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20 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

28 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.

25 Jun 2015: N π1Ki πKi (B) for all i = 1,. , r. ... This holds for all i and hence N and s(M ) agree inside the union.

9 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

21 Oct 2015: a) the set {(j, k) : 1 j < k n} of all transpositions in Sn;(b) the set {(j, j 1) : 1 j < n};(c) the set {(1, k) : 1 < k ... Find a (natural) bijection between the set of all leftcosets and the set of all right cosets of H in G.

7 Mar 2015: i.e. T(xy) = 0 for all y K implies x = 0).(b) Show that the sequence xn = T(α. ... n) is the output from a LFSR of length d. (c) The period of (xn)n>0 is the least integer r > 1 such that xnr = xn for all sufficiently largen.

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.

18 Feb 2015: Show that the sum of the residuesof f at all its poles equals zero. ... 6 Let p(z) = z5 z. Find all z such that |z| = 1 and Im p(z) = 0.

9 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.

11 Feb 2015: 4. Compute H(K(Z,k); Q) for all k. [Hint: Use PK(Z,k) K(Z,k) and the equiva-lence K(Z,k) ' K(Z,k 1).]. ... R = πi(Sn) R for all i.