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31 Jan 2023: Regarding T as a map from C into C, show that there existunique complex numbers A, B such that for every z C, T(z) = Az Bz. ... Can such a map be the restriction tothe strip of a Möbius map?

25 Jul 2017: By Lemma 1, we can find A B with|A| = 1 and colours ca (0 6 a 6 k) such that if x is in the span of A andhas pattern ... By Lemma 1, there exist a subset A B with |A| = 1 and colours cπ(π Π′) such that if x is in the span of A and has pattern π

14 Mar 2005: Proof. Define B [N]2 to be the set of all (x,y) such that xy A. ... It follows that there are functions a, b and c. taking values in [0, 1] such that.

10 Nov 2014: linearfunctional θ V such that θ(x) 6= θ(y).(b) Suppose that V is finite dimensional. ... If no such subspaceexists, show that there exists a basis 〈v,α(v),.

13 Oct 2022: 1. Calculate d = (a,b) and find integers r and s such that ra sb = d when. ... iii) Find constants c and d such that λ(a,b) 6 c log b d.

8 Oct 2021: X. (ii) Cov(Y,f(X)) = Cov(X,Y )E[f ′(X)] for suitable f.(ii) Let Q be the equivalent probability measure such that dQ. ... 7. A contingent claim with time-1 payout Y is called attainable iff there exists a scalar aand portfolio b Rd such that Y = a b>S1

9 Nov 2023: SC =2π. κ. C. [Q[ξ] ξ BH]. However, it is not possible for θ to be of the form of a totalvariation δBH on H in general because such a ... M =. [Q[t] t B]. and there is no difficulty in obtaining an M that satisfies thedesired relation.

28 Oct 2022: 2. Given (fi)iI an arbitrary collection of continuous functions [0, 1] R such thatsupiI |fi(x)| < at each x [0, 1], show that there is an interval [a,b] [0, ... 1] witha < b such that supx[a,b] supiI |fi(x)| <. 3. Let X be a closed subspace of 1. Assume

13 Feb 2017: Such a c is called a fixed point of g.Give an example of a bijection h: [0, 1] [0, 1] with no fixed point. ... Thatis, for every a < b and every y there exists x (a,b) such that f(x) = y.Can such a function be continuous at any point?

14 May 2020: a) Showthat the unique minimiser is x = 0.(b) Find constants 0 < m < M and L > 0 such that m f ′′(x) M and |f ′′′(x)| L forall x.(c) Apply ... 8. Given constants b1, b2 such that b1 eb2 0 use the Lagrangian method to.