Search

20 Nov 2015: An 6= holds for all n. Must it bethat. n=1 An 6=? ... Can S be uncountable?Is there an uncountable family T PN such that AB is finite for all distinct A, B T?

2 Nov 2015: 1. Quickies: (a) Describe all continuous functions f : [0, 1] Rn satisfying ‖ 10f‖ =. 10‖f‖. (b) Show that two norms ‖ ‖, ‖ ‖′ on a vector space V are Lipschitz equivalent if and only ... denote the vector space of all bounded Riemann

23 Feb 2015: Wii =1. aiV (µi){g′(µi)}2,. (you need not specify i(σ2), and you may assume /βjσ2 = /σ2βj for all j). ... c) How do the expressions in (a) and (b) simplify when g(µi) is the canonical linkfunction?

19 Jan 2015: 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 and let ... constituent of IndGKχ, then all |G : K| conjugates of χ are constituents of ResGKφ.

10 Feb 2015: Show that f is a polynomial, of degree k, if and only if there is a constantM for which |f(z)| < M(1 |z|)k for all z.(ii) Show that ... Show that if there existsk such that |f(z)| |z|k for all z with |z| sufficiently large, then f is a rational function

23 Nov 2015: 3. (i) Show that the function ψ(A,B) = tr(ABT ) is a symmetric positive definite bilinear form on the spaceMatn(R) of all nn real matrices. ... sn forms a basis for Pn.(iii) For all 1 k n, sk spans the orthogonal complement of Pk1 in Pk.(iv) sk is an

31 Aug 2015: for every x X, there is an open neighborhood U of x such thatf(y) = f(x) for all y U. ... Hint:first find a subsequence (fni ) such that fni (x) converges for all x Q [0, 1].)Deduce that the closure of B in (C[0, 1], d) is compact.

27 Apr 2015: Show that f is continuous ifand only if f(A) f(A) for all A X. ... Show thatthis is a topology, that all points of R are closed with respect to it, but thatthe topology is not Hausdorff.

15 Oct 2015: a) The set C of continuous functions. (b) The set {f C : |f(t)| 1 for all t [0, 1]}.(c) The set {f C : f(t) 0 as t }.(d) ... i Ui. Give an example where Ui Uj = {0} for all i 6= j, yet U1.

27 Feb 2015: continuous function g: Bε(I) Mn such that g(A)2 = A for all A Bε(I). ... b) = inf (γ), wherethe inf is taken over all continuous γ: [0, 1] E with γ(0) = a, γ(1) = b.

23 Jan 2015: Show that for all v Rn, ‖v‖2 ‖ΠWv‖2 ‖ΠV v‖2. (b) Consider the linear model (1) but where only the first p0 components of β are non-zero. ... Now prove that if A1,. ,Am and B1,. ,Bm are all independent real-valuedrandom variables and A1 B1,.

26 Nov 2015: 4. Find all bases for which 39 is an Euler pseudoprime. ... More generally, showthat if p and 2p1 are both prime numbers, then N = p(2p1) is a pseudoprimefor precisely half of all bases.

2 Jun 2015: Let Kn be the function field of X(n) over K. Let B K3 /(K3 )3 be thesubgroup generated by all rational functions on X(3) with zeros and poles only ... All computer calculations in supportof this work were carried out using Magma [2].

15 Sep 2015: Q(x1,x2,. ,xn) =. 1ijn. cijxixj, (1). where all coefficients cij lie in Z. ... Thus we have established items (a)–(c), for all n, in the case that D = U(a,b) is a constantdistribution supported on a box [a,b].

18 Feb 2015: In the best of all possible worlds one would expect the following picture. ... for all infinite primes v of F and product decompositions. G(R) =v|.

22 Sep 2015: 3. for all a and b. 41. Exercise 2.3.1. (Only part (ii).). ... for some constant B. 106. If h(t) = C exp at for all t,.

7 Aug 2015: bx,b′x′] := bb′[x,x′] b′(x′ b)x b(x b′)x′. for all b,b′ B and x,x′ L. ... bx,c(b′x′)] = c[bx,b′x′] (1 σ)(bx)(c)b′x′. for all b,b′,c B and x,x′ L.

20 May 2015: a) for all u1,u2 U, u1 u2 U;(b) for all λ F and u U, λu U;(c) 0 U. ... fn〉. Then B = P1AP. Proof. This is a special case of the change of basis formula for all linear mapsbetween f.d.

2 Dec 2015: x X is linear.(3) For all x [a,b], δx : C([a,b], R) R; f 7 f(x) is linear.(4) I : C([a,b], R) C([a,b], ... Therefore for all v ker α, φ(v) = θα(v) = θ(0) = 0.

10 Dec 2015: We also thank all the refereesfor corrections and criticisms which, we hope, have substantially improved the exposition. ... Proof. All the representations we construct will have non-zero B-invariant vectors.