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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).

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,.

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,

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†.

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 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.

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:.