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5 Oct 2013: The problem is not new, and therehave been many different approaches developed for its solution (see, for example, [2, 11, 12, 17, 18, 22,23, 24, 25, 52] and references therein).

5 Oct 2013: The second inclusion follows by Theorem VIII.24 ([RS72], p. 290)if we can show that PnAPn A in the strong resolvent sense. ... 11. Proof. Now, (i) follows from the fact that A1,n A strongly (see Theorem VIII.24 in [RS72],p.

26 Jul 2013: 24" as! 1 iDe1"!̃3!1"! 1 iDe2"#̃̇3! 1". =12. De2 De11 2iDe1. ... 32# 3#3 8#De1De2 3#3De1De2+ 24#De1. 2 4"! 20 3#2 12De12 8De1De2.

26 Jul 2013: f2x =. 2π. 0. 〈τ2xy 〉|y=0 dx, (3.24). and only the mean component of the second-order stress, 〈τ2xy 〉, contributes tothe net force.

4 Nov 2013: 8 of 21. [24] Thus, disentangling the effects of varying reflectioncoefficient and signal strength is not straightforward, espe-cially when tuning effects are important.

19 Oct 2013: 1. Submission 28 June 2012. Advice to a Young Researcher:. with reminiscences of a life in science. J. Michael T. Thompson. Honorary Fellow, Department of Applied Mathematics and Theoretical. Physics, University of Cambridge, Cambridge, CB3 0WA, UK

24 Apr 2013: models for active systems have beenproposed [2, 17–24], but most of them have yet to be tested against experiments [14, 25]. ... where. M =(. 2 A 00 0. ) (0|k|2 2|k|4 i λ0kxv0)I (24).

19 Oct 2013: Fluid. Dyn. Res., 24, 375–404.Janssen, P. A. E. M. 1982 Quasilinear approximation for the spectrum of wind-generated.

28 Dec 2013: Natural Sciences Tripos, Part IA. Lent Term 2012. 24 lectures. Tuesday, Thursdays and Saturdays at 9 am, starting 19 January.

5 Oct 2013: Thiswas later considered by Eldar et al, who extended this framework to frames in arbitrary Hilbert spaces[18, 19, 20, 24]. ... M2 =. C(γ)M11. 2. (24). Proof. Without loss of generality, in this proof, M1 and M2 will be even.