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17 Jan 2008: whenever this limit exists. Remarks. • It follows from (1.20) and the definition of a partial deriva-tive (1.14) that. ... To derive (1.52) start from the definition of a directional derivative (1.20), and use the definition of.

27 Jan 2008: 20. for sufficiently small t. We will prove this. Let (U,ϕ) be a chart containing p then, in thesecoordinates, τ will have the form.

29 Jan 2008: 502, Campus Universitaire, Orsay, F-91405, France. Received 29 June 2007; accepted 15 November 2007; published online 20 December 2007. ... s1. Most experiments took place insidea tube of diameter d = 20 mm; other diameter values.

29 Jan 2008: d = 20 mm corresponding values of Ret sin :48.5, 19.5, 0. ... Control parameter values: At = 3.9 103, d = 20 mm, = 106 m2 / s, and = 1.

29 Jan 2008: The results were staggering,especially the minimization of a periodic function F (x), x Rn, with 100 vari-ables, although problems with n = 20 were considered large at that time. ... The average values of these counts with m = 2n1 are 931, 1809,3159 and

1 Feb 2008: 0 10 20 30 40 50 600. 5. 10. 15. 20. ... 3è1V6ñEG;=1"J /. i>31Vð@6:v3EF<@> <Cç35è1ENêïé-DF;1þ:C>-A 17>1V6ð"õÂ651V;éDN35;zEF> :@>. 5. 0 10 20 30 400.

25 Feb 2008: l1 =2V 21 sin θ1 cos θ1. g=. 2eV 20 sin θ0 cos θ0g. = ... L =. n=0. enl0 =2V 20 sin θ0 cos θ0. g(1 e).

26 Feb 2008: 1 Two cyclists start 40 km apart and head towards each other, each going at a steadyrate of 20 km per hour.

26 Feb 2008: f (x1)f (x2). =f (αx1)f (αx2). 20. for any α. (If we changed our units of length from metres to centimetres, for instance, we would haveα = 100.).

18 Mar 2008: Eight points have beencomputed for each integral. 0 20 40 60 80 1000. ... w0 20 40 60 80 100. 0. 0.5. 1. 1.5. 2.