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The duality statement Ka = Kb is perhaps geometrically obvious, since Ka and Kb arespherical cones with “center” ω and half-angles θa and θb, respectively, with cos θa = a,cos θb = ... Thetriangle inequality in spherical geometry gives γ α β.

z. (. µu. z. ). , 0 = p. y, 0 =. p. z ρg cos(α),. (2.6). 4 D. Pritchard, B. R. Duffy and S. ... The solution for p is. p = pa ρg cos(α)(h z) γ2h.

The fixed direction is given by the axis c. 5. Remark 2.6.b. ... ur. um. ul. ( )E. (a). (b). u'm. ur. um. ul( )H( )E.

cos is a special case of our reparametrised PER. 3 Model Search and EvaluationAs in Duvenaud et al. ... SE cos. Both demonstrate, using Bochner’s theorem(Bochner, 1959), that these kernels can approximate anystationary covariance function.

In this case, C = A cos(kL) 6= 0, B, and. ... values satisfy,. cos(k(2)n L) = 0 k(2)n =(. n 1. 2. ).

Rev. B 54, 6227-6242 (1996). 2. I. Avrutsky, Y. Zhao, and V. ... The lattice has primitive vectors a and b, where |a| = |b| and ab = |a|2 cos 60o.

of the α-effect, αv(B = 0) = χv = sin2 θ cos θ and Cα < 0(this determines the sign value of the hydrodynamic α ef-fect, see below in Sect. ... The value |Cα| (and |D|) chosen is perhaps largerthen expected because we use the profile χv = sin2 θ

j1 =12[sin(Φ) dΘ sin(Θ) cos(Θ) cos(Φ) dΦ],. j2 =12[cos(Φ) dΘ sin(Θ) cos(Θ) sin(Φ) dΦ],. ... with (recall that a2 a b2 = 0, see (3.4)). J′1 =12sin(Θ) cos(Θ) (a dΦ b dΦ),.

W(f,v)(a) = 0 iff BC1a(f) cos(θa) BC2a(f) sin(θa) = 0, f dom (Tmax). ... that. Sf = τf,. f dom (S) ={g dom (Tmax). BC1a(g) cos(θa) BC2a(g) sin(θa) = 0}.

bottom topography considered is described by:. b(x) =. . 0.25 cos(10π(x 1.5)) 1)) m if 1.4 m x 1.6 m,0 otherwise. ... 0. 0.5. 1. 1.5. x (m). (b. b U)/. b ref.