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21 Sep 2005: 236.5 Calculation of seasonal indices. 24. 7 Fitting ARIMA models 25. ... 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

14 Mar 2005: Quasirandomness, Counting and Regularity for 3-Uniform Hypergraphs. W. T. Gowers. Abstract. The main results of this paper are regularity and counting lemmas for 3-. uniform hypergraphs. A combination of these two results gives a new proof of a

8 Aug 2005: 01. 20. 24. 2.5. 2. 1.5. 1. 0.5. 0. 0.5. 1. ... In Proc. IEEE European Conference onComputer Vision, pages 3–19, 2002. [24] G.

24 Apr 2005: 6" 4038- > 24/ 9J/1,?

24 May 2005: Part IID RIEMANN SURFACES (2004–2005): Revision Example Sheet. (a.g.kovalev@dpmms.cam.ac.uk). Note. The present 24-lecture course on Riemann Surfaces was developed from the 16-lecture

13 Jan 2005: lateral segmentation should help• Many optimisation issues! • Lateral segmentation: 1cm2 matched to RMoliere• Longitudinal segmentation: 40 layers (24 X0, 0.9λhad)• Achieves Good Energy Resolution:. ... ILCD Paris 13/1/2005 Mark Thomson 24.

9 Aug 2005: 73&)24!""K!{¢ I1[TS[Ma_XJ&:a[L5U£ [M3[TOL(¤O,Ls¥UIUL51Uc5=)0F0324!95c = c<-Cj,j,j. ... L5t$U3Ï[L$b,Ã=¿)71>'24<!'<;/"642B'0324),!"csjj&$. s"03)w!>Æ;24-),646B8% Ð!<24!$03<?71624!8732B!03!

3 Oct 2005: Sankhyā Ser. A 25,13–24. Csiszár, I. (1965). A Note on limiting distributions on topological groups.Magyar Tud.

6 Jun 2005: ii) Use the regression equation to predict the weight of an 18 to 24 year-old male who is 67 inches tall. ... iii) Use the above regression equation to predict the weight of an 18 to 24 year-old male who is 60 inches tall.

16 Aug 2005: 1 Bricklaying 2.24. 2 Masonry 2.24. 3 Carpentry 2.24. 4 Slater, tiler, thatcher 2.24. ... 5 Plasterer, painter, decorator 2.24. 6 Plumber, glazier 2.24. 7 Construction, public works, etc.

18 Jan 2005: E[. (J†δ)(J†δ)T]. = J†E[. δδT]. J†T. = (J T J)1 (24). ... 23). Analogously to Eq. (24), imagelocations are assumed corrupted by independent Gaussiannoise with σ = 1 pixel.

15 Feb 2005: 18.5. 19. 19.5. 20. mea. n D. ER. on. all. 24 d. ... RT-04f RT-04f RT-04f 23.86 24.12† Official evaluation submission. • New segmenter and clusterer resulted in 14% absolute drop in primary DER. •

25 Apr 2005: with a viewfor use ast are beingns. A widethat are notdels. Thet also that. tressing. Statistical Lifetime Predictions for Aramid FibersK. G. N. C. Alwis1 and C. J. Burgoyne2. Abstract: This paper investigates the statistical procedures that can be

8 Aug 2005: Both, centretemplateandmarginalizedtemplateshow betterclassificationperfor-mancethanthe trainedclassifier, in particular in the high detection range.At de-tectionratesof 0.99 thefalsepositive ratefor thecentretemplateis 0.24, wherasitis 0.64for

9 Aug 2005: 9 24 r24@&:$"fc!eC<<c (. %=. (.r2>< (5"< ; "9.( 0=& (!"f"5.-&86. (%= ... 7' n" (."#"<$ "< g" < (| &!ihj ( (!"|.$: ". E. "|c% } :=p &86 #.<2>.c"< # /.".)6 ><$ n"<$ 24!m (eY2K (12>< / "<$.

15 Feb 2005: 0.16 0.18 0.2 0.22 0.24 0.2612. 14. 16. 18. 20. 22. ... Cambridge UniversityEngineering Department. RT04 EARS workshop 24. Recent Developments at Cambridge in Broadcast News Transcription.

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16 Nov 2005: MICROTUBULES AND MAPs. By LINDA A. AMOS AND DANIEL SCHLIEPERMRC Laboratory of Molecular Biology, Hills Road, Cambridge CB2 2QH, UK. I. IntroductionII. Microtubule Structure A. Subunit Lattices B. Atomic Structure of the Protofilament C. Atomic

8 Aug 2005: Weizmann Horse Dataset [24]We evaluated the performance of our detector on a verychallenging dataset of side-on horse images; this dataset hasbeen used for evaluating segmentation algorithms [2], butwe do ... training images, a benefit in both collecting

9 Aug 2005: Z l <HlV6Cqh9<FH8:<98hdp. s. hZ<.hZ -;o8CEk%<o>F<H9<HieNPFWCq:u=6YH9<eNd+ s. s l <HdEeNHC 24-&V6CqCE:u7H18:<98hdp. ... d o FWCNo PeC:<9>=iHiX XeOCE:<98dpkoWCNeh9;e<HdEeNHC 24-&V6CqCE:u7 k6:e e4HCO9LwoWCE,:eN o89 oW=8HiXw.

19 Nov 2005: 24.27%) 5.66 neutralM 0.53 1.60 2.38(24.90%) 4.89 chargedM 0.30 1.63 2.40(25.10%) 1.57.

9 Aug 2005: 5 "_U#x! @#24)# $S9$M8$s!@P.6 [0!iIJ0!nU#C9$rrr. K=0!3&t(m;6tm$ 01.T$_]! 5' N-a! 3! 'T$0/k)! ... BC],0M01@"NM $A. #24) $-95l84+#24 1I{0!6A ¤!iv30M0MM@1#Y;A (An01. 2#A yAz 9$M|,U ZBC,D. ,=

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.

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

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

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

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,.

2 Jun 2005: Test. Group. Semantic dementia. Control (n 24)Experiment 1 Experiment 2. M SD M SD M SD. ... There was a significant effect of group on the large categories,F(1, 14) 9.24, p.

10 Dec 2005: construction, Procs Inst.Civ. Engrs, 21, 19-30, Nov 1943. 24 Burgoyne19. Thomas, F.G. ... structure, Concrete Constr. Engng, 43, 239-244, August 1948. 24. Anon, Prestressing applied to Strengthening a Church Tower in Staffordshire,.

19 May 2005: Systemeval03. dev04s25 fsh Avg. P2-cn HLDA 26.6 18.4 22.6 18.7. P3a-cn SAT 24.5 17.1 20.9 17.3P3c-cn SPron 24.7 ... cn SPAM 24.1 16.9 20.6 17.2P3h-cn SATSPAM 23.9 16.9 20.5 16.8P3i-cn CTRL 24.5 17.5 21.1 17.6.

7 Feb 2005: Coordinates are in. MNI space. A. Maril et al. / NeuroImage 24 (2005) 1130–11381136. ... A. Maril et al. / NeuroImage 24 (2005) 1130–1138 1137. the prefrontal cortex whose activation was linearly modulated by.

9 Aug 2005: 2 2 M# 39 D 24@ 9a3 8 C 6 6 @ C 6 6. ... r r$pqj q n,p. 8 /2!6 #(H 24@ A 2 @ 3C!3586 ( 0?

7 Feb 2005: Mathematical Tripos Part IILent 2005. Professor A. Iserles. Numerical Analysis – Lecture 121. Methods 3.25A generalν-stageRunge–Kutta methodis. kl = f. . tn clh, yn h. ν. j=1. al,j kj. .  where. ν. j=1. al,j = cl, l = 1, 2,. ,

12 Nov 2005: Find a monic polynomial over Z of degree 4 whoseGalois group is V = {e, (12)(34), (13)(24), (14)(23)}.(ii) Let f Z[X] be monic and separable of degree

4 Mar 2005: Implementation 5.24 It is quite usual to solvehyperbolic PDEs (advection equation, wave equation,Schr̈odinger equation, Euler equations of invicid compressibleflow. )

9 Aug 2005: 67 "#. @"# "5 8 ,"#0.@% #=24! #"5 $ % "#'"#! 8 24! 5"#A % "5'?"# ,"# =24. @ '?24,%. "#6. % "#'"#!( %<'" #9 =":9- 2. ... le2 22,. z q! % ,>- %<'! "! "# %<' %<?'. "#< ,"# #"#$ % "#'" 7 %?'. '! '? %<' ,"#. ) %<' 6'1 A( ), -2 01 '9;! "# & (&), - 24! '!"#:

5 Sep 2005: Phys. Commun. 162 24[20] Pask J E and Sterne P A 2005 Modelling Simul. ... Rev. Lett. 93 176403[24] Mostofi A A, Skylaris C-K, Haynes P D and Payne M C 2002 Comput.

9 Aug 2005: 2 f > f 35350 J<?L t'g<] e ]25342 gn h' [ <h'h <k% l e%f s<?]lxgk e 24<0l. ... 35 n J<?gh'l>lL2q0@ ' [ < f t_<?00 f 2' >'g#e[[ 35 e 24<0Nt e e,f / / l. (&)

7 Feb 2005: y0, n = 0, 1, 2,. 24. The following four-stage Runge–Kutta method has order four,.

9 Aug 2005: " #$ %& '& ()!,&-&. 0/ 12436587:9 ;=<?>"@BAC361D4EGF!H 24@. I8JLKNMOKQPSRUTWVXNXZY-[YT]_'aOJ=KZbO[cYd. T]_'aOJ=KZbO[cY'eGMOKNfcYJ=gLKQPSRihMO[cKZMOY-YJ=KZMO[kj Ylm]J=Pn_,Y-MoP. ... D>WH 7 m;6D1 33 24; "5 @ 5 D>"@!;m;6>W@ 9G7 58< 7.

29 Jan 2005: 20. we might take. E1,1 =. . . . 24 12. ... . , E1,2 =. . . . 24. 2. 4 02.

15 Jun 2005: α0. 24 1(1)(2). 35 = α0w = nXi=1. αsvmi yiG1(oi; ) (22). ... O; )). 375 (24)This allowsβ to be set using maximum margin training. Howeverempirically settingβ has a number of advantages.

12 Apr 2005: 302k 9k+ 16.0 13.9 24.8 –MLE 415k 9k+ 16.0 13.5 24.3 –. 398k 12k+ 16.1 13.6 24.5 –. 302k 9k+ 13.2 11.2 ... dev04 eval03 dev04f. MLEMPron 16.0 13.6 24.5SPron 15.6 13.5 24.2. MPEMPron 12.9 11.1 19.1SPron 12.7 10.8 18.8.

26 Jul 2005: LT. 0.500E-01 CONF 1 E = 62.24 ( 0.044) replaces structure 31 Minimization converged; gradient = 0.360E-01. ... LT. 0.500E-01 CONF 1 E = 62.24 ( 0.032) replaces structure 31 Minimization converged; gradient = 0.439E-01.

26 Jul 2005: 3 CONF 1 E = -201.24 ( 0.607) is unique and stored as structure 4 Minimization converged; gradient = 0.398E-01. ... 0.161) replaces structure 24 CONF 1 E = -221.33 ( 0.425) rejected by starting geometry Minimization converged; gradient = 0.456E-01.

26 Jul 2005: LT. 0.500E-01 CONF 1 E = 154.00 ( 0.049) rejected by structure 24 Minimization converged; gradient = 0.380E-01. ... LT. 0.500E-01 CONF 1 E = 154.00 ( 0.028) rejected by structure 24 Minimization converged; gradient = 0.469E-01.