Solutions to Morning Exam Mathematics

Solutions to Morning Exam Part I. Solve four of the following five problems. Problem 1. Verify that Z 2π X 3−(2n+1)/2 cos2 x dx = 6 (−1)n 2n + 1 0 n>0 by computing both sides. Solution: The left-hand side can be computed using either integration by parts (u = cos x, v = sin x), or the identity cos2 x = (1 + cos(2x))/2, the antiderivatives obtained being respectively (x+cos x sin x)/2 and x/2+sin(2x)/4 up√ to an additive constant. Thus the left-hand side is π. The right-hand side is 6f (1/ 3), where f (x) = X (−1)n n>0 x2n+1 . 2n + 1 Differentiating, we find f 0 (x) = X (−1)n x2n = n>0 1 , 1 + x2 dx = tan−1 x + C, and since f (0) = 0 we get C = 0. Thus the 1 + x2 √ right-hand side is 6 tan−1 (1/ 3) = π, which is the left-hand side. Problem 2. Suppose that y = y(t) is a differentiable function on R satisfying 2 y 0 (t) − sin(2t)y(t) = esin t . If y(0) = 0 what is y(π)? R Solution: Since − sin(2t) dt = cos2 t + C, we multiply both sides of the differ2 ential equation by ecos t , obtaining so f (x) = R 2 d (y(t)ecos t ) = e. dt 2 2 Therefore y(t)ecos t = et + C and y(t) = (et + C)e− cos t . Setting t = 0 and using 2 2 y(0) = 0, we obtain C = 0, so y(t) = ete− cos t or in other words y(t) = tesin t . Putting t = π gives y(π) = π. Problem 3. Let D be the upper half of the standard unit ball in R3 , defined by the inequalities x2 + y 2 + z 2 6 1 and z > 0. Assuming that D is of constant density, find the “centroid” or “center of mass” of D. You may use symmetry considerations and a standard volume formula to reduce the amount of calculation. Solution: Write the centroid as (x, y, z). Symmetry considerations (i. e. the invariance of D under rotation about the z-axis) give x = y = 0, and since the volume of D is (4π/3)/2 = 2π/3, we have Z Z Z 3 z= z dx dy dz. 2π D 1 2 Switching to spherical coordinates, we find that z is Z 2π Z π/2 Z 1 3 (ρ cos φ)(ρ2 sin φ) dρ dφ dθ = 3/8. 2π 0 0 0 Rx Problem 4. Let f be a continuous function on R, define F (x) = 0 f (t) dt, and suppose that a and b are real numbers with a < b. Apply the Mean Value Theorem to F on [a, b], simplifying your answer and expressing the result entirely in terms of f . Then interpret the result geometrically. Solution: The Mean Value Theorem asserts that F (b) − F (a) = F 0 (c)(b − a) for some c ∈ (a, b). By the Fundamental Theorem of Calculus, F 0 = f , so we get Z b f (x) dx = f (c)(b − a). a So the area under the graph of f from a to b is equal to the area under the horizontal line y = f (c) for some c between a and b. We can also write Z b 1 f (x) dx = f (c), b−a a and then we are saying that the average value of f from a to b is an actual value of f between a and b (the Intermediate Value Theorem for Integrals). Problem 5. Let ε(n) be the nth digit in the decimal expansion of π, so that ε(1) = 1, ε(3) = 4, ε(4) = 1, ε(5) = 5, and so on. Does the infinite series P = 3, ε(2) ε(n) (−1) (ln(1 + 1/n) − 1/n) converge? Why or why not? n>1 to see that P Solution: Since a series converges if it converges absolutely, 2it suffices 3 | ln(1 + 1/n) − 1/n| converges. Now ln(1 + x) = x − x /2 + x /3 − . . . for n>1 |x| < 1, and consequently | ln(1 + x) − x| 6 Cx2 for some constant C and all x near 0. Thus for large n we have | ln(1 + 1/n) − 1/n| 6 C/n2 . Since the series n>1 1/n2 converges (p-series with p = 2 > 1) we conclude that P n>1 | ln(1 + 1/n) − 1/n| converges and hence that the given series converges. P Part II. Solve three of the following six problems. R Problem 6. Find the value of the line integral C (y + ex )dx + (x2 − x + ey )dy, where C is the ellipse x2 /4 + y 2 /9 = 1 in the xy-plane, oriented counterclockwise. Solution: Let R be the region x2 /4 + y 2 /9 6 1. By Green’s theorem, the given line integral is Z Z Z Z ∂(x2 − x + ey ) ∂(y + ex ) − ) dx dy = (2x − 2) dx dy. ( ∂x ∂y R R The integral of 2x over R is 0 because 2x is odd and R is symmetric about the y-axis. So the integral is −2 times the area of R. The area of the ellipse x2 /a2 + y 2 /b2 = 1 is πab, so we conclude that the value of the given integral is −2(6π) = −12π. Problem 7. Let f and g be real-valued functions on R. Assume |f (x)| 6 M for some constant M > 0 and limx→0 g(x) = 0. (a) Using the formal definition of “limit,” prove that limx→0 f (x)g(x) = 0. 3 Solution: Let ε > 0 be given. Since limx→0 g(x) = 0, there exists δ > 0 such that if 0 < |x| < δ then |g(x)| < ε/M . Hence if 0 < |x| < δ then |f (x)g(x)| < M (ε/M ) = ε, and we conclude that limx→0 f (x)g(x) = 0. (b) Use (a) to show that the function ( x2 sin(1/x) if x 6= 0, r(x) = 0 if x = 0 is differentiable at 0. Solution: Let f (x) = sin(1/x) for x 6= 0, and put f (0) = 0. Also put M = 1. Then |f (x)| 6 M . Let g(x) = x, so that limx→0 g(x) = 0. We have r(0 + h) − r(0) = lim f (h)g(h) = 0 h→0 h 0 by (a), so r is differentiable at 0 and r (0) = 0. Problem 8. Find the maximum and minimum values of f (x) = xz + yz on the sphere S = {(x, y, z) ∈ R3 : x2 + y 2 + z 2 = 4}. Solution: Let g(x, y, z) = x2 + y 2 + z 2 . According to the method of Lagrange multipliers, the points where an extreme value occurs are among the points P ∈ S where ∇f (P ) = λ∇g(P ) for some λ ∈ R. Thus we seek solutions to the system  z = λ2x    z = λ2y  x + y = λ2z    2 x + y 2 + z 2 = 4. lim h→0 The first and third equations show√that √ if λ = 0 then z = 0 and x = −y, whence the fourth equation gives P = ±( 2, − 2, 0), so that f (P ) = 0. On the other hand, if λ 6= 0, then the first and second equations give x = z/(2λ) and y = z/(2λ), whence the third equation gives z/(2λ2 ) = z. Since √ z 6= 0 (else x = y = 0 also, contradicting the fourth√equation) we get λ = ±1/ 2. The first two equations then give x = y and z √= ± 2x, so the fourth equations gives 4x2 = 4. Thus x = ±1 and P = ±(1, with ε ∈ {±1}. Taking account of all possible signs, we find √ 1, ε 2) √ √ f (P ) = ±2 2. So 2 2 is the maximum value of f and −2 2 is the minimum value. Problem 9. Let {xn } be the sequence of positive real numbers defined by x1 = 1 and, for n > 1, xn+1 = 1 . xn + x−1 n Show that {xn } converges. To what number does it converge? Solution: Since xn > 0 for all n, we have xn xn+1 = 2 < xn . xn + 1 Thus {xn } is a decreasing sequence bounded below (by 0), and so it converges. If x = limn→∞ xn then xn x x = lim xn+1 = lim 2 = 2 . n→∞ n→∞ xn + 1 x +1 4 The only solution in R to x = x/(x2 + 1) is x = 0, so {xn } converges to 0. Problem 10. Define functions fn : [0, 1] → R for n > 1 by ( xn ln x if x 6= 0, fn (x) = 0 if x = 0. a) Is fn is continuous at 0? Justify your answer. Solution: Yes. By L’Hôpital’s Rule, lim+ fn (x) = lim+ x→0 x→0 xn ln x = lim = 0, −n x→0 −n x which is fn (0). b) Is {fn } a uniformly convergent sequence of functions? Justify your answer. Solution: Yes. Since ln x < 0 for x ∈ (0, 1), we see that fn (x) 6 0. Also, fn0 (x) = nxn−1 ln x + xn−1 and therefore fn0 (x) = 0 if and only if x = e−1/n . We see in fact that fn0 (x) < 0 for x < e−1/n and fn0 (x) > 0 for x > e−1/n , so the minimum value of the continuous function fn on [0, 1] is f (e−1/n ) = −1/(ne). Thus |fn (x)| 6 1/(ne) for x ∈ [0, 1]. Since the upper bound 1/(ne) is independent of x and goes to 0 as n goes to infinity, we see that {fn } is uniformly convergent to 0 on [0, 1]. RR Problem 11. Find the value of the surface integral F · dS, where the vector S field F is given by F(x, y, z) = (ey + xz, ex − yz, z), the surface S is the tetrahedron with vertices (0, 0, 0), (1, 0, 0), (0, 1, 0), and (0, 0, 1), and the normal vector points outward. Solution: By the Divergence Theorem, the given integral equals Z Z Z Z Z Z ∇ · F dV = 1 dx dy dz, D D where D is the interior of S and dV is the volume element. The right-hand side is Z 1 Z 1−z Z 1−y−z Z 1 (1 − z)2 1 dx dy dz = dz, 2 0 0 0 0 which is 1/6. Part III. Solve one of the following three problems. Pb Problem 12. Let S be the set of finite sums of the form n=a 1/n, where 1 6 a 6 b. Prove that S is dense in the set of nonnegative real numbers. Solution: Given x ∈ [0, ∞) and ε > 0, we must show that there exists s ∈ S such that |x − s| < ε. If x = 0 we choose n such that 1/n < ε and we take s = 1/n (i. e. a = b = n). Now suppose x > 0, and choose a so that 1/a < min(x, ε). The set B = {c > a : c X 1/n < x} n=a P∞ is nonempty because a ∈ B and is finite because n=a 1/n = ∞, i. e. the harmonic Pb series diverges. Put b = max(B) and s = n=a 1/n. Then 0 < s < x. On the 5 Pb+1 Pb+1 other hand, n=a 1/n > x by the definition of b. But n=a 1/n = s + 1/(b + 1), so we conclude that s < x < s + 1/(b + 1), and thus 1 1 1 0 0, and in particular 6= 0, for all (x, y). So by the Inverse Function Theorem, there are open neighborhoods V of P and and W of f (P ) such that f |V is a C ∞ -diffeomorphism, and thus in particular a homeomorphism, of V onto W . Thus N = f (U ∩ V ) is an open neighborhood of f (P ) contained in f (U ). Problem 14. Let X be a complete metric space with metric d satisfying the following condition: For every ε > 0 there is a collection of finitely many open balls of radius ε which covers X. Prove that X is compact. Solution: Given a sequence {xn } in X we will choose a subsequence {yn } such that for every N > 1, if m, n > N then d(yn , ym ) < 2/N . Since X is complete it will follow that the Cauchy subsequence {yn } converges, whence X is compact. To construct the subsequence {yn }, we proceed inductively. First, choose finitely many open balls of radius 1 which cover X. Then one of the open balls contains infinitely many terms of the sequence {xn }, and so we can choose a subsequence (1) {yn } satisfying (1) d(yn(1) , ym )<2 for all n, m > 1. (i) Now suppose that we have chosen sequences {yn } for 1 6 i 6 N such that (0) (i) (i−1) } for 1 6 i 6 N (with {yn } understood to be {yn } is a subsequence of {yn {xn }) and (i) d(yn(i) , ym ) < 2/i for all n, m > 1. Choose finitely many open balls of radius 1/(N +1) which cover X. (N ) Then one of the open balls contains infinitely many terms of the sequence {yn }, (N +1) and so we can choose a subsequence {yn } satisfying (N +1) d(yn(N +1) , ym ) < 2/(N + 1) for all n, m > 1. (ν) Finally, put yν = yν . Then for every N > 1 and all n, m > N , the terms yn (N ) and ym are terms of the sequence {yν }, and consequently they satisfy d(yn , ym ) < 2/N , as desired.