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10 Dec 2006: B. C. A. 0. cosh. sinh. cos. sin. =. -. L. ... a. a. a. a. a. a. a. a. -. -. =. 0. ). sinh. (sin. ). cosh. (cos. =. -. -. B.

23 May 2006: Solution. h1 = cos θ A(T ) cos(θ τ ) B(T ) sin(θ τ ). n>1. [An cos n(θ τ ) Bn sin n(θ τ )] eαn2(n21)τ. ... Secularity condition – resonant sin(θ τ ) and cos(θ τ ). d. dT. (. A. B. ). =. (. a b. b a. )(.

30 May 2006: 5.13 a) = cos{Ic(y-sy))-Sjsin{k(y-sy)) (5.13 b ) = ( q 2 c2) cos {Ic(y - 9 ) tan-1 (&I)} ( 5. ... 5.16). (5.1 7 a). 00. = 2 K 2 b / (7 cos K 7 csin q)2dx, (5.17 b ) --m.

25 Jan 2006: Conversely, show that for any three positive numbers a, b, c less than πsatisfying the above conditions, we have cos (b c) < cosa < cos (b c), and that thereis a spherical ... The polar triangle A′B′C′ is defined by the unit vectorsin the

14 Feb 2006: nbr = n-b, (5). where n = e, cos Be,sin /I is the unit normal to the band, the remote stress tensor 6x is. ... and. e, = log C cos(B-4-& cos (P - 6) 1 Wb) For small values of y”, C/I and 4, we can ignore the difference between fi and Do,.

22 Jun 2006: We can substitute c in each case for b and v using eq. ... Rearranging, and noting that r̂1 = (sin θ, cos θ) and r̂2 = ( sin θ, cos θ), we have:.

19 Apr 2006: a@ a(1 [ b)/[1 [ 2b(1 ] e cos f )/(1 [ e2)] , (19). ... e@ (1 [ b)1Je2 ] 2be cos f ] b2 , (20)u8 @ [ u8 f [ f @ arctan [b sin f/(b cos f ] e)].

28 Nov 2006: RAPID COMMUNICATIONS. PHYSICAL REVIEW B 68, 241312R! 2003! Raman spectroscopy of silicon nanowires. ... g. of. i-l. es. nf-l. se. uC(0,q)u25 exp(2q2d2/16p 2).6 We approximate the Si TOdispersion as v(q) 5@ A1 B cos(qp/2)#

5 Jun 2006: B Numerical Integration 108. C Derivation of Multi-Variate Gaussian to Log-Normal Mapping 111. ... case of CDHMMs, B will consist of a set of means, variances and mixture com-ponent weights.

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).

30 May 2006: 2. 1 0 4 and. 992n(& 7) = ( - )%-l sin (o cosh < cos 7 cos t ) cos (o sinh 6 sin 11 sin t ) se2,(t) dt I 0 =. { 2.10 b ) ... 4. 1 d ). x = 8% cosh [cos@, y = & sinh %sin@,.

3 Feb 2006: cosh c = cosh a cosh b sinh a sinh b cos γ. ... Hint: Reduce this to showing that (sinh a sinh b)2(cosh a cosh b cosh c)2 is symmetricin a, b and c.].

28 Nov 2006: ata-rf. ered-. ughrustr tos is. b-. C,. 6006 J. Appl. Phys., Vol. ... In ocaseuC(0,q)u25 exp(2q2d2/16p 2). We approximate the STO dispersion asv(q) 5@ A1 B cos(qp/2)# 0.51 D, with A5 1.7143 105 cm2

19 Apr 2006: cos. ; ð13Þ. where apl=a, and fs;1, fd, and fi are coefficients corre-sponding to the secular, resonant direct, and resonant indi-rect parts of the disturbing function, respectively. ... The result is. cosm2 : 1. 0:39 0:061=e ; ð26Þ. and these lines

26 Apr 2006: Fk"(2pR2)6kR2;Ph#. 0. sin h cos hAPh. h#. sin /. d3d/B dh. ... oq. j3. 6kR2:h#0. sin h cos h d3dh. :h#0. sin h cos hA:hh# sin / d3 d/B dh,.

22 May 2006: 1. 0. dy cos(py) = ddp. (sin(p). p. ). =1. p2sin(p) 1. ... X(x) = A sin(αx) B cos(αx) ,. T(t) = D exp(κα2t). •

16 Nov 2006: cosh a = cosh b cosh c sinh b sinh c cos α. ... Let G be the set of Möbius transformationsz 7 azbczd with a, b, c, d Z[i] and ad bc = 1.

30 May 2006: Howard’s technique applied along a = 9 4 3 with n = Q & (a - J)Q and H = (cos $Y)(&--)’ yields. ... There is another neutral eigen- solution. (2.13 a, b ). J J3(a) = 2 - a2 (J2 < a < 4 3 ) (2.13 c) c = 0, 4 = (cos 9y)4(’t-J)’ (sin

30 May 2006: 6 ). We note the alternative representation. sgn ($77 - t ) exp ( - ikr cos f? ... exp ( - ik< cos 8) sin (ky sin 0). (

26 Apr 2006: dd. dy 0 at y 0 ð4aÞ. and. d! H R cos ycos y. ... Wecomplete the variable transformation by definingthe horizontal coordinate x 1 cos y so that f isa positive, monotonically decreasing function of x,where x ranges between 0 and 1.Eq.