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11 Feb 2014: α =. C. A dx. α =. C. A dx =. B dS. Aϕ =g. 4πr. 1 cos θsin θ. B =g. ... 1 cos θsin θ. B =g. 4πr2̂r. References. [1] Gibbons and Rychenkova, 9608085.

21 Jan 2014: 7. Suppose γ : [a,b] R2 is a smooth curve parametrized by arc length. ... r cos z,r sin z,z).(c) U = (0, 2π) (0, 2π), σ(θ,φ) = ((a b cos φ) cos θ, (a b cos φ) sin θ,b sin φ).

29 Aug 2014: cosh ξ cos η ,. z =c sinh ξ. cosh ξ cos η , (B.1). ... UndVndη. , (B.17). ψ̂. ξ= 3. 2sinh ξ. (cosh ξ cos η)5/2.

21 Jan 2014: 7. Suppose γ : [a,b] R2 is a smooth curve parametrized by arc length. ... r cos z,r sin z,z).(c) U = (0, 2π) (0, 2π), σ(θ,φ) = ((a b cos φ) cos θ, (a b cos φ) sin θ,b sin φ).

18 Jan 2014: L =1. 2m(. ρ̇2 ρ2θ̇2). 12kρ2 q. (. Eρ cos θ B. ... A =B. 2. (. ρ sin θ, ρ cos θ, 0).

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)).

6 Feb 2014: 1. Find the General Solution with f (t) = 1 (t 0).Complementary function is y = A cos(3t) B sin(3t). ... Particular integral is y = 19General solution is y = 19 A cos(3t) B sin(3t).

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(.

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),.

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

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).

3 Feb 2014: From simple trigonometry, this leads tothe following expressions. jGxx j sin2 a; jGyy j cos2 a; jGxy j sin a cos a (1). ... a diameterwith end-points ðjSxx;jSxyÞ ðj sin2 a;j sin a cos aÞ and ðjSyy; jSxyÞ ð0; j sin a cos aÞ; see Fig.

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

25 Mar 2014: xz. -y. -xz zz. (a) (b). angle = πJt. 2Î1yÎ2z2πJ12tÎ1zÎ2z cos (πJ12t) 2Î1yÎ2z sin (πJ12t) Î1x. ... cos (τ) sin (τ)Îy sin (τ) sin (τ)Îx Îx Spin echoes 20.

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.

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.

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.

18 Oct 2014: B. Mennesson1, R. Millan-Gabet2, E. Serabyn1, M. M. Colavita1, O. Absil3, G. ... groups (Figure 4, panels B & E): stars in Group 2 have their observed distribution shifted towards higher.

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.

11 Jun 2014: 2λ cos (πλ) 1)AI AR sin (πλ) CR sin (2πλ) = 0, (40b). ( ... κ2 cos (2πλ))CRµ CIµsin (2πλ) (κ1 2λ cos (πλ))AR AI sin (πλ) = 0, (40c)(κ2.

12 Sep 2014: dσd. =dσdt. dtd. =1. 2πdt. d(cos θ)dσdt. 4 The Dirac Equation. ... 2(ceR)2 1. 4 (1 cos θ)2. ]=. 12g4Z s. 2. m4Z.

13 Jul 2014: r = b. 21 162. b2 1/4 cos. 2,. i = 1. ... s = sinn. Lx n:even. cos nL. x n:odd , 16and the critical wavenumber is.

20 Jul 2014: indy =3 cos θ sin θ. 64π η1h3. [1 1. 2γ 2 1γ 2 1. ... 62). Using the same methodology as that described in Sec. IV B 2 and Appendix B, we get that,indy =. [,indy. ]0 cos2 θ ,x x cos θ sin θ ,x

14 Jun 2014: A and B) Neonatal diaphragmwas hybridized with probes for CD24 (A) or LL5 (B). ... Figure 7. LL5 is required for AChR clustering.(A) COS cells were cotransfected with GFP-LL5 plus control duplex VIII siRNA (left), LL5-RNAi#1 (center), or LL5-RNAi#2 (right

10 Sep 2014: χn = r nn. m=0P mn (μ)(cmn cos mφ c̃mn sin mφ). ... pn = r nn. m=0P mn (μ)(amn cos mφ ãmn sin mφ).

24 Sep 2014: The product is exp iθ exp iφ =exp i(θ φ). (b) SO(2) =. {R(θ) =. (cos θ sin θsin θ cos θ. ): 0 θ 2π. }. ... 3.3.1 Lie Algebra of SO(2). g(t) =. (cos f(t) sin f(t)sin f(t) cos f(t).

10 Jan 2014: x =. Nn=1. b cos θN E[{θn}] = FbNn=1. cos θn. ... f cos θn f =Fb. kBT. =. [4π sinh(f). f. ]N.

8 Dec 2014: P is an orthogonal matrix and has the form. P =. (cos φ sin φsin φ cos φ. ), cot 2φ = e2K sinh B. ... λr+ 00 λr. )(cos 2φ sin 2φsin 2φ cos 2φ. )(λr+ 00 0. )],

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

9 May 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).

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.

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.

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.

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.