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15 Nov 2015: B,terminating in X (the x of the other numeral, also defined below). ... λa.λb. Z ( [a A] [(b B) X] ). 1.

25 Jun 2015: Then we may choose smooth functions ϕi : M [0, 1] such that supp(ϕi) int(Vi) and p(ν K) (. ϕi)1(1) and choose B =. B(A, ϕ1,. , ϕr) A ... Our choice of B guarantees that the linearcombination s =. ϕisi has s A.

22 Jan 2015: likelihood estimate of θ is (2a b)/(b 2c). 4. (a) Let X1,. ... b) For some n > 2, let X1,. ,Xn be iid with Xi Exponential(θ).

10 Jul 2015: Part a. Let R† be defined by. 〈a, b〉〈a, b〉 R 〈a, b〉 〈b, c〉. ... Consider thesetwo derivations:. 1. 〈a, e〉. 〈a, b〉 〈b, c〉 〈c, d〉 〈d, e〉.

11 Feb 2015: 5. Let p : E B be a Serre fibration, e0 E, b0 = p(e0), F = p1(b0). ... Fi //. Ep //. p. B. {b0} // B B. and show using the naturality of the Serre spectral sequence that the compositions.

14 Oct 2015: algebra. 1.2. Show that the following sets of subsets of R all generate the same σ-algebra:(a) {(a,b) : a < b}, (b) {(a,b] : a < b}, (c) {(,b] : b ... 1.8. Let B be a Borel subset of the interval [0, 1].

25 Feb 2015: iii) Show that the map. H(A)iA H(M) [M] Hn(M). (iB) Hn(B). ... w(F), [Hint: sum of vector bundles is given by pulling backE F B B along the diagonal; relate and Th(E F) to Th(E) and Th(F)].

5 Feb 2015: 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) = 0H(Y |I,X) = P(I = 0)H(Y |I = 0,X)

4 Feb 2015: 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) = 0H(Y |I,X) = P(I = 0)H(Y |I = 0,X)

20 Nov 2015: 6. Let f : X Y. Let A, B X and C, D Y. ... For each of claims (a)–(f) below, give aproof or counter-example. (a) f(AB) = f(A)f(B) (b) f1(C D) = f1(C)f1(D)(c) f(AB) =

4 Feb 2015: possible area is an arc of a circle passing through A and B. ... 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.

17 Dec 2015: possible area is an arc of a circle passing through A and B. ... 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.

7 Feb 2015: This one ishard: you will need your answer to (ii)). (v) from (A B) A into (A B) B. ... So T is a natural functionfrom (A B) A into (A B) B.) Show that something is afixpoint combinator iff it is a fixed point for T.

22 Oct 2015: λfABxA : (A B) B -int (1). λxA.(λfAB.fABxA) : A ((A B) B) -int (2). ... Bλx.x(fx) : (A B) B -int (2). λf.λx.x(fx) : ((A B) A) ((A B) B) -int (1).

28 Jul 2015: a. a. a. a a. a. b b. b. b b. ... 8 S. M. GERSTEN. every a A there is a point b B with d(a, b) H and for every b B there is apoint a A with d(a,

22 Sep 2015: f(t) dt. (iii) If a c b, b. a. f(t) dt+. ... log(xa)b = b log xa = ba log x = ab log x = log xab,.

7 Aug 2015: Proof. Write xb := σ(x)(b) and bx := bx for all x L and b B. ... bx,b′x′] := bb′[x,x′] b′(x′ b)x b(x b′)x′. for all b,b′ B and x,x′ L.

11 Sep 2015: for some a,b K. The discriminant condition is 2b(a2 4b) 6= 0. ... We have div(f/x. 21) = 2b and div(f. †/x21) = 2b† with b b†.

11 Jul 2015: i) for every X [A]ω there is b B that is a (proper) initial segment of X; or. ... ii) C [A]<ω =. Any b B [A]<ω of minimal size must belong to C, so (ii) cannot hold.

9 Jan 2015: attach(tax). a <- (y-x)/x ; b <- (z-y)/y # we compare relative increases. ... Now compare a with b, but pretending a,b independent of each other.

10 Mar 2015: Each vertex w is joined to thevertices wb for each letter b B. ... B(b,r) = {v {0, 1}N : d(b,v) < r}. Its volume is the number of points that it contains:.

10 Dec 2015: Having defined the standardapartment, the building B = B(Un,F) is the unique Un-set up to isomorphism satisfying the followingconditions (see [Tit79, 2.1]):. • ... b) The exponents of the Jacquet modules of these representations are as follows.

8 Jan 2015: Cambridge University Press.Venables, W.N. and Ripley, B.D.(2002) Modern Applied Statistics with S-plus. ... Giant armadillo 60.000 81.00. Rock hyrax-b 3.600 21.00. Raccoon 4.288 39.20.

7 Oct 2015: in (b) we have(x z)w > (x y)w = (y z)w. ... 14. D D. a’ a. vu. b b’. By following a path via a′,a,b and b′, we see that d(u, v) 6 6δ.