Sentence examples for l t for from inspiring English sources

Specially, for the case r = 2 and p = 1 β + 1, β > − 1 2, the author discusses the rate of approximation of σ 2 2 β + 1 by ϵ 2 λ 2, 1 / ( β + 1 ) under the condition E X 2 I ( | X | > t ) = O ( t − δ l ( t ) ) for some δ > 0, where l ( t ) is a slowly varying function at infinity.

If there exist positive constants C 1 and C 2 such that C 1 l ( t ) ≤ t δ T ( t ) ≤ C 2 l ( t ) for sufficiently large t and some δ > 0, where l ( t ) is a slowly varying function at infinity, then from Lemma 2.2(4) and Theorem 2.3, we have E | X | 2 + δ < ∞ ⇔ ∫ t ∞ l ( s ) s d s → 0 as  t → ∞.

The L t,NG (Table 16) is the sum of the L t for pipeline transport and the L t for tanker transport.

The volume of gas losses (L t) for pipeline transport can be calculated using Eq. (8).

According to Eq. (15), the value of L t for CSG transportation can be easily calculated.

Assuming that the recovery rate of BOG (r) is 0.7, the volume of gas losses (L t) for tanker transport can be calculated using Eq. (9).

The coefficient l t of was defined for each examination.

We shall prove that (l(t)=1) for all (t>0).

Let us prove that (l(t)=1) for all (t>0).

Since, we do not have any information about ε k and δ m, we can suppose uniform distribution in [−L r /2c L r /2c] and [−L t /2c L t /2c] for ε k and δ m, respectively.

For reflections we have L ( T ) = 2, for the rotations, L ( T ) = 0. Therefore, L ( G ) = 1.