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19 Oct 2013: "# $ $ %& ' '! (# ) (# %. , , , , -. -. - - /10324052468789:2<;-= - - - > ,@? - - - - - -. -. -. -= -= -. -.A , , , , B B(C. ... D8EGF#. % H IJIH&. D8EGF#% % IJI $ , , , , - K= -= -. LNMPO ;. - -. - QA , , , , B B RSR- - - -. -. -T. VUXW Y Y H& H H, , , ,

19 Oct 2013: " # # $ $ $ $&% $ ' ( '' ' '' ' '' ' ),.-'' ' ( '' ' ( '' ' '( ' / / ' '0 $ $ $ $ 1/ '32 '"4. 11 576 ' ' ' '98. # # : # # ; # # < = >@?BADC9A"CFEHGHIJC. > ; < ; : < =? BADC9A"CFEHGHIKC 0 $ $ $ $ ' ' ' ' / L '( / 2 M5N5PORQTSUQVSXWY[ZXS. -/ ' '0 $ $ $

19 Oct 2013: d T T T TV V V V V. b V VW V VX V V V V V VX VT e TgfV VX V V. ... h?h T T T TikjmlnjolEp [:q l VX VX b b VX Vc VX V r!s t.

6 Mar 2013: b. Thus two linearly independent solutions of Legendre’s equation can be written, in the casewhen m is not an integer (so that the exponents at the origin do not differ ... a(a 1)b(b 1). c(c 1). z2. 2! 2.

26 Jul 2013: b) Two dimers. We now consider the case of two dimersa and b. ... 482 2d3, 3a. V = ̄ab2 b1b1b2. ̄ba2 a1. a1a2 f a f b.

19 Aug 2013: Rotation of particles – rigid and dilute. Spheroid: axes a, b, b, aspect ratio r = ab. ... Averaged stress. σ = pI 2µE 2µφ[AE : 〈pppp〉. B(E 〈pp〉 〈pp〉E) CE FDrot〈pp〉].

5 Oct 2013: Toquantify this, we consider the condition number κ(B) =‖B‖‖B1‖. The following lemma is also found in [2]:. ... Technical reportNA2010/07, DAMTP, University of Cambridge, 2010. [2] B. Adcock and A.

26 Jul 2013: 3(b) denoting the 1=. ffiffiffiffiN. pisotropy. that is expected in our system. ... The case shown is for a puller swimmer (b 1) witha volume fraction / 0.01.

18 Jul 2013: B B. However, for our non-Abelian magnetic field, B B is simply a gaugetransformation. ... B. 2 ξi (k)0. (1.1). ψi =4. B. 2π. dk. 2πexp.

5 Oct 2013: Toquantify this, we consider the condition number κ(B) =‖B‖‖B1‖. The following lemma is also found in [2]:. ... Technical reportNA2010/07, DAMTP, University of Cambridge, 2010. [2] B. Adcock and A.

26 Jul 2013: dashed lines) and 3π/4 (dashed-dotted lines), with A = 1 and B = 10. ... A/B) sin zand η2 (A/B) sin(z φ) as might be expected.

17 Jul 2013: Phys. J. B 12, 589 (1999); K. D. Danov,R. Dimova, and B. ... 99, L44 (2010). [11] A. R. Honerkamp-Smith, B. B. Machta, and S.

26 Jul 2013: b c d e. ii. iii. iv. Kn = 0. Kn = 0.2 Kn = 0.4 Kn = 0.5. ... Kn Kn/Knc. 1.00 10.00. (a) (b). CD. 0.2. 0. 0.4. 0.6.

26 Jul 2013: a). (b). (b) (c). (d). (c). (d). 1. 2. FIG. 8. ... New York, 1992.10 B. Chen, P. D. Wu, and H. Gao, Proc.

7 Mar 2013: On the other hand, the demand of re-1/2 1/2 1/2 1/2B (B B ) Bsources depends on the number of cells in the colony andthe total cells ... gb Bvif 1, v p 1,dc Cv. g gb B b Bv vif!

26 Jul 2013: V2 =¡ (1) (2) ¢ ¢ ¢ (n). ¢;. taking A = ¤ , B = B, Q = I, R = 2I and ¤ cl = ¡ ¤ ¤ , and applying the aboverelationships, it follows that (3.7 ... written in the form _À = ¤ À Bu, where the diagonal matrix ¤ contains theunstable eigenvalues

9 Jan 2013: We must instead consider its logarithmic norm(Söderlind 2006) µ[B], namely the largest eigenvalue of the symmetric matrix B̃ =12 (B B. >). ... cussion yet, even if B is skew symmetric, we are not done yet!

4 Nov 2013: H = h/SV , L = l/SV , B = b/SV , X = x/SH and T = t/ST ,(2.13a, b, c, d , e). ... 366 J. A. Neufeld and H. E. Huppert. B. B. Injection point.

8 Jun 2013: a. b. c. e. f. g. d. HoloceneYDB–AODLGM. Figure 4 | Hemispheric temperatures. ... 33. Stephens, B. B. & Keeling, R. F. The influence of Antarctic sea ice on glacial-interglacial CO2 variations.

26 Jul 2013: 2x,0 = a cosx t 1 x x,0 b sinx t 1. ... ̃̇1, D12we obtain, after some algebra,. 4 2 =De1De2 De1. 1 De12 b.

26 Jul 2013: a) (b) (c). 0 1.0 0 1.0–1.0. 1.0. FIGURE 2. Trajectories, ξ(t), of surface material points for stroke A (a), stroke B (b), andstroke C ... a). (b). (c). (d ). (e). ( f ). 0 1.0 0 1.00 1.0.

4 Nov 2013: vf (x, t) = kf. μ. ρg(h b). b= φγ kf. ... xf = 11 cm. Balance Balance. PumpW = 1.1 cm. b = 2.1 cm.

26 Jul 2013: a) (b). b/a = 1/2. V (e). qΩ. V (e). qΩ. ... n 1 Anan2. n 1BnanPn1. 49. B. Efficiency and optimization of an entirelyporous sphere.

4 Nov 2013: ηN =[. 2π 1. 0. f0. s[ B f (y), z]yd zd y. ... s0( f B) B f (33a)and. F( f B) 2. 4(2 ) ( f B)3 , (33b).

4 Nov 2013: the current first reaches the sink, as illustrated in figure 2(a, b). ... B). The results of this more detailedmodel are also shown in figure 6.

5 Oct 2013: The degree ofan operator A B(H) is defined by. deg(A) = supn1. ... Let A B(H) and let {ej} be an orthonormal basis for H.

19 Oct 2013: Km F6 =I QK 6{ K! x== F6Z? ] =4-<4- 0#B6< =I ]? {? 4g :I KK 0 B"L UBJÔ< =;:I KK ITIT B XABUBJÔ<? ... B"K J5,.#G?=II:; IRÂ0==IIKmG KrKÝG= #g 4< X8 K 4- 'F,.#.=II.

5 Oct 2013: The degree ofan operator A B(H) is defined by. deg(A) = supn1. ... Let A B(H) and let {ej} be an orthonormal basis for H.

5 Jun 2013: above, i.e. compass-wise, in the northern hemisphere, corresponding to up-. ward B. ... waves. The background rotation and azimuthal magnetic field B point. toward the viewer.

21 Mar 2013: from the start of the alphabet (e.g, a,b,c,. ) take values 1,. ... dx , as before (we note here that A, B, C and D, defined by (2.9).

19 Oct 2013: ê -K5ø¥HF é öF ëLð @ é @ ò D@CK í F ë9þÁê øÒø ûK?îKpñ-K5ø ò-í K5î , ð ê @)ø¥H BED þÁé öF9ù ò øÒø òþÁë5ý B ëLé ö- ... ýí é H ëLê @pö-øö-Dî@ é K óê @pö-øF éê D ê D7üÿ/ö-DîøÒöFWKpJö-øF

19 Oct 2013: m o t i o n o f a r u b b e r m e m b r a n e at the b o t t o m , ... t l b ). O/t - 0 Z = O(a 2) forcing (4.12a).

20 Dec 2013: a µ, b ν etc.). For example, consider the quantity ηaXa in theabstract index notation. ... Aµνfν, e′µ = B. νµeν (2.25). for some matrices Aµν and Bνµ.