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18 Jan 2011: b(k) = d t cos(Á(k) µ)n (1). where b(k) is the observed intensity in the kth direction, d is the diffuse component which isindependent of insonification direction ... b(k) = d te. n. cos1. sin µcos µ cos Ψ cos µ sin Ψ.

21 Feb 2011: and a b c < 2π. Conversely, show that for any three positive numbers a,b,c lessthan π satisfying the above conditions, we have cos(b c) < cos a < cos(bc), ... Deduce theformula. sin α sin β cos c = cos γ cos α cos β.

25 Oct 2011: 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). ). Suppose f

25 Nov 2011: u(1, t) = ν(t) = eω2t cos ω b. ω2. (1 eω. ... p0(1, t) = ν(t) = eω2t cos ω b. ω2. (1 eω.

22 Jun 2011: cos x. (x2 a2)(x2 b2)dx. Also evaluate. cos x. (x2 a2)2dx. ... a)12π. eax2. eiωxdx (b)12π. sin x. xeiωxdx. (c). 12π. eiωx. x2 a2dx.

31 Oct 2011: 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). ). Suppose f

21 Feb 2011: 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. ... 10. 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). ). Suppose f

22 Nov 2011: 1! 1!cos##"". F B µµ–. e+ e–. µµ++. F B e.g. ... backward hemisphere. Prof. M.A. Thomson Michaelmas 2011 506. -1! 1!cos##"". F B!

13 Jan 2011: Algebraically,. 1. ap1 = 1, 0, 0 , 3. 1. ap2 =. b. acos , sin , 0 , 4. 1. ap3 =. c. acos , cos sin , sin sin , 5. where a, b, and c are the lengths ... 1. b/a0, 1sin , cos sin sin , 11av3 =. u1 u2u1 u2 u3.

21 Oct 2011: Freundt. where h̃ = h di tan θ and d = δd di / cos θ. ... t. (hi /g)1/2. =FT (0, φ) tan φ. (cos θ )1/2 (tan φ tan θ ). (5.2). Figure 8(b) reveals that this solution is a poor representation

2 Aug 2011: It can be shown that. D B sin cos eŁ tan. ... Dc B cos e tan. 2 cos =4 þ =2ð Þ sin =4 =2ð Þ (3).

10 Apr 2011: B s =(. η2. 4 ζ 2. )1/2 [Bx sin ω By cos ω. ] ... s = η2. [sin θ sin ωx̂ sin θ cos ωŷ cos θ ẑ. ].

10 Apr 2011: In a coordinate system moving at the speed of the colony, theradial and polar velocities are ur U[(c x3)cos A(x, )]and u U[(d x3)sin B(x, )], ... B 254, 107–113.12. Berg, H. C. & Purcell, E. M. (1977) Biophys.

26 Oct 2011: f0R(x) =1(x B(0,L cnε))cos(α(x))µ(C(L)). ,. where α(x) is the angle between π1(x) and (0,. ... f1R(x) =1(x π(C(ζ(0),ε)))cos(α(x))µ(C(ζ(0),ε)). and ξ(1) has density function given by. f1(x) =1(x

24 Jan 2011: 2. (2P1 cos(ωt ϕ) Q2. )py =. α. 2. (2P1 sin(ωt ϕ) P2. ... µ and WµWµ. The coefficients are. m2. λ[e cos α e′ sin α].

10 Apr 2011: as a Fourier series:. R(z) = R0(. 1. 0dq b(q) cos(qz φ(q)). ... Our aim is to expand up to second orderin the coefficients b(q) characterizing the deformation.

15 Jul 2011: ψ′ = Re. [. N 2H. f0 m sin(mH)eikx sin(ly) cos{m(z H)}. ]. ... ψ′ = Re[. eikx sin(ly){. B (t t0) cos(mz) C (z H) sin(mz)}]. ,.

13 Jan 2011: CP, Vi m. pbn,mVi log p b n , m V i , [5]. ... Berry RS, Elmaci N, Rose JP, Vekhter B (1997) Proc Natl Acad Sci USA.

10 Apr 2011: 2 ix1 (uku>2. ĥkeikx. 5 h̄1uĥ1uS 2 cos x1 f! 1 (uku>2. ... h x,t 5 0!5 h̄ 11 a cos x!, 0<x<2p. 58!

24 Oct 2011: 1 cos θ2 (E3/E1)(1 cos θ). ]. whereθ is the angle through which the positron has scattered. ... b) The Lorentz invariant matrix element forπ µνµ decay is. Mf i =g2W.

10 Mar 2011: The basic states B(z) and B(z) of the model obey. d. ... The geometrical functions A() and B() have the form. AðÞ 11þ7 cos, BðÞ ð1þ cosÞ2 3 cos94 cos: ð56Þ.

13 Aug 2011: 2i{(iωI A)1[et(iωA) I] (iωI A)1[et(iωA) I]}b. = etAy0 (ω2I A2)1Ab sin ωt ω(ω2I A2)1b cos ωt+ ω(ω2I A2)1etAb. ... I ω2A2)1Ab. = etAy0. r=0. (1)r. ω2r1(etA cos ωtI)A2rb sin ωt.

2 Aug 2011: r s 1 sin cos 2łð Þ. z s 1 þ sin cos 2łð Þ. ... ds cos þ 2s sin dł. þ n sr. sin cos dr þ sin2 sin.

6 Oct 2011: 0 0 0o b e i n g g i v e n b y :. R 2 = R 2 r 2 - 2 r R cos 00 (3). ... y b e t w e e n 1 a n d oo.

10 Apr 2011: VOL.2 3 7. B. 641. (Price 8s.) 5 [Published I 4Augucit I 952. ... b ) T h a t if A = A' ei$ and b = a$, where A', a, P, 41 are real, then ,%?Aebt= A' eat cos (pt 4).

1 Sep 2011: x[1,1]|f(x)|, and Tn(x) := cos n arccos x is the Chebyshev polynomial. ... otherwise. b) if νi > 0, then σi (λi,νi), c) if νi < 0, then σi (νi,µi).

14 Jul 2011: The z′ cpt of (2.1 a) is simply. i ω ŵ′ = σ̂ cos θ (2.12 a). ... and (2.1 b) reduces to. i ω σ̂ = N 2 ŵ′ cos θ (2.12 b).