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16 Feb 2008: English) s = sm1. Note that each tj and si is a segment, which is to say a string that.

19 Nov 2008: Prove the results (a = |a|),. (i) [Ti , Tj ] = iijkTk ,. (ii) (aT)3 = a2aT ,. (iii) exp(iaT) = 1 iaT sin aa (aT)2 cos a1.

22 Aug 2008: On full abstra tion for PCF:I. Models, observables and the full abstra tion problemII. Dialogue games and inno ent strategiesIII. A fully abstra t and universal game modelJ. M. E. HylandDepartment of Pure Mathemati s and Mathemati al Statisti

21 Oct 2008: XJETEO@6)9!;8ANEE0Y]!6_YU-E]E;Y(;O9(%OV?"%S(TJ M6!3#)G2574($;-)9W?I4/[LY.EE29NQ4S-%.U?)74__6KD9XTZG'

16 Feb 2008: Word alignment a = aJ1: saj tj , j = 1, 2, , J = hidden r.v.Conditional likelihood P(t, a|s) = complete dataSentence translation P(t|s) =. a P(t, a|s) = incomplete ... t1 t2 ….tj- tj tJtj-…. …. s1 ….si sI…. tj1. αj (i, φ, h) =i′,φ′,h′.

17 Jan 2008: Mathematical Tripos: IA Vector Calculus. Contents. 0 Introduction i. 0.1 Schedule. i. 0.2 Lectures. i. 0.3 Printed Notes. ii. 0.4 Examples Sheets. iii. 0.5 Previous Comments, Comments on Comments and Style of Lectures. iii. 0.6 Books. iv. 0.7

31 Jul 2008: If we consider the two following expansions:. h(t) =. j=0. aj tj, h(t) =. k=0. bk. (. t. 1 t. )k. , (8). which converge inside D, then it is true ... By using the first expansion in (8), it follows that. bk =. j=0. aj Bj,k, Bj,k =1. 2πi. Ct. tj(1 t)k1.

22 Aug 2008: On full abstra tion for PCF:I. Models, observables and the full abstra tion problemII. Dialogue games and inno ent strategiesIII. A fully abstra t and universal game modelJ. M. E. HylandDepartment of Pure Mathemati s and Mathemati al Statisti

28 Feb 2008: for each continuous function f on [a, b]. (2) Let Tj be the jth Chebychev polynomial.

12 Dec 2008: M. K. JONES, DARJ. N. POSTGATE, TJ. D. RAY, SE. ReadersC.