1 lim. f(x) sin(nx)dx = 0. n sin(nx)dx

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1 Problem A. Calculate ta(.) to 4 decimal places. Solutio: The power series for si(x)/ cos(x) is x + x 3 /3 + (2/5)x 5 +. Puttig x =. gives ta(.) =.3. Problem 2A. Let f : R R be a cotiuous fuctio. Show that lim f(x) si(x)dx = Solutio: Note that So 2 f(x) si(x)dx = f(x) si(x)dx = + f(x) si(x)dx f ( x ( ( ) si x )) dx = + f(x) si(x)dx + + f ( x ) si(x)dx ( ( f(x) f x )) si(x)dx Sice f is cotiuous ad [, 2] is compact, f it uiformly cotiuous o [, 2] ad we have f < C o [, 2] for some uiform costat C. By the uiform cotiuity, give ay ɛ > we ca fid a N such that for all > N we have Therefore, f(x) f ( x ) < ɛ 2 f(x) si(x)dx < 2 C + ɛ. As ɛ was chose arbitrarily, this proves the claim.

2 Problem 3A. Let f : R R be a smooth fuctio that satisfies the differetial equatio f (x) = g(x)f (x) for some strictly positive smooth fuctio g. Prove, i detail, that f is either strictly icreasig, strictly decreasig or costat. Solutio: If f was ot mootoe or costat, the f (y) = ad f (z) for some y, z R Set X = {x R : f (x) = }. Note that X is closed ad y X, z X. Sice X is closed, we ca fid a y X such that z y is miimal. So I = [z, y ) (if y > z) or I = (y, z] (if y < z) is disjoit from X. The o I we have (log f (x) ) = f (x) f (x) = g(x) So for all x I log f (x) = log f (z) + x z g(t)dt As g is cotiuous, it follows that log f < C o I for some uiform costat C. Therefore, f > e C o I. Sice f is cotiuous, it follows that f (y ) e C >, i cotradictio to the choice of y. Problem 4A. Fid the itegral dz ( + z 2 ) 3 Solutio: Apply Cauchy s residue theorem to the usual semicircular cotour. The residue at z = i is i3/6 ad the itegral is 2i times this which is 3/8. Problem 5A.

3 Determie all complex aalytic fuctios o the disk D = {z C : z < } which satisfy: for all positive itegers. f (/) + f(/) =, Solutio: If f is aalytic o D the g = f + f is also aalytic o D. Sice the sequece / for has a accumulatio poit at zero ad g vaishes o this sequece, it must vaish o D. Thus, the restrictio of f to the iterval (, ), call it h(x), must satisfy the secod order ODE h (x) + h(x) =, which has solutios a cos(x) + b si(x). Sice a aalytic fuctio o D is completely determied by its values o the iterval (, ), the oly possibilities for f are the fuctios a cos(z) + b si(z) for complex a, b. Problem 6A. Suppose A ad B are ivertible matrices of complex umbers. (a) Show that AB is diagoalizable if ad oly if BA is diagoalizable. (b) Give AB = SΛS, fid T such that BA = T ΛT. Solutio: (a) AB ad are similar. (b) so so T = BS will do. BA = A (AB)A BAB = BSΛS BA = BSΛS B Problem 7A. Let A, B R be two symmetric matrices with positive eigevalues. Show that there is a ivertible matrix C R such that C T AC = ( C T BC )

4 Solutio: Cosider the symmetric pairigs (u, v) A = u T Av ad (u, v) B = u T Bv, which are positive defiite. By the spectral theorem, we ca fid a basis e,..., e R that is orthoormal with respect to (, ) A ad that satisfies (e i, e j ) B = λ i δ ij. Sice B is positive defiite, we have λ i = e T i Be i >. Let ow v i = λ /4 i e i ad let C be the matrix with colum vectors v,..., v. The C T AC = diag(λ /2,..., λ /2 ) ad C T BC = diag(λ /2,..., λ /2 ). Problem 8A. Let R be the rig of complex umbers of the form m + 2 with m, itegers. Prove that if a, b R with b the we ca fid q, r R such that a = bq + r ad r < b. Give a example to show this is false for the rig of umbers m + 3. Solutio: We have to fid q R such that a/b q <. This says that the complex umbers are covered by ope discs of radius with ceters at elemet of R, which is easy to check geometrically. For the rig of umbers m + 3 this is false if we take b =, a = ( + 3)/2. Problem 9A. Fid the largest possible order of a elemet of the symmetric group o 2 poits. Solutio: The largest possible order will be give by a product of disjoit cycles, ad oe may assume that these are of prime power order for distict primes. For 2 poits some trial ad error shows that the largest order is 6, give by the product of disjoit 3, 4, ad 5 cycles. Problem B.

5 Let N be a positive iteger ad ω = e 2i/N. (a) Show that the N N matrix F with elemets F pq = ωpq N is uitary. (b) For each x R ad each iteger k with k N, fid a explicit formula for the sum of the series x pn+k E k (x) = (pn + k)!. p= Solutio: (a) Use the fiite geometric series. (b) For j N The for k N we have But e j (x) = e ωjx = N ω jk e j (x) = N j= iff is cogruet to k modulo N. Hece E k (x) = N = = N ω j( k) = N j= N j= j x ω!. N ω jk j x ω N!. j= e 2ijk/N e xe2ij/n. Problem 2B. Suppose that f is a cotiuous real fuctio o [, ]. Prove that lim α α + (The fuctio f eed ot be differetiable.) x α f(x)dx = f().

6 Solutio: This is obvious for f a costat, so by subtractig f() from both sides we ca assume f() =. Choose ay ɛ > ad choose δ so that t f(x) < ɛ wheever x δ ad choose M so that f(x) < M for all x. The while α δ δ α x α f(x)dx α x α f(x)dx α δ x α ɛdx = ɛ, x α Mdx = M( δ α ) which teds to as α teds to. So for ay ɛ > the limit is less tha ɛ i absolute value, so the limit is. (Assumig that f is differetiable allows a easier solutio by itegratig by parts.) Problem 3B. For each positive iteger defie f : R R by f (x) = si(x). Prove that the sequece of fuctios f has o subsequece that is uiformly coverget o compact subsets of R. Solutio: Assume for cotradictio that there is a subsequece j such that f j coverges uiformly o [, ] to some fuctio f. Note that f() = lim j f j () = si() =, ad f must be cotiuous sice it is a uiform limit of cotiuous fuctios. Thus, there is a eighborhood of, say ( t, t), o which f(x) < /2. Choose N large eough so that f j (x) f(x) < /2 o [, ] wheever j N. Thus, we have f j (x) < wheever x < t ad j N. But this is absurd sice f j (/2 j ) = for all j. Problem 4B. Fid = si() Solutio:

7 Method : Let f(x) = for x a ad otherwise. The the Fourier coefficiets ˆf(k) = 2 f(x)e ikx si(ka) dx = 2 k ad ˆf() = / 2. Sice f is smooth ear x =, summatio gives = 2 k Z Settig a = ad x = gives (after some algebra) = si() ˆf(k)e ikx. =. 2 Method 2: Use the Taylor series of l( z) o the circle of covergece. Problem 5B. Show that there is a complex aalytic fuctio defied o U = { z > 4} whose derivative is f(z) = Is there a fuctio o U whose derivative is: g(z) = z (z )(z 2)(z 3). z 3 (z )(z 2)(z 3)? Solutio: By Morera s theorem, f(z) has a atiderivative o U if ad oly if f(z)dz = C for every simple closed cotour cotaied i U. If C does ot ecircle the disk { z 4} the f is aalytic i the iterior of C ad by Cauchy-Goursat the itegral is zero. Otherwise the iterior of C must cotai the three poles of f at, 2, 3; by the residue theorem, for such C we have 2i C f(z) = ( 2)( 3) + 2 (2 )(2 3) + 3 (3 )(3 2) = =,

8 so ideed f(z) has a atiderivative o U. O the other had, for a cotour cotaiig the poles, 2, 3, we have g(z) = 2i ( 2)( 3) + 8 (2 )(2 3) + 27 (3 )(3 2) = , C so g(z) does ot have a atiderivative o U. Problem 6B. Show that for each iteger p the sum k= is a polyomial of degree p + i the variable. Solutio: The map takig the polyomial S() to S(+) S() is a liear map from polyomials of degree p + to polyomials of degree p. Its kerel is the costats of dimesio, so it is oto. So for ay degree p polyomial T (i particular p ) we ca fid a polyomial S with S( + ) = S() + T (). Takig S( ) = we get S() = T () + T () T (). k p Problem 7B. (a) If A is a matrix over a algebraically closed field of characteristic with A = I for some > show that A is diagoalizable. (I is the idetity matrix.) (b) Show that over ay field of characteristic p > there is a matrix with A = I for some > that is ot diagoalizable. Solutio: (a) The miimal polyomial divides z which has distict roots (i characteristic ) so the matrix is diagoalizable. (b) Take a matrix A defiig a cyclic permutatio of p basis vectors of a p-dimesioal vector space. This satisfies A p = I, but its oly eigevalues are so if it were diagoalizable it would have to be the idetity matrix, which it is ot. Problem 8B.

9 Let G be a abelia group. For ay iteger m, deote by G[m] the subgroup of elemets of G whose orders divide m. Let p be a prime umber. Assume that G[p] has order p r for some iteger r. Prove that the order of G[p ] is at most p r for ay iteger. Solutio: Cosider the homomorphism G[p ] G[p ] give by multiplicatio by p. The kerel is exactly G[p]. It follows that we have a ijectio G[p ]/G[p] G[p ]. Thus we have G[p ]/G[p] G[p ]. It implies G[p ] G[p] G[p ]. Now the result follows from iductio. Problem 9B. Fid the smallest possible order of a o-abelia group of odd order. Solutio: The smallest example is a semidirect product of a ormal cyclic subgroup of order 7 by a cyclic group of order 3 actig o-trivially o it. Groups of prime order 3, 5, 7,, 3 are cyclic, groups of prime squared order 9 are abelia, ad groups of order 5 have ormal subgroups of order 3 ad 5 by Sylow s theorems so are cyclic.