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16 Nov 2005: c) P = 2. q(. 1. q cos p). sin p and Q = log(. ... q1 = Q1 cos λ P2 sin λ , q2 = Q2 cos λ P1 sin λ.

10 May 2022: µs(θ)=. l=0. µslPl(cos θ), µp(θ)=. l=0. µplPl(cos θ). (2.8a,b). In the case of our spherical particle, the slip velocity due to either one ... s0 =sb(rc)r sR3. 2r2(sz cos θ s. y sin θ sin φ).

8 Jul 2009: B′ cos KX instead of A and B, with solution. N (0) ce(. ... A′ B′. 2sin 2T. B′. (2)2cos 2T. )cos KX. (5.10). Note that the oscillating part has to be retained in the present analysis up to the secondorder.

13 Dec 2018: b/ø (x,h) = exp{ik [xS h (x) cos 0 ] }. •ei•$[ 1 ikh(x)cos O- ( k2/2 )h2(x )cos 2 0]. (9). ... k2 •b• ø (x,zt) = - 1 ikh (x) cos 0- • cos 2 0 X [h2(x) (Zl-h) 2] )e ikxs -(za-h) •xx ikcos0.

26 Feb 2008: e1 B. (cos(βt) i sin(βt). )e2}. = eαt(a cos(βt) b sin(βt). ... J =. (0 1. (g/l) cos θ 0. ). The phase diagram has this form:.

8 Jul 2021: φ̇n = ω0 βω0 (cos χn cos χn1 ) βτ ω20 (sin χn1 sin χn ), (16). ... θ1 sin θ2 ) sin α1 sin δ cos θ1 (sin θ1 sin θ2 )]}, (46).

19 Oct 2018: µs(θ)=. l=0. µslPl(cos θ), µp(θ)=. l=0. µplPl(cos θ). (2.8a,b). In the case of our spherical particle, the slip velocity due to either one ... s0 =sb(rc)r sR3. 2r2(sz cos θ s. y sin θ sin φ).

2 Jul 2024: along the a = (sin, 0, cos) axis or along the b = (sin, 0, cos) axis. ... hEPR|Az B |EPRi = sin2. 2 cos2. 2= cos. This means that if we take the four operators A2, B1, A1 and B2 to be spin operators,.

7 Feb 2012: a magnetic moment m aligned along its axis. Let there be a magnetic field B, so that. ... each molecule has a potential energy mB cos θ (θ being the angle between the axis.

27 Mar 2022: R. cos. (. ω. (. t. -. t. 0. ). ). (4.299). with. a. ,. b. ,. R. and. t. 0. integration constants. ... A. cos. (. ω. (. t. -. t. 0. ). ). and. p. =. -. m. ω. A. sin. (. ω. (.

18 Nov 2015: 2. z2h5 B. 2. θ2h5 = 0,. (curiously A α/6, but B β/80). ... t1/4. h(π) =1 cos L. t1/4. (K ((π L) cos L sin L.

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

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).

9 Dec 2016: φjφk〉c =. dDq. (2π)Deiq(xjxk). r0 2βJD. µ=1 cos qµ. where qµ (π,π] is the µ-component of q.(b) Now take as the Hamiltonian (restoring the explicit ... G̃(q) =. [2. a2. Dµ=1. (1 cos(aqµ)) m2]1. [You need not repeat all the steps that are in

26 Jul 2013: With this in mind we let g2 = g1(z φ) with g1(z) = cos(z), andwe set equal for both sheets the bending stiffness B and the forcing amplitude A. ... B(576 B2. ) C, (3.7a)η2(z) = A. 12B [cos z cos(z φ)] [288 sin z (288 B2) sin(z φ)]B(576 B2).

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@,. -

26 Jan 2023: The general solution is. x(t) = X R sin(!Bt ) and y(t) = Y R cos(!Bt ) (1.2). ... m!B. Y = y(t) R cos(!Bt ) = y ẋ! B= y.

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.

26 Feb 2008: Tx = T sin θ cos φ. = T (r/l)(x/r). = T x/l,. ... ζ = eiωt sin λ(A cos(. g/l t) B sin(. g/l t). ).

26 Jul 2013: us(ξ,η)= u0s. 2(1 cos ξ)( cos ξ)(1 e2) (1 e2) cos(2ξ) s(ξ,η). ... Here hwis the minimal distance of the centroid to the wall without penetration, hw =(1 e2) (1 e2) cos(2θ)/2.

30 Sep 2008: General solution,. χ(x) = A sin(kx) B cos(kx). Boundary conditions from continuity of χ at x = 0 and x = a,. ... χ(x) = A cos(kx) B sin(kx). even parity condition,. χ(x) = χ(x) B = 0 χ(x) = A cos(kx).

8 Apr 2008: 4x.).ResonanceConsider ÿ ω20y = sin(ω0t). We know that yh = A sin(ω0t) B cos(ω0t) then the. ... sin. 1 κ2τ. B cos. 1 κ2τ ). This is a damped oscillator with decay time (time that takes the am-plitude to decrease by 1/e of the original value)

7 Apr 2021: x(t) = X R sin(!B(t t0)) and y(t) = Y R cos(!B(t t0)). ... AN=. g. 4r. 1 cos. sin(6.25). The resulting magnetic field is.

18 Feb 2009: E = (E0, 0, 0) cos(kz ωt), B = (0,1. cE0, 0) cos(kz ωt), (9). ... for E0 real,. (Ephys)x = 2E0 sin kz sin ωt, and (Bphys)y = (2/c)E0 cos kz cos ωt.

26 Oct 2020: Substituting y,a,b and f by theirFourier series and using (3.3)-(3.5) we obtain an infinite dimensional system of linear equationsfor the Fourier coefficients ŷn:. ... For instance: choosinga(x) = f(x) = cos πx, b(x) = sin 2πx (which incidentally

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)

26 Jul 2013: 1x, = U1ey , D4a. 1x,0 = b cos x tex a sinx t ey , D4b. ... UN =12 b. 2 2ab cos a2. D17. Allowing tangential motion does therefore reduce the swim-ming speed by the same amount, and the final results of thepapers, Eq.

24 Nov 2023: 26). 4. If a b = 0, and a, b 0 then cos θ = 0, necessarily, i.e. ... We have. a b = ab cos θ,(2, 1, 0) (3, 0, 1) =. 5. 10 cos θ,. cos θ = 6/. 50,. and thus θ = cos1(6/.

26 Oct 2015: When λ 0 then X(x) = a sin(λx) b cos(. λx) solves X′′ = λX. ... λx B cos. λx, but if we impose the boundary condition y(1) = y(1) then.

28 Sep 2020: z = r cos θ. Then compute the Christoffel symbols by directly applying (1.8). ... l2(1 e cos φ) (4.16). where e is the eccentricity, a free parameter.

14 Jun 2017: cos. (ωλ ,1(ck,t). )sin. (ελ ,1(ck,t). )sin. (ωλ ,1(ck,t). ), (29). ... 0. + cos. (ωλ ,1(ck,t). )(t ck).  t ck(t ck)21.

15 Oct 2020: S =. α β. β α. β. β α.  ,are given by λk = α 2 iβ cos kx , and wk = (i. ... λk (A) =λk (C ). λk (B )=. 1 12 iµ cos kx.

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).

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(.

1 Dec 2023: GF′′=α2F, σG+ iαA[G cos X F′ sin X]F′′=G′′α2G, (4.1a,b). where F and G are functions of X alone. ... which have the leading-order solution. F =a(Y)[sin X cos ξY X π], G=a(Y) sin X cos ξY (4.5a,b).

7 Nov 2017: u u u1, 1, 0, 1,3. 2sin 1 cos , 1r q q q q b q= = = f q( ) ( ) ( ) ( ) ( ). where θ is the azimuthal angle from the squirmer orientation e and ... From equation (1) we have that the tangential velocity on the squirmer is given byu 3 2 sin 1 cos ,

19 Jul 2019: x(t) = X R sin(!B(t t0)) and y(t) = Y R cos(!B(t t0)). ... 4r. 1 cos. sin(1.27). This again gives rise to the magnetic field (1.26).

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.

3 Jul 2024: x(t) = A cos(ωt) B sin(ωt) with ω =. k. m. ... E = mgl cos θ0. We can determine the period T of the pendulum using (2.6).

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). ]

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:.

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.

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.

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.

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: 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.

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.

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.