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12 Nov 2016: 8. By considering the Fourier transform of the function. f(x) =. {cos x for |x| < π/20 for |x| π/2 ,. David Skinner d.b.skinner@damtp.cam.ac.uk. ... is. fα(x) =2i. π(α2 x2)(α cosh α sin xx cos x sinh α).

5 Nov 2016: n=0. P(cos θ)Pn(cos θ). 3. Show that whenever r 6= r′, the function 1/|r r′| satisfies Laplace’s equation. ... y(x, 0) = A sin(πx/L) andy. t(x, 0) = 0. (b) If an extra transverse force F0 sin(πx/L) cos(πct/L) per unit length acts on the

11 Oct 2016: 8. µAµ@A. x =v! cos!t. y = v! sin!t. mẍ = eẋ B! ... x =v! cos!t. y = v! sin!t. mẍ = eẋ B!

1 Oct 2016: L1. L1{L2, s1 sin ωt,s1 cos ωt} ds1. L2. L2{s2,0,0} ds2. ... r1 r = {αs,b sin(kx ωt ),b cos(kx ωt )}, (21)r2 r = {b cos(kx ωt ),b sin(kx ωt ), αs}, (22).

1 Jun 2016: sin kθ0 =. n=0a(s)n Pn(cos θ0)=. n=1. b(s)n Vn(cos θ0) (3.4a). ... similar. Let. Qn(k)= π. 0sin θPn(cos θ) sin kθ dθ, (B 1).

12 Aug 2016: 2 2 ). ( 3. 2 3 ). (3.24). so that the normal derivative dominates the tangential terms; recall that our choice (3.1) has XR = b cos 0. ... B 3). - b/2R3. The magnetic boundary conditions then become. 0 6&=, = [ 3ih cos 812 - a ] ei@,. -

28 Jun 2016: x= a sin σcosh τ cos σ , y=. a sinh τcosh τ cos σ , (2.9a,b). ... n=1. enτ cos nσ. ). (2.10). The last equality follows from representation of the scale factor h in terms of a Fourierseries in σ.

3 Oct 2016: κ3(ω) = 2/. [1 exp. (log4(1 ω). )];. d. κ4(ω) = cos(. ... The function κ4 is a special case of. κa,b(ω) = cos(.

12 Jul 2016: 2Q v. 2Q). Expressions for A and B were obtained in [15], C (correct-. ... 10] Berloff, N. G. and Svistunov, B. V. Scenario of strongly nonequilibrated BoseEinstein condensation.

1 Oct 2016: A = 12 ln[cosh (ŷ ŷ0) cos (x̂ x̂0)], (15). B = 12 ln[cosh (ŷ ŷ0) cos (x̂ x̂0)]. ... We obtain. F H = ξH aω(1 ρ) sin cos 4.92 pN μm (54).