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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

5 Jul 2005: II | 27/170 48/146. III | 11/52 29/64. Unclassified | 24/118 49/104. ... P.M.E.Altham 24. Null deviance: 452.346 on 89 degrees of freedom. Residual deviance: 83.201 on 84 degrees of freedom.

27 Oct 2005: 24) (i) Show that H(X|Y ) 0 with equality if and only if X is a function of Y.

23 Mar 2005: DIFFERENTIAL GEOMETRY, D COURSE, 24 LECTURES. • Smooth manifolds in Rn, tangent spaces, smooth maps and the inverse func-tion theorem. ... O’Neill, Elementary Differential Geometry, Harcourt 2nd ed 1997.1. 2 DIFFERENTIAL GEOMETRY, D COURSE, 24

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,

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

21 May 2005: 24}. It follows that in dx, the coefficient of cv,w isalso zero, and so dx = 0. ... 10,. , 24 circle, we see that c10,. , c24 allrepresent the same element as c0 in H0.

24 Nov 2005: T.A.Fisher@dpmms.cam.ac.uk - 1 - 24 November 2005. 54) Alice and Bob are issued with RSA public keys (N, e1) and (N, e2), and correspondingprivate keys (N, d1) ... Show that the sequence in (ii) has period 2d 1. T.A.Fisher@dpmms.cam.ac.uk - 2 - 24

18 Oct 2005: ANALYSIS II EXAMPLES 1. Michaelmas 2005 J. M. E. Hyland. The Basic Questions are cover examinable material from the course. The Additional Questions arefor those wishing to take things a bit further. The questions are not all equally difficult; I

21 May 2005: 24. Let f : R2 R be a continuous function satisfying a Lipschitz condition|f(x,y1) f(x,y2)| K|y1 y2|.