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20 Jan 2007: φ(s,v) = α(s) r(n(s) cos v b(s) sin v)where n = n(s) and b = b(s) are the normal and binormal vectors of α. ... 9. Consider a surface of revolution parametrized by φ : (0, 2π) (a,b) R3, whereφ(u,v) = (f(v) cos u,f(v) sin u,g(v)).

21 Nov 2007: Show that f = gR f gLand hence, or otherwise, show that f is differentiable at B. ... Show that f is differentiable at (a, b). (b) Suppose instead that D1f exists and is bounded on some open ball around (a, b), and that for fixed x.

3 Jun 2007: 2. Let T : z 7 (az b)/(cz d) be a Möbius transformation.24. ... s s2 1 and s. s2 1. where. s =|a|2 |b|2 |c|2 |d|2.

5 Apr 2007: If V is a normed vector space, then B is bounded iff B B(t) forsome t > 0, where B(t) denotes the open unit ball around the origin of radius ... Tα(B(ǫ)) Tαx0 B(n). Let n0 = supα |Tαx0|. We have then.

26 Nov 2007: b, c (b,. ). a&b, c (a&b, {b}) (a&b) d, c ((a&b) d, {a&b}). ... 0 A , A. A. 0 B , B. B ( A) ( B),.

18 Sep 2007: If x, t [a, b]. f (t) =n. j=0. f (j)(x). ... b. a. f (x) dx b. a. g(x) dx. Lemma 12.6.

19 May 2007: numbers of thesepairs belonging to B be g(B) and b(B) respectively. ... Then there must exist for this r a set B such thatg(B) γ1b(B) > αrm2, which implies that |B| > αr/2m and b(B) 6 γ|B|2.

5 Oct 2007: b) Use (a) to show that every finitely presented group is the fundamental groupof some space. ... Suppose that the maps A A′, B B′, D D′, and E E′ are isomorphisms.Show that the middle map γ : C C′ must be an isomorphism, as follows.

3 Jun 2007: It corresponds to the Möbius transformation C C ; z 7 (az b)/(cz d). ... Exercises. Let T : z 7 (az b)/(cz d) be a Möbius transformation.7.

25 Oct 2007: b) Let R([0, 1]. )denote the vector space of all integrable functions on [0, 1]. ... An interval is a set taking one of the forms (, ), (, b), (, b], [a, ), (a, ), [a, b], [a, b), (a, b]or (a, b) for some a, b R with a b.

3 Jun 2007: Proof:. The theorem enables us to construct an entire function b whose zeros are poles of f. ... Thus G is a subgroup of the groupof translations: {z 7 z b : b C}.

8 Nov 2007: B+. B. I+. H. I. II. At this point, let us make the following definitions:. ... 23. The geometry of these regionsis completely different, however, in particular, the sets B = B are timelike,and this allows for the repeating structure to develop.

20 Apr 2007: ii) We say that f : A B is surjective3 if given b B there exists ana A such that f (a) = b. ... leastone solution for each b B; f is bijective if the equation f (a) = b has exactlyone solution for each b B.

5 Oct 2007: Hn1(K) Hn(A B). Hn1(f ). y yHn(f )Hn1(K′). Hn(A′ B′). (b) Show that if the maps Hi(AB) Hi(A′B′), HiA HiA′, and HiB ... is not a 2-manifold.(b) Let K be an n-dimensional pseudomanifold. Show that Hn(K) is isomorphic.

2 Mar 2007: B. Next an X-cotensor is called finite if X is finitely presentable, in the definitive. ... b) moreover the remaining locations are untouched. The monad structure is inherited from (S )S.

26 Oct 2007: 3. Two players A and B play a best of 5 set tennis match. ... Hint: Recall that H(A,B) = H(A|B) H(B).]. (b) Show that. H(I|Y,X) = 0.

3 Jun 2007: Find a continuousfunction B : R which is harmonic on and has B| = b. ... Note that the function B need no longer be uniquely determined by b.

31 Jan 2007: WbuXRxuXcX"uÜRqr6ú"û O ßßßÜ wü"sý[weYuWYXZ0úRxRx4"b fWRxu0E i XrY[CWYXrW }X } peYu]W4e|Yd efY>Ramr. ... O WZnWYXZY}YwXr|.qRqXERq}fYasd eYfvWYX"> ][w >qWYXmO YKKI|.0X}d efpcv ][w nKRaWhX}uowuqRqusoaZnÝ 4e|Yd ef_o{u"Y4Rx[w RqdQWs dexXr >XuRqvr_b.

5 Feb 2007: If R is a domain, then a b (a|b b|a). We call an elementa RR irreducible if for any factorization a = bc one of b,c is a unit ... θ =12. (a b. 7 c. 2 d. 14) ,. where a,b,c,d are rational integers.