Send all communications regarding Elementary P r o b l e m s and Solutions to P r o f e s s o r A. P. Hillman, Mathematics Department,

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1 ELEMENTARY PROBLEMS AND SOLUTIONS Edited by A.P. HILLMAN Uiversity of Sata Clara, Sata Clara, Califoria Sed all commuicatios regardig Elemetary P r o b l e m s ad Solutios to P r o f e s s o r A. P. Hillma, Mathematics Departmet, Uiversity of Sata Clara, Sata Clara, Califoria. Ay problem believed to be ew i the a r e a of r e c u r r e t sequeces ad ay ew approaches to existig problems will be welcomed. The proposer should submit each problem with solutio i legible form, preferably typed i double spacig with ame ad a d d r e s s of the proposer as a headige Solutios to problems should be submitted o separate sheets i the format used below withi two moths of publicatio. B Proposed by Verer E. Hoggatt, Jr., Sa Jose State College, Sa Jose, Califoria Show that L L,, = L~,,+(-1). where L is the -th Lucas +1 2+l umber defied by y L, = 1, L = 3, ad L l = L M + L ^ +1 B Proposed by Verer E. Hoggatt, Jr., Sa Jose State College, Sa Jose, Califoria Let u ad v be sequeces satisfyig u, J a u,,+bu = ad v. T+CV,, +dv = where a, b, c, ad d a r e costats ad let +Z +1 A % 7 (E 2 +ae+b)(e 2 +ce+d) = E 4 +pe +qe + r E + s. Show that y =u +v satisfies y j / i + p y j q y. o + r y. i + s y ~ 7 +4 r / + 3 / l J B Proposed by D.G. Mead, Uiversity of Sata Clara, Sata Clara, Califoria Fid costats p, q, r, ad s such that y p y q y r y s y = o is a 4th order r e c u r s i o relatiofor the t e r m - b y - t e r m products y =u v 3 r J of solutios of u l - u M - u = ad v, - 2 v,, -v = B Proposed by D.G. Mead, Ui versity of Sata Clara, Sata Clara, Califoria Fid the sum 1' 1 +1 * 2+2 B 5+3' F G, where F ^ = F,. +F +Z +1 ad G + 2 = 2 G G. 153

2 154 ELEMENTARY PROBLEMS AND SOLUTIONS April B Proposed by Walter W. Horer, Pittsburgh, Pesylvaia Fid expressios i t e r m s of Fiboacci umbers which will geerate itegers for the dimesios ad diagoal of a rectagular parallelopiped, i. e., solutios of 2,, 2, 2,2 a +b +c = d B Proposed by Werer E. Hoggatt, Jr., Sa Jose State College, Sa Jose, Califoria Solve the system of simultaeous equatios: x F,, +yf = x +y +1 J J x F y F + l = x x y where F is the -th Fiboacci umber. SOLUTIONS CHEBYSHEV POLYNOMIALS B Proposed by D.C. Cross, Exeter, Eglad Corrected ad restated from Vol. 1, No. 4: The Chebyshev Polyomials P (x) a r e defied by P (x) = cos(arccos x). Lettig ( f > = Arccos x, we have :os <f> - x = P (x), cos (2$) = 2cos - 1 = 2x - 1 = P? (x), cos (3<f>) = 4cos - 3cos <j> = 4x - 3x = P~(x), cos (4) = 8 c o s 4-8 c o s = 8 x 4-8x = P 4 (x), etc. It is well kow that P + 2< x > = 2 x P + l< X > " P ( x > ' Show that m P (x) = x I B. x ' 2 j j j=

3 155 ELEMENTARY PROBLEMS AND SOLUTIONS 155 where m = [/2 J, the greatest iteger ot exceedig / 2, ad (1) B = 2 ' o _ 1 ( 2 ) B.,, x l = 2 B... - B.. j + l, + l j + l, j, - l (3) I f S = B I + B B I, the S 1 = 2S,. + S ' o' ' I 1 ' m ' Solutio by Douglas Lid, Uiversity of Virgiia, Charlottesville, Va. By De Moivre's Theorem, (cos <f> + i si </>) - cos + i si rub Lettig x = cos <fi, ad expadig the left side, / 2 c o s + i s i = (x + i vl - x ).. = 2 (-l) j / 2 (Jx - j (l - x V / 2 j=o We equate real p a r t s, otig that oly the eve t e r m s of the sum a r e real, [/2] k cos < = P j x ) = h X ( - l ) k ( 2^) x " 2 k ( l - x 2 ). k= We m a y p r o v e from this (cf. Formula (22), p. 185, Higher Trascedtal Fuctios, Vol. 2 by Erdelyi et al; R. G. Buschma, "Fiboacci Numbers, Chebyshev Polyomials, Geeralizatios ad Differece Equatios, " Fiboacci Quarterly, Vol. 1, No. 4, p. 2) that (*) F r o m this, we have B. J, 11 \ - l / O \1L-J-Xf. H (-Zj): (1) B = 2 ~ l. o,

4 156 ELEMENTARY PROBLEMS AND SOLUTIONS April It is also easy to show from (*) that (2) B.,,,, = 2 B.,, - B.. j + l, + l J + l, J, -1 Now (*) implies so that (2) becomes B. = (-1) J, J B. ' j, ' ("D j + 1 1B. j + 1 l, + J l J = 2 ( - l ) j + 1 B. ' j + T l, J I + ( - D J + 1 B. ' j, -, l B.,,,, = 2 B., T + B... 1 j+1, +1 ' ' j+1, 1 j, -1 Summig both sides for j to I = J, we have (3) S,. = 2 S + S. v ' +1-1 Also solved by the proposer. A S P E C I A L C A S E B Proposed by Veter E. Hoggatt, Jr., Sa Jose State College, Sa Jose, Califoria Showthat F F ^ - F 2 = ( x - l ) + 1, where F is the -th Fib oacci umber, defied by F, = F~ = 1 ad F = F, + F Solutio by Joh L. Broiv, Jr., Pesylvaia State Uiversity, State College, Pa. Idetity XXII (Fiboacci Quarterly, Vol. 1, No. 2, April 1963, p. 68) states: F F - F. F,. = ( - l ) ~ k F, F,. m -k m+k k m+k- The proposed idetity is immediate o takig m = ad k = 2. More geerally, we have F Z - F _ F = ( - l ) " k F, 2 for < k <. -k +k k Also solved by Marjorie Bickell, Herta T. Freitag, Joh E. Homer, Jr., J.A.H. Huter, Douglas Lid, Gary C. MacDoald, Robert McGee, C.B.A. Peck, Howard Walto, Joh Wesser, Charles Ziegefus, ad the proposer.

5 1965 E L E M E N T A R Y P R O B L E M S AND S O L U T I O N S 157 S U MMING M U L T I P L E S O F S Q U A R E S B Proposed by Verer E. Hoggatt, Jr., Sa Jose State College, Sa Jose, Califoria Show t h a t (2-1 ) F? + (2 - Z)Y* F ^ _ = F!T 1 2 Z-1 2 Solutio by James D. Mooey, Uiversity of Notre Dame, Notre Dame, Idiaa Rememberig that I ^ F, 2 = F F,, k +1, k= 2 2 we m a y proceed by iductio. Clearlyfor = 1, F, = 1 = F -. Assume [2(-l) - l]? \ + [>(-l) - 2] T\ F ^ ^ ^ = = (2-3)F2 + (2-4) F F 2 _ 3 = F ^. The ( 2 - l ) F j F 2 _ x = [(2-3)Fj F 2 _ 3 ] l 2(F? v F 2 9) + F 2. = F 2? l F k F? > k k= k= F F 2-2 F 2 - l + F 2 - l F 2 = F F 2-2 F 2 - l + + F 2 - l ( F F 2-1> = F F 2-2 F 2 - l + F 2-1 = < F F 2 - l ) 2 = F 2 " Q ' E " D ' Also solved by Marforie Bickell, J.L. Brow, Jr., Douglas Lid, Joh E. Homer, Jr., Robert McGee, C.B.A. Peck, Howard Walto, David Zeitli, Charles Ziegefus, ad the proposer.

6 158 ELEMENTARY PROBLEMS AND SOLUTIONS April RECURRENCE RELATION FOR DETERMINANTS B Proposed by C.A. Church, Jr., Duke Uiversity, Durham, N. Carolia Show that the -th order determiat a x 1-1 a 2 1 () -1 a. ~ satisfies the r e c u r r e c e f() - a f(-l) + f(-z) for > 2. Solutio by Joh E. Homer, Jr., La Crosse. Wiscosi Expadig by elemets of the -th colum yields the desired relatio immediately. Also solved by Marjorie Bickell, Douglas Lid, Robert McGee, C.B.A. Peck, Charles Ziegefus, ad the proposer. AN EQUATION FOR THE GOLDEN M E A N B From a proposal by Charles R. Wall, Texas Christia Uiversity, Ft. Worth, Texas Show that x - x F - F, = has o solutio to greater tha a, -1 where a = (1 + >/ft)/2, F is the -th Fiboacci umber, ad > 1. Solutio by G.L. Alexaderso, Uiversity of Sata Clara, Califoria For > 1 let p(x, ) = x - x F - F,, g(x) = x - x - 1, ad i, x -2-3, , -k-1.. _, _ h(x, ) = x + x + 2 x F, x F x + F.. k -2-1 It is easily see that p(x, ) = g(x)h(x, ), g(x) < for - l / a < x < a, g(a) =, g(x) > for x > a, ad h(x, ) > for x >. Hece x = a is the uique positive root of p(x, ) =. Also solved by J.L. Broum, Jr., Douglas Lid, C.B.A. Peck, ad the proposer.

7 1965 ELEMENTARY PROBLEMS AND SOLUTIONS 159 GOLDEN MEAN AS A LIMIT B Proposed by Charles R. Wall, Texas Christia Uiversity, Ft. Worth, Texas Let F be the -th Fiboacci umber. Let x^ > ad defie x,, x,.. by J x... = f(x, ) where 1 Z k+1 k f(x) = \/F. + x F. -1 For > 1, prove that the limit of x, as k goes to ifiity exists ad fid the limit. (See B-43 ad B-55. ) Solutio by G.L. Alexaderso, Uiversity of Sata Clara, Sata Clara, Califoria For > 1 let p(x) = X - x F - F.. Let a = (1 + \[5)/2. As r v -1 i the proof of B-55, oe sees that p(x) > for x > a ad that p(x) < for < x < a e If x, > a, we the have (x ) > x F + F = (x ) k k -1 k+1 ad so x, > x.,,. It is also clear that x, > a implies k k+1 k (x. ^ ) = x, F + F, > a F + F. = a 11 k+1 k -1-1 ad hece x,,, > a. Thus x > a implies r x > x, > x^ >... > a. k+1 o o 1 Z Similarly, < x < a implies < x < x. < x_ <... < a. I both ' o o 1 Z cases the sequece x, x,,... is mootoic ad bouded. Hece x, o 1 k has a limit L > as k goes to ifiity. Sice L satisfies L = ^Y+rr, -1 L must be the uique positive solutio of p(x) =. Also solved by Douglas Lid ad the proposer. A F I B O N A C C I - L U C A S I N E Q U A L I T Y B Proposed by G.L. Alexaderso, Uiversity of Sata Clara, Sata Clara, Califoria Let F ad L be the -th Fiboacci ad -th Lucas um ber respectively. Prove that for all itegers > 2. ( F 4 / ) > L 2 L 6 L 1... L 4 _ 2

8 16 ELEMENTARY PROBLEMS AND SOLUTIONS April Solutio by David Zeitli, Mieapolis, Miesota Usig mathematical iductio, oe may show that F 4 = 2 L 4 k - 2 ' = 1, 2 f... k=l If we apply the well-kow a r i t h m e t i c - g e o m e t r i c iequality to the uequal positive umbers L OJ L,, L. AJ..., L., we obtai for 2 o 1 4-Z = 2, 3, 2 L 4 k k=l N / L 2 L 6 L 1... L ^ ^ which is the desired iequality. Also solved by Douglas hid ad the proposer. ACKNOWLEDGMENT XXXXXXXXXXXXXXX It is a pleasure to ackowledge the assistace furished by Prof. Verer E. Hoggatt, J r. cocerig the essetial idea of "Maximal Sets" ad the lie of proof suggested i the latter part of my article "O the Represetatios of Itegers as Distict Sums of Fiboacci Numbers. " The a r t i c l e a p p e a r e d I F e b *, s H. H. F e r s CORRECTION Volume 3, Number 1 Page 26, lie 1 from bottom of page Page 27, lies 4 ad 5 V ~+V- /)+V 7 _ = F Q - F _ = F A = 8 7, 3 7, 4 7, F ^ + F. + F, *... +F = F,, -1 ( eve) F Q +F C +F^ F = F,. -1 ( odd) ACKNOWLEDGMENT Both the papers "Fiboacci Residues" ad "O a Geeral Fiboacci Idetity, " by Joh H. Halto, were supported i part by NSF grat GP2163. CORRECTION Volume 3, Number 1 Page 4, Equatio (81), the R. H. S. should have a additioal t e r m - V 2 F V + 2