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13 Jun 2008: It focuses on:.  visual and simple.  the fact that the first 20% of effort gets 80% of the result.

1 Jul 2008: ρ0. z. (ρ0f. 20. N2. ψ′. z. )= q′ , (15).

11 Aug 2008: AdaptingEqs. (4) and (5) for a rotating solution [20] gives. r r r Rr; (6). ... 15. 20. 25. 30. 35. µ. FIG. 3 (color online). Chemical potential as a function ofpumping spot size R.

7 Oct 2008: 4.1features as a dashed line. 8. 0 0.05 0.1 0.15 0.20.82. ... 9. 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.20.82.

31 Jul 2008: 13.1.29]. U (a, c, z) = z1cU (a 1 c, 2 c, z), (20). ... 10. 9. 8. 7. 6. 5. 4. 3. 2. K0 10 20 30 40 50.

18 Mar 2008: Eight points have beencomputed for each integral. 0 20 40 60 80 1000. ... w0 20 40 60 80 100. 0. 0.5. 1. 1.5. 2.

7 May 2008: We note that in contrast to the kinetic equations pre-sented in [19] and [20] for the one-component case, Eq. ... 20] C. Connaughton, and Y. Pomeau, C.R. Physique 5 5, 91 (2004).[21] S.

1 Aug 2008: 20. 2.4. ω. 100908070. 4.2. 3.8. 50. 3.4. 3.0. 30. 2.6. ... Figure 4.1: Pointwise absolute error for two Filon-type methods. 20. at the endpoints, the second uses both function values and derivatives there.

10 Apr 2008: mπ)4. 20. (mπ)6. ]. (3.5). 128ω5[. 1. (mπ)6. 22. 3. 1. ... 3T 3. 1. 5T 5). 1. (iω)6. (. T 20. 3(T 3.

6 May 2008: They also include the bound. k1 max [ /k, γ ] k max [ δ1 , γ (M µk)1 ] (M µk) k, kK, (3.20). ... 9. the last part being obtained from the property (3.11). We replace expression(3.20) by the weaker condition.