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16 Aug 2011: Compressions and Probably Intersecting Families. Paul A. Russell. A family of sets is said to be intersecting if AB f or all A, B.

3 Feb 2011: Suppose that f ((x y)/2) (f (x) f (y))/2 for all x, y [a, b].Prove that f is continuous on (a, b). ... Which of (1)–(4) must betrue? (1) If f is increasing then f ′(x) 0 for all x (a, b).(2) If f ′(x) 0 for all x (a, b) then

15 Feb 2011: Suppose that f ((x y)/2) (f (x) f (y))/2 for all x, y [a, b].Prove that f is continuous on (a, b). ... Which of (1)–(4) must betrue? (1) If f is increasing then f ′(x) 0 for all x (a, b).(2) If f ′(x) 0 for all x (a, b) then

13 Oct 2011: Which of the following sets of functions form a vector subspace of RR?(a) The set C of continuous functions.(b) The set {f C : |f (t)| 1 for all t ... i6=j Ui = {0}. (iii) The Bi are pairwise disjoint and their union B is a basis for U =. i Ui. Give an

11 Oct 2011: 6. Show that R(s, t) 6(st2s1. )for all s, t > 2. ... Suppose that (i) |A| = 3 for each A A; and (ii) |A B| = 1for all distinct A, B A.

3 Feb 2011: 4. Find all analytic functions of the form f (z) = f (x iy) = u(x) iv(y). ... Show that there is an analyticfunction H : C C such that h(z) = exp H(z) for all z.

22 Jun 2011: The first four are maybe tricky, so makesure that you do all the integrals. ... Show that all roots lie inside the circle|z| = 3/2. (b) How many zeros does z4 12z 1 have in the annulus 2 < |z| < 3?

31 Oct 2011: Is it Lipschitz equivalent. to the uniform norm? (b) Let R[0, 1] denote the vector space of all integrable functions on [0, 1]. ... Usingthe fact that all norms on a finite-dimensional space are Lipschitz equivalent, deduce that α iscontinuous.

5 Nov 2011: c) Set pn = (2In)1p̃n. Check that pn is a polynomial of degree 2n, and that supx[a,b] |pn(x) f (x)| < 2 for all sufficiently large n. ... supx[a,b]. |xn p(x)| supx[a,b]. |xn q(x)|. for all polynomials q of degree n 1.

13 Nov 2011: Show that f is continuous at(0, 0) and that it has directional derivatives in all directions there. ... some r > 0 and a continuous functiong : B(I; r) Mn such that g(X)2 = X for all X B(I; r).Is it possible to define a square-root

25 Nov 2011: and 18t 1 are all prime numbers. Prove that N is a Carmichael number. ... More generally, show. that if p and 2p 1 are both prime numbers, then N = p(2p 1) is a pseudoprime forprecisely half of all bases.

10 Jan 2011: submodules.(b) Find all the RG-homomorphisms between the irreducible RG-modules.(c) Show that the conclusion of Schur’s Lemma (‘every homomorphism from an irre-. ... Determine all finite groups which have a faithful 2-dimensional representation over R

17 Nov 2011: 7. Let. n=1 xn be a divergent series, wherexn > 0 for all n. ... Can S be uncountable?Is there an uncountable familyT PN such thatA B is finite for all distinctA, B T?

31 Oct 2011: Is it Lipschitz equivalent. to the uniform norm? (b) Let R[0, 1] denote the vector space of all integrable functions on [0, 1]. ... Usingthe fact that all norms on a finite-dimensional space are Lipschitz equivalent, deduce that α iscontinuous.

2 Nov 2011: 3. Which of these are true for all natural numbers a , b and c? ... 6. In the sequence 41, 43, 47, 53, 61,. , each difference is two more than the previous one.Are all the numbers in the sequence prime?

24 Jan 2011: 1 Recall the character table of S4 from Sheet 2. Find all the characters of S5 induced fromthe irreducible characters of S4. ... 7 Let ρ : G GL(V ) be a representation of G of dimension d.(a) Compute the dimension of SnV and ΛnV for all n.(b) Let g G

21 Nov 2011: 13 Prove the following results:(i) If K is a field containing F2, then (a b)2 = a2 b2 for all a,b K.(ii) If P F2[X] and K ... is a field containing F2, then P(a)2 = P(a2) for all a K.(iii) Let K be a field containing F2 in which X7 1 factorises into

21 Nov 2011: Show that there is nodecodable coding c such that all codewords have length 2 or less. ... Determine whether there are optimal binary codes with (a) all but onecodeword of the same length, or (b) each codeword a different length.

6 May 2011: f) Define a subset of the integers Z to be open either if it is empty or if for some k Zthe set S contains all integers k. ... b) Show f : X Y is continuous if and only if for all A X, f (cl(A)) cl(f (A)).Deduce that if f is surjective, the continuous

16 Aug 2011: Then there existsan injection φ : I(A) I(C) such that |φ(B)| = |B| for all B I(A). ... So fix X. LetY = {X X12,(1) : 2X B for all B AX}.