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14 Sep 2015: 0.0. 0.5. e cos(ϖ - ϖplt)-0.5 0.0 0.5. A > ϖplt. ... coefficient, such that. b(j)s (α) 1. π. 2π0. cos(jψ)dψ. (1 2α cos ψ α2)s.

13 Jan 2015: repr. oduc. ed w. ithou. t per. mis. sion. In the last step, we used spherical polars with z = r cos θ. ... is. φ = a b. r. (90). Next we have solutions, like the dipole potential, cos θ,.

30 Nov 2015: c) Let f: [a,b] R2 be continuous, and differentiable on (a,b). ... Dfm(x)(h))for each h Rn.(b) Define f: R3 R2 by f(x,y,z) = (exyz, cos x2y).

8 Oct 2015: The other end,B, is fixed. The particle moves in a horizontal circle a distance d below B. ... cos θ,. where cos θ = d/l. 2 A particle rests at the top of a smooth sphere.

17 Dec 2015: Laplace coefficients aregiven by. b(ν)s (α) =1π. 2π0. cos(νx)dx(1 2α cos( x) α2)s. , ... f 2(L1 L2)2 4L1 L2],. (10). where A = (πa1/21 b.

2 Feb 2015: The inner product ab, also known as the dot or scalar product, of two vectorsa and b, is a scalar with magnitude |a||b|cos θ, where |a| and |b| are ... Here |a| (aa) 12 , sothat the expression for ab is effectively an algebraic definition of cos θ.

16 Oct 2015: 5. Let (fn) be a sequence of real-valued continuous functions on a closed, bounded interval [a,b], and. ... m=1 mam cos mx. (b) Show that f(x) =. m=1 am sin mx defines a continuous function on [π,π], but that the seriesm=1 mam cos mx need

9 Nov 2015: 87 –. while the jump condition gives. B( cos(1 ξ)) A cos ξ = 1. ... The generalsolution of this homogeneous equation c1 cos x c2 sin x and the inhomogeneous boundary.

16 Feb 2015: 16. Let 4 be a hyperbolic triangle, with angles α, β, γ, and sides of length a, b, c (theside of length a being opposite the vertex with angle α, and ... cosh c = cosh a cosh b sinh a sinh b cos γ.[For a slicker proof of this result, and of the

15 Jan 2015: Theyfound that the centre of libration is given by cos ϕ2:10 =0.39 0.061/e. ... The best fit is of the form. ϕ0 = C1 cos (δϕ0/C2).