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1 May 2014: 2 sin χ 2AD sin χ/C cos χC cos χ 4AB/C cos χ. ... See Appendix B for the full equations of motion. NB: cξ = cos ξ and sξ = sin ξ.

4 Feb 2014: φ(s,v) = α(s) r(n(s) cos v b(s) sin v)where n = n(s) and b = b(s) are the normal and binormal vectors of α. ... 9. Consider a surface of revolution parametrized by φ : (0, 2π) (a,b) R3, whereφ(u,v) = (f(v) cos u,f(v) sin u,g(v)).

31 May 2014: The weighted framebounds A,B > 0 satisfy. A 2 e2πδB1d,. B e2πδB1d < 2. ... ω B(ω̃), |f̂(ω)| |f̂(zω̃)|. Since also µω = meas(V E. ω.

6 Feb 2014: S = b cos θ tan γffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1 cos2θ tan2γ. p ; [2]. ... cos γA1cos γB1. =cos γA2cos γB2. [21]. must be satisfied to enable stacking.

4 Nov 2014: as. Θ(0, t) = Θ0 A cos (2πt/tD) B cos (2πt/tY) (4.30). – ... We can solve these as. X(x)T(t) =[a sin(. λx) b cos(.

20 Oct 2014: 4. 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

28 Jan 2014: 5. Define.  . . . . cos/cos/ 2. 2220 1. ... 2006). Acta Mat., 54, 1297-1306. Kachanov, M., Sevostianov, I. and Shafiro, B.

12 Feb 2014: f (t) = a0. n=1. (an cos(nt) bn sin(nt)). =. n=1. ... f (t) = a0. n=1. (an cos(nt) bn sin(nt)). =eπ eπ.

5 Nov 2014: When small-amplitudewaves propagate along ex, the surface of the cylinder is de-scribed by r̂ = [h(s, t) b cos φ]ey b sin φez sex and the. ... p̃ ηωh̃i cos θAK1(kr)ũr h̃ωi cos θ[αArkK1(kr) BK2(kr) CK0(kr)]ũθ h̃ωi sin θ[BK2(kr)

28 Nov 2014: f ′′cot θf ′6f =A cos θ B. (2.14)Together with the particular solution, the full solution is then ψ = cr3(at bt3 1)for t =cos θ. ... The function H(ν) is plotted for cone angles in figure 1(b).

23 Oct 2014: ui(0) = sin( πxi. b. ), u′i(0) =. π. b. gd(xi) cos. ( ... b. gd(xN ) cos. ( πxNb. ))T,. (39). where U (t) denotes the N -dimensional vector with entries ui(t) u(xi, t),.

22 Dec 2014: eπ2. sinh(π) j = 0,. j cos( jt)sin( jt)π2. j21. j Z+. ... a = 0, b = π, (t [0,π],q(t) = 20 cos(2t)), α1 6= 0, β1 6= 0, α2 = β2 = 0,.

9 Nov 2014: 2. V (t) =µπ2c2. 4L. n=1. n2[An cos. (nπct. L. ) Bn sin. ... m=1. i=1. [Cmi cos mθ Dmi sin mθ] cos(kmict) Jm(kmir). (5.32).

14 Oct 2014: obtain:. Φ(ϕ) = A sin(λϕ) B cos(λϕ),. with positve constants A, B. ... Aλrλ Bλr. λ) sin(λϕ) (Cλrλ Dλr. λ) cos(λϕ). E0 F0 ln(r).

13 Oct 2014: b) Now consider an almost ideal FJC with N monomers of type a. ... Show that these can be written as. W(n)(x) = A cos(k(n)x) B sin(k(n)x) D cosh(k(n)x) E sinh(k(n)x) ,. and

29 Aug 2014: û% |r =1 = K cos(nθ ). . 0. cos φ cos θ sin φ. ... Thus, the previous equations canbe put in the form. s û% = K cos φk.

31 May 2014: The weighted framebounds A,B > 0 satisfy. A 2 e2πδB1d,. B e2πδB1d < 2. ... ω B(ω̃), |f̂(ω)| |f̂(zω̃)|. Since also µω = meas(V E. ω.

6 Nov 2014: The Frobenius integrability condition yields. [X,Y ] = aX bY. for some a and b. ... 2.14). Let ΣA′B′ be the usual basis of SD two forms.

9 Sep 2014: fprop = (ξ‖ ξ)u sin θ cos θ ex. (13)In addition, in order to generate a net propulsion from a. ... u(x, h(x, t)) = ht. ey = bω cos(kx ωt)ey , (21).

22 Apr 2014: dE. dα BγAL. (V. B. )2/3 [ 43C1 cos(θE )α. 1 13. ] ... B. Solving the problem. We start by rearranging Eq. (12) by differentiating onceto eliminate Ft , Fn, and M and obtain a third-order equationfor ψ ,. d3ψ. ds3+ γ cos θE. EI.