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3 Feb 2011: Show that f (x) = 0 for allx. 8. Let f : [a, b] R be bounded. ... 13. Let f : R R be a function which has the intermediate value property: If f (a) < c < f (b),then f (x) = c for some x between a and b.

15 Feb 2011: Show that f (x) = 0 for allx. 8. Let f : [a, b] R be bounded. ... 13. Let f : R R be a function which has the intermediate value property: If f (a) < c < f (b),then f (x) = c for some x between a and b.

20 Feb 2011: 5. Let f be an entire function such that for some a C and r > 0, f takes no values in the discD(a, r). ... Show that |f (z)| as |z| ,and deduce that |f (z)| takes a minimum at some a C.

13 Nov 2011: Explain geometrically why this is thecase. 5. (i) Suppose that f : R2 R is such that f/x is continuous in some open ball around (a,b),and f/y exists ... 11. Show that there is a continuous square-root function on some neighbourhood of I in Mn;that is,

1 Feb 2011: Xn be independent Poisson random variables, with Xi having parameteriθ for some θ > 0. ... b) For some n > 2, let X1,. ,Xniid Exp(θ). Find a minimal sufficient statistic T , and.

10 Jan 2011: δ is abelian.(c) Assume that δ(g) = 1 for some g G. ... b) Find an example of a representation of some finite group over some field of charac-teristic p, which is not completely reducible.

12 Oct 2011: about. (ii) m n a b (n m) [(a = 1) (b = 1) (ab 6= n)] ,. 5. The sum of some (not necessarily distinct) natural numbers is 100. ... 10. Let A = {1, 2, 3} and B = {1, 2, 3, 4, 5}?

12 Oct 2011: Show that L = K(α) = {a bα |a, b K} for some α F with α2 K. ... ii) Show that if L/K is a finite extension with Q K, for which there exist only finitelymany intermediate subfields K K′ L, then L = K(α) for some α

24 Oct 2011: Show that they have the samesplitting field if and only if b = cnar for some c K and r N with gcd(r, n) = 1. ... for some divisor d > 1 of n.

2 Nov 2011: 2. Find the convergents to the fraction 5744. Prove that if x and y are integers such that 57x 44y = 1 , then x = 17 44k and y = 57k 22 for some ... 3. Which of these are true for all natural numbers a , b and c?

30 Nov 2011: limz. znf (z) = c,. prove that f (z0) = 0 for some z0 C. ... a) g(z) = z for some z S1. (b) g(z) = z for some z S1.

30 Oct 2011: Hint: argue by contradiction.). (6) Let f : B B be a continuous map from the open disc B = {(x, y) R2 : x2 y2 < 1} into it-self. ... 1. (11) Suppose f : R R has the intermediate value property that if a, b R, f (a) < c < f (b),then f (y) = c for some y

2 Dec 2011: Find this solution to some reasonable accuracy using a pocket calculator or thecalculator on your computer (remember to work in radians!), and justify the claimed accuracy ofyour approximation. ... norm. Show, however, that T is a contraction under the

12 Nov 2011: 3.17. (i) Let f K(X). Show that K(X) = K(f) if and only if f = (aX b)/(cX d)for some a, b, c, d K with ad ... Deduce that every finite abelian group is the Galois group of some Galois extension K/Q.[It is a long-standing unsolved problem (inverse Galois

16 Aug 2011: larger family Aand then randomly discard some of the sets in A to form a subfamily B. ... Then A = vf (W )Xfor some W U , X X W2 YW and W X B.

29 Jul 2011: Then Lemma 5(applied to the image of ψ/Fl) gives the desired map log : U Lie U ad V ′. Step 6: Some properties of G0(Fl). The pair ... k1γ(T(Fl)/Z0)γ1k = k′(T(Fl)/Z0)k′1. for some k′ B(Fl). Then (kk′)1γ lies in H and we deduce that Γ

25 Aug 2011: n1. maximising the number of intersecting subfamilies of size s. Theorem 2tells us that A = [n](>3) B for some B [n](2). ... Let U = {A [n](t) : B A for some B U} be the upper shadow of Uin layer t.

21 Dec 2011: So if we lift a1 to Z[X5] thena21c4 c6 = 2f for some f Z[X5]. ... Notice that if Φ′ = gΦfor some g = [A,B] with A,B GL5(K) then (Φ′) = (Φ) v(det g).

9 May 2011: First we construct some covariant columns. Let x1 = (a,b,c,d,e)T. If γ. ... Let (a : b : c : d : e) be the K-point on X(11) corresponding to (E,φ) for some.

29 Jul 2011: In the firstsection we recall some results on the Galois representations associatedto automorphic forms. ... If π is ramified, then as above we have for some t > 0.

18 Jul 2011: linear. The useof our methods to compute some elements of X(E/Q)[5] will be describedin future work. ... H1(K, E[2]) = ker(L/(L)2. NL/K K/(K)2). Therefore α has square norm, say NL/K α = α1α2α3 = b2 for some b K.

12 Jul 2011: This is easy to do later. Here, let Vbe some irreducible G-space. ... For 1 6 i 6 n, hgi = gj,some j 6= i, so ith row of [ρreg(h)]B has 0s in every place, except column j – inparticular, the (i,