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14 Feb 2012: dx log n for n N. Show that In =. 1/20. f (t/n) dt and hence that|In| < 1/9n2. By considering. ... 7. Let In = π/20. cosn x. Prove that nIn = (n 1)In2, and hence 2n2n1 I2n1/I2n 1.

25 Oct 2012: D) In lectures I mentioned some observations of Fermat and Euler.(Fermat) If primes p and q are each congruent to one of 3 or 7 modulo 20 then thereproduct is ... congruent to 3 or 7 mod 20 then 2p is so expressible.What can you say about these

21 Feb 2012: Use these to reconstruct thecharacter table of S5. Then repeat, replacing S4 by the subgroup 〈(12345), (2354)〉 of S5 of order 20.

8 Nov 2012: Use these to reconstruct thecharacter table of S5. Then repeat, replacing S4 by the subgroup 〈(12345), (2354)〉 of S5 of order 20.

16 Feb 2012: 20 and σ. 21 (> σ. 20). Show. that this test is a size α uniformly most powerful test for testing H′0 : σ2 σ20 against. ... Improved No difference WorsePlacebo 18 17 15Half dose 20 10 20Full dose 25 13 12.

14 Feb 2012: dx log n for n N. Show that In =. 1/20. f (t/n) dt and hence that|In| < 1/9n2. By considering. ... 7. Let In = π/20. cosn x. Prove that nIn = (n 1)In2, and hence 2n2n1 I2n1/I2n 1.

11 Oct 2012: 2. Between 10 and 20 there are 4 primes; does it ever happen again that there are 4 primesbetween two consecutive multiples of 10 (apart from between 0 and 10)?

21 Feb 2012: Use these to reconstruct thecharacter table of S5. Then repeat, replacing S4 by the subgroup 〈(12345), (2354)〉 of S5 of order 20.

28 Feb 2012: α (deg) 5 10 15 20 25 30 35 40 45sin 2α 0.1736 0.3420 0.5 0.6428 0.7660 0.8660 0.9397 0.9848 1range (m) 4860

27 Oct 2012: . . 1 1 00 3 20 1 0. .  ,. . . 1 1 10 3 20 1 0. .  ,. . . 1 1 11 3 11 1 1.

28 Nov 2012: Theorem 1.20. If the smooth map f : X Y is transversal to a submanifoldZ Y , then f1(Z) is submanifold of X. ... Definition 2.20. The determinant of dNp is the Gaussian curvature K(p) ofS at p.

27 Nov 2012: 20. For Riemannian geometers, on the other hand, the objects of study are fromthe beginning Riemannian manifolds.

27 Nov 2012: Dodecahedron 20 3 60Icosahedron 12 5 60. These give finite groups of orientation preserving symmetries. ... Lecture 5 19. Lecture 5 20. 5.2 Regularity. The Platonic solids are the only regular, convex polyhedra.

26 Apr 2012: 1 20 1. ). defines a representation (check). The representation R2 breaks up as 〈e1〉 〈e1 e2〉 as subreps even though thematrix is upper triangular but not diagonal. ... χρ,χρ〉G = 1|G|. gG χρ(g)χρ(g) =|N||G|. gNG/N χρ(gN)χρ(gN) =

9 Nov 2012: Use these to reconstruct thecharacter table of S5. Then repeat, replacing S4 by the subgroup 〈(12345), (2354)〉 of S5 of order 20.

28 Nov 2012: 1 20 1. ). defines a representation (check). The representation R2 breaks up as 〈e1〉 〈e1 e2〉 as subreps even though thematrix is upper triangular but not diagonal. ... Lecture 9. Recall the character table of S3. Example. S3. 20 SIMON WADSLEY.

27 Nov 2012: 20. Let G be the graph obtained by subdividing a single edge of Kn,n by a new vertex.

2 Mar 2012: 20 is smooth ifchar(k) 6= 2, 3. Find the degree and ramification degrees for (i) the projection φ = (X0 : X1) : V P1 (ii) theprojectionφ = (X0 : X2) : V P1.

14 May 2012: arX. iv:1. 102. 2606. v3 [. mat. h.R. T]. 11. May. 201. 2. ON IRREDUCIBLE REPRESENTATIONS OF COMPACT p-ADIC. ANALYTIC GROUPS. KONSTANTIN ARDAKOV AND SIMON WADSLEY. Abstract. We prove that the canonical dimension of a coadmissible repre-sentation of

23 Jun 2012: 20) Suppose p 6= 2 is prime number. Choose a Z which is not a square and notdivisible by p.

19 Nov 2012: 20 November, 2012t.yoshida@dpmms.cam.ac.uk.

5 Nov 2012: 3.20. Let ΦN Z[X] denote the N-th cyclotomic polynomial. Show that:(i) If N is odd and N ̸= 1 then Φ2N(X) = ΦN(X).(ii) If p

6 Feb 2012: In particular F• is unique up to isomorphism. Proof: See [12, Section 20.1]. ... a2 7. 13 5 20 20 1540 22 40 20 1020 35 3 15 1515 0 30 13 015 15 10 5 7.

22 Nov 2012: 31. Lecture 20. Now we define the theta function of Λ to be.

22 Nov 2012: 20. Since [Fi : Kv] = deg(fi) we get:. Corollary 5.3.

24 Jul 2012: Most readers will already be aware of the next fact. Lemma 20 If f : T C is continuously differentiable, then. ... x20x21 x. 22 x. 23)(y. 20 y. 21 y. 22 y.

28 Nov 2012: Theorem 1.20. If the smooth map f : X Y is transversal to a submanifoldZ Y , then f1(Z) is submanifold of X. ... Definition 2.20. The determinant of dNp is the Gaussian curvature K(p) ofS at p.

19 Jul 2012: u=δ. We may call these coordinate systems canonical coordinates. 1051–20. Let us note that in such coordinates the metric takes the form. ... This assumption can be made more explicit by replacing (20) with the assumption.