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PROBLEMS AND SOLUTIONS Edited by Gerald A. Edgar, Doug Hensley, Douglas B. West with the collaboration of Paul T. Bateman, Mario Benedicty, Itshak Borosh, Paul Bracken, Ezra A. Brown, Randall Dougherty, Tam ´ as Erd ´ elyi, Zachary Franco, Chris- tian Friesen, Ira M. Gessel, Jerrold Grossman, Frederick W. Luttmann, Vania Mas- cioni, Frank B. Miles, Richard Pfiefer, Cecil C. Rousseau, Leonard Smiley, John Henry Steelman, Kenneth Stolarsky, Richard Stong, Walter Stromquist, Daniel Ull- man, Charles Vanden Eynden, and Fuzhen Zhang. Proposed problems and solutions should be sent in duplicate to the MONTHLY problems address on the inside front cover. Submitted solutions should arrive at that address before May 31, 2008. Additional information, such as generaliza- tions and references, is welcome. The problem number and the solver’s name and address should appear on each solution. An asterisk (*) after the number of a problem or a part of a problem indicates that no solution is currently available. PROBLEMS 11334. Proposed by Jonathan Bober and Jeffrey Lagarias, University of Michigan, Ann Arbor, MI. Let x and y be positive integers with x 2 − 3y 2 = 1. (a) Show that if x cannot be written as 2 n ,2 n ± 1, 3 · 2 n ,or3· 2 n ± 1 for any natural number n,thenxy has more than 3 distinct prime factors. (b) Show that if x = 2 n − 1 with n > 3, then y has at least three distinct prime factors. 11335. Proposed by Juan L ´ opez Gonz ´ alez, Madrid, Spain. Let σ(n) denote the sum of the divisors of n. (a) Find, with proof, the least positive integer m such that σ(6m)<σ(6m + 1). (b) Show that there are infinitely many m such that σ(6m)<σ(6m + 1). (cf G. Martin, The Smallest Solution of φ(30n + 1)<φ(30n) Is ,thisM ONTHLY 106 (1999) 449-451.) 11336. Proposed by Donald Knuth, Stanford University, Stanford, CA. A near- deBruijn cycle of order d is a cyclic sequence of 2 d − 1 zeros and ones in which all 2 d − 1 substrings of length d are distinct. For all d > 0, construct a near-deBruijn cycle of order d + 1 such that the front and back substrings of length 2 d − 1 are both near-deBruijn cycles of order d. (Thus, for example, 1100010 is near-deBruijn of order 3, while 110 and 010 are both near-deBruijn of order 2.) 11337. Proposed by Marian Tetiva, National College “Gheorghe Ros¸ca Codreanu”, B ˆ ırlad, Romania. Suppose in triangle ABC we have opposite sides of lengths a, b,and c, respectively, with a ≤ b ≤ c.Letw a and w b be the lengths of the bisectors of the angles A and B respectively. Show that a + w a ≤ b +w b . 11338. Proposed by Ovidiu Furdui, Cluj, Romania. Let denote the classical gamma function, and let G(n) = n k=1 (1/ k).Find lim n→∞ G(n + 1) 1/(n+1) − G(n) 1/n . January 2008] PROBLEMS AND SOLUTIONS 71 11339. Proposed by Jos ´ eLuisD ´ ıaz-Barrero, Universitat Polit ` ecnica de Catalunya, Barcelona, Spain. Let F n and L n denote the nth Fibonacci and Lucas numbers, respec- tively. Prove that for all n ≥ 1, 1 2 F 1/F n n + L 1/L n n ≤ 2 − F n+1 F 2n . (The Fibonacci and Lucas numbers are given by the recurrence a n+1 = a n +a n−1 , with F 0 = 0, F 1 = 1, L 0 = 2andL 1 = 1.) 11340. Proposed by ´ Oscar Ciaurri and Luz Roncal, Universidad de la Rioja, Logro ˜ no, Spain. An umbrella of radius 1 meter is spun with angular velocity ρ in the xz-plane about an axis (call it the y-axis) parallel to the ground. It is wet, and drops of water crawl along the ribs and fly off as they reach their ends. Each drop leaves the umbrella with a velocity vector equal to the velocity of the tip of the rib at the point where it exited. It then follows a parabolic trajectory. If a drop spins off while on the downspin, then the high point in its arc will be the point of departure. Otherwise, the high point is the vertex of a parabolic arc in the xz-plane. Determine a parameterized family P ρ of polynomials in two variables such that when- ever ρ 2 > g, the various arc vertices reached by the water droplets all lie on the curve P ρ (x, z) = 0. (Here g denotes the magnitude of the downward acceleration due to gravity.) The figure shows the case ρ 2 /g = 2, with the umbrella spinning counterclock- wise. SOLUTIONS A Condition on the Commutator 11196 [2006, 79]. Proposed by Mohammad Hossein Mehrabi, Iran University of Sci- ence and Technology, Tehran, Iran. Let A and B be real n × n matrices. Show that if AB − BAis invertible and A 2 + B 2 = √ 3(AB − BA),thenn is a multiple of 6. Solution by P ´ al P ´ eter D ´ alyay, De ´ ak Ferenc High School, Szeged, Hungary. From (A + iB)( A − iB) = A 2 + B 2 − i(AB − BA) = ( √ 3 −i)(AB − BA), it follows that det[(A + iB)(A −iB)]=( √ 3 − i) n det(AB − BA).SinceAB − BA is invertible, so is (A + iB)(A −iB). We conclude that det[(A +iB)(A −iB)]=det(A +iB) det(A −iB) = det(A +iB) det(A +iB)>0. Since A and B are real, also det(AB − BA) is real; hence ( √ 3 − i ) n is real. Since √ 3 − i = 2e −πi/6 , we conclude that n is a multiple of 6. Also solved by S. Amghibech (Canada), R. Chapman (U. K.), Y. Dumont (France), O. Furdui, J P. Gri- vaux (France), C. C. Heckman, E. A. Herman, O. P. Lossers (Netherlands), M. Omarjee (France), S. Pierce, A. K. Shafie (Iran), R. Stong, M. Tetiva (Romania), J. Vinuesa (Spain), L. Zhou, NSA Problems Group, and the proposer. 72 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 115 Can One Recover a (Spherical) Triangle from Its Medial Triangle? 11201 [2006, 179]. Proposed by Robert Russell, New York, NY. Given the midpoints of the sides of a spherical triangle, provide a construction for the original triangle. Composite solution by Christoph Soland (Switzerland), William Dickenson, and the editors. There is one configuration of midpoints, namely, mutually perpendicular, for which many choices of original triangle work. Otherwise, the underlying triangle may be recovered by a construction. We take it to be part of the definition of a spherical triangle that the three vertices do not lie on a common great circle. We also assume (by definition) that the sides of a spherical triangle are less than π radians. Thus, to construct a spherical triangle, it suffices to construct the three vertices. Let O be the center of our sphere S. Without loss of generality, we take O to be the origin of R 3 and the radius of S to be 1. When A and B are not antipodal, let m(A, B) be the midpoint of the spherical line segment AB, defined as the shorter of the two portions of the great circle through A and B delimited by A and B. We first consider the exceptional case. If any side of the medial triangle M 1 M 2 M 3 is π/2, then, we claim, all of them are, and there are infinitely many triangles with the same three midpoints, so we cannot uniquely construct the “original triangle”. For the proof of this claim, suppose that M 1 ⊥ M 2 . Without loss of generality, M 1 = (1, 0, 0) and M 2 = (0, 1, 0). By assumption, there exist distinct S 1 , S 2 ,andS 3 , no two antipodal, such that for any permutation (i, j, k) of (1, 2, 3), m(S i , S j ) = M k . Note that the ends of a line segment in spherical geometry are nearer the midpoint than its antipode. Now if S 2 = (x 2 , y 2 , z 2 ),thenx 2 > 0, S 3 = (x 2 , −y 2 , −z 2 ) so that y 2 < 0, and S 1 = (−x 2 , −y 2 , z 2 ).Nowz 2 = 0sinceS 1 =−S 2 , and it follows that M 3 = (0, 0, sign(z 2 )). Thus all three axes are perpendicular. Glancing over the foregoing algebra, we see that any choice of S 2 = (x 2 , y 2 , z 2 ) with x 2 , z 2 > 0 > y 2 and x 2 2 + y 2 2 + z 2 2 = 1givesriseto S 2 , S 3 ,andS 1 for which (1, 0, 0), (0, 1, 0),and(0, 0, 1) are the vertices of the medial triangle. We now consider the composition R = H 3 ◦ H 2 ◦ H 1 ,whereH j is the bijection of S obtained by reflecting it through the axis of ±M j . Either R is the identity map on S, or it is a nontrivial rotation that fixes two antipodal points, call them ±S. We claim that the M j are mutually perpendicular if and only if R = I . In one direction this is trivial by coordinate algebra after setting M 1 = (1, 0, 0) and so on. In the other direction, suppose R = I . Then for any permutation (i, j, k) of (1, 2, 3), H i ◦ H j = H k . Thus the M j are distinct, and no two are antipodal. Now let P = H 2 M 1 and Q = H 3 M 1 .If M 1 ⊥ M 2 ,thenM 1 and M 2 determine a line L on S, P, M 1 ,andM 2 are distinct points on L,andP is not antipodal to M 1 . Furthermore, H 1 Q = P since H 1 H 3 H 2 = I . Thus Q = H 1 P.SinceP lies on L and H 1 carries any line through M 1 onto itself, Q is on L.FromQ = H 3 M 1 ,we have also that Q is on the line L through M 1 and M 3 . Thus Q = M 1 , Q =−M 1 ,or L = L .ButQ = M 1 because M 3 =±M 1 ,andQ =−M 1 because that would make P = H 1 Q =−M 1 , a contradiction. Finally, if L = L , each H j transposes the poles of S if we take L as the equator, and so R transposes those poles, contrary to the the assumption that R = I . This proves that M 1 ⊥ M 2 , and by the same logic, all three M j are mutually perpendicular. If, on the other hand, no two of the M j are perpendicular, then the underlying S j may be recovered from the M j by a construction. Henceforth, we assume that no two of the M j are orthogonal. Thus, R = I .We first show that the axis ±S of rotation for R is not perpendicular to M 1 . To see this, suppose to the contrary that H 3 H 2 H 1 S = S,andS ⊥ M 1 .TakeS = H 1 S and S = January 2008] PROBLEMS AND SOLUTIONS 73 H 2 S .ThenH 3 S = H 3 H 2 H 1 S = S.IfS and S are antipodal, then S = H 2 S = −H 2 S.SinceH 3 S = S, M 3 S, and thus M 3 ⊥ M 1 , a contradiction. If S and S are not antipodal, then, viewing these points as vectors in R 3 , from H 2 S = S it follows that M 2 (S + S ).IfalsoS and S are not antipodal then M 3 S + S = S − S . But (S + S ) ⊥ (S − S ) so M 2 ⊥ M 3 . If instead, S and S are antipodal, then S = S ,ButthenH 2 S = S so that M 2 ±S.FromM 1 ⊥ S we have M 1 ⊥ M 2 .This shows that S ⊥ M 1 .NowtakeS 2 as the nearer of ±S to M 1 .ThentakeS 3 = H 1 S 2 , and S 1 = H 2 S 3 . It remains to discuss how S 2 may be located via a construction. This can depend on what tools are permitted. Here, we assume that lines through points, intersections of lines, midpoints of line segments, and perpendicular bisectors of line segments are constructible, that we can pick points on lines that are distinct from given points or their antipodes, that we can construct reflections of a point about a given axis, that we can determine which of two points is nearer a given point if the points are not equidistant, and we can pick points not on specified lines. Consider, then, an arbitrary pair (A, B) of distinct, non-antipodal points. If either of these is fixed by R,thenwehave±S 2 and we are done. Otherwise, S 2 lies on the per- pendicular bisector of ( A, R(A)), as well as the perpendicular bisector of (B, R(B)). If these are distinct, their intersection again gives S 2 . If not, then the axis of rotation for R lies on the intersection of the lines AB and RA RB, and that gives S 2 . Editorial comment. This problem would be trivial in Euclidean geometry: draw through each midpoint a line parallel to the segment joining the other two midpoints; these lines define the original triangle. In spherical geometry there are no parallels, so this construction doesn’t make sense. There are other snags, and the somewhat narrow definition used here for the term ‘spherical triangle’ was the only means known to the editors to avoid these. Also solved by R. Stong, A. Tissier (France), and the Microsoft Research Problems Group. Differentiable and Discontinuous Densely 11221 [2006, 367]. Proposed by Paolo Perfetti, University “Tor Vergata”, Rome, Italy. Give an example of a function g from R into R such that g is differentiable everywhere, g is differentiable on one dense subset of R,andg is discontinuous on another dense subset of R. Solution by Richard Bagby, New Mexico State University, Las Cruces, NM. Define f : R → R by f (x) = x 2 sin 2 (π/x) for 0 < x < 1and f (x) = 0 otherwise. Let {q k } be an enumeration of the rationals. We will show that g(x) = ∞ k=1 2 −3k f 2 k (x − q k ) has the required properties. Note that f is differentiable everywhere: 0 ≤ f (x) ≤ x 2 shows f (0) = 0and0≤ f (x) ≤ x 2 sin 2 (π − π/x) ≤ π 2 (x − 1) 2 shows f (1) = 0. Also note that | f (x)| is bounded and f is discontinuous at 0. Since f is continuous and the series for g(x) converges uniformly, the sum g(x) exists and defines a continuous function. The derived series ∞ k=1 2 −2k f 2 k (x − q k ) (1) 74 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 115 converges uniformly, so g is differentiable and g (x) is the sum of this derived series. We claim that g is discontinuous at each rational q j . Write (1) as g (x) = 2 −2 j f 2 j (x − q j ) + k=j 2 −2k f 2 k (x − q k ) . The last series defines a function continuous at q j since each term is continuous there and the series converges uniformly. The discontinuity of f at0thenshowsthatg is discontinuous at q j . For n a positive integer, let E n = ∞ k=n (q k − 2 1−k , q k + 2 1−k ) and E = ∞ n=1 E n . Thus E n has Lebesgue measure at most 2 3−n and E has measure zero, so the set of irrationals in the complement of E is dense. We claim g is differentiable at every irrational x 0 ∈ E.Sincex 0 is neither q k nor q k +2 −k for any k, each term 2 −3k f (2 k (x − q k )) is twice differentiable at x 0 .Letn be such that x 0 ∈ E n . Then the derived series (1) reduces to a finite series, and its derivative at x 0 is n−1 k=1 2 −k f 2 k (x 0 − q k ) . (2) To prove this finite sum is g (x 0 ),let α(x) = ∞ k=n 2 −2k f 2 k (x − q k ) . The term f 2 k (x − q k ) vanishes when |x − x 0 |≤2 −k ,so α(x) − α(x 0 ) ≤ C 2 −k ≤|x −x 0 | 2 −k |x − x 0 |≤2C|x − x 0 | 2 , where C is a bound for | f |. Hence α (x 0 ) = 0, so g (x 0 ) is indeed given by the finite series (2). Also solved by J. H. Lindsey II, M. D. Meyerson, A. Stadler (Switzerland), R. Stong, Szeged Problem Solving Group “Fej ´ ental ´ altuka” (Hungary), GCHQ Problem Solving Group (U. K.), and the proposer. Snapshots for Velocities 11223 [2006, 459]. Proposed by Christopher Hillar, Texas A& M University, College Station, TX, and Lionel Levine, University of California at Berkeley, Berkeley, CA. Consider n unlabeled particles moving each at its own constant velocity along the real line. An observer is promised some number P of snapshots of the particles, to be taken at uniformly spaced intervals of time. When particles coincide, the snapshot will show how many are at a given point. (a) Show that if P = n +1 then the observer can determine the velocity of each of the particles. (b ∗ ) As a function of n, what is the minimum value of P that will suffice to ensure that the observer can determine all n velocities? Solution to (a) by S. C. Locke, Florida Atlantic University, Boca Raton, FL. Suppose the kth snapshot is taken at time t k and the positions of the n particles at that time are (x k, j ) n j=1 ,wherex k, j is not necessarily the position of the jth particle. Let G = {(t k , x k, j ) : 1 ≤ k ≤ n + 1, 1 ≤ j ≤ n}.Ifp j (t) is the position of the jth particle at time t , then the points L j ={(t, p j (t)) : t ∈ R} would constitute a line meeting G in exactly n + 1 points (not counted with multiplicity). January 2008] PROBLEMS AND SOLUTIONS 75 If a line L meets G in at least n + 1 points (and therefore exactly n + 1 points), then at least two of these points are generated by the same particle, say particle j. Thus L = L j , since two distinct points determine a line. The lines meeting G in n + 1 points are therefore exactly the lines representing the motion of the individual particles, and the velocities of the particles are the slopes of these lines. Editorial comment. No solutions to (b) were received. Several solvers noted that for n ≤ 3, n + 1 snapshots are required. For n = 3, the three snapshots (−2, 0, 2), (−1, 0, 1) and (−2, 0, 2) are ambiguous. The GCHQ Problem Solving Group verified that for n = 4, four snapshots suffice. The best lower bound was due to Petr Skovron. Let P(n) be the minimum number of snapshots required for n particles. Suppose the n trajectories { f i (t)} n i=1 are indistinguishable from {g i (t)} n i=1 with snapshots at times 1, ,P − 1. For any constant a,the2n trajectories { f i (t) + a(t − P)} n i=1 ∪{g i (t) − a(t − P)} n i=1 are indistinguishable from { f i (t) − a(t − P)} n i=1 ∪{g i (t) + a(t − P)} n i=1 with snapshots at times 1, ,P. Hence P(2n) ≥ P(n) + 1. It follows that P(n) ≥ log 2 n+2. Part (a) also solved by D. Beckwith, K. Bernstein, D. R. Bridges, P. Budney, R. Chapman (U. K.), J. H. Lindsey II, L. Pebody, R. E. Prather, F. Yang, GCHQ Problem Solving Group (U. K.), Houghton College Problem Solving Group, Szeged Problem Solving Group “Fej ´ ental ´ altuka” (Hungary), and the proposer. A Unique Solution 11226 [2006, 460]. Proposed by Franck Beaucoup, Ottawa, Canada, and Tam ´ as Erd ´ elyi, Texas A& M University, College Station, TX. Let a 1 , ,a n be real numbers, each greater than 1. For n ≥ 2, show that there is exactly one solution in the interval (0, 1) to n j=1 ( 1 − x a j ) = 1 − x. Solution by Microsoft Research Problems Group, Redmond, WA. If a ≥ 1andx ≥ 0, then x a ≥ 1 +a(x − 1) by the mean value theorem (MVT). Equality occurs only for a = 1orx = 1. If f (x) = n j=1 (1 − x a j ) for 0 ≤ x ≤ 1, then 0 ≤ f (x) ≤ (1 − x a 1 )(1 − x a 2 ) ≤ a 1 a 2 (1 − x) 2 . Hence f (x)<1 − x as x approaches 1 − . On the other hand, 1 ≥ f (x) ≥ 1 − (x a 1 + ···+x a n ) shows that f (x)>1 − x as x approaches 0 + . By the intermediate value theorem, f (x) = 1 − x has a root in (0, 1). There is only one such root. Let g(x) = log(1 − x) − log f (x). Suppose g(0) = g(x 1 ) = g(x 2 ) = 0where0< x 1 < x 2 < 1. By MVT, h(x) = (1 − x)g (x) has at least one zero in (0, x 1 ) and another in (x 1 , x 2 ). By MVT again, h has a root in (0, x 2 ).However,h(x) =−1 + n j=1 a j x a j −1 (1 − x)(1 − x a j ) −1 ,soh (x) = n j=1 a j x a j −2 (x a j − 1 − a j x + a j )(1 − x a j ) −2 and is thus positive on (0, 1). Also solved by K. F. Andersen (Canada), R. Bagby, P. P. D ´ alyay (Hungary), J. H. Lindsey II, O. P. Lossers (Netherlands), T. L. McCoy (Taiwan), R. Mortini (France), L. Pebody, J. Rooin and A. Mahmoodi (Iran), A. Stadler (Switzerland), R. Stong, BSI Problems Group (Germany), GCHQ Problem Solving Group (U. K.), and the proposers. 76 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 115 A Triangle Inequality 11228 [2006, 460]. Proposed by Marian Tetiva, B ˆ ırlad, Romania. Prove that in an acute triangle with angles A, B,andC radians, (1 − cos A)(1 − cos B)(1 − cos C) cos A cos B cos C ≥ 8(tan A + tan B + tan C) 3 27(tan A + tan B)(tan A + tan C)(tan B + tan C) . Solution by Minh Can, Irvine Valley College, Irvine, CA. (tan A + tan B + tan C) 3 cos A cos B cos C (tan A + tan B)(tan B + tan C)(tan C + tan A)(1 − cos A)(1 − cos B)(1 − cos C) = (tan A + tan B + tan C) 3 cos A cos B cos C sin C cos A cos B sin A cos B cos C sin B cos C cos A (1 − cos A)(1 − cos B)(1 − cos C) = (sin A) 2 (sin B) 2 (sin C) 2 (1 − cos A)(1 − cos B)(1 − cos C) = (1 − cos 2 A)(1 − cos 2 B)(1 − cos 2 C) (1 − cos A)(1 − cos B)(1 − cos C) = (1 + cos A)(1 +cos B)(1 +cos C) ≤ 3 + cos A + cos B + cos C 3 3 ≤ 3 + 3cos (A + B + C)/3 3 3 = 27 8 . The last inequality follows from the concavity of cos x on the interval [0,π/2].The inequality before that follows from the arithmetic–geometric mean inequality. Equality holds if and only if the triangle is equilateral. Also solved by S. Amghibech (Canada), A. Arkady, A. R. Avidon, M. Battaille (France), D. Beckwith, A. Bun- gale (India), R. Chapman (U. K.), G. H. Chung, P. P. D ´ alyay (Hungary), P. De (Ireland), O. Faynshteyn (Germany), D. Fleischman, M. Goldenberg, M. Hajja (Jordan), E. A. Herman, Y J. Kuo, J. Lee (Korea), O. P. Lossers (Netherlands), M. Mabuchi (Japan), D. J. Moore, L. Pebody, D. Perkins, C. R. Pranasachar (In- dia), J. Rooin (Iran), V. Schindler (Germany), S. Shaebani (Iran), A. Stadler (Switzerland), R. Stong, T. Tam, S. Varosanec (Croatia), J. Vinuesa (Spain), M. Vowe (Switzerland), GCHQ Problem Solving Group (U. K.), Microsoft Research Problems Group, Szeged Problem Solving Group “Fej ´ ental ´ altuka” (Hungary), and the proposer. A Function Inequality 11232 [2006, 567]. Proposed by Michael W. Botsko, Saint Vincent College, Latrobe, PA. Let f be a continuous mapping from R into R that is bounded below. Show that there exists a real number x 0 such that f (x 0 ) − f (x)<|x − x 0 | holds for all x other than x 0 . Solution by Albert Stadler, D ¨ ubendorf, Switzerland. Put g(x) := f (x) +|x|/2. Then g is continuous, bounded below, and g(x) →+∞as x →±∞. Therefore there is a real number x 0 such that g(x 0 ) = min x∈R g(x). Thus f (x) +|x|/2 ≥ f (x 0 ) +|x 0 |/2 or f (x 0 ) − f (x) +|x 0 |/2 −|x|/2 ≤ 0. We conclude that for x = x 0 , f (x 0 ) − f (x) −|x − x 0 | < f (x 0 ) − f (x) −|x − x 0 |/2 ≤ f (x 0 ) − f (x) +|x 0 |/2 −|x|/2 ≤ 0. January 2008] PROBLEMS AND SOLUTIONS 77 Also solved by 41 others, and the proposer. A Derivative Formula 11233 [2006, 568]. Proposed by Robert Downes, Mountain Lakes High School, Moun- tain Lakes, NJ. Show that for positive integer n,andforx = 0, d n dx n x n−1 sin 1 x = (−1) n x n+1 sin 1 x + nπ 2 . Solution by Jean-Pierre Grivaux, Paris, France. We will show more generally that if f is an n-times differentiable real-valued function, then d n dx n x n−1 f 1 x = (−1) n x n+1 f (n) 1 x . The proof is by induction on n. The case n = 1 is obvious. Suppose it is true for n − 1. Applying the inductive hypothesis, and taking g(x) = x n−2 f (1/x) in Leibniz’s formula d n dx n (xg(x)) = x d n g dx n + nx d n−1 g dx n−1 , we compute d n dx n x n−1 f 1 x = x d n dx n x n−2 f 1 x + nx d n−1 dx n−1 x n−2 f 1 x = x d dx (−1) n−1 x n f (n−1) 1 x + nx (−1) n−1 x n f (n−1) 1 x = x (−1) n n x n+1 f (n−1) 1 x + (−1) n x n+2 f (n) 1 x + (−1) n−1 n x n−1 f (n−1) 1 x = (−1) n x n+1 f (n) 1 x . Specializing to f (x) = sin(x) gives the desired result. Editorial comment. The more or less equivalent result (obtained by setting f (x) = e x ,) d n dx n x n−1 e 1/x = (−1) n x −n−1 e 1/x , appears as a problem in 1850 Exercices de Math ´ ematiques Pos ´ es ` a l’Oral du CAPES de Math ´ ematiques et des Concours des Grande ´ Ecoles, by L. Moisotte. The most general formula of this kind submitted was given by O. P. Lossers, who showed d k dx k x m f 1 x = k j=0 k! j! m − j k − j x m−k−j (−1) j f ( j) 1 x , from which the above formula follows by taking k = n and m = n − 1. Also solved by 74 other readers and the proposer. 78 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 115 . solutions should be sent in duplicate to the MONTHLY problems address on the inside front cover. Submitted solutions should arrive at that address before May 31, 2008. Additional information, such as. gamma function, and let G(n) = n k=1 (1/ k).Find lim n→∞ G(n + 1) 1/(n+1) − G(n) 1/n . January 2008] PROBLEMS AND SOLUTIONS 71 11339. Proposed by Jos ´ eLuisD ´ ıaz-Barrero, Universitat Polit ` ecnica. this, suppose to the contrary that H 3 H 2 H 1 S = S,andS ⊥ M 1 .TakeS = H 1 S and S = January 2008] PROBLEMS AND SOLUTIONS 73 H 2 S .ThenH 3 S = H 3 H 2 H 1 S = S.IfS and S are