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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.

30 May 2006: A ) = 1 m(sin h f - Rh cos A f ) h(6) df nrh(lR2h2) 0 For R = 0, the second term in the integrals in ( 6. ... RBO(5) [sin (C/R) i sgn k cos (C/B)], (A 1 ) where.