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26 Sep 2005: Adexp(a) = expm(ada). (2.24). 2.4. Differential equations on manifolds. We wish to return to general differential equations on manifolds, as given in (2.1).In order to construct

15 Feb 2005: 29 results in:. AArB. x. 12.575.025.01. 24.55.02. 0. %. Eq. 31. ... Technical Report of the University of Cambridge CUED/D-SOILS/TR 334 Page 24.

9 Aug 2005: Equality at. A : From (24) we find thatf 4 A 6 j 4 d 6. ... 3. Equality of the first derivative at A : From (24) we find that.

9 Aug 2005: "!$#&%')(,- /.0%1+-23! 4356078 9. :<;/=?>0=A@CBED)FHGAGI FHJLK0;9=NMOHPQR>0OSG?FH>1M. T)PVUWPXJKPY;[ZYSH]. _SaNbcdaNb"bXeXfWgihHgjajkmlnaNbbpoqpr asgigie"cutWkmfLv"kdl0bpajcSeWf/hHgjajkml tWkmfLgj_we

12 Jan 2005: 17.1/66.1 44.9/20.1/68.8 51.7/24.8/79.8bnrt04 v1.0 tg 49.7/15.4/68.6 50.2/15.0/68.5 ... 68.0 44.3/24.4/72.5 50.7/28.6/82.7bnrt04 v1.0 cl40-tg 49.1/17.1/68.3 49.4/21.2/74.6

26 Jul 2005: LT. 0.500E-01 CONF 1 E = 40.20 ( 0.043) rejected by energy CONF 1 E = 24.24 ( 0.123) is unique and stored as structure 12 CONF 1 E = ... 0.102) rejected by structure 16 CONF 1 E = -13.19 ( 0.631) rejected by starting geometry CONF 1 E = 24.24 ( 0.100)

3 Feb 2005: Accretion Discs Mathematical Tripos, Part III. Dr G. I. Ogilvie Lent Term 2005. Linear case. In the linear case, the general solution may be found as a linear superposition of elemen-. tary solutions. One may look for elementary solutions in which

5 Apr 2005: 14 Rubin149 07 24 14.40 -00 33 04.1 14.124 14.145 14.206 689. ... x RU149D 07 24 15.36 -00 32 47.9 11.435 11.445 11.469 689 syn.

30 Jan 2005: But what we need is simply a Hamilton path from 0to 54 in G3 [[0, 21] {24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60}], for which ... 0, 3, 1, 4, 7, 10, 13, 16, 19, 57, 60, 20, 17, 14, 11, 8, 5, 2, 6, 9, 12, 15, 18, 21,24, 27, 30, 33, 36, 39, 42,

25 Oct 2005: 81] 0.63 [.58,. 68] 0.58 [.52,. 64] 0.59 [.52,. 66]d2 Depth 2, source discrimination 0.32 [.24,. ... 54] 0.31 [.27,. 35] 0.36 [.31,. 41] 0.29 [.25, 33]a3 Guess depth 3, item detected 0.24 [.2,.