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13 May 2016: 7. (a) If f : [0, 1] [0, 1] is continuous, show there is some x [0, 1] withf(x) = x. ... For eachinteger n > 1, show that there is some x [0, 1] with f(x) = f(x 1.

17 Jan 2016: C/Λ2 are conformally equivalent if and only if thelattices are related by Λ2 = λΛ1 for some λ C. ... 5. Show that complex tori C/〈1,τ1〉 and C/〈1,τ2〉 are analytically isomorphic if and only if τ2 =. (aτ1 b)/(cτ1 d), for some matrix(a bc d

21 Nov 2016: 1. The square matrices A and B over the field F are congruent if B = PTAP for some invertible matrixP over F. ... 7. Let S be an nn real symmetric matrix with Sk = I for some k 1.

7 Nov 2016: a) All words w {0, 1} consisting of either the string 01 repeated some number oftimes (possibly none), or the string 010 repeated some number of times (possiblynone). ... b) All words w {a,b,c} consisting of some number of a’s (possibly none), followed

15 Feb 2016: Could there be some other Euclidean function φ making Z[3] into a Euclidean domain? ... iii) Show that if R is a PID, the gcd of elements a and b exists and can be written as rasbfor some r,s R.

11 Nov 2016: u(0,x) =1. 1 x2. 1. 5. (Cauchy-Kovalevskaya Theorem for systems of ODEs) Suppose b > 0 and F : u0 (b,b)d Rd bereal analytic in a neighbourhood of ... u′(t) = F(t,u(t)), u(0) = u0 Rd,. on (a,a) for some a > 0 with u((a,a)) u0 (b,b)d.

17 Jan 2016: 8. By considering the singularity at or otherwise, show that any injective analytic map f : C Chas the form f(z) = az b, for some a C and b C. ... conformally equivalent). 10. Define an equivalence relation on C by z w iff z = 2sw for some s Z.

27 Oct 2016: Hint: Bolzano-Weierstrass.). 3. If A and B are subsets of Rn, let A B = {a b |a A,b B}. ... vn} C such that{B(v1),. ,B(vn)} cover C. (b) Show that there exists some > 0 such that for every v C, B(v) Uα for some α.Deduce that

9 Feb 2016: 5. Let f be a function on a set A and let B A. ... some c (a,b) the vectors(f(b),g(b). )(f(a),g(a). )and. (f ′(c),g′(c). )in R2.

18 Feb 2016: f(x) = limx. g(x) = 0. Assume that for some c > 0, we have g′(x) 6= 0 for all x > c and that f′(x)g′(x). ... Show thatif f ′(a) < y < f ′(b) for some a < b in I and y R, then there exists x I witha < x < b and f ′(x) = y.