Featuring the 2005/6 British Mathematical Olympiad, Rounds 1 and ... the
Singapore Mathematical Olympiad 2004, Open Section, Special. Round;. - the
18th ...
34: No 7
November / Novembre 2008
Published by: Canadian Mathemati al So iety So iet e mathematique du Canada 577 King Edward Ottawa, ON K1N 6N5 Fax/Tele : 613 565 1539
CANADIAN
MATHEMATICAL SOCIETY 2008. ALL RIGHTS RESERVED.
SYNOPSIS
385 Skoliad: No. 113
Robert Bilinski
- National Bank of New Zealand Junior Mathemati s Competition 2005 - National Bank of New Zealand Con ours mathematiques Niveau junior 2005 - solutions to the Montmoren y Contest 2005-06, Se V. 396 Mathemati al Mayhem 396 398 404 406
Ian VanderBurgh
Mayhem Problems: M363{M368 Mayhem Solutions: M325{M331 Problem of the Month Ian VanderBurgh Adding Up Bru e Shawyer
408 The Olympiad Corner: No. 273 R.E. Woodrow Featuring the 2005/6 British Mathemati al Olympiad, Rounds 1 and 2; the Bulgarian National Olympiad, National Round, 2006; the Indian Mathemati al Olympiad 2006 Team Sele tion Problems; the South Afri an Mathemati al Olympiad 2004, Third Round; the 2006 Vietnamese Mathemati al Olympiad, Days 1 and 2; and readers' solutions to some of the problems from - the 2004 Chinese Mathemati al Olympiad; - the Singapore Mathemati al Olympiad 2004, Open Se tion, Spe ial Round; - the 18th Nordi Mathemati al Contest 2004; - the 2004 International Mathemati al Olympiad (Athens), problems proposed but not used. 422 Book Reviews 422
John Grant M Loughlin
Impossible? Surprising Solutions to Counterintuitive Conundrums
by Julian Havil Reviewed by Edward Barbeau
424
Twin Problems on Non-Periodi Fun tions
by Eugen J. Ionas u One version of L'Hospital's Rule states that if f and g are di erlim f (x) = lim g(x) = ∞ and entiable fun tions on (a, ∞) with x→∞ x→∞ f ′ (x) = L, g ′ (x)
f (x) then x→∞ lim = L. To this version of L'Hospital's g(x) Rule orresponds a Stolz-Cesaro Lemma, namely that if hxn in≥1 and y − yn hyn in≥1 , are sequen es with lim xn = ∞ and lim n+1 = L, n→∞ n→∞ x −x
lim
x→∞
y lim n then n→∞ x
n
n+1
= L.
n
In this gentle way, the author introdu es the Stolz-Cesaro Lemma and then he uses it to solve a warm-up problem. After that he pulls out some basi tools from analysis and with the Stolz-Cesaro Lemma in hand he pro eeds to solve, in a uni ed way, two problems on nonperiodi fun tions that appeared in the Ameri an Mathemati al Monthly. Enjoy!
430 Problems: 3376{3388 This month's \free sample" is:
3384.
Proposed by Mi hel Bataille, Rouen, Fran e
.
Show that, for any real number x, lim
n→∞
n−1 1 X
n−k−1 ⌊x⌋ + {x}2 k · x + = n2 k=1 n 2
,
where ⌊a⌋ is the integer part of the real number a and {a} = a − ⌊a⌋. .................................................................
3384.
Propose par Mi hel Bataille, Rouen, Fran e
.
Pour un nombre reel quel onque x, montrer que lim
n→∞
n−1 1 X
n−k−1 ⌊x⌋ + {x}2 k · x + = n2 k=1 n 2
,
ou ⌊a⌋ est la partie entiere du nombre reel a et {a} = a − ⌊a⌋. 435 Solutions: 3276{3281, 3283{3288
