Determine for which real numbers s the series n>1 (log n)s /n converges, giving reasons for your answer.

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1 Problem A. Determine for which real numbers s the series n> (log n)s /n converges, giving reasons for your answer. Solution: It converges for s < and diverges otherwise. To see this use the integral test, using the fact that log log x has derivative /x log x and (log x) s+ has derivative (s + )(log x) s /x. Problem 2A. Does there exist an infinite set I = {I n : n N} of disjoint closed intervals I n = [a n, b n ] of R (with a n < b n ) such that n N I n is compact? Solution: Yes, consider I = {[, 0]} {[, ] : n N}. Suppose that O is an open 2n+ 2n cover of I. O has a finite subcover of [, 0], which must also cover all but finitely many of the intervals in I. Each of the finitely many remaining intervals in I has a finite subcover within O. The union of these finite subcovers with the finite subcover of [, 0] provides a finite subcover of I. Problem 3A. Suppose that f(x) is a positive-valued differentiable function with domain R. Show that for all α >, lim inf x f (x) f(x) α 0. Solution: If not, then fix ɛ > 0 and a so that for all x a, f (x) f(x) b a f (x) f(x) dx = α ( α + )f(x) α which must be greater than ɛ(b a). Then, b a = ( α > (α )ɛ(b a) f(a) α > ɛ. For b > a consider f(a) α f(b) α ),

2 for all b, which is impossible. Problem 4A. (a) Find all residues of all poles of /(z 3 cos z) (b) Evaluate the sum / 3 /3 3 + /5 3 / Solution: (a) There is a pole of order 3 at 0 with residue /2, and poles of order at z = (2n + )π/2 of residue ( ) n+ ((2n + )π/2) 3 for n Z. (b) The integral over suitable large rectangles tends to 0, so the sum of all residues is 0. So / 3 /3 3 + /5 3 /7 3 + = π 3 /32. Problem 5A. Find the number of complex numbers z such that z < and e z = z 4 + 5z 3 +. Solution: The function 5z 3 has absolute value larger than that of e z z 4 on the circle z =, so by Rouche s theorem the number of solutions is the number of zeros of 5z 3 = 0 which is 3. Problem 6A. Let u, v, w R n be three vectors with the property that u = v = w = and v, w = 0 (Here, denote the Euclidean inner product and denotes the Euclidean norm on R n ). Prove that u, v 2 + u, w 2. When does equality occur? Solution: Complete v, w to an orthonormal basis e = v, e 2 = w, e 3,..., e n R n. Then = u 2 = u, v 2 + u, w 2 + u, e u, e n 2 u, v 2 + u, w 2 with equality if and only if u, e 3 =... = u, e n = 0, i.e. u span{v, w}. Problem 7A.

3 Let a, b, c > 0 and consider the symmetric matrix 0 a b A = a 0 c. b c 0 Let λ λ 2 λ 3 be the spectral values (eigenvalues) of A. (a) Show that λ, λ 2 < 0 and λ 3 > 0. (b) Show that if v R 3 then Av, v λ 3 Av, Av (c) Show that λ 3 (a + b)2 + (b + c) 2 + (c + a) 2. 2(a + b + c) Solution: (a) We first compute that λ + λ 2 + λ 3 = tra = 0, λ λ 2 λ 3 = det A = 2abc > 0. The first equation implies that λ < 0 and the second equation implies that an even number of spectral values has to be negative. So λ 2 < 0. Then λ 3 > 0 as their sum is 0. (b) Let e, e 2, e 3 R 3 be an eigenbasis of A. As A is symmetric, we can choose e, e 2, e 3 to be orthonormal. Choose u, u 2, u 3 R such that Then since λ < 0, λ 2 < 0, λ 3 > 0 v = u e + u 2 e 2 + u 3 e 3. Av, v λ 3 = (λ u 2 + λ 2 u λ 3 u 2 3)λ 3 λ 2 u 2 + λ 2 2u λ 2 3u 2 3 = Av, Av (c) follows from (b) by putting v := Problem 8A. Let f(x) be a polynomial of degree d with rational coefficients. Assume that f(n) Z for any n Z. Prove that all coefficients of d!f(x) are integers.

4 Solution: In the situation of the problem, say that f(x) is integrally valued. Write f(x) = a d x d + + a 0. Consider g(x) = f(x + ) f(x). Then deg(g) = d and g(x) is integrally valued. By induction, we can assume that (d )!g(x) has integer coefficients. In particular, its leading coefficient b = (d )! da d = d!a d is an integer. Consider ( ) x x(x ) (x d + ) h(x) = f(x) b = f(x) b. d d! Then h(x) is integrally valued, since all combinatoric numbers ( n d) with n Z are integers. Note that deg(h) d. By induction again, we have that d!h(x) = d!f(x) bx(x ) (x d + ) has integer coefficients. Then d!f(x) has integer coefficients. This finishes the proof. Problem 9A. For each of the following polynomials, say whether or not it is irreducible in Z[x], giving reasons for your answers. (a) x 6 + 6x + 3 (b) 4x 4 + (c) 3x 3 2x 2 + 2x + Solution: (a) x 6 + 6x + 3 is irreducible by Eisenstein (b) 4x 4 + factorizes as (2x 2 + 2x + )(2x 2 2x + ) (c) 3x 3 2x 2 + 2x + factorizes as (3x + )(x 2 x + ) Problem B. Find a degree 3 polynomial with integer coefficients that has cos(2π/7) as a root.

5 Solution: 8z 3 + 4z 2 4z. One method is to put cos(2π/7) = (x + x )/2 where x is a primitive 7th root of, satisfying x 6 + x 5 + x 4 + x 3 + x 2 + x + = 0. (The problem can also be done using multiple angle formulas for cos and sin.) Problem 2B. Show that the following are equivalent properties of a subset A of R.. A is countable. 2. There is a non-decreasing function f such that A is the set of x such that f is discontinuous at x. Solution: (a implies b). Let (a n : n N) be a counting of A. For x R, let f(x) = a n x /2n. (b implies a). Suppose that f is given as above and let A be the set of discontinuities of f. Since Q is countable, it is suffcient to define an injection from A to Q. Fix a counting (q n : n N) of Q. Define the injection from A to Q by sending a A to q k, where k is least such that sup x<a f(x) < q k < inf a<x f(x). Problem 3B. Show that sin(x) dx converges but does not converge absolutely. x Solution: To show convergence, use integration by parts. sin(x) x dx = cos(x) x which is finite by comparison to the integral of /x 2. It is not absolutely convergent since which diverges. sin(x) dx > x k= (2k+)π 2kπ+π/2 cos(x) dx, x 2 sin(x) x dx > (2k+)π sin(x)dx, (2k + )π k= 2kπ+π/2

6 Problem 4B. The gamma function Γ(s) = 0 e t t s dt is holomorphic for Rs > 0. (a) Prove that Γ(s + ) = sγ(s) (b) Prove that Γ can be analytically continued to a meromorphic function for all complex s. (c) Find the poles and their residues for the analytic continuation. Solution: (a) This follows by integration by parts. (b) The analytic continuation follows by repeatedly applying the functional equation Γ(s) = Γ(s + )/s, because if Γ is defined for R(s) > N then the functional equation extends it to R(s) > N. (c) The poles are at s = 0,, 2, 3,... with residues /0!, /!, /2!, /3!,... This follows from the functional equation and the fact that Γ() =. Problem 5B. Let f be a meromorphic function on C that is analytic in a neighborhood of 0. Let its Maclaurin series be a k z k. k=0 Assume that all a k are real and 0, that there are no poles at z with z < r, and that there is at least one pole at z with z = r. Show that there is a pole at z = r. Solution: Let w be a pole with w = r. Then lim f(( ɛ)w) =. ɛ 0 + Since f(( ɛ)r) = a k (( ɛ)r) k = k=0 a k ( ɛ)w k a k (( ɛ)w) k = f(( ɛ)w), k=0 k=0 we have that lim z w f(z) does not exist (as a finite value). Therefore f has a nonremovable singularity at w. Since f is meromorphic, that singularity must be a pole. Problem 6B.

7 Let A be an anti-symmetric n n matrix (A T = A) with real-valued entries. Show that all eigenvalues of A are of the form iλ for λ R. Solution: Let v C n be an eigenvector and let µ C. Let v be the vector that arises from v by complex conjugation of all its entries. For any two vectors u, w C n denote by the (linear) pairing. Then As (u, w) = u w u n w n µ(v, v) = (Av, v) = (v, A T v) = (v, A v) = (Av, v) = µ(v, v) = µ( v, v) = µ(v, v). (v, v) = v v n 2 > 0 this implies that µ = µ and therefore µ is purely imaginary. (Alternative solution: ia is Hermitean so has real eigenvalues.) Problem 7B. Let M be an n n matrix with integer coefficients. Then M defines a homomorphism M : Z n Z n by M(α) = Mα, where α Z n is viewed as a column vector. Assume that det(m) 0. Prove that the group Z n /(MZ n ) has order det(m). Solution: Note that we can write M = ADB for some diagonal matrices D M n (Z) and invertible matrices A, B GL n (Z). Using the transformations A and B, we see that M : Z n Z n is isomorphic to D : Z n Z n, where the latter is defined by left multiplication by D. Note that D is diagonal, it is easy to have that the order of Z n /(DZ n ) is det(d). By det(a) = ± and det(b) = ±, we have det(d) = det(m). This finishes the proof. Problem 8B. Find the number of automorphisms of the group G = (Z/2Z) (Z/4Z).

8 Solution: Write x = (, 0) and y = (0, ) according to the isomorphism G = (Z/2Z) (Z/4Z). An automorphism φ should make φ(y) an element of order 4, which can be ±y or ±y + x, and thus has 4 choices. Then 2φ(y) = ±2y = 2y in all these cases. Also, by x 2y, the automorphism should make φ(x) an element of order 2 and different from 2φ(y), which can be x or x + 2y, which has 2 choices. Then the total number of automorphisms is 4 2 = 8. Problem 9B. Let P d (n) be the largest number of regions into which d-dimensional Euclidean space can be divided by n hyper planes (so for example P 2 (3) = 7 as 3 lines in the plane will divide it into a maximum of 7 regions). Find P 3 (0). Solution: We have P d (0) =, P 0 (n) =, and P d (n) = P d (n ) + P d (n ), because adding a new hyperplane in d-dimensions will usually add an extra P d (n ) regions. So P d (n) = ( ) ( n d + n ) ( d + n ) d 2 +, and P 3 (0) = ( ) ( ) ( ) ( + 0 ) 0 = = 76.