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Integral domain

We replace the integral (1.7) by an integral from Si to rather than from 0 to 00. The errors due to ttiis restriction of the integration domain - the cut-off errors — can easily be estimated. [Pg.81]

We want to approximate the integral f x)dx by dividing the integration domain into n intervals of the same length h and by approximating /(x) in each interval by its value at the center of the interval. The discretization error is then... [Pg.89]

Although the wave functions employed to compute polarizability at separate internuclear distances are of very high accuracy the presence of the singularities in the integration domain lowers significantly the precision of the final results. Nevertheless, the present computation reveals that the total polarizability of the... [Pg.157]

This formula is exact, but less simple than it looks. The time ordering requires that the exponential be expanded in a series and that in each term of that series the operators B are written in chronological order. That means that the multiple integrals have to be broken up in a number of terms for different parts of the integration domain. Before proceeding, however, we collect a number of properties of the time ordering in the form of Exercises. [Pg.390]

It can be easily shown that Eqs. (2.7) and (2.10) are identical. Indeed, the three-dimensional integration domain in (2.7) can be divided into six subdomains, each corresponding to a term of (2.10) upon a proper change of time variables ... [Pg.350]

These integration domains are depicted in Fig. 3. The first three terms in (2.11) represent the third, the second, and the first terms in (2.10), respectively. A simple substitution of time variables, as specified in Table I, finally recovers the six terms of (2.10). [Pg.352]

Due to equivalence of points rj and r, both integrals in this equation are equal. Let us consider the first one. The integration domain here consists of two intervals. For the first one f 2 - rj ro and consequently... [Pg.400]

In the last section of this chapter, in Section 8.7, we present identities about roots of unity in integral domains. The results will be useful in Section 12.4 where we shall investigate Coxeter sets of cardinality 2. [Pg.154]

A commutative associative ring D with 1 is called an integral domain if the product of any two elements in D 0 is in D 0. ... [Pg.179]

Throughout this section, the letter D stands for an integral domain. [Pg.179]

If the d-d problem is not severe, this method is very efficient and runs much faster than pure real 16. In some cases, it is useful to split the integration domain into two or more parts in the manner as described above, and apply the a-p VEGAS choosing different values of to in different parts. In more difficult cases, however, pure real 16 works faster since it does not require the overhead needed in computing (46). [Pg.172]

Theorem. A closed set in k" is irreducible iff its ring of functions is an integral domain. [Pg.49]

Theorem, (a) Spec A is irreducible iff A modulo its nilradical is an integral domain. [Pg.52]

Lemma. Let A be an integral domain finitely generated over afield k. Let P be a nonzero prime ideal. The fraction field of A/P has lower transcendence degree than the fraction field of A. [Pg.106]

Theorem. Let k be a field, A B finitely generated k-algebras with A an integral domain. Then there are nonzero elements a in A and b in B such that that the map of localizations Aa -> Bb is faithfully flat. [Pg.114]

We first assume G is smooth. Let x be the idempotent for which Ax is fc[G°], an integral domain. Then Ax is a subring of Bx, and by (13.4) some AXtt Bxb is faithfully flat. In particular BA is nontrivial, so xb is not nilpo-tent and there is a maximal ideal P of B with xb 4 P- Then... [Pg.117]

Corollary. Let A B be Hopf algebra integral domains, L their fraction... [Pg.119]

Corollary. If B is a Hopf algebra integral domain and A a Hopf subalgebra with the same fraction field, then A = B. [Pg.119]

Corollary. Let B be a Hopf algebra integral domain. It is a finitely generated k-algebra iff its fraction field is a finitely generated field extension. [Pg.119]

Explicit integration of Eqn. (40) is intractable due to the form of the limit state function which defines the integration domain ([Pg.380]


See other pages where Integral domain is mentioned: [Pg.442]    [Pg.445]    [Pg.170]    [Pg.86]    [Pg.157]    [Pg.512]    [Pg.124]    [Pg.151]    [Pg.179]    [Pg.179]    [Pg.180]    [Pg.181]    [Pg.182]    [Pg.113]    [Pg.171]    [Pg.172]    [Pg.221]    [Pg.222]    [Pg.56]    [Pg.61]    [Pg.61]    [Pg.98]    [Pg.99]    [Pg.106]    [Pg.109]    [Pg.119]    [Pg.152]    [Pg.29]   
See also in sourсe #XX -- [ Pg.179 ]

See also in sourсe #XX -- [ Pg.179 ]




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