CONTENTS ARTICLES 3
Kirkman's Schoolgirls Wearing Hats and Walking through Fields of Numbers, by Ezra Brown and Keith E. Mellinger
16
When Euler Met I'Hopital, by William Dunham
25
Math Bite: A Magic Eight, by Paul and Vincent Steinfeld
26
Matroids You Have Known, by David L. Nee/ and Nancy Ann Neudauer
41
Letter to the Editor: Archimedes, Taylor, and Richardson, by Richard D. Neidinger
NOTES 42
Series that Probably Converge to One, by Thomas}. Pfaff and Max M. Tran
49
Flip the Script: From Probability to Integration, by David A. Rolls
55
Dirichletino, by lnta Bertuccioni
56
Proof Without Words: An Arctangent Identity, by Hasan Una/
57
Evil Twins Alternate with Odious Twins, by Chris Bernhardt
62
Proof Without Words: Bernoulli's Inequality, by
A ngel Plaza
PROBLEMS 63
Proposals 1811-1815
64
Quickies 987-988
64
Solutions 1786-1790
69
Answers 987-988
REVIEWS 70
NEWS AND LETTERS 72
68th Annual William Lowell Putnam Mathematical Competition
77
Letter to the Editor: Isosceles Dissections, by Roger B. Nelsen
80
New Editor of MATHEMATICS MAGAZINE
EDITORIAL POLICY Mathematics Magazine aims to provide lively and appealing mathematical exposi tion. The Magazine is not a research jour nal, so the terse style appropriate for such a journal (lemma-theorem-proof-corollary) is not appropriate for the Magazine. Articles should include examples, applications, his torical background, and illustrations, where appropriate. They should be attractive and accessible to undergraduates and would, ideally, be helpful in supplementing un dergraduate courses or in stimulating stu dent investigations. Manuscripts on history are especially welcome, as are those show ing relationships among various branches of mathematics and between mathematics and other disciplines. A more detailed statement of author guidelines appears in this Magazine, Vol. 74, pp. 75-76, and is available from the Edi tor or at www.maa.org/pubs/mathmag.html. Manuscripts to be submitted should not be concurrently submitted to, accepted for pub lication by, or published by another journal or publisher. Please submit new ma[luscripts by email to Editor-Elect Walter Stromquist at mathmag®maa.org. A brief message with an attached PDF file is preferred. Word processor and DVI files can al.so be con sidered. Alternatively, manuscripts may be mailed to Mathematics Magazine, 132 Bodine Rd., Berwyn, PA 19312-1027. If possible, please include an email address for further correspondence.
Cover image: Kirkman's Schoolgirls Walking through Fields of Numbers, by Hunter Cowdery, art student at West Val ley College, who is animating his way to San jose State University, and jason Challas, who lectures on computer graphics and fine art at West Valley.
AUTHORS Ezra (Bud) Brown grew up in New Orleans, has de grees from Rice and LSU, and has been at Virginia Tech since 1969, where he is currently Alumni Distinguished Professor of Mathematics. His re search interests include number theory and com binatorics, and he particularly enjoys discovering connections between apparently unrelated areas of mathematics and working with students who are engaged in research. In his spare time, Bud en joys singing everything from grand opera to rock and roll, playing jazz piano, and talking about his granddaughter Phoebe Rose. Under the direction of his wife jo, he has become a fairly tolerable gar dener. He occasionally bakes biscuits for his stu dents, and he once won a karaoke contest.
William Dunham is the Truman Koehler Professor of Mathematics at Muhlenberg College and a card carrying member of the Euler Fan Club. Over the years, Dunham has enjoyed grazing through Euler's work, finding nuggets of pure gold, and sharing them with the wider community. The present pa per is a case in point: here Euler resolves the Basel Problem (i.e., sums the reciprocals of the squares) by using four transcendental functions and three applications of I'H6pital's rule! It is Uncle Leon hard at his symbol-manipulating best.
Keith E. Mellinger earned his Ph.D. in finite geom etry at the University of Delaware. After graduate school he spent two years as a VIGRE postdoc at the University of Illinois at Chicago. He currently lives in Fredericksburg, VA, where he is an asso ciate professor and department chair at the Univer sity of Mary Washington. Keith's interests include many areas of discrete mathematics and he reg ularly lures unsuspecting undergraduates into his projects. This article grew out of some lively con versations between Keith and Bud about whether or not anybody would ever want to read something they would coauthor. In his spare time, Keith en joys spending time with his wife and two darling children, and plays guitar and mandolin in a local bluegrass band.
David L. Neel earned his Ph.D. at Dartmouth Col
lege under Kenneth Bogart, who started him down the path of matroids. He has since strayed occa sionally into graph theory as he migrated west to Truman State University in Missouri and then west ward again to Seattle University, his current mi lieu where he serves as an associate professor of mathematics. There he enjoys discrete math, and, discreetly, other arts like literature (Woolf, Gad dis, Wallace, Bernhard), film (Coens, P.T. Ander son, Antonioni), and music (Radiohead, Schoen berg, Monk). He believes the world would benefit from fiction written with mathematical sophistica tion.
Nancy Ann Neudauer received her Ph.D. from the University of Wisconsin. Her research interests in clude matroid theory, graph theory, enumeration, and their connections to other areas of mathemat ics. She has been involved in the MAA for as long as she can remember, at least since she gave a talk at an MAA meeting while still a high school student. She now serves on the Board of Gover nors, is active in the Pacific Northwest section, is a PNW section NExT officer, and will always be a silver dot. She enjoys finding the hidden matroid in all mathematics talks she attends. When not try ing to show her students the beauty of matroids, she races sailboats and travels where ever she can, rarely leaving the house without her passport and a packed bag, just in case.
Vol. 82,
No. 1,
February 2009
MATHEMATICS MAGAZINE EDITO R Frank Farris Santa Clara University
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Deanna B. H aunsperger Carleton College
Warren P. Johnson
Connecticut College
Elgin H . Johnston Iowa State University
Victor J. Katz University of District of Columbia
Keith M. Kendig Cleveland State University
Roger B. N elsen Lewis & Clark College
Kenneth A. Ross
University of Oregon, retired
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Harry Waldman
MAA, Washington, DC
EDITO R I A L ASSI STA N T Martha L. Giannini STUD E N T EDITO R I AL ASSI STA N T Michael V. Ryan
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ARTICLES Kirkman's Schoolgirls Wearing Hats and Walking through Fields of Numbers EZRA BROWN
Virginia Pol ytech nic I nstitute a n d State University B l acks b u rg, VA 24061 brown@math . vt.edu
KEITH E. MELLINGER University of Mary Washington Fredericksb u rg, VA 22401 kmel ling® u mw. edu
F ifteen young lad ies at school
Imagine fifteen young ladies at the Emmy Noether Boarding School-Anita, Barb, Carol, Doris, Ellen, Fran, Gail, Helen, Ivy, Julia, Kali, Lori, Mary, Noel, and Olive. Every day, they walk to school in the Official ENBS Formation, namely, in five rows of three each. One of the ENBS rules is that during the walk, a student may only talk with the other students in her row of three. These fifteen are all good friends and like to talk with each other-and they are all mathematically inclined. One day Julia says, "I wonder if it's possible for us to walk to school in the Official Formation in such a way that we all have a chance to talk with each other at least once a week?" "But that means nobody walks with anybody else in a line more than once a week," observes Anita. "I'll bet we can do that," concludes Lori. "Let's get to work." And what they came up with is the schedule in TABLE 1 . TA B L E 1 : Walking to school MON
TUE
WED
THU
FRI
SAT
a,b, e
a,c, f
a,d,h
a,g,k
a,j,m
a,n,o
c, I, o
b,m,o
b,c,g
b, f,k
d, f,m
d,g,n
b,h, I
e,j,o
c,d,j
b,d, i
g, i, j
e,h, i
f , I, n
e,m,n f, i, 0
h,k,n
j,k,l
i, k,m
SUN
a, i, I
b,j,n
c, e,k
c,h,m
d, e,l
f,h,j
d,k,o
g,h,o
g,l,m
e, f,g
c, i, n
TABLE 1 was probably what T. P. Kirkman had in mind when he posed the Fifteen Schoolgirls question in 1 850. Appearing in the unlikely-sounding Lady's and Gentle men's Diary [15], it reads as follows:
Fifteen young ladies of a school walk out three abreast for seven days in succes sion: it is required to arrange them daily so that no two shall walk abreast more than once. Kirkman's publication of this problem and solution [15, 16] is one of the starting points for what has become the vast modern field of combinatorial design theory. Its 3
4
MAT H E MAT ICS MAGAZ I N E
poser, Thomas Pennyngton Kirkman (1806-1 895), is one of the more intriguing fig ures in the history of mathematics. He published his first mathematical paper when he was 40, and was the first to describe many structures in discrete mathematics. Among these are block designs, which form the basis for the statistical design of experiments; bipartite graphs, which are essential for such problems as classroom scheduling and medical school admissions; and Hamiltonian circuits, which are at the heart of the fa mous Traveling Salesman Problem. (Biggs [2] gives more details about Kirkman's life and work.) For these achievements, combinatorialists regard him as the "Father of De sign Theory"-yet his fame outside the field rests entirely on the Schoolgirls Problem and his solution. This story is about the very problem that made Kirkman famous. His solu tion is an example of a resolvable (15, 35, 7, 3, I)-design, and we begin by ex plaining what those words and numbers mean. We describe how one of us found such a design by looking in a most unlikely place: the algebraic number field K = Ql(.J2, ,J3, v's, ,./7). This proves to be a particularly fertile field in which several other block designs grow. We talk about spreads and packings in finite geome tries, how a particular packing in the geometry P G (3, 2) answers Kirkman's question, and how the P G (3, 2) design is really the same as the number field design. Finally, we show how our design is a solution to a certain problem in recreational mathematics called the Fifteen Hats Problem. We begin by talking about block designs. Block desi gns an d Kirkman Tri ple Systems
Design theory began with Euler's studies of Latin squares in the 1 8th century, inter est in which was recently rekindled with the world-wide popularity of Sudoku. Many decades after their invention by Kirkman, block designs appeared in connection with R. A. Fisher's work [10, 11] on the statistical design of agricultural experiments, and the first comprehensive mathematical study of the field was due to R. C. Bose [4]. More recently, they have found applications in coding theory, cryptography, network design, scheduling, communication theory, and computer science. Finally, designs have al ways appealed to mathematicians because of their elegance, beauty, high degree of symmetry, and connections with many other fields of mathematics [5]. A balanced incomplete block design with parameters v, b, r, k, and A is a collection B of b subsets (or blocks) of a v-element set V of objects (or varieties) such that each block contains k varieties, each variety appears in r blocks and each pair of distinct varieties appears together in A blocks. Such a design is also called a (v, b, r, k, A) design. We say a design like this is incomplete if k v. From a combinatorial point of view, complete designs are not very interesting. However, statisticians do use them to design experiments. The five parameters in these designs are not independent. Since there are b blocks, each of size k, there are bk occurrences of varieties in the design. On the other hand, there are v varieties, each occurring in r blocks, and so a total of vr varieties appear in the design. Hence bk = vr . A similar counting argument shows that r (k - 1) = A(v - 1). Hence A(v - 1) and b = AV(v - 1) . r= k-1 k(k - 1) Because of these relations, such a design is frequently called a (v, k, A)-design. (There are more details about block designs in [5].)
P ( E; )
=
( 1 - P (£1 ) ) ( 1 - P (E )) 2
•
•
•
>
( 1 - P ( E; _ J ) )
b + (i - 1 )l . b + (l - 1 ) / + r + (l - 1 )d
.
To show that the whole sample space is covered, notice that P ( F' )
=
00 r + j d . and so i--> limoo P ( F' ) = fl fl r + j d j =O b + r + j (l + d) ' j =O b + r + j (l + d) i-l
If the the log of the product approaches
f In ( j=l
- oo ,
the product approaches 0 , so consider
r + jd
b + r + j (l + d)
) .
The integral test gives the desired result:
1 00 In ( 1
r + xd
b + r + x (l + d)
) dx =
- oo .
Intuitively, for j sufficiently large, (r + jd) / [b + r + j (l + d) ] is less than one when l =!= 0; thus the limit of the product is zero. Calculating the first few probabilities b - __ P (E 1 ) b+r'
P (E ) 3
=
�
d j (l + d) ,
r (b + l ) P (E ) 2 - (b + r ) (b + r + l + d)
which
and
r (r + d) (b + 2l)
(b + r ) (b + r + l + d) (b + r + 2l + 2d) '
leads to the general result P (E; )
=
b + (i - 1 )/
i-l
r + jd
(3)
r + (i - 1 ) d D b + r + j (l + d) ·
The two cases above, the original telescoping series and (2) , arose from (b , r, l , d) equal to ( 1 , 1 , 0 , 1) and ( 1 , 1 , 0 , 2) . In the case ( 1 , 1 , 0 , 0) and, in fact, whenever b = r and l = d, the complicated formula (3) reduces to 1 /2; and the series is just ( 1 ) , since these cases are the same as flipping a fair coin. Some other cases that work out nicely are ( 1 , 1 , 1 , 0), ( 1 , 1 , 2 , 0) , (k , 1 , 0 , 1) (k 1 ) , and (b , r, b , r ) : 1 -
� - f:r
1 = � L...., i=l
>
i
k (i - 1 ) ! k ! (k + i ) !
00
1 - ""'
(i + 1 ) ! '
,
and
2i - 1
8 2i (i ! ) , oo b ( r
1 - 2: i=l r
b+r
)i
MATH EMAT I C S MAG AZ I N E
46
As an exercise, readers can show that the first three are telescoping series and the last is a special case of the geometric series. For one more layer of complexity, let's start with a bag containing b 0 blue mar bles and r 0 red marbles. In this case, if a red marble is removed at the ith tum we add i l blue marbles and i d + 1 red marbles. Again we have P(£ 1 ) = b/(b + r) . Conditioning on the previous draw and using I:�= I k = n (n + 1 ) /2 we discover that >
>
x
x
P (E; ) =
n
2b + i (i - 1 )l 2r + J (J - l )d 2b + 2r + J (J - l ) (l + d) . 2r + i (i - 1 )d j =1
As usual we let F; be the event that i consecutive red marbles are selected. The event F; is calculated in essentially the same manner as E; except that the last draw is now a red instead of a blue. Hence, (
p F
,
; IT )-
2r + j (j - l )d 2b + 2r + j (j - 1 ) (1 + d) · j=l
In this case, lim;-+ oo P ( F; ) does not always equal 0. For instance, when l = 0 and d = r = b we discover, according to Hansen [3] and Mathematica, that lim P (F; ) i -" :xo
=
OO IT 2 + j. (j. - 1) = cosh j = I 4 + j (J - 1 )
--(.J!Jr) (v'IS:r r) 2
sech
2
�
0. 1455.
Intuitively this makes sense since we are continually adding a lot of red marbles and no blue marbles thus making it possible never to draw a blue marble. But now S F U (ur:;,1 E; ) , where F is the event that a blue marble is never drawn. Hence, =
1
=
P (S)
=
P (F) +
8 P (E; ) = cosh 00
+
On the other hand, if we let d = 0 and r then
=
( .j7[ :rr ) ( v'I5:rr ) -- sech -2 2
oo
]
i 2 + j (j - 1) 4 + 1 (j _ 1) · 2 + iu - 1) b = l, adding more blue marbles than red,
8
2
n
n
D
()
lim 4 + . 2 . 1 ) ::::; lim � ::::; lim � -+ oo P ( F; ) = i--+oo ilim i --+oo j = l 4 i --+oo 4 j (J j= l
;
= 0,
and so (4) For a slight variation we can start with b 0 blue marbles in the bag and r 0 red marbles and continue to select marbles until a blue is chosen, under the rule that if a red is chosen at the ith tum we add l · i ! blue marbles and d · i ! + 1 red marbles. In this case, calculating in the same manner, we find that P (£1 ) = b/(b + r), P(E2) = r (b + l)/(b + r ) (b + r + l + d), and for i 2 >
>
>
P (E,. ) _
� r [b + z I:�� � k ! ] (b + r) [r + d L��I� k!]
.
i- l r + d L:{=1 k! , fl . = + (l + d) I: �= I k! b + r j l
47
VO L . 8 2 , N O . 1 F E B RU A RY 2 009 I
and
i-l
r + d L... "V k =l k! . . P(F; ) = b +r r n b + r + (l + d) L:i k ! =l =l i For d = 0 and r b = l we have
i
--
=
i- 1 P(F;) = 1
and clearly lim; � 00 P(F;) = 0 and so
2
D
1 1 + Lk=o k ! '
00 L:��� k! 1 1 1 = - + - + I: . I . . (5) 2 3 i= 3 2 [1�� � [ 1 + L:i =o k!] On the other hand, for l = 0 and r = d = b, we find that i- 1 "V i k .' k =� . P(F;) = -21 n L... (6) 1 + L:i =o k ! i =l In this case, we don't necessarily expect the limit to be 0 since we are adding a lot of red marbles but no blue marbles. In fact, Mathematica suggests that the limit is around 0.23. We leave it as a challenge to the reader to calculate the limit or at least show that
it is positive (or wait until the next section to see why). In yet another variation, we start with b 0 blue marbles and r 0 red marbles in the bag, but this time if a red is chosen at the ith tum add z i blue marbles and di + 1 red marbles. Here P(E1) = bf(b + r ) , P (E2 ) = r (b + l ) / (r + b) (r + d + b + l ) , >
P (£. ) = I
>
1
" ki = l z k ] r [b + L...
in
r + L... "V ki = l d k (r + b) [r + L � = l dk] i = l r + b + L k = l dk + L k=l [ k '
and
i- 1 n
i
k
r + _"V L... k =l d . P(F) = r + b . i = l r + b + L:i = I dk + L:i = I [ k When d = 0 and b = r = l 1 , we have 1
r
--
>
h oo
. hm P(F;) =
1 . 1 = 0, 2 [1�� 1 [2 + ( 1 - ri+ l )j(l - r)]
P (E1 ) = 1 / 2 , P (E2 ) = 1 / 3 , and for i
>
2
i-l P (E;) = (r +. r 2) (r - 1 );-z -
2 [1�� 1 [ri + 2r - 3] I
'
which yields, in the specific case when r = 2, the series 00 1 1 2i-2
. (7) 1 = - + - + I: . I 2 3 i =3 TI ��� [2i + 1 1 Of course, when r = b and l = d, we again come up with the sum given in ( 1 ) since
the bag always has equal amounts of red and blue marbles. We leave it to the reader
48
MAT H EMAT I C S MAGAZ I N E
to check this if they wish. For the case l = 0 and b = = d we don't expect P ( F; ) to converge to 0 so we leave that case alone. Certainly there are other avenues to pursue since there are many possibilities of what to put back in the bag. We could also stop when we have chosen k blue marbles in a row instead of just one. We could also put more than two colors of marbles in the bag: we could have red, blue, and green marbles and stop if a red is chosen on an even draw or a blue is chosen on an odd draw, while adding marbles after each draw. r
Conn ecti ons and con clus i ons
It is well known [2, Theorem 3.20] that 2:: : 1 an and fl:1 (1 + an ) either both con verge or both diverge. Taking the log of the product the series 2:: : 1 log(l + an ) be haves just like the other two. As an application, we can show that for {an } positive, n: , an / 0 + an ) is positive if and only if I:: 1 /an converges, since I
log
(fJ � ) = f - log ( 1 n = l 1 + an
n= l
+
2_) , an
which converges to a nonzero value if and only if 2::: 1 1 /an converges. This shows that the limit in (6) must be positive, since the choice a11 = L� =O k! guarantees the convergence of 2:: : 1 l /an . To prove another corollary we need the fact that for {an } positive, which follows by replacing a; with 1 + a; positive, we conclude
an
-1
to make the sum telescope. Hence, for if
diverges ,
i=l
(8)
and 00
if L a; converges , i=l
(9) both of which follow from the relationship between 2:: � 1 a ; and n:, o + an ) . Many of our results can be derived from these facts. For example, take a; = 1 + G) to derive (4), a; = L� = ' (k - 1 ) ! to obtain (5) and a; = 2 ; to get (7). Since our methods provide a probabilistic interpretation of these special cases of (8) and (9), it is natural to wonder whether they can be used to prove (8) and (9) in general. Indeed, they can as the following shows: Given a sequence of positive numbers an , define a game where on the ith turn we throw a dart at a unit square board. The game ends if the dart lands in the region of the square below height a; I ( 1 + a; ) and continues if it lands above that height. Let E; be the event that the game ends at the ith turn and F; be the event that the game continues to the next turn. Calculation shows that P (F1 ) = 1 / ( l + a 1 ) and so P (F; )
=
P (F; - I )
1 -1 + a;
i
=
1 n --
n = l 1 + an
49
VO L . 8 2 , N O . 1, F E B R U ARY 2 009
and
P ( E; )
i 1 . . . target area on the ith throw) = a; n = P ( F; _ 1 ) P (h 1ttmg n= l 1 + an ---
Finally, with F n�1 F; and S = U� 1 E ; U F we see that (8) and (9) are equivalent to the fact that P (S ) = 1 , and that L�! a; converges if and only if n�, ( 1 + a; ) converges. Of course, there are series that we have not yet derived by these methods, such as 2:: , 6j (nn) 2 = I and l::o 1 / (n ! e ) = 1 . More generally, if we have a series of positive terms is there always a corresponding probabilistic scenario? =
Acknowledgment.
W e would like to thank Warren Johnson w h o carefully read this paper a n d provided numerous
thoughts that greatly improved it.
R E F E R E N C ES I . G.
College Math. J. 33 (2002) 3 9 1 -394. Real Infinite Series, MAA, 2006 . 3. E. R. Hansen, A Table of Series and Products, Prentice Hall, Englewood Cliffs, NJ, 1 97 5 . Berresford. Runs in coin tossing: Randomness revealed.
2 . Daniel D. B onar and Michael J . Khoury,
F l ip the Script: From Probabi l ity to I ntegration DAV I D A . RO L L S
The U n ivers i ty of Me l bo u r n e Pa r kv i l le, V I C, 3 01 0, AUSTRALIA D. Ro l l s ® m s . u nime l b . e d u . a u
Sometimes in mathematics we can reinterpret a problem or a result to advantage. The problem becomes easier to solve, or the result becomes even more useful. For an easy example from probability, imagine calculating the integral (1)
The integrand resembles f (x) = 3e- 3x , x ::::: 0, the probability density function (p.d.f.) of a rate 3 exponential probability distribution. Although they differ by a multiplying constant, that's easy to address, and we can write 1 - 3x dx 3e- 3x dx = - ( 1 ) = - .
1 0
00
e
=
-
3
1 oo
I
1
3
0
3
The key point is that the integral of a p.d.f. over its domain must be one. Compare that with the standard calculus approach using a substitution u = -3x and a limit for the improper integral
1 oo e - 3x dx = lim 1 e -3x dx = lim - l o eu du t
f -+ 00
0
=
� lim eu 3
f -+ 00
0
f -+ 00
I 3t = �3 0
-
-
1
3 - 3t
0 = �. 3
MAT H EMAT I CS MAGAZ I N E
50
Which solution would you rather use? Now, imagine calculating
1oo � 5 2xy e -2x dy dx
(2)
in your head. Think about it for a moment, then read on ! Obviously, students take calculus-based probability after a number of calculus courses where they exercise their skills of integration, including integration by parts and improper integrals. But after the course in probability, what might be said about calculus that couldn't be said before? This note illustrates a few such points, primarily by interpreting integrands as familiar probability density functions . Monte Carlo inte gration is also discussed as a technique to approximate integrals. To do this, integrands must again be interpreted in terms of probability density functions.
Defin ition s from probability Suppose X is a continuous random variable, meaning that X assigns a real number to every possible outcome of some particular experiment [10, p. 1 87] . Then there is a nonnegative function f called the probability density function (p.d.f.), or simply the probability density, that we use to compute probabilities. A key property of probability densities is
i: f (x) dx
=
1.
The expected value (also called the mean) of X and g (X) [10, pp. 1 9 1- 1 92] are defined to be E [X]
=
i: xf (x) dx
and
E [g (X)]
=
i: g (x ) f (x) dx
(3)
respectively, where g is a real-valued function. Let us review two common examples of continuous distributions. The rate A. expo nential distribution has density
j (x) =
{ Ae-icx '
0,
x �O otherwise
and mean 1 /A. . The uniform distribution on the interval [a , b] has density
f (x)
=
I
b�a'
0,
otherwise
and mean given by the midpoint (a + b) /2. (Note that inclusion of the endpoints is optional for continuous distributions since they don' t change the value of an integral.) Sometimes probability densities are specified using only the nonzero portion, with the function understood to be zero elsewhere. The univariate definitions extend to n dimensions, imagining jointly continuous random variables X I · . . . , Xn [10, Section 6. 1 ] . Then for the joint p.d.f. /x 1 . . . . . x. (X I , . . . , Xn ) we have
1
00
- 00
••
·
1-
00 00
fx 1 , ... , Xn (x i , . . . , Xn ) dx i
·
·
·
dxn
=
1
VO L . 8 2 , N O . 1, F E B R U A RY 2 009
51
and for the expected value of g(X 1 ,
E [g (X J , . . . , Xn ) ]
=
•
.
•
, Xn ) where g is real-valued, we have
i: i: g (XJ , . . . , Xn ) fx 1 ·
·
·
•...
,Xn (XJ , . . . , Xn ) dXJ
·
·
· dXn .
Expected value behaves in a linear fashion: For any constants a and b
E [a X + b] The condition that X 1 , saying for all (x , , Xn )
1
•
•
•
•
•
E
•
=
a E [X] + b .
, Xn are independent [10, Section 6 . 2] is equivalent to JRn the joint density factors as
where /x1 (x J ) , . . . , /x1 (xn ) are all univariate densities. In this case
where g1 , g2 ,
•
•
.
,
gn are real-valued functions.
Ma king q u ick work of in tegrals So, how might we reinterpret to advantage? Imagine calculating the integral
which resembles the integral in ( 1 ) but adds integration by parts into the mix. Again, we can think of a rate 3 exponential distribution, which has mean 1 /3 . By (3) we have
1 00 xe-3x dx 0
=
1
1 00 x (3e - 3x ) dx 3 0
=
1 - E [X] 3
=
1 - .
9
It takes longer to write down the integral than to compute it ! For a simple bivariate example, consider the slightly tedious but straightforward integral
/1
=
11 11 (x + y) dx dy .
To the student of probability, if X and Y are random variables with uniform distribu tions on [0, 1 ] we have h
=
E [X + Y]
=
E [X] + E [ Y ]
=
1 1 2+2
=
1
since expected value behaves in a linear fashion and the mean of a uniform distribution is the midpoint of its interval of definition. The joint p.d.f. here is the constant 1 over the interval of integration. A slightly more advanced example is the integral h
=
� 3 � 3 (x + y) dx dy .
(4)
MATH EMAT I C S MAGAZ I N E
52
Corresponding to each integral is an interval [ 1 , 3]. A uniform distribution on this interval has p.d.f. f(x ) =
{�'
1 sxs3
0, otherwise. We can form a joint p.d.f. f (x , y) here as the product f (x , y) = f (x) f (y ), so h = 2
2
1 3 1 3 (x 2� y) dx dy
II
=
4E[X + Y]
4(E[X] + E[Y]) = 4(2 + 2) = 16. This idea extends so easily. Try calculating =
1 3 1 3 (x 1 + ·
·
·
·
·
·
+ x 1 0) dx 1
·
·
·
(5)
dx 1 0.
Again, the idea is to write the integral using an integrand that is a legitimate density. How does your calculation time compare with the time just to type the iterated integrals into Maple? The intervals of integration need not be the same, and the integrand doesn't have to be a sum. In fact, the distributions need not be the same either. Consider the integral h =
100 1 5 2xy e -Zx dy dx
mentioned earlier in (2). Thinking of X as a rate 2 exponential distribution, and Y as a uniform distribution on [ 1 , 5], the integrand resembles (up to a constant) an expected value using a joint density which is the product of the two densities
{
2e- 2x '
x ;:: O
{I
1sys5
= 4> 0, otherwise and fy (y) 0, otherwise. Thus, we can treat X and Y as independent. Since we know E[X] 1 /2 and E[Y] = 3 we have 00 xy fy (y) fx (x) dy dx = 4E[XY] = 4E[X] E[Y] = 6. h =4
fx (x)
=
1 15
=
Monte Carlo integrati on
The idea of using probability to calculate integrals really shows benefits in one of the applications of Monte Carlo methods [9] called Monte Carlo integration [2, pp. 670-672], [3, pp. 2 13-221]. Monte Carlo methods are a popularization of statistical sampling techniques that started with an idea of the mathematician Stanislaw Ulam in the early 1940s. He conceived the idea as an attempt to approximate the proba bility of winning at solitaire, where combinatorics leads to an exponential explosion in the number of possible configurations of playing cards. About this he said, "In a sufficiently complicated problem, actual sampling is better than an examination of all the chains of possibilities" [11, p. 197]. The term itself seems to have been coined by Nicholas Metropolis [8, p. 1 27], a colleague and co-author of Ulam, partly in con nection to an uncle of Ulam's who would borrow money to visit the casino in Monte Carlo, one of the districts of the Principality of Monaco.
VO L . 8 2 , N O . 1 F E B R U ARY 2 009
53
I
The idea of Monte Carlo integration is to use probabilistic simulation to compute an estimate of the true value of the integral. As above, the key point is to find a density that allows interpreting the integrand in terms of probability. For the integral in (4 ), imagine simulating 2N independent random values {X 1 , . . . , X N } and { Y1 , . . . , YN } , all from a uniform distribution on [ l , 3]. Then an estimate of I2 is given by N l IN = 4- �)X;
N i =1 + YJ and by the Strong Law of Large numbers, with probability 1, as N A
(6) ---+ oo
1 N IN = 4- :�.)X; + Y; ) ---+ 4E[X + Y] = fz. N i=l Unlike with deterministic methods such as Riemann sums, 10 simulations using Monte Carlo integration can yield 10 different estimates, depending on the particular random values that have been generated. For example, five simulations using (6) and either N = 5 or N 5000 yielded the estimates in TABLE 1 . Larger N reduces the variance of the estimate-that is, a collection of estimates is more closely clustered around the true value. Such simulations can by done with a variety of computer tools and languages such as R, Maple, and C, which provide pseudorandom values using some mathematical algorithm. A measure of the quality of the estimate fN is its variance. The smaller the variance, the tighter the clustering of the estimates, thus increasing the confidence that the error is small. It is also possible several distributions, and so several densities, could be used to generate estimates of the same integral. Smaller variance for similar N is a good reason to prefer one over the other, assuming computation time is comparable. In fact, there are a number of more advanced schemes designed to reduce variance in the estimate. A
=
TA B L E 1 : Five est i m ates of ( 4 ) u s i n g N 5 (to p row) a n d N 5 000 (bot tom row) . Larger N l owers the va r i a nce-the est i m ates a re m o re c l os e l y c l u stered a ro u n d the t r u e va l ue o f s i xtee n . =
Is Isooo
=
1 6.8 1 970 14.46337 17.40961 2 14.60413 1 6.40 1 83 16.01292 15.98240 16.08013 1 5.89388 1 6.02809
Notice that no antiderivative is required for Monte Carlo integration. In fact, if you had one, why use an approximation at all? So, generally you can think of Monte Carlo integration as an alternative to a deterministic numerical integration method (such as the trapezoid rule). In one dimension, those methods are preferable. But an advantage of the Monte Carlo approach is how it scales with dimension. Notice that a Monte Carlo estimate of (4) requires N points of the form (X; , Y; ), i = I , . . . , N so 2N ran dom values. Similarly, an estimate for the integral in (5) generalized to d dimensions requires N points (X�i l, . . . , xJl), i = 1 , . . . , N so N d random values. On the other hand, a numerical approximation using a Riemann sum on a grid with M values in each dimension involves M2 points for (4) and Md points for the generalization of (5). Since Md grows exponentially in d, it is potentially a much larger number than N d, requiring more time to generate the estimate. This is sometimes called the curse of dimensionality. A rule of thumb is that Monte Carlo integration is preferable when the dimension is about eight or more. In fact, Monte Carlo integration can provide
MAT H EMAT I C S MAGAZ I N E
54
X
estimates of integrals that appear virtually impossible otherwise. Suppose 1 , , Xd are random variables, possibly not independent, with joint p.d.f. fx1 , , xd (x i , . . . , xd ) . The marginal expectation of 1 o i s the d-dimensional integral
X
E[Xd =
/_: · · · /_: x dx 1 ,
•••
•..
•
•
•
, x/XJ , . . . , xd ) dx 1 · · · dxd .
The support of f may be a d-dimensional hypercube, �d , or some more complicated volume. The dimension d might be large (the number of days in the year or obser vations in a large experiment). One can always sample from the volume and thereby create an estimate of the expected value.
Final thoughts The examples shown here use distributions whose features can be reasonably remem bered. If you can remember the density and mean of other distributions, you can calcu late more integrals similarly. Maybe you can extend to variances and higher moments too. But remember, not all distributions have finite expected mean or variance-so be careful which distributions you use ! Also, all the iterated integrals here had constant or infinite limits of integration-no variables. This allows using ideas of independence. Unfortunately, not all integrals are so simple. What we've seen here is the use of probability to reinterpret some problems from calculus. The references below give more information about probability, in general, and Monte Carlo methods, in particular. But we could equally consider any topic through the lens of a later course. What else can you reinterpret to advantage? Acknowledgment. I would like to thank Edward Boone and the anonymous referees for their helpful comments that improved this paper.
R E F E R E N C ES I . Jay
L. Devore, Probability and Statistics for Engineering and the Sciences, 6th ed., Brooks/Cole, Belmont,
CA, 2004. 2. Haym Benaroya and Seon Han, Probability Models in Engineering and Science, Taylor and Francis, New York, 2005 . 3. Michael J. Evans and Jeffrey S . Rosenthal, Probability and Statistics: The Science of Uncertainty, W. H. Free man, New York, 2004. 4. W. R. Gilks, S. Richardson and D. J. Spiegelhalter, eds. , Markov Chain Monte Carlo in Practice, Chapman & Hall, London, 1 996. 5 . Paul Hoffman, The Man Who Loved Only Numbers, Hyperion Books, New York, 1 998.
6 . Malvin H. Kalos and Paula A. Whitlock, Monte Carlo Methods: Volume 1 , John Wiley, New York, 1 986. 7 . Frances Y. Kuo and Ian H. Sloan, Lifting the curse of dimensionality, Notices Amer. Math. Soc. 52 (2005 ) 1 320-1 329. 8 . N. Metropolis, The beginning of the Monte Carlo method, Los Alamos Sci. 15 Special Issue ( 1 987) 1 25-1 30. 9 . Nicholas Metropolis and S . Ulam, The Monte Carlo method, J. Amer. Statist. Assoc. 44 ( 1 949) 335-34 1 . 1 0 . Sheldon Ross, A First Course in Probability, 6th ed. , Prentice-Hall, Upper Saddle River, NJ, 2002. 1 1 . S . M. Ulam, Adventures of a Mathematician, Charles Scribner, New York, 1 976.
VO L . 82 , N O . 1, F E B R U ARY 2 009
55
Dirichletino I N TA B E RT U CC I O N I Chem i n de Ia Raye, 1 1
C H - 1 024 E c u b l ens, SWITZE RLAN D l nta . B ertucc i o n i ®gma i l .com
In 1 837 P. G. Lejeune Dirichlet published his celebrated theorem (1], stating that any arithmetic progression a , a + b, a + 2b, a + 3b, . . . wherein a and b have no common factor contains infinitely many prime numbers. All known proofs are difficult, and the most readable ones (see, for instance, Serre [3] ) use nontrivial results of complex analysis. It can be shown [ 4, 5] that, in a very precise technical sense, elementary proofs are only possible when a 2 = 1 modulo b, in particular for a = 1 and a = - 1 . Recently, Hillel Gauchman published a simple proof [2] for the case a = 1 . We show here what we believe to be an even simpler proof, using an idea of Gauchman's [2] (the only idea we need, in fact) and a little lemma. To fix the notation, we restate what we want to prove. THEOREM . If N is any positive integer, there are infin itely many primes of the form 1 + m N, m = 1 , 2, . . . .
Proof Let q 1 , , qr be r primes of the form 1 + m N . We will find another prime of this form. Let p1 , , Ps be the distinct prime divisors of N. Consider, as in Gauch man [2] , for k = 1 , . . . , s the polynomials •
•
•
•
.f, k (X) =
•
•
XN - 1 = ( X N ! Pk ) Pk - l X N IPk - 1
+
( XN IPk ) Pk - 2 + . . . + 1 .
Fixing an index k, we can decompose fk (X) into a product of irreducible monic poly nomials in Z [X ] . Since fk ( e 2T
a - 1.
University
-ANGEL P LAZA
of Las Pa1mas
de Gran Canaria
35017-Las Pa1mas G . C . , SPAIN ap1aza @ dmat.u1pgc .es
PRO B LEMS E LG I N H . JOH N STO N,
Editor
I owa State U n ivers i ty
RAZVAN G E LCA, Texas Tech U n ivers i ty; RO B E RT G REGORAC, I owa G E RALD H E U E R, Concord i a Co l lege; VAN IA MASCIO N I , B a l l State U n iver BYRON WAL D E N, Santa C l ara U n ivers i ty; PAU L ZEITZ, The U n ivers i ty of San Fra n c i sco
Assista n t Editors: State U n ivers i ty; s i ty;
P R O PO S A L S To b e considered for publication, solutions should b e received by july 1 , 2 009.
1811.
Proposed by Emeric Deutsch, Polytechnic University, Brooklyn, NY. a connected graph G with vertices v 1 , v2 , . . . , Vn , let d;, j denote the
Given distance from v; to v j . (That is, d;, j is the minimal number of edges that must be traversed in traveling from v; to vj .) The Wiener index W (G) of G is defined by
W(G)
=
L
l �.i < j � n
di,j ·
a. Find the Wiener index for the grid-like graph
on 2n vertices. b. Find the Wiener index for the comb-like graph
on
2n vertices.
Proposed by Bob Tamper, University of North Dakota, Grand Forks, ND. Let m and n be relatively prime positive integers. Prove that 1812.
We invite readers to submit problems believed to be new and appealing to students and teachers of advanced undergraduate mathematics. Proposals must, in general, be accompanied by solutions and by any bibliographical information that will assist the editors and referees. A problem submitted as a Quickie should have an unexpected, succinct solution. Solutions should be written in a style appropriate for this
MAGAZINE.
Solutions and new proposals should be mailed t o Elgin Johnston, Problems Editor, Department of
lbT!Y( file) to on each page the reader's
Mathematics, Iowa State University, Ames lA 500 1 1 , or mailed electronically (ideally as a
ehj ohnst t!l i astat e . edu.
All communications, written or electronic, should include
name, full address, and an e-mail address and/or FAX number.
63
64
MAT H EMAT I C S MAGAZ I N E
Proposed by Elton Bojaxhiu, Albania, and Enkel Hysnelaj, Australia. Let a, b, and c be positive real numbers . Prove that 1813.
1 1 3 1 + + a ( 1 + b) b ( 1 + c) c ( l + a) � ( 1 + �) >
---
.
1814. Proposed by Michael Goldenberg and Mark Kaplan, The Ingenuity Project, Baltimore Polytechnic Institute, Baltimore, MD. Let A 1 A2A 3 be a triangle with circumcenter 0 , and let B 1 be the midpoint of A 2 A 3 , B2 be the midpoint of A 3 A � o and B3 be the midpoint of A 1 A 2 • For -oo t :S oo and k 1 , 2, 3, let Bk t be the point defined by 0 Bk :t t 0 Bk (where by Bk oo we mean the point at infinity in the direction of 0 Bk ). Prove that for any t E ( -oo, oo], the lines A k Bk t k 1 , 2, 3, are concurrent, and that the locus of all such points of concurrency is the Euler line of triangle A 1 A 2 A 3 • 1815. Proposed by Stephen J. Herschkorn, Rutgers University, New Brunswick, NJ. It is well known that if R is a subring of the ring Z of integers, then there is a unique mZ. Determine a similar unique characterization for positive integer m such that R
=
=
N and an x E D with l fn (x) - f (x) l :=:: E . Thus there is an increasing sequence { nd of positive integers and a sequence { xd of elements of D such that l !n k (xk ) - f (xk ) l :=:: E for all k. Because D is compact, there is a subse quence {xkm } of { xd with Xkm -+ x for some x E D. Now by the above observation, f (xkm ) -+ f (x) and, because f is continuous, f (xkm ) -+ f (x). But this implies that I fn km (xk,. ) - f (xkm ) I -+ 0 and contradicts the fact that I fn k (xk ) - f (xk ) I 2: E for all k. We conclude that fn -+ f uniformly on D . Also solved by Michel Bataille (France), Elton Bojaxhiu (Albania) and Enkel Hysnelaj (Australia), Paul Bud ney, Robert Calcaterra, Eugene A. Herman, Shoban Mandel (India), Nicholas C. Singer; Bob Tomper; Stuart V. Witt, and the proposer.
February 2008
A determinant and an inverse
Proposed by Harris Kwong, SUNY Fredonia, Fredonia, NY. For nonzero real numbers a 1 , a 2 , an , define s I:�= I 1 /ak and t t + a2 1789.
•
.
.
=
.
t where
t
is
a real number with
st
=1= - 1 . Find A - I and det(A ) .
VO L . 82, N O . 1, F E B R U ARY 2009
67
Solution by Robert Calcaterra, University of Wisconsin Platteville, Platteville, WI. = 1 j a j , and D be B 2 = s B and that
Let I be the n x n identity matrix, B be the n x n matrix with bij the n x n diagonal matrix with d;,; = a; . It is easy to check that A D - 1 = I + t B . It then follows that A D- 1
( I - t ( l + st ) - 1 B) = (I + tB) {I - t ( l + st ) - 1 B) = I - t ( l + st ) - 1 B + t B - t 2 ( 1 + st) - 1 B 2 = I. ( I - t (l + st)- 1 B ) .
Thus, A - 1 = D - 1 To determine the determinant, subtract the last row of A from each of the other rows of A . The result is the matrix
T=
(ri
0
t
t
.
which has the same determinant as A . For each
t a;
i,
1
S
i
S
n-
1 , add
.
( - 1 ) - (row ! of T ) to row n of T . The result is an upper triangular matrix
S with the same determinant as A
and with
1 sisn-l i = n. It follows that det(A)
= (st +
1)
TI�= I a; .
Also solved by Ricardo Alfaro, Michel Bataille (France), Elton Bojaxhiu (Albania) and Enkel Hysnelaj (Aus tralia), Hongwei Chen, Chip Cunis, Luz M. DeAlba, John Ferdinands, Fisher Problem Solving Group, Dmitry Fleischman, David Gould and Chi-Kwong Li and Victoria Perrigan, Russell Jay Hendel, Eugene A. Herman, Parviz Khalili, Kim Mcinturff, Eric Pite (France), Rob Pratt, Ken Ross, Nicholas C. Singer; A lben Stadler (Switzerland), Jeffrey Stuart, Marian Tetiva (Romania), Bob Tomper; Dave Trautman, Yarning Yu, Chris Zin and Samuel Otten, and the proposer.
February 2008
A ring of power?
1790. Proposed by Erwin Just (Emeritus), Bronx Community College of the City Uni versity ofNew York, Bronx, NY. Let R be a ring and assume that for each x E R, x + x 2 + x 3 + x 4 = x 11 + x 1 2 + x 1 3 + x 2 s . Prove that there is an integer
N
>
1
such that for each x
E
R, we have x = x N .
Solution by the proposer. = x 1 27 for each x E R. We first prove the following lemma: LEMMA . /fu E R with u 2 = 0, then u = 0. Proof. By the hypotheses we have u + u2 + u 3 + u 4 = u 1 1 + u 1 2 + u 1 3 + u 28 • Because all terms in the sum but the first are 0, we conclude u = 0. •
We show that x
Now let x
E
R and set
MAT H EMAT I C S MAGAZ I N E
68
Then v = X and it follows by induction that Next, let X
vnX
= X for all positive integers n .
Then W x 2 = x and it follows by induction that for positive integers p
1 , then f ( l js ) has the wrong denominator. Thus s ± I. =
=
=
=
=
Solution to B6. Notation: [a , b ] li mited permutations of [ I , n ] and set
n
(i) I f 2 :S
:S
2k, Pn . k
=
{a, a + 1 , . . . , b } . Let Sn .k denote the set o f k Pn .k = { a E Sn .k I a ([2, k + 1]) £: [ k + 2, 2k + I ] } .
=
Sn ,k ·
Proof 1 [2, k + l ] n [ l , n ] l
>
l [k + 2, 2k + l ] n [ 1 , n] l .
(ii) If 2k + 1 :S n and a E Sn , k \ P11, k , then a stabilizes the unique involution fixing 1 and switching i and i
[ 1 , 2k + 1 ] whereupon it induces + k for 2 :;: i :;: k + 1 .
Proof That a ( [ 1 , 2k + 1 ] ) [ 1 , 2k + 1 ] follows from the fact that a ( I ) U a ( [k + 2, 2k + 1]) must fill the positions in [ 1 , k + 1 ] . The specified form of a l l 1 . 2k + l l follows by induction on i . =
(iii)
I P11 ,k l
is even.
Proof
In fact, we exhibit a pairing of the elements of Pn , k · For a E mi nimal such that a (j ) :S k + 1 . Define a' E Sn ,k by
a ' (i) Then
a'
E
Pn,k and a"
=
=
l
a (j ) a(l) a (i )
if i if i
=
=
Pn , k . let j
:=:::
2 be
1, j,
otherwise.
a.
We now establish the required result by induction on n . For n :S 1 , I Sn,k I is I and therefore odd, while for 2 :S n :S 2k, I Sn ,k I is even by virtue o f ( i ) and (iii). For the inductive step, we note, using (ii), that when n :=::: 2k + I , there is a bijection f : Sn ,k \ Pn ,k --+ Sn - k- l , k 2 whereby f (a ) (i ) a (i + 2 k + 1 ) - 2k - 1 , and so, b y (iii), I Sn .k l = 1 Sn - 2k- 1 ,k l (mod 2 ) . =
VO L . 82, N O . 1, F E B R U A RY 2 009
77
Lette r to t h e E d i to r : I sosce l es D i ssect i o n s I enjoyed Des MacHale 's "Proof Without Words : Isosceles Dissections" i n the December 2008 issue of the MAGAZINE. However, his third claim (that a triangle can be dissected into two isosceles triangles if and only if one of its angles is three times another or if the triangle is right angled) is incorrect. There is a third case to consider. Consider an acute !::::.ABC, two of whose angles are x and 2x ( 0 < x < 45 ° ) , as il lustrated in FIGURE 1 a. Such a triangle can be dissected into two isosceles triangles as follows. As in FIGURE 1 b, draw the cevian CD equal in length to BC so that !::,.B CD is isosceles . It now follows that LACD = x , so that !::::.A CD is also isosceles . c
c
D Figure 1
A tri a n g l e w i th o n e a n g l e twi ce a nothe r
Two familiar members of this family are the triangles with x = 30° ( which makes !::::.AB C a right triangle) and x = 36° (when !::::.A B C is itself isosceles) . Mac Hale also showed how every acute triangle can be dissected into three isosceles triangles . The construction in FIGURE 1 can be extended to show that every isosceles obtuse triangle can also be dissected into three isosceles triangles, as in FIGURE 2.
Figure 2
An obtuse i sosce les tri a n g l e d i ssect i o n
-RO G E R B. N E L S E N Lewis and Clark College Portland, OR 972 1 9
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New from the Mathematical Association of America Numb er Theory Through Inquiry David C. Marshall, Edward Odell &
Michael S tarbird
Number Theory Through Inquiry is an i nnovative textbook
that leads students o n a gu ided d i scovery of i n trodu ctory n u m
b e r t h eory. The book h a s t w o equally significa n t goal s . One goal is to hel p students develop mathematical t h i n king ski l l s, particu l a r l y theorem-proving ski l l s . The other goal is to h e l p students u n d erstand s o m e of t h e wonderfu l l y r i c h i d e a s i n the m a thematical s t u d y of n u mbers. Th is book is a p p ropriate for a p roof t ransitions cou rse, for an i ndependent study experience, or for a cou rse designed as an i n trod u ction to abstract m a thematics. This text is s u i table for mathe matics or re l a ted m a j o rs or a n yone i n terested i n e x p l o r i n g m a thematical i deas on th e i r o w n .
Number Theory Through Inquiry con ta i n s a carefu l l y arranged sequence of chal
lenges t h a t l e a d students to d i scover i d eas abou t n u mbers a n d t o d i scover methods
of proof on thei r ow n . It i s designed to be u sed w i t h an i nstructional tech n i que vari ously known as g u i ded d i scovery or the Modi fied Moore Method or I n q u i ry B ased Lea rn i ng ( I B L ) . The resu l t of this approach w i l l be that students: • • •
Lea rn to t h i n k i ndependently. Lea rn to depend on t h e i r own reaso n i n g to determ i ne rig h t from w rong. Develop the cen t ral, i m porta n t ideas of i ntrod uctory nu mber t h eory on t h e i r ow n .
From that experience, they l e a rn that t h ey can perso n a l l y create i m portant i deas. They deve l op a n attitude of perso n a l rel ia nce a n d a sense that t h ey can think effec tively abo u t d i fficu l t p rob l e m s . These goa l s a re fu n d a m e n tal to the educational enterprise w i t h i n a n d beyond m a th e m a t i cs . MAA Tex tbooks
•
1 50 $51 .00
Code: N T I List:
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p p . , H a rdbou nd,
•
M A A Member:
2007 $41 .00 •
ISBN
978-0-88385-751 -9
Order your copy today www.maa.org
What Is a
Tools of
Number?
American
Mathematical Concepts
Mathematics
and Their Origins
Teaching,
Robert Tubbs
1 800-2000
Robert Tubbs examines
Peggy Aldrich
how mathematical concepts like number,
Kidwell, Amy
geometric truth, infin
Ackerberg-Hastings,
ity, and proof have been
and David Lindsay
employed by artists, theo logians, philosophers, writers, and cosmologists from ancient times to the modern era. Looking at a broad range of topics-from Pythagoras's exploration of the connection berween harmonious sounds and mathematical ratios to the understanding of time in both Western and pre-Columbian thought-Tubbs ties together seemingly disparate ideas to demon strate the relationship berween the sometimes elusive
Roberts Peggy Aldrich Kidwell, Amy Ackerberg-Hastings, and David Lindsay Roberts present the first systematic historical study of the objects used in the American mathematics classroom. Engaging and ac cessible, this volume tells the stories of how specific objects such as protractors, geometric models, slide rules, electronic calculators, and computers came to
thought of artists and philosophers and the concrete
be used in classrooms, and how some disappeared.
logic of mathematicians.
$70.00 hardcover
$27. 50 paperback
Mathematical
Adventures in
Mathematial Work. Printed in the Amcncas 1 554-1700
Works Printed in the Americas,
Group Theory Rubik's Cube, Merlin's Machine, and Other
1 5 54- 1 700
Mathematical Toys SECOND EDITION
Bruce Stanley Burdick
David Joyner
This magisterial annotated
"If you like puzzles, this
bibliography of the earli
is a somewhat fun book.
est mathematical works to be printed in the New World challenges long-held
If you like algebra, this is a fun book. If you like puzzles and algebra, this is a really fun book."-MAA On/in�
"Joyner does convey some of the excitement and
adventure in picking up knowledge of group theory by trying to understand Rubik's Cube. Enthusiastic students will learn a lot of mathematics from this book."-Ammcan Scimtist
assumptions about the earliest examples of American mathematical endeavor. Bruce Stanley Burdick brings together mathematical writings from Mexico, Lima, and the English colonies of Massachusetts, Pennsylvania, and New York. The book provides important information such as author, printer, place of publication, and location of original copies of each of the works discussed.
$27.50 paperback
$55.00 hardcover
THE JOHNS
HOPKINS UNIVERSITY PRESS 1 -800-537-5487
•
www.press.jhu.edu
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