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9 Aug 2005: £ð5M-ERDMNSéø3v s d d ( Nd û|uü | ø. T%B%V% T(! ... ERDMZSMZà EH<ZîZ["Û<-QHMZ=?SLT Û["ë3MNSLT?ÛdMZS#<ZûNûZðWÅ;[SsERD=?B fã.

18 Jan 2005: Lïiì Õ Þòð;ñ;à Õ Ð5 Ðîbª5àÏTÐbÑÑEÑEÐàÒ Â5ËE»b«»b¿ª¥bÔwb¿ ¥ÂµwEÁmY«=ª¥«'á;Sû!YE»lª=5bi=T¥«=µ¥ ðSÔw}Á'TþwEÁÁEíYE=¥i¥w bªEÁmª¥@¥Âµ(ðÔs«=ÓªEüÖwE¿ª@ªÁm¥w¥Âµ¥w¥«=

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12 Apr 2005: If B B, then fB(ω) is definedto be the value of f(φ) for any φ with φ(A) = ω(B). ... 3.1 as follows. Let K1 be the chain of all subsets of all sets B B.

16 Aug 2005: yi2/C&HJ&%U-216!7--!"3H&!721PWE'}Z>0"2 -&B&%U@H.4&!70"62 GN-%Z&H&!7-B_-B-!7H.4074#! "C-%p@H. @-!"0724GP&%p&4. ... 721Y24GucM} 4qJ@H.4-!7gY5T4070"B-B-!7H.40uB>&%PWN'%!"5%$!"Z>0"2 -&Bj!7$>.

19 Jan 2005: I AvI,so positive definiteness implies̄v>Av > 0 andv̄>(A B B>)v > 0. ... ForJacobi’s method,A B B> is the same asA except that the signs of the off-diagonal elements arereversed.

9 Dec 2005: Then Cramér’s estimate is. P {B > b} 1 νµ. κηeκb as b. ... a1eκb P {B > b} a2e. κb b 0 (4.5). for constants a1, a2 1.

27 Sep 2005: M.10 (4L, weeks 5–8). DR R. B. B. WARDYPlato Ion. F. ... Tu. Th. 12 DR R. B. B. WARDYPresocratics: Thales to Heraclitus (8L, weeks.

21 May 2005: B. C. C. B. B. C. C. B. C. C. B. ... B. C. C. B. B. A. A. A. A. A. A.

19 Oct 2005: 0.1. 0 1. a) b). c) d). e) f). Theta_max. B B Dd+ -. s-s. - π π π π π. Acc. epta. nce. Figure The geometric acceptance as a ... BsBs mixing. The general formalismof B B mixinghas beendescribed many times and is reviewed in Ref Allowing for B!

26 Aug 2005: actions are shown after 10 thermal cycles at the end of the high temperaturethe peg,x < b; (b) normal traction on the TGO/bond coat interface beyond the. ... a) Traction on plane within the peg,x1 < b; (b) traction on the TGO/bond coat interface beyond

15 Feb 2005: horizontal loading of a rigid rectangular on the surface since the ratio l/B approaches zero. ... Broms, B.B. (1964) “Lateral Resistance of Piles in Cohesive Soils,” Journal of the Soil Mechanics and Foundations Division, ASCE, Vol.

24 Nov 2005: p p. a. b. ’. b’. q=pb’ bp. Then (in Lorenz gauge) Aµ satisfies. ... p p. a. b. ’. b’. q=pb’ bp. Hence. |Mf i|2 =e2ae.

9 Aug 2005: E 9v?"(9Q"$ /Q58 Qv?"-! B!"1l Q Q } 8r? " vV( v! ... b v;X 2t ; ' v( Q?" rv! 8?": v( 3R?3Rvv v 2t < H8.

17 May 2005: . N. M.  =. . A. B. B. D. . ... It is assumed that the laminate is symmetric B = 0,and is subject to pure bending and large curvature κx,i.e.

11 Jul 2005: No! Things aren’t that simple…. Dense Jet: B=0. neutral. ve- ve. ... puri. ty. c. b. b). c (b bkgr). ECM = 91 GeV.

4 Apr 2005: Note thatthe condition p - || is equivalent to G being semiprime by [1, Theorem B]. ... By Lemma 4.4(a) and (b), XV i is finitely generated projective.

8 Dec 2005: say with active coordinate set I, such. that for all a [m]n and all b, b′ L{L+}, we have c(a, b) = c(a, b′) = c′(a),say. ... ii) A B U if and only if A U or B U;.

20 Sep 2005: 1. B AN xN = A1. B b. Why is it not fair to saythat the simplex algorithm is just a complicated way to invert a matrix? [ ... F5. A gambler at a horse race has an amount b to bet.

17 Jul 2005: As can. a. b. B. (. B. (. Figure 4. The average B-factors per residue taken from thecrystal structures of (a) ubiquitin and (b) MoaD. ... Biophys. J. 86, 2383–2391. Rasmussen, B. F., Stock, A. M., Ringe, D. &

24 Apr 2005: ijBJHGimvwKN?k2HKBI(9PZN]N?IqeHrB Q/C;MQ/bHKBI(E H KZdTP Q/GF;n2HT9A yG/ W a % HKBM/C %SILOFMN]B : /G/ W B/#b ]tHKhb = imy HKBIdTBIk4iy ... 243">$/B=6I B:, ]5pQ E; -02 (.<(E - V @L- > B:2 -1>=,B > 3 1?>< , 3@@?L-FE 0-2 - >. VGW ) : )¢ / W B/ b V (). ) :.

17 Feb 2005: mf (b, b). xkymk. ei(bκ1aκ2) mf (b, a). xkymk+ ei(aκ1aκ2). mf (a, a). ... s 1, k = 1, 2, 3, 4,. wherev1 = (b, b), v2 = (b, a), v3 = (a, a), v4 = (a, b).

14 Dec 2005: 5?/2 g à )9 $DtÄyª g à 93x?ÄVg à ;9 $D»Äª B B Ã Ä G. ... Bg à ;9 D<Ä G. ÆIg à )9 D,<?Äyª B B à 9Ä.

14 Mar 2005: Proof. Define B [N]2 to be the set of all (x,y) such that xy A. ... labelling vertices of a cube (as they would if a, b and c denoted three orthogonal vectors.

21 May 2005: BA. B. B. B. A A D. A. DCC. AA A A. ... 4, 5, 6} }. B B. AA. 2. 1. 2. 1. 3 4. 65. defines a simplicial complex that is a pseudo 2-manifold but not a 2-manifold.

9 Aug 2005: D /kb,"FiAF,"b.@b(D 5/ 5tEb?A,F,Bsk7 bBKb?4.'b. ... kD x?A,"Fb?4.'/ b?A5kB), 5cAb.D /A,8 DZt j,. 7n795j,8 bF.D /kDl/}x].'/48 b, B)b(D /}x]B,"bBEb?4.'b6.'F, /A5b68D9B9G5D

9 Aug 2005: $%! b%)(P3- % A "& 8$C(/$%B#."&TS %fN % B7 &N # )(.&B#.8 7G+ / %&&)(P%WPb%B&b%B!'&"#G6B&! & "# ... 7= # % &N (< %"! B7 <= # %. #b <b% 8. %BA -b%B!<'&B#.@(<b%B7bB! % 8 )($! %

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8 Dec 2005: Position relative to pile tip, 2h/B(b). Vol. umet. ric s. trai. ... 18. 200 2 4 6 8 10. Horizontal position relative to pile tip, 2x/B(b).

21 Sep 2005: t=1. sin2(ωt) = T σ2/2 ,. 15. cov[A(ω), B(ω)] = E. [. ... 24. 7 Fitting ARIMA models. 7.1 The Box-Jenkins procedure. A general ARIMA(p, d, q) model is φ(B)(B)dX = θ(B)ǫ, where (B) = I B.The

29 Jan 2005: an R2n2. Proposition 6 The following is true:. B>B C>C = I, (4.3). ... a>p aq = [ b>p c. >p ]. [. bqcq. ]. = (b>p bq c>p cq) = (B. >B C>C)p,q = δp,q.

21 May 2005: BA. B. B. B. A A D. A. DCC. AA A A. ... 3. (b) Now show that γ is an epimorphism (i.e., is surjective).

9 Aug 2005: N$9%9 #%>7#,- %>.- "%-. r29B!7#.- >!$&9B&B/ <B&!-B/a -.I!& 7#. -$& , ,-7#Gp! ... 213%#B>! @}(|B&B&29 s BC!,p & R7#. $.p#{S B&B&! F '- ,$7#B&.-,{S = 4$&5.

19 Apr 2005: Equilibrium un-folding data were collected under the same conditions using aPerkin-Elmer LS-50b fluorimeter with excitation and detection at280 nm and 356 nm, respectively (K.T., B.B.K., ... B.B.K., F.M.P., and K.T. acknowledge sup-port from the John and

7 Jul 2005: φ. Section B-B. sd. p. Geometry 1. 10 mm. Geometry 2. ... Geometry 2,G = 0:092: (a) unconstrained compression; (b) bonded compression; (c) constrained compression.

9 Aug 2005: YL < :k| K2 -J4)? > :B,B"E77)W(& 8-B:I; D 7? ... 7! bP : k2q5),) ) &y ,4/0? x :B,B -,)4sE: )E,B4 j H.

19 Jan 2005: by solving. (A B)x(k1) = Bx(k) b, k = 0, 1, 2,. ... If the real symmetric matrices A and A B B> areboth positive definite, then the spectral radius ofH = (A B)1B is strictly less than one.

21 May 2005: A B = {a b : a A, b B}, AB = {ab : a A, b B}. ... It shouldbe clear that. b. af > 0, and you should explain why this is so.

24 Apr 2005: 7 @?A% B DGF H J /! 78+#/! ; & 5$ ),H ; 3 £? V BN 8% $! ; ' ... N ¥ A "p ), ( " N ¥ A = & Q' 5 B u p 2@A u.

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21 May 2005: 6. If A and B are subsets of Rn, we write A B for the set {a b : a A,b B}. ... Show that if Aand B are both closed and one of them is bounded, then A B is closed.

30 Sep 2005: Ξ = {ξ(a) : ξ(a) 6= ξ(b) for a 6= b} (4.3). ... look like figure 7. B B B ( )( ( )1 0 11)0 1.

24 Apr 2005: d( 6 Á j¤ : %'=:<> iÁ Á j¤ : %'=: 'Áª B # z zE C @EF Õ$# C! ... E. E. Õ E. Õ E. Õ. Õ. [E. [E. B #. B #. f«z Ñ / ¡ @zRDR@¤H Å zRzHzRp«$ ¢ª@HRHz ¡ ¡¡ zHªD « 4Gg2 S l %' #¤z¿ ( Á @?KJ B|DzR/ #

9 Aug 2005: c7[9Z; U >M}:@ Hu<:7lM9G>K<:; S HA?B?<?B?B?<?B?<?B?C?B?B?<?B?<?B?<?B?B?<?B?C?B?<?B?B?<?B? ... B?<?B?B?<?B?<?B?C?B?B?<?B?<?B?<?B?B?<?B?C?B?<?B?B?<?B?fDQ?

21 May 2005: 6. If A and B are subsets of Rn, we write A B for the set {a b : a A, b B}. ... Show that if Aand B are both closed and one of them is bounded, then A B is closed.

9 Aug 2005: B [ b D6D y L B S B eAR [ DYALNQcR0L B y D [4T D6BCD Q LN] D SOU l K,x. ... n0LNQkQ D y<eAT B RZ]cRZe B b D Q% eAQML B y D x ePLN]c] D M b ] 2LA[.

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21 May 2005: 8. Part (b) is incorrect: In the figure, the segments labelled A are supposed to beidentified and the segments labelled B are supposed to be identified, but theresulting segments A and ... a},{1, 2, 5}, {1, 2, 6}, {1, 3, 4}, {1, 5, 6},{2, 3, 5}, {2, 4, 6}

26 Sep 2005: Let us compute the commutator of twosuch vector fields F(A,a) and F(B,b). ... Thus. [F(A,a),F(B,b). ]= F(C,c), where (C, c) = (AB BA,Ab Ba).