A Combinatorial Proof of a Theorem of Katsuura

Transcription

1 Mathematical Assoc. of America College Mathematics Joural 45:1 Jue 2, :34 p.m. TSWLatexiaTemp tex A Combiatorial Proof of a Theorem of Katsuura Bria K. Miceli Bria Miceli (bmiceli@triity.edu) is a Associate Professor of Mathematics at Triity Uiversity i Sa Atoio, TX. He received his B.S. i Mathematics from Cal Poly, SLO ad his Ph.D. i Mathematics from UC, Sa Diego. Icludig his home tow, he has always lived i a city begiig with Sa which lies southeast of his previous city of residece, leavig Sa Jua, Puerto Rico as his most likely retiremet destiatio. He eoys rootig for the 49ers, cookig, cruisig the zoo with his wife ad daughter, goig to rock cocerts, ad playig racquetball. Dece [2], i workig with Laplace trasforms ad row-sums o Pascal s triagle, provided two coectures which ca be combied ito the followig statemet. Theorem 1. Suppose k, are positive itegers with the same parity such that k. The b 1 2 c X ( 1)! ( 0 if k <, ad ( 2)k = 1 2! if k =. (1) Dece s coecture was first proved i 2009 as cosequece of followig result. Theorem 2 (Katsuura [4]). Suppose k, are positive itegers such that k ad x, y C. The! ( X 0 if k <, ad ( 1) (y x)k = (2) x! if k =. Later, usig idetities amog arbitrary complex polyomials, Fegmig, Kim, ad Yeog [3] proved a more geeral theorem from which Katsuura s result, ad hece Theorem 1, are deduced as special cases. As Katsuura remarks i [4], Equatio (2) is curious ad almost too good to be true sice it says that the left had side of the equatio is idepedet of the choice of the umber y. The goal of this paper is to give a combiatorial proof of Theorem 2, ad we hope that such a proof will shed some light as to why the choice of y does ot affect this sum. Ideed, whe oe sees such a beautiful formula with biomial coefficiets, itegral powers, ad factorials, a coutig proof surely seems plausible, but with the ( 1) term ad the fact that x ad y eed ot be positive (or eve itegers!), we eed to be a little more clever about how to use our combiatorial toolbox. As it turs out, our proof will ivolve both the Ivolutio Priciple (IP) applied to certai sets of colored words ad the Fudametal Theorem of Algebra. VOL. 45, NO. 1, JANUARY 2014 THE COLLEGE MATHEMATICS JOURNAL 1 page 1

2 Mathematical Assoc. of America College Mathematics Joural 45:1 Jue 2, :34 p.m. TSWLatexiaTemp tex page 2 For the full details of the IP method, we refer the reader to Beami ad Qui [1], but we give a quick recap here. A ivolutio is a biective fuctio g : X X such that g(g(x)) = x for every x X. Ay x X such that g(x) = x is called a fixed poit of the ivolutio, ad we see that our ivolutio partitios X ito the two sets g fix (X) = {x X g(x) = x} ad g fix (X) = {x X g(x) x}. Now, suppose our goal is to compute the value of S(A) := y A γ(y), where A is a set of obects ad γ represets a weight fuctio o the obects of A. We would like to fid a ivolutio f : A A such that γ(y) = γ(f(y)) wheever y. Here, f is called a sig-reversig ivolutio, ad give such a f, we rewrite our desired sum as follows: S(A) = γ(y) + γ(y). Because f is sig-reversig, the weights i the right sum come i pairs which cacel out. Accordigly, the sum o the right is 0, givig that S(A) = γ(y). We ca the use this formula for computig S(A), hopig that it is a easier task summig over oly the fixed poits of our ivolutio. For our combiatorial proof of Theorem 2, we will be iterested i coutig colored words over a certai alphabet. Give a fiite set W = {w 1, w 2,..., w s }, we defie a word over W to be ay strig cosistig of elemets (called letters) of W, ad we let W deote the set of all words over W. The umber of letters of w is called the legth of w, deoted by w. For example, if B = {0, 1}, the b = B ad b = 7. We say that w W is a permutatio of W if w = s ad each w i appears exactly oce i w. Give a set of colors C = {c 1, c 2,..., c m }, we defie W C to be the set of elemets of W where each elemet is colored with the colors of C, i.e., W C cotais m distict copies of each elemet of W. We the defie a C-colored word to be ay elemet of WC. Here, uv deotes that the letter u is colored with the color v. For example, if D = {1, 2, 3}, the B D = {0 1, 0 2, 0 3, 1 1, 1 2, 1 3 } ad d = BD with d = 7. Before we prove Theorem 2, we prove the followig result, i which x ad y must be positive itegers rather tha arbitrary complex umbers. Theorem 3. Suppose x, y, k, are positive itegers such that k. The { ( 1) (y x) k 0 if k <, ad = x! if k =. (3) Proof. Let Y = {b 1, b 2,..., b y }, C = {1, 2,..., x}, ad [] = {1, 2,..., }. Defie A := {(T, w) T [] ad w (Y T C ) with w = k}. 2 THE MATHEMATICAL ASSOCIATION OF AMERICA

3 Mathematical Assoc. of America College Mathematics Joural 45:1 Jue 2, :34 p.m. TSWLatexiaTemp tex page 3 For example, if = 10, k = 8, x = 3, ad y = 3, the ({1, 2, 6, 9, 10}, b b 3 b b 1 ) A. For ay (T, w) A, let the weight of (T, w) be γ((t, w)) = ( 1) T +u(w), where u(w) is the umber of letters i w that come from T C. Usig the example above, T = {1, 2, 6, 9, 10}, givig that T = 5, ad the word w = b b 3 b b 1 cotais four letters from T C, so γ(({1, 2, 6, 9, 10}, b b 3 b b 1 )) = ( 1) 5+4 = 1. For ay fixed, if T =, the there are ( ) such -elemet subsets of [], ad there are (y + x) k words of legth k cosistig of letters from Y T C. Thus, the umber of pairs (T, w) A such that T = is ( ) (y + x) k. Replacig x with ( x) weights each pair (T, w) with ( 1) u(w), ad accordigly, ( 1) (y x) k = γ((t, w)). T = Summig both side of this equatio over all possible values of gives that ( 1) (y x) k = T = γ((t, w)) = γ((t, w)). (4) Give (T, w) A, let l be the smallest elemet of [] for which o colored versio appears i w, ad defie f : A A i the followig way. i. If l T, the f((t, w)) = (T, w), where T = T {l}. ii. If l / T, the f((t, w)) = (T, w), where T = T {l}. iii. If o such l exists, the f((t, w)) = (T, w). Usig our previous example, l = 3, ad sice 3 / T = {1, 2, 6, 9, 10}, step ii. above gives that f(({1, 2, 6, 9, 10}, b b 3 b b 1 )) = ({1, 2, 3, 6, 9, 10}, b b 3 b b 1 ). This fuctio f is a sig-reversig ivolutio, sice if f((t, w)) = (T, w) with T T, the f((t, w)) = (T, w), ad T = T ± 1. Usig the IP to compute Equatio (4) gives that ( 1) (y x) k = γ((t, w)). (5) Give (T, w), it must be the case that w cotais some colored versio of every elemet of []. Sice the legth of w = k, there are o fixed poits of f i the case that k <, ad right had side of Equatio (5) is 0. Otherwise k =. I this VOL. 45, NO. 1, JANUARY 2014 THE COLLEGE MATHEMATICS JOURNAL 3

4 Mathematical Assoc. of America College Mathematics Joural 45:1 Jue 2, :34 p.m. TSWLatexiaTemp tex page 4 case, for w to cotai some colored versio of every elemet of [] gives that T must be the set [], ad = {([], w) w is a C-colored permutatio of []}. Here, if ([], w), the γ(([], w) = ( 1) [] +u(w) = ( 1) + = 1, ad so Equatio (5) reduces to ( 1) (y x) k = 1 =. (6) Sice there are! permutatios of [] ad x = x!, completig the proof. ways to color each permutatio, Remark. From this proof, we see that the umber y i our sum correspods to a set of y letters used to make words which, whe we apply our sig-reversig ivolutio f, oly appear i the o-fixed poits of f, thus ot cotributig to the fial value of the sum. To show that Theorem 3 exteds to ay x, y C, we use the Fudametal Theorem of Algebra, usually stated as follows. Theorem 4 (Fudametal Theorem of Algebra). Let be a positive iteger ad suppose that a(x) = a x + a 1 x a 1 x + a 0 such that each a i C ad a 0. The a(x) has exactly complex roots, couted with multiplicity. Remark. A cosequece of this theorem is that if p(x) is a complex polyomial of fiite degree with a ifiite umber of roots, the p(x) must be the zero polyomial. Proof of Theorem 2. First let, k, y be fixed positive itegers ad defie q k (x) := ( 1) (y x) k. Theorem 3 gives that q (x) = x! for ifiitely may values of x, ad so the polyomial p(x) := q (x) x! has a ifiite umber of distict complex roots (the positive itegers). However, p(x) is a polyomial of degree at most, ad by the Fudametal Theorem of Algebra, p(x) = 0 for all x C. A similar argumet holds i the case where k <, as q k (x) = 0, where q k (x) is a polyomial of degree k with ifiitely may roots. Now lettig, k be positive itegers ad fixig x C, a similar argumet shows that for every y C, { r k (y) := ( 1) (y x) k 0 if k <, ad = x! if k =. Thus, our desired result holds for all positive itegers, k ad all x, y C. Summary. We give a combiatorial proof of a algebraic result of Katsuura s. Moreover, we use the proof of this result to shed some light o a iterestig property of the result itself. 4 THE MATHEMATICAL ASSOCIATION OF AMERICA

5 Mathematical Assoc. of America College Mathematics Joural 45:1 Jue 2, :34 p.m. TSWLatexiaTemp tex page 5 Refereces 1. A. T. Beami & J. J. Qui, A alterate approach to alteratig sums: A method to die for, The College Mathematics Joural, 39 (2008) T. P. Dece, Some half-row sums from Pascals triagle via Laplace trasforms, The College Mathematics Joural, 38 (2007) D. Fegmig, H. W. Kim, & L. T. Yeog, A family of idetities via arbitrary polyomials, The College Mathematics Joural, 44 (2013) H. Katsuura, Summatios ivolvig biomial coefficiets, The College Mathematics Joural, 40 (2009) VOL. 45, NO. 1, JANUARY 2014 THE COLLEGE MATHEMATICS JOURNAL 5