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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: 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. (&)

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.

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

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.

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

17 May 2005: chosen to have a stiff-ener width of 0.06R = 24 mm, load angle γ = 7, hingeangle η = 14 and a stiffener thickness of 2 materialplies. ... This results in a plan stiffener width of22.6 mm, compared to that of 24 mm in Config VI.It also worth noting that

9 Sep 2005: ophthalmologist preferably within 24 hours of the injury occurring. The injured person must not drive. ... wavelength, power/energy per pulse and pulse duration. Addenbrookes Hospital: Accident and Emergency Department open 24 hours a day.

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: S1 6K (28)0. 34.1 26.0 30.2 26.4S4 9K (36). (ML)33.0 24.8 29.0 25.3. ... 32.3 22.9 27.8pMPEMPE 35.1 26.0 30.7 33.2 24.1 28.8 32.9 23.6 28.4.