weakly-increasing. If all items are larger than g ⋅ B {\displaystyle g\cdot B} , then the number of items in each group is at most k / g {\displaystyle k/g} .
Wikipedia - Karmarkar–Karp bin packing algorithms
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In each group, only the larger items are rounded up. This can be done such that: m ( K ) ≤ F O P T ( J ) / k + ln ⁡ ( 1 / g ) {\displaystyle m(K)\leq FOPT(J)/k+\ln(1/g)} . O P T ( J ) ≤ O P T ( K ) + 2 k ⋅ ( 2 + ln ⁡ ( 1 / g ) ) {\displaystyle OPT(J)\leq OPT(K)+2k\cdot (2+\ln(1/g))} .
Wikipedia - Karmarkar–Karp bin packing algorithms
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Let k > 1 {\displaystyle k>1} be an integer parameter. Put the largest k {\displaystyle k} items in group 1; the next-largest k {\displaystyle k} items in group 2; and so on (the last group might have fewer than k {\displaystyle k} items). Let J {\displaystyle J} be the original instance. Let K ′ {\displaystyle K'} be ...
Wikipedia - Karmarkar–Karp bin packing algorithms
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Then: In K ′ {\displaystyle K'} all items have the same size. In K {\displaystyle K} the number of different sizes is m ( K ) ≤ n / k + 1 {\displaystyle m(K)\leq n/k+1} . O P T ( K ) ≤ O P T ( J ) {\displaystyle OPT(K)\leq OPT(J)} - since group 1 in J {\displaystyle J} dominates group 2 in K {\displaystyle K} (all k it...
Wikipedia - Karmarkar–Karp bin packing algorithms
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Let k be a commutative ring and let λ {\displaystyle \lambda } be given. A linear operator R on a k-algebra A is called a Rota–Baxter operator of weight λ {\displaystyle \lambda } if it satisfies the Rota–Baxter relation of weight λ {\displaystyle \lambda }: R ( x ) R ( y ) = R ( R ( x ) y ) + R ( x R ( y ) ) + λ R ( x...
Wikipedia - Rota–Baxter algebra
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The terms Baxter operator algebra and Baxter algebra are also used. Let R {\displaystyle R} be a Rota–Baxter of weight λ {\displaystyle \lambda } . Then − λ I d − R {\displaystyle -\lambda Id-R} is also a Rota–Baxter operator of weight λ {\displaystyle \lambda } . Further, for μ {\displaystyle \mu } in k, μ R {\display...
Wikipedia - Rota–Baxter algebra
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Let k be a field, A an associative k-algebra, and M an A-bimodule. The enveloping algebra of A is the tensor product A e = A ⊗ A o {\displaystyle A^{e}=A\otimes A^{o}} of A with its opposite algebra. Bimodules over A are essentially the same as modules over the enveloping algebra of A, so in particular A and M can be c...
Wikipedia - Hochschild cohomology
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Let k be a field, and let n be a positive integer. Since the polynomial ring k is a unique factorization domain, the divisor class group of affine space An over k is equal to zero. Since projective space Pn over k minus a hyperplane H is isomorphic to An, it follows that the divisor class group of Pn is generated by th...
Wikipedia - Divisor class
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{\displaystyle \operatorname {Pic} _{X/k}^{0}.} For k of characteristic zero, Pic X / k 0 {\displaystyle \operatorname {Pic} _{X/k}^{0}} is an abelian variety, the Picard variety of X.For R the ring of integers of a number field, the divisor class group Cl(R) := Cl(Spec R) is also called the ideal class group of R. It ...
Wikipedia - Divisor class
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Then the line D in X defined by x = z = 0 is not principal on X near the origin. Note that D can be defined as a set by one equation on X, namely x = 0; but the function x on X vanishes to order 2 along D, and so we only find that 2D is Cartier (as defined below) on X. In fact, the divisor class group Cl(X) is isomorph...
Wikipedia - Divisor class
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Let k be an algebraically closed field and G a linear algebraic group (that is, affine algebraic group) over k. By definition, Lie(G) is the Lie algebra of all derivations of k that commute with the left action of G. As in the Lie group case, it can be identified with the tangent space to G at the identity element.
Wikipedia - Distribution on a linear algebraic group
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Let k be an algebraically closed field and let Pn be the projective n-space over k. Let f in k be a homogeneous polynomial of degree d. It is not well-defined to evaluate f on points in Pn in homogeneous coordinates. However, because f is homogeneous, meaning that f (λx0, ..., λxn) = λd f (x0, ..., xn), it does make se...
Wikipedia - Complex variety
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{\displaystyle Z(S)=\{x\in \mathbf {P} ^{n}\mid f(x)=0{\text{ for all }}f\in S\}.} A subset V of Pn is called a projective algebraic set if V = Z(S) for some S.: 9 An irreducible projective algebraic set is called a projective variety. : 10 Projective varieties are also equipped with the Zariski topology by declaring a...
Wikipedia - Complex variety
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Given a subset V of Pn, let I(V) be the ideal generated by all homogeneous polynomials vanishing on V. For any projective algebraic set V, the coordinate ring of V is the quotient of the polynomial ring by this ideal. : 10 A quasi-projective variety is a Zariski open subset of a projective variety. Notice that every af...
Wikipedia - Complex variety
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Let k be an algebraically closed field, and V be a finite-dimensional vector space over k. The symmetric algebra of the dual vector space V* is called the polynomial ring on V and denoted by k. It is a naturally graded algebra by the degree of polynomials. The projective Nullstellensatz states that, for any homogeneous...
Wikipedia - Algebraic geometry of projective spaces
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Let k be defined as an element in F, the array of Fibonacci numbers. n = Fm is the array size. If n is not a Fibonacci number, let Fm be the smallest number in F that is greater than n. The array of Fibonacci numbers is defined where Fk+2 = Fk+1 + Fk, when k ≥ 0, F1 = 1, and F0 = 1. To test whether an item is in the li...
Wikipedia - Fibonacci search technique
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There is no match; the item is not in the array. Compare the item against element in Fk−1. If the item matches, stop.
Wikipedia - Fibonacci search technique
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If the item is less than entry Fk−1, discard the elements from positions Fk−1 + 1 to n. Set k = k − 1 and return to step 2. If the item is greater than entry Fk−1, discard the elements from positions 1 to Fk−1. Renumber the remaining elements from 1 to Fk−2, set k = k − 2, and return to step 2.Alternative implementatio...
Wikipedia - Fibonacci search technique
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If K < Ki, go to Step 3; if K > Ki go to Step 4; and if K = Ki, the algorithm terminates successfully. Step 3. If q=0, the algorithm terminates unsuccessfully.
Wikipedia - Fibonacci search technique
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Otherwise set (i, p, q) ← (i - q, q, p - q) (which moves p and q one position back in the Fibonacci sequence); then return to Step 2 Step 4. If p=1, the algorithm terminates unsuccessfully. Otherwise set (i,p,q) ← (i + q, p - q, 2q - p) (which moves p and q two positions back in the Fibonacci sequence); and return to S...
Wikipedia - Fibonacci search technique
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Let k {\displaystyle k} be a field and let g {\displaystyle {\mathfrak {g}}} be a finite-dimensional Lie algebra over k {\displaystyle k} . There exists a unique maximal solvable ideal, called the radical, for the following reason. Firstly let a {\displaystyle {\mathfrak {a}}} and b {\displaystyle {\mathfrak {b}}} be t...
Wikipedia - Radical of a Lie algebra
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Then a + b {\displaystyle {\mathfrak {a}}+{\mathfrak {b}}} is again an ideal of g {\displaystyle {\mathfrak {g}}} , and it is solvable because it is an extension of ( a + b ) / a ≃ b / ( a ∩ b ) {\displaystyle ({\mathfrak {a}}+{\mathfrak {b}})/{\mathfrak {a}}\simeq {\mathfrak {b}}/({\mathfrak {a}}\cap {\mathfrak {b}})}...
Wikipedia - Radical of a Lie algebra
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Let k {\displaystyle k} be a field of characteristic zero (using Kambayashi's theorem one can reduce most results to the case k = C {\displaystyle k=\mathbb {C} } ) and let A = k {\displaystyle A=k} be a polynomial algebra.
Wikipedia - Locally nilpotent derivation
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Let k {\displaystyle k} be a field with a primitive cube root of unity. Let D {\displaystyle {\mathfrak {D}}} be the following subset of the projective plane P k 2 {\displaystyle {\textbf {P}}_{k}^{2}}: D = { , , } ⊔ { | a 3 = b 3 = c 3 } . {\displaystyle {\mathfrak {D}}=\{,,\}\sqcup \{{\big |}a^{3}=b^{3}=c^{3}\}.}...
Wikipedia - Sklyanin algebra
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Let l1 = and l2 = be a pair of distinct lines. Then the intersection of lines l1 and l2 is point a P = (x0, y0, z0) that is the simultaneous solution (up to a scalar factor) of the system of linear equations: a1x + b1y + c1z = 0 and a2x + b2y + c2z = 0.The solution of this system gives: x0 = b1c2 - b2c1, y0 = a2c1 - ...
Wikipedia - Incidence (geometry)
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Let m {\displaystyle m} be a message to be encrypted where 0 ≤ m < n {\displaystyle 0\leq m
Wikipedia - Paillier cryptosystem
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Let n 0 {\displaystyle n_{0}} be the source node on the Control-flow graph. The dominators of a node n {\displaystyle n} are given by the maximal solution to the following data-flow equations: Dom ⁡ ( n ) = { { n } if n = n 0 { n } ∪ ( ⋂ p ∈ preds ( n ) Dom ⁡ ( p ) ) if n ≠ n 0 {\displaystyle \operatorname {Dom} (n)={\...
Wikipedia - Dominator (graph theory)
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The node n {\displaystyle n} is also in the set of dominators for n {\displaystyle n} . An algorithm for the direct solution is: // dominator of the start node is the start itself Dom(n0) = {n0} // for all other nodes, set all nodes as the dominators for each n in N - {n0} Dom(n) = N; // iteratively eliminate nodes tha...
Wikipedia - Dominator (graph theory)
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Let n B z {\displaystyle nB_{z}} be the translated version of n B {\displaystyle nB} to the point z, that is, n B z = { x ∈ E | x − z ∈ n B } {\displaystyle nB_{z}=\{x\in E|x-z\in nB\}} . A shape n B z {\displaystyle nB_{z}} centered at z is called a maximal disk in a set A when: n B z ∈ A {\displaystyle nB_{z}\in A} ,...
Wikipedia - Morphological skeleton
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Let n be a nonnegative integer. The (analytical model of, or canonical) oriented (real) projective space or (canonical) two-sided projective space T n {\displaystyle \mathbb {T} ^{n}} is defined as T n = { { λ Z: λ ∈ R > 0 }: Z ∈ R n + 1 ∖ { 0 } } = { R > 0 Z: Z ∈ R n + 1 ∖ { 0 } } . {\displaystyle \mathbb {T} ^{n}=\{\...
Wikipedia - Oriented projective geometry
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Let n be a nonnegative integer. The oriented complex projective space C P S 1 n {\displaystyle {\mathbb {CP} }_{S^{1}}^{n}} is defined as C P S 1 n = { { λ Z: λ ∈ R > 0 }: Z ∈ C n + 1 ∖ { 0 } } = { R > 0 Z: Z ∈ C n + 1 ∖ { 0 } } {\displaystyle {\mathbb {CP} }_{S^{1}}^{n}=\{\{\lambda Z:\lambda \in \mathbb {R} _{>0}\}:Z\...
Wikipedia - Oriented projective geometry
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Let n be a positive integer such that n is odd, ( b 2 + 4 c n ) = − 1 {\displaystyle \left({\frac {b^{2}+4c}{n}}\right)=-1} and ( − c n ) = 1 {\displaystyle \left({\frac {-c}{n}}\right)=1} , where ( x n ) {\displaystyle \left({\frac {x}{n}}\right)} denotes the Jacobi symbol. Set B = 50000 {\displaystyle B=50000} . Then...
Wikipedia - Quadratic Frobenius test
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(2) Test whether n ∈ Z {\displaystyle {\sqrt {n}}\in \mathbb {Z} } . If yes, then stop as n is composite. (3) Compute x n + 1 2 mod ( n , x 2 − b x − c ) {\displaystyle x^{n+1 \over 2}\,{\bmod {\,}}{\big (}n,x^{2}-bx-c)} .
Wikipedia - Quadratic Frobenius test
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If x n + 1 2 ∉ Z / n Z {\displaystyle x^{n+1 \over 2}\notin \mathbb {Z} {\big /}n\mathbb {Z} } then stop as n is composite. (4) Compute x n + 1 mod ( n , x 2 − b x − c ) {\displaystyle x^{n+1}\,{\bmod {\,}}{\big (}n,x^{2}-bx-c)} . If x n + 1 ≢ − c {\displaystyle x^{n+1}\not \equiv -c} then stop as n is composite.
Wikipedia - Quadratic Frobenius test
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(5) Let n 2 − 1 = 2 r s {\displaystyle n^{2}-1=2^{r}s} with s odd. If x s ≢ 1 mod ( n , x 2 − b x − c ) {\displaystyle x^{s}\not \equiv 1{\bmod {\,}}{\big (}n,x^{2}-bx-c)} , and x 2 j s ≢ − 1 mod ( n , x 2 − b x − c ) {\displaystyle x^{2^{j}s}\not \equiv -1{\bmod {\,}}{\big (}n,x^{2}-bx-c)} for all 0 ≤ j ≤ r − 2 {\disp...
Wikipedia - Quadratic Frobenius test
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Let n denote the input size, B = B(n) denote the total number of qubits in the circuit (inputs, ancillary, output and postselection qubits), and G = G(n) denote the total number of gates. Represent the ith gate by its transition matrix Ai (a real unitary 2 B × 2 B {\displaystyle 2^{B}\times 2^{B}} matrix) and let the i...
Wikipedia - PostBQP
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S0) to be the set of basis states corresponding to P = 1, Q = 1 (resp. P = 1, Q = 0) and define the probabilities π 1 := Pr = ∑ ω ∈ S 1 Ψ ω 2 {\displaystyle \pi _{1}:={\text{Pr}}=\sum _{\omega \in S_{1}}\Psi _{\omega }^{2}} π 0 := Pr = ∑ ω ∈ S 0 Ψ ω 2 . {\displaystyle \pi _{0}:={\text{Pr}}=\sum _{\omega \in S_{0}}\Ps...
Wikipedia - PostBQP
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Let n {\displaystyle n} be a large RSA composite and let p 1 = 2 , p 2 = 3 , … {\displaystyle p_{1}=2,p_{2}=3,\ldots } the sequence of primes. Let k {\displaystyle k} , the block length, be the largest integer such that ∏ i = 1 k p i < n {\displaystyle \textstyle \prod _{i=1}^{k}p_{i}
Wikipedia - Very smooth hash
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Let n {\displaystyle n} be the number of vertices in the graph and let m {\displaystyle m} be the number of edges. Using big O notation, in his original publication Leighton states the complexity of RLF to be O ( n 3 ) {\displaystyle {\mathcal {O}}(n^{3})} ; however, this can be improved upon. Much of the expense of th...
Wikipedia - Recursive largest first algorithm
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These calculations can be performed in O ( m ) {\displaystyle {\mathcal {O}}(m)} time, meaning that the overall complexity of RLF is O ( m n ) {\displaystyle {\mathcal {O}}(mn)} .If the heuristics of Step 2 are replaced with random selection, then the complexity of this algorithm reduces to O ( n + m ) {\displaystyle {...
Wikipedia - Recursive largest first algorithm
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Let n {\displaystyle n} be | V | {\displaystyle |V|} , the number of vertices. To find all n 2 {\displaystyle n^{2}} of s h o r t e s t P a t h ( i , j , k ) {\displaystyle \mathrm {shortestPath} (i,j,k)} (for all i {\displaystyle i} and j {\displaystyle j} ) from those of s h o r t e s t P a t h ( i , j , k − 1 ) {\di...
Wikipedia - Floyd–Warshall algorithm
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Let n ∈ N {\displaystyle n\in \mathbb {N} } . It is required to prove that d d x x n = n x n − 1 . {\displaystyle {\frac {d}{dx}}x^{n}=nx^{n-1}.} The base case may be when n = 0 {\displaystyle n=0} or n = 1 {\displaystyle n=1} , depending on how the set of natural numbers is defined.
Wikipedia - Power rule
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When n = 0 {\displaystyle n=0} , d d x x 0 = d d x ( 1 ) = lim h → 0 1 − 1 h = lim h → 0 0 h = 0 = 0 x 0 − 1 . {\displaystyle {\frac {d}{dx}}x^{0}={\frac {d}{dx}}(1)=\lim _{h\to 0}{\frac {1-1}{h}}=\lim _{h\to 0}{\frac {0}{h}}=0=0x^{0-1}.} When n = 1 {\displaystyle n=1} , d d x x 1 = lim h → 0 ( x + h ) − x h = lim h → ...
Wikipedia - Power rule
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{\displaystyle {\frac {d}{dx}}x^{1}=\lim _{h\to 0}{\frac {(x+h)-x}{h}}=\lim _{h\to 0}{\frac {h}{h}}=1=1x^{1-1}.} Therefore, the base case holds either way. Suppose the statement holds for some natural number k, i.e. d d x x k = k x k − 1 . {\displaystyle {\frac {d}{dx}}x^{k}=kx^{k-1}.} When n = k + 1 {\displaystyle n=k...
Wikipedia - Power rule
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Let n ≥ 1 {\displaystyle n\geq 1} be an integer, X 1 , … , X n {\displaystyle X_{1},\ldots ,X_{n}} be TVSs (not necessarily locally convex), let Y {\displaystyle Y} be a locally convex TVS whose topology is determined by a family Q {\displaystyle {\mathcal {Q}}} of continuous seminorms, and let M: ∏ i = 1 n X i → Y {\d...
Wikipedia - Locally convex
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Let n ≥ 2 and H {\displaystyle {\mathcal {H}}} be a separable Hilbert space. Consider the C*-algebra A {\displaystyle {\mathcal {A}}} generated by a set { S i } i = 1 n {\displaystyle \{S_{i}\}_{i=1}^{n}} of isometries (i.e. S i ∗ S i = 1 {\displaystyle S_{i}^{*}S_{i}=1} ) acting on H {\displaystyle {\mathcal {H}}} sat...
Wikipedia - Cuntz algebra
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A simple C*-algebra is said to be purely infinite if every hereditary C*-subalgebra of it is infinite. O n {\displaystyle {\mathcal {O}}_{n}} is a separable, simple, purely infinite C*-algebra. Any simple infinite C*-algebra contains a subalgebra that has O n {\displaystyle {\mathcal {O}}_{n}} as a quotient.
Wikipedia - Cuntz algebra
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Let n ≥ 2 k {\displaystyle n\geq 2k} be integer. Let us take V {\displaystyle V} to be the natural permutation representation of the symmetric group S n {\displaystyle S_{n}} . This n {\displaystyle n} -dimensional representation is a sum of two irreducible representations: the standard and trivial representations, V =...
Wikipedia - Partition algebra
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Let n ≥ 2. For the free group Fn fix a "rose" Rn, that is a wedge, of n circles wedged at a vertex v, and fix an isomorphism between Fn and the fundamental group π1(Rn, v) of Rn. From this point on we identify Fn and π1(Rn, v) via this isomorphism. A marking on Fn consists of a homotopy equivalence f: Rn → Γ where Γ is...
Wikipedia - Outer space (mathematics)
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Up to a (free) homotopy, f is uniquely determined by the isomorphism f#: π1(Rn) → π1(Γ), that is by an isomorphism Fn → π1(Γ). A metric graph is a finite connected graph γ {\displaystyle \gamma } together with the assignment to every topological edge e of Γ of a positive real number L(e) called the length of e. The vol...
Wikipedia - Outer space (mathematics)
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Let of circles be described on the sides BC, CA, AB of triangle ABC whose external segments contain the two triads of angles C, A, B and B, C, A respectively. Each triad of circles determined by a triad of angles intersect at a common point thus yielding two such points. These points are called the Brocard points of tr...
Wikipedia - Modern triangle geometry
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Let one bit be assigned for each truth value: 01=T and 10=F with 00=N and 11=B.Then the subset relation in the power set on {T, F} corresponds to order ab
Wikipedia - Four-valued logic
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Let p 1 , p 2 , … , p n {\displaystyle p_{1},p_{2},\ldots ,p_{n}} be the first n primes. For a natural number a ≥ 1, define φ ( x , a ) := | { n ≤ x: p | n ⟹ p > p a } | , {\displaystyle \varphi (x,a):=\left|\left\{n\leq x:p|n\implies p>p_{a}\right\}\right|,} which counts natural numbers no greater than x with all prim...
Wikipedia - Meissel–Lehmer algorithm
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With these, we have φ ( x , a ) = ∑ k = 0 ∞ P k ( x , a ) , {\displaystyle \varphi (x,a)=\sum _{k=0}^{\infty }P_{k}(x,a),} where the sum only has finitely many nonzero terms because P k ( x , a ) = 0 {\displaystyle P_{k}(x,a)=0} when p a k > x {\displaystyle p_{a}^{k}>x} . Using the fact that P 0 ( x , a ) = 1 {\displa...
Wikipedia - Meissel–Lehmer algorithm
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Let p = p 0 + p 1 X + p 2 X 2 + ⋯ + p m − 1 X m − 1 + p m X m , {\displaystyle p=p_{0}+p_{1}X+p_{2}X^{2}+\cdots +p_{m-1}X^{m-1}+p_{m}X^{m},} be a nonzero polynomial with p m ≠ 0 {\displaystyle p_{m}\neq 0} The constant term of p is p 0 . {\displaystyle p_{0}.} It is zero in the case of the zero polynomial. The degree o...
Wikipedia - Polynomial ring
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{\displaystyle p_{m}.} In the special case of the zero polynomial, all of whose coefficients are zero, the leading coefficient is undefined, and the degree has been variously left undefined, defined to be −1, or defined to be a −∞.A constant polynomial is either the zero polynomial, or a polynomial of degree zero.
Wikipedia - Polynomial ring
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A nonzero polynomial is monic if its leading coefficient is 1. {\displaystyle 1.} Given two polynomials p and q, one has deg ⁡ ( p + q ) ≤ max ( deg ⁡ ( p ) , deg ⁡ ( q ) ) , {\displaystyle \deg(p+q)\leq \max(\deg(p),\deg(q)),} and, over a field, or more generally an integral domain, deg ⁡ ( p q ) = deg ⁡ ( p ) + deg ⁡...
Wikipedia - Polynomial ring
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{\displaystyle \deg(pq)=\deg(p)+\deg(q).} It follows immediately that, if K is an integral domain, then so is K.It follows also that, if K is an integral domain, a polynomial is a unit (that is, it has a multiplicative inverse) if and only if it is constant and is a unit in K. Two polynomials are associated if either o...
Wikipedia - Polynomial ring
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Let p be a point on the Hermitian variety H. A line L through p is by definition tangent when it is contains only one point (p itself) of the variety or lies completely on the variety. One can prove that these lines form a subspace, either a hyperplane of the full space. In the latter case, the point is singular.
Wikipedia - Hermitian variety
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Let p equal the probability of error. We claim that if A × B = C, then p = 0, and if A × B ≠ C, then p ≤ 1/2.
Wikipedia - Freivalds' algorithm
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Let p {\displaystyle p} be an odd prime number. Consider the polynomial f ( x ) = a 0 + a 1 x + ⋯ + a n x n {\textstyle f(x)=a_{0}+a_{1}x+\cdots +a_{n}x^{n}} over the field Z p {\displaystyle \mathbb {Z} _{p}} of remainders modulo p {\displaystyle p} . The algorithm should find all λ {\displaystyle \lambda } in Z p {\d...
Wikipedia - Berlekamp–Rabin algorithm
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Let p ∈ M {\displaystyle p\in M} . In what follows, we make the identification T v T p M ≅ T p M ≅ R n {\displaystyle T_{v}T_{p}M\cong T_{p}M\cong \mathbb {R} ^{n}} . Gauss's Lemma states: Let v , w ∈ B ϵ ( 0 ) ⊂ T v T p M ≅ T p M {\displaystyle v,w\in B_{\epsilon }(0)\subset T_{v}T_{p}M\cong T_{p}M} and M ∋ q := exp p...
Wikipedia - Gauss's lemma (Riemannian geometry)
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{\displaystyle \langle T_{v}\exp _{p}(v),T_{v}\exp _{p}(w)\rangle _{q}=\langle v,w\rangle _{p}.} For p ∈ M {\displaystyle p\in M} , this lemma means that exp p {\displaystyle \exp _{p}} is a radial isometry in the following sense: let v ∈ B ϵ ( 0 ) {\displaystyle v\in B_{\epsilon }(0)} , i.e. such that exp p {\displays...
Wikipedia - Gauss's lemma (Riemannian geometry)
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Let p: X → R {\displaystyle p:X\to \mathbb {R} } be a non-negative function. The following are equivalent: p {\displaystyle p} is a seminorm. p {\displaystyle p} is a convex F {\displaystyle F} -seminorm. p {\displaystyle p} is a convex balanced G-seminorm.
Wikipedia - Seminormable space
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If any of the above conditions hold, then the following are equivalent: p {\displaystyle p} is a norm; { x ∈ X: p ( x ) < 1 } {\displaystyle \{x\in X:p(x)<1\}} does not contain a non-trivial vector subspace. There exists a norm on X , {\displaystyle X,} with respect to which, { x ∈ X: p ( x ) < 1 } {\displaystyle \{x\i...
Wikipedia - Seminormable space
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Let q = (qx, qy, qz) and p = (px, py, pz) denote the position vector and momentum vector of a particle of an ideal gas, respectively. Let F denote the net force on that particle. Then the time-averaged kinetic energy of the particle is: ⟨ q ⋅ F ⟩ = ⟨ q x d p x d t ⟩ + ⟨ q y d p y d t ⟩ + ⟨ q z d p z d t ⟩ = − ⟨ q x ∂ H...
Wikipedia - Ideal gas equation
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Let q {\displaystyle \mathbf {q} } be the vector of nodal displacements of a typical element. The displacements at any other point of the element may be found by the use of interpolation functions as, symbolically: where u {\displaystyle \mathbf {u} } = vector of displacements at any point {x,y,z} of the element. N {\d...
Wikipedia - Finite element method in structural mechanics
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Let r ( t ) {\displaystyle \mathbf {r} (t)} be a parametric smooth curve. The tangent vector is given by r ′ ( t ) {\displaystyle \mathbf {r} '(t)} , where we have used a prime instead of the usual dot to indicate differentiation with respect to parameter t. The unit tangent vector is given by
Wikipedia - Tangent vectors
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Let r v = g s y e {\displaystyle r_{v}=g^{s}y^{e}} Let e v = H ( r v ∥ M ) {\displaystyle e_{v}=H(r_{v}\parallel M)} If e v = e {\displaystyle e_{v}=e} then the signature is verified.
Wikipedia - Schnorr signature
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Let s be any positive constant, let (V,+,.) be any real inner product space and let Vs={v ∈ V :|v|
Wikipedia - Gyrovector space
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The free functor C −: M o n S i g → M o n C a t {\displaystyle C_{-}:\mathbf {MonSig} \to \mathbf {MonCat} } , i.e. the left adjoint to the forgetful functor, sends a monoidal signature Σ {\displaystyle \Sigma } to the free monoidal category C Σ {\displaystyle C_{\Sigma }} it generates. String diagrams (with generators...
Wikipedia - String diagram
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Let the X := ( x i j ) ∈ R N × M {\displaystyle \mathbf {X} :=(x_{ij})\in \mathbb {R} ^{N\times M}} denote the input data matrix, M {\displaystyle M} the number of columns corresponding with the number of samples of mixed signals and N {\displaystyle N} the number of rows corresponding with the number of independent so...
Wikipedia - FastICA
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After centering, each row of X {\displaystyle \mathbf {X} } has an expected value of 0 {\displaystyle 0} .Whitening the data requires a linear transformation L: R N × M → R N × M {\displaystyle \mathbf {L} :\mathbb {R} ^{N\times M}\to \mathbb {R} ^{N\times M}} of the centered data so that the components of L ( X ) {\di...
Wikipedia - FastICA
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Let the bipartite graph be G = ( X + Y , E ) {\displaystyle G=(X+Y,E)} , where X = ( x 1 , … , x n ) {\displaystyle X=(x_{1},\ldots ,x_{n})} and Y = ( y 1 , … , y n ′ ) {\displaystyle Y=(y_{1},\ldots ,y_{n'})} and n ≤ n ′ {\displaystyle n\leq n'} . Let the given maximum matching be M = { ( x 1 , y 1 ) , … , ( x t , y t...
Wikipedia - Maximally matchable edge
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Note that there are no edges with both endpoints unsaturated by M, since this would contradict the maximality of M.Theorem: All M {\displaystyle M} -lower edges are maximally matchable. : sub.2.2 Proof: suppose e = ( x i , y j ) {\displaystyle e=(x_{i},y_{j})} where x i {\displaystyle x_{i}} is saturated and y i {\disp...
Wikipedia - Maximally matchable edge
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Hence, it remains to find the maximally matchable edges among the M-upper ones. Let H be the subgraph of G induced by the M-saturated nodes. Note that M is a perfect matching in H. Hence, using the algorithm of the previous subsection, it is possible to find all edges that are maximally matchable in H. Tassa explains h...
Wikipedia - Maximally matchable edge
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