Mathematical Association of America
Mathematical Association of America
Mathematical Association of America
Reviewed work(s):
Source: The American Mathematical Monthly, Vol. 118, No. 4 (April 2011), pp. 371-378
Published by: Mathematical Association of America
Stable URL: http://www.jstor.org/stable/10.4169/amer.math.monthly.118.04.371 .
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PROBLEMS AND SOLUTIONS
Edited by Gerald A. Edgar, Doug Hensley, Douglas B. West
with the collaboration of Mike Bennett, Itshak Borosh, Paul Bracken, Ezra A. Brown,
Randall Dougherty, Tamás Erdélyi, Zachary Franco, Christian Friesen, Ira M. Ges-
sel, László Lipták, Frederick W. Luttmann, Vania Mascioni, Frank B. Miles, Bog-
dan Petrenko, Richard Pfiefer, Cecil C. Rousseau, Leonard Smiley, Kenneth Stolarsky,
Richard Stong, Walter Stromquist, Daniel Ullman, Charles Vanden Eynden, Sam Van-
dervelde, and Fuzhen Zhang.
PROBLEMS
11564. Proposed by Albert Stadler, Herrliberg, Switzerland. Prove that
√ !
e−x (1 − e−6x )
Z ∞
3+ 5
d x = log .
0 x(1 + e−2x + e−4x + e−6x + e−8x ) 2
11570. Proposed by Kirk Bresniker, Hewlett-Packard, Granite Bay, CA, and Stan
Wagon, Macalester College, St. Paul, MN. Alice and Bob play a number game. Start-
ing with a positive integer n, they take turns changing the number; Alice goes first.
Each player in turn may change the current number k to either k − 1 or dk/2e. The
person who changes 1 to 0 wins. For instance, when n = 3, the players have no choice,
k proceeds from 3 to 2 to 1 to 0, and Alice wins. When n = 4, Alice wins if and only
if her first move is to 2. For which initial n does Alice have a winning strategy?
11571. Proposed by Finbarr Holland, University College Cork, Cork, R 1 Ireland. Let
f be a nonnegative Lebesgue-measurable function on [0, 1], with 0
f (x) d x = 1.
Let K (x, y) = (x − y) f (x) f (y), F(t) = [0,t]×[0,t] K (x, y) dy d x, and G(t) =
2
R
SOLUTIONS
An Occasional Congruence
11411 [2009, 179]. Proposed by Alun Wyn-jones, KPMG, London, U.K. For positive
integers k and n, let L k (n) = n−1j=1 (−1) j .
j k
P
(a) Show that L 1 (n) ≡ L 5 (n) (mod n) if and only if n is not a multiple of 4.
(b) Given distinct, odd, positive integers i and j with {i, j} 6 = {1, 5}, show that the set
of n such that L i (n) ≡ L j (n) (mod n) is finite.
Solution to part (a) by A. Kumar, Goleta, CA . By induction on n, one can check that
(n − 2)(n − 1)n(n + 1)
L 5 (n) − L 1 (n) = (−1)n−1 (2n − 1) .
4
The numerator contains the product of four consecutive integers. If n is not divisible
by 4, then a factor other than n is divisible by 4, and n divides L 5 (n) − L 1 (n). If n is
divisible by 4, then L 5 (n) − L 1 (n) = (−1)n−1 (2n − 1)(n − 2)(n − 1)(n + 1)(n/4) ≡
(−1)(−1)(−2)(−1)(1)(n/4) = n/2 6 ≡ 0 (mod n).
Solution to part (b) by Richard Stong, Center for Communications Research, San
Diego, CA . We prove the restriction of the claim to odd n. The functions L k (n) are
related to the Euler polynomials by L k (n) = 12 [E k (0) + (−1)n−1 E k (n)] (see 23.1.4 in
Since
∞
X tr 2 et − 1
Er (0) = t =1− t = 1 − tanh(t/2),
r =0
r! e +1 e +1
With r = 0, we obtain E 1 (0) = −1/2. Since the even terms vanish, setting r = 2m in
(??) yields a recurrence for the odd terms:
m−1
1X 2m
E 2m+1 (0) = E 2s+1 (0)E 2(m−s−1)+1 (0). (3)
2 s=0 2s + 1
Induction on m yields E 2m+1 (0) ∈ Z[ 21 ], where Z[ 12 ] is the set of rational numbers with
powers of 2 as denominators.
Let E 0 = (−1)m+1 E 2m+1 (0). Multiplying (??) by (−1)m+1 yields E 2m+1 0
=
Pm−1 2m+1
1
2 s=0 E 0
E 0
2s+1 2(m−s−1)+1 . With E 1 (0) = −1/2, inductively all E 0
2m+1 are positive.
In particular, |E 2m+1 (0)| > m2 |E 2m−1 (0)|, so |E 5 (0)| < |E 7 (0)| < · · · . Since E 1 (0) =
E 5 (0) = − 21 and E 3 (0) = 14 , the values E 2m+1 (0) for m ≥ 0 are distinct except for
E 1 (0) = E 5 (0).
Factoring the generating function (??) as e xt · et 2+1 , we obtain
∞ ∞ ∞
X tr X xr tr X tr
Er (x) = · Er (0) ,
r =0
r! r =0
r ! r =0 r!
Z[ 12 ] and constant term Er (0). For odd i and j, we can now write
1
L i (2m + 1) − L j (2m + 1) = (E i (0) + E i (2m + 1) − E j (0) − E j (2m + 1))
2
= E i (0) − E j (0) + (2m + 1)P(2m + 1),
[Editor’s note - In the original printing, “an+r ” read “an+1 ”; we regret the error.]
j
(
X a p+ j a j+1 , if j is odd;
= a p+ j a j+1 − a p a1 − a p (−1) =
s
s=1
a p+ j a j+1 − a p , if j is even.
There are bi−1 b j pairs of tilings such that one has length i − 1 in positions 2 through i
and the other has length j in positions 1 through j.
To count these another way, group the pairs by the first fault, where a fault is a
position where both tilings have a breakpoint. The first fault immediately follows the
leftmost square in the two tilings, say at position r , and there are k ways to color that
square. This determines both tilings through position r , and there are bi−r b j−r ways to
complete the two tilings.
This counts all the pairs when j is odd, but when j is even there may be no fault. In
this case both tilings begin with j/2 dominos, and there are bi− j−1 ways to complete
the longer tiling.
Editorial comment. The special case of the identity in Solution II of (a) obtained by
setting r = n − 1, t = 1, and s = n + ` was proved (for k = 1) and applied in Br.
A. Brousseau, Summation of infinite Fibonacci series, Fibonacci Quarterly 7 (1969)
143–168.
Also solved by M. Bataille (France), D. Beckwith, R. Chapman (U.K.), P. Corn, C. Curtis, P. P. Dályay (Hun-
gary), Y. Dumont (France), D. Fleischman, J. Grivaux (France), O. Kouba (Syria), H. Kwong, J. H. Lindsey II,
O. P. Lossers (Netherlands), A. Nakhash, J. P. Robertson, J. Simons (U.K.), A. Stadler (Switzerland), R. Stong,
M. Tetiva (Romania), Microsoft Research Problems Group, and the proposer. Part (b) solved by G. C. Greubel
and GCHQ Problem Solving Group (U.K.).
An Arccos Integral
11457 [2009, 747]. Proposed by M. L. Glasser, Clarkson University, Potsdam, NY. For
real numbers a and b with 0 ≤ a ≤ b, find
Z b
x
arccos √ d x.
x=a (a + b)x − ab
Solution by FAU Problem Solving Group, Florida Atlantic University, Boca Raton, FL.
ab
We integrate by parts, using x − a+b as the antiderivative of 1. Notice that
d x b(x − a) − a(b − x)
arccos √ =− √ ,
dx (a + b)x − ab 2(a + b) x − a+b ab
(b − x)(x − a)
√
and that arccos(x/ (a + b)x − ab) vanishes for x = a and x = b. Thus
Z b b
x ab x
arccos √ dx = x − arccos √
a (a + b)x − ab a+b (a + b)x − ab a
" Z √
b Z b√ #
1 x −a b−x
+ b √ dx − a √ dx
2(a + b) a b−x a x −a
" Z √
b Z b√ #
1 x −a b−x
= b √ dx − a √ dx .
2(a + b) a b−x a x −a