Then by Arzelà – Ascoli theorem the space K is compact.

Then the Zariski tangent space at a point p ∈ X is the collection of K-derivations D: O < sub > X, p </ sub >→ K, where K is the ground field and O < sub > X, p </ sub > is the stalk of O < sub > X </ sub > at p.

⟨ H ⟩, be the group generated by H. Then the word problem in H < sup >*</ sup > is solvable: given two words h, k in the generators H of H < sup >*</ sup >, write them as words in X and compare them using the solution to the word problem in G. It is easy to think that this demonstrates a uniform solution the word problem for the class K ( say ) of finitely generated groups that can be embedded in G. If this were the case the non-existence of a universal solvable word problem group would follow easily from Boone-Rogers.

* Douglas K. Smith, Robert C. Alexander, Fumbling the Future: How Xerox Invented, Then Ignored, the First Personal Computer ( William Morrow and Company, New York, 1988 ) ISBN 1-58348-266-0

Let K be a closed subset of a compact set T in R < sup > n </ sup > and let C < sub > K </ sub > be an open cover of K. Then

Then Congress Party President K. Kamaraj was instrumental in making Shastri Prime Minister on 9 June.

Theorem: Let R be a Dedekind domain with fraction field K. Let L be a finite degree field extension of K and denote by S the integral closure of R in L. Then S is itself a Dedekind domain.

Then, where K < sub > 0 </ sub >( R ) is the Grothendieck group of the commutative monoid of finitely generated projective R modules.

Then, the B < sub > m </ sub > make up a neighbourhood basis of 0 in K.

Then the action of H on Ω extends in a natural way to an action of H on the group K by

Then the unrestricted wreath product A Wr < sub > Ω </ sub > H of A by H is the semidirect product K ⋊ H. The subgroup K of A Wr < sub > Ω </ sub > H is called the base of the wreath product.

Then NA < sup > 2 </ sup >+ NB < sup > 2 </ sup >+ NC < sup > 2 </ sup >+ NH < sup > 2 </ sup > = 3R < sup > 2 </ sup > where R is the common circumradius and if PA < sup > 2 </ sup >+ PB < sup > 2 </ sup >+ PC < sup > 2 </ sup >+ PH < sup > 2 </ sup > = K < sup > 2 </ sup >, where K is kept constant, then the locus of P is a circle centered at N with a radius.

In 2008, U. K. label Stage Door Records released the retrospective collection Shirley Jones — Then & Now featuring twenty-four songs from Jones's musical career, including songs from the films Oklahoma !, Carousel and April Love.

Then, in A. D. 556 Tikal enacted a ch ’ ak ( axe ) war event, and defeated Caracol ; this caused Lord Water to enact the first know star-war event in 562 ( 9. 6. 8. 4. 2 ), and defeated Tikal's Lord Wak Chan K ’ awiil ( Double Bird ).

Then A is an algebra over K if the following identities hold for any three elements x, y, and z of A, and all elements (" scalars ") a and b of K:

Then the Black formula states the price for a European call option of maturity T on a futures contract with strike price K and delivery date T ( with ) is

We can consider this an exchange ( Margrabe ) option by considering the first asset to be and the second asset to be the riskless bond paying off $ 1 at time T. Then the call option is exercised at time T when the first asset is worth more than K riskless bonds.

Then came other Freestyle artists that were Puerto Rican such as Brenda K. Starr, Marc Anthony, Cynthia, George LaMond, La India, Judy Torres, TKA, Lil Suzy and Lissette Melendez.

Then the energy of the vacuum is exactly E < sub > 0 </ sub >.

Then, p < sup > 2 </ sup > is the fraction of the population homozygous for the first allele, 2pq is the fraction of heterozygotes, and q < sup > 2 </ sup > is the fraction homozygous for the alternative allele.

) Then X < sub > i </ sub > is the value ( or realization ) produced by a given run of the process at time i. Suppose that the process is further known to have defined values for mean μ < sub > i </ sub > and variance σ < sub > i </ sub >< sup > 2 </ sup > for all times i. Then the definition of the autocorrelation between times s and t is

Then X is reflexive if and only if each X < sub > j </ sub > is reflexive.

Then the cotangent space at x is defined as the dual space of T < sub > x </ sub > M:

Then I < sub > x </ sub > and I < sub > x </ sub >< sup > 2 </ sup > are real vector spaces and the cotangent space is defined as the quotient space T < sub > x </ sub >< sup >*</ sup > M = I < sub > x </ sub > / I < sub > x </ sub >< sup > 2 </ sup >.

Then the complex derivative of ƒ at a point z < sub > 0 </ sub > is defined by

Indeed, following, suppose ƒ is a complex function defined in an open set Ω ⊂ C. Then, writing for every z ∈ Ω, one can also regard Ω as an open subset of R < sup > 2 </ sup >, and ƒ as a function of two real variables x and y, which maps Ω ⊂ R < sup > 2 </ sup > to C. We consider the Cauchy – Riemann equations at z = 0 assuming ƒ ( z ) = 0, just for notational simplicity – the proof is identical in general case.

Then, for any given sequence of integers a < sub > 1 </ sub >, a < sub > 2 </ sub >, …, a < sub > k </ sub >, there exists an integer x solving the following system of simultaneous congruences.

Then the overall runtime is O ( n < sup > 2 </ sup >).

Then the Cartesian product set D < sub > 1 </ sub > D < sub > 2 </ sub > can be made into a directed set by defining ( n < sub > 1 </ sub >, n < sub > 2 </ sub >) ≤ ( m < sub > 1 </ sub >, m < sub > 2 </ sub >) if and only if n < sub > 1 </ sub > ≤ m < sub > 1 </ sub > and n < sub > 2 </ sub > ≤ m < sub > 2 </ sub >.

Then, dividing the units of energy ( such as eV ) by a fundamental constant that has units of velocity ( M < sup > 0 </ sup > L < sup > 1 </ sup > T < sup >-1 </ sup >), facilitates the required conversion of using energy units to describe momentum.

Then the periodic Bernoulli functions P < sub > n </ sub > are defined as

Then, in terms of P < sub > n </ sub >( x ), the remainder

Then, using the periodic Bernoulli function P < sub > n </ sub > defined above and repeating the argument on the interval, one can obtain an expression of ƒ ( 1 ).

For instance, suppose that each input is an integer z in the range 0 to N − 1, and the output must be an integer h in the range 0 to n − 1, where N is much larger than n. Then the hash function could be h

The algorithm for deciding this is conceptually simple: it constructs ( the description of ) a new program t taking an argument n which ( 1 ) first executes program a on input i ( both a and i being hard-coded into the definition of t ), and ( 2 ) then returns the square of n. If a ( i ) runs forever, then t will never get to step ( 2 ), regardless of n. Then clearly, t is a function for computing squares if and only if step ( 1 ) terminates.

Then, we can show that if the game starts with n spots, it will end in no more than 3n − 1 moves and no fewer than 2n moves.

Then a fuzzy subset s: S of a set S is recursively enumerable if a recursive map h: S × N Ü exists such that, for every x in S, the function h ( x, n ) is increasing with respect to n and s ( x ) = lim h ( x, n ).

Then the statistician must analyze the properties of and, which are viewed as random vectors since a randomly different selection of n cases to observe would have resulted in different values for them.

Then for arbitrary ε > 0 there is an embedding ( or immersion ) ƒ < sub > ε </ sub >: M < sup > m </ sup > → R < sup > n </ sup > which is

Then the n-th series coefficient c < sub > n </ sub > is given by:

Then, using a simple greedy algorithm, the easy knapsack can be solved using O ( n ) arithmetic operations, which decrypts the message.

Then n is palindromic if and only if a < sub > i </ sub > = a < sub > k − i </ sub > for all i. Zero is written 0 in any base and is also palindromic by definition.

* Then t ( n ( c )) =

Then Y = u ( X < sub > 1 </ sub >, X < sub > 2 </ sub >, ..., X < sub > n </ sub >) is a sufficient statistic for θ if and only if, for some function H,