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20 Jan 2009: Repeat the calculation in spherical geometry, i.e. for concentric spheres of radiir = a, b, b > a. ... For (iii), from a decent diagram of spherical polars, find k = er cos θ eθ sin θ. ].

13 Mar 2008: F = (3x2yz2, 2x3yz, x3z2), G = (3x2y2z, 2x3yz, x3y2),. H =(. 3x2 tan z y2exy2. sin y, (cos y 2xy sin y)exy2. , ... u = ((x cos y cos y y sin y)ex, (x sin y sin y y cos y)ex).

28 Nov 2013: X = 0 is given byF′ G = c0, (cos X σ0) G sin XF = 0, (4.4a,b). ... Θ = Ae(1κ2)τ cos κX Z, W = Ae(1κ2)τ cos κX, (5.1a,b)for any amplitude A.

19 Jul 2019: 161 –. by a sign, since this can be compensated by the cos function. ... Insisting that the two. cos functions agree, up to sign, gives us the condition.

2 Jul 2024: is given by. H(t) = eE0 x cos(!t). Our goal is to find the eigenstates of the time-dependent Hamiltonian. ... The diagonal matrixelements are particularly simple. h i|H| ii = eE0 h i|x| ii cos(!t) = 0.

11 Nov 2006: Then. z1z2 = r1r2 (cos θ1. cos θ2 sin θ1. sin θ2i (sin θ1. ... 1.30). Proof. From (1.25). cos nθ i sin nθ = ei (nθ).

2 Nov 2018: 12 e2iωγerf. (iω(γ 1). (iω)12. )(1 γ). ]. 27. Let. Ca,b(α) :=. ba. cos(αt)eiωt2dt =. π12 exp. ( iα24ω. )4(iω). 12. [erf. (a(iω). ... 2(iω)12. )],. C1,a,b(α) :=. bat cos(αt)eiωt. 2dt =. i. 2ω. [cos(aα) exp(iωa2) cos(bα) exp(iωb2). ]

3 Sep 2012: vr = rν cos φ[c(ν 1)Pν1 b(ν 1) cos θ Pν. ], ... 2.12a). vθ = rν cos φ[cP′ν1 2a csc θ Pν b.

15 Feb 2017: cos (λ,1(ck, t)) rλ,1(ck, t)sλ,1(ck, t). ... r′λ(t) = (1 q(t)λ) sin(θλ(t)) cos(θλ(t))rλ(t). and yields the seminal result:.

15 Aug 2005: Denoting by θn. a(t, yn), it is clear that. F (x) = exA(tn,yn) =. (cos xθn θ. 1n sin xθn. θn sin xθn cos xθn. ),. ... 1](x) = ϕ1(x). (cos 2θnx. θ2n sin 2θnx. 2θ3nx x. θ2n+ sin 2θnx.

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.

6 Jan 2021: Phys. Comm. 181 (2010) 232, arXiv:0903.1805. 4. B. C. Allanach, C. ... Sridhar, ‘Explaining Tevatron leptons, photons, missing ET events with. supersymmetry”, Proceedings of the IX International Symposium on Particles, Strings and Cos-.

30 Nov 2009: f (2k1)(1) f (2k1)(1)],. 1. 1f (x) sin(bx) dx cos b. ... where. F (a, b) = cos(a b) exp. (. a2. 4z2. ) [.

22 Jan 2018: b "arctan ðcos2 W þ q sin2 WÞR. 2ð1 " qÞsin W cos Wh! : ... Q2 1 þ 2hð1 " qÞsin W cos WRðcos2 W þ q sin2 WÞ!

26 Jul 2013: 4 m1=1. m2=1. 1. m12 m221 m12. 2m12 m2. 2cos 2 b m1x cos 2 b m2y , 18from which we note that large contributions to Eqs. ... cos m2w W. m1=1. 1m Cm 12m1 12 ln S sdwds , 20b.

26 Jul 2013: xm(x, t) =x 'a cos(x Re ω(t)t φ)and ym(x, t) = 'b sin(x Re ω(t)t),. ... xy|y=0 b sinx. 2ψ1. y2|y=0, (3.15). ψ2,x |(x,0) = a cos(x φ)2ψ1.

6 Feb 2019: This implies a = b,so using the formula cos θ 1 = 2 cos2(θ/2), we obtain,. ... 4.11). The antiderivative of cos(2 arctan(2x)) is arctan(2x) x, so there exist (new) realconstants a, b and c such that.

22 Nov 2017: rA A O ‘( cos u1, sin u1) ð4:1Þ. and. rB B O ‘( cos u1 þ cos u2, sin u1 þ sin u2), ð4:2Þ. ... 0 its magnitude andt̂ ( cos u2, sin u2) the unit tangent vector joining A and B.Under the assumption of creeping flow, the drag forces areFA 2 zvA

8 Jul 2021: Using spherical coordinates, wefollow ref. [12], and prescribe the tangential surface veloc-ity, us, as us = (3U0/2)(sin θ β sin θ cos θ)eθ, where U0. ... For a given value ofthe dipole strength, the maximum growth rate is obtainedfor cos k = 1, i.e.

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).