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25 Nov 2011: It considers a forcing term that is oscillatory in space. tu = 2x b sin ωx, (3.24). ... 2t cos ω xbeω2t sin ω,. 21. Figure 3.1: The solution of (3.24) for different values of ω.

27 Sep 2011: 2004) to identify missing intronsin poor gene models, and 3) Pfam (24.0) analysis (Finn et al.2008) to identify conserved functional domains in eachprotein (METE: PF01717, PF08267; METH:PF02965,PF02965,

1 Sep 2011: F0 = 1 < F2 =π2. 8< < 4. π< < F3 =. π3. 24< F1 =. π. 2. In this section we obtain a difference analogue of the Bohr-Favard inequality in theform. ... 1. 24. ]= 1. 325. 192. 1. α4. With α = 1 and α = 2, we obtain c4(πn ) =. 517192. = 2.6927, and c4

10 Apr 2011: simplifying assumptions (a), (b) above [24]. While our physical picture and conclusions aredifferent from theirs, we are indebted to them for emphasizing these issues. ... The late-time asymptotics have been studiedby Granek and Olami [24].). What is the

10 Apr 2011: f1. t+. dxdt. f1x. dpdt. f1p. (24). This symmetry is most apparent in the new variables, and translates into the fact thatequations (22) and (23) can be satisfied up to O(

10 Apr 2011: The identity22,23,24. exp[ -A3({P})]. = exp[ - A 3 ( {81M}) ]exp(f dr J(r) p (r») I J= 0' (2.22). ... f/Jc. where the N independent amplitude is. 1 (w) _ '24 [g2(X) p/2 1/2 -yk ()[ ()]1/2 gl X -g4 X.

1 Sep 2011: up to the sign we obtaina11 = a22 = k; a12 = 1:From (12.3) we have a21 = k1k1 k0k 10 = k2 1;i.e., A = (1)k1 24 k 1k2 1 k ... 24. 19. ConclusionIt is clear that the above MAPLE algorithm may help to verify de Boor's conjecture only for particularvalues m.

30 Sep 2011: CR strain CC124 [12] was grown axenically in tris-acetate-phosphate medium [11] in a growth chamber(Binder, Germany) set to 24 C and a cycle of 16 h lightat 100 ... Nikaido, J. Cell Sci. 109, 57 (1996). [24] M. A. Silverman and M.

10 Oct 2011: and. 2β.αβ.β. = 2|α||β|. cos θ Z (1-24). Taking the product we find that 4 cos2 θ Z. ... 24. The Cartan subalgebra H can be taken to be the traceless diagonal matrices; there aren(n 1) roots associated with the Ei,j for i ̸= j.

10 Apr 2011: V R! 26aa0. 5. R6. 24! Expressing this result in terms of the electric polarizabilvia. ... Thus the van deWaals potential assumes the form. V R! 24.